An air bubble of negligible weight having radius r rises steadily through a solution of density σ at speed v. The coefficient of viscosity of the solution is given by :

Solution

Physics sir · 4 steps

The bubble moves at a steady speed, so the buoyant push up exactly balances the viscous drag down.

1. Forces on bubble. The bubble's weight is negligible, so only two forces act. Buoyancy pushes it up and is equal to the weight of displaced liquid, 4/3π r³ σ g. Stokes drag acts opposite to the motion, so it points down, with size 6π η r v.
2. Steady speed. Constant velocity means zero acceleration, so the net force is zero and the two forces are equal.

   4/3π r³ σ g = 6π η r v

3. Solve for eta. η = (4π r³ σ g)/(3 × 6π r v) = (4 r² σ g)/(18 v) = (2 r² σ g)/(9 v). The π cancels and one power of r cancels.
4. So. The coefficient of viscosity is η = (2 r² σ g)/(9 v), which is option (b).

The idea to keep. Terminal or steady speed always means the forces balance. Write buoyancy, weight and 6πη r v, set the net force to zero, and the π cancels to give the familiar 2/9 factor.

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