
Which one of the following is not a unit of Young’s modulus?
A) $N{m^{ - 1}}$
B) $N{m^{ - 2}}$
C) ${\text{dyne c}}{{\text{m}}^{ - 2}}$
D) Megapascal
Answer
580.2k+ views
Hint: We know that Young’s modulus is a ratio of longitudinal stress by strain. As we apply deforming force there is a change in its configuration. Young’s modulus is a ratio of longitudinal stress to the longitudinal strain. It is denoted by a symbol $Y$.
Complete answer:
Let us Consider a wire of length L and area of cross-section A, which is suspended from a rigid support. Let $\Delta L$ be the increase in the length of the wire due to a load F applied at the other end of the wire.
The Longitudinal stress is given as $\sigma = \dfrac{F}{A}$
Where σ is the longitudinal stress, $F$ is the force applied on the wire and A is the area of the wire.
The Longitudinal strain is given as the, $\varepsilon = \dfrac{{\Delta L}}{L}$
Where ε is a longitudinal strain, $\Delta L$ is equal to change in length, L is equal to the length of the wire.
We know that young’s modulus is equal to the ratio of longitudinal stress and longitudinal strain. Thus the young’s modulus is given by the formula,
Young’ modulus, $Y = \dfrac{\sigma }{\varepsilon }$
Let's put the values of longitudinal stress and strain in the young's modulus equation, we get
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}} = \dfrac{{FL}}{{A\Delta L}}$
Now write SI units of each quantity, we get
$Y = \dfrac{{Nm}}{{{m^2}m}} = \dfrac{N}{{{m^2}}}$
Now we have the SI unit of Young’s modulus as $N{m^{ - 2}}$.
Then this $N{m^{ - 2}}$ can also be written as Pascal.
We know that the CGS unit of Young’ modulus is ${\text{dyne c}}{{\text{m}}^{ - 2}}$.
In given options, the (B),(C) and(D) represents the units of young's modulus in $N{m^{ - 2}}, $${\text{dyne c}}{{\text{m}}^{ - 2}}$ and Mega Pascal respectively.
Option (A) $N{m^{ - 1}}$ is not a unit of Young’s modulus.
$\therefore$ The correct option is (A).
Additional information:
When a body is subjected to deforming force, a restoring force is developed in the body. in the equilibrium position, this internal restoring force is equal in magnitude and opposite in direction to the applied force. The internal restoring force developed per unit area of a deformed body is called stress.
under the action of deforming force, the configuration of a body changes, and the body is said to be strained by the force. When a body is deformed the ratio of change in configuration to the original configuration is called strain. A Strain is a dimensionless quantity and has no units.
The ratio of stress by strain is called the modulus of elasticity. depending upon the type of stress applied and the resulting strain, there are three types of modulus of elasticity.
Note:
Elasticity is a property of a material that regains its original length, volume, and shape after the removal of deforming force.
When a solid is deformed by the application of external force, the atoms or molecules are displaced from its initial position.
Complete answer:
Let us Consider a wire of length L and area of cross-section A, which is suspended from a rigid support. Let $\Delta L$ be the increase in the length of the wire due to a load F applied at the other end of the wire.
The Longitudinal stress is given as $\sigma = \dfrac{F}{A}$
Where σ is the longitudinal stress, $F$ is the force applied on the wire and A is the area of the wire.
The Longitudinal strain is given as the, $\varepsilon = \dfrac{{\Delta L}}{L}$
Where ε is a longitudinal strain, $\Delta L$ is equal to change in length, L is equal to the length of the wire.
We know that young’s modulus is equal to the ratio of longitudinal stress and longitudinal strain. Thus the young’s modulus is given by the formula,
Young’ modulus, $Y = \dfrac{\sigma }{\varepsilon }$
Let's put the values of longitudinal stress and strain in the young's modulus equation, we get
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}} = \dfrac{{FL}}{{A\Delta L}}$
Now write SI units of each quantity, we get
$Y = \dfrac{{Nm}}{{{m^2}m}} = \dfrac{N}{{{m^2}}}$
Now we have the SI unit of Young’s modulus as $N{m^{ - 2}}$.
Then this $N{m^{ - 2}}$ can also be written as Pascal.
We know that the CGS unit of Young’ modulus is ${\text{dyne c}}{{\text{m}}^{ - 2}}$.
In given options, the (B),(C) and(D) represents the units of young's modulus in $N{m^{ - 2}}, $${\text{dyne c}}{{\text{m}}^{ - 2}}$ and Mega Pascal respectively.
Option (A) $N{m^{ - 1}}$ is not a unit of Young’s modulus.
$\therefore$ The correct option is (A).
Additional information:
When a body is subjected to deforming force, a restoring force is developed in the body. in the equilibrium position, this internal restoring force is equal in magnitude and opposite in direction to the applied force. The internal restoring force developed per unit area of a deformed body is called stress.
under the action of deforming force, the configuration of a body changes, and the body is said to be strained by the force. When a body is deformed the ratio of change in configuration to the original configuration is called strain. A Strain is a dimensionless quantity and has no units.
The ratio of stress by strain is called the modulus of elasticity. depending upon the type of stress applied and the resulting strain, there are three types of modulus of elasticity.
Note:
Elasticity is a property of a material that regains its original length, volume, and shape after the removal of deforming force.
When a solid is deformed by the application of external force, the atoms or molecules are displaced from its initial position.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

