The reading of pressure meter attached with a closed water pipe is `3.5 xx 10^(5) Nm^(-2)` . On opening the valve of the pipe, the reading of pressure meter is reduced to `3 xx 10^(5) Nm^(-2)`. Calculate the speed of water flowing out of the pipe.

Explanation

To solve the problem of calculating the speed of water flowing out of the pipe, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in motion. Here’s a step-by-step solution:

### Step 1: Write Bernoulli's Equation
Bernoulli's equation states:
\[ P_1 + \frac{1}{2} \rho V_1^2 + \rho g H_1 = P_2 + \frac{1}{2} \rho V_2^2 + \rho g H_2 \]

### Step 2: Identify the Variables
- \( P_1 = 3.5 \times 10^5 \, \text{N/m}^2 \) (pressure inside the pipe before the valve is opened)
- \( P_2 = 3.0 \times 10^5 \, \text{N/m}^2 \) (pressure outside the pipe after the valve is opened)
- \( \rho = 1000 \, \text{kg/m}^3 \) (density of water)
- \( g \) (acceleration due to gravity) is not needed here as the heights \( H_1 \) and \( H_2 \) are equal and will cancel out.
- \( V_1 = 0 \) (initial velocity of water inside the pipe when the valve is closed)
- \( V_2 \) (velocity of water flowing out of the pipe, which we need to find)

### Step 3: Simplify the Equation
Since \( H_1 = H_2 \) and \( V_1 = 0 \), the equation simplifies to:
\[ P_1 - P_2 = \frac{1}{2} \rho V_2^2 \]

### Step 4: Substitute the Known Values
Substituting the known values into the equation:
\[ 3.5 \times 10^5 - 3.0 \times 10^5 = \frac{1}{2} \times 1000 \times V_2^2 \]

### Step 5: Calculate the Pressure Difference
Calculate the left side:
\[ 0.5 \times 10^5 = \frac{1}{2} \times 1000 \times V_2^2 \]
\[ 0.5 \times 10^5 = 500 \times V_2^2 \]

### Step 6: Solve for \( V_2^2 \)
Rearranging gives:
\[ V_2^2 = \frac{0.5 \times 10^5}{500} \]
\[ V_2^2 = 100 \]

### Step 7: Take the Square Root
Taking the square root to find \( V_2 \):
\[ V_2 = \sqrt{100} = 10 \, \text{m/s} \]

### Final Answer
The speed of water flowing out of the pipe is:
\[ V_2 = 10 \, \text{m/s} \]

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