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


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

UGC NET Paper 2 Computer Science Mock Test - 2 for UGC NET 2024 is part of UGC NET preparation. The UGC NET Paper 2 Computer Science Mock Test - 2 questions and answers have been prepared according to the UGC NET exam syllabus.The UGC NET Paper 2 Computer Science Mock Test - 2 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 - 2 below.
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UGC NET Paper 2 Computer Science Mock Test - 2 - Question 1

If link transmits 4000 frames per second and each slot has 8 bits, then the transmission rate of circuit of this TDM is

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

Transmission rate = frame rate * no of bits in a slot.

Frame rate = 4000 frames per second

# of bits in a slot = 8 bit

TDM = 4000 × 8 bps = 32 Kbps.

So, option (C) is correct.

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

If X is a binary number which is power of 2, then the value of X & (X - 1) is

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

let X = 2= 8 = 1000

then X - 1 = 7 = 0111

now X&(X - 1) = 0000

(here & is bitwise AND = If both bits in the compared position of the bit patterns are 1, the bit in the resulting bit pattern is 1, otherwise 0)

So answer is B.

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

Which component of CRT regulate the intensity of electron beam?

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

Explanation:

In a cathode ray tube (CRT), the intensity of the electron beam is regulated by the control grid.

  • The control grid is a negatively charged wire mesh that is located between the cathode and the anode of the CRT.
  • The control grid is used to control the flow of electrons from the cathode to the anode.
  • When a high negative voltage is applied to the control grid, it attracts the electrons being emitted by the cathode and allows them to pass through to the anode. As a result, the intensity of the electron beam is increased.
  • On the other hand, when a low negative voltage is applied to the control grid, it repels the electrons and blocks their passage to the anode, decreasing the intensity of the electron beam.

Additional Information

Focusing Anode - The focusing anode is a component of the CRT that is used to focus the electron beam onto a small spot on the fluorescent screen. The focusing anode is typically shaped like a cylinder or a conical pyramid, with a hole or a depression at the tip to focus the electron beam.

Heating Filament- In a cathode ray tube (CRT), the heating filament is a small wire that is used to heat the cathode, which is a negatively charged electrode located at one end of the CRT.

Accelerating Anode- In a cathode ray tube (CRT), the accelerating anode is a positively charged electrode located at the other end of the CRT, opposite the cathode. The cathode is a negatively charged electrode that is responsible for emitting a stream of electrons.

UGC NET Paper 2 Computer Science Mock Test - 2 - Question 4

The following statements (S1 and S2) pertain to the compiler in a computer.

S1 : It is a system software to convert source language program into target language program.

S2 : It validates the input program to the source language specification – produces error messages /warnings.

Choose the correct option

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 4

Both S1 and S2 are correct statements about compilers.

S1: It is a system software to convert source language program into target language program.

  1. A compiler is a system software that is used to convert source code written in a high-level programming language like C, C++, Java, Python, etc., into machine language (binary code) that can be executed by the computer's hardware.
  2. The compiler translates the source code into an intermediate form (assembly code), and then into machine language, which can be understood by the computer's CPU.
  3. The process of translating high-level code into machine language is called compilation.

S2: It validates the input program to the source language specification – produces error messages /warnings.

  1. When a program is written in a programming language, it has to confirm to a specific set of rules and syntax.
  2. The compiler validates the input program to the source language specification and produces error messages and warnings if the program violates any of the rules or syntax.
  3. For example, if a program tries to use an undefined variable, the compiler will produce an error message indicating that the variable is undefined.
  4. Similarly, if the program contains syntax errors like missing semicolons or parentheses, the compiler will generate error messages and prevent the program from being compiled.

let's take an example to understand the role of a compiler in converting source code into machine code and detecting errors.

Suppose we have the following C program:

#include <stdio.h>

int main() {
printf("Hello, World!");
return 0;
}

This program is written in C language and is intended to display the text "Hello, World!" on the screen when executed.

When we run this program, the compiler will perform the following steps:

  • The preprocessor will include the header file stdio.h, which contains the standard input/output library functions.
  • The compiler will analyze the syntax of the program to ensure that it conforms to the rules and syntax of the C language.
  • If there are any syntax errors, the compiler will produce an error message and stop the compilation process.

For example, if we misspelled printf as print, the compiler would produce the following error message:

error: implicit declaration of function ‘print’

  • If there are no syntax errors, the compiler will translate the C code into an intermediate language called assembly language. This process is called assembly.
  • Finally, the assembler will convert the assembly code into machine code, which can be executed by the computer's hardware. This process is called linking.

If there are any errors or warnings detected during any of these steps, the compiler will report them to the programmer.

For example, if we forgot to include the stdio.h header file, the compiler would produce the following warning message:

warning: implicit declaration of function ‘printf’

  • By reporting errors and warnings,
  • the compiler helps the programmer to identify and fix problems in their code,
  • making the process of developing software more efficient and reliable.
UGC NET Paper 2 Computer Science Mock Test - 2 - Question 5
____________ is when you promote someone else's work or ideas as your own, with or without their permission, by combining it into your work without giving them full acknowledgement.
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 5

The correct answer is option 4.

Concept:

Plagiarism:

Plagiarism is when you promote someone else's work or ideas as your own, with or without their permission, by combining them into your work without giving them full acknowledgment.

If we copy some content from the Internet but do not mention the source or the original creator, then it is considered an act of plagiarism.

Hence the correct answer is Plagiarism.

Additional Information

  • Hacking is the act of unauthorized access to a computer, computer network, or any digital system. Hackers usually have the technical expertise in hardware and software.
  • Identity thieves increasingly use personal information stolen from computers or computer networks, to commit fraud by using the data gained unlawfully.
  • Cyberbullying is bullying with the use of digital technologies. It can take place on social media, messaging platforms, gaming platforms, and mobile phones.
UGC NET Paper 2 Computer Science Mock Test - 2 - Question 6
Classification and regression techniques come under which type of machine learning algorithm?
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 6

The correct answer is option 2.

Concept:

Machine learning may be categorized into four categories:

Supervised Machine Learning:

Supervised learning, often known as supervised machine learning, is artificial intelligence and machine learning subcategory. Its use of labeled datasets to train algorithms that properly categorize data or predict outcomes defines it.

Supervised Learning techniques, such as regression and classification, are used to predict in Machine Learning and operate with labeled datasets.

Unsupervised Machine Learning:

Unsupervised learning, also known as unsupervised machine learning, analyses and clusters unlabeled information using machine learning techniques. Without the need for human interaction, these algorithms uncover hidden patterns or data groupings.

Semi-Supervised Machine Learning:

Semi-supervised machine learning combines the advantages of both supervised and unsupervised machine learning techniques. In supervised machine learning, you train a machine learning algorithm using a "labeled" dataset, where each record contains the outcome information.

Reinforcement Learning:

Reinforcement learning is a machine learning training method based on rewarding desired behaviors and/or punishing undesired ones.

Hence the correct answer is Supervised Machine Learning.

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

How many tuples does the result of the following relational algebra expression contain? Assume that the schema of A∪B is the same as that of A.

(A∪B) ⋈ A.Id > 40 ∨ C.Id < 15 C

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

The correct answer is option 1.

Key Points

Apply first cross-product then apply filter. Cross product yields 10 rows, then you play filter A.ID>40 or C.ID<15 produces 7 rows.

AUB= A.Id

12,15,25,98,99

|x| is cross product followed selection and projection.

Hence the correct answer is 7.

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

Using relational schema find the total balance of the depositors at each branch of the bank separately , rather than for the sum of entire branches.

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

The attribute at the left-hand subscript of G indicates that the input relation deposit_schema must be divided into groups based on the value of branch_name.

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

What is the completion time of process P2 using shortest job first (pre-emptive)?

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

0 3 4 8 10 15

Completion time of P2 is 15.

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

What is the average turnaround time using Shortest Job first (non-pre-emptive)?

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

0 3 9 11 15

P1 is in system for: 0-3 => 3

P2 is in system for: 1-9=>8

P3 is in system for: 4-15=>11

P4 is in system for: 6-11=>5

Average turnaround time = 27/4 = 6.75

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

What is the waiting time of process P3 using shortest job first (pre-emptive)?

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

0 3 4 8 10 15

Waiting time of P3 is 0

UGC NET Paper 2 Computer Science Mock Test - 2 - Question 12

Which of the following sorting algorithms suit the given statements?

Consider the total number of elements are n.

Statement 1: Best case= zero swaps and n(n-1)/2 comparsions.

Statement 2: Worst case = n(n-1)/2 swaps and n(n-1)/2 comparsions.

Statement 3: In best and worst cases total n-1 passes it runs.

Statement 4: In the first iteration it sorts the highest element and for the next iterations sorts the next highest element.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 12

The correct answer is option 3.

Concept:

The given statement follows the bubble sort.

Bubble Sort:

Bubble Sort is the most basic sorting algorithm, which operates by exchanging neighboring elements in the wrong order repeatedly. Bubble Sort is the most basic sorting algorithm, which operates by exchanging neighboring elements in the wrong order repeatedly.

Best Case:

Bubble sort best-case swaps are zero. The list would already be sorted. So no swaps are required.

Best case example:

Input: 1 2 3 4 5

Output: 1 2 3 4 5

Pass 1:

1 2 3 4 5 compare 1, 2 and no swap.

1 2 3 4 5 compare 2, 3 and no swap.

1 2 3 4 5 compare 3, 4 and no swap.

1 2 3 4 5 compare 4, 5 and no swap.

Pass 2:

1 2 3 4 5 compare 1, 2 and no swap.

1 2 3 4 5 compare 2, 3 and no swap.

1 2 3 4 5 compare 3, 4 and no swap.

Pass 3:

1 2 3 4 5 compare 1, 2 and no swap.

1 2 3 4 5 compare 2, 3 and no swap.

Pass 4:

1 2 3 4 5 compare 1, 2 and no swap.

Here statement 1 is true, Best case= zero swaps and n(n-1)/2 comparisons.

Worst Case:

Bubble sort worst case swaps n(n-1)/2. The smallest element is at end of the list so it needs the n(n-1)/2 swaps are required.

Worst-case example:

Input: 5 4 3 2 1

Output: 1 2 3 4 5

Pass 1:

4 5 3 2 1 compare 4, 5 and swap

4 3 5 2 1 compare 3, 5 and swap

4 3 2 5 1 compare 2, 5 and swap

4 3 2 1 5 compare 1, 5 and swap

Pass 2:

3 4 2 1 5 compare 3, 4, and swap

3 2 4 1 5 compare 2, 4 and swap

3 2 1 4 5 compare 1, 4 and swap

Pass 3:

2 3 1 4 5 compare 2, 3, and swap

2 1 3 4 5 compare 1, 3, and swap

Pass 4:

1 2 3 4 5 compare 1, 2 and swap

Total swaps are 1+2+3+4 =10 =5(5-1)/2

Here statement 2 is true, Worst case = n(n-1)/2 swaps and n(n-1)/2 comparsions.

Important Points

Statement 3 and statement 4 is true.

  • In the first pass of the bubble sorts the highest element at the end. After each pass, the next highest element is stored.
  • If there are no swaps in the list then the list is stored in order but the bubble sort runs n-1 passes.

Hence the correct answer is Bubble sort.

UGC NET Paper 2 Computer Science Mock Test - 2 - Question 13

Consider the following statements regarding interrupts.

(I) If a process is interrupted during system call handling then the OS will crash.

(II) If a process is interrupted during system call handling then the process is moved to the READY queue.

(III) If an interrupt handler is interrupted, the OS will crash.

While of the above statements are true?

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

The correct answer is option 1.

Concept:

The interrupt is a signal emitted by hardware or software when a process or an event needs immediate attention. It alerts the processor to a high-priority process requiring interruption of the current working process.

Option 1: If a process is interrupted during system call handling then the OS will crash.

False, If a process is interrupted during a system call then the system call was interrupted and the system call returned an error so definitely OS will not crash.

Option 2: If a process is interrupted during system call handling then the process is moved to the READY queue.

False, When a process is interrupted, that process is transferred to the waiting queue. If the process has been completed or aborted, the process is discarded. In either case, the dispatcher then selects a process from the queue to execute.

Option 3: If an interrupt handler is interrupted, the OS will crash.

False, If an interrupt handler is interrupted, such kernel control paths may be arbitrarily nested an interrupt handler may be interrupted by another interrupt handler, thus giving rise to a nested execution of kernel threads.

Hence the correct answer is none of the above.

UGC NET Paper 2 Computer Science Mock Test - 2 - Question 14
Which one of the following well-formed formulae in predicate calculus is NOT valid?
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 14

Concept:

p ⇒ q means ¬p ∨ q

It means that p is true, and q is false, in that case this implication fails.

Explanation:

Consider, p(x) = x is a prime number

q(x) = x is a non-prime number

Option 1:

(∀x p(x) ⇒ ∀x q(x)) ⇒ (∃x ¬ p(x) ∨ ∀x q(x))

In this, case after implication it becomes true and in case implication if there is true after implication then that predicate is valid. (∃x ¬ p(x) ∨ ∀x q(x)), this will become true by the example.

Option 2:

(∃x p(x) ∨ ∃x q(x)) ⇒ ∃x (p(x) ∨ q(x))

This implication is also valid. In this, before implication it will become false. So, it is valid.

Option 3:

∃x (p(x) ∧ q(x)) ⇒ (∃x p(x) ∧ ∃x q(x))

There exists a number x, which is both prime and non-prime which is false. A false statement can imply either true or false. So, it is valid.

Option 4:

∀x (p(x) ∨ q(x)) ⇒ (∀x p(x) ∨ ∀x q(x))

Here, before implication, for all x, x is either prime or non-prime which is true.

After implication, for all x, x is prime or for all x, x is non-prime which is false. So, here a true statement implies a false statement which results in false. So, it is not valid.
UGC NET Paper 2 Computer Science Mock Test - 2 - Question 15
Which of the following is an advantage of fragmentation occurring in the memory?
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 15

Fragmentation: It is a situation or problem occurring in operating system where many processes are loaded and unloaded in the memory which results in more and more fragmented pieces of memory. And then a time comes when the total available memory is more than the required memory for a process but since it is not contiguous, it is useless.

Advantages of Fragmentation:

  1. Fast Data Writes: With fragmentation present in the memory, data writing turns out to be faster as in the other cases, reorganising the data storage to enable contiguous data writes wastes a lot of time.
  2. Fewer Failures: There has to be sufficient space in the memory so that when data writes are done, write shouldn't fail. This probability is much higher in systems which don't support fragmentation.
  3. Stable Optimisation: Sometimes, processes arrive in such a way that fragmented systems potentially make better use of a storage device by utilising every available storage block.

Disadvantages of Fragmentaion:

  1. Need of regular defragmentation: With time, the performance of the storage device will reduce significantly which will further require for time consuming defragmentation processes.
  2. Slower Read Times: If the storage device or memory becomes more and more fragmented, it will take equivalently more time to read the non-sequential data from the memory.
UGC NET Paper 2 Computer Science Mock Test - 2 - Question 16

Simplex method of solving linear programming problem uses

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 16

Simplex method:

  • The simplex method is the most popular method used for the solution of Linear Programming Problems (LPP).
  • The Simplex method is a search procedure that shifts through the set of basic feasible solutions, one at a time until the optimal basic feasible solution is identified.
  • It can be used for two or more variables as well (always advisable for more than two variables to avoid lengthy graphical procedure).
  • The simplex method is not used to examine all the feasible solutions.
  • It deals only with a small and unique set of feasible solutions, the set of vertex points (i.e. extreme points/corner points) of the convex feasible space that contain the optimal solution.
  • All the resource values or constraints should be non-negative.
  • All the inequalities of the constraint should be converted to equalities with the help of slack or surplus variables.
UGC NET Paper 2 Computer Science Mock Test - 2 - Question 17

The basic computer has three instruction code formats, each format has 2 bytes. If addressing mode has 1 bit, operand has 1 byte then what is the size of opcode in a bit?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 17

Instruction size = 2 byte = 16 bits

Addressing mode = 1 bit

operand = 1 byte = 8 bits

Instruction size = Addressing mode + opcode + operand

16 = 1 + opcode + 8

∴ opcode = 7 bits

Instruction code format:

UGC NET Paper 2 Computer Science Mock Test - 2 - Question 18

The individual pieces that make up a bitmapped image are often called _____________.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 18

The correct answer is Pixel.

Key Points

  • Pixel stands for "Picture Element."
  • Pixel is also known as "PEL."
  • Pixel is the smallest element of an image on a computer display.
  • It includes both types of display that is LCD or CRT monitors.
  • A screen is made up of a matrix of thousands or millions of pixels.
  • It is represented with a dot or a square on a computer screen.
  • Each pixel has a unique logical address.
  • It can have a single color at a time.
  • The Colour of a pixel is determined by the number of bits which is used to represent it.
  • A resolution of a computer screen depends upon the graphics card and display monitor, the quantity, size and color combination of pixels.
  • When an image has zoomed in, the surface of the image will look like filled dots and these dots are light receptors called photodiode.
  • These dots are known as pixels.
UGC NET Paper 2 Computer Science Mock Test - 2 - Question 19

Consider the 4 to 1 multiplexer with two select lines S1 and S0 given below.

The minimal sum of products form of the Boolean expression for the output F of the multiplexer is

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 19

The output is given by:

F = P'Q' + P'QR + PQR'

= P'[Q' + QR] + PQR'

= P'[Q' + Q][Q' + R] + PQR'

= P'[Q' + R] + PQR'

= P'Q' + P'R + PQR'

UGC NET Paper 2 Computer Science Mock Test - 2 - Question 20

Match the list

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 20

The correct answer is A → 4, B → 2, C→ 1, D → 3

Additional Information

  • The data stored in the database at a particular moment of time is called an instance of the database
  • A database schema is the skeleton structure that represents the logical view of the entire database
  • Metadata is the data information that provides information about the other data
  • A database model is a type of data model that determines the logical structure of a database and fundamentally determines in which manner data can be stored, organized and manipulated
UGC NET Paper 2 Computer Science Mock Test - 2 - Question 21
Which of the following statement is False in case of DMA mode?
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 21
  • During DMA the CPU is idle and it has no control over the memory buses. The DMA controller takes over the buses to manage the transfer directly between the I/O devices and the memory unit.
  • Bus Request: It is used by the DMA controller to request the CPU to relinquish the control of the buses.
  • Bus Grant: It is activated by the CPU to inform the external DMA controller that the buses are in high impedance state and the requesting DMA can take control of the buses.
  • After the completion of the DMA transfer, the processor is notified by an interrupt signal.
  • The Controller uses the registers to store the starting address, word count and the status of the operation.
UGC NET Paper 2 Computer Science Mock Test - 2 - Question 22
Pumping lemma for regular sets is used to
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 22
Key Points

Pumping Lemma is to be applied to show that certain languages are not regular. It should never be used to show a language is regular. If L is regular, it satisfies Pumping Lemma. If L does not satisfy Pumping Lemma, it is non-regular.

  • The minimum pumping length must always be greater than 0, even if there are no strings in the language.
  • Pumping lemma satisfy regular language.
  • Pumping lemma also satisfies CFLs.
  • Proves why non-regular grammar is non-regular.

Hence the correct answer is to prove that certain sets are not regular.

UGC NET Paper 2 Computer Science Mock Test - 2 - Question 23
CASE combines
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 23
Software engineering tools provide automated or semi-automated support for the process and the methods. When tools are integrated so that information created by one tool can be used by another, a system for the support of software development, called computer-aided software engineering, is established. CASE combines software, hardware, and a software engineering database which is repository containing important information about analysis, design, program construction, and testing.
UGC NET Paper 2 Computer Science Mock Test - 2 - Question 24

In the Simplex method if in pivot column all the entries are negative or zero when choosing leaving variable then

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

Simplex method: It is a step by step method in which solution is started with initial feasible solution and in the next step the initial feasible solution is improved, the steps are repeated until the optimum solution is reached.

Following are the set of variables in Simplex Method.

  • Artificial Variable
    • This variable is introduced in case of greater than or equal to type constraint to not violate the non-negativity constraint.
  • Basic Variables
    • They can be defined as the variables which can take any value other than zero.
  • Non basic Variables
    • This variable is not in the basic solution and is having the value zero.
  • Slack Variable
    • This variable is introduced in the simplex method to eliminate less-than (<) constraints.
  • Surplus Variable
    • This variable is introduced in the simplex method to eliminate greater-than (>) constraints.

A simplex table is shown below for example for the given objective function and constraints

Maximize Z = 4 x1 + 6 x2 + x3

Subject to 2 x1 - x2 + 3 x3 ≤ 5

x2 ≤ 2

x1, x2, x3 ≥ 0

Following are the special cases of simplex method:

Under Simplex Method, the existence of multiple optimal solutions is indicated by a situation under which a non-basic variable in the final simplex table showing optimal solution to a problem, has a net zero contribution.

Optimal solution: If at least one of the non-basic variable in the (Cj- Zj) row of the final simplex table has a zero value, it indicates that there is more than one optimal solution.

Unbounded solution: If all the replacement rations, bi/ai (ai – key column or pivot column coefficient are negative) then it is the solution is unbounded.

Infeasible solution: If artificial variables are present in the base, then the solution obtained is infeasible.

Degenerate solution: If some of the values in the constant column (bi) are zero, then the solution becomes degenerate.

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

The essential prime implicants of are

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

To cover 0, 10 min terms,

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

Consider the following statements:

S1: ∀x P(x) ∨ ∀xQ(x) and ∀x(P(x) ∨ Q(x)) are not logically equivalent.

S2: ∃x P(x) ∧ ∃x Q(x) and ∃x (P(x) ∧ Q(x)) are not logically equivalent

Which of the following statements is/are correct?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 26
S1: ∀x P(x) ∨ ∀xQ(x) and ∀x(P(x) ∨ Q(x)) are not logically equivalent.

This statement is correct. Consider an example: x – is a number. P(x) denotes x is even and Q(x) denotes x is odd. 

∀x P(x) ∨ ∀xQ(x) - it means for all numbers x is even or for all numbers x is odd.

∀x(P(x) ∨ Q(x)) – it means for all x, x is either odd or even.

From this, it is clear that both these are not logically equivalent.

S2: ∃x P(x) ∧ ∃x Q(x) and ∃x (P(x) ∧ Q(x)) are not logically equivalent

This statement is correct.  Consider the same example as of first case.

∃x P(x) ∧ ∃x Q(x) – It means there exists a number which is odd and there exists a number which is even also which is true.

∃x (P(x) ∧ Q(x)) - It means, there exists a number which is both even and odd which is not possible. It is false.

So, these are not logically equivalent.

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

A scheduler which selects processes from secondary storage device is called:

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 27
A scheduler which selects processes from secondary storage device is called Medium term scheduler. Medium-term scheduling is a part of swapping. It removes the processes from the memory. It reduces the degree of multiprogramming. The medium-term scheduler is in-charge of handling the swapped out-processes. Through this scheduler, processes are removed from memory. Medium-term scheduler cut down the degree of degree of multi-programming.
UGC NET Paper 2 Computer Science Mock Test - 2 - Question 28

Hive also support custom extensions written in ____________.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 2 - Question 28

Hive also supports custom extensions written in Java, including user-defined functions (UDFs) and serializer-deserializers for reading and optionally writing custom formats. Hive allows users to read, write, and manage petabytes of data using SQL. Hive is built on top of Apache Hadoop, which is an open-source framework used to efficiently store and process large datasets. As a result, Hive is closely integrated with Hadoop and is designed to work quickly on petabytes of data.

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

If the partial order of a set has at most one minimal element, then to test whether it has a non-crossing Hasse diagram its time complexity __________.

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

If the partial order has at most one minimal element, or it has at most one maximal element, then to test whether a partial order with multiple sources and sinks can be drawn as a crossing-free Hasse diagram or not it’s time complexity is NP-complete.

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

Which of the following transmission systems provides the highest data rate to an individual device?

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

Computer Bus provides the highest data rate to an individual device. The Computer Bus is a communication link used in a computer system to send the data, addresses, control signals, and power to various components in a computer system. The computer buses are used to connect the various hardware components that are part of the computer system.

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