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1.Units, Dimensions and Measurement
medium
Of the following quantities, which one has dimensions different from the remaining three
AEnergy per unit volume
BForce per unit area
CProduct of voltage and charge per unit volume
DAngular momentum per unit mass
(AIIMS-1987) (AIPMT-1989)
Solution
(d) 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.
Standard 11
Physics
