• # question_answer The speed of a boat is 5 km/h in still water. It crosses a river of width 1.0 km along the shortest possible path in 15 min. The velocity of the river water is: (in km/h) [AIPMT 1998] A) 5 B) 1                      C) 3 D) 4

 Let ${{v}_{r}}=$ velocity of river ${{v}_{br}}=$ velocity of boat in still water and $w=$ width of river Time taken to cross the river$=15\text{ }min$ $=\frac{15}{60}h=\frac{1}{4}\,h$ Shortest path is taken when ${{v}_{b}}$ is along AB. In this case $v_{br}^{2}=v_{r}^{2}+v_{b}^{2}$ Now,             $t=\frac{w}{{{v}_{b}}}=\frac{w}{\sqrt{v_{br}^{2}-v_{r}^{2}}}$ $\therefore$      $\frac{1}{4}=\frac{1}{\sqrt{{{5}^{2}}-v_{r}^{2}}}$ $\Rightarrow$   ${{5}^{2}}-v_{r}^{2}=16$ $\Rightarrow$   $v_{r}^{2}=25-16=9$ $\therefore$      ${{v}_{r}}=\sqrt{9}\,=3\,km/h$ Note:    If ${{v}_{r}}\ge {{v}_{br}},$ the boatman can never reach at point B.