One end of a horizontal pipe is closed with the help of a velve and the reading of a barometer attached in the pipe is `3xx10^(5)` pascal When the value in the pipe is opened then the reading of barometer falls to `10^(5)` pascal the velocity of water flowing through the pipe will be in m/s-

Options

A. 0.2
B. 2
C. 20
D. 200

Answer

To solve the problem, we will apply Bernoulli's principle to the horizontal pipe. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have a horizontal pipe with one end closed. Initially, the pressure in the pipe is given as \( P_1 = 3 \times 10^5 \) Pa. When the valve is opened, the pressure drops to \( P_2 = 10^5 \) Pa. We need to find the velocity of water flowing through the pipe after the valve is opened.

### Step 2: Apply Bernoulli's Equation
Bernoulli's equation for a horizontal flow can be expressed as:
\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]
Where:
- \( P_1 \) = initial pressure
- \( P_2 \) = final pressure
- \( \rho \) = density of water (approximately \( 1000 \, \text{kg/m}^3 \))
- \( v_1 \) = initial velocity (which is \( 0 \) since the pipe is closed)
- \( v_2 \) = final velocity of water flowing through the pipe

### Step 3: Substitute Known Values
Since the initial velocity \( v_1 = 0 \), the equation simplifies to:
\[ P_1 = P_2 + \frac{1}{2} \rho v_2^2 \]
Substituting the known values:
- \( P_1 = 3 \times 10^5 \, \text{Pa} \)
- \( P_2 = 10^5 \, \text{Pa} \)
- \( \rho = 1000 \, \text{kg/m}^3 \)

The equation becomes:
\[ 3 \times 10^5 = 10^5 + \frac{1}{2} \times 1000 \times v_2^2 \]

### Step 4: Simplify the Equation
Rearranging the equation gives:
\[ 3 \times 10^5 - 10^5 = \frac{1}{2} \times 1000 \times v_2^2 \]
\[ 2 \times 10^5 = \frac{1}{2} \times 1000 \times v_2^2 \]

### Step 5: Solve for \( v_2^2 \)
Multiplying both sides by 2 to eliminate the fraction:
\[ 4 \times 10^5 = 1000 \times v_2^2 \]
Now, divide both sides by 1000:
\[ v_2^2 = \frac{4 \times 10^5}{1000} = 400 \]

### Step 6: Calculate \( v_2 \)
Taking the square root of both sides:
\[ v_2 = \sqrt{400} = 20 \, \text{m/s} \]

### Final Answer
The velocity of water flowing through the pipe after the valve is opened is \( 20 \, \text{m/s} \).

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