A wire of length L and 3 identical cells of n | vedclass

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(B) Let $R$ be the resistance and $m$ be the mass of the first wire. The second wire has length $2L$,so its resistance is $2R$ and its mass is $2m$.Le

Vedclass · Accepted Answer

$6$

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(B) Let $R$ be the resistance and $m$ be the mass of the first wire. The second wire has length $2L$,so its resistance is $2R$ and its mass is $2m$.Let $E$ be the $EMF$ of each cell and $S$ be the specific heat capacity of the material.For the first wire,the current is $i_1 = \frac{3E}{R}$. The heat produced is $H_1 = i_1^2 R t = m S \Delta T$.Substituting $i_1$,we get $(\frac{3E}{R})^2 R t = m S \Delta T \implies \frac{9E^2 t}{R} = m S \Delta T$.For the second wire,the current is $i_2 = \frac{NE}{2R}$. The heat produced is $H_2 = i_2^2 (2R) t = (2m) S \Delta T$.Substituting $i_2$,we get $(\frac{NE}{2R})^2 (2R) t = 2m S \Delta T \implies \frac{N^2 E^2 t}{4R^2} \cdot 2R = 2m S \Delta T \implies \frac{N^2 E^2 t}{2R} = 2m S \Delta T$.Dividing the two heat equations: $\frac{9E^2 t / R}{N^2 E^2 t / 2R} = \frac{m S \Delta T}{2m S \Delta T} \implies \frac{18}{N^2} = \frac{1}{2} \implies N^2 = 36 \implies N = 6$.

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