NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties Fluids

NCERT solutions for class 11 Physics Chapter 10 Mechanical Properties of Fluids is a prime study resource on which you can rely on to prepare for class 11 examination as well as for the preparation of important entrance examination.

BYJU’S provides free NCERT solutions for all the classes from class 6 to 12. These NCERT solutions for class 11 Physics Chapter 10 Mechanical Properties of Fluids include the solution to every question in class 11 Physics NCERT textbook along with Exemplar problems, Worksheets, MCQ (multiple choice questions), tips and tricks.

Class 11 Physics NCERT Solutions for  Chapter 10 Mechanical Properties of Fluids

These solutions are prepared by a team of subject experts as per the latest CBSE syllabus 2018-19 in order to help students in their exam preparations and to score good marks. Students can easily access these solutions for class 11 Physics chapter 10 and practice various questions based on this topic. Those students who find difficulties in solving the problems can refer to this free class 11 Physics NCERT solution for chapter 10 PDF either by downloading the PDF files or by studying online.

Liquids and gases can flow and are therefore called fluids. It is this property that distinguishes liquids and gases from solids in a basic way. NCERT solutions are one of the best tools to prepare for class 11 Worksheets. Mechanical Properties of Fluids is a crucial chapter in CBSE class 11. Students must prepare this chapter well to score well in their examination and to understand the topic in depth.

Subtopics of Chapter 10 Mechanical Properties of Fluids

  1. Introduction
  2. Pressure
  3. Streamline flow
  4. Bernoulli’s principle
  5. Viscosity
  6. Reynolds number
  7. Surface tension.

Important questions of Class 11 Physics Chapter 10 Mechanical Properties of Fluids


Q.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 the pressure is force divided by area

Ans:

(i). The blood column to the feet is at a greater height than the head, thus the blood pressure in the feet is greater than that in the brain.

(ii). The density of the atmosphere does not decrease linearly with the increase in altitude, in fact, most of the air molecules are close to the surface. Thus there is this nonlinear variation of atmospheric pressure.

(iii). In hydrostatic pressure the force is transmitted equally in all direction in the liquid, thus there is no fixed direction of pressure making it a scalar quantity.

Q.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 disolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape
Ans:

(a) Water molecules have weak intermolecular forces and a strong force of attraction towards solids. Thus, they spill out. Whereas mercury molecules have a stronger intermolecular force of attraction and a weak attraction force towards solids, thus they form droplets.

(b) The angle of contact is the angle between the line tangent to the liquid surface at the point of contact and the surface of the liquid. It is donated by θ in the following diagram:

1

In the diagram SSL, SLA and   SSA, are the respective interfacial tensions between the liquid-solid, liquid-air, and solid-air interfaces. At the line of contact, the surface forces between the three media are in equilibrium, i.e.,

cosΘ=(SSA    SLASLA) cos \Theta = \left( \frac{ S_{ SA }\; -\; S_{ LA }}{ S_{ LA }} \right )

Thus, for mercury, the angle of contact θ, is obtuse because SSA < SLA. And for water, the angle is acute because SSL < SLA

(c) A liquid always tends to acquire minimum surface area because of the presence of surface tension. And as a sphere always has the smallest surface area for a given volume, a liquid drop will always take the shape of a sphere under zero external forces.

(d) Surface tension is independent of the area of the liquid surface because it is a force depending upon the unit length of the interface between the liquid and the other surface, not the area of the liquid.

(e) Clothes have narrow pores that behave like capillaries, now we know that the rise of liquid in a capillary tube is directly proportional to cos θ. So a soap decreases the value of θ in order to increase the value of cos θ, allowing the faster rise of water through the pores of the clothes.

Q.3: Fill in the blanks using the word(s) from the list appended with each statement:
(a) The surface tension of liquids generally … with temperatures (increases/decreases)
(b) The viscosity of gases … with temperature, whereas the viscosity of liquids … with temperature (increases/decreases)
(c) For solids with an 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)

Ans:

(i) Decreases

(ii) Shear strain, rate of shear strain

(iii) Gases, liquid

(iv) Greater

(v) Conservation of mass, Bernoulli’s principle

Q.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

 Ans:

(a) By covering the opening the area of the water outlet is reduced, this causes the velocity to increase in order to satisfy the equation; Area × Velocity = constant

(b) A spinning football would have followed a parabolic path had there been no air, but in the presence of air Magnus effect takes places causing the spinning ball to take a curved path.

(c) This backward thrust on the vessel is because of the principle of conservation of momentum. The outgoing fluid has forward momentum while the vessel attains backward momentum.

(d) According to the Bernoulli’s theorem  P + 12pv2\frac{ 1 }{ 2 } pv^{ 2 } = Constant ,  for a constant height.

In this equation, pressure has unit power while velocity has a square power. Hence the needle (as it controls the flow velocity of the liquid) controls the flow rate better than the pressure applied by the doctor.

(e) When we blow over a piece of paper the velocity of air over it increases, causing the pressure on it to decrease as according to Bernoulli’s theorem.  While blowing under it causes the pressure below it to decrease thereby making it hard for the paper to remain horizontal.

Q.5: A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter of 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

Ans:

Given:

Radius of the heel, r = d2\frac{ d }{ 2 } = 0.01 m

Mass of the lady, m = 50 kg

Diameter of the heel, d = 0.8 cm = 0.008 m

Area of the heel, A = πr2 = π (0.004)2= 5.024 × 10-5 m2

Force on the floor due to the heel: F = mg = 60 × 9.8 = 588 N

Pressure exerted by the heel on the floor:

P = FA\frac{ F }{ A }  = 588 5.024× 105\frac{ 588 }{  5.024 \times  10^{ -5 }}

P = 1.17×1071.17 \times 10^{ 7 } Nm -2

 Q-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

Ans:

We know:
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

Let the height of the French wine column = h2
Acceleration due to gravity, g = 9.8 m/s2.

We know that:
Pressure in the mercury column = Pressure in the wine column
ρ1h1 g =ρ2h2 g

\Rightarrow  h2 = ρ1h1ρ2\frac{ \rho _{ 1 } h_{ 1 } }{ \rho _{ 2 } } \Rightarrow  h2 = 13.6×103×0.76984 \frac{ 13.6 \times 10^{ 3 } \times 0.76 }{ 984 } = 10.5 m

Q-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

Ans:

Given:

The maximum stress the structure can handle, P = 109 Pa
Depth of the sea, d = 3 km = 3 × 103 m
Density of water, ρ = 103 kg/m3
Acceleration due to gravity, g = 9.8 m/s2

We know:
The pressure exerted by the seawater at depth, d = ρdg = 103 x 3 × 103 × 9.8 = 2.94 × 107 Pa

As the sea exerts a pressure lesser than the maximum stress the structure can handle, the structure can survive on the oil well in the sea.

Q-8: 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 ?

Ans:

Given:
Maximum mass 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 x 9.8 =  30.0 x 10 3 Pa
The maximum pressure  on the load carrying piston , P = F / A

P = 3000 500×104\frac{ 3000 }{  500 \times 10^{ -4 }} =  4.9 x 105 Pa

In a liquid, the pressure is transmitted equally in all directions. Therefore, the maximum pressure on the smaller is 4.9 × 105 Pa

Q-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?

Ans:

Given:

Height of the spirit column, h1 = 12.5 cm = 0.125 m

Height of the water column, h2 = 10 cm = 0.1 m

Let, A and B be the points of contact between spirit and mercury and water and mercury, respectively.
P0 = Atmospheric pressure
ρ1 = Density of spirit
ρ2 = Density of water

Pressure a point A = P0 + ρ1h 1g

Pressure at point B = P0 + ρ2h 2g

We know pressure at B and D is the same so;

P0 + ρ1h1g = P0 + ρ2h 2g

ρ1ρ2=h2h1=1012.5\frac{ \rho _{ 1 } }{ \rho _{ 2 } } = \frac{ h _{ 2 } }{ h _{ 1 } } = \frac{ 10 }{ 12.5 } = 0.8

Therefore the specific gravity of water is 0.8.

Q-10: In the previous problem, if 15.0 cm of water and spirit each is 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)

Ans:

Given:

Height of the water column, h1 = 15 + 10 = 25 cm

Height of the spirit column, h2 = 12.5 + 10 = 22.5 cm

Here density of water, ρ1 = 1 g cm-3

The density of spirit, ρ2 = 0.8 g cm-3

Density of mercury = 13.6 g cm -3

Let, h be the difference in the mercury levels of the two arms.

Pressure exerted by the mercury column of height h = hρg = h × 13.6 g  . . . . . . . . . . . . . . . . (1)

Difference between the pressures due to spirit and water = h1 ρ1 g – h2 ρ2 g

= g (20 × 1 – 22.5 × 0.8) = 2 g . . . . . . . . . . . . . . . . . . . . . . (2)

Equating (1) and (2) we have:

13.6 g × h = 2g

Therefore, h = 0.147 cm

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

Ans:

Bernoulli’s equation cannot be applied to the water flowing in a river because it is applicable only for ideal liquids in a streamlined flow and the water in a stream is turbulent.

Q-12: Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain

Ans:

Yes, one can definitely use a gauge instead of absolute pressure while applying Bernoulli’s equation as long as the atmospheric pressure on the two points where the equation is being applied is significantly different.

 Q-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].

Ans: 

Given:
Length of the horizontal tube, l = 2 m

Radius of the tube, r = 1 cm = 0.01 m

Diameter of the tube, d = 2r = 0.02 m

Glycerin is flowing at the rate of 2.0 × 10-3 kg/s

M = 2.0 × 10-3 kg/s

Density of glycerin, ρ = 1.3 × 103 kg m-3

Viscosity of glycerin, η = 0.83 Pa s

We know, volume of glycerin flowing per sec:

V = Mdensity\frac{ M }{ density } = 2  ×  1031.3  ×  103 \frac{ 2\; \times \;10^{ -3 } }{ 1.3\; \times \;10^{ 3 } } = 1.54 x 10 -6 m3 /s

Using Poiseville’s formula, we get:

V = πpr48  η  l\frac{ \pi p^{‘} r^{ 4 }}{ 8\; \eta \;l} p=V  8  η  lπr4p^{‘} = \frac{ V \;8\; \eta \;l}{ \pi r^{ 4 }}

Where p’ is the pressure difference between the two ends of the pipe.

p’ = 1.54  ×  106  ×  8  ×  0.83  ×  2π  ×  0.014\frac{1.54\; \times \;10^{-6} \;\times \;8\; \times \;0.83\; \times \;2 }{\pi \;\times \;0.01^{4} } = 6.51 x 102 Pa

And, we know:
R=4  ρ  Vπ  d  ηR = \frac{ 4 \;\rho \;V }{ \pi \;d\; \eta }, [Where R = Reynolds’s number]

R = 4×1.3×103×1.54×106π×0.83×0.02\frac{4 \times 1.3 \times 10^{3} \times 1.54 \times 10^{-6} }{\pi \times 0.83 \times 0.02 } = 0.153

Since the Reynolds’s number is 0.153   which is way smaller than 2000 , the flow of glycerin in the pipe is laminar.

Q-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 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

Ans:

Given:
Speed of wind on the upper side of the wing, V1 = 70 m/s
Speed of wind on the lower side of the wing, V2 = 63 m/s
Area of the wing, A = 2.5 m2
Density of air, ρ = 1.3 kg m -3

Using Bernoulli’s theorem, we get :
P1+12ρV12=P2+12ρV22P_{ 1 } + \frac{ 1 }{ 2 }\rho V_{ 1 }^{ 2 } = P_{ 2 } + \frac{ 1 }{ 2 }\rho V_{ 2 }^{ 2 } P2P1=12(ρV22ρV12)P_{ 2 } – P_{ 1 } = \frac{ 1 }{ 2 }\left (\rho V_{ 2 }^{ 2 } – \rho V_{ 1 }^{ 2 } \right )

Where, P1 = Pressure on the upper side of the wing
P2 = Pressure on the lower side of the wing

Now the lift on the wing = ( P2 – P1 ) x Area
= 12ρ(V12V22)×A\frac{ 1 }{ 2 }\rho \left ( V_{ 1 }^{ 2 } – V_{ 2 }^{ 2 } \right ) \times A
= 12×1.3((80)2(70)2)×2.5\\\frac{ 1 }{ 2 }\times 1.3 \left ( \left ( 80 \right )^{ 2 } – \left ( 70 \right )^{ 2 } \right ) \times 2.5 = 2.437 x 103 N

Therefore the lift experienced by the wings of the air craft is 2.437 x 103 N.

 Q-15: Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?

3.

Ans:

Figure ( b ) is the incorrect one. This is because at the kink , the velocity of the liquid increases decreasing the pressure (Bernoulli’s principle), subsequently the liquid in the tube at the kink shouldn’t have risen higher.

Q-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?

Ans:

Given:
Number of holes, n = 40
Cross-sectional area  of the spray pump, A1 = 10 cm -2= 10 × 10-4 m-2
Radius of each hole, r = 0.5 × 10-3 m
Cross-sectional area of each hole, a = πr2 = π (0.5 × 10-3 )2 m2
Total area of 50 holes, A2= n × a = 50×π(0.5×103)2 50 \times \pi \left ( 0.5 \times 10^{ -3 } \right )^{ 2 } m2 = 3.92 × 10 -5 m2
Speed of flow of water inside the tube, V1 = 1.5 m/min = 0.025 m/s
Let, the water ejected through the holes at a speed = V2

Using the law of continuity:
A1V1 = A2 V2

V2 = A1V1A2\frac{ A_{ 1 } V_{ 1 } }{ A_{ 2 } }  = 10×104×0.0253.92×105\frac{ 10 \times 10^{ -4 }\times 0.025 }{ 3.92 \times 10^{ -5 } }

Therefore, V2 = 0.638 m/s

Q-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 ?

Ans:

Given:
The maximum weight the film can support, W = 1.5 × 10-2 N
Length of the slider, l = 30 cm = 0.3 m

A soap film has two free surfaces.
Thus, total length = 2l = 2 × 0.3 = 0.6m

We know, surface tension =  Weight2l\frac{ Weight }{ 2l } = 2×1020.6\frac{ 2 \times 10^{ -2 } }{ 0.6 }
Thus the surface tension of the film = 3.3 x 10-2 N /m

Q-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.

 

1

Ans:

Case (a):

Length of the liquid film, l = 40 cm = 0.4 m
Weight supported by the film, W = 2× 10-2 N

We know, Surface tension = W2l\frac{ W }{ 2l } [As a film has two surface]
2×1020.8\frac{ 2\times 10^{ -2 }}{ 0.8 } = 2.5 x 10-2 N/m
It is given that all the three figures have the same liquid at the same temperature thus they will also have the same surface tension, i.e., 2.5 x 10-2 N/m
As all the cases have the same film lengths , the weight supported by each film will also be the same i.e. 2× 10-2 N.

Q-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.

Ans:

Given:
Surface tension of mercury, S = 4.65 × 10-1 N m-1
Radius of the mercury drop, r = 3.00 mm = 3 × 10-3 m
Atmospheric pressure, P0 = 1.01 × 105 Pa

We know:

Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure
= 2S/r + P0 = (2×4.65×1012×103)+1.01×105\left (\frac{ 2 \times 4.65 \times 10^{ -1 } }{ 2 \times 10^{ -3 } } \right ) + 1.01 \times 10^{ 5 } = 1.014 x 105 Pa

Excess pressure = 2Sr\frac{ 2S }{ r }

= (2×4.65×1012×103)\left (\frac{ 2 \times 4.65 \times 10^{ -1 } }{ 2 \times 10^{ -3 } } \right ) = 4.65 x 102 Pa

Q-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)

Ans:

Given:
Surface tension of the soap solution, S = 2.50 × 10-2 N/m
 r = 4.00 mm = 4 × 10-3 m
Density of the soap solution, ρ = 1.2 × 103 kg/m3
Relative density of the soap solution = 1.20
Air bubble is at a depth, h = 30 cm = 0.3 m
Radius of the air bubble, r = 4 mm = 4 × 10-3 m
Acceleration due to gravity, g = 9.8 m/s2
1 atmospheric pressure = 1.01 × 105 Pa
We know;
P = 4Sr\frac{ 4S } { r }

= (4×2.5×1024×103)\left (\frac{ 4 \times 2.5 \times 10^{ -2 } }{ 4 \times 10^{ -3 } } \right ) = 25 Pa

Thus the excess pressure inside the soap bubble is 25 Pa.
Now for the excess pressure inside the air bubble, P’ = 2Sr\frac{ 2S }{ r }
P’ = (2×2.5×1024×103)\left (\frac{ 2 \times 2.5 \times 10^{ -2 } }{ 4 \times 10^{ -3 } } \right ) = 12.5 Pa

Thus, the excess pressure inside the air bubble is 12.5 Pa

At the depth of 0.3 m, the total pressure inside the air bubble = Atmospheric pressure + hρg + P’
= 1.01 x 105 + 0.3 x 1.2 x 103 x 9.8 + 12.5
= 1.05 x 105 Pa.

Q-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. 

Ans:
Given:
Area of the hinged door, a = 20 cm2= 20 × 10-4 m
Base area of the given tank, A = 2 m2
Density of water, ρ1 = 103 kg/m3
Density of acid, ρ2 = 1.7 × 103 kg/m3
Height of the water column, h1 = 3 m
Height of the acid column, h2 = 3 m
Acceleration due to gravity, g = 9.8 ms-2
Pressure exerted by water , P = h1ρ1g  = 3 x 103 x 9.8  = 2.94 x 104 Pa
the pressure exerted by acid , P2 = h2ρ2g  = 3 x 1.7 x 103 x 9.8 = 5 x 104 Pa
Pressure difference between the above two :
ΔP = P2 – P1
= (5 – 2.94) X 104= 2.06 x 104Pa
Thus, the force on the door = ΔP x a
=2.06 x 104 x 20 × 10-4= 41.2 N
Hence the force required to keep the door closed is 41.2 N

Q-22: 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. 10.25 (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) is poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).
2

Ans:

For diagram (a):
Given,  Atmospheric pressure, P0 = 76 cm of Hg
The difference between the levels of mercury in the two arms is gauge pressure.
Thus, gauge pressure is 20 cm of Hg.
We know, Absolute pressure = Atmospheric pressure + Gauge pressure
        = 76 + 20 = 96 cm of Hg

For diagram (b):
Difference between the levels of mercury in the two arms = –18 cm
Hence, gauge pressure is –18 cm of Hg.
And, Absolute pressure = Atmospheric pressure + Gauge pressure
= 76 cm – 18 cm = 58 cm
(2) It is given that 13.6 cm of water is poured into the right arm of figure (b).
We know that relative density of mercury = 13.6
=> A 13.6 cm column of water is equivalent to 1 cm of mercury.
Let, h be the difference in the mercury levels of the two arms.
Now, pressure in the right arm PR = Atmospheric pressure + 1 cm of Hg

= 76 + 1 = 77 cm of Hg . . . . . . (a)

The mercury column rises in the left arm, thus the pressure in the left limb, PL = 58 + h . . . . . . (b)
Equating equations (a) and (b) we get :

77 = 58 + h

Therefore the difference in the mercury levels of the two arms, h = 19 cm

Q-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 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 ?

Ans:

As the base area is the same the pressure and thus the force acting on the two vessels will also be the same. However, force is also exerted on the walls of the vessel, which have a nonvertical component when the walls are not perpendicular to the base. The net non-vertical component on the sides of the vessel is lesser for the second vessel than the first.  Therefore, the vessels have different weights despite having the same force on the base.

Q-24: During a 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? [Use the density of whole blood from Table 10.1]

Ans:

Given:
Density of whole blood, ρ = 1.06 × 103 kg m-3
Gauge pressure, P = 2000 Pa
Acceleration due to gravity, g = 9.8 m/s2
let, the blood vessel be at a height = h
We know,
Pressure of the blood container, P = hρg
h = Pρg \frac{ P }{ \rho g } = 1800 9.8×1.06×103\frac{ 1800 }{  9.8 \times 1.06 \times 10^{ 3 }} = 0.173 m.
Thus, the blood will enter the vein if the blood bag is kept at height equal to or greater than 0.173m.

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

Ans:

(a)  Given:
Viscosity of blood , η= 2.08 x 10-3 m
Diameter of the arteriole, d = 1 × 10-3 m
Density of blood, ρ = 1.06 × 103 kg/m3
We know, Reynolds’ number for laminar flow, NR = 2000
Therefore, greatest  average velocity of blood is:
VAVG = NRηρd\frac{ N_{ R }\eta }{ \rho d }
= 2000×2.08×1031.06×103×1×103\\\frac{ 2000 \times 2.08 \times 10^{-3} }{ 1.06 \times 10^{ 3 } \times 1 \times 10^{ -3 } } = 3.924 m/s

(b). With the increase in fluid velocity, the dissipative forces become more significant because of the increase in turbulence.

Q-26: (a) What is the largest average velocity of blood flow in an artery of radius 2×10–3m if the flow must remain lanimar? (b) What is the corresponding flow rate ? (Take viscosity of blood to be 2.084 × 10–3 Pa s).

Ans:

Given:
Radius of the vein, r = 2 × 10-3 m
Diameter of the vein, d = 2 × 1 × 10-3 m = 2 × 10-3 m
Viscosity of blood ,η = 2.08 x 10-3 m
Density of blood, ρ = 1.06 × 103 kg/m3

We know, Reynolds’ number for laminar flow, NR = 2000
Therefore, greatest  average velocity of blood is:
VAVG= NRηρd\frac{ N_{ R }\eta }{ \rho d }
= 2000×2.08×1031.06×103×2×103\\\frac{ 2000 \times 2.08 \times 10^{-3} }{ 1.06 \times 10^{ 3 } \times 2 \times 10^{ -3 } }\\ = 1.962 m/s
And, flow rate R = VAVG π r2 = 1.962 x 3.14 x ( 10-3)2= 6.160 x 10-6 m3/s

Q-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. (Take air density to be 1 kg m–3).

Ans:

Given:
Total area of the wings of the plane, A = 2 × 25 = 50 m2
Velocity  of wind over the lower wing, V1 = 30 m/s
Velocity of wind over the upper wing, V2 = 50 m/s
Density of air, ρ = 1 kg m-3
Let, the pressure of air over the lower wing = P1 and the pressure of air over the upper wing= P2
According to Bernoulli’s equation :
P1+12ρV12=P2+12ρV22P_{ 1 } + \frac{ 1 }{ 2 }\rho V_{ 1 }^{ 2 } = P_{ 2 } + \frac{ 1 }{ 2 }\rho V_{ 2 }^{ 2 }\\ P2P1=12ρ(V22V12)\\P_{ 2 } – P_{ 1 } = \frac{ 1 }{ 2 }\rho \left ( V_{ 2 }^{ 2 } – V_{ 1 }^{ 2 } \right )
Now the upward force  = ( P2 – P1 ) x Area
= 12ρ(V22V12)A\frac{ 1 }{ 2 }\rho \left ( V_{ 2 }^{ 2 } – V_{ 1 }^{ 2 } \right ) A
= 12×1((50)2(30)2)50\frac{ 1 }{ 2 } \times 1 \left ( \left ( 50 \right )^{ 2 } – \left ( 30 \right )^{ 2 } \right ) 50 = 40000 N
We know, F = mg

i.e. 40000 = m x 9.8
Therefore, the mass of the plane is , m = 4081.63 kg

Q-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.

Ans:

Given:
Acceleration due to gravity, g = 9.8 m/s2
Radius of the uncharged drop, r = 2.0 × 10-5 m
Density of the uncharged drop, ρ = 1.2 × 103 kg m-3
Viscosity of air, η =  1.8 × 10-5 Pa s
We consider the density of air to be zero in order to neglect the buoyancy of air.
Therefore terminal velocity (v) is :

v =  2r2gρ9η\frac{ 2 r^{ 2 } g \rho }{ 9 \eta }

= 2(1.0×105)2×9.8×1.2×1039×1.8×105\frac{ 2 \left ( 1.0 \times 10^{ -5 } \right )^{ 2 } \times 9.8 \times 1.2 \times 10^{ 3 } }{ 9 \times 1.8 \times 10^{ -5 } } = 1.45 x 10-2 m/s
And the viscous force on the drop is :
F = 6πηrv
= 6 x 3.14 x 1.8 × 10-5x1.45 x 10-2 x  10-5
= 4.91 x 10-11 N

Q-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

Given:
Density of mercury, ρ =13.6 × 103 kg/m3
Angle of contact between mercury and soda lime glass, θ = 140°
Surface tension of mercury at that temperature , s = 0.465 N m-3
Radius of the narrow tube, r = 2/2  = 1 mm = 1 × 10-3 m
Let, the dip in the depth of mercury = h
Acceleration due to gravity, g = 9.8 m/s2
We know, surface tension S = hgρr2cosΘ\frac{ h g \rho r }{ 2 cos\Theta }
h=2ScosΘgρr\Rightarrow h = \frac{ 2 S cos\Theta }{ g \rho r } h=2×0.465×cos14013.6×9.8\Rightarrow h = \frac{ 2 \times 0.465 \times cos140^{\circ} }{ 13.6 \times 9.8 } = -5.34 mm

The negative sign indicates the falling level of mercury. Thus, the mercury dips by 5.34 mm.

 Q-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).
Ans:

Given:
Diameter of the first bore, d1 = 3.0 mm = 3 × 10-3 m
Radius of the first bore, r1 = 3/2 = 1.5 x 10-3 m.
Diameter of the second bore, d2 =6mm
Radius of the second bore, r2 = 6/2 = 3 x 10-3 mm
Surface tension of water, s = 7.3 × 10-2 N /m
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.
Thus, the difference in the height:
h1 – h2 = 2sCosΘr1ρg2sCosΘr2ρg\frac{2sCos\Theta }{r_{1}\rho g} -\frac{2sCos\Theta }{r_{2}\rho g}

Since, h = 2sCosΘrρg\frac{2sCos\Theta }{r\rho g}

h1 – h2 = 2sCosΘρg[1r11r2]\frac{2sCos\Theta }{\rho g}[\frac{1}{r_{1}} -\frac{1}{r_{2}}]
= 2×7.3×102×1103×9.8[12×10313×103]\frac{2\times 7.3\times 10^{-2}\times 1 }{10^{3}\times 9.8}\left [ \frac{1}{2\times 10^{-3}} -\frac{1}{3 \times 10^{-3}} \right ] = 2.482 mm
Therefore, the difference in the water levels of the two arms =2.482 mm.

Q-31: (a) According to the law of atmospheres density of air decreases with increase in height y as ρoeyy0\rho _{o}e^{\frac{-y}{y_{0}}}. Where ρ0 =1.25 kg m-3 is the density of air at sea level and y0 is a constant. Derive this equation/law considering that the atmosphere and acceleration due to gravity remain constant.


(b) A zeppelin of volume 1500 m3 is filled with helium and it is lifting a mass of 400 kg. Assuming that the radius of the zeppelin remains constant as it ascends. How high will it rise?[ y0 = 8000 m and ρHe =0.18 kg m-3]
.
Ans:

(a). We know that rate of decrease of density ρ of air is directly proportional  to the height y.
i.e., dρdy=  ρy0\frac{ d \rho }{ d y } = \;- \frac{ \rho }{ y_{ 0 }} . . . . . . . . . . . . . . ( 1 )
Where y is the constant of proportionality and the –ve sign indicates the decrease in density with increase in height.

Integrating equation (1) , we get :
ρ0ρdρ=  0y1y0dy\int_{\rho _{0}}^{\rho }\frac{d}{\rho } =\; – \int_{0}^{y } \frac{1}{y_{0}}dy\\
[logρ]ρ0ρ=[yy0]0y[log \rho ]_{\rho _{0}}^{\rho } = -[\frac{y}{y_{0}}]_{0}^{y}
Where ρ0 = density of air at sea level ie y =0
Or, loge0 /ρ ) = -y/y0
Therefore, ρ=ρ0  eyy0\rho =\rho _{0}\;e^{\frac{y}{y_{0}}}

(b).  Given:
Volume of zeppelin  = 1500 m3
Mass of payload, m = 400 kg
y0 = 8000 m
ρ0 =1.25 kg m-3
density of helium, ρ He =0.18 kg m-3

Density ρ = MassVolume\frac{ Mass }{ Volume } = Mass  of  payload+Mass  of  heliumVolume\frac{ Mass \;of \;payload + Mass \;of \;helium }{ Volume }

= 400+1500× 0.181500\frac{ 400 + 1500 \times  0.18 }{ 1500 }  [ Mass = volume × density]       
=  0.45 kg m-3
Using equation (1), we will get:
ρ=ρ0eyy0\rho =\rho _{0}e^{\frac{y}{y_{0}}}\\
loge(ρ0ρ)=y0y \\\Rightarrow log_{e}(\frac{\rho _{0}}{\rho })= \frac{y_{0}}{y}\\
Or, y=y0loge(ρ0ρ)\\y= \frac{y_{0}}{log_{e}(\frac{\rho _{0}}{\rho })}\\
y=8000loge(1.250.45) \\\Rightarrow y= \frac{8000}{log_{e}(\frac{1.25}{0.45 })}
Therefore,  y ≈ 8km

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CBSE is the most renowned educational board in India. The Central Board of Secondary Education follows the NCERT syllabus to conduct examinations for 10th and class 12th students. The NCERT Solutions Class 11 Physics Chapter 10 Mechanical Properties of Fluids of Matter is given so that students can understand the concepts of this chapter in depth.

The basic property of a fluid is that it can flow. The fluid does not have any resistance to change of its shape. Thus, the shape of the fluid is governed by the shape of its container. Some key points on Mechanical Properties of Fluids of Matter are given here.

Pressure is a scalar quantity. The definition of the pressure as “force per unit area” may give one false impression that pressure is a vector. The “force” in the numerator of the definition is the component of the force normal to the area upon which it is impressed.

As the temperature rises, the atoms of the liquid become more mobile and the coefficient of viscosity η falls. In a gas, the temperature rise increases the random motion of atoms and η increases.

BYJU’S is committed to providing the best study materials, NCERT notes and has resources that can make your learning interesting and interactive. BYJU’S videos and animation help you in remembering the topic for a long period and this helps you in scoring good marks in class 1 examination and entrance examinations.

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