Time period of simple pendulum at a planet is 2 sec. Length of simple pendulum is 2 meter. Find the value of g at that planet (in m/s^{2})

(1) 2$\pi$^{2}

(2) 9.8 m/s^{2}

(3) 10 m/s^{2}

(4) $\pi^2\over2$m/s^{2}

If you are commenting for the first time, please read the guidelines.

If you are answering for the first time, please read the guidelines.

Ans. (1) 2$\pi$^{2}

Sol. $T=2\pi\sqrt{\ell\over g_\text{planet}}$

$\implies2=2\pi\sqrt{2\over g_\text{planet}}$

$\implies{1\over\pi^2}={2\over g_\text{planet}}$

$\implies$ g_{planet} = 2$\pi$^{2} m/sec^{2}

Welcome to Jee Neet QnA, where you can ask questions and receive answers from other members of the community.

Join our Telegram group for live discussion.

Subscribe our YouTube channel for video solutions with explanation.

Download Jee Neet QnA Books in PDF for offline learning.

1.2k questions

842 answers

384 comments

93 users