Viscosity of a fluid with specific gravity 1.3 is measured to be 0.0034 Ns/m². Its kinematic viscosity, in m²/s, is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 1 MARCH 2017 Morning Shift
Detailed Explanation
Kinematic viscosity is determined by dividing dynamic viscosity by the fluid's mass density. Density is calculated by multiplying specific gravity by standard water density (1000 kg/m³).
Given: \(\mu = 0.0034\text{ N}\cdot\text{s/m}^2\), \(S = 1.3\):
\[ \rho = S \times \rho_{\text{water}} = 1.3 \times 1000\text{ kg/m}^3 = 1300\text{ kg/m}^3 \]
\[ \nu = \frac{\mu}{\rho} = \frac{0.0034}{1300} = \frac{34 \times 10^{-4}}{13 \times 10^2} = \frac{34}{13} \times 10^{-6} \approx 2.615 \times 10^{-6}\text{ m}^2/\text{s} \]
Why “2.6 × 10⁻⁶” Is Correct
Dividing 0.0034 by 1300 gives 2.615 × 10⁻⁶ m²/s, which rounds to 2.6 × 10⁻⁶ m²/s.
| 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. 4.4 × 10⁻⁶: 4.4 × 10⁻⁶ results from an arithmetic error in dividing 0.0034.
- C. 5.8 × 10⁻⁶: 5.8 × 10⁻⁶ is an incorrect distractor.
- D. 7.2 × 10⁻⁶: 7.2 × 10⁻⁶ corresponds to multiplying rather than dividing by 1.3.
Key Point
ν = μ / ρ; 0.0034 / 1300 ≈ 2.6 × 10⁻⁶ m²/s.
Key Concepts for Students
- Specific gravity S = 1.3 gives density ρ = 1300 kg/m³.
- Kinematic viscosity ν has dimensions [L² T⁻¹].
- 2.6 × 10⁻⁶ m²/s = 2.6 cSt.
Practice more: this is question 65 in Properties of Fluid MCQs – Part 5. Try the whole set and check every answer instantly.