Ans .

B

1. Explanation :

Piggybacking is a bi-directional data transmission technique in the network layer (OSI model). It makes the most of the sent data frames from receiver to emitter, adding the confirmation that the data frame sent by the sender was received successfully (ACK acknowledge).

Ans .

E

1. Explanation :

Encryption and Decryption is provided by presentation layer.

Ans .

D

1. Explanation :

Baud rate (signal rate)=bit rate/no of bits per signal or Bit rate=baud rate x no of bits per signal so here baud rate is given 1000 bauds(signals)per sec and no of bits per signal =4 so bit rate =4000 bits per sec.

Ans .

A

1. Explanation :

In ground propagation, radio waves travel through the lowest portion of the atmosphere. These low-frequency signals travel in all directions from the transmitting antenna.

Ans .

B

1. Explanation :

Compute φ= (p − 1)(q − 1) =12 x 30 =360
choose an int e such that 1 < e< φ and e & φ are co-prime d.
e =1 mod φ 7.
e=1 mod 360 =361 mod 360 =721 mod 360 =>7e=721 =>e=103.

Ans .

A

1. Explanation :

FAN IN =Number of components that can call or pass control to component A.
FAN OUT=Number of components that are called by component A.

Ans .

C

1. Explanation :

Cohesion - cohesion refers to the degree to which the elements of a module belong together.
Thus, cohesion measures the strength of relationship between pieces of functionality within a given module.
For example, in highly cohesive systems functionality is strongly related.
Coupling -in software engineering, coupling is the manner and degree of interdependence between software modules; a measure of how closely connected two routines or modules are; the strength of the relationships between modules.
Coupling is usually contrasted with cohesion.

Ans .

D

1. Explanation :

In software engineering, software configuration management (SCM or S/W CM) is the task of tracking and controlling changes in the software, part of the larger cross-disciplinary field of configuration management. SCM practices include revision control and the establishment of baselines. Software Configuration Management (SCM) is the the overall management of a software design project as it evolves into a software product or system. This includes technical aspects of the project, all level of communications, organization, and the control of modifications changes to the project plan by the programmers during the development phase. Software Configuration Management is also called Software Control Management.

Ans .

C

1. Explanation :

Requirement Elicitation, Requirement Analysis, Requirement Documentation and Requirement Review are the four crucial process steps of requirement engineering.Design is in itself a different phase of Software Engineering.

Ans .

B

1. Explanation :

Beta testing -performed at client side by client under actual environment with actual data It is performed in Real Time Environment.

Ans .

A

1. Explanation :

Integrity constraint is meant for "integrity" of data and mark <= 100 is part of it. By English meaning D option can be right but in Database, is there something like "Feasible Constraint"? Integrity Constraint is not meant to avoid data loss.

Ans .

B

1. Explanation :

Go bottom will take u to the last record (i.e 30th record) now skip 1 will take u to 29th skip 2 to 28th and skip 3 to 27th record.

Ans .

A

1. Explanation :

E-R Model Includes all four.

Ans .

B

1. Explanation :

Based on the cardinality ratio and participation constraint associated with a relationship type, choose either the Foreign Key Design, the Cross Referencing Design or Mutual referencing Design.

Ans .

B

1. Explanation :

a) is defined by integrity constraints c) and D) is not related to data integrity control.

Ans .

B

1. Explanation :

It is represent a pointer and it is pointing to an arra of size 10.

Ans .

D

1. Explanation :

For B, they declare ch to be a char and initialize it with the value 65. Since this is the ASCII code for 'A', that character constant can also be used to initialize ch to 65."

Ans .

D

1. Explanation :

All of the options can be overloaded.But in c++ we cannot overload this operator 1) .* 2) :: 3) ?:

Ans .

B

1. Explanation :

Inheritance allows to create classes which are derived from other classes, so that they automatically include some of its "parent's" members, plus its own.

Ans .

A

1. Explanation :

Options C and D will return the floor integer value of the given number. e.g: for 20.6 both options will return 20.
Option A is the correct answer where typecast is used after adding 0.5 to the original number.

Ans .

D

1. Explanation :

Following is algorithm for evaluation postfix expressions.
1) Create a stack to store operands (or values).
2) Scan the given expression and do following for every scanned element.
a) If the element is a number, push it into the stack
b) If the element is a operator, pop operands for the operator from stack. ​Evaluate the operator and push the result back to the stack
3) When the expression is ended, the number in the stack is the final answer
Postfix Expression=a b c d + - *
=a b ( c + d ) - *
=a ( b - ( c + d ) ) *
=( a * ( b - ( c + d ) ) )
Now Put the values a=8 ,b=4 c=2 and d=5
( 8 * ( 4 - ( 2 + 5 ) ) )
= ( 8 * ( 4 - ( 7) ) )
=( 8 * ( -3 ) )
=-24

Ans .

C

1. Explanation :

In linked list implementation of a queue, the last inserted node is always pointed by 'rear' and the first node is always pointed by 'front'.Therefore, only rear pointer changes.

Ans .

C

1. Explanation :

Correctness of recursive algorithms is often proven by mathematical induction.

Ans .

C

1. Explanation :

In B tree every internal node other than the root is at least half-full, i.e. t-1≤ #keys ≤2t-1, t≤ #children ≤2t

Ans .

D

1. Explanation :

The number of comparisons needed in the worst case by the merge sort algorithm will be m+n-1 i.e. only last element will not be compare only.

Ans .

D

1. Explanation :

Both S1 and S2 re correct.

Ans .

D

1. Explanation :

Ans .

C

1. Explanation :

If L is regular then reversal of L is also regular. hence u and v are regular Concatenation of regular languages are regular. Hence uv is regular S1 is correct To find wwR we need to remember what is w. Regular grammar is checked by finite automata, which has no memory (stack) To find wwR we need at least NPDA S2 is incorrect

Ans .

D

1. Explanation :

Relocation is the process of assigning load addresses to position-dependent code of a program and adjusting the code and data in the program to reflect the assigned addresses. Symbol resolution is the process of searching files and libraries to replace symbolic references or names of libraries with actual usable addresses in memory before running a program.

Ans .

A

1. Explanation :

Top-down parser - Leftmost derivation
Bottom-Up parser - Reverse of rightmost derivation

Ans .

C

1. Explanation :

Boolean duals are generated by simply replacing AND's with OR's , OR's with AND's , ‘1’s with ‘0’s and ‘0’s with ‘1’s. Variables and their complements are left alone.

Ans .

D

1. Explanation :

(292) 10 =(1204) x
2*102 + 9*10 + 2 = x 3 + 2*x2 + 0 + 4
200 + 90 + 2 = x3 + 2x2 + 4
x3 + 2x2 - 288 = 0
solution for x = 6
none of the option satisfy the equation.

Ans .

D

1. Explanation :

(x+y)z¯z¯
=(xz¯z¯+yz¯z¯)
=x(y+y¯)z¯+(x+x¯)yz¯x(y+y¯)z¯+(x+x¯)yz¯
=xyz¯+xy¯z¯+xyz¯+x¯yz¯xyz¯+xy¯z¯+xyz¯+x¯yz¯
=xyz¯+xy¯z¯+x¯yz¯.

Ans .

C

1. Explanation :

Statement 1 say . all integer divides all integer. Which is false.
Statement 2 say . there exists an integer which divides all integer.
Integer 1 divides all numbers. Statement 2 is true.
Hence option C.

Ans .

B

1. Explanation :

If this is as x→∞x→∞, then log(xk+a)=O(log(x))log⁡(xk+a)=O(log⁡(x)) for any fixed kk and aa.
Also, log(x)=O(xc)log⁡(x)=O(xc) for any c>0c>0. This should be enough.

Ans .

B

1. Explanation :

In each tree we have k-1 edges for k vertices. So, for t trees with total n vertices (all trees are disconnected in a forest) we would have n-t edges.

Ans .

A

1. Explanation :

f(g) = 2(3x+2)+3= 6x+7
g(f)= 3(2x+3)+2 = 6x+11

Ans .

A

1. Explanation :

Ans .

C

1. Explanation :

According to Kuratowski's theorem a finite graph is planar if and only if it does not contain a subgraph that is a subdivision of K5 (the complete graph on five vertices) or of K3,3 (complete bipartite graph on six vertices, three of which connect to each of the other three, also known as the utility graph).

Ans .

D

1. Explanation :

1.A circuit that adds two bits, producing a sum bit and carry bit is called half adder.
2.A circuit that adds two bits and a carry bit producing a sum bit and a carry bit is called full adder.
3.A device that accepts the value of a Boolean variable as input and produces its complement is called an inverter.
Ans will be (d)

Ans .

D

1. Explanation :

Starting with Internet Explorer 3.0 (1996), Microsoft added support to host ActiveX controls within HTML content. If the browser encountered a page specifying an ActiveX control via an OBJECT tag, it would automatically download and install the control with little or no user intervention.
Ans will be (d)

Ans .

D

1. Explanation :

Ans will be (d)

Ans .

B

1. Explanation :

Asynchronous transfer mode (ATM) is, according to the ATM Forum, "a telecommunications concept defined by ANSI and ITU (formerly CCITT) standards for carriage of a complete range of user traffic, including voice, data, and video signals". ATM was developed to meet the needs of the Broadband Integrated Services Digital Network, as defined in the late 1980s, and designed to unify telecommunication and computer networks. It was designed for a network that must handle both traditional high-throughput data traffic (e.g., file transfers), and real-time, low-latency content such as voice and video. The reference model for ATM approximately maps to the three lowest layers of the ISO-OSI reference model: network layer, data link layer, and physical layer. ATM is a core protocol used over the SONET/SDH backbone of the public switched telephone network (PSTN) and Integrated Services Digital Network (ISDN), but its use is declining in favour of all IP.

ATM provides functionality that is similar to both circuit switching and packet switching networks: ATM uses asynchronous time-division multiplexing, and encodes data into small, fixed-sized packets (ISO-OSI frames) called cells. This differs from approaches such as the Internet Protocol or Ethernet that use variable sized packets and frames. ATM uses a connection-oriented model in which a virtual circuit must be established between two endpoints before the actual data exchange begins. These virtual circuits may be “permanent”, i.e. dedicated connections that are usually preconfigured by the service provider, or “switched”, i.e. set up on a per-call basis using signalling and disconnected when the call is terminated.

Ans .

B

1. Explanation :

ANSI (American National Standards Institute) is the primary organization for fostering the development of technology standards in the United States. ANSI works with industry groups and is the U.S. member of the International Organization for Standardization (ISO) and the International Electrotechnical Commission (IEC).

Ans .

B

1. Explanation :

Ans .

B

1. Explanation :

T EP Priority After T
00 P1 P1[0] P2[-] P3[-] P4[-]
01 P2 P1[1] P2[0] P3[-] P4[-]
02 P1 P1[1] P2[X] P3[2] P4[-] ........P2 completed at 2
03 P3 P1[2] P2[X] P3[2] P4[2]
04 P4 P1[3] P2[X] P3[3] P4[2]
05 P1 P1[3] P2[X] P3[4] P4[X] .......P4 completed at 5
06 P3 P1[4] P2[X] P3[4] P4[X]
07 P1 P1[4] P2[X] P3[X] P4[X] .......P3 completed at 7
08 -- P1[X] P2[X] P3[X] P4[X] .......P1 completed at 8

Ans .

C

1. Explanation :

page size of 4KB ,so offset bits are 12 bits.
Remaining bits of virtual address 32 - 12 = 20 bits will be used for indexing.
number of sets =644644 = 16 (4-way set) => 4 bits.
So tag bits = 20 - 4 = 16 bits.

Ans .

A

1. Explanation :

Acc to SCAN 53 to 37 then to 14 and then to 0 (16+23+14=53) now direction will be reversed so 0 o 65 then to 67 then to 98,122,124,183 ( 65+2+31+24+2+59=183)total head movements=183+53=236

Acc to SSTF 53 to 65 then to 67 then 37 ,14,98,122,124,183 (so head movement= 12+2+30+23+84+24+2+59=236)

Ans .

A

1. Explanation :

We need a bit for each block. Number of blocks = disk size/ block size= 1.3 GB / 512 B.
This will give the size of bit map in bits. To get the byte size divide by 8 and we get 332.8 KB.
For disk size, GB is taken as 1024*1024*1024 bytes and for KB it is 1024 bytes.

Ans .

B

1. Explanation :

Fair scheduling is a method of assigning resources to jobs such that all jobs get, on average, an equal share of resources over time. When there is a single job running, that job uses the entire cluster. When other jobs are submitted, tasks slots that free up are assigned to the new jobs, so that each job gets roughly the same amount of CPU time. Unlike the default Hadoop scheduler, which forms a queue of jobs, this lets short jobs finish in reasonable time while not starving long jobs. It is also an easy way to share a cluster between multiple of users. Fair sharing can also work with job priorities - the priorities are used as weights to determine the fraction of total compute time that each job gets.