Question

A simple pendulum has a time period $T_1$ when on the earth's surface and $T_2$ when taken to a height R above the earth's surface, where R is the radius of the earth. The value of $\frac{T_2}{T_1}$ is :

• A. 1
• B.
• C. 4
• D. 2

Answer 1

The correct option is D

2

$T\propto \frac{1}{\sqrt{g}}$ i.e., $\frac{T_2}{T_1}=\sqrt{\frac{g_1}{g_2}}$

Where $g_1$ = acceleration due to gravity on earth's surface = g

$g_2$ = acceleration due to gravity at a height h = R from earth's surface = $\frac{g}{4}$

$\left[using\right(h)=\frac{g}{{(1+\frac{h}{R})}^2}]\frac{T_2}{T_1}=\sqrt{\frac{g}{\frac{g}{4}}}=2$