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UGC NET Paper 2 Computer Science Mock Test - 5 - UGC NET MCQ


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30 Questions MCQ Test UGC NET Mock Test Series 2024 - UGC NET Paper 2 Computer Science Mock Test - 5

UGC NET Paper 2 Computer Science Mock Test - 5 for UGC NET 2024 is part of UGC NET Mock Test Series 2024 preparation. The UGC NET Paper 2 Computer Science Mock Test - 5 questions and answers have been prepared according to the UGC NET exam syllabus.The UGC NET Paper 2 Computer Science Mock Test - 5 MCQs are made for UGC NET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for UGC NET Paper 2 Computer Science Mock Test - 5 below.
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UGC NET Paper 2 Computer Science Mock Test - 5 - Question 1

The octal equivalent of the binary number 1011101011 is

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 1

Explanation: First we will make pair of 3 bits from LSB :

ie- 1 011 101 011

now convert these bits into decimal and that will be

1 – 1

011 – 3

101 – 5

011 – 3

so the answer will be 1353.

So, option (B) is correct.

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 2

The process of dividing an analog signal into a string of discrete outputs, each of constant amplitude, is called

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 2
  • The process of dividing an analog signal into a string of discrete outputs, each of constant amplitude, is called Quantization.
  • The process of enhancing an analog signal is called Amplification.
  • Irregular behaviour or jerk or loss of continuity in analog signal is called Strobing.
  • Conditioning is a process of manipulating an analog signal in a way that prepares it for the next stage of processing.

So, option (D) is correct.

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UGC NET Paper 2 Computer Science Mock Test - 5 - Question 3

A fuzzy set wherein no membership function has its value equal to 1 is called

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 3

Subnormal fuzzy set: A fuzzy set wherein no membership function has its value equal to 1.

Convex fuzzy set: It is a fuzzy set which has a membership function whose membership values are strictly monotonically increasing or strictly monotonically decreasing or strictly monotonically increasing than strictly monotonically decreasing with increasing values for elements in the universe

Nonconvex fuzzy set: The membership values of the membership function are nor strictly monotonically increasing or decreasing or strictly monotonically increasing than decreasing.

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 4
The following keys 22, 28, 23, 12, and 13 are inserted into an initially empty hash table of length 9 using the hash function h(k) = k mode 9 with open addressing and linear probing. At which index is the key 13 inserted?
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 4

To insert a key into a hash table using open addressing with linear probing, we follow these steps:

  1. Calculate the hash value of the key using the hash function.
  2. If the calculated index is empty, insert the key at that index.
  3. If the calculated index is occupied, probe linearly until an empty slot is found and insert the key at that slot.

Using the given hash function h(k) = k mod 9, we can calculate the hash value of each key as follows:

  • h(22) = 22 mod 9 = 4
  • h(28) = 28 mod 9 = 1
  • h(23) = 23 mod 9 = 5
  • h(12) = 12 mod 9 = 3
  • h(13) = 13 mod 9 = 4

Steps -

  • To insert the keys into the hash table using linear probing, we start by inserting key 22 at index 4.
  • The next key, 28, is inserted at index 1 since index 4 is already occupied.
  • The third key, 23, is inserted at index 5 since index 1 is occupied and index 4 is already occupied by key 22.
  • The fourth key, 12, is inserted at index 3 since index 5 is occupied, index 1 is occupied by key 28, and index 4 is already occupied by key 22.
  • Finally, when we try to insert the key 13, we calculate its hash value as h(13) = 4, which is the same as the hash value of key 22.
  • Since index 4 is already occupied by key 22, we need to probe linearly to find the next available slot.
  • We start at index 5 (next index after 4) and find that it is occupied by key 23.
  • We continue probing linearly and find that the next available slot is index 6. Therefore, the key 13 is inserted at index 6.
  • So the key 13 is inserted at index 6 in the hash table.
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 5

Consider three floating-point numbers A, B and C stored in registers RA, RB and RC, respectively as per IEEE-754 single-precision floating point format. The 32-bit content stored in these registers (in hexadecimal form) are as follows.


Which one of the following is FALSE?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 5

The correct answer is option 2.

Concept:

IEEE single-precision floating-point:

IEEE single-precision floating-point computer numbering format is a binary computing format that takes up 4 bytes (32 bits) of memory. Binary32 is the official name for the 32-bit base 2 formats in IEEE 754-2008. IEEE 754-1985 referred to it as single.

IEEE single-precision format:

Explanation:

The given data,

Decimal value =(-1)s x 1.M x 2Base Exponent -Bias

Bias value in IEEE single-precision format is 127

RA = 1100 0001 0100 0000 0000 0000 0000 0000

RA sign= 1

RA Base Exponent =100 0001 0 = 130

RA Mantisa = 100 0000 0000 0000 0000 0000 = 1.100 0000 0000.....

Decimal value = (-1)1 x1.1 x2130-127 =-1.1x23= -1100 = (-12)10

A=-12

RB = 0100 0010 0001 0000 0000 0000 0000 0000

RA sign= 0

RA Base Exponent =100 0010 0= 132

RA Mantisa = 001 0000 0000 0000 0000 0000 = 1.001 000000.....

Decimal value = (-1)0 x1.001 x2132-127 =+1.001x25= + 100100 = (+36)10

B=+36

RC = 0100 0001 0100 0000 0000 0000 0000 0000

RA sign= 0

RA Base Exponent =100 0001 0= 130

RA Mantisa =100 0000 0000 0000 0000 0000= 1.100 0000.....

Decimal value = (-1)0 x1.1 x2130-127 =+1.1x23= + 1100 = (+12)10

C=+12

Option 1: A + C = 0

True, A+C= -12+12=0

Hence it is true.

Option 2: C = A + B

False, A+B= -12+36=+24

it not equal to C. Hence it is false.

Option 3: B = 3C

True, B=3C

=3x+12 =36 =B

it equal to B. Hence it is true.

Option 4: (B - C) > 0

True, (B-C) >0

=(36-12)=24>0

Hence it is true.

Hence the correct answer is C = A + B.

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 6
Fuzzy logic machines __________ washing load to ensure correct spinning.
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 6

Fuzzy logic machines are used in washing machines at assesses the various parameters of washing load and continuously adjusts the settings of spinning. Various system assesses and adjust some of parameters like:

i) The weight of the load

ii) Type of the fabric

iii) The amount and temperature of the water

iv) The amount of detergent

v) The dirt in the water or rinse needs

vi) The speed and duration of the spin cycle
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 7

Consider 4 matrices Q, R, S and T with dimensions 13 × 12, 12 × 30, 30 × 15 and 15 × 18 respectively. What is the least number of scalar multiplications needed to find the product QRST using the basic matrix multiplication method?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 7

Given matrices Q, R, S, and T with dimensions 13 × 12, 12 × 30, 30 × 15, and 15 × 18 respectively. From the given matrices sequence of dimensions can be seen as:

<P0, P1, P2, P3, P4> = <13, 12, 30, 15, 18>

Let c[i, j] be the minimum number of scalar multiplications needed to compute the product QRST.

Using Dynamic Programming recursive definition of c[i, j] is given as-

In the given scenario i and j can be 1, 2 or 3, and j can be 2, 3 or 4.

c[1, 2] = min {4680} = 4680


Hence the minimum number of scalar multiplications needed to compute the product QRST is 11250 using (Q(RS))T).

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 8

Which of the following not correct according to input to output, the compiler phase that is processes by it?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 8

Concept:

Phases of the compiler are:

Diagram

The Correct Answer is option 4.

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 9

The correct statement regarding WiFi and Wi-Max technology is/are -

(I) WiFi uses radio waves to create wireless connection, WiMax uses spectrum to deliver the connection.

(II) WiFi is defined under IEEE 802.11x standards, while WiMax is defined under IEEE 803.16y standards

(III) WiMax covers a comparatively larger area than WiFi.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 9

The correct answer is option 2.

Concept:

Statement 1: WiFi uses radio waves to create wireless connections, WiMax uses spectrum to deliver the connection.

​The given statement is true.

  • Wifi is an abbreviation for Wireless Fidelity. WiFi creates wireless high-speed internet and network connections by utilizing radio waves. To build hotspots, a wireless adaptor is required.
  • WiMax is an abbreviation for Wireless Interoperability for Microwave Access. WiMax uses spectrum to deliver the connection to the network and handle a larger inter-operable network.​

Statement 2: WiFi is defined under IEEE 802.11x standards, while WiMax is defined under IEEE 803.16y standards.

​The given statement is true.

  • Wifi is specified by IEEE 802.11x standards, where x denotes different WiFi versions.
  • WiMax is specified by IEEE 802.16y standards, where y denotes different WiMax versions.

Statement 3: WiMax covers a comparatively larger area than WiFi.

​The given statement is true.

  • The WiFi network has a maximum range of 100 meters.
  • The WiMax network has a maximum range of 90 kilometers.

Hence the correct answer is (I), (II), and (III).

Note:

As per the official answer key, statement two is incorrect, but as per the IEEE standards statement, two is also correct. All three statements are correct hence the correct answer is option 2 i.e (I), (II), and (III).

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 10

Given below are several usages of the anchor tag in HTML.

(a) <A href=”http://w3c.org/test.html”>Test</A>

(b) <A href=”/HTML/test.html”> Test</A>

(c) <A href=”test.html”>Test</A>

(d) <A href=”test.html#top”>Test</A>

Which of the above are valid?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 10

Concept

Anchor tag <a> is used to define a hyperlink that links one page to another. Attribute that is used with anchor tag is href which indicates the destination of the link.

Syntax of the tag:

<a href = “www.google.com”>Page</a>

Sometimes if there is need to link to another page then we can use target attribute of anchor tag. With this, link can be opened in new page.

There are three appearances in the anchor tag. These are unvisited link, visited link, active link.

All the options are correct”

NOTE:
“ ” → is not compulsory

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 11
In a software project, COCOMO (Constructive Cost Model) is used to estimate
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 11

The Constructive Cost Model (COCOMO) is a procedural software cost estimation model developed by Barry W. Boehm.

The basic COCOMO model gives an approximate estimate of the project parameters. The basic COCOMO estimation model is given by the following expressions:

Effort = a1 х (KLOC) a2 PM

Tdev = b1 x (Effort)b2 Months

Where

  • KLOC is the estimated size of the software product expressed in Kilo Lines of Code,
  • a1, a2, b1, b2 are constants for each category of software products,
  • Tdev is the estimated time to develop the software, expressed in months,
  • An effort is a total effort required to develop the software product, expressed in person months (PMs)
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 12

Asynchronous Transfer Mode (ATM) is also known as-

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 12

ATM

It is a data-link layer protocol like Ethernet, aimed at wide area networks (WANs) as well as local area networks (LANs). Whereas Ethernet is really geared towards carrying only Internet Protocol (IP) traffic, ATM is designed to integrate both data and voice needs in one network. It is efficient for call relay and it transmits all information including multiple service types such as data, video or voice which is conveyed in small fixed size packets called cells. Cells are transmitted asynchronously and the network is connection oriented.

SMDS (Switched Multimegabit Data Service)

It is a public, packet-switched service aimed at enterprises that need to exchange large amounts of data with other enterprises over the wide-area network SMDS extends the performance and efficiencies of a company's local area network (LANs) over a wide area on a switched,

X.25

It is Packet Switched networks allow remote devices to communicate with each other over private digital links without the expense of individual leased lines. Packet Switching is a technique whereby the network routes individual packets of HDLC data between different destinations based on addressing within each packet. An X.25 network consists of a network of interconnected nodes to which user equipment can connect. The user end of the network is known as Data Terminal Equipment (DTE) and the carrier’s equipment is Data Circuit-terminating Equipment (DCE)

Frame Relay

It is a simplified form of Packet Switching, similar in principle to X.25, in which synchronous frames of data are routed to different destinations depending on header information. The biggest difference between Frame Relay and X.25 is that X.25 guarantees data integrity and network managed flow control at the cost of some network delays.

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 13
White box testing is used _______
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 13

White-box testing, sometimes called glass-box testing, is a test case design method that uses the control structure of the procedural design to derive test cases.

Using white-box testing methods, the software engineer can derive test cases that

(1) guarantee that all independent paths within a module have been exercised at least once

(2) exercise all logical decisions on their true and false sides

(3) execute all loops at their boundaries and within their operational bounds

(4) exercise internal data structures to ensure their validity
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 14
Which one of the following is not related to Normal Forms (Normalization) rule with regards to the Relational Model?
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 14

The correct answer is Fields that can contain non-numeric data are to be removed and placed within their own tables with an associated Primary key.

Key Points

  • Normalization rules and normal forms in the relational model do not discriminate between data types (numeric vs. non-numeric) in their requirements. Instead, normalization focuses on the structural aspects of the database to ensure redundancy is minimized and data integrity is maximized through the organization of fields, primary keys, and relationships between tables.
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 15
The smallest integer that can be represented by an 8 - bit number in 2's complement form is
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 15

Data:

number of bits = n = 8

Formula:

Range of 2's compliment number with n bits is (–2n-1 ) to +(2n-1 – 1)

Smallest integer = –2n-1

largest integer = 2n-1 – 1

Calculation:

Smallest integer = –28-1 = –128

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 16

What is the average waiting time of processes in this schedule? (Assume Pre-emptive priority scheduling policy, further assume lower number represents higher priority).

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 16

From the last question Schedule B is Pre-emptive priority schedule, so its Gantt chart is:

P1 wait for: 4-8 => 4

P2 wait for: 1-14 => 13

P3 wait for: 2-15 => 13

P4 wait for: 4-4 => 0

Average wait time = 30/4 = 7.5

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 17

The following Schedule is an example of which scheduling policy?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 17

It is clear that it is FCFS because the order of execution is the order of arrival.

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 18

What is the average turn around time of process P2? (Assume Pre-emptive priority scheduling policy, further assume lower number represents higher priority)

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 18

Correct answer is 14

Pre-emptive priority schedule, so its Gantt chart is:

TAT = CT - AT (Completion Time (CT), Arrival Time (AT))

TAT = 15 - 1 = 14

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 19
The alternate key for Student table will be ________.
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 19

The correct answer is StuAadhar

Key PointsAn alternate key in database terminology is a candidate key that is not chosen as the primary key but is still unique and can identify each record in a table. For the Student table, the ideal alternate key would be an attribute that uniquely identifies each student, aside from the primary key (which we may assume to be something like StuID if following common conventions).

  • StuName is generally not unique because different students can share the same name.
  • StuContact could be unique, as typically each student would have a unique contact number. However, this depends on the policy of the organization (e.g., whether parents' contact numbers are used for younger students).
  • StuAadhar is a highly likely candidate for uniqueness. The Aadhar number is a unique identifier issued by the UIDAI to residents of India, and it is meant to be a unique identifier for each individual.
  • StuClass would not be unique as many students belong to the same class.

Given these options, the most definitive and universally unique across all possible scenarios would be: StuAadhar
This assumes that StuAadhar refers to the Aadhar number, which is unique to every individual in India. As such, it would serve as an excellent alternate key, assuming that uniqueness and the ability to identify each record in the table are the primary criteria.

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 20
Which of the following SQL query will fetch ID of those issued books which have not been returned?
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 20

The correct answer is Select BookID from Stulibrary where Return_date is NULL;

Key PointsThis query fetches the BookID of books from the Stulibrary table where the Return_date is NULL, indicating that the books have been issued but have not yet been returned.

  • SELECT BookID: This retrieves only the `BookID` column, focusing on book identification.
  • FROM Stulibrary: This specifies the `Stulibrary` table as the source of data.
  • WHERE Return_date is NULL: This crucial part filters the results. A `NULL` value in the `Return_date` column indicates that the book has not been returned yet.
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 21
If p and q are assigned the values 2 and 3 respectively then the statement p = q++;
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 21

The correct answer is option 3.

Concept:

Pre-increment (++i) :

Before assigning the value to the variable, the value is incremented by one.

Post-increment (i++) :

After assigning the value to the variable, the value is incremented by one.

Explanation:

The given data,

p=2

q=3

p = q++;

Here post-increment for variable q. After assigning the value to the variable, the value is incremented by one. i.e(p has 3 and q has 4)

p=3

Now q=4

Hence the correct answer is to assign a value of 3 to p.

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 22
Which one of the following is not a software process model?
Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 22

Correct answer: Option 2

Explanation:

  • A software process model is a blueprint of the flow of processes that should be followed when developing a software. This model defines each task, the input and output of each task and the sequence in which the tasks should be carried out.
  • The linear sequential model, prototype model and spiral model are examples of a software process model. Each model has a different approach to the flow of processes for developing a software.
  • The COCOMO model, on the other hand, is a cost estimation model. It identifies the amount of labour required and the schedule to be followed.

Important Points:

  • The COCOMO model, short for COnstructive COst MOdel, is used to determine the monetary cost of a software project. The cost is directly related to the manpower required (effort) and how quickly it is required (schedule).
  • Either the schedule influences the effort or the effort influences the schedule. The project manager decides which of the two factors are more important and lets it be the deciding factor.
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 23

The amount of uncertainty in a system of the symbol is called ________________.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 23
The amount of uncertainty in a system of the symbol is called entropy. The entropy of a random variable is the average level of"information", "surprise", or "uncertainty" inherent in the variable's possible outcomes. The concept of information entropy was introduced by Claude Shannon in his 1948 paper "A Mathematical Theory of Communication", and is sometimes called Shannon entropy in his honour.
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 24

Agents behaviour can be best described by ____________.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 24
An agent’s behaviour is described by the agent function that maps any given percept sequence to an action, which can be implemented by agent program. The agent function is an abstract mathematical description; the agent program is a concrete implementation, running on the agent architecture.
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 25

What is the linker?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 25
The linker is a program in the compiler which is required to create a load module. Linkers are also called transitions or discourse markers. They help us establish our ideas explicitly. Linkers make it easy for us to compare, contrast, illustrate, define, and summarize our thoughts and develop coherent paragraphs.
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 26

The automaton which allows transformation to a new state without consuming any input symbols:

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 26

The automaton which allows transformation to a new state without consuming any input symbols is NFA-l. NFA-l or e-NFA is an extension of Non-deterministic Finite Automata which are usually called NFA with epsilon moves or lambda transitions. 

Nondeterministic finite automaton with ε-moves (NFA-ε) is a further generalization to NFA. This automaton replaces the transition function with the one that allows the empty string ε as a possible input. The transitions without consuming an input symbol are called ε-transitions.

UGC NET Paper 2 Computer Science Mock Test - 5 - Question 27

Which section of a synchronous modem contains the scrambler?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 27
Transmission section  of a synchronous modem contains the scrambler. A scrambler is a device that transposes or inverts signals or otherwise encodes a message at the sender's side to make the message unintelligible at a receiver not equipped with an appropriately set descrambling device.
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 28

The transmission signal coding method for T1, the carrier is called:

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 28
The transmission signal coding method of T1 carrier is called bipolar. Bipolar encoding is a type of return-to-zero line code, where two nonzero values are used, so that the three values are +, −, and zero. Bipolar signaling, also called bipolar transmission, is a Baseband method of sending binary data over wire or cable. There are two logic states, low and high, represented by the digits 0 and 1 respectively. The bandwidth of the signal is inversely proportional to the duration of each data bit.
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 29

Recursive languages are _____________.

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 29
Recursive languages are a proper superset of CFL. When we come to recursive language it always halt whether it is accepted by the machine or not. if it is accepted it reaches at (q accept) and halt. and if not accepted by the machine it directly reach (q halt). We know CFL is a proper subset to Recursive language, this means Recursive language is a proper superset to CFL.
UGC NET Paper 2 Computer Science Mock Test - 5 - Question 30

Which is considered as the sequence of characters in a token?

Detailed Solution for UGC NET Paper 2 Computer Science Mock Test - 5 - Question 30

A Lexeme is a sequence of alphanumeric characters in a token. The term is used in both the study of language and in the lexical analysis of computer program compilation. In the context of computer programming, lexemes are part of the input stream from which tokens are identified.

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