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

Options

A. `1//9`
B. `1//3`
C. `sqrt(3)`
D. 9

Answer

B

Explanation

As `T=2pi sqrt(l)/(g)` ui.e `T prop 1 //sqrt(g)`
`therefore (T_(1))/(T_(2))=sqrt(g')/(g)=sqrt(gR^(2)//(R+2R)^(2)/(g)]^(1//2)=1/3`