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Last Updated on November 13, 2026 by XAM CONTENT
NSEP 2023 Solved Paper
Practice the complete NSEP 2023 Question Paper with Solutions for Physics. All 60 questions are arranged in the original order with the supplied answers, academic diagrams and worked solutions. Attempt each question first and then open Show Solution to check the detailed working.
Exam
NSEP 2023
Subject
Physics
Paper Code
61
Date
26 November 2023
Maximum Marks
216
Time
2 Hours
Paper Format
Questions 1–48 are single-correct questions. Questions 49–60 may have one or more correct options. The answers and multiple-answer combinations below follow the supplied NSEP 2023 solutions source.
Quick Check
NSEP 2023 Answer Key — Paper Code 61
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 | c | c | c | a | d | d | b | c | d | a | a | c | c | b |
Questions 16–30
| Question | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Answer | c | d | a | c | b | d | a | a | b | b | a | b | b | c | c |
Questions 31–45
| Question | 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Answer | b | a | d | b | b | a | a | a | b | b | d | a | c | d | c |
Questions 46–60
| Question | 46 | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Answer | d | d | d | b, d | a, d | d | a, b, c, d | c, d | a, b, d | a, c, d | b, c | a, b, c, d | a, b, c | a, b, c, d | a, b, c |
Solved Paper
NSEP 2023 Question Paper with Detailed Solutions
60 questions with source diagrams and expandable worked solutions.
60
Questions
Questions 1–15
NSEP 2023 Physics — Paper Code 61
1
A target of \({ }^{7} \mathrm{Li}\) is bombarded with a proton beam of current \(10^{-4}\) ampere for 1 hour to produce \({ }^{7} \mathrm{Be}\) of activity \(1.8 \times 10^{8}\) disintegrations per second. Assuming that bombarding of 1000 protons produces one \({ }^{7} \mathrm{Be}\) radioactive nucleus, the half-life of \({ }^{7} \mathrm{Be}\) is estimated to be approximately
Show Solution
Solution
Ans. (c)
\[
\begin{aligned}
& \Delta Q=i \Delta t \\
& =10^{-4} \times(3600) \mathrm{C} \\
& \text { Number of protons }=\frac{10^{-4} \times 3600}{1.6 \times 10^{-19}} \\
& \text { Number of Be }(\mathrm{N})=\frac{10^{-4} \times 3600}{1.6 \times 10^{-19} \times 1000} \\
& \because \quad \mathrm{~A}=\lambda N \\
& \therefore \quad \text { Half life }=\frac{\ln 2}{\lambda}=\frac{N}{A} \ln 2 \\
& =\frac{10^{-4} \times 3600 \times \ln (2)}{1.6 \times 10^{-19} \times 1000 \times 1.8 \times 10^{8}} \\
& =8665200 \text { seconds } \\
& \simeq 2407 \text { hour }
\end{aligned}
\]
2
A long straight wire carrying a current \(i=10 A\) and a rectangular metallic loop of dimensions \(b \times c\) lie in the same plane as shown in the figure. The parameters are \(a=10 \mathrm{~cm}, b=30 \mathrm{~cm}\) and \(c=50 \mathrm{~cm}\). The mutual inductance of the system is nearly
Show Solution
Solution
Ans. (c)
\[
\begin{aligned}
& a=\frac{1}{10} \mathrm{~m} \\
& b=\frac{3}{10} \mathrm{~m} \\
& c=\frac{1}{2} \mathrm{~m}
\end{aligned}
\]
\[
\begin{aligned}
& d \phi=\frac{\mu_{0} i}{2 \pi x} c \cdot d x, \phi=\int d \phi, \phi=\frac{\mu_{0} i c}{2 \pi}(\ln x)_{a}^{a+b} \\
& \phi=\frac{4 \pi \times 10^{-7} \times 10}{2 \pi} \times \frac{1}{2} \ln \left(\frac{40}{10}\right) \\
& =10^{-6} \times 2 \ln 2 \\
& \phi=1.386 \times 10^{-6} \\
& M \times 10=1.386 \times 10^{-6} \\
& M=1.386 \times 10^{-7} \\
& \simeq 139 \mathrm{nH}
\end{aligned}
\]
3
Impedance of a given series LCR circuit, fed with alternating current, is the same for two frequencies \(f_{1}\) and \(f_{2}\). The resonance frequency \(f_{R}\) of the circuit is
Show Solution
Solution
Ans. (c)
\(Z=\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}\)
\[
\begin{aligned}
& Z_{1}=\sqrt{R^{2}+\left(\frac{1}{C \omega_{1}}-L \omega_{1}\right)^{2}} \\
& Z_{2}=\sqrt{R^{2}+\left(\frac{1}{C \omega_{2}}-L \omega_{2}\right)^{2}}
\end{aligned}
\]
Here \(\mathrm{Z}_{1}=\mathrm{Z}_{2}\)
\[
\begin{aligned}
& \text { So, } \frac{1}{C \omega_{1}}-L \omega_{1}=\frac{1}{C \omega_{2}}-L \omega_{2} \\
& \frac{1}{C \omega_{1}}-\frac{1}{C \omega_{2}}=L \omega_{1}-L \omega_{2} \\
& \frac{\left(\omega_{2}-\omega_{1}\right)}{C \omega_{1} \omega_{2}}=L\left(\omega_{1}-\omega_{2}\right)
\end{aligned}
\]
\[
\begin{aligned}
& \frac{\left(\omega_{2}-\omega_{1}\right)}{C \omega_{1} \omega_{2}}=-L\left(\omega_{2}-\omega_{1}\right) \\
& -\omega_{1} \omega_{2}=\frac{1}{L C} \\
& \left|-2 \pi f_{1} \times 2 \pi f_{2}\right|=\left|(2 \pi f r)^{2}\right| \\
& f_{r}=\sqrt{f_{1} f_{2}}
\end{aligned}
\]
4
A lawn roller is a solid cylinder of mass \(M\) and radius \(R\). As shown in the figure, it is pulled at its center by a horizontal force \(F\) and rolls without slipping on a horizontal surface. Then the
Show Solution
Solution
Ans. (c)
\[
\begin{aligned}
& F-f=M a \\
& f \cdot R=\frac{M R^{2}}{2} \times\left(\frac{a}{R}\right) \\
& f=\frac{M a}{2}
\end{aligned}
\]
From (1), \(F=M a+\frac{M a}{2}=\frac{3 M a}{2}\)
\[
\begin{aligned}
& a=\frac{2 F}{3 M} \\
& f=\frac{M}{2}\left(\frac{2 F}{3 M}\right)=\frac{F}{3} \\
& f=\mu \cdot M g=\frac{F}{3} \\
& \mu=\frac{F}{3 M g}
\end{aligned}
\]
5
A hydrogen atom \(\left(M_{\mathrm{H}}=1.67 \times 10^{-27} \mathrm{~kg}\right)\), initially at rest, emits a photon and goes from the excited state \(n=5\) to the ground state. The recoil speed of the atom is nearly
Show Solution
Solution
Ans. (a)
\[
\begin{aligned}
& n_{1}=1 \\
& n_{2}=5 \\
& \text { mass of hydrogen atom }=1.67 \times 10^{-27} \mathrm{~kg} \\
& p=\frac{h}{\lambda} \\
& v=\frac{h}{m \lambda} \rightarrow \text { recoil speed } \\
& \frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\
& \frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{1^{2}}-\frac{1}{5^{2}}\right) \\
& \lambda=9.48 \times 10^{-8} \mathrm{~m}
\end{aligned}
\]
So, recoil speed \((v)=\frac{6.67 \times 10^{-34}}{1.6 \times 10^{-27} \times 9.48 \times 10^{-4}}\) \(\approx 4.2 \mathrm{~m} / \mathrm{s}\)
6
Two nuclides \(A\) and \(B\) are isotopes. The nuclides \(B\) and \(C\) are isobars. All the three nuclides \(A, B\) and \(C\) are radioactive. You may then conclude that
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Solution
Ans. (d)
\[
\begin{aligned}
& A \rightarrow(Z+N) \\
& B \rightarrow\left(Z+N^{\prime}\right) \\
& C \rightarrow Z^{\prime}+N^{\prime \prime}
\end{aligned}
\]
(a) × not same element
(b) ×
(c) \(A \rightarrow B\) decay \(p \rightarrow n \quad z\) will change
(d) \(B\) to \(C\), atomic no change
7
Numerical aperture of an optical fibre is a measure of
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Solution
Ans. (d)
Numerical aperture of optical fibre (NA) \(=\sin \alpha=\sqrt{n_{1}^{2}-n_{2}^{2}}\) It is clear that on increasing NA, higher scattering loss from greater concentrations of dopant. – It measures light-gathering ability.
8
Heavy stable nuclei have more neutrons than protons. This is because of the fact that
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Solution
Ans. (b)
In nucleus, there are two types of force, Nuclear force → attractive Force among protons → repulsive In heavy nucleus, repulsive force is more than attractive. But for stable heavy nucleus, these repulsive force is minimised by neutrons.
9
An equi-concave lens of radii of curvature of the two surfaces numerically equal to 7 cm and refractive index \(\mu=1.5\) has a small silver dot on the rear surface. As a result of this, a ray of light incident parallel to the principal axis gets reflected from its rear surface and then reflected also from the inner front surface. The ray after the second reflection emerges out of the thin lens and appears to focus at a point P on the principal axis. The point \(P\) lies
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Solution
Ans. (c)
The ray undergoes 2 reflection and 3 refractions
\[
\begin{aligned}
& \therefore \quad \frac{1}{f_{\mathrm{eq}}}=\frac{3}{f_{l}}-\frac{2}{f_{m}} \\
& \frac{1}{f_{l}}=(1.5-1) \frac{2}{R}=\frac{1}{R} \\
& f_{l}=7 \mathrm{~cm} \\
& f_{m}=\frac{R}{2}=-\frac{7}{2} \mathrm{~cm} \\
& \therefore \quad \frac{1}{f_{\mathrm{eq}}}=\frac{3}{7}+\frac{2}{7} \times 2=\frac{7}{7} \\
& f_{\mathrm{eq}}=1 \mathrm{~cm}
\end{aligned}
\]
∴ It will be focused 1 cm beyond the lens.
10
Light emerges out uniformly from a point source placed at the focus of a concave mirror to give out a spherical wave front. As a result of reflection of the paraxial rays from the concave mirror, according to Huygen’s theory the reflected light is in the form of a
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Solution
Ans. (d)
After reflection the light goes parallel to the principle axis. ∴ In case of parallel light, the shape of wavefront is plane. Option (d) is correct.
11
An equi-convex lens of focal length ‘ \(f\) is cut along a diameter, in two halves (pieces). The two identical pieces of the lens are now arranged as shown in the figure on a common axis at a separation \(f\) between the two. The image of an object \(A B\) placed at \(x=0\) cannot be formed at the distance \(x=\xi\) from the object along the axis, for the value of \(\xi\) as
Show Solution
Solution
Ans. (a)
Ans. 2f
12
During the processes of annihilation of a stationary electron of mass \(m_{0}\) with a stationary positron of equal mass, a radiation is emitted. The wavelength of the resulting radiation is
Show Solution
Solution
Ans. (a)
⇒ From conservation of momentum, photons will travel in opposite direction with equal magnitude of momentum \(\frac{h c}{\lambda}\)
⇒ From energy conservation
\[
\begin{aligned}
& \frac{h c}{\lambda}+\frac{h c}{\lambda}=m_{0} c^{2}+m_{0} c^{2} \\
& \lambda=\frac{h}{m_{0} c}
\end{aligned}
\]
13
The convex surface of a concavo-convex lens of refractive index 1.5 and radii of curvature \(R_{1}=20 \mathrm{~cm}\) and \(R_{2}=40 \mathrm{~cm}\) has been silvered so as to make it reflecting. The distance of a luminous object from the reflecting system when placed in front of it on its principal axis, so that the image coincides with the object is
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Solution
Ans. (c)
\(R_{L}=40 \mathrm{~cm}\)
The object should be at \({ }^{2} f_{\text {net }}\), so that the image coincides with it.
\[
\begin{aligned}
& \text { Now }\left(-\frac{1}{f_{\text {net }}}\right)=2\left(\frac{1}{f_{\text {lens }}}\right)+\left(-\frac{1}{f_{\text {mirror }}}\right) \\
& -\frac{1}{f_{\text {net }}}=2(1.5-1)\left(-\frac{1}{40}-\frac{1}{-20}\right)+\left(-\frac{1}{-10}\right) \\
& =\left(\frac{1}{20}-\frac{1}{40}\right)+\frac{1}{10} \\
& =\frac{1}{40}+\frac{1}{10}=\frac{10+40}{400} \\
& \Rightarrow f_{\text {net }}=-\frac{400}{50}=-8 \mathrm{~cm} \\
& \left.\therefore \quad\right|^{2} f_{\text {net }}=16 \mathrm{~cm}
\end{aligned}
\]
Option (c) is correct
14
Two balls are projected from the top of a cliff with equal initial speed \(u\). One starts at angle \(\theta\) above the horizontal while the other starts at angle \(\theta\) below. Difference in their ranges on ground is
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Solution
Ans. (c)
\[
\begin{aligned}
\Rightarrow \Delta R & =R_{1}-R_{2}=u \cos \theta\left(t_{1}-t_{2}\right) \\
& =u \cos \theta\left(\frac{2 u \sin \theta}{g}\right) \\
& =\frac{u^{2} \sin 2 \theta}{g}
\end{aligned}
\]
Option (c) is correct.
15
A solid block of mass 3 kg is suspended from the bottom of a 5 kg block with the help of a rope \(A B\) of mass 2 kg as shown in the figure. When pulled by an upward force \(F\), the whole system experiences an upward acceleration \(a=2.19 \mathrm{~ms}^{-2}\). Choose the correct option
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Solution
Ans. (b)
\[
\begin{aligned}
& F-10 \mathrm{~g}=10 \mathrm{a} \\
& \begin{array}{r}
F=21.9+100 \\
=121.9 \mathrm{~N}
\end{array}
\end{aligned}
\]
At mid-point of rope
\[
\begin{aligned}
& T-4 \mathrm{~g}=4 \times 2.19 \\
& T=47.96 \\
& \quad=48 \mathrm{~N}
\end{aligned}
\]
Questions 16–30
NSEP 2023 Physics — Paper Code 61
16
A block \(P\) of mass 0.4 kg is attached to a vertical rotating spindle by two strings \(A P\) and \(B P\) of equal length 1.0 m as shown in the figure. The period of rotation is 1.2 s. Tensions \(T_{1}\) and \(T_{2}\) in string \(A P\) and \(B P\) are
Show Solution
Solution
Ans. (c)
\[
\begin{align*}
& T=\frac{2 \pi}{\omega} \\
& \omega=\frac{2 \pi}{1.2}=\frac{5 \pi}{3} \\
& T_{1} \sin \theta=4+T_{2} \sin \theta \\
& \left(T_{1}+T_{2}\right) \cos \theta=m r \omega^{2} \\
& T_{1}=5+T_{2} \tag{i}\\
& \left(T_{1}+T_{2}\right)=10.95 \\
& T_{1}=7.94 \mathrm{~N} \\
& T_{2}=3.03 \mathrm{~N}
\end{align*}
\]
17
A particle of mass \(m\) moves in a straight line under the influence of a certain force such that the power \((P)\) delivered to it remains constant. Starting from rest, the straight-line distance travelled by the moving particle in time \(t\) is
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Solution
Ans. (d)
\[
\begin{aligned}
& P=F V \\
& F=m a \\
& \therefore \quad v a=\frac{P}{m} \\
& a=\frac{d v}{d t} \\
& \frac{v d v}{d t}=\frac{P}{m} \\
& \frac{v^{2}}{2}=\frac{P}{m} t \\
& v=\sqrt{\frac{2 P}{m}} t \\
& v=\frac{d x}{d t} \\
& \int_{0}^{x} d x=\frac{2}{3} \sqrt{\frac{2 P}{m}}(t)^{\frac{3}{2}} \\
& x=\left(\frac{8}{9} \frac{P}{m} t^{3}\right)^{\frac{1}{2}}
\end{aligned}
\]
18
A bullet is fired vertically up with half the escape speed from the surface of the Earth. The maximum altitude reached by it (ignore the effect of rotation of the Earth) in terms of radius of Earth \(R\) is
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Solution
Ans. (a)
\(V=\sqrt{\frac{G M}{2 R}}\) By conservation of energy
\[
\begin{aligned}
& -\frac{G M m}{R}+\frac{1}{2} m\left(\frac{G M}{2 R}\right)=-\frac{G M m}{R+h} \\
& -\frac{3}{4} \frac{G M m}{R}=-\frac{G M m}{R+h} \\
& h=\frac{R}{3}
\end{aligned}
\]
19
A can is a hollow cylinder of radius \(R\) and height \(h\). Its ends are sealed with circular sheets of the same material. The can is made of thin sheet metal of areal mass density \(\sigma\left(\mathrm{kg} / \mathrm{m}^{2}\right)\). Moment of inertia of this closed can about its vertical axis of symmetry is
Show Solution
Solution
Ans. (c)
\(h\)
Mass of seal \(=\pi R^{2} \sigma\) M.I. of both seal
\[
I_{1}=\frac{\left(\pi R^{2} \sigma\right) R^{2}}{2} \times 2
\]
Mass of curve surface \(=2 \pi R \cdot h \cdot \sigma\) M.I. of curved part
\[
I_{2}=2 \pi R \cdot h \sigma \cdot R^{2}
\]
\[
\begin{aligned}
\text { Total M.I. } & =l_{1}+l_{2} \\
& =\pi R^{4} \sigma+2 \pi R^{3} \sigma \cdot \mathrm{~h} \\
& =\pi R^{3} \sigma(R+2 h)
\end{aligned}
\]
20
A particle of mass \(m\) is revolving in a horizontal circle on a frictionless horizontal table with the help of a string tied to it and passing through a hole at the center of the table. Two equal masses \(M\) are attached to the other end of the string as shown. If one of the hanging masses \(M\) is removed gently, the radius of the circular motion of \(m\)
Show Solution
Solution
Ans. (b)
Initially, tension in string \(=2 M g\)
\[
\begin{equation*}
\therefore \quad 2 M g=m \omega_{1}^{2} r_{1} \tag{i}
\end{equation*}
\]
When one mass is removed
\[
\begin{equation*}
M g=m \omega_{2}^{2} r_{2} \tag{ii}
\end{equation*}
\]
By conservation of angular momentum
\[
\begin{aligned}
& \omega_{1} r_{1}^{2}=\omega_{2} r_{2}^{2} \\
& \frac{\omega_{1}}{\omega_{2}}=\left(\frac{r_{2}}{r_{1}}\right)^{2}
\end{aligned}
\]
On dividing (i) and (ii)
\[
\begin{aligned}
& 2=\left(\frac{\omega_{1}}{\omega_{2}}\right)^{2}\left(\frac{r_{1}}{r_{2}}\right) \\
& 2=\left(\frac{r_{2}}{r_{1}}\right)^{3} \\
& r_{2}=r_{1}(2)^{\frac{1}{3}} \\
& r_{2}=1.26 r_{1}
\end{aligned}
\]
21
Three stars of equal mass \(M\) rotate in a circular path of radius \(r\) about their center of mass such that the stars always remain equidistant from each other. The common angular speed \((\omega)\) of rotation of the stars can be expressed as
Show Solution
Solution
Ans. (d)
\[
\begin{aligned}
& F=\frac{G M^{2}}{\sqrt{3} r^{2}}=M \omega^{2} r \\
& \therefore \quad \omega=\left(\frac{G M}{\sqrt{3} r^{3}}\right)^{\frac{1}{3}}
\end{aligned}
\]
22
The density of a liquid is \(\rho\) at the surface. The bulk modulus of the liquid is \(B\). The increase \(\Delta \rho\) in the density of the liquid at a depth \(h\) from the surface is (with \(\Delta \rho \ll \rho\) )
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Solution
Ans. (a)
\(\rho v=\) constant
\[
\begin{array}{ll}
\frac{\Delta \rho}{\rho}+\frac{\Delta v}{v}=0 & \\
B=\left|\frac{\Delta \rho}{\Delta v}\right| & \left|\frac{\Delta v}{v}\right|=\frac{\rho g h}{B}
\end{array}
\]
\(\ldots\) (i)
…(ii) From (i) and (ii)
\[
\begin{aligned}
& \left|\frac{\Delta \rho}{\rho}\right|=\frac{\rho g h}{B} \\
& (\Delta \rho)=\frac{\rho^{2} g h}{B}
\end{aligned}
\]
23
Water flows at 1.2 m/s through a hose of diameter 1.59 cm. The time required to fill a cylindrical container of radius 2 m to a height of \(h=1.25 \mathrm{~m}\) will be nearly
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Solution
Ans. (a)
Time \(=\frac{\text { Volume }}{\text { Rate of flow of liquid }}\)
\[
\begin{aligned}
& =\frac{\pi R^{2} h}{A V} \\
& =\frac{\pi(2)^{2} \times(1.25)}{1.2 \times \pi \times\left(\frac{1.59}{2}\right)^{2} \times 10^{-4} \times 3600} \text { hours } \\
& =18.3 \text { hours }
\end{aligned}
\]
24
A police car, moving at speed of 108 km/hour, approaches a truck moving at 72 km/hour in opposite direction. The natural frequency of the siren of the car is 800 Hz and the surrounding temperature is 27°C. The frequency heard by the truck driver as the car passes him
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Solution
Ans. (b)
\[
\begin{aligned}
\text { Velocity of sound } & =\sqrt{\frac{\gamma R T}{M}} \\
& =\sqrt{\frac{1.4 \times 8.3 \times 300 \times 1000}{28.97}}=346.88 \mathrm{~m} / \mathrm{s}
\end{aligned}
\]
\[
\begin{aligned}
& f=800 \mathrm{~Hz} \\
& V_{\text {car }}=108 \times \frac{5}{18}=30 \mathrm{~m} / \mathrm{s} \\
& V_{\text {truck }}=72 \times \frac{5}{18}=20 \mathrm{~m} / \mathrm{s} \\
& \begin{aligned}
\therefore \quad f^{\prime} & =\left(\frac{V_{S}+V_{\text {truck }}}{V_{S}-V_{\text {car }}}\right) \times 800 \\
& =\frac{366.88}{316.88} \times 800=926.230
\end{aligned}
\end{aligned}
\]
When car passes
\[
\begin{aligned}
& f^{\prime \prime}=\left(\frac{V_{S}-V_{\text {truck }}}{V_{S}+V_{\text {car }}}\right) \times 800=\frac{326.88}{376.88} \times 800=693.86 \\
& \therefore \quad \Delta f=-232 \mathrm{~Hz}
\end{aligned}
\]
25
A rope of mass \(M\) and length \(L\) hangs vertically. Time needed for a transverse pulse to travel from its bottom end to the support is
Show Solution
Solution
Ans. (b)
\[
\begin{aligned}
& T=\frac{M x}{L} g \\
& \therefore \quad v=\sqrt{\frac{T}{\mu}}=\sqrt{g x} \\
& v=\frac{d x}{d t}=\sqrt{g x} \\
& \int_{0}^{L} \frac{d x}{\sqrt{x}}=\sqrt{g} \int_{0}^{t} d t \\
& 2 \sqrt{L}=\sqrt{g} t \\
& \therefore t=2 \sqrt{\frac{L}{g}}
\end{aligned}
\]
26
The figure shows a smooth tunnel \(A B\) (length \(=2 \ell\) ) in a uniform density planet (say Earth) of mass \(M\) and radius \(R\). A small ball of mass \(m\) is released from rest at the end \(A\) of the tunnel. Acceleration due to gravity at surface of the planet is \(g\). Time taken by the ball to reach the end \(B\) is
Show Solution
Solution
Ans. (a)
Time period of oscillation
\[
T=2 \pi \sqrt{\frac{R}{g}}
\]
Time taken from one extreme to another \(=\frac{T}{2}\)
\[
\therefore t=\pi \sqrt{\frac{R}{g}}
\]
27
When the speaker \(S_{1}\) is switched ON, the sound intensity at a point \(P\) in a room is 80 dB. But when the speaker \(S_{2}\) is switched ON ( \(S_{1}\) is switched OFF), the sound intensity at the same point \(P\) in the room is 85 dB. The sound intensity level (in dB) at the same point \(P\) in the room, if the two speakers \(S_{1}\) and \(S_{2}\) are simultaneously switched ON, is (consider the speakers to be incoherent)
Show Solution
Solution
Ans. (b)
Total intensity \(I=l_{1}+l_{2}\)
\[
\begin{aligned}
& 80=10 \log \left(\frac{I_{1}}{I_{0}}\right) \\
& 85=10 \log \left(\frac{I_{2}}{I_{0}}\right) \\
& L=10 \log \left(\frac{I_{1}+I_{2}}{I_{0}}\right) \\
& L=86.2 \mathrm{~dB}
\end{aligned}
\]
28
A block \(B\) of mass 0.5 kg moving, on a horizontal frictionless table at \(2.0 \mathrm{~ms}^{-1}\), collides with a massless pan \(P\) (at origin \(O\) ) and sticks to it. The pan is connected at the end of a horizontal un-stretched (relaxed) spring of force constant \(K=32 \mathrm{Nm}^{-1}\) as shown in figure. After the block collides, the displacement \(x(t)\) of the block as a function of time \(t\) is given by
Show Solution
Solution
Ans. (b)
It is a simple harmonic motion
\[
\begin{aligned}
& \omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{32}{0.5}}=8 \\
& \text { Amplitude } A=\sqrt{\frac{\frac{1}{2} m v^{2}}{\frac{1}{2} K}}=\sqrt{\frac{0.5 \times 4}{32}}=\sqrt{\frac{1}{16}} \\
& \qquad \quad=\frac{1}{4}=0.25 \mathrm{~m} \\
& \text { At } t=0, \text { block is at its mean position } \\
& \therefore x=0.25 \sin (8 t) \mathrm{m}
\end{aligned}
\]
29
Which of the following functions does not represent a traveling wave?
Show Solution
Solution
Ans. (c)
For travelling wave
\[
\frac{d^{2} y}{d t^{2}}=v^{2} \frac{d^{2} y}{d x^{2}}
\]
For option (c)
\[
\frac{d^{2} y}{d t^{2}} \neq v^{2} \frac{d^{2} y}{d x^{2}}
\]
∴ It is not a travelling wave
30
Two Carnot heat engines are connected in series such that the sink of the first engine is heat source of the second. Efficiency of the engines are \(\eta_{1}\) and \(\eta_{2}\) respectively. Net efficiency \(\eta\) of the combination is given by
Show Solution
Solution
Ans. (c)
Sol.
\[
\begin{aligned}
& \eta_{1}=1-\frac{T_{2}}{T_{1}} \\
& \eta_{2}=1-\frac{T_{3}}{T_{2}} \\
& \eta=1-\frac{T_{3}}{T_{1}} \\
& \frac{T_{3}}{T_{1}}=\left(1-\eta_{1}\right)\left(1-\eta_{2}\right) \\
& \therefore \eta=\eta_{1}+\eta_{2}-\eta_{1} \eta_{2}
\end{aligned}
\]
Questions 31–45
NSEP 2023 Physics — Paper Code 61
31
An air bubble of radius 2 mm at a depth 12 m below the surface of water at temperature of 8°C, rises to the surface where the temperature is 16°C. Neglecting the effect of Surface Tension, the radius of the bubble at the surface is estimated to be
Show Solution
Solution
Ans. (b)
\[
\begin{aligned}
& \frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \\
& P_{1}=P_{0}+\rho g h \\
& h=12 \mathrm{~m} \\
& P_{1}=2.2 \times 10^{5} \\
& P_{2}=P_{0}=10^{5} \\
& T_{1}=281 \\
& T_{2}=289 \\
& \therefore \frac{V_{2}}{V_{1}}=\frac{2.2}{1} \times \frac{289}{281}=2.26 \\
& \therefore r_{2}=r_{1}(2.26)^{\frac{1}{3}} \\
& r_{2}=2.61 \mathrm{~mm}
\end{aligned}
\]
32
Two soap bubbles of radii \(a\) and \(b\) coalesce to form a single bubble of radius \(c\) under isothermal conditions. If the external pressure is \(P_{A}\), then the Surface Tension \((T)\) of the soap solution is
Show Solution
Solution
Ans. (a)
\(P_{1} V_{1}+P_{2} V_{2}=P_{3} V_{3}\)
\[
\begin{aligned}
& \left(P_{A}+\frac{4 T}{a}\right) a^{3}+\left(P_{A}+\frac{4 T}{b}\right) b^{3}=\left(P_{A}+\frac{4 T}{c}\right) c^{3} \\
& P_{A}\left(a^{3}+b^{3}-c^{3}\right)=4 T\left(c^{2}-b^{2}-a^{2}\right) \\
& \therefore T=\frac{P_{A}}{4} \frac{\left(a^{3}+b^{3}-c^{3}\right)}{\left(c^{2}-b^{2}-a^{2}\right)}
\end{aligned}
\]
33
An open-end organ pipe 30 cm in length and a closed-end organ pipe 23 cm in length, both of equal diameter, are each sounding their first overtone and both are in unison at 1100 Hz. The speed of sound in air, is estimated to be nearly
Show Solution
Solution
Ans. (d)
For open end organ pipe for \(1^{\text {st }}\) overtone \(f=\frac{V}{\left(I_{1}+2 e\right)}\) For closed end organ pipe for \(1{ }^{\text {st }}\) overtone \(f=\frac{3 v}{4\left(I_{2}+e\right)}\)
\[
\begin{aligned}
& f=1100 \mathrm{~Hz} \\
& I_{1}+2 e=\frac{v}{f} \\
& I_{2}+e=\frac{3 v}{4 f} \\
& \therefore v=352 \mathrm{~m} / \mathrm{s}
\end{aligned}
\]
34
The figure shows a lagged bar \(X Y\) of non-uniform cross section. One end \(X\) of the bar is maintained at 100°C and the other end \(Y\) at 0°C. The variation of temperature along its length from \(X\) to \(Y\) in steady state is best represented by the curve.
A
B
C
D
Show Solution
Solution
Ans. (b)
Side of the cross-section varies linearly.
\[
\begin{aligned}
& \therefore \ell=\ell_{0}-x c \\
& a=\ell^{2}=\left(\ell_{0}-x c\right)^{2}
\end{aligned}
\]
\(Q\) is same through all cross-section
\[
\begin{aligned}
& Q=\frac{\left(T_{x}-T\right)}{x} K\left(\ell_{0}-x c\right)^{2} \\
& \therefore T=T_{x}-\frac{Q x}{K\left(\ell_{0}-x c\right)^{2}}
\end{aligned}
\]
35
An ideal gas ( \(n\) moles) is initially at pressure \(P\) and temperature \(T\). It is cooled isochorically to a pressure \(\frac{P}{4}\). The gas is then expanded at a constant pressure so as to attain back its initial temperature \(T\). Work done by gas during the entire process is
Show Solution
Solution
Ans. (b)
In Isochoric process \(W=0\) In Isobaric process \(W=n R \Delta T\)
\[
\begin{aligned}
T_{2} & =T \\
T_{1} & =\frac{T}{4} \\
\therefore W & =n R\left(T-\frac{T}{4}\right) \\
& =\frac{3}{4} n R T
\end{aligned}
\]
36
Assuming the Sun to be a spherical body (radius \(R_{S}\) ) of surface temperature \(T\), the total radiation power received by Earth (radius \(R_{E}\) ) at a distance \(r\) from Sun is
Show Solution
Solution
Ans. (a)
Power radiated by Sun \(=\sigma \varepsilon 4 \pi R_{S}^{2} T^{4}\) Intensity at earth \(=\frac{\sigma \varepsilon 4 \pi R_{S}^{2} T^{4}}{4 \pi r^{2}}\) ∴ Power received by Earth \(=\pi R_{E}^{2} \times I=\frac{\pi R_{E}^{2} \sigma \varepsilon R_{S}^{2} T^{4}}{r^{2}}\) \(\varepsilon=1\) for sun \(\therefore \quad \varepsilon=\frac{\sigma \pi R_{E}^{2} R_{S}^{2} T^{4}}{r^{2}}\)
37
The figure shows five point-charges on a straight line. Separation between successive charges is 10 cm. For what values of \(q_{1}\) and \(q_{2}\) would the net force on each of the other three charges be zero?
Show Solution
Solution
Ans. (a)
\(\begin{array}{ccccc} 2 \mu \mathrm{C} & q_{1} & 1 \mu \mathrm{C} & q_{2} & 2 \mu \mathrm{C} \bullet & \bullet & \bullet & \bullet & \bullet\end{array}\) Force on \(1 \mu \mathrm{C}\) to be zero.
\[
\begin{aligned}
& \frac{2}{(2)^{2}}+\frac{q_{1}}{(1)^{2}}-\frac{q_{2}}{(1)^{2}}-\frac{2}{(2)^{2}}=0 \\
& \therefore q_{1}=q_{2}
\end{aligned}
\]
Force on \(2 \mu \mathrm{C}\) to be zero.
\[
\begin{aligned}
& \frac{q_{1}}{1^{2}}+\frac{1}{2^{2}}+\frac{q_{2}}{3^{2}}+\frac{2}{4^{2}}=0 \\
& q_{1}\left(1+\frac{1}{9}\right)=-\left(\frac{1}{4}+\frac{2}{16}\right) \\
& q_{1}\left(\frac{10}{9}\right)=-\frac{6}{16} \\
& q_{1}=-\frac{27}{80} \mu \mathrm{C}
\end{aligned}
\]
38
Two equal blocks, each of mass \(M\), hang on either side of a frictionless light pulley with a light string. A rider of mass \(m\) is placed on one of the blocks (as shown). When the system is released, the block with rider descends a distance \(H\) till the rider is caught by a ring that allows the block to pass through. The system moves a further distance \(D\) taking time \(t\). In such a situation, the acceleration due to gravity is
Show Solution
Solution
Ans. (a)
Acceleration of blocks \(==\frac{m}{(2 M+m)} g\) ∴ velocity after descending \(H\)
\[
v=\sqrt{\frac{2 m g H}{(2 M+m)}}
\]
After that velocity is constant
\[
\begin{aligned}
& \therefore D=v t \\
& D^{2}=\frac{2 m g H}{(2 M+m)} t^{2} \\
& g=\frac{(2 M+m) D^{2}}{2 m H t^{2}}
\end{aligned}
\]
39
A very small electric dipole of dipole moment \(\vec{p}\) lies along the \(x\) axis (i.e. \(\vec{p}=p \hat{i}\) ) in a non-uniform electric field \(\vec{E}=\frac{C}{x} \hat{i}\) (where \(c\) is a constant). The force on the dipole is
Show Solution
Solution
Ans. (b)
\[
\begin{aligned}
& F=p \frac{d E}{d x} \\
& =p \frac{d}{d x}\left(\frac{c}{x}\right) \\
& F=-\frac{p c}{x^{2}} \hat{i}
\end{aligned}
\]
40
A conducting thick spherical shell of radii \(a\) and \(b(b>a)\) has been charged with uniform surface charge density \(-\sigma \mathrm{C} / \mathrm{m}^{2}\) on the inner and \(+\sigma \mathrm{C} / \mathrm{m}^{2}\) on the outer surface. Then
Show Solution
Solution
Ans. (b)
Total Charge \(=4 \pi \sigma\left(b^{2}-a^{2}\right)\) Electric field outside \(=\frac{\sigma b}{\varepsilon_{0} r^{2}}\) Charge in the cavity \(=4 \pi \sigma a^{2}\)
41
A spherical conductor is charged up to a potential of 450 V. The potential outside, at a distance 15 cm from the surface, is 300 V. Then
Show Solution
Solution
Ans. (d)
\[
\begin{aligned}
& \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R}=450 \ldots(1) \\
& \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R+0.15}=300 \\
& \Rightarrow \frac{R+0.15}{R}=1.5 \Rightarrow \frac{0.15}{R}=0.5 \Rightarrow R=30 \mathrm{~cm} \\
& \Rightarrow Q=\frac{450 \times 0.30}{9 \times 10^{9}} C=15 \mathrm{nC} \\
& E_{\text {surface }}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R^{2}}=9 \times 10^{9} \times \frac{15 \times 10^{-9}}{0.3^{2}} \\
& =1500 \mathrm{~N} / \mathrm{C}
\end{aligned}
\]
\[
\begin{aligned}
& \text { Electrical energy of the conductor }=\frac{Q^{2}}{8 \pi \varepsilon_{0} R} \\
& =3.375 \mu \mathrm{~J}
\end{aligned}
\]
42
Capacitors \(C_{1}=3 \mu \mathrm{~F}, C_{2}=6 \mu \mathrm{~F}, C_{3}=4 \mu \mathrm{~F}\) and \(C_{4}=1 \mu \mathrm{~F}\) are connected in a circuit as shown to a battery of 60 V. Now if key \(K\) is closed, the charge that will flow through \(K\) is
Show Solution
Solution
Ans. (a)
\[
\begin{aligned}
& Q_{1}=3 \times 30=90 \mu \mathrm{C} \\
& Q_{2}=6 \times 30=180 \mu \mathrm{C}
\end{aligned}
\]
∴ charge flown from switch
\[
=180-90=90 \mu \mathrm{C}
\]
From \(b\) to \(a\)
43
The electrical conductivity of a sample of semiconductor is found to increase when the electromagnetic radiation of wave length just shorter than 2480 nm is incident normally on its surface. The band gap of the semiconductor is
Show Solution
Solution
Ans. (c)
\[
\begin{aligned}
& E_{g}=\frac{h c}{\lambda} \\
& =\frac{1240 \mathrm{eV} \cdot \mathrm{~nm}}{2480 \mathrm{~nm}} \\
& =0.5 \mathrm{eV}
\end{aligned}
\]
44
A U-shaped conducting wire of mass \(m=10 \mathrm{~g}\), having length of its horizontal section as \(\ell=20 \mathrm{~cm}\), is free to move vertically up and down. The two ends of the wire are immersed in mercury for proper electrical contact. The wire is in a homogenous field of magnetic induction \(B=0.1 \mathrm{~T}\) as shown. The wire jumps up to a height \(h=3 \mathrm{~m}\) when a charge \(q\), in the form of a current pulse, is sent through the wire. Considering that the duration of the current pulse is very small compared to the time of flight, the charge \(q\) passed through the wire is estimated to be nearly
Show Solution
Solution
Ans. (d)
Sol. Magnetic force \(F=i \ell B\)
\[
\begin{aligned}
& \Rightarrow \text { impulse }=\int(i \ell B) d t \\
& =B \ell \int i d t \\
& =B \ell q=M v \\
& \Rightarrow v=\frac{B \ell q}{M}=\sqrt{2 g h} \\
& \Rightarrow q=\frac{M}{B \ell} \sqrt{2 g h} \\
& =\frac{10}{1000} \\
& 0.1 \times 0.2 \\
& =\frac{1}{2} \sqrt{60}=\sqrt{15} \simeq 3.84 \mathrm{C}
\end{aligned}
\]
45
A direct vision spectroscope has been designed to obtain dispersion without deviation by arranging alternate inverted thin prisms of crown glass (refractive index \(\mu_{1}=\sqrt{2}\) ) and flint glass \(\left(\mu_{2}=\sqrt{3}\right)\) with refracting angle \(\theta_{\text {flint }}=3^{\circ}\). The refracting angle \(\theta_{\text {crown }}\) of the crown glass prism is
Show Solution
Solution
Ans. (c)
For no deviation,
\[
\begin{aligned}
& (\mu-1) A=\left(\mu^{\prime}-1\right) A^{\prime} \\
& \Rightarrow(\sqrt{2}-1) \theta_{\text {crown }}=(\sqrt{3}-1) 3 \\
& \Rightarrow \theta_{\text {crown }}=5.3^{\circ}
\end{aligned}
\]
Questions 46–60
NSEP 2023 Physics — Paper Code 61
46
Continuous and characteristic X-rays are produced when electron beam accelerated by a high potential difference of V volt (say) is made to hit the metallic target such as Molybdenum in a modern Coolidge tube. Let \(\lambda_{\text {min }}\) be the smallest possible wavelength of continuous X-rays and \(\lambda_{L \alpha}\) be the maximum wavelength of the characteristic X-rays. Then
Show Solution
Solution
Ans. (d)
Since \(\lambda_{L \alpha}\) is characteristic, it will be independent of \(V\). Also, \(\lambda_{\text {min }}=\frac{h c}{e V}\)
47
While performing an experiment for determining the focal length of a concave mirror by u-v method, a student recorded the given sets of the positions (in cm) of the object and the corresponding image on the bench as (12, 51), (18, 54), (30, 50), (48, 34), (42, 42) and (78, 98). She used an optical bench of length 1.5 m and the mirror is fixed at the 90 cm mark on the bench. The maximum acceptable error in the location of the image is 0.2 cm. The reading (observation) that cannot be obtained from experimental measurement and has been incorrectly recorded, for a mirror of focal length = 24 cm, is
Show Solution
Solution
Ans. (d)
The co-ordinates ( \(x, y\) ) must satisfy
\[
\frac{1}{90-x}+\frac{1}{90-y}=\frac{1}{24}
\]
48
A parallel beam, of 6.0 mW radiation of wavelength 200 nm and of area of cross-section \(1.0 \mathrm{~mm}^{2}\), falls normally on a plane metallic surface. If the radiations are completely reflected, the pressure exerted by the radiations on the metallic surface is estimated to be
Show Solution
Solution
Ans. (d)
\(P=\frac{2 l}{C}\)
\[
\begin{aligned}
& =2 \times \frac{\frac{6 \times 10^{-3}}{1 \times 10^{-6}}}{3 \times 10^{8}} \mathrm{~N} / \mathrm{m}^{2} \\
& =4 \times 10^{-5} \mathrm{~Pa}
\end{aligned}
\]
A – 2 ANY NUMBER OF OPTIONS 4, 3, 2 OR 1 MAY BE CORRECT MARKS WILL BE AWARDED ONLY IF ALL THE CORRECT OPTIONS ARE BUBBLED
49
A metal rod of mass \(m\) and length \(\ell\) slides on frictionless parallel metal rails of negligible resistance. A resistance \(R\) is connected between the rails at their ends as shown in the figure. A uniform magnetic field \(B\) is directed into the plane of paper perpendicular to the plane of rails throughout the space. The rod is given an initial velocity vo (towards right). No other force acts on the rod. Then
Show Solution
Solution
Ans. (b, d)
\(\varepsilon=B \ell v\)
\[
\begin{aligned}
& \Rightarrow \text { Force on rod }=-i \ell B=-\frac{B \ell v}{R} \cdot \ell \cdot B \\
& \Rightarrow \quad-\frac{B^{2} \ell^{2}}{R} v=\frac{m d v}{d t} \\
& \Rightarrow \quad \frac{d v}{v}=-\frac{B^{2} \ell^{2}}{m R} d t \\
& \Rightarrow \quad v=v_{0} e^{\frac{-B^{2} \ell^{2} t}{m R}} \\
& \Rightarrow \quad \frac{d x}{d t}=v_{0} e^{\frac{-B^{2} \ell^{2} t}{m R}} \\
& \Rightarrow \quad d x=v_{0} e^{\frac{-B^{2} \ell^{2} t}{m R}} \\
& \Rightarrow \quad \Delta x=v_{0} \frac{1}{\frac{B^{2} \ell^{2}}{m R}}=\frac{m v_{0} R}{B^{2} \ell^{2}}
\end{aligned}
\]
Also, \(i R=B \ell v \quad[\because \varepsilon=i R]\)
\[
\begin{array}{r}
\Rightarrow \quad \int i d t=\frac{B \ell}{R} \int v d t \\
=\frac{B \ell}{R} \cdot \Delta x=\frac{m v_{0}}{B \ell}
\end{array}
\]
⇒ Options b \& d
50
The magnetic field \(\vec{B}=2 \times 10^{-5} \sin \left\{\pi\left(0.5 \times 10^{3} x+1.5 \times 10^{11} t\right)\right\} \hat{j} T\) represents a plane electromagnetic wave travelling in space with \(x\) in meter and \(t\) in second. The correct statement(s) are
Show Solution
Solution
Ans. (a, d)
\[
\begin{aligned}
& k=\pi \times 0.5 \times 10^{3} \Rightarrow \frac{2 \pi}{\lambda}=500 \pi \\
\Rightarrow & \lambda=\frac{1}{250} \mathrm{~m}=4 \mathrm{~mm}
\end{aligned}
\]
Aslo \(\omega=\pi \times 1.5 \times 10^{11}=2 \pi f \Rightarrow f=75 \mathrm{GHz}\)
Energy density \(\mu=\frac{B_{0}^{2}}{2 \mu_{0}}=\frac{4 \times 10^{-10}}{2 \times 4 \pi \times 10^{-7}} \mathrm{~J} / \mathrm{m}^{3}\)
\[
=\frac{1}{2 \pi} \times 10^{-3} \mathrm{~J} / \mathrm{m}^{3} \simeq 159 \mu \mathrm{~J} / \mathrm{m}^{3}
\]
Also, \(\vec{E}=\left[2 \times 10^{-5} \times 3 \times 10^{8}\right] \sin \left[\pi\left\{0.5 \times 10^{3} x+1.5 \times 10^{11} t\right\}\right] \hat{k} \mathrm{~N} / \mathrm{C}\) ⇒ options a \& d
51
The force \(F(x)\) acting on a body of mass m changes with position \(x\) (in meter) as shown. It is given that the potential energy \(U(x)=0\) at \(x=0\)
Choose correct option(s).
Show Solution
Solution
Ans. (d)
\(F=\frac{-d U}{d x}\)
\[
\begin{aligned}
& \Rightarrow \quad d U=-F d x \\
& \Rightarrow \quad \int_{0}^{x} d U=\int_{0}^{x}-F d x \\
& \Rightarrow \quad U(x)=\int_{0}^{x} F d x \\
& \quad[2,4]: F=-4 x+12 \\
& \begin{aligned}
\Rightarrow & U(x)=-\left[8+\int_{2}^{x}(12-4 x) d x\right] \\
& \quad=-\left[8+12(x-2)-2\left(x^{2}-4\right)\right]=2 x^{2}-12 x+8
\end{aligned} \\
& \quad[4,6]: F=3 x-18 \\
& \begin{aligned}
\Rightarrow \quad U & =-\left[8+\int_{4}^{x}(3 x-18) d x\right] \\
\quad & =-\left[8+\frac{3}{2}\left(x^{2}-16\right)-18(x-4)\right] \\
\quad & =-\frac{3}{2} x^{2}+18 x-56
\end{aligned} \\
& \begin{aligned}
U(x=3) & =2(3)^{2}-12(3)+8=18-36+8=-10 \mathrm{~J}
\end{aligned}
\end{aligned}
\]
52
A deuteron of mass \(M\) moving at speed \(v\) collides elastically with an \(\alpha\)-particle of mass \(2 M\), initially at rest. The deuteron is scattered through 90° from initial direction of its motion with speed \(V_{d}\) while the \(\alpha\)-particle is scattered with speed \(V_{\alpha}\) at an angle \(\theta\) from the initial direction of motion of deuteron. Then
Show Solution
Solution
Ans. (a, b, c, d)
\[
\begin{aligned}
& M V=2 M V_{\alpha} \cos \theta \\
& M V_{d}=2 M V_{\alpha} \sin \theta \\
& V \cos \theta=V_{\alpha}+V_{d} \sin \theta \quad[\because e=1]
\end{aligned}
\]
…(ii)
\[
\begin{aligned}
& \left.\Rightarrow \quad \begin{array}{l}
V=2 V_{\alpha} \cos \theta \\
V_{d}=2 V_{\alpha} \sin \theta
\end{array}\right] \Rightarrow 2 V_{\alpha} \cos ^{2} \theta=V_{\alpha}+2 V_{\alpha} \sin ^{2} \theta \\
& \Rightarrow 2 \cos ^{2} \theta=1+2 \sin ^{2} \theta \\
& \Rightarrow 2 \cos 2 \theta=1 \\
& \Rightarrow 2 \theta=60^{\circ} \Rightarrow \theta=30^{\circ}
\end{aligned}
\]
\[
\begin{aligned}
\Rightarrow & V_{\alpha}=\frac{V}{\sqrt{3}} \\
& V_{d}=V_{\alpha}=\frac{V}{\sqrt{3}} \\
\text { Fraction } & =\frac{\frac{1}{2} \cdot 2 M \cdot V_{\alpha}^{2}}{\frac{1}{2} \cdot M \cdot V^{2}}=2 \cdot \frac{1}{3}=\frac{2}{3}
\end{aligned}
\]
⇒ Options a, b, c, \& d
53
Two plane progressive waves travelling on a string as
\(\)
\begin{aligned}
& Y_{1}=2.5 \times 10^{-3} \sin (30 x-420 t)
& Y_{2}=2.5 \times 10^{-3} \sin (30 x+420 t)
\end{aligned}
\(\)
Superimpose to produce a standing wave. The variables \(x\) and \(y\) are in meter and \(t\) is in second. Then
Show Solution
Solution
Ans. (c, d)
\(y=y_{1}+y_{2}\)
\[
\begin{aligned}
& =2.5 \times 10^{-3}[2 \sin (30 x) \cos (420 t)] \\
& =5 \times 10^{-3} \sin (30 x) \cos (420 t)
\end{aligned}
\]
For antinode, \(\sin (30 x)= \pm 1\)
\[
\begin{array}{ll}
\Rightarrow & 30 x=(2 n-1) \frac{\pi}{2} \\
\Rightarrow & x=(2 n-1) \frac{\pi}{60} \\
& (2 n-1) \frac{\pi}{60}=\frac{1}{4} \Rightarrow n=2.88 \Rightarrow \text { For closest }, n=3 \\
\Rightarrow & x \simeq 0.262 \mathrm{~m} \\
\Rightarrow & A(x=0.17 \mathrm{~m})=\left|5 \times 10^{-3} \sin (30 \times 0.17)\right| \mathrm{m} \simeq 4.6 \mathrm{~mm} \\
\Rightarrow & \text { Options c \& d }
\end{array}
\]
54
Two moles of nitrogen in a container, of negligible thermal capacity, are initially at 17°C. The gas is compressed adiabatically from an initial volume of 120 liter to 80 liter. The correct option(s) is/are
Show Solution
Solution
Ans. (a, b, d)
\[
\begin{align*}
& P V=\text { constant } \tag{i}\\
& P V=n R T \tag{ii}\\
\Rightarrow \quad & P_{i} \times\left(120 \times 10^{-3}\right)=2 \times 8.314 \times 290 \\
\Rightarrow \quad & P_{i} \simeq 40.2 \mathrm{kPa} \\
& T V-1=c \\
\Rightarrow & (290)(120)^{2 / 5}=\left(T_{f}\right)(80)^{2 / 5} \\
\Rightarrow & T_{f}=290\left(\frac{3}{2}\right)^{2 / 5} \mathrm{~K} \simeq 68^{\circ} \mathrm{C} \\
& W=\frac{n R\left[T_{i}-T_{f}\right]}{\gamma-1}=\frac{-2 \times 8.314}{\frac{2}{5}} \times 51 \mathrm{~J} \simeq-2.12 \mathrm{~kJ} \\
\Rightarrow & \Delta U=-W=+2.12 \mathrm{~kJ} \\
\Rightarrow & \text { Options (a), (b) \& (d) }
\end{align*}
\]
55
A small dipole is placed at the origin with its dipole moment \(\vec{P}=p \hat{i}\) oriented along \(x\) axis. \(E\) and \(V\), are respectively, the electric field and potential at point \(A(x, y)\). The observations at the Point \(A(x, y)\) which is at a large distance \(r\) from the origin, show that
Show Solution
Solution
Ans. (a, c, d)
\[
\begin{aligned}
E_{2}= & \frac{1}{4 \pi \varepsilon_{0}} \frac{p \sin \theta}{r^{3}} \\
\Rightarrow & E_{\mathrm{x}}=E_{1} \cos \theta-E_{2} \sin \theta \\
& =\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\left[2 \cos ^{2} \theta-\sin ^{2} \theta\right] \\
& =\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\left[2 \cdot \frac{x^{2}}{r^{2}}-\frac{y^{2}}{r^{2}}\right] \\
& =\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{5}}\left[2 x^{2}-y^{2}\right] \\
E_{\mathrm{y}}= & E_{1} \sin \theta+E_{2} \cos \theta \\
& =\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}[2 \sin \theta \cos \theta+\sin \theta \cos \theta] \\
& =\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\left[\frac{3 x y}{r^{2}}\right] \\
& =\frac{1}{4 \pi \varepsilon_{0}} \frac{3 p x y}{r^{5}} \\
V= & \frac{1}{4 \pi \varepsilon_{0}} \frac{p \cos \theta}{r^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\vec{p} \cdot \vec{r}}{r^{2}}
\end{aligned}
\]
⇒ Option a, c, \& d
56
Two equal positive charges \(+Q\) each lie on \(y\) axis at \((0, a)\) and \((0,-a)\). The electric field strength \(E\) at a point \((x, 0)\) satisfies:
Show Solution
Solution
Ans. (b, c)
\[
\begin{aligned}
& E=2 \mathrm{E}_{0} \operatorname{Cos} \theta=2 \cdot \frac{K Q}{4 \pi \varepsilon_{0}\left(\sqrt{a^{2}+} x^{2}\right)^{2}} \\
& =\frac{2 Q x K}{\left(a^{2}+x^{2}\right)^{3 / 2}}
\end{aligned}
\]
for large \(E \propto \frac{X}{X^{3}} \propto X^{2}(\mathrm{a} \ll \mathrm{x})\) for max \(\frac{d E}{d x}=0\)
\[
\begin{aligned}
& \frac{1}{\left(a^{2}+x^{2}\right)^{3 / 2}}-\frac{3}{2} \frac{x \cdot 2 x}{\left(a^{2}+x^{2}\right)^{5 / 2}}=0 \\
& a^{2}+x^{2}-3 x^{2}=0 \\
& x=\frac{a}{\sqrt{2}} \\
\Rightarrow & (\text { b \& c })
\end{aligned}
\]
57
In the circuit shown, the current in the \(8 \Omega\) resistance across \(G\) and \(H\) is \(i=0.5\) ampere. The ammeter is ideal. The internal resistance of the cell is \(0.8 \Omega\). Choose correct option(s).
Show Solution
Solution
Ans. (a, b, c, d)
\[
\begin{aligned}
& V E G H F=12 \times 0.5=6 \text { volt } \\
& I E F=1 \mathrm{~A}
\end{aligned}
\]
\[
\begin{align*}
V_{C D} & =1.5 \times 2+1 \times 6+1.5 \times 2 \\
& =12 \text { volt } \\
& V_{A C D B}=2.5 \times 2+1 \times 12 \times 2.5 \times 2+2.5 \times 0.8 \\
& =5+12+5+2 \\
& =24 \\
V_{C F} & =2 \times 1.5+1 \times 6=9 \mathrm{~V} \tag{C}\\
V_{A H} & =2.5 \times 2+1.5 \times 2+0.5 \times 10 \\
& =5+3+5=13 \tag{B}
\end{align*}
\]
58
In an experiment with Lummer Gehrcke plate, the two coherent beams of light, caused by multiple reflections inside the transparent plate of refractive index \(\mu=1.54\), reach the points \(P\) and \(Q\) on the screen. The net path difference between the two beams reaching either at \(P\) or \(Q\) is \(\Delta x=5000\) nm. Which of the wavelengths in the visible range ( \(\lambda=390 \mathrm{~nm}\) to \(\lambda=780 \mathrm{~nm}\) ) is/are most likely to produce a constructive interference (a maximum) at the point \(P\) as well as at \(Q\) on the screen?
Show Solution
Solution
Ans. (a, b, c)
For constructive interference,
\[
\begin{aligned}
& \Delta x=n \lambda \\
& \Rightarrow 5000=n \lambda \\
& \Rightarrow \lambda=\frac{5000}{n} \mathrm{~nm}
\end{aligned}
\]
⇒ Allowed \(\lambda\) are:
\[
\frac{5000}{7} \mathrm{~nm}, \frac{5000}{8} \mathrm{~nm} \ldots \ldots, \frac{5000}{12} \mathrm{~nm}
\]
59
Two identical transparent solid cylinders, each of radius 10 cm and refractive index \(\mu=\sqrt{3}\), lie horizontally parallel to each other on a horizontal plane mirror with a separation \(x\) between their horizontal axes. A ray of light is incident horizontally on the cylinder A at a height \(h\) above the plane mirror so as to emerge from this cylinder at a height \(h_{1}=0.1 \mathrm{~m}\) above the plane mirror. The ray emerging out from the first cylinder A is reflected from the horizontal plane mirror to enter the second parallel cylinder \(B\) at a height \(h_{2}\) and then this ray emerges out of the second cylinder, parallel and in-line with the original incident ray. The correct statement(s) is/are:
Show Solution
Solution
Ans. (a, b, c, d)
\[
\begin{align*}
& \sin \alpha=\sqrt{3} \sin \theta \tag{1}\\
& \alpha=2 \theta \tag{2}\\
& \Rightarrow 2 \cos \theta=\sqrt{3} \\
& \Rightarrow \theta=30^{\circ}
\end{align*}
\]
\[
\begin{aligned}
\Rightarrow h=R+R \sin 2 \theta & =10\left[1+\frac{\sqrt{3}}{2}\right] \\
& =18.66 \mathrm{~cm}
\end{aligned}
\]
\[
\begin{aligned}
x=R+2 R \cot \alpha+R & =2 R\left[1+\frac{1}{\sqrt{3}}\right] \\
& =31.54 \mathrm{~cm}
\end{aligned}
\]
Angle of incidence for plane mirror \(=90-\alpha=30^{\circ}\)
60
In the working of a \(p-n\) junction
Show Solution
Solution
Ans. (a, b, c)
(a) Diffusion > drift because of density difference
(b) Drift > diffusion because of high electric field
(c) Depletion region decrease with increase in forward bias voltage
(d) Dopant is not ionized but because of density difference, they diffuse
