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
Answer
257.7k+ views
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.
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.
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

JEE Main Colleges 2026: Complete List of Participating Institutes

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

