Water jet issues water from a nozzle of `2 cm^(2)` cross-section with velocity 30 cm/s and strikes a plane surface placed at right angles to the jet. The force exerted on the plane is

Options

A. 200 dyne
B. 400 dyne
C. 1800 dyne
D. None of the above

Explanation

To solve the problem, we will follow these steps:

### Step 1: Find the volume flow rate of water.
The volume flow rate (Q) can be calculated using the formula:
\[ Q = A \times v \]
where:
- \( A \) is the cross-sectional area of the nozzle,
- \( v \) is the velocity of the water.

Given:
- Cross-sectional area \( A = 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 \) (since \( 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \)),
- Velocity \( v = 30 \, \text{cm/s} = 0.3 \, \text{m/s} \).

Now, substituting the values:
\[ Q = (2 \times 10^{-4} \, \text{m}^2) \times (0.3 \, \text{m/s}) \]
\[ Q = 6 \times 10^{-5} \, \text{m}^3/\text{s} \]

### Step 2: Find the mass flow rate of water.
The mass flow rate (\( \dot{m} \)) can be calculated using the formula:
\[ \dot{m} = \rho \times Q \]
where \( \rho \) is the density of water, approximately \( 1000 \, \text{kg/m}^3 \).

Now substituting the values:
\[ \dot{m} = (1000 \, \text{kg/m}^3) \times (6 \times 10^{-5} \, \text{m}^3/\text{s}) \]
\[ \dot{m} = 0.06 \, \text{kg/s} \]

### Step 3: Calculate the force exerted on the plane surface.
The force (\( F \)) exerted on the plane can be calculated using the change in momentum per unit time:
\[ F = \dot{m} \times v \]

Substituting the values:
\[ F = (0.06 \, \text{kg/s}) \times (0.3 \, \text{m/s}) \]
\[ F = 0.018 \, \text{N} \]

### Final Answer:
The force exerted on the plane surface is \( 0.018 \, \text{N} \).

---