JEE Main Practice — Physics Set 3

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 satellite revolves in a circular orbit at a height above the Earth's surface equal to the Earth's radius R=6400R=6400 km. Taking g=10 m/s2g=10\ \mathrm{m/s^2}, its period of revolution is closest to

  1. A.4 h
  2. B.2 h
  3. C.8 h
  4. D.1.4 h

Correct answer: A

Explanation: r=2Rr=2R. T=2πr3GM=2π8R3gR2=2π8Rg=2π5.12×106≈1.42×104T=2\pi\sqrt{\frac{r^3}{GM}}=2\pi\sqrt{\frac{8R^3}{gR^2}}=2\pi\sqrt{\frac{8R}{g}}=2\pi\sqrt{5.12\times10^6}\approx1.42\times10^4 s ≈3.95\approx3.95 h ≈4\approx4 h. (1.4 h is the period of a surface-grazing orbit.)

Question 2

Physics · 4 marks · −1

A satellite of mass mm is shifted from a circular orbit of radius 2R2R to one of radius 3R3R around the Earth (radius RR, surface gravity gg). The minimum energy required is

  1. A.mgR6\frac{mgR}{6}
  2. B.mgR12\frac{mgR}{12}
  3. C.mgR24\frac{mgR}{24}
  4. D.mgR3\frac{mgR}{3}

Correct answer: B

Explanation: Total energy in orbit rr: E=−GMm2rE=-\frac{GMm}{2r}. ΔE=GMm2(12R−13R)=GMm12R=mgR12\Delta E=\frac{GMm}{2}\left(\frac{1}{2R}-\frac{1}{3R}\right)=\frac{GMm}{12R}=\frac{mgR}{12} (using GM=gR2GM=gR^2).

Question 3

Physics · 4 marks · −1

At what depth below the Earth's surface is the acceleration due to gravity equal to its value at a height R/4R/4 above the surface? (RR = radius of Earth; treat Earth as uniform.)

  1. A.0.64R0.64R
  2. B.0.25R0.25R
  3. C.0.5R0.5R
  4. D.0.36R0.36R

Correct answer: D

Explanation: gh=g(5/4)2=0.64gg_h=\frac{g}{(5/4)^2}=0.64g and gd=g(1−dR)g_d=g\left(1-\frac dR\right). Setting equal: 1−dR=0.64⇒d=0.36R1-\frac dR=0.64\Rightarrow d=0.36R.

Question 4

Physics · 4 marks · −1

Two stars of masses MM and 2M2M, separated by a fixed distance dd, revolve in circular orbits about their common centre of mass under mutual gravitation. Which statement is correct?

  1. A.Period =2πd33GM=2\pi\sqrt{\frac{d^3}{3GM}}; kinetic energies (lighter : heavier) =2:1=2:1
  2. B.Period =2πd32GM=2\pi\sqrt{\frac{d^3}{2GM}}; kinetic energies (lighter : heavier) =2:1=2:1
  3. C.Period =2πd33GM=2\pi\sqrt{\frac{d^3}{3GM}}; kinetic energies (lighter : heavier) =1:2=1:2
  4. D.Period =2πd3GM=2\pi\sqrt{\frac{d^3}{GM}}; kinetic energies (lighter : heavier) =1:1=1:1

Correct answer: A

Explanation: Radii: r1=2d3r_1=\frac{2d}{3} (mass MM), r2=d3r_2=\frac d3. For MM: G(2M)Md2=Mω22d3⇒ω2=3GMd3\frac{G(2M)M}{d^2}=M\omega^2\frac{2d}{3}\Rightarrow\omega^2=\frac{3GM}{d^3}, so T=2πd33GMT=2\pi\sqrt{\frac{d^3}{3GM}} (i.e. G(M1+M2)G(M_1+M_2)). Both share ω\omega, so K=12mω2r2K=\frac12m\omega^2r^2: K1K2=M(2d/3)22M(d/3)2=2\frac{K_1}{K_2}=\frac{M(2d/3)^2}{2M(d/3)^2}=2. (Equal and opposite momenta also give K∝1/mK\propto1/m.)

Question 5

Physics · 4 marks · −1

A planet has the same mean density as the Earth but half its radius. A person who weighs 600 N on the Earth would weigh (in newton) ______ on the planet.

Correct answer: 300

Explanation: g=GMR2=43πGρR∝Rg=\frac{GM}{R^2}=\frac43\pi G\rho R\propto R for equal density. Half the radius gives g/2g/2, so weight =300=300 N.

Question 6

Physics · 4 marks · −1

The minimum energy needed to take a 10 kg body from the Earth's surface to a height equal to the Earth's radius is x×107x\times10^{7} J. Take R=6400R=6400 km and g=10 m/s2g=10\ \mathrm{m/s^2}. The value of xx is ______.

Correct answer: 32

Explanation: ΔU=−GMm2R+GMmR=GMm2R=mgR2=10×10×6.4×1062=3.2×108\Delta U=-\frac{GMm}{2R}+\frac{GMm}{R}=\frac{GMm}{2R}=\frac{mgR}{2}=\frac{10\times10\times6.4\times10^6}{2}=3.2\times10^8 J =32×107=32\times10^7 J.

Question 7

Physics · 4 marks · −1

A ball is thrown vertically upward with a speed of 20 m/s. Taking g=10 m/s2g=10\ \mathrm{m/s^2} and neglecting air resistance, the maximum height reached and the total time of flight are respectively

  1. A.20 m and 2 s
  2. B.40 m and 4 s
  3. C.20 m and 4 s
  4. D.10 m and 2 s

Correct answer: C

Explanation: H=u22g=40020=20H=\frac{u^2}{2g}=\frac{400}{20}=20 m. Time up =u/g=2=u/g=2 s, so time of flight =4=4 s.

Question 8

Physics · 4 marks · −1

Two trains, each 150 m long, move on parallel tracks in opposite directions with speeds 60 km/h and 40 km/h. The time taken by them to cross each other completely is

  1. A.5.4 s
  2. B.10.8 s
  3. C.54 s
  4. D.27 s

Correct answer: B

Explanation: Relative speed =100=100 km/h =2509=\frac{250}{9} m/s. Distance to be covered =150+150=300=150+150=300 m. t=300×9250=10.8t=\frac{300\times9}{250}=10.8 s.

Question 9

Physics · 4 marks · −1

Two projectiles are launched from the same point on level ground with the same speed at angles 30∘30^\circ and 60∘60^\circ to the horizontal. The ratio of their times of flight (T30:T60T_{30}:T_{60}) is

  1. A.1:31:3
  2. B.1:31:\sqrt3
  3. C.3:1\sqrt3:1
  4. D.1:11:1

Correct answer: B

Explanation: T=2usin⁡θg∝sin⁡θT=\frac{2u\sin\theta}{g}\propto\sin\theta: T30T60=sin⁡30∘sin⁡60∘=1/23/2=13\frac{T_{30}}{T_{60}}=\frac{\sin30^\circ}{\sin60^\circ}=\frac{1/2}{\sqrt3/2}=\frac{1}{\sqrt3}. (Their ranges are equal; their maximum heights are in the ratio 1:31:3.)

Question 10

Physics · 4 marks · −1

A particle starts from the origin at t=0t=0 and moves along the xx-axis such that its velocity is v=3xv=3\sqrt{x} (SI units). Its displacement at t=2t=2 s is

  1. A.4.5 m
  2. B.6 m
  3. C.18 m
  4. D.9 m

Correct answer: D

Explanation: a=vdvdx=3x⋅32x=4.5 m/s2a=v\frac{dv}{dx}=3\sqrt x\cdot\frac{3}{2\sqrt x}=4.5\ \mathrm{m/s^2}, constant, and v=0v=0 at x=0x=0. So x=12(4.5)(2)2=9x=\frac12(4.5)(2)^2=9 m. (Check: dxdt=3x⇒2x=3t⇒x=94t2\frac{dx}{dt}=3\sqrt x\Rightarrow 2\sqrt x=3t\Rightarrow x=\frac94t^2.)

Question 11

Physics · 4 marks · −1

A river 400 m wide flows at 3 m/s. A boat can move at 5 m/s relative to the water. If the boat crosses the river in the minimum possible time, how far downstream does it land?

  1. A.240 m
  2. B.300 m
  3. C.0 m
  4. D.160 m

Correct answer: A

Explanation: Minimum time: head perpendicular to the bank, t=4005=80t=\frac{400}{5}=80 s. Drift =3×80=240=3\times80=240 m.

Question 12

Physics · 4 marks · −1

A particle is projected from the bottom of a fixed inclined plane of inclination 30∘30^\circ, up the plane, with speed 15 m/s at an angle of 60∘60^\circ to the horizontal (i.e. 30∘30^\circ to the incline). Taking g=10 m/s2g=10\ \mathrm{m/s^2}, the distance along the incline at which it lands is

  1. A.22.5 m
  2. B.15 m
  3. C.10 m
  4. D.30 m

Correct answer: B

Explanation: Axes along/perpendicular to incline: ux=15cos⁡30∘=7.53u_x=15\cos30^\circ=7.5\sqrt3, uy=15sin⁡30∘=7.5u_y=15\sin30^\circ=7.5; ax=−gsin⁡30∘=−5a_x=-g\sin30^\circ=-5, ay=−gcos⁡30∘=−53a_y=-g\cos30^\circ=-5\sqrt3. Time of flight T=2uy∣ay∣=1553=3T=\frac{2u_y}{|a_y|}=\frac{15}{5\sqrt3}=\sqrt3 s. Range =uxT+12axT2=7.53⋅3−12(5)(3)=22.5−7.5=15=u_xT+\frac12a_xT^2=7.5\sqrt3\cdot\sqrt3-\frac12(5)(3)=22.5-7.5=15 m. (Formula: 2u2sin⁡αcos⁡(α+θ)gcos⁡2θ\frac{2u^2\sin\alpha\cos(\alpha+\theta)}{g\cos^2\theta}.)

Question 13

Physics · 4 marks · −1

The position of a particle moving along a straight line is x=t3−6t2+9tx=t^3-6t^2+9t (in metre, tt in second). The total distance travelled by the particle in the first 4 s is ______ m.

Correct answer: 12

Explanation: v=3t2−12t+9=3(t−1)(t−3)v=3t^2-12t+9=3(t-1)(t-3), zero at t=1t=1 and t=3t=3. x(0)=0x(0)=0, x(1)=4x(1)=4, x(3)=0x(3)=0, x(4)=4x(4)=4. Distance =∣4−0∣+∣0−4∣+∣4−0∣=12=|4-0|+|0-4|+|4-0|=12 m.

Question 14

Physics · 4 marks · −1

A stone dropped from rest from the top of a tower travels 25 m in the last second of its fall. Taking g=10 m/s2g=10\ \mathrm{m/s^2}, the height of the tower (in metre) is ______.

Correct answer: 45

Explanation: Distance in the last second of a fall lasting tt s: 12gt2−12g(t−1)2=g(t−12)=25⇒t=3\frac12gt^2-\frac12g(t-1)^2=g\left(t-\frac12\right)=25\Rightarrow t=3 s. Height =12×10×9=45=\frac12\times10\times9=45 m.

Question 15

Physics · 4 marks · −1

The rms speed of the molecules of an ideal gas at 300 K is vv. At 1200 K the rms speed will be

  1. A.4v4v
  2. B.v/2v/2
  3. C.2v2v
  4. D.2 v\sqrt2\,v

Correct answer: C

Explanation: vrms=3RTM∝Tv_{rms}=\sqrt{\frac{3RT}{M}}\propto\sqrt T. 1200/300=2\sqrt{1200/300}=2, so 2v2v.

Question 16

Physics · 4 marks · −1

According to the law of equipartition of energy, the average energy of a molecule of a rigid diatomic ideal gas at temperature TT is

  1. A.32kBT\frac32k_BT
  2. B.52kBT\frac52k_BT
  3. C.72kBT\frac72k_BT
  4. D.3kBT3k_BT

Correct answer: B

Explanation: A rigid diatomic molecule has 3 translational + 2 rotational = 5 degrees of freedom, each contributing 12kBT\frac12k_BT: total 52kBT\frac52k_BT.

Question 17

Physics · 4 marks · −1

A mixture contains 2 mol of helium and 3 mol of nitrogen (treated as rigid diatomic). The ratio of specific heats γ=Cp/Cv\gamma=C_p/C_v of the mixture is

  1. A.2921\frac{29}{21}
  2. B.32\frac{3}{2}
  3. C.75\frac{7}{5}
  4. D.3121\frac{31}{21}

Correct answer: D

Explanation: Cv=2⋅32R+3⋅52R5=3+7.55R=2.1RC_v=\frac{2\cdot\frac32R+3\cdot\frac52R}{5}=\frac{3+7.5}{5}R=2.1R, Cp=Cv+R=3.1RC_p=C_v+R=3.1R. γ=3.12.1=3121≈1.48\gamma=\frac{3.1}{2.1}=\frac{31}{21}\approx1.48.

Question 18

Physics · 4 marks · −1

The mean free path of the molecules of an ideal gas is λ\lambda. If the absolute temperature is halved and the pressure is doubled, the mean free path becomes

  1. A.λ\lambda
  2. B.4λ4\lambda
  3. C.λ/4\lambda/4
  4. D.λ/2\lambda/2

Correct answer: C

Explanation: λ=12πd2n\lambda=\frac{1}{\sqrt2\pi d^2n} with n=PkBTn=\frac{P}{k_BT}, so λ∝TP\lambda\propto\frac TP. New λ′=λ×1/22=λ4\lambda'=\lambda\times\frac{1/2}{2}=\frac\lambda4.

Question 19

Physics · 4 marks · −1

An ideal gas has density 1.2 kg/m3^3 at a pressure of 1.0×1051.0\times10^5 Pa. The rms speed of its molecules is

  1. A.500 m/s
  2. B.289 m/s
  3. C.250 m/s
  4. D.866 m/s

Correct answer: A

Explanation: P=13ρvrms2⇒vrms=3Pρ=3×1051.2=2.5×105=500P=\frac13\rho v_{rms}^2\Rightarrow v_{rms}=\sqrt{\frac{3P}{\rho}}=\sqrt{\frac{3\times10^5}{1.2}}=\sqrt{2.5\times10^5}=500 m/s.

Question 20

Physics · 4 marks · −1

A thermally insulated container holding nitrogen gas (molar mass 28 g/mol, treated as rigid diatomic ideal gas) moves with speed 100 m/s and is suddenly stopped. Assuming all the kinetic energy of ordered motion becomes internal energy of the gas, the rise in temperature is approximately (R=8.314R=8.314 J mol−1^{-1}K−1^{-1})

  1. A.11.2 K
  2. B.6.7 K
  3. C.4.8 K
  4. D.3.4 K

Correct answer: B

Explanation: Per mole: 12Mv2=CvΔT=52RΔT\frac12Mv^2=C_v\Delta T=\frac52R\Delta T. ΔT=Mv25R=0.028×1045×8.314=28041.57≈6.7\Delta T=\frac{Mv^2}{5R}=\frac{0.028\times10^4}{5\times8.314}=\frac{280}{41.57}\approx6.7 K. (11.2 K would follow from Cv=32RC_v=\frac32R, a monatomic gas.)

Question 21

Physics · 4 marks · −1

At what temperature (in kelvin) will oxygen molecules (M=32M=32 g/mol) have the same rms speed as hydrogen molecules (M=2M=2 g/mol) at 300 K?

Correct answer: 4800

Explanation: TM\frac{T}{M} must be equal: TO2=300×322=4800T_{O_2}=300\times\frac{32}{2}=4800 K.

Question 22

Physics · 4 marks · −1

The internal energy of 3 mol of a rigid diatomic ideal gas at 400 K, in kJ, is ______. (Take R=253R=\frac{25}{3} J mol−1^{-1} K−1^{-1}.)

Correct answer: 25

Explanation: U=n⋅52RT=3×52×253×400=25000U=n\cdot\frac52RT=3\times\frac52\times\frac{25}{3}\times400=25000 J =25=25 kJ.

Question 23

Physics · 4 marks · −1

A 5 kg block on a smooth horizontal floor is pulled by a force of 20 N acting at 60∘60^\circ above the horizontal. The acceleration of the block is

  1. A.4 m/s2^2
  2. B.232\sqrt3 m/s2^2
  3. C.2 m/s2^2
  4. D.1 m/s2^2

Correct answer: C

Explanation: Horizontal component =20cos⁡60∘=10=20\cos60^\circ=10 N (the block stays on the floor since 20sin⁡60∘<mg20\sin60^\circ<mg). a=10/5=2a=10/5=2 m/s2^2.

Question 24

Physics · 4 marks · −1

A 60 kg person stands on a weighing scale in a lift that accelerates upward at 2 m/s2^2. The reading of the scale (in newton) is (g=10 m/s2g=10\ \mathrm{m/s^2})

  1. A.600
  2. B.480
  3. C.720
  4. D.120

Correct answer: C

Explanation: N−mg=ma⇒N=m(g+a)=60×12=720N-mg=ma\Rightarrow N=m(g+a)=60\times12=720 N.

Question 25

Physics · 4 marks · −1

Blocks of mass 3 kg and 5 kg hang from the two ends of a light inextensible string passing over a light frictionless pulley. The tension in the string is (g=10 m/s2g=10\ \mathrm{m/s^2})

  1. A.37.5 N
  2. B.30 N
  3. C.50 N
  4. D.40 N

Correct answer: A

Explanation: a=(5−3)g8=2.5a=\frac{(5-3)g}{8}=2.5 m/s2^2. T=m1(g+a)=3×12.5=37.5T=m_1(g+a)=3\times12.5=37.5 N =2m1m2gm1+m2=\frac{2m_1m_2g}{m_1+m_2}.

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