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.
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is: