Q 11-07-124JEE MainJEE Main 2022 (25 Jul, Shift 2)Medium
The length of a seconds pendulum at a height $h = 2R$ from earth surface will be: (Given: $R$ = radius of earth and acceleration due to gravity at the surface of earth $g = \pi^2\ \text{m s}^{-2}$)
Answer: (D) $\dfrac19\ \text{m}$
At $h = 2R$: $g' = \dfrac{g}{(3)^2} = \dfrac{\pi^2}{9}$. A seconds pendulum has $T = 2\ \text{s}$:
$$l = \frac{g'T^2}{4\pi^2} = \frac{\pi^2}{9}\cdot\frac{4}{4\pi^2} = \frac19\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics