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An oil of specific gravity 0.9 has viscosity of 0.28 strokes at 38°C. What will be its viscosity in Ns/m²?
- a)0.2520
- b)0.0311
- c)0.0252
- d)0.0206
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
An oil of specific gravity 0.9 has viscosity of 0.28 strokes at 38°C....
Dynamic viscosity,
= (0.9 x 1000) x (0.28 x 10-4)
= 0.0252Ns/m2
Most Upvoted Answer
An oil of specific gravity 0.9 has viscosity of 0.28 strokes at 38°C....
To calculate the viscosity in Ns/m², we need to convert the given viscosity in strokes to Ns/m².
Given data:
Specific gravity of oil = 0.9
Viscosity of oil at 38°C = 0.28 strokes
First, we need to convert the specific gravity to density in kg/m³.
Density = specific gravity × density of water
Density = 0.9 × 1000 kg/m³ (density of water at 4°C is 1000 kg/m³)
Density = 900 kg/m³
Next, we need to convert strokes to Ns/m².
1 stroke = 1 cm³/g = 1 × 10⁻⁶ m³/kg
Viscosity in Ns/m² = (Viscosity in strokes) × (Density in kg/m³)
Viscosity in Ns/m² = 0.28 × 10⁻⁶ m³/kg × 900 kg/m³
Viscosity in Ns/m² = 0.252 × 10⁻⁶ Ns/m²
Therefore, the viscosity of the oil at 38°C in Ns/m² is 0.2520 Ns/m².
Hence, the correct answer is option 'A' i.e., 0.2520 Ns/m².
Given data:
Specific gravity of oil = 0.9
Viscosity of oil at 38°C = 0.28 strokes
First, we need to convert the specific gravity to density in kg/m³.
Density = specific gravity × density of water
Density = 0.9 × 1000 kg/m³ (density of water at 4°C is 1000 kg/m³)
Density = 900 kg/m³
Next, we need to convert strokes to Ns/m².
1 stroke = 1 cm³/g = 1 × 10⁻⁶ m³/kg
Viscosity in Ns/m² = (Viscosity in strokes) × (Density in kg/m³)
Viscosity in Ns/m² = 0.28 × 10⁻⁶ m³/kg × 900 kg/m³
Viscosity in Ns/m² = 0.252 × 10⁻⁶ Ns/m²
Therefore, the viscosity of the oil at 38°C in Ns/m² is 0.2520 Ns/m².
Hence, the correct answer is option 'A' i.e., 0.2520 Ns/m².
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An oil of specific gravity 0.9 has viscosity of 0.28 strokes at 38°C. What will be its viscosity in Ns/m²?a)0.2520b)0.0311c)0.0252d)0.0206Correct answer is option 'C'. Can you explain this answer?
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