Tuesday 31 July 2018

UGC NET July 2018 result released,Check Result at cbsenet.nic.in.

Steps to check UGC NET July 2018 result and and revised answer key?

The Central Board of Secondary Education  had conducted the UGC NET 2018 on July 8 (Sunday) and re-exam on July 22, declared the CBSE UGC-NET 2018 result today evening. Candidates who will clear the paper will be eligible for a JRF or can apply for the post of assistant professors. This year, there were only 2 papers for the first time. paper 1 is about research aptitude and paper 2 is subject specific. Total of 11.48 lakh candidates took the eligibility test in 84 different subjects this year. All the eligible candidates who had appeared for the UFC NET exam can check the results at cbsenet.nic.in. Here's the direct link of the result -
http://cbseresults.nic.in/UGCpxy/net_july2018.htm

Login with your Application number, your Roll Number and DoB  -



















Your marks in UGC NET 2018 with the respective percentage will be shown.

You provided with link for the  UGC NET CUT-OFF . 


Revised answer keys after verification of challenges -July 2018


Detail page for result of CBSE-UGC NET RESULT - July 2018 with
candidate details, Obtained marks, Total marks and percentages, Remarks, Subject-wise/category-wise cut off for Assistant Professor and JRF for UGC NET 2018. Remarks marks qualification in CBSE UGC NET 2018 for for Assistant Professor.

  

Thursday 26 July 2018

UGC NET July 2018:Answer key released,Check at cbsenet.nic.in

Steps to check UGC NET July 2018 Answer Key and Your Recorded Response?

Step 1 – Visit the official website and login wit your Application number, your DoB and Security pin provided -
https://jeemain.nic.in/cbsenet/keychallange/Login.aspx

After login you can see buttons:

View Recorded Response - To view your answers
View Answer Keys - To view the official answer key
Challenge Response - A call to prove or justify your response.
Challenge Answer Keys -A call to prove or justify Answer keys.

You can also find the link on the same page for image of your OMRs. 
Click Here To View Images of OMR Sheet

You will be shown page like below after successful Login:



1) View Answer Keys:
After clicking VIEW ANSWER KEYS button, you will be redirected to page where you have to select following options for UGC NET paper 1:

Select Question Paper (Examination Date) : -date of your UGC NET paper I
Select Subject : - Not Applicable (General paper)
Book No : - booklet No. for your UGC NET paper I

After submitting the above data you will be shown the answer keys for the selected Booklet No.



Tuesday 10 July 2018

NTA UGC-NET Computer Science November 2017 Solved Questions Paper-2

The NTA UGC-NET November, 2017 Examination was conducted on 5th November by CBSE( Central Board of Secondary Education). The following are the answer keys of Test Booklet Series B, Paper-1 of UGC NET / CBSE NET held on 5th November, 2017. The answer keys are as per the answer keys released by CBSE on behalf of UGC. It is to be noted that the UGC NET Examination 2017 June and 2017 December was not held. You can also Download PDF of solved computer science nta ugc-net previous year question paper with answer key from below links.

  Question paper NTA NET Computer Science Paper 2 Nov 2017: Download Free PDF Now

Question paper with Answer key NTA NET Computer Science Paper 2 Nov 2017: Download Free PDF Now

1. If the time is now 4 O’clock, what will be the time after 101 hours from now ?
(1) 9 O’clock
(2) 8 O’clock
(3) 5 O’clock
(4) 4 O’clock

Answer is: (1) 9 O’clock
101 / 24 = = 96 + 5 = reminder 5 => after 96 hour it would be 4 O’clock and then after 5 hour,
it's 9 O'clock

2. Let m=(313)4 and n=(322)4. Find the base 4 expansion of m+n.
(1) (635)4
(2) (32312)4
(3) (21323)4
(4) (1301)4

Answer is: (4) (1301)4
m+n
(313)4 + (322)4
( _ _ 5)4 =>4 (here add unit's digit, we get 5 >= 4 , so carry 1, (so one is added to the ten's digit))
( _ 41)4=> ( _ 0 1)4 (here, we sum ten's digits => (2+1+1 (carry 1 from unit's digit sum))) =4 >=4. so, carry 1 will be passed to hundreds digit.
( 7 4 1)4=> ( 3 0 1)4(here, we sum hundreds digits => (3+3+1 (carry 1 from tens digit sum))) =7 >=4. so, carry 1 will be passed to thousands digit. so, finally (1 3 0 1)4

3. 
Find the boolean product A⊙B of the two matrices.

(A) (1) (B) (2) (C) (3) (D) (4)
Answer is: (A) (1)

4. How many distinguishable permutations of the letters in the word BANANA are there ?

(1) 720
(2) 120
(3) 60
(4) 36

Answer is: (3) 60
Total permutation with 6 words are 720
A is repeated 3 times and N is repeated 2 times.
Permutations of 3 A's are 3!
Permutations of 3 N's are 2!
So, divide the total 720 permutations by 2! and 3!, gives answer 60 permutations.

5. Consider the graph given below :

Use Kruskal’s algorithm to find a minimal spanning tree for the graph. The List of the edges of the tree in the order in which they are choosen is ?
(1) AD, AE, AG, GC, GB, BF
(2) GC, GB, BF, GA, AD, AE
(3) GC, AD, GB, GA, BF, AE
(4) AD, AG, GC, AE, GB, BF

Kruskal's algorithm is a MST(minimum-spanning-tree) algo., which finds an edge with the least possible edge weight that combines any two trees in the forest (initially all vertex are considered as separate tree). Kruskal's algorithm follows the greedy algorithm approach in graph theory as it finds a MST for a connected weighted graph adding increasing cost edges at each step (Such that graph doen't form a cycle).
Sorted edges by weight
Edge  weight
AD       2
GC       2
GB       3
GA       3
BC       4
BF       4
DG      4
AE      4
ED      5
CF      5
AC     6
DB   weight not given
So, the bold edges above will form a MST of graph in the order of increasing edge weight.

6. The Boolean function with the Karnaugh map is :

(1) (A+C).D+B
(2) (A+B).C+D
(3) (A+D).C+B
(4) (A+C).B+D

Answer is: (1) (A+C).D+B

7. The Octal equivalent of the binary number 1011101011 is :
(1) 7353
(2) 1353
(3) 5651
(4) 5657

Answer is: (2) 1353
Group the three bits from the right side (from LSB) and convert to equivalent decimal.

8. Let P and Q be two propositions, ¬ (P ↔ Q) is equivalent to :
(1) P ↔ ¬ Q
(2) ¬ P ↔ Q
(3) ¬ P ↔ ¬ Q
(4) Q → P

Answer is: (1) P ↔ ¬ Q

9. Negation of the proposition ∃ x H(x) is :
(1) ∃x ¬ H(x)
(2) ∀ x ¬ H(x)
(3) ∀ x H(x)
(4) ¬ x H(x)


Ans is: (2) ∀ x ¬ H(x)
Negation of (∃ x H(x)) = ¬ ∃ x H(x) = ∀ x ¬ H(x)(as negation symbol moves from left to right it changes the symbol ∃ to ∀).

10. The output of the following combinational circuit is F.

The value of F is :
(1) P1 + P2' P3
(2) P1 + P2' P3'
(3)P1 + P2 P3'
(4)P1'+ P2 P3

Correct answer is : (2) P1+P2'P3' outputs of OR gates are:
(P1+P2'+P3'(P1+P2'+P3(P1+P2+P3')
= (P1+P1P2'+P1P3+P2'+P2'P3+P3'P1+P3'P2+0) . (P1+P2+P3')
=(P1+P1P2'+P1P3+P3'P1+P2'+P2'P3+P3'P2. (P1+P2+P3'
(P1+P2'. (P1+P2+P3'
=(P1+P1P2+P1P3'+P2'P1+0+P2'P3')
=(P1+P2'P3')

11. ‘ptrdata’ is a pointer to a data type. The expression *ptrdata++ is evaluated as (in C++) :
(1) *(ptrdata++)
(2) (*ptrdata)++
(3) *(ptrdata)++
(4) Depends on compiler

Answer is: (1) *(ptrdata++)
as the ++ has higher precedence than * operator.

12. The associativity of which of the following operators is Left to Right, in C++ ?
(1) Unary Operator
(2) Logical not
(3) Array element access
(4) addressof

Answer is: (3) Array element access
Unary Operator -- right to left associativity
Logical not -- right to left associativity
Array element access -- Array subscript - [ ] - Left to Right associativity addressof -- right to left associativity

13. A member function can always access the data in __________ , (in C++).
(1) the class of which it is member
(2) the object of which it is a member
(3) the public part of its class
(4) the private part of its class

Correct Answer is: (1) the class of which it is member

14. Which of the following is not correct for virtual function in C++ ?
(1) Must be declared in public section of class.
(2) Virtual function can be static.
(3) Virtual function should be accessed using pointers.
(4) Virtual function is defined in base class.

Answer is: (2) Virtual function can be static. A virtual function must be declared within public section of base class and is overridden by the derived-class. When we refer to a derived class object using a reference of the base class, you can call a virtual function for that class object.then, version of the function defines in derived class will execute.

15. Which of the following is not correct (in C++) ?
(1) Class templates and function templates are instantiated in the same way.
(2) Class templates differ from function templates in the way they are initiated.
(3) Class template is initiated by defining an object using the template argument.
(4) Class templates are generally used for storage classes.

OPTION 2, 3, 4 ARE CORRECT (as per UGC answer key)

16. Which of the following is/are true with reference to ‘view’ in DBMS ?
(a) A ‘view’ is a special stored procedure executed when certain event occurs.
(b) A ‘view’ is a virtual table, which occurs after executing a pre-compiled query.(1) Only (a) is true
(2) Only (b) is true
(3) Both (a) and (b) are true
(4) Neither (a) nor (b) are true

(2) Only (b) is true View is the table or the result of query after query execution.

17. In SQL, __________ is an Aggregate function.
(1) SELECT (2) CREATE (3) AVG (4) MODIFY

Correct Answer is: AVG function used to find the average of the particular attribute values. e.g., to find avg salary of employee of a company.

18. Match the following with respect to RDBMS :
(a) Entity integrity (i) enforces some specific business rule that do not fall into entity or domain.
(b) Domain integrity(ii) Rows can’t be deleted which are used by other records
(c) Referential integrity(iii) enforces valid entries for a column
(d) User defined integrity (iv) No duplicate rows in a table
Code :
(1) (iii) (iv) (i) (ii)
(2) (iv) (iii) (ii) (i)
(3) (iv) (ii) (iii) (i)
(4) (ii) (iii) (iv) (i)

Correct Answer is: (2) (iv) (iii) (ii) (i)
Entity integrity: No duplicate rows in a table. Every entity(row) in the table must be unique based on the primary key of the table. Table would not allow the same primary key twice.
Domain integrity:Defines the what type of values to be inserted to the particular (attribute)column of DB for an entity. If we define the salary attribute in employee class then, domain of such values are numeric (say double). This integrity enforces valid entries for a column. (not allow alpha-numeric values for the salary).
Referential integrity: is based on foreign keys (so, it referred by other referencing table, deletion of such entries causes the undefined value for referencing table). Rows can’t be deleted which are used by other records.
User defined integrity: enforces some specific business rule that do not fall into entity or domain.
For e.g., you defining the constraint over the values of the attribute. let's assume that you are inserting the values to the table such that max and minimum salary for the part-time employee must be in in range 10,000 and 100. (So, you are restricting any insert to DB outside this salary range).

19. In RDBMS, different classes of relations are created using __________ technique to prevent modification anomalies.
(1) Functional Dependencies
(2) Data integrity
(3) Referential integrity
(4) Normal Forms

Correct Answer is: (4) Normal Forms
Normal forms are used to reduce the redundancy and increase the data consistency.

20. __________ SQL command changes one or more fields in a record.
(1) LOOK-UP
(2) INSERT
(3) MODIFY
(4) CHANGE

Correct Answer is: (3) MODIFY
21. Consider an array representation of an n element binary heap where the elements are stored
from index 1 to index n of the array. For the element stored at index i of the array (i<=n),
the index of the parent is :
(1) floor ((i+1)/2)
(2) ceiling ((i+1)/2)
(3) floor (i/2)
(4) ceiling (i/2)

Correct Answer is: (3) floor (i/2)
As shown below the root of the 13 and 10 are floor of the 4 and 5 respectively.

22. The following numbers are inserted into an empty binary search tree in the given order :
10, 1, 3, 5, 15, 12, 16. What is the height of the binary search tree ?
(1) 3
(2) 4
(3) 5
(4) 6

Correct Answer is: (1) 3
Here, height is longest path from root to the leaf node. As shown green number in below image indicates the count of the tree edges which has max height.

23. Let G be an undirected connected graph with distinct edge weight. Let Emax be the edge
with maximum weight and Emin the edge with minimum weight. Which of the following
statements is false ?
(1) Every minimum spanning tree of G must contain Emin.
(2) If Emax is in minimum spanning tree, then its removal must disconnect G.
(3) No minimum spanning tree contains Emax.
(4) G has a unique minimum spanning tree.

Correct Answer is: (3) No minimum spanning tree contains Emax.

24. A list of n strings, each of length n, is sorted into lexicographic order using merge - sort
algorithm. The worst case running time of this computation is :
(1) O(n log n)
(2) O(n2 log n)
(3) O(n2 + log n)
(4) O(n3)

Correct Answer is: (2) O(n2 log n)

25. Postorder traversal of a given binary search tree T produces following sequence of keys :
3, 5, 7, 9, 4, 17, 16, 20, 18, 15, 14
Which one of the following sequences of keys can be the result of an in-order traversal of
the tree T ?
(1) 3, 4, 5, 7, 9, 14, 20, 18, 17, 16, 15
(2) 20, 18, 17, 16, 15, 14, 3, 4, 5, 7, 9
(3) 20, 18, 17, 16, 15, 14, 9, 7, 5, 4, 3
(4) 3, 4, 5, 7, 9, 14, 15, 16, 17, 18, 20

Correct answer is: (4) 3, 4, 5, 7, 9, 14, 15, 16, 17, 18, 20
3, 5, 7, 9, 4, 17, 16, 20, 18, 15, 14
Here, the root of the binary tree is 14 (last element), as it's post-order traversal.
Right side of the tree contains bigger elements than 14
Left side of the tree contains bigger elements than 14
So, the first number >14 on left of root (14) (check from right to left).
=>root of right sub-tree is 15
So, the first number <14 on left of root (14) (check from right to left).
=>root of left sub-tree is 4
now we have two sequences: right tree and left tree
right tree:  17, 16, 20, 18, 15,
left tree: 3, 5, 7, 9, 4
Here, also similarly  find left and right sub-tree of both tree root-right(15) and the root-left(4).
So, final binary tree is

26. Which of the following devices takes data sent from one network device and forwards it to
the destination node based on MAC address ?
(1) Hub
(2) Modem
(3) Switch
(4) Gateway

Correct answer is: (3) Switch

27. __________ do not take their decisions on measurements or estimates of the current traffic
and topology.
(1) Static algorithms
(2) Adaptive algorithms
(3) Non - adaptive algorithms
(4) Recursive algorithms

Correct answer is: (3) Non - adaptive algorithms

28. The number of bits used for addressing in Gigabit Ethernet is __________.
(1) 32 bits
(2) 48 bits
(3) 64 bits
(4) 128 bits

Correct answer is: (2) 48 bits

29. Which of the following layer of OSI Reference model is also called end-to-end layer ?
(1) Network layer
(2) Datalink layer
(3) Session layer
(4) Transport layer

Correct answer is: (4) Transport layer

30. The IP address __________ is used by hosts when they are being booted.
(1) 0.0.0.0
(2) 1.0.0.0
(3) 1.1.1.1
(4) 255.255.255.255

Correct answer is:  (1) 0.0.0.0

31. Consider the following program fragment in assembly language :
          mov ax, 0h
          mov cx, 0A h
doloop :
          dec ax
          loop doloop
What is the value of ax and cx registers after the completion of the doloop ?
(1) ax=FFF5 h and cx=0 h
(2) ax=FFF6 h and cx=0 h
(3) ax=FFF7 h and cx=0A h
(4) ax=FFF5 h and cx=0A h

Correct Answer: (2) ax=FFF6 h and cx=0 h

32. Consider the following assembly program fragment :
stc
mov al, 11010110b
mov cl, 2
rcl al, 3
rol al, 4
shr al, cl
mul cl
The contents of the destination register ax (in hexadecimal) and the status of Carry Flag (CF) after the execution of above instructions, are :
(1) ax=003CH; CF=0
(2) ax=001EH; CF=0
(3) ax=007BH; CF=1
(4) ax=00B7H; CF=1

Correct Answer: (1) ax=003CH; CF=0

33. Which of the following regular expressions, each describing a language of binary numbers (MSB to LSB) that represents non-negative decimal values, does not include even values ?
(1) 0*1+0*1* (2) 0*1*0+1* (3) 0*1*0*1+ (4) 0+1*0*1* Where {+, *} are quantification characters.


34. Which of the following statements is/are TRUE ? (a) The grammar S → SS | a is ambiguous. (Where S is the start symbol) (b) The grammar S → 0S1 | 01S | ε is ambiguous. (The special symbol ε represents the empty string) (Where S is the start symbol) (c) The grammar (Where S is the start symbol) S → T/U T → x S y | xy | e U → yT generates a language consisting of the string yxxyy.
(1) Only (a) and (b) are TRUE.
(2) Only (a) and (c) are TRUE.
(3) Only (b) and (c) are TRUE.
(4) All of (a), (b) and (c) are TRUE.

Correct Answer: (4) All of (a), (b) and (c) are TRUE.

35. Match the description of several parts of a classic optimizing compiler in List - I, with the names of those parts in List - II : List - I List - II
(a) A part of a compiler (i) Optimizer syntax. that is responsible for recognizing (b) A part of a compiler that takes as input a stream of (ii) Semantic Analysis characters and produces as output a stream of words along with their associated syntactic categories. (c) A part of a compiler that understand the meanings of (iii) Parser variable names and other symbols and checks that they are used in ways consistent with their definitions. (d) An IR-to-IR transformer that tries to improve the IR (iv) Scanner program in some way (Intermediate Representation).
Code : (a) (b) (c) (d)
(1) (iii) (iv) (ii) (i)
(2) (iv) (iii) (ii) (i)
(3) (ii) (iv) (i) (iii)
(4) (ii) (iv) (iii) (i)

Correct Answer:  (1) (iii)  (iv)  (ii)   (i)

36. In Distributed system, the capacity of a system to adapt the increased service load is called ____.
(1) Tolerance
(2) Scalability
(3) Capability
(4) Loading

Correct Answer: (2) Scalability 

37. In __________ disk scheduling algorithm, the disk head moves from one end to other end of the disk, serving the requests along the way. When the head reaches the other end, it immediately returns to the beginning of the disk without serving any requests on the return trip.
(1) LOOK
(2) SCAN
(3) C - LOOK
(4) C - SCAN

Correct Answer:  (4) C - SCAN

38. Suppose there are six files F1, F2, F3, F4, F5, F6 with corresponding sizes 150 KB, 225 KB, 75 KB, 60 KB, 275 KB and 65 KB respectively. The files are to be stored on a sequential device in such a way that optimizes access time. In what order should the files be stored ?
(1) F5, F2, F1, F3, F6, F4
(2) F4, F6, F3, F1, F2, F5
(3) F1, F2, F3, F4, F5, F6
(4) F6, F5, F4, F3, F2, F1

Correct Answer:  (2) F4, F6, F3, F1, F2, F5
If we put the small file first then, then the average time to access all the file will be minimum. Lets put the largest file in  first place and smaller after that. then, in this case to access smaller file multiple time, systems need to cross the larger file sequentially each time which is very complex.

39. Which module gives control of the CPU to the process selected by the short-term scheduler?
(1) Dispatcher
(2) Interrupt
(3) Schedular
(4) Threading

Correct Answer: (1) Dispatcher

40. Two atomic operations permissible on Semaphores are __________ and __________.
(1) wait, stop
(2) wait, hold
(3) hold, signal
(4) wait, signal

Correct Answer: (4) wait, signal

41. Software does not wear-out in the traditional sense of the term, but software does tend to deteriorate as it evolves, because :
(1) Software suffers from exposure to hostile environments.
(2) Defects are more likely to arise after software has been used often.
(3) Multiple change requests introduce errors in component interactions.
(4) Software spare parts become harder to order.

Correct Answer: (3) Multiple change requests introduce errors in component interactions.

42. Software re-engineering is concerned with :
(1) Re-constructing the original source code from the existing machine (low - level) code program and modifying it to make it more user - friendly.
(2) Scrapping the source code of a software and re-writing it entirely from scratch.
(3) Re-organising and modifying existing software systems to make them more maintainable.
(4) Translating source code of an existing software to a new machine (low - level) language.

Correct Answer: (3) Re-organising and modifying existing software systems to make them more maintainable.

43. Which of the following is not a key issue stressed by an agile philosophy of software engineering ?
(1) The importance of self-organizing teams as well as communication and collaboration between team members and customers.
(2) Recognition that change represents opportunity.
(3) Emphasis on rapid delivery of software that satisfies the customer.
(4) Having a separate testing phase after a build phase.

Correct Answer:  (4) Having a separate testing phase after a build phase.

44. What is the normal order of activities in which traditional software testing is organized ?
(a) Integration Testing
(b) System Testing
(c) Unit Testing
(d) Validation Testing
Code :
(1) (c), (a), (b), (d)
(2) (c), (a), (d), (b)
(3) (d), (c), (b), (a)
(4) (b), (d), (a), (c)

Correct Answer: (2) (c), (a), (d), (b)

45. Which of the following testing techniques ensures that the software product runs correctly after the changes during maintenance ?
(1) Path Testing
(2) Integration Testing
(3) Unit Testing
(4) Regression Testing

Correct Answer: (4) Regression Testing

46. Which of the following Super Computers is the fastest Super Computer ?
(1) Sun-way TaihuLight
(2) Titan
(3) Piz Daint
(4) Sequoia

Correct Answer:  (1) Sun-way TaihuLight

47. Which of the following statements about ERP system is true ?
(1) Most ERP software implementations fully achieve seamless integration.
(2) ERP software packages are themselves combinations of seperate applications for manufacturing, materials, resource planning, general ledger, human resources, procurement and order entry.
(3) Integration of ERP systems can be achieved in only one way.
(4) An ERP package implemented uniformly throughout an enterprise is likely to contain very flexible connections to allow charges and software variations.

Correct Answer: (2)

48. Which of the following is not a Clustering method ?
(1) K - Mean method
(2) Self Organizing feature map method
(3) K - nearest neighbor method
(4) Agglomerative method

Correct Answer:(3) K - nearest neighbor method

49. Which of the given wireless technologies used in IoT, consumes the least amount of power ?
(1) Zigbee
(2) Bluetooth
(3) Wi-Fi
(4) GSM/CDMA

Correct Answer: (2) Bluetooth

50. Which speed up could be achieved according to Amdahl’s Law for infinite number of processes if 5% of a program is sequential and the remaining part is ideally parallel ?
(1) Infinite
(2) 5
(3) 20
(4) 50

Correct Answer: (3) 20

UGC NET Computer Science December 2019 | Question 16

Question 16 In a certain coding language. 'AEIOU' is written as 'TNHDZ'. Using the same coding language. 'BFJPV' wil...

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