A cubical block of mass $M$ and edge $a$ slides down a rough inclined plane of inclination $\theta$ with a uniform velocity. Choose the correct statement(s):

The correct options are
A Torque of normal force on the block about centre is $\frac{Mga\sin \theta }{2}$
C Normal reaction force shifts by a distance $\frac{a\tan \theta }{2}$
To avoid toppling, let the normal reaction $\left(N\right)$ shift by a distance $x$ towards the right edge as shown in figure.


Applying rotational equilibrium (for just toppling) about the centre of the cube $\left(A\right)$, torques in clockwise and anticlockwise direction should be equal.
${\tau }_{Mg}=0,{\tau }_f=\frac{fa}{2},{\tau }_N=Nx$, about point $A$.
${\tau }_{\text{clockwise}}={\tau }_{\text{anticlockwise}}$
$\Rightarrow f\times \frac{a}{2}=Nx...\left(i\right)$
From translational equilibrium of cube, since it is moving with uniform velocity, acceleration $a=0$.
$f=Mg\sin \theta ...\left(ii\right)$
$N=Mg\cos \theta ...\left(iii\right)$
From Eq.$\left(i\right),\left(ii\right),\&\left(iii\right)$, we get,
$\left(Mg\sin \theta \right)\times \frac{a}{2}=\left(Mg\cos \theta \right)\times x$
$\therefore x=\frac{a\tan \theta }{2}$
Now, torque of normal reaction about its centre $\left(A\right)$ is,
${\tau }_N=Nx=Mg\cos \theta \times \frac{a\tan \theta }{2}$
$\therefore {\tau }_N=\frac{Mga\sin \theta }{2}$
Hence, option (a) and (c) are correct.