## Introduction

The NET(National Eligibility Test) is executed on behalf of the University Grants Commission (UGC) to decide the eligibility of Indian nationals for the Eligibility of Assistant Professorship, Junior Research Fellowship, or both for Indian Colleges and Universities.

This article discusses the previous year's UGC-NET computer science questions of June 2014 Paper III. We will see these questions along with their answers and explanation. Before going through the article, refer to __June 2014 Paper III - Part 1__ & __June 2014 Paper III - Part 2__.

## Questions

**Q.51 The pure object-oriented programming language with extensive metadata available and modifiable at run time is**

**Answer:** A

**Explanation:** Smalltalk was one of the first object-oriented (OO) languages (others like Simula and Eiffel) and can be said to be highly "pure" in an OO sense:

Everything is an object, and objects are only communicated via sending messages.

No primitives (no ints, booleans, etc.)

No control structures (no for, while, if, etc.).

**Q.52 Match the following interfaces of the Java Servlet package:**

**Codes : **

a b c d

(A) iii iv ii i

(B) iii ii iv i

(C) ii iii iv i

(D) iv i ii iii

**Answer:** D

**Explanation:** A servlet container uses a servlet configuration object to pass information to a servlet during initialization.

A ServletContext defines a collection of methods that a servlet utilizes to communicate with its servlet container, enabling servlets to log events.

ServletRequest Defines an object to deliver client request information to a servlet. A ServletRequest object provides data, including parameter names and values, attributes, and an input stream.

ServletResponse defines an object to help a servlet in sending a response to the client. A servlet container makes a ServletResponse object and passes it as an argument to the servlet's service method.

**Q.53 The syntax of capturing events method for document object is **

**Answer:** C

**Explanation:** It causes the window or document to capture all events of the specified type.

**Q.54 Linking to another location in the same or another webpage requires two A (Anchor) tags, first with the ________ attribute and second with the _________ attribute. **

(A) NAME & LINK

(B) LINK & HREF

(C) HREF & NAME

(D) TARGET & VALUE

**Answer:** C

**Explanation:** To make an anchor HTML link that points to another place on the same webpage, we need to create two anchors:

1. First, we create an anchor link that the person can click on. Here is the code for the anchor link.

<a href = "link"> Link Name </a>

2. Next, we need to create a named anchor in the spot you will jump to. This anchor will require the following syntax.

**Q.55 An image is given of size 1024 × 1024 pixels; each pixel's intensity is an 8-bit quality. It needs _______ of storage space if the image is not compressed. **

(A) one Terabyte

(B) one Megabyte

(C) 8 Megabytes

(D) 8 Terabytes

**Answer:** B

**Explanation:**

Each pixel requires to store 1 Byte( 8 bit)

Total pixels = 1024 x 1024

Total storage space = 1 x 1024 x 1024 = 1 x 2^{10} x 2^{10} = 2^{20} = 1 MB

**Q.56 Match the subsequent cryptographic algorithms with their design issues:**

**Codes : **

a b c d

(A) ii i iv iii

(B) iii i iv ii

(C) iii iv ii i

(D) iv i ii iii

**Answer:** C

**Explanation:**

- DES is private key encryption with a 56-bit key
- AES is private key encryption with a 128-bit key
- RSA is a public-key encryption algorithm
- Secure Hash Algorithm or SHA is a cryptographic hash function that produces a 160-bit hash value known as a message digest.

**Q.57 Suppose a code with five valid codewords of length ten:**

**Hamming distance of the code is**

(A) 5

(B) 10

(C) 8

(D) 9

**Answer:** A

**Explanation:** The Hamming distance of a code is the minimum Hamming distance between any two codewords in the code. HammingDistance({0000000000,0000011111,1111100000,1110000011,1111111111}) = 5

**Q.58 Which of the subsequent special cases does not require reformulation of the problem to obtain a solution? **

(A) Alternate optimality

(B) Infeasibility

(C) Unboundedness

(D) All of the above

**Answer:** A

**Explanation:** Infeasibility (no feasible solution) and unboundedness (unbounded solution) surely require reformulation of the issue. Only Alternate optimality does not need reformulation as one solution is already there.

An alternate optimal solution is also called alternate optima when a linear / integer programming problem has more than one optimal solution. Generally, an optimal solution is a solution to a problem that satisfies the set of conditions of the problem and the objective function, which is to minimize or maximize.

**Q.59 The given maximization assignment issue can be converted into a minimization problem by:**

**Answer:** B

**Explanation:** We can convert the given maximization assignment problem into a minimization problem by Option (B).

**Q.60 The initial basic viable solution of the following transportation problem:**

** is given as **

**Then the minimum cost is?**

(A) 76

(B) 78

(C) 80

(D) 82

**Answer:** A

**Explanation:** Solving, we get

Cost = 5x2 + 3x2 + 1x6 + 4x7 + 2x12 + 1x2

= 10 + 6 + 6 + 28 + 24 + 2

= 76

**Q.61 Given the following equalities:**

**Which of the following is true?**

**Answer:** B

**Explanation:**

For E2, find which is large:

N^{3} * 2^{N} = 6N^{2} * 3^{N}

N * 2^{N} = 6 * 3^{N} (Now take log on both sides)

log(N * 2^{N}) = log(6 * 3^{N})

logN + N log2 = log6 +N log3

logN + N = 1.5 N (log6 = CONSTANT, log3 = 1.5)

logN = 0.5N

O(0.5N) = O(6 * N^{2} * 3^{N})

So E2 is false.

For E1:

Compare N^{E} = logn

Since E > 0, N^{E} is large compared to log(n), E1 is correct.

Therefore, option (B) is correct since E1 is correct, AND E2 is not correct.

**Q.62 Consider the fractional knapsack instance where n = 4, (w1 , w2 , w3 , w4 ) = (2, 4, 6, 9)), (p1 , p2 , p3 , p4 ) = (10, 10, 12, 18) and M = 15. The maximum profit is given by (Assuming p and w denotes profit and weight of objects respectively) **

(A) 40

(B) 38

(C) 32

(D) 30

**Answer:** B

**Explanation:**

p1/w1 = 5

p2/w2 = 2.5

p3/w3 = 2

p4/w4 = 2

Now select the one which has a max(p/w) ratio

that is p1/w1 = 5, so select 10

next p2/w2 = 2.5, select 10

now p3/w3 and p4/w4 has same ratio but p4 gives maximum profit so select p4. Therefore the total weight=(2+4+9)=15 and max profit = 10+10+18 = 38

**Q.63 The solution of the recurrence relation of **** is **

(A) O(n2)

(B) O(nlog n)

(C) O(n)

(D) O(log n)

**Answer:** C

**Explanation:**

T(n) = 3T(n/2) + n

T(n) = 3(3T(n/4) + n/2) + n

T(n) = 9T(n/4) + 5n/2

T(n) = 9T(n/4) + 5/2T(n) - 15/2T(n/2)

3/2T(n) = 15/2T(n/2) - 9T(n/4)

T(n) = 5T(n/2) - 6T(n/4)

=> O(n)

**Q.64 **

(A) less than 1

(B) less than log n

(C) greater than 1

(D) greater than log n

**Answer:** A

**Explanation:** The particular key x is less than 1; this is the theorem in Universal Hashing.

**Q.65 Given the following statements:**

**Which of the following is true? **

**Answer:** B

**Explanation:** S1 is an NP problem, and S2 is an NP problem.

The subgraph isomorphism problem is in NP. This can be proved using Clique Theory. Read the NP Problems that can be solved using Clique Theory. The set-partition problem is also on NP.

Suppose the certificate y be a subset Y of the set S. The verification algorithm(A) takes the subsequent steps to verify:

1. Check if each number in Y is a number in S.

2. Compute the sum of every number in the set Y.

3. Compute the set S & Y.

4. Compute the sum of every number in the set S - Y.

5. Compare two numbers obtained in (2) and (4) to see if these two numbers are identical. If yes, then return yes.

**Q.66 Suppose the splits at each level of quicksort are in the proportion (1 - α) to α, where 0 < α < ½ is a constant. The minimum depth of a leaf in a recursion tree is approximately given by:**

**Answer:** C

**Explanation:** If we split with α, we will get minimum depth. Suppose α=1/2 means pivot is the central element. With such a case recursion tree will be with minimum depth. In other cases, it will have a maximum.

1. nα^k = 1

k = -logn/logα

2. n(1-α)^k = 1

k = -logn/log(1-α)

**Q.67 Ten signals needing 3000 Hz are multiplexed to a single channel using FDM. How much minimum bandwidth is required by the multiplexed channel? Assuming the guard bands are 300 Hz wide. **

(A) 30,000

(B) 32,700

(C) 33,000

(D) None of the above

**Answer:** B

**Explanation:**

Ten signals need frequency = 10x3000 = 30000 Hz

(10-1) Guard bands (or gaps) needs =300x(10-1) = 2700 Kz

So, the minimum bandwidth need 30000+2700 = 32700Hz

**Q.68 A terminal multiplexer contains six 1200bps terminals and ‘n’ 300 bps terminals connected. If the outgoing line is 9600bps, what is the value of n? **

(A) 4

(B) 8

(C) 16

(D) 28

**Answer:** B

**Explanation:** Since there are six 1200bps terminals, therefore, 6*1200 + n*300 = 9600

On solving this, we get n = 8.

**Q.69 Which of the subsequent is used in the options field of IPv4? **

(A) Strict source routing

(B) Loose source routing

(C) timestamp

(D) All of the above

**Answer:** D

**Explanation:** IPv4 option field used to provide:

Source routing, also known as path addressing, allows a packet sender to partially or completely specify the packet's route through the network.

Source routing can be of two types:

1. Strict source routing

2. Loose source routing

We may also include the timestamp in the option field.

**Q.70 Which OSI layers reference model are host-to-host layers? **

(A) Transport, Session, Presentation, Application

(B) Network, Transport, Session, Presentation

(C) Data-link, Network, Transport, Session

(D) Physical, Data-link, Network, Transport

**Answer:** A

**Explanation:**

**Q.71 **

(A) 1024

(B) 2048

(C) 4096

(D) 8192

**Answer:** C

**Explanation:** Subnet mask contains l's in the network ID part and o's in the host ID part.

255.255.240.0 = 11111111. 11111111. 11110000. 00000000

maximum number of hosts = 2^{12} - 2 = 4094

**Q.72 Four bits are used for packed sequence numbering in a sliding window protocol used in a computer network. What is the maximum window size? **

(A) 4

(B) 8

(C) 15

(D) 16

**Answer:** C

**Explanation:** Maximum window size = 2^{n }- 1

since, n = 4

So, Maximum window size = 2^{4} - 1 = 15

**Q.73 Given the following two grammars : **

**G1: S → AB | aaB **

**A → a | Aa **

**B → b **

**G2: S → a S b S | b S a S | λ **

**Which statement is correct? **

(A) G1 is unambiguous, and G2 is unambiguous.

(B) G1 is unambiguous, and G2 is ambiguous.

(C) G1 is ambiguous, and G2 is unambiguous.

(D) G1 is ambiguous, and G2 is ambiguous.

**Answer:** D

**Explanation:**

**Q.74 Match the following :**

**Codes:**

a b c d

(A) iv iii i ii

(B) iv iii ii i

(C) iii iv i ii

(D) iii iv ii i

**Answer:** C

**Explanation:** A context-free grammar G is in Chomsky normal form if all of its PRs (production rules) are of the form:

```
A -> BC, or
A -> a, or
S - ε,
```

where A, B, & C are nonterminal symbols, a is a terminal symbol, and ε denotes the empty string.

Greibach Normal form: A grammar is said to be in GNF if it is in the form: A-> aα where α ∈ V*.

**Q.75 Given the following two languages:**

**Which statement is correct?**

**Answer:** D

**Explanation:** L1 = {a, aabb, aaabbb...........}

In this, if the input is single "a," then we can pop it and can reach the final state, and if input contains more than one "a," then after pushing all a's when "b" comes as input, start deleting one "a" for each "b" input symbol and then reach the final stage.

Hence L1 is deterministic.

L2 is also deterministic.