Question

A swimmer crosses a river of width d flowing at velocity v. While swimming he heads himself always at an angle of ${120}^o$ with the river flow and on reaching to the other end he finds a drift of d/2 in the direction of flow of river. The speed of swimmer with respect to the river is -

• A. $(2-\sqrt{3})v$
• B. $2(2-\sqrt{3})v$
• C. $4(2-\sqrt{3})v$
• D. $(2+\sqrt{3})v$

Answer 1

The correct option is C $4(2-\sqrt{3})v$

In the given figure

$AB=\frac{d}{2}$

$V_R=\sqrt{{\left(V_m\right)}^2+V^2+2V_{m}V\cos 120}$

so,

$V_R=\sqrt{{V_m}^2+V^2-V_{m}V}.........\left(1\right)$

Let $\theta$ be the angle made by the resultant velocity

vector with the flow of the river from parallelogram law of

vector addition.

$\theta =\frac{V_m\sin 120}{V+V_m\cos 120}..........\left(2\right)$

In $\Delta ABC,$ $\angle AOB=\left(90-\theta \right)$

$\tan \left(90-\theta \right)=\frac{d/2}{d}=\frac{1}{2}$

$\cot \theta =\frac{1}{2}so\theta ={\tan }^{-1}\mathrm{2.........}\left(3\right)$

${\tan }^{-1}2=\frac{V_m\sin 120}{V+V_m\cos 120}$

from $\left(2\right)$ and $\left(3\right)$

$03.43=\frac{V_m\times \left(0.86\right)}{V-0.5v}$

$V_m=1.95v$

putting this in $\left(1\right)$ we get

$V_R=\sqrt{{\left(1.91v\right)}^2+v^2-1.95v^2}$

Hence,

Option $C$ is correct answer.