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101 | a-current-carrying-infinitely-long-wire-is-kept-along-the-diameter-of-a-circular-wire-loop-without-touching-it-the-correct-statements-isare | a-current-carrying-infinitely-long-wire-is-kept-along-the-diameter-of-a-circular-wire-loop-without-touching-it-the-correct-statements-isare-79722 | <div class="question">A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it, the correct statement(s) is(are)</div> | ['Physics', 'Electromagnetic Induction', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">The emf induced in the loop is zero if the current is constant.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">The emf induced in the loop is finite if the current is constant.</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">The emf induced in the loop is zero if the current decreases at a steady rate.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">The emf induced in the loop is infinite if the current decreases at a steady rate.</span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value">The emf induced in the loop is zero if the current is constant., The emf induced in the loop is zero if the current decreases at a steady rate.</span> </div> | <div class="solution">If the current is constant, the emf induced in the loop zero. Emf will be induced in the circular wire loop when flux through it changes with time.
<br/> <br/>$\mathrm{e}=-\frac{\Delta \phi}{\Delta \mathrm{t}}$
<br/> <br/>when the current is constant, the flux changing through it will be zero.
<br/> <br/>Also, if the current decreases at steady rate, the emf. induced in the loop is zero.
<br/> <br/>When the current is decreasing at a steady rate then the change in the flux (decreasing inwards) on the right half of the wire is equal to the change in flux (decreasing outwards) on the left half of the wire such that $\Delta \phi$ through the circular loop is zero.
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/pMb-i-hx_6TMOAXEfVyhqx49W_QA3YpqMVkpEO5sdg4.original.fullsize.png"/></div> | MarksBatch1_P1.db |
102 | a-cylindrical-cavity-of-diameter-a-exists-inside-a-cylinder-of-diameter-2-a-as-shown-in-the-figure-both-the-cylinder-and-the-cavity-are-infinity-long--1 | a-cylindrical-cavity-of-diameter-a-exists-inside-a-cylinder-of-diameter-2-a-as-shown-in-the-figure-both-the-cylinder-and-the-cavity-are-infinity-long-1-91282 | <div class="question">A cylindrical cavity of diameter a exists inside a cylinder of diameter $2 a$ as shown in the figure. Both the cylinder and the cavity are infinity long. A uniform current density $J$ flows along the length. If the magnitude of the magnetic field at the point $P$ is given by $\frac{N}{12} \mu_{0} a J$, then the value of $N$ is
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/YfqH0WzwUDjeUhuXA7otppktNta3TUFOyi7I8kErGug.original.fullsize.png"/></div> | ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">5</span> </div> | <div class="solution">Current density $J=\frac{\text { current }}{\text { area }}=\frac{I}{A} \Rightarrow I=\mathrm{JA}$ Magnetic field $\mathrm{B}_{\mathrm{R}}$ after removing cavity $(\mathrm{C})$ $\mathrm{B}_{\mathrm{R}}=\mathrm{B}_{\text {total }}-\mathrm{B}_{\text {cavity }}$
<br/> <br/>$\begin{array}{l}
<br/> <br/>\frac{\mu_{0} I_{t}}{2 \pi a}-\frac{\mu_{0} I_{c}}{2 \pi\left(\frac{3}{2} a\right)} \\
<br/> <br/>\quad=\frac{\mu_{0}}{\pi a}\left[\frac{I_{t}}{2}-\frac{I_{c}}{3}\right]\left(\text { here } I_{t}=J\left(\pi a^{2}\right) I_{c}=J\left(\frac{\pi a^{2}}{4}\right)\right) \\
<br/> <br/>=\frac{\mu_{0}}{\pi a}\left[\frac{\pi a^{2} J}{2}-\frac{\pi a^{2} J}{12}\right] \\
<br/> <br/>\text { or, } \mathrm{B}_{\mathrm{R}}=\frac{5 \mu_{0} a J}{12}
<br/> <br/>\end{array}$
<br/> <br/>Comparing it with $\frac{N}{12} \mu_{0} a J$ We get $N=5$</div> | MarksBatch1_P1.db |
103 | a-cylindrical-furnace-has-height-h-and-diameter-d-both-1-m-it-is-maintained-at-temperature-360-k-the-air-gets-heated-inside-the-furnace-at-constant-pr-1 | a-cylindrical-furnace-has-height-h-and-diameter-d-both-1-m-it-is-maintained-at-temperature-360-k-the-air-gets-heated-inside-the-furnace-at-constant-pr-1-75469 | <div class="question"><p>A cylindrical furnace has height <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>H</mi></mfenced></math> and diameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>D</mi></mfenced></math> both <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo> </mo><mi mathvariant="normal">m</mi></math>. It is maintained at temperature <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>360</mn><mo> </mo><mi mathvariant="normal">K</mi></math>. The air gets heated inside the furnace at constant pressure <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>a</mi></msub></math> and its temperature becomes <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>T</mi><mo>=</mo><mn>360</mn><mo> </mo><mi mathvariant="normal">K</mi></math>. The hot air with density <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ρ</mi></math> rises up a vertical chimney of diameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mo>=</mo><mn>0</mn><mo>.</mo><mn>1</mn><mo> </mo><mi mathvariant="normal">m</mi></math> and height <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>h</mi><mo>=</mo><mn>9</mn><mo> </mo><mi mathvariant="normal">m</mi></math> above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>ρ</mi><mi>a</mi></msub><mo>=</mo><mn>1</mn><mo>.</mo><mn>2</mn><mo> </mo><mi>kg</mi><mo> </mo><msup><mi mathvariant="normal">m</mi><mrow><mo>-</mo><mn>3</mn></mrow></msup></math>, pressure <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>a</mi></msub></math> and temperature <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>T</mi><mi>a</mi></msub><mo>=</mo><mn>300</mn><mo> </mo><mi mathvariant="normal">K</mi></math> enters the furnace. Assume air as an ideal gas, neglect the variations in <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ρ</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>T</mi></math> inside the chimney and the furnace. Also ignore the viscous effects.</p><br/><br/><p>[Given: The acceleration due to gravity <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>g</mi><mo>=</mo><mn>10</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>2</mn></mrow></msup></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>π</mi><mo>=</mo><mn>3</mn><mo>.</mo><mn>14</mn></math>]</p><br/><br/><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/3fbf9ea5-3f88-41c1-b37d-775e24fecac2.png" style="width: 357px; height: 652px;"/></p><br/><br/>When the chimney is closed using a cap at the top, a pressure difference <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∆</mo><mi>P</mi></math> develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air, due to the stoppage of air flow, are negligible then the value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∆</mo><mi>P</mi></math> is _____ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">N</mi><mo> </mo><msup><mi mathvariant="normal">m</mi><mrow><mo>−</mo><mn>2</mn></mrow></msup></math>.</div> | ['Physics', 'Mechanical Properties of Fluids', 'JEE Advanced', 'JEE Advanced 2023 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">7850</span> </div> | <div class="solution"><blockquote><p>Applying Bernoulli’s theorem between top and bottom points of furnace :</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>a</mi></msub><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><msub><mi>ρ</mi><mi>a</mi></msub><msup><mi>V</mi><mn>2</mn></msup><mo>=</mo><msub><mi>P</mi><mi>a</mi></msub><mo>+</mo><mi>ρ</mi><mi>g</mi><mi>H</mi><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>ρ</mi><msup><mi>V</mi><mn>2</mn></msup><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo> </mo><mfenced><mn>1</mn></mfenced></math></p><p>Also, since <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mi>M</mi><mo>=</mo><mi>ρ</mi><mi>R</mi><mi>T</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><msub><mi>ρ</mi><mi>a</mi></msub><mo>×</mo><mn>300</mn><mo>=</mo><mi>ρ</mi><mo>×</mo><mn>360</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi>ρ</mi><mo>=</mo><mn>1</mn><mo> </mo><mi>kg</mi><mo> </mo><msup><mi mathvariant="normal">m</mi><mrow><mo>-</mo><mn>3</mn></mrow></msup><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo> </mo><mfenced><mn>2</mn></mfenced></math></p><p>From equation(1), we can write</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mfrac><msup><mi>V</mi><mn>2</mn></msup><mn>2</mn></mfrac><mfenced><mrow><mn>0</mn><mo>.</mo><mn>2</mn></mrow></mfenced><mo>=</mo><mn>1</mn><mo>×</mo><mn>10</mn><mo>×</mo><mn>1</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi>V</mi><mo>=</mo><mn>10</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p><p>Now, let <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V</mi><mo>'</mo></math> be the speed of air in chimney.</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo></math> By continuity equation,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi mathvariant="normal">π</mi><msup><mi>D</mi><mn>2</mn></msup></mrow><mn>4</mn></mfrac><mo>·</mo><mi>V</mi><mo>=</mo><mfrac><mrow><mi mathvariant="normal">π</mi><msup><mi>d</mi><mn>2</mn></msup></mrow><mn>4</mn></mfrac><mo>·</mo><mi>V</mi><mo>'</mo></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi>V</mi><mo>'</mo><mo>=</mo><mn>100</mn><mi>V</mi><mo>=</mo><mn>1000</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p><p>Therefore, pressure difference <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∆</mo><mi>p</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mi>ρ</mi><mi>V</mi><msup><mo>'</mo><mn>2</mn></msup><mo>·</mo></math>area</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>1</mn><mo>×</mo><msup><mfenced><mn>1000</mn></mfenced><mn>2</mn></msup><mo>×</mo><mfrac><mrow><mi>π</mi><msup><mi>d</mi><mn>2</mn></msup></mrow><mn>4</mn></mfrac><mspace linebreak="newline"></mspace><mo>=</mo><mn>1</mn><mo>×</mo><msup><mfenced><mn>1000</mn></mfenced><mn>2</mn></msup><mo>×</mo><mfrac><mrow><mn>3</mn><mo>.</mo><mn>14</mn><mo>×</mo><msup><mfenced><mrow><mn>0</mn><mo>.</mo><mn>1</mn></mrow></mfenced><mn>2</mn></msup></mrow><mn>4</mn></mfrac><mspace linebreak="newline"></mspace><mo>=</mo><mn>7850</mn><mo> </mo><mi mathvariant="normal">N</mi><mo> </mo><msup><mi mathvariant="normal">m</mi><mn>2</mn></msup></math></p></blockquote></div> | MarksBatch1_P1.db |
104 | a-cylindrical-furnace-has-height-h-and-diameter-d-both-1-m-it-is-maintained-at-temperature-360-k-the-air-gets-heated-inside-the-furnace-at-constant-pr | a-cylindrical-furnace-has-height-h-and-diameter-d-both-1-m-it-is-maintained-at-temperature-360-k-the-air-gets-heated-inside-the-furnace-at-constant-pr-47059 | <div class="question"><p>A cylindrical furnace has height <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>H</mi></mfenced></math> and diameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>D</mi></mfenced></math> both <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo> </mo><mi mathvariant="normal">m</mi></math>. It is maintained at temperature <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>360</mn><mo> </mo><mi mathvariant="normal">K</mi></math>. The air gets heated inside the furnace at constant pressure <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>a</mi></msub></math> and its temperature becomes <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>T</mi><mo>=</mo><mn>360</mn><mo> </mo><mi mathvariant="normal">K</mi></math>. The hot air with density <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ρ</mi></math> rises up a vertical chimney of diameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mo>=</mo><mn>0</mn><mo>.</mo><mn>1</mn><mo> </mo><mi mathvariant="normal">m</mi></math> and height <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>h</mi><mo>=</mo><mn>9</mn><mo> </mo><mi mathvariant="normal">m</mi></math> above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>ρ</mi><mi>a</mi></msub><mo>=</mo><mn>1</mn><mo>.</mo><mn>2</mn><mo> </mo><mi>kg</mi><mo> </mo><msup><mi mathvariant="normal">m</mi><mrow><mo>-</mo><mn>3</mn></mrow></msup></math>, pressure <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>a</mi></msub></math> and temperature <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>T</mi><mi>a</mi></msub><mo>=</mo><mn>300</mn><mo> </mo><mi mathvariant="normal">K</mi></math> enters the furnace. Assume air as an ideal gas, neglect the variations in <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ρ</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>T</mi></math> inside the chimney and the furnace. Also ignore the viscous effects.</p><br/><br/><p>[Given: The acceleration due to gravity <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>g</mi><mo>=</mo><mn>10</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>2</mn></mrow></msup></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>π</mi><mo>=</mo><mn>3</mn><mo>.</mo><mn>14</mn></math>]</p><br/><br/><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/3fbf9ea5-3f88-41c1-b37d-775e24fecac2.png" style="width: 357px; height: 652px;"/></p><br/><br/>Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is _____ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>gm</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>−</mo><mn>1</mn></mrow></msup></math>.</div> | ['Physics', 'Mechanical Properties of Fluids', 'JEE Advanced', 'JEE Advanced 2023 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">7850</span> </div> | <div class="solution"><blockquote><p>Applying Bernoulli’s theorem between top and bottom points of furnace :</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>a</mi></msub><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><msub><mi>ρ</mi><mi>a</mi></msub><msup><mi>V</mi><mn>2</mn></msup><mo>=</mo><msub><mi>P</mi><mi>a</mi></msub><mo>+</mo><mi>ρ</mi><mi>g</mi><mi>H</mi><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>ρ</mi><msup><mi>V</mi><mn>2</mn></msup><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo> </mo><mfenced><mn>1</mn></mfenced></math></p><p>Also, since <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mi>M</mi><mo>=</mo><mi>ρ</mi><mi>R</mi><mi>T</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><msub><mi>ρ</mi><mi>a</mi></msub><mo>×</mo><mn>300</mn><mo>=</mo><mi>ρ</mi><mo>×</mo><mn>360</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi>ρ</mi><mo>=</mo><mn>1</mn><mo> </mo><mi>kg</mi><mo> </mo><msup><mi mathvariant="normal">m</mi><mrow><mo>-</mo><mn>3</mn></mrow></msup><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo> </mo><mfenced><mn>2</mn></mfenced></math></p><p>From equation(1), we can write</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mfrac><msup><mi>V</mi><mn>2</mn></msup><mn>2</mn></mfrac><mfenced><mrow><mn>0</mn><mo>.</mo><mn>2</mn></mrow></mfenced><mo>=</mo><mn>1</mn><mo>×</mo><mn>10</mn><mo>×</mo><mn>1</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi>V</mi><mo>=</mo><mn>10</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p><p>Required mass flow rate <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mi>V</mi><mo>×</mo><mfrac><mrow><mi>π</mi><msup><mi>D</mi><mn>2</mn></msup></mrow><mn>4</mn></mfrac><mo>×</mo><mi>ρ</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>10</mn><mo>×</mo><mfrac><mrow><mn>3</mn><mo>.</mo><mn>14</mn></mrow><mn>4</mn></mfrac><mo>×</mo><msup><mn>1</mn><mn>2</mn></msup><mo>×</mo><mn>1</mn><mo> </mo><mi>kg</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>7</mn><mo>.</mo><mn>85</mn><mo> </mo><mi>kg</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>7850</mn><mo> </mo><mi>gm</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p></blockquote></div> | MarksBatch1_P1.db |
105 | a-cylindrical-vessel-of-height-500-mm-has-an-orifice-small-hole-at-its-bottom-the-orifice-is-initially-closed-and-water-is-filled-in-it-upto-height-h--1 | a-cylindrical-vessel-of-height-500-mm-has-an-orifice-small-hole-at-its-bottom-the-orifice-is-initially-closed-and-water-is-filled-in-it-upto-height-h-1-84143 | <div class="question">A cylindrical vessel of height $500 \mathrm{~mm}$ has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it upto height $H$. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 $\mathrm{mm}$. Find the fall in height (in $\mathrm{mm}$ ) of water level due to opening of the orifice.<br/>[Take atmospheric pressure $=1.0 \times 10^5 \mathrm{Nm}^{-2}$, density of water $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$ and $g=10$ $\mathrm{ms}^{-2}$. Neglect any effect of surface tension.]</div> | ['Physics', 'Mechanical Properties of Fluids', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">6</span> </div> | <div class="solution">In this question we will have to assume that temperature of enclosed air about water is constant (or $p V=$ constant)<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/8dM8QYJxDmaHvc7TZMSYMcDZb9J5Qzf03BYUgRzDGn4.original.fullsize.png"/><br/><br/><br/>$$<br/>\begin{gathered}<br/>p=p_0-\rho g h \\<br/>p_0[A(500-H)]=p[A(200)]<br/>\end{gathered}<br/>$$<br/>Solving these two equations, we get<br/>$$<br/>\begin{aligned}<br/>H & =206 \mathrm{~mm} \\<br/>\therefore \quad \text { Level fall } & =(206-200) \mathrm{mm} \\<br/>& =6 \mathrm{~mm}<br/>\end{aligned}<br/>$$</div> | MarksBatch1_P1.db |
106 | a-decapeptide-molecular-weight-796-on-complete-hydrolysis-gives-glycine-molecular-weight-75-alanine-and-phenylalanine-glycine-contributes-470-to-the-t | a-decapeptide-molecular-weight-796-on-complete-hydrolysis-gives-glycine-molecular-weight-75-alanine-and-phenylalanine-glycine-contributes-470-to-the-t-35922 | <div class="question">A decapeptide (molecular weight 796) on complete hydrolysis gives glycine (molecular weight 75), alanine and phenylalanine. Glycine contributes $47.0 \%$ to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is</div> | ['Chemistry', 'Biomolecules', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">6</span> </div> | <div class="solution">$$<br/>\text { A decapeptide has nine peptide (amide) linkage as }<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/QRsCQ14li6NT9gp0NOovuoPzTGcmi68dMG0a74dUtUg.original.fullsize.png"/><br/><br/>Therefore, no hydrolysis, it will absorb nine water molecules.<br/>Hence, total mass of hydrolysis product $=796+18 \times 9=958$<br/>$\Rightarrow$ mass of glycine in hydrolysis product $=\frac{958 \times 47}{100}=450$<br/>$\Rightarrow$ number of glycine molecule in one molecule of decapeptide $=\frac{450}{75}=6$</div> | MarksBatch1_P1.db |
107 | a-diatomic-ideal-gas-is-compressed-adiabatically-to-32-1-of-its-initial-volume-if-the-initial-temperature-of-the-gas-is-t-i-t-i-in-kelvin-and-the-fina | a-diatomic-ideal-gas-is-compressed-adiabatically-to-32-1-of-its-initial-volume-if-the-initial-temperature-of-the-gas-is-t-i-t-i-in-kelvin-and-the-fina-30056 | <div class="question">A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_i$ $T_i$ (in kelvin) and the final temperature is $a T_i$, the value of $a$ is</div> | ['Physics', 'Thermodynamics', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">4</span> </div> | <div class="solution">In adiabatic process,<br/>$$<br/>\begin{aligned}<br/>& T V^{\gamma-1}=\text { constant } \\<br/>& \therefore T_i V_i^{0.4}=T_f V_f^{0.4} \\<br/>& \text { (as } \gamma=1.4 \text { for diatomic gas) } \\<br/>& \text { or } \quad T_i V_i^{0.4}=\left(\alpha T_i\right)\left(\frac{V_i}{32}\right)^{0.4} \\<br/>& \text { or } \quad \alpha(32)^{0.4}=4 \\<br/>&<br/>\end{aligned}<br/>$$<br/>$\therefore$ The correct answer is 4 .</div> | MarksBatch1_P1.db |
108 | a-disk-of-radius-4-a-having-a-uniformly-distributed-charge-6-c-is-placed-in-the-x-y-plane-with-its-centre-at-2-a-0-0-a-rod-of-length-a-carrying-a-unif-1 | a-disk-of-radius-4-a-having-a-uniformly-distributed-charge-6-c-is-placed-in-the-x-y-plane-with-its-centre-at-2-a-0-0-a-rod-of-length-a-carrying-a-unif-1-53825 | <div class="question">A disk of radius $\frac{a}{4}$ having a uniformly distributed charge $6 \mathrm{C}$ is placed in the $x-y$ plane with its centre at $\left(\frac{-a}{2}, 0,0\right)$. A rod of length a carrying a uniformly distributed charge $8 \mathrm{C}$ is placed on the $x$-axis from $x=\frac{a}{4}$ to $x=\frac{5 a}{4}$. Two point charges $-7 \mathrm{C}$ and $3 \mathrm{C}$ are placed at $\left(\frac{a}{4}, \frac{-a}{4}, 0\right)$ and $\left(\frac{-3 a}{4}, \frac{3 a}{4}, 0\right)$, respectively. Consider a cubical surface formed by six surfaces $x=\pm \frac{a}{2}, y=\pm \frac{a}{2}, z=\pm \frac{a}{2}$.<br/>The electric flux through this cubical surface is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/AvcnvTT5sePtQoutuhERbdjweJ91m1yd8m1dkYFaLNw.original.fullsize.png"/><br/></div> | ['Physics', 'Electrostatics', 'JEE Main'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{-2 C}{\varepsilon_0}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{2 C}{\varepsilon_0}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{10 C}{\varepsilon_0}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{12 C}{\varepsilon_0}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{-2 C}{\varepsilon_0}$<br/></span> </div> | <div class="solution">Total enclosed charge as already shown is<br/>$$<br/>q_{\text {net }}=\frac{6 C}{2}+\frac{8 C}{4}-7 C=-2 C<br/>$$<br/>From Gauss-theorem, net flux,<br/>$$<br/>\varphi_{\text {net }}=\frac{q_{\text {net }}}{\varepsilon_0}=\frac{-2 C}{\varepsilon_0}<br/>$$</div> | MarksBatch1_P1.db |
109 | a-disk-of-radius-4-a-having-a-uniformly-distributed-charge-6-c-is-placed-in-the-x-y-plane-with-its-centre-at-2-a-0-0-a-rod-of-length-a-carrying-a-unif | a-disk-of-radius-4-a-having-a-uniformly-distributed-charge-6-c-is-placed-in-the-x-y-plane-with-its-centre-at-2-a-0-0-a-rod-of-length-a-carrying-a-unif-11617 | <div class="question">A disk of radius $\frac{a}{4}$ having a uniformly distributed charge $6 \mathrm{C}$ is placed in the $x-y$ plane with its centre at $\left(\frac{-a}{2}, 0,0\right)$. A rod of length a carrying a uniformly distributed charge $8 \mathrm{C}$ is placed on the $x$-axis from $x=\frac{a}{4}$ to $x=\frac{5 a}{4}$. Two point charges $-7 \mathrm{C}$ and $3 \mathrm{C}$ are placed at $\left(\frac{a}{4}, \frac{-a}{4}, 0\right)$ and $\left(\frac{-3 a}{4}, \frac{3 a}{4}, 0\right)$, respectively. Consider a cubical surface formed by six surfaces $x=\pm \frac{a}{2}, y=\pm \frac{a}{2}, z=\pm \frac{a}{2}$.<br/>The electric flux through this cubical surface is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/AvcnvTT5sePtQoutuhERbdjweJ91m1yd8m1dkYFaLNw.original.fullsize.png"/><br/></div> | ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2009 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{-2 C}{\varepsilon_0}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{2 C}{\varepsilon_0}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{10 C}{\varepsilon_0}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{12 C}{\varepsilon_0}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{-2 C}{\varepsilon_0}$<br/></span> </div> | <div class="solution">Total enclosed charge as already shown is<br/>$$<br/>q_{\text {net }}=\frac{6 C}{2}+\frac{8 C}{4}-7 C=-2 C<br/>$$<br/>From Gauss-theorem, net flux,<br/>$$<br/>\varphi_{\text {net }}=\frac{q_{\text {net }}}{\varepsilon_0}=\frac{-2 C}{\varepsilon_0}<br/>$$</div> | MarksBatch1_P1.db |
110 | a-disk-of-radius-r-with-uniform-positive-charge-density-is-placed-on-the-x-y-plane-with-its-center-at-the-origin-the-coulomb-potential-along-the-z-axi | a-disk-of-radius-r-with-uniform-positive-charge-density-is-placed-on-the-x-y-plane-with-its-center-at-the-origin-the-coulomb-potential-along-the-z-axi-39818 | <div class="question"><p>A disk of radius <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> with uniform positive charge density <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>σ</mi></math> is placed on the <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mi>y</mi></math> plane with its center at the origin. The Coulomb potential along the <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi></math>-axis is<br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V</mi><mfenced><mi>z</mi></mfenced><mo>=</mo><mfrac><mi>σ</mi><mrow><mn>2</mn><msub><mi>ϵ</mi><mn>0</mn></msub></mrow></mfrac><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></math></p><p>A particle of positive charge <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>q</mi></math> is placed initially at rest at a point on the <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi></math>-axis with <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi><mo>=</mo><msub><mi>z</mi><mn>0</mn></msub></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>></mo><mn>0</mn></math>. In addition to the Coulomb force, the particle experiences a vertical force <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>F</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mi>c</mi><mover><mi>k</mi><mo>^</mo></mover></math> with <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>c</mi><mo>></mo><mn>0</mn></math>. Let <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>=</mo><mfrac><mrow><mn>2</mn><mi>c</mi><msub><mi>ε</mi><mn>0</mn></msub></mrow><mrow><mi>q</mi><mi>σ</mi></mrow></mfrac></math>. Which of the following statement(s) is(are) correct?</p></div> | ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2022 (Paper 2)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>=</mo><mfrac><mn>1</mn><mn>4</mn></mfrac></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>=</mo><mfrac><mn>25</mn><mn>7</mn></mfrac><mi>R</mi></math>, the particle reaches the origin.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>=</mo><mfrac><mn>1</mn><mn>4</mn></mfrac></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>=</mo><mfrac><mn>3</mn><mn>7</mn></mfrac><mi>R</mi></math>, the particle reaches the origin.</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>=</mo><mfrac><mn>1</mn><mn>4</mn></mfrac></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>=</mo><mfrac><mi>R</mi><msqrt><mn>3</mn></msqrt></mfrac></math>, the particle returns back to <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi><mo>=</mo><msub><mi>z</mi><mn>0</mn></msub></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>></mo><mn>1</mn></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>></mo><mn>0</mn></math>, the particle always reaches the origin.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value">For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>=</mo><mfrac><mn>1</mn><mn>4</mn></mfrac></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>=</mo><mfrac><mn>25</mn><mn>7</mn></mfrac><mi>R</mi></math>, the particle reaches the origin., For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>=</mo><mfrac><mn>1</mn><mn>4</mn></mfrac></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>=</mo><mfrac><mi>R</mi><msqrt><mn>3</mn></msqrt></mfrac></math>, the particle returns back to <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi><mo>=</mo><msub><mi>z</mi><mn>0</mn></msub></math>, For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>></mo><mn>1</mn></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>></mo><mn>0</mn></math>, the particle always reaches the origin.</span> </div> | <div class="solution"><p>For the particle to reach at origin, work done by the force should be greater than the increase in the potential energy of the charged particle. Therefore,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>c</mi><mi>z</mi><mo>≥</mo><mfrac><mrow><mi>σ</mi><mi>q</mi></mrow><mrow><mn>2</mn><msub><mi>ϵ</mi><mn>0</mn></msub></mrow></mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mspace linebreak="newline"></mspace><mo>⇒</mo><mfrac><mrow><mi>c</mi><mn>2</mn><msub><mi>ϵ</mi><mn>0</mn></msub></mrow><mrow><mi>σ</mi><mi>q</mi></mrow></mfrac><mo>≥</mo><mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mi>z</mi></mfrac><mspace linebreak="newline"></mspace><mo>⇒</mo><mi>β</mi><mo>≥</mo><mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mi>z</mi></mfrac></math></p><p>For option A</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mi>z</mi></mfrac><mo>=</mo><mfrac><mrow><mi>R</mi><mo>-</mo><mi>R</mi><mfenced><mrow><msqrt><mn>1</mn><mo>+</mo><mstyle displaystyle="true"><mfrac><mn>625</mn><mn>49</mn></mfrac></mstyle></msqrt><mo>-</mo><mstyle displaystyle="true"><mfrac><mn>25</mn><mn>7</mn></mfrac></mstyle></mrow></mfenced></mrow><mrow><mi>R</mi><mstyle displaystyle="true"><mfrac><mn>25</mn><mn>7</mn></mfrac></mstyle></mrow></mfrac><mo>≃</mo><mn>0</mn><mo>.</mo><mn>242</mn></math></p><p>Since, <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn><mo>.</mo><mn>25</mn><mo>></mo><mn>0</mn><mo>.</mo><mn>242</mn></math></p><p>Option A is correct.</p><p>For option B</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mi>z</mi></mfrac><mo>=</mo><mfrac><mrow><mi>R</mi><mo>-</mo><mi>R</mi><mfenced><mrow><msqrt><mn>1</mn><mo>+</mo><mstyle displaystyle="true"><mfrac><mn>9</mn><mn>49</mn></mfrac></mstyle></msqrt><mo>-</mo><mstyle displaystyle="true"><mfrac><mn>3</mn><mn>7</mn></mfrac></mstyle></mrow></mfenced></mrow><mrow><mi>R</mi><mstyle displaystyle="true"><mfrac><mn>3</mn><mn>7</mn></mfrac></mstyle></mrow></mfrac><mo>≃</mo><mn>0</mn><mo>.</mo><mn>79</mn></math></p><p>Since, <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn><mo>.</mo><mn>25</mn><mo><</mo><mn>0</mn><mo>.</mo><mn>79</mn></math></p><p>Option B is incorrect</p><p>For option C</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mi>z</mi></mfrac><mo>=</mo><mfrac><mrow><mi>R</mi><mo>-</mo><mi>R</mi><mfenced><mrow><msqrt><mn>1</mn><mo>+</mo><mstyle displaystyle="true"><mfrac><mn>1</mn><mn>3</mn></mfrac></mstyle></msqrt><mo>-</mo><mstyle displaystyle="true"><mfrac><mn>1</mn><msqrt><mn>3</mn></msqrt></mfrac></mstyle></mrow></mfenced></mrow><mrow><mi>R</mi><mstyle displaystyle="true"><mfrac><mn>1</mn><msqrt><mn>3</mn></msqrt></mfrac></mstyle></mrow></mfrac><mo>≃</mo><mn>0</mn><mo>.</mo><mn>73</mn></math></p><p>Since, <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn><mo>.</mo><mn>25</mn><mo><</mo><mn>0</mn><mo>.</mo><mn>73</mn></math></p><p>Therefore, the particle returns to <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub></math></p><p>Hence, option C is correct.</p><p>For option D,</p><p>For any value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi><mo>></mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mi>z</mi></mfrac><mo><</mo><mn>1</mn></math></p><p>Hence, option D is correct.</p></div> | MarksBatch1_P1.db |
111 | a-double-star-system-consists-of-two-stars-a-and-b-which-have-time-period-t-a-and-t-b-radius-r-a-and-r-b-and-mass-m-a-and-m-b-choose-the-correct-optio | a-double-star-system-consists-of-two-stars-a-and-b-which-have-time-period-t-a-and-t-b-radius-r-a-and-r-b-and-mass-m-a-and-m-b-choose-the-correct-optio-82746 | <div class="question">A double star system consists of two stars $A$ and $B$ which have time period $T_A$ and $T_B$. Radius $R_A$ and $R_B$ and mass $M_A$ and $M_B$. Choose the correct option.</div> | ['Physics', 'Gravitation', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>If $T_A>T_B$, then $R_A>R_B$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>If $T_A>T_B$, then $M_A>M_B$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\left(\frac{T_A}{T_B}\right)^2=\left(\frac{R_A}{R_B}\right)^3$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$T_A=T_B$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$T_A=T_B$</span> </div> | <div class="solution">In case of binary star system angular velocity and hence the time period of both the stars are equal.</div> | MarksBatch1_P1.db |
112 | a-few-electric-field-lines-for-a-system-of-two-charges-q-1-and-q-2-fixed-at-two-different-points-on-the-x-axis-are-shown-in-the-figure-these-lines-sug | a-few-electric-field-lines-for-a-system-of-two-charges-q-1-and-q-2-fixed-at-two-different-points-on-the-x-axis-are-shown-in-the-figure-these-lines-sug-34862 | <div class="question">A few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $x$-axis are shown in the figure. These lines suggest that<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/R5lzBNcidnS01bSprkXrbMp3l9bUBkdMt6YycbuULb8.original.fullsize.png"/><br/></div> | ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\left|Q_1\right|>\left|Q_2\right|$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\left|Q_1\right| < \left|Q_2\right|$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>at a finite distance to the left of $Q_1$ the electric field is zero<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>at a finite distance to the right of $Q_2$ the electric field is zero</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$\left|Q_1\right|>\left|Q_2\right|$<br/>, <br/>at a finite distance to the right of $Q_2$ the electric field is zero</span> </div> | <div class="solution">From the behaviour of electric lines, we can say that $Q_1$ is positive and $Q_2$ is negative. Further, $\left|Q_1\right|>\left|Q_2\right|$<br/>At some finite distance to the right of $Q_2$, electric field will be zero. Because electric field due to $Q_1$ is towards right (away from $Q_1$ ) and due to $Q_2$ is towards left (towards $Q_2$ ). But since magnitude of $Q_1$ is more, the two fields may cancel each other because distance of that point from $Q_1$ will also be more.<br/>$\therefore$ The correct options are (a) and (d).</div> | MarksBatch1_P1.db |
113 | a-field-line-is-shown-in-the-figure-this-field-cannot-represent | a-field-line-is-shown-in-the-figure-this-field-cannot-represent-40164 | <div class="question">A field line is shown in the figure. This field cannot represent<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/sBuAndXrVPaCS8wD88iMYjym52y85QgiPtevdojilzs.original.fullsize.png"/><br/></div> | ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>Magnetic field<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>Electrostatic field<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>Induced electric field<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>Gravitational field</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>Magnetic field<br/>, <br/>Gravitational field</span> </div> | <div class="solution">Electrostatic and gravitational field do not make closed loops.</div> | MarksBatch1_P1.db |
114 | a-gas-described-by-van-der-waals-equation | a-gas-described-by-van-der-waals-equation-24428 | <div class="question">A gas described by van der Waals' equation</div> | ['Chemistry', 'States of Matter', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>behaves similar to an ideal gas in the limit of large molar volumes<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>behaves similar to an ideal gas in the limit of large pressures<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>is characterised by van der Waals' coefficients that are dependent on the identity of the gas but are independent of the temperature<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>has the pressure that is lower than the pressure exerted by the same gas behaving ideally</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>behaves similar to an ideal gas in the limit of large pressures<br/>, <br/>is characterised by van der Waals' coefficients that are dependent on the identity of the gas but are independent of the temperature<br/>, <br/>has the pressure that is lower than the pressure exerted by the same gas behaving ideally</span> </div> | <div class="solution">The gas described by van der Waals' equation in the limit of large molar volume, in the intermediate range of pressure experience balancing by two factors.<br/>$$<br/>\left(p+\frac{a}{V^2}\right)(V-b)=R T \Rightarrow p V=\frac{a}{V}-p b+\frac{a b}{V^2}=R T<br/>$$<br/>(We can neglect the product of two very small constants, $\frac{a b}{V^2}$ )<br/>So,<br/>$$<br/>p V+\frac{a}{V}-p b=R T<br/>$$<br/>As<br/>$$<br/>\frac{a}{V}=p b<br/>$$<br/>The gas behaves ideally, $p V=R T$<br/>- The van der Waals' constants ' $a$ ' and ' $b$ ' are characteristics of the gas. The value of these constants are determined by the critical constants of the gas<br/>$$<br/>a=3 p_C V_C^2=3 p_C\left(\frac{3 R T_C}{8 p_C}\right)^2=\frac{27\left(R T_C\right)^2}{64 p_C} ; b=\frac{1}{3}\left(\frac{3 R T_C}{8 p_C}\right)=\frac{R T_C}{8 p_C}<br/>$$<br/>Question is based on description by van der Waals' equation, that's why we have to ignore the effect of temperature on $a$ and $b$. Actually, the so-called constant varies to some extent with temperature and this shows that the van der Waals' equation is not a complete solution of the behaviour of real gas</div> | MarksBatch1_P1.db |
115 | a-glass-tube-of-uniform-internal-radius-r-has-a-valve-separating-the-two-identical-ends-initially-the-valve-is-in-a-tightly-closed-position-end-1-has- | a-glass-tube-of-uniform-internal-radius-r-has-a-valve-separating-the-two-identical-ends-initially-the-valve-is-in-a-tightly-closed-position-end-1-has-30310 | <div class="question">A glass tube of uniform internal radius $(r)$ has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius $r$. End 2 has sub-hemispherical soap bubble as shown in figure. Just after opening the valve,<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/aTOnsM_ko_rZ4DTOZ1tXO4djJRRDKbR9XCHC2FJmEWk.original.fullsize.png"/><br/></div> | ['Physics', 'Mechanical Properties of Fluids', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>air from end 1 flows towards end 2. No change in the volume of the soap bubbles<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>no change occurs<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>air from end 2 flows towards end 1. Volume of the soap bubble at end 1 increases</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases<br/></span> </div> | <div class="solution">$\Delta p_1=\frac{4 T}{r_1}$ and $\Delta p_2=\frac{4 T}{r_2}$<br/>$$<br/>\begin{array}{cc} <br/>& r_1 < r_2 \\<br/>\therefore & \Delta p_1>\Delta p_2<br/>\end{array}<br/>$$<br/>Air will flow from 1 to 2 and volume of bubble at end- 1 will decrease.<br/>Therefore, correct option is (b).</div> | MarksBatch1_P1.db |
116 | a-hollow-pipe-of-length-08-m-is-closed-at-one-end-at-its-open-end-a-05-m-long-uniform-string-is-vibrating-in-its-second-harmonic-and-it-resonates-with | a-hollow-pipe-of-length-08-m-is-closed-at-one-end-at-its-open-end-a-05-m-long-uniform-string-is-vibrating-in-its-second-harmonic-and-it-resonates-with-43352 | <div class="question">A hollow pipe of length $0.8 \mathrm{~m}$ is closed at one end. At its open end a $0.5 \mathrm{~m}$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50 \mathrm{~N}$ and the speed of sound is $320 \mathrm{~ms}^{-1}$, the mass of the string is</div> | ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$5 \mathrm{~g}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$10 \mathrm{~g}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$20 \mathrm{~g}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$40 \mathrm{~g}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$10 \mathrm{~g}$<br/></span> </div> | <div class="solution">$2\left(\frac{v_1}{2 l_1}\right)=\frac{v_2}{4 l_2}$ $\therefore \quad \frac{\sqrt{T / \mu}}{l_1}=\frac{320}{4 l_2}$<br/>( $\mu=$ mass per unit length of wire)<br/>or $\frac{\sqrt{50 / \mu}}{0.5}=\frac{320}{4 \times 0.8}$<br/>Solving we get $\mu=0.02 \frac{\mathrm{kg}}{\mathrm{m}}=20 \frac{\mathrm{g}}{\mathrm{m}}$<br/>$\therefore$ Mass of string<br/>$$<br/>=\left(20 \frac{\mathrm{g}}{\mathrm{m}}\right)(0.5 \mathrm{~m})=10 \mathrm{~g}<br/>$$<br/>$\therefore$ The correct option is (b).</div> | MarksBatch1_P1.db |
117 | a-horizontal-force-f-is-applied-at-the-centre-of-mass-of-a-cylindrical-object-of-mass-m-and-radius-r-perpendicular-to-its-axis-as-shown-in-the-figure- | a-horizontal-force-f-is-applied-at-the-centre-of-mass-of-a-cylindrical-object-of-mass-m-and-radius-r-perpendicular-to-its-axis-as-shown-in-the-figure-39442 | <div class="question"><p>A horizontal force <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi></math> is applied at the centre of mass of a cylindrical object of mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> and radius <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math>, perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>μ</mi></math>. The centre of mass of the object has an acceleration <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math>. The acceleration due to gravity is <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>g</mi></math>. Given that the object rolls without slipping, which of the following statement(s) is (are) correct?</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/c40aaf47-970f-42c6-a45c-5b3be0c39e9a-image.png" style="width: 200px; height: 177px;"/></p></div> | ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2021 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">For the same <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi></math>, the value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math> does not depend on whether the cylinder is solid or hollow.</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">For a solid cylinder, the maximum possible value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mi>μ</mi><mi>g</mi></math>.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">The magnitude of the frictional force on the object due to the ground is always <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>μ</mi><mi>m</mi><mi>g</mi></math>.</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">For a thin-walled hollow cylinder, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi><mo>=</mo><mfrac><mi>F</mi><mrow><mn>2</mn><mi>m</mi></mrow></mfrac></math>.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value">For a solid cylinder, the maximum possible value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mi>μ</mi><mi>g</mi></math>., For a thin-walled hollow cylinder, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi><mo>=</mo><mfrac><mi>F</mi><mrow><mn>2</mn><mi>m</mi></mrow></mfrac></math>.</span> </div> | <div class="solution"><p>Let moment of inertia <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mi>I</mi></math></p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/f4eb0c6e-932c-407c-b3e4-d033856c2fa9-758f5244-8b39-44c0-b24d-ebcc14e0b238-image.png" style="width: 250px; height: 256px;"/></p><p>For pure rolling, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi><mo>=</mo><mi>α</mi><mi>R</mi></math></p><p>and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi><mo>−</mo><mi>f</mi><mo>=</mo><mi>m</mi><mi>a</mi><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mn>1</mn></mfenced></math></p><p>and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mi>R</mi><mo>=</mo><mi>I</mi><mi>α</mi></math></p><p>or <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mi>R</mi><mo>=</mo><mfrac><mrow><mi>I</mi><mi>a</mi></mrow><mi>R</mi></mfrac></math></p><p>or <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>=</mo><mfrac><mrow><mi>I</mi><mi>a</mi></mrow><msup><mi>R</mi><mn>2</mn></msup></mfrac></math></p><p>Therefore, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi><mo>−</mo><mfrac><mrow><mi>I</mi><mi>a</mi></mrow><msup><mi>R</mi><mn>2</mn></msup></mfrac><mo>=</mo><mi>m</mi><mi>a</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mo> </mo><mo> </mo><mi>F</mi><mo>=</mo><mfenced><mrow><mi>m</mi><mo>+</mo><mfrac><mi>I</mi><msup><mi>R</mi><mn>2</mn></msup></mfrac></mrow></mfenced><mi>a</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo></math> For the same value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi></math>, value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math> depends on <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>I</mi></math>.</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo></math> For a solid cylinder<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><mi>I</mi><mo>=</mo><mfrac><mrow><mi>m</mi><msup><mi>R</mi><mn>2</mn></msup></mrow><mn>2</mn></mfrac></mrow></mfenced></math> (to calculate maximum <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math>)</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>=</mo><mi>μ</mi><mi>m</mi><mi>g</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mi>μ</mi><mi>m</mi><mi>g</mi><mo>=</mo><mfenced><mfrac><mrow><mi>m</mi><msup><mi>R</mi><mn>2</mn></msup></mrow><mn>2</mn></mfrac></mfenced><mo>×</mo><mfrac><mi>a</mi><msup><mi>R</mi><mn>2</mn></msup></mfrac></math></p><p>Therefore, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi><mo>=</mo><mn>2</mn><mi>μ</mi><mi>g</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>f</mi></math> is dependent on applied force.</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo></math> For thin-walled hollow cylinder<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><mi>I</mi><mo>=</mo><mi>m</mi><msup><mi>R</mi><mn>2</mn></msup></mrow></mfenced></math>,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi><mo>=</mo><mfenced><mrow><mi>m</mi><mo>+</mo><mfrac><mrow><mi>m</mi><msup><mi>R</mi><mn>2</mn></msup></mrow><msup><mi>R</mi><mn>2</mn></msup></mfrac></mrow></mfenced><mi>a</mi><mo>=</mo><mn>2</mn><mi>m</mi><mi>a</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mo> </mo><mi>a</mi><mo>=</mo><mfrac><mi>F</mi><mrow><mn>2</mn><mi>m</mi></mrow></mfrac></math> </p></div> | MarksBatch1_P1.db |
118 | a-hyperbola-having-the-transverse-axis-of-length-2-sin-is-confocal-with-the-ellipse-3-x-2-4-y-2-12-then-its-equation-is | a-hyperbola-having-the-transverse-axis-of-length-2-sin-is-confocal-with-the-ellipse-3-x-2-4-y-2-12-then-its-equation-is-14840 | <div class="question">A hyperbola, having the transverse axis of length $2 \sin \theta$, is confocal with the ellipse $3 x^2+4 y^2=12$. Then, its equation is</div> | ['Mathematics', 'Ellipse', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$x^2 \sec ^2 \theta-y^2 \operatorname{cosec}^2 \theta=1$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$x^2 \sin ^2 \theta-y^2 \cos ^2 \theta=1$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$x^2 \cos ^2 \theta-y^2 \sin ^2 \theta=1$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$<br/></span> </div> | <div class="solution">The given ellipse is<br/>$$<br/>\begin{aligned}<br/>& \frac{x^2}{4}+\frac{y^2}{3}=1 \Rightarrow \alpha=2, \beta=\sqrt{3} \\<br/>& \therefore \quad 3=4\left(1-e^2\right) \Rightarrow e=\frac{1}{2} \\<br/>&<br/>\end{aligned}<br/>$$<br/>$$<br/>\therefore \quad a e=1<br/>$$<br/>Hence the eccentricity $e_1$, of the hyperbola is given by<br/>$$<br/>\begin{array}{rlrl} <br/>& & 1 & =e_1 \sin \theta \\<br/>\Rightarrow & & e_1 & =\operatorname{cosec} \theta \\<br/>\Rightarrow & b^2 & =\sin ^2 \theta\left(\operatorname{cosec}^2 \theta-1\right)=\cos ^2 \theta<br/>\end{array}<br/>$$<br/>Hence, the hyperbola is<br/>$$<br/>\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1 \text { or } x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1<br/>$$</div> | MarksBatch1_P1.db |
119 | a-lamina-is-made-by-removing-a-small-disc-of-diameter-2-r-from-a-bigger-disc-of-uniform-mass-density-and-radius-2-r-as-shown-in-the-figure-the-moment--1 | a-lamina-is-made-by-removing-a-small-disc-of-diameter-2-r-from-a-bigger-disc-of-uniform-mass-density-and-radius-2-r-as-shown-in-the-figure-the-moment-1-30622 | <div class="question">A lamina is made by removing a small disc of diameter $2 R$ from a bigger disc of uniform mass density and radius $2 R$, as shown in the figure. The moment of inertia of this lamina about axes passing though $O$ and $P$ is $I_{O}$ and $I_{P}$ respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $I_{P} / I_{O}$ to the nearest integer is
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/FCGosuZSs9bvX9TAOyMX0Db910xOh_xEi9Gnk8hO45Q.original.fullsize.png"/></div> | ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">3</span> </div> | <div class="solution">Let $\sigma$ be the surface mass density.
<br/> <br/>Moment of inertia of the lamina about axes passing through 'O'
<br/> <br/>$\begin{array}{l}
<br/> <br/>I_{O}=\frac{1}{2} \sigma\left[\pi(2 R)^{2}\right] \times(2 R)^{2}- \\
<br/> <br/>=\frac{13}{2} \pi \sigma R^{4} \quad\left[\frac{1}{2}\left(\sigma \pi R^{2}\right)^{2}+\sigma\left(\pi R^{2}\right) \times R^{2}\right]
<br/> <br/>\end{array}$
<br/> <br/>Moment of inertia of the lamina about axes passing through 'P'
<br/> <br/>$\begin{array}{l}
<br/> <br/>I_{P}=8 \pi \sigma \mathrm{R}^{4}+\sigma \pi(2 \mathrm{R})^{2} \times(2 \mathrm{R})^{2} \\
<br/> <br/>{\left[\frac{1}{2} \sigma\left(\pi R^{2}\right) R^{2}+\sigma\left(\pi R^{2}\right)\left(\sqrt{(2 R)^{2}+R^{2}}\right)^{2}\right]}
<br/> <br/>\end{array}$
<br/> <br/>$\begin{array}{l}
<br/> <br/>=24 \pi \sigma \mathrm{R}^{4}-5.5 \sigma \pi \mathrm{R}^{4}=18.5 \pi \sigma \mathrm{R}^{4} \\
<br/> <br/>\therefore \quad \frac{I_{P}}{I_{O}}=\frac{18.5 \pi \sigma R^{4}}{\frac{13}{2} \pi \sigma R^{4}}=\frac{37}{13} \approx 3
<br/> <br/>\end{array}$</div> | MarksBatch1_P1.db |
120 | a-large-glass-slab-3-5-of-thickness-8-cm-is-placed-over-a-point-source-of-light-on-a-plane-surface-it-is-seen-that-light-emerges-out-of-the-top-surfac | a-large-glass-slab-3-5-of-thickness-8-cm-is-placed-over-a-point-source-of-light-on-a-plane-surface-it-is-seen-that-light-emerges-out-of-the-top-surfac-12503 | <div class="question">A large glass slab $\left(\mu=\frac{5}{3}\right)$ of thickness 8 $\mathrm{cm}$ is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius $R \mathrm{~cm}$. What is the value of $R$ ?</div> | ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">6</span> </div> | <div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/36cY2ycowgTDhPchNs-SiYyjriRvKBG1N-mYocUGmvo.original.fullsize.png"/><br/><br/>$$<br/>\frac{R}{t}=\tan \theta_C<br/>$$<br/>or $\quad R=t\left(\tan \theta_C\right)$<br/>$$<br/>\begin{aligned}<br/>& \text { But, } \quad \sin \theta_C=\frac{1}{\mu}=\frac{3}{5} \\<br/>& \therefore \quad \tan \theta_C=\frac{3}{4} \\<br/>& \therefore \quad R=\frac{3}{4} t=\frac{3}{4}(8 \mathrm{~cm})=6 \mathrm{~cm} \\<br/>&<br/>\end{aligned}<br/>$$<br/>Hence the answer is 6 .</div> | MarksBatch1_P1.db |
121 | a-light-beam-is-travelling-from-region-i-to-region-iv-refer-figure-the-refractive-index-in-region-i-ii-iii-and-iv-are-n-0-2-n-0-6-n-0-and-8-n-0-respec | a-light-beam-is-travelling-from-region-i-to-region-iv-refer-figure-the-refractive-index-in-region-i-ii-iii-and-iv-are-n-0-2-n-0-6-n-0-and-8-n-0-respec-21573 | <div class="question">A light beam is travelling from Region I to Region IV (refer figure). The refractive index in Region I, II, III and IV are $n_0, \frac{n_0}{2}, \frac{n_0}{6}$ and $\frac{n_0}{8}$, respectively. The angle of incidence $\theta$ for which the beam just misses entering Region IV is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/XM7-nvO7wvCxwFQBPcgcs6H5x6K1A6HAofcbqGfpC6g.original.fullsize.png"/><br/></div> | ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\sin ^{-1}\left(\frac{3}{4}\right)$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\sin ^{-1}\left(\frac{1}{8}\right)$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\sin ^{-1}\left(\frac{1}{4}\right)$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\sin ^{-1}\left(\frac{1}{3}\right)$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\sin ^{-1}\left(\frac{1}{8}\right)$<br/></span> </div> | <div class="solution">Critical angle from region III to region IV<br/>$$<br/>\sin \theta_c=\frac{n_0 / 8}{n_0 / 6}=\frac{3}{4}<br/>$$<br/>Now applying Snell's law in region I and region III :<br/>$$<br/>n_0 \sin \theta=\frac{n_0}{6} \sin \theta_c<br/>$$<br/>or<br/>$$<br/>\begin{aligned}<br/>\sin \theta & =\frac{1}{6} \sin \theta_c=\frac{1}{6}\left(\frac{3}{4}\right)=\frac{1}{8} \\<br/>\theta & =\sin ^{-1}\left(\frac{1}{8}\right)<br/>\end{aligned}<br/>$$<br/>$\therefore$ correct option is (b).</div> | MarksBatch1_P1.db |
122 | a-light-inextensible-string-that-goes-over-a-smooth-fixed-pulley-as-shown-in-the-figure-connects-two-blocks-of-masses-036-kg-and-072-kg-taking-g-10-ms-1 | a-light-inextensible-string-that-goes-over-a-smooth-fixed-pulley-as-shown-in-the-figure-connects-two-blocks-of-masses-036-kg-and-072-kg-taking-g-10-ms-1-30944 | <div class="question">A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36 \mathrm{~kg}$ and $0.72 \mathrm{~kg}$. Taking $g=10 \mathrm{~ms}^{-2}$, find thework done (in Joule) by string on the block of mass $0.36 \mathrm{~kg}$ during the first second after the system is released from rest.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/WKlYQVtpL-uxteeE1iYredzRLzJ0_pwkpH4jdbu3QDI.original.fullsize.png"/><br/></div> | ['Physics', 'Work Power Energy', 'JEE Main'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">8</span> </div> | <div class="solution">$$<br/>\text { } \begin{aligned}<br/>\text { Now, } B & =\frac{\mu_0}{4 \pi} \frac{I}{12 x / 5}\left[\sin 37^{\circ}+\sin 53^{\circ}\right] \\<br/>& =7\left(\frac{\mu_0 I}{48 \pi x}\right) \\<br/>a & =\frac{\text { Net pulling force }}{\text { Total mass }} \\<br/>& =\frac{0.72 g-0.36 g}{0.72+0.36}=\frac{g}{3} \\<br/>s & =\frac{1}{2} a t^2=\frac{1}{2}\left(\frac{g}{3}\right)(1)^2=\frac{g}{6}<br/>\end{aligned}<br/>$$<br/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/BeQguzyOCEjbb25-f_J4-t2aGHS-oU6VRYL-wWQF-Ys.original.fullsize.png"/><br/><br/><br/>$$<br/>\begin{aligned}<br/>& T-0.36 g=0.36 a=0.36 \frac{g}{3} \\<br/>& \therefore \quad T=0.48 g \\<br/>& \text { Now, } W_T=T S \cos 0^{\circ} \text { (on } 3.6 \mathrm{~kg} \text { mass) } \\<br/>& =(0.48 g)\left(\frac{g}{6}\right)(1)=0.08\left(g^2\right) \\<br/>& =0.08(10)^2=8 \mathrm{~J} \\<br/>&<br/>\end{aligned}<br/>$$</div> | MarksBatch1_P1.db |
123 | a-light-inextensible-string-that-goes-over-a-smooth-fixed-pulley-as-shown-in-the-figure-connects-two-blocks-of-masses-036-kg-and-072-kg-taking-g-10-ms | a-light-inextensible-string-that-goes-over-a-smooth-fixed-pulley-as-shown-in-the-figure-connects-two-blocks-of-masses-036-kg-and-072-kg-taking-g-10-ms-66985 | <div class="question">A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36 \mathrm{~kg}$ and $0.72 \mathrm{~kg}$. Taking $g=10 \mathrm{~ms}^{-2}$, find thework done (in Joule) by string on the block of mass $0.36 \mathrm{~kg}$ during the first second after the system is released from rest.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/WKlYQVtpL-uxteeE1iYredzRLzJ0_pwkpH4jdbu3QDI.original.fullsize.png"/><br/></div> | ['Physics', 'Work Power Energy', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">8</span> </div> | <div class="solution">$$<br/>\text { } \begin{aligned}<br/>\text { Now, } B & =\frac{\mu_0}{4 \pi} \frac{I}{12 x / 5}\left[\sin 37^{\circ}+\sin 53^{\circ}\right] \\<br/>& =7\left(\frac{\mu_0 I}{48 \pi x}\right) \\<br/>a & =\frac{\text { Net pulling force }}{\text { Total mass }} \\<br/>& =\frac{0.72 g-0.36 g}{0.72+0.36}=\frac{g}{3} \\<br/>s & =\frac{1}{2} a t^2=\frac{1}{2}\left(\frac{g}{3}\right)(1)^2=\frac{g}{6}<br/>\end{aligned}<br/>$$<br/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/BeQguzyOCEjbb25-f_J4-t2aGHS-oU6VRYL-wWQF-Ys.original.fullsize.png"/><br/><br/><br/>$$<br/>\begin{aligned}<br/>& T-0.36 g=0.36 a=0.36 \frac{g}{3} \\<br/>& \therefore \quad T=0.48 g \\<br/>& \text { Now, } W_T=T S \cos 0^{\circ} \text { (on } 3.6 \mathrm{~kg} \text { mass) } \\<br/>& =(0.48 g)\left(\frac{g}{6}\right)(1)=0.08\left(g^2\right) \\<br/>& =0.08(10)^2=8 \mathrm{~J} \\<br/>&<br/>\end{aligned}<br/>$$</div> | MarksBatch1_P1.db |
124 | a-light-ray-travelling-in-glass-medium-is-incident-on-glassair-interface-at-an-angle-of-incidence-the-reflected-r-and-transmitted-t-intensities-both-a | a-light-ray-travelling-in-glass-medium-is-incident-on-glassair-interface-at-an-angle-of-incidence-the-reflected-r-and-transmitted-t-intensities-both-a-61328 | <div class="question">A light ray travelling in glass medium is incident on glass-air interface at an angle of incidence $\theta$. The reflected $(R)$ and transmitted $(T)$ intensities, both as function of $\theta$, are plotted. The correct sketch is</div> | ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/8-2OElXNEWowMDwk6QM0zKc2EYtOtP1JZE_EFnzTF0I.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/MRxmX1Tor6LkgA33aw9FVk8crkAbKnw_SgDn1-nDFDs.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OMaB27Dwg6Rl0ATVY-Af7D0a69q79QdneMKQ989T0f4.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/2SrQ0m4NG6zXfh1kZhFGBFx5biIvPp9IOPH4zj2V_6s.original.fullsize.png"/><br/></span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OMaB27Dwg6Rl0ATVY-Af7D0a69q79QdneMKQ989T0f4.original.fullsize.png"/><br/></span> </div> | <div class="solution">After critical angle reflection will be $100 \%$ and transmission is $0 \%$. Options (b) and (c) satisfy this condition. But option (c) is the correct option. Because in option (b) transmission is given $100 \%$ at $\theta=0^{\circ}$, which is not true.<br/>Analysis of Question<br/>From examination point of view question is moderately difficult. Because it takes little time to understand the given graphs clearly.</div> | MarksBatch1_P1.db |
125 | a-line-with-positive-direction-cosines-passes-through-the-point-p-2-1-2-and-makes-equal-angles-with-the-coordinate-axes-the-line-meets-the-plane-2-x-y | a-line-with-positive-direction-cosines-passes-through-the-point-p-2-1-2-and-makes-equal-angles-with-the-coordinate-axes-the-line-meets-the-plane-2-x-y-26191 | <div class="question">A line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2 x+y+z=9$ at point $Q$. The length of the line segment $P Q$ equals</div> | ['Mathematics', 'Three Dimensional Geometry', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>1<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\sqrt{2}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\sqrt{3}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>2</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\sqrt{3}$<br/></span> </div> | <div class="solution">$$<br/>\text { Since, } l=m=n=\frac{1}{\sqrt{3}}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/ich1orC4Q3FFYIKtKIixDQa3M4swfCBvT6uaMCex8bo.original.fullsize.png"/><br/><br/><br/>$\therefore$ Equation of line are<br/>$$<br/>\begin{gathered}<br/>\frac{x-2}{1 / \sqrt{3}}=\frac{y+1}{1 / \sqrt{3}}=\frac{z-2}{1 / \sqrt{3}} \\<br/>\Rightarrow \quad x-2=y+1=z-2=r \text { (say) }<br/>\end{gathered}<br/>$$<br/>$\therefore$ Any point on the line is<br/>$$<br/>Q \equiv(r+2 r-1, r+2)<br/>$$<br/>$\because Q$ lies on the plane $2 x+y+z=9$<br/>$$<br/>\begin{aligned}<br/>& \therefore \quad 2(r+2)+(r-1)+(r+2)=9 \\<br/>& \Rightarrow 4 r+5=9 \Rightarrow r=1 \Rightarrow Q(3,0,3) \\<br/>& \therefore P Q=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2} \\<br/>& =\sqrt{3}<br/>\end{aligned}<br/>$$</div> | MarksBatch1_P1.db |
126 | a-long-circular-tube-of-length-10-m-and-radius-03-m-carries-a-current-i-along-its-curved-surface-as-shown-a-wire-loop-of-resistance-0005-and-of-radius | a-long-circular-tube-of-length-10-m-and-radius-03-m-carries-a-current-i-along-its-curved-surface-as-shown-a-wire-loop-of-resistance-0005-and-of-radius-87473 | <div class="question">A long circular tube of length $10 \mathrm{~m}$ and radius $0.3 \mathrm{~m}$ carries a current I along its curved surface as shown. A wire loop of resistance $0.005 \Omega$ and of radius $0.1 \mathrm{~m}$ is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as $I=I_0 \cos 300 t$, where $I_0$ is constant. If the magnetic moment of the loop is $N \mu_0 I_0 \sin 300 t$, then $N$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/CbAZwTPq4oMzOvE4BMMDId0ZYqPEcVqy6u9nDKdL4lA.original.fullsize.png"/><br/></div> | ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">6</span> </div> | <div class="solution">Take the circular tube as a long solenoid. The wires are closely wound. Magnetic field inside the solenoid is<br/>$$<br/>B=\mu_0 n i<br/>$$<br/>Here, $n=$ number of turns per unit length<br/>$\therefore n i=$ current per unit length<br/>In the given problem,<br/><br/>$$<br/>\begin{aligned}<br/>& n i=\frac{I}{L} \\<br/>\therefore \quad B & =\frac{\mu_0 I}{L}<br/>\end{aligned}<br/>$$<br/>Flux passing through the circular coil is<br/>$$<br/>\phi=B S=\left(\frac{\mu_0 I}{L}\right)\left(\pi r^2\right)<br/>$$<br/>Induced emf, $e=-\frac{d \phi}{d t}=-\left(\frac{\mu_0 \pi r^2}{L R}\right) \cdot \frac{d I}{d t}$<br/>Induced current,<br/>$$<br/>i=\frac{e}{R}=-\left(\frac{\mu_0 \pi r^2}{L R}\right) \frac{d I}{d t}<br/>$$<br/>Magnetic moment $i A=i \pi r^2$<br/>or $\quad M=-\left(\frac{\mu_0 \pi^2 r^4}{L R}\right) \cdot \frac{d I}{d t}$<br/>Given, $\quad I=I_0 \cos 300 t$<br/>$$<br/>\therefore \quad \frac{d I}{d t}=-300 I_0 \sin (300 t)<br/>$$<br/><br/>Substituting in Eq. (i), we get<br/>$$<br/>\begin{aligned}<br/>M & =\left(\frac{300 \pi^2 r^4}{L R}\right) \mu_0 I_0 \sin 300 t \\<br/>\therefore \quad N & =\frac{300 \pi^2 r^4}{L R}<br/>\end{aligned}<br/>$$<br/>Substituting the values, we get<br/>$$<br/>\begin{aligned}<br/>N & =\frac{300(22 / 7)^2(0.1)^4}{(10)(0.005)} \\<br/>& =5.926<br/>\end{aligned}<br/>$$<br/>or $\quad N \simeq 6$<br/>Analysis of Question<br/>Question is difficult to understand inside the examination hall. But 5 to $10 \%$ question in IIT JEE are always difficult. Students should not panic. Because topper of IIT JEE scores approximately $80-90 \%$.</div> | MarksBatch1_P1.db |
127 | a-long-hollow-conducting-cylinder-is-kept-coaxially-inside-another-long-hollow-conducting-cylinder-of-larger-radius-both-the-cylinders-are-initially-e | a-long-hollow-conducting-cylinder-is-kept-coaxially-inside-another-long-hollow-conducting-cylinder-of-larger-radius-both-the-cylinders-are-initially-e-61563 | <div class="question">A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral.</div> | ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>a potential difference appears between the two cylinders when a charge density is given to the inner cylinder<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>a potential difference appears between the two cylinders when a charge density is given to the outer cylinder<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>no potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>no potential difference appears between the two cylinders when same charge density is given to both the cylinders</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>a potential difference appears between the two cylinders when a charge density is given to the inner cylinder<br/></span> </div> | <div class="solution">There will be an electric field between two cylinders (using Gauss theorem. This electric field will produce a potential difference.<br/>$\therefore$ Answer is (a).</div> | MarksBatch1_P1.db |
128 | a-long-insulated-copper-wire-is-closely-wound-as-a-spiral-of-n-turns-the-spiral-has-inner-radius-a-and-outer-radius-b-the-spiral-lies-in-the-xyplane-a | a-long-insulated-copper-wire-is-closely-wound-as-a-spiral-of-n-turns-the-spiral-has-inner-radius-a-and-outer-radius-b-the-spiral-lies-in-the-xyplane-a-37403 | <div class="question">A long insulated copper wire is closely wound as a spiral of $N$ turns. The spiral has inner radius $a$ and outer radius $b$. The spiral lies in the XY-plane and a steady current $I$ flows through the wire. The Z-component of the magnetic field at the centre of the spiral is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/WmycTUrrVf7IKN4c5VBT4TB4PBxzkeLl7dO0m0tVDr4.original.fullsize.png"/><br/></div> | ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/sUooC5bCiYfSxQLUrgtSEhqOIKhrcCw8OvgH71nhc30.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OkdpbkPeRzs32HIE0PC_5k62_Xe10HO1oaGyCgqX0PU.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/6aA8dsRAmibaeIeEGBWhgL5Hqp3JSZHy07T-JtrFE_4.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/GOAVi2UUHeYkKsJUWy0MiygIxqx6Ss8UyFnYWKY_jfk.original.fullsize.png"/><br/></span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/sUooC5bCiYfSxQLUrgtSEhqOIKhrcCw8OvgH71nhc30.original.fullsize.png"/><br/></span> </div> | <div class="solution">If we take a small strip of $d r$ at distancer from centre, then number of turns in this strip would be<br/>$$<br/>d N=\frac{N}{b-a} d r<br/>$$<br/>Magnetic field due to this element at the centre of the coil will be<br/>$$<br/>\begin{aligned}<br/>& d B=\frac{\mu_0(d N) I}{2 r}=\frac{\mu_0 N I}{2(b-a)} \frac{d r}{r} \\<br/>\therefore \quad & B=\int_{r=a}^{r=b} d B=\frac{\mu_0 N I}{2(b-a)} \ln \frac{b}{a}<br/>\end{aligned}<br/>$$<br/>$\therefore$ Correct answer is (a).<br/>Analysis of Question<br/>(i) If we see this problem independently, then I will rate this question moderately difficult. But the idea of this question is taken from question number $3.245$ of IE Irodov.<br/>(ii) Interestingly the same question was asked in IIT-JEE 2001 also.</div> | MarksBatch1_P1.db |
129 | a-long-straight-wire-carries-a-current-i-2-ampere-a-semicircular-conducting-rod-is-placed-beside-it-on-two-conducting-parallel-rails-of-negligible-res | a-long-straight-wire-carries-a-current-i-2-ampere-a-semicircular-conducting-rod-is-placed-beside-it-on-two-conducting-parallel-rails-of-negligible-res-11416 | <div class="question"><p>A long straight wire carries a current, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>I</mi><mo>=</mo><mn>2</mn></math> ampere. A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are parallel to the wire. The wire, the rod and the rails lie in the same horizontal plane, as shown in the figure. Two ends of the semi-circular rod are at distances <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo> </mo><mi>cm</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>4</mn><mo> </mo><mi>cm</mi></math> from the wire. At time <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mo>=</mo><mn>0</mn></math>, the rod starts moving on the rails with a speed <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi><mo>=</mo><mn>3</mn><mo>.</mo><mn>0</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> (see the figure).</p><p>A resistor <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi><mo>=</mo><mn>1.4</mn><mo> </mo><mi mathvariant="normal">Ω</mi></math> and a capacitor <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>0</mn></msub><mo>=</mo><mn>5.0</mn><mo> </mo><mi>μF</mi></math> are connected in series between the rails. At time <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mo>=</mo><mn>0</mn></math>, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>0</mn></msub></math> is uncharged. Which of the following statement(s) is (are) correct? [<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>μ</mi><mn>0</mn></msub><mo>=</mo><mn>4</mn><mi>π</mi><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>7</mn></mrow></msup></math> SI units. Take <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ln</mi><mn>2</mn><mo>=</mo><mn>0.7</mn></math>]</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/b7441289-f46c-471e-9ace-bb7533e4ce9e-image.png" style="width: 200px; height: 275px;"/></p></div> | ['Physics', 'Electromagnetic Induction', 'JEE Advanced', 'JEE Advanced 2021 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">Maximum current through <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1.2</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>6</mn></mrow></msup></math> ampere</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">Maximum current through <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3.8</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>6</mn></mrow></msup></math> ampere.</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">Maximum charge on capacitor <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>0</mn></msub></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>8.4</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>12</mn></mrow></msup></math> coulomb.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">Maximum charge on capacitor <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>0</mn></msub></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mo>.4</mo><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>12</mn></mrow></msup></math> coulomb.</span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value">Maximum current through <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1.2</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>6</mn></mrow></msup></math> ampere, Maximum charge on capacitor <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>0</mn></msub></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>8.4</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>12</mn></mrow></msup></math> coulomb.</span> </div> | <div class="solution"><p>Magnetic field due the long wire, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>B</mi><mo>=</mo><mfrac><mrow><msub><mi>μ</mi><mn>0</mn></msub><mi>I</mi></mrow><mrow><mn>2</mn><mi>π</mi><mi>x</mi></mrow></mfrac></math></p><p>Motional emf in small element, </p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">d</mi><mi>ε</mi><mo>=</mo><mi>B</mi><mi>v</mi><mo>d</mo><mi>x</mi><mo>=</mo><mfrac><mrow><mn>3</mn><msub><mi>μ</mi><mn>0</mn></msub><mfenced><mn>2</mn></mfenced></mrow><mrow><mn>2</mn><mi>π</mi><mi>x</mi></mrow></mfrac><mi mathvariant="normal">d</mi><mi>x</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>ε</mi><mo>=</mo><mfrac><mrow><mn>3</mn><msub><mi>μ</mi><mn>0</mn></msub></mrow><mi>π</mi></mfrac><msubsup><mo>∫</mo><mn>1</mn><mn>4</mn></msubsup><mfrac><mn>1</mn><mi>x</mi></mfrac><mi mathvariant="normal">d</mi><mi>x</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mo> </mo><mi>ε</mi><mo>=</mo><mfrac><mrow><mn>3</mn><msub><mi>μ</mi><mn>0</mn></msub></mrow><mi>π</mi></mfrac><mi>ln</mi><mn>4</mn></math><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>16.8</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>7</mn></mrow></msup></math></p><p>Maximum charge on capacitor, </p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>q</mi><mi>max</mi></msub><mo>=</mo><mi>ε</mi><mi>C</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfenced><mrow><mn>16.8</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>7</mn></mrow></msup></mrow></mfenced><mfenced><mrow><mn>5</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>6</mn></mrow></msup></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>84</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>13</mn></mrow></msup><mo> </mo><mi mathvariant="normal">C</mi><mo>=</mo><mn>8.4</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>12</mn></mrow></msup><mo> </mo><mi mathvariant="normal">C</mi></math></p><p>Maximum current in circuit, </p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>I</mi><mi>max</mi></msub><mo>=</mo><mfrac><mi>ε</mi><mi>R</mi></mfrac><mo>=</mo><mfrac><mrow><mn>16.8</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>7</mn></mrow></msup></mrow><mn>1.4</mn></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>1</mn><mo>.</mo><mn>2</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>-</mo><mn>6</mn></mrow></msup><mo> </mo><mi mathvariant="normal">A</mi></math></p></div> | MarksBatch1_P1.db |
130 | a-loop-carrying-current-i-lies-in-the-x-y-plane-as-shown-in-the-figure-the-unit-vector-k-is-coming-out-of-the-plane-of-the-paper-the-magnetic-moment-o | a-loop-carrying-current-i-lies-in-the-x-y-plane-as-shown-in-the-figure-the-unit-vector-k-is-coming-out-of-the-plane-of-the-paper-the-magnetic-moment-o-12791 | <div class="question">A loop carrying current $I$ lies in the $x-y$ plane as shown in the figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/ZQI4dW5wmYwSBqFfVnVz4R2SgFtX3OlMJYgeeFwFsz8.original.fullsize.png"/></div> | ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$a^{2} I \hat{k}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$-\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$(2 \pi+1) a^{2} I \hat{k}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value">$\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}$</span> </div> | <div class="solution">Magnetic moment of a current carrying loop $\vec{M}=N I \vec{A}$
<br/> <br/>Here $N=1, A=a^{2}+2 \pi\left(\frac{a}{2}\right)^{2}=a^{2}\left[1+\frac{\pi}{2}\right]$
<br/> <br/>From Screw law, direction of $m$ is outward or in +ve z-direction.
<br/> <br/>$\therefore \quad \vec{M}=I a^{2}\left[1+\frac{\pi}{2}\right] \hat{k}$
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/gqv5YWas2u0OrTX0h5hHA_A8CiX4RBcAhWprW1WAXoM.original.fullsize.png"/></div> | MarksBatch1_P1.db |
131 | a-magnetic-field-b-b-0-j-exists-in-the-region-a-x-2-a-and-b-b-0-j-in-the-region-2-a-x-3-a-where-b-0-is-a-positive-constant-a-positive-point-charge-mov-1 | a-magnetic-field-b-b-0-j-exists-in-the-region-a-x-2-a-and-b-b-0-j-in-the-region-2-a-x-3-a-where-b-0-is-a-positive-constant-a-positive-point-charge-mov-1-39438 | <div class="question">A magnetic field $\mathbf{B}=B_0 \hat{\mathbf{j}}$ exists in the region $a < x < 2 a$ and $\mathbf{B}=B_0 \hat{\mathbf{j}}$ in the region $2 a < x < 3 a$, where $B_0$ is a positive constant. A positive point charge moving with a velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/_N5VIaTqGyv5jcrg44Q0Bg2NBIG0WPxXf5MuRx9KtH4.original.fullsize.png"/><br/></div> | ['Physics', 'Magnetic Effects of Current', 'JEE Main'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/CuWOOBYq3jS2YpduXSHKvuhX2J8MwJbhYpKeuhCF9vw.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/mh_9iwlp6wjACA2hwq5AR2fr7YhbhDe2HfcZMMPWs4E.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5DLSkEyTqEIOhN9TjwKWqu3FgulcbE1av2dEBbNN1r0.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/uNBd5cdR_7EE6qPM2BbIMVGhSvFPMwO_gXUHVMsYF48.original.fullsize.png"/><br/></span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/CuWOOBYq3jS2YpduXSHKvuhX2J8MwJbhYpKeuhCF9vw.original.fullsize.png"/><br/></span> </div> | <div class="solution">$\mathbf{F}_m=q(\mathbf{v} \times \mathbf{B})$<br/>Correct option is (a).</div> | MarksBatch1_P1.db |
132 | a-magnetic-field-b-b-0-j-exists-in-the-region-a-x-2-a-and-b-b-0-j-in-the-region-2-a-x-3-a-where-b-0-is-a-positive-constant-a-positive-point-charge-mov | a-magnetic-field-b-b-0-j-exists-in-the-region-a-x-2-a-and-b-b-0-j-in-the-region-2-a-x-3-a-where-b-0-is-a-positive-constant-a-positive-point-charge-mov-65056 | <div class="question">A magnetic field $\mathbf{B}=B_0 \hat{\mathbf{j}}$ exists in the region $a < x < 2 a$ and $\mathbf{B}=B_0 \hat{\mathbf{j}}$ in the region $2 a < x < 3 a$, where $B_0$ is a positive constant. A positive point charge moving with a velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/_N5VIaTqGyv5jcrg44Q0Bg2NBIG0WPxXf5MuRx9KtH4.original.fullsize.png"/><br/></div> | ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/CuWOOBYq3jS2YpduXSHKvuhX2J8MwJbhYpKeuhCF9vw.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/mh_9iwlp6wjACA2hwq5AR2fr7YhbhDe2HfcZMMPWs4E.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5DLSkEyTqEIOhN9TjwKWqu3FgulcbE1av2dEBbNN1r0.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/uNBd5cdR_7EE6qPM2BbIMVGhSvFPMwO_gXUHVMsYF48.original.fullsize.png"/><br/></span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/CuWOOBYq3jS2YpduXSHKvuhX2J8MwJbhYpKeuhCF9vw.original.fullsize.png"/><br/></span> </div> | <div class="solution">$\mathbf{F}_m=q(\mathbf{v} \times \mathbf{B})$<br/>Correct option is (a).</div> | MarksBatch1_P1.db |
133 | a-man-walks-a-distance-of-3-units-from-the-origin-towards-the-northeast-n-4-5-e-direction-from-there-he-walks-a-distance-of-4-units-towards-the-northw | a-man-walks-a-distance-of-3-units-from-the-origin-towards-the-northeast-n-4-5-e-direction-from-there-he-walks-a-distance-of-4-units-towards-the-northw-56920 | <div class="question">A man walks a distance of 3 units from the origin towards the North-East $\left(\mathrm{N} 45^{\circ} \mathrm{E}\right)$ direction. From there, he walks a distance of 4 units towards the North-West ( $\mathrm{N} 45^{\circ} \mathrm{W}$ ) direction to reach a point $P$. Then, the position of $P$ in the Argand plane is</div> | ['Mathematics', 'Complex Number', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$3 e^{i \pi / 4}+4 i$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$(3-4 i) e^{i \pi / 4}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$(4+3 i) e^{i \pi / 4}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$(3+4 i) e^{i \pi / 4}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$(3+4 i) e^{i \pi / 4}$</span> </div> | <div class="solution">Let $O A=3$, so that the complex number associated with $A$ is $3 e^{i \pi / 4}$. If $z$ is the complex number associated with $P$, then<br/>$$<br/>\begin{array}{rlrl} <br/>& & \frac{z-3 e^{i \pi / 4}}{0-3 e^{i \pi / 4}} & =\frac{4}{3} e^{-i \pi / 2}=-\frac{4 i}{3} \\<br/>\Rightarrow & 3 z-9 e^{i \pi / 4} & =12 i^{i \pi / 4} \\<br/>\Rightarrow & z & z & =(3+4 i) e^{i \pi / 4}<br/>\end{array}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/qlN5mJTHFvHbuCdEp3r2OS8iZQJDWNFAAOodCbaNB6Y.original.fullsize.png"/><br/></div> | MarksBatch1_P1.db |
134 | a-massless-rod-b-d-is-suspended-by-two-identical-massless-strings-a-b-and-c-d-of-equal-lengths-a-block-of-mass-m-is-suspended-point-p-such-that-bp-is- | a-massless-rod-b-d-is-suspended-by-two-identical-massless-strings-a-b-and-c-d-of-equal-lengths-a-block-of-mass-m-is-suspended-point-p-such-that-bp-is-87850 | <div class="question">A massless rod $B D$ is suspended by two identical massless strings $A B$ and $C D$ of equal lengths. A block of mass ' $m$ ' is suspended point $P$ such that $B P$ is equal to ' $x$ ', if the fundamental frequency of the left wire is twice the fundamental frequency of right wire, then the value of $x$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/kFubdTzCAp7Sxepzvw02bmX6QUSOvEenKMtjcCm52SA.original.fullsize.png"/><br/></div> | ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$1 / 5$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$1 / 4$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$41 / 5$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$31 / 4$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$1 / 5$<br/></span> </div> | <div class="solution">$f \propto v \propto \sqrt{T}$<br/>Further $\Sigma \tau_p=0$<br/>or<br/>or<br/>$$<br/>f_{A B}=2 f_{C D} \Rightarrow T_{A B}=4 T_{C D}<br/>$$<br/>$$<br/>\begin{aligned}<br/>T_{A B}(x) & =T_{C D}(l-x) \\<br/>4 x & =l-x\left(\text { as } T_{A B}-4 T_{C D}\right) \\<br/>x & =\frac{l}{5}<br/>\end{aligned}<br/>$$</div> | MarksBatch1_P1.db |
135 | a-metal-rod-a-b-of-length-10-x-has-its-one-end-a-in-ice-at-0-c-and-the-other-end-b-in-water-at-10-0-c-if-a-point-p-on-the-rod-is-maintained-at-40-0-c--1 | a-metal-rod-a-b-of-length-10-x-has-its-one-end-a-in-ice-at-0-c-and-the-other-end-b-in-water-at-10-0-c-if-a-point-p-on-the-rod-is-maintained-at-40-0-c-1-32812 | <div class="question">A metal rod $A B$ of length $10 x$ has its one end $A$ in ice at $0^{\circ} \mathrm{C}$ and the other end $B$ in water at $100^{\circ} \mathrm{C}$. If a point $P$ on the rod is maintained at $400^{\circ} \mathrm{C}$, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is $540 \mathrm{calg}^{-1}$ and latent heat of melting of ice is $80 \mathrm{calg}^{-1}$. If the point $P$ is at a distance of $\lambda x$ from the ice end $A$, find the value of $\lambda$. [Neglect any heat loss to the surrounding.]</div> | ['Physics', 'Thermal Properties of Matter', 'JEE Main'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">9</span> </div> | <div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/YfSQoXhjZNvHl8wU_-OPaLyrnErrKwulkUaBdtVyaCo.original.fullsize.png"/><br/><br/><br/>Heat will flow both sides from point $P$.<br/>$$<br/>\begin{aligned}<br/>L_1 \frac{d m_1}{d t} & =\left(\frac{\text { Temperature difference }}{\text { Thermal resistance }}\right)_1 \\<br/>& =\frac{400}{(\lambda x) / k A} \\<br/>L_1 \frac{d m_2}{d t} & =\frac{400-100}{(100-\lambda) x / k A}<br/>\end{aligned}<br/>$$<br/>In about two equations,<br/>$$<br/>\begin{aligned}<br/>& \frac{d m_1}{d t}=\frac{d m_2}{d t} \quad \text { (given) } \\<br/>& L_1=80 \mathrm{calg}^{-1} \text { and } L_2=540 \mathrm{calg}^{-1}<br/>\end{aligned}<br/>$$<br/>Solving these two equations we get $\lambda=9$.</div> | MarksBatch1_P1.db |
136 | a-metal-rod-a-b-of-length-10-x-has-its-one-end-a-in-ice-at-0-c-and-the-other-end-b-in-water-at-10-0-c-if-a-point-p-on-the-rod-is-maintained-at-40-0-c- | a-metal-rod-a-b-of-length-10-x-has-its-one-end-a-in-ice-at-0-c-and-the-other-end-b-in-water-at-10-0-c-if-a-point-p-on-the-rod-is-maintained-at-40-0-c-70467 | <div class="question">A metal rod $A B$ of length $10 x$ has its one end $A$ in ice at $0^{\circ} \mathrm{C}$ and the other end $B$ in water at $100^{\circ} \mathrm{C}$. If a point $P$ on the rod is maintained at $400^{\circ} \mathrm{C}$, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is $540 \mathrm{calg}^{-1}$ and latent heat of melting of ice is $80 \mathrm{calg}^{-1}$. If the point $P$ is at a distance of $\lambda x$ from the ice end $A$, find the value of $\lambda$. [Neglect any heat loss to the surrounding.]</div> | ['Physics', 'Thermal Properties of Matter', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">9</span> </div> | <div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/YfSQoXhjZNvHl8wU_-OPaLyrnErrKwulkUaBdtVyaCo.original.fullsize.png"/><br/><br/><br/>Heat will flow both sides from point $P$.<br/>$$<br/>\begin{aligned}<br/>L_1 \frac{d m_1}{d t} & =\left(\frac{\text { Temperature difference }}{\text { Thermal resistance }}\right)_1 \\<br/>& =\frac{400}{(\lambda x) / k A} \\<br/>L_1 \frac{d m_2}{d t} & =\frac{400-100}{(100-\lambda) x / k A}<br/>\end{aligned}<br/>$$<br/>In about two equations,<br/>$$<br/>\begin{aligned}<br/>& \frac{d m_1}{d t}=\frac{d m_2}{d t} \quad \text { (given) } \\<br/>& L_1=80 \mathrm{calg}^{-1} \text { and } L_2=540 \mathrm{calg}^{-1}<br/>\end{aligned}<br/>$$<br/>Solving these two equations we get $\lambda=9$.</div> | MarksBatch1_P1.db |
137 | a-metal-rod-of-length-l-and-mass-m-is-pivoted-at-one-end-a-thin-disc-of-mass-m-and-radius-r-l-is-attached-at-its-centre-to-the-free-end-of-the-rod-con | a-metal-rod-of-length-l-and-mass-m-is-pivoted-at-one-end-a-thin-disc-of-mass-m-and-radius-r-l-is-attached-at-its-centre-to-the-free-end-of-the-rod-con-44195 | <div class="question">A metal rod of length $L$ and mass $m$ is pivoted at one end. A thin disc of mass $M$ and radius $R( < L)$ is attached at its centre to the free end of the rod. Consider two ways the disc is attached. Case A-the disc is not free to rotate about its centre and case $B$-the disc is free to rotate about its centre. The rod-disc system performs<br/>SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is/are true?<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/2Z9bbEPubSFDgbkK7f83upg2RB4bCyUHjrCj_uu9Q6U.original.fullsize.png"/><br/></div> | ['Physics', 'Oscillations', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>Restoring torque in case $A=$ Restoring torque in case $B$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>Restoring torque in case $A < $ Restoring torque in case $B$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>Angular frequency for case $A>$ Angular frequency for case $B$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>Angular frequency for case $A < $ Angular frequency for case $B$</span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>Restoring torque in case $A=$ Restoring torque in case $B$<br/>, <br/>Restoring torque in case $A < $ Restoring torque in case $B$</span> </div> | <div class="solution">$\tau_A=\tau_B=m g \frac{L}{2} \sin \theta+\operatorname{MgL} \sin \theta$<br/>$=$ Restoring torque about point $O$.<br/>In case $A$, moment of inertia will be more. Hence, angular acceleration $(\alpha=\tau / I)$ will be less. Therefore, angular frequency will be less.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/1s5RswcuRs38mBLtlc7w0I0vX4MgAYvlJTzjP3tzLLk.original.fullsize.png"/><br/><br/>$\therefore$ Correct options are (a) and (d).<br/>Analysis of Question<br/>Question is difficult from my point of view. Because this type of SHM is rarely taught in the class and questions of this type are not given in standard books.</div> | MarksBatch1_P1.db |
138 | a-meter-bridge-is-set-up-as-shown-in-figure-to-determine-an-unknown-resistance-x-using-a-standard-10-resistor-the-galvanometer-shows-null-point-when-t | a-meter-bridge-is-set-up-as-shown-in-figure-to-determine-an-unknown-resistance-x-using-a-standard-10-resistor-the-galvanometer-shows-null-point-when-t-36255 | <div class="question">A meter bridge is set up as shown in figure, to determine an unknown resistance $X$ using a standard $10 \Omega$ resistor. The galvanometer shows null point when tapping key is at $52 \mathrm{~cm}$ mark. The end-corrections are $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$ respectively for the ends $A$ and $B$. The determined value of $X$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/-5wAQXPr0ogSu-GX8RP1V0KuFgaPpJl2ZFwXoZMVR7U.original.fullsize.png"/><br/></div> | ['Physics', 'Current Electricity', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$10.2 \Omega$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$10.6 \Omega$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$10.8 \Omega$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$11.1 \Omega$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$10.6 \Omega$<br/></span> </div> | <div class="solution">Using the concept of balanced Wheatstone bridge, we have<br/>$$<br/>\begin{array}{rlrl} <br/>& \frac{P}{Q} & =\frac{R}{S} \\<br/>\therefore & & \frac{X}{(52+1)} & =\frac{10}{(48+2)} \\<br/>\therefore & & X & =\frac{10 \times 53}{50}=10.6 \Omega<br/>\end{array}<br/>$$<br/>$\therefore$ Correct option is (b).<br/>Analysis of Question<br/>Question is moderately tough, because normally end corrections are not taught in the coaching/schools in this type of problem.</div> | MarksBatch1_P1.db |
139 | a-mixture-of-2-moles-of-helium-gas-atomic-mass-4-amu-and-1-mole-of-argon-gas-atomic-mass-40-amu-is-kept-at-300-k-in-a-container-the-ratio-of-the-rms-s | a-mixture-of-2-moles-of-helium-gas-atomic-mass-4-amu-and-1-mole-of-argon-gas-atomic-mass-40-amu-is-kept-at-300-k-in-a-container-the-ratio-of-the-rms-s-94660 | <div class="question">A mixture of 2 moles of helium gas (atomic mass $=4 \mathrm{amu}$ ) and 1 mole of argon gas (atomic mass $=40 \mathrm{amu}$ ) is kept at $300 \mathrm{~K}$ in a container. The ratio of the rms speeds $\left(\frac{v_{\mathrm{rms}}(\text { helium })}{v_{\mathrm{rms}}(\operatorname{argon})}\right)$ is</div> | ['Physics', 'Kinetic Theory of Gases', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$0.32$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$0.45$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$2.24$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$3.16$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value">$3.16$</span> </div> | <div class="solution">Using $V_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \Rightarrow V_{r m s} \propto \frac{1}{\sqrt{M}}$ $\frac{v_{\text {rms (helium) }}}{v_{\text {rms (argon) }}}=\sqrt{\frac{M_{\text {argon }}}{M_{\text {helium }}}}=\sqrt{\frac{40}{4}}=\sqrt{10} \approx 3.16$</div> | MarksBatch1_P1.db |
140 | a-monoatomic-ideal-gas-undergoes-a-process-in-which-the-ratio-of-p-to-v-at-any-instant-is-constant-and-equals-to-1-what-is-the-molar-heat-capacity-of- | a-monoatomic-ideal-gas-undergoes-a-process-in-which-the-ratio-of-p-to-v-at-any-instant-is-constant-and-equals-to-1-what-is-the-molar-heat-capacity-of-86103 | <div class="question">A monoatomic ideal gas undergoes a process in which the ratio of $P$ to $V$ at any instant is constant and equals to 1 . What is the molar heat capacity of the gas?</div> | ['Chemistry', 'Thermodynamics (C)', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{4 R}{2}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{3 R}{2}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{5 R}{2}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>0</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{4 R}{2}$<br/></span> </div> | <div class="solution">The molar heat capacity for any process is given by a following expression<br/>$C=C_V+\frac{R}{1-\gamma}$ when $p V^\gamma=$ constant and $C_p / C_V=\gamma$<br/>Here, $\frac{p}{V}=1$ i.e., $p V^{-1}=$ constant<br/>$$<br/>\therefore \quad C=\frac{3}{2} R+\frac{R}{1-(-1)}=\frac{3}{2} R+\frac{R}{2}=\frac{4}{2} R<br/>$$</div> | MarksBatch1_P1.db |
141 | a-parallel-plate-capacitor-c-with-plates-of-unit-area-and-separation-d-is-filled-with-a-liquid-of-dielectric-constant-k-2-the-level-of-liquid-is-3-d-i-1 | a-parallel-plate-capacitor-c-with-plates-of-unit-area-and-separation-d-is-filled-with-a-liquid-of-dielectric-constant-k-2-the-level-of-liquid-is-3-d-i-1-22884 | <div class="question">A parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K=2$. The level of liquid is $\frac{d}{3}$ initially. Suppose the liquid level decreases at a constant speed $v$, the time constant as a function of time $t$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/kXR5joieLkCoSaad8_WTmvKA05ny_GxeO0ptHUUMHLY.original.fullsize.png"/><br/></div> | ['Physics', 'Capacitance', 'JEE Main'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{6 \varepsilon_0 R}{5 d+3 v t}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{(15 d+9 v t) \varepsilon_0 R}{2 d^2-3 d v t-9 v^2 t^2}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{6 \varepsilon_0 R}{5 d-3 v t}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{(15 d-9 v t) \varepsilon_0 R}{2 d^2+3 d v t-9 v^2 t^2}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{6 \varepsilon_0 R}{5 d+3 v t}$<br/></span> </div> | <div class="solution">After time $t$, thickness of liquid will remain $\left(\frac{d}{3}-v t\right)$.<br/>Now, time constant as function of time<br/>$$<br/>\begin{aligned}<br/>\tau_c & =C R \\<br/>& \left.=\frac{\varepsilon_0(1) \cdot R}{\left(d-\frac{d}{3}+v t\right)+\frac{d / 3-v t}{2}} \quad \quad \text { Applying } C=\frac{\varepsilon_0 A}{d-t+\frac{t}{k}}\right) \\<br/>& =\frac{6 \varepsilon_0 R}{5 d+3 v t}<br/>\end{aligned}<br/>$$<br/>$\therefore$ correct option is (a).</div> | MarksBatch1_P1.db |
142 | a-parallel-plate-capacitor-c-with-plates-of-unit-area-and-separation-d-is-filled-with-a-liquid-of-dielectric-constant-k-2-the-level-of-liquid-is-3-d-i | a-parallel-plate-capacitor-c-with-plates-of-unit-area-and-separation-d-is-filled-with-a-liquid-of-dielectric-constant-k-2-the-level-of-liquid-is-3-d-i-97609 | <div class="question">A parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K=2$. The level of liquid is $\frac{d}{3}$ initially. Suppose the liquid level decreases at a constant speed $v$, the time constant as a function of time $t$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/kXR5joieLkCoSaad8_WTmvKA05ny_GxeO0ptHUUMHLY.original.fullsize.png"/><br/></div> | ['Physics', 'Capacitance', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{6 \varepsilon_0 R}{5 d+3 v t}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{(15 d+9 v t) \varepsilon_0 R}{2 d^2-3 d v t-9 v^2 t^2}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{6 \varepsilon_0 R}{5 d-3 v t}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{(15 d-9 v t) \varepsilon_0 R}{2 d^2+3 d v t-9 v^2 t^2}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{6 \varepsilon_0 R}{5 d+3 v t}$<br/></span> </div> | <div class="solution">After time $t$, thickness of liquid will remain $\left(\frac{d}{3}-v t\right)$.<br/>Now, time constant as function of time<br/>$$<br/>\begin{aligned}<br/>\tau_c & =C R \\<br/>& \left.=\frac{\varepsilon_0(1) \cdot R}{\left(d-\frac{d}{3}+v t\right)+\frac{d / 3-v t}{2}} \quad \quad \text { Applying } C=\frac{\varepsilon_0 A}{d-t+\frac{t}{k}}\right) \\<br/>& =\frac{6 \varepsilon_0 R}{5 d+3 v t}<br/>\end{aligned}<br/>$$<br/>$\therefore$ correct option is (a).</div> | MarksBatch1_P1.db |
143 | a-particle-moves-in-the-x-y-plane-under-the-influence-of-a-force-such-that-its-linear-momentum-is-p-t-a-i-cos-k-t-j-sin-k-t-where-a-and-k-are-constant-1 | a-particle-moves-in-the-x-y-plane-under-the-influence-of-a-force-such-that-its-linear-momentum-is-p-t-a-i-cos-k-t-j-sin-k-t-where-a-and-k-are-constant-1-92214 | <div class="question">A particle moves in the $X-Y$ plane under the influence of a force such that its linear momentum is $\mathbf{p}(t)=A[\hat{\mathbf{i}} \cos k t-\hat{\mathbf{j}} \sin k t]$, where $A$ and $k$ are constants. The angle between the force and the momentum is</div> | ['Physics', 'Mathematics in Physics', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$0^{\circ}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$30^{\circ}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$45^{\circ}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$90^{\circ}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$90^{\circ}$</span> </div> | <div class="solution">$$<br/>\begin{aligned}<br/>& \mathbf{F}=\frac{d \mathbf{p}}{d t}=-k A \sin k t \hat{\mathbf{i}}-k A \cos k t \hat{\mathbf{j}} \\<br/>& \mathbf{p}=A \cos k t \hat{\mathbf{i}}-A \sin k \hat{\mathbf{j}}<br/>\end{aligned}<br/>$$<br/>Since,<br/>F $\cdot \mathbf{p}=0$<br/>$\therefore$ Angle between $\mathbf{F}$ and pshould be $90^{\circ}$.<br/>$\therefore$ Correct option is $(\mathrm{d})$.</div> | MarksBatch1_P1.db |
144 | a-particle-of-mass-10-3-kg-and-charge-10-c-is-initially-at-rest-at-time-t-0-the-particle-comes-under-the-influence-of-an-electric-field-e-t-e-0-sin-t--1 | a-particle-of-mass-10-3-kg-and-charge-10-c-is-initially-at-rest-at-time-t-0-the-particle-comes-under-the-influence-of-an-electric-field-e-t-e-0-sin-t-1-14819 | <div class="question">A particle, of mass <math><msup><mrow><mn>10</mn></mrow><mrow><mo>-</mo><mn>3</mn></mrow></msup><mi> kg</mi></math> and charge <math><mn>1.0</mn><mi> </mi><mn>C</mn></math>, is initially at rest. At time <math><mi>t</mi><mo>=</mo><mn>0</mn></math>, the particle comes under the influence of an electric field <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>E</mi><mo>→</mo></mover><mfenced separators="|"><mi>t</mi></mfenced><mo>=</mo><msub><mi>E</mi><mn>0</mn></msub><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mi> </mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math> where <math><msub><mrow><mi>E</mi></mrow><mrow><mn>0</mn></mrow></msub><mo>=</mo><mn>1.0</mn><mi> N</mi><mi> </mi><msup><mrow><mn>C</mn></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> and <math><mi>ω</mi><mo>=</mo><msup><mrow><mn>10</mn></mrow><mrow><mn>3</mn></mrow></msup><mi> rad</mi><mi> </mi><msup><mrow><mn>s</mn></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math>. Consider the effect of only the electrical force on the particle. Then the maximum speed, in <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>m</mn><mo> </mo><msup><mn>s</mn><mrow><mo>-</mo><mn>1</mn></mrow></msup></math>, attained by the particle at subsequent times is ____________.</div> | ['Physics', 'Laws of Motion', 'JEE Advanced', 'JEE Advanced 2018 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">2</span> </div> | <div class="solution"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mo>=</mo><msup><mn>10</mn><mrow><mo>-</mo><mn>3</mn></mrow></msup><mi> kg</mi><mo>,</mo><mi> </mi><mi> </mi><mi>q</mi><mo>=</mo><mn>1 </mn><mn>C</mn><mo>,</mo><mi> </mi><mi>t</mi><mo>=</mo><mn>0</mn></math><br/>Force on particle will be<br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi><mo>=</mo><mi>q</mi><mi>E</mi><mo>=</mo><mi>q</mi><msub><mi>E</mi><mn>0</mn></msub><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math><br/>At <math><msub><mrow><mi>v</mi></mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></msub><mo>,</mo><mi> </mi><mi>F</mi><mo>=</mo><mn>0</mn></math> <math><mo>⇒</mo><mi>q</mi><msub><mrow><mi>E</mi></mrow><mrow><mn>0</mn></mrow></msub><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>ω</mi><msub> <mi>t</mi> <mi>o</mi> </msub> <mo>=</mo><mn>0</mn></mrow> </mrow></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>ω</mi><msub><mi>t</mi><mi>o</mi></msub><mo>=</mo><mi>π</mi></math><br/><math><mi>F</mi><mo>=</mo><mi>q</mi><msub><mrow><mi>E</mi></mrow><mrow><mn>0</mn></mrow></msub><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>ω</mi><mi>t</mi></mrow></mrow></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>d</mn><mi>v</mi></mrow><mrow><mn>d</mn><mi>t</mi></mrow></mfrac><mo>=</mo><mi>q</mi><mfrac><msub><mi>E</mi><mn>0</mn></msub><mi>m</mi></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mo>∫</mo><mn>0</mn><mi>v</mi></msubsup><mn>d</mn><mi>v</mi><mo>=</mo><msubsup><mo>∫</mo><mn>0</mn><mrow><mi>π</mi><mo>∕</mo><mi>ω</mi></mrow></msubsup><mfrac><mrow><mi>q</mi><msub><mi>E</mi><mn>0</mn></msub></mrow><mi>m</mi></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mo> </mo><mn>d</mn><mi>t</mi></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi><mo>-</mo><mn>0</mn><mo>=</mo><mfrac><mrow><mi>q</mi><msub><mi>E</mi><mn>0</mn></msub></mrow><mrow><mi>m</mi><mi>ω</mi></mrow></mfrac><msubsup><mfenced close="]" open="[" separators="|"><mrow><mo>-</mo><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></mrow></mfenced><mn>0</mn><mrow><mi>π</mi><mo>∕</mo><mi>ω</mi></mrow></msubsup><mi> </mi></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi><mo>-</mo><mn>0</mn><mo>=</mo><mfrac><mrow><mi>q</mi><msub><mi>E</mi><mn>0</mn></msub></mrow><mrow><mi>m</mi><mi>ω</mi></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><mfenced separators="|"><mrow><mo>-</mo><mi>cos</mi><mfenced><mi>π</mi></mfenced></mrow></mfenced><mo>-</mo><mfenced separators="|"><mrow><mo>-</mo><mi>cos</mi><mfenced><mn>0</mn></mfenced></mrow></mfenced></mrow></mfenced></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi><mo>=</mo><mfrac><mrow><mn>1</mn><mo>×</mo><mn>1</mn></mrow><mrow><mi>m</mi><mi>ω</mi></mrow></mfrac><mo>×</mo><mn>2</mn><mo>=</mo><mn>2</mn><mo> m</mo><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></div> | MarksBatch1_P1.db |
145 | a-particle-of-mass-m-and-charge-q-moving-with-velocity-v-enters-region-ii-normal-to-the-boundary-as-shown-in-the-figure-region-ii-has-a-uniform-magnet | a-particle-of-mass-m-and-charge-q-moving-with-velocity-v-enters-region-ii-normal-to-the-boundary-as-shown-in-the-figure-region-ii-has-a-uniform-magnet-86730 | <div class="question">A particle of mass $m$ and charge $q$, moving with velocity $v$ enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field $B$ perpendicular to the plane of the paper. The length of the Region II is $l$. Choose the correct choice(s)<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/2R-9CzzQsKQiIWeWGqRQfQoFeKBfJB3YsXFFqvgUbGU.original.fullsize.png"/><br/></div> | ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>The particle enters Region III only if its velocity $v>\frac{q l B}{m}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>The particle enters Region III only if its velocity $v < \frac{q I B}{m}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>Path length of the particle in Region II is maximum when velocity $v=\frac{q I B}{m}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>Time spent in Region II is same for any velocity $v$ as long as the particle returns to Region I</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>The particle enters Region III only if its velocity $v>\frac{q l B}{m}$<br/>, <br/>Path length of the particle in Region II is maximum when velocity $v=\frac{q I B}{m}$<br/>, <br/>Time spent in Region II is same for any velocity $v$ as long as the particle returns to Region I</span> </div> | <div class="solution">$\overrightarrow{\mathbf{V}} \perp \overrightarrow{\mathbf{B}}$ in region II. Therefore, path of particle is circle in region II. Particle enters in region III if, radius of circular path, $r>1$ or $\frac{m v}{B q}>1$ or $v>\frac{B q l}{m}$<br/>If $v=\frac{B q l}{m}, r=\frac{m v}{B q}=l$, particle will turn back and path length will be maximum as shown in figure in region II. If particle returns to region I, time spent in region II will be $t=\frac{T}{2}=\frac{\pi m}{B q}$, which is independent of $v$.<br/>$\therefore$ correct options are (a), (c) and (d).<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/0u3oe3umQiRM7QrzjiMfVEpq5UCFuBEPfFEcz8oRLs4.original.fullsize.png"/><br/></div> | MarksBatch1_P1.db |
146 | a-particle-of-unit-mass-is-moving-along-the-xaxis-under-the-influence-of-a-force-and-its-total-energy-is-conserved-four-possible-forms-of-the-potentia | a-particle-of-unit-mass-is-moving-along-the-xaxis-under-the-influence-of-a-force-and-its-total-energy-is-conserved-four-possible-forms-of-the-potentia-63223 | <div class="question">A particle of unit mass is moving along the x-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column I (a and <math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></math> are constants). Match the potential energies in column I to the corresponding statement(s) in column II.<table border="1"> <tbody> <tr> <td> </td> <td><strong>Column I</strong></td> <td> </td> <td><strong>Column II</strong></td> </tr> <tr> <td>A.</td> <td><math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>1</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mfenced close="]" open="[" separators="|"><mrow><mn>1</mn><mo>-</mo><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></math></td> <td>P.</td> <td>The force acting on the particle is zero at <math><mi mathvariant="normal">x</mi><mi mathvariant="normal"> </mi><mo>=</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">a</mi></math></td> </tr> <tr> <td>B.</td> <td><math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>2</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></math></td> <td>Q.</td> <td>The force acting on the particle is zero at <math><mi mathvariant="normal">x</mi><mi mathvariant="normal"> </mi><mo>=</mo><mi mathvariant="normal"> </mi><mn>0</mn></math> .</td> </tr> <tr> <td>C.</td> <td><math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>3</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup><mrow><mrow><mi mathvariant="normal">exp</mi></mrow><mo></mo><mrow><mfenced close="]" open="[" separators="|"><mrow><mo>-</mo><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfenced></mrow></mrow></math></td> <td>R.</td> <td>The force acting on the particle is zero at <math><mi mathvariant="normal">x</mi><mo>=</mo><mi mathvariant="normal"> </mi><mo>-</mo><mi mathvariant="normal">a</mi></math></td> </tr> <tr> <td>D.</td> <td><math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>4</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac><mo>-</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>3</mn></mrow></msup></mrow></mfenced></math></td> <td>S.</td> <td>The particle experiences an attractive force towards <math><mi mathvariant="normal">x</mi><mi mathvariant="normal"> </mi><mo>=</mo><mi mathvariant="normal"> </mi><mn>0</mn></math> in the region <math><mfenced close="|" open="|" separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo><</mo><mi mathvariant="normal">a</mi></math></td> </tr> <tr> <td> </td> <td> </td> <td>T.</td> <td>The particle with total energy <math><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>4</mn></mrow></mfrac></math> can oscillate about the point <math><mi mathvariant="normal">x</mi><mo>=</mo><mi mathvariant="normal"> </mi><mo>-</mo><mi mathvariant="normal">a</mi></math></td> </tr> </tbody></table></div> | ['Physics', 'Work Power Energy', 'JEE Advanced', 'JEE Advanced 2015 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">a-s,t;b-s;c-s,t;d-t;</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">a-s;b-r,s;c-r,s,t;d-s,t;</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">a-r,s;b-q;c-q,r;d-t;</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">a-p,q,r,t;b-q,s;c-p,q,r,s;d-p,r,t;</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value">a-p,q,r,t;b-q,s;c-p,q,r,s;d-p,r,t;</span> </div> | <div class="solution"><math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>1</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mfenced close="]" open="[" separators="|"><mrow><mn>1</mn><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mo>-</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mn>2</mn><mfenced close="]" open="[" separators="|"><mrow><mn>1</mn><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced><mfenced close="]" open="[" separators="|"><mrow><mo>-</mo><mfrac><mrow><mn>2</mn><mi mathvariant="normal">x</mi></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced></math><br/><math><mo>=</mo><mfrac><mrow><mn>2</mn><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>4</mn></mrow></msup></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>-</mo><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfenced><mi mathvariant="normal">x</mi></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mfrac><mrow><mn>2</mn><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>4</mn></mrow></msup></mrow></mfrac><mi mathvariant="normal">x</mi><mfenced separators="|"><mrow><mi mathvariant="normal">a</mi><mo>-</mo><mi mathvariant="normal">x</mi></mrow></mfenced><mfenced separators="|"><mrow><mi mathvariant="normal">a</mi><mo>+</mo><mi mathvariant="normal">x</mi></mrow></mfenced></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mn>0</mn></math> , at <math><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mo>-</mo><mi mathvariant="normal">a</mi></math><br/><math><mfenced close="}" open="" separators="|"><mrow><mtable><mtr><mtd><mi mathvariant="normal">x</mi><mo>=</mo><mo>-</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">U</mi><mo>=</mo><mn>0</mn><mo>,</mo></mtd></mtr><mtr><mtd><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">U</mi><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac></mtd></mtr></mtable></mrow></mfenced><mo>â</mo></math> Oscillate when total ME is less then <math><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac></math><br/><math><mi mathvariant="normal">A</mi><mo>â</mo><mi mathvariant="normal">P</mi><mo>,</mo><mi mathvariant="normal">Q</mi><mo>,</mo><mi mathvariant="normal">R</mi><mo>,</mo><mi mathvariant="normal">T</mi></math><br/>B- <math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mo>-</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub><mi mathvariant="normal">x</mi></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mn>0</mn><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn></math><br/><math><mi mathvariant="normal">B</mi><mo>â</mo><mi mathvariant="normal">Q</mi><mo>,</mo><mi mathvariant="normal">S</mi></math><br/>C- <math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>3</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup><mrow><mrow><mi mathvariant="normal">exp</mi></mrow><mo>â¡</mo><mrow><mfenced separators="|"><mrow><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced></mrow></mrow></mrow></mfenced></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mo>-</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><mfrac><mrow><mn>2</mn><mi mathvariant="normal">x</mi></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mrow><mrow><mi mathvariant="normal">exp</mi></mrow><mo>â¡</mo><mrow><mfenced separators="|"><mrow><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced></mrow></mrow><mo>+</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mrow><mrow><mi mathvariant="normal">exp</mi></mrow><mo>â¡</mo><mrow><mfenced separators="|"><mrow><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced><mfenced separators="|"><mrow><mo>-</mo><mfrac><mrow><mn>2</mn><mi mathvariant="normal">x</mi></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced></mrow></mrow></mrow></mfenced></math><br/><math><mo>=</mo><mo>-</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mfrac><mrow><mn>2</mn><mi mathvariant="normal">x</mi></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>4</mn></mrow></msup></mrow></mfrac><mrow><mrow><mi mathvariant="normal">exp</mi></mrow><mo>â¡</mo><mrow><mfenced separators="|"><mrow><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced></mrow></mrow><mfenced close="]" open="[" separators="|"><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>-</mo><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfenced></math><br/><math><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>4</mn></mrow></msup></mrow></mfrac><msup><mrow><mi mathvariant="normal">e</mi></mrow><mrow><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></msup><mi mathvariant="normal"> </mi><mfenced close="]" open="[" separators="|"><mrow><mi mathvariant="normal">x</mi><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi><mo>+</mo><mi mathvariant="normal">a</mi></mrow></mfenced><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi><mo>-</mo><mi mathvariant="normal">a</mi></mrow></mfenced></mrow></mfenced></math><br/><math><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">x</mi><mo>=</mo><mo>-</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">x</mi><mo>=</mo><mo>+</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">F</mi><mo>=</mo><mn>0</mn></math><br/><math><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">U</mi><mo>=</mo><mn>0</mn></math> even function hence minima<br/><math><mi mathvariant="normal">C</mi><mo>â</mo><mi mathvariant="normal">P</mi><mo>,</mo><mi mathvariant="normal">Q</mi><mo>,</mo><mi mathvariant="normal">R</mi><mo>,</mo><mi mathvariant="normal">S</mi></math><br/>D- <math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>4</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow><mrow><mn>3</mn><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow></mfrac></mrow></mfenced></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mo>-</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow></mfrac></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>-</mo><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfenced><mi mathvariant="normal"> </mi></math><br/><math><mi mathvariant="normal">x</mi><mo>=</mo><mo>+</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">x</mi><mo>=</mo><mo>-</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">F</mi><mo>=</mo><mn>0</mn></math><br/><math><mi mathvariant="normal">x</mi><mo>=</mo><mo>-</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">U</mi><mo>=</mo><mo>-</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>3</mn></mrow></mfrac><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">x</mi><mo>=</mo><mo>+</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">U</mi><mo>=</mo><mo>+</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>3</mn></mrow></mfrac></math><br/>Hence oscillate about <math><mi mathvariant="normal">x</mi><mo>=</mo><mo>-</mo><mi mathvariant="normal">a</mi></math> if <math><mi mathvariant="normal">T</mi><mo>.</mo><mi mathvariant="normal">M</mi><mo>.</mo><mi mathvariant="normal">E</mi><mo><</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>3</mn></mrow></mfrac></math><br/><math><mi mathvariant="normal">D</mi><mo>â</mo><mi mathvariant="normal">P</mi><mo>,</mo><mi mathvariant="normal">R</mi><mo>,</mo><mi mathvariant="normal">T</mi></math></div> | MarksBatch1_P1.db |
147 | a-particle-p-starts-from-the-point-z-0-1-2-i-where-i-1-it-moves-first-horizontally-away-from-origin-by-5-units-and-then-vertically-away-from-origin-by | a-particle-p-starts-from-the-point-z-0-1-2-i-where-i-1-it-moves-first-horizontally-away-from-origin-by-5-units-and-then-vertically-away-from-origin-by-14396 | <div class="question">A particle $P$ starts from the point $z_0=1+2 i$, where $i=\sqrt{-1}$. It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point $z_1$. From $z_1$ the particle moves $\sqrt{2}$ units in the direction of the vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and then it moves through an angle $\frac{\pi}{2}$ in anti-clockwise direction on a circle with centre at origin, to reach a point $z_2$. The point $z_2$ is given by</div> | ['Mathematics', 'Complex Number', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$6+7 i$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$-7+6 i$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$7+6 i$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$-6+7 i$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$-6+7 i$</span> </div> | <div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/7V-wZrcaDesQMMcdHT0vrL9hcmdmdeq6EEypYnkMDJU.original.fullsize.png"/><br/><br/>$$<br/>z_2^{\prime}=\left(6+\sqrt{2} \times \cos 45^{\circ}, 5+\sqrt{2} \sin 45^{\circ}\right)=(7,6)=7+6 i<br/>$$<br/>By rotation about $(0,0)$,<br/>$$<br/>\begin{aligned}<br/>& \frac{z_2}{z_2^{\prime}}=e^{i \frac{\pi}{2}} \\<br/>& \Rightarrow \quad z_2=z_2^{\prime}\left(e^{i \frac{\pi}{2}}\right) \\<br/>& \Rightarrow \quad z_2=(7+6 i)\left(\cos \frac{\pi}{-}+i \sin \frac{\pi}{-}\right)=(7+6 i)(i)=-6+7 i<br/>\end{aligned}<br/>$$</div> | MarksBatch1_P1.db |
148 | a-person-blows-into-openend-of-a-long-pipe-as-a-result-a-high-pressure-pulse-of-air-travels-down-the-pipe-when-this-pulse-reaches-the-other-end-of-the | a-person-blows-into-openend-of-a-long-pipe-as-a-result-a-high-pressure-pulse-of-air-travels-down-the-pipe-when-this-pulse-reaches-the-other-end-of-the-49024 | <div class="question">A person blows into open-end of a long pipe. As a result, a high pressure pulse of air travels down the pipe.
<br/> <br/>When this pulse reaches the other end of the pipe,</div> | ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open.</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value">a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open., a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.</span> </div> | <div class="solution">When a sound pulse is reflected from open end of a pipe, phase changes by $180^{\circ}$. A high pressure pulse i.e., compression is reflected as a low pressure pulse i.e., rarefaction. When sound pulse is reflected through a rigid boundary (closed end of a pipe), no phase change occurs so a high pressure pulse is reflected as a high pressure pulse.</div> | MarksBatch1_P1.db |
149 | a-piece-of-ice-heat-capacity-2100-j-kg-1-c-1-and-latent-heat-336-1-0-5-j-kg-1-of-mass-m-gram-is-at-5-c-at-atmospheric-pressure-it-is-given-420-j-of-he | a-piece-of-ice-heat-capacity-2100-j-kg-1-c-1-and-latent-heat-336-1-0-5-j-kg-1-of-mass-m-gram-is-at-5-c-at-atmospheric-pressure-it-is-given-420-j-of-he-50454 | <div class="question">A piece of ice (heat capacity $=2100 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and latent heat $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ ) of mass $\mathrm{m}$ gram is at $-5^{\circ} \mathrm{C}$ at atmospheric pressure. It is given $420 \mathrm{~J}$ of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that $1 \mathrm{~g}$ of ice has melted. Assuming there is no other heat exchange in the process, the value of $m$ is</div> | ['Physics', 'Thermal Properties of Matter', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">8</span> </div> | <div class="solution">Language of question is slightly wrong. As heat capacity and specific heat are two different physical quantities. Unit $\mathrm{J}-\mathrm{kg}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}$. The heat capacity given in the question is really the specific heat. Now applying the heat exchange equation.<br/>$$<br/>\begin{array}{r}<br/>420=\left(m \times 10^{-3}\right)(2100)(5)+\left(1 \times 10^{-3}\right) \\<br/>\left(3.36 \times 10^5\right)<br/>\end{array}<br/>$$<br/>Solving this equation we get,<br/>$$<br/>m=8 \mathrm{~g}<br/>$$<br/>$\therefore$ The correct answer is 8 .</div> | MarksBatch1_P1.db |
150 | a-piece-of-wire-is-bent-in-the-shape-of-a-parabola-y-k-x-2-yaxis-vertical-with-a-bead-of-mass-m-on-it-the-bead-can-slide-on-the-wire-without-friction--1 | a-piece-of-wire-is-bent-in-the-shape-of-a-parabola-y-k-x-2-yaxis-vertical-with-a-bead-of-mass-m-on-it-the-bead-can-slide-on-the-wire-without-friction-1-90961 | <div class="question">A piece of wire is bent in the shape of a parabola $y=k x^2$ (y-axis vertical) with a bead of mass $m$ on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x$-axis with the constant acceleration $a$. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the $y$-axis is</div> | ['Physics', 'Laws of Motion', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{a}{g k}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{a}{2 g k}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{2 a}{g k}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{a}{4 g k}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{a}{2 g k}$<br/></span> </div> | <div class="solution">$N \sin \theta=m g$<br/>$$<br/>\begin{aligned}<br/>N \cos \theta & =m a \\<br/>\tan \theta & =\frac{g}{a} \\<br/>\cot \theta & =\frac{a}{g}=\tan \left(90^{\circ}-\theta\right)=\frac{d y}{d x}=2 k x<br/>\end{aligned}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/kDkPeNyYC5fMi_8lwwdUnIYuwQGNAzZla9ZDheSXGmA.original.fullsize.png"/><br/><br/>$\therefore \quad x=\frac{a}{2 k g}$</div> | MarksBatch1_P1.db |
151 | a-plane-passes-through-1-2-1-and-is-perpendicular-to-two-planes-2-x-2-y-z-0-and-x-y-2-z-4-then-the-distance-of-the-plane-from-the-point-1-2-2-is | a-plane-passes-through-1-2-1-and-is-perpendicular-to-two-planes-2-x-2-y-z-0-and-x-y-2-z-4-then-the-distance-of-the-plane-from-the-point-1-2-2-is-82024 | <div class="question">A plane passes through $(1,-2,1)$ and is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, then the distance of the plane from the point $(1,2,2)$ is</div> | ['Mathematics', 'Three Dimensional Geometry', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>0<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>1<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\sqrt{2}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$2 \sqrt{2}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$2 \sqrt{2}$</span> </div> | <div class="solution">Let the equation of plane be,<br/>$$<br/>a(x-1)+b(y+2)+c(z-1)=0<br/>$$<br/>which is perpendicular to $2 x-2 y+z=0$ and $x-y+2 z=4$<br/>$$<br/>\begin{array}{llrl}<br/>\Rightarrow & 2 a-2 b+c & =0 \text { and } a-b+2 c=0 \\<br/>\text { or } & \frac{a}{-2} 1 & =\frac{b}{2}=\frac{c}{-2} \\<br/>\Rightarrow & -12 & 1 & -1 \\<br/>\Rightarrow & \frac{a}{-3} & =\frac{b}{-3}=\frac{c}{0} \\<br/>& \frac{a}{1}=\frac{b}{1}=\frac{c}{0}<br/>\end{array}<br/>$$<br/>So, the equation of plane is,<br/>$x-1+y+2=0$ or $x+y+1=0$, its distance from the point $(1,2,2)$ is $\frac{|1+2+1|}{\sqrt{2}}=2 \sqrt{2}$.</div> | MarksBatch1_P1.db |
152 | a-point-charge-q-of-mass-m-is-suspended-vertically-by-a-string-of-length-l-a-point-dipole-of-dipole-moment-p-is-now-brought-towards-q-from-infinity-so | a-point-charge-q-of-mass-m-is-suspended-vertically-by-a-string-of-length-l-a-point-dipole-of-dipole-moment-p-is-now-brought-towards-q-from-infinity-so-57461 | <div class="question"><p>A point charge <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>q</mi></math> of mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> is suspended vertically by a string of length <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>l</mi><mo>.</mo></math> A point dipole of dipole moment <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>p</mi><mo>â</mo></mover></math> is now brought towards <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>q</mi></math> from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi><mo>Ã</mo><mo>(</mo><mi>m</mi><mi>g</mi><mi>h</mi><mo>)</mo></math>, where <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>g</mi></math> is the acceleration due to gravity, then the value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi></math> is ____________ (Note that for three coplanar forces keeping a point mass in equilibrium, <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi mathvariant="normal">F</mi><mrow><mi>sin</mi><mi>θ</mi></mrow></mfrac></math> is the same for all forces, where <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi></math> is any one of the forces and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>θ</mi></math> is the angle between the other two forces)</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/66b1f109-9a81-453a-8162-902331dfe560-image.png"/></p></div> | ['Physics', 'Work Power Energy', 'JEE Advanced', 'JEE Advanced 2020 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">2</span> </div> | <div class="solution"><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/d05922c2-b8c0-4772-8864-2f402f721035-image.png"/></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>W</mi><mo mathvariant="italic">=</mo><mi>q</mi><mi>V</mi><mo mathvariant="italic">+</mo><mi>m</mi><mi>g</mi><mi>h</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo mathvariant="italic">=</mo><mi>q</mi><mfenced><mfrac><mrow><mi>k</mi><mi>p</mi></mrow><msup><mi>x</mi><mn>2</mn></msup></mfrac></mfenced><mo mathvariant="italic">+</mo><mi>m</mi><mi>g</mi><mi>h</mi><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mn>1</mn></mfenced></math></p><p>For equilibrium</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>q</mi><mi>E</mi></mrow><mrow><mi>sin</mi><mi>β</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mrow><mi>sin</mi><mfenced><mrow><mn>90</mn><mo>-</mo><mfrac><mi>α</mi><mn>2</mn></mfrac></mrow></mfenced></mrow></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>q</mi><mi>E</mi></mrow><mrow><mi>sin</mi><mo>(</mo><mn>180</mn><mo>-</mo><mi>α</mi><mo>)</mo></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mrow><mi>cos</mi><mfenced><mfrac><mi>α</mi><mn>2</mn></mfrac></mfenced></mrow></mfrac><mo>;</mo><mo>  </mo><mfrac><mrow><mi>q</mi><mi>E</mi></mrow><mrow><mi>sin</mi><mi>α</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mrow><mi>cos</mi><mfenced><mfrac><mi>α</mi><mn>2</mn></mfrac></mfenced></mrow></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>9</mn><mi>E</mi></mrow><mrow><mn>2</mn><mi>sin</mi><mfenced><mfrac><mi>α</mi><mn>2</mn></mfrac></mfenced><mi>cos</mi><mfenced><mfrac><mi>α</mi><mn>2</mn></mfrac></mfenced></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mrow><mi>cos</mi><mfenced><mfrac><mi>α</mi><mn>2</mn></mfrac></mfenced></mrow></mfrac><mo>;</mo><mo>  </mo><mi>q</mi><mi>E</mi><mo>=</mo><mn>2</mn><mo mathvariant="italic">mg</mo><mi>sin</mi><mfenced><mfrac><mi>α</mi><mn>2</mn></mfrac></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>q</mi><mfenced><mfrac><mrow><mi>k</mi><mi>p</mi></mrow><msup><mi>x</mi><mn mathvariant="italic">2</mn></msup></mfrac></mfenced><mo mathvariant="italic">=</mo><mi>m</mi><mi>g</mi><mi>x</mi><mi>sin</mi><mstyle mathvariant="italic"><mrow><mo>(</mo><mfrac><mi>α</mi><mn>2</mn></mfrac><mo>)</mo></mrow></mstyle><mo>=</mo><mi>m</mi><mi>g</mi><mi>sin</mi><mfenced><mfrac><mi>α</mi><mn mathvariant="italic">2</mn></mfrac></mfenced><mo mathvariant="italic">·</mo><mfrac><mi>h</mi><mrow><mi>sin</mi><mfenced><mfrac><mi mathvariant="normal">α</mi><mn>2</mn></mfrac></mfenced></mrow></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>q</mi><mfrac><mrow><mi>k</mi><mi>p</mi></mrow><msup><mi>x</mi><mn>2</mn></msup></mfrac><mo>=</mo><mi>m</mi><mi>g</mi><mi>h</mi><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mn>2</mn></mfenced></math></p><p>From (1) and (2)</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>W</mi><mo>=</mo><mn>2</mn><mi>m</mi><mi>g</mi><mi>h</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>=</mo><mn>2</mn></math></p><p> </p></div> | MarksBatch1_P1.db |
153 | a-point-mass-is-subjected-to-two-simultaneous-sinusoidal-displacements-in-x-direction-x-1-t-a-sin-t-and-x-2-t-a-sin-t-3-2-adding-a-third-sinusoidal-di | a-point-mass-is-subjected-to-two-simultaneous-sinusoidal-displacements-in-x-direction-x-1-t-a-sin-t-and-x-2-t-a-sin-t-3-2-adding-a-third-sinusoidal-di-26735 | <div class="question">A point mass is subjected to two simultaneous sinusoidal displacements in $x$-direction, $x_1(t)=A \sin \omega t$ and $x_2(t)=A \sin \left(\omega t+\frac{2 \pi}{3}\right)$. Adding a third sinusoidal displacement $x_3(t)=B \sin (\omega t+\phi)$ bring the mass to a complete rest. The values of $B$ and $\phi$ are</div> | ['Physics', 'Oscillations', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\sqrt{2 A}, \frac{3 \pi}{4}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$A, \frac{4 \pi}{3}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\sqrt{3} A, \frac{5 \pi}{6}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$A, \frac{\pi}{3}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$A, \frac{4 \pi}{3}$<br/></span> </div> | <div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/aUgqUfC-c6FUMydFnyL-WX3gfqcC22f0LCjoN10xBxo.original.fullsize.png"/><br/><br/>Resultant amplitude of $x_1$ and $x_2$ is $A$ at angle $\left(\frac{\pi}{3}\right)$ from $A_1$. To make resultant of $x_1, x_2$ and $x_3$ to be zero. $A_3$ should be equal to $A$ at angle $\phi=\frac{4 \pi}{3}$ as shown in figure.<br/>$\therefore$ Correct answer is (b).<br/>Alternate Solution<br/>If we substitute,<br/>$$<br/>x_1+x_2+x_3=0<br/>$$<br/>or $A \sin \omega t+A \sin \left(\omega t+\frac{2 \pi}{3}\right)$<br/>$$<br/>+B \sin (\omega t+\phi)=0<br/>$$<br/>Then, by applying simple mathematics, we can prove that<br/>$$<br/>\begin{array}{rlrl}<br/>B & =A \\<br/>\text { and } & \phi & =\frac{4 \pi}{3}<br/>\end{array}<br/>$$<br/>Analysis of Question<br/>(i) Question is simple.<br/>(ii) Question can be solved by applying mathematics also.<br/>(iii) Amplitudes of two or more sine or cosine functions of same frequency $\omega$ can be added by vector method.</div> | MarksBatch1_P1.db |
154 | a-point-mass-of-1-kg-collides-elastically-with-a-stationary-point-mass-of-5-kg-after-their-collision-the-1-kg-mass-reverses-its-direction-and-moves-wi | a-point-mass-of-1-kg-collides-elastically-with-a-stationary-point-mass-of-5-kg-after-their-collision-the-1-kg-mass-reverses-its-direction-and-moves-wi-26616 | <div class="question">A point mass of $1 \mathrm{~kg}$ collides elastically with a stationary point mass of $5 \mathrm{~kg}$. After their collision, the $1 \mathrm{~kg}$ mass reverses its direction and moves with a speed of $2 \mathrm{~ms}^{-1}$. Which of the following statement(s) is/are correct for the system of these two masses?</div> | ['Physics', 'Center of Mass Momentum and Collision', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>Total momentum of the system is $3 \mathrm{~kg}-\mathrm{ms}^{-1}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>Momentum of $5 \mathrm{~kg}$ mass after collision is $4 \mathrm{~kg}-\mathrm{ms}^{-1}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>Kinetic energy of the centre of mass is $0.75 \mathrm{~J}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>Total kinetic energy of the system is $4 \mathrm{~J}$</span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>Total momentum of the system is $3 \mathrm{~kg}-\mathrm{ms}^{-1}$<br/>, <br/>Kinetic energy of the centre of mass is $0.75 \mathrm{~J}$<br/></span> </div> | <div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/z_PhW8UfShvxIqh556cc7MWcXwR9YnrJGM4z7hNa3Po.original.fullsize.png"/><br/><br/>$$<br/>\begin{aligned}<br/>& v_1^{\prime}=\left(\frac{m_1-m_5}{m_1+m_5}\right) v_1+\left(\frac{2 m_5}{m_1+m_5}\right) v_5 \\<br/>& -2=\left(\frac{1-5}{1+5}\right) v_1+0 \quad\left(\text { as } v_5=0\right) \\<br/>& \therefore \quad v_1=3 \mathrm{~ms}^{-1} \\<br/>& v_5^{\prime}=\left(\frac{m_5-m_1}{m_1+m_5}\right) v_5+\left(\frac{2 m_1}{m_1+m_2}\right) v_1 \\<br/>& =0+\left(\frac{2 \times 1}{6}\right)(3)=1 \mathrm{~ms}^{-1} \\<br/>& P_{\mathrm{CM}}=P_i=(1)(3) \\<br/>& =3 \frac{\mathrm{kg}-\mathrm{m}}{\mathrm{s}} \\<br/>& P_5{ }^{\prime}=(5)(1)=5 \frac{\mathrm{kg}-\mathrm{m}}{5} \\<br/>& K_{\mathrm{CM}}=\frac{P_{\mathrm{CM}}^2}{2 M_{\mathrm{CM}}} \\<br/>& =\frac{9}{2 \times 6}=0.75 \mathrm{~J} \\<br/>& K_{\text {total }}=\frac{1}{2} \times 1 \times(3)^2 \\<br/>& =4.5 \mathrm{~J} \\<br/>&<br/>\end{aligned}<br/>$$<br/>$\therefore$ correct options are (a) and (c).</div> | MarksBatch1_P1.db |
155 | a-point-object-is-placed-at-distance-of-20-cm-from-a-thin-planoconvex-lens-of-focal-length-15-cm-the-plane-surface-of-the-lens-is-now-silvered-the-ima | a-point-object-is-placed-at-distance-of-20-cm-from-a-thin-planoconvex-lens-of-focal-length-15-cm-the-plane-surface-of-the-lens-is-now-silvered-the-ima-23586 | <div class="question">A point object is placed at distance of $20 \mathrm{~cm}$ from a thin planoconvex lens of focal length $15 \mathrm{~cm}$. The plane surface of the lens is now silvered. The image created by the system is at<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/yc5nxE-emhqaMWcy037V8GiTt-GsnpNEs0_7ETm70lE.original.fullsize.png"/><br/></div> | ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$60 \mathrm{~cm}$ to the left of the system<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$60 \mathrm{~cm}$ to the right of the system<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$12 \mathrm{~cm}$ to the left of the system<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$12 \mathrm{~cm}$ to the right of the system</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$12 \mathrm{~cm}$ to the left of the system<br/></span> </div> | <div class="solution">Refraction from lens<br/>$$<br/>\begin{aligned}<br/>\frac{1}{v_1}-\frac{1}{-20} & =\frac{1}{15} \\<br/>v & =60 \mathrm{~cm}<br/>\end{aligned}<br/>$$ <img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/rBYnE5k12tvD8EtQ2q85ORZ8rV5TAkDX7qj-t3E8GmM.original.fullsize.png"/><br/><br/><br/>i.e., first image is formed at $60 \mathrm{~cm}$ to the right of lens system.<br/>Reflection from mirror<br/>After reflection from the mirror, the second image will be formed at a distance of $60 \mathrm{~cm}$ to the left of lens system.<br/>Refraction from lens<br/>$$<br/>\begin{aligned}<br/>\frac{1}{v_3}-\frac{1}{60} & =\frac{1}{15} \\<br/>v_3 & =12 \mathrm{~cm}<br/>\end{aligned}<br/>$$<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/GTmt2gH5CzNqArz52bKt-jDaojEvoq9Fy1CFfBtJbOY.original.fullsize.png"/><br/></div> | MarksBatch1_P1.db |
156 | a-police-car-with-a-siren-of-frequency-8-khz-is-moving-with-uniform-velocity-36-km-h-towards-a-tall-building-which-reflects-the-sound-waves-the-speed- | a-police-car-with-a-siren-of-frequency-8-khz-is-moving-with-uniform-velocity-36-km-h-towards-a-tall-building-which-reflects-the-sound-waves-the-speed-23247 | <div class="question">A police car with a siren of frequency $8 \mathrm{kHz}$ is moving with uniform velocity $36 \mathrm{~km} / \mathrm{h}$ towards a tall building which reflects the sound waves. The speed of sound in air is $320 \mathrm{~m} / \mathrm{s}$. The frequency of the siren heard by the car driver is</div> | ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$8.50 \mathrm{kHz}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$8.25 \mathrm{kHz}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$7.75 \mathrm{kHz}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$7.50 \mathrm{kHz}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$8.50 \mathrm{kHz}$<br/></span> </div> | <div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/N3B2m51q3dVOLcZ5J59iF2tGdvIQA5PNpL1HMwO7NI0.original.fullsize.png"/><br/><br/>Apparent frequency of sound heard by car driver (observer) reflected from the building will be<br/>$$<br/>\begin{aligned}<br/>f^{\prime} & =f\left(\frac{v+v_0}{v-v_s}\right) \\<br/>& =8\left(\frac{320+10}{320-10}\right) \\<br/>& =8.5 \mathrm{kHz}<br/>\end{aligned}<br/>$$<br/>$\therefore$ Correct option is (a).<br/>Analysis of Question<br/>(i) Question is simple.<br/>(ii) Driver will listen two sounds, direct and reflected. Direct sound will be of $8 \mathrm{kHz}$ as driver has no relative motion with the car. But reflected sound is of increased frequency because driver and image of car both are approaching towards each other.</div> | MarksBatch1_P1.db |
157 | a-positron-is-emitted-from-11-23-na-the-ratio-of-the-atomic-mass-and-atomic-number-of-the-resulting-nuclide-is | a-positron-is-emitted-from-11-23-na-the-ratio-of-the-atomic-mass-and-atomic-number-of-the-resulting-nuclide-is-87482 | <div class="question">A positron is emitted from ${ }_{11}^{23} \mathrm{Na}$. The ratio of the atomic mass and atomic number of the resulting nuclide is</div> | ['Chemistry', 'Chemical Kinetics', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>22/10<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>22/11<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>23/10<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$23 / 12$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>23/10<br/></span> </div> | <div class="solution">$$<br/>{ }_{11}^{23} \mathrm{Na} \longrightarrow{ }_{10}^{23} \mathrm{Na}+{ }_{-1}^0 e<br/>$$</div> | MarksBatch1_P1.db |
158 | a-proton-is-fired-from-very-far-away-towards-a-nucleus-with-charge-q-120-e-where-e-is-the-electronic-charge-it-makes-a-closest-approach-of-10-fm-to-th | a-proton-is-fired-from-very-far-away-towards-a-nucleus-with-charge-q-120-e-where-e-is-the-electronic-charge-it-makes-a-closest-approach-of-10-fm-to-th-57892 | <div class="question">A proton is fired from very far away towards a nucleus with charge $Q=120 e$, where $e$ is the electronic charge. It makes a closest approach of $10 \mathrm{fm}$ to the nucleus. The de Broglie wavelength (in units of $\mathrm{fm}$ ) of the proton at its start is: (take the proton mass, $m_{p}=(5 / 3) \times 10^{-27} \mathrm{~kg} ; h / e=4.2 \times$ $10^{-15} \mathrm{~J} . \mathrm{S} / \mathrm{C}$;
<br/> <br/>$\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~m} / \mathrm{F} ; 1 \mathrm{fm}=10^{-15} \mathrm{~m}$</div> | ['Physics', 'Dual Nature of Matter', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">7</span> </div> | <div class="solution">From energy conservation Loss in K.E. of proton = gain in potential energy of the proton - nucleus system
<br/> <br/>$\begin{array}{c}
<br/> <br/>\frac{1}{2} m v^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \quad \therefore \frac{p^{2}}{2 m}=\frac{1}{4 \pi \in_{0}} \frac{q_{1} q_{2}}{r} \\
<br/> <br/>\therefore \frac{1}{2 m}\left(\frac{h^{2}}{\lambda^{2}}\right)=\frac{1}{4 \pi \in_{0}} \frac{q_{1} q_{2}}{r} \quad \therefore \lambda=\sqrt{\frac{4 \pi \epsilon_{0} r \cdot h^{2}}{q_{1} q_{2}(2 m)}}
<br/> <br/>\end{array}$
<br/> <br/>Putting the values of $4 \pi \varepsilon_{0}, r, h, q_{1}, q_{2}$ and $m$ we get, deBroglie wavelength of proton, $\lambda=7 \mathrm{fm}$</div> | MarksBatch1_P1.db |
159 | a-radioactive-sample-s-1-having-an-activity-of-5-ci-has-twice-the-number-of-nuclei-as-another-sample-s-2-which-has-an-activity-of-10-ci-the-half-lives-1 | a-radioactive-sample-s-1-having-an-activity-of-5-ci-has-twice-the-number-of-nuclei-as-another-sample-s-2-which-has-an-activity-of-10-ci-the-half-lives-1-93785 | <div class="question">A radioactive sample $S_1$ having an activity of $5 \mu \mathrm{Ci}$ has twice the number of nuclei as another sample $S_2$ which has an activity of $10 \mu \mathrm{Ci}$. The half lives of $S_1$ and $S_2$ can be</div> | ['Physics', 'Nuclear Physics', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>20 yr and 5 yr, respectively<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>20 yr and 10 yr, respectively<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>10 yr each<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>5 yr each</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>20 yr and 5 yr, respectively<br/></span> </div> | <div class="solution">Activity of $S_1=\frac{1}{2}$ (activity of $S_2$ )<br/>or<br/>$$<br/>\lambda_2 N_1=\frac{1}{2}\left(\lambda_2 N_2\right) \text { or } \frac{\lambda_1}{\lambda_2}=\frac{N_2}{2 N_1}<br/>$$<br/>or<br/>$$<br/>\begin{aligned}<br/>& \frac{T_1}{T_2}=\frac{2 N_1}{N_2}\left(T=\text { half-life }=\frac{\ln 2}{\lambda}\right) \\<br/>& N_1=2 N_2 \\<br/>& \frac{T_1}{T_2}=4<br/>\end{aligned}<br/>$$<br/>Given<br/>$\therefore$ correct option is (a).</div> | MarksBatch1_P1.db |
160 | a-ray-of-light-travelling-in-water-is-incident-on-its-surface-open-to-air-the-angle-of-incidence-is-which-is-less-than-the-critical-angle-then-there-w | a-ray-of-light-travelling-in-water-is-incident-on-its-surface-open-to-air-the-angle-of-incidence-is-which-is-less-than-the-critical-angle-then-there-w-16051 | <div class="question">A ray of light travelling in water is incident on its surface open to air. The angle of incidence is $\theta$, which is less than the critical angle. Then there will be</div> | ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>only a reflected ray and no refracted ray<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>only a refracted ray and no reflected ray<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>a reflected ray and refracted ray and the angle between them would be less than $180^{\circ}-2 \theta$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>a reflected ray and a refracted ray and the angle between them would be greater than $180^{\circ}-2 \theta$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>a reflected ray and refracted ray and the angle between them would be less than $180^{\circ}-2 \theta$<br/></span> </div> | <div class="solution">Since $\theta < \theta_c$, both reflection and refraction will take place. From the figure we can see that angle between reflected and refracted rays $\alpha$ is less than $180^{\circ}-2 \theta$.<br/>$\therefore$ Option (c) is correct.<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/8tpGTJwd4YDjGKQxEU46fhTSHh_eezAmuy-VX6ti1gw.original.fullsize.png"/><br/></div> | MarksBatch1_P1.db |
161 | a-ray-op-of-monochromatic-light-is-incident-on-the-face-a-b-of-prism-a-bc-d-near-vertex-b-at-an-incident-angle-of-6-0-see-figure-if-the-refractive-ind | a-ray-op-of-monochromatic-light-is-incident-on-the-face-a-b-of-prism-a-bc-d-near-vertex-b-at-an-incident-angle-of-6-0-see-figure-if-the-refractive-ind-79168 | <div class="question">A ray $O P$ of monochromatic light is incident on the face $A B$ of prism $A B C D$ near vertex $B$ at an incident angle of $60^{\circ}$ (see figure). If the refractive index of the material of the prism is $\sqrt{3}$, which of the following is (are) correct?<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/Y0fLkA7kftBwNbLUTcG8moZnPGiATLL8kEguqIkDJW4.original.fullsize.png"/><br/></div> | ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>The ray gets totally internally reflected at face $C D$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>The ray comes out through face $A D$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>The angle between the incident ray and the emergent ray is $90^{\circ}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>The angle between the incident ray and the emergent ray is $120^{\circ}$</span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>The ray gets totally internally reflected at face $C D$<br/>, <br/>The ray comes out through face $A D$<br/>, <br/>The angle between the incident ray and the emergent ray is $90^{\circ}$<br/></span> </div> | <div class="solution">$$<br/>\begin{aligned}<br/>& \sqrt{3}=\frac{\sin 60^{\circ}}{\sin r} \\<br/>& \therefore r=30^{\circ}<br/>\end{aligned}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OqCuJa-SSDWftpM_SKULPN9o12Ag40EDBSrs9Q8ivPw.original.fullsize.png"/><br/><br/>$$<br/>\theta_C=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)<br/>$$<br/>or $\sin \theta_C=\frac{1}{\sqrt{3}}=0.577$<br/>At point $Q$, angle of incidence inside the prism is $i=45^{\circ}$.<br/>Since $\sin i=\frac{1}{\sqrt{2}}$ is greater than $\sin \theta_C=\frac{1}{\sqrt{2}}$, ray gets totally internally reflected at face $C D$. Path of ray of light after point $Q$ is shown in figure.<br/>From the figure, we can see that angle between incident ray $O P$ and emergent ray $R S$ is $90^{\circ}$.<br/>Therefore, correct options are (a), (b) and (c).</div> | MarksBatch1_P1.db |
162 | a-real-gas-behaves-like-an-ideal-gas-if-its | a-real-gas-behaves-like-an-ideal-gas-if-its-37638 | <div class="question">A real gas behaves like an ideal gas if its</div> | ['Physics', 'Kinetic Theory of Gases', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>pressure and temperature are both high<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>pressure and temperature are both low<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>pressure is high and temperature is low<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>pressure is low and temperature is high</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>pressure is low and temperature is high</span> </div> | <div class="solution">A real gas behaves like an ideal gas at low pressure and high temperature. $\therefore$ correct option is (d).</div> | MarksBatch1_P1.db |
163 | a-resistance-of-2-is-connected-across-one-gap-of-a-meterbridge-the-length-of-the-wire-is-100-cm-and-an-unknown-resistance-greater-than-2-is-connected--1 | a-resistance-of-2-is-connected-across-one-gap-of-a-meterbridge-the-length-of-the-wire-is-100-cm-and-an-unknown-resistance-greater-than-2-is-connected-1-23356 | <div class="question">A resistance of $2 \Omega$ is connected across one gap of a meter-bridge (the length of the wire is $100 \mathrm{~cm}$ ) and an unknown resistance, greater than $2 \Omega$, is connected across the other gap. When these resistances are interchanged, the balance point shifts by $20 \mathrm{~cm}$. Neglecting any corrections, the unknown resistance is</div> | ['Physics', 'Current Electricity', 'JEE Main'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$3 \Omega$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$4 \Omega$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$5 \Omega$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$6 \Omega$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$3 \Omega$<br/></span> </div> | <div class="solution">$$<br/>\begin{aligned}<br/>& R & >2 \Omega \\<br/>\therefore & 100-x & >x \\<br/>\text { Applying } & \frac{P}{Q} & =\frac{R}{S}<br/>\end{aligned}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/bj5wXDgKnTfN4Vg1PIwLyg-p-M4Vw5b76zBexKtpjvI.original.fullsize.png"/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/k0HDe0hHZT2EmygS2hd4fIRONvnb06ZJCeC2L7TIA9I.original.fullsize.png"/><br/><br/>We have<br/>$$<br/>\begin{aligned}<br/>& \frac{2}{R}=\frac{x}{100-x} \\<br/>& \frac{R}{2}=\frac{x+20}{80-x}<br/>\end{aligned}<br/>$$<br/>Solving Eqs. (i) and (ii), we get<br/>$$<br/>R=3 \Omega<br/>$$<br/>$\therefore$ Answer is (a).</div> | MarksBatch1_P1.db |
164 | a-resistance-of-2-is-connected-across-one-gap-of-a-meterbridge-the-length-of-the-wire-is-100-cm-and-an-unknown-resistance-greater-than-2-is-connected- | a-resistance-of-2-is-connected-across-one-gap-of-a-meterbridge-the-length-of-the-wire-is-100-cm-and-an-unknown-resistance-greater-than-2-is-connected-19203 | <div class="question">A resistance of $2 \Omega$ is connected across one gap of a meter-bridge (the length of the wire is $100 \mathrm{~cm}$ ) and an unknown resistance, greater than $2 \Omega$, is connected across the other gap. When these resistances are interchanged, the balance point shifts by $20 \mathrm{~cm}$. Neglecting any corrections, the unknown resistance is</div> | ['Physics', 'Current Electricity', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$3 \Omega$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$4 \Omega$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$5 \Omega$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$6 \Omega$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$3 \Omega$<br/></span> </div> | <div class="solution">$$<br/>\begin{aligned}<br/>& R & >2 \Omega \\<br/>\therefore & 100-x & >x \\<br/>\text { Applying } & \frac{P}{Q} & =\frac{R}{S}<br/>\end{aligned}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/bj5wXDgKnTfN4Vg1PIwLyg-p-M4Vw5b76zBexKtpjvI.original.fullsize.png"/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/k0HDe0hHZT2EmygS2hd4fIRONvnb06ZJCeC2L7TIA9I.original.fullsize.png"/><br/><br/>We have<br/>$$<br/>\begin{aligned}<br/>& \frac{2}{R}=\frac{x}{100-x} \\<br/>& \frac{R}{2}=\frac{x+20}{80-x}<br/>\end{aligned}<br/>$$<br/>Solving Eqs. (i) and (ii), we get<br/>$$<br/>R=3 \Omega<br/>$$<br/>$\therefore$ Answer is (a).</div> | MarksBatch1_P1.db |
165 | a-satellite-is-moving-with-a-constant-speed-v-in-a-circular-orbit-about-the-earth-an-object-of-mass-m-is-ejected-from-the-satellite-such-that-it-just- | a-satellite-is-moving-with-a-constant-speed-v-in-a-circular-orbit-about-the-earth-an-object-of-mass-m-is-ejected-from-the-satellite-such-that-it-just-78503 | <div class="question">A satellite is moving with a constant speed $v$ in a circular orbit about the earth. An object of mass $m$ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is</div> | ['Physics', 'Gravitation', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{1}{2} m v^2$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$m v^2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{3}{2} m v^2$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$2 m v^2$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$m v^2$<br/></span> </div> | <div class="solution">In circular orbit of a satellite, potential energy<br/>$$<br/>\begin{aligned}<br/>& =-2 \times(\text { kinetic energy) } \\<br/>& =-2 \times \frac{1}{2} m v^2=-m v^2<br/>\end{aligned}<br/>$$<br/>Just to escape from the gravitational pull, its total mechanical energy should be zero. Therefore, its kinetic energy should be $+m v^2$.<br/>$\therefore$ Correct answer is (b).<br/>Analysis of Question<br/>(i) Question is moderately difficult.<br/>(ii) In circular orbit of satellite $U=-\frac{G M m}{r}$ and $K=\frac{G M m}{2 r}$ or $U=-2 K$<br/>Here, $U$ is potential energy and $K$ is kinetic energy.</div> | MarksBatch1_P1.db |
166 | a-series-r-c-circuit-is-connected-to-a-c-voltage-source-consider-two-cases-a-when-c-is-without-a-dielectric-medium-and-b-when-c-is-filled-with-dielect | a-series-r-c-circuit-is-connected-to-a-c-voltage-source-consider-two-cases-a-when-c-is-without-a-dielectric-medium-and-b-when-c-is-filled-with-dielect-12486 | <div class="question">A series $R$ - $C$ circuit is connected to $A C$ voltage source. Consider two cases; $(A)$ when $C$ is without a dielectric medium and $(B)$ when $C$ is filled with dielectric of constant 4. The current $I_R$ through the resistor and voltage $V_C$ across the capacitor are compared in the two cases. Which of the following is/are true?</div> | ['Physics', 'Alternating Current', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$I_R^A>I_R^B$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$I_R^A < I_R^B$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$V_C^A>V_C^B$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$V_C^A < V_C^B$</span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$I_R^A < I_R^B$, <br/>$V_C^A>V_C^B$<br/></span> </div> | <div class="solution">$Z=\sqrt{R^2+X_C^2}=\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}$<br/>In case (b) capacitance $C$ will be more.<br/>Therefore, impedance $Z$ will be less.<br/>Hence, current will be more.<br/>$\therefore$ Option (b) is correct.<br/>Further, $\quad \begin{aligned} V_C & =\sqrt{V^2-V_R^2} \\ & =\sqrt{V^2-(I R)^2}\end{aligned}$<br/>In case (b), since current $I$ is more.<br/>Therefore, $V_C$ will be less.<br/>$\therefore$ Option (c) is correct.<br/>$\therefore$ Correct options are (b) and (c).<br/>Analysis of Question<br/>(i) Question is moderately difficult.<br/>(ii) In my opinion problems of alternating currents are not very difficult.<br/>(iii) Topic of $A C$ is small. One can feel comfortable in this topic by putting less efforts.</div> | MarksBatch1_P1.db |
167 | a-series-r-c-combination-is-connected-to-an-ac-voltage-of-angular-frequency-500-rad-s-if-the-impedance-of-the-r-c-circuit-is-r-125-the-time-constant-i | a-series-r-c-combination-is-connected-to-an-ac-voltage-of-angular-frequency-500-rad-s-if-the-impedance-of-the-r-c-circuit-is-r-125-the-time-constant-i-48889 | <div class="question">A series $R-C$ combination is connected to an $\mathrm{AC}$ voltage of angular frequency $\omega=500 \mathrm{rad} / \mathrm{s}$. If the impedance of the $R-C$ circuit is $R \sqrt{1.25}$, the time constant (in millisecond) of the circuit is</div> | ['Physics', 'Alternating Current', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">4</span> </div> | <div class="solution">$Z=\sqrt{R^2+X_C^2}=R \sqrt{1.25}$<br/>$$<br/>\therefore \quad R^2+X_C^2=1.25 R^2<br/>$$<br/>or $\quad X_C=\frac{R}{2}$<br/>or $\quad \frac{1}{\omega C}=\frac{R}{2}$<br/>$\begin{aligned} \therefore \text { Time constant } & =C R=\frac{2}{\omega} \\ & =\frac{2}{500} \mathrm{~s}=4 \mathrm{~ms}\end{aligned}$<br/>$\therefore$ Answer is 4 .<br/>Analysis of Question<br/>(i) Question is very simple.<br/>(ii) I think this is one of the simplest formula based question of this paper.</div> | MarksBatch1_P1.db |
168 | a-ship-is-fitted-with-three-engines-e-1-e-2-and-e-3-the-engines-function-independently-of-each-other-with-respective-probabilities-2-1-4-1-and-4-1-for | a-ship-is-fitted-with-three-engines-e-1-e-2-and-e-3-the-engines-function-independently-of-each-other-with-respective-probabilities-2-1-4-1-and-4-1-for-12411 | <div class="question">A ship is fitted with three engines $E_{1}, E_{2}$ and $E_{3}$. The engines function independently of each other with respective probabilities $\frac{1}{2}, \frac{1}{4}$ and $\frac{1}{4}$. For the ship to be operational at least two of its engines must function. Let $X$ denote the event that the ship is operational and let $X_{1}, X_{2}$ and $X_{3}$ denote respectively the events that the engines $E_{1}, E_{2}$ and $E_{3}$ are functioning. Which of the following is(are) true?</div> | ['Mathematics', 'Probability', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$P\left[X_{1}^{c} \mid X\right]=\frac{3}{16}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$P$ [Exactly two engines of the ship are functioning $[X]=\frac{7}{8}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$P\left[X \mid X_{2}\right]=\frac{5}{16}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$P\left[X \mid X_{1}\right]=\frac{7}{16}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value">$P$ [Exactly two engines of the ship are functioning $[X]=\frac{7}{8}$, $P\left[X \mid X_{1}\right]=\frac{7}{16}$</span> </div> | <div class="solution">Given that $P\left(X_{1}\right)=\frac{1}{2}, \mathrm{P}\left(X_{2}\right)=\frac{1}{4}, \mathrm{P}\left(X_{3}\right)=\frac{1}{4}$ $P(X)=P$ (at least 2 engines are functioning)
<br/> <br/>$\begin{aligned}=P\left(X_{1} \cap X_{2}\right.&\left.\cap X_{3}^{C}\right)+P\left(X_{1} \cap X_{2}^{C} \cap X_{3}\right) \\ &+P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}\right) \end{aligned}$
<br/> <br/>$=\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{4}$
<br/> <br/>(a) $P\left(X_{1}^{C} / X\right)=\frac{P\left(X_{1}^{C} \cap X\right)}{P(X)}=\frac{P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)}{P(X)}$
<br/> <br/>$=\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}}{\frac{1}{4}}=\frac{1}{8}$
<br/> <br/>$\therefore$ (a) is not true.
<br/> <br/>(b) $P$ [Exactly two engines are functioning $/ X]$
<br/> <br/>$=\frac{P[(\text { Exactly two engines are functioning }) \cap X]}{P(X)}$
<br/> <br/>$=\frac{P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2}^{C} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}^{C}\right)}{P(X)}$
<br/> <br/>$=\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}}=\frac{7}{8}$
<br/> <br/>$\therefore$ (b) is true.
<br/> <br/>(c) $P\left(X / X_{2}\right)=\frac{P\left(X \cap X_{2}\right)}{P\left(X_{2}\right)}$
<br/> <br/>$=\frac{P\left(X_{1} \cap X_{2} \cap X_{3}\right)+P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}^{C}\right)}{P\left(X_{2}\right)}$
<br/> <br/>$=\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}}=\frac{5}{8}$
<br/> <br/>$\therefore$ (c) is not true.
<br/> <br/>(d) $P\left(X / X_{1}\right)=\frac{P\left(X \cap X_{1}\right)}{P\left(X_{1}\right)}$
<br/> <br/>$=\frac{P\left(X_{1} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2}^{C} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}^{C}\right)}{P\left(X_{1}\right)}$
<br/> <br/>$=\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{2}}=\frac{7}{16}$
<br/> <br/>$\therefore$ (d) is true.</div> | MarksBatch1_P1.db |
169 | a-signal-which-can-be-green-or-red-with-probability-5-4-and-5-1-respectively-is-received-by-station-a-and-then-transmitted-to-station-b-the-probabilit | a-signal-which-can-be-green-or-red-with-probability-5-4-and-5-1-respectively-is-received-by-station-a-and-then-transmitted-to-station-b-the-probabilit-82953 | <div class="question">A signal which can be green or red with probability $\frac{4}{5}$ and $\frac{1}{5}$ respectively, is received by station $A$ and then transmitted to station B. The probability of each station receiving the signal correctly is $\frac{3}{4}$. If the signal received at station $B$ is green, then the probability that the original signal green is</div> | ['Mathematics', 'Probability', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{3}{5}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{6}{7}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{20}{23}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{9}{20}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{20}{23}$<br/></span> </div> | <div class="solution">From the tree-diagram it follows that<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/l4-0jmsEW2b3RYr9-D6RJ2U3VsRy7evmHsKptiune_4.original.fullsize.png"/><br/><br/>$$<br/>\begin{gathered}<br/>P\left(B_G \mid G\right)=\frac{10}{16}=\frac{5}{8} \\<br/>\therefore \quad P\left(B_G \cap G\right)=\frac{5}{8} \times \frac{4}{5}=\frac{1}{2} \\<br/>P\left(G \mid B_G\right)=\frac{\frac{1}{2}}{P\left(B_G\right)}=\frac{1}{2} \times \frac{80}{46}=\frac{20}{23}<br/>\end{gathered}<br/>$$</div> | MarksBatch1_P1.db |
170 | a-silver-sphere-of-radius-1-cm-and-work-function-47-ev-is-suspended-from-an-insulating-thread-in-freespace-it-is-under-continuous-illumination-of-200--2 | a-silver-sphere-of-radius-1-cm-and-work-function-47-ev-is-suspended-from-an-insulating-thread-in-freespace-it-is-under-continuous-illumination-of-200-2-35541 | <div class="question">A silver sphere of radius $1 \mathrm{~cm}$ and work function $4.7 \mathrm{eV}$ is suspended from an insulating thread in free-space. It is under continuous illumination of $200 \mathrm{~nm}$ wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is $A \times 10^Z$ (where $1 < A < 10$ ). The value of $Z$ is</div> | ['Physics', 'Dual Nature of Matter', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">7</span> </div> | <div class="solution">Photoemission will stop when potential on silver sphere becomes equal to the stopping potential.<br/>$\therefore \quad \frac{h c}{\lambda}-W=e V_0$<br/>Here, $\quad V_0=\frac{1}{4 \pi \varepsilon_0} \frac{n e}{r}$<br/>$$<br/>\begin{aligned}<br/>& \therefore \quad\left(\frac{1240}{1200} \mathrm{eV}\right)-(4.7 \mathrm{eV}) \\<br/>&=\frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}} \\<br/>&(6.2-4.7)=\frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}} \\<br/>& \text { or } \quad n= \frac{1.5 \times 10^{-2}}{9 \times 1.6 \times 10^{-10}}=1.04 \times 10^7<br/>\end{aligned}<br/>$$<br/>$\therefore$ Answer is 7 .<br/>Analysis of Question<br/>(i) Question is moderately difficult.<br/>(ii) In $\mathrm{eV}_0$ if we substitute the value of $e=1.6 \times 10^{-19} \mathrm{C}$, then answer is in J. If we leave it then, answer is in $e \mathrm{~V}$.</div> | MarksBatch1_P1.db |
171 | a-simple-telescope-used-to-view-distant-objects-has-eyepiece-and-objective-lens-of-focal-lengths-f-e-and-f-0-respectively-then | a-simple-telescope-used-to-view-distant-objects-has-eyepiece-and-objective-lens-of-focal-lengths-f-e-and-f-0-respectively-then-14648 | <div class="question">A simple telescope used to view distant objects has eyepiece and objective lens of focal lengths $f_e$ and $f_0$, respectively. Then<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/ijYR7qiWVf7-qQhgeKaI74NNU25wHAxcWA_wEDPcLlc.original.fullsize.png"/><br/></div> | ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>(a) p, (b) r, (c) r, (d) p,q,s<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>(a) q, (b) r, (c) s, (d) p,s<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>(a) p, (b) p, r, (c) q, (d) p,q<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>(a) q, (b) s, (c) r, (d) p,q,s</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>(a) p, (b) r, (c) r, (d) p,q,s<br/></span> </div> | <div class="solution">(a) $\rightarrow p$<br/>(b) $\rightarrow r$<br/>(c) $\rightarrow r$<br/>(d) $\rightarrow$ p, q, s</div> | MarksBatch1_P1.db |
172 | a-small-block-is-connected-to-one-end-of-a-massless-spring-of-unstretched-length-49-m-the-other-end-of-the-spring-see-the-figure-is-fixed-the-system-l-1 | a-small-block-is-connected-to-one-end-of-a-massless-spring-of-unstretched-length-49-m-the-other-end-of-the-spring-see-the-figure-is-fixed-the-system-l-1-70589 | <div class="question">A small block is connected to one end of a massless spring of un-stretched length $4.9 \mathrm{~m}$. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by $0.2 \mathrm{~m}$ and released from rest at $t=0$. It then executes simple harmonic motion with angular frequency $\omega=\pi / 3 \mathrm{rad} / \mathrm{s}$. Simultaneously at $t=0$, a small pebble is projected with speed $v$ form point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \mathrm{~m}$ from $O$. If the pebble hits the block at $t=1 \mathrm{~s}$, the value of $v$ is (take $g$ $=10 \mathrm{~m} / \mathrm{s}^{2}$ )
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/8JfHbEDRw9CzUF3KhntqqOIXAPC3H9itqgYrnpnqqJI.original.fullsize.png"/></div> | ['Physics', 'Motion In Two Dimensions', 'JEE Main'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\sqrt{50} \mathrm{~m} / \mathrm{s}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\sqrt{51} \mathrm{~m} / \mathrm{s}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\sqrt{52} \mathrm{~m} / \mathrm{s}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\sqrt{53} \mathrm{~m} / \mathrm{s}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value">$\sqrt{50} \mathrm{~m} / \mathrm{s}$</span> </div> | <div class="solution">Time of flight of projectile,
<br/> <br/>$T=\frac{2 V \sin \theta}{g}$
<br/> <br/>$\therefore \quad 1=\frac{2 V \sin 45^{\circ}}{g} \quad \therefore \quad V=\sqrt{50} \mathrm{~ms}^{-1}$
<br/> <br/>Hence pebble is projected with a speed $V=\sqrt{50} \mathrm{~ms}^{-1}$</div> | MarksBatch1_P1.db |
173 | a-small-block-is-connected-to-one-end-of-a-massless-spring-of-unstretched-length-49-m-the-other-end-of-the-spring-see-the-figure-is-fixed-the-system-l | a-small-block-is-connected-to-one-end-of-a-massless-spring-of-unstretched-length-49-m-the-other-end-of-the-spring-see-the-figure-is-fixed-the-system-l-85437 | <div class="question">A small block is connected to one end of a massless spring of un-stretched length $4.9 \mathrm{~m}$. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by $0.2 \mathrm{~m}$ and released from rest at $t=0$. It then executes simple harmonic motion with angular frequency $\omega=\pi / 3 \mathrm{rad} / \mathrm{s}$. Simultaneously at $t=0$, a small pebble is projected with speed $v$ form point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \mathrm{~m}$ from $O$. If the pebble hits the block at $t=1 \mathrm{~s}$, the value of $v$ is (take $g$ $=10 \mathrm{~m} / \mathrm{s}^{2}$ )
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/8JfHbEDRw9CzUF3KhntqqOIXAPC3H9itqgYrnpnqqJI.original.fullsize.png"/></div> | ['Physics', 'Motion In Two Dimensions', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\sqrt{50} \mathrm{~m} / \mathrm{s}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\sqrt{51} \mathrm{~m} / \mathrm{s}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\sqrt{52} \mathrm{~m} / \mathrm{s}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\sqrt{53} \mathrm{~m} / \mathrm{s}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value">$\sqrt{50} \mathrm{~m} / \mathrm{s}$</span> </div> | <div class="solution">Time of flight of projectile,
<br/> <br/>$T=\frac{2 V \sin \theta}{g}$
<br/> <br/>$\therefore \quad 1=\frac{2 V \sin 45^{\circ}}{g} \quad \therefore \quad V=\sqrt{50} \mathrm{~ms}^{-1}$
<br/> <br/>Hence pebble is projected with a speed $V=\sqrt{50} \mathrm{~ms}^{-1}$</div> | MarksBatch1_P1.db |
174 | a-small-block-of-mass-of-01-kg-lies-on-a-fixed-inclined-plane-pq-which-makes-an-angle-with-the-horizontal-a-horizontal-force-of-1-n-acts-on-the-block- | a-small-block-of-mass-of-01-kg-lies-on-a-fixed-inclined-plane-pq-which-makes-an-angle-with-the-horizontal-a-horizontal-force-of-1-n-acts-on-the-block-82917 | <div class="question">A small block of mass of $0.1 \mathrm{~kg}$ lies on a fixed inclined plane PQ which makes an angle $\theta$ with the horizontal. A horizontal force of $1 \mathrm{~N}$ acts on the block through its centre of mass as shown in the figure.
<br/> <br/>The block remains stationary if (take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/eyWmHlpMbKkU5KNCI11wped3TI_abWjiG9pQcOAesZI.original.fullsize.png"/><br/></div> | ['Physics', 'Laws of Motion', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\theta=45^{\circ}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\theta>45^{\circ}$ and a frictional force acts on the block towards P.</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$\theta>45^{\circ}$ and a frictional force acts on the block towards Q.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\theta < 45^{\circ}$ and a frictional force acts on the block towards Q.</span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value">$\theta=45^{\circ}$, $\theta>45^{\circ}$ and a frictional force acts on the block towards Q.</span> </div> | <div class="solution">The various forces acting on the block are as shown in the figure.
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/aswDGCZDDEq0RtQdOpKwJtq4gJ_Yh9XUhPtzcG6z2p4.original.fullsize.png"/><br/> <br/> <br/>When $\theta=45^{\circ}, \sin \theta=\cos \theta$
<br/> <br/>The block will remain stationary and the frictional force is zero.
<br/> <br/>When $\theta>45^{\circ}, \sin \theta>\cos \theta$
<br/> <br/>Therefore a frictional force acts towards $Q$.
<br/> <br/>When $\theta < 45^{\circ}, \cos \theta>\sin \theta$
<br/> <br/>Therefore a frictional force acts towards $P$.</div> | MarksBatch1_P1.db |
175 | a-small-circular-loop-of-area-a-and-resistance-r-is-fixed-on-a-horizontal-x-y-plane-with-the-center-of-the-loop-always-on-the-axis-n-of-a-long-solenoi-1 | a-small-circular-loop-of-area-a-and-resistance-r-is-fixed-on-a-horizontal-x-y-plane-with-the-center-of-the-loop-always-on-the-axis-n-of-a-long-solenoi-1-58555 | <div class="question"><p>A small circular loop of area <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>A</mi></math> and resistance <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> is fixed on a horizontal <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mi>y</mi></math>-plane with the center of the loop always on the axis <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math> of a long solenoid. The solenoid has <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> turns per unit length and carries current <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>I</mi></math> counter clockwise as shown in the figure. The magnetic field due to the solenoid is in <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math> direction. List-I gives time dependences of <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math> in terms of a constant angular frequency <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ω</mi></math>. List-II gives the torques experienced by the circular loop at time <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mo>=</mo><mfrac><mi>π</mi><mrow><mn>6</mn><mi>ω</mi></mrow></mfrac></math>, Let <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>α</mi><mo>=</mo><mfrac><mrow><msup><mi>A</mi><mn>2</mn></msup><msubsup><mi>μ</mi><mn>0</mn><mn>2</mn></msubsup><msup><mi>m</mi><mn>2</mn></msup><msup><mi>I</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac></math>.</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/3dbd710a-2aec-4891-a220-baef9f053b38-image.png" style="width: 220px; height: 150px;"/></p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td style="text-align: center;"> </td> <td style="text-align: center;"><strong>List-I</strong></td> <td style="text-align: center;"> </td> <td style="text-align: center;"><strong>List-II</strong></td> </tr> <tr> <td>(i)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(p)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn></math></td> </tr> <tr> <td>(ii)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(q)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mfrac><mi>α</mi><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></td> </tr> <tr> <td>(iii)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(r)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>3</mn><mi>α</mi></mrow><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></td> </tr> <tr> <td>(iv)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover><mo>+</mo><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(s)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>α</mi><mn>4</mn></mfrac><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></td> </tr> <tr> <td> </td> <td> </td> <td>(t)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mfrac><mrow><mn>3</mn><mi>α</mi></mrow><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></td> </tr> </tbody></table><p>Which one of the following options is correct?</p></div> | ['Physics', 'Magnetic Effects of Current', 'JEE Main'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">t</mi></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">t</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">p</mi></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">r</mi></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">t</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">r</mi></math></span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">r</mi></math></span> </div> | <div class="solution"><blockquote><p>Note: This question was given bonus by JEE council.</p><p>The magnetic field in a solenoid is given by, <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><msub><mi>μ</mi><mn>0</mn></msub><mi>m</mi><mi>I</mi><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math>. Therefore, the flux through the loop will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfenced><mrow><mi>A</mi><mover><mi>k</mi><mo>^</mo></mover></mrow></mfenced><mo>·</mo><mfenced><mrow><msub><mi>μ</mi><mn>0</mn></msub><mi>m</mi><mi>I</mi><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></mrow></mfenced></math></p><p>For case I:</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p>The emf will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ε</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p>Therefore, the current in the loop will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p>The magnetic moment of the loop will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>m</mi><mo>→</mo></mover><mo>=</mo><mi>i</mi><mi>A</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>=</mo><mfrac><mrow><mi>B</mi><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></math></p><p>Therefore, the torque</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mover><mi>m</mi><mo>→</mo></mover><mo>×</mo><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mfenced><mrow><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>×</mo><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced close="]" open="["><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></mfenced><msup><mi>sin</mi><mn>2</mn></msup><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced close="]" open="["><mrow><msup><mi>sin</mi><mn>2</mn></msup><mfenced><mfrac><mi>π</mi><mn>6</mn></mfrac></mfenced></mrow></mfenced><mo>=</mo><mfrac><mrow><mo>-</mo><mi>α</mi></mrow><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi mathvariant="normal">I</mi></mfenced><mo>→</mo><mi>q</mi></math></p><p>For case II</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mn>0</mn></math></p><p>Therefore, we can directly write <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>τ</mi><mo>=</mo><mn>0</mn></math></p><p>Hence <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>II</mi></mfenced><mo>→</mo><mi>p</mi></math></p><p>Following the same process as in case I for case III</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>m</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><mi>B</mi><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mover><mi>m</mi><mo>→</mo></mover><mo>×</mo><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mo>×</mo><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mi>ω</mi><mi>t</mi><mfenced><mrow><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>×</mo><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><msup><mi>sin</mi><mn>2</mn></msup><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mfrac><mi>α</mi><mn>4</mn></mfrac><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>III</mi><mo>→</mo><mi>s</mi></math></p><p>Following the same process as in case I for case IV</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>=</mo><mo>-</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>m</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><mi>B</mi><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>cos</mi><mi>ω</mi><mi>t</mi><mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mover><mi>m</mi><mo>→</mo></mover><mo>×</mo><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced><mrow><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>×</mo><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></mrow></mfenced><msup><mi>cos</mi><mn>2</mn></msup><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>+</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></mfenced><mo>·</mo><msup><mi>cos</mi><mn>2</mn></msup><mfenced><mfrac><mi>π</mi><mn>6</mn></mfrac></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>+</mo><mfrac><mn>3</mn><mn>4</mn></mfrac><mi>α</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>IV</mi></mfenced><mo>→</mo><mi>r</mi></math></p></blockquote></div> | MarksBatch1_P1.db |
176 | a-small-circular-loop-of-area-a-and-resistance-r-is-fixed-on-a-horizontal-x-y-plane-with-the-center-of-the-loop-always-on-the-axis-n-of-a-long-solenoi | a-small-circular-loop-of-area-a-and-resistance-r-is-fixed-on-a-horizontal-x-y-plane-with-the-center-of-the-loop-always-on-the-axis-n-of-a-long-solenoi-86013 | <div class="question"><p>A small circular loop of area <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>A</mi></math> and resistance <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> is fixed on a horizontal <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mi>y</mi></math>-plane with the center of the loop always on the axis <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math> of a long solenoid. The solenoid has <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> turns per unit length and carries current <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>I</mi></math> counter clockwise as shown in the figure. The magnetic field due to the solenoid is in <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math> direction. List-I gives time dependences of <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math> in terms of a constant angular frequency <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ω</mi></math>. List-II gives the torques experienced by the circular loop at time <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mo>=</mo><mfrac><mi>π</mi><mrow><mn>6</mn><mi>ω</mi></mrow></mfrac></math>, Let <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>α</mi><mo>=</mo><mfrac><mrow><msup><mi>A</mi><mn>2</mn></msup><msubsup><mi>μ</mi><mn>0</mn><mn>2</mn></msubsup><msup><mi>m</mi><mn>2</mn></msup><msup><mi>I</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac></math>.</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/3dbd710a-2aec-4891-a220-baef9f053b38-image.png" style="width: 220px; height: 150px;"/></p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td style="text-align: center;"> </td> <td style="text-align: center;"><strong>List-I</strong></td> <td style="text-align: center;"> </td> <td style="text-align: center;"><strong>List-II</strong></td> </tr> <tr> <td>(i)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(p)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn></math></td> </tr> <tr> <td>(ii)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(q)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mfrac><mi>α</mi><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></td> </tr> <tr> <td>(iii)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(r)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>3</mn><mi>α</mi></mrow><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></td> </tr> <tr> <td>(iv)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover><mo>+</mo><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(s)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>α</mi><mn>4</mn></mfrac><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></td> </tr> <tr> <td> </td> <td> </td> <td>(t)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mfrac><mrow><mn>3</mn><mi>α</mi></mrow><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></td> </tr> </tbody></table><p>Which one of the following options is correct?</p></div> | ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2022 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">t</mi></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">t</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">p</mi></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">r</mi></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">t</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">r</mi></math></span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">r</mi></math></span> </div> | <div class="solution"><blockquote><p>Note: This question was given bonus by JEE council.</p><p>The magnetic field in a solenoid is given by, <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><msub><mi>μ</mi><mn>0</mn></msub><mi>m</mi><mi>I</mi><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math>. Therefore, the flux through the loop will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfenced><mrow><mi>A</mi><mover><mi>k</mi><mo>^</mo></mover></mrow></mfenced><mo>·</mo><mfenced><mrow><msub><mi>μ</mi><mn>0</mn></msub><mi>m</mi><mi>I</mi><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></mrow></mfenced></math></p><p>For case I:</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p>The emf will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ε</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p>Therefore, the current in the loop will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p>The magnetic moment of the loop will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>m</mi><mo>→</mo></mover><mo>=</mo><mi>i</mi><mi>A</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>=</mo><mfrac><mrow><mi>B</mi><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></math></p><p>Therefore, the torque</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mover><mi>m</mi><mo>→</mo></mover><mo>×</mo><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mfenced><mrow><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>×</mo><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced close="]" open="["><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></mfenced><msup><mi>sin</mi><mn>2</mn></msup><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced close="]" open="["><mrow><msup><mi>sin</mi><mn>2</mn></msup><mfenced><mfrac><mi>π</mi><mn>6</mn></mfrac></mfenced></mrow></mfenced><mo>=</mo><mfrac><mrow><mo>-</mo><mi>α</mi></mrow><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi mathvariant="normal">I</mi></mfenced><mo>→</mo><mi>q</mi></math></p><p>For case II</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mn>0</mn></math></p><p>Therefore, we can directly write <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>τ</mi><mo>=</mo><mn>0</mn></math></p><p>Hence <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>II</mi></mfenced><mo>→</mo><mi>p</mi></math></p><p>Following the same process as in case I for case III</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>m</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><mi>B</mi><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mover><mi>m</mi><mo>→</mo></mover><mo>×</mo><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mo>×</mo><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mi>ω</mi><mi>t</mi><mfenced><mrow><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>×</mo><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><msup><mi>sin</mi><mn>2</mn></msup><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mfrac><mi>α</mi><mn>4</mn></mfrac><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>III</mi><mo>→</mo><mi>s</mi></math></p><p>Following the same process as in case I for case IV</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>=</mo><mo>-</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>m</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><mi>B</mi><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>cos</mi><mi>ω</mi><mi>t</mi><mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mover><mi>m</mi><mo>→</mo></mover><mo>×</mo><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced><mrow><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>×</mo><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></mrow></mfenced><msup><mi>cos</mi><mn>2</mn></msup><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>+</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></mfenced><mo>·</mo><msup><mi>cos</mi><mn>2</mn></msup><mfenced><mfrac><mi>π</mi><mn>6</mn></mfrac></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>+</mo><mfrac><mn>3</mn><mn>4</mn></mfrac><mi>α</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>IV</mi></mfenced><mo>→</mo><mi>r</mi></math></p></blockquote></div> | MarksBatch1_P1.db |
177 | a-small-mass-m-is-attached-to-a-massless-string-whose-other-end-is-fixed-at-p-as-shown-in-the-figure-the-mass-is-undergoing-circular-motion-in-the-x-y | a-small-mass-m-is-attached-to-a-massless-string-whose-other-end-is-fixed-at-p-as-shown-in-the-figure-the-mass-is-undergoing-circular-motion-in-the-x-y-93373 | <div class="question">A small mass $m$ is attached to a massless string whose other end is fixed at $P$ as shown in the figure. The mass is undergoing circular motion in the $x-y$ plane with centre at $O$ and constant angular speed $\omega$. If the angular momentum of the system, calculated about $O$ and $P$ are denoted by $\vec{L}_{O}$ and $\vec{L}_{P}$ respectively, then
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/8I6vSvu0X5rikcmCVJdUb7jfJ20GlgxqNY9J1KxpsgU.original.fullsize.png"/></div> | ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\vec{L}_{O}$ and $\vec{L}_{P}$ do not vary with time</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\vec{L}_{O}$ varies with time while $\vec{L}_{P}$ remains constant</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$\vec{L}_{O}$ remains constant while $\vec{L}_{P}$ varies with time</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\vec{L}_{O}$ and $\vec{L}_{P}$ both vary with time</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value">$\vec{L}_{O}$ remains constant while $\vec{L}_{P}$ varies with time</span> </div> | <div class="solution">For all locations of $\mathrm{m}$ the angular momentum of the mass $m$ about $O$ i.e., $L_{o}$ is $m r^{2} \omega$ and is directed toward $+z$ direction. The angular momentum of mass $m$ about $P$ i.e., $L_{p}$ is $m v l$ and is directed for the given location of $m$ as shown in the figure. For different location of $m$, the direction of $\vec{L}_{P}$ remains changing.
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/bwsNwamWN-cQ7HKxrMmC3itcaB_EkdPHkLoI6EWocck.original.fullsize.png"/></div> | MarksBatch1_P1.db |
178 | a-small-object-of-uniform-density-rolls-up-a-curved-surface-with-an-initial-velocity-v-it-reaches-up-to-a-maximum-height-of-4-g-3-v-2-with-respect-to--2 | a-small-object-of-uniform-density-rolls-up-a-curved-surface-with-an-initial-velocity-v-it-reaches-up-to-a-maximum-height-of-4-g-3-v-2-with-respect-to-2-54739 | <div class="question">A small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial$$<br/>\text { position. The object is }<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5hO_SeRZPfuRbMzeqNADHQSXsGxc3QBbS3m0w8QuJ4E.original.fullsize.png"/><br/></div> | ['Physics', 'Work Power Energy', 'JEE Main'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>ring<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>solid sphere<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>hollow sphere<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>disc</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>disc</span> </div> | <div class="solution">$$<br/>\begin{aligned}<br/>\frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2 & =m g\left(\frac{3 v^2}{4 g}\right) \\<br/>I & =\frac{1}{2} m R^2<br/>\end{aligned}<br/>$$<br/>So, body is disc.<br/>The correct option is (d).</div> | MarksBatch1_P1.db |
179 | a-small-object-of-uniform-density-rolls-up-a-curved-surface-with-an-initial-velocity-v-it-reaches-up-to-a-maximum-height-of-4-g-3-v-2-with-respect-to- | a-small-object-of-uniform-density-rolls-up-a-curved-surface-with-an-initial-velocity-v-it-reaches-up-to-a-maximum-height-of-4-g-3-v-2-with-respect-to-35455 | <div class="question">A small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial$$<br/>\text { position. The object is }<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5hO_SeRZPfuRbMzeqNADHQSXsGxc3QBbS3m0w8QuJ4E.original.fullsize.png"/><br/></div> | ['Physics', 'Work Power Energy', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>ring<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>solid sphere<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>hollow sphere<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>disc</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>disc</span> </div> | <div class="solution">$$<br/>\begin{aligned}<br/>\frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2 & =m g\left(\frac{3 v^2}{4 g}\right) \\<br/>I & =\frac{1}{2} m R^2<br/>\end{aligned}<br/>$$<br/>So, body is disc.<br/>The correct option is (d).</div> | MarksBatch1_P1.db |
180 | a-small-roller-of-diameter-20-cm-has-an-axle-of-diameter-10-cm-see-figure-below-on-the-left-it-is-on-a-horizontal-floor-and-a-meter-scale-is-positione | a-small-roller-of-diameter-20-cm-has-an-axle-of-diameter-10-cm-see-figure-below-on-the-left-it-is-on-a-horizontal-floor-and-a-meter-scale-is-positione-20970 | <div class="question">A small roller of diameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>20</mn><mo>Â </mo><mi>cm</mi></math> has an axle of diameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>10</mn><mo>Â </mo><mi>cm</mi></math> (see figure below on the left). It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle (see figure on the right). The scale is now pushed slowly on the axle so that it moves without slipping on the axle, and the roller starts rolling without slipping. After the roller has moved <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>50</mn><mo>Â </mo><mi>cm</mi><mo>,</mo></math> the position of the scale will look like (figures are schematic and not drawn to scale)
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/4f447cbf-b054-43c8-93b9-d22dc77ca59c-image.png"/></div> | ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2020 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/aab3bcfa-9e87-46ad-a299-0d0c725c3c1a-image.png"/></p></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/e10ff687-0ab8-40fb-8e7b-5fcfbe1e8c0a-image.png"/></p></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/dafb5633-4f65-49be-9a08-b083eee0b470-image.png"/></p></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/12b2502c-ee2b-4354-871e-c3f54559e6fc-image.png"/></span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/e10ff687-0ab8-40fb-8e7b-5fcfbe1e8c0a-image.png"/></p></span> </div> | <div class="solution"><p>Angular displacement <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>θ</mi><mo>=</mo><mfrac><msub><mi>S</mi><mn>0</mn></msub><mn>10</mn></mfrac><mo>=</mo><mn>5</mn></math></p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/e339f447-da08-4479-99ca-b33b00d22109-image.png"/></p><p>Displacement of point <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>B</mi></math> is,<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mo>=</mo><mi>r</mi><mo>.</mo><mi>θ</mi><mo>=</mo><mn>15</mn><mo>Ã</mo><mn>5</mn><mo>=</mo><mn>75</mn><mo> </mo><mi>cm</mi></math></p><p>Displacement of roller <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>50</mn><mo> </mo><mi>cm</mi></math></p><p>Relative displacement <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>75</mn><mo>-</mo><mn>50</mn><mo>=</mo><mn>25</mn><mo> </mo><mi>cm</mi></math></p><p> </p></div> | MarksBatch1_P1.db |
181 | a-solid-sphere-is-in-pure-rolling-motion-on-an-inclined-surface-having-inclination | a-solid-sphere-is-in-pure-rolling-motion-on-an-inclined-surface-having-inclination-33574 | <div class="question">A solid sphere is in pure rolling motion on an inclined surface having inclination $\theta$.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/SqpFTGNOh5r0Id1yrfTHMgHqm61n25jCGlaCyaWC2gQ.original.fullsize.png"/><br/></div> | ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>frictional force acting on sphere is $f=\mu m g \cos \theta$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$f$ is disspative force<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>friction will increase its angular velocity and decreases its linear velocity<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>If $\theta$ decreases, friction will decrease</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>friction will increase its angular velocity and decreases its linear velocity<br/>, <br/>If $\theta$ decreases, friction will decrease</span> </div> | <div class="solution">In case of pure rolling,<br/>$$<br/>\begin{aligned}<br/>& f=\frac{m g \sin \theta}{1+\frac{m R^2}{I}} \\<br/>\therefore & f \propto \sin \theta<br/>\end{aligned}<br/>$$<br/>(upwards)<br/>Therefore, as $\theta$ decreases force of friction will also decrease.</div> | MarksBatch1_P1.db |
182 | a-solid-sphere-of-radius-r-has-a-charge-q-distributed-in-its-volume-with-a-charge-density-k-r-a-where-k-and-a-are-constants-and-r-is-the-distance-from | a-solid-sphere-of-radius-r-has-a-charge-q-distributed-in-its-volume-with-a-charge-density-k-r-a-where-k-and-a-are-constants-and-r-is-the-distance-from-94247 | <div class="question">A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho=k r^a$, where $k$ and $a$ are constants and $r$ is the distance from its centre. If the electric field at $r=\frac{R}{2}$ is $\frac{1}{8}$ times that at $r=R$, find the value of $a$.</div> | ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">2</span> </div> | <div class="solution">From Gauss theorem,<br/>$$<br/>\begin{aligned}<br/>& E \propto \frac{q}{r^2} \quad(q=\text { charge enclosed }) \\<br/>& \therefore \quad \frac{E_2}{E_1}=\frac{q_2}{q_1}=\frac{r_1^2}{r_2^2} \\<br/>& 8=\frac{\int_0^R\left(4 \pi r^2\right) k r^a d r}{\int_0^{R / 2}\left(4 \pi r^2\right) k r^a d r} \times \frac{(R / 2)^2}{(R)^2} \\<br/>&<br/>\end{aligned}<br/>$$<br/>Solving this equation we get, $a=2$</div> | MarksBatch1_P1.db |
183 | a-solid-sphere-of-radius-r-has-moment-of-inertia-j-about-its-qeometrical-axis-if-it-is-m-e-lt-e-d-in-t-o-a-d-i-sco-f-r-a-d-i-u-s-r-an-d-t-hi-c-kn-ess--1 | a-solid-sphere-of-radius-r-has-moment-of-inertia-j-about-its-qeometrical-axis-if-it-is-m-e-lt-e-d-in-t-o-a-d-i-sco-f-r-a-d-i-u-s-r-an-d-t-hi-c-kn-ess-1-93161 | <div class="question">$$<br/>\text { A solid sphere of radius } R \text { has moment of inertia } J \text { about its qeometrical axis. If it is }<br/>$$<br/>melted into a disc of radius $r$ and thickness $t$. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$, then the value of $r$ is equal to<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/lSgwN8rcG18285o-EaZ9esu_dYPU3wCgkbldDfNUel8.original.fullsize.png"/><br/></div> | ['Physics', 'Rotational Motion', 'JEE Main'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{2}{\sqrt{15}} R$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{2}{\sqrt{5}} R$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{3}{\sqrt{15}} R$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{\sqrt{3}}{\sqrt{15}} R$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{2}{\sqrt{15}} R$<br/></span> </div> | <div class="solution">$\frac{2}{3} M R^2=\frac{1}{2} M r^2+M r^2$ or $\frac{2}{3} M R^2=\frac{3}{2} M r^2 \Rightarrow r=\frac{2}{\sqrt{15}} R$</div> | MarksBatch1_P1.db |
184 | a-solid-sphere-of-radius-r-has-moment-of-inertia-j-about-its-qeometrical-axis-if-it-is-m-e-lt-e-d-in-t-o-a-d-i-sco-f-r-a-d-i-u-s-r-an-d-t-hi-c-kn-ess- | a-solid-sphere-of-radius-r-has-moment-of-inertia-j-about-its-qeometrical-axis-if-it-is-m-e-lt-e-d-in-t-o-a-d-i-sco-f-r-a-d-i-u-s-r-an-d-t-hi-c-kn-ess-10772 | <div class="question">$$<br/>\text { A solid sphere of radius } R \text { has moment of inertia } J \text { about its qeometrical axis. If it is }<br/>$$<br/>melted into a disc of radius $r$ and thickness $t$. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$, then the value of $r$ is equal to<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/lSgwN8rcG18285o-EaZ9esu_dYPU3wCgkbldDfNUel8.original.fullsize.png"/><br/></div> | ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{2}{\sqrt{15}} R$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{2}{\sqrt{5}} R$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{3}{\sqrt{15}} R$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{\sqrt{3}}{\sqrt{15}} R$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{2}{\sqrt{15}} R$<br/></span> </div> | <div class="solution">$\frac{2}{3} M R^2=\frac{1}{2} M r^2+M r^2$ or $\frac{2}{3} M R^2=\frac{3}{2} M r^2 \Rightarrow r=\frac{2}{\sqrt{15}} R$</div> | MarksBatch1_P1.db |
185 | a-solution-of-a-metal-ion-when-treated-with-ki-gives-a-red-precipitate-which-dissolved-in-excess-ki-to-give-a-colourless-solution-moreover-the-solutio | a-solution-of-a-metal-ion-when-treated-with-ki-gives-a-red-precipitate-which-dissolved-in-excess-ki-to-give-a-colourless-solution-moreover-the-solutio-36957 | <div class="question">A solution of a metal ion when treated with $\mathrm{KI}$ gives a red precipitate which dissolved in excess KI to give a colourless solution. Moreover, the solution of metal ion on treatment with a solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitqte. The metal ion is</div> | ['Chemistry', 'Practical Chemistry', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\mathrm{Pb}^{2+}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\mathrm{Hg}^{2+}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\mathrm{Cu}^{2+}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\mathrm{Co}^2$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\mathrm{Hg}^{2+}$<br/></span> </div> | <div class="solution">$$<br/>\begin{aligned}<br/>& \mathrm{Hg}^{2+}+2 \mathrm{I}^{-} \longrightarrow \mathrm{HgI}_2, \mathrm{HgI}_2+\mathrm{H}^{-} \\<br/>& \text {(Excess) }\left.\longrightarrow \mathrm{HgI}_4\right]^{2-} \\<br/>& \text { Soluble }^{2-} \\<br/>& \mathrm{Hg}^{2+}+\mathrm{Co}(\mathrm{SCN})_2 \longrightarrow \mathrm{Hg}(\mathrm{SCN})_2<br/>\end{aligned}<br/>$$<br/>Blue crystalline ppt.</div> | MarksBatch1_P1.db |
186 | a-solution-of-colourless-salt-h-on-boiling-with-excess-naoh-produces-a-nonflammable-gas-the-gas-evolution-ceases-after-sometime-upon-addition-of-zn-du | a-solution-of-colourless-salt-h-on-boiling-with-excess-naoh-produces-a-nonflammable-gas-the-gas-evolution-ceases-after-sometime-upon-addition-of-zn-du-14813 | <div class="question">A solution of colourless salt $\mathrm{H}$ on boiling with excess $\mathrm{NaOH}$ produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of $\mathrm{Zn}$ dust to the same solution, the gas evolution restarts. The colourless salt (s) $\mathrm{H}$ is (are)</div> | ['Chemistry', 'd and f Block Elements', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\mathrm{NH}_4 \mathrm{NO}_3$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\mathrm{NH}_4 \mathrm{NO}_2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\mathrm{NH}_4 \mathrm{Cl}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$</span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$\mathrm{NH}_4 \mathrm{NO}_3$<br/>, <br/>$\mathrm{NH}_4 \mathrm{NO}_2$<br/></span> </div> | <div class="solution">$$<br/>\begin{aligned}<br/>& \mathrm{NH}_4 \mathrm{NO}_3+\mathrm{NaOH} \longrightarrow \mathrm{NaNO}_3+\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \\<br/>& \mathrm{NaNO}_3+8[\mathrm{H}] \mathrm{NaOH}+\mathrm{NH}_3+2 \mathrm{H}_2 \mathrm{O} \\<br/>& \mathrm{NH}_4 \mathrm{NO}_2+\mathrm{NaOH} \longrightarrow \mathrm{NaHO}_2+\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \\<br/>& \mathrm{NaNO}_2+6[\mathrm{H}] \longrightarrow \mathrm{NaOH}+\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O}<br/>\end{aligned}<br/>$$</div> | MarksBatch1_P1.db |
187 | a-solution-when-diluted-with-h-2-o-and-boiled-it-gives-a-white-precipitate-on-addition-of-excess-nh-4-cl-nh-4-oh-the-volume-of-precipitate-decreases-l | a-solution-when-diluted-with-h-2-o-and-boiled-it-gives-a-white-precipitate-on-addition-of-excess-nh-4-cl-nh-4-oh-the-volume-of-precipitate-decreases-l-42315 | <div class="question">A solution when diluted with $\mathrm{H}_2 \mathrm{O}$ and boiled, it gives a white precipitate. On addition of excess $\mathrm{NH}_4 \mathrm{Cl} / \mathrm{NH}_4 \mathrm{OH}$, the volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate which dissolves in $\mathrm{NH}_4 \mathrm{OH} / \mathrm{NH}_4 \mathrm{Cl}$.</div> | ['Chemistry', 'Practical Chemistry', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\mathrm{Zn}(\mathrm{OH})_2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\mathrm{Al}(\mathrm{OH})_3$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\mathrm{Mg}(\mathrm{OH})_2$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\mathrm{Ca}(\mathrm{OH})_2$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\mathrm{Zn}(\mathrm{OH})_2$<br/></span> </div> | <div class="solution">A solution of $\mathrm{Zn}^{2+}$ ion reacts with water which on boiling give white ppt. of $\mathrm{Zn}(\mathrm{OH})_2$. This precipitates dissolve in excess of $\mathrm{NH}_4 \mathrm{OH}$ in presence of $\mathrm{NH}_4 \mathrm{Cl}$ because $\mathrm{Zn}(\mathrm{OH})_2$ on reaction with $\mathrm{NH}_4 \mathrm{OH}$ in presence of $\mathrm{NH}_4 \mathrm{Cl}$ forms tetrammine zinc (II) complex (soluble).<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/uZilZltjS4fVUCxfYQUh2bfU6By7vtSKZ1G1L66vvg4.original.fullsize.png"/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/54xVnRJAct4YK6nXe_KVzZIA81hhc1znvIPmpFiT-TQ.original.fullsize.png"/><br/></div> | MarksBatch1_P1.db |
188 | a-sphere-is-rolling-without-slipping-on-a-fixed-horizontal-plane-surface-in-the-figure-a-is-the-point-of-contact-b-is-the-centre-of-the-sphere-and-c-i | a-sphere-is-rolling-without-slipping-on-a-fixed-horizontal-plane-surface-in-the-figure-a-is-the-point-of-contact-b-is-the-centre-of-the-sphere-and-c-i-26004 | <div class="question">A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, $A$ is the point of contact. $B$ is the centre of the sphere and $C$ is its topmost point. Then,<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/7Xsr-fGTyJjgVVwNEX_nZqDunz93zW95YxOhMA7BOEM.original.fullsize.png"/><br/></div> | ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_A=2\left(\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_C\right)$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_B=\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_A$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\left|\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_A\right|=2\left|\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_C\right|$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\left|\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_A\right|=4\left|\overrightarrow{\mathbf{v}}_B\right|$</span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_B=\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_A$<br/>, <br/>$\left|\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_A\right|=2\left|\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_C\right|$<br/></span> </div> | <div class="solution">$$<br/>v_A=0, v_B=v \text { and } v_C=2 v<br/>$$<br/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/CyOWdRHTYq2MSjWUtaeoFm7lHACpL2peoHF_5bBE4g8.original.fullsize.png"/><br/></div> | MarksBatch1_P1.db |
189 | a-spherical-metal-shell-a-of-radius-r-a-and-a-solid-metal-sphere-b-of-radius-r-b-r-a-are-kept-far-apart-and-each-is-given-charge-q-now-they-are-connec | a-spherical-metal-shell-a-of-radius-r-a-and-a-solid-metal-sphere-b-of-radius-r-b-r-a-are-kept-far-apart-and-each-is-given-charge-q-now-they-are-connec-17365 | <div class="question">A spherical metal shell $A$ of radius $R_A$ and a solid metal sphere $B$ of radius $R_B < \left(R_A\right)$ are kept far apart and each is given charge $+Q$. Now, they are connected by a thin metal wire. Then</div> | ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$E_A^{\text {inside }}=0$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$Q_A>Q_B$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{\sigma_{\mathrm{A}}}{\sigma_B}=\frac{R_B}{R_A}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$E_A^{\text {on surface }} < E_B^{\text {on surface }}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$E_A^{\text {inside }}=0$<br/>, <br/>$Q_A>Q_B$<br/>, <br/>$\frac{\sigma_{\mathrm{A}}}{\sigma_B}=\frac{R_B}{R_A}$<br/>, <br/>$E_A^{\text {on surface }} < E_B^{\text {on surface }}$</span> </div> | <div class="solution">Inside a conducting shell, electric field is always zero. Therefore, option (a) is correct. When the two are connected, their potentials become the same.<br/>$$<br/>\begin{array}{ll}<br/>\therefore & V_A=V_B \\<br/>\text { or } & \frac{Q_A}{R_A}=\frac{Q_B}{R_B} \quad\left(\because V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\right)<br/>\end{array}<br/>$$<br/>Since, $R_A>R_B$<br/>$$<br/>\therefore \quad Q_A>Q_B<br/>$$<br/>$\therefore$ Option (b) is correct.<br/>Potential is also equal to,<br/>$$<br/>\begin{array}{rlrl}<br/>V & =\frac{\sigma R}{\varepsilon_0} \\<br/>V_A & =V_B \\<br/>\therefore \quad \sigma_A R_A & =\sigma_B R_B \\<br/>\text { or } \quad \sigma_B & \frac{\sigma_A}{\sigma_B} & =\frac{R_B}{R_A}<br/>\end{array}<br/>$$<br/><br/>$\therefore$ Option (c) is correct.<br/>Electric field on surface,<br/>$$<br/>E=\frac{\sigma}{E_0} \text { or } E \propto \sigma<br/>$$<br/>or $\quad \sigma_A < \sigma_B$<br/>Since, $\quad \sigma_A < \sigma_B$<br/>$\therefore \quad E_A < E_B$<br/>$\therefore$ Option (d) is also correct.<br/>$\therefore$ Correct options are (a), (b), (c) and (d).<br/>Analysis of Question<br/>(i) Question is simple.<br/>(ii) $E=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2}=\frac{\sigma}{\varepsilon_0}$<br/>(On surface)<br/>As, $\sigma=$ surface charge density<br/>$$<br/>=\frac{Q}{4 \pi R^2}<br/>$$<br/>(iii) $V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}=\frac{\sigma R}{\varepsilon_0}$</div> | MarksBatch1_P1.db |
190 | a-spherical-portion-has-been-removed-from-a-solid-sphere-having-a-charge-distributed-uniformly-in-its-volume-as-shown-in-the-figure-the-electric-field-1 | a-spherical-portion-has-been-removed-from-a-solid-sphere-having-a-charge-distributed-uniformly-in-its-volume-as-shown-in-the-figure-the-electric-field-1-25523 | <div class="question">A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/w32mClG-c0ydnRkgWN7fvwcQKA3ZiypT8WiDkA-9iB4.original.fullsize.png"/><br/></div> | ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>zero everywhere<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>non-zero and uniform<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>non-uniform<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>zero only at its centre</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>non-zero and uniform<br/></span> </div> | <div class="solution">Inside the cavity, field at any point is uniform and non-zero. Therefore, correct option is (b).</div> | MarksBatch1_P1.db |
191 | a-spherically-symmetric-gravitational-system-of-particles-has-a-mass-density-0-for-r-r-0-for-r-r-w-h-ere-rho0-i-s-a-co-n-s-t-an-t-a-t-es-t-ma-ssc-an-u-1 | a-spherically-symmetric-gravitational-system-of-particles-has-a-mass-density-0-for-r-r-0-for-r-r-w-h-ere-rho0-i-s-a-co-n-s-t-an-t-a-t-es-t-ma-ssc-an-u-1-16691 | <div class="question">A spherically symmetric gravitational system of particles has a mass density<br/>$$<br/>\rho=\left\{\begin{array}{l}<br/>\rho_0 \text { for } r \leq R \\<br/>0 \text { for } r>R<br/>\end{array}\right.<br/>$$<br/>where $\rho_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $v$ as a function of distance $r(0 < r < \infty)$ from the centre of the system is represented by</div> | ['Physics', 'Gravitation', 'JEE Main'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/LA_vMBB7WhU2Vip7y_eiK1-zkU_SDBCOiBh6fpNcRvE.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OSf9wp55TuxJM9G6WaYwibu2xUu2N2abQyF2IXtTUak.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/9r4XEvZ2nmCJLJy-w_gGqxYumnEQB0ILHwTYuH9voJY.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/c7v7l3sB9fKCfhxFSsS39jb0MepdzAJ1UM82vwi3us8.original.fullsize.png"/><br/></span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/9r4XEvZ2nmCJLJy-w_gGqxYumnEQB0ILHwTYuH9voJY.original.fullsize.png"/><br/></span> </div> | <div class="solution">For $r \leq R$ :<br/>$$<br/>\frac{m v^2}{r}=\frac{G \cdot m m^{\prime}}{r^2}<br/>$$<br/>Here, $m^{\prime}=\left(\frac{4}{3} \pi r^3\right) \rho_0$<br/>Substituting in Eq. (i) we get, $v \propto r$<br/>i.e., $v-r$ graph is a straight line passing through origin.<br/>For $r>R$ :<br/>$$<br/>\frac{m v^2}{r}=\frac{G \cdot m\left(\frac{4}{3} \pi R^3\right) \rho_0}{r^2}<br/>$$<br/>or $\quad V \propto \frac{1}{\sqrt{r}}$<br/>The corresponding $v-r$ graph will be as shown in option (c).<br/>$\therefore$ correct option is (c).</div> | MarksBatch1_P1.db |
192 | a-spherically-symmetric-gravitational-system-of-particles-has-a-mass-density-0-for-r-r-0-for-r-r-w-h-ere-rho0-i-s-a-co-n-s-t-an-t-a-t-es-t-ma-ssc-an-u | a-spherically-symmetric-gravitational-system-of-particles-has-a-mass-density-0-for-r-r-0-for-r-r-w-h-ere-rho0-i-s-a-co-n-s-t-an-t-a-t-es-t-ma-ssc-an-u-24122 | <div class="question">A spherically symmetric gravitational system of particles has a mass density<br/>$$<br/>\rho=\left\{\begin{array}{l}<br/>\rho_0 \text { for } r \leq R \\<br/>0 \text { for } r>R<br/>\end{array}\right.<br/>$$<br/>where $\rho_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $v$ as a function of distance $r(0 < r < \infty)$ from the centre of the system is represented by</div> | ['Physics', 'Gravitation', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/LA_vMBB7WhU2Vip7y_eiK1-zkU_SDBCOiBh6fpNcRvE.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OSf9wp55TuxJM9G6WaYwibu2xUu2N2abQyF2IXtTUak.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/9r4XEvZ2nmCJLJy-w_gGqxYumnEQB0ILHwTYuH9voJY.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/c7v7l3sB9fKCfhxFSsS39jb0MepdzAJ1UM82vwi3us8.original.fullsize.png"/><br/></span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/9r4XEvZ2nmCJLJy-w_gGqxYumnEQB0ILHwTYuH9voJY.original.fullsize.png"/><br/></span> </div> | <div class="solution">For $r \leq R$ :<br/>$$<br/>\frac{m v^2}{r}=\frac{G \cdot m m^{\prime}}{r^2}<br/>$$<br/>Here, $m^{\prime}=\left(\frac{4}{3} \pi r^3\right) \rho_0$<br/>Substituting in Eq. (i) we get, $v \propto r$<br/>i.e., $v-r$ graph is a straight line passing through origin.<br/>For $r>R$ :<br/>$$<br/>\frac{m v^2}{r}=\frac{G \cdot m\left(\frac{4}{3} \pi R^3\right) \rho_0}{r^2}<br/>$$<br/>or $\quad V \propto \frac{1}{\sqrt{r}}$<br/>The corresponding $v-r$ graph will be as shown in option (c).<br/>$\therefore$ correct option is (c).</div> | MarksBatch1_P1.db |
193 | a-stationary-source-is-emitting-sound-at-a-fixed-frequency-f-0-which-is-reflected-by-two-cars-approaching-the-source-the-difference-between-the-freque-1 | a-stationary-source-is-emitting-sound-at-a-fixed-frequency-f-0-which-is-reflected-by-two-cars-approaching-the-source-the-difference-between-the-freque-1-50848 | <div class="question">A stationary source is emitting sound at a fixed frequency $f_0$, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is $1.2 \%$ of $f_0$. What is the difference in the speeds of the cars (in km per hour) to the nearest integer ? The cars are moving at constant speeds much smaller than the speed of sound which is $330 \mathrm{~ms}^{-1}$.</div> | ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">7</span> </div> | <div class="solution">Firstly, car will be treated as an observer which is approaching the source. Then, it will be treated as a source, which is moving in the direction of sound.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/D31ONv5MKojv2wN20KPcx_S2EmT-1I0UavavMug4QkI.original.fullsize.png"/><br/><br/>Hence, $\quad f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)$<br/>$$<br/>\begin{aligned}<br/>f_2 & =f_0\left(\frac{v+v_2}{v-v_2}\right) \\<br/>\therefore \quad f_1-f_2 & =\left(\frac{1.2}{100}\right) f_0 \\<br/>& =f_0\left[\frac{v+v_1}{v-v_1}-\frac{v+v_2}{v-v_2}\right] \\<br/>\text { or }\left(\frac{1.2}{100}\right) f_0 & =\frac{2 v\left(v_1-v_2\right)}{\left(v-v_1\right)\left(v-v_2\right)} f_0<br/>\end{aligned}<br/>$$<br/>As $v_1$ and $v_2$ are very very less than $v$. We can write, $\left(v-v_1\right)$ or $\left(v-v_2\right) \approx v$<br/>$$<br/>\therefore \quad\left(\frac{1.2}{100}\right) f_0=\frac{2\left(v_1-v_2\right)}{v} f_0<br/>$$<br/>or $\left(v_1-v_2\right)=\frac{v \times 1.2}{200}$<br/>$$<br/>\begin{aligned}<br/>& =\frac{330 \times 1.2}{200}=1.98 \mathrm{~ms}^{-1} \\<br/>& =7.128 \mathrm{kmh}^{-1}<br/>\end{aligned}<br/>$$<br/>$\therefore$ The nearest integer is 7 .<br/>Hence, $\quad f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)$<br/>$$<br/>\begin{aligned}<br/>f_2 & =f_0\left(\frac{v+v_2}{v-v_2}\right) \\<br/>\therefore \quad f_1-f_2 & =\left(\frac{1.2}{100}\right) f_0 \\<br/>& =f_0\left[\frac{v+v_1}{v-v_1}-\frac{v+v_2}{v-v_2}\right] \\<br/>\text { or }\left(\frac{1.2}{100}\right) f_0 & =\frac{2 v\left(v_1-v_2\right)}{\left(v-v_1\right)\left(v-v_2\right)} f_0<br/>\end{aligned}<br/>$$<br/>As $v_1$ and $v_2$ are very very less than $v$. We can write, $\left(v-v_1\right)$ or $\left(v-v_2\right) \approx v$<br/>$$<br/>\therefore \quad\left(\frac{1.2}{100}\right) f_0=\frac{2\left(v_1-v_2\right)}{v} f_0<br/>$$<br/>or $\left(v_1-v_2\right)=\frac{v \times 1.2}{200}$<br/>$$<br/>\begin{aligned}<br/>& =\frac{330 \times 1.2}{200}=1.98 \mathrm{~ms}^{-1} \\<br/>& =7.128 \mathrm{kmh}^{-1}<br/>\end{aligned}<br/>$$<br/>$\therefore$ The nearest integer is 7 .</div> | MarksBatch1_P1.db |
194 | a-steady-current-i-goes-through-a-wire-loop-pqr-having-shape-of-a-right-angle-triangle-with-pq-3-x-pr-4-x-and-qr-5-x-if-the-magnitude-of-the-magnetic--1 | a-steady-current-i-goes-through-a-wire-loop-pqr-having-shape-of-a-right-angle-triangle-with-pq-3-x-pr-4-x-and-qr-5-x-if-the-magnitude-of-the-magnetic-1-20260 | <div class="question">A steady current $I$ goes through a wire loop $P Q R$ having shape of a right angle triangle with $P Q=3 x, P R=4 x$ and $Q R=5 x$. If the magnitude of the magnetic field at $P$ due to this loop is $k\left(\frac{\mu_0 I}{48 \pi x}\right)$, find the value of $k$.</div> | ['Physics', 'Magnetic Effects of Current', 'JEE Main'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">7</span> </div> | <div class="solution">Magnetic field at point $P$ due to wires $R P$ and $R Q$ is zero. Only wire $Q R$ will produce magnetic field at $P$.<br/>$$<br/>r=3 x \cos 37^{\circ}=(3 x)\left(\frac{4}{5}\right)=\frac{12 x}{5}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/-Tnw_k9oHJNYG2G47lZ6gOLhOzmflmnXKOJwQUhvb5g.original.fullsize.png"/><br/></div> | MarksBatch1_P1.db |
195 | a-steady-current-i-goes-through-a-wire-loop-pqr-having-shape-of-a-right-angle-triangle-with-pq-3-x-pr-4-x-and-qr-5-x-if-the-magnitude-of-the-magnetic- | a-steady-current-i-goes-through-a-wire-loop-pqr-having-shape-of-a-right-angle-triangle-with-pq-3-x-pr-4-x-and-qr-5-x-if-the-magnitude-of-the-magnetic-95144 | <div class="question">A steady current $I$ goes through a wire loop $P Q R$ having shape of a right angle triangle with $P Q=3 x, P R=4 x$ and $Q R=5 x$. If the magnitude of the magnetic field at $P$ due to this loop is $k\left(\frac{\mu_0 I}{48 \pi x}\right)$, find the value of $k$.</div> | ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">7</span> </div> | <div class="solution">Magnetic field at point $P$ due to wires $R P$ and $R Q$ is zero. Only wire $Q R$ will produce magnetic field at $P$.<br/>$$<br/>r=3 x \cos 37^{\circ}=(3 x)\left(\frac{4}{5}\right)=\frac{12 x}{5}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/-Tnw_k9oHJNYG2G47lZ6gOLhOzmflmnXKOJwQUhvb5g.original.fullsize.png"/><br/></div> | MarksBatch1_P1.db |
196 | a-straight-line-l-through-the-point-3-2-is-inclined-at-an-angle-6-0-to-the-line-3-x-y-1-if-l-also-intersects-the-x-axis-then-the-equation-of-l-is-1 | a-straight-line-l-through-the-point-3-2-is-inclined-at-an-angle-6-0-to-the-line-3-x-y-1-if-l-also-intersects-the-x-axis-then-the-equation-of-l-is-1-78188 | <div class="question">A straight line $L$ through the point $(3,-2)$ is inclined at an angle $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If $L$ also intersects the $X$-axis, then the equation of $L$ is</div> | ['Mathematics', 'Straight Lines', 'JEE Main'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$y+\sqrt{3} x+2-3 \sqrt{3}=0$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$y-\sqrt{3} x+2+3 \sqrt{3}=0$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\sqrt{3} y-x+3+2 \sqrt{3}=0$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\sqrt{3} y+x-3+2 \sqrt{3}=0$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$y-\sqrt{3} x+2+3 \sqrt{3}=0$<br/></span> </div> | <div class="solution">A straight line passing through $P$ and making an angle of $\alpha=60^{\circ}$, is given by $\frac{y-y_1}{x-x_1}=\tan (\theta \pm \alpha)$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/BbwFhwlBUeWDKh1kwXK76q7lwHtU9fgCoUCgBopRkvA.original.fullsize.png"/><br/><br/>where, $\quad \sqrt{3} x+y=1$<br/>$$<br/>\Rightarrow \quad y=-\sqrt{3} x+1<br/>$$<br/>Then, $\quad \tan \theta=-\sqrt{3}$<br/>$$<br/>\begin{array}{ll}<br/>\Rightarrow & \frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha}{1 \mp \tan \theta \tan \alpha} \\<br/>\Rightarrow & \frac{y+2}{x-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})(\sqrt{3})} \\<br/>\text { and } & \frac{y+2}{x-3}=\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})}<br/>\end{array}<br/>$$<br/><br/>$$<br/>\begin{array}{ll}<br/>\Rightarrow & y+2=0 \\<br/>\text { and } & \frac{y+2}{x-3}=\frac{-2 \sqrt{3}}{1-3}=\sqrt{3} \\<br/>\Rightarrow & y+2=\sqrt{3} x-3 \sqrt{3}<br/>\end{array}<br/>$$<br/>Neglecting, $y+2=0$ as it does not intersect $Y$-axis.</div> | MarksBatch1_P1.db |
197 | a-straight-line-l-through-the-point-3-2-is-inclined-at-an-angle-6-0-to-the-line-3-x-y-1-if-l-also-intersects-the-x-axis-then-the-equation-of-l-is | a-straight-line-l-through-the-point-3-2-is-inclined-at-an-angle-6-0-to-the-line-3-x-y-1-if-l-also-intersects-the-x-axis-then-the-equation-of-l-is-70203 | <div class="question">A straight line $L$ through the point $(3,-2)$ is inclined at an angle $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If $L$ also intersects the $X$-axis, then the equation of $L$ is</div> | ['Mathematics', 'Straight Lines', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$y+\sqrt{3} x+2-3 \sqrt{3}=0$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$y-\sqrt{3} x+2+3 \sqrt{3}=0$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\sqrt{3} y-x+3+2 \sqrt{3}=0$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\sqrt{3} y+x-3+2 \sqrt{3}=0$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$y-\sqrt{3} x+2+3 \sqrt{3}=0$<br/></span> </div> | <div class="solution">A straight line passing through $P$ and making an angle of $\alpha=60^{\circ}$, is given by $\frac{y-y_1}{x-x_1}=\tan (\theta \pm \alpha)$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/BbwFhwlBUeWDKh1kwXK76q7lwHtU9fgCoUCgBopRkvA.original.fullsize.png"/><br/><br/>where, $\quad \sqrt{3} x+y=1$<br/>$$<br/>\Rightarrow \quad y=-\sqrt{3} x+1<br/>$$<br/>Then, $\quad \tan \theta=-\sqrt{3}$<br/>$$<br/>\begin{array}{ll}<br/>\Rightarrow & \frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha}{1 \mp \tan \theta \tan \alpha} \\<br/>\Rightarrow & \frac{y+2}{x-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})(\sqrt{3})} \\<br/>\text { and } & \frac{y+2}{x-3}=\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})}<br/>\end{array}<br/>$$<br/><br/>$$<br/>\begin{array}{ll}<br/>\Rightarrow & y+2=0 \\<br/>\text { and } & \frac{y+2}{x-3}=\frac{-2 \sqrt{3}}{1-3}=\sqrt{3} \\<br/>\Rightarrow & y+2=\sqrt{3} x-3 \sqrt{3}<br/>\end{array}<br/>$$<br/>Neglecting, $y+2=0$ as it does not intersect $Y$-axis.</div> | MarksBatch1_P1.db |
198 | a-straight-line-through-the-vertex-p-of-a-pqr-intersects-the-side-qr-at-the-point-s-and-the-circumcircle-of-the-pqr-at-the-point-t-if-s-is-not-the-cen | a-straight-line-through-the-vertex-p-of-a-pqr-intersects-the-side-qr-at-the-point-s-and-the-circumcircle-of-the-pqr-at-the-point-t-if-s-is-not-the-cen-98308 | <div class="question">A straight line through the vertex $P$ of a $\triangle P Q R$ intersects the side $Q R$ at the point $S$ and the circumcircle of the $\triangle P Q R$ at the point $T$. If $S$ is not the centre of the circumcircle, then</div> | ['Mathematics', 'Straight Lines', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{1}{P S}+\frac{1}{S T} < \frac{2}{\sqrt{Q S \cdot S R}}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{1}{P S}+\frac{1}{S T}>\frac{2}{\sqrt{Q S \cdot S R}}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{1}{P S}+\frac{1}{S T} < \frac{4}{Q R}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{1}{P S}+\frac{1}{S T}>\frac{4}{Q R}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$\frac{1}{P S}+\frac{1}{S T}>\frac{2}{\sqrt{Q S \cdot S R}}$<br/>, <br/>$\frac{1}{P S}+\frac{1}{S T}>\frac{4}{Q R}$</span> </div> | <div class="solution">Let a straight line through the vertex $P$ of a given $\triangle P Q R$ intersects the side $Q R$ at the point $S$ and the circumcircle of $\triangle P Q R$ at the point $T$.<br/>Points $P, Q, R, T$ are concyclic, hence<br/>$$<br/>\begin{array}{rlrl}<br/>P S \cdot S T & =Q S \cdot S R & & \\<br/>\Rightarrow \quad P_1 & \geq 2 \sqrt{P S \cdot S T} & \\<br/>& \text { Now, } \frac{P S+S T}{2} & \geq \sqrt{P S \cdot S T} & {[\because \mathrm{AM} \geq \mathrm{GM}]}<br/>\end{array}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/vrafhqNxAoheGeJLHlvdpbhAM3nv4TAWfk7gv8-lXXs.original.fullsize.png"/><br/><br/>and $\quad \frac{1}{P S}+\frac{1}{S T} \geq \frac{2}{\sqrt{P S \cdot S T}}=\frac{2}{\sqrt{Q S \cdot S R}}$<br/>Also, $\quad \frac{S Q+Q R}{2} \geq \sqrt{S Q \cdot S R}$<br/>$\Rightarrow \quad \frac{Q R}{2} \geq \sqrt{S Q \cdot S R}$<br/><br/>$$<br/>\begin{array}{ll}<br/>\Rightarrow & \frac{1}{\sqrt{S Q \cdot S R}} \geq \frac{2}{Q R} \Rightarrow \frac{2}{\sqrt{S Q \cdot S R}} \geq \frac{4}{Q R} \\<br/>\text { Hence, } & \frac{1}{P S}+\frac{1}{S T} \geq \frac{2}{\sqrt{Q S \cdot S R}} \geq \frac{4}{Q R} .<br/>\end{array}<br/>$$</div> | MarksBatch1_P1.db |
199 | a-student-is-performing-the-experiment-of-resonance-column-the-diameter-of-the-column-tube-is-4-cm-the-frequency-of-the-tuning-fork-is-512-hz-the-air- | a-student-is-performing-the-experiment-of-resonance-column-the-diameter-of-the-column-tube-is-4-cm-the-frequency-of-the-tuning-fork-is-512-hz-the-air-48462 | <div class="question">A student is performing the experiment of resonance column. The diameter of the column tube is $4 \mathrm{~cm}$. The frequency of the tuning fork is $512 \mathrm{~Hz}$. The air temperature is $38^{\circ} \mathrm{C}$ in which the speed of sound is $336 \mathrm{~m} / \mathrm{s}$. The zero of the meter scale coincides with the top end of the resonance column tube. When the first resonance occurs, the reading of the water level in the column is</div> | ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$14.0 \mathrm{~cm}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$15.2 \mathrm{~cm}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$16.4 \mathrm{~cm}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$17.6 \mathrm{~cm}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value">$15.2 \mathrm{~cm}$</span> </div> | <div class="solution">Considering the end correction $[e=0.3 D$ where $\mathrm{D}=$ diameter $]$
<br/> <br/>$f=n\left[\frac{v}{4(l+e)}\right]$ For first resonance, $n=1$
<br/> <br/>$\therefore f=\frac{v}{4(l+0.3 D)} \Rightarrow l=\frac{v}{4 f}-0.3 D$
<br/> <br/>or, $l=\left(\frac{336 \times 100}{4 \times 512}\right)-0.3 \times 4=15.2 \mathrm{~cm}$</div> | MarksBatch1_P1.db |
200 | a-student-performed-the-experiment-of-determination-of-focal-length-of-a-concave-mirror-by-u-v-method-using-an-optical-bench-of-length-15-m-the-focal- | a-student-performed-the-experiment-of-determination-of-focal-length-of-a-concave-mirror-by-u-v-method-using-an-optical-bench-of-length-15-m-the-focal-37259 | <div class="question">A student performed the experiment of determination of focal length of a concave mirror by $u-v$ method using an optical bench of length $1.5 \mathrm{~m}$. The focal length of the mirror used is 24 $\mathrm{cm}$. The maximum error in the location of the image can be $0.2 \mathrm{~cm}$.<br/>The 5 sets of $(u, v)$ values recorded by the student (in $\mathrm{cm})$ are : $(42,56),(48$, $48),(60,40),(66,33),(78,39)$. The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is(are)</div> | ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2009 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$(42,56)$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$(48,48)$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$(66,33)$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$(78,39)$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$(66,33)$<br/>, <br/>$(78,39)$</span> </div> | <div class="solution">Values of options (c) and (d) don't match with the mirror formula,<br/>$$<br/>\frac{1}{v}+\frac{1}{u}=\frac{1}{f}<br/>$$</div> | MarksBatch1_P1.db |