Water is flowing through a horizontal pipe whose one end is closed with a valve. Initially the pressure recorded by pressure gauge ia `4 xx 10^(5) N//m^2.` On opening the valve, the pressure gauge records `2 xx 10^5 N//m.` Find the speed of water flowing through the pipe.

Answer

To find the speed of water flowing through the pipe after the valve is opened, we can use Bernoulli's equation. Let's go through the solution step by step.

### Step 1: Understand the Initial Conditions
Initially, when the valve is closed, the pressure in the pipe is given as:
- \( P_1 = 4 \times 10^5 \, \text{N/m}^2 \)
- The velocity of water, \( V_1 = 0 \, \text{m/s} \) (since the water is not flowing).

### Step 2: Understand the Conditions After Opening the Valve
After opening the valve, the pressure in the pipe changes to:
- \( P_2 = 2 \times 10^5 \, \text{N/m}^2 \)
- The velocity of water, \( V_2 \) (which we need to find).

### Step 3: Apply Bernoulli's Equation
Bernoulli's equation states that for an incompressible, non-viscous fluid flowing in a streamline, the following holds:
\[ \frac{1}{2} \rho V_1^2 + P_1 + \rho gh_1 = \frac{1}{2} \rho V_2^2 + P_2 + \rho gh_2 \]
Since the pipe is horizontal, the heights \( h_1 \) and \( h_2 \) are the same, thus they cancel out. We can simplify the equation to:
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]

### Step 4: Substitute Known Values
Substituting the known values into the equation:
- \( V_1 = 0 \)
- \( P_1 = 4 \times 10^5 \, \text{N/m}^2 \)
- \( P_2 = 2 \times 10^5 \, \text{N/m}^2 \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)

The equation becomes:
\[ 4 \times 10^5 + 0 = 2 \times 10^5 + \frac{1}{2} \times 1000 \times V_2^2 \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ 4 \times 10^5 - 2 \times 10^5 = \frac{1}{2} \times 1000 \times V_2^2 \]
\[ 2 \times 10^5 = \frac{1}{2} \times 1000 \times V_2^2 \]

### Step 6: Solve for \( V_2^2 \)
Multiplying both sides by 2:
\[ 4 \times 10^5 = 1000 \times V_2^2 \]
Dividing both sides by 1000:
\[ V_2^2 = \frac{4 \times 10^5}{1000} = 400 \]

### Step 7: Calculate \( V_2 \)
Taking the square root:
\[ V_2 = \sqrt{400} = 20 \, \text{m/s} \]

### Final Answer
The speed of water flowing through the pipe after the valve is opened is:
\[ \boxed{20 \, \text{m/s}} \]

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