Question

If a copper wire is stretched to make its radius decrease by $0.1$%, then the percentage increase in its resistance is approximately:

• A. $0.1$%
• B. $0.2$%
• C. $0.4$%
• D. $0.8$%

Answer 1

The correct option is C $0.4$%

$\text{Given:}{r}^{\prime }=r-0.1\%r=r-\frac{0.1}{100}r=0.999r$

$\text{Step 1:}$

Volume of the wire remains same.

$V_i=V_f$

$lA=l^{\prime }{A}^{\prime }$ $.....\left(1\right)$

$\text{Step 2: When radius is decreased by}$ $0.1$%,

$r^{\prime }=0.999r$

$A^{\prime }=\pi r^{\prime 2}$

$\Rightarrow A^{\prime }=(0.999{)}^{2}A$

Using equation (1)

$l^{\prime }=\frac{l}{(0.999{)}^2}$

$\text{Step 3: Resistance of wire}$

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

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

Dividing equation (3) by (2)

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

$\Rightarrow \frac{R^{\prime }}{R}=\frac{1}{(0.999{)}^4}$

Using Binomial approximation

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

$\frac{1}{(0.999{)}^4}=(0.999{)}^{-4}=(1-0.001{)}^{-4}=(1+4\times 0.001)=1.004$

$\Rightarrow R^{\prime }=1.004R$

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

Thus percentage increase in resistance$=\frac{R^{\prime }-R}{R}\times 100$ % $=(1.004-1)\times 100$ %

$=0.4$%

Hence option(C) is correct.