RBSE Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids

Rajasthan Board RBSE Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Textbook Exercise Questions and Answers.

RBSE Class 11 Physics Solutions Chapter 10 Mechanical Properties of Fluids

RBSE Class 11 Physics Mechanical Properties of Fluids Textbook Questions and Answers

Question 10.1.
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?
Answer:
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
(a) 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 one surface. At a height of about 6 km, density decreases to nearly half of its value a 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.

Question 10.2.
Explain why:
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute?
(b) 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)?
(c) Surface tension of a liquid is independent of the area of the surface?
(d) Water with detergent dissolved in it should have small angles of contact?
(e) A drop of liquid under no external forces is always spherical in shape?
Answer:
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 figures.

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.,
cos θ = \(\frac{S_{s a}-S_{s l}}{S_{l a}}\)
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).
(a) Mercury molecules (which make an obtuse angle with glass) have a strong force of attraction between themselves and a weak force of attraction towards solids. Hence, they tend to form drops.

(b) On the other hand, water molecules made acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction towards 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 cos0 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 becasue 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.

Question 10.3.
Fill in the blanks using the word (s) from the list appended with each statement:
(a) Surface tension of liquids generally ................................. with temperature, (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 proportinoal 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: The surface tension of a liquid is inversely proportinoal 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 for turbulence for an actual plane. This follows from Bernoulli’s principle and different Reynolds’ numbers are associated with the motions of the two planes.

Question 10.4.
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?
Answer:
(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.
According to the equation of continuity: Area x velcoity = constant

(b) 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 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 of 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.

Question 10.5.
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:
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 x 10^-5 m²
Froce exerted by the heel on the floor:
F = mg = 50 x 9.8 = 490 N
Pressure exerted by the heel on the floor:
P = \(\frac{\text { Force }}{\text { Area }}=\frac{490}{7.85 \times 10^{-5}}\)
= 6.24 x 10⁶ N/m²
Therefore, the pressure exerted by the heel on the horizontal floor is 6.24 x 106 N/m².

Question 10.6.
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.
Answer:
Density of mercury, ρ₁ = 13.6 x 10 kg/m³
Height of the mercury column, h₁ = 0.76 m
Density fo French wine, ρ₂ = 984 kg/m³
Height of the French wine column = h₂
Acceleration due to gravity, g = 9.8m/s²
The pressure in both the columns is equal, i.e.,
Pressure in the mercury column = Pressrue in the French wine column
ρ₁h₁g = ρ₂h₂g
h₂ = \(\frac{\rho_1 h_1}{\rho_2}\)
= \(\frac{13.6 \times 10^3 \times 0.76}{984}\)
= 10.5 m
Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m.

Queston 10.7.
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:
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, ρ = 10³ kg/m³
Acceleration due to gravity, g = 9.8 m/s²
The pressure exerted because of the sea water at depth, (d) = ρdg
= 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.

Question 10.8.
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:
The maximum mass of a car that can be lifted, m = 3000 kg;
Area of cross-section of the load-carrying piston, A = 425cm² = 425 x 10^-4 m²
The maximum force exerted by the load, F = mg = 3000 x 9.8 = 29400 N
The maximum pressure exerted on load-carrying piston, P = \(\frac{F}{A}\)
= \(\frac{29400}{425 \times 10^{-4}}\) = 6.917 x 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 x 10⁵ Pa

Question 10.9.
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:
The given system of water, mercury, and methylated spirit is shown as follows:
Height of. the spirit column,
h₁ = 12.5 cm = 0.125 m

Height of the water column, h₂ = 10 cm = 0.1 m,
P₀ = Atmospheric pressure
ρ₁ = Density of spirit
ρ₂ = Density of water
Pressure at point B= P₀ + h₁ρ₁g
Pressure at point D = P₀ + h₂ρ₂g
Pressure at points B and D is the same.
P₀ + h₁ρ₁g = ρ₀h₂ρ₂g
\(\frac{\rho_1}{\rho_2}=\frac{h_2}{h_1}=\frac{10}{12.5}\) = 0.8
Therefore, the specific gravity of spirit is 0.8.

Question 10.10.
In the previous 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)
Answer:
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, ρ₁ = 1 g cm^-3
Density of spirit, ρ₂ = 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 x 13.6 g .....................(i)
Difference between the pressure exerted by water and spirit:
= h₁ρ₁g - h₂ρ₂g
= g(25 x 1 - 27.5 x 0.8) = 3g ....................(ii)
Equating equations (i) and (ii), we get:
13.6 hg = 3gh
= 0.220588 ≈ 0.221 cm
Hence, the difference between the levels of mercury in the two arms is 0.221 cm.

Question 10.11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
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.

Question 10.12.
Does it matter if one uses gauge instead of absolute pressure in applying Bernoulli’s equation? Explain.
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.

Question 10.13.
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 x 10^-3 kg/s, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 x 10³ kg/m³ 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:
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 x 10^-3 kg/s.
M = 4.0 x 10^-3 kg/s
Density of glycerine, ρ = 1.3 x 10³ kg/m³
Viscosity of glycerine, η = 0.83 Pa s
Volume of glycerine flowing per sec:
V = \(\frac{M}{\rho}\)
= \(\frac{4.0 \times 10^{-3}}{1.3 \times 10^3}\)
= 3.08 x 10^-6 m³/s
According to Poiseville’s formula, we have the relation for the rate of flow:
V = \(\frac{\pi p r^4}{8 \eta l}\)
Where, p is the pressure difference between the two ends of the tube
∴ p = \(\frac{V 8 \eta l}{\pi r^4}\)
= \(\frac{3.08 \times 10^{-6} \times 8 \times 0.83 \times 1.5}{\pi \times(0.01)^4}\)
= 9.8 x 10² Pa
Reynolds’ number is given by the relation:
R = \(\frac{4 \rho V}{\pi d \eta}\)
= \(\frac{4 \times 1.3 \times 10^3 \times 3.08 \times 10^{-6}}{\pi \times(0.02) \times 0.83}\) = 0.3
Reynolds’ number is about 0.3.
Hence, the flow is laminar.

Question 10.14.
In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m/s and 63 m/s 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/m³.
Answer:
Speed of wind on the upper surface of the wing, V₁ = 70 m/s
Speed of wind on the lower surface of the wing, V₂ = 63 m/s
Area of the wing, A = 2.5 m²
Density of air, ρ = 1.3 kg/m³
According to Bernoulli’s theorem, we have the relation:
P₁ + \(\frac{1}{2}\)ρV₁² = P₂ + \(\frac{1}{2}\)ρV₂²
P₂ - P₁ = \(\frac{1}{2}\)ρ(V₁² - V₂²)
Where,
P₁ = Pressure on the upper surface of the wing
P₂ = Pressure on the lower surface of the wing
The pressure difference between the upper and lower surface of the wing provides lift to the aeroplane. 
Lift on the wing = (P₂ - P₁) A 
= \(\frac{1}{2}\)ρ(V₁² - V₂²)A
= \(\frac{1}{2}\) x 1.3 {(70)2 - (63)2} x 2.5
= 1512.87
= 1.51 x 10³ N
Therefore, the lift on the wing of the aeroplane is 1.51 x 10³ N.

Question 10.15.
Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why? 

Answer:
(a) Take the case given in figure (b).
Where, A₁ = Area of pipe 1
A₂ = Area of pipe 2
V₁ = Speed of the fluid in pipe 1
V₂ = Speed of the fluid in pipe 2
From the law of continuity, we have:
A₁V₁ = A₂V₂
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.

Question 10.16.
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, what is the speed of ejection of the liquid through the holes?
Answer:
Area of cross-section of the spray pump,
A₁ = 8cm² = 8 x 10^-4 m²
Number of holes, n = 40
Diameter of holes, n = 40
Diameter of each hole, d = 1 mm = 1 x 10 m
Radius of each hole, r = d/2 = 0.5 x 10^-3 m
Area of cross-section of each hole,
a = πr² = π(0.5 x 10^-3)² m²
Total area of 40 holes, A₂ = n x a = 40 x π(0.5 x 10^-3)² m² = 31.41 x 10^-6 m²
Speed of flow of liquid inside the tube, V₁ = 1.5 m/min = 0.025 m/s
Speed of ejection of liquid through the holes = V₂
According to the law of continuity, we have:
A₁V₁ = A₂V₂
V₂ = \(\frac{A_1 V_1}{A_2}\)
= \(\frac{8 \times 10^{-4} \times 0.025}{31.61 \times 10^{-6}}\) = 0.633 m/s
Therefore, the speed of ejection of the liquid thorugh the holes is 0.633 m/s.

Question 10.17.
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 weight that the soap film supports, W = 1.5 x 10^-2 N
Length of the slider, l = 30 cm = 0.3 m
A soap film has two free surfaces.
∴ Total length = 2l = 2 x 0.3 = 0.6 m
Surface tension, S = \(\frac{\text { Force or Weight }}{2 l}\)
= \(\frac{1.5 \times 10^{-2}}{0.6}\) = 2.5 x 10^-2 N/m
Therefore, the surface tension of the film is 2.5 x 10^-2 N/m.

Question 10.18.
Figure (a) shows a thin liquid film supporting a small weight = 4.5 x 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:
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 x 10^-2 N
A liquid film has two free surfaces.
∴ Surface tension = \(\frac{W}{2 l}\)
= \(\frac{4.5 \times 10^2}{2 \times 0.4}\) = 5.625 x 10² N/m
In all the three figures, the liquid is same. Temperature is also 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 x 10^-2 N/m.
Since the length of the film in all the cases is 40 cm, the weight supported in each case is 4.5 x 10^-2 N.

Question 10.19.
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. The atmospheric pressure is 1.01 x 10⁵ Pa. Also give the excess pressure inside the drop.
Answer:
Radius of the mercury drop,
r = 3.00 mm = 3 x 10^-3 m
Surface tension of mercury, S = 4.65 x 10^-1 N/m
Atmospheric pressure, P₀ = 1.01 x 10⁵ Pa
Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure
= \(\frac{2 S}{r}\) + P₀
= \(\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}\) + 1.01 x 10⁵
= 1.0131 x 10⁵ = 1.01 x 10⁵ Pa
Excess pressure = \(\frac{2 S}{r}=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}\)
= 310 Pa

Question 10.20.
What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solutibn at the temperature (20°C) is 2.50 x 10^-2 N/m? If an air same 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:
Soap bubble is of radius, r = 5.00 mm = 5 x 10^-3 m
Surface tension of the soap solution,
S = 2.50 x 10^-2 N/m
Relative density of the soap solution = 1.20
∴ Density of the soap solution, ρ = 1.2 x 10³ kg/m³
Air bubble formed at a depth, h = 40cm = 0.4 m
Radius of the air bubble, r = 5 mm = 5 x 10^-3 m
1 atmospheric pressure = 1.01 x 10⁵ Pa
Acceleration due to gravity, g = 9.8 m/s²
Hence, the excess pressure inside the soap bubble is given by the relation:
P = \(\frac{4 S}{r}\)
= \(\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\)
= 20 Pa
Therefore, the excess pressure inside the soap bubble is 20 Pa.
The excess pressure inside the air bubble is given by the relation:
P' = \(\frac{2 S}{r}\)
= \(\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\)
= 10 Pa
Threfore, 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'
= 1.01 x 10⁵ + 0.4 x 1.2 x 10³ x 9.8 +10
= 1.057 x 10⁵ Pa
= 1.06 x 10⁵ Pa
Therefore, the pressure inside the air bubble is
1.6 x 10⁵ Pa.

Additional Exercise

Question 10.21.
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 (or 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:
Base area of the giveen tank, A = 1.0 m²
Area of the hinged door,
a = 20cm² = 20 x 10^-4 m²
Density of water, ρ₁ = 10³ kg/m³
Density of acid, ρ₂ =1.7 x 10³ kg/m³
Height of the water column, h₁ = 4m
Height of the acid column, h₂ = 4 m
Acceleration due to gravity, g = 9.8 pressure due to water is given as:
P₁ = h₁ρ₁g
= 4 x 10³ x 9.8
= 3.92 x 10⁴ Pa
Pressure due to acid is given as;
P₂ = h₂ρ₂g
= 4 x 1.7 x 10³ x 9.8 = 6.664 x 10⁴ Pa
Pressure difference between the water and acid columns:
∆P = P₂ - P₁
= 6.664 x 10⁴ - 3.92 x 10⁴
= 2.744 x 10⁴ Pa
Hence, the force exerted on the door = ∆P x a
= 2.744 x 10⁴ x 20 x 10^-4
= 54.88 N
Therefore, the force necessary to keep the door closed is 54.88 N. 

Question 10.22.
A manometer reads the pressure of a gas in an enclosure as shown in fig. (a). When a pump removes some of the gas, the manometer reads as in fig. (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) is poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).
Answer:
(a) For figure (a)
Atmospheric pressure, P₀ = 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
= 76cm - 18cm = 58cm
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.

(b) Let h be the difference between the levels of mercury in the two limbs.
The pressure in the right limb is given as:
PR = atmospheric pressure + 1 cm of Hg
76 + 1 = 77 cm of Hg ........................(i)
The mercury column will rise in the left limb.
P = 58 + h ...................(ii)
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.

Question 10.23.
Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water than the second vessel requires to fill upto a particular common height. Is the force exerted by the water on the base of the vessel is 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:
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 componetns 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.

Question 10.24.
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:
Gauge pressure, P = 2000 Pa
Density of whole blood, p = 1.06 x 10³ kg/m³
Acceleration due to gravity, g = 9.8 m/s²
Height of the blood container = h
Pressure of the blood container, P = hρg
∴ h = \(\frac{P}{\rho g}\)
= \(\frac{2000}{1.06 \times 10^3 \times 9.8}\)
= 0.1925 m
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.

Question 10.25.
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 m if the flow must remain lajninar? (b) Do the dissipative forces become more importndt as the fluid velocity increases? Discuss qualitatively.
Answer:
Diameter of the artery, d = 2 x 10^-3 m
Velocity of blood, η = 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 given as:
V_rg = \(\frac{N_R \eta}{\rho d}\)
= \(\frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^3 \times 2 \times 10^{-3}}\)
= 1.966 m/s
Therefore, the largest average velocity of blood is 1.966 m/s.
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.

Question 10.26.
(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 Pa s).
Answer:
Radius of the artery, r = 2 x 10 m
Diameter of the artery,
d = 2 x 2 x 10^-3 m = 4 x 10^-3 m
Viscosity of blood, η = 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 given by the relation:
V_avg = \(\frac{N_R \eta}{\rho d}\)
= \(\frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^3 \times 4 \times 10^{-3}}\)
= 0.983 m/s
Therefore, the largest average velocity of blood is 0.983 m/s.
Flow rate is given by the relation:
R = πr²V_avg
= 3.14 x (2 x 10^-3)² x 0.983 = 1.235 x 10^-5 m³s^-1
Therefore, the corresponding flow rate is 1.235 x 10^-5 m³/s.

Question 10.27.
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 ).
Answer:
The area of the wings of the plane, A = 2 x 25 = 50 m²
Speed of air over the lower wing, V₁ = 180 km/h = 50 m/s, speed of air over the upper wing, V₂ = 234 km/h = 65 m/s
Density of air, ρ = 1 kg/m³
Pressure of air over the lower wing = P₁
Pressure of air over the upper wing = P₂
The upward force on the plane can be obtained using Bernoulli’s equation as:
P₁ + \(\frac{1}{2}\)ρV₁² = P₂ + \(\frac{1}{2}\)ρV₂²
P₂ - P₁ = \(\frac{1}{2}\)ρ(V₁² - V₂²) ..........................(i)
The upward force (F) on the plane can be calculated as:
(P₁ - P₂)A
= \(\frac{1}{2}\)ρ(V₂² - V₁²)A (Using equation (i))
= \(\frac{1}{2}\) x 1 x {(65)² - (50)²} x 50
= 43125 N
Using Newton’s force equation, we can obtain the mass (m) of the plane as:
F = mg
∴ m = \(\frac{43125}{9.8}\)
= 4400.51 kg
~ 4400 kg
Hence, the mass of the plane is about 4400 kg.

Question 10.28.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 x 10^-5 m and density 1.2 x 10³ kg/m³? 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.
Answer:
Terminal speed = 5.8 cm/s; Viscous force = 3.9 x 10^-10 N;
Radius of the given uncharged drop, r = 2.0 x 10^-5 m;
Density of the uncharged drop, ρ = 1.2 x 10³ kg/m³
Viscosity of air, η = 1.8 x 10^-5 Pa s
Density of air (ρ₀) can be taken as zero in order to neglect buoyancy of air.
Acceleration due to gravity, g = 9.8 m/s
Terminal velocity (v) is given by the relation:
v = \(\frac{2 r^2 \times\left(\rho-\rho_0\right) g}{9 n}\)
= \(\frac{2 \times\left(2.0 \times 10^{-5}\right)^2\left(1.2 \times 10^3-0\right) \times 9.8}{9 \times 1.8 \times 10^{-5}}\)
= 5.807 x 10^-2 m/s
= 5.8 cm/s
Hence, the terminal speed of the drop is 5.8 cm/s.
The viscous force on the drop is given by:
F = 6πηrv
∴ F = 6 x 3.14 x 1.8 x 10^-5 x 2.0 x 10^-5 x 5.8 x 10^-2
= 3.9 x 10^-10 N
Hence, the viscous force on the drop is 3.9 x 10^-10 N.

Question 10.29.
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/m³.
Answer:
Angle of contact between mercury and soda lime glass, θ = 140°
Radius of the narrow tube, r = 1 mm = 1 x 10^-3 m
Surface tension of mercury at the given temperature, T = 0.465 N/m
Density of mercury, ρ = 13.6 x 10³ kg/m³
Dip in the height of mercury = h
Acceleration due to gravity, g = 9.8 m/s²
Surface tension is related with the angle of contact and the dip in the height as:

Hence, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by 5.34 mm.

Question 10.30.
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 temperatrue 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/m³, (g = 9.8 m/s²).
Answer:
Diameter of the first bore, d₁ = 3.0 mm = 3 x 10^-3 m
Hence, the radius of the first bore,
r₁ = \(\frac{d_1}{2}\) = 1.5 x 10^-3 m
Diameter of the second bore, d₂ = 6.0 mm
Hence, the radius of the second bore,
r₂ = \(\frac{d_2}{2}\) = 3 x 10^-3 m
Surface tension of water, T = 7.3 x 10^-2 N/m
Angle of contact between the bore surface and water, θ = 0
Density of water, ρ = 1.0 x 10³ kg/m³
Acceleration due to gravity, g = 9.8 m/s²
Let h₁ and h₂ be the heights to which water rises in the first and second tubes respectively.
These heights are given by the relations:
h₁ = \(\frac{2 T \cos \theta}{r_1 \rho g}\) ...........................(i)
h₂ = \(\frac{2 T \cos \theta}{r_2 \rho g}\) .................................(ii)
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.

Calculator/Computer based Problem

Question 10.31.
(a) It is known that density p of air decreases with height y as ρ₀e^-y/y0
Where ρ₀ = 1.25 kg/m³ is the density at sea level, and y₀ is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains constant (isothermal conditions). Also assume that the value of g remains constant.
(b) A large He balloon of volume 1425 m³ is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?
[Take y₀ = 8000 m and ρ_He = 0.18 kg/m³]
Answer:
(a) Volume of the balloon, V = 1425 m³
Mass of the payload, m = 400 kg
Acceleration due to gravity, g = 9.8 m/s²
y₀ = 8000 m
ρ_He = 0.18 kg/m³
ρ₀ = 1.25 kg/m³
Density of the balloon = ρ
Height to which the balloon rises = y
Density (ρ) of air decreases with height (y) as:
ρ = ρ₀e^-y/ya
\(\frac{\rho}{\rho_a}\) = e^-y/y0 ................................(i)
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 ρ0 to ρ.
Integrating the sides between these limits, we get;

Comparing equations (i) and (ii), we get:
k = \(\frac{1}{y_0}\) ............................(iii)
From equations (i) and (iii), we get
ρ = ρ₀e^-y/y0

From equations (ii) and (iii), we can obtain y as;
ρ = ρ₀e^-y/y0
\(\log _e \frac{\rho}{\rho_0}=-\frac{y}{y_0}\)
y = -8000 x log_e\(\frac{0.46}{1.25}\)
= -8000 x -1
= 8000 m = 8 km
Hence, the balloon will rise to a height of 8 km.