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}}=16{{R}_{1}}=16R $