The reading of pressure-meter fitted in a closed pipe is `4.5 xx 10^(5) Nm^(-2)` . On opening the value of the pipe, the reading the meter reduces to `4 xx 10^(5) Nm^(-2)`. Calculate the speed of water flowing in the pipe.

Options

A. a) 5 m/s
B. b) 10 m/s
C. c) 15 m/s
D. d) 20 m/s

Answer

To solve the problem, we will use Bernoulli's theorem, which relates the pressure, velocity, and height of a fluid in a streamline flow. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Initial pressure in the closed pipe, \( P_1 = 4.5 \times 10^5 \, \text{N/m}^2 \)
- Final pressure after opening the valve, \( P_2 = 4.0 \times 10^5 \, \text{N/m}^2 \)
- Density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)
- Initial velocity, \( v_1 = 0 \) (since the pipe was closed)

### Step 2: Apply Bernoulli's equation
In a closed pipe, Bernoulli's equation states:
\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2 \]
Since the heights \( h_1 \) and \( h_2 \) are the same, we can ignore the potential energy terms (\( \rho gh \)). Thus, the equation simplifies to:
\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]

### Step 3: Substitute known values
Substituting the known values into the equation:
\[ 4.5 \times 10^5 + \frac{1}{2} \times 10^3 \times 0^2 = 4.0 \times 10^5 + \frac{1}{2} \times 10^3 v_2^2 \]
This simplifies to:
\[ 4.5 \times 10^5 = 4.0 \times 10^5 + \frac{1}{2} \times 10^3 v_2^2 \]

### Step 4: Rearrange the equation
Rearranging the equation to solve for \( v_2^2 \):
\[ 4.5 \times 10^5 - 4.0 \times 10^5 = \frac{1}{2} \times 10^3 v_2^2 \]
\[ 0.5 \times 10^5 = \frac{1}{2} \times 10^3 v_2^2 \]

### Step 5: Solve for \( v_2^2 \)
Multiplying both sides by 2:
\[ 10^5 = 10^3 v_2^2 \]
Dividing both sides by \( 10^3 \):
\[ v_2^2 = 10^2 \]

### Step 6: Take the square root to find \( v_2 \)
Taking the square root of both sides:
\[ v_2 = \sqrt{10^2} = 10 \, \text{m/s} \]

### Final Answer
The speed of water flowing in the pipe when the valve is opened is \( 10 \, \text{m/s} \).

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