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.

When there is an operand PUSH it on the stack, when there is an operator, POP top two operands from the stack and perform the operation and push the result back.

Here are the steps followed:

Total push operations = 8
Total Pop operations = 6

ECL remains for Emitter-Coupled Logic. It is intended for greatly fast application. It is appropriate for substantial centralized server PC that requires high number of operation every second.

TRUTH TABLE:

I, II and III are all true.

Condition coverage test:

Each one of the Boolean expression with values TRUE and FALSE. For this, tester should know the code. Hence it is White box testing.

Equivalence class partitioning:

It divides the input test data into each partition at least once of equivalent data from which test cases can be derived. Only input values are needed, not the code. Hence it is black-box testing.

Volume testing:

Check system performance with increasing volumes of data in the database.

Beta Testing:

It is a kind of system testing done at the customer site to validate the usability, functionality.

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.

Semaphore: it is an integer variable which is used by different processes in a mutually exclusive manner so as to achieve synchronization.

There are two types of Semaphores:

1. Counting Semaphores: The integer values in this type of semaphore can range over more than 0s and 1s
2. Binary Semaphores: The integer values in this type of semaphore can only be binary in nature, that is, 0s and 1s.

There are two types of operations that are atomic and mutually exclusive in nature:

1. Wait (S) or P: If the semaphore value is greater than 0, decrement it. If it is less than 0, then wait for the semaphore to become greater than 0 and then decrement it.
2. Signal (S) or V: Increment the value of the semaphore. If the semaphore is of binary type, and is already 1, then again replace the value by 1.

Calculations:

Let's start by analyzing the semaphore operations one by one:

18P: This operation decrements the semaphore value by 18.
3V: This operation increments the semaphore value by 3.
7V: This operation increments the semaphore value by 7.
2P: This operation decrements the semaphore value by 2.
6V: This operation increments the semaphore value by 6.
Now, let's apply these operations in sequence starting from an initial value of S:

After 18P: S - 18
After 3V: S - 15
After 7V: S - 8
After 2P: S - 10
After 6V: S - 4.

The final value of the semaphore is S - 4.

To ensure that one process is in the suspended list, the final value of the semaphore should be negative. Therefore, we need S - 4 to be less than 0.

S - 4 < 0
S < 4

The largest initial value of S that would satisfy this condition is 3.

Therefore, the largest initial value of S to keep one process in the suspended list is 3.

The correct option is (2)

A → III, B → I, C → IV, D → II

Concept:-

Match the following:

A: Modem → III: The device used for conversion between electric signals and digital bits.

B: Ethernet card → I: A network adapter is used to set up a wired network.

C: Repeater → IV: Regenerate the signals.

D: Switch → II: Connect multiple computers.

Key Points

• Modem stands for "modulator Demodulator". It refers to a device used for conversion between analog signals and digital bits.
• An Ethernet card, also known as a Network interface card (NIC card in short) is a network adapter used to set up a wired network. It acts as an interface between a computer and the network.
• A repeater is an analog device that works with signals on the cables to which it is connected. The weakened signal appearing on the cable is regenerated and put back on the cable by a repeater.
• A switch is a networking device that plays a central role in local area network.

The correct answer is "option 4".

CONCEPT:

A planar graph is a graph that can be drawn on the plane in such a way that its edges must intersect only at their endpoints.

In a planar graph, the graph is drawn in such a way that no edges must cross each other.

The graph whose edges overlap or cross each other is known as a Non-planar graph.

A graph to be planar must satisfy the following Euler’s formula:

v - e + f = 2

where

v is the number of vertices

e is the number of edges

f is the number of faces

CALCULATION

v = 10, e = 15

According to Euler’s formula:

v - e + f = 2

10 – 15 + f = 2

f = 2 – 10 + 15

f = 7

Out of 7, there will be always one unbounded face.

So, the number of faces is 6.

Hence, the correct answer is "option 4".

AND- OR search algorithm:

AND OR search algorithm is different from other algorithms in the way that it finds goal node at each state in order to reach the final goal state. It works in non-deterministic environment.

And OR search algorithm works in the following way:

1) It has to find a solution tree of subtrees.

2) Solution tree contains the root of the subtrees.

3) If a non-terminal AND node is in subtree and solution tree then all its children are in solution tree.

4) If a non-terminal OR node is in subtree and solution tree then exactly one of its children are in solution tree.

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))

Baud rate = 6000 bits/sec

Number of bits needed to send 1 byte of data

= 1(start bit) + 8(data) + 1 (parity) + 2 (stop) = 12 bits

1 second → 500 byte

∴ 33 seconds → 16500 bytes

(a ∨ (¬ a ∧ b)) ↔ ¬ (b ∨ a)

≡ (a ∨ ¬ a) ∧ (a ∨ b) ↔ ¬ (a ∨ b)

≡ (T ∧ (a ∨ b)) ↔ ¬ (a ∨ b)

≡ (a ∨ b)) ↔ ¬ (a ∨ b)

Let a ∨ b = p

≡ p ↔ ¬ p

≡ F

Truth Table of p ↔ q:

∴ Conditional statement is contradiction

Important Points:

Satisfiable: If there exist at least one true result for the given logical statement

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.

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.

The correct answer is option 2.

Concept:

A topological sort:

Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge u v, vertex u comes before v in the ordering. A topological ordering is possible if and only if the graph has no directed cycles, that is if it is a directed acyclic graph (DAG).

The given adjacency matrix graph is,

Option 1: (h, c, a, e, f, d, i, g, b)

False, Here node e was visited before node d. Hence not a topological sort.

Option 2: (c, h, a, d, e, f, i, g, b)

True, It is one of the topological sorts of the above graph.

Option 3: (c, a, d, e, f, g, h, i, b)

False, Here node g was visited before node i. Hence not a topological sort.

Option 4: (h, e, l, f, c, d, a, b, g)

False, Here node e was visited before node d. Hence not a topological sort.

Hence the correct answer is (c, h, a, d, e, f, i, g, b).

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 =

scanf("%d",&x);

will take input 8 from console and store in variable x.

x>>2;

This statement will return right-shifted bits in x by 2 positions but does not change the bits stored in variable x.

But here no assignment exists to hold that shifted bits. and x contains 8 only ( no change )

y=x;

now y also contains 8 only.

printf("OutputA=%d",y);

this statement will print OutputA=8.

Hence None of the given options are correct.

Concept:

• The cardinality of a set A is the measure of the "number of elements" of the set. It is denoted by n(A).
  For example, a set containing 3 elements has a cardinality of 3.
• If A ⊂ B, then A ∪ B = B.

Calculations:

Since, A ⊂ B and B ⊂ C, therefore A ∪ B ∪ C = C.

⇒ n(A ∪ B ∪ C) = n(C).

∴ The cardinality (number of elements) of A ∪ B ∪ C = cardinality (number of elements) of C.

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.

Correct answer is Option3Key Points

• Informed Search algorithms have information on the goal state which helps in more efficient searching.
• Uninformed Search algorithms have no additional information on the goal node.

Uninformed Search

• Breath first search
• Depth First Search
• Uniform Cost Search
• Depth First Iterative Deepening search

Informed Search

• Hill climbing
• Best first search
• Greedy search
• Beam Search
• Algorithm A

The correct answer is A, B and C only.

As per the question,

Statement A:- No S is P.

Converse of statement A - No P is S

Obverse of statement A - All S are non-P

Hence, among the above statements, statements A, B, and C are logically equivalent.

Hence, the correct answer is A, B, and C only.

Additional InformationLogical argument:- A logical argument is a process of creating a new statement from the existing statements. It comes to a conclusion from a set of premises by means of logical implications via logical inference.

• A categorical proposition is a simple proposition containing two terms, subject(S) and predicate (P), in which the predicate is either asserted or denied of the subject.
• Every categorical proposition can be reduced to one of four logical forms, named A, E, I and O.
• The 'A' proposition, is the universal affirmative usually translated as 'All S is P'.
• In the 'E' proposition, is the universal negative is usually 'No S is P'.
• The 'I' proposition, is the particular affirmative usually translated as 'Some S are P'.
• In the 'O' proposition, is the particular negative is usually translated as 'Some S are not P'
• Obversion is the inference in which the quality of the proposition is changed and the predicate is interchanged with its complement. It is valid for all four forms. The obverse is logically equivalent to the original proposition.
• Conversion is the inference in which the subject and predicate are interchanged. In modern logic, it is only valid for the E and I propositions. The valid converse is logically equivalent to the original proposition.

Many computer-based systems must recover from faults and resume processing within a prespecified time.

If recovery is automatic reinitialization, check pointing mechanisms, data recovery, and restart are evaluated for correctness.

The limitation of dynamic programming is that it does not allow itself for machine design. The limitation of dynamic programming is that it does not allow itself to the construction of general-purpose computer programs suitable for a wide range of distinct problems and dynamic programming has been used for optimizing the shape of pin-jointed structures and for the optimal design of transmission towers by palmer and Sheppard.

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.

Every node in a full binary tree has either 0 or 2 children. A binary tree can be generated by two traversals if one of them is in-order. But, we can generate a full binary tree using post-order and pre-order traversals.

A series of statements explaining how the data is to be processed is called program. A program is a sequence of instructions, written to perform a task by computer. It requires programs to function, typically executing the program’s instructions in a central processor. A program is usually written by a computer programmer in a programming language. From the program in its human-readable form of source code, a compiler or assembler can derive machine code a form consisting of instructions that the computer can directly execute.