A river is flowing from west to east direction with speed of 9 ~km ~h^ -1 . If a boat capable of moving at a maximum speed of 27 ~km ~h^ -1 in still water, crosses the river in half a minute, while moving with maximum

Correct answer: B. 112.5 m V_ = river flow =27 × 60^ =(27)/(2) ~km / hr . Time taken =30 sec . S=Vt =(27)/(2) × (5)/(18) × 30 ~m=112.5 ~m

JEE Main 20252 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Morning Shift
Subject: Physics
Chapter: Motion in Two Dimensions

JEE Main 2025 · 2 April · Morning Shift · Physics · Motion in Two Dimensions

A river is flowing from west to east direction with speed of 9 ~km...

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 : A river is flowing from west to east direction with speed of 9 ~km ~h^ -1 . If a boat capable of moving at a maximum

A river is flowing from west to east direction with speed of 9 ~km ~h^ -1 . If a boat capable of moving at a maximum

Options

A300 m
B112.5 m
C75 m
D $112.5 \times \sqrt{3} \mathrm{~m}$

Correct answer

B. 112.5 m

Step-by-step solution

\therefore \mathrm{V}_{\perp}=

river flow

=27 \times \cos 60^{\circ}=\frac{27}{2} \mathrm{~km} / \mathrm{hr}

.
Time taken

=30 \mathrm{sec}

\therefore

\mathrm{S}=\mathrm{Vt}

=\frac{27}{2} \times \frac{5}{18} \times 30 \mathrm{~m}=112.5 \mathrm{~m}

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