UGC NET Paper 2 Computer Science Mock Test - 7 - UGC NET MCQ

Analog signal has a bit rate of 8000 bps and a baud rate of 1000. 

So, each signal will clearly carry bit rate / baud rate bits. i.e. 8000 / 1000 = 8 bits and 28= 256 signal. 

So, option (A) is correct.

LDAX – Load accumulator indirect( This instruction copies the contents of that memory location into the accumulator. )

LHLD – Load H and L register direct ( This instruction loads the contents of the 16- bit memory location into the H and L register pair. )

LXI – Load register pair immediate( The instruction loads 16-bit data in the register pair designated in the operand.)

INX – Increment register pair by 1.( It will increment the register value by 1.)

So, option (C) is correct.

(A) An instructor Id. No. cannot be null, provided Instructor Id No. being primary key. Correct by Codd’s rule 

(B) No two citizens have same Adhar-Id. Correct because Adhar is identification for citizens so it must be unique

(C) Budget of a company must be zero. We cant say or it is not necessarily true

So, option (C) is correct.

The correct answer is option 2.

Concept:

Order or SQL query evaluation:

1. From: Cartesian product of tables.
2. Where: Select the rows.
3. Group by: Divide the rows into groups.
4. Having: Select the groups.
5. Expressions: Expressions In select are evaluated.
6. Distinct: Distinct in select (Eliminate the duplicates).
7. Ser-Operations: (Union, intersect, Except)
8. Order By: Sort the rows.

The given data,

SQ1:

SELECT R.B,AVG(S.B)
FROM R, S
WHERE R.A=S.C AND S.D<7
GROUP BY R.B;

For 1st query it will select 6 tuples by joining R and S, where R.A and S.C are equal. Among that 5 are S.D<7

The relation R is,

The relation S is,

Output: Four tuples selected M=4.

SQ2:

SELECT DISTINCT S.B,MIN(S.C)
FROM S GROUP BY S.B
HAVING COUNT (DISTINCT S.D)>1;

Here we are selecting distinct value of S.D>1. In the table, we can see only one value is not distinct here. So, getting 5 tuples

The relation S is,

Output: Only two tuples selected N=2.

The correct answer is (I) and (II) are true

Concept:

Statement 1: A super key is an attribute or combination of attributes that uniquely identify records in an RDBMS table.

True, A super key, or just key, is a combination of every attribute that may be used to specifically identify rows (or tuples) in a database. This indicates that a superkey may contain additional information that isn't required for uniquely identifying table rows.

Statement 2: A candidate key is a subset of a super key.

True, A Candidate key is a subset of Super keys and is clear of any unnecessary attributes that are not essential for tuple identification. For all tuples, the Candidate key value is distinct and non-null. Additionally, each table must include at least one Candidate key.

Statement 3: All super keys are candidate keys but vice versa is not true.

False, Candidate keys are a subset of Super keys. They contain only those attributes which are required to uniquely identify tuples. All Candidate keys are Super keys. But the vice-versa is not true.

Hence the correct answer is (I) and (II) are true.

The correct answer is option 3.

Concept:

Hexadecimal Number:

Hexa stands for sixteen. It has 16 digits in the hexadecimal number system. It is made up of numbers and letters. It has 16 digits and includes the numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 as well as A, B, C, D, E, and F.

Binary Numbers:

A binary number is a number that only has the digits 0 and 1, and the data in this system is a mixture of 0s and 1s. Binary numbers are those that have only two digits.

Conversation from Hexadecimal to binary:

Every digit of Hexa-decimal can occupy the four digits in binary format. i.e If a digit 7 in Hexa decimal, then 0111 is in binary format

Explanation:

In the above table, a single-digit Hexa-decimal number can represent the maximum 4 digits and a minimum 1 digit is required in a binary number system. Hence the maximum and minimum number digits are required to represent the number X with n digits in a binary number system are 4n and 4n-3 respectively.

For instance: The F in the hexadecimal number required 4 bits in binary to be represented, and the 0 in hexadecimal required 1 bit in the binary to be represented.

So F is n and X is the number of binary digits. The F is only one hexadecimal number so n=1, therefore (4n) 4×1=4 so a maximum of 4 digits is required, and for 0 in Hexa-decimal, n=1, so (4n-3) 4×1-3=1, hence only 1 digit are required. 4 is the maximum digits in binary to represent Hexadecimal and 1 is the minimum for 1 hexadecimal number.

Hence the correct answer is 4n, 4n-3.

The precedence graph for the given schedule is:

It can be derived that:

• Read(X) in T3 is followed by Write(X) in T1.
• Read(Y) in T1 is followed by Write(Y) in T2.
• Write(X) in T3 is followed by Read(X) in T2.

Therefore, the sequence will be T3 → T1 → T2.

Block no. 3 i.e. 4th block.

Answer: Option 1

Explanation:

for Bottom-up parsing the Rules associated with a production executed at the time of reduction.

Option 4: 011332

This Option is not correct, since the 011332 can not be generated from the given Grammar.

Option 3: 033112

This option is also not correct, since the 033112 can not be generated from the given Grammar.

Option 2: 031312

This option is also not correct, since the 031312 can not be generated from the given Grammar.

Option 1: 013132

This Option is the correct answer Since 013132 can generate the output "cbbbba":

Answer: Option 2

Explanation:

Given that the PDA should accept the language { an|n ≥ 0} ∪ {anbn|n ≥ 0}

If we observe carefully the first state of PDA accepting the first part of the language given.

we just need to think about the transitions for accepting {anbn|n ≥ 0}.

In First State, all a's are already pushed to the stack and we need to start performing pop X operation for each 'b' we encounter in the string so transition A will be

(b, X/ϵ).

and for B, we have already popped all X (as per the number of a's) and if occurrences of 'b' are finished then we should go to the final state via epsilon transition.

Hence B = ( ϵ , Z / Z).

Option 2 is correct.

BFS:

Breadth first search in artificial intelligence is implemented as FIFO. Path is find out by considering the number of edges in that path. Start from the root node, it considers all the nodes. But at the end, selects the path having lowest number of edges. If b is the branching factor and d is the depth. Then overall cost of BFS is O(b^d).

DFS:

It is based on the concept of LIFO as that of stack procedure. It considers the nodes one by one from root node to the leaf as a single path. If it does not terminate it can go to the infinity. If b is the branching factor and m is the depth. Then cost of DFS is O(bm).

Depth – Limited Search

In this, depth is specified first, and all the nodes are traversed till that depth then it goes on to next level. If b is the branching factor. Then overall cost of depth limited search is O(bl).

Iterative – deepening search:

It starts from root node i.e. depth 1, expands all the nodes and then goes on to the depth 2, expand all the nodes of that depth. In this way, it expands all the nodes. If b is the branching factor and d is the depth, then overall cost is O(bd).

Index addressing mode:

In this, effective address of operand is obtained by adding the content of index register and address part of instruction.

Indirect addressing mode:

In this, address field specifies the memory location that contains effective address of operand.

Auto increment addressing mode:

In this addressing mode, effective address is the content of register and after accessing operand, content of register is incremented.

Auto decrement addressing mode:

In this addressing mode, the contents of a register specified in the instruction are first decremented and these contents are used as effective address of operand.

Therefore option 4 is correct.

Concept:

Probability of event to happen P(E) =

The product rule states that the probability of two (or more) independent events occurring together can be calculated by multiplying the individual probabilities of the event.

Calculation:

Total number of ways to select an answer for all 3 question = n(S) = 4 × 4 × 4 = 64

Total number of ways to select all 3 questions correct = 1

Total number of favorable events = n(E) = 64 - 1 = 63

Here, favorable event = selecting incorrect options.

Required probability = 63/64

∴ The probability that students will not answer all three questions correctly = 63/64.

Concept:

Data scrubbing or data cleansing deals with detecting and removing errors and inconsistencies from data in order to improve the quality of data.

Explanation:

When multiple data sources need to be integrated, the need for data cleaning increases significantly because sources often contain redundant data in different representations.

It includes ETL process:

ETL means extraction, transformation, loading in which data transformations deal with data translation and integration with filtering and aggregating data to be stored in data warehouse. Data cleaning is performed in separate staging area before loading the transformed data into warehouse.

The correct answer is option 3.

Concept:

Basic idea of syntax directed translation:
Syntax-Directed Constructing a parse tree or syntax tree and computing the values of attributes at the nodes of the tree by visiting them in some order is the definition of translation. In many circumstances, translation can be done without creating an explicit tree during parsing.

Evaluation Orders for syntax directed translation:

• Dependency Graphs.
• Ordering the Evaluation of Attributes.
• S-Attributed Definitions.
• L-Attributed Definitions.
• Semantic Rules

​

Note:

• Evaluation of the semantic rules may generate code, save information in a symbol table, issue error messages, or perform any other activities.
• Special cases of syntax-directed definitions can be implemented in a single pass by evaluating semantic rules during parsing, without explicitly constructing a parse tree or a graph showing dependencies between attributes.

Hence the correct answer is the Conceptual view of syntax-directed translation.

Network security audit is a information security audit. A network security audit should include review of following:

1) Firewalls: It should check for firewall configuration and vulnerabilities.

2) Antivirus software: Check that system is virus free. All systems should have updates antivirus software.

3) Password approach: Should check the password policies of company so that there will not be misuse of passwords.

4) Backups: There must be backing up of crucial data in organizations to overcome the loss in case of failure.

5) Active directory: All the information must be saved in a centralized directory. Directory must be managed on regular basis.

The correct answer is option 3.

Concept:

In compiler design, every phase takes inputs from its previous stage and feeds its output to the next phase of the compiler. There are 7 phases in a compiler.

• Lexical analysis
• Syntax analysis
• Semantic analysis
• Intermediate code generator
• Code independent optimizer
• Code dependent optimizer
• Code generator

Machine-dependent optimization is done after the target code has been generated and when the code is transformed according to the target machine architecture. It involves CPU registers and may have absolute memory references rather than relative references.

Machine-independent optimization attempts to improve the intermediate code to get a better target code. The part of the code which is transformed here does not involve any absolute memory location or any CPU registers.

Machine independent phases are

• Lexical analysis,
• Syntax analysis,
• Semantic analysis,
• Intermediate code generation and
• sometimes code optimization.

Hence the correct answer is A and C.

The internal representation of a string has a null character ‘\0’ at the end, so the physical storage required is one more than the number of characters written between the quotes.

Depth first search algorithm is used to traverse the vertices of a graph. It uses the concept of back tracking. Depth first search uses a data structure for its implementation which is known as stack.

Application of depth first search:

1) It is used to find the strongly connected components.

2) It helps in topological sorting.

3) to check the bipartiteness of a graph.

4) It helps in detecting cycles in a graph.

5) To find articulation points in a graph.

Example:

DFS traversal for this: 1 2 4 5 3 6 7

Min. number of nodes of height h, N(h) = N(h-1) + N(h-2) + 1

Here, 1 is added to include the root node.

N(0) = 1,

N(1) = 2,

N(2) = N(1) + N(0) + 1 = 2+1+1 = 4,

N(3) = N(2) + N(1) + 1 = 4 + 2 + 1 = 7,

N(4) = N(3) + N(2) + 1 = 7 + 4 + 1 = 12,

N(5) = N(4) + N(3) + 1 = 12 + 7 + 1 = 20.

Memory of a computer is organized for running program into three segments: the text segment, stack segment, heap segment.

Stack and heap is where storage is allocated for data storage. Stack is where memory is allocated for automatic variables within functions.

A dotted line can be displayed by generating very long dashes with spacing equal to or greater than dash size. DDA is a line drawing algorithm that is used to draw lines on the basis of the value of slop that which is equal to dy/dx. we are preparing one of the variations of the DDA algorithm where we are going to learn how to draw a dotted line using DDA. Algorithm for drawing dotted line:

Step 1:- First read the two ends point of line that is (x1,y1) and (x2,y2).

Step 2:- Find out the value of dy and dx by using following formula:

dy = y2-y1

dx = x2-x1

Step 3:- now we are going to compare the value of dx and dy in order to find out that m (slop) <=1 or m(slop) >1.

if (dx>dy)

Step = dx

else

Step =dy

Step 4:- Now we have to decide how much we have to increase in x-direction or in y-direction.

IncX = dx/Step

IncY  = dy/Step

Step 5:- Now we are going to plot first pixel of line.

x = x1

y = y1

Now we put pixel at (x,y)

Make i = 1;

Make m = 0;

Step 6:- While i is less than equal to Step.

Step 7:- Stop.

Users should be required to connect to a WAN via a VPN, including network devices that are connected to a WAN from a remote site. Additionally, SD-WAN has a key-exchange function which is used to authenticate devices on different endpoints. They can be private to connect parts of a business or they can be more public to connect smaller networks together.

Heap allocation is required for languages that support dynamic data structures.

In many situations, it is not possible to allocate data in a LIFO fashion with a stack. In this case, data is dynamically allocated from a pool of memory commonly called the heap. Heap allocation is considerably slower and more complex than stack allocation and may also result in allocations scattered all over memory. Such scattered allocations can lead to a loss of coherence and a reduction in memory access efficiency. A parallelized heap allocator should be used when writing parallel programs that use dynamic data structures.

Single ended Queue is not a type of the Queue. Queue has two ends in which one end is used for the insertion and another end is used for the deletion. Therefore, it is not possible for the Queue to have a single ended queue.

The type of queue:

• Linear Queue
• Input restricted Queue
• Double ended Queue
• Circular Queue
• Output restricted Queue
• Priority Queue

The algorithmic cost modeling utilizes the basic regression formula with such parameters that have been derived from the data of a historical project and current as well as characteristics of a future project. Algorithmic cost modelling uses a mathematical formula to predict project costs based on estimates of the project size, the number of software engineers, and other process and product factors.

The function setInteriorRepresentation (ws, fi, fs, pi, fc) can be used to set the bundled area-fill attributes. Subsequently defined fill areas are then displayed on each active workstation according to the table entry specified by the fill index parameter fi. Other fill-area attributes, such as pattern reference point and pattern size, are independent of the workstation designation and are set with the functions.