Case Study:

In many parts of India, milk adulteration is a serious concern. Small dairy cooperatives and local vendors are sometimes suspected of mixing water into milk to increase profit. To prevent this, food inspectors use a simple instrument called a lactometer.

A lactometer works on the principle of buoyancy. Pure milk has a density slightly higher than water, approximately 1.03 g/cm³. When placed in pure milk, the lactometer floats at a particular level. If water is added, the density of the mixture decreases, and the lactometer sinks deeper.

In a village dairy testing camp, an inspector tested two samples. Sample A had a density of 1.03 g/cm³, while Sample B showed a density of 1.01 g/cm³ at room temperature (30°C). The inspector concluded that Sample B might be diluted.

Further analysis showed that if 100 mL of pure milk (density = 1.03 g/cm³) is mixed with 20 mL of water (density = 1.00 g/cm³), the average density decreases noticeably.

Students are asked to analyze how changes in density affect buoyant force, how floating level indicates purity, and why temperature must also be considered during testing.

Questions:

Section A - MCQs

1. When water is added to milk, the lactometer sinks deeper because:
A. The mass of milk increases
B. The density of the mixture decreases
C. The volume of lactometer decreases
D. The buoyant force becomes zero

2. If pure milk has density 1.03 g/cm³ and water has 1.00 g/cm³, the density of their mixture will be:
A. Greater than 1.03 g/cm³
B. Equal to 1.03 g/cm³
C. Between 1.00 and 1.03 g/cm³
D. Less than 1.00 g/cm³

3. The working principle of a lactometer is based on:
A. Newton’s First Law
B. Archimedes’ Principle
C. Law of Gravitation
D. Conservation of Energy

4. If the temperature of milk increases significantly, its density will generally:
A. Increase
B. Decrease
C. Remain constant
D. Become zero

Section B - Short Answer Questions

1. Explain why a lactometer floats higher in pure milk compared to diluted milk.

2. Calculate the approximate density of a mixture formed by mixing 100 mL milk (1.03 g/cm³) with 20 mL water (1.00 g/cm³). (Show steps briefly.)

3. Why must temperature be considered while testing milk purity using density?

Section C - Long Answer Question

1. A dairy inspector finds that a milk sample has density 1.015 g/cm³ at 30°C.
(a) Suggest whether the milk is likely adulterated.
(b) Explain using buoyant force and density concepts.
(c) What if the temperature were 40°C instead of 30°C? How might this affect the result?

Answer Key:

Section A - MCQs

1. B
2. C
3. B
4. B

Section B - Short Answers

1. Pure milk has higher density, so it provides greater buoyant force. The lactometer floats higher because less volume needs to be displaced to balance its weight.

2. Mass of milk = 100 × 1.03 = 103 g
Mass of water = 20 × 1.00 = 20 g
Total mass = 123 g
Total volume = 120 mL
Density = 123/120 = 1.025 g/cm³

3. With increase in temperature, liquids expand and density decreases. This can give a false reading, making milk appear diluted even if pure.

Section C - Long Answer

(a) Since 1.015 g/cm³ is lower than 1.03 g/cm³, the milk is likely diluted.
(b) Lower density means less buoyant force; the lactometer sinks deeper, indicating adulteration.
(c) At 40°C, density naturally decreases due to thermal expansion. The reading may appear lower even without adulteration. Temperature correction is necessary for accurate conclusion.