Water flows steadily through a horizontal pipe of a variable cross-section. If the pressure of water is p at a point where the velocity of flow is v, what is the pressure at another point where the velocity of flow is 2v, `rho` being the density of water?

Options

A. `p + 2 rho v^2`
B. `p - 2 rho v^2`
C. `p + 3/2 rho v^2`
D. `p - 3/2 rho v^2`

Explanation

To solve the problem, we will use Bernoulli's equation, which relates the pressure, velocity, and height at two points in a fluid flow. Since the pipe is horizontal, the heights at both points are the same, allowing us to simplify the equation.

### Step-by-Step Solution:

1. **Write down Bernoulli's equation**:
\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \]
Since the pipe is horizontal, \( h_1 = h_2 \), so the terms involving height cancel out.

2. **Substitute known values**:
Let \( P_1 = P \), \( v_1 = v \), and \( v_2 = 2v \). The equation simplifies to:
\[ P + \frac{1}{2} \rho v^2 = P_2 + \frac{1}{2} \rho (2v)^2 \]

3. **Calculate \( (2v)^2 \)**:
\[ (2v)^2 = 4v^2 \]
Substitute this back into the equation:
\[ P + \frac{1}{2} \rho v^2 = P_2 + \frac{1}{2} \rho (4v^2) \]

4. **Rewrite the equation**:
\[ P + \frac{1}{2} \rho v^2 = P_2 + 2 \rho v^2 \]

5. **Rearrange to solve for \( P_2 \)**:
Move \( P_2 \) to one side and all other terms to the other side:
\[ P_2 = P + \frac{1}{2} \rho v^2 - 2 \rho v^2 \]

6. **Combine like terms**:
\[ P_2 = P + \frac{1}{2} \rho v^2 - \frac{4}{2} \rho v^2 \]
\[ P_2 = P - \frac{3}{2} \rho v^2 \]

7. **Final expression for \( P_2 \)**:
\[ P_2 = P - \frac{3}{2} \rho v^2 \]

### Final Answer:
The pressure at the point where the velocity of flow is \( 2v \) is:
\[ P_2 = P - \frac{3}{2} \rho v^2 \]