Ans: D
Let $w = \dfrac{k\sqrt{u}}{v^2}$, where $k$ is a non-zero constant.
Let $w = \dfrac{k\sqrt{u}}{v^2}$, where $k$ is a non-zero constant.
$\begin{array}{rcl}
w & = & \dfrac{k\sqrt{u}}{v^2} \\
w^2 & = & \dfrac{k^2 u}{v^4} \\
k^2 & = & \dfrac{w^2v^4}{u}
\end{array}$
Since $k$ is a constant, then $k^2$ is also a constant.
Hence, $\dfrac{w^2v^4}{u}$ must be a constant.
