UGC NET · Topic-wise previous year questions

Computer Science Previous Year Questions

570 Computer Science questions asked in UGC NET previous year papers, each with the correct answer and explanation. Read them free on UTpaper, or attempt the full papers as timed online tests.

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Q1UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
A Boolean function F is called self-dual if and only if F(x₁, x₂, ......., xₙ) = F(x₁', x₂', ....., xₙ') How many Boolean function of degree n are self-dual ?
  1. A2ⁿ
  2. B(2)^(2ⁿ)
  3. C(2)^(n²)
  4. D(2)^(2^(n−1))
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Correct answer: (D) (2)^(2^(n−1))

Explanation

A self-dual function is fixed by its values on one half of the input combinations: the 2ⁿ input combinations form 2ⁿ⁻¹ mutually complementary pairs, and the function's value on one member of each pair determines its value on the other. So the number of self-dual functions is 2^(2^(n−1)).

Q2UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Consider a complete bipartite graph kₘ,ₙ. For which values of m and n does this, complete graph have a Hamilton circuit
  1. Am = 3, n = 2
  2. Bm = 2, n = 3
  3. Cm = n ≥ 2
  4. Dm = n ≥ 3
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Correct answer: (C) m = n ≥ 2

Explanation

Kₘ,ₙ is Hamilton circuit if and only if m = n ≥ 2, (Circuit is a path similar to a cycle that starts and ends at the same vertex) and Kₘ,ₙ is a Hamilton circuit if m = n ≥ 2, so the correct answer is (c).

Q3UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Mergesort makes two recursive calls. Which statement is true after these two recursive calls finish, but before the merge step?
  1. AThe array elements form a heap.
  2. BElements in each half of the array are sorted amongst themselves.
  3. CElements in the first half of the array are less than or equal to elements in second half of the array.
  4. DAll of the above
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Correct answer: (B) Elements in each half of the array are sorted amongst themselves.

Explanation

After the two recursive calls in Mergesort complete, but before the merge step, elements in each half of the array are sorted amongst themselves because, Mergesort is possible when two arrays are sorted. If two unsorted arrays are submitted to Mergesort algorithm. It will sort each of them first before merging them, because Mergesort is a divide and conquer algorithm, Hence the answer is option (b).

Q4UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Searching for an element in the hash table requires O(1) time for the ______ time, whereas for direct addressing it holds for the _______ time.
  1. Aworst-case, average
  2. Bworst-case, worst-case
  3. Caverage, worst-case
  4. Dbest, average
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Correct answer: (C) average, worst-case

Explanation

In searching for an element in hash table requires O(1) time for the average time, whereas for direct it holds for the worst-case time. So the correct answer is option (c).

Q5UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
An algorithm is made up of 2 modules M₁ and M₂. If time complexity of modules M₁ and M₂ are h(n) and g(n) respectively, the time complexity of the algorithm is
  1. Amin (h(n), g(n))
  2. Bmax (h(n), g(n))
  3. Ch(n) + g(n)
  4. Dh(n) ∗ g(n)
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Correct answer: (B) max (h(n), g(n))

Explanation

When the two modules run one after the other, the total time is h(n) + g(n), which is O(max(h(n), g(n))) because the larger term dominates and lower-order terms are ignored. Hence the answer is (b).

Q6UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Big-O estimates for the factorial function and the logarithm of the factorial function i.e. n! and log n! is given by
  1. AO(n!) and O(n log n)
  2. BO(nⁿ) and O(n log n)
  3. CO(n!) and O(log n!)
  4. DO(nⁿ) and O(log n!)
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Correct answer: (B) O(nⁿ) and O(n log n)

Explanation

N! < nⁿ for all n ≥ 1, so n! is O(nⁿ). Using Stirling's approximation, log(n!) ≈ n log n − n + O(log n), so log n! is O(n log n). Therefore, the correct answer is (b) O(nⁿ) and O(n log n).

Q7UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
When we pass an array as an argument to a function, what actually gets passed?
  1. AAddress of the array
  2. BValues of the elements of the array
  3. CBase address of the array
  4. DNumber of elements of the array
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Correct answer: (C) Base address of the array

Explanation

Base address of an array is the starting address of an array. One can easily access to all the cells of an array if information about this can be achieved.

Q8UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
while (87) printf ("computer"); The above C statement will
  1. Aprint "computer" 87 times
  2. Bprint "computer" 0 times
  3. Cprint "computer" 1 times
  4. Dprint "computer" infinite times
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Correct answer: (D) print "computer" infinite times

Explanation

Parameter of a while statement in a languages takes two values true in the form of non-zero and false in the form of zero. Here 87 is a non-zero value and hence the while block gets true for infinite times.

Q9UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
A friend function can be used to
  1. Aavoid arguments between classes.
  2. Ballow access to classes whose source code is unavailable.
  3. Callow one class to access an unrelated class.
  4. Dnone of the above
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Correct answer: (C) allow one class to access an unrelated class.

Explanation

Unrelated class means the classes are not connected via inheritance. But a relation can be achieved between two unrelated classes through friend function.

Q10UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Which of the following is the correct value returned to the operating system upon the successful completion of a program?
  1. A0
  2. B1
  3. C− 1
  4. DProgram do not return a value.
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Correct answer: (A) 0

Explanation

0 is the identifier for a program to inform the operating system (OS) about proper running of a program. Each of the non-zero value signifies unique error doing running of a program.

Q11UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Manager's salary details are hidden from the employee. This is called as
  1. AConceptual level data hiding
  2. BPhysical level data hiding
  3. CExternal level data hiding
  4. DLocal level data hiding
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Correct answer: (C) External level data hiding

Explanation

External level (highest level). It is the user's view of the database and it describes part of the database for Particular group of user and provides a powerful and flexible security mechanism by hiding part of the database from certain users. The user is not aware of the existence of any attribute that are missing from the view. It permits user to access data in a way that is customized to their needs. So that the same data can be seen by different user in different ways, at the same time.

Q12UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
A clustering index is created when ______.
  1. APrimary key is declared and ordered
  2. BNo key ordered
  3. CForeign key ordered
  4. DThere is no key and no order
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Correct answer: (A) Primary key is declared and ordered

Explanation

Clustered index determines the order in which the rows of a table are stored on disk. If a table has a clustered index. Then the rows of that table will be stored on disk in the same exact order as the clustered index. In some cases, the index is created on non-primary key columns which may not be unique for each record. Clustering index is defined as an ordered data files.

Q13UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Which of the following statements is false?
  1. AAny relation with two attributes is in BCNF.
  2. BA relation in which every key has only one attribute is in 2NF.
  3. CA prime attribute can be transitively dependent on a key in 3NF relation.
  4. DA prime attribute can be transitively dependent on a key in BCNF relation.
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Correct answer: (D) A prime attribute can be transitively dependent on a key in BCNF relation.

Explanation

A relation is in BCNF, if and only if, every determinant is a candidate key i.e. the left hand side of every non trivial dependency is a super key or there is a trivial dependency or both. After 3NF, there may arise a situation that two candidate keys are having one attribute in common (overlapping) and this overlapping must be cleared to satisfy BCNF. A prime attribute (an attribute that is a part of a candidate key) can on depend on a super key. So a prime attribute cannot be transitively dependent on a key in BCNF relation. So the false statement is option (d).

Q14UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
The bit rate of a signal is 3000 bps. If each signal unit carries 6 bits, the baud rate of the signal is _____.
  1. A500 baud/sec
  2. B1000 baud/sec
  3. C3000 baud/sec
  4. D18000 baud/sec
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Correct answer: (A) 500 baud/sec

Explanation

Bit rate = number of bits per unit time; baud rate = number of signal units per unit time. Here each signal unit carries 6 bits, so baud rate = bit rate / 6 = 3000/6 = 500 baud/sec.

Q15UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
The minimum frame length for 10 Mbps Ethernet is ______ bytes and maximum is _____ bytes.
  1. A64 & 128
  2. B128 & 1518
  3. C1518 & 3036
  4. D64 & 1518
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Correct answer: (D) 64 & 1518

Explanation

The original ethernet IEEE 802.3 standard define the minimum ethernet frame size as 64 bytes and maximum as 1518 bytes. Where payload size in ethernet defined the minimum 46 bytes and maximum 1500 bytes.

Q16UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
A bridge has access to ______ address in the same network.
  1. APhysical
  2. BNetwork
  3. CDatalink
  4. DApplication
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Correct answer: (A) Physical

Explanation

A bridge functions in the data link layer, which means it cannot read IP addresses; it has access to the physical (MAC) addresses of all stations connected to it.

Q17UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
KPA in CMM stands for
  1. AKey Process Area
  2. BKey Product Area
  3. CKey Principal Area
  4. DKey Performance Area
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Correct answer: (A) Key Process Area

Explanation

Key Process Area (KPA) used in CMM identifies a cluster of related activities that, when performed collectively, achieve a set of goals considered important.

Q18UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
In function point analysis, the number of complexity adjustment factors is
  1. A10
  2. B12
  3. C14
  4. D20
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Correct answer: (C) 14

Explanation

There are 14 relative complexity adjustment factor are (1) Requirement for reliable backup and recovery (2) Require for data communication (3) Extent of distributing processing (4) Performance requirement (5) Expected operational environment (6) Extent of online data entries (7) Extent of multi-operation online data input. (8) Extent of online updating of master files. (9) Extent of complex inputs, outputs. (10) Extent of complex data processing. (11) Extent of currently developed code can be designed for reuse. (12) Extent of conversion and installation in the design. (13) Extent of multiple installation in an organization. (14) Extent of change and focus on ease of use.

Q19UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
_______ of a system is the structure or structures of the system which comprise software elements, the externally visible properties of these elements and the relationship amongst them.
  1. ASoftware construction
  2. BSoftware evolution
  3. CSoftware architecture
  4. DSoftware reuse
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Correct answer: (C) Software architecture

Explanation

Software architecture refers to the high level structure of software system, the discipline of creating such structures, and the documentation.

Q20UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Object Request Broker (ORB) is I. A software program that runs on the client as well as on the application server. II. A software program that runs on the client side only. III. A software program that runs on the application server, where most of the components reside.
  1. AI, II & III
  2. BI & II
  3. CII & III
  4. DI only
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Correct answer: (D) I only

Explanation

Object request broker (ORB) is middleware that uses the CORBA (Common Object Request Broker Architecture) specification. The ORB takes care of all of the details in routing a request from client to object and the response to its destination. ORB runs on clients as well as on the application server.

Q21UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
A software agent is defined as I. A software developed for accomplishing a given task. II. A computer program which is capable of acting on behalf of the user in order to accomplish a given computational task. III. An open source software for accomplishing a given task.
  1. AI
  2. BII
  3. CIII
  4. DAll of the above
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Correct answer: (B) II

Explanation

A software agent is a piece of software that function as a agent for a user or another program, working autonomously and continuously in a particular environment. It is inhibited by other processes and agents, but it is also able to learn from its experience in functioning in an environment over a long period of time.

Q22UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Regression testing is primarily related to
  1. AFunctional testing
  2. BDevelopment testing
  3. CData flow testing
  4. DMaintenance testing
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Correct answer: (D) Maintenance testing

Explanation

Regression testing is the process of testing changes to computer program to make sure that the older programming still works with the new changes. This happens only in case of maintenance. So it relates to the maintenance testing, hence the answer is option (d).

Q23UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Which of the following commands will output "one two three"?
  1. AFor val; do echo-n $val; done < one two three
  2. BFor one two three; do echo-n-; done
  3. CFor n in one two three; do echo-n $n; done
  4. DFor n in one two three {echo -n $ n}
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Correct answer: (C) For n in one two three; do echo-n $n; done

Explanation

The for loop "for n in one two three; do echo -n $n; done" iterates n over the words one, two and three and echoes each, so it gives the output "one two three". So the answer is (c).

Q24UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Which one of the following is not a risk management technique for managing the risk due to unrealistic schedules and budgets?
  1. ADetailed multi source cost and schedule estimation.
  2. BDesign cost
  3. CIncremental development
  4. DInformation hiding
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Correct answer: (D) Information hiding

Explanation

Risk management is the Process of identifying, controlling threats to an organization's capital and earning. Risk management can be considered as dealing with the possibility and actual occurrence of those events that are not regular expected and information hiding is the principle of segregation of the design decision in a computer program that are most like to change, thus protecting other parts of the program from extensive modification if the design decision is changed.

Q25UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Which of the following suffices to convert an arbitrary CFG to an LL(1) grammar?
  1. ARemoving left recursion alone
  2. BRemoving the grammar alone
  3. CRemoving left recursion and factoring the grammar
  4. DNone of the above
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Correct answer: (D) None of the above

Explanation

Removing left recursion and left factoring the grammar do not suffice to convert an arbitrary CFG to an LL(1) grammar. LL(1) parsing is top-down; for it the grammar must also be unambiguous and deterministic (e.g. an ambiguous grammar stays non-LL(1)), besides being free from left recursion.

Q26UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Which of the following flip-flops is free from race condition?
  1. AT flip-flop
  2. BSR flip-flop
  3. CMaster-slave JK flip-flop
  4. DNone of the above
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Correct answer: (C) Master-slave JK flip-flop

Explanation

In a JK flip-flop, when J = 1 and K = 1, Q toggles between 0 and 1 continuously as long as the clock is enabled, so the final value of Q is uncertain (race condition). The master-slave JK flip-flop is used to resolve this problem.

Q27UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
The output of a sequential circuit depends on
  1. Apresent input only
  2. Bpast input only
  3. Cboth present and past input
  4. Dpast output only
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Correct answer: (C) both present and past input

Explanation

There are 2 types of circuits: (1) Combinational and (2) Sequential. Combinational circuits are those whose outputs depend only upon the current input, like adder, subtractor, MUX, DEMUX etc. Sequential circuits are those whose output depends on both current and previous inputs. Hence the answer is (c).

Q28UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Given U = {1, 2, 3, 4, 5, 6, 7} A = {(3, 0.7), (5, 1), (6, 0.8)} then ~A will be : (where ~ → complement)
  1. A{(4, 0.7), (2, 1), (1, 0.8)}
  2. B{(4, 0.3), (5, 0), (6, 0.2)}
  3. C{(1, 1), (2, 1), (3, 0.3), (4, 1), (6, 0.2), & (7, 1)}
  4. D{(3, 0.3), (6, 0.2)}
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Correct answer: (C) {(1, 1), (2, 1), (3, 0.3), (4, 1), (6, 0.2), & (7, 1)}

Explanation

Μ~A(x) = 1 − μA(x) for all x ∈ U. The given set A can be seen as A = {(1, 0), (2, 0), (3, 0.7), (4, 0), (5, 1), (6, 0.8), (7, 0)}. So the complement of A is {(1, 1), (2, 1), (3, 0.3), (4, 1), (6, 0.2), (7, 1)} (element 5 has membership 0). Hence the answer is (c).

Q29UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
A byte addressable computer has a memory capacity of 2ᵐ K bytes and can perform 2ⁿ operations. An instruction involving 3 operands and one operator needs a maximum of
  1. A3m bits
  2. Bm + n bits
  3. C3m + n bits
  4. D3m + n + 30 bits
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Correct answer: (D) 3m + n + 30 bits

Explanation

Memory capacity is 2ᵐ Kbytes = 2^(m+10) bytes, so the whole memory can be addressed with m + 10 bits. 2ⁿ operations can be represented by n bits. An instruction of the form OP, operand₁, operand₂, operand₃ will therefore take 3(m + 10) + n = 3m + n + 30 bits. So the answer is (d).

Q30UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Consider a uniprocessor system where new processes arrive at an average of five processes per minute and each process needs an average of 6 seconds of service time. What will be the CPU utilization?
  1. A80%
  2. B50%
  3. C60%
  4. D30%
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Correct answer: (B) 50%

Explanation

New processes arrive at 5 per minute, i.e. one process every 60/5 = 12 seconds. Each process needs 6 seconds of service time. CPU utilization = (total service time required / total available time) × 100 = (6/12) × 100 = 50%. Hence the right option is (b).

Q31UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Assume that a program will experience 200 failures in infinite time. It has now experienced 100 failures. The initial failure intensity was 20 failures/CPU hr. Then the current failure intensity will be
  1. A5 failures/CPU hr.
  2. B10 failures/CPU hr.
  3. C20 failures/CPU hr.
  4. D40 failures/CPU hr.
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Correct answer: (B) 10 failures/CPU hr.

Explanation

Ν₀ = 200 failures, μ = 100 failures, λ₀ = 20 failures/CPU hr. Current failure intensity = λ₀(1 − μ/ν₀) = 20(1 − 100/200) = 20(1 − 0.5) = 10 failures/CPU hr. So the answer is (b).

Q32UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
The advantage of ––––––– is that it can reference memory without paying the price of having a full memory address in the instruction.
  1. ADirect addressing
  2. BIndexed addressing
  3. CRegister addressing
  4. DRegister Indirect addressing
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Correct answer: (D) Register Indirect addressing

Explanation

In register indirect addressing the address of the operand is held in a register, so memory can be referenced without having a full memory address in the instruction. Hence the answer is (d).

Q33UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
One of the main features that distinguish microprocessor from micro-computers is
  1. Awords are usually larger in microprocessors.
  2. Bwords are shorter in microprocessors.
  3. Cmicroprocessor does not contain I/O devices.
  4. DNone of the above.
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Correct answer: (C) microprocessor does not contain I/O devices.

Explanation

A microprocessor is a computer processor which incorporates the functions of a computer's control processing unit (CPU) on a single integrated circuit (IC), or at most a few integrated circuits. A microcomputer is small, relatively, inexpensive computer with a microprocessor as its control processing unit (CPU). It includes a microprocessor, memory, and minimal input/output (I/O) circuitry mounted on a single printed circuit board. That means, microprocessor does not contain input/output devices.

Q34UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
The reverse polish notation equivalent to the infix expression ((A + B) * C + D)/(E + F + G)
  1. AA B + C * D + EF + G +
  2. BA B + C D * + E F + G +
  3. CA B + C * D + E F G + +
  4. DA B + C* D + E + F G +
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Correct answer: (A) A B + C * D + EF + G +

Explanation

((A + B) ∗ C + D) / (E + F + G) → (AB+ ∗ C + D) / (E + F + G) → (AB+C∗ + D) / (E + F + G) → AB+C∗D+ / (EF+ + G) → AB+C∗D+ / EF+G+ → AB+C∗D+EF+G+/ Hence, the answer is (a).

Q35UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
–––––––– is used in game trees to reduce the number of branches of the search tree to be traversed without affecting the solution.
  1. ABest first search
  2. BGoal stack planning
  3. CAlpha-beta pruning procedure
  4. DMin-max search
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Correct answer: (C) Alpha-beta pruning procedure

Explanation

It is a search algorithm that seeks to decrease the number nodes that are evaluated by the Min-max algorithm in its search tree. It is an adversial search algorithm used commonly for machine playing of two-player games.

Q36UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Suppose that the splits at every level of quicksort are in the proportion (1 − α) to α, where 0 < α ≤ 1/2 is a constant. The minimum depth of a leaf in the recursion tree is approximately given by
  1. A−lg n / lg(1 − α)
  2. B−lg(1 − α) / lg n
  3. C−lg n / lg α
  4. D−lg α / lg n
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Correct answer: (C) −lg n / lg α

Explanation

In Quicksort, the depth of a leaf depends on how the input size reduces at each level. If the split is always in the ratio (1 − α) to α, the minimum depth corresponds to the path where the recursion always follows the smaller partition, i.e. of size αn. At each level the size becomes n, αn, α²n, α³n, … We stop when the size becomes 1: αᵏn = 1 ⇒ k log α + log n = 0 ⇒ k = −lg n / lg α. Hence the answer is (c).

Q37UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
If h is chosen from a universal collection of hash functions and is used to hash n keys into a table of size m, where n ≤ m, the expected number of collisions involving a particular key K is (b) less than lg n
  1. Aless than 1
  2. Bless than lg n
  3. Cgreater than 1
  4. Dgreater than lg n
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Correct answer: (A) less than 1

Explanation

This is a theorem in universal hashing: if h is chosen from a universal collection and is used to hash n keys into a table of size m, where n ≤ m, the expected number of collisions involving a particular key is less than 1. So the answer is (a).

Q38UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Code blocks allow many algorithms to be implemented with the following parameters :
  1. Aclarity, elegancy, performance
  2. Bclarity, elegance, efficiency
  3. Celegance, performance, execution
  4. Dexecution, clarity, performance
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Correct answer: (B) clarity, elegance, efficiency

Explanation

Code blocks not only allow many algorithms to be implemented with clarity, elegance and efficiency, but also help the programmer better conceptualize the true nature of the algorithm being implemented. So the answer is (b).

Q39UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Consider f (N) = g(N) + h(N) Where function g is a measure of the cost of getting from the start node to the current node N and h is an estimate of additional cost of getting from the current node N to the goal node. Then f(N) = h(N) is used in which one of the following algorithms?
  1. AA* algorithm
  2. BAO* algorithm
  3. CGreedy best first search algorithm
  4. DIterative A* algorithm
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Correct answer: (C) Greedy best first search algorithm

Explanation

The greedy algorithm work in a way that it has same estimate (called a heuristic) of how for from the goal any vertex is. Instead of selecting the vertex closet to the starting point, it selects the vertex closet to the goal, so the answer is (c).

Q40UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Match the following cryptographic algorithms with their design issues : List-I (a) DES (b) AES (c) RSA (d) SHA-1 List-II (i) Message Digest (ii) Public Key (iii) 56-bit key (iv) 128-bit key Codes : a b c d
  1. Aii i iv iii
  2. Biii i iv ii
  3. Ciii iv ii i
  4. Div i ii iii
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Correct answer: (C) iii iv ii i

Explanation

(a) DES : DES is an implementation of a fiestel cipher. It uses 16 round fiestel structure. DES has an effective key length of 56 bits. (b) AES : The more popular and widely adopted symmetric encryption algorithm is AES. The features of AES is symmetric key symmetric block cipher and it has 128 bit key. (c) RSA : RSA algorithm works on two different keys i.e public key & private key. As the name public key given to everyone & private key is kept private. (d) SHA-1 : It is message digest.

Q41UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
The solution of the recurrence relation of T(n) = 3T(floor(n/4)) + n is
  1. AO(n²)
  2. BO(n lg n)
  3. CO(n)
  4. DO(lg n)
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Correct answer: (C) O(n)

Explanation

T(n) = 3T(n/4) + n. Here a = 3, b = 4, f(n) = n and n^(log₄3) ≈ n^0.79. Since f(n) = n grows polynomially faster than n^(log₄3), case 3 of the master theorem gives T(n) = Θ(n). So T(n) = O(n) (it is also O(n²), but (c) is the tight bound and the best choice).

Q42UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Consider the fractional knapsack instance n = 4, (p₁, p₂, p₃, p₄) = (10, 10, 12, 18), (w₁, w₂, w₃, w₄) = (2, 4, 6, 9) and M = 15. The maximum profit is given by (Assume p and w denotes profit and weight of objects respectively)
  1. A40
  2. B38
  3. C32
  4. D30
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Correct answer: (B) 38

Explanation

P₁/w₁ = 5, p₂/w₂ = 2.5, p₃/w₃ = 2, p₄/w₄ = 2. Select object 1 (profit 10), then object 2 (profit 10). Objects 3 and 4 have the same ratio, but object 4 gives more profit and fits exactly: total weight = 2 + 4 + 9 = 15. Maximum profit = 10 + 10 + 18 = 38. So the answer is (b).

Q43UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Match the following with respect to I/O classes in object oriented programming : List-I (A) fopen() (B) fclose() (C) ferror() (D) feof() List-II (i) returns end of file (ii) return for any problem report (iii) returns 0 (iv) returns a file pointer Codes : A B C D
  1. Aiv i ii iii
  2. Biii i iv ii
  3. Cii iii iv i
  4. Div iii i ii
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Correct answer: (A) iv i ii iii

Explanation

Fopen() : creates a file or opens an existing file and returns a file pointer, e.g. FILE ∗fp; fp = fopen("<file-name>", "<mode>"); fclose() : closes a file; returns 0 on success and EOF otherwise. ferror() : checks the error indicator for the given file stream, returning non-zero if set and 0 otherwise. feof() : checks the end-of-file indicator for the given file stream, returning non-zero if set and 0 otherwise.

Q44UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
The pure object oriented programming language with extensive metadata available and modifiable at run time is
  1. ASmall talk
  2. BC++
  3. CJava
  4. DEiffel
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Correct answer: (A) Small talk

Explanation

Small talk was one of the earliest OOP's language (with other like simulate and Eiffel) and can be said to be extremely "Pure" in an OOP sense. In small talk everything is an object and objects are only communicated with via the sending of message. Hence the answer is (a).

Q45UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
The syntax of capturing events method for document object is
  1. ACaptureEvents()
  2. BCaptureEvents(Orgs eventType)
  3. CCaptureEvents(eventType)
  4. DCaptureEvents(eventVal)
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Correct answer: (C) CaptureEvents(eventType)

Explanation

CaptureEvents (EventType) method instructs the window to capture all events of a particular type.

Q46UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Superficially the term "object-oriented", means that, we organize software as a
  1. Acollection of continuous objects that incorporates both data structure and behaviour.
  2. Bcollection of discrete objects that incorporates both discrete structure and behaviour.
  3. Ccollection of discrete objects that incorporates both data structure and behaviour.
  4. Dcollection of objects that incorporates both discrete data structure and behaviour.
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Correct answer: (C) collection of discrete objects that incorporates both data structure and behaviour.

Explanation

Object oriented development makes software closet to abstractions the exists in the real world and the term object oriented means that software is a collection of discrete objects that incorporate both data structure and behaviour.

Q47UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Consider a program that consists of 8 pages (from 0 to 7) and we have 4 page frames in the physical memory for the pages. The page reference string is : 1 2 3 2 5 6 3 4 6 3 7 3 1 5 3 6 3 4 2 4 3 4 5 1 The number of page faults in LRU and optimal page replacement algorithms are respectively (without including initial page faults to fill available page frames with pages) :
  1. A9 and 6
  2. B10 and 7
  3. C9 and 7
  4. D10 and 6
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Correct answer: (B) 10 and 7

Explanation

Simulating LRU with 4 frames gives 10 page replacements (faults after the frames are filled), and the optimal page replacement algorithm gives 7 replacements. So the correct answer is (b).

Q48UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Which of the following statements is not true about disk-arm scheduling algorithms?
  1. ASSTF (shortest seek time first) algorithm increases performance of FCFS.
  2. BThe number of requests for disk service are not influenced by file allocation method.
  3. CCaching the directories and index blocks in main memory can also help in reducing disk arm movements.
  4. DSCAN and C-SCAN algorithms are less likely to have a starvation problem.
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Correct answer: (B) The number of requests for disk service are not influenced by file allocation method.

Explanation

Option (b) is NOT CORRECT as the number of request for disk drive can be greatly influenced by file allocation method.

Q49UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
––––––– maintains the list of free disk blocks in the Unix file system
  1. AI-node
  2. BBoot block
  3. CSuper block
  4. DFile allocation table
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Correct answer: (C) Super block

Explanation

Super block describes the state of file system, the total size of partition, the block size, pointers to a list of free blocks, the node number of root directory, magic number etc.

Q50UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
A part Windows 2000 operating system that is not portable is
  1. ADevice Management
  2. BVirtual Memory Management
  3. CProcessor Management
  4. DUser Interface
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Correct answer: (B) Virtual Memory Management

Explanation

The design of virtual memory assumes that the underlying hardware supports virtual to physical mapping.

How to use these UGC NET Computer Science questions

Every question here was asked in a real UGC NET paper; the paper and shift are shown on the card, and the number opens the question on its own page. Try each one before tapping Show answer, then attempt the full UGC NET papers as timed tests to check your speed under exam conditions.