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.
Net gravitational force at the center of a square is found to be \( F_1 \) when four particles having masses \( M, 2M, 3M \) and \( 4M \) are placed at the four corners of the square as shown in figure, and it is \( F_2 \) when the positions of \( 3M \) and \( 4M \) are interchanged. The ratio \( \dfrac{F_1}{F_2} = \dfrac{\alpha}{\sqrt{5}} \). The value of \( \alpha \) is 

In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
