Pipelining in Computer Architecture MCQ

MCQs on Pipelining in Computer Architecture for UGC-NET , GATE, SET and other examinations.

Q1. In a 4-stage pipeline with stage delays of 60 ns, 50 ns, 90 ns, and 80 ns, and a latch delay of 10 ns, what is the pipeline cycle time?
A. 90 ns
B. 100 ns
C. 110 ns
D. 80 ns
Answer: B. 100 ns
Explanation: Maximum stage delay = 90 ns; Cycle time = 90 + 10 = 100 ns

Q2. If non-pipelined execution takes 280 ns and pipelined execution takes 100 ns, what is the speed-up?
A. 2.0
B. 2.5
C. 2.8
D. 3.0
Answer: C. 2.8
Explanation: Speed-up = 280 / 100 = 2.8

Q3. A pipeline has stage delays of 150 ns, 120 ns, 160 ns, and 140 ns. If the latch delay is 5 ns, what is the pipeline cycle time?
A. 150 ns
B. 155 ns
C. 160 ns
D. 165 ns
Answer: D. 165 ns
Explanation: Maximum stage delay = 160 ns; Cycle time = 160 + 5 = 165 ns

Q4. A non-pipelined processor runs at 2.5 GHz and takes 4 cycles per instruction. A pipelined version runs at 2 GHz with 5 stages and no stalls. What is the speed-up?
A. 3.2
B. 2.0
C. 2.5
D. 4.0
Answer: A. 3.2
Explanation:
Non-pipelined time = 4 / 2.5 = 1.6 ns
Pipelined time = 1 / 2 = 0.5 ns
Speed-up = 1.6 / 0.5 = 3.2

Q5. In a 5-stage pipeline with stage delays of 2.5 ns, 1.5 ns, 2.0 ns, 1.5 ns, and 2.5 ns, and a latch delay of 0.5 ns, what is the pipeline cycle time?
A. 2.5 ns
B. 3.0 ns
C. 2.0 ns
D. 2.8 ns
Answer: B. 3.0 ns
Explanation: Maximum stage delay = 2.5 ns; Cycle time = 2.5 + 0.5 = 3.0 ns.

10 Distance Vector and Link-State Routing MCQ

 10 multiple-choice questions on Distance Vector and Link-State Routing with explanations, useful for UGC NET, SET, GATE, ISRO, and other competitive exams.

Question 1: Basic Concept of Distance Vector Routing

Q1: In a distance vector routing protocol, how do routers share routing information?

• A) By flooding the entire network with routing updates

• B) By periodically exchanging routing tables with their immediate neighbors

• C) By using a link-state database to determine the shortest path

• D) By dynamically updating paths based on link failure reports

Answer: B) By periodically exchanging routing tables with their immediate neighbors
Explanation: In Distance Vector Routing, routers share their routing tables with neighbors at regular intervals, leading to a simple but slow convergence mechanism.

Question 2: Example of Distance Vector Routing Protocol

Q2: Which of the following is an example of a Distance Vector routing protocol?

• A) OSPF

• B) RIP

• C) IS-IS

• D) BGP

Answer: B) RIP
Explanation: Routing Information Protocol (RIP) follows the Distance Vector approach, using hop count as its metric to determine the shortest path.

Question 3: Count to Infinity Problem

Q3: The "count to infinity" problem is associated with which type of routing protocol?

• A) Link-State

• B) Distance Vector

• C) Hybrid

• D) Static Routing

Answer: B) Distance Vector
Explanation: The count to infinity problem occurs in Distance Vector routing when routers continuously increase hop counts due to network changes, leading to slow convergence.

Question 4: Solution to Count to Infinity Problem

Q4: Which of the following is used to mitigate the "count to infinity" problem in Distance Vector Routing?

• A) Split Horizon

• B) Dijkstra’s Algorithm

• C) Link-State Advertisement

• D) Shortest Path Bridging

Answer: A) Split Horizon
Explanation: Split Horizon prevents routing loops by not advertising routes back to the router from which they were learned, reducing the risk of infinite loops.

Question 5: Link-State Routing Protocol

Q5: Which of the following is a Link-State routing protocol?

• A) RIP

• B) OSPF

• C) BGP

• D) IGRP

Answer: B) OSPF
Explanation: Open Shortest Path First (OSPF) is a Link-State protocol that builds a complete network topology using Link-State Advertisements (LSAs).

Question 6: Algorithm Used in Link-State Routing

Q6: Link-State Routing protocols primarily use which algorithm?

• A) Bellman-Ford Algorithm

• B) Dijkstra’s Algorithm

• C) Distance Vector Algorithm

• D) Greedy Algorithm

Answer: B) Dijkstra’s Algorithm
Explanation: Dijkstra’s Algorithm computes the shortest path from a source router to all destinations by maintaining a Link-State Database (LSDB).

Question 7: Routing Table Updates

Q7: How does a Link-State routing protocol update its routing table?

• A) By exchanging full routing tables periodically

• B) By sending incremental updates only when a topology change occurs

• C) By using static routes predefined by network administrators

• D) By relying solely on TCP connections

Answer: B) By sending incremental updates only when a topology change occurs
Explanation: Unlike Distance Vector protocols, Link-State protocols send updates only when network topology changes, reducing overhead and improving efficiency.

Question 8: Metric Used in OSPF

Q8: What metric does OSPF use to determine the best path?

• A) Hop count

• B) Bandwidth

• C) Cost

• D) Delay

Answer: C) Cost
Explanation: OSPF assigns cost to each link based on bandwidth, where higher bandwidth = lower cost, ensuring optimal path selection.

Question 9: Routing Protocol for Large Networks

Q9: Which routing protocol is most suitable for a large-scale enterprise network?

• A) RIP

• B) OSPF

• C) Distance Vector

• D) Static Routing

Answer: B) OSPF
Explanation: OSPF is scalable, supports hierarchical design using areas, and ensures faster convergence, making it ideal for large networks.

Question 10: Advantage of Link-State Over Distance Vector

Q10: Which of the following is a key advantage of Link-State routing over Distance Vector routing?

• A) Requires less memory

• B) Faster convergence

• C) Uses fewer control messages

• D) Uses hop count as a metric

Answer: B) Faster convergence
Explanation: Link-State protocols converge faster because routers maintain a full topology view, allowing immediate recalculations when a link fails.

10 Routing Protocol MCQ

10 multiple-choice questions on Routing protocol s with explanations, useful for UGC NET, SET, GATE, ISRO, and other competitive exams.

Question 1: Purpose of Routing Protocols

Q1: What is the primary purpose of a routing protocol?

• A) To assign IP addresses to devices

• B) To determine the best path for data packets to reach their destination

• C) To encrypt data during transmission

• D) To physically connect different networks

Answer: B) To determine the best path for data packets to reach their destination
Explanation: Routing protocols help routers determine the most efficient path for data transmission across a network, ensuring fast and reliable communication.

Question 2: Classification of Routing Protocols

Q2: Routing protocols can be broadly classified into which two main categories?

• A) TCP and UDP

• B) Distance Vector and Link-State

• C) IPv4 and IPv6

• D) Static and Dynamic

Answer: B) Distance Vector and Link-State
Explanation: Routing protocols are categorized as Distance Vector (e.g., RIP), which use hop count for route selection, and Link-State (e.g., OSPF), which consider network topology.

Question 3: Distance Vector Routing Protocol

Q3: Which of the following is an example of a Distance Vector routing protocol?

• A) OSPF

• B) RIP

• C) BGP

• D) IS-IS

Answer: B) RIP
Explanation: Routing Information Protocol (RIP) is a Distance Vector protocol that determines routes based on hop count, making it simple but less scalable.

Question 4: Link-State Routing Protocol

Q4: Which of the following is a Link-State routing protocol?

• A) RIP

• B) OSPF

• C) BGP

• D) EIGRP

Answer: B) OSPF
Explanation: Open Shortest Path First (OSPF) is a Link-State protocol that uses Dijkstra’s algorithm to compute the shortest path.

Question 5: Routing in Large Networks

Q5: Which routing protocol is best suited for large-scale networks like the Internet?

• A) RIP

• B) OSPF

• C) BGP

• D) EIGRP

Answer: C) BGP
Explanation: Border Gateway Protocol (BGP) is the de facto protocol used for routing between Autonomous Systems (AS) on the Internet, making it ideal for large networks.

Question 6: Hybrid Routing Protocol

Q6: Which of the following is a Hybrid routing protocol?

• A) RIP

• B) OSPF

• C) EIGRP

• D) IS-IS

Answer: C) EIGRP
Explanation: Enhanced Interior Gateway Routing Protocol (EIGRP) combines features of both Distance Vector and Link-State protocols for efficiency and scalability.

Question 7: Administrative Distance

Q7: What is the administrative distance of RIP?

• A) 120

• B) 110

• C) 90

• D) 170

Answer: A) 120
Explanation: Administrative Distance (AD) is a measure of a protocol’s trustworthiness. RIP has an AD of 120, making it less preferred than OSPF (110) and EIGRP (90).

Question 8: Routing Table Update Interval in RIP

Q8: How often does RIP update its routing table?

• A) Every 10 seconds

• B) Every 30 seconds

• C) Every 60 seconds

• D) Every 90 seconds

Answer: B) Every 30 seconds
Explanation: RIP sends updates to its neighboring routers every 30 seconds, which can lead to slower convergence compared to modern protocols.

Question 9: Protocol for Wireless Networks

Q9: Which routing protocol is commonly used in wireless ad-hoc networks?

• A) OSPF

• B) RIP

• C) AODV

• D) BGP

Answer: C) AODV
Explanation: Ad-hoc On-Demand Distance Vector (AODV) is a routing protocol specifically designed for wireless ad-hoc networks, enabling dynamic route discovery.

Question 10: Convergence Speed

Q10: Which routing protocol has the fastest convergence time?

• A) RIP

• B) OSPF

• C) BGP

• D) EIGRP

Answer: D) EIGRP
Explanation: EIGRP uses an advanced Diffusing Update Algorithm (DUAL), which allows rapid convergence by maintaining backup routes in its topology table.

10 Symbol Table MCQ

 Here are 10 multiple-choice questions on Symbol Table with explanations, useful for UGC NET, SET, GATE, ISRO, and other competitive exams.

Question 1: Purpose of Symbol Table

Q1: What is the main purpose of a Symbol Table in a compiler?

A) To store all the variables of a program
B) To store information about identifiers like variables, functions, and classes
C) To store the machine code of a program
D) To store the syntax tree of a program

Answer: B) To store information about identifiers like variables, functions, and classes

Explanation: The Symbol Table keeps track of identifiers (variables, functions, classes, objects) and their associated information like scope, type, memory location, and access rights.

Question 2: Data Structure for Symbol Table

Q2: Which of the following data structures is commonly used for implementing a Symbol Table?

A) Stack
B) Queue
C) Hash Table
D) Linked List

Answer: C) Hash Table

Explanation: A Hash Table allows fast lookups, insertions, and deletions, making it ideal for managing a Symbol Table. Other structures like linked lists can be used in case of collisions.

Question 3: Symbol Table Operations

Q3: Which of the following is NOT an operation performed on a Symbol Table?

A) Insert
B) Delete
C) Lookup
D) Sorting

Answer: D) Sorting

Explanation: A Symbol Table inserts, deletes, and looks up identifiers, but sorting is not required because the compiler only needs to retrieve and update information efficiently.

Question 4: Scope Handling

Q4: How does a compiler manage multiple symbol tables in block-structured languages?

A) Using a single global table
B) Using a stack of symbol tables
C) Using a queue
D) Using an array

Answer: B) Using a stack of symbol tables

Explanation: When entering a new block or function, a new symbol table is pushed onto the stack. When exiting, it is popped to restore the previous scope.

Question 5: Symbol Table in Different Compiler Phases

Q5: In which compiler phase is the Symbol Table created and maintained?

A) Lexical Analysis
B) Syntax Analysis
C) Semantic Analysis
D) Code Optimization

Answer: C) Semantic Analysis

Explanation: The Symbol Table is populated in the Lexical and Syntax Analysis phases but fully utilized during Semantic Analysis to perform type checking, scope resolution, and name resolution.

Question 6: Hashing in Symbol Table

Q6: Why is hashing commonly used for implementing Symbol Tables?

A) It provides constant-time average-case lookup
B) It reduces memory usage
C) It maintains the order of identifiers
D) It allows sequential access to identifiers

Answer: A) It provides constant-time average-case lookup

Explanation: Hashing ensures fast access (O(1) on average) to identifiers, making insertions and lookups efficient in a Symbol Table.

Question 7: Collision Resolution in Hashing

Q7: Which of the following is a collision resolution technique used in Symbol Tables?

A) Depth-First Search
B) Chaining
C) Sorting
D) Binary Search

Answer: B) Chaining

Explanation: Chaining handles hash collisions by maintaining a linked list of entries that hash to the same index. Linear probing and double hashing are other methods.

Question 8: Nested Scopes

Q8: In a programming language with nested scopes, how does the compiler resolve identifier references using the Symbol Table?

A) By searching only in the local Symbol Table
B) By searching the local table first, then the global table if not found
C) By using only the global Symbol Table
D) By using only the most recently defined identifier

Answer: B) By searching the local table first, then the global table if not found

Explanation: The compiler first checks the current scope’s symbol table. If the identifier is not found, it looks in enclosing scopes up to the global level.

Question 9: Semantic Analysis and Symbol Table

Q9: During Semantic Analysis, what role does the Symbol Table play?

A) Checking whether variables have been declared before use
B) Generating machine code
C) Optimizing the program
D) Parsing the input program

Answer: A) Checking whether variables have been declared before use

Explanation: The Semantic Analyzer uses the Symbol Table to verify type consistency, function calls, and variable declarations before generating code.

Question 10: Types of Symbol Tables

Q10: What is the difference between Global Symbol Table and Local Symbol Table?

A) The Global Symbol Table stores all identifiers, while the Local Symbol Table stores only function parameters
B) The Global Symbol Table is used only for built-in functions, while the Local Symbol Table is used for user-defined functions
C) The Global Symbol Table stores identifiers visible throughout the program, while the Local Symbol Table is specific to functions or blocks
D) There is no difference between them

Answer: C) The Global Symbol Table stores identifiers visible throughout the program, while the Local Symbol Table is specific to functions or blocks

Explanation: Global Symbol Table stores variables accessible anywhere, while Local Symbol Tables manage variables within a function or block to handle scoping rules properly.

10 compiler design mcq

 Here are 10 multiple-choice questions on Compiler Design for UGC NET, SET, GATE, ISRO, and other competitive exams.

Question 1: Symbol Table Management

Q1: In block-structured languages, how does the compiler manage multiple symbol tables?

A) Using a single global table
B) Using a stack of symbol tables
C) Using a queue
D) Using arrays

Answer: B) Using a stack of symbol tables

Explanation: Block-structured languages use nested scopes, so the compiler needs to manage different symbol tables dynamically. A stack is the best data structure for this because when entering a new block, a new symbol table is pushed onto the stack, and when exiting, it is popped.

Question 2: Lexical Analysis

Q2: Which of the following is the main function of a Lexical Analyzer in a compiler?

A) To generate intermediate code
B) To check syntax and semantics
C) To convert source code into tokens
D) To optimize the final machine code

Answer: C) To convert source code into tokens

Explanation: The lexical analyzer scans the source code and converts it into tokens, which are then used by the syntax analyzer. It removes whitespace, comments, and identifies keywords, identifiers, operators, and literals.

Question 3: Parsing Techniques

Q3: Which parsing technique is used in top-down parsing without backtracking?

A) LL(1) Parsing
B) LR(1) Parsing
C) Recursive Descent Parsing
D) Operator Precedence Parsing

Answer: A) LL(1) Parsing

Explanation: LL(1) parsers read input from Left to right and construct a Leftmost derivation. They work without backtracking by using a lookahead token to decide the next step.

Question 4: Intermediate Code Generation

Q4: Which of the following is commonly used as an Intermediate Representation (IR) in a compiler?

A) Parse Tree
B) Abstract Syntax Tree (AST)
C) Three-Address Code (TAC)
D) Lexical Tokens

Answer: C) Three-Address Code (TAC)

Explanation: Three-Address Code (TAC) is a widely used Intermediate Representation (IR) where each instruction has at most three operands. It simplifies the process of code optimization and translation into machine code.

Question 5: Syntax Analysis

Q5: What is the role of a Syntax Analyzer in a compiler?

A) To convert source code into machine code
B) To check the syntax of the source program using grammar rules
C) To optimize the generated code
D) To execute the program

Answer: B) To check the syntax of the source program using grammar rules

Explanation: The Syntax Analyzer (Parser) ensures that the program follows correct syntax according to a context-free grammar (CFG). It constructs a parse tree and reports syntax errors.

Question 6: Compiler Phases

Q6: In which phase of the compiler are syntax errors detected?

A) Lexical Analysis
B) Syntax Analysis
C) Semantic Analysis
D) Code Generation

Answer: B) Syntax Analysis

Explanation: The Syntax Analysis phase checks whether the input code follows the grammar rules of the programming language. Errors like missing parentheses, incorrect nesting, or misplaced operators are detected here.

Question 7: Code Optimization

Q7: Which of the following techniques is NOT used for code optimization?

A) Constant Folding
B) Loop Unrolling
C) Symbol Table Management
D) Dead Code Elimination

Answer: C) Symbol Table Management

Explanation: Code optimization improves performance by reducing execution time and space. Techniques like constant folding, loop unrolling, and dead code elimination are used to optimize code. Symbol Table Management is part of semantic analysis, not optimization.

Question 8: Error Handling

Q8: Which type of error can be detected only during runtime?

A) Syntax Error
B) Semantic Error
C) Logical Error
D) Lexical Error

Answer: C) Logical Error

Explanation: Logical errors (e.g., incorrect calculations, infinite loops) do not violate syntax or semantics but cause incorrect results. These errors cannot be detected by the compiler and only appear when the program is executed.

Question 9: Compiler vs Interpreter

Q9: Which of the following statements is true about compilers and interpreters?

A) A compiler executes code line by line, while an interpreter translates the whole program at once
B) A compiler translates the entire program before execution, while an interpreter executes line by line
C) Both compilers and interpreters translate the entire program before execution
D) Interpreters generate machine code, while compilers do not

Answer: B) A compiler translates the entire program before execution, while an interpreter executes line by line

Explanation: A compiler translates the whole source code into machine code before execution, while an interpreter translates and executes code line by line.

Question 10: Just-In-Time (JIT) Compilation

Q10: What is the main advantage of Just-In-Time (JIT) compilation?

A) Faster execution by compiling code at runtime
B) Reducing memory usage
C) Eliminating the need for an operating system
D) Improving syntax error detection

Answer: A) Faster execution by compiling code at runtime

Explanation: JIT compilation is used in Java (JVM), .NET (CLR), and Python (PyPy). It compiles bytecode into native machine code at runtime, improving execution speed compared to traditional interpretation.

10 Normal form identifying MCQ

10 multiple-choice questions related to identifying the normal form of a table based on given functional dependencies, useful for UGC NET, SET, GATE, ISRO, and other competitive exams. 

Question 1: 1NF Identification

Q1: A table with the following structure is given:

• Attributes: {A, B, C}

• Functional Dependencies: A → B, A → C

Does this table meet the requirements of 1NF?

• A) Yes, because all attributes are atomic

• B) No, because there are partial dependencies

• C) Yes, because there are no repeating groups

• D) No, because it violates 3NF

Answer: A) Yes, because all attributes are atomic

Explanation: A table is in 1NF if all its attributes contain atomic values and there are no repeating groups. The given table does not have any repeating groups or multi-valued attributes, so it satisfies the criteria of 1NF.

Question 2: 2NF Identification

Q2: A table with the following structure is given:

• Attributes: {A, B, C}

• Functional Dependencies: A → B, A → C

Is this table in 2NF?

• A) Yes, because there are no partial dependencies

• B) No, because it violates 1NF

• C) Yes, because it is in 1NF and has no partial dependencies

• D) No, because there are transitive dependencies

Answer: C) Yes, because it is in 1NF and has no partial dependencies

Explanation: A table is in 2NF if it is in 1NF and has no partial dependencies. Since A is the only key in this table, there cannot be any partial dependencies (which occur when a non-key attribute depends on part of a composite key). Thus, this table is in 2NF.

Question 3: 3NF Identification

Q3: A table with the following structure is given:

• Attributes: {A, B, C}

• Functional Dependencies: A → B, B → C

Is this table in 3NF?

• A) Yes, because there are no partial dependencies

• B) No, because there are transitive dependencies

• C) Yes, because it is in 2NF

• D) No, because it violates 1NF

Answer: B) No, because there are transitive dependencies

Explanation: A table is in 3NF if it is in 2NF and does not have transitive dependencies. Here, A → B and B → C imply that A → C indirectly through B. Since C is not fully dependent on A directly, this is a transitive dependency, violating 3NF.

Question 4: BCNF Identification

Q4: A table with the following structure is given:

• Attributes: {A, B, C, D}

• Functional Dependencies: AB → C, C → D

Is this table in BCNF?

• A) Yes, because every determinant is a candidate key

• B) No, because there are partial dependencies

• C) No, because C is not a candidate key

• D) Yes, because it is in 3NF

Answer: C) No, because C is not a candidate key

Explanation: A table is in BCNF if it is in 3NF and every determinant (an attribute that functionally determines other attributes) is a candidate key. Since C → D and C is not a candidate key, this table is not in BCNF.

Question 5: 4NF Identification

Q5: A table with the following structure is given:

• Attributes: {A, B, C}

• Functional Dependencies: A → B, A → C, B ↔ C

Is this table in 4NF?

• A) Yes, because there are no multi-valued dependencies

• B) No, because it is not in BCNF

• C) No, because B ↔ C is a multi-valued dependency

• D) Yes, because it is in BCNF

Answer: C) No, because B ↔ C is a multi-valued dependency

Explanation: A table is in 4NF if it is in BCNF and has no multi-valued dependencies (MVDs). Since B ↔ C indicates a multi-valued dependency, the table is not in 4NF.

Question 6: Partial Dependency Identification

Q6: A table with the following structure is given:

• Attributes: {A, B, C, D}

• Functional Dependencies: AB → C, A → D

Is there a partial dependency in this table?

• A) Yes, because A → D

• B) No, because AB → C

• C) Yes, because AB is a composite key

• D) No, because A is a candidate key

Answer: A) Yes, because A → D

Explanation: A partial dependency occurs when a non-key attribute depends on only a part of a composite primary key. Here, A → D means D depends only on A (not AB), which is a partial dependency.

Question 7: Transitive Dependency Identification

Q7: A table with the following structure is given:

• Attributes: {A, B, C}

• Functional Dependencies: A → B, B → C

Is there a transitive dependency in this table?

• A) Yes, because A → B and B → C

• B) No, because A → B

• C) Yes, because A is the primary key

• D) No, because there are no partial dependencies

Answer: A) Yes, because A → B and B → C

Explanation: A transitive dependency occurs when A → B and B → C implies A → C indirectly. This is a violation of 3NF.

Question 8: Candidate Key Identification

Q8: A table with the following structure is given:

• Attributes: {A, B, C, D}

• Functional Dependencies: AB → C, AB → D

Which of the following is a candidate key?

• A) A

• B) B

• C) AB

• D) CD

Answer: C) AB

Explanation: A candidate key is a minimal key that determines all attributes. Here, AB determines C and D, so AB is the candidate key.

Question 9: 2NF Violation Identification

Q9: A table with the following structure is given:

• Attributes: {A, B, C}

• Functional Dependencies: AB → C, A → B

Why is this table not in 2NF?

• A) It is in 2NF

• B) Because of partial dependency A → B

• C) Because of transitive dependency AB → C

• D) Because it is in 3NF

Answer: B) Because of partial dependency A → B

Explanation: Since A → B exists and AB is the primary key, this is a partial dependency (B depends on part of a composite key), violating 2NF.

Question 10: Normal Form Classification

Q10: A table with the following structure is given:

• Attributes: {A, B, C, D}

• Functional Dependencies: A → B, C → D, A → C, B → D

In which normal form is this table?

• A) 1NF

• B) 2NF

• C) 3NF

• D) BCNF

Answer: D) BCNF

Explanation: Since all determinants (A, C, B) are candidate keys, this table satisfies BCNF.

10 Normalization MCQ DBMS

Multiple-Choice Questions on Normalization and Normal Forms in DBMS for UGC NET, GATE, SET, ISRO and other exams 

Question 1: Normalization Definition

Q1: Normalization in a database is a process of:

• A) Reducing data redundancy and improving data integrity

• B) Increasing data redundancy for faster access

• C) Encrypting the database

• D) Creating more tables

Answer: A) Reducing data redundancy and improving data integrity

Explanation: Normalization is a systematic approach to organizing data in a database to minimize redundancy and ensure data integrity. It involves decomposing larger tables into smaller, well-structured tables while maintaining relationships among data.

Question 2: First Normal Form (1NF)

Q2: A table is in the First Normal Form (1NF) if:

• A) All values in the table are atomic and there are no repeating groups

• B) It contains no partial dependencies

• C) It contains no transitive dependencies

• D) It is free of all anomalies

Answer: A) All values in the table are atomic and there are no repeating groups

Explanation: The First Normal Form (1NF) requires that all columns in a table contain only atomic values (indivisible values) and that there are no repeating groups or arrays in a single column.

Question 3: Second Normal Form (2NF)

Q3: A table is in the Second Normal Form (2NF) if:

• A) It is in 1NF and there are no partial dependencies

• B) It is in 1NF and contains no transitive dependencies

• C) It is in 1NF and all non-key attributes are fully functionally dependent on the primary key

• D) It is in 3NF

Answer: A) It is in 1NF and there are no partial dependencies

Explanation: The Second Normal Form (2NF) builds upon 1NF by ensuring that all non-key attributes are fully dependent on the entire primary key, meaning there should be no partial dependencies (where a non-key attribute depends only on part of a composite key).

Question 4: Third Normal Form (3NF)

Q4: A table is in the Third Normal Form (3NF) if:

• A) It is in 2NF and all non-key attributes are fully functionally dependent on the primary key

• B) It is in 2NF and contains no transitive dependencies

• C) It is in 1NF and there are no repeating groups

• D) It is free of all anomalies

Answer: B) It is in 2NF and contains no transitive dependencies

Explanation: The Third Normal Form (3NF) ensures that all attributes depend only on the primary key and not on other non-key attributes. This removes transitive dependencies, where one non-key attribute is dependent on another non-key attribute.

Question 5: Boyce-Codd Normal Form (BCNF)

Q5: A table is in Boyce-Codd Normal Form (BCNF) if:

• A) It is in 3NF and every determinant is a candidate key

• B) It is in 2NF and contains no transitive dependencies

• C) It is in 1NF and contains no repeating groups

• D) It is free of all anomalies

Answer: A) It is in 3NF and every determinant is a candidate key

Explanation: BCNF is a stricter version of 3NF, ensuring that every determinant (a column that functionally determines another column) is a candidate key, eliminating potential anomalies that may still exist in 3NF.

Question 6: Functional Dependency

Q6: In a relational database, a functional dependency occurs when:

• A) One attribute uniquely determines another attribute

• B) An attribute depends on multiple attributes

• C) An attribute is independent of other attributes

• D) Multiple attributes determine a single attribute

Answer: A) One attribute uniquely determines another attribute

Explanation: Functional dependency means that if you know the value of one attribute, you can uniquely determine the value of another attribute. For example, in a student database, the student ID uniquely determines the student’s name.

Question 7: Partial Dependency

Q7: A partial dependency exists in a table when:

• A) A non-key attribute depends on part of a composite primary key

• B) A non-key attribute depends on the entire primary key

• C) A non-key attribute depends on another non-key attribute

• D) A primary key depends on a non-key attribute

Answer: A) A non-key attribute depends on part of a composite primary key

Explanation: Partial dependency occurs when a non-key attribute depends on only part of a composite primary key. This issue exists in 1NF and is eliminated in 2NF by ensuring that all attributes depend on the whole primary key.

Question 8: Transitive Dependency

Q8: A transitive dependency exists in a table when:

• A) A non-key attribute depends on another non-key attribute

• B) A non-key attribute depends on the primary key

• C) The primary key depends on a non-key attribute

• D) A primary key depends on part of a composite key

Answer: A) A non-key attribute depends on another non-key attribute

Explanation: A transitive dependency occurs when a non-key attribute depends on another non-key attribute rather than directly on the primary key. This is removed in 3NF.

Question 9: Fourth Normal Form (4NF)

Q9: A table is in the Fourth Normal Form (4NF) if:

• A) It is in BCNF and has no multi-valued dependencies

• B) It is in 3NF and contains no transitive dependencies

• C) It is in 2NF and all non-key attributes are fully functionally dependent on the primary key

• D) It is in 1NF and contains no repeating groups

Answer: A) It is in BCNF and has no multi-valued dependencies

Explanation: The Fourth Normal Form (4NF) removes multi-valued dependencies, which occur when one key determines multiple independent values for another column. This ensures that the data is stored in separate tables when necessary.

Question 10: Fifth Normal Form (5NF)

Q10: A table is in the Fifth Normal Form (5NF) if:

• A) It is in 4NF and is decomposed into smaller tables to eliminate redundancy

• B) It is in 3NF and contains no transitive dependencies

• C) It is in 2NF and all non-key attributes are fully functionally dependent on the primary key

• D) It is in BCNF and has no multi-valued dependencies

Answer: A) It is in 4NF and is decomposed into smaller tables to eliminate redundancy

Explanation: The Fifth Normal Form (5NF) deals with eliminating redundancy by decomposing tables into smaller ones while preserving the ability to reconstruct the original data without loss. It ensures that there are no join dependencies that could lead to anomalies.

10 Boot Process MCQ

10 multiple-choice questions on the boot process in operating systems for ISRO, EMRS, GATE, KV, NET, and other exams.

Question 1: Boot Process Definition

Q1: The boot process is the sequence of events that occurs:

• A) When a user logs out

• B) When the system is powered on and the operating system is loaded

• C) When an application is opened

• D) When the system goes to sleep

Answer: B) When the system is powered on and the operating system is loaded

Explanation: The boot process refers to the series of steps that occur when a computer is turned on, including hardware initialization, loading the operating system, and preparing the system for use.

Question 2: First Step in Boot Process

Q2: The first step in the boot process is typically:

• A) Loading the operating system kernel

• B) Initializing system hardware

• C) Running the BIOS or UEFI firmware

• D) Loading device drivers

Answer: C) Running the BIOS or UEFI firmware

Explanation: When a computer starts, the BIOS (or UEFI) firmware is executed first. It initializes hardware components and prepares the system for loading the bootloader.

Question 3: BIOS Function

Q3: The BIOS is responsible for:

• A) Managing user applications

• B) Initializing and testing hardware components

• C) Providing network services

• D) Managing memory allocation

Answer: B) Initializing and testing hardware components

Explanation: BIOS (Basic Input/Output System) performs the Power-On Self-Test (POST) to check hardware components and then locates the bootloader.

Question 4: Bootloader

Q4: The bootloader's primary role is to:

• A) Perform power-on self-test (POST)

• B) Load the operating system kernel into memory

• C) Manage hardware resources

• D) Provide a user interface

Answer: B) Load the operating system kernel into memory

Explanation: The bootloader (e.g., GRUB, Windows Boot Manager) is responsible for loading the OS kernel into memory and starting system initialization.

Question 5: POST (Power-On Self-Test)

Q5: During the POST process, the system:

• A) Loads the user applications

• B) Tests the hardware components for proper operation

• C) Loads the operating system kernel

• D) Initializes network connections

Answer: B) Tests the hardware components for proper operation

Explanation: POST checks the computer’s hardware (e.g., CPU, RAM, storage) to ensure they are functioning correctly before proceeding with the boot process.

Question 6: Master Boot Record (MBR)

Q6: The Master Boot Record (MBR) is located:

• A) In the CPU

• B) At the beginning of the bootable disk

• C) In the operating system kernel

• D) In the RAM

Answer: B) At the beginning of the bootable disk

Explanation: The MBR is the first sector (512 bytes) of a bootable storage device. It contains the partition table and the bootloader.

Question 7: Kernel Loading

Q7: The operating system kernel is loaded into memory by:

• A) BIOS/UEFI

• B) Bootloader

• C) POST

• D) Application software

Answer: B) Bootloader

Explanation: Once the bootloader is executed, it loads the OS kernel into memory, allowing the operating system to take control of the hardware.

Question 8: GRUB

Q8: GRUB (GRand Unified Bootloader) is a:

• A) Type of BIOS

• B) Bootloader commonly used in Linux systems

• C) Hardware initialization tool

• D) Network management utility

Answer: B) Bootloader commonly used in Linux systems

Explanation: GRUB is a multi-bootloader used in Linux that allows users to choose between different operating systems and kernel versions.

Question 9: Initial RAM Disk (initrd)

Q9: The initial RAM disk (initrd) is used to:

• A) Store the operating system permanently

• B) Provide necessary drivers and modules during the boot process

• C) Test the hardware components

• D) Manage user accounts

Answer: B) Provide necessary drivers and modules during the boot process

Explanation: The initrd or initramfs contains essential drivers and files required for the kernel to initialize and access the root filesystem.

Question 10: UEFI

Q10: UEFI (Unified Extensible Firmware Interface) is:

• A) An older firmware interface used before BIOS

• B) A replacement for BIOS with more advanced features

• C) A type of operating system kernel

• D) A network protocol

Answer: B) A replacement for BIOS with more advanced features

Explanation: UEFI is a modern firmware interface that replaces BIOS, providing faster boot times, support for larger storage devices, and security features like Secure Boot.

10 Paging (OS) MCQ

10 multiple-choice numerical questions on paging in operating systems for UGC NET, SET, ISRO and Other exam.

Question 1: Page Size Calculation

Q1: If a system uses a 32-bit address space and a page size of 4 KB, how many pages are there in the address space?

• A) 1024

• B) 4096

• C) 1048576

• D) 262144

Answer: C) 1048576

Explanation:
The number of pages is calculated as:

$$\frac{\text{Total address space}}{\text{Page size}} = \frac{2^{32} \text{ bytes}}{2^{12} \text{ bytes}} = 2^{20} = 1048576 \text{ pages}$$

Question 2: Number of Frames

Q2: If the physical memory is 1 GB and the page size is 4 KB, how many frames are available in the physical memory?

• A) 256

• B) 1024

• C) 256000

• D) 262144

Answer: D) 262144

Explanation:
The number of frames is given by:

$$\frac{\text{Physical memory}}{\text{Page size}} = \frac{1 \text{ GB}}{4 \text{ KB}} = \frac{2^{30} \text{ bytes}}{2^{12} \text{ bytes}} = 2^{18} = 262144 \text{ frames}$$

Question 3: Page Number and Offset

Q3: In a system with a 16-bit address space and a page size of 256 bytes, what are the page number and offset for the address 0x1234?

• A) Page number: 0x12, Offset: 0x34

• B) Page number: 0x123, Offset: 0x4

• C) Page number: 0x1, Offset: 0x234

• D) Page number: 0x1234, Offset: 0x0

Answer: A) Page number: 0x12, Offset: 0x34

Explanation:
Page number is obtained by dividing the address by the page size (256 bytes = 2⁸).

$$0x1234 / 256 = 0x12$$

Offset is the remainder:

$$0x1234 \mod 256 = 0x34$$

Question 4: Address Translation

Q4: If the logical address is 0x1A3 and the page size is 64 bytes, what is the physical address assuming the page table entry for the page is 0x02?

• A) 0x02A3

• B) 0x00A3

• C) 0x0223

• D) 0x0243

Answer: C) 0x0223

Explanation:
Page number: $0x1A3 / 64 = 0x6$
Offset: $0x1A3 \mod 64 = 0x23$
Physical address: The page table entry provides frame number 0x02, so the physical address is:

$$\text{Physical address} = (\text{Frame number} \times \text{Page size}) + \text{Offset} = (0x02 \times 64) + 0x23 = 0x0223$$

Question 5: Effective Access Time (EAT)

Q5: If the memory access time is 100 ns and the page fault service time is 20 ms, what is the effective access time if the page fault rate is 0.001?

• A) 20.1 ms

• B) 200 ns

• C) 20.0001 ms

• D) 200.1 ns

Answer: B) 200 ns

Explanation:

$$EAT = (1 - p) \times \text{Memory access time} + p \times \text{Page fault time}$$ $$= (1 - 0.001) \times 100 + 0.001 \times 20,000,000$$ $$= 99.9 + 20000 = 200 ns$$

Question 6: Page Table Size

Q6: For a system with a 64-bit address space and a page size of 4 KB, what is the size of the page table assuming each page table entry is 8 bytes?

• A) 64 GB

• B) 32 MB

• C) 64 MB

• D) 128 MB

Answer: D) 128 MB

Explanation:
Number of pages:

$$\frac{2^{64}}{2^{12}} = 2^{52}$$

Page table size:

$$2^{52} \times 8 \text{ bytes} = 2^{55} \text{ bytes} = 128 \text{ MB}$$

Question 7: Number of Pages

Q7: In a system with a 48-bit virtual address space and a page size of 8 KB, how many pages are there?

• A) $2^{35}$

• B) $2^{36}$

• C) $2^{34}$

• D) $2^{33}$

Answer: C) $2^{34}$

Explanation:

$$\text{Number of pages} = \frac{2^{48}}{2^{13}} = 2^{34}$$

Question 8: TLB Hit Ratio

Q8: If the TLB hit time is 20 ns, the memory access time is 100 ns, and the TLB hit ratio is 90%, what is the effective memory access time?

• A) 102 ns

• B) 108 ns

• C) 110 ns

• D) 120 ns

Answer: B) 108 ns

Explanation:

$$EAT = (\text{Hit ratio} \times (\text{TLB time} + \text{Memory time})) + (\text{Miss ratio} \times (\text{TLB time} + 2 \times \text{Memory time}))$$ $$= (0.9 \times 120) + (0.1 \times 220) = 108$$

Question 9: Page Table Levels

Q9: In a system with a 32-bit address space and 4 KB pages, what is the number of page table levels if each page table entry is 4 bytes and each level can map a maximum of 1024 entries?

• A) 2

• B) 3

• C) 4

• D) 5

Answer: B) 3

Question 10: Effective Access Time with Page Faults

Q10: If the memory access time is 200 ns, the page fault service time is 10 ms, and the page fault rate is 0.0005, what is the effective access time?

• A) 200 ns

• B) 205 ns

• C) 202 ns

• D) 502 ns

Answer: D) 502 ns