Q 11-13-109JEE MainJEE Main 2021 (25 Feb, Shift 1)Easy
If the time period of a two meter long simple pendulum is 2 s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:
Answer: (A) $2\pi^2\ \text{m s}^{-2}$
$g = \dfrac{4\pi^2L}{T^2} = \dfrac{4\pi^2\times2}{2^2} = 2\pi^2\ \text{m s}^{-2}$
Solution by Sreeraj P, M.Sc Physics