If the boat crosses the river in shortest possible time, then its drift will be:-

zero

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250 m

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125 m

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\frac{250}{\sqrt{3}} m

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Solution

The correct option is A $250 m$
$v_{b} = 4, v_{r} = 2, d = 500 m$
For shortest possible time $v_{b}$ should be along AB
$t = \frac{d}{v_{b}}$
$t = \frac{500}{4} = 125 s e c$
$d r i f t$ : horizontal displacement of the boat
$d r i f t$ $=$ $v_{r} \times t$
$= 2 \times 125$
$= 250 m$

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If the boat crosses the river in shortest possible time, then its drift will be:-

The correct option is A 250 mvb=4,vr=2,d=500mFor shortest possible time vb should be along ABt=dvbt=5004=125secdrift : horizontal displacement of the boatdrift = vr×t =2×125 =250m

The correct option is A $250 m$
$v_{b} = 4, v_{r} = 2, d = 500 m$
For shortest possible time $v_{b}$ should be along AB
$t = \frac{d}{v_{b}}$
$t = \frac{500}{4} = 125 s e c$
$d r i f t$ : horizontal displacement of the boat
$d r i f t$ $=$ $v_{r} \times t$
$= 2 \times 125$
$= 250 m$