A 2kg steel rod of length 0 . 6m is clamped on a table vertically...

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

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