JEE Main Practice — Physics Set 5

JEE Main · 2026 · with answers and explanations

questions
25
Total marks
100
Duration
60 minutes
Negative marking
−1

Subjects: Physics

Question 1

Physics · 4 marks · −1

A large open tank stands on level ground. A small hole in its side is 1.8 m above the ground, and the water surface is 5 m above the hole. The water jet strikes the ground at a horizontal distance from the tank of

  1. A.3 m
  2. B.4.2 m
  3. C.6 m
  4. D.10 m

Correct answer: C

Explanation: Efflux speed v=2gh=100=10v=\sqrt{2gh}=\sqrt{100}=10 m/s. Time to fall 1.8 m: t=2×1.810=0.6t=\sqrt{\frac{2\times1.8}{10}}=0.6 s. Range =10×0.6=6=10\times0.6=6 m (=2hH=2\sqrt{hH}).

Question 2

Physics · 4 marks · −1

Two soap bubbles of radii 3 cm and 4 cm, in vacuum, coalesce isothermally to form a single bubble. The radius of the new bubble is

  1. A.7 cm
  2. B.5 cm
  3. C.3.5 cm
  4. D.4.5 cm

Correct answer: B

Explanation: In vacuum the pressure inside a bubble is only the excess pressure 4Tr\frac{4T}{r}. Isothermal and no gas lost: ∑PV\sum PV conserved: 4Tr1⋅43πr13+4Tr2⋅43πr23=4TR⋅43πR3⇒R2=r12+r22=25\frac{4T}{r_1}\cdot\frac43\pi r_1^3+\frac{4T}{r_2}\cdot\frac43\pi r_2^3=\frac{4T}{R}\cdot\frac43\pi R^3\Rightarrow R^2=r_1^2+r_2^2=25, R=5R=5 cm.

Question 3

Physics · 4 marks · −1

100 g of ice at 0∘0^\circC is added to 300 g of water at 40∘40^\circC in an insulated vessel of negligible heat capacity. Taking latent heat of fusion 80 cal/g and specific heat of water 1 cal g−1^{-1} ∘^\circC−1^{-1}, the final temperature (in ∘^\circC) is ______.

Correct answer: 10

Explanation: Heat available from water cooling to 0∘0^\circC =12000=12000 cal >> 8000 cal needed to melt the ice, so all ice melts. 300(40−T)=100×80+100(T−0)⇒12000−300T=8000+100T⇒T=10∘300(40-T)=100\times80+100(T-0)\Rightarrow12000-300T=8000+100T\Rightarrow T=10^\circC.

Question 4

Physics · 4 marks · −1

Two rods of equal length and equal cross-section, with thermal conductivities KK and 2K2K, are joined end to end. The free end of the first rod is kept at 100∘100^\circC and that of the second at 10∘10^\circC. In the steady state, the junction temperature (in ∘^\circC) is ______.

Correct answer: 40

Explanation: Equal heat current: K(100−T)=2K(T−10)⇒100−T=2T−20⇒T=40∘K(100-T)=2K(T-10)\Rightarrow100-T=2T-20\Rightarrow T=40^\circC.

Question 5

Physics · 4 marks · −1

An object is placed 30 cm in front of a concave mirror of focal length 15 cm. The image is

  1. A.at 15 cm, real and diminished
  2. B.at 30 cm, virtual and erect
  3. C.at 60 cm, real and magnified
  4. D.at 30 cm, real, inverted and of the same size

Correct answer: D

Explanation: 1v+1u=1f\frac1v+\frac1u=\frac1f with u=−30u=-30, f=−15f=-15: 1v=−115+130=−130⇒v=−30\frac1v=-\frac1{15}+\frac1{30}=-\frac1{30}\Rightarrow v=-30 cm. m=−vu=−1m=-\frac vu=-1: real, inverted, same size (object at C).

Question 6

Physics · 4 marks · −1

The critical angle for light going from a glass of refractive index 2\sqrt2 into air is

  1. A.30∘30^\circ
  2. B.45∘45^\circ
  3. C.60∘60^\circ
  4. D.90∘90^\circ

Correct answer: B

Explanation: sin⁡C=1μ=12⇒C=45∘\sin C=\frac1\mu=\frac1{\sqrt2}\Rightarrow C=45^\circ.

Question 7

Physics · 4 marks · −1

An object is placed 30 cm from a thin convex lens of focal length 20 cm. The image distance and magnification are

  1. A.60 cm, −2-2
  2. B.12 cm, −0.4-0.4
  3. C.60 cm, +2+2
  4. D.−60-60 cm, +2+2

Correct answer: A

Explanation: 1v−1u=1f\frac1v-\frac1u=\frac1f with u=−30u=-30, f=20f=20: 1v=120−130=160\frac1v=\frac1{20}-\frac1{30}=\frac1{60}, v=+60v=+60 cm. m=vu=60−30=−2m=\frac vu=\frac{60}{-30}=-2 (real, inverted, magnified).

Question 8

Physics · 4 marks · −1

An equiconvex lens made of glass (μ=1.5\mu=1.5) has radii of curvature 20 cm each. When it is completely immersed in a liquid of refractive index 1.25, its focal length becomes

  1. A.20 cm
  2. B.40 cm
  3. C.50 cm
  4. D.−50-50 cm

Correct answer: C

Explanation: In air: 1f=(1.5−1)220⇒f=20\frac1f=(1.5-1)\frac{2}{20}\Rightarrow f=20 cm. In liquid: 1f′=(1.51.25−1)220=0.2×0.1=0.02⇒f′=50\frac1{f'}=\left(\frac{1.5}{1.25}-1\right)\frac{2}{20}=0.2\times0.1=0.02\Rightarrow f'=50 cm (still converging since μglass>μliquid\mu_{glass}>\mu_{liquid}).

Question 9

Physics · 4 marks · −1

A prism of refracting angle 60∘60^\circ has refractive index 2\sqrt2. Its angle of minimum deviation is

  1. A.45∘45^\circ
  2. B.60∘60^\circ
  3. C.15∘15^\circ
  4. D.30∘30^\circ

Correct answer: D

Explanation: μ=sin⁡A+δm2sin⁡A2⇒sin⁡60∘+δm2=2sin⁡30∘=12⇒60∘+δm2=45∘⇒δm=30∘\mu=\frac{\sin\frac{A+\delta_m}{2}}{\sin\frac A2}\Rightarrow\sin\frac{60^\circ+\delta_m}{2}=\sqrt2\sin30^\circ=\frac1{\sqrt2}\Rightarrow\frac{60^\circ+\delta_m}{2}=45^\circ\Rightarrow\delta_m=30^\circ.

Question 10

Physics · 4 marks · −1

An astronomical telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. It is adjusted so that the final image of a distant object is formed at the least distance of distinct vision (25 cm). The magnifying power and the separation between objective and eyepiece are respectively

  1. A.24 and 104.2 cm
  2. B.20 and 105 cm
  3. C.24 and 105 cm
  4. D.20 and 104.2 cm

Correct answer: A

Explanation: Objective forms the image at its focus (100 cm). For the eyepiece v=−25v=-25 cm, f=5f=5 cm: 1u=1v−1f=−125−15=−625⇒∣ue∣=4.17\frac1u=\frac1v-\frac1f=-\frac1{25}-\frac15=-\frac{6}{25}\Rightarrow|u_e|=4.17 cm. Length =100+4.17≈104.2=100+4.17\approx104.2 cm. M=fofe(1+feD)=20×1.2=24M=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=20\times1.2=24. (20 and 105 cm correspond to normal adjustment.)

Question 11

Physics · 4 marks · −1

A coin lies at the bottom of a tank containing 12 cm of water (μ=4/3\mu=4/3) with a 6 cm layer of oil (μ=1.5\mu=1.5) floating on it. Viewed from directly above (near-normal incidence), the apparent depth of the coin below the oil surface (in cm) is ______.

Correct answer: 13

Explanation: Apparent depth =∑tiμi=124/3+61.5=9+4=13=\sum\frac{t_i}{\mu_i}=\frac{12}{4/3}+\frac{6}{1.5}=9+4=13 cm.

Question 12

Physics · 4 marks · −1

A thin convex lens of focal length 20 cm is placed in contact with a thin concave lens of focal length 30 cm. The focal length of the combination (in cm) is ______.

Correct answer: 60

Explanation: 1F=120−130=160⇒F=+60\frac1F=\frac1{20}-\frac1{30}=\frac{1}{60}\Rightarrow F=+60 cm (converging).

Question 13

Physics · 4 marks · −1

The moment of inertia of a thin uniform ring of mass MM and radius RR about a tangent perpendicular to its plane is

  1. A.32MR2\frac32MR^2
  2. B.MR2MR^2
  3. C.2MR22MR^2
  4. D.12MR2\frac12MR^2

Correct answer: C

Explanation: Parallel-axis theorem: I=Icm+MR2=MR2+MR2=2MR2I=I_{cm}+MR^2=MR^2+MR^2=2MR^2. (32MR2\frac32MR^2 is about a tangent in the plane of the ring.)

Question 14

Physics · 4 marks · −1

A skater spinning at 2 rev/s with arms outstretched pulls her arms in, reducing her moment of inertia to one-third of its initial value. Her new rate of spin is

  1. A.6 rev/s
  2. B.23\frac23 rev/s
  3. C.4 rev/s
  4. D.18 rev/s

Correct answer: A

Explanation: No external torque: I1ω1=I2ω2⇒ω2=II/3×2=6I_1\omega_1=I_2\omega_2\Rightarrow\omega_2=\frac{I}{I/3}\times2=6 rev/s.

Question 15

Physics · 4 marks · −1

A solid sphere, starting from rest, rolls without slipping down an incline through a vertical height hh. Its speed at the bottom is

  1. A.2gh\sqrt{2gh}
  2. B.10gh7\sqrt{\frac{10gh}{7}}
  3. C.4gh3\sqrt{\frac{4gh}{3}}
  4. D.gh\sqrt{gh}

Correct answer: B

Explanation: mgh=12mv2(1+k2R2)mgh=\frac12mv^2\left(1+\frac{k^2}{R^2}\right) with k2R2=25\frac{k^2}{R^2}=\frac25: v2=2gh7/5=10gh7v^2=\frac{2gh}{7/5}=\frac{10gh}{7}. (4gh/3\sqrt{4gh/3} is for a disc; gh\sqrt{gh} for a ring.)

Question 16

Physics · 4 marks · −1

For a solid sphere rolling without slipping on a horizontal surface, the ratio of its rotational kinetic energy to its translational kinetic energy is

  1. A.2:72:7
  2. B.5:75:7
  3. C.1:21:2
  4. D.2:52:5

Correct answer: D

Explanation: KrotKtr=12⋅25mR2ω212mv2=25\frac{K_{rot}}{K_{tr}}=\frac{\frac12\cdot\frac25mR^2\omega^2}{\frac12mv^2}=\frac25 since v=ωRv=\omega R. (2:72:7 is rotational : total.)

Question 17

Physics · 4 marks · −1

A light string is wound on a uniform disc-shaped pulley of mass 2 kg and radius 0.1 m that can rotate freely about its fixed horizontal axis. A 1 kg block hangs from the free end of the string. The acceleration of the block is (g=10g=10 m/s2^2)

  1. A.10 m/s2^2
  2. B.5 m/s2^2
  3. C.3.3 m/s2^2
  4. D.6.7 m/s2^2

Correct answer: B

Explanation: Block: mg−T=mamg-T=ma. Pulley: TR=12MR2αTR=\frac12MR^2\alpha with a=αR⇒T=Ma2a=\alpha R\Rightarrow T=\frac{Ma}{2}. So a=mgm+M/2=101+1=5a=\frac{mg}{m+M/2}=\frac{10}{1+1}=5 m/s2^2.

Question 18

Physics · 4 marks · −1

A uniform rod of mass mm and length LL is hinged at one end and held horizontal. It is then released. At the instant of release, the magnitude of the force exerted by the hinge on the rod is

  1. A.mg4\frac{mg}{4}
  2. B.mg2\frac{mg}{2}
  3. C.3mg4\frac{3mg}{4}
  4. D.mgmg

Correct answer: A

Explanation: About the hinge: mgL2=mL23α⇒α=3g2Lmg\frac L2=\frac{mL^2}{3}\alpha\Rightarrow\alpha=\frac{3g}{2L}. Acceleration of the centre of mass a=αL2=3g4a=\alpha\frac L2=\frac{3g}{4} (vertical; ω=0\omega=0 so no centripetal part). mg−N=m3g4⇒N=mg4mg-N=m\frac{3g}{4}\Rightarrow N=\frac{mg}{4}.

Question 19

Physics · 4 marks · −1

From a uniform square plate of side 12 cm, a square of side 6 cm is cut away at one corner. Taking that corner of the original plate as origin and its two edges as the axes, the xx-coordinate (in cm) of the centre of mass of the remaining plate is ______.

Correct answer: 7

Explanation: Areas: full 144 (centre (6,6)), removed 36 (centre (3,3)). x=144×6−36×3144−36=864−108108=7x=\frac{144\times6-36\times3}{144-36}=\frac{864-108}{108}=7 cm.

Question 20

Physics · 4 marks · −1

A solid sphere of mass 2 kg and radius 0.1 m rolls without slipping on a horizontal floor with its centre moving at 5 m/s. Its total kinetic energy (in joule) is ______.

Correct answer: 35

Explanation: K=12mv2(1+25)=710mv2=710×2×25=35K=\frac12mv^2\left(1+\frac25\right)=\frac{7}{10}mv^2=\frac{7}{10}\times2\times25=35 J.

Question 21

Physics · 4 marks · −1

To obtain a p-type semiconductor, pure silicon should be doped with

  1. A.phosphorus
  2. B.arsenic
  3. C.boron
  4. D.antimony

Correct answer: C

Explanation: A trivalent impurity (acceptor) such as boron creates holes as majority carriers, giving p-type silicon. P, As and Sb are pentavalent donors (n-type).

Question 22

Physics · 4 marks · −1

A two-input logic gate gives output 1 only when both inputs are 0. The gate is

  1. A.NAND
  2. B.NOR
  3. C.OR
  4. D.AND

Correct answer: B

Explanation: NOR: Y=A+B‾Y=\overline{A+B}, which is 1 only for A=B=0A=B=0. NAND is 0 only for A=B=1A=B=1.

Question 23

Physics · 4 marks · −1

A silicon diode (cut-in 0.7 V) and a germanium diode (cut-in 0.3 V) are connected in parallel, both forward-biased, and the combination is connected in series with a 1 kΩ1\ \mathrm{k\Omega} resistor across a 12 V battery. Treating each diode as ideal beyond its cut-in voltage, the current drawn from the battery is

  1. A.11.3 mA
  2. B.11.0 mA
  3. C.23.0 mA
  4. D.11.7 mA

Correct answer: D

Explanation: The Ge diode conducts first and clamps the parallel combination at 0.3 V, so the Si diode never reaches 0.7 V and carries no current. I=12−0.31000=11.7I=\frac{12-0.3}{1000}=11.7 mA.

Question 24

Physics · 4 marks · −1

A full-wave rectifier is fed from a 60 Hz AC supply. The fundamental frequency of the ripple in the output is

  1. A.30 Hz
  2. B.60 Hz
  3. C.120 Hz
  4. D.240 Hz

Correct answer: C

Explanation: A full-wave rectifier gives one output pulse per half cycle, so the output repeats at 2f=1202f=120 Hz (a half-wave rectifier would give 60 Hz).

Question 25

Physics · 4 marks · −1

Inputs AA and BB are fed to a NOR gate. The output of the NOR gate and the input AA are then fed to an AND gate, whose output is YY. Then

  1. A.Y=0Y=0 for all inputs
  2. B.Y=AY=A
  3. C.Y=ABˉY=A\bar B
  4. D.Y=AˉBˉY=\bar A\bar B

Correct answer: A

Explanation: Y=A⋅A+B‾=A⋅Aˉ⋅Bˉ=0Y=A\cdot\overline{A+B}=A\cdot\bar A\cdot\bar B=0 for all inputs, since AAˉ=0A\bar A=0. Truth table check: (0,0)→0⋅1=0(0,0)\to0\cdot1=0; (1,0)→1⋅0=0(1,0)\to1\cdot0=0; (0,1)→0(0,1)\to0; (1,1)→0(1,1)\to0.

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