# CSIR Placement-Paper on March 2013

Discussion in 'Latest Placement papers' started by krishna, Sep 22, 2013.

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1. The output of a logic gate is 1 when all its inputs are at logic 0. The gate is either

(A) A NAND or an EX-OR

(B) An OR or an EX-NOR

(C) An AND or an EX-OR

(D) A NOR or an EX-NOR (Ans)

Hints and Solution : The output of a logic gate is 1 when all inputs are at logic 0. The gate is either a NOR or an EX- NOR.
Input Output
A B Y
0 0 1
0 1 0
1 0 0
1 1 0

Truth Table for NOR Gate
Input Output
A B Y
0 0 1
0 1 0
1 0 0
1 1 1

Truth Table for EX-NOR Gate

2. What is the simplification of the following Boolean expression in a Product-Of-Sum form?

(A) Out = (A+B+ NOT(C)) (A+NOT(C)+ D) (C+NOT(D))

(B) Out = (A+B) (A+NOT(D)) (B+NOT(C)+D)

(C) Out = (B+C+ NOT(D)) (A+C+ NOT(D)) ( NOT(C)+D) (Ans)

(D) Out = (A+NOT(B)+D) (C+D) (B+NOT(C)+D)

Hints and Solution : Transfer the seven maxterms to the map below as 0s.

Map the 0s as they appear left to right top to bottom on the map above. Now form groups of cells.

The final result is product of the three sums i.e.

Out = (B+C+ NOT(D)) (A+C+ NOT(D)) ( NOT(C)+D)

3. What is the appropriate form for the given k-map?

(A) F(W,X,Y,Z) = Σm (0,3,4,6,8,10,11,12,14)

(B) F(W,X,Y,Z) = Σm (0,2,5,6,8,10,13,14,15) (Ans)

(C) F(W,X,Y,Z) = Σm (1,2,5,6,8,9,11,14,15)

(D) F(W,X,Y,Z) = Σm (1,2,3,7,9,10,11,14,15)

F(W,X,Y,Z) = Σm (0,2,5,6,8,10,13,14,15)

4. How many fibres are required by a unidirectional and bidirectional ring respectively,to support their working traffic?

(A) 1 and 1

(B) 2 and 1

(C) 2 and 2

(D) 1 and 2 (Ans)

Hints and Solution : SONET rings can be classified by the routing principle and the SONET overhead used for triggering protection switching. A ring is called a unidirectional ring if bidirectional working signals follow opposite physical routes around a ring , while bidirectional working signals in a bidirectional ring follow the same route. Due to this routing principle, a unidirectional and a bidirectional ring, require one and two fibers respectively, to support their working traffic.

5. Find odd one out related to transmission media cables?

(A) Basic rate ISDN can transmit data at a rate of 512 kilobits per second on an existing local telephone line. (Ans)

(B) A T1 line is a dedicated telephone connection of 24 channels.

(C) A T1 channel can be configured to carry either voice or data traffic.

(D) Cable modems provide high-speed transmission over cable TV lines and are shared by many users.

Hints and Solution : Each ISDN line is made up of separate 64-Kbps "channels" for sending and receiving calls, plus a channel that is used primarily for signaling.

6. Mechanism to protect private networks from outside attack is

(A) Firewall (Ans)

(B) Antivirus

(C) Digital signature

(D) Formatting

Hints and Solution : Firewall is the mechanism to protect private networks from outside attack. It is a software or hardware used to isolate a private network from a public network.

7. What does a metric of 16 hops represent when using RIP?

(A) Number of hops to the destination

(B) Destination unreachable (Ans)

(C) Number of routers

(D) Bandwidth

Hints and Solution : Routing Information Protocol is a distance vector routing protocol that uses hop count as its metric. The maximum hop count is 15. 16 hops are considered unreachable. RIP updates are broadcast every 30 seconds by default. RIP has an administrative distance of 120.

8. If the 8085 adds 87H and 79H, specify the contents of the accumulator and the status of the S, Z, and CY flag?

(A) 10H; S =1, Z = 0, CY = 1

(B) 01H; S =0, Z = 0, CY = 1

(C) 00H; S =0, Z = 1, CY = 1 (Ans)

(D) 11H; S =1, Z = 1, CY = 0

Hints and Solution : The sum of 87H and 79H =100H. Therefore, the accumulator will have 00H, and the flags will be S =0, Z = 1, CY = 1

9. If the stack pointer is initialized with (4FEB) H, then after execution of Push operation in 8085 microprocessor, the Stack Pointer shall be

(A) 4FEA

(B) 4FEC

(C) 4FE9

(D) 4FED (Ans)

Hints and Solution : If the stack pointer is initialized with (4FEB) H, then after execution of push operation in 8085 microprocessor, stack pointer shall be 4FED.

10. If an input and output port can have the same 8-bit address how does the 8085 differentiate between the ports?

(A) The input port requires the WR and the output port requires the RD signal

(B) The input port requires the RD and the output port requires the WR signal (Ans)

(C) The input port and output port requires low I/O

(D) None of the above

Hints and Solution : The 8085 differentiates between the input and output ports of the same address by the control signal. The input port requires the RD and the output port requires the WR signal.

11. Consider the grammar given below

E → E+E | E*E | E-E | E/E | E^E | (E) | id

Assume that + and – have the same but least precedence, * and / have the next higher precedence but the same precedence and finally ^ has the highest precedence. Assume + and – associate to the left like * and / and that ^ associates to the right. Choose the correct statement with respect to relations for the ordered pairs (^,^) , (-,-) , (+,+) , (*,*) in the operator precedence table constructed for the grammar

(A) all <

(B) all >

(C) <,>,=,<

(D) <,>,>,> (Ans)

Hints and Solution : Relations for the ordered pairs (^,^),(-,-) , (+,+), (*,*) in the operator precedence table constructed for the grammar will be <,>,>,> as exponent is of right associative to itself and + , – , * and / are left associative

12. P, Q, R are three languages. If P and R are regular and if PQ=R, then

(A) Q has to be regular

(B) Q cannot be regular

(C) Q need not be regular (Ans)

(D) Q has to be a CFL

Hints and Solution : Proof as follows:

1. Let P=Q=R=EMPTY SET. The equation is satisfied. So this rules out (B).

2. Let P=R be Σ* then any Q will satisfy the equation. The demands of (A) & (D) are invalid.

3. So the answer is (C).

13. Which of the following conversion is not possible (algorithmically)?

(A) regular grammar to context-free grammar

(B) nondeterministic FSA to deterministic FSA

(C) nondeterministic PDA to deterministic PDA (Ans)

(D) nondeterministic TM to deterministic TM

Hints and Solution : It is not possible to convert nondeterministic PDA to deterministic PDA

14. The following are the set of processes with their respective CPU burst time (in milliseconds).

Processes CPU Burst time

P1 10

P2 5

P3 5

What will be the average waiting time if the process arrived in the order: P1, P2 & P3?

(A) 6.82 unit

(B) 7.56 unit

(C) 8.33 unit (Ans)

(D) 9.97 unit

Hints and Solution : CPU burst time indicates the time for which the process needs the CPU. Considering FCFS scheduling

Processes CPU Burst time

P1 10

P2 5

P3 5

For processes arriving in the order: P1 , P2 , P3

The Gantt Chart for the schedule is:.
P1 P2 P3

0 10 5 20

Waiting time for P1 = 0; P2 = 10; P3 = 15

Average waiting time: (0 + 10 + 15)/3 = 8.33 unit of time

15. Match the following:

Column I Column II

p) time sharing (1) Program first executed when a computer is turned on

q) process (2) Part of an operating system that communicates with the user

r) bootstrap (3) Technique that allows multiprocessing on a computer with a single CPU

s) shell (4) Activity of executing a program

(p) (q) (r) (s)

(A) (4), (1), (2), (3)

(B) (3), (4), (1), (2) (Ans)

(C) (2), (3), (4), (1)

(D) (1), (4), (2), (3)

Hints and Solution : p) time sharing-Technique that allows multiprocessing on a computer with a single CPU

q) process-Activity of executing a program

r) bootstrap-Program first executed when a computer is turned on

s) shell-Part of an operating system that communicates with the user

16. In Priority Scheduling a priority number is associated with each process. The CPU is allocated to the process with the highest priority. The problem of starvation is resolved by which of the following?

(A) Terminating the process.

(B) Aging (Ans)

(C) Mutual Exclusion

(D) Semaphore

Hints and Solution : Aging resolves or avoids the problem of starvation.
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17. Which of the following does not support for Page-Stealer process?

(A) It is a kernel process that makes room for the incoming pages

(B) It is created by the Kernel at the system initialization and invokes it throughout the lifetime of the system.

(C) Kernel locks a region when a process faults on a page in the region, so that page stealer cannot steal the page, which is being faulted in.

(D) All are correct. (Ans)

Hints and Solution : Option (A), (B) and (C) all are correct.

Page-Stealer process is the Kernel process that makes room for the incoming pages,

by swapping the memory pages that are not the part of the working set of a process.

It is created by the Kernel at the system initialization and invokes it throughout the

Kernel locks a region when a process faults on a page in the region, so that page

stealer cannot steal the page, which is being faulted in.

18. If you type ‘cat prog.c’ at a UNIX command prompt, which of the following sequences of system calls would be invoked?

(A) The shell calls fork(); the child process calls exec() and the parent calls wait() (Ans)

(B) The shell calls fork(); the child calls wait() and the parent calls exec()

(C) The shell calls exec() and then wait() and then fork()

(D) The shell calls wait() then fork(), creating a child which calls exec()

Hints and Solution : The shell calls fork(); the child process calls exec() and the parent calls wait(). The shell is just another process that can take a string as standard input, look for the program referenced by the string, and then run this program. Unless the program is put in the background, the shell will wait until the program has finished.

19. The 4.3BSD operating system is the version of:-

(A) UNIX (Ans)

(B) C

(C) C#

(D) C++

Hints and Solution : 4.3BSD was released in June 1986. Its main changes were to improve the performance of many of the new contributions of 4.2BSD that had not been as heavily tuned as the 4.1BSD code.

20. The mechanism of understanding what customer wants, analyzing needs, assessing

feasibility are provided by

(A) Analyzing process

(B) Requirement engineering (Ans)

(C) System modeling

(D) Checking process

Hints and Solution : The mechanism of understanding what customer wants, analyzing needs, assessing feasibility are provided by Requirement engineering

21. Which of these describe the activity of a contract review?

(A) Evaluation of the target market (Ans)

(B) Evaluation of the development risks

(C) Evaluation of the weather during the software development.

(D) Evaluation of the staff’s personal background

Hints and Solution : Evaluation of the target market is an activity of a contract review

22. Insufficient identification is a

(A) Technology-related problem

(B) Process-related problem (Ans)

(C) People-related problem

(D) Product-related problem

Hints and Solution : Insufficient identification is a process-related problem. Unidentified, partially identified, and unplanned risks pose a threat to the success of a software project. You need to intensively identify risks and evolve a risk management plan such that the project is completed successfully, on time

23. Which of the following is a formal reference point that measures system characteristics at a specific time?

(A) Benchmark

(B) Baseline (Ans)

(C) Functional

(D) Control

Hints and Solution : Baseline is a formal reference point that measures system characteristics at a specific time.

24. Five nines of reliability refers to

(A) Five software engineering practices that must be in place and assessed at level 9 in order to ensure reliability

(B) A product that is operational 99.999% of the time (Ans)

(C) A product that fails one time in 10,000 days

(D) A product which fails only five times in 99,999 days

Hints and Solution : Five nines of reliability refer to a product that is operational 99.999% of the time.

25. Which of the following is not a static testing tool?

(A) Static analyzers

(B) Code inspectors

(C) Output comparators (Ans)

(D) Standard enforces

Hints and Solution :

Static analyzers A static analyzer operates from a pre-computed database of descriptive information derived from the source text of the program.

Code inspectors A code inspector does a simple job of enforcing standards in auniform way for many programs. These can be single statement or multiple statement rules.

Output comparators These are used in dynamic testing-both single-module andmultiple-module varieties to check that predicted and actual outputs are equivalent.

Standard enforces This tool is like a code inspector. The main distribution is that afull-blown static analyzer looks at whole programs, whereas a standard enforcer looks at only single statements.

26. Match the following-

Column I Column II

p) Disparate data (i) Archive data

q) Non volatile data (ii) Level of detail

r) Data granularity (iii) Query and analysis

(iv) Production data

(p) (q) (r)

(A) (i), (iv), (ii)

(B) (iii), (ii), (iv)

(C) (ii), (i), (iii)

(D) (iv), (iii), (ii) (Ans)

Hints and Solution :

Disparate data- Production data

Non volatile data- Query and analysis

Data granularity- Level of detail

27. Select the options due to which credit theft is impossible with smart card

I. Key to unlock encrypted information required

II. No physical signature on the card

III. There is no external account number on the card

IV. Smart cards can be used with only acquainted merchants

(A) Only I and II

(B) Only II and III

(C) Only I and III (Ans)

(D) Only I and IV

Hints and Solution : Credit theft is impossible with smart card as key to unlock encrypted information is required and also due to the absence of external account number on the card.

28. Amazon.com comes under the following model-

(A) B2B

(B) B2C (Ans)

(C) C2C

(D) C2B

Hints and Solution : Amazon.com comes under B2C model where the consumer accesses the system of the supplier. It is still a two-way function but is usually done solely through the Internet. In B2C e-commerce, companies sell goods to consumers online in a dynamic environment. Each transaction under B2C represents an individual buying online.

29. Match the following:

Column I Column II

p) FAT12 (i) 1996 (Windows 95 OSR2)

q) FAT16 (ii) 1977 (Microsoft Disk BASIC)

r) FAT32 (iii) 1988 (MS-DOS 4.0)

(p) (q) (r)

(A) (iii), (i), (ii)

(B) (iii), (ii), (i)

(C) (ii), (iii), (i) (Ans)

(D) (i), (iii), (ii)

Hints and Solution :

FAT12 is a12-bit version developed by Microsoft in1977 for Microsoft Disk BASIC

FAT16 is a 16-bit version introduced in July 1988 for MS-DOS 4.0

FAT32 is a 32-bit version introduced in August 1996 for Windows 95 OSR2

30. Which macro sends an output message to the debug window of compiler:

(A) COUT

(B) FOUT

(C) TRACE (Ans)

(D) Display

Hints and Solution : TRACE macro sends an output message to the debug window of compiler. User can also see the trace messages without the debugger running using Debug View.

31. In windows95 which tool is used to kill errant hidden processes?

(B) Process Manager

(C) PVIEW (Ans)

(D) Explorer

Hints and Solution : PVIEW is used to kill errant hidden processes in windows 95. With PView, user can modify status of processes running on the system. As a result the entire system’s processes can be stopped and potentially halt.

32. Which question corresponds best to the following query?

SELECT CID, CDUR - 1,' = PRICE' FROM COURSES ORDER BY 2

(a) Select three columns from the COURSES table, of which the third one has a constant value, i.e. “ = PRICE”. Leave an empty line after every second line.

(b) Select two columns from the COURSES table, the second one gets as title “ = PRICE”. Sort the data according to the second column, in ascending order.

(c) Select three columns from the COURSES table, of which the third one has a constant value, i.e. “ = PRICE”. Sort the data according to the second column, in ascending order. (Ans)

(d) Select two columns from the COURSES table, of which the second one has a constant value, i.e. “ = PRICE”. Sort the data according to the second column, in ascending order.

Hints and Solution : Select three columns from the COURSES table, of which the third one has a constant value, i.e. “ = PRICE”. Sort the data according to the second column, in ascending order

33. . A modification to the database expressed in terms of a view must be translated to the –

(A) Actual relation in the conceptual model of the database (Ans)

(B) Queries in the actual database

(C) Relations of all the views of that database

(D) Need not be translated and the view of a database accommodates the modification

Hints and Solution : It must be translated to the actual relation in the conceptual model of the database

34. If D1, D2 ….. Dn are domains in relational model then the relation is a table which is a subset of ---

(A) { D1, D2 ….. Dn}

(B) D1 x D2 x …. Dn (Ans)

(C) D1 U D2 U …..Dn

(D) Maximum { D1, D2 ….. Dn}

Hints and Solution : Because it may have the values from all the domains.
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35. If the in order and pre order traversal of a binary tree are DBFEGHAC and ABDEFGHC respectively then, the post order traversal of that tree is

(A) DFGABCHE

(B) FHDGEBCA

(C) DFHGEBCA (Ans)

(D) CGHFEDBA

Hints and Solution : In order traversal is left->root->right and pre order is root->left-> right so it can used to find the post order traversal. first check the node in preorder and then find the root in in order, check left and right nodes and create a tree and then find the post order

36. “n” elements of a queue are to be reversed using another queue. The number of “ADD” and “REMOVE” required to do so is,

(A) 2*n

(B) 4*n

(C) n

(D) the task cannot be done (Ans)

Hints and Solution : The queue can not be reversed as it is based on the concept of FIFO.

37. Prim’s algorithm is a method available for finding out the minimum cost of a spanning tree. Its time complexity is given by:

(A) O(n*n) (Ans)

(B) O(n logn)

(C) O(n)

(D) O(1)

Hints and Solution : The time complexity of Prim’s Algorithm is O(n*n).

38. Which of the following is a collision resolution technique with open addressing in the context of hashing?

(A) Linear probing (Ans)

(B) Separate chaining

(C) Folding

(D) Mid-square method

Hints and Solution : Linear probing is a scheme in computer programming for resolving hash collisions of values of hash functions using two values - one as a starting value and one as an interval between successive values in modular arithmetic. The second value, which is the same for all keys and known as the stepsize, is repeatedly added to the starting value until either the entire table is traversed, or until a free space is found. This algorithm, which is used in openaddressed hash tables, provides good memory caching, through good locality of reference, but also results in clustering, an unfortunately high probability that where there has been one collision there will be more

39. Maximum number of children in a node in a B-tree of order “m” is:

(A) m (Ans)

(B) m/2-1

(C) m/2+1

(D) m/2

Hints and Solution : If a B-tree is created with order m then the node may have max. of m children and then it splits into two and middle child becomes the root

40. A has one share in a lottery in which there is 1 prize & 2 blanks; B has three shares in a lottery in which there are 3 prizes & 6 blanks: compare the probability of A’s success to that of B’s success as

(A) 7 : 16 (Ans)

(B) 16 : 7

(C) 6 : 14

(D) 14 : 6

Hints and Solution : A can draw a ticket in 3C1 = 3 ways.

Number of cases in which A can get a prize is 1. Probability of A’s success = 1
3

B can draw a ticket in 9C3 ways = 9.8.7 = 84 ways.
3.2.1

Number of ways in which B gets all blanks = 6C3 = 6.5.4 = 20
32.1

Number of ways of getting a prize = 84 – 20 = 64

Thus the probability of B’s success = 9 64 = 16
84 21

So A’s probability of success: B’s probability of success = 1 : 16
3 21

41. The following ‘C’ code : -

# include < stdio.h >

main ( )

{ file * FP ; FP = fopen ( “ trial “ , “ r “ ); }

FP points to :-

(A) First character in the file

(B) A structure which contains a ‘ char ‘ pointer to the first character in the file.

(C) Name of the file.

(D) None. (Ans)

Hints and Solution : Here P represents file pointer. It points to the structure, in which char pointer exists which points to the first character of file.

42. The eccentricity of node labeled 5 in the graph shown below is:

(A) 6

(B) 7 (Ans)

(C) 8

(D) 5

Hints and Solution : Eccentricity of a given node is the maximum of minimum path from other nodes to the given node.

Cost of minimum path from 1 to 5 is 7.

Cost of minimum path from 2 to 5 is 6

Cost of minimum path from 3 to 5 is 4

Cost of minimum path from 4 to 5 is 7

43. Let M be a 3 × 3 , adjacent matrix corresponding to a given graph of 3 nodes labeled 1, 2, 3. If entry (1, 3) in M3 is 2, then the graph could be

(A) (Ans)

(B)

(C)

(D)

Hints and Solution : If (1, 3) entry in M3 is 2, it means there are 2 path of length 3, connecting nodes 1 and 3. If you see the graphs in option (a) then we have two paths ( → 2 → 3 → 3 & 1 → 3 → 3 → 3)

44. Error detection at the data link level is achieved by:-

(A) Bit stuffing

(B) Cyclic redundancy check. (Ans)

(C) Hamming codes

(D) Equalization

Hints and Solution : At Data link layer there are 3 techniques for error detection:-

(1) Parity check

(2) Check summing method

(3) Cyclic redundancy check.

45. How many characters per second (7 bits + 1 parity) can be transmitted over a 2400 bps line if the transfer is synchronous (1 start and 1 stop bit)?

(A) 300 (Ans)

(B) 240

(C) 250

(D) 275

Hints and Solution : Start and stop bits are not needed in synchronous transfer of data, so it is 2400/ 8 = 300.

46. Represent (54.78)10 in normalized binary floating number.

(A) S = 1 . M = 011011000110, E = 11010011

(B) S = 0. M = 101101100011, E = 10000100 (Ans)

(C) S = 0, M = 1010011 10010, E = 10011011

(D) S =1, M = 101101100011, E = 10000100

Hints and Solution :

46.(B) (54.78)10 = (110110.1100011)2

= 1.101101100011×25

∴ M = 101101100011

E = 5 + 127 = (132)10 = (10000100)2

and s = 0 (positive)

∴ s = 0, M = 101101100011, E = 10000100

47. Which of the following statements are true?

(i) Constructors may be declared in private, public or protected section.

(ii) An object with a constructor or destructor can not be used as a member of a union.

(iii) There can be virtual constructors.

Choose the correct option:

(A) Only (i) is true

(B) Both (i) and (ii) are true

(C) Only (iii) is true

(D) Only (ii) is true (Ans)

Hints and Solution : A constructor is a special member function whose task is to initialize the objects of its class. It is special because its name is same as that of the class. A destructor is used to destroy the objects that have been created by a constructor. Constructors should be declared in the public section. An object with a constructor or destructor cannot be used as a member of a union. Constructors can never be virtual.

48. What happens when an exception is not caught?

(A) Code in the catch block is generated.

(B) An error occurs.

(C) The program is aborted. (Ans)

(D) The program executes normally.

Hints and Solution : When an exception is not caught, the program is aborted. Whenever an exception is generated (in the code which is placed in the ‘try’ block), it is thrown using a ’throw’ statement in the ‘try’ block which is caught by the ‘catch’ block where it is handled appropriately.

49. ______ is the abstraction process of introducing new characters to an existing class

of objects to create one or more new classes of objects.

(A) Specialization (Ans)

(B) Generalization

(C) Abstraction

(D) Aggregation

Hints and Solution : Abstraction is simplification mechanism used to hide superfluous details of a set of objects. It allows one to concentrate on the properties that are of interest to the application.
Generalization is the abstraction process of viewing sets of objects as a single general class by concentrating on the general characteristics of the constituent sets while suppressing or ignoring their difference.
Specialization is the abstraction process of introducing new characteristics to an existing class of objects to create one or more new classes of objects.
Aggregation is the process of compiling information on an object, thereby abstracting a higher-level object.

50. What will be the result of the following addition?

(3BCA.5078)16 + (9EBD.97F3)16 + (5FB.E2C)16

(A) (E082.CB2C)16

(B) (14916.9241)16

(C) (E083.CB2B)16 (Ans)

(D) (E916.9241)16

Hints and Solution : Addition in hexadecimal number system is same as other number system. Since hexadecimal number, upto 15(F) are defined, a carry will be generated if addition is larger than F. Another way is to convert the hexadecimal numbers into equivalent binary numbers and then add the numbers.