NCERT Solutions Class 11 Physics Chapter 10 Mechanical Properties of Fluids

S_{la =} interfacial tensions between the liquid-air,
S_sainterfacial tensions between the solid-air and
 S_sl = 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 = (S_sa – S_sl)/ (S_la)
The angle of contact θ, is obtuse if S_sa< S_la (as in the case of mercury on glass). This angle is acute if S_sl< S_la (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 T_SA = T_SL + T_LA 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))
(ρ₁/ 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?