UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 41-45)
Answer: 1). 49
Solution : Here G is 1/4, Therefore S = G x e-G = 0.195 (19.5),
this means 250 x 0.195 = 49. Only 49 frames out of 250 will probably survive.
Answer: 2). 10⁻² KHz
Solution: Explanation: We know that:
frequency(f) = 1 / Time period(T)
Given time period(T) = 100 ms
f = 1 / 100ms
f = 1000 / 100000 ms
We know that 1000 ms = 1 sec and we can write 1000 as 1 k, i.e. f = 1k / 100s
f = 10-2 kHz (1 / s = Hz) So, option (B) is correct.
Answer: 3). 129.11.11.239
Explanation: IPV4 address = 10000001 00001011 00001011 11101111
10000001 = 129, 00001011 = 11, 00001011 = 11, 11101111 = 239
IPV4 address 10000001 00001011 00001011 11101111 is equivalent to 129.11.11.239
So, option (C) is correct.
Answer: 2). (b) and (c) only
Reason:
Answer: 2). (ii) (iv) (i) (iii)
Reason:
Data Encryption Standard(DES) block cipher uses block size 64 and key size 64
International Data Encryption Algorithm (IDEA) uses block size 64 and key size 12.
BLOW FISH is the relacement for DES or IDEA, it uses block size 64 and key size ranges between 32 and 448
Advanced Encryption Standard (AES) uses block size 128 and key sizes 128, 192, 256
Q 41.
A slotted ALOHA network transmits 200-bit frames using a shared channel with a
200 Kbps bandwidth. Find the throughput of the system, if the system (all
stations put together) produces 250 frames per second :
- 49
- 368
- 149
- 151
Answer: 1). 49
Solution : Here G is 1/4, Therefore S = G x e-G = 0.195 (19.5),
this means 250 x 0.195 = 49. Only 49 frames out of 250 will probably survive.
Q 42.
The period of a signal is 100 ms. Its frequency is _________.
- 100³ Hertz
- 10⁻² KHz
- 10⁻³ KHz
- 10⁵ Hertz
Answer: 2). 10⁻² KHz
Solution: Explanation: We know that:
frequency(f) = 1 / Time period(T)
Given time period(T) = 100 ms
f = 1 / 100ms
f = 1000 / 100000 ms
We know that 1000 ms = 1 sec and we can write 1000 as 1 k, i.e. f = 1k / 100s
f = 10-2 kHz (1 / s = Hz) So, option (B) is correct.
Q 43.
The dotted-decimal notation of the following IPV4 address in binary notation is
_________.
10000001
00001011 00001011 11101111
- 111.56.45.239
- 129.11.10.238
- 129.11.11.239
- 111.56.11.239
Explanation: IPV4 address = 10000001 00001011 00001011 11101111
10000001 = 129, 00001011 = 11, 00001011 = 11, 11101111 = 239
IPV4 address 10000001 00001011 00001011 11101111 is equivalent to 129.11.11.239
So, option (C) is correct.
Q 44.
Which of the following statements are true ?
(a)
Advanced Mobile Phone System (AMPS) is a second generation cellular phone
system.
(b)
IS – 95 is a second generation cellular phone system based on CDMA and DSSS.
(c) The Third generation cellular phone system will provide universal personnel communication.
Code
:
- (a) and (b) only
- (b) and (c) only
- (a), (b) and (c)
- (a) and (c) only
Reason:
- Advanced Mobile Phone System (AMPS) is a First generation cellular phone system.False
- IS – 95 is a second generation cellular phone system based on CDMA and DSSS.True
- The Third generation cellular phone system will provide universal personnel communication.True
SO, option (B) is correct.
45.
Match the following symmetric block ciphers with corresponding block and key
sizes :
Reason:
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