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. `sqrt(2)` second
B. `2sqrt(2)` seconds
C. `(1)/(sqrt(2))` second
D. `(1)/(2sqrt(2))` second

Answer

B

Explanation

`g=(GM)/(R^(2)) & g'=(G(2M))/((2R)^(2))` so `g'=(g)/(2)`
`T'= 2pi sqrt((2l)/(g)) rArr T'=sqrt(2)T rArr T' = 2 sqrt(2)` sec.