Unit 7: Data Communication and Computer Networks

DATA COMMUNICATION & COMPUTER NETWORKS

UGC NET / JRF — Computer Science & Applications

High-Yield Study Notes & PYQ-Pattern Workbook (Unit 7)

Contents

Ch.1 — Introduction to Computer Networks

Ch.2 — Models

Ch.3 — Physical Layer

Ch.4 — Data Link Layer

Ch.5 — Network Layer

Ch.6 — Transport Layer

Ch.7 — Session, Presentation & Application Layer

Ch.8 — Network Security

Ch.9 — Mobile and Cloud Computing

One-Shot Revision

Mini Mock Test (Answer Key)

Beginner → Concept → NET-level → JRF-level. Compact by design — exam value over page count.

Chapter 1 — Introduction to Computer Networks

1.1 Types of Networks by Scale

Type	Coverage
PAN (Personal Area Network)	Very short range (e.g., Bluetooth, a few meters)
LAN (Local Area Network)	Single building/campus
MAN (Metropolitan Area Network)	A city
WAN (Wide Area Network)	Country/continent/global (e.g., the Internet)

1.2 Network Topologies

Topology	Idea	Key weakness
Bus	All nodes share one common backbone cable	Single cable failure disables the entire network
Star	All nodes connect to one central hub/switch	Central hub failure disables the entire network
Ring	Each node connects to exactly two neighbours, forming a loop	A single broken link can disrupt the whole ring (unless dual-ring)
Mesh	Every node connects to every other node (full mesh)	Very high cabling/cost — n(n−1)/2 links needed

JRF-level numericalA full mesh topology with 6 nodes needs how many links? Formula: n(n−1)/2 = 6×5/2 = 15 links. Each node also needs (n−1)=5 ports/interfaces. This link-count formula is a very frequently tested mesh-topology numerical.

1.3 Network Devices

Device	OSI Layer	Function
Repeater/Hub	Physical (1)	Regenerates/broadcasts signal to all ports (no intelligence, no filtering)
Bridge/Switch	Data Link (2)	Forwards frames based on MAC address, filters traffic between segments
Router	Network (3)	Forwards packets based on IP address, connects different networks
Gateway	All layers (up to Application)	Connects networks using DIFFERENT protocols, does protocol translation

JRF trap — hub vs switchA HUB simply broadcasts every incoming signal to ALL other ports (creates ONE large collision domain — all connected devices compete for the same bandwidth). A SWITCH learns MAC addresses and forwards frames ONLY to the intended destination port (each port is its OWN separate collision domain), dramatically reducing collisions. "Which device eliminates the collision domain problem" is a classic JRF distinguishing question.

MUST REMEMBER — Chapter 1

• Scale order: PAN < LAN < MAN < WAN.
• Mesh topology link count: n(n−1)/2; needs (n−1) ports per node.
• Hub = Physical layer, broadcasts to all (one collision domain). Switch = Data Link layer, learns MAC, forwards selectively (separate collision domains per port).
• Router = Network layer, forwards by IP, connects different networks. Gateway = protocol translation across all layers.

DON'T CONFUSE

• Hub (broadcasts, one collision domain) vs Switch (selective forwarding, separate collision domains).
• Bridge/Switch (MAC-based, same network type) vs Router (IP-based, connects different networks) vs Gateway (protocol translation).

JRF CHALLENGE ZONE — Chapter 1

1. A full mesh topology has 8 nodes. How many links are needed?
Answer: 8×7/2 = 28

2. Which device operates at the Network layer, forwarding based on IP address? (a) Hub (b) Switch (c) Router (d) Repeater
Answer: (c)

3. Which device creates a SEPARATE collision domain for each connected port? (a) Hub (b) Switch (c) Repeater (d) None of these
Answer: (b)

Practice Questions — Chapter 1 (6)

1. Order PAN, LAN, MAN, WAN from smallest to largest coverage.
   Ans: PAN < LAN < MAN < WAN
2. A full mesh topology has 5 nodes. How many links are needed?
   Ans: 5×4/2 = 10
3. What is the main weakness of a Bus topology?
   Ans: A single cable failure can disable the entire network
4. Differentiate a Hub and a Switch in terms of collision domains.
   Ans: A hub creates one shared collision domain for all ports; a switch creates a separate collision domain per port
5. At which OSI layer does a Router operate?
   Ans: Network layer (Layer 3)
6. What does a Gateway do that a Router does not?
   Ans: Translates between different protocols/network types, not just forwarding within compatible networks

---

Chapter 2 — Models

2.1 OSI Reference Model — 7 Layers

#	Layer	Key Function
7	Application	User-facing services (HTTP, FTP, SMTP)
6	Presentation	Data translation, encryption, compression
5	Session	Establishes/manages/terminates sessions between applications
4	Transport	End-to-end delivery, reliability (TCP), segmentation
3	Network	Logical addressing (IP), routing between networks
2	Data Link	Framing, MAC addressing, error detection, media access control
1	Physical	Raw bit transmission over the physical medium

Memory trickTop→Bottom (7→1): "All People Seem To Need Data Processing" = Application, Presentation, Session, Transport, Network, Data Link, Physical.

2.2 TCP/IP Model — 4 Layers

TCP/IP Layer	Corresponds to OSI Layer(s)
Application	Application + Presentation + Session (7,6,5)
Transport	Transport (4)
Internet	Network (3)
Network Access (Link)	Data Link + Physical (2,1)

JRF trapThe TCP/IP model COMBINES OSI's top 3 layers (Application, Presentation, Session) into just ONE "Application" layer, and combines OSI's bottom 2 layers (Data Link, Physical) into one "Network Access" layer — this exact layer-merging mapping is one of the most frequently tested OSI-vs-TCP/IP questions.

2.3 Encapsulation & PDU Names

Layer	PDU (Protocol Data Unit) Name
Transport	Segment (TCP) / Datagram (UDP)
Network	Packet
Data Link	Frame
Physical	Bits

JRF trapData flows DOWN the stack at the sender (each layer ADDS its own header — "encapsulation") and UP the stack at the receiver (each layer REMOVES/reads its header — "decapsulation"). Knowing the exact PDU-name-per-layer sequence (Segment→Packet→Frame→Bits) is frequently tested via "what is data called at the network layer" style questions.

MUST REMEMBER — Chapter 2

• OSI 7 layers top-to-bottom: Application, Presentation, Session, Transport, Network, Data Link, Physical.
• TCP/IP 4 layers: Application (=OSI 7+6+5), Transport, Internet (=OSI Network), Network Access (=OSI Data Link+Physical).
• PDU names: Transport=Segment/Datagram, Network=Packet, Data Link=Frame, Physical=Bits.
• Encapsulation adds headers going down the stack (sender); decapsulation removes them going up (receiver).

DON'T CONFUSE

• OSI's 7 separate layers vs TCP/IP's 4 (merged) layers — know exactly which OSI layers combine into which TCP/IP layer.
• Packet (Network layer PDU) vs Frame (Data Link layer PDU) vs Segment (Transport layer PDU).

JRF CHALLENGE ZONE — Chapter 2

1. The TCP/IP "Application" layer corresponds to which OSI layers? (a) Only Application (b) Application+Presentation (c) Application+Presentation+Session (d) All 7 layers
Answer: (c)

2. What is the PDU called at the Network layer? (a) Frame (b) Packet (c) Segment (d) Bits
Answer: (b)

Practice Questions — Chapter 2 (7)

1. List the 7 OSI layers from top to bottom.
   Ans: Application, Presentation, Session, Transport, Network, Data Link, Physical
2. List the 4 TCP/IP layers.
   Ans: Application, Transport, Internet, Network Access
3. Which OSI layers does the TCP/IP "Network Access" layer combine?
   Ans: Data Link and Physical
4. What is the PDU name at the Data Link layer?
   Ans: Frame
5. What is the PDU name at the Transport layer (for TCP)?
   Ans: Segment
6. What happens to data as it moves DOWN the protocol stack at the sender?
   Ans: Each layer adds its own header (encapsulation)
7. Which OSI layer handles encryption and compression?
   Ans: Presentation layer

---

Chapter 3 — Physical Layer

3.1 Transmission Media

Guided (Wired)	Key trait
Twisted Pair	Cheap, moderate bandwidth, susceptible to interference (twisting reduces crosstalk)
Coaxial Cable	Better shielding than twisted pair, higher bandwidth
Fiber Optic	Uses LIGHT pulses; HIGHEST bandwidth, immune to electromagnetic interference, but most expensive

JRF trapFiber optic cable is immune to electromagnetic interference because it transmits data as LIGHT, not electrical signals — this is why it's preferred in electrically noisy environments and for very long-distance, high-bandwidth links (e.g., undersea cables). Unguided media (wireless: radio, microwave, infrared) don't need a physical medium but are more susceptible to interference and eavesdropping.

3.2 Transmission Modes

Mode	Meaning
Simplex	ONE direction only, ever (e.g., keyboard→computer, TV broadcast)
Half-Duplex	BOTH directions, but only ONE at a time (e.g., walkie-talkie)
Full-Duplex	BOTH directions SIMULTANEOUSLY (e.g., telephone call)

3.3 Multiplexing

Type	Idea
FDM (Frequency Division)	Splits available bandwidth into separate FREQUENCY bands, each channel gets its own slice simultaneously
TDM (Time Division)	Each channel gets the FULL bandwidth, but only for a short, rotating TIME SLOT
WDM (Wavelength Division)	Like FDM but specifically for FIBER OPTIC — different light wavelengths carry separate channels

JRF trap — FDM vs TDMFDM divides the channel by FREQUENCY (all users transmit simultaneously, each on a different frequency band). TDM divides the channel by TIME (all users share the same frequency but take turns in short time slots). Confusing "simultaneous but different frequency" (FDM) with "same frequency but different time" (TDM) is the classic multiplexing trap.

3.4 Switching Techniques

Type	Idea
Circuit Switching	A DEDICATED physical path is reserved for the ENTIRE communication session (e.g., traditional telephone network) — guaranteed bandwidth, but wastes capacity during idle periods
Packet Switching	Data broken into PACKETS, each routed INDEPENDENTLY (may take different paths); no dedicated path, more efficient bandwidth use (e.g., the Internet)
Message Switching	Entire message stored and forwarded hop-by-hop as ONE unit ("store-and-forward") — older technique, largely replaced by packet switching

JRF trapCircuit switching has NO setup delay for DATA transfer once the circuit is established, but requires CONNECTION SETUP time upfront and wastes bandwidth if the circuit is idle. Packet switching has NO dedicated path (better bandwidth utilization, resilient to failure — packets can reroute) but individual packets may arrive OUT OF ORDER or with variable delay (jitter), requiring reassembly at the destination. This full tradeoff comparison is a very high-yield JRF topic.

MUST REMEMBER — Chapter 3

• Fiber optic: highest bandwidth, immune to EM interference (uses light, not electricity).
• Simplex=one-way always; Half-duplex=both ways, one at a time; Full-duplex=both ways simultaneously.
• FDM=simultaneous, different frequency bands. TDM=same frequency, rotating time slots. WDM=FDM for fiber optics.
• Circuit switching=dedicated path, guaranteed bandwidth, setup delay, can waste idle capacity. Packet switching=no dedicated path, efficient, but variable delay/out-of-order arrival.

DON'T CONFUSE

• FDM (divides by frequency) vs TDM (divides by time).
• Circuit switching (dedicated path, telephone-style) vs Packet switching (independent routing, Internet-style).

JRF CHALLENGE ZONE — Chapter 3

1. Which multiplexing technique allows all channels to transmit SIMULTANEOUSLY, each on a different frequency band? (a) TDM (b) FDM (c) Both equally (d) Neither
Answer: (b)

2. Which switching technique reserves a dedicated path for the ENTIRE session? (a) Packet switching (b) Message switching (c) Circuit switching (d) None
Answer: (c)

3. Why is fiber optic immune to electromagnetic interference? (a) It uses thicker cables (b) It transmits data as light, not electrical signals (c) It uses more shielding (d) It doesn't transmit data at all
Answer: (b)

Practice Questions — Chapter 3 (7)

1. Which transmission medium offers the highest bandwidth?
   Ans: Fiber optic cable
2. Differentiate Half-Duplex and Full-Duplex.
   Ans: Half-Duplex allows both directions but only one at a time; Full-Duplex allows both directions simultaneously
3. Differentiate FDM and TDM.
   Ans: FDM splits bandwidth by frequency (simultaneous transmission); TDM splits by time (rotating time slots, same frequency)
4. What is WDM, and where is it used?
   Ans: Wavelength Division Multiplexing — like FDM but for fiber optic, using different light wavelengths
5. What is the main advantage of Circuit Switching?
   Ans: Guaranteed, dedicated bandwidth for the entire session
6. What is the main advantage of Packet Switching over Circuit Switching?
   Ans: More efficient bandwidth use, since there's no dedicated (potentially idle) path
7. Give an example each of a simplex, half-duplex, and full-duplex communication.
   Ans: Simplex: TV broadcast; Half-duplex: walkie-talkie; Full-duplex: telephone call

---

Chapter 4 — Data Link Layer

4.1 Error Detection — Parity & Checksum

Parity bitAdds one extra bit to make the total number of 1s EVEN (even parity) or ODD (odd parity). Detects only SINGLE-bit errors reliably — if TWO bits flip, parity can stay unchanged and the error goes undetected.

4.2 CRC (Cyclic Redundancy Check)

JRF-level numerical — CRC computationData = 1101011011, Generator = 10011 (degree 4, so append 4 zero bits). Append 4 zeros: 11010110110000. Perform XOR-based binary division by the generator 10011, repeatedly XOR-ing wherever the leading bit is 1, until the remainder is shorter than the generator. The final remainder (CRC checksum, 4 bits) is appended to the original data to form the transmitted frame. At the RECEIVER, the same division is performed on the received frame — if the remainder is ALL ZEROS, no error is detected; any nonzero remainder signals an error. This "append zeros, XOR-divide, use remainder as checksum" procedure is THE standard CRC JRF numerical (exact remainder value varies by problem, but the METHOD is what's tested).

NET pointCRC can detect ALL single-bit errors, ALL double-bit errors (if generator has ≥2 terms), all errors with an ODD number of bit flips (if generator includes factor (x+1)), and any burst error SHORTER than the generator's degree — CRC is far more powerful than a simple parity bit.

4.3 Hamming Code (Error Correction)

IdeaHamming code can DETECT AND CORRECT single-bit errors by adding redundant parity bits at positions that are POWERS OF 2 (1,2,4,8...). Each parity bit checks a specific subset of data bit positions (determined by binary representation).

Number of redundant (parity) bits r needed for m data bits: 2^r ≥ m + r + 1

JRF-level numericalFor 7 data bits (m=7), find minimum redundant bits r. Try r=3: 2³=8 ≥ 7+3+1=11? NO (8<11). Try r=4: 2⁴=16 ≥ 7+4+1=12? YES. So minimum r = 4 — total codeword length = 7+4 = 11 bits. This "solve for minimum r satisfying 2^r ≥ m+r+1" is the standard Hamming code JRF numerical.

4.4 Flow Control

Protocol	Idea
Stop-and-Wait	Sender sends ONE frame, waits for an ACK before sending the next — simple but very inefficient (idle time waiting for each ACK)
Sliding Window	Sender can send MULTIPLE frames (up to a "window size") before needing an ACK — much more efficient use of bandwidth

JRF-level numerical — efficiency of Stop-and-WaitBandwidth=1 Mbps, Propagation delay=20ms (one-way), frame size=1000 bits (transmission time=1000/1,000,000=1ms). Efficiency = Transmission time / (Transmission time + 2×Propagation delay) = 1 / (1 + 2×20) = 1/41 ≈ 2.44%. This extremely LOW efficiency — dominated by round-trip propagation delay — is exactly why Sliding Window protocols are preferred for high bandwidth-delay-product links; this efficiency-formula numerical is a favourite JRF question.

4.5 Medium Access Control (MAC) Protocols

Protocol	Used in	Idea
CSMA/CD	Wired Ethernet (traditional)	Listen before sending; if a COLLISION is DETECTED during transmission, stop, wait a random backoff time, retry
CSMA/CA	Wireless (WiFi)	Listen before sending; tries to AVOID collisions proactively (since wireless collisions are harder to detect while transmitting)

JRF trap — CD vs CACSMA/CD (Collision DETECTION) works because a wired sender CAN listen to the medium WHILE transmitting and detect a collision in progress. CSMA/CA (Collision AVOIDANCE) is used in wireless networks because a wireless device generally CANNOT reliably detect a collision while transmitting (its own strong outgoing signal drowns out the ability to hear a weaker colliding signal) — so it must AVOID collisions in the first place, e.g., using RTS/CTS handshakes. "Why does WiFi use CA instead of CD" is a very frequently tested JRF conceptual question.

MUST REMEMBER — Chapter 4

• Parity bit detects only single-bit errors (odd number of flips); fails on 2-bit flips.
• CRC: append (generator degree) zeros, XOR-divide, remainder=checksum; receiver checks for all-zero remainder.
• Hamming code: find min r where 2^r ≥ m+r+1; parity bits at positions 1,2,4,8...; corrects single-bit errors.
• Stop-and-Wait efficiency = Tt/(Tt+2Tp) — very low for high-delay links; Sliding Window is much more efficient.
• CSMA/CD (wired, detects collision while sending) vs CSMA/CA (wireless, avoids collision proactively since detection isn't reliable).

DON'T CONFUSE

• Error DETECTION (parity, CRC — just flags an error) vs Error CORRECTION (Hamming code — fixes it too).
• CSMA/CD (wired, collision detection) vs CSMA/CA (wireless, collision avoidance).

JRF CHALLENGE ZONE — Chapter 4

1. For 4 data bits, find the minimum number of Hamming redundant bits r needed.
Answer: Try r=3: 2³=8 ≥ 4+3+1=8? YES (equal, still valid). So r=3.

2. Why can't WiFi reliably use CSMA/CD? (a) WiFi doesn't need collision handling (b) A transmitting wireless device can't reliably hear a weaker colliding signal over its own strong outgoing signal (c) WiFi has no collisions ever (d) CSMA/CD is faster for WiFi
Answer: (b)

3. Bandwidth=2Mbps, one-way propagation delay=10ms, frame=2000 bits. Find Stop-and-Wait efficiency.
Answer: Tt=2000/2,000,000=1ms. Efficiency=1/(1+2×10)=1/21≈4.76%

Practice Questions — Chapter 4 (8)

1. What type of error can a simple parity bit reliably detect?
   Ans: Single-bit errors (fails if exactly 2 bits flip)
2. What is the general procedure for computing a CRC checksum?
   Ans: Append zeros (equal to the generator's degree) to the data, XOR-divide by the generator, and use the remainder as the checksum
3. What formula gives the minimum redundant bits r for m data bits in Hamming code?
   Ans: 2^r ≥ m + r + 1
4. For 11 data bits, find the minimum Hamming redundant bits r.
   Ans: Try r=4: 2⁴=16≥11+4+1=16? Yes (equal). So r=4
5. Why is Stop-and-Wait protocol inefficient over high-delay links?
   Ans: The sender spends most of the time idly waiting for an ACK across the round-trip propagation delay rather than transmitting
6. What advantage does Sliding Window have over Stop-and-Wait?
   Ans: It allows multiple frames to be sent before waiting for an ACK, much better utilizing available bandwidth
7. Differentiate CSMA/CD and CSMA/CA.
   Ans: CSMA/CD detects collisions while transmitting (wired); CSMA/CA proactively avoids collisions since detection isn't reliable (wireless)
8. Does Hamming code perform error detection only, or detection AND correction?
   Ans: Both — it can detect and correct single-bit errors

---

Chapter 5 — Network Layer

5.1 IPv4 Addressing — Classful

Class	First bits	Range (first octet)	Default mask
A	0	0–127	/8 (255.0.0.0)
B	10	128–191	/16 (255.255.0.0)
C	110	192–223	/24 (255.255.255.0)
D (Multicast)	1110	224–239	N/A

5.2 Subnetting — Numericals

JRF-level numerical — finding subnets and hostsGiven network 192.168.1.0/26. Find number of subnets (if borrowed from a /24) and hosts per subnet. /26 means 26 network bits, so host bits = 32−26 = 6. Hosts per subnet = 2⁶−2 = 62 (subtract network address and broadcast address). Number of subnets created from the original /24 = 2^(26−24) = 2² = 4 subnets. This "given a CIDR prefix, find hosts-per-subnet and subnet-count" numerical is THE most common IP addressing question at NET/JRF level.

JRF-level numerical — finding network/broadcast addressIP = 192.168.1.100, subnet mask = /26 (255.255.255.192). Block size = 256−192 = 64. Subnet boundaries: 0,64,128,192... 100 falls in the 64–127 block. Network address = 192.168.1.64. Broadcast address = 192.168.1.64+63 = 192.168.1.127. Valid host range = 192.168.1.65 to 192.168.1.126. This "find which subnet an IP belongs to" technique using block size is a very frequently tested numerical.

TrapHosts per subnet = 2^(host bits) − 2 (subtracting network address and broadcast address) — NOT simply 2^(host bits). Forgetting the "−2" is the single most common subnetting numerical mistake.

5.3 IPv4 vs IPv6

	IPv4	IPv6
Address length	32 bits	128 bits
Address format	Dotted decimal (e.g., 192.168.1.1)	Hexadecimal, colon-separated (e.g., 2001:db8::1)
Total addresses	~4.3 billion (2³²)	~340 undecillion (2¹²⁸)
Header	Variable length, more complex	Fixed length (40 bytes), simplified for faster processing

5.4 Routing Algorithms

Type	Idea	Example
Distance Vector	Each router shares its ENTIRE routing table with DIRECTLY connected neighbours only; "routing by rumor"	RIP
Link State	Each router builds a COMPLETE map of the whole network topology (via flooding link-state info to ALL routers) and independently computes shortest paths	OSPF

JRF trap — count-to-infinity problemDistance Vector routing (e.g., RIP) suffers from the "count-to-infinity" problem — when a link fails, incorrect/outdated routing info can circulate between neighbours for many iterations before converging, since routers only trust their immediate neighbours' claims without seeing the full topology. Link State routing (e.g., OSPF) avoids this since every router has the COMPLETE topology map and can independently verify/recompute — this "why distance vector converges slowly after failure" is a favourite JRF conceptual question.

5.5 ARP & RARP

IdeaARP (Address Resolution Protocol): given an IP address, finds the corresponding MAC address (used when a device knows the destination IP but needs the physical address to actually send a frame on the LAN). RARP (Reverse ARP): given a MAC address, finds the corresponding IP address (largely obsolete now, replaced by DHCP).

MUST REMEMBER — Chapter 5

• Class A: 0-127; Class B: 128-191; Class C: 192-223 (first octet ranges).
• Hosts per subnet = 2^(host bits) − 2 (subtract network + broadcast addresses).
• Subnet count from CIDR change = 2^(new prefix − old prefix).
• Block size method: find which block an IP falls in to get network/broadcast addresses.
• IPv4=32-bit; IPv6=128-bit, fixed 40-byte header.
• Distance Vector (RIP): shares full table with neighbours only, suffers count-to-infinity. Link State (OSPF): full topology map, faster convergence.
• ARP: IP→MAC. RARP: MAC→IP (obsolete, replaced by DHCP).

DON'T CONFUSE

• Distance Vector (neighbour-only info, count-to-infinity risk) vs Link State (full topology, faster convergence).
• ARP (IP→MAC) vs RARP (MAC→IP) — opposite direction lookups.

JRF CHALLENGE ZONE — Chapter 5

1. Network 172.16.0.0/20. How many host bits, and how many usable hosts per subnet?
Answer: Host bits=32−20=12. Hosts=2¹²−2=4094

2. IP=10.0.0.70, subnet mask=/27 (block size=32). Find the network address.
Answer: Blocks: 0,32,64,96... 70 falls in 64-95 block → network address=10.0.0.64

3. Which routing protocol type suffers from the "count-to-infinity" problem? (a) Link State (b) Distance Vector (c) Both equally (d) Neither
Answer: (b)

Practice Questions — Chapter 5 (8)

1. Network 192.168.10.0/28. Find hosts per subnet.
   Ans: Host bits=32−28=4; hosts=2⁴−2=14
2. Network 10.0.0.0/24 is subnetted to /26. How many subnets are created?
   Ans: 2^(26−24)=4
3. What is the first octet range for Class B addresses?
   Ans: 128–191
4. Differentiate IPv4 and IPv6 address lengths.
   Ans: IPv4: 32 bits; IPv6: 128 bits
5. Differentiate Distance Vector and Link State routing.
   Ans: Distance Vector shares its routing table with only its immediate neighbours; Link State floods topology info to build a full network map
6. What causes the "count-to-infinity" problem in Distance Vector routing?
   Ans: Routers only trust neighbour-reported distances without a full topology view, so outdated info can circulate for many iterations after a failure
7. What does ARP resolve, and in which direction?
   Ans: It resolves an IP address to its corresponding MAC address
8. IP=172.16.5.130, subnet mask=/25 (block size=128). Find the network address.
   Ans: Blocks: 0,128... 130 falls in 128-255 block → network address=172.16.5.128

---

Chapter 6 — Transport Layer

6.1 TCP vs UDP

	TCP	UDP
Connection	Connection-oriented (3-way handshake)	Connectionless
Reliability	Reliable — guarantees delivery, ordering, retransmits lost data	Unreliable — no delivery/order guarantee
Speed/Overhead	Slower, more overhead (ACKs, sequencing)	Faster, minimal overhead (no ACKs by default)
Used for	Web (HTTP), email, file transfer — where correctness matters	Video/audio streaming, DNS, gaming — where speed matters more than occasional loss

JRF trapUDP is preferred for real-time applications (VoIP, video calls, live streaming) specifically BECAUSE waiting for TCP's retransmission of a lost/late packet would cause more disruption (stuttering/lag) than just skipping that bit of data and moving on — "why does streaming use UDP despite being unreliable" is a frequently tested conceptual JRF question.

6.2 TCP 3-Way Handshake

Client → SYN → Server Client ← SYN-ACK ← Server Client → ACK → Server (Connection established)

NET pointClosing a TCP connection typically uses a 4-way handshake (FIN, ACK, FIN, ACK) since each side must independently signal it has no more data to send — this asymmetry between the 3-way OPEN and 4-way CLOSE is occasionally tested.

6.3 Flow Control vs Congestion Control

	Flow Control	Congestion Control
Protects against	Overwhelming the RECEIVER (receiver's buffer overflow)	Overwhelming the NETWORK (router/link congestion)
Mechanism	Sliding window, receiver advertises available buffer space	Slow start, congestion avoidance, detecting packet loss/timeout as a congestion signal

JRF trapFlow control is about the RECEIVER's capacity (a fast sender overwhelming a slow receiver) — a purely END-TO-END concern between two hosts. Congestion control is about the NETWORK's capacity (too much combined traffic overwhelming routers/links in between) — a concern about the ENTIRE PATH, not just the two endpoints. Mixing up "receiver-focused" (flow) vs "network-focused" (congestion) control is a common JRF trap.

6.4 TCP Congestion Control Phases

Phase	Idea
Slow Start	Congestion window starts small, DOUBLES every RTT (exponential growth) until reaching a threshold
Congestion Avoidance	After threshold, window grows LINEARLY (by 1 per RTT) — more cautious increase
On packet loss	Threshold is halved, window resets (varies by TCP variant — Tahoe resets to 1, Reno may use fast recovery)

6.5 Port Numbers

Range	Category
0–1023	Well-known ports (e.g., HTTP=80, HTTPS=443, FTP=21, SMTP=25, DNS=53)
1024–49151	Registered ports
49152–65535	Dynamic/private ports (ephemeral, used by client-side connections)

MUST REMEMBER — Chapter 6

• TCP: connection-oriented, reliable, ordered, slower. UDP: connectionless, unreliable, faster.
• UDP preferred for real-time media (retransmission delay is worse than occasional loss).
• TCP opens with 3-way handshake (SYN, SYN-ACK, ACK); closes with 4-way handshake.
• Flow control = protects the RECEIVER (end-to-end). Congestion control = protects the NETWORK (whole path).
• Slow start = exponential window growth; Congestion avoidance = linear growth after threshold.
• Well-known ports: 0-1023 (HTTP=80, HTTPS=443, FTP=21, DNS=53).

DON'T CONFUSE

• Flow control (receiver-focused) vs Congestion control (network-focused).
• Slow Start (exponential growth) vs Congestion Avoidance (linear growth).

JRF CHALLENGE ZONE — Chapter 6

1. Why does live video streaming typically use UDP instead of TCP? (a) UDP is more reliable (b) Retransmission delay would cause worse disruption than occasional data loss (c) TCP cannot carry video data (d) UDP has connection setup
Answer: (b)

2. Flow control primarily protects against: (a) Network congestion (b) Overwhelming the receiver's buffer (c) Packet loss on the wire (d) DNS failures
Answer: (b)

3. What is the standard port number for HTTPS?
Answer: 443

Practice Questions — Chapter 6 (7)

1. List two key differences between TCP and UDP.
   Ans: Any two of: connection-oriented vs connectionless; reliable vs unreliable; slower/more overhead vs faster/less overhead
2. What are the three steps of the TCP 3-way handshake?
   Ans: SYN, SYN-ACK, ACK
3. Differentiate flow control and congestion control.
   Ans: Flow control protects the receiver from being overwhelmed; congestion control protects the network from being overwhelmed
4. What happens to TCP's congestion window during Slow Start?
   Ans: It grows exponentially (doubles every RTT) until reaching a threshold
5. What happens to TCP's congestion window during Congestion Avoidance?
   Ans: It grows linearly (by 1 per RTT)
6. What port number does HTTP use by default?
   Ans: 80
7. Why is UDP preferred for DNS queries over TCP in most cases?
   Ans: DNS queries are small and need to be fast; UDP's lower overhead suits quick request-response exchanges

---

Chapter 7 — Session, Presentation & Application Layer

7.1 Session Layer

IdeaEstablishes, manages, synchronizes, and terminates sessions (dialogues) between applications on different hosts — includes features like checkpointing/synchronization points, so a long transfer can resume from a checkpoint rather than restarting entirely if interrupted.

7.2 Presentation Layer

IdeaHandles data TRANSLATION (e.g., character encoding like ASCII/Unicode conversion between systems), ENCRYPTION/decryption, and COMPRESSION/decompression — ensures data sent by one system's format is properly interpreted by another's.

7.3 Application Layer — Common Protocols

Protocol	Port	Purpose
HTTP	80	Web page transfer (stateless)
HTTPS	443	Secure (encrypted) web transfer
FTP	21 (control), 20 (data)	File transfer
SMTP	25	Sending email
POP3	110	Retrieving email (downloads & typically deletes from server)
IMAP	143	Retrieving email (keeps synced on server, multi-device friendly)
DNS	53	Domain name → IP address resolution
DHCP	67/68	Automatic IP address assignment

JRF trap — POP3 vs IMAPPOP3 typically DOWNLOADS mail to the local device and DELETES it from the server (mail tied to one device). IMAP keeps mail SYNCHRONIZED on the server, allowing access from MULTIPLE devices consistently. "Which protocol suits a user who checks email from phone, laptop, AND desktop" → IMAP — this practical distinction is frequently tested.

7.4 DNS Resolution Process

IdeaDNS resolves a domain name to an IP address hierarchically: query goes to a Root server → TLD (Top-Level Domain, e.g., .com) server → Authoritative name server for the specific domain — each step narrows down until the final IP is returned (often cached at various points to speed up future queries).

MUST REMEMBER — Chapter 7

• Session layer: manages dialogues, checkpointing/synchronization for resumable transfers.
• Presentation layer: translation (encoding), encryption, compression.
• Key ports: HTTP=80, HTTPS=443, FTP=21/20, SMTP=25, POP3=110, IMAP=143, DNS=53, DHCP=67/68.
• POP3=downloads+deletes from server (single device); IMAP=stays synced on server (multi-device).
• DNS resolution: Root → TLD → Authoritative server (hierarchical).

DON'T CONFUSE

• POP3 (download+delete, single-device) vs IMAP (server-synced, multi-device).
• Session layer (dialogue management) vs Presentation layer (data format/encryption/compression).

JRF CHALLENGE ZONE — Chapter 7

1. Which email protocol is best for a user accessing email consistently across phone, laptop, and desktop? (a) SMTP (b) POP3 (c) IMAP (d) FTP
Answer: (c)

2. Which OSI layer handles encryption/compression, distinct from actual dialogue management? (a) Session (b) Presentation (c) Application (d) Transport
Answer: (b)

Practice Questions — Chapter 7 (6)

1. What is the main function of the Session layer?
   Ans: Establishing, managing, synchronizing, and terminating sessions between applications
2. What three tasks does the Presentation layer typically handle?
   Ans: Data translation/encoding, encryption/decryption, compression/decompression
3. What port does HTTPS use?
   Ans: 443
4. Differentiate POP3 and IMAP.
   Ans: POP3 downloads and typically deletes mail from the server (single-device); IMAP keeps mail synced on the server (multi-device access)
5. What does DNS do?
   Ans: Resolves a domain name to its corresponding IP address
6. Describe the hierarchical order of DNS resolution servers.
   Ans: Root server → TLD server → Authoritative name server

---

Chapter 8 — Network Security

8.1 Cryptography — Symmetric vs Asymmetric

	Symmetric	Asymmetric
Keys used	SAME key for encryption and decryption	DIFFERENT keys — public key encrypts, private key decrypts (or vice versa)
Speed	Fast	Much SLOWER (computationally expensive)
Key distribution problem	Sharing the secret key securely is a challenge	No shared secret needed — public key can be shared openly
Example algorithms	AES, DES	RSA

JRF trap — hybrid approach in practiceReal-world secure systems (like HTTPS/TLS) commonly use a HYBRID approach: Asymmetric cryptography is used ONLY to securely exchange a symmetric session key (solving the key-distribution problem), then Symmetric cryptography (much faster) is used for the actual BULK data encryption. "Why not use asymmetric encryption for everything, since it's more secure" → because it's too slow for bulk data; this practical reasoning is a common JRF question.

8.2 Digital Signatures

IdeaThe sender encrypts a message digest (hash) with their OWN PRIVATE key. Anyone can verify it using the sender's PUBLIC key — proving the message genuinely came from that sender (authentication) and wasn't altered (integrity). Note: this is the REVERSE key usage from normal asymmetric encryption (private key to "sign", public key to "verify" — not for confidentiality).

8.3 Firewalls

IdeaA firewall monitors and filters incoming/outgoing network traffic based on a defined set of security RULES — acting as a barrier between a trusted internal network and untrusted external networks (e.g., the Internet).

8.4 Common Network Attacks

Attack	Idea
DoS/DDoS	Overwhelms a target with excessive traffic/requests, making the service unavailable to legitimate users
Man-in-the-Middle (MITM)	Attacker secretly intercepts/relays communication between two parties, who believe they're communicating directly
Phishing	Deceptive attempt (often via fake emails/websites) to trick users into revealing sensitive information
SQL Injection	Malicious SQL code inserted into an input field to manipulate/access the backend database improperly

MUST REMEMBER — Chapter 8

• Symmetric = same key (fast, key-sharing problem); Asymmetric = public/private key pair (slow, no shared secret needed).
• Practical systems (TLS/HTTPS) use HYBRID: asymmetric to exchange a symmetric session key, then symmetric for bulk data.
• Digital signature: sign with sender's PRIVATE key, verify with sender's PUBLIC key (reverse of confidentiality use).
• Firewall: filters traffic based on rules, sits between trusted and untrusted networks.
• DoS/DDoS=traffic overload; MITM=intercepts communication; Phishing=social engineering; SQL Injection=malicious DB input.

DON'T CONFUSE

• Symmetric (one shared key) vs Asymmetric (public/private key pair).
• Digital signature key usage (private to sign, public to verify) vs normal asymmetric encryption (public to encrypt, private to decrypt) — reversed roles.

JRF CHALLENGE ZONE — Chapter 8

1. In a digital signature scheme, the sender signs the message digest using: (a) Their own public key (b) Their own private key (c) The receiver's public key (d) The receiver's private key
Answer: (b)

2. Why do systems like HTTPS use a hybrid encryption approach? (a) Asymmetric alone is too slow for bulk data (b) Symmetric alone is unbreakable (c) Hybrid is required by law (d) There's no real reason
Answer: (a)

Practice Questions — Chapter 8 (6)

1. Differentiate symmetric and asymmetric cryptography.
   Ans: Symmetric uses the same key for encryption/decryption; asymmetric uses a public/private key pair
2. Why is asymmetric cryptography typically used only for key exchange, not bulk data, in practice?
   Ans: It is computationally much slower than symmetric cryptography
3. Which key does a sender use to create a digital signature?
   Ans: Their own private key
4. What does a firewall do?
   Ans: Monitors and filters network traffic based on defined security rules
5. What is a Man-in-the-Middle attack?
   Ans: An attacker secretly intercepts and relays communication between two parties who believe they are communicating directly
6. What is the goal of a DoS/DDoS attack?
   Ans: To overwhelm a target with traffic, making its service unavailable to legitimate users

---

Chapter 9 — Mobile and Cloud Computing

9.1 Cellular Network Generations

Generation	Key characteristic
1G	Analog voice only
2G	Digital voice, basic SMS/data (GSM)
3G	Mobile broadband — video calls, mobile internet
4G/LTE	High-speed all-IP data (no separate voice circuit — voice also travels as data, VoLTE)
5G	Very high speed, low latency, supports massive IoT device density

9.2 Cloud Computing — Service Models

Model	Provider gives you	You manage	Example
IaaS (Infrastructure)	Virtual machines, storage, networking	OS, runtime, applications, data	AWS EC2
PaaS (Platform)	+ OS, runtime environment	Applications, data only	Google App Engine
SaaS (Software)	+ The complete application itself	Just your data/usage — nothing to manage	Gmail, Google Docs

JRF trap — the management-responsibility gradientMoving from IaaS → PaaS → SaaS, the CLOUD PROVIDER takes on progressively MORE responsibility, while the CUSTOMER manages progressively LESS. IaaS gives you the most control but the most management burden (you handle OS patching, scaling config, etc.); SaaS gives you the least control but zero management burden (you just use the app). "Which service model requires you to manage the OS yourself" → IaaS specifically — this responsibility-boundary distinction is the most frequently tested cloud-computing JRF question.

9.3 Cloud Deployment Models

Model	Idea
Public Cloud	Shared infrastructure, owned/operated by a third-party provider, accessible to the general public/many customers
Private Cloud	Dedicated infrastructure for a SINGLE organization — more control/security, but more cost/management overhead
Hybrid Cloud	Combination of public + private, allowing data/applications to move between them (e.g., "cloud bursting" for peak demand)
Community Cloud	Shared by SEVERAL organizations with common concerns (e.g., compliance requirements), not the general public

9.4 Cloud Characteristics

• On-demand self-service: provision resources without needing human interaction with the provider.
• Broad network access: accessible over the network via standard mechanisms (any device).
• Resource pooling: provider's resources are pooled to serve multiple customers (multi-tenancy), dynamically assigned.
• Rapid elasticity: resources can be scaled up/down quickly, often automatically, to match demand.
• Measured service: usage is monitored/controlled/billed ("pay-as-you-go").

MUST REMEMBER — Chapter 9

• Generation order: 1G(analog voice)→2G(digital voice+SMS)→3G(mobile broadband)→4G(all-IP,VoLTE)→5G(ultra-fast,low-latency,IoT).
• IaaS=infrastructure only (you manage OS+); PaaS=+OS/runtime (you manage apps+data); SaaS=complete app (you manage nothing).
• Provider responsibility increases IaaS→PaaS→SaaS; customer management burden decreases in the same direction.
• Public=shared/third-party; Private=dedicated to one org; Hybrid=mix of both; Community=shared by orgs with common needs.
• 5 key cloud characteristics: on-demand self-service, broad network access, resource pooling, rapid elasticity, measured service.

DON'T CONFUSE

• IaaS (you manage OS/apps) vs PaaS (you manage only apps/data) vs SaaS (you manage nothing — just use it).
• Private cloud (single org) vs Community cloud (multiple orgs with shared concerns, still not public).

JRF CHALLENGE ZONE — Chapter 9

1. Which cloud service model requires the customer to manage the operating system themselves? (a) SaaS (b) PaaS (c) IaaS (d) None of these
Answer: (c)

2. Which cellular generation introduced all-IP data with voice also carried as data (VoLTE)? (a) 2G (b) 3G (c) 4G (d) 1G
Answer: (c)

Practice Questions — Chapter 9 (6)

1. Order the cellular generations 1G through 5G with one key trait each.
   Ans: 1G: analog voice; 2G: digital voice/SMS; 3G: mobile broadband; 4G: all-IP data/VoLTE; 5G: ultra-fast, low-latency, IoT-scale
2. Differentiate IaaS, PaaS, and SaaS in terms of what the customer must manage.
   Ans: IaaS: OS, runtime, apps, data; PaaS: apps and data only; SaaS: nothing, just usage
3. Which cloud service model gives you the most control but the most management responsibility?
   Ans: IaaS
4. Differentiate Public and Private cloud.
   Ans: Public cloud is shared, third-party-operated infrastructure; Private cloud is dedicated to a single organization
5. What is a Hybrid cloud?
   Ans: A combination of public and private clouds, allowing data/applications to move between them
6. Name three of the five key characteristics of cloud computing.
   Ans: Any three of: on-demand self-service, broad network access, resource pooling, rapid elasticity, measured service

---

One-Shot Revision — Unit 7

Key facts across all chapters

• Scale: PAN<LAN<MAN<WAN. Mesh links=n(n−1)/2. Hub=1 collision domain; Switch=per-port collision domain. Router=IP-based; Gateway=protocol translation.
• OSI 7 layers (top-down): App,Presentation,Session,Transport,Network,DataLink,Physical. TCP/IP 4 layers merge App+Pres+Session into one, DataLink+Physical into one.
• PDU names: Segment(Transport)→Packet(Network)→Frame(DataLink)→Bits(Physical).
• Fiber=highest bandwidth, EM-immune. FDM=frequency-split,simultaneous; TDM=time-split,same frequency. Circuit switching=dedicated path; Packet switching=independent routing.
• CRC: append generator-degree zeros, XOR-divide, remainder=checksum. Hamming: min r where 2^r≥m+r+1.
• Stop-and-Wait efficiency=Tt/(Tt+2Tp) — low for high-delay links. CSMA/CD=wired,detects collision; CSMA/CA=wireless,avoids collision.
• Subnetting: hosts=2^(hostbits)−2; subnets from prefix change=2^(newprefix−oldprefix); use block-size method for network/broadcast address.
• Distance Vector(RIP)=neighbour-only,count-to-infinity risk; Link State(OSPF)=full topology,faster convergence. ARP=IP→MAC.
• TCP=connection-oriented,reliable,slower; UDP=connectionless,unreliable,faster (preferred for real-time media). 3-way handshake: SYN,SYN-ACK,ACK.
• Flow control=protects receiver; Congestion control=protects network. Slow Start=exponential; Congestion Avoidance=linear.
• POP3=download+delete(single device); IMAP=server-synced(multi-device). Key ports: HTTP80,HTTPS443,FTP21/20,SMTP25,DNS53.
• Symmetric=same key(fast); Asymmetric=public/private pair(slow). Practical systems use hybrid. Digital signature: sign with PRIVATE key, verify with PUBLIC key.
• Cellular: 1G analog→2G digital→3G broadband→4G all-IP/VoLTE→5G ultra-fast/IoT. IaaS(most mgmt)→PaaS→SaaS(least mgmt) — provider responsibility increases in that order.

Potential future exam areasPotential high-value exam area based on syllabus importance and historical question patterns: subnetting numericals (hosts/subnets/network-broadcast address); Hamming code and CRC computation numericals; Stop-and-Wait efficiency calculations; OSI-vs-TCP/IP layer mapping questions; TCP congestion control phase identification; and cloud service-model responsibility-boundary scenario questions.

---

Unit 7 — UGC NET/JRF Mini Mock Test

50 questions across all 9 chapters. NTA/UGC NET-style question patterns — mixed NET/JRF difficulty, numerical, statement-based, matching and scenario-based. Answer key with brief explanations follows each question.

Q1. A full mesh topology has 7 nodes. How many links are needed?
Ans: 7×6/2 = 21 [Ch1 | NET numerical]

Q2. Which device creates a separate collision domain per port? (a) Hub (b) Switch (c) Repeater (d) None
Ans: (b) [Ch1 | NET]

Q3. Which OSI layer handles logical (IP) addressing?
Ans: Network layer [Ch2 | NET]

Q4. The TCP/IP "Application" layer corresponds to which OSI layers? (a) 7 only (b) 7,6 (c) 7,6,5 (d) All 7
Ans: (c) [Ch2 | NET]

Q5. What is the PDU name at the Data Link layer?
Ans: Frame [Ch2 | NET]

Q6. Which transmission medium is immune to electromagnetic interference? (a) Twisted pair (b) Coaxial cable (c) Fiber optic (d) None
Ans: (c) [Ch3 | NET]

Q7. Which multiplexing divides the channel by TIME, with all users sharing the same frequency? (a) FDM (b) TDM (c) WDM (d) None
Ans: (b) [Ch3 | NET]

Q8. Which switching technique is used by the modern Internet? (a) Circuit switching (b) Packet switching (c) Message switching (d) None
Ans: (b) [Ch3 | NET]

Q9. For 8 data bits, find the minimum Hamming redundant bits r.
Ans: Try r=4: 16≥8+4+1=13? Yes → r=4 [Ch4 | JRF numerical]

Q10. Bandwidth=1Mbps, one-way propagation delay=25ms, frame=500 bits. Find Stop-and-Wait efficiency.
Ans: Tt=500/1,000,000=0.5ms. Eff=0.5/(0.5+50)=0.5/50.5≈0.99% [Ch4 | JRF numerical]

Q11. Which protocol is used by wired Ethernet for collision handling? (a) CSMA/CA (b) CSMA/CD (c) TDMA (d) None
Ans: (b) [Ch4 | NET]

Q12. A simple parity bit can reliably detect: (a) Any number of bit errors (b) Only single-bit errors (c) Only burst errors (d) Nothing
Ans: (b) [Ch4 | NET]

Q13. Network 192.168.5.0/27. Find hosts per subnet.
Ans: Host bits=32−27=5; hosts=2⁵−2=30 [Ch5 | NET numerical]

Q14. IP=192.168.1.50, mask=/26 (block size 64). Find the network address.
Ans: Blocks:0,64,128,192. 50 falls in 0-63 → network address=192.168.1.0 [Ch5 | JRF numerical]

Q15. Which routing protocol type builds a complete topology map at every router? (a) Distance Vector (b) Link State (c) Both (d) Neither
Ans: (b) [Ch5 | NET]

Q16. What does ARP resolve?
Ans: IP address to MAC address [Ch5 | NET]

Q17. How many bits are in an IPv6 address?
Ans: 128 [Ch5 | NET]

Q18. Which protocol is connection-oriented and reliable? (a) UDP (b) TCP (c) IP (d) ARP
Ans: (b) [Ch6 | NET]

Q19. What are the three steps of the TCP handshake, in order?
Ans: SYN, SYN-ACK, ACK [Ch6 | NET]

Q20. Congestion control primarily protects: (a) The receiver's buffer (b) The network/links from being overwhelmed (c) The sender's CPU (d) DNS servers
Ans: (b) [Ch6 | NET]

Q21. During TCP Slow Start, the congestion window grows: (a) Linearly (b) Exponentially (c) Not at all (d) Randomly
Ans: (b) [Ch6 | NET]

Q22. What is the well-known port for SMTP?
Ans: 25 [Ch6 | NET]

Q23. Which email protocol keeps mail synchronized across multiple devices? (a) SMTP (b) POP3 (c) IMAP (d) FTP
Ans: (c) [Ch7 | NET]

Q24. Which OSI layer handles data encryption/compression specifically?
Ans: Presentation layer [Ch7 | NET]

Q25. DNS resolution follows which hierarchical order? (a) TLD→Root→Authoritative (b) Root→TLD→Authoritative (c) Authoritative→Root→TLD (d) No hierarchy
Ans: (b) [Ch7 | NET]

Q26. Which cryptography type uses the SAME key for encryption and decryption? (a) Asymmetric (b) Symmetric (c) Both (d) Neither
Ans: (b) [Ch8 | NET]

Q27. A digital signature is created using the sender's: (a) Public key (b) Private key (c) The receiver's public key (d) A shared secret
Ans: (b) [Ch8 | NET]

Q28. Why do secure systems like HTTPS use a hybrid encryption approach?
Ans: Asymmetric encryption is too slow for bulk data; it's used only to exchange a symmetric session key [Ch8 | JRF]

Q29. What does a firewall primarily do?
Ans: Filters network traffic based on defined security rules [Ch8 | NET]

Q30. Which attack intercepts and relays communication between two unsuspecting parties?
Ans: Man-in-the-Middle (MITM) [Ch8 | NET]

Q31. Which cellular generation introduced all-IP data with voice as data (VoLTE)? (a) 2G (b) 3G (c) 4G (d) 5G
Ans: (c) [Ch9 | NET]

Q32. Which cloud service model requires the LEAST management from the customer? (a) IaaS (b) PaaS (c) SaaS (d) All equal
Ans: (c) [Ch9 | NET]

Q33. Which cloud deployment model is dedicated to a single organization?
Ans: Private cloud [Ch9 | NET]

Q34. Name the cloud characteristic describing pay-as-you-go usage tracking.
Ans: Measured service [Ch9 | NET]

Q35. Which switching technique guarantees dedicated bandwidth for the entire session? (a) Packet switching (b) Circuit switching (c) Message switching (d) None
Ans: (b) [Ch3 | NET]

Q36. A network is 10.0.0.0/22, subnetted to /24. How many subnets are created?
Ans: 2^(24−22) = 4 [Ch5 | JRF numerical]

Q37. Which of these is FALSE about UDP? (a) It is connectionless (b) It guarantees delivery order (c) It has lower overhead than TCP (d) It's used for DNS queries
Ans: (b) — UDP does NOT guarantee delivery or order. [Ch6 | NET]

Q38. Which is TRUE about Distance Vector routing? (a) It has full topology knowledge (b) It can suffer from count-to-infinity (c) It never has convergence issues (d) It's used only in link-state protocols
Ans: (b) [Ch5 | JRF]

Q39. What is the standard well-known port for DNS?
Ans: 53 [Ch7 | NET]

Q40. Which layer is responsible for MAC addressing and framing?
Ans: Data Link layer [Ch2 | NET]

Q41. RARP resolves: (a) IP to MAC (b) MAC to IP (c) Domain to IP (d) IP to Domain
Ans: (b) [Ch5 | NET]

Q42. Which cloud model gives you VMs/storage only, requiring you to manage the OS yourself?
Ans: IaaS [Ch9 | NET]

Q43. Which attack floods a target with traffic to make it unavailable?
Ans: DoS/DDoS [Ch8 | NET]

Q44. Which topology requires n(n-1)/2 links for full connectivity?
Ans: Mesh [Ch1 | NET]

Q45. What is the class of an IP address starting with 195 in its first octet? (a) A (b) B (c) C (d) D
Ans: (c) [Ch5 | NET]

Q46. Which TCP congestion control phase follows Slow Start once the threshold is reached?
Ans: Congestion Avoidance [Ch6 | NET]

Q47. Which OSI layer establishes and manages sessions between applications?
Ans: Session layer [Ch7 | NET]

Q48. Which is generally FASTER — symmetric or asymmetric cryptography?
Ans: Symmetric [Ch8 | NET]

Q49. Which cloud deployment model is shared by multiple organizations with common concerns, but not the general public?
Ans: Community cloud [Ch9 | NET]

Q50. Which of these is FALSE? (a) CRC can detect burst errors shorter than the generator's degree (b) Hamming code can correct single-bit errors (c) A parity bit can reliably detect any number of bit-flip errors (d) CRC uses XOR-based binary division
Ans: (c) — a parity bit fails to detect an even number of bit flips (e.g., 2-bit errors). [Ch4 | JRF]

— End of Mock Test — Cross-check your score, revisit the "Don't Confuse" and "JRF Challenge Zone" boxes for any topic you missed, then re-attempt after 48 hours. —

Share: