question_ID stringlengths 32 32 | module_name stringclasses 12
values | chapter_name stringclasses 106
values | question stringlengths 0 17.9k | type stringclasses 5
values | level float64 1 3 ⌀ | solution stringlengths 0 29.4k | options stringlengths 2 29.2k | correct_options stringclasses 16
values | correct_value float64 -13.84 5.04k ⌀ | tags stringclasses 2
values |
|---|---|---|---|---|---|---|---|---|---|---|
57f97d12a0308a3d1650f0cfd47c1850 | WBJEE_PHY | Current Electricity | In the circuit shown in figure potential difference between points $A$ and $B$ is $16 \mathrm{~V}$. the current passing through $2 \Omega$ resistance will be
<img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/UZbUESIx80GM6K0l50vBIuvRhaCpq4e_7uhL3_sImFs.original.fullsize.png"> | singleCorrect | 2 | $\therefore 4 \mathrm{i}_{1}+2\left(\mathrm{i}_{1}+\mathrm{i}_{2}\right)-3+4 \mathrm{i}_{1}=16 \mathrm{~V}$
Using Kirchhoff's second law in the closed loop we have
$$
9-i_{2}-2\left(i_{1}+i_{2}\right)=0
$$
Solving equations (i) and (ii), we get $\mathrm{i}_{1}=1.5 \mathrm{~A}$ and $\mathrm{i}_{2}=2 \math... | ["$2.5 \\mathrm{~A}$", "$3.5 \\mathrm{~A}$", "$4.0 \\mathrm{~A}$", "zero"] | [1] | null | PYQ |
63ae13bce05024e2b6156c532602256c | WBJEE_PHY | Current Electricity | A wire of length <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>l</mi></math> has a resistance <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math>. If half of the length is stretched to make the radius half of its original value, then the final resistance of the wire is<br /><img alt="" src="https:... | singleCorrect | 2 | <p><img alt="" src="https://cdn.quizrr.in/question-assets/bitsat/py_sjn4d5v/75337_1_.png" style="width: 319px; height: 113px;" /><br /><math> <mi>R</mi><mo>=</mo><mfrac> <mrow> <mi>ρ</mi><mi>l</mi></mrow> <mi>A</mi> </mfrac> <mo>=</mo><mfrac> <mrow> <mi>ρ</mi><mi>l</mi></mrow> <mrow> <mi>π</mi><msup> <mi>r</mi> <mn>2</... | ["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>9</mn><mi>R</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>5</mn><mi>R</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mfrac><mrow><mn>17</mn><mi>R</mi></mrow><mn>2</mn></mfrac></math>", "<math xmlns=\"http://www.w3.org/1... | [2] | null | PYQ |
f2543e85cbe1dacdc08892c6ecf7bb18 | WBJEE_PHY | Current Electricity | A wire $X$ is half the diameter and half the length of a wire $Y$ of similar material. The ratio of resistance of $\mathrm{X}$ to that of $\mathrm{Y}$ is | singleCorrect | 1 | $\mathrm{R} \propto \frac{\ell}{\mathrm{D}^{2}} \Rightarrow \frac{\mathrm{Rx}}{\mathrm{Ry}}=\frac{2}{1}$ | ["8: 1", "4: 1", "2: 1", "1: 1"] | [2] | null | PYQ |
6537a2c22008de2f6a318c920dd0be02 | WBJEE_PHY | Current Electricity | For the circuit (figure), the current is to be measured. The ammeter shown is a galvanometer with a resistance $\mathrm{R}_{\mathrm{G}}=60.00 \Omega$ converted to an ammeter by a shunt resistance $\mathrm{r}_{\mathrm{s}}=0.02 \Omega$. The value of the current is
<img src="https://cdn-question-pool.getmarks.app/pyq/b... | singleCorrect | 2 | $\mathrm{R}_{\mathrm{G}}=60.00 \Omega,$ shunt resistance, $\mathrm{r}_{\mathrm{s}}=0.02 \Omega$
Total resistance in the circuit is $\mathrm{R}_{\mathrm{G}}+3=63 \Omega$ Hence, $\mathrm{I}=3 / 63=0.048 \mathrm{~A}$
Resistance of the galvanometer converted to an ammeter is,
$$
\frac{\mathrm{R}_{\mathrm{G}} \m... | ["$0.79 \\mathrm{~A}$", "$0.29 \\mathrm{~A}$", "$0.99 \\mathrm{~A}$", "$0.8 \\mathrm{~A}$"] | [2] | null | PYQ |
8f772fe5cd46e6b80784badcd257a363 | WBJEE_PHY | Current Electricity | The resistance of a wire is $R$. It is bent at the middle by $180^{\circ}$ and both the ends are twisted together to make a shorter wire. The resistance of the new wire is | singleCorrect | 1 | Resistance of wire $(R)=\rho \frac{l}{A}$ If wire is bent in the middle then
$$
l^{\prime}=\frac{l}{2}, A^{\prime}=2 A
$$
$\therefore$ New resistance, $\mathrm{R}^{\prime}=\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}=\frac{\rho \frac{l}{2}}{2 A}=$
$\frac{\rho l}{4 A}=\frac{R}{4}$ | ["$2 R$", "$R / 2$", "$R / 4$", "$R / 8$"] | [2] | null | PYQ |
a12eb25bf85b302c1c5247d4f9ebcbbe | WBJEE_PHY | Current Electricity | In the circuit shown in the figure, find the current in $45 \Omega$.
<img src=" data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAQEAAACnCAYAAAAR3um6AAAAAXNSR0IArs4c6QAAAARnQU1BAACxjwv8YQUAAAAJcEhZcwAADsMAAA7DAcdvqGQAADOoSURBVHhe7Z0FexRJF6jvv7j3+74V3N3dJRDcgru7uyzu7u7s4rA4iwV3d5dFg7PALstu3X5PusNkMoFJMtIzU+/z9ENSk4SZ7qp... | singleCorrect | 1 | <img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/k1gYpbYmFx4xl4y5OPK-kPU12RNgkYv570kmu78bSHI.original.fullsize.png"><br>
$i=\frac{180}{90}=2 A$ | ["$4 \\mathrm{~A}$", "$2.5 \\mathrm{~A}$", "$2 \\mathrm{~A}$", "None of these"] | [2] | null | PYQ |
bb2d1e6c2b8ee7065d93b2f9f02dae32 | WBJEE_PHY | Current Electricity | The drift velocity of electrons in silver wire with cross-sectional area $3.14 \times 10^{-6} \mathrm{~m}^{2}$ carrying a current of $20 \mathrm{~A}$ is. Given atomic weight of $\mathrm{Ag}=$ $108,$ density of silver $=10.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ | singleCorrect | 2 | $\quad$ Number of electrons per kg of silver
$$
=\frac{6.023 \times 10^{26}}{108}
$$
Number of electrons per unit volume of silver
$$
\begin{array}{c}
n=\frac{6.023 \times 10^{26}}{108} \times 10.5 \times 10^{3} \\
\qquad \begin{array}{c}
v_{d}=\frac{I}{n e A} \\
=\frac{20}{6.023 \times 10... | ["$2.798 \\times 10^{-4} \\mathrm{~m} / \\mathrm{sec}$", "$67.98 \\times 10^{-4} \\mathrm{~m} / \\mathrm{sec}$", "$0.67 \\times 10^{-4} \\mathrm{~m} / \\mathrm{sec}$", "$6.798 \\times 10^{-4} \\mathrm{~m} / \\mathrm{sec}$"] | [3] | null | PYQ |
840451ac700e384f1b9ddbc72f20f03a | WBJEE_PHY | Current Electricity | Two resistances at $0^{\circ} \mathrm{C}$ with temperature coefficient of resistance $\alpha_{1}$ and $\alpha_{2}$ joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be | singleCorrect | 1 | $\begin{aligned} R_{1} &=R_{0}\left(1+\alpha_{1} t\right)+R_{0}\left(1+\alpha_{2} t\right) \\ &=2 R_{0}\left(1+\frac{\alpha_{1}+\alpha_{2}}{2} t\right) \end{aligned}$
Comparing with $R=R_{0}(1+\alpha t)$
$\alpha=\frac{\alpha_{1}+\alpha_{2}}{2}$ | ["$\\alpha_{1}+\\alpha_{2}$", "$\\frac{\\alpha_{1} \\alpha_{2}}{\\alpha_{1}+\\alpha_{2}}$", "$\\frac{\\alpha_{1}-\\alpha_{2}}{2}$", "$\\frac{\\alpha_{1}+\\alpha_{2}}{2}$"] | [3] | null | PYQ |
f76fe9829bec71bd4c6e15bce205a239 | WBJEE_PHY | Current Electricity | The resistance of a wire at <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>20</mn><mo> </mo><mo>°</mo><mi mathvariant="normal">C</mi></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>20</mn><mo> </mo><mi mathvariant="normal">Ω</mi></math> and <math xmlns="http://www.w3.org/1998/M... | singleCorrect | 1 | <p>We have, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>R</mi><mi>t</mi></msub><mo>=</mo><msub><mi>R</mi><mn>0</mn></msub><mo> </mo><mfenced><mrow><mn>1</mn><mo>+</mo><mi>α</mi><mi>t</mi></mrow></mfenced></math></p><p>Here, <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>20</mn><mo>=</mo><... | ["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>50</mn><mo> </mo><mo>°</mo><mi mathvariant=\"normal\">C</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>60</mn><mo> </mo><mo>°</mo><mi mathvariant=\"normal\">C</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/Math... | [3] | null | PYQ |
944aa6b78dffce805c56ffb9bbad8dfc | WBJEE_PHY | Current Electricity | <p>In the circuit shown, the heat produced in <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>5</mn><mo> </mo><mi mathvariant="normal">Ω</mi></math> resistor is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>10</mn><mo> </mo><mi>cal</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow... | singleCorrect | 1 | <p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>H</mi><mo>=</mo><msup><mi>I</mi><mn>2</mn></msup><mi>R</mi><mi>t</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mfrac><msub><mi>H</mi><mn>1</mn></msub><msub><mi>H</mi><mn>2</mn></msub></mfrac><mo>=</mo><msup><mfe... | ["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>1</mn><mo> </mo><mi>cal</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>2</mn><mo> </mo><mi>cal</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>3</mn><mo> </mo><mi>cal</mi></math>", "<math xmlns=\"http://... | [1] | null | PYQ |
39c081ed53f27b280a3b47bedde39bfc | WBJEE_PHY | Current Electricity | An ammeter has resistance <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>R</mi><mn>0</mn></msub></math> and range <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi></math>. What resistance should be connected in parallel with it to increase its range by <math xmlns="http://www.... | singleCorrect | 1 | Given, <math> <mrow> <msub> <mrow> <mi>i</mi><mo>⁡</mo> </mrow> <mrow> <mi>g</mi><mo>⁡</mo> </mrow> </msub> <mo>=</mo> <mi>i</mi><mo>⁡</mo> <mo>, </mo> <mi>G</mi><mo>⁡</mo> <mo>=</mo> <msub> <mrow> <mi>R</mi><mo>⁡</mo> </mrow> <mn>0</mn> </msub> </mrow> </math>; <br /><math xmlns... | ["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><msub><mi>R</mi><mn>0</mn></msub><mo>/</mo><mfenced><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></mfenced></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><msub><mi>R</mi><mn mathvariant=\"italic\">0</mn></msub><mo>/</mo><mfenced><mrow><mi>n</mi><mo>+</mo... | [2] | null | PYQ |
dc6103ac90d9a3e29173f7da2b952627 | WBJEE_PHY | Current Electricity | In the circuit given below, the value of resistance <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>X</mi></math>, when the potential difference between the points <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>B</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>D</mi></math> is zero, will... | singleCorrect | 1 | <p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mo>=</mo><mn>15</mn><mo>+</mo><mn>6</mn><mo>=</mo><mn>21</mn><mo> </mo><mi mathvariant="normal">Ω</mi><mo>,</mo><mo> </mo><mi>Q</mi><mo>=</mo><mfrac><mrow><mn>8</mn><mi>X</mi></mrow><mrow><mn>8</mn><mo>+</mo><mi>X</mi></mrow></mfrac><mo>+</mo... | ["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>9</mn><mo> </mo><mi mathvariant=\"normal\">Ω</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>8</mn><mo> </mo><mi mathvariant=\"normal\">Ω</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>4</mn><mo>... | [1] | null | PYQ |
37155d5bf822867b40bf02279725e88d | WBJEE_PHY | Current Electricity | <p>In the given circuit, it is observed that the current <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>I</mi></math> is independent of the value of resistance <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>R</mi><mn>5</mn></msub></math>. Then, the resistance value must satisfy</p><p><img src="ht... | singleCorrect | 1 | <p>From the given circuit, we can say that <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>I</mi></math> is independent of resistance <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>R</mi><mn>5</mn></msub><mo>,</mo></math> so no current flows through <math xmlns="http://www.w3.org/1998/Math/Ma... | ["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mfrac><mn>1</mn><msub><mi>R</mi><mn>5</mn></msub></mfrac><mo>+</mo><mfrac><mn>1</mn><msub><mi>R</mi><mn>6</mn></msub></mfrac><mo>=</mo><mfrac><mn>1</mn><mrow><msub><mi>R</mi><mn>1</mn></msub><mo>+</mo><msub><mi>R</mi><mn>2</mn></msub></mrow></mfrac><mo>·</mo><m... | [1] | null | PYQ |
1ea9340a848b386bef34c9046a7c45d5 | WBJEE_PHY | Current Electricity | <p>The value of current <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>I</mi></math> as shown in the given circuit diagram is</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/0f2e5a9c-67bd-4935-be12-e11e8ae36e87-image.png" style="width: 200px; height: 161px;" /></p> | singleCorrect | 2 | <p>The given circuit can be redrawn as</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/c4510ba6-fbf7-4b13-a751-dafb6ac2ade9-image.png" style="width: 180px; height: 167px;" /></p><p>Since, <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><msub><mi>R</mi><mrow><mi>A</mi><mi>B</mi></mrow></msub><m... | ["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>2</mn><mo> </mo><mi mathvariant=\"normal\">A</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>1</mn><mo>.</mo><mn>5</mn><mo> </mo><mi>A</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>0</mn><mo>.</mo><mn>9</mn>... | [2] | null | PYQ |
70f2b466a92bd7179e88b7b34d1beae9 | WBJEE_PHY | Current Electricity | An ammeter of resistance <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>.</mo><mn>5</mn><mo> </mo><mi mathvariant="normal">Ω</mi></math> can measure currents upto <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo> </mo><mi mathvariant="normal">A</mi></math>. The value of... | singleCorrect | 1 | <p>Given, for an ammeter, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>I</mi><mi>g</mi></msub><mo>=</mo><mn>1</mn><mo> </mo><mi mathvariant="normal">A</mi><mo>,</mo><mo> </mo><msub><mi>R</mi><mi>g</mi></msub><mo>=</mo><mn>1</mn><mo>.</mo><mn>5</mn><mo> </mo><mi mathvariant="normal">Ω</... | ["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>0</mn><mo>.</mo><mn>5</mn><mo> </mo><mi mathvariant=\"normal\">Ω</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>1</mn><mo> </mo><mi mathvariant=\"normal\">Ω</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML... | [0] | null | PYQ |
e4787cbfa857ba5d5b413adf4b37834f | WBJEE_PHY | Current Electricity | Two wires $\mathrm{A}$ and $\mathrm{B}$ of the same material, having radii in the ratio 1: 2 and carry currents in the ratio 4: 1 . The ratio of drift speed of electrons in $\mathrm{A}$ and $\mathrm{B}$ is | singleCorrect | 1 | Current flowing through the conductor, $\mathrm{I}=\mathrm{n} \mathrm{e} \mathrm{v} \mathrm{A}$. Hence
$\frac{4}{1}=\frac{\text { nev }_{\mathrm{d}_{1}} \pi(1)^{2}}{\operatorname{nev}_{\mathrm{d}_{2}} \pi(2)^{2}}$ or $\frac{v_{\mathrm{d}_{1}}}{\mathrm{v}_{\mathrm{d}_{2}}}=\frac{4 \times 1}{1}=\frac{16}{1}$ | ["16: 1", "1: 16", "1: 4", "4: 1"] | [0] | null | PYQ |
6838398de9f5dab671af9bb7461c5c31 | WBJEE_PHY | Current Electricity | Determine the current in $2 \Omega$ resistor.
<img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/9LHP1mZg_THupKug7G8OcWj-yVsFw0ol27BuuiaIxdQ.original.fullsize.png"> | singleCorrect | 1 | At steady state the capacitor will be fully charged and thus there will be no current in the $1 \Omega$ resistance. So the effective circuit becomes
<img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/-IM_6Ce_QQBMUduk08aOH6hQKvdi7r4EThDa5QFH4lU.original.fullsize.png"><br>
Net current from the $6 \mathrm{~... | ["$1 \\mathrm{~A}$", "$1.5 \\mathrm{~A}$", "$0.9 \\mathrm{~A}$", "$0.6 \\mathrm{~A}$"] | [2] | null | PYQ |
99ca6873c1fd657c9b8f0e1de62c9ca8 | WBJEE_PHY | Current Electricity | A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively: | singleCorrect | 1 | \(\mathrm{R}=\frac{\rho \mathrm{l}}{\mathrm{A}}\)
Now, \(1=1+\frac{1}{10}=\frac{111}{10}\)
and therefore, \(\mathrm{A}=\frac{10 \mathrm{~A}}{11}\)
So \(\mathrm{R}^{\prime}=\frac{\mathrm{p} \times\left(\frac{111}{10}\right)}{\frac{10 \mathrm{~A}}{11}}=\frac{\rho \mathrm{l}}{\mathrm{A}} \times \frac{(11)^{2}}{(1... | ["$1.2$ times, $1.3$ times", "$1.21$ times, same", "both remain the same", "$1.1$ times, $1.1$ times"] | [1] | null | PYQ |
17d845b115b419f8533bc52d64f954d8 | WBJEE_PHY | Current Electricity | A moving coil galvanometer has resistance of $10 \Omega$ and full scale deflection of $0.01 \mathrm{~A}$. It can be converted into voltmeter of $10 \mathrm{~V}$ full scale by connecting into resistance of | singleCorrect | 2 | $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}=\frac{10}{0.01}-10=990 \Omega$ in series. | ["$9.90 \\Omega$ in series", "$10 \\Omega$ in series", "$990 \\Omega$ in series", "$0.10 \\Omega$"] | [2] | null | PYQ |
ada922c5aac88ee9d56fda2b7c53e92c | WBJEE_PHY | Current Electricity | In the arrangement shown in figure, when the switch $S_2$ is open, the galvanometer, shows no deflection for $l=50 \mathrm{~cm}$ when the switch $S_2$ is closed, the galvanometer shows no deflection for $l=0.416 \mathrm{~m}$. The internal resistance $(r)$ of $6 \mathrm{~V}$ cell is
<img src="https://cdn-question-pool.... | singleCorrect | 3 | When $S_2$ open.
Assume resistance of $X Y=R$.
Resistance of wire per unit length, $x=\frac{R}{L}=R \Omega \mathrm{m}^{-1}$
$\because \quad I=E_0 / R$
Now, the potential drop across $50 \mathrm{~cm}$ length is $6 \mathrm{~V}$, So
$$
\begin{array}{cc}
& \frac{E_0}{R} \times R \times \frac{50}{100}=6 \\
\Rightar... | ["\r\n$2 \\Omega$\r\n", "\r\n$3 \\Omega$\r\n", "\r\n$5 \\Omega$\r\n", "\r\n$9 \\Omega$"] | [0] | null | PYQ |
5578aaa52312e308196889ef5b52b7c5 | WBJEE_PHY | Current Electricity | The resistance of a wire is $5 \Omega$. It's new resistance in ohm if stretched to 5 times of its original length will be : | singleCorrect | 2 | <br /><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/AHskDoEY5sab6S_juKuP1x9Cv9AcAlJ_pFWfMJGRGiM.original.fullsize.png"><br><br />Let resistance of a wire $R$ and length 1 .<br />$\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}=5 \Omega$<br />$\therefore \quad$ Volume of wire is constant in stretching<br />$V_i=V_... | ["625", "5", "125", "25"] | [2] | null | PYQ |
0f1791a04f1b8e57fe16e6a7606a16ea | WBJEE_PHY | Current Electricity | Ratio of thermal energy released in two resistor $R$ and $3 R$ connected in parallel in an electric circuit is : | singleCorrect | 1 | Thermal energy is given by<br />$\mathrm{H}=\mathrm{P} \times \mathrm{t}=\frac{\mathrm{V}^2}{\mathrm{R}} \times \mathrm{t}$<br />Here, voltage $\mathrm{V}$ is same.<br />$\therefore \quad \frac{\mathrm{H}_1}{\mathrm{H}_2}=\frac{\frac{\mathrm{V}^2 \mathrm{t}}{\mathrm{R}}}{\frac{\mathrm{V}^2 \mathrm{t}}{3 \mathrm{R}}}=3:... | ["<br />3 : 1", "<br />1 : 1", "<br />1 : 3", "<br />1 : 27"] | [0] | null | PYQ |
27cbf73ce9484c05fd283a41721b4d0d | WBJEE_PHY | Current Electricity | Five cells each of emf $E$ and internal resistance $r$ send the same amount of current through an external resistance $R$ whether the cells are connected in parallel or in series. Then the ratio $\left(\frac{R}{r}\right)$ is | singleCorrect | 2 | Given : Number of cells, $n=5$, emf of each cell $=E$
Internal resistance of each cell $=r$
In series, current through resistance $R$
$I=\frac{n E}{n r+R}=\frac{5 E}{5 r+R}$
In parallel, current through resistance $R$
$I^{\prime}=\frac{E}{\frac{r}{n}+R}=\frac{n E}{r+n R}=\frac{5 E}{r+5 R}$
According to question, ... | ["$2$", "$\\frac{1}{2}$", "$\\frac{1}{5}$", "$1$"] | [3] | null | PYQ |
382b71d361019ae0754ecbab798b3775 | WBJEE_PHY | Current Electricity | A nichrome wire $50 \mathrm{~cm}$ long and $1 \mathrm{~mm}^2$ cross-section carries a current of $4 \mathrm{~A}$. When connected to 2 volt battery. The resistivity of nichrome wire in $\Omega-\mathrm{m}$ is | singleCorrect | 2 | $\begin{aligned} l=50 \mathrm{~cm} & =50 \times 10^{-2} \mathrm{~m} \\ A & =1 \mathrm{~mm}^2=1 \times 10^{-4} \mathrm{~m}^2 \\ i & =4 \mathrm{~A}, V=2 \mathrm{volt}\end{aligned}$
$\therefore$ Resistance of wire, $R=\frac{V}{i}=\frac{2}{4}=0.5 \Omega$
$\begin{aligned} \therefore \text { Resistiviry, } \rho... | ["$1 \\times 10^{-6}$", "$4 \\times 10^{-7}$", "$3 \\times 10^{-7}$", "$2 \\times 10^{-7}$"] | [0] | null | PYQ |
8958dd58651b7abb05ad5bfb3f396f24 | WBJEE_PHY | Current Electricity | A wire of resistance $160 \Omega$ is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be | singleCorrect | 1 | Let
Initial length $=l_1$
Final length $=l_2$
Initial area $=\mathrm{A}_1$
Final area $=A_1$
$\therefore$ Volume remains same
$\therefore \mathrm{A}_1 \mathrm{l}_1=\mathrm{A}_2 \mathrm{l}_2 \Rightarrow \mathrm{~A}_1 \mathrm{l}_1=\mathrm{A}_2 \frac{\mathrm{l}_1}{4}$
$\Rightarrow 4 \mathrm{~A}_1=\mathrm{A}_2$
Ini... | ["$10 \\Omega$", "$640 \\Omega$", "$40 \\Omega$", "$16 \\Omega$"] | [0] | null | PYQ |
890443beadac097f1d436d4e496a1234 | WBJEE_PHY | Current Electricity | In the given circuit, $E_1=E_2=E_3=2 \mathrm{~V}$ and $R_1=R_2=4 \Omega$, then current flowing through the branch $A B$ is
<img src="https://cdn.quizrr.in/question-assets/bitsat/py_sjn4d5v/ezyKE-IOTRn0fQtjMrQqefXc54BO8ateC8lfj65ElqU.original.fullsize.png"/><br/> | singleCorrect | 1 | The given circuit can be redrawn as,
<img src="https://cdn.quizrr.in/question-assets/bitsat/py_sjn4d5v/XnqLnlC7H6t4kzVS6Mk9vAr4QceBEtSCi9RQdPqWDO8.original.fullsize.png"/>
where, equivalent emf of the combination,
$\begin{aligned}
E_{\text {eq }} & =\frac{E_1 R_1+E_2 R_2}{R_1+R_2} \\
& =\frac{2 \times 4+2 \times 4... | ["zero", "$2 A$ from $A$ and $B$", "$2 A$ from $B$ to $A$", "$5 A$ from $B$ to $B$"] | [1] | null | PYQ |
0d8e639349a170857e3abe1db9007c42 | WBJEE_PHY | Current Electricity | In the following circuit diagram, when $3 \Omega$ resistor is removed, then equivalent resistance of the network
<img src="https://cdn.quizrr.in/question-assets/bitsat/py_sjn4d5v/2nFV1ykWqPDmyi63jAYi7Vzr0wuo4xhg5Sb_CY6fMsw.original.fullsize.png"/><br/> | singleCorrect | 2 | The equivalent Wheatstone's bridge network of the given circuit is shown in figure. Here,
<img src="https://cdn.quizrr.in/question-assets/bitsat/py_sjn4d5v/HOCSsMPj1T3xXx6quAK4y7HgGUATTni2X4KZj1lKzOU.original.fullsize.png"/>
The points $B$ and $D$ are at the same potential as the bridge is balanced. So, the $3 \Omega... | ["increases", "decreases", "remains same", "None of these"] | [2] | null | PYQ |
2514d083281c7bbf6f3807886b90c1bc | WBJEE_PHY | Current Electricity | A conducting wire is stretched by applying a deforming force, so that its diameter decreases to $40 \%$ of original value. The percentage change in its resistance will be | singleCorrect | 1 | On stretching, volume of wire remains constant.
So,
$V=A l$ or $l=\frac{V}{A}$
$\begin{array}{ll}
\therefore & R=\rho \cdot \frac{l}{A}=\frac{\rho V}{A^2}=\frac{\rho V}{\frac{\pi^2 D^4}{16}}=\frac{16 \rho V}{\pi^2 D^4} \\
\therefore & \frac{\Delta R}{R}=-4 \frac{\Delta D}{D}=-4(-0.4)=1.6 \%
\end{array}$ | ["$0.9 \\%$", "$0.12 \\%$", "$1.6 \\%$", "$0.5 \\%$"] | [2] | null | PYQ |
1a3094864ebe859a7914e25ce1ffefe7 | WBJEE_PHY | Current Electricity | Several lamps of $50 \mathrm{~W}$ and $100 \mathrm{~V}$ rating are available. How many of them can be connected in parallel across a battery of a $120 \mathrm{~V}$ of internal resistance $10 \Omega$, so that all bulbs glow in full power? | singleCorrect | 2 | According to given condition, let $n$ number of bulbs are connected in parallel as shown below
<br><img src="https://cdn-question-pool.getmarks.app/pyq/comedk/WGYzoGYavXQI8LBwabFzXNdzlnQZXzrkCZtYLaLobog.original.fullsize.png"><br>
$\begin{aligned} \text { Resistance of each bulb, } R &=\frac{V^{2}}{P}\left[\because P=\... | ["2", "4", "6", "8"] | [1] | null | PYQ |
cd71356abee432b52f67a87111dabeac | WBJEE_PHY | Current Electricity | $I$ and $V$ are respectively the current and voltage in a metal wire of resistance $R$. The $I-V$ graph for the two different temperatures $T_{1}$ and $T_{2}$ given, then
<br><img src="https://cdn-question-pool.getmarks.app/pyq/comedk/cqeYQuLWV-WRy3PbpERI-dqD2kh8xyPDyGFfnj4mss0.original.fullsize.png"><br> | singleCorrect | 1 | From the $I-V$ graph,
$$
\begin{aligned}
&\text { Slope of the line }=\frac{1}{R} \\
&\Rightarrow \quad R=\frac{1}{\text { slope }}
\end{aligned}
$$
Since, slope of line having temperature $T_{1}$ is greater than slope of line having temperature $T_{2}$.
Hence, $\quad R_{2}>R_{1}$
Since, on increasing temperature of a ... | ["$T_{\\mathrm{I}}=T_{2}$", "$T_{1}>T_{2}$", "$T_{1} < T_{2}$", "$T_{1}=2 T_{2}$"] | [2] | null | PYQ |
4a31d4185f38dd3e056ba66498fa7b9e | WBJEE_PHY | Current Electricity | If only $2 \%$ of the main current is to be passed through a galvanometer of resistance $G$, then the resistance of shunt should be | singleCorrect | 1 | For shunt resistance connected to galvanometer, current is given by
$I_{g}=\left(\frac{S}{S+G}\right) I$
$\frac{2}{100} I=\left(\frac{S}{S+G}\right) I$
$\begin{aligned} \Rightarrow & S+G &=50 S \\ \text { or } & S &=\frac{G}{49} \end{aligned}$ | ["$G / 50$", "$G / 49$", "$50 G$", "$49 G$"] | [1] | null | PYQ |
90867e677da1bdf0e659e13b9d503415 | WBJEE_PHY | Current Electricity | Temperature coefficient of resistance of platinum is $4 \times 10^{-3} / \mathrm{K}$ at $20^{\circ} \mathrm{C}$. Temperature at which increase in resistance of platinum is $10 \%$ its value at $20^{\circ} \mathrm{C}$ is | singleCorrect | 3 | Given, $\alpha_{r}=4 \times 10^{-3} / \mathrm{K}, T_{1}=20^{\circ} \mathrm{C}$
Final resistance, $R_{2}=R+10 \% R=1.1 R$
We know that, $\Delta R=\alpha_{r} R \Delta T$
$\begin{array}{lc}
\Rightarrow & 1.1 R-R=4 \times 10^{-3} \times R \times\left(T_{2}-20\right) \\
\Rightarrow & 0.1=4 \times 10^{-3} \times\left(T_{2}-2... | ["$25^{\\circ} \\mathrm{C}$", "$70^{\\circ} \\mathrm{C}$", "$45^{\\circ} \\mathrm{C}$", "$100^{\\circ} \\mathrm{C}$"] | [2] | null | PYQ |
1a70779588113fdfaab3cd83ea635cf3 | WBJEE_PHY | Current Electricity | Ideal voltmeter connected as shown reads
<br><img src="https://cdn-question-pool.getmarks.app/pyq/comedk/NwUp8OwkhuZNFtGFuBecwsBHM78P_GpVjejgdEQSoJk.original.fullsize.png"><br> | singleCorrect | 1 | Here, $6 \Omega$ and $12 \Omega 2$ resistors are in parallel combination, so their equivalent resistance,
$R_{\mathrm{eq}}=\frac{6 \times 12}{6+12}=\frac{72}{18}=4 \Omega$
Now, $R_{\mathrm{eq}}$ and $4 \Omega$ are in series, so total resistance,
$R_{t}=4+4=8 \Omega$
Emf of battery, $E=16 \mathrm{~V}$
$\therefore$ Readi... | ["$16 \\mathrm{~V}$", "$12 \\mathrm{~V}$", "$4 \\mathrm{~V}$", "$8 \\mathrm{~V}$"] | [3] | null | PYQ |
0e4e3a5a8a2e4cc2ea14a9de93f6b69d | WBJEE_PHY | Current Electricity | The magnitude of $I$ in ampere unit is
<br><img src="https://cdn-question-pool.getmarks.app/pyq/comedk/jlYa3EDSbIs_-uJ58WToVJBLjCFBPhstIVkUj6YZ4mw.original.fullsize.png"><br> | singleCorrect | 2 | Consider the diagram with current distribution as shown below.
<br><img src="https://cdn-question-pool.getmarks.app/pyq/comedk/gdLALTrEsDWyYKyuP5c_EoBQm5V7BjwKUz_bgQh2V3I.original.fullsize.png"><br>
As voltage is same in parallel combination, so
$\begin{aligned}
&60 I=(15+5) I_{1} \\
&60 I=20 I_{1} \text { or } I_{1}=3... | ["$0.1$", "$0.3$", "$0.6$", "None of these"] | [0] | null | PYQ |
025cf4b09109f587a000816d956db644 | WBJEE_PHY | Current Electricity | If the potential difference across the internal resistance $r_{1}$ is equal to the $\operatorname{emf} E$ of the battery, then
<br><img src="https://cdn-question-pool.getmarks.app/pyq/comedk/THTLlkeXsE43Vfh6QQwBty7jtAKAHiLTUmSHoeGMNuE.original.fullsize.png"><br> | singleCorrect | 1 | Total emf of circuit, $E_{T}=E+E=2 E$
Total resistance of circuit, $R_{T}=r_{1}+r_{2}+R$
Current flowing through the circuit,
$i=\frac{E_{T}}{R_{T}}=\frac{2 E}{r_{1}+r_{2}+R}...(i)$
As per question $E=i r_{1}$
$\Rightarrow \quad i=\frac{E}{r_{1}}...(ii)$
From Eqs. (i) and (ii), we get
$\begin{array}{ll}
& \frac{E}{r_{... | ["$R=r_{1}+r_{2}$", "$R=\\frac{r_{1}}{r_{2}}$", "$R=r_{1}-r_{2}$", "$R=\\frac{r_{2}}{r_{1}^{*}}$"] | [2] | null | PYQ |
7e2601df98eff4627970021ab44f97d0 | WBJEE_PHY | Current Electricity | By using only two resistance coils-singly, in series, or in parallel-one should be able to obtain resistances of $3,4,12$ and 16 ohms. The separate resistances of the coil are | singleCorrect | 3 | The maximum and minimum resistances are $16 \Omega$ and $3 \Omega$ that are obtained by series and parallel combination of resistances. From given option only $4 \Omega$ and $12 \Omega$ resistors satisfies this condition. | ["3 and 4", "4 and 12", "12 and 16", "16 and 3"] | [1] | null | PYQ |
c2ba68c507f79543c446d618b1caac23 | WBJEE_PHY | Current Electricity | In the given circuit, the voltmeter records $5 \mathrm{~V}$. The resistance of the voltmeter (in ohms) is
<br><img src="https://cdn-question-pool.getmarks.app/pyq/comedk/d3sFmLV4AKrv53Khyri00skKT6_wJtjYOMk_lf1qjR0.original.fullsize.png"><br> | singleCorrect | 1 | If $R$ be the resistance of voltmeter, then the equivalent resistance of circuit is
$\begin{aligned}
R_{e q} &=\frac{R \times 100}{100+R}+50 \\
&=\frac{150 R+5000}{R+100}
\end{aligned}$
Total current in circuit,
$i=\frac{V}{R_{\text {eq }}}=\frac{10}{\frac{150 R+5000}{(R+100)}}$
$=\frac{10(R+100)}{150 R+5000}$
Voltage ... | ["200", "100", "10", "50"] | [1] | null | PYQ |
6f4fce4045c0ba1e3a45019f3a958677 | WBJEE_PHY | Current Electricity | A $5.0$ A current is setup in an external circuit by a $6.0 \mathrm{~V}$ storage battery for $6.0 \mathrm{~min}$. The chemical energy of the battery is reduced by | singleCorrect | 2 | Given, $I=5 \mathrm{~A}, V=6 \mathrm{~V}, t=6 \mathrm{~min}=6 \times 60=360 \mathrm{~s}$
Heat produced in a circuit =Chemical energy reduced in the battery
$\Rightarrow$
$\begin{aligned}
H &=V I t \\
&=6 \times 5 \times 360 \\
&=10800=1.08 \times 10^{4} \mathrm{~J}
\end{aligned}$ | ["$1.08 \\times 10^{4} \\mathrm{~J}$", "$1.08 \\times 10^{4} \\mathrm{~V}$", "$1.8 \\times 10^{4} \\mathrm{~J}$", "$1.8 \\times 10^{4} \\mathrm{~V}$"] | [0] | null | PYQ |
31c1bc9eed870e70282f45ecf061e1e9 | WBJEE_PHY | Current Electricity | The equivalent resistance between the points $A$ and $B$ will be (each resistance is $15 \Omega$ )<br><img src="https://cdn.quizrr.in/question-assets/comedk/pysf5e2f/8rBmNdEDTgHA2hAitD3VAqgH8P1ZykbAtywqNF2ib4o.original.fullsize.png"><br> | singleCorrect | 2 | Redrawing the given circuit network as<br><img src="https://cdn.quizrr.in/question-assets/comedk/pysf5e2f/fgRTkrGAXbgSL7C7lsCKkBIYVtxyUYPtELCpKPaSwC8.original.fullsize.png"><br>Equivalent resistance between points $D$ and $C$,<br>$$<br>R_{D C}=\frac{(15+15) \times 15}{(15+15)+15}=\frac{30 \times 15}{45}=10 \Omega<br>$$... | ["$30 \\Omega$", "$8 \\Omega$", "$10 \\Omega$", "$40 \\Omega$"] | [1] | null | PYQ |
1dd2fec1f91874464927af5c2baf0746 | WBJEE_PHY | Current Electricity | The terminals of a $18 \mathrm{~V}$ battery with an internal resistance of $1.5 \Omega$ are connected to a circular wire of resistance $24 \Omega$ at two points distant at one quarter of the circumference of a circular wire. The current through the bigger arc of the circle will be | singleCorrect | 3 | Since, the terminals of battery is connected at one quarter distance. Hence, ratio of lengths between two parts will be<br>$\begin{aligned}<br>\frac{l_{1}}{l_{2}} &=\frac{1}{3} \\<br>\Rightarrow \quad l_{1}: l_{2} &=1: 3 \\<br>\text { Since, } R \propto l \\<br>\therefore \quad R_{1}: R_{2} &=1: 3 \\<br>R_{1} &=\frac{2... | ["$0.75 \\mathrm{~A}$", "$1.5 \\mathrm{~A}$", "$2.25 \\mathrm{~A}$", "$3 \\mathrm{~A}$"] | [0] | null | PYQ |
1d432e71c6dcb0cdcad31563a6e17670 | WBJEE_PHY | Current Electricity | In the circuit shown in Fig. 22.106, cells $E_1$ and $E_2$ have emfs 4 V and 8 V and internal resistances $0.5 \Omega$ and $1 \Omega$ respectively.
<img src="https://cdn-question-pool.getmarks.app/modules/ms/wbjee/0QIfasU2VK_Tkc1ipq3NklfLwfMjyCJncybbjJICUdA.original.fullsize.png"/><br/> | multipleCorrect | 2 | Equivalent resistance of the parallel combination of $3 \Omega$ and $6 \Omega$ is $R=3 \times 6 /(3+6)=2 \Omega$. As the cells are in opposition, net emf $E=E_2-E_1=8-4=4 \mathrm{~V}$.
Therefore, current is
$I=\frac{E}{R+4.5+r_1+r_2}=\frac{4}{2+4.5+0.5+1}=0.5 \mathrm{~A}$
Potential difference across $E_1$ is $V_1=E_1+I... | ["The potential difference across $E_1$ is 4.25 V", "The potential difference across $E_1$ is 3.75 V", "The potential difference across $E_2$ is 8.5 V", "The potential difference across $E_2$ is 7.5 V ."] | [0, 3] | null | PYQ |
66a84e09af1ce190100816959b10f896 | WBJEE_PHY | Current Electricity | Which of the following statements are correct about the circuit shown in Fig. 22.101 where $1 \Omega$ and $0.5 \Omega$ are the internal resistances of the 6 V and 12 V batteries respectively?
<img src="https://cdn-question-pool.getmarks.app/modules/ms/wbjee/elq_ISYrktPRcbDV8ERIv23tesCfNb5CbJx1oZT41mQ.original.fullsize.... | multipleCorrect | 3 | Total resistance $=4+1+0.5+0.5=6 \Omega$. Net voltage in the circuit is 6 V . Current $I=\frac{6}{6}=1 \mathrm{~A}$ in the anticlockwise direction
$V_{P R}=1 \times 4=4 \mathrm{~V}$
Since $R$ is connected to earth, $V_R=0$. Hence $V_P=4 \mathrm{~V}$
$V_{S Q}=0.5 \times 1=0.5 \mathrm{~V} . S \text { is at a }$
higher po... | ["The potential at point $P$ is 6 V .", "The potential at point $Q$ is -0.5 V", "fIf a voltmeter is connected across the 6 V battery, it will read 7 V .", "If a voltmeter is connected across the 6 V battery, it will read 5 V ."] | [1, 2] | null | PYQ |
3948dc9f9a48035e0a669c2d2f4f0229 | WBJEE_PHY | Current Electricity | A galvanometer has a resistance of $100 \Omega$ and fullscale range of $50 \mu \mathrm{~A}$. It can be used as a voltmeter or an ammeter, provided a resistance is connected to it. Pick the correct range and resistance combination (s): | multipleCorrect | 2 | For voltmeter, the resistance $R$ to be connected in series with the galvanometer is given by
$R=\frac{V}{I_g}-G$
For 50 V range, $R=\frac{50}{50 \times 10^{-6}}-100$
$=10^{-6}-100 \simeq 10^6 \simeq 1000 \mathrm{k} \Omega$
Hence choice (a) is incorrect.
For 10 V range, $R=\frac{10}{50 \times 100^{-6}}-100$
$=2 \times ... | ["50 V range with $10 \\mathrm{k} \\Omega$ resistance in series", "10 V range with $200 \\mathrm{k} \\Omega$ resistance in series", "5 mA range with $1 \\Omega$ resistance in parallel", "10 mA range with $2 \\Omega$ resistance in parallel"] | [1, 2] | null | PYQ |
1e4bfc1672573e3b9614432681eb9dad | WBJEE_PHY | Current Electricity | Choose the correct statements from the following. | multipleCorrect | 3 | The correct choices are (a), (c) and (d). In choice (b) the emf of the combination $E$ and not $E / n$. | ["If $n$ identical cells, each of emf $E$ and internal resistance $r$ are connected in series, the emf of the combination is $n E$ and the internal resistance of the combination is $n r$.", "If $n$ identical cells, each of emf $E$ and internal resistance $r$ are connected in parallel, the emf of the combination is $E /... | [0, 2, 3] | null | PYQ |
278caa236648553ffd22476a42ec83d1 | WBJEE_PHY | Current Electricity | The internal resistance of the cell shown in Fig. 22.98 is negligible. On closing the key $K$, the ammeter reading changes from 0.25 A to $5 / 12 \mathrm{~A}$, then
<img src="https://cdn-question-pool.getmarks.app/modules/ms/wbjee/MUQw0vHNbKm5wQpxpTtfeXtHVAZ4kAfpxgVeHXzvRiw.original.fullsize.png"/><br/> | multipleCorrect | 1 | Before closing the key,
$I_1=\frac{E}{R} \Rightarrow 0.25=\frac{E}{10} \Rightarrow E=2.5 \mathrm{~V}$
After closing the key,
Effective resistance is $R_2=\frac{R R_1}{\left(R+R_1\right)}=\frac{10 R_1}{\left(10+R_1\right)}$.
or
$\frac{5}{12}=\frac{2.5 \times\left(10+R_1\right)}{10 R_1}$
which gives $R_1=15 \Omega$.
Befo... | ["$R_1=10 \\Omega$", "$R_1=15 \\Omega$", "the power drawn from the cell increases", "the current through $R$ decreases by $40 \\%$"] | [1, 2] | null | PYQ |
f6c46675c3555bff373177af8e727780 | WBJEE_PHY | Current Electricity | Choose the correct statements from the following. | multipleCorrect | 1 | The current $I$ drawn from a supply of voltage $E$ and internal resistance $r$ is given by $I=E / r$. So $r$ must be very small so that a high current can be drawn. Statement (b) is also correct. At such a high voltage, the current drawn from the supply will become dangerously large if its internal resistance is small.... | ["A low voltage supply of, say, 6 V must have a very low internal resistance.", "A high voltage supply of, say, 6000 V must have a very high internal resistance.", "A wire carrying current stays electrically neutral.", "A high resistance voltmeter is used to measure the emf of a cell."] | [0, 1, 2] | null | PYQ |
5aa3ded6b975f19a8d01af0748552db9 | WBJEE_PHY | Current Electricity | Two equal resistances $R_1=R_2=R$ are connected with a $30 \Omega$ resistor and a battery of terminal voltage $E$. The currents in the two branches are 2.25 A and 1.5 A as shown in Fig. 22.100. Then
<img src="https://cdn-question-pool.getmarks.app/modules/ms/wbjee/STObS_VSe_H_5014u8YbRWInIJn67k3Nlkb9qRiyR3s.original.fu... | multipleCorrect | 2 | Using Kirchhoff's Ist law current through $R_2$ is 2.25 $-1.5=0.75 \mathrm{~A}$. Also since $R_2$ is in parallel with the $30 \Omega$ resistance, $R_2$ must be $60 \Omega$ since only half the current flows through it compared to the current through $30 \Omega$ resistor. Total resistance in the circuit becomes $60+20=80... | ["$R_2=15 \\Omega$", "$R_2=60 \\Omega$", "$E=36 \\mathrm{~V}$", "$E=180 \\mathrm{~V}$"] | [1, 3] | null | PYQ |
b1bb0971739d46facbf192c0d9bea427 | WBJEE_PHY | Current Electricity | In the circuit shown in Fig. 22.102,
<img src="https://cdn-question-pool.getmarks.app/modules/ms/wbjee/z5U27nrX2W8z06rro0hM-g7deShF9pjtP6lEQ0WVvME.original.fullsize.png"/><br/> | multipleCorrect | 2 | Potential difference across $M P=$ p.d. across $N O=$ p.d. across $N P$ (see Fig. 22.104)
Current across $N P, I_{N P} \times 10=20 \times 1$ or $I_{N P}=2 \mathrm{~A}$
Across MP, $0.5 R_1=20$ or $R_1=40 \Omega$
Total current $=2+0.5+1.0=3.5 \mathrm{~A}$
$3.5=\frac{69}{R+40 / 7}$ yields $R=14 \Omega$
Hence the correct... | ["the current through $N P$ is 0.5 A", "the value of $R_1=40 \\Omega$", "the value of $R=14 \\Omega$", "the potential difference across $R=49 \\mathrm{~V}$"] | [1, 2] | null | PYQ |
a510c1b20b93116347557e27be1d1754 | WBJEE_PHY | Current Electricity | The resistance network shown in Fig. 22.103 is connected to a battery of emf 30 V and internal resistance of $1 \Omega$. Then
<img src="https://cdn-question-pool.getmarks.app/modules/ms/wbjee/g2pHxSUpSYJrU361V8Z8LV-CWdcRPZaqNSL3ye2pjg8.original.fullsize.png"/><br/> | multipleCorrect | 3 | Total resistance of parallel combination is given by
$\frac{1}{R_1}=\frac{1}{12}+\frac{1}{6}+\frac{1}{4}$
or $R_1=2 \Omega$. Total resistance in circuit is $R=R_1+2$ $=4 \Omega$. Therefore, current in the circuit is
$I=\frac{E}{R+r}=\frac{30}{4+1}=6 \mathrm{~A}$
$\therefore \quad$ Potential drop across $2 \Omega$ resis... | ["the voltage drop across the $2 \\Omega$ resistor is 12 V .", "the voltage drop across the $12 \\Omega$ resitor is 12 V .", "the terminal voltage of the battery is 24 V .", "the voltage drop across the internal resistance of the battery is 6 V ."] | [0, 1, 2, 3] | null | PYQ |
ffaf7c42c66e3533609e623089baa375 | WBJEE_PHY | Current Electricity | A current $I$ flows in the circuit shown in Fig. 22.99. Then
<img src="https://cdn-question-pool.getmarks.app/modules/ms/wbjee/JzAkWK6YBmwoGz1iC_1Noj7sp1-oS6kjX5azVYS5f-s.original.fullsize.png"/><br/> | multipleCorrect | 2 | When $R_2=R$ is connected in parallel with $R_1=R$, the resistance of the combination is $R / 2$. Therefore, the current in the circuit is $V / R / 2=2 V / R=2 I$, where $I=V / R$ was the current in the circuit when $R_2$ was not connected. Current $2 I$ divides equally among two equal parallel resistors. Hence the cur... | ["If a resistance $R_2=R$ is connected in parallel with $R_1=R$, the current through $R_1$ will remain equal to $I$.", "If a resistance $R_2=2 R$ is connected in parallel with $R_1=R$, the current through $R_1$ will remain equal to $I$.", "If a resistance $R_2=2 R$ is connected in parallel with $R_1=R$, the current thr... | [0, 1, 3] | null | PYQ |
87a6d8e3ecac56f247527292bd5d497f | WBJEE_PHY | Dual Nature of Matter | An electron in an excited state of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>L</mi><msup><mi>i</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math> ion has angular momentum <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>3</mn><mi>h</mi></mrow><mrow><mn>2</mn><mi mathvariant="normal">π</mi><... | singleCorrect | 1 | From Bohr's law<br /><math><mi mathvariant="normal">m</mi><mi mathvariant="normal">v</mi><mi mathvariant="normal">r</mi><mo>=</mo><mfrac><mrow><mi mathvariant="normal">n</mi><mi mathvariant="normal">h</mi></mrow><mrow><mn>2</mn><mi mathvariant="normal">π</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>3</mn><mi mathvaria... | ["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mi>λ</mi><mo>=</mo><mn>2</mn><mi>π</mi><msub><mi>a</mi><mn>0</mn></msub></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mi>\u03bb</mi><mo>=</mo><mn>4</mn><mi>\u03c0</mi><msub><mi>a</mi><mn>0</mn></msub></math>", "<math xmlns=\"http://www.w3.... | [0] | null | PYQ |
928580bf6bcd8b4e0eb8b071ee16b2c4 | WBJEE_PHY | Dual Nature of Matter | The threshold wavelength of the tungsten is 2300 $Ã… .$ If ultraviolet light of wavelength $1800 Ã…$ is incident on it, then the maximum kinetic energy of photoelectrons would be about - | singleCorrect | 2 | $\mathrm{K}_{\max }=\mathrm{hv}-\mathrm{hv}_{0}=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)$
$1.24 \times 10^{-6}\left(\frac{10^{8}}{18}-\frac{10^{8}}{23}\right)=1.49 \mathrm{eV}$ | ["$1.49 \\mathrm{eV}$", "$2.2 \\mathrm{eV}$", "$3.0 \\mathrm{eV}$", "$5.0 \\mathrm{eV}$"] | [0] | null | PYQ |
3c2af9c67e8ddba5da55534ae449bdbf | WBJEE_PHY | Dual Nature of Matter | The beam of light has three wavelengths $4144 Ã, 4972 Ã$ and $6216 Ã$ with a total intensity of $3.6 \times 10^{-3} \mathrm{Wm}^{2}$ equally distributed amongst the three wavelengths. The beam falls normally on the area $1 \mathrm{~cm}^{2}$ of a clean metallic surface of work function $2.3 \mathrm{eV}$. Assume that the... | singleCorrect | 2 | As we know, threshold wavelength
$$
\begin{array}{c}
\left(\lambda_{0}\right)=\frac{\mathrm{hc}}{\phi} \\
\Rightarrow \lambda_{0}=\frac{\left(6.63 \times 10^{-34}\right) \times 3 \times 10^{8}}{2.3 \times\left(1.6 \times 10^{-19}\right)}=5.404 \times 10^{-7} \mathrm{~m}
\end{array}
$$
$\Rightarrow ... | ["$2 \\times 10^{9}$", "$1.075 \\times 10^{12}$", "$9 \\times 10^{8}$", "$3.75 \\times 10^{6}$"] | [1] | null | PYQ |
590cce42164ee27564fc871e72e299e8 | WBJEE_PHY | Dual Nature of Matter | Find the number of photon emitted per second by a 25 watt source of monochromatic light of wavelength $6600 Å$. What is the photoelectric current assuming $3 \%$ efficiency for photoelectric effect? | singleCorrect | 2 | $P_{\text {in }}=25 \mathrm{~W}, \lambda=6600 \AA=6600 \times 10^{-10} \mathrm{~m}$
$n h v=P$
Number of photons emitted/sec $\begin{aligned} n &=\frac{P}{\frac{h c}{\lambda}}=\frac{P \lambda}{h c}=\frac{25 \times 6600 \times 10^{-10}}{6.64 \times 10^{-34} \times 3 \times 10^{8}} \\ &=8.28 \times 10^{19}=\frac{25}... | ["$\\frac{25}{3} \\times 10^{19} \\mathrm{~J}, 0.4 \\mathrm{amp}$", "$\\frac{25}{4} \\times 10^{19} \\mathrm{~J}, 6.2 \\mathrm{amp}$", "$\\frac{25}{2} \\times 10^{19} \\mathrm{~J}, 0.8 \\mathrm{amp}$", "None of these"] | [0] | null | PYQ |
494444bb923ce41dc5362081088eedb0 | WBJEE_PHY | Dual Nature of Matter | The de-Broglie wavelength of a proton (charge <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>1</mn><mo>.</mo><mn>6</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>-</mo><mn>19</mn></mrow></msup><mo> </mo><mi mathvariant="normal">C</mi><mo>,</mo><mi>m</mi><mo>=</mo><mn>1</mn><mo>.</mo><mn>6</mn><mo>&#... | singleCorrect | 1 | <p>de-Broglie wavelength,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>λ</mi><mo>=</mo><mfrac><mi>h</mi><mi>p</mi></mfrac><mo>=</mo><mfrac><mi>h</mi><msqrt><mn>2</mn><mi>m</mi><mi>E</mi></msqrt></mfrac><mo>=</mo><mfrac><mi>h</mi><msqrt><mn>2</mn><mi>m</mi><mi>q</mi><mi>V</mi></msqrt></mfrac></math></... | ["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>600</mn><mo> </mo><mo>Å</mo></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>0</mn><mo>.</mo><mn>9</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>-</mo><mn>12</mn></mrow></msup><mo> </mo><mi mathvariant=\"normal\">m</mi></math>", "... | [1] | null | PYQ |
3febd544c40f1037c26ce708f9dae55d | WBJEE_PHY | Dual Nature of Matter | Electrons are accelerated through a potential difference $\mathrm{V}$ and protons are accelerated through a potential difference $4 \mathrm{~V}$. The de - Broglie wavelengths are $\lambda_{e}$ and $\lambda_{p}$ for electrons and protons respectively. The ratio of $\frac{\lambda_{e}}{\lambda_{p}}$ is given by: (given $m... | singleCorrect | 1 | Energy $=$ charge $\times$ potential diff.
$\mathrm{E}_{\text {electron }}=\mathrm{q}_{\mathrm{e}} \mathrm{V}$ and $\mathrm{E}_{\text {proton }}=\mathrm{q}_{\mathrm{p}} 4 \mathrm{~V}$
de $-$ Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$
$\lambda_{e}=\frac{h}... | ["$\\sqrt{\\frac{m_{p}}{m_{e}}}$", "$\\sqrt{\\frac{m_{e}}{m_{p}}}$", "$\\frac{1}{2} \\sqrt{\\frac{m_{e}}{m_{p}}}$", "$2 \\sqrt{\\frac{m_{p}}{m_{e}}}$"] | [3] | null | PYQ |
c34332057a8a09cf0c3bb325bbaef4d6 | WBJEE_PHY | Dual Nature of Matter | The photoelectric current of voltage in a certain experiment is $1.5 \mathrm{~V}$. What is the maximum kinetic energy of photoelectrons emitted? | singleCorrect | 1 | Given, $V_{0}=15 \mathrm{~V}$
Maximum kinetic energy of emitted photoelectron,
$\begin{aligned} K_{\max } &=e V_{0} \\ &=1.6 \times 10^{-19} \times 1.5=2.4 \times 10^{-19} \mathrm{~J} \end{aligned}$ | ["$24 \\times 10^{-19} \\mathrm{~J}$", "$-24 \\times 10^{-19} \\mathrm{~J}$", "$2.4 \\times 10^{-19} \\mathrm{~J}$", "$2.04 \\times 10^{-19} \\mathrm{~J}$"] | [2] | null | PYQ |
d9ca8861d8c610eefd46a5735d99b112 | WBJEE_PHY | Dual Nature of Matter | A proton and an alpha particle are subjected to same potential difference $V$. Their de-Broglie wavelengths $\lambda_{p}, \lambda_{\alpha}$ will be in the ratio | singleCorrect | 1 | T
he de-Broglie wavelength of a particle is given by
$\lambda=\frac{h}{m v}=\frac{h}{p}$
The momentum of proton,
$p_{p}=\sqrt{2 m E}=\sqrt{2 m e V}$
Similarly, for $\alpha$ - particle,
$\begin{aligned}
p_{\alpha} &-\sqrt{2 \times 4 m \times 2 e \times V} \\
&=\sqrt{16 m e V} \\
\therefore \quad \frac{\lambda_{p}}{\lamb... | ["$2: 1$", "$2 \\sqrt{2}: 1$", "$1: 1$", "$1: 2$"] | [1] | null | PYQ |
d1b19b5e62566ade6ae0d26c43caf97b | WBJEE_PHY | Dual Nature of Matter | Work function of a photosensitive metal is $3 \mathrm{eV}$. The wavelength of incident radiations which can just eject photoelectrons from the metal is | singleCorrect | 2 | Given, work function, $\phi=3 \mathrm{eV}$
The wavelength of incident radiation is given by
$\lambda=\frac{h c}{\phi}=\frac{1242}{\phi}=\frac{1242}{3}=414 \mathrm{~nm}$ | ["$600 \\mathrm{~nm}$", "$510 \\mathrm{~nm}$", "$414 \\mathrm{~nm}$", "$378 \\mathrm{~nm}$"] | [2] | null | PYQ |
432d6f4eb0607667107dd77242190a27 | WBJEE_PHY | Dual Nature of Matter | The energy that should be added to an electron to reduce its de-Broglie wavelength from $1 \mathrm{~nm}$ to $0.5 \mathrm{~nm}$ is | singleCorrect | 1 | Given, initial value of de-Broglie wavelength,
$\lambda_{1}=1 \mathrm{~nm}=1 \times 10^{-9} \mathrm{~m}$
Final value of de-Broglie wavelength,
$\lambda_{2}=0.5 \mathrm{~nm}=0.5 \times 10^{-9} \mathrm{~m}$
de-Broglie wavelength in terms of energy $E$ is given as
$\begin{aligned}
& \lambda &=\frac{h}{\sqrt{2 m E}} \\
& \... | ["four times the initial energy", "equal to the initial energy", "twice the initial energy", "thrice the initial energy"] | [3] | null | PYQ |
8827478331aaeff5c1cb22b407e2b128 | WBJEE_PHY | Dual Nature of Matter | When a piece of metal is illuminated by a monochromatic light of wavelength $\lambda$, then stopping potential is $3 \mathrm{~V}-\mathrm{s}$. When same surface is illuminated by light of wavelength $2 \lambda$, then stopping potential becomes $V_{s}$. The value of threshold wavelength for photoelectric emission will be | singleCorrect | 2 | According to first situation wavelength of incident radiation,<br>$$<br>\lambda_{1}=\lambda<br>$$<br>Stopping potential, $V_{01}=3 V_{\mathrm{s}}$<br>According to Einstein's photoelectric equation,<br>$\begin{array}{ll} <br>& K_{\max }=h v_{1}-\phi \\<br>\Rightarrow \quad & e V_{01}=\frac{h c}{\lambda_{1}}-\frac{h c}{\... | ["$4 \\lambda$", "$8 \\lambda$", "$4 / 3 \\lambda$", "$6 \\lambda$"] | [0] | null | PYQ |
ebd452642cb0932b262b937393456611 | WBJEE_PHY | Dual Nature of Matter | Light of two different frequencies whose photons have energies \( 1 \mathrm{eV} \) and \( 2.5 \mathrm{eV} \) respectively,<br>successively illuminate a metallic surface whose work function is \( 0.5 \mathrm{eV} \). Ratio of maximum<br>speeds of emitted electrons will be | singleCorrect | 1 | Kinetic energy, \( K h=f W \)<br>where \( W \) is work function; \( f \) is frequency of incident photon.<br>For photon of energy<br>\( 1 e V, K_{1}=1 e V-0.5 e V=0.5 e V \)<br>For photon of energy<br>\( 2.5 \mathrm{eV}, K_{2}=2.5 \mathrm{eV}-0.5 \mathrm{eV}=2.0 \mathrm{eV} \)<br>Taking ratio, \( \frac{K_{1}}{K_{2}}=\f... | ["$1: 5$", "\\( 1: 4 \\)", "\\( 1: 2 \\)", "1:1"] | [2] | null | PYQ |
4dba5bc2331e0e3e04a66032a4b341ca | WBJEE_PHY | Dual Nature of Matter | Find the de-Broglie wavelength of an electron with kinetic energy of \( 120 \mathrm{eV} \). | singleCorrect | 2 | Given, kinetic energy of electron $=120 \mathrm{eV}$
de Broglie wavelength is related to $\mathrm{V}$ as
$\lambda=\frac{1.227 \mathrm{~nm}}{\sqrt{V}}=\frac{1.227 \times 10^{-9}}{\sqrt{120}}=\frac{1.227 \times 10^{-9}}{10.95}$
$=0.112 \times 10^{-9}$
Therefore,
$\lambda=112 \times 10^{-12} \mathrm{~m}=112 \mathrm{pm}$ | ["\\( 95 \\mathrm{pm} \\)", "\\( 102 \\) pm", "\\( 112 \\mathrm{pm} \\)", "\\( 124 \\mathrm{pm} \\)"] | [2] | null | PYQ |
4e48578e15571762932366ce49e28ad4 | WBJEE_PHY | Dual Nature of Matter | From the following graph of photo current against collector plate potential, for two different<br>intensities of light \( I_{1} \) and \( I_{2} \), one can conclude<br><img src="https://cdn-question-pool.getmarks.app/pyq/kcet/ques_98.png"> | singleCorrect | 3 | For a fixed frequency of incident light, photo-current is directly proportional to the intensity.<br>Thus, from graph, photo current for light of Intensity \( I_{2} \) is more than for light of intensity \( I_{1} \), Therefore, \( I_{1} < I_{2} \) | ["\\( O I_{1}=I_{2} \\)", "\\( I_{1}>I_{2} \\)", "\\( I_{1} < I_{2} \\)", "Comparison is not possible."] | [2] | null | PYQ |
87d4dc6a8363775c228e9e1669797baa | WBJEE_PHY | Dual Nature of Matter | The number of photons falling per second on a completely darkened plate to produce a force<br>of \(6.62 \times 10^{-5} \mathrm{~N}\) is ' \(n\) '. If the wavelength of the light falling is \(5 \times 10^{-7} \mathrm{~m}\), then \(n=\)_______\(\times 10^{22}\).<br>\(\left(h=6.62 \times 10^{-34}\right) \mathrm{J}_{\mathr... | singleCorrect | 2 | Given, force, $F=6.62 \times 10^{-5} \mathrm{~N} ;$ wavelength,$\lambda=5 \times 10^{-7} \mathrm{~m}$<br>Now, $C \times F=\frac{h c}{\lambda} \times \frac{n}{t}$<br>or $F=\frac{h n}{\lambda t}$<br>$\Rightarrow \frac{n}{t}=\frac{F \lambda}{h}=\frac{6.62 \times 10^{-5}}{6.62 \times 10^{-34}} \times 5 \times 10^{-7}=5 \ti... | ["\\(11\\)", "\\(05\\)", "\\(0.2\\)", "\\(3.3\\)"] | [1] | null | PYQ |
93f57ad6c96e813adbc62e325af06eb8 | WBJEE_PHY | Dual Nature of Matter | Light of certain frequency and intensity incident on a photosensitive material causes<br>photoelectric effect. If both the frequency and intensity are doubled, the photoelectric<br>saturation current becomes | singleCorrect | 1 | (D)<br>We know photoelectric current is directly proportional to the intensity of incident light provided incidentfrequency is<br>greater than threshold frequency.<br>\(\therefore\) When intensity is doubled, photoelectric saturation current doubles. | ["halved", "quadrupled", "unchanged", "doubled"] | [3] | null | PYQ |
a138c2e786a9ec1240edb369562d7235 | WBJEE_PHY | Dual Nature of Matter | A proton moving with a momentum $p_{1}$ has a kinetic energy $1 / 8$ th of its rest mass-energy. Another light photon having energy equal to the kinetic energy of the possesses a momentum $p_{2}$. Then, the ratio $\frac{p_{1}-p_{2}}{p_{1}}$ is equal to | singleCorrect | 2 | For proton, $v^{2}=\frac{c^{2}}{4}$ $\left(\because v=\frac{c}{2}\right)$<br>$\begin{aligned}<br>p_{1} &=\sqrt{2 m E_{k}} \\<br>&=\sqrt{2 \times m \times \frac{1}{8} m c^{2}} \quad\left(\because E_{k}=\frac{1}{8} m c^{2}\right) \\<br>p_{1} &=\frac{m \varepsilon}{2}<br>\end{aligned}$<br>For photon,<br>$\begin{aligned}<b... | ["1", "$1 / 4$", "$1 / 2$", "$3 / 4$"] | [3] | null | PYQ |
cefbc4d73cdc68bf2e9268ffa6c63fee | WBJEE_PHY | Dual Nature of Matter | The figure shows standing de-Broglie waves due to the revolution of electron in a certain orbit of hydrogen atom. Then, the expression for the orbit radius is (All notations have their usual meanings )<br><img src="https://cdn.quizrr.in/question-assets/kcet/py54ggf1c/llhf-kA3wU76oEdfTFxp0OIzM0r1RZGonJFJnrLzY2k.original... | singleCorrect | 3 | According to de-Broglie, the circumference of a stationary orbit must be an integral number of wavelengths.<br>$\begin{aligned}<br>&\qquad n \lambda=2 \pi r_{n} \\<br>&\text { Also, angular momentum, } m v_{n} r_{n}=\frac{n h}{2 \pi} \\<br>&\Rightarrow \quad m \frac{e^{2}}{2 n h \varepsilon_{0}} r_{n}=\frac{n h}{2 \pi}... | ["$\\frac{h^{2} \\varepsilon_{0}}{\\pi m e^{2}}$", "$\\frac{4 h^{2} \\varepsilon_{0}}{\\pi m e^{2}}$", "$\\frac{9 h^{2} \\varepsilon_{0}}{\\pi m e^{2}}$", "$\\frac{36 h^{2} \\varepsilon_{0}}{\\pi m e^{2}}$"] | [3] | null | PYQ |
5d8e897da61f464c3a6554627d90f12f | WBJEE_PHY | Dual Nature of Matter | The kinetic energy of the photoelectrons increases by $0.52 \mathrm{eV}$ when the wavelength of incident light is changed from $500 \mathrm{~nm}$ to another wavelength which is approximately | singleCorrect | 2 | Given, change in kinetic energy of photoelectrons,<br>$\begin{aligned}<br>\Delta K & =0.52 \mathrm{eV} \\<br>\lambda_1 & =500 \mathrm{~nm}, \lambda_2=?<br>\end{aligned}$<br>We know that, kinetic energy of emitted photoelectron (using Einstein's equation)<br>$\begin{gathered}<br>K=\frac{h c}{\lambda}-\phi \\<br>\therefo... | ["$400 \\mathrm{~nm}$", "$1250 \\mathrm{~nm}$", "$1000 \\mathrm{~nm}$", "$700 \\mathrm{~nm}$"] | [0] | null | PYQ |
b11ac2ae7578e2ad05b59e09654dd4d8 | WBJEE_PHY | Dual Nature of Matter | The de-Broglie wavelength of a particle of kinetic energy $K$ is $\lambda$, the wavelength of the particle, if its kinetic energy $\frac{K}{4}$ is | singleCorrect | 1 | de-Broglie wavelength $(\lambda)$ of the particle in terms of kinetic energy $K$ is given as<br>$\lambda=\frac{h}{\sqrt{2 m K}}$<br>where, $m$ is the mass of the particle and $h$ is Planck's constant.<br>Let at wavelength $\lambda^{\prime}$, kinetic energy of particle becomes $K^{\prime}=\frac{K}{4}$<br>[from Eq. (i)] | ["$2 \\lambda$", "$\\frac{\\lambda}{2}$", "$4 \\lambda$", "$\\lambda$"] | [0] | null | PYQ |
5b03d43fdf9c3376489cdd5d7cfc3e07 | WBJEE_PHY | Dual Nature of Matter | Light of energy $E$ falls normally on a metal of work function $\frac{E}{3}$. The kinetic energies $K$ of the photo electrons are | singleCorrect | 1 | Given, work function, $\phi_0=E / 3$
According to Einstein's photoelectric equation,
$K_{\max }=E-\phi_0=E-\frac{E}{3}$
$\Rightarrow \quad K_{\max }=\frac{2 E}{3}$
$\therefore$ Kinetic energy lies between 0 to $2 E / 3$. | ["$K=\\frac{2 E}{3}$", "$K=\\frac{E}{3}$", "$0 \\leq K \\leq \\frac{2 E}{3}$", "$0 \\leq K \\leq \\frac{E}{3}$"] | [2] | null | PYQ |
9170fae14e9f7560c371618d38f4f717 | WBJEE_PHY | Dual Nature of Matter | Find the frequency of light which ejects electron from a metal surface fully stopped by a retarding potential of 3 V . The photoelectric effect begins in this metal at a frequency of $6 \times 10^{15} \mathrm{~Hz}$. | singleCorrect | 1 | According to Einstein's photoelectric equation,
$E=h v-w$
If $V_s$ is retarding or stopping potential and $v_0$ is the threshold frequency, then the above equation becomes
$\begin{aligned} e V_s & =h v-h v_0 \\ h v & =e V_s+h v_0 \\ v & =\frac{e V_s}{h}+v_0\end{aligned}$
Hence, $e=1.6 \times 10^{-19} \mathrm{C}$
$V_s=3... | ["$1.324 \\times 10^{15} \\mathrm{~Hz}$", "$2.295 \\times 10^{16} \\mathrm{~Hz}$", "$3.678 \\times 10^{18} \\mathrm{~Hz}$", "$2.7 \\times 10^{14} \\mathrm{~Hz}$"] | [0] | null | PYQ |
06f06eca19a632958510a6f1100390e0 | WBJEE_PHY | Dual Nature of Matter | The wavelength of incident light falling on a photosensitive surface is changed from $2000 Å$ to $2100 Å$. The corresponding change in stopping potential is | singleCorrect | 3 | Given, $\begin{aligned} \lambda_1=2000 Å & =2000 \times 10^{-10} \mathrm{~m} \\ & =2 \times 10^{-7} \mathrm{~m}\end{aligned}$
$\lambda_2=2100 Å$
$=2.1 \times 10^{-7} \mathrm{~m}$
$\therefore \quad \frac{h c}{\lambda_1}=W+e V_0\ldots$ (i)
and$\frac{h c}{\lambda_2}=W+e V_0^{\prime}\ldots$ (ii)
Subtracting Eq. (ii) from E... | ["0.03 V", "0.3 V", "3 V", "3.3 V"] | [1] | null | PYQ |
4c57a6e297a847660d1751bf809867f3 | WBJEE_PHY | Dual Nature of Matter | In a photoemissive cell with exciting wave length $\lambda$, the fastest electron has a speed $v$. If the exciting wavelength is changed to $3 \lambda / 4$, then the speed of the fastest emitted electron will be | singleCorrect | 2 | From Einstein's photoelectric equation,
$\frac{1}{2} m v^2=\frac{h c}{\lambda}-W\ldots$ (i)
and $\quad \frac{1}{2} m v^{\prime 2}=\frac{h c}{3 \lambda / 4}-W\ldots$ (ii)
On dividing Eq. (ii) by Eq. (i), we get
$\left(\frac{v^{\prime}}{v}\right)^2=\frac{\frac{4 h c}{3 \lambda}-W}{\frac{h c}{\lambda}-W}$
$\begin{array}{l... | ["$\\left(\\frac{3}{4}\\right)^{\\frac{1}{2}}$", "$v\\left(\\frac{4}{3}\\right)^{\\frac{1}{2}}$", "less than $v\\left(\\frac{4}{3}\\right)^{\\frac{1}{2}}$", "greater than $v\\left(\\frac{4}{3}\\right)^{\\frac{1}{2}}$"] | [3] | null | PYQ |
98bde0fa95e2913f6fe3f8b24d03e41b | WBJEE_PHY | Dual Nature of Matter | The photoelectric cut-off voltage in an experiment was found to be 1.5 V . The work function for the material used in the experiment was 4.2 eV . The maximum kinetic energy of the photoelectrons that emitted was | singleCorrect | 1 | If stopping potential is $V_0$, then maximum kinetic energy of photoelectrons is given by
$E_k=e V$
Given, $V=1.5$ volt
$E_k=1.5 \mathrm{eV}$ | ["1.5 eV", "2.7 eV", "4.2 eV", "5.7 eV"] | [0] | null | PYQ |
a03e8616b03778c5e7cefc9fc0e32275 | WBJEE_PHY | Dual Nature of Matter | Photons of 5.5 eV energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 4.0 eV . The stopping voltage required for these electrons is | singleCorrect | 2 | Max. kinetic energy of photoelectrons $=4 \mathrm{eV}$
If stopping potential is $V_0$, then
$\begin{aligned}
& & e V_0 & =4 \mathrm{eV} \\
& & V_0 & =4 \mathrm{volt}
\end{aligned}$ | ["5.5 V", "1.5 V", "9.5 V", "4.0 V"] | [3] | null | PYQ |
4fb999046f871568f7055ddb8e8f47b6 | WBJEE_PHY | Dual Nature of Matter | If $\lambda$ is the incident wavelength and $\lambda_0$ is the threshold wavelength for a metal surface, photoelectric effect takes place only, if | singleCorrect | 1 | For photoelectric effect to take place,' the incident wavelength $(\lambda)$ should be equal to or smaller than the threshold wavelength $\left(\lambda_0\right)$ i.e., \(\lambda \leq \lambda_0\) | ["$\\lambda \\leq \\lambda_0$", "$\\lambda \\geq \\lambda_0$", "$\\lambda \\geq 2 \\lambda_0$", "None of these"] | [0] | null | PYQ |
d29f3e551e88889ab7e7316e9526257d | WBJEE_PHY | Dual Nature of Matter | An electron, helium ion $\left(\mathrm{He}^{++}\right)$and proton having the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda_\theta$, $\lambda_{\mathrm{He}^{++}}$and $\lambda_p$ is | singleCorrect | 1 | de-Broglie wavelength is given by
$\begin{array}{l}
\lambda =\frac{h}{p}=\frac{h}{\sqrt{2 m k}} \\
\lambda \propto \frac{1}{\sqrt{m}} \quad\left(\because k_e=k_p=k_{\mathrm{He}^{++}}\right) \\
\text {As, } m_{\mathrm{He}^{++}} \gt m_p \gt m_\theta \\
\therefore \lambda_{\mathrm{He}^{++}} \lt \lambda_p \lt \lambda_\th... | ["$\\lambda_e\\gt\\lambda_p\\gt\\lambda_{\\mathrm{He}^{++}}$", "$\\lambda_e\\gt\\lambda_{\\mathrm{He}^{++}}\\gt\\lambda_\\rho$", "$\\lambda_e \\lt \\lambda_p \\lt \\lambda_{\\mathrm{He}^{++}}$", "$\\lambda_e \\lt \\lambda_{\\mathrm{He}^{++}}=\\lambda_p$"] | [0] | null | PYQ |
80079a605f88698eb58c43ea08cc1f07 | WBJEE_PHY | Dual Nature of Matter | The maximum velocity of the photoelectron emitted by the metal surface is 'v'. Charge and mass of the photoelectron is denoted by 'e' and 'm' respectively. The
stopping potential in volt is | singleCorrect | 1 | $\frac{1}{2} m v^{2}=e v$
$v=\frac{m v^{2}}{2 e}=\frac{v^{2}}{2\left(\frac{e}{m}\right)}$ | ["$\\frac{v^{2}}{\\left(\\frac{m}{e}\\right)}$", "$\\frac{v^{2}}{\\left(\\frac{e}{m}\\right)}$", "$\\frac{v^{2}}{2\\left(\\frac{m}{e}\\right)}$", "$\\frac{v^{2}}{2\\left(\\frac{e}{m}\\right)}$"] | [3] | null | PYQ |
3120df70937529a0dbde4c8bf4778a7a | WBJEE_PHY | Dual Nature of Matter | An electron and photon are accelerated through the same potential difference. The ratio of the de-Broglie wavelength $\lambda_{\mathrm{p}}$ to $\lambda_{\mathrm{e}}$ is $\left[\mathrm{m}_{\mathrm{e}}=\right.$ mass of electron, $\mathrm{m}_{\mathrm{p}}=$ mass of proton $]$ | singleCorrect | 1 | Both will have same kinetic energy $\mathrm{k}$. Momentum of electron $\quad P_{\mathrm{e}}=\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{k}}$ Momentum of proton $P_{p}=\sqrt{2 m_{p} k}$
$\begin{array}{l}
\lambda_{p}=\frac{h}{p_{p}}, \lambda_{e}=\frac{h}{p_{e}} \\
\therefore \frac{\lambda_{p}}{\lambda_{e}}=\frac{P_{e}}{P_{p... | ["$\\left(\\frac{m_{p}}{m_{e}}\\right)^{\\frac{1}{2}}$", "$\\left(\\frac{m_{e}}{m_{p}}\\right)^{\\frac{1}{2}}$", "$\\left(\\frac{m_{e}}{m_{p}}\\right)$", "$\\left(\\frac{m p}{m}\\right)$"] | [1] | null | PYQ |
10158e69897a1babe4bbed06bf4c9454 | WBJEE_PHY | Dual Nature of Matter | In experiment of photoelectric effect, the stopping potential for a given metal is
${ }^{\prime} \mathrm{V}_{0}{ }^{\prime}$ volt, when radiation of wavelength ${ }^{\prime} \lambda_{0}{ }^{\prime}$ is used. If radiation of wavelength ${ }^{\prime} 2 \lambda_{0}{ }^{\prime}$
is used for the same metal, then the stoppin... | singleCorrect | 2 | $h v-\omega_{0}=e V_{0}$
$\frac{h c}{\lambda_{0}}-\omega_{0}=e V_{0}$
$\frac{h c}{2 \lambda_{0}}-\omega_{0}=e V_{s}$
$=\frac{h c}{\lambda_{0}}\left[1-\frac{1}{2}\right]=e\left(V_{0}-V_{s}\right)=e V_{0}-e V_{s}$
$V_{s}=V_{0}-\frac{h c}{2 \lambda_{0} e}$ | ["$\\mathrm{V}_{0}+\\frac{\\mathrm{hc}}{2 \\mathrm{e} \\lambda_{0}}$", "$\\mathrm{~V}_{0}-\\frac{\\mathrm{hc}}{2 \\mathrm{e} \\lambda_{0}}$", "$\\frac{V_{0}}{2}$", "$2 \\mathrm{~V}_{0}$"] | [1] | null | PYQ |
4fe9ecdd93c288174fee6ca2905ceed8 | WBJEE_PHY | Dual Nature of Matter | When a photosensitive surface is irradiated by lights of wavelengths $\lambda_{1}$ and $\lambda_{2}$, kinetic energies of emitted photoelectrons are $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$ respectively. The work function of the photosensitive surface is | singleCorrect | 1 | Using photoclutic effect
\(\frac{h c_{1}}{\lambda_{1}}=\phi+k_{1}-(D)\) ...(i)
and \(\frac{h_{1}}{r_{2}}=\phi k k_{2}\) ...(ii)
form (i) and (i)
\(\frac{\frac{h n_{1}}{\lambda_{1}}}{\frac{h_{2}}{\lambda_{2}}}=\frac{\min (\eta) \text { an }}{\phi+k_{1}}\)
\(\Rightarrow \frac{1}{d_{1}} \times \frac{d_{2}}{1}=\frac{\phi+... | ["$\\frac{\\lambda_{2} E_{2}-\\lambda_{1} E_{1}}{\\lambda_{2}-\\lambda_{1}}$", "$\\frac{\\lambda_{1} E_{1}+\\lambda_{2} E_{2}}{\\lambda_{2}+\\lambda_{1}}$", "$\\frac{\\lambda_{1} E_{1}-\\lambda_{2} E_{2}}{\\lambda_{2}-\\lambda_{1}}$", "$\\frac{\\lambda_{2} E_{1}+\\lambda_{2} E_{2}}{\\lambda_{2}-\\lambda_{1}}$"] | [2] | null | PYQ |
a082c0d93f977051a6d3d51aaa34efdb | WBJEE_PHY | Dual Nature of Matter | Light of incident frequency 2 times the threshold frequency is incident on a photosensitive material. If the incident frequency is made $\left(\frac{1}{3}\right)^{\text {rd }}$ and intensity is doubled then the photoelectric current will | singleCorrect | 2 | No Solution | ["decrease.", "increase.", "be halved.", "be zero."] | [3] | null | PYQ |
e076f727a816a8d9aece067f2ed88d4c | WBJEE_PHY | Dual Nature of Matter | Energy of the incident photon on the metal surface is ' $3 \mathrm{~W}$ ' and then ' $5 \mathrm{~W}$ ', where 'W' is the work function for that metal. The ratio of velocities of emitted photoelectrons is | singleCorrect | 1 | $$
\begin{aligned}
& \frac{1}{2} m v_{1}^{2}=3 W-W=2 W \\
& \frac{1}{2} m v_{2}^{2}=5 W-W=4 W \\
& \frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{2} \\
\therefore & \frac{v_{1}}{v_{2}}=\frac{1}{\sqrt{2}}
\end{aligned}
$$ | ["$1: \\sqrt{2}$", "$1: 1$", "$1: 2$", "$1: 4$"] | [0] | null | PYQ |
bbfc7694f7758a8f07977510c323c371 | WBJEE_PHY | Dual Nature of Matter | The light of wavelength $\lambda^{\prime}$ incident on the surface of metal having work function $\phi$ emits the electrons. The maximum velocity of electrons emitted is $(c=$ velocity of light, h=Planck's constant, m=mass of electron $)$ | singleCorrect | 1 | (C)
$\mathrm{hv}-\phi=\mathrm{E}_{\max }=\frac{1}{2} \mathrm{mv}^{2}$
$\frac{\mathrm{hc}}{\lambda}-\phi=\frac{1}{2} \mathrm{mv}^{2}$
$\mathrm{v}^{2}=2 \frac{(\mathrm{hc}-\lambda \phi)}{\lambda \mathrm{m}}$ | ["$\\left[\\frac{2(\\mathrm{~h} \\mathrm{C}-\\phi)}{\\mathrm{m} \\lambda}\\right]$", "$\\left[\\frac{2(\\mathrm{~h} \\mathrm{c}-\\lambda \\phi)}{\\mathfrak{m \\lambda}}\\right]^{\\frac{1}{2}}$", "$\\left[\\frac{2(h \\mathrm{C}-\\lambda)}{m \\lambda}\\right]^{\\frac{1}{2}}$", "$\\left[\\frac{2(\\mathrm{~h} \\mathrm{v}-\... | [1] | null | PYQ |
380a89335fa32ca6b02662f880d6ad20 | WBJEE_PHY | Dual Nature of Matter | The threshold wavelengths for photoelectric emission from two metals $\mathrm{A}$ and $\mathrm{B}$
are $400 \mathrm{~nm}$ and $800 \mathrm{~nm}$ respectively. The ratio of their work functions, $\phi_{A}$ to $\phi$ ?
is | singleCorrect | 2 | (D)
work function $\phi=\frac{h c}{\lambda_{0}}$
$\therefore \frac{\phi_{B}}{\phi_{A}}=\frac{\lambda_{B}}{\lambda_{A}}=\frac{800}{400}=2$ | ["$\\frac{1}{2}$", "4", "$\\frac{1}{4}$", "2"] | [3] | null | PYQ |
0e542cda0120f30c6f9b57c3c5d9eb25 | WBJEE_PHY | Dual Nature of Matter | If the maximum kinetic energy of emitted electrons in photoelectric effect is $3 \cdot 2 \times 10^{-19} \mathrm{~J}$ and the work function for metal is $6 \cdot 63 \times 10^{-19} \mathrm{~J}$, then stopping
potential and threshold wavelength respectively are
[Planck's constant $\left.\mathrm{h}=6.63 \times 10^{-34} \... | singleCorrect | 2 | (D)
$\begin{array}{l}
(\mathrm{k} \cdot \mathrm{E})_{\max }=\mathrm{eV}_{\mathrm{s}}=3.2 \times 10^{-19} \mathrm{~J} \\
\therefore \mathrm{V}_{\mathrm{s}}=\frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}}=2 \mathrm{~V}
\end{array}$
In the given options, only option (D) has $\mathrm{V}_{\mathrm{s}}=2 \mathrm{~V}$ | ["$3 \\mathrm{~V}, 4000 \u00c5$", "$4 \\mathrm{~V}, 6000 \u00c5$", "$1 \\mathrm{~V}, 1000 \u00c5$", "$2 \\mathrm{~V}, 3000 \u00c5$"] | [3] | null | PYQ |
4cc2b4cd8285590c9f53247633713a53 | WBJEE_PHY | Dual Nature of Matter | Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $\lambda_{1}=360 \mathrm{~nm}$ and $\lambda_{2}=600 \mathrm{~nm} .$ What is the ratio of work functions
for lights of wavelength ${ }^{\prime} \lambda_{1}{ }^{\prime}$ to $\lambda_{2}{ }^{\prime}$? | singleCorrect | 1 | $\frac{h c}{\lambda_{1}}-\phi_{1}=\mathrm{KE} \quad \quad \lambda_{1}=360 \mathrm{~nm}=360 \times 10^{-9} \mathrm{~m}$
$\frac{\mathrm{hc}}{\lambda_{2}}-\phi_{2}=\mathrm{KE} \quad \quad\lambda_{2}=600 \mathrm{~nm}=600 \times 10^{-9} \mathrm{~m}$
$\frac{h c}{\lambda_{1}}=\phi_{1} \quad \quad \phi \propto \frac{1}{\lambda... | ["$6: 1$", "$1: 6$", "$5: 3$", "$3: 5$"] | [2] | null | PYQ |
c00486342114c0b58bd8939128b354a7 | WBJEE_PHY | Dual Nature of Matter | Two incident radiations having energies two times and ten times of the work function of a metal surface, produce photoelectric effect. The ratio of maximum velocities of emitted photo electrons respectively is | singleCorrect | 1 | $\frac{1}{2} m v_{1}^{2}=2 \phi_{0}-\phi_{0}$
$\frac{1}{2} m v_{2}^{2}=10 \phi_{0}-\phi_{0}$
$\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{9}$
$\therefore \frac{v_{1}}{v_{2}}=\frac{1}{3}$ | ["$3: 2$", "$1: 3$", "$2: 3$", "$1: 2$"] | [1] | null | PYQ |
f64eebc1c694ff25b20e594d7b936b4a | WBJEE_PHY | Dual Nature of Matter | Photoelectrons are emitted when photons of energy $4.2 \mathrm{eV}$ are incident on a photosensitive metallic sphere of radius $10 \mathrm{~cm}$ and work function $2.4 \mathrm{eV}$. The number of photoelectrons emitted before the emission is stopped is
$$
\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text { SI unit; ... | singleCorrect | 2 | $$
\mathrm{h} v=4.2 \mathrm{eV}, \phi=2.4 \mathrm{eV},(\mathrm{KE})_{\max }=4.2-2.4=1.8 \mathrm{eV}
$$
The electrons will not be able to escape from the surface when its potential becomes $1.8 \mathrm{~V}$.
$$
\begin{aligned}
& \mathrm{V}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}} \\
& 1.8=9 \times 10^9 \... | ["$1.25 \\times 10^6$", "$1.25 \\times 10^8$", "$1.25 \\times 10^2$", "$1.25 \\times 10^4$"] | [1] | null | PYQ |
a9d66f6ab87e25ab504df930493a2fdd | WBJEE_PHY | Dual Nature of Matter | 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 | 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 $\begin{aligned} v(\text { infrared }) & =\frac{3 \times 10^8}{10,000 \times 10^{-10}} \\ & =3 \times 10^{14} \mathrm{~Hz}\end{aligned}$
Wavele... | ["50 watt infrared lamp", "100 watt red neon lamp", "60 watt sodium lamp", "5 watt ultraviolet lamp"] | [2, 3] | null | PYQ |
6c508627dcdb088ae7304a3b5b1bec0a | WBJEE_PHY | Dual Nature of Matter | 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 | 1 | 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 |
fb60ef8c3c18f60d75cd79f18d2ef9fe | WBJEE_PHY | Dual Nature of Matter | Figure 28.14 shows the stopping potential $V_0$ versus frequency $v$ for photoelectric emission from two metals $A$ and $B$. Choose the correct statement(s) from the following.
<img src="https://cdn-question-pool.getmarks.app/modules/ms/wbjee/_IPNGlq2P3K4iovHhduRktUf4RKn9hEUxAcbLz2cktc.original.fullsize.png"/><br/> | multipleCorrect | 2 | $h v=h v_0+e V_0 \text { or } V_0=\frac{h}{e} v-\frac{h}{e} v_0=\frac{h}{e}\left(v-v_0\right)$
Thus $V_0=0$ if $v=v_0$. If follow from graphs in Fig. 28.14 that the intercept $v_0$ on $v$-axis is less for metal $A$ than for metal $B$. Hence choice (d) is correct. Also since $W_0=h v_0$, the work function of metal $A$ ... | ["Work function of $A$ is greater than that of $B$.", "Work function of $B$ is greater than that of $A$.", "Threshold frequency of $A$ is greater than that of $B$.", "Threshold frequency of $B$ is greater than that of $A$"] | [1, 3] | null | PYQ |
f17e3714e497df6efb4cd021ae156769 | WBJEE_PHY | Dual Nature of Matter | Consider white light whose wavelength spread is from 400 nm to 700 nm . Its energy is uniformly distributed in this spectrum (i.e. $\frac{d E}{d \lambda}=$ constant, where $d E$ is the energy of the photons wavelengths between $\lambda$ and $\lambda+d \lambda$ ). The light is
incident on a metal $A$ of work function 1.... | multipleCorrect | 3 | Let $d n$ be number of photons from $\lambda$ to $\lambda+d \lambda$
$\begin{aligned}
& d E=d n \frac{(h c)}{\lambda} \\
& \frac{d E}{d \lambda}=k \\
& \int_0^n(d n) h c=k \int_{400}^{\lambda_{t n}} \lambda d \lambda
\end{aligned}$
For $A: \lambda_{t n}=800 \mathrm{~nm}$
For $B: \lambda_{t n}=500 \mathrm{~nm}$
$\begin... | ["Stopping potential for experiment with metal $A$ is 0.22 V", "Stopping potential for experiment with metal $B$ is 0.62 V", "Saturation photo current for metal $B$ will be 2 mA", "Saturation photo current for metal $B$ will be $\\frac{18}{11} \\mathrm{~mA}$"] | [1, 3] | null | PYQ |
fdab4097dd4fe13705b9fbb616a904de | WBJEE_PHY | Dual Nature of Matter | 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 |
2721383af5a45cff4b09001f8508032a | WBJEE_PHY | Dual Nature of Matter | Figure 28.16 shows graphs between cut-off voltage $V_0$ and $\frac{1}{\lambda}$ for three metals 1,2 and 3 , where $\lambda$ is the wavelength of the incident radiation in nm . If $W_1, W_2$ and $W_3$ are the work functions of metals 1,2 and 3 respectively, then
<img src="https://cdn-question-pool.getmarks.app/modules/... | multipleCorrect | 3 | Work function $W=h v_0=\frac{h c}{\lambda_0}$, where $\lambda_0$ is the threshold wavelength. Hence
$W_1: W_2: W_3=\frac{h c}{\left(\lambda_0\right)_1}: \frac{h c}{\left(\lambda_0\right)_2}: \frac{h c}{\left(\lambda_0\right)_3}$
$\begin{aligned} & =\frac{1}{\left(\lambda_0\right)_1}: \frac{1}{\left(\lambda_0\right)_2}:... | ["$W_1: W_2: W_3=1: 2: 4$", "$W_1: W_2: W_3=4: 2: 1$", "The graphs for metals 1,2 and 3 are parallel to each other and the slope of each graph is $h c / e$, where $h=$ Planck's contant, $c=$ speed of light and $e=$ charge of an electron.", "Ultraviolet light will eject photoelectrons from metals 1 and 2 and not from me... | [0, 3] | null | PYQ |
2a9e18beee6835f4ae7ac9a43acd755d | WBJEE_PHY | Dual Nature of Matter | When photons of energy 4.25 eV strike the surface of a metal A , the ejected photoelectrons have maximum kinetic energy, $\mathrm{T}_{\mathrm{A}}$ expressed in eV and de Broglie wavelength $\lambda_{\mathrm{A}}$. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is... | multipleCorrect | 2 | $\begin{aligned} & h v=\mathrm{K} . \mathrm{E} \cdot(T)+\text { work function }(\mathrm{W}) \\ & \Rightarrow \mathrm{hv}=\mathrm{T}+\mathrm{W} \\ & \Rightarrow 4.25 \mathrm{eV}=\mathrm{T}_{\mathrm{A}}+\mathrm{W}_{\mathrm{A}}(\text { for Metal } A) \\ & \Rightarrow 4.70 \mathrm{eV}=\mathrm{T}_{\mathrm{B}}+\mathrm{W}_{\m... | ["the work function of A is 2.25 eV", "the work function of B is 4.20 eV", "$\\mathrm{T}_{\\mathrm{A}}=2.00 \\mathrm{eV}$", "$\\mathrm{T}_{\\mathrm{B}}=2.75 \\mathrm{eV}$"] | [0, 1, 2] | null | PYQ |
dead4752059964819738b674a0295e7c | WBJEE_PHY | Dual Nature of Matter | When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are respectively 0.6 volt and 18.0 mA . If the same source is placed 0.6 m away from the photoelectric cell, then | multipleCorrect | 2 | The cut off potential remains the same as long as the frequency of the incident light remains unchanged.
The saturation current is proportion to the intensity of light. Since the distance has become 3 times its previous value, the intensity is only $\frac{1}{3^2}$ or $\frac{1}{9}$ of its previous value. Hence the new ... | ["the stopping potential will be 0.2 volt", "the stopping potential will be 0.6 volt", "the saturation current will be 6.0 mA", "the saturation current will be 2.0 mA ."] | [1, 3] | null | PYQ |
e462f331414110e1b3a15175b94a1325 | WBJEE_PHY | Dual Nature of Matter | When photoelectrons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy $\mathrm{T}_{\mathrm{A}} \mathrm{eV}$ and de-Broglie wavelength $\lambda_{\mathrm{A}}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photons of energy 4.70 eV i... | multipleCorrect | 1 | $\begin{aligned}
& \mathrm{p}_{\mathrm{A}}=\frac{\mathrm{h}}{\lambda_{\mathrm{A}}}, \mathrm{~T}_{\mathrm{A}}=\frac{\mathrm{P}_{\mathrm{A}}^2}{2 \mathrm{~m}}=\frac{\mathrm{h}^2}{2 \mathrm{~m} \ell_{\mathrm{A}}^2}, \mathrm{~T}_{\mathrm{B}}=\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda_{\mathrm{B}}^2}=\frac{\mathrm{h}^2}{8 \m... | ["work function of A is 2.25 eV", "work function of B is 4.20 eV", "$\\mathrm{T}_{\\mathrm{A}}=2.00 \\mathrm{eV}$", "$\\mathrm{T}_{\\mathrm{B}}=2.75 \\mathrm{eV}$"] | [0, 1, 2] | null | PYQ |
5a6ceba4c458a622ef3a99125b8d55eb | WBJEE_PHY | Dual Nature of Matter | Photoelectric effect supports the quantum nature of light because: | multipleCorrect | 1 | Existence of cut-off frequency and photoemission takes place even when intensity is low. | ["There is a minimum frequency of light below which no photoelectrons are emitted.", "The maximum K.E. of photoelectrons depends only on the frequency of light and not on its intensity.", "Even when the metal surface is faintly illuminated by light of wavelength less than the threshold wavelength, the photoelectrons le... | [0, 1, 2] | null | PYQ |
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