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. `sqrt(2)`
C. 4
D. 2

Answer

D

Explanation

We know that
`T_(1)=2 pi sqrt((t)/(g))`
and `T_(2) = 2 pi sqrt((l)/(g'))`
`:. (T_(2))/(T_(1)) = sqrt((g)/(g'))`
Also `g = (GM)/(R^(2))`
`:. g' = -(GM)/((2R)^(2)) =(GM)/(4 R^(2))`
`:. (g)/(g') =4 rArr (T_(2))/(T_(1)) = 2`.