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a741e607fcc048f338a4b68123607e86 | WBJEE_PHY | Alternating Current | Voltage $V$ and current $i$ in $A C$ circuit are given by $\mathrm{V}=50 \sin (50 \mathrm{t})$ volt, $\mathrm{i}=50 \sin \left(50 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{mA}$<br />The power dissipated in the circuit is | singleCorrect | 2 | Given $V=50 \sin (50$ t) $V$ Maximum voltage, $\mathrm{V}_{0}=50 \mathrm{~V}$<br />$i=\left(50 t+\frac{\pi}{3}\right) \mathrm{mA}$<br />Maximum current, $i_{0}=50 \mathrm{~mA}=50 \times 10^{-3} \mathrm{~A}$<br />Power dissipated, $\mathrm{P}=\frac{\mathrm{i}_{0}}{\sqrt{2}} \times \frac{\mathrm{V}_{0}}{\sqrt{2}}$<br />$=\frac{50 \times 50 \times 10^{-3}}{2}=\frac{2500 \times 10^{-3}}{2}=1.25 \mathrm{~W}$ | ["$5.0 \\mathrm{~W}$", "$2.5 \\mathrm{~W}$", "$1.25 \\mathrm{~W}$", "zero"] | [2] | null | PYQ |
f198f76e8eae7e808148aa6c7329f641 | WBJEE_PHY | Alternating Current | A direct current of $5 \mathrm{~A}$ is superposed on an alternating current $I=10 \sin \omega t$ flowing through the wire. The effective value of the resulting current will be | singleCorrect | 2 | Total carrent, $1=(5+10 \sin \omega \mathrm{t})$<br />$\begin{aligned}<br />\Rightarrow I_{\text {eff }} &=\left[\frac{\int_{0}^{\mathrm{T}} \mathrm{I}^{2} \mathrm{dt}}{\int_{0}^{\mathrm{T}} \mathrm{dt}}\right]^{1 / 2} \\<br />&=\left[\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}(5+10 \sin \omega \mathrm{t})^{2} \mathrm{dt}\right]^{1 / 2} \\<br />&=\left[\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}\left(25+100 \sin \omega \mathrm{t}+100 \sin ^{2} \omega \mathrm{t}\right)\right]^{1 / 2} \\<br />\text { But, } \quad \frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}} \sin \omega \mathrm{t} \mathrm{dt}=0<br />\end{aligned}$<br />and<br />$\begin{array}{l}<br />\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}} \sin ^{2} \omega \mathrm{t} \mathrm{dt}=\frac{1}{2} \\<br />\text { So, } \mathrm{I}_{\mathrm{eff}}=\left[25+\frac{1}{2} \times 100\right]^{1 / 2}=5 \sqrt{3} \mathrm{~A}<br />\end{array}$ | ["$(15 / 2) \\mathrm{A}$", "$5 \\sqrt{3} \\mathrm{~A}$", "$5 \\sqrt{5} \\mathrm{~A}$", "$15 \\mathrm{~A}$"] | [1] | null | PYQ |
4bcbb2090367607886b0745c8a02b697 | WBJEE_PHY | Alternating Current | A resistor <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math>, an inductor <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>L</mi></math>, a capacitor <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi></math> and voltmeters <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>V</mi><mn>1</mn></msub><mo>,</mo><msub><mi>V</mi><mn>2</mn></msub></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>V</mi><mn>3</mn></msub></math> are connected to an oscillator in the circuit as shown in the adjoining diagram. When the frequency of the oscillation is increased, then at the resonant frequency, the voltmeter reading is zero in the case of<br /><img src="https://cdn.quizrr.in/question-assets/bitsat/py_sjn4d5v/1676fe19-0483-4851-ac28-556e31160568-image.png" style="width: 280px; height: 201px;" /> | singleCorrect | 1 | <p>At resonance, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>X</mi><mi>L</mi></msub><mo>=</mo><msub><mi>X</mi><mi>C</mi></msub></math> or <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ω</mi><mi>L</mi><mo>=</mo><mfrac><mn>1</mn><mrow><mi>ω</mi><mi>C</mi></mrow></mfrac></math></p><p>Voltage across the series <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>L</mi><mo>-</mo><mi>C</mi></math> combination,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>V</mi><mn>2</mn></msub><mo>=</mo><mi>i</mi><mfenced><mrow><msub><mi>X</mi><mi>L</mi></msub><mo>-</mo><msub><mi>X</mi><mi>C</mi></msub></mrow></mfenced><mo>=</mo><mn>0</mn></math></p> | ["voltmeter <math xmlns=\"http://www.w3.org/1998/Math/MathML\"><msub><mi>V</mi><mn>1</mn></msub></math> only", "voltmeter <math xmlns=\"http://www.w3.org/1998/Math/MathML\"><msub><mi>V</mi><mn>2</mn></msub></math> only", "voltmeter <math xmlns=\"http://www.w3.org/1998/Math/MathML\"><msub><mi>V</mi><mn>3</mn></msub></math> only", "All the three voltmeters"] | [1] | null | PYQ |
178d2a0f58247d1c4e9142a87ed9982e | WBJEE_PHY | Alternating Current | A transformer with efficiency <math><mn>80</mn><mi>%</mi></math> works at <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>4</mn><mi mathvariant="normal"> </mi><mi>kW</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>100</mn><mi mathvariant="normal"> </mi><mi mathvariant="normal">V</mi></math>. If the secondary voltage is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>200</mn><mi mathvariant="normal"> </mi><mi mathvariant="normal">V</mi><mo>,</mo></math> then the primary and secondary currents are respectively | singleCorrect | 2 | Here, <math><mi>η</mi><mo>=</mo><mn>80</mn><mi>%</mi></math><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>i</mi></msub><mo>=</mo><mn>4</mn><mo> </mo><mi>kW</mi><mo>=</mo><mn>4000</mn><mi mathvariant="normal"> </mi><mi mathvariant="normal">W</mi></math><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>V</mi><mi>p</mi></msub><mo>=</mo><mn>100</mn><mi mathvariant="normal"> </mi><mi mathvariant="normal">V</mi><mo>,</mo><mi> </mi><msub><mi>V</mi><mi>s</mi></msub><mo>=</mo><mn>200</mn><mi mathvariant="normal"> </mi><mi mathvariant="normal">V</mi></math><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>I</mi><mi>p</mi></msub><mo>=</mo><mfrac><msub><mi>P</mi><mi>i</mi></msub><msub><mi>V</mi><mi>p</mi></msub></mfrac><mo>=</mo><mfrac><mn>4000</mn><mn>100</mn></mfrac><mo>=</mo><mn>40</mn><mi mathvariant="normal"> </mi><mi mathvariant="normal">A</mi></math><br /><math><mi>η</mi><mo>=</mo><mfrac><mrow><msub><mrow><mi>V</mi></mrow><mrow><mi>s</mi></mrow></msub><msub><mrow><mi>I</mi></mrow><mrow><mi>s</mi></mrow></msub></mrow><mrow><msub><mrow><mi>V</mi></mrow><mrow><mi>p</mi></mrow></msub><msub><mrow><mi>I</mi></mrow><mrow><mi>p</mi></mrow></msub></mrow></mfrac><mo>;</mo><mfrac><mrow><mn>80</mn></mrow><mrow><mn>100</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>200</mn><msub><mrow><mi>I</mi></mrow><mrow><mi>s</mi></mrow></msub></mrow><mrow><mn>4000</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><msub><mrow><mi>I</mi></mrow><mrow><mi>s</mi></mrow></msub></mrow><mrow><mn>20</mn></mrow></mfrac></math><br />or <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>I</mi><mi>s</mi></msub><mo>=</mo><mfrac><mrow><mn>20</mn><mo>×</mo><mn>80</mn></mrow><mn>100</mn></mfrac><mo>=</mo><mn>16</mn><mi mathvariant="normal"> </mi><mi mathvariant="normal">A</mi></math> | ["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>40</mn><mi mathvariant=\"normal\"> </mi><mi mathvariant=\"normal\">A</mi></math> and <math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>16</mn><mi mathvariant=\"normal\"> </mi><mi mathvariant=\"normal\">A</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>16</mn><mi mathvariant=\"normal\"> </mi><mi mathvariant=\"normal\">A</mi></math> and <math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>40</mn><mi mathvariant=\"normal\"> </mi><mi mathvariant=\"normal\">A</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>20</mn><mi mathvariant=\"normal\"> </mi><mi mathvariant=\"normal\">A</mi></math> and <math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>90</mn><mi mathvariant=\"normal\"> </mi><mi mathvariant=\"normal\">A</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>40</mn><mi mathvariant=\"normal\"> </mi><mi mathvariant=\"normal\">A</mi></math> and <math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>20</mn><mi mathvariant=\"normal\"> </mi><mi mathvariant=\"normal\">A</mi></math>"] | [0] | null | PYQ |
dd4b31381610186dc9bee169d1330903 | WBJEE_PHY | Alternating Current | An alternating voltage $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ is applied across a circuit. As a result, a current $\mathrm{I}=\mathrm{I}_{0} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)$ flows in it. The power consumed per cycle is | singleCorrect | 2 | The phase angle between voltage $V$ and current $I$ is $\frac{\pi}{2}$.<br />Therefore, power factor $\cos \phi=\cos \left(\frac{\pi}{2}\right)=0$.<br />Hence the power consumed is zero. | ["zero", "$0.5 \\mathrm{~V}_{0} \\mathrm{I}_{0}$", "$0.707 \\mathrm{~V}_{0} \\mathrm{I}_{0}$", "$1.414 \\mathrm{~V}_{0} \\mathrm{I}_{0}$"] | [0] | null | PYQ |
f59551d3765ca92d62b9e08d4879e757 | WBJEE_PHY | Alternating Current | A direct current of $6 \mathrm{~A}$ is superimposed on an alternating current $I=10 \sin \omega t$ flowing through a wire. The effective value of the resulting current will be | singleCorrect | 2 | Given, $I=6+10 \sin \omega t$<br />$\begin{aligned}<br />I_{\mathrm{eft}} & =\left[\frac{\left.\int_0^T I^2 d t\right]^{1 / 2}}{\int_0^T d t}=\left[\frac{1}{T} \int_0^T(6+10 \sin \omega t)^2 d t\right]^{1 / 2}\right. \\<br />& =\left[\frac{1}{T} \int_0^T\left(36+120 \sin \omega t+100 \sin ^2 \omega t\right) d t\right]^{1 / 2}<br />\end{aligned}$<br />But as, $\frac{1}{T} \int_0^T \sin \omega t d t=0$ and $\frac{1}{T} \int_0^T \sin ^2 \omega t=\frac{1}{2}$<br />$\Rightarrow \quad I_{\mathrm{eff}}=\left[36+\frac{1}{2} \times 100\right]^{1 / 2}=9.27$<br />Thus, $\quad I_{\text {eff }}=9.27 \mathrm{~A}$. | ["$5 \\sqrt{2}$<br />", "$5 \\sqrt{3}$<br />", "$9.27$<br />", "$8.37$"] | [2] | null | PYQ |
4b18691d99c8e1ee04f4828e6a94a23c | WBJEE_PHY | Alternating Current | In the circuit shown in the figure, the $\mathrm{AC}$ source gives a voltage $V=20 \cos (2000 t)$ neglecting source resistance, the voltmeter and ammeter reading will be<br /><img src="https://cdn.quizrr.in/question-assets/bitsat/py_sjn4d5v/v5y3bhAJ29Rq6lxNBnA5ho52xioZBvUCYER9Mb7lpWI.original.fullsize.png"><br> | singleCorrect | 2 | Given, $R_1=8 \Omega$<br />$\begin{aligned} R_2 & =2 \Omega \\ L & =5 \mathrm{mH} \\ C & =50 \mu \mathrm{F}\end{aligned}$<br />and $\quad V_0=20 \cos (2000 t)$<br />Impedance, $Z=\sqrt{R^2+\left(X_L-X_C\right)^2}$<br />As, here, $X_L=\omega L=2000 \times 5 \times 10^{-3}=10 \Omega$<br />Similarly, $X_L=\frac{1}{\omega C}=\frac{1}{2000 \times 50 \times 10^{-6}}=10 \Omega$ $\because \quad X_L=X_C$, Hence, $i_{\max }=\frac{V_0}{Z}=\frac{20}{(8+2)}=2 \mathrm{~A}$<br />Hence, $\quad i_{\operatorname{rms}}=\frac{i_{\max }}{\sqrt{2}}=\frac{2}{\sqrt{2}}=1.41 \mathrm{~A}$ and<br />$\begin{aligned}<br />V & =R_2 i_{\mathrm{rms}}=1.41 \times 2 \\<br />& =282 \mathrm{~V}<br />\end{aligned}$ | ["$0 \\mathrm{~V}, 0.47 \\mathrm{~A}$<br />", "$282 \\mathrm{~V}, 1.41 \\mathrm{~A}$<br />", "$1.41 \\mathrm{~V}, 0.47 \\mathrm{~A}$<br />", "$15 \\mathrm{~V}, 837 \\mathrm{~A}$"] | [1] | null | PYQ |
750c9cad9325178d34c34d8954ebe310 | WBJEE_PHY | Alternating Current | An alternating voltage $V(t)=220 \sin 100 \pi t$ volt is applied to a purely resistive load of $50 \Omega$. The time taken for the current to rise from half of the peak value to the peak value is: | singleCorrect | 2 | Rising half to peak
$t=T / 6$
$\mathrm{t}=\frac{2 \pi}{6 \omega}=\frac{\pi}{3 \omega}=\frac{\pi}{300 \pi}=\frac{1}{300}=3.3 \mathrm{~ms}$
Here $\omega=100 \pi \mathrm{rad} / \mathrm{s}$ | ["5 ms", "3.3 ms", "7.2 ms", "2.2 ms"] | [1] | null | PYQ |
e9df3dca048674df62033f68e6fe9f8b | WBJEE_PHY | Alternating Current | The instantaneous voltage of a $50 \mathrm{~Hz}$ generator giving peak voltage as $300 \mathrm{~V}$, The generator equation for this voltage is | singleCorrect | 1 | Given, frequency of $\mathrm{AC}, \mathrm{v}=50 \mathrm{~Hz}$
Peak voltage, $V_{0}=300 \mathrm{~V}$
Generator equation for the voltage is given as
$\begin{aligned}
& V=V_{0} \sin \omega t \\
=& V_{0} \sin 2 \pi v t \\
=& 300 \sin (2 \pi \times 50 \times t) \\
=& 300 \sin 100 \pi t
\end{aligned}$ | ["$V=50 \\sin 300 \\pi i$", "$V=300 \\sin 100 \\pi t$", "$V=6 \\sin 100 \\pi t$", "$V=50 \\sin 100 \\pi t$"] | [1] | null | PYQ |
e896511054761a3ee9b6ccc27d6bcb98 | WBJEE_PHY | Alternating Current | Quality factor of a series $L-C-R$ circuit decreases from 3 to 2 . Resonant frequency is $600 \mathrm{~Hz}$. Change in bandwidth is | singleCorrect | 3 | Given, $f_{0}=600 \mathrm{~Hz}, Q_{1}=3, Q_{2}=2$
The bandwidth in $L-C-R$ circuit,
$\beta=\frac{f_{0}}{Q}$
As, quality factor decreases, bandwidth increases. This increase in bandwidth is given by
$\Delta \beta=\beta_{2}-\beta_{1}=\frac{f_{0}}{Q_{2}}-\frac{f_{0}}{Q_{1}}=f_{0}\left(\frac{1}{Q_{2}}-\frac{1}{Q_{1}}\right)$
$=600\left(\frac{1}{2}-\frac{1}{3}\right)=100 \mathrm{~Hz}$ | ["zero", "$100 \\mathrm{~Hz}$ increase", "$100 \\mathrm{~Hz}$ decrease", "$300 \\mathrm{~Hz}$ increase"] | [1] | null | PYQ |
6738a4a1873b6d31b7acfc02322f1c36 | WBJEE_PHY | Alternating Current | Quality factor of a series $L-C-R$ circuit decreases from 3 to 2 . Resonant frequency is $600 \mathrm{~Hz}$. Change in bandwidth is | singleCorrect | 3 | Given, $f_{0}=600 \mathrm{~Hz}, Q_{1}=3, Q_{2}=2$
The bandwidth in $L-C-R$ circuit,
$\beta=\frac{f_{0}}{Q}$
As, quality factor decreases, bandwidth increases. This increase in bandwidth is given by
$\Delta \beta=\beta_{2}-\beta_{1}=\frac{f_{0}}{Q_{2}}-\frac{f_{0}}{Q_{1}}=f_{0}\left(\frac{1}{Q_{2}}-\frac{1}{Q_{1}}\right)$
$=600\left(\frac{1}{2}-\frac{1}{3}\right)=100 \mathrm{~Hz}$ | ["zero", "$100 \\mathrm{~Hz}$ increase", "$100 \\mathrm{~Hz}$ decrease", "$300 \\mathrm{~Hz}$ increase"] | [1] | null | PYQ |
20037d353bec400ae9b9dfb85e39e21f | WBJEE_PHY | Alternating Current | A current of $5 \mathrm{~A}$ is flowing at $220 \mathrm{~V}$ in the primary coil of a transformer. If the voltage produced in the secondary coil is $2200 \mathrm{~V}$ and $50 \%$ of power is lost, then the current in the secondary will be | singleCorrect | 1 | For a transformer,<br>$\begin{aligned} I_{P}=5 \mathrm{~A}, V_{P} &=220 \mathrm{~V} \\ V_{S}=2200 \mathrm{~V} \end{aligned}$<br>Since, $50 \%$ power is lost, hence power at secondary side,<br>$\begin{array}{cl} <br>& P_{0}=50 \% \text { of } P_{i}=\frac{50}{100} \times 1100 \\<br>\Rightarrow & P_{0}=550 \mathrm{~W} \\<br>\Rightarrow & V_{S} I_{S}=550 \\<br>\Rightarrow & 2200 \times I_{S}=550 \\<br>\Rightarrow & I_{S}=\frac{550}{2200}=\frac{1}{4}=0.25 \mathrm{~A}<br>\end{array}$ | ["$2.5 \\mathrm{~A}$", "$5 \\mathrm{~A}$", "$0.25 \\mathrm{~A}$", "$0.5 \\mathrm{~A}$"] | [2] | null | PYQ |
645d2b051afcebb38d02047a2029e471 | WBJEE_PHY | Alternating Current | For a series $L-C-R$ circuit at resonance, the statement which is not true ? | singleCorrect | 1 | In series $L-C-R$ circuit,<br>$\begin{aligned}<br>X_{L} &=X_{C} \\<br>\therefore \quad & Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}<br>\end{aligned}$<br>$\begin{aligned}<br>&=\sqrt{R^{2}+0^{2}} \\<br>Z &=R \\<br>\therefore \text { Power factor } &=\cos \phi \\<br>&=\frac{R}{Z}=\frac{R}{R}=1<br>\end{aligned}$<br>Hence, power factor is not zero.<br>Therefore option (d) is not correct.<br>In $L-C-R$ series resonance circuit, peak energy stored by a capacitor is equal to peak energy stored by an inductor.<br>Average power, $P_{\mathrm{avg}}=V_{\mathrm{rms}} \times I_{\mathrm{rms}} \times \cos 0^{\circ}$<br>$\begin{aligned}<br>&=V_{\text {rms }} \times I_{\text {rms }} \\<br>&=\text { apparent power }<br>\end{aligned}$<br>$L-C-R$ series resonance circuit is purely resistive because $X_{L}=X_{C}$, hence no wattless current exists in this circuit therefore wattless current is zero. | ["Peak energy stored by a capacitor = peak energy stored by an inductor", "Average power $=$ Apparent power", "Wattless current is zero.", "Power factor is zero."] | [3] | null | PYQ |
edd8e81f7f6c72660e785f9f6f692816 | WBJEE_PHY | Alternating Current | For a series $L-C-R$ circuit at resonance, which statement is not true? | singleCorrect | 2 | In $L-C \cdot R$ series resonance circuit,<br>$X_{L_{L}}=X_{C}$<br>$\therefore$ Impedance, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}-\sqrt{R^{2}+0^{2}}$<br>$\Rightarrow \quad Z=R$<br>$\therefore$ Power factor, cos $\phi=\frac{n}{Z}=\frac{R}{R}=1$<br>Which is not zero.<br>Since, circuit behaves like a resistive circuit. $\mathrm{So}$, power loss occurs always in this circuits.<br>i.e, wattless carrent is zero. | ["Wattless current is zero.", "Power factor is zero.", "Peak energy stored by a capacitor = peak energy stored by an inductor.", "Average power = Apparent power."] | [1] | null | PYQ |
38288258077efbe14c9b1f4890312494 | WBJEE_PHY | Alternating Current | A DC ammeter and a hot wire ammeter are connected to a circuit in series. When a direct current is passed through circuit, the DC ammeter shows $6 \mathrm{~A}$. When $\mathrm{AC}$ current flows through circuit, what is the average readings in $\mathrm{DC}$ ammeter and the $\mathrm{AC}$ ammeter, if $\mathrm{DC}$ and AC currents flow simultaneously through the circuit? | singleCorrect | 3 | Resultant current is superposition of two currents, i.e. I (instantaneous total current)<br>$$<br>=6+I_0 \sin \omega t<br>$$<br>DC ammeter will read average value<br>$\begin{aligned}<br>& =\overline{6+I_0 \sin \omega t} \\<br>& =6 \mathrm{~A} \quad\left(\because \overline{I_0 \sin \omega t}=0\right)<br>\end{aligned}$<br>AC ammeter will read average value<br>$\begin{aligned}<br>& =\sqrt{\overline{\left(6+I_0 \sin ^2 \omega t\right)^2}} \\<br>& =\sqrt{36+12 I_0 \sin \omega t+I_0^2 \sin ^2 \omega t} \quad\left(\because \overline{I_0 \sin \omega t}=0\right)<br>\end{aligned}$<br>Since, $\overline{\sin ^2 \omega t}=\frac{1}{2}$ and $I_{\mathrm{rms}}=8=\frac{I_0}{\sqrt{2}}$<br>$$<br>\therefore \text { AC reading }=\sqrt{36+\frac{I_0^2}{2}}=\sqrt{36+64}<br>$$<br>$$<br>=10 \mathrm{~A}<br>$$ | ["$\\mathrm{DC}=6 \\mathrm{~A}, \\mathrm{AC}=10 \\mathrm{~A} \\quad$", "$\\mathrm{DC}=3 \\mathrm{~A}, \\mathrm{AC}=5 \\mathrm{~A}$", "$\\mathrm{DC}=5 \\mathrm{~A}, \\mathrm{AC}=8 \\mathrm{~A} \\quad$", "$\\mathrm{DC}=2 \\mathrm{~A}, \\mathrm{AC}=3 \\mathrm{~A}$"] | [0] | null | PYQ |
5bda2d0f2b98fd6734ed4cb8528561a8 | WBJEE_PHY | Alternating Current | The instantaneous values of alternating current and voltages in a circuit given as
$\begin{aligned}
& i=\frac{1}{\sqrt{2}} \sin (100 \pi t) \mathrm{amp}, \\
& e=\frac{1}{\sqrt{2}} \sin (100 \pi t+\pi / 3) \text { volt }
\end{aligned}$
The average power (in watts) consumed in the circuit is | singleCorrect | 1 | Given,
$i=\frac{1}{\sqrt{2}} \sin (100 \pi t) \mathrm{amp}$
and $e=\frac{1}{\sqrt{2}} \sin (100 \pi t+\pi / 3)$ volt
$\Rightarrow \quad i_0=\frac{1}{\sqrt{2}}$ and $e_0=\frac{1}{\sqrt{2}}$
We know that, average power,
$\begin{aligned}
P_{\mathrm{av}}= & V_{\mathrm{rms}} \times i_{\mathrm{rms}} \cos \phi=\frac{1}{2} \times \frac{1}{2} \times \cos 60^{\circ} \\
& \left(\because i_{\mathrm{rms}}=i_0 / \sqrt{2} \text { and } V_{\mathrm{rms}}=V_0 / \sqrt{2}\right) \\
= & \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8} \mathrm{~W}
\end{aligned}$ | ["$\\frac{1}{4}$", "$\\frac{\\sqrt{3}}{4}$", "$\\frac{1}{2}$", "$\\frac{1}{8}$"] | [3] | null | PYQ |
d6244b6bff485e50acf4972058664ac3 | WBJEE_PHY | Alternating Current | In an electrical circuit $R, L, C$ and $A C$ voltage source are all connected in series. When $L$ is removed from the circuit, the phase difference between the voltage and the current in the circuit is $\pi / 3$. If instead $C$ is removed from the circuit, the phase difference is again $\pi / 3$. The power factor of the circuit is | singleCorrect | 3 | Here, phase difference
$\tan \phi=\frac{X_L-X_C}{R} \Rightarrow \tan \frac{\pi}{3}=\frac{X_L-X_C}{R}$
When $L$ is removed, $\tan \frac{\pi}{3}=\frac{X_C}{R}=\sqrt{3}$
<img src="https://cdn.quizrr.in/question-assets/comedk/pysf5e2f/xc03FX16f8itL4YY8zpmxjSVELy6sL9BIZK57UpQ0v4.original.fullsize.png"><br>Similarly, when $C$ is removed
$\begin{aligned}
\tan \frac{\pi}{3} & =\frac{X_L}{R}=\sqrt{3} \\
\Rightarrow \quad X_L & =\sqrt{3} R
\end{aligned}$<img src="https://cdn.quizrr.in/question-assets/comedk/pysf5e2f/1k5GgKX0wbFvS4wH3Ezx2l5kR1F-zP3L8hl1QR2XZrE.original.fullsize.png"><br>$\begin{array}{lc}
\text { Now, } & \tan \phi=0 \\
\Rightarrow & \phi=0^{\circ}
\end{array}$
$\therefore$ Power factor, $\cos \phi=\cos 0^{\circ}=1$ | ["$1 / 2$", "$1 \\sqrt{2}$", "1", "$\\frac{\\sqrt{3}}{2}$"] | [2] | null | PYQ |
af16760cc1294e0eb203ee07c843bb90 | WBJEE_PHY | Alternating Current | In an LCR circuit, at resonance | singleCorrect | 1 | In an LCR circuit, at resonance the inductive reactance and capacitive reactance are equal in magnitude but in opposite<br>direction therefore they cancel each other. Thus, at resonance the current and voltage are in phase.<br><img src="https://cdn.quizrr.in/question-assets/kcet/py54ggf1c/sol_136.png"> | ["the current and voltage are in phase", "the impedance is maximum", "the current is minimum", "the current leads the voltage by II / \\(2\\)"] | [0] | null | PYQ |
97719a2eca98450eeb446339c7e3294c | WBJEE_PHY | Alternating Current | A capacitor of capacitance \(10 \mu \mathrm{F}\) is connected to an AC source and an AC Ammeter. If the<br>source voltage varies as \(V=50 \sqrt{2} \sin 100 t\), the reading of the ammeter is | singleCorrect | 2 | Given, capacitance $=10 \mu F=10 \times 10^{-6} F ;$ source voltage $=50 \sqrt{2} \sin 100 t$ and $\omega=100$
We know,
$I_{\text {rms }}=\frac{V_{\text {rms }}}{X_{C}}$ and $V_{\text {rms }}=\frac{V_{\max }}{\sqrt{2}}$
Now, $V_{\max }=50 \sqrt{2} ; X_{C}=\frac{1}{\omega C}$
Therefore,
$I_{\text {rms }}=\frac{V_{\max }}{\sqrt{2}} \times \omega C=\frac{50 \sqrt{2}}{\sqrt{2}} \times 100 \times 10 \times 10^{-6}$
$I_{\text {rms }}=5 \times 10^{4} \times 10^{-6}=50 \times 10^{-3}$
Therefore, average value of ac current over a cycle is $50 \mathrm{~mA}$ | ["\\(50 \\mathrm{~mA}\\)", "\\(70.7 \\mathrm{~mA}\\)", "\\(5.0 \\mathrm{~mA}\\)", "\\(7.07 \\mathrm{~mA}\\)"] | [0] | null | PYQ |
ef65ba9fc465a5bd2c849b41b8fa08b8 | WBJEE_PHY | Alternating Current | In a series L.C.R circuit, the potential drop across \(L, C\) and \(R\) respectively are \(40 \mathrm{~V}, 120 \mathrm{~V}\) and<br>\(60 \mathrm{~V}\). Then the source voltage is | singleCorrect | 1 | Given, voltage drop across \(\mathrm{L}, V_{L}=40 \mathrm{~V} ;\) voltage dropacross \(\mathrm{C}, V_{c}=120 \mathrm{~V} ;\) voltage drop across \(R, V_{R}=60 \mathrm{~V}\)<br><img src="https://cdn.quizrr.in/question-assets/kcet/py54ggf1c/sol_224.png"><br>Then source voltage \(=\sqrt{{V_{R}}^{2}+\left(V_{L}-V_{C}\right)^{2}}\)<br>\(=\sqrt{60^{2}+(120-40)^{2}}\)<br>\(=\sqrt{3600+6400}\)<br>\(=\sqrt{10000}\)<br>\(=100 \mathrm{~V}\)<br>Therefore, source voltage \(=100 \mathrm{~V}\) | ["\\(220 \\mathrm{~V}\\)", "\\(160 \\mathrm{~V}\\)", "\\(180 \\mathrm{~V}\\)", "\\(100 \\mathrm{~V}\\)"] | [3] | null | PYQ |
90423cb6fe45ce2dea39fe6bf9a1df62 | WBJEE_PHY | Alternating Current | In a series L.C.R. circuit an alternating emf (v) and current (i) are given by the equation<br>\(v=v_{0} \sin \omega t, l=i_{0} \sin \left(\omega t+\frac{\Pi}{3}\right)\)<br>The average power discipated in the circuit over a cycle of \(A C\) is | singleCorrect | 2 | Given, emf<br>\(v=v_{0} \sin \omega t ;\) current \(i=i_{0} \sin \left(\omega t+\frac{\Pi}{3}\right)\)<br>Then, average power dissipated in circuit over a cycle of ac is \(P_{\text {avg }}=v_{\mathrm{rms}} i_{\mathrm{ms}} \cos \phi\)<br>We know, \(v_{\mathrm{ms}}=\frac{v_{0}}{\sqrt{2}} ; i_{\mathrm{rms}}=\frac{i_{0}}{\sqrt{2}} ; \phi=\frac{\Pi}{3}\) \(\therefore P_{\mathrm{avg}}=\frac{v_{0}}{\sqrt{2}} \cdot \frac{i_{0}}{\sqrt{2}} \cos \left(\frac{\Pi}{3}\right)=\frac{v_{0} i_{0}}{2} \times \frac{1}{2}=\frac{v_{0} i_{0}}{4}\)<br>\(\therefore P_{\text {avg }}=\frac{v_{0}}{\sqrt{2}} \cdot \frac{i_{0}}{\sqrt{2}} \cos \left(\frac{\Pi}{3}\right)=\frac{v_{0} i_{0}}{2} \times \frac{1}{2}\) Thus, average power dissipated \(=\frac{v_{0} i_{0}}{4}\) | ["\\(\\frac{v_{0} i_{0}}{2}\\)", "\\(\\frac{v_{0} i_{0}}{4}\\)", "\\(\\frac{\\sqrt{3}}{2} v_{0} i_{0}\\)", "Zero"] | [1] | null | PYQ |
25aab46ce018b258cd81531fb0aa7347 | WBJEE_PHY | Alternating Current | A coil of inductive reactance \(\frac{1}{\sqrt{3}} \Omega\) and resistance \(1 \Omega\) is connected to a \(200 \mathrm{~V}, 50 \mathrm{~Hz}\) ac.<br>supply. The time lag between maximum voltage and current is | singleCorrect | 1 | Given, inductive reactance, \(\omega L=\frac{1}{\sqrt{3}} \Omega\); resistance, \(R=1 \Omega\); ac voltage, \(V=200 \mathrm{~V}, 50 \mathrm{~Hz}\);<br>Now, \(\tan \phi=\frac{\omega L}{R}\)<br>\(\begin{array}{l}
\Rightarrow \tan \phi=\frac{1}{\sqrt{3}} \\
\Rightarrow \phi=30^{\circ} \text { or } \phi=\frac{\Pi}{6} \\
\Rightarrow \omega t=\frac{\Pi}{6} \\
\Rightarrow t=\frac{\Pi}{6} \times \frac{1}{\omega}
\end{array}\)<br>Since \(\omega=2 \Pi f=2 \Pi \times 50\)<br>\(t=\frac{\Pi}{6} \times \frac{1}{2 \Pi \times 50}=\frac{1}{600} s\)<br>Therefore, time lag between maximum voltage and current is \(1 / 600 \mathrm{~s}\). | ["\\(\\frac{1}{300} s\\)", "\\(\\frac{1}{600} s\\)", "\\(\\frac{1}{500} \\mathrm{~s}\\)", "\\(\\frac{1}{200} s\\)"] | [1] | null | PYQ |
341030f71432e5c177288c3c53eb2f31 | WBJEE_PHY | Alternating Current | In Karnataka, the normal domestic power supply AC is \(220 \mathrm{~V}, 50 \mathrm{~Hz} .\) Here \(220 \mathrm{~V}\) and \(50 \mathrm{~Hz}\)<br>refer to | singleCorrect | 1 | In Karnataka, the normal domestic power supply ac is \(220 \mathrm{~V}, 50 \mathrm{~Hz}\). Here, \(220 \mathrm{~V}\) and \(50 \mathrm{~Hz}\) refer to rms value of voltage<br>and frequency. | ["Peak value of voltage and frequency", "Rms value of voltage and frequency", "Mean value of voltage and frequency", "Peak value of voltage and angular frequency"] | [1] | null | PYQ |
72dc497e5c41d0f603203fedccbb1dfe | WBJEE_PHY | Alternating Current | The frequency of an alternating current is \(50 \mathrm{~Hz}\). What is the minimum time taken by current to<br>reach its peak value from rms value? | singleCorrect | 1 | (D)<br><img src="https://cdn.quizrr.in/question-assets/kcet/py54ggf1c/sol_478.png"><br>We know that \(v_{r m s}=v_{0} \sin \omega t\)<br>\(\frac{v_{0}}{\sqrt{2}}=v_{0} \sin \omega t\)<br>\(\therefore \sin \omega t=\left(\frac{1}{\sqrt{2}}\right)\)<br>i.e., \(\omega t=\frac{\Pi}{4} \Rightarrow \frac{2 \Pi}{T} \cdot t=\frac{\Pi}{4}\)<br>\(\therefore t=\frac{T}{8}\)<br>Time for current to reach from rms value to peak value is i.e., \(I_{r m s} \rightarrow I_{0}\)<br>\(\frac{T}{4}-\frac{T}{8}=\frac{T}{8}\left(\frac{I_{0}}{\sqrt{2}} \rightarrow I_{0}\right)\)<br>\(f=50 \therefore t=\frac{I}{8}\)<br>\(=\frac{1}{50 \times 8}=\frac{1}{400}=0.25 \times 10^{-2}\)<br>\(=2.5 \times 10^{-3} \mathrm{~s}\)<br>\(\therefore[\mathrm{D}]\) is correct. | ["\\(0.02 \\mathrm{~s}\\)", "\\(5 \\times 10^{-3} s\\)", "\\(10 \\times 10^{-3}\\)", "\\(2.5 \\times 10^{-3} \\mathrm{~s}\\)"] | [3] | null | PYQ |
20b51ccbce36efd64373c167bb1b286b | WBJEE_PHY | Alternating Current | In the given circuit the peak voltage across $C$, $L$ and $R$ are $30 \mathrm{~V}, 110 \mathrm{~V}$ and $60 \mathrm{~V}$, respectively. The rms value of the applied voltage is<br><img src="https://cdn.quizrr.in/question-assets/kcet/py54ggf1c/vcqTERtpY1SVz-vL6PkNqHcARkpYgWS_Oe21Bn19qTM.original.fullsize.png"><br> | singleCorrect | 1 | Given, $V_{C}=30 \mathrm{~V}, V_{L}=110 \mathrm{~V}$ and $V_{R}=60 \mathrm{~V}$ The peak voltage across series $L-C-R$ circuit is given by<br>$\begin{aligned} V_{O} &=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}} \\ &=\sqrt{(60)^{2}+(110-30)^{2}} \\ &=\sqrt{(60)^{2}+(80)^{2}}=100 \mathrm{~V} \end{aligned}$<br>$\therefore$ Rms value of voltage, $V_{\text {rms }}=\frac{100}{\sqrt{2}} V=70.7 \mathrm{~V}$ | ["$100 \\mathrm{~V}$", "$200 \\mathrm{~V}$", "$70.7 \\mathrm{~V}$", "$141 \\mathrm{~V}$"] | [2] | null | PYQ |
3f9137786853608e716e5aeace916f70 | WBJEE_PHY | Alternating Current | In the given circuit, the resonant frequency is<br><img src="https://cdn.quizrr.in/question-assets/kcet/py54ggf1c/D_Sb0flnpDSjeZkTDl4RMLVsy-wSIsf6KgviguE41YU.original.fullsize.png"><br> | singleCorrect | 2 | Given, $L=0.5 \mathrm{mH}=0.5 \times 10^{-3} \mathrm{H}$ and $C=20 \mu \mathrm{F}=20 \times 10^{-6} \mathrm{~F}$<br>The resonant frequency of $L-C$ circuit is<br>$\begin{aligned}<br>f &=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{0.5 \times 10^{-3} \times 20 \times 10^{-6}}} \\<br>&=\frac{1}{2 \pi \times \sqrt{10 \times 10^{-9}}}=\frac{10^{4}}{2 \pi}=1592.3 \simeq 1592 \mathrm{~Hz}<br>\end{aligned}$ | ["$15.92 \\mathrm{~Hz}$", "$159.2 \\mathrm{~Hz}$", "$1592 \\mathrm{~Hz}$", "$15910 \\mathrm{~Hz}$"] | [2] | null | PYQ |
4d979b156e449bf153632eba0c616354 | WBJEE_PHY | Alternating Current | In a series $L C R$ circuit, $R=300 \Omega, L=0.9 \mathrm{H}$, $C=2.0 \mu \mathrm{F}$ and $\omega=1000 \mathrm{rad} / \mathrm{s}$, then impedance of the circuit is | singleCorrect | 1 | Given, $R=300 \Omega, L=0.9 \mathrm{H}, \omega=1000 \mathrm{rad} / \mathrm{s}$ $C=2.0 \mu \mathrm{F}=2 \times 10^{-6} \mathrm{~F}$<br>Impedance of given $R-L-C$ circuit,<br>$\begin{aligned}<br>Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\<br>& =\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\left[\therefore X_L=\omega L \text { and } X_C=\frac{1}{\omega C}\right] \\<br>& =\sqrt{(300)^2+\left(1000 \times 0.9-\frac{1}{1000 \times 2 \times 10^{-6}}\right)^2} \\<br>& =\sqrt{90000+(900-500)^2} \\<br>& =\sqrt{90000+160000}=\sqrt{250000}=500 \Omega<br>\end{aligned}$ | ["$900 \\Omega$", "$500 \\Omega$", "$400 \\Omega$", "$1300 \\Omega$"] | [1] | null | PYQ |
074369def583a58fef694bed7ad51c7e | WBJEE_PHY | Alternating Current | In an $L-C-R$ series circuit, the values of $R, X_L$ and $X_C$ are $120 \Omega, 180 \Omega$ and $130 \Omega$, what is the impedance of the circuit? | singleCorrect | 1 | Given, $R=120 \Omega$
$X_L=180 \Omega, X_C=130 \Omega$
The impedance of $L-C-R$ circuit
$Z=\sqrt{R^2+\left(X_L-X_C\right)^2}$
$\begin{aligned} Z & =\sqrt{(120)^2+(180-130)^2} \\ & =130 \Omega\end{aligned}$ | ["$120 \\Omega$", "$130 \\Omega$", "$180 \\Omega$", "$330 \\Omega$"] | [1] | null | PYQ |
947ebdcb23c4bfc3f998ebb7ed034f07 | WBJEE_PHY | Alternating Current | A sinusoidal voltage of peak value 300 V and an angular frequency $\omega=400 \mathrm{rad} / \mathrm{s}$ is applied to a series $L-C-R$ circuit, in which $R=3 \Omega$, $L=20 \mathrm{mH}$ and $C=625 \mu \mathrm{~F}$. The peak current in the circuit is | singleCorrect | 1 | The impedance of the circuit is
$\begin{aligned}
Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
X_L & =\omega L=400 \times 20 \times 10^{-3}=8 \mathrm{H} \\
X_C & =\frac{1}{\omega C}=\frac{1}{400 \times 625 \times 10^{-6}}=4 \mathrm{C} \\
Z & =\sqrt{(3)^2+(8-4)^2}=5 \\
i & =\frac{E}{Z}=\frac{300}{5}=60 \mathrm{~A}
\end{aligned}$ | ["$30 \\sqrt{2} \\mathrm{~A}$", "60 A", "100 A", "$60 \\sqrt{2} \\mathrm{~A}$"] | [1] | null | PYQ |
4fac025b6e98ab62fa1504517304d40f | WBJEE_PHY | Alternating Current | A transformer of $100 \%$ efficiency has 200 turns in the primary and 40000 turns in secondary. It is connected to a 220 V main supply and secondary feeds to a $100 \mathrm{k} \Omega$ resistance. The potential difference per turn is | singleCorrect | 1 | From transformer ratio
$\begin{aligned}
\frac{V_s}{V_p} & =\frac{N_s}{N_p} \\
\Rightarrow \quad V_s & =\frac{V_p \times N_s}{N_p} \\
& =\frac{220 \times 40000}{200} \\
& =44000 \mathrm{~V}
\end{aligned}$<br/>Potential difference per turn is
$\frac{V_s}{N_s}=\frac{44000}{40000}=1.1 \mathrm{~V}$ | ["1.1 V", "25 V", "18 V", "11 V"] | [0] | null | PYQ |
b01a6a9540ab1b71a472507d749d7595 | WBJEE_PHY | Alternating Current | An alternating e.m.f. is given by e $=e_{0}$ sin wt. In what time the e.m.f. will have half its maximum value, if 'e' starts from zero? (T = Time period) $\left(\sin 30^{\circ}=\cos 60^{\circ}=0 \cdot 5, \quad \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)$ | singleCorrect | 3 | \(\frac{E_{0}}{2}=E_{0} \sin \left(\frac{2 \pi t}{T}\right) \therefore \frac{1}{2}=\frac{\sin (\pi)}{6}=\frac{\sin (2 \pi t)}{T}\)
\(\therefore \frac{2 \pi t}{T}=\frac{\pi}{6} \therefore t=\frac{T}{12}\) | ["$\\frac{T}{12}$", "$\\frac{T}{16}$", "$\\frac{T}{8}$", "$\\frac{T}{4}$"] | [0] | null | PYQ |
1900f03aab896eaaafccda6185064757 | WBJEE_PHY | Alternating Current | The resonant frequency of a series LCR circuit is ' $\mathrm{f}$ '. The circuit is now connected to the sinusoidally alternating e.m.f. of frequency ' $2 \mathrm{f}^{\prime}$. The new reactance $\mathrm{X}_{\mathrm{L}}^{\prime}$ and
$\mathrm{X}_{\mathrm{C}}^{\prime}$ are related as | singleCorrect | 2 | In a series LCR circuit, the resonant frequency is given by:
$f_0=\frac{1}{2 \pi \sqrt{L C}}$
When the circuit is connected to a sinusoidally alternating e.m.f. with a frequency $f$, the reactance of the circuit (composed of inductive reactance $X_L$ and capacitive reactance $X_C$ ) will change with frequency. At resonance, $X_L=X_C$, but off-resonance, the total reactance changes depending on the frequency of the applied e.m.f.
Answer: (1) is correct, as it shows the relationship between reactance and frequency in a series LCR circuit. | ["$\\mathrm{X}_{\\mathrm{C}}^{\\prime}=\\frac{1}{4} \\mathrm{X}_{\\mathrm{L}}^{\\prime}$", "$\\mathrm{X}_{\\mathrm{C}}^{\\prime}=2 \\mathrm{X}_{\\mathrm{L}}^{\\prime}$", "$\\mathrm{X}_{\\mathrm{C}}^{\\prime}=\\mathrm{X}_{\\mathrm{L}}^{\\prime}$", "$\\mathrm{X}_{\\mathrm{C}}^{\\prime}=\\frac{1}{2} \\mathrm{X}_{\\mathrm{L}}^{\\prime}$"] | [0] | null | PYQ |
54650187e6d508a1e1d148eae7b79d9e | WBJEE_PHY | Alternating Current | In a capactive circuit, the reactance of capacitor at frequency 'f' is ' $\mathrm{X}_{\mathrm{c}}{ }^{\prime}$. What will be
its reactance at frequency $4 \mathrm{f} ?$ | singleCorrect | 1 | Capacitive reactance $\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \pi \mathrm{fC}}$
$\begin{array}{l}
\therefore \frac{\mathrm{X}_{\mathrm{C}}^{\prime}}{\mathrm{X}_{\mathrm{c}}}=\frac{\mathrm{f}}{\mathrm{f}^{\prime}}=\frac{1}{4} \\
\therefore \mathrm{X}_{\mathrm{c}}^{\prime}=\frac{\mathrm{X}_{\mathrm{c}}}{4}
\end{array}$ | ["$\\frac{\\mathrm{X}_{\\mathrm{C}}}{2}$", "$\\frac{X_{C}}{4}$", "$\\frac{\\mathrm{X}_{\\mathrm{C}}}{8}$", "$X_{C}$"] | [1] | null | PYQ |
83b87292a646963a92040d5783e2a63c | WBJEE_PHY | Alternating Current | In series LCR circuit, resistance is $18 \Omega$ and impedance is $33 \Omega$. An r.m.s. voltage of
$220 \mathrm{~V}$ is applied across the circuit. The true power consumed in a.c. circuit is | singleCorrect | 2 | $I=\frac{V}{Z}=\frac{220}{33}=\frac{20}{3} A$
$P=I^{2} R=\left(\frac{20}{3}\right)^{2} \times 18=800 \mathrm{~W}$ | ["$400 \\mathrm{~W}$", "$600 \\mathrm{~W}$", "$800 \\mathrm{~W}$", "$200 \\mathrm{~W}$"] | [2] | null | PYQ |
5fa60e73c8bc79fa7043e21e3b1b4e69 | WBJEE_PHY | Alternating Current | In LCR circuit the inductance is changed from L to $9 \mathrm{~L}$. For same resonant
frequency the capacitance should be changed from $\mathrm{C}$ to | singleCorrect | 1 | $\omega=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$
$\therefore \quad \mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$
$\quad \mathrm{C}_{2}=\frac{\mathrm{L}_{1} \mathrm{C}_{1}}{\mathrm{~L}_{2}}=\frac{\mathrm{L}}{9 \mathrm{~L}} \times \mathrm{C}=\frac{\mathrm{C}}{9}$ | ["$9 \\mathrm{C}$", "$3 \\mathrm{C}$", "$\\frac{C}{9}$", "$\\frac{C}{3}$"] | [2] | null | PYQ |
387d68167c0888401d1ba599e4924080 | WBJEE_PHY | Alternating Current | Alternating current of peak value $\left(\frac{2}{\pi}\right)$ A flows through the primary coil of transformer. The coefficient of mutual inductance between primary and secondary coil is $1 \mathrm{H}$. The peak e.m.f. induced in secondary coil is (Frequency of a.c. $=50 \mathrm{~Hz}$ ) | singleCorrect | 3 | Given \(: \mathrm{I}_{0}=\frac{2}{\pi}\) ampere \(v=50 \mathrm{~Hz} \mathrm{~L}=1 \mathrm{H}\)
Thus \(w=2 \pi v=2 \pi(50)=100 \pi\)
Alternating current flowing through the coil is given by \(I=I_{0} \sin w t\)
Differentiating it wr.t. time we get \(\frac{\mathrm{dI}}{\mathrm{dt}}=\mathrm{I}_{0} \mathrm{w} \cos w \mathrm{t}\) \(\left.\therefore \frac{\mathrm{dI}}{\mathrm{dt}}\right|_{\max }=\mathrm{I}_{0} \mathrm{~W}=\frac{2}{\pi} \times 100 \pi=200\) ampere per second
Peak e.m.f induced \(\mathcal{E}=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}\)
\(\Longrightarrow \mathcal{E}=1 \times 200=200 \mathrm{~V}\) | ["$400 \\mathrm{~V}$", "$200 \\mathrm{~V}$", "$300 \\mathrm{~V}$", "$100 \\mathrm{~V}$"] | [1] | null | PYQ |
fc579b18dd4c8c11a1641b0f65fdddea | WBJEE_PHY | Alternating Current | A step-up transformer has 300 turns of primary winding and 450 turns of secondary winding. A primary is connected to 150 volt and the current flowing through it is $9 \mathrm{~A}$. The current and voltage in the secondary are | singleCorrect | 2 | $6.0 A, 225 V$
Given that, number of turns in primary winding, $N_{p}=300$
Number of turns in secondary winding, $N_{S}=450$
Primary voltage, $\mathrm{V}_{\mathrm{p}}=150 \mathrm{~V}$
Primary current, $I_{p}=9 \mathrm{~A}$
For step-up transformer,
$\begin{array}{l}
\frac{V_{S}}{V_{P}}=\frac{N_{S}}{N_{P}} \\
\frac{V_{S}}{150}=\frac{450}{300} \\
\Rightarrow V_{S}=\frac{450}{300} \times 150 \\
\Rightarrow V_{S}=225 \mathrm{~V}
\end{array}$
$\begin{array}{l}
\text { Again, } \mathrm{V}_{\mathrm{p}} \mathrm{I}_{\mathrm{p}}=\mathrm{V}_{\mathrm{s}} \mathrm{I}_{\mathrm{S}} \\
150 \times 9=225 \times \mathrm{I}_{\mathrm{S}} \\
\mathrm{I}_{\mathrm{s}}=\frac{1350}{225}=6.0 \mathrm{~A}
\end{array}$ | ["$13.5 \\mathrm{~A}, 225 \\mathrm{~V}$", "$13.5 \\mathrm{~A}, \\quad 100 \\mathrm{~V}$", "$4.5 \\mathrm{~A}, \\quad 100 \\mathrm{~V}$", "$6.0 \\mathrm{~A}, \\quad 225 \\mathrm{~V}$"] | [3] | null | PYQ |
5218ef4641195dd7388d657230f9378b | WBJEE_PHY | Alternating Current | Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time combination is connected to a 220 volt a.c. supply line. The power drawn by the combination in each case respectively will be: | singleCorrect | 2 | <img src="https://cdn.quizrr.in/question-assets/neet/2025/pyqsd64af1fs5/Asqere.png"/><br>
\(\begin{aligned} & \mathrm{R}_1=\frac{\mathrm{V}^2}{\mathrm{P}_1}=\frac{(220)^2}{100}=484 \Omega \\ & \mathrm{R}_2=\frac{\mathrm{V}^2}{\mathrm{P}_2}=\frac{(220)^2}{100}=484 \Omega \\ & \mathrm{I}=\frac{220}{484 \times 2}=\frac{5}{22} \\ & \mathrm{P}=\mathrm{I}^2\left(\mathrm{R}_1+\mathrm{R}_2\right)=\frac{25}{22 \times 22} \times(484 \times 2) \\ & =50 \mathrm{~W}\end{aligned}\)<br>
<img src="https://cdn.quizrr.in/question-assets/neet/2025/pyqsd64af1fs5/pBZ4NFTGgghe.png"/><br>
\(\begin{aligned} & P=\frac{V^2}{R_1}+\frac{V^2}{R_2} \\ & =\frac{2 \times(220)^2}{484}=200 \mathrm{~W}\end{aligned}\) | ["50 watt, 10 watt", "100 watt, 50 watt", "200 watt, 150 watt", "50 watt, 200 watt"] | [3] | null | PYQ |
53bd277c3d36c4abe18ee59c46158c45 | WBJEE_PHY | Alternating Current | An electric kettle has two heating coils. When one of the coils is connected to a.c. source, the water in the kettle boils in 10 minute. When the other coil is used the water boils in 40 minute. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be: | singleCorrect | 3 | <img src="https://cdn.quizrr.in/question-assets/neet/2025/pyqsd64af1fs5/_bSMx4npJAnec0qlD3o9a9-F4b7QnjU1uR_uxVgq1PQ.original.fullsize.png"/>
Here $Q=\frac{V^2}{R_1} \times t_1=\frac{V^2}{R_2} \times t_2$
$\begin{aligned}
& =\frac{V^2}{R} \times t \\
& \therefore \quad \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\
& \Rightarrow \quad \frac{Q}{V^2 t}=\frac{Q}{V^2 t_1}+\frac{Q}{V^2 t_2} \\
& \Rightarrow \quad \frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_4} \\
& \Rightarrow \quad t=\frac{t_1 t_2}{t_1+t_2} \\
& =\frac{10 \times 40}{10+40}=8 \mathrm{~min} \\
&
\end{aligned}$ | ["8 minute", "4 minute", "25 minute", "15 minute"] | [0] | null | PYQ |
077f5050b664a91c3d77d6aafe9b1ae0 | WBJEE_PHY | Alternating Current | The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux $\phi$ linked with the primary coil is given by $\phi=\phi_v+4 t$, where $\phi$ is in webers, $t$ is time in seconds and $\phi_v$ is a constant, the output voltage across the output voltage across the secondary coil is: | singleCorrect | 2 | The magnetic flux linked with the primary coil is given by<br>
\(\phi=\phi_0+4 t\)<br>
So, voltage across primary<br>
\(\begin{aligned}<br>
& V_p=\frac{d \phi}{d t}=\frac{d}{d t}\left(\phi_0+4 t\right) \\<br>
& \left.=4 \text { volt (as } \phi_0=\text { constant }\right)<br>
\end{aligned}\)<br>
Also, we have<br>
\(N_p=50 \text { and } N_s=1500\)<br>
From relation,<br>
\(\begin{aligned}<br>
& \frac{V_s}{V_p}=\frac{N_s}{N_p} \\<br>
& \text {or } V_s=V_p \frac{N_s}{N_p} \\<br>
& =4\left(\frac{1500}{50}\right) \\<br>
& =120 \mathrm{~V}<br>
\end{aligned}\) | ["120 volts", "220 volts", "30 volts", "90 volts."] | [0] | null | PYQ |
1799776e0a25c9122a6bf4550352b725 | WBJEE_PHY | Alternating Current | An alternating voltage (in volts) varies with time $t$ (in seconds) as
$V=200 \sin (100 \pi t)$ | multipleCorrect | 1 | Comparing the equation $V=200 \sin (100 \pi t)$ with $V=V_0 \sin \omega t$, we find that $V_0=200 \mathrm{~V}$ and $\omega=100 \pi \mathrm{rad} \mathrm{s}^{-1}$ or $2 \pi \nu=100 \pi$ or $v=50 \mathrm{~Hz}$.
Also $V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}=\frac{200}{\sqrt{2}}=100 \sqrt{2} \mathrm{~V}$.
Hence the correct choices are (a), (c) and (d). | ["The peak value of the voltage is 200 V .", "The rms value of the voltage is 220 V .", "The rms value of the voltage is $100 \\sqrt{2} \\mathrm{~V}$", "The frequency of the voltage is 50 Hz ."] | [0, 2, 3] | null | PYQ |
d3fb82aeb56b8f56faf30b7e1dd7257a | WBJEE_PHY | Alternating Current | Figure 25.69 shows a series $L C R$ circuit connected to a variable frequency 200 V source. $L=5 \mathrm{H}$, $C=80 \mu \mathrm{~F}$ and $R=40 \Omega$.
<img src="https://cdn-question-pool.getmarks.app/modules/ms/wbjee/5EWr2SCtOOPJwTf1mbK0OW2Eb7t5VRdW2PfUdipXyy0.original.fullsize.png"/><br/> | multipleCorrect | 3 | The resonant angular frequency is
$\begin{aligned} \omega_r & =\frac{1}{\sqrt{L C}} \\ & =\frac{1}{\left(5.0 \times 80 \times 10^{-6}\right)^{1 / 2}}=50 \mathrm{rad} \mathrm{s}^{-1}\end{aligned}$
Therefore, the resonant frequency is
$Z=\left[R^2+\left(\omega L-\frac{1}{\omega C}\right)^2\right]^{1 / 2}$
When $\omega=\omega_r=1 / \sqrt{L C}$ (i.e. at resonance), $\omega L=1 / \omega C$, and therefore
$Z=R=40 \Omega$
Current amplitude at resonance is
$\begin{aligned}
I_0=\frac{V_0}{Z}=\frac{V_0}{R} & =\frac{\sqrt{2} V_{\mathrm{rms}}}{R} \\
& =\frac{\sqrt{2} \times 200}{40}=5 \sqrt{2} \mathrm{~A} .
\end{aligned}$
The rms current in the circuit is
$I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{R}=\frac{200}{40}=5 \mathrm{~A}$
$\therefore$ The rms potential drop across $L$ is
$\begin{aligned}
& =I_{\mathrm{rms}} \times \omega_r \times L=5 \times 50 \times 5 \\
& =1250 \mathrm{~V}=1.25 \mathrm{kV}
\end{aligned}$
The rms potential drop across $C$ is
$\begin{aligned}
& =I_{\mathrm{rms}} \times \frac{1}{\omega_r C} \\
& =5 \times \frac{1}{50 \times 80 \times 10^{-6}}=1.25 \mathrm{kV}
\end{aligned}$
The rms potential drop across $R$ is
$=I_{\mathrm{rms}} \times R=5 \times 40=200 \mathrm{~V}$
Hence the correct choices are (a), (c) and (d). | ["The impedance of the circuit at resonance is $40 \\Omega$.", "The current amplitude at resonance is 5 A .", "The rms potential drop across the inductor at resonance is 1250 V .", "The rms potential drop across the resistor at resonance is 200 V ."] | [0, 2, 3] | null | PYQ |
f6c2758c45e6644530abb5ff3c8cd327 | WBJEE_PHY | Alternating Current | $L, C$ and $R$ respectively represent inductance, capacitance and resistance. Which of the following combinations have the dimensions of frequency? | multipleCorrect | 1 | The correct choices are (a), (b) and (d). The dimensions of $\omega L$ and $\frac{1}{\omega C}$ are the same as those of resistance, where $\omega=2 \pi \nu$. | ["$\\frac{R}{L}$", "$\\frac{1}{R C}$", "$\\frac{R}{\\sqrt{L C}}$", "$\\frac{1}{\\sqrt{L C}}$"] | [0, 1, 3] | null | PYQ |
fbfb41520f420b09ab926f92de6498cd | WBJEE_PHY | Alternating Current | In a transformer, the number of turns in the primary and secondary are 400 and 200 respectively. The power input to the primary is 10 kW at 200 V . The efficiency of the transformer is $90 \%$. | multipleCorrect | 2 | $\begin{aligned} \text { Output voltage } e_s=\frac{N_s}{N_p} \times e_p & =\frac{2000}{400} \times 200 \\ & =1000 \mathrm{~V} \\ \text { Output power }=90 \% \text { of } 10 \mathrm{~kW} & =9 \mathrm{~kW}\end{aligned}$
Current in primary $I_p=\frac{\text { Input power }}{e_p}=\frac{10,000 \mathrm{~W}}{200 \mathrm{~V}}$
$=50 \mathrm{~A}$
$\begin{aligned}
\text { Current in secondary } I_s & =\frac{\text { Output power }}{e_s} \\
& =\frac{9000 \mathrm{~W}}{1000 \mathrm{~V}}=9 \mathrm{~A}
\end{aligned}$
Hence the correct choices are (a), (b) and (c). | ["The output voltage is 100 V .", "The output power is 9 kW .", "The current in the primary is 50 A .", "The current in the secondary is 10 A ."] | [0, 1, 2] | null | PYQ |
e0cd8855e81f9cf9c5cca1c0fad26c08 | WBJEE_PHY | Alternating Current | A series $R-C$ circuit is connected to AC voltage source. Consider two cases; (A) when C is without a dielectric medium and $(\mathrm{B})$ 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? | multipleCorrect | 2 | Case $A: I_{\mathrm{R}}^{\mathrm{A}}=\frac{V}{Z_A}=\frac{V}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}} \qquad\mathrm{(i)}$
$V_{\mathrm{C}}^{\mathrm{A}}=\frac{I_{\mathrm{R}}^{\mathrm{A}}}{\omega \mathrm{C}}=\frac{V}{\sqrt{(R \omega C)^2}+\mathrm{I}} \qquad\mathrm{(ii)}$
Case B: $C_B=K C=4 C$. Hence
$\begin{aligned}
I_{\mathrm{R}}^{\mathrm{B}} & =\frac{V}{\sqrt{R^2+\left(\frac{1}{4 \omega C}\right)^2}} \qquad\mathrm{(iii)} \\
V_{\mathrm{C}}^{\mathrm{B}} & =\frac{I_{\mathrm{R}}^{\mathrm{B}}}{4 \omega C}=\frac{V}{\sqrt{(4 R \omega C)^2}+1} \qquad\mathrm{(iv)}
\end{aligned}$
From Eqs. (i) to (iv) it follows that
$I_{\mathrm{R}}^{\mathrm{A}} \lt I_{\mathrm{R}}^{\mathrm{B}}$ and $V_{\mathrm{C}}^{\mathrm{A}}\gtV_{\mathrm{C}}^{\mathrm{B}}$ | ["$I_{\\mathrm{R}}^{\\mathrm{A}}\\gtI_{\\mathrm{R}}^{\\mathrm{B}}$", "$I_{\\mathrm{R}}^{\\mathrm{A}} \\lt I_{\\mathrm{R}}^{\\mathrm{B}}$", "$V_{\\mathrm{C}}^{\\mathrm{A}}\\gtV_{\\mathrm{C}}^{\\mathrm{B}}$", "$V_{\\mathrm{C}}^{\\mathrm{A}} \\lt V_{\\mathrm{C}}^{\\mathrm{B}}$"] | [1, 2] | null | PYQ |
da307eaa2ee02b586913ea378c430731 | WBJEE_PHY | Alternating Current | <img src="https://cdn-question-pool.getmarks.app/modules/ms/wbjee/LmlYfJdNHCg6zoTySI0IH37LxPXqTBZucUuKuRnzHMk.original.fullsize.png"/><br/>
In the circuit shown in the figure, the A.C. source gives a voltage $V=20 \cos (2000 t)$ volt. Neglecting source resistance, the voltmeter and ammeter readings will be :- | multipleCorrect | 1 | $\begin{aligned} & \text { Here } \frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{5 \times 10^{-3} \times 50 \times 10^{-6}}}=2000 \mathrm{rad} / \mathrm{s} \\ & \Rightarrow \omega=\frac{1}{\sqrt{\mathrm{LC}}} \text { so resonance condition } \Rightarrow \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{20 / \sqrt{2}}{6+4}=\sqrt{2}=1.4 \mathrm{~A} \\ & \Rightarrow \text { Reading of ammeter }=1.4 \mathrm{~A} . \Rightarrow \text { Reading of voltmeter }=4 \sqrt{2} \mathrm{~V}=5.6 \mathrm{~V}\end{aligned}$ | ["$0 \\mathrm{~V}, 2 \\mathrm{~A}$", "$0 \\mathrm{~V}, 1.4 \\mathrm{~A}$", "$5.6 \\mathrm{~V}, 0.47 \\mathrm{~A}$", "$1.68 \\mathrm{~V}, 0.47 \\mathrm{~A}$"] | [1, 2] | null | PYQ |
1265d61a0e9d017221e624680b7c9333 | WBJEE_PHY | Alternating Current | In a certain series LCR A.C. circuit it is found that $X_L=2 X_C$ and phase difference between the current and voltage is $\pi / 4$. If now capacitance is made $1 / 4^{\text {th }} \&$ the inductance and resistance are doubled then which are not correct. | multipleCorrect | 2 | $\begin{aligned} & X_L=2 X_C \\ & X_C^{\prime}=4 X_C \\ & X_L^{\prime}=2 X_L \\ & \Rightarrow X_L^{\prime}-X_C^{\prime}=2 X_L-4 X_C=2\left[X_L-2 X_C\right]=0\end{aligned}$ | ["voltage will lead current by a phase of $\\tan ^{-1}(1 / 2)$.", "voltage will lag behind current by a phase of $\\tan ^{-1}(1 / 2)$.", "voltage will lag behind current by a phase of $\\pi / 4$.", "voltage and current will be in same phase."] | [0, 1, 2] | null | PYQ |
c69f02cbda3b2b4773642f5e85cbbf91 | WBJEE_PHY | Alternating Current | <img src="https://cdn-question-pool.getmarks.app/modules/ms/wbjee/FBjrS5CtPIal5yIc8wANTkGQi7VRVjaMIuMKf3R3vEs.original.fullsize.png"/><br/>
In a RLC series circuit shown, the readings of voltmeters $V_1$ and $V_2$ are 100 V and 120 V , respectively. The source voltage is 130 V . For this situtation mark out the correct statement (s) | multipleCorrect | 2 | $\begin{aligned} & V_1^2=V_R^2+V_L^2 \\ & V_2=V_C-V_L \\ & V^2=V_R^2+\left(V_C-V_L\right)^2 \\ & P F=\frac{V_R}{V_C-V_L}\end{aligned}$ | ["Voltage across resistor, inductor and capacitor are $50 \\mathrm{~V}, 86.6 \\mathrm{~V}$ and 206.6 V , respectively", "Voltage across resistor, inductor and capacitor are $10 \\mathrm{~V}, 90 \\mathrm{~V}$ and 30 V , respectively", "Power factor of the circuit is $\\frac{5}{13}$", "Circuit is capactivite in nature"] | [2, 3] | null | PYQ |
7d4c516f16cccd4268661a86bc03c659 | WBJEE_PHY | Alternating Current | An $L R$ series circuit consists of inductance $L=\frac{100}{\pi} \mathrm{mH}$ and a resistance $R=10 \Omega$. A sinusoidal voltage $V=V_0 \sin (2 \pi v t)$ is applied. It is given that $V_{\mathrm{rms}}=200 \mathrm{~V}$ and $v=50 \mathrm{~Hz}$. | multipleCorrect | 3 | Inductive reactance $X_L=\omega L=2 \pi v L$
$=2 \pi \times 50 \times \frac{100}{\pi} \times 10^{-3}=10 \Omega$. Therefore, impedance is
$Z=\left(X_L^2+R^2\right)^{1 / 2}=\left(10^2+10^2\right)^{1 / 2}=10 \sqrt{2} \Omega$
The amplitude of the current in the steady state is
$I_0=\frac{V_0}{Z}=\frac{V_{\mathrm{rms}} \times \sqrt{2}}{Z}=\frac{200 \times \sqrt{2}}{10 \sqrt{2}}=20 \mathrm{~A}$
The phase difference between the current and the voltage is
$\phi=\tan ^{-1}\left(\frac{X_L}{R}\right)=\tan ^{-1}\left(\frac{10}{10}\right)=\tan ^{-1}$ $\quad$(1)
which gives $\phi=\frac{\pi}{4}$.
In an $L R$ circuit, the current lags behind the voltage by a phase angle $\phi$. In the given circuit $V=V_0 \sin (2 \pi v t), I_0=20 \mathrm{~A}$ and $\phi=\pi / 4$. Hence the current $I$ varies with time $t$ as
$I=I_0 \sin \left(2 \pi v t-\frac{\pi}{4}\right)$
or $\quad I=20 \sin \left(\frac{2 \pi t}{T}-\frac{\pi}{4}\right)$ ampere $\quad$(1)
where $\quad T=\frac{1}{v}=\frac{1}{50}=0.02 \mathrm{~s}$.
At $t=0, I=20 \sin \left(-\frac{\pi}{4}\right)=-10 \sqrt{2}$ A.
From Eq. (1) it follows that $I=0$ at $t=\frac{T}{8}, \frac{5 T}{8}, \cdots$ | ["The peak value of the current in the steady state is 20 A .", "The phase difference between the current and the voltage is $\\pi / 2$", "At time $t=0$, the current in the circuit is $-10 \\sqrt{2} \\mathrm{~A}$", "The current in the circuit is zero at $t=\\frac{T}{8}$, $\\frac{5 T}{8}, \\ldots$, where $T=0.02 \\mathrm{~s}$."] | [0, 2, 3] | null | PYQ |
bd80c62e3838dedfe78f608915daa245 | WBJEE_PHY | Alternating Current | Which of the following statements are correct? | multipleCorrect | 2 | Statement (a) is incorrect. Since the capacitor offers infinite resistance to direct current, the bulb will not glow at all. Statement (b) is also incorrect. The capacitor will offer finite reactance to alternating current. Hence a current will flow in the circuit and the bulb will glow. Statement (c) is correct. The reactance of a capacitor is $1 / \omega C$. Hence if $C$ is decreases, the reactance will increase and as a result the current is decreased. Hence the brightness of the bulb is reduced.
Statement (d) is also correct, because the voltages across the different elements are not in phase. Therefore, they cannot be added algebraically; they are added vectorially. | ["A capacitor is connected in series with a bulb and this combination is connected to a variable voltage dc source. The brightness of the bulb will increase with increase in the voltage.", "A capacitor is connected in series with a bulb. When this combination is connected to an ac source, the bulb does not glow.", "a variable capacitor is connected in series with a bulb and this combination is connected to an ac source. The brightness of the bulb is reduced if the capacitance of the variable capacitor is decreased.", "In a series ac circuit, the applied voltage is not equal to the algebraic sum of voltages across the different elements of the circuit."] | [2, 3] | null | PYQ |
76210f3e68cbf78d3b286462af6a112d | WBJEE_PHY | Atomic Physics | If hydrogen atom, an electron jumps from bigger orbit to smaller orbit so that radius of smaller orbit is one-fourth of radius of bigger orbit. If speed of electron in bigger orbit was $\mathrm{v}$, then speed in smaller orbit is | singleCorrect | 2 | Radius of the orbit, $\mathrm{r}_{\mathrm{n}} \propto \mathrm{n}^{2}$<br />$\begin{aligned}<br />& \frac{\mathrm{r}_{\mathrm{nbig}}}{\mathrm{r}_{\mathrm{n} \text { small }}}=\frac{\mathrm{n}_{\text {big }}^{2}}{\mathrm{n}_{\text {small }}^{2}}=\frac{4}{1} \\<br />\Rightarrow & \frac{\mathrm{n}_{\text {big }}}{\mathrm{n}_{\text {small }}}=2 \\<br />\Rightarrow & \frac{\mathrm{n}_{\text {small }}}{\mathrm{n}_{\text {big }}}=\frac{1}{2}<br />\end{aligned}$<br />Velocity of electron in $\mathrm{n}^{\text {th }}$ orbit<br />$\begin{array}{c}<br />\quad \mathrm{v}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}} \\<br />\frac{\mathrm{v}_{\mathrm{n} \mathrm{big}}}{\mathrm{v}_{\mathrm{n} \text { small }}}=\frac{\mathrm{n}_{\text {small }}}{\mathrm{n}_{\text {big }}}=\frac{1}{2} \\<br />\Rightarrow \mathrm{v}_{\mathrm{n} \text { small }}=2\left(\mathrm{v}_{\mathrm{n} \mathrm{big}}\right)=2 \mathrm{v}<br />\end{array}$ | ["$\\frac{v}{4}$", "$\\frac{v}{2}$", "$v$", "$2 v$"] | [3] | null | PYQ |
adf798ef6ed05e9ef68fe4e2b998a4a5 | WBJEE_PHY | Atomic Physics | One of the lines in the emission spectrum of $\mathrm{Li}^{2+}$ has the same wavelength as that of the $2^{\text {nd }}$ line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is $\mathrm{n}=12 \rightarrow \mathrm{n}=\mathrm{x} .$ Find the value of $\mathrm{x}$ | singleCorrect | 1 | For $2^{\text {nd line of Balmer series in hydrogen }}$ spectrum<br />$\frac{1}{\lambda}=\mathrm{R}(1)\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3}{16} \mathrm{R}$<br />$\text {For } \mathrm{Li}^{2+}\left[\frac{1}{\lambda}=\mathrm{R} \times 9\left(\frac{1}{\mathrm{x}^{2}}-\frac{1}{12^{2}}\right)=\frac{3 \mathrm{R}}{16}\right]$<br />which is satisfied by $n=12 \rightarrow n=6$ | ["8", "6", "7", "5"] | [1] | null | PYQ |
f7127083d3d860aedb8432b31decf2be | WBJEE_PHY | Atomic Physics | If the series limit wavelength of Lyman series for the hydrogen atom is $912 Å,$ then the series limit wavelength for Balmer series of hydrogen atoms is | singleCorrect | 1 | $\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$<br />For limiting wavelength of Lyman series<br />$n_{1}=1, n_{2}=\infty \quad \frac{1}{\lambda_{L}}=R$<br />For limiting wavelength of Balmer series $n_{1}=2, n_{2}=\infty$<br />$\begin{array}{l}<br />\frac{1}{\lambda_{B}}=R\left(\frac{1}{4}\right) \Rightarrow \lambda_{B}=\frac{4}{R} \\<br />\therefore \lambda_{B}=4 \lambda_{L}=4 \times 912 Å<br />\end{array}$ | ["$912 \u00c5$", "$912 \\times 2 \u00c5$", "$912 \\times 4 \u00c5$", "$\\frac{912}{2} \u00c5$"] | [2] | null | PYQ |
ca18d214ce435596318d123bba07f806 | WBJEE_PHY | Atomic Physics | Monochromatic radiation of wavelength <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>λ</mi></math> is incident on a hydrogen sample in ground state. Hydrogen atom absorbs a fraction of light and subsequently emits radiations of six different wavelengths. The wavelength <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>λ</mi></math> is | singleCorrect | 1 | <p>As <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">H</mi></math>-atom emits <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>6</mn></math> spectral lines <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>n</mi><mfenced><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></mfenced></mrow><mn>2</mn></mfrac><mo>=</mo><mn>6</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mi>n</mi><mo>=</mo><mn>4</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">Δ</mi><mi>E</mi><mo>=</mo><msub><mi>E</mi><mn>4</mn></msub><mo>-</mo><msub><mi>E</mi><mn>1</mn></msub><mo>=</mo><mn>13</mn><mo>.</mo><mn>6</mn><mo>-</mo><mn>0</mn><mo>.</mo><mn>85</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>12</mn><mo>.</mo><mn>75</mn><mo> </mo><mi>eV</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>λ</mi><mo>=</mo><mfrac><mrow><mi>h</mi><mi>c</mi></mrow><mrow><mo>∆</mo><mi>E</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>1242</mn><mo> </mo><mi>nm</mi></mrow><mrow><mn>12</mn><mo>.</mo><mn>75</mn></mrow></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mi>λ</mi><mo>=</mo><mn>97</mn><mo>.</mo><mn>5</mn><mo> </mo><mi>nm</mi></math></p> | ["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>97</mn><mo>.</mo><mn>5</mn><mo> </mo><mi>nm</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>121</mn><mo>.</mo><mn>6</mn><mo> </mo><mi>nm</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>110</mn><mo>.</mo><mn>3</mn><mo> </mo><mi>nm</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>45</mn><mo>.</mo><mn>2</mn><mo> </mo><mi>nm</mi></math>"] | [0] | null | PYQ |
65d80a40d73020f6481aa760ff7d5afd | WBJEE_PHY | Atomic Physics | The energy of electron in the nth orbit of hydrogen atom is expressed as $\mathrm{E}_{\mathrm{n}}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}$.<br />The shortest and longest wavelength of Lyman series will be | singleCorrect | 1 | $\frac{1}{\lambda_{\max }}=\mathrm{R}\left[\frac{1}{(\mathrm{l})^{2}}-\frac{1}{(2)^{2}}\right]$<br />$\Rightarrow \lambda_{\mathrm{max}}=\frac{4}{3 \mathrm{R}} \approx 1213 Ã…$<br />and<br />$\frac{1}{\lambda_{\min }}=\mathrm{R}\left[\frac{1}{(1)^{2}}-\frac{1}{\infty}\right] \Rightarrow \lambda_{\min }=\frac{1}{\mathrm{R}} \approx 910 Ã…$ | ["$910 \u00c3, 1213 \u00c3$", "$5463 \u00c3, 7858 \u00c3$", "$1315 \u00c3, 1530 \u00c3$", "None of these"] | [0] | null | PYQ |
3f7be8e0f4f8d35984836e5d7fdd0785 | WBJEE_PHY | Atomic Physics | The frequency for a series limit of Balmer and Paschen series respectively are <math><msubsup><mrow><mi>f</mi></mrow><mrow><mn>1</mn></mrow></msubsup></math> and <math><msubsup><mrow><mi>f</mi></mrow><mrow><mn>3</mn></mrow></msubsup></math>. If the frequency of the first line of Balmer series is <math><msubsup><mrow><mi>f</mi></mrow><mrow><mn>2</mn></mrow></msubsup></math> then the relation between<math><msup><mrow><mi> </mi></mrow></msup><msubsup><mrow><mi>f</mi></mrow><mrow><mn>1</mn></mrow></msubsup><mo>,</mo><mi> </mi><msubsup><mrow><mi>f</mi></mrow><mrow><mn>2</mn></mrow></msubsup></math> and <math><msubsup><mrow><mi>f</mi></mrow><mrow><mn>3</mn></mrow></msubsup></math> is | singleCorrect | 1 | <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>c</mi><mo>=</mo><mi>f</mi><mi>λ</mi><mo>,</mo><mfrac><mn>1</mn><mi>λ</mi></mfrac><mo>=</mo><mfrac><mi>f</mi><mi>c</mi></mfrac></math><br /><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mi>λ</mi></mfrac><mo>=</mo><mi>R</mi><mfenced separators="|"><mrow><mfrac><mn>1</mn><msup><mi>P</mi><mn>2</mn></msup></mfrac><mo>-</mo><mfrac><mn>1</mn><msup><mi>n</mi><mn>2</mn></msup></mfrac></mrow></mfenced></math><br /><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>f</mi><mo>=</mo><mi>R</mi><mi>c</mi><mfenced separators="|"><mrow><mfrac><mn>1</mn><msup><mi>P</mi><mn>2</mn></msup></mfrac><mo>-</mo><mfrac><mn>1</mn><msup><mi>n</mi><mn>2</mn></msup></mfrac></mrow></mfenced></math><br /><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><msub><mi>f</mi><mn>2</mn></msub><mo>=</mo><mi>R</mi><mi>c</mi><mfenced separators="|"><mrow><mfrac><mn>1</mn><msup><mn>2</mn><mn>2</mn></msup></mfrac><mo>-</mo><mfrac><mn>1</mn><msup><mn>3</mn><mn>2</mn></msup></mfrac></mrow></mfenced><mo>=</mo><mi>R</mi><mi>c</mi><mfenced separators="|"><mrow><mfrac><mn>1</mn><mn>4</mn></mfrac><mo>-</mo><mfrac><mn>1</mn><mn>9</mn></mfrac></mrow></mfenced><mo>=</mo><mfrac><mrow><mn>5</mn><mi>R</mi><mi>c</mi></mrow><mn>36</mn></mfrac></math><br /><br /><math><msub><mrow><mi>f</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>=</mo><mi>R</mi><mi>c</mi><mfenced separators="|"><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><msup><mrow><mn>2</mn></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced><mo>=</mo><mfrac><mrow><mi>R</mi><mi>c</mi></mrow><mrow><mn>4</mn></mrow></mfrac></math><br /><br /><math><msub><mrow><mi>f</mi></mrow><mrow><mn>3</mn></mrow></msub><mo>=</mo><mi>R</mi><mi>c</mi><mfenced separators="|"><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><msup><mrow><mn>3</mn></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced><mo>=</mo><mfrac><mrow><mi>R</mi><mi>c</mi></mrow><mrow><mn>9</mn></mrow></mfrac></math><br /><br /><math><mo>∴</mo><mi> </mi><mi> </mi><mi> </mi><msub><mrow><mi>f</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>-</mo><msub><mrow><mi>f</mi></mrow><mrow><mn>3</mn></mrow></msub><mo>=</mo><mi>R</mi><mi>c</mi><mfenced separators="|"><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><mn>4</mn></mrow></mfrac><mo>-</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>9</mn></mrow></mfrac></mrow></mfenced></math><br /><br /><math><mo>∴</mo><mi> </mi><mi> </mi><msub><mrow><mi>f</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>-</mo><msub><mrow><mi>f</mi></mrow><mrow><mn>3</mn></mrow></msub><mo>=</mo><msub><mrow><mi>f</mi></mrow><mrow><mn>2</mn></mrow></msub></math><br /><br /><math><mo>∴</mo><mi> </mi><mi> </mi><mi> </mi><msub><mrow><mi>f</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>-</mo><msub><mrow><mi>f</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><msub><mrow><mi>f</mi></mrow><mrow><mn>3</mn></mrow></msub></math> | ["<math><msub><mrow><mi>f</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>-</mo><msub><mrow><mi>f</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><msub><mrow><mi>f</mi></mrow><mrow><mn>3</mn></mrow></msub></math>", "<math><msub><mrow><mi>f</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>+</mo><msub><mrow><mi>f</mi></mrow><mrow><mn>3</mn></mrow></msub><mo>=</mo><msub><mrow><mi>f</mi></mrow><mrow><mn>2</mn></mrow></msub></math>", "<math><msub><mrow><mi>f</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>+</mo><msub><mrow><mi>f</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><msub><mrow><mi>f</mi></mrow><mrow><mn>3</mn></mrow></msub></math>", "<math><msub><mrow><mi>f</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><msub><mrow><mi>f</mi></mrow><mrow><mn>3</mn></mrow></msub><mo>=</mo><mn>2</mn><msub><mrow><mi>f</mi></mrow><mrow><mn>1</mn></mrow></msub></math>"] | [0] | null | PYQ |
7e540863ec3391430da476259ae4cf21 | WBJEE_PHY | Atomic Physics | Which of the following transitions of $\mathrm{He}^{+}$ion will give rise to spectral line which has same wavelength as the spectral line in hydrogen atom? | singleCorrect | 3 | Let the transition of electron in hydrogen atom $(Z=1)$ gives wavelength $\lambda_1$ for transition from $n_2$ to $n_1$, then
$\frac{1}{\lambda}=R Z^2\left(\frac{1}{m^2}-\frac{1}{n^2}\right)$
Here, $\lambda=\lambda_1, Z=1, m=n_1, n=n_2$
$\frac{1}{\lambda_1}=R(1)^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Now, if electron makes transition in $\mathrm{He}^{+}$ion from $n_4$ to $n_3$ and gives wavelength $\lambda_2$, then
$\begin{aligned}
\frac{1}{\lambda_2} & =R(2)^2\left(\frac{1}{n_3^2}-\frac{1}{n_4^4}\right) \\
& =R\left[\frac{1}{\left(\frac{n_3}{2}\right)^2}-\frac{1}{\left(\frac{n_4}{2}\right)^2}\right]
\end{aligned}$
Given that, $\quad \lambda_1=\lambda_2$
$\Rightarrow R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=R\left[\frac{1}{\left(\frac{n_3}{2}\right)^2}-\frac{1}{\left(\frac{n_4}{2}\right)^2}\right]$
On comparing beth sides, we get
$n_1=\frac{n_3}{2} \text { and } n_2=\frac{n_4}{2}$
The value of $n_1, n_2, n_3$ and $n_4$ must be an integer, as they denote principal quantum number.
If $n_3=2$ and $n_4=4$, then the given condition is satisfied. | ["$n=4$ to $n=2$", "$n=6$ to $n=5$", "$n=6$ to $n=3$", "None of these"] | [0] | null | PYQ |
cd101eeaaae0da16aa8601c08a1d8f8e | WBJEE_PHY | Atomic Physics | Electrons are excited from $n=1$ to $n=4$ state. During downward transitions, possible number of spectral lines observed in Balmer series is | singleCorrect | 1 | In Balmer series, the electrons moves to $n=2$ state
So, possible transitions are $4 \rightarrow 2,3 \rightarrow 2$.
$\therefore$ Number of spectral lines are 2 . | ["4", "3", "2", "1"] | [2] | null | PYQ |
0e418c2de18449a256127fcb92fd7d55 | WBJEE_PHY | Atomic Physics | The electron in a hydrogen atom makes a transition from $n=n_{1}$ to $n=n_{2}$ state. The time period of the electron in the initial state $n_{1}$ is eight times that in the final state $n_{2}$. The possible values of $n_{1}$ and $n_{2}$ are | singleCorrect | 1 | Here, for hydrogen atom, $Z=1$
We know that, time period of electron in $n$th orbit is given as
$\begin{aligned} T_{n} & \propto n^{3} \\ \therefore \quad & \frac{T_{n_{1}}}{T_{n_{2}}}=\frac{n_{1}^{3}}{n_{2}^{3}} \end{aligned}$
Since, $\quad T_{n_{1}}=8 T_{n_{2}}$
$\therefore \quad \frac{8 T_{n_{2}}}{T_{n_{2}}}=\frac{n_{1}^{3}}{n_{2}^{3}} \Rightarrow 2^{3}=\left(\frac{n_{1}}{n_{2}}\right)^{3}$
$\Rightarrow \quad n_{1}=2 n_{2}$
Hence, this condition satisfies only in option (b).
i.e., $\quad n_{1}=4$ and $n_{2}=2$ | ["$n_{1}=8, n_{2}=1$", "$n_{1}=4, n_{2}=2$", "$n_{1}=2, n_{2}=4$", "$n_{1}=1, n_{2}=8$"] | [1] | null | PYQ |
1707a4c2ce4e0fd46a3e22398a477cc0 | WBJEE_PHY | Atomic Physics | The ratio of minimum wavelength of Lyman and Balmer series will be | singleCorrect | 2 | We know that, wavelength of hydrogen spectrum is given by<br>$$<br>\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)<br>$$<br>For minimum wavelength of Lyman series,<br>$\begin{aligned}<br>& n_{1}=1 \text { and } n_{2}=\infty \\<br>\therefore \quad & \frac{1}{\left(\lambda_{L}\right)_{\min }}=R\left(\frac{1}{(1)^{2}}-\frac{1}{\infty}\right)=R \\<br>\Rightarrow \quad\left(\lambda_{2}\right)_{\min }=\frac{1}{R}<br>\end{aligned}$<br>For minimum wavelength of Balmer series,<br>$n_{1}=2$ and $n_{2}=\infty$<br>$\therefore \quad \frac{1}{\left(\lambda_{B}\right)_{\min }}=R\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)$<br>$=R\left(\frac{1}{4}-0\right)=\frac{R}{4}$<br>$\Rightarrow \quad\left(\lambda_{B}\right)_{\min }=\frac{4}{R}$<br>$\therefore \quad \frac{\left(\lambda_{L}\right)_{\min }}{\left(\lambda_{B}\right)_{\min }}=\frac{\frac{1}{R}}{\frac{4}{R}}=\frac{1}{4}=0.25$ | ["$1.25$", "$0.25$", "5", "10"] | [1] | null | PYQ |
049f59c57722c1865ed72b6ab8a9a386 | WBJEE_PHY | Atomic Physics | Which state of triply ionised Beryllium $\left(\mathrm{Be}^{+++}\right)$ has the same orbital radius as that of the ground state of hydrogen? | singleCorrect | 2 | Radius of $n$ th-orbit in hydrogen like atom is given by $r=\frac{n^{2} h^{2}}{4 \pi^{2} m k Z e^{2}}$<br>i.e., $\quad r \propto \frac{n^{2}}{Z}$<br>For hydrogen, $Z=1, n=1$ in ground state $\Rightarrow \quad \frac{n^{2}}{Z}=\frac{1^{2}}{1}=1$<br>For Beryllium $\left(\mathrm{Be}^{+++}\right), Z=4$ orbital is same<br>$\therefore \quad \frac{n^{2}}{Z}=1$<br>$\Rightarrow \quad n^{2}=1 \times Z \Rightarrow n^{2}=4$<br>$\Rightarrow \quad n=2$<br>Thus, the second level of triply ionised $\mathrm{Be}^{+++}$has same radius as the ground state of hydrogen. | ["$n=1$", "$n=2$", "$n=3$", "$n=4$"] | [1] | null | PYQ |
8495b29ba8d67a23a8119e5a8ef4bf76 | WBJEE_PHY | Atomic Physics | Excitation energy of a hydrogen like ion in its first excitation state is $40.8 \mathrm{eV}$. Energy needed to remove the electron from the ion in ground state is | singleCorrect | 2 | Excitation energy of hydrogen like ion in its first excitation state, $\Delta E=40.8 \mathrm{eV}$<br>For the hydrogen like ion, energy in the ground state $(n=1)$,<br>$$<br>E=\frac{-13.6 Z^{2}}{n^{2}} \mathrm{eV}=-13.6 Z^{2} \mathrm{eV}<br>$$<br>Energy in the first excited state, $E_{2}=\frac{-13.6 Z^{2}}{n^{2}}$<br>$$<br>=\frac{-13.6 Z^{2}}{4}<br>$$<br>Excitation energy, $\Delta E=E_{2}-E_{1}$<br>$\begin{aligned}<br>&=\left(\frac{-13.6 Z^{2}}{4}+13.6 Z^{2}\right) \mathrm{eV} \\<br>&=(-3.4+13.6) Z^{2} \mathrm{eV} \\<br>\Rightarrow \quad 40.8 \mathrm{eV} &=10.2 Z^{2} \mathrm{eV} \\<br>\Rightarrow \quad \quad Z^{2} &=\frac{40.8}{10.2} \\<br>\Rightarrow \quad \quad Z^{2} &=4 \\<br>\Rightarrow \quad \quad Z &=2<br>\end{aligned}$<br>Thus, energy of the electron in the ground state is given as<br>$\begin{array}{ll}<br>\Rightarrow & E_{1}=-13.6(2)^{2} \mathrm{eV} \\<br>\Rightarrow & E_{1}=-54.4 \mathrm{eV}<br>\end{array}$<br>Hence, the energy needed to remove the electron from the ion in ground state (ionisation energy)<br>$$<br>=-(-54.4) \mathrm{eV}=54.4 \mathrm{eV}<br>$$ | ["$54.4 \\mathrm{eV}$", "$13.6 \\mathrm{eV}$", "$40.8 \\mathrm{eV}$", "$27.2 \\mathrm{eV}$"] | [0] | null | PYQ |
07cb5babb9db1e8742421a3f659ce497 | WBJEE_PHY | Atomic Physics | If an electron in hydrogen atom jumps from an orbit of level $n=3$ to an orbit at level $n=2$, emitted radiation has a frequency of ( $R=$ Rydberg's constant and $c=$ velocity of light) | singleCorrect | 1 | When an electron jumps from orbit $n_{i}$ to $n_{f}$, then the frequency of emitted photon is<br>$$<br>v=c R\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)<br>$$<br>Here, $n_{i}=3, n_{f}=2$<br>$$<br>\Rightarrow \quad v=c R\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]=\frac{5 R c}{36}<br>$$ | ["$\\frac{3 R c}{27}$", "$\\frac{R c}{25}$", "$\\frac{8 R c}{9}$", "$\\frac{5 R c}{36}$"] | [3] | null | PYQ |
631a7e82d0dedc6d0c48df30e3c8e5b2 | WBJEE_PHY | Atomic Physics | Calculate the radius of the second Bohr's orbit of electron of hydrogen atom. | singleCorrect | 1 | Radius of the $n$th Bohr's orbit of the hydrogen atom is given by<br/>$\begin{aligned}<br/>r_n & =\frac{4 \pi \varepsilon_0 n^2 h^2}{4 \pi^2 m e^2} \\<br/>r_2 & =\frac{4 \pi \times 8.85 \times 10^{-12} \times(2)^2 \times\left(6.625 \times 10^{-34}\right)^2}{4 \pi^2 \times 9.1 \times 10^{-31} \times\left(1.6 \times 10^{-19}\right)^2} \\<br/>& =\frac{8.85 \times 10^{-12} \times 4 \times\left(6.625 \times 10^{-34}\right)^2}{\pi \times 9.1 \times 10^{-31} \times\left(1.6 \times 10^{-19}\right)^2} \\<br/>& =2.12 \times 10^{-10} \mathrm{~m} \\<br/>& =2.12 <br/>\end{aligned}$ | ["$3.68 \u00c3\u2026$", "$2.12 \u00c3\u2026$", "$4.77 \u00c3\u2026$", "$2.68 \u00c3\u2026$"] | [1] | null | PYQ |
8838dc471bff1084a9d61bdcbaeb7369 | WBJEE_PHY | Atomic Physics | In a helium ion, if potential energy of electron in ground state is assumed to be zero, then its energy in first excited state is equal to | singleCorrect | 3 | Potential energy of a $\mathrm{He}^{+}$ion in ground state is<br/>$\begin{aligned}<br/>& U_1=2 E_1=2\left(-13.6 \frac{Z^2}{n^2}\right) \\<br/>& \because \quad n=1 \text { and } Z=2 \\<br/>& \Rightarrow \quad U_1=-108.8 \mathrm{eV}<br/>\end{aligned}$<br/>But PE in ground state is given zero. So, we have to increase PE (and total energy) by 108.8 eV .<br/>Hence, for given $\mathrm{He}^{+}$ion, energy of first excited state $(n=2)$ will be<br/>$\begin{aligned}<br/>E(n=2) & =-13.6 \frac{Z^2}{n^2}+108.8 \\<br/>& =-13.6 \times \frac{4}{4}+108.8 \\<br/>& =95.2 \mathrm{eV}<br/>\end{aligned}$ | ["13.6 eV", "27.2 eV", "54.4 eV", "95.2 eV"] | [3] | null | PYQ |
2aa134668792345a7c5ff8df32f13c36 | WBJEE_PHY | Atomic Physics | If an electron in hydrogen atom jumps from an orbit of level \(n=3\) to an orbit of level \(n=2\). the<br>emitted radiation has a frequency ( \(R=\) Rydberg constant, \(C=\) velocity of light) | singleCorrect | 1 | We know that<br>\(\begin{array}{l}
\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\
\text { Using } \lambda f=c \Rightarrow \frac{1}{\lambda}=\frac{f}{c} \\
\frac{f}{c}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \Rightarrow f=R c\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\
\text { Given } n_{1}=2 ; n_{2}=3 \\
\Rightarrow f=R c\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R c\left(\frac{1}{4}-\frac{1}{9}\right)=R c \frac{(9-4)}{4 \times 9}=\frac{5 R C}{36}
\end{array}\)<br>Therefore, frequency of emitted radiation is \(\frac{5}{36} \mathrm{Rc}\) | ["\\(\\frac{3 R C}{27}\\)", "\\(\\frac{R C}{25}\\)", "\\(\\frac{8 R C}{9}\\)", "\\(\\frac{5 R C}{36}\\)"] | [3] | null | PYQ |
483d46df3f0dbb09b5781eab131dc3d1 | WBJEE_PHY | Atomic Physics | The total energy of an electron revolving in the second orbit of hydrogen atom is | singleCorrect | 1 | Total energy of an electron revolving in nth orbit of hydrogen atom is given as<br>\(E_{n}=\frac{-13.6}{n^{2}} \mathrm{eV}\)<br>For second orbit, \(n=2\), energy is<br>\(E_{2}=\frac{-13.6}{2^{2}}=-3.4 e V\) | ["\\(-13.6 \\mathrm{ev}\\)", "\\(-1.51 \\mathrm{ev}\\)", "\\(-3.4 \\mathrm{ev}\\)", "Zero"] | [2] | null | PYQ |
27e1d901c41036fa5e2083328827d111 | WBJEE_PHY | Atomic Physics | A hydrogen atom in ground state absorbs \(10.2 \mathrm{eV}\) of energy. The orbital angular momentum of<br>the electron is increased by | singleCorrect | 2 | (B)<br>By absorbing \(10.2 \mathrm{ev}\), electron goes to \(2\) nd orbit as<br>\(\begin{array}{l}<br>E_{1}=-13.6 \mathrm{eV} \\<br>E_{2}=-3.4 \mathrm{eV} \\<br>E_{2}-E_{1}=10.2 \mathrm{eV} \\<br>L_{2}-L_{1}=\frac{n_{2} h}{2 \pi}-\frac{n_{1} h}{2 \pi}=\frac{2 h}{2 \Pi}-\frac{h}{2 \pi}=\frac{6.62 \times 10^{-34}}{2 \times 3.14} \\<br>=1.05 \times 10^{-34} \mathrm{Js}<br>\end{array}\) | ["\\(3.16 \\times 10^{-34} \\mathrm{Js}\\)", "\\(1.05 \\times 10^{-34} \\mathrm{Js}\\)", "\\(4.22 \\times 10^{-34} \\mathrm{Js}\\)", "\\(2.11 \\times 10^{-34} \\mathrm{~J}_{S}\\)"] | [1] | null | PYQ |
320fa59b453007aaffb1c4b77e91d2da | WBJEE_PHY | Atomic Physics | A beam of fast moving alpha particles were directed towards a thin film of gold. The parts $A, B$ and $C$ of the transmitted and reflected beams corresponding to the incident parts $A, B$ and $C$ of the beam are shown in the adjoining diagram. The number of alpha particles in<br><img src="https://cdn.quizrr.in/question-assets/kcet/py54ggf1c/En-5ZPueGGR-HvWceJwTi8SxWmjPDeeJymdbys5T3eg.original.fullsize.png"><br> | singleCorrect | 2 | According to Rutherford's $\alpha$-particles scattering experiment, following observations are made<br>(i) Most of the $\alpha$-particles passed through the gold foil undeflected.<br>(ii) Only about $0.14 \%$ of the incident $\alpha$-particles scattered by an angle greater than $1^{\circ}$.<br>(iii) About one $\alpha$-particle in every $8000 \alpha$-particles deflects by angle more than $90^{\circ}$.<br>So, from above observation, we can conclude about the number of $\alpha$-particle in given figure as,<br>$n_{A^{\prime}}>n_{C^{\prime}}>n_{B^{\prime}}$<br>i.e., number of $\alpha$-particle will be maximum in $A^{\prime}$ and minimum in $B^{\prime}$. | ["$B^{\\prime}$ will be minimum and in $C^{\\prime}$ maximum", "$A^{\\prime}$ will be maximum and in $B^{\\prime}$ minimum", "$A^{\\prime}$ will be minimum and in $B^{\\prime}$ maximum", "$C^{\\prime}$ will be minimum and in $B^{\\prime}$ maximum"] | [1] | null | PYQ |
05a4a5dfc0fa1cfbfbf073bbd3c92d1f | WBJEE_PHY | Atomic Physics | An electron in an excited state of $\mathrm{Li}^{2+}$ ion has angular momentum $\frac{3 h}{2 \pi}$. The de-Broglie wavelength of electron in this state is $p \pi a_{0}$ (where, $a_{0}=$ Bohr radius ). The value of $p$ is | singleCorrect | 2 | According to de-Broglie hypothesis,<br>$=\frac{n h}{2 \pi}=\frac{3 h}{2 \pi}=m v r$<br>$\Rightarrow \quad n=3$ <br>As, wavelength, $\lambda=\frac{h}{p}=\frac{h}{m v}=\frac{h r}{m v r}$ <br>$=\frac{h r}{3 h}=\frac{2}{3} \pi r$ <br>For Li ${ }^{2+}$ atom, radius of orbit, <br>$r=r_{0} \frac{n^{2}}{Z}=a_{0} \frac{3^{2}}{3}=3 a_{0}$ <br>$\lambda=\frac{2}{3} \pi \times a_{0} \times 3=2 \pi a_{0}=p \pi a_{0} \text { (given) }$ <br>$\quad p=2$ | ["3", "2", "1", "4"] | [1] | null | PYQ |
34a7f1b62d9bdcc9e2b8bbbcd6854a57 | WBJEE_PHY | Atomic Physics | In accordance with the Bohr's model, the quantum number that characterises the Earth's revolution around the Sun in an orbit of radius $1.5 \times 10^{11} \mathrm{~m}$ with orbital speed $3 \times 10^4 \mathrm{~ms}^{-1}$ is<br>[given, mass of Earth $=6 \times 10^{24} \mathrm{~kg}$ ] | singleCorrect | 3 | Given, $v=3 \times 10^4 \mathrm{~m} / \mathrm{s}$<br>$\begin{aligned}<br>r & =1.5 \times 10^{11} \mathrm{~m} \\<br>m_e & =6 \times 10^{24} \mathrm{~kg}<br>\end{aligned}$<br>According Bohr's atomic model,<br>Angular momentum $=m v r=\frac{n h}{2 \pi}$<br>where, $h=$ Planck's constant $=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}$<br>and $n=$ quantum number<br>$\begin{aligned}<br>\therefore \quad n & =\frac{2 \pi\left(m_e v r\right)}{h} \\<br>& =\frac{2 \times 314 \times 6 \times 10^{24} \times 3 \times 10^4 \times 1.5 \times 10^{11}}{6.62 \times 10^{-34}} \\<br>& =2.57 \times 10^{74}<br>\end{aligned}$<br>Hence, the quantam number that characterises the Earth's revolution is $2.57 \times 10^{74}$. | ["$2.57 \\times 10^{38}$", "$8.57 \\times 10^{64}$", "$2.57 \\times 10^{74}$", "$5.98 \\times 10^{86}$"] | [2] | null | PYQ |
de15b10a9000fdd0171e9ce1a809d157 | WBJEE_PHY | Atomic Physics | Three energy levels of hydrogen atom and the corresponding wavelength of the emitted radiation due to different electron transition are as shown. Then,
<img src="https://cdn.quizrr.in/question-assets/kcet/py54ggf1c/LXhBjHvrz39_eAxo8oEFEWN3gVkQ9cWUgBKjZlkDUqU.original.fullsize.png"> | singleCorrect | 2 | <img src="https://cdn.quizrr.in/question-assets/kcet/py54ggf1c/SEnCiQiNHHn8WdKXyIFobMfBFkWtEpbzPi6h73jIidI.original.fullsize.png"><br>From the given diagram,
$\begin{gathered}E_2-E_1=\frac{h c}{\lambda_1} \\ E_3-E_2=\frac{h c}{\lambda_3} \\ E_3-E_1=\frac{h c}{\lambda_2}\end{gathered}$
Adding Eq. (i) and Eq. (ii), we get
$\begin{aligned} E_3-E_1 & =\frac{h c}{\lambda_1}+\frac{h c}{\lambda_3} \\ \Rightarrow \quad \frac{h c}{\lambda_2} & =\frac{h c}{\lambda_1}+\frac{h c}{\lambda_3}\end{aligned}$
[From Eq. (iii)]
$\begin{aligned} & \Rightarrow \quad \frac{1}{\lambda_2}=\frac{1}{\lambda_1}+\frac{1}{\lambda_3} \\ & \Rightarrow \quad \lambda_2=\frac{\lambda_1 \lambda_3}{\lambda_1+\lambda_3}\end{aligned}$ | ["$\\lambda_3=\\frac{\\lambda_1 \\lambda_2}{\\lambda_1+\\lambda_2}$", "$\\lambda_1=\\frac{\\lambda_2 \\lambda_3}{\\lambda_2+\\lambda_3}$", "$\\lambda_2=\\lambda_1+\\lambda_3$", "$\\lambda_2=\\frac{\\lambda_1 \\lambda_3}{\\lambda_1+\\lambda_3}$"] | [3] | null | PYQ |
3b07a103027f48d988c7b7c039fb049c | WBJEE_PHY | Atomic Physics | In alpha particle scattering experiment, if $v$ is the initial velocity of the particle, then the distance of closest approach is $d$. If the velocity is doubled, then the distance of closest approach changes to: | singleCorrect | 2 | If $r_0$ be the distance of closest approach, then
$(\mathrm{KE})_\alpha=\frac{(Z e)(2 e)}{4 \pi \varepsilon_0 \cdot\left(r_0\right)_\alpha} \Rightarrow \frac{1}{2} m v_\alpha^2=\frac{(Z e)(2 e)}{4 \pi \varepsilon_0 \cdot\left(r_0\right)_\alpha}$
$\Rightarrow \quad v_\alpha^2 \propto \frac{1}{\left(r_0\right)_\alpha} \Rightarrow\left(r_0\right)_\alpha \propto \frac{1}{v_\alpha^2}$
$\Rightarrow \quad \frac{\left(r_0\right)_{\alpha_i}}{\left(r_0\right)_{\alpha_f}}=\frac{v_{\alpha_f}^2}{v_{\alpha i}^2}=\frac{\left(2 v_{\alpha_i}\right)^2}{v_{\alpha_i}^2}=4$
$\therefore \quad\left(r_0\right)_{\alpha_f}=\frac{\left(r_0\right) \alpha_i}{4}=\frac{d}{4} \quad\left[\because\left(r_0\right)_{\alpha_i}=d\right]$ | ["$4 d$", "$2 d$", "$\\frac{d}{2}$", "$\\frac{d}{4}$"] | [3] | null | PYQ |
9c6106bc1bd53b09dec632656c5844e0 | WBJEE_PHY | Atomic Physics | The ratio of volume of $\mathrm{Al}^{27}$ nucleus to its surfactarea is (Given, $R_0=1.2 \times 10^{-15} \mathrm{~m}$ ) | singleCorrect | 1 | $\frac{\text { Volume of } \mathrm{Al}^{27} \text { nucleus }}{\text { Surface area }}=\frac{\frac{4}{3} \pi R^3}{4 \pi R^2}=\frac{R}{3}$
$=\frac{1}{3}\left(R_0 A^{\frac{1}{3}}\right) \quad\left[\because R=R_0 A^{\frac{1}{3}}\right]$
$=\frac{1}{3} \times 1.2 \times 10^{-15}(27)^{\frac{1}{3}}$
$=1.2 \times 10^{-15} \mathrm{~m}$ | ["$2.1 \\times 10^{-15} \\mathrm{~m}$", "$1.3 \\times 10^{-15} \\mathrm{~m}$", "$0.22 \\times 10^{-15} \\mathrm{~m}$", "$1.2 \\times 10^{-15} \\mathrm{~m}$"] | [3] | null | PYQ |
2ce083c0af41c25c6f03b1d2ae70bfd5 | WBJEE_PHY | Atomic Physics | If levels 1 and 2 are separated by an energy $E_2-E_1$, such that the corresponding transition frequency falls in the middle of the visible range, calculate the ratio of the populations of two levels in the thermal equilibrium at room temperature. | singleCorrect | 2 | At thermal equilibrium, the ratio $\frac{N_2}{N_1}$ is given as
$\frac{N_2}{N_1}=\exp \left(-\frac{E_2-E_1}{k T}\right)$<br/>The middle of the visible range is taken at
$\begin{array}{ll}
\lambda=550 \mathrm{~nm} \\
\Rightarrow & E_2-E_1=\frac{h c}{\lambda}=3.16 \times 10^{-19} \mathrm{~J} \\
\Rightarrow & \frac{N_2}{N_1}=\exp \left(\frac{-3.16 \times 10^{-19} \mathrm{~J}}{\left(1.38 \times 10^{-23} 1 / \mathrm{k}\right) \cdot(300 \mathrm{k})}\right) \\
\Rightarrow & \frac{N_2}{N_1}=1.1577 \times 10^{-38}
\end{array}$ | ["$1.1577 \\times 10^{-38}$", "$2.9 \\times 10^{-35}$", "$2.168 \\times 10^{-36}$", "$1.96 \\times 10^{-20}$"] | [0] | null | PYQ |
06080bc88bb700230a9dfb4d785ee50f | WBJEE_PHY | Atomic Physics | The ionisation potential of hydrogen atom is 13.6 eV . How much energy need to be supplied to ionise the hydrogen atom in the first excited state? | singleCorrect | 1 | Ionisation energy of nth state $=-E_n$
For the first excited state, $n=2$
and energy in that state is, $\frac{13.6}{2^2} \mathrm{eV}=\frac{13.6}{4} \mathrm{eV}$
$=3.4 \mathrm{eV}$ | ["13.6 eV", "27.2 eV", "3.4 eV", "6.8 eV"] | [2] | null | PYQ |
109c904085ce93ad9a0fdb09b6dd27e0 | WBJEE_PHY | Atomic Physics | Ionisation potential of hydrogen atom is 13.6 V . Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV . The spectral lines emitted by hydrogen atoms according to Bohr's theory will be | singleCorrect | 1 | Final energy of electron
$=-13.6+12.1=-1.51 \mathrm{eV}$
This energy corresponds to third level i.e., $n=3$
Hence, number of spectral lines emitted
$=\frac{n(n-1)}{2}=\frac{3(3-1)}{2}=3$ | ["one", "two", "three", "four"] | [2] | null | PYQ |
bd3067a543f5bfee374471ba654d98ed | WBJEE_PHY | Atomic Physics | If an electron in $n=4$ orbit of hydrogen atom jumps down to $n=3$ orbit, the amount of energy released and the wavelength of radiation emitted are | singleCorrect | 1 | $\begin{aligned} E & =\frac{-13.6}{n^2} \mathrm{eV} \\ E_4-E_3 & =13.6\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right]\end{aligned}$
$\begin{aligned} & =13.6\left[\frac{7}{144}\right] \mathrm{eV}=0.66 \mathrm{eV} \\ \lambda & =\frac{1}{R\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right]}=\frac{144}{1.09 \times 10^7 \times 7} \mathrm{~m} \\ & =1.88 \times 10^{-6} \mathrm{~m}\end{aligned}$ | ["$0.66 \\mathrm{eV}, 1.88 \\times 10^{-6} \\mathrm{~m}$", "$1.89 \\mathrm{eV}, 1.98 \\times 10^{-7} \\mathrm{~m}$", "$0.29 \\mathrm{eV}, 1.78 \\times 10^{-5} \\mathrm{~m}$", "$0.98 \\mathrm{eV}, 0.93 \\times 10^{-6} \\mathrm{~m}$"] | [0] | null | PYQ |
0b203bc12c1d973a849a44e5593eb398 | WBJEE_PHY | Atomic Physics | If $^{\prime} \lambda_{1}$ ' and ' $\lambda_{2}$ ' are the wavelengths of de-Broglie waves for electrons in first and second Bohr orbits in hydrogen atom, then $\left(\frac{\lambda_{1}}{\lambda_{2}}\right)$ is equal to (Energy in $1^{\text {st }}$ Bohr
orbit $=-13.6 \mathrm{eV}$ ) | singleCorrect | 2 | The de-Broglie wavelength of an electron in the $n$-th orbit is inversely proportional to the radius of the orbit. Since the radius for the first orbit is $R$ and for the second orbit is $4 R$, the de-Broglie wavelength in the second orbit is 4 times that in the first orbit.
Thus, the wavelength ratio is:
$\frac{\lambda_1}{\lambda_2}=4$
The energy in the $n$-th Bohr orbit is given by:
$E_n=-\frac{13.6 \mathrm{eV}}{n^2}$ | ["$\\frac{1}{5}$", "$\\frac{1}{2}$", "$\\frac{1}{4}$", "$\\frac{1}{3}$"] | [2] | null | PYQ |
d7843c926976d81bac9453117f2c3247 | WBJEE_PHY | Atomic Physics | When an electron in a hydrogen atom jumps from the third orbit to the second
orbit, it emits a photon of wavelength ${\lambda} \lambda^{\prime}$. When it jumps from the fourth orbit to
third orbit, the wavelength emitted by the photon will be | singleCorrect | 2 | $\begin{aligned} \frac{1}{\lambda} &=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{R \times 5}{36} \\ \frac{1}{\lambda^{\prime}} &=R\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)=R\left(\frac{1}{9}-\frac{1}{16}\right)=R \times \frac{7}{144} \\ \frac{\lambda^{\prime}}{\lambda} &=\frac{5}{36} \times \frac{144}{7}=\frac{20}{7} \\ \therefore \lambda^{\prime} &=\frac{20}{7} \lambda \end{aligned}$ | ["$\\frac{20}{13} \\lambda$", "$\\frac{16}{25} \\lambda$", "$\\frac{9}{16} \\lambda$", "$\\frac{20}{7} \\lambda$"] | [3] | null | PYQ |
7b7cb2249f5db1b0f9c225f837bea7ee | WBJEE_PHY | Atomic Physics | An electron of mass 'm' is revolving around the nucleus in a circular orbit of radius
'r' has angular momentum 'L'. The magnetic field produced by the electron at the centre of the orbit is
(e $=$ electric charge, $\mu_{0}=$ permeability of free space) | singleCorrect | 2 | $\frac{B}{L}=\frac{\mu_{0}}{4 \pi} \frac{\frac{e V}{r^{2}}}{m V r} L=m V r$
$\frac{B}{L}=\frac{\mu_{0}}{4 \pi} \frac{e V}{e^{2}} \times \frac{1}{m V r}$
$\Rightarrow B=\frac{\mu_{0} e L}{4 \pi m r^{3}}$ | ["$\\frac{\\mu_{0} \\mathrm{eL}}{4 \\pi m r^{2}}$", "$\\frac{\\mu_{0} \\mathrm{eL}}{4 \\pi m r^{3}}$", "$\\frac{\\mu_{0} \\mathrm{eL}}{2 \\pi \\mathrm{mr}^{2}}$", "$\\frac{\\mu_{0} \\mathrm{eL}}{2 \\pi m r^{3}}$"] | [1] | null | PYQ |
351f97b1a4f2b5c15eb15bf10dee85c3 | WBJEE_PHY | Atomic Physics | The electron in hydrogen atom is moving in an orbit of radius $0.53 Å$. It takes
$1.571 \times 10^{-16} \mathrm{~s}$ to complete one revolution. The velocity of electron will be
$[\pi=3 \cdot 142]$ | singleCorrect | 1 | $r=0.53 Å, t=1.571 \times 10^{-16} \mathrm{sec}$
$v=\frac{2 \pi r}{t}=2 \times \frac{22}{7} \times \frac{53}{100} \times 10^{-10} \times \frac{10^{11}}{1.57}$
$=\frac{2 \times 22 \times 53}{700 \times 1.571 \times 10^{6}}$
$=2.12 \times 10^{6} \mathrm{~m} / \mathrm{s}$ | ["$5.3 \\times 10^{6} \\frac{\\mathrm{m}}{\\mathrm{s}}$", "$4 \\times 10^{6} \\frac{\\mathrm{m}}{\\mathrm{s}}$", "$3 \\times 10^{8} \\frac{\\mathrm{m}}{\\mathrm{s}}$", "$2 \\cdot 12 \\times 10^{6} \\frac{\\mathrm{m}}{\\mathrm{s}}$"] | [3] | null | PYQ |
5a7560ff6ce60fc060aa7f74798ab2d2 | WBJEE_PHY | Atomic Physics | Let the series limit for Balmer series be ${ }^{\prime} \lambda_{1}^{\prime}$ and the longest wavelength for Brackett series be ${ }^{\prime} \lambda_{2}{ }^{\prime} .$ Then $\lambda_{1}$ and $\lambda_{2}$ are related as | singleCorrect | 2 | We know,
\(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right)\) for Balmer series \(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{\mathrm{n}^{2}}\right)\) for Brackett series
Now,\(\frac{1}{\lambda_{1}}=\frac{R}{4}\)
\(\frac{1}{\lambda_{2}}=\mathrm{R}\left[\frac{1}{16}-\frac{1}{25}\right]=\mathrm{R}\left[\frac{25-16}{16 \times 25}\right]=\frac{9 \mathrm{R}}{16 \times 25}\)
\(\therefore \frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{R}}{4} \times \frac{16 \times 25}{9 \mathrm{R}}=\frac{100}{9}\)
\(\therefore \frac{9}{100} \times \lambda_{1}=\lambda_{2}\)
\(\therefore \lambda_{2}=0.09 \times \lambda_{1}\) | ["$\\lambda_{2}=0 \\cdot 09 \\lambda_{1}$", "$\\lambda_{1}=0 \\cdot 09 \\lambda_{2}$", "$\\lambda_{1}=1 \\cdot 11 \\lambda_{2}$", "$\\lambda_{2}=1 \\cdot 11 \\lambda_{1}$"] | [1] | null | PYQ |
80cd602d06ebe12cc0fdb08d821e8049 | WBJEE_PHY | Atomic Physics | The force acting on the electrons in hydrogen atom (Bohr's theory) is related to the principle quantum number 'n' as | singleCorrect | 1 | Now $\begin{aligned} & F=\frac{m v^{2}}{r} \\ & v \propto \frac{1}{n} \text { and } r \propto n^{2} \\ & F \propto \frac{1}{n^{2}} \times \frac{1}{n^{2}} \\ F & \propto n^{-4} \end{aligned}$
$\therefore \quad \mathrm{F} \propto \frac{1}{\mathrm{n}^{2}} \times \frac{1}{\mathrm{n}^{2}}$
$\therefore \quad \mathrm{F} \propto \mathrm{n}^{-4}$ | ["$n^{-2}$", "$n^{4}$", "$n^{-4}$", "$n^{2}$"] | [2] | null | PYQ |
850e7842b7c8222949e8f9133068a122 | WBJEE_PHY | Atomic Physics | The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) second to first energy level and (ii) highest energy level to $2^{\text {nd }}$ level is respectively | singleCorrect | 1 | The energy of photons is given by $E=\operatorname{Rhc}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
where, $\mathrm{R}$ is Rydberg constant, $\mathrm{h}$ is Planck's constant and $\mathrm{c}$ is the speed of light.
(i) Energy of photon produced from second to first energy level,
$\mathrm{E}_{1}=\operatorname{Rhc}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\mathrm{E}_{1}=\frac{3}{4} \mathrm{Rhc}$
(ii) Energy of photon produced from highest energy level (i.e., \infty) to second level,
$\begin{array}{l}
\mathrm{E}_{2}=\operatorname{Rhc}\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)=\frac{1}{4} \mathrm{Rhc} \\
\therefore \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\frac{3}{4} \mathrm{Rhc}}{\frac{1}{4} \mathrm{Rhc}}=\frac{3}{1} \text { or } 3: 1
\end{array}$ | ["$4: 1$", "$2: 1$", "$5: 1$", "$3: 1$"] | [3] | null | PYQ |
e30155a2319c41eeb40b6d5961ce626b | WBJEE_PHY | Atomic Physics | The ground state energy of hydrogen atom is $-13 \cdot 6 \mathrm{eV}$. The kinetic and potential
energy of the electron in the second excited state is respectively | singleCorrect | 1 | For second excited state $n=3$
Total energy in the third orbit
$E_{3}=\frac{-13.6}{9} \mathrm{eV}=-1.51 \mathrm{eV}$
K.E. $=-$ T.E. $\quad=1.51 \mathrm{eV}$
P.E. $=2$.T.E. $=-3.02 \mathrm{eV}$ | ["$+3 \\cdot 02 \\mathrm{eV},-1 \\cdot 51 \\mathrm{eV}$", "$1 \\cdot 51 \\mathrm{eV},-3 \\cdot 02 \\mathrm{eV}$", "$-1 \\cdot 51 \\mathrm{eV},+3 \\cdot 02 \\mathrm{eV}$", "$+3 \\cdot 02 \\mathrm{eV},+1 \\cdot 51 \\mathrm{eV}$"] | [1] | null | PYQ |
4c0fda78a678c85569c422b1cd11c5e5 | WBJEE_PHY | Atomic Physics | The energy levels with transitions for the atom are shown. The transitions corresponding to emission of radiation of maximum and minimum wavelength are respectively
<img src="https://cdn.quizrr.in/question-assets/mhtcet/py35bdsq/tZcAD141WnBfaxHzPl-28NaKHL40OwOSa1CsUqAoPig.original.fullsize.png"> | singleCorrect | 1 | Energy $\quad \mathrm{E}=\frac{\mathrm{hc}}{\lambda} \quad \therefore \mathrm{E} \propto \frac{1}{\lambda}$
In transition A, change in energy is minimum, hence wavelength is maximum. In transition D, Change in energy is maximum, hence wavelength is minimum. | ["A, D", "B, C", "C, D", "A, C"] | [0] | null | PYQ |
2d3fe4d1fb95ccf77c7656a36103e2bb | WBJEE_PHY | Atomic Physics | The shortest wavelength for Lyman series is $912 Å$. The longest wavelength in Paschen series is | singleCorrect | 3 | Shortest wavelength in Lyman series is given by
$\begin{aligned}
& \frac{1}{\lambda_{\mathrm{I}}}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\infty}\right]=\mathrm{R} \\
& \therefore \lambda_{\mathrm{I}}=\frac{1}{\mathrm{R}}
\end{aligned}$
The longest wavelength in Paschen series is given by
$\begin{aligned}
& \frac{1}{\lambda_{\mathrm{p}}}=\mathrm{R}\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right] \\
& =\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right]=\mathrm{R} \cdot \frac{7}{144} \\
& \lambda_{\mathrm{p}}=\frac{144}{7 \mathrm{R}} \\
& \therefore \frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{L}}}=\frac{144}{7} \mathrm{R}=\frac{144}{7} \\
& \therefore \lambda_{\mathrm{p}}=\frac{144}{7 \mathrm{R}} \cdot \lambda_{\mathrm{L}}=\frac{144}{7} \times 912=18760 Å
\end{aligned}$ | ["$1216 \u00c5$", "$3646 \u00c5$", "$18760 \u00c5$", "$8208 \u00c5$"] | [2] | null | PYQ |
7a2785d0fa693500a4fa5a147850f5cc | WBJEE_PHY | Atomic Physics | For wavelength of visible radiation of hydrogen spectrum Balmer gave an equation as $\lambda=\frac{\left(\mathrm{km}^2\right)}{\left(\mathrm{m}^2-4\right)}$, where $m$ is the integer value. The value of $k$ in terms of Rydberg's constant $R$ is | singleCorrect | 1 | Wavelength of visible radiation in Balmer series is given by,
$\begin{aligned} & \frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{m^2}\right) \text { for } m \geq 2 \\ & \Rightarrow \lambda=\frac{4}{R} \cdot \frac{m^2}{m^2-4}=\frac{k m^2}{m^2-4}\end{aligned}$
$\therefore k=\frac{4}{R}$ | ["$\\frac{R}{4}$", "$\\frac{4}{R}$", "$R$", "$4 R$"] | [1] | null | PYQ |
200e67810b1e8678b0bf03693fc26252 | WBJEE_PHY | Atomic Physics | The potential energy of the orbital electron in the ground state of hydrogen atoms is $-\mathrm{E}$. What is the kinetic energy? | singleCorrect | 2 | The potential energy of
$\mathrm{U}(\mathrm{n}=1)=-\mathrm{E}$
Ground state: total energy is given by
$\mathrm{TE}=\mathrm{U}+\mathrm{K}$
$\mathrm{U}=-\frac{\left(\mathrm{ze}^2\right)}{4 \pi \varepsilon_0}$
For orbit consider force balance
<img src="https://cdn.quizrr.in/question-assets/mhtcet/py35bdsq/O-qQwYhu0UVeIwxr3b_DzKINjEn2rEBbQzxAUHGPjlQ.original.fullsize.png">
$\frac{\mathrm{ze}^2}{4 \pi \varepsilon_0 \mathrm{r}^2}=\frac{\mathrm{mv}^2}{\mathrm{r}}$
$\therefore \mathrm{K}=\frac{\mathrm{mv}^2}{2}=\frac{1}{2}\left(\frac{\mathrm{ze}^2}{4 \pi \varepsilon_0 \mathrm{r}^2}\right)$
Thus, $\mathrm{K}=\frac{\mathrm{E}}{2}=\frac{\mathrm{U}}{2}$ | ["$4 \\mathrm{E}$", "$\\frac{E}{4}$", "$\\frac{E}{2}$", "$2 E$"] | [2] | null | PYQ |
aa33b256b1b9250f9b3df37ae3030d9a | WBJEE_PHY | Atomic Physics | The threshold frequency for photoelectric emission from a material is $4.5 \times 10^{14} \mathrm{~Hz}$. Photoelectrons will be emitted when this material is illuminated with monochromatic light from a | multipleCorrect | 2 | 1. Electrons will be emitted if the frequency of incident light is greater than $4.5 \times 10^{14} \mathrm{~Hz}$.
Wavelength of infrared light $\simeq 10,000 Å$, its frequency is $v($ infrared $)=\frac{3 \times 10^8}{10,000 \times 10^{-10}}$ $=3 \times 10^{14} \mathrm{~Hz}$.
Wavelength of red light is about $7800 Å$, its frequency is about $3.8 \times 10^{14} \mathrm{~Hz}$. Frequency of sodium light is about $5 \times 10^{14} \mathrm{~Hz}$ and the frequency of ultraviolet light is about $15 \times 10^{14} \mathrm{~Hz}$. Hence the correct choices are (c) and (d). | ["50 watt infrared lamp", "100 watt red neon lamp", "60 watt sodium lamp", "5 watt ultraviolet lamp"] | [2, 3] | null | PYQ |
11b0ed5abd1366aee5767a4e97641462 | WBJEE_PHY | Atomic Physics | When monochromatic light from a bulb falls on a photosensitive surface, the number of photoelectrons emitted per second is $n$ and their maximum kinetic energy is $K_{\max }$. If the distance of the lamp from the surface is halved, then | multipleCorrect | 2 | The value of $n$ is proportional to the intensity of incident light. If the distance of the lamp is halved, intensity becomes four times. But $K_{\max }$ is independent of the intensity of light. Hence the correct choices are (b) and (d). | ["$n$ is doubled", "$n$ becomes 4 times", "$K_{\\max }$ is doubled", "$K_{\\max }$ remains unchanged"] | [1, 3] | null | PYQ |
bbce49b9f2ec64b867e14e311f19d75b | WBJEE_PHY | Atomic Physics | The maximum kinetic energy of photoelectrons in a photocell depends upon | multipleCorrect | 1 | The correct choices are (a) and (b) | ["the frequency of the incident radiation", "the work function of the photosensitive material used in the cell", "the intensity of the incident radiation", "all the above parameters."] | [0, 1] | null | PYQ |
b568e6c60fa599d852eb14fa9831f176 | WBJEE_PHY | Atomic Physics | When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{\max }$. When X-rays are incident on the same cell, then | multipleCorrect | 2 | The frequency of X-rays is higher than that of ultraviolet light. Now $K_{\max }=h\left(v-v_0\right)$. Hence $K_{\text {max }}$ increases as $v$ is increased. Also $K_{\text {max }}=e V_0$, where $V_0$ is the stopping potential. Hence $V_0$ also increases with frequency. Hence the correct choices are (a) and (b). | ["$V_0$ will increase", "$K_{\\max }$ will increase", "$V_0$ will decrease", "$K_{\\max }$ will decrease"] | [0, 1] | null | PYQ |
2a7b8f60edb12867e8f695ed49d11ae9 | WBJEE_PHY | Atomic Physics | The work function of metal $A$ is greater than that for metal $B$. The two metals are illuminated with appropriate radiation of frequency $v$ so as to cause photoelectric emission in both metals. If $v_0$ is the threshold frequency and $K_{\max }$, the maximum kinetic energy of photoelectrons, then | multipleCorrect | 1 | Work function $W_0=h v_0$ and $K_{\max }=h\left(v-v_0\right)$. So the correct choices are (a) and (d). | ["$v_0$ for metal $A$ is greater than that for metal $B$.", "$v_0$ for metal $A$ is less than that for metal $B$.", "$K_{\\max }$ for metal $A$ is greater than that for metal $B$.", "$K_{\\max }$ for metal $A$ is less than that for metal $B$."] | [0, 3] | null | PYQ |
4a2875416ad2a3c249c447c88178bf12 | WBJEE_PHY | Atomic Physics | X-rays are used to cause photoelectric emission from sodium and copper. Then | multipleCorrect | 3 | The work function of sodium is smaller than that of copper. Since $W_0=h v_0$, the threshold frequency for sodium is less than that for copper. So choice (c) is correct and choice (d) is incorrect.
Since the work function of sodium is lower than that of copper, it is easier to extract electrons from sodium than from copper. Therefore, the electrons ejected from sodium will have a greater kinetic energy and will hence need a greater stopping potential. So choice (a) is incorrect and choice (b) is correct. | ["the stopping potential is more for copper than for sodium.", "the stopping potential is less for copper than for sodium.", "the threshold frequency is more for copper than for sodium.", "the threshold frequency is less for copper than for sodium."] | [1, 2] | null | PYQ |
30b1913bcbed61c988fcd691806b9ccc | WBJEE_PHY | Atomic Physics | When a point light source, of power $W$ emitting monochromatic light of wavelength $\lambda$ is kept at a distance $a$ from a photosensitive surface of work function $\phi$, and area $S$, we will have | multipleCorrect | 3 | The energy of each photon is $\frac{h c}{\lambda}$ so that the number of photons released per unit time is $W \div\left(\frac{h c}{\lambda}\right)$.
These photons are spread out in all directions over an area $4 \pi a^2$ so that the 'share' of an area $S$ is a fraction $S / 4 \pi a^2$ of the total number of photons emitted.
The maximum energy of the emitted photoelectrons is
$E_{\max }=h v-\phi=\frac{h c}{\lambda}-\phi=\frac{1}{\lambda}(h c-\lambda \phi)$
The stopping potential is given by $e V_S=E_{\max }$.
Hence $V_S=\frac{1}{e} E_{\max }=\frac{1}{e \lambda}(h c-\lambda \phi)$.
Hence choice (c) is incorrect.
For photoemission to be possible, we must have $h \nu \geq \phi$. Hence $\frac{h c}{\lambda} \geq \phi$ or $\lambda \leq h c / \phi$
Thus the permitted range of values of $\lambda$ is $0 \leq \lambda \leq h c / \phi$. Hence the correct choices are (a), (b) and (d). | ["number of photons striking the surface per unit time as $\\frac{W \\lambda S}{4 \\pi h c a^2}$", "the maximum energy of the emitted photoelectrons as $\\frac{1}{\\lambda}(h c-\\lambda \\phi)$", "the stopping potential needed to stop the most energetic emitted photoelectrons as $\\frac{e}{\\lambda}(h c-\\lambda \\phi)$.", "photoemission occurs only if $\\lambda$ lies in the range $0 \\leq \\lambda \\leq h c / \\phi$"] | [0, 1, 3] | null | PYQ |
0ff943db5e6dd30c07888034efb70178 | WBJEE_PHY | Atomic Physics | Which of the following statements are correct about photons? | multipleCorrect | 2 | The correct statements are (a), (b) and (c). | ["The rest mass of a photon is zero", "The energy of a photon of frequency $v$ is $h v$", "The momentum of a photon of frequency $v$ is $\\frac{h v}{c}$", "Photons do not exert any pressure on surface on which they are incident."] | [0, 1, 2] | null | PYQ |
9b8fcf18f506972d91f499a5cfec4a5c | WBJEE_PHY | Atomic Physics | The intensity of X-rays from a Coolidge tube is plotted against wavelength as shown in Fig. 28.15. The minimum wavelength found is $\lambda_C$ and the wavelength of $K_\alpha$ line is $\lambda_k$. If the accelerating voltage is increased
<img src="https://cdn-question-pool.getmarks.app/modules/ms/wbjee/EB0N-AyWzwrueInFuwOpCBIdcWsKDNxxfWHqHnFeo2k.original.fullsize.png"/><br/> | multipleCorrect | 2 | The minimum wavelength is given by $\lambda_C=\frac{h c}{e V}$
As $V$ increases, $\lambda_C$ decreases. Since the wavelength of $K_\alpha$ line is due to transition $n=2$ to $n=1$ in the element of the target in the tube, wavelength $\lambda_K$ remains unchanged as $V$ is increased. Hence the difference ( $\lambda_K-\lambda_C$ ) increases with increase in the accelerating voltage. Thus the correct choices are (A) and (C). | ["$\\lambda_C$ decreases", "$\\lambda_K$ increases", "$\\left(\\lambda_K-\\lambda_C\\right)$ increases", "$\\lambda_C$ and $\\lambda_K$ both decrease but $\\left(\\lambda_K-\\lambda_C\\right)$ remains unchanged"] | [0, 2] | null | PYQ |
6cc423847ae64ada7f10968856b5a5fa | WBJEE_PHY | Atomic Physics | A hydrogen atom and a $\mathrm{Li}^{2+}$ ion are both in the second excited state. If $l_{\mathrm{H}}$ and $l_{\mathrm{Li}}$ are their respective electronic angular momenta, and $E_{\mathrm{H}}$ and $E_{\mathrm{Li}}$ their respective energies, then | multipleCorrect | 3 | For a hydrogen-like atom, the energy in the $n$th excited state is
$\therefore E \propto\left(-\frac{Z^2}{n^2}\right)$
Since Z for $\mathrm{Li}^{2+}$ is greater than Z for $\mathrm{H}^{+}$, $\left|E_{\mathrm{Li}}\right|\gt\left|E_{\mathrm{H}}\right|$.
Also $l=\sqrt{n(n+1)}\left(\frac{h}{2 \pi}\right)$. Hence $l_{\mathrm{Li}}=l_{\mathrm{H}}$. Thus, the correct choices are (a) and (c). | ["$l_{\\mathrm{H}}=l_{\\mathrm{Li}}$", "$l_{\\mathrm{H}}\\gtl_{\\mathrm{Li}}$", "$E_{\\mathrm{H}} \\lt E_{\\mathrm{Li}}$", "$E_{\\mathrm{H}}\\gtE_{\\mathrm{Li}}$"] | [0, 2] | null | PYQ |
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