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

$(a)\;1 \quad (b)\;\sqrt 2 \quad (c)\; 4 \quad (d)\;2$

$\large\frac{g_2}{g_1}=\frac{{R_1}^2}{{R_2}^2}$
$R_1=R$
$R_2=2R$
$\large\frac{g_2}{g_1}=\frac{R^2}{4R^2}$
$\qquad=\large\frac{1}{4}$
$T= 2 \pi \sqrt {\large\frac{l}{g}}$
$\large\frac{T_2}{T_1}=\sqrt {\large\frac{g_1}{g_2}}$
$\qquad=\sqrt 4=2$
Hence  d is the correct answer.

edited Feb 17, 2014 by meena.p