Class 12th

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New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-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 ago

0 Follower 3 Views

A
alok kumar singh

Contributor-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 * - 2 = - 10 d B

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 ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a

length of building is given by l= 500m

And λ ~ 4l= 4 *  100 =400mf

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-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 ago

0 Follower 2 Views

A
alok kumar singh

Contributor-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 ago

0 Follower 1 View

A
alok kumar singh

Contributor-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 * 10 6 = 9 ( N m a x ) 1/2

Nmax= (5/9 * 10 6 )2= 3.086 * 1011/m3

For F2 layer 8MHz

8 * 10 6 = 9 ( N m a x ) 1/2

Nmax= (8/9 * 10 6 )2= 7.9 * 1011/m3

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a  Short Answer Type Questions as classified in NCERT Exemplar

maximun distance to cover entire surface of earth for communication is

dm2= (R+h)2+ (R+h)2= 2 (R+h)2

So dm= 2 h R + 2 h R = 2 2 h R

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 ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Option D is correct.

New answer posted

a year ago

0 Follower 71 Views

V
Vishal Baghel

Contributor-Level 10

Given,  Δ=|1xyz1yzx1zxy|

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 (y2z2) -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 ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Range = 2 h R = 2 * 20 * 6.4 * 10 6 = 16 k m

Area = π R 2 = 3.14 * 16 * 16 =803.84km2

When H= 25 m

Range = 2 h R + 2 H R = 2 * 20 * 6.4 * 10 6 + 2 * 25 * 6.4 * 10 6 = 33.9km

Area= 3.14 * 33.9 * 33.9 = 3608.52 k m 2

percentage increase = 3608.52 - 803.84 803.84 * 100 = 348.9 %

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