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

Privileged instruction cannot be the answer as system call is done from user mode and privileged instruction cannot be done from user mode.

Correct answer is (D). 

Hardware interrupts: The 8085 microprocessor has five hardware interrupt inputs (five pins). They are TRAP, RST 7.5, RST 6.5, RST 5.5 , and INTR in the decreasing order of priority. If two or more interrupts go high at the same time, the 8085 will service them on priority basis.

The correct option is option 3) 8, 20, 18, 23, 14, 45, 70, 82, 56, 38.

Concept:

• To determine the postorder traversal of a binary search tree (BST) based on the given preorder traversal, we need to understand the properties of a BST and how its nodes are arranged.
• In a BST, the preorder traversal visits nodes in the following order: root, left subtree, right subtree. Since the preorder traversal given is 38, 14, 8, 23, 18, 20, 56, 45, 82, 70, we can determine the structure of the BST as follows:
• The first element, 38, represents the root of the BST.
• Elements smaller than 38 will be in the left subtree, and elements greater than 38 will be in the right subtree.

Explanation:

Using this information, we can break down the preorder traversal step by step and build the BST:

Step 1: The root node is 38.
Step 2: The next element, 14, is smaller than 38, so it becomes the left child of 38.
Step 3: The next element, 8, is smaller than 38 and 14, so it becomes the left child of 14.
Step 4: The next element, 23, is greater than 14 and 8, so it becomes the right child of 8.
Step 5: The next element, 18, is greater than 14 and 8 but smaller than 23, so it becomes the left child of 23.
Step 6: The next element, 20, is greater than 14, 8, and 18, so it becomes the right child of 18.
Step 7: The next element, 56, is greater than 38, so it becomes the right child of 38.
Step 8: The next element, 45, is greater than 38 but smaller than 56, so it becomes the left child of 56.
Step 9: The next element, 82, is greater than 38 and 56, so it becomes the right child of 56.
Step 10: The final element, 70, is greater than 38 and 56 but smaller than 82, so it becomes the left child of 82.

Based on this process, the BST can be represented as follows:

Now that we have the BST structure, we can find the postorder traversal by visiting the nodes in the following order: left subtree, right subtree, root. Applying this to the BST above, the postorder traversal will be: 8, 20, 18, 23, 14, 45, 70, 82, 56, 38

Thus, the postorder traversal of the given binary search tree is 8, 20, 18, 23, 14, 45, 70, 82, 56, 38.

• This is the original heap as per the array.
• Swap 300 and 215.
• Swap 300 and 250.
• Swap 300 and 289.
• A total of three swaps are required to convert it to max-heap.

The correct answer is “option 1”.

EXPLANATION:

Statement P: True

Every edge in a given cycle graph will become a vertex in the new defined graph L(G) and every vertex of the cycle graph will become an edge in the new graph.

Statement R: False

Consider a planar graph with 5 vertices and 9 edges.

Also assume, the degree of one vertex is 2 and the rest is 4.

So L(G) has 9 vertices (since G has 9 edges) and 25 edges.

But, the planar graph must satisfy the condition:

|E| < = 3|V| - 6

So for 9 vertices,

|E| < = 3×9- 6

|E| < = 21

But L(G) has 25 edges & hence it is not planar.

So this statement is false.

Since Option 2 and 3 contains R, these options are eliminated.

Statement S: False

From this tree:

By drawing its line graph, a cycle graph can be formed of 3 vertices.

So this statement is wrong.

Hence, option 4 is wrong.

Hence, the correct answer is “option 1”.

NOTE THAT:

TRY TO DRAW AN NFA AND THEN OBTAIN REGULAR EXPRESSION.

Overlay: Whenever a program is running or is loaded in the RAM, it will not use the complete program, but only some part of it. Whenever the part of the program is required, it will load up and then once the part of the program will be of no use, it will unload itself. When it is needed again, it can be reloaded and run again.

Total memory requirement:

For A  BD = 2 + 3 + 9 = 14 KB

For A  BE = 2 + 3 + 8 = 13 KB

For A  CF = 2 + 4 + 7 = 13 KB

For A  CG = 2 + 4 + 5 = 11 KB

As we can see, the largest program part size is 14 KB. Hence, we need 14 KB maximum size to run the given program

Concept:

Follow(X) is the set of terminals that can appear immediately to the right of non-terminal X in some sentential form. Rules for finding the follow (X) are given here as:

1) Follow (S) = {$}, where S is the start symbol.

2) If A → bBC is the production, where b, B and C are grammar symbols, then Follow (B) is first (C).

3) If A → bB is production then follow (B) is follow (A).

4) If A → bBC is production and first(c) contains ε, then follow (B) contains {first (C) - ε} U follow (A).

Explanation:

FIRST(B) = {r, ε}

FIRST(C) = {q, ε}

FIRST(D) = {p}

FIRST(A) = FIRST (BCD) = {r, q, p}

Concept:

Source Port number, Destination Port number and Checksum is present in both TCP header and UDP header.

Diagram:

Important Point:

Header length in TCP header and Length in UDP header are different.

Explanation:

objective function z = 5x + 2y under the linear constraints

x ≥ 0, y ≥ 0, x ≥ y and 2 ≤ x + y ≤ 4

So, LPP is

Maximize z = 5x + 2y

subject to

x + y ≥ 2

x + y ≤ 4

x ≥ 0, y ≥ 0

Feasible region is ABCDA whose corner points are

A(2, 0), B(4, 0), C(0, 4), D(0, 2)

z = 5x + 2y

Z(2, 0) = 10, Z(4, 0) = 20, Z(0, 4) = 8, Z(0, 2) = 4

So, max z = 20 at (4, 0)

Hence option (2) is correct

The correct answer is option 2.

Key Points

Min-max algorithm works like a backtracking algorithm. Max will try to maximize its utility(best move). Min will try to minimize its utility(worst move).

Hence the correct answer is 2.

Formula:

Explanation:

Given:

Delay of stage S₁ = 10 ns

Stage delays are represented as :

Execution time with pipeline (T_p) = max {all stage delays}

= 10 ns

Execution time without pipeline(T_n) = sum of all stage delays

= 10 + 5 + 10 + 5 + 10 = 40 ns

The correct answer is option 1.

Concept:

Statement 1: Feasibility Analysis is used to obtain the outline of the problem and decide whether a feasible or appropriate solution exists or not.

True, A feasibility study examines all important aspects of a project to assess the viability and probability of completion. Feasibility Analysis is performed to outline the problem and determine whether or not a viable or adequate solution exists.

Statement 2: A feasibility study can be considered as the final investigation that helps the software development life cycle.

False, Feasibility studies are preliminary investigations that assist management in determining if a system study is viable for development or not.

Hence the correct answer is Only statement 1.

Fuzzy logic:

• Fuzzy logic is a concept of certain degree. Boolean logic is a subset of fuzzy logic.
• Fuzzy logic is a form of many-valued logic which deals with reasoning that is approximate rather than fixed and exact.
• Compared to traditional binary sets (where variables may take on true or false values), fuzzy logic variables may have a truth value that ranges in degree between 0 and 1.
• It is employed to handle the concept of partial truth, where the truth value may range between completely true and completely false.

Crisp Logic - In this, the values are of boolean nature and strictly precise values, that is, either 0 or 1.

Fuzzy Logic - It is a type of logic that allows for the inclusion of uncertain or ambiguous information.

Defuzzification - It is the process of producing a quantifiable result in crisp logic, given fuzzy sets and corresponding membership degrees. It is the process that maps a fuzzy set to a crisp set. It is typically needed in fuzzy control systems.

• It converts imprecise data into precise data.
• Methods of Defuzzification - Maximum membership principle, centroid method, weighted average method, centre of sums, etc.
• Complicated in nature.
• Uses centre of gravity method to find centroids of given sets.

In an "m-ary" tree. the number of total nodes (N) is given by

N=mi + 1 ----(1)

Where,

i: Number of internal nodes

Also, in a tree, N=i + L ----(2)

Where,

L=number of leaf nodes

Here m=2

From equation (1) and equation (2);

N = 2i + 1

2i + 1 = i + L

L = i + 1

The number of leaves are 1 plus the number of internal nodes in binary tree.

Here, given L=n, substitute above and we will get,

i = L - 1

Correct answer is Option C
Key Points
Following will be contents of temporary table S

Following will be contents of temporary table T

• Following will be the result of natural join of above two tables.
• The key thing to note is that the natural join happens on column name with same name which is Manager in the above example.
• “Hari” appears two times in Manager column, so there will be four entries with Manager as “Hari”.

• Dijkstra’s Shortest Path uses the greedy method to find the shortest path of a graph G(V, E).
• Floyd Warshall algorithm is based on the principle of dynamic programming. It is used to solve the all pair shortest path problem.
• Kruskal algorithm is a minimum spanning tree algorithm in which in every iteration, minimum weighted edge is found and then it is added to the construction of minimum spanning tree. Edges are added in increasing order of the edge weights. That’s why it is a greedy approach.
• Merge sort is an efficient sorting algorithm that uses a divide-and-conquer approach to order elements in an array.

As, 1 bit is fixed, we have left with 9 positions, bit used are 2(0 & 1). So, total possible strings with this are 2⁹.

Strings of length 10 that end with 00:

In this, 2 position are fixed, we are left with only 8 positions. Total possible strings with this are 2⁸.

Strings of length 10 that starts with 1 and end with 00:

In this, 3 position are already fixed, and we have only 7 positions. Total strings possible are 2⁷.

Now strings possible that starts with 1 or end with 00:

A U B = A + B - A ꓵ B

A U B = 2⁹ + 2⁸ - 2⁷ = 512 + 256 - 128 = 640

DML refers to the tools used to add, update and access the data within a database, including things like artists, albums, and so on in our music database example. DML is a categorization of existing SQL commands. DML commands include SELECT, INSERT, UPDATE, and DELETE.

Fortran is a scientific computer language.

It was originally developed by a team at IBM in 1957 for scientific calculations. It also allows the compiler to generate efficient binary code. It is a scientific computing language. It is also used in organizations such as weather forecasters, financial trading, and engineering simulations.

Model–view–controller (MVC) is a software design pattern commonly used for developing user interfaces that divide the related program logic into three interconnected elements Model, View and Controller. But the Model element is deal with the database. Model is the central component of the architecture. It is the application's dynamic data structure, independent of the user interface. It directly manages the data, logic and rules of the application.

Baselines is a software configuration management concept that helps us to control change without seriously impeding justifiable change. A baseline is a milestone and reference point in software development that is marked by completion or delivery of one or more software configuration items and formal approval of a set of predefined products is obtained through formal technical review. Baseline is a shared project database.

Dynamic binding or run-time binding or late binding is that type of binding which happens at the execution time of the program or code. Function or method overriding is the perfect example of this type of binding. Virtual functions are used to achieve the concept of function overriding.

Multiprogramming allows multiple programs to reside in separate areas of the core at the time. Multiprogramming is a rudimentary form of parallel processing in which several programs are run at the same time on a uniprocessor. Instead, the operating system executes part of one program, then part of another, and so on. To the user, it appears that all programs are executing at the same time.

Recursive descent parser is a kind of top-down parser built from a set of mutually recursive procedures (or a non-recursive equivalent) where each such procedure implements one of the nonterminal of the grammar. Thus the structure of the resulting program closely mirrors that of the grammar it recognizes.

Depth-first search implements stack operation because it always expands the deepest node in the current tree. Depth-first search (DFS) is an algorithm for traversing or searching tree or graph data structures. The algorithm starts at the root node (selecting some arbitrary node as the root node in the case of a graph) and explores as far as possible along each branch before backtracking.

Given, the post order traversal of a binary tree is OPQRST. The method is use is post order traversing is- Left, Right, Root

So, the binary tree is-

Now, the pre order traversal,

Method- Root, Left, Right.

So, traversal is TQOPSR.