Category: Class 11

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2

11th Maths Exercise 7.2 Question 1.
Without expanding the determinant, prove that
Solution:

11th Maths Exercise 7.2 Answers Question 2.
Show that
Solution:

Matrices And Determinants Class 11 Solutions Pdf Question 3.
Prove that
Solution:
LHS
Taking a from C₁, b from C₂ and c from C₃ we get

Expanding along R₁ we get
(2c) (abc) (1) [ab + ab] = abc (2c) (2ab)
1 = (abc) (4abc) = 4a²b²c²
= RHS

Matrices And Determinants Class 11 State Board Solutions Question 4.

Solution:

11th Maths Matrices And Determinants Solutions Question 5.
Prove that
Solution:

11th Maths Exercise 7.2 In Tamil Question 6.
Show that
Solution:

11th Maths Determinants Solutions Question 7.
Write the general form of a 3 × 3 skew-symmetric matrix and prove that its determinant is 0.
Solution:

11th Maths Exercise 7.2 Samacheer Kalvi Question 8.

Solution:

we get – (aα² + 2bα + c) [ac – b²]
So Δ = 0 ⇒ (aα² + 2bα + c) (ac -b²) = – 0 = 0
⇒ aα² + 2bα + c = 0 or ac – b² = 0
(i.e.) a is a root of ax² + 2bx + c = 0
or ac = b²
⇒ a, b, c are in G.P.

11th Maths Matrix Solutions Question 9.
Prove that
Solution:

Exercise 7.2 Class 11 Maths Solutions State Board Question 10.
If a, b, c are p^th, q^th and r^th terms of an A.P., find the value of
Solution:
We are given a = t_p,b = t_q and c = t_r
Let a be the first term and d be the common difference

11th Maths Matrix And Determinants Question 11.
Show that is divisible by x⁴
Solution:

Multiplying R₁ by a, R₂ by b and R₃ by c and
taking out a from C₁ b from C₂ and c from C₃ we get
==

11th Maths Ex 7.2 Question 12.
If a, b, c are all positive, and are p^th, q^th and r^th terms of a G.P., show that

Solution:

Exercise 7.2 Class 11 Maths Solutions Question 13.
Find the value of if x, y, z ≠ 1.
Solution:
Expanding the determinant along R₁

11th Std Maths Determinants Solutions Question 14.

Solution:

11th Maths Book Volume 2 Chapter 7 Question 15.
Without expanding, evaluate the following determinants:

Solution:

10th Maths Exercise 7.2 Samacheer Kalvi Question 16.
If A is a square matrix and |A| = 2, find the value of |AA^T|.
Solution:
|A| = 2 (Given) |A^T| = 2
Now |AA^T| = |A| |A^T| = 2 × 2 = 4.

12th Maths Exercise 7.2 Samacheer Kalvi Question 17.
If A and B are square matrices of order 3 such that |A| = -1 and |B| = 3, find the value of |3AB|.
Solution:
Given |A| = -1 : |B| = 3
Given A and B are square matrices of order 3.
∴ |kAB| = k³ |AB|
Here k = 3 ∴ |3AB| = 3³ |AB|
= 27 |AB|
= 27 (-1) (3)
= -81

Class 11 Maths Exercise 7.2 Solutions Question 18.
If λ = -2, determine the value of
Solution:
Given λ = -2
∴ 2λ = -4; λ² = (-2)²; 3λ² + 1 = 3 (4) + 1 = 13
6λ – 1 = 6(-2) – 1 = -13

expanding along R₁
0(0) + 4 (0 + 13) + 1 (-52 + 0) = 52 – 52 = 0
Aliter: The determinant value of a skew-symmetric matrix is zero

Matrices And Determinants Class 11 Solutions Question 19.
Determine the roots of the equation
Solution:

Given the determinant value is 0
⇒ 30(1 + x) (2 – x) = 0
⇒ 1 + x = 0 or 2 – x = 0
⇒ x = -1 or x = 2
So, x = -1 or 2.

Matrices And Determinants Class 11 Exercise Question 20.
Verify that det (AB) = (det A) (det B) for
Solution:


{(-20)(52) (-19) + (10)(38)(—49) + (2)(64)(-17)} – {(-49)(52) (2) + (-17)(38)(-20) + (-19)(64)(10)}
= (19760 – 18620 – 2176) – (-5096 + 12920 – 12160)
= (19760 + 5096 + 12160) – (18620 + 2176 + 12920)
= 37016 – 33716 = 3300 ….(3)
Now (1) × (2) = (3)
(i.e.,) (-33) (-100) = 3300
⇒ det (AB) = (det A), (det B)

Determinants Class 11 State Board Solutions Question 21.
Using cofactors of elements of second row, evaluate |A|, where
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 Additional Problems

Exercise 7.2 Class 10 Samacheer Kalvi Question 1.

Solution:

11th Maths Matrices And Determinants Pdf Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Students can Download Accountancy Chapter 5 Trial Balance Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy 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 11th Accountancy Solutions Chapter 5 Trial Balance

Samacheer Kalvi 11th Accountancy Trial Balance Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

11th Accountancy Chapter 5 Book Back Answers Question 1.
Trial balance is a ………………
(a) Statement
(b) Account
(c) Ledger
(d) Journal
Answer:
(a) Statement

Trial Balance Questions With Solutions Pdf Question 2.
After the preparation of ledger, the next step is the preparation of ………………
(a) Trading account
(b) Trial balance
(c) Journal
(d) Profit and loss account
Answer:
(b) Trial balance

11th Accountancy 5th Chapter Solutions Question 3.
The trial balance contains the balances of ………………
(a) Only personal accounts
(b) Only real accounts
(c) Only nominal accounts
(d) All accounts
Answer:
(d) All accounts

Trial Balance Questions And Answers Pdf Question 4.
Which of the following is/are the objective(s) of preparing trial balance?
(a) Serving as the summary of all the ledger accounts
(b) Helping in the preparation of final accounts
(c) Examining arithmetical accuracy of accounts
(d) (a), (b) and (c)
Answer:
(d) (a), (b) and (c)

Accounting 11 Chapter 5 Answers Question 5.
While preparing the trial balance, the accountant finds that the total of the credit column is short by ₹ 200. This difference will be ………………
(a) Debited to suspense account
(b) Credited to suspense account
(c) Adjusted to any of the debit balance
(d) Adjusted to any of the credit balance
Answer:
(b) Credited to suspense account

Trial Balance Questions And Answers Question 6.
A list which contains balances of accounts to know whether the debit and credit balances are matched is ………………
(a) Journal
(b) Day book
(c) Trial balance
(d) Balance sheet
Answer:
(c) Trial balance

Trial Balance Question Question 7.
Which of the following method(s) can be used for preparing trial balance?
(a) Balance method
(b) Total method
(c) Total and Balance method
(d) (a), (b) and (c)
Answer:
(d) (a), (b) and (c)

Trial Balance Questions Question 8.
The account which has a debit balance and is shown in the debit column of the trial balance is ………………
(a) Sundry creditors account
(b) Bills payable account
(c) Drawings account
(d) Capital account
Answer:
(c) Drawings account

Trial Balance Question And Answer Question 9.
The difference of totals of both debit and credit side of trial balance is transferred to ………………
(a) Trading account
(b) Difference account
(c) Suspense account
(d) Miscellaneous account
Answer:
(c) Suspense account

Trial Balance Sums Question 10.
Trial balance is prepared ………………
(a) At the end of the year
(b) On a particular date
(c) For a year
(d) None of the above
Answer:
(b) On a particular date

II. Very Short Answer Questions

Trial Balance Class 11 Pdf Question 1.
What is trial balance?
Answer:
Trial balance is a statement containing the debit and credit balances of all ledger accounts on a particular date. It is arranged in the form of debit and credit columns placed side by side and prepared with the object of checking the arithmetical accuracy of entries made in the books of accounts and to facilitate preparation of financial statements.

Samacheer Kalvi Guru 11th Accountancy Question 2.
Give the format of trial balance.
Answer:
Trial balance is prepared in the following format under the balance method:

Trial Balance Questions With Solutions Question 3.
What are the methods of preparation of trial balance?
Answer:

1. Balanced method
2. Total method
3. Total and balance method

Trial Balance Questions With Answers Question 4.
State whether the balance of the following accounts should be placed in the debit or the credit column of the trial balance:

1. Carriage outwards
2. Carriage inwards
3. Sales
4. Purchases
5. Bad debts vi. Interest paid
6. Interest received
7. Discount received
8. Capital
9. Drawings
10. Sales returns
11. Purchase returns

Answer:
Trial balance as on 31^st March, 2017

III. Short Answer Questions

Trial Balance Chapter Pdf Question 1.
What are the objectives of preparing trial balance?
Answer:
Trial balance is prepared with the following objectives:
1. Test of arithmetical accuracy:
Trial balance is the means by which the arithmetical accuracy of the book – keeping work is checked. When the totals of debit column and credit column in the trial balance are equal, it is assumed that posting from subsidiary books, balancing of ledger accounts, etc. are arithmetically correct. However, there may be some errors which are not disclosed by trial balance.

2. Basis for preparing final accounts:
Financial statements, namely, trading and profit and loss account and balance sheet are prepared on the basis of summary of ledger balances obtained from the trial balance.

3. Location of errors:
When the trial balance does not tally, it is an indication that certain errors have occurred. The errors may have occurred at one or more of the stages of accounting process, namely, journalising or recording in subsidiary books, totalling subsidiary books, posting in ledger accounts.

Balancing the ledger accounts, carrying ledger account balances to the trial balance and totalling the trial balance columns, etc. Hence, the errors should be located and rectified before preparing the financial statements.

4. Summarised information of ledger accounts:
The summary of ledger accounts is shown in the trial balance. Ledger accounts have to be seen only when details are required in respect of an account.

Trial Balance Exercise With Answer Pdf Question 2.
What are the limitations of trial balance?
Answer:
The following are the limitations of trial balance:

1. It is possible to prepare trial balance of an organisation, only if the double entry system is followed.
2. Even if some transactions are omitted, the trial balance will tally.
3. Trial balance may tally even though errors are committed in the books of account,
4. If trial balance is not prepared in a systematic way, the final accounts prepared on the basis of trial balance may not depict the actual state of affairs of the concern.
5. Agreement of trial balance is not a conclusive proof of arithmetical accuracy of entries made in the accounting records. This is because there are certain errors which are not : disclosed by trial balance such as complete omission of a transaction, compensating errors and error of principle.

Accountancy Class 11 Chapter 5 Solutions Question 3.
‘A trial balance is only a prima facie evidence of the arithmetical accuracy of records’. Do you agree with this statement? Give reasons.
Answer:
Yes, I agree with the statement. ‘A trial balance is only a prima facie evidence of the arithmetical accuracy of records.

Reasons:
Trial balance is the means by which the arithmetical accuracy of the book – keeping. work is checked. When the totals of debit column and credit column in the trial balance are equal, it is assumed that porting from subsidiary books, balancing of ledger accounts, etc.

IV. Exercises

Trial Balance Sums For Class 11 Question 1.
Prepare a trial balance with the following information: (3 Marks)

Answer:
Trial balance

Chapter 5 Accounts Class 11 Solutions Question 2.
Prepare the trial balance from the following information: (3 Marks)

Answer:
Trial Balance

Samacheer Kalvi Accountancy 11th Question 3.
Prepare the trial balance from the following balances of Chandramohan as on 31^st March, 2017. (3 Marks)

Answer:
Trial balance of Chandramohan as on 31^st March, 2017

Samacheer Kalvi 11th Accountancy Guide Question 4.
Prepare the trial balance from the following balances of Babu as on 31^st March, 2016. (3 Marks)

Answer:
Trial balance of Babu as on 31^st March, 2016

Question 5.
From the following balances of Aijun, prepare the trial balance as on 31^st March, 2018. (3 Marks)

Answer:
Trial balance of Arjun as on 31^st March, 2018

Question 6.
Prepare the trial balance from the following balances of Rajesh as on 31^st March, 2017. (3 Marks)

Answer:
Trial balance of Rajesh as on 31^st March, 2017

Question 7.
Prepare the trial balance from the following balances of Karthik as on 31^st March, 2017. (3 Marks)

Answer:
Trial balance of Karthik as on 31^st March, 2017

Question 8.
From the following balance of Rohini, Prepare the trial balance as on 31^st March, 2016. (3 Marks)

Answer:
Trial balance of Rohini as on 31^st March, 2017

Question 9.
Balan who has a car driving school gives you the following ledger balances. Prepare trial balance as on 31^st December, 2016. (3 Marks)

Answer:
Trial balance of Balan as on 31^st March, 2017

Question 10.
The following balances are extracted from the books of Ravichandran on 31^st December, 2016. Prepare the trial balance. (5 Marks)

Answer:
Trial balance of Rohini as on 31^st March, 2017

Question 11.
From the following balances, Prepare trial balanceof Baskar as on 31^st March 2017. Transfer the difference, if any, to suspense account. (5 Marks)

Answer:
Trial balance of Rohini as on 31^st March, 2017

Question 12.
From the following balances extracted from the books of Rajeshwari as on 31^st March, 2017, prepare the trial balance.

Answer:
Trial balance of Rajeshwari as on 31^st March, 2017

Question 13.
Correct the following trail balance:

Answer:
The Corrected Trial balance

Textbook Case Study Solved

Question 1.
Mary runs a textile store. She has prepared the following trial balance from her ledger balances. Her trial balance does not tally. She needs your help to check whether what she has done is correct.

Solution:
There are some errors in Mary’s Trial balance. The corrected trial balance has been written below:
The Corrected Trial balance of Mary

Students who are interested in learning of 11th English Reading Note-Making and Summarizing Questions and Answers can use Tamilnadu State Board Solutions of 11th English Chapter Wise Pdf. First check in which chapter you are lagging and then Download Samacheer Kalvi 11th English Book Solutions Questions and Answers Summary, Activity, Notes Chapter Wise. Students can build self confidence by solving the solutions with the help of Tamilnadu State Board English Solutions. English is the scoring subject if you improve your grammar skills. Because most of the students will lose marks by writing grammar mistakes. So, we suggest you to Download Tamilnadu State Board 11th English Solutions according to the chapters.

Tamilnadu Samacheer Kalvi 11th English Reading Note-Making and Summarizing

Check out the topics covered in Reading Note-Making and Summarizing Questions and Answers before you start your preparation. Improve your grammar skills with the help of Samacheer Kalvi 11th English Book Solutions Questions and Answers pdf links. The solutions for Tamilnadu State Board 11th English Textbook are prepared by the English experts. So, if you follow Tamilnadu State Board Solutions 11th English Textbook Solutions you can cover all the topics in Reading Note-Making and Summarizing Questons and Answers. This helps to improve your communication skills.

Notes are short written record of facts to aid the memory. Notes are usually taken to record a speech or dictation while listening to it or after reading a book, magazine or article. They are referred back whenever needed and may be reproduced in the desired way.

 

A Good Business Letter Note Making The Necessity Of Note-Making

Knowledge is vast and unlimited, but our memory is limited. We cannot remember all the information all the time. Hence note-making is necessary. With the help of notes we can recall the entire information read/heard months ago. It is quite useful to students preparing for many subjects. At the time of examinations, it is not possible to go through voluminous books. At such critical times, notes are quite handy. Hence note-making fulfils three useful functions:
(i) It keeps a lot of information at our disposal for ready reference.
(ii) It helps us reconstruct what was said or written and thus accelerates the process of remembering/recall.
(iii) It comes in handy in delivering a speech, participation in a debate/discussion, writing an essay and revising lessons before an examination.

How Note-Making Helps Us

While making notes we do not simply read the passage/listen to speech but consider various points made by the writer/speaker and draw our own inferences about what is being presented. Thus note-making helps us in understanding the passage in a better way and organising our thoughts systematically.

 

Characteristics Of Good Notes

• Short and Compact: Good notes must be short and compact.
• Complete Information: They must contain all the important information.
• Logical: They must be presented in a logical way.
• Understandable: They should be understandable when consulted at a later stage.

Mechanics Of Note-Making

While making notes we follow certain standard practices. These may be listed as follows:
(a) Heading and Sub-headings
(b) Abbreviation and Symbols
(c) Note-form
(d) Numbering and Indentation

Heading And Sub-Headings
The heading reflects the main theme whereas the sub-headings point out how it has been developed. The selection of proper heading and sub-heading reveals the grasp of the passage by the students. In the absence of proper assimilation of main ideas and subsidiary points it is impossible to make notes.

 

Abbreviations And Symbols
They are used for precision and economy of words and hence quite helpful in note-making. At least four recognisable abbreviations are to be used in note-making in your board examination. These are essential components of note-making. Students often make use of abbreviations and symbols in doing their written work.

Conversation Is Indeed The Most Easily Solved Questions
Read the following passages carefully:

Conversation Is Indeed The Most Easily Teachable Passage Answers Passage 1
1. The conversation is indeed the most easily teachable of all arts. All you need to do in order to become a good conversationalist is to find a subject that interests you and your listeners. There are, for example, numberless hobbies to talk about. But the important thing is that you must talk about other fellow’s hobby rather than your own. Therein lies the secret of your popularity. Talk to your friends about the things that interest them, and you will get a reputation for good fellowship, charming wit, and a brilliant mind. There is nothing that pleases people so much as your interest in their interest.

 

2. It is just as important to know what subjects to avoid and what subjects to select for good conversation. If you don’t want to be set down as a wet blanket or a bore, be careful to avoid certain unpleasant subjects. Avoid talking about yourself, unless you are asked to do so. People are interested in their own problems, not in yours. Sickness or death bores everybody. The only one who willingly listens to such talk is the doctor, but he gets paid for it.

3. To be a good conversationalist you must know not only what to say, but how also to say it. Be mentally quick and witty. But don’t hurt others with your wit. Finally, try to avoid mannerisms in your conversation. Don’t bite your lips or click your tongue, or roll your eyes or use your hands excessively as you speak.

4. Don’t be like that Frenchman who said, “How can I talk if you hold my hand?”

11th English Summary Writing Questions:
A. On the basis of your understanding of the above passage make notes on it using headings and sub-headings. Use recognizable abbreviations wherever necessary. Give an appropriate title.
B. Write a summary of the above passage in about 80 words.
Answers:
A. TITLE: The Art of Conversation NOTES:
1. Conv’n—most easily tch’ble art
(i) Reqd. interest’g subject – hobbies
(ii) Talk abt other fellow’s int./hobby
(iii) Win’g reptn. as good conversationalist
(a) good f ’ship
(b) charm’g wit
(c) brl. mind

 

2. Fit subs, for conversationalist
(i) What subs, to avoid/select?
(ii) Avoid unpl’nt subs.
(a) sickness
(b) death
(iii) Avoid talk’g abt self

3. Qualities A a good conversationalist
(i) What to say & how to say it
(ii) ment’y quick & witty
(iii) pleasant & unhurt’g
(iv) avoid mannerisms

Key to Abbreviations and Symbols used

• Conv’n – Conversation
• tch’ble – teachable
• Reqd. – Required
• interest’g – interesting
• abt – about
• int. – interest
• Win’g – Winning
• reptn. – reputation
• f’ship – friendship
• charm’g – charming
• brl. – brilliant
• subs. – subjects
• unpl’nt – unpleasant
• talk’g – talking
• A – of
• ment’y – mentally
• & – and
• unhurt’g – unhurting

 

B. Summary
Conversation is the easiest and the most effective tool than other arts. To have such attractive quality, you need to pick a subject that interests your listeners more than you. Talk to your friends on topics that can indulge your friends in the conversation for a longer period of time. Being a good conversationalist, you have to be quick and witty. You should have a pleasant and unhurting quality. Mannerism should be avoided.

Passage 2
1. A good business letter is one that gets results. The best way to get results is to develop a letter that, in its appearance, style and content, conveys information efficiently. To perform this function, a business letter should be concise, clear and courteous.

2. The business letter must be concise: don’t waste words. Little introduction or preliminary chat is necessary. Get to the point, make the point, and leave it. It is safe to assume that your letter is being read by a very busy person with all kinds of papers to deal with. Re-read and revise your message until the words and sentences you have used are precise. This takes time, but is a necessary part of a good business letter. A short business letter that makes its point quickly has much more impact on a reader than a long-winded, rambling exercise in creative writing. This does not mean that there is no place for style and even, on occasion, humour in the business letter. While it conveys a message in its contents, the letter also provides the reader with an impression of you, its author: the medium is part of the message.

 

3. The business letter must be clear. You should have a very firm idea of what you want to say, and you should let the reader know it. Use the structure of the letter—the paragraphs, topic sentences, introduction and conclusion—to guide the reader point by point from your thesis, through your reasoning, to your conclusion. Paragraph often, to break up the page and to lend an air of organisation to the letter. Use an accepted business-letter format. Re-read what you have written from the point of view of someone who is seeing it for the first time, and be sure that all explanations are adequate, all information provided (including reference numbers, dates, and other identification). A clear message, clearly delivered, is the essence of business communication.

4. The business letter must be courteous. Sarcasm and insults are ineffective and can often work against you. If you are sure you are right, point that out as politely as possible. Explain why you are right, and outline what the reader is expected to do about it. Another form of courtesy j is taking care in your writing and typing of the business letter. Grammatical and spelling errors (even if you call them typing errors) tell a reader that you don’t think enough of him or can > lower the reader’s opinion of your personality faster than anything you say, no matter how idiotic. There are excuses for ignorance; there are no excuses for sloppiness.

 

5. The business letter is your custom-made representative. It speaks for you and is a permanent record of your message. It can pay big dividends on the time you invest in giving it a concise message, a clear structure, and a courteous tone.

Note Making Conversation Is Indeed Questions:
A. On the basis of your understanding of the above passage make notes on it using headings and sub-headings. Use recognizable abbreviations wherever necessary. Give an appropriate title.
B. Write a summary of the above passage in about 80 words.
Answers:
A. TITLE: Writing a Business Letter
Notes:
1. Features of a gd. busns. letter
(i) conveys info efficiently to get results
(ii) is concise
(iii) is clear
(iv) is courteous

 

2. How to write a gd. busns. letter
(i) Making letter concise
(a) Intro shd be brief
(b) make ur pt in precise words and sent’s
(c) short letr more effective
(d) style is imp.—may ocasnly have hum’
(ii) Achieving clarity
(a) Have a clear idea of what you wish to say
(b) structr the letter—intro & conclsn.
(c) use accepted format; para, topic, sent’s
(d) check facts, expl’ns, refs.
(iii) Bejng courteous
(a) Expin. ur pt. politely—avoid sarcasm/insults.
(b) careful wrtg & typg.
(c) gram. & spel’g errors to be avoided

3. Importance of busns. letr
(i) a representative
(ii) permanent rec. message

 

Key to Abbreviations and Symbols used

• gd – good
• busns – business
• info – information
• shd – should
• letr – letter
• pt – point
• sent’s – sentences
• ur – your
• imp – important
• ocasnly – occasionally
• hum’r – humour
• & – and
• structr – structure
• intro – introduction
• conclsn – conclusion
• para – paragraphs
• expl’ns – explanations
• refs – references
• Expln – Explain
• wrtg – writing
• typg – typing
• gram – grammar
• spel’g – spelling
• rec – record

 

B. Summary
A good business letter is that lends you positive and quality results. To get such results, a business letter should be effective in appearance, style and content. Apart from this a letter should be concise, clear and courteous. The business letter should be to the point as the message can be clear to the reader with an impression of you. The structure of letter should have topic sentence, introduction, paragraphs to conclusion. Re-read the points you have written to avoid sarcasm and insults that can work against your motive. Further more grammar and spelling errors need to be avoided.

The main aim is to share the knowledge and help the students of 11th English to secure the best score in their final exams. Use the concepts of Samacheer Kalvi 11th English Book Solutions Reading Note-Making and Summarizing Questions and Answers in Real time to enhance your skills. If you have any doubts you can post your comments in the comment section, We will clarify your doubts as soon as possible without any delay.

Students can Download Computer Science Chapter 5 Working with Typical Operating System (Windows & Linux) Questions and Answers, Notes Pdf, Samacheer Kalvi 11th 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 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux)

Samacheer Kalvi 11th Computer Science Working with Typical Operating System (Windows & Linux) Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

11th Computer Science Chapter 5 Book Back Answers Question 1.
From the options given below, choose the operations managed by the operating system …………….
(a) Memory
(b) Processor
(c) I/O devices
(d) all of the above
Answer:
(d) all of the above

11th Computer Science 5th Lesson Book Back Answers Question 2.
Which is the default folder for many Windows Applications to save your file?
(a) My Document
(b) My Pictures
(c) Documents and Settings
(d) My Computer
Answer:
(a) My Document

Chapter 5 Computer Science Class 11 Question 3.
Under which of the following OS, the option Shift + Delete – permanently deletes a file or folder?
(a) Windows 7
(b) Windows 8
(c) Windows 10
(d) all of the OS
Answer:
(d) all of the OS

Samacheer Kalvi 11th Computer Application Question 4.
What is the meaning of “Hibernate” in Windows XP/Windows 7?
(a) Restart the Computer in safe mode
(b) Restart the Computer in hibernate mode
(c) Shutdown the Computer terminating all the running applications
(d) Shutdown the Computer without closing the running applications
Answer:
(c) Shutdown the Computer terminating all the running applications

11th Computer Application Samacheer Kalvi Question 5.
Which of the following OS is not based on Linux?
(a) Ubuntu
(b) Redhat
(c) Cent OS
(d) BSD
Answer:
(d) BSD

Question 6.
Which of the following in Ubuntu OS is used to view the options for the devices installed?
(a) Settings
(b) Files
(c) Dash
(d) VBox_GAs_5.2.2
Answer:
(d) VBox_GAs_5.2.2

Question 7.
Identify the default email client in Ubuntu …………….
(a) Thunderbird
(b) Firefox
(c) Internet Explorer
(d) Chrome
Answer:
(a) Thunderbird

Question 8.
Which is the default application for spreadsheets in Ubuntu? This is available in the software launcher ……………..
(a) LibreOffice Writer
(b) LibreOffice Calc
(c) LibreOffice Impress
(d) LibreOffice Spreadsheet
Answer:
(b) LibreOffice Calc

Question 9.
Which is the default browser for Ubuntu?
(a) Firefox
(b) Internet Explorer
(c) Chrome
(d) Thunderbird
Answer:
(a) Firefox

Question 10.
Where will you select the option to log out, suspend, restart, or shut down from the desktop of Ubuntu OS?
(a) Session Indicator
(b) Launcher
(c) Files
(d) Search
Answer:
(a) Session Indicator

PART – 2
II. Short Answers

Question 1.
Differentiate cut and copy options.
Answer:
Cut:

• The file is moved to the destination location.
• Edit → cut, Edit → paste is used for cut.
• Ctrl + X, Ctrl + V is used.
• Right click ? cut, Right click ? paste is used.

Copy:

• File will be present in both source destination places.
• Edit → copy, Edit → paste is used for copy.
• Ctrl + C, Ctrl + V is used.
• Right click → copy, Right click → paste is used.

Question 2.
What is the use of a file extension?
Answer:
File Extension is the second part of the file name. It succeeds the decimal point in the file name. It is used to identify the type of file and it is normally upto 3 to 4 characters long.
Example: exe, html.

Question 3.
Differentiate Files and Folders.
Answer:
The basic difference between the two is that files store data, while folders store files and other folders.
Files:
Files, on the other hand, can range from a few bytes to several gigabytes. They can be documents, programs, libraries, and other compilations of data.

Folders:
The folders, often referred to as directories, are used to organize files on the computer. The folders themselves take up virtually no space on the hard drive.

Question 4.
Differentiate Save and save As option.
Answer:
Save:
This will save the document without asking for a new name or location. It will over – write the original.

Save as:
This will prompt to save the document using a dialog box. You will have the ability to change the file name and/or location. If you choose the same file name and location it will over – write the original. Your working document will be the one you just saved.

Question 5.
What is Open Source?
Answer:
Open source refers to a program or software in which the source code is available in the web to all public, without any cost.

Question 6.
What are the advantages of open source?
Answer:
Advantages:

1. It is free of cost
2. Accessible to everybody.
3. Not belongs to particular vendor.
4. It can be modified, for requirements.

Question 7.
Mention the different server distributions in Linux OS.
Answer:
The server distribution in Linux OS:

1. Ubuntu Linux
2. Linux Mint
3. Arch Linux
4. Deepin
5. Fedora
6. Debian
7. Centos

Question 8.
How will you log off from Ubuntu OS?
Answer:
By choosing an option logout through the session indicator on the far right side of the top panel.

PART – 3
III. Explain in Brief

Question 1.
Analyse: Why the drives are segregated?
Answer:

1. It saves space and increases system performance.
2. Sharing and protection.
3. Segment and segregate the data to defend it from cyber attack.

Question 2.
If you are working on multiple files at a time, sometimes the system may hang.
What is the reason behind it. How can you reduce it?
Answer:
When we open multiple applications, each application opened on the system takes some internal and hardware resources to keep it running. Memory is used by all the applications. System processing speed will be shared and it will become slow finally the system will hang.

Methods of reducing it:

1. It is advisable to run one program at a time.
2. Upgrade RAM
3. Clean the cache Memories
4. Defragmentation can be done

Question 3.
Are drives such as hard drive and floppy drives represented with drive letters? If so why, if not why?
Answer:
Yes, hard drives and floppy drives are represented with drive letters.
A: drive is used for floppy disk of 3.5 inches and storage capacity of 1.44 MB.
B: drive is for floppy of size 5.25 inches and of storage capacity of 1.2 MB
So, we can say like
A: First Floppy Drive
B: Second Floppy Drive
C: D : E: …………….. Z: Hard Disk Drives, CD/DVD

Question 4.
Write the specific use of Cortana.
Answer:
Uses of Cortana:

1. Cortana is a voice activated personal assistant.
2. Give the reminders based on time, places, or people.
3. Track packages, teams, interests, and flights.
4. Send emails and texts.
5. Manage the calendar and keeps up to date.
6. Create and manage lists.
7. Chit chat and play games.
8. Find facts, files, places, and info.
9. Open any app on the system.

Question 5.
List out the major differences between Windows and Ubuntu OS.
Answer:

1. Ubuntu supports high security than windows.
2. Windows is more user friendly than Ubuntu
3. Ubuntu is a free and open source based Linux OS whereas windows is a closed source and it is not free.
4. Ubuntu uses Linux kernel whereas windows uses a hybrid kernel.

Question 6.
Are there any difficulties you face while using Ubuntu? If so, mention it with reasons.
Answer:

1. Ubuntu has less hardware support for commercial / industrial / medical/ logistical therefore its less votted for use in big time backends.
2. Ubuntu doesn’t support middlewares such as C panel, cloud linux and a plethora of other infrastructure or monitoring tools.
3. Hard to install graphic drivers especially for old hardwares. It is not possible to play the modern games, because of poor graphics quality.
4. The user switching from windows will not like the user experience on Ubuntu and will have difficulty in operating the OS.
5. Ubuntu is not capable of playing MP3 files by default.

Question 7.
Differentiate Thunderbird and Firefox in Ubuntu OS.
Answer:
Difference between Thunderbird and Firefox in Ubuntu:

1. Thunderbird is inbuilt email software of ubuntu whereas Firefox is a web browser.
2. Thunderbird is used to access email such as Gmail, hotmail whereas Firefox is used to open web pages including e-mail.

Question 8.
Differentiate Save, Save As and Save a Copy in Ubuntu OS.
Answer:

PART – 4
IV. Explain in Detail

Question 1.
Explain the versions of Windows Operating System.
Answer:

Question 2.
Draw and compare the icon equivalence in Windows and Ubuntu.
Answer:

Elements of Ubuntu:
Search your Computer Icon : This icon is equal to search button in Window’s OS. Here, you have to give the name of the File or Folder for searching them.

Files : This icon is equivalent to My Computer icon. From here, you can directly go to Desktop, Documents and so on.

Firefox Web Browser : By clicking this icon, you can directly browse the internet. This is equivalent to clicking the Web Browser in Task bar in Windows.

Libre Office Writer : This icon will directly take you to document preparation application like MS Word in Windows.

Libre Office Calc: This icon will open LibreOffice Calc application. It is similar to MS Excel in Windows.

LibreOffice Impress : By clicking this icon, you can open LibreOffice Impress to prepare any presentations in Ubuntu like MS PowerPoint.

Ubuntu Software Icon : This icon will let you add any additional applications you want. This can be done by clicking the Update option at the top right comer of that screen.

Online Shopping icon : Using this icon user can buy and sell any products online.

System Settings Icons : This icon is similar to the Control panel in the Windows Operating System. But here, you need to authenticate the changes by giving your password. You cannot simply change as you do in Windows.

Trash : This icon is the equivalent of Recycle bin of windows OS. All the deleted Files and Folders are moved here.

Question 3.
Complete the following matrix.

Answer:

Question 4.
Observe the figure and mark all the window elements. Identify the version of the Windows OS.
Answer:

Question 5.
Write the procedure to create, rename, delete and save a file in Ubuntu OS. Compare it with Windows OS.
Answer:

Samacheer kalvi 11th Computer Science Working with Typical Operating System (Windows & Linux) Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
………………. is Open source Operating System for desktop and server.
(a) Windows series
(b) Android
(c) iOS
(d) Linux
Answer:
(d) Linux

Question 2.
The most common way of opening a file or a Folder is to click on it.
(a) left
(b) right
(c) double
(d) single
Answer:
(c) double

Question 3.
If you want to select multiple files or folders, use ……………….
(a) Ctrl + shift
(b) Ctrl + click
(c) shift + click
(d) Ctrl + shift + click
Answer:
(b) Ctrl + click

Question 4.
………………. is a special folder to keep the files or folders deleted by the user, which means you still have an opportunity to recover them.
(a) My computer
(b) Documents
(c) Recycle bin
(d) Pictures
Answer:
(c) Recycle bin

Question 5.
………………. is one of the popular Open Source versions of the UNIX Operating System.
(a) Windows 7
(b) Windows 8
(c) Linux
(d) Android
Answer:
(c) Linux

Question 6.
………………. icon is equivalent to My Computer icon. From here, you can directly go to Desktop, Documents and so on.
(a) Files
(b) Documents
(c) Downloads
(d) Computer
Answer:
(a) Files

Question 7.
………………. icon is the equivalent of Recycle bin of windows OS. All the deleted Files and Folders are moved here.
(a) Trash
(b) Files
(c) Online shopping
(d) Libre Office Impress
Answer:
(a) Trash

Question 8.
The vertical bar of icons on the left side of the desktop is called the ………………..
(a) Search
(b) Libre office calc
(c) Launcher
(d) Files
Answer:
(c) Launcher

Question 9.
………………. manages network connections, allowing you to connect to a wired or wireless network.
(a) Toolbar
(b) Title bar
(c) Session indicator
(d) Network indicator
Answer:
(d) Network indicator

Question 10.
To permanently delete a file or folder (i.e. to avoid sending a file or folder to the Recycle Bin), hold down the SHIFT key, and press on the keyboard.
(a) restore
(b) delete
(c) send to
(d) cut
Answer:
(b) delete

Question 11.
Clock is available in ……………….
(a) system tray
(b) Files
(c) start
(d) My documents
Answer:
(a) system tray

Question 12.
………………. command should be typed in the Run dialog bar to open Calculator?
(a) Calculator
(b) Calc
(c) Arithmetic
(d) Calculator open
Answer:
(b) Calc

Question 13.
The menu bar is present below the ……………….
(a) Task bar
(b) Scroll bar
(c) Title bar
(d) Function bar
Answer:
(c) Title bar

Question 14.
Which of the following OS bar plug and play feature ……………….
(a) Window XP
(b) Windows 98
(c) Windows 95
(d) Windows me
Answer:
(b) Windows 98

Question 15.
………………. has the task for frequently used applications?
(a) Quick Launch Tool bar
(b) Settings
(c) My pc
(d) This pc
Answer:
(a) Quick Launch Tool bar

Question 16.
The winkey combination used to display desktop is ……………….
(a) winkey + dt
(b) winkey + T
(c) winkey + alt + D
(d) winkey + D
Answer:
(d) winkey + D

Question 17.
SSD stands for ……………….
(a) Solid State Devices
(b) Simple Stage Driver
(c) Single State Drivers
(d) Synchronized State Devices
Answer:
(a) Solid State Devices

Question 18.
The mouse pointer becomes ………………. when it is positioned over a border or a comer of a window.
(a) +
(b) arrow
(c) single headed arrow
(d) double headed arrow
Answer:
(d) double headed arrow

Question 19.
What is the name given to the document window to enter or type the text?
(a) Work space
(b) Work Area
(c) Typing Area
(d) Space
Answer:
(a) Work space

Question 20.
………………. enables alternate method of opening an application.
(a) Running
(b) Searching
(c) Run
(d) Open
Answer:
(c) Run

Question 21.
The disk drives mounted in the system can be seen by clicking ……………….
(a) Disk drive Icon
(b) Drive Icon
(c) Device Driver Icon
(d) My Computer Icon
Answer:
(d) My Computer Icon

Question 22.
What is the name given to the rectangular area in an application or a document?
(a) Document
(b) Window
(c) Application
(d) Desktop
Answer:
(b) Window

Question 23.
Windows 10 was developed in the year ……………….
(a) 2009
(b) 2012
(c) 2015
(d) 2018
Answer:
(c) 2015

Question 24.
Match the following:

(a) (3) (4) (2) (1)
(b) (1) (2) (3) (4)
(c) (4) (3) (2) (1)
(d) (4) (2) (1) (3)
Answer:
(a) (3) (4) (2) (1)

Question 25.
The Rulers are used to set ……………….
(a) Orientations
(b) Header
(c) Footer
(d) Margins
Answer:
(d) Margins

Question 26.
Which one of the following boots faster, mns apps faster compared to HDD.
(a) FDD
(b) Cache
(c) SSD
(d) DVD
Answer:
(c) SSD

Question 27.
Which functional key is used to bring the focus on, the first menu of the menu bar?
(a) F5
(b) F10
(c) F11
(d) F7
Answer:
(b) F10

Question 28.
How many disk drive icon options are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) 5

Question 29.
Which one of the following is used to open search results dialog box?
(a) search
(b) See more results
(c) search more results
(d) searching web
Answer:
(b) See more results

Question 30.
Which icon is used to check whether one system is connected to another system?
(a) Network
(b) System
(c) Control panel
(d) Hard drive
Answer:
(a) Network

Question 31.
The keyboard shortcut to save a file is ……………….
(a) alt + s
(b) Ctrl + s
(c) Ctrl + alt + s
(d) winkey + s
Answer:
(b) Ctrl + s

Question 32.
Which command is used to create new folder?
(a) File → folder
(b) File → New folder
(c) New → folder
(d) File → New → folder
Answer:
(d) File → New → folder

Question 33.
Applications or files or folders are opened using related shortcut icons by ……………….
(a) Click and drag
(b) double click
(c) click
(d) drag and drop
Answer:
(b) double click

Question 34.
In windows 7, which option is used from file menu to quit an application?
(a) Exit
(b) Close
(c) Quit
(d) Exit window
Answer:
(b) Close

Question 35.
Which option is used to save the file?
(a) Ctrl + s
(b) Save
(c) File + save
(d) All the above
Answer:
(d) All the above

Question 36.
Which is inbuilt Word Processor application to creat and manipulate text documents?
(a) Word pad
(b) MS – word
(c) Staroffice writer
(d) Notepad
Answer:
(a) Word pad

Question 37.
ubuntu supports office suite called ……………….
(a) Open Office
(b) Star Office
(c) Libre Office
(d) MS – Office
Answer:
(c) Libre Office

Question 38.
The in – built e – mail software facility in ubuntu is ……………….
(a) Debain
(b) Thunderbird
(c) Gmail
(d) Firefox
Answer:
(b) Thunderbird

Question 39.
Which one of the following is a server distribution of Linux?
(a) Deepin
(b) Firefox
(c) MS.word
(d) Files
Answer:
(a) Deepin

Question 40.
The popular Linux distributions is ……………….
(a) Deepin
(b) Centos
(c) Debian
(d) All the above
Answer:
(d) All the above

Question 41.
Which option is used to delete all files in the Recycle bin?
(a) Remove the Recycle bin
(b) Empty the Recycle bin
(c) Clear the Recycle bin
(d) Clean the Recycle bin
Answer:
(b) Empty the Recycle bin

Question 42.
The search text box in the computer disk drive screen will appear at ……………….
(a) Bottom right comer
(b) Top left comer
(c) Bottom left comer
(d) Top right comer
Answer:
(d) Top right comer

Question 43.
Which key is used to access the menu’s in the menu bar?
(a) shift
(b) control
(c) alt
(d) Tab
Answer:
(c) alt

Question 44.
Which one of the following is the open source for desktop and server?
(a) Linux
(b) MS – DOS
(c) BASIC
(d) COBOL
Answer:
(a) Linux

Question 45.
Which mouse actions is used to display popup menu?
(a) right click
(b) click
(c) Double click
(d) drag and drop
Answer:
(a) right click

Question 46.
What is used to interact with windows by clicking icons?
(a) Mouse
(b) Keyboard
(c) Monitor
(d) Printer
Answer:
(a) Mouse

Question 47.
Which menu contains layout options?
(a) option
(b) view
(c) organize
(d) Menu bar
Answer:
(c) organize

Question 48.
V Box stands for ……………….
(a) Virtual Image Box
(b) Virtual box
(c) Validity box
(d) Vat Box
Answer:
(b) Virtual box

Question 49.
Session Indicator is present in which comer?
(a) top left
(b) top right
(c) bottom left
(d) bottom right
Answer:
(b) top right

Question 50.
Hardware settings is used in which option?
(a) Monitor
(b) Display
(c) Theme
(d) My Computer
Answer:
(b) Display

Question 51.
In Text entry settings En, Fr, Ku are ……………….
(a) Desktop Layouts
(b) Keyboard Layouts
(c) Message Layouts
(d) Data Entry Layouts
Answer:
(b) Keyboard Layouts

Question 52.
Which one in ubuntu is similar to task bar of windows?
(a) Launcher
(b) Desktop
(c) start
(d) Notification area
Answer:
(a) Launcher

Question 53.
Notification area of ubuntu desktop is otherwise called as ……………….
(a) Task bar
(b) Desktop
(c) status bar
(d) Indicator area
Answer:
(d) Indicator area

Question 54.
The indicator that helps to access instant messenger and email client is ……………….
(a) Messaging
(b) Network
(c) Email
(d) Text entry
Answer:
(a) Messaging

Question 55.
The application in Ubuntu similar to MS – Excel in windows ……………….
(a) Spread sheet
(b) Libre Office calc
(c) Open Office calc
(d) Star Office calc
Answer:
(b) Libre Office calc

Question 56.
Which dropdown list box is used to select the location where you want to save the file?
(a) Location
(b) Look – in
(c) Peep in
(d) Search for
Answer:
(b) Look – in

Question 57.
Which menu has the rename option?
(a) File
(b) Edit
(c) View
(d) Window
Answer:
(a) File

Question 58.
How will you rename the file?
(a) Edit → Rename
(b) press F2
(c) right click → rename
(d) All the above
Answer:
(d) All the above

Question 59.
In which panel of disk drive window, the files and folders are displayed in tree like structures?
(a) Top
(b) Centre
(c) Left
(d) Right
Answer:
(c) Left

Question 60.
Delete option is present in which menu?
(a) File
(b) Edit
(c) View
(d) Tools
Answer:
(a) File

Question 61.
Which option reboot the computer?
(a) Restart
(b) Boot
(c) Reboot
(d) Reselect
Answer:
(a) Restart

Question 62.
Identify the menu item which is not present is the keyboard indicator menu?
(a) Character Map
(b) Keyboard Layout
(c) Keyboard Layout Chart
(d) Text entry settings
Answer:
(b) Keyboard Layout

Question 63.
There are ………………. types of indicators in the Menu bar.
(a) 5
(b) 6
(c) 1
(d)8
Answer:
(b) 6

Question 64.
What is the name of the default Ubuntu 16.04 theme?
(a) Ambiance
(b) V Box
(c) Trash
(d) Files
Answer:
(a) Ambiance

Question 65.
Who developed Ubuntu OS?
(a) Mark Shuttleworth
(b) Ricki Mascitti
(c) Dan Bricklin
(d) Bob Franston
Answer:
(a) Mark Shuttleworth

Question 66.
Which year ubuntu OS was developed?
(a) 2003
(b) 2004
(c) 2005
(d) 2006
Answer:
(b) 2004

Question 67.
Identify the statement which is given wrong?
(a) Open source code is available free of cost.
(b) ubuntu is a linux based OS.
(c) ubuntu has smart searching facility
(d) The horizontal bar of icons on the left side of the desktop is called Launcher.
Answer:
(d) The horizontal bar of icons on the left side of the desktop is called Launcher.

Question 68.
How many ways of creating files are there is windows?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 69.
How many sets of scroll bars are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 70.
How many versions of windows 2000 are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 71.
Which one of the following is used for high traffic computer networks in windows 2000?
(a) Professional
(b) Server
(c) Advanced server
(d) Data centre server
Answer:
(d) Data centre server

Question 72.
How many types of Icons are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 73.
What is the name given to the larger window?
(a) Work window
(b) Document window
(c) Application window
(d) Desktop
Answer:
(c) Application window

Question 74.
Which symbol is used to indicate that sub menu is attached to this option?
(a) +
(b) ▷
(c) ☐
(d) ◁
Answer:
(b) ▷

Question 75.
What is the keyboard shortcut for Exit option?
(a) Ctrl + E
(b) Alt + E
(c) Ctrl + Q
(d) Alt + Q
Answer:
(c) Ctrl + Q

Question 76.
How many methods of Renaming file are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 77.
Which one of the following is not a method of pasting the contents?
(a) Edit → paste
(b) Ctrl + V
(c) alt + V
(d) Right click → paste
Answer:
(c) alt + V

Question 78.
Which option is used as a part of installing new software or windows update?
(a) Lock
(b) Restart
(c) Sleep
(d) Hibernate
Answer:
(b) Restart

Question 79.
Which option is found only on Laptop?
(a) Lock
(b) Restart
(c) Sleep
(d) Hibernate
Answer:
(d) Hibernate

PART – 2
II. Short Answers

Question 1.
Name some distributions of Linux.
Answer:

1. Fedora
2. Ubuntu
3. BOSS
4. RedHat
5. Linux Mint.

Question 2.
What is Ubuntu?
Answer:
It is a Linux – based operating system. Designed for computers, smart phones and network servers.

Question 3.
What is Open source Operating system?
Answer:
It is the software in which the source code is available to the general public for use and modification from its original design, free of charge.

Question 4.
Mention any 2 significant features of Ubuntu?
Answer:

1. It supports the office suite called Libreoffice.
2. Ubuntu has in built email software called Thunderbird, which gives the user access to email such as Exchange, Gmail, Hotmail, etc.

Question 5.
What are the similarities between Ubuntu and other operating systems?
Answer:
All are based on the concepts of Graphical User Interface.

Question 6.
What is Multitasking?
Answer:
Multiple applications which can be executed simultaneously in Windows is known as Multitasking.

Question 7.
What is a workspace of a window?
Answer:
The workspace is the area in the document window to enter or type the text of your document. The workspace is the element of a window.

Question 8.
What is a Wordpad?
Answer:
It is an in – built word processor application in windows OS to create and manipulate text windows.

Question 9.
What is Folder?
Answer:
It is a container of files.

Question 10.
What is Firefox?
Answer:
It is a browser.

PART – 3
III. Explain in Brief

Question 1.
How will you renaming the files or folders?
Answer:
Using the FILE Menu

1. Select the File or Folder you wish to Rename.
2. Click File → Rename.
3. Type in the new name.
4. To finalise the renaming operation, press Enter

Question 2.
Draw the diagram of overview of an operating system
Answer:

Question 3.
List out the important functions of an OS?
Answer:
Following are some of the important functions of an Operating System:

1. Memory Management
2. Process Management
3. Device Management
4. File Management
5. Security Management
6. Control overall system performance
7. Error detecting aids
8. Coordination between other software and users

Question 4.
How will you create a shortcuts on the desktop?
Answer:
Creating Shortcuts on the Desktop:
Shortcuts to your most often used folders and files may be created and placed on the Desktop to help automate your work.

1. Select the file or folder that you wish to have as a shortcut on the Desktop.
2. Right click on the file or folder.
3. Select Send to from the shortcut menu, then select Desktop (create shortcut) from the submenu.
4. A shortcut for the file or folder will now appear on your desktop and you can open it from the desktop in the same way as any other icon.

Question 5.
What are the elements of Ubuntu?
Answer:

1. Search your Computer Icon
2. Files
3. Firefox Web browser
4. Libre Office Writer
5. Libre Office Calc
6. Libre office Impress
7. Ubuntu software icon
8. Online shopping icon
9. System settings icon
10. Trash

Question 6.
Write note on windows XP.
Answer:

1. Windows XP was developed in the year 2001
2. It introduced 64 bit processor.
3. Improved windows appearance with themes and offered a stable version.

Question 7.
What are the four versions of windows 2000.
Answer:

1. Professional – business desktop and laptop systems.
2. Server – used as both web server and an office server.
3. Advanced server – for line of business applications.
4. Data center server – for high-traffic computer networks.

Question 8.
What are the advantages of using windows 8 OS?
Answer:

1. Windows 8 is faster than previous versions of windows.
2. It has multi – core processing, Solid State Drivers (SSD), touch screens and other alternate input methods.
3. It served as common platform for mobile and computer.

Question 9.
What is a Window?
Answer:
Window is a typical rectangular area in an application or a document. It is an area on the screen that displays information for a specific program. The two types of windows are application window and document window.

Question 10.
Differentiate Application Window and Document Window?
Answer:
Application Window:

• It is the larger window.
• This window helps the user to communicate with the application program.

Document Window:

• It is the smaller window and appears inside the Application Windows.
• This window is used for typing, editing, drawing and formatting the text and graphics.

Question 11.
How will you start an application?
Answer:

1. Click the start button and then point to all programs. The program menu appears. Point to the group that contains the application you want to start and then click the application name.
2. Click Run on the start menu and type the name of the application and click ok.

Question 12.
How will you close an application?
Answer:

1. To quit a application, click the close button in the upper right comer of the application window.
2. Click File → Exit
3. Click File → Close option

Question 13.
How will you open the word pad?
Answer:

1. Click start → All programs → Accessories → Wordpad
2. Run → type wordpad, click ok

Question 14.
How will you delete a file or folder?
Answer:
Select the file or folder you wish to delete.

1. Right click the file or folder, select delete option from the popup menu or click file → delete (or) Press delete key from the keyboard.
2. The file will be deleted and moved to the Recycle bin.

Question 15.
Write note on Recycle bin.
Answer:
Recycle bin is a special folder to keep the files or folders deleted by the user. From the recycle bin, files can be restored back.

PART – 4
IV. Explain in Detail

Question 1.
Discuss about the mouse actions.
Answer:

Action	Reaction
Point to an item	Move the mouse pointer over the item.
Click	Point to the item on the screen, press and release the left mouse button.
Right click	Point to the item on the screen, press and release the right mouse button. Clicking the right mouse button displays a pop up menu with various options.
Double-click	Point to the item on the screen, quickly press twice the left mouse button.
Drag and drop	Point to an item then hold the left mouse button as you move the pointer press and you have reached the desired position, release the mouse button.

Question 2.
Discuss the following Icons.
Answer:
Shortcut Icons:
Shortcut icons can be created for any application or file or folder. By double clicking the icon, the related application or file or folder will open. This represents the shortcut to open a particular application.

Disk drive icons:
The disk drive icons graphically represent five disk drive options.

1. Hard disk.
2. CD – ROM/DVD Drive.
3. Pen drive.
4. Other removable storage such as mobile, smart phone, tablet etc.
5. Network drives if your system is connected with other system.

Question 3.
How will you copying files and folders to removable disk?
Answer:
Copying Files and Folders to removable disk:
There are several methods of transferring files to or from a removable disk.

1. Copy and Paste
2. Send To

METHOD I – Copy and Paste:

1. Plug the USB flash drive directly into an available USB port.
2. If the USB flash drive or external drive folder does NOT open automatically, follow these steps:
3. Click Start → Computer.
   
4. Double Clicking on the Removable Disk associated with the USB flash drive.
   
5. Navigate to the folders in your computer containing files you want to transfer. Right – click on the file you want to copy, then select Copy.
   
6. Return to the Removable Disk window, right – click within the window, then select Paste.
   

METHOD II – Send To:

1. Plug the USB flash drive directly into an available USB port.
2. Navigate to the folders in your computer containing files you want to transfer.
3. Right – click on the file you want to transfer to your removable disk.
4. Click Send To and select the Removable Disk associated with the USB flash drive.
   

Question 4.
Explain the various elements of a window.
Answer:
Elements of a window:

Title Bar : The title bar will display the name of the application and the name of the document opened. It will also contain minimize, maximize and close button.

Menu Bar : The menu bar is seen under the title bar. Menus in the menu bar can be accessed by pressing Alt key and the letter that appears underlined in the menu title. Additionally, pressing Alt or F10 brings the focus on the first menu of the menu bar. In Windows 7, in the absence of the menu bar, click Organise and from the drop down menu, click the Layout option and select the desired item, from that list.

The Workspace : The workspace is the area in the document window to enter or type the text of your document.

Scroll bars : The scroll bars are used to scroll the workspace horizontally or vertically.

Corners and borders : The comers and borders of the window helps to drag and resize the windows. The mouse pointer changes to a double headed arrow when positioned over a border or a comer. Drag the border or comer in the direction indicated by the double headed arrow to the desired size as shown in the Figure. The window can be resized by dragging the comers diagonally across the screen.

Question 5.
Explain various methods of creating Files and Folders.
Answer:
Creating files and Folders:
Creating Folders:
You can store your files in many locations – on the hard disk or in other devices. To better organise your files, you can store them in folders.
There are two ways in which you can create a new folder:

Method I:
Step 1: Open Computer Icon.
Step 2: Open any drive where you want to create a new folder. (For example select D:)
Step 3: Click on File → New → Folder.
Step 4: A new folder is created with the default name “New folder”.
Step 5: Type in the folder name and press Enter key.

Method II:
In order to create a folder in the desktop:
Step 1: In the Desktop, right click → New → Folder.
Step 2: A Folder appears with the default name “New folder” and it will be highlighted.
Step 3: Type the name you want and press Enter Key.
Step 4: The name of the folder will change.

Question 6.
Explain how will you rename files and folders.
Answer:
Renaming Files or Folders:
There are number of ways to rename files or folders. You can rename using the File menu, left mouse button or right mouse button.

Method I:
Using the FILE Menu

1. Select the File or Folder you wish to Rename.
2. Click File → Rename.
3. Type in the new name.
4. To finalise the renaming operation, press Enter.

Method II:
Using the Right Mouse Button

1. Select the file or folder you wish to rename.
2. Click the right mouse button over the file or folder.
3. Select Rename from the pop-up menu. ’
4. Type in the new name.
5. To finalise the renaming operation, press Enter.
6. The folder “New Folder” is renamed as C++.

Method III:
Using the Left Mouse Button

1. Select the file or folder you wish to rename.
2. Press F2 or click over the file or folder. A surrounding rectangle will appear around the name.
3. Type in the new name.
4. To finalise the renaming operation, press Enter.

Question 7.
How will you move the file or folder from one place to another?
Answer:
Moving/Copying Files and Folders:
You can move your files or folders to other areas using variety of methods.

Moving Files and Folders:
Method I – CUT and PASTE
To move a file or folder, first select the file or folder and then choose one of the following:

1. Click on the Edit → Cut or Ctrl + X Or right click → cut from the pop – up menu.
2. To move the file(s) or folder(s) in the new location, navigate to the new location and paste it using Click Edit → Paste from edit menu or Ctrl + V using keyboard.
3. Or Right click → Paste from the popup menu. The file will be pasted in the new location.

Method II – Drag and Drop:

In the disk drive window, we have two panes called left and right panes. In the left pane, the files or folders are displayed like a tree structure. In the right pane, the files inside the specific folders in the left pane are displayed with various options.

1. In the right pane of the Disk drive window, select the file or folder you want to move.
2. Click and drag the selected file or folder from the right pane, to the folder list on the left pane.
3. Release the mouse button when the target folder is highlighted (active).
4. Your file or folder will now appear in the new area.

Question 8.
How will you copy files and folders.
Answer:
Copying Files and Folders:
There are variety of ways to copy files and folders:

Method I – COPY and PASTE:
To copy a file or folder, first select the file or folder and then choose one of the following:

1. Click Edit → Copy or Ctrl + C or right click → Copy from the pop – up menu.
2. To paste the file(s) or folder(s) in the new location, navigate to the target location then do one of the following:
3. Click Edit → Paste or Ctrl + V.
4. Or Right click → Paste from the pop – up menu.

Method II – Drag and Drop:

1. In the RIGHT pane, select the file or folder you want to copy.
2. Click and drag the selected file and/or folder to the folder list on the left, and drop it where you want to copy the file and/or folder.
3. Your file(s) and folder(s) will now appear in the new area.

Question 10.
Explain various menu bar Indicator.
Menu bar:
The menu bar is located at the top of the screen. The menu bar incorporates common functions used in Ubuntu. The frequently used icons in the menu bar are found on the right. The most common indicators in the Menu bar are located in the indicator or notification area.

Network indicator:
This manages network connections, allowing you to connect to a wired or wireless network. Text entry settings:
This shows the current keyboard layout (such as Eli, Fr, Ku, and so on). If more than one keyboard layout is shown, it allows you to select a keyboard layout out of those choices. The keyboard indicator menu contains the following menu items: Character Map, Keyboard Layout Chart, and Text Entry Settings.

Messaging indicator:
This incorporates your social applications. From here, you can access instant messenger and email clients.

Sound indicator:
This provides an easy way to adjust the volume as well as access your music player.

Clock:
This displays the current time and provides a link to your calendar and time and date settings.

Question 11.
Explain various elements of Ubuntu.
Answer:

Search your Computer Icon : This icon is equal to search button in Windows OS. Here, you have to give the name of the File or Folder for searching them.

Files : This icon is equivalent to My Computer icon. From here, you can directly go to Desktop, Documents and so on.

Firefox Web Browser : By clicking this icon, you can directly browse the internet. This is equivalent to clicking the Web Browser in Task bar in Windows.

Libre Office Writer : This icon will directly take you to document preparation application like MS Word in Windows.

Libre Office Calc : This icon will open LibreOffice Calc application. It is similar to MS Excel in Windows.

Libre Office Impress : By clicking this icon, you can open LibreOffice Impress to prepare any presentations in Ubuntu like MS PowerPoint.

Ubuntu Software Icon : This icon will let you add any additional applications you want. This can be done by clicking the Update option at the top right comer of that screen.

Online Shopping icon : Using this icon user can buy and sell any products online.

System Settings Icons : This icon is similar to the Control panel in the Windows Operating System. But here, you need to authenticate the changes by giving your password. You cannot simply change as you do in Windows.

Trash : This icon is the equivalent of Recycle bin of windows OS. All the deleted Files and Folders are moved here.

Students can Download Accountancy Chapter 7 Subsidiary Books – II Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy 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 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Samacheer Kalvi 11th Accountancy Subsidiary Books – II Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

11th Accountancy Chapter 7 Book Back Answers Question 1.
Cash book is a ………………
(a) Subsidiary book
(b) Principal book
(c) Journal proper
(d) Both subsidiary book and principal book
Answer:
(d) Both subsidiary book and principal book

Subsidiary Books Questions And Answers Pdf Question 2.
The cash book records ………………
(a) All cash receipts
(b) All cash payments
(c) Both (a) and (b)
(d) All credit transactions
Answer:
(c) Both (a) and (b)

11th Accountancy 7th Chapter Solutions Question 3.
When a firm maintains a simple cash book, it need not maintain ………………
(a) Sales account in the ledger
(b) Purchases account in the ledger
(c) Capital account in the ledger
(d) Cash account in the ledger
Answer:
(d) Cash account in the ledger

11th Accountancy Book Back Answers Question 4.
A cash book with discount, cash and bank column is called ………………
(a) Simple cash book
(b) Double column cash book
(c) Three column cash book
(d) Petty cash book
Answer:
(c) Three column cash book

11th Accountancy – Book Back Answers Question 5.
In Triple column cash book, the balance of bank overdraft brought forward will appear in ………………
(a) Cash column debit side
(b) Cash column credit side
(c) Bank column debit side
(d) Bank column credit side
Answer:
(d) Bank column credit side

Samacheer Kalvi 11th Accountancy Chapter 7 Question 6.
Which of the following is recorded as contra entry?
(a) Withdrew cash from bank for personal use
(b) Withdrew cash from bank for office use
(c) Direct payment by the customer in the bank account of the business
(d) When bank charges interest
Answer:
(b) Withdrew cash from bank for office use

Samacheer Kalvi Guru 11th Accountancy Question 7.
If the debit and credit aspects of a transaction are recorded in the cash book, it is ………………
(a) Contra entry
(b) Compound entry
(c) Single entry
(d) Simple entry
Answer:
(a) Contra entry

Accounting 11 Chapter 7 Answer Key Question 8.
The balance in the petty cash book is ………………
(a) An expense
(b) A profit
(c) An asset
(d) A liability
Answer:
(c) An asset

Class 11 Accountancy Chapter 7 Solutions Question 9.
Petty cash may be used to pay ………………
(a) The expenses relating to postage and conveyance
(b) Salary to the Manager
(c) Purchase of furniture and fixtures
(d) Purchase of raw materials
Answer:
(a) The expenses relating to postage and conveyance

Accounting Chapter 7 Answer Key Question 10.
Small payments are recorded in a book called ………………
(a) Cash book
(b) Purchase book
(c) Bills payable book
(d) Petty cash book
Answer:
(d) Petty cash book

II. Very Short Answer Questions

11th Accountancy Book Answers Question 1.
What is cash book?
Answer:
Cash book is the book in which only cash transactions are recorded in the chronological order. The cash book is the book of original entry or prime entry as cash transactions are recorded for the first time in it. Cash transactions here may include bank transactions also. Cash receipts are recorded on the debit side while cash payments are recorded on the credit side.

Class 11 Accounts Subsidiary Books Solutions Question 2.
What are the different types of cash book?
Answer:
The main cash book may be of various types and following are the three most common types.

1. Simple or single column cash book (only cash column)
2. Cash book with cash and discount column (double column cash book)
3. Cash book with cash, discount and bank columns (three column cash book)

Accountancy Class 11 Chapter 7 Solutions Question 3.
What is simple cash book?
Answer:
Single column cash book or simple cash book, like a ledger account has only one amount column, i.e., cash column on each side. Only cash transactions are recorded in this book. All cash receipts and payments are recorded systematically in this book.

Chapter 7 Accountancy Class 11 Question 4.
Give the format of ‘Single column cash book’.
Answer:
Simple Cash Book

Subsidiary Books Class 11 Notes Question 5.
What is double column cash book?
Answer:
It is a cash book with cash and discount columns. As there are two columns, i.e., discount and cash columns, both on debit and credit sides, this cash book is known as ‘double column cash book’.

Question 6.
Give the format of ‘Double column cash book’.
Answer:
Cash book with cash and discount columns

Question 7.
What is three column cash book?
Answer:
A three column cash book includes three amount columns on both sides, i.e., cash, bank and discount. This cash book is prepared in the same way as simple and double column cash books are prepared. The transactions which increase the cash and bank balance are recorded on the debit side of the cash and bank columns respectively. Opening balance of cash and favourable bank balance appear as the first item on the debit side of the three column cash book in case of existing business. If the business is a new one, capital contributed in cash and/or bank deposit appear as the first item on the debit side.

Question 8.
What is cash discount?
Answer:
Cash discount is allowed to the parties making prompt payment within the stipulated period of time or early payment. It is discount allowed (loss) for the creditor and discount received (gain) for the debtor who makes payment. The discount is allowed when payment is received or made and hence, the entry for discount is also passed with the entry of payment. The earlier the payment, the more may be the discount. Cash discount motivates the debtor to make the payment at an earlier date to avail discount facility.

Question 9.
What is trade discount?
Answer:
Trade discount is a deduction given by the supplier to the buyer on the list price or catalogue price of the goods. It is given as a trade practice or when goods are purchased in large quantities. It is shown as a deduction in the invoice. Trade discount is not recorded in the books of accounts. Only the net amount is recorded.

Question 10.
What is a petty cash book?
Answer:
Business entities have to pay various small expenses like taxi fare, bus fare, postage, carriage, stationery, refreshment and other sundry items. These are small payments and repetitive in nature. If all these small payments are recorded in the main cash book, it will be loaded with lot of entries. Hence, all petty payments of the business may be recorded in a separate book, which is called as petty cash book and the person who maintains the petty cash book is called the petty cashier.

III. Short Answer Questions

Question 1.
Explain the meaning of imprest system of petty cash book.
Answer:
Under this system, a fixed amount necessary or sufficient to meet petty payments determined on the basis of past experience is paid to the petty cashier on the first day of the period. (It may be a week or fortnight or month). The amount given to the petty cashier in advance is known as “Imprest Money”.

The word imprest means payment in advance. The petty cashier makes payments from this amount and records them in petty cash book. At the end of a particular period, the petty cashier submits the petty cash book to the head cashier.

The head cashier scrutinises the petty payments and gives amount equal to the amount spent by petty cashier so that the total amount with the petty cashier is now equal to the amount he had received in the beginning as advance. Under the system, the total cash with the petty cashier never exceeds the imprest at any time during the period.

Question 2.
Bring out the differences between cash discount and trade discount.
Answer:
Following are the difference between cash discount and trade discount:

Question 3.
Write the advantages of maintaining petty cash book.
Answer:
Following are the advantages of maintaining petty cash book:

1. There can be better control over petty payments.
2. There is saving of time of the main cashier.
3. Cash book is not loaded with many petty payments.
4. Posting of entries from main cash book and petty cash book is comparatively easy.

Question 4.
Write a brief note on accounting treatment of discount in cash book.
Answer:
In the discount columns, cash discount, i.e., cash discount allowed and cash discount received are recorded. The net amount received is entered in the amount column on the debit side and the net amount paid is entered in the amount column on the credit side. For the seller who allows cash discount, it is a loss and hence it is debited and shown on the debit side of the cash book. For the person making payment, discount received is a gain because less payment is made and it is credited and shown on the credit side of the cash book.

Question 5.
Briefly explain about contra entry with examples.
Answer:
To denote that there are contra entries, the alphabet ‘C’ is written in L.F. column on both sides. Contra means that particular entry is posted on the other side (contra) of the same book, because Cash account and Bank account are there in the cash book only and there are no separate ledger accounts needed for this purpose. The alphabet ‘C’ indicates that no further posting is required and the relevant account is posted on the opposite side.

IV. Exercises

Question 1.
Enter the following transactions in a single column cash book of Seshadri for May, 2017.

Answer:
In the books of Seshadri
Single column cash book

Question 2.
Enter the following transactions in a single column cash book of Pandeeswari for the month ofJune, 2017

Answer:
In the books of Pandeeswari
Single column cash book

Question 3.
Enter the following transactions in a single column cash book of Ramalingam for month of July, 2017.

Answer:
In the books of Ramalingam
Single column cash book

Question 4.
Enter the following transaction in Ahamad’s cash book with discount and cash columns.

Answer:
In the books of Ahamad’s
Cash book with discount and cash columns

Question 5.
Enter the following transaction in Chandran’s cash book with cash and discount column.

Answer:
In the books of chandran
Cash book with cash and discount columns

Question 6.
Enter the following transactions in cash book with discount and cash column of Anand

Answer:
In the books of Anand
Cash book with cash and discount and cash columns

Question 7.
Write out a cash book with discount, cash and bank columns in the books of Mahendran.

Answer:
In the books of Mahendran
There – column cash book

Question 8.
Enter the following transactions in the three column cash book of Kalyana Sundaram

Answer:
In the books of Kalyana Sundaram
There – column cash book

Question 9.
Enter the following transactions of Fathima in the cash book with cash, bank and discount columns for the month of May, 2017.

Answer:
In the books of Fathima
Three – column cash book with cash, bank and discount

Question 10.
Enter the following transactions in the three column cash book of Chozhan.

Answer:
In the books of Fathima
Three – column cash book with cash, bank and discount

Question 11.
Enter the following transactions in a cash book with cash, bank and discount columns of Sundari.

Answer:
In the books of Sundari
Three – column cash book

Question 12.
Record the following transactions in the three column cash book of Rajeswari for the month of June, 2017.

Answer:
In the books of Rajeswari
Three – column cash book

Question 13.
Record the following transactions in three column cash book of Ramachandran.

Answer:
In the books of Ramachandran
Three – column cash book

Question 14.
Record the following transactions in the three column cash book of John Pandian.
In the books of John Pandian
Three – column cash book

Answer:
In the books of John Pandian
Three – column cash book

Question 15.
Prepare a triple column cash book of Rahim from the following transactions:

In the hooks of Rahim
Three – column cash book
Answer:

Question 16.
Prepare analytical petty cash book from the following particulars under imprest system:

Answer:
Analytical Petty Cash Book (in ₹)

Question 17.
From the following information prepare an analytical petty cash book under imprest system:

Answer:
Analytical Petty Cash Book (in ₹)

Question 18.
Record the following transactions in an analytical petty cash book and balance the same. On 1^st November, 2017, the petty cashier started with imprest cash ₹ 2,000.

Answer:
Analytical Petty Cash Book (in ₹)

Question 19.
Enter the following transactions in Iyyappan’s petty cash book with analytical columns under imprest system.

Answer:
Analytical Petty Cash Book (in ₹)

Textbook Case Study Solved

Vetri is a sole trader selling food products. He maintains a simple cash book. He sells and purchases goods both on cash and credit. He maintains the cash book by himself. He allows discount and receives discount. He has his personal bank account. He also has so many petty expenses. Now, he wants to establish his business. But he wants to maintain the cash book all by himself.
Now, discuss on the following points:

Question 1.
What could be the reason that Vetri maintains the cash book by himself?
Answer:
He is a sole trader, he need not show the accounts to anybody else, he wants to know about profit or loss for himself only. So he maintains cash book only.

Question 2.
Is it convenient for him to record all the cash transactions in the simple cash book?
Answer:
No, it is not convenient for him to record all the cash transactions in the simple cash book.

Question 3.
Will his personal bank account serve the purpose of his business transactions?
Answer:
No, his personal bank account will not serve the purpose of his business transactions.

Question 4.
Suggest him some better ways of maintaining the cash transactions.
Answer:
Suggestions:

He may maintain triple column cash book, because he can know all cash transactions is a same account.
Instead of personal bank account he can open business bank account (i.e.) current account.

Question 5.
When his business becomes large, what other books will he be maintaining?
Answer:
He will be maintaining the following other books:

1. Triple column cash book.
2. Petty cash book (Analytical).
3. Purchases book for credit purchases
4. Sales book for credit sales.
5. Purchases returns book.
6. Sales returns book.
7. Business book account (i.e. current account to be maintained).
8. Proper journal for other assets maintaining:
   • All cash transactions recorded in cash book.
   • All petty expenses are to recorded in analytical petty cash book.
   • All credit transactions to recorded in special purpose books (i.e. purchases book, sales book, purchases return book and sales returns book and proper journal.

Samacheer Kalvi 11th Accountancy Subsidiary Books – II Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
Cash receipts are recorded on the ……………. of the cash book.
(a) debit side
(b) credit side
(c) journal
(d) ledger
Answer:
(a) debit side

Question 2.
Cash payments are recorded on the ……………. of the cash book.
(a) debit side
(b) credit side
(c) contra
(d) journal
Answer:
(b) credit side

Question 3.
All cash transactions are straightway recorded in the …………….
(a) cash book
(b) ledger
(c) journal
(d) contra
Answer:
(a) cash book

Question 4.
……………. serves the purpose of a journal and a ledger.
(a) cash book
(b) purchase book
(c) sales book
(d) petty cash book
Answer:
(a) cash book

Question 5.
There are ……………. types of cash book (common).
(a) three
(b) four
(c) five
(d) two
Answer:
(a) three

Question 6.
R.N. expands for …………….
(a) Receipts number
(b) Roll number
(c) Route number
(d) Rank number
Answer:
(a) Receipts number

Question 7.
L.F. expands for ……………..
(a) Long file
(b) Ledger folio
(c) Line folio
(d) Link file
Answer:
(b) Ledger folio

Question 8.
V.N. exapnds for …………….
(a) Value number
(b) Voucher number
(c) Vendor number
(d) Volume number
Answer:
(b) Voucher number

Question 9.
As both the debit and credit aspects of a transaction are recorded in the cash book, such entries are called ……………. entries.
(a) Contra
(b) Journal
(c) Compound
(d) Opening
Answer:
(a) Contra

Question 10.
The word ……………. means payment in advance.
(a) imprest
(b) loan
(c) money
(d) petty cash
Answer:
(a) imprest

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Download the Samacheer Kalvi 11th Physics Book Solutions Questions and Answers Notes Pdf, for all Chapter Wise through the direct links available and take your preparation to the next level. The Tamilnadu State Board Solutions for 11th Physics Chapter 1 Nature of Physical World and Measurement Questions and Answers covers all the topics and subtopics within it. Practice using these Samacheer Kalvi 11th Physics Book Solutions Questions and Answers for Chapter 1 Nature of Physical World and Measurement PDF and test your preparation level and bridge the knowledge gap accordingly.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

If you have any queries take the help of the Tamilnadu State Board Solutions for 11th Physics Chapter 1 Nature of Physical World and Measurement Questions and Answers learn all the topics in it effectively. We have everything covered and you can Practice them often to score better grades in your exam. By going through the Samacheer Kalvi 11th Physics Book Solutions Questions and Answers you can attempt the actual exam with utmost confidence.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement TextBook Questions Solved

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Multiple Choice Questions

Samacheer Kalvi 11th Physics Solution Chapter 1 Question 1.
One of the combinations from the fundamental physical constants is \(\frac{h c}{\mathrm{G}}\). The unit of this expression is
(a) Kg²
(b) m³
(c) S^-1
(d) m
Answer:
(a) Kg²

Nature Of Physical World And Measurement Question 2.
If the error in the measurement of radius is 2%, then the error in the determination of volume of the sphere will be …….
(a) 8%
(b) 2%
(c) 4%
(d) 6%
Answer:
(d) 6%

Nature Of Physical World And Measurement Questions And Answers Question 3.
If the length and time period of an oscillating pendulum have errors of 1 % and 3% respectively
then the error in measurement of acceleration due to gravity is …… [Related to AMPMT 2008]
(a) 4%
(b) 5%
(c) 6%
(d) 7%
Answer:
(d) 7%

11th Physics Nature Of Physical World And Measurement Question 4.
The length of a body is measured as 3.51 m, if the accuracy is 0.01mm, then the percentage error in the measurement is ……
(a) 351%
(b) 1%
(c) 0.28%
(d) 0.035%
Answer:
(c) 0.28%

Nature Of Physical World And Measurement Class 11 Question 5.
Which of the following has the highest number of significant figures?
(a) 0.007 m²
(b) 2.64 × 1024 kg
(c) 0.0006032 m²
(d) 6.3200 J
Answer:
(d) 6.3200 J

Nature Of Physical World And Measurement Class 11 Notes Question 6.
If π = 3.14, then the value of π² is …..
(a) 9.8596
(b) 9.860
(c) 9.86
(d) 9.9
Answer:
(c) 9.86

Samacheer Kalvi Guru 11th Physics Question 7.
Which of the following pairs of physical quantities have same dimension?
(a) force and power
(b) torque and energy
(c) torque and power
(d) force and torque
Answer:
(b) torque and energy

11th Physics Unit 1 Question 8.
The dimensional formula of Planck’s constant h is [AMU, Main, JEE, NEET]
(a) [ML²T^-1]
(b) [ML²T^-3]
(c) [MLTT^-1]
(d) [MLTT^3-3]
Answer:
(a) [ML²T^-1]

Physics Class 11 Chapter 1 Question 9.
The velocity of a particle v at an instant t is given by v = at + bt². The dimensions of b is ……
(a) [L]
(b) [LT^-1]
(c) [LT^-2]
(d) [LT^-3]
Answer:
(d) [LT^-3]

Samacheerkalvi.Guru 11th Physics Question 10.
The dimensional formula for gravitational constant G is [Related to AIPMT 2004]
(a) [ML^-3T^-2]
(b) [M^-1L³T^-2]
(c) [M^-1L^-3T^-2]
(d) [ML^-3T²]
Answer:
(b) [M^-1L³T^-2]

Nature Of Physical World And Measurement In Tamil Question 11.
The density of a material in CGS system of units is 4 g cm^-3. In a system of units in which unit of length is 10 cm and unit of mass is 100 g, then the value of density of material will be ……
(a) 0.04
(b) 0.4
(c) 40
(d) 400
Answer:
(c) 40

Samacheer Kalvi 11th Physics Question 12.
If the force is proportional to square of velocity, then the dimension of proportionality constant
is [JEE-2000] ……
(a) [MLT⁰]
(b) [MLT^-1]
(c) [ML^-2T]
(d) [ML^-1T⁰]
Answer:
(d) [ML^-1T⁰]

Samacheer Kalvi 11th Physics Solution Book Question 13.
[MainAIPMT2011]
(a) length
(b) time
(c) velocity
(d) force
Answer:
(c) velocity

11 Th Physics Samacheer Kalvi Guide Question 14.
Planck’s constant (h), speed of light in vaccum (c) and Newton’s gravitational constant (G) are taken as three fundamental constants. Which of the following combinations of these has the dimension of length? [NEET 2016 (Phase II)]

Answer:

Samacheer Kalvi 11th Physics Guide Question 15.
A length-scale (l) depends on the permittivity (ε) of a dielectric material, Boltzmann constant (k_B), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression for l is dimensionally correct?. [JEE (advanced) 2016]

Answer:

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Short Answer Questions

11th Physics Chapter 1 Numerical Problems Question 1.
Briefly explain the types of physical quantities.
Answer:
Physical quantities are classified into two types. There are fundamental and derived quantities. Fundamental or base quantities are quantities which cannot be expressed in terms of any other physical quantities. These are length, mass, time, electric current, temperature, luminous intensity and amount of substance.
Quantities that can be expressed in terms of fundamental quantities are called derived quantities. For example, area, volume, velocity, acceleration, force.

11th Physics Samacheer Kalvi Question 2.
How will you measure the diameter of the Moon using parallax method?
Answer:
Let θ is the angular diameter of moon
d – is the distance of moon from earth, from figure, θ = \(\frac{\mathrm{D}}{d}\)

Diameter of moon D = d.θ
by knowing θ, d, diameter of moon can be calculated

Samacheer Kalvi Physics 11th Question 3.
Write the rules for determining significant figures.
Answer:
Rules for counting significant figures:

Note: 1 Multiplying or dividing factors, which are neither rounded numbers nor numbers representing measured values, are exact and they have infinite numbers of significant figures as per the situation.
For example, circumference of circle S = 2πr, Here the factor 2 is exact number. It can be written as 2.0, 2.00 or 2.000 as required.
Note: 2 The power of 10 is irrelevant to the determination of significant figures.
For example x = 5.70 m = 5.70 × 10² cm = 5.70 × 10³ mm = 5.70 × 10^-3 km.
In each case the number of significant figures is three.

Physics Class 11 Samacheer Kalvi Question 4.
What are the limitations of dimensional analysis?
Answer:
Limitations of Dimensional analysis
1. This method gives no information about the dimensionless constants in the formula like 1, 2, ……… π, e, etc.
This method cannot decide whether the given quantity is a vector or a scalar.
This method is not suitable to derive relations involving trigonometric, exponential and logarithmic functions.
It cannot be applied to an equation involving more than three physical quantities.
It can only check on whether a physical relation is dimensionally correct but not the correctness of the relation. For example, using dimensional analysis, is dimensionally correct whereas the correct relation is

Samacheer Kalvi Guru 11 Physics Question 5.
Define precision and accuracy. Explain with one example.
Answer:
The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity. Precision of a measurement is a closeness of two or more measured values to each other.
The true value of a certain length is near 5.678 cm. In one experiment, using a measuring instrument of resolution 0.1 cm, the measured value is found to be 5.5 cm. In another experiment using a measuring instrument of greater resolution, say 0.01 cm, the length is found to be 5.38 cm. We find that the first measurement is more accurate as it is closer to the true value, but it has lesser precision. On the contrary, the second measurement is less accurate, but it is more precise.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Long Answer Questions

Question 1.
(i) Explain the use of screw gauge and vernier caliper in measuring smaller distances.
Answer:
(i) Measurement of small distances: screw gauge and vernier caliper Screw gauge:
The screw gauge is an instrument used for measuring accurately the dimensions of objects up to a maximum of about 50 mm. The principle of the instrument is the magnification of linear motion using the circular motion of a screw. The least count of the screw gauge is 0.01 mm. Vernier caliper: A vernier caliper is a versatile instrument for measuring the dimensions of an object namely diameter of a hole, or a depth of a hole. The least count of the vernier caliper is 0.1 mm.

(ii) Write a note on triangulation method and radar method to measure larger distances. Triangulation method for the height of an accessible object;
Let AB = h be the height of the tree or tower to be measured. Let C be the point of observation at distance x from B. Place a range finder at C and measure the angle of elevation, ACB =
θ as shown in figure.
From right angled triangle ABC,


(or) height h = x tan θ
Knowing the distance x, the height h can be determined.
RADAR method
The word RADAR stands for radio detection and ranging. A radar can be used to measure accurately the distance of a nearby planet such as Mars. In this method, radio waves are sent from transmitters which, after reflection from the planet, are detected by the receiver. By measuring, the time interval (t) between the instants the radio waves are sent and received, the distance of the planet can be determined as where v is the speed of the radio wave. As the time taken (t) is for the distance covered during the forward and backward path of the radio waves, it is divided by 2 to get the actual distance of the object. This method can also be used to determine the height, at which an aeroplane flies from the ground.

Question 2.
Explain in detail the various types of errors.
Answer:
The uncertainty in a measurement is called an error. Random error, systematic error and gross error are the three possible errors.
(i) Systematic errors: Systematic errors are reproducible inaccuracies that are consistently *, in the same direction. These occur often due to a problem that persists throughout the experiment. Systematic errors can be classified as follows
(1) Instrumental errors: When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.
(2) Imperfections in experimental technique or procedure: These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied
(3) Personal errors: These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.
(4) Errors due to external causes: The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity, or pressure during measurements may affect the result of the measurement.
(5) Least count error: Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument’s
resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.
(ii) Random errors: Random errors may arise due to random and unpredictable variations in
experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer who performs the experiment. Random errors are sometimes called “chance error”. When different readings are obtained by a person every time he repeats the experiment, personal error occurs. For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.

Question 3.
What do you mean by propagation of errors? Explain the propagation of errors in addition and multiplication.
Answer:
A number of measured quantities may be involved in the final calculation of an experiment. Different types of instruments might have been used for taking readings. Then we may have to look at the errors in measuring various quantities, collectively.
The error in the final result depends on
(i) The errors in the individual measurements
(ii) On the nature of mathematical operations performed to get the final result. So we should know the rules to combine the errors.
The various possibilities of the propagation or combination of errors in different mathematical operations are discussed below:
(i) Error in the sum of two quantities
Let ∆A and ∆B be the absolute errors in the two quantities A and B respectively. Then, Measured value of A = A ± ∆A
Measured value of B = B ± ∆B
Consider the sum, Z = A + B
The error ∆Z in Z is then given by

The maximum possible error in the sum of two quantities is equal to the sum of the absolute errors in the individual quantities.
Error in the product of two quantities: Let ∆A and ∆B be the absolute errors in the two quantities A, and B, respectively. Consider the product Z = AB
The error ∆Z in Z is given by Z ± ∆Z = (A ± ∆A) (B ± ∆B)
= (AB) ± (A ∆ B) ± (B ∆ A) ± (∆A • ∆B)
Dividing L.H.S by Z and R.H.S by AB, we get,

As ∆A/A, ∆B/B are both small quantities, their product term can be neglected.
The maximum fractional error in Z is

Question 4.
Write short note on the following:
(a) Unit
(b) Rounding – off
(c) Dimensionless quantities
Answer:
(a) Unit: An arbitrarily chosen standard of measurement of a quantity, which is accepted internationally is called unit of the quantity.
The units in which the fundamental quantities are measured are called fundamental or base units and the units of measurement of all other physical quantities, which can be obtained by a suitable multiplication or division of powers of fundamental units, are called derived units.
(b) Rounding – off: In no case should the result have more significant figures than die figures involved in the data used for calculation. The result of calculation with numbers containing more than one uncertain digit should be rounded off. The rules for rounding off are given below.

(c) Dimensionless quantities: On the basis of dimension, dimensionless quantities are classified into two categories.
(i) Dimensionless variables:
Physical quantities which have no dimensions, but have variable values are called dimensionless variables. Examples are specific gravity, strain, refractive index etc.
(ii) Dimensionless Constant:
Quantities which have constant values and also have no dimensions are called dimensionless constants. Examples are π, e, numbers etc.

Question 5.
Explain the principle of homogeniety of dimensions. What are its uses? Give example.
Answer:
The principle of homogeneity of dimensions states that the dimensions of all the terms in a physical expression should be the same. For example, in the physical expression v² = u² + 2as, the dimensions of v², u² and 2 as are the same and equal to [L²T^-2].
(i) To convert a physical quantity from one system of units to another: This is based on the fact that the product of the numerical values (n) and its corresponding unit (u) is a constant, i.e, = constant n₁[u₁] = constant (or) n₁[u₁] = n₂[u₂].
Consider a physical quantity which has dimension ‘a’ in mass, ‘b’ in length and ‘c’ in time. If the fundamental units in one system are M₁, L₁ and T₁ and the other system are M₂, L₂ and T₂ respectively, then we can write,
We have thus converted the numerical value of physical quantity from one system of units into the other system.
Example: Convert 76 cm of mercury pressure into Nm^-2 using the method of dimensions. Solution: In cgs system 76 cm of mercury pressure = 76 × 13.6 × 980 dyne cm^-2
The dimensional formula of pressure P is [ML^-1T^-2]

(ii) To check the dimensional correctness of a given physical equation:
Example: The equation \(\frac{1}{2} m v^{2}\) = mgh can be checked by using this method as follows.
Solution: Dimensional formula for

Dimensional formula for

Both sides are dimensionally the same, hence the equations is dimensionally correct.
(iii) To establish the relation among various physical quantities:
Example: An expression for the time period T of a simple pendulum can be obtained by using this method as follows.
Let true period T depend upon
(i) mass m of the bob
(ii) length l of the pendulum and
(iii) acceleration due to gravity g at the place where the pendulum is suspended. Let the constant involved is K = 2π.
Solution:

Here k is the dimensionless constant. Rewriting the above equation with dimensions.

Comparing the powers of M, L and T on both sides, a – 0, b + c = 0, -2c = 1
Solving for a, b and c a = 0, b = 1/2, and c = -1/2
From the above equation

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Numerical Problems

Question 1.
In a submarine equipped with sonar, the time delay between the generation of a pulse and its echo after reflection from an enemy submarine is observed to be 80 sec. If the speed of sound in water is 1460 ms^-1. What is the distance of enemy submarine?
Answer:
Given:
Speed of sound in water = 1460 ms^-1
Time delay = 80s
Distance of enemy ship = ?
Solution:
Total distance covered = speed × time
= 1460 ms^-1 × 80s = 116800 m
Time taken is for forward and backward path of sound waves.

= 58400 m (or) 58.4 km

Question 2.
The radius of the circle is 3.12 m. Calculate the area of the circle with regard to significant figures.
Answer:
Given: radius : 3.12 m (Three significant figures)
Solution:
Area of the circle = πr² = 3.14 × (3.12 m)² = 30.566
If the result is rounded off into three significant figure, area of the circle = 30.6 m²

Question 3.
Assuming that the frequency v of a vibrating string may depend upon
(i) applied force (F)
(ii) length (l)
(iii) mass per unit length (m), prove that using dimensional analysis. [Related to JIPMER 2001]
Answer:
Given: The frequency v of a vibrating string depends
(i) applied force (F)
(ii) length (l)
(iii) mass per unit length (m)
Solution:

Substitute the dimensional formulae of the above quantities

Comparing the powers of M, L, T on both sides,
x + z = 0, x + y – z = 0, -2x = -1
Solving for x, y, z, we get

Substitute x, y, z values in equ(1)

Question 4.
Jupiter is at a distance of 824.7 million km from the Earth. Its angular diameter is measured to be 35.72″. Calculate the diameter of Jupiter.
Answer:
Given,
Given Distance of Jupiter = 824.7 × 10⁶ km = 8.247 × 10¹¹ m
angular diameter = 35.72 × 4.85 × 10^-6rad = 173.242 × 10^-6 rad
= 1.73 × 10^-4 rad
∴ Diameter of Jupiter D = D × d = 1.73 × 10^-4 rad × 8.247 × 10¹¹ m
= 14.267 × 1o⁷ m = 1.427 × 10⁸ m (or) 1.427 × 10^{5</sup km}

Question 5.
The measurement value of length of a simple pendulum is 20 cm known with 2 mm accuracy. The time for 50 oscillations was measured to be 40 s within 1 s resolution. Calculate the percentage of accuracy in the determination of acceleration due to gravity ‘g’ from the above measurement.
Answer:
Given,
Length of simple pendulum (l) = 20 cm
absolute error in length (∆l) = 2 mm = 0.2 cm
Time taken for 50 oscillation (t) = 40 s
error in time ∆T = 1 s
Solution: Time period for one oscillation (T)

Hence, the percentage error in g is

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Conceptual Questions

Question 1.
Why is it convenient to express the distance of stars in terms of light year (or) parsec rather than in km?
Answer:
A parsec is 206, 265 AU and is roughly the distance to the nearest stars. If we were to view a giant star with a diameter of 1 AU at a distance of one parsec, it would appear to be just 1/3600^th of a degree in angular size. For comparison, the sun and moon are both half a degree in angular size when viewed from Earth.

Question 2.
Show that a screw gauge of pitch 1 mm and 100 divisions is more precise than a vernier caliper with 20 divisions on the sliding scale.
Answer:

As shown, the least count of screw gauge is lesser then vernier caliper, hence screw gauge is more precise.

Question 4.
Having all units in atomic standards is more useful. Explain.
Answer:
An atomic mass unit (symbolized AMU or amu) is defined as precisely 1/12 the mass of an atom of carbon-12. The carbon-12 (C-12) atom has six protons and six neutrons in its nucleus.
In imprecise terms, one AMU is the average of the proton rest mass and the neutron rest mass. This is approximately 1.67377 × 10^-27 kilogram (kg), or 1.67377 × 10^-24 gram (g). The mass of an atom in AMU is roughly equal to the sum of the number of protons and neutrons in the nucleus.
The AMU is used to express the relative masses of, and thereby differentiate between, various isotopes of elements. Thus, for example, uranium-235 (U-235) has an AMU of approximately 235, while uranium-238 (U-238) is slightly more massive. The difference results from the fact that U-238, the most abundant naturally occurring isotope of uranium, has three more neutrons than U-235, an isotope that has been used in nuclear reactors and atomic bombs.

Question 5.
Why dimensional methods are applicable only up to three quantities?
Answer:
Understanding dimensions is of utmost importance as it helps us in studying the nature of physical quantities mathematically. The basic concept of dimensions is that we can add or subtract only those quantities which have same dimensions. Also, two physical quantities are equal if they have same dimensions. these basic ideas help us in deriving the new relation between physical quantities, it is just like units.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Additional Questions Solved

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Multiple Choose Questions

Question 1.
The unit of surface tension ……
(a) MT^-2
(b) Nm^-2
(c) Nm
(d) Nm^-1
Answer:
(d) Nm^-1

Question 2.
One atomus equal to ……

Answer:
(c) 160 ms

Question 3.
One light year is ……
(a) 3.153 × 10⁷ m
(b) 1.496 × 10⁷ m
(c) 9.46 × 10¹² km
(d) 3.26 × 10¹⁵ km
Answer:
(c) 9.46 × 10¹² km

Question 4.
One Astronomical unit is
(a) 3.153 × 10⁷ m
(b) 1.496 × 10⁷ m
(c) 9.46 × 10¹² m
(d) 3.26 × 10¹⁵ m
Answer:
(b) 1.496 × 10⁷ m

Question 5.
One parsec is …..
(a) 3.153 × 10⁷ m
(b) 3.26 × 10¹⁵ m
(c) 30.84 × 10¹⁵ m
(d) 9.46 × 10¹⁵ m
Answer:
(c) 30.84 × 10¹⁵ m

Question 6.
One Fermi is …..
(a) 10^-9 m
(b) 10^-10 m
(c) 10^-12 m
(d) 10^-15 m
Answer:
(d) 10^-15 m

Question 7.
One Angstrom is ………
(a) 10^-9 m
(b) 10^-10m
(c) 10^-12 m
(d) 10^-15 m
Answer:
(b) 10^-10 m

Question 8.
One solar mass is ….
(a) 2 × 10³⁰ kg
(b) 2 × 10³⁰ g
(c) 2 × 10³⁰ mg
(d) 2 × 10³⁰ tonne
Answer:
(d) 2 × 10³⁰ tonne

Question 9.
\(\frac{1}{12}\) of the mass of carbon 12 atom is …..
(a) 1 TMC
(b) mass of neutron
(c) 1 amu
(d) mass of hydrogen
Answer:
(d) mass of hydrogen

Question 10.
The word physics is derived from the word …..
(a) scientist
(b) fusis
(c) fission
(d) fusion
Answer:
(b) fusis

Question 11.
The study of forces acting on bodies whether at rest or in motion is …..
(a) classical mechanics
(b) quantum mechanics
(c) thermodynamics
(d) condensed matter physics
Answer:
(a) classical mechanics

Question 12.
Mass of observable universe …..
(a) 10³¹ kg
(b) 10⁴¹ kg
(c) 10⁵⁵ kg
(d) 9.11 × 10³¹ kg
Answer:
(c) 10⁵⁵ kg

Question 13.
Mass of an electron …….

Answer:
(b) 9.11 × 10^-31 kg

Question 14.
The study of production and propagation of sound waves …..
(a) Astrophysics
(b) Acoustics
(c) Relativity
(d) Atomic physics
Answer:
(b) Acoustics

Question 15.
The study of the discrete nature of phenomena at the atomic and subatomic levels.
(a) Quantum mechanics
(b) High energy physics
(c) Acoustics
(d) Classical mechanics
Answer:
(a) Quantum mechanics

Question 16.
The techniQuestion used to study the crystal structure of various rocks are …….
(a) diffraction
(b) interference
(c) total internal reflection
(d) refraction
Answer:
(a) diffraction

Question 17.
The astronomers used to observe distant points of the universe by …….
(a) Electron telescope
(b) Astronomical telescope
(c) Radio telescope
(d) Radar
Answer:
(c) Radio telescope

Question 18.
The comparison of any physical quantity with its standard unit is known as ……..
(a) fundamental quantities
(b) measurement
(c) dualism
(d) derived quantities
Answer:
(b) measurement

Question 19.
Fundamental quantities can also be known as …… quantities.
(a) original
(b) physical
(c) negative
(d) base
Answer:
(d) base

Question 20.
Which one of the following is not a fundamental quantity?
(a) length
(b) luminous intensity
(c) temperature
(d) water current
Answer:
(d) water current

Question 21.
The system of unit not only based on length, mass, and time is
(a) FPS
(b) CGS
(c) MKS
(d) SI
Answer:
(d) SI

Question 22.
The coherent system of units …..
(a) CGS
(b) SI
(c) FPS
(d) MKS
Answer
(b) SI

Question 23.
The triple point temperature of water is ……
(a) -273.16 K
(b) 0K
(c) 273.16 K
(d) 100 K
Answer:
(d) 100 K

Question 24.
Which of the following is a unit of distance?
(a) Light year
(b) Leap year
(c) Dyne-sec
(d) Pauli
Answer:
(a) Light year

Question 25.
The unit of moment of force ……
(a) Nm²
(b) Nm
(c) N
(d) NJ rad
Answer:
(b) Nm

Question 26.
1 radian is ……
(a) 2.91 × 10^-4 m
(b) 57.27°
(c) 180°
(d) \(\frac{\pi}{180}\)
Answer:
(b) 57.27°

Question 27.
One degree of arc is …….
(a) 1″
(b) 60″
(c) 60′
(d) 60°
Answer:
(c) 60′

Question 28.
One degree of arc is equal to …….
(a) 1.457 × 10² rad
(b) 1.457 × 10^-2 rad
(c) 1.745 × 10² rad
(d) 1.745 × 10^-2 rad
Answer:
(b) 1.457 × 10^-2 rad

Question 29.
1 minute of arc is equal to …….
(a) 1.745 × 10^-2 rad
(b) 2.91 × 10^-4 rad
(c) 2.91 × 10⁴ rad
(d) 4.85 × 10^-6 rad
Answer:
(b) 2.91 × 10^-4 rad

Question 30.
1 second of arc is equal to ………

Answer:

Question 31.
1 second of arc is equal to ….
(a) 0.00027°
(b) 1.745 × 10^-2 rad
(c) 2.91 × 10^-4 rad
(d) 4.85 × 10^-6 rad
Answer:
(a) 0.00027°

Question 32.
Unit of impulse ….
(a) NS²
(b) NS
(c) Nm
(d) Kgms^-2
Answer:
(b) NS

Question 33.
The ratio of energy and temperature is known as ……
(a) Stefen’s constant
(b) Boltzmann constant
(c) Plank’s constant
(d) Kinetic constant
Answer:
(b) Boltzmann constant

Question 34.
The range of distance can be measured by using direct methods is …..
(a) 10^-2 to 10^-5 m
(b) 10^-2 to 10² m
(c) 10² to 1(T5 m {d) 10″2 to 105 m
Answer:
(b) 10^-2 to 10² m

Question 35.
Which of the following is in increased order?
(a) exa, tera, hecto
(b) tera, exa, hecto
(c) giga, tera, exa
(d) hecto, exa, giga
Answer:
(c) giga, tera, exa

Question 36.
10^-18 is called as ……
(a) nano
(b) pico
(c) femto
(d) atto
Answer:
(d) atto

Question 37.
A radio signal sent towards the distant planet, returns after “t” s. If “c” is the speed of radio
waves then the distance of the planet and from the earth is …….

Answer:

Question 38.
Find odd one out ….
(a) Newton
(b) metre
(c) candela
(d) Kelvin
Answer:
(a) Newton

Question 39.
The shift in the position of an object when viewed with two eyes, keeping one eye closed at a
time is known as …
(a) basis
(b) fundamental
(c) parallax
(d) pendulum
Answer:
(c) parallax

Question 40.
Chandrasekar limit is ….. times the mass of the sun.
(a) 1.2
(b) 1.4
(c) 1.6
(d) 1.8
Answer:
(b) 1.4

Question 41.
The smallest physical unit of time is
(a) second
(b) minute
(c) microsecond
(d) shake
Answer:
(d) shake

Question 42.
Size of atomic nucleus is …..
(a) 10^-10 m
(b) 10^-12 m
(c) 10^-14 m
(d) 10^-18 m
Answer:
(c) 10^-14 m

Question 43.
Time interval between two successive heart beat is in the order of …….
(a) 10° s
(b) 10 s
(c) 10² s
(d) 10^-3 s
Answer:
(a) 10° s

Question 44.
Half life time of a free neutron is in the order of ……
(a) 10°
(b) 10¹ s
(c) 10² s
(d) 10³ s
Answer:
(d) 10³ s

Question 45.
The uncertainty contained in any measurement is ……
(a) rounding off
(b) error
(c) parallax
(d) gross
Answer:
(b) error

Question 46.
Zero error of an instrument is a ……
(a) Systematic error
(b) Random error
(c) Gross error
(d) Both (a) and (b)
Answer:
(a) Systematic error

Question 47.
Error in the measurement of radius of a sphere is 2%. Then error in the measurement of surface
area is ….
(a) 1%
(b) 2%
(c) 3%
(d) 4%
Answer:
(d) 4%

Question 48.
Imperfections in experimental procedure gives ….. error.
(a) random
(b) gross
(c) systematic
(d) personal
Answer:
(c) Systematic

Question 49.
Random error can also be called as ….
(a) personal error
(b) chance error
(c) gross error
(d) system error
Answer:
(b) chance error

Question 50.
To get the best possible true value of the quantity has to be taken.
(a) rms value
(b) net value
(c) arithmetic mean
(d) mode
Answer:
(c) arithmetic mean

Question 51.
The error caused due to the shear carelessness of an observer is called as …… error.
(a) Systematise
(b) Gross
(c) Random
(d) Personal
Answer:
(b) Gross

Question 52.
The uncertainty in a measurement is called as ….
(a) error
(b) systematic
(c) random error
(d) gross error
Answer:
(a) error

Question 53.
The difference between the true value and the measured value of a quantity is known as …..
(a) Absolute error
(b) Relative error
(c) Percentage error
(d) Systemmatic error
Answer:
(a) Absolute error

Question 54.
If a₁, a₂, a₃ …. a_n are the measured value of a physical quantity “a” and a_m is the true value then absolute error …..

Answer:
(d) \(\Delta a_{n}=a_{m}-a_{n}\)

Question 55.
If ‘a_m‘ and ‘∆a_m ‘ are true value and mean absolute error respectively, then the magnitude of the quantity may lie between …..

Answer:
(b) \(a_{m}-\Delta a_{m} \text { to } a_{m}+\Delta a_{m}\)

Question 56.
The ratio of the mean absolute error to the mean value is called as ……
(a) absolute error
(b) random error
(c) relative error
(d) percentage error
Answer:
(c) Relative error

Question 57.
Relative error can also be called as ……
(a) fractional error
(b) absolute error
(c) percentage error
(d) systematic error
Answer:
(a) fractional error

Question 58.
A measured value to be close to targeted value, percentage error must be close to
(a) 0
(b) 10
(c) 100
(d) ∝
Answer:
(a) 0

Question 59.
The maximum possible error in the sum of two quantities is equal to …….
(a) Z = A + B
(b) ∆Z = ∆A + ∆B
(c) ∆Z = ∆A/∆B
(d) ∆Z = ∆A – ∆B
Answer:
(b) ∆Z = ∆A + ∆B

Question 60.
The maximum possible error in the difference of two quantities is ……

Answer:

Question 61.
The maximum fractional error in the division of two quantities is ….

Answer:

Question 62.
The fractional error in the n^th power of a quantity is …..

Answer:

Question 63.
A physical quantity is given as y = \(\frac{a b^{3}}{c^{2}}\). If ∆a, ∆b, ∆c are absolute errors, the possible fractional error in y is …..

Answer:

Question 64.
Number of significant digits in 3256 …
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 65.
Number of significant digits in 32005 ……
(a) 1
(b) 2
(c) 5
(d) 2
Answer:
(c) 5

Question 66.
Number of significant digits in 2030 …
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 67.
Number of significant digits in 2030N …..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 68.
Number of significant digits in 0.0342 …..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c)3

Question 69.
Number of significant digit in 20.00 …..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 70.
Number of significant digit in 0.030400
(a) 6
(b) 5
(c) 4
(d) 3
Answer:
(b) 5

Question 71.
The force acting on a body is measured as 4.25 N. Round it off with two significant figure ..
(a) 4.3
(b) 4.2
(c) both
(a) or (b)
(d) 4.25
Answer:
(b) 4.2

Question 72.
The quantities a, b, c are measured as 3.21, 4.253, 7.2346. The sum (a + b + c) with proper
significant digits is ……
(a) 14.6976
(b) 14.697
(c) 14.69
(d) 14.6
Answer:
(c) 14.69

Question 73.
The dimensions of gravitational constant G are …

Answer:
(b) \(\mathbf{M}^{-1} \mathbf{L}^{3} \mathbf{T}^{-2}\)

Question 74.
The ratio of one nanometer to one micron is
(a) 10^-3
(b) 10³
(c) 10^-9
(d) 10^-6
Answer:
(b) 10³

Question 75.
Which of the following pairs does not have same dimension?
(a) Moment of inertia and moment of force
(b) Work and torque
(c) Impulse and momentum
(d) Angular momentum and Plank’s constant
Answer:
(a) Moment of inertia and moment of force

Question 76.
Two quantities A and B have different dimensions. Which of the following is physically meaningful?
(a) A + B
(b) A – B
(c) A /B
(d) None
Answer:
(c) A /B

Question 77.
The dimensional formula for moment of inertia ……

Answer:
(d) \(\mathbf{M} \mathbf{L}^{2} \mathbf{T}^{\mathbf{0}}\)

Question 78.
Which of the following is having same dimensional formula?
(a) Work and power
(b) Radius of gyration and displacement
(c) Impulse and force
(d) Frequencies and wavelength
Answer:
(b) Radius of gyration and displacement

Question 79.
Which of the following quantities is expressed as force per unit area?
(a) Pressure
(b) Stress
(c) Both (a) and (b)
(d) None
Answer:
(c) Both (a) and (b)

Question 80.
In equation of motion the dimensional formula for K is …..

Answer:
(b) \(\left[\mathbf{L} \mathbf{T}^{-2}\right]\)

Question 81.
The dimensional formula for heat capacity ……

Answer:
(d) \(\left[\mathbf{M} \mathbf{L}^{2} \mathbf{T}^{2} \mathbf{K}^{-1}\right]\)

Question 82.
The product of Avogadro constant and elementary charge is known as …… constant.
(a) Planck’s
(b) Avagadro
(c) Boltzmann
(d) Faraday
Answer:
(d) Faraday

Question 83.
The force F is given by F = at + bt² where t is time. The dimensions of ‘a’ and ‘b’ respectively are

Answer:
(d) \(\left[\mathrm{MLT}^{-2}\right]\) and \(\left[\mathrm{MLT}^{-0}\right]\)

Question 84.
Dimensions of impulse are …..

Answer:
(c) \(\left[\mathrm{MLT}^{-1}\right]\)

Question 85.
If speed of light (c), acceleration due to gravity (g) and pressure (P) are taken as fundamental
units, the possible relation to gravitational constant (G) is ….

Answer:
(c) \(c^{0} g^{2} p^{-1}\)

Question 86.
Equivalent of one joule is ……
(a) Nm²
(b) kg m² s^-2
(c) kg m s^-1
(d) N kg m²
Answer:
(b) kg m² s^-2

Question 87.
Pick out the dimensionless quantity …..
(a) force
(b) specific gravity
(c) planck’s constant
(d) velocity
Answer:
(b) specific gravity

Question 88.
Odd one out ………
(a) strain
(b) refractive index
(c) numbers
(d) stress
Answer:
(d) stress

Question 89.
A wire has a mass 0.3 ± 0.003g, radius 0.5 ± 0.005 mm and length 6 + 0.06 cm. The maximum percentage error in the measurement of its density is …….
(a) 1%
(b) 2%
(c) 3%
(d) 4%
Answer:
(d) 4%

Question 90.
The dimensions of planck constant equals to that of …..
(a) energy
(b) momentum
(c) angular momentum
(d) power
Answer:
(c) angular momentum

Question 91

(a) The unit of λ is same as that of x and A
(b) The unit of λ is same as that of x but not of A
(c) The unit of c is same as that of 2π/λ
(d) The unit of (ct – x)is same as that 2π/λ
Answer:
(a) The unit of λ is same as that of x and A

Question 92.
The number of significant figures in 0.06900 is …….
(a) 2
(b) 2
(c) 4
(d) 5
Answer:
(c) 4

Question 93.
The numbers 3.665 and 3.635 on rounding off to 3 significant figures will give
(a) 3.66 and 3.63
(b) 3.66 and 3.64
(c) 3.67 and 3.63
(d) 3.67 and 3.64
Answer:
(b) 3.66 and 3.64

Question 94.
Which of the following measurements is most precise?
(a) 4.00 mm
(b) 4.00 cm
(c) 4.00 m
(d) 4.00 km
Answer:
(a) 4.00 mm

Question 95.
The mean radius of a wire is 2 mm. Which of the following measurements is most accurate? (a) 1.9 mm
(b) 2.25 mm
(c) 2.3 mm
(d) 1.83 mm
Answer:
(a) 1.9 mm

Question 96.
If error in measurement of radius of sphere is 1%. What will be the error in measurement of volume?
(a) 1%
(b) \(\frac{1}{3}\)%
(c) 3%
(d) 10%
Answer:
(c) 3%

Question 97.
Dimensions [M L^-1 T^-1] are related to …….
(a) torque
(b) work
(c) energy
(d) Coefficient of viscosity
Answer:
(d) Coefficient of viscosity

Question 98.
Heat produced by a current is obtained a relation H = I²RT. If the errors in measuring these quantities current, resistance, time are 1%, 2%, 1% respectively then total error in calculating the energy produced is
(a) 2%
(b) 4%
(c) 5%
(d) 6%
Answer:
(c) 5%

Question 99.
Length cannot be measured by ….
(a) fermi
(b) angstrom
(c) parsec
(d) debye
Answer:
(d) debye

Question 100.
The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate by using the formula p = \([/late\frac{\mathrm{F}}{l^{2}}x]. If the maximum errors in the measurement
of force and length are 4% and 2% respectively, then the maximum error in the measurement of pressure is ……..
(a) 1%
(b) 2%
(c) 8%
(d) 10%
Aswer:
(c) 8%

Question 101.
Which of the following cannot be verified by using dimensional analysis?
1 mv²

Answer:
(b) y = a sin wt

Question 102.
Percentage errors in the measurement of mass and speed are 3% and 2% respectively. The error in the calculation of kinetic energy is …….
(a) 2%
(b) 3%
(c) 5%
(d) 7%
Answer:
(d) 7%

Question 103.
More number of readings will reduce …….
(a) random error
(b) systematic error
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(a) random error

Question 104.
If the percentage error in the measurement of mass and momentum of a body are 3% and
2%respectively, then maximum possible error in kinetic energy is
(a) 2%
(b) 3%
(c) 5%
(d) 7%
Answer:
(d) 7%

Question 105.
In a vernier caliper, n divisions of vernier scale coincides with (n – 1) divisions of main scale. The least count of the instrument is ………

Answer:

Question 106.
The period of a simple pendulum is recorded as 2.56s, 2.42s, 2.71s and 2.80s respectively.
The average absolute error is
(a) 0.1s
(b) 0.2s
(c) 1.0s
(d) 0.11s
Answer:
(d) 0.11s

Question 107.
In a system of units, if force (F), acceleration (A) and time (T) are taken as fundamental units
then the dimensional formula of energy is
(a) [FA²T]
(b) [FAT²]
(c) [F²AT]
(d) [FAT]
Answer:
(b) [FAT²]

Question 108.
The random error in the arithmetic mean of 50 observations is ‘a’, then the random error in the
arithmetic mean of 200 observations a would be

Answer:
(c) [latex]\frac{a}{4}\)

Question 109.
Which of the following is not dimensionless?
(a) Relative permittivity
(b) Refractive index
(c) Relative density
(d) Relative velocity
Answer:
(d) Relative velocity

Question 110.
If V-velocity, K – kinetic energy and T – time are chosen as the fundamental units, then what is the dimensional formula for surface tension?

Answer:
(a) [K V^-2 T^-2]

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Short Answer Questions (1 Mark)

Question 1.
A new unit of length is chosen such that the speed of light in vaccum is unity. What is the distance between the sun and the earth in terms of the new unit if light takes 8 min and 20 s to cover this distance.
Answer:
Speed of light in vacuum, c = 1 new unit of length s^-1
t = 8 min. 20 sec, = 500 s
x = ct= 1 new unit of length s^-1 × 500s
x = 500 new unit of length

Question 2.
If x = a + bt + ct², where x is in metre and t in seconds, what is the unit of c ?
Answer:
The unit of left hand side is metre so the units of ct² should also be metre.
Since t² has unit of s², so the unit of c is m/s².

Question 3.
What is the difference between mN, Nm and nm ?
Answer:
mN means milli newton, 1 mN = 10^-3 N, Nm means Newton meter, nm means nano meter.

Question 4.
The radius of atom is of the order of 1 A° & radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of the atom as compared to the volume of nucleus ?
Answer:

Question 5.
How many kg make 1 unified atomic mass unit ?
Answer:
1u = 1.66 × 10^-27 kg

Question 6.
Name some physical quantities that have same dimension.
Answer:
Work, energy and torque.

Question 7.
Name the physical quantities that have dimensional formula [ML ^-1T^-2].
Answer:
Stress, pressure, modulus of elasticity.

Question 8.
Give two examples of dimensionless variables.
Answer:
Strain, refractive index.

Question 9.
State the number of significant figures in
(i) 0.007 m²
(ii) 2.64 × 1024 kg
(iii) 0.2370 g cm^-3
(iv) 0.2300m
(v) 86400
(vi) 86400 m
Answer:
(i) 1,
(ii) 3,
(iii) 4,
(iv) 4,
(v) 3,
(vi) 5 since it comes from a measurement the last two zeros become significant.

Question 10.
Given relative error in the measurement of length is 0.02, what is the percentage error ?
Answer:
2%.

Question 11.
A physical quantity P is related to four observables a, b, c and d as follows :

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage error in the quantity P?
Answer:
Relative error in P is given by

Question 12.
A boy recalls the relation for relativistic mass (m) in terms of rest mass (m₀) velocity of particle V, but forgets to put the constant c (velocity of light). He writes correct the equation by putting the missing ‘c’.
Answer:
Since quantities of similar nature can only be added or subtracted, v² cannot be subtracted from 1 but v²/c² can be subtracted from 1.

Question 13.
Name the technique used in locating.
(a) an under water obstacle
(b) position of an aeroplane in space.
Answer:
(a) SONAR ➝ Sound Navigation and Ranging.
(b) RADAR ➝ Radio Detection and Ranging.

Question 14.
Deduce dimensional formulae of—
(i) Boltzmann’s constant
(ii) mechanical equivalent of heat.
Answer:
(i) Boltzmann Constant:

Question 15.
Give examples of dimensional constants and dimensionless constants.
Answer:
Dimensional Constants : Gravitational constant, Plank’s constant. Dimensionless Constants : it, e.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Short Answer Questions (2 Marks)

Question 16.
The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm. Calculate the minimum inaccuracy in the measurement of distance.
Answer:
Minimum inaccuracy = Vernier constant

Question 17.
If the unit of force is 100N, unit of length is 10m and unit of time is 100s. What is the unit of Mass in this system of units ?
Answer:

Question 18.
State the principle of homogeneity. Test the dimensional homogeneity of equations

Answer:

as Dimensions of L.H.S. = Dimensions of R.H.S.
∴ The equation to dimensionally homogeneous.

(ii) S_n = Distance travelled in n^th sec that is (S_n – S_{n – 1})

Hence this is dimensionally correct.

Question 19.

Answer:
Since dimensionally similar quantities can only be added

Question 20.
Magnitude of force experienced by an object moving with speed v is given by F = kv². Find dimensions of k.
Answer:

Question 21.
A book with printing error contains four different formulae for displacement. Choose the
correct formula/formulae

Answer:
The arguments of sine and cosine function must be dimensionless so (a) is the probable correct formulae. Since

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Numericals Questions

Question 22.
Determine the number of light years in one metre.
Answer:

Question 23.
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces 20.15 g and 20.17 g are added to the box.
(i) What is the total mass of the box ?
(ii) The difference in masses of the pieces to correct significant figures.
Answer:
(i) Mass of box = 2.3 kg
Mass of gold pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg.
Total mass = 2.3 + 0.04032 = 2.34032 kg
In correct significant figure mass = 2.3 kg (as least decimal)
(ii) Difference in mass of gold pieces = 0.02 g
In correct significant figure (2 significant fig. minimum decimal) will be 0.02 g.

Question 24.
5.74 g of a substance occupies 1.2 cm³. Express its density to correct significant figures.
Answer:

Here least significant figure is 2, so density = 4.8 g/cm³.

Question 25.
If displacement of a body s = (200 ± 5) m and time taken by it t = (20 + 0.2) s, then find the percentage error in the calculation of velocity.
Answer:

Question 26.
If the error in measurement of mass of a body be 3% and in the measurement of velocity be 2%. What will be maximum possible error in calculation of kinetic energy.
Answer:

Question 27.
The length of a rod as measured in an experiment was found to be 2.48 m, 2.46 m, 2.49 m, 2.50 m and 2.48 m. Find the average length, absolute error and percentage error. Express the result with error limit.
Answer:

Question 28.
A physical quantity is measured as Q = (2.1 ± 0.5) units. Calculate the percentage error in (1) Q² (2) 2Q.
Answer:

Question 29.
When the planet Jupiter is at a distance of 824.7 million km from the earth, its angular diameter is measured to be 35.72″ of arc. Calculate diameter of Jupiter.
Answer:

Question 30.
A laser light beamed at the moon takes 2.56s and to return after reflection at the moon’s surface. What will be the radius of lunar orbit?
Answer:
t = 2.54 s

Question 31.
Convert
(i) 3 m.s^-2 to km h^-2
(ii) G = 6.67 × 10^-11 N m² kg^-2 to cm³ g^-1 s^-2
Answer:

Question 32.
A calorie is a unit of heat or energy and it equals 4.2 J where 1J = 1 kg m²s^-2. Suppose we employ a system of units in which unit of mass is α kg, unit of length is β m, unit of time γs. What will be magnitude of calorie in terms of this new system.
Answer:

Question 33.
The escape velocity v of a body depends on—
(i) the acceleration due to gravity ‘g’ of the planet,
(ii) the radius R of the planet. Establish dimensionally the relation for the escape velocity.
Solution:

Question 34.
The frequency of vibration of a string depends of on,
(i) tension in the string
(ii) mass per unit length of string,
(iii) vibrating length of the string. Establish dimensionally the relation for frequency.
Answer:

Question 35.
One mole of an ideal gas at STP occupies 22.4 L. What is the ratio of molar volume to atomic volume of a mole of hydrogen? Why is the ratio so large? Take radius of hydrogen molecule to be 1°A.
Answer:

This ratio is large because actual size of gas molecule is negligible in comparison to the inter molecular separation.

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Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 4 Theoretical Concepts of Operating System

Samacheer Kalvi 11th Computer Science Theoretical Concepts of Operating System Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

11th Computer Science Chapter 4 Book Back Answers Question 1.
Operating system is a ……………….
(a) Application Software
(b) Hardware
(c) System Software
(d) Component
Answer:
(c) System Software

Theoretical Concepts Of Operating System Question 2.
Identify the usage of Operating Systems ……………….
(a) Easy interaction between the human and computer
(b) Controlling input & output Devices
(c) Managing use of main memory
(d) All the above
Answer:
(d) All the above

Computer Operating System In Tamil Pdf Question 3.
Which of the following is not a function of an Operating System?
(a) Process Management
(b) Memory Management
(c) Security management
(d) Compiler Environment
Answer:
(d) Compiler Environment

Samacheer Kalvi Advantages And Disadvantages Question 4.
Which of the following OS is a Commercially licensed Operating system?
(a) Windows
(b) UBUNTU
(c) FEDORA
(d) REDHAT
Answer:
(a) Windows

Samacheer Kalvi Guru 11th Computer Science Question 5.
Which of the following Operating systems support Mobile Devices?
(a) Windows 7
(b) Linux
(c) BOSS
(d) iOS
Answer:
(d) iOS

Question 6.
File Management manages ……………….
(a) Files
(b) Folders
(c) Directory systems
(d) All the Above
Answer:
(d) All the Above

Question 7.
Interactive Operating System provides ……………….
(a) Graphics User Interface (GUI)
(b) Data Distribution
(c) Security Management
(d) Real Time Processing
Answer:
(a) Graphics User Interface (GUI)

Question 8.
Android is a ……………….
(a) Mobile Operating system
(b) open Source
(c) Developed by Google
(d) All the above
Answer:
(d) All the above

Question 9.
Which of the following refers to Android operating system’s version?
(a) JELLY BEAN
(b) UBUNTU
(c) OS/2
(d) MITTIKA
Answer:
(a) JELLY BEAN

PART – 2
II. Short Answers

Question 1.
What are the advantages of memory management in Operating System?
Answer:

1. Allocating memory is easy and cheap
2. Any free page is ok, OS can take first one out of.list it keeps
3. Eliminates external fragmentation
4. Data (page frames) can be scattered all over PM
5. Pages are mapped appropriately anyway
6. Allows demand paging and pre – paging
7. More efficient swapping
8. No need for considerations about fragmentation
9. Just swap out page least likely to be used

Question 2.
What is the multi – user Operating system?
Answer:
A Multi – user Operating system is a computer operating system (OS) that allows multiple users on different computers or terminals to access a single system with one OS on it.

Question 3.
What is a GUI?
Answer:
GUI – Graphical User Interface – It allows the use of icons or other visual indicators to interact with electronic devices, rather than using only text via the command line. For example, all versions of Microsoft Windows utilize a GUI, whereas MS – DOS does not.

Question 4.
List out different distributions of Linux operating system.
Answer:
Different server distributions in Linux OS:

1. Ubuntu
2. Linux Mint
3. Debian
4. Fedora
5. RedHa.

Question 5.
What are the security management features available in Operating System?
Answer:
Security Management features:

1. File access level security
2. System level security
3. Network level securny.

Question 6.
What is multi – processing?
Answer:
Multiprocessing is one of the features of operating system. It has two or more processors for a single running process. Each processor works on different parts of the same task or two or more different tasks.

Question 7.
What are the different Operating Systems used in computer?
Answer:
Different operating system used:

1. Single user, single Task Operating system
2. Multi user operating system
3. Multiprocessing operating system
4. Distributed Operating system

PART – 3
III. Explain in Brief

Question 1.
What are the advantages and disadvantages of Time – sharing features?
Answer:
Advantages and disadvantages of Time Sharing Option (TSO)
Advantages:

• Each task and each user get its time.
• Systems have to give time to these application individual tasks and other applications also, so that all system behave correctly.
• Reduces the CPU ideal time

Disadvantages:

• Problem in Reliability
• It consumes many resources so it need special operating systems.
• Need High specification hardware

Question 2.
Explain and List out examples of mobile operating system.
Answer:
A Mobile Operating System (or mobile OS) is an operating system that is specifically designed to run on mobile devices such as phones, tablets, smart watches, etc.
Example: Apple IOS

Google android : Android is a mobile OS developed by Google, based on Linux and designed for smart phones and tabs. iOS was developed by Apple.
Example : Android, ColorOS, LGUX, MIUI.

Question 3.
What are the differences between Windows and Linux Operating system?
Answer:
Windows:

• Microsoft windows is a proprietary OS which is commercial.
• Windows can be modified only by the company that owns it.
• Difficult to customize.
• Vulnerable to virus and malware attacks.

Linux:

• Linux is open source, i.e., free licensed.
• Linux can be modified by anyone.
• Easy to customize.
• More secure.

Question 4.
Explain the process management algorithms in Operating System.
Answer:
The following algorithms are mainly used to allocate the job (process) to the processor: FIFO, SJF, Round Robin, Based on priority.

1. First In First Out (FIFO) Scheduling – This algorithm is based on queuing technique. Technically, the process that enters the queue first is executed first by the CPU, followed by the next and so on. The processes are executed in the order of the queue.
2. Shortest Job First (SJF) Scheduling – This algorithm works based on the size of the job being executed by the CPU.
3. Round Robin Scheduling – Round Robin (RR) Scheduling algorithm is designed especially -for time sharing systems, jobs are assigned and processor time in a circular method.
4. Based on priority – The given job (process) is assigned on a priority. The job which has higher priority is more important than ether jobs.

PART – 4
IV. Explain in Detail

Question 1.
Explain the concept of a Distributed Operating System.
Answer:
The data and application that are stored and processed on multiple physical locations across the world over the digital network (intemet/intranet). The Distributed Operating System is used to access shared data and files that reside in any machine around the world. The user can handle the data from different locations. The users can access as if it is available on their own computer.

The advantages of distributed Operating System are as follows:

1. A user at one location can make use of all the resources available at another location over the network.
2. Many computer resources can be added easily in the network
3. Improves the interaction with the customers and clients.
4. Reduces the load on the host computer.

Question 2.
Explain the main purpose of an operating system.
Operating system has become essential to enable the users to design applications without the knowledge of the computer’s internal structure of hardware.
Operating system manages all the software and hardware. Most of the time there are many different computer programmes running at the same time, they all need to access the computers, CPU, memory and storage. The need of operating system is basically – an interface between the user and the hardware.

Operating System works as translator, while it translates the user request into machine language(Binary language), processes it and then sends it back to Operating System. Operating System converts processed information into user readable form.

Uses of Operating Systems:

1. to ensure that a computer can be used to extract what the user wants it do.
2. Easy interaction between the users and computers.
3. Starting computer operation automatically when power is turned on (Booting).
4. Controlling Input and Output Devices
5. Manage the utilisation of main memory.
6. Providing security to user programs.

Question 3.
Explain advantages and disadvantages of open source operating systems.
Answer:
The benefits of open source are tremendous and have gained huge popularity in the IT field in recent years. They are as follows:

1. Open source (OS) is free to use, distribute and modify.
2. Open source is independent of the company as author who originally created it.
3. It is accessible to everyone. Anyone can debug the coding.
4. It doesn’t have the problem of incompatible formats that exits in proprietary software.
5. It is easy to customize as per our needs.
6. Excellent support will be provided by programmers who will assist in making solutions.

Some of the disadvantages are:

1. Latest hardware are incompatible, i.e. lack of device drivers.
2. It is less user friendly and not as easy to use.
3. There may be some indirect costs involved such as paying for external support.
4. Malicious users can potentially view it and exploit any vulnerability.

Samacheer kalvi 11th Computer Science Theoretical Concepts of Operating System Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
Software is classified into ………………. types.
(a) five
(b) two
(c) four
(d) six
Answer:
(b) two

Question 2.
A computer consists of a collection of processes, they are classified as ………………. categories.
(a) 7
(b) 3
(c) 8
(d) 2
Answer:
(d) 2

Question 3.
Which one of the following is not an algorithm?
(a) NTFS
(b) FIFO
(c) SJE
(d) Round Robin
Answer:
(a) NTFS

Question 4.
The operating system provides ………………. levels of securities to the user end.
(a) three
(b) five
(c) seven
(d) ten
Answer:
(a) three

Question 5.
Which one of the following is not a prominent operating system?
(a) UNIX
(b) IOS
(c) GUI
(d) Android
Answer:
(c) GUI

Question 6.
………………. is a family of multitasking.
(a) LINUX
(b) Microsoft Windows
(c) UNIX
(d) iOS
Answer:
(a) LINUX

Question 7.
Which one of the following comes under proprietary licence?
(a) Apple Mac OS
(b) Google’s Android
(c) UNIX
(d) LINUX
Answer:
(a) Apple Mac OS

Question 8.
The LINUX operating system was originated in ……………….
(a) 1996
(b) 1998
(c) 2000
(d) 1991
Answer:
(d) 1991

Question 9.
………………. is the second most popular mobile operating system globally after Android.
(a) Microsoft Windows
(b) iOS
(c) UNIX
(d) LINUX
Answer:
(b) iOS

Question 10.
Which one of the following is an application software to play audio and video files?
(a) Audio Player
(b) Media Player
(c) VLC Player
(d) All of these
Answer:
(c) VLC Player

Question 11.
Which one of the following is a System software?
(a) Operating System
(b) Language Processor
(c) Both a & b
(d) none of these
Answer:
(c) Both a & b

Question 12.
Which one of the following is a set of instructions that perform specific tasks?
(a) Hardware
(b) Software
(c) Processor
(d) I/O devices
Answer:
(b) Software

Question 13.
Hardware and software are managed by ……………….
(a) GUI
(b) OS
(c) Bootstrap
(d) keyboard
Answer:
(b) OS

Question 14.
The process of starting computer operation automatically when the power is turned on is called ……………….
(a) Booting
(b) Compiling
(c) executing
(d) Storing
Answer:
(a) Booting

Question 15.
An OS that allows only a single user to perform a task at a time is called as ……………….
(a) Single user os
(b) Single task os
(c) Both a & b
(d) Multi tasking os
Answer:
(c) Both a & b

Question 16.
Identify the single user and single task OS?
(a) MS – DOS
(b) UNIX
(c) LINUX
(d) iOS
Answer:
(a) MS – DOS

Question 17.
Identify the multi-user OS?
(a) Windows
(b) Linux
(c) UNIX
(d) All of these
Answer:
(d) All of these

Question 18.
To build a cheap computer, ………………. os is used.
(a) Windows
(b) Raspbion OS
(c) iOS
(d) None of these
Answer:
(b) Raspbion OS

Question 19.
GUI stands for ……………….
(a) Geo User Interact
(b) Global User Inter Change
(c) Graphical User Interface
(d) Global User Interface
Answer:
(c) Graphical User Interface

Question 20.
A ………………. is the unit of work or program in a computer.
(a) Process
(b) Code
(c) Concept
(d) Log file
Answer:
(a) Process

Question 21.
The operating system processes are executed by ……………….
(a) User code
(b) System code
(c) Task
(d) Program
Answer:
(b) System code

Question 22.
System level security is provided by ………………. in a multi user environment.
(a) Permission
(b) execute
(c) Password
(d) Security code
Answer:
(c) Password

Question 23.
NTFS is a ……………….
(a) game
(b) file management technique
(c) os
(d) System level security
Answer:
(b) file management technique

Question 24
………………. os is used to access shared data that resides in any machine around the world.
(a) Time sharing
(b) fixed
(c) MS – Dos
(d) distributed
Answer:
(d) distributed

Question 25.
Unix was developed in the year ……………….
(a) 1970
(b) 1980
(c) 1990
(d) 1960
Answer:
(a) 1970

Question 26.
Unix was developed by?
(a) Ken Thompson
(b) Dennis Ritchie
(c) Both a & b
(d) Ricki Mascitti
Answer:
(c) Both a & b

Question 27.
………………. is a windows alternative open source operating system.
(a) React OS
(b) Boss
(c) Redhat
(d) Fedora
Answer:
(a) React OS

Question 28.
Google has developed ………………. for wrist watches.
(a) Android wear
(b) Android wrist
(c) Android wrist watches
(d) Android watches
Answer:
(a) Android wear

Question 29.
Which among the following is not an android moblie open source versions?
(a) Dotnut
(b) Froyo
(c) Nougat
(d) Alpha
Answer:
(a) Dotnut

PART – 2
II. Short Answers

Question 1.
What is an operating system?
Answer:
An operating system is a software which serves as the interface between a user and a computer.

Question 2.
What are the different types of Operating System?
Answer:
Single user and single task operating system, Multi user operating system, Multi – Processing Operating system, Time sharing Operating system.

Question 3.
What is Real Time operating system?
Answer:
It is a multi – tasking and multi user operating system designed for real time based applications such as robotics, weather and climate prediction software etc.,

Question 4.
What is meant by Distributed Operating system?
Answer:
The data and applications are stored and processed on multiple locations in an around world over the digital network.

Question 5.
What are the advantages of Distributed Operating system?
Answer:
Resources can be used in different locations. Improves interaction with customers and clients. Reduces load on host computers. The data can be exchanged via email and chat.

Question 6.
What is an Interactive Operating system?
Answer:
This is the operating system that provides a GUI through which the user can navigate and interact.

Question 7.
Explain Round Robin Scheduling.
Answer:
This type of scheduling is also known as Time sharing scheduling process. In this, each program is given a fixed amount of time to execute.

Question 8.
What is Memory Management?
Answer:
It is the main functionality of an operating system which handles or manages primary memory and moves processes back and forth between main memory and disk during execution.

Question 9.
Mention different management techniques?
Answer:
Single continuous allocation, Partitioned allocation, Paged memory management, Segmented memory management.

Question 10.
What is Linux?
Answer:
Linux is a family of open-source operating systems.

Question 11.
What is an Android?
Answer:
Android is a mobile operating system developed by Google, based on the Linux and designed primarily for touch screen mobile devices such as smart phones and tablets.

Question 12.
What are the types of software?
Answer:
Application software and System software

Question 13.
What is Process Management?
Answer:
Process management is a function that includes creating and deleting processes and providing mechanisms for processes to communicate and synchronize with each other.

Question 14.
What is Round Robin Scheduling?
Answer:
This algorithm is designed especially for time sharing systems.

Question 15.
What are the 3 levels of security?
Answer:

1. File Access Level
2. System Level
3. Network Level.

Question 16.
Name an OS which is Multitasking and Multi – user operating system.
Answer:
UNIX.

PART – 3
III. Explain in Brief

Question 1.
Write a short note on Android.
Answer:
Android:
Android is a mobile operating system developed by Google, based on Linux and designed primarily for touch screen mobile devices such as smart phones and tablets. Google has further developed Android TV for televisions, Android Auto for cars and Android Wear for wrist watches, each with a specialized user interface. Variants of Android are also used on game consoles, digital cameras, PCs and other electronic gadgets.

Question 2.
Why the operating system is needed?
Answer:
Need for operating system:
Operating System has become essential to enable the users to design applications without the knowledge of the computer’s internal structure of hardware. Operating System manages all the Software and Hardware. Most of the time there are many different computer programmes running at the same time, they all need to access the Computers, CPU, Memory and Storage. The need of Operating System is basically – an interface between the user and hardware.

Operating System works as translator, while it translates the user request into machine language(Binary language), processes it and then sends it back to Operating System. Operating System converts processed information into user readable form.

Question 3.
Write a note on OS for mobile devices.
Answer:
Operating systems for mobile devices:
Mobile devices such as phones, tablets and MP3 players are different from desktop and laptop computers and hence they need special Operating Systems. Examples of mobile Operating Systems are Apple iOS and Google Android. The iOS running on an iPad is Operating systems for mobile devices generally are not as fully featured as those made for desktop and laptop computers and they are not able to run all software.

Question 4.
Write a note on iOS – iphone OS.
Answer:
iOS (formerly iPhone OS) is a mobile Operating System created and developed by Apple Inc., exclusively for its hardware. It is the Operating System that presently powers many of the company’s mobile devices, including the iPhone, iPad and iPod Touch. It is the second most popular mobile Operating System globally after Android.

Question 5.
What is a software? Explain its types in detail.
Answer:
A software is a set of instructions that perform specific task. It interacts basically with the hardware to generate the desired output.
Types of Software:
Software is classified into two types:

1. Application Software
2. System Software

Application Software:
Application software is a set of programs to perform specific task. For example MS-word is an application software to create text document and VLC player is familiar application software to play audio, video files and many more.

System Software:
System software is a type of computer program that is designed to run the computer’s hardware and application programs. For example Operating System and Language Processor.

Question 6.
Write note on single user OS?
Answer:
An os allows only a single user to perform a task at a time. It is called as a single user and single task os.
Example : MS – DOS.

Question 7.
Write note on Raspbion os?
Answer:
Raspbion os is a platform that is designed to teach how to build a computer, working principle of every part of a circuit board, write code apps or games. The platform is available in pre – designed kits.

Question 8.
Explain the memory management activities done by os?
Answer:

1. Keeping track of which portion of memory are currently being used and who is using them.
2. Determining which processes and data to more in and out of memory.
3. Allocation and de – allocation of memory blocks as needed by the program in main memory. (Garbage collection).

Question 9.
Define Process?
Answer:

1. A process is the unit of work (program) in a computer.
2. A word processing program being run by an individual user on a computer is a process.
3. A system task, such as sending output to printer or screen can also be called as a process.

Question 10.
How are the processes classified on process management?
Answer:
A computer consists of a collection of process they are classified as two categories:

1. Operating system process which is executed by system code.
2. User processes which is executed by user code.

Question 11.
Name the activities done by os related with the process management?
Answer:

1. Scheduling processes and threads on the cpu.
2. Creating and deleting both user and system processes.
3. Suspending and resuming processes.
4. Providing mechanisms for process synchronization.
5. Providing mechanisms for process communication.

Question 12.
Write note an Fault Tolerance?
Answer:

1. The operating systems should be robust.
2. When there is a fault, the operating system should not crash, instead the os have fault tolerance capabilities and retain the existing state of system.

Question 13.
Write note on File Allocations Table (FAT).
Answer:

1. Any type of data in a computer is stored in the form of flies and directories / folders through File Allocation Table (FAT).
2. The FAT stores general information about files like file name, type (text or binary), size, starting address and access mode (sequential / indexed / indexed – sequential / direct / relative).

Question 14.
Give an example for Time – Sharing concept?
Answer:
Let us assume that there are three processes called P_1, P_2, P₃ and time allocated for each process 30,40, 50 minutes. If the process P₁ completes with in 20 minutes then processor takes the next Process P₁ for the execution. If the process P₂ could not complete within 40 minutes, then the current process P₂ will be paused and switch over to the next process P₃.

Question 15.
Write note on React OS.
React os is a windows – alternative open source os which is being developed on the principles of windows – without using any of the microsoft code.

Question 16.
List any 6 Android Mobile Open Source OS.
Answer:

1. Honey comb
2. Jelly Bean
3. Kitkat
4. Lollipop
5. Marshmallow
6. Nougat

PART – 4
IV. Explain in Detail

Question 1.
Write six uses of operating system.
Answer:
The main use of Operating System is

1. to ensure that a computer can be used to extract what the user wants it do.
2. Easy interaction between the users and computers.
3. Starting computer operation automatically when power is turned on (Booting).
4. Controlling Input and Output Devices
5. Manage the utilization of main memory.
6. Providing security to user programs.

Question 2.
Draw the diagram for the key features of the operating system.
Answer:

Question 3.
Explain User Interface?
Answer:
User Interface:
User interface is one of the significant feature in Operating System. The only way that user can make interaction with a computer. If the computer interface is not user – friendly, the user slowly reduces the computer usage from their normal life. This is the main reason for the key success of GUI (Graphical User Interface) based Operating System. The GUI is a window based system with a pointing device to direct I/O, choose from menus, make selections and a keyboard to enter text. Its vibrant colours attract the user very easily. Beginners are impressed by the help and pop up window message boxes. Icons are playing vital role of the particular application.

Now Linux distribution is also available as GUI based Operating System. The following points are considered when User Interface is designed for an application.

1. The user interface should enable the user to retain this expertise for a longer time.
2. The user interface should also satisfy the customer based on their needs.
3. The user interface should save user’s precious time. Create graphical elements like Menus,Window,Tabs, Icons and reduce typing work will be an added advantage of the Operating System.
4. The ultimate aim of any product is to satisfy the customer. The User Interface is also designed to satisfy the customer.
5. The user interface should reduce number of errors committed by the user with a little practice the user should be in a position to avoid errors (Error Log File)

Question 4.
Explain Memory Management?
Answer:
Memory Management:
Memory Management is the process of controlling and coordinating computer’s main memory and assigning memory block (space) to various running programs to optimize overall computer performance. The Memory management involves the allocation of specific memory blocks to individual programs based on user demands. At the application level, memory management ensures the availability of adequate memory for each running program at all times.

The objective of Memory Management process is to improve both the utilization of the CPU and the speed of the computer’s response to its users via main memory. For these reasons the computers must keep several programs in main memory that associates with many different Memory Management schemes.

The Operating System is responsible for the following activities in connection with memory management:

1. Keeping track of which portion of memory are currently being used and who is using them.
2. Determining which processes (or parts of processes) and data to move in and out of memory.
3. Allocation and de – allocation of memory blocks as needed by the program in main memory. (Garbage Collection)

Question 5.
Explain process Management.
Answer:
Process Management:
Process management is function that includes creating and deleting processes and providing mechanisms for processes to communicate and synchronize with each other. A process is the unit of work (program) in a computer. A word processing program being run by an individual user on a computer is a process. A system task, such as sending output to a printer or screen, can also be called as a Process.

A computer consists of a collection of processes, they are classified as two categories:

1. Operating System processes which is executed by system code.
2. User Processes which is execute by user code.

All these processes can potentially execute concurrently on a single CPU. A process needs certain resources including CPU time, memory, files and I/O devices to finish its task.

The Operating System is responsible for the following activities associated with the process management:

1. Scheduling processes and threads on the CPUs.
2. Creating and deleting both user and system processes.
3. Suspending and resuming processes.
4. Providing mechanisms for process synchronization.
5. Providing mechanisms for process communication.

The following algorithms are mainly used to allocate the job (process) to the processor.

1. FIFO
2. SJF
3. Round Robin
4. Based on Priority

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Samacheer Kalvi 11th Physics Kinematics Textual Questions Solved

Samacheer Kalvi 11th Physics Kinematics Multiple Choice Questions

Samacheer Kalvi 11th Physics Solution Chapter 2 Question 1.
Which one of the following Cartesian coordinate systems is not followed in physics?

Answer:

11th Physics Chapter 2 Book Back Answers Question 2.
Identify the unit vector in the following:
(a) \(\hat{i}+\hat{j}\)
(b) \(\frac{\hat{i}}{\sqrt{2}}\)
(c) \(\hat{k}-\frac{\hat{j}}{\sqrt{2}}\)
(d) \(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\)
Answer:
(d) \(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\)

11th Physics Kinematics Book Back Answers Question 3.
Which one of the following physical quantities cannot be represented by a scalar?
(a) Mass
(b) length
(c) momentum
(d) magnitude of acceleration
Answer:
(c) momentum

11th Physics Lesson 2 Book Back Answers Question 4.
Two objects of masses m₁ and m₂, fall from the heights h₁ and h₂ respectively. The ratio of the magnitude of their momenta when they hit the ground is [AIPMT 20121]
(a) \(\sqrt{\frac{h_{1}}{h_{2}}}\)
(b) \(\sqrt{\frac{m_{1} h_{1}}{m_{2} h_{2}}}\)
(c) \(\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)
(d) \(\frac{m_{1}}{m_{2}}\)
Answer:
(c) \(\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)

11th Physics 2nd Chapter Book Back Answers Question 5.
If a particle has negative velocity and negative acceleration, its speed
(a) increases
(b) decreases
(c) remains same
(d) zero
Answer:
(a) increases

Physics Class 11 Unit 2 Kinematics Question 6.
If the velocity is\(\overrightarrow{\mathrm{v}}\) – 2\(\hat{i}\) +t²\(\hat{j}\) – 9\(\overrightarrow{\mathrm{k}}\) , then the magnitude of acceleration at t = 0.5 s is
(a) 1 m s^-2
(b) 1 m
(c) zero
(d) -1 m s s^-2
Answer:
(a) 1 m s^-2

11th Physics 2nd Lesson Book Back Answers Question 7.
If an object is dropped from the top of a building and it reaches the ground at t = 4 s, then the height of the building is (ignoring air resistance) (g = 9.8 m s^-2).
(a) 77.3 m
(b) 78.4 m
(c) 80.5 m
(d) 79.2 m
Answer:
(b) 78.4 m

Samacheer Kalvi 11th Physics Solution Chapter 1 Question 8.
A ball is projected vertically upwards with a velocity v. It comes back to ground in time t. Which v -1 graph shows the motion correctly?[NSEP 00 – 01]

Answer:

Physics Chapter 2 Kinematics Question 9.
If one object is dropped vertically downward and another object is thrown horizontally from the same height, then the ratio of vertical distance covered by both objects at any instant is
(a) 1
(b) 2
(c) 4
(d) 0.5
Answer:
(a) 1

11th Physics Unit 2 Question 10.
A ball is dropped from some height towards the ground. Which one of the following represents the correct motion of the ball?

Answer:

Samacheer Kalvi Guru 11th Physics Question 11.
If a particle executes uniform circular motion in the xy plane in clockwise direction, then the angular velocity is in
(a) +y direction
(b) +z direction
(c) -z direction
(d) -x direction
Answer:
(c) -z direction

Samacheer Kalvi 11th Physics Question 12.
If a particle executes uniform circular motion, choose the correct statement [NEET 2016]
(a) The velocity and speed are constant.
(b) The acceleration and speed are constant.
(c) The velocity and acceleration are constant.
(d) The speed and magnitude of acceleration are constant.
Answer:
(d) The speed and magnitude of acceleration are constant.

Samacheer Kalvi 11th Physics Solution Book Question 13.
If an object is thrown vertically up with the initial speed u from the ground, then the time taken by the object to return back to ground is
(a) \(\frac{u^{2}}{2 g}\)
(b) \(\frac{u^{2}}{g}\)
(c) \(\frac{u}{2 g}\)
(d) \(\frac{2 u}{g}\)
Answer:
(d) \(\frac{2 u}{g}\)

Samacheer Kalvi Physics 11th Question 14.
Two objects are projected at angles 30° and 60° respectively with respect to the horizontal direction. The range of two objects are denoted as R_30° and R_60°– Choose the correct relation from the following:
(a) R_30° = R_60°
(b) R_30° = 4R_60°
(c) \(\mathrm{R}_{30^{\circ}}=\frac{\mathrm{R}_{60^{\circ}}}{2}\)
(d) R_30° = 2R_60°
Answer:
(a) R_30° = R_60°

11th Physics Samacheer Kalvi Question 15.
An object is dropped in an unknown planet from height 50 m, it reaches the ground in 2 s. The acceleration due to gravity in this unknown planet is
(a) g = 20 m s^-2
(b) g = 25 m s^-2
(c) g = 15 m s^-2
(d) g = 30 m s ^-2
Answer:
(a) g = 25 m s^-2

Samacheer Kalvi 11th Physics Kinematics Short Answer Questions

11th Physics Samacheer Kalvi Question 1.
Explain what is meant by Cartesian coordinate system?
Answer:
At any given instant of time, the frame of reference with respect to which the position of the object is described in terms of position coordinates (x, y, z) is called Cartesian coordinate system.

Physics Class 11 Samacheer Kalvi Question 2.
Define a vector. Give examples.
Answer:
Vector is a quantity which is described by the both magnitude and direction. Geometrically a vector is directed line segment.
Example – force, velocity, displacement.

Class 11 Physics Samacheer Kalvi Question 3.
Define a scalar. Give examples.
Answer:
Scalar is a property which can be described only by magnitude.
Example – mass, distance, speed.

Samacheer Kalvi 11th Physics Solution Chapter 3 Question 4.
Write a short note on the scalar product between two vectors.
Answer:
The scalar product (or dot product) of two vectors is defined as the product of the magnitudes of both the vectors and the cosine of the angle between them. Thus if there are two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) having an angle 0 between them, then their scalar product is defined as \(\overrightarrow{\mathrm{A}}\) • \(\overrightarrow{\mathrm{B}}\) = AB cos 0. Here, AB and are magnitudes of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\).

11th Samacheer Kalvi Physics Book Back Answers Question 5.
Write a short note on vector product between two vectors.
Answer:
The vector product or cross product of two vectors is defined as another vector having a magnitude equal to the product of the magnitudes of two vectors and the sine of the angle between them. The direction of the product vector is perpendicular to the plane containing the two vectors, in accordance with the right hand screw rule or right hand thumb rule. Thus, if\(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are two vectors, then their vector product is written as \(\overrightarrow{\mathrm{A}}\) × \(\overrightarrow{\mathrm{B}}\) which is a vector C defined by \(\overrightarrow{\mathrm{c}}\) = \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = (AB sin 0) \(\hat{n}\)
The direction \(\hat{n}\) of \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) , i.e., \(\overrightarrow{\mathrm{c}}\) is perpendicular to the plane containing the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\).

Samacheer Kalvi 11 Physics Solutions Question 6.
How do you deduce that two vectors are perpendicular?
Answer:
If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are perpendicular to each other than their scalar product \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) = 0 because cos 90° = 0. Then he vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are said to be mutually orthogonal.

Samacheer Kalvi 11th Physics Book Back Answers Question 7.
Define displacement and distance.
Answer:
Distance is the actual path length traveled by an object in the given interval of time during the motion. It is a positive scalar quantity. Displacement is the difference between the final and initial positions of the object in a given interval of time. It can also be defined as the shortest distance between these two positions of the object. It is a vector quantity.

Samacheer Kalvi 11th Physics Guide Pdf Question 8.
Define velocity and speed.
Answer:
Speed is defined as the ratio of total distance covered to the total time taken, it is a scalar quantity and always it is positive. Velocity is defined as the ratio of the displacement vector to the corresponding time interval. It is a vector quantity or it can also be defined as rate of change of displacement.

Question 9.
Define acceleration.
Answer:
Acceleration of a particle is defined as the rate of change of velocity or it can also be defined as the ratio of change in velocity to the given interval of time.

Question 10.
What is the difference between velocity and average velocity.
Answer:

Question 11.
Define a radian?
One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle.
1 rad = 57.295°

Question 12.
Define angular displacement and angular velocity.
Answer:
1. Angular displacement:
The angle described by the particle about the axis of rotation in a given time is called angular displacement.

2. Angular velocity:
The rate of change of angular displacement is called angular velocity.

Question 13.
What is non uniform circular motion?
If the speed of the object in circular motion is not constant, then we have non-uniform circular motion. For example, when the bob attached to a string moves in vertical circle, the speed of the bob is not the same at all time Whenever the speed is not same in circular motion, the particle will have both centripetal and tangential acceleration.

Question 14.
Write down the kinematic equations for angular motion.
Answer:
Kinematic equations for circular motion are –

1. \(\omega=\omega_{0}+\alpha t\)
2. \(\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}\)
3. \(\omega^{2}=\omega_{o}^{2}+2 \alpha \theta\)
4. \(\theta=\frac{\left(\omega_{0}+\omega\right)}{2} t\)

Here,
ω₀  = initial angular velocity
ω = final angular velocity
θ = angular displacement
α = angular acceleration
t = time.

Question 15.
Write down the expression for angle made by resultant acceleration and radius vector in the non uniform circular motion.
Answer:
The angle made by resultant acceleration and radius vector in the non uniform circular motion is –
\(\tan \theta=\frac{a_{t}}{\left(\frac{V^{2}}{r}\right)}\) or \(\theta=\tan ^{-1}\left(\frac{a_{t}}{\left(\frac{V^{2}}{r}\right)}\right)\)

Samacheer Kalvi 11th Physics Kinematics Long Answer Questions

Question 1.
Explain in detail the triangle law of addition.
Answer:
Let us consider two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) as shown in figure. To find the resultant of the two vectors we apply the triangular.

Law of addition as follows:
present the vectors A and by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.


To explain further, the head of the first vector \(\overrightarrow{\mathrm{A}}\) is connected to the tail of the second vect \(\overrightarrow{\mathrm{B}}\) Let O he the angle between\(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\). Then \(\overrightarrow{\mathrm{R}}\) is the resultant vector connecting the tail of the first vector \(\overrightarrow{\mathrm{A}}\) to the head of the second vector \(\overrightarrow{\mathrm{B}}\) The magnitude of \(\overrightarrow{\mathrm{R}}\). (resultant) given geometrically by the length of (OQ) and the direction of the resultant vector is the angle between \(\overrightarrow{\mathrm{R}}\). and \(\overrightarrow{\mathrm{A}}\). Thus we write
\(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{A}}\) +\(\overrightarrow{\mathrm{B}}\) \(\overrightarrow{\mathrm{OQ}}\) = \(\overrightarrow{\mathrm{OP}}\) + \(\overrightarrow{\mathrm{PQ}}\)

1. Magnitude of resultant vector:
The magnitude and angle of the resultant vector ar determined by using triangle law of vectors as follows.From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

From figure, let R is the magnitude of the resultant of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\).
cos θ = \(\frac { AN}{ B }\) ∴ AN = B cos θ and sinθ = \(\frac { BN}{ B }\) ∴BN = B sinθ
For ∆ OBN, we have OB² = ON² + BN²
⇒ R² = (A + B cos θ)² + (B sinθ)²
⇒ R² = A² + B² cos²θ + 2ABcosθ B² sin²θ
⇒ R² = A² + B²(cos²θ + sin²θ) + 2AB cos θ
⇒ R² = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

2. Direction of resultant vectors:
If 0 is the angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) then,
\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}\)
If R makes an angle α with \(\overrightarrow{\mathrm{A}}\) , then in AOBN,
tan α = \(\frac { BN}{ ON }\) = \(\frac { BN}{ OA + AN }\)
tan α = \(\frac { B sinθ }{ A + B cosθ}\) ⇒ α = \(\tan ^{-1}\left(\frac{B \sin \theta}{A+B \cos \theta}\right)\)

Question 2.
Discuss the properties of scalar and vector products.
Answer:
Properties of scalar product of two vectors are:
(1) The product quantity \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\) is always a scalar. It is positive if the angle between the vectors is acute (i.e., < 90°) and negative if the angle between them is obtuse (i.e. 90°<0< 180°).

(2) The scalar product is commutative, i.e. \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) ≠ \(\overrightarrow{\mathrm{B}}\). \(\overrightarrow{\mathrm{A}}\)

(3) The vectors obey distributive law i.e. \(\overrightarrow{\mathrm{A}}\)(\(\overrightarrow{\mathrm{B}}\) + \(\overrightarrow{\mathrm{C}}\)) = \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\) + \(\overrightarrow{\mathrm{A}}\) .\(\overrightarrow{\mathrm{C}}\)
(4) The angle between the vectors θ = \(\cos ^{-1}\left[\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{\mathrm{AB}}\right]\)

(5) The scalar product of two vectors will be maximum when cos θ = 1, i.e. θ = 0°, i.e., when the vectors are parallel;
\((\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}})_{\max }=\mathrm{AB}\)

(6) The scalar product of two vectors will be minimum, when cos θ = -1, i.e. θ = 180°.
\((\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}})_{\min }=-\mathrm{AB}\) when the vectors are anti-parallel.

(7) If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are perpendicular to each other than their scalar product \(\overrightarrow{\mathrm{A}}\) .\(\overrightarrow{\mathrm{B}}\) = 0, because cos 90° 0. Then the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are said to be mutually orthogonal.

(8) The scalar product of a vector with itself is termed as self-dot product and is given by (\(\overrightarrow{\mathrm{A}}\))² = \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{A}}\) = AA cos 0 = A². Here angle 0 = 0°.
The magnitude or norm of the vector \(\overrightarrow{\mathrm{A}}\) is |\(\overrightarrow{\mathrm{A}}\)| = A = \(\sqrt{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}}\).

(9) In case of a unit vector \(\hat{n}\)
\(\hat{n}\) . \(\hat{n}\) = 1 x 1 x cos 0 = 1. For example, \(\hat{i}\) – \(\hat{i}\) = \(\hat{j}\) . \(\hat{j}\) = \(\hat{k}\) . \(\hat{k}\) = 1.

(10) In the case of orthogonal unit vectors, \(\hat{i}\),\(\hat{j}\) and \(\hat{k}\),
\(\hat{i}\) . \(\hat{j}\) = \(\hat{j}\).\(\hat{k}\) = \(\hat{k}\) . \(\hat{i}\)= 1.1 cos 90° = 0

(11) In terms of components the scalar product of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) can be written as
\(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\) = (A_x\(\hat{i}\) + A_y\(\hat{j}\) + A_z\(\hat{k}\)).(B_x\(\hat{i}\) + B_y\(\hat{j}\) + B_z\(\hat{k}\))
= A _xB_x + A_yB_y+ A_zB_z, with all other terms zero.
The magnitude of vector | \(\overrightarrow{\mathrm{A}}\) | is given by
| \(\overrightarrow{\mathrm{A}}\) | = A = \(\sqrt{\mathrm{A}_{x}^{2}+\mathrm{A}_{y}^{2}+\mathrm{A}_{z}^{2}}\)

Properties of vector product of two vectors are:
(1) The vector product of any two vectors is always another vector whose direction is perpendicular to the plane containing these two vectors, i.e., orthogonal to both the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\), even though the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) may or may not be mutually orthogonal.

(2) The vector product of two vectors is not commutative, i.e., \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) ≠ \(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\). But,
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)=-\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\).
Here it is worthwhile to note that |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| =
|\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)| = AB sin 0 i.e., in the case of the product vectors \(\overrightarrow{\mathrm{B}}\)=-\(\overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\), the magnitudes are equal but directions are opposite to each other.

(3) The vector product of two vectors will have maximum magnitude when sin 0 = 1, i.e., 0 = 90° i.e., when the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are orthogonal to each other.
\((\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})_{\mathrm{max}}=\mathrm{AB} \hat{n}\) = AB \(\hat{n}\)

(4) The vector product of two non-zero vectors will be minimum when sin θ = 0, i.e θ = 0° or 180°
\((\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})_{\min }=0\)
i. e., the vector product of two non – zero vectors vanishes, if the vectors are either parallel or anti parallel.

(5) The self – cross product, i.e., product of a vector with itself is the null vector
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{A}}\) = AA sin 0° \(\hat{n}\) = \(\overrightarrow{\mathrm{0}}\) In physics the null vector 0 is simply denoted as zero.

(6) The self – vector products of unit vectors are thus zero.
\(\hat{i}\) x \(\hat{i}\) = \(\hat{ j}\) x \(\hat{j}\) = \(\hat{k}\) x \(\hat{k}\) = 0

(7) In the case of orthogonal unit vectors, \(\hat{i}\), \(\hat{j}\). \(\hat{k}\) , in accordance with the right hand screw rule:
\(\hat{i}\) x \(\hat{j}\) = \(\hat{k}\), \(\hat{j}\) x \(\hat{k}\) = \(\hat{i}\) and \(\hat{k}\) x \(\hat{i}\) = \(\hat{j}\)

Also, since the cross product is not commutative,
\(\hat{j}\) x \(\hat{i}\) = –\(\hat{k}\), \(\hat{k}\) x \(\hat{j}\) = –\(\hat{i}\) and \(\hat{i}\) x \(\hat{k}\) = \(\hat{j}\)

(8) In terms of components, the vector product of two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is –

Note that in the \(\hat{j}^{\mathrm{th}}\) component the order of multiplication is different than \(\hat{i}^{\mathrm{th}}\) and \(\hat{k}^{\mathrm{th}}\) components.

(9) If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) form adjacent sides in a parallelogram, then the magnitude of |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| will give the area of the parallelogram as represented graphically in figure.

(10) Since we can divide a parallelogram into two equal triangles as shown in the figure, the area of a triangle with \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) as sides is \(\frac { 1 }{ 2 }\) |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| . This is shown in the Figure. A number of quantities used in Physics are defined through vector products. Particularly physical quantities representing rotational effects like torque, angular momentum, are defined through vector products.

Question 3.
Derive the kinematic equations of motion for constant acceleration.
Answer:
Consider an object moving in a straight line with uniform or constant acceleration ‘a’. Let u be the velocity of the object at time t = 0, and v be velocity of the body at a later time t.

Velocity – time relation:
(1) The acceleration of the body at any instant is given by the first derivative of the velocity with respect to time,
a = \(\frac {dv}{dt}\) or dv = a dt
Integrating both sides with the condition that as time changes from 0 to t, the velocity changes from u to v. For the constant acceleration,

(2) The velocity of the body is given by the first derivative of the displacement with respect to time.
v = \(\frac {ds}{dt}\) or ds = vdt
and since v = u + at,
we get ds = (u+ at ) dt
Assume that initially at time t = 0, the particle started from the origin. At a later time t, the particle displacement is s. Further assuming that acceleration is time-independent, we have

Velocity – displacement relation:

(3) The acceleration is given by the first derivative of velocity with respect to time.
a = \(\frac {dv}{dt}\) = \(\frac {dv}{ds}\) = \(\frac {ds}{dt}\) = \(\frac {dv}{ds}\) v [since ds/dt = v] where s is displacement traverse
This is rewritten as a = \(\frac{1}{2} \frac{d v^{2}}{d s}\) or ds = \(\frac{1}{2 a} d\left(v^{2}\right)\) Integrating the above equation, using the fact when the velocity changes from u² to v², displacement changes from 0 to s, we get

We can also derive the displacement 5 in terms of initial velocity u and final velocity v. From equation we can
write,
at = v – u
Substitute this in equation, we get

Question 4.
Derive the equations of motion for a particle (a) falling vertically (b) projected vertically.
Answer:
’Equations of motion for a particle falling vertically downward from certain height. Consider an object of mass m falling from a height h. Assume there is no air resistance. For convenience, let us choose the downward direction as positive y – axis as shown in the figure. The object experiences acceleration ‘g’ due to gravity which is constant near the surface of the Earth. We can use kinematic equations to explain its motion. We have The acceleration \(\overrightarrow{\mathrm{a}}\) = g \(\hat{i}\)
By comparing the components, we get,
Equations of motion for a particle thrown vertically upwards,
a_x = 0, a_x = 0, a_y = g Let us take for simplicity, a_y = a = g

If the particle is thrown with initial velocity ‘u’ downward which is in negative y – axis, then velocity and position at of the particle any time t is given by
v = u + gt
v = ut + \(\frac {1}{2}\) – gt²
The square of the speed of the particle when it is at a distance y from the hill – top, is v² = u² + 2 gy
Suppose the particle starts from rest.
Then u = 0
Then the velocity v, the position of the particle and v² at any time t are given by (for a point y from the hill – top)
v = gt …………(i)
y = \(\frac {1}{2}\) – gt² …………(ii)
v² = 2gy …………(iii)
The time (t = T) taken by the particle to reach the ground (for which y = h), is given by using equation (ii),
h = \(\frac {1}{2}\) – gT² …………(iv)
T = \(\sqrt{\frac{2 h}{g}}\) …………(v)
The equation (iv) implies that greater the height (h), particle takes more time (T) to reach the ground. For lesser height (h), it takes lesser time to reach the ground. The speed of the particle when it reaches the ground (y = h) can be found using equation (iii), we get,
\(v_{\text {ground }}=\sqrt{2 g h}\) …………(vi)
The above equation implies that the body falling from greater height (h) will have higher velocity when it reaches the ground. The motion of a body falling towards the Earth from a small altitude (h<<R), purely under the force of gravity is called free fall. (Here R is radius of the Earth).

case (ii):
A body thrown vertically upwards:
Consider an object of mass m thrown vertically upwards with an initial velocity u. Let us neglect the air friction. In this case we choose the vertical direction as positive y axis as shown in the figure, then the acceleration a = -g (neglect air friction) and g points towards the negative y axis. The kinematic equations for this motion are,
The velocity and position of the object at any time t are,

v = u – gt ……………(vii)
s = ut – \(\frac {1}{2}\) – gt² …………..(viii)
The velocity of the object at any position y (from the point where the object is thrown) is
v² = u² – 2gy …………..(ix)

Question 5.
Derive the equation of motion, range and maximum height reached by the particle thrown at an oblique angle 9 with respect to the horizontal direction.
Answer:
This projectile motion takes place when the initial velocity is not horizontal, but at some angle with the vertical, as shown in Figure.
(Oblique projectile)
Examples:

• Water ejected out of a hose pipe held obliquely.
• Cannon fired in a battle ground.

Consider an object thrown with initial velocity at an angle θ with the horizontal.
Then,
\(\overrightarrow{\mathrm{u}}\) = u_x î + u_y\(\hat{j}\) .
where u_x = u cos θ is the horizontal component and u_y = u sin θ the vertical component of velocity. Since the acceleration due to gravity is in the direction opposite to the direction of vertical component u_y , this component will gradually reduce to zero at the maximum height of the projectile. At this maximum height, the same gravitational force will push the projectile to move downward and fall to the ground. There is no acceleration along the x direction throughout the motion. So, the horizontal component of the velocity (u_x = u cos θ) remains the same till the object reaches the ground. Hence after the time t, the velocity along horizontal motion v_x = u_x + a_xt = u_x = u cos θ. The horizontal distance travelled by projectile m time t is s_x = \(u_{x} t+\frac{1}{2} a_{x} t^{2}\)
Here, s_x = x, u_x = u cos θ, a_x = 0

Thus, x = u cos θ or t = \(\frac {x}{u cos θ}\) ……..(i)
Next, for the vertical motion v_y= u_y + a_yt
Here u_y = u sin θ, a_y = -g (acceleration due to gravity acts opposite to the motion).
Thus, v_y= u sin θ – gt
The vertical distance traveled by the projectile in the same time t is
Here, s_y = y, u_y = u sin θ, a_x = -g. Then
y = u sinθ t – \(\frac {1}{2}\) – gt² ………..(ii)
Substitute the value of t from equation (i) in equation (ii), we have .

Thus the path followed by the projectile is an inverted parabola Maximum height (h_max): The maximum vertical distance travelled by the projectile during the journey is called maximum height. This is determined as follows:
For the vertical part of the motion.
\(v_{y}^{2}=u_{y}^{2}+2 a_{y} s\)
Here, u_y = u sin θ, a = -g, s = h_max, and at the maximum height v_y = 0
Hence, (0)² = u² sin² θ = 2 gh_max or \(h_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}\)

Time of flight (T_f):
The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight. This time of flight is the time taken by the projectile to go from point O to B via point A as shown
we know that s_y = y = 0 (net displacement in y-direction is zero),
u_y = u sin θ, a_y = -g , t = T_f Then

Horizontal range (R):
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). This is found easily since the horizontal component of initial velocity remains the same. We can write Range R = Horizontal component of velocity x time of flight = u cos θ x \(\mathrm{T}_{f}=\frac{u^{2} \sin 2 \theta}{g}\) The horizontal range directly depends on the initial speed (u) and the sine of angle of projection (θ). It inversely depends on acceleration due to gravity ‘g’.
For a given initial speed u, the maximum possible range is reached when sin 2θ is maximum, sin 2θ = 1. This implies 2θ = π/2 or θ = \(\frac {π}{4}\) This means that if the particle is projected at 45 degrees with respect to horizontal, it attains maximum range, given by.
\(\mathrm{R}_{\max }=\frac{u^{2}}{g}\) ………..(vi)

Question 6.
Derive the expression for centripetal acceleration.
Answer:
In uniform circular motion the velocity vector turns continuously without changing its magnitude (speed), as shown in figure.

Note that the length of the velocity vector is not changed during the motion, implying that the speed remains constant. Even though the velocity is tangential at every point in the circle, the acceleration is acting towards the center of the circle. This is called centripetal acceleration. It always points towards the center of the circle. This is shown in the figure.

The centripetal acceleration is derived from a simple geometrical relationship between position and velocity vectors.

Let the directions of position and velocity vectors shift through the same angle θ in a small interval of time ∆t, as shown in figure. For uniform circular motion, r = \(\left|\vec{r}_{1}\right|\) = \(\left|\vec{r}_{2}\right|\) and v = \(\left|\vec{v}_{1}\right|\) = \(\left|\vec{v}_{2}\right|\). If the particle moves from position vector \(\vec{r}_{1}\) to \(\vec{r}_{2}\), the displacement is given by ∆\(\overrightarrow{\mathrm{r}}\) = \(\vec{r}_{2}\) – \(\vec{r}_{1}\) and the change in velocity from \(\vec{v}_{1}\) to\(\vec{v}_{2}\) is given by ∆\(\overrightarrow{\mathrm{v}}\) = \(\vec{v}_{2}\) – \(\vec{v}_{1}\),. The magnitudes of the displacement ∆r and of ∆v satisfy the following relation. \(\frac {∆r}{r}\) = \(\frac {-∆v}{v}\) = θ Here the negative sign implies that ∆v points radially inward, towards the center of the circle.

For uniform circular motion v = cor, where co is the angular velocity of the particle about center. Then the centripetal acceleration can be written as.
a = -ω²r

Question 7.
Derive the expression for total acceleration in the non uniform circular motion.
Answer:
If the speed of the object in circular motion is not constant, then we have non-uniform circular motion. For example, when the bob attached to a string moves in vertical circle, the speed of the bob is not the same at all time. Whenever the speed is not same in circular motion, the particle will have both centripetal and tangential acceleration as shown in the figure.

The resultant acceleration is obtained by vector sum of centripetal and tangential acceleration Since centripetal acceleration is \(\frac{v^{2}}{r}\), the magnitude of this resultant acceleration is given by –\(\dot{a}_{\mathrm{R}}=\sqrt{a_{t}^{2}+\left(\frac{v^{2}}{r}\right)^{2}}\)
This resultant acceleration makes an angle 0 with the radius vector as shown in figure.
This angle is given by tan θ = \(\frac{a_{t}}{\left(v^{2} / r\right)}\)

Samacheer Kalvi 11th Physics Kinematics Numerical Questions

Question 1.
The position vector of the particle has length 1 m and makes 30° with the x-axis. What are the lengths of the x and y – components of the position vector?
Answer:
Given,
Length of position vector = 1 m
Angle made with x axis = 30
Solution:
Length of X component (OB) = OA cos θ
= 1 x cos 30°
= \(\frac{\sqrt{3}}{2}\) (or) 0.87 m
Length of Y component (AB) = OA sin θ = 1 x sin 30° = \(\frac { 1 }{ 2 }\) = 0.5 m.

Question 2.
A particle has its position moved from \(\vec{r}_{1}\) = 3\(\hat{i}\) + 4\(\hat{j}\) to r2 = \(\hat{i}\) + 2\(\hat{i}\). Calculate the displacement vector (∆\(\overrightarrow{\mathrm{r}}\) ) and draw the \(\vec{r}_{1}\), \(\vec{r}_{2}\) and ∆\(\overrightarrow{\mathrm{r}}\) vector in a two dimensional Cartesian coordinate system.
Answer:
Given,
Position vectors \(\vec{r}_{1}\) = 3\(\hat{i}\) + 4\(\hat{j}\)
\(\vec{r}_{1}\) = \(\hat{i}\) + 2\(\hat{j}\)
Solution:
Displacement vector:
∆r= \(\vec{r}_{2}\) – \(\vec{r}_{1}\) = (1 – 3)\(\hat{i}\) + (2 – 4) \(\hat{j}\)
∆r = -2\(\hat{i}\) -2\(\hat{j}\) = -2(\(\hat{i}\) + \(\hat{j}\))

Question 3.
Calculate the average velocity of the particle whose position vector changes from \(\vec{r}_{1}\) = 5\(\hat{i}\) + 6\(\hat{j}\) to \(\vec{r}_{2}\) = 2\(\hat{i}\) + 3 \(\hat{j}\) in a time 5 seconds.
Answer:
Given,
Position vectors of a particle
\(\vec{r}_{1}\) = 5\(\hat{i}\) + 6\(\hat{j}\),
\(\vec{r}_{2}\) = 2\(\hat{i}\) + 35\(\hat{j}\)
time(t) = 5s
Solution:

Question 4.
Convert the vector \(\overrightarrow{\mathrm{r}}\) = 3\(\hat{i}\) + 2\(\hat{j}\) into a unit vector.
Answer:
Given:
Position vector\(\hat{r}\) = 3\(\hat{i}\) + 2\(\hat{j}\)
Solution:

Question 5.
What are the resultants of the vector product of two given vectors given by \(\overrightarrow{\mathrm{A}}\) = 4\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\) and \(\overrightarrow{\mathrm{B}}\) = 5\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\) ?
Answer:
Given,
Vectors \(\overrightarrow{\mathrm{A}}\) = 4\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\)
\(\overrightarrow{\mathrm{B}}\) = 5\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\)
Solution:

Question 6.
object at an angle such that the horizontal range is 4 times of the maximum height. What is the angle of projection of the object?
Answer:
Give,
Horizontal range = 4H_max
Solution:

Question 7.
The following graphs represent velocity – time graph. Identity what kind of motion a particle undergoes in each graph.

Answer:
(a) At all the points, slope of the graph is constant.
∴ \(\overrightarrow{\mathrm{a}}\) = constant

(b) No change in magnitude of velocity with respect to time
∴ \(\overrightarrow{\mathrm{v}}\) = constant

(c) Slope of this graph is greater than graph (a) but constant
∴ \(\overrightarrow{\mathrm{a}}\) = constant but greater than the graph (a)

(d) At each point slope of the curve increases.
∴ \(\overrightarrow{\mathrm{a}}\) is a variable and object is in accelerated motion.

Question 8.
The following velocity – time graph represents a particle moving in the positive x-direction. Analyse its motion from 0 to 7 s. Calculate the displacement covered and distance travelled by the particle from 0 to 2 s.

Answer:
As per graph,
(a) From 0 to 1.5 s the particle moving in a opposite direction.

• From 1.5 s to 2 s the particle is moving with increasing velocity.
• From 2 s to 5 s velocity of the particle is constant of magnitude 1 ms ^-1
• From 5 s to 6 s velocity of the particle is decreasing.
• From 6 s to 7 s the particle is at rest.

(b) Distance covered by the particle – Area covered under (v -t) graph

Displacement of the particle

Question 9.
A particle is projected at an angle of θ with respect to the horizontal direction. Match the following for the above motion.
(a) v_x – decreases and increases
(b) v_y  – remains constant
(c) Acceleration – varies
(d) Position vector – remains downward
Answer:
(a) v_x = remains constant
(b) v_y = decreases and increases
(c) a = remains downward
(d) r = varies

Question 10.
A water fountain on the ground sprinkles water all around it. If the speed of the water coming out of the fountain is v, calculate the total area around the fountain that gets wet.
Answer:
Given,
Speed of water = v
Solution:
Water comes from a fountain can be taken as projectile and the distance covered is maximum range of projectile i.e. θ = 45°.
Range of the particle (R_max) = \(\frac{v^{2}}{g}\) sin 2θ = \(\frac{v^{2}}{g}\)
here, R_max is radius of the area covered.

Question 11.
The following table gives the range of a particle when thrown on different planets. All the particles are thrown at the same angle with the horizontal and with the same initial speed. Arrange the planets in ascending order according to their acceleration due to gravity, (g value).

Answer:
Range = \(\frac{v^{2}}{g}\) sin 2θ ∴ g α \(\frac { 1 }{ range }\)
Ascending order of the planet with respect to their “g” is Mercury, Mars, Earth, Jupiter.

Question 12.
The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B. Then the angle between A and B is
(a) 30°
(b) 45°
(c) 150°
(d) 120°
Answer:
Given:
Resultant of \(\overrightarrow{\mathrm{A}}\) & \(\overrightarrow{\mathrm{B}}\) is perpendicular to \(\overrightarrow{\mathrm{A}}\) and magnitude of resultant (C) = \(\frac { 1 }{ 2 }\) \(\overrightarrow{\mathrm{B}}\) and α = 90°
Solution:
(i) Magnitude of resultant:

(ii) direction of resultant:

Question 13.
Compare the components for the following vector equations
(a) T\(\hat{j}\) -mg\(\hat{j}\) = ma\(\hat{j}\)
(b) \(\overrightarrow{\mathrm{T}}\) + \(\overrightarrow{\mathrm{F}}\) = \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
(c) \(\overrightarrow{\mathrm{T}}\) – \(\overrightarrow{\mathrm{F}}\) = \(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)
(d) T\(\hat{j}\) + mg\(\hat{j}\)= ma\(\hat{j}\)
Answer:
Components of the vectors
(a) T – mg = ma
(b) \(\overline{\mathrm{T}}_{x}+\overline{\mathrm{F}}_{x}\) = \(\overline{\mathrm{A}}_{x}+\overline{\mathrm{B}}_{x}\) (or) \(\overline{\mathrm{T}}_{y}+\overline{\mathrm{F}}_{y}=\overline{\mathrm{A}}_{y}+\overline{\mathrm{B}}_{y}\)
(c) \(\overline{\mathrm{T}}_{x}-\overline{\mathrm{F}}_{x}=\overline{\mathrm{A}}_{x}+\overline{\mathrm{B}}_{x}\)  (or) \(\overline{\mathrm{T}}_{y}-\overline{\mathrm{F}}_{y}=\overline{\mathrm{A}}_{y}+\overline{\mathrm{B}}_{y}\)
(d) T + mg = ma

Question 14.
Calculate the area of the triangle for which two of its sides are given by the vectors A = 5\(\hat{i}\) – 3\(\hat{j}\), B = 4\(\hat{i}\) + 6\(\hat{j}\) .
Answer:
Solution:
Area of the triangle = \(\frac { 1 }{ 2 }\) |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{A}}\)|

Question 15.
If Earth completes one revolution in 24 hours, what is the angular displacement made by Earth in one hour? Express your answer in both radian and degree.
Answer:
Given,
time period of earth = 24 hours
Solution:
Earth covers 360° in 24 hours
∴Angular displacement m 1 hour = \(\frac { 360° }{ 24 }\) = 15° (or) \(\frac { π }{ 12 }\)
Angular displacement in radian = \(\frac { 15° }{ 57.295° }\) = 0.262 rad

Question 16.
An object is thrown with initial speed 5 ms ^-1 with an angle of projection 30°. What is the height and range reached by the particle?
Answer:
Given,
Initial speed (u) = 5 ms^-1
Angle of projection θ = 30°
Solution:

Question 17.
A foot – ball player hits the ball with speed 20 ms^-1 with angle 30° with respect to horizontal direction as shown in the figure. The goal post is at a distance of 40 m from him. Find out whether ball reaches the goal post.

Answer:
Given:
Initial speed (u) = 20 ms^-1
Angle of projection (θ) = 30°
The distance of the goal post = 40 m
Solution:
Range of the projectile

The distance of goal post is 40 m. But the range of the ball is 35.35 m only. So ball will not reach the goal post.

Question 18.
If an object is thrown horizontally with an initial speed 10 ms ^-1 from the top of a building of height 100 m. What is the horizontal distance covered by the particle?
Answer:
Given,
Initial speed =10 ms^-1
Height of the building (h) = 100 m
Range = ?
Solution:
Range of the object = R = \(u \sqrt{\frac{2 h}{g}}\) = 10\(\sqrt{\frac{200}{9.8}}\) = 45.1 m
R = 45 m.

Question 19.
An object is executing uniform circular motion with an angular speed of \(\frac { π }{ 12 }\) radian per second. At t = 0 the object starts at angle θ = 0. What is the angular displacement of the particle after 4 s?
Answer:
Given:
Angular speed = \(\frac { π }{ 12 }\) rad/ sec
Solution:
Angular speed = \(\frac { Angular displacement}{ time taken }\)
Angular displacement = \(\frac { π }{ 12 }\) x 4 = \(\frac { π }{ 12 }\) = 60°

Question 20.
Consider the x-axis as representing east, the v – axis as north and z – axis as vertically upwards. Give the vector representing each of the following points.
(a) 5 m north east and 2 m up
(b) 4 m south east and 3 m up
(c) 2 m north west and 4 m up
Answer:
Given,
Solution:
(a) Length along X – axis = 5 cos 45° = \(\frac{5}{\sqrt{2}}\)m
Length along Y- axis = 5 sin 45° = \(\frac{5}{\sqrt{2}}\)m
Length along Z – Axis = 2 m
In vector rotation = \(\frac{5}{\sqrt{2}}\)\(\hat{i}\) + \(\frac{5}{\sqrt{2}}\)\(\hat{j}\) + 2\(\hat{k}\) = \(\frac{5(\hat{i}+\hat{j})}{\sqrt{2}}\) + 2\(\hat{k}\)

(b) Length along X = 4 cos 45° = \(\frac{4}{\sqrt{2}}\)m
Length along Y = 4 sin 45° = \(\frac{4}{\sqrt{2}}\)m
Length along Z-axis = 3 m
In vector rotation = \(\frac{4}{\sqrt{2}}\)\(\hat{i}\) – \(\frac{4}{\sqrt{2}}\)\(\hat{i}\) + 3k = 4(\(\hat{i}\) – \(\hat{j}\)) \(\sqrt{2}+3 \hat{k}\)

(c) Length along X = – 2 cos 45° = \(\sqrt{2}+3 \hat{k}\) = \(\frac{2}{\sqrt{2}} m=\sqrt{2} m\)
Length along Y = 2 sin 45° = \(\frac{2}{\sqrt{2}} m=\sqrt{2} m\)
length along Z = 4 m
∴ In vector rotation = \(-\sqrt{2} \hat{i}+\sqrt{2} \hat{j}+4 \hat{k}\)

Question 21.
The Moon is orbiting the Earth approximately once in 27 days, what is the angle transformed by the Moon per day?
Answer:
Given,
period of moon = 27 days
Solution:
i.e. in 27 days moon covers 360°
In one day angle traversed by moon = \(\frac { 360° }{ 2H }\) = 13.3°

Question 22.
An object of mass m has angular acceleration a = 0.2 rad s2. What is the angular displacement covered by the object after 3 second? (Assume that the object started with angle zero with zero angular velocity).
Answer:
Given,
Angular acceleration = α = 0.2 rad s^-2
Time = 3s
Initial velocity = 0
Solution:

Samacheer Kalvi 11th Physics Kinematics Additional Questions Solved

Samacheer Kalvi 11th Physics Kinematics Multiple Choice Questions 

Question 1.
The radius of the earth was measured by –
(a) Newton
(b) Eratosthenes
(c) Galileo
(d) Ptolemy
Answer:
(b) Eratosthenes

Question 2
The branch of mechanics which deals with the motion of objects without taking force into account is –
(a) kinetics
(b) dynamics
(c) kinematics
(d) statics
Answer:
(c) kinematics

Question 3.
If the coordinate axes (x, y, z) are drawn in anticlockwise direction then the co-ordinate system is known as –
(a) Cartesian coordinate system
(b) right handed coordinate system
(c) left handed coordinate system
(d) cylindrical coordinate system
Answer:
(b) right handed coordinate system

Question 4.
The dimension of point mass is –
(a) 0
(b) 1
(c) 2
(d) kg
Answer:
(a) 0

Question 5.
If an object is moving in a straight line then the motion is known as –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(a) linear motion

Question 6.
An athlete running on a straight track is an example for the whirling motion of a stone attached to’a string is a –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(a) linear motion

Question 7.
The whirling motion of a stone attached to a string is a –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(b) circular motion

Question 8.
Spinning of the earth about its own axis is known as –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(d) rotational motion

Question 9.
If an object executes a to and fro motion about a fixed point, is an example for –
(a) rotational motion
(b) vibratory motion
(c) circular motion
(d) curvilinear motion
Answer:
(b) vibratory motion

Question 10.
Vibratory motion is also known as –
(a) circular motion
(b) rotational motion
(c) oscillatory motion
(d) spinning
Answer:
(c) oscillatory motion

Question 11.
The motion of satellite around the earth is an example for –
(a) circular motion
(b) rotational motion
(c) elliptical motion
(d) spinning
Answer:
(a) circular motion

Question 12.
An object falling freely under gravity close to earth is –
(a) one dimensional
(b) circular motion
(c) rotational motion
(d) spinning motion
Answer:
(a) one dimensional

Question 13.
Motion of a coin on a carrom board is an example of –
(a) one dimensional motion
(b) two dimensional motion
(c) three dimensional motion
(d) none
Answer:
(b) two dimensional motion

Question 15.
A bird flying in the sky is an example of –
(a) one dimensional motion
(b) two dimensional motion
(c) three dimensional motion
(d) none
Answer:
(c) three dimensional motion

Question 16.
Example for scalar is –
(a) distance
(b) displacement
(c) velocity
(d) angular momentum
Answer:
(a) distance

Question 17.
Which of the following is not a scalar?
(a) Volume
(b) angular momentum
(c) Relative density
(d) time
Answer:
(b) angular momentum

Question 18.
Vector is having –
(a) only magnitude
(b) only direction
(c) bot magnitude and direction
(d) either magnitude or direction
Answer:
(c) both magnitude and direction

Question 19.
“norm” of the vector represents –
(a) only magnitude
(b) only direction
(c) both magnitude and direction
(d) either magnitude or direction
Answer:
(a) only magnitude

Question 20.
If two vectors are having equal magnitude and same direction is known as –
(a) equal vectors
(b) col-linear vectors
(c) parallel vectors
(d) on it vector
Answer:
(a) equal vectors

Question 21.
The angle between two collinear vectors is / are –
(a) 0°
(b) 90°
(c) 180°
(d) 0° (or) 180°
Answer:
(d) 0° (or) 180°

Question 22.
The angle between parallel vectors is –
(a) 0°
(b) 90°
(c) 180°
(d) 0° (or) 180°
Answer:
(a) 0°

Question 23.
The angle between anti parallel vectors is –
(a) 0°
(b) 90°
(c) 180°
(d) 0° (or) 180°
Answer:
(c) 180°

Question 24.
Unit vector is –
(a) having magnitude one but no direction
(b) \(A \widehat{A}\)
(c) \(\frac{\widehat{A}}{A}\)
(d) |A|
Answer:
(c) \(\frac{\widehat{A}}{A}\)

Question 25.
A unit vector is used to specify –
(a) only magnitude
(b) only direction
(c) either magnitude (or) direction
(d) absolute value
Answer:
(b) only direction

Question 26.
The angle between any two orthogonal unit vectors is –
(a) 0°
(b) 90°
(c) 180°
(d) 360°
Answer:
(b) 90°

Question 27.
If \(\hat{n}\) is a unit vector along the direction of \(\overrightarrow{\mathrm{A}}\), the \(\hat{n}\) is-
(a) \(\overrightarrow{\mathrm{A}}\) A
(b) n x A
(c) \(\overrightarrow{\mathrm{A}} / \mathrm{A}\)
(d \(\overrightarrow{\mathrm{A}}\) |A|
Answer:
(c) \(\overrightarrow{\mathrm{A}} / \mathrm{A}\)

Question 28.
The magnitude of a vector can not be-
(a) positive
(b) negative
(e) zero
(cl) 90
Answer:
(b) negative

Question 29.
If R = P + Q, then which of the following is true?
(a) P > Q
(b) Q >P
(c) P = Q
(d) R > P, Q
Answer:
(d) R > P, Q

Question 30.
A force of 3 N and 4 N are acting perpendicular to an object, the resultant force is-
(a) 9 N
(b) 16 N
(c) 5 N
(d) 7 N
Answer:
(c) 5 N

Question 31.
Torque is a-
(a) scalar
(b) vector
(c) either scalar (or) vector
(d) none
Answer:
(6) vector

Question 32.
The resultant of \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\) acts along x – axis. If A = 2\(\hat{i}\) – 3 \(\hat{j}\) + 2\(\hat{k}\) then B is-
(a) -2\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)
(b) 3\(\hat{j}\) – 2\(\hat{k}\)
(c) -2\(\hat{i}\) -3 \(\hat{j}\)
(d) -2\(\hat{i}\) – 2\(\hat{k}\)
Answer:
(b) 3\(\hat{j}\) – 2\(\hat{k}\)

Question 33.
The angle between (\(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)) and (\(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)) can be –
(a) only 0°
(b) only 90°
(c) between 0° and 90°
(d) between 0° and 180°
Answer:
(d) between 0° and 180°

Question 34.
If a vector \(\overrightarrow{\mathrm{A}}\) = 3\(\hat{i}\) + 2\(\hat{j}\) then what is 4 A-
(a) 12\(\hat{i}\) + 8\(\hat{j}\)
(b) 0.75\(\hat{i}\) + 0.5\(\hat{j}\)
(c) 3\(\hat{i}\) + 2\(\hat{j}\)
(d) 7\(\hat{i}\) + 6\(\hat{j}\)
Answer:
(a) 12\(\hat{i}\) + 8\(\hat{j}\)

Question 35.
If P = mV then the direction of P along-
(a) m
(b) v
(c) both (a) and (b)
(d) neither m nor v
Answer:
(b) v

Question 36.
The scalar product \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\) is equal to-
(a) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
(b) AB sin θ
(c) AB cos θ
(d) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
Answer:
(c) AB cos θ

Question 37.
The scalar product \(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\)is equal to-
(a) \(\overrightarrow{\mathrm{A}}\) +\(\overrightarrow{\mathrm{B}}\)
(b) \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\)
(c) AB sin θ
(d) (\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)
Answer:
(b) \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\)

Question 38.
The scalar product of two vectors will be maximum when θ is equal to –
(a) 0°
(b) 90°
(c) 180°
(d) 270°
Answer:
(a) 0°

Question 39.
The scalar product of two vectors will be minimum. When θ is equal to –
(a) 0°
(b) 45°
(c) 180°
(d) 60°
Answer:
(c) 180°

Question 40.
The vectors A and B to be mutually orthogonal when –
(a) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\) = 0
(b) \(\overrightarrow{\mathrm{A}}\) –\(\overrightarrow{\mathrm{B}}\) = 0
(c) \(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\) = 0
(d) \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = 0
Answer:
(c) \(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\) = 0

Question 41.
The magnitude of the vector is –
(a) A²
(b) \(\sqrt{\mathrm{A}^{2}}\)
(c) \(\sqrt{\mathrm{A}}\)
(d) \(\sqrt[3]{\mathrm{A}}\)
Answer:
(b) \(\sqrt{\mathrm{A}^{2}}\)

Question 42.
\(\hat{i}\) .\(\hat{j}\) is –
(a) 0
(b) I
(c) ∞
(d) none
Answer:
(a) 0

Question 43.
If \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are two vectors, which are acting along x, y respectively, then \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) lies along –
(a) x
(b) y
(c) z
(d) none
Answer:
(c) z

Question 44.
The direction of \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) is given by-
(a) right hand screw rule
(b) right hand thumb rule
(c) both (a) and (b)
(d) neither (a) and (b)
Answer:
(c) both (a) and (b)

Question 45.
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) is –
(a) AB cos θ
(b) AB sin θ
(c) AB tan θ
(d) AB sec θ
Answer:
(b) AB sin θ

Question 46.
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) isequal to –
(a) \(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)
(b) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
(c) –\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)
(d) \(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)
Answer:
(c) –\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)

Question 47.
The vector product of any two vectors gives a –
(a) vector
(b) scalar
(e) tensor
(d) col-linear
Answer:
(a) vector

Question 48.
|\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| is equal to –
(a) -|\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)|
(b) |\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)|
(c) -|\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)|
(d) \( \frac{\overline{\mathrm{A}} \times \overline{\mathrm{B}}}{|\overline{\mathrm{A}} \times \overline{\mathrm{B}}|}\)
Answer:
(b) |\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)|

Question 49.
The vector product of two vectors will have maximum magnitude when θ is equal to –
(a) 0°
(b) 90°
(c) 180°
(d) 360°
Answer:
(b) 90°

Question 50.
The vector product of two non-zero vectors will be minimum when O is equal to –
(a) 0°
(b) 180°
(e) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(e) both (a) and (b)

Question 51.
The product of a vector with itself is equal to –
(a) 0
(b) 1
(c) ∞
(d) A²
Answer:
(a) 0

Question 52.
\(\hat{i}\) x \(\hat{i}\) is –
(a) 0
(b) 1
(c) ∞
(d) \(\hat{j}\)
Answer:
(a) 0

Question 53.
\(\hat{i}\) x \(\hat{j}\) is –
(a) \(\hat{i}\)
(b) \(\hat{j}\)
(c) \(\hat{k}\)
(d) \(\overrightarrow{\mathrm{z}}\)
Answer:
(c) \(\hat{k}\)

Question 54.
\(\hat{j}\) x \(\hat{i}\) is –
(a) –\(\hat{i}\)
(b) –\(\hat{j}\)
(c) –\(\hat{k}\)
(d) \(\overrightarrow{\mathrm{z}}\)
Answer:
(c) –\(\hat{k}\)

Question 55.
If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) form adjacent sides of parallelogram, then the magnitude of |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| will give of parallelogram –
(a) length
(b) area
(c) volume
(d) diagonal
Answer:
(b) area

Question 56.
If \(\overrightarrow{\mathrm{P}}\) – \(\overrightarrow{\mathrm{Q}}\) then which of the following is incorrect. –
(a) \(\hat{P}\) = \(\hat{Q}\)
(b) |\(\hat{P}\)| = |\(\hat{Q}\)|
(c) P\(\hat{Q}\) = Q\(\hat{A}\)
(d) \(\hat{P}\) \(\hat{Q}\) = PQ
Answer:
(d) \(\hat{P}\) \(\hat{Q}\) = PQ

Question 57.
The momentum of a particle is \(\overrightarrow{\mathrm{P}}\) = cos θ \(\hat{i}\) + sin θ \(\hat{j}\) . The angle between momentum and the force acting on a body is –
(a) 0°
(b) 45°
(c) 90°
(d) 180°
Answer:
(c) 90°

Question 58.
A and B are two vectors, if A and B are perpendicular to each other then –
(a) \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = 0
(b) \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = l
(c) \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) = 0
(d) \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) = \(\overrightarrow{\mathrm{A}}\)\(\overrightarrow{\mathrm{B}}\)
Answer:
(c) \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) = 0

Question 59.
The angle between two vectors -3\(\hat{i}\) + 6\(\hat{k}\) and 2\(\hat{i}\) + 3\(\hat{j}\) + \(\hat{k}\) is –
(a) 0°
(b) 45°
(c) 60°
(d) 90°
Answer:
(d) 90°

Question 60.
The radius vector is 2\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) while linear momentum is 2\(\hat{i}\) + 3\(\hat{j}\) + \(\hat{k}\) Then the angular momentum is
(a) -2\(\hat{i}\) + 4\(\hat{k}\)
(b) 4\(\hat{i}\) – 8\(\hat{k}\)
(c) 2\(\hat{k}\) – 4\(\hat{j}\) + 2\(\hat{k}\)
(d) 4\(\hat{i}\) – 8\(\hat{j}\)
Answer:
(a) -2\(\hat{i}\) + 4\(\hat{k}\)

Question 61.
Which of the following cannot be a resultant of two vectors of magnitude 3 and 6?
(a) 3
(b) 6
(c) 10
(d) 7
Answer:
(c) 10

Question 62.
Twelve forces each of magnitude 10 N acting on a body at an angle of 30° with other forces then their resultant is-
(a) 10 N
(b)120 N
(c) \(\frac{10}{\sqrt{3}}\)
(d) zero

Question 63.
Two forces are in the ratio of 3 : 4. The maximum and minimum of their resultants are in the ratio is –
(a) 4 : 3
(b) 3 : 4
(c) 7 : 1
(d) 1 : 7
Answer:
(c) 7 : 1

Question 64.
If | \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) | = |\(\overrightarrow{\mathrm{P}}\) | + |\(\overrightarrow{\mathrm{Q}}\)|. The angle between the vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) is –
(a) 0°
(b) 180°
(c) 60°
(d) 90°
Answer:
(a) 0°
|\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) | = |\(\overrightarrow{\mathrm{P}}\) | + |\(\overrightarrow{\mathrm{Q}}\)|
Square on both sides and the resultant becomes
P² + Q² + 2PQ cos θ = P² + Q² + 2PQ cos θ = 1
θ = 0

Question 65.
If |\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = |\(\overrightarrow{\mathrm{P}}\) | — |\(\overrightarrow{\mathrm{P}}\)|, then the angle between the vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\)
(a) 0°
(b) 90°
(c) 180°
(d) 360°
Answer:
(c) |\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\)| = |\(\overrightarrow{\mathrm{P}}\) | |\(\overrightarrow{\mathrm{P}}\)| ‘
Square on both side, and the resultant becomes
P² + Q² + 2PQ cos θ = P² + Q² – 2PQ .
cos θ = -1
θ = 180°

Question 66.
If |\(\overrightarrow{\mathrm{P}}\) x \(\overrightarrow{\mathrm{Q}}\)| = |\(\overrightarrow{\mathrm{P}}\) . \(\overrightarrow{\mathrm{Q}}\)| then angle between \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) then angle between P and Q will be –
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Answer:
(c) |\(\overrightarrow{\mathrm{P}}\) x \(\overrightarrow{\mathrm{Q}}\)| = |\(\overrightarrow{\mathrm{P}}\) . \(\overrightarrow{\mathrm{Q}}\)| Expand the terms
PQ sinθ = PQ cos θ
tan θ = 1
θ = 45°

Question 67.
If | \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) | = |\(\overrightarrow{\mathrm{P}}\) | |\(\overrightarrow{\mathrm{Q}}\)|, then angle between \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) will be –
(a) 0°
(b) 45°
(c) 90°
(d) 180°
Answer:
(c) | \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) | = |\(\overrightarrow{\mathrm{P}}\)| |\(\overrightarrow{\mathrm{Q}}\) |
Square on both side, and the resultants become,
P² + Q² + 2PQ cos 0 = P² + Q² – 2PQ cos θ 4PQ cos θ = 0
θ = 90°

Question 68.
If A and B are the sides of triangle, then area of triangle –
(a) \(\frac{1}{2}|\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}|\)
(b) \(\frac{1}{2}|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|\)
(c) AB sin θ
(d) AB cos θ
Answer:
(b) \(\frac{1}{2}|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|\)

Question 69.
A particle moves in a circular path of radius 2 cm. If a particle completes 3 rounds, then the distance and displacement of the particle are –
(a) 0 and 37.7
(b) 37.7 and 0
(c) 0 and 0
(d) 37.7 and 37.7
Answer:
(b) Radius = 2 cm
Circumference of the circle = 2nr = 4n cm
Distance covered in 3 rounds = 127r cm = 37.7 cm
Initial and final positions are same
∴ Displacement = 0

Question 70.
If rx and r2 are position vectors, then the displacement vector is –
(a) \(\vec{r}_{1} \times \vec{r}_{2}\)
(b) \(\vec{r}_{1} \cdot \vec{r}_{2} \)
(c) \(\vec{r}_{1}+\vec{r}_{2}\)
(d) \(\vec{r}_{2}+\vec{r}_{1} \)
Answer:
(d) \(\vec{r}_{2}+\vec{r}_{1} \)

Question 71.
The ratio of the displacement vector to the corresponding time interval is –
(a) average speed
(b) average velocity
(c) instantaneous speed
(d) instantaneous velocity
Answer:
(b) average velocity

Question 72.
The ratio of total path length travelled by the particle in a time interval –
(a) average speed
(b) average velocity
(c) instantaneous speed
(d) instantaneous velocity
Answer:
(a) average speed

Question 73.
The product of mass and velocity of a particle is –
(a) acceleration
(b) force
(c) torque
(d) momentum
Answer:
(d) momentum

Question 74.
The area under the force, displacement curve is –
(a) potential energy
(b) work done.
(c) impulse
(d) acceleration
Answer:
(b) work done

Question 75.
The area under the force, time graph is –
(a) momentum
(b) force
(c) work done
(d) impulse
Answer:
(d) impulse

Question 76.
The unit of momentum is –
(a) kg ms^-1
(b) kg ms^-2
(c) kg m²s^-1
(d) kg^-1 m² s^-1
Answer:
(b) kg ms^-2

Question 77.
The slope of the position – time graph will give –
(a) displacement
(b) velocity
(c) acceleration
(d) force
Answer:
(d) force

Question 78.
The area under velocity-time graph gives-
(a) positive
(b) negative
(c) either positive (or) negative
(d) zero
Answer:
(c) either positive (or) negative

Question 79.
The magnitude of distance is always-
(a) positive
(b) negative
(c) either positive (or) negative
(d) zero
Answer:
(a) positive

Question 80.
If two objects A and B are moving along a straight line in the same direction with the velocities v_A and v_B respectively, then the relative velocity is-
(a) v_A + v_B
(b) v_A – v_B
(c) v_A v_B
(d) v_A / v_B
Answer:
(b) V_A – V_B

Question 81.
If two objects A and B are moving along a straight line in the opposite direction with the velocities V_A and V_B respectively, then relative velocity is-
(a) V_A + V_B
(b) V_A – V_B
(c) V_{A .} V_B
(d) V_A / V_B
Answer:
(a) V_A + V_B

Question 82.
If two objects moving with a velocities of V_A and V_B at an angle of 0 between them, the relative velocity is –

Answer:

Question 83.
A person moving horizontally with velocity \(\overrightarrow{\mathrm{V}_{m}}\) The relative velocity of rain with respect to the person is –
(a) \(\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{\mathrm{m}}\)
(b) \(\sqrt{\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{m}}\)
(c) \(\mathrm{V}_{\mathrm{R}}-\mathrm{V}_{m}\)
(d) \(\sqrt{\mathrm{v}_{\mathrm{R}}^{2}+\mathrm{V}_{m}^{2}}\)
Answer:
(d) \(\sqrt{\mathrm{v}_{\mathrm{R}}^{2}+\mathrm{V}_{m}^{2}}\)

Question 84.
A person moving horizontally with velocity \(\overrightarrow{\mathrm{V}_{m}}\) . Rain falls vertically with velocity \(\overrightarrow{\mathrm{V}_{R}}\) To save himself from the rain, he should hold an umbrella with vertical at an angle of –
(a) \(\tan ^{-1}\left(\frac{V_{R}}{V_{m}}\right)\)
(b) \(\tan ^{-1}\left(\frac{V_{m}}{V_{R}}\right)\)
(c) \(\tan \theta=\mathrm{V}_{m}+\mathrm{V}_{\mathrm{R}}\)
(d) \(\tan ^{-1}\left(\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{m} / \mathrm{V}_{\mathrm{R}}-\mathrm{V}_{m}\right)\)
Answer:
(b) \(\tan ^{-1}\left(\frac{V_{m}}{V_{R}}\right)\)

Question 85.
A car starting from rest, accelerates at a constant rate x for sometime after which it decelerates at a constant rate v to come to rest. If the total time elapsed is t, the maximum velocity attained by the car is given by –

Answer:

Question 86.
A car covers half of its journey with a speed of 10 ms^-1 and the other half by 20 ms^-1. The average speed of car during the total journey is –
(a) 70 ms^-1
(b) 15 ms^-1
(c) 13.33 ms^-1
(d) 7.5 ms^-1
Answer:
(c) Let x is the total distance
Time to cover 1st half = \(\frac{x / 2}{10}\)
Time to cover 2nd half = \(\frac{x / 2}{20}\)
Average speed =

Question 87.
A swimmer can swim in still water at of 10 ms^-1 While crossing a river his average speed is 6 ms^-1. If he crosses the river in the shortest possible time, what is the speed of flow of water?
(a) 16 ms^-1
(b) 4 ms^-1
(c) 60 ms^-1
(d) 8 ms^-1
Answer:
(d) The resultant velocity of swimmer must be perpendicular to speed of water to cross the river in a shortest time
∴ \(v_{s}^{2}=v^{2}+v_{w}^{2}\)
\(v_{w}^{2}=v_{s}^{2}-v^{2}\) = 100 – 36 = 64
∴ V = 8 m/s^-1

Question 88.
A 100 m long train is traveling from North to South at a speed of 30 ms^-1. A bird is flying from South to North at a speed of 10^-1. How long will the bird take to, cross the train?
(a) 3 s
(b) 2.5 s
(c) 10 s
(d) 5 s
Answer:
(b) Length of train = 100 m
Relative velocity = 30 + 10 = 40 ms^-1
Time taken to cross the train (t) = \(\frac {distance}{ R.velocity }\) = \(\frac { 100 }{ 40 }\) = 2.5 s

Question 89.
The first derivative of position vector with respect to time is –
(a) velocity
(b) acceleration
(c) force
(d) displacement
Answer:
(a) velocity

Question 90.
The second derivative of position vector with respect to time is –
(a) velocity
(b) acceleration
(c) force
(d) displacement
Answer:
(b) acceleration

Question 91.
The slope of displacement-time graph gives –
(a) velocity
(b) acceleration
(c) force
(d) displacement
Answer:
(a) velocity

Question 92.
The slope of velocity-time graph gives –
(a) velocity
(b) acceleration
(c) force
(d) displacement
Answer:
(b) acceleration

Question 93.
The position vector of a particle is \(\vec{r}\) = 4t²\(\hat{i}\) + 2t\(\hat{j}\) + 3t\(\hat{k}\) The acceleration of a particle is having only –
(a) X – component
(b) Y – component
(c) Z – component
(d) X – Y component
Answer:
(a) X – component
4t²\(\hat{i}\) + 2t\(\hat{j}\) + 3t\(\hat{k}\)
\(\vec{v}\) = \(\frac{d \vec{r}}{dt}\) = 8t\(\hat{i}\) + 2\(\hat{j}\)
a = \(\frac{d^{2} r}{d t^{2}}\) = 8\(\hat{i}\) a is having only X-component.

Question 94.
The position vector of a particle is \(\vec{r}\) = 4t²\(\hat{i}\) + 2t\(\hat{j}\) + 3t\(\hat{k}\). The speed of the particle at t = 5 s is –
(a) 42 ms^-1
(b) 3s
(c) 3 ms^-1
(d) 40 ms^-1
Answer:
(a) 42 ms^-1
\(\vec{r}\) = 4t²\(\hat{i}\) + 2t\(\hat{j}\) + 3t\(\hat{k}\)
Speed v = — = \(\frac{d \vec{r}}{dt}\) = 8t\(\hat{i}\) + 2\(\hat{j}\)
at t = 5 s v = 40 + 2 = 42

Question 95.
An object is moving in a straight line with uniform acceleration a, the velocity-time relation is –
(a) u = v + at
(b) v = u + at
(c) v² = u² + a²t²
(d) v² – u² = at
Answer:
(b) v = u + at

Question 96.
An object is moving in a straight line with uniform acceleration, the displacement-time relation is –
(a) S = \(u t^{2}+\frac{1}{2} a t^{2}\)
(b) S = \(u t-\frac{1}{2} a t^{2}\)
(c) S = \(u t+\frac{1}{2} a t^{2}\)
(d) S = \(u t-a t^{2}\)
Answer:
(c) S = \(u t+\frac{1}{2} a t^{2}\)

Question 97.
An object is moving in a straight line with uniform acceleration, the velocity-displacement reflation is –
(a) V = u + 2as
(b) S = ut + -at
(c) V² = u² – 2as
(d) V² = u² + 2as
Answer:
(d) V² = u² + 2as

Question 98.
For free-falling body, its initial velocity is –
(a) 0
(b) 1
(c) ∞
(d) none
Answer:
(a) 0

Question 99.
An object falls from a height h (h<<R), the speed of the object when it reaches the ground is –
(a) \(\frac{1}{2} g t^{2}\)
(b) \(\sqrt{g t}\)
(c) gh
(d) \(\sqrt{2 g h}\)
Answer:
(d) \(\sqrt{2 g h}\)

Question 100.
An object falls from a height h (h<< R) the time taken by an object to reaches the ground is –
(a) \(\frac{1}{2} g t^{2}\)
(b) \(\sqrt{2 g h}\)
(c) \(\sqrt{\frac{2 h}{g}}\)
(d) \(\sqrt{\frac{2 g}{h}}\)
Answer:
(d) \(\sqrt{\frac{2 g}{h}}\)

Question 101.
In the absence of air resistance, horizontal velocity of the projectile is –
(a) always negative
(b) equal to ‘g’
(c) directly proportional to g
(d) a constant
Answer:
(d) a constant

Question 102.
In the horizontal projection, the range of the projectile is –
(a) \(\sqrt{\frac{2 h}{g}}\)
(b) \(u \sqrt{\frac{h}{g}}\)
(c) \(u \sqrt{\frac{2 h}{g}}\)
(d) \(u \sqrt{\frac{g}{2 h}}\)
Answer:
(c) \(u \sqrt{\frac{2 h}{g}}\)

Question 103.
In oblique projection, maximum height attained by the projectile is –
(a) \(\frac { t }{ u cos θ }\)
(b) \(\frac { u cos θ }{ 2g }\)
(c) \(\frac { 2g }{ u cos θ }\)
(d) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
Answer:
(d) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)

Question 104.
In oblique projection time of flight of a projectile is –
(a) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
(b) \(\frac { 2u cos θ }{ g }\)
(c) \(\frac{u^{2} \sin 2 \theta}{g}\)
(d) \(\frac{u^{2}}{g}\)
Answer:
(b) \(\frac { 2u cos θ }{ g }\)

Question 105.
In oblique projection horizontal range of the projectile is –
(a) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
(b) \(\frac { 2u cos θ }{ g }\)
(c) \(\frac{u^{2} \sin 2 \theta}{g}\)
(d) \(\frac{u^{2}}{g}\)
Answer:
(a) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)

Question 106.
In oblique projection, maximum horizontal range of the projectile is –
(a) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
(b) \(\frac { 2u cos θ }{ g }\)
(c) \(\frac{u^{2} \sin 2 \theta}{g}\)
(d) \(\frac{u^{2}}{g}\)
Answer:
(d) \(\frac{u^{2}}{g}\)

Question 107.
One radian is equal to –
(a) \(\frac {π}{ 180 }\) degree
(b) 60°
(c) 57.295°
(d) 53.925°
Answer:
(c) 57.295°

Question 108.
In relation between linear and angular velocity is –
(a) ω = vr
(b) ω = \(\frac {v }{ r }\)
(c) ω = \(\frac { r}{ v }\)
(d) ω = \(\frac { r }{ ω }\)
Answer:
(b) ω = \(\frac {v }{ r }\)

Question 109.
Centripetal acceleration is given by –
(a) \(\frac{v^{2}}{r}\)
(b) \(-\frac{v^{2}}{r}\)
(c) \(\frac{r}{v^{2}}\)
(d) \(-\frac{r}{v^{2}}\)
Answer:
(b) \(-\frac{v^{2}}{r}\)

Question 110.
In uniform circular motion –
(a) Speed changes but velocity constant
(b) Velocity changes but speed constant
(c) both speed and velocity are constant
(d) both speed and velocity are variable
Answer:
(b) Velocity changes but speed constant

Question 111.
In non – uniform circular motion, the resultant acceleration is given by –

Answer:

Question 112.
In non – uniform circular motion, the resultant acceleration makes an angle with the radius vector is –

Answer:

Question 113.
A compartment of an uniformly moving train is suddenly detached from the train and stops after covering some distance. The distance covered by the compartment and distance covered by the train in the given time –
(a) both will be equal
(b) second will be half of first
(c) first will be half of second
(d) none
Answer:
(c) first will be half of second

Question 114.
An object is dropped from rest. Its v – t graph is –

Answer:

Question 115.
When a ball hits the ground as free fall and renounces but less than its original height? Which is represented by –

Answer:

Question 116.
Which of the following graph represents the equation y = mx – C?

Answer:

Question 117.
Which of the following graph represents the equation y = mx + C?

Answer:

Question 118.
Which of the following graph represents the equation y = mx?

Answer:

Question 119.
Which of the following graph represents the equation y = -mx + C?

Answer:

Question 120.
Which of the following graph represents the equation y = kx²?

Answer:

Question 121.
X = -ky² is represented by –

Answer:

Question 122.
X = ky² is represented by –

Answer:

Question 123.
y = kx² is represented by –

Answer:

Question 124.
X °∝\(\frac { 1 }{ Y }\) (or) XY = constant is represented by –

Answer:

Question 125.
y = e^-kx is represented by –

Answer:

Question 126.
Y = 1 – e^-kx is represented by –

Answer:

Question 127.
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 is represented by –

Answer:

Question 128.
Let y =f(x) is a function. Its maxima (or) minima can be obtained by –
(a) y = 0
(b) f(x) = 0
(c) \(\frac {dy}{dx}\) = 0
(d) \(\frac{d^{2} y}{d x^{2}}\) = 0
Answer:
(c) \(\frac {dy}{dx}\) = 0

Question 129.
A particle at rest starts moving in a horizontal straight line with uniform acceleration. The ratio of the distance covered during the fourth and the third second is –
(a) \(\frac {4}{ 3 }\)
(b) \(\frac { 26 }{ 9 }\)
(e) \(\frac { 7}{ 5 }\)
(d) 2
Answer:
(c) The distance travelled during nth second –
S_n  = u + \(\frac { 1 }{ 2 }\) a (2n -1)
Distance travelled during 4th second S₁ = \(\frac { 1 }{ 2 }\) (8 – 1)
Distance travelled during 3rd second S₂ = \(\frac { 1 }{ 2 }\) a(6 – 1)
\(\frac{\mathrm{S}_{1}}{\mathrm{S}_{2}}\) = \(\frac { 7 }{ 5 }\)

Question 130.
The distance travelled by a body, falling freely from rest in t = 1s, t = 2s and t = 3s are in the ratio of –
(a) 1 : 2 : 3
(h) 1 : 3 : 5
(c) 1 : 4 : 9
(d) 9 : 4 : 1
Answer:
(c) The distance travelled by a free falling body S = \(\frac { 1 }{ 2 }\) gt²
∴ S α t²
∴ S₁ : S₂  : S₃ : 1² : 2² : 3² = 1 : 4 : 9.

Question 131.
The displacement of the particle along a straight line at time ¡ is given by X = a + ht + ct² where a, b, c are constants. The acceleration of the particle is-
(a) a
(b) b
(c) c
(d) 2c
Answer:
(d) X = a + bt + ct²
\(\frac { dX }{ dt }\) = v = b + 2ct
Acceleration = \(\frac{d^{2} X}{d t^{2}}\) = 2c.

Question 132.
Two bullets are fired at an angle of θ and (90 – θ) to the horizontal with same speed. The ratio of their times of flight is –
(a) 1 : 1
(b) 1: tan θ
(c) tan θ : 1
(d) tan² θ : 1
Answer:
(c) Time of flight t_f =  \(\frac { 2x sinθ}{ 9 }\)
t_f α sinθ
∴ \(\frac{t_{f_{1}}}{t_{f_{2}}}\) = \(\frac { sinθ }{sin (90 – θ) }\) = \(\frac { sinθ }{cos θ }\) = tanθ
\(t_{f_{1}}: t_{f_{2}}\) = tan θ : 1

Question 133.
A particle moves along a circular path under the action of a force. The work done by the force is –
(a) positive and non zero
(b) zero
(c) negative and non-zero
(d) none
Answer:
(b) zero

Question 134.
For a particle, revolving in a circle with speed, the acceleration of the particle is –
(a) along the tangent
(b) along the radius
(c) along its circumference
(d) zero
Answer:
(b) along the radius

Question 135.
A gun fires two bullets with same velocity at 60° and 30° with horizontal. The bullets strike at the same horizontal distance. The ratio of maximum height for the two bullets is in the ratio of –
(a) 1 : 2
(b) 3 : 1
(c) 2 : 1
(d) 1 : 3
Answer:
(b) 3 : 1
Max height attained h_max = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
∴ h_max α sin² θ i.e h_max α \(\frac { 1-cos2θ}{ 2 }\)
\(\frac{h_{\max 1}}{h_{\max } 2}\) = \(\frac{3 / 2}{1 / 2}\) = 3

Question 136.
A ball is thrown vertically upward. it is a speed of lo m/s. When it has reached one half of its maximum height. I-low high does the ball rise? (g = 10 ms^-2)
(a) 5 m
(b) 7 m
(c) 10 m
(d) 12 m
Answer:
(c) 10 m

Question 137.
A car moves from X to Y with a uniform speed V_n  and returns to Y with a uniform speed V_d The average speed for this round trip is –

Answer:

Question 138.
Two projectiles of same mass and with same velocity are thrown at an angle of 60° and 30° with the horizontal then which of the following will remain same?
(a) time of flight
(b) range of projectile
(c) maximum height reached
(d) all the above
Answer:
(b) range of projectile

Question 139.
A n object of mass 3 kg is at rest. Now a force of \(\overrightarrow{\mathrm{F}}\) = 6 t²\(\hat{i}\) + 4t\(\hat{j}\) is applied on the object, then the velocity of object at t = 3 second is –
(a) 18\(\hat{i}\) + 3 \(\hat{j}\)
(b) 18\(\hat{i}\) + 6\(\hat{j}\)
(c) 3 \(\hat{i}\) + 18\(\hat{j}\)
(d) 18 \(\hat{i}\) + 4\(\hat{j}\)
Answer:
(b) F = 6 t²\(\hat{i}\) + 4t\(\hat{j}\)

Question 140.
The angle for which maximum height and horizontal range are same for a projectile is –
(a) 32°
(b) 48°
(c) 76°
(d) 84°
Answer:
(c) 76°
H_max = horizontal range
\(\frac{u^{2} \sin ^{2} \theta}{2 g}\) = \(\frac{u^{2} \sin 2 \theta}{g}\)
\(\frac{\sin ^{2} \theta}{2}\) = 2 sin θ cos θ = sin θ = 4 cos θ tan θ = 4 θ = 76°

Question 141.
A bullet is dropped from some height, when another bullet is fired horizontally from the same height. They will hit the ground –

(a) depends upon mass of bullet
(b) depends upon the observer
(c) one after another
(d) simultaneously
Answer:
(d) simultaneously

Question 142.
From this velocity – time graph, which of the following is correct?
(a) Constant acceleration
(b) Variable acceleration
(c) Constant velocity
(d) Variable acceleration
Answer:
(b) Variable acceleration

Question 143.
When a projectile is at its maximum height, the direction of its velocity and acceleration are –
(a) parallel to each other
(b) perpendicular to each other
(c) anti – parallel to each other
(d) depends on its speed
Answer:
(b) perpendicular to each other

Question 144.
At the highest point of oblique projection, which of the following is correct?
(a) velocity of the projectile is zero
(b) acceleration of the projectile is zero
(c) acceleration of the projectile is vertically downwards
(d) velocity of the projectile is vertically downwards
Answer:
(c) acceleration of the projectile Is vertically downwards

Question 145.
The range of the projectile depends –
(a) The angle of projection
(b) Velocity of projection
(c) g
(d) all the above
Answer:
(d) all the above

Question 146.
A constant force is acting on a particle and also acting perpendicular to the velocity of the particle. The particle describes the motion in a plane. Then –
(a) angular displacement is zero
(b) its velocity is zero
(c) it velocity is constant
(d) it moves in a circular path
Answer:
(d) it moves in a circular path

Question 147.
If a body moving in a circular path with uniform speed, then –
(a) the acceleration is directed towards its center
(b) velocity and acceleration are perpendicular to each other
(c) speed of the body is constant but its velocity is varying
(d) all the above
Answer:
(d) all the above

Question 148.
A body is projected vertically upward with the velocity y = 3\(\hat{i}\) + 4\(\hat{j}\) ms-¹. The maximum height attained by the body is (g 10 ms^-2).
(a) 7 m
(b) 1.25 m
(c) 8 m
(d) 0.08 m
Answer:
(b) v = 3\(\hat{i}\) + 4\(\hat{j}\)
H_max = \(\frac{v^{2} \sin ^{2} \theta}{2 g}\) = \(\frac{v^{2}}{2 g}\) [ θ = 90 ]
v = \(\sqrt{9+16}\)  = \(\sqrt{25}\)
v² = 25
H_max = \(\frac{25}{20}\)  = \(\frac { 5 }{ 4 }\) = 1.25 m

Samacheer Kalvi 11th Physics Kinematics Short Answer Questions – I (1 Mark)

Question 1.
What is frame of reference?
Answer:
In a coordinate system, the position of an object is described relative to it, then such a coordinate system is called as frame of reference.

Question 2.
What are the types of motion?
Answer:

• Linear motion
• Circular motion
• Rotational motion
• Vibratory motion.

Question 3.
What is linear motion? Give example.
Answer:
An object is said to be in linear motion if it moves in a straight line.
Example – an athlete running on a straight track.

Question 4.
What is circular motion? Give example.
Answer:
Circular motion is defined as a motion described by an object traversing a circular path.
Example – The whirling motion of a stone attached to a string.

Question 5.
What is rotational motion? Give example.
Answer:
During a motion every point in the object traverses a circular path about an axis except the points located on the axis, is called as rotational motion.
Example – Spinning of the earth about its own axis.

Question 6.
What is vibratory motion? Give example.
Answer:
If an object or particle executes a to and fro motion about a fixed point, it is said to be in vibratory motion.
Example – Vibration of a string on a guitar.

Question 7.
What is one dimensional motion? Give example.
Answer:
One dimensional motion is the motion of a particle moving along a straight line.
Example –  Motion of a train along a straight railway track.

Question 8.
What is two dimensional motion? Give example.
Answer:
If a particle moving along a curved path in a plane, then it is said to be in two dimensional motion.
Example – Motion of a coin on a carrom board.

Question 9.
What is three dimensional motion? Give example.
Answer:
If a particle moving in used three dimensional space, then the particle is said to be in three dimensional motion.
E.g. A bird flying in the sky.

Question 10.
Write about the properties of components of vectors.
Answer:
If two vectors \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{B}}\) are equal, then their individual components are also equal. then their individual components are also equal.
Let\(\overline{\mathrm{A}}\) = \(\overline{\mathrm{B}}\)
then A_x  \(\hat{i}\) + A_y \(\hat{j}\) + A_z \(\hat{k}\) = B_x \(\hat{i}\) + B_y \(\hat{j}\) + B_z \(\hat{k}\)
i.e. A_x = B_x, A_y =  B_y  = A_z = B_z

Question 11.
Give an example for scalar product of two vectors.
Answer:
The work done by a force \(\overrightarrow{\mathrm{F}}\) to move an object through a small displacement \(\overrightarrow{\mathrm{dr}}\) then
Work done W = \(\overrightarrow{\mathrm{F}}\) .\(\overrightarrow{\mathrm{dr}}\) (or) W = F dr cos θ

Question 12.
Give any three example for vector product of two vectors.
Answer:

1. Torque \(\overrightarrow{\mathrm{t}}\) = \(\overrightarrow{\mathrm{r}}\) x \(\overrightarrow{\mathrm{F}}\). Where i is force and \(\overrightarrow{\mathrm{F}}\) is force and \(\overrightarrow{\mathrm{r}}\) position vector of a particle.
2. Angular momentum \(\overrightarrow{\mathrm{L}}\) = \(\overrightarrow{\mathrm{r}}\) x \(\overrightarrow{\mathrm{P}}\) where \(\overrightarrow{\mathrm{P}}\) is the linear momentum.
3. Linear velocity \(\overrightarrow{\mathrm{v}}\) = \(\overrightarrow{\mathrm{ω}}\) x \(\overrightarrow{\mathrm{r}}\) where \(\overrightarrow{\mathrm{ω}}\) is angular velocity.

Question 13.
What is position vector?
Answer:
It is vector which denotes the position of a particle at any instant of time, with respect to some reference frame or coordinate system.
The position \(\overrightarrow{\mathrm{r}}\) vector of the particle at a point P is given by
\(\overrightarrow{\mathrm{r}}\) = x\(\hat{i}\) + y\(\hat{j}\) + z\(\hat{k}\)
where x, y and z are components of \(\overrightarrow{\mathrm{r}}\).

Question 14.
Write a note an momentum.
Answer:
Momentum of a particle is defined as product of mass with velocity. It is denoted as \(\overrightarrow{\mathrm{p}}\) Momentum is also a vector quantity
\(\overrightarrow{\mathrm{r}}\) = m\(\overrightarrow{\mathrm{v}}\)
The direction of momentum is also in the direction of velocity, and the magnitude of momentum is equal to product of mass and speed of the particle.
p = mv
In component form the momentum can be written as
p_x\(\hat{i}\) + p_y\(\hat{j}\)+ p_z\(\hat{k}\) = mv_x\(\hat{i}\) + mv_y\(\hat{j}\) + mv_z\(\hat{k}\)
Here,
p_x = x component of momentum and is equal to mv_x
P_x = y component of momentum and is equal to mv_y
P_x = z component of momentum and is equal to mv_z

Question 15.
“Displacement vector is basically a position vector”. Comment on it.
Answer:
This statement is almost correct only. Because the displacement vector also gives the position of a point just like a position vector. The difference between these two vectors is p. The displacement vector gives the position of a point with respect to a point other than origin but position vector gives the position of a point with respect to origin.

Question 16.
Will two dimensional motion with an acceleration only in one dimension?
Answer:
Yes. In oblique projection, the acceleration is acting vertically downward but the object follows a parabolic path.

Question 17.
A foot ball is kicked by a player with certain angle to the horizontal. Is there any point at which velocity is perpendicular to its acceleration.
Answer:
Yes. At its maximum height in the parabolic path vertical velocity is zero but due to horizontal component, velocity acts along horizontally, but acceleration of the

Question 18.
Give any two examples for parallelogram law of vectors.
Answer:

• the flight of a bird
• working of a sling.

Question 19.
Why does rubber ball bounce greater heights on hills than in plains?
Answer:
The maximum height attained by the projectile is inversely proportional to acceleration due to gravity. At greater height, acceleration due to gravity will be lesser than plains. So ball can bounce higher in hills than in plains.

Question 20.
Is it possible for body to have variable velocity but constant speed? Give example.
Answer:
Yes, it is possible. In horizontal circular motion the speed of a particle is always constant but due to the variation in direction continuously, the velocity of a particle varies.

Question 21.
What is relative velocity?
Answer:
When two objects are moving with different velocities, then the velocity of one object with respect to another object is called relative velocity of an object with respect to another.

Question 22.
What is average acceleration?
Answer:
The average acceleration is defined as the ratio of change in velocity over the time interval
a_avg = \(\frac{\Delta \overrightarrow{\mathrm{v}}}{\Delta t}\) It is a vector quantity.

Question 23.
Write a note an instantaneous acceleration.
Answer:
Instantaneous acceleration or acceleration of a particle at time ‘t’ is given by the ratio of change in velocity over ∆t, as ∆t approaches zero.
Acceleration
In other words, the acceleration of the particle at an instant t is equal to rate of change of velocity

(1) Acceleration is a vector quantity. Its SI unit is ms^-2 and its dimensional formula is [M°L¹ T^-2]

(2) Acceleration is positive if its velocity is increasing, and is negative if the velocity is decreasing. The negative acceleration is called retardation or deceleration.

Question 24.
Write an acceleration in terms of its component?
(Or)
Show that the acceleration is the second derivative of position vector with respect to time.
Answer:
in terms of components, we can write,


are the components of instantaneous acceleration. Since each component of velocity is the derivative of the corresponding coordinate, we can express the components a_x, a_y, and a_z as

Then the acceleration vector \(\overrightarrow{\mathrm{a}}\) it self is

Thus acceleration is the second derivative of position vector with respect to time.

Question 25.
What are the examples of projectile motion?
Answer:

1. An object dropped from window of a moving train.
2. A bullet fired from a rifle.
3. A ball thrown in any direction.
4. A javelin or shot put thrown by an athlete.
5. A jet of water issuing from a hole near the bottom of a water tank.

Question 26.
Explain projectile motion.
Answer:
A projectile moves under the combined effect of two velocities.

• A uniform velocity in the horizontal direction, which will not change provided there is no air resistance.
• A uniformly changing velocity (i.e., increasing or decreasing) in the vertical direction.

There are two types of projectile motion:

• Projectile given an initial velocity in the horizontal direction (horizontal projection)
• Projectile given an initial velocity at an angle to the horizontal (angular projection)

To study the motion of a projectile, let us assume that,

• Air resistance is neglected.
• The effect due to rotation of Earth and curvature of Earth is negligible.
• The acceleration due to gravity is constant in magnitude and direction at all points of the motion of the projectile.

Question 27.
What is time of flight?
Answer:
The time taken for the projectile to complete its trajectory or time taken by the projectile to hit the ground is called time of flight.

Question 28.
Under what condition is the average velocity equal the instantaneous velocity?
Answer:
When the body is moving with uniform velocity.

Question 29.
Draw Position time graph of two objects, A & B moving along a straight line, when their relative velocity is zero.

Question 30.
Suggest a situation in which an object is accelerated and have constant speed.
Answer:
Uniform Circular Motion.

Question 31.
Two balls of different masses are thrown vertically upward with same initial velocity. Maximum heights attained by them are h₁ and h₂ respectively what is h₁/h₂?
Answer:
Same height,
∴  h₁/h₂ = 1

Question 32.
A car moving with velocity of 50 kmh^-1 on a straight road is ahead of a jeep moving with velocity 75 kmh^-1 would the relative velocity be altered if jeep is ahead of car?
Answer:
No change.

Question 33.
Which of the two – linear velocity or the linear acceleration gives the direction of motion of a body?
Answer:
Linear velocity.

Question 34.
Will the displacement of a particle change on changing the position of origin of the coordinate system?
Answer:
Will not change.

Question 35.
If the instantaneous velocity of a particle is zero, will its instantaneous acceleration be necessarily zero?
Answer:
No. (highest point of vertical upward motion under gravity).

Question 36.
A projectile is fired with Kinetic energy 1 KJ. If the range is maximum, what is its Kinetic energy, at the highest point?
Answer:
Here \(\frac { 1 }{ 2 }\) mv² =1kJ=1000 J, θ = 45°
At the highest point, K.E. = \(\frac { 1 }{ 2 }\) m(v cos 0)² = \(\frac { 1 }{ 2 }\)\(\frac{m v^{2}}{2}\) = \(\frac {1000}{ 2 }\) = 500 J.

Question 37.
Write an example of zero vector.
Answer:
The velocity vectors of a stationary object is a zero vectors.

Question 38.
State the essential condition for the addition of vectors.
Answer:
They must represent the physical quantities of same native.

Question 39.
When is the magnitude of (\(\overline{\mathrm{A}}\) + \(\overline{\mathrm{B}}\)) equal to the magnitude of (\(\overline{\mathrm{A}}\) – \(\overline{\mathrm{B}}\))?
Answer:
When \(\overline{\mathrm{A}}\) is perpendicular to \(\overline{\mathrm{B}}\).

Question 40.
What is the maximum number of component into which a vector can be resolved?
Answer:
Infinite.

Question 41.
A body projected horizontally moves with the same horizontal velocity although it moves under gravity Why?
Answer:
Because horizontal component of gravity is zero along horizontal direction.

Question 42.
What is the angle between velocity and acceleration at the highest point of a projectile motion?
Answer:
90°.

Question 43.
When does

• height attained by a projectile maximum?
• horizontal range is maximum?

Answer:

• Height is maximum at θ = 90
• Range is maximum at θ = 45.

Question 44.
What is the angle between velocity vector and acceleration vector in uniform circular motion?
Answer:
90°.

Question 45.
A particle is in clockwise uniform circular motion the direction of its acceleration is radially inward. If sense of rotation or particle is anticlockwise then what is the direction of its acceleration?
Answer:
Radial in ward.

Question 46.
A train is moving on a straight track with acceleration a. A passenger drops a stone. What is the acceleration of stone with respect to passenger?
Answer:
\(\sqrt{a^{2}+g^{2}}\) where g = acceleration due to gravity.

Question 47.
What is the average value of acceleration vector in uniform circular motion over one cycle?
Answer:
Null vector.

Question 48.
Does a vector quantity depends upon frame of reference chosen?
Answer:
No.

Question 49.
What is the angular velocity of the hour hand of a clock?
Answer:
ω = \(\frac {2π}{ 12 }\) = \(\frac { π }{6}\) rad h^-1

Question 50.
What is the source of centripetal acceleration for earth to go round the sun?
Answer:
Gravitation force of sun.

Question 51.
What is the angle between (\(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{A}}\) ) and (\(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{A}}\)) ?
Answer:
90°

Samacheer Kalvi 11th Physics Kinematics Short Answer Questions – II (2 Marks)

Question 1.
What are positive and negative acceleration in straight line motion?
Solution:
If speed of an object increases with time, its acceleration is positive. (Acceleration is in the direction of motion) and if speed of an object decreases with time its acceleration is negative (Acceleration is opposite to the direction of motion).

Question 2.
Can a body have zero velocity and still be accelerating? If yes gives any situation.
Solution:
Yes, at the highest point of vertical upward motion under gravity.

Question 3.
The displacement of a body is proportional to t³, where t is time elapsed. What is the nature of acceleration –  time graph of the body?
Solution:
As a α t³ ⇒ s = kt³
Velocity, V = \(\frac { ds }{ dt }\) = 3 kt³
Acceleration, a = \(\frac { dv }{ dt }\) = 3 kt³
i.e., a α t
⇒ motion is uniform, acceleration motion, a – t graph is straight-line.

Question 4.
Suggest a suitable physical situation for the following graph.

Solution:
A ball thrown up with some initial velocity rebounding from the floor with reduced speed after each hit.

Question 5.
An object is in uniform motion along a straight line, what will be position time graph for the motion of object, if (i) x₀ = positive, v = negative is constant.
(i) x₀ = positive, v = negative is |\(\vec{v}\) | constant.
(ii) both x₀ and v are negative |\(\vec{v}\) | is constant.
(iii) x₀ = negative, v = positive |\(\vec{v}\) | is constant.
(iv) both x₀ and v are positive |\(\vec{v}\) | is constant where x₀ is position at t = 0.
Solution:

Question 6.
A cyclist starts from centre O of a circular park of radius 1 km and moves along the path OPRQO as shown. If he maintains constant speed of 10 ms^-1. What is his acceleration at point R in magnitude & direction?

Solution:
Centripetal acceleration, a_c = \(\frac{v^{2}}{r}\) = \(\frac{10^{2}}{1000}\) = 0.1 m/s² along RO.

Question 7.
What will be the effect on horizontal range of a projectile when its initial velocity is doubled keeping angle of projection same?
Solution:
\(\frac{u^{2} \sin 2 \theta}{g}\) ⇒ R α u²
Range comes four times.

Question 8.
The greatest height to which a man can throw a stone is h. What will be the greatest distance upto which he can throw the stone?
Solution:
Maximum height:
H = \(\frac{u^{2} \sin ^{2} \theta}{g}\) ⇒ H_max = \(\frac{u^{2}}{2g}\) = h (at θ = 90)
Maximum range R_max = \(\frac{u^{2}}{g}\) = 2h

Question 9.
A person sitting in a train moving at constant velocity throws a ball vertically upwards. How will the ball appear to move to an observer.

• Sitting inside the train
• Standing outside the train

Solution:

• Vertical straight line motion
• Parabolic path.

Question 10.
A gunman always keep his gun slightly tilted above the line of sight while shooting. Why?
Solution:
Because bullet follow Parabolic trajectory under constant downward acceleration.

Question 11.
Is the acceleration of a particle in circular motion not always towards the center. Explain.
Solution:
No acceleration is towards the center only in case of uniform circular motion.

Samacheer Kalvi 11th Physics Kinematics Short Answer Questions – III (3 Marks)

Question 1.
Draw
(a) acceleration – time
(b) velocity – time
(c) Position – time graphs representing motion of an object under free fall. Neglect air resistance.
Solution:
The object falls with uniform acceleration equal to ‘g’

Question 2.
The velocity time graph for a particle is shown in figure. Draw acceleration time graph from it.

Solution:

Question 3.
For an object projected upward with a velocity v₀, which comes back to the same point after some time, draw
(i) Acceleration – time graph
(ii) Position – time graph
(iii) Velocity time graph
Solution:

Question 4.
The acceleration of a particle in ms² is given by a = 3t² + 2t + 2, where time t is in second. If the particle starts with a velocity v = 2 ms^-1 at t = 0, then find the velocity at the end of 2s.
Solution:

Question 5.
At what angle do the two forces (P + Q) and (P – Q) act so that the resultant is \(\sqrt{3 P^{2}+Q^{2}}\)?
Solution:
Use
R = \(\sqrt{3 P^{2}+Q^{2}}\)
R = \(\sqrt{3 \mathrm{P}^{2}+\mathrm{Q}^{2}}\)
A = P + Q
B = P – Q
solve, θ = 60°

Question 6.
A car moving along a straight highway with speed of 126 km h 1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?
Solution:
Initial velocity of car,
u = 126 kmh^-1 = 126 x \(\frac {5}{18}\) ms^-1 = 35 ms^-1 ………(i)
Since, the car finally comes to rest, v = 0
Distance covered, s = 200 m, a = ?, t = ?
v² = u² – 2as
or a = \(\frac{v^{2}-u^{2}}{2 s}\) ………..(ii)
substituting the values from eq. (i) in eq . (ii), we get
a = \(\frac{0-(35)^{2}}{2 \times 200}\) = \(\frac{0-(35)^{2}}{2 \times 200}\)
= \(-\frac{46}{16}\) ms^-2 = -3.06 ms^-2
Negative sign shows that acceleration in negative which is called retardation, i.e., car is uniformly retarded at – a = 3.06 ms^-2.
To find t, let us use the relation
v = u + at
t = \(\frac {v-u}{ a }\)
use a = -3.06 ms^-2, v = 0, u = 35 ms^-1
∴ t = \(\frac {v-u}{ a }\) = \(\frac {0-35}{ -3.06 }\) = 11.44 s
∴ t = 11.44 sec

Samacheer Kalvi 11th Physics Kinematics Long Answer Questions
Question 1.
Explain the types of motion with example.
Answer:
(a) Linear motion:
An object is said to be in linear motion if it moves in a straight line.
Examples:

• An athlete running on a straight track
• AA particle falling vertically downwards to the Earth.

(b) Circular motion:
Circular motion is defined as a motion described by an object traversing a circular path.
Examples:

• The whirling motion of a stone attached to a string.
• The motion of a satellite around the Earth.
• These two circular motions are shown in figure.

(c) Rotational motion:
If any object moves in a rotational motion about an axis, the motion is called ‘rotation’. During rotation every point in the object transverses a circular path about an axis, (except the points located on the axis).
Examples:

• Rotation of a disc about an axis through its center
• Spinning of the Earth about its own axis.
• These two rotational motions are shown in figure

(d) Vibratory motion:
If an object or particle executes a to-and-fro motion about a fixed point, it is said to be in vibratory motion. This is sometimes also called oscillatory motion.
Examples:

• Vibration of a string on a guitar
• Movement of a swing
• These motions are shown in figure

Other types of motion like elliptical motion and helical motion are also possible.

Question 2.
What are the different types of vectors? ,
Answer:
1. Equal vectors:
Two vectors A and B are said to be equal when they have equal magnitude and same direction and represent the same physical quantity

(a) Collinear vectors:
Col-linear vectors are those which act along the same line. The angle between them can be 0° or 180°.

(i) Parallel vectors:
If two vectors A and B act in the same direction along the same line or on parallel lines, then the angle between them is 0°. Geometrical representation of parallel vectors.

(ii) Anti-parallel vectors:
Two vectors A and B are said to be anti – parallel when they are in opposite directions along the same line or on parallel lines. Then the angle between them is 180°.

2. Unit vector:
A vector divided by its magnitude is a unit vector. The unit vector for \(\overrightarrow{\mathrm{A}}\) is denoted by \(\widehat{A}\) . It has a magnitude equal to unity or one.
Since, \(\widehat{A}\) = \(\frac{\bar{A}}{A}\) we can write \(\overrightarrow{\mathrm{A}}\) = A\(\widehat{A}\)
Thus, we can say that the unit vector specifies only the direction of the vector quantity.

3. Orthogonal unit vectors:
Let \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) be three unit vectors which specify the directions along positive x-axis, positive y-axis and positive z-axis respectively. These three unit vectors are directed perpendicular to each other, the angle between any two of them is 90°.\(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) and are examples of orthogonal vectors. Two vectors which are perpendicular to each other are called orthogonal vectors as shown in the figure.

Question 3.
Explain the concept of relative velocity in one and two dimensional motion.
Answer:
When two objects A and B are moving with different velocities, then the velocity of one object A with respect to another object B is called relative velocity of object A with respect to B.

Case I:
Consider two objects A and B moving with uniform velocities V_A and V_B, as shown, along straight tracks in the same direction \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\), \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) with respect to ground.
The relative velocity of object A with respect to object B is \(\overrightarrow{\mathrm{V}}_{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\).
The relative velocity of object B with respect to object A is \(\overrightarrow{\mathrm{V}}_{\mathrm{BA}}\) = \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) Thus, if two objects are moving in the same direction, the magnitude of relative velocity of one object with respect to another is equal to the difference in magnitude of two velocities.

Case II.
Consider two objects A and B moving with uniform velocities \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) along the same straight tracks but opposite in direction.

The relative velocity of an object A with respect to object B is –
\(\overrightarrow{\mathrm{V}}_{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) – (-\(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\)) = \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) + \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\)

The relative velocity of an object B with respect to object A is
\(\overrightarrow{\mathrm{V}}_{\mathrm{AB}}\) = –\(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) = – (\(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) + \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\))
Thus, if two objects are moving in opposite directions, the magnitude of relative velocity of one object with respect to other is equal to the sum of magnitude of their velocities.

Case III.
Consider the velocities \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) at an angle θ between their directions. The relative velocity of A with respect to B, \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\)
Then, the magnitude and direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) is given by \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = \(\sqrt{\vec{v}_{\mathrm{A}}^{2}+\vec{v}_{\mathrm{B}}^{2}-2 v_{\mathrm{A}} v_{\mathrm{B}} \cos \theta}\) and tan β = \(\frac{v_{\mathrm{B}} \sin \theta}{v_{\mathrm{A}}-v_{\mathrm{B}} \cos \theta}\) (Here β is angle between (\(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) and \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\))
\(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) cos θ .

(i) When θ = 0, the bodies move along parallel straight lines in the same direction, We have \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = (\(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\)) in the direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) . Obviously \(\overrightarrow{\mathrm{v}}_{\mathrm{BA}}\) = (\(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) + \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\)) in the direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\).

(ii) When θ = 180°, the bodies move along parallel straight lines in opposite directions,
We have \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = (\(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\)+ \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\)) in the direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) . Similarly, vBA = (vB + vA) in the direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) .

(iii) If the two bodies are moving at right angles to each other, then 0 = 90°. The magnitude of the relative velocity of A with respect to B = \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = \(\sqrt{v_{\mathrm{A}}^{2}+v_{\mathrm{B}}^{2}}\).

(iv) Consider a person moving horizontally with velocity \(\overrightarrow{\mathrm{V}}_{\mathrm{M}}\) . Let rain fall vertically with velocity \(\overrightarrow{\mathrm{V}}_{\mathrm{R}}\) . An umbrella is held to avoid the rain. Then the relative velocity of the rain with respect to the person is,

which has magnitude
\(\overrightarrow{\mathrm{V}}_{\mathrm{RM}}\) = \(\overrightarrow{\mathrm{V}}_{\mathrm{R}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{M}}\)
And direction 0 = tan^-1\(\left(\frac{V_{\mathrm{M}}}{\mathrm{V}_{\mathrm{R}}}\right)\) with the vertical as shown in figure.

Question 4.
Shows that the path of horizontal projectile is a parabola and derive an expression for
1. Time of flight
2. Horizontal range
3. resultant relative and any instant
4. speed of the projectile when it hits the ground?
Answer:
Consider a projectile, say a ball, thrown horizontally with an initial velocity \(\vec{u}\) from the top of a tower of height h. As the ball moves, it covers a horizontal distance due to its uniform horizontal velocity u, and a vertical downward distance because of constant acceleration due to gravity g. Thus, under the combined effect the ball moves along the path OPA. The motion is in a 2 – dimensional plane. Let the ball take time t to reach the ground at point A, Then the horizontal distance travelled by the ball is x(t) = x, and the vertical distance travelled is y(t) = y.

We can apply the kinematic equations along the x direction and y direction separately. Since this is two-dimensional motion, the velocity will have both horizontal component u_x and vertical component u_y.

Motion along horizontal direction:
The particle has zero acceleration along x direction. So, the initial velocity ux remains constant throughout the motion. The distance traveled by the projectile at a time t is given by the equation x = u_xt +\(\frac { 1 }{ 2 }\) at² . Since a = 0 along x direction, we have x = u_xt ……….(1)

Motion along downward direction:
Here u_y = 0 (initial velocity has no downward component), a = g (we choose the + ve y – axis in downward direction), and distance y at time t.
From equation, y = u_xt +\(\frac { 1 }{ 2 }\) at² we get
y = \(\frac { 1 }{ 2 }\) at² …………..(2)

Substituting the value oft from equation (i) in equation (ii) we have

y = Kx²
where K = \(\frac{g}{2 u_{x}^{2}}\) is constant
Equation (iii) is the equation of a parabola. Thus, the path followed by the projectile is a parabola.

1. Time of Flight:
The time taken for the projectile to complete its trajectory or time taken by the projectile to hit the ground is called time of flight. Consider the example of a tower and projectile. Let h be the height of a tower. Let T be the time taken by the projectile to hit the ground, after being thrown horizontally from the tower.

We know that s_y = u_yt + \(\frac { 1 }{ 2 }\) at² for vertical motion. Here .s_y = h, t = T, u_y = 0 (i.e., no initial vertical velocity). Then h = \(\frac { 1 }{ 2 }\) gt² or T = \(\sqrt{\frac{2 h}{g}}\) Thus, the time of flight for projectile motion depends on the height of the tower, but is independent of the horizontal velocity of projection. If one ball falls vertically and another ball is projected horizontally with some velocity, both the balls will reach the bottom at the same time. This is illustrated in the Figure

2. Horizontal range:
The horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground is called horizontal range.
For horizontal motion, we have
s_x = u_xt + \(\frac { 1 }{ 2 }\) at²
Here,s_x = R (range), u_x = u, a = 0 (no horizontal acceleration) T is time of flight. Then horizontal range = uT
Since the time of flight T = \(\sqrt{\frac{2 h}{g}}\) we substitute this and we get the horizontal range of the particle as R = u \(\sqrt{\frac{2 h}{g}}\)
The above equation implies that the range R is directly proportional to the initial velocity u and inversely proportional to acceleration due to gravity g.

3. Resultant Velocity (Velocity of projectile at any time):
At any instant t, the projectile has velocity components along both x-axis and y-axis. The resultant of these two components gives the velocity of the projectile at that instant t, as shown in figure. The velocity component at any t along horizontal (x-axis)
is V_x = U_x + a_xt
Since, u_x = u, ax = 0 , we get
u_x = u a_x = 0 we get
v_x = u
The component of velocity along vertical direction (y – axis) is v_y = u_y + a_yt
Since, u_y= 0, a_y = g, we get
V_y = gt
Hence the velocity of the particle at any instant is –
v = u\(\hat{i}\) + g\(\hat{j}\)
The speed of the particle at any instant t is given by
v = \(\sqrt{v_{x}^{2}+v_{y}^{2}}\)
v= \(\sqrt{u^{2}+g^{2} t^{2}}\)

4. Speed of the projectile when it hits the ground:
When the projectile hits the ground after initially thrown horizontally from the top of tower of height h, the time of flight is –
t = \(\sqrt{\frac{2 h}{g}}\)
The horizontal component velocity of the projectile remains the same i.e v_x = u.
The vertical component velocity of the projectile at time T is
v = gT = g \(\sqrt{\frac{2 h}{g}}\) = \(\sqrt{\frac{2gh}\)
The speed of the particle when it reaches the ground is
v = \(\sqrt{u^{2}+2 g h}\).

Question 5.
Derive the relation between Tangential acceleration and angular acceleration.
Answer:
Consider an object moving along a circle of radius r. In a time ∆t, the object travels in an arc distance As as shown in figure. The corresponding angle subtended is ∆θ
The ∆s can be written in terms of ∆θ
∆s = r∆θ ………(i)
in a time ∆t, we have
\(\frac { ∆s }{ ∆t}\) = t \(\frac { ∆θ }{ ∆t}\) …………(ii)
¡n the limit ∆t – 0, the above equation becomes
\(\frac { ds }{ dt}\) = rω …………….(iii)
Here \(\frac { ds }{ dt}\) is linear speed (y) which is tangential to the circle and co is angular speed.
So equation (iii) becomes.
v _r = rω …….(iv)

which gives the relation between linear speed and angular speed.
Eq. (iv) is true only for circular motion. In general the relation between linear and angular velocity is given by
\(\vec{v}\) = \(\vec{\omega} \times \vec{r}\) ………..(v)
For circular motion eq. (y) reduces to eq. (iv) since \(\overrightarrow{\mathrm{ω}}\) and \(\overrightarrow{\mathrm{r}}\)are perpendicular to each other.
Differentiating the eq. (iv) with respect to time, we get (since r is constant)
\(\frac { dv }{ dt}\) = \(\frac { rdv }{ dt}\) = rα
Here \(\frac { dv }{ dt}\) Is the tangential acceleration and is denoted as a_t = \(\frac {dω}{ dt}\)is the angular acceleration
α. Then eq. (v) becomes
a_t = rα ………..(vii)

Samacheer Kalvi 11th Physics Kinematics Numerical Questions

Question 1.
The V – t graphs of two objects make angle 30° and 60° with the time axis. Find the ratio of their accelerations.
Solution:
\(\frac{a_{1}}{a_{2}}\) = \(\frac{tan 30}{tan 60}\) = \(\frac{1 / \sqrt{3}}{\sqrt{3}}\) = \(\frac{1}{3}\) = 1 : 3.

Question 2.
When the angle between two vectors of equal magnitudes is 2n/?>, prove that the magnitude of the resultant is equal to either.
Solution:
R = \(\left(\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cos \theta\right)^{1 / 2}\) = \(\left(p^{2}+p^{2}+2 p \cdot p \cos \frac{2 \pi}{3}\right)\) = \(\left[2 p^{2}+2 p^{2}\left(\frac{-1}{2}\right)\right]^{1 / 2}\) = p.

Question 3.
If \(\overline{\mathrm{A}}\) = 3\(\hat{i}\) + 4\(\hat{j}\) and \(\overline{\mathrm{B}}\) = 7\(\hat{i}\) + 24 \(\hat{j}\), find a vector having the same magnitude as \(\overline{\mathrm{B}}\) and parallel to \(\overline{\mathrm{A}}\).
Solution:
\(|\overrightarrow{\mathrm{A}}|=\sqrt{3^{2}+4^{2}}=5\)
also \(|\overrightarrow{\mathrm{B}}|=\sqrt{7^{2}+24^{2}}=25\)
desired vector = \(|\overrightarrow{\mathrm{B}}|\) \(\widehat{A}\) = 25 x \(\frac{3 \hat{i}+4 \hat{j}}{5}\) = 5(3\(\hat{j}\) + 4\(\hat{j}\)) = 15 \(\hat{i}\) + 20\(\hat{j}\).

Question 4.
What is the vector sum of n coplanar forces, each of magnitude F, if each force makes an angle of \(\frac {2 π}{ n }\) with the preceding force?
Solution:
Resultant force is zero.

Question 5.
A car is moving along X- axis. As shown in figure it moves from O to P in 18 seconds and return from P to Q in 6 second. What are the average velocity and average speed of the car in going from

• O to P
• From O to P and hack to Q

Solution:

• O to P, Average velocity =20 ms^-1
• O to P and back to Q

Average velocity = 10 ms^-1
Average speed = 20 ms^-1

Question 6.
On a 60 km straight road, a bus travels the first 30 km with a uniform speed of 30 kmh^-1 . How fast must the bus travel the next 30 km so as to have average speed of 40 kmh^-1 for the entire trip?
Solution:

Question 7.
The displacement x of a particle varies with time as x = 4t² – 15t + 25. Find the position, velocity and acceleration of the particle at t = O.
Solution:
Position, x = 25 m
Velocity = \(\frac { dx}{dt}\)8t – 15
t = 0, v = 0 – 15 = -15 m/s
acceleration, a = \(\frac { dx}{dt}\) = 8 ms^-2

Question 8.
A driver takes 0.20 second to apply the breaks (reaction time). If he is driving car at a speed of 54 kmh^-1 and the breaks cause a deceleration of 6.0 ms^-2. Find the distance travelled by car after he sees the need to put the breaks.
Solution:
(distance covered during 0.20 s) + (distance covered until rest)
= (15 x 0.25) + [18.75] = 21.75 m.

Question 9.
From the top of a tower 100 m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velocity of 25 m/s. Find when and where the two balls will meet? (g = 9.8 m/s)
Solution:
For the ball dropped from the top
x = 4.9 t²…….(i)
For the ball thrown upwards
100 – x =25t – 4.9 t² …….(ii)
From eq. (i) and (ii)
t = 4s  x = 78.4 m.

Question 10.
A ball thrown vertically upwards with a speed of 19.6 ms^-1 from the top of a tower returns to the earth in 6s. Find the height of the tower, (g = 9.8 m/s)
Solution:
s = ut + \(\frac { 1 }{ 2 }\) at²
-h = 19. 6 x 6 + \(\frac { 1 }{ 2 }\) x (-9.8) x (6)²
h = 58.8 m.

Question 11.
Two town A and B are connected by a regular bus service with a bus leaving in either direction every T min. A man cycling with a speed of 20 kmh^-1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed do the buses ply of the road?
Solution:
V =40 krnh^-1 and T = 9 min.

Question 12.
A motorboat is racing towards north at 25 kmh^-1 and the water current in that region is 10 kmh^-1 in the direction of 60° east of south. Find the resultant velocity of the boat.
Solution:
V= 21.8 kmh^-1
angle with north, θ = 23.4°.

Question 13.
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft position 10 second apart is 30°, what is the speed of the aircraft?
Solution:
Speed = 182.2 ms^-1

Question 14.
A boat is moving with a velocity (3\(\hat{i}\) -4\(\hat{j}\)) with respect to ground. The water in river is flowing with a velocity (-3\(\hat{i}\) – 4\(\hat{j}\)) with respect to ground. What is the relative velocity of boat with respect to river?
Solution:
\(\overrightarrow{\mathrm{V}}_{\mathrm{BW}}\)= \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\).

Question 15.
A hiker stands on the edge of a clift 490 m above the ground and throws a stone horizontally with an initial speed of 15 ms^-1. Neglecting air resistance, find the time taken by the stone to reach the ground and the speed with which it hits the ground. (g 9.8 ms^-2)
Solution:
time = 10 seconds
V = \(\sqrt{\mathrm{V}_{s}^{2}+\mathrm{V}_{Y}^{2}}\) = \(\sqrt{15^{2}+98^{2}}\) = 99.1 m/s^-1

Question 16.
A bullet fired at an angle of 30° with the horizontal hits the ground 3 km away. By adjusting the angle of projection, can one hope to hit the target 5 km away ? Assume that the muzzle speed to be fixed and neglect air resistance.
Solution:
Maximum Range = 3.46 km
So it is not possible.

Question 17.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 seconds, what is the magnitude and direction of acceleration of the stone?
Solution:
\(\frac { 88 }{ 25 }\) rad s^-1, \(\frac { 2π}{ T}\) = \(\frac { 2πN}{t}\)
a = 991.2 cm s^-2

Question 18.
A cyclist is riding with a speed of 27 kmh^-1 . As he approaches a circular turn on the road of radius 30 m^-2, he applies brakes and reduces his
speed at the constant rate 0.5 ms^-2. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Solution:
a^c = \(\frac{v^{2}}{r}\) = 0.7ms^-2
a^t = 0.5 ms^-2
a = \(\sqrt{a^{2}-a^{2}}\) = 0.86 ms^-2
If O is the angle between the net acceleration and the velocity of the cyclist, then
0= tan^-1 \(\left(\frac{a_{c}}{a_{\mathrm{T}}}\right)\) = tan^-1 = 54°28′

Question 19.
If the magnitude of two vectors are 3 and 4 and their scalar product is 6, find angle between them and also find \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\)|
Solution:
\(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\) = AB cos θ
|\(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\)| = AB sin θ
or 6 = (3 x 4) cos θ = 3 x 4 x 60°
or θ = 60°
= 3 x 4 x \(\frac{\sqrt{3}}{2}\) = \(6 \sqrt{3}\)

Question 20.
Find the value ofA so that the vector \(\overrightarrow{\mathrm{A}}\) = 2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) and \(\overrightarrow{\mathrm{B}}\) = 4\(\hat{i}\) -2 \(\hat{j}\) +2\(\hat{k}\) are perpendicular to each other.
Solution:
\(\overrightarrow{\mathrm{A}} \perp \overrightarrow{\mathrm{B}}\)
⇒ \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}\)
⇒ t = 3

Question 21.
The velocity time graph of a particle is given by –

•  Calculate distance and displacement of particle from given v – t graph.
• Specify the time for which particle undergone acceleration, retardation and moves with constant velocity.
• Calculate acceleration, retardation from given v – t graph.
• Draw acceleration-time graph of given v – t graph.

Solution:
(i) distance = area of ∆OAB + area of trapezium BCDE = 12 + 28 = 40 m
(ii) displacement area of ∆OAB – area of trapezium BCDE = 12 – 28 = – 16 m

• times acc. \((0 \leq t \leq 4) \) and \((12\leq t \leq 16) \)
• retardation \((4\leq t \leq 8) \)
• constant velocity \((8\leq t \leq 12) \)
  

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