UGC NET Computer Science 2024 Previous Year Questions (PYQs) – Page 3 of 20

Transmission delay

$= \frac{\text{Frame size}}{\text{Bandwidth}}$

$= \frac{10\times10^6}{6\times10^6}$

$= 1.6667\ s$

Propagation delay

Distance $=5000\ km =5\times10^6\ m$

$= \frac{5\times10^6}{2\times10^8}$

$=0.025\ s$

Router delay

Each router delay $=2\mu s+1\mu s=3\mu s$

Total routers $=20$

Total delay $=20\times3\mu s=60\mu s=0.00006\ s$

Total latency

$=1.6667+0.025+0.00006$

$\approx1.691726\ s$

First bit is fixed as 0, so remaining 3 bits are free.

Possible strings:

0000

0001

0010

0011

0100

0101

0110

0111

Total outcomes $=8$

Strings with at least two consecutive 0's:

0000

0001

0010

0011

0100

Favourable outcomes $=5$

Probability

$=\frac{5}{8}$

In LR parsing, shift-reduce conflicts occur when the parser can either shift the next input symbol or reduce using a rule.

These conflicts are resolved using operator precedence and associativity rules.

Backpropagation computes error at the output layer (sink) and propagates the error backwards through hidden layers toward the input layer (source) to update weights.

Speed hierarchy of memory/storage:

CPU Registers → fastest

SRAM → cache memory

DRAM → main memory

Hard Disk → secondary storage

Magnetic Tape → slowest storage

Order:

C → B → A → E → D

A → False (only owner or DBA can grant privileges)

B → True

C → False (system error is not a malicious attack)

D → True

False statements:

A and C

Regular languages:

Closed under union, concatenation, intersection, complement, kleene star.

Context-free languages:

Closed under union, concatenation, kleene star.

Not closed under intersection and complement.

Thus incorrect statements:

C and D

Effective address:

$EA = ADR + XR$

This is indexed addressing, where the address field is added to the index register.

For 8-bit signed integer (two’s complement):

Range = $-2^{7}$ to $2^{7}-1$

Least value = $-128$

For 8-bit unsigned integer:

Range = $0$ to $2^{8}-1$

Least value = $0$

Cardinality means number of elements in the set.

$|A_1| = 2$

$|A_2| = 4$

$|A_3| = 1$

$|A_4| = 3$

$|A_5| = 5$

Increasing order:

$1 < 2 < 3 < 4 < 5$

So order becomes:

$A_3, A_1, A_4, A_2, A_5$