A uniform thin rod of mass M and length L is hinged at one end so that it can rotate freely in a vertical plane. It is released from rest from the…

About the hinge $I=\frac{ML^2}{3}$. At release the torque is $Mg\frac L2$, so $\alpha=\frac{MgL/2}{ML^2/3}=\frac{3g}{2L}$, hence (A) is false (it would be double). The centre of mass has acceleration $\alpha L/2=\frac{3g}{4}$ downward and no horizontal acceleration, so $Mg-N=M\frac{3g}{4}$ and $N=\frac{Mg}{4}$, which makes (B) true. When vertical, energy conservation gives $\frac12\cdot\frac{ML^2}{3}\omega^2=Mg\frac L2$, so $\omega^2=\frac{3g}{L}$. There is no tangential force there ($\alpha=0$), and the centripetal requirement is $N-Mg=M\omega^2\frac L2=\frac{3Mg}{2}$, so $N=\frac{5Mg}{2}$. Option (D) is true and (C) is false. Note that the hinge force has both a vertical and, in general, a horizontal component; here the horizontal component is zero at both instants considered because the centre of mass has no horizontal acceleration.