Q 11-13-023NEETJEE MainEasy
The length of a seconds pendulum (period $2$ s) at a place where $g = \pi^2\ \text{m/s}^2$ is
Answer: (A) $1$ m
$$T = 2\pi\sqrt{\frac{l}{g}} \;\Rightarrow\; l = \frac{gT^2}{4\pi^2} = \frac{\pi^2 \times 4}{4\pi^2} = 1\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics