Q. The acceleration due to gravity at a place is $\pi^{2}\ m/s^{2}.$ Then the time period of a simple pendulum of length one $metre$ is
A) $\frac{2}{\pi}s$
B) $2\pi\ s$
C) $2\ s$
D) $\pi\ s$
Explanation
(c) $T = 2\pi\sqrt{l/g} = 2\pi\sqrt{\frac{1}{\pi^{2}}} = 2\ s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.