A fluid is flowing through a horizontal pipe of varying cross - section, with speed `v ms^(-1)` at a point where the pressure is P Pascal .
At another point where pressure is `P/2`
Pascal its speed is `V ms^(-1)`. If the density of the fluid is `rho " kg "m^(-3)` and the flow is streamline, then V is equal to :

Options

A. `sqrt(P/rho+v)`
B. `sqrt((2P)/rho+v^2)`
C. `sqrt(P/(2rho) + v^2)`
D. `sqrt(P/rho+v^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 streamline manner. Since the pipe is horizontal, the height terms can be ignored.

### Step-by-step Solution:

1. **Write down Bernoulli's Equation**:
\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]
Here, \( P_1 \) and \( P_2 \) are the pressures at points 1 and 2, and \( v_1 \) and \( v_2 \) are the velocities at those points.

2. **Assign values from the problem**:
- At point 1: \( P_1 = P \), \( v_1 = v \)
- At point 2: \( P_2 = \frac{P}{2} \), \( v_2 = V \)

3. **Substitute these values into Bernoulli's equation**:
\[ P + \frac{1}{2} \rho v^2 = \frac{P}{2} + \frac{1}{2} \rho V^2 \]

4. **Rearrange the equation**:
\[ P - \frac{P}{2} + \frac{1}{2} \rho v^2 = \frac{1}{2} \rho V^2 \]
This simplifies to:
\[ \frac{P}{2} + \frac{1}{2} \rho v^2 = \frac{1}{2} \rho V^2 \]

5. **Multiply the entire equation by 2 to eliminate the fractions**:
\[ P + \rho v^2 = \rho V^2 \]

6. **Rearrange to solve for \( V^2 \)**:
\[ \rho V^2 = P + \rho v^2 \]
\[ V^2 = \frac{P}{\rho} + v^2 \]

7. **Take the square root to find \( V \)**:
\[ V = \sqrt{\frac{P}{\rho} + v^2} \]

### Final Result:
Thus, the speed \( V \) at the second point is given by:
\[ V = \sqrt{\frac{P}{\rho} + v^2} \]