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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.
Base area of the given tank, A = 1.0 m2
Area of the hinged door, a = 20 cm2 = 20 × 10–4 m2
Density of water, ρ1 = 103 kg/m3
Density of acid, ρ2 = 1.7 × 103 kg/m3
Height of the water column, h1 = 4 m
Height of the acid column, h2 = 4 m
Acceleration due to gravity, g = 9.8
Pressure due to water is given as:
Pressure due to acid is given as:
Pressure difference between the water and acid columns:
Hence, the force exerted on the door = ΔP × a
= 2.744 × 104 × 20 × 10–4
= 54.88 N
Therefore, the force necessary to keep the door closed is 54.88 N.
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. (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).
(a) 96 cm of Hg & 20 cm of Hg; 58 cm of Hg & –18 cm of Hg
(b) 19 cm
(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
= 76 cm – 18 cm = 58 cm
(b) 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.
Let h be the difference between the levels of mercury in the two limbs.
The pressure in the right limb is given as:
Atmospheric pressure + 1 cm of Hg
= 76 + 1 = 77 cm of Hg … (i)
The mercury column will rise in the left limb.
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.
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 upto 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?
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 components 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.
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?
Gauge pressure, P = 2000 Pa
Density of whole blood, ρ = 1.06 × 103 kg m–3
Acceleration due to gravity, g = 9.8 m/s2
Height of the blood container = h
Pressure of the blood container, P = hρg
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.
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.
(a) 1.966 m/s (b) Yes
(a) Diameter of the artery, d = 2 × 10–3 m
Viscosity of blood, 
Density of blood, ρ = 1.06 × 103 kg/m3
Reynolds’ number for laminar flow, NR = 2000
The largest average velocity of blood is given as:
Therefore, the largest average velocity of blood is 1.966 m/s.
(b) 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.
(a) What is the largest average velocity of blood flow in an artery of radius 2 × 10–3 m if the flow must remain laminar? (b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10–3 Pa s).
(a)Radius of the artery, r = 2 × 10–3 m
Diameter of the artery, d = 2 × 2 × 10–3 m = 4 × 10– 3 m
Viscosity of blood, 
Density of blood, ρ = 1.06 × 103 kg/m3
Reynolds’ number for laminar flow, NR = 2000
The largest average velocity of blood is given by the relation:
Therefore, the largest average velocity of blood is 0.983 m/s.
(b) Flow rate is given by the relation:
R = π r2
Therefore, the corresponding flow rate is
.
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).
The area of the wings of the plane, A = 2 × 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 m–3
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:
The upward force (F) on the plane can be calculated as:
Using Newton’s force equation, we can obtain the mass (m) of the plane as:
∼ 4400 kg
Hence, the mass of the plane is about 4400 kg.
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.
Terminal speed = 5.8 cm/s; Viscous force = 3.9 × 10–10 N
Radius of the given uncharged drop, r = 2.0 × 10–5 m
Density of the uncharged drop, ρ = 1.2 × 103 kg m–3
Viscosity of air, 
Density of air 
can be taken as zero in order to neglect buoyancy of air.
Acceleration due to gravity, g = 9.8 m/s2
Terminal velocity (v) is given by the relation:
Hence, the terminal speed of the drop is 5.8 cm s–1.
The viscous force on the drop is given by:
Hence, the viscous force on the drop is 3.9 × 10–10 N.
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.
Angle of contact between mercury and soda lime glass, θ = 140°
Radius of the narrow tube, r = 1 mm = 1 × 10–3 m
Surface tension of mercury at the given temperature, s = 0.465 N m–1
Density of mercury, ρ =13.6 × 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:
Here, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by 5.34 mm.
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).
Diameter of the first bore, d1 = 3.0 mm = 3 × 10–3 m
Diameter of the second bore, 
= 6.0 mm
Surface tension of water, s = 7.3 × 10–2 N m–1
Angle of contact between the bore surface and water, θ= 0
Density of water, ρ =1.0 × 103 kg/m–3
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:
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.
(a) It is known that density ρ of air decreases with height y as 
Where 
= 1.25 kg m–3 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 a constant (isothermal conditions). Also assume that the value of gremains 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 
= 0.18 kg m–3].
(a) Volume of the balloon, V = 1425 m3
Mass of the payload, m = 400 kg
Acceleration due to gravity, g = 9.8 m/s2
Density of the balloon = ρ
Height to which the balloon rises = y
Density (ρ) of air decreases with height (y) as:
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 
to ρ.
Integrating the sides between these limits, we get:
(b)
Hence, the balloon will rise to a height of 8 km.
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