Q 11-07-187JEE MainJEE Main 2019 (11 Jan, Shift 2)Easy
The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is $2$ s. The period of oscillation of the same pendulum on the planet would be:
Answer: (D) $2\sqrt3$ s
$g = \dfrac{GM}{R^2}$, so on the planet $g' = \dfrac{3}{3^2}g = \dfrac g3$.
$$T' = T\sqrt{\frac{g}{g'}} = 2\sqrt3\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics