Question

If an air bubble of diameter 2 mm rises steadily through a liquid of density 2000 kg/m³ at a rate of 0.5 cm/s, then the coefficient of viscosity of the liquid is _______ Poise.

• A. 0.88
• B. 8.8
• C. 88.8
• D. 0.088

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When a bubble rises steadily, it has reached its terminal velocity.
At this point, the upward buoyant force is perfectly balanced by the downward weight of the bubble and the downward viscous drag force (Stokes' Law).
Since the density of air is negligible compared to the liquid, we can safely ignore the bubble's weight.
Step 2: Key Formula or Approach:
Force balance equation: \(F_{buoyancy} = F_{weight} + F_{viscous}\).
\(\frac{4}{3} \pi r^3 \rho_{liquid} g = \frac{4}{3} \pi r^3 \rho_{air} g + 6 \pi \eta r v\).
Ignoring \(\rho_{air}\), the equation simplifies to:
\(\eta = \frac{2}{9} \frac{r^2 \rho_{liquid} g}{v}\).
Remember to convert the final answer from SI units (\(\text{Pa}\cdot\text{s}\)) to CGS units (Poise), where \(1 \text{ Pa}\cdot\text{s} = 10 \text{ Poise}\).
Step 3: Detailed Explanation:
Given values:
Radius \(r = \frac{2 \text{ mm}}{2} = 1 \text{ mm} = 10^{-3} \text{ m}\).
Density of liquid \(\rho = 2000 \text{ kg/m}^3\).
Terminal velocity \(v = 0.5 \text{ cm/s} = 0.005 \text{ m/s}\).
Gravity \(g = 10 \text{ m/s}^2\).
Substitute these values into the derived formula:
\[ \eta = \frac{2}{9} \frac{(10^{-3})^2 \times 2000 \times 10}{0.005} \] \[ \eta = \frac{2}{9} \frac{10^{-6} \times 20000}{0.005} \] \[ \eta = \frac{2}{9} \frac{0.02}{0.005} = \frac{2}{9} (4) = \frac{8}{9} \text{ Pa}\cdot\text{s} \] Calculate the value in SI units:
\[ \eta \approx 0.888 \text{ Pa}\cdot\text{s} \] Convert \(\text{Pa}\cdot\text{s}\) to Poise by multiplying by 10:
\[ \eta_{Poise} = 0.888 \times 10 = 8.88 \text{ Poise} \] This matches the \(8.8\) option closely depending on the \(\pi\) or \(g\) strict rounding, but strictly \(8.88 \approx 8.8\).
Step 4: Final Answer:
The coefficient of viscosity is \(8.8 \text{ Poise}\).