Question

pls ans above question ​

Answer 1

Answer:

The answer is 9.77 m/s^2

Explanation:

We know that,

Time period of a simple pendulum = $2\pi\sqrt{\frac{l}{g} }$

∴$t^{2} = 4 \pi^{2} \frac{l}{g}$

⇒ $g = \frac{4 \pi ^ {2} l}{t^{2}}$

⇒ $g = \frac{4 \pi^{2} * 1}{2.01^{2}}$

On solving we  get

g = 9.77 m/s^2