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A man is trying to cross a river flowing at a speed of $5 m s^{- 1}$ to have least possible displacement by swimming at an angle of $143 °$ to the stream. The drift he suffers when he is crossing the same river in least possible time is (width of river $=$ 1 km)

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800 m

200 m

160 m

400 m

answer is B.

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Detailed Solution

$\sin 50^{0} = \frac{V_{r}}{V_{b r}}$ , $\frac{4}{5} = \frac{5}{V_{b r}} \Rightarrow V_{b r} = \frac{25}{4}$

For shortest time :

$t = \frac{d}{V_{b r}} =$ 160 seconds

Drift BC $= V_{r} \times t = 5 \times 160 = 800 m$

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