A 2   kg steel rod of length 0 . 6   m is clamped on a table vertically at its lower end and is…

In the above case the torque due to weight of the rod, here use the concept of translation work done and rotational kinetic energy, 

$\Rightarrow W=K.E.\Rightarrow mgl=\frac{1}{2}I{\omega }^2$, now the moment of inertia of the rod when the axis of rotation passes through one end of the rod, $I=\frac{1\times m{l}^2}{3}$,
$\Rightarrow mgl=\frac{1}{2}\times \frac{m{l}^2}{3}{\omega }^2$
$\Rightarrow \omega =\sqrt{\frac{6\mathrm{g}}{l}}$, now use the relation between linear velocity and angular velocity, 
$\Rightarrow v=l\omega =\sqrt{3.6g}=6 \mathrm{m} {\mathrm{s}}^{-1}$