MCQ
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$0.2$
- ✓$0.4$
- D$0.8$
✓
Answer
Correct option: C.
$0.4$
c
${\rm{V}} = \pi {{\rm{r}}^2}l$
${\rm{V}} = \pi {{\rm{r}}^2}l$
$\frac{{\Delta V}}{V} = 2\frac{{\Delta r}}{r} + \frac{{\Delta l}}{l}$
$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 \% $
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