NCERT Exemplar Class 11 Physics Solutions Chapter 10 Mechanical Properties of Fluids

Vishal kumarUpdated on 10 Dec 2025, 01:24 PM IST

Fluids such as air and water show remarkable behaviour! Students will learn concepts such as fluid flow, fluid pressure, and buoyancy in this chapter. You will know why big ships sail, how aeroplanes take off, and even how the circulation of fluids such as blood takes place in living organisms.

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NCERT Exemplar Class 11 Physics Solutions Chapter 10: MCQI
NCERT Exemplar Class 11 Physics Solutions Chapter 10: MCQII
NCERT Exemplar Class 11 Physics Solutions Chapter 10: Very Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 10: Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 10: Long Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 10: Important Concepts and Formulas
Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 10 Mechanical Properties of Fluids
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
NCERT Exemplar Class 11th Solutions

NCERT Exemplar Class 11 Physics Chapter 10 Mechanical Properties of Fluids

The NCERT Exemplar Class 11 Physics Solutions Chapter 10 Mechanical Properties of Fluids is also easier to learn due to its detailed solutions, objective questions, and numerical problems, which are solved, making learning these concepts of fluid mechanics easier and exam-oriented. The NCERT Exemplar Class 11 Solutions Physics Chapter 10 Mechanical Properties of Fluids states how gases and liquids are able to move freely as opposed to rigid solids. The most important concepts, like viscosity, pressure change within fluids and intermolecular forces within fluids, are represented clearly. The NCERT Exemplar solutions are presented in a clear and understandable manner, with their derivations and practice being based on formulas so that students can build their conceptual knowledge and get high marks in tests.

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NCERT Exemplar Class 11 Physics Solutions Chapter 10: MCQI

NCERT Exemplar Class 11 Physics Chapter 10 Multiple-choice questions I (MCQs) have been developed to assess your knowledge about the behaviour of fluids under various conditions. These questions include pressure, buoyancy, the Bernoulli law and viscosity. Answering the MCQs can assist students in revising the concepts faster and in studying well in time to take exams.

Question:10.1

A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plot shown in figure, indicate the one that represents the velocity (v) of the pebble as a function of time (t).
v-t graph

Answer:

The answer is option (c)
When a cylinder is filled with viscous oil and a pebble is dropped from the top, it falls under the effect of gravity and acceleration due to this force. When it enters the oil, a dragging force $f = 6 \pi r n v$ acts opposite, and hence the acceleration of the pebble decreases till it becomes zero and the velocity becomes constant. Hence, option c is correct.

Question:10.2

Which of the following diagrams does not represent a streamline flow?

Answer:

The answer is option (d)
The velocity remains constant for a liquid in the case of a streamlined flow. So, the layers do not cross each other. And hence, d is the correct choice.

Question:10.3

Along a streamline
a) the velocity of a fluid particle remains constant
b) the velocity of all fluid particles crossing a given position is constant
c) the velocity of all fluid particles at a given instant is constant
d) the speed of a fluid particle remains constant

Answer:

The answer is option (b), the velocity of all fluid particles crossing a given position is constant.

The speed at any point or any cross-section for a liquid is constant in the case of a streamlined flow.

Question:10.4

An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5 cm and 3.75 cm. The ratio of the velocities in the two pipes is
a) 9:4
b) 3:2
c) $\sqrt{3}:\sqrt{2}$
d) $\sqrt{2}:\sqrt{3}$

Answer:

The answer is option (a) 9:4
a₁ v₁ = a₂ v₂
$\frac{v_1}{v_2}= \frac{a_2}{a_1}$
$\frac{v_1}{v_2} =\frac{ d_{2}^{2}}{d_{1}^{2}}$
putting values,
$\frac{v_1}{v_2} = \frac{\left ( 3.75 \times 3.75 \right )}{2.5 \times 2.5} =\frac{9}{4}$
so, $v_1: v_2$ = 9:4. Hence option a is correct.

Question:10.5

The angle of contact at the interface of water-glass is $0^{\circ}$, ethylalcohol-glass is $0^{\circ}$, mercury-glass is $140^{\circ}$, and methyl iodide-glass is $30^{\circ}$. A glass capillary is put in a trough containing one of these four liquids. It is observed that the meniscus is convex. The liquid in the trough is
a) water
b) ethyl alcohol
c) mercury
d) methyliodide

Answer:

The answer is option (c) Mercury.
In the case of mercury glass, the meniscus of liquid is convex, which makes the angle of contact obtuse. Hence, option c is the correct choice.

NCERT Exemplar Class 11 Physics Solutions Chapter 10: MCQII

The behaviour of fluids is also an important branch of physics and is significant in solving problems dealing with pressure, flow and buoyancy. The MCQ-II part of the NCERT Exemplar Class 11 Physics Chapter 10 assists students in studying fluid concepts more profoundly and enhancing conceptual awareness for competitive exams. These are the properly organised solutions of the MCQ-II that simplify the preparation.

Question:10.6

For a surface molecule
a) the net force on it is zero
b) there is a net downward force
c) the potential energy is less than that of a molecule inside
d) the potential energy is more than that of a molecule inside

Answer:

The answer is option (b) and (d)
The force exerted on a molecule by all other neighbouring molecules is equal and opposite in all directions. Hence, the horizontal component is zero.
A downward force acts vertically, which is not cancelled and hence b is the correct option.
The potential energy of the molecules on the surface is negative as compared to the molecules in the subsequent layers. But the magnitude of this PE is higher than the following molecules. Hence, option d is also correct.

Question:10.7

Pressure is a scalar quantity because
a) it is the ratio of force to area and both force and area are vectors
b) it is the ratio of the magnitude of the force to area
c) it is the ratio of component of the force normal to the area
d) it does not depend on the size of the area chosen

Answer:

The answer is option (c)
The pressure is nothing but a component of some force acting along a particular area. The directions here are fixed for both these quantities, which makes pressure a scalar quantity. Hence, option c is the right choice.

Question:10.8

A wooden block with a coin placed on its top, floats in water as shown in the figure:
principle of floatation
The distances l and h are shown in the figure. After some time, the coin falls into the water. Then
a) l decreases
b) h decreases
c) l increases
d) h increases

Answer:

The answers are options (a) and (b)
The amount of displaced fluid is equal to the weight of the floating body; this is as per the flotation principle. When the coin is dropped, the net force acting on the block decreases as the coin displaces water in an equivalent amount to its weight. This is the reason that the block rises up. As the block rises, l decreases. The height h will also decrease, as the amount of water getting displaced now will be less. Hence, options a and b are correct.

Question:10.9

With increase in temperature, the viscosity of
a) gases decreases
b) liquids increases
c) gases increases
d) liquids decreases

Answer:

The answers are options (c) and (d)
If n is the coefficient of viscosity of a liquid, $n\propto \frac{1}{\sqrt{T}}$
In case of the viscosity of gases, $n\propto \sqrt{T}$
As the temperature increases, kinetic energy in the gas molecules also increases and hence, due to the collisions between the molecules, the viscosity increases.
Whereas in the case of a liquid, the kinetic energy increases with temperature, but the relative speed of the layers of liquid remains fairly the same as before. Hence, options c and d are correct.

Question:10.1

Streamline flow is more likely for liquids with
a) high density
b) high viscosity
c) low density
d) low viscosity

Answer:

The answers are options (b) and (c)
Liquids which have a lower density are more subjected to a streamline flow. The velocity gradient of a liquid increases with its viscosity. Hence, the line of flows can be differentiated accordingly.
Reynold number, $R =\frac{ pvD}{n}$
As we increase n and decrease p, R decreases. This decrease in R makes the flow more streamlined. Hence, options b and c are correct.

NCERT Exemplar Class 11 Physics Solutions Chapter 10: Very Short Answer

The NCERT Exemplar Class 11 Physics Chapter 10 VSAQs serve to reinforce the fundamental conceptual knowledge in the students. These questions test understanding of concepts such as pressure, buoyancy, viscosity, Pascal's law, and streamline flow. They are even perfect for revision within a short time before exams.

Question:10.11

Is viscosity a vector?

Answer:

Viscosity: the dragging force between two layers of a liquid per unit area, where their velocity gradient is called viscosity. This quantity just has a magnitude and no direction, and hence it is not a vector quantity.

Question:10.12

Is surface tension a vector?

Answer:

Surface tension = work done/ change in the surface area
These are scalar quantities, and hence surface tension is a scalar and not a vector quantity.

Question:10.13

Iceberg floats in water with part of it submerged. What is the fraction of the volume of iceberg submerged if the density of ice is $\rho_i = 0.917 \frac{g}{cm^{3}}$?

Answer:

The weight of the iceberg is equal to the weight of the displaced liquid, as the iceberg is floating on the surface of the sea.
$V \rho_{i} g = V' \rho_{w} g$
The volume of an iceberg submerged in water/ vol of iceberg = $\frac{V^{'}}{V}=\frac{\rho_i}{\rho_w}=\frac{0.917}{1}$
Hence, the portion of the iceberg which is submerged in water is 0.917

Question:10.14

A vessel filled with water is kept on a weighing pan and the scale adjusted to zero. A block of mass M and density $\rho$ is suspended by a massless spring of spring constant k. This block is submerged inside into the water in the vessel. What is the reading of the scale?

Answer:

The given figure has a beaker filled with water. The beaker has been placed on a weighing scale, which is adjusted to zero.

As we submerge the block into the water, a force equal to the weight of the block acts upward in the form of a buoyant force. So, the downward force applied by the block equals V $\rho_w$ g
V = the water volume which is displaced by the block
$\rho_w$ = density of water
Mass of block = M = V
$V=\frac{M}{\rho}$
So, the reading on the weighing scale reads:
$\frac{M}{\rho}. \rho_w g = \frac{\rho_w}{\rho}Mg$

Question:10.15

A cubical block of density $\rho$ is floating on the surface of water. Out of its height L, fraction x is submerged in water. The vessel is in an elevator accelerating upwards with acceleration a. What is the fraction immersed?

Answer:

Through the principle of flotation, when the block gets submerged in water
$V\rho g$ = $V^{'} \rho_{w} g$
Where V’ = volume of water displaced by block
V’ = base area of block x height of block inside the water
$V^{'} = L^{2}x$
Let V be the volume of the block and $\rho_b$ be the density of the block
Hence,
$V= L^{2}\times \rho_w$
$\frac{\rho_b}{\rho_w }=\frac{x}{L}$ ---------(1)
$x =\frac{\rho_b}{\rho_w} L$
As the block lifts upward, the acceleration = (g + a)
Weight of L³ $\rho b$ block = m(g+a) = V $\times \rho_b$(g+a) =$ L^{3} \rho_b (g+a)$
Let x₁ be the part of the block which is submerged.
We know that the buoyant force will be equal and opposite to the weight of the block.
$L^{3} \rho_b (g+a) $= x₁ $L^{2} \rho_w (g+a)$
So,
$\frac{\rho_b}{\rho_w}=\frac{x_1}{L}$
So, $x_1=\frac{\rho_b}{\rho_w}L$ ------------(2)
From (1) and (2), we can infer that$\frac{\rho_b}{\rho_w}L$ is the part of the block which is submerged in water. This is independent of the acceleration of the block (upwards, downwards or at rest)

NCERT Exemplar Class 11 Physics Solutions Chapter 10: Short Answer

The behaviour of fluids can be used to explain a large number of natural and everyday phenomena, including floating ships and flying aeroplanes. The Mechanical Properties of Fluids Class 11 NCERT Exemplar gives properly structured short questions which enhance clarity and examination preparation.

Question:10.16

The sap in trees, which consists mainly of water in summer, rises in a system of capillaries of radius $r = 2.5 \times 10^{-5} m$. The surface tension of sap is $T = 7.28 \times 10^{-2} N/m$ and the angle of contact is $0^{\circ}$. Does surface tension alone account for the supply of water to the top of all trees?

Answer:

r = radius of the capillary
$r = 2.5 \times 10^{-5} m$
$S=T = 7.28 \times 10^{-2} N/m$
$g = 9.8 \frac{m}{s^{2}}$
$\theta =0$, $\rho$ = 1000 kg/m³
now, $h =\frac{2 s \cos \theta}{r\rho g} $=$ 2 \times 7.28 \times 10^{-2} \frac{\cos 0}{2.5 \times 10^{-5}\times 1000 \times 9.8}$

$h = \frac{104}{175}\times \frac{1000}{1000} = \frac{104}{175} = 0.594 \approx 0.6m$

Question:10.17

The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts accelerating the free surface will be tilted by an angle $\theta$. If the acceleration is a m/s², what will be the slope of the free surface?

Answer:

Let us assume the tanker is pulled by a force F, and it produces an acceleration in the forward direction. Let $\delta$m be the mass of a small element. As the tanker is pulled forward and accelerates, the small element also experiences a force in the same direction. The inertia of rest makes the element remain at rest, and hence, as the tank moves forward, the element moves backwards as a result of inertia.
Force acting on $\delta$m, F₁ = $\delta$ma in the backward direction
F₂ = $\delta$mg, which is in a vertically downward direction.
The normal reaction gets balanced by various components: $\delta$$ma \sin \theta$ from F₁
The situation when there is a maximum angle inclination by the surface,
$\delta ma \cos \theta = \delta mg \sin \theta$
So required slope, $\frac{\sin \theta }{\cos \theta}=\frac{a}{g}$
So, $\tan \theta=\frac{a}{g}$

Question:10.18

Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury $T = 435.5 \times 10^{-3} \frac{N}{m}$.

Answer:

Surface tension energy, $E=\sigma \Delta A$
As we can infer from the law of conservation of mass, the volume of a drop V = v₁ + v₂
$r_1 = 0.1 cm = 0.001 m$
$r_2 = 0.2 cm = 0.002 m$
$\Delta A = 4 \pi r_1^{2} + 4 \pi r_2^{2} - 4\pi R^{2}$
$\Delta A = 4 \pi \left [r_1^{2} + r_2^{2}- R^{2} \right ]$
The new, bigger drop has a radius of R,
$\frac{4}{3}\pi R^{3}=\frac{4}{3}\pi \left [ r_1^{3}+r_2^{3} \right ]$
$R^{3}= r_1^{3}+r_2^{3}$
$R^{3}= 9 \times 10^{-9}$
$R^{3}= 2.1 \times 10^{-3}m$
Now, $E=\sigma \Delta A$
$E = 4 \times 3.14 [(10^{-3})^{2} + (2 \times 10^{-3})^{2} - (2.1 \times 10^{-3})^{2}] \times 435.5 \times 10^{-3}$
$E = 1742 \times 3.14 \times 10^{-9} [0.59] = 5469.88 \times 0.59 \times 10^{-9}$
$E = 32.27 \times 10^{-7} J$
Finally, the area will be smaller when energy is released as the bigger drop is formed from smaller drops.

Question:10.19

If a drop of liquid breaks into smaller droplets, it results in a lowering of the temperature of the droplets. Let a drop of radius R, break into N small droplets, each of radius r. Estimate the drop in temperature.

Answer:

now, $\Delta E=\sigma$
(final area - initial area) of surface
$\Delta E=m s \Delta t$
As per the law of conservation of mass,
Final volume =initial volume of one drop, which has a radius ,R which is then split into N drops r
So, $\frac{4}{3}\pi R^{3}=N.\frac{4}{3}\pi r^3$
$N.\frac{4}{3}\pi r^{3}ps\Delta t=4\pi\sigma \left [ Nr^{2}-R^{2} \right ]$
$\Delta t=\frac{4\pi \sigma \times 3}{N.4\pi r^{3}ps}\left [ Nr^{2}-R^{2} \right ]$
$\Delta t=\frac{3\sigma}{Nps}\left [ \frac{Nr^{2}}{r^{3}}-\frac{R^{2}}{r^{3}} \right ]$
$\Delta t=\frac{3\sigma N}{Nps}\left [ \frac{1}{r}-\frac{1}{R} \right ]$
$\Delta t=\frac{3\sigma}{ps}\left [ \frac{1}{r}-\frac{1}{R} \right ]$ as R>r
Hence, we can say that $\Delta t$ will be positive and therefore, $\frac{1}{R} < \frac{1}{r}$
So, the temperature increases as the formation of smaller drops takes place, and this change takes place through absorption of energy from the surroundings.

Question:10.20

The surface tension and vapour pressure of water at $20^{\circ}C$ is $7.28 \times 10^{-2} \frac{N}{m}$ and $2.33 \times 10^{3}$ Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at $20^{\circ}C$?

Answer:

In case the water pressure in the liquid is greater than the surface pressure above the liquid, the drop will evaporate. We assume R to be the radius of the droplet before evaporation.
We know that the pressure of vapour = the pressure which becomes excess inside the drop
$P=\frac{2\sigma}{R}$
$R = 2 \times 7.28 \times 10^{-2}$$/P$ $=\frac{2 \times 7.28 \times 10^{-2}}{2.33 \times 1000 }= \frac{14560}{233}\times 100000$
Hence, $R= 6.25 \times 10^{-5} m$

NCERT Exemplar Class 11 Physics Solutions Chapter 10: Long Answer

Mechanical Properties of Fluids Class 11 NCERT Exemplar Long answer question helps to assess the conceptual clarity and numerical problem-solving ability in the long answer questions. These questions are concerned with the practical uses of fluid mechanics, which include the calculation of pressure, the Bernoulli principle, viscosity and the flow of fluid. Explanations in detail make students understand the reasoning behind the solutions and prepare them properly for exams.

Question:10.21

a) Pressure decreases as one ascends the atmosphere. If the density of air is $\rho$, what is the change in pressure dp over a differential height dh?
b) Considering the pressure p to be proportional to the density, find the pressure p at a height h if the pressure on the surface of the earth is $\rho _{0}$.
c) If $p_{0}=1.03 \times 10^{5}N/m^{2}$, $\rho _{0}=1.29\frac{kg}{m^{3}}$ and g is $9.8 \frac{m}{s^{2}}$ at what height will the pressure drop to (1/10) the value at the surface of the earth?
d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.

Answer:

a) Let us assume that there is a packet which has a thickness of dh. In a liquid, at any point, the pressure is working in all directions in an equal manner. The force as a result of the pressure is opposed by the buoyant force of the liquid.
(P + dP) A - P.A = - V$\rho g$
$dP A = - Adh\rho g$
$dP = - \rho gdh$ -----------(1) {since pressure decreases as height increases, the negative sign is present here}

b) Let us assume P_o to be the density of air on Earth.
At any point p, the pressure is directly proportional to the density
Hence,
$P = \frac{P}{P_{0}}p_0$
now, $dP = \rho gdh$
$dP = \frac{P}{P_{0}}p_{0}gdh$
$\frac{dP}{P} = -p_{0}\frac{g}{P_{0}}dh$
On integrating both sides, we get,
$\log \frac{P}{P_{0}} = \frac{-p_{0}}{P_0}gh$ -----------(3)
$\frac{P}{P_{0}} =e ^{\frac{-p_{0}}{P_0}gh}$
c)
we know that, $\log \frac{P}{P_{0}} = \frac{-p_{0}}{P_0}gh$
but, $P=\frac{P_{0}}{10}$ (given)
$\log \frac{\frac{P_{0}}{10}}{P_{0}}=\frac{-p_0gh}{P_{0}}$
$\log \frac{1}{10}=\frac{-p_0gh}{P_{0}}$
$h=\frac{P_{0}}{p_0g}\log 10$
$h = \frac{1.013 \times 10^5 \times 2.303}{1.29 \times 9.8} = 18.4 km$
d)
The temperature does not remain constant at a greater height. It only remains constant near the surface of the Earth.

Question:10.22

Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at the boiling point. Given that the latent heat of vaporization for water L _v = 540 k cal $kg^{-1}$, the mechanical equivalent of heat $J = 4.2 J cal^{-1}$, density of water $\rho_w = 103 kg l^{-1}$, Avagadro’s number $N_{A} = 6.0 \times 10^{26} k mole ^{-1}$ and the molecular weight of water $M_{A} = 18 kg$ for 1 k mole.
(a) Estimate the energy required for one molecule of water to evaporate.
(b) show that the intermolecular distance for water is
$d=\left [ \frac{M_{A}}{N_{A}}\times \frac{1}{\rho _{w}} \right ]^{\frac{1}{3}}$
(c) 1 g of water in the vapor state at 1 atm occupies 1601 cm³. Estimate the intermolecular distance at boiling
point, in the vapour state.
(d) During vaporisation a molecule overcomes a force F, assumed constant, to go from an inter-molecular distance d to d ′ . Estimate the value of F where d=3.1x10^-10m
(e) Calculate $\frac{F}{d}$, which is a measure of the surface tension.

Answer:

a) $L_v = 540 k cal kg^{-1}$ $= 540 \times 4.2 \times 1000 J/Kg$
For the evaporation of 1 kg of water, the energy required = L_v K Cal
For evaporation of M_A kg of water, L_A M_A K Cal
Number of molecules in M_A kg of water = N_A
Hence, for the evaporation of 1 molecule, the energy required (U)= (L_v M_A/N_A) K. Cal
$U = \frac{M_A L_v \times 1000 \times 4.2}{N_A}$ = $\frac{18 \times 540 \times 1000 \times 4.2}{6 \times 10 ^{26}}$ =$ 6.8 \times 10 ^{-20} J$

b)
$\text{water molecules are assumed to be of a point size and have distance d between each other}$
total volume of Na molecules of water = mass/density
$=\frac{M_{A}}{\rho}$
volume occupied by 1 molecule =
$d^{3} = \frac{M_{A}}{N_{A}}\times p$
$d = \left [ \frac{M_{A}}{N_{A}} \times p \right ]^{\frac{1}{3}}$
Hence, the volume occupied by 1kg vapour $= 1601 cm^3 = 1601 \times 10^{-6} m^3$
c)
the volume occupied by 18kg water vapour $= 18\times 1601 \times 10^{-6} cm^3$
number of molecules in 18 kg water =$6\times 10^{26}$
so, the volume occupied by 1 molecule = $\frac{18 \times 1601 \times 10^{-6} cm^{3}}{6 \times 10 ^{26}}$
hence $d^{'3}=\frac{18 \times 1601 \times 10^{-6} cm^{3}}{6 \times 10 ^{26}}$
$d^{'} = 36.3 \times 10 ^{-10} m$
d) work done to change the distance to d’ from d = F (d’-d)
$6.8 \times 10^{-20}= F (36.3 \times 10^{-10} - 3.1 \times 10^{-10}) = F \times 33.2 \times 10^{-10}$
Hence, $F = \frac{6.8 \times 10^{-20}}{33.2 \times 10^{-10}}$
$F = 2.05 \times 10 ^{-11} N$

e)
From the formula of surface tension, we know that surface tension = F/d
Surface tension = $\frac{2.05 \times 10 ^{-11} N}{3.1 \times 10^{-10}}$
Surface tension =$6.6 \times 10^{-2} N/m$

Question:10.23

A hot air balloon is a sphere of radius 8 m. The surface inside is at a temperature of $60^{\circ}C$. How large a mass can the balloon lift when the outside temperature is $20^{\circ}C$?

Answer:

The amount of pressure outside a balloon is less than the amount of pressure inside the balloon.

$P_{i}-P_{0}=\frac{2\sigma}{R}$ {$\sigma$: surface tension, R: balloon’s radius}

Air Balloon

$P_i V = ni R T_i$

Where V is the volume of the balloon and n_i is the number of moles of gas in the balloon

R is the gas constant, T_i is the temperature

$n_i = \frac{P_i V}{R T_i }$= Mass of balloon (Mi)/Molecular mass (Ma)

$Mi = \frac{P_i V Ma}{R T_i }$-------------(1)

In the same way, $n_{0} = \frac{P_{0} V}{R T_{0}}$

According to principal of floatation, $W + M_i g = M_{0} g$, here W is the weight which the balloon lifts

$W= (M_{0} - M_{i}) g$

If the mass displaced by the balloon is M_0, and M_a is the mass inside or outside of the balloon

$n_{0} = \frac{M_{0}}{M_{a}} = \frac{P_{0} V}{R T_{0}}$

$M_{0} =\frac{P_{0} V M_{a}}{R T_{0}}$---------------(2)

Hence, from (1) and (2) we can write,

$W = \left [ \frac{P_{0} V Ma}{R T_{0} } -\frac{P_{i} V M_{a}}{R T_{i}}\right ] g$

$W = \frac{V M_{a}}{R}\left [ \frac{P_{0}}{T_{0}}-\frac{P_{i}}{T_{i}} \right ] g$

Ma = 79% N₂ + 21%O₂

$Ma = 0.79 \times 28 + 0.21 \times 32 = 28.84 g = 0.2884 kg$

$P_{i}-P_{0}=\frac{2\sigma}{R}$

$P_{i} = P_{o}$ + Pressure due to ST of membrane

$Pi = \left [1.1013 \times 10^{5} + 2\times \frac{5}{8} \right ] = \sim 1.1013 \times 10^{5}$

$W = \frac{4}{3} \times \frac{3.14 \times 8 \times 8 \times 8\times 0.02884}{3\times 8.314} \left [\frac{1.013 \times 10^{5}}{293} - \frac{1.013 \times 10^{5}}{333} \right ]g$

$W = \frac{4}{3} \times \frac{3.14 \times 8 \times 8 \times 8\times 0.02884\times 1.013 \times 10^{5} }{3\times 8.314} \left [\frac{1}{293} - \frac{1}{333} \right ]g$

$W = 3044.2 N$

NCERT Exemplar Class 11 Physics Solutions Chapter 10: Important Concepts and Formulas

The study of the mechanical behaviour of fluids is fundamental in many processes, including engineering to biology. Class 11 Chapter 10 describes such concepts as the pressure within fluids, the buoyant force, the Bernoulli theorem, and the movement of liquids. The following are the main formulas and concepts in their simplified version:

1. Pressure (P):

Pressure is defined as force per unit area. Fluids exert pressure in all directions.

$
P=\frac{F}{A}
$
Pressure depends on depth in a fluid column due to the weight of the liquid above.

2. Pascal’s Law:

A change in pressure applied to an enclosed fluid is transmitted equally throughout the fluid. This is used in hydraulic machines like car brakes and lifts.

3. Pressure at Depth:

The pressure increases with depth in a liquid.

$
P=h \rho g
$

where
$h=$ depth,
$\rho=$ density of fluid,
$g$ = acceleration due to gravity.

4. Buoyant Force (Archimedes' Principle):

A body immersed in fluid experiences an upward force equal to the weight of fluid displaced.

$
F_b=\rho g V
$

5. Streamline Flow:

A flow in which every particle of the fluid follows a smooth path, not crossing others. The velocity, pressure and density remain constant at a point.

6. Bernoulli’s Principle:

The total mechanical energy in streamline flow remains constant.

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

7. Viscosity:

It is the property of fluids that opposes motion between adjacent layers.
The SI unit is poise (Pa·s).

8. Stokes’ Law:

Used for calculating viscous drag on a spherical body moving in a fluid.

$
F=6 \pi \eta r v
$

9. Coefficient of Viscosity (η):

$\eta=\frac{F}{6 \pi r v}$

10. Terminal Velocity (vₜ):

The maximum velocity of a body falling through a viscous fluid:

$
v_t=\frac{2 r^2(\rho-\sigma) g}{9 \eta}
$

where
$\rho=$ density of sphere,
$\sigma=$ density of fluid.

11. Surface Tension (T):

Force acting per unit length on the surface of a liquid.

$
T=\frac{F}{L}
$

12. Capillarity (Capillary Rise/Drop):

Liquids rise or fall in thin tubes due to surface tension.

$
h=\frac{2 T \cos \theta}{r \rho g}
$

Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 10 Mechanical Properties of Fluids

The NCERT Exemplar Class 11 Physics Chapter 10 offers a good conceptual base on the behaviour and properties of fluids like viscosity, buoyancy and pressure. Such solutions aid students in solving key numerical problems and building up their skills of analysis. They are also useful with students going through competitive exams such as JEE and NEET.

By explaining basic concepts, e.g. pressure, density, fluid flow, Pascal's law, and buoyancy in an elaborate and simplified manner, they assist students to gain a better insight into core concepts.
The problem-solving method proposed in a step-by-step manner assists the students in knowing how to approach the numerical questions with confidence during an examination.
Students also get to see most of the key question formats, such as MCQ, short answers, and the short essay format as well, a fact that enhances their general preparedness for the exam.
This chapter enhances analytical and reasoning abilities because students will have an opportunity to apply the laws of fluid mechanics to real-life processes, such as ship floating, blood flow, and hydraulic systems.
These model solutions are developed by experts in the subject, making them accurate and clear, enabling the students to score better grades in school and competitive exams.

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

NCERT Exemplar Class 11 Physics solutions are very handy in the exam preparation because they contain conceptual questions, numericals and application-based problems of all the chapters. These solutions can help students who are preparing not only for CBSE exams but also such competitive exams as JEE or NEET. Chapter links enable students to have a good and orderly material in a single place, thus making the learning process more systematic.

Chapter 2 Units and Measurement

Chapter 3 Motion in a Straight Line

Chapter 4 Motion in a Plane

Chapter 5 Laws of Motion

Chapter 6 Work, Energy, and Power

Chapter 7 System of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solids

Chapter 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 12 Thermodynamics

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Chapter 15 Waves

NCERT Exemplar Class 11th Solutions

Check Class 11 Physics Chapter-wise Solutions

NCERT solutions for Class 11 Physics Chapter 1 Units and Measurement

NCERT solutions for Class 11 Physics Chapter 2 Motion in a straight line

NCERT solutions for Class 11 Physics Chapter 3 Motion in a Plane

NCERT solutions for Class 11 Physics Chapter 4 Laws of Motion

NCERT solutions for Class 11 Physics Chapter 5 Work, Energy and Power

NCERT solutions for Class 11 Physics Chapter 6 System of Particles and Rotational Motion

NCERT solutions for Class 11 Physics Chapter 7 Gravitation

NCERT solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids

NCERT solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

NCERT solutions for Class 11 Physics Chapter 10 Thermal Properties of Matter

NCERT solutions for Class 11 Physics Chapter 11 Thermodynamics

NCERT solutions for Class 11 Physics Chapter 12 Kinetic Theory

NCERT solutions for Class 11 Physics Chapter 13 Oscillations

NCERT solutions for Class 11 Physics Chapter 14 Waves

Also, Read NCERT Solution subject-wise -

Check NCERT Notes subject-wise -

Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

Q: How are these solutions helpful for students?

Students can get a clear and better idea about the topics, and the theorems explained in the chapter and how they have to answer the questions in the exam.

Q: What questions are solved from this chapter?

From this chapter, all the questions mentioned in the MCQ and short and long answer type questions are solved in complete detail in NCERT exemplar class 11 physics chapter 10 solutions.

Q: Are these solutions useful for entrance exams?

Yes, these NCERT exemplar class 11 physics solutions chapter 10 are highly useful for those who are preparing for the entrance exams. One can make use of these solutions to understand questions in more depth.

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