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UGC NET Paper 2 Computer Science Mock Test - 9 - UGC NET MCQ


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30 Questions MCQ Test - UGC NET Paper 2 Computer Science Mock Test - 9

UGC NET Paper 2 Computer Science Mock Test - 9 for UGC NET 2024 is part of UGC NET preparation. The UGC NET Paper 2 Computer Science Mock Test - 9 questions and answers have been prepared according to the UGC NET exam syllabus.The UGC NET Paper 2 Computer Science Mock Test - 9 MCQs are made for UGC NET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for UGC NET Paper 2 Computer Science Mock Test - 9 below.
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UGC NET Paper 2 Computer Science Mock Test - 9 - Question 1

Which of the following arguments are not valid?

(a) 'If Gora gets the job and works hard, then he will be promoted. If Gora gets promotion, then he will be happy. He will not be happy, therefore, either he will not get the job or he will not work hard'.

(b) 'Either Puneet is not guilty or Pankaj is telling the truth. Pankaj is not telling the truth, therefore, Puneet is not guilty'.

(c) If n is a real number such that n > 1, then n2 > 1. Suppose that n2 > 1, then n > 1.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 1

Option A let

  • p: Gora get the job
  • q: he works hard
  • r: he will be promoted  
  • s: he will be happy

So we have  which will give us  (by contrapositive law) and . Now ¬S is given so it implies ¬r and so we have  i.e either he does not get the job or he does not work hard. So it is valid

Option B valid as EXACTLY one of two statements must be true. Given one statement is not true so other must be true 

Option C not valid. Here we have a ⇒ b. But this does not always mean b ⇒a. 

Hence A and B are valid  C is no  valid so Option A is correct . 

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 2

If an image has a height of 4 inches and an aspect ratio of 3 : 2, then what will be the width of the image?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 2

Key Points
3 : 2 aspect ratio means that for every 3 units of width, there are 2 units of height.

The same applies to the 6 : 4 aspect ratio - for every 6 units of width, there are 4 units of height.

The formula for finding aspect ratio is:

AR = W / H

Where:

AR is the aspect ratio

W is the width of the displayed image

H is the height of the displayed image

Note that W and H are measured in inches or millimeters.

Hence the correct answer is 6 inches.

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UGC NET Paper 2 Computer Science Mock Test - 9 - Question 3

The code for C:

# include <stdio.h>

int main ()

{

int i = 0;

while (i < 3)

i++;

printf (" In while loop\n");

}

How many times 'i' value is checked?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 3

Concept:

The correct answer is: 4
Explanation:

In the provided code, 'i' is initially set to 0. The condition of the while loop, i < 3, is checked each time before the loop is executed.

  • On the first iteration, i is 0, so the condition is true (0 < 3), and 'i' is incremented to 1.
  • On the second iteration, i is 1, so the condition is again true (1 < 3), and 'i' is incremented to 2.
  • On the third iteration, i is 2, so the condition is still true (2 < 3), and 'i' is incremented to 3.
  • On the fourth check, i is 3, so the condition is now false (3 < 3), and the loop is exited.
  • So, the condition i < 3 is checked 4 times in total before the loop is exited.
UGC NET Paper 2 Computer Science Mock Test - 9 - Question 4
Let E1 and E2 be two entities in an E/R diagram with simple single-valued attributes. R1 and R2 are two relationships between E1 and E2, where R1 is one-to-many and R2 is many-to-many. R1 and R2 do not have any attributes of their own. What is the minimum number of tables required to represent this situation in the relational model?
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 4

The correct answer is option 1.

Concept:

ER Diagram stands for Entity Relationship Diagram, also known as ERD is a diagram that displays the relationship of entity sets stored in a database. In other words, ER diagrams help to explain the logical structure of databases. ER diagrams are created based on three basic concepts: entities, attributes and relationships.

Note:

Minimum number of tables between two entities. Suppose R is relation between two entities E1 and E2 with some cardinality ratio like,

cardinality ratio one to one:

2 tables,

  • E1R, E2 or
  • E1, E2R

cardinality ratio one to many:

2 tables,

  • E1, E2R

cardinality ratio many to one:

2 tables,

  • E1R, E2

cardinality ratio many to many:

3 tables,

  • E1, E2, R

Given data,

Only three table are requried,

  • E2 R1
  • E1
  • R2

Hence the correct answer is 3.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 5

Which of the following basic steps are needed in order to perform a memory read operation?

1. The address of the location from which the word is to be read is loaded into the memory address register (MAR).

2. A signal, called read (issued by the CPU indicating that the word whose address is in the MAR) is to be read into the MDR.

3. Corresponding to the memory delay in reading the specified word, the required word will be loaded by the memory into the MDR ready for use by the CPU.

Select the correct answer using the code given below.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 5

Memory Address Register:

  • Memory Address Register is the CPU register that either stores the memory address from which data will be fetched to the CPU or the address to which data will be sent and stored. (statement 1 is correct)
  • In other words, MAR holds the memory location of data that needs to be accessed.
  • MAR holds the address of the current instruction that is to be fetched from memory or the address in the memory to which data is to be transferred.

Memory Data Register:

  • Memory Data Register is the register in a computer's CPU that stores the data being transferred to and from the immediate access storage.
  • In other words, MDR is the register used for holding information that is in the process of being transferred from the memory to the central processor or vice versa.
  • MDR holds the contents found at the address held in the MAR or data which is to be transferred to primary memory.
UGC NET Paper 2 Computer Science Mock Test - 9 - Question 6

Find the correct match for terms in Column I to those in Column II:

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 6

The correct answer is a - S, b - P, c - R, d - Q

Key Points

  • a. Topological sorting - S (Serializable order): Topological sorting is a method used in concurrency control to achieve a serializable order of transactions. Therefore, the match for "topological sorting" is "Serializable order."
  • b. Privileges - refer to permissions or rights granted to users or roles in databases or operating systems, such as the ability to read, write, or execute files or data. The correct match in Column II is "P. grant," since granting is how privileges are typically assigned or modified.
  • c. Multivalued dependency - is a concept from database theory, particularly in the normalization of databases. A table is said to have a multivalued dependency if, for two attributes, A and B, for each value of A, there are multiple values of B that are independent of other attributes in the table. This concept leads directly to "R. 4NF" (Fourth Normal Form), which specifically deals with removing non-trivial multivalued dependencies to prevent certain types of anomalies.
  • d. Improved throughput - refers to the increased rate of processing work or tasks, often through optimiztion of processes or technology. In the context provided, it is best matched with "Q. Concurrency," as concurrency mechanisms enable systems to handle more tasks simultaneously, thereby potentially improving throughput.

The closest match based on key terms is: a - S, b - P, c - R, d - Q

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 7

Consider the following table of arrival time and burst time for three processes P0, Pl and P2.


The pre-emptive shortest job first scheduling algorithm is used. Scheduling is carried out only at arrival or completion of processes. What is the average waiting time for the three processes?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 7

The correct answer is option 1.

Concept:

Shortest Remaining Time First is a preemptive shortest job first scheduling algorithm, in which the process with the smallest amount of burst time is executed in the CPU.

Formula:

TAT = CT - AT

WT = TAT - BT

Calculation:

GANTT CHART:


Average waiting time =

Hence the correct answer is5.

NOTE:

TAT = Turn Around Time

CT = Completion Time

WT = Waiting Time

BT = Burst Time

AT = Arrival Time

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 8

Which of the following languages over the alphabet ∑ = {0, 1} is regular?

I. 0n1m

II. 0n1m where m = n + 1

III. 0m1n where m ≡ n (mod 3).

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 8

The correct answer is option 4.

Concept:

A regular language is a language that can be expressed with a regular expression or deterministic or non-deterministic finite automata or state machines. A language is a set of strings that are made up of characters from a specified alphabet, or set of symbols.

Option 1: 0n1m

True, It is a regular language. It accepts language like 0*1*(any number of 0's followed by any number of 1's). The strings like {ε, 0, 1, 00, 11, 01, 000, 001, 011, 111,...}. There is no use of the stack.

Option 2: 0n1m where m = n + 1

False, It is not a regular language because number 0's are depends on the number of 1's. So we can't compare the number of 0's to the number of 1's.

It accepts the strings like {1, 011, 00111, ...}. There is no use of the stack.

Option 3: 0m1n where m ≡ n (mod 3)

True, It is a regular language. The strings where m is the number of digits 0 in the string, n is the number of digits 1 in the string, and m = n modulo 3. Here no one symbol is depending on the other. There is no use of the stack.

Hence the correct answer is I and III.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 9
LL(1) parser is also known as:
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 9

The correct answer is option 3.

Concept:

Non-recursive descent parser or table-driven is also known as LL(1) parser. This parser follows the leftmost derivation (LMD). The main problem during predictive parsing is that of determining the production to be applied for a non-terminal.

LL(1) Parsing:

Here the 1st L represents that the scanning of the Input will be done from Left to Right manner and the second L shows that in this parsing technique we are going to use Left most Derivation Tree. And finally, the 1 represents the number of look-ahead, which means how many symbols are you going to see when you want to make a decision.

Hence the correct answer is a Non-recursive descent parser.

Additional Information

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 10

There are n jobs are waiting in the ready queue and the CPU uses a non-preemptive shortest job first scheduling algorithm. what is the time complexity of the shortest job first algorithm when jobs are non-preemptive in nature?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 10

Concept:

Shortest Job First (SJF):

Shortest Job First (SJF) is an algorithm in which the process having the smallest execution time is chosen for the next execution. This scheduling method can be preemptive or non-preemptive. It significantly reduces the average waiting time for other processes awaiting execution. The full form of SJF is Shortest Job First.

Non-Preemptive SJF:

In non-preemptive scheduling, once the CPU cycle is allocated to process, the process holds it till it reaches a waiting state or is terminated.

Algorithm:

  • Start with sorting all the processes based on their time of arrival at the processor.
  • Then select the earliest arriving process with the shortest burst time.
  • While executing the current process, all the waiting processes are pushed in the waiting queue, and when the current process ends, choose the process with the shortest burst time among the waiting processes.

Analysis:

Shortest Job First uses Insert and Extract-Min operations. Both the Insert and the Extract-Min operations have a running time of O(log n). There are n tasks and each task has to be inserted into the heap and later extracted from the heap, resulting in a running time of O(n 2 log n).

Insert has an O(log n) running time because we first add the new job to the heap's final position when inserting.

  • This does not guarantee that the heap property is preserved, because the priority of the new task may be higher than the priority of its parent. This is why the Insert operation includes the Heapify-Up method, which restores heap order. The Heapify-Up operation takes O minutes to complete (log n).

The Extract-Min operation takes O(log n) time because it deletes the heap's root first and then moves the final task to the first slot.

  • The Extract-Min process includes the execution of the Heapify-Down procedure, which may violate the heap property. The Heapify-Down operation takes O seconds to complete (log n).

The time complexity of this algorithm is O(N* 2* log(N)).

Hence the correct answer is O(n log n).

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 11

Which of the following statement is False about inter-process communication?

I. The shared memory method can allow access to a common data area even amongst the processes that are not related.

II. In message based IPC at the receiver end, messages are put in a stack.

III. In direct message passing, the communication link get established automatically.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 11

I. True: For shared memory communication, one process would write into a certain commonly accessed area and another process would read subsequently from that area.

II. False: In message based IPC at the receiver end, messages are put in a queue.

III. True: In direct message passing, the communication link get established automatically, which can be either unidirectional or bidirectional, but one link can be used between one pair of the sender and receiver and one pair of sender and receiver should not possess more than one pair of link.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 12
A validation testing is achieved through a series of _________
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 12

Software validation is achieved through a series of black-box tests that demonstrate conformity with requirements.

A test plan outlines the classes of tests to be conducted and a test procedure defines specific test cases that will be used to demonstrate conformity with requirements
UGC NET Paper 2 Computer Science Mock Test - 9 - Question 13

Which of the following is NOT a characteristic of an SRS document?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 13

Concept:

Characteristics of SRS Document

  • Correctness: User review is used to provide the accuracy of requirements stated in the SRS. SRS is said to be perfect if it covers all the needs that are truly expected from the system.
  • Completeness: The SRS is complete if, and only if, all essential requirements, whether relating to functionality, performance, design, constraints, attributes, or external interfaces.
  • Consistency: The SRS is consistent if, and only if, no subset of individual requirements described in its conflict
  • Unambiguousness: SRS is unambiguous when every fixed requirement has only one interpretation. This suggests that each element is uniquely interpreted. In case there is a method used with multiple definitions, the requirements report should determine the implications in the SRS so that it is clear and simple to understand.
  • Ranking for importance and stability: The SRS is ranked for importance and stability if each requirement in it has an identifier to indicate either the significance or stability of that particular requirement.
  • Modifiability: SRS should be made as modifiable as likely and should be capable of quickly obtain changes to the system to some extent. Modifications should be perfectly indexed and cross-referenced.
  • Verifiability: SRS is correct when the specified requirements can be verified with a cost-effective system to check whether the final software meets those requirements. The requirements are verified with the help of reviews.
  • Traceability: The SRS is traceable if the origin of each of the requirements is clear and if it facilitates the referencing of each condition in future development or enhancement documentation.

Hence Option 4 is correct

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 14

_____ register Connected to the Processor bus is a single-way transfer capable.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 14
Z register Connected to the Processor bus is a single-way transfer capable.

The Z register is a special register that can interact with the processor BUS only. Z register is an 8-bit temporary register of 8085 microprocessor, which is not accessible to the user. This register is used either to store 8-bit information in Z registers. Z register pairs with lower-order 8-bits in Z.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 15

The reason for software bugs and failures is due to ____________.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 15
The reason for software bugs and failures is due to Software Developers and Software companies.

Software companies are responsible for making policies and providing working atmosphere for the software development, so in turn these companies become a part of software development process. Bugs from developers side is no new thing, even the slightest mistake in the program causes a bug.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 16

Which of the following is not a characteristic of High-level languages?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 16
Machine code is not a characteristic of High-level languages.

A high-level language (HLL) is a programming language such as C, Fortan, or Pascal that enables a programmer to write programs that are more or less independent of a particular type of computer. Such languages are considered high-level because they are closer to human languages and further from machine languages.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 17

The register used to store the flags is called as _________.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 17
The register used to store the flags is called as Status register. The status register stores the condition codes of the system. The status register is a hardware register that contains information about the state of the processor. Individual bits are implicitly or explicitly read and/or written by the machine code instructions executing on the processor. Some CPU architectures, such as the MIPS and Alpha, do not use a dedicated flag register.
UGC NET Paper 2 Computer Science Mock Test - 9 - Question 18

The assembler stores all the names and their corresponding values in ______.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 18
The table where the assembler stores the variable names along with their corresponding memory locations and values is Symbol Table. The symbol table contains information to locate and relocate symbolic definitions and references. The assembler creates the symbol table section for the object file. It makes an entry in the symbol table for each symbol that is defined or referenced in the input file and is needed during linking.
UGC NET Paper 2 Computer Science Mock Test - 9 - Question 19

The computer architecture aimed at reducing the time of execution of instructions is ________.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 19
The computer architecture aimed at reducing the time of execution of instructions is RISC. The RISC stands for Reduced Instruction Set Computer. A Reduced Instruction Set Computer is a type of microprocessor architecture that utilizes a small, highly-optimized set of instructions rather than the highly-specialized set of instructions typically found in other architectures. RISC processors can be designed more quickly than CISC processors due to its simple architecture.
UGC NET Paper 2 Computer Science Mock Test - 9 - Question 20

The primary aim of structural optimization is to determine the __________.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 20
The primary aim of structural optimization is to determine the most suitable combination of design variables, so as to achieve the satisfactory performance of the structures subjected to the behavior and geometric constraints imposed with the goal of optimality being defined by the conditions and the three basic features of the structural optimization problem are:
  • The design variables
  • The objective function
  • The constraints
UGC NET Paper 2 Computer Science Mock Test - 9 - Question 21

How the new states are generated in genetic algorithm?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 21
New states are generated by mutation and by crossover, which combines a pair of states from the population. Crossover and mutation perform two different roles. Crossover (like selection) is a convergence operation which is intended to pull the population towards a local minimum/maximum. In an interesting aside, the crossover is not one of the original genetic operators; Holland's original thesis used only selection and probabilistic mutation.

Mutation is a divergence operation. It is intended to occasionally break one or more members of a population out of a local minimum/maximum space and potentially discover a better minimum/maximum space.

Since the end goal is to bring the population to convergence, selection/crossover happens more frequently (typically every generation). Mutation, being a divergence operation, should happen less frequently, and typically only affects a few members of a population (if any) in any given generation.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 22

Physical memory is divided into sets of finite size called as __________.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 22

Physical memory is divided into sets of finite size called as Frames. A frame refers to a storage frame or central storage frame. In terms of physical memory, it is a fixed-sized block in physical memory space or a block of central storage. In computer architecture, frames are analogous to logical address space pages.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 23

______________ can be defined as the duplication of another creator’s or developer’s product trailing a thorough examination of its production or development.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 23

Reverse engineering can be made functional to diverse aspects of software development and hardware improvement activities. This practice absorbs how the system or the application works and what concepts have to implement in order to crack or duplicate it.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 24

Which of the following is not of the same group?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 24

Apple talk is not of the same group. Apple talk is a discontinued proprietary suite of networking protocols developed by Apple Inc. for their Macintosh computers. AppleTalk includes a number of features that allow local area networks to be connected with no prior setup or the need for a centralized router or server of any sort.

Bus topology uses one main cable to which all nodes are directly connected. The main cable acts as a backbone for the network.

Ring topology, the computers in the network are connected in a circular fashion, and the data travels in one direction.

The Internet is a vast network that connects computers all over the world.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 25

What will happen when a process terminates?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 25

When a process terminates, it removes from all queues. All allocated resources to that particular process are deallocated and all those resources are returned back to OS.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 26

_________ hides the limitations of Java behind a powerful and concise Clojure API for Cascading.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 26

Cascalog hides the limitations of Java behind a powerful and concise Clojure API for Cascading. Cascalog also adds Logic Programming concepts inspired by Datalog. Hence the name “Cascalog” is a contraction of Cascading and Datalog. Cascalog is a Clojure-based query language for Hadoop inspired by Datalog.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 27

Does the Load instruction do the following operation/s?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 27

The load instruction is basically used to load the contents of a memory location onto a register. Load instructions move data from memory into a register. The address for the load is the sum of a register specified in the instruction and a constant value that is coded into the instruction. LOAD is used to load data from memory to the accumulator. STORE is used to store the accumulator result in memory.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 28
The document listing all procedures and regulations that generally govern an organization is the:
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 28

The document listing all procedures and regulations that generally govern an organization is the organization manual. This manual gives a detailed account of the organization. The authority and responsibility of every person is given in detail. It avoids confusion and conflicts among various persons. The extent of authority and the relationship of executives is explained in this manual. The extent of span of management and delegation of authority are facilitated by a properly drawn manual.

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 29

During the transfer of data between the processor and memory, we use ________.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 29

During the transfer of data between the processor and memory, we use registers. Registers are a type of computer memory used to quickly accept, store, and transfer data and instructions that are being used immediately by the CPU. The registers used by the CPU are often termed Processor registers. A processor register may hold an instruction, a storage address, or any data (such as bit sequence or individual characters).

UGC NET Paper 2 Computer Science Mock Test - 9 - Question 30

Solve for x: log2(x2-3x)=log2(5x-15).

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 9 - Question 30

By using the property if logax = logay

then x=y, gives x2-3x=5x-15.

Now, to solve the equation x2-3x-5x+15=0

⇒ x(x-3)-5(x-3)=0

⇒ (x-3)(x-5)=0

⇒ x=3, x=5

For x=3: log2(32-33) = log2(53-15) ⇒ true

For x=5: log2(52-35) = log2(55-15) ⇒ true

The solutions to the equation are : x=3 and x=5.

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