Question

A swimmer crosses a river with minimum possible time $10$ second. And, when he reaches the other end starts swimming in the direction 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 $\frac{2}{\sqrt{5}}$

Let $v$ be the velocity of man w.r.t river, $u$ be the velocity of flow in river and $d$ be the width of river.

Tme taken by man while going from A to B,
$\frac{d}{v}=10.....\left(i\right)$
Tme taken by man while going from B to A,
$\frac{d}{v\cos \theta }=15\dots .\left(ii\right)$
$\tan \theta =\frac{u}{v}.\dots \left(iii\right)$
Solving the above 3 equations, we get
$\frac{v}{\sqrt{u^2+v^2}}=\frac{2}{3}$
$\frac{v}{u}=\frac{2}{\sqrt{5}}$