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NCERT Exemplar Class 12 Physics Solutions Chapter 15 Communication System

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

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.

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

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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:

52719_1
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:

52556_1
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 dB = 10 log10 (Po /Pi ).
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 ‘ht’ is equal to ‘hr’ 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: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:

52548
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
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NCERT Exemplar Class 12 Physics Chapter Wise Links

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.

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Frequently Asked Question (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. 

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hello,

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I hope this was helpful!

Good Luck

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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

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