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?

Explanation

Time period of second.s pendulum is given by
`T = 2pi sqrt((l)/(g))`
On the Earth, T = 2, l =1
`therefore 2 = 2pi sqrt((1)/(g_(e)))" "…(i)`
On the moon, T = 2, l = l., `g_(m) = (g_(e))/(6)`
`therefore 2 = 2pi sqrt((l.)/(g_(m))) = 2pi sqrt((6l)/(g_(e))) " "...(ii)`
From (i) and (ii)
`6l. =l`
or `l. = (l)/(6) = (1)/(6) m = 0.17m`