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The ends $\mathrm{Q}$ and $\mathrm{R}$ of two thin wires, $\mathrm{PQ}$ and $RS$, are soldered (joined) togetker. Initially each of the wires has a length of $1 \mathrm{~m}$ at $10^{\circ} \mathrm{C}$. Now the end $\mathrm{P}$ 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 $RS$ and the coefficient of linear thermal expansion of $P Q$ is $1.2 \times 10^{-5} \mathrm{~K}^{-1}$, the change in length of the wire $\mathrm{PQ}$ is
$0.78 \mathrm{~mm}$
$0.90 \mathrm{~mm}$
$1.56 \mathrm{~mm}$
$2.34 \mathrm{~mm}$
Solution
Let the temperature of the junction be $T.$
$\therefore \text { rate of heat transfer }=\frac{d Q}{d t}=\frac{2 K A(T-10)}{L}=\frac{K A(400-T)}{L}$
$\Rightarrow 2(T-10)=400-T$
$\text { or, } T =140^{\circ} C$
Now, for the wire $P Q$, let us imagine a small length $\Delta x$ at a distance $x$ from the junction.
$\therefore \frac{\Delta T }{\Delta x }=\frac{140-10}{1}=130$
So, temperature at distance $x :$
$T=10+130 x$
or, $T -10=130 x$
Increase in length of the small element $\Delta x$ is expressed as:
$\frac{d y}{d x}=\alpha \Delta T =\alpha( T -10)$
$\text { or, } \frac{ dy }{ dx }=\alpha \times 130 x$
Integrating both sides, we get:
$\int_0^{\Delta L } dy =13 \circ \alpha \int_0^{ L } xdx$
$\because L =1 m \text { (Given) }$
$\text { or, } \Delta L =\frac{13 \circ ax ^2}{2}=\frac{130 \times 1.2 \times 10^{-5} \times 1}{0.78} m=0.78 \ mm$
