Question

A horizontal cylindrical water jet of diameter $D_1 = 2 \text{ cm}$ strikes a vertical solid plate with a hole of diameter $D_2 = 1 \text{ cm}$, as shown in the figure. A part of the jet passes through the hole and the rest is deflected along the plate. The density of water is $1000\ \text{kg m}^{-3}$. If the jet speed is $20\ \text{m s}^{-1}$, the magnitude of the horizontal force required to hold the plate stationary is

• A. \(30\pi\)
• B. \(10\pi\)
• C. \(20\pi\)
• D. \(5\pi\)

Hint:

For force caused by fluid jets, always use momentum change. Only the mass flow whose direction changes contributes to force.

Answer 1

The Correct Option is A

Consider a horizontal water jet striking a plate with a central hole. A part of the flow passes straight through, while the remaining flow is deflected symmetrically upward and downward. The force on the plate equals the net change in horizontal momentum of the flow.

Step 1: Compute mass flow rate of the incoming jet.
Jet diameter: \(D_1 = 2\ \text{cm} = 0.02\ \text{m}\)
Incoming area: \[ A_1 = \frac{\pi D_1^2}{4} = \frac{\pi (0.02)^2}{4} = 0.0001\pi\ \text{m}^2 \] Mass flow rate: \[ \dot{m}_1 = \rho A_1 V = 1000 \cdot (0.0001\pi) \cdot 20 = 2\pi\ \text{kg s}^{-1} \]

Step 2: Compute mass flow rate passing through the hole.
Hole diameter: \(D_2 = 1\ \text{cm} = 0.01\ \text{m}\)
Hole area: \[ A_2 = \frac{\pi (0.01)^2}{4} = 0.000025\pi\ \text{m}^2 \] Mass flow rate: \[ \dot{m}_2 = 1000 \cdot (0.000025\pi) \cdot 20 = 0.5\pi\ \text{kg s}^{-1} \]

Step 3: Mass flow rate deflected by the plate.
\[ \dot{m}_{\text{deflected}} = \dot{m}_1 - \dot{m}_2 = 2\pi - 0.5\pi = 1.5\pi\ \text{kg s}^{-1} \]

Step 4: Horizontal momentum change.
Water passing through hole suffers no change in horizontal momentum. Water deflected upward/downward loses all horizontal momentum: \[ F = \dot{m}_{\text{deflected}} \cdot V = (1.5\pi)(20) = 30\pi\ \text{N} \] Thus, the horizontal force needed to hold the plate is: \[ F = 30\pi\ \text{N} \]

Final Answer: \(30\pi\)