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 (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":

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).

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.

Calculation:

The number of input lines in a available multiplexer is 2ⁿ.

The number of input lines in a desired multiplexer is 2^N.

The number of individual multiplexer required to construct the desired multiplexers circuit is equal to the ratio of The number of input lines in a desired multiplexer to the number of input lines in a available multiplexer.

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.

Correct answer is <a href="http://google.com”>MCQ Questions</a>

Key Points
In computer science, a syntax error is an error in the syntax of a sequence of characters or tokens that is intended to be written in a particular programming language. These are detected by compiler. A program will not compile until all syntax errors are corrected.

Concept:

Topological sort:

• Topological sort order will be A B because here A is not dependent on any vertex but B is dependent on A so A will resolved first then remove vertex A, see next vertex which is not dependent on any other node or say which is dependent on less number of nodes that node or vertex will resolved next.

Explanation:

• h and c are not dependent on any other nodes but other are dependent on at least one vertex so topological sort will start from either h or c.
• If we select h then next will be c because c is not dependent of any other node, option 4 is false because e is dependent on d.
• If we select c then next we can select among (a,d,h) here a is selected then d is selected but next selected e is dependent upon h so option 3 is false.
• If we select h then next we can select only c after selected c next select among (d,a) here a is selected after a here e is selected but we can’t select e because e is dependent on d so option 1 is false.
• Option 2 is the correct answer.

LRU page replacement algorithm:

Hit miss table:

∴ number of miss = page faults = 8

Important Points:

M is miss which corresponds to a page fault

H is hit which corresponds to page hit (no page fault)

Following are the various examples of row matrix:

1. [6 5 4]: The order of this matrix 1 × 3, i.e., 1 row and three columns.

2. [0]: The order of this matrix is 1 × 1, i.e., 1 row and 1 column.

3. [1 2 0 6 8 9]: The order of this matrix is 1 × 6, i.e., 1 row and 6 column.

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.

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.

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.

In the Least Recently Used (LRU) page replacement policy, the page that is used least recently will be replaced. Add a register to every page frame - contain the last time that the page in that frame was accessed. Use a "logical clock" that advances by 1 tick each time a memory reference is made. The page that has not been used for the longest time in the main memory will be selected for replacement.

The extended notation would be (0+1)⁴ but however, we may allow some or all the factors to be ε. Thus ε needs to be included in the given regular expression.

Expression (E)= Length 4 or less means, length 0 or length (1) or length (2) or length (3) or length 4.