NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves

NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves

Vishal kumarUpdated on 23 Sep 2025, 12:49 AM IST

Have you ever wondered how your phone can pick up radio signals or how the sun can travel millions of kilometres to hit the Earth? The solution is in electromagnetic waves, which constitute the basis of Chapter 8 of Class 12 Physics. This chapter discusses the properties, behaviour and uses of the electromagnetic waves in real life, and the NCERT Solutions of Class 12 Physics Chapter 8 - Electromagnetic Waves provide these simple and step-by-step explanations, which help students to understand them well.

This Story also Contains

  1. NCERT Solutions for Class 12 Physics Chapter 8 - Electromagnetic Waves: Download PDF
  2. Class 12 Physics Chapter 8 - Electromagnetic Waves: Exercise Question
  3. Class 12 Physics Chapter 8 - Electromagnetic Waves: Additional Questions
  4. Class 12 Physics Chapter 8 - Electromagnetic Waves: Higher Order Thinking Skills (HOTS) Questions
  5. Class 12 Physics Chapter 8 - Electromagnetic Waves: Topics
  6. Approach to Solve Questions of Class 12 Physics Chapter 8 - Electromagnetic Waves
  7. What Extra Should Students Study Beyond NCERT For JEE/NEET?
  8. NCERT Solutions for Class 12 Physics Chapter-Wise
NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves
Electromagnetic Wave Solutions

These NCERT Solutions for Class 12 Physics Chapter 8 - Electromagnetic Waves are very useful in the preparation of CBSE Class 12 board examination and competitive examinations such as JEE Main and NEET because they contain not only solved exercises, but also extra practice questions and an understanding of the concept. These NCERT solutions, which are designed by the subject experts, not only conserve the precious study time but also simplify the revision of the complex concepts. It is also free of cost, and the NCERT Solutions for Class 12 Physics Chapter 8 - Electromagnetic Waves PDF can be downloaded by the student, allowing one to revise anytime and anywhere conveniently. These solutions are well constructed, and thus learners are able to reinforce their fundamentals and build confidence in exams.

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NCERT Solutions for Class 12 Physics Chapter 8 - Electromagnetic Waves: Download PDF

The Class 12 Physics Chapter 8 - Electromagnetic Waves question answers help students gain a clear understanding of the concepts and reinforce their knowledge by giving out detailed and stepwise answers to all the questions present in the chapter. The solutions are crafted according to the new NCERT syllabus and are also very beneficial in preparation for the board exams and competitive exams like JEE and NEET. You may as well download the free PDF to read continuously and revise successfully.

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Class 12 Physics Chapter 8 - Electromagnetic Waves: Exercise Question

Electromagnetic Waves class 12 question answers of textbook exercises offer detailed and easy-to-understand answers to all in-text and exercise problems. These solutions help students grasp key concepts, practice effectively, and prepare confidently for board exams and competitive tests like JEE and NEET.

Q1.(a) Figure 8.5 shows a capacitor made of two circular plates, each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A. Calculate the capacitance and the rate of change of potential difference between the plates.

Answer:

Radius of the discs(r) = 12cm

Area of the discs(A) = πr2=π(0.12)2=.045m2

Permittivity, ϵ0 = 8.85×1012C2N1m2

Distance between the two discs = 5cm=0.05m

Capacitance=ε0Ad

= 8.85×1012C2N1m2×.045m20.05m

= 8.003×1012F

=8.003 pF

Rate of change of potential=dVdt

But

V=QC

Therefore,

Rate of change of potential=d(QC)dt=dQCdt=iC

=0.15A8.003pF

= 1.87×1010Vs1

Q1.(c) Figure 8.5 shows a capacitor made of two circular plates, each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A. Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

Answer:

Yes, Kirchhoff's First rule (junction rule) is valid at each plate of the capacitor. This might not seem like the case at first, but once we take into consideration both the conduction and displacement current, Kirchhoff's first rule will hold good.

Q2.(a) A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300rads1 . What is the rms value of the conduction current?

Answer:

Capacitance(C) of the parallel plate capacitor = 100 pF

Voltage(V) = 230 V

Angular Frequency (ω) = 300 rad s1

Rms Current(I)=VoltageCapacitive Reactance

Capacitive Reactance(Xc)=1Cω=1100 pF×300 rad s1

Xc=3.33×107Ω

I=VXc=230V3.33×107Ω=6.9×106A

RMS value of conduction current is 6.9μA

Q2.(b) A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300rads1 Is the conduction current equal to the displacement current?

Answer:

Yes, conduction current is equal to the displacement current. This will be the case because otherwise, we will get different values of the magnetic field at the same point by taking two different surfaces and applying Ampere–Maxwell Law.

Q2.(c) A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300rads1 Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

Answer:

We know Ampere-Maxwell's Law,

$\oint \vec{B} \cdot \vec{d l}=\mu_0\left(i_c+\varepsilon_0 \frac{d \phi_E}{d t}\right)$

Between the plates conduction current ic=0.

For a loop of radius r smaller than the radius of the discs,

μ0ε0dϕEdt=μ0idπr2πR2=μ0idr2R2

B(2πr)=μ0idr2R2

B=μ0idr2πR2

Since we have to find the amplitude of the magnetic field, we won't use the RMS value but the maximum value of the current.

imax=2×irms
imax=2×6.9μA
imax=9.76μA

Bamp=μ0imaxr2πR2
Bamp=4π×107×9.33×106×(.03)2π×(0.06)2
Bamp=1.63×1011T

The amplitude of B at a point 3.0 cm from the axis between the plates is 1.63×1011T.

Q3. What physical quantity is the same for X-rays of wavelength 1010 m, red light of wavelength 6800 A˙ and radiowaves of wavelength 500m?

Answer:

The speed with which these electromagnetic waves travel in a vacuum will be the same and will be equal to 3×108ms1 (speed of light in vacuum).

Q4. A plane electromagnetic wave travels in a vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?

Answer:

Since the electromagnetic wave travels along the z-direction, its electric and magnetic field vectors are lying in the x-y plane as they are mutually perpendicular.

Frequency of wave = 30 MHz

Wavelength = Speed of lightFrequency
=3×10830×106

=10m.

Q5. A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Answer:

Frequency range = 7.5 MHz to 12 MHz

Speed of light = 3×108ms1

Wavelength corresponding to the frequency of 7.5 MHz
=3×108ms17.5×106Hz

=40m

Wavelength corresponding to the frequency of 12 MHz =

3×108ms112×106Hz

=25m

The corresponding wavelength band is 25m to 40m.

Q6. A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Answer:

The frequency of the electromagnetic waves produced by the oscillation of a charged particle about a mean position is equal to the frequency of the oscillation of the charged particle. Therefore, the electromagnetic waves produced will have a frequency of 109 Hz.

Q7. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0=510nT. What is the amplitude of the electric field part of the wave?

Answer:

Magnetic Field (B0 )=510 nT = 510 × 109 T

Speed of light(c) = 3 × 108 ms1

Electric Field = B0 × c

= 510 × 109 T × 3 × 108 ms1

= 153 NC1

Q8. Suppose that the electric field amplitude of an electromagnetic wave is E0=120N/C and that its frequency is ν=50.0 MHz (a) Determine, B0,ω, k,λ (b) Find expressions for E and B.

Answer:

E0 = 120 NC -1

ν=50.0 MHz

(a)

Magnetic Field amplitude(B0)=E0c

= 1203×108

=400nT

Angular frequency ( ω ) = 2 πν

=2 ×π×50 × 106

=3.14 × 108 rad s1

Propagation constant(k)

= 2πλ

= 2πνλν

= ωc

= 3.14×1083×108

=1.05 rad m1

Wavelength( λ ) = cν

= 3×10850×106 =6m

(b) Assuming the electromagnetic wave propagates in the positive z-direction, the Electric field vector will be in the positive x-direction and the magnetic field vector will be in the positive y-direction, as they are mutually perpendicular and E0×B0 gives the direction of propagation of the wave.

E=E0sin(kxωt)i^

=120sin(1.05x3.14×108t)NC1i^

B=B0sin(kxωt)j^

=400sin(1.05x3.14×108t) nTj^

Q.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

Answer:

E=hν=hcλ

E=hν=6.6×1034×3×108λ×1.6×1019eV=12.375×107λeV

Now substitute a different range of wavelengths in the electromagnetic spectrum to obtain the energy

EM waveOne wavelength is taken from the rangeEnergy in eV
Radio1 m1.2375×106
Microwave1 mm1.2375×103
Infra-red1000 nm1.2375
Light500 nm2.475
Ultraviolet1nm1237.5
X-rays0.01 nm123750
Gamma rays0.0001 nm12375000

10(a) In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 20×1010Hz and amplitude 48Vm1

What is the wavelength of the wave?

Answer:

Frequency( ν ) =20 × 1010 Hz

E0 = 48 Vm1

Wavelength(λ)=Speed of light(c)Frequency(ν)

= 3×10820×1010

=1.5mm

Q10.(c) In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0×1010Hz and amplitude 48Vm1 .

Show that the average energy density of the E field equals the average energy density of the B field. [c=3×108ms1.]

Answer:

The average energy density of the Electric field(uE)

= 12ϵE2

= 12ϵ(Bc)2 (as E=Bc)

= 12ϵB2μϵ (c=1μϵ)

=UB

Therefore, the average energy density of the electric field is equal to the average energy density of the Magnetic field.

Class 12 Physics Chapter 8 - Electromagnetic Waves: Additional Questions

Electromagnetic Waves NCERT Solutions of additional questions offer additional work beyond school textbook questions to learners to allow them to solidify their concept knowledge, such as the propagation of waves, energy transfer, and uses of electromagnetic waves. These are solutions which are capable of raising exam preparedness and problem-solving abilities, both on the board, as well as competitive examinations.

Q.1 (a) Suppose that the electric field part of an electromagnetic wave in vacuum is E={(3.1N/C)cos[(1.8rad/m)y+(5.4×106rad/s)t]}i^. What is the direction of propagation?

Answer:

E={(3.1N/C)cos[(1.8rad/m)y+(5.4×106rad/st)]}i^

E={(3.1N/C)sin[(π2((1.8rad/m)y+(5.4×106rad/st))]}i^

The electric field vector is in the negative x-direction, and the wave propagates in the negative y-direction.

Q.1 (b) Suppose that the electric field part of an electromagnetic wave in vacuum is E={(3.1N/C)cos[(1.8rad/m)y+(5.4×106rad/s)t]}i^. What is the wavelength?

Answer:

From the equation of the wave given, we can infer k = 1.8 rad m1

Wavelength(λ)=2πk

=2π1.8

=3.49m

Q.1 (c) Suppose that the electric field part of an electromagnetic wave in vacuum is E={(3.1N/C)cos[(1.8rad/m)y+(5.4×106rad/s)t]}i^. What is the frequency?

Answer:

From the given equation of the electric field, we can infer angular frequency( ω ) = 5.4 × 108 rad s1

Frequency( ν ) = ω2π

= 5.4×1082π

=8.6 × 107 Hz

=86 MHz

Q1 (d) Suppose that the electric field part of an electromagnetic wave in vacuum is E={(3.1N/C)cos[(1.8rad/m)y+(5.4×106rad/s)t]}i^. What is the amplitude of the magnetic field part of the wave?

Answer:

From the given equation of the electric field, we can infer Electric field amplitude (E 0 ) =3.1NC1

Magneticfieldamplitude(B0)=E0c

=3.13×108

=1.03 × 107 T

Q1(e) Suppose that the electric field part of an electromagnetic wave in vacuum is E={(3.1N/C)cos[(1.8rad/m)y+(5.4×106rad/s)t]}i^. Write an expression for the magnetic field part of the wave.

Answer:

As the electric field vector is directed along the negative x-direction and the electromagnetic wave propagates along the negative y-direction, the magnetic field vector must be directed along the negative z-direction. ( i^×k^=j^ )

Therefore, B={B0cos[(1.8rad/m)y+(5.4×106rad/s)t]}k^

B={1.03×107cos[(1.8rad/m)y+(5.4×106rad/s)t]}Tk^

Q2.(a) About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation at a distance of 1m from the bulb? Assume that the radiation is emitted isotropically and neglect reflection.

Answer:

Total power which is converted into visible radiation = 5% of 100W = 5W

The above means 5J of energy is passing through the surface of a concentric sphere(with the bulb at its centre) per second.

Intensity for a sphere of radius 1m

=54π(1)2

=0.398 Wm2

Q2.(b) About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation at a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection.

Answer:

Total power which is converted into visible radiation = 5% of 100W = 5W

The above means 5J of energy is passing through the surface of a concentric sphere(with the bulb at its centre) per second.

Intensity for a sphere of radius 10 m

=54π(10)2

=3.98 × 103 Wm2

Q3) Use the formula λmT=0.29cmK to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?

Answer:

EM waveOne wavelength is taken from the rangeTemperature T=0.29λ
Radio100 cm2.9×103K
Microwave0.1cm2.9 K
Infra-red100000ncm2900K
Light50000 ncm5800K
Ultraviolet100ncm2.9×106K
X-rays1 ncm2.9×108K
Gamma rays0.01 nm2.9×1010K

These numbers indicate the temperature ranges required for obtaining radiation in different parts of the spectrum

Answer the following questions

Q5. (a) Long-distance radio broadcasts use short-wave bands. Why?

Answer:

Long-distance radio broadcasts use short-wave bands as these are refracted by the ionosphere.

Answer the following questions

Q5. (b) It is necessary to use satellites for long-distance TV transmission. Why?

Answer:

As TV signals are of high frequencies, they are not reflected by the ionosphere and therefore satellites are used to reflect them.

Answer the following questions

Q5 (c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?

Answer:

X-rays are absorbed by the atmosphere, and therefore, the source of X-rays must lie outside the atmosphere to carry out X-ray astronomy. Therefore, satellites orbiting the earth are necessary, but radio waves and visible light can penetrate through the atmosphere, and therefore optical and radio telescopes can be built on the ground.

Answer the following questions

Q5. (d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?

Answer:

The small ozone layer on top of the stratosphere is crucial for human survival as it absorbs the ultraviolet radiation coming from the sun, which is very harmful to humans.

Answer the following questions

Q5. (e) If the Earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?

Answer:

If the Earth did not have an atmosphere, its average surface temperature would be lower than what it is now, as in the absence of an atmosphere, there would be no greenhouse effect.

Answer the following questions

Q5 f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?

Answer:

The use of nuclear weapons would cause the formation of smoke clouds, preventing the light from the sun from reaching Earth's surface, and it would also deplete the atmosphere and therefore stop the greenhouse effect and thus doubling the cooling effect.

Class 12 Physics Chapter 8 - Electromagnetic Waves: Higher Order Thinking Skills (HOTS) Questions

Chapter 8 Higher Order Thinking Skills (HOTS) Questions provide students with an opportunity to think beyond ordinary exercises and think critically in understanding electromagnetic waves. These Electromagnetic Waves class 12 question answers also assist students to apply concepts in real life and prepare themselves well for both board exams as well as competitive ones like JEE and NEET.

Q1: A red LED emits light at 0.2 watts uniformly around it. The amplitude of the electric field of the light at a distance of 3 m from the diode is ___ V/m.

Answer:

I=P/4πr2 and pavg=12ε0Eo2
P/4πr2=(12)E0E02C

Or E0=2P4πε0r2C

Here p=0.2w,r=3 m,c=3×108 ms214πε0=9×109

E0=2×0.2×9×1099×3×108E0=0.4×9×10927×108E0=3627

E0=1.15 V/m


Q2: The following figure shows a capacitor made of two circular plates. The capacitor is being charged by an external source which supplies a constant current equal to 0.15 A. What is the displacement current (in amperes) across plates?

Answer:

As we know,

Displacement current

Id=ε0dϕcdt

But ϕE=EA=qAε0A=qε0

Id=ε0ddt(qε0)=ε0ε0dqdt=dqdt=I

Here I=0.15 AId=0.15 A


Q3: A radiation of 200 W is incident on a surface which is 60% reflecting and 40% absorbing. The total force on the surface is

Answer:
Force exerted by incident radiation is F=PC where P is the power of radiation and C is the velocity of light. For absorbed radiations are 40% then exerted force is Fabs=0.4Pc and 60% part of radiations is reflected by the surface, so force exerted by theses radiations is Fref=2×0.6PC=1.2PC Then total force exerted by radiations on the surface is Ftotal =Fref +Fabs =1.2Pc+0.4Pc=1.6Pc=1.6×2003×108=1.07×106 N


Q4: The magnetic field of a plane electromagnetic wave is given by :

B=B0[cos(kzωt)]i^+B1cos(kz+ωt)j^

where B0=3×105T and B1=2×106 T.
The rms value of the force (in Newton) experienced by a stationary charge Q=104C at Z=0 is closest to (up to one decimal).

Answer:

B=B0cos(kxωt)i^+B1cos(kz+ωt)j^B0=3×105TB1=2×106T


Amplitude of resultant magnetic field =B=Bo2+B12

Q=104CFrms=QErms=Q(cBrms)Frms=Q(CB02)2+(CB12)2=104×3×1082(3×105)2+(2×106)2


=104×3×10829×1010+4×1012=104×3×1082900×1012+4×1012=3×1042904×106=0.6 N


Q5: Two light waves are given by, E1=2sin(100πtkx+30) and E2=3cos(200πtkx+60). The ratio of the intensity of the first wave to that of the second wave is :
Answer:
As we learned
Intensity of EM wave -

I=12ϵoEo2c

where
ϵo= Permittivity of free space
Eo= Electric field amplitude
c= Speed of light in vacuum
so

IA2

I1I2=2232=4/9


Class 12 Physics Chapter 8 - Electromagnetic Waves: Topics

Class 12 Physics Chapter 8 - Electromagnetic Waves is to explain the oscillation of electric and magnetic fields in order to transmit the energy across the space. The chapter identifies the source, characteristics and the uses of electromagnetic waves, which is the backbone of the modern communication systems. These topics are not just some good preparation in board exams, but also a good way of learning the concepts that are helpful in JEE and NEET.

8.1 Introduction
8.2 Displacement current
8.3 Electromagnetic waves
8.3.1 Sources of electromagnetic waves
8.3.2 Nature of electromagnetic waves
8.4 Electromagnetic spectrum
8.4.1 Radio waves
8.4.2 Microwaves
8.4.3 Infrared waves
8.4.4 Visible rays
8.4.5 Ultraviolet rays
8.4.6 X-rays
8.4.7 Gamma rays

Approach to Solve Questions of Class 12 Physics Chapter 8 - Electromagnetic Waves

The electromagnetic waves are the foundation of the modern world of communication, medical science and even transfer of energy in nature. During the process of answering questions in this chapter, the students are required to concentrate on the properties of waves and their equations and application in real life rather than memorization. The systematic approach will assist in addressing the conceptual as well as the numerical issues without any problems.

  • Revise Basics of Maxwell's Equations - Learn how electric and magnetic fields that vary with time produce electromagnetic waves.
  • Memorise Important Formulas -Speed of EM waves, relationship between electric field (E) and magnetic field (B) energy density.
  • Know the Spectrum - Learn the frequency, wavelength ranges, and other uses of various components of the electromagnetic spectrum (radio, microwave, infrared, visible, UV, X-ray, gamma).
  • Concentrate on Wave Properties: Polarization, reflection, refraction, and interaction of EM waves with objects.
  • Solve Conceptual Problems -Use reasoning to solve problems in the real world such as communication device, satellite, medical imaging.
  • Step-by-Step approach to solving Numerical Problems - It is always important to begin with the known quantities (E, B, c, f) and use relations in a systematic manner.
  • Practice Past Year questions -Work on JEE/ NEET and CBSE board questions to be acquainted with concepts that are commonly asked.
  • ake Short Notes - Summarize formulas, spectrum chart, and key concepts for quick revision.
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What Extra Should Students Study Beyond NCERT For JEE/NEET?

Beyond NCERT, students preparing for JEE/NEET should study additional concepts like Energy Density and Intensity of Electromagnetic Waves, which explain how energy is distributed in electric and magnetic fields and how it propagates through space. Understanding these topics enhances conceptual clarity and equips students to solve advanced numerical and application-based problems in competitive exams.


NCERT Solutions for Class 12 Physics Chapter-Wise

NCERT Solutions for Class 12 Physics Chapter-wise links provide a complete guide to all chapters, making it easier for students to access solved exercises, important formulas, and key concepts in one place. These solutions are designed to help students prepare effectively for board exams, JEE, NEET, and other competitive tests.


Also Check NCERT Books and NCERT Syllabus here:

NCERT solutions subject wise

Frequently Asked Questions (FAQs)

Q: How can students complete electromagnetic waves ncert solutions for chapter 8 physics class 12 with a perfect score?
A:

Experts at Careers360 have created the chapter 8 physics class 12 ncert solutions after completing extensive research on each concept. To assist students perform well on class assignments and board exams, every small element is well addressed. Additionally, it enables students to successfully complete their tasks on time.

Q: What is the prerequisite of the NCERT chapter electromagnetic waves?
A:

Before starting the preparation of NCERT class 12 chapter 8 it is better to have an idea of chapters 1 to 7. Chapter 1 to 7 is interconnected.

Q: How many questions are expected from Class 12 Physics Chapter 8 NCERT solutions for NEET?
A:

One question from chapter 8 electromagnetic waves can be expected for the NEET exam. The questions can be theoretical or numerical.

Q: What is the weightage of the NCERT class 12 chapter electromagnetic wave for JEE Main?
A:

From the chapter on electromagnetic waves, one question can be expected for JEE Main. More than one question can also be asked.

Q: How many marks are allotted to electromagnetic wave for CBSE board exams?
A:

3 to 5 marks questions are asked from class 12 chapter electromagnetic waves for CBSE board exams.

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For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
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