The correct options are
A Torque of normal force on the block about centre is
Mgasinθ2 C Normal reaction force shifts by a distance
atanθ2To avoid toppling, let the normal reaction
(N) shift by a distance
x towards the right edge as shown in figure.
![](https://search-static.byjusweb.com/question-images/byjus/ckeditor_assets/pictures/950655/original_10.1.png)
Applying rotational equilibrium (for just toppling) about the centre of the cube
(A), torques in clockwise and anticlockwise direction should be equal.
τMg=0, τf=fa2, τN=Nx, about point
A.
τclockwise=τanticlockwise ⇒f×a2=Nx ...(i) From translational equilibrium of cube, since it is moving with uniform velocity, acceleration
a=0.
f=Mgsinθ ...(ii) N=Mgcosθ ...(iii) From Eq.
(i), (ii), & (iii), we get,
(Mgsinθ)×a2=(Mgcosθ)×x ∴x=atanθ2 Now, torque of normal reaction about its centre
(A) is,
τN=Nx=Mgcosθ×atanθ2 ∴τN=Mgasinθ2 Hence, option (a) and (c) are correct.