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. 0.5
B. 2
C. `(1)/(sqrt(2))`
D. `sqrt(2)`

Answer

A

Explanation

`T_(1)=2pi sqrt((l)/(g))`
and `(g)/((1+(h)/(R ))^(2))=(g)/((1+(R )/(R ))^(2))=(g)/(4)`
`therefore T_(2)=2pi sqrt((4l)/(g)) therefore (T_(1))/(T_(2))=sqrt((l)/(g)xx(g)/(4l))=(1)/(2)`