1. Use Bernoulli’s equation for streamline flow in a horizontal pipe:

$P + \frac{1}{2}\rho v^2 = \text{constant}$

Since the pipe is horizontal, the gravitational term $\rho g h$ is same at both points and cancels out.

2. Apply Bernoulli’s equation at the two points:

At point 1:

• Pressure $= P$
• Speed $= v$

At point 2:

• Pressure $= \dfrac{P}{2}$
• Speed $= V$

So,

$P + \frac{1}{2}\rho v^2 = \frac{P}{2} + \frac{1}{2}\rho V^2$

3. Rearrange to find $V$:

Subtract $\dfrac{P}{2}$ from both sides:

$\frac{P}{2} + \frac{1}{2}\rho v^2 = \frac{1}{2}\rho V^2$

Multiply by $2$:

$P + \rho v^2 = \rho V^2$

Divide by $\rho$:

$V^2 = \frac{P}{\rho} + v^2$

Hence,

$V = \sqrt{\frac{P}{\rho} + v^2}$

4. Match with options:

This corresponds to Option B.