The time period of a simple pendulum on the surface of the earth is 4s. Its time period on the surface of the moon is

Options

A. 4s
B. 8s
C. 10s
D. 12s

Answer

C

Explanation

Acceleration due to gravity on the surface of the moon is `(1)/(6)` that of surface of the earth.
`thereforeg_(m)=(1)/(6)g_(e)` . . . (i)
On earth, `T_(e)=2pisqrt((L)/(g_(e)))`
On moon, `T_(m)=2pisqrt((L)/(g_(m)))`
`therefore(T_(m))/(T_(e))=sqrt((g_(e))/(g_(m)))=sqrt(6)`
`T_(m)=sqrt(6)T_(e)=sqrt(6)xx4s=10s`