An incompressible, ideal fluid flows steadily through a horizontal...

For the steady flow of an incompressible fluid, the volume flow rate $Q$ is constant.

According to the equation of continuity:
$Q = A v = \pi r^2 v$
$\Rightarrow v = \dfrac{Q}{\pi r^2}$

The kinetic energy per unit volume ($K$) is given by:
$K = \dfrac{1}{2}\rho v^2$

Substituting the expression for $v$:
$K = \dfrac{1}{2}\rho \left(\dfrac{Q}{\pi r^2}\right)^2 = \dfrac{\rho Q^2}{2\pi^2 r^4}$

Since $\rho$ and $Q$ are constants for a given steady flow, we get:
$K \propto \dfrac{1}{r^4}$ or $K \propto r^{-4}$

Thus, the graph between $K$ and $r$ will be a non-linear curve representing an inverse fourth-power relationship.

Answer: A non-linear curve representing $K \propto r^{-4}$