**This article is formulated according to the 10th chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.**

## Stokes’ law

Stokes proved that the viscous drag (𝐹) acting on a spherical body of radius 𝑎 moving with velocity 𝑣 in a fluid of co-efficient of viscosity 𝜂 is given by,

𝑭 = 𝟔𝝅𝜼𝒂𝒗

The viscous drag increases with increases velocity of the body, but it is found that the body after attaining certain velocity starts moving with a constant velocity called terminal velocity.

## Terminal velocity (𝒗_{𝒕})

When a body is dropped in a viscous fluid, it is first accelerated and then its acceleration becomes zero and it attains a constant velocity. This constant velocity is called terminal velocity.

## Expression for Terminal velocity

Consider a spherical body of radius 𝑎 falling through a viscous fluid having density 𝜎 and co-efficient of viscosity 𝜂. Let 𝜌 be the density of the material of the body. The viscous forces acting on that spherical body are, (i) its weight (𝑚𝑔) in downward direction. (ii) upward thrust 𝑇 equal to the weight of the displaced fluid. (iii) viscous drag 𝐹 in a direction opposite to the direction of motion of the body. Net downward force acting on that body = 𝑚𝑔 − 𝑇 – 𝐹. When the body attains terminal velocity, acceleration, 𝑎 = 0.

From Newton’s I law, Net force acting on the body is zero.

𝑚𝑔 − 𝑇 − 𝐹 = 0

Now,

(i) 𝑚𝑔 = (𝑉 × 𝜌)𝑔 = (4/3)𝜋𝑎^{3}𝜌𝑔

(ii) 𝑇 = 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 = (𝑉 × 𝜎)𝑔 = (4/3)𝜋𝑎^{3}𝜎𝑔

(iii) According to Stoke’s law, 𝐹 = 6𝜋𝜂𝑎𝑣_{𝑡}

Therefore, (4/3)𝜋𝑎^{3}𝜌𝑔 – (4/3)𝜋𝑎^{3}𝜎𝑔 − 6𝜋𝜂𝑎𝑣_{𝑡} = 0

(4/3)𝜋𝑎^{3}𝑔(𝜌 − 𝜎) = 6𝜋𝜂𝑎𝑣_{𝑡}

𝑣_{𝑡} = [(4/3)𝜋𝑎^{3}𝑔(𝜌 − 𝜎)/6𝜋𝜂𝑎]

𝒗_{𝒕} = (𝟐/𝟗)[𝒂^{𝟐}𝒈(𝝆 − 𝝈)/𝜼]

**Note:**

(i) 𝑣_{𝑡} ∝ 𝑎^{2}, Terminal velocity depends on square of the radius of the sphere.

(ii) 𝑣_{𝑡} ∝ 1/𝜂, Terminal velocity depends inversely on viscosity of the medium. (iii) Falling of rain drops through air and the descent of a parachute can be explained using Stoke’s law.

## Reynolds number

When the rate of flow of a fluid becomes very large, the flow loses its orderliness and becomes turbulent. In turbulent flow, the velocity of particles of the fluid at any point varies randomly with time.

**Example:**

(i) An obstacle placed in the path of the fast-moving fluid causes turbulence.

(ii) The smoke rising from burning wood.

(iii) Air left by cars, aeroplanes boat etc. Osborne Reynolds defined a dimension less number which gives an approximate idea about whether the flow would be turbulent or not. This number is called Reynolds number denoted by 𝑅_{𝑒}. If 𝜂 is the viscosity and 𝜌 is the density of fluid flowing with a speed 𝑣 in a pipe of diameter 𝑑, the value of 𝑅𝑒 is given by,

𝑹_{𝒆} = 𝝆𝒗𝒅/𝜼

## Classification of flow based on Reynolds number

(i) For laminar or streamline flow, 𝑅_{𝑒} < 1000.

(ii) For the turbulent flow, 𝑅_{𝑒} > 2000.

(iii) For 1000 < 𝑅_{𝑒} < 2000, the flow becomes unsteady.

## Variation of Reynolds number

For a given fluid, the density and co-efficient of viscosity and constant and for a given pipe its diameter is constant. Then, 𝑅_{𝑒} ∝ 𝑣. As the velocity as increases 𝑅_{𝑒} also increases. If the velocity of the fluid increases beyond a limiting value the flow becomes turbulent.

## Critical Reynolds number

The value of Reynolds number at which the turbulence just occurs is called critical Reynolds number.

## Critical velocity

The maximum velocity of a fluid in a tube for which the flow remains streamline is called Critical velocity.

## Importance of Reynolds number

The value of Reynolds number is very much useful in designing of ships, submarines, race cars, aeroplanes etc.

**Note:** Turbulence promotes mixing and increases the rates of transfer of mass, momentum, and energy.

## Surface Tension

The property of a liquid at rest by virtue of which its free surface behaves like a stretched membrane under tension and tries to minimise the surface area is called surface tension.

**Explanation: **Consider three spheres indicating the spheres of influence of the molecule at 𝐴 , 𝐵 and 𝐶. The surface 𝑋𝑌 is the free surface of the liquid. The molecule 𝐴 is well inside the liquid surface is attracted equally in all directions by the neighbouring molecules. Hence the resultant force on 𝐴 is zero. The molecule 𝐵 has a part of its sphere of influence above the liquid surface. Hence 𝐵 experiences a net force vertically downwards in the liquid.

The molecule 𝐶 is just on the surface of the liquid, hence it experiences maximum force pulling into the liquid as a result the surface of a liquid behaves as a stretched membrane and the attractive inter molecular forces on the surface of the liquid tend to compress the liquid surface, so that the liquid tries to minimise its surface area.

## Surface energy

The potential energy of the surface molecules per unit area of the surface is called surface energy.

𝑺𝒖𝒓𝒇𝒂𝒄𝒆 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝒑𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚/𝒂𝒓𝒆𝒂

Unit of surface energy is 𝑗𝑜𝑢𝑙𝑒/𝑚𝑒𝑡𝑟𝑒^{2} and dimensions are 𝑀𝐿^{0}𝑇^{−2}.

## Surface energy and surface tension

Consider a thin horizontal film of liquid ending in bar, free to slide over parallel guide as shown. Let the bar be moved along horizontal through a small distance 𝑑 to increase the surface area. Some work must be done against the internal force. If the internal force is 𝐹, then work done by the internal force is 𝐹𝑑. This work done is stored as additional energy (potential energy) in the film. The increase in the area of the film is 𝑙𝑑, since the film has two surfaces, the increase in the area of the film should be 2𝑙𝑑. If 𝑆 is the surface energy of the film per unit area, the extra energy supplied to the film is, 𝑆(2𝑙𝑑) = 𝐹𝑑

𝑺 = 𝑭𝒅/𝟐𝒍𝒅 = 𝑭/𝟐𝒍

The quantity 𝑆 is the magnitude of surface tension and it is equal to the surface energy per unit area of the liquid surface.

**Note: **Surface tension can also be defined as, force per unit length acting on the surface of the liquid. Unit of surface tension is 𝑁𝑚^{−1} and dimensions are 𝑀𝐿^{0}𝑇^{−2}.

## Measurement of surface tension

A flat vertical glass plate, below which a vessel of some liquid is kept, forms one arm of the balance. The plate is balanced by weights on the other side. The vessel is raised slowly till the liquid just touches the glass plate and pulls it down a little because of surface tension. Weights are added till the plate just clears water. If the additional weight required is 𝑊 then the surface tension of the liquid air interface is, 𝑆_{𝑙𝑎} = 𝑊/2𝑙 = 𝑚𝑔/2𝑙 where 𝑙 →length of the plate.

## Angle of contact

The angle between tangent to the liquid surface at the point of contact and solid surface inside the liquid is called as angle of contact. It is denoted by 𝜃. The value of 𝜃 determines whether a liquid will spread on the surface of a solid (Wetting liquid) or it will form droplet on it (non-wetting liquid).

Now let us consider three interfacial tensions at all the three interfaces, liquid-air, solid-air and solid-liquid denoted by 𝑆_{𝑙𝑎}, 𝑆_{𝑠𝑎} and 𝑆_{𝑠𝑙} respectively. Since the surface is at rest, at the line of contact the surface forces between the three media must in equilibrium. Then, 𝑆_{𝑙𝑎}𝑐𝑜𝑠 𝜃 + 𝑆_{𝑠𝑙}_{ }= 𝑆_{𝑠𝑎}

If 𝑆_{𝑠𝑙} > 𝑆_{𝑙𝑎}, the angle of contact is obtuse. When 𝜃 is obtuse angle, the molecules of the liquid are attracted strongly to themselves and weakly to those of solids and the liquid will not wet the surface.

Ex:

(i) water-leaf interface.

(ii) water in waxy or oily surface.

(iii) Mercury on any surface. etc.

If 𝑆_{𝑠𝑙} < 𝑆_{𝑙𝑎}, the angle of contact is acute. When 𝜃 is acute angle, the molecules of the liquid are attracted strongly to those of solids and the liquid will wet the surface.

Ex: (i) Water on glass. (ii) Water on plastic sheet.

## Drops and Bubbles

Due to surface tension, surface of liquid always has a tendency to have least surface area. For a given volume a sphere has the minimum surface. Therefore, a small volume of liquid take the shape of a sphere, that is why drop of a liquid and bubbles assumes the shape of a sphere.

## Pressure difference across a liquid surface

**(i) For concave surface of liquid:** If the surface of the liquid is concave the resultant force 𝑅 due to surface tension on a molecule on the surface act vertically upwards. To balance this, there should be an excess of pressure acting downward on the concave side.

**(ii) For convex surface of liquid:** If the surface of the liquid is convex, the resultant force 𝑅 acts downwards and there must be an excess of pressure on the concave side acting upwards. Thus, there is always an excess pressure on the concave side of a curved liquid surface over the pressure on its convex side due to surface tension.

## Excess pressure inside a liquid drop and Bubble

**(i) For a liquid drop: **

Consider a liquid drop of radius 𝑟 having surface tension 𝑆. Let 𝑃_{𝑖} be the pressure inside the drop and 𝑃_{𝑜} be the pressure outside the drop. The excess pressure inside the drop = (𝑃_{𝑖} − 𝑃_{𝑜})

Outward force acting on the drop = pressure × surface area of the drop Outward force = (𝑃_{𝑖} − 𝑃_{𝑜}) × 4𝜋𝑟^{2}

Due to this force, the drop expands. Let its radius increases by 𝑑𝑟. Hence work done = force × change in radius

work done = (𝑃_{𝑖}_{ }− 𝑃_{𝑜}) × 4𝜋𝑟^{2} × 𝑑𝑟

The work done against the force of surface tension is stored inside the drop in the form of its potential energy.

Increase in potential energy = surface tension × increase in surface area = 𝑆 × [4𝜋(𝑟 + 𝑑𝑟)^{2} − 4𝜋𝑟^{2}] = 𝑆 × [4𝜋(𝑟^{2} + 𝑑𝑟^{2} + 2𝑟𝑑𝑟) − 4𝜋𝑟^{2}] = 𝑆 × [4𝜋𝑟^{2} + 4𝜋𝑑𝑟^{2} + 8𝜋𝑟𝑑𝑟 − 4𝜋𝑟^{2}] = 𝑆[4𝜋𝑑𝑟^{2} + 8𝜋𝑟𝑑𝑟]

Since 𝑑𝑟 is very small 𝑑𝑟^{2} can be neglected.

Increase in potential energy = 8𝜋𝑆𝑟𝑑𝑟

Therefore, work done = 8𝜋𝑆𝑟𝑑𝑟

(𝑃_{𝑖} − 𝑃_{𝑜}) × 4𝜋𝑟^{2} × 𝑑𝑟 = 8𝜋𝑆𝑟𝑑𝑟

(𝑃_{𝑖} − 𝑃_{𝑜}) = 8𝜋𝑆𝑟𝑑𝑟/(4𝜋𝑟^{2} × 𝑑𝑟)

(𝑷_{𝒊} − 𝑷_{𝒐}) = 𝟐𝑺/𝒓

**(ii) For a bubble:** In the case of a liquid bubble, there are two surfaces-inner and outer. Consider a liquid bubble of radius 𝑟 having surface tension 𝑆. Let 𝑃_{𝑖 }be the pressure inside the bubble and 𝑃𝑜 be the pressure outside the bubble. The excess pressure inside the bubble = (𝑃_{𝑖} − 𝑃_{𝑜}).

The outward force acting on the bubble = pressure × surface area of the bubble

Outward force = (𝑃_{𝑖} − 𝑃_{𝑜}) × 4𝜋𝑟^{2}

Due to this force, the bubble expands. Let its radius increases by 𝑑𝑟.

Hence work done = force × change in radius work done = (𝑃_{𝑖} − 𝑃_{𝑜}) × 4𝜋𝑟^{2} × 𝑑𝑟.

The work done against the force of surface tension is stored inside the bubble in the form of its potential energy. Increase in potential energy = surface tension × increase in surface area = 𝑆 × 2[4𝜋(𝑟 + 𝑑𝑟)^{2} − 4𝜋𝑟^{2}] = 𝑆 × 2[4𝜋(𝑟^{2} + 𝑑𝑟^{2} + 2𝑟𝑑𝑟) − 4𝜋𝑟^{2}] = 𝑆 × 2[4𝜋𝑟^{2} + 4𝜋𝑑𝑟^{2} + 8𝜋𝑟𝑑𝑟 − 4𝜋𝑟^{2}] = 2𝑆[4𝜋𝑑𝑟^{2} + 8𝜋𝑟𝑑𝑟] Since 𝑑𝑟 is very small 𝑑𝑟^{2} can be neglected.

Increase in potential energy = 16𝜋𝑆𝑟𝑑𝑟

Therefore, work done = 16𝜋𝑆𝑟𝑑𝑟

(𝑃_{𝑖}_{ }− 𝑃_{𝑜}) × 4𝜋𝑟^{2} × 𝑑𝑟 = 16𝜋𝑆𝑟𝑑𝑟

(𝑃_{𝑖} − 𝑃_{𝑜}) = 16𝜋𝑆𝑟𝑑𝑟/(4𝜋𝑟^{2} × 𝑑𝑟)

(𝑷_{𝒊} − 𝑷_{𝒐}) = 𝟒𝑺/𝒓

**Note:** Thus, excess pressure in a bubble is two times the excess pressure in a liquid drop.

## Capillarity

When a capillary tube is dipped in water, the water rises in the tube. This rise of liquid in a capillary tube is known as capillarity.

## Capillary rise

The surface of the water in the capillary is concave. Then there is a pressure difference between the two sides of the top surface, which is given by,

(𝑃_{𝑖} − 𝑃_{𝑜}) = 2𝑆𝑐𝑜𝑠 𝜃/𝑟 where 𝑆 cos 𝜃 → vertical component of 𝑆

Thus, the pressure of the water inside the tube, just at the meniscus is less than the atmospheric pressure. Consider the two points 𝐴 and 𝐵(as in figure). They must be at same height. Then,

𝑃_{𝑜} + 𝜌𝑔ℎ = 𝑃_{𝑖}

𝑃_{𝑖}_{ }− 𝑃_{𝑜} = 𝜌𝑔ℎ

where 𝜌 is the density of water and ℎ is called capillary rise

2𝑆 𝑐𝑜𝑠 𝜃/𝑟 = 𝜌𝑔ℎ

𝒉 = 𝟐𝑺 𝒄𝒐𝒔 𝜽/𝝆𝒈𝒓

Practical applications of capillarity

(i) The oil in the lamp rises in the wick to its top by capillary action.

(ii) Sap and water rise up to the top of the leaves of the tree by capillary action.

(iii) Ink is absorbed by the blotting paper due to capillarity.

(iv) The moisture rises in the capillaries of the soil to the surface, where it evaporates. To prevent this and preserve moisture in the soil, capillaries must be destroyed by ploughing and levelling fields.

## Detergents and Surface tension

We clean dirty clothes containing grease and oil stains by adding detergents or soap to water. Washing only with water does not remove grease stains. This is because water does not wet grease dirt, because there is very little area of contact between them. If the water wet the grease, the flow of water could carry some grease away. This is achieved by adding detergents. The molecules of detergents are hairpin shaped, with one end attached to water and the other to molecules of grease, oil, or wax. Thus, they tend to form water-oil interfaces. In other words, the addition of detergent reduces drastically the surface tension of water. This kind of process using surface active detergent or surfactants is important not only in cleaning but also in recovering oil, mineral ores etc.

## Important Questions for Exams.

**One mark.**

1. Define Pressure at a point inside a liquid.

2. Write equation of continuity.

3. State Pascal’s law.

4. What is principle behind the uplift of an aeroplane?

5. State Stokes law.

6. Give one example for turbulent flow.

7. What is streamline flow?

8. Define surface energy.

9. Define angle of contact.

10. What is value of angle of contact for water-leaf interface?

11. What is capillarity?

**Two marks.**

1. State and explain Bernoulli’s theorem.

2. State Bernoulli’s theorem and mention any two applications.

3. What is the order of Reynold’s number for streamline flow and turbulent flow?

4. What is venture-meter? On which principle it works.

5. State and explain Pascal’s law for transmission of fluid pressure.

6. What is terminal velocity? Mention the expression for it.

7. Mention the factors on which terminal velocity depends.

8. How the viscosity of liquids and gases does vary with temperature.

9. Mention three applications of capillarity.

**Three Marks.**

1. State Bernoulli’s theorem with an example.

2. Derive an expression for liquid pressure at a point inside the liquid. or Arrive at an expression for pressure at a point due to liquid.

3. Mention the applications of Pascal’s law.

4. Mention the limitations of Bernoulli’s Principle.

5. Distinguish between Streamline motion and turbulent motion.

6. What is surface tension? Mention its SI unit. Mention expression for it