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If a copper wire is stretched to make its radius decrease by 0.15% then the percentage increase in resistance is approximately

If a copper wire is stretched to make its radius decrease by 0.15% then the percentage increase in resistance is approximately

Grade:12th pass

1 Answers

Kshitij Sharma
33 Points
4 years ago
0.6% increase.Volume is const. so, dV=d(LA)=AdL+LdA=0\Rightarrow​AdL= – LdA......(1)
Formula is AR=\rho L\Rightarrow \log A+\log R=\log \rho +\log L​.On partial differentiation, \frac{dA}{A}+\frac{dR}{R}=\frac{d\rho }{\rho }+\frac{dL}{L}​ ,         d\rho = 0  So,\frac{dR}{R}=\frac{AdL-LdA}{AL}​  Using (1)\Rightarrow \frac{dR}{R}=\frac{-2dA}{A}=\frac{-2d(\pi r^{2})}{\pi r^{2}}=\frac{-4dr}{r}​.Putting \frac{dr}{r}=-0.15/100​, we get the answer as 0.6%.​

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