Question

A wire of length L is stretched such that its radius becomes half. Then its resistance becomes

• A. 2 times
• B. 4 times
• C. $\frac{1}{4}times$
• D. 16 times

Answer 1

The correct option is D 16 times

Resistance(R) of a wire of length L, area of cross-section A and resistivity $\rho$ is:
$R=\rho \frac{L}{A}$
$=\rho \frac{V}{A^2}[\because Volume,V=A\times L]$
$=\rho \frac{V}{(\pi r^2{)}^2}=\rho \frac{V}{{\pi }^2{r}^4}[\because A=\pi r^2]$
$\Rightarrow R\propto \frac{1}{r^4}$
Let $R_1$ be the initial resistance of the wire of radius $r_1$ and $R_2$ be the resistance of the wire when its radius $r_2$ becomes half after being stretched.
$\Rightarrow \frac{R_2}{R_1}={\left(\frac{r_1}{r_2}\right)}^4$
$\Rightarrow R_2=2^4{R}_1=16R_1[\because r_2=\frac{r_1}{2}]$
Thus, the resistance of the wire after being stretched becomes 16 times.
Hence, the correct answer is option (d).