Question

When a wire is stretched and its radius becomes $r/2$, then its resistance will be

• A. 16R
• B. 2R
• C. 4R
• D. 0

Answer 1

Answer: (A)

Initial radius of wire, $r_1=r$ and
final radius of wire, $r_2=r/2=0.5r$.
Since volume of the wire after stretch remains constant,
therefore $l_1{A}_1=l_2{A}_2$
or $\frac{l_1}{l_2}=\frac{A_2}{A_1}=\frac{r_2^2}{r_1^2}$
$={\left(\frac{0.5r}{r}\right)}^2=\frac{1}{4}.$
The resistance $\left(R\right)=\rho \frac{l}{A}\propto \frac{l}{A}.$
Therefore $\frac{R_1}{R_2}=\frac{l_1}{l_2}\times \frac{A_2}{A_1}$
$=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
or $R_2=16R_1=16R.$