JEE Physics 2025 Question with Solution

Given: River speed is $9 \, \text{km/hr}$. Boat speed in still water is $27 \, \text{km/hr}$. The boat moves at $150^\circ$ to the direction of river flow. Time taken to cross is half a minute, i.e. $30 \, \text{s} = \frac{1}{120} \, \text{hr}$.

Find: The width of the river.

The width depends on the component of the boat's velocity perpendicular to the river flow.

Resolve the boat velocity:

$$\vec{v}_{br} = 27(\cos 150^\circ \, \hat{i} + \sin 150^\circ \, \hat{j})$$

Using $\sin 150^\circ = \frac{1}{2}$, the perpendicular component is

$$v_{y} = 27 \times \frac{1}{2} = 13.5 \, \text{km/hr}$$

Now width of river is

$$W = v_{y} \times t$$ $$W = 13.5 \times \frac{1}{120} \, \text{km}$$ $$W = \frac{13.5}{120} \, \text{km} = \frac{9}{80} \, \text{km}$$

Converting into metres,

$$W = \frac{9}{80} \times 1000 = 112.5 \, \text{m}$$

Therefore, the width of the river is $112.5 \, \text{m}$. The correct option is B.

Given: Take river flow from west to east as the positive $x$-axis. The perpendicular direction across the river is the positive $y$-axis.

• River velocity: $\vec{v}_r = 9\hat{i} \, \text{km/hr}$
• Boat speed relative to river: $|\vec{v}_{br}| = 27 \, \text{km/hr}$
• Direction of boat: $150^\circ$ to river flow
• Crossing time: $0.5 \, \text{min} = \frac{1}{120} \, \text{hr}$

Find: The width of the river.

First resolve the boat velocity relative to the river into components:

$$\vec{v}_{br} = 27(\cos 150^\circ \, \hat{i} + \sin 150^\circ \, \hat{j})$$

Since $\cos 150^\circ = -\frac{\sqrt{3}}{2}$ and $\sin 150^\circ = \frac{1}{2}$,

$$\vec{v}_{br} = 27\left(-\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right)$$ $$\vec{v}_{br} = -13.5\sqrt{3}\hat{i} + 13.5\hat{j} \, \text{km/hr}$$

Now the boat velocity with respect to ground is

$$\vec{v}_b = \vec{v}_{br} + \vec{v}_r$$ $$\vec{v}_b = (-13.5\sqrt{3}\hat{i} + 13.5\hat{j}) + 9\hat{i}$$ $$\vec{v}_b = (9 - 13.5\sqrt{3})\hat{i} + 13.5\hat{j} \, \text{km/hr}$$

The component responsible for crossing the river is the $y$-component:

$$v_{by} = 13.5 \, \text{km/hr}$$

Hence,

$$W = v_{by}t = 13.5 \times \frac{1}{120} \, \text{km}$$ $$W = \frac{9}{80} \, \text{km} = 112.5 \, \text{m}$$

Therefore, the width of the river is $112.5 \, \text{m}$.