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

So, option (C) is correct answer.

ans must be 1)SCRIPT

<script> 

document.getElementById("demo").innerHTML = "Hello JavaScript!"; 

</script>

rest options are part of a normal HTML document

The 8087 is divided into two sections namely control unit and numeric extension unit in which the numeric extension unit executes all the numeric processor instructions.

Hence the correct option is (C).

Answer --> a - iii, b - iv, c - i, d - ii

Explanation -

FTP (File Transfer Protocol) -> File Transfer
FTP is a protocol that is used for transferring files over a network. It is a standard protocol that is used for uploading and downloading files from a server to a client or vice versa.

TFTP (Trivial File Transfer Protocol) -> Trivial File Transfer
TFTP is a simplified version of FTP that is used for transferring small files over a network. It is a lightweight protocol that does not provide authentication or encryption.

Telnet -> Remote Login
Telnet is a protocol that is used for remote login to a server or a networking device. It enables a user to access and control a remote computer over a network connection.

UDP (User Datagram Protocol) -> Send Datagram
UDP is a protocol that is used for sending datagrams over a network. It is a connectionless protocol that does not provide any reliability or error checking. It is commonly used for applications that require the fast transmission of data, such as online gaming and video streaming.

Concept:

The given circuit is Multiplexer.

Multiplexers:

It is a combinational circuit that has many data inputs and single output depending on control or select inputs.​ For N input lines, log n (base2) selection lines, or we can say that for 2ⁿ input lines, n selection lines are required. Multiplexers are also known as “Data n selector, parallel to serial convertor, many to one circuit, universal logic circuit​”.

Hence the correct answer is a⊙b⊙c.

Answer: Option 3

Explanation:

Fm = 5 kHz

carrier frequency Fc = 1 MHz ≡ 1000 kHz
(fc+fm) and(fc−fm) are called sideband frequencies.

(1000+ 5) and (1000 - 5 ) are sideband frequencies.

1005 kHz and 995 kHz

and 1.005 Mhz and 0.995 Mhz

• Verification and validation techniques are very similar since both these techniques are designed to help remove errors in a software
• In spite of the apparent similarity between their objectives, the underlying principles of these two bug detection techniques and their applicability are very different
• Verification is the process of determining whether the output of one phase of software development conforms to that of its previous phase
• Validation is the process of determining whether a fully developed software conforms to its requirements specification

A self-organizing map:

• It is a type of artificial neural network (ANN) that is trained using unsupervised learning to produce a low-dimensional (typically two-dimensional), discretized representation of the input space of the training samples, called a map, and is therefore a method to do dimensionality reduction.
• Self-organizing maps differ from other artificial neural networks as they apply competitive learning as opposed to error-correction learning (such as backpropagation with gradient descent), and in the sense that they use a neighbourhood function to preserve the topological properties of the input space.

The correct answer is 1600

EXPLANATION:

In a CSMA/CD (Carrier Sense Multiple Access with Collision Detection) network, after a collision, the station needs to wait for a certain amount of time before attempting to transmit again. This waiting time is determined using the exponential backoff algorithm.

The formula for calculating the backoff time is as follows:

where

In Ethernet, the slot time is typically considered to be twice the maximum one-way signal propagation time.

Given:

• Transmission rate = 8 Mbps
• One-way signal propagation time = 25 μs

Let's calculate the slot time:

• 
• Slot time=2×25μs=50μs

Now, after 5 successive collisions, the value of K would be in the range of 0 to  (as per Ethernet standard).

So, the maximum backoff time would be :

Therefore, the answer is 4) 1600.

The correct answer is in both 2 NF and 3 NF

Key PointsIn the given schema R = (S, T, U, V) with dependencies S → T, T → U, U → V, and V → S, every attribute is functionally dependent on another, creating a cyclic dependency. This construct forms an unusual scenario not typically encountered in straight-forward normalization discussions.

• Analysis Based on Normalization Forms:
  • 2NF (Second Normal Form) checks for partial dependencies, meaning that non-prime attributes (those not part of a candidate key) should not depend on just a part of any candidate key. In this case, since every attribute appears to depend on another in a circular manner, and no single attribute can be considered partially dependent on any candidate key (assuming each attribute by itself forms a single attribute candidate key due to the cyclic dependency), the schema by a strict interpretation could be considered in 2NF.
  • 3NF (Third Normal Form) concerns itself with transitive dependencies, where a non-prime attribute is dependent on another non-prime attribute. Given the direct dependency of each attribute on another in a cyclic form, there are no clear transitive dependencies in the traditional sense where one non-prime attribute depends on another non-prime attribute through a third attribute. Each attribute’s dependency is direct and not through another non-prime attribute.
• Conclusion: In a simplified explanation, considering the unique cyclic dependency scenario provided, it can be reasonably argued that the schema is in both 2NF and 3NF because it does not exhibit the partial or transitive dependencies that 2NF and 3NF respectively seek to eliminate. Each attribute’s dependency is direct and forms a closed loop, not easily fitting into the typical scenarios of violation for these normalization forms.

Note: The assumption here is based on classical definitions of 2NF and 3NF, and the unique circular dependency pattern doesn't squarely fit traditional examples of normalization form violations, leading to the conclusion of being in both 2NF and 3NF (Option 4).

Reflexive Relation :A Relation R defined on set A is said to be reflexive, if (x, x) ∈ R, x ∈ A ,i.e xRx, ∀ x,y ∈ A

Here define Relation p = {(a, a), (b, b), (c, c), (d, d), (a, b), (b, a), (b, c), (c, b).

If a relation is reflexive, then (a, a) ∈ R for every a ∈ {a, b, c, d}

Since (a, a) ∈ R ,(b, b) ∈ R,(c, c) ∈ R, (d, d) ∈ R

So, R is reflexive.

Check Symmetric

Symmetric Relation: A Relation R is defined on set A is said to be symmetric, if (x, y)  R ⇒ (y, x)R

To check symmetric,

If (a, b) ∈ R, Then (b, a) ∈ R

Since(a, b) ∈ R and (b, a) ∈ R also (b ,c) ∈ R and (c, b) ∈ R

So, R is reflexive and symmetric.

Additional Information
Transitive Relation: A Relation R defined on set A is said to be transitive if (x, y) ∈ R , (y, z) ∈ R ⇒ (x, z) ∈ R ∀ x,y,z ∈ A.

If (a, b) ∈ R, (b, a) ∈ R since (a, c) ∉ R

so, R is not transitive Relation.

If R is satisifed all the condition of Equivalence then the R Relation be the Equivalence Otherwise not.

Key Points

NP-Complete problem:

Following are some NP-Complete problems, for which no polynomial-time algorithm is known. Determining whether a graph has a Hamiltonian cycle Determining whether a Boolean formula is satisfiable, etc.

• Hamiltonian-Cycle problem
• The Travelling Salesman Problem
• The Clique problem

The above algorithms are not feasible under polynomial-time algorithm problems.

For Shortest Path Problem you can use Dijkstra's algorithm. With a normal binary heap, this gives you a time complexity of O((E + V) log V). With a Fibonacci heap, this can be improved to O(E + V log V), which is faster for dense graphs.

Hence the correct answer is Shortest Path Problem.

The correct answer is option 4.

Concept:

Count():

It takes one set of values and gives one output. The COUNT() function returns the number of records returned by a select query.

Note: NULL values are not counted.

Syntax:

COUNT(expression)

Explanation:

The given table is,

Consider the following student relation.

Student relation has Name, Subject1, and Subject2 attributes.

Query1:

Select count (*)

From student;

Explanation:

The query1 has count(*) prints the number of rows that are not null. Here no complete row is null. So it prints the 6 as output.

Query2:

Select count (10)

From student;

Explanation:

Note:

Count(constant) prints the number of rows in a relation.

The query2 has count(10) prints the number of rows that are not null. Here no complete row is null. So it prints the 6 as output.

Hence the correct answer is Q1=Q2.

The correct answer is option 1.

Concept:

Distinct Clause:

By default, SQL shows all the data retrieved through query as output. However, there can be duplicate values. The SELECT statement when combined with the DISTINCT clause returns records without repetition (distinct records).

The given data,

Student(Rno, name, Branch, Year)

Query:

Select distinct branch

from student ;

Step 1:

It selects all the rows in the student table.

Step 2:

DISTINCT clause returns records without repetition Which means distinct records in the attribute branch so it prints only CSE, IT, and ME as output.

Hence the correct answer is option 1.

In R(ABCDEFGH), left-hand side of FD has attributes: {G, B, C, F, H, A, E, G}

Since it doesn't contain D we cannot derive it unless it is the present right-hand side.

(AD)+ = {A,D, B, C, F, H, G, E }

Since all the 8 attributes are present, AD is a candidate key

(BD)+ = {B, D, C, F, H, G, E, A }

Since all the 8 attributes are present, BD is a candidate key

(ED)+ = {E, D, A, B, C, F, H, G }

Since all the 8 attributes are present, ED is a candidate key

(FD)+ = {F, D, E, G, A, B, C, H }

Since all the 8 attributes are present, FD is a candidate key

The non-prime attributes are dependent on a partial candidate key.

A → BC ---(A → C)
B → CFH ---((B → H)
F → EG ----(F → G)

Therefore it is in 1NF but not in 2 NF

FIRST and FOLLOW:

Parsing Table:

Hence the answer is

E1: S → aAbB, S → ε

The correct answer is option 3.

Concept:

Statement 1: A single deque can support both stack and queue operations.

True, The word deque, usually pronounced deck, is short for double-ended queue. A deque is a list that supports insertion and removal at both ends. Thus, a deque can be used as a queue or as a stack.

• A deque elements insertion and deletion are done from only one end so it follows the stack property(First in Last out).
• A deque elements insertions are done from only one end and deletion is done from another end. so it follows the queue property(Last in first out).

Statement 2: A single stack can support both deque and queue operations.

False, A stack can not supports the deque and queue operations because it has the restriction that insertion and deletion are done from only one end.

Statement 3: A single queue can support both stack and queue operations.

False, A queue can not supports the deque operations because it has the restriction that insertion on one end and deletion is done from another end.

Hence the correct answer is only statement 2 and statement 3.

The correct answer is option 1.

Concept:

The given data,

The time taken for 600 characters = 1 minute.

i.e 60 seconds =600 characters.

1 second= 600/ 60 characters per second.

1 second =10 characters per second.

Note:

The baud rate is the rate at which information is transferred in a communication channel.

If one character stores in eleven bit then the transfer time of 11 bit is=?

11 bits = 11x time taken each bit

11 bits= 11 x 10 bauds

11 bits=110 bauds.

Hence the correct answer is 110 bauds.

After reducing the cost of link D - E to 1, a routing table for Router E:

Calculating minimum and maximum number of nodes from height:

If binary search tree has height h, minimum number of nodes is h+1.

If binary search tree has height h, maximum number of nodes will be when all levels are completely full. Total number of nodes will be = 2^(h+1) - 1.

Now,

maximum number of nodes = 2^(h+1)-1 = 26-1 = 63

minimum number of nodes = h+1 = 5+1 = 6

So, the maximum and the minimum number of nodes in a binary tree of height 5 are respectively 63 and 6.

AI assistants, like Alexa and Siri, are examples of intelligent agents as they use sensors to perceive a request made by the user and automatically collect data from the internet without the user's help. They can be used to gather information about its perceived environment such as weather and time.

If the referred page is not present in the main memory, then there will be a miss, and the concept is called Page miss or page fault. The CPU has to access the missed page from the secondary memory. If the number of page fault is very high, then the effective access time of the system will become very high.

• Binary images are images whose pixels have only two possible intensity values. They are normally displayed as black and white. Numerically, the two values are often 0 for black, and either 1 or 255 for white.
• Binary images are often produced by thresholding a grayscale or color image, in order to separate an object in the image from the background.
• The color of the object (usually white) is referred to as the foreground color. The rest (usually black) is referred to as the background color.
• However, depending on the image which is to be thresholded, this polarity might be inverted, in which case the object is displayed with 0 and the background is with a non-zero value.

If the angle between 2 connected line segments is very small then, Miter join can generate a long spike that distorts the appearance of the polyline. It causes the outside corner of the stroke to come to a sharp point. In computer graphics, a point of intersection between two strokes in which the joining comes to a point, but an angle is limited, as at the top of some typefaces, such as the capital letter "A" in Palatino: a

This is used to check the overflow that occurs in the operation. In computer processors, the overflow flag (sometimes called the V flag) is usually a single bit in a system status register used to indicate when an arithmetic overflow has occurred in an operation, indicating that the signed two's-complement result would not fit in the number of bits used for the result. Some architectures may be configured to automatically generate an exception on an operation resulting in overflow.

Tree algorithm is used to solve any kind of problem because specific variants of the algorithm embed different strategies. Much of the work on search in artificial intelligence deals with trees. These are usually defined implicitly by a so-called problem representation, and the process of searching for a solution of the given problem can be represented by a search tree (more generally an acyclic graph, because of transpositions).

To sort the array firstly create a min-heap with first k+1 elements and a separate array as resultant array. 2) because elements are at most k distance apart from original position so, it is guranteed that the smallest element will be in this K+1 elements. 3) remove the smallest element from the min-heap(extract min) and put it in the result array. 4) Now,insert another element from the unsorted array into the mean-heap, now,the second smallest element will be in this, perform extract min and continue this process until no more elements are in the unsorted array.finally, use simple heap sort for the remaining elements Time Complexity ------------------------ 1) O(k) to build the initial min-heap 2) O((n-k)logk) for remaining elements... 3) 0(1) for extract min so overall O(k) + O((n-k)logk) + 0(1) = O(nlogk).