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Did you ever imagine how radio signals are relayed to your cell phone or even how sunlight transports energy across millions of kilometres to the planet? The answer is given in the electromagnetic waves, which are the foundation of the Physics Class 12 Chapter 8. This chapter discusses the characteristics, behaviour and use of electromagnetic waves in our everyday lives. The Solutions of NCERT by Careers360 breaks down these concepts by explaining them in an easy-to-understand manner, which will prove very helpful to students seeking a clear understanding.
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These NCERT solutions are not only important in preparation for board exams conducted by CBSE, but also in the preparation of competitive exams like JEE Main and NEET exams, as they give step-by-step answers to all the exercise questions. All the topics are described in a way to help students and make their studying time shorter and more efficient. The students also have a chance to check their knowledge by solving the PDF-based solved questions in the NCERT solutions for Class 12 Physics, which can be easily found online for free of cost so they may revise their knowledge at any time and from any place.
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The NCERT Solutions to class 12 Physics chapter 8 Electromagnetic waves 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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NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves: Exercise Questions 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.
Answer:
Radius of the discs(r) = 12cm
Area of the discs(A) = $\pi r^{2}=\pi (0.12)^{2}=.045m^{2}$
Permittivity, $\epsilon _{0}$ = $8.85\times 10^{-12}C^{2}N^{-1}m^{2}$
Distance between the two discs = 5cm=0.05m
$Capacitance= \frac{\varepsilon _{0}A}{d}$
= $\frac{8.85\times 10^{-12}C^{2}N^{-1}m^{2}\times .045m^{2}}{0.05m}$
= $8.003\times 10^{-12}F$
=8.003 pF
$Rate\ of \ change\ of \ potential =\frac{dv}{dt}$
But
$V =\frac{Q}{C}$
Therefore,
$Rate\ of \ change\ of \ potential=\frac{d(\frac{Q}{C})}{dt}=\frac{dq}{Cdt}=\frac{i}{C}$
$=\frac{0.15A}{8.003pF}$
= $1.87\times 10^{10}Vs^{-1}$
Answer:
The value of the displacement current will be the same as that of the conduction current, which is given to be 0.15A. Therefore displacement current is also 0.15A.
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, the Kirchhoff's first rule will hold good.
Answer:
Capacitance(C) of the parallel plate capacitor = 100 pF
Voltage(V) = 230 V
Angular Frequency (ω) = $300\ rad\ s^{-1}$
$Rms\ Current(I) =\frac{Voltage}{Capacitive\ Reactance}$
$Capacitive \ Reactance(X_{c})= \frac{1}{C\omega }=\frac{1}{100\ pF\times 300\ rad\ s^{-1}}$
$X_{c}= 3.33\times 10^{7} \Omega$
$I=\frac{V}{X_{c}}=\frac{230V}{3.33\times 10^{7}\Omega }= 6.9\times 10^{-6}A$
RMS value of conduction current is $6.9\mu A$
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.
Answer:
We know Ampere-Maxwell's Law,
$\oint B\cdot \vec{dl} = \mu _{0}(i_{c}+\varepsilon _{0}\frac{d\phi _{E}}{dt})$
Between the plates conduction current $i_{c}=0$.
For a loop of radius r smaller than the radius of the discs,
$\mu _{0}\varepsilon _{0}\frac{\mathrm{d} \phi _{E}}{\mathrm{d} t}=\mu _{0}i_{d}\frac{\pi r^{2}}{\pi R^{2}}=\mu _{0}i_{d}\frac{ r^{2}}{ R^{2}}$
$B(2\pi r)=\mu _{0}i_{d}\frac{r^{2}}{R^{2}}$
$B=\mu _{0}i_{d}\frac{r}{2\pi R^{2}}$
Since we have to find the amplitude of the magnetic field, we won't use the RMS value but the maximum value of current.
$\\i_{max}=\sqrt{2}\times i_{rms}\\$
$ i_{max}=\sqrt{2}\times6.9\mu A\\$
$ i_{max}=9.76\mu A$
$\\B_{amp}=\mu _{0}i_{max}\frac{r}{2\pi R^{2}}\\$
$ B_{amp}=\frac{4\pi\times 10^{-7}\times 9.33\times 10^{-6}\times (.03) }{2\pi\times (0.06)^{2} }\\$
$ B_{amp}=1.63\times 10^{-11}T$
The amplitude of B at a point 3.0 cm from the axis between the plates is $1.63\times 10^{-11}T$ .
Answer:
The speed with which these electromagnetic waves travel in a vacuum will be the same and will be equal to $3\times 10^{8}ms^{-1}$ (speed of light in vacuum).
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 = $\frac{Speed\ of\ light}{Frequency}$
$=\frac{3\times 10^{8}}{30\times 10^{6}}$
$=10m$ .
Answer:
Frequency range = 7.5 MHz to 12 MHz
Speed of light = $3\times 10^{8}ms^{-1}$
Wavelength corresponding to the frequency of 7.5 MHz
=$\frac{3\times 10^{8}ms^{-1}}{7.5\times 10^{6}Hz}$
$=40m$
Wavelength corresponding to the frequency of 12 MHz =
$\frac{3\times 10^{8}ms^{-1}}{12\times 10^{6}Hz}$
$=25m$
The corresponding wavelength band is 25m to 40m.
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, electromagnetic waves produced will have a frequency of $10^9$ Hz.
Answer:
Magnetic Field ($B_0$ )=510 nT = 510 $\times$ $10^{-9}$ T
Speed of light(c) = 3 $\times$ $10^{8}$ $ms^{-1}$
Electric Field = $B_0$ $\times$ c
= 510 $\times$ $10^{-9}$ T $\times$ 3 $\times$ $10^{8}$ $ms^{-1}$
= 153 $NC^{-1}$
Answer:
$E_0$ = 120 NC -1
$\nu =50.0\ MHz$
(a)
$Magnetic\ Field \ amplitude(B0) =\frac{E_{0}}{c}$
= $\frac{120}{3\times 10^{8}}$
$=400 nT$
Angular frequency ( $\omega$ ) = 2 $\pi \nu$
$=2$ $\times \pi \times 50$ $\times$ $10^{6}$
=3.14 $\times$ $10^{8}$ rad $s^{-1}$
Propagation constant(k)
= $\frac{2\pi }{\lambda }$
= $\frac{2\pi \nu }{\lambda\nu }$
= $\frac{\omega }{c}$
= $\frac{3.14\times 10^{8} }{3\times 10^{8}}$
=1.05 rad $m^{-1}$
Wavelength( $\lambda$ ) = $\frac{c}{\nu }$
= $\frac{3\times 10^{8}}{50\times 10^{6}}$ $= 6 m$
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 $E_{0}\times B_{0}$ gives the direction of propagation of the wave.
$\vec{E}=E_{0}sin(kx-\omega t)\hat{i}$
$= 120sin(1.05x-3.14\times 10^{8} t)NC^{-1}\hat{i}$
$\vec{B}=B_{0}sin(kx-\omega t)\hat{j}$
$= 400sin(1.05x-3.14\times 10^{8} t)\ nT\hat{j}$
Answer:
$E=h\nu=\frac{hc}{\lambda}$
$E=h\nu=\frac{6.6\times 10^{-34}\times3\times10^8}{\lambda \times1.6\times10^{-19}}eV=\frac{12.375\times 10^{-7}}{\lambda}eV$
Now substitute a different range of wavelengths in the electromagnetic spectrum to obtain the energy
EM wave | One wavelength is taken from the range | Energy in eV |
Radio | 1 m | $1.2375\times10^{-6}$ |
Microwave | 1 mm | $1.2375\times10^{-3}$ |
Infra-red | 1000 nm | 1.2375 |
Light | 500 nm | 2.475 |
Ultraviolet | 1nm | 1237.5 |
X-rays | 0.01 nm | 123750 |
Gamma rays | 0.0001 nm | 12375000 |
What is the wavelength of the wave?
Answer:
Frequency( $\nu$ ) =20 $\times$ $10^{10}$ Hz
$E_0$ = 48 V$m^{-1}$
$Wavelength(\lambda) =\frac{Speed\ of\ light(c)}{Frequency (\nu )}$
= $\frac{3\times 10^{8}}{20\times 10^{10}}$
$=1.5 mm$
What is the amplitude of the oscillating magnetic field?
Answer:
The amplitude of the oscillating magnetic field(B 0 ) = $\frac{E_{0}}{c}$
$=$ $\frac{48}{3\times 10^{8}}$
$=160 nT$
Answer:
The average energy density of the Electric field(U E )
= $\frac{1}{2}\epsilon E^{2}$
= $\frac{1}{2}\epsilon (Bc)^{2}$ (as E=Bc)
= $\frac{1}{2}\epsilon \frac{B^{2}}{\mu \epsilon }$ $(c=\frac{1}{\sqrt{\mu \epsilon }})$
=$U_B$
Therefore, the average energy density of the electric field is equal to the average energy density of the Magnetic field.
Additional Questions of Electromagnetic Waves Class 12 NCERT 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.
Answer:
$E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i$
$E = \left \{ ( 3.1 N/C )\sin [ (\frac{\pi }{2}-(( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t ))] \right \} \hat i$
The electric field vector is in the negative x-direction and the wave propagates in the negative y-direction.
Answer:
From the equation of the wave given we can infer k = 1.8 rad $m^{-1}$
$Wavelength(\lambda )=\frac{2\pi }{k}$
$=\frac{2\pi }{1.8}$
$=3.49 m$
Answer:
From the given equation of the electric field, we can infer angular frequency( $\omega$ ) = 5.4 $\times$ 108 rad $s^{-1}$
Frequency( $\nu$ ) = $\frac{\omega }{2\pi }$
= $\frac{5.4\times10^{8}}{2\pi }$
=8.6 $\times$ $10^{7}$ Hz
=86 MHz
Answer:
From the given equation of the electric field, we can infer Electric field amplitude (E 0 ) =3.1$NC^{-1}$
$Magnetic field amplitude (B_0) =\frac{E_{0}}{c}$
$=\frac{3.1}{3\times 10^{8}}$
=1.03 $\times$ $10^{-7}$ T
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. ( $-\hat{i}\times -\hat{k}=-\hat{j}$ )
Therefore, $\vec{B}= \left \{ B_{0}\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s )t] \right \} \hat k$
$\vec{B}= \left \{ 1.03\times 10^{-7}\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s )t] \right \}T \hat k$
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
$=\frac{5}{4\pi (1)^{2}}$
=0.398 $Wm^{-2}$
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
$=\frac{5}{4\pi (10)^{2}}$
=3.98 $\times$ $10^{-3}$ $Wm^{-2}$
Answer:
EM wave | One wavelength is taken from the range | Temperature $T=\frac{0.29}{\lambda}$ |
Radio | 100 cm | $2.9\times 10^{-3}K$ |
Microwave | 0.1cm | 2.9 K |
Infra-red | 100000ncm | 2900K |
Light | 50000 ncm | 5800K |
Ultraviolet | 100ncm | $2.9\times10^{6}K$ |
X-rays | 1 ncm | $2.9\times10^8K$ |
Gamma rays | 0.01 nm | $2.9\times10^{10}K$ |
These numbers indicate the temperature ranges required for obtaining radiation in different parts of the spectrum
Answer:
Frequency( $\nu$ )=1057 MHz
Wavelength( $\lambda$ ) $=\frac{c}{\nu }$
$=\frac{3\times 10^{8}}{1057\times 10^{6} }$
=0.283 m
=28.3 cm
Radio waves
Answer:
Using formula, $\lambda _{m}T=0.29\ cmK$
$\lambda =\frac{0.29}{2.7}$
= 0.107cm
Microwaves.
Answer:
E=14.4 keV
Wavelength( $\lambda$ ) = $\frac{hc}{E}$
= $\frac{6.6\times 10^{-34}\times 3\times 10^{8}}{14.4\times 10^{3}\times 1.6\times 10^{-19}}$
= 0.85 $\dot{A}$
X-rays
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
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
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 atmosphere, there would be no greenhouse effect.
Answer the following questions
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.
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 questions 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:
$I=P / 4 \pi r^2$ and $p_{avg}=\frac{1}{2} \varepsilon_0 E_{o}^2$
$\therefore P / 4 \pi r^2=\left(\frac{1}{2}\right) \mathcal{E}_0 E_0^2 C$
Or $E_0=\sqrt{\frac{2 P}{4 \pi \varepsilon_0 r^2 C}}$
$\begin{aligned} & \text { Here } p=0.2 \mathrm{w}, r=3 \mathrm{~m}, c=3 \times 10^8 \mathrm{~ms}^{-2} \\ & \frac{1}{4} \pi \varepsilon_0=9 \times 10^9\end{aligned}$
$\begin{aligned} & E_0=\sqrt{\frac{2 \times 0.2 \times 9 \times 10^9}{9 \times 3 \times 10^8}} \\ & E_0=\sqrt{\frac{0.4 \times 9 \times 10^9}{27 \times 10^8}} \\ & E_0=\sqrt{\frac{36}{27}}\end{aligned}$
$E_0=1.15 \mathrm{~v} / \mathrm{m}$
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
$I_d=\frac{\varepsilon_0 d \phi_c}{d t}$
But $\phi_E=E A=\frac{q}{A \varepsilon_0} A=\frac{q}{\varepsilon_0}$
$\therefore \quad I_d=\varepsilon_0 \frac{d}{d t}\left(\frac{q}{\varepsilon_0}\right)=\frac{\varepsilon_0}{\varepsilon_0} \frac{d q}{d t}=\frac{d q}{d t}=I$
Here $I=0.15 \mathrm{~A} \quad \therefore I_d=0.15 \mathrm{~A}$
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 $F=\frac{P}{C}$ where $P$ is the power of radiation and $C$ is the velocity of light. For absorbed radiations are $40 \%$ then exerted force is $F_{a b s}=\frac{0.4 P}{c}$ and $60 \%$ part of radiations is reflected by the surface, so force exerted by theses radiations is $ F_{r e f}=2 \times \frac{0.6 P}{C}=\frac{1.2 P}{C} $ Then total force exerted by radiations on the surface is $ \begin{aligned} & F_{\text {total }}=F_{\text {ref }}+F_{\text {abs }} \\ & =\frac{1.2 P}{c}+\frac{0.4 P}{c}=\frac{1.6 P}{c} \\ & =\frac{1.6 \times 200}{3 \times 10^8}=1.07 \times 10^{-6} \mathrm{~N} \end{aligned} $
Q4: The magnetic field of a plane electromagnetic wave is given by :
$
\vec{B}=B_0[\cos (k z-\omega t)] \hat{i}+B_1 \cos (k z+\omega t) \hat{j}
$
where $B_0=3 \times 10^{-5} T$ and $B_1=2 \times 10^{-6} \mathrm{~T}$.
The rms value of the force (in Newton) experienced by a stationary charge $Q=10^{-4} C$ at $Z=0$ is closest to (up to one decimal).
Answer:
$
\begin{aligned}
& B=B_0 \cos (k x-\omega t) \hat{i}+B_1 \cos (k z+\omega t) \hat{j} \\
& B_0=3 \times 10^{-5} T \\
& B_1=2 \times 10^{-6} T
\end{aligned}
$
Amplitude of resultant magnetic field $=B^{\prime}=\sqrt{B_o^2+B_1^2}$
$
\begin{aligned}
Q & =10^{-4} C \\
F_{r m s} & =Q E_{r m s}=Q\left(c B_{r m s}^{\prime}\right) \\
F_{r m s} & =Q \sqrt{\left(\frac{C B_0}{\sqrt{2}}\right)^2+\left(\frac{C B_1}{\sqrt{2}}\right)^2} \\
& =10^{-4} \times \frac{3 \times 10^8}{\sqrt{2}} \sqrt{\left(3 \times 10^{-5}\right)^2+\left(2 \times 10^{-6}\right)^2}
\end{aligned}
$
$
\begin{aligned}
& =10^{-4} \times \frac{3 \times 10^8}{\sqrt{2}} \sqrt{9 \times 10^{-10}+4 \times 10^{-12}} \\
& =10^{-4} \times \frac{3 \times 10^8}{\sqrt{2}} \sqrt{900 \times 10^{-12}+4 \times 10^{-12}} \\
& =\frac{3 \times 10^4}{\sqrt{2}} \sqrt{904} \times 10^{-6} \\
& =0.6 \mathrm{~N}
\end{aligned}
$
Q5: Two light waves are given by, $E_1=2 \sin (100 \pi t-k x+30)$ and $E_2=3 \cos (200 \pi t-k x+60)$. 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=\frac{1}{2} \epsilon_o E_o^2 c
$
where
$\epsilon_o=$ Permittivity of free space
$E_o=$ Electric field amplitude
$\mathrm{c}=$ Speed of light in vacuum
so
$
I \propto A^2 \therefore \frac{I_1}{I_2}=\frac{2^2}{3^2}=4 / 9
$
NCERT Solutions for Class 12 Physics Chapter 8: Topics cover the key concepts of electromagnetic waves, including their propagation, characteristics, and applications in daily life. These solutions help students understand each topic clearly and provide a structured approach to learning for exams like CBSE, 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
To answer questions on Electromagnetic Waves in Class 12 Physics, one should read the question carefully and be able to pick the known and unknown data as well as what is needed, which could be wave properties, speed, wavelength, its frequency, energy or intensity. Learn how speed, frequency, and wavelength are related to one another, how the energy of a photon depends on the related quantities, and what the intensity of the wave is. A labelled diagram of wave propagation and the direction of electric and magnetic fields may be helpful in visualising them. Choose the right formulas, such as $c=f \lambda, E=h f$, or $I=\frac{c \epsilon_0 E^2}{2}$. Input the given values replacing them and maintain consistency in the units. Solve the problem step by step as a series of smaller calculations, and ensure the final answer makes physical sense. Solidify conceptual understanding, derivations, and energy density and intensity of EM waves especially on advanced or HOTS questions. This systematic method offers certainty, precision, and clear understanding of how to answer board and competitive exam questions.
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.
Concept Name | JEE | NCERT |
Displacement Current | ✅ | ✅ |
Maxwell's 4 Equations | ✅ | ✅ |
Nature Of Electromagnetic Waves | ✅ | ✅ |
Energy Density And Intensity Of EM Waves | ✅ | ❌ |
Electromagnetic Spectrum | ✅ | ✅ |
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.
Frequently Asked Questions (FAQs)
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.
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.
One question from chapter 8 electromagnetic waves can be expected for the NEET exam. The questions can be theoretical or numerical.
From the chapter on electromagnetic waves, one question can be expected for JEE Main. More than one question can also be asked.
3 to 5 marks questions are asked from class 12 chapter electromagnetic waves for CBSE board exams.
On Question asked by student community
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Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.
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Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.
From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .
If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.
The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.
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