Question

A boat which has a speed of $15\text{km/hr}$ in still water crosses a river of width $3\text{km}$ along the shortest possible path in $20\text{minutes}$ . The velocity of the river water in $\text{km/hr}$ is

• A. $4.5$
• B. $11$
• C. $13$
• D. $12$

Answer 1

The correct option is D $12$

It is given that,

Displacement $=3\text{km}$
$t=\frac{20}{60}=\frac{1}{3}\text{hr}$

Velocity along $\text{AB}=\frac{\text{AB}}{t}$
i.e $v_{br}\cos \theta =\frac{3}{\frac{1}{3}}=9\text{km/hr}$
$\Rightarrow \cos \theta =\frac{9}{v_{br}}=\frac{3}{5}$
$\Rightarrow \sin \theta =\frac{4}{5}$
From triangle $\text{ABC}$,
$\sin \theta =\frac{v_r}{v_{br}}=\frac{v_r}{15}$
($\because$ Horizontal component of velocity of boat must become zero)
$\therefore v_r=12\text{km/hr}$

Hence option D is the correct answer