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Non-Rationalised Science NCERT Notes and Solutions (Class 6th to 10th)
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Non-Rationalised Science NCERT Notes and Solutions (Class 11th)
Physics Chemistry Biology
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Physics Chemistry Biology

Class 11th (Physics) Chapters
1. Physical World 2. Units And Measurements 3. Motion In A Straight Line
4. Motion In A Plane 5. Laws Of Motion 6. Work, Energy And Power
7. System Of Particles And Rotational Motion 8. Gravitation 9. Mechanical Properties Of Solids
10. Mechanical Properties Of Fluids 11. Thermal Properties Of Matter 12. Thermodynamics
13. Kinetic Theory 14. Oscillations 15. Waves



Chapter 10 Mechanical Properties Of Fluids



Introduction

This chapter explores the mechanical properties of liquids and gases. Both liquids and gases have the ability to flow, and for this reason, they are collectively known as fluids. This property is what fundamentally distinguishes them from solids.

Fluids play a crucial role in our world and biological systems. The Earth is enveloped by air and its surface is largely covered by water. All life processes, in both plants and animals, are mediated by fluids.

Differences between Solids, Liquids, and Gases

Unlike solids, fluids do not have a definite shape of their own; they take the shape of their container.



Pressure

The impact of a force depends not only on its magnitude but also on the area over which it is applied. This concept is quantified by pressure.

When a fluid is at rest, it exerts a force on any surface in contact with it. This force is always directed normal (perpendicular) to the surface. If there were a tangential component of the force, the fluid would flow, which contradicts the condition that it is at rest.

Force exerted by a fluid on a submerged object is always normal to the surface at every point.

Definition of Pressure

Pressure (P) is defined as the normal force (F) acting per unit area (A).

$P = \frac{F}{A}$

Pressure is a scalar quantity. It has magnitude but no direction. The force exerted by a fluid on a surface is always perpendicular to that surface, regardless of the surface's orientation.


Density

Density ($\rho$) is another crucial property of fluids, defined as mass (m) per unit volume (V).

$\rho = \frac{m}{V}$


Pascal’s Law

This fundamental principle of fluid statics was discovered by Blaise Pascal.

Pascal's Law states that the pressure in a fluid at rest is the same at all points if they are at the same height.

Another form of the law states: Whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions.

Hydraulic Machines

Pascal's law is the principle behind hydraulic machines like the hydraulic lift and hydraulic brakes. These devices use an incompressible fluid to transmit force.

In a hydraulic lift with two pistons of areas $A_1$ and $A_2$ ($A_2 > A_1$), a small force $F_1$ applied to the smaller piston creates a pressure $P = F_1/A_1$. This pressure is transmitted to the larger piston, producing a large upward force $F_2 = P \times A_2$.

$F_2 = \left(\frac{F_1}{A_1}\right) A_2 = F_1 \left(\frac{A_2}{A_1}\right)$

The force is magnified by a factor of $A_2/A_1$, which is the mechanical advantage of the lift.

Schematic of a hydraulic lift, where a small force F1 on a small piston A1 generates a large force F2 on a large piston A2.

Variation of Pressure with Depth

In a fluid under gravity, pressure increases with depth. Consider a fluid of constant density $\rho$ at rest. The pressure difference between two points separated by a vertical height $h$ is given by:

$P_2 - P_1 = \rho g h$

If the surface of the liquid (at height $h=0$) is open to the atmosphere, the pressure at the surface is atmospheric pressure, $P_a$. The absolute pressure $P$ at a depth $h$ below the surface is:

$P = P_a + \rho g h$

Gauge Pressure

The excess pressure above atmospheric pressure, $P - P_a$, is called the gauge pressure. So, Gauge Pressure $= \rho g h$.


Example 1. What is the pressure on a swimmer 10 m below the surface of a lake?

Answer:

Given: depth $h = 10$ m, density of water $\rho = 1000 \text{ kg/m}^3$. Let $g = 9.8 \text{ m/s}^2$ and atmospheric pressure $P_a = 1.01 \times 10^5 \text{ Pa}$.

The absolute pressure is $P = P_a + \rho g h$.

$P = (1.01 \times 10^5 \text{ Pa}) + (1000 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 \times 10 \text{ m})$

$P = 1.01 \times 10^5 + 0.98 \times 10^5 = 1.99 \times 10^5 \text{ Pa}$.

This is approximately $2 \times 10^5$ Pa, or about 2 atmospheres.



Streamline Flow

The study of fluids in motion is called fluid dynamics. We often analyze an idealized type of motion called steady or streamline flow.

The flow of a fluid is said to be steady if, at any given point, the velocity of each passing fluid particle remains constant over time. In steady flow, the path taken by a fluid particle is called a streamline. A streamline is a curve whose tangent at any point gives the direction of the fluid velocity at that point. In steady flow, streamlines do not cross each other.

A region of streamline flow showing smooth, non-crossing paths of fluid particles.

Equation of Continuity

For an incompressible fluid (density $\rho$ is constant) in steady flow, the mass of fluid entering a section of a pipe must equal the mass leaving it. This is a statement of the conservation of mass.

If the fluid flows with speed $v_1$ through an area of cross-section $A_1$ and speed $v_2$ through an area $A_2$, then:

Mass flow rate = $\rho A_1 v_1 = \rho A_2 v_2$

Since $\rho$ is constant, we get the equation of continuity:

$A_1 v_1 = A_2 v_2 \quad \text{or} \quad Av = \text{constant}$

The product $Av$ is the volume flow rate. This equation implies that where the pipe is narrower (smaller A), the fluid speed is greater, and where the pipe is wider (larger A), the speed is smaller.

Turbulent Flow

At low speeds, flow is typically steady (or laminar). Above a certain critical speed, the flow becomes unsteady and chaotic. This irregular flow is called turbulent flow and is characterized by the formation of eddies and whirlpools.



Bernoulli’s Principle

Bernoulli's principle is a fundamental principle in fluid dynamics that relates pressure, velocity, and height for a moving fluid. It is essentially a statement of the conservation of energy for an ideal fluid (incompressible and non-viscous) in steady flow.

Bernoulli's Equation

Consider an ideal fluid flowing through a pipe of varying cross-section and height. The work done on a volume of fluid by pressure differences goes into changing its kinetic and potential energy.

Diagram showing an ideal fluid flowing through a pipe with changing height and cross-sectional area, used for deriving Bernoulli's equation.

The relationship derived from the work-energy theorem is Bernoulli's equation:

$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$

This means that along a streamline, the sum of the pressure ($P$), the kinetic energy per unit volume ($\frac{1}{2}\rho v^2$), and the potential energy per unit volume ($\rho g h$) remains constant.

$P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$

A key consequence for horizontal flow ($h$=constant) is that where the speed is higher, the pressure is lower, and vice versa.


Applications of Bernoulli's Principle

1. Speed of Efflux: Torricelli’s Law

Consider a tank filled with a liquid of density $\rho$, open to the atmosphere, with a small hole at a depth $h$ below the surface. Applying Bernoulli's equation between the surface of the liquid (point 2) and the hole (point 1), we find the speed of the liquid exiting the hole (efflux) is:

$v_1 = \sqrt{2gh}$

This is Torricelli's law. The speed of efflux is the same as the speed an object would acquire by freely falling through the same height $h$.

2. Venturi-meter

A Venturi-meter is a device used to measure the flow speed of a fluid. It consists of a tube with a narrow constriction (a "throat"). According to the equation of continuity, the fluid speeds up in the throat. According to Bernoulli's principle, the pressure in the throat is therefore lower than in the wider section. By measuring this pressure difference (often with a U-tube manometer), the flow speed can be calculated.

3. Dynamic Lift (Aerofoil and Magnus Effect)

The shape of an aeroplane wing (an aerofoil) is designed so that the air flows faster over its curved upper surface than its flatter lower surface. This difference in speed creates a pressure difference: the pressure above the wing is lower than the pressure below it. This pressure difference results in a net upward force called dynamic lift, which counteracts the weight of the aircraft.

A similar effect explains the curve of a spinning ball (Magnus effect). A spinning ball drags air with it, causing the air speed to be higher on one side and lower on the other, creating a pressure difference and a net force that makes the ball swerve.



Viscosity

Ideal fluids are non-viscous, but real fluids offer resistance to flow. This internal friction in a fluid is called viscosity. It arises from the frictional forces between adjacent layers of the fluid that are in relative motion.

Consider a fluid between two parallel plates, with the bottom plate fixed and the top plate moving at a constant velocity $v$. The layer of fluid in contact with the top plate moves with velocity $v$, while the layer in contact with the bottom plate is at rest. The velocity of intermediate layers varies linearly from 0 to $v$. This type of flow, where layers slide over one another, is called laminar flow.

Laminar flow of a viscous fluid between a fixed and a moving plate, showing the velocity profile of the fluid layers.

The shearing stress ($F/A$) in the fluid is found to be proportional to the rate of change of strain (strain rate), which is given by $v/l$. The constant of proportionality is the coefficient of viscosity ($\eta$).

$\frac{F}{A} = \eta \frac{v}{l} \implies F = \eta A \frac{v}{l}$


Stokes' Law and Terminal Velocity

When a solid object moves through a viscous fluid, it experiences a drag force that opposes its motion. For a small sphere of radius $a$ moving at a low speed $v$ through a fluid of viscosity $\eta$, the viscous drag force $F_v$ is given by Stokes' Law:

$F_v = 6\pi \eta a v$

When an object falls through a fluid, it initially accelerates due to gravity. As its speed increases, the viscous drag force also increases. Eventually, the upward drag force plus the buoyant force will balance the downward gravitational force. At this point, the net force is zero, the acceleration ceases, and the object falls with a constant maximum velocity called the terminal velocity ($v_t$).



Surface Tension

The free surface of a liquid behaves like a stretched elastic membrane. This phenomenon is called surface tension. It is responsible for many effects, such as the spherical shape of small liquid drops, the ability of insects to walk on water, and capillary action.

Microscopic Origin and Surface Energy

Molecules inside a liquid are attracted equally in all directions by their neighbors. However, molecules at the surface experience a net inward pull from the molecules below them. This net inward force pulls the surface molecules closer together and causes the surface to contract to the minimum possible area.

To bring a molecule from the interior to the surface, work must be done against these cohesive forces. This work is stored as potential energy in the surface. Thus, the molecules at the surface have a higher potential energy than those in the bulk. This extra energy per unit area of the surface is called surface energy.

Forces on a molecule deep inside a liquid are balanced, while forces on a surface molecule have a net inward pull.

Definition of Surface Tension

Surface tension (S) is defined as the force per unit length acting in the plane of the liquid surface, perpendicular to a line drawn on the surface. It is also numerically equal to the surface energy per unit area.

$S = \frac{\text{Force}}{\text{Length}} = \frac{\text{Surface Energy}}{\text{Area}}$


Angle of Contact

When a liquid is in contact with a solid surface, the liquid surface is generally curved near the point of contact. The angle of contact ($\theta$) is the angle between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid.


Pressure Difference across a Curved Surface (Laplace Pressure)

Due to surface tension, there is a pressure difference across any curved liquid-air interface. The pressure is greater on the concave side of the surface.


Capillary Action

The rise or fall of a liquid in a narrow tube (a capillary) is called capillary action or capillarity. This is a result of surface tension and the angle of contact.

The height $h$ to which a liquid of density $\rho$ and surface tension $S$ rises in a capillary of radius $a$ is given by:

$h = \frac{2S \cos\theta}{\rho g a}$

The rise is greater for narrower tubes.



What is Blood Pressure?

The circulatory system of upright animals, including humans, has evolved to handle the demands of moving blood against gravity. The venous system, which returns blood from the lower parts of the body to the heart, is particularly adapted for this task. Without these adaptations, blood would pool in the lower extremities, as seen in some animals like snakes or rats, which cannot survive being held in an upright position for long.


Pressure Variation in the Human Body

The pressure in the arteries varies significantly at different points in the body due to gravity. We can understand this variation using a simplified form of Bernoulli's principle. For blood flow in major arteries, the velocity is relatively small and constant, so the kinetic energy term ($\frac{1}{2}\rho v^2$) can be ignored. Bernoulli's equation then reduces to the hydrostatic pressure relationship:

$P + \rho g y = \text{constant}$

This means the pressure difference between any two points depends on their vertical separation. Let's denote the gauge pressures at the brain, heart, and foot as $P_B$, $P_H$, and $P_F$ respectively. The relationship between them when a person is standing is:

$P_F = P_H + \rho g h_H = P_B + \rho g h_B$

where $\rho$ is the density of blood, $h_H$ is the height of the heart above the feet, and $h_B$ is the height of the brain above the feet.

Schematic diagram showing the gauge pressures in kPa in the arteries of a human at the brain, heart, and foot, both while standing and lying down.

Using typical values for an adult ($\rho \approx 1.06 \times 10^3 \text{ kg/m}^3$, $h_H \approx 1.3 \text{ m}$, and the height from heart to brain being 0.4 m, so $h_B \approx 1.7 \text{ m}$), and an average heart pressure of $P_H = 13.3 \text{ kPa} \approx 100 \text{ mm of Hg}$, we can estimate the pressures:

When the person is lying down, all these points are at roughly the same horizontal level, so the pressures become nearly equal.


Biological Adaptations for Blood Circulation

The human body has remarkable mechanisms to assist blood return to the heart against gravity:

This is why a soldier standing at attention for a long time may faint; the lack of muscle movement reduces blood return to the heart, leading to insufficient blood supply to the brain. Lying the person down equalizes the pressure and restores consciousness.


Measuring Blood Pressure

Blood pressure is typically measured using an instrument called a sphygmomanometer. The measurement is taken at the upper arm because it is at the same level as the heart, providing a pressure reading close to that at the heart.

Systolic and Diastolic Pressure

Blood pressure is not constant; it fluctuates with each heartbeat.

The Measurement Process

The sphygmomanometer works by temporarily stopping and then releasing the blood flow in the brachial artery of the upper arm and listening to the resulting sounds with a stethoscope.

A diagram showing blood pressure measurement with a sphygmomanometer cuff on the upper arm, a pressure gauge, and a stethoscope.
  1. An inflatable cuff is wrapped around the upper arm and inflated to a pressure high enough to completely close the brachial artery, stopping blood flow.
  2. The pressure in the cuff is then slowly released. A stethoscope is placed over the artery just below the cuff.
  3. When the cuff pressure drops just below the systolic pressure, the artery opens briefly during the peak pressure phase of each heartbeat. The blood rushes through the constricted opening, creating turbulent flow, which is heard as a distinct tapping sound. The pressure reading at which this first sound is heard is recorded as the systolic pressure.
  4. As the cuff pressure continues to decrease, the tapping sound continues.
  5. When the cuff pressure drops to the diastolic pressure, the artery remains open throughout the entire heart cycle. The flow is still turbulent but the sound changes from a tapping to a continuous, dull roar or disappears altogether. The pressure reading at which the tapping sound ceases is recorded as the diastolic pressure.

The blood pressure reading is given as a ratio of these two values, for example, 120/80 mm of Hg for a healthy resting adult. Readings consistently above 140/90 mm of Hg are considered high blood pressure (hypertension) and require medical attention.



Exercises



Question 10.1. 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 pressure is force divided by area.

Answer:

Question 10.2. 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

Answer:

Question 10.3. 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:

Question 10.4. 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:

Question 10.5. 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:

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:

Question 10.7. A vertical off-shore structure is built to withstand a maximum stress of $10^9$ 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:

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 cm$^2$. What maximum pressure would the smaller piston have to bear ?

Answer:

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:

Question 10.10. 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:

Question 10.11. Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river ? Explain.

Answer:

Question 10.12. Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation ? Explain.

Answer:

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 $4.0 \times 10^{–3} \text{ kg s}^{–1}$, what is the pressure difference between the two ends of the tube ? (Density of glycerine = $1.3 \times 10^3 \text{ kg m}^{–3}$ 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:

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$^{–1}$ and 63 m s$^{-1}$ respectively. What is the lift on the wing if its area is 2.5 m$^2$ ? Take the density of air to be 1.3 kg m$^{–3}$.

Answer:

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 ?

Two figures showing fluid flow through pipes of varying cross-section. In (a), the pipe narrows, and the streamlines get closer. In (b), the pipe narrows, but the streamlines spread farther apart.

Answer:

Question 10.16. The cylindrical tube of a spray pump has a cross-section of 8.0 cm$^2$ 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:

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 $1.5 \times 10^{–2}$ 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:

Question 10.18. Figure 10.24 (a) shows a thin liquid film supporting a small weight = $4.5 \times 10^{–2}$ 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.

Three diagrams showing a liquid film stretched on frames of different shapes and sizes. (a) is 40 cm long. (b) is 40 cm long but curved. (c) is a circle of 40 cm diameter. All support a weight.

Answer:

Question 10.19. 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 $4.65 \times 10^{–1} \text{ N m}^{–1}$. The atmospheric pressure is $1.01 \times 10^5$ Pa. Also give the excess pressure inside the drop.

Answer:

Question 10.20. 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 $2.50 \times 10^{–2} \text{ N m}^{–1}$ ? 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 $1.01 \times 10^5$ Pa).

Answer:

Additional Exercises

Question 10.21. A tank with a square base of area 1.0 m$^2$ is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm$^2$. 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.

Answer:

Question 10.22. 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.

Two U-tube manometers. In (a), the mercury level in the arm connected to the gas is lower than the open arm. In (b), the mercury level in the arm connected to the gas is higher than the open arm.

(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) are poured into the right limb of the manometer ? (Ignore the small change in the volume of the gas).

Answer:

Question 10.23. 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 ?

Answer:

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:

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 $2 \times 10^{–3}$ m if the flow must remain laminar ? (b) Do the dissipative forces become more important as the fluid velocity increases ? Discuss qualitatively.

Answer:

Question 10.26. (a) What is the largest average velocity of blood flow in an artery of radius $2 \times 10^{–3}$m if the flow must remain lanimar? (b) What is the corresponding flow rate ? (Take viscosity of blood to be $2.084 \times 10^{–3}$ Pa s).

Answer:

Question 10.27. A plane is in level flight at constant speed and each of its two wings has an area of 25 m$^2$. 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:

Question 10.28. In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius $2.0 \times 10^{–5}$ m and density $1.2 \times 10^3 \text{ kg m}^{–3}$. Take the viscosity of air at the temperature of the experiment to be $1.8 \times 10^{–5}$ Pa s. How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop due to air.

Answer:

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 = $13.6 \times 10^3 \text{ kg m}^{–3}$.

Answer:

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 $7.3 \times 10^{–2} \text{ N m}^{–1}$. Take the angle of contact to be zero and density of water to be $1.0 \times 10^3 \text{ kg m}^{–3}$ (g = 9.8 m s$^{–2}$) .

Answer:

Calculator/Computer – Based Problem

Question 10.31. (a) It is known that density $\rho$ of air decreases with height y as

$\rho = \rho_0 e^{-y/y_0}$

where $\rho_0 = 1.25 \text{ kg m}^{–3}$ is the density at sea level, and $y_0$ 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 g remains constant.

(b) A large He balloon of volume 1425 m$^3$ 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 $y_0 = 8000$ m and $\rho_{He} = 0.18 \text{ kg m}^{–3}$].

Answer: