# Fluid Mechanics Questions and Answers – Surface Tension

This set of Fluid Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Surface Tension”.

1. Which of the following contribute to the reason behind the origin of surface tension?
a) only cohesive forces
c) neither cohesive forces nor adhesive forces
d) both cohesive forces and adhesive forces

Explanation: The molecules on the surface of a liquid experience cohesive forces due to surrounding liquid molecules acting downward and adhesive forces due to surrounding gaseous molecules acting upwards. Surface tension orginates due to this unbalanced force on the surface molecules.

2. A soap film is trapped between a frame and a wire of length 10 cm as shown.

If the surface tension is given as 0.0049 N/m, what will be the value of m (in mg) such that the wire remains in equilibrium?
a) 0.1
b) 1
c) 10
d) 100

Explanation: For the wire to be in equilibrium, Force exerted by the film on the wire due to surface tension (acting upwards) must be equal to the downward force due to the weight of the wire (acting downwards). If σ=surface tension, l=length of the wire
2σl = mg
Substituting all the values, m = 2σl/g = 2 * 0.0049 * 0.01 ⁄ 9.81 = 99.9mg.

3. What will be the diameter (in mm) of a water droplet, the pressure inside which is 0.05 N/cm2 greater than the outside pressure? (Take surface tension as 0.075 N/m)
a) 3
b) 0.3
c) 0.6
d) 6

Explanation: p = 4σ/d
where p = pressure difference between the liquid droplet and the surrounding medium, σ = surface tension and d = diameter of the droplet. Substituting all the values,

4. A soap bubble of d mm diameter is observed inside a bucket of water. If the pressure inside the bubble is 0.075 N/cm2, what will be the value of d? (Take surface tension as 0.075 N/m)
a) 0.4
b) 0.8
c) 1.6
d) 4

Explanation: p = 8σ/d
where p = pressure difference between the bubble and the surrounding medium, σ = surface tension and d = diameter of the bubble. Substituting all the values,

5. A liquid jet of 5 cm diameter has how much pressure difference (in N/m2). (Take surface tension as 0.075 N/m)
a) 12
b) 6
c) 3
d) 1.5

Explanation: p = 2σ/r = 4σ/d
where p = pressure difference across the interface of a liquid jet to the surface tension and the radius of the jet, σ = surface tension and d = diameter of the jet. Substituting all the values,
p = (4*0.075) / 5 * 10-2 = 6 N/m2.
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6. The rise in the level of a liquid in a tube is h. What will be the rise in the level if the same amount of liquid is poured into a tube of half the diameter.
a) 0
b) h/2
c) h
d) 2h

Explanation:
where h = rise in liquid height in the tube, S = surface tension, θ = the angle of contact, d = diameter of the tube, ρ = density of liquid and g = acceleration due to gravity. All other factors remaining constant, h α d. Thus, if d is halved, h will be doubled.

7. The ratio of the surface tension S and density ρ of liquid 1 and 2 are 1:2 and 1:4 respectively. Equal amount of the two liquids is poured into two identical tubes. what will be the ratio of the rise in the liquid level in the two tubes? (Assume the angle of contact to be same)
a) 1:2
b) 2:1
c) 8:1
d) 1:8

Explanation:
where h = rise in liquid height in the tube, S = surface tension, θ = the angle of contact, d = diameter of the tube, ρ = density of liquid and g = acceleration due to gravity.
Given, S1 / ρ1 = 1 : 2 and S2 / ρ2 = 1 : 4.

8. The rise in the level of a liquid in a tube is h. If half the amount is poured outside, what will be the new rise in liquid level?
a) 0
b) h/2
c) h
d) 2h

Explanation: The rise in liquid level for a liquid is independent of the amount of liquid present in the tube. Since, same tube is used and same liquid is considered, the rise in the liquid level will remain the same.

9. If a glass tube of 10 mm diameter is immersed in water, what will be the rise or fall in capillary?
(Take surface tension = 0.075 N/m, g = 10 m/s2 and angle of contact = 0)
a) 0.75
b) 1.5
c) 3
d) 6

Explanation:
where h = rise in liquid height in the tube, S = surface tension, θ = the angle of contact, d = diameter of the tube, ρ = density of liquid and g = acceleration due to gravity. Substituting all the values,

10. A water drop of diameter 1 cm breaks into 1000 similar droplets of same diameter. What will be the gain or loss in the surface energy? (Take surface tension as 0.075 N/m)
a) gain of 0.424 mJ
b) gain of 0.212 mJ
c) loss of 0.212 mJ
d) loss of 0.424 mJ

Explanation: According to the Principle of Conservation of mass, M = 1000 * m, where M = mass of the big drop, m = mass of each droplet. Assuming density to be constant, D3 = 1000 * d3, i.e. D = 10d, where D = diameter of big drop, d = diameter of a droplet.
Change in surface energy = Surface tension * Change in surface area = 0:075*(1000 * πd2 – πD2) = 0:075 * (10 * πD2 – πD2) = 0:075 * 9π * (10-2)2 = 0:212 mJ Since, the change is positive, there will be a gain in the surface energy.

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