Question

A wire of radius $r$ has resistance $R$. If it is stretched to a radius of $r/2$ , its resistance becomes

• A. 0.5 R
• B. 2 R
• C. 4 R
• D. 16 R

Answer 1

Answer: (D)

Given : Initial radius of wire $\left(r_1\right)=r$ ; Initial resistance $=R$ and final radius of wire $\left(r_2\right)=$ $0.5r$. Let $L_1$ and $A_1$ be the length and cross-sectional area of original wire and $L_2,A_2$ be the values for the stretched wire
Theretore, $L_1\times \pi r_1^2=L_2\times \pi r_2^2$ or
$L_{1}r=L_2\times (0.5r{)}^2$ or $\frac{L_1}{L_2}=\frac{1}{4}$ and $\frac{A_1}{A_2}=\frac{\pi (r{)}^2}{\pi (0.5r{)}^2}=4$.
Resistance $(R)=\rho \frac{L}{A}\propto \frac{L}{A}$ or
$\frac{R_1}{R_2}=\frac{L_1}{L_2}\times \frac{A_2}{A_1}=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$ or $R_2=16R_1$