Question

In an o/w emulsion, mineral oil (specific gravity - 0.9) dispersed in aqueous phase having specific gravity of 1.05. If oil particles have average diameter of 5 micrometer, the external phase has viscosity of 0.5 poise and gravity constant is 981 \(cm/sec^{2}\), what is the velocity of creaming in cm/day?

• A. 0.35
• B. 35.0
• C. 4.1
• D. $4.1 \times 10^{-6}$

Hint:

Always ensure units are consistent (CGS or SI). Here, $1 \mu m = 10^{-4} cm$.

Answer 1

The Correct Option is A

Step 1: Concept
The velocity of creaming ($v$) is given by Stokes' Law: $v = \frac{d^{2} (\rho_{s} - \rho_{0}) g}{18\eta}$.

Step 2: Meaning
$d = 5 \times 10^{-4}$ cm ($5 \mu m$), $\rho_{s} = 0.9$ g/cm$^3$, $\rho_{0} = 1.05$ g/cm$^3$, $g = 981$ cm/sec$^2$, $\eta = 0.5$ poise. Note: A negative velocity indicates creaming (upward movement) since oil is less dense than water.

Step 3: Analysis
$v = \frac{(5 \times 10^{-4})^{2} \times (0.9 - 1.05) \times 981}{18 \times 0.5}$ $v = \frac{25 \times 10^{-8} \times (-0.15) \times 981}{9} \approx -4.0875 \times 10^{-6}$ cm/sec. To convert to cm/day: $v \times 60 \times 60 \times 24 = 4.0875 \times 10^{-6} \times 86400 \approx 0.353$ cm/day.

Step 4: Conclusion
The velocity of creaming is approximately 0.35 cm/day.

Final Answer: (A)