Question

At what depth from the surface of earth the time period of a simple pendulum is $5\%$ more than that on the surface of the Earth? $($Radius of earth is $6400km)$

• A. $32km$
• B. $64km$
• C. $96km$
• D. $595km$

Answer 1

The correct option is D $595km$

Time period of simple pendulum is inversely proportional to root of g

$\Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{g_2}{g_1}}$

Given, time period is increased by 5%, therefore $T_2=1.05T_1$

let the depth at which it happens be 'd'

$\Rightarrow \frac{T_1}{1.05T_1}=\sqrt{\frac{g(1-\frac{d}{R})}{g}}$

$1-\frac{d}{R}=\frac{1}{{1.05}^2}\Rightarrow d=0.093R=0.093\times 6400=595km$