(A) The formula for Young's modulus is $Y = \frac{F L}{A l}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $l$

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

${l_s} = 0.75 \ cm, \ {l_c} = 1.25 \ cm$

(A) The formula for Young's modulus is $Y = \frac{F L}{A l}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $l$ is the elongation.Since the wires are similar,$F, L,$ and $A$ are the same for both. Thus,$l \propto \frac{1}{Y}$.Therefore,$\frac{l_s}{l_c} = \frac{Y_c}{Y_s} = \frac{1.2 \times 10^{11}}{2.0 \times 10^{11}} = \frac{1.2}{2.0} = \frac{3}{5}$.This implies $l_c = \frac{5}{3} l_s$ ... $(i)$.Given the difference in elongations is $l_c - l_s = 0.5 \ cm$ ... (ii).Substituting $(i)$ into (ii): $\frac{5}{3} l_s - l_s = 0.5 \ cm$.$\frac{2}{3} l_s = 0.5 \ cm \Rightarrow l_s = 0.5 \times \frac{3}{2} = 0.75 \ cm$.Then,$l_c = 0.75 + 0.5 = 1.25 \ cm$.

Class 11 Physics Mechanical Properties of Solids Young’s Modulus

Medium

Two exactly similar wires of steel and copper are stretched by equal forces. If the difference in their elongations is $0.5 \ cm$,find the elongation $(l)$ of each wire. Given: ${Y_s} = 2.0 \times {10^{11}} \ N/m^2$ and ${Y_c} = 1.2 \times {10^{11}} \ N/m^2$.

A
${l_s} = 0.75 \ cm, \ {l_c} = 1.25 \ cm$
B
${l_s} = 1.25 \ cm, \ {l_c} = 0.75 \ cm$
C
${l_s} = 0.25 \ cm, \ {l_c} = 0.75 \ cm$
D
${l_s} = 0.75 \ cm, \ {l_c} = 0.25 \ cm$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Mechanical Properties of Solids

Subtopic

Young’s Modulus

$A$ copper wire of length $1.0\, m$ and a steel wire of length $0.5\, m$ having equal cross-sectional areas are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by $1\, mm$. If the Young's moduli of copper and steel are respectively $1.0 \times 10^{11}\, N/m^2$ and $2.0 \times 10^{11}\, N/m^2$,the total extension of the composite wire is ........ $mm$.

MediumJEE MAIN 2013

View Solution

The force required to stretch a steel wire of area of cross-section $1 \,mm^2$ to double its length is (Young's modulus of steel $= 2 \times 10^{11} \,N \,m^{-2}$)

MediumAP EAMCET 2025

View Solution

$A$ uniformly tapering conical wire is made from a material of Young's modulus $Y$ and has a normal,unextended length $L$. The radii at the upper and lower ends of this conical wire have values $R$ and $3R$,respectively. The upper end of the wire is fixed to a rigid support and a mass $M$ is suspended from its lower end. The equilibrium extended length of this wire would be:

DifficultJEE MAIN 2016

View Solution

Two uniform rods $AB$ and $BC$ have Young's moduli $1.2 \times 10^{11} \, N/m^2$ and $1.5 \times 10^{11} \, N/m^2$ respectively. If the coefficient of linear expansion of $AB$ is $1.5 \times 10^{-5} /^{\circ}C$ and both have equal area of cross-section,then the coefficient of linear expansion of $BC$,for which there is no shift of the junction at all temperatures,is ............. $\times 10^{-5} /^{\circ}C$.

Difficult

View Solution

$A$ force $F$ is applied on a wire of radius $r$ and length $L$,and the change in the length of the wire is $l$. If the same force $F$ is applied on a wire of the same material with radius $2r$ and length $2L$,then what is the change in the length of the second wire?

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Two exactly similar wires of steel and copper are stretched by equal forces. If the difference in their elongations is $0.5 \ cm$,find the elongation $(l)$ of each wire. Given: ${Y_s} = 2.0 \times {10^{11}} \ N/m^2$ and ${Y_c} = 1.2 \times {10^{11}} \ N/m^2$.

Similar Questions

The force required to stretch a steel wire of area of cross-section $1 \,mm^2$ to double its length is (Young's modulus of steel $= 2 \times 10^{11} \,N \,m^{-2}$)

$A$ force $F$ is applied on a wire of radius $r$ and length $L$,and the change in the length of the wire is $l$. If the same force $F$ is applied on a wire of the same material with radius $2r$ and length $2L$,then what is the change in the length of the second wire?

Vedclass Test Series

Exam Paper Generator

Online Exam Module