Question

If the length of a seconds pendulum on a planet is 2m then the acceleration due to gravity on the surface of the surface of that planet is

Answer 1

$\small{\underline{\sf{\orange{Given:\rightarrow}}}}$

• $\sf{Lenght= 2m}$

• $\sf{T=2s}$

$\small{\underline{\sf{\blue{Find:\rightarrow}}}}$

acceleration due to gravity on the surface of the surface of that planet is

$\boxed{\sf{ g_p}}$

$\small{\underline{\sf{\green{SoLution:\rightarrow}}}}$

$\boxed{\sf{g_p=4\pi^2 \dfrac{l}{T} }}$

Putting values in the formula we get,

$\sf{g_p= 4\pi^2 \times\dfrac{2}{2^2} }$

$\sf{g_p = 2\pi^2 m/s^2}$

$\sf{\underline {\therefore Acceleration= 2\pi^2 m/s^2 }}$