COMMUNICATION SYSTEMS MCQs for NEET UG PHYSICS — Mixed difficulty

1 Easy

Understanding +4 | -1

In the communication systems, AM is used for broadcasting because

View answer and explanation

Correct option: A

Its use avoids receiver complexity

Option A (Its use avoids receiver complexity) is the reason AM is used for broadcasting, as it simplifies the design and cost of receivers.

Solution approach

Correct Answer + Core Reason: AM is used for broadcasting because its use avoids receiver complexity.

Step-by-Step Solution:

Step 1: Consider the requirements for broadcasting: simple and inexpensive receivers.

Step 2: AM receivers are simpler and cheaper to manufacture than FM or other modulation schemes.

Step 3: Simplicity in receiver design is crucial for widespread adoption in broadcasting.

Why other options are wrong:
- Option B: AM is actually more susceptible to noise than FM.
- Option C: AM requires more transmitting power than some other modulation techniques.
- Option D: Other modulation systems can provide the necessary bandwidth, but at the cost of increased receiver complexity.

Key Concepts:
- Amplitude Modulation (AM)
- Receiver Complexity
- Broadcasting Requirements

Memory Aids:
- AM = Affordable Modulation (simple receivers)

Quick Tips:
- Remember that AM's main advantage is the simplicity of its receivers.

2 Easy

2 of 44Question permalink

Application +4 | -1

The maximum and minimum voltage of an amplitude modulated signal are

60 \mathrm{~V}

and

20 \mathrm{~V}

respectively. The percentage modulation index will be:

View answer and explanation

Correct option: B

50 \%

Option B (50%) is correct because the percentage modulation index is calculated as ((Vmax - Vmin) / (Vmax + Vmin)) * 100.

Solution approach

Correct Answer + Core Reason: 50 %. The percentage modulation index is calculated using the formula based on maximum and minimum voltages.

Step-by-Step Solution:

Step 1: Identify the given information: Maximum voltage ( $V_{max}$ ) = 60 V, Minimum voltage ( $V_{min}$ ) = 20 V.

Step 2: Recall the formula for modulation index ( $m$ ) in AM modulation: $m = \frac{{V_{max} - V_{min}}}{{V_{max} + V_{min}}}$

Step 3: Calculate the modulation index: $m = \frac{{60 - 20}}{{60 + 20}} = \frac{40}{80} = 0.5$

Step 4: Calculate the percentage modulation index: Percentage modulation = $m \times 100 = 0.5 \times 100 = 50 \%$

Step 5: Final Answer: The percentage modulation index is 50%.

Why other options are wrong:
- Option A (0.5 %): This is incorrect because it represents the modulation index without converting it to a percentage.
- Option C (2 %): This is incorrect as it doesn't result from the correct application of the percentage modulation index formula.
- Option D (30 %): This is incorrect as it doesn't result from the correct application of the percentage modulation index formula.

Key Concepts:
- Amplitude Modulation (AM)
- Modulation Index
- Percentage Modulation

Memory Aids:
- Remember to multiply the modulation index by 100 to get the percentage.

Quick Tips:
- Always remember to convert the modulation index to percentage when asked for percentage modulation.

3 Easy

3 of 44Question permalink

Remembering +4 | -1

Depth of modulation

\left( {{m_a}} \right)

in terms of

{{\text{E}}_{{\text{max}}}}\,{\text{and}}\,{{\text{E}}_{{\text{min}}}}

is

View answer and explanation

Correct option: C

{{\text{m}}_{\text{a}}}{\text{ = }}{{\text{E}}_{{\text{max}}}}{\text{ - }}{{\text{E}}_{{\text{min}}}}{\text{/}}{{\text{E}}_{{\text{max}}}} + {{\text{E}}_{{\text{min}}}}

Option C,

{{\text{m}}_{\text{a}}}{\text{ = }}{{\text{E}}_{{\text{max}}}}{\text{ - }}{{\text{E}}_{{\text{min}}}}{\text{/}}{{\text{E}}_{{\text{max}}}} + {{\text{E}}_{{\text{min}}}}

, is the correct formula for calculating the depth of modulation (

m_a

) in terms of the maximum and minimum amplitudes of the modulated wave.

Solution approach

Correct Answer + Core Reason: ${{\text{m}}_{\text{a}}}{\text{ = }}{{\text{E}}_{{\text{max}}}}{\text{ - }}{{\text{E}}_{{\text{min}}}}{\text{/}}{{\text{E}}_{{\text{max}}}} + {{\text{E}}_{{\text{min}}}}$ . This is the standard formula for depth of modulation.

Step-by-Step Solution:

Step 1: Define depth of modulation ( $m_a$ ). Depth of modulation is a measure of the extent to which the amplitude of the carrier wave is varied by the message signal.

Step 2: Recall the formula for depth of modulation in terms of $E_{max}$ and $E_{min}$ :
$m_a = \frac{{E_{max} - E_{min}}}{{E_{max} + E_{min}}}$

Step 3: Compare the given options with the correct formula.

Why other options are wrong:
- Option A: ${{\text{m}}_{\text{a}}}{\text{ = }}{{\text{E}}_{{\text{max}}}}{\text{ + }}{{\text{E}}_{{\text{min}}}}{\text{/}}{{\text{E}}_{{\text{min}}}}$ is incorrect. The numerator should be the difference, not the sum.
- Option B: ${{\text{m}}_{\text{a}}}{\text{ = }}{{\text{E}}_{{\text{max}}}}{\text{ - }}{{\text{E}}_{{\text{min}}}}{\text{/}}{{\text{E}}_{{\text{max}}}}$ is incorrect. The denominator should be the sum, not just the maximum.
- Option D: ${{\text{m}}_{\text{a}}}{\text{ = }}{{\text{E}}_{{\text{max}}}}{\text{ + }}{{\text{E}}_{{\text{min}}}}{\text{/}}{{\text{E}}_{{\text{max}}}} + {{\text{E}}_{{\text{min}}}}$ is incorrect. The numerator should be the difference, not the sum.

Key Concepts:
- Amplitude Modulation (AM)
- Depth of Modulation

Memory Aids:
- Remember the formula: $m_a = \frac{{E_{max} - E_{min}}}{{E_{max} + E_{min}}}$

Quick Tips:
- Understand the relationship between $E_{max}$ , $E_{min}$ , and the amplitude of the message signal and carrier wave.

4 Easy

4 of 44Question permalink

Remembering +4 | -1

The process of superimposing signal frequency (i.e., audio wave) on the carrier wave is known as

View answer and explanation

Correct option: C

Modulation

Option C (Modulation) is correct because modulation is the process of superimposing a signal onto a carrier wave.

Solution approach

Correct Answer + Core Reason: Modulation is the process of superimposing a signal onto a carrier wave.

Step-by-Step Solution:

Step 1: Define the terms.

Step 2: Modulation is the process of varying one or more properties of a carrier wave with a modulating signal.

Step 3: Transmission is the process of sending a signal over a medium.

Step 4: Reception is the process of receiving a signal.

Step 5: Detection is the process of extracting the original signal from the modulated wave.

Step 6: Therefore, the correct answer is modulation.

Why other options are wrong:
- Option A: Transmission is the process of sending the modulated wave, not the superimposition itself.
- Option B: Reception is the process of receiving the modulated wave, not the superimposition itself.
- Option D: Detection is the process of extracting the original signal, not the superimposition itself.

Key Concepts:
- Modulation
- Transmission
- Reception
- Detection

Memory Aids:
- Modulation: Think of \"modifying\" the carrier wave.

Quick Tips:
- Understand the definitions of each term in communication systems.

5 Hard

5 of 44Question permalink

Application +4 | -1

The antenna current of an AM transmitter is 8 A when only the carrier is sent, but it increases to 8.93 A when the carrier is modulated by single sine wave. Find the percentage modulation.

View answer and explanation

Correct option: B

0.701

Option B (0.701) is correct because it represents the modulation index calculated from the antenna currents with and without modulation.

Solution approach

Correct Answer + Core Reason: 0.701. The percentage modulation is calculated using the formula $m = \sqrt{2((\frac{I_t}{I_c})^2 - 1)} \times 100$ .

Step-by-Step Solution:

Step 1: Identify the given information.\nCarrier current, $I_c = 8 \text{{ A}}$ \nTotal current with modulation, $I_t = 8.93 \text{{ A}}$

Step 2: Recall the formula for the total current in an AM transmitter:
$I_t = I_c \sqrt{1 + \frac{m^2}{2}}$ , where $m$ is the modulation index.

Step 3: Rearrange the formula to solve for $m$ :
$\frac{I_t}{I_c} = \sqrt{1 + \frac{m^2}{2}}$
$(\frac{I_t}{I_c})^2 = 1 + \frac{m^2}{2}$
$\frac{m^2}{2} = (\frac{I_t}{I_c})^2 - 1$ \nm² = 2(( $\frac{I_t}{I_c}$ )² - 1) $\nm = \sqrt{2((\frac{I_t}{I_c})^2 - 1)}$

Step 4: Substitute the given values into the formula:\nm = $\sqrt{2((\frac{8.93}{8})^2 - 1)}$ $\n\n**Step 5:** Calculate the value:\nm = \sqrt{2((1.11625)^2 - 1)}$ \nm = $\sqrt{2(1.246 - 1)}$ $\nm = \sqrt{2(0.246)}$ \nm = $\sqrt{0.492}$ ≈ 0.701 $\n\n**Step 6:** Convert to percentage:\nPercentage modulation =$ 0.701 × 100 = 70.1 \% $or 0.701\n\n**Why other options are wrong:**\n- **Option A:** 0.601: This is incorrect. It might be a result of a miscalculation in the formula.\n- **Option C:** 0.801: This is incorrect. It might be a result of a miscalculation in the formula.\n- **Option D:** 0.501: This is incorrect. It might be a result of a miscalculation in the formula.\n\n**Key Concepts:**\n- Amplitude Modulation\n- Modulation Index\n- Antenna Current\n\n**Memory Aids:**\n- Remember the relationship between total current, carrier current, and modulation index.\n\n**Quick Tips:**\n- Be careful with the calculations and square roots.\n- Remember to convert the modulation index to percentage if required.$

6 Easy

6 of 44Question permalink

Application +4 | -1

An audio signal of amplitude one half the carrier amplitude is used in amplitude modulation. Calculate the modulation index?

View answer and explanation

Correct option: A

0.5

Option A (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 the ratio of the amplitude of the modulating signal to the amplitude of the carrier signal.

Step-by-Step Solution:

Step 1: Define the variables: Let $A_m$ be the amplitude of the modulating signal and $A_c$ be the amplitude of the carrier signal.

Step 2: Given that the amplitude of the audio signal is one half the carrier amplitude, we have $A_m = \frac{1}{2} A_c$ .

Step 3: The modulation index ( $m$ ) is defined as: $m = \frac{{A_m}}{{A_c}}$

Step 4: Substitute the given relationship: $m = \frac{{\frac{1}{2} A_c}}{{A_c}}$

Step 5: Simplify: $m = \frac{1}{2} = 0.5$

Step 6: Final Answer: The modulation index is 0.5.

Why other options are wrong:
- Option B (0): This is incorrect because the modulating signal has a non-zero amplitude.
- Option C (0): This is incorrect because the modulating signal has a non-zero amplitude.
- Option D (0): This is incorrect because the modulating signal has a non-zero amplitude.

Key Concepts:
- Amplitude Modulation (AM)
- Modulation Index

Memory Aids:
- Remember the formula for modulation index: $m = \frac{{A_m}}{{A_c}}$

Quick Tips:
- Always identify the amplitudes of the modulating and carrier signals correctly.

7 Easy

7 of 44Question permalink

Understanding +4 | -1

When electromagnetic waves enter the ionised layer of ionosphere, then the relative permittivity

ie

, dielectric constant dielectric constant of the ionised layer

View answer and explanation

Correct option: C

Appears to decrease

The correct answer is C (Appears to decrease) because the refractive index of the ionosphere is related to the electron density, and as electromagnetic waves enter, the effective dielectric constant decreases.

Solution approach

Correct Answer + Core Reason: Appears to decrease. The relative permittivity of the ionized layer decreases when electromagnetic waves enter.

Step-by-Step Solution:

Step 1: Recall the relationship between relative permittivity and refractive index in the ionosphere.

Step 2: The relative permittivity (dielectric constant) of the ionosphere is related to the electron density. As the electromagnetic waves interact with the ionized layer, the effective dielectric constant decreases.

Step 3: Therefore, the relative permittivity appears to decrease.

Why other options are wrong:
- Option A: Does not change - The permittivity changes due to the interaction of the electromagnetic waves with the ionized particles.
- Option B: Appears to increase - The permittivity decreases, not increases.
- Option D: Sometimes appears to increase and sometimes to decrease - The permittivity consistently decreases as the waves enter the ionized layer.

Key Concepts:
- Ionosphere
- Relative Permittivity (Dielectric Constant)
- Electromagnetic Waves

Memory Aids:
- Remember that the ionosphere is ionized, and the interaction with EM waves reduces the effective permittivity.

Quick Tips:
- Understand the basic properties of the ionosphere and its interaction with electromagnetic waves.

8 Medium

8 of 44Question permalink

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.

9 Easy

9 of 44Question permalink

Understanding +4 | -1

In frequency modulated wave

View answer and explanation

Correct option: A

Frequency varies with time

The correct answer is A (Frequency varies with time) because, by definition, in frequency modulation, the frequency of the carrier wave is varied according to the message signal.

Solution approach

Correct Answer + Core Reason: Frequency varies with time. In frequency modulation (FM), the frequency of the carrier signal changes with the message signal.

Step-by-Step Solution:

Step 1: Understand the concept of Frequency Modulation (FM).

Step 2: In FM, the instantaneous frequency of the carrier wave is varied in proportion to the message signal's amplitude.

Step 3: Therefore, the frequency varies with time.

Why other options are wrong:
- Option B: Amplitude varies with time - This is the principle of Amplitude Modulation (AM), not FM.
- Option C: Both frequency and amplitude vary with time - Only frequency varies in FM.
- Option D: Both frequency and amplitude are constant - This is true for an unmodulated carrier wave.

Key Concepts:
- Frequency Modulation (FM)
- Amplitude Modulation (AM)
- Carrier Wave
- Message Signal

Memory Aids:
- FM: Frequency changes with the message.

Quick Tips:
- Remember the basic definitions of AM and FM.

10 Medium

10 of 44Question 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.

11 Medium

11 of 44Question 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.

12 Medium

12 of 44Question 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.

13 Hard

13 of 44Question permalink

Application +4 | -1

Calculate the phase velocity of electromagnetic wave having electron density and frequency of D-layer, N=400 electron

c{c^{ - 1}},v = 200\,kHz

View answer and explanation

Correct option: C

6.9 \times {10^8}\mathrm{m}{\mathrm{s}^{ - 1}}

Option C (

6.9 \times {10^8}\mathrm{m}{\mathrm{s}^{ - 1}}

) is correct because it is the calculated phase velocity using the given electron density and frequency.

Solution approach

Correct Answer + Core Reason: $6.9 \times {10^8}\mathrm{m}{\mathrm{s}^{ - 1}}$ . The phase velocity is calculated using the formula $v_p = \frac{c}{\sqrt{1 - \frac{81N}{f^2}}}$ .

Step-by-Step Solution:

Step 1: Identify the given information.\nElectron density, $N = 400 \text{{ electrons/cc}} = 400 \times 10^6 \text{{ electrons/m}}^3$ \nFrequency, $f = v = 200 \text{{ kHz}} = 200 \times 10^3 \text{{ Hz}}$ \nSpeed of light, $c = 3 \times 10^8 \text{{ m/s}}$

Step 2: Recall the formula for phase velocity ( $v_p$ ) of an electromagnetic wave in an ionized medium:
$v_p = \frac{c}{\sqrt{1 - \frac{81N}{f^2}}}$

Step 3: Substitute the given values into the formula:\nv_p = \frac{3 × 10⁸}{ $\sqrt{1 - \frac{81 × 400 × 10^6}{(200 × 10^3)^2}}$ } $\n\n**Step 4:** Simplify the expression inside the square root:\n$ $\frac{81 × 400 × 10^6}{(200 × 10^3)^2}$ = $\frac{81 × 400 × 10^6}{40000 × 10^6}$ = $\frac{81 × 400}{40000}$ = $\frac{81}{100}$ = 0.81 $\n\n**Step 5:** Calculate the square root:\n$ $\sqrt{1 - 0.81}$ = $\sqrt{0.19}$ ≈ 0.436 $\n\n**Step 6:** Calculate the phase velocity:\nv_p = \frac{3 \times 10^8}{0.436} \approx 6.88 \times 10^8 \text{{ m/s}} \approx 6.9 \times 10^8 \text{{ m/s}}$

Why other options are wrong:
- Option A: $3 \times {10^8}\mathrm{m}{\mathrm{s}^{ - 1}}$ : This is the speed of light in a vacuum. The phase velocity in an ionized medium is greater than the speed of light.
- Option B: $3.5 \times {10^8}\mathrm{m}{\mathrm{s}^{ - 1}}$ : This is incorrect. It might be a result of a miscalculation in the formula.
- Option D: $1.1 \times {10^9}\mathrm{m}{\mathrm{s}^{ - 1}}$ : This is incorrect. It might be a result of a miscalculation in the formula.

Key Concepts:
- Phase velocity of electromagnetic waves
- Ionospheric propagation

Memory Aids:
- Remember the formula for phase velocity in an ionized medium.

Quick Tips:
- Pay attention to units and conversions.
- Be careful with the calculations inside the square root.

14 Medium

14 of 44Question 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.

15 Medium

15 of 44Question 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.

16 Easy

16 of 44Question permalink

Application +4 | -1

In AM modulation, a signal is modulated on a carrier wave such that maximum and minimum amplitudes are found to be

6 \mathrm{~V}

and

2 \mathrm{~V}

respectively. The modulation index is

View answer and explanation

Correct option: D

0.5

Option D (0.5) is correct because the modulation index is calculated as (Amax - Amin) / (Amax + Amin).

Solution approach

Correct Answer + Core Reason: 0.5. The modulation index is calculated using the formula based on maximum and minimum amplitudes.

Step-by-Step Solution:

Step 1: Identify the given information: Maximum amplitude ( $A_{max}$ ) = 6 V, Minimum amplitude ( $A_{min}$ ) = 2 V.

Step 2: Recall the formula for modulation index ( $m$ ) in AM modulation: $m = \frac{{A_{max} - A_{min}}}{{A_{max} + A_{min}}}$

Step 3: Substitute the given values into the formula: $m = \frac{{6 - 2}}{{6 + 2}}$

Step 4: Calculate the modulation index: $m = \frac{4}{8} = 0.5$

Step 5: Final Answer: The modulation index is 0.5.

Why other options are wrong:
- Option A (1): This is incorrect because it would imply that the minimum amplitude is zero, which is not the case here.
- Option B (0.8): This is incorrect as it doesn't result from the correct application of the modulation index formula.
- Option C (0.6): This is incorrect as it doesn't result from the correct application of the modulation index formula.

Key Concepts:
- Amplitude Modulation (AM)
- Modulation Index

Memory Aids:
- Remember the formula: $m = \frac{{A_{max} - A_{min}}}{{A_{max} + A_{min}}}$

Quick Tips:
- Always identify the maximum and minimum amplitudes correctly before applying the formula.

17 Easy

17 of 44Question permalink

Application +4 | -1

Find the modulation index of an AM wave having 8

\mathrm{V}

variation where maximum amplitude of the AM wave is

9 \mathrm{~V}

.

View answer and explanation

Correct option: A

0.8

Option A (0.8) is correct because the modulation index is calculated as variation/ (2 * A_carrier), where A_carrier is A_max - variation/2.

Solution approach

Correct Answer + Core Reason: 0.8. The modulation index is calculated using the formula based on the variation and maximum amplitude.

Step-by-Step Solution:

Step 1: Identify the given information: Variation = 8 V, Maximum amplitude ( $A_{max}$ ) = 9 V.

Step 2: Determine the carrier amplitude ( $A_c$ ). Since the variation is 8V, the minimum amplitude is $9 - 8 = 1$ V. The carrier amplitude is the average of the max and min amplitudes: $A_c = (9 + 1)/2 = 5$ V.

Step 3: Calculate the modulation index ( $m$ ) using the formula: $m = \frac{{\text{{Variation}}}}{{2A_c}} = \frac{8}{2*5} = \frac{8}{10} = 0.8$

Step 4: Final Answer: The modulation index is 0.8.

Why other options are wrong:
- Option B (0.5): This is incorrect as it doesn't result from the correct application of the modulation index formula with the given variation and maximum amplitude.
- Option C (0.2): This is incorrect as it doesn't result from the correct application of the modulation index formula with the given variation and maximum amplitude.
- Option D (0.1): This is incorrect as it doesn't result from the correct application of the modulation index formula with the given variation and maximum amplitude.

Key Concepts:
- Amplitude Modulation (AM)
- Modulation Index

Memory Aids:
- Remember the relationship between variation, maximum amplitude, and modulation index.

Quick Tips:
- Ensure you correctly determine the carrier amplitude from the given variation and maximum amplitude.

18 Medium

18 of 44Question 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.

19 Medium

19 of 44Question 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.

20 Easy

20 of 44Question permalink

Remembering +4 | -1

In earth's atmosphere, for E-layer, the virtual height and critical frequency are

View answer and explanation

Correct option: D

110 km and 4 MHz

Option D, 110 km and 4 MHz, is the closest to the typical values for the E-layer virtual height and critical frequency in Earth's atmosphere.

Solution approach

Correct Answer + Core Reason: 110 km and 4 MHz. These are the approximate values for the E-layer's virtual height and critical frequency.

Step-by-Step Solution:

Step 1: Recall the different layers of the ionosphere: D, E, F1, and F2 layers.

Step 2: Remember the approximate altitudes and critical frequencies for each layer:
- D layer: 60-90 km, absorbs medium and high frequencies during the day.
- E layer: 90-150 km, reflects high frequencies during the day, disappears at night. Virtual height is around 110 km and critical frequency is around 4 MHz.
- F1 layer: 150-250 km, exists during the day, merges with F2 layer at night.
- F2 layer: 250-400 km, most important layer for long-distance radio communication.

Step 3: Identify the option that matches the E-layer characteristics.

Why other options are wrong:
- Option A: 80 km is too low for the E-layer, and 3 MHz is a lower critical frequency than typical.
- Option B: 90 km is slightly low for the E-layer, and 3.5 MHz is a lower critical frequency than typical.
- Option C: 120 km is within the E-layer range, but 4.5 MHz is a slightly higher critical frequency than typical.

Key Concepts:
- Ionosphere layers (D, E, F1, F2)
- Virtual height
- Critical frequency

Memory Aids:
- Remember the order of the layers: D-E-F (lowest to highest).

Quick Tips:
- Memorize the approximate altitudes and critical frequencies of the ionospheric layers.

21 Medium

21 of 44Question 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.

22 Medium

22 of 44Question 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.

23 Medium

23 of 44Question 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).

24 Hard

24 of 44Question permalink

Understanding +4 | -1

In a radio receiver, the short wave and medium wave stations are tuned by using the same capacitor but coils of different inductance

{L_s}

and

{L_m}

respectively then

View answer and explanation

Correct option: B

{L_s} < {L_m}

Option B (

{L_s} < {L_m}

) is correct because for the same capacitor, a lower inductance is required for tuning to higher frequencies (short waves) and a higher inductance is required for tuning to lower frequencies (medium waves).

Solution approach

Correct Answer + Core Reason: ${L_s} < {L_m}$ . Short waves have higher frequency, requiring lower inductance.

Step-by-Step Solution:

Step 1: Recall the formula for the resonant frequency of an LC circuit: $f = \frac{1}{2\pi \sqrt{LC}}$

Step 2: For a given capacitor C, the resonant frequency f is inversely proportional to the square root of the inductance L: $f \propto \frac{1}{\sqrt{L}}$

Step 3: Short waves have higher frequencies than medium waves: $f_s > f_m$

Step 4: Therefore, for the same capacitor, the inductance for short waves must be smaller than the inductance for medium waves: $L_s < L_m$

Why other options are wrong:
- Option A: ${L_s} > {L_m}$ is incorrect because higher inductance corresponds to lower frequency.
- Option C: ${L_s} = {L_m}$ is incorrect because short and medium waves have different frequencies.
- Option D: Option B is the correct relationship between the inductances.

Key Concepts:
- Resonant Frequency
- LC Circuit
- Inductance
- Capacitance

Memory Aids:
- Low Inductance, High Frequency: Remember that lower inductance corresponds to higher frequency in an LC circuit.

Quick Tips:
- Use the formula $f = \frac{1}{2\pi \sqrt{LC}}$ to relate frequency and inductance.

25 Hard

25 of 44Question permalink

Application +4 | -1

Two electrons are moving in opposite direction with speeds 0.8 c and 0.4 c, where c the speed of light in vacuum. Then the relative speed is about

View answer and explanation

Correct option: C

0.9 c

Option C (0.9 c) is correct because it represents the relative speed calculated using the relativistic velocity addition formula.

Solution approach

Correct Answer + Core Reason: 0.9 c. The relative speed is calculated using the relativistic velocity addition formula.

Step-by-Step Solution:

Step 1: Identify the given information.
$v_1 = 0.8c$
$v_2 = 0.4c$ \nSince the electrons are moving in opposite directions, we can consider one velocity positive and the other negative.

Step 2: Recall the relativistic velocity addition formula:\nv = $\frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}$ $\n\n**Step 3:** Substitute the given values into the formula. Since they are moving in opposite directions, let$ v_1 = 0.8c $and$ v_2 = -0.4c $.\nv = \frac{0.8c - 0.4c}{1 + \frac{(0.8c)(-0.4c)}{c^2}}$

Step 4: Simplify the expression:\nv = $\frac{0.4c}{1 - 0.32}$ = $\frac{0.4c}{0.68}$ ≈ 0.588c $\n\nHowever, the question asks for the *relative* speed, which means we want the speed of one electron relative to the other. So, we should use$ v_2 = 0.4c $(positive).\nv = \frac{0.8c + 0.4c}{1 + \frac{(0.8c)(0.4c)}{c^2}}$ \nv = $\frac{1.2c}{1 + 0.32}$ = $\frac{1.2c}{1.32}$ ≈ 0.909c ≈ 0.9c $\n\n**Why other options are wrong:**\n- **Option A:** 0.4 c: This is simply one of the given speeds and doesn't account for the relative motion or relativistic effects.\n- **Option B:** 0.8 c: This is simply one of the given speeds and doesn't account for the relative motion or relativistic effects.\n- **Option D:** 1.2 c: This is the sum of the speeds, but it's greater than the speed of light and doesn't account for relativistic effects. The relative speed can never exceed c.\n\n**Key Concepts:**\n- Relativistic Velocity Addition\n\n**Memory Aids:**\n- Remember the relativistic velocity addition formula.\n\n**Quick Tips:**\n- Pay attention to the signs of the velocities when they are in opposite directions.\n- Remember that the relative speed cannot exceed the speed of light.$

26 Medium

26 of 44Question 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.

27 Easy

27 of 44Question permalink

Remembering +4 | -1

The sky wave propagation is suitable for radio-waves of frequency

View answer and explanation

Correct option: C

From 2 MHz to 30 MHz

Option C, From 2 MHz to 30 MHz, is correct because sky wave propagation, which relies on ionospheric reflection, is most effective within this frequency range.

Solution approach

Correct Answer + Core Reason: From 2 MHz to 30 MHz. Sky wave propagation is suitable for radio waves in this frequency range.

Step-by-Step Solution:

Step 1: Understand sky wave propagation. Sky wave propagation (also known as ionospheric propagation) involves the transmission of radio waves that are reflected or refracted by the ionosphere back to Earth.

Step 2: Recall the frequency ranges suitable for different propagation methods:
- Ground wave propagation: Low frequencies (up to 2 MHz), follows the curvature of the Earth.
- Sky wave propagation: Medium and high frequencies (2 MHz to 30 MHz), reflected by the ionosphere.
- Space wave propagation: Very high frequencies (above 30 MHz), travels in a straight line.

Step 3: Identify the option that corresponds to sky wave propagation.

Why other options are wrong:
- Option A: Upto 2 MHz is suitable for ground wave propagation.
- Option B: From 2 MHz to 20 MHz is a subset of the correct range, but not the complete range.
- Option D: From 2 MHz to 50 MHz includes frequencies that are more suitable for space wave propagation.

Key Concepts:
- Sky wave propagation
- Ionosphere
- Ground wave propagation
- Space wave propagation

Memory Aids:
- Sky wave: Think of waves bouncing off the sky (ionosphere).

Quick Tips:
- Remember the frequency ranges for different radio wave propagation methods.

28 Medium

28 of 44Question 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.

29 Easy

29 of 44Question permalink

Application +4 | -1

The audio signal used to modulate

60\sin \left( {2\pi \times {{10}^6}\;t} \right)

is 

15\,\,\sin \,\,300\pi t.

The depth of modulation is

View answer and explanation

Correct option: C

0.25

Option C (0.25) is correct because the depth of modulation is calculated as the ratio of the amplitude of the modulating signal to the amplitude of the carrier signal.

Solution approach

Correct Answer + Core Reason: 0.25. Depth of modulation is the ratio of modulating signal amplitude to carrier signal amplitude.

Step-by-Step Solution:

Step 1: Identify the carrier signal: $60\sin(2\pi \times 10^6 t)$ . Its amplitude is $A_c = 60$ .

Step 2: Identify the modulating signal: $15\sin(300\pi t)$ . Its amplitude is $A_m = 15$ .

Step 3: Calculate the depth of modulation ( $\mu$ ): $\mu = \frac{A_m}{A_c}$

Step 4: Substitute the values: $\mu = \frac{15}{60}$

Step 5: Simplify: $\mu = 0.25$

Why other options are wrong:
- Option A: 0.5 is incorrect. It might be obtained by incorrectly dividing 30 by 60.
- Option B: 0.4 is incorrect. There is no direct calculation that leads to this value.
- Option D: 0.15 is incorrect. It might be obtained by incorrectly dividing 9 by 60 (15% of 60 is 9, not related to depth of modulation).

Key Concepts:
- Amplitude Modulation (AM)
- Depth of Modulation
- Carrier Signal
- Modulating Signal

Memory Aids:
- Modulation = Message / Carrier: Remember that the depth of modulation is the ratio of the message signal amplitude to the carrier signal amplitude.

Quick Tips:
- Always identify the carrier and modulating signals correctly before calculating the depth of modulation.

30 Medium

30 of 44Question 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.

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