Physics

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a

length of building is given by l= 500m

And $\lambda ~$ 4l= 4 $*$  100 =400mf

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 

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 ω_c, (ω_c – ω_m) and (ω_c + ω_m)

So to minimise cost of radiation we use both (ω_c – ω_m) and (ω_c + ω_m)

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

Frequency, fmax= 9 (N_max)^1/2

For F1 layer frequency is 5MHz

So 5 $*{10}^6=9\left(Nmax\right)$ ^1/2

Nmax= (5/9 $*{10}^6$ )²= 3.086 $*$ 10¹¹/m³

For F2 layer 8MHz

8 $*{10}^6=9\left(Nmax\right)$ ^1/2

Nmax= (8/9 $*{10}^6$ )²= 7.9 $*$ 10¹¹/m³

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

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

Range = $\sqrt[]{2hR}=\sqrt[]{2*20*6.4*{10}^6}=16km$

Area = $\pi R^2=3.14*16*16$ =803.84km²

When H= 25 m

Range = $\sqrt[]{2hR}+\sqrt[]{2HR}=\sqrt[]{2*20*6.4*{10}^6}+\sqrt[]{2*25*6.4*{10}^6}$ = 33.9km

Area= 3.14 $*33.9*33.9=3608.52km$ ²

percentage increase = $\frac{3608.52-803.84}{803.84}*100=348.9$ %

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

Total distance = 5km and loss is 2 dB/km

So total loss = 5 (2)= 10 dB

Total gain in amplifier  10+20= 30dB and gain in signal is 20dB

So by the formula 20= 10log_{10 $\frac{p_o}{p_i}$}

log_{10 $\frac{p_o}{p_i}$} = 2

so po/pi= 10²

so Po= Pi (100)= 101mW

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

In modulating signal we add career wave or noise signal to send it to receiver end and this wave is varied from time to time that is why more noise appear.

But in frequency modulation frequency is not varied so less noise appear.

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

Frequency tuned amplifier is $\frac{1}{2\pi \sqrt[]{LC}}=1MHz$

$\sqrt[]{LC}$ =1/2 $\pi *{10}^6$

$LC=2.54*{10}^{-14}s$

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

Maximum amplitude = Am+Ac =12 while minimum amplitude is Ac-Am=3

Using elimination method = 2Ac= 18

Ac=9 and Am= 6V

So modulating index m = 6/9= 2/3