The horizontal speed of a jet of water is 100 cm/sec and `50 cm^3` of water hits the plate each second. Assume that the water moves parallel to the plate after striking it. The force exerted on the stationary plate if it is held perpendicular to the jet of water is :

Options

A. `5 xx 10^(-2) N`
B. `5 xx 10^(2)` N
C. `5 xx 10^(-1)` N
D. `5 N`

Answer

To solve the problem, we need to calculate the force exerted on the stationary plate by the jet of water. We can use the principle of momentum change to find the force.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Horizontal speed of the water jet, \( v = 100 \, \text{cm/sec} = 1 \, \text{m/sec} \) (convert to meters)
- Volume of water hitting the plate per second, \( V = 50 \, \text{cm}^3 = 50 \times 10^{-6} \, \text{m}^3 \) (convert to cubic meters)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)

2. **Calculate the Mass Flow Rate:**
The mass flow rate \( \frac{dm}{dt} \) can be calculated using the formula:
\[ \frac{dm}{dt} = \rho \times \frac{dV}{dt} \]
Here, \( \frac{dV}{dt} \) is the volume flow rate, which is given as \( 50 \times 10^{-6} \, \text{m}^3/s \).
\[ \frac{dm}{dt} = 1000 \, \text{kg/m}^3 \times 50 \times 10^{-6} \, \text{m}^3/s = 0.05 \, \text{kg/s} \]

3. **Calculate the Change in Momentum:**
The change in momentum per second (which is the force) can be calculated using:
\[ F = \frac{dm}{dt} \times v \]
Substituting the values:
\[ F = 0.05 \, \text{kg/s} \times 1 \, \text{m/s} = 0.05 \, \text{N} \]

4. **Conclusion:**
The force exerted on the stationary plate by the jet of water is:
\[ F = 0.05 \, \text{N} \]

### Final Answer:
The force exerted on the stationary plate is **0.05 N**.