Viscosity and Surface Tension: Important Solved Problems

IMPORTANT SOLVED PROBLEMS

1. Calculate the terminal velocity of a rain drop of diameter 10^‒3m. Density of air relative to water is 1.3 × 10^‒3; η = 1.81 x 10^‒5 Nsm^‒2.

Solution

Given data

Diameter (d) = 1 x 10^‒3m

∴ Radius r = 10^‒3 / 2 = 5 × 10^‒4 m

Density of water ρ = 1000 kg m^‒3

Density of air ρ' = Density of air relative to water × Density of water

ρ' = 1.3 x 10^‒3 × 1000

ρ' = 1.3kg m^‒3

 η = 1.81 × 10^‒5 Nsm^‒2 and g = 9.8 ms^‒2

Formula

Terminal velocity v =

 (or) v = 30ms^‒1

Terminal velocity v = 30ms^‒1

2. Find the radius of the drop of water falling through air, if the terminal velocity of the drop is 1.2 x 10^‒2 ms^‒1. Coefficient of viscosity for air 18 x 10^‒6 Nsm^‒2 and density of air = 1.21 kg m^‒3.

Solution

Given data

 v = 1.2 x 10^‒2 ms^‒1;

 η = 18 x 10^‒6 Nsm^‒2

Density of water drop (ρ) = 1000 kg m^‒3

Density of air (ρ')= 1.21 kg m^‒3;

g = 9.8 ms^‒2,

Formula

From Stokes' formula,

Terminal velocity of the drop of water (v) = 

Radius of the drop of water (r) = 9.968 μm

3. A glass tube of external diameter 2 mm is used and 100 drops of water are collected. The mass of these drops is 2.8 gms. Find the surface tension using drop weight method.

Solution

Given data

Diameter = 2mm (or) 2× 10^‒3m

Radius r = 1 × 10⁻³ m.

Mass m = 2.8×10^‒3 / 100 = 2.8 ×10^‒5 kg

Formula

 σ = m.g / 3.8r

 σ = [ (2.8×10^‒5)9.8 ] / [3.8×10^‒3 ]

∴ σ = 0.07221 Nm^‒1

∴ Surface tension (σ) = 0.07221 Nm^‒1

ASSIGNMENT PROBLEMS

1. Two drops of water of same size are falling through air with terminal velocites of 0.1ms^‒1. If the two drops combine to form a single drop, what will be the new terminal velocity? [Ans : V₁ = 0.159 ms^‒1]

2. Castor oil at 20°C has a coefficient of viscosity 2.42 Nsm^‒2 and a density 940 kgm^‒3. Calculate the terminal velocity of a steel ball of radius 2 x 10^‒2 m falling under gravity in the oil, taking the density of steel as 7800 kg m^‒3. [Ans : 0.025 ms^‒1]

3. A small steel ball bearing falling through glycerine has a terminal velocity of 2 x 10^‒2 ms^‒1. Find the terminal velocity of an air bubble of the same size rising through glycerine at the same temperature. Neglect the weight of air in the bubble and assume that the viscous force on each sphere is proportional to its terminal velocity. (Density of steel 7700 kgm^‒3; Density of glycerine = 1260 kg m^‒3)  [Ans: 39 x 10^‒3 ms^‒1]

4. The difference of pressure between the inside and outside of a soap bubble of diameter 3 x 10^‒3 m is 10^‒2 m of liquid of density 800 kg m^‒3. Calculate the surface tension of the soap solution. (g = 9.8ms^‒2). [Ans 29.4 x 10^‒3 Nm^‒1]

5. A liquid drop of radius R breaks into 64 tiny drops. find the resulting change in energy. Surface tension of the liquid is σ. [Аns: 12πσ R²]