JEE Main 2025 · 22 January · evening · Single correct · +4 / −1
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet.
Reason (R) : The mass of the pendulum remains unchanged at Earth and the other planet.
In the light of the above statements, choose the correct answer from the options given below :
- (a) Both (A) and (R) are true and (R) is the correct explanation of (A)
- (b) (A) is false but (R) is true
- (c) (A) is true but (R) is false
- (d) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
Answer and solution
Answer (d) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
Solution
Physics sir · 4 steps
Find g on the planet first: mass and radius pull in opposite directions here, and they cancel exactly.
- Gravity on the planet. Surface gravity is g = GM/R². The planet has mass 4M and radius 2R, so put these in.
g' = G(4M)/(2R)² = 4GM/4R² = g
- Check the Assertion. The period is T = 2π√(L/g). The length is the same and g' = g, so T is the same on both. (A) is true.
- Check the Reason. Mass is the amount of matter in the bob, so it does not change from Earth to the planet. (R) is true. But the bob's mass does not appear in T = 2π√(L/g), so it plays no part in keeping T the same.
- So. Both statements are true, but (A) holds because g is unchanged, not because the mass is unchanged. The answer is (d).
The idea to keep. A pendulum's period depends only on L and g, never on the bob's mass. On a new planet, find g = GM/R² and compare that.
Answer from the NTA final answer key. Solution by PrepWiser.