DBMS Notes Series-1: Tricky Questions from Previous Papers

Hello Friends!

Let's discuss questions on DBMS from previous year papers. To begin with, some of the multiple choice type questions are listed below:

Q. Location transparency allows :
 I. Users to treat the data as if it is done at one location.
 II. Programmers to treat the data as if it is at one location.
 III. Managers to treat the data as if it is at one location.
 Which one of the following is correct ?
 (A) I, II and III (B) I and II only (C) II and III only (D) II only

The right answer is (A). 
Reason: Location transparency means the location of data must not matter to the person who accesses/manipulates the data. This is a feature of distributed databases, which applies to every kind of database user. According to a definition on Wikipedia, "The location of a resource doesn't matter to either the software developers or the end-users. This creates the illusion that the entire system is located in a single computer, which greatly simplifies software development."

The I and II are database users. The III is a component of distributed databases. Database Manager components are responsible for providing seamless data access to users without regards to its location. Hence, this covers all 3 choices.

Q. Which of the following is correct?
I. Two phase locking is an optimistic protocol.
II. Two phase locking is pessimistic protocol
III. Time stamping is an optimistic protocol.
IV. Time stamping is pessimistic protocol.

(A) I and III (B) II and IV (C) I and IV (D) II and III

The right answer is (D).
Reason: Optimistic Vs. Pessimistic approach: The optimistic concurrency control approach doesn't actually lock anything. It is based on the assumption that conflicts of database operations are very less. Means, when when oner transaction is executing, other transactions will not access the same data item being accessed by the executing one. It lets transactions run to completion and only checks for conflicts when they are about to commit. Thus, a transaction is executed without any restrictions until it is committed.

The pessimistic approach believes that some other transaction might try to access the same piece of data. So, in order to prevent any conflict, a transaction will first acquire all the required locks, then perform all the operations. It has two phases:
1. Growing Phase, where a transaction must first acquire all the locks.
2. Shrinking Phase, where a transaction releases all the locks one-by-one.(It cannot issue lock requests here.)

Q. Data warehousing refers to
(A) storing data offline at a separate site (B) backing up data regularly
(C) is related to data mining (D) uses tape as opposed to disk

The right answer is (C)

Reason: NOT A because: Not all data of Data warehouse stored offline as it depends on nature and usage of data.

NOT B because: A data warehouse typically stores a lot of historical data, that is often not subject to change. Data that does not change only needs to be backed up once.

NOT D because: Data may be stored using a proper mix of disks, tapes, or near-line storage.

Common data warehouse models include a data warehouse that is subject oriented, time variant, non-volatile, and integrated.

Q. The "PROJECT' operator of a relational algebra creates a new table that has always

(A) More columns than columns in original table

(B) More rows than original table

(C) Same number of rows as the original table

(D) Same number of columns as the original table

The right answer is (A) 

Reason: The number of tuples in the result of PROJECT  <list> (R) is always less or equal to the number of tuples in R.

Now, a few questions with descriptive answers:

Q. Show that 2-phase locking ensures serializability?

A. In databases and transaction processing two-phase locking, (2PL) is a concurrency control method that guarantees serializability. A transaction is said to follow the two-phase locking protocol if all locking operations (read_lock, write_lock) precede the first unlock operation in the transaction. Such a transaction can be divided into two phases:

Phase 1: Growing Phase
i)  transaction may obtain locks
ii)  transaction may not release locks

Phase 2: Shrinking Phase
i)  transaction may release locks
ii)  transaction may not obtain locks

If lock conversion is allowed, then upgrading of locks (from read-locked to write-locked) must be done during the expanding phase, and downgrading of locks (from write-locked to read-locked) must be done in the shrinking phase. Hence, a read_lock(X) operation that downgrades an already held write lock on X can appear only in the shrinking phase.

The protocol assures serializability. It can be proved that the transactions can be serialized in the order of their lock points  (i.e. the point where a transaction acquired its final lock). Two-phase locking does not ensure freedom from deadlocks.

Q. How to determine candidate key(s) for a relation?

A. Finding candidate keys is just as simple as applying some algorithm here and there.
In the first example, there are five attributes:
W H O S E
WH -> S
HOS -> E
Steps:
1. Find the attributes that are neither on the left and right side
> (none)
2. Find attributes that are only on the right side
> E
3. Find attributes that are only on the left side
> WHO
4. Combine the attributes on step 1 and 3
> since step 1 has no attributes, it’s just WHO
5. Test if the closures of attributes on step 4 are all the attributes
> in this case, it is true. Because with WH we can get S, and by HOS, we can get E.
So we have only one candidate key that is WHO.

Q. What are steps of a Database design?

A. Major Steps in Database Design are:

1. Requirements Analysis: Talk to the potential users! Understand what data is to be stored, and what operations and requirements are desired.
2. Conceptual Database Design: Develop a high-level description of the data and constraints (we will use the ER data model)
3. Logical Database Design: Convert the conceptual model to a schema in the chosen data model of the DBMS. For a relational database, this means converting the conceptual to a relational schema (logical schema).
4. Schema Refinement: Look for potential problems in the original choice of schema and try to redesign.
5. Physical Database Design: Direct the DBMS into choice of underlying data layout (e.g., indexes and clustering) in hopes of optimizing the performance.
6. Applications and Security Design: It defines how the underlying database will interact with surrounding applications.

Q. Differentiate b/w Specialisation and Generalisation in ER Model?

A. Specialisation:
Top-down design process; we designate subgroupings within an entity set that are distinctive from other entities in the set.
These subgroupings become lower-level entity sets that have attributes or participate in relationships that do not apply to the higher-level entity set.
Depicted by a triangle component labeled ISA (E.g. customer “is a” person).
Attribute inheritance – a lower-level entity set inherits all the attributes and relationship participation of the higher-level entity set to which it is linked.

Generalisation:
A bottom-up design process – combine a number of entity sets that share the same features into a higher-level entity set.
Specialization and generalization are simple inversions of each other; they are represented in an E-R diagram in the same way.
The terms specialization and generalization are used interchangeably.

Cracking NET: No Instant Recipe but Strategy

Dear Friends!

I have been sharing my views on strategy for succeeding in NET exam on this blog as well as in Facebook groups. Most of times, I am thrown the queries on "Instant Recipe" for success in NET....like:

"Do you have study material for NET exam?"

"How I can prepare in next 2-3 months so that I crack the next exam?"

I start wondering why people look for "Instant Formula" to get success in competitive exams like UGC NET? Perhaps the "successful" formulas used by them in academic exams of their degree made them to think the same way for UGC NET exam also, but sorry, it is not that easy.

Before the pattern of NET examination was changed to 100% objective, Paper-III was subjective. You were supposed to secure just passing marks (40%) in paper III and 50% in total of Paper I and Paper II. It means, there was no criteria of falling in top 15% of eligible candidates. Still, the rate of success in Computer Science was dismally low all over India.

Now, the pattern has been changed and the high ratio of successful candidates is an indicator that cracking NET is far easier now. But, still, there is no "Instant Recipe" for success in NET.

NET preparation takes planning as well as patience. Unless you are born genius and covered all important topics of NET exam with utmost sincerity as a student and started preparing while you were in 1st semester of Masters degree....it definitely takes time to qualify. How much of time? It all depends on 3 factors, which we will discuss at the end.

Previously I posted two articles: One-> Getting success in NET exam, Two-> Securing great marks. In this post, I am sharing some more ideas on planning for your goal to qualify the Exam. Your success depends much on the dedicated execution of these steps.

First:

Select the topics you can study and prepare well for the next NET exam. You better focus on only those subjects you know are important to cover and also, you are confident that you can understand and can study them in short time. One strategy to choose the topics is to select 7-8 old topics, means, the topics you have already studied in your Masters. Additionally, select one or two new topics which you have never studied but are important for NET exam. For example, I had never heard about Theory of Computation (TOC) while doing MCA. The questions from this subject are always asked in paper II and/or paper III. You will need to select around 10 subjects.
For your old subjects, make a list of books you need to study. Follow the standard and famous books.

For the new topics, it would be better either you watch tutorial videos on NPTEL website, Youtube or find a good book for beginners on such topic. In my case, to learn Automata, I started with the book by Ullman. But that was a mistake, since the language of this book is not suitable for beginners. So, I shifted to online Videos and other web-based tutorials.  I later came to know that books on Automata by Peter Linz and Mishra were better for beginners.

For now, in first step, Do not start reading/re-reading books. Do this after following the next step.

Second:

Now, download all previous UGC papers from cbsenet.nic.in as well as ugcnetonline.in. For saving your time, here are the URLs to old question papers:

1. December 2015
2. June 2015

3. December 2014

4. June 2009 to June 2014

Now take printout of all the papers if you prefer to study offline. In my case, I took printout of all question papers and binded all papers in order like an MCQ book. Study the questions of all 3 papers. Try to understand the pattern and the types of questions asked. (Right now, do not focus on solutions to the problems, just develop an understanding on the types of the questions). 

To do this exercise in a planned manner, select any one topic and study questions from that topic only - in all question papers. For example, if you choose to begin with OS, make a list of only OS related questions from previous papers. On average, you will find 5-6 questions on OS from each exam session. It means, you will have more than 60 questions in hand on OS. If you have just began to prepare for NET, it is for sure that you will find that more than 50% questions are challenging. Highlight such questions. Do the same for all other topics. (Right now, we are not focusing on the solutions to the questions, but simply developing understanding of the topics). Solve  the questions as mentioned in third step.

Third:

Now solve the questions (which you selected in previous step) in three ways:

1. Find the topics in the text book, for example Galvin, and read the text. Make notes of important concepts. If you are short of time, simply use highlighter to mark in the book itself (Don't do this in borrowed books :-)

2. If  you could not find useful information in the book, take help of Google to research these questions. You will find Tens of useful links to study.

3. Still, if you fail to find the solution, post your question if forums like Facebook groups you have joined, online blogs related to your topic etc. Someone will definitely post answer to your question in any of the forums.

Remember, solving each challenging question can take time. Sometimes, it may take even 2 hours just to find the right answer and concept behind the asked question. (DO NOT CRAM ANS KEYS GIVEN BY UGC. I too made this mistake. It consumes time but is least helpful.) Why I have asked to do this step? You are not supposed to remember the given answers, but study the concept behind each such question. If you get the concept right, you will find many other questions easy to answer which are based on the same concept. If later you read the same questions or other questions on the concept you learnt, you will realise that you can now understand the problem in a better manner before you covered the topics.

Fourth:

After covering most of previous NET questions, research more on MCQ from other resources (both offline and online). Start solving questions from MCQ book by authors like Timothy Williams and any other you consider as good book. 

You may also use web sites some of which I am listing below:

http://www.geeksforgeeks.org/

http://nptel.ac.in/course.php

http://www.indiabix.com/engineering/

Last thing! There are three factors which will decide the direction of your exam preparation:
#1. (Most Important) Your focus on your goal (means how much serious you are to crack the NET)

#2. Your previous academics, means how much and which subjects you have learnt good by the time you complete your masters.

#3. Your perseverance, means "Never Say Die" spirit even if it is taking longer than you thought to qualify the exam. I stressed on this attitude in one of my previous posts also.

These 3 factors will decide when and how you will succeed.

Readers! The only objective of this post is to make you realise that it is better to equip yourself with all necessary Armour for hitting the exam. Give yourself at least 6 months of dedicated preparation. After 6 months, you will be confident on the concepts as well as on how to solve MCQs.

Best of Luck for July 2016 exam!

UGC NET/SET Exam: Some Ideas for success

Friends,

I want to share some of my ideas on NET/SET preparation. I have done MCA and qualified UGC NET exam twice and SET(H.P.)-2013 exam.

The very first time I appeared in NET exam simply to know the pattern of the exam and I didn't prepare much. At that time, exam had subjective pattern. I was able to clear Paper I and Paper II but I could not clear paper III. Still, I was very happy that I cleared first 2 papers. Then, I appeared again, then again. In total 5 of my efforts, I was not able to clear Paper III. In my 4th and 5th attempt, pattern of the paper had become totally objective. Still, I was not able to clear paper III.

I had started losing my faith in myself and I decided to opt out of Dec' 13 NET Exam and preferred to sit in another exam on the same date. I believed that the same story will get repeated. But one good thing I did was that before Dec' 13 exam, Himachal Pradesh Public Service commission had announced SET/SLET exam for the state universities/colleges. I applied for the exam and had appeared in the exam on November 17, 2013. The result of the exam was announced on  March 14, 2014 and I was declared qualified! WOW!!

The success in SET gave me much needed high and faith in myself. The point here is, we throw seeds on the land without bothering about which seed will bore fruits and when. Our efforts in one direction may show result somewhere else. The time I devoted in preparation of NET helped me to crack SET exam. Further, the success in SET exam enthused me and started preparing in a more determined manner. SET is limited in scope and a bit easier than NET. But success in this exam rekindled my faith in myself that NET too is not far and can be cracked with some more planning.

And the result was that I was able to qualify December 2014 exam. For the first time, I Cracked the "hard NUT" called UGC NET.

If one is failing 3 or more times in NET, it is quite possible that one will get disappointed and will lose one's confidence. To uplift one's spirits, focusing on some other exam/objective (for time being) of your chosen career will help you to forget the stress you might have been facing due to rejection in NET. You may choose to focus on some other competition exams for computer teacher/lecturers like CTET, SET, KV schools exams etc. whichever is announced in the time frame. Believe me, these exams can be cracked more easily than NET.

Your sole objective may be NET only, but if you succeed in other exams I have mentioned, it will lift your morale and will give birth to an enthusiastic mindset which is very crucial for cracking exams like NET.

PLEASE post your ideas (If whatever I have mentioned is right or wrong)..

Useful Q & A on Networking - Part 1

Q 1. What happens at every layer of the OSI model when we type "www.google.com" in the browser? 

Ans. The following sequence of steps takes place every time we type the name of a website in a browser->
1) Your PC will need to resolve "www.google.com" to an IP address. It will therefore need to send a 
query to your DNS server ( usually your ISP's DNS).

2) The PC constructs a DNS query packet with a destination IP address equal to your DNS server 
and a destination MAC address of your router (gateway).

3) The DNS query process begins to resolve the IP address of the given URL.

4) Once your PC (web browser application) receives the IP address of the destination web server, it will construct a HTTP GET request and send it to Google's web server. (As per OSI model at application layer [layer 7] application software which is client web browser now know the IP address of google.com). After that it pass that information down to presentation layer.

5) Presentation layer[ layer 6] converts that HTTP request into a standard format which is HTTP format so that the other layers can understand.and pass that information down to session layer.

6) Session layer[ layer 5] at client side creates session for www.google.com which is HTTP session so that it can be separated from other sessions.and pass that information down to transport Layer.

7) Transport Layer[ layer 4]  at client side chooses TCP for every HTTP session which is reliable connection which creates virtual connection by using three way hand-shake before sending actual data [HTTP Request]. This layer also prepares segments by adding source and destination port number. Source port is chosen by upper layer which is random number range from 1024 to 65535  and destination port number is here 80, which is well known port number used for HTTP services.

Layer 4 assign source port number to distinguish the web browser application from every other program running on computer and it also used to identify which application should receive return traffic. The destination port number is used to make sure that messages coming from web browser gets and sent to web server program running on the server and is not grabbed by other program. After preparing segment, layer 4 at client side pass that information down to network layer.

8) Network Layer [ layer 3] at client side  prepares packet by adding source and destination IP address. It also check whether destination IP address is on local network or it is on remote network. If it is on local network it checks ARP cache to find mac address of local device. If cache is empty it sends ARP message to resolve IP address into MAC address. Here the device is on remote network so client PC sends ARP to find MAC address of default gate way [192.168.1.1]. Every devices on that network segment gets that ARP request because as it layer 2 broadcast message destined for all devices on that broadcast domain but only the default gateway, which is 192.168.1.1, replies with its MAC address. Then network layer passes this information down to data link layer.

9) Data link layer[ layer 2]  prepares frame by adding source and destination MAC address of default gate way[192.168.1.1]. It also rund CRC which simply checks the data and frame header bits and add that result in to frame check sequence (FCS) field. Then, passes this information down to Physical layer.

10) Physical layer[ layer 1]: converts all this information in 1's and 0's and sends it to destination device using Ethernet cable.

11) Google's Web server will reply and start sending your application the necessary data using TCP session.

12) The application will start to draw and present the website on your screen.

At the server side, i.e. Google:

1) At Physical layer, server receives the bits. Layer 2 of server builds frames and runs CRC and checks answer with FCS field. If answer didn't match then the frame is discarded. If it matches then the destination MAC address is checked. Here, destination MAC address is right, so it checks Ethernet type field to find the protocol used at network layer (which is IP). It retrieves the packet from the frame and gives to IP at network layer.

2) Network layer checks destination IP address and in our case it matches so it checks protocol type field to find the protocol used at Layer 4 (which is TCP). Now network layer of server sends all information up to TCP at layer 4.

3) At transport layer, destination port number is checked which is well know port number 80, which is destined to web server application running on that particular server i.e. in our case www.google.com. So it passes all those information up to google web server. Google web server sends acknowledgement message to client to ensure that it has received its request. Finally, it sends the web page in the form of packet by packet over the network to client. This information gives google web page to our client web browser.

Notes on Computer Networking:Part II

IP address classes:

Class	Leftmost bits	Start address	Finish address
A	0xxx	0.0.0.0	127.255.255.255
B	10xx	128.0.0.0	191.255.255.255
C	110x	192.0.0.0	223.255.255.255
D	1110	224.0.0.0	239.255.255.255
E	1111	240.0.0.0	255.255.255.255

IP address range for Intranets (Private Networks):

Class	Private start address	Private finish address
A	10.0.0.0	10.255.255.255
B	172.16.0.0	172.31.255.255
C	192.168.0.0	192.168.255.255

IP packets addressed by them cannot be transmitted onto the public Internet. If such a private network needs to connect to the Internet, it must use either a network address translator (NAT) gateway, or a proxy server.

Some good facts on IP V6:

In IPV6, The address block fc00::/7 has been reserved for private networks.

IP officially reserves the entire range from 127.0.0.0 through 127.255.255.255 for loopback purposes.

Very Imp: IPv6 does not use classes. IPv6 supports the following three IP address types: 

1. Unicast 

2. Multicast 

3. Anycast

IPv6 does not support broadcast. Multicast addresses in IPv6 start with 'FF' (255) just like IPv4 addresses. Unicast addresses have 3 defined scopes, including link-local, site-local and global; and multicast addresses have 14 scopes.

The number of IPv6 addresses is 10^28. There is no ARP in V6. Currently, DHCP, FTP, PPP, RIP, SNMP, VPN, L2TP and Telnet do not support IPv6.

IPv6 does not require NAT. NAT, too, doesn't support V6. Currently, IPv6 packets are not forwarded.

IPv6 reserves just two special addresses: 0:0:0:0:0:0:0:0 and 0:0:0:0:0:0:0:1. IPv6 uses 0:0:0:0:0:0:0:0 internal to the protocol implementation, so nodes cannot use it for their own communication purposes. IPv6 uses 0:0:0:0:0:0:0:1 as its loopback address, equivalent to 127.0.0.1 in IPv4. The minimum size of an IP datagram is 28 bytes, including 20 bytes of header.

Anycast is a network addressing and routing methodology in which datagrams from a single sender are routed to the topologically nearest node in a group of potential receivers, though it may be sent to several nodes, all identified by the same destination address. On the Internet, anycast is usually implemented by using BGP.

In denial-of-service attacks (DoS), a rogue network host may advertise itself as an anycast server for a vital network service, to provide false information or simply block service.

6to4 is an Internet transition mechanism for migrating from IPv4 to IPv6, a system that allows IPv6 packets to be transmitted over an IPv4 network. 6to4 does not facilitate interoperation between IPv4-only hosts and IPv6-only hosts, but simply a transparent mechanism used as a transport layer between IPv6 nodes.

The network requests supporting DNS lookups run over TCP and UDP, port 53 by default.

Some Questions-Answers with explanation

In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges. Why is the spanning tree algorithm used for bridge-routing? (GATE 2005)

(a) For shortest path routing between LANs         (b) For avoiding loops in the routing paths

(c) For fault tolerance                                      (d) For minimizing collisions

SOLUTION: Spanning tree is a protocol that allows the bridges to exchange information so that only one of them will handle a given message that is being sent between two computers within the network. This  protocol prevents the condition known as a BRIDGE LOOP. It is typical to add a second bridge between two network segments as a backup in case the primary bridge fails (both bridges need to have some way to understand which bridge is the primary one). To do this, they have a separate path connection just between the bridges in which they exchange information, using bridge protocol data units (BPDUs).

The program in each bridge that allows it to determine how to use the protocol is known as the spanning tree algorithm. The algorithm is specifically constructed to avoid bridge loops (for a bridge using only the most efficient path when faced with multiple paths). If the best path fails, the algorithm recalculates the network and finds the next best route.

How many 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits and one parity bit?
(1) 600         (2) 800         (3) 876           (4) 1200

For 9600 baud, 1 bit=1/9600=0.104mS. Each char would require 11 bits. That means, to transmit one char, it would take 1.144 mS. So, applying the formula 1000/1.144 (mS in one Sec/transmission time for one char)=approx 875. Nearest answer is (3)

The single stage network is also called

A) one sided network                             B) two sided network

C) recirculating network                      D) pipeline network

Single-Stage Network is a single stage of switching elements (SEs) existing between the inputs and the outputs of the network. Data is circulated a number of times around the network.

 If a class B network on the Internet has a subnet mask of 255.255.248.0, what is the maximum number of hosts per subnet?
(a) 1022 (b) 1023 (c) 2046 (d) 2047

Explanation: Convert the subnet mask into binary format.

255.255.248.0 = 11111111.11111111.11111000.00000000

Number of 1's in the subnet mask indicates the Network-ID and the Subnet-ID part. Number of 0's in the subnet mask indicates the Host-ID part. Maximum number of Hosts per subnet = 2¹¹ = 2048, where 11 = Number of 0's in the Subnet Mask. Out of 2048 values, 2 addresses are reserved, hence we remove them (2048-2 = 2046). Note: In the host part of the address:- all bits as 1 is reserved as broadcast address and all bits as 0 is used as network address of subnet.

Which of the following system calls results in the sending of SYN packets?
(a) socket (b) bind (c) listen (d) connect

The connect system call is normally called by the client process to connect to the server process. The socket system call creates a new socket and assigns the protocol and resources to the created socket descriptor. The bind system call associates a local network transport address with a socket. For a client process, it is not mandatory to issue a bind call. The kernel takes care of doing an implicit binding when the client process issues the connect system call. It is often necessary for a server process to issue an explicit bind request before it can accept connections or start communication with clients. The listen call indicates to the protocol that the server process is ready to accept any new incoming connections on the socket. There is a limit on the number of connections that can be queued up, after which any further connection requests are ignored.

Paper-1 Notes: Series I

I have compiled some notes for Research and Communication Topic. I will post some more notes in near future. Also, I plan to share many important questions with explanation.
Please give your valuable comments on this post. It will help to customise the contents according to what is desired.

Some notes on Research and Communication

The following five phases outline a simple and effective strategy for conducting effective research:

I.   The conceptual phase

II. Phase of construction of research design

III. Empiric phase

IV. Analytic phase

V.  Disseminative phase

DIFF. TYPES OF RESEARCHES

Applied research refers to scientific study and research that seeks to solve practical problems. Applied research is used to find solutions to everyday problems, cure illness, and develop innovative technologies.

Action research is an applied research design used by practitioners (i.e., teachers, administrators, and other school personnel) to solve problems or supply useful information regarding educational policy making and practice at the local level.These types of researches are applied to solve immediate problems.

However, pure science is something with a lab component. Examples are biology, chemistry, physics, anatomy and physiology. Also, an applied science is a science that incorporates many sciences such as fire science or nutrition.

Experimental research is commonly used in sciences such as sociology and psychology, physics, chemistry, biology and medicine etc. It is a collection of research designs which use manipulation and controlled testing to understand causal processes. Generally, one or more variables are manipulated to determine their effect on a dependent variable.

The term descriptive research refers to the type of research question, design, and data analysis that will be applied to a given topic. Descriptive statistics tell what is, while inferential statistics try to determine cause and effect. Descriptive research can be either quantitative or qualitative. It involves gathering data that describe events and then organizes, tabulates, depicts, and describes the data collection. It often uses visual aids such as graphs and charts to aid the reader in understanding the data distribution. The intent of some descriptive research is to produce statistical information about aspects of education that interests policy makers and educators

A deductive argument is an argument in which it is thought that the premises provide a guarantee of the truth of the conclusion. In a deductive argument, the premises are intended to provide support for the conclusion that is so strong that, if the premises are true, it would be impossible for the conclusion to be false.

An inductive argument is an argument in which it is thought that the premises provide reasons supporting the probable truth of the conclusion. In an inductive argument, the premises are intended only to be so strong that, if they are true, then it is unlikely that the conclusion is false.

Phenomenology refers to an approach that concentrates on the study of consciousness and the objects of direct experience. OR The discipline of phenomenology may be defined initially as the study of structures of experience, or consciousness. Literally, phenomenology is the study of “phenomena”: appearances of things, or things as they appear in our experience, or the ways we experience things, thus the meanings things have in. The discipline of phenomenology may be defined initially as the study of structures of experience, or consciousness. Literally, phenomenology is the study of “phenomena”: appearances of things, or things as they appear in our experience, or the ways we experience things, thus the meanings things have in our experience.

Probability Sampling:  is any method of sampling that utilizes some form of random selection. In order to have a random selection method, you must set up some process or procedure that assures that the different units in your population have equal probabilities of being chosen. Humans have long practiced various forms of random selection, such as picking a name out of a hat, or choosing the short straw. These days, we tend to use computers as the mechanism for generating random numbers as the basis for random selection.

In this sampling technique, the researcher must guarantee that every individual has an equal opportunity for selection and this can be achieved if the researcher utilizes randomization.

Analysis of Variance: In statistics, analysis of variance (ANOVA) is a collection of statistical models, and their associated procedures, in which the observed variance in a particular variable is partitioned into components attributable to different sources of variation.

Correlational Study:

Sociogram: a sociometric diagram representing the pattern of relationships between individuals in a group, usually expressed in terms of which persons they prefer to associate with.

Sample Questions:

Q An example of asynchronous medium is:

(A) Radio (B) Television    (C) Film (D) Newspaper

Answer is: Newspaper

Explanation:  Asynchronous communication is communication other than in `real-time'-feedback is significantly delayed rather than potentially immediate. This feature ties together the presence or absence of the producer (s) of the text and the technical features of the medium. Asynchronous interpersonal communication is primarily through verbal text (e.g. letters, fax, e-mail). Asynchronous mass communication is primarily through verbal text, graphics and/or audio-visual media (e.g. film, television, radio, newspapers, magazines etc.).

Q In analog mass communication, stories are

(A) static      (B) dynamic          (C) interactive       (D) exploratory

Ans.static ( newspapers, letters, board game etc. are analog while e-mail, web newspaper etc are digital)

Q Which of the following is not an example of a continuous variable?

(A) Family size                     (B) Intelligence                     (C) Height             (D) Attitude

Ans. A

Continuous variables can have an infinite number of different values between two given points. Discrete variables can have only a certain number of different values between two given points. For example, in a family, there can be one, two, or three children, but there cannot be a continuous scale of 1.1, 1.5, or 1.75 children. A variable such as a person's height can take on any value in a range.

Q In the process of conducting research "Formulation of Hypothesis” is followed by

(A) Statement of Objectives               (B) Analysis of Data

(C) Selection of Research Tools        (D) Collection of Data

Q Transforming thoughts, ideas and messages into verbal and non-verbal signs is referred to as

(A) channelisation               (B) mediation        (C) encoding      (D) decoding

At their most basic, transmission models consist of three parts:

1. source  2. channel and  3. receiver

A sender encodes a message, which is transmitted through an appropriate channel (in the case of speech, in a face-to-face interaction, this is air), to a receiver who subsequently decodes the message. A source (a person with a reason for communicating) first accesses his or her communication encoder (a device that manipulates the source’s thoughts into some kind of code) in order to formulate a message. When messages are spoken in face-to-face interactions the channel is the air between the speaker and the listener. In the same way that a source requires an encoder to render his or her thoughts into messages, so a receiver requires a decoder to decipher the message. The receiver is, self-evidently, the person(s) at the end of the channel. Put another way, a person (source) formulates an idea – a concept – and encodes this concept linguistically, i.e. into strings of sounds, syllables and words, then transmits this encoded thought as a sound wave, whereupon another person (receiver) decodes the sound wave back into the original concept.