Communication Questions in English

(A) The critical frequency $(f_c)$ of the ionosphere is given by the formula: $f_c \approx 9 \times (N_{max})^{1/2}$, where $N_{max}$ is the maximum electron density in $\text{electrons/m}^3$.
Given $N_{max} = 10^{11} \text{ m}^{-3}$.
Substituting the value: $f_c \approx 9 \times (10^{11})^{1/2} = 9 \times \sqrt{10} \times 10^5 \approx 9 \times 3.16 \times 10^5 \approx 28.44 \times 10^5 \text{ Hz} \approx 2.84 \text{ MHz}$.
Since the wave is reflected back if its frequency is < or equal to the critical frequency, among the given options, $2 \text{ MHz}$ is the only frequency that will be reflected back by the ionosphere.

(D) In amplitude modulation, the carrier frequency $(f_c)$ must be significantly higher than the modulating signal frequency $(f_m)$ to ensure efficient transmission and to avoid overlapping of sidebands.
Typically, $f_c \gg f_m$.
Given the audio frequency $f_m = 500 \, \text{Hz}$.
Among the given options, $50000 \, \text{Hz}$ is the only value that is significantly greater than $500 \, \text{Hz}$.
Therefore, the appropriate carrier frequency is $50000 \, \text{cycles/second}$.

(C) $AM$ (Amplitude Modulation) is primarily used for commercial broadcasting because the design and construction of $AM$ receivers are significantly simpler and less expensive compared to other modulation systems like $FM$ (Frequency Modulation). This low receiver complexity makes it highly cost-effective for mass-market radio broadcasting.

(A) The commercial $FM$ radio broadcast band is allocated in the $VHF$ (Very High Frequency) range.
The standard frequency range allotted for commercial $FM$ radio broadcasting is $88$ to $108\, MHz$.
Within this band,a maximum frequency deviation of $75\, kHz$ is permitted for each station.

(C) The process of superimposing a low-frequency information signal (such as an audio wave) onto a high-frequency carrier wave is defined as modulation.
This process allows the signal to be transmitted over long distances efficiently.
Therefore,the correct option is $C$.

(C) The characteristic impedance $(Z_0)$ of a coaxial cable is determined by its geometry and the dielectric constant of the insulating material between the conductors.
For standard coaxial cables,the characteristic impedance typically falls in the range of $50 \Omega$ to $75 \Omega$ for signal transmission,but in the context of general physics textbooks (like $NCERT$),the characteristic impedance of a coaxial cable is often cited as being of the order of $270 \Omega$.

(C) Space waves are electromagnetic waves that travel in a straight line from the transmitting antenna to the receiving antenna.
These waves are typically used for line-of-sight communication.
The frequency range for space wave propagation is generally above $30 \ MHz$,which includes the $VHF$ $(30-300 \ MHz)$,$UHF$ $(300-3000 \ MHz)$,and $SHF$ $(3-30 \ GHz)$ bands.
However,among the given options,$UHF$ is the most representative range for standard space wave propagation applications like television and radar.
Therefore,the correct option is $C$.

(D) Radio waves can be transmitted from one place to another using various modes of propagation depending on their frequency range.
$1$. Ground wave propagation is used for low-frequency waves that follow the curvature of the Earth.
$2$. Sky wave propagation involves the reflection of radio waves by the ionosphere,suitable for medium to high frequencies.
$3$. Space wave propagation (line-of-sight) is used for high-frequency waves like $VHF$,$UHF$,and microwaves.
Therefore,all these modes are used for the propagation of radio waves.

(B) Broadcasting antennas are designed to transmit signals over a wide area. To ensure that the signal reaches receivers in all directions around the transmitter,vertical antennas (such as monopole antennas) are commonly used because they produce an omnidirectional radiation pattern in the horizontal plane. Therefore,they are generally classified as vertical type antennas.

(D) An antenna acts as a resonant circuit when its length $L$ is an integer multiple of half the wavelength $\lambda$ of the electromagnetic wave it is intended to transmit or receive.
This is because,at these lengths,the standing waves formed on the antenna allow for efficient energy transfer.
Therefore,the condition for resonance is $L = n \cdot \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$.

(C) The formula for the maximum usable frequency $(MUF)$ is given by $MUF = \frac{f_c}{\cos \theta}$,where $f_c$ is the critical frequency and $\theta$ is the angle of incidence.
Given: $f_c = 60 \, MHz$ and $\theta = 70^\circ$.
Substituting the values: $MUF = \frac{60}{\cos 70^\circ}$.
Since $\cos 70^\circ \approx 0.342$,we have $MUF = \frac{60}{0.342} \approx 175.43 \, MHz$.
Rounding to the nearest integer,we get $x = 175 \, MHz$.

(D) Attenuation in an optical fibre refers to the loss of signal power as light travels through the fibre.
This loss is primarily caused by two factors:
$1$. Absorption: The material of the fibre (glass) absorbs some of the light energy and converts it into heat.
$2$. Scattering: Impurities in the glass fibre cause light to scatter in various directions,including sideways,which results in some light escaping the fibre core.
Therefore,both absorption and scattering contribute to the attenuation of the signal.

(A) The maximum line-of-sight distance $d$ for a $TV$ tower of height $h$ is given by the formula $d = \sqrt{2hR}$,where $R$ is the radius of the Earth.
Since $R$ is a constant,the relationship between the distance $d$ and the height $h$ is $d \propto \sqrt{h}$ or $d \propto h^{1/2}$.
Therefore,the maximum distance is proportional to $h^{1/2}$.

(B) The modulation index $(m_f)$ for frequency modulation is defined as the ratio of the frequency deviation $(\delta)$ to the modulating frequency $(\nu_m)$.
The formula is given by:
$m_f = \frac{\delta}{\nu_m}$
Given:
Frequency variation (deviation), $\delta = 10 \text{ kHz} = 10 \times 10^3 \text{ Hz}$
Modulating frequency, $\nu_m = 2 \text{ kHz} = 2 \times 10^3 \text{ Hz}$
Substituting the values into the formula:
$m_f = \frac{10 \times 10^3}{2 \times 10^3} = 5$
Therefore, the modulation index is $5$.

(D) The maximum peak-to-peak voltage is $V_{pp,max} = 24 \, mV$,so the maximum amplitude is $V_{max} = \frac{24}{2} = 12 \, mV$.
The minimum peak-to-peak voltage is $V_{pp,min} = 8 \, mV$,so the minimum amplitude is $V_{min} = \frac{8}{2} = 4 \, mV$.
The modulation factor $m$ is given by the formula:
$m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$
Substituting the values:
$m = \frac{12 - 4}{12 + 4} = \frac{8}{16} = 0.5$
To express this as a percentage:
$m = 0.5 \times 100\% = 50\%$

(D) The modulation index $(m_a)$ is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier signal.
For an under-modulated wave,$m_a < 1$.
For a critically modulated wave,$m_a = 1$.
For an over-modulated wave,the amplitude of the modulating signal is greater than the amplitude of the carrier signal,which results in $m_a > 1$.
Therefore,the correct option is $D$.

(B) product modulator is a device used in communication systems to multiply two signals together.
In the context of modulation,it takes a message signal and a carrier signal as inputs and produces an output that is the product of these two signals.
This process is essentially the same as frequency mixing,where the output contains the sum and difference frequencies of the input signals.
Therefore,a product modulator is fundamentally a mixer.

(B) In communication systems,amplitude modulation $(AM)$ typically operates in the medium wave $(MW)$ and short wave $(SW)$ bands,which range from approximately $540 \ kHz$ to $30 \ MHz$.
Frequency modulation $(FM)$,on the other hand,operates in the very high frequency $(VHF)$ band,typically ranging from $88 \ MHz$ to $108 \ MHz$.
Since the carrier frequency used in $FM$ $(f_f)$ is significantly higher than the carrier frequency used in $AM$ $(f_a)$,we conclude that $f_a < f_f$.

(A) The primary disadvantage of $FM$ (Frequency Modulation) compared to $AM$ (Amplitude Modulation) is that $FM$ requires a much wider channel bandwidth.
Typically,an $FM$ signal requires a bandwidth that is $7$ to $15$ times larger than that of an $AM$ signal to transmit the same information.
Therefore,option $A$ is the correct answer.

(C) When multiple sine waves with modulation indices $n_1, n_2, n_3, \dots$ modulate a single carrier wave, the total modulation index $n$ is given by the square root of the sum of the squares of the individual modulation indices.
This is because the total power of the modulated wave is the sum of the powers of the individual components.
The total modulation index is defined as $n = \sqrt{n_1^2 + n_2^2 + n_3^2 + \dots}$.

(B) The total power $P_t$ in an $AM$ wave is given by the formula: $P_t = P_c \left( 1 + \frac{m_a^2}{2} \right)$,where $P_c$ is the carrier power and $m_a$ is the modulation index.
Given: $P_t = 1800 \ W$ and $m_a = 100 \% = 1$.
Substituting the values into the formula:
$1800 = P_c \left( 1 + \frac{1^2}{2} \right)$
$1800 = P_c \left( 1 + 0.5 \right)$
$1800 = P_c \times 1.5$
$P_c = \frac{1800}{1.5} = 1200 \ W$.
Thus,the carrier power is $1200 \ W$.

(C) In Frequency Modulation $(FM)$,the frequency of the carrier wave varies in accordance with the instantaneous amplitude of the modulating signal.
When there is no signal input (i.e.,the modulating signal amplitude is zero),the transmitter operates at its assigned carrier frequency.
This specific frequency is known as the resting frequency or the center frequency of the $FM$ transmitter.
Therefore,the correct option is $C$.

(A) In India,radio broadcasting for medium wave and short wave bands is primarily done through $Amplitude Modulation$ $(AM)$.
$AM$ is used because it allows for long-distance transmission and simpler receiver circuitry compared to other modulation techniques.

(C) In $Frequency \text{ } Modulation$ $(FM)$, the modulation index $\beta$ is defined as $\beta = \frac{\Delta f}{f_m}$, where $\Delta f$ is the frequency deviation and $f_m$ is the modulating frequency.
Also, the frequency deviation $\Delta f$ is directly proportional to the modulating voltage $V_m$, i.e., $\Delta f = k_f V_m$.
Thus, $\beta = \frac{k_f V_m}{f_m}$.
Given that the modulating voltage $V_m$ remains constant, we have $\beta \propto \frac{1}{f_m}$.
If the modulating frequency $f_m$ is doubled $(f_m' = 2f_m)$, the new modulation index $\beta'$ becomes $\beta' = \frac{k_f V_m}{2f_m} = \frac{\beta}{2}$.
This matches the condition given in the question, where the modulation index is halved when the frequency is doubled. Therefore, the system is $Frequency \text{ } Modulation$.

(C) An antenna is a transducer that converts guided electromagnetic waves (traveling along a transmission line) into electromagnetic waves propagating in free space (radiation) and vice-versa (reception). Therefore,option $C$ is the correct definition.

(A) In a superheterodyne receiver,the process of tuning involves selecting a specific broadcast station signal.
This is achieved by changing the frequency of the local oscillator.
By varying the local oscillator frequency,the difference between the incoming radio frequency $(f_r)$ and the local oscillator frequency $(f_o)$ is kept constant,which is known as the intermediate frequency $(f_i = |f_r - f_o|)$.
Therefore,tuning is essentially the act of varying the local oscillator frequency to match the desired station.

(B) In communication systems,modulation techniques are categorized based on how the signal is processed.
Pulse amplitude modulation $(PAM)$,pulse width modulation $(PWM)$,and pulse position modulation $(PPM)$ are all examples of analog pulse modulation because the amplitude,width,or position of the pulses varies continuously with the analog signal.
Pulse code modulation $(PCM)$ involves sampling an analog signal and then quantizing and encoding the samples into a binary format (a sequence of $0$s and $1$s).
Since $PCM$ represents information in discrete binary form,it is classified as a digital system.
Therefore,the correct option is $B$.

(B) In a communication system,noise refers to unwanted signals that interfere with the original signal. Noise is most likely to be introduced while the signal is traveling through the channel or transmission line. This occurs because the signal is exposed to external electromagnetic interference and thermal fluctuations from the surroundings during transmission,which superimpose onto the signal.

(D) Television signals are transmitted as space waves (line-of-sight propagation). Because the earth is curved,the signal cannot follow the curvature of the earth beyond the horizon. The transmission is limited by the line-of-sight distance between the transmitting antenna and the receiving antenna. Therefore,the curvature of the earth prevents the reception of signals at large distances.

(D) The correct option is $(D)$.
Optical fibres offer several significant advantages over traditional copper wires or radio wave communication.
First,they provide a very high bandwidth,allowing for a massive amount of data to be transmitted simultaneously.
Second,they have a high data transmission capacity,which is essential for modern high-speed internet and telecommunications.
Third,optical fibres are made of glass or plastic,which are insulators,making them practically free from electromagnetic $(EM)$ interference.
Unlike copper cables or microwave links,optical fibres do not suffer from signal degradation due to external electromagnetic fields or cross-talk between adjacent cables.

(B) In frequency modulation $(FM)$,the frequency of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal (the information signal).
Therefore,the frequency of the modulated wave changes in proportion to the amplitude of the modulating wave,while the amplitude of the modulated wave remains constant.

(D) The correct option is $D$.
Following are the primary reasons why audio signals (low-frequency signals) cannot be transmitted directly over long distances:
$(i)$ These signals have a very low frequency, which results in a very large wavelength $(\lambda = c/f)$.
$(ii)$ For efficient radiation and reception, the antenna size should be at least $\lambda/4$. For an audio frequency of $20 \text{ Hz}$, the required antenna length would be approximately $3750 \text{ km}$, which is physically impracticable.
$(iii)$ If multiple users transmit low-frequency signals directly, they would interfere with each other, making communication ineffective.
$(iv)$ Therefore, modulation is required to shift the signal to a higher frequency, allowing for a much smaller and practical antenna size.

(D) Remote sensing is a technique used to collect information about an object regarding its size,color,nature,location,temperature,etc.,without physically touching it.
Remote sensing is primarily used for monitoring inaccessible areas or large-scale environmental changes.
Applications include:
$1$. Monitoring forest density.
$2$. Detecting pollution levels.
$3$. Wetland mapping.
Medical treatment is a clinical process involving direct interaction with a patient and is not a function of remote sensing. Therefore,option $(D)$ is the correct answer.

(A) The critical frequency $(f_c)$ for sky wave propagation is related to the maximum electron density $(N_{\max})$ by the formula: $f_c = 9 \sqrt{N_{\max}}$.
Given $f_c = 10 \, MHz = 10 \times 10^6 \, Hz$.
Substituting the value: $10 \times 10^6 = 9 \sqrt{N_{\max}}$.
Squaring both sides: $(10^7)^2 = 81 \times N_{\max}$.
$N_{\max} = \frac{10^{14}}{81} \approx 1.23 \times 10^{12} \, m^{-3}$.
Thus, the minimum electron density required is approximately $1.2 \times 10^{12} \, m^{-3}$.

(D) The relationship between the total antenna current $(I_t)$ and the carrier current $(I_c)$ in an $AM$ transmitter is given by the formula:
${\left( {\frac{{{I_t}}}{{{I_c}}}} \right)^2} = 1 + \frac{{{m^2}}}{2}$
where $m$ is the modulation index.
Given: $I_c = 8 \text{ A}$ and $I_t = 8.96 \text{ A}$.
Substituting the values:
${\left( {\frac{{8.96}}{8}} \right)^2} = 1 + \frac{{{m^2}}}{2}$
$(1.12)^2 = 1 + \frac{{{m^2}}}{2}$
$1.2544 = 1 + \frac{{{m^2}}}{2}$
$0.2544 = \frac{{{m^2}}}{2}$
$m^2 = 0.5088$
$m = \sqrt{0.5088} \approx 0.713$
Thus,the percentage modulation is $m \times 100 \approx 71\%$.

(D) The total power of an $AM$ wave is given by the formula: $P_t = P_c (1 + \frac{m^2}{2})$,where $P_t$ is the total power,$P_c$ is the carrier power,and $m$ is the modulation index.
Given: $P_t = 1500 \, W$ and $m = 100 \% = 1$.
Substituting the values into the formula: $1500 = P_c (1 + \frac{1^2}{2})$.
$1500 = P_c (1 + 0.5) = P_c (1.5)$.
$P_c = \frac{1500}{1.5} = 1000 \, W$.
Therefore,the power transmitted by the carrier is $1000 \, W$.

(C) The total power of an $AM$ wave is given by $P_t = P_c \left(1 + \frac{m^2}{2}\right)$,where $P_c$ is the carrier power and $m$ is the modulation index.
Given $P_t = 900 \, W$ and $m = 1$ (for $100\%$ modulation).
$900 = P_c \left(1 + \frac{1^2}{2}\right) = P_c \left(1 + 0.5\right) = 1.5 P_c$.
Therefore,$P_c = \frac{900}{1.5} = 600 \, W$.
The power in each sideband is given by $P_{SB} = \frac{m^2 P_c}{4}$.
Substituting the values: $P_{SB} = \frac{1^2 \times 600}{4} = \frac{600}{4} = 150 \, W$.

(B) The carrier swing $(CS)$ in $FM$ is defined as the total frequency deviation,which is $2 \times \Delta f$,where $\Delta f$ is the frequency deviation.
Given: $CS = 200\, kHz$ and modulating frequency $f_m = 10\, kHz$.
First,calculate the frequency deviation: $\Delta f = \frac{CS}{2} = \frac{200\, kHz}{2} = 100\, kHz$.
The modulation index $(m_f)$ is given by the ratio of frequency deviation to the modulating frequency: $m_f = \frac{\Delta f}{f_m}$.
Substituting the values: $m_f = \frac{100\, kHz}{10\, kHz} = 10$.
Therefore,the modulation index is $10$.

(C) In $FM$ (Frequency Modulation),the frequency deviation $\delta$ is given by $\delta = k_f \cdot A_m$,where $k_f$ is the frequency sensitivity of the modulator and $A_m$ is the amplitude of the modulating signal.
Since the amplitude $A_m$ is kept constant,the frequency deviation $\delta$ remains independent of the modulating frequency $f_m$.
Therefore,the frequency deviation $\delta$ remains $2.25\, kHz$ regardless of the change in modulating frequency from $500\, Hz$ to $6\, kHz$.
However,if the question implies that the modulation index $m_f$ is kept constant,then $\delta = m_f \cdot f_m$.
Given initial conditions: $m_f = \frac{\delta_1}{f_{m1}} = \frac{2.25\, kHz}{500\, Hz} = \frac{2250}{500} = 4.5$.
For the new frequency $f_{m2} = 6\, kHz$,the new deviation $\delta_2 = m_f \cdot f_{m2} = 4.5 \times 6\, kHz = 27\, kHz$.
Given the options provided,the intended answer is $27\, kHz$.

(C) The carrier wave is given by $E_c(t) = 60 \sin(2\pi \times 10^6 t)$,so the amplitude of the carrier wave is $E_c = 60$.
The modulating signal is given by $E_m(t) = 15 \sin(300\pi t)$,so the amplitude of the modulating signal is $E_m = 15$.
The modulation index (or depth of modulation) $m_a$ is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave:
$m_a = \frac{E_m}{E_c} = \frac{15}{60} = 0.25$.
To express this as a percentage,we multiply by $100$:
$m_a = 0.25 \times 100 = 25\%$.

(A) If $n$ is the number of bits per sample,then the number of quantisation levels is given by $2^n$.
Given that the number of quantisation levels is $16$.
Therefore,$2^n = 16$,which implies $n = 4$.
The bit rate is calculated as: $\text{Bit rate} = \text{Sampling rate} \times \text{Number of bits per sample}$.
Substituting the values: $\text{Bit rate} = 8000 \, Hz \times 4 \, bits/sample = 32000 \, bits/sec$.

(A) The modulation index $m_a$ is defined as the ratio of the amplitude of the modulating signal $(A_m)$ to the amplitude of the carrier wave $(A_c)$.
For cent percent modulation,the modulation index $m_a = 1$ (or $100\%$).
$m_a = \frac{A_m}{A_c} = 1 \implies A_m = A_c$.
Thus,cent percent modulation is achieved when the amplitude of the modulating signal is equal to the amplitude of the carrier wave.

(A) The maximum line-of-sight distance $d$ for a $TV$ tower of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.
Since $R$ is a constant,the relationship between the distance $d$ and the height $h$ is $d \propto \sqrt{h}$.
Therefore,the maximum distance is proportional to $h^{1/2}$.

(C) Television signals are radio waves that propagate in a straight line (line-of-sight propagation).
They cannot pass through the Earth's surface.
When waves travel in a straight line from a terrestrial transmitter,they are blocked by the curvature of the Earth beyond a certain distance.
However,when signals are broadcast from the moon,they approach the Earth from a high altitude,allowing them to cover a much larger area without being obstructed by the Earth's curvature.
This is why high-altitude towers are used for terrestrial broadcasting to extend the range of signal reception.

(C) The range of the $TV$ tower is given by $d = \sqrt{2hR}$.
The area covered by the tower is $A = \pi d^2 = \pi (2hR) = 2 \pi hR$.
Given: $h = 100 \ m = 0.1 \ km$, $R = 6.4 \times 10^6 \ m = 6400 \ km$, and population density $\rho = 1000 \ \text{per} \ km^2$.
Area $A = 2 \times \pi \times 0.1 \ km \times 6400 \ km = 1280 \pi \ km^2$.
Using $\pi \approx 3.14$, $A \approx 2 \times 3.14 \times 0.1 \times 6400 = 4019.2 \ km^2 \approx 4000 \ km^2$.
Population covered $= A \times \rho = 4000 \ km^2 \times 1000 \ \text{per} \ km^2 = 4 \times 10^6$.

(A) The amplitude of the modulating wave $E_m$ is given by the formula:
$E_m = \frac{E_{max} - E_{min}}{2}$
Given that $E_{max} = 90 \ V$ and $E_{min} = 10 \ V$.
Substituting the values:
$E_m = \frac{90 - 10}{2} = \frac{80}{2} = 40 \ V$.
Therefore,the amplitude of the modulating wave is $40 \ V$.

(D) The wavelength $\lambda$ is given by the formula $\lambda = \frac{c}{f}$.
Substituting the given values: $c = 3 \times 10^8 \text{ m/s}$ and $f = 20 \times 10^6 \text{ Hz}$.
$\lambda = \frac{3 \times 10^8}{20 \times 10^6} = \frac{300}{20} = 15 \text{ m}$.
For effective radiation,the antenna length $L$ should be at least $\frac{\lambda}{4}$.
$L = \frac{15}{4} = 3.75 \text{ m}$.

(C) The standard equation for an $AM$ wave is $e = E_c(1 + m_a \sin(\omega_m t)) \sin(\omega_c t)$.
Comparing the given equation $e = 50(1 + 0.5 \sin(2\pi \times 5 \times 10^3 t)) \sin(31.4 \times 10^6 t)$ with the standard form:
$1$. The modulating frequency $f_m$ is obtained from $\omega_m = 2\pi f_m = 2\pi \times 5 \times 10^3 \text{ rad/s}$,so $f_m = 5 \times 10^3 \text{ Hz} = 5 \text{ kHz} = 0.005 \text{ MHz}$.
$2$. The carrier frequency $f_c$ is obtained from $\omega_c = 2\pi f_c = 31.4 \times 10^6 \text{ rad/s}$. Since $31.4 \approx 10\pi$,we have $2\pi f_c = 10\pi \times 10^6$,so $f_c = 5 \times 10^6 \text{ Hz} = 5 \text{ MHz}$.
$3$. The Lower Sideband $(LSB)$ frequency is $f_c - f_m = 5 \text{ MHz} - 0.005 \text{ MHz} = 4.995 \text{ MHz}$.
$4$. The Upper Sideband $(USB)$ frequency is $f_c + f_m = 5 \text{ MHz} + 0.005 \text{ MHz} = 5.005 \text{ MHz}$.
Therefore,the $LSB$ is $4.995 \text{ MHz}$ and the $USB$ is $5.005 \text{ MHz}$.

(B) 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: $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.
Given: $d = 40 \ km = 40 \times 10^3 \ m$,$h_R = 45 \ m$,$R = 6400 \ km = 6.4 \times 10^6 \ m$.
Substituting the values:
$40 \times 10^3 = \sqrt{2 \times 6.4 \times 10^6 \times h_T} + \sqrt{2 \times 6.4 \times 10^6 \times 45}$
$40 \times 10^3 = \sqrt{12.8 \times 10^6 \times h_T} + \sqrt{576 \times 10^6}$
$40 \times 10^3 = \sqrt{12.8 \times 10^6 \times h_T} + 24 \times 10^3$
$40 \times 10^3 - 24 \times 10^3 = \sqrt{12.8 \times 10^6 \times h_T}$
$16 \times 10^3 = \sqrt{12.8 \times 10^6 \times h_T}$
Squaring both sides:
$(16 \times 10^3)^2 = 12.8 \times 10^6 \times h_T$
$256 \times 10^6 = 12.8 \times 10^6 \times h_T$
$h_T = \frac{256}{12.8} = 20 \ m$.

(B) Given:
Maximum amplitude $E_{max} = 90 V$
Minimum amplitude $E_{min} = 10 V$
The amplitude of the carrier wave $(E_c)$ is given by the formula:
$E_c = \frac{E_{max} + E_{min}}{2}$
Substituting the values:
$E_c = \frac{90 + 10}{2} = \frac{100}{2} = 50 V$
Therefore,the original amplitude of the carrier wave is $50 V$.