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 dfrac-xT-0-12- then x is -

Question

Toppr Admin · Accepted Answer

Time period of a simple pendulum is given by
$T = 2 π \sqrt{\frac{L}{g}}$
When we go to another planet with different mass, the value of $g$ changes accordingly.
we have
$g = \frac{G M}{R^{2}}$ ; 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.
$M = d e n s i t y \times v o l u m e = ρ \times \frac{4}{3} π R^{3}$
$g = \frac{G \times ρ \frac{4}{3} π R^{3}}{R^{2}} = G \times ρ \frac{4}{3} π R$
here $R^{'} = 4 R$
$\Rightarrow g^{'} = G \times ρ \frac{4}{3} π (4 R) = 4 g$
The new time period
$T^{'} = 2 π \sqrt{\frac{L}{4 g}} = T / 2$
hence $\frac{x}{12} = \frac{1}{2} \Rightarrow x = 6$

A simple pendulum has a time period of T0 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 xT012 then x is :

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 :