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

Answer

B

Explanation

Gravity, `g = (GM)/(R^(2))` (G is constant)
`therefore" "(g_("earth"))/(g_("planet"))=(M_(e))/(M_(p))xx(R_(p)^(2))/(R_(e)^(2)) 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) = 2 sqrt(2) s`