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 $$g_d=g(1-\frac{d}{R})$$ $\Rightarrow \frac{g_d}{g}=(1-\frac{d}{R})...\left(1\right)$

The time period of a simple pendulum is given by $T=2\pi {\left(\frac{l}{g}\right)}^{\frac{1}{2}}$.

Hence, $T\propto \sqrt{\frac{1}{g}}$,

Let, the time period of pendulum at the surface is $T$ and time period at depth $d$ is $T_d$.
$\Rightarrow \frac{T_d}{T}={\left(\frac{g}{g_d}\right)}^{\frac{1}{2}}...\left(2\right)$

According to the problem,
$T_d=T+0.5\%ofT=1.005T$

Using equation $\left(1\right)$ and $\left(2\right)$,

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

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

$\Rightarrow 1.005=\frac{1}{{(1-\frac{d}{6400})}^{\frac{1}{2}}}$

$\Rightarrow {\left(\frac{1}{1.005}\right)}^2=1-\frac{d}{6400}$

$\Rightarrow \frac{d}{6400}=0.0099$

$\Rightarrow d=63.99\approx 64\text{km}$

Hence, option (b) is correct answer.