Coupon Accepted Successfully!



Explain why
(a) A balloon filled with helium does not rise in air density but halts after a certain height (Neglect winds)

(b) The force required by a man to raise his limbs immersed in water is smaller than the force for the same movement in air.

(c) The blood pressure in humans is greater at the feet than at the brain.

(d) Atmospheric pressure at a height of about 6 km decreases to nearly half its value at the sea level, though the height of the atmosphere is more than 100 km.

(e) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area and force is a vector.

(a) A balloon filled with helium goes on rising in air so long as the weight of the air displaced by it is greater than the weight of the filled balloon. We know that the density of air decreases with height. Therefore, the balloon halts after attaining a height at which density of air is such that the weight of air displaced just equals the weight of filled balloon.

(b) The upthrust on the limbs of a man in water is much more than when immersed in air. Hence, net weight of the limbs in water is much less than in air. Therefore, it is easier for him to move his limbs in water than in air.

(c) The height of the blood column is quite large at feet than at the brain. Consequently, the blood pressure in humans is greater at the feet than at the brain.

(d) Density of air decreases very rapidly with increase in height and reduces to nearly half its value at the sea level at a height of about 60 km. After 6 km, the density of air decreased but rather very slowly. This is the reason that at a height of about 6 km, atmosphere pressure reduces to nearly half its value at the sea level.

(e) Force is a vector quantity but pressure is a scalar quantity. According to Pascal's law pressure is transmitted equally in all directions i.e. direction is not associated with the pressure. Hence it is not a scalar quantity.


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 the glass while mercury does not).

(c) Surface tension of a liquid is independent of the area of the surface.

(d) Detergents should have small angles of contact.

(e)  A drop of liquid under no external forces is always spherical in shape.

(a) The cohesive force between mercury molecules is greater than the adhesive force between mercury and glass. So, the surface is convex and hence the angle is obtuse. For water-glass the adhesive force is greater so the meniscus is concave and hence angle of contact is acute.

(b) The adhesive force between water molecules and glass molecules is greater than cohesive force between water molecule. So it spreads. For mercury, cohesive force between molecules is greater. So they form drops.

(c) Surface tension = Force/length.
Thus surface tension depends on force, length and temperature and not on area.

(d) If detergents do not have small angle of contact, then they will not be soluble in water.

(e) When there is no external force, the only force on the drop is surface tension, which tries to make the drop to have the least surface area. Hence the drop become spherical.


Fill in the blanks using the word Cs0 from list appended with each statement:
(a) Surface tension of liquids generally -------- with temperature (increase/decrease).

(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 liquids it is proportional to ----------- (shear strain/rate of shear strain).

(d) For a fluid in steady flow, the increases in flow speed at a constriction follows from --------- (conservation of mass/Bernoulli equation).

(e) For the model of a plane in a wind tunnel, turbulence occurs at a -------- speed than the critical speed for turbulence for an actual plane (greater/smaller).

(a) decreases.

ηof the gases increases, ηof liquid decreases with temperature.

(c) shear strain, rate of shear strain.

(d) conservation of mass, Bernoulli's equation.

(e) greater.


Explain why
(a) To keep a 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.

(a) When we blow over the paper, the velocity of air increases and hence the pressure of the air decreases, where as pressure of air below the paper is atmospheric. Hence the paper stays horizontal.

(b) According to the equation of continuity i.e. av = a constant, the area of the water that jets out is reduced, so the velocity of the water increases.

(c) Here, the size of the needle controls the velocity of flow and thumb pressure controls the water pressure. 

According to Bernoulli's theorem, P =
ρgh + 1/2
                                            Pv2 = a constant
This shows that P occurs with powers of 1 and v occurs with powers of 2, so the velocity has more influence. Hence the needle has a better control over the flow.

(d) When a fluid is flowing out of a small hole in a vessel, it acquires a large velocity and hence possesses large momentum. Since no external force is acting on the system, a backward velocity must be attained by the vessel. As a result of it, an impulse is experienced by the vessel.

(e) The spinning cricket ball displaces air and as a result the speed of air becomes different at upper and lower portion of the ball. Thus a pressure difference is created due to which the trajectory of cricket ball does not remain parabolic. The parabolic trajectory remains only when gravitational forces act on the ball.


A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

Force applied, F = 50 kg f = 50 × 10 N
Diameter of the heel, d = 1.0 cm = 0.01 m
Radius of the heel, r = 0.5 cm = 0.005 m
Area of the heel, A =
πr2 = 3.14 × (0.005)2 
                                     = 7.85
× 10-5 m2
Pressure = = 6.2 × 106 Pa.


Torcelli'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.

Atmospheric pressure hρg = 1.013 × 105 Nm-2
Height h = m
             = m
             = 10.5 m.


A vertical offshore 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 Bombay high? Take the depth of the sea to be roughly 3km and ignore the ocean currents.

Depth of the sea h = 3km = 3 × 103 m
Maximum stress = 109 Pa
Therefore the density ρ = 103 kg/m3
Pressure exerted by a water column of depth 3 km
                                   P = h ρ g
                                      = 3 × 103 × 103 × 9.8
                                      = 2.94 × 107 Pa
This pressure is less than the maximum stress of 109 Pa;  so the structure is suitable.


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?

Pressure on the piston P = F/A
Force F = m × a
          = 3000 × 9.8
          = 35280N
Area of cross section A = 425 × 10-4 sq m

Therefore the pressure P = = 6.92 × 10-5 Pa.


A U tube contains water and methylated spirit separated by mercury. The mercury columns in two arms are in level with 10 cm of water in one arm and 12.5 cm of spirit in the other. What is the relative density of spirit?


Since the mercury columns in the two arms are at the same level,
Pressure due to water column = pressure due to spirit column.
ρ1g = h2ρ2g
ρ2 = h1ρ1 / h2 
where h1 = 10 cm
                     h2 = 12.5 cm
ρ1 = 1 g cm-3 
ρ2 = = 0.8 g cm-3 = 800 kg cm-1
Relative density of spirit = = 800 / 1000 = 0.8.


In the previous problem, if 15 cm of water and spirit each are further poured in to the respective arms of the tube, what is the difference in the levels of mercury in the 2 arms? ( relative density of mercury = 13.6).


The two points A and B lying in the same horizontal plane. So, applying Pascal's law, we get
Pressure at A = Pressure at B

p0+ hw ρwg = p0 + (hsρsg + hmρmg)
ρ0 is the atmospheric pressure.
Now,      hw
ρw = hsρs + hmρm

25 × 1 = 27.5 × 0.8 + hm × 13.6
× 13.6 = 25 - 27.5 × 0.8
hm = cm = 0.2206 cm.


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

No. Bernoulli's principle can be applied only to stream line flow.


Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli's equation? Explain.

One cannot use gauge instead of absolute pressures in applying Bernoulli equation because the atmospheric pressure at the two points where the Bernoulli's equation is applied are significantly different.


Glycerine flows steadily through a horizontal tube of length 1.5m and radius 1cm. If the amount of glycerine collected per second at one end is 4 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 = 0.83 Pa s).

Length of the horizontal tube l = 1.5 m
Radius of the horizontal tube r = 1 cm
Mass of the liquid flowing per sec = = 4
× 10-3 kg s-1
Viscosity of the liquid
η = 0.83 Nsm-2
Density of glycerin
ρ = 1.3 × 1000 kg/cubic metre
We know that, Q = =
                    Q = =
Therefore the pressure difference P = =  = 9.755
× 102 Pa.


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 70m/s and 63m/s respectively. What is the lift on the wing if its area is 2.5 m2? Take the density of air to be 1.3kg / m3.

Volume in the upper surface of the wing V1 = 70 ms-1
Volume in the lower surface of the wing V2 = 63 ms-1
Area of the wing A = 2.5 m2
Density of air
ρ = 1.3 kgm-3
By Bernoulli's theorem P1 +
ρv12 + h1ρg = P2 +ρv22 + h2ρg
Here the gravitational potential energy is same, so h1
ρg = h2ρg
Therefore P2 - P1          =
ρv12 - ρv22  
ρ (v12 -v22 )
                                 = (4900 - 3969)
                                 = 605.15
Force = Pressure
× Area
        = (P2 - P1)
× A
        = 605.15
× 2.5
        = 1.5
× 103 N
        = 1500 N.


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

Figure (a) is incorrect. The reason is, at a constriction (i. e,  where the area of cross section of the tube is smaller) the flow speed is larger due to mass conservation. Consequently, pressure there is smaller accordance with Bernoulli's Equation. The assumption is that the fluid is incompressible.


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 1.0mm. If the liquid flow inside the tube is 1.5 m per minute, what is the speed of ejection of the liquid through the holes.

The cross-section area of the cylindrical tube A1 = 8 cm2
                                                                      = 8 x 10-4 m2
The speed of the liquid flow inside the tube v1 = 1.5 m per minute
                                                                   = ms-1
Area of each hole =
π (0.5 x 10-3)2 m2
Area of 40 hole A2 = 40
π (0.5 x 10-3)2 m2
                   A1v1 = A2v2
                      v2 =  
                          =  = 0.636 ms-1.


The U shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and a 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?

Weight mg = 1.5 × 10-2 N
Length of slider l = 30 × 10-2 m
Force due to surface tension = weight
We know that σ × 2l = mg
Therefore surface tension σ =
= = 2.5 × 10-2 Nm-1.


Figure (a) below 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 ?


The length of the slider is same in all figures. The weight supported by liquid film in all the cases will be the same i.e., 4.5 × 10-2 N.


What is the pressure inside a drop of mercury of radius 3mm at room temperature? Surface tension of mercury at that temperature (20 °C) is 4.65 x 10-1 N/m. The atmospheric pressure is 1.013 × 105 Pa. Also give the excess pressure inside the drop?

Radius of the mercury drop r = 3 × 10-3m
Surface tension of Hg
σ         = 4.65 × 10-1 Nm-1
Excess pressure P                =  2
                                          =   = 310 Pa
Total pressure = atmospheric pressure + excess pressure
                     = 1.013
× 105 Pa + 310 Pa
                     = 101300 pa + 310 pa
                     = 101610 pa
                     = 1.016
× 105 pa.


What is the excess pressure inside a bubble of soap solution of radius 5 mm, given that the surface tension of soap solution at the temperature ( 20° C) is 2.5 x 10-2 N/m? If an air bubble of the same dimension were formed at a depth of 40 cm inside a container containing the soap solution (of relative density 1.2), what would be the pressure inside the bubble? (1 atm = 1.01 ×105 Pa)

Radius of the bubble r = 5mm = 5 × 10-3m
Surface tension of the soap solution
σ = 2.5 × 10-2 Nm-1
Pressure inside the bubble P = =  = 20 Pa
Excess pressure inside an air bubble under soap solution P
                                                                                  = = 10 Pa
Density of soap solution = relative density
× 1000 = 1.2 × 1000 = 1200 kgm-3.
Pressure inside the bubble = atmospheric pressure + pressure due to soap solution of height 40cm
                                       = 1.01
×105 + (0.4 x 1.2 × 1000 × 9.8)
                                       = 1.05
× 105 Pa
The excess pressure is so small that up to 3 significant figures, total pressure inside the air bubble = 1.05
× 105 Pa.


A tank with a square base of area 1 sq m is divided by a vertical partition in the middle. The bottom of the partition has a small hinged door of area 20 sq cm. 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 4m. compute the force necessary to keep the door closed?

Height of water = Height of acid = h = 4.00 m
Density of water ρ = 103 kgm-3
Density of acid ρ = relative density of acid × density of water
                         = 1.7 × 103 kgm-3
Area of the door A = 20cm2  
                          = 20 × 10-4 m2
Pressure at the bottom of water compartment = P × A
                                                                = 39200A N
Similarly, force on the door from acid compartment = h ρ g A
                                                                       = 4 × 1.7 × 103 × 9.8A N
                                                                       = 66640A N
The force on the door from acid compartment is large
                     Difference in force = 66640A – 39200A
                                               = A(66640 – 39200)
                                               = 27440 × 20 × 10-4 = 54.88 N.


A manometer reads the pressure of a gas in an enclosure as shown in figure (a). When some of the gas is removed by a pump, the manometer reads as in (b). The liquid used in the manometer is mercury and the atmospheric pressure is 76cm of mercury.
(i) Give the absolute and gauge pressure of the gas in the enclosure for cases
(a) and (b) in units of cm of mercury.

(ii) How would the levels change in case (b) if 13.6 cm of water are poured in to the right limb of the manometer? ( ignore the small change in the volume of the gas) 


(i) In this case in fig (a) the absolute pressure is equal to 96 cm of Hg and the gauge pressure = 20 cm of Hg and in figure (b) the absolute pressure = 58 cm of Hg and gauge pressure = -18 cm of Hg.
P = P0 + h1
P = 76 cm of Hg + 20 cm of Hg
P = 96 cm of Hg
Gauge pressure is the difference between the absolute pressure and atmospheric pressure.
Gauge pressure = 96 cm of Hg - 76 cm of Hg = 20 cm of Hg.
In the fig. (b), the absolute pressure P of the gas is less than atmospheric pressure.
P = P0 - h2
   = 76 cm of Hg - 18 cm of Hg
   = 58 cm of Hg
Gauge pressure = -58 cm of Hg - 76 cm of Hg
                      = -18 cm of Hg

(ii) After 13.6 cm of water is poured in the right limb. 
Adding 13.6 cm of water in the right limb is equivalent to adding 1 cm of mercury. So, mercury will rise in the left limb such that the difference in its levels in the two limbs becomes 19 cm. It may be noted that the absolute and gauge pressure remain unchanged.


A spring balance reads 10 kg when a bucket of water is suspended from it. What is the reading in the spring balance when

(i) An ice cube of mass 1.5 kg is put into the bucket.

(ii) An iron piece of mass 7.8 kg suspended by another string is immersed with half the volume inside the water in the bucket (relative density of iron = 7.8).

(i) If an ice cube of 1.5 kg is added the reading on the spring = 10 + 1.5 = 11.5 kg
The spring balance will show the reaction of the above force = 11.5 kg

(ii) If an iron piece of mass 7.8 kg is immersed such that half its volume is inside the water then
Density of iron = 7.8
× 1000 kg m-3
Therefore, Volume of iron = = = 0.001 m3
Since only half of iron is immersed,
Volume of water displaced = m3
Up thrust = weight of water displaced = (volume
× density) kg = 0.5 kg
The total upward reaction = 10 + 0.5 = 10.5 kg
The reading on the spring balance = 10.5 kg.


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?

Density of the whole blood, ρ = 1.06 × 103 kg m-3
Pressure P = h
                = 2000 Pa
        2000 = h
× 1.06 × 103 × 9.8
             h = = 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) How does the pressure change as the fluid moves along the tube if dissipative forces are present?

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

(a) If the dissipative forces are present then the pressure drop is greater.

(b) Yes, the dissipative forces become more important with increasing flow velocity.


(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
× 103 Pa).

(a) Radius of the artery r = 2 × 10-3 m
Diameter, D = 2
× 2 × 10-3 m = 4 × 10-3 m
Viscosity of the blood
η= 2.084 × 103 Pa
Density of the blood
ρ = 1.06 × 103 kg m-3
Maximum value of Reynold number of flow to be laminar, NR = 20 N
Average velocity vc =  
                            = = 0.98 m/s.

(b) Flow rate = av =
                          = 3.14
× (2 × 10-3)2 × 9.8
                          = 1.24
× 10-5 m3s-1.


A plane is in level flight at constant speed and each of its 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).

Area of each wing = 25 m2
Speed of the air over the lower wingv1 = 234 km/h =
                                                                       = 65 m/s

Speed of the air over the upper wing v2 = 180 km/h =
                                                                        = 50 m/s

Let P1 and P2 be the pressures of air at the upper and lower wings of plane respectively, then
                            P1 - P2 =
                            P1 - P2 =
                                      = = 862.5
Pressure =

Force = Pressure × Area
Force = 862.5
× 25
         = 21562.5
Mass = 2156.25 kg
Mass of the plane = 2156.25
× 2 = 4312 kg.


In Millikans oil drop experiment, what is the terminal speed of a drop of radius 2 × 10-5m and the density 1.2 × 103 kg/m3 ? Take the viscosity of air at the temperature of the experiment to be 1.8 × 10-5 Nsm-2. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

Radius of the drop a = 2 × 10-5 m
Density of the oil drop = 1.2
× 103 kgm-3
Coefficient of viscosity
η = 1.8 × 10-5 Nsm-2
The terminal velocity can be calculated using the formula
                              v =
                                = 5.8 cm/s
                                = 5.8
× 10-2 ms-1
The viscous force on the drop at that speed can be calculated using the formula
                              F = 6
                                = 6
× 3.14 × 2 ×105 × 1.8 ×10-5 ×5.8 ×10-2 
                                = 3.9
× 10-10N.


Mercury has an angle of contact equal to 140o with soda-lime glass. A narrow tube of radius 1mm 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 × 103 kg/ m3.

Surface tension σ =
                  0.465 =
                  0.465 = -x (0.086) - 2.838 x 10-5
                        x = -5.4 mm
Negative sign shows that the mercury gets depressed by 5.4 mm.


Two narrow bores of diameters 3mm and 6 mm are joined together to form a U-shaped tube open at both the 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 x 10-2 N/m. Take the angle of contact to be zero and density of water to be = 1000 kg/ m3.

For 1st bore: 
Radius of the bore r = 1.5 x 10-3 m
Surface tension
σ = 73 × 10-3 Nm-1
... Pressure difference =
                                 = 97.3 Pa
For 2nd bore:
Radius of the bore R = 3
× 10-3 m
Surface tension    
σ = 73 × 10-3 Nm-1
Pressure difference =
                          P = 48.7 Pa
Difference in pressure = 97.3 - 48.7
                                 = 48.6 Pa
                   Pressure = h
                              h = P/
                                 = (48.7
× 10-5) / (9.8 × 1000)
                                 = 5 mm
The level in the narrower bore is higher.


It is known that density ρ of air decreases with height y (in metres) as ρ = ρ0e-y/y0 where ρ0 = 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 constant (isothermal conditions). Also assume that the value of g remains constant.

We know thatρ = ρ0 e-αh
where ρ0 = 1.25 kg m-3, α = 1.2 ×10-4 m-1
Pressure due to small air column of length dh at height h is
dp = (dh) dg
    = (dh) (
ρ0 e-αh)g
ρ0g e-αh dh
Total atmospheric pressure is

   p =
  p =
    = 1.06
×104 Nm-2.

Test Your Skills Now!
Take a Quiz now
Reviewer Name