9.1 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.
Solution:
(a) $P = h \rho g$. The height (h) of the blood column in the human body is more at the feet than at the brain. Therefore , blood pressure in humans is greater at the feet than at the brain.
(b) Atmospheric pressure $P = h \rho g$ where $\rho$ is the density of air. As we go up , the density of air decrease rapidly and at a height of about 6 km , it decreases to nearly half of its value at the sea level. Due to this reason , atmospheric pressure at a height of about 6 Km decreases to nearly half of its value at the sea level.
(C) When force is applied on a liquid , hydrostatic pressure is transmitted equally in all directions in the liquid. Therefore, there is no sense to assign direction to hydrostatic pressure and it is considered a scalar quantity.
9.2 Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
Solution:
a) The adhesive force between glass and water molecules is greater as compared to the cohesive force between water molecules. This makes a concave meniscus for the surface of water. Thus, water makes an acute angle from the glass surface. The adhesive force between glass and mercury molecules is less as compared to the cohesive force between mercury molecules. This makes a convex meniscus for the surface of mercury. Thus, mercury makes an obtuse angle from the glass surface.
(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.)
Solution:
(c) Surface tension of a liquid is independent of the area of the surface.
Solution: Surface tension is defined as the force acting per unit length at the interface between the plane of a liquid and any other surface. It depends upon the nature of liquid not on the area of contact.
$$S = \frac{F}{l}$$
(d) Water with detergent dissolved in it should have small angles of contact.
Solution:
Water with detergent dissolved in it has small angles of contact θ. 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. If θ is small, then cos θ will be large and the rise of the detergent water in the cloth will be fast.
(e) A drop of liquid under no external forces is always spherical in shape.
Solution:
A liquid tends to acquire the minimum surface area. Since, the surface area of a sphere is minimum for a given volume therefore, under no external forces, liquid drops always take spherical shape
9.3 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).
Solution:
(a) Decreases
(b) Increase; decrease
(c) Shear strain; rate of shear strain
(d) Conservation of mass and Bernoulli's principle
(e) Greater
9.4 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.
Solution:
(a) When we blow over the piece of paper, then the velocity of air increases. As a result From the Bernoulli’ theorem, the pressure over it decreases whereas the atmospheric pressure below remains the same. Thus, to keep the paper horizontal, we should blow over the paper not under it.
(b) From the equation of continuity, Area×velocity=constant , av=constant .
The above equation shows that, for an opening with smaller area velocity is more as compared to the opening with larger area. Thus, when we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
(c) For a constant height, the Bernoulli’s theorem is expressed as,
$$P+ \rho gh+ \frac{1}{2} \rho v^{2} = constant$$
In this equation, the pressure P occurs with a single power whereas the velocity occurs with a square power. Therefore, the velocity has more effect compared to the pressure. For this reason needle of the syringe controls flow rate better than the thumb pressure exerted by the doctor.
(d) This is because of principle of conservation of momentum. While the flowing fluid carries forward momentum, the vessel gets a backward momentum.
(e) A spinning cricket ball would have followed a parabolic trajectory if there been no air. But because of air, the Magnus effect takes place. Due to the Magnus effect the spinning cricket ball deviates from its parabolic trajectory. Thus, a spinning cricket ball does not follow a parabolic trajectory.
9.5 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 ?
Solution:
Mass of girl , m = 50 Kg , Radius of heel , r = 1/2 cm = 1/200 m
$$P = \frac{Force}{Area} = \frac{mg}{\pi r^{2}}$$
$$P= \frac{50 \times 9.8}{ \pi \times (1/200)^{2}}$$
$$P = 6.24 \times 10^{6}Nm^{-2}$$
9.6 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.
Solution :
Density of Mercury $\rho_{1} = 13.6 \times 10^{3} Kg/m^{3}$
Height of the mercury columns, $h_{1} = 0.76 m$
Density of French wine , $\rho_{2} = 984 kg/m^{3}$
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
$$\rho_{1}h_{1}g = \rho_{2}h_{2}g$$
$$h_{2} = \frac{\rho_{1}h_{1}}{\rho_{2}}$$
$$h_{2} =\frac{13.6 \times 10^{3} \times 0.76}{984}$$
$$h_{2} = 10.5 m$$
Hence , The height of the French wine column for normal atmospheric pressure is 10.5 m
9.7 A vertical off-shore structure is built to withstand a maximum stress of $10^{9}$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.
Solution:
Here h = 3 km = 3000 m
$$\rho = 1000\frac{kg}{m^{3}}$$
$$g = 9.8 \frac{m}{s^{2}}$$
Pressure
$$P = h \rho g$$
$$P = 3000\times 1000 \times 9.8$$
$$P = 2.94 \times 10^{7}Pa$$
Since the pressure exerted by sea water is less than the maximum stress the structure can withstand the structure is suitable
9.8 A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000kg. The area of cross-section of the piston carrying the load is $425 cm^{2}$. What maximum pressure would the smaller piston have to bear ?
Solution:
Area of cross section of bigger piston (carrying load ) ,
$$A= 425 cm^{2} = 425 \times 10^{-4} m^{2}$$
Max load bigger piston can bear ,
$$F = m \times g$$
$$F = 3000 \times 9.8N$$
$\therefore$ Max pressure on bigger piston ,
$$P = \frac{F}{A}$$
$$P = \frac{3000 \times 9.8 }{425 \times 10^{-4}}$$
$$P= 6.92 \times 10^{5}Nm^{-2}$$
Since liquids transmit pressure equally in all directions, the smaller piston can bear maximum pressure= $6.92 \times 10^{5}Nm^{-2}$
9.9 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 ?
Solution:
Figure shows the conditions of the problem. The pressure at the interfaces marked D and A is the same. It means pressure exerted by water at D and Pressure exerted by spirit at A are the same i.e;
$$P_{1} = P_{2}$$
$$h_{1} \rho_{1}g = h_{2} \rho_{2}g$$
$$10 \times 1 \times g = 12.5 \times \rho_{2} \times g$$
$$\rho_{2} = \frac{10}{12.5} = 0.8 g cm^{-3}$$
$$relative Density = \frac{Spirit Density}{Water Density}$$
$$relative Density = \frac{0.8}{1}$$
$\therefore$ Specific gravity of Relative density of spirit = 0.8
9.10 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)
Solution:
when we pour 15 cm of water and spirit in the respective arms ,
$$h_{1} = 10+15= 25 cm$$
$$h_{2} = 12.5 + 15 = 27.5 cm$$
Pressure at the mercury - water interface is
$$P_{1} = h_{1} \rho_{1}g = 25 \times 1 \times g$$
$$P_{1} = 25g$$
Pressure at the mercury - spirit interface is
$$P_{2} = h_{2} \rho_{2}g = 27.5 \times 0.8 \times g$$
$$P_{2} = 22g$$
Since pressure at water - mercury interface is more than the pressure at spirit-mercury interface , the mercury will rise in the spirit arm If h is the difference in the levels of mercury in the two arms , then ,
$$P_{1}-P_{2} = h \rho g$$
$$25g - 22g = h \times 13.6 \times g$$
$$3g=h \times 13.6 \times g$$
$$h = \frac{3}{13.6} cm$$
$$h = 0.221 cm$$
9.11 Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river ? Explain.
Solution:
No, Bernoulli's equation can be applied only when the flow of liquid is streamline.
$$P_{1} + \rho gh_{1} +\frac{1}{2}\rho v_{1}^{2} = P_{2} + \rho gh_{2} +\frac{1}{2}\rho v_{2}^{2}$$
9.12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation ? Explain.
Solution:
Using Bernoulli's theorem :
$$P + \rho gh +\frac{1}{2}\rho v^{2}=constant$$
Where P is the absolute pressure at a point , $\rho$ is the density of the fluid , h is the height of the point above a reference point and v is the velocity of fluid at that point.
$$P_{1} + \rho gh_{1} +\frac{1}{2}\rho v_{1}^{2} = P_{2} + \rho gh_{2} +\frac{1}{2}\rho v_{2}^{2}$$
Subtracting atmospheric pressure from both sides
We get ;
$$P_{1}-P_{0} + \rho gh_{1} +\frac{1}{2}\rho v_{1}^{2}= P_{2} -P_{0} + \rho gh_{2} +\frac{1}{2}\rho v_{2}^{2}$$
Gauge Pressure $$P_{i}^{'} = P_{i}-P_{0}(i = 1,2)$$
$$P^{'}_{1} + \rho gh_{1} +\frac{1}{2}\rho v_{1}^{2} = P^{'}_{2} + \rho gh_{2} +\frac{1}{2}\rho v_{2}^{2}$$
Thus the Bernoulli's equation remains in the same form.
Hence , it does not matter if one uses gauge instead of absolute pressure in applying Bernoulli's equation.
9.13 Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0cm. If the amount of glycerine collected per second at one end is $4.0 \times 10^{-3} kg s^{-1}$,what is the pressure difference between the two ends of the tube ? (Density of glycerine= $1.3 \times 10^{3} kg m^{-3}$ and viscosity of glycerine = 0.83 Pa s).
Solution:
Length of the horizontal tube, l = 1.5 m
Radius of the tube, r = 1 c m = 0.01 m
Diameter of the tube, d = 2 r = 0.02 m
Glycerine is flowing at a rate of $4 \times 10^{-3} \frac{kg}{s}$
$$M = 4 \times 10^{-3} \frac{kg}{s}$$
Density of glycerine ,
$$\rho = 1.3 \times 10^{3} \frac{kg}{m^{3}}$$
Viscosity of glycerine ,
$$\eta = 0.83 Pas$$
Volume of glycerine flowing per sec :
$$V = \frac{m}{\rho}$$
$$V = \frac{4 \times 10^{-3}}{1.3 \times 10^{3}}$$
$$V = 3.08 \times 10^{-6} m^{3}/s$$
According to Poisevilles 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
$$p = \frac{V 8 \eta l}{\pi r^{4}}$$
$$p = \frac{3.08 \times 10^{-6} \times 8 \times 0.83 \times 1.5}{\pi \times 0.01^{4}}$$
$$p = 9.8 \times 10^{2} Pa$$
9.14 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 $70 ms^{-1}$ and $63ms^{-1}$ respectively. What is the lift on the wing if its area is $2.5 m^{2}$ ? Take the density of air to be $1.3 kg m^{–3}$.
Solution:
Let $v_{1}$ and $v_{2}$ be the speeds on the upper and lower surfaces the wing of aeroplane and $P_{1}$ and $P_{2}$ the corresponding pressures. According to Bernoulli's equation for horizontal flow ,
$$P_{1}+\frac{1}{2} \rho v_{1}^{2} = P_{2}+\frac{1}{2} \rho v_{2}^{2}$$
$$P_{2} - P_{1} = \frac{1}{2} \rho (v_{1}^{2} - v_{2}^{2} )$$
Given : $\rho = 1.3 kg m^{-3}$ , $v_{1} = 70 ms^{-1}$ , $v_{2} = 63 ms^{-1}$
$$P_{2} - P_{1} = \frac{1}{2} \times 1.3 (70^{2} - 63^{2} )$$
$$P_{2} - P_{1} = 605.15 Nm^{-2}$$
$$\therefore Lifting \ force = (P_{2} - P_{1}) \times \ Area \ of \ wing$$
$$Lifting \ force= 605.15 \times 2.5 = 1.51 \times 10^{3}N$$
9.15 Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
Solution:
Fig. 9.20 (a) is incorrect. According to equation of continuity and Bernoulli's equation for horizontal flow , we have ,
av = Constant
$$P + \frac{1}{2} \rho v^{2} = constant$$
According to equation of continuity the velocity of flow is more at the construction (where area of X-section is less) and according to bernoulli's equation where velocity is more, pressure should be less. Fig 9.20 (a) shows that pressure is more ( higher level of liquid) at the construction and hence is incorrect.
9.16 The cylindrical tube of a spray pump has a cross-section of $8.0 cm^{2}$ 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 ?
Solution:
According to equation of continuity ,
$$a_{1}v_{1} = a_{2}v_{2}$$
Area of X-section of tube ,
$$a_{1} = 8.0 cm^{2} =8.0 \times (10^{-2}m)^{2}$$
$$a_{1} = 8.0 \times 10^{-4}m^{2}$$
speed of liquid in tube ,
$$v_{1} = \frac{1.5}{60} ms^{-1}$$
Area of cross-section of 40 holes ,
$$a_{2} = 40 \times \pi r^{2}$$
Given :
$$D= 1mm = 0.1 cm $$
$$r = \frac{D}{2} \times 10^{-2}m$$
$$r = \frac{0.1}{2} \times 10^{-2}m$$
$$r = 0.50 \times 10^{-2}m$$
$$r = 50 \times 10^{-4}m$$
$$a_{2}= \pi \times 40 \times (50 \times 10^{-4})^{2} m^{2} $$
$$v_{2} = ?$$
$\therefore$ Speed of ejection of liquid through the holes is
$$v_{2} = \frac{a_{1} v_{1}}{a_{2}}$$
$$v_{2} = \frac{(8 \times 10^{-4}) \times \frac{1.5}{60} }{40 \times \pi (5 \times 10^{-4})^2}$$
$$v_{2} =0.636m/sec $$
9.17 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 $1.5 \times 10^{-2}N$ (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film ?
Solution:
The force due to surface tension acts upward on the slider. since the soap film has the two surface, the total length of the film to be supported = 2l = 2×30 = 60cm = 0.6 m
$\therefore$ Total upward force on the slider due to surface tension is
$$F = T \times 2l $$
$$F= T \times 0.6 = 0.6 T Newton$$
Since the slider is in equilibrium, the total upwards force F on the slider must be equal to its weight i.e
$$F = mg$$
$$0.6T = 1.5 \times 10^{-2}$$
$$T = \frac{1.5 \times 10^{-2}}{0.6}$$
$$T = 0.025 N m^{-1}$$
9.18 Figure 9.21 (a) shows a thin liquid film supporting a small weight = $4.5 \times 10^{–2} N$.What is the weight supported by a film of the same liquid at the same temperature in Fig. 9.21 (b) and (c) ? Explain your answer physically.
Solution:
The length of the slider is l = 40 cm = 0.4 m. Since the shop film has two free surfaces , the total length of the film to be supported
$$2l = 2 \times 0.4 = 0.8 m.$$
Total upward force on the slider due to surface tension is
$$F = T \times 2l = T \times 0.8$$
and weight supported by the film is $$F=4.5 \times 10^{-2}N$$
$$ T \times 0.8 = 4.5 \times 10^{-2}$$
$$T = \frac{4.5 \times 10^{-2}}{0.8}$$
$$T= 5.625 \times 10^{-2}Nm^{-1}$$
As the liquid and the temperature in the three cases are the same , therefore , surface tension for cases b and c will also be the same. Again the length of the slider is the same in the three cases , the weight supported by the liquid film in cases b and c will be the same i.e , 4.5 × $10^{-2}N$
9.19 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 $4.65 \times 10^{–1} N m^{–1}$. The atmosphere pressure is $1.01 \times 10^{5} Pa)$. Also give the excess pressure inside the drop.
Solution:
The excess pressure inside a liquid drop is given by :
$$P_{2} - P_{1} = \frac{2T}{R}$$
Here , $P_{2} - P_{1}$ = excess pressure;
$$T = 4.65 \times 10^{-1} Nm^{-1}$$
$$R = 3.0 mm = 3.0 \times 10^{-3}m$$
$$P_{2} - P_{1} = \frac{2 \times (4.65 \times 10^{-1})}{3.0 \times 10^{-3}}$$
$$P_{2} - P_{1} = 3.1 \times 10^{2} N/m^{2}$$
Pressure inside the drop ,
$$P_{2} = P_{1} + 3.1 \times 10^{2} N/m^{2}$$
$$P_{2} = 1.01 \times 10^{5} + 3.1 \times 10^{2}$$
$$P_{2} = 1.0131 \times 10^{5}Pa$$
9.20 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 $2.50 \times 10^{-2} N m^{–1}$ ? 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 $1.01 \times 10^{5}$Pa).
Solution:
The excess pressure inside a soap bubble is given by :
$$P_{2} - P_{1} = \frac{4T}{R}$$
$$T = 2.5 \times 10^{-2} Nm^{-1}$$
$$R = 5.0 mm = 5.0 \times 10^{-3}m$$
$$P_{2} - P_{1} = \frac{4 \times (2.5 \times 10^{-2})}{5.0 \times 10^{-3}}$$
$$P_{2} - P_{1} = 20 Pa$$
Excess pressure inside an air bubble in soap solution :
$$P_{2} - P_{1} = \frac{2T}{R} = \frac{2 \times ( 2.5 \times 10^{-2})}{5.0 \times 10^{-3}}$$
$$= 10 Pa$$
Total pressure inside the air bubble= $$p_{a} + h \rho g + \frac{2T}{R}$$
$$= 1.01 \times 10^{5} + 0.4 \times 1200 \times 9.8 + 10$$
$$= 1.06 \times 10^{5} Pa$$