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Q 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 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).

For the angle of contact , radius of the tube r, surface tension t, the density of fluid  the rise in the column is given by The radii of the two limbs r1 and r2 are 3.0 mm and 1.5 mm respectively  The level in the limb of diameter 6.0 mm is The level in the limb of diameter 3.0 mm is The difference in the heights is h2 - h1 = 4.96 mm

Q 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-1. Density of mercury = 13.6 × 103 kg m-3.

Since the angle of contact is obtuse the Pressure will be more on the Mercury side. This pressure difference is given as The dip of mercury inside the narrow tube would be equal to this pressure difference The mercury dips down in the tube relative to the liquid surface outside by an amount of 5.34 mm.

Q 10.28 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.

Neglecting buoyancy due to air the terminal velocity is Viscous Force Fv at this speed is

Q 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.

Speed of the wind above the upper wing surface is v1 = 234 km h-1 Speed of the wind over the lower wing is v2 = 180 km h-1 Let the pressure over the upper and lower wing be P1 and P2 Let the plane be flying at a height of h The density of air is  Applying Bernoulli's Principle at two points over the upper and lower wing we get Area of each wing is a = 25 m2 The net upward force on the plane...

Q 10.26 (b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10-3 Pa s).

Volumetric flow rate is given as

Q 10.26 (a) What is the largest average velocity of blood flow in an artery of radius 2×10-3m if the flow must remain laminar?

The diameter of the artery is d The viscosity of blood is  The density of blood is  The average velocity is given by  Taking the Maximum value of Reynold's Number ( NRe = 2000) at which Laminar Flow takes place we have

Q 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.

(b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.

As the fluid velocity increases the dissipative forces become important as turbulence rises due to which drag due to friction forces increases.

Q 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 × 10–3 m if the flow must remain laminar?

The diameter of the artery is  The viscosity of blood is  The density of blood is  The average velocity is given by  Taking the Maximum value of Reynold's Number ( NRe = 2000) at which Laminar Flow takes place we have

Q 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?

The density of whole blood  Gauge Pressure  Height at which the blood container must be placed so that blood may just enter the vein

Q 10.23 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?

Since the height of the water level in the vessels is the same the Pressure at the bottom would be equal. As the area of the bottom is also the same the Force exerted by the water on the bottom would be the same. The difference in the reading arises due to the fact that the weight depends on the volume of the water inside the container which is more in the first vessel. The vertical component...

Q 10.22 A manometer reads the pressure of a gas in an enclosure as shown in Figure (a) When a pump removes some of the gas, the manometer reads as in Figure (b) The liquid used in the manometers is mercury and the atmospheric pressure is 76 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? As we know Specific Gravity of Mercury is 13.6 therefore 13.6 cm of water column would be equal to  1 cm of Mercury column. The pressure at the Mercury Water interface in the right column = Atmospheric Pressure + 1 cm of Mercury = 77 cm of Mercury The difference in Pressure due to the level of the Mercury column = Pressure at the Mercury Water interface - Absolute Pressure of the Glass...

Q 10.22 A manometer reads the pressure of a gas in an enclosure as shown in Figure (a) When a pump removes some of the gas, the manometer reads as in Figure (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. In figure (a) Gauge Pressure = 20 cm of Mercury Absolute Pressure = Atmospheric Pressure + Gauge Pressure Absolute Pressure = 76 + 20= 96 cm of Mercury In figure (b) Gauge Pressure = -18 cm of Mercury Absolute Pressure = Atmospheric Pressure + Gauge Pressure Absolute Pressure = 76 + (-18)= 58 cm of Mercury

Q 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 (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

Pressure in the waterside at the bottom is Pressure in the acid side at the bottom is The pressure difference across the door is Area of the door, a = 20 cm2  The force necessary to keep the door closed is Note: The dimensions of the door are small enough to neglect pressure variations near it.

Q 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 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 × 105Pa).

Excess Pressure inside a bubble is given by             (It's double the usual value because of the presence of 2 layers in case of soap bubble) where T is surface tension and r is the radius of the bubble Atmospheric Pressure is  The density of soap solution is  The pressure at a depth of 40 cm (h) in the soap solution is  Total Pressure inside an air bubble at that depth

Q 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 × 10-1 N m-1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.

Surface Tension of Mercury is  The radius of the drop of Mercury is r = 3.00 mm Excess pressure inside the Mercury drop is given by Atmospheric Pressure is  Total Pressure inside the Mercury drop is given by

Q 10.18 Figure (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. As the liquid and the temperature is the same in all three the surface tension will also be the same. Since the length is also given to be equal (40 cm) in all three cases the weight being supported is also the same and equal to .

Q 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 × 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?

Total weight supported by the film   Since a soap film has two surfaces, the total length of the liquid film is 60 cm. Surface Tension is T

Q 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-1, what is the speed of ejection of the liquid through the holes?

Cross-sectional area of cylindrical tube is a1 = 8.0 cm2 The total area of the 40 fine holes is a2 Speed of liquid inside the tube is v1 = 1.5 m min-1 Let the speed of ejection of fluid through the holes be v2 Using the continuity equation

Q 10.15 Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why? By the continuity equation, the velocity of the non-viscous liquid will be large at the kink than at the rest of the tube and therefore pressure would be lesser here by Bernoulli's principle and the air column above it, therefore, should be of lesser height. Figure (a) is therefore incorrect.

Q 10.14 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 m s-1 and 63 ms-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

The speed of air above and below the wings are given to be v1 = 70 m s-1 and v2 = 63 m s-1 respectively.  Let the pressure above and below the wings be p1 and p2 and let the model aeroplane be flying at a height h from the ground. Applying Bernoulli's Principle on two points above and below the wings we get The pressure difference between  the regions below and above the wing is 605.15 Pa The...
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