JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 21)
The speed of a swimmer is $$4 \mathrm{~km} \mathrm{~h}^{-1}$$ in still water. If the swimmer makes his strokes normal to the flow of river of width $$1 \mathrm{~km}$$, he reaches a point $$750 \mathrm{~m}$$ down the stream on the opposite bank.
The speed of the river water is ___________ $$\mathrm{km} ~\mathrm{h}^{-1}$$
Answer
3
Explanation
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Time to cross the River width $\omega=1000 \mathrm{~m}$ is $=\frac{1 \mathrm{~km}}{4 \mathrm{~km} / \mathrm{h}}$
Drift $\mathrm{x}=\mathrm{Vm} / \mathrm{g} \times \mathrm{t}$
Where $\mathrm{Vm} / \mathrm{g}$ is velocity of River w.r. to ground.
$$ \begin{aligned} & \mathrm{x}=\mathrm{Vm} / \mathrm{g} \times \frac{1}{4}=750 \mathrm{~m}=\frac{3}{4} \mathrm{~km} \\\\ & \mathrm{Vm} / \mathrm{g}=3 \mathrm{~km} / \mathrm{hr} \end{aligned} $$
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