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JEE SECTION - A [PHYSICS MCQ]

JEE Main 3-April-2025 Paper - Shift 1 · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 20 questions.

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SECTION - A [PHYSICS MCQ]

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20 questions · 6 auto-graded MCQ + 14 self-marked written.

MCQ 14 Marks
Consider following statements for refraction of light through prism, when angle of deviation is minimum.
(A) The refracted ray inside prism becomes parallel to the base.
(B) Larger angle prisms provide smaller angle of minimum deviation.
(C) Angle of incidence and angle of emergence becomes equal.
(D) There are always two sets of angle of incidence for which deviation will be same except at minimum deviation setting.
(E) Angle of refraction becomes double of prism angle.
Choose the correct answer from the options given below.
  • ✓
    A, C and D Only
  • B
    B, C and D Only
  • C
    A, B and E Only
  • D
    B, D and E Only
Answer
Correct option: A.
A, C and D Only
(A) A, C and D Only
$
\delta=I+e-A
$
For $\delta_{\min } \Rightarrow I = e$ and refracted ray is parallel to base $A , C , D$ are correct
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MCQ 54 Marks
A particle is released from height S above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.
  • A
    $\frac{ S }{2}, \sqrt{\frac{3 gS }{2}}$Q
  • B
    $\frac{ S }{2}, \frac{3 gS }{2}$
  • C
    $\frac{ S }{4}, \frac{3 gS }{2}$
  • ✓
    $\frac{ S }{4}, \sqrt{\frac{3 gS }{2}}$
Answer
Correct option: D.
$\frac{ S }{4}, \sqrt{\frac{3 gS }{2}}$
(D) $\frac{ S }{4}, \sqrt{\frac{3 gS }{2}}$
$
\begin{aligned}
V^2 & =0+2 g(S-x) \\
V^2 & =2 g(S-x)
\end{aligned}
$
At $B$, Potential energy $=m g x$
$
\begin{array}{l}
mgx=3 \times \frac{1}{2} mv^2 \\
gx=\frac{3}{2} \times 2 g(S-x) \\
4 x=S \\
x=\frac{S}{4} \\
\Rightarrow V=\sqrt{2 g \times \frac{3 S}{4}}=\sqrt{\frac{3 gS}{2}}
\end{array}
$
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MCQ 74 Marks
The electrostatic potential on the surface of uniformly charged spherical shell of radius $R =10 cm$ is 120 V . The potential at the centre of shell, at a distance $r =5 cm$ from centre, and at a distance $r =15 cm$ from the centre of the shell respectively, are :
  • A
    $120 V, 120 V, 80 V$
  • B
    $40 V, 40 V, 80 V$
  • C
    $0 V, 0 V, 80 V$
  • D
    $0 V, 120 V, 40 V$
Answer
(A) $120 V, 120 V, 80 V$
Potential inside shell is equal to potential on surface
$
\begin{array}{l}
V_{\text {in }}=V_{\text {surface }}=\frac{kQ}{R}=120 V \\
\text { at } r=15 cm \\
V=\frac{kQ}{r}=\frac{120 \times 10}{15}=80 V
\end{array}
$
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MCQ 94 Marks
A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of $800 cm^3$ and temperature $27^{\circ} C$. The change in temperature when the gas is adiabatically compressed to $200 cm^3$ is :
(Take $\gamma=1.5: \gamma$ is the ratio of specific heats at constant pressure and at constant volume)
  • A
    327 K
  • B
    600 K
  • C
    522 K
  • ✓
    300 K
Answer
Correct option: D.
300 K
(D) $300k$
$
\begin{array}{l}
V_1=800 cm^3 \quad V_2=200 cm^3 \\
T_1=300 K
\end{array}
$
for adiabatic
$
\begin{array}{l}
TV^{\gamma-1}=\text { const. } \\
(300)(800)^{1.5-1}=T_2(200)^{1.5-1} \\
T_2=300\left[\frac{800}{200}\right]^{0.5}=300 \times\left(2^2\right)^{1 / 2} \\
T_2=600 K \\
\Delta T=600-300=300 K
\end{array}
$
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MCQ 104 Marks
A piston of mass M is hung from a massless spring whose restoring force law goes as $F =- kx ^3$, where k is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with ' $n$ ' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature T ) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height $L_0$ to $L_1$, the total energy delivered by the filament is (Assume spring to be in its natural length before heating)
Image
  • A
    $3 n R T \ln \left(\frac{L_1}{L_0}\right)+2 Mg \left(L_1-L_0\right)+\frac{k}{3}\left(L_1^3-L_0^3\right)$
  • B
    $n R T \ln \left(\frac{L_1^2}{L_0^2}\right)+\frac{M g}{2}\left(L_1-L_0\right)+\frac{k}{4}\left(L_1^4-L_0^4\right)$
  • C
    $n R T \ln \left(\frac{L_1}{L_0}\right)+M g\left(L_1-L_0\right)+\frac{k}{4}\left(L_1^4-L_0^4\right)$
  • D
    $n R T \ln \left(\frac{L_1}{L_0}\right)+M g\left(L_1-L_0\right)+\frac{3 k}{4}\left(L_1^4-L_0^4\right)$
Answer
(C) $n R T \ln \left(\frac{L_1}{L_0}\right)+M g\left(L_1-L_0\right)+\frac{k}{4}\left(L_1^4-L_0^4\right)$
Using WET
Total energy supplied $=$ gravitational potential energy + spring potential energy + work done by gas
$
Mg\left(L_1-L_0\right)+\int_{L_0}^{L_1} kx^3 dx+nRT \ell n
$
$\begin{array}{l}{\left[\frac{L_1 A}{L_0 A}\right]+W_{e x t}=0} \\ \frac{K}{4}\left[x^4\right]_{L_0}^{L_1}+M g\left(L_1-L_0\right)+\int_{L_0}^{L_1} k x^3 d x+n R T \ell n \\ {\left[\frac{L_1}{L_0}\right]+W_{e x t}=0}\end{array}$
$\begin{array}{l}\frac{ k }{4}\left(L_1^4- L _0^4\right)+ Mg \left( L _1- L _0\right)+ nRT \ell n \\ {\left[\frac{ L _1}{L_0}\right]+ W _{ ext }=0} \\ W_{ ext }=\frac{ k }{4}\left(L_1^4- L _0^4\right)+ Mg \left( L _1- L _0\right)+ nRT \ell n \left[\frac{ L _1}{L_0}\right]\end{array}$
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MCQ 114 Marks
  • A
    $3.5 m / s ^2$
  • B
    $0.35 m / s ^2$
  • C
    $2.5 m / s ^2$
  • D
    $0.25 m / s ^2$
Answer
(A) $3.5 m / s ^2$
Image
Torque about bottom point
$
\begin{array}{l}
F \times 2 r=I \alpha \\
49 \times 2 r=\frac{7}{5} mr^2 \alpha \\
14=4 r \alpha
\end{array}
$
As sphere rolls without slipping
$
\begin{array}{l}
a=r \alpha \\
a=\frac{14}{4}=\frac{7}{2}=3.5 m / s^2
\end{array}
$
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MCQ 124 Marks
  • A
    A-IV, B-III, C-II, D-I
  • B
    A-III, B-II, C-I, D-IV
  • C
    A-II, B-IV, C-III, D-I
  • D
    A-I, B-III, C-IV, D-II
Answer
(A) A-IV, B-III, C-II, D-I
$
\begin{array}{l}
\text { (A) } G=\frac{Fr^2}{m^2} \\
{[G]=\frac{\left[MLT^{-2}\right]\left[L^2\right]}{\left[M^2\right]}=\left[M^{-1} L^3 T^{-2}\right](IV)}
\end{array}
$
(B)
$
\begin{array}{l}
\text { P.E. }=mgh=\left[MLT^{-2} L\right] \\
=\left[ML^2 T^{-2}\right] \text { (III) }
\end{array}
$
$
\begin{array}{l}
\text { (C) Gravitational Potential }=\frac{GM}{r} \\
=\frac{\left[M^{-1} L^3 T^{-2}\right][M]}{[L]}=\left[M^0 L^2 T^{-2}\right]=\left[L^2 T^{-2}\right] \text { (II) }
\end{array}
$
(D) Acceleration due to gravity $=[ g ]=\left[ LT ^{-2}\right]$ (I)
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MCQ 134 Marks
A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $\varepsilon_1$ and $\varepsilon_2$, as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are $C_1$ and $C_2$ respectively, then $\frac{C_1}{C_2}$ is :
Image
Image
  • A
    $\frac{\varepsilon_1 \varepsilon_2^2}{\left(\varepsilon_1+\varepsilon_2\right)^2}$
  • B
    $\frac{4 \varepsilon_1 \varepsilon_2}{\left(\varepsilon_1+\varepsilon_2\right)^2}$
  • C
    $\frac{\varepsilon_1 \varepsilon_2}{\varepsilon_1+\varepsilon_2}$
  • D
    $\frac{\varepsilon_0\left(\varepsilon_1+\varepsilon_2\right)}{2}$
Answer
(B) $\frac{4 \varepsilon_1 \varepsilon_2}{\left(\varepsilon_1+\varepsilon_2\right)^2}$
Image
Area of plate is A.
then
$
\begin{array}{l}
C=\frac{\varepsilon_2 \varepsilon_0 A}{d / 2}=\frac{2 \varepsilon_2 \varepsilon_0 A}{d} \\
C^{\prime}=\frac{\varepsilon_1 \varepsilon_0 A}{d / 2}=\frac{2 \varepsilon_1 \varepsilon_0 A}{d}
\end{array}
$
Let $C _0=\frac{\varepsilon_0 A}{ d }$
$
C=2 \varepsilon_2 C_0
$
$
C^{\prime}=2 \varepsilon_1 C_0
$
$C \& C ^{\prime}$ are in series
$
\begin{array}{l}
C_1=\frac{CC^{\prime}}{C+C^{\prime}}=\frac{4 \varepsilon_2 \varepsilon_1 C_0^2}{2 C_0\left(\varepsilon_2+\varepsilon_1\right)} \\
=\frac{2 \varepsilon_2 \varepsilon_1 C_0}{\left(\varepsilon_2+\varepsilon_1\right)}
\end{array}
$
Image
Here
$
C=\frac{\varepsilon_1 \varepsilon_0 A}{2 d}=\frac{\varepsilon_1 C_0}{2}
$
$
C^{\prime}=\frac{\varepsilon_2 C_0}{2}
$
$C \& C ^{\prime}$ are inparallel
$
C_2=C^{\prime}+C=\left(\varepsilon_1+\varepsilon_2\right) \frac{C_0}{2}
$
Thus $\frac{ C _1}{ C _2}=\frac{2 \varepsilon_2 \varepsilon_1 C _0}{\left(\varepsilon_2+\varepsilon_1\right)} \times \frac{2}{\left(\varepsilon_1+\varepsilon_2\right) C _0}$
$
=\frac{4 \varepsilon_2 \varepsilon_1}{\left(\varepsilon_2+\varepsilon_1\right)^2}
$
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MCQ 154 Marks
Two blocks of masses m and $M ,( M > m )$, are placed on a frictionless table as shown in figure. A massless spring with spring constant k is attached with the lower block. If the system is slightly displaced and released then
( $\mu=$ coefficient of friction between the two blocks)
Image
(A) The time period of small oscillation of the two blocks is $T =2 \pi \sqrt{\frac{(m+ M )}{ k }}$
(B) The acceleration of the blocks is $a =\frac{ kx }{ M + m }$ ( $x =$ displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is $\frac{ m \mu| x |}{ M + m }$
(D) The maximum amplitude of the upper block, if it does not slip, is $\frac{\mu(M+m) g}{k}$
Choose the correct answer from the options given below :
  • A
    A, B, D Only
  • B
    B , C, D Only
  • C
    C, D, E Only
  • D
    A, B, C Only
Answer
A
(A) As both blocks moving together so Time period $=2 \pi \sqrt{\frac{m}{ K }} ; \quad$ where $m = M + m$
$
T=2 \pi \sqrt{\frac{M+m}{K}}
$
(B) Let block is displaced by x in $(+ ve )$ direction so force on block will be in(-ve) direction
$
\begin{array}{l}
F=-Kx \\
(M+m) a=-Kx \\
a=-\frac{Kx}{(M+m)}
\end{array}
$
(C) As upper block is moving due to friction thus
$
f=ma=\frac{mKx}{(M+m)}
$
(D) This option is like two block problem in friction for maximum amplitude, force on block is also maximum, for which both blocks are moving together.
Image
Image
$\begin{array}{l} KA =( M + m ) a \\ a =\frac{ KA }{( M + m )} \\ f = ma =\frac{ mKA }{( M + m )} \\ f _{\max }= f _{ L }=\mu mg \\ f =\mu mg \\ \frac{m K A}{( M + m )}=\mu mg \\ A =\frac{\mu( M + m ) g }{ K }\end{array}$
(E) Maximum friction can be $\mu mg$ as force is acting between blocks & normal force here is mg.
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MCQ 164 Marks
A wire of length 25 m and cross-sectional area $5 mm^2$ having resistivity of $2 \times 10^{-6} \Omega m$ is bent into a complete circle. The resistance between diametrically opposite points will be
  • A
    $12.5 \Omega$
  • B
    $50 \Omega$
  • C
    $100 \Omega$
  • D
    $2.5 \Omega$
Answer
(D) $25 \Omega$
Image
$
\begin{array}{l}
L=25 m, A=5 mm^2=5 \times 10^{-6} m^2 \\
\rho=2 \times 10^{-6} \Omega m \\
R_{\text {wixe }}=\frac{\rho L}{A}=\frac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}}=10 \\
R_{eq}=\frac{R}{4}=\frac{10}{4}=2.5 \Omega
\end{array}
$
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MCQ 174 Marks
  • A
    $1.5 \times 10^{-8}$ Pascals
  • B
    0
  • C
    $6 \times 10^{-8}$ Pascals
  • D
    $3 \times 10^{-8}$ Pascals
Answer
(C). $6 \times 10^{-8}$ Pascals
$
P_{rad}=\frac{2 I}{C}
$
Where $I =$ intensity at surface
$
C=\text { Speed of light }
$
$
\begin{array}{l}
I=\frac{\text { Power }}{\text { Area }}=\frac{450}{4 \pi r^2} \\
=\frac{450}{4 \pi \times 4}=\frac{450}{16 \pi}
\end{array}
$
$
\begin{array}{l}
P_{rad}=\frac{2 \times 450}{16 \pi \times 3 \times 10^8}=\frac{150}{8 \pi \times 10^8} \\
=5.97 \times 10^{-8} \approx 6 \times 10^{-8} \text { Pascals }
\end{array}
$
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MCQ 194 Marks
Consider a completely full cylindrical water tank of height 1.6 m and cross-sectional area $0.5 m^2$. It has a small hole in its side at a height 90 cm from the bottom. Assume, the cross-sectional area of the hole to be negligibly small as compared to that of the water tank. If a load 50 kg is applied at the top surface of the water in the tank then the velocity of the water coming out at the instant when the hole is opened is : $\left(g=10 m / s ^2\right)$
  • A
    $3 m / s$
  • B
    $5 m / s$
  • C
    $2 m / s$
  • D
    $4 m / s$
Answer
D $4 m/s$
Image
Apply Bernouli equation between points 1 $\&$ 2
$
\begin{array}{l}
P_1+\frac{1}{2} \rho v_1^2+\rho g h=P_2+\frac{1}{2} \rho v_2^2+0 \\
P_0+\frac{mg}{A}+\rho g \frac{70}{100}=P_0+\frac{1}{2} \rho v_2^2 \\
\frac{5000}{0.5}+10^3 \times 10 \frac{70}{100}=\frac{1}{2} \times 10^3 v_2^2 \\
10^3+10^3 \times 7=\frac{10^3}{2} v_2^2
\end{array}
$
$
v_2^2=16
$
$
v_2=4 m / s
$
As the tank area is large $v _1$ is negligible compared to $v _2$
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MCQ 204 Marks
During the melting of a slab of ice at 273 K at atmospheric pressure :
  • A
    Internal energy of ice-water system remains unchanged.
  • B
    Positive work is done by the ice-water system on the atmosphere.
  • C
    Internal energy of the ice-water system decreases.
  • ✓
    Positive work is done on the ice-water system by the atmosphere.
Answer
Correct option: D.
Positive work is done on the ice-water system by the atmosphere.
(D) Positive work is done on the ice-water system by the atmosphere.
Volume decreases during melting of ice so positive work is done on ice water system by atmosphere
Heat absorbed by ice water so $\Delta Q$ is positive, work done by ice water system is negative
Hence by first law of thermodynamics
$
\Delta U=\Delta Q+\Delta W=\text { Positive }
$
So internal energy increases
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