
Which of the following is not a unit of young’s modulus?
A. ${\text{N }}{{\text{m}}^{ - 1}}$
B. ${\text{N }}{{\text{m}}^{ - 2}}$
C. ${\text{Dyne c}}{{\text{m}}^{ - 2}}$
D. ${\text{Mega pascal}}$
Answer
217.5k+ views
Hint: In this question, we have to find out the unit which does not belong to the units of young’s modulus. Using the definition of young’s modulus, we frame the formula of it and after that, we write the units of young’s modulus in SI units, and CGS units and check the given options which one does not belong to it.
Complete step by step solution:
As we know, young’s modulus is defined as the ratio of stress and strain. It is written as;
$Y = \dfrac{{{\text{Stress}}}}{{{\text{Strain}}}} - - - - (1)$
To frame the unit of young’s modulus, let us find the units of stress and strain, we get;
Stress is defined as when we apply the external force to the object per unit area of the object. It is written as;
$\sigma ({\text{stress) = }}\dfrac{F}{A}$
Here, $F$ is the force, and $A$ is the cross-section area of an object.
As we know that the unit of force is newton$(N)$and the unit of area is ${m^2}$. Therefore unit of stress is written as,
${\text{Stress(}}\sigma {\text{) = }}\dfrac{{\text{N}}}{{{{\text{m}}^{\text{2}}}}} \\
\Rightarrow {\text{Stress(}}\sigma {\text{) = N}}{\text{.}}{{\text{m}}^{ - 2}} \\ $
Strain is defined as the force applied per unit area to deform the object. It is dimensionless quantity.Now, in equation $(1)$ we can put the stress and strain unit we get that,
$Y = {\text{N}}{\text{.}}{{\text{m}}^{ - 2}}$
Now ${\text{N}}{\text{.}}{{\text{m}}^{ - 2}}$ is the unit of young’s modulus in SI units and CGS unit we have,
$Y = {\text{Dyne}}{\text{.c}}{{\text{m}}^{ - 2}}$or ${\text{Mega pascal}}$.
Therefore, we are not having the unit ${\text{N}}{\text{.}}{{\text{m}}^{ - 1}}$ in young’s modulus.
Hence, Option A is correct.
Note: Sometimes students get confused between stress and strain. Stress is defined as the force per unit area and strain is defined as the ratio of change in length to the original length. It is a dimensionless quantity and always takes care of taking the unit in the SI unit system or CGS unit system.
Complete step by step solution:
As we know, young’s modulus is defined as the ratio of stress and strain. It is written as;
$Y = \dfrac{{{\text{Stress}}}}{{{\text{Strain}}}} - - - - (1)$
To frame the unit of young’s modulus, let us find the units of stress and strain, we get;
Stress is defined as when we apply the external force to the object per unit area of the object. It is written as;
$\sigma ({\text{stress) = }}\dfrac{F}{A}$
Here, $F$ is the force, and $A$ is the cross-section area of an object.
As we know that the unit of force is newton$(N)$and the unit of area is ${m^2}$. Therefore unit of stress is written as,
${\text{Stress(}}\sigma {\text{) = }}\dfrac{{\text{N}}}{{{{\text{m}}^{\text{2}}}}} \\
\Rightarrow {\text{Stress(}}\sigma {\text{) = N}}{\text{.}}{{\text{m}}^{ - 2}} \\ $
Strain is defined as the force applied per unit area to deform the object. It is dimensionless quantity.Now, in equation $(1)$ we can put the stress and strain unit we get that,
$Y = {\text{N}}{\text{.}}{{\text{m}}^{ - 2}}$
Now ${\text{N}}{\text{.}}{{\text{m}}^{ - 2}}$ is the unit of young’s modulus in SI units and CGS unit we have,
$Y = {\text{Dyne}}{\text{.c}}{{\text{m}}^{ - 2}}$or ${\text{Mega pascal}}$.
Therefore, we are not having the unit ${\text{N}}{\text{.}}{{\text{m}}^{ - 1}}$ in young’s modulus.
Hence, Option A is correct.
Note: Sometimes students get confused between stress and strain. Stress is defined as the force per unit area and strain is defined as the ratio of change in length to the original length. It is a dimensionless quantity and always takes care of taking the unit in the SI unit system or CGS unit system.
Recently Updated Pages
Ionisation Energy and Ionisation Potential Explained

Addition of Three Vectors: Methods & Examples

Addition of Vectors: Simple Guide for Students

Algebra Made Easy: Step-by-Step Guide for Students

Relations and Functions: Complete Guide for Students

Analytical Method of Vector Addition Explained Simply

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

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

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

Understanding Atomic Structure for Beginners

