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. 200%
• B. 100%
• C. 50%
• D. 300%

Answer 1

Answer: (D)

Given: $ l=l+100%l=2l $ 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 $ $ %\Delta R=\frac{3R}{R}\times 100% $ $ =300% $