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A solids sphere (diameter 6 mm) is rising through oil (mass density 900 kg/m3, dynamic viscosity 0.7 kg/ms) at a constant velocity of 1 cm/s. What is the specific weight of the material from which the sphere is made? (Take g = 9.81 m/s2)
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- 4.3 kN/m3
- 5.3 kN/m3
- 8.7 kN/m3
- 12.3 kN/m3
Correct Option: B
In dynamic equilibrium conditions, Upward force = Downward force
⇒ Buoyant force = weight of sphere + Drag force.
| ρoil. v.g = ρsphere. v.g + | cd.ρsphere.A V2 | |
| 2 |
ρoil = 900 kg/m3
V = Surface volume of sphere
| = | πR3 | |
| 3 |
| R = | = 0.03 m | |
| 2 |
| ∴ V = | π × (0.003)3 = 1.31 × 10-7 m3 | |
| 3 |
| Re = | ||
| υoil |
| = | = | |||||
 |  | (0.7 / 900) | ||||
| e | ||||||
= 0.0771 < 1
| ∴ CD = | ||
| Re |
A = πR2
= π × (0.003)2
= 0.283 × 10-4 m2
V = 0.01 m / s
| ∴ 900 × 1.131 × 10-4 = ρsphere × 1.131 × 10-7 × 9.81 + | | × 311.28 × 900 × 0.283 × 10-4 × (0.01)2 | | |
| 2 |
∴ ρsphere = 543.37 kg / m3
⇒ γsphere = 543.4 × 9.81 N / m3 ≈ 5.33 kN / m3
