Question

An aluminium wire is stretched to make its length, $04\%$ larger. Then percentage change in resistance is:

• A. $0.4\%$
• B. $0.2\%$
• C. $0.8\%$
• D. $0.6\%$

Answer 1

Answer: (C)

$R=\frac{\rho \ell }{A}$
$\frac{\Delta R}{R}=\frac{\Delta \ell }{\ell }-\frac{\Delta A}{A}$
$\ell A=k$
$\frac{\Delta \ell }{\ell }+\frac{\Delta A}{A}=0$
$\frac{\Delta R}{R}=\frac{2\Delta \ell }{\ell }$
$\frac{\Delta R}{R}=2\times 0.4=0.8\%$