Properties of Matter and Fluid Mechanics
What is Fluid?
The substances which can flow easily are called fluids. All liquids and gases are fluids . example – water, oil, air etc.
Why Liquid is incompressible?
A liquid is considered incompressible because its molecules are tightly packed together, leaving very little space between them, so applying pressure cannot significantly decrease their volume; essentially, there’s no room to push the molecules closer together as they are already very close to one another.
Liquids are always considered to be incompressible fluids, as density changes caused by pressure and temperature are small. While intuitively gases may always seem to be incompressible fluids if the gas is permitted to move, a gas can be treated as being incompressible if its change in density is small.
Pressure
A sharp needle when pressed against our skin pierces it. Our skin, however, remains intact when a blunt object with a wider contact area (say the back of a spoon) is pressed against it with the same force. Such everyday experiences convince us that both the force and its coverage area are important. Smaller the area on which the force acts, greater is the impact. This impact is known as pressure.
Normal force acting per unit area of a surface is called pressure. It is generally represented by \(p\).
If surface area is \(A\) and normal force acting on the surface is \(F\) (shown in figure), then pressure on the surface is
\(
p=\frac{\text { Normal force }(F)}{\text { Area }(A)}
\)
If area is very small, then \(p=\lim _{\Delta A \rightarrow 0} \frac{\Delta F}{\Delta A}=\frac{d F}{d A}\)
It is clear from the above formula that pressure acting on a surface depends upon the normal force acting on the surface and its area. Thus, force of same magnitude exerts different pressure on different area that is for the same force, if area is lesser, then pressure will be greater.
SI unit of pressure is \(\mathrm{Nm}^{-2}\) or pascal (Pa). Pressure is a scalar quantity and its dimensional formula is \(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\).
Pressure in a fluid
When a fluid (either liquid or gas) is at rest, it exerts a force perpendicular to any surface in contact with it, such as a container wall or a body immersed in the fluid. While the fluid as a whole is at rest, the molecules or atoms that make up the fluid are in motion. These molecules or atoms collide continuously with the walls of container. Pressure of a fluid is due to molecules colliding with the container walls. In Figure below, we can see the pressure acting on all sides of the vessel.
The pressure applied by a fluid can be defined as the magnitude of the normal force (applied by fluid) acting per unit surface area.
If the pressure is same at all points of a finite plane surface with area \(A\), then
\(
p=\frac{F_{\perp}}{A}
\)
where, \(F_{\perp}\) is the normal force on the surface.
Unit of fluid pressure French scientist Pascal contributed many researches in the field of fluid pressure. Thus, in his honour, SI unit of pressure is known as pascal which is denoted by \(\mathrm{Pa} .1 \mathrm{~Pa}=1 \mathrm{Nm}^{-2}\)
One unit used practically in meterology is the bar which is equal to \(10^5 \mathrm{~Pa} .1\) bar \(=10^5 \mathrm{~Pa}\)
When an air filled balloon is immersed inside the water in a vessel it immediately comes up and floats on water. This shows that water (or liquid) exerts pressure in the upward direction. It is shown in Figure below.
Similarly, liquid pressure acts in lateral sides also. When a bottle having water is pierced on the sides we can see water coming out with a speed as in Figure below. This is because liquid exerts lateral pressure on the walls the container.
Example 1: The two thigh bones (femurs), each of cross-sectional area \(10 \mathrm{~cm}^2\) support the upper part of a human body of mass 40 kg. Estimate the average pressure sustained by the femurs.
Solution: Total cross-sectional area of the femurs is \(A=2 \times 10 \mathrm{~cm}^2=20 \times 10^{-4} \mathrm{~m}^2\). The force acting on them is \(F=40 \mathrm{~kg}\) wt \(=400 \mathrm{~N}\) (taking \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\) ). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is
\(
P_{a v}=\frac{F}{A}=2 \times 10^5 \mathrm{Nm}^{-2}
\)
Atmospheric pressure \(\left(p_{a}\right)\)
It is the pressure of the earth’s atmosphere. This changes with weather and elevation. Normal atmospheric pressure at sea level (an average value) is \(1.013 \times 10^5 \mathrm{~Pa}\). Thus,
\(
1 \mathrm{~atm}=1.013 \times 10^5 \mathrm{~Pa}
\)
Other practical units of atmospheric pressure are bar and torr \((\mathrm{mm}\) of Hg\() . \quad 1 \mathrm{~atm}=1.01 \mathrm{bar}=760\) torr
In the experiment it is found that the mercury column in the barometer has a height of about 76 cm at sea level equivalent to one atmosphere ( 1 atm ). A common way of stating pressure is in terms of cm or mm of mercury \((\mathrm{Hg})\). A pressure equivalent of 1 mm is called a torr (after Torricelli).
1 torr \(=133 \mathrm{~Pa}\).
The mm of Hg and torr are used in medicine and physiology. In meteorology, a common unit is the bar and millibar.
\(1 \mathrm{bar}=10^5 \mathrm{~Pa}\)
Absolute pressure and gauge pressure
The excess pressure above atmospheric pressure is usually called gauge pressure and the total pressure is called absolute pressure. Thus,
Gauge pressure = Absolute pressure – Atmospheric pressure
Absolute pressure is always greater than or equal to zero. While gauge pressure can be negative also.
Density of a liquid
Density ( \(\rho\) ) of a liquid or any substance is defined as the mass per unit volume.
\(
\rho==\frac{\text { Mass }}{\text { Volume }}=\frac{m}{V}
\)
Where mass of the fluid is \(m\) occupying volume \(V\).
If the volume is small we can write,
Density, \(\rho=\frac{\text { Mass }}{\text { Volume }}=\lim _{\Delta V \rightarrow 0} \frac{\Delta m}{\Delta V}=\frac{d m}{d V}\)
SI unit of density is \(\mathrm{kgm}^{-3}\) and its dimensions are \(\left[\mathrm{ML}^{-3} \mathrm{~T}^0\right]\) and CGS unit is \(\mathrm{gcm}^{-3}\) with \(1 \mathrm{gcm}^{-3}=10^3 \mathrm{~kg} \mathrm{~m}^{-3}\).
Density of a mixture of two or more liquids
Case I: Suppose two liquids of densities \(\rho_1\) and \(\rho_2\) having masses \(m_1\) and \(m_2\) are mixed together. Then, the density of the mixture will be
\(
\begin{aligned}
\rho & =\frac{\text { Total mass }}{\text { Total volume }} \\
& =\frac{\left(m_1+m_2\right)}{\left(V_1+V_2\right)}=\frac{\left(m_1+m_2\right)}{\left(\frac{m_1}{\rho_1}+\frac{m_2}{\rho_2}\right)}
\end{aligned}
\)
\(
\text { If } m_1=m_2 \text {, then } \rho=\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2} \text {. }
\)
Case II: If two liquids of densities \(\rho_1\) and \(\rho_2\) having volumes \(V_1\) and \(V_2\) are mixed, then the density of the mixture,
If \(\quad V_1=V_2\), then \(\rho=\frac{\rho_1+\rho_2}{2}\)
Example 2: Two liquids of densities \(\rho\) and \(3 \rho\) having volumes \(3 V\) andV are mixed together. Find density of the mixture.
Solution: Given, density of first liquid, \(\rho_1=\rho\)
Density of second liquid, \(\rho_2=3 \rho\)
Volume of first liquid, \(V_1=3 \mathrm{~V}\)
Volume of second liquid, \(V_2=V\)
\(\therefore\) Density of the mixture,
\(
\begin{aligned}
\rho_m & =\frac{\rho_1 V_1+\rho_2 V_2}{V_1+V_2} \\
& =\frac{\rho \times 3 V+3 \rho \times V}{3 V+V}=\frac{6 \rho}{4}=\left(\frac{3}{2}\right) \rho
\end{aligned}
\)
Effect of temperature on density
When the temperature of a liquid is increased, the mass remains the same while the volume is increased and hence, the density of the liquid decreases \(\left(\right.\) as \(\left.\rho \propto \frac{1}{V}\right)\). Thus, \(\frac{\text { Density after increase of temperature by } \Delta \theta}{\text { Initial density }}=\frac{\rho^{\prime}}{\rho}\)
\(
=\frac{V}{V^{\prime}}=\frac{V}{V+d V}=\frac{V}{V+V \gamma \Delta \theta} \quad \text { or } \quad \frac{\rho^{\prime}}{\rho}=\frac{1}{1+\gamma \Delta \theta}
\)
Here, \(\quad \gamma=\) thermal coefficient of volume expansion and \(\Delta \theta=\) rise in temperature.
Therefore,
\(
\rho^{\prime}=\frac{\rho}{1+\gamma \Delta \theta}
\)
Effect of pressure on density
As pressure is increased, volume decreases and hence, density will increase. Thus, \(\rho \propto \frac{1}{V}\)
\(
\begin{array}{r}
\therefore \quad \frac{\text { Density after increase of pressure }}{\text { Initial density }}=\frac{\rho^{\prime}}{\rho} \\
=\frac{V}{V^{\prime}}=\frac{V}{V+d V}=\frac{V}{V-\left(\frac{d p}{B}\right) V}
\end{array}
\)
\(
\frac{\rho^{\prime}}{\rho}=\frac{1}{1-\frac{d p}{B}}
\)
Here, \(d p=\) change in pressure and \(\quad B=\) bulk modulus of elasticity of the liquid.
Therefore,
\(
\rho^{\prime}=\frac{\rho}{1-\frac{d p}{B}}
\)
Illustration 1: The thermal coefficient of volume expansion of a liquid is \(5 \times 10^{-4} \mathrm{~K}^{-1}\). If its temperature is increased by \(30^{\circ} \mathrm{C}\), find the ratio of new density to the previous one.
Solution: Given, thermal coefficient of volume expansion,
\(
\gamma=5 \times 10^{-4} \mathrm{~K}^{-1}
\)
Rise in temperature, \(\Delta \theta=30^{\circ} \mathrm{C}\)
\(
\therefore \quad \quad \rho^{\prime}=\frac{\rho}{1+\gamma \Delta \theta}
\)
\(
\begin{aligned}
\frac{\rho^{\prime}}{\rho} & =\frac{1}{1+\gamma \Delta \theta} \\
& =\frac{1}{1+\left(5 \times 10^{-4}\right)(30)} \simeq 0.98
\end{aligned}
\)
Illustration 2: The bulk modulus of a liquid is \(8 \times 10^9 \mathrm{Nm}^{-2}\) and its density is \(11 \mathrm{~g} \mathrm{~cm}^{-3}\). What will be the density of liquid under a pressure of \(20,000 \mathrm{Ncm}^{-2}\) ?
Solution: Given, bulk modulus of liquid, \(B=8 \times 10^9 \mathrm{Nm}^{-2}\) Density of liquid, \(\rho=11 \mathrm{~g} \mathrm{~cm}^{-3}\) and pressure, \(p=20000 \mathrm{Ncm}^{-2}\)
\(\therefore\) New density of liquid, \(\rho^{\prime}=\frac{\rho}{\left(1-\frac{d p}{B}\right)}\) or \(\frac{\rho}{\left(1-\frac{p}{B}\right)}\)
\(
\begin{aligned}
& =\frac{11}{1-\frac{20000}{8 \times 10^9 \times 10^{-4}}} \\
& =\frac{11 \times 40}{39}=\frac{440}{39} \mathrm{gcm}^{-3}
\end{aligned}
\)
Relative density (RD)
In case of a liquid (or a substance), sometimes an another term relative density (RD) or specific gravity is defined. It is the ratio of density of the liquid (or a substance) to the density of water at \(4^{\circ} \mathrm{C}\).
Hence, \(\quad R D=\frac{\text { Density of liquid or substance }}{\text { Density of water at } 4^{\circ} \mathrm{C}}\)
Relative density is a pure ratio, so it has no units. Density of water at \(4^{\circ} \mathrm{C}\) in CGS is \(1 \mathrm{gcm}^{-3}\). Therefore, numerically the relative density and density of a substance in CGS are equal. In SI unit, the density of water at \(4^{\circ} \mathrm{C}\) is \(1000 \mathrm{kgm}^{-3}\).
Pressure due to liquid column
A tall beaker is filled with liquid so that it forms a liquid column. The area of cross section at the bottom is \(A\). The density of the liquid is \(\rho\). The height of the liquid column is \(h\). In other words the depth of the water from the top level surface is ‘ \(h\) ‘ as shown in Figure below.
We know that thrust at the bottom of the column \(({F})=\) weight of the liquid.
Therefore, \({F}={mg} \dots(1)\)
We can get the mass of the liquid by multiplying the volume of the liquid and its density.
Mass, \({m}=\rho {V} \dots(2)\)
Volume of the liquid column, \({V}=\) Area of cross section \(({A}) \times\) Height \(({h})={Ah} \dots(3)\)
Substituting (3) in (2)
Hence, mass, \(m=\rho A h \dots(4)\)
Substituting (4) in (1)
Force \(=m g=\rho A h g\)
Pressure, \(P=\frac{\text { Thrust }({F})}{\operatorname{Area}({A})}=\frac{{mg}}{{A}}=\frac{\rho({Ah}) {g}}{{A}}=\rho {hg}\)
\(\therefore\) Pressure due to a liquid column, \({P}={h} \rho {g}\)
This expression shows that pressure in a liquid column is determined by depth, density of the liquid and the acceleration due to gravity. Pressure exerted by a liquid at a point is determined by,
(i) depth (\(h\))
(ii) density of the liquid ( \(\rho\) )
(iii) acceleration due to gravity (\(g\)).
Interestingly, the final expression for pressure does not have the term area \(A\) in it. Thus pressure at a given depth does not depend upon the shape of the vessel containing the liquid or the amount of liquid in the vessel. It only depends on the depth. In Figure below, the pressure is the same even though the containers have different amounts of liquid in them, and are of different shapes.
Variation of fluid pressure with depth
Consider a fluid of density \(\rho\) is kept at rest, in a cylindrical vessel of height \(h\) as shown in figure.
In Figure above point 1 is at height \(h\) above a point 2. The pressures at points 1 and 2 are \(p_1\) and \(p_2\) respectively. Consider a cylindrical element of fluid having area of base \(A\) and height \(h\). As the fluid is at rest the resultant horizontal forces should be zero and the resultant vertical forces should balance the weight of the element. The forces acting in the vertical direction are due to the fluid pressure at the top \(\left(p_1 A\right)\) acting downward, at the bottom \(\left(p_2 A\right)\) acting upward. If \(m g\) is weight of the fluid in the cylinder we have
\(
\left(p_2-p_1\right) A=m g
\)
Now, if \(\rho\) is the mass density of the fluid, we have the mass of fluid to be \(m=\rho V=\rho h A\) so that
\(
p_2-p_1=\rho g h \dots(i)
\)
From the above equation we can see the Pressure difference depends on the vertical distance \(h\) between the points ( 1 and 2), mass density of the fluid \(\rho\) and acceleration due to gravity \(g\).
If the point 1 under discussion is shifted to the top of the fluid (say, water), which is open to the atmosphere, \({p}_1\) may be replaced by atmospheric pressure \(\left({p}_{{a}}\right)\) and we replace \({p}_2\) by \(p\). Then Eq. (i) gives
\(
p=p_{\mathrm{a}}+\rho g h \dots(ii)
\)
Thus, the pressure \(p\), at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount \(\rho g h\). The excess of pressure, \(p-p_{{a}}\), at depth \(h\) is called a gauge pressure at that point.
According to Eq. (ii), pressure increases linearly with depth, if \(\rho\) and \(g\) are uniform.
A graph between \(p\) and \(h\) is shown below
The area of the cylinder is not appearing in the expression of absolute pressure in Eq. (ii). Thus, the height of the fluid column is important and not cross-sectional or base area or the shape of the container. The liquid pressure is the same at all points at the same horizontal level (same depth).
Hydrostatic paradox
Consider three vessels A, B and C [Figure below] of different shapes. They are connected at the bottom by a horizontal pipe. On filling with water, the level in the three vessels is the same, though they hold different amounts of water. This is so because water at the bottom has the same pressure below each section of the vessel.
Example 3: What is the pressure on a swimmer 10 m below the surface of a lake?
Solution: Here
\(h=10 \mathrm{~m}\) and \(\rho=1000 \mathrm{~kg} \mathrm{~m}^{-3}\). Take \(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\)
From Eq. (ii)
\(
\begin{aligned}
p & =p_{\mathrm{a}}+\rho g h \\
& =1.01 \times 10^5 \mathrm{~Pa}+1000 \mathrm{~kg} \mathrm{~m}^{-3} \times 10 \mathrm{~m} \mathrm{~s}^{-2} \times 10 \mathrm{~m} \\
& =2.01 \times 10^5 \mathrm{~Pa} \\
& \approx 2 \mathrm{~atm}
\end{aligned}
\)
This is a \(100 \%\) increase in pressure from surface level. At a depth of 1 km , the increase in pressure is 100 atm ! Submarines are designed to withstand such enormous pressures.
Important points related to fluid pressure
Important points related to fluid pressure are given below
- At a point in the liquid column, the pressure applied on it is same in all directions.
- In a liquid, pressure will be same at all points at the same level (height).
- The pressure exerted by a liquid depends only on the height of fluid column and is independent of the shape of the containing vessel.
- Consider following shapes of vessels: Pressure at the base of each vessel,
\(
\begin{aligned}
& p_x=p_y=p_z=p_0+\rho g h \text { but weight, } w_x \neq w_y \neq w_z \\
& \text { where, } \rho=\text { density of liquid in each vessel, } \\
& \qquad h=\text { height of liquid in each vessel }
\end{aligned}
\)
and \(\quad p_0=\) atmospheric pressure.
- In the figure, a block of mass \(m\) floats over a fluid surface, If \(\rho=\) density of the liquid,
\(A=\) area of the block, then
Pressure at the base of the vessel is
\(
p=p_0+\rho g h+\frac{m g}{A}
\)
Example 4: Find the pressure exerted below a column of water, open to the atmosphere, at depth 10 m
(Take, density of water \(=1 \times 10^3 \mathrm{kgm}^{-3}, g=10 \mathrm{~ms}^{-2}\) )
Solution: Pressure at depth of 10 m ,
\(
\begin{aligned}
p & =p_a+\rho g h \\
& =1.013 \times 10^5 \mathrm{~Pa}+\left(1 \times 10^3 \mathrm{kgm}^{-3}\right)\left(10 \mathrm{~ms}^{-2}\right)(10 \mathrm{~m}) \\
& =1.013 \times 10^5 \mathrm{~Pa}+1 \times 10^5 \mathrm{~Pa} \\
& =2.013 \times 10^5 \mathrm{~Pa} \\
& \approx 2 \mathrm{~atm}
\end{aligned}
\)
Example 5: For the arrangement shown in the figure, what is the density of oil?
Solution: According to the figure, for the equilibrium of liquid, Net downward force \(=\) Net upward force
\(
\begin{aligned}
p_0+\rho_w g l & =p_0+\rho_{\mathrm{oil}}(l+d) g \\
\Rightarrow \quad & \rho_{\mathrm{oil}}=\frac{\rho_w l}{l+d}=\frac{1000 \times(135)}{(135+12.3)}=916 \mathrm{kgm}^{-3}
\end{aligned}
\)
Pascal’s law
Pascal’s law states that whenever external pressure is applied to any part of an enclosed liquid. This pressure gets distributed in all directions equally. The French scientist Blaise Pascal observed that the pressure in a fluid at rest is the same at all points if they are at the same height.
Consider an example in the figure above which a flask fitted with a piston is filled with a liquid. Let an external force \(F\) is applied on the piston. If the cross-sectional area of the piston is \(A\), the pressure just below the piston is increased by \(F / A\). By Pascal’s law, the pressure applied by piston will be transmitted equally at all points in the flask. If we make some holes in the flask, then the liquid from all the holes will emerge out with same intensity. The hydraulic lift is also an application of this law.
Hydraulic Machines
Let us now consider what happens when we change the pressure on a fluid contained in a vessel. Consider a horizontal cylinder with a piston and three vertical tubes at different points [Figure below]. The pressure in the horizontal cylinder is indicated by the height of liquid column in the vertical tubes. It is necessarily the same in all. If we push the piston, the fluid level rises in all the tubes, again reaching the same level in each one of them.
This indicates that when the pressure on the cylinder was increased, it was distributed uniformly throughout. We can say whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions. This is another form of the Pascal’s law and it has many applications in daily life.
Hydraulic lift
A well known application of Pascal’s law is the hydraulic lift used to support or lift heavy objects. A number of devices, such as hydraulic lift and hydraulic brakes, are based on the Pascal’s law. In these devices, fluids are used for transmitting pressure. It is schematically illustrated in figure.
In a hydraulic lift, as shown in Figure above, two pistons are separated by the space filled with a liquid. A piston of small cross-section \(A_1\) is used to exert a force \(F_1\) directly on the liquid. The pressure \(P=\frac{F_1}{A_1}\) is transmitted throughout the liquid to the larger cylinder attached with a larger piston of area \(A_2\), which results in an upward force of \(P \times A_2\). Therefore, the piston is capable of supporting a large force (large weight of, say a car, or a truck, placed on the platform)
\(F_2=P A_2=\frac{F_1 A_2}{A_1}\).
By changing the force at \(A_1\), the platform can be moved up or down. Thus, the applied force has been increased by a factor of \(\frac{A_2}{A_1}\) and this factor is the mechanical advantage of the device.
Now, since \(A_2 \gg A_1\), therefore \(F_2 \gg F_1\). Thus, hydraulic lift is a force multiplying device with a multiplication factor equal to the ratio of the areas of the two pistons. Dentist’s chairs, car lifts and jacks, hydraulic elevators and hydraulic brakes all use this principle.
Pascal’s Law Derivation
Let us consider a right-angled triangle(with sides p, q, and r) prism (height s) submerged in the liquid of density \(\rho\), also assume the size of the submersed element is negligible with compare to the volume of the liquid, and all the points on the element experience the same gravitational force.
Now, the area of the faces PQRS, PSUT, and QRUT of the prism is ps, qs, and rs respectively. Also, assume the pressure applied by the liquid on these faces is \(\mathrm{P}_1\), \(P_2\), and \(P_3\) respectively.
Exerted force by this pressure to the faces in the perpendicular inward direction is \(F_1, F_2\), and \(F_3\).
Thus, \(F_1=P_1 \times\) Area of PQRS \(=P_1 \times p s\)
\(F_2=P_2 \times\) Area of PSUT \(=P_2 \times q s\)
\(F_3=P_3 \times\) Area of QRUT \(=P_3 \times r s\)
Now, in triangle PQT,
\(\sin \theta=p / r\) and \(\cos \theta=q / r\)
The net force on the prism will be zero since the prism is in equilibrium.
\(F_3 \sin \theta=F 1\) and \(F_3 \cos \theta=F_2\) (putting values of \(F_1, F_2\), and \(F_3\) from the above values)
\(
\begin{aligned}
& \Rightarrow P_3 \times r s \times p / r=P_1 \times p s \text { and } P_3 \times r s \times q / r=P_2 \times q s \\
& \Rightarrow P_3=P_1 \text { and } P_3=P_2
\end{aligned}
\)
Thus, \(\mathrm{P}_1=\mathrm{P}_2=\mathrm{P}_3\)
Therefore, pressure throughout the liquid remains the same.
Example 5: Two pistons of a hydraulic machine have diameters 20 cm and 2 cm . Find the force exerted on the larger piston when 50 kg -wt is placed on the smaller piston. When the smaller piston moves in through 5 cm , by what distance, the other piston moves out?
Solution: For smaller piston, area, \(A_1=\pi \times(1)^2\)
For larger piston, area, \(A_2=\pi \times(10)^2\)
\(\therefore\) Force exerted on the larger piston,
\(
\begin{aligned}
F_2 & =\frac{A_2}{A_1} \times F_1 \\
& =\frac{\pi(10)^2}{\pi(1)^2} \times 50 \times 9.8 \\
& =100 \times 50 \times 9.8 \\
& \simeq 5 \times 10^4 \mathrm{~N}
\end{aligned}
\)
This is the force exerted on the larger piston. The liquids are considered incompressible. Therefore, volume covered by movement of smaller piston inwards equal to the outward movement of larger piston.
\(
\begin{aligned}
\therefore \quad L_1 A_1 & =L_2 A_2 \\
\Rightarrow \quad L_2 & =\left(\frac{A_1}{A_2}\right) L_1 \\
& =\frac{\pi(1 \mathrm{~cm})^2}{\pi(10 \mathrm{~cm})^2} \times 5 \mathrm{~cm} \\
& =\frac{1}{100} \times 5 \mathrm{~cm} \\
& =0.05 \mathrm{~cm}
\end{aligned}
\)
So, the distance moved out by the larger piston is 0.05 cm.
Example 6: In a car lift compressed air exerts a force \(F_1\) on a small piston having a radius of 5.0 cm . This pressure is transmitted to a second piston of radius 15 cm (Fig 9.7). If the mass of the car to be lifted is 1350 kg , calculate \(F_1\). What is the pressure necessary to accomplish this task? \(\left(g=9.8 \mathrm{~ms}^{-2}\right)\).
Solution: Since pressure is transmitted undiminished throughout the fluid,
\(
\begin{aligned}
F_1=\frac{A_1}{A_2} F_2= & \frac{\pi\left(5 \times 10^{-2} \mathrm{~m}\right)^2}{\pi\left(15 \times 10^{-2} \mathrm{~m}\right)^2}\left(1350 \mathrm{~kg} \times 9.8 \mathrm{~m} \mathrm{~s}^{-2}\right) \\
& =1470 \mathrm{~N} \\
& \approx 1.5 \times 10^3 \mathrm{~N}
\end{aligned}
\)
The air pressure that will produce this force is
\(
P=\frac{F_1}{A_1}=\frac{1.5 \times 10^3 \mathrm{~N}}{\pi\left(5 \times 10^{-2}\right)^2 \mathrm{~m}}=1.9 \times 10^5 \mathrm{~Pa}
\)
This is almost double the atmospheric pressure.
Measurement of pressure
Pressure can be measured by using following two devices
(i) Barometer:
It is a device used to measure atmospheric pressure. In principle, any liquid can be used to fill the barometer, but mercury is the substance of choice because its high density makes possible an instrument of reasonable size. A barometer is an inverted evacuated tube, put over a mercury volume. Outside pressure pushes mercury into tube till the weight of liquid column equalises the force due to external pressure.
In given diagram, in equilibrium,
\(p_1=p_2\)
\(
\begin{array}{ll}
\text { Here, } & p_1=\text { atmospheric pressure }\left(p_0\right) [\text { same as } \left(p_a\right)\\
\text { and } & p_2=0+\rho g h=\rho g h
\end{array}
\)
where, \(\rho=\) density of mercury
\(
\therefore \quad p_0=\rho g h
\)
Thus, the mercury barometer reads the atmospheric pressure \(\left(p_0\right)\) directly from the height of the mercury column.
e.g. If the height of mercury in a barometer is 760 mm , then atmospheric pressure will be
\(
\begin{aligned}
p_0 & =\rho g h \\
& =\left(13.6 \times 10^3\right)(9.8)(0.760) \\
& =1.01 \times 10^5 \mathrm{Nm}^{-2}
\end{aligned}
\)
(ii) Manometer:
It is a device used to measure the pressure of a gas inside a container.
The U-shaped tube often contains mercury. Let 1 and 2 are points on same horizontal level, then
\(
p_1=p_2
\)
Here, \(\quad p_1=\) pressure of the gas in the container \((p)\)
and \(\quad p_2=\) atmospheric pressure \(\left(p_0\right)+\rho g h\)
\(
\therefore \quad p=p_0+\rho g h
\)
This can also be written as
\(
p-p_0=\text { gauge pressure }=\rho g h
\)
Here, \(\rho\) is the density of the liquid used in U-tube. Thus, by measuring \(h\), we can find absolute (or gauge) pressure in the vessel.
Example 7: What will be the length of mercury column in a barometer tube, when the atmospheric pressure is 76 cm of mercury and the tube is inclined at an angle of \(30^{\circ}\) with the horizontal direction?
Solution: Here, \(h=76 \mathrm{~cm}, \theta=30^{\circ}\)
If \(l\) is the length of mercury column in a barometer tube, then
\(
\frac{h}{l}=\sin 30^{\circ}
\)
\(
\begin{array}{ll}
\Rightarrow & \frac{76 \mathrm{~cm}}{l}=\frac{1}{2} \\
\Rightarrow & l=2 \times 76=152 \mathrm{~cm}
\end{array}
\)
Example 8: A manometer tube contains a liquid of density \(3 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\). When connected to a vessel containing a gas, the liquid level in the other arm of the tube is higher by 20 cm . When connected to another sample of enclosed gas, the liquid level in the other arm of the manometer tube falls 8 cm below the liquid level in the first arm. Which of the two samples exerts more pressure and by what amount?
Solution:
For Sample 1:
Difference in level of liquids, \(h_1=20 \mathrm{~cm}=0.2 \mathrm{~m}\)
Pressure of the gas in the left arm, \(p_1=p_a+\rho g h_1 \dots(i)\)
For sample 2:
In this case, level of the liquid in the left arm is higher than that in the right arm by 8 cm .
\(\therefore\) Atmospheric pressure \(p_a\) is greater than the pressure exerted by the sample, i.e.
\(
p_a=p_2+\rho g h_2 \Rightarrow p_2=p_a-\rho g h_2 \dots(ii)
\)
Comparing Eqs. (i) and (ii), it is clear that \(p_1>p_2\).
Therefore, the gas in sample 1 exerts greater pressure than that in sample 2.
The difference in the two pressures is
\(
\begin{aligned}
p_1-p_2 & =\left(p_a+\rho g h_1\right)-\left(p_a-\rho g h_2\right) \\
& =\rho g\left(h_1+h_2\right)=\rho g(28 \mathrm{~cm}) \\
& =\left(3 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\right) \times\left(9.8 \mathrm{~ms}^{-2}\right)(0.28 \mathrm{~m}) \\
& =8.23 \times 10^3 \mathrm{~Pa} \approx 8 \mathrm{kPa}
\end{aligned}
\)
Archimedes’ principle
If a heavy object is immersed in water, it seems to weigh less than when it is in air. This is because the water exerts a net upward normal force called buoyant force.
The Archimedes principle gives the magnitude of buoyant force on a body. It states that, “a body wholly or partially submerged in a fluid experiences an upward force which is equal to the weight of the displaced fluid.”
Thus, the magnitude of buoyant force \((F)\) which is also called upthrust is given by
\(
F=V_i \rho_l g
\)
where, \(\quad V_i=\) immersed volume of solid,
\(\rho_l=\) density of liquid
and \(\quad g=\) acceleration due to gravity.
The upthrust act vertically upwards through the centre of gravity of displaced fluid.
Apparent weight of a body inside a liquid
If a body is completely immersed in a liquid, its effective weight gets decreased. The decrease in its weight is equal to the upthrust on the body.
Hence, apparent weight of body \(=w_{\text {app }}=w_{\text {actual }}-\) upthrust or
\(
w_{\mathrm{app}}=V \rho_s g-V \rho_l g
\)
Here, \(\quad V=\) total volume of the body,
\(\rho_s=\) density of body and \(\rho_l=\) density of liquid.
Thus, \(w_{\text {app }}=V g\left(\rho_s-\rho_l\right)\)
\(
\begin{array}{ll}
\text { or } & w_{\text {app }}=V \rho_s g\left(1-\frac{\rho_l}{\rho_s}\right) \\
\therefore & w_{\text {app }}=w_{\text {actual }}\left(1-\frac{\rho_l}{\rho_s}\right)
\end{array}
\)
Special cases Three possibilities may now arise
(i) If \(\rho_s<\rho_l\), then in this condition, the upthrust applied by the liquid will be greater than the weight of the body. That means, if the body is completely immersed in liquid, it will experience a net upward force. When released, the body comes up to the fluid surface till the upthrust becomes equal to the weight of the body, at this point, the body floats partially immersed in the fluid.
(ii) If \(\rho_s=\rho_l\), then \(w_{\mathrm{app}}=0\) in this condition, upthrust of the liquid balances the weight of the body. The body floats completely submerged just below the surface of the fluid.
(iii) If \(\rho_s>\rho_l\), then in this condition, the upthrust of the liquid is less than the weight of the liquid, i.e. it is not sufficient to balance the weight of the body, so the body sinks.
Example 9: An ornament weighing 50 g in air weights only 46 g in water. Assuming that some copper is mixed with gold to prepare the ornament. Find the amount of copper in it. Specific gravity of gold is 20 and that of copper is 10 .
Solution: Let \(m\) be the mass of the copper in ornament.
Then, mass of gold in it is \((50-m)\).
Volume of copper, \(V_1=\frac{m}{10} \quad\left(\because \text { Volume }=\frac{\text { Mass }}{\text { Density }}\right)\)
and \(\quad\) volume of gold, \(V_2=\frac{50-m}{20}\)
When immersed in water ( \(\rho_w=1 \mathrm{gcm}^{-3}\) ),
decrease in weight \(=\) upthrust
\(
\therefore \quad(50-46) g=\left(V_1+V_2\right) \rho_w g
\)
\(
4=\frac{m}{10}+\frac{50-m}{20} \text { or } 80=2 m+50-m
\)
\(\therefore \quad m=30 \mathrm{~g}\)
Example 10: Density of ice is \(900 \mathrm{~kg} \mathrm{~m}^{-3}\). A piece of ice is floating in water (of density \(1000 \mathrm{~kg} \mathrm{~m}^{-3}\) ). Find the fraction of volume of the piece of ice outside the water.
Solution: Let \(V\) be the total volume and \(V_i\) the volume of ice piece immersed in water. For equilibrium of ice piece,
weight \(=\) upthrust
\(
\therefore \quad V \rho_i g=V_i \rho_w g \dots(i)
\)
Here,\(\quad \rho_i=\) density of ice \(=900 \mathrm{kgm}^{-3}\)
and \(\quad \rho_w=\) density of water \(=1000 \mathrm{kgm}^{-3}\)
Substituting in Eq. (i), we get
\(
\frac{V_i}{V}=\frac{\rho_i}{\rho_w}=\frac{900}{1000}=0.9
\)
i.e. The fraction of volume outside the water, \(f=1-0.9=0.1\)
Example 11: A piece of ice is floating in a glass vessel filled with water. How will the level of water in the vessel changes when the ice melts?
Solution: Let \(m\) be the mass of ice piece floating in water.
In equilibrium, weight of ice piece \(=\) upthrust
\(
m g=V_i \rho_w g \text { or } V_i=\frac{m}{\rho_w} \dots(i)
\)
Here, \(V_i\) is the volume of ice piece immersed in water.
When the ice melts, let \(V\) be the volume of water formed by \(m\) mass of ice.
Then,
\(
V=\frac{m}{\rho_w} \dots(ii)
\)
From Eqs. (i) and (ii), we see that, \(V_i=V\)
Hence, the level will not change.
Example 12: A piece of ice having a stone frozen in it, floats in a glass vessel filled with water. How will the level of water in the vessel changes when the ice melts?
Solution: Let \(m_1=\) mass of ice, \(m_2=\) mass of stone,
\(\rho_S=\) density of stone and \(\rho_w=\) density of water.
In equilibrium, when the piece of ice floats in water,
\(
\begin{array}{rlrl}
& \text { weight of (ice }+ \text { stone }) & =\text { upthrust } \\
\left(m_1+m_2\right) g & =V_i \rho_w g \dots(i) \\
\therefore \quad V_i & =\frac{m_1}{\rho_w}+\frac{m_2}{\rho_w}
\end{array}
\)
Here, \(V_i=\) volume of ice immersed.
When the ice melts, \(m_1\) mass of ice converts into water and stone of mass \(m_2\) is completely submerged.
Volume of water formed by \(m_1\) mass of ice, \(V_1=\frac{m_1}{\rho_w}\)
Volume of stone (which is also equal to the volume of water displaced),
\(
V_2=\frac{m_2}{\rho_S}
\)
Since, \(\rho_S>\rho_w\), so \(V_1+V_2<V_i\) or the level of water will decrease.
Buoyant force in accelerating fluids
If a body is dipped inside a liquid of density \(\rho_L\) placed in an elevator moving upward with an acceleration \(a\) as shown in figure.
Now, consider the force acting on the liquid replaced by the body. For upward motion of the replaced liquid, we can write,
\(
F-w=m a
\)
where, \(w=\) weight of the displaced liquid \(F=\) buoyant force acting on the bodly
\(\therefore \quad F=w+m a=m(g+a)=\rho_L V(g+a)\)
\(
F=V \rho_L g_{\mathrm{eff}}
\)
Here, \(g_{\mathrm{eff}}=|g+a|\)
e.g. If the lift is moving upwards with an acceleration \(a\), the value of \(g_{\text {eff }}\) is \(g+a\) and if it is moving downwards with acceleration \(a\), the \(g_{\text {eff }}\) is \(g-a\). In a freely falling lift, \(\boldsymbol{g}_{\text {eff }}\) is zero (as \(a=g\) ) and hence, net buoyant force is zero. This is why, in a freely falling vessel filled with some liquid, the air bubbles do not rise up (which otherwise move up due to buoyant force).
Example 13: The tension in a string holding a solid block below the surface of a liquid (of density greater than that of solid) as shown in figure is \(T_0\) when the system is at rest. What will be the tension \(T\) in the string, if the system has an upward acceleration \(a\)?
Solution: Let \(m\) be the mass of block.
Initially for the equilibrium of block, \(F=T_0+m g \dots(i)\)
Here, \(F\) is the upthrust on the block.
When the lift is accelerated upwards, \(g_{\text {eff }}\) becomes \(g+a\) instead of \(g\).
Hence, \(\frac{F^{\prime}}{F}=\frac{m(g+a)}{mg}\)
\(
F^{\prime}=F\left(\frac{g+a}{g}\right) \dots(ii)
\)
From Newton’s second law,
\(
F^{\prime}-T-m g=m a \dots(iii)
\)
Solving Eqs. (i), (ii) and (iii), we get
\(
T=T_0\left(1+\frac{a}{g}\right)
\)
Flow of Fluids
So far we have studied fluids at rest. The study of the fluids in motion is known as fluid dynamics. In this section, we will study fluids in motion. When a fluid is in flow, its motion can either be smooth (steady flow which is known as also streamline flow) or irregular (turbulent flow) depending on its velocity of flow.
Streamline Flow (Laminar Flow)
When a water tap is turned on slowly, the water flow is smooth initially but loses its smoothness when the speed of the outflow is increased. In studying the motion of fluids, we focus our attention on what is happening to various fluid particles at a particular point in space at a particular time. The flow of the fluid is said to be steady if at any given point, the velocity of each passing fluid particle remains constant in time. This does not mean that the velocity at different points in space is same. The velocity of a particular particle may change as it moves from one point to another. That is, at some other point the particle may have a different velocity, but every other particle which passes the second point behaves exactly as the previous particle that has just passed that point. Each particle follows a smooth path, and the paths of the particles do not cross each other. Streamline flow typically occurs at lower velocities. When the speed of the fluid increases too much, the flow can become turbulent. Steamline flow has following characterstics (a stream line flow is shown below):
Parallel Paths:
In streamlined flow, fluid particles move in straight or gently curving paths that never intersect with one another.
Consistent Speed:
Each particle moves at a constant speed along its path. This means that if you were to watch a particle, it would keep moving steadily without speeding up or slowing down.
Orderly Movement:
There’s no mixing or swirling in streamlined flow. The fluid layers slide over one another smoothly, like sheets of paper sliding over each other.
Predictable:
Because the flow is so orderly, it’s easy to predict where a particle will go. This makes calculations and designs based on streamlined flow very reliable.
Low Velocity:
Streamline flow typically occurs at lower velocities. When the speed of the fluid increases too much, the flow can become turbulent.
If the velocity of fluid particles at any point does not vary with time, the flow is said to be steady. Steady flow is also called streamlined or laminar flow. The velocity at different points may be different. Hence, in the figure, we can say that velocity of fluid particles at different points 1,2 and 3 remains same with time.
i.e. \(\mathbf{v}_1=\) constant, \(\mathbf{v}_2=\) constant, \(\mathbf{v}_3=\) constant, but
\(
\mathbf{v}_1 \neq \mathbf{v}_2 \neq \mathbf{v}_3
\)
The path followed by a fluid particle in steady flow is called streamline. Velocity of fluid particle at any point of the streamline is along tangent to the curve at that point. Streamline flow is possible only if the liquid velocity does not exceed a limiting value called critical velocity.
Streamline flow Equation of Continuity
The Equation of Continuity asserts that for an incompressible fluid, the product of area and fluid speed at all points all along the pipe is constant. The Equation of Continuity asserts that if no leaks exist, the volume of fluid entering one end of a tube in a particular time interval equals the volume leaving the other end of the tube within the same time interval.
This equation helps us understand how fluids behave when they move through different shapes of pipes or channels. It’s crucial for designing systems like water supply networks, air conditioning ducts, and even for understanding blood flow in arteries. Think of a river. In some places, it’s wide and the water flows gently. In other places, the river narrows, and the water rushes through quickly. The Equation of Continuity explains this behavior – the water speeds up where the river is narrow because the area decreases, so the velocity must increase to maintain the balance.
The Equation of Continuity is based on the principle of conservation of mass, which states that mass cannot be created or destroyed within a closed system.
Imagine a pipe (shown above in diagram) with varying cross-sectional areas through which an incompressible fluid (like water) is flowing. Let’s take two different sections of the pipe, section 1 with area \(\left({A}_1\right)\) and section 2 with area \(\left({A}_2\right)\).
In a small time interval \((\Delta {t})\), the fluid will move a distance \(\left(\Delta {x}_1\right)\) at section 1 with velocity \(\left({v}_1\right)\) and \(\left(\Delta {X}_2\right)\) at section 2 with velocity \(\left({v}_2\right)\).
The volume of fluid passing through section 1 in time \((\Delta t)\) is \(\left({A}_1 \Delta {x}_1\right)\), and through section 2 is \(\left(A_2 \Delta x_2\right)\).
The distance of each section of fluid travels is related to its velocity and time:
\(
\begin{aligned}
& \Delta x_1=v_1 \Delta t \\
& \Delta x_2=v_2 \Delta t
\end{aligned}
\)
Since the fluid is incompressible, its density ( \(\rho\) ) remains constant. The mass of fluid passing through each section is the product of density and volume (\(\rho=\frac{m}{V}\)):
\(
\begin{aligned}
& m_1=\rho A_1 \Delta x_1 \\
& m_2=\rho A_2 \Delta x_2
\end{aligned}
\)
The mass entering section 1 must equal the mass leaving section 2 in the time \((\Delta t)\) :
\(
\begin{aligned}
m_1 & =m_2 \\
\rho A_1 \Delta x_1 & =\rho A_2 \Delta x_2
\end{aligned}
\)
Replace ( \(\Delta {x}_1\) ) and ( \(\Delta {x}_2\) ) with their respective expressions involving velocity and time:
\(
\rho A_1 v_1 \Delta t=\rho A_2 v_2 \Delta t
\)
The density ( \(\rho\) ) and time ( \(\Delta t\) ) are the same on both sides of the equation, so they cancel out:
\(
A_1 v_1=A_2 v_2
\)
This equation tells us that the product of the cross-sectional area and the velocity at any two points along the pipe is constant for an incompressible fluid. It implies that where the pipe narrows (small (\(A\)), the velocity (\(v\)) increases, and where the pipe widens (large \(A\)), the velocity (\(v\)) decreases.
Example 14: Water is flowing through a horizontal tube of non-uniform cross-section. At a place, the radius of the tube is 1.0 cm and the velocity of water is \(2 \mathrm{~ms}^{-1}\). What will be the velocity of water, where the radius of the pipe is 2.0 cm ?
Solution: Using equation of continuity,
\(
\begin{gathered}
A_1 v_1=A_2 v_2 \quad \text { or } \quad v_2=\left(\frac{A_1}{A_2}\right) v_1 \\
v_2=\left(\frac{\pi r_1^2}{\pi r_2^2}\right) v_1=\left(\frac{r_1}{r_2}\right)^2 v_1
\end{gathered}
\)
Given, \(r_1=1 \mathrm{~cm}=1 \times 10^{-2} \mathrm{~m}, r_2=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}\), \(v_1=2 \mathrm{~ms}^{-1}\)
Substituting the above values, we get
\(
v_2=\left(\frac{1.0 \times 10^{-2}}{2.0 \times 10^{-2}}\right)^2(2) \text { or } v_2=0.5 \mathrm{~ms}^{-1}
\)
Turbulent Flow
Turbulent flow is characterized by chaotic changes in pressure and flow velocity. It is the opposite of streamlined flow and occurs at high velocities or with large obstacles. Turbulent flow is a type of fluid motion that is chaotic and unpredictable. Unlike streamlined flow, where the fluid moves in orderly layers, turbulent flow is characterized by irregular fluctuations and mixing.
In rivers and canals, where speed of water is quite high or the boundary surfaces cause abrupt changes in velocity of the flow, then the flow becomes irregular. Such flow of liquid is known as turbulent flow. Thus, the flow of fluid in which velocity of all particles crossing a given point is not same and the motion of the fluid becomes irregular or disordered. This is called turbulent flow.
A few examples of turbulent flow are
- a jet of air striking a flat plate.
- the smoke rising from a burning stock of wood.
A jet of air striking a flat plate placed perpendicular to it. This is an example of turbulent flow.
Reynolds number
Reynolds number \(\left(R_e\right)\) is a dimensionless number, whose value gives us an idea whether the flow would be laminar (streamlined) or turbulent.
\(
R_e=\frac{\text { Inertial force }}{\text { Viscous force }}
\)
Thus, \(R_e\) represents the ratio of inertial force of moving fluid to viscous force offered by the fluid.
The expression for \(R_e\) is
\(
R_e=\frac{\rho v D}{\eta}
\)
Here, \(\rho=\) density of fluid,
\(v=\) speed of fluid,
\(D=\) diameter of tube
and \(\quad \eta=\) viscosity of fluid (later on it is discussed in detail)
For \(R_e<1000\), flow is streamline or laminar.
For \(R_e \rightarrow 1000\) to 2000 , flow is unsteady.
For \(R_e>2000\), flow is turbulent.
Example 15: Water is flowing in a pipe of diameter 6 cm with an average velocity \(7.5 \mathrm{cms}^{-1}\) and its density is \(10^3 \mathrm{~kg} \mathrm{~m}^{-3}\). What is the nature of flow? Given, coefficient of viscosity of water is \(10^{-3} \mathrm{kgm}^{-1} \mathrm{~s}^{-1}\).
Solution: Reynold’s number for the given situation is given as
\(
R_e=\frac{\rho v D}{\eta}
\)
Here, density, \(\rho=10^3 \mathrm{kgm}^{-3}\)
Coefficient of viscosity, \(\eta=10^{-3} \mathrm{kgm}^{-1} \mathrm{~s}^{-1}\)
Average velocity of water, \(v=7.5 \mathrm{cms}^{-1}\)
\(
=0.075 \mathrm{~ms}^{-1}
\)
Diameter of pipe, \(D=6 \mathrm{~cm}=0.06 \mathrm{~m}\)
Hence, \(\quad \begin{aligned} R_e & =\frac{10^3 \times 0.075 \times 0.06}{10^{-3}} \\ & =10^6 \times 0.0045=4500\end{aligned}\)
\(
\because \quad R_e>2000
\)
Therefore, the flow is turbulent.
Bernoulli’s Principle
Bernoulli’s theorem is based on the law of conservation of energy and applied to ideal fluids.
It states that the sum of pressure energy per unit volume, kinetic energy per unit volume and potential energy per unit volume of an incompressible, non-viscous fluid in a streamlined irrotational flow remains constant at every cross-section throughout the liquid flow.
Mathematically, it can be expressed as
\(
p+\frac{1}{2} \rho v^2+\rho g h=\text { constant }
\)
where, \(p\) represents the pressure energy per unit volume (or pressure), \(\frac{1}{2} \rho v^2\) the kinetic energy per unit volume and \(\rho g h\) the potential energy per unit volume.
Proof: Consider an ideal fluid having streamline flow through a pipe of varying area of cross-section as shown in figure.
Assumptions: The density of the incompressible fluid remains constant at both points. The energy of the fluid is conserved (no loss of fluid) as there are no viscous forces in the fluid.
If \(p_1\) and \(p_2\) are the pressures at two ends of the tube respectively, the work done by pressure difference in pushing the volume \(\Delta V\) of fluid from the points \(1\) to \(2\) through the tube is given by:
\(
\begin{aligned}
& W=F_1 d x_1-F_2 d x_2 \\
& W=p_1 A_1 d x_1-p_2 A_2 d x_2 \\
& W=p_1 d V-p_2 d V=\left(p_1-p_2\right) \Delta V \dots(i)
\end{aligned}
\)
We know that the work done on the fluid was due to the conservation of change in gravitational potential energy and change in kinetic energy.
The change in potential energy of mass \(\Delta m\) (of volume \(\Delta V\) ),
\(
\Delta U=\Delta m g\left(h_2-h_1\right) \dots(ii)
\)
The change in kinetic energy of the fluid is given as:
\(
\Delta K=\frac{1}{2} \Delta m\left(v_2^2-v_1^2\right) \dots(iii)
\)
By conservation of energy, \(W=\Delta K+\Delta U\)
Putting the values from Eqs. (i), (ii) and (iii), we get
\(
\left(p_1-p_2\right) \Delta V=\frac{1}{2} \Delta m\left(v_2^2-v_1^2\right)+\Delta m g\left(h_2-h_1\right) \dots(iv)
\)
We know that \(\rho=\frac{\Delta m}{\Delta V}\)
Replacing \(\Delta m=\rho \times \Delta V\) in the equation (iv) we see\(\Delta V\) gets cancelled and finally, we get
\(
p_1+\frac{1}{2} \rho v_1^2+\rho g h_1=p_2+\frac{1}{2} \rho v_2^2+\rho g h_2
\)
\(
p+\frac{1}{2} \rho v^2+\rho g h=\text { constant } \dots(v)
\)
This is called Bernoulli’s equation.
On dividing both sides of Eq. (v) by \(\rho g\), we get
\(
\frac{p}{\rho g}+h+\frac{v^2}{2 g}=\frac{\text { constant }}{\rho g}=\text { new constant } \dots(vi)
\)
Here, \(p / \rho g\) is called pressure head, \(h\) is called gravitational head and \(v^2 / 2 g\) is called velocity head.
Note: Bernoulli’s equation for the fluid at rest When a fluid is at rest, i.e. the velocity is zero everywhere, then the Bernoulli’s equation becomes
\(
p_1+\rho g h_1=p_2+\rho g h_2 \Rightarrow p_1-p_2=\rho g\left(h_2-h_1\right)
\)
Example 16: Calculate the rate of flow of glycerine of density \(1.25 \times 10^3 \mathrm{kgm}^{-3}\) through the conical section of a horizontal pipe, if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length is \(10 \mathrm{Nm}^{-2}\).
Solution: According to the question, we draw the following diagram.
From the continuity equation,
\(
\begin{aligned}
& A_1 v_1=A_2 v_2 \text { or } \frac{v_1}{v_2}=\frac{A_2}{A_1}=\frac{\pi r_2^2}{\pi r_1^2} \\
& \frac{v_1}{v_2}=\left(\frac{r_2}{r_1}\right)^2=\left(\frac{0.04}{0.1}\right)^2=\frac{4}{25} \dots(i)
\end{aligned}
\)
\(
\text { From Bernoulli’s equation, } p_1+\frac{1}{2} \rho v_1^2=p_2+\frac{1}{2} \rho v_2^2\left(\text { As } h_1=h_2\right)
\)
\(
\begin{aligned}
v_2^2-v_1^2 & =\frac{2\left(p_1-p_2\right)}{\rho} \\
v_2^2-v_1^2 & =\frac{2 \times 10}{1.25 \times 10^3}=1.6 \times 10^{-2} \mathrm{~m}^2 \mathrm{~s}^{-2} \dots(ii)
\end{aligned}
\)
Solving Eqs. (i) and (ii), we get
\(
v_2=\sqrt{\frac{1.6 \times 10^{-2} \times 625}{(625-16)}} \approx 0.128 \mathrm{~ms}^{-1}
\)
\(\therefore\) Rate of volume flow through the tube,
\(
\begin{aligned}
Q & =A_2 v_2=\left(\pi r_2^2\right) v_2=\pi(0.04)^2(0.128) \\
& =6.43 \times 10^{-4} \mathrm{~m}^3 \mathrm{~s}^{-1}
\end{aligned}
\)
Applications based on Bernoulli’s theorem
Speed of Efflux: Torricelli’s Law
The word efflux means fluid outflow. Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body. Consider a tank containing a liquid of density \(\rho\) with a small hole in its side at a height \(y_l\) from the bottom (see Figure below). The air above the liquid, whose surface is at height \(y_2\), is at pressure \(P\). From the equation of continuity, we have
\(
\begin{aligned}
& v_1 A_1=v_2 A_2 \\
& v_2=\frac{A_1}{A_2} v_1
\end{aligned}
\)
If the cross-sectional area of the tank \(A_2\) is much larger than that of the hole \(\left(A_2 \gg A_1\right)\), then we may take the fluid to be approximately at rest at the top, i.e., \(v_2=0\). Now, applying the Bernoulli equation at points 1 and 2 and noting that at the hole \(P_1=P_a\), the atmospheric pressure, we have from Eqn (\(P_1+\frac{1}{2} \rho V_1^2+\rho g h_1=P_2+\frac{1}{2} \rho V_2^2+\rho g h_2\))
\(
P_a+\frac{1}{2} \rho v_1^2+\rho g y_1=P+\rho g y_2
\)
Taking \(y_2-y_1=h\) we have
\(
v_1=\sqrt{2 g h+\frac{2\left(P-P_a\right)}{\rho}}
\)
When \(P \gg P_a\) and \(2 g h\) may be ignored, the speed of efflux is determined by the container pressure. Such a situation occurs in rocket propulsion. On the other hand, if the tank is open to the atmosphere, then \(P=P_a\) and
\(
v_1=\sqrt{2 g h} \dots(1)
\)
This is also the speed of a freely falling body. Equation (1) represents Torricelli’s law.
From the above formula, it is clear that “The velocity of efflux of a liquid issuing out of an orifice is the same as it would attain, if allowed to fall freely through the vertical height between the liquid surface and orifice.” The above statement is also known as Torricelli’s theorem.
Horizontal range of liquid (R)
The outflow of a fluid is called efflux and the speed of the liquid coming out is called speed of efflux.
Consider a closed vessel filled with a liquid upto height \(H\) and a small hole is made in the wall of the vessel at a depth \(h\) below the surface of liquid.
Horizontal range: The escaping liquid flows in the form of a parabola. Let \(t\) be the time taken by the liquid to fall through a height \(h^{\prime}=(H-h)\).
Now, \({h}^{\prime}=0 \times {t}+\frac{1}{2} {gt}^2, \quad\) Initial velocity in the vertically downward direction is zero.
\(
{h}^{\prime}=\frac{1}{2} {gt}^2
\)
\(
{t}=\sqrt{\frac{2 {~h}^{\prime}}{{g}}}
\)
Let \(R\) be the horizontal range. In order to calculate \(R\), we shall consider the horizontal motion of the projectile. The horizontal motion takes place with constant velocity \(v\).
\(
\therefore {R}=v \times {t} \quad[\text { Distance }=\text { Speed } \times \text { Time }]
\)
\(
\text { or } {R}=\sqrt{2 g h} \times \sqrt{\frac{2 h^{\prime}}{g}}=2 \sqrt{{hh}^{\prime}}=2 \sqrt{h(H-h)}
\)
\(
R^2=4\left(H h-h^2\right)
\)
For \(R\) to be maximum, \(\frac{d R^2}{d h}=0\)
or \(H-2 h=0\) or \(h=\frac{H}{2}\)
i.e. \(R\) is maximum at \(h=\frac{H}{2}\)
\(
\begin{aligned}
& R_{\max }=2 \sqrt{\frac{H}{2}\left(H-\frac{H}{2}\right)} \\
& R_{\max }=H
\end{aligned}
\)
i.e. The maximum horizontal distance covered by liquid coming out of a hole is equal to the height of the liquid column.
Time taken to empty a tank is given by the formula
Let A: Area of the tank’s cross-section.
\(a\) : Area of the orifice.
\(\boldsymbol{H}\) : Initial height of the liquid in the tank.
\(g:\) Acceleration due to gravity.
Torricelli’s Law: \(v=\sqrt{2 g h}\), where \(v\) is the velocity of efflux and \(h\) is the height of the fluid.
Volume flow rate: \(Q=A v\), where \(Q\) is the flow rate, \(A\) is the area, and \(v\) is the velocity.
The volume flow rate out of the orifice is given by:
\(Q_{\text {out }}=a v=a \sqrt{2 g h}\)
The rate of change of volume in the tank is:
\(Q_{\text {in }}=A \frac{d h}{d t}\)
Equating the outflow and the change in volume:
\(A \frac{d h}{d t}=-a \sqrt{2 g h}\)
The negative sign indicates that the height is decreasing.
Integrate both sides:
\(
\int_H^0 \frac{d h}{\sqrt{h}}=-\frac{a}{A} \sqrt{2 g} \int_0^t d t
\)
Evaluate the integrals:
\([2 \sqrt{h}]_H^0=-\frac{a}{A} \sqrt{2 g}[t]_0^t\)
\(0-2 \sqrt{H}=-\frac{a}{A} \sqrt{2 g} t\)
Rearrange the equation to solve for \(t\) :
\(
t=\frac{2 \sqrt{H}}{\frac{a}{A} \sqrt{2 g}}
\)
\(
t=\frac{A}{a} \sqrt{\frac{2 H}{g}}
\)
The time taken to empty the tank is \(t=\frac{A}{a} \sqrt{\frac{2 H}{g}}\)
Example 17: If the water emerge from an orifice in a tank in which the gauge pressure is \(4 \times 10^5 \mathrm{Nm}^{-2}\) before the flow starts, then what will be the velocity of the water emerging out? (Take, density of water is \(1000 \mathrm{kgm}^{-3}\) )
Solution: Here, \(p=4 \times 10^5 \mathrm{Nm}^{-2}\) and \(\rho=1000 \mathrm{~kg} \mathrm{~m}^{-3}, g=10 \mathrm{~m} \mathrm{~s}^{-2}\)
Apply, \(p=h \rho g \Rightarrow h=\frac{p}{\rho g}=\frac{4 \times 10^5}{1000 \times 10}\)
Velocity of efflux, \(v=\sqrt{2 g h}=\sqrt{\frac{2 \times 10 \times 4 \times 10^5}{1000 \times 10}}\)
\(=\sqrt{800}=28.28 \mathrm{~ms}^{-1}\)
Dynamic Lift
Dynamic lift is the force that acts on a body, such as airplane wing, a hydrofoil or a spinning ball, by virtue of its motion through a fluid. In many games such as cricket, tennis, baseball, or golf, we notice that a spinning ball deviates from its parabolic trajectory as it moves through air. This deviation can be partly explained on the basis of Bernoulli’s principle.
- Ball moving without spin: Figure (a) shows the streamlines around a non-spinning ball moving relative to a fluid. From the symmetry of streamlines, it is clear that the velocity of fluid (air) above and below the ball at corresponding points is the same resulting in zero pressure difference. The air, therefore, exerts no upward or downward force on the ball.
- Ball moving with spin: A ball which is spinning drags air along with it. If the surface is rough more air will be dragged. Figure(b) shows the streamlines of air for a ball which is moving and spinning at the same time. The ball is moving forward and relative to it the air is moving backwards. Therefore, the velocity of air above the ball relative to the ball is larger and below it is smaller. The stream lines, thus, get crowded above and rarified below. This difference in the velocities of air results in the pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spining is called Magnus effect.
- Aerofoil or lift on aircraft wing: Figure (c) shows an aerofoil, which is a solid piece shaped to provide an upward dynamic lift when it moves horizontally through air. The cross-section of the wings of an aeroplane looks somewhat like the aerofoil shown in Figure (c) with streamlines around it. When the aerofoil moves against the wind, the orientation of the wing relative to flow direction causes the streamlines to crowd together above the wing more than those below it. The flow speed on top is higher than that below it. There is an upward force resulting in a dynamic lift of the wings and this balances the weight of the plane. The following example illustrates this.
Example 18: A fully loaded Boeing aircraft has a mass of \(3.3 \times 10^5 \mathrm{~kg}\). Its total wing area is \(500 \mathrm{~m}^2\). It is in level flight with a speed of \(960 \mathrm{~km} / \mathrm{h}\). (a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is \(\rho\) \(\left.=1.2 \mathrm{~kg} \mathrm{~m}^{-3}\right]\)
Solution: (a) The weight of the Boeing aircraft is balanced by the upward force due to the pressure difference
\(
\begin{aligned}
& \Delta P \quad A=3.3 \times 10^5 \mathrm{~kg} \times 9.8 \\
& \begin{aligned}
\Delta P & =\left(3.3 \times 10^5 \mathrm{~kg} \times 9.8 \mathrm{~m} \mathrm{~s}^{-2}\right) / 500 \mathrm{~m}^2 \\
& =6.5 \times 10^3 \mathrm{Nm}^{-2}
\end{aligned}
\end{aligned}
\)
(b) We ignore the small height difference between the top and bottom sides in Bernoulli Eqn. The pressure difference between them is then
\(
\Delta P=\frac{\rho}{2}\left(v_2^2-v_1^2\right)
\)
where \(V_2\) is the speed of air over the upper surface and \(V_1\) is the speed under the bottom surface.
\(
\left(v_2-v_1\right)=\frac{2 \Delta P}{\rho\left(v_2+v_1\right)}
\)
Taking the average speed
\(
V_{\mathrm{av}}=\left(v_2+v_1\right) / 2=960 \mathrm{~km} / \mathrm{h}=267 \mathrm{~m} \mathrm{~s}^{-1}
\)
we have
\(
\left(v_2-v_1\right) / v_{\mathrm{av}}=\frac{\Delta P}{\rho v_{\mathrm{av}}^2} \approx 0.08
\)
The speed above the wing needs to be only 8 % higher than that below.
Viscocity
Most of the fluids are not ideal ones and offer some resistance to motion. This resistance to fluid motion is like an internal friction analogous to friction when a solid moves on a surface. It is called viscosity.
A real fluid flowing in a pipe experiences frictional forces. There is friction with the walls of the pipe, and there is friction within the fluid itself, converting some of its kinetic energy into thermal energy. The frictional forces that try to prevent different layers of fluid from sliding past each other are called viscous forces. Viscosity is a measure of a fluids resistance to relative motion within the fluid. We can measure the viscosity of a fluid by measuring the viscous drag between two plates.
Lets us assume layer-1 is located at a distance \(x\) from fixed surface and its velocity is \(v\). Layer-2 is located at a distance of \(x+dx\) and its velocity is \( v+dv\). According to Newton, the frictional force \(F\) (or viscous force) between two layers depends upon the following factors:
(i) Force \(F\) is directly proportional to the area \((A)\) of the layers in contact, i.e. \(F \propto A\).
(ii) Force \(F\) is directly proportional to the velocity gradient \(\left(\frac{d v}{d x}\right)\) between the layers.
Combining these two, we have
\(
F \propto A \frac{d v}{d x}
\)
\(
F=-\eta A \frac{d v}{d x}
\)
Here, \(\eta\) is a constant of proportionality and is called coefficient of viscosity. Its value depends on the nature of the fluid.
The negative sign in the above equation shows that the direction of viscous force \(F\) is opposite to the direction of the relative velocity of the layer.
The SI unit of \(\eta\) is \(\mathrm{Nsm}^{-2}\). It is also called decapoise or pascal second.
Thus, \(\quad 1\) decapoise \(=1 \mathrm{Nsm}^{-2}=1 \mathrm{~Pa}-\mathrm{s}=10\) poise
Dimensions of \(\eta\) are \(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]\).
Coefficient of viscosity of water at \(10^{\circ} \mathrm{C}\) is \(\eta=1.3 \times 10^{-3} \mathrm{Nsm}^{-2}\). Experiments show that the coefficient of viscosity of a liquid decreases as its temperature rises.
Liquids have different viscosities depending on their properties (shown in the figure below). If you have a range of liquids at the same temperature some like water will have a low viscosity of around 1, while others like honey will have a higher viscosity of over 2000.
Example 19: A plate of area \(2 m^2\) is made to move horizontally with a speed of \(2 \mathrm{~ms}^{-1}\) by applying a horizontal tangential force over the free surface of a liquid. The depth of the liquid is 1 m and the liquid in contact with the bed is stationary. Coefficient of viscosity of liquid is 0.01 poise. Find the tangential force needed to move the plate.
Solution:
Velocity gradient \(=\frac{2-0}{1-0}=2 \mathrm{~s}^{-1}\)
From Newton’s law of viscous force,
\(
\begin{aligned}
|F| & =\eta A \frac{\Delta v}{\Delta y}=\left(0.01 \times 10^{-1}\right) \\
& =4 \times 10^{-3} \mathrm{~N}
\end{aligned}
\)
So, to keep the plate moving a force of \(4 \times 10^{-3} \mathrm{~N}\) must be applied.
Alternate Way:
This force exists when there is relative motion between layers of the liquid. Suppose we consider a fluid like oil enclosed between two glass plates as shown in Figure below. The bottom plate is fixed while the top plate is moved with a constant velocity \({v}\) relative to the fixed plate. If oil is replaced by honey, a greater force is required to move the plate with the same velocity. Hence we say that honey is more viscous than oil. The fluid in contact with a surface has the same velocity as that of the surfaces. Hence, the layer of the liquid in contact with top surface moves with a velocity \({v}\) and the layer of the liquid in contact with the fixed surface is stationary. The velocities of layers increase uniformly from bottom (zero velocity) to the top layer (velocity \(v\)). For any layer of liquid, its upper layer pulls it forward while lower layer pulls it backward. This results in force between the layers. This type of flow is known as laminar. The layers of liquid slide over one another as the pages of a book do when it is placed flat on a table and a horizontal force is applied to the top cover. When a fluid is flowing in a pipe or a tube, then velocity of the liquid layer along the axis of the tube is maximum and decreases gradually as we move towards the walls where it becomes zero.
On account of this motion (as shown with arrow in diagram below), a portion of liquid, which at some instant has the shape \(A B C D\), take the shape of AEFD after short interval of time ( \(\Delta t\) ). During this time interval the liquid has undergone a shear strain of \(\Delta x / l\). Since, the strain in a flowing fluid increases with time continuously. Unlike a solid, here the stress is found experimentally to depend on ‘rate of change of strain’ or ‘strain rate’ i.e. \(\Delta x /(l \Delta t)\) or \(v / l\) instead of strain itself.
The coefficient of viscosity (pronounced ‘eta’) for a fluid is defined as the ratio of shearing stress to the strain rate.
\(
\eta=\frac{\text { shearing stress }}{ \text { strain rate }}=\frac{F / A}{v / l}=\frac{F l}{v A}
\)
Note: Strain rate is a measure of how quickly a material deforms or changes shape over time. Shear strain is the deformation of an object that occurs when a force acts parallel to its surface.
Strain rate = (Change in strain) / (Change in time)=\(\Delta x / l / (\Delta t)=\Delta x /(l \Delta t)=v/l\)
Strain rate is represented by the formula “\(v/l\)“, where ” \(v\) ” is the velocity and “\(l\)” is the length, meaning it calculates the rate of deformation by dividing the change in length by the original length per unit time; essentially, how fast a material is changing shape relative to its original size.
Example 20: A metal block of area \(0.10 \mathrm{~m}^2\) is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless) as shown below. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released the block moves to the right with a constant speed of \(0.085 \mathrm{~m} \mathrm{~s}^{-1}\). Find the coefficient of viscosity of the liquid.
Solution: Answer The metal block moves to the right because of the tension in the string. The tension \(T\) is equal in magnitude to the weight of the suspended mass \(m\). Thus, the shear force \(F\) is
\(
F=T=m g=0.010 \mathrm{~kg} \times 9.8 \mathrm{~m} \mathrm{~s}^{-2}=9.8 \times 10^{-2} \mathrm{~N}
\)
Shear stress on the fluid \(=F / A=\frac{9.8 \times 10^{-2}}{0.10} \mathrm{~N} / \mathrm{m}^2\)
\(
\begin{aligned}
& \text { Strain rate }=\frac{v}{l}=\frac{0.085}{0.30 \times 10^{-3}} \\
& h=\frac{\text { stress }}{\text { strain rate }} \mathrm{s}^{-1} \\
& =\frac{\left(9.8 \times 10^{-2} \mathrm{~N}\right)\left(0.30 \times 10^{-3} \mathrm{~m}\right)}{\left(0.085 \mathrm{~m} \mathrm{~s}^{-1}\right)\left(0.10 \mathrm{~m}^2\right)} \\
& =3.46 \times 10^{-3} \mathrm{~Pa} \mathrm{~s}
\end{aligned}
\)
Stokes’ Law
When an object moves through a fluid, it experiences a viscous force which acts in opposite direction of its velocity. It is seen that the viscous force is proportional to the velocity of the object and is opposite to the direction of motion. The other quantities on which the force \(F\) depends are viscosity \(\eta\) of the fluid and radius \(r\) of the sphere. Sir George G. Stokes (1819-1903), an English scientist enunciated clearly the viscous drag force \(F\) as
\(F=6 \pi \eta r v\) (where, \(\eta=\) coefficient of viscosity)
This law is called Stokes’ law.
Terminal velocity \(\left(v_t\right)\)
This law is an interesting example of retarding force, which is proportional to velocity. We can study its consequences on an object falling through a viscous medium. We consider a raindrop in air. It accelerates initially due to gravity. As the velocity increases, the retarding force also increases. Finally, when viscous force plus buoyant force becomes equal to the force due to gravity, the net force becomes zero and so does the acceleration. The sphere (raindrop) then descends with a constant velocity. Thus, in equilibrium, this velocity is known as terminal velocity \(v_{\mathrm{t}}\).
Consider a small sphere falling from rest through a large column of viscous fluid.
The forces acting on the sphere are
(i) weight \(w\) of the sphere acting vertically downwards.
(ii) upthrust buoyant force \(F_t\) acting vertically upwards.
(iii) viscous force \(F_v\) acting vertically upwards, i.e. in a direction opposite to velocity of the sphere.
Initially,
\(
\begin{aligned}
F_v & =0 \quad (\because v=0) \\
w & >F_t
\end{aligned}
\)
and the sphere accelerates downwards. As the velocity of the sphere increases, \(F_v\) increases. Eventually, a stage is reached when
\(
w=F_t+F_v \dots(i)
\)
After this net force on the sphere is zero and it moves downwards with a constant velocity called terminal velocity \(\left(v_t\right)\).
It can be defined as the maximum constant velocity acquired by a body while falling through a viscous medium.
Substituting proper values in Eq. (i), we get
\(
\frac{4}{3} \pi r^3 \rho g=\frac{4}{3} \pi r^3 \sigma g+6 \pi \eta r v_t \dots(ii)
\)
Here, \(\quad \rho=\) density of sphere, \(\sigma=\) density of fluid and \(\quad \eta=\) coefficient of viscosity of fluid.
From Eq. (ii), we get
\(
v_t=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}
\)
The figure shows the variation of the velocity \(v\) of the sphere with time
Note: From the above expression, we can see that the terminal velocity of a spherical body is directly proportional to the difference in the densities of the body and the fluid \((\rho-\sigma)\). If the density of the fluid is greater than that of body (i.e. \(\sigma>\rho\) ), the terminal velocity is negative. This means that the body instead of falling, moves upward. This is why air bubbles rise up in water.
Example 21: With what terminal velocity will an air bubble 0.8 mm in diameter rise in a liquid of viscosity \(0.15 \mathrm{Nsm}^{-2}\) and specific gravity \(0.9 \mathrm{kgm}^{-3}\)? (Take, the density of air is \(1.293 \mathrm{~kg} \mathrm{~m}^{-3}\) )
Solution: The terminal velocity of the bubble is given by
\(
v_t=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}
\)
Here,
\(
\begin{aligned}
& r=0.4 \times 10^{-3} \mathrm{~m}, \sigma=0.9 \times 10^3 \mathrm{kgm}^{-3} \\
& \rho=1.293 \mathrm{kgm}^{-3}, \eta=0.15 \mathrm{Nsm}^{-2}
\end{aligned}
\)
and \(\quad g=9.8 \mathrm{~ms}^{-2}\)
Substituting the values, we get
\(
\begin{aligned}
v_t & =\frac{2}{9} \times \frac{\left(0.4 \times 10^{-3}\right)^2\left(1.293-0.9 \times 10^3\right) \times 9.8}{0.15} \\
& =-0.0021 \mathrm{~ms}^{-1} \text { or } v_t=-0.21 \mathrm{cms}^{-1}
\end{aligned}
\)
Note Here, a negative sign implies that the bubble will rise up.
Example 22: Two spherical raindrops of equal size are falling vertically through air with a terminal velocity of \(1 \mathrm{~ms}^{-1}\). What would be the terminal speed, if these two drops were to coalesce to form a large spherical drop?
Solution: As,
\(
v_t \propto r^2
\)
Let \(r\) be the radius of small raindrops and \(R\) the radius of large drop.
Equating the volumes, we get
\(
\begin{aligned}
& & \frac{4}{3} \pi R^3 & =2\left(\frac{4}{3} \pi r^3\right) \\
& \therefore & R & =(2)^{1 / 3} \cdot r \text { or } \frac{R}{r}=(2)^{1 / 3} \\
& \because & \frac{v_t^{\prime}}{v_t} & =\left(\frac{R}{r}\right)^2=(2)^{2 / 3} \\
& \therefore & v_t^{\prime} & =(2)^{2 / 3}(1.0) \mathrm{ms}^{-1}=1.587 \mathrm{~ms}^{-1}
\end{aligned}
\)
Surface Tension
Surface tension is measured as the energy required to increase the surface area of a liquid by a unit of area. The surface tension of a liquid results from an imbalance of intermolecular attractive forces, the cohesive forces between molecules:
- A molecule in the bulk liquid experiences cohesive forces with other molecules in all directions.
- A molecule at the surface of a liquid experiences only net inward cohesive forces.
A microscopic view of water illustrates the difference between molecules at the surface of a liquid and water molecules within a liquid. A small drop of liquid is spherical in shape, because for given volume, sphere has minimum surface area. This stretched behaviour is due to net downward force acting on a molecule on the surface of liquid as shown in the figure below. The cohesive forces between molecules down into a liquid are shared with all neighboring atoms. Those on the surface have no neighboring atoms above and exhibit stronger attractive forces upon their nearest neighbors on the surface. This enhancement of the intermolecular attractive forces at the surface is called surface tension.
For molecules at the surface, the cohesive forces are unbalanced as there are no molecules above them. This means molecules at the surface of a liquid are pulled towards their neighbours below the surface. This creates a “skin” at the surface known as surface tension.
A molecule well inside the liquid experiences no net force (the one that is submerged).
Now, consider a line \(A B\) on the free surface of the liquid. The small elements of the surface on this line are in equilibrium because they are acted upon by equal and opposite forces, acting perpendicular to the line from either side as shown in the figure below.
The force acting on this line is proportional to the length of this line. If \(l\) is the length of imaginary line and \(F\) the total force on either side of the line,
\(F \propto l\)
\(
S=T=\frac{F}{l}
\)
where:
\(S\) is the surface tension,
\(F\) is the force acting perpendicular to the line, and \(l\) is the length of the line.
Surface tension, \(S=\frac{\text { Force }}{\text { Length }}\)
From this expression, surface tension can be defined as the force acting per unit length of an imaginary line drawn on the liquid surface, the direction of force being perpendicular to this line and tangential to the liquid surface.
It is denoted by \(S\) and it is a scalar quantity.
Units and dimensions of surface tension
SI unit of surface tension \(=\mathrm{Nm}^{-1}\), CGS unit of surface tension \(=\) dyne \(\mathrm{cm}^{-1}\),
Dimensions of surface tension
\(
=\frac{\text { Force }}{\text { Length }}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]}=\left[\mathrm{ML}^0 \mathrm{~T}^{-2}\right]
\)
Examples of Surface tension Shown below:
Water strider, which are small insects, can walk on water as their weight is considerably less to penetrate the water surface. Like this, there are various examples of surface tension which are found in nature. Some cases are provided below:
Example 23: A liquid is kept in a beaker of radius 4 cm . Consider a diameter of the beaker on the surface of the water. Find the force by which the surface on one side of the diameter pulls the surface on the other side. (Take, surface tension of liquid \(=0.075 \mathrm{Nm}^{-1}\) )
Solution: The length of the diameter, \(l=2 r=8 \mathrm{~cm}=0.08 \mathrm{~m}\)
The surface tension is \(S=F / l\).
Thus, \(F=S l=\left(0.075 \mathrm{Nm}^{-1}\right) \times(0.08 \mathrm{~m})=6 \times 10^{-3} \mathrm{~N}\)
Surface energy
The free surface of a liquid always has a tendency to contract and possess minimum surface area. To increase the surface area of the liquid work has to be done. This work done is stored in the surface film of the liquid as its potential energy. This potential energy per unit area of the surface film is called the surface energy.
Hence, the surface energy may be defined as the amount of work done in increasing the area of the surface film through unity. Thus,
\(
\text { Surface energy }=\frac{\text { Work done in increasing the surface area }}{\text { Increase in surface area }}=\frac{W}{\Delta A}
\)
The SI unit of surface energy is \(\mathrm{Jm}^{-2}\).
Relation between surface energy and surface tension
Consider a rectangular frame \(P Q R S\). Here, wire \(Q R\) is movable. A soap film is formed on the frame. The film pulls the movable wire \(Q R\) inward due to surface tension.
As, \(\quad\) surface tension \(=\frac{\text { force }}{\text { length }}=\frac{F^{\prime}}{2 l} \Rightarrow F^{\prime}=S \times 2 l\) (A film has two sides and the liquid in between so length is \(2l\))
Suppose that we move the bar by a small distance \(d\) as shown. Since the area of the surface increases, the system now has more energy, this means that some work has been done against an internal force.
If \(Q R\) is moved through a distance \(d\) by an external force \(F\) very slowly, then some work has to be done against this force.
\(\therefore\) External work done \(=\) Force \(\times\) Distance
\(=S \times 2 l \times d \quad\left(\because F^{\prime}=F\right)\)
Increase in surface area of film \(=2 l \times d\) (As soap film has two sides)
Surface energy \(=\frac{\text { Work done }}{\text { Surface area }}=\frac{S 2 l d}{2 l d}=S\)
So, value of surface energy of liquid is numerically equal to the value of surface tension.
Note:
(i) Work done in forming a drop, \(W=S \times 4 \pi r^2\)
\(
\left(A s, \Delta A=4 \pi r^2-0=4 \pi r^2\right)
\)
(ii) Work done in forming a bubble, \(W=S \times 4 \pi r^2 \times 2\) (As, bubble in air has two surfaces)
Example 24: How much work will be done in increasing the diameter of a soap bubble from 2 cm to 5 cm . (Take, surface tension of soap solution is \(3.0 \times 10^{-2} \mathrm{Nm}^{-2}\) )
Solution: The work done in increasing the diameter of a soap bubble from 2 cm to 5 cm .
What’s given in the problem?
Initial diameter of the soap bubble: \(\boldsymbol{d}_{\mathbf{1}}=2 \mathrm{~cm}=0.02 \mathrm{~m}\)
Final diameter of the soap bubble: \(d_2=5 \mathrm{~cm}=0.05 \mathrm{~m}\)
Surface tension of soap solution: \(S=3.0 \times 10^{-2} \mathrm{Nm}^{-1}\)
The surface area of a sphere is given by \(A=4 \pi r^2\), where \(r\) is the radius.
A soap bubble has two surfaces, so the total surface area is \(2 A=8 \pi r^2\).
The work done in changing the surface area of a soap bubble is given by \(W=S \Delta A\), where \(\Delta A\) is the change in surface area.
Calculate the initial surface area \(A_1\)
\(
r_1=\frac{d_1}{2}
\)
\(
A_1=4 \pi r_1^2
\)
Calculate the final surface area \(\boldsymbol{A}_2\)
\(
r_2=\frac{d_2}{2}
\)
\(
A_2=4 \pi r_2^2
\)
\(
\Delta A=2(A_2-A_1) \text { Soap bubble has two surfaces. }
\)
\(
\begin{aligned}
\Delta A & =2\left[4 \pi\left\{\left(2.5 \times 10^{-2}\right)^2-\left(1.0 \times 10^{-2}\right)^2\right\}\right] \\
& =1.32 \times 10^{-2} \mathrm{~m}^2
\end{aligned}
\)
\(\begin{aligned} \therefore \text { Work done, } W & =\left(3.0 \times 10^{-2}\right)\left(1.32 \times 10^{-2}\right) \mathrm{J} \\ & =3.96 \times 10^{-4} \mathrm{~J}\end{aligned}\)
Example 25: Calculate the energy released when 1000 small water drops each of radius \(10^{-7} \mathrm{~m}\) coalesce to form one large drop. (Take, surface tension of water is \(7.0 \times 10^{-2} \mathrm{Nm}^{-1}\) )
Solution: Let \(r\) be the radius of smaller drops and \(R\) of bigger one. Equating the initial and final volumes, we get
\(
\frac{4}{3} \pi R^3=(1000)\left(\frac{4}{3} \pi r^3\right)
\)
\(
R=10 r=(10)\left(10^{-7}\right) \mathrm{m} \text { or } R=10^{-6} \mathrm{~m}
\)
Further, the water drops have only one free surface.
Therefore,
\(
\begin{aligned}
\Delta A & =4 \pi R^2-(1000)\left(4 \pi r^2\right) \\
& =4 \pi\left[\left(10^{-6}\right)^2-\left(10^3\right)\left(10^{-7}\right)^2\right] \\
& =-36 \pi\left(10^{-12}\right) \mathrm{m}^2
\end{aligned}
\)
Here, negative sign implies that surface area is decreasing. Hence, energy released in the process,
\(
\begin{aligned}
U & =T|\Delta A|=\left(7 \times 10^{-2}\right)\left(36 \pi \times 10^{-12}\right) \mathrm{J} \\
& =7.9 \times 10^{-12} \mathrm{~J}
\end{aligned}
\)
Measuring Surface Tension
A fluid will stick to a solid surface if the surface energy between fluid and the solid is smaller than the sum of surface energies between solid-air, and fluid-air. Now there is attraction between the solid surface and the liquid. It can be directly measured experimentaly as schematically shown in Figure below. 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, with its horizontal edge just over water. The vessel is raised slightly 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.
Suppose the additional weight required is \(W\). Then the surface tension of the liquid-air interface is
\(
S_{l a}=(W / 2 l)=(m g / 2 l)
\)
where \(m\) is the extra mass and \(l\) is the length of the plate edge. The subscript (\(la\)) emphasises the fact that the liquid-air interface tension is involved.
Angle of Contact
When the free surface of a liquid comes in contact with a solid, then the surface of the liquid becomes curved at the point of contact. Whenever the liquid surface becomes a curve, then the angle between the two medium (solid-liquid interface) comes in the picture. Surfaces that repel water, like the leaves of the Colocasia plant (as shown below) , are called hydrophobic. “Hydro-” is a Greek root word that means water, “Phobia” means fear.
When a water drop lands on the Colocasia leaf, it beads up into a sphere and simply rolls away. Why doesn’t the water make the surface wet? Because this leaf is superhydrophobic !
The angle between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid is known as the angle of contact between the solid and the liquid. It is denoted by \(\theta\) (Read it as “theta” which is Greek alphabet small letter).
Its value is different at interfaces of different pairs of solids and liquids. In fact, it is the factor which decides whether a liquid will spread on the surface of a chosen solid or it will form droplets on it.
Let us consider three interfaces such as liquid-air, solid-air and solid-liquid with reference to the point of contact ‘ O ‘ and the interfacial surface tension forces \(\mathrm{T}_{\mathrm{sa}}, \mathrm{T}_{\mathrm{sl}}\) and \(\mathrm{T}_{\mathrm{la}}\) on the respective interfaces as shown in Figure below.
Since the liquid is stable under equilibrium, the surface tension forces between the three interfaces must also be in equilibrium. Therefore,
\(
T_{s a}=T_{l a} \cos \theta+T_{s l} \Rightarrow \cos \theta=\frac{T_{s a}-T_{s l}}{T_{l a}}
\)
From the above equation, there are three different possibilities which can be discussed as follows.
(i) If \(\mathrm{T}_{\mathrm{sa}}>\mathrm{T}_{\mathrm{sl}}\) and \(\mathrm{T}_{\mathrm{sa}}-\mathrm{T}_{\mathrm{sl}}>0\) (water-plastic interface) then the angle of contact \(\theta\) is acute angle ( \(\theta\) less than \(90^{\circ}\) ) as \(\cos \theta\) is positive.
(ii) If \(\mathrm{T}_{\mathrm{sa}}<\mathrm{T}_{\mathrm{sl}}\) and \(\mathrm{T}_{\mathrm{sa}}-\mathrm{T}_{\mathrm{ls}}<0\) (water-leaf interface) then the angle of contact is obtuse angle ( \(\theta\) less than \(180^{\circ}\) ) and as \(\cos \theta\) is negative.
(iii) If \(T_{\mathrm{sa}}>\mathrm{T}_{\mathrm{la}}+\mathrm{T}_{\mathrm{sl}}\) then there will be no equilibrium and liquid will spread over the solid.
Therefore, the concept of angle of contact between the solid-liquid interface leads to some practical applications in real life. For example, soaps and detergents are wetting agents. When they are added to an aqueous solution, they will try to minimize the angle of contact and in turn penetrate well in the cloths and remove the dirt. On the other hand, water proofing paints are coated on the outer side of the building so that it will enhance the angle of contact between the water and the painted surface during the rainfall.
For example, water forms droplets on lotus leaf as shown in Figure below(a) while spreads over a clean plastic plate as shown in Figure below (b).
We consider the three interfacial tensions at all the three interfaces, liquid-air, solid-air and solid-liquid denoted by \(S_{\mathrm{la}}, S_{\mathrm{sa}}\) and \(S_{\mathrm{sl}}\), respectively as given in Fig. (a) and (b). At the line of contact, the surface forces between the three media must be in equilibrium. From the Fig. (b) the following relation is easily derived.
\(
S_{\mathrm{la}} \cos \theta+S_{\mathrm{sl}}=S_{\mathrm{sa}}
\)
The following cases arise:
Case-I: If the surface tension at the solid-liquid \(S_{\mathrm{sl}}\), interface is greater than the surface tension at the liquid-air \(S_{\mathrm{la}}\) interface, i.e. \(S_{\mathrm{sl}}>S_{\mathrm{la}}\), then \(\theta>90^{\circ}\) (the angle of contact is obtuse angle as shown in Fig. (a)).
The molecules of a liquid are attracted strongly to themselves and weakly to those of solid. It costs a lot of energy to create a liquid-solid surface. The liquid then does not wet the solid.
e.g. Water-leaf or glass-mercury interface.
Case-II: If the surface tension at the solid-liquid \(S_{\mathrm{sl}}\) interface is less than the surface tension at the liquid-air \(S_{\text {la }}\) interface, i.e. \(S_{\mathrm{sl}}<S_{\mathrm{la}}\), then \(0<90^{\circ}\) (the angle of contact is acute angle as shown in Fig. (b)).
The molecules of the liquid are strongly attracted to those of solid and weakly attracted to themselves. It costs less energy to create a liquid-solid surface and liquid wets the solid.
e.g. When soap or detergent is added to water, the angle of contact becomes small.
Excess pressure (inside a bubble or drop)
Due to surface tension, a drop or bubble tends to contract and so compresses the matter enclosed. This in turn increases the internal pressure which prevents further contraction and equilibrium is achieved. So, in equilibrium, the pressure inside a bubble or drop is greater than outside and the difference of pressure between two sides of the liquid surface is called excess pressure.
Following ways are used to calculate the excess pressure inside a bubble or drop.
A soap bubble consists of two spherical surface films with a thin layer of liquid between them. Because of surface tension, the film tend to contract in an attempt to minimize their surface area. But as the bubble contracts, it compresses the inside air, eventually increasing the interior pressure to a level that prevents further contraction.
Another interesting consequence of surface tension is that the pressure inside a spherical drop Figure above (a) is more than the pressure outside. Suppose a spherical drop of radius \(r\) is in equilibrium. If its radius increase by \(\Delta r\). The extra surface energy is
\(
\left[4 \pi(r+\Delta r)^2-4 \pi r^2\right] S_{\mathrm{la}}=8 \pi r \Delta r S_{\mathrm{la}}
\)
If the drop is in equilibrium this energy cost is balanced by the energy gain due to expansion under the pressure difference ( \(P_i-P_o\) ) between the inside of the bubble and the outside. The work done is
\(
W=\left(P_i-P_{{o}}\right) 4 \pi r^2 \Delta r
\)
so that
\(
\left(P_{\mathrm{i}}-P_{\mathrm{o}}\right)=\left(2 S_{\mathrm{la}} / r\right)
\)
In general, for a liquid-gas interface, the convex side has a higher pressure than the concave side. For example, an air bubble in a liquid, would have higher pressure inside it. See Figure (b).
A bubble Figure (c) differs from a drop and a cavity; in this it has two interfaces. Applying the above argument we have for a bubble
\(
\Delta P=\left(P_1-P_{\mathrm{o}}\right)=\left(4 S_{\mathrm{la}} / r\right)=\frac{4 T}{r}
\)
This is probably why you have to blow hard, but not too hard, to form a soap bubble. A little extra air pressure is needed inside!
Example 26: 0 .04 cm liquid column balances the excess pressure inside a soap bubble of radius 6 mm . Evaluate density of the liquid. (Take, surface tension of soap solution \(=0.03 \mathrm{Nm}^{-1}\) )
Solution: The excess pressure inside a soap bubble,
\(
\begin{aligned}
\Delta p & =4 T / R \\
& =\frac{4 \times 0.03 \mathrm{Nm}^{-1}}{6 \times 10^{-3} \mathrm{~m}}=20 \mathrm{Nm}^{-2}
\end{aligned}
\)
The pressure due to 0.04 cm of the liquid column,
\(
\Delta p=h \rho g=\left(0.04 \times 10^{-2} \mathrm{~m}\right) \rho\left(10 \mathrm{~ms}^{-2}\right)
\)
Thus, \(20 \mathrm{Nm}^{-2}=\left(0.04 \times 10^{-2} \mathrm{~m}\right) \rho\left(10 \mathrm{~ms}^{-2}\right)\)
\(\therefore\) Density of the liquid, \(\rho=5 \times 10^3 \mathrm{kgm}^{-3}\)
Example 27: Two separate soap bubbles (radii 0.004 m and 0.002 m ) formed of the same liquid (surface tension \(0.07 \mathrm{Nm}^{-1}\) ) come together to form a double bubble. Find the radius and the sense of curvature of the internal film surface common to both the bubbles.
Solution: Excess pressure inside first soap bubble, \(p_1=p_0+\frac{4 T}{r_1}\)
Excess pressure inside second soap bubble, \(p_2=p_0+\frac{4 T}{r_2}\)
\(\because \quad r_2<r_1\), therefore \(p_2>p_1\)
i.e. Pressure inside the smaller bubble will be more. The excess pressure,
\(
p=p_2-p_1=4 T\left(\frac{r_1-r_2}{r_1 r_2}\right) \dots(i)
\)
This excess pressure acts from concave to convex side, the interface will be concave towards smaller bubble and convex towards larger bubble. Let \(R\) be the radius of interface, then
\(
p=\frac{4 T}{R} \dots(ii)
\)
From Eqs. (i) and (ii), we get
\(
R=\frac{r_1 r_2}{r_1-r_2}=\frac{(0.004)(0.002)}{(0.004-0.002)}=0.004 \mathrm{~m}
\)
Note: Excess pressure inside a liquid drop A liquid drop has only one surface film, so excess pressure inside the liquid drop is given by
\(
\Delta p=\frac{2 T}{R}
\)
Some Important points regarding excess pressure
Case-I: If we have an air bubble inside a liquid, a single surface is formed. There is air on the concave side and liquid on the convex side. The pressure in the concave side (i.e. in the air) is greater than the pressure in the convex side (i.e. in the liquid) by an amount \(2 T / R\).
The above expression has been written by assuming \(p_1\) to be constant from all sides of the bubble. For small size bubbles this can be assumed.
Case II: From the above discussion, we can make a general statement. The pressure on the concave side of a spherical liquid surface is greater than the convex side by \(2 T / R\) as shown below.
Case-III: Excess pressure inside a soap bubble can also be understood in terms of excess pressure inside a curved surface as shown below.
\begin{aligned}
&\begin{array}{r}
p_1-p^{\prime}=\frac{2 T}{R} \\
\Rightarrow \quad p^{\prime}-p_2=\frac{2 T}{R}
\end{array}\\
&\text { Excess pressure, } \Delta p=p_1-p_2=\frac{4 T}{R}
\end{aligned}
\)
Case-IV: If two bubbles of different sizes are connected with each other through a thin tube (as shown in figure), then the air will rush from smaller to larger bubble (pressure inside smaller bubble will be greater than pressure inside larger bubble), so that the smaller will shrink while the larger will expand till the smaller bubble reduces to a droplet.
Example 28: What should be the pressure inside a small air bubble of 0.1 mm radius situated just below the water surface? (Take, surface tension of water \(=7.2 \times 10^{-2} \mathrm{Nm}^{-1}\) and atmospheric pressure \(=1.013 \times 10^5 \mathrm{Nm}^{-2}\) )
Solution: Surface tension of water,
\(
T=7.2 \times 10^{-2} \mathrm{Nm}^{-1}
\)
Radius of air bubble,
\(
\begin{aligned}
R & =0.1 \mathrm{~mm} \\
& =10^{-4} \mathrm{~m}
\end{aligned}
\)
The excess pressure inside the air bubble is given by
\(
p_2-p_1=\frac{2 T}{R}
\)
\(\therefore\) Pressure inside the air bubble, \(p_2=p_1+\frac{2 T}{R}\)
Substituting the values, we get
\(
\begin{aligned}
p_2 & =\left(1.013 \times 10^5\right)+\frac{\left(2 \times 7.2 \times 10^{-2}\right)}{10^{-4}} \\
& =1.027 \times 10^5 \mathrm{Nm}^{-2}
\end{aligned}
\)
Shape of liquid surface
The curved surface of the liquid is called meniscus. The shape of the meniscus (convex or concave) is determined by the relative strengths of cohesive and adhesive forces. The force between the molecules of the same material is known as cohesive force and the force between the molecules of different kinds of material is called adhesive force.
When the adhesive force between solid and liquid molecules is more than the cohesive force between liquid-liquid molecules (as with water and glass), shape of the meniscus is concave and the angle of contact \(\theta\) is less than \(90^{\circ}\). In this case, the liquid wets or adheres to the solid surface.
When the adhesive force between solid and liquid molecules is less than the cohesive force between liquid-liquid molecules than shape of meniscus is convex.
Note The angle of contact between water and clean glass is zero and that between mercury and clean glass is \(137^{\circ}\).
Capillarity
The term capilla means hair which is Latin word. A tube of very fine (hair-like) bore is called a capillary tube.
If a capillary tube of glass is dipped in liquid like water, the liquid rises in the tube, but when the capillary tube is dipped in a liquid like mercury, the level of liquid falls in the tube.
This phenomenon of rise or fall of a liquid in the capillary is called capillarity.
Some examples are
(i) Small capillaries in fibres of towels soaks water from our skin.
(ii) In trees sap rises in stem due to capillary action.
Capillary Rise
One consequence of the pressure difference across a curved liquid-air interface is the wellknown effect that water rises up in a narrow tube in spite of gravity. The word capilla means hair in Latin; if the tube were hair thin, the rise would be very large. To see this, consider a vertical capillary tube of circular cross section (radius a) inserted into an open vessel of water (Figure above). The contact angle between water and glass is acute. Thus the surface of water in the capillary is concave. This means that there is a pressure difference between the two sides of the top surface. This is given by
\(
\begin{aligned}
& \left(P_i-P_o\right)=(2 S / r)=2 S /(a \sec \theta) \\
& =(2 S / a) \cos \theta \dots(i)
\end{aligned}
\)
Thus the pressure of the water inside the tube, just at the meniscus (air-water interface) is less than the atmospheric pressure. Consider the two points A and B in Figure (a) above. They must be at the same pressure, namely
\(
P_o+h \rho g=P_i=P_A \dots(ii)
\)
where \(\rho\) is the density of water and \(h\) is called the capillary rise [Figure (a)]. Using Eq. (i) and (ii) we have
\(
h \rho g=\left(P_i-P_o\right)=(2 S \cos \theta) / a \dots(iii)
\)
\(
h=\frac{2 S \cos \theta}{r \rho g}
\)
The above equations make it clear that the capillary rise is due to surface tension. It is larger, for a smaller a. Typically it is of the order of a few cm for fine capillaries.
Important Points on Formula of Capillary Rise:
When a capillary tube is dipped in a liquid, then the level of liquid in capillary tube rises or falls w.r.t. free surface of liquid outside the capillary.
This phenomena of rises or fall of liquid is called capillary action.
The formula in capillary motion is
\(
h=\frac{2 T \cos \theta}{r \rho g}
\)
where, \(h=\) height of liquid column rises or falls,
\(r=\) radius of capillary tube,
\(\rho=\) density of liquid,
\(g=\) acceleration due to gravity,
\(\theta=\) angle of contact and \(T=\) surface tension.
The result has following notable features
(i) If the contact angle \(\theta\) is greater than \(90^{\circ}\), the term \(\cos \theta\) becomes negative and hence, \(h\) is negative. The result, then gives the depression of the liquid in the tube.
(ii) Suppose a capillary tube is held vertically in a liquid which has a concave meniscus, then capillary rise is given by
\(
\begin{aligned}
& h=\frac{2 T \cos \theta}{r \rho g}=\frac{2 T}{R \rho g} \quad\left(\because R=\frac{r}{\cos \theta}\right) \\
& h R=\frac{2 T}{\rho g}
\end{aligned}
\)
Note:
(i) If the tube is of a length \(l\) less than \(h(l<h)\), the liquid does not overflow. The angle made by liquid surface with the tube adjusts in such a way that rise will stop when it is equal to length of tube.
(ii) If the tube makes an angle \(\phi\) units with vertical, rise will be
\(
h^{\prime}=\frac{h}{\cos \phi}
\)
Example 29: A capillary tube whose inside radius is 0.5 mm is dipped in water having surface tension \(7.0 \times 10^{-2} \mathrm{Nm}^{-1}\). To what height is the water raised above the normal water level? Angle of contact of water with glass is \(0^{\circ}\). Density of water is \(10^3 \mathrm{~kg} \mathrm{~m}^{-3}\) and \(g=9.8 \mathrm{~ms}^{-2}\).
Solution: Height raised, \(h=\frac{2 T \cos \theta}{r \rho g}\) Substituting the proper values, we get
\(
\begin{aligned}
h & =\frac{(2)\left(7.0 \times 10^{-2}\right) \cos 0^{\circ}}{\left(0.5 \times 10^{-3}\right)\left(10^3\right)(9.8)}=2.86 \times 10^{-2} \mathrm{~m} \\
& =2.86 \mathrm{~cm}
\end{aligned}
\)
Example 30: A glass tube of radius 0.4 mm is dipped vertically in water. Find upto what height, the water will rise in the capillary. If the tube is inclined at an angle of \(60^{\circ}\) with the vertical, how much length of the capillary is occupied by water? (Take, surface tension of water \(=7.0 \times 10^{-2} \mathrm{Nm}^{-1}\), density of water \(=10^3 \mathrm{kgm}^{-3}\) )
Solution: For glass-water, angle of contact, \(\theta=0^{\circ}\).
Now, height of water in capillary,
\(
\begin{aligned}
h & =\frac{2 T \cos \theta}{r \rho g}=\frac{(2)\left(7.0 \times 10^{-2}\right) \cos 0^{\circ}}{\left(0.4 \times 10^{-3}\right)\left(10^3\right)(9.8)} \\
& =3.57 \times 10^{-2} \mathrm{~m} \\
& =3.57 \mathrm{~cm}
\end{aligned}
\)
Length of capillary occupied after tilting it is,
\(
\begin{aligned}
l & =\frac{h}{\cos 60^{\circ}}=\frac{3.57}{1 / 2} \\
& =7.14 \mathrm{~cm}
\end{aligned}
\)
Example 31: The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at temperature of the experiments is \(7.30 \times 10^{-2} \mathrm{Nm}^{-1}\). 1 atmospheric pressure \(=\) \(1.01 \times 10^5 \mathrm{~Pa}\), density of water \(=1000 \mathrm{~kg} / \mathrm{m}^3\), \(\mathrm{g}=9.80 \mathrm{~m} \mathrm{~s}^{-2}\). Also calculate the excess pressure.
Solution: The excess pressure in a bubble of gas in a liquid is given by \(2 S / r\), where \(S\) is the surface tension of the liquid-gas interface. You should note there is only one liquid surface in this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure in that case is \(4 S / r\).) The radius of the bubble is \(r\). Now the pressure outside the bubble \(P_o\) equals atmospheric pressure plus the pressure due to 8.00 cm of water column. That is
\(
\begin{aligned}
P_{\mathrm{o}}=\left(1.01 \times 10^5 \mathrm{~Pa}+0.08 \mathrm{~m}\right. & \times 1000 \mathrm{~kg} \mathrm{~m}^{-3} \\
& \left.\times 9.80 \mathrm{~m} \mathrm{~s}^{-2}\right)
\end{aligned}
\)
\(
=1.01784 \times 10^5 \mathrm{~Pa}
\)
Therefore, the pressure inside the bubble is
\(
\begin{aligned}
P_{\mathrm{i}} & =P_{\mathrm{o}}+2 \mathrm{~S} / \mathrm{r} \\
& =1.01784 \times 10^5 \mathrm{~Pa}+\left(2 \times 7.3 \times 10^{-2} \mathrm{~Pa} \mathrm{~m} / 10^{-3} \mathrm{~m}\right) \\
& =(1.01784+0.00146) \times 10^5 \mathrm{~Pa} \\
& =1.02 \times 10^5 \mathrm{~Pa}
\end{aligned}
\)
where the radius of the bubble is taken to be equal to the radius of the capillary tube, since the bubble is hemispherical! (The answer has been rounded off to three significant figures.) The excess pressure in the bubble is 146 Pa .