Physics

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

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

As frequency of B is more than A so it has more refractive index also and if a wave have higher refractive index then it has less angle of refraction. So wave B travel more in ionosphere.

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

When we send a signal to be transmitted through sky waves must have a frequency range of 1710 kHz to 40 MHz.

But, the frequency of TV signals are 60 MHz which is beyond the required range.

So, sky waves will not be suitable for transmission of TV signals of 60 MHz frequency.

Frequency of career wave is 20MHz

Bandwidth require for modulation is 3kHz/2= 1.5kHz

For demodulating we need to reciprocal it

1/f= 1/20MHz= 0.5 $*$ 10^-7s

For modulation = 1/1.5KHz= 0.7 $*$ 10³s

According to first option =RC= 1 $*$ 0.01= 10^-5s

So it will demodulated

According to 2^nd option- RC= 10^-4s

So it can also be demodulated

According to third option – RC= 10^-8s

So it cannot be modulated

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

in first case crowding spectrum is visible

If we want more wave to be modulated then more crowding will occur and more mixing up of signal.

But we want to accommodate this we use higher band width and frequency career wave

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

Maximum voltage = 100/2 = 50V

And minimum voltage = 20/2 = 10V

Percentage modulation= max voltage -min voltage/ max voltage + min voltage $*100$

= $\frac{50-10}{50+10}*100$ = 66.67%

Peak career voltage= max voltage+ min voltage/2= 50+10/2=30V

Peak value of information voltage= 66.67/100 $*$  30= 20V

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

height of satellite hs= 600km
As we know velocity = distance/time

 2x/4.0410^-3 = 3⁸

So x=606 km after solving 
According to Pythagoras theorem d²=x²-h²= 606²-600²= 7236

So d= 85.06km so total distance will be double from receiver to transmitter = 170.1km
And d = √2Rh
 h=7236/2*6400 = 565m

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

as we know that I= I_{0 $e^{-\propto x}$}

And I= 25%of I₀=

I=I₀/4

I₀/4= I_{0 $e^{-\propto x}$}

I₀ cancel from both sides

¼= $e^{-\propto x}$

Taking log on both sides log1 -log4= - $\propto x$ loge

X= log4/ $\propto$