Physics - Gravitation Question with Solution | TestHub

Time period of a simple pendulum at Earth's surface is given by

$T=2\pi \sqrt{\frac{l}{g}}$.

When distance from centre of Earth, i.e., at $h=R$, acceleration due to gravity is $g\text{'}=\frac{g}{{\left(1+{\displaystyle \frac{h}{R}}\right)}^2}=\frac{g}{{\left(2\right)}^2}=\frac{g}{4}$

Now, time period is $xT=2\pi \sqrt{\frac{l}{\left({\displaystyle \frac{g}{4}}\right)}}$

$\Rightarrow xT=2\times 2\pi \sqrt{\frac{l}{g}}$

$$\begin{gathered} \Rightarrow xT=2T \\ \Rightarrow \mathrm{x}=2 \end{gathered}$$

Thus, $x=2$.