Question

A river is flowing from west to east direction with speed of \(9\) km/hr. If a boat capable of moving at a maximum speed of \(27\) km/hr 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. 300 m
• B. 112.5 m
• C. 75 m
• D. \( 112.5 \times \sqrt{3} \) m

Hint:

When solving river crossing problems, decompose the boat's velocity into components: one parallel to the river flow and one perpendicular to it. The perpendicular component gives the speed for crossing the river.

Answer 1

The Correct Option is B

The speed of the boat relative to the river is \( 27 \, \text{km/hr} \), and the boat crosses the river at an angle of \( 150^\circ \) to the direction of the river flow. 
Using the formula for the effective speed component of the boat in the direction perpendicular to the flow of the river: \[ V_{L} = 27 \, \text{km/hr} \times \cos 60^\circ = \frac{27}{2} = 13.5 \, \text{km/hr} \] The time taken to cross the river is \( 30 \, \text{seconds} \) or \( \frac{1}{2} \) minute. Using the formula for distance: \[ S = V_t \times t = 13.5 \, \text{km/hr} \times \frac{30}{60} \, \text{hr} = 13.5 \times \frac{1}{2} = 112.5 \, \text{m} \] Thus, the width of the river is \( 112.5 \, \text{m} \).