JEE Physics 2025 Question with Solution

Given: A simple pendulum is taken to a planet whose mass is $4M_E$ and radius is $2R_E$.

Find: Whether Assertion (A) and Reason (R) are true, and whether (R) explains (A).

For a simple pendulum,

$$T = 2\pi \sqrt{\frac{L}{g}}$$

and acceleration due to gravity is

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

For the new planet,

$$M_P = 4M_E, \quad R_P = 2R_E$$

So,

$$g_P = \frac{GM_P}{R_P^2} = \frac{G(4M_E)}{(2R_E)^2} = \frac{4GM_E}{4R_E^2} = g_E$$

Hence,

$$T_P = 2\pi \sqrt{\frac{L}{g_P}} = 2\pi \sqrt{\frac{L}{g_E}} = T_E$$

Therefore, the time period remains the same, so Assertion (A) is true.

The mass of the pendulum does remain unchanged, but the time period of a simple pendulum does not depend on the bob's mass. So this statement does not explain the assertion.

The solution contains a discrepancy: one approach states the correct option is C, while the detailed working shows that both (A) and (R) are true, but (R) is not the correct explanation.

Therefore, the correct option is D.

Given: The planet has mass $4$ times that of Earth and radius $2$ times that of Earth.

Find: The correct relation between Assertion (A) and Reason (R).

The pendulum time period depends on $g$, not on the mass of the pendulum bob.

Using

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

we compare the two planets:

$$\frac{g_P}{g_E} = \frac{M_P/M_E}{(R_P/R_E)^2} = \frac{4}{2^2} = 1$$

Thus,

$$g_P = g_E$$

Now,

$$T = 2\pi \sqrt{\frac{L}{g}}$$

Since $g$ is unchanged and the pendulum length $L$ is unchanged, the time period also remains unchanged.

So Assertion (A) is true.

Reason (R) says the mass of the pendulum remains unchanged. That physical fact is true, but it is not the reason for the same time period, because $T$ is independent of pendulum mass.

Therefore, both statements are true, but (R) is not the correct explanation of (A). Hence, the correct option is D.