Ten identical cells connected in series are needed to heat a wire of length one meter and radius ' $

Question

Ten identical cells connected in series are needed to heat a wire of length one meter and radius ' $r$ ' by $10^{\circ} \mathrm{C}$ 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 ' $t$ '?\

Professor PC Thomas Classes · Answer

(a) : Let $ho$ be resistivity of the material of the wire and $r$ be radius of the wire.\ Therefore, resistance of 1 m wire is\ $R=\frac{ho(1)}{\pi r^{2}}=\frac{ho}{\pi r^{2}}$\ Let $\varepsilon$ be emf of each cell. In first case, 10 cells each of emf $\varepsilon$ are connected in series to heat the wire of length 1 m by $\Delta T\left(=10^{\circ} \mathrm{C}ight)$ in time $t$. $$ \left(\because R=\frac{ho l}{A}ight) $$ $	herefore \quad \frac{(10 \varepsilon)^{2}}{R} t=m s \Delta T$\ In second case, resistance of same wire of length 2 m is\ $R^{\prime}=\frac{ho(2)}{\pi r^{2}}=\frac{2 ho}{\pi r^{2}}=2 R$\ Let $n$ cells each of emf $\varepsilon$ are connected in series to heat the same wire of length 2 m , by the same temperature $\Delta T\left(=10^{\circ} \mathrm{C}ight)$ in the same time $t$.\ $	herefore \quad \frac{(n \varepsilon)^{2} t}{2 R}=(2 m) s \Delta T$\ Divide (ii) by (i), we get\ $\frac{n^{2}}{200}=2 \Rightarrow n^{2}=400 \quad 	herefore \quad n=20$\