UGC NET Computer Science 2025 Previous Year Questions (PYQs) – Page 2 of 9
Study Stuff for MCA Examinations
Solution
Solution
Since the cards are drawn “one after the other” and order matters, we use permutations:
Number of ways = \( ^{42}P_3 = 42 \times 41 \times 40 \).
Compute: \( 42 \times 41 = 1722 \), and \( 1722 \times 40 = 68880 \).
Solution
| List I | List II |
|---|---|
| A. Address Space | I. Associative Mapping |
| B. Memory Space | II. Logical address |
| C. Cache Memory | III. Physical address |
| D. Segmented Program | IV. Virtual address |
Solution
- A. Address Space → Virtual Address (IV)
Address space refers to the range of virtual addresses a process can use in memory. - B. Memory Space → Physical Address (III)
Memory space corresponds to the actual physical memory locations in RAM. - C. Cache Memory → Associative Mapping (I)
Cache memory uses mapping techniques such as associative mapping to store data efficiently. - D. Segmented Program → Logical Address (II)
In segmentation, programs are divided into segments, each referenced by logical addresses.
Consider the following table defining the sample inputs and corresponding target values for a perceptron model.
| Sample No | x1 | x2 | target | w1 | w2 |
|---|---|---|---|---|---|
| S1 | 0 | 0 | 0 | 0 | 0 |
| S2 | 0 | 1 | 1 | ||
| S3 | 1 | 0 | 1 | ||
| S4 | 1 | 1 | 1 |
What shall be the value of updated weights after applying all the samples S1 to S4 (in the order S1, S2, S3, S4) to this model. Given that the initial weights w1=0, w2=0, learning rate=0.1 and no bias is involved in the perceptron. The activation function for this perceptron is given below:
yobserved =
{ 1 if yin > 0
0 if yin < 0 }
- w1 = 0.1, w2 = 0.1
- w1 = 0.0, w2 = 0.2
- w1 = 0.0, w2 = 0.1
- w1 = 0.2, w2 = 0.2
Solution
Using perceptron learning rule: Δw = η (t − y) x, where y = 1 if yin > 0 else 0.
S1 (0,0,t=0): yin=0 ⇒ y=0 ⇒ no change ⇒ (w1,w2)=(0,0)
S2 (0,1,t=1): yin=0 ⇒ y=0 ⇒ Δw=0.1(1−0)(0,1) ⇒ (w1,w2)=(0,0.1)
S3 (1,0,t=1): yin=0 ⇒ y=0 ⇒ Δw=0.1(1−0)(1,0) ⇒ (w1,w2)=(0.1,0.1)
S4 (1,1,t=1): yin=0.1+0.1=0.2>0 ⇒ y=1 ⇒ no change
Final Weights: w1=0.1, w2=0.1
Answer: (1)
Arrange the following in the increasing order of coupling from lowest coupling to highest coupling.
A. Common Coupling
B. Stamp Coupling
C. Control Coupling
D. External Coupling
E. Content Coupling
Choose the correct answer from the options given below:
Solution
Increasing coupling (lower → higher):
-
Stamp coupling (B): modules share a composite data structure but use only parts of it.
-
Control coupling (C): one module controls another via flags/controls.
-
External coupling (D): modules depend on externally imposed formats/protocols/devices.
-
Common coupling (A): modules share global data.
-
Content coupling (E): one module relies on or modifies another’s internal logic (worst).
Solution
Solution
Match List I with List II
| List I | List II |
|---|---|
| A. Clustered Page Table | III. Useful for sparse address spaces. |
| B. Hierarchical Page Table | I. Generally considered inappropriate for 64-bit architectures. |
| C. Segmentation | IV. Supports a user view of the system. |
| D. Inverted Page Table | II. Has only entry for each real page (or frame) of memory. |
Choose the correct answer from the options given below:
Solution
Explanation:
- $A \rightarrow III$ → Clustered Page Table is useful for sparse address spaces.
- $B \rightarrow I$ → Hierarchical Page Table is inappropriate for 64-bit architectures.
- $C \rightarrow IV$ → Segmentation supports user’s logical view.
- $D \rightarrow II$ → Inverted Page Table has only one entry per real page frame.
Solution
Why: The LCS length for the two sequences is $4$. $2,3,2,1$ is a subsequence of both: $\{1,\underline{2},3,\underline{2},4,1,\underline{2}\}$ and $\{\underline{2},4,\underline{3},1,\underline{2},\underline{1}\}$. Option (4) has length $5$ (too long), and (1), (2) are not subsequences of both.
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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