Calculate the length of the second's pendulum on the surface of moon where acceleration due to gravity is one sixth as that on earth.

Explanation

For second pendulum, t = 2s
As `T = 2pisqrt((l)/(g_(m)))`
`therefore (T^(2)g_(m))/(4pi^(2)) = l`
`((2)^(2) xx g_(e))/(6 xx 4 xx (3.14)^(2)) = l [because g_(m) (g_(e))/(6)]`
` (4xx 9.8)/(24 xx 3.14 xx 31.4) = l`
l = 0.165 m