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

Explanation

For a second pendulum `2=2pisqrt((l)/(g))`
On earth, `l=1m, g=g_(e), ` therefore, `2=2pisqrt((l)/(g_(e)))` …(i)
On moon `l=l^(')(say), g=g_(e)//6, ` then `2=2pisqrt((l^('))/(g_(e)//6)) =2pisqrt((6l^('))/(g_(e)))` …(ii)
From (i) and (ii), we have `6l^(')=1 or l^(')=(1)/(6)m=0.17m`