Density scales inversely with volume under constant mass

Steps:

• Original volume $V = L^3$, mass $m = \rho V = \rho L^3$.
• New side length $L/2$, so new volume $V' = (L/2)^3 = L^3/8 = V/8$.
• Mass remains constant at $m = \rho L^3$.
• New density $\rho' = m / V' = (\rho V) / (V/8) = 8\rho$.

Why D is correct:

• Density is mass per volume; with mass conserved, volume reduction by factor of 8 increases density by 8, per the formula $\rho = m/V$.

Why the others are wrong:

• A: Assumes volume scales with area ($(1/2)^2 = 1/4$), ignoring 3D scaling.
• B: Assumes volume halves with linear halving, but volume scales cubically.
• C: Assumes volume quarters ($(1/2)^2$), confusing 2D with 3D.

Final answer: D