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Which of the following is dimensionally correct?
(A) Pressure = Energy per unit volume
(B) Pressure = Energy per unit area
(C) Pressure = Momentum per unit volume per unit time
(D) Pressure = Force per unit volume

Last updated date: 11th Jun 2024
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Hint: We can use the definition of terms like Pressure, Energy, Momentum to define it in terms of fundamental units and then compare the two sides. We can also compare the units to check whether the statement is correct or not.
Formula Used:
\[\Pr essure = \dfrac{{Force}}{{Area}}\]

Complete step by step answer
Pressure is defined as the force per unit area, so we have the dimensions of pressure as
\[\left[ {\Pr essure} \right] = \dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]\], this is the dimension of left-hand side of the options. Now we will see the dimensions of the right hand-side of each option.
For option A $\left[ {\dfrac{{Energy}}{{Volume}}} \right] = \left[ {\dfrac{{M{L^2}{T^{ - 2}}}}{{{L^3}}}} \right] = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$which is same as the dimension of pressure hence it is correct.
For option B $\left[ {\dfrac{{Energy}}{{Area}}} \right] = \left[ {\dfrac{{M{L^2}{T^{ - 2}}}}{{{L^2}}}} \right] = \left[ {M{L^0}{T^{ - 2}}} \right]$which is not the dimension of pressure but the dimension of surface tension and surface energy, hence it is incorrect.
For option C $\left[ {\dfrac{{Momentum}}{{Volume \times time}}} \right] = \left[ {\dfrac{{ML{T^{ - 1}}}}{{{L^3} \times T}}} \right] = \left[ {M{L^{ - 2}}{T^{ - 2}}} \right]$which is not the same as the dimension of pressure hence it is incorrect.
For option D $\left[ {\dfrac{{Force}}{{Volume}}} \right] = \left[ {\dfrac{{ML{T^{ - 2}}}}{{{L^3}}}} \right] = \left[ {M{L^{ - 2}}{T^{ - 2}}} \right]$which is not same as the dimension of pressure hence it is incorrect.

The correct answer is A. Pressure = Energy per unit volume

Additional information
Dimensions can be used for many purposes like checking the correctness of an equation (by principle of homogeneity of dimensions), converting the units from one system to another, forming relations among quantities. Few shortcomings of dimensional analysis are we cannot analyze constants (dimensionless) or trigonometrical terms or exponential terms.

NoteIf the option had been Pressure is equal to momentum per unit area per unit time, then the statement would have been correct.