UGC NET Computer Science 2022 Previous Year Questions (PYQs) – Page 1 of 30
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Solution
I. $a-b$
II. $d-f$
III. $b-f$
IV. $d-c$
V. $d-e$
Choose the correct answer from the options given below:
Solution
Using Kruskal's algorithm, edges are selected in increasing order of weight.
From the graph:
$a-b = 1$
$d-f = 1$
$b-f = 2$
$d-c = 2$
$d-e = 3$
So, the valid order must follow:
Weight $1$ edges first:
$I, II$ in any order
Then weight $2$ edges:
$III, IV$ in any order
Then weight $3$ edge:
$V$
Now check option (c):
II, I, III, V, IV
Here, edge $V = d-e$ has weight $3$, but edge $IV = d-c$ has weight $2$.
So, selecting $V$ before $IV$ violates Kruskal's increasing weight rule.
Therefore, this order cannot be used.
What is the safest order while simplifying Context Free Grammar?
Solution
While simplifying a CFG, the safest order is:
First remove $\epsilon$-productions.
Then remove unit productions.
Finally remove useless symbols and productions.
Reason:
After removing $\epsilon$-productions, new unit productions may be created.
After removing unit productions, some symbols may become useless.
So, useless symbols should be removed at the end.
Correct order is:
$\epsilon$-productions → Unit productions → Useless symbols and productions
Rearrange the following sequence in the context of OSI Layers:
I. Transforming the raw bits in the form of frame for transmission
II. Transmission of raw bits over communication channel
III. Handling user interfaces
IV. Control and monitoring of subnet
V. Transmission data through connection oriented or connection less using datagrams
Choose the correct answer from the options given below:
Solution
Physical layer:
Transmission of raw bits over communication channel.
So, II comes first.
Data Link layer:
Transforms raw bits into frames.
So, I comes second.
Network layer:
Controls and monitors subnet.
So, IV comes third.
Transport layer:
Provides connection-oriented or connection-less data transmission.
So, V comes fourth.
Application layer:
Handles user interfaces.
So, III comes last.
Choose the correct option describing the features of Artificial neural network
I. It is essentially machine learning algorithm.
II. It is useful when solving the problems for which the data set is very large.
III. They are able to extract features without input from the programmer.
IV. These are systems modeled on the human brain and nervous system
Choose the correct answer from the options given below:
Solution
Statement I:
Artificial Neural Network is a machine learning technique.
So, statement I is correct.
Statement II:
ANN is useful for large datasets because it can learn complex patterns from data.
So, statement II is correct.
Statement III:
Neural networks can automatically learn important features from data without manual feature programming.
So, statement III is correct.
Statement IV:
ANN is inspired by the human brain and nervous system.
So, statement IV is correct.
Therefore, all statements are correct.
Four persons: P, Q, R and S are in police custody and one of them has committed a crime. They confess as follows:
A. Person P: Q did it.
B. Person Q: S did it.
C. Person R: I did not do it.
D. Person S: Q lied.
If exactly one of the statements is false, which of the following is the guilty person.
Solution
Check each case one by one.
Case 1: Suppose P is guilty.
P says: Q did it → False
Q says: S did it → False
R says: I did not do it → True
S says: Q lied → True
Here, two statements are false. So, P is not guilty.
Case 2: Suppose Q is guilty.
P says: Q did it → True
Q says: S did it → False
R says: I did not do it → True
S says: Q lied → True
Here, exactly one statement is false.
So, Q can be guilty.
Case 3: Suppose R is guilty.
P says: Q did it → False
Q says: S did it → False
R says: I did not do it → False
S says: Q lied → True
Here, three statements are false. So, R is not guilty.
Case 4: Suppose S is guilty.
P says: Q did it → False
Q says: S did it → True
R says: I did not do it → True
S says: Q lied → False
Here, two statements are false. So, S is not guilty.
Therefore, the guilty person is Q.
Assertion A:
A Raster scan device is a CRT graphic device and can use a television monitor for display.
Reason R:
In Raster scan display the picture is composed of a series of dots. These dots are traced out as a series of horizontal lines. Television works in a similar fashion.
In the light of the above statements, choose the correct answer from the options given below:
Solution
Raster scan display forms an image using many small dots called pixels.
These pixels are scanned line by line from left to right and top to bottom.
Television display also works in a similar raster scanning manner.
So, Assertion A is true.
Reason R correctly explains why raster scan devices can use television monitors.
So, Reason R is also true and it is the correct explanation of Assertion A.
Solution
In Round Robin scheduling, each process gets CPU for one time quantum.
Given:
Time quantum $=4$ ms
Number of processes $=20$
Maximum waiting time occurs for the last process in the ready queue.
Before the last process gets CPU, the other $19$ processes will execute.
So,
Maximum waiting time $=19 \times 4$
$=76$ ms
Statement 1:
Given a graph $G=(V,E)$ in which each vertex $v \in V$ has an associated positive weight $w(v)$, we can use linear programming to find the lower bound on the weight of the minimum-weight vertex cover.
Statement 2:
The lower bound can be found by maximizing the following
$ \sum_{v \in V} w(v)x(v) $
subject to
$ x(u)+x(v) \ge 1 $ for each $(u,v) \in E$
$ x(v) \le 1 $ for each $v \in V$
$ x(v) \ge 0 $ for each $v \in V$
In the light of the above statements, choose the most appropriate answer from the options given below:
Solution
For minimum-weight vertex cover, we can write an integer linear programming formulation.
Its LP relaxation can be used to find a lower bound.
So, Statement 1 is correct.
But for minimum-weight vertex cover, the objective should be minimized, not maximized.
Correct objective should be:
$ \text{Minimize } \sum_{v \in V} w(v)x(v) $
Statement 2 says maximizing the objective, so Statement 2 is incorrect.
A TCP Server application is programmed to listen on port P on Host S. A TCP Client is connected to the TCP Server over the network. Considered that while TCP Connection is active the server is crashed and rebooted. Assume that the client does not use TCP keepalive timer. Which of the following behaviours is/are possible?
Statement I:
If client is waiting to receive a packet, it may wait indefinitely.
Statement II:
If the client sends a packet after the server reboot, it will receive the FIN segment.
In the light of the above statements, choose the correct answer from the options given below.
Solution
When the server crashes and reboots, the server loses the old TCP connection state.
Statement I:
If the client is only waiting to receive data and TCP keepalive timer is not used, then the client may not immediately know that the server has crashed.
So, the client may wait indefinitely.
Therefore, Statement I is true.
Statement II:
After reboot, if the client sends a packet for the old TCP connection, the server will not recognize that old connection.
In this case, the server sends a reset segment, that is $RST$, not a $FIN$ segment.
$FIN$ is used for normal connection termination.
$RST$ is used when the connection is invalid or not recognized.
Therefore, Statement II is false.
UGC NET Computer Science
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UGC NET Computer Science
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
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