$\therefore \text { rate of heat transfer }=\frac{d Q}{d t}=\frac{2 K A(T-10)}{L}=\frac{K A(400-T)}{L}$

$\Rightarrow 2(T-10)=400-T$

$\text { or, } T =140^{\circ} C$

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:

$\int_0^{\Delta L } dy =13 \circ \alpha \int_0^{ L } xdx$

$\because L =1 m \text { (Given) }$

$\text { or, } \Delta L =\frac{13 \circ ax ^2}{2}=\frac{130 \times 1.2 \times 10^{-5} \times 1}{0.78} m=0.78 \ mm$