
The increase in energy of a metal bar of length ‘L’ and cross sectional area ‘A’ when compressed with a load ‘M’ along its length is: (Y= young modulus of the material of metal bar)
A) $\dfrac{{FL}}{{2AY}}$
B) $\dfrac{{{F^2}L}}{{2AY}}$
C) $\dfrac{{FL}}{{AY}}$
D) $\dfrac{{{F^2}{L^2}}}{{2AY}}$
Answer
170.4k+ views
Hint: For the above question Young’s modulus concept is as follows:
Young Modulus is given by the formula:
$Y = \dfrac{\text{longitudinal stress}}{{strain}}$
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}$ , F is the force, A is the area and l is the length and $\Delta L$ is the change in length.
Energy, U is given by $ = \dfrac{1}{2} \times stress \times strain \times volume$
Using the above relations we will calculate the energy of the metal bar.
Complete step by step solution:
Young’s Modulus is a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression .Young’s modulus is the ratio of longitudinal stress to strain.SI unit of Young’s modulus is Pascal.
Now let’s derive the equation for the energy of the metal bar
U$ = \dfrac{1}{2} \times stress \times strain \times volume$(equation for energy)
Young’s modulus is given by:
$stress = \dfrac{F}{A}$ (F is force and A is the area)
Then, strain is given by:
$strain = \dfrac{{stress}}{Y} = \dfrac{F}{{AY}}$ (Y is young modulus here)
Volume of the metal bar is given: AL.
Therefore, energy of the metal rod is given as:
$U = \dfrac{1}{2} \times \dfrac{F}{A} \times (\dfrac{F}{{AY}})AL$ .....................1
Now, we will cancel the common terms in equation 1 to bring out the formula for energy.
$
\Rightarrow U = \dfrac{1}{2} \times \dfrac{F}{A} \times (\dfrac{F}{Y})L \\
\Rightarrow U = \dfrac{1}{2} \times \dfrac{{{F^2}L}}{{AY}} \\
$
Note: Application of Young’s Modulus in our day to day life is in designing a bridge ,one has to keep in mind the traffic load, weight of the bridge ,force of wind .Bridge should also not bend too much nor break. Another application is heavy load lifted by the crane, which depends on the weight of the load and the stress and strain applied by the crane because pressure is equal to the force upon the area.
Young Modulus is given by the formula:
$Y = \dfrac{\text{longitudinal stress}}{{strain}}$
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}$ , F is the force, A is the area and l is the length and $\Delta L$ is the change in length.
Energy, U is given by $ = \dfrac{1}{2} \times stress \times strain \times volume$
Using the above relations we will calculate the energy of the metal bar.
Complete step by step solution:
Young’s Modulus is a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression .Young’s modulus is the ratio of longitudinal stress to strain.SI unit of Young’s modulus is Pascal.
Now let’s derive the equation for the energy of the metal bar
U$ = \dfrac{1}{2} \times stress \times strain \times volume$(equation for energy)
Young’s modulus is given by:
$stress = \dfrac{F}{A}$ (F is force and A is the area)
Then, strain is given by:
$strain = \dfrac{{stress}}{Y} = \dfrac{F}{{AY}}$ (Y is young modulus here)
Volume of the metal bar is given: AL.
Therefore, energy of the metal rod is given as:
$U = \dfrac{1}{2} \times \dfrac{F}{A} \times (\dfrac{F}{{AY}})AL$ .....................1
Now, we will cancel the common terms in equation 1 to bring out the formula for energy.
$
\Rightarrow U = \dfrac{1}{2} \times \dfrac{F}{A} \times (\dfrac{F}{Y})L \\
\Rightarrow U = \dfrac{1}{2} \times \dfrac{{{F^2}L}}{{AY}} \\
$
Note: Application of Young’s Modulus in our day to day life is in designing a bridge ,one has to keep in mind the traffic load, weight of the bridge ,force of wind .Bridge should also not bend too much nor break. Another application is heavy load lifted by the crane, which depends on the weight of the load and the stress and strain applied by the crane because pressure is equal to the force upon the area.
Recently Updated Pages
Molarity vs Molality: Definitions, Formulas & Key Differences

Preparation of Hydrogen Gas: Methods & Uses Explained

Polymers in Chemistry: Definition, Types, Examples & Uses

P Block Elements: Definition, Groups, Trends & Properties for JEE/NEET

Order of Reaction in Chemistry: Definition, Formula & Examples

Hydrocarbons: Types, Formula, Structure & Examples Explained

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Instantaneous Velocity - Formula based Examples for JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
NCERT Solution for Class 11 Physics Chapter 1 Units and Measurements - 2025-26

NCERT Solution for Class 11 Physics Chapter 2 Motion In A Straight Line - 2025-26

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
