The acceleration due to gravity on the moon is `(1)/(6)`th the acceleration due to gravity on the surface of the earth. If the length of a second's pendulum is 1 m on the surface of the earth, then its length on the surface of the moon will be

Options

A. `(1)/(2)m`
B. 6 m
C. `(1)/(6)m`
D. `(1)/(4)m`

Answer

C

Explanation

`T=2p sqrt((l)/(g)) therefore 2=2pisqrt((l)/(g)) therefore l=(g)/(pi^(2))`
`therefore l prop g therefore " when g becomes " (g)/(6)`
l will be `(l)/(6)=(1)/(6) m`