![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
The coefficient of linear expansion of brass and steel are ${\alpha _1}$ and ${\alpha _2}$. If we take a brass rod of length ${l_1}$ and steel rod of length ${l_2}$ at $0^\circ C$ and their difference in length $(\,{l_2}\, - \,{l_{1\,}})$ will remain the same at all temperatures if
(A) \[{\alpha _1}{l_2} = {\alpha _2}{l_1}\]
(B) ${\alpha _1}{l_2}^2 = {\alpha _2}{l_1}^2$
(C) ${\alpha _1}^2{l_2} = {\alpha _2}^2{l_1}$
(D) ${\alpha _1}{l_1} = {\alpha _2}{l_2}$
Answer
124.8k+ views
Hint: Lengths of rods made up of brass and steel are ${l_1}$ and ${l_2}$ and their coefficient of linear expansion are ${\alpha _1}$ and ${\alpha _2}$ at $0^\circ C$ temperature. The length of a metal rod will increase from ${l_1}$ to ${l_1}\,(\,1 + {\alpha _1}\,)$. Here $\alpha $ is the coefficient of linear expansion. We will then have to check each option to find out which equation is correct.
Complete step by step answer
Coefficient of linear expansion $(\alpha )$:
For $1^\circ $ rise in temperature, the ratio of the increase in length of wire to its actual length of wire is known as coefficient of linear expansion.
$\alpha = \dfrac{{\Delta l}}{{\Delta T}}$
Reason for expansion:
When heat is supplied to the object, intermolecular force of attraction between the molecules decreases and molecules start to separate. This results in increase in spacing between the molecules and therefore results in increase in size and change in shape of substance. This is the reason why length, area or volume of substance expands.
In this case we will study about length expansion only.
At temperature T,
Length of brass rod $ = {l_1}\,(\,1 + {\alpha _1}\,T\,)$
Length of steel rod $ = {l_2}\,(\,1 + {\alpha _2}\,T\,)$
Difference in lengths $ = {l_1}\,(\,1 + {\alpha _1}\,T\,) - {l_2}\,(\,1 + {\alpha _2}\,T\,)$
$ = (\,{l_2}\, - \,{l_{1\,}}) + T\,(\,{l_1}{\alpha _1} - \,{l_2}{\alpha _2}\,)$
Length will be independent of temperature only when coefficient of temperature will be equal to zero.
$(\,{l_1}{\alpha _1} - \,{l_2}{\alpha _2}\,)\, = \,0$
${l_1}{\alpha _1} = \,{l_2}{\alpha _2}\,$
Since D is a satisfying option. Hence it is the correct relation.
Note
If $(\,{l_2}\, - \,{l_{1\,}})$ equals to zero then both the lengths will become equal which cannot be possible. So, it is a wrong understanding.
According to the concept coefficient of linear expansion is multiplied by length of same material used then option A, B and C are not possible.
Hence, we are left with only one option I.e. D. so the correct solution is D.
Complete step by step answer
Coefficient of linear expansion $(\alpha )$:
For $1^\circ $ rise in temperature, the ratio of the increase in length of wire to its actual length of wire is known as coefficient of linear expansion.
$\alpha = \dfrac{{\Delta l}}{{\Delta T}}$
Reason for expansion:
When heat is supplied to the object, intermolecular force of attraction between the molecules decreases and molecules start to separate. This results in increase in spacing between the molecules and therefore results in increase in size and change in shape of substance. This is the reason why length, area or volume of substance expands.
In this case we will study about length expansion only.
At temperature T,
Length of brass rod $ = {l_1}\,(\,1 + {\alpha _1}\,T\,)$
Length of steel rod $ = {l_2}\,(\,1 + {\alpha _2}\,T\,)$
Difference in lengths $ = {l_1}\,(\,1 + {\alpha _1}\,T\,) - {l_2}\,(\,1 + {\alpha _2}\,T\,)$
$ = (\,{l_2}\, - \,{l_{1\,}}) + T\,(\,{l_1}{\alpha _1} - \,{l_2}{\alpha _2}\,)$
Length will be independent of temperature only when coefficient of temperature will be equal to zero.
$(\,{l_1}{\alpha _1} - \,{l_2}{\alpha _2}\,)\, = \,0$
${l_1}{\alpha _1} = \,{l_2}{\alpha _2}\,$
Since D is a satisfying option. Hence it is the correct relation.
Note
If $(\,{l_2}\, - \,{l_{1\,}})$ equals to zero then both the lengths will become equal which cannot be possible. So, it is a wrong understanding.
According to the concept coefficient of linear expansion is multiplied by length of same material used then option A, B and C are not possible.
Hence, we are left with only one option I.e. D. so the correct solution is D.
Recently Updated Pages
JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Algebra Important Concepts and Tips for Exam Preparation
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
States of Matter Chapter For JEE Main Chemistry
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Uniform Acceleration - Definition, Equation, Examples, and FAQs
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Mass and Weight
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The ratio of the diameters of two metallic rods of class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Class 11 JEE Main Physics Mock Test 2025
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
![arrow-right](/cdn/images/seo-templates/arrow-right.png)