A) The aluminium wire is 3.9 times longer
B) The aluminium wire is 1.3 times longer
C) The aluminium wire is 2.6 times longer
D) The steel wire is 3.9 times longer
Correct Answer: A
Solution :
\[{{R}_{S}}={{R}_{Al}}\] \[{{\rho }_{1}}=\frac{{{l}_{1}}}{{{A}_{1}}}={{\rho }_{2}}\frac{{{l}_{2}}}{{{A}_{2}}}\] \[{{\rho }_{1}}=\frac{l_{1}^{2}}{{{A}_{1}}{{l}_{1}}}={{\rho }_{2}}\frac{l_{2}^{2}}{{{A}_{2}}{{l}_{2}}}\] \[{{A}_{1}}{{l}_{1}}\] and \[{{A}_{2}}{{l}_{2}}\] will be the volumes of the wires. \[\therefore \] \[{{\rho }_{1}}=\frac{l_{1}^{2}}{{{V}_{1}}}={{\rho }_{2}}\frac{l_{2}^{2}}{{{V}_{2}}}\] \[{{\rho }_{1}}=\frac{{{d}_{1}}l_{1}^{2}}{{{m}_{1}}}={{\rho }_{2}}=\frac{{{d}_{2}}l_{2}^{2}}{{{m}_{2}}}\] Both wires have equal masses, ie, \[{{m}_{1}}={{m}_{2}}\] \[\therefore \] \[{{\rho }_{1}}l_{1}^{2}{{d}_{1}}={{\rho }_{2}}l_{2}^{2}{{d}_{2}}\] \[0.15\,l_{1}^{2}\times 7.8\times {{10}^{3}}=0.028\,l_{2}^{2}\times 2.7\times {{10}^{3}}\] \[{{\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)}^{2}}=\frac{0.028\times 2.7\times {{10}^{3}}}{0.15\times 7.8\times {{10}^{3}}}\] \[\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)=0.254\] \[\therefore \] \[{{l}_{2}}=3.9{{l}_{1}}\] Aluminium wire is 3.9 times longer.
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