(B) The energy stored in a stretched wire (elastic potential energy) is given by the formula:$U = \frac{1}{2} \times \text{Force} \times \text{Extensi

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

(B) The energy stored in a stretched wire (elastic potential energy) is given by the formula:$U = \frac{1}{2} \times \text{Force} \times \text{Extension}$Here, the force $F$ is the weight hung, $F = mg = 10 \, kg \times 10 \, m/s^2 = 100 \, N$.The extension $l$ is $1 \, mm = 1 \times 10^{-3} \, m$.Substituting the values:$U = \frac{1}{2} \times 100 \, N \times 1 \times 10^{-3} \, m$$U = 50 \times 10^{-3} \, J$$U = 0.05 \, J$.

Class 11 Physics Mechanical Properties of Solids Work Done in Stretching a Wire

Medium

$A$ $5 \, m$ long wire is fixed to the ceiling. $A$ weight of $10 \, kg$ is hung at the lower end and is $1 \, m$ above the floor. The wire is elongated by $1 \, mm$. The energy stored in the wire due to stretching is ......... $J$.

A
$0$
B
$0.05$
C
$100$
D
$500$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Mechanical Properties of Solids

Subtopic

Work Done in Stretching a Wire

$A$ wire of Young's modulus $1.6 \times 10^{12} \,N/m^{2}$ is stretched by a force so as to produce a strain of $2 \times 10^{-4}$. The energy density of the wire is

MediumMHT CET 2020

View Solution

Calculate the work done, if a wire is loaded by $Mg$ weight and the increase in length is $l$.

Medium

View Solution

If the work done in stretching a wire by $1 \ mm$ is $2 \ J$,the work necessary for stretching another wire of the same material but with double the radius of cross-section and half the length by $1 \ mm$ is:

EasyAP EAMCET 2024

View Solution

$A$ metallic rod of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$. If the rod is elongated by an amount $y$,then the work done is proportional to ......

Easy

View Solution

When a $8 \,m$ long wire is stretched by a load of $10 \,kg-wt$, it is elongated by $1.5 \,mm$. The energy stored in the wire in this process is $\left(g=10 \,ms^{-2}\right)$ (in $\,J$)

MediumAP EAMCET 2021

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

$A$ $5 \, m$ long wire is fixed to the ceiling. $A$ weight of $10 \, kg$ is hung at the lower end and is $1 \, m$ above the floor. The wire is elongated by $1 \, mm$. The energy stored in the wire due to stretching is ......... $J$.

Similar Questions

$A$ wire of Young's modulus $1.6 \times 10^{12} \,N/m^{2}$ is stretched by a force so as to produce a strain of $2 \times 10^{-4}$. The energy density of the wire is

Calculate the work done, if a wire is loaded by $Mg$ weight and the increase in length is $l$.

If the work done in stretching a wire by $1 \ mm$ is $2 \ J$,the work necessary for stretching another wire of the same material but with double the radius of cross-section and half the length by $1 \ mm$ is:

$A$ metallic rod of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$. If the rod is elongated by an amount $y$,then the work done is proportional to ......

When a $8 \,m$ long wire is stretched by a load of $10 \,kg-wt$, it is elongated by $1.5 \,mm$. The energy stored in the wire in this process is $\left(g=10 \,ms^{-2}\right)$ (in $\,J$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module