- (a) The pressure of a liquid is given by the relation:
P= hρg
Where,
P= Pressure
h= Height of the liquid column
ρ= Density of the liquid
g = Acceleration due to the gravity
It can be inferred that pressure is directly proportional to height. Hence, the blood pressure in human vessels depends on the height of the blood column in the body. The height of the blood column is more at the feet than it is at the brain. Hence, the blood pressure at the feet is more than it is at the brain.
(b) Density of air is the maximum near the sea level. Density of air decreases with increase in height from the surface. At a height of about 6 km, density decreases to nearly half of its value at the sea level. Atmospheric pressure is proportional to density. Hence, at a height of 6 km from the surface, it decreases to nearly half of its value at the sea level.
(c) When force is applied on a liquid, the pressure in the liquid is transmitted in all directions. Hence, hydrostatic pressure does not have a fixed direction and it is a scalar physical quantity. - (a) The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact (θ), as shown in the given figure.
S_{la},S_{sa}, and S_{sl} are the respective interfacial tensions between the liquid-air, solid-air, and solid-liquid interfaces. At the line of contact, the surface forces between the three media must be in equilibrium, i.e.,
The angle of contact θ ,is obtuse if S_{sa} < S_{la} (as in the case of mercury on glass). This angle is acute if S_{sl}< S_{la} (as in the case of water on glass).
(b) Mercury molecules (which make an obtuse angle with glass) have a strong force of attraction between themselves and a weak force of attraction toward solids. Hence, they tend to form drops.
On the other hand, water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction toward solids. Hence, they tend to spread out.
(c) Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface. This force is independent of the area of the liquid surface. Hence, surface tension is also independent of the area of the liquid surface.
(d) Water with detergent dissolved in it has small angles of contact (θ). This is because for a small θ,there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportional to the cosine of the angle of contact (θ). If θ is small, then cosθ will be large and the rise of the detergent water in the cloth will be fast.
(e) A liquid tends to acquire the minimum surface area because of the presence of surface tension. The surface area of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take spherical shape. - (a) decreasesThe surface tension of a liquid is inversely proportional to temperature.
(b) increases; decreases
Most fluids offer resistance to their motion. This is like internal mechanical friction, known as viscosity. Viscosity of gases increases with temperature, while viscosity of liquids decreases with temperature.
(c) Shear strain; Rate of shear strain
With reference to the elastic modulus of rigidity for solids, the shearing force is proportional to the shear strain. With reference to the elastic modulus of rigidity for fluids, the shearing force is proportional to the rate of shear strain.
(d) Conservation of mass/Bernoulli’s principle
For a steady-flowing fluid, an increase in its flow speed at a constriction follows the conservation of mass/Bernoulli’s principle.
(e) Greater
For the model of a plane in a wind tunnel, turbulence occurs at a greater speed than it does for an actual plane. This follows from Bernoulli’s principle and different Reynolds’ numbers are associated with the motions of the two planes.
- (a) When air is blown under a paper, the velocity of air is greater under the paper than it is above it. As per Bernoulli’s principle, atmospheric pressure reduces under the paper. This makes the paper fall. To keep a piece of paper horizontal, one should blow over it. This increases the velocity of air above the paper. As per Bernoulli’s principle, atmospheric pressure reduces above the paper and the paper remains horizontal.
(b) According to the equation of continuity:
Area × Velocity = Constant
For a smaller opening, the velocity of flow of a fluid is greater than it is when the opening is bigger. When we try to close a tap of water with our fingers, fast jets of water gush through the openings between our fingers. This is because very small openings are left for the water to flow out of the pipe. Hence, area and velocity are inversely proportional to each other.
(c) The small opening of a syringe needle controls the velocity of the blood flowing out. This is because of the equation of continuity. At the constriction point of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure applied.
(d) When a fluid flows out from a small hole in a vessel, the vessel receives a backward thrust. A fluid flowing out from a small hole has a large velocity according to the equation of continuity:
Area × Velocity = Constant
According to the law of conservation of momentum, the vessel attains a backward velocity because there are no external forces acting on the system.
(e) A spinning cricket ball has two simultaneous motions –rotatory and linear. These two types of motion oppose the effect of each other. This decreases the velocity of air flowing below the ball. Hence, the pressure on the upper side of the ball becomes lesser than that on the lower side. An upward force acts upon the ball. Therefore, the ball takes a curved path. It does not follow a parabolic path. - Mass of the girl, m= 50 kg
Diameter of the heel, d= 1 cm = 0.01 m
Radius of the heel, r
Area of the heel
= π (0.005)^{2}
= 7.85 × 10^{–5} m^{2}
Force exerted by the heel on the floor:
F = mg
= 50 × 9.8
= 490 N
Pressure exerted by the heel on the floor:
= 6.24 × 10^{6}N m^{–2}
Therefore, the pressure exerted by the heel on the horizontal floor is
6.24 × 10^{6}Nm^{–2}. - 10.5 m
Density of mercury, ρ_{1}= 13.6 × 10^{3} kg/m^{3}
Height of the mercury column, h_{1} = 0.76 m
Density of French wine,ρ_{2} = 984 kg/m^{3}
Height of the French wine column = h_{2}
Acceleration due to gravity, g = 9.8 m/s^{2}
The pressure in both the columns is equal, i.e.,
Pressure in the mercury column = Pressure in the French wine column
= 10.5 m
Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m. - Answer: Yes
The maximum allowable stress for the structure, P = 10^{9} Pa
Depth of the ocean, d= 3 km = 3 × 10^{3} m
Density of water, ρ= 10^{3} kg/m^{3}
Acceleration due to gravity, g = 9.8 m/s^{2}
The pressure exerted because of the sea water at depth, d = ρdg
= 3 × 10^{3} × 10^{3} × 9.8
= 2.94 × 10^{7} Pa
The maximum allowable stress for the structure (10^{9} Pa) is greater than the pressure of the sea water (2.94 × 10^{7} Pa). The pressure exerted by the ocean is less than the pressure that the structure can withstand. Hence, the structure is suitable for putting up on top of an oil well in the ocean.
The maximum mass of a car that can be lifted, m = 3000 kg
Area of cross-section of the load-carrying piston, A = 425 cm^{2 }= 425 × 10^{–4} m^{2}
The maximum force exerted by the load, F = mg
= 3000 × 9.8
= 29400 N
The maximum pressure exerted on the load-carrying piston,
= 6.917 × 10^{5} Pa
Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is 6.917 × 10^{5} Pa.- The given system of water, mercury, and methylated spirit is shown as follows:
Height of the spirit column, h_{1} = 12.5 cm = 0.125 m
Height of the water column, h_{2} = 10 cm = 0.1 m
P_{0} = Atmospheric pressure
ρ_{1}= Density of spirit
ρ_{2}= Density of water
Pressure at point B =
Pressure at point D =
Pressure at points B and D is the same.
Therefore, the specific gravity of spirit is 0.8. - Height of the water column, h_{1} = 10 + 15 = 25 cm
Height of the spirit column, h_{2} = 12.5 + 15 = 27.5 cm
Density of water, ρ_{1} = 1 g cm^{–3}
Density of spirit, ρ_{2} = 0.8 g cm^{–3}
Density of mercury = 13.6 g cm^{–3}
Let h be the difference between the levels of mercury in the two arms.
Pressure exerted by height h, of the mercury column:
= hρg
= h × 13.6g … (i)
Difference between the pressures exerted by water and spirit:
= g(25 × 1 – 27.5 × 0.8)
= 3g … (ii)
Equating equations (i) and (ii), we get:
13.6 hg = 3g
h = 0.220588 ≈ 0.221 cm
Hence, the difference between the levels of mercury in the two arms is 0.221 cm. - Answer: No
Bernoulli’s equation cannot be used to describe the flow of water through a rapid in a river because of the turbulent flow of water. This principle can only be applied to a streamline flow.
Answer: No
It does not matter if one uses gauge pressure instead of absolute pressure while applying Bernoulli’s equation. The two points where Bernoulli’s equation is applied should have significantly different atmospheric pressures.- Answer: 9.8 × 10^{2} Pa
Length of the horizontal tube, l = 1.5 m
Radius of the tube, r= 1 cm = 0.01 m
Diameter of the tube, d= 2r = 0.02 m
Glycerine is flowing at a rate of 4.0 × 10^{–3} kg s^{–1}.
M = 4.0 × 10^{–3} kg s^{–1}
Density of glycerine, ρ= 1.3 × 10^{3} kg m^{–3}
Viscosity of glycerine,η = 0.83 Pa s
Volume of glycerine flowing per sec:
= 3.08 × 10^{–6} m^{3} s^{–1}
According to Poiseville’s formula, we have the relation for the rate of flow:
Where, p is the pressure difference between the two ends of the tube
= 9.8 × 10^{2} Pa
Reynolds’ number is given by the relation:
Reynolds’ number is about 0.3. Hence, the flow is laminar. - Speed of wind on the upper surface of the wing, V_{1} = 70 m/s
Speed of wind on the lower surface of the wing, V_{2} = 63 m/s
Area of the wing, A= 2.5 m^{2}
Density of air, ρ= 1.3 kg m^{–3}
According to Bernoulli’s theorem, we have the relation:
Where,
P_{1} = Pressure on the upper surface of the wing
P_{2} = Pressure on the lower surface of the wing
The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane.
Lift on the wing
= 1512.87
= 1.51 × 10^{3} N
Therefore, the lift on the wing of the aeroplane is 1.51 × 10^{3} N.
Answer: (a)
Take the case given in figure (b).
Where,
A_{1} = Area of pipe1
A_{2} = Area of pipe 2
V_{1} = Speed of the fluid in pipe1
V_{2} = Speed of the fluid in pipe 2
From the law of continuity, we have:
When the area of cross-section in the middle of the venturimeter is small, the speed of the flow of liquid through this part is more. According to Bernoulli’s principle, if speed is more, then pressure is less.
Pressure is directly proportional to height. Hence, the level of water in pipe 2 is less.
Therefore, figure (a) is not possible.- Area of cross-section of the spray pump, A_{1} = 8 cm^{2 }= 8 × 10^{–4} m^{2}
Number of holes, n= 40
Diameter of each hole,d = 1 mm = 1 × 10^{–3} m
Radius of each hole, r= d/2 = 0.5 × 10^{–3} m
Area of cross-section of each hole, a = πr^{2}= π (0.5 × 10^{–3})^{2} m^{2}
Total area of 40 holes,A_{2 }= n × a
= 40 × π(0.5 × 10^{–3})^{2} m^{2}
= 31.41 × 10^{–6} m^{2}
Speed of flow of liquid inside the tube, V_{1 }= 1.5 m/min = 0.025 m/s
Speed of ejection of liquid through the holes = V_{2}
According to the law of continuity, we have:
= 0.633 m/s
Therefore, the speed of ejection of the liquid through the holes is 0.633 m/s.
The weight that the soap film supports, W = 1.5 × 10^{–2} N
Length of the slider, l= 30 cm = 0.3 m
A soap film has two free surfaces.
∴Total length = 2l = 2 × 0.3 = 0.6 m
Surface tension,
=
Therefore, the surface tension of the film is 2.5 × 10^{–2} N m^{–1}.- Take case (a):
The length of the liquid film supported by the weight, l = 40 cm = 0.4 cm
The weight supported by the film, W = 4.5 × 10^{–2} N
A liquid film has two free surfaces.
∴Surface tension
In all the three figures, the liquid is the same. Temperature is also the same for each case. Hence, the surface tension in figure (b) and figure (c) is the same as in figure (a), i.e., 5.625 × 10^{–2}N m^{–1}.
Since the length of the film in all the cases is 40 cm, the weight supported in each case is 4.5 × 10^{–2} N. - Answer:
Radius of the mercury drop, r= 3.00 mm = 3 × 10^{–3}m
Surface tension of mercury, S= 4.65 × 10^{–1}N m^{–1}
Atmospheric pressure, P_{0}= 1.01 × 10^{5}Pa
Total pressure inside the mercury drop
= Excess pressure inside mercury + Atmospheric pressure
= 1.0131 × 10^{5}
= 1.01 ×10^{5}Pa
Excess pressure
Excess pressure inside the soap bubble is 20 Pa;
Pressure inside the air bubble is
Soap bubble is of radius, r= 5.00 mm = 5 × 10^{–3}m
Surface tension of the soap solution, S = 2.50 × 10^{–2}Nm^{–1}
Relative density of the soap solution = 1.20
∴Density of the soap solution, ρ= 1.2 × 10^{3}kg/m^{3 }
Air bubble formed at a depth, h= 40 cm = 0.4 m
Radius of the air bubble, r= 5 mm = 5 × 10^{–3}m
1 atmospheric pressure = 1.01 × 10^{5}Pa
Acceleration due to gravity, g = 9.8 m/s^{2}
Hence, the excess pressure inside the soap bubble is given by the relation:
Therefore, the excess pressure inside the soap bubble is 20 Pa.
The excess pressure inside the air bubble is given by the relation:
Therefore, the excess pressure inside the air bubble is 10 Pa.
At a depth of 0.4 m, the total pressure inside the air bubble
= Atmospheric pressure + hρg + P’
Therefore, the pressure inside the air bubble is- Base area of the given tank, A = 1.0 m^{2}
Area of the hinged door, a = 20 cm^{2 }= 20 × 10^{–4} m^{2}
Density of water, ρ_{1} = 10^{3} kg/m^{3}
Density of acid, ρ_{2} = 1.7 × 10^{3} kg/m^{3}
Height of the water column, h_{1} = 4 m
Height of the acid column, h_{2} = 4 m
Acceleration due to gravity, g = 9.8
Pressure due to water is given as:
Pressure due to acid is given as:
Pressure difference between the water and acid columns:
Hence, the force exerted on the door = ΔP × a
= 2.744 × 10^{4} × 20 × 10^{–4}
= 54.88 N
Therefore, the force necessary to keep the door closed is 54.88 N. - Answer: (a) 96 cm of Hg & 20 cm of Hg; 58 cm of Hg & –18 cm of Hg
(b) 19 cm
(a) For figure (a)
Atmospheric pressure, P_{0 }= 76 cm of Hg
Difference between the levels of mercury in the two limbs gives gauge pressure
Hence, gauge pressure is 20 cm of Hg.
Absolute pressure = Atmospheric pressure + Gauge pressure
= 76 + 20 = 96 cm of Hg
For figure (b)
Difference between the levels of mercury in the two limbs = –18 cm
Hence, gauge pressure is –18 cm of Hg.
Absolute pressure = Atmospheric pressure + Gauge pressure
= 76 cm – 18 cm = 58 cm
(b) 13.6 cm of water is poured into the right limb of figure (b).
Relative density of mercury = 13.6
Hence, a column of 13.6 cm of water is equivalent to 1 cm of mercury.
Leth be the difference between the levels of mercury in the two limbs.
The pressure in the right limb is given as:
Atmospheric pressure + 1 cm of Hg
= 76 + 1 = 77 cm of Hg … (i)
The mercury column will rise in the left limb.
Hence, pressure in the left limb,
Equating equations (i) and (ii), we get:
77 = 58 + h
∴h= 19 cm
Hence, the difference between the levels of mercury in the two limbs will be 19 cm. - Answer:Yes
Two vessels having the same base area have identical force and equal pressure acting on their common base area. Since the shapes of the two vessels are different, the force exerted on the sides of the vessels has non-zero vertical components. When these vertical components are added, the total force on one vessel comes out to be greater than that on the other vessel. Hence, when these vessels are filled with water to the same height, they give different readings on a weighing scale. - Gauge pressure, P= 2000 Pa
Density of whole blood,ρ= 1.06 × 10^{3}kg m^{–3}
Acceleration due to gravity, g = 9.8 m/s^{2}
Height of the blood container = h
Pressure of the blood container,P= hρg
The blood may enter the vein if the blood container is kept at a height greater than 0.1925 m, i.e., about 0.2 m. - Answer: (a)1.966 m/s (b) Yes
(a) Diameter of the artery, d= 2 × 10^{–3}m
Viscosity of blood,
Density of blood,ρ=1.06 × 10^{3}kg/m^{3}
Reynolds’number for laminar flow,N_{R}= 2000
The largest average velocity of blood is given as:
Therefore, the largest average velocity of blood is 1.966 m/s.
(b) As the fluid velocity increases, the dissipative forces become more important. This is because of the rise of turbulence. Turbulent flow causes dissipative loss in a fluid. - (a)Radius of the artery, r= 2 × 10^{–3}m
Diameter of the artery, d= 2 × 2 × 10^{–3}m = 4 × 10^{–3 }m
Viscosity of blood,
Density of blood,ρ= 1.06 × 10^{3}kg/m^{3}
Reynolds’number for laminar flow, N_{R}= 2000
The largest average velocity of blood is given by the relation:
Therefore, the largest average velocity of blood is 0.983 m/s.
(b) Flow rate is given by the relation:
R= π r^{2}
Therefore, the corresponding flow rate is. - The area of the wings of the plane, A= 2 × 25 = 50 m^{2}
Speed of air over the lower wing, V_{1}= 180 km/h = 50 m/s
Speed of air over the upper wing, V_{2}= 234 km/h = 65 m/s
Density of air, ρ= 1 kg m^{–3}
Pressure of air over the lower wing = P_{1}
Pressure of air over the upper wing=P_{2}
The upward force on the plane can be obtained using Bernoulli’s equation as:
The upward force (F) on the plane can be calculated as:
Using Newton’s force equation, we can obtain the mass (m) of the plane as:
∼ 4400 kg
Hence, the mass of the plane is about 4400 kg. - Terminal speed = 5.8 cm/s; Viscous force = 3.9 × 10^{–10 }N
Radius of the given uncharged drop, r = 2.0 × 10^{–5} m
Density of the uncharged drop, ρ =1.2 × 10^{3} kg m^{–3}
Viscosity of air,
Density of air can be taken as zero in order to neglect buoyancy of air.
Acceleration due to gravity, g = 9.8 m/s^{2}
Terminal velocity (v) is given by the relation:
Hence, the terminal speed of the drop is 5.8 cm s^{–1}.
The viscous force on the drop is given by:
Hence, the viscous force on the drop is 3.9 × 10^{–10 }N.
Angle of contact between mercury and soda lime glass, θ= 140°
Radius of the narrow tube, r = 1 mm = 1 × 10^{–3} m
Surface tension of mercury at the given temperature, s = 0.465 N m^{–1 }
Density of mercury, ρ=13.6 × 10^{3} kg/m^{3}
Dip in the height of mercury = h
Acceleration due to gravity, g = 9.8 m/s^{2}
Surface tension is related with the angle of contact and the dip in the height as:
Here, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by 5.34 mm.- Diameter of the first bore, d_{1} = 3.0 mm = 3 × 10^{–3} m
Diameter of the second bore, = 6.0 mm
Surface tension of water, s = 7.3 × 10^{–2}N m^{–1}
Angle of contact between the bore surface and water, θ=0
Density of water, ρ =1.0 × 10^{3} kg/m^{–3}
Acceleration due to gravity, g = 9.8 m/s^{2}
Let h_{1} and h_{2 }be the heights to which water rises in the first and second tubes respectively. These heights are given by the relations:
The difference between the levels of water in the two limbs of the tube can be calculated as:
Hence, the difference between levels of water in the two bores is 4.97 mm.
(a) Volume of the balloon, V = 1425 m^{3}
Mass of the payload, m = 400 kg
Acceleration due to gravity, g = 9.8 m/s^{2}
Density of the balloon = ρ
Height to which the balloon rises = y
Density (ρ) of air decreases with height (y) as:
This density variation is called the law of atmospherics.
It can be inferred from equation (i) that the rate of decrease of density with height is directly proportional to ρ, i.e.,
Where, k is the constant of proportionality
Height changes from 0 to y, while density changes from to ρ.
Integrating the sides between these limits, we get:
(b)
Hence, the balloon will rise to a height of 8 km.
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Monday, October 29, 2012
Mechanical Properties Of Fluids
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