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Table of contents
1.
Introduction
2.
FAQs
2.1.
What is the UGC NET exam?
2.2.
What is the exam format for the UGC NET?
2.3.
Is it possible to take the UGC NET after completing a postgraduate degree?
2.4.
How can solving PYQs help in my exam preparation?
2.5.
How many papers are there in the UGC NET exam?
3.
Conclusion
Last Updated: Mar 27, 2024
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Dec 2015 Paper-III Part-1

Author Aditya Singh
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Introduction

UGC NET Exam is a very popular exam in India for people interested in research. Solving Previous Year Questions are an excellent way to learn about the exam pattern and practice for it. By solving the PYQs, you will get a basic idea about your preparation. In this article, we have given the questions of UGC NET 2015 Dec Paper-III. We have also explained every problem adequately to help you learn better.

Note: This article contains Question No. 1 to Question No. 25 out of the 75 questions asked in UGC NET Dec 2015 paper III. You can find Question No 26 to Question No 50  in part-2 and Question No 51 to Question No 75 in part-3.
 

Question 1: The three outputs x1x2x3 from the 8x3 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?

  1. BC
  2. A4
  3. BD
  4. AC

Solution: (B) A4

Explanation:

The 8 X 3 priority encoder will have the following priority preference:

000 First 

001 Second 

010 Third

011 Fourth

100 Fifth

101 Sixth

110 Seventh 

111 Eighth

The second highest priority vector address, according to the question, will be 10100100, or 1010 0100.

It will be A4 if we convert it to hexadecimal.

As a result, option (B) is the correct answer.

 

Question 2: What will be the output at PORT1 if the following program is executed?

MVI B, 82H

MOV A, B

MOV C, A

MVI D, 37H

OUT PORT1

HLT

  1. 37H
  2. 82H
  3. B9H
  4. 00H

Solution: (B) 82H

Explanation:

The outcome of an operation is saved in the accumulator in 8085 programmings.

As a result, the output is 82H.
 

Question 3: Which of the following 8085 microprocessor hardware interrupts has the lowest priority?

  1. RST 6.5
  2. RST 7.5
  3. TRAP
  4. INTR

Solution: (D) INTR

Explanation:

There are five hardware interrupts on the 8085 microprocessor. TRAP, RST 7.5, RST 6.5, RST 5.5, and INTR are the acronyms for "Trap, RST 7.5, RST 6.5, RST 5.5, and INTR."

The preceding order is diminishing in importance.

As a result, option (D) is the correct answer.

 

Question 4: A dynamic RAM has a 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?

  1. 0.64
  2. 0.96
  3. 2.00
  4. 0.32

Solution: (D) 0.32

Explanation:

In 1 msec, refresh=32 times

memory cycle= 1msec/250nsec= 4000 times

Therefore % refresh time:

(32*100nsec)/(4000*250 nsec) 

= 3200/1000000=0.0032 

= 0.32%

 

Question 5: 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?

  1. 0.06%
  2. 0.12%
  3. 1.2%
  4. 2.5%

Solution: (B) 0.12%

Explanation: 

The DMA combines four consecutive characters (bytes) into one word, resulting in 4800 chars/s = 4800 bytes/s = 1200 words/s (one word = 32 bits = 4 bytes).

If each CPU instruction is one word long, 1 million instructions per second equal 1 million words per second or 106 words per second.

So, in one second, 1200 words are received and (106-1200) words are 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 performing DMA transfers, the CPU only processes 106 - 1200 = 998800 instructions [998800 / 106]. * 100 percent = 99.88 percent

Slowdown = 100 minus 99.88 = 0.12%

The processor will be slowed by 0.12 percent.
 

Question 6:   A CPU handles interrupt by executing interrupt service subroutine.................

  1. by checking interrupt register after execution of each instruction
  2. by checking interrupt register at the end of the fetch cycle
  3. whenever an interrupt is registered
  4. by checking interrupt register at regular time interval

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

Explanation: 

After each instruction is executed, the interrupt service subroutine is executed, and the interrupt register is checked.

As a result, option (A) is the correct answer.

 

Question 7:  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?

  1. 1
  2. 2
  3. 3
  4. 4

Solution: Marks to all

Explanation:

Father (X, Jane) is true for those values of which make subgoals male(X) and parents (Jane, Y, X) true. Recursively apply this reasoning until reach rules that are facts; called backward chaining only "Alice" is matched.

 

Question 8:   Forward chaining systems are ............. whereas backward chaining systems are ................

  1. Data driven, Data driven   
  2. Goal driven, Data driven
  3. Data driven, Goal-driven
  4. Goal driven, Goal-driven

Solution: (C) Data driven, Goal-driven

Explanation: 

Backward chaining systems are motivated by goals, whereas forward chaining systems are driven by data.

 

Question 9:  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)

Solution: Marks to all

Explanation

  • JAVA is statically object oriented language.
  • JPython is a dynamic object oriented language.
  • AVProlog is a non-object oriented language.
  • A is statically object oriented language   

So, no option is correct.

 

Question 10: The combination of an IP address and a port number is known as...................

  1. network number     
  2. socket address
  3. subnet mask number
  4. MAC address

Solution:(B) socket address

Explanation: 

  • Socket address is the combination of an IP address and a port number.
  • Subnet mask is a 32-bit value that conceals an IP address.
  • Your computer's MAC (Media Access Control) address is a unique hardware number.
  • The number that is logical AND between a 32-bit IP address and a Network mask is known as the Network Number.

So, option (B) is correct.
 

Question 11:    A network with a 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?

  1. 2 Mbps    
  2. 60 Mbps
  3. 120 Mbps    
  4. 10 Mbps

Solution: (A)  2 Mbps 

Explanation: 

Throughput is the quantity of data successfully transferred from one location to another in a particular amount of time, and is commonly measured in bits per second (bps), megabits per second (Mbps), or gigabits per second (GBPS) (Gbps).

Throughput = 15000 x 8000/60 = 2 Mbps in this case.

 

Question 12:    Consider a subnet with 720 routers. If a three-level hierarchy is chosen with eight clusters, each containing 9 regions of 10 routers, then the total number of entries in the routing table is ...................

  1. 25             
  2. 27
  3. 53             
  4. 72

Solution: (A) 25 

Explanation:

The amount of data transmitted successfully from one location to another in a particular period of time is measured in bits per second (bps), megabits per second (Mbps), or gigabits per second (GBPS) in data transmission (Gbps).

Throughput equals 15000 x 8000/60 = 2 Mbps in this case.

 

Question:13   In classful addressing, the IP addresses with 0 (zero) as network number:

  1. refers to the current network
  2. refers to broadcast on the local network
  3. refers to broadcast on a distant network
  4. refers to loopback testing

Solution: (A)  refers to the current network

Explanation: 

The IP addresses with 0 (zero) as the network number correspond to the present network in classful addressing.

 

Question: 14   In electronic mail, which of the following protocols allows the transfer of multimedia messages?

  1. IMAP        
  2. SMTP
  3. POP 3     
  4. MIME

Solution: (D)  MIME

Explanation:

  • Emails are retrieved from a mail server using the Internet Message Access Protocol (IMAP).
  • The Simple Mail Transfer Protocol (SMTP) is a standard for email communication on the Internet.
  • The Multipurpose Internet Mail Extensions (MIME) standard allows multimedia messages to be sent.
  • The most recent version of a standard protocol for receiving email is POP3 (Post Office Protocol 3).

So, option (D) is correct.

 

Question: 15    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?

  1. 50             
  2. 200
  3. 400          
  4. 800

Solution: (C)  400  

Explanation:

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

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

 

Question: 16 In the Activity-Selection problem, each activity i has a start time si and a finish time fi where si≤fi. Activities i and j are compatible if:

  1. si≥fj                       
  2. sj≥fi
  3. si≥fj or sj≥fi           
  4. si≥fj and sj≥fi

Solution: (C) si≥fj or sj≥fi 

Explanation:

If two tasks can be accomplished in some order and do not overlap in time, they are compatible. To complete two activities, one activity's START time must be larger than or equal to the other's FINISH time, i.e. one activity must begin only after the other has finished. 

So, If Si >= Fj, then, Sj >= Fi.

 

Question: 17  Given two sequences X and Y:

X = <a, b, c, b, d, a, b>

Y = <b, d, c, a, b, a>

The longest common subsequence of X and Y is:

  1. <b, c, a>           
  2. <c, a, b>
  3. <b, c, a, a> 
  4. <b, c, b, a>

Solution: (D)  <b, c, b, a>

Explanation:

A subsequence is a sequence that can be created by eliminating some elements from another sequence while keeping the order of the remaining components the same.

If Z is a subsequence of both X and Y, G is said to be a common subsequence of X and Y.

In the case of X = a, b, c, b, d, a, b>, the sequences b,c,a>, c,a,b>, and b,c,b,a> are X subsequences.

If Y = b, d, c, a, b, a> is a second sequence of symbols, then b,c,a>, c,a,b>, b,c,b,a> are common subsequences to both X and Y.

The longest common subsequence of X and Y, however, is b,c,b,a>.

 

Question:18   If there are n integers to sort, each integer has d digits and each digit is in the set {1,2, ..., k}, radix sort can sort the numbers in:

  1. O(d n k)               
  2. O(d nk)
  3. O((d+n)k)            
  4. O(d(n+k))

Solution: (D) O(d(n+k))

Explanation:

When there are n integers to sort, each with d digits and each digit in the range 1, 2,..., k, radix sort can sort the numbers in O(d (n + k)) time.

 

Question: 19 The solution of the recurrence relation is:

  1. O(lg n)     
  2. O(n)
  3. O(n lg n) 
  4. None of the above

Solution: (D) None of the above

Explanation:

Assume T(n) = O(1) for small ns 80.

T(n) <= T(n/5)+T((7n/10) +6) +O(n)

Inductively verify that T(n) <= cn for some constant o

T(n) <= c(n/5)+c(7n/10+6)+0(n)

<= 9cn/10 + 6c + O(n)

<= cn

In above, choose c so that c((n/10) -6) beats the function O(n) for all n.

Note: They given 's' instead of 5.

 

Question: 20  Floyd-Warshall algorithm utilises ............... to solve the all-pairs shortest paths problem on a directed graph in ................ time.

  1. Greedy algorithm, θ(V3)            
  2. Greedy algorithm, θ(V2 lgn)
  3. Dynamic programming, θ(V3) 
  4. Dynamic programming, θ(V2 lgn)

Solution: (C)  Dynamic programming, θ(V3) 

Explanation:

Floyd-Warshall algorithm solves the all-pairs shortest routes issue on a directed network in (V3) time using dynamic programming.

 

Question: 21 Let n=4 and (a1, a2, a3, a4) = (do, if, int, while). Let p(1:4) = (3/8, 3/8, 1/8, 1/8) and q(1:4) = (2/8, 3/8, 1/8, 1/8, 1/8) where p(i) and q(i) denotes the probability with which we search ai and the identifier x being searched satisfy ai < x < ai+1 respectively. The optimal search tree is given by:

Solution:(B)

Explanation: 

(b) Search success nodes are

(do, if, int, while) in ascending order. 

Therefore, option (c) and (d) cannot be true.

 Now the minimum cost of a given tree in option (a) and (b).

            

      Cost = (⅛)*(1)+(⅜ +⅛)*2*(⅜+⅛+⅛+⅛)*3+ (2/8+⅜)*4 = 5.875

                                                    

Cost =  (⅜ )*(1) + (⅜ + ⅛) * 2 * (2/8 + ⅜ + ⅛ + ⅛)*3+ (⅛ + ⅛ )*4 = 5

 

Question: 22 The family of context-sensitive languages is ................. under union and ................. under reversal.

  1. closed, not closed                     
  2. not closed, not closed
  3. closed, closed                
  4. not closed, closed

Solution:(C) closed, closed        

Explanation:

Context-sensitive languages are closed under union, intersection, Kleen closer, concatenation and reversal.

 

Question: 23  Match the following :

                        List - I  

List - II

 

(a) {an bn|n > 0} is a deterministic        

context free language        

(i) but not recursive language

 

(b) The complement of {an bn an|n > 0} 

is a context free language

(ii) but not context free language
(c) {an bn an} is context sensitive language         (iii) but can not be accepted by a  deterministic pushdown automaton
(d) L is a recursive language  (iv) but not regular

  Codes :

  (a)   (b)   (c)   (d)

A. (i)   (ii)   (iii)   (iv)

B. (i)   (ii)   (iv)   (iii)

C. (iv) (iii)  (ii)    (i)

D. (iv) (iii)  (i)    (ii)

Solution:(Marks to all)

Explanation: 

  • Anbn | n > 0 is a context-free deterministic language, but it is not a regular language.
  • The complement of anbn | n > 0 is a context-free language, but deterministic pushdown automata do not accept it.
  • The language anbn n is context-sensitive but not recursive.
  • L is a recursive language, although it is not context-free.

 

Question: 24   The language of all non-null strings of a’s can be defined by a context-free grammar as follow:

S→a S|S a| a

The word a3 can be generated by ................ different trees.

  1. Two          
  2. Three
  3. Four         
  4. Five

Solution: (C)  Four

Explanation:

Question: 25  Which one of the following non-functional quality attributes is not highly affected by the architecture of the software?

  1. Performance      
  2. Reliability
  3. Usability              
  4. Portability

Solution: (C) Usability

Explanation: 

(C) “Usability” should be the answer. It should not be heavily influenced by the software's design because it is a core feature; however, the rest of the options are optional features provided by the software and maybe heavily influenced by the product's architecture.

 

FAQs

What is the UGC NET exam?

UGC NET is a national-level exam organized by UGC to determine the eligibility of the candidates for lectureship and JRF.

What is the exam format for the UGC NET?

The UGC NET exam is conducted entirely online.

Is it possible to take the UGC NET after completing a postgraduate degree?

You are qualified to take the exam if you have finished a PG Diploma from a recognized university.

How can solving PYQs help in my exam preparation?

Solving PYQs will give you a good idea about the exam pattern and help you identify your weak topics to prepare them better for the examination.

How many papers are there in the UGC NET exam?

There are two papers, and the candidates get 3 hours for both papers. There are 150 questions in UGC NET combining both papers.

 

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Conclusion

We have extensively discussed the Dec 2015 paper-III. We hope that this blog has helped you understand the UGC pattern. 

This article contains Question No. 1 to Question No. 25 out of the 75 questions asked in NET Dec 2015 paper III. You can find Question No 26 to Question No 50  in part-2 and Question No 51 to Question No 75 in part-3.

You can refer to this article for more details on UGC NET 2022. 

You can refer to other Previous Year papers using the links given below.

Refer to our carefully curated articles and videos and code studio library if you want to learn more. Refer to our guided paths on Coding Ninjas Studio to learn more about DSA, Competitive Programming, JavaScript, System Design, etc. Enrol in our courses and refer to the mock test and problems available. Take a look at the interview experiences and interview bundle for placement preparations.

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