Join now for JEE/NEET and also prepare for Boards Learn Science & Maths Concepts for JEE, NEET, CBSE @ Rs. 99! Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
How do Detergents Clean Dirty Clothes?
Properties of Surface Tension
Surface Energy
Simulation for Size of Bubble as a Comparison With Depth
Force Approach
Angle of Contact
Case I: When q < 90^{o}
Case II: When q > 90^{o}
Capillarity
Rise of Liquid in a tube of Insufficient Length
Energy Required to Raise a Liquid in a Capillary tube
Stokes' Law and Terminal Velocity
Related Resources
“Surface tension is the property of a liquid by virtue of which its free surface behaves like a stretched membrane and supports, comparatively heavier objects placed over it”. It is measured in terms of force of surface tension.
The free surface of a liquid contracts so that its exposed surface area is a minimum, i.e., it behaves as if it were under tension, somewhat like a stretched elastic membrane. This property is known as surface tension. The surface tension of a liquid varies with temperature as well as dissolved impurities, etc. When soap is mixed with water, the surface tension of water decreases. Also, the surface tension decreases with increase in temperature.
Consider a wire frame (see the adjacent figure) equipped with a sliding wire AB. It is dipped in a soapy water. A film of liquid is formed on it. A force F has to be applied to hold the wire in place. Since the soap film has two surfaces attached to the wire, the total length of the film in contact with the wire is 2L.
T (surface tension) = F/2L.
Surface tension of a liquid is measured by the normal force acting per unit length. On either side of an imaginary line drawn on the free surface of a liquid, the direction of this force is perpendicular to the line and tangential to the free surface of liquid.
Scalar quantity.
Temperature sensitive.
Impurity sensitive.
Depends only n the nature of the liquid.
Unit of surface tension, N/m.
Dimension of surface tension, ML^{0}T^{-2.}
If the area of the liquid surface has to be increased, work has to be done against the force of surface tension. The work done to from a film is stored as potential energy in the surface and the amount of this energy per unit area of this surface under isothermal condition is the "intrinsic surface energy" or free surface energy density.
Work done in small displacement dx
dW = F × dx = 2TL dx
= 2TLx = TA
As A = 2Lx (area of both sides),
W/A = T (intrinsic surface energy)
Question:-
What is the surface energy of a soap bubble of radius r?
Answer:-
E = TA = T × 4Πr^{2} × 2 (as it has two surfaces)
= 8Πr^{2 }T.
_____________________________________________________________________________________________
What is the surface energy of an air bubble inside a soap solution?
E = T × A = 4Πr^{2}T, as it has only one surface
The pressure inside a soap bubble and that outside it are not identical due to surface tension of the soap bubble. To calculate this pressure difference, let's first consider an air bubble inside a liquid. If the pressure difference is Δp, then the work done to increase the radius of the bubble from r to (r + Δr) is given by:
W = FΔr = 4Πr^{2} Δp Δr
Change in area
ΔS = 4Π (r + Δr)^{2} - 4Πr^{2} = 8ΠrΔr.
From the definition of surface tension
T = W/ΔS = (4Πr^{2} Δp Δr)/(8ΠrΔr) or Δp = 2T/r
For a soap bubble in air, there are two surfaces.
So, Δp = 2 × 2T/r = 4T/r
Bubble bubble boil and trouble. This animaition is used to show size as a comparison with depth. The user move the bubble around while making observations of size versus depth. Then the student can make inferences as to why they get this behavior.
Consider the equilibrium of a hemispherical portion of a liquid bubble of radius R and surface tension T as shown in the figure. For the equilibrium of the liquid bubble.
F_{0} - F_{i }+ F_{T} = 0 = 0 where F_{0} = force due to the outside pressure.
F_{i} = force due to the inside pressure
F_{T} = force due to the surface tension
P_{0}ΠR^{2} - P_{i}ΠR^{2} + 2(2ΠRT) = 0
=> P_{i }- P_{0} = 4T/R
where R is the radius of the bubble.
Two soap bubbles of different radii (r, R: r < R) are connected by means of a tube. What will happen to the larger bubble?
1. Angle of contact, for a solid and a liquid, is defined as the angle between tangent to the liquid surface drawn at the point of contact and the solid surface inside the liquid.
2. The angle of contact of a liquid surface on a solid surface depends on the nature of the liquid and the solid.
The liquid surface curves up towards the solid. This happens when the force of cohesion between two liquid molecules is less than force of adhesion between the liquid and the solid. If such a liquid is poured into a solid tube, it will have a concave meniscus. For example, a glass rod dipped in water or water inside a glass tube.^{}
The liquid surfaces get curved downward in contact with a solid. In this case, the force of cohesion is greater than the force of adhesion. In such cases, solids do not get "wet". When such liquids are put into a solid tube, a convex meniscus is obtained.
For example, a glass rod dipped in mercury or mercury within a solid glass tube.
When a piece of chalk is dipped into water, it is observed that water rises through the pores of the chalk and wets it.
Consider a glass capillary of radius R dipped in water as shown in the figure. The pressure below the meniscus will be (p0 - 2T/r). To compensate for this pressure difference, water in the capillary rises so that
2T/r = rgh or h = 2T/rgρ
where r in the radius of meniscus,
r = R/cos θ
where q is the angle of contact.
Thus,
If q < 90^{o}, the meniscus will be concave, for illustration: at a water-glass interface.
If q < 90^{o}, the meniscus will be convex, for illustration: at a mercury-glass interface.
We have seen, how a liquid rises up into a capillary tube, dipped into it, until the weight of the liquid in the tube is just balanced by the force due to its surface tension. If q be the angle of contact between the liquid and the tube, and R, the radius of liquid meniscus in the tube, we have r = R cos q, where r is the radius of the tube; so that above relation now becomes,
T = (Rcos θ.hρ.g)/(2 cos θ )=(R.h.ρ.g)/2
where h is the height of the liquid column in the tube
Here, clearly
R.h = 2T/ρg, a constant
Now with the tube sufficiently longer than h, it is the value of h alone that changes to satisfy the above relation for T. But if the tube be smaller than the calculated value of h, the only variable in the above relation is R, because now h = l, the length of the tube (a constant) and so is q a constant for the given liquid and the tube. The liquid thus just spreads over the walls of the tube at the top and its meniscus acquires a new radius of curvature R', such that R'l = 2T/ρg, or that R'l = R.h = a constant. And since l is smaller than R' > R, i.e., the meniscus becomes less curved.
We have seen above how when a capillary tube is dipped vertically into a liquid which wets the walls of the tube, there is rise of the liquid inside the tube. Due to the rise, the liquid, gains in potential energy. The question, therefore arises as to where does it get this increase in its potential energy from. The explanation is, however, simple.
We have three surface of separation to consider when a capillary tube is immersed in a liquid viz., (i) an air-liquid surface (ii) an air-glass surface and (iii) a glass-liquid surface each having its own surface tension, difference from the others, and equal to its free surface energy per unit area.
Now, as the plane liquid surface in the tube acquires a curvature, (i.e. become concave), the air-liquid surface increases and, as the liquid rises in the tube, the glass-liquid surface increases, the air-glass surface decreasing by an equal amount. Thus, the surface energy of the air-liquid and the glass-liquid surface increases while that of the air-glass surface decreases by the same amount. In other words, the energy required to raise the liquid in the capillary tube is obtained from the surface energy of the air-glass surface.
On the other hand, a liquid, which does not wet the walls of the tube, get depressed inside it, below its level outside the tube. In this case, obviously the glass-liquid surface decreases, whereas the air-glass surface increases by an equal amount, resulting in a net increase in the surface energy of the whole system. This energy is derived from the depression of the liquid inside the tube, whose gravitational potential energy is thus decreased by an equal amount.
A meniscus drop of radius 1 cm is sprayed into 10^{6} droplets of equal size. Calculate the energy expended if surface tension of mercury is 435 × 10^{-3} N/m.
Solution:-
Energy expended will be the work done against the increase in surface area, i.e.
n(4Πr^{2}) -4ΠR^{2}
E = W = T DS
= T.4Π (nr^{2} - R^{2})
But the total volume remains constant
i.e. 4/3 ΠR^{3} = n 4/3 Πr^{3} or r = R/(n)^{1/3}
=> E = 4 ΠR^{2}T (n^{1/3}- 1) = 4 × 3.14 × (1 × 10^{-2})^{2} × 435 × 10^{-3}(10^{2}- 1)
= 54.1 × 10^{-3} J.
Problem 2 :-
If a number of little droplets of water, all of the same radius, coalesce to form a single drop of radius R, show that the rise in temperature of water will be given by θ = 3T/J (1/r - 1/R) × 10^{-3} where T is the surface tension of water and J, the mechanical equivalent of heat. (in Joule per calorie)
Le the number of little droplets be n and the radius of each droplet r. Then the surface area of all the droplets = n4Πr^{2} and surface area of the single drop formed by their coalescing together = 4ΠR^{2}.
And therefore, decrease in surface area = n4Πr^{2} - 4ΠR^{2} and
=> decrease in surface energy = (n.4Πr^{2} - 4ΠR^{2})
Hence, heat produced = (n.4Πr^{2} - 4ΠR^{2}) T/J
This heat is obviously taken up by the single drop formed of volume 4/3 ΠR^{3} and hence of mass 4/3 ΠR^{3} × 1 taking density of water to be 1000 kg/m^{3}
If therefore, θ^{o}C be the rise in temperature, we have
(4/3 ΠR^{3} × 1) × 1000 × θ = (n.4Πr^{2} - 4ΠR^{2})T/J
or, θ = 3T/J (nr^{3}/R^{3} -1/R) × 10^{-3}
Now, volume of the single drop = volume of n droplets
i.e., 4/3 ΠR^{3} = n.4/3 r^{3}, where R^{3} = nr^{3}
so that, θ = 3T/J (1/r - 1/R) × 10^{-3} [substituting the value of R^{3}]
When a fluid such that a velocity gradient is set up within it, forces act within the fluid so as to prevent the velocity gradient from existing. This force is due to a property called viscosity.
Suppose that a glass plate in contact with a water column of height h is moved with constant velocity v. Forces of viscosity appear between the solid surface and the layer in contact.
i.e. F = ηA dv/dx.
where h is a constant called co-efficient of viscosity. The CGS unit of coefficient of viscosity y is poise, its dimension is ML^{-1}T^{-1}. The SI units of viscosity equal 10 poise.
Problem 3 :-
A metal plate 0.04 m^{2} in area is lying on a liquid layer of thickness 10^{-3} m and co-efficient of viscosity 140 poise. Calculate the horizontal force needed to move the plate with a speed of 0.040 m/s.
Area of the plate, A = 0.04 m^{2}
Thickness,Δx = 10^{-3} m
Δx is the distance of the free surface with respect to the fixed surface
Velocity gradient, Δv/Δx = 22.4 N
Problem 4(JEE Main) :-
A small air bubble of radius r in water is at a depth h below the water surface. If P is atmospheric pressure, d and T are density and surface tension of water respectively, the pressure inside the bubble will be,
(a) P+hdg – (4T/r) (b) P+hdg+(2T/r)
(c) P+hdg – (2T/r) (d) P+hdg+(4T/r)
P_{in} – P_{out} = 2T/r
So, P_{in} = P_{out} + (2T/r)
_{ }= (P+hdg)+(2T/r)
From the above observation we conclude that, option (b) is correct.
When a smooth sphere of radius r moves with a velocity v through a fluid of viscosity η , the viscous force opposing the motion of the sphere is
F = 6Πηrv
If, for a sphere, viscous force becomes equal to the net weight acting downward, the velocity of the body becomes constant and is known as termination velocity.
6Πηrv_{T} = 4/3 Πr^{3} (ρ - σ)g
So,
(a) the liquid wets the solids.
(b) shape of meniscus is concave upwards.
(c) liquid rises up into a capillary tube made of that solid.
(a) the liquid does not wet the solids.
(b) shape of meniscus is convex upwards.
(c) the liquid is depressed in a capillary tube made of that solid.
Problem 5 (JEE Main):-
The radii of the two columns is U-tube are r_{1} and r_{2}. When a liquid of density ρ (angle of contact is 0º) is filled in it, the level difference of liquid in two arms is h. Find out the surface tension of liquid.
We know that, h = 2T/rρg
So, h_{1} = 2T/r_{1}ρg
h_{2} = 2T/r_{2}ρg
h_{1}- h_{2} = h = 2T/ρg(1/r_{1} – 1/r_{2})
Thus, T = hρgr_{1}r_{2}/2(r_{2 }- r_{1})
Problem 6 (JEE Main):-
A metal ball A (density 3.2 g/cc) is dropped in water, while another metal ball B (density 6.0 g/cc) is dropped in a liquid of density 1.6 g/cc. If both the balls have the same diameter and attain the same terminal velocity, find out the ratio of viscosity of water to that of the liquid.
Terminal velocity is given by,
v_{T} = 2/9 [r^{2}(ρ_{1} - ρ_{2})/η]g
(v_{T})_{A} = 2/9 [r^{2}(32 - 1)/η_{w}]g
(v_{T})_{B} = 2/9 [r^{2}(6.0 - 1.6)/η_{l}]g
Given, (v_{T})_{A} = (v_{T})_{B}
Equating these two we get, η_{w}/η_{l} = 0.5
Question 1 :-
Two molecules are separated an appreciable distance apart. What is the nature of the force between them:
(a) attractive (b) repulsive
(c) both attractive and repulsive (d) none of them
Question 2 :-
The force of surface tension acts in such a direction that the curvature of the surface should:
(a) increase (b) decrease
(c) remain the same (d) none of these
Question 3 :-
When the temperature is increased, the angle of contact of a liquid:
(a) first increases and then decreases (b) decreases
(c) remains the same (d) increases
Question 4:-
When a capillary tube is dipped in a liquid, the level of the liquid inside the tube rises because of:
(a) viscosity (b) surface tension
(c) osmosis (d) diffusion
a
d
b
You might like to refer Solved Examples on Fluid Mechanics.
For getting an idea of the type of questions asked, refer the Previous Year Question Papers.
Click here to refer the most Useful Books of Physics.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Fluid Statics Table of Content Fluid Fluid...
Fluid Dynamics Table of Content Bernoulli's...
Archimedes Principle Table of Content Refer this...
Solved Examples on Fluid Mechanics Problem 1 :-...