Communication Questions in English

(C) Sky wave propagation is a mode of radio wave propagation that uses the ionosphere to reflect radio waves back towards the Earth.
This mode is typically used for long-distance communication beyond the horizon.
The frequency range suitable for sky wave propagation is generally between $3 \ MHz$ and $30 \ MHz$.
Among the given options,$10^7 \ Hz$ (which is $10 \ MHz$) falls within this range.
Therefore,the correct option is $C$.

(A) The ionosphere is divided into different layers based on their altitude ranges:
$1$. $D$-layer: $65-75 \ km$ (Matches $IV$)
$2$. $E$-layer: $95-120 \ km$ (Matches $III$)
$3$. $F_1$-layer: $170-190 \ km$ (Matches $II$)
$4$. $F_2$-layer: $250-400 \ km$ (Matches $I$)
Therefore,the correct matching is $A-IV, B-III, C-II, D-I$.

(B) Speech signals are human voice signals which typically have a frequency range from $20 \ Hz$ to $20 \ kHz$. However,for commercial telephonic communication,it is not necessary to transmit the entire audible range. To ensure clarity and efficiency in transmission,the frequency range is restricted. The standard frequency range adequate for speech signals in commercial telephony is $300 \ Hz$ to $3100 \ Hz$.

(C) coaxial cable is a type of transmission line that consists of an inner conductor surrounded by a tubular insulating layer,surrounded by a tubular conducting shield.
It is widely used for cable television,broadband internet,and other high-frequency signal transmissions.
The frequency bandwidth of a coaxial cable is typically in the range of up to $750 \text{ MHz}$.
Therefore,the correct option is $C$.

(B) The process of the loss of strength of a signal while propagating through a medium is called attenuation. As a signal travels through a communication channel,its energy is absorbed by the medium,leading to a decrease in its amplitude and power,which is referred to as attenuation.

(C) Attenuation is the gradual loss of intensity or strength of a signal as it travels through a transmission medium. It occurs in all types of signals,whether digital or analog,due to factors like absorption,reflection,and scattering.

(C) Carrier frequency $= 20 GHz = 20 \times 10^9 Hz$.
Bandwidth available for transmission $= 20 \% \text{ of } 20 GHz = 0.20 \times 20 \times 10^9 Hz = 4 \times 10^9 Hz$.
Bandwidth required per channel $= 5 kHz = 5 \times 10^3 Hz$.
Number of channels $= \frac{\text{Total bandwidth available}}{\text{Bandwidth per channel}} = \frac{4 \times 10^9}{5 \times 10^3} = 0.8 \times 10^6 = 8 \times 10^5$.

(B) The frequencies of the side bands in amplitude modulation are given by the carrier frequency $\nu_c$ plus and minus the message signal frequency $f_m$.
Given: $f_m = 15 kHz$,Upper Side Band $(USB)$ $= 1515 kHz$,Lower Side Band $(LSB)$ $= 1485 kHz$.
The side bands are $\nu_c + f_m$ and $\nu_c - f_m$.
Therefore,$\nu_c + 15 kHz = 1515 kHz$ or $\nu_c - 15 kHz = 1485 kHz$.
Solving for $\nu_c$: $\nu_c = 1515 kHz - 15 kHz = 1500 kHz$.
Converting to $MHz$: $1500 kHz = 1.5 MHz$.

(A) Standard $AM$ (Amplitude Modulation) broadcast radio operates in the Medium Frequency $(MF)$ band.
The frequency range allocated for standard $AM$ broadcasting is $540 \text{ kHz}$ to $1600 \text{ kHz}$.
Therefore,the correct option is $A$.

(D) In radio communications,a side-band is a band of frequencies higher than or lower than the carrier frequency,which are the result of the modulation process.
The side-bands carry the information transmitted by the radio signal.
The message signal is superimposed on the carrier wave.
Thus,the side-bands produced have frequencies given by: $\text{Side-band frequencies} = \text{Carrier frequency} \pm \text{Message frequency}$.
Given: $\text{Carrier frequency} = 3 MHz = 3000 kHz$ and $\text{Message frequency} = 20 kHz$.
Therefore,the side-band frequencies are: $3000 kHz + 20 kHz = 3020 kHz$ and $3000 kHz - 20 kHz = 2980 kHz$.

(D) Television signals typically operate in the frequency range of $40 MHz$ to $900 MHz$.
Sky-wave propagation relies on the reflection of electromagnetic waves by the ionosphere,which is only effective for frequencies in the range of $3 MHz$ to $30 MHz$.
Since television signal frequencies are much higher than $30 MHz$,they penetrate the ionosphere and are not reflected back to Earth.
Therefore,television signals cannot be received through sky-wave propagation; they require line-of-sight communication or satellite communication.
Thus,the Assertion $(A)$ is false,while the Reason $(R)$ is a true statement regarding the ionospheric reflection range.

(D) Low frequency signals possess very low energy; hence,they cannot be transmitted over long distances without modulation.
In modulation,the characteristics of a carrier wave are varied according to the amplitude of the low-frequency (message) signal.
Carrier waves are high-frequency signals that are used to transmit low-frequency signals over long distances through the process of modulation.

(A) Given:
Number of channels $(N)$ = $5 \times 10^5$
Central frequency $(f_c)$ = $25 \text{ GHz} = 25 \times 10^9 \text{ Hz}$
Bandwidth per channel $(\Delta f)$ = $2 \text{ kHz} = 2 \times 10^3 \text{ Hz}$
The total bandwidth required for the network is $N \times \Delta f = (5 \times 10^5) \times (2 \times 10^3) = 10 \times 10^8 = 10^9 \text{ Hz}$.
The percentage of the link used is given by the ratio of the total bandwidth required to the central frequency (total available bandwidth capacity):
Percentage = $\frac{\text{Total Bandwidth Required}}{\text{Central Frequency}} \times 100$
Percentage = $\frac{10^9 \text{ Hz}}{25 \times 10^9 \text{ Hz}} \times 100$
Percentage = $\frac{1}{25} \times 100 = 4 \%$.
Therefore,$4 \%$ of the link is used for the network.

(A) The frequency of the carrier wave is $f_c = 1 \text{ MHz} = 1000 \text{ kHz}$.
The frequency of the message signal is $f_m = 28 \text{ kHz}$.
The frequencies of the side bands are given by the upper side band $(USB)$ and lower side band $(LSB)$ formulas:
$USB = f_c + f_m = 1000 \text{ kHz} + 28 \text{ kHz} = 1028 \text{ kHz}$.
$LSB = f_c - f_m = 1000 \text{ kHz} - 28 \text{ kHz} = 972 \text{ kHz}$.
Therefore,the side band frequencies are $1028 \text{ kHz}$ and $972 \text{ kHz}$.

(C) The maximum line-of-sight distance $d$ between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by the formula: $d = \sqrt{2 R h_T} + \sqrt{2 R h_R}$.
Given: $h_T = 33.8 \ m = 33.8 \times 10^{-3} \ km$,$h_R = 64.8 \ m = 64.8 \times 10^{-3} \ km$,and $R = 6400 \ km$.
Substituting the values:
$d = \sqrt{2 \times 6400 \times 33.8 \times 10^{-3}} + \sqrt{2 \times 6400 \times 64.8 \times 10^{-3}}$
$d = \sqrt{12800 \times 0.0338} + \sqrt{12800 \times 0.0648}$
$d = \sqrt{432.64} + \sqrt{829.44}$
$d = 20.8 \ km + 28.8 \ km = 49.6 \ km$.

(C) In amplitude modulation,the amplitude of the side bands is given by the formula:
$A_{SB} = \frac{\mu A_c}{2}$
Where $\mu$ is the modulation index and $A_c$ is the carrier amplitude.
However,in the context of a standard signal where the modulation index $\mu$ is assumed to be $1$ (maximum efficiency) and the message signal amplitude $A_m$ is equal to the product $\mu A_c$,the amplitude of each side band is:
$A_{SB} = \frac{A_m}{2}$
Given the peak voltage of the message signal $A_m = 12 \ V$:
$A_{SB} = \frac{12 \ V}{2} = 6 \ V$
Thus,the amplitude of each side band is $6 \ V$.

(D) For an amplitude modulated wave,the maximum amplitude $A_{\max} = 10 \ V$ and the minimum amplitude $A_{\min} = 2 \ V$.
The modulation index $\mu$ is defined by the formula:
$\mu = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}}$
Substituting the given values:
$\mu = \frac{10 - 2}{10 + 2} = \frac{8}{12}$
Simplifying the fraction:
$\mu = \frac{2}{3}$

(B) Given: Message signal frequency,$f_m = 10 kHz = 0.01 MHz$.
Carrier wave frequency,$f_c = 6 MHz = 6000 kHz$.
The sideband frequencies are given by $(f_c + f_m)$ and $(f_c - f_m)$.
Upper sideband frequency $(f_{USB}) = f_c + f_m = 6000 kHz + 10 kHz = 6010 kHz$.
Lower sideband frequency $(f_{LSB}) = f_c - f_m = 6000 kHz - 10 kHz = 5990 kHz$.
Therefore,the sideband frequencies are $5990 kHz$ and $6010 kHz$.

(C) Modulation is the process of superimposing a low-frequency baseband signal onto a high-frequency carrier wave.
This process is essential because low-frequency signals cannot be transmitted efficiently over long distances due to the requirement of impractically large antenna sizes and high signal attenuation.
By modulating the signal,it can be transmitted over large distances with minimal loss and smaller antenna requirements.
Therefore,the correct option is $C$.

(D) The standard equation for an amplitude modulated wave is given by $c_m(t) = A_c [1 + \mu \sin(\omega_m t)] \sin(\omega_c t)$.
Comparing the given equation $c_m(t) = 10[1 + 0.6 \sin(1250 t)] \sin(10^8 t)$ with the standard equation:
Here,$A_c = 10$,$\mu = 0.6$,$\omega_m = 1250 \text{ rad/s}$,and $\omega_c = 10^8 \text{ rad/s}$.
Thus,the modulation index $\mu$ is $0.6$.

(B) The power $P$ radiated by a linear antenna is given by the relation $P \propto \frac{l^2}{\lambda^2}$,where $l$ is the length of the antenna and $\lambda$ is the wavelength.
Since $P \propto l^2$,statement $A$ is true.
Since $\lambda = \frac{c}{f}$,we have $P \propto \frac{l^2}{(c/f)^2} \propto f^2$. This means power radiated increases with the square of the frequency.
Therefore,statement $B$ is false because it claims power decreases with increasing frequency.
For efficient radiation,the antenna size $l$ must be comparable to the wavelength $\lambda$ (typically $l \geq \frac{\lambda}{4}$),so statement $C$ is true.
Since long wavelength signals correspond to low frequencies,the radiated power $P \propto f^2$ is very small,making statement $D$ true.

(A) The modulation index $m$ is defined as the ratio of the peak voltage of the modulating signal $(A_m)$ to the peak voltage of the carrier wave $(A_c)$.
Given:
Carrier wave peak voltage,$A_c = 10 \,V$
Modulation index,$m = 80 \% = 0.8$
Formula:
$m = \frac{A_m}{A_c}$
Substituting the values:
$0.8 = \frac{A_m}{10 \,V}$
$A_m = 0.8 \times 10 \,V = 8 \,V$
Therefore,the peak voltage of the modulating signal should be $8 \,V$.

(B) The modulation index $m$ for an amplitude modulated wave is given by the formula:
$m = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}}$
Given $m = 0.5$ and $A_{\max} = 10.0 \ V$.
Substituting the values into the formula:
$0.5 = \frac{10 - A_{\min}}{10 + A_{\min}}$
$0.5(10 + A_{\min}) = 10 - A_{\min}$
$5 + 0.5 A_{\min} = 10 - A_{\min}$
$1.5 A_{\min} = 5$
$A_{\min} = \frac{5}{1.5} = \frac{50}{15} = 3.33 \ V$.

(C) Given that,maximum amplitude,$A_{\max} = 10 \,V$.
Minimum amplitude,$A_{\min} = 4 \,V$.
The modulation index $\mu$ is defined as the ratio of the difference of maximum and minimum amplitudes to the sum of maximum and minimum amplitudes.
$\mu = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}}$.
Substituting the given values:
$\mu = \frac{10 - 4}{10 + 4} = \frac{6}{14} = \frac{3}{7}$.

(B) Initially, the modulation index $\mu_1 = 0.5$ and the message signal amplitude $A_m = 10 \,V$.
Using the formula for modulation index, $\mu = \frac{A_m}{A_c}$, we find the initial carrier amplitude $A_{c_1} = \frac{A_m}{\mu_1} = \frac{10}{0.5} = 20 \,V$.
When the modulation index is changed to $\mu_2 = 0.8$ while keeping $A_m = 10 \,V$ constant, the new carrier amplitude is $A_{c_2} = \frac{A_m}{\mu_2} = \frac{10}{0.8} = 12.5 \,V$.
The change in carrier amplitude is $\Delta A_c = A_{c_1} - A_{c_2} = 20 \,V - 12.5 \,V = 7.5 \,V$.
Since the carrier amplitude decreased from $20 \,V$ to $12.5 \,V$, the peak voltage of the carrier signal must be reduced by $7.5 \,V$.

(A) The modulation index $\mu$ is defined as the ratio of the amplitude of the message signal $(A_m)$ to the amplitude of the carrier wave $(A_c)$.
Given:
Amplitude of message signal,$A_m = 15 V$
Amplitude of carrier wave,$A_c = 30 V$
Formula:
$\mu = \frac{A_m}{A_c}$
Calculation:
$\mu = \frac{15 V}{30 V} = 0.5$
Therefore,the modulation index is $0.5$.

(D) The standard form of an amplitude-modulated signal is given by:
$C_m(t) = A_c \sin(\omega_c t) + \frac{\mu A_c}{2} \sin((\omega_c + \omega_m)t) - \frac{\mu A_c}{2} \sin((\omega_c - \omega_m)t)$.
Comparing this with the given equation $C_m(t) = A_1 \sin(\omega_1 t) + A_2 \sin(\omega_2 t) - A_2 \sin(\omega_3 t)$:
We identify the carrier amplitude $A_c = A_1$ and carrier frequency $\omega_c = \omega_1$.
The sideband frequencies are $\omega_c + \omega_m = \omega_3$ and $\omega_c - \omega_m = \omega_2$.
Subtracting these two equations: $(\omega_c + \omega_m) - (\omega_c - \omega_m) = \omega_3 - \omega_2$,which gives $2\omega_m = \omega_3 - \omega_2$,so $\omega_m = \frac{\omega_3 - \omega_2}{2}$.
Comparing the amplitudes of the sidebands: $\frac{\mu A_c}{2} = A_2$.
Substituting $A_c = A_1$,we get $\frac{\mu A_1}{2} = A_2$,which implies $\mu = \frac{2 A_2}{A_1}$.
Thus,the modulation index is $\frac{2 A_2}{A_1}$ and the angular frequency of the message signal is $\frac{\omega_3 - \omega_2}{2}$.

(C) In the demodulation process,the detector performs the following functions:
$(a)$ Rectification: This process converts all negative-going peaks into positive-going peaks,effectively allowing the envelope of the modulated wave to be extracted.
$(b)$ Filtering: This step involves the separation of the high-frequency carrier components from the low-frequency message signal using a low-pass filter,leaving only the original information signal.

(D) The formula for the maximum line-of-sight $(LOS)$ distance $d_m$ between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by $d_m = \sqrt{2Rh_T} + \sqrt{2Rh_R}$, where $R$ is the radius of the Earth.
Given: $h_T = 20 \,m$, $d_m = 40 \,km = 40,000 \,m$, and $R = 6400 \,km = 6.4 \times 10^6 \,m$.
Substituting the values:
$40,000 = \sqrt{2 \times 6.4 \times 10^6 \times 20} + \sqrt{2 \times 6.4 \times 10^6 \times h_R}$
$40,000 = \sqrt{256 \times 10^6} + \sqrt{12.8 \times 10^6 \times h_R}$
$40,000 = 16,000 + \sqrt{12.8 \times 10^6 \times h_R}$
$24,000 = \sqrt{12.8 \times 10^6 \times h_R}$
Squaring both sides:
$(24,000)^2 = 12.8 \times 10^6 \times h_R$
$576,000,000 = 12,800,000 \times h_R$
$h_R = \frac{576,000,000}{12,800,000} = 45 \,m$.
Thus, the height of the receiving antenna is $45 \,m$.

(D) Given: Carrier frequency $f_c = 5 MHz$,carrier peak voltage $V_c = 40 V$,modulation index $\mu = 0.75$.
$1$. Calculation of message signal peak voltage $(V_m)$:
The modulation index is defined as $\mu = \frac{V_m}{V_c}$.
Substituting the values: $0.75 = \frac{V_m}{40 V}$.
$V_m = 40 V \times 0.75 = 30 V$.
$2$. Calculation of message signal frequency $(f_m)$:
The frequency separation between the two side-bands (upper side-band and lower side-band) is equal to the bandwidth,which is $2 f_m$.
Given: $2 f_m = 40 kHz$.
$f_m = \frac{40 kHz}{2} = 20 kHz$.
Therefore,the peak voltage and frequency of the message signal are $30 V$ and $20 kHz$ respectively.

$(C)$ The power $P$ of a signal radiated by an antenna is proportional to the square of the frequency $v$, i.e., $P \propto v^2$.
Given frequencies are $v_1 = 10^7 \,Hz$ and $v_2 = 10^6 \,Hz$.
The ratio of powers is $\frac{P_1}{P_2} = \left(\frac{v_1}{v_2}\right)^2$.
Substituting the values, we get $\frac{P_1}{P_2} = \left(\frac{10^7}{10^6}\right)^2 = (10)^2 = 100$.
This implies $P_1 = 100 P_2$.
Since the power radiated at $10^7 \,Hz$ is $100$ times the power radiated at $10^6 \,Hz$ for the same input signal strength, the broadcaster must increase the signal strength at $10^6 \,Hz$ by $100$ times to achieve equal strength at the receiver.

(B) The maximum line-of-sight distance $d_{\max}$ between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by the formula:
$d_{\max} = \sqrt{2Rh_T} + \sqrt{2Rh_R}$
where $R$ is the radius of the Earth,approximately $6400 \,km = 6.4 \times 10^6 \,m$.
Given: $h_T = 49 \,m$,$h_R = 64 \,m$,$R = 6.4 \times 10^6 \,m$.
Substituting the values:
$d_{\max} = \sqrt{2 \times 6.4 \times 10^6 \times 49} + \sqrt{2 \times 6.4 \times 10^6 \times 64}$
$d_{\max} = \sqrt{12.8 \times 10^6 \times 49} + \sqrt{12.8 \times 10^6 \times 64}$
$d_{\max} = 10^3 \times \sqrt{12.8} \times (7 + 8)$
$d_{\max} = 10^3 \times 3.577 \times 15$
$d_{\max} \approx 53660 \,m = 53.66 \,km$.
Thus,the maximum distance is approximately $53.6 \,km$.

(B) Given:
Frequency of message signal $(f_m) = 1 \text{ kHz} = 1 \times 10^3 \text{ Hz}$.
Peak voltage of message signal $(E_m) = 5 \text{ V}$.
Carrier frequency $(f_c) = 1 \text{ MHz} = 1 \times 10^6 \text{ Hz}$.
Peak voltage of carrier $(E_c) = 15 \text{ V}$.
The equation of an amplitude modulated wave is given by:
$e(t) = E_c [1 + \mu \sin(\omega_m t)] \sin(\omega_c t)$
where $\mu = \frac{E_m}{E_c}$ is the modulation index.
$\mu = \frac{5}{15} = \frac{1}{3}$.
Substituting the values:
$e(t) = 15 [1 + \frac{1}{3} \sin(2 \pi f_m t)] \sin(2 \pi f_c t)$
$e(t) = 15 [1 + \frac{1}{3} \sin(2 \pi \times 10^3 t)] \sin(2 \pi \times 10^6 t)$.

(C) The modulation index $m_a$ is defined as the ratio of the peak voltage of the message signal $(E_m)$ to the peak voltage of the carrier wave $(E_c)$.
Given:
Peak voltage of message signal,$E_m = 20 V$
Peak voltage of carrier wave,$E_c = 30 V$
Using the formula:
$m_a = \frac{E_m}{E_c}$
$m_a = \frac{20}{30} = \frac{2}{3} \approx 0.67$
Therefore,the modulation index is $0.67$.

(C) The modulation index $\mu$ is defined as the ratio of the peak voltage of the modulating signal $(E_m)$ to the peak voltage of the carrier wave $(E_c)$.
$\mu = \frac{E_m}{E_c}$
Given:
Carrier wave peak voltage $E_c = 12 \,V$
Modulation index $\mu = 75 \% = 0.75 = \frac{3}{4}$
Substituting the values into the formula:
$0.75 = \frac{E_m}{12}$
$E_m = 0.75 \times 12$
$E_m = 9 \,V$
Therefore, the peak voltage of the modulating signal is $9 \,V$.

(B) The maximum line-of-sight distance $d$ for a transmitting antenna of height $h_T$ and a receiving antenna at ground level is given by the formula: $d = \sqrt{2 R_e h_T}$.
Given:
$R_e = 6.4 \times 10^6 \ m$
$h_T = 128 \ m$
Substituting the values:
$d = \sqrt{2 \times (6.4 \times 10^6) \times 128}$
$d = \sqrt{12.8 \times 10^6 \times 128}$
$d = \sqrt{128 \times 10^5 \times 128}$
$d = 128 \times \sqrt{10^5} \ m$
$d = 128 \times \sqrt{10 \times 10^4} \ m$
$d = 128 \times 100 \sqrt{10} \ m$
$d = 12800 \sqrt{10} \ m$
Converting to kilometers $(1 \ km = 1000 \ m)$:
$d = \frac{12800 \sqrt{10}}{1000} \ km = 12.8 \sqrt{10} \ km$
Wait,checking the calculation again:
$d = \sqrt{2 \times 6.4 \times 10^6 \times 128} = \sqrt{12.8 \times 10^6 \times 128} = \sqrt{1638.4 \times 10^6} = \sqrt{16.384 \times 10^8} = 40477 \ m \approx 40.5 \ km$.
Re-evaluating the provided options: $128/\sqrt{10} \approx 128/3.16 \approx 40.5 \ km$. Thus,option $B$ is correct.

(B) The range $d$ of a transmitting antenna of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.
Given: $h = 980 \ m = 0.98 \ km$ and $R = 6400 \ km$.
Substituting the values into the formula:
$d = \sqrt{2 \times 6400 \times 0.98}$
$d = \sqrt{12800 \times 0.98}$
$d = \sqrt{12544}$
$d = 112 \ km$.
Therefore,the range of the transmitting antenna is $112 \ km$.

(D) The layer of the atmosphere that reflects radio waves,including high-frequency waves,is the ionosphere,which is a part of the thermosphere.
During the night,the ionosphere becomes more stable and effective at reflecting these waves,allowing for long-distance communication.
Therefore,the correct layer is the thermosphere.

(C) The radio horizon distance $d$ for a transmitting antenna of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.
Given: $h = 39.2 \ m = 39.2 \times 10^{-3} \ km$ and $R = 6400 \ km$.
Substituting the values into the formula:
$d = \sqrt{2 \times 6400 \times 39.2 \times 10^{-3}}$
$d = \sqrt{12800 \times 0.0392}$
$d = \sqrt{501.76}$
$d = 22.4 \ km$.
Therefore,the correct option is $C$.

(B) The process of the loss of strength of a signal while propagating through a medium is known as attenuation.
As a signal travels through a medium,its energy is absorbed or scattered,leading to a decrease in its amplitude or intensity,which is termed attenuation.

(B) The range $(d)$ of a $TV$ transmission antenna of height $h$ is given by the formula: $d = \sqrt{2hR}$,where $R$ is the radius of the Earth.
Given: $h = 40 \ m$ and $R = 6400 \ km = 6400 \times 10^3 \ m$.
The service area covered by the antenna is the area of a circle with radius $d$,given by: $Area = \pi d^2$.
Substituting the value of $d^2 = 2hR$ into the area formula:
$Area = \pi (2hR)$
$Area = \pi \times 2 \times 40 \times (6400 \times 10^3)$
$Area = \pi \times 80 \times 6400 \times 10^3$
$Area = 512000 \times 10^3 \pi \ m^2$
$Area = 512 \times 10^6 \pi \ m^2$.

(B) The range $d$ of a $TV$ transmission tower of height $h$ is given by the formula $d = \sqrt{2hR}$,where $R$ is the radius of the Earth.
Initially,$d = \sqrt{2hR}$.
When the height is increased to $h' = 3/2 h$,the new range $d'$ becomes:
$d' = \sqrt{2h'R} = \sqrt{2(3/2 h)R} = \sqrt{3/2} \sqrt{2hR} = \sqrt{3/2} d$.
The change in range is $\Delta d = d' - d$.
Substituting the values,we get $\Delta d = \sqrt{3/2} d - d = (\sqrt{3/2} - 1) d$.

(C) For satisfactory communication in Line-of-Sight $(LOS)$ mode, the maximum distance $d_{\text{max}}$ between a transmitting antenna of height $h_1$ and a receiving antenna of height $h_2$ is given by:
$d_{\text{max}} = \sqrt{2Rh_1} + \sqrt{2Rh_2}$
Given that $h_1 = h_2 = d \text{ meters}$ and $d_{\text{max}} = 2d \text{ kilometers} = 2d \times 1000 \text{ meters}$.
Substituting these values into the formula:
$2d \times 1000 = \sqrt{2Rd} + \sqrt{2Rd}$
$2000d = 2\sqrt{2Rd}$
$1000d = \sqrt{2Rd}$
Squaring both sides:
$(1000d)^2 = 2Rd$
$1,000,000 d^2 = 2 \times 6400 \times 1000 \times d$
Dividing both sides by $d$ (assuming $d \neq 0$):
$1,000,000 d = 2 \times 6400 \times 1000$
$1000 d = 2 \times 6400$
$d = \frac{12800}{1000} = 12.8 \text{ m}$

(A) The ionosphere is divided into several layers based on their altitude ranges:
$D$-layer: $65-75 \ km$
$E$-layer: $95-120 \ km$
$F_1$-layer: $170-190 \ km$
$F_2$-layer: $250-400 \ km$
Comparing these with the given table:
$A$ ($D$-layer) matches with $IV$ $(65-75 \ km)$
$B$ ($E$-layer) matches with $III$ $(95-120 \ km)$
$C$ ($F_1$-layer) matches with $II$ $(170-190 \ km)$
$D$ ($F_2$-layer) matches with $I$ $(250-400 \ km)$
Therefore,the correct matching is $A-IV, B-III, C-II, D-I$.

(C) The formula for the maximum line-of-sight distance $d$ between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$,where $R$ is the radius of the Earth.
Given: $h_T = 45 \ m = 0.045 \ km$,$d = 40 \ km$,and $R = 6400 \ km$.
Substituting the values into the formula:
$40 = \sqrt{2 \times 6400 \times 0.045} + \sqrt{2 \times 6400 \times h_R}$
$40 = \sqrt{12800 \times 0.045} + \sqrt{12800 \times h_R}$
$40 = \sqrt{576} + \sqrt{12800 \times h_R}$
$40 = 24 + \sqrt{12800 \times h_R}$
$16 = \sqrt{12800 \times h_R}$
Squaring both sides:
$256 = 12800 \times h_R$
$h_R = \frac{256}{12800} \ km = 0.02 \ km$
Converting to meters: $h_R = 0.02 \times 1000 \ m = 20 \ m$.

(C) The modulation index $\mu$ is defined as the ratio of the peak voltage of the modulating signal $(A_m)$ to the peak voltage of the carrier wave $(A_c)$.
$\mu = \frac{A_m}{A_c}$
Given,modulation index $\mu = 60 \% = 0.6$ and carrier peak voltage $A_c = 20 V$.
Substituting the values in the formula:
$0.6 = \frac{A_m}{20}$
$A_m = 0.6 \times 20 = 12 V$
Therefore,the peak voltage of the modulating signal is $12 V$.

(A) In amplitude modulation,the sideband frequencies are given by $f_1 = f_c - f_m$ and $f_2 = f_c + f_m$.
Adding these two equations: $f_1 + f_2 = (f_c - f_m) + (f_c + f_m) = 2f_c$,which implies $f_c = \frac{f_1 + f_2}{2}$.
Subtracting the first from the second: $f_2 - f_1 = (f_c + f_m) - (f_c - f_m) = 2f_m$,which implies $f_m = \frac{f_2 - f_1}{2}$.
Therefore,the ratio $\frac{f_c}{f_m} = \frac{(f_1 + f_2)/2}{(f_2 - f_1)/2} = \frac{f_1 + f_2}{f_2 - f_1}$.
Taking the absolute value to ensure a positive ratio,we get $\left|\frac{f_1+f_2}{f_2-f_1}\right|$.