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 $ T_2/T_1 $ is

• A. 1
• B. $\sqrt 2$
• C. 4
• D. 2

Answer 1

Answer: (D)

$ 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 =g/4
$ \, \, \, \, \bigg[ Using\: g(h) = \frac{g}{\bigg( 1+\frac{h}{R}\bigg)^2} \bigg] $
$\Rightarrow \, \, \, \, \, \, \, \, \, \frac{T_2}{T_1} =\sqrt{\frac{g}{g/4}} =2$