A) \[16{}^\circ C\]
B) \[10{}^\circ C\]
C) \[20{}^\circ C\]
D) \[12{}^\circ C\]
Correct Answer: C
Solution :
| [c] 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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