Question

A simple pendulum has a time period $T_1$ on the surface
of earth of radius $R$. When taken to a height of $R$ above
the earth's surface, its time period is $T_2$. Then the ratio $\frac{T_2}{T_1}is$

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. 2
• D. 4
• E. 1/2

Answer 1

Answer: (C)

$T\propto \sqrt{l/g}$
or $\frac{T_1}{T_2}=\sqrt{\frac{g_2}{g_1}}$... (i)
$g_1=g,g_2=\frac{g}{(1+\frac{h}{R}{)}^2}$
$=\frac{g}{(1+\frac{R}{R}{)}^2}(ash=R)$
$or{g}_2=\frac{g}{4}$
Putting these values in Eq. (i), we obtain
$\frac{T_1}{T_2}=\sqrt{\frac{g/4}{g}}=\sqrt{\frac{1}{4}}=\frac{1}{2}or\frac{T_2}{T_1}=2$