When water is taken in a glass vessel, the free surface of the water near the walls is curved concave upward. If mercury is taken in a glass vessel, the free surface of mercury near the walls is convex upwards.

When the liquid is in contact with solid, the angle between the solid surface and the tangent to the free surface of the liquid at the point of contact, measured from inside the liquid is called the angle of contact.

When the liquid surface is curved concave upwards, the angle of contact is acute and when the liquid surface is curved convex upwards, the angle of contact is obtuse.

Characteristics of the Angle of Contact:

• The angle of contact is constant for a given liquid-solid pair.
• When the angle of contact between the liquid and a solid surface is small (acute), the liquid is said to wet the surface. Thus water wets glass.
• If the angle of contact is large the surface is not wetted. Mercury does not wet glass.
• If there are impurities in liquid, then they alter the values of the angle of contact.
• The angle of contact decreases with an increase in temperature.
• For a liquid which completely wets the solid, the angle of contact is equal to zero.

Notes:

• The smearing of glycerine on a glass increases the angle of contact of water with glass from acute to obtuse, hence a glass window is smeared with glycerine.
• Smaller the angle of contact, better is the detergent.
• Teflon coating is done on the surface of the non-stick pan because Teflon increases the angle of contact from acute to obtuse.

The Shape of a Liquid Drop on Solid Surface:

When a small quantity of liquid is poured on a plane solid surface, a force of surface tension acts along a surface separating the two media. There is a formation of the liquid drop.

When the drop is in equilibrium, its shape depends on the forces of interface media. Thus there is surface tension along a surface between a) liquid and air b) solid and air and c) liquid and solid.

Let, θ  = the angle of contact of given solid-liquid pair.

T1 = Surface tension at the liquid-solid interface

T2 = Surface tension at the air-solid interface

T3 = Surface tension at the air-liquid interface

For equilibrium of the drop

T₂ = T₁ + T₃ cos θ

Case -1:

If T₂ > T₁, and T₂ -T₁ < T₃, then cos θ is positive and the angle of contact is acute. e.g. kerosene on the glass surface

Case – 2:

If T₂ < T₁, and T₁ -T₂ < T₃, then cos θ is negative and the angle of contact is obtuse. e.g. mercury on the glass surface Mercury does not wet the glass and thus it forms a drop on the glass.

Case – 3:

If T₂ -T₁ = T₃, then cos θ is 1 and the angle of contact is zero. For pure water, cos θ is almost 1 i.e. water wets the glass and thus water spreads on the glass.

Case – 4:

If T₂ -T₁ > T₃, then cos  θ > 1 which is not possible. Thus liquid will spread over the solid and drop shall not be formed.

Notes:

For the small drop, gravitational potential energy is very small and hence can be neglected. Hence for stability, the surface tension of the drop should be least i.e. the surface area should be least. It can be achieved by forming a spherical drop because the surface area of a sphere is least for the given volume.

For the large drop, gravitational potential energy dominates to make the potential energy minimum. It does so by bringing the centre of gravity as low as possible, due to which the drop flattens.

The Formation of Concave Surface of Liquid (Explanation of Acute Angle of Contact):

When impure water or kerosene is taken in a glass vessel, it is found that the surface near the walls is curved concave upwards.

Consider a molecule of water M on the free surface very close to the wall of the glass vessel. The force of cohesion C due to other water molecules is as shown in the figure. In addition to this, a force of adhesion A acts due to the glass molecule as shown in the figure. The net adhesive force between water molecules and air molecules is negligible. The gravitational force on the molecule is also negligible.

The magnitude of A is greater than the magnitude of C and resultant of the two molecular forces of attraction R is directed towards the glass or outside the liquid. Hence the molecule A is attracted towards the walls of the glass vessel.

The free surface of water adjusts itself at right angles to the resultant R. Therefore molecules like M creep upward on the solid surface. Thus the water surface is curved concave upwards and the angle of contact is acute.

The Formation of Convex Surface of Liquid (Explanation of Obtuse Angle of Contact):

When mercury is taken in a glass vessel, it is found that the surface near the walls is curved convex upwards.

Consider a molecule of mercury M on the free surface very close to the wall of the glass vessel. The force of cohesion C due to other mercury molecules is as shown in the figure. In addition to this, a force of adhesion A acts due to the glass molecule as shown in the figure. The net adhesive force between mercury molecules and air molecules is negligible. The gravitational force on the molecule is also negligible.

The magnitude of A is very less than the magnitude of C and resultant of the two molecular forces of attraction R is directed inside the liquid. Hence the molecule A is attracted towards other molecules of mercury.

The free surface adjusts itself at right angles to the resultant R. The molecule A creeps downwards on the glass surface. Thus the surface of the mercury in the glass is curved convex upwards and the angle of contact is obtuse.

The Formation of Flat Surface of Liquid:

When pure is taken in a glass vessel, it is found that the surface of the water is perfectly flat.

Consider a molecule of water M on the free surface very close to the wall of the glass vessel. The force of cohesion C due to other water molecules is negligible. The force of adhesion A acts due to the glass molecule as shown in the figure. The gravitational force on the molecule is also negligible.

The magnitude of A is large and resultant of the two molecular forces of attraction R is along the direction of the adhesive force. Hence the molecule A is attracted towards glass molecules.

The free surface adjusts itself at right angles to the wall. Thus the surface of the pure water is perfectly flat.

Mercury Drops Coalesce Together When Brought Together:

The attractive force between the two molecules of the same substance is called as a cohesive force and the attractive force between the two molecules of the different substance is called an adhesive force.

In the case of mercury, the cohesive force between its molecules is very strong compared to negligible adhesive force between any surface and mercury molecules. The net adhesive force between mercury molecules and air molecules is negligible. The gravitational force on the molecule is also negligible. Thus there is a strong attraction between mercury molecules and almost no attractive force between surface and mercury molecules. Hence mercury drops coalesce together when brought together.

Water on a clean glass surface tends to spread out, while mercury on the same surface tends to form a drop. 

The shape of the drop depends on the angle of contact of the liquid with glass. In the case of a glass water interface, the angle of contact is acute. Thus to achieve this angle water spreads on the glass. In the case of a glass mercury interface, the angle of contact is obtuse. Thus to achieve this angle mercury forms a drop as shown.

Previous Topic: Problems on Change in Surface Energy

Next Topic: Laplace’s Law of Spherical Membrane

Science > Physics > Surface Tension > Angle of Contact

The post Angle of Contact appeared first on The Fact Factor.

In this article, we shall study Laplace’s Law about the excess of pressure inside a drop or a bubble of a liquid.

Excess of Pressure Inside a Drop of a Liquid and Soap Bubble:

Due to surface tension, free liquid drops and bubbles are spherical, if the effect of gravity and air resistance are negligible. Due to the spherical shape, the inside pressure P_i is always greater than the outside pressure P_o. The excess of pressure is P_i– P_o.

Laplace’s Law of a Spherical Membrane for a Liquid Drop:

Due to the spherical shape, the inside pressure P_i is always greater than the outside pressure P_o. The excess of pressure is P_i– P_o.

Let the radius of the drop increases from r to r + Δr, where Δr is very very small, hence the inside pressure is assumed to be constant.

Initial surface area = A₁ = 4 π r²

Final surface area = A₂ = 4 π (r + Δr)²

Final surface area = 4 π (r² + 2r.Δr + Δr²)

Final surface area =  4 πr² + 8 πr.Δr + 4 πΔr²

Δr is very very small, hence Δr² still smsll hence the term 4 πΔr² can be neglected.

Final surface area = A₂ = 4 πr² + 8 πr.Δr

Hence Change in area = A₂ – A₁ = 4 πr² + 8 πr.Δr  –  4 πr²

Change in area = dA = 8 πr.Δr

Now, work done in increasing the surface area is given by

dW = T. dA = T.  8 πr.Δr    …………… (1)

By definition of work in mechanics we have

dW = Force ∴ displacement = F .Δr  …………… (2)

But P = F /A, Hence F = Excess pressure × Area

F = (P_i– P_o) × 4 πr²

Substituting in equation (2) we have

dW = (P_i– P_o) × 4 πr².Δr  …………… (3)

From equations (3) and (4) we have

(P_i– P_o) × 4 πr².Δr = T.  8 πr.Δr

(P_i– P_o) = 2T / r

This relation is known as Laplace’s law for the spherical membrane for a liquid drop.

Laplace’s Law of a Spherical Membrane for a Liquid Bubble:

Due to the spherical shape, the inside pressure P_i is always greater than the outside pressure P_o. The excess of pressure is P_i– P_o.

Let the radius of the bubble increases from r to r + Δr, where Δr is very very small, hence the inside pressure is assumed to be constant.

Bubble has two free surfaces, one inside and another outside

Initial surface area = A₁ = 2 × 4 π r² = 8 π r²

Final surface area = A₂ = 2 × 4 π (r + Δr)²

Final surface area = 8 π (r² + 2r.Δr + Δr²)

Final surface area =  8 πr² + 16 πr.Δr + 8 πΔr²

Δr is very very small, hence Δr² still smsll hence the term 8 πΔr² can be neglected.

Final surface area = A₂ = 8 πr² + 16 πr.Δr

Hence Change in area = A₂ – A₁ =  8 πr² + 16 πr.Δr  –  8 πr²

Change in area = dA = 16 πr.Δr

Now, work done in increasing the surface area is given by

dW = T. dA = T.  16 πr.Δr    …………… (1)

By definition of work in mechanics we have

dW = Force × displacement = F .Δr  …………… (2)

But P = F /A, Hence F = Excess pressure × Area

F = (P_i– P_o) × 4 πr²

Substituting in equation (2) we have

dW = (P_i– P_o) × 4 πr².Δr  …………… (3)

From equations (3) and (4) we have

(P_i– P_o) × 4 πr².Δr = T.  16 πr.Δr

(P_i– P_o) = 4T / r

This relation is known as Laplace’s law for spherical membrane for a liquid bubble.

Difference in Pressure Across the Surface Film:

Rise and fall of liquid in a capillary tube can be explained by knowing the fact that a pressure difference exists across a curved free surface of the liquid.

When the liquid surface is flat:

In this case, the surface tension is horizontal and hence has no normal component to the horizontal surface. Due to which there is no extra pressure on outside or inside. Thus the pressure on the liquid side is equal to the pressure on the vapour side.

When the liquid surface is concave:

In this case, the component of surface tension acts vertically upward. Due to which pressure inside liquid decreases. Thus for the equilibrium of concave surface, the pressure on the vapour side should be more than that on the liquid side. Hence liquid rises in the capillary tube. In this case, the adhesion is greater than the cohesion.

When the liquid surface is convex:

In this case, the component of surface tension acts vertically downward. Due to which pressure inside liquid increases. Thus for the equilibrium of convex surface, the pressure on the vapour side should be less than that on the liquid side. Hence liquid dips in the capillary tube. In this case, the cohesion is greater than the adhesion.

Conclusion:

In case 2 and 3 we can see that, when liquid surface is curved, the surface tension give rises to a pressure which is directed towards the centre of curvature of the surface, to balance this there is excess pressure acting on the surface. Thus, there is always excess of pressure on the concave side of a curved liquid surface over the pressure on its convex side due to surface tension.

Numerical Problems on Excess of Pressure:

Example – 1:

A raindrop of diameter 4 mm is about to fall on the ground. Calculate the pressure inside the raindrop. Surface tension of water = 0.072 N/m. Atmospheric pressure = 1.013 x 10⁵ N/m².

Given: Diameter of soap bubble = 4 mm, Radius of raindrop = r = 4/2 = 2 mm = 2 × 10^-3 m, Surface tension = T = 0.072 N/m, Outside pressure = P_o = Atmospheric pressure = 1.013 × 10⁵ N/m².

To Find: Inside pressure = P_i =?

Solution:

By Laplace’s law of spherical membrane for a drop

(P_i– P_o) = 2T / r

∴    P_i   = (2T / r) + P_o

∴    P_i   = (2 × 0.072 / 2 × 10^-3) + 1.013 × 10⁵

∴    P_i   =72 + 1.013 × 10⁵

∴    P_i   = 1.01372 × 10⁵ N/m²

Ans: The pressure inside the raindrop is 1.01372 × 10⁵ N/m²

Example – 2:

Find the excess pressure inside the soap bubble of diameter 3 cm. The surface tension of the soap solution is 3 × 10^-2 N/m. 

Given: Diameter of soap bubble = 3 cm, Radius of soap bubble = r = 3/2 =1.5 cm = 1.5 × 10^-2 m, Surface tension = T = 3 × 10^-2 N/m

To Find: excess pressure = (P_i– P_o) =?

Solution:

By Laplace’s law of spherical membrane for a bubble

(P_i– P_o) = 4T / r

∴   (P_i– P_o) = 4 × 3 × 10^-2 / 1.5 × 10^-2

∴   (P_i– P_o) = 8 N/m²

The excess pressure inside the soap bubble is 8N/m²

Example – 3:

What should be the diameter of a soap bubble, in order that the excess pressure inside it is 51.2 N/m². The surface tension of the soap solution is 3.2 × 10^-2 N/m. 

Given: excess pressure = (P_i– P_o) = 51.2 N/m², Surface tension = T = 3.2 × 10^-2 N/m

To Find: Diameter of soap bubble =?

Solution:

By Laplace’s law of spherical membrane for a bubble

(P_i– P_o) = 4T / r

∴ r = 4T / (P_i – P_o)

∴ r = 4 × 3.2 × 10^-2 / 51.2

∴ r = 2.5 × 10^-3 m = 2.5 mm

Diameter of soap bubble = 2 × 2.5 mm = 5 mm

Ans: The diameter of soap bubble is 5 mm

Previous Topic: Concept of Angle of Contact

Next Topic: The Concept of Capillary Action

Science > Physics > Surface Tension > Laplace’s Law of Spherical Membrane

The post Laplace’s Law of Spherical Membrane appeared first on The Fact Factor.

In this article, we shall study to solve problems on capillary action.

Important Formulae:

and Jurin’s law, hr = constant

where h  =  height of liquid level in the capillary

T  =  Surface tension

ρ =  Density of liquid

r  =  Radius of the bore of the capillary tube

g  =  Acceleration due to gravity

θ =  Angle of contract

Example – 1:

A capillary tube of a uniform bore is dipped vertically in water which rises by 7 cm in the tube. Find the radius of the capillary if the surface tension is 70 dynes/cm g= 9.8 m/s².

Given: Rise in the tube = h = 7 cm = 7× 10^-2 m, Surface tension = T = 70 dynes/cm = 70 × 10^-3 N/m, Acceleration due to gravity = g  = 9.8 m/s², Density of water = ρ = 1 ×10³ kg/m³, Angle of contact = θ  = 0°

To Find: Radius of the tube = r =?

Solution:

The rise in the tube is given by

∴ r = (2 × 70 × 10^-3 × cos 0°) / (7× 10^-2 × 1 ×10³ × 9.8)

∴ r = 2 × 10^-4 m = 0.2 × 10^-3 m = 0.2 mm

Ans: The radius of the capillary is 0.2 mm

Example – 2:

Water rises to a height of 4.5 cm in a capillary tube of radius r. Find r assuming the surface tension of water is 72 dynes/cm. Take the angle of contact of water in the glass as 15°. Density of water=1000 kg/m³, g = 9.8 m/s².

Given: Rise in the tube = h =4.5 cm = 4.5 × 10^-2 m, Surface tension = T = 72 dynes/cm = 72 × 10^-3 N/m, Acceleration due to gravity = g = 9.8 m/s², Density of water = ρ = 10³ kg/m³, Angle of contact = θ = 15°

To Find: Radius of the tube = r =?

Solution:

The rise in the tube is given by

∴ r = (2 × 72 × 10^-3 × cos 15°) / (4.5× 10^-2× 1 ×10³ × 9.8)

∴ r = (2 × 72 × 10^-3 × 0.9659) / (4.5× 10^-2× 1 ×10³ × 9.8)

∴ r = 3.15 × 10^-4 m = 0.315 × 10^-3 m = 0.315 mm

Ans: The radius of the tube is 0.315 mm

Example – 3:

A liquid rises to a height of 9 cm in a glass capillary of radius 0.02 cm. What will be the height of the liquid column in a similar glass capillary of radius 0.03 cm?

Given: Rise in first tube = h₁ = 9 cm, Radius of first tube = r₁ = 0.02 cm, Radius of second tube = r₂ = 0.03 cm,

To Find: Radius in second tube = h₂ =?

Solution:

For a particular liquid-solid interface by Jurin’s law, hr = constant

∴    h₁ r₁ = h₂ r₂

∴    h₂   = h₁ r₁ / r₂

∴    h₂ = 9 × 0.02 / 0.03 = 6 cm

Ans: The height of the liquid column in a similar glass capillary is 6 cm

Example – 4:

A liquid rises to a height of 4.5 cm in a glass capillary of radius 0.01 cm. What will be the height of the liquid column in a similar glass capillary of radius 0.02 cm?

Given: Rise in first capillary tube = h₁ = 4.5 cm, Radius of first capillary tube = r₁ = 0.01 cm, Radius of second capillary tube = r₂ = 0.02 cm,

To Find: Radius in second capillary tube = h₂ =?

For a particular liquid-solid interface by Jurin’s law, hr = constant

∴ h₁ r₁ = h₂ r₂

∴ h₂   = h₁ r₁ / r₂

∴ h₂ = 4.5 × 0.01 / 0.02 = 2.25 cm

Ans: The height of the liquid column in a similar glass capillary is 2.25 cm

Example – 5:

Water rises to height 3.2 cm in glass capillary tube. Find height to which same water rises in another capillary having half area of cross-section.

Given: Rise in first capillary tube = h₁ = 3.2 cm, Area of cross-section of first capillary tube = A₁,  Area of cross-section of second capillary tube = A₂ , A₂ = ½ A₁,

To Find: Rise in second capillary tube = h₂ =?

Solution:

Given A₂ = ½ A₁

∴    π r₂²  = ½ π r₁²

∴    r₂² = ½ r₁²

∴    r₂ = 0.707 r₁

For a particular liquid-solid interface by Jurin’s law hr = constant

∴    h₁ r₁ = h₂ r₂

∴    h₂   = h₁ r₁ / r₂

∴    h₂ = 3.2.5 × r₁ / 0.707  r₁  = 4.525 cm

Ans: The height of the liquid column in the second glass tube is 4.525 cm

Example – 6:

A liquid of density 900 kg/m³ rises to a height of 7 mm in a capillary tube of 2 mm internal diameter. If the angle of contact of the liquid in the glass is 25°, find the surface tension of the liquid. g=9.8 m/s².

Given: Rise in capillary tube = h = 7 mm = 7× 10^-3 m, Internal diameter = 2 mm, Radius of capillary tube = r = 2/2 = 1 mm = 1 × 10^-3 m, Acceleration due to gravity = g = 9.8 m/s², Density of liquid = ρ =900 kg/m³, Angle of contact = θ = 25°

To Find: Surface tension = T =?

Solution:

The rise in the capillary tube is given by

∴ T = (7× 10^-3 × 1 × 10^-3 × 900 × 9.8) / (2 × cos 25°)

∴ T = (7× 10^-3 × 1 × 10^-3 × 900 × 9.8) / (2 × 0.9063)

∴ T = 0.064 N/m

Ans: The surface tension of the liquid is 0.064 N/m

Example – 7:

When a capillary tube of radius 0.45 mm is dipped into water, the level inside the capillary tube rises to height 3 cm above the surface of the water. Calculate the surface tension of water, if its angle of contact is zero and its density is 1 g/cm³, g = 9.8 m/s².

Given: Rise in capillary tube = h =3 cm = 3 × 10^-2 m, Radius of capillary tube = r = 0.45 mm = 0.45 × 10^-3 m, Acceleration due to gravity = g = 9.8 m/s², Density of water =  1 g/cm³ = 1 × 10³ kg/m³, Angle of contact = θ  = 0°

To Find: Surface tension = T =?

Solution:

The rise in the capillary tube is given by

∴ T = (3× 10^-3 × 0.45 × 10^-3 × 1 × 10³ × 9.8) / (2 × cos 0°)

∴ T = (3× 10^-3 × 0.45 × 10^-3 × 1 × 10³ × 9.8) / (2 ×1)

∴ T = 0.006615 N/m

Ans: The surface tension of the liquid is 0.066615 N/m

Problem – 8:

The capillary rise when a glass capillary tube of diameter 0.4 mm is dipped vertically in olive oil is 3.82 mm. If the angle of contact of olive oil with glass is 0°and the density of olive oil is 920 kg/m³, find the surface tension of olive oil.

Given: Rise in capillary tube = h =3.82 mm = 3.82 × 10^-3 m, Radius of capillary tube = r = 0.4 mm = 0.4 × 10^-3 m, Acceleration due to gravity = g  = 9.8 m/s², Density of olive oil = 920kg/m³, Angle of contact = θ  = 0°

To Find: Surface tension = T =?

Solution:

The rise in the tube is given by

∴ T =(3.82× 10^-3  × 0.4 × 10^-3 × 920 × 9.8) / (2 × cos 0°)

∴ T =(3.82 × 10^-3  × 0.4 × 10^-3 × 920 × 9.8) / (2 ×1)

∴ T = 6.889 × 10^-3 N/m

Ans: The surface tension of olive oil is 6.889 × 10^-3 N/m

Example – 9:

A mercury barometer tube is 2.5 mm in internal radius. Find the error introduced in the observed reading because of surface tension. g = 9.8 m/s², T=0.540 N/m. Density of mercury = 13600 kg /m³, angle of contact of mercury in glass = 135°.

Given: Radius of capillary tube = r =2.5 mm = 2.5 × 10^-3 m, Surface tension = T = 0.540N/m, Acceleration due to gravity = g = 9.8 m/s², Density of mercury = ρ = 13600 kg/m³, Angle of contact = θ  = 135°

To Find: Error introduced in the observation = h =?

Solution:

The rise in capillary tube is given by

∴ h= (2 × 0.540× cos 135°) / (2.5× 10^-3× 13600 × 9.8)

∴ h= (2 × 0.540× (-0.707)) / (2.5× 10^-3× 13600 × 9.8)

∴ h= – 2.29 × 10^-3 m = – 2.29 mm

Ans: The error introduced in the observation is – 2.29 mm

Hence 2.29 mm should be added to height of mercury measurement.

Example – 10:

The tube of a mercury barometer is 3 mm in diameter. What error is introduced in the reading because of surface tension? Angle of contact of mercury in glass = 135° and S.T. of mercury = 0.460 N/m. Density of mercury = 13.6 g/cc, g=9.8 m/s²

Given: Diameter of capillary tube = 3 mm, Radius of capillary tube = r =3/2 = 1.5 mm = 1.5 × 10^-3 m, Surface tension = T = 0.460N/m, Acceleration due to gravity = g  = 9.8 m/s², Density of mercury = ρ = 13.6 g/cc= 13.6 × 10³ kg/m³, Angle of contact = θ  = 135°

To Find: Radius of capillary tube = r =?

Solution:

The rise in capillary tube is given by

∴ h= (2 × 0.460× cos 135°) / (1.5× 10^-3× 13.6 × 10³ × 9.8)

∴ h= (2 × 0.460× (-0.707)) / (1.5× 10^-3× 13.6 × 10³ × 9.8)

∴ h= – 3.25 × 10^-3 m = – 3.25 mm

Ans: The error introduced in the observation is – 3.25 mm

Hence 3.25 mm should be added to height of mercury measurement.

Example – 11:

A mercury barometer tube is 1 cm in diameter. Find the error introduced in the observed reading because of surface tension. g = 9.8 m/s², T=435.5 dyne/cm. The density of mercury = 13600 kg /m³, angle of contact of mercury in glass = 140°.

Given: Diameter of capillary tube = 1 cm, Radius of capillary tube = r =1/2 = 0.5 cm = 0.5 × 10^-2 m, Surface tension = T =435.5 dyne/cm = 435.5 × 10^-3 N/m, Acceleration due to gravity = g = 9.8 m/s², Density of mercury = ρ = 13600 kg/m³, Angle of contact = θ  = 140°

To Find: Correction due to capillary action = h =?

Solution:

The rise in capillary tube is given by

∴ h= (2 × 435.5 × 10^-3 × cos 140°) / (0.5× 10^-2× 13600 × 9.8)

∴ h= (- 2 × 435.5 × 10^-3 × sin 140°) / (0.5× 10^-2× 13600 × 9.8)

∴ h= (- 2 × 435.5 × 10^-3 × 0.766) / (2.5× 10^-3× 13600 × 9.8)

∴ h= – 1.01 × 10^-3 m = – 1.01 mm

Ans: The correction due to capillary action is – 1.01 mm

Hence 1.01 mm should be added to height of mercury measurement.

Example – 12:

A capillary tube of radius 0.5 mm is dipped vertically in a liquid of surface tension 0.04 N/m and density 0.8 g/cc. Calculate the height of capillary rise, if the angle of contact is 10°.

Given: Radius of capillary tube = r =0.5 mm = 0.5 × 10^-3 m, Surface tension = T =0.04 N/m, Acceleration due to gravity = g  = 9.8 m/s², Density = ρ = 0.8 g/cc = 0.8  × 10³  kg/m³, Angle of contact = θ  = 10°

To Find: Height of capillary rise = h =?

Solution:

The rise in capillary tube is given by

∴ h= (2 × 0.04× cos 10°) / (0.5× 10^-3× 0.8 × 10³ × 9.8)

∴ h= (2 × 0.04× 0.9848) / (0.5× 10^-3× 0.8 × 10³ × 9.8)

∴ h= 2.01 × 10^-2 m = 2.01 cm

Ans: The height of capillary rise is 2.01 cm

Problem – 13:

A capillary tube 0.12 mm in diameter has its lower end immersed in a liquid of surface tension 0.054 N/m. If the density of the liquid is 860 kg/m³, find the height to which the liquid rises in the tube. Given the angle of contact 30° and g = 9.8 m/s².

Given: Radius of capillary tube = r =0.12 mm = 0.12 × 10^-3 m, Surface tension = T =0.054 N/m, Acceleration due to gravity = g  = 9.8 m/s², Density = ρ = 860 kg/m³, Angle of contact = θ  = 30°

To Find: Height of capillary rise = h =?

Solution:

The rise in the tube is given by

∴ h= (2 × 0.054× cos 30°) / (0.12× 10^-3×860 × 9.8)

∴ h= (2 × 0.054× 0.5) / (0.12× 10^-3 × 860 × 9.8)

∴ h= 5.34 × 10^-2 m = 5.34 cm

Ans: The height of capillary rise is 5.34 cm

Example – 14:

A capillary tube of radius 0.5 mm is immersed in a beaker of mercury. The mercury level inside the tube is found to be 0.80 cm below the level of the reservoir. Determine the angle of contact between mercury and glass. Surface tension of mercury = 0.465 N/m and density is 13.6 × 10³ kg/m³, g = 9.8 m/s^2.

Given: Radius of capillary tube = r =0.5 mm = 0.5 × 10^-3 m, Height of capillary rise = h = – 0.80 cm = – 0.80 × 10^-2 m, Surface tension = T =0.465 N/m, Acceleration due to gravity = g = 9.8 m/s², Density = ρ =13.6  × 10³  kg/m³,

To Find:  Angle of contact = θ =?

Solution:

The rise in capillary tube is given by

∴ cos θ =   (- 0.80 × 10^-2 × 0.5 × 10^-3 × 13.6  × 10³ × 9.8)/ (2× 0.465)

∴ cos θ =   – 0.5732

∴ cos (π – θ) =  0.5732

∴ (π – θ) = cos^-1 (0.5732)

∴   (180° – θ) =55°2′

∴ θ = 180° – 55°2′ = 124°58′

Ans: The angle of contact is 124°58′

Example – 15:

Calculate the density of paraffin oil, if glass capillary of diameter 0.25 mm dipped in a paraffin oil of the surface tension 0.0245 N/m rises to a height of 4 cm. Angle of contact of paraffin oil with glass is 28°, g  = 9.8 m/s²,

Given: Diameter of capillary tube = 0.25 mm, Radius of capillary tube = r = 0.25/2 = 0.125 mm = 0.125 × 10^-3 m, Height of capillary rise = h = 4 cm = 4 × 10^-2 m , Surface tension = T =0.0245 N/m, Acceleration due to gravity = g  = 9.8 m/s², Angle of contact = θ  = 28°

To Find:   Density = ρ =?

Solution:

The rise in capillary tube is given by

∴ ρ = (2 × 0.0245 × cos 28°) / (0.125 × 10^-3 × 4 × 10^-2 × 9.8)

∴ ρ = (2 × 0.0245 × 0.8829) / (0.125 × 10^-3 × 4 × 10^-2 × 9.8)

∴ ρ = 882.9 kg/m³

Ans: The density of paraffin oil is 882.9 kg/m³

Example – 16:

Water rises to a height of 4 cm in a certain capillary tube. If the same capillary tube is dipped in mercury, the level of mercury decreases to 3 cm. Compare the surface tension of water and mercury, if densities of mercury and water are 13.6 × 10³ kg/m³ and 10³ kg/m³ respectively. The angle of contact for water and mercury are 0° and 135° respectively.

Given:   

For water: The rise in tube = h_w =4 cm, Density = ρ_w = 10³ kg/m³, Angle of contact = θ_w =0°, Radius of capillary = r_w= r.

For mercury: The rise in tube = h_m = – 3 cm, Density = ρ_m = 13.6 × 10³ kg/m³, Angle of contact = θ_m =135°, Radius of capillary = r_m = r

To Find: T_w / T_m =?

Solution:

The rise in the tube is given by

Ans: The ratio of the surface tension of water to the surface tension of mercury 0.06932:1

Example – 17:

A U tube has one limb of radius 0.5 cm and the other limb of radius 0.1 cm. Water is poured in the tube. Find the difference in the level of water in the two tubes. S.T. of water = 0.070 N/m. g = 9.8 m/s².

Given: Radius of first limb = r₁ =0.5 cm =  0.5 × 10^-2 m ,  Radius of second limb = r₂ =0.1 cm =  0.1 × 10^-2 m , Surface tension = T = 0.070 N/m, Acceleration due to gravity = g  = 9.8 m/s², Density = ρ = 1  × 10³  kg/m³, Angle of contact = θ  = 0°

To Find: Differene in levels   = |h₂ – h₁| =?

Solution:

The rise in the capillary tube is given by

Hence h ∝ 1/r

Now r₁ > r₂, hence  h₂ > h₁

Ans: The difference in the level of water in the two tubes is 1.142 cm

Previous Topic: The Concept of Capillary Action (Capillarity)

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A capillary tube is a tube with a very fine bore. The phenomenon of a rise and fall of a liquid inside a capillary tube when it is dipped in the liquid is called capillary action or capillarity. When a glass capillary is immersed in water, the water rises inside the tube. When a glass tube is immersed in mercury, the level of mercury inside the tube falls below the level in the outer vessel.

Examples of capillarity: Blotting paper absorbs ink or water, ink rises in a pen, oil rises in wicks of a lamp.

When chalk piece is immersed in water, bubbles are emitted:

Chalk is porous in nature. In these pores, the air is trapped. When chalk piece is dropped in water, due to capillary action water enters the pores of the chalk. Due to which the trapped air comes out of the pores in the form of bubbles.

Everyday Examples of Capillarity:

• Due to capillary action oil rises through wicks of lamps.
• Due to capillary action water rises through the sap of trees.
• Due to capillary action ink is absorbed by blotting paper.
• Due to capillary action liquids are absorbed by sponges.
• Bricks and mortar, which are porous, permit the rise of soil water through them by capillary action. To avoid it, the base is made using damp-proof cement.
• Soil water rises to the surface by capillary action and evaporates. To conserve soil water, the farmers plough their fields in summer. The loosening of the top layer of the soil breaks up the fine capillaries through which the deeper down water is sucked up. Thus the loss of water in the soil is prevented.
• Due to depression in the level of mercury in a capillary tube, a correction is to be applied for mercury barometer.

Capillary Action in Water:

 We know that when a liquid is taken in a vessel, the free surface of the liquid near the walls becomes curved either concave upwards or convex upwards.

In the case of the capillary tube, its bore is very small hence all the points on the free surface are very close to the walls, and hence, the entire free surface becomes curved. It is called a meniscus.

When a glass capillary immersed in water, the meniscus is concave upwards. The air pressure on the upper or concave side of the meniscus is the atmospheric pressure P. We know that the pressure on the convex side of the free surface is less than that on the concave side. It means that the pressure just below the meniscus i.e. on the convex side of the meniscus is less than atmospheric pressure.

The pressure at points just below the flat surface of the water in the surrounding vessel is atmospheric pressure. The system is, therefore not in equilibrium and the liquid rises in the capillary tube till the equilibrium is reached.

Capillary Action in case of Mercury:

We know that when a liquid is taken in a vessel, the free surface of the liquid near the wall becomes curved either concave upwards or convex upwards.

In case of a capillary tube, its bore is very small hence all the points on the free surface are very close to the walls, and hence, the entire free surface becomes curved. It is called a meniscus.

When a glass capillary immersed in mercury, the meniscus is convex upwards. The air pressure on the upper or convex side of the meniscus is the atmospheric pressure P. We know that the pressure on the concave side of the free surface is greater than that on the convex side. It means that the pressure just below the meniscus i.e. on the concave side is more than atmospheric pressure.

The pressure of points just below the flat surface of the mercury in the surrounding vessel is atmospheric pressure.  The system is therefore not in equilibrium and the liquid descends in the capillary tube till equilibrium is reached.

Expression for the Rise in the Liquid in a Capillary Tube:

If a glass tube of a smaller bore (capillary tube) is immersed in a liquid which wets the glass (water), then the liquid level inside the tube rises. If the tube is immersed in a liquid which does not wet the glass (mercury), then the liquid level inside the tube decreases. This phenomenon of the rise or fall of liquid in a capillary tube is called capillary action or capillarity.

Consider a capillary tube immersed in a liquid that wets it. The liquid will rise in the capillary tube. The surface of the liquid will be concave.

The surface tension ‘T’ acts along the tangent to the liquid surface at the point of contact as shown. Let θ be the angle of contact. The force of surface tension is resolved into two components vertical Tcos θ and horizontal T sin θ. The components Tsinθ cancel each other as they are equal in magnitude and radially outward (opposite to each other). The unbalanced component T cos θ will push the liquid up into the capillary tube. This explains the rise in the liquid layer in the capillary tube.

If ‘r’ is the radius of the bore of the capillary tube, the length along which the force of surface tension acts is 2πr. Hence total upward force is 2πr T cos θ.

Due to this force the liquid rise up in the tube. The weight of liquid acts vertically downward. The liquid goes on rising till the force of surface tension is balanced by the weight of the liquid column.

Total upward force = Weight of liquid in the capillary tube.

2πr T cos θ =   mg

Where ‘m’ is the mass of liquid in the capillary tube.

2πr T cos θ  =  V ρ g

Where ‘V’ is the volume of liquid in capillary and ρ is the density of the liquid in the capillary tube.

2πr T cos θ   =    π r²h ρ g

Where ‘h’ is the height of the liquid column in the capillary tube, then

This is an expression for the rise in the liquid in the capillary tube.

Factor’s Affecting the Rise in a Capillary Tube:

The height to which the liquid rises in the capillary tube is

• inversely proportional to the radius of the bore of the capillary tube.
• inversely proportional to the density of the liquid
• directly proportional to the surface tension of the liquid.

Important Observations:

• For those liquids whose angle of contact is acute, the liquid rises in the capillary tube.
• For those liquids whose angle of contact is obtuse, the liquid depresses in the capillary tube.
• For a particular liquid-solid interface, h r = constant. This relation is known as Jurin’s Law.

Determination of the Surface Tension of Liquid:

A glass capillary tube is dipped in a beaker containing liquid (say water) whose surface tension is to be measured. Due to surface tension, the liquid level rises up inside the capillary. The phenomenon is known as capillary action or capillarity.

Using a travelling microscope, the microscope is focused on the lower meniscus of the curved concave surface. The reading is taken using the microscope. The microscope is lowered and it is focused on the water level inside the beaker. The reading is taken again. The difference between the two readings gives the height of the water column inside capillary (h). The inner diameter of the capillary tube is measured using the travelling microscope. Hence we calculate the radius of the bore (r).

Using the following formula Surface tension (T) can be calculated.

where h  =  height of liquid level in the capillary

T  =  Surface tension

ρ =  Density of liquid

r  =  Radius of the bore of the capillary tube

g  =  Acceleration due to gravity

θ =  Angle of contract

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