Mechanical Properties Of Fluids
Class 11th Physics Part II CBSE Solution
Exercise
Question 1.Explain why
The blood pressure in humans is greater at the feet than at the brain
Answer:
Human body is filled with blood which constitutes for the hydrostatic pressure. Hydrostatic pressure depends on the height of the liquid column.
P = ρgh
Where, P = Pressure
h = height of the liquid column
g = Acceleration due to gravity = 9.8 m/s
Height of the blood column at feet is greater than the height of blood column at the brain.
Therefore blood pressure in humans is greater at the feet than at the brain.
Question 2.
Explain why
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
Answer:
The air at sea level is compressed by the atmospheric air column above that level. At higher altitudes the air column above decreases. So the air will expand and its density decreases. Also at higher altitudes gravity is less because of which air molecules can spread freely. Hence density of air decreases with increasing altitude. So the density of air is less at a height of about 6 km from the ground than the density of air at sea level.
We know that hydrostatic pressure is directly proportional to density of the fluid and the height of the liquid column above that level.
P = ρgh
As both density of air and height of the air column 6 km above sea level are less compared to that at the sea level , Atmospheric pressure decreases to nearly half of its value at 6 lm above the sea level.
Question 3.
Explain why
Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.
Answer:
Pascal's law states that pressure exerted anywhere on a confined incompressible fluid is transmitted equally and undiminished throughout the entire fluid. So when force is applied on a liquid, pressure is transmitted in all directions. A vector has only one direction but hydrostatic pressure does not have a specific direction . So it is considered a scalar quantity.
Figure showing the pictorial representation of Pascal's law
Question 4.
Explain why
The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
Answer:
The angle between the glass surface and tangent to the liquid surface at the point of contact with the glass surface is known as angle of contact (θ)
Figure showing various tensions at the interfaces
Let Sla = Interfacial tension between liquid-air
Ssa = Interfacial tension between solid-air
Ssl = Interfacial tension between solid-liquid
At the line of contact, the surface forces between the three media should be in equilibrium.
SlaCosθ = Ssa - Ssl
cosθ = (Ssa - Ssl)/Sla
We also know that surface force is directly proportional to the density of the substance.
In case of mercury, the density of mercury is greater than that of density of glass, therefore cohesive forces exerted by mercury molecules are greater than adhesive forces between the glass and the mercury molecules .But density of air is less that of density of glass; so adhesive forces are greater than the cohesive forces. The surface tension of a liquid results from the imbalance of the cohesive forces between molecules. Therefore interfacial tension between the glass and mercury is greater than the tension between solid-air surface.
So, Sla>Ssa.
cosθ = (Ssa - Ssl)/Sla is negative. Hence θ is an obtuse.
In case of water, the density of water is lesser than that of density of glass molecules, therefore cohesive forces exerted by water are lesser than adhesive forces between the glass and water molecules. Therefore the interfacial tension between glass-water is less than tension between air-glass surface.
So, Sla<Ssa
cosθ = (Ssa - Ssl)/Sla is positive. Hence θ is acute.
Question 5.
Explain why
Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)
Answer:
Mercury molecules have strong forces of attraction between themselves (cohesive forces) and the forces of attraction between molecules of mercury and glass are weak (Adhesive forces).So they form a droplets on the glass surface.
Water molecules have weak forces of attraction between themselves (cohesive forces) and the forces of attraction between the water molecules and glass molecules is strong (adhesive forces). So water spreads on glass.
Question 6.
Explain why
Surface tension of a liquid is independent of the area of the surface
Answer:
The force acting per unit length at the interface between the plane of a liquid and any other surface is known as surface tension. The force acting does not depend on the area of the surface. Therefore surface tension is independent of area of the surface.
Question 7.
Explain why
Water with detergent dissolved in it should have small angles of contact.
Answer:
If angle of contact (θ) is small, cosθ will be large. We know that the capillary rise is directly proportional to cosθ
As water enters into the pores of our clothes by capillary action, the capillary rise of a detergent dissolved in a liquid will be more if the angle of contact is acute.
Question 8.
Explain why
A drop of liquid under no external forces is always spherical in shape
Answer:
The most stable state of any system is that which possesses the least amount of energy. A liquid tends to have minimum surface area because it has less surface energy. Surface area of a sphere is minimum for a given volume. Hence under no external pressure drop of a liquid always forms a spherical surface.
Question 9.
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)
Answer:
(a) decreases
Explanation: Surface tension decreases with increase in temperature because cohesive forces in the liquid decrease with increasing molecular thermal activity. As surface tension depends on cohesive forces of the liquid, it's value decreases at higher temperatures.
(b) increases, decreases
Explanation: Usually gases are free flowing in the atmosphere. When temperature increases collisions between the molecules increase and as a result the their free flow is resisted. Resistance to free flow is known as viscosity. Therefore viscosity increases in gases due to increase in temperature.
(c) shear strain, rate of shear strain
Explanation: In solids 
Therefore in solids, shear stress is directly proportional to shear strain.
In fluids according to continuum mechanics, in a Newtonian fluid shear stress is directly proportional to rate of shear strain.
(d) Conservation of mass, Bernoulli's principle
Explanation: A steady flowing fluid's increase in flow speed at a constriction depends on both conservation of mass and Bernoulli's principle.
(e) greater
Explanation: For the model of plane in a wind tunnel turbulence occurs at a greater speed than for an actual plane. This is due to the difference in Reynolds's number associated with the planes.
Question 10.
Explain why
To keep a piece of paper horizontal, you should blow over, not under, it
Answer:
When we blow over a paper velocity of air above the plane of the paper increases. As a result Kinetic energy above the plane of paper increases.
According to Bernoulli's principle
As K.E increases, Pressure energy above the plane of paper decreases below the atmospheric pressure. The pressure below the paper is atmospheric pressure As a result pressure above the paper is less than pressure below the paper. So the paper remains horizontal and does not fall.
Question 11.
Explain why
When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers
Answer:
We know that according to equation of continuity
Area × velocity = constant
By closing a water tap with our fingers, the area of outlet of water jet is reduced. As a result velocity of the outlet jet increases according to the above equation of continuity.
Question 12.
Explain why
The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection
Answer:
From Bernoulli's theorem, we know that
We know that diameter of the needle is very less. So flow velocity through the needle will be very high according to equation of continuity; Area × velocity = constant
So needle controls the velocity of the fluid flow
By using our thumb we apply more pressure to the flow of fluid . So thumb controls pressure.
The power of pressure is 1 and the power of velocity is 2. Thus size of the needle has more control over the flow rate than the thumb pressure.
Question 13.
Explain why
A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel
Answer:
The area through which water is flowing out of the hole in the vessel is small. According to equation of continuity ,
Area × velocity = constant
So the velocity of water flowing through the hole is large.
According to law of conservation of momentum,
m1v1 = m2v2
Due to this the vessel attains a backward velocity as there are no external forces acting on the system. So we can that water exerts a backward thrust on the vessel.
Question 14.
Explain why
A spinning cricket ball in air does not follow a parabolic trajectory
Answer:
A spinning cricket ball has velocity both due to linear(v) and rotational motion(u). when a ball is moved forward, the air occupies the space left by the ball with a velocity v (say). When the ball spins, the layer of air around it also moves with the ball, say with the velocity 'u'. So the resultant velocity of air above the ball becomes (v-u) and below the ball becomes (v + u). Hence, the Due to this the velocity of air above the ball decreases and below the ball increases. According to Bernoulli's principle decrease in velocity increases pressure on the upper side of the ball.
So there is a downward force on the ball and because of this and it fails to follow parabolic trajectory.
Question 15.
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:
Given:
Mass of the girl (m) = 50kg
Diameter of the heel (d) = 1.0cm
Radius of the heel (r) = d/2 = 0.5cm
Area of the heel(A) = π × r2
= π × (0.5 × 10-2 m)2
= 7.8539 × 10-5 m2
We know that Pressure (P) = F/A
F = Force exerted by heel on the floor
A = Area of the heel
F = m × g where m = mass of the girl
= 50 kg × 9.8 m/s2 g = acceleration due to gravity
= 490 N
P = F/A
= 490 N/(7.8539 × 10-5) m2
= 6.24 × 106 N/m2
Pressure exerted by heel on the horizontal floor is 6.24 × 106N/m2
Question 16.
Toricelli’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.
Answer:
In Toricelli's barometer mercury is used
Density of mercury (ρ1) = 13.6 × 103 kg/m3
Height of mercury column at atmospheric pressure (h1) = 0.76 m
Density of French wine (ρ2) = 984 kg/m3
Height of the French wine column at atmospheric pressure = h2
We know that pressure in a column of height h and density ρ = ρgh
where, g = acceleration due to gravity = 9.8 m/s2
Pressure in both the columns is the same i.e. atmospheric pressure
ρ1gh1 = ρ2gh2
ρ1h1 = ρ2h2
(13.6 × 103) × 0.76 = (984) × h2
h2 = 10.504 m
Hence the height of French white column at normal atmospheric pressure is 10.504 m.
Question 17.
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:
Given:
Maximum stress the structure can withstand is (P) = 109 Pa
Depth of the ocean (d) = 3 km = 3 × 103 m
Density of water (ρ) = 1000 kg/m3
We know that pressure due to depth d = ρgd
Pressure due to depth d of water = ρgd
= 1000 kg/m3 × 9.8 m/s2 × (3 × 103m)
= 2.94 × 107 Pa
The maximum allowable stress for the structure(109Pa ) is greater than the pressure of the sea (2.94 × 107Pa). Hence the structure can be put on top of an oil well in the ocean.
Question 18.
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear?
Answer:
Maximum mass of a car that can be lifted (m) = 3000 kg
Area of cross section of piston carrying the load (A1) = 452 cm2
= 0.0452 m2
Maximum force exerted by the car due to its weight (F) = mg
= 3000kg × 9.8m/s2
= 29400 kgm/s2
Pressure on the piston lifting the car P = F/A
= 
= 6.504 × 105 Pa
Pressure is transmitted equally in all the directions of a fluid. Therefore the maximum pressure on the smaller piston is 6.504 × 105 Pa.
Question 19.
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?
Answer:
Figure showing heights of liquids in an U-Tube
Given:
Height of spirit column (h1) = 12.5 cm
Height of water column (h2) = 10.0 cm
Let P0 = Atmospheric pressure
ρ1 = density of the spirit
ρ2 = density of the water column
Pressure due to spirit column at B = P0 + ρ1gh1
Pressure due to water column at D = P0 + ρ2gh2
Pressure is equal at the points having the same height from the ground i.e. at points B and D.
P0 + ρ1gh1 = P0+ ρ2gh2
ρ1h1 = ρ2h2
We know that 
Therefore specific gravity of spirit is 0.8.
Question 20.
In the previous problem, 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:
Given:
Height of the spirit column (h1) = 12.5 + 15 = 27.5 cm
Height of the water column (h2) = 10 + 15 = 25 cm
Density of spirit (ρ1) = 1 gm/cc
Density of spirit (ρ2) = 0.8 gm/cc
Density of mercury (ρ) = 13.6 gm/cc
Mercury rises because of the differences in pressure exerted by the columns of water and spirit.
Pressure due to height h of mercury column = ρgh
= (13.6gm/cc) × (980 cm/s) × (h cm)
Differences in pressures due to water and spirit = ρ1gh1-ρ2gh2
= (1 × 980 × 25)-(0.8 × 980 × 27.5)
= 2940 gm/cm2s
Pressure due to height h of mercury = Differences in pressures due to water and spirit
13.6 × 980 × h = 2940
h = 0.2205 cm
Difference in the levels of mercury in two columns is 0.2205.
Question 21.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
Answer:
The flow of water through a rapid in a river consist of turbulent flow of water. The Bernoulli’s equation cannot be applied to the turbulent flow. It can only be applied to the streamline flow.
Thus, The Bernoulli’s equation cannot be used to describe the flow of water through a rapid in a river.
Question 22.
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
Answer:
The excess of pressure, P − Pa, at depth h is called a gauge pressure at that point.
The bernoulli’s equation works when the atmospheric pressure at the two points at which it is applied is significantly different. Using gauge pressure instead of absolute pressure in bernoulli’s equation not matter as we are using the difference only.
Question 23.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 103 kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].
Answer:
The Poiseuille’s law states the relation:
Where,
‘p’ is the difference in the pressure between the two ends.
‘r’ is the radius of the tube.
‘ɳ’ is the viscosity of the fluid.
‘l’ is the horizontal length of the tube.
Given,
Length of the tube, l = 1.5 m
Radius of the tube, r = 1 cm=0.01 m
Mass of glycerine flowing per second, M = 4.0 × 10-3 kg/s
Density of the glycerine, ρ = 1.3 × 103 kg/m3
Viscosity of the glycerine, ɳ = 0.83 Pa s
Volume of glycerine flowing per second is given as,
⇒V= 
⇒ V=3.08 × 10-6 m3/sec
Applying Poiseuille’s law we get the relation,
⇒ 
⇒ p=9.8 × 102 Pa
To check whether the assumption of laminar flow in the tube is correct we use the Reynolds’s number relation given as,
Where,
and if the Reynolds number is less than 2000, the flow is laminar.
Where,
‘ρ’ is the density of the fluid.
‘V’ is the volume of fluid flowing per sec
‘d’ is the diameter of the pipe
‘ɳ’ is the viscosity of the fluid
Thus,
⇒ 
⇒ R =0.3
Since, the Reynolds’s number is less than 2000. Hence, the flow is laminar.
The pressure difference between the two ends of the tube is9.8 × 102Pa
Question 24.
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–1and 63 m s-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:
The Bernoulli’s theorem states the relation,
Pa + 
Va2= Pb + Vb2
Thus we get,
Pa – Pb =
(va2 – vb2) …..(i)
Where,
‘Pa’ is the pressure on the upper surface of the wing
‘Pb’ is the pressure on the lower surface of the wing.
‘v1’ is the speed on the upper surface of the wing.
‘v2’ is the speed on the lower surface of the wing.
‘ρ’ is the density of the air.
The reason for the lift of the airplane is the difference between the pressure differences between the upper and lower surface,
Thus,
The force which is responsible for the airplane lift, F = (Pb-Pa) A
Replacing the value of Pressure difference from equation (i),
⇒F = (va2– vb2) A
Given,
Speed of the wind on the upper surface, v1 = 70 m/s
Speed of the wind on the lower surface , v2 = 63 m/s
Area of the wing, A = 2.5 m2
Density of the air, ρ = 1.3 kg / m3
⇒ 
⇒ F = 1512.87 N = 1.512 × 103 N
the lift on the wing of the airplane is 1.512 × 103 N.
Question 25.
Figures 10.23(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?
Answer:
The law of continuity states that,
A v= A’ v’
As taken in the figure,
Where,
Area of ‘Pipe a’ is A
Area of ‘Pipe b’ is A’
Speed of the fluid in the ‘pipe a’ = v
Speed of the fluid in the ‘pipe b’ = v’
The law thus states that if the area of cross section in the Venturimeter is small, the speed of the flowing liquid through the part will be more.
And the Bernoulli’s principle states that if the speed is more around a part, the pressure will be less.
The pressure is directly proportional to the height.
In the figure, the area A’ is less than area A, Thus, the speed in the is higher and thus the height of the in ‘pipe b’ must be higher.
The figure ‘a’ is the incorrect option.
Question 26.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min–1, what is the speed of ejection of the liquid through the holes?
Answer:
The law of continuity states that,
A1v1 = A2v2
⇒ v2 = A1v1 / A2 ……….(i)
Where,
‘A1’ is the area of cross section of the spray pump
‘A2’ is the area of cross section of the end with the 40 holes
‘v1’is the speed of flow of fluid inside the tube
‘v2’ is the speed of flow of the fluid through the holes
Given,
Area of cross section of spray pump, A1 = 8 cm2 = 8 × 10-4 m2
Diameter of each hole, d = 1 mm = 1 × 10-3 m
Radius of each hole, r = d/2 = 
= 0.5 × 10-3 m
Velocity of the liquid flow inside the tube, v1 = 1.5 m min-1
⇒ v1 = 1.5/60 m sec-1 = 0.025 m/s
Thus,
Area of cross section of each hole, a = πr2
⇒ a = 3.14 × (0.5 × 10-3 m)2
⇒ a = 0.785 × 10-6 m2
Total area of cross section of 40 holes, A2 = 40 × a
⇒ A2 = 40 × 0.785 × 10-6 m2
⇒ A2 = 31.41 × 10-6 m2
Therefore, from equation (i),
⇒ v1 =
⇒ 0.633 m/s
Therefore, the speed of ejection of the liquid through the holes is 0.633 m/s.
Question 27.
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 × 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 file?
Answer:
Surface tension, S is written as
S= 
Where,
‘F’ is the Force exerted or weight supported between the wire and the slider
‘l’ is the length of the slider.
Given,
The force exerted, F =1.5 × 10-2 N
The length of the slider, l =30 cm = 0.30 m
The soap film has two free surfaces thus,
Total length = 2l = 2 × 0.30 m = 0.6 m
Therefore, the surface tension, S =
⇒ S = 2.5 × 10-2 N / m
The surface tension of the file is 2.5 × 10-2 N /m
Question 28.
Figure 10.24 (a) shows a thin liquid film supporting a small weight = 4.5 × 10–2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.
Answer:
The surface tension = 
Where,
‘F’ is the weight supported by the films,
‘l’ is the length of the liquid film.
From the equation, we can see that the surface tension only depends on the weight and the length of the film. The liquid is same and at the same temperature in all the three cases (a), (b), (c).
Thus, the surface tension will be same in all the three cases and thus, the weight supported by will be same in all the three cases i.e 4.5 × 10-2 N
Question 29.
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 × 10–1 N m–1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.
Answer:
Total pressure inside a mercury drop is given as ‘P’,
P = Excess pressure inside the mercury drop + Atmospheric pressure
Excess pressure inside the mercury drop, P1 =
Where,
‘S’ is the surface tension of the mercury
‘r’ is the radius of the mercury drop.
Thus,
P = P1 + P0
Given,
Surface tension of the mercury, S = 4.65 × 10-1 N
Radius of the mercury drop, r = 3.0 mm = 3.0 × 10-3 m
Atmospheric pressure, P0 = 1.01 × 105 Pa
⇒ P1 =
⇒ P1 = 3.1 × 102 Pa = 0.0031 × 105 Pa
Therefore,
P = 0.0031 × 105 Pa + 1.01 × 105 Pa
⇒ P = 1.0131 × 105 Pa
The pressure inside the drop of mercury is 1.0131 × 105 Pa and the excess pressure inside the drop is 310 Pa.
Question 30.
What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 × 10–2 N m–1? 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 × 105 Pa).
Answer:
Excess pressure inside the soap bubble:
Excess pressure inside the air bubble:
Where,
S is the surface tension of the soap solution.
r is the radius of the soap bubble.
r’ is the radius of the air bubble.
Total Pressure inside the air bubble at a depth of 0.4 m= P1
P1 = P0 + hρg + P’
Where,
‘P0’ is the atmospheric pressure
Given,
Radius of the soap bubble, r = 5.0 mm = 5 × 10-3 m
Surface tension of the soap solution, S = 2.50 × 10-2 N / m
Relative density to the air of the soap solution, ρ = 1.2 × 103 kg /m3
The height at which the air bubble is formed, h = 40 cm = 0.4 m
Radius of the air bubble, r’ = r = 5 × 10-3 m
Acceleration due to gravity = 9.8 m/s2
The atmospheric pressure, P0 = 1.01 × 105 Pa
Therefore,
Excess pressure inside the soap bubble
⇒ 
⇒ P = 20 Pa
Excess pressure inside the air bubble,
⇒ 
⇒ P’ = 10 Pa
The total pressure inside an air bubble at the depth of 0.4 m, P1
P1 = 1.01 × 105 Pa + (0.4 m × 1.2 × 103 kg/m3 × 9.8 m/s2 ) + 10 Pa
P1 = 1.057 × 105 Pa
The excess pressure inside the soap bubble is 20 Pa, the excess pressure inside the air bubble 10 Pa and the total pressure inside the air bubble is 1.057 × 105 Pa
Question 31.
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 4.0 m. compute the force necessary to keep the door close.
Answer:
Given,
Base area of the tank = 1 m2
Area of the hinged door = 20 cm2
Height of both water and acid columns, h = 4 m
Density of the water, ρw = 1000 kg/m3
Density of the acid, ρa = 1.7 × ρw = 1700 kg/m3
Pressure, P = ρgh
Where,
ρ = density
g = acceleration due to gravity
h = height of the column
∴ Pressure exerted by water, Pw = 1000 × 9.81 × 4
= 3.92 × 104 pa
Pressure exerted by the acid, Pa = 1700 × 9.81 × 4
= 6.664 × 104 pa
∴ Pressure difference, ΔP = Pa – Pw
= (6.664-3.92) × 104 pa
= 2.744 × 104 pa
Thus, Force exerted on the hinged plate, 
⇒ 
= 54.88 N
Question 32.
A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (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 (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) Local atmospheric pressure, Patm = 76 cm of Hg
Gauge pressure = mercury height in (Right limb – Left limb)
Absolute pressure = Atmospheric pressure + Gauge pressure
For figure a:
Gauge pressure = 20 cm of Hg
Absolute pressure = 76 + 20 = 96 cm of Hg
For figure b:
Gauge pressure = -18 cm of Hg
Absolute pressure = 76 - 18 = 56 cm of Hg
(b) Relative density of the mercury = 13.6
∴ 13.6 cm of water exerts same pressure as 1cm of mercury exerts.
Thus,
Pressure in the right limb, Pr = Atmospheric + 1 cm of Hg
= 76 + 1 = 77 cm of Hg
Pressure in the left limb, Pl = 58 + h cm of Hg
∴ Height difference between limbs, h = 77 - 58 = 19 cm
Question 33.
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 up to 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?
Answer:
Given,
Two vessels are having same base area.
Two vessels are filled with water.
We have,
Pressure, P = ρgh
Where,
ρ = density
g = acceleration due to gravity
h = height of the column
Thus, for given two vessels pressure only depends on height of the water column only.
Force at the base, 
Hence, Force exerted on the base of two vessels is same.
Weight = mg
Where,
m = mass
g = acceleration due to gravity
Different volumes of two vessels have different mass. So, on weighing scale they show different readings.
Question 34.
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].
Answer:
From the table 10.1,
Density of the blood, ρ = 1.06 × 103 kg/m3
Gauge pressure, P = 2000 pa
We have,
Pressure, P = ρgh
Where,
ρ = density
g = acceleration due to gravity
h = height of the column
It is clear that, pressure exerted by the column of the blood should at least equal to inside vein pressure to enter blood into the vein from bottle.
Thus,
Required height, 
= 
= 0.1925 m
So, h≈0.2m required to make blood enter into the vein.
Question 35.
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 × 10–3 m if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.
Answer:
(a) Reynolds’s number, 
Where,
ρ = density of the fluid
v = largest average velocity of the fluid
D = diameter of the duct
μ = viscosity of the fluid
We have,
Density of the blood, ρ = 1.06 × 103 kg/m3
Reynolds’s number for laminar flow in duct, Re = 2000
Viscosity of the blood, μ = 2.084 × 10-3 Nm/s
Thus,
Largest average velocity, 
= 1.962 m/s
(b) Dissipative forces are more important when flow velocity is increasing because increase in flow leads to turbulence.
Question 36.
What is the largest average velocity of blood flow in an artery of radius 2 × 10–3m if the flow must remain lanimar?
Answer:
Reynolds’s number,
Where,
ρ = density of the fluid
v = largest average velocity of the fluid
D = diameter of the duct
μ = viscosity of the fluid
We have,
Density of the blood, ρ = 1.06 × 103 kg/m3
Reynolds’s number for laminar flow in duct, Re = 2000
Viscosity of the blood, μ = 2.084 × 10-3 Nm/s
Thus,
Largest average velocity,
= 0.982 m/s
Question 37.
What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10–3 Pa s).
Answer:
Flow rate, Q = Av
Where,
A = Area of the duct
v = velocity of the fluid
Thus, flow rate Q = πr2v
= π × (2 × 10-3)2 m2 × 0.982 m/s
= 1.235 × 10-5 m3/s
Question 38.
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 surface, determine the plane’s mass. (Take air density to be 1 kg m–3).
Answer:
Given,
Area of the two wings, A = 2 × 25 = 50 m2
Air Velocity over wings, v1 = 234 km/h = 65 m/s
Air Velocity below wings, v2 = 180 km/h = 50 m/s
Density of the air, ρ = 1 kg/m3
Bernoulli’s equation,
(∵ neglecting potential energy)
Let, P1 = pressure over wings
P2 = pressure below wings
⇒ 
⇒ 
= 862.5 pa
We have,
Where,
ΔP = change in pressure = P2 – P1
A = area
∴ F = 862.5 pa × 50 m2
= 43125 N
As per Newton’s second law,
F = mg
Where,
F = force
g = acceleration due to gravity
⇒ mass of the plane, 
Thus,
Mass of the plane, m ≈ 4400 kg
Question 39.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10–5 m and density 1.2 × 103 kg m–3. Take the viscosity of air at the temperature of the experiment to be 1.8 × 10–5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
Answer:
Radius of the drop = 2 × 10-5 m
Density of the drop = 1.2 × 103 kg/m3
Viscosity of the air, μ = 1.8 × 10-5 Pa.s
*Consider density of air to be zero in order to neglect buoyancy force create by air on drop.
Terminal velocity is given by the relation,
Where,
r = radius
ρ = highest density = density of drop
ρ0 = Lowest density = density of air
g = acceleration due to gravity = 9.81 m/s
μ = viscosity
∴ 
= 5.8 × 10-2 m/s
= 5.8 cm/s
Hence, the terminal speed of the drop = 5.8 cm/s
The viscous force on the drop is given by,
Where,
μ = viscosity
r = radius
v = velocity
∴ 
= 3.9 × 10-10 N
Hence, the viscous force on drop = 3.9 × 10-10 N
Question 40.
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–1. Density of mercury = 13.6 × 103 kg m–3.
Answer:
Surface tension is related with both angle of contact and dip in the height as, 
Where,
h = dip height
ρ = density = 13.6 × 103 kg/m3
g = acceleration due to gravity = 9.81 m/s
r = radius of the dipped tube = 1 × 10-3 m
θ = contact angle = 1400
and, Surface Tension S = 0.465 N/m
Thus, height of dip down, 
= -0.00534 m
= -5.34 mm
Negative sign indicates decrease in level of mercury in tube. Hence the mercury level is dipped by 5.34 mm.
Question 41.
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 × 10–2 N m–1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kg m–3 (g = 9.8 m s–2).
Answer:
Diameter of the first tube = 3 mm
Diameter of the second tube = 6 mm
Thus, Radius of the first tube, r1 = 1.5 mm = 1.5 × 10-3 m
Radius of the first tube, r2 = 3 mm = 3 × 10-3 m
Surface tension of water, S = 7.3 × 10-2 N/m
Density of water, ρ = 1000 kg/m3
Angle of contact between bore surface and water, θ = 00
Acceleration due to gravity, g = 9.8 m/s
Height of water rise,
Where,
S = Surface tension
θ = Contact angle
ρ = Density
g = acceleration due to gravity
r = radius
Let h1 and h2 be the height rise of water in tube1 and tube2 respectively. Then the height are given by,
; 
Thus, the difference between heights is given by,
Δh = h1 – h2
= 4.996 × 10-3 m
≈ 4.97 mm
∴ The difference between the water levels in two bores = 4.97 mm