A manometer connected to a closed tap reads `3.5xx10^(5)N//m^(2)`.When the value is opened, the reading of manometer fall is `3.0xx10^(5)N//m^(2)`, then velocity of flow of water is

Options

A. `100 ms^( -1)`
B. `10 ms^( -1)`
C. `1 ms^( -1)`
D. `10sqrt(10) ms^( -1)`

Answer

To solve the problem step-by-step, we will use Bernoulli's equation, which relates the pressure and velocity of a fluid in a streamline flow.

### Step-by-Step Solution:

1. **Identify Given Values:**
- When the tap is closed (initial state), the pressure \( P_1 = 3.5 \times 10^5 \, \text{N/m}^2 \).
- When the tap is opened (final state), the pressure \( P_2 = 3.0 \times 10^5 \, \text{N/m}^2 \).
- The density of water \( \rho = 1000 \, \text{kg/m}^3 \).

2. **Apply Bernoulli's Equation:**
Bernoulli's equation states that:
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]
Here, \( V_1 \) is the velocity of water when the tap is closed, and \( V_2 \) is the velocity when the tap is opened. Since the tap is closed, \( V_1 = 0 \).

3. **Rearranging the Equation:**
Since \( V_1 = 0 \), the equation simplifies to:
\[ P_1 = P_2 + \frac{1}{2} \rho V_2^2 \]
Rearranging gives:
\[ \frac{1}{2} \rho V_2^2 = P_1 - P_2 \]

4. **Substituting Values:**
Substitute the values of \( P_1 \) and \( P_2 \):
\[ \frac{1}{2} \rho V_2^2 = (3.5 \times 10^5) - (3.0 \times 10^5) \]
\[ \frac{1}{2} \rho V_2^2 = 0.5 \times 10^5 \, \text{N/m}^2 \]
\[ \frac{1}{2} \rho V_2^2 = 5 \times 10^4 \, \text{N/m}^2 \]

5. **Solving for \( V_2^2 \):**
Now, substituting the density of water:
\[ \frac{1}{2} (1000) V_2^2 = 5 \times 10^4 \]
\[ 500 V_2^2 = 5 \times 10^4 \]
\[ V_2^2 = \frac{5 \times 10^4}{500} \]
\[ V_2^2 = 100 \]

6. **Finding \( V_2 \):**
Taking the square root:
\[ V_2 = \sqrt{100} = 10 \, \text{m/s} \]

### Final Answer:
The velocity of flow of water when the tap is opened is \( V_2 = 10 \, \text{m/s} \).