The mass and diameter of a planet are twice those of earth. What will be the period of oscillation of a pendulum on this plenet. If it is a 2 second's pendulum on earth?

Options

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

Answer

B

Explanation

As we know `g = (GM)/(R^(2))`
`rArr (k_("earth"))/g_("planet") = (M_(e))/(M_(p)) rArr(g_(e))/(g_(p)) = (2)/(1)`
Also `T prop (1)/(sqrt(g)) rArr (T_(e))/(T_(p)) = sqrt((g_(p))/(g_(e))) rArr (2)/(T_(p)) = sqrt((1)/(2))`
`T_(p) = 2sqrt(2)sec`