Question

The resistance of a wire is $10\Omega$. Its length increased by $10\%$ by stretching. The new resistance will now be?

• A. $12.22\Omega$
• B. $1.2\Omega$
• C. $13\Omega$
• D. $11\Omega$

Answer 1

Answer: (A)

$12.22\Omega$

Given resistance of the wire=10$\Omega$ and change in length=10%

But we know total volume of the wire will remain same

Initial volume = Final volume after stretching

Now volume of cylinder$=\pi r^{2}L$

As the length increases its radius decreases thus the new area of the wire will be A$=\frac{V}{1.1L}=.9A$

Also let the resistance of the wire $\rho \frac{L}{A}$

Thus the new resistance of the wire $=\rho \frac{1.1L}{.9A}=1.1\times \rho \frac{L}{A}=\frac{1.1\times 10}{.9}=12.22\Omega$