The length of a second's pendulum on the surface of earth is 1 m. What will be the length of a second's pendulum on the moon?

Options

A. `(1)/(3)m`
B. 6 m
C. `(1)/(6)m`
D. 3 m

Answer

C

Explanation

For a second's pendulum `T=2 s and g_("moon")=(g_(e ))/(6)`
`T_("earth")=2pi sqrt((L_(e ))/(g_(e )))`
`T_("moon")=2pi sqrt((L_("moon"))/(g_(m)))`
It is given that `T_(e )=T_(m)=2s`
`therefore` From (1) and (2), `sqrt((L_(e ))/(g_(e )))=sqrt((L_(m))/(g_(m)))`
`therefore (L_(m))/(L_(e ))=(g_(m))/(g_(e ))=(1)/(6) therefore L_(m)=(L_(e ))/(6)=(1)/(6)m`