Updated By SaraNextGen

On April 21, 2024, 11:35 AM

Question 10.24:

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? [Use the density of whole blood from Table 10.1].

Answer:

Gauge pressure, P = 2000 Pa

Density of whole blood, ρ = 1.06 × 10³ kg m^–3

Acceleration due to gravity, g = 9.8 m/s²

Height of the blood container = h

Pressure of the blood container, P = hρg

The blood may enter the vein if the blood container is kept at a height greater than 0.1925 m, i.e., about 0.2 m.

Question 10.25:

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.

Answer:

Answer: (a) 1.966 m/s (b) Yes

(a) Diameter of the artery, d = 2 × 10^–3 m

Viscosity of blood, 

Density of blood, ρ = 1.06 × 10³ kg/m³

Reynolds’ number for laminar flow, N_R = 2000

The largest average velocity of blood is given as:

Therefore, the largest average velocity of blood is 1.966 m/s.

(b) As the fluid velocity increases, the dissipative forces become more important. This is because of the rise of turbulence. Turbulent flow causes dissipative loss in a fluid.

Question 10.26:

(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 × 10^–3 Pa s).

Answer:

(a)Radius of the artery, r = 2 × 10^–3 m

Diameter of the artery, d = 2 × 2 × 10^–3 m = 4 × 10^{– 3} m

Viscosity of blood, 

Density of blood, ρ = 1.06 × 10³ kg/m³

Reynolds’ number for laminar flow, N_R = 2000

The largest average velocity of blood is given by the relation:

Therefore, the largest average velocity of blood is 0.983 m/s.

(b) Flow rate is given by the relation:

R = π r² 

Therefore, the corresponding flow rate is .

Question 10.27:

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).

Answer:

The area of the wings of the plane, A = 2 × 25 = 50 m²

Speed of air over the lower wing, V₁ = 180 km/h = 50 m/s

Speed of air over the upper wing, V₂ = 234 km/h = 65 m/s

Density of air, ρ = 1 kg m^–3

Pressure of air over the lower wing = P₁

Pressure of air over the upper wing= P₂

The upward force on the plane can be obtained using Bernoulli’s equation as:

The upward force (F) on the plane can be calculated as:

Using Newton’s force equation, we can obtain the mass (m) of the plane as:

∼ 4400 kg

Hence, the mass of the plane is about 4400 kg.

Question 10.28:

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.

Answer:

Terminal speed = 5.8 cm/s; Viscous force = 3.9 × 10^–10 N

Radius of the given uncharged drop, r = 2.0 × 10^–5 m

Density of the uncharged drop, ρ = 1.2 × 10³ kg m^–3

Viscosity of air, 

Density of air   can be taken as zero in order to neglect buoyancy of air.

Acceleration due to gravity, g = 9.8 m/s²

Terminal velocity (v) is given by the relation:

Hence, the terminal speed of the drop is 5.8 cm s^–1.

The viscous force on the drop is given by:

Hence, the viscous force on the drop is 3.9 × 10^–10 N.

Question 10.29:

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.

Answer:

Angle of contact between mercury and soda lime glass, θ = 140°

Radius of the narrow tube, r = 1 mm = 1 × 10^–3 m

Surface tension of mercury at the given temperature, s = 0.465 N m^–1

Density of mercury, ρ =13.6 × 10³ kg/m³

Dip in the height of mercury = h

Acceleration due to gravity, g = 9.8 m/s²

Surface tension is related with the angle of contact and the dip in the height as:

Here, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by 5.34 mm.

Question 10.30:

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).

Answer:

Diameter of the first bore, d₁ = 3.0 mm = 3 × 10^–3 m

Diameter of the second bore,  = 6.0 mm

Surface tension of water, s = 7.3 × 10^–2 N m^–1

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. These heights are given by the relations:

The difference between the levels of water in the two limbs of the tube can be calculated as:

Hence, the difference between levels of water in the two bores is 4.97 mm.

Question 10.31:

(a) It is known that density ρ of air decreases with height y as 

Where  = 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 a constant (isothermal conditions). Also assume that the value of gremains constant.

(b) A large He balloon of volume 1425 m³ is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?

[Take y₀= 8000 m and  = 0.18 kg m^–3].

Answer:

(a) Volume of the balloon, V = 1425 m³

Mass of the payload, m = 400 kg

Acceleration due to gravity, g = 9.8 m/s²

Density of the balloon = ρ

Height to which the balloon rises = y

Density (ρ) of air decreases with height (y) as:

This density variation is called the law of atmospherics.

It can be inferred from equation (i) that the rate of decrease of density with height is directly proportional to ρ, i.e.,

Where, k is the constant of proportionality

Height changes from 0 to y, while density changes from  to ρ.

Integrating the sides between these limits, we get:

(b)

Hence, the balloon will rise to a height of 8 km.