Fluid:

A substance having no definite shape and can flow from one place to another.

Fluid statics:

The branch of physics which deals with the study of fluid in the state of rest is known as known as fluid statics.

Density (Absolute density):

The density of a body is defined as the ratio of its mass and volume.
Density, $\rho = \frac{\text{mass}}{\text{volume}}$
Its unit is kgm⁻³.

Relative Density (Specific gravity):

It is defined as the ratio of density of a body to the density of water at 4°C. It is the measure of lightness or heaviness of a body as compared to the water. Mathematically, it is given as.
Specific gravity $= \frac{\text{density of substance}}{\text{density of water at 4°C}}$
or, S.G. = $\frac{\text{mass of certain volume of substance}}{\text{mass of equal volume of water at 4°C}}$
or, S.G. = $\frac{\text{wt. of certain vol. of substance}}{\text{wt. of equal vol. of water at 4°C}}$
It has no unit.
Note: Specific gravity is numerically equal to density in CGS.

Pascal’s Law of Pressure:

It states that when external pressure is applied to an isolated and enclosed liquid, the pressure is equally transmitted in all directions.

According to Pascal’s Law of Pressure,
$P_1 = P_2$
Or, $\frac{F_1}{A_1} = \frac{F_2}{A_2}$
Or, $\frac{F_2}{F_1} = \frac{A_2}{A_1} > 1$ [since $A_2 > A_1$]
Or, $F_2 > F_1$
Thus using Pascal’s Law of Pressure, a small force can be converted into large force.

Applications of Pascal’s Law of Pressure

(i) Hydraulic press
(ii) Hydraulic jack
(iii) Hydraulic brakes

(iv) Teeth cleaning / scaling by water jet
(v) High pressure water jet cutting etc

Expression for pressure exerted by liquid at any depth ‘h’

Liquid pressure, $P = \frac{\text{Force}}{\text{Area}}$
$P = \frac{\text{Weight of liquid}}{\text{Area of liquid}}$
$P = \frac{mg}{A}$
$P = \frac{\nu \rho g}{A}$ [since $\rho = \frac{m}{\nu}$]
$P = \frac{A \times h \rho g}{A}$ where, h is depth of liquid
$P = \rho gh$

Upthrust (force of buoyancy):

The upward force exerted by fluid on the immersing body is called upthrust or force of buoyancy. It is denoted by ‘U’. It depends so the volume of body and the density of fluid. It is also the loss of weight of object in fluid, i.e.
Upthrust = wt. of object in air – wt. of object in fluid.
∴ $U = W_{\text{air}} – W_{\text{fluid}}$.

Archimedes’ Principle:

It states that when a body is fully or partially immersed in a fluid, it loses its weight, which is equal to the weight of the fluid displaced by the body (or upthrust), i.e.
Wt. loss = Upthrust = $W_{\text{(air)}} – W_{\text{(fluid)}} = \text{Wt. of displaced fluid}$

Condition for floating:

If $\rho$ be the overall density of object and $\sigma$ be the density fluid then
(i) If $\rho > \sigma$ then the object will sink and lies at the bottom.

(ii) If $\rho = \sigma$ then the object will float with its total part inside the liquid.

(iii) If $\rho < \sigma$ then the object will float with its some part outside the liquid.

Principle of Floatation:

It states that the weight of a floating body is equal to weight of the liquid displaced. i.e floating object displaces the fluid of its own weight.
Wt. of floating body = Wt. of fluid displaced
∴ Wt. of floating body = Upthrust.

Centre of gravity (C.G.), Centre of buoyancy (C.B.) and Meta centre (M.C.):

Centre of gravity (C.G.):

The centre of gravity of a object is defined as the point at which the whole weight of the body is supposed to be acted.

Centre of buoyancy (C.B.):

Centre of buoyancy (C.B.) of a floating body is defined as the point at which the centre of gravity of the displaced liquid lie. In other words it is the point where the upthrust caused by a displaced fluid acts.
Note: The wt. of a body acts vertically downward at C.G. and the upthrust (force of buoyancy) acts vertically upward at C.B.

Meta centre (M.C.):

When a floating body is in stable equilibrium the C.G. and C.B. lie on the same vertical line (fig: a) but when the floating body gets tilted, the C.B. displaces to a new position (fig: b). Then the point of intersecting of the vertical line through the new C.B. and the original vertical line. The position of metacentre determines whether the floating body is stable or not.

In heavy bottom floating object C.G. lies below the M.C. so when tilted it regains its original stable equilibrium (fig:b). But in heavy top floating object C.G. lies above the M.C. so when tilted, a torque is produced and cannot regain its original stable equilibrium (fig: c).

Numerical

1. According to Archimedes’ principle,
   Weight loss = Wt of the liquid displaced = Upthrust = $V \sigma g$
2. Weight loss = Weight in air – weight in water = Wt. of the liquid displaced
3. According to principle of floatation
   Wt. of floating object = Wt. of liquid displaced

Type: 1

1. A string supports a solid iron object of mass 200gm totally immersed in a liquid of specific gravity 0.9. Calculate the tension in the string it the density of iron is 8000 kg/m⁻³.
   Ans: 1.775N

Here,
Mass of iron, $m_i = 200 \text{ gm} = 0.2\text{kg}$
Sp. gravity of liquid = 0.9
Density of liquid, $\sigma = 0.9 \text{ gm/cc} = 900 \text{ kgm}^{-3}$
Density of iron, $\rho = 8000 \text{ kgm}^{-3}$
Tension, $T = ?$
At equilibrium,
$T + U = W$
$T = W – U$
$= mg – \text{weight of liquid displaced}$
$= mg – V_i \sigma g$
$= mg – \frac{m_i}{\rho} \sigma g$ [Volume of displaced liquid = vol. of ion]
$= 0.2 \times 10 – \frac{0.2}{8000} \times 900 \times 10$
$T = 1.775\text{N}$

2. A string supports a solid iron object of mass 200 gm totally immersed in a liquid of density 800kgm⁻³. The density of iron is 8000 kgm⁻³. Calculate the tension in the string.
   Ans: 1.8 N
3. A string supports a solid iron of mass 200 gm totally immersed in a liquid of density 900 kgm⁻³. Calculate the tension in the string if the density of iron is 8000 kgm⁻³.
   Ans: 1.775 N
4. A string supports a solid iron object of mass 180g totally immersed in a liquid of density 800 kg/m³. The density of iron is 8000 kg/m³. Calculate the tension in the string.
   Ans: 1.62 N

Type: 2 (mixed type)

5. A boy can lift a maximum load of 250 N of water. How many litre of mercury of density 13600 kgm⁻³ he can lift in an identical vessel?
   Ans: 1.83 litre

Here,
Maximum weight, $W = 250\text{N}$
Volume of mercury that can be lifted, $V_1 = ?$
Maxm wt. of mercury that he can lift,
$W = 250 \text{ N}$
$mg = 250$
$V \sigma g = 250$
$V = \frac{250}{\sigma g} = \frac{250}{13600 \times 10} = 1.83 \times 10^{-3} \text{ m}^3$
∴ Volume in litre, $V = 1.83 \times 10^{-3} \times 1000 = 1.83 \text{ lit.}$

6. A geologist finds that a moon rock whose mass is 7.2 kg has an apparent mass 5.88 kg when submerged in water. What is the density of the rock?
   Ans: 5454.54 kg m⁻³

Here,
Mass of rock, $m = 7.2 \text{ kg}$
Apparent mass of rock in water = $5.88 \text{ kg}$
Density of rock, $\rho = ?$
According to Archimedes’ principle,
Weight loss = Wt of the liquid displaced
Weight in air – weight in water = Wt of the liquid displaced
$m_{\text{air}} \times g – m_{\text{water}} \times g = V \sigma g$
$\frac{7.2 – 5.88}{\sigma} = V$
$\frac{m}{\rho} = \frac{7.2 – 5.88}{1000}$
$\frac{7.2}{\rho} = 1.32 \times 10^{-3}$
∴ Density of rock, $\rho = 5454.54 \text{ kgm}^{-3}$

7. A piece of gold- Aluminium alloy weights 100g in air and 80 g in water. What is the weight of the gold in the alloy if the relative density of gold is 19.3 and that of aluminum is 2.5?
   Ans: 0.057 kg

Here,
Mass of alloy in air = 100g
Mass of alloy in water = 80g
Mass of gold, $m_{\text{g}} = ?$
Relative density of gold = 19.3 i.e. $\rho_{\text{(gold)}} = 19.3 \text{ gcm}^{-3}$
Relative density of Aluminum = 2.5 i.e. $\rho_{\text{(Al)}} = 2.5 \text{ gcm}^{-3}$
According to Archimedes’ principle
Weight loss = Weight of the water displaced
$(100-80) \times g = V \sigma g$
$V_{\text{alloy}} = 20$
$V_{\text{g}} + V_{\text{Al}} = 20$
$\frac{m_{\text{g}}}{\rho_{\text{g}}} + \frac{m_{\text{Al}}}{\rho_{\text{Al}}} = 20$
$\frac{m_{\text{g}}}{19.3} + \frac{100-m_{\text{g}}}{2.5} = 20$
$\frac{m_{\text{g}}}{19.3} + \frac{100-m_{\text{g}}}{2.5} = 20$

8. An alloy of mass 588 g and volume 100 c.c. is made of iron of density 8.0 gm/cc and aluminum of density 2.7 gm/cc. Calculate the proportion by (i) volume (ii) by mass of the constituents of the alloy. Ans: (i) 60 cc & 40 cc (ii) 480 g & 108 g.

Here,
Mass of alloy, $M = 588 \text{ g}$
Mass of alloy, $V = 100 \text{ cc}$
Density of iron, $\rho_{\text{i}} = 8.0 \text{ gm/cc}$
Density of aluminum, $\rho_{\text{al}} = 2.7 \text{ gm/cc}$
Mass of iron & mass of aluminum = ?
Volume of iron & volume of aluminum = ?

Now,
Mass of alloy = mass of iron + mass of aluminum
i.e. $M = m_{\text{i}} + m_{\text{al}}$
Or, $m_{\text{i}} + m_{\text{al}} = 588 \text{ g} \dots (i)$
Also, $V = V_{\text{i}} + V_{\text{al}}$
Or, $V = \frac{m_{\text{i}}}{\rho_{\text{i}}} + \frac{m_{\text{al}}}{\rho_{\text{al}}}$
Or, $100 = \frac{588-m_{\text{al}}}{8} + \frac{m_{\text{al}}}{2.7}$
Or, $100 \times 8 \times 2.7 = 1587.6 – 2.7 m_{\text{al}} + 8 m_{\text{al}}$
Or, $2160 – 1587.6 = 5.3 m_{\text{al}}$
Or, $m_{\text{al}} = \frac{2160 – 1587.6}{5.3} = 108 \text{ g}$
From equation (i)
$m_{\text{i}} = 588 – 108 = 480 \text{ gm}$
Now $V_{\text{al}} = \frac{m_{\text{al}}}{\rho_{\text{al}}}$
Or, $V_{\text{al}} = \frac{108}{2.7} = 40 \text{ cc}$
Similarly, $V_{\text{i}} = \frac{m_{\text{i}}}{\rho_{\text{i}}} = \frac{480}{8} = 60 \text{ cc}$

Type: 3

9. An iceberg having volume 2.06 litre floats in sea-water of density 1.03 gmcm⁻³ with a portion of 224 CC above the surface. Calculate the density of ice.
   Ans: 918 kg/m³

Here,
Volume of ice, $V = 2.06 \text{ l} = 2.06 \times 1000 \text{ cc} = 2060 \text{ cc}$
Density of sea-water, $\sigma = 1.03 \text{ gm/cm}^3$
Volume of ice above water surface, $V_{\text{above}} = 224 \text{ cc}$
Density of ice, $\rho = ?$
According to principle of floatation
Wt. of ice = Wt. of sea water displaced
Or, $m_{\text{i}} g = V_{\text{sub}} \sigma g$
Or, $V_{\text{i}} \rho = (V_{\text{i}} – V_{\text{above}}) \sigma$
Or, $\rho = \frac{(V_{\text{i}} – V_{\text{above}}) \times \sigma}{V_{\text{i}}}$
Or, $\rho = \frac{(2060 – 224) \times 1.03}{2060}$
Or, $\rho = 0.918 \text{ gm/cc}$
∴ $\rho = 918 \text{ kg/m}^3$

10. An iceberg having a volume of 2060 cc floats in sea-water of density 1030 kg m⁻³ with a portion of 224 cc above the surface. Calculate the density of ice.
    Ans: 918 Kg/m³
11. The density of ice is 971 km m⁻³, and the approximate density of seawater in which an iceberg floats is 1025 kg m⁻³. What fraction of the iceberg, is beneath the water surface? Ans: 0.95
12. A 25 cm thick block of ice floating on fresh water can support an 80 kg man standing on it, what is the smallest area of the ice block? (sp. gravity of ice = 0.917). Ans: 3.85 m².

Here,
Thickness of ice block, $t = 25\text{cm} = 0.25\text{m}$
Mass of man, $m = 80 \text{ kg}$
Minimum area, $A_{\text{(min)}} = ?$
From the principle of floatation,
Wt. of (man + ice) = $V_{\text{w}} \sigma g$
Or, $80 \times g + m_{\text{i}} g = V_{\text{w}} \sigma g$
Or, $80 + m_{\text{i}} = V_{\text{w}} \sigma$
Or, $80 + V_{\text{i}} \rho = V_{\text{i}} \sigma$
Or, $80 = V_{\text{i}} \sigma – V_{\text{i}} \rho$
Or, $80 = V_{\text{i}} (\sigma-\rho)$
Or, $80 = A \times t (\sigma-\rho)$
Or, $A = \frac{80}{t (\sigma-\rho)} = \frac{80}{0.25 \times (1000-917)}$
Or, $A = 3.85 \text{ m}^2$

13. A slab of ice floats on fresh water lake. What minimum volume must the slab have for a 45 kg girl to be able to stand on it without getting her feet wet? Sp. Gravity of ice is 0.920.
    Ans: 0.563 m³
14. A cube of wood floating in water supports 1 kg mass resting on the centre of its top surface. When the mass is removed, the cube rises 2.5 cm. Find the length of the cube.
    Ans: 20 cm

Surface tension (T)

The property of liquid in rest by virtue of which its free surface behaves like a stretched membrane and tries to occupy minimum possible surface area is called surface tension. It is due to molecular cohesive force. It is denoted by ‘T’ and mathematically defined as the force acting per unit length on an imaginary line drawn on the surface of liquid acting normally to the line and tangentially to the liquid surface.
i.e. $T = \frac{f}{L}$
Where ‘L’ is the total length in contact or total peripheral length. Its unit is Nm⁻¹ and dimension = $[MLT^{-2}]$

1. Effect of temperature

The surface tension of liquid decreases with increase in temperature. The temperature at which surface tension of liquid becomes zero is called critical temperature. Also, the surface tension of liquid is zero at its boiling point.

2. Effect of impurities

a) The surface tension of liquid increases if the impurities added are highly soluble.
b) The surface tension of liquid decreases if the impurities added are less soluble or insoluble.

3. On electrification

The surface tension of liquid decreases on electrification.

Some examples of surface tension.

1. Small water drops are spherical but larger water drops …
2. Soap bubbles are spherical
3. Hot soup is tasty…
4. Washing clothes in hot water…
5. Washing powder/ detergent…
6. Watches in street market…
7. Kerosene oil….
8. Painting brush…
9. Wet fly cannot fly…..
10. Antiseptic is used in cut…
11. Small insects can easily walk on the surface of water …
12. A metallic needle can float on water surface

Molecular theory of surface tension

Fig: Molecular force in liquid molecules

The molecules of liquid attract each other with a force of cohesion. The molecules of liquid on surface experiences different net force than that of molecules well below the surface. Let us consider a molecule A of liquid lying well below the free surface of a liquid it has a sphere of influence of radius of order 10⁻⁹m. As this molecule is attracted by neighboring molecules lying within the sphere of influence as shown in figure. The resultant force due to all molecules on A is zero. Again let us consider other molecule B on the surface of liquid. There are few molecules of liquid in vapour state above the free surface. The molecule experiences force of attraction only due to molecules lying in the lower half of the sphere of influence. The resultant of all these forces is downward. So the molecules on the surface experience maximum downward force. If we wish to bring a molecule from interior of the liquid to the liquid surface some work has to be done against the for this force. This work done on the molecules is stored as potential energy. For equilibrium system must have minimum potential energy so there must be minimum number of molecules on the liquid surface which is possible only when the liquid surface is minimum. That’s why the liquid surface contracts like a stretched stressed elastic membrane.

Surface energy ($\sigma$):

The surface energy of a liquid is defined as the work done to increase its surface area by Unity.
It is denoted by ($\sigma$) and mathematically given by
Surface energy $= \frac{\text{work done to increase surface area}}{\text{Increased in surface area}}$
The unit of surface energy ($\sigma$) is Nm⁻¹.

Expression for surface energy

Or

Relation between surface energy and surface tension (T).

Fig: Soap film in a rectangular film

Let us consider a rectangular frame of War ABCD such that hat the side AB is movable is dipped completely in the soap solution of surface tension (T) and then taken out of solution. On doing so, a thin soap film is formed on the rectangular frame. Due to surface tension of soap film, it tries to occupy minimum surface area and hence, pulls AB towards left. Therefore,
$T = \frac{F}{L}$
Where, ‘L’ is the total length in contact with the free surface.
or, $T = \frac{F}{2l}$ Here, factor ‘2’ shows that soap film has two free surfaces.

Now,
Let us consider of force (F) is applied to side AB to displace it to a new position A’B’ through a small displacement dx.
From the definition of surface energy we have,
Surface energy $(\sigma) = \frac{\text{work done to increase surface area (w)}}{\text{Increased in surface area(AA)}}$
$\sigma = \frac{F.dx}{l.dx \times 2}$
From equation (i) we have,
$\sigma = \frac{T.2l.dx}{l.dx \times 2}$
$\sigma = T$
Which is the required expression for surface energy.
Above relation shows that surface energy of liquid is numerically equal to its surface tension.

Angle of contact:

It is defined as the angle made by the tangent drawn on the surface of liquid at the point of contact with solid surface inside the liquid.
It is denoted by ‘$\theta$’ and depends upon nature of liquid and surface of solid.

Capillarity (capillary action)

The rise or fall of liquid in tube of very fine bore is known as capillarity or capillary action.
It depends upon angle of contact.
i) If $\theta < 90°$, the liquid rises of in the capillary tube. ii) If $\theta > 90°$, the liquid falls down in the capillary tube.
iii) If $\theta = 90°$, the liquid neither rises not falls down.

Fig: (a) Angle of contact in water is acute angle (b) and obtuse in mercury

Expression for the rise of liquid in a capillary tube:

Fig: Rise of liquid in a capillary tube due to surface tension

Let us consider a capillary tube of sufficient height and very fine bore of radius ‘r’ is partially dipped in a liquid of surface tension ‘T’ and density ‘$\rho$’. On doing so, if the angle of contact is less than 90° the liquid rises in the capillary tube to a height at as shown in the figure above. Here ‘T’ is the surface tension drawn on liquid surface at the point of contact. The meniscus of liquid is concave upward. Here ‘R’ is the reaction of surface tension which has equal magnitude to that of surface tension. Resolving ‘T’ into its constituent components we get,
horizontal component = $T \sin \theta$ and vertical component = $T \cos \theta$

Fig: Horizontal components of T

Here all the horizontal components ($T \sin \theta$) being equal in magnitude and opposite in direction cancel each other, i.e. there is no net horizontal force.

Fig: Vertical components of T

Here all the vertical components ($T \cos \theta$) being equal in magnitude and same in direction do not cancel each other.
So, the net vertical upward force = $T \cos \theta \cdot 2\pi r$
Which is responsible for rise of liquid in a capillary tube.
In equilibrium condition,
Total downward force = total upward force
The total downward force is weight of the liquid raised in a capillary tube above the reference level.
i.e. $T \cos \theta \cdot 2\pi r = mg$
or, $T \cos \theta \cdot 2\pi r = V \rho g \dots (i)$
where V is the volume of liquid raised in the capillary tube
Now
Volume of liquid raised $V = \pi r^2 h + \pi r^2 r – \frac{2}{3} \pi r^3$
$= \pi r^2 h + \pi r^3 – \frac{2}{3} \pi r^3$
$= \pi r^2 h + \pi r^3 \left(1 – \frac{2}{3}\right)$
$= \pi r^2 h + \pi r^3 \left(\frac{1}{3}\right)$
$= \pi r^2 \left(h + \frac{r}{3}\right)$
Since r is very small, $\frac{r}{3}$ can be neglected.
So, $V = \pi r^2 h$
∴ Equation (i) becomes
$T \cos \theta \cdot 2\pi r = \pi r^2 h \rho g$
$h = \frac{2T \cos \theta}{r \rho g}$
which is the required expression for the rise of liquid in a capillary tube.

Surface Tension

Numerical

Type: 1

1. A capillary tube or 0.4 mm diameter is placed vertically inside a liquid of density 800 kgm⁻³, surface tension $5 \times 10^{-2}$ Nm⁻¹ and angle of contact 30°. Calculate the height to which liquid rises in the capillary tube.
   Ans: 0.054 m

Here,
d = 0.40 mm
r = 0.2mm = $0.2 \times 10^{-3}$ m
$\rho = 800 \text{ kg/m}^3$
$T = 5 \times 10^{-2} \text{ N/m}$
$\theta = 30°$
h = ?
We have,
$h = \frac{2T \cos \theta}{r \rho g}$
$h = \frac{2 \times 5 \times 10^{-2} \cos 30°}{0.2 \times 10^{-3} \times 800 \times 10}$
$h = 0.055 \text{ m}$
∴ The height to which the liquid raised in the capillary tube in 0.054m.

2. Angle of contact of mercury with glass is 135°. A narrow tube of glass having diameter 2 mm is dipped in a beaker containing mercury. By what height does the mercury go down in the tube relative to the level of mercury outside? [Density of mercury = 13600 kgm⁻³, Surface tension = 0.547 Nm⁻¹]
   Ans: 5.7 mm
3. A capillary tube of 0.3 m diameter is placed vertically inside a liquid of density 800 Kgm⁻³, surface tension $5 \times 10^{-4}$ Nm⁻¹ and angle of contact 30°. Calculate the height to which the liquid rises in the capillary tube
   Ans: $7.2 \times 10^{-1}$ m

Type: 2

4. A rectangular plate of dimensions 6cm by 4 cm and thickness 2 mm is placed with its largest face flat on the surface of water. Calculate the downward force on the plate due to surface tension assuming zero angle of contact. What is the downward force if the plate is placed vertical so that its longest side just touches the water?
   Ans: $1.4 \times 10^{-2}$ N, $8.92 \times 10^{-3}$ N

Solution:
Here, length (l) = 6 cm = 0.06 m
Breadth (b) = 4 cm = 0.04m
Thickness (t) = 2 mm = $2 \times 10^{-3}$m
$\theta = 0$
$T = 7.2 \times 10^{-2} \text{ N/m}$
For the largest surface,
$T = \frac{F}{L} = \frac{F}{2(l+b)} = \frac{F}{0.2}$
or, $F = T \times 0.2 = 0.0144\text{N}$
∴ The downward force on the largest face of the plate in 0.0144N.
For the longest side,
$L = 2 (l+t) = 0.1424\text{m}$
Again, $F = T \times L = (7.2 \times 10^{-2}) (0.124) = 8.928 \times 10^{-3}\text{N}$
∴ The downward force on the largest side of the plate in $8.928 \times 10^{-3}\text{N}$.

5. A rectangular plate of dimensions 6cm by 4 cm and thickness 2mm is placed with its largest face flat on the surface of water. Calculate the downward force on the plate due to the surface tension assuming zero, angle of contact. (Surface tension of water = $7 \times 10^{-2}$ Nm⁻¹)
   Ans: $1.4 \times 10^{-2}$ N
6. A rectangular plate of dimensions 6cm by 4 cm and thickness 2mm is placed vertical so that its largest side just touches the surface of the water. Calculate the downward force on the plate due to surface tension. (Surface tension of water = $7.0 \times 10^{-2}$ Nm⁻¹)
   Ans: $8.92 \times 10^{-3}$ N

Type:-3

7. Find the work done required to break up a drop of water of radius $5 \times 10^{-3}$ m into eight drops of water, assuming isothermal condition. [Surface tension of water = $72 \times 10^{-3}$ N/m]
   Ans: $2.26 \times 10^{-5}$ J
8. Calculate the work done in breaking a drop of water of 2mm diameter into million droplets, of same size. The surface tension of water is $72 \times 10^{-3}$ Nm⁻¹.
   Ans: $8.9 \times 10^{-5}$ J

Solution:
Here,
W = ?
d = 2mm = $2 \times 10^{-3}$m
r = $1 \times 10^{-3}$m
n = $1 \times 10^6$
$T = 72 \times 10^{-3}$ N/m
We know,
$\sigma = \frac{W}{\Delta A}$
or, $W = \sigma \times \Delta A$
or, $W = T \times \Delta A$
Since $\Delta A = 10^6 \times \text{surface area of smaller droplet} – \text{surface area of larger drop}$
or, $W = (72 \times 10^{-3}) (10^6 \times 4\pi r^2 – 4\pi R^2) \dots (i)$
Now,
volume of lager drop = $10^6 \times \text{volume of small droplets}$
$\frac{4}{3} \pi R^3 = 10^6 \left(\frac{4}{3} \pi r^3\right)$
Or, $R^3 = 10^6 r^3$
Or, $R = (10^6 r^3)^{1/3} = 10^2 r$
Or, $R = 100 r$
Or, $r = \frac{R}{100} = \frac{1 \times 10^{-3}}{100} = 10^{-5}\text{m}$
Equation (i) becomes,
$W = (72 \times 10^{-3}) [10^6 \times 4\pi(10^{-5})^2 – 4\pi(10^{-3})^2]$
$W = 8.95 \times 10^{-5}\text{J}$

9. Calculate the amount of energy evolved, when 8 droplets of water (Surface tension of water = 72 dyne cm⁻¹) of radius 0.5 mm each combine into one. Ans: 9.05erg
10. Calculate the force required to take away a flat plate of radius 5 cm from the surface of water. Given, (Surface tension of water = 72 dyne cm⁻¹) Ans: 2261.95dyne
11. What correction is to be made in a barometer reading if the barometer has a glass tube of internal diameter 3mm? [Surface tension of mercury = 0.545N/m, density of mercury = $13.6 \times 10^3$ kg/m³, angle of contact = 140°]
    Ans: 4.1 mm must be added to the observed value

Viscosity

Extra Numerical problems

17. A liquid is flowing through a horizontal pipeline of varying cross-section. At a certain cross-section, the diameter of a pipe is 5 cm and the velocity of flow of the liquid is 25 cm/s. Calculate the velocity of flow at another cross-section, where the diameter is 1 cm.
    Ans: 625cm/s.
18. The water entering a house flows with a speed of 0.1 m/s through a pipe of 21 mm in diameter. What is the speed of water at a point, where pipe tapers to a diameter of 7 mm?
    Ans: 0.9m/s
19. Air of density 1.3kg/m³ blows horizontally with a speed of 108km/hr. A house has a plane roof of area 40 m². Find the magnitude of aerodynamic lift on the roof.
    Ans: 23400N
20. Water flowing in a horizontal main of uniform bore has a velocity of 100 cm/s at appoint, where the pressure is 1/10th of the atmospheric pressure. What will be the velocity at a point, where the pressure is one half that at the first point?
    Ans: 333.16 cm/s

Viscosity

1. Viscosity is the property of
   (a) liquid only
   (b) gases only
   (c) solid only
   (d) liquid and gas both
   Ans: d
2. When a body acquires terminal velocity, which of the following quantity gets terminated
   (a) velocity
   (b) acceleration
   (c) momentum
   (d) energy
   Ans: b
3. If two ping pong balls are suspended near each other and a fast stream of air is produced within the space between balls. The balls,
   (a) come closer
   (b) move farther
   (c) remain at original position
   (d) fall down
   Ans: a
4. Bernoulli’s theorem is based on the principle of
   (a) conservation of charge
   (c) conservation of mass
   (b) conservation of momentum
   (d) conservation of energy
   Ans: d
5. On increasing the temperature, the viscosity of gas
   (a) increases
   (b) decreases
   (c) remain constant
   (d) none of the above
   Ans: a
6. On increasing the temperature, the viscosity of the liquid
   (a) increases
   (b) decreases
   (c) remains constant
   (d) none of the above
   Ans: b
7. Water is flowing at 12m/s in horizontal pipe. If the pipe widens to twice its original diameter, the flow speed in the wider section is
   (a) 6m/s
   (b) 9 m/s
   (c) 2 m/s
   (d) 3 m/s
   Ans: d
8. Water falls along a horizontal pipe of cross-sectional area 48cm² which has a constriction of cross-sectional area 12 cm² at one place. If the speed of water at constriction is 8m/s then the speed at wider part is
   (a) 1 m/s
   (b) 2 m/s
   (c) 3 m/s
   (d) 4 m/s
   Ans: b
9. The terminal velocity of a drop of water is 2 m/s. If 27 such droplets coalesce then the terminal velocity of single drop will be
   (a) 2 cm/s
   (b) 6 cm/s
   (c) 12 cm/s
   (d) 18 cm/s
   Ans: d
10. The difference in pressure between two ends of pipe of 20 m long and 5 mm radius if the flow of water is 800 cc/s (viscosity of water = $10^{-3}$ Nsm⁻²).
    (a) $3.5 \times 10^4$ Nm⁻²
    (b) $6.5 \times 10^4$ Nm⁻²
    (c) $6.5 \times 10^5$ Nm⁻²
    (d) $13 \times 10^4$ Nm⁻²
    Ans: b
11. A rain drop of radius 0.3mm has terminal velocity 1 m/s in air. The viscosity of air is $18 \times 10^{-5}$ Poise then viscous force is
    (a) $1.01 \times 10^{-7}$ N
    (b) $1.01 \times 10^{-8}$ N
    (c) $1.6 \times 10^{-9}$ N
    (d) $1.6 \times 10^{-8}$ N
    Ans: a
12. A shower head has 20 circular openings, each with radius 1.0mm. The shower head is connected to a pipe with radius 0.8 cm. If the speed of water in the pipe is 3.0 m/s, what is its speed as it exists the shower head openings?
    (a) 60 m/s
    (b) 36 m/s
    (c) 24 m/s
    (d) 9.6 m/s
    Ans: d