A cubical block of mass M and edge a slides down a rough inclined plane of inclination θ with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude
(a) zero

(b) Mga

(c) Mga sinθ

(d) $\frac{1}{2}Mga\mathrm{sin\theta }$.

(d) $\frac{1}{2}Mga\mathrm{sin\theta }$

Let N be the normal reaction on the block.



From the free body diagram of the block, it is clear that the forces N and mgcosθ pass through the same line. Therefore, there will be no torque due to N and mgcosθ. The only torque will be produced by mgsinθ.
$$\begin{gathered} \therefore \overrightarrow{\tau }=\overrightarrow{F}\times \overrightarrow{r} \\ \mathrm{Since}a\mathrm{is}\mathrm{the}\mathrm{edge}\mathrm{of}\mathrm{the}\mathrm{cube},r=\frac{a}{2}. \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ \tau =mg\sin \theta \times \frac{a}{2} \\ =\frac{1}{2}mga\sin \theta \end{gathered}$$