An air bubble of diameter 6 mm rises steadily through a solution of density 1750 kg / m³ at the rate of 0.35 cm / s. The co-efficient of viscosity of the solution (neglect density of air) is ___________ Pas (given, g=10 ms⁻² ).

Solution

Physics sir · 4 steps

A bubble rising steadily has no net force, so buoyancy exactly balances Stokes drag. Solve that for η.

1. Balance the forces. The bubble moves at a constant speed, so it is at terminal velocity and the net force on it is zero. The weight of the air is neglected. That leaves the upward buoyant force 4/3π r³ ρ g balanced by the downward viscous drag 6πη r v.

   6πη r v = 4/3π r³ ρ g ⇒ η = (2 r² ρ g)/(9 v)

2. Convert to SI. The radius is half the diameter, so r = 3 mm = 3 × 10⁻³ m. The speed is v = 0.35 cm s⁻¹ = 3.5 × 10⁻³ m s⁻¹.
3. Put in numbers. The numerator is 2 × 9 × 10⁻⁶ × 1750 × 10 = 0.315. The denominator is 9 × 3.5 × 10⁻³ = 0.0315. So η = 0.315/0.0315 = 10 Pa s.
4. So. The coefficient of viscosity of the solution is 10 Pa s.

The common slip: 40. Using the diameter of 6 mm as r makes r² four times too large and gives 40, not 10.

The idea to keep. Steady motion through a fluid means terminal velocity, so the net force is zero. Set the drag 6πη rv equal to the net of buoyancy and weight, and always use the radius in SI units.

Answer from the NTA final answer key. Solution by PrepWiser.