NCERT Solutions Class 11 Physics Chapter 10 Mechanical Properties of Fluids

Class 11 - Physics
Chapter 10 - Mechanical Properties of Fluids

NCERT Solutions Class 11 Physics Textbook

Question :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 than100 km
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

Answer :

  • The height of the blood capillaries is larger at feet than at brain. As a result blood pressure is greater at feet than the brain.
  • Density of air decreases with increase in height and reduces to half its value at the sea level at a height of about 60km.
  • After 6km, the density of air decreases but rather very slowly. That is why at a height of about 6km, atmosphere pressure reduces to nearly half of its value at the sea level.
  • Force is a vector quantity but pressure is a scalar quantity .
  • According to Pascal law pressure is transmitted equally in all directions i.e. direction is not associated with the pressure. Hence it is not a scalar quantity.

Question : 2.
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 glass while mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent dissolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape

Answer :
The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact (θ), as shown in the given figure.

Sla = interfacial tensions between the liquid-air,
Ssainterfacial tensions between the solid-air and
 Ssl = interfacial tensions between the liquid-air
At the line of contact, the surface forces between the three media must be in equilibrium, i.e.
cos θ = (Ssa – Ssl)/ (Sla)
The angle of contact θ, is obtuse if Ssa< Sla (as in the case of mercury on glass). This angle is acute if Ssl< Sla (as in the case of water on glass).

  • Explanation: – Mercury molecules (which make an obtuse angle with glass) have a strong force of attraction between themselves and a weak force of attraction toward solids.
  • Hence, they tend to form drops.
  • Explanation: – For equilibrium of a liquid drop on the surface of a solid, the equation TSA = TSL + TLA cos θ … (i) must be true.

For mercury-glass, angle of contact is obtuse. In order to achieve this obtuse value of angle of contact, the mercury tends to form a drop.

  • Explanation: – Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface.
  • This force is independent of the area of the liquid surface. Hence, surface tension is also independent of the area of the liquid surface.
  • Explanation: – As the cloth has narrow spaces in the form of fine capillaries. The rise of liquid in a capillary tube is given by:-

 h = (2Scos (θ).)/ (rρ g); 
=> h ∝  cos (θ).  It shows that if θ is small, cos θ will be large and as a result detergent raises more in fine capillaries in the cloth.
As detergents having small angles of contact, therefore they can penetrate more, so will remove dirt from the cloth.
(e)  Explanation: – In the absence of external forces, only force acting on the liquid drop is due to surface tension.
A drop of liquid tends to acquire minimum surface area due to the property of surface tension.
Since for a given volume of liquid, surface area is least for a sphere. So the liquid drop will always assume a spherical shape.


Question : 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)

Answer :

  • Surface tension of liquids generally decreases with temperatures.

Explanation: – The surface tension is inversely proportional to temperature.
(b) Viscosity of gases increases with temperature, whereas viscosity of liquids decreases with temperature.
Explanation: – The resistance offered by fluids to their motion is known as viscosity.
Viscosity of gases increases with temperature, while viscosity of liquids decreases with temperature.
(c)   For solids with elastic modulus of rigidity, the shearing force is proportional to shear strain while for fluids it is proportional to rate of shear strain.
Explanation: – The elastic modulus of rigidity of solids, the shearing force is proportional to the shear strain.
And the elastic modulus of rigidity of fluids, the shearing force is proportional to the 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’.
Explanation: – For a fluid in a steady flow,
the increase in flow speed at a constriction follows from ‘ conservation of mass’ while the decrease of pressure there follows from ‘ Bernoulli’s principle’.
 (e) For the model of a plane in a wind tunnel, turbulence occurs at a greater speed for turbulence for an actual plane.
Explanation: – For the model of a plane in a wind tunnel, turbulence occurs at a greater speed then it does for an actual plane.
This follows from Bernoulli’s principle and different Reynolds’ numbers are associated with the motions of the two planes.


Question : 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

Answer :

  • Explanation: – If we blow over a piece of paper, velocity of air above the paper becomes more than that below it.
  • As per Bernoulli’s principle, atmospheric pressure reduces under the paper.
  • This makes the paper fall. To keep a piece of paper horizontal, we should blow over it.
  • This increases the velocity of air above the paper. As per Bernoulli’s principle, atmospheric pressure reduces above the paper and the paper remains horizontal.
  • Explanation: – According to the equation of continuity:

Area × Velocity = Constant.
When we try to close a tap of water with our fingers, fast jets of water gush through the openings between our fingers.
This is because very small openings are left for the water to flow out of the pipe. Hence, area and velocity are inversely proportional to each other.

  • Explanation: The small opening of a syringe needle controls the velocity of the blood flowing out.
  • This is because of the equation of continuity.
  • At the constriction point of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure applied.
  • Explanation: When a fluid flows out from a small hole in a vessel, the vessel receives a backward thrust.
  • A fluid flowing out from a small hole has a large velocity according to the equation of continuity:

 Area × Velocity = Constant
According to the law of conservation of momentum, the vessel attains a backward velocity because there are no external forces acting on the system.

  • Explanation: A spinning cricket ball has two simultaneous motions – rotatory and linear.
  • These two types of motion oppose the effect of each other.
  • This decreases the velocity of air flowing below the ball.
  • Hence, the pressure on the upper side of the ball becomes lesser than that on the lower side.
  • An upward force acts upon the ball. Therefore, the ball takes a curved path. It does not follow a parabolic path.

Question : 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 ?

Answer :
Given:
Mass of the girl, m = 50 kg
Diameter of the heel, d = 1 cm = 0.01 m
Radius of the heel, r = (d/2) = 0.005m
Area of the heel = πr2
= (π (0.005)2) = (7.85 × 10–5) m2
Force exerted by the heel on the floor:
F = mg
 = (50 × 9.8) = 490 N
Pressure exerted by the heel on the floor:
P= (Force/Area)
 = ((490)/ (7.85×10-5))
= (6.24 × 106) N m–2
Therefore, the pressure exerted by the heel on the horizontal floor is
(6.24 × 106) Nm–2


Question : 6.
Torricelli’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.

Answer :
Given:
Density of mercury, ρ1 = (13.6 × 103) kg/m3
Height of the mercury column, h1 = 0.76 m
Density of French wine, ρ2 = 984 kg/m3
Height of the French wine column = h2
Acceleration due to gravity, g = 9.8 m/s2
The pressure in both the columns is equal, i.e.
Pressure in the mercury column = Pressure in the French wine column
ρ1h1g = ρ2h2g
h2= (ρ1h1)/ (ρ2)
= (13.6×103x0.76)/ (984)
= 10.5 m
Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m.


Question : 7.
A vertical off-shore 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 the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.

Answer :
Given:
The maximum allowable stress for the structure, P = 109 Pa
Depth of the ocean, d = 3 km = 3 × 103 m
Density of water, ρ = 103 kg/m3
Acceleration due to gravity, g = 9.8 m/s2
The pressure exerted because of the sea water at depth, d = ρdg
= (3 × 103 × 103 × 9.8) = 2.94 × 107 Pa
The maximum allowable stress for the structure (109 Pa) is greater than the pressure of the sea water (2.94 × 107 Pa).
The pressure exerted by the ocean is less than the pressure that the structure can withstand.
Hence, the structure is suitable for putting up on top of an oil well in the ocean.


Question : 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 cm2. What maximum pressure would the smaller piston have to bear?

Answer :
Given:
The maximum mass of a car that can be lifted, m = 3000 kg
Area of cross-section of the load-carrying piston, A = 425 cm2 = (425 × 10–4) m2
The maximum force exerted by the load,
F = mg = (3000 × 9.8) = 29400 N
The maximum pressure exerted on the load-carrying piston, P = (F/A)
= ((29400)/ (425×105)
= (6.917 × 105) Pa
Pressure is transmitted equally in all directions in a liquid.
Therefore, the maximum pressure that the smaller piston would have to bear is (6.917 × 105) Pa.


Question : 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?

Answer : 

 


Given:
Height of the spirit column, h1 = 12.5 cm = 0.125 m
Height of the water column, h2 = 10 cm = 0.1 m
P0 = Atmospheric pressure
ρ1 = Density of spirit
ρ2 = Density of water
Pressure at point B = (P0 + (h1 ρ1g))
Pressure at point D = (P0 + (h2 ρ2g))
Pressure at points B and D is the same.
(P0 + (h1 ρ1g)) = (P0 + (h2 ρ2g))
1/ ρ2) = (h2/ h1)
= (10/12.5)
= 0.8
Therefore, the specific gravity of spirit is 0.8.


Question : 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)

Answer :
Given:
Height of the water column, h1 = (10 + 15) = 25 cm
Height of the spirit column, h2 = (12.5 + 15) = 27.5 cm
Density of water, ρ1 = 1 g cm–3
Density of spirit, ρ2 = 0.8 g cm–3
Density of mercury = 13.6 g cm–3
Let h = difference between the levels of mercury in the two arms.
Pressure exerted by height h, of the mercury column:
= hρg = (h × 13.6g)………………. (i)
Difference between the pressures exerted by water and spirit:
= (h1ρ1g) – (h2ρ2g)
= g ((25 × 1) – (27.5 × 0.8)) = 3g………… (ii)
Equating equations (i) and (ii), we get:
13.6 hg = 3g h
= 0.220588 ≈ 0.221 cm
Hence, the difference between the levels of mercury in the two arms is 0.221 cm.


Question : 11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.

Answer :
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 :
No. It does not matter if one uses gauge pressure instead of absolute pressure while applying Bernoulli’s equation.
Unless the atmospheric pressure at the two points where Bernoulli’s equation is applied are significantly different.


Question : 13.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is
4.0 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube?
(Density of glycerine = 1.3 × 103 kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].

Answer :
Given:
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) kg s–1
Density of glycerine, ρ = (1.3 × 103) kg m–3
Viscosity of glycerine, η = Pa s
Volume of glycerine flowing per sec:
V= (M/ ρ)
= (4.0 × 10–3)/ (1.3 × 103)
V= (3.08 × 10–6) m3 s–1
According to Poiseville’s formula, the rate of flow:
V = (π pr4)/ (8 η l)
Where, p = pressure 0.83 difference between the two ends of the tube
Therefore p = (V 8 η l)/ (π r4)
= ((3.08 × 10–6) x 8 x 0.83 x 1.5)/ (π x (0.01)4)
= 9.8 × 102 Pa
Reynolds’ number R = (4 ρV)/ (π x (0.02) x0.83)
= (4 x (1.3 × 103) x (3.08 × 10–6))/ (π x (0.02) x0.83))
R =0.3
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, the flow speeds on the upper and lower
surfaces of the wing are 70 m s–1and 63 m s-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.

Answer :
Given:
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+ (1/2) (ρ V12)) = (P2+ (1/2) (ρV22))
(P2-P1) = ((1/2) ρ (V12 – V22))
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 of the wing provides lift to the aeroplane.
Dynamic Lift on the wing = (P2-P1) A
= ((1/2) ρ (V12 – V22) A)
= (1.3((70)2 – (63)2) x2.5)
= 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 :

  • Consider figure(b):-

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:

A1 V1 = A2 V2
When the area of cross-section in the middle of the venturimeter is small, the speed of the flow of liquid through this part is more.
According to Bernoulli’s principle, if speed is more, then pressure is less.
Pressure is directly proportional to height. Hence, the level of water 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 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?

Answer :
Given:
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:
A1V1=A2V2
V2= ((A1V1)/ A2)
= ((8 × 10–4x0.025)/ (31.61×10-6))
= 0.633 m/s
Therefore, the speed of ejection of the liquid through the holes is 0.633 m/s.


Question : 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?

Answer :
Given:
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.
Total length = 2l = (2 × 0.3) = 0.6 m
Surface tension, S = (Force or Weight)/2l
= (1.5×10-2)/ (0.6) = (2.5×10-2) N/m
Therefore, the surface tension of the film is (2.5 × 10–2) N m–1.
 


Question : 18.
Figure 10.24 (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.

Answer :

  • 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.
Therefore, Surface tension = (W/2l) = (4.5 x 10-2)/ (2×0.4)
=5.625 x 10-2 Nm-1
In all the three figures, the liquid is the same. Temperature is also the same for each case.
Hence, the surface tension in figure (b) and figure (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 weight supported in each case is (4.5 × 10–2) N.


Question : 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.

Answer :
Given:
Radius of the mercury drop, r = 3.00 mm = 3 × 10–3 m
Surface tension of mercury, S = (4.65 × 10–1) N m–1
Atmospheric pressure, P0 = (1.01 × 105) Pa
Total pressure inside the mercury drop
= (Excess pressure inside mercury + Atmospheric pressure)
= (2S/r) + P0
= (2×4.65×10-1)/ (3×10-3)
= (1.0131 × 105) = (1.01 ×105) Pa
Excess pressure = (2S/r)
 = (2×4.65×10-1)/ (3×10-3)
= 310 Pa


Question : 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 × 105 Pa).

Answer :
Given:
Excess pressure inside the soap bubble = 20 Pa;
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
Density of 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 the relation:
P= (4S/r)
= (4×2.5×10-2)/ (5×10-3)
=20Pa
Therefore, the excess pressure inside the soap bubble is 20 Pa.
The excess pressure inside the air bubble is given by the relation:
P’= (2S/r)
= (2×2.5×10-2)/ (5×10-3)
=10Pa
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’)
= (1.01×105+0.4×1.2×103x9.8+10)
= (1.057×105) Pa
= (1.06×105) Pa
Therefore, the pressure inside the air bubble is (1.06×105) Pa.

Click here to visit Official CBSE website

Click here for NCERT solutions

Click here to visit Official Website of NCERT

Click here to download NCERT Textbooks

Leave a Reply