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

We have to make string which have sum = 7.There are:

2,2,1,1,1 = 5! / 2! × 3! = 10

2,2,2,1,0 = 5! / 3!      = 20

3,1,1,1,1 = 5! / 4!      = 5

3,2,1,1,0 = 5! / 2!      = 60

3,2,2,0,0 = 5! / 2! × 2! = 30

3,3,1,0,0 = 5! / 2! × 2! = 30

4,2,1,0,0 = 5! / 2!      = 60 

4,3,0,0,0 = 5! / 3!      = 20

4,1,1,1,0 = 5! / 3!      = 20

5,1,1,0,0 = 5! / 2! × 2! = 30

5,2,0,0,0 = 5! / 3!      = 20

6,1,0,0,0 = 5! / 3!      = 20

7,0,0,0,0 = 5! / 4!      = 5 

total = 10 + 20 + 5 + 60 + 30 + 30 + 60 + 20 + 20 + 30 + 20 + 20 + 5 = 330.

So, option (B) is correct.

Alternative method –

Lets digits are a, b, c, d, and e. Therefore,

a + b + c + d + e = 7 

Total number of combinations are ^(n-1+r)C_(r) = ^(5-1+7)C₇ = ¹¹C₇ = ¹¹C₄ = 330.

Decorator is used to enhancing the functionality of an object at run-time, i.e., it allows you to add new behaviour to other objects at run-time.

So, option (B) is correct.

 In 3G network, W-CDMA is also known as UMTS. The minimum spectrum allocation required for W-CDMA is 5 MHz.

So, option (D) is correct.

TRUTH TABLE:

I, II and III are all true.

Here, both statements are correct because,

• The correlation between the two statements lies in the concept of pipelining in computer architecture.
• Here, statement 1 has multiple stages of instruction execution can occur simultaneously, allowing for a more efficient use of resources and faster overall processing.
• Here, statement 2 aligns with the idea that pipelining involves breaking down the execution of an instruction into smaller stages that can be performed parallel.

Option 1 is correct.

The number of nodes present in a graph will be equal to the number of principal nodes present in an electric circuit.

The number of branches present in a graph will be less than or equal to the number of branches present in an electric circuit.

In a graph there is one and only one path between every pair of vertices.

Branches are the connections between nodes. A branch is an element(resistor, capacitor, source).

The number of branches in a circuit is equal to the number of elements.

Waterfall Model:

Moving forward compulsorily, it is impossible to go back to previous project phase in waterfall model. Hence, this method is inflexible.

Evolutionary:

The model keeps changing with time and according to requirements. Hence, it is incremental in nature.

Component based:

This model relies on reuse-based approach to defining, implementing and composing loosely coupled independent components into systems.

Spiral:

This model is the most advanced. It includes four phases - Planning, Risk Analysis, Engineering and Evaluation.

Concept:

Total participation: It specifies that each entity in the entity set must compulsorily participate in at least one relationship instance in that relationship set.

Partial participation: It specifies that each entity in the entity set may or may not participate in the relationship instance in that relationship set.

Explanation:

In one to many or many to one relation, the relation between two entities is merged on the many side with total participation. As, it is given that relationship R doesn’t have its own attributes. So, it must be combined with entity A. So, the relation must be many to one and there should be total participation of A in R.

Answer: option 4

Explanation:

Consider the string "00111"

on 0 transition, q₀ → q₁

on 0 transition, q₁ → q₁

on 1 transition, q₁ → q₂

on 1 transition, q₂ → q₂

on 1 transition, q₂ → q₂

on Y transition, q₂ → q₃

Hence "00111" is accepted, now we can directly rule out options 1, 2, and 3.

The only option left is Option 4.

Even we dont use the above elimination strategy, It can be clearly understood with help of rules given in the Question that if the input contains 0 in starting then only TM will proceed further q0 → q1 (First Point)

Now if there more 0's in the input it will stay in the same state i.e. q₁ and if 1 in the input then it will make a transition from q1 → q₂ and now if again 0 in the input then it will make the transition back to q₁. (Second point) (this indicate complete language )

Now if there more 0's in the input it will stay in the same state i.e. q₂.

In this instance, we observed that we will reach state q₂ only if we last see a 1 in input and if input finishes TM then halts.

So the language accepted is all the string which starts with 1 and ends with 0.

The correct answer is ((1, 0, 1), (a, a, a), (0, 0, 0))

Key PointsThe sets A and B are defined as follows:

1. A = {0, 1} × {0, 1} × {0, 1}, which produces all possible combinations of three digits where each digit can be either 0 or 1.
2. B = {a, b, c} × {a, b, c} × {a, b, c}, which produces all possible combinations of three letters where each letter can be either a, b, or c.

Given the order of elements, A × B × A in lexicographic order means:

• First, we order by the first element of A,
• Then, by elements of B,
• Finally, by the second element of A.

Given the current element ((1, 0, 0), (c, c, c), (1, 1, 1)), we need to find the next element.

For the first A, "(1, 0, 0)" is fixed in this step because to move to the next element in the sequence, we look into B first as it can vary next before the second A gets incremented. Since "(c, c, c)" is the last possibility in B's lexicographic order, we need to move to the next possibility in the second A and then cycle B back to its first.

For the second A, "(1, 1, 1)" is already at the last possibility. So, we increment the first A from "(1, 0, 0)" to "(1, 0, 1)" as the next logical step, given that the elements of A and B are considered in a combined lexicographic progression.

Since the element from B was "(c, c, c)" and we're rolling over, B should reset to its first element "(a, a, a)".

Then, the final component (the second A) should also reset back to the beginning, "(0, 0, 0)", as you effectively cycle through every combination of B with each element of A before moving to the next element of A.

Therefore, the correct answer is: ((1, 0, 1), (a, a, a), (0, 0, 0))

Since wait(R) can pass 3 time without blocking which will make 6 signal(R), but if P3 process will consume all 6 signal(R) then P2 wouldn’t get the chance to unblock wait(R) and printf(“A”) wouldn’t able to execute.

There are only one wait(L) and there are no signal(L) and initial value of L is 3, so wait(L) can pass three times before P1 process get block so there are three “C” can print after that P1 will be blocked.

Concept:-

Entity integrity constraint:- The entity integrity rule is designed to assure that every relation has a primary key and that the data values for that primary key are all valid. Entity integrity guarantees that every primary key attribute is non-null.

Key Points

• The primary key performs the unique identification function in a relational model.
• An entity that cannot be identified is a contradiction in terms, hence the name entity integrity.
• A null is a value that is assigned to an attribute when no other value applies, or when the applicable value is unknown.

Additional InformationReferential integrity constraint:- A referential integrity constraint is a rule that maintains consistency among the rows of two tables(relations). The rule states that if there is a foreign key in one relation, either each foreign key value must match a primary key value in the other table, or else the foreign key must be null.

Domain constraint:- All the values that appear in a column of a relation (table) must be taken from the same domain. A domain is a set of values that may be assigned to an attribute.

Key constraints:- The term 'key' has too many meanings in the database world. In the relational model alone, there are candidate keys, primary keys, alternate keys, foreign keys, search keys, and so on.

Data:

Packet size = 1500 bytes = 12000 bits

Bandwidth = 100 Mbps

Propagation speed = 2 × 10⁸ m/s

Calculation:

Transmission time =

Length of cable needed to exactly contain packet = 120 × 2 × 10⁸

= 24000 m

In 24000 m, bits are = 12000

1 m = 12000/24000 = ½

100 m = 50 bits

But it is given that there is an delay of 10 bits in each 100 m, so we have total 60 bits/100 m or 0.6 bits/m.

So, 12000-bit packet can fill =

The correct answer is option 3

Concept:

Comment Tag:

We can add comments to our HTML source code by using the following syntax

<!-- This line is a comment -->

Comments are not displayed by the browser, but comments are helpful to understand our HTML source code.

So the option (3) <!-- Testing--> is correct

• A configuration management plan is written during the project planning phase; It lists all controlled objects
• The managers who develop the plan must strike a balance between controlling too much and controlling too little; If too much is controlled, overheads due to configuration management
• Configuration management is carried out through two principal activities:

1. Configuration identification: It involves deciding which parts of the system should be kept track of.
2. Configuration control: It ensures that changes to a system happen smoothly

The correct answer is option 2.

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.

Option 1:If L is a regular language, then L2 is also a regular language.

True, If L is a regular language, then L2 means L.L is also regular. Regular languages are closed under concatenation. Hence it L2 is also a regular language.

Option 2: The language L = {an: n ≥ 4} is NOT regular.

False, The language L = {an : n ≥ 4} is regular. It accepts the strings like {a⁴, a⁵,a⁶,a⁷, a⁸,....}.

Here minimum length string is "aaaa" So only 5 states are sufficient to construct DFA. Hence The language L = {an:: n ≥ 4} is regular.

Option 3: A language L is regular if and only if there exists a corresponding DFA to accept it.

True, A Regular expression by definition is the language accepted by a nondeterministic finite automaton (NFA) or deterministic finite automaton (DFA) it can be generated by a regular grammar.

Regular language = Type-3 language = Regular expressions = Regular grammar = DFA =NFA = ε-NFA = Right linear grammar = Left linear grammar

Option 4: If L is regular, then L- {λ} is also regular.

True, A language is regular, by definition, if you can create a DFA for it. Then I need to prove that if L is regular, then so is L− {λ} for any λ ∈ L.

Here L − {λ} λ ∈ L.

λ ∈ L means it accepts by the given regular language Hence λ is also regular.

= L− {Regular}

= L ∩ {Regular}^c { By set rule A - B = A∩B^c}

=L ∩ {Regular} {By closed properties Complement of regular is regular}

= Regular ∩ Regular {By closed properties intersetion of regular is regular }

= Regular

Hence the correct answer is The language L = {an:: n ≥ 4} is NOT regular.

• A page fault occurs when a page is not found in the memory and needs to be loaded from the disk
• If a page fault occurs and all memory frames have been already allocated, then the replacement of a page in memory is required at the request of a new page; This is referred to as demand-paging
• The choice of which page to replace is specified by a page replacement algorithm
• The commonly used page replacement algorithms are FIFO, LRU, optimal page replacement algorithms, etc
• Generally, on increasing the number of frames to a process’ virtual memory, its execution becomes faster as less number of page faults occur; Sometimes the reverse happens, that is, more numbers of page faults occur when more frames are allocated to a process; This most unexpected result is termed as Belady’s Anomaly
• FIFO and random replacement algorithms suffer from Belady’s Anomaly

The size of int type data is 4 bytes.

The array stores 9 elements.

So, the size of the array will 9 × 4 = 36 bytes.

Grammar:

S → XY

X → aX | a

Y → aYb | ϵ

X generates at least one a. Generates a string of type {a, aa, aaa, aaaa……}

a^mbⁿ, if n = 0 and m = 0 ∴ string = ϵ which cannot be generated from the given grammar and hence option (B) is incorrect.

Y generates an equal number of a and b { ϵ, ab, aabb, ……} with a comes before b.

Since at least one a is generated by X and an equal number of a’s and b’s generated by Y ∴ m> n and hence option (A) is incorrect

The smallest length string generated by Y is ϵ. In a^mbⁿ, n ≥ 0 and hence option (D) is incorrect.

{ a, aab, aaabb,………….} which is equivalent to:

L = {a^mbⁿ | m > n, n ≥ 0}

CRAY BLITZ was a computer chess program written by Robert Hyatt, Harry L. Nelson, and Albert Gower to run on the Cray supercomputer. It was derived from "Blitz" a program that Hyatt started to work on as an undergraduate. "Blitz" played its first move in the fall of 1968 and was developed continuously from that time until roughly 1980 when Cray Research chose to sponsor the program. Cray Blitz participated in computer chess events from 1980 through 1994 when the last North American Computer Chess Championship was held in Cape May, New Jersey. Cray Blitz won several ACM computer chess events, and two consecutive World Computer Chess Championships, the first in 1983 in New York City, and the second in 1986 in Cologne, Germany.

Oracle is an RDBMS (Relational Database Management System). It is known as Oracle Database, Oracle DB, or Oracle Only. The first database for enterprise grid computing is the Oracle database. Therefore, Oracle is not an operating system.

We consider a hypothetical state space where every state has b successors. The root of the search tree generates b nodes at the first level, each of which generates b more nodes, for a total of b2 at the second level. Each of these generates b more nodes, yielding b3 nodes at the third level, and so on. Now suppose that the solution is at depth d. In the worst case, we would expand all but the last node at level d (since the goal itself is not expanded), generating bd+1- b nodes at level d+1.

The Aspect source flag function specifies the aspect source flags for all bundled line attributes. The bundled line attributes are line type, line width, and line colour. For each of the bundled line attributes, the value defined by the Aspect source flag function can be INDIVIDUAL or BUNDLED. The default for all line aspect source flags is INDIVIDUAL. The Aspect source flag function sets the Aspect source flag entries in the General Attributes and Output Control State List. . We can chose any one of the above attributes by setting switch for each of the attributes.