If `T_(1)` and `T_(2)` are the time-periods of oscillation of a simple pendulum on the surface of earth (of radius R) and at a depth d, the d is equal to

Options

A. `(1-(T_(1)^(2))/(T_(2)^(2)))R`
B. `(1-(T_(2)^(2))/(T_(1)^(2)))R`
C. `(1-(T_(1))/(T_(2)))R`
D. `(1-(T_(2))/(T_(1)))R`

Answer

A

Explanation

`T_(1)=(k)/(sqrt(g))" and "T_(2)=(k)/(sqrt(g(1-(d)/( R ))))`
So, `(T_(1))/(T_(2))=sqrt(1-(d)/( R ))=((T_(1))/(T_(2)))^(2)=1-(d)/( R )`
`d=[1-((T_(1))/(T_(2)))^(2)]R`