In amplitude modulation, the instantaneous voltage of carrier waves is varied by the instantaneous voltage of modulating waves. On transmission through a channel noise signals can also be added, resulting a change in the amplitude of modulated waves. Due to it, the receiver receives the modulating signal having a part of noise. However, in frequency modulation, the frequency of carrier waves is changed as per instantaneous voltage of modulating waves. This can only be done at the mixing or modulating stage and not while signal is transmitted in channel. Hence noise does not effect FM signal.

Frequency of tuned amplifier is; \(v=\frac { 1 }{ 2\pi \sqrt { LC } } \)
As per question, \(v=1MHz={ 10 }^{ 6 }Hz=\frac { 1 }{ 2\pi \sqrt { LC } } \) or  \(LC=\frac { 1 }{ 4{ \pi }^{ 2 }\times \left( { 10 }^{ 6 } \right) ^{ 2 } } =\frac { 7\times 7 }{ 4\times \left( 22\times 22 \right) \times { 10 }^{ 12 } } \)
= 2.5 x 10^-14 s

We know that refractive index \(\mu\) of a medium is related with wavelength of wave according to relation
\(\mu =A+\frac { B }{ { \lambda }^{ 2 } } +\frac { C }{ { \lambda }^{ 4 } } =\frac { \sin { i } }{ \sin { r } } \ and \ \lambda =\frac { c }{ v } \)
Therefore, the value of refractive index \(\mu\) of a medium increases with the decrease in wavelength or increase in frequency of wave travelling through medium. For higher frequency wave, angle of refraction r is less, i.e., bending of wave is less in medium. Due to it, the condition for total internal reflection is attained after travelling longer distance by higher frequency wave. Hence 3 MHz wave will travel longer distance in the ionosphere before suffering total internal reflection.

No, because the TV signal of frequency 60 MHz is greater than 30 MHz (i.e., the upper limit of signal frequency for sky wave propagation). Therefore, T.V. signal will not be reflected by the ionosphere but will penetrate through the ionosphere.

(i) analog  
(ii) analog  
(iii) digital  
(iv) digital