JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Elasticity & SHM

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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 but (R) is NOT the correct explanation of (A)✓ Correct
(B)(A) is true but (R) is false
(C)(A) is false but (R) is true
(D)Both (A) and (R) are true and (R) is the correct explanation of (A)

Answer: (A)

Solution

The time period of a simple pendulum is

T = 2 π ℓ / g

, which depends on g but not on the mass of the bob.

Acceleration due to gravity at a planet's surface is

g = \frac{GM}{R ^{2}}

. For a planet with mass

M^{'} = 4 M

and radius

R^{'} = 2 R

g^{'} = \frac{G ( 4 M )}{( 2 R ) ^{2}} = \frac{4 GM}{4 R ^{2}} = \frac{GM}{R ^{2}} = g

Since g is unchanged, the period is unchanged — Assertion (A) is true.

The mass of the pendulum bob does stay the same on both bodies — Reason (R) is also true, but it is irrelevant to why the period is unchanged, since T does not depend on the bob's mass at all. The real reason is that g happens to come out the same on the new planet.

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

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Elasticity & SHM

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