Q: 1 Which of the following is not range of frequency band operate in Industrial Scientific and Medical (ISM)?
902 – 928 MHz
2.4 – 4.835 GHz
5.725 – 5.850 GHz
1.5 – 1.6 GHz
[ Option D ]
Industrial, Scientific, and Medical (ISM) bands are portions of the radio spectrum reserved internationally for non-commercial purposes like industrial heating, medical equipment, and scientific experimentation. Over time, they have also been widely used for unlicensed wireless communication, such as Wi-Fi and Bluetooth.
Q: 2 Find the incorrect statement about Manchester encoding in computer networks
It is used in physical layer
Main advantage is signal synchronization
Signal transitions always occur at the bit boundary
It is also called as biphase code
[ Option C ]
Manchester Encoding is a digital line‑coding scheme used at the physical layer to transmit data along with clock information. Each bit period is split into two phases, and there is always a transition in the middle of the bit, which provides self‑clocking and helps the receiver stay synchronized.
So, Manchester Encoding is used in the Physical Layer and provides signal synchronization through frequent transitions. It is called a Biphase Code because each bit has two phases, and its required transition occurs at the center of the bit, not necessarily at the bit boundary.
Q: 3 The primary use of scramblers in Line Coding Schemes is to—
Make the data look random
Compress transmission data
Decrease signal bandwidth
Increase signal bandwidth
[ Option A ]
A scrambler is used in line coding schemes to modify the original data before transmission.
Q: 4 Construction of a coaxial cable can be briefly given as
A solid inner conductor surrounded by an insulator
A solid inner conductor surrounded by an insulator and an overall conductive sheath
A solid inner conductor surrounded by a conductive sheath
A solid inner conductor twisted with an insulation
[ Option B ]
A coaxial cable consists of multiple layers designed to transmit signals efficiently with minimal interference. Its construction includes:
Q: 5 During transmission digital data is more secure because of _____________.
High Frequency Transmission
Low Frequency Transmission
Power Consumption Less
It Remain Encryption
[ Option D ]
Digital Data Transmission is considered more secure because the data can be encrypted before transmission. Encryption converts the original information into a coded form so that unauthorized users cannot easily understand or access the data.
In digital communication systems, techniques such as encryption, encoding, and authentication are commonly used to improve security during data transfer.
Q: 6 In Ethernet 1000BASE-X, the encoding scheme used is ________ that encodes ________ bit data into ________ bit data groups.
5B/4B, 5, 4
4B/5B, 4, 5
8B/10B, 8, 10
10B/8B, 10, 8
[ Option C ]
Encoding schemes in Ethernet convert data bits into transmittable symbols over physical media (fiber/copper). They ensure clock recovery (frequent transitions), DC balance (equal 0s/1s), and error detection by adding overhead bits.
In Gigabit Ethernet (1000BASE-X), data transmission uses an encoding scheme to ensure reliable communication, proper synchronization, and error detection. The scheme used is 8B/10B encoding.
In this technique, 8 bits of actual data are converted into 10-bit code groups before transmission. This extra 2 bits help in:
Therefore, 1000BASE-X uses 8B/10B encoding, where 8-bit data is converted into 10-bit groups for reliable high-speed transmission.
Q: 7 The frequency which is used for transmission of signal from the satellite to earth station receive is known as ___________.
Uplink Frequency
Backlink Frequency
Forward Link Frequency
Downlink Frequency
[ Option D ]
In satellite communication, signals travel between the earth station and the satellite using different frequencies. There are two main communication links:
The frequency used for sending signals from the satellite to the earth station receiver is called the Downlink Frequency.
For example, a TV broadcast satellite sends signals to home dish antennas using downlink frequency.
Q: 8 In digital modulation schemes using two carriers, a constellation diagram is particularly useful because it helps to
Measure only the frequency variations in the signal.
Visualize the amplitude and phase of each signal element.
Encode binary data directly without modulation.
Eliminate noise from the transmitted signal.
[ Option B ]
In digital communication, Modulation is the process of varying a carrier signal (sine wave) to transmit data. Digital modulation schemes often use two or more carriers, which can vary in amplitude, phase, or frequency to represent binary information.
A Constellation Diagram is a graphical representation of these modulated signals on a 2D plane, where the x-axis represents the in-phase component (I) and the y-axis represents the quadrature component (Q). Each point in the diagram corresponds to a unique amplitude and phase combination, allowing engineers to visualize the transmitted symbols and detect errors.
| TOOL | PURPOSE | OBSERVED PARAMETERS |
|---|---|---|
| Constellation Diagram | Visualization of digital modulation. | Amplitude and Phase. |
| Eye Diagram | Time-domain signal analysis. | Signal timing, Jitter, ISI. |
| Spectrum Analyzer | Frequency analysis. | Frequency, bandwidth, harmonics. |
Q: 9 Which of the following uses simplex mode of data transfer?
Print from Computer to Printer
Internet Browsing
Telephone Communication
None of the above
[ Option A ]
In data communication, Simplex Mode is a type of transmission where data flows in only one direction with no possibility of data flowing back, like traffic on a one-way street. One device acts as the sender, and the other only receives.
In Computer to Printer communication, data flows only from the computer to the printer. The printer does not send data back to the computer. This is a perfect example of simplex communication.
| Term | Explanation |
|---|---|
| Internet Browsing | Uses two-way (full-duplex) communication (request and response). |
| Telephone Communication | Uses full-duplex communication because both users speak and listen simultaneously. |
Q: 10 In a Walkie—Dalkie which data communication method is used?
Simplex Method
Half—Duplex Method
Full—Duplex Method
Bi—Flex Method
[ Option B ]
A Walkie-Talkie uses the Half-Duplex communication method. In half-duplex communication, data can travel in both directions, but not at the same time.
In a walkie-talkie:
Q: 11 In digital transmission the error performance is quantified in terms of ___________.
Baud Rate
Data Rate
BER
Bit Rate
[ Option C ]
In digital communication, the quality or error performance of transmission is measured using BER (Bit Error Rate).
BER represents the number of bits received incorrectly compared to the total number of bits transmitted.
For example, if 1000 bits are transmitted and 10 bits are received incorrectly, then:
BER = 10/1000 = 0.01
A lower BER means better transmission quality.
| Term | Description | Example |
|---|---|---|
| Baud Rate | Number of signal changes (symbols) transmitted per second in a communication channel. One baud may carry one or more bits. | If a modem changes its signal 2400 times in one second, then baud rate is 2400 baud. |
| Symbol Rate | Number of symbols transmitted per second. It is another name for Baud Rate in digital communication. | If 5000 symbols are transmitted each second, Symbol Rate is 5000 baud. |
| Bit Rate | Total number of bits transmitted per second, including data and control bits. | If 1,000,000 bits are transmitted in one second, Bit Rate = 1 Mbps. |
| Data Rate | Amount of actual useful data transferred per second. It may be less than bit rate because of extra control/error bits. | Out of 1200 transmitted bits, if only 1000 bits are actual data, then data rate is 1000 bps. |
| BER (Bit Error Rate) | Measurement of errors in digital transmission. It shows how many bits are received incorrectly. | If 10 bits are wrong out of 10,000 transmitted bits, BER = 10/10000 = 0.001. |
| Bandwidth | Range of frequencies used for communication or maximum data carrying capacity of a channel. | A channel with 20 MHz bandwidth can carry more data than a 5 MHz channel. |
| Throughput | Actual amount of data successfully delivered over a network in real conditions. | A 100 Mbps network may provide only 80 Mbps throughput because of traffic and delays. |
| Latency | Time taken by data to travel from sender to receiver. | In online gaming, lower latency gives faster response. |
Q: 12 Identify the diagram and write the type of conversion the system is performing.

Digital to Analog
Digital to Digital
Analog to Digital
Analog to Analog
[ Option B ]
Q: 13 Which of the following statement(s) is/are true regarding Data Communication?
I. Half duplex communication is multidirectional communication.
II. Full duplex communication is bidirectional communication.
Only I
Only II
Both I and II
Neither I Nor II
[ Option B ]
Data communication is the process of transferring data between two or more devices through communication channels such as cables, optical fibers, or wireless networks.
Communication modes define the direction of data flow between sender and receiver. The three main communication modes are:
In Simplex Communication, data flows only in one direction. One device only sends data and the other device only receives data.
In Half Duplex Communication, data can flow in both directions, but not at the same time. Only one device can transmit data at a particular moment. Therefore, half duplex communication is bidirectional, not multidirectional.
In Full Duplex Communication, data can flow in both directions simultaneously. Both devices can send and receive data at the same time. Therefore, full duplex communication is bidirectional communication.
Q: 14 If the channel band is limited to 6 kHz and signal to noise ratio is 16, what would be the capacity of the channel?
15.15 kbps
24.74 kbps
30.12 kbps
52.18 kbps
[ Option B ]
Q: 15 Which type of channel does not represent any correlation between input and output symbols?
Noiseless channel
Lossless channel
Useless channel
Deterministic channel
[ Option C ]
A Useless Channel is a communication channel where there is no correlation between input and output symbols. This means the output is completely independent of the input, the received symbol gives no information about what was sent. In this, the channel capacity is zero because the output does not carry any meaningful information about the input.
A noiseless channel, output is exactly the same as input, so there is a perfect correlation between input and output.
A lossless channel, input can be uniquely determined from the output, so correlation still exists.
A deterministic channel, each input symbol corresponds to exactly one output symbol, hence strong correlation.
Q: 16 A receiver using Manchester decoding expects a transition in every bit period. If the line goes from high to high during one bit period, what does the receiver most likely interpret?
Bit 0
Bit 1
Bit error or synchronization loss
Bit 1 followed by Bit 0
[ Option C ]
In Manchester Encoding, each bit period must have a transition in the middle, Low to High for bit 1 and High to Low for bit 0. This transition helps maintain synchronization between sender and receiver. If the line stays high throughout a bit period (High to High), it means no transition occurred, violating the Manchester rule. Therefore, the receiver interprets it as a bit error or loss of synchronization.
Q: 17 We want to digitize a human voice signal. Assuming that each audio sample is represented using 8 bits and the human voice normally contains frequencies from 0 to 4000 Hz. What is the resulting bit rate of the digitized signal?
128 kbps
32 kbps
64 kbps
256 kbps
[ Option C ]
To digitize a human voice signal, the Pulse Code Modulation (PCM) method is used, which requires sampling the analog signal.
According to the Nyquist theorem, the sampling rate must be at least twice the highest frequency present in the signal.
Since the human voice ranges up to 4000 Hz, the signal must be sampled at 8000 samples per second. If each sample is represented by 8 bits, the total bit rate becomes 8000 × 8 = 64,000 bits per second, or 64 kbps.
Q: 18 If a signal consists of V discrete levels and channel bandwidth is B, then according to Nyquist's theorem, the maximum data rate is:
4B log2V bits/sec
2V log2B bits/sec
2B log2V bits/sec
V log2B bits/sec
[ Option C ]
According to Nyquist’s theorem, the maximum data rate of a noiseless channel depends on its bandwidth (B) and the number of discrete signal levels (V) used in transmission.
Maximum Data Rate = 2B log2V bits per second
Q: 19 Four channel are multiplexed using TDM. If each channel sends 100 bytes/seconds, then the bit rate for the link is __________.
400 bps
800 bps
1600 bps
3200 bps
[ Option D ]
In Time Division Multiplexing (TDM), multiple channels share a single communication link, and the total bit rate is the sum of the bit rates of all channels.
Each channel sends 100 bytes/second. Since 1 byte = 8 bits, we first convert bytes to bits:
One channel: 100*8 = 800 bits/second
Now, there are 4 channels, so total bit rate: 800*4 = 3200 bits/second (3200 bps)
Q: 20 In the ________ encoding scheme, each 24 bits become four 6-bit chunks, and eventually are sent as 32 bits.
8 bits
Binary
Base 64
Quoted-Printable
[ Option C ]
In computer networks, binary data cannot always be transmitted directly over text-based protocols like email (SMTP) or web (HTTP). To solve this problem, Base64 encoding is used. Base64 takes every 24 bits of input data (3 bytes) and divides it into four groups of 6 bits each. Each 6-bit value is then mapped to a printable ASCII character.
Since each Base64 output is an ASCII character, it requires 8 bits of storage. Therefore, four characters × 8 bits = 32 bits of output are generated from the original 24 bits. This explains why Base64 increases the size of the encoded data by about 33%.
When encoding the string "SRK" into Base64, the process begins by converting each character into its ASCII binary form. The character S has an ASCII value of 83, which is 01010011 in binary. The character R has an ASCII value of 82, represented as 01010010. Finally, the character K has an ASCII value of 75, represented as 01001011. Together, the string "SRK" produces a 24-bit binary stream:
01010011 01010010 01001011
Next, these 24 bits are divided into groups of 6 bits each:
010100 110101 001001 001011
Each 6-bit group is then converted into its decimal equivalent:
Using the Base64 index table, these decimal values are mapped to their respective characters:
Thus, the final Base64 encoded output of the string "SRK" is: U1JL
Q: 21 For a noiseless channel, the ___________ formula defines the theoretical maximum bit rate.
Nyquist bit rate
Shannon bit rate
Ethernet bit rate
None bit rate
[ Option A ]
For a Noiseless communication channel, the theoretical maximum bit rate is determined by the Nyquist bit rate formula. The Nyquist formula calculates the highest possible data rate based on the channel bandwidth (B) and the number of discrete signal levels (M) used in transmission.
Maximum bit rate = 2 × B × log2(M)
In contrast, the Shannon bit rate formula considers noisy channels and accounts for signal-to-noise ratio (SNR), making it applicable when noise is present.
Q: 22 The transmission media that can carry huge data to long distances with less delay/latency is _____________.
Optical Fiber Cables
Low Noise, High Quality, Underwater Coaxial Cables
Microwave / RF/ Wireless
Shielded Twisted Pair Cables
[ Option A ]
To carry huge data over long distances with low delay (Latency), a transmission medium must combine high bandwidth, low signal loss (Attenuation), and little interference.
Optical Fiber Cables use light signals to transmit data, which allows them to carry very large amounts of data over long distances with minimal loss and very low latency.
Q: 23 While transmission of data from a sender to receiver using Modem on Internet, the data is first converted to ________ from ________.
Analog, Analog
Analog, Digital
Digital, Analog
Digital, Digital
[ Option B ]
A Modem (Modulator-Demodulator) is a device that allows computers to communicate over telephone lines or other analog transmission media.
This process is called Modulation (Digital to Analog) and Demodulation (Analog to Digital), which is why the device is called a MODEM (Modulator-Demodulator).
Q: 24 Quantizing noise occurs in–
PCM
TDM
FDM
PPM
[ Option A ]
Quantizing noise occurs during the quantization process in Pulse Code Modulation (PCM). PCM converts an analog signal into a digital signal by sampling the amplitude of the analog signal at regular intervals and then quantizing these samples to the nearest value among a finite set of levels. The difference between the actual analog value and the quantized digital value is called quantization error or quantizing noise.
Q: 25 What is the maximum binary signal transmission rate of a noiseless 3-kHz channel?
12000 bps
6000 bps
3000 bps
1500 bps
[ Option B ]
The maximum data rate of a noiseless channel is determined by the Nyquist Theorem. It defines the theoretical upper limit for data transmission rate in an noiseless channel.
Maximum bit rate = 2×B×log2(M)
Where:
B = Bandwidth of the channel.
M = Number of discrete signal levels (symbols) used.
Given Values:
B=3000 Hz, M=2 (says binary signal and binary means two levels, 0 and 1)
Maximum bit rate = 2 × 3000 × log2(2) and log2(2)=1
So, Maximum bit rate = 2 × 3000 × 1 = 6000 bps.
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