Question

Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R): \begin{itemize} \item[(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. \item[(R)] The mass of the pendulum remains unchanged at Earth and the other planet. \end{itemize} In light of the above statements, choose the correct answer from the options given below:

• A. (A) is false, but (R) is true.
• B. Both (A) and (R) are true and (R) is the correct explanation of (A).
• C. (A) is true but (R) is false.
• D. Both (A) and (R) are true, but (R) is NOT the correct explanation of (A).

Hint:

The time period of a simple pendulum depends on the acceleration due to gravity \( g \). Gravity is determined by the mass and radius of the planet.

Answer 1

The Correct Option is C

- The time period \( T \) of a simple pendulum is given by: \[ T = 2\pi \sqrt{\frac{L}{g}}, \] where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. The time period depends on \( g \), which is given by \( g = \frac{GM}{R^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is its radius. - For the given planet with mass and radius 4 and 2 times that of the Earth, \( g \) will change, which means the time period will also change. Thus, Assertion (A) is false. - The mass of the pendulum does indeed remain the same, so Reason (R) is true. Thus, the correct answer is \( \boxed{(3) (A) \text{ is true but } (R) \text{ is false.}} \).

Answer 2

Step 1: Formula for the time period of a simple pendulum.
\[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( g \) is the acceleration due to gravity.

Step 2: Relation of \( g \) with planet’s mass and radius.
\[ g = \frac{GM}{R^2}. \] Let \( g_E \) be the acceleration due to gravity on Earth and \( g_P \) on the new planet.

Given: \[ M_P = 4M_E, \quad R_P = 2R_E. \] Thus, \[ g_P = \frac{G M_P}{R_P^2} = \frac{G(4M_E)}{(2R_E)^2} = \frac{4GM_E}{4R_E^2} = g_E. \]

Step 3: Compare time periods.
Since \( g_P = g_E \), \[ T_P = 2\pi \sqrt{\frac{L}{g_P}} = 2\pi \sqrt{\frac{L}{g_E}} = T_E. \] Hence, **time period remains the same.**

Step 4: Analyze statements.
- Assertion (A): “Time period remains same” → ✅ True (since \( g_P = g_E \)). - Reason (R): “Mass of the pendulum remains unchanged” → although true as a physical fact, **it is not the reason** why the time period remains same — time period depends on \( g \), not on the pendulum’s mass.

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Final Answer:

\[ \boxed{\text{(A) is true but (R) is false.}} \]