To mop-clean a floor, a cleaning machine presses a circular mop of radius $$R$$ vertically down with a total force $$F$$ and rotates it with a constant angular speed about its axis. If the force $$F$$ is distributed uniformly over the mop and if coefficient of friction between the mop and the floor is $$\mu$$, the torque, applied by the machine on the mop is:

We have a flat, circular mop of radius $$R$$ pressed against the floor by a total normal force $$F$$. The force is stated to be distributed uniformly, so the normal pressure on the floor is the same at every point of the mop.

First, we write the formula for pressure. If a total force $$F$$ is spread uniformly over an area $$A$$, the pressure $$P$$ on the surface is

$$P=\dfrac{F}{A}\;.$$

The area of a circle of radius $$R$$ is $$A=\pi R^{2}$$, so the uniform pressure exerted by the mop on the floor is

$$P=\dfrac{F}{\pi R^{2}}\;.$$

Now we examine a small ring‐shaped element of the mop. Choose a ring of radius $$r$$ and infinitesimal width $$dr$$. The area of this thin ring is

$$dA=2\pi r\,dr\;.$$

The normal force acting on this small ring is the pressure multiplied by the ring’s area,

$$dF_{\text{normal}}=P\,dA =\dfrac{F}{\pi R^{2}}\,(2\pi r\,dr) =\dfrac{2F\,r}{R^{2}}\,dr\;.$$

Because the coefficient of kinetic friction between the mop and the floor is $$\mu$$, the friction force acting on that same ring element is

$$dF_{\text{friction}}=\mu\,dF_{\text{normal}} =\mu\left(\dfrac{2F\,r}{R^{2}}\right)\!dr =\dfrac{2\mu F\,r}{R^{2}}\,dr\;.$$

Friction acts tangentially, and the torque contributed by this ring about the central axis is the friction force times its lever arm (which is simply the radius $$r$$ of the ring). Thus, for this element,

$$d\tau = r\,dF_{\text{friction}} = r\left(\dfrac{2\mu F\,r}{R^{2}}\right)\!dr =\dfrac{2\mu F\,r^{2}}{R^{2}}\,dr\;.$$

To obtain the total torque $$\tau$$ on the entire mop, we integrate $$d\tau$$ from the centre of the mop ($$r=0$$) to its rim ($$r=R$$):

$$\tau=\int_{0}^{R}d\tau =\int_{0}^{R}\dfrac{2\mu F\,r^{2}}{R^{2}}\,dr\;.$$

We now carry out the integral. The constant factors $$\dfrac{2\mu F}{R^{2}}$$ come outside the integral:

$$\tau=\dfrac{2\mu F}{R^{2}}\int_{0}^{R}r^{2}\,dr\;.$$

We recall the standard integral formula $$\int r^{n}\,dr=\dfrac{r^{n+1}}{n+1}+C$$. Applying it with $$n=2$$, we get

$$\int_{0}^{R}r^{2}\,dr=\left[\dfrac{r^{3}}{3}\right]_{0}^{R} =\dfrac{R^{3}}{3}-0 =\dfrac{R^{3}}{3}\;.$$

Substituting this result back, we find

$$\tau=\dfrac{2\mu F}{R^{2}}\left(\dfrac{R^{3}}{3}\right)=\dfrac{2\mu F R}{3}\;.$$

Hence the torque that the machine must supply to keep the mop rotating is

$$\boxed{\tau=\dfrac{2\mu F R}{3}}\;.$$

Comparing with the given options, this matches Option A.

Hence, the correct answer is Option A.