A boat which has a speed of $5km{h}^{-1}$ in still water, crosses a river of width 1 km along the shortest possible path in 15 min. The velocity of the river water (in $km{h}^{-1}$) is ________

Shortest possible path comes when absolute velocity of boatman comes perpendicular to river current as shown in the figure.
  
$t=\frac{\omega }{v_b}=\frac{\omega }{\sqrt{v_{br}^2-v_r^2}}$ 
$\therefore \frac{1}{4}=\frac{1}{\sqrt{25-v_r^2}}$ 
Solving this equation, we get $v_r=3km{h}^{-1}$