If the length of second pendulum becomes `(1)/(3)` what will be its periodic time?

Explanation

`implies (2)/(sqrt(3)) s`
Time period of second pendulum `T_(1) = 2s`
In equation `T= 2pi sqrt((l)/(g)), 2pi, g` are constant
`therefore T propto sqrt(l)`
`therefore (T_2)/(T_1)= sqrt((l_2)/(l_1)) = sqrt((l)/(3xx l_(1)))= (1)/(sqrt(3))" "[therefore l_(2)= (l_1)/(3)]`
`therefore T_(2) = (T_1)/(sqrt(3))= (2)/(sqrt(3))s`.