Gravitation — NEET UG practice

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Sample questions with solutions

Q1 · 2026

The amount of work done to raise a mass 'mm' from the surface of the Earth to a height equal to the radius of the Earth 'RR' will be

  • A.

    2mgR2mgR

  • B.

    mgR4mg\dfrac{R}{4}

  • C.

    mgRmgR

  • D.

    mgR2mg\dfrac{R}{2}

Answer: D

  1. Recall the formula for gravitational potential energy. The gravitational potential energy of a mass mm at distance rr from the centre of Earth (mass MM) is U=−GMmrU = -\dfrac{GMm}{r}

  2. Write the initial and final potential energies. At the surface, r=Rr = R, and at height RR above the surface, r=R+R=2Rr = R + R = 2R, so U1=−GMmR,U2=−GMm2RU_1 = -\dfrac{GMm}{R}, \quad U_2 = -\dfrac{GMm}{2R}

  3. Find the work done. Since work done equals the change in potential energy, W=U2−U1=−GMm2R+GMmR=GMm2RW = U_2 - U_1 = -\dfrac{GMm}{2R} + \dfrac{GMm}{R} = \dfrac{GMm}{2R} Using g=GMR2g = \dfrac{GM}{R^2}, this simplifies to W=mgR2W = \dfrac{mgR}{2}

Hence, the answer is D (mgR2mg\dfrac{R}{2}).

Q2 · 2026

Two planets P1P_1 and P2P_2 with equal mass have radii R1R_1 and R2R_2, respectively, where R2=R12R_2=\dfrac{R_1}{2}. The escape speeds of P1P_1 and P2P_2 are v1v_1 and v2v_2, respectively. Then v2v1\dfrac{v_2}{v_1} is:

  • A.

    22

  • B.

    12\dfrac{1}{\sqrt{2}}

  • C.

    11

  • D.

    2\sqrt{2}

Answer: D

  1. Recall the escape velocity formula. Since escape velocity depends on the mass MM and radius RR of a planet, it can be written as Ve=2GMRV_e = \sqrt{\dfrac{2GM}{R}}

  2. Apply the condition of equal mass. Since P1P_1 and P2P_2 have the same mass MM, the escape velocity depends only on the radius, so Ve∝1RV_e \propto \dfrac{1}{\sqrt{R}}

  3. Find the ratio of escape velocities. Using the proportionality above for the two planets, v2v1=R1R2\dfrac{v_2}{v_1} = \sqrt{\dfrac{R_1}{R_2}} Given R2=R12R_2 = \dfrac{R_1}{2}, therefore v2v1=2\dfrac{v_2}{v_1} = \sqrt{2}

Hence, the answer is D (2\sqrt{2}).

Q3 · 2026

In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius RR is proportional to:

  • A.

    R3R^3

  • B.

    R1/2R^{1/2}

  • C.

    R3/2R^{3/2}

  • D.

    R2R^2

Answer: C

  1. Apply Kepler's Third Law. Kepler's third law states that the square of the orbital time period is proportional to the cube of the orbital radius, so T2∝R3T^2 \propto R^3

  2. Solve for T in terms of R. Taking the square root of both sides gives the proportionality between TT and RR, T∝R3/2T \propto R^{3/2}

Hence, the answer is C (R3/2R^{3/2}).

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