Maharashtra Board Class 11 Physics Chapter 13 Electromagnetic Waves and Communication System Solutions

1. Choose The Correct Option.

Question 1. The EM wave emitted by the Sun and responsible for heating the Earth's atmosphere due to green house effect is
(A) Infra-red radiation
(B) X ray
(C) Microwave
(D) Visible light
Answer: (A) Infra-red radiation
In simple words: Infrared radiation from the Sun gets trapped by greenhouse gases, warming the Earth's atmosphere.

🎯 Exam Tip: Remember the different types of EM waves and their primary interactions with the Earth's atmosphere, especially regarding the greenhouse effect.

Question 2. Earth's atmosphere is richest in
(A) UV
(B) IR
(C) X-ray
(D) Microwaves
Answer: (B) IR
In simple words: The Earth's atmosphere contains a significant amount of infrared radiation due to absorption and emission processes.

🎯 Exam Tip: Understand the composition of the electromagnetic spectrum present in the Earth's atmosphere and the factors influencing it.

Question 3. How does the frequency of a beam of ultraviolet light change when it travels from air into glass?
(C) decreases
(D) remains same
Answer: (D) remains same
In simple words: When light moves from one medium to another, its speed and wavelength change, but its frequency remains constant as it's determined by the source.

🎯 Exam Tip: A key principle of wave propagation is that the frequency of a wave is determined by its source and does not change when it passes into a different medium.

Question 4. The direction of EM wave is given by
(A) \( \vec{E} \times \vec{B} \)
(B) \( \vec{E} \cdot \vec{B} \)
(C) along \( \vec{E} \)
(D) along \( \vec{B} \)
Answer: (A) \( \vec{E} \times \vec{B} \)
In simple words: The direction of propagation for an electromagnetic wave is perpendicular to both the electric and magnetic field vectors, given by their cross product.

🎯 Exam Tip: Remember that electromagnetic waves are transverse, with electric and magnetic fields oscillating perpendicular to each other and to the direction of propagation, which is represented by the cross product.

Question 5. The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to
(A) \( h^{1/2} \)
(B) h
(C) \( h^{3/2} \)
(D) \( h^2 \)
Answer: (A) \( h^{1/2} \)
In simple words: The maximum distance for line-of-sight communication from a TV tower is directly proportional to the square root of the tower's height due to Earth's curvature.

🎯 Exam Tip: Understand the formula for the line-of-sight communication range, \(d = \sqrt{2Rh}\), which clearly shows the direct proportionality to the square root of the height \(h\).

Question 6. The waves used by artificial satellites for communication purposes are
(A) Microwave
(B) AM radio waves
(C) FM radio waves
(D) X-rays
Answer: (A) Microwave
In simple words: Microwaves are used for satellite communication because they can easily penetrate the Earth's atmosphere with minimal attenuation and are suitable for high-bandwidth data transmission.

🎯 Exam Tip: Be aware of the specific applications of different parts of the electromagnetic spectrum, especially for communication technologies like satellite transmission.

Question 7. If a TV telecast is to cover a radius of 640 km, what should be the height of transmitting antenna?
(A) 32000 m
(B) 53000 m
(C) 42000 m
(D) 55000 m
Answer: (A) 32000 m
In simple words: This question involves calculating the required antenna height for a given communication range, using the line-of-sight formula considering the Earth's radius.

🎯 Exam Tip: For problems involving communication range and antenna height, always remember the formula \(d = \sqrt{2Rh}\) and ensure consistent unit conversion (e.g., km to m) for accurate results.

2. Answer Briefly.

Question 1. State two characteristics of an EM wave.
Answer:
(i) The electric and magnetic fields, \( \vec{E} \) and \( \vec{B} \) are always perpendicular to each other and also to the direction of propagation of the EM wave. Thus, the EM waves are transverse waves.
(ii) The cross product \( (\vec{E} \times \vec{B}) \) gives the direction in which the EM wave travels. \( (\vec{E} \times \vec{B}) \) also gives the energy carried by EM wave.
In simple words: EM waves are transverse waves where electric and magnetic fields oscillate perpendicular to each other and to the direction of wave travel. Their direction of propagation is given by the cross product of the electric and magnetic fields.

🎯 Exam Tip: Clearly state that EM waves are transverse, define the orthogonality of their fields, and mention that the Poynting vector (or the cross product \( \vec{E} \times \vec{B} \)) indicates the direction of energy flow.

Question 2. Why are microwaves used in radar?
Answer: Microwaves are used in radar systems for identifying the location of distant objects like ships, aeroplanes etc.
In simple words: Microwaves are effective for radar because their short wavelengths allow for good resolution and they can travel long distances without significant atmospheric absorption.

🎯 Exam Tip: Focus on the properties of microwaves (e.g., short wavelength, ability to penetrate atmosphere, suitable for detection) that make them ideal for radar applications.

Question 3. What are EM waves?
Answer: Waves that are caused by the acceleration of charged particles and consist of electric and magnetic fields vibrating sinusoidally at right angles to each other and to the direction of propagation are called EM waves or EM radiation.
In simple words: Electromagnetic waves are disturbances created by accelerating charges, featuring mutually perpendicular oscillating electric and magnetic fields that propagate through space.

🎯 Exam Tip: Define EM waves by their origin (accelerating charges) and their fundamental structure (oscillating, perpendicular electric and magnetic fields, transverse propagation).

Question 4. How are EM waves produced?
Answer:
(1) According to quantum theory, an electron, while orbiting around the nucleus in a stable orbit does not emit EM radiation even though it undergoes acceleration.
(2) It will emit an EM radiation only when it falls from an orbit of higher energy to one of lower energy.
(3) EM waves (such as X-rays) are produced when fast moving electrons hit a target of high atomic number (such as molybdenum, copper, etc.).
(4) An electric charge at rest has an electric field in the region around it but has no magnetic field.
(5) When the charge moves, it produces both electric and magnetic fields.
(6) If the charge moves with a constant velocity, the magnetic field will not change with time and hence, it cannot produce an EM wave.
(7) But if the charge is accelerated, both the magnetic and electric fields change with space and time and an EM wave is produced.
(8) Thus, an oscillating charge emits an EM wave which has the same frequency as that of the oscillation of the charge.
In simple words: Electromagnetic waves are fundamentally produced by accelerating charged particles, which create changing electric and magnetic fields that propagate as waves. They can also be produced by electron transitions in atoms or by high-speed electrons striking a target.

🎯 Exam Tip: The core concept is that accelerated or oscillating charges produce EM waves. Also, briefly mention atomic transitions and X-ray production as specific examples.

Question 5. Can we produce a pure electric or magnetic wave in space? Why?
Answer: No. In vacuum, an electric field cannot directly induce another electric field so a "pure” electric field wave cannot exist and same can be said for a "pure" magnetic wave.
In simple words: No, pure electric or magnetic waves cannot exist independently because a changing electric field induces a magnetic field, and a changing magnetic field induces an electric field; they are inherently coupled in an EM wave.

🎯 Exam Tip: Emphasize the interdependence of electric and magnetic fields in an EM wave, as described by Maxwell's equations, making a "pure" field wave impossible.

Question 6. Does an ordinary electric lamp emit EM waves?
Answer: Yes, ordinary electric lamp emits EM waves.
In simple words: Yes, an ordinary electric lamp emits electromagnetic waves, primarily in the form of visible light and infrared radiation, due to the heated filament.

🎯 Exam Tip: Remember that hot objects, like lamp filaments, emit thermal radiation, which is a form of electromagnetic waves spanning various wavelengths, including visible light.

Question 7. Why light waves travel in vacuum whereas sound wave cannot?
Answer: Light waves are electromagnetic waves which can travel in vacuum whereas sound waves travel due to the vibration of particles of medium. Without any particles present (like in a vacuum) no vibrations can be produced. Hence, the sound wave cannot travel through the vacuum.
In simple words: Light waves are electromagnetic and can propagate without a medium, unlike sound waves, which are mechanical waves requiring a material medium for particle vibrations to transmit energy.

🎯 Exam Tip: Distinguish clearly between electromagnetic waves (like light) that do not require a medium and mechanical waves (like sound) that do.

Question 8. What are ultraviolet rays? Give two uses.
Answer:
Production:
(1) Ultraviolet rays can be produced by the mercury vapour lamp, electric spark and carbon arc lamp.
(2) They can also be obtained by striking electrical discharge in hydrogen and xenon gas tubes.
(3) The Sun is the most important natural source of ultraviolet rays, most of which are absorbed by the ozone layer in the Earth's atmosphere.
Uses:
(1) Ultraviolet rays destroy germs and bacteria and hence they are used for sterilizing surgical instruments and for purification of water.
(2) Used in burglar alarms and security systems.
(3) Used to distinguish real and fake gems.
In simple words: Ultraviolet rays are a part of the EM spectrum beyond violet light, produced by sources like the Sun or mercury lamps. They are widely used for sterilization and detection due to their germicidal properties and interaction with certain materials.

🎯 Exam Tip: Know the sources and at least two distinct applications of each major type of electromagnetic radiation, especially UV rays, for common exam questions.

Question 9. What are radio waves? Give its two uses.
Answer:
(1) Radio waves are produced by accelerated motion of charges in a conducting wire. The frequency of waves produced by the circuit depends upon the magnitudes of the inductance and the capacitance.
(2) Thus, by choosing suitable values of the inductance and the capacitance, radio waves of desired frequency can be produced.
Uses:
(1) Radio waves are used for wireless communication purpose.
(2) They are used for radio broadcasting and transmission of TV signals.
(3) Cellular phones use radio waves to transmit voice communication in the ultra high frequency (UHF) band.
In simple words: Radio waves are the lowest frequency EM waves, produced by oscillating electric circuits, and are primarily used for various forms of wireless communication, including broadcasting and mobile phone signals.

🎯 Exam Tip: Understand that radio waves are generated by oscillating electrical circuits and their primary applications are in wireless communication, emphasizing their role in broadcasting and mobile technology.

Question 10. Name the most harmful radiation entering the Earth's atmosphere from the outer space.
Answer: Ultraviolet radiation.
In simple words: Ultraviolet (UV) radiation from outer space is the most harmful type that reaches Earth's surface, posing risks like skin damage and cancer if not filtered by the ozone layer.

🎯 Exam Tip: Identify ultraviolet radiation as a key harmful EM radiation from space and recall the protective role of the ozone layer.

Question 11. Give reasons for the following:
(i) Long distance radio broadcast uses short wave bands.
(ii) Satellites are used for long distance TV transmission.
Answer:
(i) Long distance radio broadcast uses short wave bands because electromagnetic waves only in the frequency range of short wave bands only are reflected by the ionosphere.
(ii) a. It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere.
b. Hence, satellites are helpful in long distance TV transmission.
In simple words: Short waves are used for long-distance radio because they reflect off the ionosphere, allowing them to travel over the horizon. Satellites are necessary for long-distance TV because high-frequency TV signals pass through the ionosphere, requiring satellite relays to reach distant receivers.

🎯 Exam Tip: Differentiate between ionospheric reflection (for shortwave radio) and ionospheric penetration (for high-frequency TV signals), which necessitates satellite communication for TV.

Question 12. Name the three basic units of any communication system.
Answer: Three basic (essential) elements of every communication system are transmitter, communication channel and receiver.
In simple words: Every communication system fundamentally consists of a transmitter to send information, a channel to carry it, and a receiver to interpret it.

🎯 Exam Tip: Memorize the three fundamental components of any communication system: Transmitter, Communication Channel, and Receiver. This is a basic and frequently asked question.

Question 13. What is a carrier wave?
Answer: The high frequency waves on which the signals to be transmitted are superimposed are called carrier waves.
In simple words: A carrier wave is a high-frequency electromagnetic wave used to transport a low-frequency information signal over long distances by modifying its properties through modulation.

🎯 Exam Tip: Define a carrier wave as a high-frequency wave used to carry information, highlighting its role in making long-distance signal transmission possible.

Question 14. Why high frequency carrier waves are used for transmission of audio signals?
Answer: An audio signal has low frequency (<20 kHz) and low frequency signals cannot be transmitted over large distances. Because of this, a high frequency carrier waves are used for transmission.
In simple words: High-frequency carrier waves are used for audio signal transmission because low-frequency audio signals cannot travel far efficiently and require impractically large antennas; carrier waves overcome these limitations.

🎯 Exam Tip: Explain that low-frequency audio signals have poor radiation efficiency and would require impractically large antennas, making high-frequency carrier waves essential for effective transmission.

Question 15. What is modulation?
Answer: The signals in communication system (e.g. music, speech etc.) are low frequency signals and cannot be transmitted over large distances. In order to transmit the signal to large distances, it is superimposed on a high frequency wave (called carrier wave). This process is called modulation.
In simple words: Modulation is the process of superimposing a low-frequency information signal onto a high-frequency carrier wave to enable efficient long-distance transmission.

🎯 Exam Tip: Clearly define modulation as superimposing a low-frequency signal onto a high-frequency carrier wave for long-distance transmission, and briefly mention why it's necessary (e.g., antenna size, mixing of signals).

Question 16. What is meant by amplitude modulation?
Answer: When the amplitude of carrier wave is varied in accordance with the modulating signal, the process is called amplitude modulation.
In simple words: Amplitude modulation (AM) is a technique where the strength (amplitude) of a high-frequency carrier wave is changed in proportion to the instantaneous amplitude of the information signal.

🎯 Exam Tip: For amplitude modulation, specify that the amplitude of the carrier wave is varied, while its frequency and phase remain constant, according to the modulating signal.

Question 17. What is meant by noise?
Answer:
(1) A random unwanted signal is called noise.
(2) The source generating the noise may be located inside or outside the system.
(3) Efforts should be made to minimize the noise level in a communication system.
In simple words: Noise refers to any unwanted random electrical signal that interferes with the desired signal in a communication system, degrading its quality.

🎯 Exam Tip: Define noise as an unwanted random signal that interferes with communication, and briefly mention its sources and the need for minimization.

Question 18. What is meant by bandwidth?
Answer: The bandwidth of an electronic circuit is the range of frequencies over which it operates efficiently.
In simple words: Bandwidth is the range of frequencies a communication channel or circuit can effectively transmit or process without significant loss.

🎯 Exam Tip: Define bandwidth as the range of frequencies a system can effectively handle, emphasizing its importance in determining the amount of information that can be transmitted.

Question 19. What is demodulation?
Answer: The process of regaining signal from a modulated wave is called demodulation. This is the reverse process of modulation.
In simple words: Demodulation is the inverse process of modulation, where the original information signal is extracted from the modulated carrier wave at the receiver.

🎯 Exam Tip: Clearly state that demodulation is the reverse of modulation, focusing on its function of recovering the original information signal from the modulated wave.

Question 20. What type of modulation is required for television broadcast?
Answer: Amplitude modulation is required for television broadcast.
In simple words: Television broadcasts typically use amplitude modulation for transmitting the video signal, while frequency modulation is used for the audio component.

🎯 Exam Tip: For television, remember that amplitude modulation is primarily used for the video signal, although FM is used for the accompanying audio.

Question 21. How does the effective power radiated by an antenna vary with wavelength?
Answer:
(1) To transmit a signal, an antenna or an aerial is needed.
(2) Power radiated from a linear antenna of length l is, \( P \propto \left(\frac{l}{\lambda}\right)^2 \) where, \( \lambda \) is the wavelength of the signal.
In simple words: The effective power radiated by a linear antenna is inversely proportional to the square of the wavelength of the signal, meaning shorter wavelengths radiate more power for a given antenna length.

🎯 Exam Tip: Remember the relationship \( P \propto \left(\frac{l}{\lambda}\right)^2 \). This implies that for efficient radiation, the antenna length should be comparable to or greater than the wavelength.

Question 22. Why should broadcasting programs use different frequencies?
Answer: If broadcasting programs run on same frequency, then the information carried by these waves will get mixed up with each other. Hence, different broadcasting programs should run on different frequencies.
In simple words: Broadcasting programs must use different frequencies to prevent their signals from interfering with each other, ensuring that listeners or viewers receive clear, distinct content from each source.

🎯 Exam Tip: The primary reason for using different frequencies in broadcasting is to avoid signal interference and ensure clear, distinct reception of various programs.

Question 23. Explain the necessity of a carrier wave in communication.
Answer:
(1) Without a carrier wave, the input signals could be carried by very low frequency electromagnetic waves but it will need quite a bit of amplification in order to transmit those very low frequencies.
(2) The input signals themselves do not have much power and need a fairly large antenna in order to transmit the information.
(3) Hence, it is necessary to impose the input signal on carrier wave as it requires less power in order to transmit the information.
In simple words: Carrier waves are essential because low-frequency information signals cannot be transmitted efficiently over long distances due to large antenna size requirements and poor radiation power; carrier waves effectively 'carry' this information.

🎯 Exam Tip: Highlight the three main reasons for using carrier waves: reducing antenna size, increasing transmission range, and enabling multiplexing (allowing multiple signals on different frequencies).

Question 24. Why does amplitude modulation give noisy reception?
Answer:
(i) In amplitude modulation, carrier is varied in accordance with the message signal.
(ii) The higher the amplitude, the greater is magnitude of the signal. So even if due to any reason, the magnitude of the signal changes, it will lead to variation in the amplitude of the signal. So its easy for noise to disturb the amplitude modulated signal.
In simple words: Amplitude modulation is prone to noise because noise primarily affects the amplitude of a signal. Since the information in AM is encoded in the amplitude, noise easily corrupts the signal, leading to poorer reception quality.

🎯 Exam Tip: Explain that in AM, information is carried by amplitude variations, making it highly susceptible to external noise, which also causes amplitude changes, thus degrading the signal quality.

Question 25. Explain why is modulation needed.
Answer: Modulation helps in avoiding mixing up of signals from different transmitters as different carrier wave frequencies can be allotted to different transmitters. Without the use of these waves, the audio signals, if transmitted directly by different transmitters, would get mixed up.
In simple words: Modulation is crucial because it allows different signals to be transmitted simultaneously on distinct high frequencies without interfering with each other, overcomes limitations of antenna size, and enables long-distance transmission of low-frequency signals.

🎯 Exam Tip: Summarize the key necessities of modulation: avoiding signal mixing, reducing antenna size, and achieving long-distance transmission for low-frequency signals.

3. Solve The Numerical Problem.

Question 1. Calculate the frequency in MHz of a radio wave of wavelength 250 m. Remember that the speed of all EM waves in vacuum is \( 3.0 \times 10^8 \) m/s.
Answer:
Given: \( \lambda = 250 \text{ m, } c = 3 \times 10^8 \text{ m/s} \)
To find: Frequency (v)
Formula: \( c = v \lambda \)
Calculation: From formula,

\( v = \frac{c}{\lambda} = \frac{3 \times 10^8}{250} = 1.2 \times 10^6 \text{ Hz} \)
\( = 1.2 \text{ MHz} \)
In simple words: This problem uses the fundamental wave equation, speed = frequency × wavelength, to calculate the frequency of a radio wave given its wavelength and the speed of light.

🎯 Exam Tip: Always remember the formula \( c = v \lambda \) for electromagnetic waves in a vacuum. Pay close attention to unit conversions, especially converting Hz to MHz for the final answer.

Question 2. Calculate the wavelength in nm of an X-ray wave of frequency \( 2.0 \times 10^{18} \) Hz.
Answer:
Given: \( c = 3 \times 10^8, v = 2 \times 10^{18} \text{ Hz} \)
To find: Wavelength (\( \lambda \))
Formula: \( c = \nu \lambda \)
Calculation. From formula,

\( \lambda = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^{18}} = 1.5 \times 10^{-10} \)
\( = 0.15 \text{ nm} \)
In simple words: This calculation uses the speed of light and the given frequency to find the X-ray's wavelength, converting the result to nanometers.

🎯 Exam Tip: Ensure you correctly rearrange the formula \( c = v \lambda \) to find wavelength. Remember to convert the final answer to the required unit (nanometers) accurately.

Question 3. The speed of light is \( 3 \times 10^8 \) m/s. Calculate the frequency of red light of wavelength of \( 6.5 \times 10^{-7} \) m.
Answer:
Given: \( c = 3 \times 10^8 \text{ m/s, } \lambda = 6.5 \times 10^{-7} \text{ m} \)
To find: Frequency (v)
Formula: \( c = \nu \lambda \)
Calculation: From formula,

\( v = \frac{c}{\lambda} = \frac{3 \times 10^8}{6.5 \times 10^{-7}} = 4.6 \times 10^{14} \text{ Hz} \)
In simple words: This problem calculates the frequency of red light by dividing the speed of light by its given wavelength.

🎯 Exam Tip: Practice working with scientific notation for calculations. Double-check that your wavelength unit matches the speed of light unit (meters) before calculation.

Question 4. Calculate the wavelength of a microwave of frequency 8.0 GHz.
Answer:
Given: \( v = 8 \text{ GHz} = 8 \times 10^9 \text{ Hz, } c = 3 \times 10^8 \text{ m/s} \)
To find: Wavelength (\( \lambda \))
Formula: \( c = \nu \lambda \)
Calculation: From formula,

\( \lambda = \frac{c}{v} = \frac{3 \times 10^8}{8 \times 10^9} = 3.75 \times 10^{-2} \)
\( = 3.75 \text{ cm} \)
In simple words: This problem determines the wavelength of a microwave by dividing the speed of light by its frequency, ensuring proper unit conversion from GHz to Hz and finally to centimeters.

🎯 Exam Tip: Crucially, convert GigaHertz (GHz) to Hertz (Hz) before applying the formula \( \lambda = \frac{c}{v} \). Also, be ready to express the final wavelength in common units like centimeters.

Question 5. In a EM wave the electric field oscillates sinusoidally at a frequency of \( 2 \times 10^{10} \) What is the wavelength of the wave?
Answer:
Given: \( v = 2 \times 10^{10} \text{ Hz, } c = 3 \times 10^8 \text{ m/s} \)
To find: Wavelength (\( \lambda \))
Formula: \( c = \nu \lambda \)
Calculation: From formula,

\( \lambda = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^{10}} = 1.5 \times 10^{-2} \)
In simple words: This calculation determines the wavelength of an electromagnetic wave by using the given oscillation frequency and the speed of light in a vacuum.

🎯 Exam Tip: For EM waves in vacuum, assume the speed of light \( c = 3 \times 10^8 \text{ m/s} \) unless otherwise specified. Accurately handle powers of ten during calculations.

Question 6. The amplitude of the magnetic field part of a harmonic EM wave in vacuum is \( B_0 = 5 \times 10^{-7} \) T. What is the amplitude of the electric field part of the wave?
Answer:
Given: \( B_0 = 5 \times 10^{-7} \text{ T, } c = 3 \times 10^8 \)
To find: Amplitude of electric field (E0)
Formula: \( c = \frac{E_0}{B_0} \)
Calculation /From formula,

\( E_0 = c \times B_0 \)
\( = 3 \times 10^8 \times 5 \times 10^{-7} \)
\( = 150 \text{ V/m} \)
In simple words: This problem uses the fundamental relationship between the amplitudes of the electric and magnetic fields in an electromagnetic wave and the speed of light to find the electric field amplitude.

🎯 Exam Tip: Remember the relationship \( c = \frac{E_0}{B_0} \) which connects the peak electric field \( E_0 \), peak magnetic field \( B_0 \), and the speed of light \( c \). This is a crucial formula for EM waves.

Question 7. A TV tower has a height of 200 m. How much population is covered by TV transmission if the average population density around the tower is 1000/km²? (Radius of the Earth = \( 6.4 \times 10^6 \) m)
Answer:
Given: \( h = 200 \text{ m, } \)
Population density \( n = 1000/\text{km}^2 = 1000 \times 10^{-6}/\text{m}^2 = 10^{-3}/\text{m}^2 \)
\( R = 6.4 \times 10^6 \text{ m} \)
To find: Population covered
Formulae: (i) \( A = \pi d^2 = \pi (\sqrt{2Rh})^2 = 2\pi Rh \)
(ii) Population covered = \( nA \)
Calculation /From formula (i),

\( A = 2\pi Rh \)
\( = 2 \times 3.142 \times 6.4 \times 10^6 \times 200 \)
\( \approx 8 \times 10^9 \text{ m}^2 \)
From formula (ii),

Population covered = \( nA \)
\( = 10^{-3} \times 8 \times 10^9 \)
\( = 8 \times 10^6 \)
In simple words: This problem calculates the population covered by a TV tower by first finding the transmission area using the tower's height and Earth's radius, then multiplying by the given population density.

🎯 Exam Tip: For such problems, ensure consistent units (e.g., convert km² to m² or vice versa). Accurately calculate the area of coverage using \( A = 2\pi Rh \) before multiplying by population density.

Question 8. Height of a TV tower is 600 m at a given place. Calculate its coverage range if the radius of the Earth is 6400 km. What should be the height to get the double coverage area?
Answer:
Given: \( h = 600 \text{ m, } R = 6.4 \times 10^6 \text{ m} \)
To find: Range (d)
Height to get the double coverage (h')
Formula: \( d = \sqrt{2hR} \)
Calculation: From formula,

\( d = \sqrt{2 \times 600 \times 6.4 \times 10^6} = 87.6 \times 10^3 = 87.6 \text{ km} \)
Now, for \( A' = 2A \)

\( \pi (d')^2 = 2 (\pi d^2) \)

\( \therefore (d')^2 = 2d^2 \)
From formula,

\( h' = \frac{(d')^2}{2R} \)

\( = \frac{2d^2}{2R} \)

\( = 2 \times h \quad \left( h = \frac{d^2}{2R} \right) \)
\( = 2 \times 600 \)
\( = 1200 \text{ m} \)
In simple words: This problem first calculates the initial coverage range of a TV tower. Then, by understanding that doubling the coverage area means doubling the square of the range, it determines the new height required for that increased coverage.

🎯 Exam Tip: When doubling the coverage area, remember that the range \(d\) increases by a factor of \( \sqrt{2} \). Since \(h \propto d^2\), doubling the area means doubling \(h\). Be careful with unit conversions, especially for Earth's radius (km to m).

Question 9. A transmitting antenna at the top of a tower has a height 32 m and that of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight mode? Given radius of Earth is \( 6.4 \times 10^6 \) m.
Answer:
Given: \( h_t = 32 \text{ m, } h_r = 50 \text{ m, } R = 6.4 \times 10^6 \text{ m} \)
To find: Maximum distance or range (d)
Formula: \( d = \sqrt{2Rh} \)
Calculation: From formula,

\( d_t = \sqrt{2Rh_t} = \sqrt{2 \times 6.4 \times 10^6 \times 32} \)
\( = 20.238 \times 10^3 \text{ m} \)
\( = 20.238 \text{ km} \)

\( d_r = \sqrt{2Rh_r} \)
\( = \sqrt{2 \times 6.4 \times 10^6 \times 50} \)
\( = 25.298 \times 10^3 \text{ m} \)
\( = 25.298 \text{ km} \)
Now, \( d = d_t + d_r \)
\( = 20.238 + 25.298 \)
\( = 45.536 \text{ km} \)
In simple words: This problem calculates the total line-of-sight communication range between two antennas by summing the individual ranges from each antenna, considering their heights and the Earth's curvature.

🎯 Exam Tip: For line-of-sight communication between two antennas, calculate the range for each antenna separately using \( d = \sqrt{2Rh} \) and then add them to get the total maximum distance. Pay attention to units.

Can You Recall? (Textbookpage No. 229)

Question 1.
(i) What is a wave?
Answer: Wave is an oscillatory disturbance which travels through a medium without change in its form.
In simple words: A wave is a disturbance that transfers energy through a medium or space without transferring matter.

🎯 Exam Tip: A fundamental definition, remember that a wave is a disturbance that propagates and transfers energy, not matter.

Question 1.
(ii) What is the difference between longitudinal and transverse waves?
Answer:
a. Transverse wave: A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of wave is called transverse wave.
b. Longitudinal wave: A wave in which particles of the medium vibrate in a direction parallel to the direction of propagation of wave is called longitudinal wave.
In simple words: In transverse waves, particles oscillate perpendicular to wave direction (like ocean waves), while in longitudinal waves, particles oscillate parallel to wave direction (like sound waves).

🎯 Exam Tip: Clearly differentiate transverse (oscillation perpendicular to propagation) and longitudinal (oscillation parallel to propagation) waves, perhaps with simple examples like light vs. sound.

Question 1.
(iii) What are electric and magnetic fields and what are their sources?
Answer:
a. Electric field is the force experienced by a test charge in presence of the given charge at the given distance from it.
b. A magnetic field is produced around a magnet or around a current carrying conductor.
In simple words: An electric field is a region where electric charges experience force, originating from charges; a magnetic field is a region where magnetic materials or moving charges experience force, originating from magnets or electric currents.

🎯 Exam Tip: Define electric and magnetic fields in terms of the forces they exert on charges/magnets and their respective sources (static/moving charges, magnets, current-carrying conductors).

Question 1.
(iv) By which mechanism heat is lost by hot bodies?
Answer: Hot bodies lose the heat in the form of radiation.
In simple words: Hot bodies lose heat primarily through thermal radiation, which involves the emission of electromagnetic waves, in addition to conduction and convection.

🎯 Exam Tip: Remember the three modes of heat transfer: conduction, convection, and radiation. For heat loss from hot bodies, radiation is a key mechanism, especially in the absence of a medium.

Question 2. What are Lenz's law, Ampere's law and Faraday's law?
Answer:
Lenz's law: Whereas, Lenz's law states that, the direction of the induced emf is such that the change is opposed.
Ampere's law: Ampere's law describes the relation between the induced magnetic field associated with a loop and the current flowing through the loop.
Faraday's law: Faraday's law states that, time varying magnetic field induces an electromotive force (emf) and an electric field.
In simple words: Lenz's law describes the direction of induced current opposing the change that caused it. Ampere's law relates magnetic fields to the electric currents producing them. Faraday's law describes how a changing magnetic field induces an electromotive force and an electric field.

🎯 Exam Tip: Memorize the core statement of each law. Lenz's law is about opposition, Ampere's law about current-magnetic field relation, and Faraday's law about changing magnetic flux inducing EMF/electric field.

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