Physics · chapter PYQs

Gravitation, on the exam screen.

4 questions from NEET 2026, Re-NEET 2026, NEET 2020, 4 minutes, marked +4 / −1. See them with solutions.

Gravitation · PYQs

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Question 1 · NEET 2026 · Q13 · +4 / −1
The amount of work done to raise a mass ' $m$ ' from the surface of the Earth to a height equal to the radius of the Earth ' $R$ ' will be
Answer and explanation
Answer: (4)
The work done equals the change in gravitational potential energy: $W = U_2 - U_1 = \left(-\frac{GMm}{2R}\right) - \left(-\frac{GMm}{R}\right) = \frac{GMm}{2R}$ . Using $g = \frac{GM}{R^2}$ , this becomes $\frac{mgR}{2}$ , option (4). It is not the naive $mgR$ because gravity weakens with height, so less work is needed than if g stayed constant.
Question 2 · Re-NEET 2026 · Q13 · +4 / −1
Two planets $P_1$ and $P_2$ with equal mass have radii $R_1$ and $R_2$ , respectively, where $R_2 = \frac{R_1}{2}$ . The escape speeds of $P_1$ and $P_2$ are $v_1$ and $v_2$ , respectively. Then $\frac{v_2}{v_1}$ is:
Answer and explanation
Answer: (3)
Escape speed is $v_e = \sqrt{\frac{2GM}{R}}$ . With the masses equal, $v_e \propto R^{-1/2}$ , so $\frac{v_2}{v_1} = \sqrt{\frac{R_1}{R_2}} = \sqrt{2}$ . The denser, smaller planet is harder to escape from, which is the physical sense check. NCERT Class 11, Gravitation (escape speed).
Question 3 · Re-NEET 2026 · Q14 · +4 / −1
In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius $R$ is proportional to:
Answer and explanation
Answer: (2)
Kepler's third law states $T^2 \propto R^3$ for orbits about a common central mass, so $T \propto R^{3/2}$ . It also follows directly from equating gravitational and centripetal force: $\frac{GMm}{R^2} = m\omega^2 R$ with $\omega = 2\pi/T$ . NCERT Class 11, Gravitation (Kepler's laws).
Question 4 · NEET 2020 · Q144 · +4 / −1
A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?
Answer and explanation
Answer: (1)
The gravitational force at a height h from the Earth's surface is given by $F = \frac{mgR^2}{(R + h)^2}$ . For $h = \frac{R}{2}$ , $F = \frac{mgR^2}{(R + \frac{R}{2})^2} = \frac{mgR^2}{\left(\frac{3R}{2}\right)^2} = \frac{4}{9} \times mg$ . Since $mg = 72\ N$ , $F = \frac{4}{9} \times 72 = 32\ N$ . Chapter: Gravitation.