Allied & ICAR Exam Bank • ALLIED_0976NAT • 2 Marks
Section 7: Dairy and Food Engineering
Dairy Processing Equipment
Subtopic: Homogenization and Separation
Question Statement:
In a USPHS homogenization test, the fat content of the top $100\text{ mL}$ is $F_{top} = 4.4\%$ and the remainder of the container has $F_{rem} = 4.0\%$. The percentage difference relative to top fat is $\frac{4.4 - 4.0}{4.4} \times 100 = \frac{0.4}{4.4} \times 100$. This value in percentage (rounded off to one decimal place) is:
Answer Key:
9.1
Step-by-Step Solution & Explanation
Difference $= (0.4 / 4.4) \times 100 = 9.09\% \approx 9.1\%$ (conforms to the $\le 10\%$ USPHS standard).