![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
A $5m$ long aluminium wire ($Y = 7 \times {10^{10}}\;N/{m^2}$ ) of diameter $3\;mm$ supports a $40\;kg$ mass. In order to have the same elongation in the copper wire ($Y = 12 \times {10^{10}}\;N/{m^2}$) of the same length under the same weight, the diameter should now be (in mm):
A) 1.75
B) 1.5
C) 2.5
D) 5.0
Answer
116.7k+ views
Hint: Stress and Strain are the two terms in Physics that describe the forces causing the deformation of the objects. Deformation is known as the change of the shape of an object by applications of force. Very small forces can also cause deformation. The object experiences it due to external forces; for example, the forces might be like squeezing, squashing, twisting, shearing, ripping, or pulling the objects apart.
Complete step by step answer:
We will now write the relation between $d$ and $E$.
${d^2} \propto \dfrac{1}{E}$
Here, d is the diameter and E is the modulus of elasticity.
We can write the above equations for different states as:
$d_1^2{E_1} = d_2^2{E_2}$
Here subscript 1 represents aluminium and subscript 2 represents copper.
Substitute the known values in the above equation:
$\begin{array}{l}
{3^2} \times 7 \times {10^{10}} = d_2^2 \times 12 \times {10^{10}}
\end{array}$
${d_2} = 2.29\;mm$
Additional Information: The relation between the stress and the strain can be found experimentally for a given material under tensile stress. In a standard test or experiment of tensile properties, a wire or test cylinder is stretched by an external force. The fractional change in length or what is referred to as strain and the external force required to cause the strain are noted. The applied external force is gradually increased step by step and the change in length is again noted. Then, a graph is plotted between the stress (equal in magnitude to the external force per unit area) and the strain.
Note: Strain is the ratio of the amount of deformation experienced by the body in the direction of force applied to the initial sizes of the body. Hooke's law expresses the relationship between stress and strain; it states that the strain in an object is proportionate to the stress applied within the range of elastic limit of that object.
Complete step by step answer:
We will now write the relation between $d$ and $E$.
${d^2} \propto \dfrac{1}{E}$
Here, d is the diameter and E is the modulus of elasticity.
We can write the above equations for different states as:
$d_1^2{E_1} = d_2^2{E_2}$
Here subscript 1 represents aluminium and subscript 2 represents copper.
Substitute the known values in the above equation:
$\begin{array}{l}
{3^2} \times 7 \times {10^{10}} = d_2^2 \times 12 \times {10^{10}}
\end{array}$
${d_2} = 2.29\;mm$
Additional Information: The relation between the stress and the strain can be found experimentally for a given material under tensile stress. In a standard test or experiment of tensile properties, a wire or test cylinder is stretched by an external force. The fractional change in length or what is referred to as strain and the external force required to cause the strain are noted. The applied external force is gradually increased step by step and the change in length is again noted. Then, a graph is plotted between the stress (equal in magnitude to the external force per unit area) and the strain.
Note: Strain is the ratio of the amount of deformation experienced by the body in the direction of force applied to the initial sizes of the body. Hooke's law expresses the relationship between stress and strain; it states that the strain in an object is proportionate to the stress applied within the range of elastic limit of that object.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How to find Oxidation Number - Important Concepts for JEE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How Electromagnetic Waves are Formed - Important Concepts for JEE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electrical Resistance - Important Concepts and Tips for JEE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Average Atomic Mass - Important Concepts and Tips for JEE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Chemical Equation - Important Concepts and Tips for JEE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Login 2045: Step-by-Step Instructions and Details
![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 Chemistry Question Paper with Answer Keys and 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
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
![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)
NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids
![arrow-right](/cdn/images/seo-templates/arrow-right.png)