Question

Two copper wires have their masses in the ratio $ 2:3 $ and the lengths in the ratio $ 3:4 $ . The ratio of their resistance is

• A. $ 4:9 $
• B. $ 27:32 $
• C. $ 16:9 $
• D. $ 27:128 $
• E. $ 1:2 $

Answer 1

Answer: (B)

The resistance of one wire
$ {{R}_{1}}=\rho \frac{{{l}_{1}}}{{{A}_{1}}} $
and the resistance of second wire
$ {{R}_{2}}=\rho \frac{{{l}_{2}}}{{{A}_{2}}} $
Ratio of their resistances $ \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{A}_{1}}}\times \frac{{{A}_{2}}}{{{l}_{2}}} $
$ \because $ $ mass=density\times volume $
$ \because $ $ mass=density\times area\times length $
Or $ \frac{{{R}_{1}}}{{{R}_{2}}}={{\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)}^{2}}\times \frac{\rho {{A}_{2}}{{l}_{2}}}{\rho {{A}_{1}}\times {{l}_{1}}} $
Or $ \frac{{{R}_{1}}}{{{R}_{2}}}={{\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)}^{2}}\times \frac{{{m}_{2}}}{{{m}_{1}}} $
Or $ \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{9}{16}\times \frac{3}{2} $
Or $ \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{27}{32} $
$ {{R}_{1}}:{{R}_{2}}=27:32 $