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Recent exams in UGC NET

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Exam Type Full Length
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1
UGCNET CSE June 2015 Paper ll
No ..just keep ur basics clear in c and try to be strong in logic thats enough
posted May 29, 2020 in UGC NET Full Length soujanyareddy13 24 takes
2
UGCNET CSE June 2015 Paper III
Dis maths:kenneth rosen really its a bible and you wl also learn many new things apart from syllabus Aptitude:r.s agarwal standard book
posted May 29, 2020 in UGC NET Full Length soujanyareddy13 9 takes
3
UGCNET CSE December 2014 Paper II
Why not option d is correct ?
posted May 26, 2020 in UGC NET Full Length soujanyareddy13 1 take
4
UGCNET CSE December 2014 Paper III
150 Marks, 150 Minutes, 75 Questions
posted May 26, 2020 in UGC NET Full Length soujanyareddy13 2 takes
5
UGCNET CSE December 2015 Paper II
I feel option c is correct because there must be some vertex from which 7 edges are coming out else our given condition cantbe satisfied
posted Aug 13, 2018 in UGC NET Full Length soujanyareddy13 3 takes
6
UGCNET CSE December 2015 Paper III
In a $\text{JK}$ flip flop the output toggles when both $\text{J}$ and $\text{K}$ inputs are $1.$ So, we must ensure that with each clock the output from the previous stage reaches the current stage. Setup time is defined as the minimum amount of time before the clock's active edge ... is $10 + \max(0,10,20) = 10 + 20 = 30ns$ which would mean a maximum clock frequency of $1/30 GHz = 33.33 MHz$
posted Aug 12, 2018 in UGC NET Full Length soujanyareddy13 2 takes
7
UGCNET CSE June 2014 Paper II
100 Marks, 75 Minutes, 50 Questions
posted Sep 12, 2016 in UGC NET Full Length Arjun 178 takes
8
UGCNET CSE June 2014 Paper III
Because microprocessor can also have hardwired control unit. Microprogrammed control unit is just a design option for control unit in a microprocessor. This "micro" is not related to the "micro" in microprocessor.
posted Sep 11, 2016 in UGC NET Full Length soujanyareddy13 1 take
9
UGCNET CSE December 2013 Paper II
100 Marks, 75 Minutes, 50 Questions
posted Sep 2, 2016 in UGC NET Full Length soujanyareddy13 3 takes
10
UGCNET CSE December 2013 Paper III
No. Its not because of stronger answer. Option B is not correct. We can prove as follows: Lets take $n = 10$. So, total number of incoming edges $=7 \times 10 = 70$ $= $ total number of outgoing edges. So, we have 70 ... is indeed a valid outgoing degree sequence as it satisfies Havel-Hakimi theorem. http://www.maths.unp.ac.za/coursework/Math236/2012/Lecture%20Slides/Math236%20Lecture%2031.pdf
posted Sep 1, 2016 in UGC NET Full Length soujanyareddy13 1 take
11
UGCNET CSE September 2013 Paper II
100 Marks, 75 Minutes, 50 Questions
posted Aug 12, 2016 in UGC NET Full Length soujanyareddy13 2 takes
12
UGCNET CSE September 2013 Paper III
150 Marks, 150 Minutes, 75 Questions
posted Aug 11, 2016 in UGC NET Full Length soujanyareddy13 2 takes
13
UGCNET CSE June 2013 Paper II
100 Marks, 75 Minutes, 50 Questions
posted Aug 2, 2016 in UGC NET Full Length soujanyareddy13 4 takes
14
UGCNET CSE June 2013 Paper III
150 Marks, 150 Minutes, 75 Questions
posted Aug 1, 2016 in UGC NET Full Length soujanyareddy13 2 takes
15
UGCNET CSE December 2012 Paper II
Answer is $O(2^n)$ The last $cn$ is for the $n$ times the for loop is executing. T(0) = 1 T(1) = 2 T(2) = 5 T(3) = 11 T(4) = 23 T(5) = 47 So, $T(n) = 2^n + 2^{n-1} - 1 = O(2^n)$
posted Jul 12, 2016 in UGC NET Full Length soujanyareddy13 3 takes
16
UGCNET CSE December 2012 Paper III
double foo(int n) { int i; double sum; if(n == 0) { return 1.0; } else { sum = 0.0; for(i = 0; i < n; i++) { sum += foo(i); } return sum; } } The time complexity of the above code is?
posted Jul 11, 2016 in UGC NET Full Length soujanyareddy13 5 takes
17
UGCNET CSE June 2012 Paper II
100 Marks, 75 Minutes, 50 Questions
posted Jul 8, 2016 in UGC NET Full Length Arjun 105 takes
18
UGCNET CSE June 2012 Paper III
As arjun sir explained dats the reason c is the answer
posted Jul 7, 2016 in UGC NET Full Length soujanyareddy13 6 takes
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