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