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(B) The heat produced is given by $H = mS\Delta T = I^2Rt$.For Case $I$: Length is $L$,so resistance is $R$ and mass is $m$. The total $EMF$ is $3E$.

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The heat produced is given by $H = mS\Delta T = I^2Rt$.For Case $I$: Length is $L$,so resistance is $R$ and mass is $m$. The total $EMF$ is $3E$. The current is $I_1 = \frac{3E}{R}$.Heat $H_1 = I_1^2 R t = \left(\frac{3E}{R}ight)^2 R t = \frac{9E^2 t}{R}$.For Case $II$: Length is $2L$,so resistance is $2R$ and mass is $2m$. The total $EMF$ is $NE$. The current is $I_2 = \frac{NE}{2R}$.Heat $H_2 = I_2^2 (2R) t = \left(\frac{NE}{2R}ight)^2 (2R) t = \frac{N^2 E^2 t}{4R} 	imes 2 = \frac{N^2 E^2 t}{2R}$.Since the temperature increase $\Delta T$ and time $t$ are the same,and the material is the same ($S$ is constant),the heat produced must be proportional to the mass: $H \propto m$.Thus,$\frac{H_1}{H_2} = \frac{m}{2m} = \frac{1}{2}$.Substituting the expressions:$\frac{9E^2 t / R}{N^2 E^2 t / 2R} = \frac{1}{2}$$\frac{18}{N^2} = \frac{1}{2}$$N^2 = 36$$N = 6$.

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