A simple pendulum is taken at a place where its separation from the earth's surface is equal to the radius of the earth. Calculate the time period of small oscillation if the length of the string is `1.0m`. Take `g = pi^(2) m//s^(2)` at the surface of the earth.

Options

A. `15 s`
B. `10 s`
C. `7 s`
D. `4 s`

Answer

D

Explanation

`g' = (g)/(1 + (h)/( R))^(2)`
Putting `h = R`
`g' = (g)/(4)`
`T = 2pi sqrt ((l)/((g//4)))`
`= 4pi sqrt((l)/(g))`
`= 4sqrt(l) = 4sqrt(1.0)` (as `g = pi^(2)`)
`= 4 s`