## Introduction

UGC NET Exam is one of the popular exams in India for people interested in research. Previous Year Questions are an excellent option to learn about the exam pattern. By solving the PYQs, you will get a basic idea about your preparation. You can evaluate your weak areas and work on them to perform better in the examination. In this article, we have given the questions of UGC NET 2016 July Paper-II. We have also explained every problem adequately to help you learn better.

**Note: **This article contains Q.No. 51 to Q.No. 75 out of the 75 questions asked in NET july 2016 paper II. The solution to Q.No. 01 to Q.No. 25 can be found in the july 2016 Paper-III Part-1 article and the solutions of Q.No. 26 to Q.No. 50 can be found in the july 2016 Paper III Part-2.

### Questions 51 to 75

**(1) Which of the following information about the UNIX file system is not correct?**

**(A) Super block contains the number of i-nodes, the number of disk blocks, and the start of the list of free disk blocks.**

**(B) An i-node contains accounting information as well as enough information to locate all the disk blocks that holds the file’s data.**

**(C) Each i-node is 256-bytes long.**

**(D) All the files and directories are stored in data blocks.**

**Answer: C**

In UNIX, each i-node is 256-bytes long. This statement is incorrect because it is 128 byte long.

**(2) Which of the following option with reference to UNIX operating system is not correct?**

**(A) INT signal is sent by the terminal driver when one types <Control-C> and it is a request to terminate the current operation.**

**(B) TERM is a request to terminate execution completely. The receiving process will clean up its state and exit.**

**(C) QUIT is similar to TERM, except that it defaults to producing a core dump if not caught.**

**(D) KILL is a blockable signal.**

**Answer: D**

In UNIX Operating System, KILL is not a blockable signal.

**(3) A Multicomputer with 256 CPUs is organized as 16x16 grid. What is the worst case delay (in hops) that a message might have to take?**

**(A) 16 (B) 15**

**(C) 32 (D) 30**

**Answer: D**

We have 16 x 16 grid, There will be (M – 1) * (N – 1) hops for M * N grid.

In the given question M = N.

# of hops = (M – 1) * (M – 1)

= 2 * (M – 1)

= 2 * (16 – 1)

= 2 * 15

= 30 hops.

**(4) Suppose that the time to do a null remote procedure call (RPC) (i.e, 0 data bytes) is 1.0 msec, with an additional 1.5 msec for every 1K of data. How long does it take to read 32 K from the file server as 32 1K RPCs?**

**(A) 49 msec (B) 80 msec**

**(C) 48 msec (D) 100 msec**

**Answer: B**

According to given question

1K RPC takes – 1.5 * 32 + 1.0 = 49.0 msec

32 1K RPC will take:

1.5 * 32 + 1.0 * 32

= 80.0 msec

**(5) Let L be the language generated by regular expression 0*10* and accepted by the deterministic finite **__Automata__** M. Consider the relation RM defined by M. As all states are reachable from the start state, RM has ................ equivalence classes.**

**(A) 2 (B) 4**

**(C) 5 (D) 6**

**Answer: D**

Here M is a DFA for L. It is not mentioned that M is a minimal DFA.

R_{M} is defined by M as the set of states reachable from the start state.

This means that all reachable states form 1 class and others form another class. So, in this case the answer must be 2.

**(6) Let L = {0n1n | n≥0} be a context free language.**

**Which of the following is correct?**

**(A) L’ is context free and Lk is not context free for any k≥1**

**(B) L’ is not context free and Lk is context free for any k≥1**

**(C) Both L’ and Lk is for any k≥1 are context free.**

**(D) Both L’ and Lk is for any k≥1 are not context free.**

**Answer: C**

Complements of L1 will also have strings like L = 1^{n}0^{n }.

In general we can make a CFG of complement of L = {0^{n}1^{n} | n≥0} as below:

S → T | aU | V b

T → aT | T a | bT | T b | ba

U → aU | W

V → V b | W

W → aW b | λ

and L = {0^{n}1^{n} | n≥0} is also context free (push 0 and pop 1 into stack and at last we have empty string). o both L1 and L2 are context free.

**(7) Given a Turing Machine**

**M = ({q0,q1,q2,q3}, {a,b}, {a,b,B}, δ, B, {q3})**

**Where δ is a transition function defined as**

**δ(q0,a) = (q1,a,R)**

**δ(q1,b) = (q2,b,R)**

**δ(q2,a) = (q2,a,R)**

**δ(q2,b) = (q3,b,R)**

**The language L(M) accepted by the Turing Machine is given as:**

**(A) aa*b (B) abab**

**(C) aba*b (D) aba***

**Answer: C**

According to given question, we have transition:

δ(q0, a) = (q1, a, R)

δ(q1, b) = (q2, b, R)

δ(q2, a) = (q2, a, R)

δ(q3, b) = (q3, b, R)

**(8) Consider a discrete memoryless channel and assume that H(x) is the amount of information per symbol at the input of the channel; H(y) is the amount of information per symbol at the output of the channel; H(x|y) is the amount of uncertainty remaining on x knowing y; and I(x;y) is the information transmission.**

**Which of the following does not define the channel capacity of a discrete memoryless channel?**

**(A) max I(x;y)**** p(x)**

**(B) max [(H(y) – H(y|x))]**** p(x)**

**(C) max [(H(x) – H(x|y))]**** p(x)**

**(D) max H(x|y)**** p(x)**

**Answer: D**

**(9) Consider a source with symbols A, B, C, D with probabilities 1/2, 1/4, 1/8, 1/8 respectively. What is the average number of bits per symbol for the Huffman code generated from above information?**

**(A) 2 bits per symbol (B) 1.75 bits per symbol**

**(C) 1.50 bits per symbol (D) 1.25 bits per symbol**

**Answer: B**

Bits required per symbol:

A – 0 (1 bit)

B – 10 (2 bit)

C – 110 (3 bit)

D – 111 (3 bit)

Average number of bits per symbol = 1 * 1 / 2 + 2 * 1 / 4 + 3 * 1 / 8 + 3 * 1 / 8 = 7 / 4 = 1.75.

**(10) Which of the following is used for the boundary representation of an image object?**

**(A) Quad Tree (B) Projections**

**(C) Run length coding (D) Chain codes**

** Answer: D**

Chain codes is used for the boundary representation of an image object.

**(11) The region of feasible solution of a linear programming problem has a ............... property in geometry, provided the feasible solution of the problem exists.**

**(A) concavity (B) convexity**

**(C) quadratic (D) polyhedron**

**Answer: B**

The region of feasible solution of a linear programming problem has a convexity property in geometry, provided the feasible solution of the problem exists.

**(12) Consider the following statements:**

**(a) Revised simplex method requires lesser computations than the simplex method.**

**(b) Revised simplex method automatically generates the inverse of the current basis matrix.**

**(c) Less number of entries are needed in each table of revised simplex method than usual simplex method.**

**Which of these statements are correct?**

**(A) (a) and (b) only (B) (a) and (c) only**

**(C) (b) and (c) only (D) (a), (b) and (c)**

**Answer: D**

Revised simplex method requires lesser computations than the simplex method. Revised simplex method automatically generates the inverse of the current basis matrix. Less number of entries are needed in each table of revised simplex method than usual simplex method.

All these statements are correct.

**(13) The following transportation problem:**

**The above solution of a given transportation problem is**

**(A) infeasible solution (B) optimum solution**

**(C) non-optimum solution (D) unbounded solution**

**Answer: B**

The above solution is an optimum solution.

**(14) Let R and S be two fuzzy relations defined as:**

**Then, the resulting relation, T, which relates elements of universe x to elements of universe z using max-min composition is given by**

**Answer: C**

Since x is related to y and y is related to z, to relate universe x and universe z we have to compute max-min composition:

x1z1= max(min(0.7, 0.9), min(0.5, 0.1))

= max(0.7 0.1)

= 0.7

x1z2= max(min(0.7, 0.6), min(0.5, 0.7))

= max(0.6, 0.5)

= 0.6

x1z3= max(min(0.7, 0.2), min(0.5, 0.5))

= max(0.2, 0.5)

= 0.5

x2z1= max(min(0.8, 0.9), min(0.4, 0.1))

= max(0.8, 0.1)

= 0.8

x2z2= max(min(0.8, 0.6), min(0.4, 0.7))

= max(0.6, 0.4)

= 0.6

x3z3= max(min(0.8, 0.2), min(0.4, 0.5))

= max(0.2, 0.4)

= 0.4

**(15) Compute the value of adding the following two fuzzy integers:**

**A = {(0.3,1), (0.6,2), (1,3), (0.7,4), (0.2,5)}**

**B = {(0.5,11), (1,12), (0.5,13)}**

**Where fuzzy addition is defined as**

**μA+B(z) = maxx+y=z (min(μA(x), μB(x)))**

**Then, f(A+B) is equal to**

**(A) {(0.5,12), (0.6,13), (1,14), (0.7,15), (0.7,16), (1,17), (1,18)}**

**(B) {(0.5,12), (0.6,13), (1,14), (1,15), (1,16), (1,17), (1,18)}**

**(C) {(0.3,12), (0.5,13), (0.5,14), (1,15), (0.7,16), (0.5,17), (0.2,18)}**

**(D) {(0.3,12), (0.5,13), (0.6,14), (1,15), (0.7,16), (0.5,17), (0.2,18)}**

**Answer: D**

A={(0.3, 1), (0.6, 2), (1, 3), (0.7, 4), (0.2, 5)}

B={(0.5,11), (1, 12), (0.5, 13)}

first add the numbers(x + y = z) and write the min membership value since function is min((μA(x),μB(x)) u will get following 15 terms:

{(0.3, 12), (0.3, 13), (0.3, 14), (0.5, 13), (0.6, 14), (0.5, 15), (0.5, 14), (1, 15), (0.5, 16), (0.5, 15), (0.7, 16), (0.5, 17), (0.2, 16), (0.2, 17), (0.2, 18)

f(A + B) is equal to {(0.3, 12), (0.5, 13), (0.6, 14), (1, 15), (0.7, 16), (0.5, 17), (0.2, 18)}

**(16) A perceptron has input weights W1 = -3.9 and W2 = 1.1 with threshold value T = 0.3. What output does it give for the input x1 = 1.3 and x2 = 2.2?**

**(A) -2.65 (B) -2.30**

**(C) 0 (D) 1**

**Answer: C**

According to given question w1 = – 3.9 and w2 = 1.1 and x1 = 1.3 and x2 = 2.2

weighted sum = w1 * x1 + w2 * x2 + . . . wn * xn

i.e. -3.9 * 1.3 + 1.1 * 2.2

= -5.07 + 2.42

= -2.65

Now we will compare the weighted sum -2.65 to the threshold 0.3.

-2.65 < 0.3

Then output will be zero.

**(17) What is the function of following UNIX command?**

**WC – l<a >b&**

**(A) It runs the word count program to count the number of lines in its input, a, writing the result to b, as a foreground process.**

**(B) It runs the word count program to count the number of lines in its input, a, writing the result to b, but does it in the background.**

**(C) It counts the errors during the execution of a process, a, and puts the result in process b.**

**(D) It copies the ‘ l ’ numbers of lines of program from file, a, and stores in file b.**

**Answer: B**

It runs the word count program to count the number of lines in its input, a, writing the result to b, but does it in the background.

**(18) Which of the following statement is not correct with reference to cron daemon in UNIX O.S?**

**(A) The cron daemon is the standard tool for running commands on a pre-determined schedule.**

**(B) It starts when the system boots and runs as long as the system is up.**

**(C) Cron reads configuration files that contain list of command lines and the times at which they invoked.**

**(D) Crontab for individual users are not stored.**

**Answer: D**

Crontab for individual users are not stored. Only this statement is incorrect as crontab is stored for individual user.

**(19) In Unix, files can be protected by assigning each one a 9-bit mode called rights bits. Now, consider the following two statements:**

**I. A mode of 641 (octal) means that the owner can read and write the file, other members of the owner’s group can read it, and users can execute only.**

**II. A mode of 100 (octal) allows the owner to execute the file, but prohibits all other access.**

**Which of the following options is correct with reference to above statements?**

**(A) Only I is correct (B) Only II is correct**

**(C) Both I and II are correct (D) Both I and II are incorrect**

**Answer: C**

In UNIX, a mode of 641 (octal) means that the owner can read and write the file, other members of the owner’s group can read it, and users can execute only.

A mode of 100 (octal) allows the owner to execute the file, but prohibits all other access.

Both statements are correct regarding UNIX.

**(20) Consider the statement,**

**"Either -2 ≤ x ≤ -1 or 1 ≤ x ≤ 2"**

**The negation of this statement is**

**(A) x<-2 or 2<x or -1<x<1 (B) x<-2 or 2<x**

**(C) -1<x<1 (D) x ≤ -2 or 2 ≤ x or -1<x<1 **

**Answer: A**

“Either – 2 ≤ x ≤ – 1 or 1 ≤ x ≤2”.

i.e. Eiter x ≥ -2 or x ≤ -1 or 1 ≤ x or x ≤ 2”

We have to find negation of above statement:

Negation of x ≥ -2 is x < 2. Negation of x ≤ -1 is x > -1.

Negation of 1 ≤ x is x < 1 Negation of x ≤ 2 is x > 2.

i.e. x < – 2 or 2 < x or – 1 < x < 1.

**(21) Which of the following is characteristic of an MIS?**** (A) Provides guidance in identifying problems, finding an evaluating alternative solutions, and selecting or comparing alternatives.**

**(B) Draws on diverse yet predictable data resources to aggregate and summarize data.**

**(C) High volume, data capture focus.**

**(D) Has as its goal the efficiency of data movement and processing and interfacing different TPS.**

**Answer: B**

Characteristic of an Management Information System(MIS) is draws on diverse yet predictable data resources to aggregate and summarize data.

**(22) How does randomized hill-climbing choose the next move each time?**

**(A) It generates a random move from the moveset, and accepts this move.**

**(B) It generates a random move from the whole state space, and accepts this move.**

**(C) It generates a random move from the moveset, and accepts this move only if this move improves the evaluation function.**

**(D) It generates a random move from the whole state space, and accepts this move only if this move improves the evaluation function.**

**Answer: C**

In randomized hill-climbing choose the next move each time it generates a random move from the moveset, and accepts this move only if this move improves the evaluation function.

**(23) Consider the following game tree in which root is a maximizing node and children are visited left to right. What nodes will be pruned by the alpha-beta pruning?**

**(A) I (B) HI**

**(C) CHI (D) GHI**

**Answer: B**

We have to maximize A and minimize B and C. Minimum of B is 3 and C is 2 But max from B is 12 and max from C is 15. So, there is no need of further expanding H and I. Because whatever value their successor will produce will be immaterial. Hence H I are pruned.

**(24) Consider a 3-puzzle where, like in the usual 8-puzzle game, a tile can only move to an adjacent empty space. Given the initial state**

**which of the following state cannot be reached?**

**Answer: C**

Only in option C, we can’t reach the Goal State.

**(25) A software program that infers and manipulates existing knowledge in order to generate new knowledge is known as:**

**(A) Data Dictionary (B) Reference Mechanism**

**(C) Inference Engine (D) Control Strategy**

**Answer: C**

A software program that infers and manipulates existing knowledge in order to generate new knowledge is known as Inference engine.

**“**Must Recommended Topic, __procedure call in compiler design__“

## 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 maximum number of attempts for the UGC NET examination?**

There is no bar on the number of attempts of this examination. Candidates can appear for the examination as long as they are eligible.

**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.