The kinetic energy at the point Q is given by
= $ \frac{1}{2} 1 \omega^2 = \frac{1}{2} \frac{ ml^2 }{ 3 } \frac{ v^2 }{ l^2 } $
= $ \frac{1}{2} \times \frac{1}{3 } mv^2 $ $ \hspace20mm$ ..(i)
The potential energy at G
$ = \frac{1}{2} mgl $...(ii)
From Eqs. (i) and (ii), we get
$ \frac{1}{2} \frac{mv^2 }{ 3} = \frac{1}{2} mgl $
v = $ \sqrt{ 3gl }$