A simple pendulum has a time period T1 when on the earth’s surface and T2 when taken to a height R above…

Q: A simple pendulum has a time period T1 when on the earth’s surface and T2 when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of T2/T1 is

(a)1

(b)√2

(c)4

(d)2

Sol: (d)

$\large T_1 = 2\pi \sqrt{\frac{l}{g}}$

Acceleration due to gravity at a height h=R above the surface of earth

$\large g’ = \frac{g}{(1+\frac{h}{R})^2} = \frac{g}{4}$

$\large T_2 = 2\pi \sqrt{\frac{l}{g’}}$

$\large T_2 = 2\pi \sqrt{\frac{l}{g/4}}$

$\large T_2 = 2 T_1 $