Question

A wire has a resistance of $10\Omega$. It is stretched by $1/10$ of its original length. Then its resistance will be

• A. $9\Omega$
• B. $10\Omega$
• C. $11\Omega$
• D. $12.1\Omega$

Answer 1

The correct option is D $12.1\Omega$

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

$R=10\Omega$

$l_{new}=\frac{1}{10}l+l=\frac{11l}{10}$

area=

volume will remain

$l\times A=\frac{11}{10}l\times A^{\prime }\Rightarrow A^{\prime }=\left(\frac{10}{11}\Omega \right)$

$R=\rho \left(\frac{l^{\prime }}{A^{\prime }}\right)$

$=\rho \left(\frac{11l}{10}\right)\times \left(\frac{11}{10A}\right)$

$=\rho \frac{121l}{100A}$

$R_{new}=1.21\left(\rho \frac{A}{l}\right)$

$R_{new}=1.21\times 10$

$R_{new}=12.1\Omega$