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 2011 December Paper-II. We have also explained every problem adequately to help you learn better.
Note: This article contains Q.No. 01 to Q.No. 25 out of the 50 questions asked in NET December 2011 paper II. The solution to Q.No. 25 to Q.No. 50 can be found in the December 2011 Paper-III Part-2 article.
Questions 01 to 25
(1) Which of the following data structure is Non-linear type ?
(A) Strings (B) Lists
(C) Stacks (D) None of the above
Answer: D
The Data structure is said to be Non-Linear Data Structures if it's elements do not form a sequence or a linear series but form a hierarchical format.
For Example:- Graph, Tree, Binary Search Tree etc.
(2) The total number of comparisons in a bubble sort is
(A) 0(log n) (B) 0(n log n)
(C) 0(n) (D) None of the above
Answer:B
First iteration of Outer loop → n-1 comparisions
Second iteration of Outer loop → n-2 comparisions (Since largest element has gone to its proper place)
Third iteration of Outer loop → n-3 comparisions (Since two largest elements has gone to their proper places)
Total number of comparisions = (n-1)+(n-2)+(n-3)+............+1 = n(n-1) / 2 = O(n^2)
(3) Which of the following is a bad example of recursion ?
(A) Factorial (B) Fibonacci numbers
(C) Tower of Hanai (D) Tree traversal
Answer:B
Fibonacci numbers is a bad example of recursion.
(4) Domain and Range of the function 
is
(A) x ≥ 3/2 , y ≥ 0 (B) x > 3/2 , y ≤ 0
(C) x ≥ 3/2 , y ≤ 0 (D) x ≤ 3/2 , y ≤ 0
Answer:D
-2x+3 >= 0 (As Square Root cannot have negative value )
-2x >= -3 implies x <= 3/2
for x = 3/2 , y=0
for any value of x , y < 0
The possible values are x <= 3/2 , y <= 0
(5) Maximum number of edges in a n Node undirected graph without self loop is
(A) n2 (B) n(n – 1)
(C) n(n + 1) (D) n(n – 1)/2
Answer: D
Total no of pairs = Total no of edges (max) = (n/2) = n(n−1)/2
(6) A hash table has space for 75 records, then the probability of collision before the table is 6% full.
(A) .25 (B) .20
(C) .35 (D) .30
Answer: B
To make the table 6% full, we need to insert at least ( 75∙6 100) = 4.5 round up to 5 values.
Probability of collision during first insertion is 1/75
Probability of collision during third insertion is 2/75
Probability of collision during fourth insertion is 3/75
Probability of collision during fifth insertion is 4/75
Probability of collision during sixth insertion is 5/75
Total probability of collision to make the table 6% full is (1 + 2 + 3 + 4 + 5) /75 = 0.2
(7) BCC in the internet refers to
(A) Black carbon copy (B) Blind carbon copy
(C) Blank carbon copy (D) Beautiful carbon copy
Answer: B
On .75th insertion probability of collision = 1/75
On 1.5th insertion probability of collision = 2/75
On 2.25th insertion probability of collision = 3/75
On 3th insertion probability of collision = 4/75
On 3.75th insertion probability of collision = 5/75
So the required probability is 1+2+3+4+5/75 = .20
(8) Hub is a term used with
(A) A Star Networks (B) A Ring Networks
(C) A Router (D) A Bridge
Answer: A
A hub is a common connection point for devices in star topology.
(9) The amount of uncertainty in a system of symbol is called
(A) Bandwidth (B) Entropy
(C) Loss (D) Quantum
Answer: B
Entropy measures the amount of uncertainty in information or data sources.
(10) Which of the following network access standard disassembler is used for connection station to a packet switched network ?
(A) X.3 (B) X.21
(C) X.25 (D) X.75
Answer: C
X.25 is a mechanism for transferring transparent data. Standards exist for encapsulation of higher-level data such as TCP/IP or SNA traffic over X.25. One of the largest uses of X.25 is the transfer of asynchronous data streams such as those produced by simple terminals like credit card readers. These devices connect to a Packet Assembler/Disassembler (PAD) that organises the async data streams into X.25 packets for transmission across the network. PAD design is based on ITU standards X.28, X.29 and X.3.
(11) A station in a network forward incoming packets by placing them on its shortest output queue. What routing algorithm is being used ?
(A) Hot potato routing (B) Flooding
(C) Static routing (D) Delta routing
Answer: A
Hot potato routing is a routing technique enabling packet routing without storing them in buffers. Instead, this technique continuously transfers data packets until they reach their destination without the packets having to wait or be stored in a buffer.
(12) Start and stop bits are used in serial communications for
(A) Error detection (B) Error correction
(C) Synchronization (D) Slowing down the communication
Answer: C
Start and stop bits are used in serial communications for synchronization to tell the receiver that ignoring these bits of data is acceptable.
(13) For a data entry project for office staff who have never used computers before (user interface and userfriendliness are extremely important), one will use
(A) Spiral model (B) Component based model
(C) Prototyping (D) Waterfall model
Answer: C
Correct answer should be C (prototyping) as only user friendliness is needed so the prototyping model is best suited.
(14) An SRS
(A) establishes the basis for agreement between client and the supplier.
(B) provides a reference for validation of the final product.
(C) is a prerequisite to high quality software.
(D) all of the above.
Answer:D
All the options are purposes of a good SRS.
(15) McCabe’s cyclomatic metric V(G) of a graph G with n vertices, e edges and p connected component is
(A) e (B) n
(C) e – n + p (D) e – n + 2p
Answer:C
Cyclomatic complexity (C) = E - N + 2P
Where:
P = number of disconnected parts of the flow graph (e.g. a calling program and a subroutine)
E = number of edges (transfers of control)
N = number of nodes (sequential group of statements containing only one transfer of control)
(16) Emergency fixes known as patches are result of
(A) adaptive maintenance (B) perfective maintenance
(C) corrective maintenance (D) none of the above
Answer:C
Corrective maintenance is a maintenance task performed to identify, isolate, and rectify a fault so that the failed equipment, machine, or system can be restored to an operational condition within the tolerances or limits established for in-service operations.
(17) Design recovery from source code is done during
(A) reverse engineering (B) re-engineering
(C) reuse (D) all of the above
Answer: D
Reverse engineering, also called back engineering, is the processes of extracting knowledge or design information from anything man-made and re-producing it or re-producing anything based on the extracted information
(18) Following is used to demonstrate that the new release of software still performs the old one did by rerunning the old tests :
(A) Functional testing (B) Path testing
(C) Stress testing (D) Regression testing
Answer: D
Regression Testing: Whenever software is corrected, some aspect of the software configuration (the program, its documentation, or the data that support it) is changed.
Regression testing ensures that changes do not introduce additional errors.
(19) The post order traversal of a binary tree is DEBFCA. Find out the preorder traversal.
(A) ABFCDE (B) ADBFEC
(C) ABDECF (D) ABDCEF
Answer: C
Step 1: Construct binary tree from the given post order traversal : debfca.
1. From the given post order traversal will identify that “a” is the root.
2. “a” being the root has a left and right side. Since left side is considered first, it is expected that it will have two sides. Hence, “a” will have “deb” on the left node and “fc” on the right.
3. Further on the left side, we understand that “b” is the root, which will have left side as “d” and right one as “e”.
4. On the right side of “a”, c will be the root which will only have a left side with “f”
5. Hence, below binary tree is constructed:
Step 2: Perform pre-order traversal of the constructed binary tree.
After traversing the above tree in the pre-order, we get
: abdecf.
(20) B + tree are preferred to binary tree in database because
(A) Disk capacities are greater than memory capacities
(B) Disk access much slower than memory access
(C) Disk data transfer rates are much less than memory data transfer rate
(D) Disk are more reliable than memory
Answer: B
The major advantage of B+ tree is in reducing the number of last level access which would be from disk in case of large data size.
(21) What deletes the entire file except the file structure ?
(A) ERASE (B) DELETE
(C) ZAP (D) PACK
Answer: C
Zap command removes all records from a table, leaving just the table structure.
The ZAP command unconditionally removes all records from a table, but leaves the table structure intact.
(22) Which command classes text file, which has been created using “SET ALTERNATIVE” “Command” ?
(A) SET ALTERNATE OFF (B) CLOSE DATABASE
(C) CLOSE ALTERNATE (D) CLEAR ALL
Answer: A
SET ALTERNATE controls the process of recording the inputs and outputs in a different text file.
(23) Data security threats include
(A) privacy invasion (B) hardware failure
(C) fraudulent manipulation of data (D) encryption and decryption
Answer: C
Fraudulent manipulation of data is related to Data Security Threats.
(24) Which of the following statements is true, when structure of database file with 20 records is modified ?
(A) ? EOF ( ) Prints. T
(B) ? BOF ( ) Prints F
(C) ? BOF ( ) Prints T
(D) ? EOF ( ) Prints F
Answer: A
EOF stands for End of File and as soon as the structure of database file that has 20 records is modified, ?EOF will print T i.e., TRUE.
(25) The SQL Expression Select distinct T. branch name from branch T, branch S where T. assets > S. assets and S. branch-city = DELHI, finds the name of
(A) all branches that have greater asset than any branch located in DELHI.
(B) all branches that have greater assets than allocated in DELHI.
(C) the branch that has the greatest asset in DELHI.
(D) any branch that has greater asset than any branch located in DELHI.
Answer: A
SELECT distinct TT. branch name
From branch TT, branch SS
Where TT. assets >SS.assets and SS.branch-city = DELHI
Return : Any branch that has greater assets than any(any one ) branch located in DELHI.
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.
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