
Q1 Explain why (a) The blood pressure in humans is greater at the feet than at the brain (b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km (c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area. Ans: (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.
Q3 Fill in the blanks using the word(s) from the list appended with each statement: (a) Surface tension of liquids generally . . . with temperatures (increases / decreases) (b) Viscosity of gases. .. with temperature, whereas viscosity of liquids . . . with temperature (increases / decreases) (c) For solids with elastic modulus of rigidity, the shearing force is proportional to . . . , while for fluids it is proportional to . .. (shear strain / rate of shear strain) (d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli's principle) (e) For the model of a plane in a wind tunnel, turbulence occurs at a ... speed for turbulence for an actual plane (greater / smaller) Ans: (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 steadyflowing 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.
Q4 Explain why (a) To keep a piece of paper horizontal, you should blow over, not under, it (b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers (c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection (d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel (e) A spinning cricket ball in air does not follow a parabolic trajectory Ans: (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.
Q5 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? Ans: 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 = πr^{2}
= 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:
P = Force / Area
= 490 / 7.85 x 10^{5}
= 6.24 × 10^{6} N m^{2}
Therefore, the pressure exerted by the heel on the horizontal floor is 6.24 × 106 Nm^{2}.
Q6 Toricelli's barometer used mercury. Pascal duplicated it using French wine of density 984 kg m3. Determine the height of the wine column for normal atmospheric pressure. Ans: Density of mercury, p_{1} = 13.6 × 10^{3} kg/m^{3}
Height of the mercury column, h_{1} = 0.76 m
Density of French wine, p_{2} = 984 kg/m3
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 =
p_{1 } h_{1 }g = p_{2 } h_{2 }g_{ }
= 10.5 m
Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m.
Q7 A vertical offshore 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. Ans: Yes
The maximum allowable stress for the structure, P = 10^{9} Pa
Depth of the ocean, d = 3 km = 3 x 10^{3} m
Density of water, p = 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 =pdg
= 3 x 10^{3} x 10^{3} x 9.8 = 2.94 x 10^{7} Pa
The maximum allowable stress for the structure (10^{9} Pa) is greater than the pressure of the sea water (2.94 x 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.
Q8 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? Ans: The maximum mass of a car that can be lifted, m = 3000 kg
Area of crosssection of the loadcarrying 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 loadcarrying piston, P = F / A
= 29400 / 425 x 10^{4}
= 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 × 105 Pa.
Q10 In problem 10.9, 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) Ans: 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, p_{1} = 1 g cm^{3}
Density of spirit, p_{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:
= hpg
= h × 13.6g … (i)
Difference between the pressures exerted by water and spirit:
= h_{1}p_{1}g
= 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.
Q11 Can Bernoulli's equation be used to describe the flow of water through a rapid in a river? Explain. Ans: 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.
Q12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli's equation? Explain. Ans: 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.
Q17 A Ushaped 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 102 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? Ans: 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}.
Q23 Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale? Ans: 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 nonzero 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.
Q24 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? [Use the density of whole blood from Table 10.1]. Ans: Gauge pressure, P = 2000 Pa
Density of whole blood, p = 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 = hpg
∴ h = P / pg
= 2000 / 1.06 x 10^{3} 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.
Q25 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 103 m if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively. Ans: (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/m^{3}
Density of blood, p = 1.06 × 10^{3} kg/m^{3}
Reynolds' number for laminar flow, N_{R} = 2000
The largest average velocity of blood is given as:
V_{ arg} = N_{R}n / 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.