NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory

Vishal kumarUpdated on 19 Sep 2025, 11:38 PM IST

Have you ever questioned yourself as to why gases expand or put pressure on the sides of a container? The scientific explanation is given by the Kinetic Theory of Gases, which assumes that gases are made of small particles in everlasting, rapid movement. This theory is the work of great scientists such as Boyle, Newton, Maxwell, and Boltzmann, and it is used to explain how gases act in response to changes in pressure and temperature and how they diffuse or mix.

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  1. NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory: Download PDF
  2. NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory- Exercise Questions
  3. NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory - Additional Questions
  4. NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory: Higher Order Thinking Skills (HOTS) Questions
  5. Class 11 Physics Chapter 12 - Kinetic Theory: Topics
  6. Class 11 Physics Chapter 12 - Kinetic Theory: Important Formulae
  7. Approach to Solve Questions of the Class 11 Physics Chapter 12 - Kinetic Theory
  8. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  9. NCERT Solutions for Class 11 Physics Chapter-Wise

The NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory are very valuable not only in the preparation for the CBSE board but also for competitive exams such as JEE and NEET. The NCERT solutions are the work of physics experts who make complex concepts simple and offer a step-by-step approach to resolving numerical and theoretical problems. These NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory are prepared as per the recent CBSE syllabus, making them conceptually clear and helping the students to develop their ability to solve problems. Through such NCERT solutions, students will be able to have confidence in the chapter as well as score highly in exams.

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NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory: Download PDF

The Class 11 Physics Chapter 12 - Kinetic Theory question answers include step-by-step solutions of the textbook questions, which help the students to have a clear picture of the behaviour of gases at the molecular scale. One can access these solutions online and download them as a PDF solution without any payment, making practice and revision convenient.

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NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory- Exercise Questions

Kinetic Theory class 11 question answers provide all the detailed solutions to the exercises, which allow students to understand the behavior of gases easily at the microscopic level. The solutions make things easy to understand, such as explain concepts such as pressure, mean free path, gas laws, and the kinetic interpretation of temperature, developing competitive fundamentals to prepare students for exams and other competitive tests like the JEE or NEET.

Q12.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.

Answer:

The diameter of an oxygen molecule, d = 3 Å.

The actual volume of a mole of oxygen molecules V actual is

Vactual=NA43π(d2)3
Vactual=6.023×1023×43π×(3×10102)3
Vactual=8.51×106m3
Vactual=8.51×103litres

The volume occupied by a mole of oxygen gas at STP is V molar = 22.4 litres

VactualVmolar=8.51×10322.4

VactualVmolar=3.8×104

Q12.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 00C ). Show that it is 22.4Litres .

Answer:

As per the ideal gas equation

PV=nRTV=nRTP

For one mole of a gas at STP we have

V=1×8.314×2731.013×105
V=0.0224m3

V=22.4 litres

Q12.3 Figure 13.8 shows plot of PV/T versus P for 1.00×103 kg of oxygen gas at two different temperatures.

PV/T versus P graph

(a) What does the dotted plot signify?
(b) Which is true: T1>T2orT1<T2 ?
(c) What is the value of PV/T where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00×103 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high-temperature region of the plot)? (Molecular mass of H2=2.02μ , of O2=32.0μ, R=8.31Jmol1K1.)

Answer:

(a) The dotted plot corresponds to the ideal gas behaviour.

(b) We know the behaviour of a real gas tends close to that of ideal gas as its temperature increases and since the plot corresponding to temperature T 1 is closer to the horizontal line that the one corresponding to T 2 we conclude T 1 is greater than T 2.

(c) As per the ideal gas equation

PVT=nR

The molar mass of oxygen = 32 g

n=132

R = 8.314

nR=132×8.314
nR=0.256JK1

(d) If we obtained similar plots for 1.00×103 kg of hydrogen, we would not get the same value of PV/T at the point where the curves meet on the y-axis as 1 g of Hydrogen would contain more moles than 1 g of Oxygen because of havinga smaller molar mass.

Molar Mass of Hydrogen M = 2 g

mass of hydrogen

m=PVTMR=0.256×28.314=5.48×105Kg

Q12.4 An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 270C . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 170C . Estimate the mass of oxygen taken out of the cylinder ( R=8.31Jmol1K1 , molecular mass of O2=32μ ).

Answer:

Initial volume, V 1 = Volume of Cylinder = 30 l

Initial Pressure P 1 = 15 atm

Initial Temperature T 1 = 27 o C = 300 K

The initial number of moles n 1 inside the cylinder is

n1=P1V1RT1
n1=15×1.013×105×30×1038.314×300
n1=18.28

Final volume, V 2 = Volume of Cylinder = 30 l

Final Pressure P 2 = 11 atm

Final Temperature T 2 = 17 o C = 290 K

The final number of moles n 2 inside the cylinder is

n2=P2V2RT2
n2=11×1.013×105×30×1038.314×290
n2=13.86

Moles of oxygen taken out of the cylinder = n 2 -n 1 = 18.28 - 13.86 = 4.42

Mass of oxygen taken out of the cylinder, m is

m=4.42×32
m=141.44g

Q12.5 An air bubble of volume 1.0cm3 rises from the bottom of a lake 40m deep at a temperature of 120C. To what volume does it grow when it reaches the surface, which is at a temperature of 350C?

Answer:

Initial Volume of the bubble, V 1 = 1.0 cm 3

Initial temperature, T 1 = 12 o C = 273 + 12 = 285 K

The density of water is ρw=103 kg m3

Initial Pressure is P 1

Depth of the bottom of the lake = 40 m

P1= Atmospheric Pressure + Pressure due to water P1=Patm+ρwghP1=1.013×105+103×9.8×40P1=4.93×105 Pa

Final Temperature, T 2 = 35 o C = 35 + 273 = 308 K

Final Pressure = Atmospheric Pressure =1.013×105Pa

Let the final volume be V 2

As the number of moles inside the bubble remains constant, we have

P1V1T1=P2V2T2

V2=P1T2V1P2T1

V2=4.93×105×308×11.013×105×285

V2=5.26 cm3

Q12.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0m3 at a temperature of 270C and 1atm pressure.

Answer:

The volume of the room, V = 25.0 m 3

Temperature of the room, T = 27 o C = 300 K

The pressure inside the room, P = 1 atm

Let the number of moles of air molecules inside the room be n

n=PVRTn=1.013×105×258.314×300n=1015.35

Avogadro's Number, NA=6.022×1023
The number of molecules inside the room is N

N=nNAN=1015.35×6.022×1023N=6.114×1026

Q12.7 Estimate the average thermal energy of a helium atom at (i) room temperature ( 270C ), (ii) the temperature on the surface of the Sun ( 6000K ), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).

Answer:

The average energy of a Helium atom is given as 3kT2 since it is monoatomic

(i)

E=3kT2

E=3×1.38×1023×3002

E=6.21×1021 J

(ii)

E=3kT2

E=3×1.38×1023×60002

E=1.242×1019 J

(iii)

E=3kT2

E=3×1.38×1023×1072

E=2.07×1016 J

Q12.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain an equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v rms the largest?

Answer:

As per Avogadro's Hypothesis, under similar conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules. Since the volume of the vessels is the same and all vessels are kept at the same conditions of pressure and temperature, they would contain an equal number of molecules.

Root mean square velocity is given as

vrms=3kTm

As we can see, vrms is inversely proportional to the square root of the molar mass; the root mean square velocity will be maximum in the case of Neon, as its molar mass is the least.

Q12.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at 200C? (atomic mass of Ar=39.9μ , of He=4.0μ ).

Answer:

As we know root mean square velocity is given as vrms=3RTM

Let at temperature T, the root mean square speed of an atom in an argon cylinder equal to the rms speed of a helium gas atom at 200C

3R×T39.9=3R×2534

T=2523.7 K

Q12.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0atm and temperature 170C. Take the radius of a nitrogen molecule to be roughly 1.0A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2=28.0μ ).

Answer:

Pressure, P = 2atm

Temperature, T = 17 o C

The radius of the Nitrogen molecule, r=1 Å.

The molecular mass of N 2 = 28 u

The molar mass of N 2 = 28 g

From the ideal gas equation

PV=nRT

nV=PRT

The above tells us about the number of moles per unit volume; the number of molecules per unit volume would be given as

n=NAnV=6.022×1023×2×1.013×1058.314×(17+273)

n=5.06×1025

The mean free path λ is given as

λ=12πnd2

λ=12×π×5.06×1025×(2×1×1010)2

λ=1.11×107 m

The root mean square velocity v rms is given as

vrms=3RTM

vrms=3×8.314×29028×103

vrms=508.26 m s1

The time between collisions T, is given as

T=1Collision Frequency

T=1ν

T=λvrms

T=1.11×107508.26

T=2.18×1010s

Collision time T' is equal to the average time taken by a molecule to travel a distance equal to its diameter

T=dvrms

T=2×1×1010508.26

T=3.935×1013s

The ratio of the average time between collisions to the collision time is

TT=2.18×10103.935×1013TT=554
Thus, we can see that the time between collisions is much larger than the collision time.

NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory - Additional Questions

The Class 11 Physics Chapter 12 - Kinetic Theory Additional Questions have been provided to boost the problem solving skills by discussing extra problems besides the NCERT textbook. The questions enhance an appreciation of such concepts as molecular motion, pressure of gases, mean free path, equipartition of energy, and are relevant in both board examinations and competitive entrance examinations such as JEE and NEET.

Q 1) A metre-long narrow bore held horizontally (and closed at one end) contains a 76cm long mercury thread, which traps a 15cm column of air. What happens if the tube is held vertically with the open end at the bottom?

Answer:

Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P 1 = 1 atm = 76 cm of Mercury

Let the crossectional area of the tube be x cm 2

The initial volume of the air column, V 1 = 15x cm 3

Let's assume that once the tube is held vertical, y cm of Mercury flows out of it.

The pressure of the air column after y cm of Mercury has flown out of the column P 2 = 76 - (76 - y) cm of Mercury = y cm of mercury

Final volume of air column V 2 = (24 + y)x cm 3

Since the temperature of the air column does not change

P1V1=P2V2

76×15x=y×(24+y)x

1140=y2+24y

y2+24y1140=0

Solving the above quadratic equation, we get y = 23.8 cm or y = -47.8 cm

Since a negative amount of mercury cannot flow out of the column, y cannot be negative. Therefore, y = 23.8 cm.

Length of the air column = y + 24 = 47.8 cm.

Therefore, once the tube is held vertically, 23.8 cm of Mercury flows out of it, and the length of the air column becomes 47.8 cm

Q 2) From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7cm3s1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2cm3s1. Identify the gas.

Answer:

As per Graham's Law of diffusion, if two gases of Molar Mass M 1 and M 2 diffuse with rates R 1 and R 2 respectively, their diffusion rates are related by the following equation

R1R2=M2M1

In the given question

R 1 = 28.7 cm 3 s -1

R 2 = 7.2 cm 3 s -1

M 1 = 2 g

M2=M1(R1R2)2

M2=2×(28.77.2)2

M2=31.78g

The above Molar Mass is close to 32, therefore, the gas is Oxygen.

Q 3) A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n2=n1exp[mg(h2h1)/kbT] where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for the sedimentation equilibrium of a suspension in a liquid column: n2=n1exp[mgNA(ρρ)(h2h1)/(ρRT)] where ρ is the density of the suspended particle, and ρ , that of the surrounding medium. [ NA s Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes' principle to find the apparent weight of the suspended particle.]

Answer:

n2=n1exp[mg(h2h1)/kbT] (i)

Let the suspended particles be spherical and have a radius r

The gravitational force acting on the suspended particles would be

FG=43πr3ρg

The buoyant force acting on them would be

FB=43πr3ρg

The net force acting on the particles become

Fnet=FGFB

Fnet=43πr3ρg43πr3ρg

Fnet=43πr3g(ρρ)

Replacing mg in equation (i) with the above equation, we get

n2=n1exp[43πr3g(ρρ)(h2h1)/kbT]

n2=n1exp[43πr3g(ρρ)(h2h1)RTNA]

n2=n1exp[mgNA(ρρ)(h2h1)RTρ]

The above is the equation to be derived

Q 4) Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Substance
Atomic Mass (u)
Density (10 3 Kg m-3 )
Carbon (diamond)
12.01
2.22
Gold
197
19.32
Nitrogen (liquid)
14.01
1
Lithium
6.94
0.53
Fluorine
19
1.14

Answer:

Let one mole of a substance of atomic radius r and density ρ have molar mass M

Let us assume the atoms to be spherical

Avogadro's number is NA=6.022×1023

NA43πr3ρ=Mr=(3M4NAπρ)13

For Carbon

NA43πr3ρ=Mr=(3×12.014×6.022×1023×π×2.22×103)13r=1.29 Å.

For gold

NA43πr3ρ=Mr=(3×197.004×6.022×1023×π×19.32×103)13r=1.59 Å.

For Nitrogen

NA43πr3ρ=Mr=(3×14.014×6.022×1023×π×1.00×103)13r=1.77 Å.

For Lithium

NA43πr3ρ=Mr=(3×6.944×6.022×1023×π×0.53×103)13r=1.73 Å.

For Fluorine

NA43πr3ρ=Mr=(3×19.004×6.022×1023×π×1.14×103)13r=1.88 Å.


NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory: Higher Order Thinking Skills (HOTS) Questions

The HOTS questions of Class 11 Physics Chapter 12 - Kinetic Theory aim at engaging students in the real life application of the concept and solve the problems that challenge them beyond the core of the topic. These high order questions develop thinking, analytical and good conceptual clarity which are the requirements of board examination as well as competitive examinations, such as JEE and NEET.

Q1: Two moles of an ideal gas with CpCv=53 are mixed 3 moles of another ideal gas with CpCv=43. The value of CpCv for the mixture is:-

Answer:

For ideal gas:- CpCv=R
For the first case:-

Cp1Cv1=53 and Cp1Cv1=R


Cp1=53Cv1 and 53Cv1Cv1=R

23Cv1=R

Cv1=32R

So, Cp1=52R
For the second case:-

Cp2Cv2=43 and Cp2Cv2=R

Cp2=43Cv2 and 43Cv2Cv2=R

Cv2=3R and Cp2=4R

Ymix =n1Cp1+n2Cp2n1Cv1+n2Cv2=2×52R+3×4R2×32R+3×3R=1.417=1.42


Q2: If temperature of the atmosphere varies with height as T=(T0ah), where a and T0 are positive constants, then the pressure as a function of height h is (assume atmospheric pressure at sea level ( h=0 ) is P0 and molecule mass M of the air and acceleration due to gravity g be constant)

Answer:

dPdh=ρg=(PMRT)g=[PMR( T0ah)]g


dPP=(MgR)dh T0ah


P0PdPP=(MgR)0hdh( T0ah)


P=P0( T0ah T0)MsRa


Q3: In two jars A and B, the pressure, volume and temperature in jar A are respectively P, V, and T, and those of B are 2P, V/4 and 2T. Then the ratio of the number of molecules in jars A and B will be,

Answer:

NANB=PAVAPBPB×TBTANANB=P×V×(2T)2P×V4×T=41


Q4: The temperature of an open room of volume 30 m3 increases from 17oC to 27oC due to the sunshine. The atmospheric pressure in the room remains 1×105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nfni will be :

Answer:

PV = nRT

ni=PVRTi,nf=PVRTfnfni=PVR(1Tf1Ti)=105×308.31×(13001290)nfni=105×308.31×10300×290=105290×8.31 Change in the Number of molecules =105×6.023×1023290×8.31=2.5×1025


Q5: A gas mixture contains 3 moles of oxygen and x moles of monoatomic gas at temperature T. Considering only translational and rotational but not vibrational modes, the total energy of the system is 15 RT then the value of x is.

Answer:

U=nf2RT

According to the question,

15RT=3×52RT+x×32RT15=152+3x2x=5


Class 11 Physics Chapter 12 - Kinetic Theory: Topics

The Class 11 Physics Chapter 12 - Kinetic Theory examines how the behaviour of gases can be considered at the molecular level, in which the behaviour at the microscopic level is related to the macroscopic behaviour in the form of pressure, volume and temperature. The chapter addresses all the basic assumptions of kinetic theory, derivations and applications that describe real-life phenomena like diffusion and pressure of gases. Knowledge of these topics does not only enhance clarity in concepts but also equips students to board examinations and other competitive exams such as JEE and NEET.

12.1 Introduction
12.2 Molecular Nature Of Matter
12.3 Behaviour Of Gases
12.4 Kinetic Theory Of An Ideal Gas
12.4.1 Pressure Of An Ideal Gas
12.4.2 Kinetic Interpretation Of Temperature
12.5 Law Of Equipartition Of Energy
12.6 Specific Heat Capacity
12.6.1 Monatomic Gases
12.6.2 Diatomic Gases
12.6.3 Polyatomic Gases
12.6.4 Specific Heat Capacity Of Solids
12.7 Mean Free Path

Class 11 Physics Chapter 12 - Kinetic Theory: Important Formulae

Kinetic Theory NCERT Solutions- Important Formulae contain important equations that describe the behavior of the molecular motion in relation to the macroscopic properties of gas such as pressure, temperature and volume. With the help of these formulas, students can solve numerical problems and see a correlation between the behaviour of microscopic particles and the observed physical phenomena.

1. Ideal Gas Equation

PV=nRT

Where:

P= pressure ,V= volume ,n= number of moles ,R= universal gas constant ,T= temperature (in Kelvin)

2. Average Kinetic Energy per Molecule

KEavg=32kT
Where k= Boltzmann constant, T= temperature in Kelvin

3. Total Kinetic Energy of Gas

KEtotal =32nRT

4. Pressure of an Ideal Gas from Kinetic Theory

P=13ρv¯2
Where ρ= mass per unit volume, v¯2= mean square speed

5. Root Mean Square Speed

vrms=3kTm=3RTM
Where:
m= mass of one molecule,
M= molar mass

7. Relation Between Different Speeds

vrms >vavg >vmp
Where:
- vrms=3RTM
- vavg=8RTπM
- vmp=2RTM

8. Degrees of Freedom and Energy

E=f2kT per molecule
Where f= degrees of freedom

Etotal =f2nRT for n moles

Approach to Solve Questions of the Class 11 Physics Chapter 12 - Kinetic Theory

The Kinetic Theory of Gases has several assumptions, equations and derivations that must be understood clearly to solve some of its questions. Students are expected to learn formulas not by memorization, but to work out the logical application of formulas and interpret physical meaning underlying a formula. Organized planning assists in dealing with numerical and conceptual issues with ease.

  • Get a clear understanding of the Assumptions: Start with the revision of the fundamental postulates of the kinetic theory, including the following: gases consist of small particles in a state of perpetual random motion, the volume is insignificant, and collisions are perfectly elastic. These assumptions assist you in relating concepts to formulas.

  • Determine the Type of Question: Determine whether the problem is based on the derivation of pressure, mean free path, root mean square (rms) velocity or laws, such as Boyle and Charles. Saving time is achieved through categorizing and guides to the correct formula.

  • Link Equations with Physical Meaning: Don’t just apply formulas like $P=\frac{1}{3} \rho c^2$ or $v_{r m s}=\sqrt{\frac{3 k T}{m}} ;$ consider what they are, relation of pressure with molecular movement or temperature with speed of particles.

  • Divide Numerical Problems into Steps: Begin with familiar values and change all the units into SI and then carefully replace them in the equation. Always verify dimensional consistency- it means that it is accurate.

  • Apply Graphical Interpretation to Theory based questions: Reasoning is involved in many questions such as pressure-temperature or volume-temperature graphs. Explanations are better understood and remembered through graphs.

  • Practice Derivations Comprehensively: Derivations such as the pressure exerted by the gas molecules, mean free path and degrees of freedom are common questions. The conceptual power is developed through writing them step by step.


What Extra Should Students Study Beyond NCERT for JEE/NEET?

NCERT textbooks are almost a must-have in studying and preparing competitive exams such as JEE and NEET, but are definitely lacking in preparing the numericals, tricky applications and deeper conceptual advancement. This is a list of what NCERT does not cover and what additional students are expected to learn in the chapter Kinetic Theory of Gases in preparation for JEE/NEET:

NCERT Solutions for Class 11 Physics Chapter-Wise

Chapter-wise links of NCERT Solutions Class 11 provides an easy way to access all chapters at one stop. These solutions are aimed at allowing students to easily get explanations, solved exercises, vital formulas, and additional questions to each chapter and therefore making exam preparation and concept reviewing easier and organized.

NCERT Solutions for Class 11 Subject Wise

Also, check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

Q: Will the NCERT solutions be available in PDF format?
A:

Yes, NCERT chapter-wise solutions can be found in PDF format and can be easily studied offline anytime, anywhere.

Q: Are all chapters of Class 11 Physics covered in these solutions?
A:

They contain all chapters of the Mechanics to the Kinetic Theory with solved exercises, HOTS questions and additional questions and formulas.

Q: What are chapter wise solutions of NCERT Class 11 Physics?
A:

These are step-by-step solutions accompanying every exercise, HOTS, and other questions in the NCERT textbooks that allow students to comprehend new information in a proper way.

Q: Why is NCERT chapter wise solutions useful in exam preparation?
A:

They mainly consist of step-wise descriptions, crucial formulas and illustrated solutions that make studying easier, which makes it easy to revise and obtain a good score both in the board as

Q: Can these solutions be of use in JEE and NEET preparation?
A:

They do reinforce the essential ideas in Physics, upon which the questions are set in JEE, NEET and other such competitive tests.

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