If a copper wire is stretched to make its radius decrease by 0.1% , then percentage increase in resi

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1.Units, Dimensions and Measurement

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If a copper wire is stretched to make its radius decrease by $0.1\%$ , then percentage increase in resistance is approximately .......... $\%$

A

$0.1$

B

$0.2$

C

$0.4$

D

$0.8$

Solution

${\rm{V}} = \pi {{\rm{r}}^2}l$

$\frac{{\Delta V}}{V} = 2\frac{{\Delta r}}{r} + \frac{{\Delta l}}{l}$

$0 = \frac{{2\Delta r}}{r} + \frac{{\Delta l}}{l} \ldots $ $(1)$

${\rm{R}} = \frac{{\rho l}}{{\pi {{\rm{r}}^2}}}$

$\frac{{\Delta {\rm{R}}}}{{\rm{R}}} = \frac{{\Delta l}}{l} – 2\frac{{\Delta {\rm{r}}}}{{\rm{r}}}$

Now from equation $(1)$

$\text { Hence, } \frac{\Delta \mathrm{R}}{\mathrm{R}} =-4 \frac{\Delta \mathrm{r}}{\mathrm{r}} $

$=0.4 \% $

Standard 11

Physics

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