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

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

(a) 0.1%      (b) 0.2%      (c) 0.4%      (d) 0.8%  

Concept: As wire is stretched the volume is assumed to be constant. So L can be written in terms of $r$. $R =\rho \cdot \frac{L}{A}$ Differentiating R w.r.t. $r$ is the way to find the change in resistance due to percentage change in $r$. 

\[V=\pi r^2 \cdot L \]

\[L=\frac{V}{\pi r^2}\]

\[R=\frac{\rho L} {A}\]

\[R=\frac{\rho L }{\pi r^2}\]

\[R=\frac{\rho}{\pi} \frac{V}{\pi r^2} \frac{1}{r^2} \]

\[R=\frac {k}{r^4}\]

\[\frac{dR}{dr}=\frac{-4}{r^5}\cdot k\]

dr=-0.1% r (decrease in radius 0.1%)

\[\frac{dR}{-0.1\% \;r}=k \cdot \frac{-4}{r^5}\]

\[dR=\frac{k}{r^4} \cdot 0.4\%\]

\[dR=0.4\% \; R\]

So 0.4% increase in resistance.

Note: The same question can be changed to find the change in resistance for a given change in length of the wire instead of radius. So the above method can be followed by replacing $r$ in terms of L. $\frac {dR}{dL}=k. 2L$