Work done in increasing length of a metre long wire of cross-sectional area ........ J. 1,mm^2 throu

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8.Mechanical Properties of Solids

hard

The work done in increasing the length of a metre long wire of cross-sectional area ........ $J.$ $1\,mm^2$ through $1\,mm$ will be $(Y = 2 \times 10^{11}\,Nm^{-2})$

A

$0.1$

B

$5$

C

$10$

D

$250$

Solution

Young's modulus $Y=\frac{F / A}{l / L} \Rightarrow F=\frac{Y A}{L} l$ Work done

$d W=F \times d l=\int_{o}^{I} \frac{Y A}{L} l d l=\frac{1}{2} Y A \frac{l^{2}}{L}$ Given $, A=1 m m^{2}=10^{-6}$

$l=1 m m=10^{-3} m, Y=2 \times 10^{11} N m^{-2}, L=1 m=$

$W=\frac{1}{2} \times 2 \times 10^{11} \times 10^{-6} \times\left(10^{-3}\right)^{2} \mathrm{W}=0.1 \mathrm{J}$

Standard 11

Physics

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(JEE MAIN-2022)