A boat which has a speed of $5\,\text{km/hr}$ in still water crosses a river of width $1\,\text{km}$ along the shortest possible path in $15$ minutes. The velocity of the river water in km/hr is

Step 1: Identify the given parameters.

Convert the crossing time of $15$ minutes into hours.

$$v_{\text{boat}} = 5\,\text{km/hr}$$ $$d = 1\,\text{km}$$ $$t = \frac{1}{4}\,\text{hr}$$

Step 2: Determine the velocity of the boat relative to the ground.

For the shortest possible path, the boat must travel perpendicular to the river banks.

$$v_{\text{ground}} = \frac{d}{t}$$ $$v_{\text{ground}} = \frac{1\,\text{km}}{\frac{1}{4}\,\text{hr}}$$ $$v_{\text{ground}} = 4\,\text{km/hr}$$

Step 3: Relate the velocities using the vector addition triangle.

The velocity of the boat in still water forms the hypotenuse of a right-angled triangle with the ground velocity and river velocity.

$$v_{\text{boat}}^2 = v_{\text{ground}}^2 + v_{\text{river}}^2$$

Step 4: Calculate the velocity of the river water.

Substitute the known values into the equation.

$$v_{\text{river}} = \sqrt{v_{\text{boat}}^2 - v_{\text{ground}}^2}$$ $$v_{\text{river}} = \sqrt{5^2 - 4^2}$$ $$v_{\text{river}} = \sqrt{9}$$ $$v_{\text{river}} = 3\,\text{km/hr}$$

Answer: The velocity of the river water is $3\,\text{km/hr}$.