A wire of length L and 3 identical cells of negligible internal resistances are connected is series. Due to the current, the temperature of the wire is raised by $\mathrm{\Delta T}$ in time t.  N number of similar cells is now connected in series with a wire of the same material and cross section but of length 2L. The temperature of the wire is raised by the same amount $\mathrm{\Delta T}$ in the same time t. The value of N is:

Let, the terminal potential difference of each cell is V and resistance of wire is R.

then heat developed in the wire in time t is

$\mathrm{H}=\left(\frac{(3 \mathrm{V}{)}^2}{\mathrm{R}}\right)=\mathrm{ms}\Delta \mathrm{T}$

where m is the mass of wire

s is the specific heat of its material and 

ΔT is rise in its temperature

When N such identical cells are connected in series the effective terminal 

potential is NV volts and if the length of wire is doubled its resistance and 

mass also doubled then heat developed in the wire is

$\mathrm{H}=\left(\frac{(\mathrm{NV}{)}^2}{2\mathrm{R}}\right)\mathrm{t}=2 \mathrm{ms}\Delta \mathrm{T}$

Dividing both equations we get

$$\begin{gathered} \frac{{\mathrm{N}}^2}{2\times 9}=2 \\ \therefore N=6 \end{gathered}$$