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Students can Download Tamil Nadu 12th Commerce Model Question Paper 1 English Medium Pdf, Tamil Nadu 12th Commerce Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Commerce Model Question Paper 1 English Medium

Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II. III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about 50 words.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in about 150 words.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in about 250 words. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 90

Part – I

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
With a wider span, there will be ………… hierarchical levels.
(a) more
(b) less
(c) multiple
(d) additional
Answer:
(b) less

Question 2.
Spot market is a market when the delivery of the financial instrument and payment of cash occurs.
(a) Immediately
(b) In the future
(c) Uncertain
(d) After one mouth
Answer:
(a) Immediately

Question 3.
How many times a security can be sold in a secondary market?
(a) only one time
(b) two time
(c) three times
(d) multiple times
Answer:
(d) multiple times

Question 4.
Money market provides ……………
(a) medium term funds
(b) short term funds
(c) long term funds
(d) shares
Answer:
(b) short term funds

Question 5.
Jobbers transact in a stock exchange.
(a) For their Clients
(b) For their Own Transactions
(c) For other Brokers
(d) For other Members
Answer:
(b) For their Own Transactions

Question 6.
Registering and controlling the functioning of collective investment schemes as ………..
(a) Mutual funds
(b) Listing
(c) Rematerialisation
(d) Dematerialisation
Answer:
(d) Dematerialisation

Question 7.
Human resource management determines the …… relationship.
(a) internal, external
(b) employer, employee
(c) owner, servant
(d) principle, agent
Answer:
(b) employer, employee

Question 8.
Advertisement is a ………. source of recruitment.
(a) internal
(b) external
(c) agent
(d) outsourcing
Answer:
(b) external

Question 9.
Selection is usually considered as a,process.
(a) positive
(b) negative
(c) natural
(d) none of these
Answer:
(b) negative

Question 10.
Improves skill levels of employees to ensure better job performance.
(a) Training
(b) Selection
(c) Recruitment
(d) Performance appraisal
Answer:
(d) Performance appraisal

Question 11.
Which one of the market deals in the purchase and sale of shares and debentures?
(a) stock exchange market
(b) Manufactured goods market
(c) local market
(d) family market
Answer:
(a) stock exchange market

Question 12.
In the following variables which one is not the variable of marketing mix?
(a) place variable
(b) product variable
(c) program variable
(d) price variable
Answer:
(c) program variable

Question 13.
Social marketing deals with
(a) Society
(b) Social clan
(c) Social change
(d) Social evil
Answer:
(c) Social change

Question 14.
The main objective of all business enterprises is …………….
(a) Providing service
(b) Providing better standard of life
(c) Providing necessities to the society
(d) Earn profit
Answer:
(d) Earn profit

Question 15.
As the consumer is having the rights, they are also having
(a) measures
(b) Promotion
(c) responsibilities
(d) duties
Answer:
(c) responsibilities

Question 16.
Which of the following is not a consumer right summed up by John.F.Kennedy.
(a) Right to safety
(b) Right to choose
(c) Right to consume
(d) Right to be informed
Answer:
(a) Right to safety

Question 17.
The chairman of the District Forum is
(a) District Judge
(b) High court Judge
(c) Supreme court Judge
(d) None of the above
Answer:
(a) District Judge

Question 18
ownership makes bold management decisions due to their strong foundation in the intimation level.
(a) Private
(b) Public
(c) Corporate
(d) MNC’s
Answer:
(a) Private

Question 19.
The property in the goods means the
(a) Possession of goods
(b) Custody of goods
(c) Ownership of goods
(d) Both (a) and (b)
Answer:
(c) Ownership of goods

Question 20.
Number of parties is a bill of exchange are
(a) 2
(b) 6
(c) 3
(d) 4
Answer:
(c) 3

Part – II

Answer any seven in which question No. 30 is compulsory. [7 x 2 = 14]

Question 21.
What are the objectives of MBO?
Answer:
Management by objectives is intended primarily:

• To measure and judge performance.
• To relate individual performance to organisational goals.
• To clarify both the job to be done and the expectations of accomplishment.
• To foster the increasing competence and growth of the subordinates.

Question 22.
Write a note on OTCEI.
Answer:
The OTCEI was set up by a premier financial institution to allow the trading of securities across the electronic counters throughout the country.

Question 23.
Define Stock Exchange.
Answer:
According to Husband and Dockerary, “Stock exchanges are privately organized markets which are used to facilitate trading in securities.”

Question 24.
Give the meaning of Human Resource.
Answer:
In an organisation, the human resource are the employees who are inevitable for the survival t and success of the enterprise.

Question 25.
What is service marketing?
Answer:
Service marketing denotes the processing of selling service goods like telecommunication, banking, insurance, car rentals, healthcare, tourism, professional services, repairs etc.

Question 26.
Write short notes on: “Right to be informed.”
Answer:
Consumers should be given all the relevant facts about the product. This implies that the manufacturer and the dealer are expected to disclose all the material facts relevant and relating to the product.

Question 27.
What is GST?
Answer:
GST is the indirect tax levied on goods and services across the country. It is a comprehensive, multi-stage, destination-based tax that is levied on every value addition.

Question 28.
What is a contract of sale of goods?
Answer:
Contract of sale of goods is a contract whereby the seller transfers or agrees to transfer the property (ownership) of the goods to the buyer for a price.

Question 29.
What is the other name of business entrepreneur?
Answer:
The term entrepreneur means the person who takes steps for commencing the business. So he is otherwise called as organiser or proprietor.

Question 30.
Define Director.
Answer:
The Companies Act 2013 section 2 (34) defines a director appointed to the board of a Company is “A Person who is appointed or elected member of the Board of Directors of a company and has the responsibility of determining and implementing policies along with others in the board”.

Part – III

Answer any seven in which question No. 31 is compulsory. [7 x 3 = 21]

Question 31.
What are the process involved in MBO?
Answer:

• Defining Organisational Objectives
• Goals of Each Section
• Fixing Key Result Areas
• Setting subordinate objectives or targets
• Matching Resources with objective
• Periodical Review meetings
• Appraisal of Activities
• Reappraisal of objectives

Question 32.
Differentiate spot market from future market.
Answer:

Question 33.
What are the documents required for a Demat account?
Answer:
For opening a demat account, we submit proof of identity and address along with a passport size photo and the account opening form.

• Documents for Identity: Pan card, Voters ID, Passport, Driver’s License, IT Returns, Electricity and Telephone Bills are the Identity documents.
• Documents for Address: Ration card, Passport, Voter’s ID card, Driving License, Telephone Bills, Electricity bills are the address documents.

Question 34.
Mention any three Role of Marketer.
Answer:
(1) Instigator: As an instigator, marketer keenly watches the developments taking place in the market and identifies marketing opportunities emerging in the ever changing market.

(2) Integrator: Marketer plays a role of integrator in the sense that he collects feedback or vital inputs from channel members and consumers.

(3) Implemented Marketer plays a role of implementer when he/she actually converts marketing opportunities into marketable product.

Question 35.
Discuss the objectives E-Marketing.
Answer:
The following are the objectives of E-Marketing:

• Expansion of market share
• Reduction of distribution and promotional expenses
• Achieving higher brand awareness
• Strengthening database

Question 36.
What do you understand by “Right to redressal”?
Answer:
The complaints and protests are not just to be heard: but the aggrieved party is to be granted compensation within a reasonable time period . There should be prompt settlement of complaints and claims lodged by the aggrieved customers.

Question 37.
Explain the concept of Privatisation.
Answer:
Privatisation means permitting the private sector to set up industries which were previously reserved for the public, sector. Under this policy many Public Sector Units (PSUs) were sold to private sector. The main reason for privatisation was that PSUs were running in losses due to mismanagement and political interference.

Question 38.
What are the characteristics of a bill of exchange?
Answer:
Characteristics of a Bill of Exchange:

• A bill of exchange is a document in writing.
• The document must contain an order to pay.
• The order must be unconditional.
• The instrument must be signed by the person who draws it.
• The name of the person on whom the bill is drawn must be specified in the bill itself.

Question 39.
Distinguish between Entrepreneur and Manager.

Question 40.
When are alternative directors appointed?
Answer:
Alternate director is appointed by the Board of Directors, as a substitute to a director who may be absent from India, for a period which is not less than three months.

Part – IV

Answer all the following questions. [7 x 5 = 35]

Question 41
(a) Explain the various functions of management.
Answer:

• Planning:- Planning is the primary function of management. Planning is a constructive reviewing of future needs so that present actions can be adjusted in view of the established goal.
• Organising:- Organising is the process of establishing harmonious relationship among the members of an organisation and the creation of network of relationship among them.
• Staffing:- Staffing function comprises the activities of selection and placement of competent personnel.
• Directing:- Directing denotes motivating, leading, guiding and communicating with subordinates on an ongoing basis in order to accomplish pre-set goals.
• Controlling:- Controlling is performed to evaluate the performance of employees and deciding increments and promotion decisions.
• Co-ordination:- Co-ordination is the synchronization of the actions of all individuals, working in the enterprise in different capacities.
• Motivating:- The goals are achieved with the help of motivation. Motivation includes increasing the speed of performance of a work and developing a willingness on the part of workers.

Subsidiary Functions:

• Innovation:- Innovation includes developing new material, new products, new techniques in production, new package, new design of a product and cost reduction.
• Representation:- A manager has to act as representative of a company. It is the duty of every manager to have good relation with others.
• Decision-making:- Every employee of an organisation has to take a number of decisions every day. Decision-making helps in the smooth functioning of an organisation.
• Communication:- Communication is the transmission of human thoughts, views or opinions from one person to another person. Communication helps the regulation of job and co-ordinates the activities.

[OR]

(b) Compare the concept of social marketing with service marketing.
Answer:
Social marketing: Social marketing is a new marketing tool.
It is the systematic application of marketing philosophy to achieve social good.
The primary aim of social marketing is ‘social good’ such as anti-tobacco, anti-drug, anti-pollution, anti-dowry, road safety, protection of girl child.

• Service marketing: A service is any activity or benefit that one party can offer to another which is essentially intangible.
• Service marketing is a specialized branch of marketing.
• Service marketing denotes the process of selling service goods like telecommunication, banking, insurance, car rentals, healthcare, tourism and repairs.

Question 42
(a) Explain the various disadvantages of MBO.
Answer:

• MBO fails to explain the philosophy; most of the executives do not know how MBO works? what is MBO? and why is MBO necessary? and how participants can benefit by MBO.
• MBO is a time consuming process. Much time is needed by senior people for framing the MBO. Next, it leads to heavy expenditure and also requires heavy paper work.
• MBO emphasises only on short-term objectives and does not consider the long-term objectives.
• The status of subordinates is necessary for proper objectives setting. But, this is not possible in the process of MBO.
• MBO is rigid one. Objectives should be changed according to the changed circumstances, external or internal. If it is not done, the planned results cannot be obtained.

[OR]
(b) Explain the duties of consumers.
Answer:
Apart from rights, there are certain duties imposed on the consumer. The following are the duties of consumers.

• Buying Quality Products at Reasonable Price: It is the duty of a consumer to purchase a product after gaining a thorough knowledge of its price, quality and other terms and conditions.
• Ensure the Weights and Measurement before Purchase: The consumer should ensure that he/she is getting the product of exact weight and. measure.
• Reading the Label Carefully: It is the duty of the consumer to read the label of the product thoroughly.
• Beware of False and Attractive Advertisements: It is the prime duty of the consumer about the genuineness of the advertisement, before purchasing the product.
• Ensuring the Receipt of Cash Bill: It is a legitimate duty of consumers to get the cash receipt and warranty card supplied along with the bill.

Question 43.
(a) Explain the Instruments of Money Market.
Answer:
(i) Treasury Bills: Treasury bills are very popular and enjoy a higher degree of liquidity since they are issued by the Government. A Treasury bill is nothing but a promissory note issued for a specified period stated therein. The Government promises to pay the specified amount mentioned therein to the bearer of the instrument on the due date. The period does not exceed a period of one year.

(ii) Certificate of Deposits: Certificate of Deposits are short-term deposit instruments issued by banks and financial institutions to raise large sums of money. The Certificate of Deposit is transferable from one party to another. Due to their negotiable feature, they are also known as negotiable certificate of deposit.

(iii) Commercial Bills: The Commercial Bill is an instrument drawn by a seller of goods on a buyer of goods. It possesses the advantages like self-liquidating in nature, recourse to two parties, knowing exact date of transactions, transparency of transactions etc.

[OR]

(b) Explain the overall performance of State Commission.
The State Commission is to be appointed by the State Government. The person who is a judge or retired judge of high court is the president of the state commission.

Performances:

• The compensation of the value should not exceed Rs.20 lakhs and below Rs.1 crore.
• The state commission has the power to call for the records and pass orders in the district forum.
• To furnish the information which is required for the purpose of the Act to any officer.

Question 44.
(a) Explain the powers of SEBI.
Answer:
The various powers of a SEBI are explained below:

• Powers Relating to Stock Exchanges and Intermediaries: SEBI has wide powers to get the information from the stock exchange and intermediaries regarding their business transactions for inspection.
• Power to Impose Monetary Penalties: SEBI has power to impose monetary penalties on capital market intermediaries for violations.
• Power to Initiate Actions in Functions Assigned
• Power to Regulate Insider Trading: SEBI has power to regulate insider trading or can regulate the functions of merchant banker.
• Powers Under Securities Contracts Act: For the regulation of stock exchange, the Ministry of Finance issued a notification, for delegating several of its powers under the securities contract Act.

[OR]

(b) Explain the impact of LPG on Indian Economy.
Impact of Liberalization:

• Liberalization has opened up new business opportunities abroad and increased foreign direct investment.
• New market for various goods came into existence and resulted not only in urban but also in rural development.
• It became very easy to obtain loans from banks for business expansion.

Impact of Privatization:

• Privatization has a positive impact on the financial growth by decreasing the deficits and debts.
• Increase in the efficiency of government undertakings.
• Provide better goods and services to the consumers.

Impact of Globalization:

• Multinational corporations can manufacture, buy and sell goods worldwide.
• Globalization has led to a boom in consumer products market.
• The advent of foreign companies and growth in economy has led to job creation.

Question 45
(a) Describe the significance of External source of recruitment.
Answer:
External sources of recruitment include sources that lie outside the organisation. This provides a wider collection of potential employees with the necessary skillset, especially for managerial and technical positions. Existing employees can recommend suitable candidates, which may lead to a higher level of teamwork and synchronisation among employees. Hiring new employees can lead to the introduction of new blood and thus the introduction of a new set of skills and ideas. External sources of recruitment offer jobs to unskilled, semi-skilled and skilled workers.

[OR]

(b) Discuss in detail the features of a cheque.
Answer:
A cheque is a negotiable instrument drawn on a particular banker.
Features:
(i) Instrument in Writings
A cheque or a bill or a promissory note must be an instrument in writing. Though the law does not prohibit a cheque being written in pencil, bankers never accept it because of risks involved. Alternation is quite easy but detection is impossible in such cases.

(ii) Unconditional Orders
The instrument must contain an order to pay money. It is not necessary that the word ‘order’ or its equivalent must be used to make the document a cheque. It does not cease to be a cheque just because the world ‘please’ is used before the word pay. Further the order must be unconditional.

(iii) Drawn on a Specified Banker Only
The cheque is always drawn on a specified banker. A cheque vitally differs from a bill in this respect as latter can be drawn on any person including a banker. The customer of a banker can draw the cheque only on the particular branch of the bank where he has an account.

(iv) A Certain Sum of Money Only
The order must be for payment of only money. If the banker is asked to deliver securities, the document cannot be called a cheque. Further, the sum of money must be certain.

(v) Payee to be Certain
The cheque must be made payable to a certain person or to the order of a certain person . or to the bearer of the instrument. The word, person includes corporate bodies, local authorities, associations, holders of the office of an institution etc.

(vi) Signed by the Drawer
The cheque is to be signed by the drawer. Further, it should tally with specimen signature furnished to the bank at the time of opening the account

Question 46.
(a) What are the differences between on the job training and off the job training?
Answer:

[OR]

(b) What are the various kinds of Debentures?
Answer:
Debenture is a document issued by the company for acknowledging the loan from the public. Debentures are classified into different categories on the basis of:

• Convertibility of the Instrument
• Security of the Instrument;
• Redemption ability; and
• Registration of Instrument.

On the basis of convertibility:

• Non-Convertible Debentures: These instruments cannot be converted into equity shares.
• Partly Convertible Debentures: A part of these instruments are converted into equity shares.
• Fully Convertible Debentures: These are fully convertible into equity shares.
• Optionally Convertible Debentures: The investor can have the option to either convert the debentures at a price decided by the issuer or agreed upon at the time of issue.

On the basis of Security:

• Secured Debentures: These instruments are secured by a charge on the fixed assets of the issuer company.
• Unsecured Debentures: These instruments are unsecured against the assets.

On the basis of Redeemability:

• Redeemable Debentures: It refers to the debentures which will be redeemed in the future.
• Irredeemable Debentures: It is a debenture, in which no specific time is specified by the companies to pay back the money.

On the basis of Registration:

• Registered Debentures: These are issued in the name of a particular person, who is registered by the company.
• Bearer Debentures: These are issued to the bearer and are negotiable instruments, and are transferred by mere delivery.

Question 41.
(a) Why the marketing is important to the society and individual firm? Explain.
Answer:
Importance of Marketing:
To the Society:

• Marketing is a connecting link between the consumer and the producer.
• Marketing helps in increasing the living standard of people.
• Marketing helps to increase the nation’s income.
• Marketing process increases employment opportunities.
• Marketing helps to maintain economic stability and rapid development in various countries.

To the Individual Firms:

• Marketing generates revenue to firms.
• Marketing gives information to the top management for taking overall decisions on production.
• Marketing and innovation are the two basic functions of all businesses.

[OR]

(b) State the qualification of directors.
Qualifications of Director: As regards to the qualification of directors, there is no direct provision in the Companies Act, 2013. In general, a director shall possess appropriate skills, experience and knowledge in the fields of finance, law, management, sales, marketing, research and other disciplines related to the business.

The following are the qualifications:

• A director must be a person of sound mind.
• A director must hold share qualification if the article of association provides such.
• A director must be an individual.
• A director should be a solvent person.
• A director should not be convicted by the Court for any offence, etc.

Students can Download Computer Science Chapter 14 Importing C++ Programs in Python Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Computer Science Solutions Chapter 14 Importing C++ Programs in Python

Samacheer Kalvi 12th Computer Science Importing C++ Programs in Python Text Book Back Questions and Answers

PART – I
I. Choose The Best Answer

Question 1.
Which of the following is not a scripting language?
(a) JavaScript
(b) PHP
(c) Perl
(d) HTML
Answer:
(d) HTML

Question 2.
Importing C++ program in a Python program is called …………………………
(a) wrapping
(b) Downloading
(c) Interconnecting
(d) Parsing
Answer:
(a) wrapping

Question 3.
The expansion of API is ……………………….
(a) Application Programming Interpreter
(b) Application Programming Interface
(c) Application Performing Interface
(d) Application Programming Interlink
Answer:
(b) Application Programming Interface

Question 4.
A framework for interfacing Python and C++ is ………………………….
(a) Ctypes
(b) SWIG
(c) Cython
(d) Boost
Answer:
(d) Boost

Question 5.
Which of the following is a software design technique to split your code into separate parts?
(a) Object oriented Programming
(b) Modular programming
(c) Low Level Programming
(d) Procedure oriented Programming
Answer:
(b) Modular programming

Question 6.
The module which allows you to interface with the Windows operating system is ………………………….
(a) OS module
(b) sys module
(c) csv module
(d) getopt module
Answer:
(a) OS module

Question 7.
getopt( ) will return an empty array if there is no error in splitting strings to …………………………..
(a) argv variable
(b) opt variable
(c) args variable
(d) ifile variable
Answer:
(c) args variable

Question 8.
Identify the function call statement in the following snippet.
if_name ==’_main_’:
main(sys.argv[l:])
(a) main(sys.argvfl:])
(b) _name_
(c) _main_
(d) argv
Answer:
(b) _name_

Question 9.
Which of the following can be used for processing text, numbers, images, and scientific data?
(a) HTML
(b) C
(c) C++
(d) PYTHON
Answer:
(d) PYTHON

Question 10.
What does name contains?
(a) C++ filename
(b) main( ) name
(c) python filename
(d) os module name
Answer:
(c) python filename

PART – II
II. Answer The Following Questions

Question 1.
What is the theoretical difference between Scripting language and other programming language?
Answer:
The theoretical difference between the two is that scripting languages do not require the 228 compilation step and are rather interpreted. For example, normally, a C++ program needs to be compiled before running whereas, a scripting language like JavaScript or Python need not be compiled. A scripting language requires an interpreter while a programming language requires a compiler.

Question 2.
Differentiate compiler and interpreter?
Answer:
Compiler:

1. It converts the whole program at a time
2. It is faster
3. Error detection is difficult. Eg. C++

Interpreter:

1. line by line execution of the source code.
2. It is slow
3. It is easy Eg. Python

Question 3.
Write the expansion of

1. SWIG
2. MinGW

Answer:
SWIG (Simplified Wrapper Interface Generator. Both C and C++)
MinGW (Minimalist GNU for Windows)

Question 4.
What is the use of modules?
Answer:
We use modules to break down large programs into small manageable and organized files. Furthermore, modules provide reusability of code. We can define our most used functions in a module and import it, instead of copying their definitions into different programs.

Question 5.
What is the use of cd command. Give an example?
Answer:
The syntax to change from c:\> to the folder where Python is located is
cd <absolute path>
where “cd” command refers to change directory and absolute path refers to the complete path where Python is installed.

In this Example to go to the folder where Python is located we should type the following command “cd C:\Program Files\OpenOffiice 4\Program”:

PART – III
III. Answer The Following Questions

Question 1.
Differentiate PYTHON and C++?
Answer:
PYTHON:

1. Python is typically an “interpreted” language
2. Python is a dynamic-typed language
3. Data type is not required while declaring variable
4. It can act both as scripting and general purpose language

C++:

1. C++ is typically a “compiled” language
2. C++ is compiled statically typed language
3. Data type is required while declaring variable
4. It is a general purpose language

Question 2.
What are the applications of scripting language?
Answer:
Applications of Scripting Languages

1. To automate certain tasks in a program
2. Extracting information from a data set
3. Less code intensive as compared to traditional programming language
4. can bring new functions to applications and glue complex systems together

Question 3.
What is MinGW? What is its use?
Answer:
MinGW refers to a set of runtime header files, used in compiling and linking the code of C, C++ and FORTRAN to be run on Windows Operating System.

MinGw-W64 (versionofMinGW) is the best compiler for C++ on Windows. To compile and execute the C++ program, you need ‘g++’ for Windows. MinGW allows to compile and execute C++ program dynamically through Python program using g++.

Python program that contains the C++ coding can be executed only through minGW-w64 project run terminal. The run terminal open the command-line window through which Python program should be executed.

Question 4.
Identify the module, operator, definition name for the following
welcome.display( )
Answer:
welcome – module name
– dot operator
display( ) – function name

Question 5.
What is sys.argv? What does it contain?
Answer:
sys.argv is the list of command-line arguments passed to the Python program, argv contains all the items that come along via the command-line input, it’s basically an array holding the command-line arguments of the program.
To use sys.argv, you will first have to import sys. The first argument, sys.argv[0], is always the name of the program as it was invoked, and sys.argv[l] is the first argument you pass to the program (here it is the C++ file).
For example:
main(sys.args[1]) Accepts the program file (Python program) and the input file (C++ file) as a list(array). argv[0] contains the Python program which is need not to be passed because by default _main_ contains source code reference and argv[l] contains the name of the C++ file which is to be processed.

PART – IV
IV. Answer The Following Questions

Question 1.
Write any 5 features of Python?
Answer:

1. Python uses Automatic Garbage Collection
2. Python is a dynamically typed language.
3. Python runs through an interpreter.
4. Python code tends to be 5 to 10 times shorter than that written in C++.
5. In Python, there is no need to declare types explicitly
6. In Python, a function may accept an argument of any type, and return multiple values without any kind of declaration beforehand.

Question 2.
Explain each word of the following command?
Answer:
The syntax to execute the Python program is
Python <filename.py> -i <C++filename without cpp extension>
Where,

Question 3.
What is the purpose of sys,os,getopt module in Python. Explain?
Answer:
1. Python’s sys module
This module provides access to some variables used by the interpreter and to functions that interact strongly with the interpreter.

sys.argv:
sys.argv is the list of command-line arguments passed to the Python program, argv contains all the items that come along via the command-line input, it’s basically an array holding the command-line arguments of the program.

To use sys.argv, you will first have to import sys. The first argument, sys.argv[0], is always the name of the program as it was invoked, and sys.argv) 1] is the first argument you pass to the program (here it is the C++ file). For example

main(sys.argv[ 1 ]) Accepts the program file (Python program) and the input file (C++ file) as a list(array). argv[0] contains the Python program which is need not to be passed because by default main contains source code reference and argv[l] contains the name of the C++ file which is to be processed.

2. Python’s OS Module
The OS module in Python provides a way of using operating system dependent functionality. The functions that the OS module allows you to interface with the Windows operating system where Python is running on.
os.system( ): Execute the C++ compiling command (a string contains Unix, C command which also supports C++ command) in the shell (Here it is Command Window). For Example to compile C++ program g++ compiler should be invoked. To do so the following command is used.
os.system (g++’ + <varciiable_namel> ‘-<mode>’+ <variable_name2>
where,

os.system(‘g++ ‘+ cpp_file + ‘-o ‘+ exe_file) g++ compiler compiles the file_cpp_file and -o (output) send to exe file
Note
‘+’ in os.system( ) indicates that all strings are concatenated as a single string and send that as a List.

3. Python getopt module
The getopt module of Python helps you to parse (split) command-line options and arguments. This module provides two functions to enable command-line argument parsing, getopt.getopt method
This method parses command-line options and parameter list. Following is the syntax for this method
<opts>,<args>=getopt.getopt(argv, options, {long_options])
Here is the detail of the parameters –
argv – This is the argument list of values to be parsed (splited). In our program the complete command will be passed as a list.
options – This is string of option letters that the Python program recognize as, for input or for output, with options (like ‘i’ or ‘o’) that followed by a colon (:). Here colon is used to denote the mode.
long options – This parameter is passed with a list of strings. Argument of Long options should be followed by an equal sign (‘=’). In our program the C++ fde name will be passed as string and ‘i’ also will be passed along with to indicate it as the input file.
getopt( ) method returns value consisting of two elements. Each of these values are stored separately in two different list (arrays) opts and args. Opts contains list of splitted strings like mode, path and args contains any string if at all not splitted because of wrong path or mode, args will be an empty array if there is no error in splitting strings by getopt( ).
For example The Python code which is going to execute the C++ file p4 in command line will have the getopt( ) method like the following one. opts, args = getopt.getopt (argv, “i:”,[‘ifile=]) where opto contains [(‘-i’, ‘c:\ \pyprg\\p4’)]
-i: – option nothing but mode should be followed by:
‘c: \ \pyprg\\p4’ value nothing but the absolute path of C++ file.
In our examples since the entire command line commands are parsed and no leftover argument, the second argument args will be empty [ ]. If args is displayed using print( ) command it displays the output as [ ].
>>>print(args)
[ ]

Question 4.
Write the syntax for getopt( ) and explain its arguments and return values?
Answer:
Python getopt module
The getopt module of Python helps you to parse (split) command-line options and arguments. This module provides two functions to enable command-line argument parsing, getopt.getopt method
This method parses command-line options and parameter list. Following is the syntax for this method –
<opts>,<args>=getopt.getopt(argx>, options, {.long_options])
Here is the detail of the parameters –

getopt( ) method returns value consisting of two elements. Each of these values are storec separately in two different list (arrays) opts and args .Opts contains list of splitted strings like mode, path and args contains any string if at all not splitted because of wrong path or mode, args will be an empty array if there is no error in splitting strings by getopt( ).
For example The Python code which is going to execute the C++ file p4 in command line will have the getopt( ) method like the following one. opts, args = getopt.getopt (argv, “i:”,[‘ifile=’])

In our examples since the entire command line commands are parsed and no leftover argument, the second argument args will be empty [ ]. If args is displayed using print( ) command it display the output as [ ].
>>>print(args)
[ ]
Some more command for wrapping C++ code
if_name_==’_main_’
main(sys.argv[1:])
_name_(A Special variable) in Python
Since there is no main() function in Python, when the command to run a Python program is given to the interpreter, the code that is at level 0 indentation is to be executed. However, before doing that, interpreter will define a few special variables. _name_ is one such special variable which by default stores the name of the file. If the source file is executed as the main program, the interpreter sets the _name_ variable to have a value as “_main_” name is a built-in variable which evaluates to the name of the current module. Thus it can be used to check whether the current script is being run on its own.
For example consider the following
if_name_ ==’_ main_’
main(sys.argv[1:])
if the command line Python program itself is going to execute first, then _main_ contains the name of that Python program and the Python special variable _name_ also contain the Python program name. If the condition is true it calls the main which is passed with C++ file as argument.

Question 5.
Write a Python program to execute the following C++ coding?
Answer:
#include <iostream>
using namespace std;
int main( )
{ cout«“WELCOME”;
return (0);
}
The above C++ program is saved in a file welcome.cpp
Python program
Type in notepad and save as welcome.cpp
#include<iostream>
using namespace std;
int main( )
{
cout<<“WELCOME”;
retum(0);
}
Open Notepad and type python program and save as welcome.py
import sys,os,getopt
def main(argv):
cppfile =”
exefile = ”
opts, args = getopt.getopt(argv, “i:”, [ifile = ‘])
for o,a in opts:
if o in (“_i”,” ifile “):
cpp_file = a+ ‘.cpp’
exe_file = a+ ‘.exe’
run(cpp_file, exe_file)
def run(cpp_file, exe_file):
print(“compiling” +cpp_file)
os.system(‘g++’ +cpp_file + ‘_o’ + exe_file)
print(“Running” + exe_file)
print(“………………………….”)
print
os.system(exe_file)
print
if — name — == ‘–main –‘;
main(sys.argv[1:])
Output:
——————-
WELCOME
——————-

Practice Programs

Question 1.
Write a C++ program to create a class called Student with the following details?
Answer:
Protected member
Rno integer
Public members
void Readno(int); to accept roll number and assign to Rno
void Writeno( ); To display Rno.
The class Test is derived Publically from the Student class contains the following details
Protected member
Mark1 float
Mark2 float
Public members
void Readmark(float, float); To accept mark1 and mark2
void Writemark( ); To display the marks
Create a class called Sports with the following detail
Protected members
score integer
Public members
void Readscore(int); To accept the score
void Writescore( ); To display the score
The class Result is derived Publically from Test and Sports class contains the following – details
Private member
Total float
Public member
void display( ) assign the sum of mark1, mark2, score in total
invokeWriteno( ), Writemark( ) and Writescore( ). Display the total also.
Save the C++ program in a file called hybrid. Write a python program to execute the hybrid.cpp
Answer:
In Notepad, type the C++ program
#include<iostream>
using namespace std;
class student
{
protected:
int mo;
public:
void readno(int rollno)
{
mo = rollno;
}
void writeno( )
{
cout<< “\n Roll no:” <<rno;
}};
class test: public student
{
protected:
float mark1,mark2;
public:
void readmark(float m1, float m2)
{
mark1 = m1;
mark2 = m2;
}
void writemark( )
{
cout<< “\n mark1 ” << mark1;
cout<< “\n mark2 ” <<mark2;
}};
class sports
{
protected:
int score;
public:
void readscore(int s)
{
score = s;
}
void writescore( )
{
cout<< “SCORE : ” <<score;
}};
class result: public test, public sports
{
private: float total; public:
void display( )
{
total = mark1 + mark2;
cout<< “TOTAL MARKS: ” <<total;
}};
int main( )
{
result r;
r.readno(5);
r.readmark(100,100);
r.readscore(200);
r.writeno( );
r.writemark( );
r.display( );
r.writescore( );
}
save this file as hybrid.cpp
Now type the python program in New Notepad file.
# python hybrid.py -i hybrid.cpp
import sys,os,getopt def main(argv): cpp Jile =” exe_file =”
opts, args = getopt.getopt(argv, “i:”,[‘ifile-])
for o, a in opts:
if o in(“-i”, “–file”):
cpp_file =”a+ ‘.cpp’
exe_file = a+ ‘.exe’
run(cpp_file, exefile)
def run(cpp_file, exe_file):
print(“Compiling” + cpp_file)
os.system(‘g++ ‘+ cpp_file + ‘-o ‘+ exe_file)
print(“Running” + exefile)
print(“———————“)
print
os.system(excfile)
print
if name ==’ main
main(sys.argv[1:])
Output:
Rollno : 5
Mark1 : 100
Mark2 : 100
TOTAL MARKS : 200
SCORE : 200

Question 2.
Write a C++ program to print boundary elements of a matrix and name the file as Border.cpp. Write a python program to execute the Border.cpp?
Answer:
Select File → New in Notepad and type the C++ program.
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
const int MAX =100;
void printBoundary(int a[][max], int m, int n)
{
for (int i=0; i < m; i++)
{
for(int j=0; j < n; j++)
{
if(i= =0 || j= =0 || i= =n-l ||j==n-l)
cout <<a[i][j]
<< ” else
cout <<“”
cout<<”
}
cout <<” \n”;
}}
int main( )
{
int a[4][MAX] = { {1,2,3,4}, {5,6,7,8}, {1,2,3,4}, {5,6,7,8}};
print Boundary(a,4,4);
return 0;
}
save it as Border.cpp
open a New notepad file and type the python program to execute border.cpp
#python border.py -i border.cpp
import sys,os,getopt
def main(argv):
cpp_file =”
exe_file =”
opts, args = getopt.getopt(argv, “i:”,[‘ifile=’])
for o, a in opts;
if o in(“-i”, “-ifile”):
cpp_file =a+ ‘.cpp’
exe_file = a+ ‘.exe’
run(cpp_file, exe_file)
def run(cpp_file, exe_file):
print(“Compiling” + cpp_file)
os.system(‘g++ ‘+ cpp_file + ‘-o ‘+ exe_file)
print(“Running” + exe_file)
print(“———————“)
print
os.system(exe_file)
print
if_name_==’_main_’:
main(sys.argv[1:])
Output:

Samacheer kalvi 12th Computer Science Importing C++ Programs in Python Additional Questions and Answers

I. Choose The Correct Answer

Question 1.
Which one of the following language act as both scripting and general purpose language?
(a) python
(b) c
(c) C++
(d) html
Answer:
(a) python

Question 2.
Find the correct statement.
(a) C++ is a dynamic typed language
(b) python is a dynamic typed language
Answer:
(b) python is a dynamic typed language

Question 3.
Identify the script language from the following.
(a) Html
(b) Java
(c) Ruby
(d) C++
Answer:
(c) Ruby

Question 4.
Pick the odd one out Perl, Ruby, ASP, Tel, Java
Answer:
Java

Question 5.
Read the statement given below and choose the correct option.
(I) All scripting languages are progamming languages.
(II) A scripting language requires an interpreter
(III) Programming language requires a compiler
(a) (I) – False, (II), (III) – True
(b) (I), (II) – False
(c) All are true
(d) All are false
Answer:
(c) All are true

Question 6.
How many values can be returned by a function in python?
(a) 1
(b) 2
(c) 3
(d) many
Answer:
(d) many

Question 7.
The process by which python periodically frees and reclaims block of memory that no longer are in use is called ……………………………..
Answer:
Garbage Collection

Question 8.
………………………. is a python like language for writing c extensions.
Answer:
cython

Question 9.
Which of the following language codes are linked by minGW on windows OS?
(a) C
(b) C++
(c) FORTRAN
(d) all of these
Answer:
(d) all of these

Question 10.
……………………. is the best compiler for C++ on windows.
(a) MinGW-w64
(b) MinGW-w63
(c) MinGW-w62
(d) MinGW-w60
Answer:
(a) MinGW-w64

Question 11.
MinGW allows to compile and execute C++ program dynamically through python program using ……………………….
(a) C++
(b) g++
(c) a++
(d) x++
Answer:
(b) g++

Question 12.
g++ is a program that calls GNU C compiler called …………………………….
(a) GCC
(b) CCC
(c) GGC
(d) GGG
Answer:
(a) GCC

Question 13.
The command to change directory is …………………………….
(a) cc
(b) cd
(c) dc
(d) changed
Answer:
(b) cd

Question 14.
………………………. keyword is used to execute the python program from command line
Answer:
Python

Question 15.
The input mode in python command is given by ………………………….
(a) -i
(b) o
(c) -p
Answer:
(a) -i

Question 16.
Which command is used to clear the screen?
(a) clear
(b) clean
(c) cls
(d) clrscr
Answer:
(c) cls

Question 17.
……………………… refer to a file containing python statements and definitions.
Answer:
Modules

Question 18.
pythons’ …………………………… provides access to same variables and functions that interact with the interpreter.
Answer:
sys module

Question 19.
Which is the list of command-line arguments passed to the python program.
(a) sys.ar
(b) sys.argv
(c) sys.sys
(d) sys.opt
Answer:
(b) sys.argv

Question 20.
Match the following related to OS module.
1. g++ – (i) Name of C++file
2. variable namel – (ii) General compiler
3. mode – (iii) input output mode
4. variable_name2 – (iv) Name of exe file
(a) 1-i, 2-ii, 3-iii, 4-iv
(b) 1-ii, 2-i, 3-iii, 4-iv
(c) 1-iv, 2-iii, 3-ii, 4-i
(d) 1-iv, 2-i, 3-iii, 4-ii
Answer:
(b) 1-ii, 2-i, 3-iii, 4-iv

Question 21.
Which in os.system( ) indicates that all strings are concatenated?
(a) +
(b) –
(c) #
(d) *
Answer:
(a) +

Question 22.
Which module helps to parse(split) command line options and arguments?
(a) getopt
(b) argopt
(c) get
(d) putopt
Answer:
(a) getopt

Question 23.
How many functions are there in getopt module to enable command line argument parsing?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 24.
getopt mode is given by ………………………….
(a) ;
(b) =
(c) #
(d) :
Answer:
(d) :

Question 25.
Argument of long options in the getopt method is followed by ………………………….
(a) ‘
(b) =
(c) #
(d) :
Answer:
(b) =

Question 26.
When the command to run a python program is given to interpreter, code at …………………………. indentation is executed.
(a) level 0
(b) level 1
(c) level 2
(d) level 3
Answer:
(a) level 0

Question 27.
Which one is a special variable which by default stores the name of the file.
(a) main
(b) none
(c) _name_
(d) get
Answer:
(c) _name_

Question 28.
The line number starts from …………………………… script.
Answer:
python

Question 29.
The …………………………… definition invoke the g++ compiler and creates the exe file?
Answer:

Question 30.
Which command of os module executes the exe file?
(a) run
(b) system( )
(c) main
(d) name
Answer:
(a) run

Question 31.
Which opeator is used to access the functions of a imported value?
(a) +
(b) *
(c) .
(d) /
Answer:
(c) .

Question 32.
………………………….. refers to a set of runtime header files used in compiling and linking the code of c, C++, Fortran to run on window os
(a) MaxGW
(b) CountGW
(c) MinGW
(d) AvgGW
Answer:
(c) MinGW

PART – II
II. Answer The Following Questions

Question 1.
Write note on scripting language?
Answer:
A scripting language is a programming language designed for integrating and communicating with other programming languages. Some of the most widely used scripting languages are JavaScript, VBScript, PHP, Perl, Python, Ruby, ASP and Tel.

Question 2.
Write note on Modular programming?
Answer:
Modular programming is a software design technique to split your code into separate parts. These parts are called modules.The focus for this separation should have modules with no or just few dependencies upon other modules.

PART – III
III. Answer The Following Questions

Question 1.
Name some commonly used Interfaces for importing C++ files on python?
Answer:
Importing C++ Files in Pythona:
Importing C++ program in a Python program is called wrapping up of C++ in Python. Wrapping or creating Python interfaces for C++ programs are done in many ways. The commonly used interfaces are

1. Python-C-API (API-Application Programming Interface for interfacing with C programs)
2. Ctypes (for interfacing with c programs)
3. SWIG (Simplified Wrapper Interface Generator.Both C and C++)
4. Cython (Cython is both a Python-like language for writing C-extensions)
5. Boost. Python (a framework for interfacing Python and C++)
6. MinGW (Minimalist GNU for Windows)

Question 2.
How will you import modules in python?
Answer:
We can import the definitions inside a module to another module. We use the import keyword to do this. To import the module factorial we type the following in the Python prompt.
>>> import factorial
Using the module name we can access the functions defined inside the module. The dot(.)
operator is used to access the functions. The syntax for accessing the functions from the module is
<module name>.<function name>
For example:
>>> factorial.fact(5)
120

Question 3.
Give the syntax for getopt module?
Answer:
The syntax for this method
<opts> ,<args>=getopt.getopt(argv, options, [long_options])
Here is the detail of the parameters –

Question 4.
Write the steps for executing the C++ program to check whether a given number is palindrome or not?
Answer:

PART – IV
IV. Answer The Following Questions

Question 1.
Explain Pythons OS Module?
Answer:
Python’s OS Module:
The OS module in Python provides a way of using operating system dependent functionality. The functions that the OS module allows you to interface with the Windows operating system where Python is running on.
os.system( ): Execute the C++ compiling command (a string contains Unix, C command which also supports C++ command) in the shell (Here it is Command Window). For Example to – compile C++ program g++ compiler should be invoked. To do so the following command is used.
os.system (g++’ + <varaiable_namel> ‘-<mode>’+ <variable_name2>
where,

For example the command to compile and execute C++ program is given below
os.system(’g++ ’+ cpp_file + ‘-o ‘+ exe_file) g++ compiler compiles the file cpp_file and -0(output) send to exe_file

Question 2.
Write a C++ program to check whether the given number is palindrome or not. Write a program to execute it?
Answer:
Example:- Write a C++ program to enter any number and check whether the number is palindrome or not using while loop.
/*. To check whether the number is palindrome or not using while loop.*/
//Now select File-> New in Notepad and type the C++ program
#include <iostream>
using namespace std; intmain( )
{
int n, num, digit, rev = 0;
cout << “Enter a positive number:”;
cin>>num;
n = num;
while(num)
{ digit=num% 10;
rev= (rev* 10) +digit;
num = num/10;
cout << “The reverse of the number is:”<<rev <<end1;
if (n ==rev)
cout<< “The number is a palindrome”;
else
cout<< “The number is a palindrome”;
return 0;
}
//save this file as pali_cpp.cpp
#Now select File → New in Notepad and type the Python program
#Save the File as pali.py.Program that complies and executes a .cpp file
import sys, os, getopt
def main(argv);
cpp_file=”
exe_file=”
opts.args = getopt.getopt(argv, “i:”,[‘ifile-])
for o, a in opts:
if o in (“-i”, “—file”):
cpp_file =a+ ‘.cpp’
exe_file =a+ ‘.exe’
run(cpp_file, exe_file)
def run(cpp_file, exe_file):
print(“Compiling” + eppfile)
os.system(‘g++’ + cpp_file +’ -o’ + exe file)
print(“Running” + exe_file)
print(“——————“)
print
os.system(exefile)
print
if_name_==’_main_’: #program starts executing from here
main(sys.argv[1:])
Output of the above program
C:\Program Files\OpenOffice 4\program>Python c:\pyprg\pali.py -i c:\pyprg\pali_cpp
Compiling c:\pyprg\pali_cpp.cpp
Running c:\pyprg\pali_cpp.exe
———————————–
Enter a positive number: 56765
The reverse of the number is: 56765
The number is a palindrome
C:\Program Files\OpenOffice 4\program>Python c:\pyprg\pali.py -i c:\pyprg\pali_cpp Compiling c:\pyprg\pali_cpp.cpp Running c:\pyprg\pali_cpp.exe
Enter a positive number: 56756
The reverse of the number is: 65765
The number is not a palindrome
————————

Question 3.
Write a C++ program to check whether the given number is palindrome or not. Write a program to execute it?
Write a C++ program to enter any number and check whether the number is palindrome or not using while loop?
Answer:
//Now select File-> New in Notepad and type the C++ program
#include <iostream>
using namespace std;
intmain( )
{
int a[3][3], i, j;
for(i=o; j<3;i++)
{
for(j=0;j<3;j++)
{count<<“enter the value for array [”<<i+1<<“]<<”<<“[“<<j+1<<“]”;
cin>>a[i][j];
}
}
system(“cls”); .
cout<<“\n \nOriginal Array\n”;
for(i=0; i<3; i++)
{
for(j=0; j<3; j++)
cout<<a[i] [j]
cout<<end1;
}
cout<<end1;
}
return 0;
}
// Save this file as trans_cpp.cpp
// Now select File → New in Notepad and type the Python program
#Save the File as transpose.py.Program that compiles and executes a .cpp file
#Python tanspose.py -i trans_cpp
import sys, os, getopt
def main(argv):
cpp_file =”
exefile =”
opts, args = getopt.getopt(argv, “i:”,[‘ifile=’])
for o, a in opts:
if o in (“-i”,–ifile”):
cpp_file =a+ ‘.cpp’
exe_file = a+ ‘.exe’
run(cpp_file, exe_file)
def run(cpp_file, exe file):
print(“Compiling” + cppfile)
os.system(‘g++’ + cpp_file + ‘-o ‘+ exe_file)
print(“Running” + exefile)
print(“———————-“)
print
os.system(exe_file)
print
if_name_==’_main_’:
main(sys.argv[1:])
Output of the above program
Original Array
1 2 3
4 5 6
7 8 9
The Transpose of Matrix
1 4 7
2 5 8
3 6 9

Question 4.
Write a C++ program to find cube of a number. Write a python program to execute it?
Answer:
/* Write a C++program using a user definedfunction to function cube of a number. */
//Now select File → New in Notepad and type the C++program
#include <iostream>
using namespace std;
//Function declaration
int cube(int num); int main( )
{
int num; int c;
cout<<“Enter any number: “<<end1;
cin>>num;
c = cube(num);
cout<<“Cube of’ <<num<<” is “<<c;
return 0;
}
//Function to find cube of any number
int cube(int num)
{
return (num * num * num);
}
//Save this file as cube_file.cpp
#Now select File → New in Notepad and type the Python program
#Save the File as fun.py
#Program that compiles and executes a .cpp file
#Python fun.py -i c:\pyprg\cuheJile.cpp
import sys, os, getopt
def main(argv):
cpp_Jile =”
exe_file =”
opts, args = getopt.getopt(argv, “i:”,[‘ifile-])
for o, a in opts:
if o in(“-i”, —ifile”):
cpp_file =a+ ‘.cpp’
exe_file = a+ ‘.exe’
run(cpp_file, exe_file)
def run(cpp_file, exe file):
print(“Compiling” + cppfile)
os.system(‘g++’+ cpp file + ‘-o ‘+ exe file)
print(“Running” + exe_file)
print(“——————–“)
print
os. sy stem(exe_file)
print
if_name_==’_main_’:
main(sys.argv[1:])
Output of the above program
Compiling c:\pyprg\cube_file.cpp
Running c:\pyprg\cube_file.exe
———————-
Enter any number:
5
Cube of 5 is 125

Question 5.
Write a C++ program to implement multilevel inheritance. Write a python program to execute it?
Answer:
// C++program to implement Multilevel Inheritance
//Now select File → New in Notepad and type the C++program
#include <iostream>
using namespace std;
// base class
class Vehicle
{
public:
Vehicle( )
{
cout<< “This is a Vehicle” <<end1;
}
};
class threeWheeler: public Vehicle
{public:
threeWheeler( )
{
cout<<“Objects with 3 wheels are vehicles”<<end1;
}
};
// sub class derivedfrom two base classes
class Auto: public threeWheeler
{
public:
Auto( )
{
cout<<“Auto has 3 Wheels”<<end1;
}
// main function
int main( )
{
//creating object of sub class will invoke the constructor of base classes
Auto obj;
return 0;
}
// Save this file as inheri_cpp.cpp
//Now select File —* New in Notepad and type the Python program
#Save the File as classpy.py
#Python classpy.py -i inheri_cpp command to execute C++ program
import sys, os, getopt
def main (argv):
cpp_file =”
exe file= 11
opts, args = getopt.getopt (argv, 11i: 11 ,[‘ifile=’])
for o, a in opts:
if o in (“-i”, —ifile”):
cpp_file =a+ ‘.cpp’
exe_file = a+ ‘.exe’
run (cpp file, exe_file)
def run(cpp_file, exe_file):
print (”Compiling 11 + cpp file)
os.system (‘g++’ + cpp_file +’ -o ‘+ exe_file)
print (“Running”+ exe file)
print(“———————“)
print
os.system (exefile) print
if_name_==’_main_’:
main (sys.argv[1:])
Output of the above program
Compiling c:\pyprg\class_file.cpp
Running c:\pyprg\class_file.exe
This is a Vehicle
Objects with 3 wheels are vehicles
Auto has 3 Wheels

Students can Download Tamil Nadu 12th Maths Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Maths Model Question Paper 2 English Medium

Instructions:

1.  The question paper comprises of four parts.
2.  You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. questions of Part I, II. III and IV are to be attempted separately
4. Question numbers 1 to 20 in Pan I are objective type questions of one -mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 90

Part-I

I. Choose the correct answer. Answer all the questions.  [20 × 1 = 20]

Question 1.
If A = \(\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]\) then 9I₂ – A =
(a) A^-1
(b) \(\frac{A^{-1}}{2}\)
(c) 3A^-1
(d) 2A^-1
Answer:
(d) 2A^-1

Question 2.
If (1+ i) (1+ 2i) (1+ 3i)…(1+ ni) = x + iy, then 2-5T0…(1 + n² ) is ..
(a) 1
(b) i
(c) x² + y²
(d) 1 + n²
Answer:
(c) x² + y²

Question 3.
If p + iq = (2 – 3i) (4 + 2i) then q is
(a) 14
(b) -14
(c) -8
(d) 8
Answer:
(c) -8

Question 4.
A zero of x³ + 64 is
(a) 0
(b) 4
(c) 4i
(d) -4
Answer:
(d) -4

Question 5.
sin^-1(2COS²A – 1) + cos^-1(1 – 2 sin²x) = .

(a) π/2
(b) π /3
(c) π/4
(d) π/6
Answer:
(a) π/2

Question 6.
If cot^-1 x = \(\frac{2 \pi}{5}\) for some x ∈ R, the value of tan^-1 x is
(a) \(-\frac{\pi}{10}\)
(b) \(\frac{\pi}{5}\)
(c) \(\frac{\pi}{10}\)
(d) \(-\frac{\pi}{5}\)
Answer:
(c) \(\frac{\pi}{10}\)

Question 7.
An ellipse has OB as semi minor axes, F and F’ its foci and the angle FBF’ is a right angle. Then the eccentricity of the ellipse is

Answer:
(a) \(\frac{1}{\sqrt{2}}\)

Question 8.
The focus of the parabola x² = 20 y is
(a) (0, 0)
(b) (5, 0)
(c) (0, 5)
(d) (-5, 0)
Answer:
(c) (0, 5)

Question 9.
If the planes \(\vec{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})=3\) and \((4 \hat{i}+\hat{j}-\mu \hat{k})=5\) are parallel, then the value of λ and μ are

Answer:
(c) \(-\frac{1}{2}\) , -2

Question 10.
The unit normal vectors to the plane 2x – y + 2z = 5 are

Question 11.
The slope of the line normal to the curve f(x) – 2cos 4x at x = \(\frac{\pi}{2}\) is ……………….
(a) -4√3
(b) -4
(c) √3/12
(d) 4√3
Answer:
(c) √3/12

Question 12.
The curve y² = x² ( 1 – x) has
(a) only one loop between x = -1 and x = 0
(b) only one loop between x = 0 and x = 1
(c) two loops between x = -1 and x = 1
(d) no loop
Answer:
(b) only one loop between x = 0 and x = 1

Question 13.
If w (x, y) = x^y, x > 0, then \(\frac{\partial w}{\partial x}\) is equal to
(a) x^y logx
(b) y logx
(c) yx^y-1
(d) xlogy
Answer:
(c) yx^y-1

Question 14.
If (x, y, z) = xy + yz + zx, then f_x – f_z is equal to
(a) z – x
(b) y – z
(c) x – z
(d) y – x
Answer:
(a) z – x

Question 15.
If \(\frac{\Gamma(n+2)}{\Gamma(n)}\) = 90 then n is
(a) 10
(b) 5
(c) 8
(d) 9
Answer:
(d) 9

Question 16.
If n is even then \(\int_{0}^{\pi / 2} \sin ^{n} x d x\) is ………………

Answer:
(b)

Question 17.
The slope at any point of a curve y =/(x) is given by \(\frac{d y}{d x}=3 x^{2}\) and it passes through (-1,1).
Then the equation of the curve is
(a) y = x³ + 2
(b) y = 3x² + 4
(c) y = 3x³ + 4
(d) y = x³ + 5
Answer:
(a) y = x³ + 2

Question 18.
The solution of the differential equation \(\frac{d y}{d x}+\frac{1}{\sqrt{1-x^{2}}}=0\) is
(a) y + sin^-1x = c
(b) x + sin^-1y = 0
(c) y² + 2sin^-1x = c
(d) x² + 2sin^-1y = 0

Question 19.
Suppose that x takes on one of the values 0, 1, and 2. If for some constant k,
P(X = i) = k P(x = i – 1) for i = 1, 2 P(x = 0) = 1/7. Then the value of k is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 20.
Determine the truth value of each of the following statements:
(a) 4 + 2 = 5 and 6 + 3 = 9
(b) 3 + 2 = 5 and 6 + 1 = 7
(c) 4 + 5 = 9 and 1 + 2 = 4
(d) 3 + 2 = 5 and4 + 7 = 11

Answer:
(1) (a) F
(b) T
(c) F
(d) T

Part – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
If A = \(\left[\begin{array}{ll}
4 & 3 \\
2 & 5
\end{array}\right]\) , find x and y such that A² + xA + yI₂ = O₂. Hence , find A^-1
Answer:

So, we get 22 + 4x + y = 0, 31+ 5x + y – 0, 27 + 3x = 0 and 18 + 2x = 0.
Hence x = -9 and y = 14. Then, we get A² – 9A + 14I₂ = O₂
Post-multiplying this equation by A^-1, we get A – 9I + 14A^-1 – I₂ = O₂. Hence, we get

Question 22.
Find the principal argument arg z, when z = \(\frac{-2}{1+i \sqrt{3}}\)
Answer:

This implies that one of the values of arg z is \(\frac{2 \pi}{3}\)
Since \(\frac{2 \pi}{3}\) lies between -π and π , the principal argument arg z is \(\frac{2 \pi}{3}\)

Question 23.
Verify whether the line \(\frac{x-3}{-4}=\frac{y-4}{-7}=\frac{z+3}{12}\) lies in the plane 5x – y + z = 8 .
Answer:
Here, (x₁, y₁, z₁) = (3,4, -3) and direction ratios of the given straight line are (a, b, c) – (-4, -7,12).
Direction ratios of the normal to the given plane are (A, B, C) = (5, -1,1).
We observe that, the given point (x₁, y₁, z₁) = (3,4, -3) satisfies the given plane 5x – y + z = 8
Next, aA+ bB + cC = (-4)(5) + (-7)(-l) + (12)( 1) = -1 ≠ 0. So, the normal to the plane is not perpendicular to the line. Hence, the given line does not lie in the plane.

Question 24.
Evaluate \(\begin{aligned}
&\lim\\
&x \rightarrow 0^{+} \quad x^{2} \log e^{x}
\end{aligned}\)
Answer:

Question 25.
Find a linear approximation for the function at the indicated points.
g(x) = \(\sqrt{x^{2}+9}\), x₀ = -4

The required linear approximation L(x) = g (x₀) + g’ (x₀) (x – x₀)

Question 26.
Evaluate: \(\int_{0}^{2 \pi} x^{2} \sin n x d x\) , Where n is a positive integer.
Answer:
Taking u = x² and v = sin nx, and applying the Bernoulli’s formula, we get

Question 27.
Form the differential equation from the equation y² = 4a (x – a)
Answer:
y²= 4 a(x-a) …(1)
Differentiating, 2yy’ = 4a …(2)
Eliminating a between (1) and (2) we get

(yy ‘)² – 2xyy’ + y² = 0

Question 28.
Compute P(X = k) for the binomial distribution, B(n,p) where n = 6, p = 1/3, k = 3
Answer:
Given n = 6, p = 1/3,k = 3
∴ q = 1 – p = \(1-\frac{1}{3}=\frac{2}{3}\)
P(X = x) = nC_xp^xq^n-x, x = 0, 1,2,…………n
P (X = k) = P (X = 3)

Question 29.
Let A = {a + √5b : a, b∈Z} . Check whether the usual multiplication is a binary operation on A.
Answer:
Let A = a + √5 b and B = c + √5 d, where a, b,c,d∈ R.
Now A * B = (a + √5 b) (c + √5 d)
= ac + √5 ad + √5 bc + √5 b √5 d
= (ac + 5 bd) + 45 (ad + bc) ∈ A .
Where a, b,c,d ∈ Z.
So * is a binary operation.

Question 30.
Find the value of \(\sin ^{-1}\left(\sin \frac{5 \pi}{9} \cos \frac{\pi}{9}+\cos \frac{5 \pi}{9} s \quad \frac{\pi}{9}\right)\)
Answer:

Part – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 x 3 = 21]

Question 31.
Solve the following system of linear equations, using matrix inversion method:
5x + 2y – 3, 3x + 2y – 5.
Answer:
The matrix form of the system is AX = B, where A :\(\left[\begin{array}{ll}
5 & 2 \\
3 & 2
\end{array}\right]\) , X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\), B = \(\left[\begin{array}{l}
3 \\
5
\end{array}\right]\)

We find |A| = \(\left|\begin{array}{ll}
5 & 2 \\
3 & 2
\end{array}\right|\) = 10 – 6 = 4 ≠ 0
So A^-1 exists A^-1 = \(\frac{1}{4}\left[\begin{array}{rr}
2 & -2 \\
-3 & 5
\end{array}\right]\)
Then, applying the formula X = A 1B, we get

So the solution is (x = -1 ,y = 4).

Question 32.
If z₁ = 2 – i and z₂ = -4 + 3i, find the inverse of z₁ z₂ and \(\frac{z_{1}}{z_{2}}\)
Answer:
z₁ = 2 – i, z₂ = -4 + 3i

Question 33.
If k is real, discuss the nature of the roots of the polynomial equation 2x² + kx + k = 0, in terms of k.
Answer:
The given quadratic equation is 2x² + kx + k = 0
a = 2,b = k,c = k
∆ = b² – 4ac = k² – 4(2) k = k² – 8k
(i) If the roots are equal
k² – 8k = 0 ⇒ k(k -8) = 0
k = 0, k = 8
(it) If the roots are real
k² – 8k > 0
k (k – 8) > 0
k ∈ (-∞, 0) ∪ (8, ∞)
(iii) ‘If this roots are imaginary
k² – 8k < 0 ⇒ f ∈ (0, 8)

Question 34.
Find the number of solution of the equation tan^-1 (x – 1) + tan^-1 x + tan^-1 (x + 1) = tan^-1(3x).
Answer:
tan^-1(x – 1) + tan^-1x + tan^-1(x + 1) = tan^-1(3x)
tan^-1(x – 1) + tan^-1(x + 1) = tan^-1 3x – tan^-1 x

So, the equation has 2 solutions.

Question 35.
Find the non-parametric form of vector equation, and Cartesian equations of the plane \(\vec{r}=(6 \hat{i}-\hat{j}+\hat{k})+s(-\hat{i}+2 \hat{j}+\hat{k})+t(-5 \hat{i}-4 \hat{j}-5 \hat{k})\)
Answer:

Non-parametric form of vector equation

Question 36.
Find two positive numbers whose sum is 12 and their product is maximum.
Answer:
Let the two numbers be x, 12 – x.
Their product p = x (12 – x) – 12x – x²
To find the maximum product.
p’ (x) = 12 – 2x
p”(x) = -2
p’ (x) = 0 ⇒ 12 – 2x = 0 2x = 12 => x = 6
at x = 6, p”(x) = -2 = -ve
⇒ p is maximum at x = 6
when x = 6, 12-x = 12 – 6 = 6
So the two numbers are 6, 6

Question 37.
Find the differential equation that will represent family of all circles having centres on the x-axis and the radius is unity.
Answer:
Equation of a circle with centre on x-axis and radius 1 unit is
(x-a)² + y² = 1 …(1)
Differentiating with respect to x,
2 (x-a) + 2yy’ = 0
⇒ 2 (x – a) = – 2yy’
(or) x-a = -yy’ …(2)
Substituting (2) in (1), we get,
(-yy’)² + y² = 1
(i.e. y² (y’)² + y² = 1 ⇒ y² [1 + (y’)²] = 1

Question 38.
The probability that a certain kind of component will survive a electrical test is \(\frac{3}{4}\). Find
the probability that exactly 3 of the 5 components tested survive.
Answer:
Given n = 5
Probability that a component survive in a test =p = \(\frac{3}{4}\)
∴ q = 1 – p = \(1-\frac{3}{4}=\frac{1}{4}\)
Let ‘X’ be the random variable denotes the number of components survived in a test.
Probability of ‘x’ successes in V trials is
P (X = x) = nC_xp^xq^n-x, x = 0, 1,2,………n
Probability that exactly 3 components survive

Question 39.
Fill in the following table so that the binary operation * on A = {a,b,c} is commutative.

Answer:
Given that the binary operation * is Commutative.
To find a * b :
a * b = b * a (∵ * is a Commutative)
Here b * a = c. So a * b = c
To find a * c :
a * c = c * a (∵ * is a Commutative)
c * a = a. (Given)
So a * c = a
To find c * b :
c * b = b * c
Here b * c = a. So c * b = a

Question 40.
Find the area of the region enclosed by y² = x and y = x – 2.
Answer:
The points of intersection of the parabola y² = x and the line y = x – 2 are (1, – 1) and (4, 2)
To compute the region [shown in the figure] by integrating with respect to x, we would have to split the region into two parts, because the equation of the lower boundary changes at x = 1. However if we integrate with respect to y no splitting is necessary.

Part – IV

IV. Answer all the questions. [ 7 x 5 = 35]

Question 41.
(a) Use product \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\left[\begin{array}{rrr}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]\)
to solve the system of equations x – y + 2z = 1, 2y – 3z = 1, 3x – 2y + 4z = 2.
Answer:

The system of equations can be written in the form AX = C, where

Thus x = 0, y = 5 and z = 3

(b) Find, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown in the Figure.
Answer:

Question 42.
(a) Simplify (-√3 + 3i)³¹
Answer:
Let – √3 + 3i = r (cos θ + z sin θ). Then, we get

(b) Solve the following equation x⁴ – 10x³ + 26x² – 10x + 1 = 0.
Answer:
This equation is Type I even degree reciprocal equation. Hence it can be rewritten as


[(y² – 2) – 10y + 26] = 0 ⇒ (y² – 10y + 24) = 0 ⇒ (y – 6) {y – 4) = 0 ⇒ y = 6 (or) 4

Hence the roots are 3 ± 2 √2, 2 ± √3

Question 43.
(a) Solve

Answer:

(b) Suppose a person deposits 10,000 Indian rupees in a bank account at the rate of 5% per annum compounded continuously. How much money will be in his bank account 18 months later?
Answer:
Let P be the principal
Rate of interest 5 %

Given, when t = 0, P = 10000
⇒ c = 10000

Question 44.
(a) Find the equation of the circle through the points (1, 0), (-1, 0), and (0, 1).
Answer:
Let the required circle be
x² + y² + 2gx + 2fy + c = 0 ……(A)
The circle passes through (1, 0), (-1, 0) and (0, 1)
(1,0) ⇒ 1 +0 + 2g(1) + 2f(0) + c = 0 -(A)
2g + c = -1
(-1, 0) ⇒ 1 + 0 + 2g (-1) + 2/(0) + c = 0 …(1)
-2g + c = -1
(0,1)⇒ 0+ 1 +2g(0) + 2f(l) + c = 0 …(2)
2 + c = -1
Now solving (1), (2) and (3) …(3)
2g + c = -1 …(1)
-2g + c = -1 (2)
(1) + (2) ⇒ 2c = -2 ⇒ c = -1
Substituting c = -1 in (1) we get
2g – 1 = -1 2g = -1 + 1= 0 ⇒ g = 0
Substituting c = -1 in (3) we get
2f – 1 = -1 ⇒ 2f = -1 + 1 = 0 ⇒ f = 0
So we get g = 0,f= 0 and c = -1
So the required circle will be
x² + y² + 2(0)x + 2(0)y – 1 = 0
(i.e) x² + y² – 1 = 0 ⇒ x² + y² = 1

[OR]

(b) Find the area of the loop of the curve 3ay² = x(x – a)².
Answer:
Put y = 0; we get x = 0, a
It meets the x – axis at x = 0 and x = a
∴ Here a loop is formed between the points (0, 0) and (a, 0) about x-axis. Since the curve is symmetrical about x-axis, the area of the loop is twice the area of the portion above the x – axis.

Question 45.
Show that the lines \(\frac{x-3}{3}=\frac{y-3}{-1}, z-1=0\) and \(\frac{x-6}{2}=\frac{z-1}{3}, y-2=0\) interest. Also find the point of intersection.
Answer:

(x₁, y₁, z₁) = (3, 3,1) and (x₂, y₂, z₂) = (6,2,1)
(b₁ b₂, b₃) = (3, -1, 0) and (d₁ d₂, d₃) = (2, 0, 3)
Condition for intersection of two lines

Given two lines are intersecting lines.
Any point on the first line

(3λ + 3, -λ + 3,1)
Any point on the Second line

(2µ + 6, 2, 3µ + 1)
∴ 3µ + 1 = 1
3µ = 0
µ = 0

-λ + 3 = 2
-λ = -1
λ = 1
∴ The required point of intersection is (6, 2, 1)

(b) Prove that \(p \rightarrow(\neg q \vee r) \equiv \neg p \vee(\neg q \vee r)\) using truth table.
Answer:

The entries in the column corresponding to \(p \rightarrow(\neg q \vee r) \text { and } \neg p \vee(\neg q \vee r)\) are identical.
Hence they are equivalent.

Question 46.
(a) A water tank has the shape of an inverted circular cone with base radius 2 metres and height 4 metres. If water is being pumped into the tank at a rate of 2m3/min, find the rate at which the water level is rising when the water is 3m deep.
Answer:
We first sketch the cone and label it as in diagram. Let V, r and h be respectively the volume of the water, the radius of the cone and the height at time t, where t is measured in minutes.
We are given that \(\frac{d \mathrm{V}}{d t}\) = 2m³/min and we are asked to find
\(\frac{d h}{d t}\) where h is 3m
The quantities V and h are related by the equation \(\mathrm{V}=\frac{1}{3} \pi r^{2} h\). But it is very useful to express V as function of h alone.
In order to eliminate r we use similar triangles in diagram to write \(\frac{r}{h}=\frac{2}{4} \Rightarrow r=\frac{h}{2}\) and the expression for V becomes \(\mathrm{V}=\frac{1}{3} \pi\left(\frac{h}{2}\right)^{2} h=\frac{\pi}{12} h^{3}\)
Now we can differentiate each side with respect to t and we have

(b) Let U (x, y) = e^x sin y, where x = st², y = s²t s, t e R. Find \(\frac{\partial \boldsymbol{U}}{\partial \boldsymbol{s}}, \frac{\partial \boldsymbol{U}}{\partial t}\) and evaluate them at s = t = 1.
Answer:

Question 47.
(a) The probability density function of X is given by f(x) = \(\begin{array}{c}
f(x)=\left\{\begin{array}{cc}
k e^{-\frac{x}{3}} & \text { for } x>0 \\
0 & \text { for } x \leq 0
\end{array}\right.
\end{array}\)
Find (i) the value of k
(ii) the distribution function
(iii) P(X < 3)
(iv) P(5 ≤ X) (v) P(X ≤ 4) .
Answer:

[OR]

(b) Solve [y(1 – xtanx) + x²cosx]dx – xdy = 0.

Students can Download Tamil Nadu 12th Biology Model Question Paper 1 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Biology Model Question Paper 1 English Medium

General Instructions:

1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Identify plant species which is popularly called as “Terror of Bengal”.
(a) Eichornia Crassipes
(b) Vallisneria spiralis
(c) Pistia stratiotes
(d) Zostera marina
Answer:
(a) Eichornia Crassipes

Question 2.
‘Gametes are never hybrid’ is concluded by __________.
(a) Law of dominance
(b) Law of segregation
(c) Law of lethality
(d) Law of independent assortment
Answer:
(b) Law of segregation

Question 3.
Match list I with list II

(a) A-i, B-iii, C-ii, D-iv
(b) A-ii, B-iii, C-iv, D-i
(c) A-ii, B-iii, C-i, D-iv
(d) A-iii, B-ii, C-i, D-iv
Answer:
(c) A-ii, B-iii, C-i, D-iv

Question 4.
EcoRI cleaves the DNA at _______.
(a) AGGGTT
(b) GTATATC
(c) GAATTC
(d) TATAGC
Answer:
(c) GAATTC

Question 5.
________ is the climax community of hydrosere.
(a) Reed Swamp stage
(b) Marsh meadow stage
(c) Shrub stage
(d) Forest stage
Answer:
(d) Forest stage

Question 6.
People’s movement for the protection of environment in Sirsi of Karnataka is ________.
(a) Chipko movement
(b) Appiko movement
(c) Amirtha Devi Bishwas movement
(d) None of the above
Answer:
(b) Appiko movement

Question 7.
Which of the following is incorrectly paired?
(a) Wheat – Himgiri
(b) Rice – Ratna
(c) Milch breed – sahiwal
(d) Pusa komal – Brassica
Answer:
(d) Pusa komal – Brassica

Question 8.
Tectona grandis is coming under the family _______.
(a) Lamiaceae
(b) Fabaceae
(c) Dipterocaipaceae
(d) Ebenaceae
Answer:
(a) Lamiaceae

Part – II

Answer any four of the following questions. [4 x 2 = 8]

Question 9.
Draw and label the structure of a mature embryo sac of angiosperm.
Answer:

Question 10.
What are plasmogenes?
Answer:
Plasmogenes are independent, self-replicating, extra-chromosomal units located in cytoplasmic organelles, chloroplast and mitochondrion.

Question 11.
Define the terms
(a) Bioventing
(b) Bioaugmentation
Answer:
(a) Bioventing is the process that increases the oxygen or air flow to accelerate the degradation of environmental pollutants.
(b) Bioaugmentation is the addition of selected microbes to speed up degradation process.

Question 12.
Differentiate between Eurythermal animals and Stenothermal animals.
Answer:
Eurythermal: Organisms which can tolerate a wide range of temperature fluctuations. Example: Zostera.
Stenothermal: Organisms which can tolerate only small range of temperature variations. Example: Mango.

Question 13.
What are Blue carbon ecosystems?
Answer:
Sea grasses and mangroves of Estuarine and coastal ecosystems are the most efficient in carbon sequestration. Hence, these ecosystems are called as “ Blue carbon ecosystems”.

Question 14.
What is silvopasture system? How it helps economy?
Answer:
The production of woody plants combined with pasture is referred to silvopasture system. The trees and shrubs may be used primarily to produce fodder for livestock or they may be grown for timber, fuel wood and fruit or to improve the soil.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Given an account an king of spices and its uses.
Answer:
King of Spices:
Pepper is one of the most important Indian spices referred to as the “King of Spices” and also termed as “Black Gold of India”. Kerala, Karnataka and Tamil Nadu are the top producers in India. The characteristic pungency of the pepper is due to the presence of alkaloid Pipeline. There are two types of pepper available in the market namely black and white pepper.

Uses:
It is used for flavouring in the preparation of sauces, soups, curry powder and pickles. It is used in medicine as an aromatic stimulant for enhancing salivary and gastric secretions and also as a stomachic. Pepper also enhances the bio-absorption of medicines.

Question 16.
Distinguish between primary production and secondary production.
Answer:

Question 17.
What is co-evolution? Explain with example.
Answer:
The interaction between organisms, when continues for generations, involves reciprocal changes in genetic and morphological characters of both organisms. This type of evolution is called Co-evolution. It is a kind of co-adaptation and mutual change among interactive species.

Examples:

• Corolla length and proboscis length of butterflies and moths (Habenaria and Moth).
• Bird’s beak shape and flower shape and size.

Question 18.
What do you mean by Germplasm conservation? Describe it.
Answer:
Germplasm conservation refers to the conservation of living genetic resources like pollen, seeds or tissue of plant material maintained for the purpose of selective plant breeding, preservation in live condition and used for many research works.

Germplasm conservation resources is a part of collection of seeds and pollen that are stored in seed or pollen banks, so as to maintain their viability and fertility for any later use such as hybridization and crop improvement. Germplasm conservation may also involve a gene bank, DNA bank of elite breeding lines of plant resources for the maintenance of biological diversity and also for food security.

Question 19.
Point out the reasons for Mendels’ success in his breeding experiment.
Answer:

1. He applied mathematics and statistical methods to biology and laws of probability to his breeding experiments.
2. He followed scientific methods and kept accurate and detailed records that include quantitative data of the outcome of his crosses.
3. His experiments were carefully planned and he used large samples.
4. The pairs of contrasting characters which were controlled by factor (genes) were present on separate chromosomes.
5. The parents selected by Mendel were pure breed lines and the purity was tested by self crossing the progeny for many generations.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Explain how Nicotiana exhibit self-compatibility in detail.
Answer:

Self-sterility means that the pollen from a plant is unable to germinate on its own stigma and will not be able to bring about fertilization in the ovules of the same plant. East (1925) observed multiple alleles in Nicotiana which are responsible for self-incompatibility or self-sterility. The gene for self-incompatibility can be designated as S, which has allelic series S₁, S₂, S₃, S₄ and S₅

The cross-fertilizing tobacco plants were not always homozygous as S₁S₁ or S₂S₂, but all plants were heterozygous as S₁S₂, S₃S₄ and S₅S₆. When crosses were made between different S₁S₂ plants, the pollen tube did not develop normally. But effective pollen tube development was observed when crossing was made with other than S₁S₂ for example S₃S₄.

When crosses were made between seed parents with S₁S₂ and pollen parents with S₂S₃, two kinds of pollen tubes were distinguished. Pollen grains carrying S₃ were not effective, but the pollen grains carrying S₃ were capable of fertilization. Thus, from the cross S₁, S₂ X S₃, S₄, all the pollens were effective and four kinds of progency resulted: S₁S₃, S₁S₄, S₂S₃ and S₂S₄.

[OR]

(b) Point out the applications of plant tissue culture.
Answer:
Plant tissue culture techniques have several applications such as:

• Improved hybrids production through somatic hybridization.
• Somatic embryoids can be encapsulated into synthetic seeds (synseeds). These encapsulated seeds or synthetic seeds help in conservation of plant biodiversity.
• Production of disease resistant plants through meristem and shoot tip culture.
• Production of stress resistant plants like herbicide tolerant, heat tolerant plants.
• Micro-propagation technique to obtain large numbers of plantlets of both crop and tree species useful in forestry within a short span of time and all through the year.
• Production of secondary metabolites from cell culture utilized in pharmaceutical, cosmetic and food industries.

Question 21.
(a) Describe the various stages of decomposition process.
Answer:
(1) Fragmentation: The breaking down of detritus into smaller particles by detritivores like bacteria, fungi and earth worm is known as fragmentation. These detritivores secrete certain substances to enhance the fragmentation process and increase the surface area of detritus particles.

(2) Catabolism: The decomposers produce some extracellular enzymes in their surroundings to break down complex organic and inorganic compounds in to simpler ones. This is called catabolism.

(3) Leaching or Eluviation: The movement of decomposed, water soluble organic and inorganic compounds from the surface to the lower layer of soil or the carrying away of the same by water is called leaching or eluviation.

(4) Humification: It is a process by which simplified detritus is changed into dark coloured amorphous substance called humus. It is highly resistant to microbial action, therefore decomposition is very slow. It is the reservoir of nutrients.

(5) Mineralisation: Some microbes are involved in the release of inorganic nutrients from the humus of the soil, such process is called mineralisation.

[OR]

(b) Explain the steps involved in hybridization.
Answer:
Steps involved in hybridization are as follows:

• Selection of Parents: Male and female plants of the desired characters are selected. It should be tested for their homozygosity.
• Emasculation: It is a process of removal of anthers to prevent self pollination before anthesis (period of opening of a flower).
• Bagging: The stigma of the flower is protected against any undesirable pollen grains, by covering it with a bag.
• Crossing: Transfer of pollen grains from selected male flower to the stigma of the female emasculated flower.
• Harvesting seeds and raising plants: The pollination leads to fertilization and finally seed formation takes place. The seeds are grown into new generation which are called hybrid.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Assertion (A): In bee society, all the members are diploid, except drones.
Reason (R): Drones are produced by parthenogenesis.
(a) A and R are true, R is the correct explanation for A
(b) A and R are true, R is not the correct explanation for A
(c) A is true, R is false
(d) Both A and R are false
Answer:
(a) A and R are true, R is the correct explanation for A

Question 2.
Colotrum is rich in ______.
(a) IgE
(b) IgA
(c) IgD
(d) IgM
Answer:
(b) IgA

Question 3.
Match column I with column II

(a) A-iv, B-ii, C-i, D-iii
(b) A-iv, B-i, C-ii, D-iii
(c) A-iv, B-i, C-ii, D-iii
(d) A-i, B-iv, C-iii, D-ii
Answer:
(b) A-iv, B-i, C-ii, D-iii

Question 4.
Which of the following is the correct sequence of event with reference to central dogma?
(a) Transcription, Translation, Replication
(b) Transcription, Replication, Translation
(c) Duplication, Transcription, Translation
(d) Replication, Transcription, Translation
Answer:
(d) Replication, Transcription, Translation

Question 5.
Spread of cancerous cells to distant sites is termed as ______.
(a) Metastasis
(b) Oncogenes
(c) Proto-oncogenes
(d) Malignant neoplasm
Answer:
(a) Metastasis

Question 6.
How many amino acids are arranged in the two chains of Insulin?
(a) Chain A has 12 and Chain B has 13 amino acids
(b) Chain A has 21 and Chain B has 30 amino acids
(c) Chain A has 20 and chain B has 30 amino acids
(d) Chain A has 12 and chain B has 20 amino acids
Answer:
(b) Chain A has 21 and Chain B has 30 amino acids

Question 7.
Who introduced the term biodiversity?
(a) Edward Wilson
(b) Walter Rosen
(c) Norman Myers
(a) Alice Norman
Answer:
(b) Walter Rosen

Question 8.
What is the name of the action plan for sustainable development framed in Rio conference in 1992?
(a) Action 21
(b) Agenda 21
(c) Declaration 21
(d) Protocol 21
Answer:
(b) Agenda 21

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
How is polyspermy avoided in humans?
Answer:
Once fertilization is accomplished, cortical granules from the cytoplasm of the ovum form a barrier called the fertilization membrane around the ovum preventing further penetration of other sperms. Thus polyspermy is prevented.

Question 10.
What are holandric genes?
Answer:
The genes present in the differential region of Y chromosome are called Y- linked or holandric genes. The Y linked genes have no corresponding allele in X chromosome.

Question 11.
What are connecting links? Give example.
Answer:
The organisms which possess the characters of two different groups (transitional stage) are called connecting links. Example Peripatus (connecting link between Annelida and Arthropoda) Archaeopteryx (connecting link between Reptiles and Aves).

Question 12.
Saccharomyces cerevisiae is called as brewer’s yeast. Justify.
Answer:
Saccharomyces cerevisiae commonly called brewer’s yeast is used for fermenting malted cereals and fruit juices to produce various alcoholic beverages. Wine and beer are produced without distillation, whereas whisky, brandy and rum are obtained by fermentation and distillation.

Question 13.
Give the diagnostic characters of a Biome.
Answer:

• Location, Geographical position (Latitude and Longitude)
• Climate and physiochemical environment
• Predominant plant and animal life
• Boundaries between biomes are not always sharply defined. Transition or transient zones are seen.

Question 14.
What would Earth be like without the greenhouse effect?
Answer:
Greenhouse effect is vital for the sustenance of life. Greenhouse gases like CO₂, water vapour etc absorb some of the reflected sun’s radiation and radiate back it to the Earth surface, thus maintaining the Earth’s warm condition. Without this effect, life on Earth would be difficult or rather impossible for existence or become hostile to most living organisms.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Write a short note on phases of life cycle.
Answer:

• Juvenile phase – Period of growth between birth of an individual and reproductive maturity.
• Reproductive phase – Period of growth when an organism attain reproductive maturity and produces new offsprings.
• Senescent plane – Period of growth when the structure and functioning of body starts degenerating.

Question 16.
What is MTP? Add a note on it.
Answer:
Medical Termination of Pregnancy (MTP): Medical method of abortion is a voluntary or intentional termination of pregnancy in a non-surgical or non-invasive way. Early medical termination is extremely safe upto 12 weeks (the first trimester) of pregnancy and generally has no impact on a women’s fertility. Abortion during the second trimester is more risky as the foetus becomes intimately associated with the maternal tissue.

Question 17.
Genetic code is ‘universal’. Give reason.
Answer:
The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However similarities are more common than differences.

Question 18.
Autoimmunity is a misdirected immune response. Justify.
Answer:
Autoimmune diseases : Autoimmunity is due to an abnormal immune response in which the immune system fails to properly distinguish between self and non-self and attacks its own body. Our body produces antibodies (auto antibodies) and cytotoxic T cells that destroy our own tissues. If a disease-state results, it is referred to as auto-immune disease. Thus, autoimmunity is a misdirected immune response.

Question 19.
PCR is a useful tool for early diagnosis of an Infectious disease. Elaborate.
Answer:
The specificity and sensitivity of PCR is useful for the diagnosis of inherited disorders (genetic diseases), viral diseases, bacterial diseases, etc., The diagnosis and treatment of a particular disease often requires identifying a particular pathogen. Traditional methods of identification involve culturing these organisms from clinical specimens and performing metabolic and other tests to identify them.

The concept behind PCR based diagnosis of infectious diseases is simple – if the pathogen is present in a clinical specimen its DNA will be present. Its DNA has unique sequences that can be detected by PCR, often using the clinical specimen (for example, blood, stool, spinal fluid, or sputum) in the PCR mixture.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Give an detailed account on various natural methods of contraception.
Answer:
Natural method is used to prevent meeting of sperm with ovum, i.e., Rhythm method (safe period), coitus interruptus, continuous abstinence and lactational amenorrhoea.

1. Periodic abstinence/rhythm method: Ovulation occurs at about the 14th day of the menstrual cycle. Ovum survives for about two days and sperm remains alive for about 72 hours in the female reproductive tract. Coitus is to be avoided during this time.

2. Continuous abstinence is the simplest and most reliable way to avoid pregnancy is not to have coitus for a defined period that facilitates conception.

3. Coitus interruptus is the oldest family planning method. The male partner withdraws his penis before ejaculation, thereby preventing deposition of semen into the vagina.

4. Lactational amenorrhoea : Menstrual cycles resume as early as 6 to 8 weeks, from parturition. However, the reappearance of normal ovarian cycles may be delayed for six months during breast-feeding. This delay in ovarian cycles is called lactational amenorrhoea. It serves as a natural, but an unreliable form of birth control. Suckling by the baby during breast-feeding stimulates the pituitary to secrete increased prolactin hormone in order to increase milk production.

This high prolactin concentration in the mother’s blood may prevent menstrual cycle by suppressing the release of GnRH (Gonadotropin Releasing Hormone) from hypothalamus and gonadotropin secretion from the pituitary.

[OR]

(b) Explain the three major categories in which fossilization occur.
Answer:
(1) Actual remains is the most common method of fossilization. When marine animals die, their hard parts such as bones and shells, etc. are covered with sediments and are protected from further deterioration. They get preserved as such as they are preserved in vast ocean; the salinity in them prevents decay.

The sediments become hardened to form definite layers or strata. For example, Woolly Mammoth that lived 22 thousand years ago were preserved in the frozen coast of Siberia as such. Several human beings and animals living in the ancient city of Pompeii were preserved intact by volcanic ash which gushed out from Mount Vesuvius.

(2) Petrifaction – When animals die the original portion of their body may be replaced . molecule for molecule by minerals and the original substance being lost through disintegration. This method of fossilization is called petrifaction. The principle minerals involved in this type fossilization are iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

(3) Natural moulds and casts – Even after disintegration, the body of an animal might leave indelible impression on the soft mud which later becomes hardened into stones. Such impressions are called moulds. The cavities of the moulds may get filled up by hard minerals and get fossilized, which are called casts. Hardened faecal matter termed as coprolites occur as tiny pellets. Analysis of the coprolites enables us to understand the nature of diet, the prehistoric animals thrived.

Question 21.
(a) Explain in detail about stem cell therapy.
Answer:
Stem cells are undifferentiated cells found in most of the multi cellular animals. These cells maintain their undifferentiated state even after undergoing numerous mitotic divisions.

Stem cell research has the potential to revolutionize the future of medicine with the ability to regenerate damaged and diseased organs. Stem cells are capable of self renewal and exhibit ‘cellular potency’. Stem cells can differentiate into all types of cells that are derived from any of the three germ layers ectoderm, endoderm and mesoderm.

In mammals there are two main types of stem cells – embryonic stem cells (ES cells) and adult stem cells. ES cells are pluripotent and can produce the three primary germ layers ectoderm, mesoderm and endoderm. Embryonic stem cells are multipotent stem cells that can differentiate into a number of types of cells. ES cells are isolated from the epiblast tissue of the inner cell mass of a blastocyst. When stimulated ES can develop into more than 200 cells types of the adult body. ES cells are immortal i.e. they can proliferate in a sterile culture medium and maintain their undifferentiated state.

Adult stem cells are found in various tissues of children as well as adults. An adult stem cell or somatic stem cell can divide and create another cell similar to it. Most of the adult stem cells are multipotent and can act as a repair system of the body, replenishing adult tissues.The red bone marrow is a rich source of adult stem cells.

The most important and potential application of human stem cells is the generation of cells and tissues that could be used for cell based therapies. Human stem cells could be used to test new drugs.

[OR]

(b) Explain in detail about various types of extinctions.
Answer:
There are three types of Extinctions
(1) Natural extinction: It is a slow process of replacement of existing species with better adapted species due to changes in environmental conditions, evolutionary changes, predators and diseases. A small population can get extinct sooner than the large population due to inbreeding depression (less adaptivity and variation)

(2) Mass extinction: The Earth has experienced quite a few mass extinctions due to environmental catastrophes. A mass extinction occurred about 225 million years ago during the Permian, where 90% of shallow water marine invertebrates disappeared.

(3) Anthropogenic extinctions: These are abetted by human activities like hunting, habitat destruction, over exploitation, urbanization and industrialization. Some examples of extinctions are Dodo of Mauritius and Steller’s sea cow of Russia. Amphibians seem to be at higher risk of extinction because of habitat destruction. The most serious aspect of the loss of biodiversity is the extinction of species. The unique information contained in its genetic material (DNA) and the niche it possesses are lost forever.

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TN State Board 12th Chemistry Model Question Paper 5 English Medium

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III and IV are to be attempted separately
4. Question numbers 1 to 15 in Part I are Multiple choice Questions of one mark each.  These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
Which one of the following ore is best concentrated by froath – floatation method?
(a) Magnetite
(b) Haematite
(c) Galena
(d) Cassiterite
Answer:
(c) Galena

Question 2.
Which compound is used as flux in metallurgy?
(a) Boric acid
(b) Borax
(c) Diborane
(d) BF₃
Answer:
(b) Borax

Question 3.
The shape of XeOF₄ is
(a) T Shaped
(b) Pyramidal
(c) Square planar
(d) Square pyramidal
Answer:
(d) Square pyramidal

Question 4.
How many moles of acidified KMnO₄ required to oxidise one mole of oxalic acid?
(a) 5
(b) 0.6
(c) 1.5
(s) 0.4
Answer:
(b) 0.6

Question 5.
The type of isomerism exhibited by [Pt(NH₃)₂ Cl₂] ?
(a) coordination isomerism
(b) linkage isomerism
(c) optical isomerism
(d) geometrical isomerism
Answer:
(d) geometrical isomerism

Question 6.
The fraction of the total volume occupied by the atoms in a fcc is

Answer:
(a) \(\frac{\pi \sqrt{2}}{6}\)

Question 7.
The half life period of a radioactive element is 140 days. After 280 days 1g of element will be
reduced to which amount of the following?
(a) 1/4
(b) 1/16
(c) 1/8
(d) 1/2
Answer:
(a) 1/4
Solution:

Question 8.
Which is not a Lewis base?
(a) BF₃
(b) PF₃
(c) CO
(d) F
Answer:
(a) BF₃

Question 9.
During electrolysis of molten copper chloride, the time required to produce 0.2 mole of chlorine gas using a current of 2A is ………..
(a) 32.66 min
(b) 321.66 min
(c) 378 min
(d) 260 min
Solution:
m = ZIt (mass of 1 mole of Cl2 gas = 71) m
t = \(\frac{\mathrm{m}}{\mathrm{Zl}}\) (mass of 0.2 mole of Cl2 gas = 14.2 g)

Question 10.
Smoke is a colloidal solution of …………
(a) Solid in gas
(b) Gas in gas
(c) Liquid in gas
(d) Gas in liquid
Answer:
(c) Liquid in gas

Question 11.
Iso propyl benzene on oxidation in presence of air and dilute acid gives
(a) C₆H₅COOH
(b) C₆H₅COCH₃
(c) C₆H₅COC₆H₅
(d) C₆H₅OH
Answer:
(d) C₆H₅OH

Question 12.
But-2 ene on ozonolyis followed by subsequent cleavage with Zn and water gives …………….
(a) ethanal
(b) Propanal
(c) Propanone
(d) Methanal
Answer:
(a) ethanal

Question 13.
3(i)
This reaction is known as
(a) Friedal- craft’s reaction
(b) HVZ reaction
(c) Schotten – Baumann reaction
(d) Cannizaro reaction
Answer:
(c) Schotten – Baumann reaction

Question 14.
The pyrimidine bases present in DNA are
(a) Cytosine and Adenine
(b) Cytosine and Guanine
(c) Cytosine and Thiamine
(d) Cytosine and Uracil
Answer:
(c) Cytosine and Thiamine

Question 15.
Nylon is an example of
(a) Polyamide
(b) Polythene
(c) Polyester
(d) Polysaccharide
Answer:
(a) Polyamide

Part – II

Answer any six questions. Question No. 24 is compulsory. [6 × 2 = 12]

Question 16.
Write a test to identify borate radical?
Answer:
When boric acid or borate salt is heated with ethyl alcohol in presence of concentrated H₂SO₄, an ester triethyl borate is formed. The Vapour of this ester bums with a green edged flame and this reaction is used to identify the presence of borate.

Question 17.
How is pure phosphine prepared from phosphorous acid ?
Answer:
Phosphine is prepared in pure form by heating phosphorous acid.

Question 18.
What are ionisation isomers? Explain with an example.
Answer:
1. Ionisation isomerism arises when an ionisable counter ion (simple ion) itself can act as a ligand.

2. The exchange of such counter ions with one or more ligands in the coordination entity will result in ionisation isomers. These isomers will give different ions in solution.

3. For example, consider the coordination compound [Pt (en)₂ Cl₂]Br₂. In this compound, both Br^– and Cl^– have the ability to act as a ligand and the exchange of these two ions result in a different isomer [Pt(en)₂Br₂]Cl₂. In solution, the first compound gives Br^– ions while the later gives Cl ions and hence these compounds are called ionization isomers.

Question 19.
What is pseudo first order reaction? Give one example.
Answer:
A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, such reaction is called pseudo first order reaction.
Let us consider the acid hydrolysis of an ester,

If the reaction is carried out with the large excess of water, there is no significant change in the concentration of water during hydrolysis, i.e., concentration of water remains almost a constant.
Now we can define k [H₂O] = k’
.’. The above rate equation becomes, Rate = k’ [CH₃COOCH₃]
Thus it follows first order kinetics.

Question 20.
State Faraday’s second law of electrolysis.
Answer:
When the same quantity of charge is passed through the solutions of different electrolytes, the , amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents.

Question 21.
How will you convert glycerol into acrolein?
Answer:

Question 22.
Give any four differences between DNA and RNA.
DNA

1. It is mainly present in nucleus, mitochondria and chloroplast
2. It contains deoxyribose sugar
3. It’s life time is high
4. It is stable and not hydrolysed easily by alkalis
5. It can replicate itself

RNA

1. It is mainly present in cytoplasm, nucleolus and ribosomes
2. It contains ribose sugar
3. It is Short lived
4. It is unstable and hydrolyzed easily by alkalis
5. It cannot replicate itself. It is formed from DNA.

Question 23.
Write short notes on Antioxidants.
Answer:

• Antioxidants are substances which retard the oxidative deteriotations of food. Food containing fats and oils is easily oxidised and turn rancid.
• To prevent the oxidation of fats and oils, chemical BHT (butyl hydroxy toluene), BHA (butylated hydroxy anisole) are added as antioxidants.
• These materials readily undergo oxidation by reacting with free radicals generated by the oxidation of oils there by stop the chain reaction of oxidation of food.
• Sulphur dioxide, sulphites are also used as antioxidant and also act as antimicrobial agents and
  enzyme inhibitors.

Question 24.
50ml of 0.05M HNO₃ is added to 50ml of 0.025M KOH. Calculate the pH of the resultant solution.
Answer:
Number of moles of HNO₃ = 0.05 × 50 × 10^-3 = 2.5 × 10^-3
Number of moles of KOH = 0.025 × 50 × 10^-3 = 1.25 × 10^-3
Number of moles of HNC₃ after mixing = 2.5 × 10^-3 – 1.5 × 10 ^-3 = 1.25 × 10^-3

After mixing, total volume = 100 ml = 100 × 10^-3 L

pH = – log [H⁺]
pH = -log(l.25 × 10^-2) = 2 – 0.0969
= 1.9031

Part – III

Answer any six questions. Question No. 33 is compulsory. [6 × 3 = 18]

Question 25.
Explain the electro metallurgy of aluminium.
Answer:
Electrochemical extraction of aluminium -Hall-Herold process: In this method, electrolysis is carried out in an iron tank lined with carbon, which acts as a cathode. The carbon blocks immersed in the electrolyte acts as a anode. A 20% solution of alumina, obtained from the bauxite ore is mixed with molten cyrolite and is taken in the electrolysis chamber. About 10% calcium chloride is also added to the solution. Here calcium chloride helps to lower the melting point of the mixture. The fused mixture is maintained at a temperature of above 1270 K. The chemical reactions involved in this process are as follows:

Ionisation of alumina : Al₂O₃ → 2Al³⁺ + 3O^2-
Reaction at cathode : 2Al³⁺ (melt) + 6e^– → 2Al _(l)
Reaction at anode : 6O^2- (melt) → 3O₂ + 12e^–
Since carbon acts as anode the following reaction also takes place on it.
C (s) + O^2- (melt) → CO + 2e^– ; C_(s) + 2O^2- (melt) → CO₂ + 4e^–

Due to the above two reactions, anodes are slowly consumed during the electrolysis. The pure aluminium is formed at the cathode and settles at the bottom. The net electrolysis reaction can be written as follows:

4Al³⁺ (melt) + 6O^2- (melt) + 3C_(s) → 4Al_(l) + 3CO_2(g)

Question 26.
Give the uses of helium.
Answer:

• Helium and oxygen mixture is used by divers in place of air oxygen mixture. This prevents the painful dangerous condition called bends.
• Helium is used to provide inert atmosphere in electric arc welding of metals
• Helium has lowest boiling point hence used in cryogenics (low temperature science).
• It is much less denser than air and hence used for filling air balloons.

Question 27.
Explain chromyl chloride Test.
Answer:
(i) When potassium dichromate is heated with any chloride salt in the presence of Conc.H₂SO₄, orange red vapours of chromyl chloride (CrO₂Cl₂) is evolved. This reaction is used to confirm the presence of chloride ion in inorganic qualitative analysis.

(ii) The chromyl chloride vapours are dissolved in sodium hydroxide solution and then acidified with acetic acid and treated with lead acetate. A yellow precipitate of lead chromate is obtained. Cr02Cl2 + 4NaOH -► Na2Cr04

Question 28.
A face centred cubic solid of an element (atomic mass 60 g mol ) has a cube edge of 4A. Calculate its density.
Answer:
For FCC unit cell n 4
Edge length(a) = 4A = 4 × 10^-8cm
Mass (M) = 60 g mol^-1
Dcnsity(ρ) = ?

Question 29.
Describe the construction of Daniel cell and write its cell reaction.
Answer:
The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.

Oxidation half cell: The metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker.
Reduction half cell: A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.
Joining the half cell : The zinc and copper strips are externally connected using a wire through a switch (k) and a load (example: volt meter). The electrolytic solution present in the cathodic and anodic compartment are connected using an inverted U tube containing a agar-agar gel mixed with an inert electrolyte such as KCl, Na₂SO₄ etc., The ions of inert electrolyte do not react with other ions present in the half cells and they are not either oxidised (or) reduced at the electrodes. The solution in the salt bridge cannot get poured out, but through which the ions can move into (or) out of the half cells.

When the switch (k) closes the circuit, the electrons flows from zinc strip to copper strip. This is due to the following redox reactions which are taking place at the respective electrodes.

Anodic oxidation:- The electrode at which the oxidation occur is called the anode. In Daniel cell, the oxidation take place at zinc electrode, i.e., zinc is oxidised to Zr²⁺ ions and the electrons. The Zn²⁺ ions enters the solution and the electrons enter the zinc metal, then flow through the external wire and then enter the copper strip. Electrons are liberated at zinc electrode and hence it is negative (- ve).
Zn(s) → Zn²⁺+(aq) + 2e^–
(loss of electron-oxidation)

Cathodic reduction:
As discussed earlier, the electrons flow through the circuit from zinc to copper, where the Cu ions in the solution accept the electrons, get reduced to copper and the same get deposited on the electrode. Here, the electrons are consumed and hence it is positive (+ve).

cu²⁺ 2e^– cu_(s)( gain of electron – reduction)

Salt bridge: -The electrolytes present in two half cells are connected using a salt bridge. We have learnt that the anodic oxidation of zinc electrodes results in the increase in concentration of Zn²⁺ in solution, i.e., the solution contains more number of Zn²⁺ ions as compared to SO^2-₄ and hence the solution in the anodic compartment would become positively charged. Similarly, the solution in the cathodic compartment would become negatively charged as the Cu²⁺ ions are reduced to copper i.e., the cathodic solution contain more number of SO^2-₄ ions compared to Cu²⁺ .

Completion of circuit:- Electrons flow from the negatively charged zinc anode into the positively charged copper electrode through the external wire, at the same time, anions move towards anode and cations move towards the cathode compartment. This completes the circuit.

Consumption of Electrodes :- As the Daniel cell operates, the mass of zinc electrode gradually decreases while the mass of the copper electrode increases and hence the cell will function until the entire metallic zinc electrode is converted into Zn²⁺ or the entire Cu²⁺ ions are converted into mettalic copper.
Daniel cell is represents as

Question 30.
Write short notes on (i) Negative catalyst (ii) Phase transfer catalyst
Answer:
(i) Negative catalyst:
(i) In certain reactions, presence of certain substances decreases the rate of the reaction. Such
substances are called negative catalyst and the process is called negative catalysis.

(ii) In oxidation of chloroform, ethanol decreases the rate of the reaction and ethanol act as negative catalyst.

(iii) Decomposition of H202 rate is decreased by glycerol and it act as negative catalyst

(ii) Phase transfer catalyst:

(i) Consider the reactant of a reaction is present in one solvent and the other reactant is present in an another solvent. The reaction between them is very slow, if the solvents are immisible.

(ii) As the solvents form separate phases, the reactants have to migrate across the boundary to react. But migration of reactants across the boundary is not easy. For such situation, a third solvent is added which is miscible with both. So, the phase boundary is eliminated the reactants freely mix and react fast.

(iii) But for large scale preparation of any product, use of a third solvent is not convenient as it . may be expensive. For such problems, phase transfer catalysis provides a simple solution, which avoids the use of solvents.

(iv) It directs the use of a phase transfer catalyst (a phase transfer reagent) to facilitate transport of a reactant in one solvent to other solvent where the second reactant is present. As the reactants are now brought together, they rapidly react and form the product.

(v) Example : Substitution of Cl^– and CN^– in the following reaction.

R Cl = 1- chloro octane
R CN = 1- cyano octane
(vi) By direct heating of two phase mixture of organic 1- chloro octane with aqueous sodium cyanide for several days, 1- cyano octane is not obtained. However, if a small amount of quartemary ammonium salt like tetra alkyl ammonium cation which has hydrophobic and hydrophilic ends, transports CN- from the aqueous phase to the organic phase using its hydrophilic end and facilitates the reactions with 1- chloro octane as shown below

(vii) So phase transfer catalyst, speeds up the reaction by transporting one reactant from one phase to another.

Question 31.
Explain the mechanism of Aldol condensation of acetaldehyde.
Answer:
In presence of dilute base NaOH, or KOH, two molecules of an aldehyde or ketone having α – hydrogen add together to give β- hydroxyl aldehyde (aldol) or β – hydroxyl ketone (ketol). The reaction is called aldol condensation reaction.

Mechanism
The mechanism of aldol condensation Of acetaldehyde takes place in three steps.
Step 1: The carbanion is formed as the a – hydrogen atom is removed as a proton by the base.

Step 2: The carbanion attacks the carbonyl carbon of another unionized aldehyde to form an alkoxide ion.

Step 3: The alkoxide ion formed is protonated by water to form aldol.

The aldol rapidly undergoes dehydration on heating with acid to form a, p unsaturated aldehyde.

Question 32.
Explain the preparation of Nylon – 6,6 and Buna- S.
Answer:
Nylon 6,6 can be prepared by mixing equimolar adipic acid and hexamethylene diamine. With the elimination of water to form amide bonds.

Buna – S is prepared by the polymerisation of buta -1,3 – diene and styrene in the ratio of 3:1 in the presence of sodium.

Question 33.
Identify A to C in the following sequence.

Part – IV

Answer all the questions. [5 × 5 = 25]

Question 34.
(a) (i) Explain how gold ore is leached by cyanide process
(ii) Explain the classification of lnosilicates

[OR]

(b) (i) What are interhalogen compounds? Give examples
(ii) Explain the preparation of KMn04.
Answer:
(a) (i) 4Au (s) + 8CN^– (aq) + 2H₂O(aq) + O₂(g)→ 4Au(CN)₂]^–(aq) + 4OH^–
2[Au(CN)₂]^–(aq) + Zn (s) → 2Au (s) + [Zn(CN)₄]^-2(aq)
In the first reaction Au changes into Au+, i.e. its oxidation takes place. In the second reaction: Au⁺ → Au° (i.e. ) reduction takes place.

(ii) Ino silicones: Silicates which contain V number of silicate units liked by sharing two or more oxygen atoms are called inosilicates. They are further classified as chain silicates and double chain silicates.

Chain silicates (or pyroxenes): These silicates contain [(SiO₃)_n]^2n- ions formed by linking ‘ri number of tetrahedral [SiO₄]^4- units linearly. Each silicate unit shares two of its oxygen atoms with other units.

Example: Spodumene – LiAl(SiO₃)₂ Double chain silicates (or amphiboles): These silicates contains [Si₄O₁₁] _n^6n-. ions. In these silicates there are two different types of tetrahedra : (i) Those sharing 3 vertices (ii) those sharing only 2 vertices.

Examples:
Asbestos: These are fibrous and noncombustible silicates. Therefore they are used for thermal insulation material, brake linings, construction material and filters. Asbestos being carcinogenic silicates, their applications are restricted.

[OR]

(b) (i) Each halogen combines with other halogens to form a series of compounds called interhalogen compounds, e.g., ClF₃, IF₅, IF₇
(ii) Potassium permanganate is prepared from pyrolusite (MnO₂) ore. The preparation involves the following steps.

Conversion of MnO₂ to potassium manganate:
Powdered ore is fused with KOH in the presence of air or oxidising agents like KNO₃ A green coloured potassium manganate is formed.

Oxidation of potassium manganate to potassium permanganate:
Potassium manganate can be oxidised in two ways, either by chemical oxidation or electrolytic oxidation.

Chemical oxidation: In this method potassium manganate is treated with ozone (O₃) or chlorine to get potassium permanganate.

Electrolytic oxidation: In this method aqueous solution of potassium manganate is electrolyzed in the presence of little alkali.
K₂MnO₄ ⇌ 2K⁺ + MnO₄^2-
H₂O ⇌ H⁺ + OH^–

Manganate ions are converted into permanganate ions at anode.

The purple coloured solution is concentrated by evaporation and forms crystals of potassium permanganate on cooling.

Question 35.
(a) (i) Explain [Fe(CN)₆]^3- is paramagnetic, using Crystal Field theory
(ii) What is schottky detect?

[OR]

(A) (0 Derive Henderson – Hassel balch equation
(ii) What is kohlraush’s law?
Answer:
(a) (i)

(ii) Schottky defect arises due to the missing of equal number of cations and anions from the crystal lattice. This effect does not change the stoichiometry of the crystal.Ionic solids in which the cation and anion are of almost of similar size show schottky defect. Example: NaCl.

Presence of large number of schottky defects in a crystal, lowers its density.
For example, the theoretical density of vanadium monoxide (VO) calculated using the edge length of the unit cell is 6.5 g cm^-3 , but the actual experimental density is 5.6 g cm^-3 .It indicates that there is approximately 14% Schottky defect in VO crystal. Presence of Schottky defect in the crystal provides a simple way by which atoms or ions can move within the crystal lattice.

[OR]

(b) (i) Henderson – Hasselbalch equation:
1. The concentration of hydronium ion in acidic buffer solution depends on the ratio of concentration of the weak acid to the concentration of its conjugate base present in the solution, i.e.,

2. The weak acid is dissociated only to a small extent. Moreover due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.

3. [Acid] and [Salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution.

4. Taking logarithm on both sides of the equation

5. Reverse the sign on both sides

Equation (6) & (7) are called Henderson – Hasselbalch equations.

(ii) Kohlrauselrs law: It is defined as, at infinite dilution the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions.

Question 36.
(a) (i) Explain Intermediate compound formation theory.
(ii) Write short notes on ultra filtration.

[OR]

(b) How the following conversions are effected?
(i) Phenol → Salicylaldehvde
(ii) Phenol → Phenolphthalein
(iii) Glycol → 1.4dioxane
Answer:
(a) (i) The intermediate compound formation theory:
A catalyst acts by providing a new path with low energy of activation.. In homogeneous catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the
catalyst is regenerated.
Consider the reactions:
A+B → AB (1)
A+C →AC (intermediate) (2)
C is the catalyst
AC+B → AB+C (3)

Activation energies for the reactions (2) and (3) are lowered compared to that of (1). Hence the formation and decomposition of the intermediate accelerate the rate of the reaction.

(ii) Ultrafiltration:

• The pores of ordinary filter papers permit the passage of colloidal solutions. In ultrafilteration, the membranes are made by using collodion, cellophane or visiking.
• When a colloidal solution is filtered using such a filter, colloidal particles are separated on the filter and the impurities are removed as washings.
• This process is quickened by application of pressure. The separation of sol particles from electrolyte by filtration through an ultrafilter is called ultrafiltration.
• Collodian is 4% solution of nitrocellulose in a mixture of alcohol and water.

[OR]

(b) (i) Phenol → Salicylaldehyde:

(ii) Phenol → Phenolphthalcin:
On heating phenol with phthalic anhydride in presence of con.H2S04, phenolphthalein
is obtained.

(iii) Glycol →1,4 dioxane:
When distilled with Conc.112S04, glycol forms 1, 4-dioxane

Question 37.
(a) Write short notes on
(i) Mustard oil reactions
(ii) Carbylamine reaction
(iii) Gabriel pathalimide synthesis

[OR]

(b) Explain the structure of Fructose.
Answer:
(a) (i) Mustard oil reactions:
When primary amines are treated with carbon disulphide (CS₂), N – alkyldithio carbonic acid is formed which on subsequent treatment with HgCl₂, give an alkyl isothiocyanate.

(ii) Carbylamine reaction:
Aliphatic (or) aromatic primary amines react with chloroform and alcoholic KOH to give isocyanides (carbylamines), which has an unpleasant smell. This reaction is known as carbylamines test. This test is used to identify the primary amines.

(iii) Gabriel pathalimide synthesis:
Gabriel synthesis is used for the preparation of Aliphatic primary amines. Phthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine.

[OR]

Structure of Fructose:

• Elemental analysis and molecular weight determination of fructose show that it has the molecular formula C₆H₁₂O₆.
• Fructose on reduction with HI and red phosphorus gives a mixture of n – hexane (major product) and 2 – iodohexane (minor product). This reaction indicates that the six carbon atoms in fructose are in a straight chain.
  
• Fructose reacts with NH₂OH and HCN. It shows the presence of carbonyl groups in the molecules of fructose.
• Fructose reacts with acetic anhydride in the presence of pyridine to form penta acetate. This reaction indicates the presence of five hydroxyl groups in a fructose molecule.
• Fructose is not oxidized by bromine water. This rules out the possibility of presence of an aldehyde (-CHO) group.
• Partial reduction of fructose with sodium amalgam and water produces mixtures of sorbitol and mannitol which are epimers at second carbon. New asymmetric carbon is formed at C-2. This confirms the presence of keto group.
  

On oxidation with nitric acid, it gives glycolic acid and tartaric acids which contain smaller number of carbon atoms than in fructose.

This shows that a keto group is present in C-2. It also shows the presence of 10 alcoholic groups at C- 1 and C- 6. From the above reaction the structure of fructose is

Question 38.
(a) (i) A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?
(ii) K_sp of Ag₂CrO₄ is 1.1 x 10^-12. What is the solubility of Ag₂CrO₄ in 0.1M K₂CrO₄?
[OR]
(b) Compound A of molecular formula C₇H₆O reduces Tollen’s reagent when A reacts with 50% NaOH gives compound B of molecular formula C₇H₈O and C of molecular formula C₇H₅O₇Na. Compound C on treatment with dil.HCl gives compound D of molecular formula C₇H₆O₂. When D is heated with sodalime gives compound E.
Identify A,B,C,D & E. Write the corresponding equations.
Answer:
(a) (i) 1. For the first order reaction k =\(\frac{2.303}{t} \log \frac{a}{(a-x)}\)
Assume, a = 100 %, x = 40%, t = 50 minutes
Therefore, a – x = 100 – 40 = 60.
k = (2.303/ 50) log (100/ 60)
k = 0.010216 min^-1
Hence the value of the rate constant is 0.010216 min^-1
2. t = ?, when x = 80%
Therefore, a – x = 100 – 80 = 20
From above, k = 0.010216 min^-1
t = (2.303/0.010216) log (100/20)
t = 157.58 min
The time at which the reaction will be 80% complete is 157.58 min.

(ii)

[OR]

(b)

Students can Download Tamil Nadu 12th Physics Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Physics Model Question Paper 3 English Medium

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III and IV are to be attempted separately
4. Question numbers 1 to 15 in Part I are Multiple choice Questions of one mark each.  These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

PART -1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 × 10^-2 C and 5 × 10^-2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
(a) 3 × 10^-2 C
(b) 4 × 10^-2 C
(c) 1 × 10^-2 C
(d) 2 × 10^-2 C
Answer:
(a) 3 × 10^-2 C

Question 2.
What is the value of resistance of the following resistor?

(a) 100 k Ω
(b) 10 kΩ
(c) 1 k Ω
(d) 1000 k Ω
Answer:
(a) 100 k Ω

Question 3.
A flow of 10⁷ electrons per second in a conduction wire constitutes a
(a) 1.6 × 10^-26A
(b) 1.6 × 10¹²A
(c) 1.6 × 10^-12A
(d) 1.6 × 10²⁶A
Hint:

Answer:
(c) 1.6 x 10^-12A

Question 4.
The force experienced by a particle having mass m and charge q accelerated through a potential difference V when it is kept under perpendicular magnetic field B is ……………….. .
(a) \(\sqrt{\frac{2 q^{3} \mathrm{BV}}{m}}\)
(b) \(\sqrt{\frac{q^{3} \mathrm{B}^{2} V}{2 m}}\)
(c) \(\sqrt{\frac{2 q^{3} \mathbf{B}^{2} \mathbf{V}}{m}}\)
(d) \(\sqrt{\frac{2 q^{3} B V}{m^{3}}}\)
Answer:
(c) \(\sqrt{\frac{2 q^{3} \mathbf{B}^{2} \mathbf{V}}{m}}\)

Question 5.
A magnetic needle is kept in a non-uniform magnetic field. It experiences
(a) a force and a torque
(b) a force but not a torque
(c) a torque but not a force
(d) neither a force nor a torque
Answer:
(a) a force and a torque

Question 6.
The instantaneous values of alternating current and voltage in a circuit are i = \(\frac{1}{\sqrt{2}}\) sin( 100πt) A and v = \(\frac{1}{\sqrt{2}} \sin \left(100 \pi t+\frac{\pi}{3}\right) \mathrm{V}\) The average power in watts consumed in the circuit is
(a) \(\frac{1}{4}\)
(b) \(\frac{\sqrt{3}}{4}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{8}\)
Answer:
(d) \(\frac{1}{8}\)

Question 7.
Which of the following is an electromagnetic wave’?
(a) α – rays
(b) β – rays
(c) γ – rays
(d) all of them
Answer:
(c) γ – rays

Question 8.
When light is incident on a soap film of thickness 5 × 10^-5 cm, the wavelength of light reflected maximum in the visible region is 5320 Å. Refractive index of the film will be.
(a) 1.22
(b) 1.33
(c) 1.51
(d) 1.83
Hint. The condition for constructive interference, (for reflection)

Answer:
(b) 1.33

Question 9.
A plane glass is placed over a various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is,
(a) red
(b) yellow
(c) green
(d) violet
Hint. Letters appear to be raised depending upon the refractive index of the material. Since violet has a higher refractive index than red (the index increases with frequency), red will be the lowermost.
Answer:
(d) violet

Question 10.
In photoelectric emission, a radiation whose frequency is 4 times threshold frequency of a certain metal is incident on the metal. Then the maximum possible velocity of the emitted electron will be
(a) \(\sqrt{\frac{h v_{0}}{m}}\)
(b) \(\sqrt{\frac{6 h u_{0}}{m}}\)
(c) \(2 \sqrt{\frac{h v_{0}}{m}}\)
(d) \(\sqrt{\frac{h v_{0}}{2 m}}\)

Hint: From Einstein’s photoelectric equation
K_max = hυ – hυ₀ [υ = 4υ₀]
\(\frac { 1 }{ 2 }\) mV²_max = 4hυ₀ – hυ₀
V²_max = \(\frac{6 h v_{0}}{m}\)
V_max = \(\sqrt{\frac{6 h v_{0}}{m}}\)
Answer:
(b) \(\sqrt{\frac{6 h u_{0}}{m}}\)

Question 11.
The number of photo-electrons emitted for light of a frequency u (higher than the threshold frequency υ₀) is proportional to
(a) Threshold frequency (υ₀)
(b) Intensity of light
(c) Frequency of light (υ)
(d) υ – υ₀
Hint: Photoelectric current of Intensity of incident light
Answer:
(b) Intensity of light

Question 12.
Atomic number of H-like atom with ionization potential 122.4 V for n = 1 is
(a) 1
(b) 2
(c) 3
(d) 4
Hint: The ionisation energy of a hydrogen atom is, IE = \(\frac{13.6 z^{2}}{n^{2}}\)

Answer:
(c) 3

Question 13.
The given electrical network is equivalent to ………….. .
(а) AND gate
(b) OR gate
(c) NOR gate
(d) NOT gate

Answer:
(c) NOR gate

Question 14.
The frequency range of 3 MHz to 30 MHz is used for
(a) Ground wave propagation
(b) Space wave propagation
(c) Sky wave propagation
(d) Satellite communication
Answer:
(c) Sky wave propagation

Question 15.
Which one of the following is the natural nanomaterial?
(a) Peacock feather
(b) peacock beak
(c) Grain of sand
(d) Skin of the wale
Answer:
(a) Peacock feather

PART – II

Answer any six questions in which Q. No 21 is compulsory. [6 × 2 = 12]

Question 16.
When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.
Answer:
Charge produced in each object q = 50 nC
q = 50 × 10^-9C
Charge of electron (e) = 1.6 × 10^-19C
Number of electron transferred, n = \(\frac{q}{e}=\frac{50 \times 10^{-9}}{1.6 \times 10^{-19}}\) = ,n =31.25 × 10^-9 × 10¹⁹
n = 31.25 × 1o¹⁰ electrons

Question 17.
State Kirchhoff’s voltage rule.
Answer:
It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system.

Question 18.
Define magnetic dipole moment.
Answer:
The magnetic dipole moment is defined as the product of its pole strength and magnetic length.
\(\overrightarrow{\mathrm{P}}\) = q_m\(\overrightarrow{\mathrm{d}}\)

Question 19.
What is meant by ‘Wattful current’?
Answer:
The component of current (I_RMS cos Φ) which is in phase with the voltage is called active component. The power consumed by this current = V_RMSI_RMS cos Φ . So that it is also known as ‘WattfuT current.

Question 20.
What are electromagnetic waves?
Answer:
Electromagnetic waves are non-mechanical waves which move with speed equals to the speed of light (in vacuum).

Question 21.
Two light sources have intensity of light as 10. What is the resultant intensity at a point where the two light waves have a phase difference of π/3?
Answer:
Let the intensities be I₀.
The resultant intensity is, I = 4 I₀ cos² (φ/2)
Resultant intensity when, Φ = π / 3, is I = 4I₀ cos² (π / 6)
I = 4I₀(√3/2)² = 3I₀

Question 22.
State de Broglie hypothesis.
Answer:
De Broglie hypothesis, all matter particles like electrons, protons, neutrons in motion are associated with waves.

Question 23.
Calculate the energy equivalent of 1 atomic mass unit.
Answer:
We take, m = 1 amu = 1.66 × 10²⁷ kg
c = 3 × 10⁸ ms^-1
Then, E = mc² = 1.66 × 10²⁷ × (3 × 1o⁸)²⁷ J

E ≈ 981 MeV
∴ 1 amu = 931 MeV

Question 24.
hat do you mean by doping?
Answer:
The process of adding impurities to the intrinsic semiconductor is called doping.

PART III

Answer any six questions in which Q.No. 32 is compulsory. [6 × 3 = 18]

Question 25.
What are the differences between Coulomb force and gravitational force?
Answer:

• The gravitational force between two masses is always attractive but Coulomb force between two charges can be attractive or repulsive, depending on the nature of charges.
• The value of the gravitational constant G = 6.626 × 10¹¹ N m² kg^-2. The value of the constant k in Coulomb law is k = 9 × 10⁹ N m² C^-2.
• The gravitational force between two masses is independent of the medium. The electrostatic force between the two charges depends on nature of the medium in which the two charges are kept at rest.
• The gravitational force between two point masses is the same whether two masses are at rest or in motion. If the charges are in motion, yet another force (Lorentz force) comes into play in addition to coulomb force.

Question 26.
The resistance of a nichrome wire at 0 °C is 10 Ω. If its temperature coefficient of resistance is 0.004/°C, find its resistance at boiling point of water. Comment on the result.
Answer:
Resistance of a nichrome wire at 0°C, R₀ = 10 Ω
Temperature co-efficient of resistance, α = 0.004/°C
Resistance at boiling point of water, R_T = ?
Temperature of boiling point of water, T = 100 °C
R_T = R₀ ( 1 + α T) = 10[1 + (0.004 × 100)]
R_T = 10(1 + 0.4) = 10 × 1.4
R_T = 14Ω
As the temperature increases the resistance of the wire also increases.

Question 27.
What is meant by magnetic induction?
Answer:
The magnetic induction (total magnetic held) inside the specimen \(\overrightarrow{\mathrm{B}}\) is equal to the sum of the magnetic field \(\overrightarrow{\mathrm{B}}\)₀ produced in vacuum due to the magnetising field and the magnetic field \(\overrightarrow{\mathrm{B}}\)_m due to the induced magnetisation of the substance.

Question 28.
An inductor blocks AC but it allows DC. Why? and How?
Answer:
An inductor L is a closely wound helical coil. The steady DC current flowing through L produces uniform magnetic field around it and the magnetic flux linked remains constant. Therefore there is no self-induction and self-induced emf (back emf). Since inductor behaves like a resistor, DC flows through an inductor.

The AC flowing through L produces time-varying magnetic field which in turn induces self- induced emf (back emf)- This back emf, according to Lenz’s law, opposes any change in the current. Since AC varies both in magnitude and direction, its flow is opposed in L. For an ideal inductor of zero ohmic resistance, the back emf is equal and opposite to the applied emf. Therefore L blocks AC.

Question 29.
Derive the relation between/and R for a spherical mirror.
Answer:
Let C be the centre of curvature of the mirror. Consider a light ray parallel to the principal axis is incident on the mirror at M and passes through the principal focus F after reflection. The geometry of reflection of the incident ray is shown in figure. The line CM is the normal to the mirror at M. Let i be the angle of incidence and the same will be the angle of reflection.
If MP is the perpendicular from M on the principal axis, then from the geometry, The angles

∠MCP = i and ∠MFP = 2i From right angle triangles ∆MCP and ∆MFP,
tan i = \(\frac{P M}{P C}\) and tan 2i = \(\frac{P M}{P F}\)
As the angles are small, tan i ≈ i, i = \(\frac{P M}{P C}\) and 2i = \(\frac{P M}{P F}\)
Simplifying further, 2 \(\frac{\mathrm{PM}}{\mathrm{PC}}=\frac{\mathrm{PM}}{\mathrm{PF}}\) ; 2PF = PC
PF is focal length f and PC is the radius of curvature R.
2f = R (or) f = \(\frac{R}{2}\)
f = \(\frac{R}{2}\) is the relation between/and R.

Question 30.
What is a photo cell? Mention the different types of photocells.
Answer:
photocells: Photo electric cell or photo cell is a device which converts light energy into electrical energy. It works on the principle of photo electric effect.
Types:

• Photo emissive cell
• Photo voltaic cell
• Photo conductive cell

Question 31.
Show that nuclear density is almost constant for nuclei with Z > 10.
Answer:

The expression shows that the nuclear density is independent of the mass number A. In other words, all the nuclei (Z >10) have the same density and it is an important characteristics of the nuclei.

Question 32.
Calculate the range of the variable capacitor that is to be used in a tuned-collector oscillator which has a fixed inductance of 150 pH. The frequency band is from 500 kHz to 1500 kHz.
Answer:
Resonant frequency f₀ = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)
On simplifying we get C = \(\frac{1}{4 \pi^{2} f_{0}^{2} L}\)
When frequency is equal to 500 kHz

When frequency is equal to 1500 kHz

There fore, the capacitor range is 75 – 676 pF

Question 33.
Distinguish between wireline and wireless communication.
Answer:

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34.
(a) Explain in detail the construction and working of a Van de Graaff generator.
Answer:
Principle: Electrostatic induction and action at points.
Construction: A large hollow spherical conductor is fixed on the insulating stand. A pulley B is mounted at the center of the hollow sphere and another pulley C is fixed at the bottom. A belt made up of insulating materials like silk or rubber runs over both pulleys. The pulley C is driven continuously by the electric motor. Two comb shaped metallic conductors E and D are ‘ fixed near the pulleys.
The comb D is maintained at a positive potential of 10⁴ V by a power supply. The upper comb E is connected to the inner side of the hollow metal sphere.

Working: Due to the high electric field near comb D, air between the belt and comb D gets ionized. The positive charges are pushed towards the belt and negative charges are attracted towards the comb D. The positive charges stick to the belt and move up. When the positive charges reach the comb E, a large amount of negative and positive charges are induced on either side of comb E due to electrostatic induction. As a result, the positive charges are pushed away from the comb E and they reach the outer surface of the sphere. Since the sphere is a conductor, the positive charges are distributed uniformly on the outer surface of the hollow sphere. At the same time, the negative charges nullify the positive charges in the belt due to corona discharge before it passes over the pulley.

When the belt descends, it has almost no net charge. At the bottom, it again gains a large positive charge. The belt goes up and delivers the positive charges to the outer surface of the sphere. This process continues until the outer surface produces the potential difference of the order of 107 V which is the limiting value. We cannot store charges beyond this limit since the extra charge starts leaking to the surroundings due to ionization of air. The leakage of charges can be reduced by enclosing the machine in a gas filled steel chamber at very high pressure. Uses: The high voltage produced in this Van de Graaff generator is used to accelerate positive ions (protons and deuterons) for nuclear disintegrations and other applications.

[OR]

Question 34.
(b) Explain the equivalent resistance of a series resistor network.
Answer:
Resistors in series: When two or more resistors are connected end to end, they are said to be in series. The resistors could be simple resistors or bulbs or heating elements or other devices. Fig. (a) shows three resistors R₁, R₂ and R₃ connected in series.


The amount of charge passing through resistor R₁ must also pass through resistors R₂ and R₃ since the charges cannot accumulate anywhere in the circuit. Due to this reason, the current I passing through all the three resistors is the same. According to Ohm’s law, if same current pass through different resistors of different values, then the potential difference across each resistor must be different. Let V₁, V₂ and V₃ be the potential difference (voltage) across each of the resistors R₁, R₂ and R₃ respectively, then we can write V₁ = IR₁ V₂ = IR₂ and V₃ = IR₃. But the total voltage V is equal to the sum of voltages across each resistor.
V = V₁ + V₂ + V₃
= IR₁ + IR₂ + IR₃ ….. (1)
V = I(R₁ + R₂ + R₃)
V = I.R_s ….. (2)
where R_s is the equivalent resistance,
R_s = R₁ + R₂ + R₃ ….. (3)
When several resistances are connected in series, the total or equivalent resistance is the sum of the individual resistances as shown in fig. (b).
Note: The value of equivalent resistance in series connection will be greater than each individual resistance.

Question 35.
(a) Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.
Answer:
Magnetic field due to long straight conductor carrying current: Consider a long straight wire NM with current I flowing from N to M. Let P be the point at a distance a from point O. Consider an element of length dl of the wire at a distance l from point O and \(\vec{r}\) be the vector joining the element dl with the point P. Let θ be the angle between dl and \(\vec{r}\) Then, the magnetic field at P due to the element is
\(d \overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{Id} \vec{l}}{4 \pi r^{2}} \sin \theta\) (unit vector perpendicular to \(d \vec{l}\) and \(\vec{r}\) )-(1)
The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by taking the cross product between two vectors \(d \vec{l}\) and \(\vec{r}\) (let it be n̂). The net magnetic field can be determined by integrating equation with proper limits.
\(\vec{B}\) = ∫d\(\vec{B}\)
From the figure, in a right angle triangle PAO,

tan (π – θ) = \(\frac{a}{l}\)
l = \(-\frac{a}{\tan \theta}\) (since tan (π – θ) = – tan θ) ⇒ \(\frac{1}{\tan \theta}\) = cot θ
l = -a cot θ and r = a cosecθ

Differentiating,
dl = a cosec²θ dθ

This is the magnetic field at a point P due to the current in small elemental length. Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. θ. Therefore, the net magnetic field at the point P which can be obtained by integrating \(d \vec{B}\) by varying the angle from θ = φ₁ to θ = φ₂ is

For a an infinitely long straight wire, l = 0 and 2 = , the magnetic field is
\(\overrightarrow{\mathrm{B}}=\frac{\mu_{0} I}{2 \pi a} \hat{n}\) …… (3)
Note that here n̂ represents the unit vector from the point O to P.

[OR]

Question 35.
(b) Show that the mutual inductance between a pair of coils is same (M₁₂ = M₂₁).
Answer:
Mutual induction: When an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil. This phenomenon is known as mutual induction and the emf is called mutually induced emf.

Consider two coils which are placed close to each other. If an electric current q is sent through coil i₁ the magnetic field produced by it is also linked with coil 2.
Let Φ₂₁ be the magnetic flux linked with each turn of the coil 2 of N₂ turns due to coil 1, then the total flux linked with coil 2 (N₂Φ₂₁) is proportional to the current i₁ in the coil 1.
N₂Φ₂₁ ∝ i₁
N₂Φ₂₁ = M₂₁i₁ or M₂₁ = \(\frac{N_{2} \Phi_{21}}{i_{1}}\)
The constant of proportionality M₂₁ is the mutual inductance of the coil 2 with respect to coil 1. It is also called as coefficient of mutual induction. If i₁ = 1 A, then M₂₁ = N₂Φ₂₁.
Therefore, the mutual inductance M₂₁ is defined as the flux linkage of the coil 2 when 1A current flows through coil 1.
When the current i₁ changes with time, an emf ξ₂ is induced in coil 2. From Faraday’s law of electromagnetic induction, this mutually induced emf ξ₂₁ is given by

The negative sign in the above equation shows that the mutually induced emf always opposes the change in current with respect to time. If \(\frac{d i_{1}}{d t}\) = 1 As^-1, then M₂₁ = -ξ₂.
Mutual inductance M₂₁ is also defined as the opposing emf induced in the coil 2 when the rate of change of current through the coil I is l As^-1.
Similarly, if an electric current i₂ through coil 2 changes with time, then emf ξ₁ is induced in coil 1. Therefore, N

where M₁₂ is the mutual inductance of the coil I with respect to coil 2. It can be shown that for a given pair of coils, the mutual inductance is same. i.e., M₂₁ = M₁₂ = M.
In general, the mutual induction between two coils depends on size, shape, the number of turns of the coils, their relative orientation and permeability of the medium.

Question 36.
(a) Derive the equation for refraction at single spherical surface. Equation for refraction at single spherical surface:
Answer:
Let us consider two transparent media having refractive indices n₁ and n₂ are separated by a spherical surface. Let C be the centre of curvature of the spherical surface. Let a point object O be in the medium n₁. The line OC cuts the spherical surface at the pole P of the surface. As the rays considered are paraxial rays, the perpendicular dropped for the point of incidence to the principal axis is very close to the pole or passes through the pole itself.

Light from O falls on the refracting surface at N. The normal drawn at the point of incidence passes through the centre of curvature C. As n₂ > n₁, light in the denser medium deviates towards the normal and meets the principal axis at I where the image is formed. Snell’s law in product form for the refraction at the point N could be written as,
n₁ sin i = n₂ sin r …… (1)
As the angles are small, sin of the angle could be approximated to the angle itself.
n₁i = n₂r ……… (2)
Let the angles
∠NOP = α ∠NCP = β ∠NIP = γ
tan α = \(\frac{P N}{P O}\) tan β = \(\frac{P N}{P C}\) tan γ = \(\frac{P N}{P I}\)
As these angles are small, tan of the angle could be approximated to the angle itself.
α = \(\frac{P N}{P O}\) ; β = \(\frac{P N}{P C}\) ; γ = \(\frac{P N}{P I}\) …… (3)
For the triangle, ∆ONC,
i = α + β ….. (4)
For the triangle, ∆INC,
β = r + γ (or) r = β – γ …… (5)
Substituting for i and r from equations (4) and (5) in the equation (2).
n₁ (α + β) = n₂ (β – γ)
Rearranging,
n₁α + n₂γ = (n₂ – n₁)β
Substituting for α, β and γ from equation (3)
\(n_{1}\left(\frac{\mathrm{PN}}{\mathrm{PO}}\right)+n_{2}\left(\frac{\mathrm{PN}}{\mathrm{PI}}\right)=\left(n_{2}-n_{1}\right)\left(\frac{\mathrm{PN}}{\mathrm{PC}}\right)\)
Further simplifying by cancelling method
\(\frac{n_{1}}{\mathrm{PO}}+\frac{n_{2}}{\mathrm{PI}}=\frac{n_{2}-n_{1}}{\mathrm{PC}}\) ….. (6)
Following sign conventions, PO = -υ , PI = +ν and PC = +R in equation (6),
\(\frac{n_{1}}{-u}+\frac{n_{2}}{v}=\frac{\left(n_{2}-n_{1}\right)}{\mathrm{R}}\)
After rearranging, finally we get,
\(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{\left(n_{2}-n_{1}\right)}{\mathrm{R}}\) ……… (7)
Equation (7) gives the relation among the object distance υ, image distance ν , refractive indices of the two media (nl and n2) and the radius of curvature R of the spherical surface. It holds for any spherical surface.
If the first medium is air then, n₁ = 1 and the second medium is taken just as n2 = n, then the equation is reduced to,
\(\frac{n}{v}-\frac{1}{u}=\frac{(n-1)}{\mathrm{R}}\) ……… (8)

[OR]

Question 36.
(b) What do you mean by electron emission? Explain briefly various methods of electron emission.
Answer:
Electron emission:

• Free electrons possess some kinetic energy and this energy is different for different electrons. The kinetic energy of the free electrons is not sufficient to overcome the surface barrier.
• Whenever an additional energy is given to the free electrons, they will have sufficient energy to cross the surface barrier. And they escape from the metallic surface.
• The liberation of electrons from any surface of a substance is called electron emission. There are mainly four types of electron emission which are given below.

(i) Thermionic emission:
When a metal is heated to a high temperature, the free electrons on the surface of the metal get sufficient energy in the form of thermal energy so that they are emitted from the metallic surface. This type of emission is known as thermionic emission.

The intensity of the thermionic emission (the number of electrons emitted) depends on the metal used and its temperature.

Examples: cathode ray tubes, electron microscopes, X-ray tubes etc.

(ii) Field emission:
Electric field emission occurs when a very strong electric field is applied across the metal. This strong field pulls the free electrons and helps them to overcome the surface barrier of the metal.

Field emission
Examples: Field emission scanning electron microscopes, Field-emission display etc.

(iii) Photo electric emission:
When an electromagnetic radiation of suitable frequency is incident on the surface of the metal, the energy is transferred from the radiation to the.free electrons. Hence, the free electrons get sufficient energy to cross the surface barrier and the photo electric emission takes place. The number of electrons emitted depends on the intensity of the incident radiation.

Examples: Photo diodes, photo electric cells etc.

(iv) Secondary emission:
When a beam of fast moving electrons strikes the surface of the metal, the kinetic energy of the striking electrons is transferred to the free electrons on the metal surface. Thus the free electrons get sufficient kinetic energy so that the secondary emission of electron occurs.

Examples: Image intensifires, photo multiplier tubes etc.

Question 37.
(a) Discuss the spectral series of hydrogen atom.
Answer:
The spectral lines of hydrogen are grouped in separate series. In each series, the distance of separation between the consecutive wavelengths decreases from higher wavelength to the lower wavelength, and also wavelength in each series approach a limiting value known as the series limit. These series are named as Lyman series. Balmer series, Paschen series, Brackett series, Pfund series, etc. The wavelengths of these spectral lines perfectly agree with the equation derived from Bohr atom model.
\(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)=\bar{v}\) …… (1)
where ν̂ is known as wave number which is inverse of wavelength, R is known as Rydberg constant whose value is 1.09737 × 10⁷ m^-1 and m and n are positive integers such that m > n. The various spectral series are discussed below:

(a) Lyman series:
Put n = 1 and m = 2,3,4 in equation (I). The wave number or wavelength of spectral lines of Lyman series which lies in ultra-violet region is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{m^{2}}\right)\)

(b) Balmer series:
Put n = 2 and m = 3,4,5 in equation (I). The wave number or wavelength of spectral lines of Balmer series which lies in visible region is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{m^{2}}\right)\)

(c) Paschen series:
Put n = 3 and m = 4,5,6 in equation (I). The wave number or wavelength of spectral lines of Paschen series which lies in infra-red region (near IR) is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{3^{2}}-\frac{1}{m^{2}}\right)\)

(d) Brackett series:
Put n = 4 and m = 5,6,7 in equation (I). The wave number or wavelength of spectral lines of Brackett series which lies in infra-red region (middle IR) is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{m^{2}}\right)\)

(e) Pfund series:
Put n = 5 and m = 6,7,8 in equation (I). The wave number or wavelength of spectral lines of Pfund series which lies in infra-red region (far IR) is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{5^{2}}-\frac{1}{m^{2}}\right)\)

[OR]

Question 37.
(b) State and prove De Morgan’s First and Second theorems.
Answer:
De Morgan’s First Theorem:
The first theorem states that the complement of the sum of two logical inputs is equal to the product of its complements.

Proof:
The Boolean equation for NOR gate is Y = \(\overline{\mathrm{A}+\mathrm{B}}\) . The Boolean equation for a bubbled AND gate is Y = \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\) Both cases generate same outputs for same inputs. It can be verified using the following truth table.

From the above truth table, we can conclude \(\overline{\mathrm{A}+\mathrm{B}}=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\). Thus De Morgan’s First Theorem is proved. It also says that a NOR gate is equal to a bubbled AND gate. The corresponding logic circuit diagram is shown in figure.

De Morgan’s Second Theorem:
The second theorem states that the complement of the product of two inputs is equal to the sum of its complements.

Proof
The Boolean equation for NAND gate is Y = \(\overline{\mathrm{AB}}\)
The Boolean equation for bubbled OR gate is Y = \(\overline{\mathrm{A}}+\overline{\mathrm{B}}\) . A and B are the inputs and Y is the output. The above two equations produces the same output for the same inputs. It can be verified by using the truth table.

From the above truth table we can conclude \(\overline{\mathrm{A} . \mathrm{B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}\). Thus De Morgan’s Second Theorem is proved. It also says, a NAND gate is equal to a bubbled OR gate. The corresponding logic circuit diagram is shown in figure.

Question 38.
(a) What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation: For long distance transmission, the low frequency baseband signal (input signal) is superimposed onto a high frequency radio signal by a process called modulation.
There are 3 types of modulation based on which parameter is modified. They are
(i) Amplitude modulation,
(ii) Frequency modulation, and
(iii) Phase modulation.

(i) Amplitude Modulation (AM):
If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal, then it is called amplitude modulation. Here the frequency and the phase of the carrier signal remain constant. Amplitude modulation is used in radio and TV broadcasting.

The signal shown in figure (a) is the message signal or baseband signal that carries information, figure (b) shows the high frequency carrier signal and figure (c) gives the amplitude modulated signal. We can see clearly that the carrier wave is modified in proportion to the amplitude of the baseband signal.
Amplitude Modulation

(ii) Frequency Modulation (FM):
The frequency of the camer signal is modified according to the instantaneous amplitude of the baseband signal in frequency modulation. Here the amplitude and the phase of the carrier signal remain constant. Increase in the amplitude of the baseband signal increases the frequency of the carrier signal and vice versa. This leads to compressions and rarefactions in the frequency spectrum of the modulated wave. Louder signal leads to compressions and relatively weaker signals to rarefactions. When the amplitude of the baseband signal is zero in Figure (a), the frequency of the modulated signal is the same as the carrier signal. The frequency of the modulated wave increases when the amplitude of the baseband signal increases in the positive direction
(A, C). The increase in amplitude in the negative half cycle (B,D) reduces the frequency of the modulated wave (figure (c)).

(iii) Phase Modulation (PM):
The instantaneous amplitude of the baseband signal modifies the phase of the camer signal keeping the amplitude and frequency constant is called phase modulation. This modulation is used to generate frequency modulated signals. It is similar to frequency modulation except that the phase of the carrier is varied instead of varying frequency. The carrier phase changes according to increase or decrease in the amplitude of the baseband signal. When the modulating signal goes positive, the amount of phase
lead increases with the amplitude of the modulating signal. Due to this, the carrier signal is compressed or its frequency is increased.

On the other hand, the negative half cycle of the baseband signal produces a phase lag in the carrier signal. This appears to have stretched the frequency of the carrier wave. Hence similar to frequency modulated wave, phase modulated wave also comprises of
compressions and rarefactions. When the signal voltage is zero (A, C and E) the carrier frequency is unchanged.

[OR]

Question 38.
(b) Elaborate any two types of Robots with relevant examples.
Answer:
(i) Human Robot:
Certain robots are made to resemble humans in appearance and replicate the human activities like walking, lifting, and sensing, etc.

• Power conversion unit: Robots are powered by batteries, solar power, and hydraulics.
• Actuators: Converts energy into movement. The majority of the actuators produce rotational or linear motion.
• Electric motors: They are used to actuate the parts of the robots like wheels, arms, fingers, legs, sensors, camera, weapon systems etc. Different types of electric motors are used. The most often used ones are AC motor, Brushed DC motor, Brushless DC motor, Geared DC motor, etc.
• Pneumatic Air Muscles: They are devices that can contract and expand when air is pumped inside. It can replicate the function of a human muscle. They contract almost 40% when the air is sucked inside them.
• Muscle wires: They are thin strands of wire made of shape memory alloys. They can contract by 5% when electric current is passed through them.
• Piezo Motors and Ultrasonic Motors: Basically, we use it for industrial robots.
• Sensors: Generally used in task environments as it provides information of real-time knowledge.
• Robot locomotion: Provides the types of movements to a robot. The different types are (a) Legged (b) Wheeled (c) Combination of Legged and Wheeled Locomotion (d) Tracked slip/skid

(ii) Industrial Robots:
Six main types of industrial robots:

1. Cartesian
2. SCARA (Selective Compliance Assembly Robot Arm)
3. Cylindrical
4. Delta
5. Polar
6. Vertically articulated

Six-axis robots are ideal for:

• Arc Welding
• Spot Welding
• Material Handling
• Machine Tending
• Other Applications

Students can Download Tamil Nadu 12th Biology Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Biology Model Question Paper 3 English Medium

General Instructions:

1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Match the following:

(a) A – (iv), B – (iii), C – (i), D – (ii)
(b) A – (ii), B – (iv), C – (i), D – (iii)
(c) A – (iii), B – (iv), C – (ii), D – (i)
(d) A – (iii), B – (i), C – (iv), D – (i)
Answer:
(b) A – (ii), B – (iv), C – (i), D – (iii)

Question 2.
Changing the codon AGC to AGA represents ___________.
(a) Mis-sense mutation
(b) Non-sense mutation
(c) Frame-shift mutation
(d) Deletion mutation
Answer:
(a) Mis-sense mutation

Question 3.
Restriction enzymes are ______.
(a) Not always required in genetic engineering
(b) Essential tools in genetic engineering ‘
(c) Nucleases that cleave DNA at specific sites
(d) Both b and c
Answer:
(d) Both b and c

Question 4.
Totipotency refers to _________.
(а) capacity to generate genetically identical plants
(b) capacity to generate a whole plant from any plant cell/explant
(c) capacity to generate hybrid protoplasts
(d) recovery to healthy plants from diseased plants
Answer:
(b) capacity to generate a whole plant from any plant cell/explant

Question 5.
In soil water available for plants is ________.
(a) gravitational water
(b) Chemically bound water
(c) Capillary water
(d) hygroscopic water
Answer:
(c) Capillary water

Question 6.
Which one is in descending order of a food chain?
(a) Producers → Secondary consumers → Primary consumers → Tertiary consumers
(b) Tertiary consumers → Primary consumers → Secondary consumers → Producers
(c) Tertiary consumers → Secondary consumers → Primary consumers → Producers
(d) Tertiary consumers → Producers → Primary consumers → Secondary consumers
Answer:
(c) Tertiary consumers → Secondary consumers → Primary consumers → Producers

Question 7.
Clean Development Mechanism (CDM) is defined is _______.
(a) Copenhagen Acord
(b) Montreal protocol
(c) Paris Agreement
(d) Kyoto protocol
Answer:
(d) Kyoto protocol

Question 8.
Pick out the odd pair.
(a) Man selection – Morphological characters
(b) Pureline selection – Repeated self pollinartion
(c) Clonal selection – Sexually propagated
(d) Natural selection – Involves nature
Answer:
(a) Man selection – Morphological characters

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Write a short note on pollen kitt.
Answer:
Pollenkitt is contributed by the tapetum and coloured yellow or orange and is chiefly made of carotenoids or flavonoids. It is an oily layer forming a thick viscous coating over pollen surface. It attracts insects and protects damage from UV radiation.

Question 10.
What is back cross?
Answer:
Back cross is a cross of F₁ hybrid with any one of the parental genotypes. The back cross is of two types; they are dominant back cross and recessive back cross. It involves the cross between the F₁ off spring with either of the two parents.

Question 11.
Productivity of profundal zone will be low. why?
Answer:
The producers of the pond ecosystem depends on phytoplankton through photosynthesis. Profundal zone lies below the limnetic zone with no effective light penetration, hence productivity rate is very low.

Question 12.
Explants for tissue culturing has to be surface sterilized. How?
Answer:
The explants are surface sterilized by first exposing the material is running tap water and then treating it in surface sterilizing agents like 0.1 % Mercuric chloride, 70% ethanol under aseptic condition inside the laminar air flow chamber.

Question 13.
Define heterosis, through which way does this condition can be maintained for generation.
Answer:
The superiority of F₁ hybrid in performance over its parents is called heterosis or hybrid vigour. Vigour refers to increase in growth, yield, adaptability resistance to disease, pest, etc., Vegetative propogation is the best suited method to maintain hybrid vigour.

Question 14.
Discuss which wood is better for making furniture.
Answer:
Teak wood is the ideal type of wood for making household furnitures because, it is highly durable and shows great resistance against the attack of termites and fungi. Moreover it does not split or crack and is a carpenter friendly wood.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Give the names of the scientists who rediscovered mendel’s work.
Answer:

• Hugo de Vries of Holland
• Carl Correns of Germany
• Erich von Tschermark of Austria.

Question 16.
Write the advantages and disadvantages of Bt cotton.
Answer:
The advantages of Bt cotton are:

• Yield of cotton is increased due to effective control of bollworms.
• Reduction in insecticide use in the cultivation of Bt cotton
• Potential reduction in the cost of cultivation.

Bt cotton has some limitations:

• Cost of Bt cotton seed is high.
• Effectiveness upto 120 days after that efficiency is reduced.
• Ineffective against sucking pests like jassids, aphids and whitefly.
• Affects pollinating insects and thus yield.

Question 17.
How cryopreservation works?
Answer:
Cryopreservation, also known as Cryo-conservation, is a process by which protoplasts, cells, tissues, organelles, organs, extracellular matrix, enzymes or any other biological materials are subjected to preservation by cooling to very low temperature of -196°C using liquid nitrogen. At this extreme low temperature any enzymatic or chemical activity of the biological material will be totally stopped and this leads to preservation of material in dormant status. Later these materials can be activated by bringing to room temperature slowly for any experimental work.

Question 18.
What is Co-evolution? Give examples.
Answer:
The interaction between organisms, when continues for generations, involves reciprocal changes in genetic and morphological characters of both organisms. This type of evolution is called Co-evolution. It is a kind of co-adaptation and mutual change among interactive species.

Examples:

• Corolla length and proboscis length of butterflies and moths (Habenaria and Moth).
• Bird’s beak shape and flower shape and size.

Question 19.
Write a short note on Chipko Movement.
Answer:
Chipko Movement:
The tribal women of Himalayas protested against the exploitation of forests in 1972. Later on it transformed into Chipko Movement by Sundarlal Bahuguna in Mandal village of Chamoli district in 1974. People protested by hugging trees together which were felled by a sports goods company. Main features of Chipko movement were,

• This movement remained non political
• It was a voluntary movement based on Gandhian thought.
• It was concerned with the ecological balance of nature
• Main aim of Chipko movement was to give a slogan of five F’s – Food, Fodder, Fuel, Fibre and Fertilizer, to make the communities self sufficient in all their basic needs.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Name the tissue that nourishes the embryo in angiospermic seeds. Explain its types.
Answer:
Structure of ovuIe (Megasporangium)
Endosperm: The primary endosperm nucleus (PEN) divides immediately after fertilization but before the zygote starts to divide, into an endosperm. The primary endosperm nucleus is the result of triple fusion (two polar nuclei and one sperm nucleus) and thus has 3n number of chromosomes. It is a nutritive tissue and regulatory structure that nourishes the developing embryo. Depending upon the mode of development three types of endosperm are recognized in angiosperms. They are nuclear endosperm, cellular endosperm and helobial endosperm.

Nuclear endosperm: Primary Endosperm Nucleus undergoes several mitotic divisions without cell wall formation thus a free nuclear condition exists in the endosperm.
Examples: Coccinia, Capsella and Arachis.

Cellular endosperm:
Primary endosperm nucleus divides into 2 nuclei and it is immediately followed by wall formation. Subsequent divisions also follow cell wall formation.
Examples: Adoxa, Helianthus and Scoparia.

Helobial endosperm: Primary Endosperm Nucleus moves towards base of embryo sac and divides into two nuclei. Cell wall formation takes place leading to the formation of a large micropylar and small chalazal chamber. The nucleus of the micropylar chamber undergoes several free nuclear division whereas that of chalazal chamber may or may not divide. Examples : Hydrilla and Vallisneria.

Ruminate endosperm: The endosperm with irregularity and unevenness in its surface forms ruminate endosperm. Examples : Areca catechu, Passiflora and Myristica.

[OR]

(b) Describe the basic steps involved in recombinant DNA technology.
Answer:

• Isolation of a DNA fragment containing a gene of interest that needs to be cloned. This is called an insert.
• Generation of recombinant DNA (rDNA) molecule by insertion of the DNA fragment into a carrier molecule called a vector that can self-replicate within the host cell.
• Selection of the transformed host cells that is carrying the rDNA and allowing them to multiply thereby multiplying the rDNA molecule.
• The entire process thus generates either a large amount of rDNA or a large amount of protein expressed by the insert.
• Wherever vectors are not involved the desired gene is multiplied by PCR technique. The multiple copies are injected into the host cell protoplast or it is shot into the host cell protoplast by shot gun method.
  

Question 21.
(a) Derive the protocol for micro propagation of banana.
Answer:

[OR]

(b) Expand NBT and Explain how it is involved in developing new traits in plant breeding.
Answer:
New Breeding Techniques (NBT) are a collection of methods that could increase and accelerate the development of new traits in plant breeding. These techniques often involve genome editing, to modify DNA at specific locations within the plants to produce new traits in crop plants. The various methods of achieving these changes in traits include the following,

• Cutting and modifying the genome during the repair process by tools like CRISPR /Cas.
• Genome editing to introduce changes in few base pairs using a technique called Oligonucleotide-directed mutagenesis (ODM).
• Transferring a gene from an identical or closely related species (cisgenesis).
• Organising processes that alter gene activity without altering the DNA itself (epigenetic methods).

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Assertion (A): Organisms show three phases in their life cycle.
Reason (R): Juvenile phase is a degenerative phase.
(a) A is correct. R is incorrect
(b) Both A and R are incorrect
(c) R is the correct explantation for A
(d) A is not correct but R is correct.
Answer:
(a) A is correct. R is incorrect

Question 2.
Find the wrongly matched pair
(a) Bleeding phase – fall in oestrogen and progesterone
(b) Follicular phase – rise in oestrogen
(c) Luteal phase – rise in FSH level
(d) Ovulatory phase – LH surge
Answer:
(c) Luteal phase – rise in FSH level

Question 3.
Identify the correct statements from the following.
(a) chlamydiasis is a viral disease
(b) Gonorrhoea is caused by a spirochaete bacterium, Treponema palladium
(c) The incubation period for syphilis is 2 to 14 days is males and 7 to 21 days in males.
(d) Both syphilis and gonorrhoea are easily cured with antibiotics.
Answer:
(d) Both syphilis and gonorrhoea are easily cured with antibiotics.

Question 4.
Klinefelter’s syndrome is characterized by a karyotype of _______.
(a) XYY
(b) XO
(C) XXX
(d) XXY
Answer:
(d) XXY

Question 5.
The golden age of repetails was _______.
(a) Mesozoic era
(b) Cenozoic era
(c) Paleozoic era
(d) Proterozoic era
Answer:
(a) Mesozoic era

Question 6.
Match list I with list II

(a) A – 2, B – 4, C – 3, D – 1
(b) A – 4, B – 3, C – 2, D – 1
(c) A – 2, B – 3, C – 4, D – 1
(d) A – 3, B – 1, C – 4, D – 2
Answer:
(b) A – 4, B – 3, C – 2, D – 1

Question 7.
The relation ship between sucker fish and shark is ________.
(a) Competition
(b) Commensalism
(c) Predation
(d) Parasitism
Answer:
(b) Commensalism

Question 8.
Who is called as the forest man of India?
(a) Sunderlal Bahuguna
(b) M.S. Swamination
(c) Dr. V. Kurier
(d) Jadav Payeng
Answer:
(d) Jadav Payeng

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Mention the importance of the position of the testes in humans.
Answer:
The testes are positioned in such a way hanging out from the body in scrotal sac that provides optimal temperature 2°C to 3°C lower than internal body temperature for effective sperm production.

Question 10.
Type A blood should not be injected to a person having B-blood group. Why?
Answer:
When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells.

Question 11.
Name the anticodons required to recognize the following codons. AAU, CGA, UAU, GCA.
Answer:
UUA, GCU, AUA and CGU.

Question 12.
What is diapedesis?
Tissue damage and infection induce leakage of vascular fluid, containing chemotactic signals like serotonin, histamine and prostaglandins. They influx the phagocytic cells into the affected area. This phenomenon is called diapedesis.

Question 13.
How was Insulin obtained before the advent of rDNA technology? What were the problems encountered?
Answer:
Conventionally, Insulin was isolated and refined from the pancreas of pigs and cows to treat diabetic patients. Though it is effective, due to minor structural changes, the animal insulin caused allergic reactions in few patients.

Question 14.
How many hotspots are there is India? Name them.
Answer:
India encloses 4 biodiversity hotspots. They are

1. Himalayan
2. Indo-Burma
3. Western ghats
4. Sundalands

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Define surrogacy.
Answer:
Surrogacy is a method of assisted reproduction or agreement whereby a woman agrees to carry a pregnancy for another person, who will become the newborn child’s parent after birth. Through In Vitro Fertilization (IVF), embryos are created in a lab and are transferred into the surrogate mother’s uterus.

Question 16.
Differentiate between divergent evolution and convergent evolution with one example for each.
Answer:

Question 17.
Under which conditions does a bacterium develops resistance against antibiotics?
Answer:
Antibiotic resistance occurs when bacteria develop the ability to defeat the drug designed to kill or inhibit their growth. It is one of the most acute threat to public health. Antibiotic resistance is accelerated by the misuse and over use of antibiotics, as well as poor infection prevention control.

Antibiotics should be used only when prescribed by a certified health professional. When the bacteria become resistant, antibiotics cannot fight against them and the bacteria multiply. Narrow spectrum antibiotics are preferred over broad spectrum antibiotics. They effectively and accurately target specific pathogenic organisms and are less likely to cause resistance.

Question 18.
Expand (i) CFC (ii)AQI (w) PAN
Answer:
(i) CFC – Chloro fluro carbon
(ii) AQI – Air Quality Index
(iii) PAN – Peroxyacetyl nitrate .

Question 19.
Mention the number of primers required in each cycle of PCR. Write the role of primers arid DNA polymerase in PCR. Name the source organism of DNA polymerase used in PCR.
Answer:

• For each cycle of PCR two primers are required.
• Primers are the small fragments of single stranded DNA or RNA which serves as template for initiating DNA polymerization.
• DNA polymerase is an enzyme that synthesize DNA molecules by pairing the Deoxyribo Nucleotides leading to formation of new strands.
• DNA polymerase used in PCR is Taq polymerase which is isolated from a thermophilic bacteria called Thermus aquatics. Taq polymerase will remain active ever at very high temperature (80°C) and hence used in PCR amplification technique.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Menstruation – a cyclic event occurring in every normal woman throughout her fertile period. Name the various phases of the menstruation and explain it.
Answer:
Menstrual cycle: The menstrual or ovarian cycle occurs approximately once in every 28/29 days during the reproductive life of the female from menarche (puberty) to menopause except during pregnancy. The cycle of events starting from one menstrual period till the next one is called the menstrual cycle during which cyclic changes occurs in the endometrium every month. Cyclic menstruation is an indicator of normal reproductive phase.

Menstrual cycle comprises of the following phases:

1. Menstrual phase
2. Follicular or proliferative phase
3. Ovulatory phase
4. Luteal or secretory phase

1. Menstrual phase: The cycle starts with the menstrual phase when menstrual flow occurs and lasts for 3-5 days. Menstrual flow is due to the breakdown of endometrial lining of the uterus, and its blood vessels due to decline in the level of progesterone and oestrogen. Menstruation occurs only if the released ovum is not fertilized. Absence of menstruation may be an indicator of pregnancy. However it could also be due to stress, hormonal disorder and anaemia.

2. Follicular or proliferative phase: The follicular phase extends from the 5th day of the cycle until the time of ovulation. During this phase, the primary follicle in the ovary grows to become a fully mature Graafian follicle and simultaneously, the endometrium regenerates through proliferation. These changes in the ovary and the uterus are induced by the secretion of gonadotropins like FSH and LH, which increase gradually during the follicular phase. It stimulates follicular development and secretion of oestrogen by the follicle cells.

3. Ovulatory phase: Both LH and FSH attain peak level in the middle of the cycle (about the 14th day). Maximum secretion of LH during the mid cycle called LH surge induces the rupture of the Graafian follicle and the release of the ovum (secondary oocyte) from the ovary wall into the peritoneal cavity. This process is called as ovulation.

4. Luteal or secretory phase: During luteal phase, the remaining part of the Graafian follicle is transformed into a transitory endocrine gland called corpus luteum. The corpus luteum secretes large amount of progesterone which is essential for the maintenance of the endometrium. If fertilization takes place, it paves way for the implantation of the fertilized ovum. The uterine wall secretes nutritious fluid in the uterus for the foetus. So, this phase is also called as secretory phase. During pregnancy all events of menstrual cycle stop and there is no menstruation.

In the absence of fertilization, the corpus luteum degenerates completely and leaves a scar tissue called corpus albicans. It also initiates the disintegration of the endometrium leading to menstruation, marking the next cycle.

[OR]

(b) Deoxy Ribo Nucleic Acid the life thread which acts as a genetic material for majority of living organism. Enlist the properties of DNA that makes it an ideal genetic material.
Answer:
(1) Self Replication: It should be able to replicate. According to the rule of base pairing and complementarity, both nucleic acids (DNA and RNA) have the ability to direct duplications. Proteins fail to fulfill this criteria.

(2) Stability: It should be stable structurally and chemically. The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism. Stability as one of property of genetic material was clearly evident in Griffith’s transforming principle. Heat which killed the bacteria did not destroy some of the properties of genetic material. In DNA the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided.

Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

(3) Information storage: It should be able to express itself in the form of ‘Mendelian characters’. RNA can directly code for protein synthesis and can easily express the characters. DNA, however depends on RNA for synthesis of proteins. Both DNA and RNA can act as a genetic material, but DNA being more stable stores the genetic information and RNA transfers the genetic information.

(4) Variation through mutation: It should be able to mutate. Both DNA and RNA are able to mutate. RNA being unstable, mutates at a faster rate. Thus viruses having RNA genome with shorter life span can mutate and evolve faster. The above discussion indicates that both RNA and DNA can function as a genetic material. DNA is more stable, and is preferred for storage of genetic information.

Question 21.
(a) Explain the structure of Immunoglobulin molecule with a suitable diagram.
Answer:
An antibody molecule is Y shaped structure that comprises of four polypeptide chains, two identical light chains (L) of molecular weight 25,000 Da (approximately 214 amino acids) and two identical heavy chains (H) of molecular weight 50,000 Da (approximately 450 amino acids). The polypeptide chains are linked together by di-sulphide (S-S) bonds. One light chain is attached to each heavy chain and two heavy chains are attached to each other to form a Y shaped structure. Hence, an antibody is represented by H2 L2. The heavy chains have a flexible hinge region at their approximate middles.

Each chain (L and H) has two terminals. They are C – terminal (Carboxyl) and amino or N-terminal. Each chain (L and H) has two regions. They have variable (V) region at one end and a much larger constant (C) region at the other end.

Antibodies responding to different antigens have very different (V) regions but their (C) regions are the same in all antibodies. In each arm of the monomer antibody, the (V) regions of the heavy and light chains combines to form an antigen – binding site shaped to ‘fit’ a specific antigenic determinant.

Consequently each antibody monomer has two such antigen – binding regions. The (C) regions that forms the stem of the antibody monomer determine the antibody class and serve common functions in all antibodies.

[OR]

(b) List out the uses of Transgenesis.
Answer:
(1) Transgenesis is a powerful tool to study gene expression and developmental processes in
higher organisms.

(2) Transgenesis helps in the improvement of genetic characters in animals. Transgenic animals serve as good models for understanding human diseases which help in the investigation of new treatments for diseases. Transgenic models exist for many human diseases such as cancer, Alzheimer’s, cystic fibrosis, rheumatoid arthritis and sickle cell anemia.

(3) Transgenic animals are used to produce proteins which are important for medical and pharmaceutical applications.

(4) Transgenic mice are used for testing the safety of vaccines.

(5) Transgenic animals are used for testing toxicity in animals that carry genes which make them sensitive to toxic substances than non-transgenic animals exposed to toxic substances and their effects are studied.

(6) Transgenesis is important for improving the quality and quantity of milk, meat, eggs and wool production in addition to testing drug resistance.

Students can Download Tamil Nadu 12th Biology Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Biology Model Question Paper 2 English Medium

General Instructions:

1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
In majority of plants, pollen is liberated at _______.
(a) 1 celled stage
(b) 2 celled stage
(c) 3 celled stage
(d) 4 celled stage
Answer:
(b) 2 celled stage

Question 2.
Which one of the following is an example for polygenic inheritance?
(a) Flower color in Mirabilis jalapa
(b) Pod shape in garden pea
(c) Production of male honey bee
(d) Skin color in humans
Answer:
(d) Skin color in humans

Question 3.
Assertion (A): Complete linkage is noticed in male species of Drosophila.
Reason (R): Completely linked genes show some crossing over.
(a) A is true, R is false
(b) Both A and R are false
(c) A is true, R is not correct explanation of A
(d) R explains A
Answer:
(a) A is true, R is false

Question 4.
Virus free germ plants are developed from _______.
(a) Organ culture
(b) Meristem culture
(c) Protoplast culture
(d) Cell suspension culture
Answer:
(b) Meristem culture

Question 5.
The unit of measuring ozone thickness is ______.
(a) Joule
(b) Kilos
(c) Dobson
(d) Watt
Answer:
(c) Dobson

Question 6.
If 1200 Joules of solar energy is trapped by producers, how much of joules of energy does the organism in the third tropic level will receive?
(a) 120 joules
(b) 12 joules
(c) 1.2 joules
(d) 0.12 joules
Answer:
(c) 1.2 joules

Question 7.
Dwarfing gene of wheat is ______.
(a) Pal 1
(b) Atomita 1
(c) Norin 10
(d) Pelita 2
Answer:
(c) Norin 10

Question 8.
Which one of the following match is correct.
(a) Palmyra – Native of Brazil
(b) Saccharum – Abundant in Kanyakumari
(c) Steveocide – Natural sweetner
(d) Palmyra sap – fermented to give ethanol
Answer:
(c) Steveocide – Natural sweetner

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Write the origin and area of cultivation of green gram and red gram.
Answer:

Question 10.
State the law of independent assortment.
Answer:
When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent to the other pair of characters. Genes that are located in different chromosomes assort independently during meiosis.

Question 11.
Expand (i) PEG (ii) PHB
Answer:
(i) PEG – Poly Ethylene Glycol
(ii) PHB – Poly Hydroxy Butyrate

Question 12.
Based on the materials used, how will you classify the culture technology. Explain.
Answer:
Based on explants used culture technology are of following types:
Organ culture – Embryos, anthers, root and shoot part are used.
Meristem culture – Meristematic tissues are used.
Protoplast culture – Protoplasts are used.
Cell culture – Single cells or aggregate of cells from callus are used.

Question 13.
Give four examples of plants cultivated in commercial agroforestry.
Answer:
Casuarina, Eucalyptus, Teak, Malaivembu

Question 14.
How does an orchid ophrys ensures its pollination by bees?
Ahnswer:
The plant, Ophrys an orchid, the flower looks like a female insect to attract the male insect to get pollinated by the male insect and it is otherwise called ‘floral mimicry’.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
List out the objectives of plant breeding.
Answer:

• To increase yield, vigour and fertility of the crop.
• To increase tolerance to environmental condition, salinity, temperature and drought.
• To prevent the premature falling of buds and fruits, etc.
• To improve synchronous maturity.
• To develop resistance to pathogens and pests.
• To develop photosensitive and thermos-sensitive varieties.

Question 16.
Spindle shaped pyramid of number is noticed in forest ecosystem. Give Reasons.
Answer:
In a forest ecosystem the pyramid of number is spindle in shape, it is because the base (T₁) of the pyramid occupies large sized trees (Producer) which are lesser in number. Herbivores (T₂) (Fruit eating birds, elephant and deer) occupying second trophic level, are more in number than the producers. In final trophic level (T₄), tertiary consumers (lion) are lesser in number than the secondary consumer (T₃) (fox and snake).

Question 17.
Distinguish between mound layering and air layering.
Answer:

Question 18.
Explain the sex determination mechanism in Carica papaya.
Answer:

Carica papaya, 2n = 36 (Papaya) has 17 pairs of autosomes and one pair of sex chromosomes. Male papaya plants have XY and female plants have XX. Unlike human sex chromosomes, papaya sex chromosomes look like autosomes and it is evolved from autosome. The sex chromosomes are functionally distinct because the Y chromosome carries the genes for male organ development and X bears the female organ developmental genes. In papaya sex determination is controlled by three alleles. They are m, M₁ and M₂ of a single gene.

Question 19.
Write the protocol for artificial seed preparation.
Answer:

Later these seeds are grown in vitro medium and converted into piantiets. These piantiets require a hardening period (either green house or hardening chamber) and then shifted to normal environment condition.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Bring out the inheritance of chloroplast gene with on example.
Answer:
Chloroplast Inheritance:
It is found in 4 O’ Clock plant (Mirabilis jalapa). In this, there are two types of variegated leaves namely dark green leaved plants and pale green leaved plants. When the pollen of dark green leaved plant (male) is transferred to the stigma of pale green leaved plant (female) and pollen of pale green leaved plant is transferred to the stigma of dark green leaved plant, the F, generation of both the crosses must be identical as per Mendelian inheritance.
Chloroplast inheritance.

But in the reciprocal cross the F₁ plant differs from each other. In each cross, the F₁ plant reveals the character of the plant which is used as female plant.

This inheritance is not through nuclear gene. It is due to the chloroplast gene found in the ovum of the female plant which contributes the cytoplasm during fertilization since the male gamete contribute only the nucleus but not cytoplasm.

[OR]

(b) Explain in detail about various types of direct gene transfer method.
Answer:

(1) Chemical mediated gene transfer:
Certain chemicals like polyethylene glycol (PEG) and dextran sulphate induce DNA uptake into plant protoplasts.

(2) Microinjection: The DNA is directly injected into the nucleus using Electroporation Methods of Gene Transfer fine tipped glass needle or micro pipette to transform plant cells. The protoplasts are immobilised on a solid support (agarose on a microscopic slide) or held with a holding pipette under suction.

(3) Electroporation Methods of Gene Transfer: A pulse of high voltage is applied to protoplasts, cells or tissues which makes transient pores in the plasma membrane through which uptake of foreign DNA occurs.

(4) Liposome mediated method of Gene Transfer: Liposomes the artificial phospholipid vesicles are useful in gene transfer. The gene or DNA is transferred from liposome into vacuole of plant cells. It is carried out by encapsulated DNA into the vacuole. This technique is advantageous because the liposome protects the introduced DNA from being damaged by the acidic pH and protease enzymes present in the vacuole. Liposome and tonoplast of vacuole fusion resulted in gene transfer. This process is called lipofection.

(5) Biolistics: The foreign DNA is coated onto the surface of minute gold or tungsten particles (1-3 mm) and bombarded onto the target tissue or cells using a particle gun (also called as gene gun/micro projectile gun/shotgun). Then the bombarded cells or tissues are cultured on selected medium to regenerate plants from the transformed cells.

Question 21.
(a) Explain the steps involved in protoplast culture.
Answer:
Protoplasts are cells without a cell wall, but bounded by a cell membrane or plasma membrane. Using protoplasts, it is possible to regenerate whole plants from single cells and also develop somatic hybrids. The steps involved in protoplast culture are:

(1) Isolation of protoplast: Small bits of plant tissue like leaf tissue are used for isolation of protoplast. The leaf tissue is immersed in 0.5% Macrozyme and 2% Onozuka cellulase enzymes dissolved in 13% sorbitol or mannitol at pH 5.4. It is then incubated over-night at 25°C. After a gentle teasing of cells, protoplasts are obtained, and these are then transferred to 20% sucrose solution to retain their viability. They are then centrifuged to get pure protoplasts as different from debris of cell walls.

(2) Fusion of protoplast: It is done through the use of a suitable fusogen. This is normally PEG (Polyethylene Glycol). The isolated protoplast are incubated in 25 to 30% concentration of PEG with Ca++ ions and the protoplast shows agglutination (the formation of clumps of cells) and fusion.

(3) Culture of protoplast: MS liquid medium is used with some modification in droplet, plating or micro-drop array techniques. Protoplast viability is tested with fluorescein diacetate before the culture. The cultures are incubated in continuous light 1000-2000 lux at 25 °C. The cell wall formation occurs within 24-48 hours and the first division of new cells occurs between 2-7 days of culture.

(4) Selection of somatic hybrid cells: The fusion product of protoplasts without nucleus of different cells is called a hybrid. Following this nuclear fusion happen. This process is called somatic hybridization.

[OR]

(b) Write a note on Henna.
Answer:
Botanical name: Lawsonia inermis.
Family: Lythraceae.
Origin and Area of cultivation: It is indigenous to North Africa and South-west Asia. It is grown mostly throughout India, especially in Gujarat, Madhya Pradesh and Rajasthan.

Uses:
An orange dye ‘Henna’ is obtained from the leaves and young shoots of Lawsonia inermis. The principal colouring matter of leaves ‘lacosone’ is harmless and causes no irritation to the skin. This dye has long been used to dye skin, hair and finger nails. It is used for colouring leather, for the tails of horses and in hair-dyes.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Animals giving birth to young ones are ______.
(a) Oviparous
(b) Ovoviviparous
(c) Viviparous
(d) Both a and b
Answer:
(c) Viviparous

Question 2.
Messelson and Stahl’s experiment proved ________.
(a) Transduction
(b) Transformation
(c) DNA is the genetic material
(d) Semi-conservative nature of DNA reflication.
Answer:
(d) Semi-conservative nature of DNA reflication.

Question 3.
Match column I with column II

(a) A – (iv), B – (ii), C – (i), D – (iii)
(b) A – (iv), B – (i), C – (iii), D – (ii)
(c) A – (i), B – (iv), C – (ii), D – (iii)
(d) A – (iv), B – (i), C – (ii), D – (iii)
Answer:
(d) A – (iv), B – (i), C – (ii), D – (iii)

Question 4.
Choose the correctly matched pair.
(a) Amphetamines – Stimulant
(b) LSD – Narcotic
(c) Heroin – Psychotropic
(d) Benzodiazepine – Pain killer
Answer:
(a) Amphetamines – Stimulant

Question 5.
The first clinical gene therapy was done for the treatment of ________.
(a) AIDS
(b) Cancer
(c) Cystic fibrosis
(d) SCID
Answer:
(d) SCID

Question 6.
Some organisms are able to maintain homeostasis by physical means ________.
(a) Conform
(b) Regulate
(c) Migrate
(d) Suspend
Answer:
(b) Regulate

Question 7.
Select the correct linear equation describing the species area relationship?
(a) log C = log S + Z log A
(b) Z log A = log S + log C
(c) log S = log C + Z log A
(d) log C = log S ± Z log C
Answer:
(b) Z log A = log S + log C

Question 8.
Oil strains in laundry can be removed using _______.
(a) Peptidane
(b) Protease
(c) Amylase
(d) Lipase
Answer:
(d) Lipase

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
The unicellular organisms which reproduce by binary fission are considered immortal. Justify.
Answer:
In unicellular organisms during binary fission, the entire cell (organism) divides completely to form two daughter cells which later on develop into adult and the process goes on repeatedly during each division leading to immortality of cell (organism). Hence unicellular organisms like amoeba are ‘biologically immortal’.

Question 10.
What is colostrum? Write its significance.
Answer:
The mammary glands secrete a yellowish fluid called colostrum during the initial few days after parturition. It has less lactose than milk and almost no fat, but it contains more proteins, vitamin A and minerals. Colostrum is also rich in IgA antibodies. This helps to protect the infant’s digestive tract against bacterial infection.

Question 11.
Define haplodiploidy.
Answer:
In haplodiploidy, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid).

Question 12.
Mention the main objections to Darwinism.
Answer:
Some objections raised against Darwinism were –

• Darwin failed to explain the mechanism of variation.
• Darwinism explains the survival of the fittest but not the arrival of the fittest.
• He focused on small fluctuating variations that are mostly non-heritable.
• He did not distinguish between somatic and germinal variations.
• He could not explain the occurrence of vestigial organs, over specialization of some organs like large tusks in extinct mammoths and over sized antlers in the extinct Irish deer, etc.

Question 13.
Differentiate between cell mediated Immunity and Antibody Mediated Immunity.
Answer:

Question 14.
What are attenuated recombinant vaccines?
Answer:
Attenuated recombinant vaccines includes genetically modified pathogenic organisms (bacteria or viruses) that are made nonpathogenic and are used as vaccines. Such vaccines are referred to as attenuated recombinant vaccines.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Alien species invasion is a threat to endemic species – substantiate this statement.
Answer:
Exotic species are organisms often introduced unintentionally or deliberately for commercial purpose, as biological control agents and other uses. They often become invasive and drive away the local species and is considered as the second major cause for extinction of species. Tilapia fish (Jilabi kendai) (Oreochromis mosambicus) introduced from east coast of South Africa in 1952 for its high productivity into Kerala’s inland waters, became invasive, due to which the native species such as Puntius dubius and Labeo kontius face local extinction.

Amazon sailfin catfish is responsible for destroying the fish population in the wetlands of Kolkata. The introduction of the Nile Perch, a predatory fish into Lake Victoria in East Africa led to the extinction of an ecologically unique assemblage of more than 200 nature species of cichlid fish in the lake.

Question 16.
In what way Peyang conserved the forests?
Answer:
The ‘Forest man of India’, Jadav Payeng who created 1,360 acres of dense and defiant forest was born in Arunasapori (a river island on the Brahmaputra). He had just completed his Class X exams in 1979 when he started to sow the seeds and shoots on the eroded island covered with sand and silt. Thirty-six years later he had converted the once unproductive land into a forest.

Payeng’s forest is now home to five Royal Bengal tigers, over a hundred deer, wild boar, vultures, and several species of birds. For his remarkable initiative, the Jawaharlal Nehru University invited Payeng in 2012 on Earth Day and honoured him with the title of the ‘ Forest Man of India ’.

Question 17.
Amniocentesis, the foetal sex determination test, is banned in our country. Is it necessary comment?
Answer:
Amniocentesis is a prenatal technique used to detect any chromosomal abnormalities in the foetus and it is being often misused to determine the sex of the foetus. Once the sex of the foetus is known, there may be a chance of female foeticide. Hence, a statutory ban on amniocentesis is imposed.

Question 18.
Write the objectives of Human Genome project.
Answer:
The main goals of Human Genome Project are as follows:

• Identify all the genes (approximately 30000) in human DNA.
• Determine the sequence of the three billion chemical base pairs that makeup the human DNA
• To store this information in databases.
• Improve tools for data analysis.
• Transfer related technologies to other sectors, such as industries.
• Address the ethical, legal and social issues (ELSI) that may arise from the project.

Question 19.
Explain the role of cry-genes in genetically modified crops.
Answer:
Bacillus thuringiensis is a soil dwelling bacterium which is commonly used as a biopesticide and contains a toxin called cry toxin. Scientists have introduced this toxin producing genes into cotton and have raised genetically engineered insect resistant cotton plants.

During sporulation Bacillus thuringiensis produces crystal proteins called Delta-endotoxin which is encoded by cry genes. Delta-endotoxins have specific activities against the insects of the orders Lepidoptera, Diptera, Coleoptera and Hymenoptera. When the insects ingest the toxin crystals their alkaline digestive tract denatures the insoluble crystals making them soluble. The cry toxin then gets inserted into the gut cell membrance and paralyzes the digestive tract. The insect then stops eating and starves to death.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Describe the structure of human spermatozoa with a labelled diagram.
Answer:

The human sperm is a microscopic, flagellated and motile gamete. The whole body of the sperm is enveloped by plasma membrane and is composed of a head, neck and a tail. The head comprises of two parts namely acrosome and nucleus. Acrosome is a small cap like pointed structure present at the tip of the nucleus and is formed mainly from the Golgi body of the spermatid.

It contains hyaluronidase, a proteolytic enzyme, popularly known as sperm lysin which helps to penetrate the ovum during fertilization. The nucleus is flat and oval. The neck is very short and is present between the head and the middle piece. It contains the proximal centriole towards the nucleus which plays a role in the first division of the zygote and the distal centriole gives rise to the axial filament of the sperm. The middle piece possesses mitochondria spirally twisted around the axial filament called mitochondrial spiral or nebenkem.

It produces energy in the form of ATP molecules for the movement of sperms. The tail is the longest part of the sperm and is slender and tapering. It is formed of a central axial filament or axoneme and an outer protoplasmic sheath. The lashing movements of the tail push the sperm forward.

[OR]

(b) Explain the Mechanism of ‘lac’ – operon of the E-coli.
Answer:
The Lac (Lactose) operon: The metabolism of lactose in E.coli requires three enzymes-permease, P-galactosidase (β-gal) and transacetylase. The enzyme permease is needed for entry of lactose into the cell, P-galactosidase brings about hydrolysis of lactose to glucose and galactose, while transacetylase transfers acetyl group from acetyl Co A to β-galactosidase.

The lac operon consists of one regulator gene (‘i’ gene refers to inhibitor) promoter sites (p), and operator site (o). Besides these, it has three structural genes namely lac z, y and lac a. The lac ‘z’ gene codes for β-galactosidase, lac ‘y’ gene codes for permease and ‘a’ gene codes for transacetylase. Jacob and Monod proposed the classical model of Lac operon to explain gene expression and regulation in E.coli.

In lac operon, a polycistronic structural gene is regulated by a common promoter and regulatory gene. When the cell is using its normal energy source as glucose, the ‘i’ gene transcribes a repressor mRNA and after its translation, a repressor protein is produced. It binds to the operator region of the operon and prevents translation, as a result, β-galactosidase is not produced. In the absence of preferred carbon source such as glucose, if lactose is available as an energy source for the bacteria then lactose enters the cell as a result of permease enzyme. Lactose acts as an inducer and interacts with the repressor to inactivate it.

The repressor protein binds to the operator of the operon and prevents RNA polymerase from transcribing the operon. In the presence of inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to bind to the promotor site and transcribe the operon to produce lac mRNA which enables formation.

Question 21.
(a) Explain stabilizing, directional and disruptive selection with examples.
Answer:
(1) Stabilising selection (centipetal selection): This type of selection operates in a stable environment. The organisms with average phenotypes survive whereas the extreme individuals from both the ends are eliminated. There is no speciation but the phenotypic stability is maintained within the population over generation. For example, measurements of sparrows that survived the storm clustered around the mean, and the sparrows that failed to survive the storm clustered around the extremes of the variation showing stabilizing selection.

(2) Directional Selection: The environment which undergoes gradual change is subjected to directional selection. This type of selection removes the individuals from one end towards the other end of phenotypic distribution. For example, size differences between male and female sparrows. Both male and female look alike externally but differ in body weight. Females show directional selection in relation to body weight.

(3) Disruptive selection: (centrifugal selection) When homogenous environment changes into heterogenous environment this type of selection is operational. The organisms of both the extreme phenotypes are selected, whereas individuals with average phenotype are eliminated. This results in splitting of the population into sub population/species.

This is a rare form of selection but leads to formation of two or more different species. It is also called adaptive radiation. (e.g:) Darwin’s finches beak size in relation to seed size inhabiting Galapagos islands. Group selection and sexual selection are other types of selection. The two major group selections are Altrusim and Kin selection.

Operation of natural selection on different traits (a) Stablishing (b) Directional and (c) Disruptive

[OR]

(b) Tabulate the various types of innate immunity and their action mechanism.
Answer:

Students can Download Tamil Nadu 12th Physics Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Physics Model Question Paper 4 English Medium

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III and IV are to be attempted separately
4. Question numbers 1 to 15 in Part I are Multiple choice Questions of one mark each.  These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks : 70

PART -1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
Two identical conducting balls having positive charges q₁ and q₂ are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be …………………… .
(a) less than before
(b) same as before
(c) more than before
(d) zero
Answer:
(c) more than before

Question 2.
Two plates are 1 cm apart and the potential difference between them is 10 V. The electric field between the plates is …………………. .
(a) 10 NC^-1
(b) 250 NC^-1
(c) 500 NC^-1
(d) 1000 NC^-1
Hint: E = \(\frac{V}{d}=\frac{10}{1 \times 10^{-2}}\) = 1000 NC^-1
Answer:
(d) 1000 NC^-1

Question 3.
A toaster operating at 240 V has a resistance of 120 Ω. The power is ………………… .
(a) 400 W
(b) 2 W
(c) 480 W
(d) 240 W
Answer:
(c) 480 W

Question 4.
Three wires of equal lengths are bent in the form of loops. One of the loops is circle, another is a semi-circle and the third one is a square. They are placed in a uniform magnetic field and same electric current is passed through them. Which of the following loop configuration will experience greater torque ?
(a) circle
(b) semi-circle
(c) square
(d) all of them
Answer:
(a) circle

Question 5.
A bar magnet of magnetic moment M is cut into two parts of equal length. The magnetic moment of either part is ………………….. .
(a) M
(b) 2M
(c) \(\frac{M}{2}\)
(d) Zero
Answer:
(c) \(\frac{M}{2}\)

Question 6.
The current flowing in a coil varies with time as shown in ® the figure. The variation of induced emf with time would be …………….. .


Answer:

Question 7.
Which of the following electromagnetic radiation is used for viewing objects through fog?
(a) microwave
(b) gamma rays
(c) X- rays
(d) infrared
Answer:
(d) infrared

Question 8.
Stars twinkle due to …………………. .
(a) reflection
(b) total internal reflection
(c) refraction
(d) polarisation
Answer:
(c) refraction

Question 9.
When a plane electromagnetic wave enters a glass slab, then which of the following will not change?
(a) Wavelength
(b) Frequency
(c) Speed
(d) Amplitude
Hint: Only the frequency of the electromagnetic wave remains unchanged.
Answer:
(b) Frequency

Question 10.
In an electron microscope, the electrons are accelerated by a voltage of 14 kV. If the voltage is changed to 224 kV, then the de Broglie wavelength associated with the electrons would …………………… .
(a) increase by 2 times
(b) decrease by 2 times
(c) decrease by 4 times
(d) increase by 4 times
Hint: At Voltage, V = 14 kV
de-Broglie wavelength of electron, λ = \(\frac{12.3}{\sqrt{14000}}\) Å = 0.104 Å
At voltage, V = 224 kV
λ’ = \(\frac{12.3}{\sqrt{224000}}\) Å = 0.026 Å
\(\frac{\lambda}{\lambda^{\prime}}=\frac{0.104}{0.0260}\) = 4 ⇒ λ = 4λ’ ⇒ λ’ = \(\frac{\lambda}{4}\)
Answer:
(c) decrease by 4 times

Question 11.
The charge of cathode rays is …………………. .
(a) positive
(b) negative
(c) neutral
(d) not defined
Answer:
(a) positive
(b) negative

Question 12.
The primary use of a zener diode is …………………… .
(a) Rectifier
(b) Amplifier
(c) Oscillator
(d) Voltage regulator
Answer:
(d) Voltage regulator

Question 13.
For a common base circuit if \(\frac{I_{C}}{I_{E}}\) = 0.98, then current gain for common emitter circuit will be ………………. .
(a) 49
(b) 98
(c) 4.9
(d) 25.5
Ir a 0 98

Answer:
(a) 49

Question 14.
The internationally accepted frequency deviation for the purpose of FM broadcasts.
(a) 75 kHz
(b) 68 kHz
(c) 80 kHz
(d) 70 kHz
Answer:
(a) 75 kHz

Question 15.
“Sky wax” is an application of nano product in the field of …………………….. .
(a) Medicine
(b) Textile
(c) Sports
(d) Automotive industry
Answer:
(c) Sports

PART – II

Answer any six questions. Question No. 23 is compulsory. [6 × 2 = 12]

Question 16.
The electric field lines never intersect. Justify.
Answer:
As a consequence, if some charge is placed in the intersection point, then it has to move in two different directions at the same time, which is physically impossible. Hence, electric field lines do not intersect.

Question 17.
A potential difference across 24 ft resistor is 12 V. What is the current through the resistor?
Answer:
V = 12 V
and R = 24 Ω
Current, I = ?
From Ohm’s law, \(I=\frac{V}{R}=\frac{12}{24}=0.5 \mathrm{A}\)

Question 18.
State Coulomb’s inverse law.
Answer:
The force of attraction or repulsion between two magnetic poles is directly proportional to the product of their pole strengths and inversely proportional to the square of the distance between ‘ them.
\(\overrightarrow{\mathrm{F}} \propto \frac{q_{m_{\mathrm{A}}} q_{m_{b}}}{r^{2}} \hat{r}\)

Question 19.
State Lenz’s law.
Answer:
Lenz’s law states that the direction of the induced current is such that it always opposes the cause responsible for its production.

Question 20.
What is angle of deviation due to reflection?
Answer:
The angle between the incident and deviated light ray is called angle of deviation of the light ray. It is written as, d = 180 – (i + r). As, i = r in reflection, we can write angle of deviation in  reflection at plane surface as. d = 180 – 2i

Question 21.
What is photoelectric effect?
Answer:
The ejection of electrons from a metal plate when illuminated by light or any other electromagnetic radiation of suitable wavelength (or frequency) is called photoelectric effect.

Question 22.
Define impact parameter.
Answer:
The impact parameter is defined as the perpendicular distance between the centre of the gold nucleus and the direction of velocity vector of alpha particle when it is at a large distance.

Question 23.
In a transistor connected in the common base configuration, α = 0.95, I_E = mA. Calculate the values of I_c and I_B
Answer:
α = \(\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{E}}}\)
I_c = αI_E = 0.95 × 1 = 0.95 mA
I_E = I_B + I_c
∴ I_B = I_c – I_E = 1 – 0.95 = 0.05 mA

Question 24.
What do you mean by Internet of Things?
Answer:
Internet of Things (IoT), it is made possible to control various devices from a single device. Example: home automation using a mobile phone.

PART-III

Answer any six questions. Question No. 28 is compulsory. [6 × 3 = 18]

Question 25.
Write down Coulomb’s law in vector form and mention what each term represents.
Answer:
The force on a charge q₁ exerted by a point charge q₂ is given by
\(\overrightarrow{\mathrm{F}}_{12}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \hat{r}_{21}\)
Here r̂₂₁ is the unit vector from charge q₂ to q₁
But r̂₂₁ = -r̂₂₁

Therefore, the electrostatic force obeys Newton’s third law.

Question 26.
Write down the various forms of expression for power in electrical circuit.
Answer:
The electric power P is the rate at which the electrical potential energy is delivered,
\(P=\frac{d U}{d t}=\frac{1}{d t}(V \cdot d Q)=V \cdot \frac{d Q}{d t}\)
[dU = V.dQ]
The electric power delivered by the battery to any electrical system.
P = VI
The electric power delivered to the resistance R is expressed in other forms.
P = VI = I(IR) = I²R
P = IV = \(\left(\frac{V}{R}\right) V=\frac{V^{2}}{R}\)

Question 27.
The repulsive force between two magnetic poles in air is 9 × 10^-3 N. If the two poles are equal in strength and are separated by a distance of 10 cm, calculate the pole strength of each pole.
Answer:
The force between two poles are given by \(\overrightarrow{\mathrm{F}}=k \frac{q_{m_{\mathrm{A}}} q_{m_{\mathrm{B}}}}{r^{2}}\)
The magnitude of the force is F = \(F=k \frac{q_{m_{A}} q_{m_{B}}}{r^{2}}\)
Given : F = 9 × 10^-3 N, r = 10 cm = 10 × 10^-2 m
Therefore 9 × 10^-3 = 10^-7 × \(\frac{q_{m}^{2}}{\left(10 \times 10^{-2}\right)^{2}}\) ⇒ q_m = 30NT^-1

Question 28.
A 200 turn coil of radius 2 cm is placed co-axially within a long solenoid of 3 cm radius. If the turn density of the solenoid is 90 turns per cm, then calculate mutual inductance of the coil.
Answer:
Numbef of turns of the solenoid, N₂ = 200
Radius of the solenoid, r cm = 2 × 10^-2 m
Area of the solenoid, A = πr² = 3.14 × (2 × 10^-2)^-2 = 1.256 × 10^-3 m²
Turn density of long solenoid per cm, N₁ = 90 × 10²

= 283956.48 × 10^-8 ⇒ M = 2.84 mH

Question 29.
Compute the speed of the electromagnetic wave in a medium if the amplitude of electric and magnetic fields are 3 × 10⁴ N C^-1 and 2 × 10^-4 T, respectively.
Answer:
The amplitude of the electric field, E₀ = 3 × 10⁴ N C^-1.
The amplitude of the magnetic field, B_o = 2 × 10^-4 T. Therefore, speed of the electromagnetic wave in a medium is
\(v=\frac{3 \times 10^{4}}{2 \times 10^{-4}}=1.5 \times 10^{8} \mathrm{ms}^{-1}\)

Question 30.
State the laws of refraction.
Answer:
Law of refraction is called Snell’s law.
Snell’s law states that,
(a) The incident ray, refracted ray and normal to the refracting surface are all coplanar (i.e. lie in the same plane).
(b) The ratio of angle of incident i in the first medium to the angle of reflection r in the second medium is equal to the ratio of refractive index of the second medium n₂ to that of the refractive index of the first medium n₁.
\(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)

Question 31.
A proton and an electron have same de Broglie wavelength. Which of them moves faster and which possesses more kinetic energy?
Answer:
We know that λ = \(\frac{h}{\sqrt{2 m K}}\)
Since proton and electron have same de Broglie wavelength, we get

Since m_e < m_p , K_p < K_e, the electron has more kinetic energy than the proton.

Since m_e < m_p ν_p < ν_e , the electron moves faster than the proton.

Question 32.
In alpha decay, why the unstable nucleus emits ⁴₂He nucleus? Why it does not emit four separate nucleons?
Answer:
After all ⁴₂He consists of two protons and two neutrons. For example, if ²³⁸₉₂U nucleus decays into ²³⁴₉₀Th by emitting four separate nucleons (two protons and two neutrons), then the disintegration energy Q for this process turns out to be negative. It implies that the total mass of products is greater than that of parent (²³⁸₉₂U) nucleus. This kind of process cannot occur in nature because it would violate conservation of energy. In any decay process, the conservation of energy, conservation of linear momentum and conservation of angular momentum must be obeyed.

Question 33.
Distinguish between intrinsic and extrinsic semiconductors.
Answer:

PART – IV

Answer all the questions. [5 × 1 = 5]

Question 34.
(a) Derive an expression for electrostatic potential due to an electric dipole.
Answer:
Electrostatic potential at a point due to an electric dipole: Consider two equal and opposite charges separated by a small distance 2a. The point P is located at a distance r from the midpoint of the dipole. Let θ be the angle between the line OP and dipole axis AB.
Let r₁ be the distance of point P from +q and r₂ be the distance of point P from -q.
Potential at P due to charge +q = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{1}}\)
Potential at P due to charge -q = \(-\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{2}}\)
Total potential at the point P,
V = \(\frac{1}{4 \pi \varepsilon_{0}} q\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)\) ……… (1)
Suppose if the point P is far away from the dipole, such that r>>a, then equation can be expressed in terms of r. By the cosine law for triangle BOP,

r²₁ = r² + a² – 2ra cos θ = r²\(\left(1+\frac{a^{2}}{r^{2}}-\frac{2 a}{r} \cos \theta\right)\)

Since the point P is very far from dipole, then r >> a. As a result the term \(\frac{a^{2}}{r^{2}}\) is very small and can be neglected. Therefore

since \(\frac{a}{r}\)<< 1, we can use binominal theorem and retain the terms up to first order r
\(\frac{1}{r_{1}}=\frac{1}{r}\left(1+\frac{a}{r} \cos \theta\right)\) …… (2)
Similarly applying the cosine law for triangle AOP,
r₂² = r² + a² – 2ra cos(180 – θ) since cos (180 – θ) = -cos θ we get
r₂² = r² + a² + 2ra cos θ
Neglecting the term \(\frac{a^{2}}{r^{2}}\) (because r >> a)

Using Binomial theorem, we get
\(\frac{1}{r_{2}}=\frac{1}{r}\left(1-a \frac{\cos \theta}{r}\right)\) ………. (3)
Substituting equations (3) and (2) in equation (1)

But the electric dipole moment p = 2qa and we get,
\(\mathrm{V}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{p \cos \theta}{r^{2}}\right)\)
Now we can write p cos θ = \(\vec{r}\) . r̂ where r̂ is the unit vector from the point O to point P.
Hence the electric potential at a point P due to an electric dipole is given by
V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\vec{p} \cdot \hat{r}}{r^{2}}\) (r >> a) …(4)
Equation (4) is valid for distances very large compared to the size of the dipole. But for a point dipole, the equation (4) is valid for any distance.

Special cases:
Case (1) If the point P lies on the axial line of the dipole on the side of +q, then θ = 0. Then the electric potential becomes
\(V=\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{2}}\)

Case (ii) If the point Plies on the axial line of the dipole on the side of -q. then θ = 180°, then
\(\mathrm{V}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{2}}\)

Case (iii) If the point P lies on the equatorial line of the dipole, then θ = 90°. Hence, V = 0.

Question 34.
(b) Explain the determination of the internal resistance of a cell using voltmeter.
Answer:
Determination of internal resistance: The emf of cell ξ is measured by connecting a high resistance voltmeter across it without connecting the external resistance R. Since the voltmeter draws very little current for deflection, the circuit may be considered as open. Hence, the voltmeter reading gives the emf of the ceíl. Then, external resistance R is included in the circuit and current I is established in the circuit. The potential difference across R is equal to the potential difference across the cell (V).

The potential drop across the resistor R is
V = IR …….. (1)

Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell ξ. It is because, certain amount of voltage (Ir) has dropped across the internal resistance r.
Then V = ξ – Ir
Ir = ξ – V ……… (2)
Dividing equation (2) by equation (1), we get
\(\frac{I r}{I R}=\frac{\xi-V}{V}\)
\(r=\left|\frac{\xi-V}{V}\right| R\) ……… (3)
since ξ, V and R are known, internal resistance r can be determined

Question 35.
(a) Calculate the magnetic induction at a point on the axial line of a bar magnet.
Answer:
Magnetic field at a point along the axial line of the magnetic dipole (bar magnet): Consider a bar magnet NS. Let N be the North Pole and S be the south pole of the bar magnet, each of pole strength q, and separated by a distance of 2l. The magnetic field at a point C (lies along the axis of the magnet) at a distance from the geometrical center O of the bar magnet
can be computed by keeping unit north pole (q_MC = 1 A m) at C. The force experienced by the unit north pole at C due to pole strength can be computed using Coulomb’s law of magnetism as follows:
The force of repulsion between north pole of the bar magnet and unit north pole at point C (in free space) is
\(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=\frac{\mu_{0}}{4 \pi} \frac{q_{m}}{(r-l)^{2}} \hat{l}\) …. (1)
where r – l is the distance between north pole of the bar magnet and unit north pole at C.
The force of attraction between South Pole of the bar magnet and unit North Pole at point C (in free space) is
\(\overrightarrow{\mathrm{F}}_{\mathrm{S}}=-\frac{\mu_{0}}{4 \pi} \frac{q_{m}}{(r+l)^{2}} \hat{l}\) ……. (2)
where r + l is the distance between South pole of the bar magnet and unit north pole at C.

From equation (1) and (2), the net force at point C is \(\vec{F}\) = \(\vec{F}\)_N + \(\vec{F}\)_s . From definition, this net force is the magnetic field due to magnetic dipole at a point C (\(\vec{F}\) = \(\vec{B}\))

Since, magnitude of magnetic dipole moment is |\(\vec{P}\)_m| = P_m = q_m .2l the magnetic field point C equation (3) can be written as
\(\vec{B}\)_axial = \(\frac{\mu_{0}}{4 \pi}\left(\frac{2 r p_{m}}{\left(r^{2}-l^{2}\right)^{2}}\right) \hat{i}\) …… (4)
If the distance between two poles in a bar magnet are small (looks like short magnet) compared to the distance between geometrical centre O of bar magnet and the location of point C i.e.,
r >> l then (r² – l²)² ≈ r⁴ ……. (5)
Therefore, using equation (5) in equation (4), we get

[OR]

Question 35.
(b) Show mathematically that the rotation of a coil in a magnetic field over one rotation induces an alternating emf of one cycle.
Answer:
Induction of emf by changing relative orientation of the coil with the magnetic field:
Consider a rectangular coil of N turns kept in a uniform magnetic field \(\vec{B}\) figure (a). The coil rotates in anti-clockwise direction with an angular velocity co about an axis, perpendicular to the field.

At time = 0, the plane of the coil is perpendicular to the field and the flux linked with the coil has its maximum value Φ_m = BA (where A is the area of the coil).

In a time t seconds, the coil is rotated through an angle θ (= ωt) in anti-clockwise direction. In this position, the flux linked is Φ_m cos ωt., a component of Φ_m normal to the plane of the coil (figure (b)). The component parallel to the plane (Φ_m sinωt) has no role in electromagnetic induction. Therefore, the flux linkage at this deflected position is NΦ_B = NΦ_m cos ωt. According to Faraday’s law, the emf induced at that instant is


ε = \(-\frac{d}{d t}\) (NΦ_B) = \(-\frac{d}{d t}\) (NΦ_m cos ωt)
= -NΦ_m (-sin ωt)ω = NΦ_m ω sin ωt
When the coil is rotated through 90° from initial position, sin ωt = 1. Then the maximum value of induced emf is
ε_m = NΦ_m ω = NBAω since Φ_m = BA
Therefore, the value of induced emf at that instant is then given by
ε = ε_m sin ωt
It is seen that the induced emf varies as sin ωt function of the time angle ωt. The graph between induced emf and time angle for one rotation of coil will be a sine curve and the emf varying in this manner is called sinusoidal emf or alternating emf.

Question 36.
(a) Write down the properties of electromagnetic waves.
Answer:
Properties of electromagnetic waves:
1. Electromagnetic waves are produced by any accelerated charge.

2. Electromagnetic waves do not require any medium for propagation. So electromagnetic wave is a non-mechanical wave.

3. Electromagnetic waves are transverse in nature. This means that the oscillating electric field vector, oscillating magnetic field vector and propagation vector (gives direction of propagation) are mutually perpendicular to each other.

4. Electromagnetic waves travel with speed which is equal to the speed of light in vacuum or free space, c = \(\frac{1}{\sqrt{\varepsilon_{0} \mu_{0}}}\) = 3 × 10⁸ ms^-1

5. The speed of electromagnetic wave is less than speed in free space or vacuum, that is, v < c. In a medium of refractive index,

Question 36.
(b) Derive the equation for acceptance angle and numerical aperture, of optical fiber.
Answer:
Acceptance angle in optical fibre:
To ensure the critical angle incidence in the core-cladding boundary inside the optical fibre, the light should be incident at a certain angle at the end of the optical fiber while entering in to it. This angle is called acceptance angle. It depends on the refractive indices of the core n₁ cladding n₂ and the outer medium n₃. Assume the light is incident at an angle i_a called acceptance angle i at the outer medium and core boundary at A.
The Snell’s law in the product form, equation for this refraction at the point A.

n₃ sin i_a = n₁ sin r_a …(1)
To have the total internal reflection inside optical fibre, the angle of incidence at the core-cladding interface at B should be atleast critical angle i_c . Snell’s law in the product form, equation for the refraction at point B is,
n₁ sin i_c = n₂ sin 90° …(2)
n₁ sin i_c = n₂ ∵ sin 90° = 1
∴ sin i_c = \(\frac{n_{2}}{n_{1}}\) …(3)
From the right angle triangle ∆ABC,
i_c = 90° – r_a
Now, equation (3) becomes, sin (90° – r_a) = \(\frac{n_{2}}{n_{1}}\)
Using trigonometry, cos r_a = \(\frac{n_{2}}{n_{1}}\) …(4)
sin r_a = \(\sqrt{1-\cos ^{2} r_{a}}\)
Substituting for cos r_a
sin r_a = \(\sqrt{1-\left(\frac{n_{2}}{n_{1}}\right)^{2}}=\sqrt{\frac{n_{1}^{2}-n_{2}^{2}}{n_{1}^{2}}}\) …… (5)
Substituting this in equation (1)
n₃ sin i_a = \(n_{1} \sqrt{\frac{n_{1}^{2}-n_{2}^{2}}{n_{1}^{2}}}=\sqrt{n_{1}^{2}-n_{2}^{2}}\) ….. (6)
On further simplification,

If outer medium is air, then n₃ = 1. The acceptance angle i_a becomes,
i_a = sin^-1\((\sqrt{n_{1}^{2}-n_{2}^{2}})\) …… (9)

Light can have any angle of incidence from 0 to i_a with the normal at the end of the optical fibre forming a conical shape called acceptance cone. In the equation (6), the term (n₃ sin i_a) is called numerical aperture NA of the optical fibre.
NA = n₃ sin i_a \((\sqrt{n_{1}^{2}-n_{2}^{2}})\) ….. (10)

If outer medium is air, then n₃ = 1. The numerical aperture NA becomes,
NA = sin i_a \(\sqrt{n_{1}^{2}-n_{2}^{2}}\) ….. (11)

Question 37.
(a) Explain the effect of potential difference on photoelectric current.
Answer:
Effect of potential difference on photoelectric current:
To study the effect of potential difference V between the electrodes on photoelectric current, the frequency and intensity of the incident light are kept constant. Initially the potential of A is kept positive with respect to C and the cathode is irradiated with the given light.

Now, the potential of A is increased and the corresponding photocurrent is noted. As the potential of A is increased, photocurrent is also increased. However a stage is reached where photocurrent reaches a saturation value (saturation current) at which all the photoelectrons from C are collected by A. This is represented by the flat portion of the graph between potential of A and photocurrent.

When a negative (retarding) potential is applied to A with respect to C, the current does not immediately drop to zero because the photoelectrons are emitted with some definite and different kinetic energies. .

The kinetic energy of some of the photoelectrons is such that they could overcome the retarding electric field and reach the electrode A.

When the negative (retarding) potential of A is gradually increased, the photocurrent starts to decrease because more and more photoelectrons are being repelled away from reaching the electrode A. The photocurrent becomes zero at a particular negative potential V₀ , called stopping or cut-off potential.

Stopping potential is that the value of the negative (retarding) potential given to the collecting electrode A which is just sufficient to stop the most energetic photoelectrons emitted and make the photocurrent zero.

At the stopping potential, even the most energetic electron is brought to rest. Therefore, the initial kinetic energy of the fastest electron (K_max) is equal to the work done by the stopping potential to stop it (eV₀).
K_max = \(\frac { 1 }{ 2 }\) mv²_maxK = eV₀ ……… (1)
where v_max is the maximum speed of the emitted photoelectron.

From equation (1),
K_max = eV₀ (in joule) (or) K_max – V₀ (in eV)

From the graph, when the intensity of the incident light alone is increased, the saturation current also increases but the value of V₀ remains constant.

Thus, for a given frequency of the incident light, the stopping potential is independent of intensity of the incident light. This also implies that the maximum kinetic energy of the photoelectrons is independent of intensity of the incident light.

[OR]

Question 37.
(b) Explain the J.J. Thomson experiment to determine the specific charge of electron.
Answer:
In 1887, J. J. Thomson made remarkable improvement in the scope of study of gases in discharge tubes. In the presence of electric and magnetic fields, the cathode rays are deflected. By the variation of electric and magnetic fields, mass normalized charge or the specific charge (charge per unit mass) of the cathode rays is measured.

A highly evacuated discharge tube is used and cathode rays (electron beam) produced at cathode are attracted towards anode disc A. Anode disc is made with pin hole in order to allow only a narrow beam of cathode rays. These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage. Further, this gas discharge tube is kept in between pole pieces of magnet such that both electric and magnetic fields are perpendicular to each other. When the cathode rays strike the screen, they produce scintillation and hence bright spot is observed. This is achieved by coating the screen with zinc sulphide.

(i) Determination of velocity of cathode rays
For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O. This means that the magnitude of electric force is balanced by the magnitude of force due to magnetic field. Let e be the charge of the cathode rays, then
eE eBv
⇒ ν = \(\frac{E}{B}\) …… (1)

(ii) Determination of specific charge:
Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode. Let V be the potential difference between anode and cathode, then the potential energy is eV. Then from law of conservation of energy,
eV = \(\frac{1}{2}\)mV² ⇒ \(\frac{e}{m}=\frac{v^{2}}{2 \mathrm{V}}\)
Substituting the value of velocity from equation (1), we get
\(\frac{e}{m}=\frac{1}{2 \mathrm{V}} \frac{\mathrm{E}^{2}}{\mathrm{B}^{2}}\) ….. (2)
Substituting the values of E, B and V, the specific charge can be determined as
\(\frac{e}{m}\) = 1.7 × 10¹¹ C kg^-1

(iii) Deflection of charge only due to uniform electric field:
When the magnetic field is turned off, the deflection is only due to electric field. The deflection in vertical direction is due to the electric force.

F_e = eE …(3)
Let m be the mass of the electron and by applying Newton’s second law of motion, acceleration of the electron is
a_e = \(\frac{1}{m}\) F_e ……. (4)
Substituting equation (4) in equation (3),
a_e = \(\frac{1}{m} e E=\frac{e}{m} E\)

Let y be the deviation produced from Deviation of path by applying uniform electric field original position on the screen. Let the initial upward velocity of cathode ray be u = 0 before entering the parallel electric plates. Let t be the time taken by the cathode rays to travel in electric field. Let l be the length of one of the plates, then the time taken is
t = \(\frac{1}{v}\) …(5)
Hence, the deflection y’ of cathode rays is (note: u = 0 and a_e = \(\frac{e}{m}\)E)

Therefore, the deflection y on the screen is
y ∝ y’ ⇒ y = Cy’
where C is proportionality constant which depends on the geometry of the discharge tube and substituting y’ value in equation (6), we get y = C \(\frac{1}{2} \frac{e}{m} \frac{l^{2} \mathrm{B}^{2}}{\mathrm{E}}\) ……. (7)
Rearranging equation (7) as \(\frac{e}{n}=\frac{2 y \mathrm{E}}{\mathrm{Cl}^{2} \mathrm{B}^{2}}\) …… (8)
Substituting the values on RHS, the value of specific charge is calculated as
\(\frac{e}{m}\) = 1.7 × 10¹¹Ckg^-1m

(iv) Deflection of charge only due to uniform magnetic field
Suppose that the electric field is switched off and only the magnetic field is switched on. Now the deflection occurs only due to magnetic field. The force experienced by the electron in uniform magnetic field applied perpendicular to its path is
F_m = evB (in magnitude)
Since this force provides the centripetal force, the electron beam undergoes a semicircular path. Therefore, we can equate F_m to centripetal force \(\frac{m v^{2}}{R}\)
F_m = evB = \(m \frac{v^{2}}{R}\)
where v is the velocity of electron beam at the point where it enters the magnetic field and R is the radius of the circular path traversed by the electron beam.
eB = \(m \frac{v}{\mathrm{R}} \Rightarrow \frac{e}{m}=\frac{v}{\mathrm{BR}}\) ……. (9)
Further, substituting equation (1) in equation (9), we get
\(\frac{e}{m}=\frac{E}{B^{2} R}\) …….. (10)
By knowing the values of electric field, magnetic field and the radius of circular path, the value of specific charge \(\left(\frac{e}{m}\right)\) can be calculated

Question 38.
(a) Draw the circuit diagram of a half wave rectifier and explain its working
Answer:
Half wave rectifier circuit:
The half wave rectifier circuit. The circuit consists of a transformer, a p-n junction diode and a resistor. In a half wave rectifier circuit, either a positive half or the negative half of the AC input is passed through while the other half is blocked. Only one half of the input wave reaches the output. Therefore, it is called halfwave rectifier. Here, a p-n junction diode acts as a rectifying diode.

During the positive half cycle:
When the positive half cycle of the ac input signal passes through the circuit, terminal A becomes positive with respect to terminal B. The diode is forward biased and hence it conducts. The current flows through the load resistor R_L and the AC voltage developed across R_L constitutes the output voltage V₀ and the waveform of the diode current.

During the negative half cycle:
When the negative half cycle of the ac input signal passes through the circuit, terminal A is negative with respect to terminal B. Now the diode is reverse biased and does not conduct and hence no current passes through R_L. The reverse saturation current in a diode is negligible. Since there is no voltage drop across R_L, the negative half cycle of ac supply is suppressed at the output.

The output of the half wave rectifier is not a steady dc voltage but a pulsating wave. This pulsating voltage is not sufficient for electronic equipments. A constant or a steady voltage is required which can be obtained with the help of filter circuits and voltage regulator circuits. Efficiency (η) is the ratio of the output dc power to the ac input power supplied to the circuit. Its value for half wave rectifier is 40.6 %

[OR]

Question 38.
(b) Fiber optic communication is gaining popularity among the various transmission media – justify.
Answer:
The method of transmitting information from one place to another in terms of light pulses through an optical fiber is called fiber optic communication. It is in the process of replacing wire transmission in communication systems.
Light has very high frequency (400 THz – 790 THz) than microwave radio systems. The fibers are made up of silica glass or silicon dioxide which is highly abundant on Earth.
Now it has been replaced with materials such as chalcogenide glasses, fluoroaluminate crystalline materials because they provide larger infrared wavelength and better transmission capability.
As fibers are not electrically conductive, it is preferred in places where multiple channels are to be laid and isolation is required from electrical and electromagnetic interference.

Applications
Optical fiber system has a number of applications namely, international communication, inter¬city communication, data links, plant and traffic control and defense applications.

Merits

• Fiber cables are very thin and weight lesser than copper cables.
• This system has much larger bandwidth. This means that its information carrying capacity is larger.
• Fiber optic system is immune to electrical interferences.
• Fiber optic cables are cheaper than copper cables.

Demerits

• Fiber optic cables are more fragile when compared to copper wires.
• It is an expensive technology.

Students can Download Tamil Nadu 12th Physics Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers  helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Physics Model Question Paper 2 English Medium

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III and IV are to be attempted separately
4. Question numbers 1 to 15 in Part I are Multiple choice Questions of one mark each.  These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

PART -1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
(a) Capacitance
(b) Charge
(c) Voltage
(d) Energy density
Answer:
(d) Energy density

Question 2.
A carbon resistance has colour bands in order yellow, brown, red. Its resistance is
(a) 41 Ω
(b) 41 × 10² Ω
(c) 4 × 10³ Ω
(d) 4.2 Ω
Answer:
(b) 41 × 10² Ω

Question 3.
The magnetic field at the center O of the following
(a) \(\frac{\mu_{0} \mathbf{I}}{4 r} \otimes\)
(b) \(\frac{\mu_{0} I}{4 r} \odot\)
(c) \(\frac{\mu_{0} I}{2 r} \otimes\)
(d) \(\frac{\mu_{0} I}{2 r} \odot\)

Answer:
(a) \(\frac{\mu_{0} \mathbf{I}}{4 r} \otimes\)

Question 4.
The horizontal component of earth’s magnetic field at a place is 3.6 × 10^-5T. If the angle of dip at this place is 60°, the vertical components of earth’s field at this place is
(a) 1.2 × 10^-5T
(b) 2.4 × 10^-5T
(c) 4 × 10^-5T
(d) 6.2 × 10^-5T
Hint: B_v = B_H tan δ = 3.6 × 10^-5 × tan 60°
B_v = 6.2 × 10^-5 T
Answer:
(d) 6.2 × 10^-5T

Question 5.
In a series RL circuit, the resistance and inductive reactance are the same. Then the phase difference between the voltage and current in the circuit is
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{6}\)
(d) zero
Answer:
(a) \(\frac{\pi}{4}\)

Question 6.
Alternating current can be measured by
(a) moving coil galvanometer
(b) hot wire ammeter
(c) tangent galvanometer
(d) none of the above
Answer:
(b) hot wire ammeter

Question 7.
Let E = E₀ sin[10⁶ × -ωt] be the electric field of plane electromagnetic wave, the value of ω is
(a) 0.3 × 10^-14 rad s^-1
(b) 3 × 10^-14 rad s^-1
(c) 0.3 × 10¹⁴ rad s^-1
(d) 3 × 10¹⁴ rad s^-1
Answer:
(d) 3 × 10¹⁴ rad s^-1

Question 8.
Two coherent monochromatic light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beam are
(A) 51 and I
(b) 51 and 31
(c) 91 and I
(d) 91 and 31
Answer:
(c) 91 and I

Question 9.
The transverse nature of light is shown in,
(a) interference
(b) diffraction
(c) scattering
(d) polarisation
Answer:
(d) polarisation

Question 10.
Emission of electrons by the absorption of heat energy is called emission.
(a) photoelectric
(b) field
(c) thermionic
(d) secondary
Answer:
(c) thermionic

Question 11.
Proton and α – particle have the same de-Broglie wavelength. What is same for both of them?
(a) Time period
(b) Energy
(c) Frequency
(d) Momentum
Hint: λ = h/p, when wavelength λ is same, momentum p is also same.
Answer:
(d) Momentum

Question 12.
The ratio of the wavelengths for the transition from n = 2 to n = 1 in Li⁺⁺, He⁺ and H is
(a) 1:2: 3
(b) 1:4: 9
(c) 3:2:1
(d) 4: 9: 36
Answer:
(d) 4: 9: 36

Question 13.
The principle in which a solar cell operates
(a) Diffusion
(b) Recombination
(c) Photovoltaic action
(d) Carrier flow
Answer:
(c) Photovoltaic action

Question 14.
The output transducer of the communication system converts the radio signal into
(a) Sound
(b) Mechanical energy
(c) Kinetic energy
(d) None of the above
Answer:
(a) Sound

Question 15.
The alloys used for muscle wires in Robots are
(a) Shape memory alloys
(b) Gold copper alloys
(c) Gold silver alloys
(d) Two dimensional alloys
Answer:
(a) Shape memory alloys

PART – II

Answer any six questions in which Q. No 22 is compulsory. [6 × 2 = 12]

Question 16.
What is Polarisation?
Answer:
Polarisation \(\overrightarrow{\mathrm{P}}\) is defined as the total dipole moment per unit volume of the dielectric.
\(\overrightarrow{\mathrm{P}}\) = χ_e \(\overrightarrow{\mathrm{P}}\)_ext

Question 17.
Why current is a scalar?
Answer:
The current I is defined as the scalar product of current density and area vector in which the charges cross.
I = \(\overrightarrow{\mathrm{j}}\) . \(\overrightarrow{\mathrm{A}}\)
The dot product of two vector quantity is a scalar form. Hence, current is called as a scalar quantity.

Question 18.
The horizontal component and vertical components of Earth’s magnetic field at a place are 0.15 G and 0.26 G respectively. Calculate the angle of dip and resultant magnetic field.
Answer:
B_H = 0.15 G and B_v = 0.26 G
tan I = \(\frac{0.26}{0.15}\) ⇒ I = tan^-1 (1.732) = 60°
The resultant magnetic field of the Earth is
\(\mathrm{B}=\sqrt{\mathrm{B}_{\mathrm{H}}^{2}+\mathrm{B}_{\mathrm{H}}^{2}}=0.3 \mathrm{G}\)

Question 19.
State Fleming’s right hand rule.
Answer:
The thumb, index finger and middle finger of right hand are stretched out in mutually perpendicular directions. If the index finger points the direction of the magnetic field and the thumb indicates the direction of motion of the conductor, then the middle finger will indicate the direction of the induced current.

Question 20.
The wavelength of a light is 450 nm. How much phase it will differ for a path of 3 mm?
Answer:
The wavelength is, λ = 450 nm = 450 × 10^-9 m
Path difference is, δ = 3 mm = 3 × 10^-3 m
Relation between phase difference and path difference is, Φ = \(\frac{2 \pi}{\lambda} \times \delta\)
Substituting, Φ = \(\frac{2 \pi}{450 \times 10^{-9}} \times 3 \times 10^{-3}=\frac{\pi}{75} \times 10^{6}\)
Φ = \(\frac{\pi}{75} \times 10^{6} \mathrm{rad}\)

Question 21.
How will you define threshold frequency?
Answer:
For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.

Question 22.
Calculate the number of nuclei of carbon-14 undecayed after 22,920 years if the initial number of carbon-14 atoms is 10,000. The half-life of carbon-14 is 5730 years.
Answer:
To get the time interval in terms of half-life, n = \(\frac{t}{\mathrm{T}_{1 / 2}}=\frac{22,920 \mathrm{yr}}{5730 \mathrm{yr}}=4\)
The number of nuclei remaining undecayed after 22,920 years
\(\mathrm{N}=\left(\frac{1}{2}\right)^{n} \mathrm{N}_{0}=\left(\frac{1}{2}\right)^{4} \times 10,000 \Rightarrow \mathrm{N}=625\)

Question 23.
A diode is called as a unidirectional device. Explain
Answer:
Diode is called as a unidirectional device, i.e., current flows in only one direction (anode to cathode internally) when a forward voltage is applied, the diode conducts and when reverse voltage is applied, there is no conduction. A mechanical analogy is a rat chat, which allows motion in one direction only.

Question 24.
Give the factors that are responsible for transmission impairments.
Answer:

• Attenuation
• Distortion (Harmonic)
• Noise

PART-III

Answer any six questions ¡n which Q.No. 26 ¡s compulsory. (6 × 3 = 18)

Question 25.
A sample of HO gas ¡s placed in a uniform electric field of magnitude 3 × 10⁴ N C^-1. The dipole moment of each HCI molecule is 3.4 × 10^-30 Cm. Calculate the maximum torque experienced by each HCl molecule.
Answer:
The maximum torque experienced by the dipole is when it is aligned perpendicular to the applied field.
\(\tau_{\max }\) = pE sin90° = 3.4 × 10^-30 × 3 × 10⁴Nm
\(\tau_{\max }\) =10.2 × 10^-26Nm

Question 26.
The resistance ola wire is 20 Ω . What will be new resistance, ¡fit is stretched uniformly 8 times its original length?
Answer:
R₁ = 20 Ω, R₂ = ?
Let the original length (l₁) be 1.
The new length, l₂ = 8l₁ (i.,e) l₂ =8l
The original resistance,


Though the wire is stretched, its volume is unchanged.
Initial volume = Final volume
A₁l₁ = A₂l₂ , A₁l =A₂8l
\(\frac{A_{1}}{A_{2}}=\frac{8 l}{l}=8\)
By dividing equation R₂ by equation R₁, we get

Substituting the value of \(\frac{A_{1}}{A_{2}}\) we get
\(\frac{R_{1}}{R_{2}}\) = 8 × 8 = 64 ⇒ R₂ = 64 × 20 = 1280 Ω
Hence, strecthing the length of the wire has increased its resistance.

Question 27.
State Biot-Savart’s law.
The magnitude of magnetic field \(d \vec{B}\) at a point P at a distance r from the small elemental length taken on a conductor carrying current varies

• directly as the strength of the current I
• directly as the magnitude of the length element \(d \vec{l}\)
• directly as the sine of the angle (say,0) between d\(d \vec{l}\) and r̂ .
• inversely as the square of the distance between the point P and length element \(d \vec{l}\).
  This is expressed as
  \(d \mathrm{B} \propto \frac{\mathrm{I} d l}{r^{2}} \sin \theta\)

Question 28.
Give the principle of AC generator.
Answer:
Alternators work on the principle of electromagnetic induction. The relative motion between a conductor and a magnetic field changes the magnetic flux linked with the conductor which in turn, induces an emf. The magnitude of the induced emf is given by Faraday’s law of electromagnetic induction and its direction by Fleming’s right hand rule.

Question 29.
A transformer is used to light a 140 W, 24 V bulb from a 240 V AC mains. The current in the main cable is 0.7 A. Find the efficiency of the transformer.
Answer:

Question 30.
What are the Cartesian sign conventions for a spherical mirror?
Answer:

• The Incident light is taken from left to right (i.e. object on the left of mirror).
• All the distances are measured from the pole of the mirror (pole is taken as origin).
• The distances measured to the right of pole along the principal axis are taken as positive.
• The distances measured to the left of pole along the principal axis are taken as negative,
• Heights measured in the upward perpendicular direction to the principal axis are taken as positive.
• Heights measured in the downward perpendicular direction to the principal axis, are taken as negative.

Question 31.
Write the relationship of de Broglie wavelength k associated with a particle of mass m in terms of its kinetic energy K.
Answer:
Kinetic energy of the particle, K = \(\frac{1}{2}\) mv² = \(\frac{p^{2}}{2 m}\)
p = \(\sqrt{2 m \mathrm{K}}\)
de-Broglie wavelength of the particle λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m \mathrm{K}}}\)

Question 32.
What is binding energy of a nucleus? Give its expression.
Answer:
when Z protons and N neutrons combine to form a nucleus, mass equal to mass defect disappears and the corresponding energy is released. This is called the binding energy of the nucleus (BE) and is equal to (Δm)c² .
BE = (Zm_p + Nm_n – M ) c²

Question 33.
Distinguish between avalanche and zener breakdown.
Answer:

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34.
(a) Calculate the electric field due to a dipole on its axial line and equatorial plane.
Answer:
Case (i) Electric field due to an electric dipole at points on the axial line. Consider an electric dipole placed on the x-ax is as shown in figure. A point C is located at a distance of r from the midpoint O of the dipole along the axial line. line
The electric field at a point C due to +q is


Since the electric dipole moment vector \(\vec{p}\) is from -q to +q and is directed along BC, the above equation is rewritten as

where P̂ is the electric dipole moment unit vector from -q to +q.
The electric field at a point C due to -q is

Since +q is located closer to the point C than -q, \(\vec{E}\)_– \(\vec{E}\)₊ us stronger than \(\vec{E}\)_–. Therefore, the length of the \(\vec{E}\)₊ vector is drawn large than that of \(\vec{E}\)_– vector.
The total electric field at point C is calculated using the superposition principle of the electric field.

Note that the total electric field is along \(\vec{E}\)₊ since +q is closer to C than -q.

The direction of \(\vec{E}\)_tot is shown in Figure
If the point C is very far away from the dipole then (r >> a).
Under this limit the term(r² – a²) ≈ r⁴ Substituting this into equation, we get

If the point C is chosen on the left side of the dipole, the total electric field is still in the direction of \(\vec{p}\).

Case (ii) Electric field due to an electric dipole at a point on the equatorial plane

Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure. Since the point C is quite-distant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of E is along BC and the direction of \(\vec{E}\)₊ is along and the direction of \(\vec{E}\)_– CA. \(\vec{E}\)₊ and \(\vec{E}\)_– are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. The perpendicular components \(\left|\overrightarrow{\mathrm{E}}_{+}\right|\) sin θ and \(\left|\overrightarrow{\mathrm{E}}_{-}\right| sin θ\) are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of \(\vec{E}\)₊ and \(\vec{E}\)_– and its direction is along -P̂

The magnitudes \(\vec{E}\)₊ and \(\vec{E}\)_– are the same and are given by


By substituting equation (1) into equation (2), we get

Question 34.
(b) Obtain the condition for bridge balance in Wheatstone’s bridge.
Answer:
An important application of Kirchhoff’s rules is the Wheatstone’s bridge. It is used to compare resistances and also helps in determining the unknown resistance in electrical network. The bridge consists of four resistances P, Q, R and S connected, A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is I_G and its resistance is G.

Applying Kirchhoff’s current rule to junction B,
I₁ – I_G – I₃ = 0 ……. (1)
Applying Kirchhoff’s current rule to junction D,
I₂ + I_G – I₄ = 0 ……. (2)
Applying Kirchhoff’s voltage rule to loop ABDA,
I₁p + I_GG – I₂R = 0 ……. (3)
Applying Kirchhoff’s voltage rule to loop ABCDA,
I₁p – I₃Q – I₄S – I₂R = 0 ……. (4)
When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B and D, no current flows through galvanometer (I_G = 0).
Substituting I_G = O in equation, (I), (2) and (3), we get
I₁ = I₃ …… (5)
I₂ = I₄ ……… (6)
I₁p = I₂R ………. (7)
Substituting the equation (5) and (6) in equation (4)
I₁P + I₁Q – I₂S – I₂R= 0
I₁(P + Q) = I₂(R+S) …… (8)
Dividing equation (8) By equation (7), we get

This is the bridge balance condition. Only under this condition, galvanometer shows null deflection. Suppose we know the values of two adjacent resistances, the other two resistances can be compared. If three of the resistances are known, the value of unknown resistance (fourth one) can be determined.

Question 35.
(a) Show the time period of oscillation when a bar magnet is kept in a uniform magnetic field is T = 2π\(\sqrt{\frac{1}{p_{m} \mathrm{B}}}\). in second, where I represents moment of interia of the bar magnet, p<sub<m is the magnetic moment and is the magnetic field.
Answer:
The magnitude of deflecting torque (the torque which makes the object rotate) acting on the bar magnet which will tend to align the bar magnet parallel to the direction of the uniform magnetic field \(\overrightarrow{\mathrm{B}}\) is
\(\vec{\tau}\) = p_m B sin θ
The magnitude of restoring torque acting on the bar magnet can be written as
\(|\vec{\tau}|=\mathrm{I} \frac{d^{2} \theta}{d t^{2}}\)
Under equilibrium conditions, both magnitude of deflecting torque and restoring torque will be equal but act in the opposite directions, which means
\(\mathrm{I} \frac{d^{2} \theta}{d t^{2}}=-p_{m} \mathrm{B} \sin \theta\)
The negative sign implies that both are in opposite directions. The above equation can be written as
\(\frac{d^{2} \theta}{d t^{2}}=-\frac{p_{m} B}{I} \sin \theta\)
This is non-linear second order homogeneous differential equation. In order to make it linear, we use small angle approximation, i.e., sin θ ≈ θ, we get
\(\frac{d^{2} \theta}{d t^{2}}=-\frac{p_{m} \mathrm{B}}{\mathrm{I}} \theta\)
This linear second order homogeneous differential equation is a Simple Harmonic differential equation. Therefore,
Comparing with Simple Harmonic Motion (SHM) differential equation \(\frac{d^{2} x}{d t^{2}}=-\omega^{2} x\)
where ω is the angular frequency of the oscillation.

where, B_H is the horizontal component of Earth’s magnetic field.

[OR]

Question 35.
(b) An inductor of inductance L carries an electric current i. How much energy is stored while establishing the current in it?
Answer:
Energy stored in an inductor: Whenever a current is established in the circuit, the inductance opposes the growth of the current. In order to establish a current in the circuit, work is done against this opposition by some external agency. This work done is stored as magnetic potential energy. Let us assume that electrical resistance of the inductor is negligible and inductor effect alone is considered. The induced emf eat any instant t is
ℰ = \(-\mathrm{L} \frac{d i}{d t}\) …(1)
Let dW be work done in moving a charge dq in a time dt against the opposition, then
dW = -edq = -ℰdq = -ℰidi [ ∵ dq = idt]
Substituting for e value from equation (1)
= \(-\left(-\mathrm{L} \frac{d i}{d t}\right) i d t\)
dW = Lidt ……. (2)
Total work done in establishing the current i is
This work done is stored as magnetic potential energy.

W = \(\frac{1}{2}\) Li² ………. (3)
This work done is stored as magnetic potential energy.
∴ U_B = \(\frac{1}{2}\) Li²

Question 36.
(a) Using Faraday’s law of electromagnetic induction, derive an equation for motional emf.
Answer:
Motional emf from Faraday’s law: Let us consider a rectangular conducting loop of width l in a uniform magnetic field \(\overrightarrow{\mathrm{B}}\) which is perpendicular to the plane of the loop and is directed inwards. A part of the loop is in the magnetic field while the remaining part is outside the field.

When the loop is pulled with a constant velocity \(\overrightarrow{\mathrm{v}}\) to the right, the area of the portion of the loop within the magnetic field will decrease. Thus, the flux linked with the loop will also decrease. According to Faraday’s law, an electric current is induced in the loop which flows in a direction so as to oppose the pull of the loop.
Let x be the length of the loop which is still within the magnetic field, then its area is lx. The magnetic flux linked with the loop is
Φ_B = \(\int_{A} \vec{B} \cdot d \vec{A}=B A \cos \theta\)
Here θ = 0° and cos 0° = 1
= BA
Φ_B = Blx ……… (1)

As this magnetic flux decreases due to the movement of the loop, the magnitude of the induced emf is given by
\(\varepsilon=\frac{d \Phi_{\mathrm{B}}}{d t}=\frac{d}{d t}(\mathrm{B} l x)\)
Here, both B and l are constants. Therefore,
\(\varepsilon=\mathrm{B} l \frac{d x}{d t}=\mathrm{B} l v\) …… (2)
where v = \(\frac{d x}{d t}\) is the velocity of the loop. This emf is known as motional emf since it is produced due to the movement of the loop in the magnetic field.

[OR]

Question 36.
(b) Derive the mirror equation and the equation for lateral magnification. The mirror equation:
Answer:
The mirror equation establishes a relation among object distance u, image distance v and focal length/for a spherical mirror. An object AB is considered on the principal axis of a concave mirror beyond the center of curvature C.
Let us consider three paraxial rays from point B on the object.
The first paraxial ray BD travelling parallel to principal axis is incident on the concave mirror at D, close to the pole P. After reflection the ray passes through the focus F. The second paraxial ray BP incident at the pole P is reflected along PB’. The third paraxial ray BC passing through centre of curvature C, falls normally on the mirror at E is reflected back along the same path.
The three reflected rays intersect at the point B’. A perpendicular drawn as A’ B’ to the principal axis is the real, inverted image of the object AB.

As per law of reflection, the angle of incidence ∠BPA is equal to the angle of reflection ∠B’PA’ . The triangles ∆BPA and ∆B’PA’ are similar. Thus, from the rule of similar triangles,
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{P A^{\prime}}{P A}\) …… (1)
The other set of similar triangles are, ∆DPF and ∆ BA.’ F. (PD is almost a straight vertical line)
\(\frac{A^{\prime} B^{\prime}}{P D}=\frac{A^{\prime} F}{P F}\)
As, the distances PD = AB the above equation becomes,
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{A^{\prime} F}{P F}\) …….. (2)
From equations (1) and (2) we can write,
\(\frac{P A^{\prime}}{P A}=\frac{A^{\prime} F}{P F}\)

As, A’ F PA’ – PF, the above equation becomes,
\(\frac{P A^{\prime}}{P A}=\frac{P A^{\prime}-P F}{P F}\)
We can apply the sign conventions for the various distances in the above equation.
PA= -u. PA’= -v, PF = -f
All the three distances are negative as per sign convention, because they are measured to the left of the pole. Now, the equation (3) becomes.

The above equation (4) is called mirror equation

Lateral magnification in spherical mirrors:
The lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object. The height of the object and image are measured perpendicular to the principal axis.

m = \(\frac{h^{\prime}}{h}\) …… (5)
Applying proper sign conventions for equation (1), A’B’ PA’
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{P A^{\prime}}{P A}\)
A’B’ = -h, AB = h, PA’ = -v, PA = -u
\(\frac{-h^{\prime}}{h}=\frac{-v}{-u}\)
On simplifying we get,
\(m=\frac{h^{\prime}}{h}=-\frac{v}{u}\) …….(6)
Using mirror equation, we can further write the magnification as,
\(m=\frac{h^{\prime}}{h}-\frac{f-v}{f}=\frac{f}{f-u}\) ……… (7)

Question 37.
(a) Obtain Einstein’s photoelectric equation with necessary explanation.
Answer:
Einstein’s explanation of photoelectric equation:
When a photon of energy hv is incident on a metal surface, it is completely absorbed by a single electron and the electron is ejected.

In this process, a part of the photon energy is used for the ejection of the electrons from the metal surface (photoelectric work function Φ₀) and the remaining energy as the kinetic energy of the ejected electron. From the law of conservation of energy,
hυ = Φ₀ + \(\frac { 1 }{ 2 }\) mv² …… (1)
where m is the mass of the electron and u its velocity

If we reduce the frequency of the incident light, the speed or kinetic energy of photo electrons is also reduced. At some frequency v0 of incident radiation, the photo electrons are ejected with almost zero kinetic energy. Then the equation (1) becomes
hυ₀ = Φ₀
where v₀ is the threshold frequency. By rewriting the equation (1), we get
hυ = hυ₀ + \(\frac { 1 }{ 2 }\) mv² …….(2)
The equation (2) is known as Einstein’s Photoelectric equation.
If the electron does not lose energy by internal collisions, then it is emitted with maximum kinetic energy K_max. Then
K_max = \(\frac { 1 }{ 2 }\) mv²_max
where υ_max is the maximum velocity of the electron ejected. The equation (1) is rearranged as follows:
K_max = hυ – Φ₀

[OR]

Question 37.
(b) Derive the energy expression for hydrogen atom using Bohr atom model.
Answer:
The energy of an electron in the nth orbit
Since the electrostatic force is a conservative force, the potential energy for the orbit is

This implies that U_n = -2 KE_n. Total energy in the «th orbit is
E_n = KE_n + U_n = KE_n – 2KE_n = -KE_n
E_n = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}} \frac{Z^{2}}{n^{2}}\)
For hydrogen atom (Z = 1),
E_n = \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}} \frac{1}{n^{2}} \text { joule }\) ……. (1)
where n stands for principal quantum number. The negative sign in equation (1) indicates that the electron is bound to the nucleus.
Substituting the values of mass and charge of an electron (m and e), permittivity of free space s0 and Planck’s constant h and expressing in terms of eW. we get
E_n = -13.6\(\frac{1}{n^{2}}\)eV
For the first orbit (ground state), the total energy of electron is E₁ = – 13.6 eV. For the second orbit (first excited state), the total energy of electron is E₂ = -3.4 eV. For the third orbit (second excited state), the total energy of electron is E₃ =1.51 eV and so on.

Notice that the energy of the first excited state is greater than the ground state, second excited state is greater than the first excited state and so on. Thus, the orbit which is closest to the nucleus (r₁) has lowest energy (minimum energy compared with other orbits). So, it is often called ground state energy (lowest energy state). The ground state energy of hydrogen (-13.6 eV ) is used as a unit of energy called Rydberg (1 Rydberg = -13.6 eV ).

The negative value of this energy is because of the way the zero of the potential energy is defined. When the electron is taken away to an infinite distance (very far distance) from nucleus, both the potential energy and kinetic energy terms vanish and hence the total energy also vanishes.

Question 38.
(a) Explain the construction and working of a full wave rectifier.
Full wave rectifier:
The positive and negative half cycles of the AC input signal pass through the full wave rectifier circuit and hence it is called the full wave rectifier. It consists of two p-n junction diodes, a center tapped transfonner, and a load resistor (RL). The centre is usually taken as the ground or zero voltage reference point. Due to the centre tap transformer, the output voltage rectified by each diode is only one half of the total secondary voltage.

During positive half cycle:
When the positive half cycle of the ac input signal passes through the circuit, terminal M is positive, G’ is at zero potential and N is at negative potential. This forward biases diode D₁ and reverse biases diode D₂ Hence, being forward biased, diode D₁ conducts and current flows along the path MD₁ AGC . As a result, positive half cycle of the voltage appears across R_L in the direction G to C.

During negative half cycle:
When the negative half cycle of the ac input signal passes through the circuit, terminal N is positive, G is at zero potential and M is at negative potential. This forward biases diode D₂ and reverse biases diode D₁. Hence, being forward biased, diode D₂ conducts and current flows along the path ND₂ BGC . As a result, negative half cycle of the voltage appears across R_L in the same direction from G to C

Hence in a full wave rectifier both positive and negative half cycles of the input signal pass through the circuit in the same direction as shown in figure (b). Though both positive and negative half cycles of ac input are rectified, the output is still pulsating in nature.

The efficiency ( η) of full wave rectifier is twice that of a half wave rectifier and is found to be 81.2 %. It is because both the positive and negative half cycles of the ac input source are rectified.

[OR]

Question 38.
(b) Give the applications of ICT in mining and agriculture sectors.
Answer:
(i) Agriculture
The implementation of information and communication technology (ICT) in agriculture sector enhances the productivity, improves the living standards of farmers and overcomes the challenges and risk factors.
(a) ICT is widely used in increasing food productivity and farm management.
(b) It helps to optimize the use of water, seeds and fertilizers etc.
(c) Sophisticated technologies that include robots, temperature and moisture sensors, aerial images, and GPS technology can be used.
(d) Geographic information systems are extensively used in farming to decide the suitable place for the species to be planted.

(ii) Mining
(a) ICT in mining improves operational efficiency, remote monitoring and disaster locating system.
(b) Information and communication technology provides audio-visual warning to the trapped underground miners.
(c) It helps to connect remote sites.