Ten identical cells connected in series are needed to heat a wire of length one meter and radius ‘r’ by ${10}^{o}C$ in time ‘t’. How many cells will be required to heat the wire of length two meter of the same radius by the same temperature in time ‘t’ ?

Solution:

Let $\rho$ be resistivity of the material of the wire and r be radius of the wire. Therefore, resistance of 1 m wire is

$R=\frac{\rho \left(1\right)}{\pi r^2}=\frac{\rho }{\pi r^2} \left(\because R=\frac{\rho l}{A}\right)$

Let $\epsilon$ be emf of each cell. 

In first case,

10 cells each of emf $\epsilon$ are connected in series to heat the wire of length 1 m by $\Delta T(={10}^{o}C)$in time t.

$\therefore \frac{(10\epsilon {)}^2}{R}t=ms\Delta T ........\left(i\right)$

In second case, Resistance of same wire of length 2 m is

$R^{\text{'}}=\frac{\rho \left(2\right)}{\pi r^2}=\frac{2\rho }{\pi r^2}=2R$

Let n cells each of emt $\epsilon$ are connected in series to heat the same wire of length 2 m, in the same temperature $\Delta T(={10}^{o}C)$ in the same time t

$\therefore \frac{(n\epsilon {)}^{2}t}{2R}=(2m{)}_{s\Delta T} .....(ii)$

Divide (ii) by (i), we get

$\frac{n^2}{200}=2\Rightarrow n^2=400\therefore n=20$