Question

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

• A. zero
• B. 2 R
• C. 8 R
• D. 16 R

Answer 1

Answer: (D)

For a wire of length $l,$ area A and specific resistance $\rho ,$ resistance (R) is given by $R=\rho \frac{l}{A}$ ?(i) Also volume = length x area which remains constant on stretching the wire, hence $l_1{A}_1=l_2{A}_2$ If $r_1$ and $r_2$ are radii of wire then, $\frac{l_1}{l_2}=\frac{A_2}{A_1}$ $=\frac{\pi r_2^2}{\pi r_1^2}$ $=\left(\frac{0.5\times r}{r}\right)=\frac{1}{4}$ Using Eq. (i), we have $\frac{R_1}{R_2}=\frac{l_1}{l_2}\times \frac{A_2}{A_1}$ $=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$ $\Rightarrow$ $R_2=16R_1=16R$