A wheel of radius ‘r’ rolls without slipping with a speed ‘v’ on a horizontal road. When it is at point ‘A’ on the road, a small lump of mud separates from the wheel at its highest point B and drops at point ‘C’ on the road. The distance AC will be

The mud blob would start out at a height of 2r and will have the instantaneous tangential velocity as of point B. The velocity would be 2v. 

 $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$is used to calculate the time of travel (t).

$2\mathrm{r}=\frac{1}{2}{\mathrm{gt}}^2$

$\mathrm{t}=2\sqrt{\frac{\mathrm{r}}{\mathrm{g}}}$

Distance AC is given as $2\mathrm{vt}=4\mathrm{v}\sqrt{\frac{\mathrm{r}}{\mathrm{g}}}$

Hence the correct answer is $4\mathrm{v}\sqrt{\frac{\mathrm{r}}{\mathrm{g}}}.$