Question

If a copper wire is stretched to make it $0.1$% longer, the percentage change in its resistance is

• A. $0.2$% increase
• B. $0.2$% decrease
• C. $0.1$% increase
• D. $0.1$% decrease

Answer 1

The correct option is A $0.2$% increase

$\text{Step 1: New Area of the wire}$

The volume of the wire remains the same even after stretching.

$V_i=V_f$

$Al=A^{\prime }{l}^{\prime }$

Given that , $l^{\prime }=l+0.1\%l=l+\frac{0.1}{100}l=1.001l$

$\Rightarrow A^{\prime }=\frac{1}{1.001}A$

$\text{Step 2: Resistance of wire}$

$R=\rho \frac{l}{A}$ $.....\left(2\right)$

$R^{\prime }=\rho \frac{l^{\prime }}{A^{\prime }}$ $.....\left(3\right)$

Dividing equation $\left(3\right)$ by $\left(2\right)$

$\frac{R^{\prime }}{R}=\frac{l^{\prime }A}{l{A}^{\prime }}$

$\Rightarrow \frac{R^{\prime }}{R}=(1.001{)}^2$

Using Binomial approximation

$(1+x{)}^a=(1+ax)$ when $ax<<1$

$(1.001{)}^2=(1+0.001{)}^2=(1+2\times 0.001)=1.002$

$\Rightarrow R^{\prime }=1.002R$

$\text{Step 4: Percentage increase in resistance of wire}$

Thus percentage increase in resistance$=\frac{R^{\prime }-R}{R}\times 100$ % $=(1.002-1)\times 100$ %

$=0.2$%

Hence option(A) is correct.