A uniform rod of mass M and length L is hinged at one end to a smooth hinge and released from rest in the horizontal position.

About the hinge $I=ML^2/3$. (A) Just after release the torque is $MgL/2$, so $\alpha=\dfrac{MgL/2}{ML^2/3}=\dfrac{3g}{2L}$: true. (B) The centre of mass accelerates down at $\alpha L/2=3g/4$, so $Mg-N=3Mg/4$ and $N=Mg/4$, not $Mg/2$: false. (C) Energy conservation, $MgL/2=\tfrac12I\omega^2$, gives $\omega^2=3g/L$. At the vertical position the centripetal requirement is $N-Mg=M\omega^2L/2=3Mg/2$, so $N=5Mg/2$: true. (D) At a general angle $\theta$ below the horizontal the centre of mass has a nonzero horizontal acceleration (for example $-9g/8$ at $45^\circ$) from both tangential and radial components, so the hinge exerts a horizontal force: false.