JEE MAIN - Physics (2025 - 2nd April Morning Shift - No. 4)
A river is flowing from west to east direction with speed of $9 \mathrm{~km} \mathrm{~h}^{-1}$. If a boat capable of moving at a maximum speed of $27 \mathrm{~km} \mathrm{~h}^{-1}$ in still water, crosses the river in half a minute, while moving with maximum speed at an angle of $150^{\circ}$ to direction of river flow, then the width of the river is :
112.5 m
75 m
300 m
$112.5 \times \sqrt{3} \mathrm{~m}$
Explanation
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$$\begin{aligned} &\therefore \mathrm{V}_{\perp}=\text { river flow }=27 \times \cos 60^{\circ}=\frac{27}{2} \mathrm{~km} / \mathrm{hr}\\ &\text { Time taken }=30 \mathrm{sec} \text {. }\\ &\therefore \mathrm{S}=\mathrm{Vt}=\frac{27}{2} \times \frac{5}{18} \times 30 \mathrm{~m}=112.5 \mathrm{~m} \end{aligned}$$
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