NCERT Exemplar Class 12 Physics Solutions Chapter 15 Communication System 2021

Q: How can students take help from these NCERT exemplar Class 12 Physics chapter 15 solutions?

Students can understand the topic and also use the solutions as a way to refer to their own answers while practicing before the exam.

Q: How many questions are solved in this chapter?

Our guides have solved all 30 questions from the NCERT exemplar Class 12 Physics chapter 15 exercise

Q: How to download NCERT exemplar Class 12 physics solutions chapter 15?

One can easily go to the solutions page, click on the link and the solutions will be downloaded in PDF format.

Q: Why is this chapter important?

This chapter is one of the base chapters for those who want to become electrical and communication engineers.

NCERT Exemplar Class 12 Physics Solutions Chapter 15 Communication System

Edited By Safeer PP | Updated on Sep 14, 2022 01:37 PM IST | #CBSE Class 12th

Upcoming Event

CBSE Class 12th Exam Date : 15 Feb' 2025 - 15 Feb' 2025

NCERT Exemplar Class 12 Physics solutions chapter 15 introduces the wonder of transmission of information, also called communication. As humans, we process millions of information fragments every second by receiving them through a multitude of sources- that is where communication comes in, to make sure the efficient transfer and delivery of information. How is efficient communication possible? How can one share his thoughts or ideas? NCERT Exemplar Class 12 Physics chapter 15 solutions given on this page would help you articulate your ideas properly and adequately communicate them, be it in your academics orally. They would also aid in giving a good performance in the 12 boards and other competitive exams. NCERT Exemplar Class 12 Physics solutions chapter 15 PDF download can also be brought into use by the students.

Latest: JEE Main high scoring chapters | JEE Main 10 year's papers

Recommended: NEET high scoring chapters

This Story also Contains

NCERT Exemplar Class 12 Physics Solutions Chapter 15 MCQI
NCERT Exemplar Class 12 Physics Solutions Chapter 15 MCQII
NCERT Exemplar Class 12 Physics Solutions Chapter 15 Very Short Answer
NCERT Exemplar Class 12 Physics Solutions Chapter 15 Short Answer
NCERT Exemplar Class 12 Physics Solutions Chapter 15 Long Answer
Main Subtopics Covered in NCERT Exemplar Class 12 Physics Solutions Chapter 15 Communication System
NCERT Exemplar Class 12 Physics Chapter Wise Links
NCERT Exemplar Class 12 Physics Solutions Chapter 15 Important Topics For Exams

NCERT Exemplar Class 12 Physics Solutions Chapter 15 MCQI

Question:1

Three waves A, B and C of frequencies 1600 kHz, 5 MHz and 60 MHz respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communication?

(a) A is transmitted via space wave while B and C are transmitted via sky wave
(b) A is transmitted via ground wave, B via sky wave and C via space wave
(c) B and C are transmitted via ground wave while A is transmitted via sky wave
(d) B is transmitted via ground wave while A and C are transmitted via space wave

Answer:

The correct option is (b) A is transmitted via ground wave, B via skywave and C via space wave. Because the ground wave propagation is only suitable for up to 2MHz frequencies, and the space wave propagation is for the frequencies more than 40 MHz, whereas the skywave propagation is used for frequencies ranging from 30 to 40 MHz.

Question:2

100m long antenna is mounted on a 500m tall building. The complex can become a transmission tower for waves with λ
(a) ~ 400 m
(b) ~ 25 m
(c) ~ 150 m
(d) ~ 2400 m

Answer:

It can easily be calculated as
Length of the building $= 500\: m (l)$
Length of the antenna $= 100\: m (L)$
And thus the wavelength of the wave transmitted would be $= 4L= 4(100) = 400 \: m$

Question:3

A1 kW signal is transmitted using a communication channel which provides attenuation at the rate of -2dB per km. If the communication channel has a total length of 5 km, the power of the signal received is [gain in dB =10 log(p0/pi)]
(a) 900 W (b) 100 W (c) 990 W (d) 1010 W

Answer:

To calculate the power of the signal transmitted,

The rate of attenuation of the signal $= -2(dB/km)$

The given total length of the path $= 5 km$

The formula used $\Rightarrow dB = 10\log(P_o/P_i)$

After putting all he values,

$P_o = 100W$

Question:4

A speech signal of 3 kHz is used to modulate a carrier signal of frequency 1 MHz, using amplitude modulation. The frequencies of the side bands will be

(a) 1.003 MHz and 0.997 MHz
(b) 3001 kHz and 2997 kHz
(c) 1003 kHz and 1000 kHz
(d) 1 MHz and 0.997 MHz

Answer:

The frequencies of the sidebands would be (a) 1.003 MHz and 0.997 MHz

Question:5

A message signal of frequency $\omega _m$ is superposed on a carrier wave of frequency $\omega _c$ to get an amplitude modulated wave (AM). The frequency of the AM wave will be

(a) $\omega _m$

(b) $\omega _c$

(c) $\frac{\omega _c +\omega _m}{2}$

(d) $\frac{\omega _c -\omega _m}{2}$

Answer:

Since the frequencies of both the carrier wave and the amplitude-modulated wave are similar, the frequency of the AM wave would be (b) ωc

Question:6
I-V characteristics of four devices are shown in the figure.
52719
Identify devices that can be used for modulation:
(a) ‘i’ and ‘iii’
(b) only ‘iii’
(c) ‘ii’ and some regions of ‘iv’
(d) All the devices can be used.

Answer:

From the above-given figures, option ii) represents the square-law diode modulation and the option iv) represents some of its parts as the square law modulation there for the correct option is (c) ‘ii’ and some regions of ‘iv’

Question:7

A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to
(a) poor selection of modulation index (selected 0 < m < 1)
(b) poor bandwidth selection of amplifiers
(c) poor selection of carrier frequency
(d) loss of energy in transmission

Answer:

When the selection of bandwidth for transmission is improper, the frequency of the modulated signal becomes more though the male voice frequency is less than female; thus, the voice sounds feminine. Therefore, the answer is (b) poor bandwidth selection of amplifiers

Question:8

A basic communication system consists of
(A) transmitter
(B) information source
(C) user of information
(D) channel
(E) receiver
Choose the correct sequence in which these are arranged in a basic communication system:
(a) ABCDE
(b) BADEC
(c) BDACE
(d) BEADC

Answer:

As clearly stated by the order explained in the given below chain –
Information source $\rightarrow$ Transmitter $\rightarrow$ Channel $\rightarrow$ Receiver $\rightarrow$ user of information
The correct order is option (b) BADEC

Question:9

Identify the mathematical expression for amplitude modulated wave:

$(a) Ac \sin\left [ \left \{ \omega_c +1k_{1Vm}\left ( t \right ) \right \} t+\phi \right ]$
$(b) Ac \sin { \omega_c\: t+\phi + k_{2Vm} (t)}$
$(c) \left \{ Ac +k_{2}V_m(t) \right \}\sin\left ( \omega _ct+\phi \right )$
$(d )A_{c\: Vm}(t)\sin(\omega _ct+\phi )$

Answer:

The correct mathematical expression for amplitude modulated wave is

$(c) \left \{ Ac +k_{2}V_m(t) \right \}\sin\left ( \omega _ct+\phi \right )$

NCERT Exemplar Class 12 Physics Solutions Chapter 15 MCQII

Question:10

An audio signal of 15 kHz frequency cannot be transmitted over long distances without modulation, because
(a) the size of the required antenna would be at least 5 km which is not convenient
(b) the audio signal cannot be transmitted through sky waves
(c) the size of the required antenna would be at least 20 km, which is not convenient
(d) effective power transmitted would be very low, if the size of the antenna is less than 5 km

Answer:

The correct answer is the options (a, b, d)
The audio frequency signals transmitted are of extremely low-frequency waves, and due to the absorption of sky waves by the atmosphere, they cannot transmit the radio waves. And the size of the antenna used must not be less than 5 km as it would degrade the quality of transmission significantly. This would be due to the resonance of wavelength of wave and antenna length. So, the correct options are (a), (b), and (d).

Question:11

Audio sine waves of 3 kHz frequency are used to amplitude modulate a carrier signal of 1.5 MHz Which of the following statements are true?
(a) The sideband frequencies are 1506 kHz and 1494 kHz
(b) The bandwidth required for amplitude modulation is 6kHz
(c) The bandwidth required for amplitude modulation is 3 MHz
(d) The sideband frequencies are 1503 kHz and 1497 kHz.

Answer:

The types of frequencies under the AM wave are of three types, and the sideband frequencies in AM waves are close to the carrier waves. So according to the given criteria,
The bandwidth required for amplitude modulation is 6kHz and The sideband frequencies are 1503 kHz and 1497 kHz
So, the correct options are (b) and (d)

Question:12

A TV transmission tower has a height of 240 m. Signals broadcast from this tower will be received by LOS communication at a distance of (assume the radius of the earth to be 6.4 × 10⁶ m)
(a) 100 km
(b) 24 km
(c) 55 km
(d) 50 km

Answer:

To calculate the maximum signal transmission from any TV towers, the formula is –
$d=\sqrt{2rh}=\sqrt{2\times 6.4\times 10^6\times 240}=55\: km$
So, the distances up to which the signal can be received are (b) 24 km, (c) 55 km, (d) 50 km

Question:13

The frequency response curve in the figure for the filter circuit used for production of AM wave should be
52556
(a) (i) followed by(ii)
(b) (ii) followed by (i)
(c) (iii)
(d) (iv)

Answer:

The bandwidth for the production of amplitude modulated wave is given by the relation =
à Frequency of upper band – frequency of a lower band
$=\left ( \omega _c +\omega _m \right )-\left ( \omega _c -\omega _m \right )$
Therefore, the correct options are (a) (i) followed by(ii), (b) (ii) followed by (i), (c) (iii)

Question:14

In amplitude modulation, the modulation index m is kept less than or equal to 1 because
(a) m > 1, will result in interference between the carrier frequency and message frequency, resulting in distortion
(b) m > 1 will result in overlapping of both sidebands resulting in loss of information
(c) m > 1 will result in a change in phase between the carrier signal and the message signal
(d) m > 1 indicates the amplitude of message signal greater than the amplitude of the carrier signal resulting in distortion

Answer:

Due to the occurrence of overmodulation $u>1$ , this leads to higher chances of sideband overlapping and fading (loss of information). If the value of u is approximately equal to 1, it results in a distortion f carrier wave.
Thus option (b) and (d) are the right answers.

Question:15

Which of the following would produce analogue signals and which would produce digital signals?
(i) A vibrating tuning forks
(ii) Musical sound due to a vibrating sitar string
(iii) Light pulse
(iv) Output of NAND gate

Answer:

Analogue signals will be produced by option i) and ii), i.e. a vibrating tuning fork and musical sound due to a vibrating sitar string.
Whereas option iii) and iv) will produce digital signals which are light pulse and output of NAND gate.

NCERT Exemplar Class 12 Physics Solutions Chapter 15 Very Short Answer

Question:16

Would sky waves be suitable for transmission of TV signals of 60 MHz frequency?

Answer:

As the frequency of TV signals is beyond 60 MHz, the skywaves are not reliable for transmission of 60 MHz frequency TV signals. The waves preferred for this type of transmission are space waves.

Question:17

Two waves A and B of frequencies 2 MHz and 3 MHz respectively are beamed in the same direction for communication via sky wave. Which one of these is likely to travel a longer distance in the ionosphere before suffering total internal reflection?

Answer:

Frequency and refractive index are directly proportional; thus, due to wave B having a higher frequency than wave A, the refractive index of wave B is higher than the refractive index of wave A.

Question:18

The maximum amplitude of an A.M. wave is found to be 15 V while its minimum amplitude is found to be 3 V. What is the modulation index?

Answer:

The change in the ratio of the amplitude of the carrier wave to the amplitude of the original carrier wave is defined as the modulation index.
The given maximum amplitude of the AM wave is $= Amax = Ac + Am = 15 V$

The given minimum amplitude of the AM wave is $= Amin = Ac - Am = 3V$
Thus, the modulation index, m, would be $= Am/Ac = 2/3$

Question:19

Compute the LC product of a tuned amplifier circuit required to generate a carrier wave of 1 MHz for amplitude modulation.

Answer:

The formula for the frequency of the tuned amplifier is,
$v=\frac{1 }{2 \pi\sqrt{LC}}$
As given in the question
$v=1MHz=10^6Hz$
After substituting the values, the answer would be
$LC = 2.54 \times 10^{-14} s^2$

Question:20

Why is an AM signal likely to be noisier than an FM signal upon transmission through a channel?

Answer:

AM signal has the characteristic feature of being noisy when broadcasted on channels as compared to the FM signal because its carrier waves’ instantaneous voltage varies by the modulating wave voltage in the AM.

NCERT Exemplar Class 12 Physics Solutions Chapter 15 Short Answer

Question:21

Figure shows a communication system. What is the output power when the input signal is of 1.01mW? (gain in d_B = 10 log₁₀ (P_o /P_i ).
52372

Answer:

The loss of signal during the transmission $= 2 \: \: dB/km$

The total path covered by the signal $= 5 \: Km$

Collective loss suffered $= -10\: dB$

Total amplifier gain $= 30 \: dB$

Total gain in signal $= 20 \: dB$

Gain in dB $= 2$

Input power $= 1.01 mW$

Thus, output power would be,

$P_o = 1.01 \times 100$

Hence, the output power is $101\: mW.$

Question:22

A TV transmission tower antenna is at the height of 20 m. How much service area can it cover if the receiving antenna is (i) at ground level, (ii) at the height of 25 m? Calculate the percentage increase in area covered in case (ii) relative to the case (i).

Answer:

As given in the question,
For the first part
i) Range of the signals, it = 16 km
Area covered by the waves = 803.84 km2
ii) Height of antenna from ground level = 25 m
Range of the antenna = dm = 33.9 km
Area covered in transmission = 3608.52 km
Thus, the percentage increase in area covered in case (ii) relative to the case (i) = 348.9 %

Question:23

If the whole earth is to be connected by LOS communication using space waves (no restriction of antenna size or tower height), what is the minimum number of antennas required? Calculate the tower height of these antennas in terms of earths radius?

Answer:

In order to connect earth with LOS using space waves,
Let the height of the antenna receiving the signals be ‘ht’
The ‘h_t’ is equal to ‘h_r’ where $R= radius \: o\! f\: the\: earth$
The line of sight between transmission and receiving antennas be ‘dm’
The maximum distance would be $R = h_t$

Question:24

The maximum frequency for reflection of sky waves from a certain layer of the ionosphere is found to be f_max= 9(N_max)^1/2, where N_max is the maximum electron density at that layer of the ionosphere.
On a certain day it is observed that signals of frequencies higher than 5 MHz are not received by reflection from the F₁layer of the ionosphere while signals of frequencies higher than 8 MHz are not received by reflection from the F₂ layer of the ionosphere. Estimate the maximum electron densities of the F₁and F₂ layers on that day.

Answer:

The values given in the question are

$F_{1}\: max = 5 MHz\\F_{2} \: max = 8 MHz$
The formula used is Maximum frequency $=> f_{max }= 9(Nmax)^{1/2}$
So, the Nmax | $F_{1 }= 3.086 \times10^{11} m^{3}$

Again, after value substitution,

$N_{max} | F_{2} = 7.9 \times 10^{^{11}}/m^{3 }$

Question:25

On radiating (sending out) and AM modulated signal, the total radiated power is due to the energy carried by $\omega c,\left ( \omega c-\omega m \right )\: and\:\left ( \omega c+\omega m \right )$ . Suggest ways to minimise the cost of radiation without compromising on the information.

Answer:

The overall power radiated because of energy carried by $\omega c,\left ( \omega c-\omega m \right )\: and\:\left ( \omega c+\omega m \right )$
Since the information is only transmitted by the amplitude modulated signals that are $\left ( \omega c-\omega m \right )\: and\:\left ( \omega c+\omega m \right )$ as these sideband frequencies are mainly closer to the carrier frequency, the cost can easily be minimised by transmitting both these signals.

NCERT Exemplar Class 12 Physics Solutions Chapter 15 Long Answer

Question:26

The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation $I=I_oe^{-ex}$ , where I0 is the intensity at $x = 0 \: and\: \alpha$ is the attenuation constant.
(a) Show that the intensity reduces by 75% after a distance of $( In \: 4/\alpha ).$
(b) Attenuation of a signal can be expressed in decibel (dB) according to the relation $dB=10 \log_{_{10}} (I/I_0)$ .What is the attenuation in dB/km for an optical fibre in which the intensity falls by 50% over a distance of 50 km?

Answer:

The given components in the question,
$I = Ioe^{-\alpha x }$
$\log _e. I/I_o = -\alpha x \: or \: \log_e . I_o/I = \alpha x$
Now, to calculate the % decrease in intensity, the formula used would be
% derease in intensity
$=\left ( \frac{I_0 -I}{I_0} \right )\times 100 =\left ( 1-\frac{I}{I_0} \right )\times 100 =\left ( 1-\frac{1}{4} \right )\times 100=75\%$
(ii) Let $\alpha$ be the attenuation in dB/km. If x is the distance travelled by signal, the
$10 \log_{10}.\frac{I}{I_0}=-\alpha x$
Given
$\frac{I}{I_0}= \frac{1}{2},x=50\: km \therefore 10 \log_{10}\left ( \frac{1}{2} \right )=- \alpha \times 50 \: or\: 10 \log_{10} \; 2 =50 \alpha\\or \log _{10}\:2=5 \alpha \; or \\\: \alpha= \frac{1}{5} \log_{10} \: 2=\frac{0.3010}{5}=0.0602 \:dB/km$

So, the final answers for both the parts are,
i) The decrease in the intensity is reduced by 75%.
ii) The attenuation of the given optical fibre = 0.0602 dB/km

Question:27

A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km above earth’s surface. Assuming re-transmission time by satellite negligible, find the distance between source and receiver. If communication between the two was to be done by Line of Sight (LOS) method, what should size and placement of receiving and transmitting antenna be?

Answer:

The values given in the question are,

Velocity of waves $= 3 \times 10^{8} m/s$

Time to reach a receiver $= 4.04 \times 10^{-3} s$

Height of satellite, $h = 600\: km$

Radius of earth $= 6400\: km$

Size of transmitting antenna $= hT$

By using the formula = distance travelled by wave/time = velocity

The final velocity of waves $= 606\: km$

After using the Pythagoras theorem,

$d = 85.06\: km$

Thus, the distance between source and receiver $= 2d = 170\: km$

So, the maximum distance from the transmitter to the ground would by the EM waves would be

$= h_t = 565\: m$

Question:28

An amplitude modulated wave is as shown in the figure. Calculate
(i) the percentage modulation
(ii) peak carrier voltage and
(iii) peak value of information voltage
52546

Answer:

52546_1

As shown in the figure,

Maximum voltage, $Vmax = 100/2 = 50 V$

Minimum voltage, $Vmin = 20/2 = 10 V$

i) Percentage modulation $= (Vmax - Vmin)/(Vmax + Vmin) \times 100$

After substituting the values of Vmax and Vmin, the percentage modulation = 66.67%

ii) the formula of peak carrier voltage is $Vc =( Vmax + Vmin)/2$

Thus, $Vc = 30 V$

iii) Peak value of information voltage $= Vm = uVc = 20V$

Question:29

(i) Draw the plot of amplitude versus ω for an amplitude modulated wave whose carrier wave is carrying two modulating signals,
(ii) Is the plot symmetrical about $\omega _c$ ? Comment especially about plot in region $\left ( \omega <\omega _c \right )$
(iii) Extrapolate and predict the problems one can expect if more waves are to be modulated.
(iv) Suggest solutions to the above problem. In the process can one understand another advantage of modulation in terms of bandwidth?

Answer:

ii) No, the plot of the frequency spectrum is not symmetrical due to the crowding for $\left ( \omega <\omega _c \right )$ the present in the spectrum
iii) The probability of mixing of signals is very high due to the crowding at $\left ( \omega <\omega _c \right )$
iv) An easy way to accommodate more signals would be to increase the bandwidth and frequency of the carrier waves.

Question:30

An audio signal is modulated by a carrier wave of 20MHz such that the bandwidth required for modulation is 3kHz. Could this wave be demodulated by a diode detector which has the values of R and C as
$\\(i) R = 1 k\Omega , C = 0.01\mu F\\ (ii) R = 10 k\Omega, C = 0.01\mu F\\ (iii) R = 10 k\Omega, C = 0.1\mu F$

Answer:

Carrier wave frequency, $f_c =20 MHz = 20 \times 10^{6} Hz$

Bandwidth $= 2fm = 3 \times 10^3 Hz$

fm $= 1.5 \times 10^3 Hz$

$\\ \frac{1}{f_c} = 0.5 \times 10^{-7}\\ \\ \frac{1}{f_m} = 0.7 \times 10^{-3}\\ i)\frac{1}{f_c}<<RC< \frac{1}{f_m}$

Therefore, it can be demodulated

$ii)\frac{1}{f_c}<<RC<< \frac{1}{f_m}$

Therefore, it can be demodulated

$iii)\frac{1}{f_c}<RC$

Therefore, it cannot be demodulated

Main Subtopics Covered in NCERT Exemplar Class 12 Physics Solutions Chapter 15 Communication System

Introduction
Elements Of A Communication System
Basic Terminology Used In Electronic Communication Systems
Bandwidth Of Signals
Bandwidth Of Transmission Medium
Propagation Of Electromagnetic Waves
Ground wave
Skywaves
Space wave
Modulation And Its Necessity
Size of the antenna or aerial
Adequate power radiated by an antenna Ex 15.7.3 – Mixing up of signals from different transmitters.
Amplitude Modulation
Production Of Amplitude Modulated Wave
Detection Of Amplitude Modulated Wave

Pearson | PTE

Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 31st DEC'24! Trusted by 3,500+ universities globally

Apply

NCERT Exemplar Class 12 Physics Chapter Wise Links

Chapter 1 Electric Charges and Fields

Chapter 2 Electrostatic Potential and Capacitance

Chapter 3 Current Electricity

Chapter 4 Moving Charges and Magnetism

Chapter 5 Magnetism and Matter

Chapter 6 Electromagnetic Induction

Chapter 7 Alternating Current

Chapter 8 Electromagnetic Waves

Chapter 9 Ray Optics and Optical Instruments

Chapter 10 Wave Optics

Chapter 11 Dual Nature of Radiation and Matter

Chapter 12 Atoms

Chapter 13 Nuclei

Chapter 14 Semiconductor Electronics

NCERT Exemplar Class 12 Physics Solutions Chapter 15 Important Topics For Exams

· A communication system makes the successful transmission of an idea or any other important information among individuals possible.

· NCERT Exemplar solutions For Class 12 Physics chapter 15 moves through communication basics, from a simple voice instruction from one person to another to a transfer of files and pictures using a server. Communication is used in telecommunication networks, relay stations, telephones, computers, mobile, telegraph, etc.

· NCERT Exemplar Class 12 Physics solutions chapter 15 also discusses the types of communication systems, the transmission and altering of signals, terms like bandwidth, attenuation, receiver and modulation, and modulation types.

NCERT Exemplar Class 12 Solutions

NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Mathematics Solutions

NCERT Exemplar Class 12 Biology Solutions

Also, check the NCERT solutions of questions given in book

Also read NCERT Solution subject wise

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download EBook

Must read NCERT Notes subject wise

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. How can students take help from these NCERT exemplar Class 12 Physics chapter 15 solutions?

Students can understand the topic and also use the solutions as a way to refer to their own answers while practicing before the exam.

2. How many questions are solved in this chapter?

Our guides have solved all 30 questions from the NCERT exemplar Class 12 Physics chapter 15 exercise

3. How to download NCERT exemplar Class 12 physics solutions chapter 15?

One can easily go to the solutions page, click on the link and the solutions will be downloaded in PDF format.

4. Why is this chapter important?

This chapter is one of the base chapters for those who want to become electrical and communication engineers.

Also Read

NCERT Class 12 Maths Notes - Download Chapter wise Free PDFs

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

NCERT Class 12 Physics Chapter 8 Notes Electromagnetic Waves - Download PDF

NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

NCERT Class 12 Physics Chapter 6 Notes Electromagnetic Induction - Download PDF

NCERT Class 12 Physics Chapter 5 Notes Magnetism and Matter - Download PDF

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

Articles

CBSE Class 12 English 2025: Expected Article Writing Questions to Prepare

Dec 17, 2024

RBSE Model Paper 2024-25 Class 12, Download RBSE Model Papers

Dec 17, 2024

WBCHSE 2025 Syllabus Revised: Key Updates for West Bengal Class 11 & 12 Students

Dec 17, 2024

Score Above 90% in CBSE Class 12 Board Exam 2025: 2-month Preparation Plan

Dec 17, 2024

TS Inter 2nd Year Exam Dates 2025 Out, Manabadi Telangana Intermediate Time Table PDF

Dec 16, 2024

CBSE Exam Rules for Class 10 and 12: New Guidelines For CBSE Exams 2024-25

Dec 13, 2024

Most Repeated Questions in Physics Class 12 CBSE Board

Dec 11, 2024

CBSE 2025 Latest News and Updates: Exam Schedule by CBSE Board

Dec 10, 2024

Upcoming School Exams

National Institute of Open Schooling 10th examination

Application Date:21 November,2024 - 20 December,2024

Application Process

Exam Pattern

Mock Test

National Institute of Open Schooling 12th Examination

Application Date:21 November,2024 - 20 December,2024

Application Process

Exam Pattern

Admit Card

National Means Cum-Merit Scholarship

Admit Card Date:27 November,2024 - 22 December,2024

Eligibility Criteria

Application Process

Exam Pattern

West Bengal Board 12th Examination

Exam Date:02 December,2024 - 20 December,2024

Result

Exam Pattern

Admit Card

Gujarat Board Secondary School Certificate Examination

Late Fee Application Date:11 December,2024 - 20 December,2024

Exam Pattern

Admit Card

Result

View All School Exams

Certifications By Top Providers

My SQL Basics

Via Great Learning

Full-Stack Web Development

Via Masai School

Business Analytics Foundations

Via PW Skills

Digital Banking Business Model

Via State Bank of India

Full-Stack Android Development

Via Masai School

Data Science for Beginners

Via Board Infinity

1115 courses

816 courses

394 courses

295 courses

Harvard University, Cambridge

98 courses

University of Michigan, Ann Arbor

83 courses

IT & Software

Teaching and Academics

Programming and Development

Health, Fitness and Medicine

Business and Management

Engineering and Architecture

General Management

Financial Management

Environmental Studies

History

Psychology

Artificial Intelligence

Explore Top Universities Across Globe

University of Essex, Colchester

Wivenhoe Park Colchester CO4 3SQ

University College London, London

Gower Street, London, WC1E 6BT

The University of Edinburgh, Edinburgh

Old College, South Bridge, Edinburgh, Post Code EH8 9YL

University of Bristol, Bristol

Beacon House, Queens Road, Bristol, BS8 1QU

University of Delaware, Newark

Newark, DE 19716

University of Nottingham, Nottingham

University Park, Nottingham NG7 2RD

Aeronautical Engineering Salary Abroad - Country Wise Top Recruiters List

7 min ⋆ Dec 17, 2024 12:12 PM IST

ETH Zurich Scholarship For International Students 2025: Check Requirements and Deadlines

5 min ⋆ Dec 17, 2024 11:12 AM IST

GMAT Focus Edition Syllabus 2025: Section Wise Syllabus QR, VR and DI

8 min ⋆ Dec 04, 2024 11:12 AM IST

Post Study Work Options in France for International Students in 2025

7 min ⋆ Dec 17, 2024 11:12 AM IST

Engineering in New Zealand Fees 2024: Living Cost and Food Expenses

8 min ⋆ May 22, 2024 12:05 PM IST

Related E-books & Sample Papers

CBSE Class 12 Accountancy Question Papers 2024 (Code-PQ1RS/1) – All Sets (1, 2, 3)

27+ Downloads

CBSE Class 12 Maths Question Papers 2024 (Code-QSS4R/4) – All Sets (1, 2, 3)

36+ Downloads

CBSE Class 12 Maths Question Papers 2024 (Code-Q5QPS/5) – All Sets (1, 2, 3)

13+ Downloads

CBSE Class 12 Maths Question Papers 2024 (Code- PQ3RS/3) – All Sets (1, 2, 3)

5+ Downloads

CBSE Class 12 Maths Question Papers 2024 (Code-PQ2RS/2) – All Sets (1, 2, 3)

4+ Downloads

CBSE Class 12 Maths Question Papers 2024 (Code-PQ1RS/1) – All Sets (1, 2, 3)

5+ Downloads

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Read Complete Answer

Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer

Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer

Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer

kumardurgesh1802 10 July,2024

View All

Stay up-to date with CBSE Class 12th News

Download Careers360 App

All this at the convenience of your phone

Regular Exam Updates
Best College Recommendations
College & Rank predictors
Detailed Books and Sample Papers
Question and Answers

Scan and download the app

Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 12 Physics Solutions Chapter 15 Communication System

NCERT Exemplar Class 12 Physics Solutions Chapter 15 MCQI

NCERT Exemplar Class 12 Physics Solutions Chapter 15 MCQII

NCERT Exemplar Class 12 Physics Solutions Chapter 15 Very Short Answer

NCERT Exemplar Class 12 Physics Solutions Chapter 15 Short Answer

NCERT Exemplar Class 12 Physics Solutions Chapter 15 Long Answer

Main Subtopics Covered in NCERT Exemplar Class 12 Physics Solutions Chapter 15 Communication System

Pearson | PTE

NCERT Exemplar Class 12 Physics Chapter Wise Links

NCERT Exemplar Class 12 Physics Solutions Chapter 15 Important Topics For Exams

NCERT Exemplar Class 12 Solutions

Also, check the NCERT solutions of questions given in book

Frequently Asked Questions (FAQs)

Also Read

Articles

Upcoming School Exams

Certifications By Top Providers

Explore Top Universities Across Globe

Related E-books & Sample Papers

Questions related to CBSE Class 12th

Popular CBSE Class 12th Questions

Colleges After 12th

JEE Main Important Physics formulas

JEE Main Important Chemistry formulas

TOEFL ® Registrations 2024

Pearson | PTE

JEE Main high scoring chapters and topics

JEE Main Important Mathematics Formulas

Stay up-to date with CBSE Class 12th News

Explore on Careers360

NCERT Solutions

NCERT Exemplars

NCERT Books

NCERT Notes

CBSE

CBSE Class 10

CBSE Class 12th

CISCE Board 10th

IMO

IEO

NSO

NMMS

NVS

JNVST

Schools By Medium

By Ownership

By City

By State

Download Careers360 App

All this at the convenience of your phone

Scan and download the app