Step 1: Define a base problem.
Let Problem~$P_1$ be the plate with side AB at $100^\circ$C and the other three sides at $0^\circ$C; let the center temperature be $T_0^\circ$C.
Step 2: Use symmetry to create four problems.
By symmetry, if instead we heat any single side (AB, BC, CD, or DA) to $100^\circ$C with the others at $0^\circ$C, the center temperature is the same $T_0^\circ$C for each case.
Step 3: Superposition.
Superimpose the four single-side-heated solutions. On the boundary, each side receives one contribution of $100^\circ$C and three of $0^\circ$C, so every side becomes $100^\circ$C.
Thus the superposed boundary condition is all four sides at $100^\circ$C.
Step 4: Center temperature for the superposed problem.
For Laplace's equation with all boundaries at $100^\circ$C, the steady-state solution is uniform: $T(x,y)\equiv 100^\circ$C throughout the plate. Hence the center temperature is $100^\circ$C.
Step 5: Relate to $T_0$.
Temperatures add under superposition, so the center temperature of the superposed problem equals $T_{\text{center}}=T_0+T_0+T_0+T_0=4T_0$.
Therefore, $4T_0 = 100 \Rightarrow T_0 = 25^\circ$C.
\[
\boxed{T_0 = 25.00^\circ\text{C}}
\]
The figures, I, II, and III are parts of a sequence. Which one of the following options comes next in the sequence as IV?
For the beam and loading shown in the figure, the second derivative of the deflection curve of the beam at the mid-point of AC is given by \( \frac{\alpha M_0}{8EI} \). The value of \( \alpha \) is ........ (rounded off to the nearest integer).