Q 11-13-064JEE MainJEE Main 2024 (30 Jan, Shift 2)Medium
A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $4\ \text{m}$, then the time period of small oscillations will be ______ s. [take $g = \pi^2\ \text{m s}^{-2}$]
Numerical value type. Enter your answer.
Answer: 8
At height $R$, $g' = \dfrac{g}{(1 + 1)^2} = \dfrac{\pi^2}{4}$.
$$T = 2\pi\sqrt{\frac{L}{g'}} = 2\pi\sqrt{\frac{4\times4}{\pi^2}} = 2\pi\cdot\frac4\pi = 8\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics