An ideal fluid flows in a non-uniform cross-sectional tube AB as shown in the figure. If $A_{1}$ and $A_{2}$ is the area of cross-section and $E_{1}$ and $E_{2}$ is the kinetic energy per unit volume of the fluid at A and B, respectively, then the ratio $E_{1} / E_{2}$ is

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TestHub · Accepted Answer

From the equation of continuity for an incompressible ideal fluid, the volume flow rate is constant:

$A_{1} v_{1} = A_{2} v_{2}$

$\Rightarrow \frac{v _{1}}{v _{2}} = \frac{A _{2}}{A _{1}}$

The kinetic energy per unit volume of the fluid is given by $E = \frac{1}{2} ρ v^{2}$ , where $ρ$ is the density of the fluid.

The ratio of kinetic energy per unit volume at points A and B is:

$\frac{E _{1}}{E _{2}} = \frac{\frac{1}{2} ρ v _{1}^{2}}{\frac{1}{2} ρ v _{2}^{2}} = (\frac{v _{1}}{v _{2}})^{2}$

Substituting the ratio of velocities from the continuity equation:

$\frac{E _{1}}{E _{2}} = (\frac{A _{2}}{A _{1}})^{2} = \frac{A _{2}^{2}}{A _{1}^{2}}$

Answer: $\frac{A _{2}^{2}}{A _{1}^{2}}$

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