A manometer connected to a closed tap reads `4.5 xx 10^(5)` pascal. When the tap is opened the reading of the monometer 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 and velocity of a fluid in a streamline flow. The equation is given as:

\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \]

In this case, since the height (h) is the same at both points, we can ignore the potential energy terms. Thus, the equation simplifies to:

\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]

### Step 1: Identify the given values
- Initial pressure \( P_1 = 4.5 \times 10^5 \, \text{Pa} \)
- Final pressure \( P_2 = 4.0 \times 10^5 \, \text{Pa} \)
- Initial velocity \( v_1 = 0 \, \text{m/s} \) (since the tap is closed)

### Step 2: Calculate the change in pressure
The change in pressure when the tap is opened is:

\[ \Delta P = P_1 - P_2 = 4.5 \times 10^5 \, \text{Pa} - 4.0 \times 10^5 \, \text{Pa} = 0.5 \times 10^5 \, \text{Pa} \]

### Step 3: Substitute values into Bernoulli's equation
Since \( v_1 = 0 \), the equation simplifies to:

\[ P_1 = P_2 + \frac{1}{2} \rho v_2^2 \]

Rearranging gives:

\[ P_1 - P_2 = \frac{1}{2} \rho v_2^2 \]

Substituting the change in pressure:

\[ 0.5 \times 10^5 = \frac{1}{2} \rho v_2^2 \]

### Step 4: Use the density of water
The density of water \( \rho \) is approximately \( 1000 \, \text{kg/m}^3 \).

### Step 5: Substitute the density into the equation
Now substituting \( \rho \):

\[ 0.5 \times 10^5 = \frac{1}{2} \times 1000 \times v_2^2 \]

### Step 6: Solve for \( v_2^2 \)
Multiplying both sides by 2:

\[ 1.0 \times 10^5 = 1000 \times v_2^2 \]

Now divide both sides by 1000:

\[ v_2^2 = \frac{1.0 \times 10^5}{1000} = 100 \]

### Step 7: Take the square root to find \( v_2 \)
Taking the square root of both sides gives:

\[ v_2 = \sqrt{100} = 10 \, \text{m/s} \]

### Final Answer
The velocity of flow of water when the tap is opened is:

\[ v_2 = 10 \, \text{m/s} \]

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