Page No 269: - Chapter 10 Mechanical Properties Of Fluids class 11 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 10.6:

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.

Answer:

10.5 m

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

= 10.5 m

Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m.

Question 10.7:

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:

Answer: Yes

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 10.8:

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?

Answer:

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, 

= 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 10.9:

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:

The given system of water, mercury, and methylated spirit is shown as follows:

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 = 

Pressure at point D = 

Pressure at points B and D is the same.

Therefore, the specific gravity of spirit is 0.8.

Question 10.10:

In problem 10.9, 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:

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 be the 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:

= 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 10.11:

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

Answer:

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 10.12:

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

Answer:

Answer: No

It does not matter if one uses gauge pressure instead of absolute pressure while applying Bernoulli’s equation. The two points where Bernoulli’s equation is applied should have significantly different atmospheric pressures.

Question 10.13:

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:

Answer: 9.8 × 10² Pa

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 = 0.83 Pa s

Volume of glycerine flowing per sec:

= 3.08 × 10^–6 m³ s^–1

According to Poiseville’s formula, we have the relation for the rate of flow:

Where, p is the pressure difference between the two ends of the tube

= 9.8 × 10² Pa

Reynolds’ number is given by the relation:

Reynolds’ number is about 0.3. Hence, the flow is laminar.

Question 10.14:

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:

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 = 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.

Lift on the wing 

= 1512.87

= 1.51 × 10³ N

Therefore, the lift on the wing of the aeroplane is 1.51 × 10³ N.

Question 10.15:

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:

Answer: (a)

Take the case given in figure (b).

Where,

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:

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 10.16:

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:

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:

= 0.633 m/s

Therefore, the speed of ejection of the liquid through the holes is 0.633 m/s.

Question 10.17:

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:

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,

 = 

Therefore, the surface tension of the film is 2.5 × 10^–2 N m^–1.

Question 10.18:

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:

Take case (a):

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.

∴Surface tension

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.