Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant. Reason (R): For a central force field the angular momentum is a constant. In the light of the above statements, choose the most appropriate answer from the options given below:

Both (A) and (R) are correct and (R) is the correct explanation of (A)

is not correct but (R) is correct

Both (A) and (R) are correct but (R) is not the correct explanation of (A)

is correct but (R) is not correct

JEE Main · Physics · Chapter 8

Gravitation JEE Main PYQs (2020–2026)

161 questions from Gravitation have appeared in JEE Main between 2020 and 2026, including 112 in the last five papers (2022–2026). The most-asked topic is Acceleration due to Gravity and its Variation (44 questions). Practise them all with instant answers and step-by-step solutions.

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Numerical

Solved Gravitation PYQs

3 recent questions with answers. The full set of 161 is free to practise.

JEE Main 2026EasyGravitational Potential Energy
Q1
A body of mass $m$ is taken from the surface of earth to a height equal to twice the radius of earth $(R_{e})$ . The increase in potential energy will be $____$ .

( $g$ is acceleration due to gravity at the surface of earth)
- A $\frac{1}{2} m g R_{e}$
- B $\frac{3}{4} m g R_{e}$
- C $\frac{1}{4} m g R_{e}$
- D $\frac{2}{3} m g R_{e}$
Show answer & solution
Correct answer: (D)
The potential energy $(U)$ of a mass $m$ at a distance $r$ from the centre of the Earth is:

$U = - \frac{GM _{e} m}{r}$

As we know that $g = \frac{GM _{e}}{R _{e}^{2}}$ , so $GM_{e} = g R_{e}^{2}$ .

$U = - \frac{m g R _{e}^{2}}{r}$

At the surface of the Earth, the distance from the centre is $r_{1} = R_{e}$ :

So, the initial gravitational potential energy of body is;

$U_{i} = - \frac{m g R _{e}^{2}}{R _{e}} = - m g R_{e}$

The body is taken to a height $h = 2 R_{e}$ above the surface. The distance from the centre is $r_{2} = R_{e} + h = R_{e} + 2 R_{e} = 3 R_{e} :$

$U_{f} = - \frac{m g R _{e}^{2}}{3 R _{e}} = - \frac{1}{3} m g R_{e}$

So, the increase in potential energy is;

$Δ U = U_{f} - U_{i}$

$\Rightarrow Δ U = (- \frac{1}{3} m g R_{e}) - (- m g R_{e})$

$\Rightarrow Δ U = m g R_{e} - \frac{1}{3} m g R_{e}$

$\Rightarrow Δ U = \frac{2}{3} m g R_{e}$

Therefore, the increase in potential energy is $\frac{2}{3} m g R_{e}$ .

Thus, the correct option is (D).
JEE Main 2026MediumAcceleration due to Gravity and its Variation
Q2
When one moves from a point 16 km below the earth’s surface to a point 16 km above the earth’s surface. The change in g is approximately $%$ . The value of $α$ is $____$ .

(Take radius of the earth $= 6400 km$ .)
- A0.12
- B0.25
- C0.50
- D0.75
Show answer & solution
Correct answer: (B)
The acceleration due to gravity at a depth d below the Earth’s surface is given by the formula :

$g_{d} = g (1 - \frac{d}{R})$

Where $g$ is the acceleration at the surface, $d$ is the depth, and $R$ is the radius of the Earth.

The depth is given as $d = 16 km$ and $R = 6400 km$ :

$g_{d} = g (1 - \frac{16}{6400})$

The acceleration due to gravity at a height $h$ above the Earth’s surface (for $h ≪ R$ ) is given by the approximation :

$g_{h} = g (1 - \frac{2 h}{R})$

Where h is the height above the surface.

The height is given as $h = 16 km$ and $R = 6400 km$ :

$g_{h} = g (1 - \frac{2 \times 16}{6400}) = g (1 - \frac{32}{6400})$

The change in $g$ as we move from depth to height is :

$Δ g = g_{h} - g_{d}$

$\Rightarrow Δ g = g (1 - \frac{32}{6400}) - g (1 - \frac{16}{6400})$

$\Rightarrow Δ g = g (1 - \frac{32}{6400} - 1 + \frac{16}{6400})$

$\Rightarrow Δ g = g (- \frac{16}{6400})$

The magnitude of the change is $∣Δ g ∣ = g (\frac{16}{6400}) = g (\frac{1}{400})$ .

So, the percentage change is,

$α = (\frac{∣Δ g ∣}{g}) \times 100$

$\Rightarrow α = (\frac{1}{400}) \times 100 = \frac{1}{4} = 0.25$

Therefore, the correct option is (B).
JEE Main 2025EasyKepler's Laws of Planetary Motion
Q3
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant.

Reason (R): For a central force field the angular momentum is a constant.

In the light of the above statements, choose the most appropriate answer from the options given below:
- Ais not correct but (R) is correct
- BBoth (A) and (R) are correct but (R) is not the correct explanation of (A)
- CBoth (A) and (R) are correct and (R) is the correct explanation of (A)
- Dis correct but (R) is not correct
Show answer & solution
Correct answer: (C)
The assertion (A) states that the “radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time” which is just another way of stating Kepler’s second law. This implies that the areal velocity is constant.

The reason (R) states that “For a central force field the angular momentum is a constant.” In a central force system (like that of a planet orbiting the Sun under gravity), the force always acts along the line joining the two bodies. This means there is no torque acting on the planet, and thus, its angular momentum is conserved.

The connection between angular momentum and areal velocity is given by:

$Areal velocity = \frac{1}{2} r^{2} \dot{θ} = \frac{L}{2 m}$

where $L$ is the angular momentum and $m$ is the mass of the planet. Since the angular momentum $L$ is constant, the areal velocity must also remain constant.

Therefore, both the assertion and the reason are correct, and the reason correctly explains why the areal velocity (as stated in the assertion) is constant.

So, the most appropriate answer is:

Option C: Both (A) and (R) are correct and (R) is the correct explanation of (A).