Step-by-step Solution:

1. Analyze the physical setup and identify knowns:

   • We have two wires, PQ and RS, joined end-to-end at QR.
   • Initial length of each wire, $L_{PQ} = L_{RS} = L = 1$ m.
   • Initial temperature of both wires, $T_{initial} = 10^\circ$C.
   • Final temperature at end P, $T_P = 10^\circ$C.
   • Final temperature at end S, $T_S = 400^\circ$C.
   • Thermal conductivity relation: $k_{PQ} = 2k_{RS}$.
   • Coefficient of linear expansion for PQ: $\alpha_{PQ} = 1.2 \times 10^{-5}$ K$^{-1}$.
   • The system is thermally insulated, implying heat flows only along the wires and reaches a steady state.

2. Determine the temperature of the junction (QR):

   • In the steady state, the rate of heat flow through wire PQ must be equal to the rate of heat flow through wire RS.
   • The formula for the rate of heat flow (heat current) is $H = \frac{kA(T_{hot} - T_{cold})}{L}$.
   • Let $T_j$ be the temperature of the junction QR. Heat flows from S to P, so $T_S > T_j > T_P$.
   • Rate of heat flow through RS: $H_{RS} = \frac{k_{RS}A(T_S - T_j)}{L_{RS}}$.
   • Rate of heat flow through PQ: $H_{PQ} = \frac{k_{PQ}A(T_j - T_P)}{L_{PQ}}$.
   • Equating the heat currents ($H_{RS} = H_{PQ}$): $\frac{k_{RS}A(T_S - T_j)}{L_{RS}} = \frac{k_{PQ}A(T_j - T_P)}{L_{PQ}}$
   • Since $L_{PQ} = L_{RS} = 1$ m and assuming the wires have the same cross-sectional area $A$, we can simplify the equation: $k_{RS}(T_S - T_j) = k_{PQ}(T_j - T_P)$
   • Substitute the given values: $k_{PQ} = 2k_{RS}$, $T_S = 400^\circ$C, and $T_P = 10^\circ$C. $k_{RS}(400 - T_j) = 2k_{RS}(T_j - 10)$
   • Cancel $k_{RS}$ from both sides: $400 - T_j = 2(T_j - 10)$ $400 - T_j = 2T_j - 20$ $420 = 3T_j$ $T_j = \frac{420}{3} = 140^\circ\text{C}$
   • So, the temperature of the junction QR is $140^\circ$C.

3. Calculate the change in length of wire PQ:

   • The temperature of wire PQ varies linearly from $T_P = 10^\circ$C at one end to $T_j = 140^\circ$C at the other end.
   • The change in length of a rod with a non-uniform temperature distribution can be found using its average temperature change. The initial temperature of the entire wire was $T_{initial} = 10^\circ$C.
   • The final temperature at end P is $10^\circ$C, so the change in temperature at this end is $\Delta T_P = 10 - 10 = 0^\circ$C.
   • The final temperature at end Q is $140^\circ$C, so the change in temperature at this end is $\Delta T_Q = 140 - 10 = 130^\circ$C.
   • The average change in temperature over the length of the wire PQ is: $\Delta T_{avg} = \frac{\Delta T_P + \Delta T_Q}{2} = \frac{0 + 130}{2} = 65^\circ\text{C}$
   • Alternatively, the final average temperature is $T_{avg} = (10+140)/2 = 75^\circ$C. The average temperature change is $\Delta T_{avg} = T_{avg} - T_{initial} = 75 - 10 = 65^\circ$C.
   • Now, we use the formula for linear thermal expansion, $\Delta L = \alpha L \Delta T_{avg}$.
   • For wire PQ: $\Delta L_{PQ} = \alpha_{PQ} L_{PQ} \Delta T_{avg}$
   • Substitute the values: $\Delta L_{PQ} = (1.2 \times 10^{-5} \text{ K}^{-1}) \times (1 \text{ m}) \times (65 \text{ K})$ (Note: A change in temperature in Celsius is equal to a change in Kelvin) $\Delta L_{PQ} = 1.2 \times 65 \times 10^{-5} \text{ m}$ $\Delta L_{PQ} = 78 \times 10^{-5} \text{ m}$
   • To convert the result to millimeters (mm), we multiply by 1000: $\Delta L_{PQ} = 78 \times 10^{-5} \times 10^3 \text{ mm} = 78 \times 10^{-2} \text{ mm} = 0.78 \text{ mm}$

4. Conclusion:

   • The change in length of the wire PQ is 0.78 mm. This matches option A.