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?

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

1.	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?
Answer» Solution :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`.