Question

A simple pendulum has a time period of $T_0$ on the surface of the earth. On another planet whose density is same that of earth but radius is 4 times then time period of this simple pendulum is $\frac{x{T}_0}{12}$ then x is :

Answer 1

Time period of a simple pendulum is given by
$T=2\pi \sqrt{\frac{L}{g}}$
When we go to another planet with different mass, the value of $g$ changes accordingly.
we have
$g=\frac{GM}{R^2}$ ; where $M$ is the mass and $R$ is radius of the body(planet).
Given, for another planet density is same but the radius is $4$ times.

$M=density\times volume=\rho \times \frac{4}{3}\pi R^3$

$g=\frac{G\times \rho \frac{4}{3}\pi R^3}{R^2}=G\times \rho \frac{4}{3}\pi R$
here $R^{\prime }=4R$

$\Rightarrow g^{\prime }=G\times \rho \frac{4}{3}\pi \left(4R\right)=4g$
The new time period

$T^{\prime }=2\pi \sqrt{\frac{L}{4g}}=T/2$
hence $\frac{x}{12}=\frac{1}{2}\Rightarrow x=6$