At what depth from the surface of earth the time period of a simple pendulum is `0.5%` more than that on the surface of the Earth? (Radius of earth is `6400 km`)

Options

A. `32 km`
B. `64 km`
C. `96 km`
D. `128 km`

Answer

B

Explanation

`T prop = (1)/(sqrtg)`
`(T_(2))/(T_(1)) = sqrt((g_(1))/(g_(2))` or `(1.05 T)/(T) = sqrt((g)/(g(1 - (h)/(R))`
Solving this equation we get, `h = 64 km`