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

Explanation

The time period of a simple pendulum `T= 2pi sqrt((l)/(g))`
`therefore T propto sqrt((l)/(g))" "2pi` is constant
`therefore (T_m)/(T_e)= sqrt((l_m)/(g_m)xx (g_e)/(l_e))`
where `T_(e ), T_(m)` are the periodic time on earth and moon
Here, `T_(e )= T_(m) = 2s`
`l_(m), l_(e )` are the length of second.s pendulum on the surface of moon and earth respectively.
Now, `g_(m), g_(e )` are the acceleration due to gravity on moon and earth respectively.
`therefore (2)/(2) = sqrt((g_e)/(g_m)xx(l_m)/(l_e))`
by squaring, `1= (g_e)/(g_m)xx(l_m)/(l_e)`
but `g_(m)= (g_e)/(6)" and "l_(e )= 1m`
`therefore 1= (g_e)/((g_e)/(6))xx (l_m)/(1)`
`therefore 1= 6xx l_(m)`
`therefore l_(m)= (1)/(6)m`.