A plate at a distance of 0.03 mm from a fixed plate moves at 0.8 m/s and requires a force of 1.50 N/m² area of plate. Determine the dynamic viscosity of liquid between the plates.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 15.11.2022 SHIFT-I
Detailed Explanation
Couette flow between parallel plates creates a linear velocity profile under laminar conditions. Newton's law of viscosity allows direct computation of dynamic viscosity from shear stress, plate velocity, and gap spacing.
Given: gap \(dy = 0.03\text{ mm} = 0.03 \times 10^{-3}\text{ m}\), velocity \(du = 0.8\text{ m/s}\), shear stress \(\tau = 1.50\text{ N/m}^2\):
\[ \tau = \mu \frac{du}{dy} \;\implies\; \mu = \frac{\tau \cdot dy}{du} \]
\[ \mu = \frac{1.50 \times (0.03 \times 10^{-3})}{0.8} = \frac{0.045 \times 10^{-3}}{0.8} = 0.05625 \times 10^{-3}\text{ N}\cdot\text{s/m}^2 = 56.25 \times 10^{-6}\text{ N}\cdot\text{s/m}^2 \]
Why “56.25 × 10⁻⁶ N·s/m²” Is Correct
Evaluating μ = (1.50 × 0.03 × 10⁻³) / 0.8 yields 56.25 × 10⁻⁶ N·s/m², matching option (a).
| Quantity | SI unit | CGS unit | Conversion |
|---|---|---|---|
| Dynamic viscosity μ | Pa·s (N·s/m²) | Poise (P) = dyne·s/cm² | 1 P = 0.1 Pa·s |
| Kinematic viscosity ν = μ/ρ | m²/s | Stoke (St) = cm²/s | 1 St = 10⁻⁴ m²/s |
| Surface tension T | N/m | dyne/cm | 1 dyne/cm = 10⁻³ N/m |
Why the Other Options Are Wrong
- B. 6.25 × 10⁻⁶ N·s/m²: 6.25 × 10⁻⁶ N·s/m² results from missing a factor of 9 in the numerator.
- C. 66.25 × 10⁻⁶ N·s/m²: 66.25 × 10⁻⁶ N·s/m² is an incorrect arithmetic distractor.
- D. 50.25 × 10⁻⁶ N·s/m²: 50.25 × 10⁻⁶ N·s/m² is an erroneous estimation.
Key Point
Newton's shear relation: μ = τ / (du/dy) = τ·dy / du.
Key Concepts for Students
- Linear velocity profile assumes laminar flow between narrowly spaced plates.
- 1 N·s/m² = 1 Pa·s = 10 poise.
- 0.03 mm = 3 × 10⁻⁵ m; careful unit conversion to SI meters is essential.
Practice more: this is question 25 in Properties of Fluid MCQs – Part 2. Try the whole set and check every answer instantly.