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 ~cm / s . Coefficient of viscosity of the solution is ______ (Assume mass of the bubble to be negligible) (Answer in Poise to the nearest integer)

Step-by-step solution

As the bubble is rising steadily the net force acting on it will be zero. (Because of density of air is negligible, so the value of m g can be neglected) So, the force of buoyancy B is equal to the weight of the displaced liquid acting upward is equal to the viscous force acting downward, i.e., B = F . ⇒ 4 π 3 R 3 ρ g = 6 π η R v Putting R = 1   mm = 10 - 3   m , ρ = 1 . 75 × 10 3   kg   m - 3 , g = 10   m   s - 2 and v = 0 . 35 × 10 - 2