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. 6m
C. `(1)/(6)m`
D. `(1)/(4)m`

Answer

C

Explanation

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