A cylindrical jet of water (density = 1000 kg/m³ ) impinges at the center of a flat, circular plate and spreads radially outwards, as shown in the figure. The plate is resting on a linear spring with a spring constant 𝒌 = 𝟏 kN/m. The incoming jet diameter is 𝑫 = 𝟏 cm.

If the spring shows a steady deflection of 1 cm upon impingement of jet, then the velocity of the incoming jet is ____________ m/s (round off to one decimal place).

The correct answer is 11.28.

Step 1: Understand the Physical System from the Image
As shown in the provided diagram, a vertical cylindrical water jet of diameter D is directed downwards, impinging on the center of a flat, horizontal circular plate. After impingement, the water spreads radially outwards in the horizontal plane. The plate is supported by a linear spring with a spring constant k. The force of the jet compresses the spring downward by a steady deflection distance, which we will denote as x.

Step 2: Identify the Given Parameter Values
- Density of water (ρ) = 1000 kg/m³
- Diameter of the incoming jet (D) = 1 cm = 0.01 m
- Spring constant (k) = 1 kN/m = 1000 N/m
- Steady deflection of the spring (x) = 1 cm = 0.01 m

Step 3: Formulate the Momentum Equation
Let V be the velocity of the incoming jet in the vertical direction. By applying the linear momentum equation in the vertical direction for a control volume enclosing the plate and the region of fluid impingement:
The vertical force exerted by the water jet on the plate, F, is equal to the rate of change of vertical momentum of the water jet:

$$F=\dot{m}\times (V_{\text{in,y}}-V_{\text{out,y}})$$

Since the water spreads radially outward along the flat plate horizontally, the final velocity component in the vertical direction is zero ($V_{\text{out,y}}=0$).
The mass flow rate of the incoming jet is:

$$\dot{m}=\rho AV$$

where A is the cross-sectional area of the cylindrical jet:
$$A=\frac{\pi }{4}{D}^2$$

Substituting the mass flow rate and velocities, we get the force exerted by the water jet on the plate:

$$F=\rho A{V}^2$$

Step 4: Relate the Jet Force to the Spring Deflection
At steady state, the downward force exerted by the water jet is balanced by the restoring upward force of the compressed spring:

$$F=kx$$

Equating the two forces yields:

$$kx=\rho A{V}^2$$

Step 5: Solve for the Velocity of the Incoming Jet (V)
Let us substitute the known values into the equation:

$$1000\times 0.01=1000\times \frac{\pi }{4}\times {\left(0.01\right)}^2\times V^2$$

Simplifying the equation:
$$10=1000\times \frac{\pi }{4}\times 0.0001\times V^2$$

$$10=0.1\times \frac{\pi }{4}\times V^2$$

Multiply both sides by 10:
$$100=\frac{\pi }{4}\times V^2$$

$$V^2=\frac{400}{\pi }$$

$$V^2\approx 127.324$$

Taking the square root of both sides:
$$V=\sqrt{127.324}\approx 11.2838\text{ m/s}$$

Thus, the velocity of the incoming jet is 11.28 m/s (or 11.3 m/s when rounded to one decimal place).