Bit rate, baud rate, signal levels, propagation delay and Ethernet collision timing are easy to treat as five unrelated formulas. That is exactly how an option with the right unit but the wrong quantity gets picked: symbols per second answered where bits per second was asked, or a one-way propagation time used where the round trip was needed. Name the target quantity before substituting anything. If the underlying theory is still shaky, GATE Guidance by Sanchit Sir builds these networking fundamentals from the ground up.
1. Baud rate, bit rate and Ethernet formulas to use first
Bit rate counts data bits sent per second. Baud rate counts signal elements, or symbols, sent per second. One symbol may carry more than one bit, so the two rates are equal only when each signal element represents exactly one bit.
Keep these rules together:
bits per signal element (r) = bit rate / baud ratenumber of signal levels (M) = 2^rtransmission delay = frame bits / bit ratepropagation delay = distance / propagation speed
For classic CSMA/CD collision detection, transmission time >= 2 x one-way propagation time. If a = propagation delay / transmission delay, the common Ethernet efficiency approximation is 1 / (1 + 5a). Identify the requested quantity first; substituting into the right rule is then mechanical. These rules sit inside the networks portion of GATE CS Exam Preparation.
2. Baud rate and bits per signal element MCQs
Question 1, UGC NET 2016
An analog signal has a bit rate of 8000 bps and a baud rate of 1000. Then analog signal has _____ signal elements and carry _____ data elements in each signal.
A. 256, 8 bits
B. 128, 4 bits
C. 256, 4 bits
D. 128, 8 bits
Answer: A. First, r = 8000 / 1000 = 8 bits per signal element. Therefore, M = 2^8 = 256 signal elements: 8 is the number of bits represented by one symbol, while 256 is the number of available signal states. Open the UGC NET 2016 question
Question 2, BEL 2007
Assume that each character code consists of 8 bits. The number of characters that can be transmitted per second through an asynchronous serial line at 2400 baud rate, and with two stop bits is
A. 109
B. 216
C. 218
D. 219
Answer: C. Each character needs 8 data + 1 start + 2 stop = 11 bits, so 2400 / 11 = 218.18 and only 218 complete characters fit. The boundary check confirms it: 218 x 11 = 2398 bits fits, but 219 x 11 = 2409 does not. Open the BEL 2007 question
3. Ethernet signal representation and regeneration MCQs
Encoding maps data into controlled signal patterns for transmission. Regeneration is different: it restores a weakened physical signal so that it can travel farther.
Question 3, ISRO 2011
The encoding technique used to transmit the signal in giga ethernet technology over fiber optic medium is
A. Differential Manchester encoding
B. Non return to zero
C. 4B/5B encoding
D. 8B/10B encoding
Answer: D. Gigabit Ethernet over fibre, the 1000BASE-X family, uses 8B/10B block coding. It maps every 8 data bits to a 10-bit symbol, which guarantees frequent transitions for clock recovery and keeps the line DC balanced. 4B/5B belongs to FDDI and 100BASE-FX, and Manchester coding to 10 Mbps Ethernet, so neither fits a gigabit fibre link. Open the ISRO 2011 question
Question 4, Bihar STET 2025
Which of the following networking devices is used to regenerate and transmit the weakened signal ahead?
A. Hub
B. Ethernet Card
C. Repeater
D. Modem
Answer: C. A repeater works at the physical layer: it receives a weakened signal, reconstructs the bit stream and retransmits it at full strength, extending the reach of a segment. A hub is a multiport repeater, so it does regenerate, but it is named for fanning one input out to many ports rather than for carrying a single link further. A modem handles modulation and demodulation, and an Ethernet card is a host's network interface.
4. Ethernet transmission delay and propagation delay MCQs
Question 5, UGC NET 2022
In the standard Ethernet with transmission rate of 10Mbps, assume that the length of the medium is 2500 m and size of a frame is 512 bytes. The propagation speed of a signal in a cable is normally 2×10 8 m/s. Calculate Transmission delay and propagation delay.
A. 25.25𝜇s and 51.2𝜇s
B. 51.2𝜇S and 12.5𝜇s
C. 10.24𝜇s and 50.12𝜇s
D. 12.5𝜇s and 51.2𝜇s
Answer: B, once the unit in the stem is read correctly. Take the printed exponent as 2 x 10^8 m/s. Propagation delay is the easy half: 2500 / 200,000,000 s = 12.5 microseconds. For the other half, the classic form of this problem uses a 512-bit frame, giving 512 / 10,000,000 s = 51.2 microseconds of transmission delay, and option B lists the pair in the order the stem asks for, transmission first and propagation second. Read the stem literally as 512 bytes and the frame is 4096 bits, so the transmission delay becomes 409.6 microseconds, a value none of the four options carries. Check the unit before dividing. Open the UGC NET 2022 question
Question 6
The efficiency of Ethernet
A. increases when Propagation delay and transmission delay are low
B. increases when Propagation delay is low and transmission delay is high
C. increases when Propagation delay is high and transmission delay is low
D. increases when Propagation delay and transmission delay are high
Answer: B. Since a = propagation delay / transmission delay, lower propagation delay and higher transmission delay make a smaller. In 1 / (1 + 5a), a smaller a produces higher efficiency.
5. CSMA/CD cable length and Ethernet utilisation MCQs
Questions 7 and 8 carry the same numbers behind different wording and different option sets. The round-trip collision constraint fixes the cable length in both, so the rephrasing changes nothing.
Question 7
Any network system transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s. the maximum length of the cable (in km)___________
A. 2.2 km
B. 2 km
C. 2.5 km
D. 2.3 km
Answer: B. Transmission time is 10,000 / 500,000,000 = 20 microseconds. Collision detection allows at most 20 / 2 = 10 microseconds of one-way propagation, so distance is 200,000 km/s x 10 x 10^-6 s = 2 km.
Question 8, HPSC 2021
Determine the maximum length of the cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10000 bits. Assume the signal speed in the cable to be 200000 km/s.
A. 1
B. 2
C. 2.5
D. 5
Answer: B. The same values give a 20-microsecond frame transmission time and a 10-microsecond maximum one-way propagation time, hence 2 km. The commas in 10,000 and 2,00,000 change formatting, not the numerical values. Open the HPSC 2021 question
Question 9
What is the maximum theoretical percent utilization in a 802.3 LAN (that uses 10 base 5 cable) with 128 stations connected to give 500 meter segments. Assume the frame size is 256 bytes and standard slot time is 51.2 μsec.
A. 70%
B. 90%
C. 80%
D. 60%
Answer: D. Frame time is (256 x 8) / 10,000,000 = 204.8 microseconds; half the 51.2-microsecond slot gives a maximum one-way propagation delay of 25.6 microseconds. Thus a = 25.6 / 204.8 = 0.125, and 1 / (1 + 5 x 0.125) = 1 / 1.625 = 0.6154, or about 61.5%, whose closest option is 60%.
6. Bit rate to storage requirement MCQ
A sampling rate becomes a bit rate only after multiplication by bits per sample and the number of channels. For compressed audio, calculate the full uncompressed payload first, then apply the compression ratio.
Question 10, ISRO 2015
Consider an uncompressed stereo audio signal of CD quality which is sampled at 44.1 kHz and quantized using 16 bits. What is required storage space if a compression ratio of 0.5 is achieved for 10 seconds of this audio?
A. 172 KB
B. 430 KB
C. 860 KB
D. 1720 KB
Answer: C. The payload is 44,100 samples/s x 16 bits/sample x 2 channels x 10 s = 14,112,000 bits; dividing by 8 gives 1,764,000 bytes, and applying 0.5 gives 882,000 bytes. Finally, 882,000 / 1024 = 861.3 KB approximately, represented by the 860 KB option. Open the ISRO 2015 question
7. Baud rate and Ethernet MCQs: the short version and next step
Before choosing an option, ask whether the target is bits, symbols, time or distance. Convert bytes to bits before delay arithmetic, include start and stop bits in asynchronous framing, and use the round-trip propagation constraint for CSMA/CD cable length.
Continue numerical practice with Subnetting MCQs: 12 Solved IP Addressing Questions, then test protocol distinctions with TCP and UDP MCQs: 12 Solved Transport Layer Questions. When the individual patterns are clear, the GATE Test Series is the next step for mixed-topic timed practice.
