UGCNET Computer Science June 2013 Paper II Re test
Ans .
(B) 6.72 Mbits
Explanation :
56kbps means 56kb of data is downloaded per second. So in 2 minutes is equal to120 seconds. the size of data downloaded will be 56*120=6720kb convert kb to mb. divide by 1000. 6720kb=6720/1000=6.72Mb.
Ans .
(D) p-persistent
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Ans .
(B) Polyalphabetic
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Ans .
(A) 100 meters
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Ans .
(D) Common Gateway Interface
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Ans .
(C) Distance vector routing algorithm
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Ans .
(C) 32 bits and 64 bits
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Ans .
(C) 4851
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Ans .
(D) Search, Insert and Delete
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Ans .
(D) – (Ig n)/Ig (1–β)
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Ans .
(C) d + 1
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Ans .
(C) Doubly linked list
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Ans .
(C) 5
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Ans .
(D) (1 + 01)* (0 + λ)
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Ans .
(A) L1 is regular language and L2 is not regular language.
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Ans .
(C) S → aSc | S1| λ ; S1 → bS1c| λ
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Ans .
(A) S1 is correct and S2 is not correct.
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Ans .
(A) 64
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Ans .
(B) 1,66,320
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Ans .
(D) All of the above
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Ans .
(C) a b e j k n o p f c d g l m h i
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Ans .
(B) more than (n – 1) (n – 2)/2 edges
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Ans .
(C) EX-OR, EX-NOR
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Ans .
(C) ∃ x P(x) ∧ ∀ x∀ y ((P(x) ∧ P(y)) → x = y)
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Ans .
(A) F ≤ F + G and F G ≤ F
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Ans .
(A) iii   iv   i   ii  
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Ans .
(C) CMM
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Ans .
(C) Software architecture
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Ans .
(A) Waterfall model
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Ans .
(B) Black box
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Ans .
(D) (i), (ii), (iii) and (iv)
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Ans .
(A) Binary Search
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Ans .
(B) 4
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Ans .
(C) 15
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Ans .
(B) Computes GCD of two numbers
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Ans .
Marks given to all
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Ans .
(B) Deadlock situation
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Ans .
(A) XSTL
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Ans .
(A) Hiding
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Ans .
(A) Classification
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Ans .
(B) Two dimensions
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Ans .
(D) PGP
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Ans .
(B) Timestamp ordering protocol
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Ans .
(A) iii i ii iv
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Ans .
(B) 140 ns
Explanation :
Hit ratio is the percentage of time that a frame number is found in the TLB – Associative lookup time is a time units – Memory cycle time is m time units – Hit ratio is r – Effective Access Time (EAT) = (m + a) r + (2m + a)(1 – r) Example If a = 20 ns, m = 100 ns, and r = 80%, then EAT = (100 + 20) 0.8 + (2(100) + 20)(1 - 0.8) = 140 ns For r = 98%, then EAT = 122 ns
Ans .
(B) 143 and 123 milliseconds
Explanation :
ans is B 143 ms and 123 ms at clock time 160 millisecond only request number 1 2 3 4 are there to be serviced
request number 5 (with track no 75) will be ready after 15 millisecond starting head position 65
now for SSTF 12 40 75 85 100 it will serve 85(since 75 is not come yet) first then 75 then 100 then 40 and finally 12
so total movements =20+10+25+60+28=143 total time 143×1 ms=143ms
now for look 12 40 75 85 100 direction towards highest no of tracks
it will serve 85(since 75 is not come yet) first then 100 then 75 then 40 and finally 12
so total movements =20+15+25+35+28=123 total time 123×1 ms=123ms
Ans .
Marks given to all
Explanation :
Ans .
(C) title = WELCOME; export title ; sh –c “echo $title”