A rod of mass 5 kg is connected to the string at point B. The span of rod is along horizontal. The other end
of the rod is hinged at point A If the string is massless, then the reaction of hinge at the instant when string
is cut, is (Take, 9 = 10 ms^-2) [UP CPMT 2015]

When string is cut, the weight of the rod constitutes a torque about the hinge point A.

So,  ${\tau }_A=mg\frac{l}{2}$          …(i)

Also, from Newton's law,   ${\tau }_A=I\alpha$              …(ii)

where, $\alpha$ = angular acceleration of the rod
and I = moment of inertia of rod.
From Eqs. (i) and (ii), we get

$I\alpha =mg\frac{l}{2}\Rightarrow \alpha =mg\frac{l}{2}/I$

As,   $I=m\frac{l^2}{3}$

So,    $\alpha =\frac{mg(l/2)}{\frac{m{l}^2}{3}}=\frac{3g}{2l}$

Now, acceleration of centre of mass of rod,  $a_{\mathrm{CM}}=\alpha \cdot r$

Here, r = distance between hinge point A and centre of the rod.

$\Rightarrow a_{\mathrm{CM}}=\frac{3}{2}\frac{g}{l}\frac{l}{2}=\frac{3g}{4}$

$\Rightarrow mg-R_A=m{a}_{\mathrm{CM}}$

where, R_A = reaction at A.

$\Rightarrow mg-R_A=m\times \frac{3g}{4}$

$\Rightarrow R_A=\frac{mg}{4}=\frac{5\times 10}{4}=12.5\mathrm{N}$