Question

A river $5\sqrt{3}km$ wide flows with a velocity of $5kmph$. A boat which can move with a velocity of $10kmph$ in still water crosses the river, without any drift. For this, the boat has to tilt its velocity by $\theta$ with the straight path in the right direction (see figiure). If the boat's velocity is tilted by the same angle $\theta$ in the wrong direction, the drift will be

• A. $8.1km$
• B. $9.1km$
• C. $10km$
• D. $12km$

Answer 1

The correct option is C $10km$

The velocities of the river and the boat can be represented vectorially as shown.

Let $\overrightarrow{OA}$ reperesent boat velocity and $\overrightarrow{AP}$ represent river velocity.

For zero drift, resultant $\overrightarrow{OP}$ should be normal to $\overrightarrow{AP}$.
Hence $\theta$ is given by
$\sin \theta =\frac{|\overrightarrow{AP}|}{|\overrightarrow{OA}|}=\frac{5}{10}\Rightarrow \theta ={30}^{\circ }$
Time to cross the river is
$\frac{d}{|\overrightarrow{OP}|}=\frac{5\sqrt{3}}{10\cos {30}^{\circ }}=1\text{hour}$

When tilted wrongly, the boat velocity will be along $OB$.
Hence, drift velocity is $10\sin {30}^{\circ }+5=10\text{kmph}$
Time to cross the river will remain same because $10\cos {30}^{\circ }$ is the normal component.
$\therefore$ Total drift $=10\times 1=10\text{km}$