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

Options

A. 1
B. `sqrt2`
C. `4`
D. 2

Answer

D

Explanation

We know that `T_1=2psqrt(1//g)` and `T_2=2psqrt(1//g')`
`therefore T_2/T_1 =sqrt(g/g')`…(i)
Also `g=(GM)/R^2 therefore g'=(GM)/(2R)^2 =(GM)/(4R^2) therefore g/(g')=4 rArr T_2/T_1 =2`