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Q.
A wire with $15 \,\Omega$. resistance is stretched by one tenth of its original length and volume of wire is kept constant. Then its resistance will be
Current Electricity
Solution:
If the wire is stretched by $(1/10)^{th}$ of its original length then the new length of wire become
$l_{2}=l+\frac{l}{10}=\frac{11\,l}{10}\quad\ldots\left(i\right)$
As the volume of wire remains constant then
$\pi r^{2}_{1}l=\pi r^{2}_{2}l_{2}=\pi r^{2}_{2}\left(\frac{11l}{10}\right)\quad$ (using $(i)$)
$\Rightarrow r^{2}_{2}=\frac{10}{11}r^{2}\quad\ldots\left(ii\right)$
Now the resistance of stretched wire.
$R_{2}=\frac{\rho\left(\frac{11}{10}l\right)}{\pi r^{2}_{2}}=\frac{\left(\frac{11}{10}\right)\rho l}{\pi \times \frac{10}{11}r^{2}_{1}}$
$=\left(\frac{11}{10}\right)^{2} \times \frac{\rho l}{\pi r^{2}_{1}}$
$\left(\because R_{1}=\frac{\rho l}{\pi r^{2}_{1}}=15\,\Omega\right)$
$\therefore R_{2}=\left(\frac{11}{10}\right)^{2} \times 15=18.15\,\Omega$