Write dimensions of each quantity

We use:

• Moment of inertia: $[I] = M L^2$
• Force: $[F] = M L T^{-2}$
• Velocity: $[v] = L T^{-1}$
• Work: $[W] = M L^2 T^{-2}$
• Length: $[L] = L$

Given quantity

The quantity is $x = \frac{IFv^2}{WL^4}$

Find dimensions of numerator

$[IFv^2] = [I][F][v]^2$

Now, $[v]^2 = (L T^{-1})^2 = L^2 T^{-2}$

So, $[IFv^2] = (M L^2)(M L T^{-2})(L^2 T^{-2})$ $= M^2 L^5 T^{-4}$

Find dimensions of denominator

$[WL^4] = [W][L]^4$ $= (M L^2 T^{-2})(L^4)$ $= M L^6 T^{-2}$

Find dimensions of $x$

$[x] = \frac{M^2 L^5 T^{-4}}{M L^6 T^{-2}}$ $= M^{2-1} L^{5-6} T^{-4-(-2)}$ $= M L^{-1} T^{-2}$

Compare with given options

A: Coefficient of viscosity

$[\eta] = M L^{-1} T^{-1}$ Not same.

B: Force constant

From $F = kx$, $[k] = \frac{[F]}{[x]} = \frac{M L T^{-2}}{L} = M T^{-2}$ Not same.

C: Energy density

Energy density = energy/volume $\frac{M L^2 T^{-2}}{L^3} = M L^{-1} T^{-2}$ Same.

D: Planck's constant

$[h] = \text{energy} \times \text{time} = M L^2 T^{-1}$ Not same.

Final answer

The dimensional formula of $x$ matches energy density.

Hence, the correct option is C.