Solids And Fluids of Class 11

When a solid body slides over another solid body, a frictional-force begins to acts between them. This force opposes the relative motion of the bodies. Similarly, when a layer of a liquid slides over another layer of the same liquid, a frictional-force acts between them which opposes the relative motion between the layers. This force is called ‘internal frictional-force’.


Suppose a liquid is flowing in streamlined motion on a fixed horizontal surface AB. The layer of the liquid which is in contact with the surface is at rest, while the velocity of other layers increases with distance from the fixed surface. In the figure, the lengths of the arrows represent the increasing velocity of the layers. Thus there is a relative motion between adjacent layers of the liquid. Let us consider three parallel layers a, b and c. Their velocities are in the increasing order. The layer a tends to retard the layer b, while b tends to retard c.

 Thus each layer tends to decrease the velocity of the layer above it. Similarly, each layer tends to increase the velocity of the layer below it. This means that in between any two layers of the liquid, internal tangential forces act which try to destroy the relative motion between the layers. These forces are called ‘viscous forces’. If the flow of the liquid is to be maintained, an external force must be applied to overcome the dragging viscous forces. In the absence of the external force, the viscous forces would soon bring the liquid to rest.

The property of the liquid by virtue of which it opposes the relative motion between its adjacent layers is known as viscosity.

The property of viscosity is seen in the following examples:

(i) A stirred liquid, when left, comes to rest on account of viscosity. Thicker liquids like honey, coaltar, glycerine, etc. have a larger viscosity than thinner ones like water. If we pour coaltar and water on a table, the coaltar will stop soon while the water will flow upto quite a large distance.

(ii) If we pour water and honey in separate funnels, water comes out readily from the hole in the funnel while honey takes enough time to do so. This is because honey is much more viscous than water. As honey tends to flow down under gravity, the relative motion between its layers is opposed strongly.

(iii) We can walk fast in air, but not in water. The reason is again viscosity which is very small for air but comparatively much larger for water.

(iv) The cloud particles fall down very slowly because of the viscosity of air and hence appear floating in the sky.

Viscosity comes into play only when there is relative motion between the layers of the same material. This is why it does not act in solids.

Flow of liquid in Tube: Critical Velocity

When a liquid flows in a tube, the viscous forces oppose the flow of the liquid. Hence a pressure difference is applied between the ends of the tube which maintains the flow of the liquid. If all particles of the liquid passing through a particular point in the tube move along the same path, the flow of the liquid is called ‘stream-lines flow’. This occurs only when the velocity of flow of the liquid is below a certain limiting value called ‘critical velocity’. When the velocity of flow exceeds the critical velocity, the flow is no longer stream-lined but becomes turbulent. In this type of flow, the motion of the liquid becomes zig-zag and eddy-currents are developed in it.

Reynold proved that the critical velocity for a liquid flowing in a tube is vc = kη/ρa, where ρ is density and η is viscosity of the liquid, a is radius of the tube and k is ‘Reynold’s number’ (whose value for a narrow tube and for water is about 1000). When the velocity of flow of liquid is less than the critical velocity, then the flow of the liquid is controlled by the viscosity, the density having no effect on it. But when the velocity of flow is larger than the critical velocity, then the flow is mainly governed by the density, the effect viscosity becoming less important. It is because of this reason that when a volcano erupts, then the lava coming out of it flows speedly inspite of being very thick (of large viscosity).

Velocity Gradient and Coefficient of Viscosity

Suppose a liquid is flowing in stream-lined motion on a horizontal surface OX. The liquid layer in contact with the surface is at rest while the velocity of other layers increases with increasing distance from the surface OX. The highest layer flows with maximum velocity. Let us consider two parallel layers PQ and RS at distance z and z + Δz from OX. Let vx and vx + Δvx be their velocities in the direction OX. Thus the change in velocity in a perpendicular distance Δz is Δvx. That is, the rate of change of velocity with distance perpendicular to the direction of flow is Δvx/Δz. This is called velocity-gradient.

Now let us consider a liquid layer of area A at a height z above OX. The layer of the liquid immediately above it tends to accelerate it with a tangential viscous force F, while the layer immediately below it tends to retard it backward with the same tangential viscous force F. According to Newton, the viscous force F acting between two layers of a liquid flowing in stream-lined motion depend upon two factors:

(i) It is directly proportional to the contact-area A of the layers (F ∝ A).

(ii) It is directly proportional to the velocity-gradient Δvx/Δz between the layers (F ∝ Δvx/Δz).

Combining both these laws, we have

F ∝ Viscosity


where η is a constant called ‘coefficient of viscosity’ of the liquid. In this formula if
A = 1 and Δvx/Δz = 1, then η = ±F. Thus, the coefficient of viscosity of a liquid is defined as the viscous force per unit area of contact between two layers having a unit velocity gradient between them.

In the above formula, ± signs indicate that the force F between two layers is a mutual-interaction force. On the layer A, the layer above it exerts a force in the forward direction while the layer below it exerts an equal force in the backward direction.

Dimensions and Unit of Coefficient of Viscosity: From the above formula, we have


∴ dimensions of η = Viscosity = [ML-1T-1]

Its unit is kg/(meter-second).

Effect of Temperature on Viscosity: The viscosity of liquids decreases with rise in temperature. On the other hand, the viscosity of gases increases with rise in temperature.

Steady Flow of Liquid through a Capillary Tube: Poiseuille’s Formula

Poiseuille gave a formula for the volume of a liquid flowing per second through or a capillary tube of uniform cross-section under a constant pressure difference across the ends of the tube. For this, Poiseuille assumed that the flow of liquid in the tube is streamlined and parallel to the axis of the tube and the pressure difference across the tube is just enough to overcome the viscous force. Also, the liquid in contact with the walls of the tube is rest.

If l and a be the length and radius of the tube and p the pressure difference across its ends, then the volume of the liquid flowing per second through the tube is given by


Where η is the coefficient of viscosity of the liquid. This is known as ‘Poiseuille’s formula’.

Effective Force on a Body Falling in a Liquid: Stoke’s Law

When a body falls through a liquid, ordinarily two forces act upon the body: weight of the body and the upthrust force due to the buoyancy of the liquid. As an example, suppose we drop a lead ball in a long jar filled with water. Let V be the volume and ρ the density of the ball. Then the mass of the ball is

m = V × ρ

Now, two forces act upon the ball:

(i) Weight of the ball, mg = V × ρ × g

(ii) Upthrust of water = V × σ × g

where σ is the density of water. The effective downward force on the ball is

Vρg – Vσg = V(ρ - σ)g.

The acceleration of the ball in water will be

g′ = Viscosity(1)

For lead, ρ = 11 and for water σ = 1.

∴g′ = Viscosity

If we drop marble (ρ = 2.5) ball, instead of lead ball, in water, then the acceleration of the marble in water is


The acceleration of the lead ball in water is nearly one and half times the acceleration of the marble ball. In the above calculation we have not considered the effect of viscosity of water. Clearly, the difference in densities of the bodies affects the motion of bodies falling in liquid.

If ρ < σ, as air bubble in water, the observed acceleration g′ of the bubble will be negative, (1), that is, the bubble will rise.

Now, we consider the viscosity of water. The velocity of the falling ball goes on increasing due to the effective force V(ρ - σ)g acting downward on the ball. The layer of water in contact with the ball tends to move with the velocity of the ball. The layer of water in contact with the ball; while the layer far away from the ball remains at rest. Thus as a result of the motion of the ball, there is a relative motion between adjacent layers of water. Hence a viscous force is developed in water which increases with increasing velocity of the ball, and a stage is reached when the opposite viscous force becomes equal to the effective force driving the ball. The net force on the ball is now zero and the ball attains a constant velocity called the ‘terminal velocity’. Thus viscous force limits the speed of the body falling in a liquid.

Stokes showed that if a small sphere of radius r is moving with a terminal velocity v through a perfectly homogeneous medium (liquid or gas) of infinite extension, then the viscous force acting on the sphere is


where η is the coefficient of viscosity of that medium.

Calculation of Terminal Velocity:

Let us consider a small ball, whose radius is r and density is ρ, falling freely in a liquid (or gas), whose density is σ and coefficient of viscosity η. When it attains a terminal velocity v, it is subjected to two forces:

(i) Effective force acting downward

= V(ρ - σ)g = Viscosity

(ii) Viscous force acting upward

= 6πηrv


Since the ball is moving with a constant velocity v i.e., there is no acceleration in it, the net force acting on it must be zero. That is


Thus, terminal velocity of the ball is directly proportional to the square of its radius.

Example 12.23

A drop of water of mass m = 0.2 g is placed between two clean glass plates, the distance between which is 0.01 cm. Find the force of attraction between the plates. Surface tension of water = 0.07 Nm-1.


Let R be the radius of the circular layer of water. Then πR2d × ρ = m

Pressure at A = Viscosity (meniscus is cylindrical in shape)

Thus pressure between the plates is less than the atmospheric pressure and so the plates are pressed together as though attracted towards each other.


F, force of attraction = Δp × area ⇒ F = Viscosity

Viscosity N

Example 12.24

A glass capillary sealed at the upper end is of length 0.11 m and internal diameter 2 × 10-5 m. The tube is immersed vertically into a liquid of surface tension 5.06 × 10-2 N/m. To what length has the capillary to be immersed so that the liquid level inside and outside the capillary becomes same? What will happen to the water level inside the capillary if the seal is now broken?


If A is the cross-sectional area of the tube and L its length, the initial volume of air inside it will be V1 = AL while pressure p1 = po = atmospheric pressure.

Now when the tube is immersed in water with its length x in water, the level of water inside and outside is same, so the volume of air in the tube will be V2 = A(L - x). Further if p2 is the pressure of gas in the tube,


Now if temperature is constant,




If the seal is broken the pressure inside the capillary will become atmospheric, i.e. po while capillarity will take place and the rise will be

Viscosity m

However, the length of the tube outside the water is 0.11 - 0.01 = 0.1 m; so the tube will be of insufficient length and so the liquid will rise to the top of the tube and will stay with radius of meniscus,


Example 12.25

A conical glass capillary tube of length 0.1 m has diameters 10-3 and 5 × 10-4 m at the ends. When it is just immersed in a liquid at 0oC with larger diameter in contact with it, the liquid rises to 8 × 10-2 m in the tube. If another cylindrical glass capillary tube B is immersed in the same liquid at 0oC, the liquid rises to 6 × 10-2 m height. The rise of liquid in the tube B is only 5.5 × 10-2 m when the liquid is at 50oC. Find the rate at which the surface tension changes with temperature considering the change to be linear. The density of the liquid is (1/14) × 104 kg/m3 and angle of contact is zero. Effect of temperature on density of liquid and glass is negligible.


If r is the radius of the meniscus in the conical tube, then as shown in figure.

tan θ = Viscosity


i.e.,r × 104 - 2.5 = 0.2 × 2.5 i.e., r = 3 × 10-4 m

Now as capillarity is independent of the shape of tube so at same temperature θ = 0oC.



so m

Now as from h = (2T/rρg) for cylindrical tube,

= 8.4 × 10-2 N/m

Now as for a given tube and liquid T ∝ h (as T = hρgr/2)

Viscosity × 8.4 × 10-2 = 7.7 × 10-2 N/m

So rate of change of surface tension with temperature assuming linearity,

−1.4 × 10-2 N/moC

Negative sign shows that with rise in temperature surface tension decreases.

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