The diameter of an air bubble which was initially 2 mm, rises steadily through a solution of density 1750 kg m⁻³ at the rate of 0.35 cms⁻¹. The coefficient of viscosity of the solution is _________ poise (in nearest integer). (the density of air is negligible).

Solution

Physics sir · 4 steps

A steady bubble means no net force, so set the upthrust equal to the Stokes drag and solve for η.

1. Balance the forces. The bubble rises at a steady speed, so it is at terminal velocity and the net force on it is zero. Air has negligible density, so its weight is negligible. The upthrust 4/3π r³ ρ g must equal the viscous drag 6πη r v.

   η = (2 r² ρ g)/(9 v)

2. Put in numbers. The radius is half the diameter, so r = 1 mm = 10⁻³ m. The speed is v = 0.35 cm s⁻¹ = 3.5 × 10⁻³ m s⁻¹. Then η = (2 × 10⁻⁶ × 1750 × 9.8)/(9 × 3.5 × 10⁻³) = 0.0343/0.0315 ≈ 1.09 Pa s.
3. Convert to poise. 1 Pa s = 10 poise, so η ≈ 10.9 poise. Taking g = 10 m s⁻² gives 11.1 poise, which rounds to the same answer.
4. So. The coefficient of viscosity is about 11 poise.

The common slip: 44. Using the 2 mm diameter as the radius makes r² four times too big and gives about 44 poise instead of 11.

The idea to keep. Terminal velocity means zero net force. For a bubble, the upthrust balances the drag. Halve the diameter before you square it, and remember that 1 Pa s is 10 poise.

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