Physics - Current Electricity Question with Solution | TestHub

Let $\rho =$ resistivity of the material of wire
$r=$ radius of wire $\therefore$ Resistance of $1\mathrm{m}$ wire is $R=\frac{\rho }{\pi r^2}$ $$\left\{\because R=\frac{\rho l}{A}\right\}$$ Let $\epsilon =emf$ of the cell in first case. All the cells are connected in series. So heat is $ms\Delta T=\frac{(106{)}^2}{R}t$ In $2^{\mathrm{nd}}$ case Resistance of some wire of length 2 $\mathrm{m}$ is : $$R^{\prime }=\frac{\rho (2)}{\pi r^2}=\frac{2\rho }{\pi r^2}=2R$$ and then heat is : $$(2\mathrm{m})\mathrm{S}\Delta \mathrm{T}=\frac{(\mathrm{n}\epsilon {)}^{2}t}{2R}$$ From eq. (i) and (ii) $$\begin{array}{rl}\frac{n^2}{200}& =2\\ \Rightarrow n^2& =400\Rightarrow n=20\end{array}$$