BEL Trainee Engineer CS Mock Test- 4 - Computer Science Engineering (CSE) MCQ

Routers are used to connect two or more segments of the network. They operate under the network layer of OSI model. They are capable of finding the best path to transfer the data. Thus, the services of the network layer are dependent on router technology.
The services of network layer to the transport layer are designed such that the network address which are made available to the transport layer should use a uniform numbering plan across local-area network as well as wide-area network.

• SRTF( Shortest Remaining Timing First)scheduling may cause starvation since the processes which require large CPU burst periods would have to wait.
• Generally preemptive scheduling doesn't cause starvation but in some cases, it may cause starvation when one process doesn't block due to I/O so others have to wait.
• In case of response time, it is quite obvious due to involvement of preemption, round robin is better than FCFS, .

Hence, all the statements are true.

According to the definition of the Algorithm,

• An algorithm is a sequence combination of finite steps to solve a particular problem. 
• It is a finite step-by-step procedure to achieve a required result.
• An algorithm always produces at least one output.
• It always terminates an infinite amount of time.
• An algorithm is independent of the programming language.

Hence, the correct option is (B).

Selection sort algorithm is used to sort the elements of an arrays by finding the maximum or minimum element of the array and then putting that element into its correct position to sort the array in ascending or descending order.
In the array given above, to sort it into an ascending array, the first step is to find the minimum element which is 11 and replace it with 64.
After the first step array becomes: 11 25 12 22 64
Hence, option 1 is correct.

P₁ decidable
P₂ undecidable
A decision problem is said to be decidable if there exists an effective algorithm that returns a correct yes or no answer to that problem.
A decision problem is said to be undecidable if there does not exist a single algorithm that always leads to a correct yes or no solution.
In terms of reducibility, P₂ ≤p P₃ denotes that P₂ is a decision problem that is reducible to P₃ in polynomial time p. This simply means that the instance of P₂ can be transformed into P₃ instance; and following the solution of P₃, we can get a solution for the problem P₂. Hence, if P₁ or P₂ is reducible to P₃, then P₃ also has the same properties as P₁ and P₂. So, if P₂ is reducible to P₃, then P₃ is undecidable.

Let A be the event in which an item has been produced by machine A and so on.

D = The event of the item being defective.

P(A) = 1/2

P(B) = 1/4

P(D/A) = P = 2/100 (An item is defective, given that A has produced it)

P(D/B) = 2/100

P(D/C) = 4/100

P(C) = 1/4

By theorem of total productivity,

Required probability = Chosen a defective bolt from Machine A+ Chosen a defective bolt from Machine B+ Chosen a defective bolt from Machine C.

P(D) = P(A) × P(D/A) + P(B) × P(D/B) + P(C) × P(D/C)

=0.025

Hence, the correct option is (B).

The simplified SOP (Sum of Product) form of the boolean expression by the two methods:

Method ( 1 ): By Minimization

Method ( 2 ): By Truth Table

Hence, the correct option is (B).

Binary Tree - Tree in which each node has at most two child nodes.
Binary search tree - A binary tree in which the left child contains only nodes with values less than the parent node, and the right child contains only nodes with values greater than or equal to the parent. Binary search tree is used for efficient searching. The binary search tree for the above problem is given below.

Given,

T(n) = 1 if n = 1
T(n) = T(n/2) + n

Now,

= n/2^k

= 1

As we know,

⇒ log n = k

= n(2) + 1

= 2n

∴T(n) = O(n)

Hence, the correct option is (C).

There are various routes possible from P to U.

First route from P to U is:

1. P → S → U: COST = 4 + 2 = 6

2. P → R → U:  COST = 1 + 2 = 3

3. P → R → S → U: COST = 1 + 1 + 2 =  4

4. P → R → T → U: COST = 1+ 8 + 3 = 12

5. P → Q → T → U: COST = 3 + 6 + 3 = 12

6. P → Q → R → U: COST = 3 + 7 + 2 = 12

7. P → Q → R → T → U: COST = 3 + 7 + 8 + 3 = 21

So, minimum cost is 3. Shortest route from P to U is 3.

Hence, the correct option is (A).

It is given that, each 5 computer needs directly connected with each router. 

Let c1 to c5 be the 5 computer needs directly connected with each router r1 to r5. Now,

c1 → r1, r2, r3, r4, r5

c2 → r1, r2, r3, r4, r5

c3 → r1, r2, r3, r4, r5

c4 → r1, r2, r3, r4, r5

c5 → 1, r2, r3,r 4, r5

c6 → 1

c7 → r2

c8 → r3

c9 → r4

c10 → r5

So, 25 connections + now remaining 5 computer, each connected to 5 different routers, so 5 connections =30 connections.

Now, any pick of 5 computers will have a direct connection to all the 5 routers.

Hence, the correct option is (C).

A Byte of data is read into the DMA Controller's buffer (During this time, the CPU has control of the bus, so we don't take this time into consideration).

Take the bus from CPU = 300 ns

Transfer the Byte = 750 ns

Give the control back to the processor = 300 ns

The block size = 160 bytes

Data transfer rate = 32 KB/s

Now,

So the total time = Time to start DMA + time to end DMA.

The time required for initiate,

= 300 + 300

= 600 ns

Since the device is slow the DMA speed is governed by DMA speed.

Time taken by the device is,

= 5 ms

Total time required:

= 750 ns + 5 ms

Total time required:

= 750 × 10⁻⁶ + 5

= 5 ms

Hence, the correct option is (A).

We are implementing a stack using a priority queue. Stack implementation is always Last In First Out (LIFO) order.
As given, POP is implemented as DELETEMIN (Q), that means the stack returns minimum element always.
So, we need to implement PUSH (C) using INSERT (Q, C, K), where K is the key chosen from a strictly-decreasing order (only this order will ensure that the stack will return minimum element when we POP an element). It will satisfy the Last In First Out (LIFO) property of stack. That is, answer option (1) is true.
Option (4) (non-increasing order) cannot be true because two same (identical) numbers cannot have same priority as priority should be distinguishable for each number.

Finite Automata(FA):

The first phase of the compiler is called lexical analysis or scanning. The lexical analyzer reads the stream of characters making up the source program and groups the characters into a meaningful sequence called lexemes.

For each lexeme, the lexical analyzer produces tokens as output for the parser and tokens are expressed in regular expressions.

So, a simple finite automaton is sufficient for it.

Phases in a compiler:

Hence, the correct option is (A).

Let n = 2^k

The meaning of word Appreciation (Noun) is: Thankfulness. It’s antonym should be: Aspersions, e.g, to criticize harshly.

One word substitute is Imposter.

Imposter: a person who pretends to be someone else in order to deceive others, especially for fraudulent gain.
Liar: a person who tells lies.
Rogue: a dishonest or unprincipled man.
Magician: a person with magical powers.

Suffrage means "the right to vote in political elections"

She also was involved in the national temperance and women's suffrage movements
The system of universal suffrage did not exist in the United Kingdom until 1928

The problem of contamination of water is being faced by the people. The author says that there is poor quality of water.

Reciprocate: To behave or feel towards somebody in the same way as we behave or feel towards you ; give in return

Sagacity means acuteness of mental discernment and soundness of judgment

The King, however, had the sagacity to secure a preemption-right as early as 1620
Abdur Rahman's attitude at this critical juncture is a good example of his political sagacity

User-level threads

User-level threads are very effective because a mode switch is not required to switch from one thread to another. Here is a detailed explanation:

• Definition: User-level threads are managed by the application without the intervention of the operating system. They are created and scheduled by the application itself.


• Advantages: User-level threads offer several advantages, including:




• No mode switch: User-level threads do not require a mode switch from user mode to kernel mode when switching between threads. This makes the context switch faster and more efficient.


• Thread management control: The application has full control over thread management, including scheduling, synchronization, and resource allocation.


• Portability: User-level threads can be easily ported across different operating systems, as they are independent of the underlying kernel.




• Disadvantages: User-level threads also have some limitations:




• Blocking system calls: If a user-level thread performs a blocking system call, it will block the entire process, as the kernel is unaware of the individual threads.


• Limited parallelism: User-level threads may not fully utilize the available hardware resources, as the operating system schedules threads at the process level rather than at the thread level.

In conclusion, user-level threads are very effective because they eliminate the need for a mode switch and provide greater control over thread management. However, they also have some limitations that need to be considered in certain scenarios.

The correct answer is c​alculate active/selected worksheets in the active workbook.

Key Points

• Shift+F9 calculates the active worksheet.
• Ctrl+Alt+F9 calculates all worksheets in all open workbooks, regardless of whether they have changed since the last calculation.

Additional Information

Answer:
The time taken by internet packets can vary depending on various factors. Let's explore these factors in detail:
1. Network Congestion:
- Internet traffic and congestion can affect the time taken by packets to reach their destination.
- During peak hours or in heavily congested networks, packets may experience delays.
2. Packet Size:
- The size of the packet can also impact the time taken.
- Larger packets may take longer to transmit compared to smaller packets.
3. Routing:
- The route the packets take from the source to the destination can also affect the time taken.
- Different routing paths may have varying latencies, causing differences in packet delivery time.
4. Network Quality:
- The quality and stability of the network infrastructure can impact packet delivery time.
- Unstable or unreliable networks may result in packet loss or increased latency.
5. Packet Priority:
- Certain packets, such as real-time audio or video packets, may be given higher priority and transmitted faster.
- However, this does not mean that the time taken by packets is irrelevant for audio packets.
In conclusion, the time taken by internet packets can vary based on network congestion, packet size, routing, network quality, and packet priority. Therefore, option B is the correct answer, while option C is incorrect as the time taken is relevant for all types of packets, including audio packets.

Answer:
To play and hear sound on a computer, one needs:
A sound card:
- A sound card is a hardware component that allows the computer to produce and process audio.
- It is responsible for converting digital audio signals into analog signals that can be played through speakers or headphones.
- The sound card also provides various audio input and output ports for connecting external devices.
Speakers:
- Speakers are output devices that convert electrical signals into sound waves.
- They are connected to the sound card and produce audio output based on the signals received.
- Speakers come in different types and configurations, such as stereo speakers, surround sound speakers, and portable speakers.
Microphone (not required for playing sound):
- A microphone is an input device that converts sound waves into electrical signals.
- While a microphone is not necessary for playing sound on a computer, it is required for recording audio or for voice communication in applications such as video conferencing or gaming.
In summary, to play and hear sound on a computer, a sound card and speakers are required. A microphone is not necessary unless audio recording or voice communication is needed.

The correct option is (2)

Buffer

Key Points

• A memory area known as a buffer is used to store data momentarily while it is being transferred from one location to another.
• When transferring data between computer processes, a buffer is employed. The vast majority of buffers are used in the software.
• The main memory (RAM buffer)'s memory serves as a short-term storage space for data being sent between two or more devices or between an application and a device.

Additional Information

•  Its acronym is ALU or arithmetic-logic unit. It is a component of a computer processor (CPU) that performs mathematical and logical processes.
• A computer's CPU contains hardware called a control unit, or CU, which controls operations. It provides instructions on how to process commands from the program for the computer's memory, logic module, and input and output devices.

Checking Statement I:

Here, as Kirti correctly remembers that her birthday is in the second half of the year, so her birthday is in July or August or September or October or November or December.

Hence data in statement I is not sufficient to answer the question.

Checking Statement II:

Here, as Kirti’s father correctly remembers that her birthday is after 31st October but before 1st December, so we can say that Kirti’s birthday is in November.

Hence data in statement II is sufficient to answer the question.

Hence, Option B is correct.

The above information can be summarized below diagram,

Therefore, the code for “was” will be “3”. 

Hence, option 4 is the correct answer.

 

The main triangle shown is in the given figure and this the total no. of triangle is 15. remaing triangle we can find out in the drawing the triangle in the image.