Question

A swimmer crosses a river with minimum possible time $10$ second. And when he reaches the other end starts swimming in the direction towards the point from where he started swimming. Keeping the direction fixed the swimmer crosses the river in $15$ sec. The ratio of speed of swimmer with respect to water and the speed of river flow is:

(Assume constant speed of river & swimmer)

• A. $\frac{3}{2}$
• B. $\frac{9}{4}$
• C. $\frac{2}{\sqrt{5}}$
• D. $\frac{\sqrt{5}}{2}$

Answer 1

The correct option is C $\frac{2}{\sqrt{5}}$

A swimmer crosses
$v$ $=$ velocity of man w.r.t . river
$u$ $=$velocity of river
$A\stackrel{t}{\to }B=\frac{d}{v}\Rightarrow 10=\frac{d}{v}\Rightarrow d=10V\to \left(1\right)$
$B\stackrel{t}{\to }C=\frac{d}{vcos\theta }\Rightarrow 15=\frac{d}{vcos\theta }$
$\Rightarrow d=15vcos\theta \to \left(2\right)$
(1) & (2)$\Rightarrow cos\theta =\frac{2}{3}\Rightarrow sec\theta =\frac{3}{2}.$
$\because tan\theta =\frac{u}{v}\therefore \sqrt{se{c}^2\theta -1}=\frac{u}{v}$
$\Rightarrow \frac{u}{v}=\sqrt{\frac{9}{4}-1}=\frac{\sqrt{5}}{2}\Rightarrow \frac{v}{u}=\frac{2}{\sqrt{5}}$