UGC NET Paper 2 Computer Science Mock Test - 10 - UGC NET MCQ

70% tlb hit ratio 30 ns to search tlb

(0.7 ×30) in hit case: to find the page

0.3×(90+30) in miss case: to find the page

90 ns to access the data from main memory

so total time:

(0.7 × 30)+ (0.3 × (90+30)) + 90 = 147

P ↔ Q

(P → Q) ∧ (Q → P)

(~ P ∨ Q) ∧ (~ Q ∨ P)

B is answer.

Correct answer is (D). 

Hardware interrupts: The 8085 microprocessor has five hardware interrupt inputs (five pins). They are TRAP, RST 7.5, RST 6.5, RST 5.5 , and INTR in the decreasing order of priority. If two or more interrupts go high at the same time, the 8085 will service them on priority basis.

• This is the original heap as per the array.
• Swap 300 and 215.
• Swap 300 and 250.
• Swap 300 and 289.
• A total of three swaps are required to convert it to max-heap.

The correct answer is option 3.

Key Points

Option 1: abaabaaabaa

The combination of {ab,aa, baa} forms abaabaaabaa. And this is correct combination.

"ab aa baa ab aa"

Option 2: aaaabaaaa

The combination of {ab,baa} forms aaaabaaaa. And this is correct combination.

“aa aa baa aa”

Option 3: baaaaabaaaab

This is not a correct combination of {ab,aa,baa}. Ending with 'b' can not get from given strings.

"baa aa ab aa aab"

Option 4: baaaaabaa

The combination of {ab,aa, baa} forms baaaaabaa. And this is the correct combination.

"baa aa ab aa"

Hence the correct answer is 1, 2 and 4.

NOTE THAT:

TRY TO DRAW AN NFA AND THEN OBTAIN REGULAR EXPRESSION.

The correct answer is option 1.

Concept:

Statement A: The union of two regular expressions is also a regular expression.

True, The union of two regular expressions is closed it means the union of two regular expressions is gives a regular expression.

Regular expression 1 ∪ Regular expression 2 = Regular expression.

Example:

RE1 = a(aa)^*

RE2 = (aa)^*
So, L1 = {a, aaa, aaaaa,.....}
and L2 ={ ε, aa, aaaa, aaaaaa,.......}
L1 ∪ L2 = { ε, a, aa, aaa, aaaa, aaaaa, aaaaaa,.......}
RE = (L1 ∪ L2) = a* which is a regular expression.

Statement B: The concatenation of two regular expressions is also a regular expression.

True, The concatenation of two regular expressions is closed which means the concatenation of two regular expressions gives a regular expression.

Regular expression 1 . Regular expression 2 = Regular expression.

RE1 = (0+1)*0

RE2 = 01(0+1)*
L1 = {0, 00, 10, 000, 010, ......}
L2 = {01, 010,011,.....}
Then, L1.L2 = {001,0010,0011,0001,00010,00011,1001,10010,.............}

L1.L2 =(0+1)*0 . 01(0+1)*

which is a regular expression.

Statement C: The iteration of a regular expression is also a regular expression.

False, Iteration is the process of repeating a process in order to generate a series of outcomes. Each iteration of the process is a single repetition of the process, and the outcome of each iteration serves as the starting point for the next iteration.

So iteration can perform many operations like Union, intersection, Subset, prefix, suffix, and many more. So a subset of any regular expressions is not closed under regular grammar. Hence iteration is not closed under regular expressions.

Example:

(a+b)^* are regular expressions it accepts all strings.

(aⁿbⁿ) is subset of above regular expressions and it is not a regular expression.

Hence the correct answer is Only C.

Additional Information
Close Properties:

The correct answer is 0.

Stack-based CPU organization uses zero address instruction.

Key Points

• The computers which use Stack-based CPU Organization are based on a data structure called a stack.
• It makes use of the Last In First Out (LIFO) access technique, which is the most common in most CPUs.
• The address of the highest member of the stack, known as the Stack pointer, is stored in a register (SP).
• Push and Pop are the two most common operations done on the stack's operators. These two operations are only carried out from one end.
• This instruction contains the opcode only with no address field. It pops the two top data from the stack, subtracting the data, and pushing the result into the stack at the top.
• PDP-11, Intel’s 8085, and HP 3000 are some examples of stack-organized computers.

The advantages of Stack-based CPU organization –

1. Efficient computation of complex arithmetic expressions.
2. Execution of instructions is fast because operand data are stored in consecutive memory locations.
3. The length of instruction is short as they do not have an address field.

The disadvantages of Stack-based CPU organization –

• The size of the program increases.

Z-buffer:

• Z-buffer is generally used for hidden surface detection. It is also known as the Depth-buffer method.
• The time complexity is the number of pixels times the number of objects and the space complexity is two times the number of pixels because two arrays of pixels are required, one for the frame buffer and the other for the depth buffer.
• These two buffers are used to store depth values and intensity of color respectively.

CALCULATION:

Memory required = 2 × 512 × 512 × 24 = 2 × 2⁹ × 2⁹ × 8 × 3 bits

= 2 × 2 × 3 × 2²⁰ bits = 12 M bits = 1.5 MB

Since 1 M = 2²⁰ and 1 byte = 8 bits

Concept:

Source Port number, Destination Port number and Checksum is present in both TCP header and UDP header.

Diagram:

Important Point:

Header length in TCP header and Length in UDP header are different.

Explanation:

objective function z = 5x + 2y under the linear constraints

x ≥ 0, y ≥ 0, x ≥ y and 2 ≤ x + y ≤ 4

So, LPP is

Maximize z = 5x + 2y

subject to

x + y ≥ 2

x + y ≤ 4

x ≥ 0, y ≥ 0

Feasible region is ABCDA whose corner points are

A(2, 0), B(4, 0), C(0, 4), D(0, 2)

z = 5x + 2y

Z(2, 0) = 10, Z(4, 0) = 20, Z(0, 4) = 8, Z(0, 2) = 4

So, max z = 20 at (4, 0)

Hence option (2) is correct

Concept

Recursion tree of the given function fib(n) will contain n levels hence will contain almost O(2ⁿ) function points thus will have O(2ⁿ) time complexity, assuming each function point takes constant time.

Example: fib(5)

Avoid overlapping sub-problems to solve this problem in O(n) time.

Use a buffer (array) to store the recently solved sub-problem and check every time we encounter a new problem.

Code using dynamic programming:

int fib[n];

fib[0] = 0; fib[0] = 1;

for(int i =2; i < n; i++)

fib[i] = fib[i-1] + fib[i-2];

Time complexity is O(n)

Fuzzy logic:

• Fuzzy logic is a concept of certain degree. Boolean logic is a subset of fuzzy logic.
• Fuzzy logic is a form of many-valued logic which deals with reasoning that is approximate rather than fixed and exact.
• Compared to traditional binary sets (where variables may take on true or false values), fuzzy logic variables may have a truth value that ranges in degree between 0 and 1.
• It is employed to handle the concept of partial truth, where the truth value may range between completely true and completely false.

In an "m-ary" tree. the number of total nodes (N) is given by

N=mi + 1 ----(1)

Where,

i: Number of internal nodes

Also, in a tree, N=i + L ----(2)

Where,

L=number of leaf nodes

Here m=2

From equation (1) and equation (2);

N = 2i + 1

2i + 1 = i + L

L = i + 1

The number of leaves are 1 plus the number of internal nodes in binary tree.

Here, given L=n, substitute above and we will get,

i = L - 1

Data:

number of pages = 128

size of page/frame = 1024 word

number of frames = 64

Calculation:

size of logical address = number of pages × size of page

size of logical address = 128 × 1024 words = 2¹⁷ words

number of bits = log₂217 = 17 bits

size of physical address = number of frames × size of a frame

size of physical address = 64 × 1024 words = 2¹⁶ word

number of bits = log2216 = 16 bits

Effective Memory Access Time = Cache hit ratio*Cache access time + Cache miss*( Cache miss penalty + Memory access time)

= 0.8*(30) + (1 - 0.8)*(30 + 150) ns

= 24 + 0.2*(180) ns = 60 ns

• Dijkstra’s Shortest Path uses the greedy method to find the shortest path of a graph G(V, E).
• Floyd Warshall algorithm is based on the principle of dynamic programming. It is used to solve the all pair shortest path problem.
• Kruskal algorithm is a minimum spanning tree algorithm in which in every iteration, minimum weighted edge is found and then it is added to the construction of minimum spanning tree. Edges are added in increasing order of the edge weights. That’s why it is a greedy approach.
• Merge sort is an efficient sorting algorithm that uses a divide-and-conquer approach to order elements in an array.

Programming Languages which are designed and developed by Microsoft are:

• C#.NET
• VB.NET
• C++.NET
• J#.NET
• F#.NET
• JSCRIPT.NET
• WINDOWS POWERSHELL
• IRON RUBY
• IRON PYTHON
• C OMEGA
• ASML(Abstract State Machine Language)

• Python is an interpreted, high-level, general-purpose programming language.
• Python can be used on a server to create web applications.
• Python lets user work quickly and integrate systems more efficiently.

Therefore, python is treated as a back-end language.

DML refers to the tools used to add, update and access the data within a database, including things like artists, albums, and so on in our music database example. DML is a categorization of existing SQL commands. DML commands include SELECT, INSERT, UPDATE, and DELETE.

setLineWidthScaleFactor (lw) function can be used to set line-width attribute. Line-width parameter (lw) is assigned a positive number to indicate the relative width of the line to be displayed. A value of 1 specifies a .standard-width line. On n pen plotter, for instance, a user could set lw to a value of 0.5 to plot a line whose width is half that of the standard line. Values greater than 1 produce lines thicker than the standard.

O(bm) is the space complexity of Depth-first search. Where b is the branching factor and m is the maximum depth of the search tree. For DFS, which goes along a single 'branch' all the way down and uses a stack implementation, the height of the tree matters. DFS algorithm needs to store only a single path from the root node, so, the space complexity of DFS is equivalent to the size of the fringe set, which is O(bm).

In a linear queue, there are chances of wastage of memory because if the rear is pointing to the last element whereas the front is pointing to the element other than the first element; it means that spaces allocated before the front are free, but it cannot be reused as rear cannot be incremented. In contrast, the last element is connected to the first element in a circular queue; if initial spaces are vacant, then rear can be incremented by using the statement (rear+1) mod max where max is the size of the array. Therefore, we conclude that the circular queue avoids wastage of memory.

Configuration management policies and processes define how to record, and process proposed system changes, how to decide what system components to change, how to manage different versions of the system and its components, and how to distribute changes to customers.

So, overflow can be detected by checking Most Significant Bit(MSB) of two operands and answer. But Instead of using 3-bit Comparator Overflow can also be detected using 2 Bit Comparator just by checking Carry-in(C-in) and Carry-Out(C-out) from MSB’s. Consider N-Bit Addition of 2’s Complement number.

MSB

C-out

Overflow Occurs when C-in is not equal to C-out. The above expression for overflow can be explained from the below Analysis.

In the first Figure, the MSB of the two numbers are 0 which means they are positive. Here if C-in is 1 we get the answer’s MSB as 1 means the answer is negative (Overflow) and C-out as 0. C-in is not equal to C-out So, overflow.

In the second Figure, the MSB of two numbers are 1 which means they are negative. Here if C-in is 0 we get the answer MSB as 0 means the answer is positive(Overflow) and C-out as 1. C-in is not equal to C-out. So, overflow.

Readers can also try out other combinations of C-in C-out and MSB’s to check overflow.

So, Carry-in and Carry-out at MSB’s are enough to detect Overflow.

Above XOR Gate can be used to detect overflow.

We can check the attribute values by stating the name of the attribute in the inquiry function. For example, the functions:

• inquirePolyLineIndex (last li)
• inquireInteriorColourIndex (last fc)

Copy the current values tor line index and fill color into parameters last li and last fc.

Depth-first search always expands the deepest leaf node in the current fringe of the search tree. DFS traverses the tree“deepest node first”, it would always pick the deeper branch until it reaches the solution (or it runs out of nodes, and goes to the next branch).

Path: S -> A -> B -> C -> G

ZooKeeper applications run on thousands of machines, and it performs best where reads are more common than writes, at ratios of around 10:1. ZooKeeper is replicated over a set of hosts (called an ensemble) and the servers are aware of each other. Zookeeper in Hadoop can be viewed as a centralized repository where distributed applications can put data and get data out of it. It is used to keep the distributed system functioning together as a single unit, using its synchronization, serialization and coordination goals.