A copper wire having cross-sectional area of $0.5 \mathrm{~mm}^{2}$ and a length of 0.1 metre is ini

Question

A copper wire having cross-sectional area of $0.5 \mathrm{~mm}^{2}$ and a length of 0.1 metre is initially at $25^{\circ} \mathrm{C}$ and is thermally insulated from the surrounding. If a current of 10 ampere is set up in this wire, (The change of resistance with the temperature of the wire may be neglected.) \ Melting point of copper $=1075^{\circ} \mathrm{C}$.\ Specific resistance of copper $=1.6 	imes 10^{-8} \mathrm{Wm}$\ Density of copper $=9 	imes 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$\ Specific heat of copper $=9 	imes 10^{-2} \mathrm{cal} / \mathrm{kg}^{\circ} \mathrm{C}$\

Professor PC Thomas Classes · Answer

Mass of wire ( $m$ )\ $=$ area $	imes$ length $	imes$ density\ $=0.5 	imes 10^{-6} 	imes 0.1 	imes 9 	imes 10^{3}=45 	imes 10^{-5} \mathrm{~kg}$\ Rise in temperature $(\Delta T)$\ $=1075^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}=1050^{\circ} \mathrm{C}$\ (i) If a current $I$ flows in a wire of resistance $R$ for a time $t$, the heat energy generated is calorie\ $Q=\frac{I^{2} R t}{4.2}$ calorie\ If $s$ is the specific heat of copper, the wire will melt in time $t$ given by $$\begin{equation*} \frac{I^{2} R t}{4.2}=m s \Delta T 	ag{i} \end{equation*}$$ The resistance $R$ is given by $$ R=\frac{ho L}{A} $$ where $ho$ is the specific resistance. Using this in the equation (i), we have\ $\frac{I^{2} ho L t}{4.2 \mathrm{~A}}=m s \Delta T$ or $t=\frac{4.2 \mathrm{~ms} \Delta T A}{I^{2} ho L}$ $$ \begin{aligned} t= & \frac{4.2 	imes\left(45 	imes 10^{-5}ight) 	imes\left(9 	imes 10^{-2}ight) 	imes 1050}{(10)^{2} 	imes\left(1.6 	imes 10^{-8}ight)(0.1)} \ & =558 \mathrm{~s}=9 \mathrm{~min} 18 \mathrm{~s} \end{aligned} $$ (ii) When the length of the wire is doubled, its resistance $R$ as well as its mass $m$ are doubled. From equation (i) $$ t=\frac{4.2 \mathrm{~ms} \Delta T}{I^{2} R} $$ As $t$ is proportional to $m / R$ and as both $m$ and $R$ are doubled, it follows that $t$ will remain the same, i.e. the wire will start melting in the same time.\