Kerala Engineering Entrance 2012 (KEAM Engg. 2012) Questions on Surface Tension

“It doesn't matter how beautiful your theory is; it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong.”

– Richard Feynman

Two questions on surface tension were included in the KEAM (Engineering) 2012 question paper. Here are the questions with solution:

(1) If two capillary tubes of radii r₁ and r₂ in the ratio 1 : 2 are dipped vertically in water, then the ratio of capillary rises in the respective tubes is

(a) 1 : 4

(b) 4 : 1

(c) 1 : 2

(d) 2 : 1

(e) 1 : √2

The capillary rise h due to surface tension is relate to the surface tension S as

            S = hrρg/2 cosθ

where r is the radius of the capillary tube, ρ is the density of the liquid, g is the acceleration due to gravity and θ is the angle of contact.

Evidently h is inversely proportional to r.

Therefore, h₁/h₂ = r₂/r₁ = 2 : 1

(2) If the excess pressure inside a soap bubble of radius r₁ in air is equal to the excess pressure inside air bubble of radius r₂ inside the soap solution, then r₁ : r₂ is         

(a) 2 : 1

(b) 1 : 2

(c) 1 : 4

(d) √2 : 1

(e) 1 : √2

The excess pressure inside a soap bubble in air is 4S/r where as the excess pressure inside an air bubble in the soap solution is 2S/r where S is the surface tension (of soap solution) and r is the radius of the bubble.

[In the case of the air bubble in the soap solution there is one liquid surfaoe only and that is why the excess pressure is 2S/r and not 4S/r. Remember that in the case of a soap bubble in air there are two liquid surfaces]. 

As given in the question, we have

            4S/r₁ = 2S/r₂

This gives r₁/r₂ = 2

Or, r₁ : r₂ = 2 : 1

EAMCET 2010 Questions (MCQ) on Surface Tension

Today we will discuss two questions on surface tension which appeared in EAMCET 2010 question papers.

The following multiple choice question appeared in EAMCET 2010 (Engineering) question paper:

The excess pressure inside a spherical soap bubble of radius 1 cm is balance by a column of oil (Sp. gravity 0.8) 2 mm high. The surface tension of oil is

(1) 3.92 N/m

(2) 0.0392 N/m

(3) 0. 392 N/m

(4) 0.00392 N/m

The excess pressure ∆P insie a spherical bubble of radius r is given by

∆P = 4 T/r where T is the surface tension.

The pressure p exerted by a liquid column of height h is given by

p = hρg where ρ is the density of the liquid.

Therefore we have

4 T/r = hρg from which T = rhρg/4

Substituting for known values, T = (1×10^–2×2×10^–3×0.8×10³×10)/4 = 0.004 N/m [Option (4)].

[If you substitute g = 9.8 ms^–2 instead of 10 ms^–2 (as we did above for convenience), you will get the answer as 0.00392 N/m].

Here is the question which appeared in EAMCET 2010 (Agriculture and Medicine) question paper:

A spherical liquid drop of diameter D breaks up to n identical spherical drops. If the surface tension of the drop is ‘σ’, the change in energy in this process is

(1) πσD²(n^1/3 – 1)

(2) πσD²(n^2/3 – 1)

(3) πσD²(n – 1)

(4) πσD²(n^4/3 – 1)

The surface energy (E₁) of a drop of radius R is given by

E₁ = 4πR²σ

Or E₁ = πD²σ where D is the diameter of the drop.

When a drop of radius R breaks into n identical droplets, the radius r of each droplet is given by (on equating the volumes)

(4/3) πR³ = n×(4/3) πr³

Therefore, r = R/n^1/3

The total surface energy (E₂) of all the n droplets is given by

E₂ = n×4πr²σ = n×4π (R/n^1/3) ²σ = 4πR²σn^1/3

Or, E₂ = πD²σn^1/3

The change in energy is E₂ – E₁ = πσD²(n^1/3 – 1)

Surface Tension–Questions involving excess of pressure inside a bubble

Here is a question which is popular among question setters:

Two spherical soap bubbles of radii r₁ and r₂ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to

(a) (r₁+r₂)/2 (b) √(r₁+r₂) (c) r₁r₂/(r₁+r₂) (d) (r₁+r₂)/√2 (e) √(r₁²+r₂²)

As the temperature is constant, we have P₁V₁+ P₂V₂ =.PV where P₁ and P₂ are the pressures inside the separate bubbles, V₁ and V₂ are their volumes, P is the pressure inside the combined bubble and V is its volume.

Since the bubbles are located in vacuum, the pressure inside the bubble is equal to the excess of pressure 4T/r, where T is the surface tension and ‘r’ is the radius so that we have (4T/r₁)×[(4/3)πr₁³]+ (4T/r₂)×[(4/3)πr₂³] = (4T/R)×[(4/3)πR³] where R is the radius of the combined bubble.

This yields R = √(r₁²+r₂²).

[You can work out this problem by equating the surface energies: 4πr₁²T+4πr₂²T= 4πR²T, from which R = √(r₁²+r₂²)]

Now, consider the following MCQ:

Excess of pressure inside one soap bubble is four times that inside another. Then the ratio of the volume of the first bubble to that of the second is

(a) 1:16 (b) 1:32 (c) 1:64 (d) 1:2 (e) 1:4

If r₁ and r₂ are the radii of the bubbles and T is the surface tension of soap solution, we have (4T/r₁)/(4T/r₂) = 4.

Therefore, r₁/r₂ = ¼. Since the volume is directly proportional to the cube of the radius, the ratio of volumes V₁/V₂ = (r₁/r₂)³ = (¼)³ = 1/64 [Option (d)].

The following MCQ appeared in AIEEE 2004 question paper. This question is popular among question setters and has appeared in other entrance test papers as well:

If two soap bubbles of different radii are connected by a tube,

(a) air flows from the bigger bubble to the smaller bubble till the sizes become equal

(b) air flows from the bigger bubble to the smaller bubble till the sizes are interchanged

(c) air flows from the smaller bubble to the bigger

(d) there is no flow of air

The excess of pressure inside the smaller bubble is greater than that inside the bigger bubble. (Remember, P = 4T/r and hence excess of pressure ‘P’ is inversely proportional to the radius ‘r’). Therefore, air will flow from the smaller bubble to the bigger bubble [Option (c)].

Properties of Fluids

In any Medical and Engineering entrance test paper you will find a few questions based on the properties of fluids. Let us discuss some typical questions. The following simple question is taken from the A.F.M.C.2004 test paper:
Application of Bernoulli’s theorem can be seen in
(a) dynamic lift of aeroplane (b) hydraulic press (c) helicopter (d) none of these
The correct option is (a). The shape of the aeroplane wings is such that when it moves forward, the air molecules at the top of the wings have a greater velocity (relative to the wings) compared to the air molecules at the bottom. Therefore, in accordance with Bernoulli’s principle, the pressure at the top of the wings is less than that at the bottom. This results in a dynamic lift of the wings which balances the weight of the plane.
Let us now consider another question involving Bernoulli’s principle:
Water is flowing steadily through two horizontal pipes of radii 3cm and 6cm connected in series. The speed of water in the first pipe is 2m/s and the pressure of water in it is 2×10⁴ pascal. The pressure of water in the second pipe will be nearly
(a) 2×10⁴ pascal (b) 2.2×10⁴ pascal (c) 2.4×10⁴ pascal (d) 2.6×10⁴ pascal (e) 3×10⁴ pascal
The speed v₂ of water in the second pipe is given (from the equation of continuity, a₁v₁ = a₂v₂ ) by v₂ = 2×a₁/a₂ = 2×1/4 = 0.5 m/s. (The ratio of cross section areas is ¼ since the radius of the second pipe is twice that of the first).
As per Bernoulli’s principle, we have P₁ + (ρv₁² )/2 + ρgh = P₂ + (ρv₂²)/2 + ρgh with usual notations. Therefore, P₂ = P₁ + ρ(v₁²- v₂²)/2 =2×10⁴ + 1000(2² – 0.5²)/2 = 2.1875×10⁴ pascal. The correct option therefore is (b).
The following question involving surface tension may appear to be difficult for some of you. See how simple it is:
Two soap bubbles of radii ‘R’ and ‘r’ (R > r) are touching each other. The radius of curvature of the common surface at the region of contact is
(a) (R+r)/2 (b) R – r (c) (R – r)/2 (d) Rr/(R+r) (e) Rr/(R-r)
At the contact region, the radius of curvature (R') of the soap film is governed by the net excess of pressure 4T/r – 4T/R where T is the surface tension. Therefore, we have 4T/r – 4T/R = 4T/R' from which R' = Rr/R-r [option (e)].
Consider now the following M.C.Q.:
Two rain drops are falling through air with the same terminal velocity of 8cm/s. If they coalesce, the terminal velocity of the combined single drop will be (in cm/s)
(a) 8/√2 (b) 16 (c) 8×4^1/3 (d) 8×2^1/3 (e) 8√2
On equating the apparent weight of a rain drop to the viscous force (air resistance) opposing the downward motion of the drop, we have, 4/3πr³(ρ-σ)g = 6πrηv, with usual notations. Therefore, the velocity, v α r²…………(1)
Since the drops have the same terminal velocity, it follows that they have the same radius. The radius (r₁ ) of the single large drop (when the two drops coalesce) is given by
2× 4/3πr³ = 4/3 πr₁³. Therefore, r₁=2^1/3 r.
The velocity of the combined single drop, v₁ α [2^1/3 r]² -------------- (2)

From equations (1)&(2), v₁= 4^1/3v = 4^1/3 × 8, since v=8cm/s. The correct option therefore is (c).

Questions on Surface Tension

God used beautiful mathematics in creating the world.

– P. A. M. Dirac

 
You might be remembering the expression for the excess of pressure ‘p’ inside a liquid drop: p = 2T/r where ‘T’ is the surface tension of the liquid and ‘r’ is the radius of the drop. The excess pressure is due to surface tension by which the free surface of a liquid behaves like a stretched membrane. The excess of pressure inside a bubble is 4T/r, which is twice that inside a drop because a bubble has two free surfaces (inner and outer).
Consider the following question which often finds a place in Medical and Engineering Entrance test papers:
Two soap bubbles A and B are obtained at the ends of a glass tube with a stop cock at its middle. The stop cock is closed initially and the bubble B is larger in size. If the stop cock is opened what happens to the bubbles?
(a) Nothing
(b) The size of A decreases and the size of B increases
(c) The size of A increases and the size of B decreases
(d) Both A and B will decrease in size
(e) Both A and B will increase in size.The excess pressure inside the smaller bubble A is greater and hence air from it will rush to B through the tube. The size of A will decrease further while the size of B will increase. The correct option therefore is (b).
You should note that the pressure on the concave side of a liquid surface is always greater by 2T/r. Consider the following question:
A tall metallic jar has a small hole of radius 0.07mm at its bottom. What is the approximate depth up to which the jar can be lowered vertically in water before any water penetrates in to it through the hole? (Surface tension of water = 0.073N/m)
(a) 7cm
(b) 11cm
(c) 16cm
(d) 21cm
(e) 27cmWater will begin to enter the jar through the hole when the hydrostatic pressure exceeds the allowed excess pressure of 2T/r on the concave side of the water surface at the hole. Therefore, the depth ‘h’ up to which the jar can be lowered safely is given by
hdg = 2T/r so that h = 2T/dgr = 2×0.073/(1000×9.8×0.00007) = 0.21m =21cm.

Let us consider one more question involving surface tension:
If T is the surface tension of soap solution, the amount of work done in increasing the radius of a soap bubble from r to 2r is
(a) (64πr²)T
(b) (32πr²)T
(c) (24πr²)T
(d) (16πr²)T
(e) (8πr²)T

Work done = Increase in surface area ×T = 2 [4π(2r)² - 4πr² )]T = (24πr²)T. So, the correct option is (c). Note that the bubble has two surfaces.

The following MCQ appeared in Kerala Engineering Entrance 2006 test paper:Water rises in a capillary tube to a height ‘h’. Choose FALSE statement regarding capillary rise from the following:
(a) On the surface of Jupiter, height will be less than h.
(b) In a lift moving up with constant acceleration, height is less than h.
(c) On the surface of moon, height is more than h.
(d) In a lift moving down with constant acceleration, height is less than h.
(e) At the poles, height is less than that at equatorThe expression for surface tension (T) in terms of capillary rise in a capillary tube is
T = hrdg/2cosθ where ‘h’ is the capillary rise (or fall), ‘r’ is the radius of the tube, ‘d’ is the density of the liquid, ‘g’ is the acceleration due to gravity and ‘θ’ is the angle of contact. When ‘g’ is less, the capillary rise ‘h’ is more and vice versa. The value of ‘g’ is smaller in options (c) and (d) only so that the value of ‘h’ is greater in these two cases. So, option (d) is false. [Note that when a lift moves down with acceleration ‘a’, the effective value of acceleration due to gravity is (g-a)]