Question

A man wants to cross the river to an exactly opposite point on the other bank. If he can row his boat with $\frac{2}{\sqrt{3}}$ times the velocity of the current, then at what angle to the current he must keep the boat pointed ?

• A. ${60}^{\circ }$
• B. ${90}^{\circ }$
• C. ${120}^{\circ }$
• D. ${150}^{\circ }$

Answer 1

The correct option is D ${150}^{\circ }$


Let,
$V_R=$ Velocity of the river
$V_B=$ Velocity of the river

In order to cross the river with the shortest path, the net horizontal component of velocity of boat w.r.t ground must be $0$.

$\therefore V_B\sin \theta =V_R$

$\Rightarrow \sin \theta =\frac{V_R}{V_B}$

According to question, $V_B=\frac{2}{\sqrt{3}}{V}_R$

$\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}$

$\therefore \theta ={60}^{\circ }$

Hence, required angle will be
${90}^{\circ }+{60}^{\circ }={150}^{\circ }$.

So, option $\left(d\right)$ is correct. $$\begin{array}{c}\text{Why this question ?}\\ \text{For crossing the river with the shortest path,}\\ \text{the velocity of boat w.r.t ground must be}\\ \text{perpendicular to river.}\end{array}$$