
A material has Poisson’s ratio 0.3. If a uniform rod of it suffers a longitudinal strain of \[3 \times {10^{ - 3}}\], what will be the percentage increase in volume?
Answer
126.9k+ views
Hint: In this question we have to find the value of increase in the volume of the material. The Poisson’s ratio and longitudinal strain of the material are given. To find the increase in volume we will use the formula for Poisson’s ratio,
${\text{Poisson's ratio(}}\sigma {\text{)}} = - \left( {\dfrac{{\Delta R/R}}{{\Delta L/L}}} \right)$
Complete step by step solution:
Given,
$\Rightarrow \dfrac{{\Delta L}}{L} = 3 \times {10^{ - 3}}$
$\Rightarrow {\text{Poisson's ratio(}}\sigma {\text{)}} = - \left( {\dfrac{{\Delta R/R}}{{\Delta L/L}}} \right)$
$\Rightarrow 0.3 = - \left( {\dfrac{{\Delta R/R}}{{3 \times {{10}^{ - 3}}}}} \right)$
$\Rightarrow \dfrac{{\Delta R}}{R} = - 0.3 \times 3 \times {10^{ - 3}}$
$\Rightarrow \dfrac{{\Delta R}}{R} = - 0.9 \times {10^{ - 3}}$
Volume of rod is $V = \pi {R^2}L$
To find an increase in volume we will convert it in this form.
$\Rightarrow \dfrac{{\Delta V}}{V} = 2\dfrac{{\Delta R}}{R} + \dfrac{{\Delta L}}{L}$
$\Rightarrow \dfrac{{\Delta V}}{V} = \left( { - 2 \times 0.9 \times {{10}^{ - 3}} + 3 \times {{10}^{ - 3}}} \right)$
$\Rightarrow \dfrac{{\Delta V}}{V} = 1.2 \times {10^{ - 3}}$
Now, we will find the percentage increase in volume
$\Rightarrow \dfrac{{\Delta V}}{V} \times 100 = 1.2 \times {10^{ - 3}} \times 100$
$\Rightarrow \dfrac{{\Delta V}}{V} \times 100 = 0.12\% $
Hence, from the above calculation we have found the value of percentage increase in volume and it comes out to be, $\dfrac{{\Delta V}}{V} \times 100 = 0.12\% $.
Additional Information:
Strain is a measure of how much a body has been deformed or stretched. Strain in a body occurs when a force is applied on it. This is a unit less quantity. Strain in a body is given by following formula;
$strain = \dfrac{{extension}}{{length}}$
If the extension in length is $\Delta L$and the total length of the body is $L$, then strain is given by following formula;
$strain = \dfrac{{\Delta L}}{L}$
There are three types of stain-
Longitudinal strain $\dfrac{{\Delta L}}{L}$
Shearing strain $\dfrac{{\Delta L}}{L}$
Volumetric strain $\dfrac{{\Delta V}}{V}$
Note: Poisson’s ratio is a measure of deformation of a material in different directions perpendicular to the direction of the force applied. In other words, a Poisson’s ratio is the ratio of the transverse strain to the longitudinal strain. It is represented by $\sigma $. Since, it is a ratio so it does not have any dimension. It is a scalar quantity.
${\text{Poisson's ratio(}}\sigma {\text{)}} = - \left( {\dfrac{{Transverse{\text{ strain}}}}{{Longitudinal{\text{ strain}}}}} \right)$
The deformation in the material in different directions perpendicular to the direction of force applied on the material is also known as Poisson’s effect.
${\text{Poisson's ratio(}}\sigma {\text{)}} = - \left( {\dfrac{{\Delta R/R}}{{\Delta L/L}}} \right)$
Complete step by step solution:
Given,
$\Rightarrow \dfrac{{\Delta L}}{L} = 3 \times {10^{ - 3}}$
$\Rightarrow {\text{Poisson's ratio(}}\sigma {\text{)}} = - \left( {\dfrac{{\Delta R/R}}{{\Delta L/L}}} \right)$
$\Rightarrow 0.3 = - \left( {\dfrac{{\Delta R/R}}{{3 \times {{10}^{ - 3}}}}} \right)$
$\Rightarrow \dfrac{{\Delta R}}{R} = - 0.3 \times 3 \times {10^{ - 3}}$
$\Rightarrow \dfrac{{\Delta R}}{R} = - 0.9 \times {10^{ - 3}}$
Volume of rod is $V = \pi {R^2}L$
To find an increase in volume we will convert it in this form.
$\Rightarrow \dfrac{{\Delta V}}{V} = 2\dfrac{{\Delta R}}{R} + \dfrac{{\Delta L}}{L}$
$\Rightarrow \dfrac{{\Delta V}}{V} = \left( { - 2 \times 0.9 \times {{10}^{ - 3}} + 3 \times {{10}^{ - 3}}} \right)$
$\Rightarrow \dfrac{{\Delta V}}{V} = 1.2 \times {10^{ - 3}}$
Now, we will find the percentage increase in volume
$\Rightarrow \dfrac{{\Delta V}}{V} \times 100 = 1.2 \times {10^{ - 3}} \times 100$
$\Rightarrow \dfrac{{\Delta V}}{V} \times 100 = 0.12\% $
Hence, from the above calculation we have found the value of percentage increase in volume and it comes out to be, $\dfrac{{\Delta V}}{V} \times 100 = 0.12\% $.
Additional Information:
Strain is a measure of how much a body has been deformed or stretched. Strain in a body occurs when a force is applied on it. This is a unit less quantity. Strain in a body is given by following formula;
$strain = \dfrac{{extension}}{{length}}$
If the extension in length is $\Delta L$and the total length of the body is $L$, then strain is given by following formula;
$strain = \dfrac{{\Delta L}}{L}$
There are three types of stain-
Longitudinal strain $\dfrac{{\Delta L}}{L}$
Shearing strain $\dfrac{{\Delta L}}{L}$
Volumetric strain $\dfrac{{\Delta V}}{V}$
Note: Poisson’s ratio is a measure of deformation of a material in different directions perpendicular to the direction of the force applied. In other words, a Poisson’s ratio is the ratio of the transverse strain to the longitudinal strain. It is represented by $\sigma $. Since, it is a ratio so it does not have any dimension. It is a scalar quantity.
${\text{Poisson's ratio(}}\sigma {\text{)}} = - \left( {\dfrac{{Transverse{\text{ strain}}}}{{Longitudinal{\text{ strain}}}}} \right)$
The deformation in the material in different directions perpendicular to the direction of force applied on the material is also known as Poisson’s effect.
Recently Updated Pages
JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key

Classification of Elements and Periodicity in Properties Chapter For JEE Main Chemistry

JEE Main 2023 (January 25th Shift 1) Maths Question Paper with Answer Key

JEE Main 2023 (January 24th Shift 2) Chemistry Question Paper with Answer Key

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

JEE Main Login 2045: Step-by-Step Instructions and Details

Class 11 JEE Main Physics Mock Test 2025

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
