Question

At what depth from the surface of earth, the time period of a simple pendulum is $0.5\%$ more than that on the surface of the earth?
$($Radius of earth is $6400\text{km})$

• A. $32\text{km}$
• B. $64\text{km}$
• C. $96\text{km}$
• D. $128\text{km}$

Answer 1

The correct option is B $64\text{km}$

The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by

$g^{\prime }=g(1-\frac{d}{R})$

The time period of a simple pendulum is given by

$T=2\pi {\left(\frac{l}{g}\right)}^{1/2}$

Hence, $T$ is inversely proportional to $g^{1/2}$ for the same pendulum.

$\therefore \frac{T^{\prime }}{T}={\left(\frac{g}{g^{\prime }}\right)}^{1/2}.....\left(1\right)$

From question,

$T^{\prime }=T+\frac{0.5}{100}T=1.005T$

putting all data in equation $\left(1\right)$, we get

$\frac{1.005T}{T}={\left(\frac{g}{g(1-\frac{d}{R})}\right)}^{1/2}$

$\Rightarrow 1.005=\frac{1}{{(1-\frac{d}{R})}^{1/2}}$

Substituting, $R=6400\text{km}$, we get

$d=63.5\approx 64\text{km}$

Hence, option $\left(b\right)$ is correct.