Plus One

Kerala State Board New Syllabus Plus One Accountancy Chapter Wise Previous Questions and Answers Chapter 8 Financial Statements – I & II.

Kerala Plus One Accountancy Chapter Wise Previous Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II

Plus One Accountancy Financial Statements Questions And Answers Question 1.
Outstanding salaries are shown as _______ (March 2010)
a) Assets
b) Liabilities
c) Capital
d) Expense
Answer:
b) Liabilities

Question 2.
Goods given as samples should be credited to ________ (March 2010)
a) Advertisement account
b) Sales account
c) Purchase account
d) None of these
Answer:
c) Purchase Account

Question 3.
State whether the following is “true” or “false”. If false, correct the same. (March 2010)
Loss of stock on account of fire should be debited to the Trading account.
Answer:
False. It should be debited to P/L a/c or credited to Trading A/c.

Question 4.
Sales are equal to _________ (March 2010)
a) cost of goods sold + profit
b) cost of goods sold – gross profit
c) gross profit – the cost of goods sold
Answer:
a) cost of goods sold + profit

Question 5.
Income tax paid by a sole proprietor on his business income should be ________ (March 2010)
a) debited to the trading account.
b) debited to the profit and loss account.
c) deducted from the capital account in the balance sheet
Answer:
c) Deducted from the capital account in the balance sheet

Question 6.
If a closing stock appears in the Trial Balance, it will appear in the _______ only. (March 2010)
Answer:
Balance Sheet

Profit And Loss Account Items List Question 7.
a) The following is the Trial Balance of Mr. Laxmi Narayan as of 30th June 2007. (March 2010)

Adjustments:
a) Stock in hand on 30th June, 2007 was Rs. 1,28,960.
b) Write off half of Mohan’s cheque.
c) Create 5% provision on debtors.
d) Wages include Rs. 1,200 for erection of machine purchased last year.
e) Depreciate plant and machinery by 5% and fixtures and fitting by 10% per year.
f) Commission accrued Rs. 600.
g) Interest on loan for the last two months is not paid.
You are required to prepare the Trading and Profit & Loss Account and also a Balance Sheet as on 30.6.2007.
Answer:
a) Trading and Profit and Loss Account of Mr. Lijin for the year ended 31st Dec. 2005

Balance Sheet as on 30/6/2007

Question 8.
Following is the Trial Balance of M/s. Kasthuri Agencies on 31st March, 2008. Prepare the Trading and Profit and Loss Account for the year ended 31st March, 2008 and a Balance Sheet on that date, after considering the adjustments. (March 2010)

Adjustments:
a) The value of closing stock on 31st March, 2008 was Rs. 32,000.
b) Outstanding wages Rs. 500.
c) Pre-paid insurance Rs. 300.
d) Commission received in advance Rs. 800.
e) Allow interest on capital @ 10% p.a.
f) Depreicate building by 2.5%, furniture and fittings by 10%.
g) Charge interest on drawing Rs. 500. Motor van by 10%.
h) Balance of interest due on loan is also to be provided for.
Answer:
Trading and Profit and Loss Account for the year ended 31.03.08

Balance Sheet as on 31.03.2008

Question 9.
_______ can be computed by deducting the cost of goods sold from sales. (March 2011)
a) Net profit
b) Gross profit
c) Net sales
d) Operating profit
Answer:
b) Gross profit

Question 10.
State whether the following is “true” or “false”. If false, correct the same. (March 2011)
Sales = cost of goods sold + profit
Answer:
True

Plus One Accountancy Chapter Wise Previous Year Questions And Answers Question 11.
State whether the following is “true” or “false”. If false, correct the same. (March 2011)
If closing stock appears in the Trial Balance, it will appear in the Trading account only.
Answer:
False, If closing stock appears in the Trial Balance, it will appear in the “Balance sheet” only.

Question 12.
The following is the Trial Balance of Mr. B. Govind on 31st March, 1990. (March 2011)

Taking into account the following adjustments, prepare the Trading and Profit and Loss account and the Balance Sheet for the year ended 31 st March, 1990.
a) Stock in hand on 31st March, 1990 was Rs. 26,800.
b) Machinery is to be depreciated at the rate of 10% and patent at the rate of 20%.
c) Salaries for the month of March, 1990 amounting to Rs. 1,500 were unpaid.
d) Insurance includes a premium of Rs. 170 on a policy expiring on 30th September, 1990.
e) A provision for bad and doubtful debts is to be credited to the extent of 5% on sundry debtors.
Answer:
Trading and Profits & Loss Account of Mr. B. Govind for the year ended 31/03/1990

Note: Insurance prepaid = 170 × 6/12 = 85. (April 1990 to September 30th)
Balance Sheet as on 31/03/1990

Question 13.
The following is the Trial Balance of Mr. Arun as on 31st December, 2009. (March 2011)

He furnishes the following additional information:
a) Closing stock on 31-12-2009 is Rs. 16,000.
b) Provide for Rs. 1,500 and Rs.2,000 for outstanding salaries and wages respectively.
c) Insurance was prepaid to the extent of Rs. 75.
d) Interest accrued Rs. 500.
e) Machinery and patents are to be depreciated @ 10% p.a. and 20% p.a. respectively.
You are required to prepare the Trading account, Profit & Loss account and the Balance Sheet of Mr. Arun as on 31.12.2009.
Answer:
Trading & Profit and Loss A/c for the year ended 31/12/2009

Balance Sheet as on 31/12/2009

Question 14.
Interest on investment received in advance is shown on the ________ (March 2012)
a) asset side of the balance sheet
b) liability side of the balance sheet
c) credit side of the revenue account
d) none of these
Answer:
b) liability side of the balance sheet

Question 15.
Contingency liabilities are shown on the footnote in the balance sheet as per the _______ accounting principle. (March 2012)
Answer:
Disclosure

Question 16.
The details of stationery for a manufacturing unit is given below: (March 2012)
Opening stock of stationery Rs. 500/-
Stationery purchase Rs. 1,200/-
Closing stock of stationery Rs.700/-
Determine the amount of stationery to be debited in the profit and loss account.
Answer:

Question 17.
Calculate the amount of operating profit from the following: (March 2012)
Sales less return – 1,30,000
Cost of goods sold – 80,000
Administration expenses – 25,000
Loss on sale of furniture – 20,000
Answer:
Operating Profit = Gross Profit – Operating expenses
Gross Profit = Net sales – cost of goods sold
= 1,30,000 – 80,000
= 50,000
Operating Profit = 50,000 – 25,000 = 25,000
Note: Loss on sales of furniture is non-operating or purely financial nature. So it is excluded while computing the operating profit.

Question 18.
A balance sheet of a sole trader is arranged in the liquidity order. Answer the following, if the trader has all typical assets. (March 2012)
a) First item is shown on the asset side of the Balance Sheet.
b) Last item is shown on the asset side of the Balance Sheet.
Answer:
a) Cash-in-hand is the first item shown on the asset side of the Balance Sheet in order of liquidity,
b) Goodwill is the last item shown on the asset side of the Balance Sheet in the order of liquidity.

Question 19.
Pass the necessary adjusting entries to record the following transactions in .the books of account on 31st March 2010. (March 2012)
a) Rent due but not paid during the year ended Rs. 500/-
b) Prepaid insurance Rs. 1,250/-
c) Depreciation on plant and machinery Rs. 5,000/-
d) Loss on sale of machinery Rs. 1,250/-
Answer:

Class 12 Accountancy Chapter Wise Questions Question 20.
From the following Trial Balance of M/s Usha Agencies on 31st March, 2010, prepare the Trading, Profit and Loss account for the year ended 31st March, 2010 and the Balance Sheet as on that date, making necessary adjustments. (March 2012)

Adjustments:
i) Closing stock Rs. 20,000.
ii) Write off further bad debts Rs. 3,000.
iii) Provide 5% for provision for bad and doubtful debts.
Answer:
Trading and Profit & Loss Account for the year ended 31/3/2010

Note:
a) New provision for bad & doubtful debts = (15000 – 3000) × 5/100 = 600
b) Interest on fixed deposit = 25,000 × 10/100 = 2500
c) Salary outstanding for one month = 1,000
Balance sheet as 31/3/2010

Question 21.
A.Mr. Koya has submitted to you the following Trial Balance as on 31st March, 2011 wherein the totals of the debit and credit balances are not equal. (March 2012)
Trial Balance as at 31st March, 2011

You are required to:
a) Redraft the Trial Balance correctly as at 31st March, 2011.
b) Prepare Trading and Profit and Loss Account for the year ended 31st March, 2011 and Balance Sheet as on that date after taking into account the following additional information.
Additional information:
i) Stock as on 31st March, 2011 worth Rs. 10,000.
ii) Wages for the month of March, 2011 amounted to Rs. 2,400 were unpaid.
iii) Depreciate building at 10% per annum.
iv) Bad debts written off Rs. 600.
v) A provision for bad debts is to be created to the extent of 10% on Sundry Debtors.
vi) Patents are written off by Rs. 2,500.
Answer:
Trial Balance as on 31.3.2011

Trading & Profit and Loss A/c for the year ended 31/12/2009

Balance Sheet as on 31/3/2011

Question 22.
The statement showing the financial position of a business is known as ______ (Say 2012)
a) Profit and Loss Account
b) Balance Sheet
c) Income and Expenditure
d) Receipts and Payments Account
Answer:
b) balance sheet

Question 23.
The purpose of preparing the ______ account is to ascertain the financial result. (Say 2012)
Answer:
Profit and Loss A/c or Income and Expenditure a/c

Question 24.
If closing balance of stock appears in the trial balance, it will appear in ______ only. (Say 2012)
Answer:
Assets side of the Balance sheet

Question 25.
Write the treatment of interest on capital, while preparing final account. (Say 2012)
Answer:
Interest on capital debited to Profit and Loss A/c and it is shown in the liability side of the Balance Sheet by adding the same to capital.

Question 26.
A firm had purchased goods worth Rs. 70,000 and it had an opening stock for Rs. 10,000. It sold a portion of goods for Rs. 75,000 at a Gross Profit of 20%. Find out its closing stock. (Say 2012)
Answer:
Closing Stock = (Opening Stock + Purchases + Gross Profit) – Sales
= (10,000 + 70,000 + 15,000) – 75,000
= 20,000
Gross Profit = 75,000 × 20/100 = 15000
OR
Another Method
Trading A/c.

Question 27.
Ram Ltd. gives the following items. As a commerce student classify them into revenue expense, capital expenditure and deferred revenue expense. (Say 2012)
Answer:
i) Computer purchased
ii) Salary paid
iii) Huge advertisement expenses
iv) Goods purchased
Answer:
i) Computer Purchased – Capital Expenditure
ii) Salary paid – Revenue Expenditure
iii) Huge Advertisement Expenses – Deferred Revenue Expenditure
iv) Goods purchased – Revenue Expenditure

Question 28.
The following Trial Balance is extracted from the books of D.Das on 31st December 2009. (Say 2012)
Trial Balance

The following adjustments are to be made:
a) Provide interest on capital at 5% per annum.
b) Salaries outstanding ₹ 1,200.
c) Insurance prepaid to the extend of ₹ 700.
d) Write off ₹ 1300 as bad debts.
e) The provision for doubtful debts is to be maintained at 5% on debtors.
f) Depreciate motor vehicles at 15% and furniture at 10%.
g) Stock in hand on 31.12.2009 was ₹ 30000.
h) A fire occurred on 26.12.2009 in the godown and the stock of the value of ₹ 6000 fully damaged. It was fully insured and the Insurance company admitted the claim in full.
You are required to prepare a Trading, Profit and Loss Account for the year ended 31 st December 2009 and a Balance sheet on that date after making the above adjustments.
Answer:
Trading and Profit and Loss A/c for the year ended 31.12.2009

Balance Sheet as on 31/12/2009

Working note
New provision for bad debt = (38500 – 1300) × 5% = 1860.

Question 29.
Prepare Trading and Profit and Loss account and Balance Sheet from the Trial Balance of Shri. Govind as on 31st March 2010. (Say 2012)

Adjustments:
1) Stock is valued at Rs. 16,000
2) Allow Interest on Capital @ 6% p.a.
3) Depreciate Furniture by 5% and Plant by 10%
4) Wages are unpaid Rs. 400.
Answer:
Trading & Profit and Loss A/c for the year ended 31/3/2010

Balance Sheet as on 31/3/2010

Question 30.
While preparing final accounts, outstanding salary is added to the salary account. The accounting principle relevant to this context is _______ (March 2013)
a) Matching
b) Business entity
c) Materially
d) Going concerned
Answer:
a) Matching

Question 31.
Which of the following is not a purpose of preparing a Balance Sheet? (March 2013)
a) To know about the sources and application of funds.
b) To know the financial position of the business.
c) To ascertain the nature and value of the asset owned.
d) To evaluate the profitability of the business
Answer:
d) To evaluate the profitability of the business

Question 32.
The Trial Balance of a trader has the following information. (March 2013)
a) Bank loan @ 12 %, Rs. 10,000
b) Interest paid Rs. 800 bank loan
Calculate the outstanding amount of interest on the bank loan.
Answer:
Interest on loan = 10,000 × 12/100 = 1200
The outstanding amount of interest on loan = 1200 – 800 = 400

Question 33.
Interest on capital is _________ (March 2013)
a) expenditure for the business
b) expense for the business
c) gain for the business
d) loss of business
Answer:
b) Expenses for the business

Question 34.
From the following information, ascertain the number of debtors to be posted in the asset side of the Balance sheet. (March 2013)
Given in Trial Balance
Bad debts – Rs. 700
Debtors – Rs. 7,000
Provision for doubtful debts – Rs. 500
Given in Adjustments:
a) Provide 5% for further bad debts.
b) Create provision for doubtful debts @ 10% on debtors.
c) Provision for discount on debtors is to be created @ 10%.
Answer:

The amount to be posted in the assets side of the B/S = 5386

Question 35.
On 31.12.2011, the balance of sundry debtors is Rs. 30,300 and further bad debts are estimated as Rs. 300. 5% of the debtors are expected to be doubtful. What are the adjustments to be required at the time of preparing the Profit and Loss Account? (March 2013)
Answer:
Rs. 300 debited to Profit and Loss Account as bad debts and Rs. 1500 (30300 – 300 = 30000 × 5/100 = 1500) debited to Profit and Loss A/c as provision for bad and doubtful debts.

Question 36.
The Trial Balance of M/s. Arathy as on 31st March, 2012 is given below. Prepare the final accounts. (March 2013)

Adjustments:
a) Closing stock was valued at Rs. 9000.
b) Insurance prepaid is Rs. 100
c) Wages outstanding is Rs. 700
d) Depreciate machinery and furniture by 10%.
Answer:
Trading and Profit & Loss A/c for the year ended 31/3/2012.

Balance Sheet as on 31/3/2012

Question 37.
Cost of goods sold is equal to sales minus _______ (March 2014)
a) net profit
b) net loss
c) gross profit
d) gross loss
Answer:
c) gross profit

Question 38.
From the Balance sheet given below, calculate: (March 2014)
a. Long term liabilities
b. Current liabilities
c. Fixed assets
Balance Sheet

Answer:
a) Long term liabilities = Debentures + capital
= 20000 + 94400
= 114400
b) Current liabilities = Sundry creditors + Loan (6 months)
= 42000 + 8000
= 50000
c) Fixed Assets = Furniture + Patent
= 30000 + 50000
= 80000

Question 39.
Given below is an extract taken from the Trial Balance of a trader as on 31.12.2012. (March 2014)
Debtors – Rs. 9100
Bad debts – Rs. 1000
Provision for bad debts – Rs. 900
Additional Information:
a) Further bad debts to be written off this year were Rs. 100.
b) 10% is reserved for further bad debts.
How much amount has to be debited to the Profit and Loss account?
Answer:

Question 40.
The following are the balances taken on 31st December, 2012, from the books of Mr.Rajendran. Prepare the Trading and Profit and Loss accounts and the Balance Sheet as on that date. (March 2014)

Other information:
1) Stock on 31st December, 2012 was Rs. 10,000.
2) Create a reserve for doubtful debts at 10%.
3) Depreciate furniture by 10%.
4) Wages unpaid were Rs. 1000.
Answer:
Trading and Profit and Loss account for the year ended December 31.12.2012

Balance Sheet as on 31/12/12

Plus One Accountancy Chapter Wise Questions And Answers Question 41.
Prepare the financial statements of Rajeevan Stores for the year 2012-13 from the information given below. (March 2014)
Trial Balance as on 31.03.2013

Adjustments:
a) Closing stock was valued at Rs. 7000.
b) Wages yet to be paid Rs. 300.
c) Rent includes advance payments of Rs. 600 for the next year.
d) Bad debts written off Rs. 400.
e) Depreciate building at 10% per annum
Answer:
Trading and Profit and Loss Account for the year ended 31.3.2013

Balance Sheet as on 31.03.2013

Question 42.
Salary of Rs.8,000 which remains unpaid during the current year is called a/an _________ (March 2015)
a) asset
b) liability
c) capital
d) income
Answer:
b) liability

Question 43.
“Under the liquidity approach, assets which are more liquid are presented at the bottom of the balance sheet”. (March 2015)
a) Is it correct? Justify your answer.
b) Write the following assets in the order of liquidity.
i) Cash in hand
ii) Cash at bank
iii) land and building
iv) Bills receivable
v) Closing stock
vi) Goodwill
Answer:
a) No, Under the liquidity approach, assets are entered up in the balance sheet following the order in which they can be converted into cash.
b) i) Cash in hand
iii) Bills Receivable
iv) Closing stock
v) Land and building
vi) Goodwill

Question 44.
Beena started the business with the following: (March 2015)
a) Classify the assets under appropriate headings.
b) Ascertain her capital.
c) Arrange them in the order of permanence by adding an intangible asset of your choice.
Cash Rs. 20,000, Building Rs. 1,00,000, Bank balance Rs. 10,000, Furniture Rs. 8,000 and Debtors Rs. 3,000.
Answer:
a) Cash – CurrentAsset
Building – Fixed Asset
Bank Balance – CurrentAsset
Furniture – Fixed Asset
Debtors – CurrentAsset
b) Capital = 20,000 + 1,00,000 +10,000 + 8,000 + 3,000 = Rs. 1,41,000
c) In the order of Permanence
1) Goodwill
2) Building
3) Furniture
4) Debtors
5) Bank
6) Cash

Question 45.
Prepare the trading, Profit, and Loss account and Balance Sheet from the Trial Balance of Mr. Shibu as on 31st March, 2013. (March 2015)

Adjustments:
1) Closing stock was valued at Rs. 20,000
2) Wages outstanding Rs. 500
3) Commission receivable Rs. 400
4) Insurance prepaid Rs. 100
Answer:
Trading and Profit and Loss A/c for the year ended 31/03/2013

Balance Sheet as on 31/03/2013

Question 46.
The following balances are taken from the books of Mr. Biswas. (March 2015)

After making the following adjustments, prepare the Trading and Profit and Loss accounts for the year ended 31st March 2014 and a Balance Sheet as on that date.
a) Stock as on 31st March 2014 worth Rs. 20,000.
b) Depreciate furniture @ 20%
c) Salary due but not paid Rs. 1,000.
d) Write-off Rs. 500 as bad debts and create a reserve for bad and doubtful debts @ 10%.
(Hint: Interest on investment accured but not received has to be considered).
Answer:
Trading and Profit and Loss A/c for the year ended 31/03/2014

Balance Sheet as on 31/3/2014

Question 47.
In which of the following items, shall commission paid by a business in advance during the year be included? (Say 2015)
a) Current expenditure
b) Current liability
c) Current asset
d) Current income
Answer:
c) Current Asset

Question 48.
The information given below is extracted form ABC Stores during the year 2013-14. (Say 2015)
Purchases – Rs. 4,000
Opening stock – Rs. 6000
Closing stock – Rs. 20,000
Gross profit – Rs. 10,000
Wages – Rs. 4,000
Find the sales during the year.
Answer:
Trading A/c for the year ended ________

Question 49.
Find the ‘sale’ of Mr. Binesh from the following information provided by him for the year ending 31st March 2013. (March 2016)

Answer:
Trading A/c for the year ended 31/3/13

Question 50.
Some adjustments are given below. (March 2016)
i) Wages outstanding Rs. 500
ii) Rent received in advance Rs. 450
iii) Further bad debts Rs. 300
a) How these adjustments are treated in the final account?
b) Write the adjustment entry for wages outstanding.
Answer:
a) Wages outstanding: It must be added to the wages account in the trading and profit and loss A/c. It will be shown on the liability side of the balance sheet.
Rent Received in advance: It will be deducted from the rent in profit and loss a/c. It will be shown on the liability side of the balance sheet.
Further Baddebts: It will be added to the baddebt in profit and loss a/c and deduct further bad debts from debtors in assets side of the balance sheet.
b) Wages a/c Dr 500
To wages outstanding a/c 500

Question 51.
a) Explain the term, current liability with an example. (March 2016)
b) Classify the following items into current liabilities and long term liabilities.
(Loan for five years, Creditors, Debentures, Bankover draft)
Answer:
a) Liabilities are payable within one year are called current liabilities
eg. creditors, Bills payable etc.
b) Current liabilities → Creditors, bank overdraft
Long term liabilities → Loan for 5 years, debentures.

Question 52.
The following balances have been extracted form the books of Kerala Traders for the year ended 31st December, 2014. (March 2016)

Adjustments:
a) Make a provision on debtors @ 6%
b) Salaries outstanding Rs.200. Repairs include Rs. 300 for the year 2015.
c) Depreciate building @ 5% and furniture @ 10%
d) Closing stock Rs. 24,000
Prepare Profit and Loss A/c and Balance Sheet for the year ending 31st December, 2014.
Answer:
Trading and Profit and Loss A/c for the year ended 31/12/2014

Balance Sheet as on 31/12/2014

Question 53.
The following is the trial balance of Ammu Associates for the year ending 31st December, 2014 and other information relating are given to you. Find the working result of business and also show the financial position of them for the year. (March 2016)
Trial Balance of Ammu Associates on 31-12-2014

Additional information:
a) Closing stock is valued at Rs. 25,000
b) Salary of an employee is not paid Rs. 500
c) Further bad debt incurred Rs. 200
d) Provision for bad debt is created at 5% on debtors
Answer:
Trading and profit and Loss A/c for the year ended 31/12/14

Balance sheet as on 31/12/2014

Question 54.
Find the odd one out from the following. State the reason. (Say 2016)
a) Rent
b) Salary
c) Wages
d) General Expenses
Answer:
c) wages
Reason: wage is a direct expenses, but all others are indirect expenses.

Question 55.
Ascertain the amount to be debited to prodit and loss.account under the head ‘Salaries’ based on the following information. (Say 2016)
Given in Trial balance:
Salary (Dr) Rs. 12,000
Salary outstanding (Cr) Rs. 2,000
Given in adjustments:
Salary includes Rs. 1,500 paid in advance
Answer:
Profit and Loss A/c

Question 56.
The following balances are extracted form the books of a sole trader as on 31/12/2015. (Say 2016)

Additional information
a) Closing stock on 31/12/2015 Rs. 33,000
b) Salary prepaid Rs. 2,000
c) Write off Rs. 1,000 as bad debts and provide 3% for provision for bad and doubtful debts
d) Wages outstanding Rs. 1,000
Prepare a trading and Profit and Loss account and Balance Sheet as on 31/12/2015.
Answer:
Trading and profit and Loss A/c for the year ended 31/12/2015

Balance sheet as on 31/12/2015

Question 57.
Athul, a commerce student prepared the following diagram showing examples of direct expenses for exhibiting in the class room. Identify the wrong examples appearing in the diagram. (March 2017)

Answer:
Rent
Repairs

Question 58.
“A Balance Sheet is more reliable than statement of affairs”. Explain. (March 2017)
Answer:
A balance sheet is prepared on the basis of those books which are maintained under double entry system. But statement of affairs is prepared on the ba¬sis of information from incomplete records. So a bal¬ance sheet is more reliable than statement of affairs.

Question 59.
Calculate the value of cost of goods sold and gross profit. (March 2017)

Answer:
Cost of goods sold = (Opening stock + Purchases + Direct exp.) – Closing stock
= 18,000 + 60,000 + 4000 – 20,000
= 62,000
Gross profit = Net sales – cost of goods sold
Net sales = 1,32,000 – 7000 = 1,25,000
Gross Profit = 1,25,000 – 62,000 = 63,000

Question 60.
Opening Stock Rs. 18,000
Purchases Rs. 22,000
Wages Rs. 5,000
Closing stock Rs. 15,000
a) Ascertain cost of goods sold.
b) How much is gross profit, if sales is Rs. 37,500? (March 2017)
Answer:
a) Cost of goods sold = Opening stock + purchases+ Direct expenses – Closing stock
= (18,000 + 22,000 + 5000) – 15,000
= 30,000
b) Gross profit = Sales – Cost of goods sold
= 37500 – 30,000
= 7,500

Question 61.
“Outstanding Expenses are to be accounted for while preparing the financial statements”. (March 2017)
a) Which concept of accounting is mentioned in the above?
b) Complete the following table.

Answer:
a) Matching Principle
b) i) Prepaid insurance – Prepaid amount deducted from insurance on the debit side of P/L a/c
ii) Depreciation on furniture – Amount of depreciation debited to P/L a/c.
iii) Bad debts given outside the Trial Balance – Deducted from Sundry debtors on the asset side of the B/s

Question 62.
Following is the trail balance of R.K. Traders on 31st December, 2014. Prepare Trading and Profit & Loss account and Balance Sheet taking into account, the adjustments given. (March 2017)

Adjustments:
a) Closing stock was valued at Rs. 2,400.
b) Commission received in advance Rs. 800.
c) Salaries due but not paid Rs. 1,300
d) Depreciate furniture by Rs. 1,500
e) 1/4 of insurance is unexpired
Answer:
Trading and profit and Loss A/c for the year ended 31/12/14

Balance Sheet as on 31/12/14

Question 63.
The following is the Trial Balance of Kiran, a trader as on 31st December, 2016. (March 2017)
Trial Balance as on 31/12/2016

Additional Information:
a) Closing stock was valued at Rs. 10,000.
b) Carriage outstanding Rs.500
c) Commission received in advance Rs.700
d) Provide 5% of debtors for bad debts.
Prepare Trading and Profit and Loss account for the year ended 31st December, 2016 and a Balance Sheet as on that date.
Answer:
Trading and Profit and Loss A/c for the year ended 31/12/2016

Balance Sheet As on 31/12/2016

Students can Download Chapter 3 Trigonometric Functions Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Plus One Maths Trigonometric Functions Three Mark Questions and Answers

Plus One Maths Chapter Wise Questions And Answers Pdf Question 1.
Prove the following

Answer:
i) LHS

ii) LHS

iii) LHS = sin 2x + 2 sin 4x + sin 6x
= 2 sin 4xcos2x + 2sin 4x
= 2 sin 4x(cos2x + 1) = 4 cos² x sin 4x

iv) LHS

v) LHS

vi) LHS

vii) LHS = sin² 6x – sin² 4x

= 2 sin 10x sin(-2x)
= 2 sin 10x sin2x

viii) LHS

Plus One Maths Trigonometric Functions Question 2.
Find the general solution of the following equations.

1. cos4x = cos2x
2. sin 2x +cosx = 0
3. cos3x + cosx – cos2x = 0

Answer:
1. Given; cos 4x = cos 2x
⇒ cos4x – cos 2x = 0
⇒ -2 sin 3x sin x = 0
General solution is
⇒ sin3x = 0; ⇒ 3x = nπ ⇒ x = \(\frac{n \pi}{3}\), ∈ Z
Again we have;
⇒ sinx = 0; ⇒ x = nπ; n ∈ Z

2. Given; sin 2x + cosx = 0
⇒ 2sin xcosx + cosx = 0
⇒ cosx(2sin x + 1) = 0
General solution is
⇒ cosx = 0 ⇒ x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z
Again we have; 2sin x + 1 = 0

3. Given; cos3x +cosx – cos2x = 0
⇒ 2 cos2x cosx – cos2x = 0
⇒ cos2x(2cosx – 1) = 0
General solution is

Again we have; 2cosx -1 = 0

Plus One Maths Trigonometry Equations Question 3.
In Triangle ABC, if a = 25, b = 52 and c = 63, find cos A and sin A.
Answer:

Plus One Maths Text Book Questions And Answers Question 4.
For any ΔABC, prove that a(b cosC – c cosB) = b² – c²
Answer:
LHS = ab cos C – ac cos B

Hsslive Maths Textbook Answers Plus One Question 5.
For any ΔABC, prove that, \(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^{2}-c^{2}}{a^{2}}\).
Answer:

Plus One Chapter Wise Questions And Answers Question 6.

1. Convert \(\frac{2 \pi}{3}\) radian measure into degree measure. (1)
2. Prove that \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x\) (2)

Answer:
1. \(\frac{2 \pi}{3}=\frac{2 \pi}{3} \times \frac{180}{\pi}=120^{\circ}\)

2. LHS

Plus One Maths Trigonometric Functions Four Mark Questions and Answers

Plus One Maths Chapter Wise Questions And Answers Question 1.
For any ΔABC, prove that

Answer:

Plus One Maths Questions And Answers Question 2.
For any ΔABC, prove that \(\sin \frac{B-C}{2}=\frac{b-c}{a} \cos \frac{A}{2}\).
Answer:

Plus One Trigonometry Questions And Answers Question 3.
(i) Which of the following is not possible. (1)
(a) sin x = \(\frac{1}{2}\)
(b) cos x = \(\frac{2}{3}\)
(c) cosec x = \(\frac{1}{3}\)
(d) tan x = 8
(ii) Find the value of sin 15°. (2)
(iii) Hence write the value of cos 75° (1)
Answer:
(i) (c) cosec x = \(\frac{1}{3}\)

(ii) sin 15° = sin(45° – 30°)
= sin45°cos30°- cos45°sin30°

(iii) sin 15° = sin(90° – 75°) = cos 75°

Plus One Maths Trigonometric Functions Six Mark Questions and Answers

Kerala Sslc Maths Chapter Wise Questions And Answers Question 1.
The angle of elevation of the top point P of the vertical tower PQ of height h from a point A is 45° and from a point B, the angle of elevation is 60°, where B is a point at a distance d from the point A measured along the line AB which makes angle 30° with AQ. Prove that d = h(\(\sqrt{3}\) – 1).
Answer:

From the figure we have ∠PAQ = 45°, ∠BAQ = 30°and ∠PBH = 60°
in right ∆AQP
Clearly ∠APQ = 45°, ∠BPH = 30° , giving ∠APB = 15° ⇒ ∠PAB = 15°
In ∆APQ ,PQ = AQ = h
AP² = h² + h² = 2h² ⇒ AP = \(\sqrt{2}\)h
From ∆ABP,

Important Questions For Class 11 Maths Trigonometry Question 2.
A tree stands vertically on a hill side which makes an angle of 15° with the horizontal. From a point on the ground 35m down the hill from the base of the tree, the angle of elevation of the top of the tree is 60°. Find the height of the tree.
Answer:

Let BC represent the tree, A be the point 35m down the hill from the base of the tree and h be the height of the tree.
Clearly in ∆ABC
∠BAC = 60°- 15° =45°;
∠ACB = 30°; ∠ABC = 105°

Trigonometric Functions Class 11 Pdf State Board Question 3.
(i) If sin x = cos x, x ∈ [0, π] then is
(a) 0
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{3}\)
(d) π
(ii) Write the following in ascending order of tits values, sin 100°, sin 0°, sin 50°, sin 200°
(iii) Solve: sin2x – sin4x + sin6x = 0
Answer:
(i) (b) \(\frac{\pi}{4}\)

(ii) sin 100° = sin(l 80 – 80) = sin 80°
sin 200° = sin(l 80° + 20°) = -sin 20°
The ascending order is
sin 200°, sin 0°, sin 50°, sin 100°

(iii) sin2x + sin6x – sin4x = 0
⇒ 2sin 4x cos2x – sin 4x = 0
⇒ sin 4x(2 cos 2x – 1) = 0
⇒ sin4x = 0 or (2cos2x – 1) = 0
⇒ 4x = nπ or cos2x = \(\frac{1}{2}\)

Plus One Maths Trigonometric Functions Practice Problems Questions and Answers

Question 1.
Convert the following degree measure into radian measure.
i)  45°
ii) 25°
iii) 240°
iv) 40°20′
v) -47°30′
Answer:

Question 2.
Convert the following radian measure into degree measure,
i)   6
ii) -4
iii) \(\frac{5 \pi}{3}\)
iv) \(\frac{7 \pi}{6}\)
v) \(\frac{11}{16}\)
Answer:

Question 3.
The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use π = 3.14)
Answer:
60 minutes = 360 degrees.
1 minutes = 6 degrees.
40 minutes = 240 degrees.
240° = 240 × \(\frac{\pi}{180}=\frac{4 \pi}{3}\)
The required distance travelled = l = rθ
= 1.5 × \(\frac{4 \pi}{3}\) = 2 × 3.14 = 6.28 cm

Question 4.
In a circle of diameter 40 cm, the length of a cord is 20 cm. Find the length of minor arc of the chord.
Answer:
The radius and chord join to form a equilateral triangle. Therefore
l = rθ = 20 × \(\frac{\pi}{3}\)
= 20 × \(\frac{3.14}{3}\) = 20.933.

Question 5.
If the arcs of the same lengths in the two circles subtend angles 65° and 110° at the centre, find the ratio of their radii.
Answer:
We have l = rθ, the radius and angle are inversely proportional. Therefore;

Question 6.
Find the values of the other five trigonometric functions in the following; (2 score each)

1. cos x = \(-\frac{3}{5}\), x lies in the third quadrant.
2. cot x = \(-\frac{5}{12}\), x lies in the second quadrant.
3. sin x = \(\frac{1}{4}\), x lies in the second quadrant.

Answer:
1. Given;
cos x = \(-\frac{3}{5}\)

2. Given;
cot x = \(-\frac{5}{12}\)

3. Given;
sin x = \(\frac{1}{4}\); cosecx = 4

Question 7.
Find the value of the trigonometric functions. (2 score each)

Answer:

Question 8.
Find the value of the following.

iv) sin 75°
v) tan 15°
Answer:

iv) sin 75° = sin(45° + 35°)
= sin 45° cos30° + cos45° sin 30°

v) tan 15° = tan(45° – 30°) = \(\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}\)

Question 9.
Find the principal and general solution of the following.

1. sin x = \(\frac{\sqrt{3}}{2}\)
2. cosx = \(\frac{1}{2}\)
3. tan x = \(\sqrt{3}\)
4. cos ecx = -2

Answer:
1. Given; sin x = \(\frac{\sqrt{3}}{2}\) = sin \(\frac{\pi}{3}\)
General solution is; x = nπ + (-1)ⁿ\(\frac{\pi}{3}\),
n ∈ Z
Put n = 0, 1 we get principal solution; x = \(\frac{\pi}{3} ; \frac{2 \pi}{3}\).

2. Given; cosx = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
General solution is; x = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z
Put n = 0, 1 we get principal solution;
n = 0 ⇒ x = \(\frac{\pi}{3}\); n = 1 ⇒ x = 2π – \(\frac{\pi}{3}\) = \(\frac{5\pi}{3}\).

3. Given; tan x = \(\sqrt{3}\) = tan\(\frac{\pi}{3}\)
General solution is; ⇒ x = nπ + \(\frac{\pi}{3}\), n ∈ Z
Put n = 0, 1 we get principal solution;
n = 0 ⇒ x = \(\frac{\pi}{3}\); n = 1 ⇒ x = π + \(\frac{\pi}{3}\) = \(4\frac{\pi}{3}\).

4. Given; cosecx = -2
⇒ sin x = \(-\frac{1}{2}\) = – sin \(\frac{\pi}{6}\) = sin(-\(\frac{\pi}{6}\) )
General solution is; x = nπ – (-1)ⁿ \(\frac{\pi}{6}\), n ∈ Z
Put n = 1, 2 we get principal solution;

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 2 Anukampa Text Book Questions and Answers, Summary, Notes.

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Samkramanam Questions and Answers

Samkramanam Summary

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Vasanavikrithi Questions and Answers

Vasanavikrithi Summary

You can Download Writing and City Life Questions and Answers, Notes, Plus One History Chapter Wise Questions and Answers Kerala Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Plus One History Chapter Wise Questions and Answers Chapter 2 Writing and City Life

Writing And City Life Extra Question And Answer Chapter 2 Question 1.
For recreating Mesopotamian history the most important sources are archaeological evidences. How do you evaluate this statement?
Answer:
Archaeological excavations were started, in Mesopotamia in the 1840s. Excavations in sites like Uruk and Mari continued for decades. The hundreds of buildings, statues, tombs, tools and seals in Mesopotamia and the thousands of written documents are useful sources in recreating Mesopotamian history.

Class 11 History Chapter 2 Important Questions And Answers Question 2.
Iraq is a land with different physical features. Explain.
Answer:
Iraq is a land with widely different physical features. In its North-Eastern region, there are green plains, rivers with their banks full of trees and hills covered with all sorts of flowers. There is enough rain for cultivating different crops. Agriculture started here between BC 7000 and 6000. On the Northern side of Iraq there is a grassy plain. This place was suitable for grazing cattle. Naturally, the people here got a better life than those of mere agricultural farmers.

In the Eastern side there were the tributaries of River Tigris and they offered excellent travel facilities towards the hilly regions. The southern side is a desert. It is in Iraq that the first writings and urban life started. The fertile silt deposited by Euphrates and Tigris Rivers made this area very suitable for cultivation and therefore it could nurture urban life.

Class 11 History Chapter 2 Extra Questions And Answers Question 3.
Agriculture was a main occupation of Mesopotamian people. What were their other occupations?
Answer:
Cattle herding and fishing.

Writing And City Life Important Questions And Answers Chapter 2 Question 4.
Urban life is very important. Explain the importance of urban life in Mesopotamia.
Answer:

1. Mutual dependence (between cities & cities, between cities & villages).
2. Division of labour
3. Social organizations.

Important Questions Of Writing And City Life Chapter 2  Question 5.
In Mesopotamia minerals were rare. Examine the validity of this statement.
Answer:
Mesopotamia was rich in food supplies. But minerals were rate there. In many parts of southern Mesopotamia, there weren’t enough stones to make work tools, seals and ornaments. The wood of the date palms and poplar trees of Iraq was not capable of being used in carts and wheels as it was too soft and brittle.

There weren’t minerals for making work tools, seals, pots, pans, and ornaments. Therefore the Mesopotamians got their clothes, timber, copper, lead, silver, gold, shells and different types of stones from Turkey and Iran.

Writing And City Life Question And Answer Chapter 2 Question 6.
Prepare a seminar paper on the Mesopotamian writing technique.
Answer:
Areas to be considered:

1. The development of writing
2. System of writing
3. Uses of writing

The development of writing: AH societies have languages. Certain sounds in the language give certain meanings. This is verbal exchange. When speech sounds are represented in visible forms, we have writing or script. Mesopotamians wrote on clay slates. The writer kneads clay and makes it into a size that he can hold in one hand, The surface would be smoothened. Using a special kind of sharp stiletto he makes wedge-shaped letter marks on the smooth surface. This is called cuneiform writing. After that, the clay-slate is dried in the sun. This way the clay slates become permanent like clay pots. These slates couldn’t be used again for writing other things.

System of Writing: A cuneiform symbol does not represent a mere consonant or vowel sound, but a number of letters. Therefore a Mesopotamian script writer had to leam hundreds of symbols. He should have had the competence to handle a wet clay-slate and write on it before it goes dry. Thus writing on the clay-slate was a highly skilled job. It was an intellectual exercise which translated the sound system of a language into a visible format.

Uses of Writing: Writing has given man invaluable contributions. There was a close tie between the writing of Mesopotamia and its urban life and trade. It was the kings that linked writing with trade.

Writing was used to store information and to exchange messages. Many saw Mesopotamian writing as a sign proclaiming the dominance of Mesopotamian urban culture.

The writing helped in communicating with other regions culturally and economically and to do the buying and selling on the basis of written agreements. In short, writing made trading easier. Writing helped in maintaining accounts and in keeping laws recorded. It was also useful for literary creations.

Writing And City Life Class 11 Important Questions Chapter 2 Question 7.
In South Mesopotamia, there were three kinds of cities. Which were they?
Answer:

1. Cities that grew around temples.
2. Cities that grew into commercial and trade centres.
3. Imperial Cities.

Important Question Of Writing And City Life Chapter 2 Question 8.
It was the control over the temples that helped kings to exercise their authority over people. Do you agree with this view? Justify.
Answer:
In due course, new institutions and traditions developed in the society. Powerful nobles began to work for their own welfare and also for the welfare of their community. The successful nobles attacked the weaker ones and looted precious things from them and gave them to the gods, beautifying the temples. They proclaimed themselves to be kings. They sent people around and got.stones and minerals for the prosperity of the community and also their gods. They also took steps to distribute the temple property efficiently. This helped the kings to get high status and authority over the people. In short, it was their control over the temples that helped them to have power over the people.

Writing And City Life Important Questions Chapter 2 Question 9.
There used to be confrontations between the shepherds and farmers of Mari. Why?
Answer:
When the shepherds passed with their flocks through the sown fields, the plants got destroyed. This destruction of the crops caused conflicts between the shepherds and farmers. Sometimes the nomadic shepherds attacked agricultural villages and forcefully took away things which the villagers had stored. Farmers, on their side, often refused to let the shepherds and their flock move.to the river through their farms. They also refused to give water from their canals to the shepherds and their flocks causing friction.

Class 11 History Chapter 2 Important Questions  Question 10.
For the Mesopotamians, urban life was very important. Examine the validity of this statement in the background of the Epic of Gilgamesh.
Answer:
Epic of Gilgamesh is an epic poem that shows the love and pride the Mesopotamians felt for their cities. This poem has 12 parts and it is at the end of the poem that their pride for their cities has been shown. Gilgamesh was the king of the city of Uruk. He was a great warrior and he conquered even faraway places.

The death of his close friend was a turning point in the fife of Gilgamesh. f he shock inflicted on him by v the death of his friend persuaded him to embark on a journey seeking the mystery of immortality. He crossed mountains and seas, but his journey was a failure and he returned to his city Uruk. He tried to console and comfort himself by walking on the city wall this way and that way.

The wall was built with baked bricks and he looked at the base of the wall with admiration. He wrote the epic poem sitting on the wall. He was able to find consolation by unburdening his load of sorrow bn the wall which his people had built. The pride he felt helped him to overcome his sorrow. The epic of Gilgamesh makes it clear that the city was like life-giving oxygen to the Mesopotamians.

Important Question Of History Class 11 Chapter 2 Question 11.
Describe the power of the Mesopotamian writing in the background of discoveries in the field of mathematics and astrology.
Answer:
The Mesopotamians gave great contributions in the realm of science. In fact, their contributions in the scientific area can be ascribed to their writing. For science, written material is necessary. Only then future generations of scholars can read it, understand it and improve it. The Mesopotamians have made great contributions in calendar-making, to fix the time of things and mathematics.

In Mathematics they discovered multiplication, division, square, square root and compound interest. Some clay slates where these things are recorded have been discovered. The square root they discovered differs only very slightly from the actual one.

Based on the rotation of the moon around the earth, a year was divided into 12 months, a month was divided into 4 weeks, and a day was divided into 24 hours, and an hour was divided into 60 minutes. This was a Mesopotamian discovery. Thus the calendar which was based on the lunar movement has been approved and accepted by the whole world.

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Students can Download Chapter 2 Units and Measurement Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 2 Units and Measurement

Plus One Physics Units and Measurement One Mark Questions and Answers

Plus One Physics Chapter 2 Questions And Answers Question 1.
How many seconds are there in a light fermi?
(a) 10^-15
(b) 3.0 × 10⁸
(c) 3.33 × 10^-24
(d) 3.3 × 10^-7
Answer:
(c) 3.33 × 10^-24
One light fermi is time taken by light to travel a distance of 1 fermi ie. 10^-15m
1 light fermi = \(\frac{10^{-15}}{3 \times 10^{8}}\) = 3.33 × 10^-24s.

Units And Measurements Questions And Answers Pdf Hsslive Question 2.
Which of the following pairs have same dimensional formula for both the quantities?

1. Kinetic energy and torque
2. Resistance and Inductance
3. Young’s modulus and pressure

(a) (1)only
(b) (2) only
(c) (1) and (3) only
(d) All of three
Answer:
(c) (1) and (3) only

Plus One Physics Units And Measurements Questions And Answers Question 3.
Give four dimensionless physical quantities.
Answer:
Angle, Poisson’s ratio, strain, specific gravity.

Plus One Physics Units And Measurements Questions Question 4.
The dimensions of plank constant are the same as those of______.
Answer:
Angular momentum

Plus One Physics Chapter Wise Questions And Answers Question 5.
A physical quantity P = \(\frac{\sqrt{a b c^{2}}}{d^{3}}\) measuring a, b, c and d separately with the percentage error of 2% , 3%, 2% and 1% respectively. Minimum amount of error is contributed by the measurement of
(a) b
(b) a
(c) d
(d) c
Answer:
(b) a
P = \(\frac{\sqrt{a b c^{2}}}{d^{3}}\)

The minimum amount of error is contributed by the measurement of a.

Plus One Physics Second Chapter Questions And Answers Question 6.
The number of significant figures in 11.118 × 10^-6 is
(a) 3
(b) 6
(c) 5
(d) 4
Answer:
As per rules, number of significant figures in 11.118 × 10^-6 is 5.

Class 11 Physics Ch 2 Important Questions Question 7.
What is the number of significant figures in 0.06070?
Answer:
4.

Plus One Physics Important Questions And Answers Pdf Question 8.
If f = x², What is the relative error in f?
Answer:
\(\frac{2 \Delta x}{x}\).

Hsslive Plus One Physics Chapter Wise Questions And Answers Question 9.
Which of the following measurement is more accu¬rate?
(i) 7000m
(ii) 7 × 10²m
(iii) 7 × 10³m
Answer:
(i) 7000 m

Units And Measurements Questions And Answers Pdf Question 10.
Which of the following measurements is most, accurate?
(a) 5.0 cm
(b) 0.005 cm
(c) 5.00 cm
Answer:
(c) Is most accurate because it has three significant figures. Greater is number of significant figures, more accurate is the measurement.
(a) has 2 significant figures
(b) has 1 significant figure.

Units And Measurements Class 11 Numericals With Solutions Question 11.
Name three physical quantities having same dimension.
Answer:
Work, Energy, and Torque.

Plus One Physics Units and Measurement Tw0 Mark Questions and Answers

Units And Measurements Class 11 Questions Answers Question 1.
Using dimensional analysis derive the relation F = ma. Where the symbols have the usual meaning.
Answer:
Force on a body depends on mass(m), acceleration (a) an
F α m^aa^bt^c
M¹L¹T^-2 = M^a(LT^-2)^bT^c
M¹L¹T^-2 = M^aL^bT^-2a+c
Equating the powers, we get a = 1 ,b = 1, -2b + c = -2, c = 0
F = m¹a¹t⁰ = ma.

Plus One Physics Chapter 2 Previous Year Questions Question 2.
Use your definition to explain how simple harmonic motion can be represented by the equation y = a sin ωt
(a) Show that the above equation is dimensionally correct
Answer:
Y = a sin ωt
sin ωt has no dimensions. Hence we get L = L
Hence this equation is dimensionaly correct.

Questions On Error Analysis Class 11 Question 3.
Fill in the blanks.

1. The curved surface area of a solid cylinder of radius 2 cm and height 20 cm is_____m² (Write answer in 3 significant digits)
2. Im = ______ ly

Answer:
1. Curved area = 2πl
= 2 × 3.14 (2 × 10²) × 20 × 10²
= 2.51 × 10^-6m²

2. l ly= 9.46 × 10¹⁵ m
lm = \(\frac{l \mathrm{ly}}{9.46 \times 10^{15}}\) ≈ 10^-6ly.

Physics Chapter 2 Class 11 Numericals Question 4.

1. Give a physical quantity with a unit and no dimension.
2. Arrange the following in the descending order.
   1 light year, 1 parsec, 1 astronomical unit

Answer:

1. Angle has no dimension. But it has unit.
2. 1 parsec, 1 light year, 1 astronomical unit.

Dimensional Analysis Questions And Answers Pdf Question 5.
Magnitude of force F experienced by a certain object moving with speed V is given by F = KV². Where K is a constant. Find the dimensions of K.
Answer:
F = KV²

Class 11 Physics Chapter 2 Important Questions With Answers Question 6.
What is the maximum percentage error in the measurement of kinetic energy if percentage errors in mass and speed are 2% and 3% respectively?
Answer:
E = \(\frac{1}{2}\)v²

% error in KE = % error in mass + 2 × % error in speed
= 2% + 2 × 3% = 8%.

Units And Measurements Class 11 Important Questions Pdf Question 7.
Solve the following with regard to significant figures.

1. 5.8 + 0.125
2. 3.9 × 10⁵ – 2.5 × 10⁴

Answer:
1. 5.8 + 0.125 = 5.925
Rounding to first decimal point, we get 5.9

2. 3.9 × 10⁵ – 2.5 × 10⁴
= 3.5 × 10⁵ – 0.25 × 10⁴
= 3.65 × 10⁵
Rounding to first decimal place, we get 3.6 × 10⁵.

Question 8.
What is maximum fractional error in
i) (a + b)
ii) a – b
iii) ab
iv) \(\frac{a}{b}\)
Given ∆ a and ∆ b are absolute errors in measurements a and b.
Answer:

Question 9.

1. What is the fractional error in aⁿ? (Given absolute error in a is ∆ a)
2. What is absolute error in the measurements according to least count?
   • 3.0 kg
   • 25 s
   • 5.62 cm

Answer:
1. n\(\frac{\Delta a}{a}\)

2. The measurements according to least count:

• 0.1 kg
• 1 s
• 0.01 cm

Plus One Physics Units and Measurement Three Mark Questions and Answers

Question 1.
A stone is thrown upwards from the ground with a velocity ‘u’.

1. What is the maximum height attained by the stone?
2. Check the correctness of the equation obtained in (a) using the method of dimensional analysis.

Answer:
1. H = \(\frac{u^{2}}{2 g}\) ___(1)
u = u, v = o, a = -g, h = ?
We can find maximum height using the equation
u² = u² + 2as
0 = u² + 2 × -g × H
2gh = u²
H = \(\frac{u^{2}}{2 g}\)

2. Dimension of H = L
Dimension of u = (LT^-1)
Dimensions of time (t) = T
Dimension of g = (LT^-2)
substituting these values in eq(1) we get
L = \(\frac{\left(L T^{-1}\right)^{2}}{\left(L T^{-2}\right)}\)
L = L.

Question 2.
Derive an empirical relationship for the force experienced on the car in terms of mass of the car m, velocity v, and radius of the track r using dimensional analysis.
Answer:
Centripetal force may depends on mass (m),radius(r) and velocity(v)
F α m^ar^bv^c
M¹L¹T^-2 = M^aL^b(LT^-1)^c
M¹L¹T^-2 = M^aL^bL^cT^-c
M¹L¹T^-2 = M^aL^b+cT^-c
Equating we get a = 1, b + c = 1, c = 2, b = -1
Substituting these values in eq(1),we get
F = \(\frac{M V^{2}}{r}\).

Question 3.
Dimensional formula of a physical quantity indicate how many times fundamental quantity is involved in the measurement of the quantity.

1. What is the dimensional formula of coefficient of viscosity?
2. Write any two drawbacks of dimensional analysis.

Answer:
1. F = ηA\(\frac{d V}{d x}\)

2. The method of dimensional analysis has the following drawbacks:

• It gives no information about the dimensionless constant involved in the equation.
• The method is not applicable to equations involv¬ing trigonometric and exponential functions.
• This method cannot be employed to derive the • exact form of the relationship if it contains sum
  of two, or more terms.
• If the given physical quantity depends on more than three unknown quantities, the method fails.

Question 4.
Principle of homogeneity is based on the fact that two quantities of same nature can be added.

1. What do you mean by principle of homogeneity?
2. Velocity V depends on the time t as V = at² + bt + c. Find dimension of constants a, b, and c.

Answer:
1. For the correctness of an equation, the dimensions on either side must be the same. This is known as the principle of homogeneity of dimensions.

2. V = at² + bt + c
M⁰L¹T^-1 = aT² + bT + c
According to principle of homogenity, we get
aT² = M⁰L¹T^-1
a = \(\frac{\mathrm{M}^{0} \mathrm{L}^{1} \mathrm{T}^{-1}}{\mathrm{T}^{2}}\)
= M⁰L¹T^-3.

Question 5.
If x = a + bt + ct² where x is in meter and t in second.

1. Find the dimensional formula of ‘b’.
2. If error in the measurement of time is 2%. What will be the error in x?

Answer:
1. According to principle of homogeneity, the dimensions of both sides must be same.
ie. L = a + bT + cT²
ie : L = bT, b = L/T

2.

% error in x = 3 × %. error in ‘t’ = 3 × 2% = 6%.

Question 6.
A physical quantity P is related to four observables a, b, c as P = \(\frac{a^{3} b^{2}}{\sqrt{c d}}\). The % error in the measurement of a, b, c, and d are 1%, 3%, 4%, 2% are respectively.

1. What do you mean by error in a measurement?
2. What is the % error in the measurement of P?

Answer:
1. The result of every measurement by any measuring instrument contains some uncertainty. This uncertainty is called error.

2.

% error in
P = 3 × 1 + 2 × 3 + 1/2 × 4 + 1/2 × 2
= 3 + 6 + 2+ 1
P = 12%

Question 7.
Rahul measured the height of Ramesh in different trials as 1.67m, 1.65m 1.64m, and 1.63m.

1. Find the mean absolute error?
2. Find the percentage error?

Answer:
1. Arithametic mean,

a_mean = 1.645m = 1.65
absolute error,
∆a₁ = a_mean – a₁
∆a₁ = 1.65 – 1.67 = -0.02
∆a₂ = 1.65 – 1.65 = 0
∆a₃ = 1.65 – 1.64 = 0.01
∆a₄ = 1.65 – 1.63 = 0.02
Mean absolute error

= 0.012

2. percentage error = \(\frac{\Delta \mathrm{a}_{\text {mean }}}{\mathrm{a}_{\text {mean }}}\) × 100
= \(\frac{0.012}{1.65}\) × 100
= 0.75%.

Question 8.
In a particular experiment Ramu used the relation F = AB + (P + Q) Y to calculate force.

1. Which principle is used to check the correctness of the equation (1)
2. If the dimensional formula of Y is M⁰L¹T^-1, then find the dimensional formula of P

Answer:
1. Principle of homogenity

2. F = AB + (P+Q)Y
F = AB + PY + QY
MLT^-2 = AB + PY+ QY
According to principle of homogeneity
MLT^-2 = PY
M¹L¹T^-2 = P M⁰L¹T^-1
ie. P = \(\frac{M^{\prime} L^{1} T^{-2}}{M^{0} L^{1} T^{-1}}\) = M¹T^-3

Question 9.

1. Which of the following is precise
   • A vernier calliperse with 40 divisions on sliding scale
   • An optical instrument that can measure length of the order of wavelength of light.
2. Is it possible to increase the accuracy of screw gauge by increasing the number of divisions on the head scale?

Answer:
1. (i) L.C of vernier caliperse = \(\frac{1}{40}\) = 0.025mm
= 0.025 × 10^-3m
= 2.5 × 10^-5m.

(ii) L.C of optical instrument = 6000A°
= 6000 × 10^-10m
(Taking λ of visible light = 6000°A)= 6 × 10^-7m

2. Yes. Because L.C proportional to number of division on the headscale. So with the increase in number of divisions, the least count will increase. This leads to increase the accuracy of above screw guage.

Plus One Physics Units and Measurement Four Mark Questions and Answers

Question 1.
In an experiment with common balance the mass of a body is found to 2.52g, 2.53g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate

1. The mean value of the body
2. Mean absolute error
3. Percentage error

Answer:
1. Mean value, M_mean
= \(\frac{2.52+2.53+2.51+2.49+2.54}{5}\)
= 2.5g

2. Absolute error,
Absolute error ∆m₁ = |2.52 – 2.52| = 0
∆m₂ = |2.52 – 2.53| = 0.01
∆m₃ = |2.52 – 2.51| = 0.01
∆m₄ = |2.52 – 2.49| = 0.03
∆m₅ = |2.52 – 2.54| = 0.02
∴ Mean absolute error
\(\frac{0+0.01+0.01+0.03+0.02}{5}\)
∆m_mean = 0.014g

3. Percentage error = \(\frac{\Delta \mathrm{m}_{\text {mean }}}{\mathrm{m}_{\text {mean }}}\) × 100
= \(\frac{0.014}{2.52}\) × 100 = 0.556.

Question 2.
While discussing the period of a pendulam, one of the student argued that period depends on the mass of the bob.

1. What is your opinion?
2. How will you prove your argument dimensionally?

Answer:

1. Period is independent of mass of the bob
2. The principle of homogeneity of dimensions also helps to derive a relationship between the different physical quantities involved; This method is known as dimensional analysis.

The period of the simple pendulum may possibly depend upon:

• The mass of the bob, m
• The length of the pendulum, I
• Acceleration due to gravity, g
• The angle of swing, q

Let us write the equation for the time period as t = k m^a l^b g^c q^d
where, k is a constant having no dimensions; a, b, care to be found out.
The dimensions of, t = T¹
Dimensions of. m = M¹
Dimensions of, l = L¹
Dimensions of, g = L¹T^-2
Angle q has no dimensions (since, q = arc/radius = L/L)
Equating the dimensions of both sides of the equation, we get,
T¹ = M^aL^b (L¹T^-2)^c
ie. T¹ = M^aL^b+c+ T^-2c.
The dimensions of the terms on both sides must be the same. Equating the powers of M, L and T.

a = 0; b + c = 0; -2c = 1
∴ c = ^–\(\frac{1}{2}\), b = ^– c = \(\frac{1}{2}\)
Hence, the equation becomes,
t = kl^1/2g^-1/2
ie, t = k\(\sqrt{1 / g}\)
Experimentally, the value of k is found to be 2p.

Plus One Physics Units and Measurement Five Mark Questions and Answers

Question 1.
In an experiment with a common balance the mass of a ring found to be 2.52g, 2.5g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate

1. The mean value of the mass of the ring
2. The absolute error in each measurement
3. Mean absolute error
4. Relative error
5. Percentage error

Answer:
1. The mean value of the mass of the ring.
M_mean = \(\frac{2.52+2.53+2.51+2.49+2.54}{5}\) = 2.52g.

2. The absolute error in each measurement.
∆m₁ = M_mean – m₁ = 2.52 – 2.52 = 0.00
∆m₂ = M_mean – m₂ = 2.52 – 2.53 = -0.01
∆m₅ = M_mean – m₅ = 2.52 – 2.54 = -0.02

3. mean absolute error = |∆m₁| + |∆m₂|………..+|∆m₅|
= 0.014

4. Relative error = δm = \(\frac{\Delta \mathrm{m}_{\text {mean }}}{\mathrm{m}_{\text {mean }}}=\frac{.014}{2.52}\) = 0.00555

5. Percentage error δm × 100 = 0.555%.

Plus One Physics Units and Measurement NCERT Questions and Answers

Question 1.
Fill in the blanks:

1. The volume of a cube of side 1 cm is equal to______m³.
2. The surface area of a solid cylinder of radius 2.0 cm and height 10.0cm is equal to____(mm)².
3. A vehicle moving with a speed of 18km h^-1 covers_____m in 1s.
4. The relative density of lead is 11.3.Its density is_____g cm^-3or_____kgm^-3.

Answer:
1. V = (1 cm)³
= (10^-2m)³
= 10^-6m³
So, answer is 10^-6.

2. Surface area = 2 πrh + 2 × πr²
= 2πr(h + r)
= 2 × \(\frac{22}{7}\) × 2 × 10(10 × 10 + 2 × 10)mm²
= 1 .5 × 10⁴mm²
So, answer is 1.5 × 10⁴

3. 18kmh^-1 = \(\frac{18 \times 1000}{3600}\)ms^-1
= 5ms^-1
So, answer is 5.

4. 11.3, 11.3 × 10³ or 1.13 × 10⁴.

Question 2.
Fill in the blanks by suitable conversion of units:

1. 1 kgm²s^-2 = _____g cm²s^-2
2. 1 m =_____1 y
3. 3.0 ms² =______kmh^-2
4. G = 6.67 × 10^-11 Nm² (kg)^-2 =_____(cm)³s^-2g^-1

Answer:

1. 10⁷
2. 10^-16
3. 3.888 × 10⁴
4. 6.67 × 10^-8

Question 3.
A calorie is a unit of heat or energy and it equals about 4.2 J, where 1 J = 1 kgm²S^-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β metere, the unit of time is second, show that a calorie has a magnitude 4.2 α^-1 β^-2 γ² in terms of the new units.
Answer:
1 cal = 4.2kg m²s^-2

Dimensional formula of energy is [ML²T^-2]. Comparing with [M^aL^bT^c], we find that
a = 1, b = 2, c = -2

Question 4.
Which of the following is the most precise device for measuring length?

1. A vernier callipers with 20 divisions on the sliding scale.
2. A screw guage of pitch 1 mm and 100 divisions on the circular scale
3. An optical instrument that can measure length to within a wavelength of light?

Answer:
The most precise device is one whose least count is the least.
1. Least count = 1SD – 1 VD = 1 SD – \(\frac{19}{20}\) SD

2. Least count

3. Wavelength = 10^-5 cm = 0.00001 cm
Clearly, the optical instrument is the most precise.

Question 5.
State the number of significant figures in the following:

1. 0.007m²
2. 2.64 × 10²⁴kg
3. 0.2370gcm³
4. 6.320 J
5. 6.032 Nm^-2
6. 0.0006032 m²

Answer:

1. 1
2. 3
3. 4
4. 4
5. 4
6. 4

Question 6.
The length, breadth, and thickness of a rectangular sheet of metal are 4.234m, 1.005m and 2.01cm respectively. Give the area and volume of the sheet to correct significant figures.
Answer:
Area of the sheet = 2(l × b + b × t + t × l)
= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)m²
= 2 (4.255 + 0.0202 + 0.0851)m²
= 2 × 4.3603m²
= 8.7206m²
= 8.72m²
Volume = lbt
4.234 × 1.005 × 0.0201m³
= 0.0855m³

Question 7.
A Physical qunatity P is related to four observables a, b, c and d as follows:
P = \(\frac{\mathrm{a}^{3} \mathrm{b}^{2}}{\sqrt{\mathrm{c}} \mathrm{d}}\). 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? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?
Answer:
P = \(\frac{\mathrm{a}^{3} \mathrm{b}^{2}}{\sqrt{\mathrm{c}} \mathrm{d}}\)

% error in P = 3% + 6% + 2%+2% = 13%
3.763 should be rounded off to 3.8.

Question 8.
A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m₀ of a particle in terms of its speed v and the speed of light c. (This relation first arose as a consequence of special relativity due to Albert Einstein). Boy recalls the relation almost correctly but forgets where to put the constant c. He writes:
\(\frac{m_{0}}{\left(1-v^{2}\right)^{1 / 2}}\)
Guess where to put the missing e.
Answer:
From the given equation, \(\frac{m_{0}}{m}=\sqrt{1-v^{2}}\)
Since left hand side is dimensionless therefore right hand side should be also dimensionless.

The correct formula is m = m₀ \((\sqrt{1-\frac{v^{2}}{c^{2}}})^{-1 / 2}\).

Students can Download Chapter 2 Units and Measurement Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 2 Units and Measurement

Plus One Physics Notes Chapter 2 Summary
Introduction

a. Fundamental or base quantities:
Physics is based on measurement of physical quantities. Certain physical quantities are chosen as fundamental or base quantities. Length, mass, time, electric current thermodynamic temperature, amount of substance and luminous intensity are such base quantities.

b. Units: Fundamental Units and Derived Units Unit:
Measurement of any physical quantity is made by comparing it with a standard. Such standard of measurement are known as unit. If length of rod is 5 m, it means that the length of rod is 5 times the standard unit ‘metre’.

Fundamental Unit:
The unit of fundamental or base quantities are called fundamental or base units. The base units are listed in table.

Derived Unit
The units of other physical quantities can be expressed as combination of base units. Such units are called derived units.
Example: Unit of force is kgms^-2 (or Newton). Unit of velocity is ms^-1.

The International System Of UnitsDerived Unit
System of Units: A complete set of fundamental and derived units is called a system of unit.

a. Different system of units:
The different systems of units are CGS system FPS (or British) system, MKS system and SI system. A comparison of these systems of unit is given in the table below, (for length, mass and time)

Note: The first three systems of units were used in earlier time. Presently we use SI system.

b. International System Of Unit (Si Unit):
The internationally accepted system of unit for measurement is system international d’ unites (French for International System of Units). It is abbreviated as SI.

The SI system is based on seven fundamental units and these units have well defined and internationally accepted symbols, (given in table – 2.1)

c. Solid Angle and Plane Angle:
Other than the seven base units, two more units are defined.
1. Plane angle (dq): It is defined as ratio of length of arc (ds) to the radius, r.


The unit of plane angle is radian. Its symbol is rad.

2. Solid Angle (dW): It is defined as the ratio of the intercepted area (dA) of spherical surface, to square of its radius.


The unit of solid angle is steradian. The symbol is Sr.

Unit And Measurement Class 11 Notes Pdf Chapter 2 Measurement Of Length
Two methods are used to measure length

• direct method
• indirect method.

The metre scale, Vernier caliper, screwgauge, spherometer are used in direct method for measurement of length. The indirect method is used if range of length is beyond the above ranges.

1. Measurement Of Large Distances:
Parallax Method:
Parallax method is used to find distance of planet or star from earth. The distance between two points of observation (observatories) is called base. The angle between two directions of observation at the two points is called parallax angle or parallactic angle (q).

Plus One Physics Chapter 2 Notes Pdf Parallax Method
The planet ‘s’ is at a distance ‘D’ from the surface of earth. To measure D, the planet is observed from two observatories A and B (on earth). The distance between A and B is b and q be the parallax angle between direction of observation from A and B.

AB can be considered as an arch A h B of length ‘b’ of a circle of radius D with its center at S. (Because q is very small, \(\frac{b}{D}\)<<1], Thus from arch-radius relation.

Thus by measuring b and q distance to planet can be determined. The size of planet or angular diameter of planet can be measured using the value of D. If the angle a (angle between two directions of observation of two diametrically opposite points on planet) is measured using a


Where d is diameter of planet.

2. Estimation Of Very Small Distances:
Size Of Molecule
Electron microscope can measure distance of the order of 0.6A0 (wavelength of electron).

3. Range Of Lengths:
The size of the objects in the universe vary over a very wide range. The table (given below) gives the range and order of lengths and sizes of some objects in the universe.

Units for short and large lengths
1 fermi = 1f = 10^-15m
1 Angstrom = 1A° = 10^-10m
1 astronomical unit = 1AU = 1.496 × 10¹¹m
1 light year = 1/y = 9.46 × 10¹⁵m
(Distance that light travels with velocity of 3 × 10⁸ m/s in 1 year)
1 par sec = 3.08 × 10¹⁶m = 3.3 light year
(par sec is the distance at which average radius of earth’s orbit subtends an angle of 1 arc second).

Units And Measurements Questions And Answers Pdf Hsslive Measurement Of Mass
Mass is basic property of matter. The S.l. unit of mass is kg. While dealing with atoms and molecules, the kilogram •is an inconvenient unit. In this case there is an important standard unit called the unified atomic mass unit( u).
1 unified atomic mass unit = lu
= (1/12)^th of the mass of carbon-12

1. Range Of Masses:
The masses of the objects in the universe vary over a very wide range which is given in the table.

Units And Measurements Class 11 Chapter 2 Measurement Of Time
To measure any time interval we need a clock. We now use an atomic standard of time, which is based on the periodic vibrations produced in a cesium atom. This is the basis of the cesium clock sometimes called atomic clock.

Plus One Physics Note Chapter 2 Definition of second:
One second was defined as the duration of 9, 192, 631, 770 internal oscillations between two hyperfine levels of Cesium-133 atom in the ground state.
Range and Order of time intervals

Class 11 Physics Chapter 2 Notes Accuracy, Precision Of Instruments And Errors In Measurement
Error:
The result of every measurement by any measuring instrument contains some uncertainty. This uncertainty is called error.

Systematic errors:
The systematic errors are those errors that tend to be in one direction, either positive or negative.

Sources of systematic errors

1. Instrumental errors
2. Imperfection in experimental technique or procedure
3. personal errors

1. Instrumental errors:
Instrumental error arise from the errors due to imperfect design or calibration of the measuring instrument.
eg: In Vernier Callipers, the zero mark of vernier scale may not coincide with the zero mark of the main scale.

2. Imperfection in experimental technique or procedure:
To determine the temperature of a human body, a thermometer placed under the armpit will always give a temperature lower than the actual value of the body temperature. Other external conditions (such as changes in temperature, humidity, velocity……..etc) during the experiment may affect the measurement.

3. Personal Errors:
Personal error arise due to an individual’s bias, lack of proper setting of the apparatus or individual carelessness etc.

Random errors
The random errors are those errors, which occur irregularly and hence are random with respect to sign and size. These can arise due to random and unpredictable fluctuations in experimental conditions (eg. unpredictable fluctuations in temperature, voltage supply, etc.)

Unit And Measurement Notes Pdf Chapter 2 Least Count Error
The smallest value that can be measured by the measuring instrument is called its least count. The least count error is the error associated with the resolution of the instrument. By using instruments of higher precision, improving experimental technique etc, we can reduce least count error.

1. Absolute Error, Relative Error And Percentage Error:
The magnitude of the difference between the true value of the quantity and the measured value is called absolute error in the measurement. Since the true value of the quantity is not known, the arithmetic mean of the measured values may be taken as the true value.

Physics Class 11 Chapter 2 Notes Explanation:
Suppose the values obtained in several measurements are a₁, a₂, a₃,………,a_n. Then arithmetic mean can be written as

The absolute error,
∆a₁ = a_mean – a₁
∆a₂ = a_mean – a₂
∆a_n = a_mean – a_n

a. Mean absolute error:
The arithmetic mean of all the absolute errors is known as mean absolute error. The mean absolute error in the above case,

b. Relative error:
The relative error is the ratio of the mean absolute error (Da_mean) to the mean value (a_mean).

c. Percentage error:
The relative error expressed in percent is called the percentage error (da).

Example:
Units And Measurements Class 11 Notes Pdf Download Question 1.
When the diameter of a wire is measured using a screw gauge, the successive readings are found to be 1.11 mm, 1.14mm, 1.09mm, 1.15mm and 1.16mm. Calculate the absolute error and relative error in the measurement.
Answer:
The arithmetic mean value of the measurement is

The absolute errors in the measurements are
1.13 – 1.14 = 0.02mm
1.13 – 1.14 = -0.01mm
1.13 – 1.09 = 0.04mm
1.13 – 1.15 =-0.02 mm
1.13 – 1.16 = 0.03mm
The arithmetic mean of the absolute errors

Percentage of relative error

2. Combination Of Errors:
When a quantity is determined by combining several measurements, the errors in the different measurements will combine in some way or other.

a. Error of a sum or a difference:
Rule: when two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute errors in the individual quantities.
Explanation:
Let two quantities A and B have measured values A ± DA and B ± DB respectively. DA and DB are the absolute errors in their measurements. To find the error Dz that may occur in the sum z = A + B,
Consider
z + ∆z = (A ± ∆A) + B ± ∆B = (A + B) ± ∆A ± ∆B
The maximum possible error in the value of z is given by,

Similarly, it can be shown that, the maximum error in the difference.
Z = A – B is also given by

b. Error of product ora quotient:
Rule: When two quantities are multiplied or divided, the relative error in the result is the sum of the relative errors in the multipliers.
Explanation:
Suppose Z=AB and the measured values of A and B are A + DA and B + DB. They
Z + DZ = (A + DA) (B + DB)
= AB ± BDA ± ADB ± DADB
Dividing LHS by Z and RHS by AB, we get

c. Errors in case of a measured quantity raised to a power:
Suppose Z = A²

Hence, the relative error in A² is two time the error in A.
In general, if \(Z=\frac{A^{P} B^{q}}{C^{T}}\)
Then

Hence the rule: The relative error in a physical quantity raised to the power K is the K times the relative error in the individual quantity.

Chapter 2 Physics Class 11 Notes Significant Figures
Every measurement involves errors. Hence the result of measurement should be reported in a way that indicates the precision of measurement.

Normally, the reported result of measurement is a number that includes all digits in the number that are known reliable plus the first digit that is uncertain. The reliable digits plus the first uncertain digit are known as significant digits or significant figures.
Example:

• The length of a rod measured is 3.52cm. Here there are 3 significant figures. The digits 3 and 5 are reliable and the last digit 2 is uncertain.
• The mass of a body measured as 3.407g. Here there are four significant figures. The figure 7 is uncertain.

When the measurement becomes more accurate, the number of significant figure is increased.
Rules to find significant figures:
1. All the non zero digits are significant.
Example:
Physics Notes For Class 11 Kerala Syllabus Chapter 2 Question 1.
Find significant figure of

• 2500
• 263.25

Answer:

• In this case, there are two nonzero numbers. Hence significant figure is 2.
• In this, there are 5 nonzero numbers. Hence significant figure is 5.

2. All the zeros between two nonzero digits are significant, no matter where the decimal point is,
Example:
Units And Measurements Class 11 Notes Chapter 2 Question 2.
Find the significant figure

• 2.05
• 302.005
• 2000145

Answer:

• Significant figure is 3
• Significant figure is 6
• Significant figure is 7

3. If the number is less than 1, the zeros on the right of decimal point but to the left to the first nonzero digits are not significant.
Example:
Class 11 Physics Notes Units And Measurements Chapter 2 Question 1.
Find the significant figure of

• 0.002308
• 0.000135

Answer:

• 4 significant figures
• 3 significant figures

4. The terminal zeros in a number without a deci¬mal point are not significant.
Example:
Physics Class 11 Chapter 2 Question 1.
Find the significant figure of

• 12300
• 60700

Answer:

• 3
• 3

Note: But if the number obtained is on the basis of actual measurement, all zeros to the right of last non zero digit are significant.
Example: If distance is measured by a scale as 2010m. This contain 4 significant figures.

5. The terminal zeros in a number with a decimal point are significant.
Example:
Gvhss Payyoli Physics Notes Chapter 2 Question 1.
Find the significant figure of

• 3.500
• 0.06900
• 4.7000

Answer:

• 4
• 4
• 5

Method to find significant figures through scientific notation:
In this notation, every number is expressed as a × 10^b, where a is a number between 1 and 10 and b is any positive or negative power. In this method, we write the decimal after the first digit.
Example:
4700m =4.700 × 10³m
The power of 10 is irrelevant to the determination of significant figures. But all zeros appearing in the base number in the scientific notation are significant. Hence each number in this case has 4 significant figures.
Significant figures in numbers:-

a. Rules for Arithmetic operations with significant figures:
1. Rules for multiplication or division:
In multiplication or division, the computed result should not contain greater number of significant digits than in the observation which has the fewest significant digits.
Examples:
(i) 53 × 2.021 =107.113
The answer is 1.1 × 10² since the number 53 has only 2 significant digits.

(ii) 3700 10.5 = 352.38
The answer is 3.5 × 10² since the minimum number of significant figure is 2 (in the number 3700)

2. Rules for Addition and Subtraction:
In addition or substraction of given numbers, the same number of decimal places is retained in the result as are present in the number with minimum number of decimal places.
Examples:
(i) 76.436 +
12.5
88.936
The answer is 88.9, since only one decimal place is found in the number 12.5.

(ii) 43.6495 +
4.31
47.9595
The answer is 47.96 since only two decimal places are to be retained.

(iii) 8.624 –
3.1726
5.4514
The answer is 5.451

(iv) 6.5 × 10^-5 – 2.3 × 10^-6 = 6.5 × 10^-5 – 0.23 × 10^-5
= 6.27 × 10^-5
The answer is = 6.3 × 10^-5

Units And Measurements Class 11 Notes Pdf Chapter 2 Dimensions And Dimensional Analysis
All physical quantities can be expressed in terms of seven fundamental quantities. (Mass, length, time, temperature, electric current, luminous intensity and amount of substance). These seven quantities are called the seven dimensions of the physical world.

The dimensions of the three mechanical quantities mass, length and time are denoted by M, L and T. Other dimensions are denoted by K (for temperature), I (for electric current), cd (for luminous intensity) and mol (for the amount of substance).

The letters [L], [M], [T] etc. specify only the nature of the unit and not its magnitude. Since area may be regarded as the product of two lengths, the dimensions of area are represented as [L] × [L] = [L]².

Similarly, volume being the product of three lengths, its dimensions are represented by [L]³. Density being mass per unit volume, its dimensions are M/L³ or M¹L³.

Thus, the dimensions of a physical quantity are the powers to which the fundamental units of length, mass, time must be raised to represent it.
Note: The dimensions of a physical quantity and the dimensions of its unit are the same.

Class 11 Physics Notes Chapter 2 Dimensional Formula And Dimensional Equations
An equation obtained by equating a quantity with its dimensional formula is called dimensional equations of the physical quantities.
Examples:
Consider for example, the dimensions of the following physical quantities.
1. Velocity: Velocity = distance/ time = L/T = L¹T^-1 \The dimension of velocity are, zero in mass, 1 in length and-1 in time.

2. Acceleration:
Acceleration = \(\frac{\text { Change in velocity }}{\text { time }}=\frac{L^{1} T^{-1}}{T}=L^{1} T^{-2}\)

3. Force: Force = mass × acceleration
Dimensions of force = M × L¹T^-2 = M¹L¹T ^-2
That is, the dimensions of force are 1 in mass, 1 in length and -2 in time.

4. Momentum: Momentum = mass × velocity
Dimensions of momentum = M × L¹T^-1 = M¹L¹T ^-1

5. Moment of a force: Moment = force × distance
Dimensions of moment = M¹L¹T^-2 × L = M¹L²T ^-2

6. Impulse: Impulse = force × time
Dimensions of impulse = M¹L¹T^-2 × T = M¹L¹T ^-1

7. Work: Work = force × distance
Dimensions of work = M¹L¹T^-2 × L = M¹L²T ^-2

8. Energy: Energy = Work done
Dimensions of energy = dimensions of work = M¹L²T^-2.

9. Power: Power = work/time
Dimensions of power \(=\frac{M^{2} L^{2} T^{-2}}{T}p\) = M¹L²T^-3

Explain Parallax Method Class 11 Chapter 2 Dimensional Analysis And Its Applications
The important uses of dimensional equations are:

1. To check the correctness of an equation.
2. To derive a correct relationship between different physical quantities.
3. To convert one system of units into another.

1. Checking the correctness of an equation:
For the correctness of an equation, the dimensions on either side must be the same. This ‘ is known as the principle of homogeneity of dimensions.

If an equation contains more than two terms, the dimensions of each term must be the same. Thus, if x = y + z, Dimensions of x = dimensions of y = dimensions of z
Example :
Units And Dimensions Class 11 Chapter 2 Question 1.
Check the correctness of the equation s = ut + 1/2at² by the method of dimensions.
Dimensions of, s = L¹
Dimensions of, u = L¹T^-1
Dimensions of, ut = L¹T^-1 × T¹ = L¹
Dimensions of, a = L¹T^-2
Dimensions of, at² = L¹T^-2 × T² = L¹
The constant 1/2 has no dimensions. Each term has dimension L¹.
Therefore, dimensions of, ut + 1/2 at² = ¹
Thus, either side of the equation has the same dimen¬sion L1 and hence the equation is dimensionally correct.
Note: Even though the equation is dimensionally correct, it does not mean that the equation is necessarily correct. For instance the equation s = ut + at² is also dimensionally correct, though the correct equation, s = ut + 1/2 at².

2. Deriving the correct relationship between different physical quantities:
The principle of homogeneity of dimensions also helps to derive a relationship between the different physical quantities involved. This method is known as dimensional analysis.
Example :
Question 1.
Deduce an expression for the period of oscillation of a simple pendulum.
The period of the simple pendulum may possibly depend upon

• The mass of the bob, m
• The length of the pendulum, I
• Acceleration due to gravity, g
• The angle of swing, q

Let us write the equation for the time period as t = km^a l^b g^c θ^d
where, k is a constant having no dimensions; a, b, c are to be found out. ’
The dimensions of, t = T¹
Dimensions of m = M¹
Dimensions of, l = L¹
Dimensions of, g = L¹T^-2
Angle q has no dimensions (since, q = arc/radius = L/L) Equating the dimensions of both sides of the equation, we get,
T¹ = M^aL^b (L¹T^-2)^c
ie. T¹ = M^aL^b+cT^-2c
The dimensions of the terms on both sides must be the same. Equating the powers of M, L and T.
a = 0; b + c = 0; -2c = 1
∴ c = \(\frac{-1}{2}\), b = ^–c = \(\frac{1}{2}{/latex]
Hence, the equation becomes,
t = kl^1/2, 2g^-1/2
ie, t = k[latex]\sqrt{l/g}\)
Experimentally, the value of k is found to be 2p.
Limitations of Dimensional Analysis:
The method of dimensional analysis has the following limitations:

• It gives no information about the dimensionless constant involved in the equation.
• The method is not applicable to equations involving trigonometric and exponential functions.
• This method cannot be employed to derive the exact form of the relationship, if it contains sum
  of two, or more terms.
• If the given physical quantity depends on more than three unknown quantities, the method fails.

3. Conversion of one system of units to another:
Suppose we have a physical quantity of dimensions a, b and c in mass, length and time. The dimensional formula for the quantity is therefore, M^aL^bT^c. Let its numerical value be n, in one system in which the fundamental units of mass, length and time are M₁, L₁ and T₁ respectively. Then, the magnitude of the physical quantity
= n₁ M₁^aL₁^bT₁^c
Also, let the numerical value of the same quantity be n₂ in another system where the fundamental units of mass, length and time are M₂, L₂ and T₂respectively. Then the magnitude of the quantity
= n₂ M₂^aL₂^bT₂^c
Equating, n₂ M₂^aL₂^bT₂^c =
n₁ M₁^aL₁^bT₁^c

Example :
Question 1.
Find the number of dynes in one newton.
Answer:
Dyne is the unit of force in the C.G.S. system and newton is the S.I.unit. The dimensional formula for force is M¹L¹T^-2. In eqn. (1) let the suffix 1 refer to quantities in S.I and 2 those in the C.G.S. system.
Here, a = 1, b = 1 and c = 2

and n₁ = 1 (ie. one Newton)
By eqn. (1),
n₂ = 1 (1000)¹ (100)¹ (1)^-2 = 10⁵
ie. 1 newton = 10⁵ dynes.