The length of a second’s pendulum at the surface of earth is 1 m. The length of second’s pendulum at the surface of moon where g is 1/6th that at earth’s surface is

Explanation

For second's pendulum at the surface of earth
`2= 2pi sqrt((l_e)/(g_3))…...........(A)`
For second's pendulum at the surface of moon
`2= 2pi sqrt((l_m)/(g_m))…...........(B)`
From (A) and (B)
`(l_e)/(g_e)= (l_m)/(g_m) rArr l_m = ((g_m)/(g_e)) l_e`
`rArr l_m = (l_e)/(6) [ because g_m= (g_e)/(6)]`