Question

The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter the change in the resistance of the wire will be

• A. 2
• B. 1
• C. 0.5
• D. 3

Answer 1

Answer: (D)

3

Answer 2

Given: $l=l+100$ Initial volume = final volume $ie,$ $\pi {{r}^{2}}l=\pi r{{}^{2}}l$ $\Rightarrow$ $r{{}^{2}}=\frac{{{r}^{2}}l}{l}={{r}^{2}}\times \frac{l}{2l}$ $\Rightarrow$ $r{{}^{2}}=\frac{{{r}^{2}}}{2}$ $\therefore$ $R=\rho \frac{l}{A}=\rho \frac{2l}{\pi r{{}^{2}}}$ $\left( \because R=\frac{\rho l}{A} \right)$ Thus, $\Delta R=R-R=4R-R=3R$ $\therefore$ $$ $=300$