(B) Given:Area of cross-section $A = 2 \, cm^2$Force $F = 2 \times 10^5 \, dynes$Initial length $= L$Final length $= 2L$Change in length $\Delta L = 2

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

(B) Given:Area of cross-section $A = 2 \, cm^2$Force $F = 2 \times 10^5 \, dynes$Initial length $= L$Final length $= 2L$Change in length $\Delta L = 2L - L = L$Strain $= \frac{\Delta L}{L} = \frac{L}{L} = 1$Young's modulus $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$Substituting the values:$Y = \frac{(2 \times 10^5) / 2}{1} = 10^5 \, dyne/cm^2$Therefore,the correct option is $B$.

Class 11 Physics Mechanical Properties of Solids Young’s Modulus

Easy

An area of cross-section of a rubber string is $2 \, cm^2$. Its length is doubled when stretched with a linear force of $2 \times 10^5 \, dynes$. The Young's modulus of the rubber in $dyne/cm^2$ will be:

A
$4 \times 10^5$
B
$1 \times 10^5$
C
$2 \times 10^5$
D
$1 \times 10^4$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Mechanical Properties of Solids

Subtopic

Young’s Modulus

If $\rho$ is the density of the material of a wire and $\sigma$ is the breaking stress, what is the greatest length of the wire that can hang freely without breaking?

Medium

View Solution

$A$ copper wire and a steel wire of the same diameter and length are connected end to end and a force is applied,which stretches their combined length by $1 \ cm$. The two wires will have

Easy

View Solution

The area of cross-section of a wire of length $1.1 \, m$ is $1 \, mm^2$. It is loaded with $1 \, kg$. If Young's modulus of copper is $1.1 \times 10^{11} \, N/m^2$,then the increase in length will be ......... $mm$ (Take $g = 10 \, m/s^2$)

Medium

View Solution

Which one is more elastic,steel or plastic? Why?

Medium

View Solution

Young's modulus of rubber is $10^4 \ N/m^2$ and the area of cross-section is $2 \ cm^2$. If a force of $2 \times 10^5 \ dynes$ is applied along its length,then its initial length $L$ becomes:

Medium

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

An area of cross-section of a rubber string is $2 \, cm^2$. Its length is doubled when stretched with a linear force of $2 \times 10^5 \, dynes$. The Young's modulus of the rubber in $dyne/cm^2$ will be:

Similar Questions

If $\rho$ is the density of the material of a wire and $\sigma$ is the breaking stress, what is the greatest length of the wire that can hang freely without breaking?

$A$ copper wire and a steel wire of the same diameter and length are connected end to end and a force is applied,which stretches their combined length by $1 \ cm$. The two wires will have

The area of cross-section of a wire of length $1.1 \, m$ is $1 \, mm^2$. It is loaded with $1 \, kg$. If Young's modulus of copper is $1.1 \times 10^{11} \, N/m^2$,then the increase in length will be ......... $mm$ (Take $g = 10 \, m/s^2$)

Which one is more elastic,steel or plastic? Why?

Young's modulus of rubber is $10^4 \ N/m^2$ and the area of cross-section is $2 \ cm^2$. If a force of $2 \times 10^5 \ dynes$ is applied along its length,then its initial length $L$ becomes:

Vedclass Test Series

Exam Paper Generator

Online Exam Module