
Assertion: Identical springs of steel and copper are equally stretched. More work will be done on the steel spring.
Reason: Steel is more elastic than copper.
A. Both assertion and reason are correct and the reason is the correct explanation of the assertion.
B. Both assertion and reason are true but the reason is not the correct explanation of the assertion.
C. The assertion is correct but the reason is incorrect.
D. Both the assertion and reason are incorrect
Answer
220.8k+ views
Hint: Stress is defined as the force per unit area. And the strain is nothing but a change in volume to the original volume and it is a dimensionless quantity.
Complete step by step solution:
We know that the work done is given by,
\[W = \dfrac{1}{2} \times stress \times strain\]
Since, \[Y = \dfrac{\text{stress}}{\text{strain}}\]
Then, \[ \Rightarrow \text{stress} = \left( {Y \times \text{strain}} \right)\]
Now, substituting the value of stress in above equation we get,
\[W = \dfrac{1}{2} \times \left( {Y \times strain} \right) \times strain\]
\[\Rightarrow W = \dfrac{1}{2} \times Y \times {\varepsilon ^2}\]
Here, you can see that the work done depends only on the young’s modulus and \[\varepsilon \] is the same because the strain is the same for both. The work done will be more if the strain is more for a material.
But we know that young’s modulus of steel is greater than young’s modulus of copper. That is, the work has to be done more on steel to stretch. This shows that the assertion is true.
Since the young’s modulus of steel is greater than the young’s modulus of copper, by this we can say that the steel is more elastic than copper which confirms the reason is also true in this case and the correct explanation of assertion.
Hence, option A is the correct answer.
Note:Young's modulus is the property of a material that tells how easy it can stretch and deform and is defined as the ratio of tensile stress to tensile strain.
Complete step by step solution:
We know that the work done is given by,
\[W = \dfrac{1}{2} \times stress \times strain\]
Since, \[Y = \dfrac{\text{stress}}{\text{strain}}\]
Then, \[ \Rightarrow \text{stress} = \left( {Y \times \text{strain}} \right)\]
Now, substituting the value of stress in above equation we get,
\[W = \dfrac{1}{2} \times \left( {Y \times strain} \right) \times strain\]
\[\Rightarrow W = \dfrac{1}{2} \times Y \times {\varepsilon ^2}\]
Here, you can see that the work done depends only on the young’s modulus and \[\varepsilon \] is the same because the strain is the same for both. The work done will be more if the strain is more for a material.
But we know that young’s modulus of steel is greater than young’s modulus of copper. That is, the work has to be done more on steel to stretch. This shows that the assertion is true.
Since the young’s modulus of steel is greater than the young’s modulus of copper, by this we can say that the steel is more elastic than copper which confirms the reason is also true in this case and the correct explanation of assertion.
Hence, option A is the correct answer.
Note:Young's modulus is the property of a material that tells how easy it can stretch and deform and is defined as the ratio of tensile stress to tensile strain.
Recently Updated Pages
Two discs which are rotating about their respective class 11 physics JEE_Main

A ladder rests against a frictionless vertical wall class 11 physics JEE_Main

Two simple pendulums of lengths 1 m and 16 m respectively class 11 physics JEE_Main

The slopes of isothermal and adiabatic curves are related class 11 physics JEE_Main

A trolly falling freely on an inclined plane as shown class 11 physics JEE_Main

The masses M1 and M2M2 M1 are released from rest Using class 11 physics JEE_Main

Trending doubts
Understanding Uniform Acceleration in Physics

Understanding Atomic Structure for Beginners

Understanding Entropy Changes in Different Processes

Common Ion Effect: Concept, Applications, and Problem-Solving

What Are Elastic Collisions in One Dimension?

Understanding Charging and Discharging of Capacitors

Other Pages
NCERT Solutions for Class 11 Physics Chapter 5 Work Energy And Power 2025-26

Free Radical Substitution and Its Stepwise Mechanism

Understanding Geostationary and Geosynchronous Satellites

NCERT Solutions For Class 11 Physics Chapter 12 Kinetic Theory - 2025-26

Define thermal expansion for alpha beta and gamma A class 11 physics JEE_Main

JEE Advanced 2026 Revision Notes for Practical Organic Chemistry

