Class 12th
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New answer posted
a year 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
a year 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
a year 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
a year 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
a year 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
a year 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
New answer posted
a year agoContributor-Level 10
This is a Short Answer Type Questions as classified in NCERT Exemplar

dm2= (R+h)2+ (R+h)2= 2 (R+h)2
So dm=
8hR= R2+2Rh+h2
R-h=0
R=h so frequency of space wave Is less than height of tower. So it is also keep in consideration. for 3 dimension coverage 2 antennna each side means total 6 antenna required.
New answer posted
a year agoContributor-Level 10
Given,
Co-factor of elements of third column

∴ Δ = a13A13 + a23A23 + a33A33
= yz (z-y) + zx [- (z-x)] + xy (y-x)
= y z2-y2z-z2x+ zx2 + xy2-x2y.
= yz2-y2z+ (xy2-xz2) + (zx2-x2y)
= yz (z-y) + x (y2 – z2) -x2 (y-z)
= -yz (y-z) + x (y + z) (y-z) -x2 (y-z)
= (y-z) [-yz + x (y + z) -x2]
= (y-z) [-yz + xy + xz-x2]
= (y-z) [-y (z-x) + x (z-x)]
= (y-z) (z-x) (x -y)
New answer posted
a year agoContributor-Level 10
This is a Short Answer Type Questions as classified in NCERT Exemplar
Range =
Area = =803.84km2
When H= 25 m
Range = = 33.9km
Area= 3.14 2
percentage increase = %
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