A rod of length $\mathcal{l}$ is sliding such that one of its ends is always in contact with a vertical wall and its other end is always in contact with horizontal surface. Just after the rod is released from rest, the magnitude of acceleration of end points of the rod is $a$ and $b$ respectively. The angular acceleration of rod at this instant will be

At the initial moment, angular velocity of rod is zero. Acceleration of end B of rod with respect to end A is shown in figure. Centripetal acceleration of point B with respect to A is zero $(\because {\omega }^{2}l=0).$ So at the initial moment , acceleration of end B with respect  to end A is perpendicular to the rod which is equal to $\sqrt{a^2+b^2}$

$a_{rel}=\mathcal{l}\alpha , \frac{\sqrt{a^2+b^2}}{\mathcal{l}}=\alpha$ where $\alpha$ is angular acceleration.