Mixture density from volume fractions Steps:

• Volume of fat: $V_f = 0.040 \times 1.00 \times 10^{-3} = 4.00 \times 10^{-5}$ m³.
• Volume of fat-free milk: $V_m = 0.96 \times 1.00 \times 10^{-3} = 9.60 \times 10^{-4}$ m³.
• Mass of fat-free milk: $m_m = 1.04 \times 10^3 \times 9.60 \times 10^{-4} = 0.9984$ kg.
• Mass of fat: $m_f = 1.035 - 0.9984 = 0.0366$ kg.
• Density of fat: $\rho_f = \frac{0.0366}{4.00 \times 10^{-5}} = 9.15 \times 10^2$ kg m^{-3}.

Why B is correct:

• Matches the calculated density using the volume-weighted mass balance formula for mixtures.

Why the others are wrong:

• A: Overestimates by assuming smaller fat volume fraction.
• C: Overestimates by ignoring fat-free milk volume adjustment.
• D: Underestimates by treating percentage as mass fraction.

Final answer: B