Chapter 10 Mechanical Properties Of Fluids Solutions
Question - 11 : - Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
Answer - 11 : -
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 - 12 : - Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
Answer - 12 : -
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 - 13 : - Glycerine flows steadily through a horizontal tube of length 1.5 mand radius 1.0 cm. If the amount of glycerine collected per second at one endis 4.0 × 10–3 kg s–1, what is the pressuredifference between the two ends of the tube? (Density of glycerine = 1.3 × 103 kgm–3 and viscosity of glycerine = 0.83 Pa s). [You may also liketo check if the assumption of laminar flow in the tube is correct].
Answer - 13 : -
Answer: 9.8 × 102 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 kgs–1
Density of glycerine, ρ = 1.3 × 103 kgm–3
Viscosity of glycerine, η = 0.83 Pa s
Volume of glycerine flowing per sec:
= 3.08 × 10–6 m3 s–1
According to Poiseville’s formula, we have the relation for therate of flow:
Where, p is the pressure difference between thetwo ends of the tube
= 9.8 × 102 Pa
Reynolds’ number is given by the relation:
Reynolds’ number is about 0.3. Hence, the flow is laminar.
Question - 14 : - In a test experiment on a model aeroplane in a wind tunnel, theflow speeds on the upper and lower surfaces of the wing are 70 m s–1and63 m s–1 respectively. What is the lift on the wing if its areais 2.5 m2? Take the density of air to be 1.3 kg m–3.
Answer - 14 : -
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 m–3
According to Bernoulli’s theorem, we have the relation:
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 surfaces ofthe wing provides lift to the aeroplane.
Lift on the wing 
= 1512.87
= 1.51 × 103 N
Therefore, the lift on the wing of the aeroplane is 1.51 × 103 N.
Question - 15 : - 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 - 15 : -
Answer: (a)
Take the case given in figure (b).
Where,
A1 =Area of pipe1
A2 =Area of pipe 2
V1 =Speed of the fluid in pipe1
V2 =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 venturimeteris small, the speed of the flow of liquid through this part is more. Accordingto Bernoulli’s principle, if speed is more, then pressure is less.
Pressure is directly proportional to height. Hence, the level ofwater in pipe 2 is less.
Therefore, figure (a) is not possible.
Question - 16 : - The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 oneend of which has 40 fine holes each of diameter 1.0 mm. If the liquid flowinside the tube is 1.5 m min–1, what is the speed of ejection of theliquid through the holes?
Answer - 16 : -
Area of cross-section of the spray pump, A1 =8 cm2 = 8 × 10–4 m2
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 = πr2 =π (0.5 × 10–3)2 m2
Total area of 40 holes, A2 = n × a
= 40 × π (0.5 × 10–3)2 m2
= 31.41 × 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:
= 0.633 m/s
Therefore, the speed of ejection of the liquid through the holesis 0.633 m/s.
Question - 17 : - A U-shaped wire is dipped in a soap solution, and removed. Thethin soap film formed between the wire and the light slider supports a weightof 1.5 × 10–2 N (which includes the small weight of theslider). The length of the slider is 30 cm. What is the surface tension of thefilm?
Answer - 17 : -
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.
∴Totallength = 2l = 2 × 0.3 = 0.6 m
Surface tension,
= 
Therefore, the surface tension of the film is 2.5 × 10–2 Nm–1.
Question - 18 : - Figure 10.24 (a) shows a thin liquid film supporting a smallweight = 4.5 × 10–2 N. What is the weight supported by a filmof the same liquid at the same temperature in Fig. (b) and (c)? Explain youranswer physically.
Answer - 18 : -
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.
∴Surfacetension
In all the three figures, the liquid is the same. Temperature isalso the same for each case. Hence, the surface tension in figure (b) andfigure (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 weightsupported in each case is 4.5 × 10–2 N.
Question - 19 : - What is the pressure inside the drop of mercury of radius 3.00 mmat room temperature? Surface tension of mercury at that temperature (20°C) is4.65 × 10–1 N m–1. The atmospheric pressure is 1.01× 105 Pa. Also give the excess pressure inside the drop.
Answer - 19 : -
Answer: 
Radius of the mercury drop, r = 3.00 mm = 3 × 10–3 m
Surface tension of mercury, S = 4.65 × 10–1 Nm–1
Atmospheric pressure, P0 = 1.01 × 105 Pa
Total pressure inside the mercury drop
= Excess pressure inside mercury + Atmospheric pressure
= 1.0131 × 105
= 1.01 ×105 Pa
Excess pressure 
= 310 Pa
Question - 20 : - What is the excess pressure inside a bubble of soap solution ofradius 5.00 mm, given that the surface tension of soap solution at thetemperature (20 °C) is 2.50 × 10–2 N m–1? If an airbubble of the same dimension were formed at depth of 40.0 cm inside a containercontaining the soap solution (of relative density 1.20), what would be thepressure inside the bubble? (1 atmospheric pressure is 1.01 × 105 Pa).
Answer - 20 : -
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
∴Densityof the soap solution, ρ = 1.2 × 103 kg/m3
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 × 105 Pa
Acceleration due to gravity, g = 9.8 m/s2
Hence, the excess pressure inside the soap bubble is given by therelation:
Therefore, the excess pressure inside the soap bubble is 20 Pa.
The excess pressure inside the air bubble is given by therelation:
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 