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Step-by-Step Solution
Step 1: Write down the given expression
We have the quantity
$$
x = \frac{I \, F \, v^2}{W \, L^4}.
$$
Step 2: Note the dimensional formulas of each symbol
Moment of inertia $I$ has dimensions $[M L^2]$ (since $I = m r^2$ for mass $m$ and radius $r$).
Force $F$ has dimensions $[M L T^{-2}]$ (from Newton’s second law: $F = ma$).
Velocity $v$ has dimensions $[L T^{-1}]$.
Work $W$ has dimensions $[M L^2 T^{-2}]$ (same as energy, since $W = F \cdot s$).
Length $L$ obviously has dimensions $[L]$.
Step 3: Substitute the dimensions into the expression
Substituting the dimensions of $I$, $F$, $v$, $W$, and $L$ into
$$
x = \frac{I \, F \, v^2}{W \, L^4}
$$
we get:
$$
[x]
= \frac{[M L^2] \times [M L T^{-2}] \times ([L T^{-1}])^2}{[M L^2 T^{-2}] \times [L]^4}.
$$
Step 4: Simplify the dimensions
Let us simplify step by step:
1. Dimensions in the numerator:
$$
[M L^2] \times [M L T^{-2}] \times [L^2 T^{-2}]
= [M \cdot M] \, [L^2 \cdot L \cdot L^2] \, [T^{-2} \cdot T^{-2}]
= [M^2] \, [L^5] \, [T^{-4}].
$$
2. Dimensions in the denominator:
$$
[M L^2 T^{-2}] \times [L^4]
= [M] \, [L^2 \cdot L^4] \, [T^{-2}]
= [M] \, [L^6] \, [T^{-2}].
$$
3. Hence,
$$
[x]
= \frac{[M^2] \, [L^5] \, [T^{-4}]}{[M] \, [L^6] \, [T^{-2}]}
= [M^{2-1}] \, [L^{5-6}] \, [T^{-4 + 2}]
= [M^1] [L^{-1}] [T^{-2}].
$$
Thus
$$
[x] = [ M L^{-1} T^{-2} ].
$$
Step 5: Identify the physical quantity
The dimensions $[M L^{-1} T^{-2}]$ match those of energy density, because energy density is energy per unit volume. Energy itself has dimensions $[M L^2 T^{-2}]$, and dividing by volume ($[L^3]$) yields $[M L^{-1} T^{-2}]$.
Final Answer
The dimensional formula for $x$ is the same as that of energy density.
