Question

A man crosses a $320\, m$ wide river perpendicular to the current in $4$ min. If in still water he can swim with a speed $5/3$ times that of the current, then the speed of the current, in $ m{{m}^{-1}} $ in is

• A. 30
• B. 40
• C. 50
• D. 60

Answer 1

Answer: (D)

$v_{r}^{2}=v_{m}^{2}-v^{2}$
$v=\frac{320}{4} \,m \min ^{-1}=80 \,m \min ^{-1}$
$v_{m}=\frac{5}{3} v_{r}$
$v_{r}^{2}=\left[\frac{5}{3} v_{r}\right]^{2}-(80)^{2}$
$\Rightarrow \frac{16}{9} v_{r}^{2}=(80)^{2}$
$v_{r}=60 \,m \min ^{-1}$