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
Concept. $T = 2\pi\sqrt{\dfrac{l}{g}} \Rightarrow l = \dfrac{gT^2}{4\pi^2}$.
Step 1. Substitute. $l = \dfrac{\pi^2 \times 4}{4\pi^2} = 1$ m.
Answer. $1$ m (Option A).
Solution by Sreeraj P, M.Sc Physics