A streamline may be defined as the path, straight or curved, the tangent to which at any point gives the direction of the flow of liquid at that point.
The two streamlines cannot cross each other and the greater is the crowding of streamlines at a place, the greater is the velocity of liquid particles at that place.
Path ABC is streamline as shown in the figure and \[\Delta U=mg({{h}_{2}}-{{h}_{1}})\], \[{{v}_{2}}\] and \[W={{P}_{1}}V-{{P}_{2}}V=({{P}_{1}}-{{P}_{2}})V\] are the velocities of the liquid particles at A, B and C point respectively.
(2) Laminar flow : If a liquid is flowing over a horizontal surface with a steady flow and moves in the form of layers of different velocities which do not mix with each other, then the flow of liquid is called laminar flow. In this flow, the velocity of liquid flow is always less than the critical velocity of the liquid. The laminar flow is generally used synonymously with streamlined flow.
(3) Turbulent flow : When a liquid moves with a velocity greater than its critical velocity, the motion of the particles of liquid becomes disordered or irregular. Such a flow is called a turbulent flow.
In a turbulent flow, the path and the velocity of the particles of the liquid change continuously and haphazardly with time from point to point. In a turbulent flow, most of the external energy maintaining the flow is spent in producing eddies in the liquid and only a small fraction of energy is available for forward flow. For example, eddies are seen by the sides of the pillars of a river bridge.
If density of body is greater than that of liquid \[\rho >\sigma \]
Weight will be more than upthrust so the body will sink
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If density of body is equal to that of liquid \[\rho =\sigma \]
Weight will be equal to upthrust so the body will float fully submerged in neutral equilibrium with its top surface in it just at the top of liquid
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If density of body is lesser than that of liquid \[\rho <\sigma \]
Weight will be less than upthrust so the body will, move upwards and in equilibrium will float and partially immersed in the liquid Such that,
\[W={{V}_{in}}\sigma \,g\Rightarrow V\rho \,g={{V}_{in}}\sigma \,g\]
\[V\rho ={{V}_{in}}\sigma \] Where \[{{V}_{in}}\] is the volume of body in the liquid
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However, if meta-centre goes below CG, the couple due to forces at G and \[{B}'\] tends to topple the floating body.
That is why a wooden more... 
As \[{{h}_{2}}>{{h}_{1}},\,{{F}_{2}}\] will be greater than \[{{F}_{1}}\], so the body will experience a net upward force
\[F={{F}_{2}}-{{F}_{1}}=A\sigma g({{h}_{2}}-{{h}_{1}})\]
If L is the vertical height of the body \[F=A\sigma gL=V\sigma g\]
[As \[V=AL=A({{h}_{2}}-{{h}_{1}})]\]
i.e., F = Weight of fluid displaced by the body.
This force is called upthrust or buoyancy and acts vertically upwards (opposite to the weight of the body) through the centre of gravity of displaced fluid (called centre of buoyancy). Though we have derived this result for a body fully submerged in a fluid, it can be shown to hold good for partly submerged bodies or a body in more than one fluid also.
(1) Upthrust is independent of all factors of the body such as its mass, size, density etc. except the volume of the body inside the fluid.
(2) Upthrust depends upon the nature of displaced fluid. This is why upthrust on a fully submerged body is more in sea water than in fresh water because its density is more than fresh water.
(3) Apparent weight of the body of density \[(\rho )\] when immersed in a liquid of density \[(\sigma )\].
Apparent weight = Actual weight - Upthrust \[=W-{{F}_{up}}\]
\[=V\rho g-V\sigma g=V(\rho -\sigma )g\]\[=V\rho g\left( 1-\frac{\sigma }{\rho } \right)\]
\[\therefore \] \[{{W}_{APP}}=W\left( 1-\frac{\sigma }{\rho } \right)\]
(4) If a body of volume V is immersed in a liquid of density \[\sigma \] then its weight reduces.
\[{{W}_{1}}\] = Weight of the body in air, \[{{W}_{2}}\] = Weight of the body in water
Then apparent (loss of weight) weight \[{{W}_{1}}-{{W}_{2}}=V\sigma g\]
\[\therefore \] \[V=\frac{{{W}_{1}}-{{W}_{2}}}{\sigma g}\]
(5) Relative density of a body
(R.D.) =\[\frac{\text{density of body}}{\text{density of water}}\] \[=\frac{\text{Weight of body}}{\text{Weight of equal volume of water}}\]= \[\frac{\text{Weight of body}}{\text{Water thrust }}\]
\[=\frac{\text{Weight of body}}{\text{Loss of weight in water}}\]
= \[\frac{\text{Weight of body in air}}{\text{Weight in air--weight in water}}\] = \[\frac{{{W}_{1}}}{{{W}_{1}}-{{W}_{2}}}\]
(6) If more... 
\[F=P\,A=\frac{f}{a}\,A=f\left( \frac{A}{a} \right)\]
As \[A>>a\], therefore \[F>>f\]. So heavy load placed on the larger piston is easily lifted upwards by applying a small force. 
(6) Hydrostatic pressure depends on the depth of the point below the surface (h), nature of liquid (\[\rho \]) and acceleration due to gravity (g) while it is independent of the amount of liquid, shape of the container or cross-sectional area considered. So if a given liquid is filled in vessels of different shapes to same height, the pressure at the base in each vessel's will be the same, though the volume or weight of the liquid in different vessels will be different.
\[{{P}_{A}}={{P}_{B}}={{P}_{C}}\] but \[{{W}_{A}}<{{W}_{B}}<{{W}_{C}}\]
(7) In a liquid at same level, the pressure will be same at all points, if not, due to pressure difference the liquid cannot be at rest. This is why the height of liquid is the same in vessels of different shapes containing different amounts of the same liquid at rest when they are in communication with each other.
(8) Gauge pressure : The pressure difference between hydrostatic pressure P and atmospheric pressure \[{{P}_{0}}\] is called gauge pressure.
\[P-{{P}_{0}}=h\rho g\] 
As excess pressure acts from concave to convex side, the interface will be concave towards the smaller bubble and convex towards larger bubble and if r is the radius of interface.
\[\Delta P=\frac{4T}{r}\] ...(ii)
From (i) and (ii) \[\frac{1}{r}=\frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}}\]
\[\therefore \] Radius of the interface \[r=\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{2}}-{{r}_{1}}}\]
(2) Formation of a single bubble
(i) Under isothermal condition two soap bubble of radii \['a'\] and \['b'\] coalesce to form a single bubble of radius \['c'\].
If the external pressure is P0 then pressure inside bubbles
\[{{P}_{a}}=\left( {{P}_{0}}+\frac{4T}{a} \right)\], \[{{P}_{b}}=\left( {{P}_{0}}+\frac{4T}{b} \right)\] and \[{{P}_{c}}=\left( {{P}_{0}}+\frac{4T}{c} \right)\]
and volume of the bubbles
\[{{V}_{a}}=\frac{4}{3}\pi {{a}^{3}}\], \[{{V}_{b}}=\frac{4}{3}\pi {{b}^{3}}\], \[{{V}_{c}}=\frac{4}{3}\pi {{c}^{3}}\]
Now as mass is conserved \[{{\mu }_{a}}+{{\mu }_{b}}={{\mu }_{c}}\]
\[\Rightarrow \] \[\frac{{{P}_{a}}{{V}_{a}}}{R{{T}_{a}}}+\frac{{{P}_{b}}{{V}_{b}}}{R{{T}_{b}}}=\frac{{{P}_{c}}{{V}_{c}}}{R{{T}_{c}}}\] \[\left[ \text{As }PV=\mu RT,\,\,\,i.e.,\,\mu =\frac{PV}{RT} \right]\]
\[\Rightarrow \] \[{{P}_{a}}{{V}_{a}}+{{P}_{b}}{{V}_{b}}={{P}_{c}}{{V}_{c}}\] ...(i) [As temperature is constant, i.e., \[{{T}_{a}}={{T}_{b}}={{T}_{c}}\]]
Substituting the value of pressure and volume
\[\Rightarrow \] \[\left[ {{P}_{0}}+\frac{4T}{a} \right]\,\left[ \frac{4}{3}\pi {{a}^{3}} \right]+\left[ {{P}_{0}}+\frac{4T}{b} \right]\,\left[ \frac{4}{3}\pi {{b}^{3}} \right]\]\[=\left[ {{P}_{0}}+\frac{4T}{c} \right]\,\left[ \frac{4}{3}\pi {{c}^{3}} \right]\]
\[\Rightarrow \] \[4T({{a}^{2}}+{{b}^{2}}-{{c}^{2}})={{P}_{0}}({{c}^{3}}-{{a}^{3}}-{{b}^{3}})\]
\[\therefore \] Surface tension of the liquid \[T=\frac{{{P}_{0}}({{c}^{3}}-{{a}^{3}}-{{b}^{3}})}{4({{a}^{2}}+{{b}^{2}}-{{c}^{2}})}\]
(ii) If two bubble coalesce in vacuum then by substituting \[{{P}_{0}}=0\] in the above expression we get
\[{{a}^{2}}+{{b}^{2}}-{{c}^{2}}=0\] \[\therefore \] \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}\]
Radius of new bubble \[=c=\sqrt{{{a}^{2}}+{{b}^{2}}}\] or can be expressed as \[r=\sqrt{r_{1}^{2}+r_{2}^{2}}\].
(3) The difference of levels of liquid column in two limbs of U-tube of unequal radii \[{{r}_{1}}\] and \[{{r}_{2}}\] is
\[h={{h}_{1}}-{{h}_{2}}=\frac{2T\cos \theta }{dg}\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]\]
(4) A large force (F) is required to draw apart two glass plate normally enclosing a thin water film because the thin water film formed between the two glass plates will have concave surface all around. Since on the concave side of a liquid surface, pressure is more, work will have to be done in drawing the plates apart.
\[F=\frac{2AT}{t}\] where T= surface tension of water film, t= thickness of film, A = area of film.
(5) When a soap bubble is charged, then its size increases due to outward force on the bubble.
(6) The materials, which when coated on a surface and water does not enter through that surface are known as water proofing agents. For example wax etc. Water proofing agent increases the angle of contact.
(7) Values of surface tension of some liquids.
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Whether the liquid will be in equilibrium in the form of a drop or it will spread out; depends on the relative strength of the force due to surface tension at the three interfaces.
\[{{T}_{LA}}=\] surface tension at liquid-air interface, \[{{T}_{SA}}=\] surface tension at solid-air interface.
\[{{T}_{SL}}=\] surface tension at solid-liquid interface, \[\theta =\] angle of contact between the liquid and solid.
For the equilibrium of molecule
\[{{T}_{SL}}+{{T}_{LA}}\cos \theta ={{T}_{SA}}\] or \[\cos \theta =\frac{{{T}_{SA}}-{{T}_{SL}}}{{{T}_{LA}}}\]
Special Cases
When one end of capillary tube of radius r is immersed into a liquid of density d which wets the sides of the capillary tube (water and capillary tube of glass), the shape of the liquid meniscus in the tube becomes concave upwards.
R = radius of curvature of liquid meniscus.
T = surface tension of liquid
P = atmospheric pressure
Pressure at point A = P, Pressure at point B = \[P-\frac{2T}{R}\]
Pressure at points C and D just above and below the plane surface of liquid in the vessel is also P (atmospheric pressure). The points B and D are in the same horizontal plane in the liquid but the pressure at these points is different.
In order to maintain the equilibrium the liquid level rises in the capillary tube upto height h.
Pressure due to liquid column = pressure difference due to surface tension
\[\Rightarrow \] \[hdg=\frac{2T}{R}\]
\[\therefore \] \[h=\frac{2T}{Rdg}\]\[=\frac{2T\cos \theta }{rdg}\] \[\left[ \text{As }R=\frac{r}{\cos \theta } \right]\]
(i) The capillary rise depends on the nature of liquid and solid both i.e. on T, d, \[\theta \] and R.
(ii) Capillary action for various liquid-solid pair.
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