NEET Physics Class 11 Chapter 5 Fluid Mechanics Notes

Fluid Mechanics Definition Of Fluid

The term fluid refers to a substance that can flow and does not have a shape of its own. For example liquids and gases.

Fluid includes properties → (1) Density (2) Viscosity (3) Bulk modulus of elasticity (4) pressure (5) specific gravity

Pressure In A Fluid

The pressure p is defined as the magnitude of the normal force acting on a unit surface area.

P = \(\frac{\Delta F}{\Delta A}\)

ΔF = normal force on a surface area ΔA.

The pressure is a scalar quantity. This is because hydrostatic pressure is transmitted equally in all directions when force is applied, which shows that a definite direction is not associated with pressure.

Thrust. The total force exerted by a liquid on any surface in contact with it is called the thrust of the liquid.

Note:
The normal force exerted by liquid at rest on a given surface in contact with it is called the thrust of liquid on that surface.

The normal force (or thrust) exerted by liquid at rest per unit area of the surface in contact with it, is called pressure of liquid or hydrostatic pressure.

If F is the normal force acting on a surface of area A in contact with liquid, then the pressure exerted by the liquid on this surface is P = F/A

1. Units : N / m² or Pascal (S.I) and Dyne/cm²(C.G.S)
2. Dimension : (P) = \(\frac{[\mathrm{F}]}{[\mathrm{A}]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)
3. At a point pressure acts in all directions and a definite direction is not associated with it. So pressure is a tensor quantity.
4. Atmospheric pressure: The gaseous envelope surrounding the earth is called the earth’s atmosphere and the pressure exerted by the atmosphere is called atmospheric pressure its value on the surface of the earth at sea level is nearly 1.013 × 10⁵ N/m² or Pascal in S.I. Other practical units of pressure are atmosphere, bar and torr (mm of Hg) 1 atm = 1.01 × 10⁵ Pa = 1.01 bar = 760 torr.
   1. The atmospheric pressure is maximum at the surface of the earth and goes on decreasing as we move up into the earth’s
5. If P₀ is the atmospheric pressure then for a point at depth h below the surface of a liquid of density ρ. hydrostatic pressure P is given by P = P₀+ hρg.

6. Hydrostatic pressure depends on the depth of the point below the surface (h). nature of liquid (ρ) and acceleration due to gravity (g) while it is independent of the amount of liquid, the shape of the container, or the cross-sectional area considered.

• So if a given liquid is filled in vessels of different shapes to the same height, the pressure at the base in each vessel will be the same, though the volume or weight of the liquid in different vessels will be different.

7. In a liquid at the same level, the pressure will be the same at all points, if not, due to the pressure difference the liquid cannot be at rest. This is why the height of the 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.

Gauge pressure: The pressure difference between hydrostatic pressure P and atmospheric pressure P0 is called gauge pressure.

P – P₀= hρg

Consequences Of Pressure

Railway tracks are laid on large-sized wooden or iron sleepers. This is because the weight (force) of the train is spread over a large area of the sleeper.

1. This reduces the pressure acting on the ground and hence prevents the yielding of ground under the weight of the train.
2. A sharp knife is more effective in cutting objects than a blunt knife. The pressure exerted = Force/area. The sharp knife transmits force over a small area as compared to the blunt knife. Hence the pressure exerted in the case of a sharp knife is more than in the case of a blunt knife.
3. A camel walks easily on sand but a man cannot in spite of the fact that a camel is much heavier than a man.

This is because the area of camel’s feet is large as compared to man’s feet. So the pressure exerted by the camel on the sand is very small as compared to the pressure exerted by man. Due to large pressure, sand under the feet of man yields, and hence he cannot walk easily on sand.

Variation Of Pressure With Height

Assumptions:

1. Unaccelerated liquid
2. Uniform density of liquid
3. Uniform gravity

The weight of the small element dh is balanced by the excess pressure. It means

⇒ \(\frac{d p}{d h}=\rho g\)

⇒ \(\int_{P_a}^P d p=\rho g \int_0^h d h\)

⇒ P = Pa + ρgh

Pascal’S Law

if the pressure in a liquid is changed at a particular, point the change is transmitted to the entire liquid without being diminished in magnitude. In the above case if P_a is increased by some amount then P must increase to maintain the difference (P – P_a) = hρg. This is Pascal’s Law which states that Hydraulic lift is a common application of Pascal’s Law.

1. Hydraulic press.

⇒ \(p=\frac{f}{a}=\frac{W}{A} \text { or } f=\frac{W}{A} \times a\)

as A >> a then f<<W…

This can be used to lift a heavy load placed on the platform of a larger piston or to press the things placed between the piston and the heavy platform. The work done by applied force is equal to the change in the potential energy of the weight in a hydraulic press.

Density: In a fluid, at a point, density ρ is defined as :

⇒ \(\rho=\lim _{\Delta V \rightarrow 0} \frac{\Delta m}{\Delta V}=\frac{d m}{d V}\)

1. In the case of a homogenous isotropic substance, it has no directional properties, so is a scalar.
2. It has dimensions (ML–and S.I. unit kg/m³ while C.G.S. unit g/cc with 1g /cc = 10³ kg/m³
3. The density of a substance means the ratio of the mass of a substance to the means the ratio of mass of a body to the volume of the body. So for a solid body. Density of body = Density of substance While for a hollow body, the density of the body is lesser than that of substance [As V_body > V_sub.]
4. When immiscible liquids of different densities are poured into a container, the liquid of the highest density will be at the bottom while that of the lowest density at the top, and interfaces will be plane.
5. Sometimes instead of density, we use the term relative density or specific gravity which is defined as:

⇒ \(\mathrm{RD}=\frac{\text { Density of body }}{\text { Density of water }}\)

6. If m₁ mass of liquid of density ρ₁ and m₂ mass of density ρ₂ are mixed. then as
m = m₁+ m₂ and V = (m₁/ ρ+ (m₂/ ρ₂)

[As V = m/ρ]

7. If  V₁ volume of liquid of density ρ₁ and V₂ volume of liquid of density ρ2are mixed, then as m = ρ₁V₁+ ρ₂V₂ and V = V₁+ V₂[As ρ = m/V]

If V₁= V₂= V ρ₁(ρ₁ + ρ₂)/2 = Arithmetic Mean

Question 1. A body of one kg is placed on two objects of negligible mass. Calculate pressure due to force on its bottom.

Answer:

⇒ \(P=\frac{F}{A}=\frac{m g}{A}\)

⇒ \(P_1=\frac{1 \times 10 \mathrm{~N}}{10 \times 10^{-2} \mathrm{~m}^2}=10^4 \mathrm{~N} / \mathrm{m}^2\)

⇒ \(P_2=\frac{1 \times 10 \mathrm{~N}}{2 \times 10^{-4}}=5 \times 10^4\left(=5 p_1\right)\)

Question 2. For a hydraulic system, A car of mass 2000 kg standing on the platform of Area 10m² while the area other side platform 10 cm² finds the mass required to balance the car
Answer:

According to the Pascal’s Law

P₁ = P₂ \(\frac{m_{\text {car }} g}{A_{\text {car }}}=\frac{\mathrm{mg}}{\mathrm{A}}\)

⇒ \(\mathrm{m}=\left(\frac{\mathrm{A}}{\mathrm{A}_{\text {car }}}\right) \times \mathrm{m}_{\text {car }}=\frac{10 \mathrm{~cm}^2}{10 \mathrm{~m}^2} \times 2000 \mathrm{~kg}\)

= 0.2 kg

= 200 gm

Question 3. If two liquids of the same masses but densities of P₁ and P₂ respectively are mixed, then the density of the mixture is given by

1. \(\rho=\frac{\rho_1+\rho_2}{2}\)
2. \(\rho=\frac{\rho_1+\rho_2}{2 \rho_1 \rho_2}\)
3. \(\rho=\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2}\)
4. \(\rho=\frac{\rho_1 \rho_2}{\rho_1+\rho_2}\)

Answer:

ρ = \(\frac{\text { Total mass }}{\text { Totalvolume }}\)

⇒ \(\frac{2 m}{V_1+V_2}=\frac{2 m}{m\left(\frac{1}{\rho_1+\rho_2}\right)}\)

∴ \(\rho=\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2}\)

Question 4. If two liquids of the same masses but different densities ρ₁ and ρ₂ are mixed, then the density of the mixture is given by

1. \(\rho=\frac{\rho_1+\rho_2}{2}\)
2. \(\rho=\frac{\rho_1+\rho_2}{2 \rho_1 \rho_2}\)
3. \(\rho=\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2}\)
4. \(\rho=\frac{\rho_1 \rho_2}{\rho_1+\rho_2}\)

Answer:

ρ = \(\frac{\text { Total mass }}{\text { Total volume }}\)

⇒ \(\frac{m_1+m_2}{2 V} \frac{V\left(\rho_1+\rho_2\right)}{2 V}\)

⇒ \(\frac{\rho_1+\rho_2}{2}\)

Question 5. If pressure at the half depth of a lake is equal to 2/3 of pressure at the bottom of the lake, then the depth of the lake [ρ_water = 10³ kg m^-3, P₁= 10⁵ N/m²]

1. 10 m
2. 20 m
3. 60 m
4. 30 m

Answer: Pressure at the bottom of the lake = P₀+hpg

Pressure at half the depth of a lake = \(P_0+h \rho g\)

According to given condition \(P_0+\frac{1}{2} h \rho g=\frac{2}{3}\left(P_0+h \rho g\right)\)

⇒ \(\frac{1}{3} P_0=\frac{1}{6} h \rho g\)

⇒ \(h=\frac{2 P_o}{\rho g}=\frac{2 \times 10^5}{10^3 \times 10}=20 \mathrm{~m}\)

Question 6. A uniform tapering vessel is filled with a liquid of uniform density 900 kg/m³. The force that acts on the base of the vessel due to the liquid is (g = 10 ms^-2)

1. 3.6 N
2. 7.2 N
3. 9.0 N
4. 14.4 N

Answer: Force acting on the base

F = P × A = hdgA = 0.4 × 900 × 10 × 2 × 10^-3 = 7.2N

Question 7. The area of the cross-section of the two arms of a hydraulic press is 1 cm² and 10 cm² respectively (figure). A force of 5 N is applied to the water in the thinner arm. What force should be applied to the water in the thicker arms so that the water may remain in equilibrium?

Answer:

In equilibrium, the pressures at the two surfaces should be equal as they lie in the same horizontal level. If the atmospheric pressure is P and a force F is applied to maintain the equilibrium, the pressures are

⇒ \(P_0+\frac{5 \mathrm{~N}}{1 \mathrm{~cm}^2}\) and \([P_0+\frac{F}{10 \mathrm{~cm}^2}\) respectively. This gives F = 50 N.

Hydraulic Brake: Hydraulic brake system is used in automobiles to retard the motion.

Hydrostatic Paradox

Pressure is directly proportional to depth and by applying Pascal’s law it can be seen that pressure is independent of the size and shape of the containing vessel. (In all three cases the heights are the same).

P_A= P_B= P_C

Atmospheric Pressure

Atmospheric Pressure Definition: The atmospheric pressure at any point is numerically equal to the weight of a column of air of a unit cross-sectional area extending from that point to the top of the atmosphere.

At 0ºC, the density of mercury = 13.595 g cm^-3, and at sea level, g = 980.66 cm s^-2

Now P = hρg.

Atmospheric pressure = 76 × 13.595 × 980.66 dyne cm^-2 = 1.013 × 10^-5 N-m²(p_a)

Height of Atmosphere

The standard atmospheric pressure is 1.013 × 10⁵ Pa (N m^-2). If the atmosphere of the earth has a uniform density ρ = 1.30 kg m^-3, then the height h of the air column which exerts the standard atmospheric pressure is given by

⇒ hρg = 1.013 × 10⁵

h = \(\frac{1.013 \times 10^5}{\rho g}\)

⇒ \(\frac{1.013 \times 10^5}{1.13 \times 9.8}\mathrm{~m}\)

m = 7.95 × 10³ m ~ 8 km. 1.13 9.8

In fact, the density of air is not constant but decreases with height. The density becomes half at about 6 km high,\(\frac{1}{4}\)th at about 12 km, and so on.

Therefore, we can not draw a clear-cut line above which there is no atmosphere. Anyhow the atmosphere extends upto 1200 km. This limit is considered for all practical purposes.

Measurement Of Atmospheric Pressure

1. Mercury Barometer.

To measure the atmospheric pressure experimentally, Torricelli invented a mercury barometer in 1643.

p_a= hρg

The pressure exerted by a mercury column of 1mm

high is called 1 Torr.

1 Torr = 1 mm of mercury column

2. Open tube Manometer

An open-tube manometer is used to measure the pressure gauge. When equilibrium is reached, the pressure at the bottom of a left limb is equal to the pressure at the bottom of the right limb.

i.e. p + y₁ ρg = p_a+ y₂ ρg

p – p_a= ρg (y₂– y₁)= ρgy

p – p_a= ρg (y₂– y₁)= ρgy

p = absolute pressure, p – p_a= gauge pressure.

Thus, knowing y and ρ (density of liquid), we can measure the gauge pressure.

Question 1. A barometer tube reads 76 cm of mercury. If the tube is gradually inclined at an angle of 60° vertically, keeping the open end immersed in the mercury reservoir, the length of the mercury column will be

1. 152 cm
2. 76 cm
3. 38 cm
4. \(38 \sqrt{3} \mathrm{~cm}\)

Answer: cos 60º = \(\frac{\mathrm{h}}{\ell}\)

⇒ \(\ell=\frac{\mathrm{h}}{\cos 60^{\circ}}=\frac{76}{1 / 2}\)

∴ l = 152 cm

Question 2. When a large bubble rises from the bottom of a lake to the surface. Its radius doubles. If atmospheric pressure is equal to that of a column of water height H, then the depth of the lake is

1. H
2. 2H
3. 7H
4. 8H

Answer:

P₁V₁= P₂V₂

⇒ \(\left(P_{\circ}+h \rho g\right) \times \frac{4}{3} \pi r^2=P_0 \times \frac{4}{3} \pi(2 r)^3\)

Question 3. A beaker containing liquid is kept inside a big closed jar If the air inside the jar is continuously pumped out, the pressure in the liquid near the bottom of the liquid will

1. Increase
2. Decreases
3. Remain constant
4. First decrease and then increase

Answer: Total pressure at (near) bottom of the liquid

P = P₀+ hρg

As air is continuously pumped out from the jar (container), P0 decreases and hence P decreases.

Question 4. Write the pressure inside the tube

Answer:

⇒ \(P_A=P_0+\rho g \frac{8}{100}=P_B\)…(i)

⇒ \(P_{\text {tube }}=P_B-\rho g \frac{6}{100}\)…

(1) and (2)

⇒ \(P_{\text {tube }}=\left(P_0+\rho g \frac{8}{100}\right)-\rho g \frac{6}{100}\)

⇒ \(P_{\text {tube }}=\left(P_0+\frac{\rho g}{50}\right)\)

⇒ \(P_0+\frac{\rho g}{50}=P\)

Question 5. Find the pressure inside the tube

Answer:

⇒ \(P_A=\left(P_0+\rho g h\right)+3 \rho g(2 h)=P_B\)

⇒ \(P_{\text {tube }}=P_B-8 \rho g\left(\frac{2 h}{3}-\frac{h}{4}\right)\)

⇒ \(P_0+\rho g h+6 \rho g h-8 \rho g\left(\frac{5}{12} h\right)\)

⇒ \(P_0+\rho g h+6 \rho g h-\frac{10}{3} \rho g h\)

⇒ \(P_0+\frac{(21-10) g g h}{3}=P_0+\frac{11}{3} \rho g h\)

⇒ \(P_0+\frac{11}{3} \rho g h\)

Question 6. The manometer shown below is used to measure the difference in water level between the two tanks. Calculate this difference for the conditions indicated.

Answer:

p_a+ h₁ρg – 40ρ₁g + 40ρg = p_a+ h₂ρg

h₂ρg – h₁ρg = 40 ρg – 40 ρ₁g

as ρ₁= 0.9ρ

(h₂– h₁)ρg = 40ρg – 36ρg

h₂– h₁= 4 cm

3. Water Barometer.

Let us suppose water is used in the barometer instead of mercury.

hρg = 1.013 × 10⁵ or h = \(\mathrm{h}=\frac{1.013 \times 10^5}{\rho \mathrm{g}}\)

The height of the water column in the tube will be 10.3 m. Such a long tube cannot be managed easily, thus, a water barometer is not feasible.

Question 7. In a given U-tube (open at one end) find out the relation between p and p_a. Given d₂= 2 × 13.6 gm/cm³ d₁= 13.6 gm/cm³

Answer: Pressure in a liquid at the same level is the same i.e. at A – A–, p_a+d₂yg+xd₁g = p

In C.G.S.

p_a+ 13.6 × 2 × 25 × g + 13.6 × 26 × g = p

p_a+ 13.6 × g [50 + 26] = p

2p_a= p   [p_a= 13.6 × g × 76]

Question 8. The truck starts from rest with an acceleration of 2.5 ms^-2 then the angle (acute) between the vertical and surface at the liquid, In equilibrium (assume that liquid is at with respect to the truck)

1. sinθ = \(\frac{4}{\sqrt{17}}\)
2. cosθ = \(\frac{1}{\sqrt{17}}\)
3. tanθ = 4
4. None of these

Answer :

Consider a particle on the liquid surface

mg cos θ = ma cos θ

gcosθ = a sinθ

tanθ = \(\frac{\mathrm{g}}{\mathrm{a}}\)

tanθ = \(\frac{10}{2.5}\) = 4

ABC

Question 9. In the previous question pressure at points A, B, and C

1. P_A = P_B = P_C
2. P_A > P_B > P_C
3. P_A < P_B < P_C
4. None of these

Answer: 2. P_A > P_B > P_C

Question 10. In the previous question, three different points, above the points A, B, and C of an accelerated liquid surface in equilibrium called A’ B’, C’ then the pressure at the points A’ B’ and C’

1. P_A‘ = P_B‘ = P_C‘
2. P_A‘ > P_B‘ > P_C‘
3. P_A‘ < P_B‘ < P_C‘
4. Po = atmosphere pressure

Answer: 2. P_A‘ > P_B‘ > P_C‘

Question 11. Highest pressure at the point inside the liquid :

1. A
2. C
3. Pressure at A, B, and C are equal, and the highest
4. None of these

Answer: 1. A

Question 12. The slope of the line on which pressure is the same considers the direction of acceleration of the truck as the X-axis

1. –4
2. –0.25
3. –2.5
4. \(-\frac{1}{4}\)

Answer: 4. \(-\frac{1}{4}\)

Archimedes’ Principle

According to this principle, when a body is immersed wholly or partially in a fluid, it loses its weight which is equal to the weight of the fluid displaced by the body.

Up thrust = buoyancy = Vρ_lg

V = volume submerged

ρ_l= density of liquid.

Relation between the density of solid and liquid

weight of the floating solid = weight of the liquid displaced

V₁ρ₁g = V₂ρ₂ g

⇒ \(\frac{\rho_1}{\rho_2}=\frac{V_2}{V_1}\)

or \(\frac{\text { Density of solid }}{\text { Density of liquid }}=\frac{\text { Volume of the immersed portion of the solid }}{\text { Total Volume of the solid }}\)

This relationship is valid in accelerating fluid also. Thus, the forces acting on the body are :

1. Its weight is Mg which acts downward and
2. Net upward thrust on the body or the buoyant force (mg)

Hence the apparent weight of the body = Mg – mg = weight of the body – weight of the displaced liquid. Or Actual Weight of body – Apparent weight of body = weight of the liquid displaced.

• The point through which the upward thrust or the buoyant force acts when the body is immersed in the liquid is called its center of buoyancy. This will coincide with the center of gravity if the solid body is homogeneous.
• On the other hand, if the body is not homogeneous, then the center of gravity may not lie on the line of the upward thrust and hence there may be a torque that causes rotation in the body.
• If the center of gravity of the body and the center of buoyancy lie on the same straight line, the body is in equilibrium.
• If the center of gravity of the body does not coincide with the center of buoyancy (i.e., the line of upthrust), then torque acts on the body. This torque causes the rotational motion of the body.

Floatation

1. Translatory equilibrium: When a body of density p and volume V is immersed in a liquid of density σ, the forces acting on the body are

Weight of body W = mg = Vρg, acting vertically downwards through the center of gravity of the body. Upthrust force = Vσg acting vertically upwards through the center of gravity of the displaced liquid i.e., the center of buoyancy.

If the density is greater than that of liquid p>σ

Weight will be more than upthrust so the body will sink

If density is equal to that of liquid p>σ

Weight will be equal to upthrust so the body will float fully submerged in neutral equilibrium with its top surface just at the top of a liquid

If the density of the body is lesser than that of liquid p>σ

Weight will be less than upthrust so the body will, move upwards and in equilibrium will float and be partially immersed in the liquid such that,

W = V_in σg

⇒ Vρg = V_in σg

Vρ = V_in σ

Where V_in is the volume of a body in the liquid

1. A body will float in liquid only and only if ρ ≤ σ
2. In the case of floating the weight of the body = upthrust
3. So W_App = Actual weight – upthrust = 0
4. In the case of floating Vρg = V_inσg

So the equilibrium of floating bodies is unaffected by variations in g though both thrust and weight depend on g.

Rotatory Equilibrium: When a floating body is slightly tied from the equilibrium position, the center of buoyancy B shifts. The vertical line passing through the new center of buoyancy B’ and the initial vertical line meet at a point M called meta-center.

If the meta-center M is above the center of gravity the couple due to forces at G (weight of body W) and at B’ (upthrust) tends to bring the body back body the meta the center must always be higher than the center of gravity of the body.

• However, if the meta-center goes CG, the couple due to forces at G and B’ tends to topple the floating body.
• That is why a wooden log cannot be made to float vertically in water or a boat is likely to capsize if the sitting passengers stand on it. In these situations, CG becomes higher than MG and so the body will topple if slightly tilted.

Question 1. A concrete sphere of radius R has a cavity of radius r which is packed with sawdust. The specific gravities of concrete and sawdust are respectively 2.4 and 0.3 for this sphere to float with its entire volume submerged under water. A ratio of the mass of concrete to the mass of sawdust will be

1. 8
2. 4
3. 3
4. Zero

Answer: Let specific gravities of concrete and sawdust be ρ₁ and ρ₂ respectively.

According to the principle of floatation weight of the whole sphere = upthrust on the sphere

⇒ \(\frac{4}{5} \pi\left(R^3-r^3\right) \rho_1 g+\frac{4}{3} \pi r^3 \rho_2 g=\frac{4}{3} \pi R^3 \times 1 \times g\)

⇒ \(R^3 \rho_1-r^3 \rho_1+r^3 \rho_2=R^3\)

⇒ \(R^3\left(\rho_1-1\right)=r^3\left(\rho_1-\rho_2\right)=\frac{R^3}{r^3}=\frac{\rho_1-\rho_2}{\rho_1-1}\)

⇒ \(\frac{R^3-r^3}{r^3}=\frac{\rho_1-\rho_2-\rho_1+1}{\rho_1-1}\)

⇒ \(\frac{\left(R^3-r^3\right) \rho_1}{r^3 \rho_2}=\left(\frac{1-\rho_2}{\rho_1-1}\right) \frac{\rho_1}{\rho_2}\)

⇒ \(\frac{\text { Mass of concrete }}{\text { Mass of saw dust }}\)

⇒ \(\left(\frac{1-0.3}{2.4-1}\right) \times \frac{2.4}{0.3}=4\)

Question 2. A metallic block of density 5 gm cm^-3 and having dimensions 5 cm × 5 cm × 5cm is weighed in water. Its apparent weight will be

1. 5 × 5× 5 × 5 gf
2. 4 × 4 × 4 × 4 gf
3. 5 × 4× 4 × 4 gf
4. 4 × 5× 5 × 5 gf

Answer: Apparent weight

= V (ρ – σ) g = 1 × b × h × (5 –1) × g

= 5 × 5 × 5 × 4 × g

Dyne = 4 × 5 × 5 × 5 gf

Question 3. A cubical block is floating in a liquid with half of its volume immersed in the liquid. When the whole system accelerates upwards with an acceleration of g/3, the fraction of volume immersed in the liquid will be

1. \(\frac{1}{2}\)
2. \(\frac{3}{8}\)
3. \(\frac{2}{3}\)
4. \(\frac{3}{4}\)

Answer: Fraction of volume immersed in the liquid \(V_{\text {in }}=\left(\frac{\rho}{\sigma}\right) V\)

i.e. it depends upon the densities of the block and liquid. So there will be no change in it if the system moves upward or downward with constant velocity or some acceleration.

Question 4. A silver ingot weighing 2.1 kg is held by a string so as to be completely immersed in a liquid of relative density 0.8. The relative density of silver is 10.5. The tension in the string in kg-wt is

1. 1.6
2. 1.94
3. 3.1
4. 5.25

Answer: T = Apparent weight = V(ρ – σ) g = \(\frac{M}{\rho}(\rho-\sigma) g\)

T = \(M\left(1-\frac{\sigma}{\rho}\right) g=2.1\left(1-\frac{0.8}{10.5}\right) g=1.94 \mathrm{gN}\)

T = 1.94 Kg-wt

Question 5. A sample of metal weighs 210 gm in air, 180 gm in water and 120 gm in liquid. The density (RD) of

1. Metal is 3
2. Metal is 7
3. Liquid is 3
4. Liquid is \(\frac{1}{3}\)

Answer: Density of metal = ρ. Density of liquid = σ

If V is the volume of the sample then according to the problem

210 = Vρg ……(1)

180 = V (ρ – 1)g ……(2)

120 = V (ρ – σ)g ……(3)

By solving (1),(2) and (3) we get ρ = 7 and σ = 3.

Question 6. A cubical block of wood of edge 3 cm floats in water. The lower surface of the cube just touches the free end of a vertical spring fixed at the bottom of the pot. Find the maximum weight that can be put on the block without wetting it. The density of wood = 800 kg/m³ and the spring constant of the spring = 50 N/m. Take g = 10 m/s².

Answer:

The specific gravity of the block = 0.8. Hence the height inside water = 3 cm × 0.8 = 2.4 cm. The height outside after = 3 cm – 2.4 = 0.6 cm. Suppose the maximum weight that can be put without wetting it is W. The block in this case is completely immersed in the water. The volume of the displaced water

= volume of the block = 27 × 10^-6 m³.

Hence, the force of buoyancy

= (27 × 10^-6 m³)× 1(1000 kg/m³)× (10 m/s²)= 0.27 N.

The spring is compressed by 0.6 cm and hence the upward force exerted by the spring = 50 N/m × 0.6 cm = 0.3 N.

The force of buoyancy and the spring force taken together balance the weight of the block plus the weight W put on the block. The weight of the block is

W′ = (27 × 10^-6 m) × (800 kg/m× (10 m/s= 0.22 N.

Thus, W = 0.27 N + 0.3 N – 0.22 N

= 0.35 N.

Pressure In Case Of Accelerating Fluid

Liquid Placed In Elevator:

When the elevator accelerates upward with acceleration a0 then the pressure in the fluid, at depth ‘h’ may be given by,

p = hρ [g + a₀]

and force of buoyancy, B = m (g + a₀)

The free surface of a liquid in horizontal acceleration :

tan θ = \(\frac{a_0}{\mathrm{~g}}\)

p₁– p₂= lρ a₀ where p₁ and p₂ are pressures at point 1 and 2. Then h₁– h₂ 

= \(\frac{\ell \mathrm{a}_0}{\mathrm{~g}}\)

Question 1. An open rectangular tank 1.5 m wide 2m deep and 2m long is half filled with water. It is accelerated horizontally at 3.27 m/sec² in the direction of its length. Determine the depth of water at each end of the tank. [g = 9.81 m/sec²]

Answer: tan θ = \(\frac{\mathrm{a}}{\mathrm{g}}=\frac{1}{3}\)

Depth at the corner ‘A’

= 1 – 1.5 tanθ

= 0.5 m

Depth at corner ‘B’

= 1 + 1.5 tan θ = 1.5 m

The free surface of a liquid in the case of a rotating cylinder.

h = \(\frac{v^2}{2 g}=\frac{\omega^2 r^2}{2 g}\)

Streamline Flow

The path taken by a particle in flowing fluid is called its line of flow. In the case of steady flow, all the particles passing through a given point follow the same path and hence we have a unique line of flow passing through a given point which is also called streamline.

Characteristics Of Streamline

1. A tangent at any point on the streamline gives the direction of the velocity of the fluid particle at that point.

2. Two streamlines never intersect each other.

1. Laminar Flow: If the liquid flows over a horizontal surface in the form of layers of different velocities, then the flow of liquid is called Laminar flow. The particles of one layer do not go to another layer. In general, Laminar flow is a streamlined flow.
2. Turbulent Flow: The flow of fluid in which the velocity of all particles crossing a given point is not the same and the motion of the fluid becomes disorderly or irregular is called turbulent flow.

Reynold’S Number

According to Reynold, the critical velocity (vc) of a liquid flowing through a long narrow tube is

1. Directly proportional to the coefficient of viscosity (η) of the liquid.
2. Inversely proportional to the density ρ of the liquid and
3. Inversely proportional to the diameter (D) of the tube.

That is \(v_c \propto \frac{\eta}{\rho D} \quad \text { or } \quad v_c=\frac{R \eta}{\rho D} \quad \text { or }=\frac{v_c \rho D}{\eta}\) ……………(1)

where R is the Reynold number.

If R < 2000, the flow of liquid is streamlined or laminar. If R > 3000, the flow is turbulent. If R lies between 2000 and 3000, the flow is unstable and may change from streamlined flow to turbulent flow.

Equation Of Continuity

The equation of continuity expresses the law of conservation of mass in fluid dynamics.

a₁v₁ = a₂v₂

In general av = constant. This is called an equation of continuity and states that as the area of the cross-section of the tube of flow becomes larger, the liquid’s (fluid) speed becomes smaller and vice-versa.

Illustrations –

1. The velocity of the liquid is greater in the narrow tube as compared to the velocity of the liquid in a broader tube.
2. Deep waters run slowly can be explained by the equation of continuity i.e., av = constant. Where water is deep the area of cross-section increases hence velocity decreases.

Energy Of A Liquid

A liquid can possess three types of energies :

Kinetic Energy: The energy possessed by a liquid due to its motion is called kinetic energy. The kinetic energy of a liquid of mass m moving with speed v is \(\frac{1}{2} m v^2\)

∴ K.E. per unit mass = \(\frac{\frac{1}{2} m v^2}{m}=\frac{1}{2} v^2\)

Potential Energy: The potential energy of a liquid of mass m at a height h is m g h.

∴ P.E. per unit mass = \(\frac{\mathrm{mgh}}{\mathrm{m}}=\mathrm{gh}\)

Pressure Energy: The energy possessed by a liquid by virtue of its pressure is called pressure energy. Consider a vessel fitted with a piston at one side (figure).

• Let this vessel is filled with a liquid. Let ‘A’ be the area of the cross-section of the piston and P be the pressure experienced by the liquid. The force acting on the piston = PA
• If dx is the distance moved by the piston, then work done by the force = PA dx = PdV where dV = Adx, the volume of the liquid swept.

This work is equal to the pressure energy of the liquid.

∴ Pressure energy of liquid in volume dV = PdV.

The mass of the liquid having volume dV = ρdV,

ρ is the density of the liquid.

∴ Pressure energy per unit mass of the liquid = \(\frac{P d V}{\rho d V}=\frac{P}{\rho}\)

Bernoulli’s Theorem

It states that the sum of pressure energy, kinetic energy, and potential energy per unit mass or per unit volume or per unit weight is always constant for an ideal (i.e. incompressible and non-viscous) fluid having stream-line flow.

i.e. \(\frac{P}{\rho}+\frac{1}{2} v^2+g h\)= constant.

Question 1. An incompressible liquid flows through a horizontal tube as shown in the following fig. Then the velocity υ of the fluid is

1. v = 2v₁ + v₂
2. v = v₁+ v₂
3. \(v=\frac{v_1 v_2}{v_1+v_2}\)
4. \(\mathrm{v}=\sqrt{\mathrm{v}_1^2+\mathrm{v}_2^2}\)

Answer: 1. v = 2v₁+ v₂

m = m₁+ m₂

ρV = ρV₁+ ρV₂

ρAv = ρ2Av₁+ ρAv₂

v = 2v₁+ v₂

Question 2. Water enters through end A with speed υ₁ and leaves through end B with speed υ₂ of a cylindrical tube AB. The tube is always completely filled with water. In case I it is horizontal and in case II it is vertical with end A upwards and in case III it is vertical with end B upwards. We have υ₁= υ₂

1. Case 1
2. Case 2
3. Case 3
4. Each case

Answer: This happens in accordance with an equation of continuity and this equation was derived on the principle of conservation of mass and it is true in every case, either tube remains horizontal or vertical.

Question 3. Water flows in a horizontal tube as shown in the figure. The pressure of water changes by 600 N/m² between A and B where the areas of cross-section are 30cm² and 15cm² respectively. Find the rate of flow of water through the tube.

Answer:

Let the velocity at A = v_A and that at B = v_B.

By the equation of continuity, \(\frac{v_B}{v_A}=\frac{30 \mathrm{~cm}^2}{15 \mathrm{~cm}^2}=2\)

By Bernoulli’s equation,

⇒ \(P_A+\frac{1}{2} \rho v_A^2=P_B+\frac{1}{2} \rho v_B^2\)

or \(P_A-P_B=\frac{1}{2} \rho\left(2 v_A\right)^2-\frac{1}{2} \rho v_A^2=\frac{3}{2} \rho v_A^2\)

or \(600 \frac{\mathrm{N}}{\mathrm{m}^2}=\frac{3}{2}\left(1000 \frac{\mathrm{kg}}{\mathrm{m}^3}\right) \mathrm{v}_{\mathrm{A}}^2\)

or \(v_{\mathrm{A}}=\sqrt{0.4 \mathrm{~m}^2 / \mathrm{s}^2}\)

=0.63 m/s

The rate of flow = (30 cm²)(0.63 m/s) = 1800 cm³/s.

Application Of Bernoulli’s Theorem

1. Bunsen burner
2. Lift of an airfoil.
3. Spinning of a ball (Magnus effect)
4. The sprayer.
5. A ping-pong ball in an air jet
6. Torricelli’s theorem (speed of efflux)

At point A, P₁= P, v₁= 0 and h₁= h

At point B, P₂= P, v₂= v (speed of efflux) and h = 0

Using Bernoulli’s theorem \(\frac{P_1}{\rho}+g h_1+\frac{1}{2} v_1^2=\frac{P_2}{\rho}+g h_2=\frac{1}{2} v_2^2\)

we have \(\frac{P}{\rho}+g h+0=\frac{P}{\rho}+0+\frac{1}{2} v^2\)

⇒ \(\frac{1}{2} v^2=g h \text { or } v=\sqrt{2 g h}\)

Venturi meter: 

It is a gauge put on a flow pipe to measure the flow speed of a liquid. Let the liquid of density ρ be flowing through a pipe of area of cross-section A₁.

Let A₂ be the area of the cross section at the throat and a manometer is attached as shown in the figure. Let v₁ and P₁ be the velocity of the flow and pressure at point A, and v₂ and P₂ be the corresponding quantities at point B.

Using Bernoulli’s Theorem:

⇒ \(\frac{P_1}{\rho}+g h_1+\frac{1}{2} v_1^2=\frac{P_2}{\rho}+g h_2+\frac{1}{2} v_2^2\), we get

⇒ \(\frac{P_1}{\rho}+g h+\frac{1}{2} v_1^2=\frac{P_2}{\rho}+g h+\frac{1}{2} v_2^2\) (Since h₁=h₂=h)

or \(\left(P_1-P_2\right)=\frac{1}{2} \rho\left(v_2^2-v_1^2\right)\) ….

According to the continuity equation, A₁v₁= A₂v₂

or \(v_2=\left(\frac{A_1}{A_2}\right) v_1\)

Substituting the value of v₂ in the equation we have

\(\left(P_1-P_2\right)=\frac{1}{2} \rho\left[\left(\frac{A_1}{A_2}\right)^2 v_1^2-v_1^2\right] \frac{1}{2} \rho v_1{ }^2\left[\left(\frac{A_1}{A_2}\right)^2-1\right]\)

Since A₁> A₂, therefore, P₁> P₂

or \(v_1^2=\frac{2\left(P_1-P_2\right)}{\rho\left[\left(\frac{A_1}{A_2}\right)^2-1\right]}\)

⇒ \(\frac{2 A_2^2\left(P_1-P_2\right)}{\rho\left(A_1^2-A_2^2\right)}\)

where (P₁– P₂)= ρ_mgh and h is the difference in heights of the liquid levels in the two tubes.

⇒ \(v_1=\sqrt{\frac{2 \rho_{\mathrm{m}} g h}{\rho\left[\left(\frac{A_1}{A_2}\right)^2-1\right]}}\)

The flow rate (R) i.e., the volume of the liquid flowing per second is given by R = v₁A₁.

During A wind storm: The velocity of air just above the roof is large so according to Bernoulli’s theorem, the pressure just above the roof is less than the pressure below the roof. Due to this pressure difference an upward force acts on the roof which is blown off without damaging other parts of the house.

When a fast-moving train crosses a person standing near a railway track, the person has a tendency to fall towards the train.

This is because a fast-moving train produces a large velocity in the air between a person and the train and hence pressure decreases according to Bernoulli’s theorem. Thus the excess pressure on the other side pushes the person towards the train.

Question 1. Water flows through a horizontal tube of variable cross-section (figure). The area of a cross-section at A and B are 4 mm² and 2 mm² respectively. If 1 cc of water enters per second through A, find

1. The speed of the water at A,
2. The speed of the water at B and
3. The pressure difference P_A– P_B.

Answer: A₁v₁= A₂v₂

⇒ \(\rho_1+\frac{1}{2} \rho v_1^2+0\)

⇒ \(r_2+\frac{1}{2} \rho v_2^2+\rho g h\)

1. 25 cm/s,
2. 50 cm/s
3. 94 N/m²

Question 2. The velocity of the liquid coming out of a small hole of a large vessel containing two different liquids of densities 2ρ and ρ as shown in the figure is

1. \(\sqrt{6 g h}\)
2. \(2 \sqrt{g h}\)
3. \(2 \sqrt{2 \mathrm{gh}}\)
4. \(\sqrt{g h}\)

Answer: Pressure at : P₁= Patm + ρ g (2h)

Applying Bernoulli’s theory between points and

⇒ \(\left[P_{a t m}+2 \rho g h\right]+\rho g(2 h)+\frac{1}{2}(2 \rho)(0)^2\)

⇒ \(P_{\mathrm{atm}}+(2 \rho) g(0)+\frac{1}{2}(2 \rho) v^2\)

⇒ \(v=2 \sqrt{g h}\)

Question 3. A horizontal pipeline carries water in a streamlined flow. At a point along the pipe where the cross-sectional area is 10 cm², the water velocity is 1 ms^-1 and the pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area is 5 cm² will be:

[Density of water = 103 kg. m^-3 ]

Answer:

From continuity equation

A₁v₁= A₂v₂

∴ \(v_2=\left(\frac{A_1}{A_2}\right) v_1=\left(\frac{10}{5}\right)(1)=2 \mathrm{~m} / \mathrm{s}\)

Applying Bernoulli’s theorem at 1 and 2

⇒ \(P_2+\frac{1}{2} \rho v_2{ }^2=P_1+\frac{1}{2} \rho v_1{ }^2\)

⇒ \(P_2=P_1+\frac{1}{2} \rho\left(v_1^2-v_2^2\right)\)

⇒ \(\left(2000+\frac{1}{2} \times 10^3(1-4)\right)\)

500 Pa

Question 4. Equal volumes of two immiscible liquids of densities ρ and 2ρ are filled in a vessel as shown in the figure. Two small holes are punched at depths h/2 and 3h/2 from the surface of a lighter liquid. If v₁ and v₂ are the velocities of efflux at these two holes, then v₁/v₂ is:

1. \(\frac{1}{2 \sqrt{2}}\)
2. 0.5
3. 0.25
4. \(\frac{1}{\sqrt{2}}\)

Answer: for hole (1)

⇒ \(P_0+\rho \frac{V_1^2}{2}=P_0+\rho g \frac{h}{2}\)

⇒ \(V_1=\sqrt{g h}\)

for hole (2)

⇒ \(P_0+\rho \frac{V_2^2}{2}\)

⇒ \(P_0+\rho g h+2 \rho h\left(\frac{h}{2}\right)\)

⇒ \(\frac{\rho V_2^2}{2}=2 \rho g h\)

⇒ \(V_2=2 \sqrt{g h}\)

⇒ \(\frac{V_1}{V_2}=\frac{1}{2}\)

= 0.5