UGC NET Paper 2 Computer Science Mock Test - 8 - UGC NET MCQ

This function is automatically called when no catch handler can be found for a thrown exception, or for some other exceptional circumstance that makes impossible to continue the exception handling process.

Dependency Preserving Decomposition:

Decomposition of R into R₁ and R₂ is a dependency preserving decomposition if closure of functional dependencies after decomposition is same as closure of of FDs before decomposition.

A simple way is to just check whether we can derive all the original FDs from the FDs present after decomposition.

In the above question R(A, B, C, D) is decomposed into R₁ (A, B) and R₂(C, D) and there are only two FDs A -> B and C -> D. So, the decomposition is dependency preserving

Lossless-Join Decomposition:

Decomposition of R into R₁ and R₂ is a lossless-join decomposition if at least one of the following functional dependencies are in F+ (Closure of functional dependencies)

R₁ ∩ R₂ → R₁

OR

R₁ ∩ R₂ → R₂

In the above question R(A, B, C, D) is decomposed into R₁(A, B) and R₂(C, D), and R₁ ∩ R₂ is empty. So, the decomposition is not lossless.

The correct answer is option 1.

Key Points

• The degree to which the elements within a module belong together is referred to as cohesion.
• The degree of interdependence between software modules is referred to as coupling. A measure of how closely two routines or modules are linked.

Coupling between M1 and M2 =

The cohesion of M1 =

After moving method m to M2, we get the following graph as

Hence the correct answer is There is no change.

The correct answer is option 4.

Concept:

Statement 1: Depth First Search uses the stack data structure.

True, When a dead end occurs in any iteration, the Depth First Search (DFS) method traverses a graph in a depth ward motion and utilizes a stack to remember to acquire the next vertex to start a search.

Statement 2: Breadth First Search uses the queue data structure.

True, When a dead end occurs during any iteration, the Breadth First Search (BFS) algorithm uses a queue to keep track of where to get the next vertex to begin a search.

Statement 3: Depth First Search uses a backtracking technique.

True, Depth First Search uses a backtracking technique. Backtracking is a method for problem-solving based on an algorithm. It constructs a solution step by step, increasing values over time, and then employs recursive calling to discover the answer. Based on the limitations provided to answer the problem, it eliminates solutions that don't lead to the solution.

Statement 4: Breadth-first search uses the Branch-and-Bound traverse technique.

True, Breadth-first search uses the Branch-and-Bound traverse technique. Branch and bound is an algorithm design paradigm that is generally used for solving combinatorial optimization problems.

Hence the correct answer is only 1, 2, 3, and 4.

Three address code: In this a type of intermediate code which is easy to generate and can be easily converted to machine code uses at most three addresses and one operator to represent an expression and the value computed at each instruction is stored in temporary variable generated by compiler.

Calculation:

The correct answer is option 4.

Key Points

S1: Regular languages are a subset of the set of languages accepted by TMs which do not write anything on the tape.

True, Regular languages are a subset of the set of languages accepted by TMs Which do not write anything on the tape. The subset of regular languages is not closed under regular grammar.

S2: The set of languages accepted by halting TMs with a bidirectional infinite tape is a proper superset of decidable language.

False, The set of languages accepted by halting TMs with a bidirectional infinite tape is a proper superset of decidable language. Halting Turing machine accepts recursive languages. Proper superset of recursive languages is an undecidable language.

S3: Every decidable language can be accepted by a DFA with a priority queue.

True, Every decidable language can be accepted by a DFA with a priority queue.

Essential prime implicant means implicant which can not be replaced by any another implicant.

Key Points

In RSA encryption, the ciphertext (C) is computed using the public key (e, n) and the original message (M) as follows:

Here, e is the public exponent, and n is the product of the two prime numbers (p and q) used to generate the keys.

Given that the participant's public key is e = 11, and the product of the two prime numbers n = pq = 5 times 13 = 65, and the ciphertext C = 8, we can set up the equation:

Now, we need to find the value of M.

To solve this equation, we can use the Chinese Remainder Theorem or simply try different values of M until we find one that satisfies the congruence.

Let's check each option:

1)
2)
3)
4)

Therefore, the correct answer is 4) 57.

Here, longest common subsequence is PQPS.

The correct answer is 2 KB.

The steps to find the page table size are:

Euler formula:-

It is written F + V - E = 2, where F is the number of faces, V the number of vertices, and E the number of edges.

This formula is true for sphere as well.

If we create one latitude and one longitude on a sphere. We'll get 2 vertices, 4 edges and 4 faces.

Here we can take any number of longitude and latitudes, it will satisfy the euler formula.

Concept Used:

P(A) represents the Powerset of A i.e., the set of all subsets of A.

|P(A)| = 2^|A|

A - B = {x| x ∈ A and x ∉ B}

Calculation:

|A| = 50, |A ∩ B| = 45 and |B| = 48

⇒ A and B have 45 elements in common.

Hence, A - B must have only 5 elements i.e., |A - B| = 5

Hence, |P(A − B)| = 2^{|(A − B)|} = 2⁵

The correct answer is 4 - 20 mA

Concept:

HART Protocol

There are several reasons for using HART Communication to increase data flow between hosts and field devices. These include reconfiguration, diagnostic and troubleshooting instruments, reading the additional measurements provided by the device, and much more.

• It can provide many benefits, including playing a key role in improving plant performance, increasing asset availability, reducing maintenance costs.
• The HART Protocol provides two simultaneous communication channels on the same wire: 4-20mA “current loop” analog and a HART digital signal.
• The analog signal continues to provide primary values to and from field instruments, the digital signal provides additional device information.
• This is originally superimposed a digital communication signal “on top of” the 4-20mA current loop to bring Caller ID technology to the field of telephony.

Communication data

There are many ways to integrate HART data and leverage intelligence in your smart field devices.

• Point-to-Point Integration is the most common method, enabling users to interrogate a device using a remote host by connecting anywhere on the current loop.
• The next step is HART-to-Analog Integration, which replaces analog control components with HART protocol signal extractors to convert real-time instrument data 4-20 mA signals for input into an existing analog control system.
• Next, HART-plus – Analog Integration employs HART technology multiplexers to replace existing I/O termination panels.

Conclusion:

Option 3 is correct.

A)182.25.110.68 -> 10110110.00011001.1101110.1000100

For 182.20.0.0/ 13 MASK is (first 13 bits are 1 and remaining are 0) 11111111 11111000 00000000 00000000

Bitwise AND result -> 10110110.00011000.0.0 -> 182.24.0.0

B) 182.25.110.68 -> 10110110.00011001.1101110.1000100

For 182.22.0.0/ 12 MASK is (first 12 bits are 1 and remaining are 0) 11111111 11110000 00000000 00000000

Bitwise AND result -> 10110110.00010000.0.0 -> 182.16.0.0

C) 182.25.110.68 -> 10110110.00011001.1101110.1000100

For 182.24.0.0/ 14 MASK is (first 14 bits are 1 and remaining are 0) 11111111 11111100 00000000 00000000

Bitwise AND result -> 10110110.00011000.0.0 -> 182.24.0.0

Longest prefix matching will be used to decide among interface 2 and 4.

The interface 2 has the longest matching prefix with the input IP address.

Calculation:

(A ∩ B' ∩ C')' = A' U B U C

∴ (A U B U C) ∩ (A ∩ B' ∩ C')'

= (A U B U C) ∩ (A' U B U C)

= (A ∩ A') U (B U C) [By distributivity of U over ∩]

= ϕ U (B U C) = B U C

Hence,

(A U B U C) ∩ (A ∩ B' ∩ C')' U C'

⇒ (B U C) ∩ C' = (B ∩ C') U (C ∩ C') = B ∩ C'

∴ Correct answer is option 1.

Statement 1: Greedy technique solves the problem correctly and always provides an optimized solution to the problem.

This Statement is False. Since the Greedy technique does not always solve a problem correctly.

Statement 2: Bellman ford, Floyd-warshal, and Prim’s algorithms use the Dynamic Programming technique to solve the Path problems.

This Statement is also False Since Prim's algorithm does not use the Dynamic Programming technique.

Prim's algorithm is a greedy algorithm that finds a minimum spanning tree for a weighted undirected graph.

Key Points

Bit plane slicing is a method of representing an image with one or more bits of the byte used for each pixel. One can use only MSB to represent the pixel, which reduces the original gray level to a binary image. The three main goals of bit plane slicing are:

• Converting a gray level image to a binary image.
• Representing an image with fewer bits and corresponding the image to a smaller size.
• Enhancing the image by focussing.

The given image is,

And each pixel is represented with 4-bits means 4-bit plans come,

So,

• 0 can represent 0000,
• 1 can represent 0001,
• 6 can represent 0110,
• 7 can represent 0111,
• 2 can represent 0010,
• 4 can represent 0100,
• 6 can represent 0110,
• 5 can represent 0101,
• 3 can represent 0011.

MSB plan 0 =

plan 1 =

plan 2 =

plan 3 =

Option 4 is not bit plan among 4-bit plan because the third row can't be 1 1 1.

Hence the correct answer is

for a)

for b)

So, 2 3 5 4 1

for c)

so, 4 3 5 2 1

for d)

So, the permutation 3, 4, 1, 5, 2 is invalid

A procedural abstraction is a named sequence of instructions that has a specific and limited function.

A data abstraction is a named collection of data that describes a data object.

Concept:

Fork system call is used to create a new process also called as child processes.

Formula:

With n fork system calls, number of child processes created are 2ⁿ – 1.

Printf statement will be executed 2ⁿ times with n system calls.

Explanation:

Here there are 3 fork system calls in the given program.

So, 2³ times printf statement will be executed.

Given:

abelian groups of order 108

Concept:

the number of abelian groups of order n is p(n) ( no of partition of n )

Calculation:

108 = 2²⋅3³.

Now, p(2) = 2 because 2 = 2 and 2 = 1 + 1,

whereas p(3) = 3 because 3 = 3, 3 = 1 + 2, and 3 = 1 + 1 + 1.

Hence, the number of abelian groups of order 108 up to isomorphism is p(2) × p(3) = 2 × 3 = 6.

Hence the option (3) is correct.

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

Where, N(0) = 1 and N(1) = 2

So,

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.

Therefore, to find the minimum number of nodes of an AVL tree of height 5, the values of N(2), N(3) and N(4) must be found. So, the values of 3 nodes have to be found.

Push 6 into the stack, Push 2 into the stack, Push 3 into the stack.

Now we encounter an operator. So, pop top 2 elements and apply the operator between them, it should be like:

( second_pop operator first_pop ) ==> 2 * 3 but not 3 * 2

result is 6 ==> push into the stack.

Now we encounter an operator. So, pop top 2 elements and apply the operator between them, it should be like:

( second_pop operator first_pop ) ==> 6 / 6

result is 1 ==> push into the stack.

Push 4 into the stack, Push 2 into the stack.

Now we encounter an operator. So, pop top 2 elements and apply the operator between them, it should be like:

( second_pop operator first_pop ) ==> 4 * 2

result is 8 ==> push into the stack.

question is asking about till evaluate up to 2nd * evaluated, ==> our task complete.

The top of the two elements of our stacks is 8,1 in the order from top to bottom.

The best way to solve game playing problem is Heuristic approach. You use a Heuristic approach, as it will find out brute force computation, looking at hundreds of thousands of positions. e.g Chess competition between Human and AI-based Computers. A heuristic, or a heuristic technique, is an approach to problem-solving that uses a practical method or various shortcuts in order to produce solutions that may not be optimal but are sufficient given a limited timeframe or deadline.

A language is regular if and only if it can be accepted by a finite automaton. Secondly, It supports no concept of auxiliary memory as it loses the data as soon as the device is shut down.

Hasse diagrams can be described as the transitive reduction as an abstract directed acyclic graph. This graph drawing techniques are constructed by Helmut Hasse(1948). The diagrams are named after Helmut Hasse (1898–1979); according to Garrett Birkhoff (1948), they are so-called because of the effective use Hasse made of them.