At Saralstudy, we are providing you with the solution of Class 11 Physics Mechanical Properties of Fluids according to the latest NCERT (CBSE) Book guidelines prepared by expert teachers. Here we are trying to give you a detailed answer to the questions of the entire topic of this chapter so that you can get more marks in your examinations by preparing the answers based on this lesson. We are trying our best to give you detailed answers to all the questions of all the topics of Class 11 Physics Mechanical Properties of Fluids so that you can prepare for the exam according to your own pace and your speed.
(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) decreases
The 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 x 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 x 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 = d / 2 = 0.005 m
Area of the heel = πr2
= π (0.005)2
= 7.85 × 10-5 m2
Force exerted by the heel on the floor:
F = mg
= 50 × 9.8 = 490 N
Pressure exerted by the heel on the floor:
P = Force / Area
= 490 / 7.85 x 10-5
= 6.24 × 106 N m-2
Therefore, the pressure exerted by the heel on the horizontal floor is 6.24 × 106 Nm-2.
Density of mercury, p1 = 13.6 × 103 kg/m3
Height of the mercury column, h1 = 0.76 m
Density of French wine, p2 = 984 kg/m3
Height of the French wine column = h2
Acceleration due to gravity, g = 9.8 m/s2
The pressure in both the columns is equal, i.e.,
Pressure in the mercury column = Pressure in the French wine column =
p1 h1 g = p2 h2 g
= 10.5 m
Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m.
Yes
The maximum allowable stress for the structure, P = 109 Pa
Depth of the ocean, d = 3 km = 3 x 103 m
Density of water, p = 103 kg/m3
Acceleration due to gravity, g = 9.8 m/s2
The pressure exerted because of the sea water at depth, d =pdg
= 3 x 103 x 103 x 9.8 = 2.94 x 107 Pa
The maximum allowable stress for the structure (109 Pa) is greater than the pressure of the sea water (2.94 x 107 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 cm2 = 425 × 10-4 m2
The maximum force exerted by the load, F = mg
= 3000 × 9.8 = 29400 N
The maximum pressure exerted on the load-carrying piston, P = F / A
= 29400 / 425 x 10-4
= 6.917 × 105 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 × 105 Pa.
Height of the water column, h1 = 10 + 15 = 25 cm
Height of the spirit column, h2 = 12.5 + 15 = 27.5 cm
Density of water, p1 = 1 g cm-3
Density of spirit, p2 = 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:
= hpg
= h × 13.6g … (i)
Difference between the pressures exerted by water and spirit:
= h1p1g
= 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.
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.
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.
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, S = Force of Weight / 2l
= 1.5 × 10-2 / 0.6 = 2.5 x 10-2 N/m
Therefore, the surface tension of the film is 2.5 x 10-2 N m-1.
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, p = 1.06 × 103 kg m-3
Acceleration due to gravity, g = 9.8 m/s2
Height of the blood container = h
Pressure of the blood container, P = hpg
∴ h = P / pg
= 2000 / 1.06 x 103 x 9.8
= 0.1925
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.
(a) 1.966 m/s (b) Yes
(a) Diameter of the artery, d = 2 × 10-3 m
Viscosity of blood, n = 2.084 x 10-3 kg/m3
Density of blood, p = 1.06 × 103 kg/m3
Reynolds' number for laminar flow, NR = 2000
The largest average velocity of blood is given as:
V arg = NRn / pd
= 2000 x 2.084 x 10-3 / 1.06 x 103 x 2 x 10-3
= 1.966 m/s
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.