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).
To find the coefficient of viscosity (η), we use Stokes' Law for a sphere moving through a viscous fluid: Fd = 6πηrv, where Fd is the drag force, r is the radius, and v is the velocity. For a sphere moving at terminal velocity, this force equals the buoyant force minus the gravitational force acting on the sphere. Since the density of air is negligible, the buoyant force dominates.
1. Calculate the radius of the bubble: r = diameter/2 = 2 mm / 2 = 1 mm = 0.001 m.
2. Convert the velocity to m/s: v = 0.35 cm/s = 0.0035 m/s.
3. Calculate the buoyant force using the fluid density (ρ) and the volume of the sphere (V):
4. Using Stokes' Law, equate the buoyant force to the viscous drag force to find η:
Fd = Fb = 6πηrv.
7.229 × 10–5 N = 6π(η)(0.001 m)(0.0035 m/s).
Solve for η: η ≈ (7.229 × 10–5 N) / (6π × 0.001 m × 0.0035 m/s).
η ≈ 11 Poise (1 Poise = 0.1 Pa.s).
The calculated viscosity is approximately 11 Poise, which fits within the specified range of 11,11. Hence, the coefficient of viscosity is conclusively 11 Poise.
