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. 1
B. `sqrt(2)`
C. 4
D. 2

Answer

B

Explanation

The acceleration due to gravity at a height h above the surface of the earth is `g_(h)=g((R)/(R+h))^(2)`
where g is the value at the surface of the earth. Now
`T_(2)=2pisqrt((l)/(g_(h)))andT_(1)=2pisqrt((l)/(g))`
`therefore (T_(2))/(T_(1))=sqrt((g)/(g_(h)))=(R+h)/(R)=(R+R)/(R)=2" "(because h=R)`