A tube has two area of cross-sections as shown in figure. The diameters of the tube are $8\,\text{mm}$ and $2\,\text{mm}$. Find range of water falling on horizontal surface, if piston is moving with a constant velocity of $0.25\,\text{m/s}$, $h = 1.25\,\text{m}$ ($g = 10\,\text{m/s}^{2}$)

Step 1: Identify the given parameters.

The diameters of the tube, the velocity of the piston, the height of the nozzle, and the acceleration due to gravity are given.

$$D_1 = 8\,\text{mm}$$ $$D_2 = 2\,\text{mm}$$ $$v_1 = 0.25\,\text{m/s}$$ $$h = 1.25\,\text{m}$$ $$g = 10\,\text{m/s}^{2}$$

Step 2: Calculate the ratio of the cross-sectional areas.

The area of a circular cross-section is proportional to the square of its diameter.

$$\frac{A_1}{A_2} = \left(\frac{D_1}{D_2}\right)^2$$ $$\frac{A_1}{A_2} = \left(\frac{8}{2}\right)^2 = 16$$

Step 3: Find the exit velocity of the water.

Apply the equation of continuity for an incompressible fluid.

$$A_1 v_1 = A_2 v_2$$ $$v_2 = v_1 \left(\frac{A_1}{A_2}\right)$$ $$v_2 = 0.25 \times 16 = 4\,\text{m/s}$$

Step 4: Calculate the time of flight of the water jet.

The water exits horizontally, so its vertical motion is a free fall from height $h$.

$$t = \sqrt{\frac{2h}{g}}$$ $$t = \sqrt{\frac{2 \times 1.25}{10}}$$ $$t = 0.5\,\text{s}$$

Step 5: Calculate the horizontal range.

The horizontal range is the product of the horizontal exit velocity and the time of flight.

$$R = v_2 t$$ $$R = 4 \times 0.5 = 2\,\text{m}$$

Answer: The range of the water falling on the horizontal surface is $2\,\text{m}$.