Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen
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New answer posted
7 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer- b
Explanation- it is represented by a diagram given below

New answer posted
7 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer- b
Explanation- Here, in this question, the frequency of modulated signal received becomes more, which is possible with the poor bandwidth selection of amplifiers.
This happens because bandwidth in amplitude modulation is equal to twice the
Frequency of modulating signal. But, the frequency of male voice is less than that of a female.
New answer posted
7 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer- c
Explanation- The device which follows square law is used for modulation purpose. Characteristics shown by (i) and (iii) corresponds to linear devices.
Characteristics shown by (ii) corresponds to square law device. Some part of (i) also
Follow square law.
Hence, (ii)and (iv) can be used for modulation.
New answer posted
7 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer- b
Explanation- frequency of career wave and frequency of amplitude modulated wave is same which is wc .
New answer posted
7 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer- a
Frequency of career signal = 1MHz and frequency = 3KHz= 0.003MHz
So frequency of side bands = 1 0.003
So 1.003MHz and 0.997MHz
New answer posted
7 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer-b
p = 1kW= 1000W
Attenuation of signal = -2dB/km and total path length = 5km
Gain in dB= 5
Gain in dB= 10 log (p0/pi)……. (i)
-10=10 log (p0/pi)= -10log (pi/po)
logPi/po=1 = log (pi/po)= log10
pi/p0=10= 1000W= 10p0
p0=100W
New answer posted
7 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer- a
length of building is given by l= 500m
And 4l= 4 100 =400mf
New answer posted
7 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer- b
Ground wave propagation – 530KHz to 1710KHz
Sky wave propagation- 1710KHz- 40MHz
Space wave propagation- 54MHz to 42GHz
New answer posted
7 months agoContributor-Level 10
This is a Short Answer Type Questions as classified in NCERT Exemplar
In amplitude modulated wave side frequency contain the information and from the question the power will be ωc, (ωc – ωm) and (ωc + ωm)
So to minimise cost of radiation we use both (ωc – ωm) and (ωc + ωm)
New answer posted
7 months agoContributor-Level 10
This is a Short Answer Type Questions as classified in NCERT Exemplar
Frequency, fmax= 9 (Nmax)1/2
For F1 layer frequency is 5MHz
So 5 1/2
Nmax= (5/9 )2= 3.086 1011/m3
For F2 layer 8MHz
8 1/2
Nmax= (8/9 )2= 7.9 1011/m3
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