The shear stress at a point in a liquid is found to be 0.02 N/m². The velocity gradient at this point is 0.20 s⁻¹. What will be the viscosity of the liquid (in poise)?
Try it first: pick an option, then read the explanation.
Asked in SSC JE 07/06/2024 (09am-11am)
Detailed Explanation
The dynamic viscosity of a fluid measures its internal resistance to shear flow. Given the local shear stress and velocity gradient, dynamic viscosity is calculated using Newton's law of viscosity and then converted from SI units to poise.
\[ \tau = \mu \frac{du}{dy} \;\implies\; \mu = \frac{\tau}{du/dy} = \frac{0.02\text{ N/m}^2}{0.20\text{ s}^{-1}} = 0.1\text{ N}\cdot\text{s/m}^2 = 0.1\text{ Pa}\cdot\text{s} \]
Convert to poise (CGS unit):
\[ 1\text{ Pa}\cdot\text{s} = 10\text{ poise} \;\implies\; \mu = 0.1 \times 10 = 1\text{ poise} \]
Why “1” Is Correct
Substituting \(\tau = 0.02\text{ N/m}^2\) and \(\frac{du}{dy} = 0.20\text{ s}^{-1}\) yields \(\mu = 0.1\text{ Pa}\cdot\text{s}\). Multiplying by the conversion factor of 10 gives exactly 1 poise.
| 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
- A. 0.4: 0.4 poise results from an incorrect arithmetic operation (such as 0.02 × 0.20 × 100).
- B. 2: 2 poise results from failing to divide 0.02 by 0.20 correctly.
- C. 10: 10 poise confuses the conversion factor 10 with the calculated viscosity.
Key Point
μ = τ / (du/dy); 1 Pa·s (N·s/m²) = 10 poise.
Key Concepts for Students
- Poise is the CGS unit of dynamic viscosity: 1 poise = 1 dyne·s/cm² = 0.1 Pa·s.
- 1 centipoise (cP) = 10⁻² poise = 1 mPa·s (approximate viscosity of water at 20°C).
- Velocity gradient has dimensions [T⁻¹] (s⁻¹).
Practice more: this is question 2 in Properties of Fluid MCQs – Part 1. Try the whole set and check every answer instantly.