A diode detector is used to detect an amplitude modulated wave of modulation by using a condenser of capacity picofarad in parallel with a load resistance . Find the maximum modulated frequency which could be detected by it.
1
2
3
4
Official Solution
Correct Option:
(2)
The higher frequency which can be detected with tolerable distortion is
02
PYQ 2014
medium
physicsID: jee-main
A transmitting antenna at the top of a tower has a height and the height of the receiving antenna is . What is the maximum distance between them for satisfactory communication in line of sight (LOS) mode ?
1
55.4 km
2
45.5 km
3
54.5 km
4
455 km
Official Solution
Correct Option:
(2)
03
PYQ 2014
medium
physicsID: jee-main
For sky wave propagation, the radio waves must have a frequency range in between :
1
1 MHz to 2 MHz
2
5 MHz to 25 MHz
3
35 MHz to 40 MHz
4
45 MHz to 50 MHz
Official Solution
Correct Option:
(2)
Sky wave propagation is suitable for frequency range to
04
PYQ 2016
medium
physicsID: jee-main
A modulated signal has the form sin (cos cos ). The carrier frequency , the modulating frequency (message frequency) , and the modulation index are respectively given by :
1
2
3
4
Official Solution
Correct Option:
(2)
05
PYQ 2018
medium
physicsID: jee-main
A carrier wave of peak voltage is used for transmitting a message signal. The peak voltage of modulating signal given to achieve a modulation index of will be:
1
2
3
4
Official Solution
Correct Option:
(3)
Amplitude of carrier wave is Index is Amplitude of modulating signal is given by
06
PYQ 2018
medium
physicsID: jee-main
A telephonic communication service is working at carrier frequency of . Only of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of ?
1
2
3
4
Official Solution
Correct Option:
(3)
Frequency of carrier Available bandwidth
Bandwidth for each telephonic channel Number of channels
07
PYQ 2018
medium
physicsID: jee-main
The carrier frequency of a transmitter is provided by a tank circuit of a coil of inductance and a capacitance of . It is modulated by an audio signal of . The frequency range occupied by the side bands is:
1
13482 kHz - 13494 kHz
2
442 kHz - 466 kHz
3
63 kHz - 75 kHz
4
18 kHz - 30 kHz
Official Solution
Correct Option:
(2)
The tank circuit of an inductance and a capacitance is know that Given carrier frequency is modulated by audio signal of . Therefore,
08
PYQ 2018
medium
physicsID: jee-main
The value closest to the thermal velocity of a Helium atom at room temperature in is :
1
2
3
4
Official Solution
Correct Option:
(2)
We know that . Given, . Therefore, or
09
PYQ 2019
easy
physicsID: jee-main
A message signal of frequency and peak voltage is used to execute amplitude modulation on a carrier wave of frequency and peak voltage . The modulation index and difference between the two side band frequencies are :
1
2
3
4
Official Solution
Correct Option:
(4)
Modulaton Index Upper band frequency (UBF) = Lower band frequency (LBF) = UBF - LBF =
10
PYQ 2019
medium
physicsID: jee-main
An amplitude modulated signal is plotted below: Which one of the following best describes the above signal ?
1
2
3
4
Official Solution
Correct Option:
(4)
Analysis of graph says (1) Amplitude varies as or (2) Two time period as (signal wave) & (carrier wave) Hence signal is
11
PYQ 2019
medium
physicsID: jee-main
In a line of sight radio communication, a distance of about is kept between the transmitting and receiving antennas. If the height of the receiving antenna is , then the minimum height of the transmitting antenna should be : (Radius of the Earth = ).
1
40 m
2
51 m
3
32 m
4
20 m
Official Solution
Correct Option:
(3)
Range =
by solving
12
PYQ 2019
medium
physicsID: jee-main
An amplitude modulated signal is given by . Here t is in seconds. The sideband frequencies (in kHz) are, [Given ]
1
1785 and 1715
2
892.5 and 857.5
3
89.25 and 85.75
4
178.5 smf 171.5
Official Solution
Correct Option:
(3)
Side band frequency are
13
PYQ 2020
medium
physicsID: jee-main
An amplitude modulated wave is represented by the expression volts. The minimum and maximum amplitudes of the amplitude modulated wave are, respectively :
1
2
3
4
Official Solution
Correct Option:
(3)
14
PYQ 2021
medium
physicsID: jee-main
Given below are two statements : Statement I : A speech signal of 2 kHz is used to modulate a carrier signal of 1 MHz. The bandwidth requirement for the signal is 4 kHz. Statement II : The side band frequencies are 1002 kHz and 998 kHz. In the light of the above statements, choose the correct answer :
1
Both Statement I and Statement II are true
2
Both Statement I and Statement II are false
3
Statement I is true but Statement II is false
4
Statement I is false but Statement II is true
Official Solution
Correct Option:
(1)
Step 1: Bandwidth of AM signal kHz kHz. Statement I is true.
Step 2: Sidebands are and . Step 3: kHz, kHz. Sidebands kHz and kHz. Statement II is true.
15
PYQ 2021
medium
physicsID: jee-main
An antenna is mounted on a 400 m tall building. What will be the wavelength of signal that can be radiated effectively by the transmission tower upto a range of 44 km ?
1
37.8 m
2
75.6 m
3
302 m
4
605 m
Official Solution
Correct Option:
(4)
Step 1: Understanding the Question:
The question asks for the wavelength ( ) of a signal that can be effectively radiated from an antenna on a 400 m building with a given range of 44 km. This question combines concepts of antenna theory and signal propagation. Step 2: Key Formula or Approach:
There are two main physical principles mentioned: effective radiation and range.
1. Effective Radiation: For an antenna to radiate effectively, its physical size (L) should be comparable to the wavelength of the signal. A common relationship for a monopole antenna (like one on a building) is or . Assuming the building height acts as the antenna length, .
2. Range of Transmission: For line-of-sight communication, the range is related to the height of the transmitting antenna and receiving antenna by , where R is the radius of the Earth. This formula does not directly involve wavelength. Step 3: Detailed Explanation:
Let's analyze the given data using the principles above.
- Antenna height, m.
- Range, km. Analysis of Effective Radiation:
If we assume is the antenna length, m.
- If , then m.
- If , then m.
Neither of these values matches the options closely. The option 302 m is somewhat close to if the effective length was different, and 605 m is not immediately obvious. Analysis of Range:
The line-of-sight range for a 400 m tower to the horizon is
km.
The given range of 44 km is well within the line-of-sight distance, so this information doesn't seem to constrain the wavelength in a standard way. Conclusion based on Official Answer Key:
The question is ambiguously worded and does not seem to correspond to a standard, simple physics formula connecting height, range, and wavelength. The information provided appears either insufficient or contradictory. However, the official answer key indicates that the correct answer is 605 m. This result may be based on specific assumptions about the type of antenna, propagation mode, or may stem from an error in the question's data. Without further context or a specific non-standard formula, it's not possible to derive this answer from first principles. Step 4: Final Answer:
Based on the provided official answer key, the wavelength is 605 m.
16
PYQ 2021
medium
physicsID: jee-main
An amplitude modulated wave is represented by volts. The modulating frequency in kHz will be _________.
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept:
An amplitude modulated (AM) signal is expressed as . The term inside the cosine represents the modulating signal (message signal). Step 2: Key Formula or Approach:
The modulating frequency is calculated from the angular modulating frequency using the relation:
Step 3: Detailed Explanation:
Comparing the given equation with the standard AM equation:
Calculating linear frequency :
Using :
Converting to kHz:
Step 4: Final Answer:
The modulating frequency is 2 kHz.
17
PYQ 2021
medium
physicsID: jee-main
A transmitting antenna has a height of 320 m and that of receiving antenna is 2000 m. The maximum distance between them for satisfactory communication in line of sight mode is 'd'. The value of 'd' is _________ km. (Radius of Earth = 6400 km)
Official Solution
Correct Option:
(1)
Step 1: Understanding the Question:
This problem deals with line-of-sight (LOS) communication between two antennas of given heights. We need to find the maximum possible separation between them for a signal to be received, taking into account the curvature of the Earth. Step 2: Key Formula or Approach:
The maximum line-of-sight distance ( ) between a transmitting antenna of height and a receiving antenna of height is the sum of their individual radio horizons. The formula is:
where is the radius of the Earth. It's important to use consistent units for all quantities. Since the answer is required in km, it's convenient to use R in km and the heights in km. Step 3: Detailed Explanation:
Given values:
Height of transmitting antenna, .
Height of receiving antenna, .
Radius of Earth, . Now, let's calculate the radio horizon for each antenna.
Radio horizon for transmitting antenna:
Radio horizon for receiving antenna:
The maximum line-of-sight distance is the sum of these two distances:
Step 4: Final Answer:
The value of 'd' is 224 km.
18
PYQ 2021
medium
physicsID: jee-main
A transmitting antenna at top of a tower has a height of 50 m and the height of receiving antenna is 80 m. What is the range of communication for Line of Sight (LoS) mode ?
[use radius of earth = 6400 km]
1
45.5 km
2
80.2 km
3
144.1 km
4
57.28 km
Official Solution
Correct Option:
(4)
Step 1: Understanding the Question:
We need to calculate the maximum line-of-sight (LoS) distance for communication between a transmitting antenna and a receiving antenna of given heights. Step 2: Key Formula or Approach:
The maximum LoS distance between two antennas of height (transmitter) and (receiver) is given by the sum of their individual horizon distances:
where R is the radius of the Earth. Step 3: Detailed Explanation:
Given values are:
Height of transmitting antenna, m.
Height of receiving antenna, m.
Radius of Earth, km = m = m. First, calculate the horizon distance for the transmitting antenna ( ):
Next, calculate the horizon distance for the receiving antenna ( ):
The total range of communication is the sum of these distances: Step 4: Final Answer:
The calculated range is approximately 57.3 km, which matches closely with option (D) 57.28 km.
19
PYQ 2021
medium
physicsID: jee-main
The logic circuit shown above is equivalent to :
1
A
2
B
3
C
4
D
Official Solution
Correct Option:
(3)
Step 1: Analyze the standard configuration. If the circuit shows two inputs and going through an OR gate followed by a NOT gate (or a NAND where inputs are inverted), we apply De Morgan’s Laws.
Step 2: For example, if the diagram is , it is a NOR gate. If it is , it is also equivalent to a NOR gate.
20
PYQ 2021
medium
physicsID: jee-main
A bandwidth of 6 MHz is available for A.M. transmission. If the maximum audio signal frequency used for modulating the carrier wave is not to exceed 6 kHz. The number of stations that can be broadcasted within this band simultaneously without interfering with each other will be ________.
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: In Amplitude Modulation (AM), the bandwidth occupied by a single station is twice the frequency of the modulating audio signal ( ). Step 2: Key Formula or Approach: 1. Bandwidth per station = . 2. Number of stations = . Step 3: Detailed Explanation: Given: Total BW = 6 MHz = Hz. Max audio signal frequency Hz. BW required per station:
Total number of stations:
Step 4: Final Answer: The number of stations is 500.
21
PYQ 2021
medium
physicsID: jee-main
If a message signal of frequency ' ' is amplitude modulated with a carrier signal of frequency ' ' and radiated through an antenna, the wavelength of the corresponding signal in air is:
1
2
3
4
Official Solution
Correct Option:
(3)
In amplitude modulation (AM), a low-frequency message signal ( ) modulates a high-frequency carrier signal ( ). The resulting AM signal contains three frequency components: the carrier frequency ( ), the upper sideband ( ), and the lower sideband ( ). Typically, the carrier frequency is much higher than the message frequency ( ). The signal that is radiated by the antenna is an electromagnetic wave. The vast majority of the power in an AM signal is concentrated at the carrier frequency. The wavelength ( ) of an electromagnetic wave is related to its frequency ( ) and the speed of light ( ) by the formula . Since the dominant frequency being transmitted is the carrier frequency, the wavelength of the radiated signal is determined by . Therefore, the wavelength is .
22
PYQ 2021
medium
physicsID: jee-main
If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at , then the maximum range of LOS communication is km. (Take radius of Earth )
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: The maximum line-of-sight (LOS) distance between two antennas of heights and is given by the sum of their individual horizons. To maximize this distance for a fixed sum of heights, the heights should be equal. Step 2: Key Formula or Approach: 1. LOS Distance: . 2. Constraint: . Step 3: Detailed Explanation: 1. Maximization condition: The function is maximized when for a constant sum . Thus, . 2. Calculate Range d:
3. Convert to km:
Step 4: Final Answer: The maximum range is 64 km.
23
PYQ 2021
medium
physicsID: jee-main
In Amplitude Modulation, the message signal volts and Carrier signal volts. The modulated signal now contains the message signal with lower side band and upper side band frequency, therefore the bandwidth of modulated signal is . The value of is :
1
200 kHz
2
100 kHz
3
50 kHz
4
0
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept:
In Amplitude Modulation (AM), the bandwidth is the difference between the maximum and minimum frequency components of the modulated signal.
The frequency components are the carrier frequency ( ), the upper sideband ( ), and the lower sideband ( ). Step 2: Key Formula or Approach:
Bandwidth ( ) of an AM signal = .
Where is the frequency of the message signal. Step 3: Detailed Explanation:
The given message signal is .
The standard form is .
Comparing the two, we get message frequency .
Convert Hz to kHz:
The bandwidth of the modulated signal is:
Given bandwidth is , so . Step 4: Final Answer:
The value of is 200.
24
PYQ 2021
medium
physicsID: jee-main
An audio signal amplitude modulates a carrier . The value of percent modulation is ________
Official Solution
Correct Option:
(1)
Step 1: Modulation index .
Step 2: Here, amplitude of message signal and amplitude of carrier .
Step 3: .
Step 4: Percent modulation .
25
PYQ 2021
medium
physicsID: jee-main
Match List - I with List - II.List - I
(a) 10 km height over earth's surface
(b) 70 km height over earth's surface
(c) 180 km height over earth's surface
(d) 270 km height over earth's surface List - II
(i) Thermosphere
(ii) Mesosphere
(iii) Stratosphere
(iv) Troposphere
(a)-(iv), (b)-(ii), (i) and (i) belong to higher regions.
Official Solution
Correct Option:
(1)
Step 1: **Troposphere**: 0 to 12 km (Matches 10 km).
Step 2: **Stratosphere**: 12 to 50 km.
Step 3: **Mesosphere**: 50 to 80 km (Matches 70 km).
Step 4: **Thermosphere**: Above 80 km (Matches 180 km and 270 km).
26
PYQ 2021
medium
physicsID: jee-main
The maximum amplitude for an amplitude modulated wave is found to be 12 V while the minimum amplitude is found to be 3 V. The modulation index is 0.6x where x is ________ .
Official Solution
Correct Option:
(1)
The formula for the modulation index ( ) in amplitude modulation is given by the maximum ( ) and minimum ( ) amplitudes of the modulated wave. . We are given the values:
V.
V. Substitute these values into the formula:
. Simplifying the fraction gives:
. The problem states that the modulation index is .
So, we have the equation . Solving for x, we get:
.
27
PYQ 2021
medium
physicsID: jee-main
A signal of 0.1 kW is transmitted in a cable. The attenuation of the cable is −5 dB per km and the cable length is 20 km. The power received at the receiver is W. The value of is __________.
Official Solution
Correct Option:
(1)
Step 1: Total loss .
Step 2: .
Step 3: .
Step 4: .
Correction: If input is , output is . If output is , .
(Note: Often in these problems, attenuation refers to current/voltage or the power values are different. Based on the logic of standard dB loss, would be 8.)
28
PYQ 2021
medium
physicsID: jee-main
A TV transmission tower antenna is at a height of 20 m. Suppose that the receiving antenna is at. (i) ground level (ii) a height of 5 m. The increase in antenna range in case (ii) relative to case (i) is n%. The value of n, to the nearest integer, is __________.
Official Solution
Correct Option:
(1)
Step 1: The range of communication is given by .
Step 2: Case (i): .
Step 3: Case (ii): m.
Note that .
Step 4: Calculate percentage increase.
29
PYQ 2021
medium
physicsID: jee-main
The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulated by a 7.7 kHz sine wave of 5 V amplitude are V and V respectively. Then the value of is _________.
Official Solution
Correct Option:
(1)
In standard Amplitude Modulation (AM), the amplitudes of the upper sideband (USB) and the lower sideband (LSB) are equal. The amplitude of each sideband is given by the formula:
where is the amplitude (peak voltage) of the carrier wave and is the modulation index. First, we calculate the modulation index, :
where is the amplitude of the modulating wave. Given: Carrier amplitude, V. Modulating wave amplitude, V. . Now, we calculate the amplitude of the sidebands: V. The amplitude of the upper sideband is 2.5 V, and the amplitude of the lower sideband is also 2.5 V. We are given that the amplitudes are V and V. Amplitude of upper sideband: . Amplitude of lower sideband: . The value of the ratio is: .
30
PYQ 2022
medium
physicsID: jee-main
A radio can tune to any station in to band The value of corresponding wavelength bandwidth will be :
1
4 m
2
20 m
3
30 m
4
50 m
Official Solution
Correct Option:
(2)
The correct option is(B): 20 m.
31
PYQ 2022
medium
physicsID: jee-main
At a particular station, the TV transmission tower has a height of 100 m. To triple its coverage range, height of the tower should be increased to
1
200 m
2
300 m
3
600 m
4
900 m
Official Solution
Correct Option:
(4)
To solve this question, we need to understand the relationship between the height of a TV transmission tower and its coverage range. The range of coverage of a transmission tower is directly proportional to the square root of its height. Mathematically, this can be expressed as:
Where is the range and is the height of the tower.
Given that initially, the tower has a height of 100 m, and we want to triple its coverage range, the calculation can be done as follows:
(for initial height where is a proportionality constant)
We want the new range to be triple the initial range, so
For the new range, the formula becomes
Equating,
Divide both sides by (assuming ),
Square both sides to solve for , we get
Thus, to triple the coverage range, the height of the tower should be increased to 900 m.
Correct Answer: 900 m
The options provided were 200 m, 300 m, 600 m, and 900 m, and based on our calculations, the correct option is indeed 900 m.
Tip: Remember that the relationship between tower height and coverage range follows the formula . This can help quickly solve similar problems.
32
PYQ 2022
medium
physicsID: jee-main
In the case of amplitude modulation to avoid distortion the modulation index should be:
1
2
3
4
Official Solution
Correct Option:
(1)
From the given options the correct answer is option (A):
33
PYQ 2022
medium
physicsID: jee-main
A FM Broad cast transmitter, using modulating signal of frequency 20 kHz has a deviation ratio of 10. The Bandwidth required for transmission is
1
220 kHz
2
180 kHz
3
360 kHz
4
440 kHz
Official Solution
Correct Option:
(4)
The modulating frequency is
The deviation ratio is
Frequency deviation
The bandwidth is given by
34
PYQ 2022
medium
physicsID: jee-main
The height of a transmitting antenna at the top of a tower is 25 m and that of receiving antenna is, 49 m. The maximum distance between them, for satisfactory communication in LOS (Line-Of-Sight) is . The value of K is ______ . (Assume radius of Earth is ) [Calculate upto nearest integer value]
Official Solution
Correct Option:
(1)
The maximum distance between them, for satisfactory communication in LOS (Line-Of-Sight) is . Then,
So, the answer is .
35
PYQ 2023
medium
physicsID: jee-main
Match List I with List II:
List I
List II
A.
Attenuation
I.
Combination of a receiver and transmitter
B.
Transducer
II.
Process of retrieval of information from the carrier wave at received
C.
Demodulation
III.
Converts one form of energy into another
D.
Repeater
IV.
Loss of strength of a signal while propagating through a medium
Choose the correct answer from the options given below :
1
A-II, B-III, C-IV, D-I
2
A-IV, B-III, C-II, D-I
3
A-I, B-II, C-III, D-IV
4
A-IV, B-III, C-I, D-II
Official Solution
Correct Option:
(2)
Let's analyze the terms from List I and match them with their corresponding terms in List II:
A: Attenuation
Attenuation refers to the loss of strength of a signal while propagating through a medium. This matches with option IV in List II: Loss of strength of a signal while propagating through a medium.
B: Transducer
A transducer is a device that converts one form of energy into another, typically used in sensors and receivers. This matches with option III in List II: Converts one form of energy into another.
C: Demodulation
Demodulation is the process of retrieval of information from the carrier wave at the receiver. This matches with option II in List II: Process of retrieval of information from the carrier wave at received.
D: Repeater
A repeater is a device that is a combination of a receiver and a transmitter, used to extend the range of communication signals. This matches with option I in List II: Combination of a receiver and transmitter.
Correct Matches:
A – IV: Attenuation – Loss of strength of a signal while propagating through a medium
B – III: Transducer – Converts one form of energy into another
C – II: Demodulation – Process of retrieval of information from the carrier wave at received
D – I: Repeater – Combination of a receiver and transmitter
The correct answer is (B) A-IV, B-III, C-II, D-I
36
PYQ 2023
medium
physicsID: jee-main
In AM modulation, a signal is modulated on a carrier wave such that maximum and minimum amplitude are found to be and respectively The modulation index is
1
2
3
4
Official Solution
Correct Option:
(4)
modulation index =Vmax+Vmin.Vmax−Vmin.×100% =6+26−2×100%=50%
37
PYQ 2023
easy
physicsID: jee-main
A sinusoidal carrier voltage is amplitude modulated The resultant amplitude modulated wave has maximum and minimum amplitude of and respectively The amplitude of each sideband is:
1
2
3
4
Official Solution
Correct Option:
(1)
Ac+Am=120 Ac−Am=80 ∴Ac=100 Am=20
Modulation index =10020=51
Amplitude of each sideband =Ac2( mod ulation index ) =100×101=10volt
So, the correct ans is (A): 10 V.
38
PYQ 2023
easy
physicsID: jee-main
The height of the antenna is 98 m. The radius of Earth is 6400 km. The area up to which it will transmit signal is:
1
3642 km2
2
3942 km2
3
11200 km2
4
22400 km2
Official Solution
Correct Option:
(2)
Step 1: Use the formula for surface area covered.- Maximum distance covered d = .- Surface area A = πd2 ≈ 2πRhT.
Step 2: Substitute the values.
A=2πRhT
A = 2 · 3.14 · 6400 · 98 · 10-3.
A ≈ 3942 km2.
Final Answer: Surface area covered = 3942km2.
39
PYQ 2023
medium
physicsID: jee-main
Which of the following frequencies does not belong to FM broadcast?
1
2
3
4
Official Solution
Correct Option:
(1)
The range of to is chosen for FM broadcasting because it provides a wide range of frequencies that can carry a diverse range of audio content while avoiding interference with other radio services operating in different frequency bands. This frequency range also allows for a reasonable balance between signal coverage and bandwidth utilization. Then, the frequency that does not belong to FM broadcast is because it falls outside the typical FM broadcast frequency range.
So, the correct option is (A):
40
PYQ 2023
medium
physicsID: jee-main
In satellite communication, the uplink frequency band used is :
1
76 – 88 MHz
2
3.7 – 4.2 GHz
3
5.925 – 6.425 GHz
4
420 – 890 MHz
Official Solution
Correct Option:
(3)
In satellite communication:
The uplink frequency band is used for transmitting signals from an earth station to a satellite.
The downlink frequency band is used for transmitting signals from a satellite to an earth station.
Uplink frequencies are generally higher than downlink frequencies to minimize interference.
Common Standard Frequency Bands:
Uplink:
Downlink:
Analysis:
The frequency band falls within the standard uplink frequency range for satellite communication.
Final Answer:
The uplink frequency band used in satellite communication is .
41
PYQ 2023
medium
physicsID: jee-main
If the height of the tower used for L.D.S is increased by 21% then percentage change in range is?
1
10%
2
21%
3
19%
4
42%
Official Solution
Correct Option:
(1)
The range of the L.D.S. (Laser Distance Sensor) is given by: Range ∝ This means that the range of the L.D.S. is directly proportional to the square root of the height of the tower. If the height of the tower is increased by 21%, then the new height is: New Height = Old Height + 0.21 x Old Height = 1.21 x Old Height So the percentage change in height is 21%. Now, using the formula above, we can write: New Range ∝ New Range ∝ New Range ∝ New Range ∝ 1.1 x Therefore, the new range is 1.1 times the square root of the old range, which is an increase of: Percentage change in range = (New Range - Old Range) / Old Range x 100% Percentage change in range = [(1.1 x ) - ] / x 100% Percentage change in range = (0.1 / ) x 100% Substituting the value of the percentage change in height, we get: Percentage change in range = ( ) x 100% = ( ) x 100% Percentage change in range = 10% Therefore, the percentage change in range is 10%. Answer. A
42
PYQ 2023
easy
physicsID: jee-main
To radiate EM signal of wavelength with high efficiency, the antennas should have a minimum size equal to:
1
2
2
3
4
Official Solution
Correct Option:
(2)
For efficient radiation of electromagnetic (EM) waves, the minimum length of an antenna should be proportional to the wavelength of the signal being transmitted. The minimum length of the antenna is given by:
where is the wavelength of the EM signal. Thus, the minimum size of the antenna for efficient radiation is .
43
PYQ 2023
medium
physicsID: jee-main
Height of the receiving and transmitting antenna in communication of a signal are 245 m and 180 m respectively. Find the maximum distance between the two antennas for proper communication.
1
104 km
2
208 km
3
52 km
4
96 km
Official Solution
Correct Option:
(1)
Understanding the Problem
We need to find the maximum distance in the line of sight between two antennas using the formula:
where:
(radius of the Earth)
(height of the transmitting antenna)
(height of the receiving antenna)
Corrected Solution
1. Convert Units:
Since is in kilometers, we need to convert and to kilometers.
2. Substitute Values into the Formula:
3. Calculate the Distance:
Error in Your Calculation
You directly added the heights in meters and multiplied them by the radius in kilometers without converting the heights to kilometers. This led to a significantly larger and incorrect result.
Final Answer
The maximum distance in the line of sight between the two antennas is approximately .
44
PYQ 2023
medium
physicsID: jee-main
A modulating signal is a square wave, as shown in the figure If the carrier wave is given as volts, the modulation index is:
1
2
1
3
4
Official Solution
Correct Option:
(3)
Given:
The carrier wave is given as (in volts).
Therefore, the amplitude of the carrier wave .
From the figure, the amplitude of the modulating signal .
Therefore, the modulation index is given by:
45
PYQ 2023
easy
physicsID: jee-main
If the height of the tower used for LOS communication is increased by 21%, the percentage change in range is?
1
5%
2
10%
3
15%
4
12%
Official Solution
Correct Option:
(2)
Percentage change in d = 10%
So, the correct option is (B): 10%
46
PYQ 2023
medium
physicsID: jee-main
By what percentage will the transmission range of a TV tower be affected when the height of the tower is increased by 21% ?
1
12%
2
15%
3
14%
4
10%
Official Solution
Correct Option:
(4)
TV Tower Transmission Range Problem
Step 1: Transmission Range Formula
The transmission range ( ) of a TV tower is given by:
where is the radius of the Earth and is the height of the tower.
Step 2: New Height
If the height is increased by 21%, the new height ( ) is:
Step 3: New Transmission Range
The new transmission range ( ) is:
Step 4: Percentage Increase
Since , the new range is:
The percentage increase in the transmission range is:
Conclusion:
The transmission range increases by 10% (Option 4).
47
PYQ 2023
medium
physicsID: jee-main
A transmitting antenna is kept on the surface of the earth. The minimum height of receiving antenna required to receive the signal in line of sight at 4 km distance from it is m. The value of is:
1
125
2
12.5
3
1250
4
1.25
Official Solution
Correct Option:
(1)
Let be the radius of the Earth, and be the height of the antenna. The distance between the two antennas is given by: We are given that , and the radius of the Earth . Substituting these values into the equation: Squaring both sides: Now, using the formula for the signal range: Thus, the value of is 125.
48
PYQ 2023
medium
physicsID: jee-main
The required height of a TV tower which can cover the population of lakh is If the average population density is per square and the radius of earth is , then the value of will be ______
Official Solution
Correct Option:
(1)
The correct answer is 150.
49
PYQ 2023
medium
physicsID: jee-main
Match List I with List I:
List I
List II
A
AM Broadcast
I
88-108 MHz
B
FM Broadcast
II
540-1600 kHz
C
Television
III
3.7-4.2 GHz
D
Satellite Communication
IV
54 MHz-890 MHz
Choose the correct answer from the options given below:
1
A-II, B-III, C-I, D-IV
2
A-II, B-I, C-IV, D-III
3
A-IV, B-III, C-I, D-II
4
A-I, B-III, C-II, D-IV
Official Solution
Correct Option:
(2)
AM Broadast 540−1600 KHz FM Broadcast 88−108 MHz Television 54−890 MHz Salellite communication 3.7−4.2 GHz
So, the correct option is (B): A-II, B-I, C-IV, D-III.
50
PYQ 2026
hard
physicsID: jee-main
Find the correct combination of A, B, C and D inputs which can cause the LED to glow.
1
0100
2
1000
3
0011
4
1101
Official Solution
Correct Option:
(2)
Step 1: Analyze the logic gates in the circuit. The circuit consists of NOR gates at the inputs and output. The LED glows only when the final output is logic HIGH. Step 2: Evaluate the logic condition. For the LED to glow, the final NOR gate must receive all LOW inputs. This occurs when:
Step 3: Write the input combination. Final Answer:
51
PYQ 2026
medium
physicsID: jee-main
Identify the correct truth table of the given logic circuit.
1
A
2
B
3
C
4
D
Official Solution
Correct Option:
(2)
Step 1: Identify the logic gates used in the circuit.
From the diagram, the upper gate connected to input is an AND gate with both inputs same, hence its output is
The lower left gate is a NAND gate with inputs and , producing output
Step 2: Analyze the middle gate.
The output of the NAND gate is fed into an AND gate whose both inputs are the same, so its output remains
Step 3: Write expression for final output.
The final gate is an AND gate combining the two signals, hence
Step 4: Simplify the Boolean expression.
Using Boolean algebra,
Step 5: Construct the truth table from the expression .
This matches exactly with Option (B).
52
PYQ 2026
hard
physicsID: jee-main
Select correct truth table.
1
2
3
4
Official Solution
Correct Option:
(4)
Step 1: Writing Boolean expression from logic circuit. From the given logic circuit, the output is:
Step 3: Matching with given options. The above truth table matches option (4).
About Communication Systems - JEE-MAIN
Communication Systems is a vital chapter for JEE-MAIN aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.
By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.
Frequently Asked Questions
Why focus on Communication Systems PYQs?
Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.
How to best use this analysis?
Review the topic breakdown to see which sub-topics within Communication Systems carry the most weight. Then, tackle the questions iteratively to solidify your understanding.
