A) \[10{}^\circ C\]
B) \[20{}^\circ C\]
C) \[5{}^\circ C\]
D) \[25{}^\circ C\]
Correct Answer: B
Solution :
Heat produced in a wire of resistance R in time \[t\]is \[Q={{I}^{2}}Rt\] Also heat required to raise the temperature of a wire by \[\Delta T\]is \[Q=ms\Delta T\] Here \[Q=Q\] \[\Rightarrow \] \[ms\Delta T={{I}^{2}}Rt\] Since, m, S,R and tare constant. So, \[\Delta T\propto {{I}^{2}}\] \[\Rightarrow \] \[\frac{\Delta {{T}_{2}}}{\Delta {{T}_{1}}}={{\left( \frac{{{I}_{2}}}{{{I}_{1}}} \right)}^{2}}={{\left( \frac{2I}{I} \right)}^{2}}=4\] \[\therefore \] \[\Delta {{T}_{2}}=4\times 5=20{}^\circ C\]
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