A quantity $$x$$ is given by $$(1Fv^2/WL^4)$$ in terms of moment of inertia $$I$$, force $$F$$, velocity $$v$$, work $$W$$ and length $$L$$. The dimensional formula for $$x$$ is same as that of:

We are told that the physical quantity $$x$$ is expressed as

$$x=\dfrac{I\,F\,v^{2}}{W\,L^{4}}$$

where the symbols have the following usual meanings:

$$I=$$ moment of inertia,   $$F=$$ force,   $$v=$$ velocity,   $$W=$$ work,   $$L=$$ length.

Our task is to obtain the dimensional formula of $$x$$ and then compare it with the dimensions of the physical quantities listed in the options.

First, we write down the standard dimensional formulas (using $$[M]$$ for mass, $$[L]$$ for length and $$[T]$$ for time):

• Moment of inertia $$I$$ is mass × length$$^{2}$$, so $$[I]=[M^{1}L^{2}T^{0}]$$.

• Force $$F$$ follows Newton’s second law $$F=ma$$, giving $$[F]=[M^{1}L^{1}T^{-2}]$$.

• Velocity $$v$$ is length per time, therefore $$[v]=[M^{0}L^{1}T^{-1}]$$. Consequently $$[v^{2}]=[M^{0}L^{2}T^{-2}]$$.

• Work $$W$$ equals force × displacement, hence $$[W]=[M^{1}L^{2}T^{-2}]$$.

• Length $$L$$ of course has $$[L]=[M^{0}L^{1}T^{0}]$$, so $$[L^{4}]=[M^{0}L^{4}T^{0}]$$.

Now we substitute each of these into the defining expression for $$x$$:

Numerator dimensions:

$$[I\,F\,v^{2}]=[I]\,[F]\,[v^{2}]= \big[M^{1}L^{2}T^{0}\big]\, \big[M^{1}L^{1}T^{-2}\big]\, \big[M^{0}L^{2}T^{-2}\big].$$

Multiplying the like powers of $$M,\,L,\,T$$ one by one, we obtain

• For mass: $$M^{1}\times M^{1}\times M^{0}=M^{2}$$,

• For length: $$L^{2}\times L^{1}\times L^{2}=L^{5}$$,

• For time: $$T^{0}\times T^{-2}\times T^{-2}=T^{-4}$$.

So, $$[I\,F\,v^{2}]=[M^{2}L^{5}T^{-4}]$$.

Denominator dimensions:

$$[W\,L^{4}]=[W]\,[L^{4}]= \big[M^{1}L^{2}T^{-2}\big]\, \big[M^{0}L^{4}T^{0}\big] = [M^{1}L^{6}T^{-2}].$$

We now divide the numerator by the denominator to get the dimensions of $$x$$:

$$[x]=\dfrac{[M^{2}L^{5}T^{-4}]}{[M^{1}L^{6}T^{-2}]}.$$

Carrying out the division term by term, we subtract the powers:

• Mass exponent: $$2-1=1 \;\Rightarrow\; M^{1},$$

• Length exponent: $$5-6=-1 \;\Rightarrow\; L^{-1},$$

• Time exponent: $$-4-(-2)=-2 \;\Rightarrow\; T^{-2}.$$

Hence, $$[x]=[M^{1}L^{-1}T^{-2}].$$

Finally, we match this result with the dimensions of the quantities in the options:

(A) Planck’s constant $$h$$ has $$[M^{1}L^{2}T^{-1}]$$ - does not match.

(B) Force constant (spring constant $$k=F/\Delta L$$) has $$[M^{1}T^{-2}]$$ - does not match.

(C) Energy density (energy per unit volume) is $$\dfrac{\text{energy}}{\text{volume}}$$.

The dimensional formula of energy is $$[M^{1}L^{2}T^{-2}]$$ and of volume is $$[L^{3}]$$. Therefore

$$\dfrac{[M^{1}L^{2}T^{-2}]}{[L^{3}]}=[M^{1}L^{-1}T^{-2}],$$

which exactly matches $$[x]$$.

(D) Coefficient of viscosity $$\eta$$ has $$[M^{1}L^{-1}T^{-1}]$$ - does not match.

So the only option with the same dimensional formula as $$x$$ is energy density.

Hence, the correct answer is Option C.