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).

(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.

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

(a) 7cm

(b) 11cm

(c) 16cm

(d) 21cm

(e) 27cm

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) (64πr

(b) (32πr

(c) (24πr

(d) (16πr

(e) (8πr(a) (64πr

^{2})T(b) (32πr

^{2})T(c) (24πr

^{2})T(d) (16πr

^{2})T(e) (8πr

^{2})T
Work done = Increase in surface area ×T = 2 [4π(2r)

^{2}- 4πr^{2})]T = (24πr^{2})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:

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)]

**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(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 equator

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)]

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