UGC NET DECEMBER 2015 PAPER 3 SOLVED PART – 1


1.       The three outputs x1x2x3 from the 8×3 priority encoder are used to provide a vector address of the form 101x1x2x300. What is the second highest priority vector address in hexadecimal if the vector addresses are starting from the one with the highest priority?

(A) BC                (B) A4

(C) BD               (D) AC

Answer: B

Explanation:

Starting from the one has the highest priority i.e. 000. Thus the second highest priority is 001.
Thus the number is 101 001 00 = A4


  1. What will be the output at PORT1 if the following program is executed?

MVI B, 82H

MOV A, B

MOV C, A

MVI D, 37HUGC net study material

OUT PORT1

HLT

(A) 37H              (B) 82H

(C) B9H             (D) 00H

Answer: B

Explanation:

In 8085 programming, the result of an operation is stored in the accumulator.

So output is 82H.


  1. Which of the following 8085 microprocessor hardware interrupt has the lowest priority?

(A) RST 6.5                   (B) RST 7.5

(C) TRAP                      (D) INTR

Answer: D

Explanation:

TRAP
RST7.5
RST6.5
RST5.5
INTR
INTA is not an interrupt. INTA is used by the Microprocessor for sending the acknowledgement. TRAP has highest priority and RST 7.5 has second highest priority and so on.

  1. A dynamic RAM has refresh cycle of 32 times per msec. Each refresh operation requires 100 nsec and a memory cycle requires 250 nsec. What percentage of memory’s total operating time is required for refreshes?

(A) 0.64              (B) 0.96

(C) 2.00             (D) 0.32

Answer: D

Explanation:

in 1ms :  refresh = 32 times

Memory cycle = 1ms/250ns   = 106ns/250ns = 4000 times

Therefore, % of refresh time  = (32 x 100ns)/(4000 x 250ns)

= 3200ns/1000000 x 100% = 0.32%


  1. A DMA controller transfers 32-bit words to memory using cycle Stealing. The words are assembled from a device that transmits characters at a rate of 4800 characters per second. The CPU is fetching and executing instructions at an average rate of one million instructions per second. By how much will the CPU be slowed down because of the DMA transfer?

(A) 0.06%          (B) 0.12%

(C) 1.2%            (D) 2.5%

Answer: B

Explanation:

The DMA combines one word from four consecutive characters (bytes) so we get

4800 chars/s = 4800 bytes/s = 1200 words/s (one word = 32 bits = 4 bytes)

If we assume that one CPU instruction is one word wide then

1 million instructions/s = 1 million words/s = 106 word/s

So we have 1200 words received during one second and (106-1200) words processed by the CPU (while DMA is transferring a word, the CPU cannot fetch the instruction so we have to subtract the number of words transferred by DMA).

While DMA transfer CPU executes only 106 – 1200 = 998800 instructions

[998800 / 106] * 100 = 99.88 %

Slowdown = 100 – 99.88 = 0.12%

The CPU will be slowed down by 0.12%.


  1. A CPU handles interrupt by executing interrupt service subroutine……………..

(A) by checking interrupt register after execution of each instruction

(B) by checking interrupt register at the end of the fetch cycle

(C) whenever an interrupt is registered

(D) by checking interrupt register at regular time interval

Answer: A


  1. Given the following set of prolog clauses:

father(X, Y):

parent(X, Y),

male(X),

parent(Sally, Bob),

parent(Jim, Bob),

parent(Alice, Jane),

parent(Thomas, Jane),

male(Bob),

male(Jim),

female(Salley),

female(Alice).

How many atoms are matched to the variable ‘X’ before the query

father(X, Jane) reports a Result?

(A) 1       (B) 2

(C) 3       (D) 4

Answer: X


  1. Forward chaining systems are …………. where as backward chaining systems are …………….

(A) Data driven, Data driven              (B) Goal driven, Data driven

(C) Data driven, Goal driven              (D) Goal driven, Goal driven

Answer: C

Explanation:

Backward Chaining is a form of reverse engineering, which is very applicable in situations where there are so many rules that could be applied to a single problem, the system could be there all day sifting through rules before it gets anywhere.

Forward Chaining is a problem solving technique used by Expert Systems when they are faced with a scenario and have to give a solution or conclusion to this scenario. The system will work its way through the rules, finding which ones fit and which leads to which using Deductive Reasoning.


  1. Match the following w.r.t. programming languages:

List – I                                     List – II

(a) JAVA                        (i) Dynamically object oriented

(b) Python                     (ii) Statically Non-object oriented

(c) Prolog                      (iii) Statically object oriented

(d) ADA                         (iv) Dynamically non-object oriented

Codes:

(a)  (b)  (c)  (d)

(A) (iii)  (i)   (ii)  (iv)

(B) (i)   (iii)  (ii)  (iv)

(C) (i)   (iii)  (iv) (ii)

(D) (ii)  (iv)  (i)   (iii)

Answer: X


  1. The combination of an IP address and a port number is known as ……………….

(A) network number                (B) socket address

(C) subnet mask number       (D) MAC address

Answer: B

Explanation:

Socket address is a combination of IP address and Port address.


11.       A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network ?

(A) 2 Mbps        (B) 60 Mbps

(C) 120 Mbps    (D) 10 Mbps

Answer: A

Explanation:

In data transmission, throughput is the amount of data moved successfully from one place to another in a given period of time, and typically measured in bits per second (bps), or in megabits per second (Mbps) or gigabits per second (Gbps).

Here, Throughput = 15000 x 8000/60 = 2 Mbps


  1. Consider a subnet with 720 routers. If a three-level hierarchy is choosen with eight clusters, each containing 9 regions of 10 routers, then total number of entries in the routing table is ……………….

(A) 25                 (B) 27

(C) 53                 (D) 72

Answer: A

Explanation:

Each router needs 10 entries for local routers, 8 entries for routing to other regions within its own cluster, and 7 entries for distant clusters, for a total of 25 entries.


  1. In a classful addressing, the IP addresses with 0 (zero) as network number:

(A) refers to the current network

(B) refers to broadcast on the local network

(C) refers to broadcast on a distant network

(D) refers to loopback testing

Answer: B


  1. In electronic mail, which of the following protocols allows the transfer of multimedia

messages?

(A) IMAP            (B) SMTP

(C) POP 3         (D) MIME

Answer: D

Explanation:

MIME stands for Multi-purpose Internet Mail Extensions or Multimedia Internet Mail Extensions . It is a encoding protocol like BinHex in Mac and UUEncode in UNIX. At first it was used as a way of sending more than just text via email. Later the protocol was extended to manage file typing by Web servers.

Email attachment

Let’s talk about email first. By using MIME, you can enclose the following types of binary file to your text-based e-mail message:

  • character sets other than ASCII
  • enriched text
  • images
  • sounds
  • other messages (reliably encapsulated)
  • tar files
  • PostScript
  • FTPable file pointers

The easiest way to send and receive MIME attachment is to use a MIME-aware Email client such as Netscape Mail or Microsoft Outlook. The procedure is self-explanatory. When a MIME ready e-mail system receives a MIME encoded file, the binary file should show up as an attachment and the proper software in your computer should be capable of reading the file. For example, if the attached file is a PostScript file, your LaserWriter Utilities should be able to download it to the printer. If the file is a graphic, DeBabelizer or PhotoShop should be able to open the file.

However, you must pay attention to the file extension. For instance, if someone sends you a JPEG image from a Mac with an extension as “JPEG,” you would not be able to open it in a PC unless you change the extension to “JPG,” and vice versa.


  1. A device is sending out data at the rate of 2000 bps. How long does it take to send a file of 1,00,000 characters ?

(A) 50                 (B) 200

(C) 400              (D) 800

Answer: C

Explanation:

1,00,000 characters = 1,00,000 x 8 bits = 8,00,000 bits

8,00,000 bits/2000 bps = 400 seconds


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