A) \[{{16}^{\text{o}}}C\]
B) \[{{10}^{\text{o}}}C\]
C) \[{{20}^{\text{o}}}C\]
D) \[{{12}^{\text{o}}}C\]
Correct Answer: C
Solution :
Key Idea: When current is passed through a conductor, electric energy is absorbed by the conductor through collisions between its atomic lattice and the charge carriers causing its temperature to rise. Energy loss in conductor \[Q={{i}^{2}}RT\] Heat developed \[=ms\,\Delta \theta \] \[\therefore ms\,\Delta \theta ={{i}^{2}}Rt\] or \[\Delta \theta \,\propto \,\,{{i}^{2}}\] \[\frac{\Delta {{\theta }_{2}}}{\Delta {{\theta }_{1}}}=\frac{i_{2}^{2}}{i_{1}^{2}}\] or \[\Delta {{\theta }_{2}}={{\left( \frac{{{i}_{2}}}{{{i}_{1}}} \right)}^{2}}\,\Delta {{\theta }_{1}}\] Here \[{{i}_{2}}=2{{i}_{1}},\,\,\Delta {{\theta }_{1}}={{5}^{o}}C\] From Eq. (i) \[\therefore \Delta {{\theta }_{2}}={{\left( \frac{2{{i}_{1}}}{{{i}_{1}}} \right)}^{2}}\times 5\] \[=4\times 5={{20}^{o}}C\]
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