2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
A 25 m long antenna is mounted on an antenna tower. The height of the antenna tower is 75 m. The wavelength (in meter) of the signal transmitted by this antenna would be :
A.
200
B.
400
C.
100
D.
300
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
If a message signal of frequency 'fm' is amplitude modulated with a carrier signal of frequency 'fc' and radiated through an antenna, the wavelength of the corresponding signal in air is :
A.
${c \over {{f_c} + {f_m}}}$
B.
${c \over {{f_c}}}$
C.
${c \over {{f_c} - {f_m}}}$
D.
${c \over {{f_m}}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
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 from the options given below :
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 from the options given below :
A.
Statement I is false but Statement II is true
B.
Both Statement I and Statement II are false
C.
Statement I is true but Statement II is false
D.
Both Statement I and Statement II are true
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
An amplitude modulated wave is represented
by the expression
vm = 5(1 + 0.6 cos 6280t) sin(211 × 104 t) volts
The minimum and maximum amplitudes of the amplitude modulated wave are, respectively
vm = 5(1 + 0.6 cos 6280t) sin(211 × 104 t) volts
The minimum and maximum amplitudes of the amplitude modulated wave are, respectively
A.
${3 \over 2}$ V, 5 V
B.
${5 \over 2}$ V, 8 V
C.
5 V, 8 V
D.
3 V, 5 V
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
In an amplitude modulator circuit, the carrier wave is given by, C(t) = 4 sin(20000 $\pi $t) while modulating signal
is given by, m(t) = 2 sin (2000 $\pi $t). The values of modulation index and lower side band frequency are :
A.
0.3 and 9 kHz
B.
0.5 and 10 kHz
C.
0.4 and 10 kHz
D.
0.5 and 9 kHz
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
A message signal of frequency 100 MHz and
peak voltage 100 V is used to execute
amplitude modulation on a carrier wave of
frequency 300 GHz and peak voltage 400 V.
The modulation index and difference between
the two side band frequencies are :
A.
0.25; 1 × 108 Hz
B.
4; 2 × 108 Hz
C.
4; 1 × 108 Hz
D.
0.25; 2 × 108 Hz
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
Given below in the the left column are different
modes of communication using the kinds of
waves given the right column.
| A | Optical Fibre communication |
P | Ultrasound |
|---|---|---|---|
| B | Radar | Q | Infrared Light |
| C | Sonar | R | Microwaves |
| D | Mobile Phones | S | Radio Waves |
A.
A-Q, B-S, C-R, D-P
B.
A-Q, B-S, C-P, D-R
C.
A-S, B-Q, C-R, D-P
D.
A-R, B-P, C-S, D-Q
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
The physical sizes of the transmitter and
receiver antenna in a communication system
are :-
A.
proportional to carrier frequency
B.
inversely proportional to carrier frequency
C.
inversely proportional to modulation
frequency
D.
independent of both carrier and modulation
frequency
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
A signal Acoswt is transmitted using v0 sin $\omega $0t
as carrier wave. The correct amplitude
modulated (AM) signal is
A.
v0 sin $\omega $0t + Acos$\omega $t
B.
(v0 + A)cos$\omega $t sin$\omega $0 t
C.
v0 sin[$\omega $0 (1+ 0.01Asin$\omega $t)t]
D.
${v_0}\sin {\omega _0}t + {{\rm A} \over 2}\sin ({\omega _0} - \omega )t + {A \over 2}\sin ({\omega _0} + \omega )t$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
In a line of sight radio communication, a
distance of about 50 km is kept between the
transmitting and receiving antennas. If the
height of the receiving antenna is 70m, then the
minimum height of the transmitting antenna
should be :
(Radius of the Earth = 6.4 × 106 m).
(Radius of the Earth = 6.4 × 106 m).
A.
40 m
B.
20 m
C.
51 m
D.
32 m
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
The wavelength of the carrier waves in a
modern optical fiber communication network
is close to :
A.
1500 nm
B.
2400 nm
C.
600 nm
D.
900 nm
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
To double the covering range of a TV transmittion tower, its height should be multiplied by :
A.
4
B.
2
C.
$\sqrt 2 $
D.
${1 \over {\sqrt 2 }}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index ?
A.
0.5
B.
0.6
C.
0.4
D.
0.3
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
An amplitude modulated signal is plotted below :
Which one of the following best describes the above signal ?
Which one of the following best describes the above signal ?
A.
(1 + 9sin (2$\pi $ $ \times $ 104 t)) sin(2.5$\pi $ $ \times $ 105t) V
B.
(9 + sin (4$\pi $ $ \times $ 104 t)) sin(5$\pi $ $ \times $ 105t) V
C.
(9 + sin (2$\pi $ $ \times $ 104 t)) sin(2.5$\pi $ $ \times $ 105t) V
D.
(9 + sin (2.5$\pi $ $ \times $ 104 t)) sin(2$\pi $ $ \times $ 104t) V
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
An amplitude modulated signal is given by V(t) = 10[1 + 0.3cos(2.2 $ \times $ 104
t)] sin(5.5 $ \times $ 105
t). Here t is in
seconds. The sideband frequencies (in kHz) are, [Given $\pi $ = 22/7]
A.
892.5 and 857.5
B.
89.25 and 85.75
C.
1785 and 1715
D.
178.5 and 171.5
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot ?
A.
2900 kHz
B.
2750 kHz
C.
2250 kHz
D.
2000 kz
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Morning Slot
A TV transmission tower has a height of 140 m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower is LOS (Line of Sight) mode ? (Given : radius of earth = 6.4 × 106 m).
A.
40 km
B.
65 km
C.
48 km
D.
80 km
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwith. The number of channels accomodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c = 3 $ \times $ 108m/s, h = 6.6 $ \times $ 10$-$34 J-s)
A.
3.75 $ \times $ 106
B.
3.86 $ \times $ 106
C.
6.25 $ \times $ 105
D.
4.87 $ \times $ 105
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 16th April Morning Slot
A carrier wave of peak voltage 14 V is used for transmitting a message signal. The peak voltage of modulating signal given to achieve a modulation index of 80% will be :
A.
7 V
B.
28 V
C.
11.2 V
D.
22.4 V
2018
JEE Mains
MCQ
JEE Main 2018 (Offline)
A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized
for transmission. How many telephonic channels can be transmitted simultaneously if each channel
requires a bandwidth of 5 kHz ?
A.
2 $\times$ 106
B.
2 $\times$ 103
C.
2 $\times$ 105
D.
2 $\times$ 104
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Evening Slot
The carrier frequency of a transmitter is provided by a tank circuit of a coil of inductance 49 $\mu $H and a capacitance of 2.5 nF. It is modulated by an audio signal of $12$ $kHz.$ The frequency range occupied by the side bands is :
A.
13482 kHz $-$ 13494 kHz
B.
442 kHz $-$ 466 kHz
C.
63 kHz $-$ 75 kHz
D.
18 kHz $-$ 30 kHz
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Morning Slot
The number of amplitude modulted broadcast stations that can be accomodated in a $300$ $kHz$ band width for the highest modulating frequency $15$ $kHz$ will be :
A.
$20$
B.
$15$
C.
$10$
D.
$8$
2017
JEE Mains
MCQ
JEE Main 2017 (Online) 9th April Morning Slot
A signal is to be transmitted through a wave of wavelength $\lambda $, using a linear antenna. The length l of the antenna and effective power radiated Peff will be given respectively as :
(K is a constant of proportionality)
(K is a constant of proportionality)
A.
$\lambda $, Peff = K $\left( {{1 \over \lambda }} \right)$2
B.
${{\lambda \over 8}}$, Peff = K $\left( {{1 \over \lambda }} \right)$
C.
${{\lambda \over 16}}$, Peff = K $\left( {{1 \over \lambda }} \right)$3
D.
${{\lambda \over 5}}$, Peff = K $\left( {{1 \over \lambda }} \right)$${\left( {{1 \over \lambda }} \right)^{{1 \over 2}}}$
2017
JEE Mains
MCQ
JEE Main 2017 (Online) 8th April Morning Slot
A signal of frequency 20 kHz and peak voltage of 5 Volt is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 Volts. Choose the correct statement.
A.
Modulation index = 5, side frequency bands are at 1400 kHz and 1000 kHz
B.
Modulation index = 5, side frequency bands are at 21.2 kHz and 18.8 kHz
C.
Modulation index = 0.8, side frequency bands are at 1180 kHz and 1220 kHz
D.
Modulation index = 0.2, side frequency bands are at 1220 kHz and 1180 kHz
2017
JEE Mains
MCQ
JEE Main 2017 (Offline)
In amplitude modulation, sinusoidal carrier frequency used is denoted by ${\omega _c}$ and the signal frequency is
denoted by ${\omega _m}$. The bandwidth ($\Delta {\omega _m}$) of the signal is such that $\Delta {\omega _m}$ < < $\omega _c$. Which of the following frequencies is not contained in the modulated wave?
A.
${\omega _m}$
B.
${\omega _c}$
C.
${\omega _m}$ + ${\omega _c}$
D.
${\omega _c}$ - ${\omega _m}$
2016
JEE Mains
MCQ
JEE Main 2016 (Online) 10th April Morning Slot
A modulated signal Cm(t) has the form Cm(t) = 30 sin 300 $\pi $t + 10 (cos 200 $\pi $t −cos 400 $\pi $t). The carrier frequency c, the modulating frequency (message frequency) f$\omega $, and the modulation index $\mu $ are respectively given by :
A.
fc = 200 Hz; f$\omega $ = 50 Hz; $\mu $ = ${1 \over 2}$
B.
fc = 150 Hz; f$\omega $ = 50 Hz; $\mu $ = ${2 \over 3}$
C.
fc = 150 Hz; f$\omega $ = 30 Hz; $\mu $ = ${1 \over 3}$
D.
fc = 200 Hz; f$\omega $ = 30 Hz; $\mu $ = ${1 \over 2}$
2016
JEE Mains
MCQ
JEE Main 2016 (Online) 9th April Morning Slot
An audio signal consists of two distinct sounds : one a human speech signal in the frequency band of 200 Hz to 2700 Hz, while the other is a high frequency music signal in the frequency band of 10200 Hz to 15200 Hz. The ratio of the AM signal bandwidth required to send both the signals together to the AM signal bandwidth required to send just the human speech is :
A.
3
B.
5
C.
6
D.
2
2016
JEE Mains
MCQ
JEE Main 2016 (Offline)
Choose the correct statement :
A.
In amplitude modulation the amplitude of the high frequency carrier wave is make to vary in proportion to the amplitude of the audio signal.
B.
In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
C.
In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
D.
In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal.
2015
JEE Mains
MCQ
JEE Main 2015 (Offline)
A signal of $5$ $kHz$ frequency is amplitude modulated on a carrier wave of frequency $2$ $MHz.$ The frequencies of the resultant signal is/are :
A.
$2005$ $kHz, 2000$ $kHz$ and $1995$ $kHz$
B.
$2000$ $kHz$ and $1995$ $kHz$
C.
$2$ $MHz$ only
D.
$2005$ $kHz$ and $1995$ $kHz$
2012
JEE Mains
MCQ
AIEEE 2012
A radar has a power of $1kW$ and is operating at a frequency of $10$ $GHz.$ It is located on a mountain top of height $500$ $m.$ The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth $ = 6.4 \times {10^6}m$) is :
A.
$80$ $km$
B.
$16$ $km$
C.
$40$ $km$
D.
$64$ $km$
2011
JEE Mains
MCQ
AIEEE 2011
This question has Statement - $1$ and Statement - $2$. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement -$1$ : Sky wave signals are used for long distance radio communication. These signals are in general, less stable then ground wave signals.
Statement -$2$ : The state of ionosphere varies from hour to hour, day to day and season to season.
Statement -$1$ : Sky wave signals are used for long distance radio communication. These signals are in general, less stable then ground wave signals.
Statement -$2$ : The state of ionosphere varies from hour to hour, day to day and season to season.
A.
Statement - $1$ is true, Statement - $2$ is true, Statement - $2$ is the correct explanation of Statement - $1$.
B.
Statement - $1$ is true, Statement - $2$ is true, Statement - $2$ is not the correct explanation of Statement - $1$.
C.
Statement - $1$ is false, Statement - $2$ is true
D.
Statement - $1$ is true, Statement - $2$ is false.
2003
JEE Mains
MCQ
AIEEE 2003
Consider telecommunication through optical fibres. Which of the following statements is not true?
A.
Optical fibres can be of graded refractive index
B.
Optical fibres are subject to electromagnetic interference from outside
C.
Optical fibres have extremely low transmission loss
D.
Optical fibres may have homogeneous core with a suitable cladding.
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 29th July Evening Shift
A modulating signal $2 \sin \left(6.28 \times 10^{6}\right) t$ is added to the carrier signal $4 \sin \left(12.56 \times 10^{9}\right) t$ for amplitude modulation. The combined signal is passed through a non-linear square law device. The output is then passed through a band pass filter. The bandwidth of the output signal of band pass filter will be _________ $\mathrm{MHz}$.
Correct Answer: 2
Explanation:
$\mathrm{W}_{\mathrm{C}}=12.56 \times 10^{9}$
$ W_{m}=6.25 \times 10^{6} $
After amplitude modulation
Bandwidth frequency
$ =\frac{2 W_{m}}{2 \pi}=\frac{2 \times 6.28}{2 \pi} \times 10^{6}=2 \mathrm{MHz} $
$ W_{m}=6.25 \times 10^{6} $
After amplitude modulation
Bandwidth frequency
$ =\frac{2 W_{m}}{2 \pi}=\frac{2 \times 6.28}{2 \pi} \times 10^{6}=2 \mathrm{MHz} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th July Morning Shift
The required height of a TV tower which can cover the population of $6.03$ lakh is $h$. If the average population density is 100 per square $\mathrm{km}$ and the radius of earth is $6400 \mathrm{~km}$, then the value of $h$ will be _____________ $m$.
Correct Answer: 150
Explanation:
$r = \sqrt {{{(h + R)}^2} - {R^2}} \cong \sqrt {2hR} $
$A = {{6.03 \times {{10}^5}} \over {100}}$
$\pi {r^2} = 6.03 \times {10^3}$
$\pi 2Rh = 6.03 \times {10^3}$
$h = {{6.03 \times {{10}^3}} \over {2 \times \pi \times R}} = 0.015 \times 10 \times {10^3}$ m
$=150$ m
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th June Morning Shift
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 K$\sqrt5$ $\times$ 102 m. The value of K is ___________.
(Assume radius of Earth is 64 $\times$ 10+5 m) [Calculate upto nearest integer value]
Correct Answer: 192
Explanation:
$d = \sqrt {2{h_t}{R_e}} + \sqrt {2 \times {h_R}{R_e}} $
$ = \sqrt {2 \times 25 \times 64 \times {{10}^5}} + \sqrt {2 \times 49 \times 64 \times {{10}^5}} $
$ = 8000\sqrt 5 + 11200\sqrt 5 $ m
$ = 19200\sqrt 5 $ m
$ = 192\sqrt 5 + {10^2}$ m
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 24th June Evening Shift
An antenna is placed in a dielectric medium of dielectric constant 6.25. If the maximum size of that antenna is 5.0 mm, it can radiate a signal of minimum frequency of __________ GHz.
(Given $\mu$r = 1 for dielectric medium)
Correct Answer: 6
Explanation:
We know that v = f$\lambda$
Putting the values,
${{3 \times {{10}^8}} \over {\sqrt {6.25} }} = f \times 20 \times {10^{ - 3}}$
$ \Rightarrow f = 6 \times {10^9}$ Hz
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 1st September Evening Shift
A carrier wave with amplitude of 250 V is amplitude modulated by a sinusoidal base band signal of amplitude 150V. The ratio of minimum amplitude to maximum amplitude for the amplitude modulated wave is 50 : x, then value of x is ____________.
Correct Answer: 200
Explanation:
Given, the amplitude of the carrier wave, Ac = 250 V
The amplitude of the message wave, Am = 150 V
We know that,
The maximum amplitude, Amax = Ac + Am
Substituting the values in the above equation, we get
Amax = 250 + 150 = 400 V
We know that,
The minimum amplitude, Amin = Ac $-$ Am
Substituting the values in the above equation, we get
Amin = 250 $-$ 150 = 100 V
Thus, the ratio of the minimum amplitude to the maximum amplitude of the modulated wave is
${{{A_{\min }}} \over {{A_{\max }}}} = {{100} \over {400}} = {1 \over 4}$
Comparing with, 1 : x
The value of the x = 4.
The amplitude of the message wave, Am = 150 V
We know that,
The maximum amplitude, Amax = Ac + Am
Substituting the values in the above equation, we get
Amax = 250 + 150 = 400 V
We know that,
The minimum amplitude, Amin = Ac $-$ Am
Substituting the values in the above equation, we get
Amin = 250 $-$ 150 = 100 V
Thus, the ratio of the minimum amplitude to the maximum amplitude of the modulated wave is
${{{A_{\min }}} \over {{A_{\max }}}} = {{100} \over {400}} = {1 \over 4}$
Comparing with, 1 : x
The value of the x = 4.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
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 __________.
Correct Answer: 500
Explanation:
Signal bandwidth = 2 fm = 12 kHz
$\therefore$ N = ${{6MHZ} \over {12kHZ}} = {{6 \times {{10}^6}} \over {12 \times {{10}^3}}} = 500$
$\therefore$ N = ${{6MHZ} \over {12kHZ}} = {{6 \times {{10}^6}} \over {12 \times {{10}^3}}} = 500$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at 160 m, then the maximum range of LOS communication is ___________ km. (Take radius of Earth = 6400 km)
Correct Answer: 64
Explanation:
hT = hR = 160 ........ (i)
$d = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $
$d = \sqrt {2R} \left[ {\sqrt {{h_T}} + \sqrt {{h_R}} } \right]$
$d = \sqrt {2R} \left[ {\sqrt x + \sqrt {160 - x} } \right]$
${{d(d)} \over {dx}} = 0$
${1 \over {2\sqrt x }} + {{1( - 1)} \over {2\sqrt {160 - x} }} = 0$
${1 \over {\sqrt x }} = {1 \over {\sqrt {160 - x} }}$
x = 80 m
${d_{\max }} = \sqrt {2 \times 6400} \left[ {\sqrt {{{80} \over {1000}}} + \sqrt {{{20} \over {1000}}} } \right]$
$ = {{80\sqrt 2 \times 2\sqrt {80} } \over {10\sqrt {10} }}$
$ = 8 \times 2 \times \sqrt 2 \times 2\sqrt 2 = 64$ km
$d = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $
$d = \sqrt {2R} \left[ {\sqrt {{h_T}} + \sqrt {{h_R}} } \right]$
$d = \sqrt {2R} \left[ {\sqrt x + \sqrt {160 - x} } \right]$
${{d(d)} \over {dx}} = 0$
${1 \over {2\sqrt x }} + {{1( - 1)} \over {2\sqrt {160 - x} }} = 0$
${1 \over {\sqrt x }} = {1 \over {\sqrt {160 - x} }}$
x = 80 m
${d_{\max }} = \sqrt {2 \times 6400} \left[ {\sqrt {{{80} \over {1000}}} + \sqrt {{{20} \over {1000}}} } \right]$
$ = {{80\sqrt 2 \times 2\sqrt {80} } \over {10\sqrt {10} }}$
$ = 8 \times 2 \times \sqrt 2 \times 2\sqrt 2 = 64$ km
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Morning Shift
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.
Correct Answer: 224
Explanation:
${d_m} = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $
${d_m} = \left( {\sqrt {2 \times 6400 \times {{10}^3} \times 320} + \sqrt {2 \times 6400 \times {{10}^3} \times 2000} } \right)$m
dm = 224 km
${d_m} = \left( {\sqrt {2 \times 6400 \times {{10}^3} \times 320} + \sqrt {2 \times 6400 \times {{10}^3} \times 2000} } \right)$m
dm = 224 km
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Morning Shift
An amplitude modulated wave is represented by
Cm(t) = 10(1 + 0.2 cos 12560t) sin(111 $\times$ 104t) volts. The modulating frequency in kHz will be ................. .
Cm(t) = 10(1 + 0.2 cos 12560t) sin(111 $\times$ 104t) volts. The modulating frequency in kHz will be ................. .
Correct Answer: 2
Explanation:
Wm = 12560 = 2$\pi$fm
${f_m} = {{12560} \over {2\pi }}$
= 2000 Hz
${f_m} = {{12560} \over {2\pi }}$
= 2000 Hz
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
The maximum amplitude for an amplitude modulated wave is found to be 12V while the minimum amplitude is found to be 3V. The modulation index is 0.6x where x is ____________.
Correct Answer: 1
Explanation:
Amax = Ac + Am = 12
Amin = Ac $-$ Am = 3
$\Rightarrow$ Ac = ${{15} \over 2}$ & Am = ${9 \over 2}$
modulation index = ${{{A_m}} \over {{A_c}}} = {{9/2} \over {15/2}} = 0.6$
$\Rightarrow$ x = 1
Amin = Ac $-$ Am = 3
$\Rightarrow$ Ac = ${{15} \over 2}$ & Am = ${9 \over 2}$
modulation index = ${{{A_m}} \over {{A_c}}} = {{9/2} \over {15/2}} = 0.6$
$\Rightarrow$ x = 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
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 5V amplitude are ${a \over {10}}V$ and ${b \over {10}}V$ respectively. Then the value of ${a \over b}$ is ____________.
Correct Answer: 1
Explanation:
${a \over {10}} = {b \over {10}} = {{\mu {A_C}} \over 2}$
$ \Rightarrow {a \over b} = 1$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
A message signal of frequency 20 kHz and peak voltage of 20 volt is used to modulate a carrier wave of frequency 1 MHz and peak voltage of 20 volt. The modulation index will be :
Correct Answer: 1
Explanation:
Modulation index
$\mu = {{{A_m}} \over {{A_c}}} = {{20} \over {20}} = 1$
$\mu = {{{A_m}} \over {{A_c}}} = {{20} \over {20}} = 1$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
A carrier wave Vc(t) = 160 sin(2$\pi$ $\times$ 106t) volts is made to vary between Vmax = 200 V and Vmin = 120 V by a message
signal Vm(t) = Am sin(2$\pi$ $\times$ 103t) volts. The peak voltage Am of the modulating signal is ___________.
signal Vm(t) = Am sin(2$\pi$ $\times$ 103t) volts. The peak voltage Am of the modulating signal is ___________.
Correct Answer: 40
Explanation:
Given, VC(t) = 160 sin(2$\pi$ $\times$ 106t) V
Vmax = 200 V, Vmin = 120 V and Vm(t) = Am sin(2$\pi$ $\times$ 103t) V.
Since, maximum amplitude,
Amax = Am + Ac
$\Rightarrow$ Vmax = Vm + Vc $\Rightarrow$ 200 = Vm + 160
$\Rightarrow$ Vm = 200 $-$ 160 $\Rightarrow$ Vm = 40
$\therefore$ Peak voltage, Am = 40
Vmax = 200 V, Vmin = 120 V and Vm(t) = Am sin(2$\pi$ $\times$ 103t) V.
Since, maximum amplitude,
Amax = Am + Ac
$\Rightarrow$ Vmax = Vm + Vc $\Rightarrow$ 200 = Vm + 160
$\Rightarrow$ Vm = 200 $-$ 160 $\Rightarrow$ Vm = 40
$\therefore$ Peak voltage, Am = 40
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
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 ___________.
(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 ___________.
Correct Answer: 50
Explanation:
For calculation of Range from tower.
$d = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $
$ \therefore $ hT = height of tower
& hR = height of receiver
For 1st case : hT = 20m, hR = 0
$ \because $ ${d_1} = \sqrt {2 \times 6400 \times {{10}^3} \times 20} $ = 16 km
For 2nd case : hT = 20m, hR = 5m
$ \because $ ${d_2} = \sqrt {2 \times 6400 \times {{10}^3} \times 20} + \sqrt {2 \times 6400 \times {{10}^3} \times 5} $
= 16 + 8 = 24 km
$ \therefore $ % change in range
$ = {{{d_2} - {d_1}} \over {{d_1}}} \times 100$
$ = {{24 - 16} \over {16}} \times 100$
$ = {8 \over {16}} \times 100$ = 50%
$d = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $
$ \therefore $ hT = height of tower
& hR = height of receiver
For 1st case : hT = 20m, hR = 0
$ \because $ ${d_1} = \sqrt {2 \times 6400 \times {{10}^3} \times 20} $ = 16 km
For 2nd case : hT = 20m, hR = 5m
$ \because $ ${d_2} = \sqrt {2 \times 6400 \times {{10}^3} \times 20} + \sqrt {2 \times 6400 \times {{10}^3} \times 5} $
= 16 + 8 = 24 km
$ \therefore $ % change in range
$ = {{{d_2} - {d_1}} \over {{d_1}}} \times 100$
$ = {{24 - 16} \over {16}} \times 100$
$ = {8 \over {16}} \times 100$ = 50%
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
For VHF signal broadcasting, ___________ km2 of maximum service area will be covered by an antenna tower of height 30 m, if the receiving antenna is placed at ground. Let radius of the earth be 6400 km. (Round off to the Nearest Integer) (Take $\pi$ as 3.14)
Correct Answer: 1206
Explanation:
$d = \sqrt {2hR} $
area = $\pi$d2
Area = $\pi$(2hR) = 3.14 $\times$ 2 $\times$ 30 $\times$ 6400 $\times$ 103 . m2
= 1205.76 km2
$ \approx $ 1206 km2
area = $\pi$d2
Area = $\pi$(2hR) = 3.14 $\times$ 2 $\times$ 30 $\times$ 6400 $\times$ 103 . m2
= 1205.76 km2
$ \approx $ 1206 km2
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
If the highest frequency modulating a carrier is 5 kHz, then the number of AM broadcast stations accommodated in a 90 kHz bandwidth are _________.
Correct Answer: 9
Explanation:
B. W. (Bandwidth) = 2 $\times$ maximum frequency at modulating signal
= 2 $\times$ 5kHz
= 10 kHz
$ \therefore $ No of stations accommodate
= ${{90} \over {10}}$ = 9
= 2 $\times$ 5kHz
= 10 kHz
$ \therefore $ No of stations accommodate
= ${{90} \over {10}}$ = 9
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
The maximum and minimum amplitude of an amplitude modulated wave is 16V and 8V respectively. The modulation index for this amplitude modulated wave is x $\times$ 10$-$2. The value of x is __________.
Correct Answer: 33
Explanation:
${A_m} = {{{A_{\max }} - {A_{\min }}} \over 2}$
${A_c} = {{{A_{\max }} + {A_{\min }}} \over 2}$
Modulation index (mi) = ${{{A_m}} \over {{A_c}}} = {{{{{A_{\max }} - {A_{\min }}} \over 2}} \over {{{{A_{\max }} + {A_{\min }}} \over 2}}} = {{{A_{\max }} - {A_{\min }}} \over {{A_{\max }} + {A_{\min }}}}$
mi = ${{16 - 8} \over {16 + 8}} = {8 \over {24}} = {1 \over 3} = 0.33$
mi = 33 $\times$ 10$-$2
$ \therefore $ x = 33
${A_c} = {{{A_{\max }} + {A_{\min }}} \over 2}$
Modulation index (mi) = ${{{A_m}} \over {{A_c}}} = {{{{{A_{\max }} - {A_{\min }}} \over 2}} \over {{{{A_{\max }} + {A_{\min }}} \over 2}}} = {{{A_{\max }} - {A_{\min }}} \over {{A_{\max }} + {A_{\min }}}}$
mi = ${{16 - 8} \over {16 + 8}} = {8 \over {24}} = {1 \over 3} = 0.33$
mi = 33 $\times$ 10$-$2
$ \therefore $ x = 33
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is $-$5 dB per km and cable length is 20 km. The power received at receiver is 10$-$x W. The value of x is ________.
[Gain in $dB = 10{\log _{10}}\left( {{{{P_o}} \over {{P_i}}}} \right)$]
[Gain in $dB = 10{\log _{10}}\left( {{{{P_o}} \over {{P_i}}}} \right)$]
Correct Answer: 8
Explanation:
Given, power of transmitted signal, Pi = 0.1 kW = 0.1 $\times$ 103W = 102 W
Rate of attenuation, R = $-$5 dB/km
Length of cable, l = 20 km
Power received at receiver, Px = 10$-$x W
Total loss, $\beta$ = R $\times$ l = $-$5 $\times$ 20 = $-$ 100 dB
$\because$ Gain ($\beta$) = $10{\log _{10}}{{{P_0}} \over {{P_i}}}$
$\therefore$ $\beta = - 100 = - 10{\log _{10}}{{{P_0}} \over {{P_i}}}$
$ \Rightarrow - 10 = {\log _{10}}{{{P_0}} \over {{P_i}}} \Rightarrow {10^{ - 10}} = {{{P_0}} \over {{P_i}}}$
$ \Rightarrow {P_0} = {10^{ - 10}}{P_i} = {10^{ - 10}} \times {10^2} = {10^{ - 8}} \Rightarrow {P_0} = {10^{ - 8}}V$
Hence, x = 8
Rate of attenuation, R = $-$5 dB/km
Length of cable, l = 20 km
Power received at receiver, Px = 10$-$x W
Total loss, $\beta$ = R $\times$ l = $-$5 $\times$ 20 = $-$ 100 dB
$\because$ Gain ($\beta$) = $10{\log _{10}}{{{P_0}} \over {{P_i}}}$
$\therefore$ $\beta = - 100 = - 10{\log _{10}}{{{P_0}} \over {{P_i}}}$
$ \Rightarrow - 10 = {\log _{10}}{{{P_0}} \over {{P_i}}} \Rightarrow {10^{ - 10}} = {{{P_0}} \over {{P_i}}}$
$ \Rightarrow {P_0} = {10^{ - 10}}{P_i} = {10^{ - 10}} \times {10^2} = {10^{ - 8}} \Rightarrow {P_0} = {10^{ - 8}}V$
Hence, x = 8