Communication Questions in English

(B) The modulation index $\mu$ is given by the formula:
$\mu = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}}$
Given,$V_{\max} = 14\,mV$ and $V_{\min} = 6\,mV$.
Substituting the values into the formula:
$\mu = \frac{14 - 6}{14 + 6}$
$\mu = \frac{8}{20}$
$\mu = 0.4$
Therefore,the modulation index is $0.4$.

(B) Given,maximum amplitude $A_{max} = A_c + A_m = 120 \, V$ and minimum amplitude $A_{min} = A_c - A_m = 80 \, V$.
Adding these two equations: $2A_c = 200 \, V \implies A_c = 100 \, V$.
Subtracting these two equations: $2A_m = 40 \, V \implies A_m = 20 \, V$.
The modulation index $\mu = \frac{A_m}{A_c} = \frac{20}{100} = 0.2$.
The amplitude of each sideband is given by $\frac{\mu A_c}{2}$.
Substituting the values: $\text{Amplitude} = \frac{0.2 \times 100}{2} = \frac{20}{2} = 10 \, V$.

(D) $1$. Attenuation $(A)$ is the loss of strength of a signal while propagating through a medium,so $A-IV$.
$2$. Transducer $(B)$ is a device that converts one form of energy into another,so $B-III$.
$3$. Demodulation $(C)$ is the process of retrieval of information from the carrier wave at the receiver,so $C-II$.
$4$. Repeater $(D)$ is a combination of a receiver and a transmitter,which picks up the signal from the transmitter,amplifies it,and retransmits it to the receiver,so $D-I$.
Thus,the correct matching is $A-IV, B-III, C-II, D-I$.

(B) The carrier wave is given by $V_c(t) = 15 \sin(1000 \pi t)$.
Comparing this with $V_c(t) = A_c \sin(2 \pi f_c t)$,we get $2 \pi f_c = 1000 \pi$,so $f_c = 500 \text{ Hz}$.
The modulating signal is given by $V_m(t) = 10 \sin(4 \pi t)$.
Comparing this with $V_m(t) = A_m \sin(2 \pi f_m t)$,we get $2 \pi f_m = 4 \pi$,so $f_m = 2 \text{ Hz}$.
An amplitude modulated signal consists of the carrier frequency $f_c$ and two sideband frequencies $(f_c - f_m)$ and $(f_c + f_m)$.
Sideband frequencies are:
Lower sideband: $f_c - f_m = 500 - 2 = 498 \text{ Hz}$.
Upper sideband: $f_c + f_m = 500 + 2 = 502 \text{ Hz}$.
Thus,the frequencies present are $500 \text{ Hz}$,$498 \text{ Hz}$,and $502 \text{ Hz}$,which correspond to options $A, D,$ and $E$.

(D) Statement $I$ is correct: For efficient transmission and reception of electromagnetic waves,the antenna length must be comparable to the wavelength of the signal,typically $l = \frac{\lambda}{4}$.
Statement $II$ is incorrect: In amplitude modulation $(AM)$,the amplitude of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal (information signal). Therefore,the amplitude of the carrier wave does not remain constant.

(B) The standard frequency range for $FM$ (Frequency Modulation) radio broadcasting is $88\,MHz$ to $108\,MHz$.
Comparing the given options:
$A) 106\,MHz$ is within the range.
$B) 64\,MHz$ is outside the range.
$C) 99\,MHz$ is within the range.
$D) 89\,MHz$ is within the range.
Therefore,$64\,MHz$ does not belong to the $FM$ broadcast band.

(C) The modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal $(A_m)$ to the amplitude of the carrier signal $(A_c)$: $\mu = \frac{A_m}{A_c}$.
In the first case,the modulating signal amplitude is $A_m = X$ and the carrier signal amplitude is $A_c = Y$. Thus,$\mu_1 = \frac{X}{Y}$.
In the second case,the modulating signal amplitude is $A_m = X$ and the carrier signal amplitude is $A_c = 2Y$. Thus,$\mu_2 = \frac{X}{2Y}$.
The ratio of the modulation indices is $\frac{\mu_1}{\mu_2} = \frac{X/Y}{X/2Y} = \frac{2}{1}$.
Therefore,the ratio is $2:1$.

(C) The transmission range $d$ of a $TV$ tower of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.
Let the initial height be $h_1$ and the initial range be $d_1 = \sqrt{2Rh_1}$.
The new height $h_2$ is increased by $21 \%$,so $h_2 = h_1 + 0.21h_1 = 1.21h_1$.
The new range $d_2$ is given by $d_2 = \sqrt{2Rh_2} = \sqrt{2R(1.21h_1)} = \sqrt{1.21} \times \sqrt{2Rh_1} = 1.1 \times d_1$.
The percentage increase in the range is given by $\frac{d_2 - d_1}{d_1} \times 100 \%$.
Substituting the values: $\frac{1.1d_1 - d_1}{d_1} \times 100 \% = (1.1 - 1) \times 100 \% = 0.1 \times 100 \% = 10 \%$.

(B) Given: Height of transmitting antenna $h_T = 98\,m = 0.098\,km$. Height of receiving antenna $h_R = 0\,m$. Radius of the earth $R = 6400\,km$.
The maximum line-of-sight distance $d$ is given by the formula $d = \sqrt{2 R h_T} + \sqrt{2 R h_R}$.
Since $h_R = 0$,the formula simplifies to $d = \sqrt{2 R h_T}$.
Substituting the values: $d = \sqrt{2 \times 6400 \times 0.098} = \sqrt{12800 \times 0.098} = \sqrt{1254.4} \approx 35.417\,km$.
The surface area $A$ covered by the antenna is given by $A = \pi d^2$.
$A = 3.14159 \times (35.417)^2 \approx 3.14159 \times 1254.4 \approx 3941.07\,km^2$.
Rounding to the nearest integer,the area is approximately $3942\,km^2$.

(B) Given,carrier wave amplitude $A_c = 15\,V$.
Baseband signal amplitude $A_m = 3\,V$.
The maximum amplitude of the amplitude modulated wave is given by $A_{\max} = A_c + A_m = 15 + 3 = 18\,V$.
The minimum amplitude of the amplitude modulated wave is given by $A_{\min} = A_c - A_m = 15 - 3 = 12\,V$.
The ratio of maximum amplitude to minimum amplitude is $\frac{A_{\max}}{A_{\min}} = \frac{18}{12} = \frac{3}{2}$.

(D) The bandwidth of an amplitude modulated $(AM)$ wave is given by the formula:
Bandwidth $= 2 f_{m}$
where $f_{m}$ is the frequency of the message signal.
Given,$f_{m} = 3\,kHz$.
Therefore,Bandwidth $= 2 \times 3\,kHz = 6\,kHz$.

(A) The distance $d$ for line-of-sight communication between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by $d = \sqrt{2 R h_T} + \sqrt{2 R h_R}$.
Given that the transmitting antenna is on the surface of the earth,$h_T = 0$.
Therefore,the formula simplifies to $d = \sqrt{2 R h_R}$.
Squaring both sides,we get $d^2 = 2 R h_R$.
Solving for $h_R$,we get $h_R = \frac{d^2}{2 R}$.
Given $d = 4\,km = 4000\,m$ and $R = 6400\,km = 6400000\,m$.
Substituting the values: $h_R = \frac{(4000)^2}{2 \times 6400000} = \frac{16000000}{12800000} = \frac{160}{128} = 1.25\,m$.
We are given $h_R = x \times 10^{-2}\,m$.
So,$1.25 = x \times 10^{-2} \implies x = 1.25 \times 10^2 = 125$.

(B) In satellite communication,the frequency band used for the uplink (transmission from Earth station to satellite) is typically in the $C$-band range of $5.925-6.425\,GHz$.
Conversely,the downlink frequency band (transmission from satellite to Earth station) is $3.7-4.2\,GHz$.
Therefore,the correct option is $B$.

(D) The carrier wave is given by $c(t) = 15 \sin(1000 \pi t)$.
The carrier frequency $f_c$ is calculated as $f_c = \frac{\omega_c}{2\pi} = \frac{1000\pi}{2\pi} = 500\,Hz$.
The modulating signal is given by $m(t) = 10 \sin(4 \pi t)$.
The modulating frequency $f_m$ is calculated as $f_m = \frac{\omega_m}{2\pi} = \frac{4\pi}{2\pi} = 2\,Hz$.
An amplitude modulated $(AM)$ wave contains the carrier frequency and two sideband frequencies: $(f_c - f_m)$ and $(f_c + f_m)$.
Sideband frequencies are $(500 - 2)\,Hz = 498\,Hz$ and $(500 + 2)\,Hz = 502\,Hz$.
Therefore,the frequencies present in the $AM$ signal are $500\,Hz, 498\,Hz,$ and $502\,Hz$,which correspond to items $(1), (4),$ and $(5)$.

(A) The atmospheric layers and their approximate heights are as follows:
$1$. The $F_1$-layer of the ionosphere is located at an approximate height of $170-190\,km$.
$2$. The $D$-layer of the ionosphere is located at an approximate height of $65-75\,km$.
$3$. The Troposphere is the lowest layer of the atmosphere,extending up to approximately $10\,km$ from the Earth's surface.
$4$. The $E$-layer of the ionosphere is located at an approximate height of $100\,km$.
Matching these values:
$(A) \rightarrow (II)$
$(B) \rightarrow (IV)$
$(C) \rightarrow (I)$
$(D) \rightarrow (III)$
Therefore,the correct sequence is $(A)-(II), (B)-(IV), (C)-(I), (D)-(III)$.

(D) 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:
$d_{\max} = \sqrt{2Rh_t} + \sqrt{2Rh_r}$
Given:
$R = 6400\,km = 64 \times 10^5\,m$
$h_t = 180\,m$
$h_r = 245\,m$
Substituting the values:
$d_{\max} = \sqrt{2 \times 64 \times 10^5 \times 180} + \sqrt{2 \times 64 \times 10^5 \times 245}$
$d_{\max} = \sqrt{128 \times 180 \times 10^5} + \sqrt{128 \times 245 \times 10^5}$
$d_{\max} = \sqrt{23040 \times 10^5} + \sqrt{31360 \times 10^5}$
$d_{\max} = \sqrt{2304 \times 10^6} + \sqrt{3136 \times 10^6}$
$d_{\max} = (48 \times 10^3) + (56 \times 10^3)\,m$
$d_{\max} = 48\,km + 56\,km = 104\,km$.

(C) repeater is a combination of a receiver and a transmitter.
It receives the signal from the transmitter,amplifies it,and retransmits it to the destination.
This process compensates for the signal loss (attenuation) that occurs over long distances,thereby extending the range of the communication system.

(A) Digital signals are signals that represent data as a sequence of discrete values at any given time. Unlike analog signals,which are continuous,digital signals consist of discrete stepwise values,typically represented by binary digits ($0$ and $1$). Therefore,they are characterized by discrete stepwise values.

(B) In a standard superheterodyne receiver block diagram,the signal from the receiving antenna first passes through an amplifier (often an $RF$ amplifier) to boost the weak signal.
Then,it goes to the Intermediate Frequency $(IF)$ stage,where the frequency is converted to a lower,fixed intermediate frequency.
After the $IF$ stage,the signal passes through a detector (demodulator) to extract the original message signal.
Finally,the signal is passed through an audio amplifier to increase its power for the output device.
Looking at the provided diagram:
- The first box '$Y$' represents an amplifier ($RF$ amplifier).
- The second box '$X$' represents the $IF$ stage.
- The third box is the detector.
- The fourth box '$Y$' represents an amplifier (audio amplifier).
Therefore,'$X$' represents the $IF$ stage and '$Y$' represents an amplifier.

(B) repeater is a device that receives a signal,amplifies it,and then retransmits it to extend the range of communication. Therefore,it essentially consists of a receiver to pick up the signal and a transmitter to send it further. Thus,it is a combination of a receiver and a transmitter.

(A) In a frequency modulated wave,only the frequency of the carrier wave varies with time in accordance with the instantaneous amplitude of the modulating signal,while the amplitude of the carrier wave remains constant.
This wave is represented by the provided image.
Frequency modulation is widely used in $FM$ broadcasting,telemetry,and $RADAR$ systems.

(B) The ionosphere consists of different layers $(D, E, F_1, F_2)$.
The $E$ layer of the ionosphere is responsible for reflecting medium-frequency radio waves.
This layer is formed due to the ionization caused by solar radiation during the daytime.
At night,the source of ionization (solar radiation) is absent,causing the $E$ layer to disappear or become ineffective.
Therefore,the $E$ layer is the correct answer.

(C) The process of superimposing a low-frequency message signal (baseband signal) onto a high-frequency carrier wave is called $Modulation$. This is necessary because low-frequency signals cannot travel long distances efficiently,and high-frequency waves act as carriers to transmit information over large distances.

(C) The critical frequency $(f_c)$ of the ionosphere is the maximum frequency of radio waves that can be reflected back to the Earth by the ionospheric layers.
It is given by the formula $f_c = 9 \sqrt{N_{max}}$, where $N_{max}$ is the maximum electron density of the ionosphere in electrons per cubic meter $(m^{-3})$.
In this context, the constant $9$ is derived from physical constants, and the expression is $9 \sqrt{N}$.

(A) In radio communication,the signal is transmitted in the form of electromagnetic waves.
These electromagnetic waves propagate through the atmosphere or vacuum without the need for any physical medium like wires or cables.
Therefore,the medium through which these waves travel is known as free space.
Thus,free space acts as the communication channel for radio communication.

(B) Given:
Frequency of the modulating signal,$f_m = 3 \text{ kHz} = 0.003 \text{ MHz}$.
Frequency of the carrier wave,$f_c = 2.5 \text{ MHz} = 2500 \text{ kHz}$.
The upper sideband frequency $(f_{USB})$ is given by $f_c + f_m = 2500 \text{ kHz} + 3 \text{ kHz} = 2503 \text{ kHz}$.
The lower sideband frequency $(f_{LSB})$ is given by $f_c - f_m = 2500 \text{ kHz} - 3 \text{ kHz} = 2497 \text{ kHz}$.
Thus,the sideband frequencies are $2503 \text{ kHz}$ and $2497 \text{ kHz}$.

(D) The range $d$ for line-of-sight propagation is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth and $h$ is the height of the antenna.
From this formula,we can see that $d \propto \sqrt{h}$.
Let the initial range be $d_1 = \sqrt{2Rh}$ and the initial height be $h_1 = h$.
Let the new height be $h_2 = 2h$.
The new range $d_2$ is given by $d_2 = \sqrt{2Rh_2} = \sqrt{2R(2h)}$.
$d_2 = \sqrt{2} \times \sqrt{2Rh} = \sqrt{2} d_1$.
Therefore,if the height is doubled,the new range becomes $\sqrt{2} d$.

(D) In amplitude modulation,the amplitude of the carrier wave is varied in accordance with the instantaneous value of the modulating signal (information signal),while the frequency and phase of the carrier wave remain constant.

(D) The modulation index $(m_a)$ is defined as the ratio of the peak voltage of the modulating signal $(V_m)$ to the peak voltage of the carrier wave $(V_c)$.
Formula: $m_a = \frac{V_m}{V_c}$
Given:
Peak voltage of carrier wave $(V_c)$ = $16 \text{ V}$
Modulation index $(m_a)$ = $75 \% = 0.75$
Calculation:
$V_m = m_a \times V_c$
$V_m = 0.75 \times 16 \text{ V}$
$V_m = 12 \text{ V}$
Therefore, the peak voltage of the modulating signal is $12 \text{ V}$.

(C) Low frequency audio signals cannot be transmitted directly over long distances because they require antennas of impractical size and suffer from high attenuation. To overcome this,the low frequency audio signal (baseband signal) is superposed on a high frequency wave,known as the carrier wave. This process is called modulation.

(A) In amplitude modulation,the modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave. If $\mu > 1$,it leads to over-modulation,which causes severe distortion in the signal. Therefore,to avoid distortion,the modulation index $\mu$ is always kept less than or equal to $1$ (i.e.,$\mu \le 1$). Thus,the statement that $\mu$ is kept greater than one is incorrect.

(A) In amplitude modulation,the modulation index is defined as $\mu = \frac{A_m}{A_c}$.
If $\mu > 1$,the amplitude of the modulating signal exceeds the amplitude of the carrier wave,which leads to over-modulation.
Over-modulation causes the envelope of the carrier wave to become distorted,resulting in the loss of information.
Therefore,to avoid distortion,the ratio $\frac{A_m}{A_c}$ (modulation index) is kept less than or equal to $1$.

(D) In amplitude modulation,the amplitude of the high-frequency carrier wave is varied in accordance with the instantaneous amplitude of the modulating (information) signal,while the frequency and phase of the carrier wave remain constant.

(A) modem stands for Modulator-Demodulator.
In the process of space communication,at the transmitting end,the modem acts as a modulator to convert digital signals into analog signals for transmission.
At the receiving end,the modem acts as a demodulator to convert the received analog signals back into digital signals.
Therefore,the modem acts as a modulator during transmission and as a demodulator during reception.

(B) The ionosphere acts as a reflecting medium for electromagnetic waves in the frequency range of $3 \ MHz$ to $30 \ MHz$. These waves are known as sky waves. When these waves are transmitted towards the sky,they are reflected back to the earth by the ionospheric layers,enabling long-distance communication.

(D) The primary advantages of optical fibres include:
$1$. They possess a very high bandwidth,allowing for the transmission of a large amount of data.
$2$. They have a high data transmission capacity,significantly exceeding that of copper wires or radio waves.
$3$. They are practically free from electromagnetic $(EM)$ interference and crosstalk,which are common issues in ordinary cables and microwave links.

(B) Space waves are used for line-of-sight $(LOS)$ communication.
Line-of-sight is a type of communication that can transmit and receive data only when transmitter and receiver stations are in view of each other with no obstacle between them.
This mode of propagation is typically used for high-frequency signals like $VHF$,$UHF$,and microwave signals.

(B) The wavelength $\lambda$ of the electromagnetic wave is given by the formula $\lambda = \frac{c}{f}$,where $c$ is the speed of light $(3 \times 10^8 \text{ m/s})$ and $f$ is the frequency $(5 \times 10^6 \text{ Hz})$.
Substituting the values: $\lambda = \frac{3 \times 10^8}{5 \times 10^6} = 0.6 \times 10^2 = 60 \text{ m}$.
For an antenna to work efficiently,its length should be at least $\frac{\lambda}{4}$.
Therefore,the required size of the antenna is $\frac{60}{4} = 15 \text{ m}$.

(C) Given: Radius of the Earth, $R = 6400 \text{ km} = 6400 \times 10^3 \text{ m}$.
Height of the antenna, $h = 500 \text{ m} = 0.5 \text{ km}$.
The formula for the range $(d)$ of an antenna is given by $d = \sqrt{2Rh}$.
Substituting the values:
$d = \sqrt{2 \times 6400 \text{ km} \times 0.5 \text{ km}}$
$d = \sqrt{6400 \times 1} \text{ km}$
$d = \sqrt{6400} \text{ km} = 80 \text{ km}$.
Thus, the range of the antenna is $80 \text{ km}$.

(D) The process of superimposing a message signal on a high-frequency carrier wave is called modulation.
It is usually applied to electromagnetic signals.
Common modulation methods are amplitude modulation,frequency modulation,and phase modulation.

(A) basic communication system is designed to transfer information from one point to another.
The process follows a specific sequence:
$1$. Information source: The origin of the message.
$2$. Transmitter: Converts the information into a signal suitable for transmission.
$3$. Channel: The medium through which the signal travels.
$4$. Receiver: Extracts the original signal from the channel.
$5$. User of information: The final destination or recipient of the message.
Therefore,the correct sequence is: Information source $\rightarrow$ Transmitter $\rightarrow$ Channel $\rightarrow$ Receiver $\rightarrow$ User of information,which corresponds to $(b)$ $\rightarrow$ $(a)$ $\rightarrow$ $(d)$ $\rightarrow$ $(e)$ $\rightarrow$ $(c)$.

(A) In amplitude modulation,the amplitude of the carrier wave is varied in accordance with the instantaneous value of the modulating signal.
Mathematically,the modulated wave consists of the carrier frequency $(f_c)$ and two sideband frequencies: the upper sideband $(f_c + f_m)$ and the lower sideband $(f_c - f_m)$,where $f_m$ is the frequency of the modulating signal.
Therefore,amplitude modulation consists of one carrier frequency and two sideband frequencies.

(B) The maximum range $d$ for line-of-sight communication is given by the formula: $d = \sqrt{2Rh_t} + \sqrt{2Rh_r}$,where $R$ is the radius of the Earth,$h_t$ is the height of the transmitting antenna,and $h_r$ is the height of the receiving antenna.
Given: $d = 57.6 \ km = 57.6 \times 10^3 \ m$,$h_r = 80 \ m$,$R = 6.4 \times 10^6 \ m$.
Substituting the values:
$57.6 \times 10^3 = \sqrt{2 \times 6.4 \times 10^6 \times h_t} + \sqrt{2 \times 6.4 \times 10^6 \times 80}$
$57.6 \times 10^3 = \sqrt{12.8 \times 10^6 \times h_t} + \sqrt{1024 \times 10^6}$
$57.6 \times 10^3 = \sqrt{12.8 \times 10^6 \times h_t} + 32000$
$57.6 \times 10^3 - 32 \times 10^3 = \sqrt{12.8 \times 10^6 \times h_t}$
$25.6 \times 10^3 = \sqrt{12.8 \times 10^6 \times h_t}$
Squaring both sides:
$(25.6 \times 10^3)^2 = 12.8 \times 10^6 \times h_t$
$655.36 \times 10^6 = 12.8 \times 10^6 \times h_t$
$h_t = \frac{655.36}{12.8} = 51.2 \ m$.

(A) Optical fibers use light waves for communication,which have very high frequencies (in the range of $THz$). Due to this high carrier frequency,the bandwidth available for signal transmission is extremely large,typically exceeding $100 GHz$. This allows for the transmission of a massive amount of data compared to traditional copper wires.

(B) digital signal represents data as a sequence of discrete values,typically $0$ and $1$,which correspond to the binary code system. Unlike analog signals that are continuous (like sine or cosine waves),digital signals are discontinuous and rely on binary logic.

(B) coaxial cable,also known as $TV$ aerial cable or coax,is primarily used to carry video and data signals from an antenna to a device such as a satellite dish or television.
This is due to the well-insulated conductor wire,which prevents frequency interference.
Coaxial cables can transmit a high range of frequencies due to low power losses.
The maximum number of $TV$ signals that can be transmitted simultaneously along a single coaxial cable is $125$.

(A) The $UHF$ (Ultra High Frequency) range typically spans from $300 \ MHz$ to $3000 \ MHz$.
Due to their high frequency,these waves cannot be reflected by the ionosphere (sky waves) and are heavily attenuated by the ground (ground waves).
Therefore,they propagate primarily through line-of-sight communication,which is achieved by means of space waves.

(A) In satellite communication,the uplink frequency is the frequency at which the ground station transmits signals to the satellite.
This frequency is kept high (in the microwave range) to ensure that the signals can penetrate the ionosphere without being reflected back to Earth.
The standard frequency band used for uplink satellite communication is $5.9-6.4 \text{ GHz}$.

(B) Given,the highest modulating frequency,$f_m = 5 kHz = 5 \times 10^3 Hz$.
For an amplitude-modulated wave,the bandwidth required for one station is $BW = 2 f_m$.
Therefore,$BW = 2 \times 5 kHz = 10 kHz = 10^4 Hz$.
The total available bandwidth is $f = 150 kHz = 150 \times 10^3 Hz$.
The number of stations that can be accommodated is given by the ratio of the total bandwidth to the bandwidth per station.
Number of stations $= \frac{f}{BW} = \frac{150 kHz}{10 kHz} = 15$.

(C) Given: Wavelength $\lambda = 800 \,nm = 800 \times 10^{-9} \,m$. Speed of light $c = 3 \times 10^8 \,m/s$.
Frequency of the source $f = c / \lambda = (3 \times 10^8) / (800 \times 10^{-9}) = 3.75 \times 10^{14} \,Hz$.
Available signal bandwidth is $1 \%$ of the source frequency:
Bandwidth $= 0.01 \times 3.75 \times 10^{14} = 3.75 \times 10^{12} \,Hz$.
Bandwidth required for one $TV$ channel $= 6 \,MHz = 6 \times 10^6 \,Hz$.
Number of channels $= \text{Total bandwidth} / \text{Bandwidth per channel} = (3.75 \times 10^{12}) / (6 \times 10^6) = 0.625 \times 10^6 = 6.25 \times 10^5$.
Expressing this in the form of the options: $6.25 \times 10^5 = (1/16) \times 10^7 = 0.0625 \times 10^7$.