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Last Updated on November 13, 2026 by XAM CONTENT
NSEP 2025 Solved Paper
Practice the complete NSEP 2025 Question Paper with Solutions for Physics. All 60 questions are arranged in the original order with the supplied answer key, diagrams and worked solutions. The paper begins with equation-based reasoning such as the condition \(\mathrm{V}=0\), so attempt each question first and then open Show Solution.
Exam
NSEP 2025
Subject
Physics
Paper Code
62
Date
23 November 2025
Maximum Marks
216
Time
2 Hours
Paper Format
Questions 1–45 are single-correct questions. Questions 46–60 may have more than one correct option. The answer choices and solution order below follow the supplied NSEP 2025 source.
Quick Check
NSEP 2025 Answer Key — Paper Code 62
Use the key for quick checking before opening the detailed solutions.
Questions 1–15
| Question | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Answer | c | b | c | b | a | b | c | b | d | d | c | c | d | c | c |
Questions 16–30
| Question | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Answer | a | d | d | b | d | b | a | d | b | d | c | a | c | b | a |
Questions 31–45
| Question | 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Answer | a or d | b | a | b | c | c | c | b | d | c | d | c | d | d | b |
Questions 46–60
| Question | 46 | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Answer | a | a | c | a,b,d | a,c,d | a,c | a,b,c | a,b,c,d | b,d | b,c | a,d | a,d | b,c,d | a,b,c | a,b,d |
Solved Paper
NSEP 2025 Question Paper with Detailed Solutions
60 questions with source diagrams and expandable worked solutions.
60
Questions
Questions 1–15
NSEP 2025 Physics — Paper Code 62
1
Two electric charges, +q at the origin O \((0,0)\) and -2q at the point A \((6,0)\) are placed on x axis. The locus of the point P in x-y plane where the potential vanishes \((\mathrm{V}=0)\) is
Show Solution
Solution
Ans. (c)
\(\mathrm{V}_{(\mathrm{x}, \mathrm{y})}=0=\frac{\mathrm{kq}}{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}}-\frac{\mathrm{k} 2 \mathrm{q}}{\sqrt{(\mathrm{x}-6)^{2}+\mathrm{y}^{2}}}\)
\[
\begin{aligned}
& \Rightarrow(\mathrm{x}-6)^{2}+\mathrm{y}^{2}=4\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) \\
& \Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+36-12 \mathrm{x}=4 \mathrm{x}^{2}+4 \mathrm{y}^{2} \\
& \Rightarrow 3 \mathrm{x}^{2}+3 \mathrm{y}^{2}+12 \mathrm{x}-36=0 \\
& \Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+4 \mathrm{x}-12=0
\end{aligned}
\]
\(\Rightarrow(\mathrm{x}+2)^{2}+\mathrm{y}^{2}=16 \ldots \ldots\). Equation of circle.
2
In the circuit shown, the Zener diode is an ideal one with breakdown voltage of 5.0 volt. The values of the resistance are \(\mathrm{R}_{\mathrm{S}}=10 \mathrm{k} \Omega\) and \(\mathrm{R}_{\mathrm{L}}=1 \mathrm{k} \Omega\). The current through the resistances, when the supply voltage is 11.0 V, is
Show Solution
Solution
Ans. (b)
Let Zener doesn’t breakdown
\[
\begin{aligned}
& \therefore \mathrm{i}=\frac{11}{11} \mathrm{~mA}=1 \mathrm{~mA} \\
& \therefore \mathrm{~V}_{\text {Zener }}=1 \mathrm{~V}<\mathrm{V}_{\text {breakdown }}
\end{aligned}
\]
∴ Zener diode will behave like open switch \& current in \(\mathrm{R}_{\mathrm{L}}=\) current in \(\mathrm{R}_{\mathrm{S}}=1 \mathrm{~mA}\)
3
In an accelerator the electrons are accelerated up to an energy of 50 MeV. The electrons do not emerge continuously from the accelerator rather they come in pulses at time interval of 5.0 milliseconds. Each pulse has a much shorter duration of 200 nanoseconds. Electron current during the pulse is 100 mA, while the current is zero between the two successive pulses (see figure), then
Show Solution
Solution
Ans. (c)
\(\mathrm{V}_{\text {acc }}=50 \mathrm{MV}\)
\(\mathrm{i}=100 \mathrm{~mA}\) during the time in which \(\mathrm{e}^{-}\)emerges.
\[
\begin{aligned}
& \therefore \mathrm{i}_{\text {avg }}=\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{200 \times 10^{-9} \times 100 \times 10^{-3}}{5 \times 10^{-3}}=4 \mu \mathrm{~A} \\
& \mathrm{P}_{\text {avg }}=\mathrm{V}_{\text {i avg }}=50 \times 10^{6} \times 4 \times 10^{-6}=200 \mathrm{~W} \\
& \mathrm{P}_{\text {peak }}=\mathrm{V}_{\text {i max }}=50 \times 10^{6} \times 100 \times 10^{-3}=5 \mathrm{MW}
\end{aligned}
\]
4
Two infinitely long straight parallel wires perpendicular to the plane of the paper are 5 m apart. One of the wires, P carries current I out of the plane of the paper and the other, Q carries the current I into the plane of paper. The magnetic field B at the origin O of the coordinate system with x and y axes as perpendicular and parallel to PQ, respectively, is [Given \(\mathrm{OP}=4 \mathrm{~m}\) and \(\mathrm{OQ}=3 \mathrm{~m}\) ]
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Solution
Ans. (b)
\[
\mathrm{PM}=\frac{16}{5} \mathrm{~m} \& \mathrm{QM}=\frac{9}{5} \mathrm{~m}
\]
\[
\begin{aligned}
& \vec{B}_{P}=\frac{\mu_{0} I}{8 \pi}\left[\frac{4}{5} \hat{i}+\frac{3}{5} \hat{j}\right] \\
& \vec{B}_{Q}=\frac{\mu_{0} I}{6 \pi}\left[\frac{3}{5} \hat{i}-\frac{4}{5} \hat{j}\right] \\
& \therefore \vec{B}_{\text {net }}=\frac{\mu_{0} I}{\pi}\left[\frac{1}{10} \hat{i}+\frac{3}{40} \hat{j}+\frac{1}{10} \hat{i}-\frac{2}{15} \hat{j}\right] \\
& =\frac{\mu_{0} I}{\pi}\left[\frac{1}{5} \hat{i}-\frac{7}{120} \hat{j}\right]=\frac{\mu_{0} I}{5 \pi}\left[\hat{i}-\frac{7}{24} \hat{j}\right]
\end{aligned}
\]
5
Charge q is uniformly distributed over the surface of a thin non-conducting annular disc of inner radius \(\mathrm{R}_{1}\) and outer radius \(\mathrm{R}_{2}\). The disc is made to rotate with constant frequency f, about an axis passing through the center of the annular disc and perpendicular to its plane. The magnetic moment of the disc is
Show Solution
Solution
Ans. (a)
\[
\sigma=\frac{\mathrm{q}}{\pi\left(\mathrm{R}_{2}^{2}-\mathrm{R}_{1}^{2}\right)}
\]
For a small ring of radius x and width ‘dx’
\[
\begin{aligned}
\mathrm{dM} & =\mathrm{IA} \\
& =f\,dq \mathrm{~A} \\
& =\mathrm{f} \sigma \mathrm{dA} \times \pi \mathrm{x}^{2} \\
& =\frac{\mathrm{q}}{\mathrm{R}_{2}^{2}-\mathrm{R}_{1}^{2}} \int_{\mathrm{R}_{1}^{2}}^{\mathrm{R}_{2}} \mathrm{x}^{3} \mathrm{dx} \\
& =\frac{2 \pi f q}{\mathrm{R}_{2}^{2}-\mathrm{R}_{1}^{2}} \times\left(\frac{\mathrm{R}_{2}^{4}}{4}-\frac{\mathrm{R}_{1}^{4}}{4}\right) \\
& =\frac{2 \pi f q\left(\mathrm{R}_{2}^{2}+\mathrm{R}_{1}^{2}\right)}{4\left(\mathrm{R}_{2}^{2}-\mathrm{R}_{1}^{2}\right)}\left(\mathrm{R}_{2}^{2}-\mathrm{R}_{1}^{2}\right) \\
& =\frac{\pi f q\left(\mathrm{R}_{2}^{2}+\mathrm{R}_{1}^{2}\right)}{2}
\end{aligned}
\]
(a) is correct
6
For a resistance R and capacitance C in series, the impedance is twice that of a parallel combination of the same elements when used with an AC voltage frequency f. The frequency f of the applied emf is
Show Solution
Solution
Ans. (b)
in series
\[
Z_{s}=\sqrt{R^{2}+\left(\frac{1}{\omega C}\right)^{2}} \tag{i}
\]
In parallel
\[
\frac{1}{\mathrm{Z}_{\mathrm{p}}}=\sqrt{\frac{1}{\mathrm{R}^{2}}+(\omega \mathrm{C})^{2}} \tag{i}
\]
here \(\mathrm{Z}_{\mathrm{S}}=2 \mathrm{Z}_{\mathrm{P}}\)
\[
\begin{aligned}
& \mathrm{R}^{2}+\frac{1}{4 \pi \mathrm{f}^{2} \mathrm{C}^{2}}=4 \frac{1}{\left(\frac{1}{\mathrm{R}^{2}}+4 \pi^{2} \mathrm{f}^{2} \mathrm{C}^{2}\right)} \\
& \left(\mathrm{R}^{2}+\frac{1}{4 \pi^{2} \mathrm{f}^{2} \mathrm{C}^{2}}\right)\left(\frac{1}{\mathrm{R}^{2}}+4 \pi^{2} \mathrm{f}^{2} \mathrm{C}^{2}\right)=4 \\
& 1+1+4 \pi \mathrm{f}^{2} \mathrm{R}^{2} \mathrm{C}^{2}+\frac{1}{4 \pi^{2} \mathrm{f}^{2} \mathrm{C}^{2} \mathrm{R}^{2}}=4 \\
& 4 \pi \mathrm{f}^{2} \mathrm{R}^{2} \mathrm{C}^{2}+\frac{1}{4 \pi^{2} \mathrm{f}^{2} \mathrm{C}^{2} \mathrm{R}^{2}}=2
\end{aligned}
\]
On solving
\[
\begin{aligned}
& 4 \pi^{2} f^{2} R^{2} C^{2}=1 \\
& f=\frac{1}{2 \pi R C}
\end{aligned}
\]
(b) is correct.
7
In an experiment on photoelectric effect on a metal surface, one finds a stopping potential of 1.8 V for the wavelength of 300 nm and a stopping potential of 0.9 V for the wavelength of 400 nm . The cutoff wavelength \(\lambda_{0}\) (the maximum wavelength that can produce photoelectric effect) for the metal is
Show Solution
Solution
Ans. (c)
from \(\mathrm{eV}_{0}=\mathrm{hC}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)\)
\[
1.8 \mathrm{eV}=\mathrm{hC}\left(\frac{1}{300 \mathrm{~nm}}-\frac{1}{\lambda_{0}}\right) \tag{i}
\]
\[
0.9 \mathrm{eV}=\mathrm{hC}\left(\frac{1}{400 \mathrm{~nm}}-\frac{1}{\lambda_{0}}\right) \tag{ii}
\]
from (i) \& (ii)
\[
\begin{aligned}
& 2\left(\frac{1}{400 \mathrm{~nm}}-\frac{1}{\lambda_{0}}\right)=\frac{1}{300 \mathrm{~nm}}-\frac{1}{\lambda_{0}} \\
& \frac{1}{200 \mathrm{~nm}}-\frac{1}{300 \mathrm{~nm}}=\frac{2}{\lambda_{0}}-\frac{1}{\lambda_{0}} \\
& \frac{3-2}{600}=\frac{1}{\lambda_{0}}
\end{aligned}
\]
\(\lambda_{0}=600 \mathrm{~nm}\)
(c) is correct
8
Given that the power dissipated in \(5 \Omega\) resistance is 7.2 W in the circuit shown. Statement (1) : Power dissipated in \(6 \Omega\) resistance is 6W. Statement (2) : Potential difference \(\mathrm{V}_{\mathrm{AB}}\) between A and B is \(\mathrm{V}_{\mathrm{AB}}=12.4 \mathrm{~V}\) Then
Show Solution
Solution
Ans. (b)
in parallel \(\mathrm{I} \propto \frac{1}{\mathrm{R}}\) so current
in \(5 \Omega\) will be 3 times that of current in \(15 \Omega\).
\[
\begin{aligned}
& \mathrm{P}_{5 \Omega}=(3 \mathrm{I})^{2} \times 5 \\
& 7.2=9 \times 5 \times \mathrm{I}^{2} \\
& \mathrm{I}^{2}=\frac{7.2}{9 \times 5}=\frac{0.8}{5}=0.16 \\
& \mathrm{I}=0.4 \mathrm{~A} \\
& \mathrm{P}_{6 \Omega}=\mathrm{I}^{2} \times 6 \Omega=0.16 \times 6=0.96 \mathrm{~W}
\end{aligned}
\]
Statement (1) is wrong.
\[
\begin{aligned}
\mathrm{V}_{\mathrm{AB}} & =4 \times 4 \mathrm{I}+5 \times 3 \mathrm{I} \\
& =31 \mathrm{I} \\
& =31 \times 0.4 \mathrm{~V} \\
& =12.4 \mathrm{Volt}
\end{aligned}
\]
Statement (2) is correct
(b) is correct.
9
A transparent and homogeneous sphere of glass of radius r is immersed in water (refractive indices of glass and water being \({ }_{\mathrm{a}} \mu_{\mathrm{g}}=\frac{3}{2}\) and \({ }_{\mathrm{a}} \mu_{\mathrm{w}}=\frac{4}{3}\) ). The image of a point object O, located at distance d on its axis in front of the sphere, is formed at point I at the same distance d from the sphere on the opposite side as shown. The distance d is equal to
Show Solution
Solution
Ans. (d)
By symmetry
\[
\begin{aligned}
& \frac{\mu_{r}}{v}-\frac{u_{i}}{v}=\frac{\mu_{r}-\mu_{i}}{R} \\
& \frac{3}{2(\infty)}-\frac{4}{3(-d)}=\frac{\frac{3}{2}-\frac{4}{3}}{r} \\
& \frac{4}{3 d}=\frac{1}{6 r} \\
& d=8 r
\end{aligned}
\]
option (d)
10
A certain substance, with a dielectric constant \(\mathrm{k}=2.5\) and the dielectric strength \(\mathrm{E}=1.8 \times 10^{7} \mathrm{~N} / \mathrm{C}\), completely fills the space between the plates of a parallel plate capacitor (with circular plates) of capacitance \(\mathrm{C}=72.0 \mathrm{nF}\). The minimum diameter of the circular plates, to ensure that the capacitor can withstand a potential difference of \(\mathrm{V}=4.0 \mathrm{kV}\), is
Show Solution
Solution
Ans. (d)
\[
\begin{aligned}
& \mathrm{k}=2.5 \\
& \mathrm{E}=1.8 \times 10^{7} \\
& \mathrm{C}=72 \mathrm{nF} \\
& \mathrm{~V}=4000 \mathrm{~V} \\
& \mathrm{E}=\frac{\mathrm{V}}{\mathrm{~d}} \\
& \mathrm{~d}=\frac{4000}{1.8 \times 10^{7}}
\end{aligned}
\]
\[
\begin{aligned}
& \frac{\mathrm{k} \epsilon_{0} \mathrm{~A}}{\mathrm{~d}}=72 \times 10^{-9} \\
& \frac{\mathrm{k} 4 \pi \epsilon_{0} \mathrm{~A}}{4 \pi \mathrm{~d}}=72 \times 10^{-9} \\
& \frac{2.5 \times \mathrm{A} \times 1.8 \times 10^{7}}{9 \times 10^{9} \times 4 \pi \times 4000}=72 \times 10^{-9} \\
& \frac{\pi \mathrm{~d}^{2}}{4}=\frac{9 \times 4 \pi \times 4 \times 4 \times 10^{-3}}{2.5} \\
& \mathrm{~d}=96 \mathrm{~cm} \\
& \text { option (d) }
\end{aligned}
\]
11
A uniform solid sphere of radius R rolls without slipping on a rough horizontal surface with a forward velocity v of its center. On its way, it suddenly encounters a small step of height 0.2 R as shown. The angular velocity of the sphere just after the impact is [given that the sphere does not bounce back, rather it goes ahead up the step]
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Solution
Ans. (c)
\[
\begin{aligned}
& \text { About P } \\
& L_{i}=L_{F} \\
& M V(0.8 R)+\frac{2}{5} M R^{2}\left(\frac{V}{R}\right)=\left(\frac{2}{5} M R^{2}+M R^{2}\right) \omega \\
& 0.8 V R+0.4 V R=1.4 R^{2} \omega \\
& \frac{1.2 V}{1.4 R}=\omega \\
& \omega=\frac{6 V}{7 R} \\
& \text { option (c) }
\end{aligned}
\]
12
The magnetic field (B) produced by the current i flowing through the sides of a square loop of side \(\ell\), at a point P at distance x from the center of the square, on the axis perpendicular to the plane of the square loop and passing through its center, is
Show Solution
Solution
Ans. (c)
B due to AB
\[
\begin{aligned}
& \mathrm{dB}=\frac{\mu_{0} \mathrm{I}}{4 \pi} \frac{\overrightarrow{\mathrm{~d} \ell} \times \overrightarrow{\mathrm{r}}}{\mathrm{r}^{3}} \\
& \overrightarrow{\mathrm{~d} \ell}=\mathrm{dz} \hat{\mathrm{k}}
\end{aligned}
\]
Position of element \(\left(0, \frac{\ell}{2}, \mathrm{z}\right)\)
\[
\begin{aligned}
& \vec{r}=\vec{r}_{p}-\vec{r}_{\text {element }} \\
& \vec{r}=x \hat{i}-\left(\frac{\ell}{2} \hat{j}+z \hat{k}\right) \\
& \vec{r}=x \hat{i}-\frac{\ell}{2} \hat{j}-z \hat{k} \\
& |\vec{r}|=\sqrt{x^{2}+\frac{\ell^{2}}{4}+z^{2}} \\
& d \vec{B}=\frac{\mu_{0} I}{4 \pi} \frac{d z \hat{k} \times\left(x \hat{i}-\frac{\ell}{2} \hat{j}-z \hat{k}\right)}{\left(x^{2}+\frac{\ell^{2}}{4}+z^{2}\right)^{\frac{3}{4}}} \\
& d \vec{B}=\frac{\mu_{0} I}{4 \pi} \frac{\left[x d z \hat{j}+\frac{\ell}{2} d z \hat{i}\right]}{\left(x^{2}+\frac{\ell^{2}}{4}+z^{2}\right)^{\frac{3}{4}}}
\end{aligned}
\]
Taking x-component
\[
\begin{aligned}
& |\mathrm{dB}|_{\mathrm{x}}=\frac{\mu_{0} \mathrm{I}}{4 \pi} \frac{\ell}{2} \frac{\mathrm{dz}}{\left(\mathrm{x}^{2}+\frac{\ell^{2}}{4}+\mathrm{z}^{2}\right)^{\frac{3}{4}}} \\
& \mathrm{~B}_{\mathrm{x}}=\frac{\mu_{0} \mathrm{I} \ell}{8 \pi} \int_{-\frac{\ell}{2}}^{\frac{\ell}{2}} \frac{\mathrm{dz}}{\left(\mathrm{x}^{2}+\frac{\ell^{2}}{4}+\mathrm{z}^{2}\right)^{\frac{3}{4}}}
\end{aligned}
\]
Consider \(\mathrm{x}^{2}+\frac{\ell^{2}}{4}=\mathrm{r}^{2}\)
\[
\begin{aligned}
B_{x} & =\frac{\mu_{0} I \ell}{8 \pi} \int_{-\frac{\ell}{2}}^{\frac{\ell}{2}} \frac{d z}{\left(r^{2}+z^{2}\right)^{\frac{3}{4}}} \\
B_{x} & =\frac{\mu_{0} I \ell^{2}}{8 \pi\left(x^{2}+\frac{\ell^{2}}{4}\right) \sqrt{x^{2}+\frac{\ell^{2}}{2}}}
\end{aligned}
\]
Net \(\mathrm{B}=4 \mathrm{~B}_{\mathrm{x}}=\frac{4 \times 4 \mu_{0} I \ell^{2} \times \sqrt{2}}{8 \pi\left(4 \mathrm{x}^{2}+\ell^{2}\right) \sqrt{2 \mathrm{x}^{2}+\ell^{2}}}\)
\[
\mathrm{B}=\frac{8 \sqrt{2} \mu_{0} \mathrm{I} \ell^{2}}{4 \pi\left(4 \mathrm{x}^{2}+\ell^{2}\right) \sqrt{2 \mathrm{x}^{2}+\ell^{2}}}
\]
Option (c)
13
A linear positive charge distribution, with linear charge density \(\lambda\) coulomb per meter, extends along +x-axis from \(\mathrm{x}=0\) to \(\mathrm{x}=\infty\). The electric field \(\overrightarrow{\mathrm{E}}\) at any point (0, y) on the y-axis
Show Solution
Solution
Ans. (d)
Option (d)
14
Imagine a situation, in which an infinite sheet with positive charge \(+\sigma\) per unit area lies in the xyplane and a second infinite sheet with negative charge \(-\sigma\) per unit area lies in the yz -plane. The net electric field E at any point (x, y, z) [that does not lie on either of these planes xy or yz] can be expressed as
Show Solution
Solution
Ans. (c)
E due to \(\infty\) sheet \(=\frac{\sigma}{2 \epsilon_{0}}\)
due to \(+\sigma: \overrightarrow{\mathrm{E}}=\frac{\sigma}{2 \epsilon_{0}} \frac{\mathrm{z}}{|\mathrm{z}|} \hat{\mathrm{k}}\)
due to \(-\sigma: \overrightarrow{\mathrm{E}}=\frac{-\sigma}{2 \epsilon_{0}} \frac{\mathrm{x}}{|\mathrm{x}|} \hat{\mathrm{i}}\)
option (c)
15
For the electric field E, in a region of space where a non-uniform, but spherically symmetric distribution of charge has a charge density \(\rho(\mathrm{r})\) as \(\rho(\mathrm{r})=\rho_{0}\left(1-\frac{\mathrm{r}}{\mathrm{R}}\right)\) for r r R one can say that \(\rho(\mathrm{r})=0 \quad\) for \(\quad \mathrm{r} \geq \mathrm{R}\)
Show Solution
Solution
Ans. (c)
\(\mathrm{E}_{\text {in }} 4 \pi \mathrm{r}^{2}=\frac{\int_{0}^{\mathrm{r}} \rho_{0}\left(1-\frac{\mathrm{r}}{\mathrm{R}}\right) 4 \pi \mathrm{r}^{2} \mathrm{dr}}{\varepsilon_{0}}\)
\(\mathrm{E}_{\text {in }}=\frac{\rho_{0}}{\varepsilon_{0} \mathrm{r}^{2}}\left[\frac{\mathrm{r}^{3}}{3}-\frac{\mathrm{r}^{4}}{4 \mathrm{R}}\right]\)
\(\mathrm{E}_{\text {in }}=\frac{\rho}{\varepsilon_{0}}\left[\frac{\mathrm{r}}{3}-\frac{\mathrm{r}^{2}}{4 \mathrm{R}}\right]\)
\[
\begin{aligned}
& \frac{\mathrm{dE}}{\mathrm{dR}}=0 \Rightarrow \frac{1}{3}-\frac{\mathrm{r}}{2 \mathrm{R}} \Rightarrow \mathrm{r}=\frac{2 \mathrm{R}}{3} \\
& \frac{\mathrm{~d}^{2} \mathrm{E}}{\mathrm{dR}^{2}}<0 \text { at } \mathrm{r}=\frac{2 \mathrm{R}}{3} \\
& \mathrm{E}_{\max }=\frac{\rho}{\varepsilon_{0}}\left[\frac{2 \mathrm{R}}{9}-\frac{4 \mathrm{R}^{2}}{9 \times 4 \mathrm{R}}\right] \\
& \frac{\rho}{\varepsilon_{0}} \frac{\mathrm{R}}{9}
\end{aligned}
\]
∴ option (c)
Questions 16–30
NSEP 2025 Physics — Paper Code 62
16
A typical network of resistance \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) shown below extends to infinity towards the right. The total resistance \(\mathrm{R}_{\text {effective }}\) of this network between points A and B is :
Show Solution
Solution
Ans. (a)
\[
\begin{aligned}
& \frac{R_{2} x}{R_{2}+x}+2 R_{1}=x \\
& \Rightarrow R_{2} x+2 R_{1} R_{2}+2 R_{1} x=R_{2} x+x^{2} \\
& \Rightarrow x^{2}-2 R_{1} x-2 R_{1} R_{2}=0 \\
& \Rightarrow x=\frac{2 R_{1} \pm \sqrt{4 R_{1}^{2}+8 R_{1} R_{2}}}{2} \\
& x=R_{1}+\sqrt{R_{1}^{2}+2 R_{1} R_{2}}
\end{aligned}
\]
option (a)
17
A cylindrical cavity of diameter ‘a’ exists inside a long solid cylinder of diameter ‘2a’ as shown in figure. Both the cylinder and the cavity are taken to be infinitely long. The axis of the cavity is parallel to the axis of the cylinder and is at distance \(\frac{\mathrm{a}}{2}\) from it. A uniform current of current density \(\mathrm{J}\left(\mathrm{Am}^{-2}\right)\) flows through the cylinder along its length and not through the cavity. The magnitude of the magnetic field at a point P on the surface of the cylinder lying farther from the axis of the cavity, is :
Show Solution
Solution
Ans. (d)
We can treat cavity as superposition of a full solid cylinder carrying current density as +J and a smaller cylinder (the cavity) carrying current density -J.
\[
\begin{aligned}
& \left(\vec{B}_{\text {net }}\right)_{\text {at point } P}=\vec{B}_{\text {Big cylinder }}+\vec{B}_{\text {small cylinder }} \\
& =\frac{\mu_{0} I_{1}}{2 \pi R}+\frac{\mu_{0} I_{2}}{2 \pi r^{\prime}} \\
& =\frac{\mu_{0} J \pi a^{2}}{2 \pi a}-\frac{\mu_{0} J\left(\frac{\pi a^{2}}{4}\right)}{2 \pi\left(\frac{3 a}{2}\right)} \\
& =\frac{\mu_{0} J a}{2}-\frac{\mu_{0} J a}{12} \\
& =\frac{5 \mu_{0} J a}{12}
\end{aligned}
\]
Option (d) is correct answer
18
A thin uniform rod, length \(\ell=0.200 \mathrm{~m}\) with negligible mass, is attached to the floor by a frictionless hinge at a fixed point P. A horizontal spring connects the other end of the rod to a vertical wall. The rod is in a uniform magnetic field \(\mathrm{B}=0.500\) tesla directed into the plane of paper. There is a current i = 10.0 A in the rod in the direction shown. Force constant of the spring is 5 N/m. The rod is in equilibrium at \(\theta=\tan ^{-1} \frac{4}{3}\). Statement-1: Torque on the rod due to magnetic force is 0.1 Nm clockwise. Statement-2 : In equilibrium the energy stored in the spring is 0.039 J.
Show Solution
Solution
Ans. (d)
(a) Magnetic force on rod
\[
\mathrm{F}_{\mathrm{m}}=\mathrm{I}(\vec{\ell} \times \overrightarrow{\mathrm{B}})=10 \times 0.2 \times 0.5 \sin 90^{\circ}
\]
\(=1\) Newton
Torque due to magnetic field
\(=(\mathrm{Fm})\) at centre of mass of rod \(\times \mathrm{r}_{\perp}\)
\[
\begin{aligned}
& =1 \times \frac{\mathrm{L}}{2}=1 \times \frac{0.2}{2} \\
& =0.1 \mathrm{Nm}
\end{aligned}
\]
clockwise
(b) Toque due to spring
\[
\begin{aligned}
& \mathrm{T}_{\text {spring }}=\left(\mathrm{F}_{\mathrm{s}}\right) \mathrm{L} \sin \theta \\
& =\mathrm{kx} \times 0.2 \times \frac{4}{5}
\end{aligned}
\]
Torque of spring must become torque due to magnetic field
\[
\mathrm{kx} \times 0.2 \times \frac{4}{5}=0.1
\]
\[
\begin{aligned}
& \mathrm{x}=\frac{0.1 \times 5}{4 \times 0.2 \times 5}=\frac{1}{8}=0.125 \mathrm{~m} \\
& \therefore \text { Energy stored in spring }=\frac{1}{2} \mathrm{kx}^{2} \\
& =\frac{1}{2} \times 5 \times(0.125)^{2} \\
& =0.039 \text { Joule } \\
& \text { (d) option is correct }
\end{aligned}
\]
Both statement (1) and (2) are correct
19
The electric flux though a certain of a dielectric medium is \(\phi=\left(8.00 \times 10^{3}\right) \mathrm{t}^{4}\) in SI units. The displacement current through that area is 12.5 pA at a time \(\mathrm{t}=20 \mathrm{~ms}\). The dielectric constant of the dielectric medium is :
Show Solution
Solution
Ans. (b)
From formula
\[
\begin{aligned}
& I_{d}=\frac{\epsilon_{0} \epsilon_{r} d \phi_{E}}{d t} \\
& \because \frac{d \phi_{E}}{d t}=\frac{d}{d t}\left(8 \times 10^{3} \times t^{4}\right) \\
& =32000 t^{3}
\end{aligned}
\]
At \(\mathrm{t}=20 \times 10^{-3}\) second
\[
\frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}=32000 \times\left(20 \times 10^{-3}\right)^{3}=0.256
\]
Using all the values in formula of displacement current
\[
\begin{aligned}
& I_{d}=\epsilon_{0} \epsilon_{r} \frac{d \phi_{E}}{d t} \\
& 12.5 \times 10^{-12}=\left(8.85 \times 10^{-12}\right) \epsilon_{r} \times 0.256 \\
& \epsilon_{r}=5.52
\end{aligned}
\]
Correct option is (b) 5.52
20
A thin equi-convex lens of flint glass (refractive index \(\mu_{1}\) ) is kept coaxially in contact with another thin equi-concave lens of crown glass (refractive index \(\mu_{2}\) ). The system is completely immersed in water \(\left({ }_{\mathrm{a}} \mu_{\mathrm{w}}=\frac{4}{3}\right)\). Parallel rays of light incident parallel to the principal axis in water are focused by this system at a distance of 24 cm beyond the system. The thickness of the system is negligible. If the radius of curvature of each surface is \(\mathrm{R}=20 \mathrm{~cm}\), the difference \(\left(\mu_{1}-\mu_{2}\right)\) is :
Show Solution
Solution
Ans. (d)
For equiconvex lens
\[
\frac{1}{f_{1}}=\left(\frac{\mu_{1}}{\mu_{\omega}}-1\right)\left(\frac{2}{R}\right)
\]
For equiconcave lens
\[
\frac{1}{\mathrm{f}_{2}}=\left(\frac{\mu_{2}}{\mu_{\omega}}-1\right)\left(-\frac{2}{\mathrm{R}}\right)
\]
Combined focal length
\[
\begin{aligned}
& \frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}} \\
& =\frac{2}{\mathrm{R}}\left[\left(\frac{\mu_{1}}{\mu_{\mathrm{w}}}-1\right)-\left(\frac{\mu_{2}}{\mu_{\omega}}-1\right)\right] \\
& \frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{2}{\mathrm{R} \mu_{\omega}}\left(\mu_{1}-\mu_{2}\right) \\
& \frac{1}{24}=\frac{2}{20 \times\left(\frac{4}{3}\right)}\left(\mu_{1}-\mu_{2}\right) \\
& \mu_{1}-\mu_{2}=\frac{80}{144}=\frac{5}{9}
\end{aligned}
\]
Answer is option (d)
21
Two identical large thin metal plates carrying charges \(+\mathrm{q}_{1}\) and \(+\mathrm{q}_{2}\left(\mathrm{q}_{1}>\mathrm{q}_{2}\right)\), respectively, are kept close at a distance d apart and parallel to each other to form a parallel plate capacitor of capacitance C. The potential difference between the plates is :
Show Solution
Solution
Ans. (b)
\[
\begin{aligned}
& q_{3}=q_{1}-\frac{q_{1}+q_{2}}{2}=\left(\frac{q_{1}-q_{2}}{2}\right) \\
& Q=C V \\
& V=\frac{Q}{C} \\
& V=\left(\frac{q_{1}-q_{2}}{2}\right) \times \frac{1}{C} \\
& V=\frac{q_{1}-q_{2}}{2 C}
\end{aligned}
\]
22
A point mass m moves in a straight line under a retardation \(\mathrm{kv}^{2}\) [where k is a positive constant and v is the instantaneous velocity]. The initial velocity of the point mass is u . The displacement of the point mass at time t is :
Show Solution
Solution
Ans. (a)
\(\mathrm{a}=-\mathrm{kv}^{2}\)
\[
\begin{aligned}
& \frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{kv}^{2} \\
& \int_{\mathrm{u}}^{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{v}^{2}}=-\mathrm{k} \int_{0}^{\mathrm{t}} \mathrm{dt} \\
& -\left.\frac{1}{\mathrm{v}}\right|_{0} ^{\mathrm{v}}=-k t \\
& \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=k t \\
& \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{u}}+k t \\
& \int_{0}^{\mathrm{x}} \mathrm{dx}=\int_{0}^{\mathrm{t}} \frac{\mathrm{u}}{1+u k t} \mathrm{dt} \\
& \mathrm{x}=\frac{\mathrm{u}}{1+u k t} \frac{\left.\mathrm{u}(1+u k t)\right|_{0} ^{\mathrm{t}}}{u k} \\
& \mathrm{x}=\frac{1}{\mathrm{k}} \ln (1+u k t)
\end{aligned}
\]
23
In the arrangement shown in figure, ‘a’ represent the magnitude of acceleration of small blocks A and B while ‘T’ is the tension in the massless string passing over the frictionless and massless pulley. The sum of the masses of blocks A and B is constant. For this system, a linear relationship can be obtained between
Show Solution
Solution
Ans. (d)
\(a=\frac{\left(m_{A}-m_{B}\right) g}{\left(m_{A}+m_{B}\right)} \quad T=\frac{2 m_{A} m_{B} g}{\left(m_{A}+m_{B}\right)}\)
\[
\begin{aligned}
& a^{2}=\frac{g^{2}}{\left(m_{A}+m_{B}\right)^{2}} \times\left(m_{A}^{2}+m_{B}^{2}-2 m_{A} m_{B}\right) \\
& a^{2}=\frac{\left(m_{A}^{2}+m_{B}^{2}\right) g^{2}}{\left(m_{A}+m_{B}\right)^{2}}-\frac{2 m_{A} m_{B}(g)(g)}{\left(m_{A}+m_{B}\right)\left(m_{A}+m_{B}\right)}
\end{aligned}
\]
\[
\begin{aligned}
& a^{2}=\frac{\left[\left(m_{A}^{2}+m_{B}^{2}\right)-2 m_{A} m_{B}+2 m_{A} m_{B}\right] g^{2}}{\left(m_{A}+m_{B}\right)^{2}}-\frac{T g}{\left(m_{A}+m_{B}\right)} \\
& a^{2}=\frac{\left[\left(m_{A}^{2}+m_{B}^{2}\right)-2 m_{A} m_{B}\right] g^{2}}{\left(m_{A}+m_{B}\right)^{2}}-\frac{T g}{\left(m_{A}+m_{B}\right)} \\
& a^{2}=\left[1-\frac{2 m_{A} m_{B}}{\left(m_{A}+m_{B}\right)^{2}}\right] g^{2}-\frac{T g}{\left(m_{A}+m_{B}\right)} \\
& a^{2}=g^{2}-\frac{2 m_{A} m_{B} g^{2}}{\left(m_{A}+m_{B}\right)^{2}}-\frac{T g}{\left(m_{A}+m_{B}\right)} \\
& a^{2}=g^{2}-\left[\frac{2 g}{\left(m_{A}+m_{B}\right)^{2}}\right] T \\
& Y=C-m x
\end{aligned}
\]
Linear relation between \(\mathrm{a}^{2} \& \mathrm{~T}\)
24
A thin uniform circular ring of mass m is rolling without slipping down an inclined plane of inclination 30° with the horizontal. The coefficient of friction between the ring and the surface is \(\mu\). The correct statement is :
Show Solution
Solution
Ans. (b)
\(\mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{mR}^{2}}}\)
\[
\begin{aligned}
& \mathrm{a}=\frac{\mathrm{g}\left(\frac{1}{2}\right)}{1+\frac{\mathrm{mR}^{2}}{\mathrm{mR}^{2}}}=\frac{\mathrm{g}}{4} \\
& \mathrm{f}=\frac{\mathrm{mg} \sin \theta^{\mathrm{mR}^{2}}}{1+\frac{\mathrm{I}}{\mathrm{I}}} \\
& \mathrm{f}=\frac{\mathrm{mg}\left(\frac{1}{2}\right)}{1+1} \\
& \mathrm{f}=\frac{\mathrm{mg}}{4} \rightarrow \text { ans (b) }
\end{aligned}
\]
\[
\begin{aligned}
& \frac{\mathrm{mg}}{4} \leq \mu \mathrm{mg} \frac{\sqrt{3}}{2} \\
& \mu_{\min } \geq \frac{1}{2 \sqrt{3}}
\end{aligned}
\]
25
A bullet of mas m can penetrate a target (a heavy block of mass M) up to a distance S, when the target M is held stationary by a stopper P (shown in figure). Up to what distance S’ the bullet will penetrate if the block of mass M is free to move (i.e. when the stopper P is removed) on the frictionless surface T.
Show Solution
Solution
Ans. (d)
Case-1 : When block is fixed
\[
\begin{aligned}
& \omega_{\mathrm{Net}}=\Delta \mathrm{KE} \\
& \omega_{\mathrm{Res}}=\mathrm{K}_{\mathrm{F}}-\mathrm{K}_{\mathrm{I}} \\
& -\mathrm{F}_{\mathrm{R}} \cdot \mathrm{~S}=\mathrm{O}-\frac{1}{2} \mathrm{mv}_{0}^{2} \\
& \mathrm{~F}_{\mathrm{R}}=\frac{1}{2} \frac{\mathrm{mv}_{0}^{2}}{\mathrm{~S}} \tag{i}
\end{aligned}
\]
\(\mathrm{F}_{\mathrm{R}}=\) Constant resistance f offered by block.
\(\mathrm{V}_{0}=\) Initial speed of bullet
Case-2 : When block is free to move
Stepe-1 : Linear momentum conservation of bullet + block system, as no consntant force is acting
\[
\mathrm{mv}_{0}=(\mathrm{m}+\mathrm{M}) \mathrm{v} \Rightarrow \mathrm{v}=\frac{\mathrm{mv}_{0}}{\mathrm{~m}+\mathrm{M}} \tag{ii}
\]
Step-2 : Work energy theorem
\[
\begin{aligned}
& W_{\text {net }}=k_{F}-K_{I} \\
& -F_{R} \cdot S^{\prime}=\frac{1}{2}(m+M) v^{2}-\frac{1}{2} m v_{0}^{2} \tag{iii}
\end{aligned}
\]
On solving equation (i), (ii) and (iii)
\[
\begin{aligned}
& -F_{R} \cdot S^{\prime}=\frac{1}{2}(m+M)\left[\frac{m v_{0}}{m+M}\right]^{2}-\frac{1}{2} m v_{0}^{2} \\
& -\frac{1}{2} \frac{m v_{0}^{2}}{S} \cdot S^{\prime}=\frac{1}{2} \frac{m^{2} v_{0}^{2}}{m+M}-\frac{1}{2} m v_{0}^{2} \\
& -\frac{S^{\prime}}{S}=\frac{m}{m+M}-1 \\
& \Rightarrow \frac{S^{\prime}}{S}=1-\frac{m}{m+M}=\frac{M}{m+M} \\
& \Rightarrow S^{\prime}=\frac{M s}{m+M}
\end{aligned}
\]
26
Knowing that the atomic masses of Al and Mg are respectively \({ }_{13}^{25} \mathrm{AI}=24.990432 \mathrm{u}\) and \({ }_{12}^{25} \mathrm{Mg}=\) 24.985839 u while electron mass is often expressed as \(\mathrm{m}_{\mathrm{e}}=0.511 \mathrm{MeV}\), the Q value (energy liberated) of the \(\beta\) decay nuclear reaction \({ }^{25} \mathrm{Al} \rightarrow{ }^{25} \mathrm{Mg}+\mathrm{e}^{+}+v\) in MeV is :
Show Solution
Solution
Ans. (c)
B
\[
\begin{aligned}
& \text { For } \beta+\text { decay process } \\
& \mathrm{Q}=\left[\mathrm{m}_{\text {parent }}-\mathrm{m}_{\text {daughter }}-2 \mathrm{~m}_{\mathrm{e}}\right] \times 931.5 \mathrm{MeV} \\
& \mathrm{Q}=\left[\mathrm{m}_{\text {parent }}-\mathrm{m}_{\text {daughter }}\right] \times 931.5-2\left[\mathrm{~m}_{\mathrm{e}} \times 931.5 \mathrm{MeV}\right] \\
& \mathrm{Q}=[24.990432-24.985839] \times 931.5-2[0.511] \\
& \mathrm{Q}=3.256 \mathrm{MeV}
\end{aligned}
\]
27
A block of mass m, lying on a rough horizontal plane, is acted upon by a horizontal force P and and simultaneously by another force Q acting at an angle \(\theta\) from the vertical as shown. The block will remain in equilibrium if the coefficient of friction between the block and the surface S is :
Show Solution
Solution
Ans. (a)
\[
\xrightarrow[\mathrm{mg}]{\stackrel{\mathrm{f}_{\mathrm{s}}}{\longleftrightarrow}} \underset{\downarrow \mathrm{Q} \cos \theta}{\stackrel{\mathrm{~N}}{\mathrm{Q} \sin \theta}} \mathrm{P}
\]
\[
\begin{aligned}
& \Sigma \mathrm{F}_{\mathrm{y}}=0 \\
& \mathrm{~N}=\mathrm{mg}+\mathrm{Q} \cos \theta \tag{i}\\
& \Sigma \mathrm{~F}_{\mathrm{x}}=0 \\
& \mathrm{Fs}=\mathrm{P}+\mathrm{Q} \sin \theta \tag{ii}\\
& \because \mathrm{fs} \leq(\mathrm{fs})_{\max } \\
& \mathrm{P}+\mathrm{Q} \sin \theta \leq \mu \mathrm{N} \\
& \mathrm{P}+\mathrm{Q} \sin \theta \leq \mu(\mathrm{mg}+\mathrm{Q} \cos \theta) \\
& \mu \geq \frac{\mathrm{P}+\mathrm{Q} \sin \theta}{\mathrm{mg}+\mathrm{Q} \cos \theta}
\end{aligned}
\]
28
Knowing that the acceleration due to gravity on the Earth surface is g and the radius of the Earth is R, a small body of mass m falls on the Earth from a height \(\mathrm{h}=\frac{\mathrm{R}}{5}\) above the Earth’s surface. During the freefall, the potential energy of the falling body decreases by:
Show Solution
Solution
Ans. (c)
\[
\begin{aligned}
& U_{I}=-\frac{G M_{e} m}{R+h} \\
& U_{F}=-\frac{G M_{e} m}{R} \\
& D U=U_{I}-U_{F} \\
& \Delta U=G M_{e} m\left[\frac{1}{R}-\frac{1}{R+h}\right] \\
& \Delta U=G M_{e} m\left[\frac{1}{R}-\frac{5}{6 R}\right] \\
& \Delta U=G M_{e} m\left[\frac{1}{6 R}\right]=\frac{G M e m}{R} \times \frac{1}{6} \\
& =m\left[\frac{G M_{e}}{R^{2}}\right] \times \frac{R}{6} \\
& \Delta U=\frac{m g R}{6}=\frac{5 m g h}{6}
\end{aligned}
\]
29
At some instant, a motor car is moving on a circular path of radius 600 m , with a speed \(u=30 \mathrm{~ms}^{-1}\). If its speed is increased at a rate of \(2 \mathrm{~ms}^{-2}\), the magnitude of the acceleration of the car at that instant is :
Show Solution
Solution
Ans. (b)
\[
\begin{aligned}
& a_{C}=a_{R}=\frac{v^{2}}{R}=\frac{(30)^{2}}{600}=\frac{900}{600}=\frac{3}{2} \text { [Centripital acceleration] } \\
& a_{t}=2 \\
& a_{N e t}=\sqrt{a_{t}^{2}+a_{c}^{2}} \\
& =\sqrt{(2)^{2}+(3 / 2)^{2}}=\sqrt{\frac{25}{4}}=\frac{5}{2} \\
& a_{n e t}=2.5 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\]
30
A cricket ball, thrown across a field, is at heights of \(\mathrm{h}_{1}\) and \(\mathrm{h}_{2}\) above the point of projection, at time \(\mathrm{t}_{1}\) and time \(\mathrm{t}_{2}\) after the throw, respectively. It is then caught by the wicket keeper at the same height as that from which it was thrown. The Time of Flight (T) of the ball is :
Show Solution
Solution
Ans. (a)
\[
\begin{aligned}
& \mathrm{h}_{1}=\mathrm{u}_{\mathrm{y}} \mathrm{t}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2} \\
& \mathrm{~h}_{1}+\frac{1}{2} \mathrm{gt}_{1}^{2}=\mathrm{u}_{\mathrm{y}} \mathrm{t}_{1} \tag{i}\\
& \mathrm{~h}_{2}=\mathrm{u}_{\mathrm{y}} \mathrm{t}_{2}-\frac{1}{2} \mathrm{gt}_{2}^{2} \\
& \mathrm{~h}_{2}+\frac{1}{2} \mathrm{gt}_{2}^{2}=\mathrm{u}_{\mathrm{y}} \mathrm{t}_{2} \tag{ii}
\end{aligned}
\]
Equation (i)/(ii)
\[
\begin{aligned}
& \frac{h_{1}+\frac{1}{2} g t_{1}^{2}}{h_{2}+\frac{1}{2} g t_{2}^{2}}=\frac{t_{1}}{t_{2}} \\
& h_{1} t_{2}-h_{2} t_{1}=\frac{g}{2}\left(t_{1} t_{2}^{2}-t_{1}^{2} t_{2}\right) \\
& g=2\left[\frac{h_{1} t_{2}-h_{2} t_{1}}{t_{1} t_{2}^{2}-t_{1}^{2} t_{2}}\right]
\end{aligned}
\]
Time of flight
\[
\begin{aligned}
& T=\frac{2 \mu g}{g} \\
& T=\frac{2}{g}\left[\frac{h_{1}+\frac{1}{2} g t_{1}^{2}}{t_{1}}\right] \\
& T=\frac{2}{t_{1}}\left[\frac{h_{1}}{g}+\frac{t_{1}^{2}}{2}\right] \\
& T=\frac{2}{t_{1}}\left[\frac{h_{1}\left(t_{1} t_{2}^{2}-t_{1}^{2} t_{2}\right)}{2\left[h_{1} t_{2}-h_{2} t_{1}\right]}+\frac{t_{1}^{2}}{2}\right] \\
& T=\left[\frac{h_{1} t_{1} t_{2}\left(t_{2}-t_{1}\right)}{t_{1}\left(h_{1} t_{2}-h_{2} t_{1}\right)}+t_{1}\right] \\
& T=\frac{h_{1} t_{2}\left(t_{2}-t_{1}\right)+t_{1} h_{1} t_{2}-t_{1}^{2} h_{2}}{h_{1} t_{2}-h_{2} t_{1}} \\
& T=\frac{h_{1} t_{2}^{2}-h_{2} t_{1}^{2}}{h_{1} t_{2}-h_{2} t_{1}}
\end{aligned}
\]
Questions 31–45
NSEP 2025 Physics — Paper Code 62
31
A plate of mass M is placed on a horizontal frictionless surface S. A block of mass m is placed on the plate. The coefficient of dynamic friction between the block and the plate is \(\mu\). If a horizontal force \(\mathrm{F}=2 \mu \mathrm{mg}\) is applied to the block (as shown), the acceleration of the plate will be :-
Show Solution
Solution
Ans. (a or d)
\[
\mathrm{F}=2 \mu \mathrm{mg}
\]
Suppose together :
\[
\mathrm{a}=\left(\frac{\mathrm{F}}{\mathrm{~m}+\mathrm{M}}\right)=\frac{2 \mu \mathrm{mg}}{\mathrm{~m}+\mathrm{M}}
\]
\[
\begin{aligned}
& \mathrm{f}=\mathrm{M} \cdot \mathrm{a}=\frac{2 \mathrm{Mm} \mu \mathrm{~g}}{\mathrm{~m}+\mathrm{M}} \\
& \mathrm{f}=\mu \mathrm{mg}\left[\frac{2 \mathrm{M}}{\mathrm{~m}+\mathrm{M}}\right] \\
& \left(\mathrm{f}_{\mathrm{s}}\right)_{\max }=\mu \mathrm{mg}
\end{aligned}
\]
If m > M :
then \(\mathrm{f} \leq\left(\mathrm{f}_{\mathrm{s}}\right)_{\text {max }}\)
block move together
\[
\mathrm{a}=\frac{2 \mu \mathrm{mg}}{\mathrm{~m}+\mathrm{M}}
\]
If m < M :
then \(\mathrm{f}>\left(\mathrm{f}_{\mathrm{s}}\right)_{\text {max }}\)
our assumption is wrong.
\[
\mathrm{a}=\frac{\mu \mathrm{mg}}{\mathrm{M}}
\]
32
A simple pendulum, with a bob of mass m, oscillates in a vertical plane, with an angular amplitude \(\theta_{0}\). The tension in its string when it passes through the mean position is 2mg. Neglecting the effect of air friction and the viscosity of air, the angular amplitude \(\theta_{0}\) is :-
Show Solution
Solution
Ans. (b)
\[
\begin{aligned}
& \mathrm{T}-\mathrm{mg}=\frac{\mathrm{mv}^{2}}{\ell} \\
& 2 \mathrm{mg}-\mathrm{mg}=\frac{\mathrm{mv}^{2}}{\ell} \tag{1}\\
& \mathrm{mg} \ell=\mathrm{mv}^{2} \\
& \mathrm{~h}=\ell(1-\cos \theta)
\end{aligned}
\]
Work done by gravity \(=\Delta \mathrm{K}\)
\[
\begin{aligned}
& m g \ell(1-\cos \theta)=\frac{1}{2} m v^{2}-0 \\
& \text { From }(1): \\
& m g \ell(1-\cos \theta)=\frac{1}{2}(m g \ell) \\
& \frac{m g \ell}{2}=m g \ell \cos \theta \\
& \cos \theta=\frac{1}{2} \\
& \theta=60^{\circ}
\end{aligned}
\]
33
Because of their mutual gravitational attraction, four identical planets each of mass m are orbiting in a circular path of radius r in the same sense (angular direction). The magnitude of the velocity of each planet is :-
Show Solution
Solution
Ans. (a)
Net force on planet A is :
\[
\begin{aligned}
& F=F_{1} \sqrt{2}+F_{0} \\
& F=\frac{G m^{2}}{(r \sqrt{2})^{2}} \sqrt{2}+\frac{G m^{2}}{(2 r)^{2}} \\
& F=\frac{G m^{2}}{r^{2}}\left[\frac{1}{\sqrt{2}}+\frac{1}{4}\right]=\frac{m v^{2}}{r} \\
& v=\sqrt{\frac{G m}{r} \frac{(1+2 \sqrt{2})}{4}}
\end{aligned}
\]
34
A rigid square sheet of size 2 m × 2 m is hinged at the middle of the vertical edges to serve as a door which can turn about the horizontal axis OO’. A fluid of density \(\rho\) fills the space to the left of the sheet up to its top. The horizontal force F required (to be applied at the lower edge) to hold the sheet vertical is :-
Show Solution
Solution
Ans. (b)
\[
\begin{aligned}
\tau_{\text {upper }} & =\int_{0}^{1} \rho \mathrm{~g}(1-\mathrm{x})(2 \mathrm{dx})(\mathrm{x}) \\
= & \rho \mathrm{g} 2\left[\frac{1}{2}-\frac{1}{3}\right]=\frac{\rho \mathrm{g}}{3}(\mathrm{cw})
\end{aligned}
\]
\[
\begin{aligned}
& \tau_{\text {lower }}=\int_{0}^{1} \rho \mathrm{~g}\left(1+\mathrm{x}^{\prime}\right)\left(2 \mathrm{dx}^{\prime}\right)\left(\mathrm{x}^{\prime}\right) \\
& \quad=\rho \mathrm{g} 2\left[\frac{1}{2}+\frac{1}{3}\right]=\rho \mathrm{g} \times \frac{5}{3}(\text { Acw. }) \\
& \tau_{\text {net }}=\rho \mathrm{g}\left(\frac{5}{3}-\frac{1}{3}\right) \\
& \quad=\frac{4}{3} \rho \mathrm{~g}=\mathrm{F} \times 1 \\
& \mathrm{~F}=\frac{4}{3} \rho \mathrm{~g}
\end{aligned}
\]
35
A major artery in human body, with radius 0.4 cm, carries blood at a flow rate of 5.0 cubic centimeters per second. The pressure difference of blood per meter length of the artery is nearly :- [Given that the coefficient of viscosity \((\eta)\) of blood at body temperature is \(4.0 \times 10^{-3} \mathrm{~Pa} . \mathrm{s}\) and the density of mercury is \(13.6 \mathrm{~g} / \mathrm{cm}^{3}\) ]
Show Solution
Solution
Ans. (c)
Rate of liquid flow is given by poiseuilles equation
\[
\begin{aligned}
& \frac{\mathrm{dQ}}{\mathrm{dt}}=\left(\frac{\pi}{8}\right) \frac{(\Delta \mathrm{P}) \mathrm{r}^{4}}{\eta \mathrm{~L}} \\
& 5 \times 10^{-6}=\frac{\pi}{8} \times \frac{\Delta \mathrm{P} \times\left(4 \times 10^{-3}\right)^{4}}{4 \times 10^{-3} \times \ell} \\
& \begin{aligned}
\frac{\Delta \mathrm{P}}{\ell} & =\frac{5 \times 10^{-6} \times 4 \times 10^{-3}}{\left(4 \times 10^{-3}\right)^{4}} \times \frac{8}{\pi} \\
& =0.2 \times 10^{3} \mathrm{~Pa} / \mathrm{m} \\
& =1.5 \mathrm{~mm} \text { of } \mathrm{Hg} / \mathrm{m}
\end{aligned}
\end{aligned}
\]
36
If P represents radiation pressure, E represents radiation energy striking per unit area per unit time and c represents speed of light then the possible values of non-zero integers x, y and z such that \(\mathrm{P}^{\mathrm{x}} \mathrm{E}^{\mathrm{y}} \mathrm{c}^{\mathrm{z}}\) is dimensionless, may be :-
Show Solution
Solution
Ans. (c)
Dimension of
\[
\begin{aligned}
& \quad \mathrm{P}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \\
& \quad \mathrm{E}=\left[\mathrm{MT}^{-3}\right] \\
& \quad \mathrm{c}=\left[\mathrm{LT}^{-1}\right] \\
& \mathrm{P}^{\mathrm{x}} \mathrm{E}^{\mathrm{y}} \mathrm{c}^{\mathrm{z}}=\left[\mathrm{M}^{\mathrm{x}+\mathrm{y}} \mathrm{~L}^{-\mathrm{x}+\mathrm{z}} \mathrm{~T}^{-2 \mathrm{x}-3 \mathrm{y}-\mathrm{z}}\right] \\
& {\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]=\left[\mathrm{M}^{\mathrm{x}+\mathrm{y}} \mathrm{~L}^{-\mathrm{x}+\mathrm{z}} \mathrm{~T}^{-2 \mathrm{x}-3 \mathrm{y}-\mathrm{z}}\right]} \\
& \Rightarrow \quad \mathrm{x}+\mathrm{y}=0 ; \mathrm{x}=\mathrm{z} \& 2 \mathrm{x}+3 \mathrm{y}+\mathrm{z}=0 \\
& 2 \mathrm{x}+3(-\mathrm{x})+\mathrm{x}=0 \\
& \mathrm{x}=0 ; \mathrm{y}=0 ; \mathrm{z}=0
\end{aligned}
\]
But question says x, y \& z must be non-zero, so only option ‘c’ satisfy this.
37
A large tank, open at the top, has two small holes in the vertical wall. One is a square hole of side ‘s’ at a depth h below the top and the other is a circular hole of radius r at a depth 4h below the top (given that \(\mathrm{s} \ll \mathrm{h} ; \mathrm{r} \ll \mathrm{h}\) ). When the tank is completely filled up to the brim with water, the quantity of water flowing out per second from each hole is the same, then r is equal to :-
Show Solution
Solution
Ans. (c)
\[
\begin{array}{ll}
v_{s}=\sqrt{2 g h} & v_{c}=\sqrt{2 g(4 h)} \\
Q_{s}=\text { Area × velocity } & v_{c}=\sqrt{8 g h} \\
Q_{s}=s^{2} \sqrt{2 g h} & Q_{c}=\left(\pi r^{2}\right) \sqrt{8 g h}
\end{array}
\]
According to given condition
\[
\begin{aligned}
& Q_{s}=Q_{c} \\
& s^{2} \sqrt{2 g h}=\pi r^{2} \sqrt{8 g h} \\
& s^{4}(2 g h)=\pi^{2} r^{4}(8 g h) \\
& s^{4}=4 \pi^{2} r^{4} \\
& r^{4}=\frac{s^{4}}{4 \pi^{2}} \\
& r=\frac{s}{\left(4 \pi^{2}\right)^{1 / 4}} \\
& r=\frac{s}{\sqrt{2 \pi}}
\end{aligned}
\]
38
A pendulum consists of a heavy but very small bob of mass M suspended at the end of a rigid rod of mass m and length L. The time period of small oscillations in the vertical plane, about a horizontal axis through the upper end of the rod is :-
Show Solution
Solution
Ans. (b)
\[
\begin{aligned}
& \tau=\mathrm{I} \alpha \\
& -\left\{(\mathrm{mg} \sin \theta) \frac{\mathrm{L}}{2}+(\mathrm{Mg} \sin \theta) \mathrm{L}\right\}=\left(\mathrm{ML}^{2}+\frac{\mathrm{mL}^{2}}{3}\right) \alpha
\end{aligned}
\]
\(\theta\) is very small
So, \(\sin \theta \approx \theta\)
\[
\begin{aligned}
& -\left(\frac{\mathrm{m}}{2}+\mathrm{M}\right) \mathrm{g} \theta=\left(\frac{3 \mathrm{M}+\mathrm{m}}{3}\right) \mathrm{L} \alpha \\
& -\left(\frac{\mathrm{m}+2 \mathrm{M}}{2}\right) \mathrm{g} \theta=\left(\frac{3 \mathrm{M}+\mathrm{m}}{3}\right) \mathrm{L} \alpha \\
& \alpha=-\left(\frac{\mathrm{m}+2 \mathrm{M}}{\mathrm{~m}+3 \mathrm{M}}\right) \frac{3 \mathrm{~g}}{2 \mathrm{~L}} \theta
\end{aligned}
\]
By comparing standard equation
\[
\begin{aligned}
& \alpha=-\omega^{2} \theta \\
& \omega=\sqrt{\left(\frac{m+2 M}{m+3 M}\right) \frac{3 g}{2 L}} \\
& T=\frac{2 \pi}{\omega} \\
& T=2 \pi \sqrt{\left(\frac{m+3 M}{m+2 M}\right) \frac{2 L}{3 g}}
\end{aligned}
\]
39
A transverse wave is travelling along a long stretched string from left to right (along +ve x direction). The snapshot of a small part of the string at any moment t is shown in the figure. At this particular instant :-
Show Solution
Solution
Ans. (d)
\(\mathrm{v}_{\mathrm{P}}=-\mathrm{v}_{\omega}\) (slope)
Wave is in +x direction, so \(\mathrm{v}_{\omega}=+\mathrm{ve}\)
Positive slope : A, B, H, I
Negative slope : D, E, F
Zero slope : C, G
From given option slope of B and H are positive, so at the given moment B \& H are moving downward.
40
Two tuning forks, with natural frequency 700 Hz each, move relative to a stationary observer. Fork (1) moves towards the observer while the fork (2) moves away from the observer. Both the forks move with same velocity v on the same line. The observer, standing between the two forks, hears 4 beats per sec. Using the speed of sound in air as \(\mathrm{v}_{\mathrm{s}}=350 \mathrm{~ms}^{-1}\), the speed of each tuning fork is :-
Show Solution
Solution
Ans. (c)
\(\mathrm{f}_{1}=\) frequency heared by observer due to tunning fork 1
\(\mathrm{f}=\) frequency heared by observer due to tunning fork 2
\[
\begin{aligned}
& f_{1}=f_{0}\left[\frac{v_{\omega} \pm v_{0}}{v_{\omega} \pm v_{s}}\right] \\
& \Rightarrow f_{1}=700\left[\frac{350+0}{350-v}\right] \\
& f_{2}=700\left[\frac{350}{350+v}\right]
\end{aligned}
\]
According to given condition,
\[
\begin{aligned}
& f_{\text {beat }}=f_{1}+f_{2} \\
& 4=700\left[\frac{350}{350-v}\right]-700\left[\frac{350}{350+v}\right] \\
& 4=700 \times 350\left[\frac{(350+v)-(350-v)}{(350)^{2}-v^{2}}\right] \\
& (350)^{2}-v^{2}=\frac{1}{2} \times 350 \times 350 \times 2 v \\
& (350)^{2}-v^{2}=(350)^{2} v \\
& v^{2}+(350)^{2} v-(350)^{2}=0 \\
& v^{2}+122500 v-122500=0
\end{aligned}
\]
From the given option the most appropriate choice for speed of tunning fork will be
\[
\mathrm{v} \approx 1 \mathrm{~m} / \mathrm{s}
\]
41
The speed of sound in a mixture of 1 mole of Helium (molar mass = 4 g) and 2 moles of oxygen (molar mass \(=32 \mathrm{~g}\) ) at 27 °C is nearly :-
Show Solution
Solution
Ans. (d)
\(\mathrm{V}=\sqrt{\frac{\gamma \mathrm{RT}}{\mathrm{M}}}\)
\[
\gamma_{\operatorname{mix}}=\frac{C_{P_{\operatorname{mix}}}}{C_{V_{\operatorname{mix}}}}
\]
\[
\begin{aligned}
& \quad=\frac{\mathrm{n}_{1} \mathrm{C}_{\mathrm{P}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{P}_{2}}}{\mathrm{n}_{1} \mathrm{C}_{\mathrm{V}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{V}_{2}}} \\
& \quad=\frac{1\left(\frac{5 \mathrm{R}}{2}\right)+2\left(\frac{7 \mathrm{R}}{2}\right)}{1\left(\frac{3 \mathrm{R}}{2}\right)+2\left(\frac{5 \mathrm{R}}{2}\right)} \\
& \quad=\frac{19}{13} \approx 1.46 \\
& \mathrm{M}_{\mathrm{mix}}=\frac{\mathrm{n}_{1} \mathrm{M}_{1}+\mathrm{n}_{2} \mathrm{M}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}} \\
& \quad=\frac{1(4)+2(32)}{1+2}=\frac{68}{3} \mathrm{gm} / \mathrm{mol} \\
& \quad=\frac{68 \times 10^{-3}}{3} \mathrm{~kg} / \mathrm{mol} \\
& \Rightarrow \mathrm{~V}=\sqrt{\frac{\gamma_{\mathrm{mix}} \times \mathrm{RT}}{\mathrm{M}_{\mathrm{mix}}}}=401 \mathrm{~m} / \mathrm{s}
\end{aligned}
\]
42
Three identical large metal plates are kept parallel and close to each other. Each plate can be treated as an ideal black body and has very high thermal conductivity. The first and third plates are maintained at high temperature \(\mathrm{T}_{1}=3 \mathrm{~T}\) and \(\mathrm{T}_{3}=2 \mathrm{~T}\). The temperature \(\mathrm{T}_{2}\) of the middle (i.e. second) plate under steady state condition is :-
Show Solution
Solution
Ans. (c)
For equilibrium,
\[
\begin{aligned}
& \mathrm{P}_{12}=\mathrm{P}_{23} \\
& \sigma \mathrm{eA}\left(\mathrm{~T}_{1}^{4}-\mathrm{T}_{2}^{4}\right)=\sigma \mathrm{eA}\left(\mathrm{~T}_{2}^{4}-\mathrm{T}_{3}^{4}\right)
\end{aligned}
\]
\[
\begin{aligned}
& \Rightarrow \quad \mathrm{T}_{2}^{4}=\frac{1}{2}\left(\mathrm{~T}_{1}^{4}+\mathrm{T}_{3}^{4}\right) \\
& \Rightarrow \quad \mathrm{T}_{2}=\left(\frac{97}{2}\right)^{1 / 4} \mathrm{~T}
\end{aligned}
\]
43
A thin uniform circular disc of radius ‘a’ is placed in XY plane with its center at origin (0,0). A small circular disc of radius b with center at (c, 0) is cut and taken out to create a hole. The center of mass of the remaining disc is at :-
Show Solution
Solution
Ans. (d)
\(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{M}_{1} \mathrm{x}_{1}-\mathrm{M}_{2} \mathrm{x}_{2}}{\mathrm{M}_{1}-\mathrm{M}_{2}}\)
\[
\begin{aligned}
\mathrm{M} & =\sigma \mathrm{A}=\sigma \pi \mathrm{r}^{2} \\
\mathrm{M}_{1} & =\sigma \pi \mathrm{a}^{2} \text { at } \mathrm{x}_{1}=0 \& \mathrm{M}_{2}=\sigma \pi \mathrm{b}^{2} \text { at } \mathrm{x}_{2}=\mathrm{c} \\
\mathrm{x}_{\mathrm{cm}} & =\frac{\left(\sigma \mathrm{A}_{1}\right)\left(\mathrm{x}_{1}\right)-\left(\sigma \mathrm{A}_{2}\right)\left(\mathrm{x}_{2}\right)}{\sigma \mathrm{A}_{1}-\sigma \mathrm{A}_{2}} \\
& =\frac{\left(\pi \mathrm{a}^{2}\right)(0)-\left(\pi \mathrm{b}^{2}\right)(\mathrm{c})}{\pi \mathrm{a}^{2}-\pi \mathrm{b}^{2}} \\
& =-\left(\frac{\mathrm{b}^{2} \mathrm{c}}{\mathrm{a}^{2}-\mathrm{b}^{2}}\right)
\end{aligned}
\]
\(\mathrm{y}_{\mathrm{cm}}=0\), because of symmetry along x-axis.
44
One mole of an ideal monoatomic gas, contained in a cylinder fitted with movable piston, is originally at \(\mathrm{P}_{1}, \mathrm{~V}_{1}\) and \(\mathrm{T}_{1}=27^{\circ} \mathrm{C}\). The gas is slowly heated. Initially 8.31 watt-hour of energy is added to it; at the same time it is allowed to expand at constant pressure to a new state \(\mathrm{P}_{1}, \mathrm{~V}_{2}\) and \(\mathrm{T}_{2}\). The correct option is :-
Show Solution
Solution
Ans. (d)
\(\mathrm{n}=1, \mathrm{~T}_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}\)
Heat added, \(\mathrm{Q}=8.31\) watt-hr
\[
\begin{aligned}
& =8.31 \times 3600 \mathrm{~J} \\
\mathrm{C}_{\mathrm{P}}= & \frac{5 \mathrm{R}}{2} ; \quad \mathrm{C}_{\mathrm{V}}=\frac{3 \mathrm{R}}{2}
\end{aligned}
\]
For an isobaric process
\[
\begin{aligned}
& \mathrm{Q}=\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{~T} \\
& 8.31 \times 3600=(1)\left(\frac{5 \mathrm{R}}{2}\right) \Delta \mathrm{T} \\
& \mathrm{R}=8.31
\end{aligned}
\]
\[
\begin{aligned}
& \Rightarrow \Delta \mathrm{T}=1440 \mathrm{~K} \\
& \Rightarrow \text { Final temperature } \\
& \quad \mathrm{T}_{2}=\mathrm{T}_{1}+\Delta \mathrm{T}=1740 \mathrm{~K} \text { not } 1740^{\circ} \mathrm{C} \\
& \text { Work done }=\mathrm{P} \Delta \mathrm{~V}=\mathrm{nR} \Delta \mathrm{~T} \\
& \quad=1440 \mathrm{R}
\end{aligned}
\]
Change in internal energy
\[
\begin{aligned}
\Delta \mathrm{U} & =\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{~T} \\
& =(1)\left(\frac{3 \mathrm{R}}{2}\right)(1440) \\
& =2160 \mathrm{R}
\end{aligned}
\]
At constant pressure
\[
\begin{aligned}
& \mathrm{V} \propto \mathrm{~T} \\
& \frac{\mathrm{~V}_{2}}{\mathrm{~V}_{1}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{1740}{300}=5.8
\end{aligned}
\]
45
A non-conducting solid sphere, of radius R, with its center at A, has a spherical cavity of diameter R with center at B as shown. There is no charge in the cavity while the solid part has a uniform volume charge density \(\rho\). Electric potential at the center of the sphere (at point A) is \(\mathrm{V}=\frac{\mathrm{k} \rho R^{2}}{12 \varepsilon_{0}}\) (in SI units) where the value of k is :-
Show Solution
Solution
Ans. (b)
\(\mathrm{Q}_{\text {total }}=\rho \times \frac{4}{3} \pi \mathrm{R}^{3}\)
\[
\begin{aligned}
& \mathrm{Q}_{\text {removed }}=\rho \times \frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)^{3} \\
& \begin{aligned}
\mathrm{V}_{\mathrm{A}} & =\frac{3}{2} \times \frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}_{\mathrm{T}}}{\mathrm{R}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}_{\mathrm{R}}}{\mathrm{R} / 2} \\
& =\frac{1}{4 \pi \varepsilon_{0} \mathrm{R}}\left[\frac{3}{2} \rho \times \frac{4}{3} \pi \mathrm{R}^{3}-2 \rho \times \frac{4}{3} \pi \frac{\mathrm{R}^{3}}{8}\right] \\
& =\frac{5 \rho \mathrm{R}^{2}}{12 \varepsilon_{0}} \\
\mathrm{~K} & =5
\end{aligned}
\end{aligned}
\]
Questions 46–60
NSEP 2025 Physics — Paper Code 62
46
Energy from the Sun falls on the Earth surface at the rate of \(1400 \mathrm{~W} / \mathrm{m}^{2}\), which is known as solar constant. The respective rms values \(\mathrm{E}_{\text {rms }}\) and \(\mathrm{B}_{\text {rms }}\) of electric and magnetic fields in the sunlight (electromagnetic radiation) reaching Earth surface are (Take speed of light \(\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\) ) :-
Show Solution
Solution
Ans. (a)
Intensity of light
\[
\begin{aligned}
\mathrm{I}= & \frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2} \mathrm{c} \\
\mathrm{E}_{0} & =\sqrt{\frac{2 \times 1400}{3 \times 10^{8} \times 8.85 \times 10^{-12}}} \\
& =\mathrm{E}_{\mathrm{rms}} \sqrt{2} \\
\mathrm{E}_{\mathrm{rms}} & =726 \mathrm{v} / \mathrm{m} \\
\mathrm{~B}_{\mathrm{rms}} & =\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{c}} \\
& =2.42 \mu \mathrm{~T}
\end{aligned}
\]
47
The figure below depicts the voltage wave forms of binary input signals A and B and the output signal C of a certain logic gate. The logic gate is
Show Solution
Solution
Ans. (a)
Truth table of given signal
| A | B | C(Output) |
| 0 | 0 | 0 |
| 1 | 1 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 0 | 0 | 0 |
| 1 | 1 | 1 |
Given gate is AND gate
48
In a certain electrical network, the three nodes A, B and C are each at a potential of 1.0 volt while the node D is at a potential 2.0 volt. The potential at the Node O in volt is
Show Solution
Solution
Ans. (c)
Apply KCL at point O
\[
\begin{aligned}
& \frac{V_{A}-V_{O}}{10}+\frac{V_{B}-V_{O}}{10}+\frac{V_{C}-V_{O}}{10}+\frac{V_{D}-V_{O}}{10}=0 \\
& V_{O}=\frac{V_{A}+V_{D}+V_{C}+V_{D}}{4}=\frac{5}{4} \text { volt }
\end{aligned}
\]
49
In young’s double slit experiment, a fine beam of coherent monochromatic light of wavelength \(\lambda=\) 600 nm is incident on identical slits \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) at separation d. The intensity at the central maximum formed at O is \(\mathrm{I}_{\text {max }}\) and the angular fringe width is \(\beta=0.1^{\circ}\). When a thin transparent film is placed in front of the slit \(\mathrm{S}_{2}\), the intensity at O changes. It is found that the smallest thickness of the film. for which the intensity at O becomes half the maximum intensity (i.e. \(\frac{\mathrm{I}_{\text {max }}}{2}\) ), is 250 nm. Neglecting the absorption of light by the film, the zero order fringe earlier at O now forms at \(\mathrm{O}^{\prime}\) where \(\mathrm{OO}^{\prime}=0.5 \mathrm{~mm}\). Choose correct option(s) :-
Show Solution
Solution
Ans. (a,b,d)
\(\lambda=600 \mathrm{~nm} \mathrm{I}_{\text {max }}=4 \mathrm{I}_{0} \quad \beta=\frac{\lambda}{\mathrm{d}}=0.1^{\circ}\)
\[
\begin{aligned}
& \mathrm{I}=4 \mathrm{I}_{0} \cos ^{2} \frac{\theta}{2} \\
& 2 \mathrm{I}_{0}=4 \mathrm{I}_{0} \cos ^{2} \frac{\theta}{2} \\
& \cos \frac{\theta}{2}= \pm \frac{1}{\sqrt{2}} \\
& \therefore \theta=\frac{\pi}{2}=\frac{2 \pi}{\lambda} \Delta \mathrm{x} \\
& \Delta \mathrm{x}=\frac{\lambda}{4} \\
& (\mu-1) \mathrm{t}=\frac{\lambda}{4} \\
& \mu-1=\frac{1}{4} \times \frac{600 \times 10^{-9}}{250 \times 10^{-9}}=0.6 \\
& \therefore \mu=1.6 \quad \text { Option (a) } \\
& \mathrm{OO}=\mathrm{d} \sin \theta \approx \text { dtan } \theta=\frac{\lambda}{4} \\
& \mathrm{~d}=\frac{0.5 \times 10^{-3}}{\mathrm{D}}=\frac{\lambda}{4} \\
& \omega=\frac{\lambda \mathrm{D}}{\mathrm{~d}}=\text { fringe width }=4 \times 0.5 \times 10^{-3}=2 \mathrm{~mm} \\
& \omega=2 \mathrm{~mm} \quad \text { Option (b) } \\
& \omega=\beta \mathrm{D} \\
& \mathrm{D}=\frac{\omega}{\beta}=\frac{2 \times 10^{-3}}{0.1 \times \pi} \times 180 \approx 1.15 \mathrm{~m} \quad \text { Option (d) }
\end{aligned}
\]
50
An insulated non-conducting solid sphere of radius ‘a’, carrying a positive charge +4Q uniformly distributed throughout its volume, is surrounded by a concentric thick conducting spherical shell of inner radius b and outer radius c. This thick shell carries a negative charge -2Q (see figure). The correct option (s) is/are :-
Show Solution
Solution
Ans. (a,c,d)
\[
\overrightarrow{\mathrm{E}}_{\mathrm{A}}=\frac{\mathrm{k}(4 \mathrm{Q}) \overrightarrow{\mathrm{r}}}{\mathrm{R}^{3}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{4 \mathrm{Q}}{\mathrm{a}^{2}} \cdot \overrightarrow{\mathrm{r}} \quad \text { Option (a) }
\]
Option (b) is incorrect
Option (c) is correct
Energy stored inside volume of a solid non conducting sphere is
\[
\mathrm{U}=\frac{1}{10} \frac{\mathrm{kQ}^{2}}{\mathrm{R}}=\frac{1}{10} \times \frac{1}{4 \pi \epsilon_{0}} \times \frac{(4 \mathrm{Q})}{\mathrm{a}}=\frac{2}{5} \frac{\mathrm{Q}^{2}}{\pi \epsilon_{0} \mathrm{a}} \quad \text { Option (d) }
\]
51
One mole of an ideal monoatomic gas of molecular mass M undergoes a cyclic process (ABCA) shown in the figure as a density \((\rho)\) versus pressure (P) curve. The correct option (s) is/are :-
Show Solution
Solution
Ans. (a,c)
A → B
\(\mathrm{P} \propto \rho\)
\(\therefore \mathrm{PV}=\) Constant \(\\) Isothermal pressure
\[
\begin{aligned}
& \mathrm{W}_{\mathrm{AB}}=\mathrm{nR} T \ln \frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\mathrm{P}_{0} \frac{\mathrm{M}}{\rho_{0}} \ln \frac{\rho_{0}}{2 \rho_{0}}=-\frac{\mathrm{P}_{0} \mathrm{M}}{\rho_{0}} \ln 2 \\
& \mathrm{~W}_{\mathrm{AB}}=-\frac{\rho_{0} \mathrm{M}}{\rho_{0}} \ln 2 \quad \text { Option (a) }
\end{aligned}
\]
B → C Isobaric pressure
\[
\begin{aligned}
& W=P \Delta V=2 P_{0}\left(\frac{M}{\rho_{0}}-\frac{M}{2 \rho_{0}}\right)=\frac{P_{0} M}{\rho_{0}} \\
& Q_{A B}=-\frac{P_{0} M}{\rho_{0}} \ell n 2
\end{aligned}
\]
\[
\begin{aligned}
& Q_{B C}=n C_{P} \Delta T=\frac{5}{2}\left(P_{2} V_{2}-P_{1} V_{1}\right)=\frac{5}{2}\left[2 P_{0} \frac{M}{\rho_{0}}-2 P_{0} \frac{M}{2 \rho_{0}}\right] \\
& =\frac{5}{2} P_{0} \frac{M}{\rho_{0}} \\
& Q_{C A}=n C_{V} \Delta T=\frac{3}{2}\left(P_{2} V_{2}-P_{1} V_{1}\right)=\frac{3}{2}\left(P_{0} \frac{M}{\rho_{0}}-2 P_{0} \frac{M}{\rho_{0}}\right) \\
& =-\frac{3}{2} \frac{P_{0} M}{\rho_{0}} \\
& Q_{a}=Q_{B C}=\frac{5}{2} \frac{P_{0} M}{\rho_{0}} \\
& W=\Sigma W=\Sigma Q=\frac{P_{0} M}{\rho_{0}}(1-\ell n 2) \\
& \eta=\frac{W}{Q_{a}}=\frac{2}{5}(1-\ell n 2) \quad \text { Option }(c) \\
& Q_{\text {rej }}=Q_{A B}+Q_{C A}=-\frac{P_{0} M}{\rho_{0}}\left(\ell n 2+\frac{3}{2}\right)
\end{aligned}
\]
52
Two ideal inductors \(\mathrm{L}_{1}=\mathrm{L}_{2}=\mathrm{L}\) and the three identical resistors \(\mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}_{3}=\mathrm{R}\) have been connected to a DC source of emf E as shown in the circuit. When the key K is kept pressed (closed) for a long time, the current through the resistance \(\mathrm{R}_{1}\) on the extreme right is measured to be I. Immediately after releasing (switching off) the key, the current through the resistors is :-
Show Solution
Solution
Ans. (a,b,c)
At steady state : inductor behave as an ideal inductor
Just after opening switch inductor do not allow its correct to change
53
A particle of mass m moves along x axis with its potential energy as \(\mathrm{U}(\mathrm{x})=\frac{\alpha}{\mathrm{x}^{2}}-\frac{\beta}{\mathrm{x}}\) where \(\alpha\) and \(\beta\) are positive constants. The particle is released from rest at \(\mathrm{x}_{0}=\frac{\alpha}{\beta}\). Then
Show Solution
Solution
Ans. (a,b,c,d)
(a) \(\mathrm{U}(\mathrm{x})=\frac{\alpha}{\mathrm{x}^{2}}-\frac{\beta}{\mathrm{x}}=\frac{\alpha}{\mathrm{x}_{0}^{2}}\left[\frac{\mathrm{x}_{0}^{2}}{\mathrm{x}^{2}}-\frac{\beta \mathrm{x}_{0}^{2}}{\mathrm{x} \alpha}\right]\)
\[
\mathrm{U}(\mathrm{x})=\frac{\alpha}{\mathrm{x}_{0}^{2}}\left[\left(\frac{\mathrm{x}_{0}}{\mathrm{x}}\right)^{2}-\frac{\mathrm{x}_{0}}{\mathrm{x}}\right] \quad \because \mathrm{x}_{0}=\frac{\alpha}{\beta}
\]
(b) \(\mathrm{K}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{K}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}\)
\[
\begin{aligned}
& O+\left(\frac{\alpha}{x_{0}^{2}}-\frac{\beta}{x_{0}}\right)=\frac{1}{2} m v^{2}+\frac{\alpha}{x_{0}^{2}}\left(\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right) \\
& \frac{1}{2} m v^{2}=\frac{\alpha}{x_{0}^{2}}\left[\frac{x_{0}}{x}-\left(\frac{x_{0}}{x}\right)^{2}\right]+\left(\frac{\alpha}{x_{0}^{2}}-\frac{\beta}{x_{0}}\right) \\
& =\frac{\alpha}{x_{0}^{2}}\left[\frac{x_{0}}{x}-\left(\frac{x_{0}}{x}\right)^{2}\right]+\left(\frac{\alpha}{x_{0}^{2}}-\frac{\alpha}{x_{0}^{2}}\right) \because \beta=\frac{\alpha}{x_{0}} \\
& v=\left[\frac{2 \alpha}{m x_{0}^{2}}\left\{\frac{x_{0}}{x}-\left(\frac{x_{0}}{x}\right)^{2}\right\}\right]^{1 / 2}
\end{aligned}
\]
(c) for \(\mathrm{V}_{\text {max }} \rightarrow \mathrm{U}_{\text {min }}\)
\[
\begin{aligned}
& \frac{d U}{d x}=0=\alpha \frac{(-2)}{x^{3}}+\frac{\beta}{x^{2}}=0 \\
& \Rightarrow-\frac{2 \alpha}{x}+\beta=0 \\
& x=\frac{2 \alpha}{\beta}=2 x_{0} \\
& \text { so } v_{\max }=\left\{\frac{2 \alpha}{m x_{0}^{2}}\left[\frac{x_{0}}{2 x_{0}}-\left(\frac{x_{0}}{2 x_{0}}\right)^{2}\right]\right\}^{1 / 2} \\
& v_{\max }=\left\{\frac{2 \alpha}{m x_{0}^{2}}\left[\frac{1}{2}-\frac{1}{4}\right]\right\}^{1 / 2}=\sqrt{\frac{\alpha}{2 m x_{0}^{2}}}
\end{aligned}
\]
(d) T.E. \(=\mathrm{K}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=0+\left(\frac{\alpha}{\mathrm{x}_{0}^{2}}-\frac{\beta}{\mathrm{x}_{0}}\right)\) \(=0+\frac{\alpha}{\mathrm{x}_{0}^{2}}-\frac{\alpha}{\mathrm{x}_{0}^{2}}=0\)
54
Two blocks A and B, of masses M and 2M, respectively, are connected by a massless spring of natural length \(\mathrm{L}_{0}\) and spring constant K. The blocks are initially at rest on a smooth horizontal floor with spring at its natural length \(\mathrm{L}_{0}\). A third block C of mass M, identical to that of block A, moves on the floor with speed v along the line joining A and B and collides with a elastically. In the subsequent motion :-
Show Solution
Solution
Ans. (b,d)
After collision between C \& A
(a) at maximum compression A, B move together
Energy conservation \(\mathrm{K}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{K}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}\)
\[
\frac{1}{2} M v^{2}+0=\frac{1}{2}(M+2 M) v_{1}^{2}+\frac{1}{2} k x_{\max }^{2} \tag{1}
\]
momentum conservation
\[
\begin{aligned}
& \mathrm{P}_{\mathrm{i}}=\mathrm{P}_{\mathrm{f}} \\
& \mathrm{MV}=(\mathrm{M}+2 \mathrm{M}) \mathrm{V}_{\mathrm{i}} \\
& \mathrm{~V}_{\mathrm{i}}=\frac{\mathrm{V}}{3} \ldots \ldots(2) \tag{2}
\end{aligned}
\]
Using (1) and (3)
\[
\begin{aligned}
& \frac{1}{2} M V^{2}=\frac{1}{2} 3 M\left(\frac{v^{2}}{9}\right)+\frac{1}{2} K x_{\max }^{2} \\
& x_{\max }=V \sqrt{\frac{2}{3} \frac{M}{K}}
\end{aligned}
\]
(b) K.E. \(=\frac{1}{2}(\mathrm{M}+2 \mathrm{M}) \mathrm{v}_{1}^{2}=\frac{1}{2}(3 \mathrm{M}) \frac{\mathrm{v}^{2}}{9}=\frac{1}{6} \mathrm{Mv}^{2}\)
(d) \(\frac{\mathrm{T}}{4}=\frac{2 \pi}{4} \sqrt{\frac{\mu}{\mathrm{k}}}=\frac{2 \pi}{4} \sqrt{\frac{\mathrm{M}(2 \mathrm{M})}{3 \mathrm{Mk}}}=\frac{\pi}{2} \sqrt{\frac{2 \mathrm{M}}{3 \mathrm{k}}}\)
55
A small block B of mass \(\mathrm{m}=0.25 \mathrm{~kg}\), lying on a frictionless horizontal table, is attached to a massless cord (breaking strength 40 N) passing through a narrow hole C at the centre of the table. Initially when the block is revolving in a circle of radius \(\mathrm{r}_{0}=0.80 \mathrm{~m}\) about a vertical axis through the hole, with a tangential speed of \(\mathrm{v}_{0}=4.00 \mathrm{~m} / \mathrm{s}\); the tension in the string is \(\mathrm{T}_{0}\) and the kinetic energy of the block is \(\mathrm{K}_{0}\). The string is then pulled down slowly from below, decreasing the radius of circular path from \(\mathrm{r}_{0}\) to r so that the kinetic energy of the block is now K and the tension in the string is T. As a result :-
Show Solution
Solution
Ans. (b,c)
(a) angular momentum conservation
\[
\begin{aligned}
& m v_{0} r_{0}=m v r \\
& v=\frac{v_{0} r_{0}}{r} \\
& T_{0}=\frac{m v_{0}^{2}}{r_{0}} \\
& T=\frac{m v^{2}}{r}=\frac{m v_{0}^{2} r_{0}^{2}}{r^{3}} \\
& \Rightarrow T=\frac{T_{0} r_{0}^{3}}{r^{3}} \\
& \text { (b) } k=\frac{1}{2} m v^{2}=\frac{1}{2} m \frac{v_{0}^{2} r_{0}^{2}}{r^{2}}=\frac{k_{0} r_{0}^{2}}{r^{2}} \\
& \text { (c) } T=\frac{T_{0} r_{0}^{3}}{r^{3}}=40 \\
& \Rightarrow \frac{m v_{0}^{2} r_{0}^{2}}{r^{3}}=40 \\
& \Rightarrow \frac{0.25 \times 16 \times 0.64}{40}=r^{3} \\
& \Rightarrow r=0.4 m \\
& \text { (d) } W=\Delta K=k_{f}-k_{i} \\
& =\frac{k_{0} r_{0}^{2}}{\left(\frac{r_{0}}{2}\right)^{2}}-k_{0}=3 k_{0}
\end{aligned}
\]
56
A thin uniform metallic rod, of length \(\ell=1.0 \mathrm{~m}\) and area of cross section \(\mathrm{A}=2 \mathrm{~mm}^{2}\), is made to rotate with angular velocity \(\omega=400 \mathrm{rad} / \mathrm{s}\) in a horizontal plane about a vertical axis through one of its ends. The density and the Young’s modulus of the material of the rod are \(\rho=10^{4} \mathrm{~kg} \mathrm{~m}^{-3}\) and \(\mathrm{Y}=2.0 \times 10^{11}\) \(\mathrm{Nm}^{-2}\). Taking r as the distance of a point on the rod from the axis of rotation, the
Show Solution
Solution
Ans. (a,d)
(a)
\[
\begin{aligned}
& \mathrm{T}=\frac{\mathrm{m}}{2} \omega^{2} \mathrm{r}=\frac{\mathrm{m}}{2} \omega^{2}\left(\frac{\ell}{2}+\frac{\ell}{4}\right) \\
& =\frac{3 \mathrm{~m} \omega^{2} \ell}{8}=\frac{3\left[\rho(\mathrm{~A} \ell) \omega^{2} \ell\right]}{8}=\frac{3 \times 10^{4} \times 2 \times 10^{-6} \times 1 \times 16 \times 10^{4}}{8}=1200 \mathrm{~N}
\end{aligned}
\]
(b)
\[
\begin{aligned}
& \mathrm{T}=\left[\frac{\mathrm{m}}{\ell}(\ell-\mathrm{r})\right] \omega^{2}\left[\mathrm{r}+\frac{\ell-\mathrm{r}}{2}\right] \\
& =\frac{\mathrm{m}(\ell-\mathrm{r})}{\ell} \omega^{2}\left(\frac{\mathrm{r}+\ell}{2}\right)=\frac{\mathrm{m} \omega^{2}\left(\ell^{2}-\mathrm{r}^{2}\right)}{2 \ell}=\frac{\rho \mathrm{A} \ell \omega^{2}\left(\ell^{2}-\mathrm{r}^{2}\right)}{2 \ell} \\
& \mathrm{~T}=\frac{10^{4} \times 2 \times 10^{-6} \times 16 \times 10^{4}\left(1-\mathrm{r}^{2}\right)}{2}=1600\left(1-\mathrm{r}^{2}\right)
\end{aligned}
\]
(c) stress \(=\frac{\mathrm{T}}{\mathrm{A}}=\frac{1200}{2 \times 10^{-6}}=6 \times 10^{8}\)
(d)
\[
\begin{aligned}
& \frac{T}{A}=Y \frac{d \ell}{d r} \\
& \int \frac{T}{A Y} d r=\int d \ell \\
& \int_{0}^{\ell} \frac{1600\left(1-r^{2}\right)}{2 \times 10^{-6} \times 2 \times 10^{11}} d r=\Delta \ell
\end{aligned}
\]
(a) \& (d)
\[
\begin{aligned}
& 4 \times 10^{-3}\left(\ell-\frac{\ell^{3}}{3}\right)=\Delta \ell \\
& \Delta \ell=\frac{8}{3} \times 10^{-3} \mathrm{~m}=\frac{8}{3} \mathrm{~mm}
\end{aligned}
\]
57
One end of a long and thin rope, stretched horizontally with a tension \(\mathrm{T}=8 \mathrm{~N}\), along x axis, is supporting a weight after passing over a pulley fixed on a vertical pole (see figure). At the other end, a simple harmonic oscillator (a clamped iron rod along the axis of a solenoid fed with AC voltage and oscillating between north and south poles) at \(\mathrm{x}=0\), generates a transverse wave of frequency 100 Hz and an amplitude of 2cm, in the rope. The wave propagates along the rope. The mass per unit length of the rope is 20 g/m. Ignoring the effect of gravity (on the rope), the correct option (s) is/are :-
Show Solution
Solution
Ans. (a,d)
Velocity of wave in rope \(=\sqrt{\mathrm{T} / \mu}=\sqrt{\frac{8}{20 \times 10^{-3}}}=20 \mathrm{~m} / \mathrm{s}\)
(a) Wavelength \(=\frac{\mathrm{v}}{\mathrm{f}}=\frac{20}{100} \mathrm{~m}=20 \mathrm{~cm}\)
(b) Max. transverse acceleration \(=\omega^{2} \mathrm{~A}\)
\[
=(2 \pi \times 100)^{2} \times \frac{2}{100}=800 \pi^{2} \mathrm{~m} / \mathrm{s}^{2}
\]
(c) Equation of wave \(=\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx}+\phi)=0.02 \sin \left(200 \pi-\frac{2 \pi}{0.2} \mathrm{x}+\phi\right)\) at \(\mathrm{t}=0, \mathrm{x}=0 \mathrm{y}=-\mathrm{A}\)
\[
\begin{aligned}
& -\mathrm{A}=\mathrm{A} \sin \phi \\
& \phi=-90^{\circ}
\end{aligned}
\]
so equation is : \(\mathrm{y}=-0.02 \cos (200 \pi-10 \pi \mathrm{x})\)
(d) if \(\mathrm{f}=200 \mathrm{~Hz} \lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{20}{200}=10 \mathrm{~cm}\)
58
Nuclei of a radioactive element A are being produced at a constant rate \(\alpha\). The element A has a decay constant \(\lambda\). If there are \(\mathrm{N}_{0}\) nuclei at \(\mathrm{t}=0\), then :-
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Solution
Ans. (b,c,d)
(a) \(\frac{\mathrm{dN}}{\mathrm{dt}}=+\alpha-\lambda \mathrm{N}\)
\[
\begin{aligned}
& \int \frac{d N}{\alpha-\lambda N}=\int d t \\
& \left.\frac{\ell n(\alpha-\lambda N)}{-\lambda}\right|_{N_{0}} ^{N}=t \\
& \alpha-\lambda N=\left(a-\lambda N_{0}\right) e^{-\lambda t} \\
& N=\frac{\alpha}{\lambda}\left(1-e^{-\lambda t}\right)+N_{0} e^{-\lambda t}
\end{aligned}
\]
(b) \(\alpha=\lambda \mathrm{N}_{0}\)
\[
\begin{aligned}
& \mathrm{N}=\mathrm{N}_{0}\left(1-\mathrm{e}^{-\lambda t}\right)+\mathrm{N}_{0} \mathrm{e}^{-\lambda t} \\
& \mathrm{~N}=\mathrm{N}_{0}=\text { constant }
\end{aligned}
\]
(c) \(\alpha=2 \lambda \mathrm{~N}_{0} \quad \mathrm{t} \rightarrow \infty \Rightarrow \mathrm{e}^{-\lambda \mathrm{t}}=0\)
\(\mathrm{N}=2 \mathrm{~N}_{0}\)
(d) \(\alpha=2 \lambda \mathrm{~N}_{0} \quad \mathrm{t}=\frac{\ln 2}{\lambda} \Rightarrow \mathrm{e}^{-\lambda \mathrm{t}}=\frac{1}{2}\)
\[
\mathrm{N}=2 \mathrm{~N}_{0}\left(1-\frac{1}{2}\right)+\mathrm{N}_{0} \frac{1}{2}=\frac{3 \mathrm{~N}_{0}}{2}
\]
59
A single electron orbits around a stationary nucleus of charge +Ze in a hydrogen-like atom, where Z is the atomic number and e is the magnitude of the charge on an electron. It requires 47.25 eV to excite the electron from second Bohr orbit to the third Bohr orbit. Ionization energy of hydrogen atom is 13.6 eV. Then
Show Solution
Solution
Ans. (a,b,c)
(a) \(\mathrm{E}_{3}-\mathrm{E}_{2}=47.25 \mathrm{eV}\)
\[
\begin{aligned}
& \Rightarrow-13.6 Z^{2}\left(\frac{1}{9}-\frac{1}{4}\right)=47.25 \\
& \Rightarrow Z^{2}=25 \\
& \Rightarrow Z=5
\end{aligned}
\]
(b) \(\mathrm{E}_{4}-\mathrm{E}_{3}=-13.6 \times 25 \times\left(\frac{1}{16}-\frac{1}{9}\right)\)
\[
=16.53 \mathrm{eV}
\]
(c) \(\mathrm{n}=1 \rightarrow \mathrm{n}=\infty\)
\[
\mathrm{E}=-13.6 \times 25\left(\frac{1}{\infty}-\frac{1}{1}\right)=340 \mathrm{eV}
\]
Wavelength required \(=\frac{12400}{340} \AA=36.56 \AA\)
(d) \(\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}=\frac{2 \times 6.62 \times 10^{-34}}{2 \times 3.14}=2.19 \times 10^{-34}\)
60
A circular coil of thin insulated copper wire ( \(\mathrm{N}=2000\) turns), wrapped around an iron cylinder of cross-section area \(\Delta \mathrm{S}=0.001 \mathrm{~m}^{2}\), is connected to a suspended type moving coil ballistic galvanometer. The suspended rectangular coil of the galvanometer is of mass \(\mathrm{m}=80 \mathrm{~g}\), length \(\ell=5\) cm, breadth \(\mathrm{b}=3 \mathrm{~cm}\) and has \(\mathrm{n}=100\) turns of fine copper wire wound on a non-metallic frame of ivory. This rectangular coil of the galvanometer is free to execute torsional oscillations in a radial magnetic field \(\mathrm{B}=0.1\) tesla.The galvanometer is being used to measure the charge by employing the formula \(\mathrm{q}=\frac{\mathrm{T}}{2 \pi} \frac{\mathrm{c}}{\mathrm{nAB}} \theta\) [Given that the moment of inertia of the oscillating coil about the vertical axis is \(\mathrm{I}=2.7 \times 10^{-6} \mathrm{kgm}^{2}\) and the torsional constant (torsional rigidity) of the suspension fiber is \(\mathrm{c}=3.0 \times\) \(10^{-3} \mathrm{Nm} /\) radian : \(\mathrm{A}=\ell \times \mathrm{b}\) is the area of the coil] When the magnetic induction of 1.0 weber per meter \({ }^{2}\), perpendicular to the plane of the circular coil, is reversed (in opposite direction), a deflection of 40 mm is observed on a scale placed 1.0 meter away in front of the reflecting mirror attached with the suspension fiber of the rectangular coil. The correct statement(s) is/are
Show Solution
Solution
Ans. (a,b,d)
(a) \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{~T}}{\mathrm{C}}}=2 \pi \sqrt{\frac{2.7 \times 10^{-6}}{3 \times 10^{-3}}}=0.19 \mathrm{~s}\)
(b) \(\Delta \phi=\mathrm{N}\left(\mathrm{BA} \cos 0^{\circ}-\mathrm{BA} \cos 180^{\circ}\right)=2 \mathrm{BAN}\)
\[
=2 \times 1 \times 0.001 \times 200=4 \mathrm{wb}
\]
(c)
Mirror doubles the beam deflection so \(\theta=\frac{\mathrm{S}}{2 \mathrm{~L}}=0.02 \mathrm{rad}\)
\[
\begin{aligned}
& \mathrm{T}=0.19 \mathrm{~s} \\
& \text { so } \mathrm{q}=\frac{\mathrm{T}}{2 \pi} \frac{\mathrm{C}}{\mathrm{nAB}} . \theta \\
& =0.03 \times \frac{0.003}{100 \times 15 \times 10^{-4} \times 0.1} \times 0.02=120 \mu \mathrm{C}
\end{aligned}
\]
(d) \(\mathrm{R}=\frac{\mathrm{N} \Delta \phi}{\mathrm{q}}=\frac{4}{120 \times 10^{-6}}=33.3 \mathrm{k} \Omega\)
Practice Tip
Try each question before opening its solution. For numerical questions, compare not only the final option but also the physical reasoning and intermediate steps.
FAQs
NSEP 2025 Question Paper with Solutions — FAQs
How many questions are there in the NSEP 2025 Physics paper?
The supplied NSEP 2025 paper contains 60 questions, and all 60 are included above with answers and solutions.
What is the NSEP 2025 paper code?
The supplied answer key identifies the paper as Paper Code 62.
Are multiple-correct questions included?
Yes. Questions 46–60 may have more than one correct option, and the answer labels above preserve the multiple-answer combinations from the supplied source.
Can I check the answer before reading the full solution?
Yes. A complete answer key appears near the top of the page. Each question also has a Show Solution control for the detailed working.
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