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 speed at an angle of 150° to direction of river flow, then the width of the river is:

The river flows due east with speed $$v_r = 9 \text{ km h}^{-1}$$. The boat can move in still water with maximum speed $$v_b = 27 \text{ km h}^{-1}$$ and is steered at an angle $$150^{\circ}$$ to the river flow (angle measured counter-clockwise from east).

Take east as the $$+x$$-axis and north as the $$+y$$-axis.
Components of the boat’s velocity relative to water:

$$v_{bx} = v_b \cos 150^{\circ} = 27 \cos 150^{\circ} = 27 \left(-\frac{\sqrt{3}}{2}\right) = -\frac{27\sqrt{3}}{2} \text{ km h}^{-1}$$ $$-(1)$$
$$v_{by} = v_b \sin 150^{\circ} = 27 \sin 150^{\circ} = 27 \left(\frac{1}{2}\right) = \frac{27}{2} \text{ km h}^{-1}$$ $$-(2)$$

The river velocity has only an $$x$$-component:
$$v_{rx} = +9 \text{ km h}^{-1}, \qquad v_{ry}=0$$

Ground (resultant) velocity of the boat:
$$v_{gx} = v_{bx} + v_{rx} = -\frac{27\sqrt{3}}{2} + 9 \text{ km h}^{-1}$$ $$-(3)$$
$$v_{gy} = v_{by} + v_{ry} = \frac{27}{2} \text{ km h}^{-1}$$ $$-(4)$$

To cross the river we need the northward component $$v_{gy}$$ only. Hence the time taken $$t$$ and river width $$D$$ are related by
$$D = v_{gy}\, t$$ $$-(5)$$

The boat crosses in half a minute:
$$t = \frac{1}{2} \text{ min} = \frac{1}{2}\times\frac{1}{60} \text{ h} = \frac{1}{120} \text{ h}$$ $$-(6)$$

Substituting $$v_{gy}$$ from $$(4)$$ into $$(5)$$:
$$D = \frac{27}{2} \times \frac{1}{120} \text{ km}$$

Simplify:
$$D = \frac{27}{2 \times 120} \text{ km} = \frac{27}{240} \text{ km} = 0.1125 \text{ km}$$

Convert kilometres to metres:
$$D = 0.1125 \times 1000 = 112.5 \text{ m}$$

Therefore, the width of the river is $$112.5 \text{ m}$$, which corresponds to Option B.