A quantity x is given by (IF v^{2} / WL^{4}) | vedclass

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(C) Given the expression: $x = \frac{I F v^{2}}{W L^{4}}$Dimensions of the quantities are:$I = [M L^{2}]$$F = [M L T^{-2}]$$v = [L T^{-1}]$$W = [M L^{

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Energy density

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(C) Given the expression: $x = \frac{I F v^{2}}{W L^{4}}$Dimensions of the quantities are:$I = [M L^{2}]$$F = [M L T^{-2}]$$v = [L T^{-1}]$$W = [M L^{2} T^{-2}]$$L = [L]$Substituting these into the expression for $x$:$[x] = \frac{[M L^{2}] [M L T^{-2}] [L T^{-1}]^{2}}{[M L^{2} T^{-2}] [L]^{4}}$$[x] = \frac{[M^{2} L^{3} T^{-2}] [L^{2} T^{-2}]}{[M L^{6} T^{-2}]}$$[x] = \frac{[M^{2} L^{5} T^{-4}]}{[M L^{6} T^{-2}]}$$[x] = [M L^{-1} T^{-2}]$Now,checking the dimensions of Energy density:Energy density = $\frac{	ext{Energy}}{	ext{Volume}} = \frac{[M L^{2} T^{-2}]}{[L^{3}]} = [M L^{-1} T^{-2}]$Since the dimensions of $x$ match the dimensions of Energy density,the correct option is $C$.

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

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