Young’s modulus of brass and steel are \[10 \times {10^{10}}N/m\] and \[2 \times {10^{11}}N/m\], respectively. A breasts wire and a steel wire of the same length are estimated by \[1mm\] under the same force. The radii of the brass and steel wires are \[{R_s}\] and \[{R_b}\]. Then:
(A) \[{R_s} = \sqrt 2 {R_b}\]
(B) \[{R_s} = \dfrac{{{R_b}}}{{\sqrt 2 }}\]
(C) \[{R_s} = 4{R_b}\]
(D) \[R = \dfrac{{{R_b}}}{4}\]
Answer
249.9k+ views
Hint: Try to recall the relation of Young’s modulus, change in length, force and area. From that equation you can solve the question by equating the force as force applied on both the wires are equal and further extensions and length both wires are also same. And we all know the basic relation between area and radius of the wire.
Complete step-by-step answer:
Young’s relation:
\[E = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}\]
Where:
\[E\] is the Young's modulus (modulus of elasticity)
\[F\] is the force exerted on an object under tension;
\[A\] is the actual cross-sectional area, which equals the area of the cross-section perpendicular to the applied force;
\[\Delta L\] is the amount by which the length of the object changes (ΔL is positive if the material is stretched, and negative when the material is compressed);
\[L\] is the original length of the object.
Or
\[F = \dfrac{{AE\Delta L}}{L}\]
As force applied is equal so on equating forces for both the wires with respective quantities we get:
\[\dfrac{{{A_b}{E_b}\Delta {L_b}}}{{{L_b}}} = \dfrac{{{A_s}{E_s}\Delta {L_s}}}{{{L_s}}}\]
Given:
\[\Delta {L_b} = \Delta {L_s}\]
\[{L_b} = {L_s}\]
On simplifying
\[{A_b}{E_b} = {A_s}{E_s}\]
\[{A_b} \times 10 \times {10^{10}} = {A_s} \times 2 \times {10^{11}}\]
And we \[A\] is proportional to \[{R^2}\]
\[\dfrac{{{A_b}}}{{{A_s}}} = \dfrac{{R_b^2}}{{R_s^2}} = 2\]
\[\dfrac{{{R_b}}}{{{R_s}}} = \sqrt 2 \]
\[{R_b} = \sqrt 2 {R_s}\]
Hence option A is correct.
Note: Young's modulus can be helpful in the calculation of the change in the dimension of a bar made of an isotropic elastic material under tensile or compressive loads. For example, it can predict how much a material sample will be extended under tension or shortens on compression. Young modulo is a mechanical property that measures the stiffness of a solid material.
Complete step-by-step answer:
Young’s relation:
\[E = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}\]
Where:
\[E\] is the Young's modulus (modulus of elasticity)
\[F\] is the force exerted on an object under tension;
\[A\] is the actual cross-sectional area, which equals the area of the cross-section perpendicular to the applied force;
\[\Delta L\] is the amount by which the length of the object changes (ΔL is positive if the material is stretched, and negative when the material is compressed);
\[L\] is the original length of the object.
Or
\[F = \dfrac{{AE\Delta L}}{L}\]
As force applied is equal so on equating forces for both the wires with respective quantities we get:
\[\dfrac{{{A_b}{E_b}\Delta {L_b}}}{{{L_b}}} = \dfrac{{{A_s}{E_s}\Delta {L_s}}}{{{L_s}}}\]
Given:
\[\Delta {L_b} = \Delta {L_s}\]
\[{L_b} = {L_s}\]
On simplifying
\[{A_b}{E_b} = {A_s}{E_s}\]
\[{A_b} \times 10 \times {10^{10}} = {A_s} \times 2 \times {10^{11}}\]
And we \[A\] is proportional to \[{R^2}\]
\[\dfrac{{{A_b}}}{{{A_s}}} = \dfrac{{R_b^2}}{{R_s^2}} = 2\]
\[\dfrac{{{R_b}}}{{{R_s}}} = \sqrt 2 \]
\[{R_b} = \sqrt 2 {R_s}\]
Hence option A is correct.
Note: Young's modulus can be helpful in the calculation of the change in the dimension of a bar made of an isotropic elastic material under tensile or compressive loads. For example, it can predict how much a material sample will be extended under tension or shortens on compression. Young modulo is a mechanical property that measures the stiffness of a solid material.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

