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. 2.4 s
B. 2.8 s
C. 3.2 s
D. Infinity

Answer

B

Explanation

`g = (Gm)/(R^(2)) rArr (g_(1))/(g_(2)) = (M_(1))/(M_(2)) xx ((R_(2))/(R_(1)))^(2) = (2)/(1)`
`(T_(1))/(T_(2))=sqrt((g_(2))/(g_(1)))=(2)/(T_(2)) = sqrt((1)/(2)) = T = 2 sqrt(2) 2 xx 1.4 = 2.8s`