Chapter 9 Mechanical Properties of Fluids

NCERT Solutions Class 11 Physics Chapter 9 – Mechanical Properties of Fluids

NCERT Solutions for Class 11 Physics Chapter 9 (Mechanical Properties of Fluids) dive into the fascinating world of substances that can flow—liquids and gases. Unlike solids, fluids do not have a fixed shape and offer little resistance to shear stress. This chapter explores how fluids behave at rest (Hydrostatics) and in motion (Hydrodynamics).

The Allen NCERT Solutions for Class 11 Physics Chapter 9 are designed by expert faculty to make fluid mechanics easy to understand with the help of lucid theory and numerical examples. The solutions are based on a systematic approach in accordance with the CBSE, JEE and NEET exam pattern to develop a strong conceptual understanding.

1.0Class 11 Physics Chapter 9 Mechanical Properties of Fluids: Key Concepts

This chapter examines the behavior of fluids under various conditions of pressure and flow. Key lessons include:

Pressure and Density: Understanding that pressure is a scalar quantity and increases with depth ( $P = P_{a} + ρ g h$ ).
Pascal’s Law:The law that the pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid.
Applications: Hydraulic lifts and hydraulic brakes.
Archimedes’ Principle and Buoyancy: Explaining why objects feel lighter in water and the conditions for floating.
Streamline and Turbulent Flow: Differentiating between steady flow and the chaotic motion that occurs at high velocities.
Equation of Continuity: The principle of conservation of mass in fluid flow ( $A_{1} v_{1} = A_{2} v_{2}$ ).
Bernoulli’s Principle: The fundamental law relating pressure, velocity, and height in a moving fluid:
$P + \frac{1}{2} ρ v^{2} + ρ g h = constant$ .
Applications: Venturi-meter, Torricelli’s Law (Efflux), and the dynamic lift of an airplane wing.
Viscosity: The internal friction of a fluid. Mastering Stoke’s Law ( $F = 6 π ηr v$ ) and the concept of Terminal Velocity.
Reynolds Number (Re): A dimensionless parameter used to predict whether a flow will be laminar or turbulent.
Surface Tension: Understanding why raindrops are spherical and the energy required to increase surface area.
Capillarity: Analyzing the rise or fall of liquid in a narrow tube due to surface tension forces.

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 9

SOLVED EXAMPLES

The two thigh bones (femurs), each of cross-sectional area $10 cm^{2}$ support the upper part of a human body of mass 40 kg. Estimate the average pressure sustained by the femurs. Sol. Total cross-sectional area of the femurs is $A = 2 \times 10 cm^{2} = 20 \times 1 0^{- 4} m^{2}$ . The force acting on them is $F = 40 kg wt = 400 N$ (taking $g = 10 m s^{- 2}$ ). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is

$P_{av} = \frac{F}{A} = 2 \times 1 0^{5} Nm^{- 2}$

What is the pressure on a swimmer 10m below the surface of a lake? Sol. Here $h = 10 m$ and $ρ = 1000 kg m^{- 3}$ . Take $g = 10 m s^{- 2}$ pressure

$P = P_{a} + ρ gh = 1.01 \times 1 0^{5} + 1000 \times 10 \times 10 = 2.01 \times 1 0^{5} Pa \approx 2 atm$

This is a 100% increase in pressure from surface level. At a depth of 1 km, the increase in pressure is 100 atm! Submarines are designed to withstand such enormous pressures.

The density of the atmosphere at sea level is $1.29 kg / m^{3}$ . Assume that it does not change with altitude. Then how high would the atmosphere extend? Sol. We use

$ρ g h ∴ = 1.29 \times 9.8 \times h = 1.01 \times 1 0^{5} Pa h = 7989 m \approx 8 km$

In reality the density of air decreases with height. So does the value of g. The atmospheric cover extends with decreasing pressure over 100 km. We should also note that at the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 10 mm or more is a sign of an approaching storm.

At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure? (c) Find the force acting on the window of area 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sea level atmospheric pressure. (The density of sea water is $1.03 \times 1 0^{3} kg m^{- 3}$ , $g = 10 m s^{- 2}$ .)

Sol. Here $h = 1000 m$ and $ρ = 1.03 \times 1 0^{3} kg m^{- 3}$ .

(a) absolute pressure

$P = P_{a} + ρ gh = 1.01 \times 1 0^{5} + 1.03 \times 1 0^{3} \times 10 \times 1000 = 104.01 \times 1 0^{5} Pa \approx 104 atm$

(b) Gauge pressure is $P - P_{a} = ρ gh = P_{g}$

$P_{g} = 1.03 \times 1 0^{3} \times 10 \times 1000 = 103 \times 1 0^{5} Pa \approx 103 atm$

(c) The pressure outside the submarine is $P = P_{a} + ρ gh$ and the pressure inside it is $P_{a}$ . Hence, the net pressure acting on the window is gauge pressure, $P_{g} = ρ gh$ . Since the area of the window is $A = 0.04 m^{2}$ , the force acting on it is

$F = P_{g} A = 103 \times 1 0^{5} \times 0.04 = 4.12 \times 1 0^{5} N$

Two syringes of different cross-sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively. (a) Find the force exerted on the larger piston when a force of 10 N is applied to the smaller piston. (b) If the smaller piston is pushed in through 6.0 cm, how much does the larger piston move out?

Sol. (a) Since pressure is transmitted undiminished throughout the fluid,

$F_{2} = \frac{A _{2}}{A _{1}} F_{1} = \frac{π ( \frac{3}{2} \times 1 0 ^{- 2} m ) ^{2}}{π ( \frac{1}{2} \times 1 0 ^{- 2} m ) ^{2}} \times 10 N = 90 N$

(b) Water is considered to be perfectly incompressible. Volume covered by the movement of smaller piston inwards is equal to volume moved outwards due to the larger piston.

$L_{1} A_{1} = L_{2} A_{2} L_{2} = \frac{A _{1}}{A _{2}} L_{1} = \frac{π ( \frac{1}{2} \times 1 0 ^{- 2} m ) ^{2}}{π ( \frac{3}{2} \times 1 0 ^{- 2} m ) ^{2}} \times 6 \times 1 0^{- 2} m ≃ 0.67 \times 1 0^{- 2} m = 0.67 cm$

Note, atmospheric pressure is common to both pistons and has been ignored.

In a car lift compressed air exerts a force $F_{1}$ on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm. If the mass of the car to be lifted is 1350 kg, calculate $F_{1}$ . What is the pressure necessary to accomplish this task? $(g = 9.8 ms^{- 2})$ . Sol. Since pressure is transmitted undiminished throughout the fluid,

$F_{1} = \frac{A _{1}}{A _{2}} F_{2} = \frac{π ( 5 \times 1 0 ^{- 2} m ) ^{2}}{π ( 15 \times 1 0 ^{- 2} m ) ^{2}} (1350 kg \times 9.8 ms^{- 2}) = 1470 N = 1.5 \times 1 0^{3} N$

The air pressure that will produce this force is

$P = \frac{F _{1}}{A _{1}} = \frac{1.5 \times 1 0 ^{3} N}{π ( 5 \times 1 0 ^{- 2} ) ^{2} m} = 1.9 \times 1 0^{5} Pa$

This is almost double the atmospheric pressure.

A fully loaded Boeing aircraft has a mass of $3.3 \times 1 0^{5} kg$ . Its total wing area is $500 m^{2}$ . It is in level flight with a speed of 960 km/h. (a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is $ρ = 1.2 kg m^{- 3}$ ] Sol. (a) The weight of the Boeing aircraft is balanced by the upward force due to the pressure difference

$Δ P \times A = 3.3 \times 1 0^{5} \times 9.8 Δ P = (3.3 \times 1 0^{5} \times 9.8) /500 = 6.5 \times 1 0^{3} Nm^{- 2}$

(b) We ignore the small height difference between the top and bottom sides. The pressure difference between them is then

$Δ P = \frac{ρ}{2} (v_{2}^{2} - v_{1}^{2})$

where $v_{2}$ is the speed of air over the upper surface and $v_{1}$ is the speed under the bottom surface.

$(v_{2} - v_{1}) = \frac{2Δ P}{ρ ( v _{2} + v _{1} )}$

Taking the average speed

$v_{av} = \frac{( v _{2} + v _{1} )}{2} = 960 km / h = 267 m s^{- 1},$

we have

$\frac{( v _{2} - v _{1} )}{v _{av}} = \frac{Δ P}{ρ v _{av}^{2}} \approx 0.08$

The speed above the wing needs to be only 8 % higher than that below.

A metal block of area $0.10 m^{2}$ is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in figure. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released, the block moves to the right with a constant speed of $0.085 m s^{- 1}$ . Find the coefficient of viscosity of the liquid.

Sol. The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. Thus, the shear force F is

$F = T = mg = 0.010 kg \times 9.8 m s^{- 2} = 9.8 \times 1 0^{- 2} N$

Shear stress on the fluid $= \frac{F}{A} = \frac{9.8 \times 1 0 ^{- 2}}{0.10} N / m^{2}$

$Strain rate = \frac{v}{t} = \frac{0.085}{0.30 \times 1 0 ^{- 3}} h = \frac{stress}{strain rate} s^{- 1} = \frac{( 9.8 \times 1 0 ^{- 2} N ) ( 0.30 \times 1 0 ^{- 3} m )}{( 0.085 m s ^{- 1} ) ( 0.10 m ^{2} )} = 3.46 \times 1 0^{- 3} Pas$

The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20°C is $6.5 cm s^{- 1}$ . Compute the viscosity of the oil at 20°C. Density of oil is $1.5 \times 1 0^{3} kg m^{- 3}$ , density of copper is $8.9 \times 1 0^{3} kg m^{- 3}$ . Sol. We have $v_{t} = 6.5 \times 1 0^{- 2} ms^{- 1}, a = 2 \times 1 0^{- 3} m$ ,

$g = 9.8 ms^{- 2}, ρ = 8.9 \times 1 0^{3} kg m^{- 3}, σ = 1.5 \times 1 0^{3} kg m^{- 3} . η = \frac{2}{9} \times \frac{a ^{2} ( ρ - σ ) g}{v _{t}} η = \frac{2}{9} \times \frac{( 2 \times 1 0 ^{- 3} ) ^{2} \times 9.8}{6.5 \times 1 0 ^{- 2}} \times 7.4 \times 1 0^{3} = 9.9 \times 1 0^{- 1} kg m^{- 1} s^{- 1} .$

The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at temperature of the experiments is $7.30 \times 1 0^{- 2} Nm^{- 1} .1$ atmospheric pressure $= 1.01 \times 1 0^{5} Pa$ , density of water $= 1000 kg / m^{3}, g = 9.80 m s^{- 2}$ . Also calculate the excess pressure. Sol. The excess pressure in a bubble of gas in a liquid is given by $\frac{2 S}{r}$ , where S is the surface tension of the liquid-gas interface. You should note there is only one liquid surface in this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure in that case is $\frac{4 S}{r}$ ). The radius of the bubble is r. Now the pressure outside the bubble $P_{0}$ equals atmospheric pressure plus the pressure due to 8.00 cm of water column. That is

$P_{0} = P_{a} + h ρ g P_{0} = 1.01 \times 1 0^{5} + 0.08 \times 1000 \times 9.80 = 1.01784 \times 1 0^{5} Pa$

Therefore, the pressure inside the bubble is

$P_{i} = P_{0} + \frac{2 S}{r} = 1.01784 \times 1 0^{5} + \frac{( 2 \times 7.3 \times 1 0 ^{- 2} )}{1 0 ^{- 3}} = (1.01784 + 0.00146) \times 1 0^{5} Pa = 1.02 \times 1 0^{5} Pa$

where the radius of the bubble is taken to be equal to the radius of the capillary tube, since the bubble is hemispherical ! (The answer has been rounded off to three significant figures.) The excess pressure in the bubble is 146 Pa.

EXERCISE QUESTION WITH SOLUTIONS

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. Sol. (a) The pressure of a liquid is given by the relation: $P = h ρ g$ Where. $P =$ Pressure $h =$ Height of the liquid column $ρ =$ Density of the liquid $g =$ Acceleration due to the gravity It can be inferred that pressure is directly proportional to height. Hence, the blood pressure in human vessels depends on the height of the blood column in the body. The height of the blood column is more at the feet than it is at the brain. Hence, the blood pressure at the feet is more than it is at the brain. (b) Density of air is the maximum near the sea level. Density of air decreases with increase in height from the surface. At a height of about 6 km, density decreases to nearly half of its value at the sea level. Atmospheric pressure is proportional to density. Hence, at a height of 6 km from the surface, it decreases to nearly half of its value at the sea level. (c) When force is applied on a liquid, the pressure in the liquid is transmitted in all directions. Hence, hydrostatic pressure does not have a fixed direction and it is a scalar physical quantity.
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 dissolved in it should have small angles of contact. (e) A drop of liquid under no external forces is always spherical in shape. Sol. (a) The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact $(θ)$ , as shown in the given figure.

$S_{la}, S_{sa}$ and $S_{sl}$ are the respective interfacial tensions between the liquid-air, solid-air, and solid-liquid interfaces. At the line of contact, the surface forces between the three media must be in equilibrium, i.e., $cos θ = \frac{S _{sa} - S _{sl}}{S _{la}}$ 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).

(b) 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. On the other hand, water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction toward solids. Hence, they tend to spread out. (c) 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. (d) Water with detergent dissolved in it has small angles of contact $(θ)$ .. This is because for a small $θ$ , there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportional to the cosine of the angle of contact $(θ)$ .If $θ$ is small, then $cos θ$ will be large and the rise of the detergent water in the cloth will be fast. (e) A liquid tends to acquire the minimum surface area because of the presence of surface tension. The surface area of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take spherical shape.

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 ____ temperature, whereas viscosity of liquids ____ 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)

Sol. (a) decreases The surface tension of a liquid is inversely proportional to temperature.

(b) increases; decreases Most fluids offer resistance to their motion. This is like internal mechanical friction, known as viscosity. Viscosity of gases increases with temperature, while viscosity of liquids decreases with temperature. (c) Shear strain; Rate of shear strain With reference to the elastic modulus of rigidity for solids, the shearing force is proportional to the shear strain. With reference to the elastic modulus of rigidity for fluids, the shearing force is proportional to the rate of shear strain. (d) Bernoulli's principle For a steadyflowing fluid, an increase in its flow speed at a constriction follows the Bernoulli's principle. (e) Greater For the model of a plane in a wind tunnel, turbulence occurs at a greater speed than 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.

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. Sol. (a) When air is blown under a paper, the velocity of air is greater under the paper than it is above it. As per Bernoulli's principle, atmospheric pressure reduces under the paper. This makes the paper fall. To keep a piece of paper horizontal, one 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. (b) According to the equation of continuity: Area × Velocity = Constant For a smaller opening, the velocity of flow of a fluid is greater than it is when the opening is bigger. 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. (c) 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. (d) 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 back ward velocity because there are no external forces acting on the system. (e) 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. [Magnus effect]
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 ?

Sol. Mass of the girl, $m = 50 kg$ Diameter of the heel, $d = 1 cm = 0.01 m$ Radius of the heel, $r = \frac{d}{2} = 0.005 m$ Area of the heel $= π r^{2} = π (0.005)^{2}$

$= 7.85 \times 1 0^{- 5} m^{2}$

Force exerted by the heel on the floor:

$F = mg = 50 \times 9.8 = 490 N$

Pressure exerted by the heel on the floor:

$P = \frac{Force}{Area} = \frac{490}{7.85 \times 1 0 ^{- 5}} = 6.24 \times 1 0^{6} N m^{- 2}$

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. Sol. Density of mercury, $ρ_{1} = 13.6 \times 1 0^{3} kg / m^{3}$ Height of the mercury column, $h_{1} = 0.76 m$ Density of French wine, $ρ_{2} = 984 kg / m^{3}$ Height of the French wine column $= h_{2}$ Acceleration due to gravity, $g = 9.8 m / s^{2}$ The pressure in both the columns is equal i.e., Pressure in the mercury column = Pressure in the French wine column

$ρ_{1} h_{1} g = ρ_{2} h_{2} g h_{2} = \frac{ρ _{1} h _{1}}{ρ _{2}} = \frac{13.6 \times 1 0 ^{3} \times 0.76}{984} = 10.5 m$

A vertical off-shore structure is built to withstand a maximum stress of $1 0^{9} Pa$ . 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. Sol. The maximum allowable stress for the structure, $P = 1 0^{9} Pa$ Depth of the ocean, $d = 3 km = 3 \times 1 0^{3} m$ Density of water, $ρ = 1 0^{3} kg / m^{3}$ Acceleration due to gravity, $g = 9.8 m / s^{2}$ The pressure exerted because of the sea water at depth, $p = ρ dg$

$= 3 \times 1 0^{3} \times 10 \times 9.8 = 2.94 \times 1 0^{7} Pa$

The maximum allowable stress for the structure (10 $^{9} Pa$ ) is greater than the pressure of the sea water ( $2.94 \times 1 0^{7} Pa$ ). 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.

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 ? Sol. The maximum mass of a car that can be lifted, $m = 3000 kg$ Area of cross-section of the load-carrying piston, $A = 425 cm^{2} = 425 \times 1 0^{- 4} m^{2}$ The maximum force exerted by the load,

$F = mg = 3000 \times 9.8 = 29400 N$

The maximum pressure exerted on the load-carrying piston, $P = \frac{F}{A} = \frac{29400}{425 \times 1 0 ^{- 4}}$

$= 6.917 \times 1 0^{5} Pa$

Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is $6.917 \times 1 0^{5} Pa$

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

Height of the spirit column, $h_{1} = 12.5 cm$ $= 0.125 m$

Height of the water column, $h_{2} = 10 cm = 0.1 m$ $P_{0} =$ Atmospheric pressure $ρ_{1} =$ Density of spirit $ρ_{2} =$ Density of water Pressure at point $B = P_{0} + h_{1} ρ_{1} g$ Pressure at point $D = P_{0} + h_{2} ρ_{2} g$ Pressure at points B and D is the same.

$P_{0} + h_{1} ρ_{1} g = ρ_{0} + h_{2} ρ_{2} g \Rightarrow \frac{ρ _{1}}{ρ _{2}} = \frac{h _{2}}{h _{1}} = \frac{10}{12.5} = 0.8$

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 $= 13.6$ ) Sol. Height of the water column, $h_{1} = 10 + 15 = 25$ cm Height of the spirit column,

$h_{2} = 12.5 + 15 = 27.5 cm$

Density of water, $ρ_{1} = 1 g cm^{- 3}$ Density of spirit, $ρ_{2} = 0.8 g cm^{- 3}$ 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:

$= h ρ g = h \times 13.6 g$

Difference between the pressures exerted by water and spirit:

$= h_{1} ρ_{1} g - h_{2} ρ_{2} g = g (25 \times 1 - 27.5 \times 0.8) = 3 g$

Equating equations (i) and (ii), we get:

$13.6 hg = 3 g \Rightarrow h = \frac{3}{13.6} h = 0.220588 = 0.221 cm$

Can Bernoulli's equation be used to describe the flow of water through a rapid in a river? Sol. 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.
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli's equation ? Explain. Sol. 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 same atmospheric pressures.
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 $4.0 \times 1 0^{- 3} kg s^{- 1}$ , what is the pressure difference between the two ends of the tube ? (Density of glycerine $= 1.3 \times 1 0^{3} kg$ $m^{- 3}$ 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]. Sol. Length of the horizontal tube, $l = 1.5 m$ Radius of the tube, $r = 1 cm = 0.01 m$ Diameter of the tube, $d = 2 r = 0.02 m$ Glycerine is flowing at a rate of $4.0 \times 1 0^{- 3} kg s^{- 1}$ .

$M = 4.0 \times 1 0^{- 3} kg s^{- 1}$

Density of glycerine, $ρ = 1.3 \times 1 0^{3} kg m^{- 3}$ Viscosity of glycerine, $η = 0.83 Pa s$ Volume of glycerine flowing per sec:

$V = \frac{M}{ρ} = \frac{4.0 \times 1 0 ^{- 3}}{1.3 \times 1 0 ^{3}} = 3.08 \times 1 0^{- 6} m^{3} s^{- 1}$

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

$V = \frac{π pr ^{4}}{8 η l}$

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

$∴ p = \frac{V 8 η l}{π r ^{4}} = \frac{3.08 \times 1 0 ^{- 6} \times 8 \times 0.83 \times 1.5}{π \times ( 0.01 ) ^{4}}$

Reynolds number is given by the relation:

$R = \frac{4 ρ V}{π d η} = \frac{4 \times 1.3 \times 1 0 ^{3} \times 3.08 \times 1 0 ^{- 6}}{π \times ( 0.02 ) \times 0.83} = 0.3$

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 $70 ms^{- 1}$ and $63 ms^{- 1}$ respectively. What is the lift on the wing if its area is $2.5 m^{2}$ ? Take the density of air to be $1.3 kg m^{- 3}$ . Sol. Speed of wind on the upper surface of the wing. $V_{1} = 70 m / s$ Speed of wind on the lower surface of the wing. $V_{2} = 63 m / s$ Area of the wing, $A = 2.5 m^{2}$ Density of air, $ρ = 1.3 kg m^{- 3}$ According to Bernoulli's theorem, we have the relation:

$P_{1} + \frac{1}{2} ρ V_{1}^{2} = P_{2} + \frac{1}{2} ρ V_{2}^{2} P_{2} - P_{1} = \frac{1}{2} ρ (V_{1}^{2} - V_{2}^{2})$

Where, $P_{1} =$ Pressure on the upper surface of the wing $P_{2} =$ 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 $= (P_{2} - P_{1}) A$

$= \frac{1}{2} ρ (V_{1}^{2} - V_{2}^{2}) A = \frac{1}{2} 1.3 ((70)^{2} - (63)^{2}) \times 2.5 = 1512.87 = 1.51 \times 1 0^{3} N$

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

Sol. (a) (Figure is incorrect) Take the case given in figure (b).

Where,

$A_{1} = Area of pipe 1 A_{2} = Area of pipe 2 V_{1} = Speed of the fluid in pipe 1 V_{2} = Speed of the fluid in pipe 2$

From the law of continuity, we have:

$A_{1} V_{1} = A_{2} V_{2}$

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.

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 ?

Sol. Area of cross-section of the spray pump,

$A = 8 cm^{2} = 8 \times 1 0^{- 4} m^{2}$

Number of holes, $n = 40$ Diameter of each hole, $d = 1 mm = 1 \times 1 0^{- 3} m$ Radius of each hole, $r = \frac{d}{2} = 0.5 \times 1 0^{- 3} m$ Area of cross-section of each hole, $a = π r^{2}$

$= π (0.5 \times 1 0^{- 3}) m^{2}$

Total area of 40 holes, $A_{2} = n \times a$

$= 40 \times π (0.5 \times 1 0^{- 3}) m^{2} = 31.41 \times 1 0^{- 6} m^{2}$

Speed of flow of liquid inside the tube,

$V_{1} = 1.5 m / min = 0.025 m / s$

Speed of ejection of liquid through the holes

$= V_{2}$

According to the law of continuity, we have:

$A_{1} V_{1} = A_{2} V_{2} V_{2} = \frac{A _{1} V _{1}}{A _{2}} = \frac{8 \times 1 0 ^{- 4} \times 0.025}{31.61 \times 1 0 ^{- 6}} = 0.633 m / s$

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 $1.5 \times 1 0^{- 2} N$ . (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film? Sol. The weight that the soap film supports,

$W = 1.5 \times 1 0^{- 2} N$

Length of the slider, $ℓ = 30 cm = 0.3 m$ A soap film has two free surfaces.

$∴ Total length = 2 ℓ = 2 \times 0.3 = 0.6 m$

Surface tension,

$S = \frac{Force or Weight}{2 ℓ} = \frac{1.5 \times 1 0 ^{- 2}}{0.6} = 2.5 \times 1 0^{- 2} N / m .$

Figure (a) shows a thin liquid film supporting a small weight $= 4.5 \times 1 0^{- 2} N$ . What is the weight supported by a film of the same liquid at the same temperature in Figure (b) and (c) ? Explain your answer physically.

Sol. 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,

$W = 4.5 \times 1 0^{- 2} N$

A liquid film has two free surfaces.

$∴ Surface tension = \frac{W}{2 ℓ} = \frac{4.5 \times 1 0 ^{- 2}}{2 \times 0.4} = 5.625 \times 1 0^{- 2} Nm^{- 1}$

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. $.5.625 \times 1 0^{- 2} Nm^{- 1}$ . Since the length of the film in all the cases is 40 cm, the weight supported in each case is $4.5 \times 1 0^{- 2} N$ .

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 $4.65 \times 1 0^{- 1} N m^{- 1}$ . The atmospheric pressure is $1.01 \times 1 0^{5} Pa$ . Also give the excess pressure inside the drop.

Sol. Radius of the mercury drop,

$r = 3.00 = 3 \times 1 0^{- 3} m$

Surface tension of mercury,

$S = 4.65 \times 1 0^{- 1} Nm^{- 1}$

Atmospheric pressure, $P_{0} = 1.01 \times 1 0^{5} Pa$ Total pressure inside the mercury drop.

$= Excess pressure inside mercury + Atmospheric pressure = \frac{2 S}{r} + P_{0} = \frac{2 \times 4.65 \times 1 0 ^{- 1}}{3 \times 1 0 ^{- 3}} + 1.01 \times 1 0^{5} = 1.0131 \times 1 0^{5} = 1.01 \times 1 0^{5} Pa$

Excess pressure $= \frac{2 S}{r} = \frac{2 \times 4.65 \times 1 0 ^{- 1}}{3 \times 1 0 ^{- 3}} = 310 Pa$

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 $2.50 \times 1 0^{- 2} N m^{- 1}$ ? 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 $1.01 \times 1 0^{5} Pa$ ). Sol. Excess pressure inside the soap bubble is 20 Pa: Pressure inside the air bubble is $1.06 \times 1 0^{5} Pa$ Soap bubble is of radius,

$r = 5.00 mm = 5 \times 1 0^{- 3} m$

Surface tension of the soap solution,

$S = 2.50 \times 1 0^{- 2} Nm^{- 1}$

Relative density of the soap solution $= 1.20$ ∴ Density of the soap solution,

$ρ = 1.2 \times 1 0^{3} kg / m^{2}$

Air bubble formed at a depth, $h = 40 cm = 0.4 m$ Radius of the air bubble, $r = 5 mm = 5 \times 1 0^{- 3} m 1$ atmospheric pressure $= 1.01 \times 1 0^{5} Pa$ . Acceleration due to gravity, $g = 9.8 m / s^{2}$ Hence, the excess pressure inside the soap bubble is given by the relation:

$P = \frac{4 S}{r} = \frac{4 \times 2.5 \times 1 0 ^{- 2}}{5 \times 1 0 ^{- 3}} = 20 Pa$

Therefore, the excess pressure inside the soap bubble is 20 Pa. The excess pressure inside the air bubble is given by the relation:

$P = \frac{2 S}{r} = \frac{2 \times 2.5 \times 1 0 ^{- 2}}{5 \times 1 0 ^{- 3}} = 10 Pa$

Therefore, the excess pressure inside the air bubble is 10 Pa. At a depth of 0.4 m, the total pressure inside the air bubble

$= Atmospheric pressure + h ρ g + P^{'} = 1.01 \times 1 0^{5} + 0.4 \times 1.2 \times 1 0^{3} \times 9.8 + 10 = 1.057 \times 1 0^{5} Pa = 1.06 \times 10 Pa$

Therefore, the pressure inside the air bubble is $= 1.06 \times 1 0^{5} Pa$

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Find chapter-wise NCERT Solutions for Class 11 Physics with simple explanations for all textbook questions. Learn important concepts, understand each solution step, and build a strong foundation in Physics for effective study and revision.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 9 (Mechanical Properties of Fluids)

Clear Understanding of Fluid Pressure: Learn how fluid pressure depends on depth, density, and gravitational acceleration and understand the hydrostatic paradox.
Step-by-Step Bernoulli’s Equation: Understand the derivation and application of Bernoulli’s equation using the work-energy principle, along with the conditions for its use.
Easy Understanding of Viscosity: Learn viscosity, terminal velocity, Stokes’ law, and viscous force through clear explanations and step-by-step numerical solutions.
Surface Tension Explained: Understand cohesive and adhesive forces, angle of contact, capillary rise, and wetting with simple explanations and relevant examples.
Prepared by ALLEN Subject Experts: Get accurate and detailed NCERT Solutions for Class 11 Physics Chapter 9, prepared according to the latest NCERT syllabus.
Simple and Concept Based Explanations: Understand Mechanical Properties of Fluids, pressure, fluid flow, viscosity, surface tension and Bernoulli’s principle with clear language and systematic solutions.
Complete NCERT Exercise Solutions: Get detailed solutions of NCERT Class 11 Physics Chapter 9 Examples and Exercise Questions. Important concepts of fluid mechanics for CBSE, JEE and NEET examination.

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