A fluid is flowing through a horizontal pipe of varying cross-section, with speed$vm{s}^{-1}$ at a point where the pressure is $Ppascal$. At another point where pressure is $P/2pascal$ its speed is $Vm{s}^{-1}$. If the density of the fluid is $\rho$ in $kg{m}^{-3}$ and the flow is streamlined, then $V$ is equal to:

The correct option is B

$\sqrt{\left[\right(P/\rho )+v^2]}$

Step 1: Given Data

Speed$=vm{s}^{-1}$

Pressure $=Ppascal$

Density of the fluid $=\rho kg{m}^{-3}$

At another point

Pressure $=P/2pascal$

Speed$=Vm{s}^{-1}$

Step 2: Find $V$

Bernoulli's equation formula is a relation between pressure, kinetic energy, and gravitational potential energy of a fluid in a container.

By Bernoulli’s equation,

$P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$

Here,

$\rho =$density of fluid

$g=$acceleration due to gravity

$P_1=$pressure at one point

$P_2=$pressure at other point

$h_1=$elevation of one point

$h_2=$elevation of other point

$v_1=$velocity at one point

$v_2=$velocity at other point

Since, pipe is horizontal, elevation of both point is same and potential energy will be same at both point and it will cancel out upon equating.

$$\begin{gathered} P+(1/2)\rho v^2=[P/2]+[1/2]\rho V^2 \\ V=\sqrt{\left[\right(P/\rho )+v^2]} \end{gathered}$$

Hence, option B is correct.