The cylindrical tube of a spray pump has a radius R, one end of which has n fine holes, each of radius r. If the speed of flow of the liquid in the tube is V, the speed of ejection of the liquid through the holes is

Cross-sectional area of tube $\left(A\right)=\pi R^2$. Cross-sectional area of each hole $=\pi r^2$.

Therefore, cross-sectional area of n holes $\left(a\right)=\pi n{r}^2$.

If v is the speed of ejection of the liquid through the holes, then from the continuity of flow, we have
$$\begin{gathered} av=AV \\ v=\frac{AV}{a}=\frac{\pi R^{2}V}{\pi n{r}^2}=\frac{V}{n}{\left(\frac{R}{r}\right)}^2 \end{gathered}$$
Hence the correct choice is (4).