UGC NET Computer Science 2023 Previous Year Questions (PYQs) – Page 2 of 10

A contains numbers of the form:

$4n + 2$

So, numbers in A are:

$6, 10, 14, 18, 22, 26, 30, ...$

B contains multiples of 3:

$3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ...$

Common elements are:

$6, 18, 30, 42, ...$

These can be written as:

$12n - 6$

In an unclustered B+ tree, retrieving records in sorted order requires scanning index entries and then fetching records from data pages.

Cost of scanning leaf level entries:

$F \times N$

Number of records:

$P \times N$

So, total cost:

$FN + PN = (F + P)N$

Final Answer: (b) (F + P) * N

Assertion A is true because FIFO page replacement may show Belady’s anomaly.

In Belady’s anomaly, increasing the number of page frames can sometimes increase the number of page faults.

Reason R is true, but it is not the correct explanation of Assertion A.

Initial semaphore value = 7

Each P operation decreases the value by 1.

Each V operation increases the value by 1.

So,

$7 - 20 + x = 5$

$x - 13 = 5$

$x = 18$

Using Amdahl’s law:

Fraction improved = $0.30$

Speedup of improved part = $2$

Overall speedup:

$S = \frac{1}{(1 - 0.30) + \frac{0.30}{2}}$

$S = \frac{1}{0.70 + 0.15}$

$S = \frac{1}{0.85}$

$S = 1.176 \approx 1.18$

Scaling transformation is used to increase or decrease the size of a 2-dimensional object.

For a 3D array $A[a][b][c]$ stored in row major order:

Address of $A[r][s][t]$ is

Base address + $w[(r \times b \times c) + (s \times c) + t]$

So,

Address = $ A[0][0][0] + w(bcr + c*s + t)$

Cost Performance Index is calculated as:

$CPI = \frac{BCWP}{ACWP}$

That means:

Budgeted cost of work performed ÷ Actual cost of work performed

After reboot, the TCP server can again start listening on port P.

The old TCP connection state is lost after reboot. So, if the client sends a packet for the old connection, the server does not recognize it and sends an RST segment.

Number of transactions per second:

$ \frac{20000}{120} = 166.67 $

Size of each article:

$3k$ bytes $= 3000$ bytes

$3000$ bytes $= 3000 \times 8 = 24000$ bits

Throughput:

$166.67 \times 24000 = 4000000$ bits per second

$= 4$ megabits per second