UGC NET Paper 2 Computer Science Mock Test - 6 - UGC NET MCQ

Number of channels = 4

1 byte is multiplexed with each channel i.e. each frame carries 1 byte and we have 4 channel, so the size of each frame is 1 × 4 byte = 4 bytes 32 bits.

Channel sends 100 frames per seconds. So, the bit rate will be 100 × 4 bytes i.e. 3200 bits/seconds.

So, option (D) is correct.

The translator which performs macro calls expansion is called Macro pre – processor. A Macro processor is a program that copies a stream of text from one place to another, making a systematic set of replacements as it does so. ... It is called micro pre-processor because it allows us to add macros.

Hexagonal cells means each cell has six neighbors.

If the central cell uses frequency  A, its six neighbors can use B, C, B, C, B, and C  respectively.

In other words, only 3 unique cells are needed. 

Consequently, each cell can have 840/3= 280 frequencies

We can generate more than 1 parse tree for a single string from the grammars. For statement (iii)

S → U

U → yT

 yT → yxSy

 yxSy → yxTy

 yxTy → yxxyy

All statement are correct.

So, option (D) is correct.

Deletion of the last node of the linked list, we need address of second last node of single linked list to make NULL of its next pointer. Since we can not access its previous node in singly linked list, so need to traverse entire linked list to get second last node of linked list.

So, option (D) is correct.

As I7 is active therefore y2, y1, y0. Where y2, y1, y0 = 1.

The solution is A + B̅ i.e. option (2)

Explanation:

5x + 9y = 45 ⇒  = 1...(i)

x + y = 2 ⇒

y = 4 ......(iii)

The feasible region is ABCDEA

A(2, 0), B(9, 0), C(9/5, 4), D(0, 4), E(0, 2)

Now, Z = x + y

Z(2, 0) = 2

Z(9, 0) = 9

Z(9/5, 4) = 9/5 + 4 = 1.8 + 4 = 5.8

Z(0, 4) = 4

So minimum value is 2 which is attained at (2, 0) and (0, 2) i.e., at the line x + y = 2

(3) is correct

The correct answer is option 3.

Concept:

Internal Fragmentation:

When memory is divided into fixed-sized chunks, internal fragmentation occurs. Internal fragmentation occurs when the memory allotted to the process is slightly bigger than the memory needed. The difference between allocated and demanded memory is known as the allocated and demanded memory difference.

External fragmentation:

External fragmentation occurs when a dynamic memory allocation mechanism allocates some memory while leaving a small quantity of useless memory. When there is excessive external fragmentation, the overall amount of memory accessible is significantly reduced. As a result, there is enough memory space to finish a request, but it is not continuous. As a result, it is referred to as exterior fragmentation.

Statement 1: Using a larger block size in a fixed partition scheme may increase internal fragmentation.

The given statement is true, As the size of the block in fixed partition scheme increases, the chances of internal fragmentation also increase. Because with a larger block size, if a smaller process arrives; the remaining block will be wasted. This is because one block can be allocated to only one process at a time.

Statement 2: Variable partition and segmentation suffer from external fragmentation.

The given statement is true, Variable partition scheme suffers from external fragmentation, and segmentation is one of the variable partition schemes.

Hence the correct answer is Both statement 1 and statement 2.

Answer: Option 1

Concept:

Fractional knapsack uses Greedy approach.

An item with maximum profit/weight is selected first

Explanation:

Profit = Item 1 + Item 3 + Item 5 + fraction part of 4

Profit = 90 + 60 + 42 +  = 269

The correct answer is option 1.

Concept:

Infix expression: The expression of the form ( a operator b). When an operator is in-between every pair of operands.

Postfix expression: The expression of the form ( a b operator). When an operator is followed for every pair of operands.

To convert infix expression to postfix expression, we will use the stack data structure. By scanning the infix expression from left to right, when we will get any operand, simply add them to the postfix form, and for the operator and parenthesis, add them in the stack maintaining the precedence of them.

Analysis:

Rules:

• Print operands as they arrive.
• If the stack is empty or contains a left Parenthesis on top, push the incoming operator onto the stack.
• If the incoming symbol is '(', push it onto the stack.
• If the incoming symbol is ')', pop the stack & Print the operators until the left parenthesis is found.
• If the incoming symbol has higher precedence than the top of the stack, push it on the stack.
• If the incoming symbol has lower precedence than the top of the stack, pop & print the top. Then test the incoming operator against the new top of the stack.
• If the incoming operator has equal precedence with the top of the stack, use associativity.
• At the end of the expression, pop & print all operators of the stack.
• Associativity Left to Right then pop & print the top of the stack & then push the incoming operator. Right to Left then pushes the incoming operator.

Hence the correct answer is - ( ( ×.

The correct answer is option 2.

Distance-vector protocols :

Distance-vector protocols update the routing tables of routers and determine the route on which a packet will be sent by the next hop which is the exit interface of the router and the IP address of the interface of the receiving router. Distance is a measure of the cost to reach a certain node.

A distance-vector routing (DVR) protocol requires that a router inform its neighbors of topology changes periodically. Historically known as the old ARPANET routing algorithm (or known as the Bellman-Ford algorithm).

Bellman-Ford Basics – Each router maintains a Distance Vector table containing the distance between itself and ALL possible destination nodes. Distances, based on a chosen metric, are computed using information from the neighbors’ distance vectors.

The given diagram is,

if link V1-V4 is removed then routing table.

Hence the correct answer is (0, 3, 8, 5, 11).

The correct answer is option 1.

Key Points

T O

+ G O

-----------------

OUT

__________

Here O goes to carry. So it must be O = 1

T = O + O = 1 + 1 = 2 Hence, T = 2

U = T + G (here T and G must produce carry) so only two number produce carry.

U = 2 + G (Here G can be 8 or 9)

if G = 8, U = 2 + 8 = 10 so U = 0 and G = 8

if G = 9, U = 2 + 9 = 11 here U = 1 and G = 9 not possible because we already give 1 to O. So G = 9 incorrect.

T + O + G + U = 11

Hence the correct answer is 11.

The characteristic equation of the flip – flop is given by

The characteristic equation of a JK flip – flop is

Both the above equation will be same if

The correct answer is option 2.

Concept:

In order to build a binary tree from given traversal sequences, one of the traversal sequences must be Inorder. The other traversal sequence might be either Preorder or Postorder. We know that the Inorder traversal of a Binary Search Tree is always in ascending order.

Preorder traversal sequence of a binary search tree,

25, 15, 10, 4, 12, 22, 18, 24, 50, 35, 31, 44, 70, 66, 90

Inorder traversal sequence of a binary search tree,

4, 10, 12, 15, 18, 22, 24, 25, 31, 35, 44, 50, 66, 70, 90

Postorder traversal is a type of traversal in which we first visit the left subtree in postorder, then the right subtree in postorder, and then visit the root node.

Postorder is= 4, 12, 10, 18, 24, 22, 15, 31, 44, 35, 66, 90, 70, 50, 25

Hence the correct answer is 4, 12, 10, 18, 24, 22, 15, 31, 44, 35, 66, 90, 70, 50, 25.

Additional Information

• Main memory Access Time (Time required to access page in main memory) = m
• Page Tables are stored in the main memory.
• Effective Memory Access Time (without TLB) = 3m
• Now, TLB (Translation Look-aside Buffer) is added to improve performance. It contains frequently referred page numbers and corresponding frame numbers.
• TLB Access Time (Time required to access page present in TLB) = c
• TLB Hit Ratio (Hit ratio means the probability of the page missing in main memory to be found in TLB) = x
• Effective Memory Access Time (with TLB)

Final write:

A-T1

B-T3

C: T3

Initial read

Write read sequence

T3 T2 T1 ------not valid

T2 T3 T1-------not valid

T2 T1T3------valid sequence

The correct answer is option 3.

Concept:

Context-free language:

A context-free language is a language generated by context-free grammar. Context-free languages have many applications in programming languages, in particular, most arithmetic expressions are generated by context-free grammars.

Union, Prefix, Suffix, Substring, Concatenation, Reversal, Kleen closure, positive closure, substitution, Homomorphism, and Inverse Homomorphism are closed under context-free language.

Close Properties:

Hence the correct answer is Intersection.

Concept:

The total number of simple graphs possible with ‘n’ vertices = 2^n(n-1)/2.

Explanation:

Simple Graph is a graph with no loops and no parallel edges.

The maximum number of edges possible in a single graph with ‘n’ vertices is ⁿC₂ where ⁿC₂ = n(n – 1)/2.

Hence, The maximum number of simple graphs possible with ‘n’ vertices = 2^n(n-1)/2.

The correct answer is option 4.

Concept:

Artificial intelligence (AI) is a broad field of computer science that focuses on creating intelligent computers that can accomplish activities that would normally need human intelligence.

There are five Logical connectives used in Artificial Intelligence (A.I.) and are;

Conjunction:

A sentence that has ∧ a connective such as, P ∧ Q is called a conjunction.

Negotiation:

A sentence such as ¬ P is called the negation of P. A literal can be either Positive literal or negative literal.

Implication:

A sentence such as P → Q is called an implication. Implications are also known as if-then rules.

Disjunction:

A sentence that has ∨ a connective, such as P ∨ Q. is called disjunction, where P and Q are the propositions.

Biconditional:

A sentence such as P⇔ Q is a Biconditional sentence.

Hence the correct answer is All of the above.

The RAD model is a high-speed adaptation of the linear sequential model in which rapid development is achieved by using component-based construction.

The RAD approach encompasses the following phases:

Business modeling, Data modeling, Process modeling, Application generation, Testing

CPU-scheduling decisions may take place under the following four circumstances:

1. When a process switches from the running state to the waiting state.

2. When a process switches from the running state to the ready state.

3. When a process switches from the waiting state to the ready state.

4. When a process terminates. If no process is ready, a system-supplied idle process is normally run.

• For situations 1 and 4, there is no choice in terms of scheduling. A new process (if one exists in the ready queue) must be selected for execution.
• There is a choice, however, for situations 2 and 3. When scheduling takes place only under circumstances 1 and 4, we say that the scheduling scheme is nonpreemptive or cooperative; otherwise, it is pre-emptive.

The correct answer is SELECT * FROM STUDENT WHERE CLASS IN ('12A', '12B');

Key Points

• The IN operator allows you to check if a value belongs to a specified set of values.
• In this case, we want to find students where CLASS is either '12A' or '12B'.
• Option (1) achieves the same result but uses the OR operator which is a common alternative.
• Option (3) uses the LIKE operator which is for pattern matching, not suitable here.
• Option (4) uses the BETWEEN operator which is typically used for ranges of numbers, not applicable here.

The correct answer is SELECT NAME FROM STUDENT GROUP BY NAME HAVING COUNT(*) > 1;

Key Points

• To determine if there are any students with duplicate names in the table, you can use a SQL query that specifically looks for names that occur more than once. From the given options, the query that efficiently finds duplicate names is: SELECT NAME FROM STUDENT GROUP BY NAME HAVING COUNT(*) > 1;
• This query groups the results by the NAME column and then filters the groups to only those with a count greater than 1, indicating duplicates.

The core to reverse engineering is an activity called extract abstractions. In abstraction activity, the engineer must evaluate older program and extract information about procedures, interface, data structure or database used. The engineer must evaluate the old program and from the source code, extract a meaningful specification of the processing that is performed, the user interface that is applied, and the program data structures or database that is used.

DNS can obtain the IP address of the host if its domain name is known and vice versa. DNS automatically converts between the names we type in our Web browser address bar to the IP addresses of Web servers hosting.

Syntax Analysis is a second phase of the compiler design process in which the given input string is checked for the confirmation of rules and structure of the formal grammar. It analyses the syntactical structure and checks if the given input is in the correct syntax of the programming language or not.

The first part of the compiler (lexical analysis) is also known as a scanner. Lexical analysis is the first stage of a three-part process that the compiler uses to understand the input program. The role of the lexical analysis is to split program source code into substrings called tokens and classify each token to their role (token class). The program that performs the analysis is called scanner or lexical analyzer. It reads a stream of characters and combines them into tokens using rules defined by the lexical grammar, which is also referred to as lexical specification.