A boat which has a speed of 5 kmh^-1 in still water, crosses a river of width 1 km along the shortest possible path in 15 min. The velocity of the river water is

Shortest possible path comes when absolute velocity of boat is perpendicular to river current as shown in the figure.

$\begin{array}{l}\mathrm{t}=\frac{\mathrm{\omega }}{{\mathrm{v}}_{\mathrm{b}}}=\frac{\mathrm{\omega }}{\sqrt{{\mathrm{v}}_{\mathrm{br}}^2-{\mathrm{v}}_{\mathrm{r}}^2}}\\ \therefore \frac{1}{4}=\frac{1}{\sqrt{25-{\mathrm{v}}_{\mathrm{r}}^2}}\end{array}$

Solving this equation, we get ${\mathrm{v}}_{\mathrm{r}}=3 {\mathrm{kmh}}^{-1}$