Pivot-to-CG Distance Along Diagonal

Steps:

• Square's center of gravity (CG) is at geometric center, intersection of diagonals.
• Hanging from corner, board rotates so diagonal from pivot is vertical, aligning CG below pivot.
• Small board diagonal: $x\sqrt{2}$; pivot-to-CG distance: $\frac{x\sqrt{2}}{2} = \frac{x}{\sqrt{2}}$.
• Large board diagonal: $2x\sqrt{2}$; pivot-to-CG distance: $\frac{2x\sqrt{2}}{2} = x\sqrt{2}$.
• Vertical distance between CGs (nails at same height): $x\sqrt{2} - \frac{x}{\sqrt{2}} = x \left( \sqrt{2} - \frac{1}{\sqrt{2}} \right) = \frac{x}{\sqrt{2}}$.

Why B is correct:

• Matches difference in half-diagonal lengths, per uniform density and equilibrium principle.

Why the others are wrong:

• A: $\sqrt{x/2}$ misapplies square root to half-side, ignoring diagonal.
• C: $x$ overestimates by assuming side-length drop, not half-diagonal.
• D: Duplicate of B, but question specifies B as correct.

Final answer: B