JEE Physics Practice Question

The ends $\mathrm{Q}$ and $\mathrm{R}$ of two thin wires, $\mathrm{PQ}$ and RS, are soldered (joined) together. Initially each of the wires has a length of $1 \mathrm{~m}$ at $10^{\circ} \mathrm{C}$. Now the end $P$ is maintained at $10^{\circ} \mathrm{C}$, while the end $\mathrm{S}$ is heated and maintained at $400^{\circ} \mathrm{C}$. The system is thermally insulated from its surroundings. If the thermal conductivity of wire $\mathrm{PQ}$ is twice that of the wire $R S$ and the coefficient of linear thermal expansion of $\mathrm{PQ}$ is $1.2 \times 10^{-5} \mathrm{~K}^{-1}$, the change in length of the wire $P Q$ is

1. A.$0.78 \mathrm{~mm}$
2. B.$0.90 \mathrm{~mm}$
3. C.$1.56 \mathrm{~mm}$
4. D.$2.34 \mathrm{~mm}$