An incompressible fluid (ideal fluid) is flowing through non uniform cross-sectional tube PQ as shown in the figure from end P to end Q. If $K_{P}$ and $K_{Q}$ are the kinetic energy per unit volume of the fluid at end P and end Q respectively, then

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From the given figure, the cross-sectional area of the tube at end P is less than the cross-sectional area at end Q, so $A_{P} < A_{Q}$ .

According to the equation of continuity for an incompressible fluid, the product of cross-sectional area and velocity remains constant:

$A_{P} v_{P} = A_{Q} v_{Q}$

Since $A_{P} < A_{Q}$ , the velocity of the fluid at P is greater than the velocity at Q, which gives $v_{P} > v_{Q}$ .

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

Since $v_{P} > v_{Q}$ , it follows that $v_{P}^{2} > v_{Q}^{2}$ .

Therefore, $\frac{1}{2} ρ v_{P}^{2} > \frac{1}{2} ρ v_{Q}^{2}$ , which implies $K_{P} > K_{Q}$ .

Answer: $K_{P} > K_{Q}$

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