Question

A boat which has a speed of $5\, km/h$ 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 $km/h$ is

• A. 1
• B. 3
• C. 4
• D. $\sqrt {41}$

Answer 1

Answer: (B)

3

Answer 2

Shortest possible path comes when absolute velocity of boatman comes perpendicular to river current as shown in figure. \hspace20mm t = \frac{\omega}{v_b}=\frac{\omega}{\sqrt{v^2_{br}-v^2_r}} \therefore \hspace15mm = \frac{1}{4}=\frac{1}{\sqrt{25-v^2_r}} Solving this equation, we get $v_r=3\, km/h$ $\therefore$ Correct answer is (b).