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Grade 12th passElectric Current

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

Munaza

9 Years agoGrade 12th pass

1 Answer

Kshitij Sharma

9 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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