Question

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

• A. $12.1\Omega$
• B. $1.21\Omega$
• C. $1.13\Omega$
• D. $11\Omega$

Answer 1

The correct option is A $12.1\Omega$

Given: $R=10\Omega$ and length increased by $10$

So, $L^{{}^{\prime }}=L+10\%\text{of}L=L+\frac{10}{100}L$

$L^{{}^{\prime }}=\left(\frac{110}{100}\right)L$

Resistance of a conductor is given by,

$R=\frac{\rho L}{A}$

Where $\rho =$ resistivity of the material.

$L=\text{Length}$

$A=\text{cross-sectional area}$

When the conductor is snatched the volume of conductor will remain same.

Volume of a cylinder,

$V=AL=\pi r^{2}h(A=\pi r^2,h=L)$

So, $V=V^{{}^{\prime }}$

$AL=A^{{}^{\prime }}{L}^{{}^{\prime }}$

$A^{{}^{\prime }}=\frac{AL}{L^{{}^{\prime }}}=A\left(\frac{100}{110}\right)$

Now new resistance is $R^{{}^{\prime }}=\rho \frac{L^{{}^{\prime }}}{A^{{}^{\prime }}}$

Or,$R^{{}^{\prime }}=\rho \frac{L}{A}\left(\frac{110\times 110}{100\times 100}\right)=10\times \frac{121}{100}$

$R^{\prime }=12.1\Omega$