 UGC NET Computer Science Question Solution DEC- 2019 -Paper -II

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Explanation:
(a) False: The running time depends on the number of subproblems multiplied the time taken by each subproblem. This would only be true if the time per subproblem is O(1).
(b) True: Cyclic dependency is a relation between two or more modules which either directly or indirectly depend on each other to function properly. Such modules are also known as mutually recursive. So, we can’t get correct solution when we are using dynamic programming.
(c) False: A bottom up implementation must go through all of the sub-problems and spend the time per subproblem for each. Using recursion and memoization only spends time on the subproblems that it needs. In fact, the reverse may be true: using recursion and memoization may be asymptotically faster than a bottom-up implementation.
(d) False: If a problem X can be reduced to a known NP-hard problem, then X must be NP-Complete, not NP – Hard.

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Explanation:
For any context-free grammar G=(V,T,P,S) without any λ-productions or unit-productions.
The maximum number of production rules are: (k−1)|P|+|T|
where k is the maximum number of the symbol on the right of production P.

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Explanation:
Time required to execute 100 tasks without pipeline = 100* 30 = 3000 ns
Time required to execute 100 tasks with 4-stages pipeline = Time taken by 1st task + Time taken by
= 1 * 4 clock cycle + 99 * 1 clock cycle
= 4 * 10 + 99 * 10
= 40 + 990
= 1030 ns
Speedup = Time required without pipeline / Time required with pipeline
= 3000 / 1030
= 2.91
Or
Speedup = Time required without pipeline / Time required with pipeline
= n *tn / (n + k – 1) tp
where, t¬n = clock cycle without pipeline,
n = total number of tasks,
k = number of segment,
tp = clock cycle with pipeline
Speedup = n *tn / (n + k – 1) tp
= 100 * 30 / (100 + 4 – 1) * 10
= 3000 / 1030
= 2.91 ns

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Explanation:
Frame attribute:
Frame attribute has capability to specify the side of the table frame that display boarder.
Frame is a part of a web page or browser window which displays content independent of its container, with the ability to load content independently.
Using Frame attribute we can divide the web browser window into multiple sections where each section can contain separate data and load independently    #### Show Answer With Best Explanation

Explanation:
Given,
Bandwidth = 10 Mbps = 10 x 106bps
Frame per minutes = 12,000 Frame per min
Frame size = 10,000 bits

Throughput = ?

Throughput::
In data Transmission, network throughput tells how much data was transferred from a source at any given time & typically measured in bps, as in Mbps or Gbps.

Caculation:
Throughput = Frames per minute * each Frame size
= 12,000 per minute* 10,000 bits
= 12,000 * 10,000 bits per minute
= (12,000 * 10,000 / 60) bits per second
= 2,000,000 bps
= 2 Mbps

So, Option(ii) is correct.

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Explanation:
By Handshaking theorem,
Sum of the Degree of vertices = 2 * Number of Edges
2n*1 + 3n*2 + n3 = 2E
2n + 6n + 3n = 2E
E = 11n/2 —————— equation (1)
We know , In Tree
Number of vertices = Number of edges + 1
2n + 3n + n = 11n/2 + 1
6n = 11n/2 +1
12n = 12n + 2
n = 2
put n =2 in eqution (1),we got
E = 11n/2
E = 11*2 / 2
E = 11
Number of vertices = number of edges + 1
V = E + 1
V = 11 + 1
V = 12

Explanation:

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Explanation:
The worst case time to insert an element in unsorted array is θ(1), because simply append an element at the end of unsorted array.
In the case of extract min, there is no specific algorithm for it. So, the only way is to do the linear search on the unsorted array. The worst case complexity of extract_min is θ(n).

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Explanation:
S1) True: These exists no algorithm for deciding if any two Turning machine M1 and M2 accept the same language.
Because “Equivalence Problem” of recursive enumerable language is undecidable.
S2) True: Let M1 and M2 be arbitrary Turing machines. The problem to determine L(M1)⊆ L(M2)is undecidable.
Because “subset problem” of recursive Enumerable language is undecidable.

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Explanation:
A microinstruction format has microoperation field which is divided into 2 subfields F1 and F2.
Each having 15 distinct micro operations.
Bits required = log2 15 = 3 bits each
Condition field CD have four status bits.
Bits required = log2 4 = 2 bits
Branch field BR having four options used in conjunction with address field AD.
Bits required = log2 4 = 2 bits
The address space is of 128 memory words.
Bits required = log2 128 = 7 bits
Micro instruction field Micro instruction size = 3+3+2+2+7 = 17 bits

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Explanation:
It is given that ,
Block pointer = 4 byte
Record pointer = 0 , {consider it as 0 because it is not given in the question}
Search Key value = 32 byte
 Size of non-leaf node in B Tree = m * Block pointer + (m – 1)*(search key value + record pointer)
= m * 4 + (m-1)*(32 + 0)
= 4m + 32m -32
= 36m – 32

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Explanation:
The SQL Server Query Optimizer is a cost-based optimizer. The query optimizer attempts to determine the most efficient way to execute a given query by considering the possible query plans. Database users do not typically interact with a query optimizer, which works in the background.
Query Optimizer does rearrangement and possible ordering of operations, eliminate redundancy in query and use efficient algorithms and indexes during the execution of a query.

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Explanation:
A) int *P = &44; Illegal Statements:
B) int *P = &r; Legal Statements:
C) int P= &a; Illegal Statements:
D) int P = a; Legal Statements:

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Explanation:
Self complementing codes are the codes that have the property that 9’s complement of a decimal no. is obtained directly by interchanging 1’s by 0’s and 0’s by 1’s i.e. by complementing each bit in the pattern.

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Explanation:
it is given that ,
TLB access time = 20ns
Memory access time = 80ns
hit ratio = 80% (or 0.8)
miss ratio = 1 – 0.8 = 0.2(or 20%)
effective memory access = hit ratio * (TLB access time + memory access time) + miss ratio * (TLB access time + memory access time)
= 0.8*(20+80) + 0.2(20+2*80)
= 0.8 * 100 + 0.2*180
= 80 + 36
= 116 ns

Explanation:

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Explanation:
Preemtive CPU scheduling is used. when,
=> A process switches from Running state to Ready state.
=> A process switches from Waiting state to Ready state.

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Explanation:
A process can be of two types:
Independent process: running process does not affected by the execution of other processes
Co-operating process: running process can be affected by other executing processes.
There are 2 different approaches are used for communication among the processes.
Shared memory.
Message passing.

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Explanation:

Confidentiality
• Confidentiality means that only the authorized individuals/systems can view sensitive or classified information.
• The data being sent over the network should not be accessed by unauthorized individuals.
Confidentiality means that the sender and the receiver expect privacy.

Integrity
• Data integrity means maintaining and assuring the accuracy and completeness of data over its entire lifecycle.
Integrity means that the data must arrive at the receiver exactly as they were sent.

Authentication
• Authentication is the process of determining whether someone is who it is declared to be.
Authentication means the receiver is ensured that the message is coming from the intended sender.

So, Option(IV) is correct.

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Explanation:
aspect ratio = width/ height
Aspect ration of 1024 × 768 pixels image = 1024/768
= 4/3
Aspect ration of modified pixels image = 640/height
aspect ratio of both the images are same
4/3 = 640/height
height = 640*3/4
= 480 pixels

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Explanation:
The time complexity to multiply two polynomials is : O(n2)
but,
The time complexity to multiply two polynomials of degree n using Fast Fourier transform method is : θ(n lg n)

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Explanation:
There are spelling mistake in Linear repression.This has to be Linear Regression.
List of Common Machine Learning Algorithms are:
• Back propagation
• Logistic Regression
• Decision Tree
• SVM
• Naive Bayes
• kNN
• K-Means
• Random Forest
• Dimensionality Reduction Algorithms

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Software-Engineering Software-process-models

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Explanation:
• A piconet is a wireless set of devices make of use Bluetooth technology protocols.
• A piconet is connection of two or more devices occupying the same physical.
• A piconet is a combination of one master device( any device whis share the resource) and seven active slave devices(devices connected to the master device like phone,printer,speaker,scanner,mic etc) which has maximum 255 further slave devices can be inactive.

Other topic related to Piconet is Scatternet.

Scatternet -> sactternet is combination of Piconet.

So the correct answer is Option(II). (One Master and Seven Slave Devices )

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Explanation:
T(n)= 4T(√n)+lg2 n
T(n)= 4T(n1/2)+lg2 n
Suppose , n = 2m
Log2 n = m
T(2m) = 4 T (2m/2) + log2 2m
T(2m) = 4 T (2m/2) + log 2m * log 2m
Suppose, T(2m) = S(m)
S(m) = 4 S (m/2) + m*m
a = 4, b= 2, k= 2, p = 0

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Explanation:

ICANN (Internet Corporation for Assigned Name and Numbers)is non-profit organization that was formed in 1998. ICANN maintains the central repository for IP Addresses & help coordinate the supply the IP Addresses. It also manages the domain name system & root servers.

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Explanation:
Let’s try to understand the basic,
In distance vector & link state routing, router have routing table & router have to store information about other router. As networks grow in size, number of router increases in network. Due to this, router can’t handle network traffic efficiently. So, routing will have to done hierarchically.

In three level hierarchically routing,
Group of routers are called region & Group of regions are called cluster.

Given:
Number of cluster = 8 cluster
Each cluster have 9 regions.
Each region have 10 routers.

Total number of entries in hierarchically table of each router
=10 entries for router + 8 entries for routing to other regions within own cluster + 7 entries for distant clusters
=10 + 8 + 7
=25

or

Total number of entries in hierarchically table of each router
= Total nmber of router in a region + (Number of regions – 1) +(Number of clusters -1)
=10 + 8 + 7
=25

So, Option(I) is correct.

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Answer: This Question Is Dropped Marks for all

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As we know .
TRUE: Dijkstra’s algorithm always find shortest path with in the same graph data structure. It uses a greedy technique to identify shortest path.
TRUE: Every time a new node is visited, we choose the node with the smallest known distance/cost (weight) to visit first
FALSE: It is false because this algorithm applicable any number of vertices.
TRUE: Dijkstra’s algorithm will support only positive weight edges.

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