A) \[3{}^\circ C\]
B) \[9{}^\circ C\]
C) \[12{}^\circ C\]
D) \[15{}^\circ C\]
Correct Answer: C
Solution :
Using the relation, \[W=JQ\] or \[{{i}^{2}}RT=Jmc\,\,\Delta \theta \] or \[\Delta \theta \,\propto \,{{i}^{2}}\] Hence, \[\frac{\Delta {{\theta }_{2}}}{\Delta {{\theta }_{1}}}={{\left( \frac{{{i}_{2}}}{{{i}_{1}}} \right)}^{2}}={{(2)}^{2}}=4\] So, \[\Delta {{\theta }_{2}}=4\times \Delta {{\theta }_{1}}=4\times 3={{12}^{o}}C\]
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