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

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(B) Let $E$ be the $EMF$ of each cell and $R$ be the resistance of the wire of length $L$. The mass of the wire is $m = \rho A L$,where $\rho$ is dens

Vedclass · Accepted Answer

$6$

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(B) Let $E$ be the $EMF$ of each cell and $R$ be the resistance of the wire of length $L$. The mass of the wire is $m = ho A L$,where $ho$ is density and $A$ is cross-sectional area.In the first case,the heat produced is equal to the heat absorbed by the wire:$\frac{(3E)^2}{R} t = m s \Delta T$ ....................... $(i)$When the length is doubled to $2L$,the new resistance becomes $2R$ and the new mass becomes $2m$.In the second case,with $N$ cells in series:$\frac{(NE)^2}{2R} t = (2m) s \Delta T$ ....................... $(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{(NE)^2 / 2R}{(3E)^2 / R} = \frac{2m s \Delta T}{m s \Delta T}$$\frac{N^2}{2 	imes 9} = 2$$N^2 = 36$$N = 6$

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