A rectangular plate of mass $20\text{kg}$ is suspended from points A and B as known, if the pin B is suddenly removed, the angular acceleration in ${\text{rad/s}}^2$(to the nearest integer) of the plate is $\alpha$.
Then $\frac{\alpha }{8}$ has value $(g=10{\text{m/s}}^2$)

When plate starts to rotate about point A, then torque. $\tau =\frac{mgl}{2}$
$I\alpha =\frac{mgl}{2}$
Here $l=0.2\text{m}$ and $b=0.15\text{m}$
Distance of COM from axis of rotation is
$d^2=\frac{l^2}{4}+\frac{b^2}{4}=\frac{0.0625}{4}$
Moment of inertia
$=\frac{m}{12}(l^2+b^2)+m{d}^2$
$=\frac{20}{12}(0.2{)}^2+(0.15{)}^2+20\left(\frac{0.0625}{4}\right)$
$=\frac{5}{3}\left[0.0625\right]+5\left[0.0625\right]$
$=0.1041+0.3125=0.416{\text{kg m}}^2$
now $I\alpha =\frac{mgl}{2}\Rightarrow \alpha =\frac{mgl}{2I}$
$\alpha =\frac{20\times 10\times 0.2}{2\times 0.416}\approx 48{\text{rad/s}}^2$
so, $\frac{\alpha }{8}=\frac{48}{8}=6$