Water flows steadily through a horizontal pipe of a variable cross-section. If the pressure of the water is p at a point , where the speed of the flow is v. What is the pressure at another point , where the speed of the flow is 2 v ? Let the density of water be `1 g cm ^(-3)`.

Options

A. `p+(3/2)rhov^2`
B. `p-2rhov^2`
C. `p - 3rhov^2`
D. `p- (3/2) rhov^2`

Answer

To solve the problem, we will use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid flowing in a horizontal pipe. The equation is given by:

\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \]

### Step 1: Identify the known values
- At point 1:
- Pressure \( P_1 = P \)
- Velocity \( v_1 = v \)

- At point 2:
- Pressure \( P_2 = ? \)
- Velocity \( v_2 = 2v \)

- Density of water \( \rho = 1 \, \text{g/cm}^3 \) (which is equivalent to \( 1000 \, \text{kg/m}^3 \) in SI units)
- Since the pipe is horizontal, the heights \( h_1 \) and \( h_2 \) are equal, so \( h_1 = h_2 \).

### Step 2: Simplify Bernoulli's equation
Since \( h_1 = h_2 \), the terms involving height cancel out:

\[ P + \frac{1}{2} \rho v^2 = P_2 + \frac{1}{2} \rho (2v)^2 \]

### Step 3: Substitute the known values
Substituting \( v_2 = 2v \):

\[ P + \frac{1}{2} \rho v^2 = P_2 + \frac{1}{2} \rho (4v^2) \]

### Step 4: Rearrange the equation to solve for \( P_2 \)
Rearranging gives:

\[ P_2 = P + \frac{1}{2} \rho v^2 - \frac{1}{2} \rho (4v^2) \]

This simplifies to:

\[ P_2 = P + \frac{1}{2} \rho v^2 - 2 \rho v^2 \]

\[ P_2 = P - \frac{3}{2} \rho v^2 \]

### Step 5: Substitute the density value
Now substituting \( \rho = 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 \):

\[ P_2 = P - \frac{3}{2} (1) v^2 \]

### Final Result
Thus, the pressure at the second point where the speed of the flow is \( 2v \) is:

\[ P_2 = P - \frac{3}{2} v^2 \]