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