The mass and diameter of a planet are twice those of the earth. What will be the time period of that pendulum on this planet which is a seconds pendulum on the earth.

Options

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

Answer

A

Explanation

On earth
`g=(GM)/(R^(2))`
On planet,
`g'=(Gxx2M)/((2R)^(2))=(g)/(2)`
`(T)/(T)= sqrt((g)/(g'))= sqrt((g)/(g//2))= sqrt(2)`
`T'= sqrt(2)xxT= sqrt(2)xx 2=2 sqrt(2)`