MODEL SOLVED PAPER FOR UGC NET JUNE 2014
1. A chip having 200 gates will be classified
as
Ans:-B
Note:
Small Scale Integration(SSI) - <10
Medium Scale Integration(MSI) - between 10 to 1000
Large Scale Integration(LSI) – Greater than 1000
Very Large Scale Integration(VLSI) – greater than 100000.
Small Scale Integration(SSI) - <10
Medium Scale Integration(MSI) - between 10 to 1000
Large Scale Integration(LSI) – Greater than 1000
Very Large Scale Integration(VLSI) – greater than 100000.
2. Consider the following statements :
I. Recursive languages are closed under complementation.
II. Recursively enumerable languages are closed under union.
III. Recursively enumerable languages are closed under complementation.
Which of the above statements are true ?
(A) I only
(B) I and II
(C) I and III
(D) II and III
I. Recursive languages are closed under complementation.
II. Recursively enumerable languages are closed under union.
III. Recursively enumerable languages are closed under complementation.
Which of the above statements are true ?
(A) I only
(B) I and II
(C) I and III
(D) II and III
Ans:-B
Very often asked
question in UGC.
Explanation:-
Recursive languages are closed under the following operations.
Recursive languages are closed under the following operations.
1.
Concatenation
2.
Kleene Star
3.
Union
4.
Intersection
5.
Complement
6.
Set difference
Recursively enumerable languages are closed under the following operations.
1.
Kleene star
2.
Concatenation
3.
Union
4.
Intersection
3. Consider the following page trace : 4,3, 2, 1, 4, 3, 5, 4, 3,
2, 1, 5 Percentage of page fault that would occur if FIFO page replacement
algorithm is used with number of frames for the JOB m = 4 will be
(A) 8
(B) 9
(C) 10
(D) 12
(A) 8
(B) 9
(C) 10
(D) 12
Ans:-C
Opearating system
based question.
Explanation:-
The reference string is 4,3,2,1,4,3,5,4,3,2,1,5
The number of frames m = 4
The first 4 references 4,3,2,1 cause page faults and brought into the empty frames.
The reference string is 4,3,2,1,4,3,5,4,3,2,1,5
The number of frames m = 4
The first 4 references 4,3,2,1 cause page faults and brought into the empty frames.
So there will be 4 page faults initially.
The next reference 4 is already available in the frame and so there will be no page fault.
Similarly, The next reference 3 is also already available and so there is no page fault.
The next reference (5) replaces page 4 which was brought in to first frame.
The next reference 4 is already available in the frame and so there will be no page fault.
Similarly, The next reference 3 is also already available and so there is no page fault.
The next reference (5) replaces page 4 which was brought in to first frame.
So, No of page faults=5. Because we already
had 4 page faults.
Next reference (4) replaces page 3 which is the next to come in.
Next reference (4) replaces page 3 which is the next to come in.
No of page faults=6.
Next reference (3) replaces 2.No of page faults till now=7.
Next reference (2) replaces 1 which was the last of the pages to come in.
No of page faults till now=8.
Next reference (1) replaces 5 which was the first to come in in the second cycle. No of page faults will increase to 9.
The last reference in the reference string is 5 which will replace 4. No of page faults will be finally 10.
Next reference (1) replaces 5 which was the first to come in in the second cycle. No of page faults will increase to 9.
The last reference in the reference string is 5 which will replace 4. No of page faults will be finally 10.
So the answer is 10.
4. Evaluate the following postfix
expression 579*+493/+–
Ans.:61.
Explanation:
1.
7*9=63
63+5=68 Part
1
2.
9/3=3
3+4=7 Part
2
Part
1 – Part 2
Result:
68-7=61
5. Which of the following strings is in the language defined by
grammar
S → 0A,
S → 0A,
A → 1A/0A/1
(A) 01100
(B) 00101
(C ) 10011
(D) 11111
(A) 01100
(B) 00101
(C ) 10011
(D) 11111
Ans:-B
Hint: Any production
needs to be started from S, which is the starting point.
From the above given
symbols, we can substitute A with 1A or 0A or 1 as per the requirements of the
result.
Explanation:- Non terminals in
the above grammar are S and A.
Non terminals are generally denoted by capital
letters.
Terminals in the given
problem are 0 and 1.
The start symbol of
the grammar is S, which is the starting point of the production. S->0A. So
rule out strings beginning with 1. so that leaves us with two options A and B.
Start of production:
S->0A
->00A Hint: In place of A, we are substituting 0A
->001A Hint: In place of A, we are substituting 1A
->0010A Hint: In place of A, we are substituting 0A
->00101 Hint: In place of A, we are substituting 1
So the option B is correct.
->00A Hint: In place of A, we are substituting 0A
->001A Hint: In place of A, we are substituting 1A
->0010A Hint: In place of A, we are substituting 0A
->00101 Hint: In place of A, we are substituting 1
So the option B is correct.
6.
Interner layer provides
Quality
of service
Routing
Addressing.
7.
The search algorithm that helps to find optimal solution.
Breadth
First search.
8. The program used to determine the
round-trip delay between a workstation and a destination address is
Trace
route
9.
Some of the data link layer standards are.
Ethernet
Frame
relay
10
base-T
Token
ring.
10.
The topology with highest reliasbility is
Mesh
topology.
11. What is the maximum number of nodes in a
B-tree of order 10 of depth 3 (root at depth 0) ?
(A) 111
(B) 999
(C) 9999
(D) None of the above
(A) 111
(B) 999
(C) 9999
(D) None of the above
Ans:-C
Explanation:-
Formula for Calculating maximum number of
nodes in a B-tree
=>mh+1-1
=>mh+1-1
Where m is the order
h is the depth
Here, m=10 and h=3. So the formula becomes
104-1 which is 10000-1=9999 and therefore C is the correct option (Hint:10^4=10*10*10*10=10000)
Here, m=10 and h=3. So the formula becomes
104-1 which is 10000-1=9999 and therefore C is the correct option (Hint:10^4=10*10*10*10=10000)
12.
What is used to regulate the flow of electrons in CRT.
Control
grid.
13.
The thin or nonexistent layer in broadcast network is
Network
layer
14.
The most powerful parsing method is
Canonical
LR
15.
Shift reduce parsers are also called as
Bottom
up parsers.
really helpful.......
ReplyDeletegood collection of questions.Thaks a lot.
ReplyDeleteSimply superb hats of u sir
ReplyDelete