Q 11-07-056JEE MainJEE Main 2024 (1 Feb, Shift 1)Easy
If $R$ is the radius of the earth and the acceleration due to gravity on the surface of the earth is $g = \pi^2\ \text{m s}^{-2}$, then the length of the seconds pendulum at a height $h = 2R$ from the surface of the earth will be
Answer: (B) $\dfrac{1}{9}\ \text{m}$
At height $2R$ the distance from the centre is $3R$, so
$$g' = \frac{g}{9} = \frac{\pi^2}{9}$$
A seconds pendulum has $T = 2\ \text{s}$:
$$T = 2\pi\sqrt{\frac{l}{g'}} \Rightarrow l = \frac{g'T^2}{4\pi^2} = \frac{\pi^2}{9}\times\frac{4}{4\pi^2} = \frac{1}{9}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics