A) both the length and the radius of the wire are doubled
B) both the length and the radius of the wire are halved
C) the radius of the wire is doubled
D) the length of the wire is doubled
Correct Answer: A
Solution :
: Heat produced\[\propto \]resistance of wire Heat \[=\frac{{{V}^{2}}}{R}t\] \[R=\rho \frac{l}{\pi {{r}^{2}}}\] Hence, heat \[(H)=\frac{{{V}^{2}}t\times \pi {{r}^{2}}}{\rho l}\] \[\therefore \]\[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{l}_{2}}r_{1}^{2}}{{{l}_{1}}r_{2}^{2}}\] or \[\frac{{{H}_{1}}}{2{{H}_{2}}}=\left( \frac{{{l}_{2}}}{{{l}_{1}}} \right)\left( \frac{r_{1}^{2}}{r_{2}^{2}} \right)\] Or \[\frac{1}{2}=\left( \frac{{{l}_{2}}}{{{l}_{1}}} \right){{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\] Option , RHS \[=\left( \frac{2{{l}_{1}}}{{{l}_{1}}} \right){{\left( \frac{{{r}_{1}}}{2{{r}_{1}}} \right)}^{2}}=\frac{2}{4}=\frac{1}{2}=\]L.H.S. This option agrees with the condition.
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