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).

Question

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). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

As the bubble is rising steadily the net force acting on it will be zero

(Because of density of air the value of mg can be neglected)
So

B = F \Rightarrow \frac{4 π}{3} R^{3} ρ g = 6 π η R v

Putting

R = 1 mm = 1 0^{- 3} m

ρ = 1.75 \times 1 0^{3} k g / m^{3}

g = 10 m / s^{2}

v = 0.35 \times 1 0^{- 2} m / s

η = \frac{10}{9} ≃ 1 \cdot 11 SI unit = 11 poise (CGS)

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

As the bubble is rising steadily the net force acting on it will be zero (Because of density of air the value of mg can be neglected) So B=F⇒34π​R3ρg=6πηRv Putting R=1mm=10−3m ρ=1.75×103kg/m3 g=10m/s2 v=0.35×10−2m/s η=910​≃1⋅11 SI unit =11 poise (CGS)

Q. The diameter of an air bubble which was initially
$2 mm$ , rises steadily through a solution of density $1750 k g m^{- 3}$ at the rate of $0.35 c m s^{- 1}$ . The coefficient of viscosity of the solution is poise (in nearest integer). (the density of air is negligible).