A simple pendulum has a time period `T_(1)` when on the earth's surface and `T_(2)` when taken to a height 2R above the earth's surface where R is the radius of the earth. The value of `(T_(1)//T_(2))` is :

Options

A. 2
B. `1/3`
C. `sqrt(3)`
D. 3

Answer

D

Explanation

The periodic time of a simple pendulum is given by,
`" "T=2pisqrt(l/g)`
Where taken to height 2R.
`g.=g(1+h/R_(e))^(2)=g(1+(2R)/R)^(-2)=g(3)^(-2)`
`therefore" "T_(1)/T_(2)=sqrt(1/3^(2))`
`rArr" "T_(2)=3T_(1)`
`rArr" "T_(2)/T_(1)=3`