Metacentre
When the body floating in liquid is slightly tilt from its equilibrium position, the centre of gravity (CG) of the body and centre of buoyancy (CB) of the displaced liquid do not lie on the same vertical line as in equilibrium position.
The point of intersection of the vertical line passing through CB of the tilted position & the original vertical line in the equilibrium position is called metacentre (MC) of the floating body. The Floating body can regain its original position or fall into the liquid when it is tilted which depends upon the position of CG of and MC of the floating body.

Fluid dynamics
The branch of physics which deals with the study of nature of fluid when it is in motion is called fluid dynamics.

Fluid
Fluid can be either liquid or gases.

Relation between Surface Tension and Surface Energy

Let us consider, a rectangular frame of wire ABCD in which the wire BC is movable as shown in figure. If we dip the frame in a soap solution, a thin film is formed which pulls the wire BC towards left due to surface tension.

Let, T = Surface tension of film

[latex]l[/latex] = BC = length of the wire

Then, the force F on BC due to T is

[latex]F = T\times 2l[/latex]

Since, the film has both upper and lower surface, so total length of contact = [latex]2l[/latex]

Suppose, the wire BC moves to B’C’ through a distance x at constant temperature, work done in increasing the surface area against force F is given by,

[latex]W = F\times x = T\times 2l\times x[/latex]

Now, increase in surface area [latex](\Delta A) = 2l\times x[/latex]

 [latex]\therefore[/latex] Surface energy [latex](\sigma) = \frac{W}{\Delta A} = \frac{T\times 2l\times x}{2l\times x}[/latex]

Or, [latex]\sigma = T[/latex]

So, surface tension (T) is equal to surface energy [latex](\sigma)[/latex] under isothermal condition.

Viscosity of fluid:
The nature of fluid by virtue of which an internal resistance comes into play when it is in motion which opposes the relative motion between its different layers is called viscosity of fluid.

Newton’s law of viscous force:

Let us consider, a steady flow of viscous fluid through the horizontal solid surface as shown in the figure above.
The layers of liquid which is in contact with solid surface is at rest while the velocity of layers of fluid goes on increasing as we move upwards from the horizontal surface.
Consider, two such layers having area of contact ‘A’ such that lower surface which is at a distance of ‘x’ from the horizontal surface has velocity ‘v’ and the layer of fluid which is at a distance x+dx has velocity v+dv as shown in figure above, then, according to Newton’s law of viscous force,
i. Viscous force depends upon area of layers in contact i.e. viscous force (F) ∝ A. ………… (i)
ii. Viscous force is directly proportional to the velocity gradient which is acting normal to the direction of flow of fluid. i.e.
F ∝  [latex]\frac{dv}{dx}[/latex]…………. (ii)
Combining (i) and (ii), we get,
F = – [latex]\eta A \frac{dv}{dx}[/latex] A , where, [latex]\eta[/latex] = coefficient of viscosity of the fluid.
The negative sign is due to the fact that viscous force is directed opposite to the direction of flow of fluid.

Coefficient of viscosity ([latex]\eta[/latex]) of fluid:
From Newton’s law of viscous force (F), we have,
F = [latex]\eta A \frac{dv}{dx}[/latex]  (in magnitude)
∴ [latex]\eta = \frac{F}{A(\frac{dv}{dx})}[/latex]
Hence, the coefficient of viscosity ([latex]\eta[/latex]) of fluid is the viscous force acting on fluid per unit area of contact between the layers of fluid having a unit velocity gradient acting normal to the direction of flow of fluid.
Dimension of [latex]\eta[/latex] = [ML^-1T^-1]
SI unit: Nm^-2S Or, Kgm^-1S^-1 or decapoise.
CGS unit: dyne cm^-2S Or, gmcm^-1S^-1 or poise.
1 decapoise = 1Nm^-2S = 10⁵ dyne (100cm)^-2 = 10 dyne cm^-2S.
                                                            = 10 Poise.

Streamline flow and turbulent flow:

(i) Streamline flow of fluid:
The flow of fluid is said to be streamline flow if every particle crossing a given point during the flow of fluid follows exactly the same path and has the same velocity (both in magnitude and direction).

The path of the flow of particles in the steady and streamline flow is called streamline.
It is a curve, the tangent drawn at any point on the curve gives the direction of velocity of particle of fluid crossing that point. No two streamlines can intersect each other because if they intersect each other then, at the point of intersection, the velocity of the particle has two different values which is impossible.
Laminar flow of fluid in which the velocity of the given layer of fluid is same is also a streamline flow.

(ii) Turbulent flow of fluid:
When the velocity of flow of fluid increases above the certain limit called critical velocity of fluid, the velocity of the particles of fluid changes continuously and haphazardly with time, such flow of fluid is called turbulent flow of fluid.
Flow of fluid behind the ship or boat in motion, flow of air behind moving bus are examples of turbulent fluid.

Poiseuille’s formula to determine the volume of fluid:
 Poiseullie studied a streamline flow of viscous fluid flowing through the capillary tube and found that the volume of fluid flowing per second (V) through the capillary tube depends on:
(i) pressure gradient (P/L)
Where, P = pressure difference between ends of the capillary tube
And, L = length of the tube.
i.e. V ∝ [latex](\frac{P}{L})[/latex] …………… (i)
(ii) radius of the capillary tube (r):
i.e. V ∝ (r)^b ………….. (ii)
(iii) coefficient of viscosity [latex]\eta[/latex] of the fluid,
i.e. V ∝ ([latex]\eta[/latex])^c …………….. (iii)
combining eq^n. (i), (ii) and (iii), we get,
V = k[latex](\frac{P}{L})^a(r)^b(\eta)^c[/latex] ……………………… (iv), where, k = dimensionless proportionality constant.
The dimensional formula of volume of fluid per second is = [M^oL³T^-1].
The dimensional formula of pressure gradient is: [latex]\frac{P}{L} = [ML^{-2}T^{-2}][/latex].
The dimensional formula of radius (r) = [M^oLT^o].
The dimensional formula of coefficient of viscosity ([latex]\eta[/latex]) is = [ML^-1T^-1].
Writing eq^n. (iv) in dimensional form, we get,
M^oL³T^-1 = [ML^-2T^-2]^a [M^oL¹T^o]^b [ML^-1T^-1]^c
Using principle of homogeneity of dimension, we get,
a + c =0 …………… (v)
-2a + b – c = 3…………. (vi)
-2b-c = -1……….. (vii)
Adding eq^n. (vi) and (vii), we get,
-a = -1.
∴ a = 1.
Putting the value of a in eq^n. (v), we get,
a + c = 0.
∴ c = -1.
Putting the value of a and c in eq^n. (iv), we get,
-2 x 1 + b + 1 = 3.
∴ b=4.
Substituting the values of a, b and c in eq^n. (iv), we get,
V = k[latex](\frac{P}{L})r^4\eta^{-1}[/latex]
From experiment, the value of k is found to be [latex]\frac{\pi}{8}[/latex]. So,
V = [latex]\frac{\pi}{8}\frac{P}{L}\frac{r^4}{\eta}[/latex] …………. (viii)
Eq^n. (viii) gives the required expression for the volume of fluid to the capillary tube.

Stroke’s law to determine the viscous force:
Terminal velocity:
When a small spherical ball falls on the viscous fluid, its velocity goes on increasing at first. A stage is reached when downward force i.e., weight of the ball is balanced by upward force acting on the ball i.e., viscous force and Upthrust force. Then, the spherical ball moves with constant velocity which is known as Terminal velocity of the quantity.

After performing various experiments by making fall of small spherical balls of different radii on the different fluids, Stroke found that the viscous force (F_v) acting on the small spherical ball during the fall depends upon,
(i) radius of the spherical ball i.e., F_v ∝ r^a …………. (i)
(ii) coefficient of viscosity of the fluid ([latex]\eta[/latex]) of the fluid. i.e., F_v = [latex]\eta[/latex]^b…………. (ii)
(iii) terminal velocity of the spherical ball. i.e., F_v = v_t^c …………. (iii)
Combining eq^n. (i), (ii) and (iii), we get,
F_v = kr^a[latex]\eta[/latex]^bv_t^c …….. (iv), where, k is dimensionless proportionality constant.
Where, a, b and c are scalars to be determined.
Dimensional formula of viscous force (F_v) = [MLT^-2]
Dimensional formula of radius (r) = [M^oLT^o]
Dimensional formula of coefficient of viscosity ([latex]\eta[/latex]) = [ML^-1T^-1]
Dimensional formula of terminal velocity (v_t) = [M^oLT^-1]
Writing eq^n. (iv) in dimensional form,
[MLT^-2] = [M^oLT^o]^a [ML^-1T^-1]^b [M^oLT^-1]^c
Using principle of homogeneity of dimension,
b = 1 ….. (v)
a – b+c = 1………….. (vi)
-b-c = -2……… (vii)
Using eq^n. (v) in eq^n. (vii), we get,
-1-c = -2.
c = 1.
Putting the values of b and c in eq^n. (vi), we get,
a-b+c = 1.
∴ a = 1.
Making the substitution of a, b and c in eq^n. (iv),
F_v = krhv_t
From experiment, the value of k is found to be 6π.
So, F_v = 6πr[latex]\eta[/latex]v_t
This is the required expression of the viscous force acting on the small spherical ball falling on viscous fluid.

Determination of coefficient of viscosity of fluid using Stroke’s law
Let us consider, a small spherical solid ball of density (ρ) at radius (r) is falling on the liquid having coefficient of viscosity ([latex]\eta[/latex]) and density ([latex]\sigma[/latex]). At first, the velocity of the spherical ball goes on increasing.

When it attains a certain velocity, the upward force acting on the ball i.e., Viscous force (F_v) and Upthrust force (U) is balanced by the weight of the ball i.e., mg. After that, the ball moves with constant velocity known as terminal velocity i.e., v_t.
When the ball starts moving with terminal velocity, then, W = U + F_v
Here,
Weight of spherical ball (W) = mg = [latex](\frac{4}{3}\pi r^3)\rho g[/latex]
Upthrust = [latex](\frac{4}{3}\pi r^3)\sigma g[/latex]
Viscous force (F_v) = 6π[latex]\eta[/latex]rv_t.
∴ Total Upward force = 6π[latex]\eta[/latex]rv_t + [latex]\frac{4}{3}\pi r^3 \sigma g[/latex]
∴ Total Downward force = weight of spherical ball
            = [latex](\frac{4}{3}\pi r^3)\rho g[/latex]
Since, upward force and downward force acting on ball gets balanced when it starts falling with terminal velocity, so,
Total Upward force = Total Downward force.
Or, 6π[latex]\eta[/latex]rv_t + [latex]\frac{4}{3}\pi r^3 \sigma g[/latex] = [latex]\rho \frac{4}{3}\pi r^3 g[/latex]
[latex]\eta = \frac{4}{18}\frac{\pi r^3 g(\rho – \sigma)}{\pi r v_t}[/latex]
∴ [latex]\eta = \frac{2}{9}\frac{r^2 g(\rho – \sigma)}{vt}[/latex].
This is the required expression for the coefficient of viscosity of the fluid as given by Stroke’s formula.

Continuity equation

The equation of continuity states that, ”For the streamline flow of an ideal liquid (non-viscous and incompressible) through a tube of varying cross section, the mass of liquid entering per second is equal to mass of liquid leaving per second, if there is no loss of liquid from the walls of the tube.”

If [latex]\rho[/latex] represents the density of the liquid, v₁ and v₂ represents the velocity of the flow of liquid where the fluid enters and leaves. Then, A₁ and A₂ represents the cross-sectional area of the segment of the liquid when it enters and leaves. Then, from continuity equation,

Mass of liquid entering per second = mass of liquid leaving per second.

[latex]\rho[/latex] x A₁ x v₁ = [latex]\rho[/latex] x A₂ x v₂

A₁ x v₁ = A₂ x v₂

i.e., the volume of liquid entering per second = volume of liquid leaving per second.

i.e., Then product of cross-sectional area and velocity of liquid at any cross section during the streamline flow of ideal liquid is constant.

Bernoulli’s Equation

Bernoulli’s equation states that, ”For the streamline flow of an ideal fluid (non – viscous and incompressible), the sum of pressure energy, kinetic energy and potential energy per unit mass at every cross section during the flow of fluid is always constant. i.e., [latex]\frac{P}{\rho}+\frac{1}{2}v^2 + gh[/latex] = constant. where, the symbols carry their usual meaning.

Proof

Let us consider, a streamline flow of ideal fluid through a tube of varying cross section as in figure. At any instant of time, let v₁, P₁, A₁, h₁ and v_2, P₂, A₂, h₂ be the velocity, pressure, cross sectional area and height of liquid at ends X and Y respectively. Also, [latex]\rho[/latex] be the density of fluid.

Now, workdone in moving a fluid through a distance (d) is: (W) = force x distance moved

            = F x d = (P.A) x d

            = P.V [latex][\because A.d = Volume (V)][/latex]

Or, [latex]\frac{W}{V} = P[/latex]

Or, workdone per unit volume = pressure …………… (i)

 [latex]\therefore[/latex] Net workdone on fluid per unit volume = P₁ – P₂

K.E. per unit volume = [latex]\frac{K.E.}{V}=\frac{1}{2}\times \frac{mass}{volume}\times (velocity)^2[/latex]

            = [latex]\frac{1}{2}\times density(\rho)\times v^2[/latex]

If v₁ and v₂ be the initial & final velocity, then,

K.E. gained per unit volume = [latex]\frac{1}{2}\rho (v_2^2 – v_1^2)[/latex]

Now, P.E. gained per unit volume = [latex]\rho g(h_2 – h_1)[/latex]

We know, from conservation of energy,

P₁ – P₂ = [latex]\frac{1}{2}\rho (v_2^2 – v_1^2) + \rho g(h_2 – h_1)[/latex]

            = [latex]\frac{1}{2}\rho v_2^2 – \frac{1}{2}\rho v_1^2 + \rho gh_2 – \rho gh_1^2[/latex]

Or, P₁ + [latex]\frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2[/latex]

[latex]P + \frac{1}{2}\rho v^2 + \rho gh[/latex] = Constant

This verifies Bernoulli’s theorem.

Liquid meniscus:

Usually, the surface of the liquid is curve when it is in contact with it. This curve surface of the liquid in contact with solid is called Liquid Meniscus. The shape of the liquid meniscus is determined by the relative strength of adhesive force that exist between the molecules of liquid and solid and cohesive force that exist between molecules of liquid. If adhesive force is greater than cohesive force, the liquid meniscus is concave and if cohesive is greater than adhesive force in strength, then, the liquid meniscus is convex. The concave and convex meniscus is shown in figure above. If the strength of both cohesive & adhesive force is equal, the liquid meniscus is flat which is shown in figure alongside.

Angle of contact:

The angle of contact between a pair of solid and liquid is defined as the angle that the tangent drawn to the surface of liquid at the point of contact of solid and liquid that makes with solid inside the liquid.

The angle of contact for a given pair of solid and liquid is acute if the liquid wets the solid and if the liquid does not wet the solid, the angle of contact is obtuse.

If the L.M. is flat, the angle of contact is 90°. The angle of contact lies between 0 to 180°.

The angle of contact is shown in figure above.

Capillary action or Capillarity:

A tube with very fine bore is called capillary tube. One end of the capillary tube open at both ends is dipped into the liquid contained in the vessel, there is either rise or fall of liquid inside the capillary tube. This phenomenon is known as capillarity.

Determination of surface tension of liquid by capillary rise method

Let us consider, one end of a glass capillary tube with internal radius ‘r’ and having both ends open is dipped into the liquid like water which wets the capillary tube then, there is rise of liquid in the capillary tube in comparison to the liquid surface outside. The liquid meniscus is concave and the angle of contact ([latex]\theta[/latex]) is acute.

Let, [latex]\rho[/latex], T and h represents density and surface Tension of liquid and height of rise of liquid in the capillary tube. At point of contact, A the liquid exerts a force to the wall of the capillary tube which acts tangent to the curve i.e., surface tension (T) [force in terms of per unit length].

From Newton’s third law of motion, the capillary tube also exert equal and opposite reaction force, R = T as in figure.

Resolving the reaction force, we get, TSin[latex]\theta[/latex] as horizontal component and TCos[latex]\theta[/latex] as vertical component. The same follows at the point of contact B as well as all points of contact on the circumference 2πr.

The horizontal component cancels each other while the vertical component is added to give rise of the liquid in the capillary tube.

Therefore, the total vertical upward force on the liquid column that rises in the capillary tube = 2πr x TCos[latex]\theta[/latex].

Volume of liquid column that rises in the capillary tube (V) = Volume of hemisphere of radius r.

V = πr³ (h+r) – [latex]\frac{2}{3}\pi r^3[/latex]

V = πr² (h +[latex]\frac{r}{3}[/latex])

∴ Weight of liquid column that rises in the capillary tube = V[latex]\rho[/latex]g

            = πr² [latex](h+\frac{r}{3})\rho g[/latex]

At equilibrium, vertically upward force acting on liquid column = wt. of liquid column

2πr x TCos[latex]\theta[/latex] = πr² [latex](h+\frac{r}{3})\rho g[/latex]

T = [latex]\frac{\rho g(h+\frac{r}{3})r}{2Cos\theta}[/latex]

For a capillary tube of a very fine bore, [latex]\frac{r}{3}[/latex] can be neglected in comparison to h.

T = [latex]\frac{\rho ghr}{2cos\theta}[/latex]. This is the required expression for the surface tension of liquid.

h = 2TCos[latex]\theta[/latex].