The problem deals with understanding the behavior of a simple pendulum under different gravitational conditions. The time period \( T \) of a simple pendulum is given by the formula:
\( T = 2\pi \sqrt{\frac{L}{g}} \)
where:
On Earth, the gravitational acceleration \( g \) is \( \frac{GM_e}{R_e^2} \), where \( G \) is the gravitational constant, \( M_e \) is the Earth's mass, and \( R_e \) is the Earth's radius.
For the other planet, with mass \( 4M_e \) and radius \( 2R_e \), the gravitational acceleration \( g' \) is:
\( g' = \frac{G \cdot 4M_e}{(2R_e)^2} = \frac{4GM_e}{4R_e^2} = \frac{GM_e}{R_e^2} = g \)
Thus, the time period \( T' \) on the planet is:
\( T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g}} = T \)
This confirms that the time period on both Earth and the planet is the same, supporting assertion (A).
Now, consider the reason (R). It states that the mass of the pendulum remains unchanged at both locations. While this is true, the mass of the pendulum does not affect the time period \( T \), as evidenced by the formula. Therefore, (R) does not explain (A).
Thus, the correct choice is: (A) is true but (R) is false.
Match the LIST-I with LIST-II
\[ \begin{array}{|l|l|} \hline \text{LIST-I} & \text{LIST-II} \\ \hline \text{A. Gravitational constant} & \text{I. } [LT^{-2}] \\ \hline \text{B. Gravitational potential energy} & \text{II. } [L^2T^{-2}] \\ \hline \text{C. Gravitational potential} & \text{III. } [ML^2T^{-2}] \\ \hline \text{D. Acceleration due to gravity} & \text{IV. } [M^{-1}L^3T^{-2}] \\ \hline \end{array} \]
Choose the correct answer from the options given below:
A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is: 