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Last Updated on November 13, 2026 by XAM CONTENT
NSEP 2024 Solved Paper
Practice the complete NSEP 2024 Question Paper with Solutions for Physics. All 60 questions are arranged in the original order with the supplied answer key, academic diagrams and worked solutions. Attempt each question first and then open Show Solution to check the detailed working.
Exam
NSEP 2024
Subject
Physics
Paper Code
63
Date
24 November 2024
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, NA entries and multiple-answer combinations below follow the supplied NSEP 2024 answer key.
Quick Check
NSEP 2024 Answer Key — Paper Code 63
Use the key for quick checking before opening the detailed solutions. “NA” means the supplied key marks the listed options as not correct.
Questions 1–15
| Question | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Answer | a | c | b | d | NA | b | c | a | d | NA | b | b | b | a | b |
Questions 16–30
| Question | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Answer | d | c | c | d | b | c | a | d | d | d | a | c | a or b | b | a |
Questions 31–45
| Question | 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Answer | b | a | b | b | d | a | b | c | b | b | a | c | b | d | d |
Questions 46–60
| Question | 46 | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Answer | c | b | d | a,b,c,d | a,b,c | b,c | a,c,d | a,b,d | b,c,d | a,c,d | a,b,c | a,b,d | b,c | b,c,d | a,b,c |
Solved Paper
NSEP 2024 Question Paper with Detailed Solutions
60 questions with source diagrams and expandable worked solutions.
60
Questions
Questions 1–15
NSEP 2024 Physics — Paper Code 63
1
OABC is a regular tetrahedron, each side of which is made of uniform wire of resistance \(4 \Omega / \mathrm{m}\). The length of each side is 2 m . The point M is the midpoint of the side BC. The resistance between O and M is
Show Solution
Solution
Ans. (a)
By folding symmetry the circuit can be reduced to
\[
\therefore \mathrm{R}_{\mathrm{OM}}=5 \Omega
\]
Resistance of each side \(=2 \times 4=8 \Omega\)
2
A box weighing W is placed on a rough horizontal floor. The coefficient of static friction between the box and the floor is \(\mu\). To move the box, a pulling force F is applied along the rope joined to the box at an angle \(\theta\) with horizontal. By suitable choice of \(\theta\), the minimum value of F that can make the box move is
Show Solution
Solution
Ans. (c)
For block to move
\[
\begin{aligned}
& F \cos \theta-\mu N \geq 0 \\
& F \sin \theta+N-W=0 \\
& \therefore N=W-F \sin \theta \\
& \therefore F \cos \theta-\mu(W-F \sin \theta) \geq 0
\end{aligned}
\]
For \(\mathrm{F}_{\text {min }}\)
\[
\begin{aligned}
& \mathrm{F}(\cos \theta+\mu \sin \underline{\theta})=\mu \mathrm{W} \\
& \therefore \mathrm{~F}_{\min }=\frac{\mu \mathrm{W}}{\sqrt{1+\mu^{2}}}\left[\text { Maximum value of } \sqrt{\mathrm{a} \sin \theta+\mathrm{b} \cos \theta}=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}\right]
\end{aligned}
\]
3
The external wall of a room measuring 2 m × 3 m consists of a layer of white pine of thickness \(\mathrm{d}_{\text {pine }}=2.0 \mathrm{~cm}\) and a layer of rock wool in succession. The external temperature is 36 K below the indoor temperature (Given the thermal conductivity coefficient of white pine \(\mathrm{K}_{\mathrm{P}}=0.10 \mathrm{Wm}^{-1} \mathrm{kelvin}^{-1}\) and that of rock wool \(\mathrm{K}_{\mathrm{w}}=0.04 \mathrm{Wm}^{-1} \mathrm{kelvin}^{-1}\) ). The thickness of the layer of the rock wool, so that the thermal conduction rate \(\mathrm{P}_{\text {cond }}=\frac{\mathrm{dQ}}{\mathrm{dt}}\) across the wall does not exceed 120 watt (assuming no loss of heat during conduction and no other way of heat transfer other than conduction), is
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Solution
Ans. (b)
Let the thickness of wool be t.
\[
\begin{aligned}
& \therefore R_{\text {pine }}=\frac{\ell}{K A}=\frac{2 \times 10^{-2}}{2 \times 3 \times 0.1} \frac{K}{W} \\
& R_{\text {wool }}=\frac{t}{K A}=\frac{t}{2 \times 3 \times 0.04} \frac{K}{W} \\
& \therefore \frac{d Q}{d t}=\frac{\Delta T}{R_{\text {net }}}=\frac{36}{\frac{2 \times 10^{-2}}{0.6}+\frac{t}{0.24}}
\end{aligned}
\]
Given \(\left.\frac{\mathrm{dQ}}{\mathrm{dt}}\right|_{\text {max }}=120\) Watt
\[
\therefore \frac{36}{\frac{2 \times 10^{-2}}{0.6}+\frac{\mathrm{t}}{0.24}}=120 \Rightarrow \mathrm{t}=0.064 \mathrm{~m}=6.4 \mathrm{~cm}
\]
4
A 240 kg block is suspended from a fixed point O, at the end of a long \((\mathrm{L}=13 \mathrm{~m})\) massless rope. A horizontal force F slowly pushes the block to move it a horizontal distance \(\mathrm{d}=5 \mathrm{~m}\) sideways, to a position B where it remains stationary as shown in the figure. Statement-1 : Force F at position B is 980 N. Statement-2 : Work done by the force to bring the box from A to B is 2352 J
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Solution
Ans. (d)
At B, block is in equilibrium
\[
\begin{aligned}
& \mathrm{F}-\mathrm{T} \sin \theta=0 \\
& \mathrm{~T} \cos \theta-240 \mathrm{~g}=0 \\
& \therefore \tan \theta=\frac{\mathrm{F}}{240 \mathrm{~g}}
\end{aligned}
\]
From geometry, \(\sin \theta=\frac{5}{13}\)
\[
\therefore \frac{5}{12}=\frac{\mathrm{F}}{240 \mathrm{~g}}=\mathrm{F}=980 \mathrm{~N}
\]
From work-energy theorem, since block is moved slowly \(\Rightarrow \Delta \mathrm{KE}=0\) \(\therefore \mathrm{W}_{\mathrm{mg}}+\mathrm{W}_{\mathrm{T}}^{0}+\mathrm{W}_{\mathrm{F}}=0 \quad[\mathrm{~T} \perp\) displacement \(]\)
\[
\begin{aligned}
& \therefore W_{F}=-W_{m g} \\
& \therefore W_{F}=-[-240 g 1(1-\cos \theta)]=240 \times 9.8 \times 1 J=2352 J .
\end{aligned}
\]
5
The number of radioactive nuclei N of a radioactive sample is experimentally measured as a function of time t. At \(\mathrm{t}=0, \mathrm{~N}(\mathrm{t}=0)=50,000\) and at \(\mathrm{t}=10 \mathrm{~s}, \mathrm{~N}(\mathrm{t}=10 \mathrm{~s})=5000 \pm 100\). The half-life of the sample is estimated from these measurement. The error in the estimation of the half-life is approximately [note that for small values of \(\mathrm{x}, \mathrm{x} \rightarrow 0, \ln(1+\mathrm{x}) \approx \mathrm{x}\) ]
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Solution
Ans. (NA)
\(\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}\)
\[
\begin{aligned}
& 5000=50000 \mathrm{e}^{-\frac{\ln 2}{\mathrm{~T}_{1 / 2}}(10)} \\
& \frac{1}{10}=\mathrm{e}^{-\frac{10 \ln 2}{\mathrm{~T}_{1 / 2}}} \\
& \ln 10=\frac{10 \ln 2}{\mathrm{~T}_{1 / 2}} \\
& \mathrm{~T}_{1 / 2}=\frac{10 \ln 2}{\ln 10}=\frac{10 \times 0.7}{2.30}=3.04 \mathrm{sec} \\
& \frac{\mathrm{~N}}{\mathrm{~N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}} \\
& \ln N-\ln N_{0}=-\lambda \mathrm{t} \\
& \frac{1}{\mathrm{~N}} \cdot \Delta \mathrm{~N}=+(\Delta \lambda)(10)=+\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)^{2}} \times \Delta \mathrm{T}_{1 / 2}(10) \\
& \frac{100}{5000}=+\frac{\ln 2}{(3.04)^{2}} \times 10 \times \Delta \mathrm{T}_{1 / 2} \\
& \Delta \mathrm{~T}_{1 / 2}=\frac{100}{5000} \times \frac{(3.04)^{2}}{\ln 2} \times \frac{1}{10} \\
& =0.026 \mathrm{sec}
\end{aligned}
\]
6
An amount of heat equal to 10.61 J is given to an ideal gas at constant pressure of one atm \(\left(1.01 \times 10^{5}\right.\) Pa). As a result the volume of the gas increases by \(30.0 \mathrm{~cm}^{3}\). The gas is
Show Solution
Solution
Ans. (b)
Since pressure is constant.
\[
\begin{aligned}
& \therefore \mathrm{dQ}=\mathrm{nC}_{\mathrm{P}} \mathrm{dT}=(\mathrm{ndT}) \mathrm{C}_{\mathrm{p}}=\frac{(\mathrm{PdV}) \mathrm{C}_{\mathrm{P}}}{\mathrm{R}}[\mathrm{PdV}=\mathrm{nRdT}] \\
& \therefore 10.61=\frac{1.01 \times 10^{5} \times 30 \times 10^{-6}}{\mathrm{R}} \cdot\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R} \\
& \therefore \mathrm{f}=5
\end{aligned}
\]
Thus gas is diatomic
7
A ball is projected from point O on the ground with a certain velocity \(\overrightarrow{\mathrm{u}}\) at angle \(\theta\) from horizontal. When it reaches point P located at a horizontal distance L from O and is at a height h above the ground, the angle \(\alpha\) subtended by the velocity vector \(\overrightarrow{\mathrm{v}}\) with horizontal at this point P is expressed as
Show Solution
Solution
Ans. (c)
\(\mathrm{y}=\mathrm{x} \tan \theta-\frac{\mathrm{g}}{2 \mathrm{u}^{2}} \mathrm{x}^{2} \sec ^{2} \theta\)
\[
\begin{aligned}
& \frac{d y}{d x}=\tan \theta-\frac{g}{2 u^{2}} \cdot 2 x \cdot \sec ^{2} \theta \\
& -\tan \alpha=\tan \theta-\frac{g}{2 u^{2}} \cdot 2 L \sec ^{2} \theta \\
& -\tan \alpha-\tan \theta=-\frac{g}{u^{2}} \cdot L \sec ^{2} \theta \\
& \tan \alpha=\frac{g L}{u^{2}} \sec ^{2} \theta-\tan \theta \\
& h=L \tan \theta-\frac{g}{2 u^{2}} L^{2} \sec ^{2} \theta \\
& \frac{h}{L}=\tan \theta-\frac{g}{2 u^{2}} L \sec ^{2} \theta \\
& \frac{g}{2 u^{2}} L \sec { }^{2} \theta=\tan \theta-\frac{h}{L} \\
& \tan \alpha=2\left(\tan \theta-\frac{h}{L}\right)-\tan \theta=\tan \theta-\frac{2 h}{L}
\end{aligned}
\]
8
A flexible chain, of length L and uniform mass per unit length \(\lambda\), slides off the edge of a frictionless table (sec figure). Initially a length \(\mathrm{y}=\mathrm{y}_{0}\) of the chain hangs over the edge, with the chain held at rest. Now the chain is let free. The velocity, of the chain when the chain becomes completely vertical (i.e. when the chain is just to leave the edge) is
Show Solution
Solution
Ans. (a)
∴ Using energy conservation,
\[
\begin{aligned}
& \mathrm{U}_{\mathrm{i}}+\mathrm{K}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{K}_{\mathrm{f}} \\
& \Rightarrow-\lambda \mathrm{y}_{0} \mathrm{~g} \frac{\mathrm{y}_{0}}{2}+0=-\lambda \operatorname{Lg} \frac{\mathrm{L}}{2}+\frac{1}{2} \lambda \mathrm{Lv}^{2} \\
& \Rightarrow \mathrm{v}=\sqrt{\mathrm{g}\left(\mathrm{~L}-\frac{\mathrm{y}_{0}^{2}}{\mathrm{~L}}\right)}
\end{aligned}
\]
9
A Keplerian telescope is adjusted in its normal setting for parallel rays. Mounting of the objective has diameter D and the diameter of the image of that mounting formed by the eye piece of the telescope is d. Magnifying power of the telescope is
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Solution
Ans. (d)
\(\mathrm{m}=\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}\) Since the telescope is in normal setting ⇒ Distance between objective & eyepiece \(=\mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}\).
\[
\therefore \mathrm{m}=\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}=\frac{\mathrm{D}}{\mathrm{~d}}
\]
10
In the circuit shown below, the resistance \(\mathrm{R}=\sqrt{3} \times 10^{3} \Omega\), the inductance L = 2 H and the capacitance \(\mathrm{C}=1 \mu \mathrm{~F}\) have been connected to an AC supply of the peak voltage of \(\mathrm{V}_{\text {max }}=5\) volt at a frequency \(\omega\). Either the switch \(\mathrm{S}_{1}\) or the switch \(\mathrm{S}_{2}\) is closed at a time. In either case, same maximum current \(\left(\mathrm{i}_{\text {max }}\right)\) is recorded in the circuit. The frequency of the AC source is nearly
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Solution
Ans. (NA)
\(\mathrm{z}_{1}=\mathrm{z}_{2}\)
\[
\begin{aligned}
& R^{2}+X_{C}^{2}=(2 R)^{2}+X_{L}^{2} \\
& R^{2}+\left(\frac{1}{\omega C}\right)^{2}=4 R^{2}+(\omega L)^{2} \\
& \left(\frac{1}{\omega C}\right)^{2}-(\omega L)^{2}=3 R^{2} \\
& \left(\frac{10^{+6}}{\omega}\right)^{2}-[\omega(2)]^{2}=3\left(3 \times 10^{6}\right)=9 \times 10^{6} \\
& \frac{10^{12}}{\omega^{2}}-4 \omega^{4}=9 \times 10^{6} \\
& 10^{12}-4 \omega^{4}=9 \times 10^{6} \omega^{2} \\
& 4 \omega^{4}+9 \times 10^{6} \omega^{2}-10^{12}=0 \\
& 4(2 \pi f)^{4}+9 \times 10^{6}(2 \pi f)^{2}-10^{12}=0 \\
& 4 \times 16 \times 100 f^{4}+9 \times 10^{6} \times 4 \times 10 f^{2}-10^{12}=0 \\
& 6400 f^{4}+36 \times 10^{7} f^{2}-10^{12}=0 \\
& f^{4}+56250 f^{2}-\frac{10^{12}}{6400}=0 \\
& f^{2}=\frac{-56250 \pm \sqrt{3789062500}}{2} \\
& f^{2}=\frac{61555.36-56250}{2}=2652.68 \\
& f=51.50 \mathrm{~Hz}
\end{aligned}
\]
11
There are two given physical quantities \(\mathrm{A}=\frac{\mathrm{F}}{\rho v^{2} \mathrm{D}}\) and \(\mathrm{B}=\frac{\rho v \mathrm{D}}{\eta}\) where F is force, \(\rho\) is density, D is diameter, v is velocity and \(\eta\) is the Coefficient of viscosity. Which of the two A and B is/are dimensionless ?
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Solution
Ans. (b)
\([\mathrm{F}]=\mathrm{MLT}^{-2}\)
\[
\begin{aligned}
& {[\rho]=\mathrm{ML}^{-3}} \\
& {[\mathrm{v}]=\mathrm{LT}^{-1}} \\
& {[\mathrm{D}]=\mathrm{L}} \\
& {[\eta]=\mathrm{ML}^{-1} \mathrm{~T}^{-1}} \\
& \therefore[\mathrm{~A}]=\frac{\mathrm{MLT}^{-2}}{\mathrm{ML}^{-3} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~L}}=\mathrm{L} \\
& {[\mathrm{~B}]=\frac{\mathrm{ML}^{-3} \mathrm{LT}^{-1} \mathrm{~L}}{\mathrm{ML}^{-1} \mathrm{~T}^{-1}}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \rightarrow \text { Dimensionless. }}
\end{aligned}
\]
12
A uniform inextensible string of length L and mass M is suspended vertically from a rigid support. A transverse pulse is allowed to propagate down through the string from the support. At the same time, a ball of mass m is dropped from the rigid support. The ball will pass the pulse at a distance of (from the top of the string i.e. from the support)
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Solution
Ans. (b)
Position of pulse at any time, t is given by
\[
x=t \sqrt{\operatorname{Lg}}-\frac{1}{2} \frac{g}{2} t^{2}
\]
Position of ball at any time t is given by
\[
\mathrm{x}=\frac{1}{2} \mathrm{gt}^{2}
\]
∴ They will cross each other if
\[
\begin{aligned}
& \mathrm{t} \sqrt{\mathrm{Lg}}-\frac{1}{4} \mathrm{gt}^{2}=\frac{1}{2} \mathrm{gt}^{2} \\
& \sqrt{\mathrm{Lg}}=\frac{3}{4} \mathrm{gt} \Rightarrow \mathrm{t}=\frac{4}{3} \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}} \\
& \therefore \mathrm{x}=\frac{1}{2} \mathrm{~g} \frac{16}{9} \frac{\mathrm{~L}}{\mathrm{~g}}=\frac{8}{9} \mathrm{~L}
\end{aligned}
\]
13
In a positron decay, the radionuclide \({ }^{11} \mathrm{C}\) decays according to \({ }^{11} \mathrm{C} \rightarrow{ }^{11} \mathrm{~B}+\mathrm{e}^{+}+v\) [the respective atomic masses are \({ }^{11} \mathrm{C}=11.01142 \mathrm{u},{ }^{11} \mathrm{~B}=11.00931 \mathrm{u}\) and the mass of positron = the mass of electron \(=\mathrm{m}_{\mathrm{e}}=0.00055 \mathrm{u}\) ]. The disintegration energy (i.e. the Q value) is approximately
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Solution
Ans. (b)
\({ }^{11} \mathrm{C} \rightarrow{ }^{11} \mathrm{~B}+\mathrm{e}^{+}+v\)
\[
\begin{aligned}
& \therefore \Delta \mathrm{m}=(11.01142-11.00931-2(0.00055) \mathrm{u} \\
& =0.00101 \mathrm{u} \\
& \therefore \theta=0.00101 \times 931.5 \mathrm{MeV} \approx 0.94 \mathrm{MeV}
\end{aligned}
\]
14
Two mechanical waves, given by \(\mathrm{y}_{1}=\mathrm{A} \sin (8 \mathrm{x}-50 \mathrm{t})\) and \(\mathrm{y}_{2}=\mathrm{A} \sin \left(8 \mathrm{x}+50 \mathrm{t}+\frac{\pi}{3}\right)\) travelling in opposite directions along x-axis, superpose. The position of the node (for \(\mathrm{x}>0\) ) nearest to the origin is (the displacements \(\mathrm{y}_{1}\) and \(\mathrm{y}_{2}\) are in meter)
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Solution
Ans. (a)
\(\mathrm{y}_{1}=\mathrm{A} \sin (8 \mathrm{x}-50 \mathrm{t})\)
\[
\begin{aligned}
& y_{2}=A \sin \left(8 x+50 t+\frac{\pi}{3}\right) \\
& \therefore y_{R}=y_{1}+y_{2}=2 A \sin \left(8 x+\frac{\pi}{6}\right) \cos \left(50 t+\frac{\pi}{6}\right) \\
& \therefore \text { For node, } \sin \left(8 x+\frac{\pi}{6}\right)=0 \\
& \therefore 8 x+\frac{\pi}{6}=n \pi \\
& \Rightarrow 8 x=n \pi-\frac{\pi}{6} \\
& \therefore x=\frac{n \pi-\frac{\pi}{6}}{8}=0.327 m=32.7 \mathrm{~cm}[n=1 \text { for minimum distance from origin }]
\end{aligned}
\]
15
At a certain location on the Earth surface, the intensity of sunlight is \(1.00 \mathrm{~kW} / \mathrm{m}^{2}\). A perfectly reflecting concave mirror, of radius of curvature R and aperture radius r, is facing the Sun to produce the light intensity of \(100 \mathrm{~kW} / \mathrm{m}^{2}\) at the image. Knowing that the disc of the Sun subtends an angle of 0.01 radian at the Earth surface, the relation between R and r is
Show Solution
Solution
Ans. (b)
Total energy incident on the mirror \(=\mathrm{I}_{0} \pi \mathrm{r}^{2}\)
\[
\text { Radius of image formed }=\mathrm{f} \theta=\frac{\mathrm{R}}{2} \theta
\]
∴ Given,
\[
\begin{aligned}
& \frac{\mathrm{I}_{0} \pi \mathrm{r}^{2}}{\pi(\mathrm{f} \theta)^{2}}=100 \Rightarrow \frac{4 \mathrm{r}^{2}}{\mathrm{R}^{2} \theta^{2}}=100 \\
& \therefore \frac{2 \mathrm{r}}{\mathrm{R}}=100 \Rightarrow \frac{\mathrm{r}}{\mathrm{R}}=\frac{10 \times 0.01}{2}=\frac{1}{20} \\
& \therefore \mathrm{R}=20 \mathrm{r}
\end{aligned}
\]
Questions 16–30
NSEP 2024 Physics — Paper Code 63
16
A charge Q is uniformly distributed throughout the volume of a non-conducting sphere of radius R. The total electrostatic energy of electric field inside the sphere is \(\mathrm{U}=\alpha \times \frac{\mathrm{Q}^{2}}{4 \pi \varepsilon_{0} \mathrm{R}}\). The value of \(\alpha\) is:
Show Solution
Solution
Ans. (d)
\(\mathrm{U}_{\mathrm{M}}=\mathrm{U}_{\text {solid }}-\mathrm{U}_{\text {hollow }}\)
\[
=\frac{3}{5} \frac{\mathrm{KQ}^{2}}{\mathrm{R}}-\frac{1}{2} \frac{\mathrm{KQ}^{2}}{\mathrm{R}}=\frac{\mathrm{KQ}^{2}}{10 \mathrm{R}}=\frac{1}{10} \frac{\mathrm{Q}^{2}}{4 \pi \varepsilon_{0} \mathrm{R}}
\]
17
A long straight vertical wire of circular cross section of radius R contains n conduction electrons per unit volume. A current I flows upward in the wire. The expression for magnetic force on an electron at the surface of the wire is [Assume all the conduction electrons are moving with drift velocity]
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Solution
Ans. (c)
\[
\begin{aligned}
& I=n A e V d \\
& \therefore V_{d}=\frac{I}{n e A}=\frac{I}{\pi n e R^{2}} \\
& B=\frac{\mu_{0} R}{2 \pi R} \\
& F=e V_{d} B=e \frac{\mu_{0} I}{2 \pi R} \times \frac{I}{\pi n e R^{2}} \\
& =\frac{\mu_{0} I^{2}}{2 \pi^{2} R^{3}} \text { inward }
\end{aligned}
\]
18
In Young’s double slit experiment, a bright Fringe is observed at \(\mathrm{y}=1.5 \mathrm{~cm}\) from the center of the fringe pattern when monochromatic light of wavelength 612 nm is used. The screen is at 1.4 m from the plane of the two slits, whose separation is 0.4 mm. The number of dark fringes between the center and the said bright fringe at \(\mathrm{y}=1.5 \mathrm{~cm}\) is:
Show Solution
Solution
Ans. (c)
\[
\begin{aligned}
& \mathrm{d}=0.4 \mathrm{~mm} \\
& \tan \theta=\frac{1.5 \times 10^{-2}}{1.4}=\frac{15}{14} \times 10^{-2} \approx \sin \theta
\end{aligned}
\]
For maxima.
\[
\begin{aligned}
& \Delta \mathrm{x}=\mathrm{d} \sin \theta=\mathrm{n} \lambda \\
& \mathrm{n}=\frac{4 \times 10^{-4} \times \frac{15}{14} \times 10^{-2}}{612 \times 10^{-9}}=7 \\
& \therefore 7 \text { dark fringes }
\end{aligned}
\]
19
An illuminated point object is placed in front of an equi-convex lens of refractive \(\mu=1.5\) and focal length \(\mathrm{f}=\) 40 cm, at a distance of \(\mathrm{u}=1.20 \mathrm{~m}\) in front of the lens on its principal axis. Water \(\left(\mu=\frac{4}{3}\right)\) fills the space behind the lens up to a distance of 40 cm from the lens. The final image is formed on the principal axis beyond the lens at a distance v from the lens. The value of v and the nature of image are
Show Solution
Solution
Ans. (d)
\[
\begin{aligned}
& \mu=1.5, \mathrm{f}=40 \mathrm{~cm} \\
& \frac{1}{40}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right)=\frac{1}{\mathrm{R}} \\
& \therefore \mathrm{R}=40 \mathrm{~cm}
\end{aligned}
\]
\(\mathbf{I}_{\mathbf{s t}}\) Surface
\[
\begin{aligned}
& \frac{1.5}{\mathrm{~V}_{1}}-\frac{1}{(-1.2)}=\frac{1.5-1}{0.4}=\frac{1}{0.8} \\
& \frac{1.5}{\mathrm{~V}_{1}}-\frac{1}{0.8}=\frac{1}{1.2}=\frac{1}{2.4} \\
& \therefore \mathrm{~V}_{1}=3.6 \mathrm{~m}
\end{aligned}
\]
\(2{ }^{\text {nd }}\) surface
\[
\begin{aligned}
& \mathrm{u}=+3.6 \mathrm{~m} \\
& \frac{\frac{4}{3}}{\mathrm{v}_{2}}-\frac{1.5}{3.6}=\frac{\frac{4}{3}-1.5}{-0.4}=+\frac{1}{2.4} \\
& \frac{4}{3 \mathrm{v}_{2}}=\frac{1}{2.4}+\frac{1.5}{3.6}=\frac{1}{1.2} \\
& \therefore \mathrm{v}_{2}=\frac{4}{3} \times 1.2=1.6 \mathrm{~m}
\end{aligned}
\]
Last plane surface
\[
\begin{aligned}
& \mathrm{S}_{\mathrm{O}}=1.6-0.4=1.2 \mathrm{~m} \\
& \frac{\mathrm{~S}_{\mathrm{i}}}{1.2}=\frac{1}{\frac{4}{3}} \\
& \therefore \mathrm{~S}_{\mathrm{i}}=\frac{3}{4} \times 1.2=0.9 \mathrm{~m}=90 \mathrm{~cm} \\
& \therefore 130 \mathrm{~cm} \text { behind lens, real image. }
\end{aligned}
\]
20
A paramagnetic gas at room temperature 27 °C is paced in an external uniform magnetic field of magnitude \(\mathrm{B}=1.5\) tesla. The atoms of the gas have magnetic dipole moment \(\mu=1.0 \mu_{\mathrm{B}}\) where \(1 \mu_{\mathrm{B}}=\frac{\mathrm{eh}}{4 \pi \mathrm{~m}}\) is Bohr magneton, and m is the mass of electron. The energy difference \(\Delta \mathrm{U}_{\mathrm{B}}\) between the parallel alignment and the antiparallel alignment of the atom’s magnetic dipole moment, with respect to the external field B, is \(\times \times 10^{-4} \mathrm{eV}\), where the value of x is:
Show Solution
Solution
Ans. (b)
Paramagnetic gas
\[
\begin{aligned}
& \mathrm{T}=300 \mathrm{~K} \\
& \mathrm{~B}=1.5 \mathrm{~T} \\
& \mathrm{U}=-\mathrm{MB}\left(\cos \theta_{2}-\cos \theta_{1}\right), \mathrm{M}=\text { magnetic moment, } \mathrm{B}=\text { Magnetic field } \\
& \theta_{2}=180^{\circ}, \theta_{1}=0^{\circ} \\
& \text { Then } \mathrm{U}=2 \mathrm{MB}=\frac{2 \mathrm{ehB}}{4 \pi \mathrm{~m}} \\
& \mathrm{U}=\frac{2 \times 1.6 \times 10^{-19} \times 6.62 \times 10^{-34} \times 1.5}{4 \times 3.14 \times 9.1 \times 10^{-31}}=2.78 \times 10^{-23} \mathrm{~J} \\
& \mathrm{U}=\frac{2.78 \times 10^{-23}}{1.6 \times 10^{-19}}=1.74 \times 10^{-4} \mathrm{eV}
\end{aligned}
\]
21
A glass capillary with inner diameter of 0.40 mm is vertically submerged in water so that the length of its part protruding above the water is \(\mathrm{h}=25 \mathrm{~mm}\). Surface tension of water is \(\mathrm{T}=0.073 \mathrm{Nm}^{-1}\). Since the water wets the glass completely, it may be concluded that the:
Show Solution
Solution
Ans. (c)
\(\mathrm{d}=0.4 \mathrm{~mm}, \mathrm{r}=0.2 \mathrm{~mm}, \mathrm{~h}=25 \mathrm{~mm}, \mathrm{~T}=0.073 \mathrm{~N} / \mathrm{m}\) Free rise
\[
\begin{aligned}
& \mathrm{h}^{\prime}=\frac{2 \mathrm{~T}}{\mathrm{r} \rho \mathrm{~g}}=\frac{2 \times 0.073}{0.2 \times 10^{-3} \times 1000 \times 10}=73 \mathrm{~mm} \\
& \mathrm{~h}^{\prime}>25 \mathrm{~mm} \\
& \rho \mathrm{gh}=\frac{2 \mathrm{~T}}{\mathrm{R}} \\
& 10^{3} \times 10 \times 25 \times 10^{-3}==\frac{2 \times 73 \times 10^{-3}}{\mathrm{R}} \\
& R=\frac{2 \times 73}{25 \times 10^{4}}=5.84 \times 10^{-4} \mathrm{~m}=0.6 \mathrm{~mm}
\end{aligned}
\]
22
The container A in the figure holds an ideal gas at a pressure of \(5.0 \times 10^{5} \mathrm{~Pa}\) at 27 °C. It is connected to the container B by a thin tube fitted with a closed valve. Container B with volume four times the volume of A holds the same ideal gas at a pressure 1.0 × 105 Pa at 127 °C The valve is now opened to allow the pressure to equalize, but the temperature of each container is maintained as before. The common pressure in the two containers (in kPa) is
Show Solution
Solution
Ans. (a)
Initial Condition
\[
\begin{array}{ll}
\mathrm{P}=5 \times 10^{5} \mathrm{~Pa} & \mathrm{P}=1 \times 10^{5} \mathrm{~Pa} \\
\mathrm{~T}=300 \mathrm{~K} & \mathrm{~T}=400 \mathrm{~K}
\end{array}
\]
\[
\begin{aligned}
& n_{1}+n_{2}=n_{1}^{\prime}+n^{\prime}{ }_{2} \\
& \frac{5 \times 10^{5} \times V}{R \times 300}+\frac{1 \times 10^{5} \times 4 V}{R \times 400}=\frac{P V}{R \times 300}+\frac{P(4 V)}{R \times 400}
\end{aligned}
\]
On Solving
\[
\mathrm{P}=2 \times 10^{5} \mathrm{~Pa}=200 \mathrm{kPa}
\]
23
A thin plastic disc, which is a quarter of a circle of radius \(\mathrm{R}=0.6 \mathrm{~m}\), lies in the first quadrant of x-y plane, with the center of curvature at the origin O as shown. It is charged uniformly on one side (one face) with surface charge density \(\sigma\). Electric potential at point P (0, 0, 0.8 m) is:
Show Solution
Solution
Ans. (d)
Potential by quarter disc \(=\frac{1}{4}\) potential of disc
\[
\begin{aligned}
& \mathrm{V}=\frac{\frac{\sigma}{2 \varepsilon_{0}}\left[\sqrt{\mathrm{R}^{2}+\mathrm{z}^{2}}-\mathrm{z}\right]}{4} \\
& =\frac{\sigma}{8 \varepsilon_{0}}\left[\sqrt{0.6^{2}+0.8^{2}}-0.8\right]=\frac{\sigma}{40 \varepsilon_{0}}
\end{aligned}
\]
24
A spring is compressed under the action of a constant force F. The compression of the spring is \(\xi\). Suppose the direction of the force is reversed suddenly as well as its magnitude is doubled. The maximum extension of the spring beyond its natural length will now be (the spring obey’s Hook’s law).
Show Solution
Solution
Ans. (d)
When force F is applied then at equilibrium position
\[
\mathrm{F}=\mathrm{k} \xi
\]
Work energy theorem
\[
\begin{aligned}
& W_{F}+W_{S P}=0 \\
& 2 F(x+\xi)-\frac{1}{2} k\left(x^{2}-\xi^{2}\right)=0 \\
& 2 F(x+\xi)=\frac{1}{2} k\left(x^{2}-\xi^{2}\right) \\
& 2 F=\frac{1}{2} \frac{F}{\xi}(x-\xi) \\
& 4 \xi=x-\xi \\
& \therefore x=5 \xi
\end{aligned}
\]
25
The charge on a small point object at A produces an electric potential \(\mathrm{V}=3\) volt at a point P. Because of an unavoidable situation, leakage of the charge from the object at starts at \(\mathrm{t}=0\) at a constant rate of \(3 \mu \mathrm{Cs}^{-1}\). To maintain the potential of 3 volt at the point P, the object is made to move towards P with a certain velocity v. When the point object crosses the point D shown in the figure, it is found to have a charge of \(10 \mu \mathrm{C}\) and the direction of its velocity is perpendicular to OD. The resulting magnetic field B, at this instant, at the loacation O (such that \(\mathrm{OD}=1.00 \mathrm{~m}\) ) is:
Show Solution
Solution
Ans. (d)
Potential at point P is given by
\[
\begin{aligned}
& V_{P}=\frac{K q}{r}=\text { constant }=3 \\
& \frac{d q}{d t} r-\frac{q d r}{d t}=0 \\
& \therefore-\frac{d q}{d t} r=-q \frac{d r}{d t} \\
& V=-\frac{d r}{d t}=\frac{r}{q}\left(-\frac{d q}{d t}\right)=\frac{k}{V_{P}}\left(-\frac{d q}{d t}\right) \\
& =\frac{9 \times 10^{9}}{3} \times 3 \times 10^{-6}=9 \times 10^{3} \mathrm{~m} / \mathrm{s}
\end{aligned}
\]
\[
\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{qv}}{\mathrm{r}^{2}}=\frac{10^{-7} \times 10 \times 10^{-6} \times 9 \times 10^{3}}{1^{2}}=9 \mathrm{nT}
\]
26
In the Bohr model of hydrogen atom, the force between the nucleus and the electron is modified as \(\mathrm{F}=\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\left(\frac{1}{\mathrm{r}^{2}}+\frac{\delta}{\mathrm{r}^{3}}\right)\), where \(\delta\) is a small constant. Using the Bohr radius the radius \(\mathrm{a}_{0}=\frac{\varepsilon_{0} \mathrm{~h}^{2}}{\pi \mathrm{me}^{2}}\), the radius of \(\mathrm{n}^{\text {th }}\) orbit is:
Show Solution
Solution
Ans. (a)
\(\mathrm{F}=\frac{\mathrm{e}^{2}}{4 \pi \epsilon_{0}}\left(\frac{1}{\mathrm{r}^{2}}+\frac{\delta}{\mathrm{r}^{3}}\right)=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\)
\[
\begin{aligned}
& m v^{2}=\frac{e^{2}}{4 \pi \varepsilon_{0}}\left(\frac{1}{r}+\frac{\delta}{r^{2}}\right) \& m v r=\frac{n h}{2 \pi} \\
& m \times \frac{n^{2} h^{2}}{4 \pi^{2} m^{2} r^{2}}=\frac{e^{2}}{4 \pi \epsilon_{0}}\left(\frac{1}{r}+\frac{\delta}{r^{2}}\right) \\
& \frac{n^{2} h^{2}}{\pi m r^{2}} \times \frac{\epsilon_{0}}{e^{2}}=\frac{1}{r}+\frac{\delta}{r^{2}} \\
& \frac{n^{2} a_{0}}{r^{2}}=\frac{1}{r}+\frac{\delta}{r^{2}} \\
& \frac{\left(n^{2} a_{0}-\delta\right)}{r^{2}}=\frac{1}{r} \therefore r=\left(n^{2} a_{0}-\delta\right)
\end{aligned}
\]
27
An infinite number of conducting rings having increasing radius \(\mathrm{r}_{0}, \mathrm{r}_{1}, \mathrm{r}_{2}, \mathrm{r}_{3}\) and so no, such that \(\mathrm{r}_{0}=\mathrm{r}\), \(\mathrm{r}_{1}=2 \mathrm{r}, \mathrm{r}_{2}=2^{2} \mathrm{r}, \mathrm{r}_{3}=2^{3} \mathrm{r}\) and so on ………. Upto \(\infty\), have been placed concentrically on a plane. All the rings carry the same current I but the current in consecutive rings is in opposite direction shown. The magnetic field produced at the common centre of the rings is:
Show Solution
Solution
Ans. (c)
We know that magnetic field due to circular loop at its centre is given by \(\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}}\)
\[
\begin{aligned}
& B=\frac{\mu_{0} i}{2 r}\left[1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots \ldots \ldots\right] \\
& =\frac{\mu_{0} i}{2 r} \frac{1}{\left(1-\left(-\frac{1}{2}\right)\right)}=\frac{\mu_{0} i}{2 r} \cdot \frac{2}{3}=\frac{\mu_{0} i}{3 r}
\end{aligned}
\]
28
One mole of an ideal monoatomic gas, initially at temperature T, is heated such a way that its molar heat capacity during the process of heating is \(\mathrm{C}=2 \mathrm{R}\). The volume of the gas gets tripled (at constant pressure) during the process. The final temperature attained by the gas is:
Show Solution
Solution
Ans. (a or b)
\(\mathrm{C}=\mathrm{C}_{\mathrm{v}}+\frac{\mathrm{R}}{1-\mathrm{x}}\)
\[
\begin{aligned}
& 2 \mathrm{R}=\frac{3}{2} \mathrm{R}+\frac{\mathrm{R}}{1-\mathrm{x}} \\
& \therefore \mathrm{x}=-1 \\
& \mathrm{PV}^{-1}=\mathrm{constant} \\
& \therefore \mathrm{P} \propto \mathrm{~V}
\end{aligned}
\]
As per heat capacity
\[
\begin{gathered}
\mathrm{P}_{\mathrm{f}}=3 \mathrm{P} \\
\therefore \mathrm{~T}_{\mathrm{f}}=9 \mathrm{~T}
\end{gathered}
\]
At constant pressure
\[
\mathrm{T}_{\mathrm{f}}=3 \mathrm{~T}
\]
29
Two small positively charged spherical balls are suspended from a common point at the ceiling by non-conducting massless string of equal length \(\ell\). The first ball has mass \(\mathrm{m}_{1}\) and charge \(\mathrm{q}_{1}\) while the second ball has mass \(\mathrm{m}_{2}\) and charge \(\mathrm{q}_{2}\). If the two strings subtend angles \(\theta_{1}\) and \(\theta_{2}\) with the vertical as shown then
Show Solution
Solution
Ans. (b)
\[
\frac{\mathrm{F}}{\sin \left(\pi-\theta_{1}\right)}=\frac{\mathrm{m}_{1} \mathrm{~g}}{\sin (\pi-\alpha)}
\]
\[
\begin{align*}
& \frac{F}{\sin \theta_{1}}=\frac{m_{1} g}{\sin \alpha} \ldots \ldots \text { (i) } \tag{i}\\
& \frac{F}{\sin \left(\pi-\theta_{2}\right)}=\frac{m_{2} g}{\sin (\pi-\beta)} \\
& \frac{F}{\sin \theta_{2}}=\frac{m_{2} g}{\sin \beta} \ldots \ldots \text { (ii) } \tag{ii}
\end{align*}
\]
(i)/(ii)
\[
\begin{aligned}
& \frac{\sin \theta_{2}}{\sin \theta_{1}}=\frac{m_{1}}{m_{2}} \cdot \frac{\sin \beta}{\sin \alpha} \\
& \frac{\sin \theta_{2}}{\sin \theta_{1}}=\frac{m_{1}}{m_{2}} \text { as } \alpha=\beta \\
& \frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{m_{1}}{m_{2}}
\end{aligned}
\]
30
An air filled parallel plate capacitor, with plate area A and plate separation d, is connected to a battery of emf V volt having negligible internal resistance. One of the plates of the capacitor vibrates with amplitude ‘a’ (a << d) and angular frequency \(\omega\). If the instantaneous current in the circuit reaches a maximum value \(\mathrm{I}_{0}\), the amplitude of the vibrations is ‘a’ equal to:
Show Solution
Solution
Ans. (a)
\[
\begin{aligned}
& C=\frac{A \epsilon_{0}}{d+x} \\
& q=C V=\frac{A \epsilon_{0} V}{d+x} \\
& i=\frac{d q}{d t}=\frac{-A \epsilon_{0} V}{(d+x)^{2}} \cdot \frac{d x}{d t} \\
& i=\frac{-A \epsilon_{0} V}{(d+x)^{2}} v \\
& i_{\max }=\frac{A \epsilon_{0} V}{d^{2}}(a \omega)=\frac{A \epsilon_{0} a \omega V}{d^{2}}=I_{0} \\
& \therefore a=\frac{I_{0} d^{2}}{A \epsilon_{0} \omega V}
\end{aligned}
\]
Questions 31–45
NSEP 2024 Physics — Paper Code 63
31
A small ball of mass m is attached to one end of a massless un-stretchable string of length \(\ell\) and is held at the point P. The other end of the string is fixed to a support at O such that OP is horizontal. The minimum downward speed u, that should be imparted to the ball at the point P so that the ball can complete the vertical circle without any slack in the string is
Show Solution
Solution
Ans. (b)
\[
\begin{aligned}
& \frac{1}{2} m u^{2}+m g \ell=\frac{1}{2} m v_{\mathrm{H}}^{2}+\operatorname{mg}(2 \ell) \\
& \frac{1}{2} m u^{2}=\frac{3 m g \ell}{2} \\
& \mathrm{u}=\sqrt{3 g \ell}
\end{aligned}
\]
32
An illuminated point object is placed on the principal axis, in front of an equi-convex glass lens of focal length \(\mathrm{f}=40 \mathrm{~cm}\), at a distance of \(\mathrm{u}=1.20 \mathrm{~m}\) from the lens. A reflecting plane mirror has been placed behind the lens perpendicular to the principal axis and facing the lens. The nature of the final image and the distance of the plane mirror from the lens so as to form the final image at the plane mirror itself, is :-
Show Solution
Solution
Ans. (a)
for \(1^{\text {st }}\) refraction from lens
\[
\begin{aligned}
& \frac{1}{\mathrm{v}}-\frac{1}{-120}=\frac{1}{40} \\
& \frac{1}{\mathrm{v}}=\frac{1}{40}-\frac{1}{120}=\frac{2}{120}
\end{aligned}
\]
\[
\mathrm{v}=60 \mathrm{~cm}
\]
Now reflection from plane mirror
\[
\begin{aligned}
& \mathrm{u}_{2}=-(\mathrm{x}-60) \\
& \mathrm{v}_{2}=\mathrm{x}-60 \text { (Behind mirror) }
\end{aligned}
\]
Now final refraction from lens.
\[
\begin{aligned}
& u_{3}=-(2 x-60) f=40 v_{3}=-40 \\
& \frac{1}{-40}+\frac{1}{+(2 x-60)}=\frac{1}{40} \\
& \frac{1}{2 x-60}=\frac{1}{20} \\
& x=40 \mathrm{~cm}
\end{aligned}
\]
33
A ball is kicked horizontally from the top C of a hemispherical rock ACB of radius R on a horizontal ground, with a velocity v, so as not to hit the rock at any point during its flight. Choose the correct statement (see the figure)
Show Solution
Solution
Ans. (b)
\[
\begin{aligned}
& \mathrm{t}=\sqrt{\frac{2 \mathrm{R}}{\mathrm{~g}}} \mathrm{v}_{\min }=\sqrt{\mathrm{gR}} \\
& \mathrm{R}_{\min }=\sqrt{2} \mathrm{R} \\
& \mathrm{BD}=\sqrt{2} \mathrm{R}-\mathrm{R}=(\sqrt{2}-1) \mathrm{R}
\end{aligned}
\]
34
The ratio of length, radii and Young’s moduls of steel to brass wires in the figure are \(\alpha, \beta\) and \(\gamma\), respectively. The corresponding ratio of the increase in their length is :-
Show Solution
Solution
Ans. (b)
\(\mathrm{T}_{\mathrm{B}}=3 \mathrm{~g}\)
\[
\begin{align*}
& \frac{3 \mathrm{~g}}{\pi \mathrm{R}_{\mathrm{B}}^{2}}=\mathrm{Y}_{\mathrm{B}} \frac{\Delta \ell_{\mathrm{B}}}{\ell_{\mathrm{B}}} \tag{1}\\
& \frac{5 \mathrm{~g}}{\pi \mathrm{R}_{\mathrm{S}}^{2}}=\mathrm{Y}_{\mathrm{S}} \frac{\Delta \ell_{\mathrm{S}}}{\ell_{\mathrm{S}}} \tag{2}
\end{align*}
\]
eq. (1) divides by eq. (2)
\[
\begin{aligned}
& \frac{3 \mathrm{R}_{\mathrm{S}}^{2}}{5 \mathrm{R}_{\mathrm{B}}^{2}}=\frac{\mathrm{Y}_{3}}{\mathrm{Y}_{5}} \frac{\Delta \ell_{\mathrm{B}}}{\Delta \ell_{\mathrm{S}}} \cdot \frac{\ell_{\mathrm{S}}}{\ell_{\mathrm{B}}} \\
& \frac{3}{5} \beta^{2}=\frac{1}{\gamma} \frac{\Delta \ell_{\mathrm{B}}}{\Delta \ell_{\mathrm{S}}} \cdot \alpha \\
& \frac{\Delta \ell_{\mathrm{S}}}{\Delta \ell_{\mathrm{B}}}=\frac{5 \alpha}{3 \gamma \beta^{2}}
\end{aligned}
\]
35
A uniform beam of light of intensity \(60 \mathrm{~mW} / \mathrm{m}^{2}\) is incident on a totally absorbing sphere of radius 2.0 \(\mu \mathrm{m}\). The density of the material of the sphere is \(\rho=5.0 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). The sphere is placed in a region of space whre gravitational force can be neglected/ignored. The magnitude of acceleration of the sphere due to the incidence of the light is :-
Show Solution
Solution
Ans. (d)
Radiation force \(=\frac{\mathrm{I}}{\mathrm{C}} \pi \mathrm{R}^{2}=\mathrm{ma}\)
\[
\begin{aligned}
& \frac{I}{C} \cdot \pi R^{2}=\rho \cdot \frac{4}{3} \pi R^{3} a \\
& a=\frac{3 I}{4 R \rho C}=\frac{3 \times 60 \times 10^{-3}}{4\left(2 \times 10^{-6}\right) 5 \times 10^{3} \times 3 \times 10^{8}} \\
& a=\frac{180 \times 10^{-3}}{12 \times 10^{6}}=1.5 \times 10^{-8} \mathrm{~m} / \mathrm{s}^{2} \approx 0
\end{aligned}
\]
36
The fuse, in the upper branch of the circuit shown, is an ideal 4.0 A fuse. The fuse has zero resistance as long as current through it remains less than 4.0 A. The fuse blows out when the current reaches 4.0 A. Needless to say that the resistance becomes infinite thereafter. Switch S is closed at time t = 0. The fuse blows out at time :-
Show Solution
Solution
Ans. (a)
initially \(\mathrm{i}_{\mathrm{R}}=0\) (as fuse is resistance less)
\[
\begin{aligned}
& V-L \frac{d i}{d t}=0 \\
& i=\frac{V}{L} \cdot t \\
& 4=\frac{10}{5} t \\
& t=2 \sec
\end{aligned}
\]
37
The bent wire PQR shown in the figure lies in a uniform magnetic field \(\overrightarrow{\mathrm{B}}=3.0 \hat{\mathrm{i}}+4.0 \hat{\mathrm{k}} \mathrm{T}\). The direction \(\hat{\mathrm{k}}\) being normal to the plane of the paper and directed towards the viewer. The two straight sections PQ and QR of the wire each have length 2.0 m and the wire carries a current of 2.5 A. Net force on the wire due to magnetic field \(\overrightarrow{\mathrm{B}}\) is :-
Show Solution
Solution
Ans. (b)
\[
\begin{aligned}
& \mathrm{i}=2.5 \mathrm{~A} \\
& \overrightarrow{\mathrm{~F}}=\mathrm{i} \overrightarrow{\mathrm{~d} \ell} \times \overrightarrow{\mathrm{B}}=5 \hat{\mathrm{i}} \times(3 \hat{\mathrm{i}} \times 4 \hat{\mathrm{k}}) \\
& \overrightarrow{\mathrm{F}}=-20 \hat{\mathrm{j}} . \mathrm{N} .
\end{aligned}
\]
38
A charged spherical capacitor consists of two concentric spherical shells of radii a and b (b > a). Half of the stored electrical energy of this system lies within a spherical region of radius r if :-
Show Solution
Solution
Ans. (c)
\[
\mathrm{E}=\frac{\mathrm{kQ}}{\mathrm{r}^{2}}
\]
Energy stored in capactor \(\mathrm{E}_{1}=\int \frac{1}{2} \epsilon_{0} \mathrm{E}^{2} . \mathrm{dV}\)
\[
\begin{aligned}
& E_{1}=\int \frac{1}{2} \epsilon_{0} \frac{k^{2} Q^{2}}{r^{4}} \cdot 4 \pi r^{2} d r \\
& E_{1}=\frac{k Q^{2}}{2} \cdot\left(\frac{1}{a}-\frac{1}{b}\right) \\
& E_{2}=\frac{1}{2} E_{1}=\frac{k Q^{2}}{2} \cdot \int_{a}^{R} r^{-2} d r \\
& \frac{k Q^{2}}{4}\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{k Q^{2}}{2}\left(\frac{1}{a}-\frac{1}{r}\right) \\
& \frac{1}{2 a}-\frac{1}{2 b}=\frac{1}{a}-\frac{1}{r} \\
& \frac{1}{r}=\frac{1}{2 a}+\frac{1}{2 b} \\
& r=\frac{2 a b}{a+b}
\end{aligned}
\]
39
Two stars of masses \(\mathrm{M}_{1}=\mathrm{M}_{\mathrm{s}}\) and \(\mathrm{M}_{2}=15 \mathrm{M}_{\mathrm{s}}\) (where \(\mathrm{M}_{\mathrm{s}}=\) mass of the Sun) form a binary system. The stars are revolving round each other, always being at a separation of \(\mathrm{d}=4 \mathrm{AU}\) between them (1 AU is distance of the Earth from the Sun), move in circular orbits about their center of mass. The period of revolution of each star is :-
Show Solution
Solution
Ans. (b)
\[
\begin{aligned}
& \mathrm{r}=1 \mathrm{Au} \\
& \omega=\sqrt{\frac{\mathrm{G}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)}{\mathrm{d}^{3}}}=\sqrt{\frac{\mathrm{G}(16 \mathrm{Ms})}{64 \mathrm{r}^{3}}} \\
& \omega=\frac{1}{2} \sqrt{\frac{\mathrm{GMs}}{\mathrm{r}^{3}}}=\frac{\omega_{\text {earth }}}{2} \\
& \mathrm{~T}=2 \mathrm{~T}_{\text {earth }}=2 \text { years. }
\end{aligned}
\]
40
The total energy released in \(\alpha\) decay of a stationary Radium nucleus \({ }^{226} \mathrm{Ra}\) (mass 116 u) is \(\mathrm{Q}=4.9\) MeV (mass of \(\alpha\) particle \(\mathrm{m}_{\alpha}=4 \mathrm{u}\) ) then the :-
Show Solution
Solution
Ans. (b)
\(\mathrm{k}_{\alpha}=\left(\frac{\mathrm{A}-4}{\mathrm{~A}}\right) \mathrm{Q}=\left(\frac{226-4}{226}\right) 4.9 \mathrm{MeV}\)
\[
=\frac{222}{226}(4.9 \mathrm{MeV})=4.81 \mathrm{MeV}
\]
41
A listener at rest (with respect to the air and the ground) hears a sound of frequency \(\mathrm{f}_{1}\) from a source moving towards him with a velocity of \(15 \mathrm{~ms}^{-1}\), towards East. If the listener now moves towards the approaching source with a velocity of \(25 \mathrm{~ms}^{-1}\), towards West, he hears a frequency \(\mathrm{f}_{2}\) that differs from \(\mathrm{f}_{1}\) by 40 Hz. The frequency f of the sound produced by the source is (speed of sound in the air is 340 \(\mathrm{ms}^{-1}\) )
Show Solution
Solution
Ans. (a)
Case-1
\[
f_{1}=\left(\frac{v}{v-v_{1}}\right) f_{0}
\]
Case-2
\[
\begin{aligned}
& f_{2}=\left(\frac{v+v_{2}}{v-v_{2}}\right) f_{0} \\
& f_{2}-f_{1}=\left(\frac{v+v_{2}}{v-v_{1}}\right) f_{0}-\left(\frac{v}{v-v_{1}}\right) f_{0}=\left(\frac{v_{2}}{v-v_{1}}\right) f_{0} \\
& 40=\left(\frac{25}{340-15}\right) f_{0} \\
& f_{0}=\frac{40 \times 325}{25} \\
& f_{0}=520 \mathrm{~Hz}
\end{aligned}
\]
42
A uniform rod of length \(\ell\) swings from a pivot as a physical pendulum. The position of the pivot can be varied along the length of the rod. The minimum time period with which the rod can oscillate with an appropriate position of the pivot is \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}}\) where L is equal to
Show Solution
Solution
Ans. (c)
for minimum time period in physical pendulum. \(\ell=\mathrm{k}(\mathrm{k}=\) Radius of Gyration \()\) so for uniform rod.
\[
\begin{aligned}
& \mathrm{K}=\frac{\mathrm{L}}{\sqrt{12}}=\frac{\mathrm{L}}{2 \sqrt{3}} \\
& \ell_{\text {eff. }}=\frac{\mathrm{K}^{2}}{\ell}+\ell=\frac{\ell^{2}}{\ell}+\ell \Rightarrow \ell_{\text {eff }}=2 \ell=\frac{\ell}{\sqrt{3}}
\end{aligned}
\]
43
The number density of conduction electrons in pure silicon at room temperature is about \(10^{16} \mathrm{~m}^{-3}\). The number density of conduction electrons is increased by a factor of \(10^{6}\) by doping the silicon lattice with phosphorus. Assume that at room temperature every phosphorus atom contributes one electron to the conduction band. The fraction of silicon atoms replaced by phosphorus atoms is (Given that the density of silicon is \(2.33 \mathrm{gm} / \mathrm{cm}^{3}\) and that the molar mass of silicon \(\mathrm{M}=28.1 \mathrm{gm}\) )
Show Solution
Solution
Ans. (b)
\(\mathrm{n}_{0}=10^{16}\)
\[
\begin{aligned}
& 10^{6} \mathrm{n}_{0}=\mathrm{n}_{0}+\mathrm{n}_{\mathrm{p}} \\
& \left(10^{6}-1\right) \mathrm{n}_{0}=\mathrm{n}_{\mathrm{p}} \\
& \mathrm{n}_{\mathrm{p}}=10^{22}
\end{aligned}
\]
No. of silicon in 1gm is
\[
\begin{aligned}
& \mathrm{N}_{\mathrm{si}}=\frac{\mathrm{m}_{\mathrm{si}} \mathrm{~N}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{si}}}=\frac{1 \times 6.023 \times 10^{23}}{28.1} \\
& \mathrm{~N}_{\mathrm{Si}}=2.143 \times 10^{22} \text { atoms } / \mathrm{gm}
\end{aligned}
\]
No. of phosphorous required is \(=10^{6}\) atoms \(/ \mathrm{m}^{3}\) No. of si atoms \(/ \mathrm{m}^{3}=\mathrm{Nsi} \times \rho\)
\[
\begin{aligned}
& =2.143 \times 10^{22} \times 2.33 \times 10^{+6} \\
& =4.99 \times 10^{28} \approx 5 \times 10^{28} \\
& \frac{\mathrm{n}_{\mathrm{p}}}{\mathrm{n}_{\mathrm{si}}}=\frac{10^{22}}{5 \times 10^{28}}=2 \times 10^{-7}
\end{aligned}
\]
44
A soap bubble 10 cm in radius, with a film thickness of \(\frac{10}{3} \times 10^{-6} \mathrm{~cm}\), is charged to a potential of 80 V. The bubble bursts and converts into a single spherical drop. Assuming that the soap solution is a good conductor, the potential at the surface of the drop is :-
Show Solution
Solution
Ans. (d)
\(\mathrm{v}=\frac{\mathrm{kQ}}{\mathrm{R}}\)
\[
\mathrm{Qk}=\mathrm{vR}
\]
Volume conservation
\[
\begin{aligned}
& 4 \pi \mathrm{R}^{2} \mathrm{t}=\frac{4}{3} \pi \mathrm{r}^{3} \\
& \mathrm{r}=10^{-3} \mathrm{~m} \\
& \mathrm{v}=\frac{\mathrm{kQ}}{\mathrm{r}}=\frac{\mathrm{vR}}{\mathrm{r}}=80\left(\frac{10^{-1}}{10^{-3}}\right) \\
& =8000 \mathrm{v}=8 \mathrm{kv}
\end{aligned}
\]
45
Three resistance, each of \(\mathrm{R}=4 \Omega\) rated 16 W, are connected across A and B as shown. Potential difference of V volt is applied between points A and B. Statement 1 : Maximum potential difference V that can be applied is 12 volt. Statement 2 : Maximum power that can be dissipated is 24 watt.
Show Solution
Solution
Ans. (d)
\[
\begin{aligned}
& \mathrm{i}_{\max }^{2} \mathrm{R}=16 \\
& \mathrm{i}_{\max }^{2}=4 \\
& \mathrm{i}_{\max }=2 \mathrm{amp} . \\
& \mathrm{v}_{\max }=\frac{\mathrm{iR}}{2}+\mathrm{iR}
\end{aligned}
\]
\[
\begin{aligned}
& =\mathrm{i}\left(\frac{3 \mathrm{R}}{2}\right)=12 \mathrm{~V} \\
& \mathrm{P}_{\max }=\frac{\mathrm{i}^{2}}{4} \mathrm{R}+\frac{\mathrm{i}^{2}}{4} \mathrm{R}+\mathrm{i}^{2} \mathrm{R} \\
& =\frac{3 \mathrm{i}^{2} \mathrm{R}}{2}=\frac{3}{2}(4) \cdot(4)^{2}=24
\end{aligned}
\]
Questions 46–60
NSEP 2024 Physics — Paper Code 63
46
Water is filled in a vertical cylinder up to a certain height h. The cylinder is made to rotate with an angular velocity \(\omega\) about a vertical axis coinciding with the axis of the cylinder. The water surface seen from top appears as a/an
Show Solution
Solution
Ans. (c)
Surface of water is seen as rotating parabola so it is paraboloid \(\left(\mathrm{y}=\frac{\omega_{0}^{2}}{2 \mathrm{~g}} \mathrm{x}^{2}\right)\)
47
A uniformly charged non-conducting sphere with its center at C carries positive charge with uniform charge density \(+\rho\), except in a spherical cavity (inside the sphere) with center O. The electric field E at any point inside the cavity is :-
Show Solution
Solution
Ans. (b)
\(\overrightarrow{\mathrm{F}}=\frac{\rho}{3 \epsilon_{0}} \overrightarrow{\mathrm{CO}}\) (uniform electric field in the cavity)
48
A cylindrical tank with base area \(\mathrm{A}=0.05 \mathrm{~m}^{2}\) is filled with water up to a height \(\mathrm{H}=50 \mathrm{~cm}\). There is a small hole of area \(\mathrm{a}=0.001 \mathrm{~m}^{2}(\mathrm{a} \ll \mathrm{A})\) in the bottom of the tank. It takes time t to empty the tank up to a height \(\frac{\mathrm{H}}{2}\) (i.e. to empty half of the water volume). The additional time required to empty the tank completely is :-
Show Solution
Solution
Ans. (d)
\(\quad \mathrm{T}=\frac{\mathrm{A}}{\mathrm{A}_{0}} \sqrt{\frac{2}{\mathrm{~g}}}(\sqrt{\mathrm{H}}-\sqrt{\mathrm{h}}) \quad\) ( \(\mathrm{H}=\) initial height, \(\mathrm{h}=\) final height)
\[
\begin{aligned}
\mathrm{h} & =\frac{\mathrm{H}}{2} \\
\mathrm{t} & =\frac{\mathrm{A}}{\mathrm{~A}_{0}} \sqrt{\frac{2}{\mathrm{~g}}}\left(\sqrt{\mathrm{H}}-\sqrt{\frac{\mathrm{H}}{2}}\right) \\
\mathbf{h} & =\mathbf{0}, \mathrm{H}=\frac{\mathrm{H}}{2}
\end{aligned}
\]
\[
\begin{aligned}
& t_{1}=\frac{A}{A_{0}} \sqrt{\frac{2}{g}}\left(\sqrt{\frac{H}{2}}-0\right) \\
& \frac{t}{t_{1}}=\frac{\sqrt{H}-\sqrt{\frac{H}{2}}}{\sqrt{\frac{H}{2}}} \\
& \frac{t}{t_{1}}=\sqrt{2}-1 \\
& t_{1}=\frac{t}{\sqrt{2}-1} \\
& t_{1}=t(\sqrt{2}+1)
\end{aligned}
\]
49
According to Bohr theory, in the ground state of hydrogen atom, an electron revolves in circular orbit of radius r with velocity v and circulation frequency f. The magnetic dipole moment of the electronic orbit is \(\mathrm{p}_{\mathrm{m}}\). The magnetic field produced by the circulating electron at the center of the atom is B. Then for a \(\mathrm{He}^{+}\)ion in the state \(\mathrm{n}=2\) (electron in the \(2^{\mathrm{nd}}\) orbit)
Show Solution
Solution
Ans. (a, b, c, d)
(a) \(\mathrm{r}=\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{\pi \mathrm{mZe}^{2}}=2 \mathrm{r}_{0}\)
(b) \(\mathrm{M}=\frac{\mathrm{q}}{2 \mathrm{~m}} \times \mathrm{L}=\frac{\mathrm{q}}{2 \mathrm{~m}} \times \frac{2 \mathrm{~h}}{2 \pi}=2 \mu_{0}\) (c) \(\mathrm{f}=\frac{\mathrm{v}}{\mathrm{r}}=\frac{\frac{\mathrm{Ze}^{2}}{2 \epsilon_{0} \mathrm{nh}}}{\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{\pi \mathrm{mZe}^{2}}} \propto \frac{\mathrm{Z}^{2}}{\mathrm{n}^{3}} \Rightarrow \frac{\mathrm{f}_{0}}{2}\) (d) \(\mathrm{B}=\frac{\mu_{0} \mathrm{q} \omega}{4 \pi \mathrm{r}} \propto \frac{\mathrm{f}}{\mathrm{r}}=\frac{\mathrm{B}_{0}}{4}\)
50
A body of mass \(\mathrm{m}=0.25 \mathrm{~kg}\) is moving along x axis under the action of a conservative force. Its potential energy as a function of position \(x\) is given by \(\mathrm{U}(\mathrm{x})=-\frac{100 \mathrm{x}}{\mathrm{x}^{2}+4} \mathrm{~J}\) (x in m). Then
Show Solution
Solution
Ans. (a, b, c)
\(\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dx}}=100 \frac{\left[\left(\mathrm{x}^{2}+4\right)-\mathrm{x} \times 2 \mathrm{x}\right]}{\left(\mathrm{x}^{2}+4\right)^{2}}\)
\[
\begin{aligned}
& =100 \frac{\left[4-x^{2}\right]}{\left(x^{2}+4\right)^{2}}=0 \Rightarrow x= \pm 2(\text { equilibrium position }) \\
& F_{0}=\frac{400}{16}=25 N \\
& \frac{d^{2} U}{d x^{2}}=-100 \frac{\left[-2 x\left(x^{2}+4\right)^{2}-\left(4-x^{2}\right) \times 2\left(x^{2}+4\right) \times 2 x\right]}{\left(x^{2}+4\right)^{4}} \\
& =\frac{-100[-4 \times 64]}{64^{2}}=+ \text { ve for } x=2 \Rightarrow \text { stable unstable for } x=-2 \\
& 0.25 a=F=100\left[\frac{4-(2+\Delta)^{2}}{(2+4)^{2}}\right] \Rightarrow a=\frac{400 \times 4}{64} \Rightarrow \omega=5 \mathrm{rad} / \mathrm{sec}
\end{aligned}
\]
51
A uniform rod OA of length L is freely pivoted at one end O. Let C be the midpoint (i.e. \(\mathrm{OC}=\frac{\mathrm{L}}{2}=\) CA). Choose the correct statement (s) :-
Show Solution
Solution
Ans. (b, c)
\[
\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{mg} \ell / 2}}=2 \pi \sqrt{\frac{\frac{\mathrm{~m} \ell^{2}}{3}}{\frac{\mathrm{~m} \ell}{2}}}=2 \pi \sqrt{\frac{2 \ell}{3 \mathrm{~g}}} \\
& \mathrm{mg} \frac{\mathrm{~L}}{2}=\frac{1}{2} \times \frac{\mathrm{mL}^{2} \omega^{2}}{3} \\
& \Rightarrow \omega=\sqrt{\frac{3 \mathrm{~g}}{\mathrm{~L}}}
\end{aligned}
\]
\[
\begin{aligned}
& -\frac{\mathrm{mg}}{2} \times \frac{\mathrm{L}}{4}(1-\cos \theta)=\mathbf{0}-\frac{1}{2} \times \frac{\mathrm{m}}{2} \frac{(\mathrm{~L} / 2)^{2}}{3} \times \frac{3 \mathrm{~g}}{\mathrm{~L}} \\
& -\frac{\mathrm{mgL}(1-\cos \theta)}{8}=-\frac{\mathrm{mL}^{2}}{48} \times \frac{3 \mathrm{~g}}{\mathrm{~L}} \\
& 1-\cos \theta=\frac{1}{2} \\
& \theta=60^{\circ}
\end{aligned}
\]
Lower part has vertical circular motion.
52
A boy stands on a stationary ice boat on a frictionless horizontal flat iceberg. The boy and the boat have a combined mass of \(M=60 \mathrm{~kg}\). Two balls of masses \(\mathrm{m}_{1}=10 \mathrm{~kg}\) and \(\mathrm{m}_{2}=20 \mathrm{~kg}\) are placed on the boat. In order to get the boat moving, the boy throws the balls backward horizontally either in succession or both together. In each case the balls are thrown backward with a certain speed \(\mathrm{v}_{\text {rel }}=6\) \(\mathrm{ms}^{-1}\) relative to the boat just when the ball is being thrown. The resulting speed of the system of the boat and the boy is
Show Solution
Solution
Ans. (a, c, d)
\_\_\_\_
\[
\begin{aligned}
& 60 v_{1}=P_{1 \mathrm{C}}=\mu v_{\text {rel }}=\frac{60 \times 30}{90} \times 6 \\
& v_{1}=2 \mathrm{~m} / \mathrm{s} \\
& v_{1}^{\prime}=\frac{80 \times 10}{90} \times 6 \\
& 60\left(v_{1}-\frac{2}{3}\right)=\frac{60 \times 20}{80} \times 6 \\
& v_{1}=\frac{2}{3}+\frac{3}{2}=\frac{13}{6}=2.17 \mathrm{~m} / \mathrm{s}
\end{aligned}
\]
\[
\begin{aligned}
& (60+10) v_{1}^{\prime}=\frac{70 \times 20}{90} \times 6 \\
& 60\left(v_{1}-\frac{4}{3}\right)=\frac{60 \times 10}{70} \times 6 \\
& v_{1}=\frac{4}{3}+\frac{6}{7}=\frac{46}{21}=2.19 \mathrm{~m} / \mathrm{s}
\end{aligned}
\]
53
An electric circuit consists of a battery of emf E, an inductance L and a resistance R in series. The switch S is closed at \(\mathrm{t}=0\). The current in the circuit grows exponentially with time as depicted by curve (1). The values of the circuit parameters (E, L or R) are now somehow changed. The circuit is closed second time, the growth of current I follows curve (2). The following conclusions(s) may be drawn :-
Show Solution
Solution
Ans. (a, b, d)
\(\mathrm{I}_{0}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{\mathrm{E}^{\prime}}{\mathrm{R}^{\prime}}\)
\[
\begin{aligned}
& I=\frac{E}{R}\left(1-e^{-\frac{R t}{L}}\right) \\
& \frac{d I}{d t}=\frac{E}{R} \times \frac{R}{L} e^{-\frac{R t}{L}} \\
& =\frac{E}{L} e^{-\frac{R t}{L}} \\
& L_{2}>L_{1}
\end{aligned}
\]
54
The Earth is revolving around the Sun (mass M) in a circular orbit of radius r with a period of revolution \(\mathrm{T}=1\) year. Suppose that during the revolution of the Earth, at any instant, the mass of the Sun instantaneously becomes double i.e. 2M. The correct alternative (s) is/are [Assume that the Sun and the Earth are point masses].
Show Solution
Solution
Ans. (b, c, d)
\[
\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}
\]
\[
\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\mathrm{a}^{3}}{\mathrm{G} \times 2 \mathrm{M}}}
\]
\[
\mathrm{R}=\mathrm{a}(1+\mathrm{e})
\]
\[
\mathrm{E}=\frac{1}{2} \mathrm{~m} \times \mathrm{v}^{2}-\frac{\mathrm{GMm}}{\mathrm{R}}
\]
\[
\begin{aligned}
& -\frac{2 \mathrm{GMm}}{2 \mathrm{a}}=\frac{1}{2} \mathrm{~m} \times \frac{\mathrm{GM}}{\mathrm{R}}-\frac{2 \mathrm{GMm}}{\mathrm{R}} \\
& \frac{1}{\mathrm{a}}=\frac{3}{2}
\end{aligned}
\]
\[
a=\frac{2}{3} R
\]
\[
\mathrm{R}=\mathrm{a}(1+\mathrm{e})
\]
\[
=\frac{2 \mathrm{R}}{3}(1+\mathrm{e})
\]
\[
e=\frac{1}{2}
\]
\[
R_{\min }=a(1-e)=\frac{2 R}{3} \times \frac{1}{2}=\frac{R}{3}
\]
\[
v_{\max }=\sqrt{\frac{\mathrm{G} \times 2 \mathrm{M}}{\frac{2 \mathrm{R}}{3}} \frac{\left(1+\frac{1}{2}\right)}{(1-\mathrm{e})}}=\sqrt{\frac{2 \mathrm{GM} \times \frac{3}{2}}{\frac{2 \mathrm{R}}{3} \frac{1}{2}}}=3 \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}
\]
55
An optical fibre consists of a glass core (refractive index \(\mathrm{n}_{1}\) ) surrounded by a clidding (refractive index \(\mathrm{n}_{2}<\mathrm{n}_{1}\) ). A beam of light enters one end of the fibre at P, from air at angle \(\theta\), with the axis of fibre as shown in the figure. Choose the correct option(s) :-
Show Solution
Solution
Ans. (a, c, d)
\[
\begin{aligned}
& v=\frac{c}{n_{1}}=\frac{L}{t_{1}} \Rightarrow t_{1}=\frac{n_{1} L}{c} \\
& n_{1} \sin i=n_{2} \sin 90^{\circ} \\
& \sin i=\frac{n_{2}}{n_{1}} \\
& 1 \sin \theta=n_{1} \sin (90-i) \\
& =n_{1} \sqrt{1-\frac{n_{2}^{2}}{n_{1}^{2}}}=\sqrt{n_{1}^{2}-n_{2}^{2}} \\
& \frac{1}{2}=\sqrt{\frac{9}{4}-n_{2}^{2}} \\
& \frac{1}{4}=\frac{9}{4}-n_{2}^{2} \\
& n_{2}=\sqrt{2}
\end{aligned}
\]
(d)
\[
\begin{aligned}
& \frac{\mathrm{c}}{\mathrm{n}_{1}}=\frac{\mathrm{L}}{\sin \theta_{\mathrm{c}} \mathrm{t}_{2}} \\
& \mathrm{t}_{2}=\frac{\mathrm{n}_{1}^{2} \mathrm{~L}}{\mathrm{c} \times \mathrm{n}_{2}} \\
& \Delta \mathrm{t}=\frac{\mathrm{n}_{1} \mathrm{~L}}{\mathrm{c}}\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}-1\right)
\end{aligned}
\]
56
A series LCR circuit fed with AC has resonant angular frequency of \(\omega=2.0 \times 10^{4} \mathrm{rad} / \mathrm{s}\). When the same circuit is driven at an angular frequency of \(2.5 \times 10^{4} \mathrm{rad} / \mathrm{s}\), it has an impedance of \(1.0 \mathrm{k} \Omega\) and phase constant of \(\phi=45^{\circ}\). The value of L, C and R for this circuit may be :-
Show Solution
Solution
Ans. (a, b, c)
\(\omega_{0}=2 \times 10^{4}=\frac{1}{\sqrt{\mathrm{LC}}}\) \(\omega>\omega_{\text {res }} \Rightarrow\) Inductive
\[
\begin{aligned}
& \tan \phi=\frac{\omega L-\frac{1}{\omega C}}{R}=+1 \\
& Z=\sqrt{\left(\omega L-\frac{1}{\omega C}\right)^{2}+R^{2}} \\
& =\sqrt{2} R=10^{3} \\
& R=\frac{1000}{\sqrt{2}}=707 \Omega \\
& \left(\omega L-\frac{1}{\omega C}\right)=+\frac{1000}{\sqrt{2}}
\end{aligned}
\]
\[
\begin{aligned}
& \mathrm{LC}=\frac{1}{4 \times 10^{8}} \\
& \frac{2.5 \times 10^{4}}{4 \times 10^{8} \mathrm{C}}-\frac{1}{2.5 \times 10^{4} \mathrm{C}}=\frac{1000}{\sqrt{2}} \\
& \mathrm{C}=\frac{\sqrt{2}}{1000}\left[\frac{5}{8}-\frac{2}{5}\right] \times 10^{-4} \\
& =\sqrt{2}\left[\frac{4}{40}\right] \times 10^{-8}=31.8 \mathrm{nF} \\
& \mathrm{~L}=\frac{10^{9}}{4 \times 10^{8} \times 31.8}=78.61 \mathrm{mH}
\end{aligned}
\]
57
A long straight wire, having a radius greater than 4.0 mm, carries a current that is uniformly distributed over its cross-section. The magnitude of magnetic field B due to the current is 0.28 mT at r \(=4.0 \mathrm{~mm}\) and 0.20 mT at \(\mathrm{r}=10.0 \mathrm{~mm}\), respectively, from the axis of the wire then the :-
Show Solution
Solution
Ans. (a, b, d)
\(\mathrm{B}=\mu_{0} \frac{\mathrm{Jx}}{2}\)
\[
\begin{aligned}
& 0.28 \times 10^{-3}=4 \pi \times 10^{-7} \times \mathrm{J} \times \frac{4 \times 10^{-3}}{2} \\
& \mathrm{~J}=\frac{0.28}{8 \pi} \times 10^{7}=1.11 \times 10^{5} \mathrm{~A} / \mathrm{m}^{2}
\end{aligned}
\]
if \(\mathrm{r}>10 \mathrm{~mm}\)
\[
0.2 \times 10^{-3}=4 \pi \times 10^{-6} \times \frac{0.28}{8 \pi} \times 10^{3} \times \frac{10^{-2}}{2}
\]
does not match
\[
\begin{aligned}
& \Rightarrow r<10 \mathrm{~mm} \\
& 0.2 \times 10^{-3}=\frac{\mu_{0} i}{2 \pi \times 0.01}=\frac{2 \times 10^{-7} \times i}{10^{-2}} \\
& i=10 \mathrm{~A} \\
& \frac{0.28 \times 10^{7}}{8 \pi}=\frac{10}{\pi \mathrm{x}^{2}} \\
& x^{2}=\frac{8 \times 10^{-6}}{0.28} \\
& x=5.34 \mathrm{~mm} \\
& B_{5}=\frac{\mu_{0} \times 1.11 \times 10^{5} \times 5 \times 10^{-3}}{2}=277.5 \mu_{0}
\end{aligned}
\]
\[
\mathrm{B}_{6}=\frac{\mu_{0} \times 10}{2 \pi \times 6 \times 10^{-3}}=265.25 \mu_{0}
\]
58
In the given combination of capacitors \(\mathrm{C}_{1}=4 \mu \mathrm{~F}, \mathrm{C}_{2}=2 \mu \mathrm{~F}, \mathrm{C}_{3}=2 \mu \mathrm{~F}, \mathrm{C}_{4}=4 \mu \mathrm{~F}\) and \(\mathrm{C}_{5}=3 \mu \mathrm{~F}\), a source of 12 volt is connected across the points A and B. Then the :-
Show Solution
Solution
Ans. (b, c)
\[
\begin{aligned}
& 4(x-12)+2(x-0)+3(2 x-12)=0 \\
& 12 x=84 \\
& x=7 \\
& Q=20+14=C_{\text {eq }} \times 12 \\
& C_{\text {eq }}=\frac{34}{12}=\frac{17}{6} \\
& U=\frac{1}{2} \times \frac{17}{6} \times 12^{2}=204
\end{aligned}
\]
59
One mole of an ideal monoatomic gas, initially at temperature T, is compressed to \(\frac{1}{8}\) of its voluem by a piston in a cylinder such that the heat dissipated into the environment is equal to the change in the internal energy of the gas. Then the
Show Solution
Solution
Ans. (b, c, d)
\(\mathbf{d Q}=+\mathbf{d U}=+\frac{3}{2} \mathrm{nRdT}=\mathrm{nCdT}\)
\[
\Rightarrow \mathbf{C}=+\frac{3}{2} R
\]
but it is not isochoric
\[
\begin{aligned}
& \Rightarrow d Q=-d U \Rightarrow C=-\frac{3}{2} R \\
& -\frac{3 R}{2}=C=\frac{3}{2} R-\frac{R}{m-1} \\
& 3 R=\frac{R}{m-1} \\
& m=\frac{4}{3} \\
& \mathrm{TV}^{1 / 3}=\mathrm{T}_{0} \mathrm{~V}_{0}^{1 / 3} \\
& \mathrm{~T}\left(\frac{\mathrm{~V}_{0}}{8}\right)^{1 / 3}=\mathrm{T}_{0} \mathrm{~V}_{0}^{1 / 3} \\
& \mathrm{~T}=2 \mathrm{~T}_{0} \\
& \mathrm{~W}=\mathrm{Q}-\Delta \mathrm{U} \\
& =-\frac{3}{2} \mathrm{nR} \times \mathrm{T}-\frac{3}{2} \mathrm{nR} \times \mathrm{T}=-3 \mathrm{nRT}
\end{aligned}
\]
60
A solid ball of mass M and radius R (see figure) is lying on a horizontal table. The ball experiences a short horizontal impulse which imparts a momentum p to the ball. The height of point of impact above the center line is \(\mathrm{h}=\alpha \mathrm{R}(0 \leq \alpha \leq 1)\). Choose the correct option/option(s).
Show Solution
Solution
Ans. (a, b, c)
\[
\begin{aligned}
& \mathrm{K}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}} \\
& \alpha \mathrm{R} \times \phi=\frac{2}{5} \mathrm{mR}^{2} \times \omega \\
& \omega=\frac{\alpha \mathrm{pR}}{\mathrm{I}} \\
& \mathrm{~K}=\frac{1}{2} \times \frac{2}{5} \mathrm{mR}^{2} \times \frac{\alpha^{2} \mathrm{p}^{2} \mathrm{R}^{2}}{\frac{4}{25} \mathrm{~m}^{2} \mathrm{R}^{4}} \\
& =\frac{5}{4} \frac{\alpha^{2} \mathrm{p}^{2}}{\mathrm{~m}} \\
& \omega=\frac{\mathrm{p}}{\mathrm{mR}}=\frac{\alpha \mathrm{pR}}{\frac{2}{5} \mathrm{mR}^{2}} \Rightarrow \alpha=\frac{2}{5}
\end{aligned}
\]
Practice Tip
Try each question before opening its solution. For numerical questions, compare both the final option and the intermediate physical reasoning.
FAQs
NSEP 2024 Question Paper with Solutions — FAQs
How many questions are there in the NSEP 2024 Physics paper?
The supplied NSEP 2024 paper contains 60 questions, and all 60 are included above with answers and solutions.
What is the NSEP 2024 paper code?
The supplied answer key identifies the paper as Paper Code 63.
Does the NSEP 2024 answer key contain NA or multiple-answer entries?
Yes. The supplied key includes NA entries as well as questions with more than one accepted option. These are preserved above exactly as part of the answer key.
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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