Of the following quantities, which one has dimensions different from the remaining three
Energy per unit volume
Force per unit area
Product of voltage and charge per unit volume
Angular momentum per unit mass
Of the following quantities, which one has dimensions different from the remaining three
Energy per unit volume = $\frac{{[M{L^2}{T^{ - 2}}]}}{{[{L^3}]}} = [M{L^{ - 1}}{T^{ - 2}}]$
Force per unit area = $\frac{{[ML{T^{ - 2}}]}}{{[{L^2}]}} = [M{L^{ - 1}}{T^{ - 2}}]$
Product of voltage and charge per unit volume $ = \frac{{V \times Q}}{{{\rm{Volume}}}} = \frac{{VIt}}{{{\rm{Volume}}}} = \frac{{{\rm{Power}} \times {\rm{Time}}}}{{{\rm{Volume}}}}$
$ \Rightarrow $ $\frac{{[M{L^2}{T^{ - 3}}]\,[T]}}{{[{L^3}]}} = [M{L^{ - 1}}{T^{ - 2}}]$
Angular momentum per unit mass = $\frac{{[M{L^2}{T^{ - 1}}]}}{{[M]}} = [{L^2}{T^{ - 1}}]$
So angular momentum per unit mass has different dimension.
Force per unit area = $\frac{{[ML{T^{ - 2}}]}}{{[{L^2}]}} = [M{L^{ - 1}}{T^{ - 2}}]$
Product of voltage and charge per unit volume $ = \frac{{V \times Q}}{{{\rm{Volume}}}} = \frac{{VIt}}{{{\rm{Volume}}}} = \frac{{{\rm{Power}} \times {\rm{Time}}}}{{{\rm{Volume}}}}$
$ \Rightarrow $ $\frac{{[M{L^2}{T^{ - 3}}]\,[T]}}{{[{L^3}]}} = [M{L^{ - 1}}{T^{ - 2}}]$
Angular momentum per unit mass = $\frac{{[M{L^2}{T^{ - 1}}]}}{{[M]}} = [{L^2}{T^{ - 1}}]$
So angular momentum per unit mass has different dimension.
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