A manometer connected to a closed tap reads `4.5 xx 10^(5)` pascal. When the tap is opened the reading of the manometer falls is `4 xx 10^(5)` pascal. Then the velocity of flow of water is

Options

A. `7 ms^(-1)`
B. `8 ms^(-1)`
C. `9 ms^(-1)`
D. `10 ms^(-1)`

Answer

To solve the problem, we will use Bernoulli's equation, which relates the pressure, velocity, and height in a flowing fluid.

### Step-by-Step Solution:

1. **Identify the Initial and Final Pressures:**
- When the tap is closed, the pressure \( P_1 = 4.5 \times 10^5 \) Pa.
- When the tap is opened, the pressure \( P_2 = 4.0 \times 10^5 \) Pa.

2. **Calculate the Pressure Difference:**
\[ \Delta P = P_1 - P_2 = (4.5 \times 10^5) - (4.0 \times 10^5) = 0.5 \times 10^5 \text{ Pa} \]

3. **Apply Bernoulli's Equation:**
- According to Bernoulli's principle, the difference in pressure between two points in a fluid can be expressed as:
\[ P_1 - P_2 = \frac{1}{2} \rho v^2 \]
where \( \rho \) is the density of the fluid (for water, \( \rho = 1000 \, \text{kg/m}^3 \)) and \( v \) is the velocity of the fluid.

4. **Rearranging the Equation to Solve for Velocity:**
\[ v^2 = \frac{2(P_1 - P_2)}{\rho} \]
\[ v = \sqrt{\frac{2(P_1 - P_2)}{\rho}} \]

5. **Substituting the Values:**
\[ v = \sqrt{\frac{2 \times (0.5 \times 10^5)}{1000}} \]
\[ v = \sqrt{\frac{1 \times 10^5}{1000}} = \sqrt{100} = 10 \, \text{m/s} \]

6. **Conclusion:**
The velocity of the flow of water when the tap is opened is \( 10 \, \text{m/s} \).