Calculate the kinematic viscosity (stoke) of the fluid, if the dynamic viscosity of fluid is 0.5 poise and specific gravity is 0.4?
Try it first: pick an option, then read the explanation.
Asked in SSC JE 24.01.2018 (Morning Shift)
Detailed Explanation
Kinematic viscosity is the quotient of dynamic viscosity and mass density. Performing the calculation directly in consistent CGS units simplifies conversion.
Given: dynamic viscosity \(\mu = 0.5\text{ poise} = 0.5\text{ g}/(\text{cm}\cdot\text{s})\), specific gravity \(S = 0.4\):
\[ \rho = S \times \rho_{\text{water, CGS}} = 0.4 \times 1\text{ g/cm}^3 = 0.4\text{ g/cm}^3 \]
\[ \nu = \frac{\mu}{\rho} = \frac{0.5\text{ poise}}{0.4\text{ g/cm}^3} = \frac{0.5}{0.4} = 1.25\text{ cm}^2/\text{s} = 1.25\text{ stokes} \]
Why “1.25” Is Correct
Dividing 0.5 by 0.4 yields 1.25 stokes, corresponding to option (c).
| 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. 0.95: 0.95 is an incorrect numerical distractor.
- B. 1: 1 stoke results if density is assumed to be 0.5 g/cm³.
- D. 1.5: 1.5 stokes results from using 0.6 / 0.4.
Key Point
Kinematic viscosity ν = μ / ρ; 0.5 poise / 0.4 g/cm³ = 1.25 stokes.
Key Concepts for Students
- In CGS units, water density is exactly 1 g/cm³.
- Specific gravity equals mass density in g/cm³ numerically.
- 1.25 stokes = 1.25 × 10⁻⁴ m²/s in SI.
Practice more: this is question 45 in Properties of Fluid MCQs – Part 3. Try the whole set and check every answer instantly.