1st PUC Physics Question Bank Chapter 10 Mechanical Properties of Fluids

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Karnataka 1st PUC Physics Question Bank Chapter 10 Mechanical Properties of Fluids

Question 1.
What are fluids?
Answer:
The materials that can flow are called fluids.

Question 2.
How are fluids different from solids?
Answer:
Fluid has no definite shape of its own but solids have definite shape.

Question 3.
Define thrust of a fluid.
Answer:
The total normal force exerted by a fluid on any surface in contact with it is called thrust of a liquid.

Question 4.
Define liquid pressure or average pressure.
Answer:
Liquid pressure is defined as the normal force acting per unit area.
P_ave = \(\frac{F}{A}\)

Question 5.
Distinguish between liquid thrust and pressure.
Answer:
Liquid thrust is the total normal force exerted by a fluid on walls of container where as pressure is fluid thrust per unit area.

Question 6.
Name the S.I. unit of fluid pressure and write the dimensional formula.
Answer:
S.I. unit is Nm’2 and dimensions are ML^-1T².
Note: Pressure is scalar quantity.

Question 7.
What is gauge pressure?
Answer:
The gauge pressure is the difference of the actual pressure and the atmospheric pressure.
Gauge pressure = P-P_a

Question 8.
What is the difference between atmospheric pressure and gauge pressure?
Answer:
The pressure of the atmosphere at any point is known as atmosphere pressure and gauge pressure is the difference of the actual pressure and the atmospheric pressure.

Question 9.
Define density of the fluid. Give its dimensions and S.I. unit.
Answer:
Density is defined as mass per unit volume of the fluid. m
ρ = m/V
where m is the mass of fluid and V is the volume of fluid.
Dimension: [ML^-3]
S.I. unit: kg m^-3

Question 10.
What is relative density of a substance?
Answer:
It is the ratio of density of substance to the density of water at 4°C.
Note: Relative density is dimensionless positive scalar quantity.

Question 11.
What is the density of water at 4°C (277 K)?
Answer:
1.0 x 10³ kg m^-3

Question 12.
If density of aluminium is 2.7 x 103 kg m 3, what is its relative density?
Answer:

Question 13.
What is the pressure exerted by the atmosphere at sea level.
Answer:
1 atm = 1.013 x 10⁵ Pa

Question 14.
Derive an expression for pressure inside a liquid.
Answer:
Consider a fluid at rest in a container as shown in the Figure.
The point 1 is at a height h above a point 2 and the pressures at point 1 and 2 are Pi and P2 respectively. Consider a ‘ cylindrical element of fluid having area of base A and height h. As the fluid is at rest the resultant vertical forces should balances the weight of the element. The forces acting in the vertical direction are due to the fluid pressure at the top (P₁A)( ∴ P = \(\frac{F}{A}\) ) acting downward at the bottom (P2A) acting upward. If mg is weight of the fluid in the cylindrical we have
(P₂ – P₁)A = F but F = mg
∴ (P₂ – P₁)A = mg …(1)
If p is the density of the fluid then mass of fluid is m = ρgA (∴ V = hA )
Equation (1) becomes
(P₂ – P₁)A = ρhAg
P₂ – P₁ = ρgh …(2)
If point 1 is shifted to the top of the fluid which is open to the atmosphere then P₁ = P_a atmospheric pressure and replace P₂ by P then equation (2) becomes P – P_a = ρgh
or P = P_a + ρgh

Question 15.
Who discovered the mercury barometer?
Answer:
Evangelista Torricelli

Question 16.
Write a note on mercury barometer. What is the use of it?
Answer:
Mercury barometer is a device used to measure atmosphere pressure. It consists of long glass tube closed at one end and filled with mercury and is inverted into a trough containing mercury as shown.
I The space above the mercury column in the tube contains only mercury vapour
whose pressure P and is so small that it may be neglected. The pressure inside , the column at point A must equal to the pressure at point B.
∴ Pressure at B = atmospheric pressure = P_a
∴ P_a = ρgh
where ρ is the density of mercury and h is the height of the mercury column in the tube.

Question 17.
What is a torr?
Answer:
A pressure equivalent of 1 mm is called a torr (after Torricelli).
Note: (i) 1 torr = 133 Pa
(ii) 1 bar = 10⁵ Pa

Question 18.
Write a note on open tube monometer. What is the use of it?
Answer:
Open tube monometer is a device used for measuring pressure differences.
Open tube monometer consists of U-tube containing a suitable liquid. Low density liquid such as oil for measuring small pressure differences and high density liquid such as mercury for large pressure differences are commonly used. One end of the tube is open to the atmosphere and other end is connected to the system whose pressure to be measured.
The pressure P at point A is equal to pressure P at point B. Therefore gauge pressure is P – Pa and is proportional to monometer height h.

Question 19.
State Pascal’s law.
Answer:
Whenever external pressure is applied on any part of a fluid contained in a vessel it is transmitted undiminished and equally in all directions.

Question 20.
Mention any two application of Pascal’s law.
Answer:
Hydraulic brake, Hydraulic lift.

Question 21.
Explain how Pascal’s law is applied in a hydraulic lift?
Answer:
For a confined static liquid pressure applied at any point in the liquid is transmitted equally and undiminished in all direction throughout the liquid. At piston P the force F₁ acts over the area A₁.
∴ P₁ = \(\frac{F_{1}}{A_{1}}\)
At piston Q the force F2 acts over area A2
∴ P₂ = \(\frac{F_{2}}{A_{2}}\)
But according to Pascal’s law the pressure Pi is transmitted equally to piston Q

Since
Thus a small force applied on the smaller piston appears as a larges piston.

Question 22.
Explain stream flow.
Answer:
The path taken by a fluid particle under a steady flow is a stream line. It is defined as a curve whose tangent at any point is in the direction of the fluid velocity at that point. This can be illustrated as follows.
(a) A typical trajectory of a fluid particle

(b) Region of streamline flow

Question 23.
What is streamline motion?
Answer:
The flow is said to be stream line if at a given point, the velocity of every liquid particle remains constant with time.

Question 24.
Can two streamlines intersect each other in a flowing liquid?
Answer:
No, two streamlines cannot intersect each other.

Question 25.
How is velocity of stream line flow of liquid is related to area of cross section of the tube?
Answer:
Velocity of liquid in a streamline flow is inversely proportional to the area of cross section of the tube.

Question 26.
How is pressure of liquid depends on the area of cross section of tube of flow?
Answer:
Pressure of liquid is directly proportional to the area of cross section.

Question 27.
How are the pressure of liquid and speed of flow related?
Answer:
Pressure of liquid is inversely proportional to the square of the speed of flow of the liquid.

Question 28.
What is an ideal fluid?
Answer:
Non viscous and incompressible liquid is called ideal fluid.

Question 29.
When does the flow of the liquid become turbulent?
Answer:
Flow of the liquid become turbulent if the velocity of a liquid at a point changes and the rate of flow exceeds the critical velocity.

Question 30.
The blood pressure in human is greater at the feet than at brain. Why?
Answer:
This is because the height of the blood column is quite large at the feet than at the brain.

Question 31.
Mention the types of flow of fluid.
Answer:
Streamline flow and turbulent flow.

Question 32.
Distinguish between stream line flow and turbulent flow.
Answer:
Streamline flow :

1. It is a regular and orderly flow of liquid.
2. In stream line flow velocity of all the liquid particles is the same at a given point.
3. The motion of liquid particles is parallel to each other.
4. Every liquid particles moves with a velocity less than the critical velocity.

Turbulent flow :

1. It is a irregular and disorderly flow of liquid.
2. In turbulent flow the velocity of all the liquid particles is different at a given point.
3. The motion of liquid particle is not parallel to each other.
4. Every’ liquid particles moves with a velocity greater than the critical velocity.

Question 33.
State and explain equation of continuity.
Answer:
It states that the product of area of cross section and the speed of a liquid remains the same at all points of a tube of flow.
If ‘A’ is the area of cross section of the tube at a point and v is the velocity of liquid in that region then 1
V ∝ 1/A
∴ vA = constant
Let A₁ and A₂ be the areas of cross sections at sections A and B respectively. Let the liquid enter with the velocity V₁ at A and leave with velocity v₂ at B.
Volume of liquid entering the tube = A₁v₁
Volume of tube leaving the tube = A₂v₂
Since the volume of liquid flows across each section of the tube, we have A₁v₁ = A₂v₂

Question 34.
State Bernoulli’s principle. Explain the Bernoulli’s equation for the flow of an ideal fluid in stream line motion. Mention any two applications.
Answer:
Statement: Along the stream line of an ideal fluid, the sum of the potential energy, kinetic energy and pressure energy per unit mass remains constant.
Explanation: Bernoulli’s theorem relates the speed of a fluid at a given point, the pressure at that point and the height of that point above a reference level. Consider a liquid contained between cross sections A and B of the tube as shown in Fig. The height of A and B are h₁ and h₂ respectively from a reference level.
Let the pressure at A and B be P₁ and P₂. The velocities at A and B be V₁ and v₂ and the density of the liquid is ρ. According to Bernoulli’s theorem.

This is known as Bernoulli’s equation. Thus for incompressible non viscous fluid in steady state flows, the sum of pressure energy, kinetic energy and potential energy per unit volume is constant.
Application: Venturimeter, Atomisers and sprayers

Question 35.
Derive the expression for Bernoulli’s equation.
Answer:
Consider a liquid to be flowing inside a tube in a streamline motion. Let P₁,A₁,v₁,h₁ be the pressure, area of cross-section, velocity and height of the liquid at A and P₂,A₂,v₂,h₂ be the pressure, area of cross-section, velocity and height of the liquid at B respectively. Then, the change in gravitational potential energy is given by
U₂ – U₁ = ∆Vρgh₂ – ∆Vρgh₁
∆U = ∆Vρg{h₂-h₁)
If the density of fluid is p, the mass passing through the pipe in time At is given by
∆m = ρA₁v₁∆t = P∆V.
The change its kinetic energy is
∆K = \(\frac{1}{2}\) ∆Vρ(v₂² -v₁²)
By applying work-energy theorem, we get
(P₁– P₂)∆v = \(\frac{1}{2}\) ρ∆v{v₂² -v₁²) + pg∆v(h₂ – h₁)
Divide each term by AV to obtain,
(P₁– P₂) = \(\frac{1}{2}\) ρ{v₂² -v₁²) + pg(h₂ – h₁)
By rearranging the above equation to get Bernoulli’s equation
P₁ + \(\frac{1}{2}\) ρv₁² + ρgh₁ = P₂ + \(\frac{1}{2}\) ρv₂² + ρgh₂
Since 1 and 2 refer to any two point locations along the pipe line,
P + \(\frac{1}{2}\) ρv₁² + ρgh = constant

Question 36.
What is the basis on which Bernoulli theorem is states?
Answer:
Bemoullis’ theorem is based on the law of conservation of energy.

Question 37.
Explain Torricelli’s law (speed of efflux)
Answer:
The word efflux means fluids out flow. Torricell’s discovered that speed of efflux from an open tank is given by a formula identical to that of a free falling body. Consider a tank containing a liquid of density p with a small hole in its side at a height yi from the bottom. The air above the liquid is at height y2 is at a pressure P. From the equation of continuity we have
V₁A₁=V₂A₂
∴ V₂ = \(\frac{A_{1}}{A_{2}} v_{1}\)
If the cross sectional area of the tank A₂ is much higher than that of the hole, the fluid to be approximately at rest at the top.
A₂ >> A₁
∴ v₂ = 0
Now applying Bernoulli’s equation at points 1 and 2 and nothing that at the whole P₁ = P_a, the atmospheric pressure we have

In the tank is open to atmosphere then P = P_a
Then v₁ = \(\sqrt{2 g h}\) . This the speed of the free falling body.
This is known as Torricelli’s law.

Question 38.
Describe the working of venture meter.
Answer:
Venturimetre is a device to measure the flowspeed of incompressible fluid. It consists of a tube with a broad diameter and a small constriction at the middle as shown in figure.
Let A be area of cross section at point (1) and ‘a’ be the area of cross-section at point (2).
According to equation of continuity.
A₁v₁ = A₂v₂
A₁ =A, A₂ =a
∴ Av₁ = av₂
∴ v₂ = \(\frac{A v_{1}}{a}\)
Since the height remains the same in the tube potential energy also remains the same. According to Bernoulli’s principle.


We know that the difference in pressure energy gives potential energy.

Question 39.
Give any three application of Venturi meter.
Answer:

1. The carburetor of automobile has a Venturi Channel through which air flows with large speed.
2. Filter pumps or aspirators
3. Bunsen burner
4. Atomisers and spayers are used under the principle of Venturi meter.

Question 40.
Explain blood flow and heart attack.
Answer:
The blood flow through artery can be explained using Bemoullis’ principle. The artery may get constricted due to accumulation of plaque (cholesterol) on its inner walls. To derive the blood through this constricted portion, a greater demand is placed on the activity of the heart. The speed of the blood in this region increases and hence the pressure inside the artery decreases. Due to external pressure the artery may collapse. Now the heart exerts further pressure to open the artery and forces the blood through. As the blood rushes with more velocity the pressure further decreases. This may results in heart attack.

Question 41.
What is dynamic lift?
Answer:
Dynamic lift is the force that acts on a body by virtue its motion through a fluid. Examples: air plane wing, hydrofoil or a spinning ball.

Question 42.
Explain dynamic lift of the ball moving with spin.
Answer:
When the ball is moving and spinning in the air it drags the air along with it. If the surface of the ball is rough the velocity of the air above the ball will be more than the velocity of the air below the ball. As a result the pressure above the ball is less compared to the pressure below the ball. Since due to high pressure and low velocity below the ball there is a dynamic lift in the ball. This is also called Magnus effect.

Question 43.
Explain the uplift of an aircraft.
Answer:
The wings of the aircraft is designed in such way that when the aircraft moves the velocity of the air above the wings is greater than the velocity of the air below the wings. As a result the pressure above the wing will be less compared to the pressure below the wing. Since the velocity of wind is less below the wings and pressure is more which helps in the uplift of the aircraft.

Question 44.
What is Magnus effect?
Answer:
The difference in the velocities of air results in the pressure difference between the lower and upper surfaces of the spinning object and net upward force on the object. This dynamic lift due to spinning is called magnus effect.

Question 45.
Write the equation connected to hydraulic lift.
Answer:
Applying Pascal’s law, P = \(\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}\)

Viscosity

Question 1.
Define viscosity of a liquid.
Answer:
The property of a liquid by virtue of which it tends to resist relative motion between different layers of it is called viscosity.

Question 2.
What is viscous force or viscous drag?
Answer:
The internal opposite force between adjacent liquid layers in motion is called viscous force.

Question 3.
Define co-efficient of viscosity.
Answer:
It is defined as the ratio of shearing stress to the strain rate.


when A = 1, l = 1, v = 1 then η = F.
Therefore, it can also be defined as tangential viscous force acting per unit area between two liquid layers with unit velocity gradient.

Question 4.
Give the dimensions and SI unit of viscosity.
Answer:
Dimension of viscosity = [ML^-1T^-1]
S.I. unit of viscosity = poiseiulle (PI)
Other units are Nsm^-2or Pa s.
Note: The viscosity of liquid decreases with increase in temperature while it increases in the case of gases.

Question 5.
State Strokes law or Give strokes formula and explain the symbols used.
Answer:
Stokes law states that “viscous drag force on a small spherical object falling inside a fluid is directly proportional to its radius and terminal speed.
F = 6πrηav
This equation is known as Stokes law.
where v = terminal speed of liquid,
η = co-efficient of viscosity
a = radius of the spherical object
F – viscous drag force.

Question 6.
Give an expression for the terminal speed of a small spherical object in a homogeneous surrounding fluid.
Answer:
V_t = \(\frac{2 a^{2}(\rho-\sigma) g}{9 \eta}\)
v_t = terminal speed of the object
ρ = density of the small spherical object
σ = density of the surrounding fluid
η = co-efficient of viscosity
a = radius of the spherical object

Question 7.
Hotter liquid flows faster than cold ones. Why?
Answer:
The coefficient of viscosity of a liquid decreases with rise in temperature. As a result the hotter Liquid will flow faster than the cold one.

Question 8.
How is terminal velocity depends on the radius of the sphere and viscosity of the medium?
Answer:
Terminal velocity depends on the square of the radius of the sphere and inversely on the viscosity of the medium.

Question 9.
What is Reynold’s number? Write its expression.
Answer:
Reynold’s number is a dimensionless number which determines the type of flow of liquid through a pipe. It is given by the expression
R_e = \(\frac{\rho v d}{\eta}\)
where v = the speed of flow of a liquid through a pipe
d = diameter of the tube
η = co-efficient of viscosity of the liquid
Note: It is found that
If R_e > 2000, the flow is turbulent
If R_e < 1000, the flow is laminar or stream line.
If R_e is between 1000 and 2000, the flow is unsteady.

Question 10.
How is Reynold’s number related to inertial force and force of viscosity?
Answer:
Reynold s number = \(\frac{\text { Initial force }}{\text { Force of viscosity }}\)

Surface Tension

Question 1.
What is surface tension?
Answer:
Surface tension is the force per unit length acting in the plane of the interface between the plane of the liquid and any other substance.

Question 2.
What is surface energy?
Answer:
The extra potential energy of the molecules in the surface layer of a liquid is called the surface energy.
Note: Surface tension is equal to the surface energy per unit surface area.

Question 3.
Write the expression of magnitude of surface tension, by the fluid on the movable bar. Explain the symbols.
Answer:
Magnitude of surface tension, S = \(\frac{F}{2 l}\)
where F is the force exerted by the fluid on the movable bar and l is the length of the plate edge.

Question 4.
Write the expression of the surface tension of the liquid – air interface.
Answer:
S_la = \(\frac{m g}{2 l}\)
where m = extra mass required
l = length of the plate edge

Question 5.
What is angle of contact? What is its importance?
Answer:
The angle between tangent to the liquid surface at the point of contact and solid surface inside the liquid is called angle of contact. It is denoted by θ.
The value of angle of contact (θ) determines whether a liquid will spread on the surface of a solid or it will form droplets on it.

Question 6.
What is meant by sphere of influence of a liquid molecule?
Answer:
A sphere of radius equal to the range of the molecular force about a given molecule as centre is called its sphere of influence.

Question7.
Give an example of a liquid which wets glass.
Answer:
Water. Because angle of contact between water and glass is zero.

Question 8.
Give an example of a liquid which does not wet glass.
Answer:
Mercury. The angle of contact between mercury and glass is 135°.

Question 9.
Small drop of liquid is spherical. Give reason.
Answer:
The surface tension exceeds the force of gravity due to less surface area in the drop. Hence drop becomes spherical.
Note: (i) For a spherical liquid drop, the pressure inside the liquid is more than the pressure out side.
(ii) For liquid – gas interface, the convex side has a higher pressure than the concave side.

Question 10.
What is capillary tube?
Answer:
A tube of very fine bore is called capillary tube.

Question 11.
The surface of water in the capillary is concave. Give reason.
Answer:
Because the angle of contact between water and glass is acute.

Question 12.
The surface of water in the capillary is concave. What does it mean?
Answer:
This means that there is a pressure difference between the two sides of the top surface.

Question 13.
Write the expression to show the surface of water in the capillary is concave.
Answer:
(P_i – P₀) =\(\left(\frac{2 S}{a}\right)\) cosθ

Question 14.
Give reason in Capillary rise of liquid in capillary tube.
Answer:
Capillary rise is due to surface tension.

Question 15.
How is surface energy related to surface tension of a liquid?
Answer:
Surface tension = surface energy x change in the area.
i.e., surface tension is equal to work done in increasing unit area of the surface film.

Question 16.
What effect does temperature have on the surface tension of a liquid?
Answer:
The surface tension of a liquid decreases with increase of temperature.

Question 17.
What is the effect of adding detergents to water?
Answer:
Surface tension of water decreases with addition of detergents.

Question 18.
What are the factors on which angle of contact depends?
Answer:
The angle of contact depends upon

1. The nature of the liquid.
2. The nature of the material of the container.
3. The temperature of the liquid.

Question 19.
What happens to angle of contact, if the temperature of the liquid increase?
Answer:
The angle of contact of a liquid increases with the increase of temperature.

Question 20.
What is capillarity?
Answer:
The rise or fall of a liquid in a capillary tube immersed partially in the liquid is called capillarity.

Numerical Problems

Question 1.
The thigh bones (femurs), each of cross – sectional area 10 cm² support the upper part of a human body of mass 40 kg. Estimate the average pressure sustained by the by the femurs.
Answer:
Total cross-sectional area of the femurs, A = 2 x 10 cm² = 20 x 10^-4m².
The force acting on them , F = 40 kg wt = 400 N (taking g = 10 ms^-2).
The average pressure, P_av= F/A = 2 x 10⁵ Nm^-2

Question 2.
What will be the effect of the angle of contact of a liquid? If the temperature increases.
Answer:
The angle of contact of a liquid increases with increase of temperature.

Question 3.
What is the pressure on a swimmer 10 m below the surface of a lake?
Answer:
Here h = 10 m and ρ = 1000 kg m^-3and g = 10 ms^-2
P = P_a+ρgh
= 1.01 x 10⁵ Pa + 1000 kg m^-3 x 10 ms^-2 x 10 m = 2.01 x 10⁵Pa = 2 atm

Question 4.
The density of the atmosphere at sea level is 1.29 kg / m3. Assume that it does not change with altitude. Then how high would the atmosphere extend?
Answer:
P = ρgh
1.05 x 10⁵ Pa = 1.29 kg m^-3 x 9.8 ms² x h m
∴ h = 7989 m ≈ 8 km
In reality the density of air decreases with height. So does the value of g.

Note: The atmospheric cover extends with decreasing pressure over 100 km. We should also note that the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 10 mm or more is a sign of an approaching storm.

Question 5.
At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure?
(d) Find the force acting on the window of area 20 cm x 20 cm of a submarine at this depth, the interior of which is maintained at sea – level atmospheric pressure. (The density of sea water is 1.03 x 10³ kg m^-3, g = 10 ms^-2).
Answer:
Here h = 1000 m and ρ = 1.03 x 10³ kg m^-3,
(a) From equation of absolute pressure,
P = P_a + ρgh
P = 1.01 x 10⁵Pa + 1.03 x 10³kg m^-3 x 10 ms^-2 x 1000 m
P = 104.01 x 10⁵Pa = 104 atm

(b) Gauge pressure is P-P_a = ρgh = P_g
P_g = 1.03 x 10³ kg m^-3 x 10 ms² x 1000 m = 103 x 10⁵ Pa = 103 atm

(c) The pressure outside the submarine is P = P_a+ ρgh and the pressure inside it is P_a.
Hence, the net pressure acting on the window is gauge pressure, P_g = ρgh.
Since the area of the window is A = 0.04 m², the force acting on it is
F = P_gA = 103 x 10⁵Pa x 0.04 m²= 4.12 x 10⁵N

Question 6.
Two syringes of different cross-sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively, (a) Find the force exerted on the larger piston when a force of 10 N is applied to the smaller piston, (b) If the smaller piston is pushed in through 6.0 cm, how much does the larger piston move out?
Answer:
(a) Since pressure is transmitted undiminished throughout the fluid,

(b) Water is considered to be perfectly incompressible. Volume covered by the movement of smaller piston inwards is equal to volume moved outwards due to the larger piston.
L₁A₁ = L₂A₂

Note: atmospheric pressure is common to both pistons and has been ignored.

Question 7.
In a car lift compressed air exerts a force F₁ on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm. If the mass of the car to be lifted is 1350 kg. Calculate F₁. What is the pressure necessary to accomplish this task? (g = 9.8 ms^-2)
Answer:
Since pressure is transmitted undiminished throughout the fluid,

The air pressure that will produce this force is

This is almost double the atmospheric pressure.

Question 8.
The flow of blood in a large artery of an anesthetized dog is diverged through a Venturi meter. The wider part of the meter has a cross-sectional area equal to that of the artery, A = 8 mm². The narrower part has an area a = 4 mm². The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery?
Answer:
The density of blood = 1.06 x 10³ kg m^-3.
The pressure drop in the artery = ρ_mgh = 24 pa

Question 9.
A fully loaded Boeing aircraft has a mass of 3.3 x 10⁵ kg. Its total wing area is 500 m². It is in level flight with a speed of 960 km/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 p = 1.2 kg m^-3]
Answer:
(a) The weight of the Boeing aircraft is balanced by the upward force due to the pressure difference = ∆P x A
3.3 x 10⁵kg x 9.8 = ∆P x 500m²

= 6.5 x 10³ Nm^-2
According to the Bernoul’s equation

We ignore the small height difference between the top and bottom sides in the above equation. The pressure difference between them is then

Where v₂. is the speed of air over the upper surface and v₁. is the speed under the bottom surface.

Taking the average speed

Question 10.
A metal block of area 0.10 m² is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. 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 ms^-1. Find the co-efficient of viscosity of the liquid.
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 = mg = 0.010 kg x 9.8ms^-2 = 9.8 x 10^-2N

Question 11.
The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20°C is 6.5 cm s^-1. Compute the viscosity of the oil at 20°C. Density of oil is 1.5 x 10³ kg m^-3, density of copper is 8.9 x 10³ kg m^-3.
Answer:
We have v_t =6.5 x 10^-3 ms^-1, a = 2 x 10^-3 m, g = 9.8 ms^-2,ρ = 8.9 x 10³ kg m^-3, σ = 1.5 x 10³ kg m^-3,

Question 12.
The flow rate of water from a tap of diameter 1.25 cm is 0.48 L/min. The co-efficient of viscosity of water is 10^-3 Pa s. After sometime the flow rate is increased to 3 L / min. Characterise the flow for both the flow rates.
Answer:
Let the speed of the flow be v and the diameter of the tap be d = 1.25 cm.
The volume of the water flowing out per second is

We then estimate the Reynolds number to be R_e = \(\frac{4 p Q}{\pi d \eta}\)

R_e = 1.019 x 10³m^-3s x Q
Since initially Q = 0.48 L / min = 8 cm³ / s = 8 x 10^-6 m³ s^-1, we obtain
R_e = 1.019 x 10³ m^-3s x 8 x 10⁸m³s^-1 = 815
Since this is below 1000, the flow is steady.
After some time when
Q = 3 L / min = 50 cm³ / s = 5 x 10^-5 m³ s^-1,
we obtain, R_e = 1.019 x 10⁸ x 5 x 10^-5 = 5095
The flow will be turbulent.

Question 13.
The lower end of a capillary tube of diameter 2.90 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 and in water? The surface tension of water at temperature of the experiments is 7.30 x 10^-2 Nm^-1. 1 atmospheric pressure = 1.01 x 10⁸ Pa. density of water = 1000 kg/m³, g = 9.80 ms^-2. Also calculate the excess pressure.
Answer:
The excess pressure in a bubble of gas in a liquid is given by 2Sir, where S is the surface tension of the liquid- gas interface. The radius of the bubble is r.
The pressure outside the bubble P₀ equals atmospheric pressure plus the pressure due to 8.00 cm of water column.
P₀ = (1.01 x 10⁵ Pa + 0.08 m x 1000 kg m^-2 x 9.80 ms^-2)
= 1.01784 x 10⁵ Pa
Therefore, the pressure inside the bubble is
P₁ = P₀ + \(\frac{2 S}{r}\)
= 1.01784 x 10⁵ Pa + (2 x 7.3 x 10^-2 Pa m / 10^-3 m)
= (1.01784 + 0.00146) x 10⁵ Pa = 1.02 x 10⁵ Pa
where the radius of the bubble is taken to be equal to the radius of the capillary tube, since the bubble is hemispherical. The excess pressure in the bubble is 146 Pa.

Note: For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure in that case is 4 Sir.

Question 14.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter
1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Answer:
Force exerted by the heel due to the weight of the girl,
F = mg = 50 x 10 = 500 N
Diameter of the heel’s circular are, d = 1cm = 10^-2 m

Question 15.
Toricelll’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^-3. Determine the height of the wine column for normal atmospheric pressure. (Assume g = 9.8 m/s²)
Answer:
Normal atmospheric pressure = 1 atm
∴ Gauge pressure of wine = 1 atm = 1.013 x 10⁵ pa
P =ρgh
1.013 x 10⁵ Pa = 984 kg m^-3 x 9.8 m/s² x h
h = \(\frac{1 \cdot 013 \times 10^{5}}{984 \times 9 \cdot 8}\) = 10.5m

Question 16.
A vertical off-shore structure is built to withstand a maximum stress of 10⁹ Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.
Answer:
Maximum stress that can be taken by the off-share structure, P_max = 10⁹ Pa.
Density of water = 10³ kg /m³, g = 9.8 m/s² and h = 3 km (depth of the sea)
∴ Pressure exerted by sea W’ater, P = ρ gh
= 10³ x 9.8 x 3 x 10³
P = 2.94 x 10⁷ Pa

Note that, P < P_max
Since the pressure exerted by the sea water is less than the maximum pressure the structure can withstand, the structure is suitable for putting on the top of the oil well.

Question 17.
8 A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of crosssection of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear?
Answer:
Let F_max be the maximum force experienced by the bigger piston.
F_max = mg = 3000 x 9.8 = 29400 N
Area of the bigger piston = 425 cm² = 425 x 10^-4 m²
Maximum presure on the bigger piston p_max = \(\frac{F_{\max }}{A}\)
P = \(\frac{29400}{425 \times 10^{-4}}\) = 6.92 x 10⁵
Since pressure is transmitted uniformly throughout the liquid the smaller piston will also bear a pressure of 6.92 x 10⁵ Pa.

Question 18.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit? If 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)
Answer:
ρ₁ → density of water
ρ₂ → density of spirit
h₁ → height of water column
h₂ → height of spirit column
Let P₀ be the atmospheric pressure, P_A = P₀ + p₁gh₁ .. .(1)
P_B = P₀ + p₂gh₂ .. .(2)
Since the mercury levels are point A and B we can say that, P_A= P_B
P₀ + p₁gh₁ = P₀ + p₂gh₂
p₁gh₁ = p₂gh₂


Since we are putting equal amount of water and spirit and specific gravity of spirit is < 1 we can assume that the mercury level will rise on the spirit side.
ρ₁ → density of water,
ρ₂ → density of spirit
ρ₃ → density of mercury,
Since A and B are at the same height,
P_A= P_B

where \(\frac{\rho_{2}}{\rho_{1}}\) ⇒ specific gravity of spirit and \(\frac{\rho_{3}}{\rho_{1}}\) ⇒ specific gravity of mercury.
Thus x = \(\frac{25-0.8 \times 27.5}{13.6}\) = 0.221 cm
Therefore the difference in the mercury level in the arms is 0.221 cm.

Question 19.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1 cm. If the amount of glycerine collected per second at one end is 4.0 x 10^-3 kg s^-1, what is the pressure difference between the two ends of the tube?, (Density of glycerine = 1.3 x 10³ kg m^-3 and viscosity of glycerine = 0.83 Pas).
Answer:
The volume of liquid flowing out per second is given by,
Q = \(\frac{\pi R^{4}\left(p_{1}-p_{2}\right)}{8 h l}\)
Pressure difference at tube ends, P₁ = P₂ = \(\frac{8 \mathrm{Qn}}{\pi r^{4}}\)
where Q is the volume of liquid flowing per second.

A pressure difference of 9.76 x 10²N/m² is obtained.
Reynolds number,

Since the Reynolds number is about 0.3, the flaw is laminar.

Question 20.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s^-1 and 63 ms^-1 respectively. What is the lift on the wing if its area is 2.5 m²? Take the density of air to be 1.3 kg m^-3.
Answer:
According to Bermolli’s theorem,

P₁ = pressure at the lower surface of wings
P₂ = pressure at the lower surface of wings
p = Density of air
v₁ = Speed of wind at the lower surface = 63 m/s.
v₂ = speed of wind at upper surface = 70m/s
P₁ – P₂ = \(\frac{1}{2}\) p (v₁² -v ₂²)
P₁ – P₂ = \(\frac{1}{2}\) x 1.3 x(70² – 63²) = 605.15 pa
Force on wings = (P₁ – P₂) x area
= 605.15 x2.5 = 1512.87 N
∴ The lift on the wings is about 1.512 x 10³ N

Question 21.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm² one end of which has 40 line holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min, what is the speed of ejection of the liquid through the holes?
Answer:
Number of holes = 40
Diameter of each hole, d = 10^-3 m

speed of liquid inside the tube, v₁ = 1.5m/ min = \(\frac{1 \cdot 5}{60}\) m / s
Area of cross section of tube, a₁ = 8.0 cm² = 8 x 10^-4 m²
If v₂ is the velocity of ejection of liquid through the holes then from equation of continuity,
a₁v₁ = a₂v₂

v₂ =0.637 m / s

Question 22.
A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 x 10^-2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
Answer:
The soap film has two free surfaces.
∴ Total length of the film to the supported, l = 2 x 30 = 60 cm = 0.6
Total force on the slider due to surface tension, F = S x 2l = S x 0.6N
At equilibrium, the force F on slider due to surface tension should balance the weight mg
∴ F = mg = 1.5 x 10^-2 N
S x 0.6 = 1.5 x 10^-2
S = 2.5 x 10^-2 N/m

Question 23.
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20 °C) is 4.65 x 10^-1 N m^-1. The atmospheric pressure is 1.1 x 10⁵ Pa. Also give the excess pressure inside the drop.
Answer:
Radius of the mercury drop, r = 3 mm = 3 x 10^-3 m
Surface tension of mercury, S = 4.65 x 10^-1 Nm^-1
Atmospheric pressure, P₀ = 1.01 x 10⁵ Pa
Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure

Question 24.
What is the excess pressure inside a bubble of soap solution of radius 5 mm, given that the surface
tension of soap solution at the temperature (20 °C) is 2.50 x 10^-2 N/rn? If an air bubble of the same
dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative
density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 x 10⁵ Pa).
Answer:
Radius of the soap bubble, r = 5mm= 5 x 10^-3 m
Surface tension of the soap solution, S = 2.5 x 10^-2 N/m
Since the soap Bubble has two surfaces,
Excess pressure. P = \(\frac{4 S}{r}\)
= \(\frac{4 \times 2 \cdot 5 \times 10^{-2}}{5 \times 10^{-3}}\)
P = 20 Pa
Excess pressure inside the soap bubble is 20 Pa
Radius of air bubble, r = 5 mm = 5 x 10^-3
Depth, h = 40 t = 0.4 m
Density of soap solution. p = 1.2 x 10^-3 kg/m²
Atmospheric pressure, P0 = 1.0.1 x 10⁵Pa
g = 9.1 m/s²
Excess pressure inside air bubble,
P₁ = 25/r
P₁ = \(\frac{2 \times 2 \cdot 5 \times 10^{-2}}{5 \times 10^{-3}}\) = 10Pa
At a depth of 0.4 m. thc toa1 pressure inside the air bubble.
P_total = P₀ + pgh +P₁
P_total =101 x 10⁵ + 1.2 x 10³x 9.8 x 0.4 +10
P_total =1.057 x 10⁵Pa
∴ the pressure inside the ait bubble is 1.06 x 10⁵ Pa.

Question 25.
A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm². The tank is filled with water in one compartment, and an acid (of relative density 1.7) In the other, both to a height of 4 m. Compute the force necessary to keep the door closed.
Answer:
Assume g = 9.8 m/s²
Density of water, pi = 10³ kg/m³ height of water column, h₁ = 4m
∴ Pressure due to water, P₁ = p₁ gh₁
= 10³ X 9.8 X 4 = 3.92 X 10⁴ Pa
Density of acid = Relative density x density of water .
= 1.7 x 10³kg/m³
Height of acid column, h₂ = 4m
∴ Pressure due to acid, P₂ = p₂ gh₂
= 1.7 x 10³ x 9.8 x 4
P₂ = 6.664 x 10⁴ Pa
Pressure difference between the water and acid columns:
∆P = P₂ – P₁
∆P = 6.664 x 10⁴ – 3.92 x 10⁴
∆P = 2.744 x 10⁴ Pa
Area of the door, a = 20 cm² = 20 x 10⁴ m²
∴ Force exerted on the door, F = ∆P x a
F = 2.144 x 10⁴ x 20 x 10^-4 = 54.88 N
Therefore, the force necessary to keep the door closed is 54.88 N.

Question 26.
A manometer reads the pressure of a gas in an enclosure as shown in Figure (a) When a pump removes some of the gas, the manometer reads as in Figure (b) The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury, (b) How would the levels change in case of (b) If 13.6 cm of water (Immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).
Answer:
(a) Difference between the mercury levels in the two limbs gives gauge pressure
.’. Gauge pressure, Pa = 20 cm of mercury.
Atmospheric pressure, PQ = 76 cm of mercury
Absolute pressure = Atmospheric pressure + Gauge pressure = 76 cm + 20 cm = 96 cm of mercury

(b) Difference between the levels of mercury in the two limbs = – 18 cm Gauge pressure = – 18cm of mercury.
Absolute pressure = Atmospheric pressure + Gauge pressure = 76 cm – 18 cm = 58 cm of mercury.

Question 27.
During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein?
Answer:
Gauge pressure, P = 2000 Pa
Density of blood, ρ = 1.06 x 10³ kg/m³
g = 9,8 m/s²
Let the height of the blood container be h
∴ pressure of the blood container, P = ρgh
2000 = pgh
h = \(\frac{2000}{1 \cdot 06 \times 10^{3} \times 9 \cdot 8}\) = 0.1925m
The blood may just enter the vein if the blood container is kept at a height slightly greater than 0.1925 m.

Question 28.
In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy.
(a) What is the largest average velocity of blood flow in an artery of diameter 2 x 10^-3 remain laminar?
(b) How does the pressure change as the fluid moves along the tube if dissipative forces are present?
(c) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.
Answer:
(a) Diameter of artery = 2 x 10^-3 m
Viscosity of blood, = 2.084 x 10^-3 Pa s
Density of blood, ρ = 1.06 x 10³ kg/m³
Reynolds number for laminar flow, N_R = 2000
The largest average velocity of blood is

The largest average velocity of blood is 1.966 m/s.

(b) If dissipative forces are present, then some forces in liquid flow due to pressure difference is spent against dissipative forces due to which the pressure drop becomes large.

(c) The dissipative forces become more important with increasing flow velocity. This is because of the rise of turbulence. Turbulent flow causes dissipative losses in fluid.

Question 29.
(a) What is the largest average velocity of blood flow in an artery of radius 2 x 10“3 m if the flow must remain laminar? (b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 x 10-3 Pas).
Answer:
(a) Diameter of artery, d = 2 x 2 x 10^-3 m = 4 x 10^-3m
Viscosity of blood, h – 2.084 x 10^-3 m Pas
Density of blood, ρ = 1.06 x 10^-3 kg/m³
Reynold’s number for laminar flow, N_R = 2000
The largest average velocity of blood is,

Hence, the largest average velocity of blood is 0.983 m/s.

(b) Flow rate is given by
R = πr²v_avg
R = 3.14 x (2 x 10^-3)² x 0.983 = 1.235 x 10^-5 m³/s
Therefore, the corresponding flow rate is 1.235 x 10^-5 m³/s

Question 30.
A plane is in level flight at constant speed and each of its two wings has an area of 25 m². If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surfaces determine the plane’s mass. (Take air density to be 1 kg m^-3).
Answer:
Total area of wings, A = 2 x 25 = 50m²
Speed of air lover the longer wing, V₁= 180 km/h = 50 m/s
Speed of air over the upper wings, v₂ = 243 km/h = 65 m/s
Density of air, p = 1 kg/m³
Let P₁ and P₂ be the pressure of air over the lower wing and upper wing respectively.
From Bemoull’s equation,
P₁ + \(\frac{1}{2}\)ρv₁² = P₂ + \(\frac{1}{2}\)ρv₂²)
P₁ – P₂ = \(\frac{1}{2}\)ρ(v₂² – v₁²)
∴ Pressure difference, ∆P
= \(\frac{1}{2}\)ρ(v₂² – v₁²)
= \(\frac{1}{2}\)x 1 x (65² – 50 ²) = 862.5pa
The upward force on the plane, F = (∆P)A
= 862.5 x 50 = 43125 N
We know that the upward force balances the weight of the plane
∴ F = mg
43125 = m x 9.8
m = 4400.51 kg

Question 31.
In Mallikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 x 10^-6 m and density 1.2 x 10³ kg m^-3? Take the viscosity of air at the temperature of the experiment to be 1.8 x 10^-5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air. (Take g = 9.8 ms^-2).
Answer:
Radius of the given unchanged drop, r = 2 x 10^-5 m
Density of the uncharged drop, p = 1.2 x 10³ kg/m^-3
Viscosity of air, η = 1.8 x 10^-5 Pa S
Density of air (p₀) can be taken as zero in order to neglect buoyancy of air.
Terminal velocity (r;) is given by:

The terminal speed of the drop is 5.8 cm/s
Viscous force, F = π6hrv
F = 6 x 3.14 x 1.8 x 10^-5 x 2 x 10^-5 x 5.8 x 10^-2 =3.9 x 10^-10N
The viscous force on the drop is = 3.9 x 10^-10 N

Question 32.
Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 N m. Density of mercury = 13.6 x 10³ kg m^-3.
Answer:
Angie of contact 1θ= 140°
Radius of the tube, r = 1 mm = 10^-3 m
Surface tension of mercury S = 0.465 N/m
Density of mercury, p = 13.6 x 10³ kg/m³
g = 9.8 m/s²
Let the Dip in height of mercury be h
Surface tension is given by,

Negative sign indicates that mercury level dips by 5.34 mm.

Question 33.
Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what, is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 x 10^-2 N/m. Take the angle of contact to be zero and density of water to be 1.0 x 10³ kg m^-3(g = 9.8 ms^-2).
Answer:
Radius of the first bore, r₁ = 3mm / 2= 1.5 x 10^-3m
Radius of the second bore, r₂ = 6mm / 2 = 3 x 10 ^-3m
Surface tension of water, S = 7.3 x 10^-2 N/m
Angle of contact, θ = 0° (give in the question)
Density of water, p = 1 x 10³ kg/m³
Let h₁ and h₂ he the heights to which the water rises in the 3 mm and 6 mm diameter bores respectively. We have,
h₁ = \(\frac{2 S \cos \theta}{r_{1} \rho g}\) h₂ = \(\frac{2 S \cos \theta}{r_{2} \rho g}\)
Difference in water level = h₁ = h₂


The difference in the water levels in the cores is 4.97 mm.