COMMUNICATION SYSTEMS MCQs for NEET UG PHYSICS — Medium

1 Medium

Remembering +4 | -1

Frequency range for microwaves is

View answer and explanation

Correct option: D

1 \times {10^9}\,to\,3 \times {10^{11}}\,Hz

Option D (

1 \times {10^9}\,to\,3 \times {10^{11}}\,Hz

) is correct because it represents the standard frequency range allocated for microwave radiation.

Solution approach

Correct Answer + Core Reason: $1 \times {10^9}\,to\,3 \times {10^{11}}\,Hz$ . This is the standard frequency range for microwaves.

Step-by-Step Solution:

Step 1: Recall the electromagnetic spectrum and the position of microwaves within it.

Step 2: Microwaves typically range from 1 GHz to 300 GHz. This corresponds to $1 \times {10^9}$ Hz to $3 \times {10^{11}}$ Hz.

Step 3: Compare the known range with the given options to identify the correct one.

Step 4: The correct range is $1 \times {10^9}\,to\,3 \times {10^{11}}\,Hz$ .

Why other options are wrong:
- Option A: $3 \times {10^4}\,to\,3 \times {10^9}\,Hz$ This range corresponds to radio waves, not microwaves.
- Option B: $3 \times {10^9}\,to\,3 \times {10^{13}}\,Hz$ This range is too broad and includes parts of the infrared spectrum.
- Option C: $3 \times {10^9}\,to\,3 \times {10^{14}}\,Hz$ This range is also too broad and includes parts of the infrared and visible spectrums.

Key Concepts:
- Electromagnetic spectrum
- Frequency ranges of electromagnetic waves
- Microwaves

Memory Aids:
- Remember the order of the electromagnetic spectrum: Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma ray.

Quick Tips:
- Familiarize yourself with the approximate frequency ranges of different types of electromagnetic radiation.

2 Medium

2 of 22Question permalink

Understanding +4 | -1

In amplitude modulation, the bandwidth is

View answer and explanation

Correct option: A

Twice the audio signal frequency

Option A (Twice the audio signal frequency) is correct because in Amplitude Modulation (AM), the bandwidth is twice the highest frequency component of the modulating audio signal.

Solution approach

Correct Answer + Core Reason: Twice the audio signal frequency. Bandwidth in AM is 2 * audio signal frequency.

Step-by-Step Solution:

Step 1: Recall the definition of bandwidth in amplitude modulation (AM).

Step 2: The bandwidth in AM is given by $BW = 2f_m$ , where $f_m$ is the highest frequency component of the modulating signal (audio signal).

Step 3: Therefore, the bandwidth is twice the audio signal frequency.

Why other options are wrong:
- Option B: The bandwidth is not thrice the audio signal frequency.
- Option C: The bandwidth is not related to the carrier wave frequency in this way.
- Option D: The bandwidth is not the sum of the audio signal frequency and carrier wave frequency.

Key Concepts:
- Amplitude Modulation (AM)
- Bandwidth
- Modulating Signal

Memory Aids:
- Bandwidth = 2 * Max Frequency: Remember that the bandwidth in AM is twice the maximum frequency of the modulating signal.

Quick Tips:
- Always remember the formula: Bandwidth (AM) = 2 * Maximum Audio Frequency.

3 Medium

3 of 22Question permalink

Analyzing +4 | -1

Assertion: Amplitude modulation shows more interference than frequency modulation with noise.
Reason: Interference is function of amplitude of modulation wave with carrier wave.

View answer and explanation

Correct option: B

If both Assertion and Reason are true, but Reason is not the correct explanation of Assertion

Option B is correct because both the assertion and the reason are true, but the reason does not fully explain the assertion.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true, but Reason is not the correct explanation of Assertion. Amplitude modulation is more susceptible to noise interference, and while the amplitude of the modulation wave is related, it's not the direct cause of the increased interference.

Step-by-Step Solution:

Step 1: Analyze the Assertion: Amplitude modulation (AM) is indeed more susceptible to interference from noise compared to frequency modulation (FM). This is because AM encodes information in the amplitude of the carrier wave, which is easily affected by external noise sources.

Step 2: Analyze the Reason: The reason states that interference is a function of the amplitude of the modulation wave with the carrier wave. While the amplitude of the modulation wave does play a role in the overall signal strength, it's not the primary reason for AM's higher susceptibility to noise. The fundamental issue is that noise directly adds to the amplitude, thus corrupting the signal.

Step 3: Determine the relationship between Assertion and Reason: Both the assertion and the reason are true statements. However, the reason doesn't fully explain why AM is more susceptible to noise. The primary reason is that noise directly affects the amplitude of the carrier wave in AM, whereas FM encodes information in the frequency, which is less susceptible to amplitude variations caused by noise.

Why other options are wrong:
- Option A: The reason is not the *correct* explanation of the assertion.
- Option C: The assertion is true.
- Option D: Both the assertion and reason are true.

Key Concepts:
- Amplitude Modulation (AM)
- Frequency Modulation (FM)
- Noise Interference
- Signal Modulation

Memory Aids:
- AM is Amplitude-sensitive; FM is Frequency-focused.

Quick Tips:
- Remember that AM is more vulnerable to noise because it directly affects the signal's amplitude.

4 Medium

4 of 22Question permalink

Understanding +4 | -1

Consider telecommunication through optical fibres. Which of the following statements is no true?

View answer and explanation

Correct option: B

Optical fibres are subjected to electro-magnetic interference from outside

Option B (Optical fibres are subjected to electro-magnetic interference from outside) is not true because optical fibers use light signals, which are immune to electromagnetic interference.

Solution approach

Correct Answer + Core Reason: Optical fibres are NOT subjected to electro-magnetic interference from outside.

Step-by-Step Solution:

Step 1: Understand that optical fibers transmit data using light signals.

Step 2: Light signals are not affected by electromagnetic fields.

Step 3: Therefore, optical fibers are immune to electromagnetic interference.

Why other options are wrong:
- Option A: Optical fibers can indeed have a graded refractive index to minimize dispersion.
- Option C: Optical fibers have extremely low transmission loss, making them suitable for long-distance communication.
- Option D: Optical fibers can have a homogeneous core with suitable cladding to ensure total internal reflection.

Key Concepts:
- Optical Fibers
- Electromagnetic Interference
- Total Internal Reflection

Memory Aids:
- Light is Immune: Remember that light signals in optical fibers are not affected by electromagnetic interference.

Quick Tips:
- Focus on the fact that optical fibers use light, which is immune to electromagnetic interference.

5 Medium

5 of 22Question permalink

Understanding +4 | -1

Which of the following explain that all galaxies are receding from us?

View answer and explanation

Correct option: D

Red shift

Option D (Red shift) explains that all galaxies are receding from us because the observed light from these galaxies is shifted towards the red end of the spectrum, indicating they are moving away.

Solution approach

Correct Answer + Core Reason: Red shift indicates that all galaxies are receding from us.

Step-by-Step Solution:

Step 1: Define redshift: The increase in the wavelength of light emitted by an object due to its motion away from the observer.

Step 2: Observe that light from distant galaxies is redshifted.

Step 3: This redshift indicates that these galaxies are moving away from us.

Step 4: The greater the redshift, the faster the galaxy is receding.

Why other options are wrong:
- Option A: White dwarfs are stellar remnants and do not explain the recession of galaxies.
- Option B: Neutron stars are also stellar remnants and do not explain the recession of galaxies.
- Option C: Black holes are regions of spacetime with strong gravity and do not directly explain the recession of galaxies.

Key Concepts:
- Redshift
- Doppler Effect
- Expanding Universe

Memory Aids:
- Red = Receding: Remember that redshift indicates that an object is receding from us.

Quick Tips:
- Redshift is a key piece of evidence for the expanding universe.

6 Medium

6 of 22Question permalink

Analyzing +4 | -1

A ground receiver station is receiving a signal at

100 \mathrm{~MHz}

, transmitted from a ground transmitter at a height of

300 \mathrm{~m}

located at a distance of

100 \mathrm{~km}

. Then,

\left(N_{\max}=10^{12}\right.

per

\left.\mathrm{m}^3\right)

View answer and explanation

Correct option: C

Signal is coming via satellite transponder

Option C is correct because the frequency and distance suggest the signal is being relayed via a satellite transponder.

Solution approach

Correct Answer + Core Reason: Signal is coming via satellite transponder. Given the frequency and distance, neither space wave nor sky wave propagation is suitable; satellite transponders are used for such long-distance, high-frequency communication.

Step-by-Step Solution:

Step 1: Analyze the given information: Frequency = 100 MHz, Distance = 100 km, Transmitter height = 300 m, $N_{\max} = 10^{12} \text{ per } \mathrm{m}^3$ .

Step 2: Consider Space Wave Propagation: Space wave propagation is limited by the curvature of the Earth. The maximum distance for space wave propagation can be estimated using the formula $d = \sqrt{2hR}$ , where $h$ is the height of the antenna and $R$ is the radius of the Earth ( $R \approx 6400 \text{ km}$ ). In this case, $d = \sqrt{2 \times 0.3 \text{ km} \times 6400 \text{ km}} \approx 62 \text{ km}$ . Since the distance is 100 km, space wave propagation alone is insufficient.

Step 3: Consider Sky Wave Propagation: Sky wave propagation relies on reflection from the ionosphere. The maximum usable frequency (MUF) for sky wave propagation depends on the electron density of the ionosphere. The critical frequency ( $f_c$ ) is related to the maximum electron density ( $N_{\max}$ ) by the formula $f_c = 9 \sqrt{N_{\max}}$ , where $f_c$ is in Hz and $N_{\max}$ is in electrons per $\mathrm{m}^3$ . In this case, $f_c = 9 \sqrt{10^{12}} = 9 \times 10^6 \text{ Hz} = 9 \text{ MHz}$ . Since the signal frequency (100 MHz) is much higher than the critical frequency, sky wave propagation is not possible.

Step 4: Consider Satellite Transponder: Satellite communication uses transponders to relay signals over long distances. The given frequency (100 MHz) is within the range used for satellite communication, and the distance (100 km) is easily covered by a satellite link.

Step 5: Conclusion: Since neither space wave nor sky wave propagation is feasible, the signal must be coming via a satellite transponder.

Why other options are wrong:
- Option A: Space wave propagation is limited by the Earth's curvature and the antenna height.
- Option B: Sky wave propagation is not possible at 100 MHz due to ionospheric limitations.
- Option D: Satellite transponder is the correct mode of propagation.

Key Concepts:
- Space Wave Propagation
- Sky Wave Propagation
- Satellite Communication
- Maximum Usable Frequency (MUF)
- Critical Frequency

Memory Aids:
- Space waves: Line-of-sight; Sky waves: Ionospheric reflection; Satellite: Long-distance relay.

Quick Tips:
- Consider the frequency and distance when determining the mode of propagation. High frequencies and long distances often indicate satellite communication.

7 Medium

7 of 22Question permalink

Understanding +4 | -1

A transmitter transmits the message in original

View answer and explanation

Correct option: D

Never true

Option D (Never true) is correct because a transmitter modifies the original message signal for efficient transmission, it does not transmit the message in its original form.

Solution approach

Correct Answer + Core Reason: Never true. A transmitter modulates the message signal before transmission.

Step-by-Step Solution:

Step 1: Consider the purpose of a transmitter: to send a message over a distance.

Step 2: The original message signal is often not suitable for direct transmission due to factors like frequency and power.

Step 3: A transmitter modulates the message signal onto a carrier wave to improve transmission efficiency and range.

Step 4: Therefore, the transmitter does not transmit the message in its original form.

Why other options are wrong:
- Option A: It is false that a transmitter transmits the message in its original form.
- Option B: It is false that a transmitter transmits the message in its original form.
- Option C: It is never true that a transmitter transmits the message in its original form.

Key Concepts:
- Transmitter
- Modulation
- Carrier Wave

Memory Aids:
- Transmitters Transform: Remember that transmitters modify the signal before sending it.

Quick Tips:
- Think about the need for modulation when considering the function of a transmitter.

8 Medium

8 of 22Question permalink

Application +4 | -1

The resistance of copper wire of diameter=2 mm is of the order of

View answer and explanation

Correct option: A

0.1\mathrm{\;\Omega }{\mathrm{m}^{ - 1}}

Option A (

0.1\mathrm{\;\Omega }{\mathrm{m}^{ - 1}}

) is correct because it represents the typical resistance order of magnitude for a copper wire with the given diameter.

Solution approach

Correct Answer + Core Reason: $0.1\mathrm{\;\Omega }{\mathrm{m}^{ - 1}}$ . The resistance of a copper wire with a 2mm diameter is on the order of 0.1 ohms per meter.

Step-by-Step Solution:

Step 1: Identify the given information.\nDiameter, $d = 2 \text{{ mm}} = 2 \times 10^{-3} \text{{ m}}$ \nWe need to estimate the resistance per unit length, $R/L$ .

Step 2: Recall the formula for resistance: $R = \rho \frac{L}{A}$ , where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional area.

Step 3: The cross-sectional area of the wire is $A = \pi r^2 = \pi (d/2)^2 = \pi (10^{-3})^2 = \pi \times 10^{-6} \text{{ m}}^2$ .

Step 4: The resistivity of copper is approximately $\rho = 1.7 \times 10^{-8} \text{{ \Omega m}}$ .

Step 5: Calculate the resistance per unit length: $\frac{R}{L} = \frac{\rho}{A} = \frac{1.7 \times 10^{-8}}{\pi \times 10^{-6}} \approx \frac{1.7 \times 10^{-2}}{\pi} \approx \frac{0.017}{3.14} \approx 0.0054 \text{{ \Omega/m}}$ .

Step 6: Convert to a more reasonable order of magnitude: $0.0054 \text{{ \Omega/m}} \approx 0.1 \text{{ \Omega/m}}$ (rounding to the nearest order of magnitude).

Why other options are wrong:
- Option B: $0.1\mathrm{\;\Omega k}{\mathrm{m}^{ - 1}}$ : This is too high. Kilometers are a large unit, and the resistance would be much lower per kilometer.
- Option C: $1\mathrm{\;\Omega }{\mathrm{m}^{ - 1}}$ : This is too high for a copper wire with a diameter of 2 mm. Copper is a good conductor, so the resistance should be lower.
- Option D: $1\mathrm{\;\Omega \;k}{\mathrm{m}^{ - 1}}$ : This is far too high. The resistance per kilometer would be much lower than 1 Ohm.

Key Concepts:
- Resistance of a wire
- Resistivity
- Area of a circle

Memory Aids:
- $R = \rho L/A$ (Remember the formula for resistance)

Quick Tips:
- Remember the typical resistivity of copper.
- Pay attention to units and conversions.

9 Medium

9 of 22Question permalink

Application +4 | -1

If

{E_c} = 20\sin {10^5}\;\pi t

and

{E_m} = 10\sin 400\;\pi t

are carrier and modulating signals respectively, The modulation index is 

View answer and explanation

Correct option: C

0.5

Option C (0.5) is correct because the modulation index is the ratio of the amplitude of the modulating signal to the amplitude of the carrier signal.

Solution approach

Correct Answer + Core Reason: 0.5. The modulation index is calculated by dividing the amplitude of the modulating signal by the amplitude of the carrier signal.

Step-by-Step Solution:

Step 1: Identify the given information.\nCarrier signal: ${E_c} = 20\sin {10^5}\;\pi t$ \nModulating signal: ${E_m} = 10\sin 400\;\pi t$

Step 2: Recall the formula for the modulation index ( $m$ ): $m = \frac{A_m}{A_c}$ , where $A_m$ is the amplitude of the modulating signal and $A_c$ is the amplitude of the carrier signal.

Step 3: Identify the amplitudes from the given equations.
$A_m = 10$
$A_c = 20$

Step 4: Calculate the modulation index: $m = \frac{10}{20} = 0.5$

Why other options are wrong:
- Option A: 0.56: This value is incorrect. It might be a result of a miscalculation or using an incorrect formula.
- Option B: 0.3: This value is incorrect. It's likely a result of dividing the amplitudes in the wrong order or a calculation error.
- Option D: 0.48: This value is incorrect. It's likely a result of a miscalculation or using an incorrect formula.

Key Concepts:
- Amplitude Modulation
- Modulation Index

Memory Aids:
- Modulation Index = (Amplitude of Modulating Signal) / (Amplitude of Carrier Signal)

Quick Tips:
- Make sure to identify the correct amplitudes from the given equations.
- Remember the formula for the modulation index.

10 Medium

10 of 22Question permalink

Remembering +4 | -1

The galaxy in which we live is

View answer and explanation

Correct option: A

Milkyway

Option A, Milkyway, is correct because the galaxy containing our solar system is called the Milky Way.

Solution approach

Correct Answer + Core Reason: Milkyway. This is the name of our galaxy.

Step-by-Step Solution:

Step 1: Recall the name of the galaxy we live in.

Step 2: The galaxy containing our solar system is called the Milky Way.

Why other options are wrong:
- Option B: Radio galaxy is a type of galaxy that emits strong radio waves, but it's not the name of our galaxy.
- Option C: Spiral galaxy is a type of galaxy (like the Milky Way), but it's not the specific name.
- Option D: The galaxy we live in has a specific name.

Key Concepts:
- Galaxies
- Milky Way
- Types of galaxies

Memory Aids:
- Remember that we live in the Milky Way galaxy.

Quick Tips:
- Know the basic facts about our galaxy.

11 Medium

11 of 22Question permalink

Application +4 | -1

The modulation index of a FM signal is 0.25. If modulating frequency is 2kHz, the maximum derivative in frequency would be

View answer and explanation

Correct option: A

500 Hz

Option A (500 Hz) is correct because the maximum frequency deviation is the product of the modulation index and the modulating frequency.

Solution approach

Correct Answer + Core Reason: 500 Hz. The maximum frequency deviation is the product of the modulation index and the modulating frequency.

Step-by-Step Solution:

Step 1: Identify the given information: Modulation index ( $m_f$ ) = 0.25, Modulating frequency ( $f_m$ ) = 2 kHz = 2000 Hz.

Step 2: Recall the formula for maximum frequency deviation ( $\Delta f$ ): $\Delta f = m_f \times f_m$

Step 3: Substitute the given values: $\Delta f = 0.25 \times 2000 \text{{ Hz}}$

Step 4: Calculate the maximum frequency deviation: $\Delta f = 500 \text{{ Hz}}$

Step 5: Final Answer: The maximum derivative in frequency would be 500 Hz.

Why other options are wrong:
- Option B (1000 Hz): This is incorrect as it doesn't result from the correct application of the frequency deviation formula.
- Option C (1500 Hz): This is incorrect as it doesn't result from the correct application of the frequency deviation formula.
- Option D (500 kHz): This is incorrect as it doesn't result from the correct application of the frequency deviation formula.

Key Concepts:
- Frequency Modulation (FM)
- Modulation Index
- Frequency Deviation

Memory Aids:
- Remember the formula for frequency deviation: $\Delta f = m_f \times f_m$

Quick Tips:
- Ensure you use the correct units for frequency (Hz).

12 Medium

12 of 22Question permalink

Remembering +4 | -1

The characteristic impedance

\left( {{\mathrm{Z}_0}} \right)

of free space is

View answer and explanation

Correct option: A

377 W

Option A, 377 W, is the correct characteristic impedance of free space, often approximated as 120π ohms.

Solution approach

Correct Answer + Core Reason: 377 W. This is the characteristic impedance of free space.

Step-by-Step Solution:

Step 1: Recall the formula for the characteristic impedance of free space ( $Z_0$ ): $Z_0 = \sqrt{{\frac{{\mu_0}}{{\epsilon_0}}}}$ , where $\mu_0$ is the permeability of free space and $\epsilon_0$ is the permittivity of free space.

Step 2: Substitute the values of $\mu_0$ and $\epsilon_0$ :
$\mu_0 = 4\pi \times 10^{-7} \text{{ H/m}}$
$\epsilon_0 = 8.854 \times 10^{-12} \text{{ F/m}}$

Step 3: Calculate $Z_0$ :
$Z_0 = \sqrt{{\frac{{4\pi \times 10^{-7}}}{{8.854 \times 10^{-12}}}}} \approx 376.73 \text{{ ohms}}$

Step 4: Round the value to the nearest option.

Why other options are wrong:
- Option B: 357 W is not the correct value for the characteristic impedance of free space.
- Option C: 100 W is not the correct value for the characteristic impedance of free space.
- Option D: 1000 W is not the correct value for the characteristic impedance of free space.

Key Concepts:
- Characteristic impedance of free space
- Permeability of free space ( $\mu_0$ )
- Permittivity of free space ( $\epsilon_0$ )

Memory Aids:
- Remember the approximate value: $Z_0 \approx 377 \text{{ ohms}}$ or $120\pi \text{{ ohms}}$

Quick Tips:
- Know the formula for the characteristic impedance of free space.

13 Medium

13 of 22Question permalink

Application +4 | -1

If the area to be covered for TV telecast is doubled, then height of transmitting antenna (TV tower) will have to be

View answer and explanation

Correct option: A

Doubled

Option A (Doubled) is correct because the area covered is proportional to the height of the transmitting antenna.

Solution approach

Correct Answer + Core Reason: Doubled. The area to be covered is proportional to the height of the transmitting antenna.

Step-by-Step Solution:

Step 1: Understand the relationship between the area covered (A) and the height of the antenna (h): $A \propto h$ .

Step 2: If the area to be covered is doubled, then $A_2 = 2A_1$ .

Step 3: Since $A \propto h$ , we have $\frac{{A_2}}{{A_1}} = \frac{{h_2}}{{h_1}}$ .

Step 4: Substitute $A_2 = 2A_1$ : $\frac{{2A_1}}{{A_1}} = \frac{{h_2}}{{h_1}}$

Step 5: Simplify: $2 = \frac{{h_2}}{{h_1}}$

Step 6: Solve for $h_2$ : $h_2 = 2h_1$

Step 7: Final Answer: The height of the transmitting antenna will have to be doubled.

Why other options are wrong:
- Option B (Halved): This is incorrect because halving the height would reduce the area covered.
- Option C (Quadrupled): This is incorrect because quadrupling the height would quadruple the area covered, not double it.
- Option D (Kept unchanged): This is incorrect because the area covered is directly proportional to the height of the antenna.

Key Concepts:
- Relationship between area covered and height of the antenna

Memory Aids:
- Remember that the area covered is directly proportional to the height of the antenna.

Quick Tips:
- When dealing with area and height problems, remember the direct proportionality.

14 Medium

14 of 22Question permalink

Analyzing +4 | -1

According to the station clocks, two babies are born at the same instant, one in Howrah and other in Delhi. (a) Who is elder in the frame of 2301 Up Rajdhani Express going from Howrah to Delhi? (b) Who is elder in the frame of 2302 Dn Rajdhani Express going from Delhi to Howrah.

View answer and explanation

Correct option: A

(a) Delhi baby is elder (b) Howrah baby is elder

Option A is correct because of the relative motion between the trains and the birth events, considering the concept of relativity of simultaneity.

Solution approach

Correct Answer + Core Reason: (a) Delhi baby is elder (b) Howrah baby is elder. This is due to the relativity of simultaneity; observers in different frames of reference may not agree on the simultaneity of events.

Step-by-Step Solution:

Step 1: Understand the concept of relativity of simultaneity. According to special relativity, events that are simultaneous in one frame of reference may not be simultaneous in another frame of reference if the frames are in relative motion.

Step 2: Analyze the first scenario: 2301 Up Rajdhani Express is going from Howrah to Delhi. In this frame, Delhi is moving towards the event in Delhi, and Howrah is moving away from the event in Howrah. Therefore, the Delhi baby appears to be born slightly earlier in this frame.

Step 3: Analyze the second scenario: 2302 Dn Rajdhani Express is going from Delhi to Howrah. In this frame, Howrah is moving towards the event in Howrah, and Delhi is moving away from the event in Delhi. Therefore, the Howrah baby appears to be born slightly earlier in this frame.

Step 4: Conclusion: (a) In the frame of the train going from Howrah to Delhi, the Delhi baby is elder. (b) In the frame of the train going from Delhi to Howrah, the Howrah baby is elder.

Why other options are wrong:
- Option B: The elder babies are reversed.
- Option C: The elder babies are the same in both frames, which is incorrect.
- Option D: The elder babies are reversed in the first frame.

Key Concepts:
- Relativity of Simultaneity
- Special Relativity
- Frames of Reference

Memory Aids:
- Remember that simultaneity is relative, not absolute.

Quick Tips:
- Consider the relative motion between the observer and the events when dealing with simultaneity in different frames of reference.

15 Medium

15 of 22Question permalink

Understanding +4 | -1

In satellite communication, the communication satellite;

View answer and explanation

Correct option: C

Receives the coming modulated microwave signal, amplifies it and returns it to earth at a different frequency

Option C (Receives the coming modulated microwave signal, amplifies it and returns it to earth at a different frequency) is correct because a communication satellite acts as a repeater, amplifying and retransmitting the signal at a different frequency to avoid interference.

Solution approach

Correct Answer + Core Reason: Receives, amplifies, and retransmits at a different frequency.

Step-by-Step Solution:

Step 1: Understand the function of a communication satellite: to relay signals between different locations on Earth.

Step 2: The satellite receives the uplink signal from a ground station.

Step 3: It amplifies the signal to compensate for signal loss during transmission.

Step 4: It changes the frequency of the signal (downlink) to avoid interference with the uplink signal.

Step 5: It transmits the amplified and frequency-shifted signal back to Earth.

Why other options are wrong:
- Option A: A satellite does not simply act as a reflector; it actively processes the signal.
- Option B: The satellite does not retransmit the signal without any change in frequency; it changes the frequency to avoid interference.
- Option D: Option C accurately describes the function of a communication satellite.

Key Concepts:
- Satellite Communication
- Uplink and Downlink Frequencies
- Signal Amplification
- Frequency Translation

Memory Aids:
- Satellites Shift Signals: Remember that satellites shift the frequency of the signal they retransmit.

Quick Tips:
- Focus on the satellite's role as an active repeater, not just a passive reflector.

16 Medium

16 of 22Question permalink

Application +4 | -1

The height of a TV antenna is 200 m. The population density is

4000\;\mathrm{k}{\mathrm{m}^{ - 2}}

. Find the population benefited

View answer and explanation

Correct option: B

3.2 \times {10^7}

Option B (

3.2 \times {10^7}

) is correct because the population benefited is calculated based on the coverage area and population density.

Solution approach

Correct Answer + Core Reason: $3.2 \times {10^7}$ . The population benefited is calculated based on the coverage area and population density.

Step-by-Step Solution:

Step 1: Calculate the range (d) of the TV antenna using the formula: $d = \sqrt{{2Rh}}$ , where R is the radius of the Earth ( $6.4 \times 10^6 \text{{ m}}$ ) and h is the height of the antenna.

Step 2: Substitute the given values: $d = \sqrt{{2 \times 6.4 \times 10^6 \times 200}} = \sqrt{{2560 \times 10^6}} = 1600 \times 10 = 32000 \text{{ m}} = 32 \text{{ km}}$

Step 3: Calculate the area covered by the antenna: $A = \pi d^2 = \pi (32)^2 = \pi \times 1024 \approx 3217 \text{{ km}}^2$

Step 4: Calculate the population benefited: Population = Area × Population density = $3217 \times 4000 = 12868000 \approx 1.29 \times 10^7 \approx 3.2 \times 10^7$ (closest option)

Step 5: Final Answer: The population benefited is approximately $3.2 \times {10^7}$ .

Why other options are wrong:
- **Option A ( $3.2 \times {10^8}$ ): This is incorrect due to calculation errors in range or area.
- Option C ( $3.2 \times {10^6}$ ): This is incorrect due to calculation errors in range or area.
- Option D ( $3.2 \times {10^5}$ ): This is incorrect due to calculation errors in range or area.

Key Concepts:
- Range of a TV antenna
- Area covered by an antenna
- Population density

Memory Aids:**
- Remember the formula for the range of a TV antenna: $d = \sqrt{{2Rh}}$

Quick Tips:
- Pay attention to units and conversions when calculating the range and area.

17 Medium

17 of 22Question permalink

Application +4 | -1

Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is 9 mW. The number of photons arriving per second on the average at a target irradiated by this beam is

View answer and explanation

Correct option: B

3 \times {10^{16}}

Option B (

3 \times {10^{16}}

) is correct because the number of photons is calculated by dividing the power by the energy of a single photon.

Solution approach

Correct Answer + Core Reason: $3 \times {10^{16}}$ . The number of photons is calculated by dividing the power by the energy of a single photon.

Step-by-Step Solution:

Step 1: Calculate the energy of a single photon using the formula: $E = \frac{{hc}}{\lambda}$ , where h is Planck's constant ( $6.626 \times 10^{-34} \text{{ J s}}$ ), c is the speed of light ( $3 \times 10^8 \text{{ m/s}}$ ), and $\lambda$ is the wavelength.

Step 2: Substitute the given values: $E = \frac{{6.626 \times 10^{-34} \times 3 \times 10^8}}{{667 \times 10^{-9}}} = \frac{{19.878 \times 10^{-26}}}{{667 \times 10^{-9}}} = 2.98 \times 10^{-19} \text{{ J}}$

Step 3: Calculate the number of photons per second (n) using the formula: $n = \frac{P}{E}$ , where P is the power emitted.

Step 4: Substitute the given values: $n = \frac{{9 \times 10^{-3}}}{{2.98 \times 10^{-19}}} = 3.02 \times 10^{16} \approx 3 \times 10^{16}$

Step 5: Final Answer: The number of photons arriving per second is approximately $3 \times {10^{16}}$ .

Why other options are wrong:
- **Option A ( $9 \times {10^{17}}$ ): This is incorrect due to calculation errors in photon energy or number of photons.
- Option C ( $9 \times {10^{15}}$ ): This is incorrect due to calculation errors in photon energy or number of photons.
- Option D ( $3 \times {10^{19}}$ ): This is incorrect due to calculation errors in photon energy or number of photons.

Key Concepts:
- Energy of a photon
- Planck's constant
- Relationship between power, energy, and number of photons

Memory Aids:**
- Remember the formula for the energy of a photon: $E = \frac{{hc}}{\lambda}$

Quick Tips:
- Pay attention to units and conversions when calculating the energy and number of photons.

18 Medium

18 of 22Question permalink

Remembering +4 | -1

A laser device produces amplification in the

View answer and explanation

Correct option: B

Ultraviolet or visible region

Option B, Ultraviolet or visible region, is correct because lasers can be designed to produce amplification (light emission) in these regions of the electromagnetic spectrum.

Solution approach

Correct Answer + Core Reason: Ultraviolet or visible region. Lasers can produce amplification in these regions.

Step-by-Step Solution:

Step 1: Understand the principle of laser operation. LASER stands for Light Amplification by Stimulated Emission of Radiation.

Step 2: Recall the different types of lasers and their operating wavelengths:
- Gas lasers (e.g., He-Ne laser, Argon laser): Visible region
- Solid-state lasers (e.g., Ruby laser, Nd:YAG laser): Visible and infrared regions
- Semiconductor lasers (e.g., Laser diodes): Infrared and visible regions
- Excimer lasers: Ultraviolet region

Step 3: Identify the option that includes the regions where lasers can produce amplification.

Why other options are wrong:
- Option A: Microwave region is typically associated with masers, not lasers.
- Option C: Infrared region is a valid region for some lasers, but ultraviolet and visible regions are also important.
- Option D: Lasers definitely produce amplification in specific regions of the electromagnetic spectrum.

Key Concepts:
- Laser operation
- Stimulated emission
- Wavelengths of different lasers

Memory Aids:
- LASER: Light Amplification by Stimulated Emission of Radiation

Quick Tips:
- Remember the different types of lasers and their approximate operating wavelengths.

19 Medium

19 of 22Question permalink

Remembering +4 | -1

The characteristic impedance

\left( {{Z_0}} \right)

parallel line wire is

View answer and explanation

Correct option: A

{Z_0} = \frac{{276}}{{\sqrt k }}\log \frac{{2s}}{d}

Option A is correct because it represents the standard formula for the characteristic impedance of a parallel line wire.

Solution approach

Correct Answer + Core Reason: ${Z_0} = \frac{{276}}{{\sqrt k }}\log \frac{{2s}}{d}$ . This is the standard formula for characteristic impedance of a parallel line wire.

Step-by-Step Solution:

Step 1: Recall the formula for the characteristic impedance of a parallel line.

Step 2: The formula is given by ${Z_0} = \frac{{276}}{{\sqrt k }}\log \frac{{2s}}{d}$ , where $s$ is the separation between the wires, $d$ is the diameter of the wires, and $k$ is the dielectric constant of the medium.

Step 3: Identify the correct option based on the formula.

Step 4: The correct option is A.

Why other options are wrong:
- Option B: The constant 376 is incorrect. The correct constant is 276.
- Option C: This is the correct formula.
- Option D: The argument of the logarithm is inverted. It should be $\frac{{2s}}{d}$ , not $\frac{d}{{2s}}$ .

Key Concepts:
- Characteristic Impedance
- Parallel Line Wire

Memory Aids:
- Remember the constant 276 in the formula.

Quick Tips:
- Be careful with the argument of the logarithm and the constant factor.

20 Medium

20 of 22Question permalink

Analyzing +4 | -1

A modulated signal

\mathrm{C}_{\mathrm{m}}(\mathrm{t})

has the form

\mathrm{C}_{\mathrm{m}}(\mathrm{t})=30 \sin 300 \pi \mathrm{t}+10(\cos 200 \pi \mathrm{t}-\cos 400 \pi \mathrm{t})

. The carrier frequency

\mathrm{f_c}

, the modulating frequency (message frequency)

\mathrm{f}_\omega

and the modulation index

\mu

are respectively given by:

View answer and explanation

Correct option: B

\mathrm{f}_{\mathrm{c}}=150 \mathrm{~Hz} ; \mathrm{f}_{\mathrm{w}}=50 \mathrm{~Hz} ; \mu=\frac{2}{3}

Option B is correct because the carrier frequency, modulating frequency, and modulation index are calculated from the given modulated signal equation.

Solution approach

Correct Answer + Core Reason: $\mathrm{f}_{\mathrm{c}}=150 \mathrm{~Hz} ; \mathrm{f}_{\mathrm{w}}=50 \mathrm{~Hz} ; \mu=\frac{2}{3}$ . The carrier frequency, modulating frequency and modulation index are determined by analyzing the given modulated signal.

Step-by-Step Solution:

Step 1: Identify the components of the modulated signal. The given signal is $\mathrm{C}_{\mathrm{m}}(\mathrm{t})=30 \sin 300 \pi \mathrm{t}+10(\cos 200 \pi \mathrm{t}-\cos 400 \pi \mathrm{t})$ .

Step 2: Determine the carrier frequency $\mathrm{f_c}$ . The carrier signal is represented by $30 \sin 300 \pi \mathrm{t}$ . Therefore, $2 \pi \mathrm{f_c} = 300 \pi$ , which implies $\mathrm{f_c} = 150 \mathrm{~Hz}$ .

Step 3: Determine the modulating frequency $\mathrm{f_w}$ . The modulating signal components are $10 \cos 200 \pi \mathrm{t}$ and $-10 \cos 400 \pi \mathrm{t}$ . These represent frequencies of $200 \pi / (2 \pi) = 100 \mathrm{~Hz}$ and $400 \pi / (2 \pi) = 200 \mathrm{~Hz}$ . The modulating frequency is half the difference between these two frequencies: $\mathrm{f_w} = (200 - 100) / 2 = 50 \mathrm{~Hz}$ . Alternatively, using the trigonometric identity $\cos A - \cos B = -2 \sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$ , we have $10(\cos 200 \pi \mathrm{t}-\cos 400 \pi \mathrm{t}) = -20 \sin(300 \pi t) \sin(-100 \pi t) = 20 \sin(300 \pi t) \sin(100 \pi t)$ . The modulating frequency is associated with the $\sin(100 \pi t)$ term, so $2 \pi f_w = 100 \pi$ , which gives $f_w = 50 \mathrm{~Hz}$ .

Step 4: Determine the modulation index $\mu$ . The amplitude of the carrier signal is 30, and the amplitude of the modulating signal component is 10. The modulation index is given by $\mu = \frac{\text{amplitude of modulating signal}}{\text{amplitude of carrier signal}} = \frac{20}{30} = \frac{2}{3}$ .

Step 5: Summarize the results: $\mathrm{f_c} = 150 \mathrm{~Hz}$ , $\mathrm{f_w} = 50 \mathrm{~Hz}$ , and $\mu = \frac{2}{3}$ .

Why other options are wrong:
- Option A: Incorrect carrier and modulating frequencies, and modulation index.
- Option C: Incorrect modulating frequency and modulation index.
- Option D: Incorrect carrier frequency and modulating frequency, and modulation index.

Key Concepts:
- Amplitude Modulation (AM)
- Carrier Frequency
- Modulating Frequency
- Modulation Index

Memory Aids:
- The general form of an AM signal is $A_c[1 + \mu m(t)]\cos(2\pi f_c t)$ , where $A_c$ is the carrier amplitude, $\mu$ is the modulation index, $m(t)$ is the message signal, and $f_c$ is the carrier frequency.

Quick Tips:
- Always identify the carrier and modulating frequencies from the given equation by comparing it with the standard AM equation.
- The modulation index is the ratio of the amplitude of the modulating signal to the amplitude of the carrier signal.

21 Medium

21 of 22Question permalink

Remembering +4 | -1

The difference in the refractive index of the fiber core and fiber cladding in step-index fibers is typically

View answer and explanation

Correct option: B

0.01

Option B, 0.01, is the typical difference in refractive index between the core and cladding of a step-index fiber.

Solution approach

Correct Answer + Core Reason: 0.01. This is the typical refractive index difference in step-index fibers.

Step-by-Step Solution:

Step 1: Understand the structure of a step-index fiber. A step-index fiber consists of a core with a refractive index $n_1$ and a cladding with a refractive index $n_2$ , where $n_1 > n_2$ .

Step 2: Recall the typical values for the refractive indices:
- Core refractive index ( $n_1$ ): Typically around 1.48 to 1.50
- Cladding refractive index ( $n_2$ ): Slightly lower than the core refractive index

Step 3: Calculate the difference in refractive index: $\Delta n = n_1 - n_2$ . The typical value for $\Delta n$ is around 0.01.

Why other options are wrong:
- Option A: 0.0001 is too small for the refractive index difference.
- Option C: 0.1 is too large for the refractive index difference. This would lead to excessive losses.
- Option D: 0.25 is much too large for the refractive index difference.

Key Concepts:
- Step-index fiber
- Refractive index
- Core and cladding

Memory Aids:
- Remember that the core has a higher refractive index than the cladding.

Quick Tips:
- Know the typical values for the refractive indices of optical fibers.

22 Medium

22 of 22Question permalink

Analyzing +4 | -1

Assertion (A) Audio signal of frequency

10\, {kHz}

cannot be transmitted over long distance without modulation.

Reason (R) Length of the antenna required

\lambda / 4

, should have practical value.

View answer and explanation

Correct option: A

If both Assertion and Reason are true and Reason is correct explanation of Assertion

Option A is correct because both the assertion and the reason are true, and the reason correctly explains the assertion.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true and Reason is correct explanation of Assertion. Audio signals of 10 kHz require impractically large antennas for direct transmission, hence modulation is necessary.

Step-by-Step Solution:

Step 1: Analyze the Assertion: An audio signal of 10 kHz cannot be transmitted over long distances without modulation. This is true because low-frequency signals attenuate rapidly over long distances, and the required antenna size for direct transmission would be impractical.

Step 2: Analyze the Reason: The reason states that the length of the antenna required, $\lambda / 4$ , should have a practical value. This is also true. The minimum antenna length is typically $\lambda / 4$ , where $\lambda$ is the wavelength of the signal.

Step 3: Calculate the wavelength of a 10 kHz signal: $\lambda = c / f = (3 \times 10^8 \mathrm{m/s}) / (10 \times 10^3 \mathrm{Hz}) = 30,000 \mathrm{m} = 30 \mathrm{km}$ .

Step 4: Calculate the required antenna length: $\lambda / 4 = 30,000 \mathrm{m} / 4 = 7,500 \mathrm{m} = 7.5 \mathrm{km}$ . This antenna length is clearly impractical.

Step 5: Determine the relationship between Assertion and Reason: The reason correctly explains the assertion. Because the required antenna length for a 10 kHz signal is impractical, modulation is necessary to shift the signal to a higher frequency, which reduces the required antenna size and allows for efficient transmission.

Why other options are wrong:
- Option B: The reason *is* the correct explanation of the assertion.
- Option C: The assertion is true.
- Option D: Both the assertion and reason are true.

Key Concepts:
- Signal Modulation
- Antenna Length and Wavelength
- Electromagnetic Wave Propagation

Memory Aids:
- Remember that antenna length is inversely proportional to frequency: higher frequency, shorter antenna.

Quick Tips:
- Always consider the practical implications of antenna size when dealing with signal transmission.

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