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

Students can Download Chapter 8 Financial Statements – I & Financial Statements – II Questions and Answers, Plus One Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

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

Plus One Accountancy Financial Statements – I & Financial Statements – II One Mark Questions and Answers

Question 1.
Closing stock is valued at
(a) Market Price
(b) Cost Price
(c) Market Price or Cost Price whichever is less.
Answer:
(c) Market Price or Cost Price whichever is less.

Question 2.
Carriage inward is debited to …………… account.
(a) Trading account
(b) Profit and Loss Account
(c) Cash Account
Answer:
(a) Trading Account

Question 3.
Balance Sheet is prepared to find out ……………
(a) Capital
(b) Net Profit
(c) Financial position
Answer:
(c) Financial Position

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

Question 4.
Gross profit is the difference between ……………..
(a) Sales and Purchases
(b) Sales and total cost
(c) Sales and cost of goods sold
Answer:
(c) Sales and cost of goods sold

Question 5.
Interest on capital is ……………… to the business.
(a) Income
(b) Expenses
(c) Asset
Answer:
(b) Expenses

Question 6.
Wages paid before it has become due is shown in the balance sheet as ……………..
(a) Asset
(b) Liability
(c) Expenses
Answer:
(a) Asset

Question 7.
Ameer’s trial balance contains the following information.

  • Bad debt Rs.400
  • Provision for bad debts Rs. 1600

It is desired to maintain a provision for bad debts at Rs. 1500. The amount to be debited to profit and loss a/c is ………….
(a) Rs. 3500
(b) Rs. 100
(c) Rs. 300
Answer:
(c) Rs. 300

Question 8.
Profit from Profit and Loss account is transferred to …………… account.
(a) Asset account
(b) Capital account
(c) Liability account
Answer:
(b) Capital Account

Question9.
The Financial statements consist of:
(a) Trial Balance
(b) Profit and Loss account
(c) Balance sheet
(d) Both b and c
Answer:
(d) Both b and c

Question 10.
Choose the correct order of ascertainment of the fol¬lowing profits from the profit and loss account:
(a) Operating profit. Net Profit, Gross profit
(b) Operating profit, Gross Profit, Net Profit
(c) Gross Profit, Operating Profit, Net Profit
(d) Gross Profit, Net Profit, Operating Profit
Answer:
(c) Gross profit, Operating profit, Net Profit

Question 11.
Which of the following are not taken into account at the time of calculation of operating profit?
(a) Normal transactions
(b) Abnormal items
(c) Expenses of a purely financial nature
(d) Both a and c
Answer:
(c) Expenses of a purely financial nature

Question 12.
If the insurance premium paid Rs. 1,000 and prepaid insurance of Rs. 400. The amount of insurance premium shown in profit and loss a/c will be:
(a) Rs. 1400
(b) Rs. 1000
(c) Rs. 400
(d) Rs. 600
Answer:
(d) Rs. 600

Question 13.
When the Manager is entitled to a commission of 10% on profits after charging such commission, it is calculated on profits before charging commission by the formula ………….
(a) 10/90
(b) 10/100
(c) 10/110
(d) 90/100
Answer:
(c) 10/110

Question 14.
Profit and Loss Account is an account prepared to find out ………………
Answer:
Net Profit or Net Loss.

Question 15
Income tax paid for the trader from the business is treated as …………..
Answer:
Drawings.

Question 16.
Adjustments are made in the final accounts to satisfy ………. principle of accounting
Answer:
Matching Principle.

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

Question 17.
Free samples distributed among customer’s will be credited to ……….. account.
Answer:
Advertisement Account.

Question 18.
…………….. is an expenditure of the revenue nature in the current year, the benefit of which accrues gradually or lasts for more than one accounting year.
Answer:
Deferred Revenue Expenditure

Question 19.
EBIT stands for …………………
Answer:
Earnings Before Interest and Tax

Plus One Accountancy Financial Statements – I & Financial Statements – II Two Mark Questions and Answers

Question 1.
Find the odd one and state the reason.

  1. Wages, carriage, Trade expenses, Freight
  2. Income tax paid, Salary paid, Rent paid, Wages paid

Answer:

  1. Trade expense – It is an indirect expense, all others are direct expenses.
  2. Income tax paid – It is a personal expense of the Proprietor, all others are business expenses.

Question 2.
The sales and cost of goods sold by John Brothers are Rs. 40,000 and Rs. 28,000 respectively. What is the amount of his gross profit?
Answer:
Gross profit = Sales – Cost of goods sold
= 40,000 – 28,000 = Rs. 12,000

Question 3.
What are the financial statements?
Answer:
The term Financial Statements generally refers to two statements prepared at the end of an accounting period for an enterprise. These are Trading and Profit and Loss Account, showing profitability of the business operations and Balance sheet, showing the financial position of the enterprises.

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

Question 4.
A new manufacturer incurred huge expenditure in advertisement. Explain the nature of expenditure.
Answer:
Deferred Revenue Expenditure:
“Expenses incurred today, and the benefit of which accrue gradually in subsequent year is called Deferred Revenue Expenditure. For example Expenses like advertisement, which may be incurred in one lump sum but the benefit of which will be received only within a series of years. The proportionate amount due the current year must be arrived at and debited to the P/L account. The balance must be shown on the assets side of the Balance sheet.

Question 5.
Calculate the value of cost of goods sold.

  • Opening stock – 10,000
  • Net purchases – 5,000
  • Direct expenses – 2,000
  • Closing stock – 10,500

Answer:
Cost of goods sold = Opening stock + Net Purchases + Direct Expenses – Closing stock
= (10,000 + 5000 + 2000) – 10500 = Rs. 6,500

Question 6.
What is Operating profit?
Answer:
Operating profit is the profit earned through the normal operations and activities of the business. Operating profit can be calculated as follows:
Operating profit = (Gross profit – Operating expense) + Operating income.

Or

= Net Profit + Non-operating expenses – Non-operating income.
Examples of non-operating expenses are loss on sale of assets, interest paid etc. Non-operating incomes are dividend received, Profit on sale of assets.

Question 7.
Operating profit earned by M/s. Asoka & Sons in 2010-11 was Rs. 17,00,000. Its non-operating incomes were Rs. 1,50,000 and non-operating expenses were Rs.3,75,000. Calculate the amount of net profit earned by the firm.
Answer:
Net profit = Operating profit – Non-operating expenses + Non operating income = 1700000 – 375000 + 150000 Net Profit =Rs. 1475000

Question 8 .
Explain the treatment of goods distributed as sample to customers.
Answer:
In order to increase sales, goods may be distributed among customers free of cost. Such free goods are in the form of samples for test use. It is in the form of advertisement expense. The amount is to be deducted from the purchases in the trading account. The cost of the sample is to be shown as an expenses in the profit and loss account.

Question 9.
To tally a balance sheet, one has to strictly adhere to the basic accounting equation ‘Asset = Liabilities + Captial’. Do you agree with this statement? Substantiate.
Answer:
Every transaction has two aspects which are equal in value but opposite in nature. One of the aspects will form part of liability and the other one asset. Be¬cause of this feature of accounting equation, a balance sheet is always tallied.

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

Question 10.

  • Closing stock – Rs. 30,000
  • Sales – Rs. 1,00,000

Gross Profit 20% on sale

  • Purchases – Rs. 60,000
  • Direct Wages – Rs. 7000

Find out opening stock.
Answer:
Opening stock = Sales + Closing stock – (Purchases + Direct Expense + Gross Profit)
= 1,00,000 + 30,000 – (60,000 + 7,000 + 20,000)
= 1,30,000 – 87,000 = 43,000

Question 11.
Provide provision for bad debt @ 10% on debtors from the particulars given below.

  • Good Debts Rs. 10,000
  • Doubtful debts Rs. 50,000
  • Bad debts Rs. 1000

Answer:
Provision for doubtful debts = Doubtful debts × % of provision for bad debt
= 50000 × 10/100 = Rs. 5000

Question 12.
Payment received from debtors does not result a change in the total assets. Comment.
Answer:
When the payment is received from debtors the asset, cash is increased and also a corresponding decrease happens on the asset, Debtors. Therefore, there won’t be any change in the value of total assets in effect.

Question 13.
What is meant by provision for discount on debtors?
Answer:
This is a discount which is being allowed by an enterprise to its debtors to encourage prompt payments. Discount likely to be allowed to customers in an accounting year can be estimated and provided for by creating a provision for discount on debtors. Here, it is to be remembered that provision for discount made on good debtors which are arrived at by deducting further bad debts and the provision for doubtful debts.

Plus One Accountancy Financial Statements – I & Financial Statements – II Three Mark Questions and Answers

Question 1.
Distinguish between capital and revenue expenditure.
Answer:
1. Capital expenditure increases earning capacity of business whereas revenue expenditure is incurred to maintain the earning capacity.

2. Capital expenditure is incurred to acquire fixed assets for operation of business whereas revenue expenditure is incurred on the day-to-day conduct of business.

3. Revenue Expenditure is recurring expenditure but capital expenditure is non-recurring by nature.

Question 2.
Differentiate between Direct and Indirect Expenses.
Answer:
1. Direct Expenses:
Means all expenses directly connected with the manufacture, purchase of goods and bringing them to the point of sale. Direct expenses include carriage inward, freight inwards, octroi, clearing charges, wages, factory lighting, coal, water, gas, fuel, import duty, cotton waste, royalty on production, heating, dock dues, customs duty etc.

2. Indirect expenses:
Are those expenses which are incurred after the manufacturing of goods. In other words, Indirect expenses are those expenses that are incurred to operate a business as a whole. Indirect expense include – carriage outward, rent, rates and tax, office expenses, selling and distribution expenses.

Question 3.
Mention the difference between a Balance Sheet and a Trial Balance. (Any two points).
Answer:

Trial Balance Balance Sheet
1. It is a list of all account balances 1. It is prepared with the balances of real and personal accounts only.
2. It is prepared to check the arithmetical accuracy of books of accounts. 2. It is prepared to ascertain the financial position of the firm.
3. It is prepared frequently. 3. It is usually prepared annually.

Question 4.
Calculate closing stock from the following

  • Sales – 20,000
  • Purchases – 12,300
  • Return inwards – 500
  • Carriage inwards – 400
  • Return outwards – 1,000
  • Gross Profit – 8000

Answer:
Trading Account

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

Question 5.
On 1st January 2008, a firm had a stock of goods valued at Rs. 10,000. During the year the following transactions took place.

  • Sales – 2,00,000
  • Purchases – 1,20,000
  • Carriage inwards – 500
  • Sales returns – 2,000
  • Purchase returns – 1,000
  • Find out the amount of Gross Profit.

Answer:

  1. Gross Profit = Net Sales – Cost of goods sold
  2. Net sales = 2,00,000 – 2,000 = 1,98,000
  3. Net Purchase = 1,20,000 -1000 = 1,19,000.
  4. Cost of goods sold = 10,000 + 1,19,000 + 500 = 1,29,500
  5. Gross Profit = 1,98,000 – 1,29,500 = Rs. 68,500

Question 6.
What do you mean by Profit and Loss Account.
Answer:
Profit and Loss a/c is an account prepared to find out the net profit earned or net loss incurred by a business during an accounting period. It is debited with all operating expenses and losses and credited with incomes and profit. This account begins with the Gross Profit or Gross Loss brought down from Trading A/c.

If the total of the credit side of this account is more than the total of the debit side, the difference is net profit. If the total of the debit side exceeds the total of the credit side, the difference is net loss. The amount of net profit or net loss transferred to capital account.

Question 7.
From the following information calculate operating profit.

  • Cost of goods sold – Rs. 5,00,000
  • Administrative expenses – 25,000
  • Selling & Distribution expenses – 35,000
  • Net sales – 7,50,000

Answer:

  • Operating Profit = Gross Profit – (Operating expenses + Operating income)
  • Gross Profit = Net sales – Cos of goods sold = 7,50,000-5,00,000 = 2,50,000
  • Operating Expenses = Administration expense + Selling and Distribution exp.
  • Operating Profit = 2,50,000 – (25,000+ 35,000) = Rs. 1,90,000

Question 8.
What do you mean by Deferred Revenue Expenditure? Can you present it with a suitable example?
Answer:
“Deferred Revenue Expenditure is an expenditure of the revenue nature in the current year, the benefit of which accrues gradually or lasts for more than one accounting year.”

Examples are advertisements of usually high amounts, cost of shifting business to a more convenient location, etc. The amounts spend on such expenses would be spread over the period for which the benefit arises. Suppose if a concern spends Rs. 5 lakh for advertisement and it is expected that the benefit of its lasts for 5 years, the amount to be treated as current year’s expenditure is only one fifth (1/5) of Rs. 5 lakhs ie. Rs. 1 lakh. The balance Rs. 4 lakhs would be treated as an asset.

Question 9.
State the reasons why the following items appear or not in the Profit and Loss A/c.

  1. Bad debts
  2. Drawings
  3. Provision for bad debts.

Answer:

  1. Bad debts is an indirect expenditure and hence it appears in the Profit and Loss A/c.
  2. Drawings will reduce the capital a/c. Therefore, it cannot be shown in the P/L a/c.
  3. Provision for bad debt is a charge against profit. Hence it will appear in the P/LA/c.

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

Question 10.
Mr. Narayanan, an accountant of Samay Ltd wrote off Rs. 2000 as bad debts in the year 2004. The total
Sundry debtors for the year 2005 is Rs. 26,000. During this year half of the bad debts written off in the last year were recovered. Give Journal Entry for the recovery of bad debts and also show how it will be dealt in the financial statement for the year 2005.
Answer:
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 2

Question 11.
What are adjusting entries? Why are they necessary for preparing final accounts?
Answer:
Entries which are given outside the trial balance are called adjustment entries, to record those entries a proper treatment is required according to the double-entry system. Here it is to be remembered that all the adjustments given outside the Trial Balance are posted at two places.

Adjustment is generally done for those items which are omitted or entered with the wrong amount and/or recorded under wrong heads. The following are reasons for recording or. incorporating these adjustment entries in preparation of final account.

  1. Through these adjustment entries, we come to know the actual figure of profit or loss.
  2. Because of these adjusting entries, we can assess the true financial position of an organisation based on accrual basis of accounting.
  3. These adjustment entries enable us to records the omitted entries and help in rectifying all those errors.
  4. These adjusting entries help in providing depreciation and making different provisions, such as Bad Debts and Depreciation.

Question 12.
What is meant by provision for doubtful debts? How are ‘the relevant accounts prepared and what journal entries are recorded in final accounts? How is the amount for provision for doubtful debts calculated?
Answer:
Provision for doubtful debts is a kind of arrangement about the expect bad debts from the debtors. Generally it is provided after deducting the amount of bad debts from the debtors. As provision for doubtful debts is made after preparing the trial balance, to record it we need a kind of adjustment entry in this regard we prepare debtors account and provision for doubtful debts account. For recording bad debts, the following journal entry is passed.

  • Profit and Loss A/c Dr
    To Provision for Doubtful Debts A/c (Being provision for doubtful debts is created out of current year profits)

Computation of the Amount of Provision for Doubtful Debts As it is given at the end of a trial balance as an adjustment, little another related adjustment may be there for instance bad debts and discount on debtors. In this case provision for doubtful debt will be created after deducting the figure for bad debts out of the debtor figure.

Plus One Accountancy Financial Statements – I & Financial Statements – II Four Mark Questions and Answers

Question 1.
What is Trading Account? Explain its purpose.
Answer:
The trading account is an account which shows the results of buying and selling of goods or services. It contains summarized form of all the transactions occurring during a trading period. This account is credited with direct incomes and debited with direct expenses.

Trading account is prepared to ascertain the gross result of the business. The gross result of the business is either gross profit or gross loss. If the net sales exceeds cost of goods sold then there is gross profit and if the opposite takes place, there is a gross loss.

Gross profit = Net sales – Cost of goods sold Gross Loss = Cost of goods sold – Net Sales Purposes of a trading account are:

  1. To ascertain the gross profit or gross loss.
  2. To enable the management to make a comparison of gross profit or gross loss with that of the previous year.
  3. To ascertain different ratios such as gross profit ratio, ratio of cost of goods sold to sales etc.

Question2.
From the following details, prepare Trading Account.

  • Opening stock – Rs. 12500
  • Purchases – Rs. 22,000
  • Purchase returns – Rs.2000
  • Wages – Rs. 2000
  • Carriage inwards – Rs. 500
  • Sales – Rs. 57,000
  • Sales returns – Rs. 3,000
  • Closing stock – Rs. 15,000

Answer:
Trading Account for the year ended ………..
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 3

Question 3.
What is a Balance Sheet? Explain the needs for preparing Balance Sheet.
Answer:
A Balance Sheet is a statement prepared to ascertain the true position of assets and liabilities as on a particular date. It is prepared at the end of the accounting period, after the preparation of Trading and Profit and Loss account. It is called Balance sheet, as it is a statement prepared with the balance of accounts left after the preparation of Trading and Profit and Loss account.

It gives clear picture of the financial position of the concern. Accounts of Assets, liabilities and Owner’s equity are shown in the Balance Sheet. Items of liabilities and capital are shown on the left side, known as “liabilities” side and the item of assets are shown on the right-hand side, known as “Assets” side of the balance sheet. Balance sheet is prepared with the following objectives.

  1. To ascertain the financial position of the concern.
  2. To ascertain the nature of assets and liabilities of the firm.
  3. To know about the source and application of funds.
  4. To ascertain working capital as on the date of Balance sheet.
  5. To ascertain the excess of assets over external liabilities.

Question 4.
What do you mean by Grouping and Marshalling of assets and liabilities?
Answer:
1. Grouping:
The term grouping means putting together items of similar nature under a common heading in the Balance sheet.

2. Marshaling:
Marshaling denotes the order in which the assets and liabilities are shown in the Balance sheet. They are arranged in the following two different ways.

a. In the order of liquidity:
Liquidity means the capacity to raise cash, Under these approach assets are presented in the order of their liquidity. ‘Cash’ being the most liquid item, it is shown as the first item whereas the least liquid item such as ‘Goodwill’ is shown as the last one. The most urgent liability is shown first and the least urgent to pay is shown last.

b. In the order of permanence:
Linder this approach, permanent assets, and liabilities are shown first followed by current assets and liabilities. Joint Stock Companies have to prepare their Balance sheet in the order of permanence. This is just the reverse of the order of liquidity.

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

Question 5.
Show the treatment in financial statements in respect of the following:

  1. Outstanding expenses
  2. Managers commission
  3. Interest on capital

Answer:
1. Outstanding expenses:
Expenses that have been incurred during the current year, but the payment has not been made is called outstanding expenses. It must be added to respective expense account in the Trading and Profit and Loss account. It will be shown on the ‘liability’ side of the Balance sheet.

2. Managers Commission:
Commission on net profit, at a specific percentage, may be allowed to the manager of a business concern. The commission as a percentage of the net profit may be ‘before’ or ‘after’ charging such commission. In the absence of any special instructions, it is assumed that commission is allowed as a percentage of the net profit before charging such commission.

  • If the commission is on the net profit before charging such commission, the formula is
    Profit before commission × \(\frac{\text { rate of commission }}{100}\)
  • If the commission is on the net profit after charging such commission the formula is
    Profit × rate/(100 + rate)
  • The amount of commission must be debited to Profit and Loss account and it must be shown as a liability in the Balance sheet.

3. Interest on capital:
Sometimes interest is paid on the Proprietor’s capital. Interest is allowed at a certain rate on the capital at the beginning of the year. Such interest is an expense to the business and is debited to Profit and Loss Account. It is shown in the liability side by adding the same to capital.

Plus One Accountancy Financial Statements – I & Financial Statements – II Five Mark Questions and Answers

Question 1.
From the following figures, prepare profit and loss account of M/s. Thomas and Sons for the year ended 31.03.2008.

  • Gross Profit – 15000
  • Printing charges – 750
  • Salaries – 5000
  • Carriage outwards – 500
  • Interest received – 2000
  • Bad debts – 500
  • Insurance charges – 400
  • Discount allowed 650
  • Discount received – 750
  • Advertisement – 440

Answer:
Profit and Loss Account of M/s. Thomas & Sons for the year ended 31.03.08
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 4

Question 2.
Arrange the following assets and liabilities in the order of liquidity and in the order of permanence.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 5
Answer:
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 6
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 7

Question 3.
Write the adjustment entries for the following:-
a) Salary outstanding Rs. 5000.
b) Insurance prepaid Rs. 5100.
c) Depreciation of Machinery Rs. 4000
d) The commission received in advance Rs. 1000
e) Interest on drawing Rs. 400
Answer:
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 8

Question 4.
What are the closing entries? Give examples.
Answer:
The preparation of trading and profit and loss account requires that the balances of accounts of all concerned items are transferred to it for its compilation. For transferring the balance of all the ledger account to concerned head is done through closing entries.
For examples:
1. Opening stock account, purchase account, wages account, carriage inward account, and direct expense account are closed by transferring to the debit side of the trading and profit and loss account.
The journal entry is:

  • Trading A/c Dr
  • To opening stock A/c
  • To Purchase A/c
  • To Wages A/c
  • To Carriage inward A/c
  • To Direct Expense A/c

2. The purchase return account is closed by transferring its balance to the purchase account. The journal entry is:

  • Purchase Return A/c Dr
  • To Purchase A/c

3. The sales return account is closed by transferring its balance to the sales account as:

  • Sales A/c Dr
  • To Sales Return A/c

4. The Sales account is closed by transferring its balance to the credit side of the trading and profit and loss account.
The Journal entry is:

  • Sales A/c Dr
  • To Trading A/c

Plus One Accountancy Financial Statements – I & Financial Statements – II Six Mark Questions and Answers

Question 1.
The following are the extracts from Trial Balance of a business.

  • Sundry Debtors = 40,000
  • Bad debts = 4,000
  • Provision for bad debts = 5,000

Additional Information:

  1. Provide further bad debts Rs. 2000
  2. Create 10% provision for bad debts.

Pass Journal entries and show how these items will appear in the final accounts.
Answer:
Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 9
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 10

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

Question 2.
Show the treatment of prepaid expenses,
depreciation, closing stock at the time of preparation of final accounts.

  1. When given inside the trial balance.
  2. When given outside the trial balance.

Answer:
Treatment of prepaid expenses, depreciation and closing stock at the time of preparing the final account.
1. When Given Inside the Trial Balance Prepaid Expenses:
When prepaid expenses are given in the trial balance itself it will be treated as current assets only and will be posted in the Assets Side of the balance sheet. No further adjustment will be required in this case.

Depreciation:
When depreciation is given in the trial balance it will be treated as an expenditure and will be shown in the debit side of the Profit and Loss Account. No further adjustment will be required in this case.

Closing Stock:
When closing stock is given in the trial balance it will purely be treated as assets and will be shown only in the Assets Side of the Balance sheet. No further adjustment will be required in this case.

2. When Given Outside the Trial Balance Prepaid Expenses:
When prepaid expenses are given outside the trial balance it will be treated as an Adjustment and will be posted at two places, first of all, it will be deducted from the concerned expenses in the debit side of Profit and Loss Account and after that, it will be treated as current assets and will be posted in the Assets side of the Balance Sheet.

Depreciation:
When depreciation is given outside the trial balance be treated as an adjustment and will be posted at two places to comply with the rules of the double-entry bookkeeping system. First of all the amount of depreciation will be shown in the debit side of Profit and Loss account as an expenditure and the amount of depreciation will be deducted from the concerned assets in the assets side of Balance Sheet.

Closing Stock:
When closing stock is given outside the trial balance it will purely be treated as an adjustment and will be posted at two places first of all the amount of closing stock will be shown at the credit side of Trading Account and after that it will be shown as an assets in the Assets side of the Balance Sheet.

Plus One Accountancy Financial Statements – I & Financial Statements – II Eight Mark Questions and Answers

Question 1.
The following is the extract takes from the Trial Balance of Vimal.
Trial Balance as on 31.03.2008
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 11
Adjustments:

  1. Salary outstanding has not been recorded – Rs.8000
  2. Prepaid insurance was meant for Proprietor’s son.
  3. Write off further Rs. 200 as bad debts and make a provision for doubtful debts @ 5%.
  4. Depreciate furniture @ 10%.

You are required to prepare Trading and Profit and Loss Account for the year ended 31.03.2008 and Balance Sheet as on the date.
Answer:
Trading and Profit and Loss a/c for the year ended 31.03.2008
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 12
Balance sheet as on 31.03.2008
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 13

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

Question 2.
Prepare Trading and Profit and Loss a/c for the year ended 31.03.05 and Balance sheet as on that date from the following balance.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 14
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 15
Answer:
Trading and Profit and Loss A/c of Mr. Babu for the year ended 31.03.05
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 16
Balance Sheet of Mr. Babu as on 31.03.05
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 17

Question 3.
The following is the Trial Balance of Vineeth as on 30.06.2008.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 18
Additional Information:

  1. Closing stock is valued at Rs. 8750.
  2. Provide 5% of debtors for bad debts and 2% of debtors and creditors for discount.
  3. Provide interest on capital at 10% and charge interest on drawings at 5%.
  4. Depreciate delivery van by 20%.
  5. Only one-fifth of advertisement is to be treated as expenses of current year.

Prepare Trading and Profit and Loss A/c for the year ended 30.06.2008 and also a Balance Sheet as on that date.
Answer:
Trading and Profit and Loss A/c for the year ended 30.06.08
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 19
Balance Sheet as on 30.06.2008
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 20
Note: Discount on Debtors
= (20000-1000) × 2/100 = 380
Discount on Creditors = 10500 × 2/100 = 210

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

Question 4.
Prepare Trading and Profit and Loss Account for the year ended 31.03.2005 and a Balance Sheet as on that date from the following balances.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 21
Adjustments:

  1. Stock on 31.03.05 Rs. 72,600.
  2. Depreciate Plant and Machinery by 33 1/3%, Furniture by 10% and Freehold property by 5%.
  3. Loose tools valued at Rs. 1,760 on 31.3.05.
  4. Of the Sundry debtors Rs. 660 are bad and should be written off.
  5. Maintain a provision of 5% debtors for doubtful debts.
  6. The Manager is entitled to a commission of 10% of the net profits after charging such commission.

Answer:
1. Trading and Profit and Loss A/c for the year ended 31.03.2005
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 22
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 23
Balance Sheet as on 31.03.2005
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 24
Note: Manager’s Commission:
Net Profit before commission = 49280 (111870-62590)
Therefore, Managers commission @ 10% of Net Profit after charging commission = 49280 × 1°/110 = 4480

Question 5.
The following is the Trial Balance of Balu as on 31/12/2011.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 25
Adjustments:

  1. Closing stock Rs. 9,500
  2. Depreciate plant and machinery at 10% p.a
  3. Interest on loan at 12% is due for the whole year.
  4. Write off further Rs.400 as bad debts and provision for bad debts is to be made equal to 5% on debtors.
  5. Provide 2% for discount on debtors.

Prepare Trading and Profit and Loss A/c for the year ended and Balance sheet as on 31/12/2011.
Answer:
Trading and Profit and Loss Account for the year ended 31/12/2011
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 26
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 27
Balance sheet as on 31/12/2011
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 28
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 29

Question 6.
The following is the Trial Balance of Mahesh as on 31/3/11.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 30
Adjustments:

  1. Closing stock Rs. 65,000
  2. Insurance prepaid Rs. 400
  3. Commission earned but not received amounts to Rs. 300
  4. Write off 1/5 of leasehold premises and provide a depreciation of 10% on plant and machinery.
  5. Debtors include goods of the cost of Rs. 4,000 sent on sale or return basis at an invoice price of Rs. 5,000. The goods are likely to be returned.
  6. Transfer 10% of net profit to reserve. Prepare Trading and Profit and Loss A/c and Balance Sheet.

Answer:
Trading and Profit and Loss A/c for the year ended 31/3/2011.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 31
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 32
Balance Sheet as on 31/03/2011
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 33
Note:
Reserve = 10% of net profit
79,500 -16,800 = 62,700 x 10% = 6,270

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

Question 7.
From the following balance extracted from the books of M/s. Hariharan Brother, you are required to prepare the trading and profit and loss account and a balance sheet as on December 31, 2005.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 34
Adjustments:

  1. Closing stock Rs. 14000
  2. Wages outstanding Rs.600, Salaries outstandings Rs. 1,000, Rent outstanding Rs. 200.
  3. Fire Insurance premium includes Rs.1200 paid in July 01, 2005, to run for one year from July 01, 2005, to June 30, 2006.
  4. Apprenticeship Premium is for three years paid in advance on January 01, 2005.
  5. Stationery bill for Rs. 60 remain unpaid.
  6. Depreciation on Premises @ 5%, furniture @ 10%, Machinery @ 10%.
  7. Interest on loan given accrued for one year @ 7%.
  8. Interest on investment @ 5% for half year to December 31, 2005, has accrued.
  9. Interest on capital to be allowed at 5% for one year.
  10. Interest on drawings to be charged to him ascertained for the year Rs. 160.

Answer:
Books of Hariharan Bros.
Trading and Profit and Loss account for the year ended December 31, 2005
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 35
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 36
Balance Sheet as at December 31, 2005
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 37

Question 8.
Prepare the trading and profit and loss account of M/s.Roni Plastic Ltd. from the following trial balance and a balance sheet as at March 31,2006.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 38
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 39
Adjustments:

  1. Depreciation on land and building at 5% and Motor vehicle at @ 15%.
  2. Interest on loan is @ 5% taken on April 01, 2005
  3. Goods costing Rs. 1200 were sent to a customer on sale on return basis for Rs.1400 on March 30, 2006, and has been recorded in the books as actual sales.
  4. Salaries amounting to Rs. 1400 and Rates amounting to Rs. 800 are due.
  5. The bad debts provision is to be brought up to @ 5% on Sundry debtors.
  6. The closing stock was Rs. 13,700.
  7. Goods costing Rs. 1,000 were taken away by the proprietor for his personal use but not entry has been made in the books of account.
  8. Insurance prepaid Rs. 350.
  9. Provide the Manager’s commission at @ 5% on Net profit after charging such commission.

Answer:
Books of Roni’s Plastic Ltd.
Trading and Profit and Loss account for the year ended December 31, 2006
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 40
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 41
Note: New provision for bad debt = (38200 -1400) 5/100 = 1840
Balance Sheet as on 31.3.2006
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 42

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

Question 9.
From the following balances extracted from the books of Raga Ltd. prepare a trading and profit and loss account for the year ended December 31, 2011, and a balance sheet as on that date.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 43
The additional information is as under

  1. The closing stock was valued at the end of the year 20,000.
  2. Depreciation on plant and machinery charged @ 5% and land and building @ 10%.
  3. Discount on debtors @ 3%.
  4. Make a provision @ 5% on debtors for bad debts.
  5. Salary outstanding was? 100 and wages prepaid was? 40.
  6. The manager is entitled a commission of 5% on net profit after charging such commission.

Answer:
Trading and Profit and Loss Account as on 31st December 2011
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 44
Balance Sheet as on 31st December 2011
Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II 45

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Students can Download Chapter 11 Thermal Properties of Matter 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 11 Thermal Properties of Matter

Plus One Physics Thermal Properties of Matter One Mark Questions and Answers

Question 1.
In which of the following processes, convection does nottake place primarily?
(a) Sea and land breeze
(b) Boiling of water
(c) Warming of glass of bulb due to filament
(d) Heating air around a furnace
Answer:
(c) Warming of glass of bulb due to filament
In convection process, the heat is transferred by the bodily motion of the heated particles. It is not so in case of warming of glass bulb due to filament heating. In fact, warming of glass bulb is due to radiation.

Question 2.
\(\frac{\text { Watt }}{\text { Kelvin }}\) a unit of
(a) Stefan’s constant
(b) Wien’s constant
(c) Cooling’s constant
(d) Thermal conductance
Answer:
(d) Thermal resistance
Plus One Physics Thermal Properties of Matter One Mark Questions and Answers 1

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 3.
For measuring temperatures in the range of 2000°C, we should employ
(a) gas thermometer
(b) platinum-rhodium thermometer
(c) barometer
(d) pyrometer
Answer:
(d) pyrometer

Question 4.
There is a hole in metal disc. What happens to the size of metal disc if the metal disc is heated?
Answer:
The size of hole increases.

Question 5.
Which has more specific heat capacity, water, and sand?
Answer:
Water.

Question 6.
Two solid spheres of the same material have the same radius but one is hollow while the other is solid. Both spheres are heated to same temperature. Then
(a) the solid sphere expands more
(b) the hollow sphere expands more
(c) expansion is same for both
(d) nothing can be solid about their relative expansion if their masses are not given
Answer:
(c) expansion is same for both

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 7.
The sprinkling of water reduces slightly the temperature of a closed room because
(a) temperature of water is less than that of the room.
(b) specific heat of water is high
(c) water has large latent heat of vaporisation
(d) water is a bad conductor of heat
Answer:
(c) water has large latent heat of vaporisation
When water is sprinkled over a large area, evaporation takes place. As the latent heat of vaporisation is large cooling takes place.

Question 8.
Why specific heat of gas at constant pressure (Cp) is greater than specific heat at constant volume?
Answer:
More heat is required to raise the temperature of gas at constant pressure than at constant volume.

Question 9.
A body is heated. But there is no change in its temperature. Is it possible?
Answer:
Yes. During change of state, there will be no increase in temperature even when heat is supplied.

Question 10.
When boiling water is put in glass tumbler, the tumbler cracks. Why?
Answer:
Glass is poor conductor of heat. So inner and outer surfaces of tumbler suffer uneven expension. Hence it breaks.

Question 11.
A small space is left between two rails on railway track. Why?
Answer:
If no space is left, the rails would bend due to thermal expansion in summer. So small space is left between two rails to allow thermal expansion.

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 12.
The pendulum of clock is made of invar, Why?
Answer:
Invar has low value of coefficient of linear expansion. So length of pendulum remains almost same in all seasons. (The change in length affects time period of pendulum).

Question 13.
Tea gets cooled, when sugar is added to it. Why?
Answer:
When sugar is added, heat content of tea gets shared with sugar & hence’ temperature decreases.

Question 14.
Ice covered in gunny bag does not melt for a long time. Why?
Answer:
A gunny bag is poor conduct of heat & hence it does not allow external heat to enter.

Question 15.
Why two layers of cloth of equal thickness provide warmer covering than a single layer of cloth of double the thickness?
Answer:
Because air between two layers of clothes is a bad ‘ conductor of heat.

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 16.
On winter nights, we feel warmer when clouds cover the sky than the sky is clear. Why?
Answer:
The clouds are bad conductor of heat. So heat of earth’s atmosphere is not conducted out.

Plus One Physics Thermal Properties of Matter Two Mark Questions and Answers

Question 1.
1. Two bodies at different temperatures T1K and T2K are brought in contact with each other

  • Is the resultant temperature be necessarily (T1 + T2)/2? If not, Why?
  • Should the resultant temperature be between T1 and T1 only? If not, Why?

Answer:
1. Two bodies at different temperatures:

  • The resultant temperature may not be necessarily \(\left(\frac{T_{1}+T_{2}}{2}\right)\). Because specific heat capacity is different for all substances.
  • If heat is not lost to the surroundings, resultant temperature must lie in between T1 and T2.

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 2.
A Patient is admitted to hospital. The temperature of the patient is measured by nurse and is found to be 97.6°F

  1. What is meant by temperature?
  2. Convert the temperature (97.6°F) is to centigrade

Answer:
1. Temperature is the degree of hotness.

2.
Plus One Physics Thermal Properties of Matter Two Mark Questions and Answers 2

Question 3.
Why iron rims are heated red hot before being put on the cart wheels?
Answer:
The radius of iron rim is smaller than radius of cart wheel. When iron rim is heated, its radius increases due to thermal expansion. After rim has planted on the wheel, iron rim is allowed to cool. Then it fits tightly on the wheel due to thermal contraction.

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 4.
How woolen clothes helps us in winter against cold?
Answer:
Wool is a heat insulator. Moreover, it contains a layer of dry air in it. This air, which is bad conductor of heat does not allow the body heat to escape & it also does not allow external cold to come in. The rough surface of woollen clothes absorbs more and reflects less heat falling on it.

Plus One Physics Thermal Properties of Matter Three Mark Questions and Answers

Question 1.
A Solid material is supplied with heat at constant rate. The temperature of the material is changing with the heat input as shown in figure.

  1. What do the horizontal region AB and CD represent?
  2. What does the slope of DE represent?
  3. The slope of OA is greater than the slope of BC. What does this indicate?

Answer:
Plus One Physics Thermal Properties of Matter Three Mark Questions and Answers 3

  1. Latent heat of fusion, Latent heat of vaporization
  2. It indicates that, the material is in vapour state
  3. It indicates that latent heat of vaporization of the material is greater than the latent heat of fusion.

Plus One Physics Thermal Properties of Matter Four Mark Questions and Answers

Question 1.
A copper block of mass 2.5kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39Jg-1 K-1; latent heat of fusion of water = 335 Jg-1).
Answer:
Mass, m = 2.5kg
= 2.5 × 103 g;
Change in temperature, ∆T = 500°C
Specific heat, c = 0.39 Jg-1K-1;
Latent heat of fusion, L = 335Jg-1
If m’ be the mass of ice melted, then m’L = mc ∆T
or m’ × 335 = 2.5 × 103 × 0.39 × 500 2.5 × 103 × 0.39 × 500
or m’ = \(\frac{2.5 \times 10^{3} \times 0.39 \times 500}{335} \mathrm{g}\)
= 1.5 kg.

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 2.
Heat from the sun reaches the earth.

  1. Write the method of transmission of heat.
  2. Thermal conductivity of Aluminium is 205 Jm-1s-1deg-1. What do you mean by this?
  3. Birds swell feathers in winter. Why?

Answer:
1. Radiation, conduction, convenction.

2. The coefficient of thermal conductivity of a substance is defind as the quauitity of heat conducted normally persecond through unit area of the substance per unit temperature gradient when the substance attains steady state.

3. By doing so the birds enclose air between the feathers. Air being a poor conductor, prevents the loss of heat from the body of the bird to the cold surroundings.

Question 3.
Two accidents are happened. The first one with water at 100°C and the second one with steam at 100°C.

  1. Which is dangerous burn due to water at 100°C, and bum due to steam at 100°C? Why?
  2. Latent heat of vapourisation of water to 536 cal/g. Explain the idea of latent heat of vopourisation.
  3. Find the heat required to convert 1g of ice at 0°C to steam at 100°C is

Answer:
1. Burn due to steam is more dangerous, because heat content in steam is very high compared to 100°C water.

2. Latent heat of vapourisation is the amount of heat required to change the state of 1 kg water in to vapour.

3. Q = ML + MC∆Q + ML1
Plus One Physics Thermal Properties of Matter Four Mark Questions and Answers 4
= 716 cal.

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 4.
When a fluid is heated, the particle rises up.

  1. Name the phenomonon behind it.
  2. Explain the formation of land breeze and sea breeze.

Answer:
1. Convection

2. During the day, land heats up more quickly than, water in lake (due to high specific heat capacity of water). The air on the surface of earth gets heated, expands, becomes less dense and rises up. The colder air (wind) replaces the space created by hot air.

It creates a sea breeze. At night the land loses its heat very quickly than water. So water remains more warmer at night.

Plus One Physics Thermal Properties of Matter NCERT Questions and Answers

Question 1.
A 10kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = 0.19Jg-1 K-1.
Answer:
Power, P = 10 kW
= 10 × 103W
= 104W
Mass, m = 8 kg; Time, t = 2.5min = 150s
Specific heat, c = 0.91Jg-1K-1
= 0.91 × 103Jkg-1K-1
Energy, Q = pt = 104 × 150J = 1.5 × 106J
It is given that 50% of energy is lost to the. surroundings. So, energy absorbed by the block is given by
Q = \(\frac{1}{2}\) × 1.5 × 106J = 0.75 × 106J
But Q = mc∆T
∴ ∆T
Plus One Physics Thermal Properties of Matter NCERT Questions and Answers 5

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 2.
A brass boiler has a base area of 0.15m2 and thickness 1.0cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s-1 m-1K-1;
Heat of vaporisation of water = 2256 × 103 J kg-1.
Answer:
A = 0.15m2, d = 10-2m,
\(\frac{m}{t}=\frac{6}{60}\) kgs-1 = 0.1kgs-1,
K = 109J s-1 m-1 K-1, L = 2256 × 103 Jkg-1,
θ1 = ?,
θ2 = 100°C
Plus One Physics Thermal Properties of Matter NCERT Questions and Answers 6

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 3.
Explain why?
(a) A body with large reflectivity is a poor emitter.
(b) A brass tumbler feels much colder than a wooden tray on a chilly day.
(c) An optical pyrometer (for measuring high temperature) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.
(d) The earth without its. atmosphere would be inhospitably cold?
(e) Heating systems based on circulation of steam are more efficient in warming a building than heating system based on circulation of hot water.
Answer:
(a) a body whose reflectivity is large would naturally absorb less heat. So, a body with large reflectivity is a poor.

(b) The thermal conductivity of brass is high i.e., brass „ is a good conductor of heat. So, when a brass tumbler is touched, heat quickly flows from human body to tumbler. Consequently, the tumbler appears colder. On the other hand, wood is a bad conductor, so, heat does not flow from the human body to the wooden tray in this case. Thus, it appears comparatively hotter.

(c) Let T the temperature of the hot iron in the furnace. Heat radiated per second per unit area, E = σT4

Kerala Syllabus 9th Standard Hindi Solutions Unit 3 Chapter 3 अंदर के और बाहर के

You can Download अंदर के और बाहर के Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Hindi Solutions Unit 3 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Hindi Solutions Unit 3 Chapter 3 अंदर के और बाहर के (व्यंग्य लेख)

अंदर के और बाहर के Summary in Malayalam and Translation

Kerala Syllabus 9th Standard Hindi Solutions Unit 3 Chapter 3 अंदर के और बाहर के 1
Kerala Syllabus 9th Standard Hindi Solutions Unit 3 Chapter 3 अंदर के और बाहर के 2
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Kerala Syllabus 9th Standard Hindi Solutions Unit 3 Chapter 3 अंदर के और बाहर के 4
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Kerala Padavali Malayalam Standard 10 Solutions Unit 4 Chapter 3 Aswamedham

Students can Download Kerala Padavali Unit 4 Chapter 3 Aswamedham Questions and Answers, Summary, Notes Pdf, Activity, Kerala Padavali Malayalam Standard 10 Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Padavali Malayalam Standard 10 Guide Unit 4 Chapter 3 Aswamedham

Aswamedham Questions and Answers, Summary, Notes

Kerala Padavali Malayalam Standard 10 Solutions Unit 4 Chapter 3 Aswamedham 1

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Kerala Padavali Malayalam Standard 10 Solutions Unit 4 Chapter 3 Aswamedham 2
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Kerala Padavali Malayalam Standard 10 Solutions Unit 4 Chapter 3 Aswamedham 10

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Kerala Padavali Malayalam Standard 10 Solutions Unit 4 Chapter 3 Aswamedham 29
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Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Students can Download Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits NCERT Text Book Questions and Answers

Question 1.
In an n-type silicon, which of the following statement is true.
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c)  Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer:
(c) Holes are minority carriers and pentavalent atoms are the dopants.

Question 2.
Carbon, silicon, and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)c, (Eg)Si and (Eg)Ge. Which of the following statements is true?
(a) (Eg)Si < (Eg)Ge < (Eg)c
(b) (Eg)c < (Eg)Ge < (Eg)Si
(c) (Eg)c > (Eg)Si > (Eg)Ge
(d) (Eg)c = (Eg)Si = (Eg)Ge
Answer:
(c) (Eg)c > (Eg)Si > (Eg)Ge

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 3.
When a forward bias is applied to a p-n junction, it.
(a) raise the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above
Answer:
(c) lowers the potential barrier.

Question 4.
For transistor action, which of the following statements are correct:
(a) Base, emitter and collector regior should have similar size and dopin concentrations.
(b) The base region must be very thin are lightly doped.
(c) The emitter junction is forward biase and collector junction is reverse biased.
(d) Both the emitter junction as well as to collector junction are forward biased.
Answer:
(b) The base region must be very thin are lightly doped,
(c) The emitter junction is forward biase and collector junction is reverse biased.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 5.
For transistor amplifier, the volta gain.
(a) remains constant for all frequencies.
(b) is high at high and low frequencies.
(c) is low at high and low frequencies a constant in the middle frequency large.
(d) None of the above.
Answer:
(c) is low at high and low frequencies a constant in the middle frequency large.

Question 6.
In half-ware rectification, what is the output frequency if the input frequency is 50Hz. What is the output frequency of a fullwave rectifier for the same input frequency.
Answer:
Given Input frequency = 50Hz
Output frequency
For Halfwave rectifier = 50 Hz
For full wave rectifier = 50 × 2 = 100 Hz.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 7.
A p-n photodiode is fabricated from a semiconductor with band gap of 2.8eV. Can it detect a wavelength of 6000 nm?
Answer:
Band gap Eg = 2.8 eV
Energy band gap corresponding to wavelength 6000 nm is given by
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 1
Since Eg, < Eg, so it can not detect a wave length of 6000 nm.

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits One Mark Questions and Answers

Question 1.
The zenerdiode works in______bias.
Answer:
reverse bias.

Question 2.
A transistor is operated in common emitter configuration at Vc = 2 V, such that a change in the base current from 100 µA to 200 µA produces a change in the collector current from 5 mA to 10 mA. The current gain is
(a) 100
(b) 150
(c) 75
(d) 50
Answer:
(d) 50
Explanation: Current gain,
 Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 2

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 3.
The electrical conductivity of an intrinsic semiconductor at 0 K is
(a) less than that an insulator
(b) equal to zero
(c) equal to infinity
(d) more than that of an insulator
Answer:
(b) equal to zero.

Question 4.
The voltage between the terminals A and B is 17 V and Zener breakdown voltage is 9 V. Find the potential across R is
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 3
Answer:
The potential across R = 17V – 9V = 8V.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 5.
Hole is
(a) an antiparticle of electron.
(b) a vacancy created when an electron leaves a covalent bond.
(c) absence of free electrons.
(d) an artificially created particle.
Answer:
(b) a vacancy created when an electron leaves a covalent bond.

Question 6.
A circuit is constructed by using certain gates is given below
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 4
1. Each gate is a…….. gate
2. Complete the truth table of above circuit
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 5
Answer:
1. NAND gate

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 6

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 7.
In both p and n-type semiconductor, actually electrons are flowing. What difference do you observe in the motion of electrons in these semiconductors?
Answer:
Electrons in valence band are flowing. Electrons in conduction band are flowing.

Question 8.
Unidirectional property of diode; Rectification. Then the break down action of Zener diode:…….
Answer:
Voltage regulation.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 9.
For an input frequency 50Hz, the output frequency……..of hall wave rectifier is and the output of full
wave rectifier for the same input frequency is………
Answer:
50 Hz, 100 Hz

Question 10.
The following questions consists of two statements. Assertion: Zener diode works as a voltage regulator Reason: Zener voltage is independent of the Zener current variations and change of load resistance. Write the correct response from the following.
(a) Both assertion and reason are true and the rea¬son is not a correct explanation of the assertion.
(b) Assertion is true, but reason is false.
(c) Both assertion and reason are true and reason is correct explanation for the assertion.
(d) Both assertion and reason are false.
Answer:
(c) Both assertion and reason are true and reason is correct explanation for the assertion.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 11.
Assertion: Semiconductors have -ve temperature co-efficient of resistance.
Reason: As temperature of a semiconductor increases, number density of charge carriers also increases.
(a) Both assertion and reason are correct, but reason is not proper explanation.
(b) Both assertion and reason are correct and reason is proper explanation.
(c) Assertion is correct but reason is wrong.
(d) Assertion is correct, and reason also is correct.
Answer:
(a) Assertion js correct, but reason is incorrect

Question 12.
Correct the following CE amplifier circuit.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 7
Answer:
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 8

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Two Mark Questions and Answers

Question 1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 9
Answer:
Semiconductor – P-type, n-type.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 2.
Fill in the blanks with appropriate word given below. (Base, collector, emitter, bias-collector junction, collector-emitter junction, emitter bias junction) Structurally, a bipolar junction transfer consists of emitter, base and………Out of these regions……….is the most heavily doped. For proper functioning of a transistor………..is forward biased and………….is reverse biased.
Answer:

  1. Collector
  2. Emitter
  3. EB junction
  4. CB junction

Question 3.
Classify the following into conductors, insulators, and semiconductors.
Ga, As, Ni, Calcite, Graphite
Answer:

  1. Conductor – Graphite, Ni
  2. Insulator – Calcite
  3. Semiconductor – Ga,As.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 4.
Construct truth table for following logic circuit.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 10
Answer:
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 11

Question 5.
State whether true or false and justify.

  1. Zener diode are used under forward bias.
  2. In n-p-n transistor current conduction is primarily due to electrons.
  3. Transistor amplifier do not strictly obey law of conservation of energy since output power is greater than input power.
  4. In a transistor amplifier all the frequency will have exactly equal gain.

Answer:

  1. False, Zener diodes are used under reverse bias
  2. True
  3. False
  4. False

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 6.
Match the following
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 12
Answer:
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 13

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Three Mark Questions and Answers

Question 1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 14
An electric circuit containing a battery, a bulb, and two switches is give above
1. Identify the gate an alogues to the above electric circuit
2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 15
If the above two input signal are applied to the gate what will be the shape of out put wave Draw the out put wave.
Answer:
1. OR gate

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 16

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 2.
A boy designs a circuit to study the input and output characteristics of an npn transistor
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 17

  1. Identify the transistor configuration, input current and output current
  2. By keeping the output voltage constant, the boy measures the input current by varying the input voltage. If a graph is drawn, what is the nature of the input characteristic? Justify your answer.

Answer:
1. Common Emitter, ib, and ic

2. Input Characteristics (CE configuration):
The graph connecting base current with base emitter voltage (at constant VCE) is the input characteristics of the transistor.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 18
To study the input characteristics, the collector to emitter voltage (VCE) is kept at constant. The base current IB against VBE is plotted in a graph. The ratio ∆ VBE /∆IB at constant VCE is called the input resistance.
i.e.,Input resistance = \(r_{i}=\frac{\Delta V_{B E}}{\Delta I_{B}}\).

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 3.
Truth table of a logic gate is given below:
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 19
1. Identify the logic gate.
2. Explain the working of this gate using diode and battery.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 20
3. If these input signals are applied across the gate, what will be the shape of the output wave form.
Answer:
1. OR gate

2. Out of syllabus
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 21

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 4.
A circuit using two switches (A and B), cell and bulb is shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 22
1. The above circuit is equivalent to

  • OR gate
  • AND gate
  • NOT gate
  • NOR gate

2. Give symbol and truth table of the above gate
Answer:
1. OR gate

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 23

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Four Mark Questions and Answers

Question 1.
Forward biased pn junction diodes are shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 24

  1. Identify the figure, which shows the correct direction of flow of charges.
  2. What do you mean by barrier potential and depletion region of a pn junction?
  3. When forward bias is applied to a p-n junction, what happens to the potential barrier and the width of depletion region.

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 25

2. The potential developed across the junction, which tends to prevent the movement of electron from the n region into the p region, in semiconductor is called a barrier potential.

The space-charge region on either side of the junction at which there is no free charge carriers is known as depletion region.

3. The potential barrier decreases and depletion region gets reduced.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 2.
The symbol of a n-p-n transistor is shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 26

  1. Redraw the symbol and mark emitter, collector and base of the transistor.
  2. Arrange the doping concentration and width of emitter, collector and base regions in ascending order.
  3. What happens when both the emitter and the collector of a transistor are forward biased?

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 27
2. Doping concentration of base < doping concentration of collector < doping concentration of emitter. Width of base < Width of emitter < Width of collector.

3. The transistor will work as two p-n junctions with common base terminals.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 3.
A green house has an electric system, which automatically switches ON a heater if the air temperature in the green house drops too low. A manual switch is included so that the automatic system can be switched off.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 28

  1. What is meant by 1 and 0 in digital circuit?
  2. Name logic gate X. Why is it used?
  3. Name the logic gate Y?
  4. Construct a truth table of this electronic system by taking A and B as inputs and D as output.

Answer:
1. 1 – means maximum, 0 – means minimum voltage

2. NOT gate

3. AND gate

4.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 29

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 4.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 30
The forward-biased diode is wrongly given above.

  1. Redraw the above circuit correctly.
  2. Draw the graph of current I with voltage v in forward bias.
  3. Classify the following circuit into forward bias, reverse-bias, unbias.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 31
Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 32
2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 33

3. circuit into forward bias, reversebias, unbias:

  • Reverse bias
  • Forward bias
  • Reverse bias
  • Unbias

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 5.
A car stereo working at stabilized voltage supply of 9 v DC and has a zener diode of 9V, 0.25W. But the voltage supply inside the car is 12V DC.

  1. Which mode of bias will you suggest to connect zener diode voltage regulator?
  2. Draw a circuit diagram of voltage regulation to help the boy.
  3. Which device is essential for circuit diagram? Find the value of that device.

Answer:
1. Reversebias

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 34

3. Resistance: R can be calculated using the equation
Vs = IR + Vz , 12 = \(\left(\frac{0.25}{9}\right)\)R + 9, R = 108Ω.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 6.
Electric current is the flow of charges along a definite direction and take place through metals as well as semiconductors.

  1. Mention the charge carriers in the above cases.
  2. Give the sketch of graph with V along X-axis and I along Y-axis for a metal at room temperature.
  3. Give the physical significance of the slope of the graph.
  4. If the above graph is drawn at 100°C, compared the nature of the graph with the graph at room temperature.

Answer:

  1. Metals – electrons
    Semiconductor – Electrons and holes
  2. Straight line graph
  3. Slope gives conductance
  4. When the temperature increases, resistance is also increased and hence slope decreases.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 7.
A transistor in the common-emitter mode can be used as an amplifier

  1. Design a circuit to amplify an ac signal given in the input region
    [Hint: Give forward biasing to input region, reverse biasing to output region and take output across a resistor]
  2. Derive expressions for voltage gain, current gain and power gain in the above transistor configuration.

Answer:
1.

2. When we apply an AC signal as input, we get an AC base current denoted by iB. Hence input AC voltage can be written as
Vi = iBr ……..(1)
where ‘r’ is the effective input resistance.
This AC input current produces an AC output current (ic) which can flow through a capacitor. Hence the output voltage can be written as
V0 = ic × output resistance
If we take output resistance as RL then v0 becomes
V0 = ic RL
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 36
Substituting eq(1) and eq(2),in the above equation we get
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 37

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits
Power gain:
The power gain Ap can be expressed as the product of the current gain and voltage gain.
ie. power gain A = βac × Av.

Question 8.
The circuit diagram of a full wave rectifier is shown
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 38

  1. Explain how its works? Also draw the output wave form
  2. If the frequency of a.c. at the input is 50Hz what will be the output frequency?

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 39
Full wave rectifier consists of transformer two diodes and a load resistance RL. Input a.c signal is applied across the primary of the transformer. Secondary of the transformer is connected to D1, and D2. The output is taken across RL.
Working:
During the +ve half cycle of the a.c signal at secondary, the diode D1 is forward biased and D2 is reverse biased. So that current flows through D1 and RL.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

During the negative half cycle of the a.c signal at secondary, the diode D1 is reverse biased and D2 is forward biased. So that current flows through D2 and RL.

Thus during both the half cycles, the current flows through RL in the same direction. Thus we get a +ve voltage across RL for +ve and -ve input. This process is called full wave rectification.

2. 100Hz

Question 9.
Forward biased pn junction diodes are shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 40
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 41

  1. Identify the figure, which shows the correct direction of flow of charges. (1)
  2. What do you mean by barrier potential and depletion region of a pn junction? (2)
  3. When forward bias is applied to a p-n junction, what happens to the potential barrier and the width of depletion region. (1)

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 42

2. The potential developed across the junction .which tends to prevent the movement of electron from the n region into the p region, in semiconductor is called a barrier potential.

The space charge region on either side of the junction at which there is no free charge carriers is known as depletion region.

3. The potential barrier decreases and depletion region gets reduced.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 10.
LEDs that can emit red, yellow, orange, etc. commercially available.

  1. How these colours are obtained in a LED. (1)
  2. Write any two uses of LED. (1)
  3. What are its advantages over ordinary bulbs? (2)

Answer:
1. Different colours are obtained by changing the concentration of arsenic and phosphors in Gallium Arsenide Phosphide.

2. LEDs find extensive use in remote controls, burglar alarm systems, optical communication, etc.

3. LEDs have the following advantages over conventional incandescent low power lamps:

  • Low operational voltage and less power.
  • Fast action and no warm-up time required.
  • The bandwidth of emitted light is 100A° to 500A° or in other words it is nearly (but not exactly) monochromatic.
  • Long life and ruggedness Fast on-off switching capability.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 11.
The symbol of a n-p-n transistor is shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 43

  1. Redraw the symbol and mark emitter, collector and base of the transistor. (1)
  2. Arrange the doping concentration and width of emitter, collector and base regions in ascending order. (2)
  3. What happens when both the emitter and the collector of a transistor are forward biased? (1)

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 44

2. Doping concentration of base < doping concentration of collector < doping concentration of emitter. Width of base < Width of emitter < Width of collector.

3. The transistor will work as two p-n junctions with common base terminals.

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Five Mark Questions and Answers

Question 1.
The transfer characteristic of n-p-n transistor in CE configuration is shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 45

  1. Identify the cut-off region, active region, saturation region from the figure.
  2. In which of these regions, a transistor is said to be switched off.
  3. For a CE transistor amplifier, the audio signal voltage across collector resistance of 2.0 kΩ is 2.0V. Suppose the current amplification factor of the transistor is 100. What should be the value of RB in series with VBB supply of 2.0V, if DC base current has to be 10 times the signal current?
  4. In the working of a transistor, the emitter-base (EB) junction is forward biased while collector-base (CB) junction is reverse biased. Why?

Answer:
1.

  • I – cut off region
  • II – active region
  • III-saturation region

2. Region I

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 46
Idc =10-5 × 10 = 10-4A
Vbb = Vbe + IbRb
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 47

4. Only forward biased emitter-base junction can send the majority charge carriers from emitter to base and only reverse biased collector can collect these majority charge carriers form the base region.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 2.
The basic building blocks of digital electronic circuits are called Logic Gates. Some logic gates and their names are given in the table.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 48
1. Match the symbols of logic gates with their names.
2. Draw the output wave form, from the given input wave form of a NAND gate as shown in figure (input terminals are A and B)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 49
3. Write the truth table forthe given circuit.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 50
Answer:
1. Match the symbols of logic gates with their names:

  • A – 5
  • B – 1
  • C – 2
  • D – 3
  • E – 4

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 51

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 52

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 3.
Following figure is an incomplete circuit of common emitter transistor in CE configuration with the input forward biased.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 53

  1. Identify the transistor an NPN or PNP.
  2. Complete the above circuit diagram by giving proper bias in the output and connect load resistance of 4 KΩ.
  3. When the base current changes by 20µ Afor VBE = .02 V. What is the voltage gain of the amplifier, if Ic = 2mA
  4. npn transistors are preferred in devices with very high-frequency source. Why?

Answer:
1. npn transistor

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 54

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 55

4. The resistance of a semiconductor decreases with rise in temperature.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 4.
Diodes are one of the building elements of electronic circuits. Some type of diods are shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 56

  1. Identify rectifier diode from the figure.
  2. Draw the circuit diagram of a forward biased rectifier diode by using a battery.
  3. Draw the forward and reverse bias characteristics of a rectifier diode and mark threshold voltage or cut in voltage.
  4. What happens to the resistance of a semiconductor on heating?

Answer:
a.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 57

b.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 58

c. The forward and reverse characteristics of a silicon diode is as shown in the fig.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 59

d. The resistance of a semiconductor decreases with rise in temperature.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 5.
A full wave rectifier circuit is shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 60
1. Draw the output wave form of the rectifier.
2. If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be

  • 50 Hz
  • 70.7 Hz
  • 100 Hz
  • 25 Hz

3. In a zener regulated power supply, a Zener diode with Vz = 6.0V is used for regulation. The load current is to be 4.0 mA and the unregulated input 10.0V. What is the value of series resistor R?
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 61
Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 62

2. 100 Hz

3. Input Voltage = 10V, Zener voltage Vz = 6V The voltage drop on the resistor should be 4V
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 63

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 6.
A P N junction diode is connected to a cell as a shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 64

  1. Name the type of biasing used here
  2. Design a circuit diagram to draw the characteristics of the diode in above biasing.
  3. Trace the characteristics curve if the polarity of battery is reversed

Answer:
1. forward biasing

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 65

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 66
In forward bias, current first increases very slowly up to a certain value of bias voltage. After this voltage, diode current increases rapidly. The diode offers low resistance in forward bias.

In reverse bias, current is very small. It remains almost constant upto break down voltage (called reverse saturation current). After this voltage reverse current increases sharply.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 7.
The circuit diagram of a full wave rectifier is shown.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 67
1. Explain how its works? Also draw the output wave form.
2. If another diode is connected in series with D2, as shown below what will happen to the out put wave form?
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 68
3. If the frequency of a.c. at the input is 50Hz what will be the output frequency of full wave rectifier?
Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 69
Full wave rectifier consists of transformer two diodes and a load resistance RL. Input a.c signal is applied across the primary of the transformer. Secondary of the transformer is connected to D1, and D2. The output is taken across RL.
Working:
During the +ve half cycle of the a.c signal at secondary, the diode D1 is forward biased and D2 is reverse biased. So that current flows through D1 and RL.

During the negative half cycle of the a.c signal at secondary, the diode D1 is reverse biased and D2 is forward biased. So that current flows through D2 and RL.

Thus during both the half cycles, the current flows through RL in the same direction. Thus we get a +ve voltage across RL for +ve and -ve input. This process is called full wave rectification.

2. Out put will be halfwave.

3. 100 Hz.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 8.
Diodes are one of the building elements of electronic circuits. Some type of diods are shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 70

  1. Identify rectifier diode from the figure. (1)
  2. Draw the circuit diagram of a forward biased rectifier diode by using a battery. (1)
  3. Draw the forward and reverse bias characteristics of a rectifier diode and mark threshold voltage or cut in voltage. (2)
  4. What happens to the resistance of a semiconductor on heating? (1)

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 71

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 72

3. The forward and reverse characteristics of a silicon diode is as shown in the fig.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 73
4. The resistance of a semiconductor decreases with rise in temperature.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 9.
A full wave rectifier circuit is shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 74
1. Draw the output wave form of the rectifier. (1)
2. If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be (2)

  • 50 Hz
  • 70.7 Hz
  • 100 Hz
  • 25 Hz (2)

3. In a zener regulated power supply, a zenerdiode with Vz = 6.0V is used for regulation. The load current is to be 4.0 mA and the unregulated input 10.0V. What is the value of series resistor R?(2)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 75
Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 76

2. 100 Hz

3. Input Voltage = 10V, Zener voltage Vz = 6V The voltage drop on the resistor should be
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 77

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 10.
The following diagram shows energy bands in a semiconductor.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 78
(a) Which diagram shows energy band positions at OK? (1)
(b) What do you mean by energy gap? Match the elements /compounds with their respective energy gap values. (1)

Diamond 6 eV
Aluminium 0.03 eV
Germanium 1.1 eV
Silicon 0.71 eV

(c) Classify solids into conductors, semiconductors, and insulators by drawing energy diagram. (3)
Answer:
(a)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 79

(b) The gap between the top of the valence band and bottom of the conduction band is called the energy band gap.

Column I Column II
A. Diamond 1. 1.1 eV
B.  Aluminium 2. 0.71 eV
C. Germanium 3. 0.03 eV
D. Silicon 4. 6 eV

(c) For conductors
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 80
For Insulators
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 81
For Semiconductors
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 82

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 11.
The given figure shows an npn transistor.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 83

  1. Redraw the figure and show the biasing voltage, direction of current and direction of flow of electrons and holes. (2)
  2. Draw the input and output characteristics of transistor connected in common emitter configuration. (2)
  3. In a transistor, a change of 7.9mA is observed in the collector current for a change of 7.99mA in the emitter current. Determine the change in base current. (1)

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 84

2. Input characteristic
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 85
Output characteristic
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 86

3. ∆IE = ∆Ic + ∆IB, ∆Ic = 7.9mA, ∆IE = 7.99mA
∆IB = ∆IE – ∆Ic
= 7.99 – 7.9 = 0.09mA.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 12.
The transfer characteristic of n-p-n transistor in CE configuration is shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 87

  1. Identify the cut-off region, active region, saturation region from the figure. (1)
  2. In which of these regions, a transistor is said to be switched off. (1)
  3. For a CE transistor amplifier, the audio signal voltage across collector resistance of 2.0 kΩ is 2.0V. Suppose the current amplification factor of the transistor is 100. What should be the value of RB in series with VBB supply of 2.0V, if DC base current has to be 10 times the signal current? (2)
  4. In the working of a transistor, the emitter-base (EB) junction is forward biased while collector-base (CB) junction is reverse biased. Why? (1)

Answer:
1. cut-off region, active region, saturation region from the figure:

  • I – cut off region
  • II – active region
  • III – saturation

2. Region I

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 88

4. Only forward biased emitter-base junction can send the majority charge carriers from emitter to base and only reverse biased collector can collect these majority charge carriers form the base region.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 13.
The basic building blocks of digital electronic circuits are called Logic Gates. Some logic gates and their names are given in the table.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 89
1. Match the symbols of logic gates with their names. (2)
2. Draw the output wave form, from the given input wave form of a NAND gate as shown in figure (input terminals are A and B) (1)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 90
3. Write the truth table for the given circuit. (2)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 92
Answer:
1. Match the symbols of logic gates with their names:

  • A – 5
  • B – 1
  • C – 2
  • D – 3
  • E – 4

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 93

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 94

Question 14.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 95

  1. According to energy gap, Classify them as metal, Insulator, and semiconductor.
  2. From which of the above material we can eject electrons with minimum effort Explain
  3. In Photo electric effect, while we are measuring photo current by varying retarding potential the variations is as shown in graph.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 96
Answer:

  1. Insulator, semiconductor, and conductor
  2. Metals
  3. This graph shows that photocurrent increases and reaches saturation with anode potential. This increase in current shows that electrons are emitted from the metal surface with different energies.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 15.
A diode can be properly doped at the time of its manufacture, so that it have a shape break down voltage
1. The above diode is called

  • Zener diode
  • Photo diode
  • Light emitting diode
  • Solar cell

2. Compare V-l Characteristics of above diode with that of an ordinary diode

3. Explain how the above diode can be used as an voltage regulator.

Answer:
1. Zener diode

2. Zener diode has sharp breakdown voltage than ordinary diode

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 97
The zener diode is connected to a fluctuating voltage supply through a resistor Rz. The out put is taken across RL.
Working:
When ever the supply voltage increases beyond the breakdown voltage, the current through zener increases (and also through Rz).

Thus the voltage across Rz increases, by keeping the voltage drop across zener diode as a constant value. (This voltage drop across Rz is proportional to the input voltage)

Similarly, when supply voltage decreases beyond a certain value, the current through the zener diode decreases. Thus the voltage across Rz decreases, by keeping the voltage drop across zener diode as constant (Zener diode as a voltage regulator).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Students can Download Chapter 11 Three Dimensional Geometry Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Plus Two Maths Three Dimensional Geometry Three Mark Questions and Answers

Question 1.
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 1

  1. Write the Cartesian equation. (1)
  2. Find the angle between the line. (2)

Answer:
1.
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 2

2. cosθ
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 3

Question 2.
Find the vector equation of the plane passing through the intersection of the planes \(\bar{r}\).(i + j + k) = 6 and \(\bar{r}\).(2i + 3 j + 4k) = -5 at the point (1,1,1).
Answer:
The Cartesian equation of the planes are x + y + z = 6 and 2x + 3y + 4z = – 5. Therefore the equation of the plane passing through the intersection of these planes is
x + y + z – 6 + k(2x + 3y + 4z + 5) = 0
Since it pass through (1, 1, 1) we get,
1 + 1 + 1 – 6 + k(2 + 3 + 4 + 5) = 0 ⇒ -3 + k14 ⇒ k = \(\frac{3}{14}\)
∴ the equation is
x + y + z + -6 + \(\frac{3}{14}\) (2x + 3 y + 4z + 5) = 0
14x + 14y + 14z – 84 + 6x + 9y + 12z + 15 = 0
20x + 23y + 26z = 69
Vector equation is \(\bar{r}\). (20i + 23j + 26k) = 69.

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 3.
Find the equation of the plane passing through the intersection of the planes x + y + 4z + 5 = 0 and 2x – y + 3z + 6 = 0 and contains the point (1, 0, 0).
Answer:
The equation of the planes passing through the intersection of the planes
x + y + 4z + 5 = 0 and 2x – y + 3z + 6 = 0 is
x + y + 4z + 5 + k(2x – y + 3z + 6) = 0 ____(1)
Since (1) pass through (1, 0, 0)
⇒ 1 + 0 + 0 + 5 + k(2 – 0 + 0 + 6) = 0
⇒ 6 + 8k = 0 ⇒ k = –\(\frac{3}{4}\); Then (1)
⇒ x + y + z + 5 + \(\frac{3}{4}\)(2x – y + 3z + 6) = 0
⇒ 4x + 4y + 16z + 20 – 6x + 3y – 9z – 18 = 0
⇒ 2x – 7y – 7z = 2.

Plus Two Maths Three Dimensional Geometry Four Mark Questions and Answers

Question 1.
Consider the point (-1, -2, -3).

  1. In which octant, the above point lies.(1)
  2. Find the direction cosines of the line joining (-1, -2, -3) and (3, 4, 5). (1)
  3. If P is any point such that OP = \(\sqrt{50}\) and direction cosines of OP are \(\frac{3}{\sqrt{50}}\), \(\frac{4}{\sqrt{50}}\) and \(\frac{5}{\sqrt{50}}\), then find the co-ordinate of P. (2)

Answer:
1. The point lies in the octant X’OY’Z’.

2. Direction ratios of the line joining (-1, -2, -3) and (3, 4, 5) are (3 + 1), (4 + 2), (5 + 3) ⇒ 4, 6, 8 ⇒ 2, 3, 4.
Therefore direction cosines are
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 4

3. Given, OP = \(\sqrt{50}\)
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 5
Therefore the point is (3, 4, 5).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 2.
Consider a cube of side ‘a’ unit has one vertex at the origin O.
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 6

  1. Write down the co-ordinate of 0, 0′, A and A’ (1)
  2. Find the direction ratios of OO’ and AA’. (2)
  3. Show that the angle between the main diagonals of the above cube is cos-1\(\left(\frac{1}{3}\right)\) (1)

Answer:
1. O(0, 0, 0), O'(a, a, a), A(a, 0, 0) and A'(0, a, a).

2. Direction ratios along OO’ is a – 0, a – 0, a – 0
⇒ a, a ,a ⇒ 1, 1, 1

3. cosθ
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 7

Question 3.
Consider two points A and B and a line L as shown in the figure.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 8

  1. Find \(\overline{A B}\) (1)
  2. Find the Cartesian equation of the line L.  (1)
  3. Find the foot of the perpendicular drawn from ( 2, 3, 4 ) to the line L. (2)

Answer:
1. \(\overline{A B}\) = (3 – 1)i + (- 3 – 3)j + (3 – 0)k = 2i – 6j + 3k.

2. The Cartesian equation of a line passing through the point (4, 0, -1 ) and parallel to the vector \(\overline{A B}\) is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 9

3. We take, \(\frac{x-4}{2}=\frac{y}{-6}=\frac{z+1}{3}\) = r then any point of the line can be taken as (2r + 4, -6r, 3r – 1). Assume that this point be the foot of the perpendicular drawn from (2, 3, 4 ). The dr’s of the line is 2 : – 6 : 3 and dr’s of the perpendicular line L is
2r + 4 – 2 : -6r – 3 : 3r – 1 – 4 ⇒ 2r + 2: -6r – 3 : 3r – 5
Since perpendicular,
2(2 r + 2) – 6(-6 r -3) + 3(3r – 5) = 0
49r = -7 ⇒ r = \(\frac{-7}{49}\) = –\(\frac{1}{7}\). Therefore the foot of the
perpendicular is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 10

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 4.
Cartesian equation of two lines are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 11
(i) Write the vector equation of the lines. (2)
(ii) Shortest distance between the lines. (2)
Answer:
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 12
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 13
Question 5.
Consider the points (1, 3, 4) & (-3, 5, 2)

  1. Find the equation of the line through P and Q. (1)
  2. At which point that the above line cuts the plane 2x + y + z + 3 = 0. (3)

Answer:
1. Equation of a line passing through( 1, 3, 4) and (-3, 5, 2) is given by,
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 14
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 15

2. Let \(\frac{x-1}{-2}=\frac{y-3}{1}=\frac{z-4}{-1}\) = λ
Then any point on the line is (-2λ + 1, λ + 3, -λ + 4)
Since the plane 2x + y + z + 3 = 0 cuts the aboveline. We have,
⇒ 2(-λ + 1) + λ + 3 – λ + 4 + 3 = 0
⇒ -2λ + 2 + λ + 3 – λ + 4 + 3 = 0
⇒ -2λ = -12 ⇒ λ = 6
∴ point of intersection is (-2 × 6 + 1, 6 + 3, -6 + 4)
⇒ (-11, 9, -2).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 6.
Let the equation of a plane be \(\bar{r}\). (2i – 3j + 5k) = 7, then

  1. Find the Cartesian equation of the plane. (1)
  2. Find the equation of a plane passing through the point (3, 4, -1) and parallel to the given plane. (2)
  3. Find the distance between the parallel planes. (1)

Answer:
1. Given, \(\bar{r}\).(2i – 3j + 5k) = 7 and if we substitute \(\bar{r}\) = xi + yj + zk Then we get the Cartesian equation as 2x – 3y + 5z – 7 = 0.

2. The equation of a plane parallel to the above plane differ only by a constant, therefore let the equation be 2x – 3y + 5z + k = 0.
⇒ 6 – 12 – 5 + k = 0 ⇒ k = 11
Therefore the equation is 2x – 3y + 5z + 11 = 0

3. The distance between the parallel planes
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 16

Question 7.

  1. State the condition for the line \(\bar{r}\) = \(\bar{a}\) + λ \(\bar{b}\) is parallel to the plane \(\bar{r}\).\(\bar{n}\) = d. (2)
  2. Show that the line \(\bar{r}\) = i + j + λ(2i + j + 4k) is parallel to the plane \(\bar{r}\). (-2i + k) = 5. (1)
  3. Find the distance between the line and The Plane in (ii). (1)

Answer:
1. The line \(\bar{r}\) = \(\bar{a}\) + λ \(\bar{b}\) is parallel to the plane \(\bar{r}\).\(\bar{n}\) = d, if the normal of the plane is perpendicularto the line.
∴ \(\bar{b}\).\(\bar{n}\) = 0.

2. Given,
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 17
The line \(\bar{r}\) = i + j + λ(2i + j + 4k) is parallel to the plane \(\bar{r}\). (-2i + 4k) = 5 ⇒ -2x + 4y = 5.

3. Distance = Distance between – 2x + 4y = 5 and point (1, 1, 0) on the line
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 19

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 8.
Choose the correct answer from the bracket,
(i) If a line in the space makes angle α, β and γ with the coordinates axes, then cos2α + cos2β + cos2γ is equal to (1)
(a) 1
(b) 2
(c) 0
(d) 3
(ii) The direction ratios of the line are \(\frac{x-6}{1}=\frac{2-y}{2}=\frac{z-2}{2}\) (1)
(a) 6, -2, -2
(b) 1, 2, 2
(c) 6, 1, -2
(d) 0, 0, 0
(iii) If the vector equation of a line is \(\bar{r}\) = i + j + k + µ(2i – 3j – 4k), then the Cartesian equation of the line
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 20
(iv) If the Cartesian equation of a plane is x + y + z =12, then the vector equation of the line is (1)
(a) \(\bar{r}\).(2i + j + k) = 12
(b) \(\bar{r}\).(i + j + k) = 12
(C) \(\bar{r}\).(i + y + 2k) = 12
(d) \(\bar{r}\).(i + 3j + k) = 12
Answer:
(i) (a) 1

(ii) (b) 1, 2, 2

(iii) (b) \(\frac{x-I}{2}=\frac{y-1}{-3}=\frac{z-1}{-4}\)

(iv) (b) \(\bar{r}\).(i + j + k) = 12.

Question 9.
Consider the lines \(\bar{r}\) = (i + 2j – 2k) + λ(i + 2 j) and \(\bar{r}\) = (i + 2j – 2k) + µ(2j – k)

  1. Find the angle between the lines.
  2. Find a vector perpendicular to both the lines.
  3. Find the equation of the line passing through the point of intersection of lines and perpendicular to both the lines.

Answer:
1. \(\bar{b}_{1}\) = i + 2j; \(\bar{b}_{2}\) = 2j – k
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 21

2. Perpendicular vector = \(\bar{b}_{1}\) × \(\bar{b}_{2}\)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 22
= i(-2 – 0) -j(-1 – 0) + k(2 – 0)
= -2i + j + k.

3. Equation of line is \(\bar{r}\) = (i + 2j – k) + µ(-2i + j + 2k).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 10.
Consider the line \(\bar{r}\) = (2i – j + k) + λ(i + 2j + 3k)

  1. Find the Cartesian equation of the line.
  2. Find the vector equation of the line passing through A (1, 0, 2) and parallel to the above line.
  3. Write two points on the line obtained in (ii) which are equidistant from A.

Answer:
1. \(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-1}{3}\).

2. \(\bar{r}\) = (i + 2k) + λ(i + 2j + 3k).

3. Put λ = a and λ = -a for any real value ‘a’.
Let us put λ = 1 and λ = -1
\(\bar{r}\) = (i + 2k) + 1(i + 2j + 3k) = 2i + 2j + 5k
⇒ (2, 2, 5)
\(\bar{r}\) = (i + 2k) – 1(i + 2j + 3k) = 0i – 2j – k
⇒ (0, -2, -1)
The equidistant points are(2, 2, 5) and (0, -2, -1).

Question 11.

  1. Find the equation of the plane through the point(1, 2, 3) and perpendicular to the plane x – y + z = 2 and 2x + y – 3z = 5 (2)
  2. Find the distance between the planes x – 2y + 2z – 8 = 0 and 6y – 3x – 6z = 57 (2)

Answer:
1. Required equation is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 23
(x – 1)(3 – 1) – (y – 2)(-3 – 2) + (z – 3)(1 + 2) = 0
2(x – 1) + 5(y – 2) + 3(z – 3) = 0
2x + 5y + 3z – 2 – 10 – 9 = 0
2x + 5y + 3z – 21 = 0

2. The planes are
x – 2y + 2z – 8 = 0 and 3x – 6y + 6z + 57 = 0
ie, 3x – 6y + 6z – 24 = 0 and 3x – 6y + 6z + 57 = 0
Distance
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 24

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 12.
Consider the Cartesian equation of a line \(\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-5}{-2}\)

  1. Find the vector equation of the line. (1)
  2. Find its intersecting point with the plane 5x + 2y – 6z – 7 = 0 (2)
  3. Find the angle made by the line with the plane 5x + 2y – 6z – 7 = 0 (1)

Answer:
1. The vector equation is \(\bar{r}\) = (3i – j + 5k) + λ(2i + 3 j – 2k).

2. Any point on the line is
\(\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-5}{-2}\) = λ
x = 2λ + 3, y = 3, λ – 1, z = -2λ + 5
Since this lies on the plane ,it satisfies the plane
5(2λ + 3) + 2(3λ – 1) -6(-2λ + 5) – 7 = 0
10λ + 6λ + 12λ + 15 – 2 – 30 – 7 = 0
28λ = 24
λ = 6/7
The point of intersection is \(\left[\frac{33}{7}, \frac{11}{7}, \frac{23}{7}\right]\).

3. Let θ be the angle between the line and the plane. The direction of the line and the plane
\(\bar{b}\) = 2i + 3j + k; \(\bar{m}\) = 5i + 2j – 6k
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 25

Question 13.
From the following figure
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 26

  1. Find \(\overline{A B}\). (1)
  2. Find the vector equation of line L. (1)
  3. Find a point on line L other than C. (2)

Answer:
1. P.v of A = i – j + 4k,
P.v. of B = 2i + j + 2k
\(\overline{A B}\) = p. v. of B – p. v. of A
= 2i + j + 2k -(i – j + 4k) = i + 2j – 2k.

2. The line L passes through (1, -2, -3) and parallel to \(\overline{A B}\)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 28
∴ Vector equation of line L is \(\bar{r}=\bar{a}+\lambda \bar{m}\)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 29

3. From (1) of part (ii), we have
xi + yj + zk = (l + λ)i + (-2 + 2λ)j + (-3 – 2λ)k
Put λ = 1
⇒ xi +yj + zk = (1 +1)i + (-2 + 2)j + (-3 – 2 )k
⇒ xi + yj + zk = 2i + 0j – 5k
Therefore a point on line L is (2, 0, -5).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 14.
Find the vector equation of the plane which is at a distance of \(\frac{6}{\sqrt{29}}\) from the origin with perpendicular vector 2i – 3j + 4k. Convert into Cartesian form. Also, find the foot of the perpendicular drawn from the origin to the Plane.
Answer:
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 30
Perpendicular distance from origin = d = \(\frac{6}{\sqrt{29}}\)
The equation of the Plane is \(\bar{r}\).\(\hat{n}\) = d
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 31
Cartesian equation is 2x – 3y + 4z = 6
The direction cosines perpendicular to the Plane is \(\frac{2}{\sqrt{29}},-\frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}\).
Perpendicular distance to the Plane is as \(\frac{6}{\sqrt{29}}\)
Hence the foot of the perpendicular is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 32

Question 15.
Consider the Plane \(\bar{r}\).(-6i -3j – 2k) + 1 = 0, find the direction cosines perpendicular to the Plane and perpendicular distance from the origin.
Answer:
Convert the equation of the plane into normal form
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 33
Direction cosines perpendicular to the Plane is \(\frac{6}{7}, \frac{3}{7}, \frac{2}{7}\)
Perpendicular distance from the origin is \(\frac{1}{7}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 16.
Consider three points (6, -1, 1), (5, 1, 2) and (1, – 5, -4) on space.

  1. Find the Cartesian equation of the plane passing through these points. (2)
  2. Find direction ratios normal to the Plane.(1)
  3. Find a unit vector normal to the Plane. (1)

Answer:
1. Equation of a plane passing through the points (6, -1, 1),(5, 1, 2) and (1, -5, 4)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 34
⇒ (x – 6)(-10 + 4) – (y + 1)(5 + 5) + (z – 1)(4 + 10) = 0
⇒ (x – 6)(-6) – (y + 1)(10) + (z – 1)(14) = 0
⇒ -6x + 36 – 10y – 10 + 14z – 14 = 0
⇒ 6x +10y – 14z -12 = 0.

2. Dr’s normal to the plane are 6 : 10 : -14 ⇒ 3 : 5 : -7.

3. Since the dr’s normal to the plane are 3 : 5 : -7, a unit vector in this direction is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 35

Question 17.
Consider a straight line through a fixed point with position vector 2i – 2j + 3k and parallel to i – j + 4k.

  1. Write down the vector equation of the straight line. (1)
  2. Show that the straight line is parallel to the plane \(\bar{r}\).(i + 5y + k) = 5 (1)
  3. Find the distance between the line and plane. (2)

Answer:
1. Vector equation of a straight line is \(\bar{r}=\bar{a}+\lambda \bar{b}\) where a is \(\bar{a}\) fixed point and \(\bar{b}\) is a vector parallel to the line. Here \(\bar{a}\) = 2i – 2y + 3 it and \(\bar{b}\) = i – j + 4k. Therefore vector equation of the line \(\bar{r}\) = 2i – 2j + 3k + λ(i – j + 4k).

2. The vector parallel to the line is i – j + 4k and vector normal to the plane is i + 5j + k.
Then, (i – j + 4k). (i + 5j + k) = 1 – 5 + 4 = 0
implies that straight line and plane are parallel.

3. A point on the line is 2i – 2j + 3k. Then the distance of 2i – 2j + 3k to the given plane
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 36

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 18.
Consider the vector equation of two planes \(\bar{r}\).(2i + j + k) = 3, \(\bar{r}\).(i – j – k) = 4

  1. Find the vector equation of any plane through the intersection of the above two planes. (2)
  2. Find the vector equation of the plane through the intersection of the above planes and the point (1, 2, -1 ) (2)

Answer:
1. The cartesian equation are 2x + y + z – 3 = 0 and x – y – z – 4 = 0 Required equation of the plane is
(2x + y + z – 3) + λ(x – y – z – 4) = 0
(2+ λ)x + (1 – λ)y + (1 – λ)z + (-3 – 4λ) = 0.

2. The above plane passes through (1, 2, -1)
(2+ λ)1 + (1 – λ)2 + (1 – λ)(-1) + (-3 – 4λ) = 0
3 – 3 + 4λ = 0
λ = 0
Equation of the plane is 2x + y + z – 3 = 0
\(\bar{r}\).(2i + j + k) = 3.

Question 19.
(i) Distance of the point(0, 0, 1) from the plane x + y + z = 3
(a) \(\frac{1}{\sqrt{3}}\) units
(b) \(\frac{2}{\sqrt{3}}\) units
(c) \(\sqrt{3}\) units
(d) \(\frac{\sqrt{3}}{2}\) units
(ii) Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to x – y + z = 0 (3)
Answer:
(i) (b) \(\frac{2}{\sqrt{3}}\) units.

(ii) Equation of the plane passing through the intersection is of the form
x + y + z – 1 + λ(2x + 3y + 4z – 5) = 0 _____(1)
(1 + 2λ)x + (1 + 3λ)j + (1 + 4λ)z – 1 – 5λ = 0
Thr Dr’s of the required plane is
(1 + 2λ), (1 + 3λ), (1 + 4λ)
Thr Dr’s of the Perpendicular plane is 1, -1, 1
⇒ (1 + 2λ)(1) + (1 + 3λ)(-1) + (1 + 4λ)(1) = 0
⇒ 1 + 2λ – 1 – 3λ + 1 + 4λ = 0
⇒ 3λ + 1 = 0 ⇒ λ = \(\frac{-1}{3}\)
(1) ⇒ x + y + z – \(\frac{1}{3}\)(2x + 3y + 4z – 5) = 0
⇒ 3x + 3y + 3z – 2x – 3y – 4z + 5 = 0
⇒ x – z + 2 = 0.

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 20.
Consider a plane \(\bar{r}\).(6i – 3j – 2k) + 1 = 0

  1. Find dc’s perpendicular to the plane. (2)
  2. Find a vector of magnitude 14 units perpendicular to given plane. (1)
  3. Find the equation of a line parallel to the above vector and passing through the point (1, 2, 1 ). (1)

Answer:
1. Given, \(\bar{r}\).(6i – 3j – 2k) + 1 = 0 ____(1)
Now, |6i – 3j – 2k| = \(\sqrt{36+9+4}\) = 7
∴ \(\frac{6}{7} i-\frac{3}{7} j-\frac{2}{7} k\) is a unit perpendicular to the plane (1)
⇒ the dc’s perpendicular to the plane (1) are \(\frac{6}{7},-\frac{3}{7},-\frac{2}{7}\).

2. We have, \(\frac{6}{7} i-\frac{3}{7} j-\frac{2}{7} k\) is a unit perpendicular to the Plane (1). Therefore, a vector of magnitude 14 units perpendicular to the Plane (1) is 14(\(\frac{6}{7} i-\frac{3}{7} j-\frac{2}{7} k\))
⇒ 12i – 6j – 4k.

3. Equation of a line parallel to the vector 12i – 6j – 4k and passing through the point (1, 2, 1 )is given by
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 37

Plus Two Maths Three Dimensional Geometry Six Mark Questions and Answers

Question 1.
Consider the pair of lines whose equations are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 38

  1. Write the direction ratios of the lines. (1)
  2. Find the shortest distance between the above skew lines. (4)
  3. Find the angle between these two lines. (1)

Answer:
1. The direction ratios are 2, 5, – 3 and – 1, 8, 4.

2. The given lines are \(\bar{r}\) = (2i + j – 3k) + λ(2i + 5j – 3k)
i.e. \(\bar{r}\) = \(\overline{a_{1}}+\lambda \overline{b_{1}}\),
where \(\overline{a_{1}}\) = 2i + j – 3k) + λ(2i + 5j – 3k)
and \(\bar{r}\) =(-i + 4j + 5k) + µ(-i + 8j + 4k)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 39

3. cosθ
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 40

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 2.
Consider the pair of lines \(\bar{r}\) = 3i + 4j – 2k + λ(-i + 2j + k) ——L1, \(\bar{r}\) = i – 7j – 2k + µ(i + 3j + 2k) ——L2

  1. Find one point each on lines L1 and L2. (1)
  2. Find the distance between those points. (2)
  3. Find the shortest distance between L1 and L2. (3)

Answer:
1. By putting λ = 0 in line L1 and µ = 0 in L2 we get the required points. L1 ⇒ \(\bar{r}\) = 3i + 4j – 2k
∴ Co-ordinate is (3, 4, -2)
L2 ⇒ \(\bar{r}\) = i – 7j – 2k
∴ Co-ordinate is (1, -7, -2).

2. Distance between (3, 4, -2) and (1, -7, -2)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 41

3. Let L1 ⇒ \(\bar{r}\) = 3i + 4j – 2k + λ(-i + 2j + k) is of the form \(\bar{r}=\overline{a_{1}}+\lambda \overline{b_{1}}\) where
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 42

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 3.
Consider the points A (2, 2, -1), B (3, 4, 2) and C (7, 0, 6).

  1. Are A, B, and C collinear? Explain.
  2. Find the vector and Cartesian equation of the plane passing these three points. (2)
  3. Find the angle between the above plane and the line \(\bar{r}\) = (i + 2j – k) + λ(i – j + k) (2)

Answer:
1. Direction ratios along A and B is 3 -2, 4 -2, 2 + 1 ⇒ 1, 2, 3
Direction ratios along B and C is
7 -3, 0 -4, 6 -2 ⇒ 4, -4, 4
Since the direction ratios are not proportional they are not collinear.

2. Cartesian equation of the Plane is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 43
⇒ (x – 2)(14 + 6) -(y – 2)(7 – 15) + (z + 1)(-2 -10) = 0
⇒ 20(x – 2) + 8(y – 2) – 12(z + 1) = 0
⇒ 20x – 40 + 8y – 16 – 12z – 12 = 0
⇒ 20x + 8y – 12z = 68
⇒ 5x + 2y – 3z = 17
Vector Equation is \(\bar{r}\).(5i + 2j – 3k) = 17.

3. Angle between the Plane and the Line
\(\bar{r}\) = (i + 2j – k) + λ(i – j + k)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 44

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 4.
Consider three points on space ( 2, 1, 0 ), (3, -2, -2)and(3, 1, 7)

  1. Find the Cartesian equation of the plane passing through the above points. (2)
  2. Convert the above equation into vector form.
  3. Hence, find a unit vector perpendicular to the above plane and also find the perpendicular distance of the plane from the origin. (2)

Answer:
1. Equation of the plane is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 45
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 46
⇒ (x – 2)(-21) – (y – 1)(7 + 2) + z(0 + 3) = 0
⇒ 21x + 42 – 9y + 9 + 3z = 0 ⇒ -21x – 9y + 3z + 51 = 0
⇒ 7x + 3y – z = 17.

2. Vector form is \(\bar{r}\).(7i + 3j – k) = 17 _____(1)

3. Now, |7i + 3j – k| = \(\sqrt{49+9+1}=\sqrt{59}\)
Dividing equation (1) by \(\sqrt{59}\), we get
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 47
Therefore the above equation is the normal form of the plane. Then \(\frac{7 i+3 j-k}{\sqrt{59}}\) is the unit vector perpendicular to the plane and \(\frac{17}{\sqrt{59}}\) is the perpendicular distance from the origin.

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 5.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 48
\(\overline{O A}\) = i + 2j + 3k
\(\overline{O B}\) = i – 2j + 4k
\(\overline{O C}\) = 2i + 3j + k
are adjacent sides of the parallelopiped.

  1. Find the base area of the parallelopiped. (2)
    (Base determined by \(\overline{O A}\) and \(\overline{O B}\))
  2. Find the volume of the parallelopiped. (2)
  3. Find the height of the parallelopiped. (2)

Answer:
1. \(\overline{O A}\) × \(\overline{O B}\) =
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 49
= 14i – j – 4k
Base area = |l4i – j – 4k|
\(=\sqrt{196+1+16}=\sqrt{213}\)

2. Volume of the parallelopiped is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 50
= (14i – j – 4k).(2i + 3 j + k)
= 28 – 3 – 4 = 21.

3. Height = \(\frac{\text {volume}}{\text {base area}}=\frac{21}{\sqrt{213}}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 6.

  1. Find the equation of the line passing through the point (2, 1, 0) and (3, 2, -1) (3)
  2. Find the shortest distance of the above line from the line \(\bar{r}\) = (i – j + 2k) + λ(2i + j – 3k) (3)

Answer:
1.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 51

2.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 52
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 53
= i(-3 + 1) – j(-3 + 2) + k(1 – 2)
= -2i + j – k
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 54

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 7.
The equation of two lines are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 55

  1. Find the dr’s of the given lines. (2)
  2. Find the angle between the given lines. (2)
  3. Find the equation of the line passing through (2, 1, 3) and perpendicular to the given lines. (2)

Answer:
1. The given lines are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 56
The dr’s of (1) are 2, 2, 3 and dr’s of (2) are -3, 2, 5.

2. The angle between (1) and (2) is given by
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 57

3. Let a, b, c be the dr’s of the line perpendicular to lines (1) and (2).
∴ 2a + 2b + 3c = 0, -3a + 2b + 5c = 0
Solving by the rule of cross-multiplication, we get
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 58
∴ dr’s of the required line are 4, -19, 10 and the line passes through (2, 1, 3).
∴ Equation of the required line is
\(\frac{x-2}{4}=\frac{y-1}{-19}=\frac{z-3}{10}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 8.

  1. Find the direction cosines of the vector 2i + 2j – k. (1)
  2. Find the distance of the point (2, 3, 4) from the plane \(\bar{r}\).(3i – 6j + 2 k) = -11. (2)
  3. Find the shortest distance between the lines \(\bar{r}\) = (2i – j – k)+ λ(3i – 5 j + 2k) an \(\bar{r}\) = (i+ 2 j + k)+ µ(i – j + k) (3)

Answer:
1. Direction ratios of the vector 2i + 2j – k is 2, 2, -1
Direction cosines of the vector 2i + 2j – k is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 59

2. The equation of the plane in the Cartesian form is 3x – 6y + 2z + 11 = 0 . Then distance from the point (2, 3, 4) is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 60

3. The given lines are \(\bar{r}\) = (2i – j – k) + λ(3i – 5j + 2k)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 61

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services

Students can Download Chapter 4 Business Services Questions and Answers, Plus One Business Studies Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services

Plus One Business Services One Mark Questions and Answers

Question 1.
Name the type of account opened in a bank with an over draft facility
Answer:
Current Account

Question 2.
DTH services are provided by
Answer:
Cellular companies

Question 3.
CWC Stands for
Answer:
Central warehousing corporation

Question 4.
Mr. James running a software development business, incur heavy expenditure for replacement of computers due to updation in technology. To avoid this expenditure he may opt for
(a) Factoring
(b) Venture capital
(c) Leasing
(d) Merchant banking
Answer:
(c) Leasing

Question 5.
Name the type of account opened in the bank with overdraft facility
Answer:
Current account

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services

Question 6.
Classify the following into appropriate categories SBI, Federal Bank, American Express Bank.
Answer:

  • SBI – Public Bank
  • Federal Bank – Private bank
  • American Express bank – Foreign bank

Question 7.
Which among the following helps in issue management of shares,
(a) Mutual fund
(b) Merchant banks
(c) Venture Capital
(d) Factoring
Answer:
(b) Merchant banks

Question 8.
Classify the following bank into Appropriate Categories,

  1. IDBI
  2. HDFC
  3. Axis Bank
  4. Canara Bank
  5. HSFC
  6. CITI Bank

Answer:

  1. IDBI – Private Bank
  2. HDFC – Private Bank
  3. Axis Bank – Public Bank
  4. Canara Bank – Public Bank
  5. HSFC – Private Bank
  6. CITI Bank – Foreign Bank

Question 9.
Which facility allows a current account holder to withdraw more amount than his deposit?
Answer:
Overdraft

Question 10.
Which of the following is not applicable in life insurance contract?
(a) Conditional contact
(b) Unilateral contract
(c) Indemnity contract
(d) None of the above
Answer:
(c) Indemnity contract

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services

Question 11.
When the same subject matter is insured with more than one insurer, it is called
Answer:
Double insurance

Question 12.
________ is a contract of insurance entered into by the insurer with another insurer to reduce the risk.
Answer:
Reinsurance

Question 13.
Rajiv insures his shop for ₹5,00,000 against fire. His shop is completely damaged by a mob attack. Rajiv claims ₹5,00,000/- from the insurance company.

  1. Will the company admit the claim of Rajiv?
  2. Which principle of insurance is applicable here?

Answer:

  1. No.
  2. Causa Proxima

Question 14.
Principle of indemnity does not apply to __________
(a) Life insurance
(b) Health insurance
(c) Fire insurance
(d) Marine insurance
Answer:
(a) Life insurance

Question. 15
I will give security against fraudulent practices of employees who deal with cash or other variables. Can you identify me? Explain its features?
Answer:
Fidelity insurance

Question 16.
Govind insures his shop for ₹6,00,000. in a fire accident. The shop incurred a loss of ₹3,00,000. But he claimed for the assured sum. Will the insurance co admit the claim in full? Identify the principle of insurance which is applicable here.
Answer:
No. Principle of indemnity

Question 17.
Life insurance is a contract of ________
Answer:
Guarantee

Question 18.
_______ principle states, the insurer step into the shoes of insured.
Answer:
Subrogation.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services

Question 19.
Jacob purchased a new car for ₹15,00,000 and insured it with National insurance company and Oriental Insurance Company for ₹15,00,000 each. On an accident, the car was damaged completely and he claimed compensation from both companies.

  1. Can he get compensation from both companies?
  2. Identify relevant principle of insurance.

Answer:

  1. No
  2. Principle of Contribution

Question 20.
According to _________ principle, the insured is not allowed to make any profit out of his loss.
Answer:
Indemnity

Question 21.
Rajan wants to take an insurance policy covering the risk arising from the dishonesty of employees. Which type to policy close he have to take?
Answer:
Fidelity insurance

Question 22.
__________ is a contract whereby the insurer, in consideration of premium, undertakes to compensate the insured for the loss or damage suffered due to fire.
Answer:
Fire insurance

Plus One Business Services Two Mark Questions and Answers

Question 1.
Define service and goods. (2)
Answer:
Services are those identifiable and intangible activities that provide satisfaction of wants. Goods is a physical product which can be delivered to a purchaser and involves the transfer of ownership from the seller to customer.

Question 2.
Explain various telecom services available for enhancing business. (2)
Answer:
The various types of telecom services are:

1. Cellular mobile services:
Mobile communication device including voice and non-voice messages, data services and PCO services utilising any type of network equipment within their service area.

2. Radio paging services:
It means of transmitting information to persons even when they are mobile.

3. Fixed line services:
It includes voice and non-voice messages and data services to establish linkage for long distance traffic.

4. Cable services:
Linkages and switched services within a licensed area of operation to operate media services which are essentially one way entertainment related services.

5. VSAT services (Very small Aperture Terminal):
It is a Satellite-based communication service. It offers government and business agencies a highly flexible and reliable communication solution in both urban and rural areas.

6. DTH services (Direct to Home):
It is a Satellite based media services provided by cellular companies with the help of small dish antenna and a set up box.

Question 3.
Mr. Dino is running a textile shop. He insured the shop for ₹1,00,000 against natural calamities. After this the shop was completely destroyed by fire due to electrics short circuit. He claimed for compensation. But the insurance company refused to admit the claim on the ground that the shop was insured against natural calamities. (2)

  1. Can you justify the decision of insurance company.
  2. Which principle of insurance is applicable here.

Answer:

  1. Yes.
  2. Causa proxima. It states that insurance company will admit the claim only if mishap has resulted directly by an event covered by under insurance.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services

Question 4.
Mr. Sabu is running a provision store. He insured the shop for ₹ 50,000 against fire. One day during course of business the shop caught fire and Mr. Sabu purposefully did not take any efforts to minimize the loss. He claimed for compensation. In this situation. (2)

  1. What will be your decision assuming that you are the officer in charge of admitting the claims for insurance company?
  2. Which principle is applicable here?

Answer:

  1. The claim can not be admitted because the insurer did not take any reasonable steps to minimize the loss.
  2. Principle of mitigation of loss

Question 5.
Why do we say that the principle of indemnity is not applicable to life insurance. Illustrate your answer. (2)
Answer:
The loss due to death of the insured cannot be measured in terms of money. So principle of indemnity is not applicable to Life insurance.

Question 6.
Mr. Sibi purchased a new car and it was completely damaged in an accident. Subsequently, he lodged a claim with the insurance company and the same was admitted. Later, the insurance company recovered the destroyed property. (2)

  1. Why did the company take ownership of the car?
  2. State the relevant principle. Explain.

Answer:

  1. Ownership of the damaged car is vested with the insurance company.
  2. According to the principle of subrogation, once the claim of the insured has been settled, the right of ownership of such property passes to .the insurer.

Question 7.
Mr. Anand your friend has started a new business. He decided to open a bank account with SBI. He has only a vague idea about the type of accounts available to the business people. So he seeks your help to choose the most favourable account for him. Give reason. (2)
Answer:
Mr. Anand can open current account. There is no restriction on the frequency and volume of deposits or withdrawals of money.

Question 8.
Match the following

A B
Warehousing Mutual Funds
Unit Trust Of India Hindrance Of Time
Insurance External Comunication
Telecom Causa proxima

Answer:

A B
Warehousing Hindrance Of Time
Unit Trust Of India Mutual Funds
Insurance Causa proxima
Telecom External Comunication

Plus One Business Services Three Mark Questions and Answers

Question 1.
Mr. Janardhanan, a business man, has a current a/c of ₹1,00,000 in the S.B.I. He is in need of another ₹50,000 for the purpose of expansion of his business. By availing which facility, can he have money from the bank? Explain its features. (3)
Answer:
Overdraft: Overdraft is an arrangement where customers are allowed to withdraw up to a level over and above their deposits in the current account. Interest is to be paid only on the actual amount of overdraft availed by him. OD is granted to businessmen against their current account.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services

Question 2.
Write a detailed note on various facilities offered by Indian Postal Department. (3)
Answer:
1. Postal services:
Indian post and telegraph department provides various postal services across India. Various facilities provided by postal department are
(a) Financial facilities:
They provide postal banking facilities to the general public and mobilise their savings through the saving schemes like public provident fund (PPF), Kisan Vikas Patra, National Saving Certificate, Recurring Deposit Scheme and Money Order facility

(b) Mail facilities:
1. Mail services consist of

  • Parcel facilities that is trans-mission of articles from one place to another
  • Registration facility to provide security of the transmitted articles
  •  Insurance facility to provide insurance cover for all risks in the course of transmission by post.

2. Allied Postal Services

  • Greetings Post: Greetings card can be sent through post offices.
  •  Media Post: Corporate can advertise their brands through post cards, envelops etc.
  •  Speed Post: It allows speedy transmission of articles to people in specified cities.
  •  e-bill post: The post offices collect payment of telephone, electricity, and water bills from the consumers.
  •  Courier Services: Letters, documents, parcels etc. can be sent through the courier service.

(2) Telecom Services:
The various types of telecom services are:

1. Cellular mobile services:
Mobile communication device including voice and non-voice messages, data services and PCO services utilising any type of network equipment within their service area.

2. Radio paging services:
It means of transmitting information to persons even when they are mobile.

3. Fixed line services:
It includes voice and non-voice messages and data services to establish linkage for long distance traffic.

4. Cable services:
Linkages and switched services within a licensed area of operation to operate media services which are essentially one way entertainment related services.

5. VSAT services (Very small Aperture Terminal):
It is a Satellite based communication service. It offers government and business agencies a highly flexible and reliable communication solution in both urban and rural areas.

6. DTH services (Direct to Home):
It is a Satellite based media services provided by cellular companies with the help of small dish antenna and a set up box.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services

Question 3.
What is e-banking? What are the advantages of e-banking? (3)
Answer:]
e-banking is a service provided by many banks, that allows a customer to conduct banking transactions, such as managing savings, checking accounts, applying for loans or paying bills over the internet using a personal computer, mobile telephone. Eg. Electronic Fund Transfer (EFT), ATM etc.

Question 4.
Babu insures his shop for ₹3,00,000. In a fire accident the shop burned and incurred a loss of ₹1,50,000. Babu claimed for the assured sum i.e, ₹3,00,000. (3)

  1. Will the insurance company admit the claim in full. Why?
  2. Which principle of insurance is applicable here.

Answer:

  1. No. According to the principle of indemnity, in the event of occurrence of loss, the insured will be indemnified to the extent of actual value of his loss or the sum insured which ever is less,
  2. Principle of Indemnity

Question 5.
Mrs. Anjali has a medical shop and it is insured with two insurers for ₹50,000 each against natural calamities. During the rainy season, the subject matter was partly destroyed by flood. The total amount of loss was estimated at ? 30,000. She claimed compensation from both insurance companies for ₹30,000 each. (3)

  1. Can she get the claim from both the insurers?
  2. Justify your answer with regard to the relevant principle of insurance.

Answer:

  1. No. She cannot get claim from both the insurance.
  2. According to the principle of contribution, the insured is eligible to receive a claim only up to the amount of actual loss suffered by her.

Question 6.
Mr. Joy keep stock of inflammable materials in his shop premises, takes an insurance policy for the shop against fire. But he did not disclose anything about the inflammable materials to the insurance company. Later, the inflammable goods caught fire and the shop was completely destroyed by fire. The insurance company refused to admit the claim of Mr. Joy on the ground that he did not disclose information about the inflammable goods. (3)

  1. Can you justify the decision of the insurance company?
  2. What principle of insurance is applicable here ?

Answer:

  1. Yes.
  2. Principle of utmost good faith. The insured must disclose all material facts about the subject matter to the insurer. Otherwise the insurer can cancel the contract. The insurer must disclose all the terms and conditions in the insurance contract to the insured.

Plus One Business Services Four Mark Questions and Answers

Question 1.
Mr. Surendran is running an electrical shop which is insured against fire for ₹50,000. Due to heavy rainfall his stock was fully damaged and he claimed for compensation. The insurance company refused to admit the claim as goods were not insured against natural calamities. Can you justify the decision of the insurance company? Which principle of insurance is applicable here? (4)
Answer:
Yes. The decision taken by the insurance company is correct. Mr. Surendran has no right to claim compensation from the insurance company. The principle of insurance applicable here is causa proxima. It states that insurance company will admit the claim only if the mishap has resulted directly by an event covered under the insurance. An event covered under the insurance.

Question 2.
Draw a chart showing types of insurance. (4)
Answer:
Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services 1

Question 3.
Explain the functions of Transport. (4)
Answer:
Functions of Transport:

  1. Helps in the movement of goods and materials from one place to another
  2. Helps in the stabilisation of prices.
  3. Helps in the social, economic and cultural development of the country
  4. Helps in national and international trade
  5. Facilitates large scale production
  6. Generates employment opportunities
  7. Increases growth of towns and cities
  8. Connects all part of the world

Plus One Business Services Five Mark Questions and Answers

Question 1.
Explain the functions of warehousing. (5)
Answer:
Functions of Warehousing:

  1. Warehouse helps in supplying the goods to the customers when it is needed.
  2. By maintaining a balance of supply of goods warehousing leads to price stabilization.
  3. By keeping the goods in the warehouse, the trader can relieve himself of the responsibility of keeping of goods.
  4. The warehouse performs the function of dividing the bulk quantity of goods into smaller quantities.
  5. Warehousing helps in the seasonal storage of goods to select businesses.
  6. The functions of grading, branding and packing of goods can be done in warehouses.
  7. The warehousing receipt can be used as a collateral security for obtaining loans.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services

Question 2.
Name the principle of insurance violated in the following cases. (5)

1. ‘A’does not own the building but is trying to get it insured as a party to the insurance policy.

2.  ‘A’ enters into life insurance contract with LIC of India. ‘A’ was ailing with heart decease but he did not reveal this at the time of entering the contract. ‘

3. ‘B’ enters into insurance contract with two companies ‘C’ and ‘D’. The subject matter is a building worth ₹ 5 lakh. The building caught fire and properties worth ₹3 lakhs were damaged. ‘C’ paid the entire claim and asked ‘D’ to share the claim. ‘D’ denies.

4. Goods worth ₹50,000 are damaged and the insurance company pays the claim to ‘Z’ for the loss. ‘Z’ not only took the compensation for loss but also claims the damaged goods.

5. ‘P’ takes an insurance policy for ₹1 lakh with ‘Q’ company. The goods of ‘P’ are damaged due to fire the loss incurred is ₹25,000. ‘Q’ shall restore the actual loss but ‘P’ claims full amount of the policy.

Answer:

  1. Principle of insurable interest
  2. Principle of utmost good faith
  3. Principle of contribution
  4. Principle of subrogation
  5. Principle of Indemnity

Question 3.
Match A with B and C.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services 2

Answer:

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services 3

Plus One Business Services Six Mark Questions and Answers

Question 1.
Distinguish between life insurance and general insurance. (6)
Answer:

Life Insurance Fire insurance Marine Insurance
The subject matter of insurance is human life The subject matter is any asset. The subject matter is a ship, cargo or freight.
it has the elements of protection and investment It has only the element of Protection It has only the element of Protection.
insurable interest exists at the time of contract alone. Insurable interest exists at the time of contract and at the time of loss. Insurable interest exists at the time of contract
Period of coverage is long period Period of coverage is up to one year Marine insurance policy is for one or period of voyage or mixed.
Life insurance is not a contract of indemnity Fie insurance is a contract of indemnity. Marine insurance is a contract of indemnity
Loss Is not measurable. Loss is measurabie. Loss is measurable.
Surrender of policy is possible Surrender of policy is not possible Surrender of policy is not possible
One can insure for any amount in life insurance. The amount of the policy cannot be more than the value of the subject matter. The amount of the policy can be the market value of the ship or cargo.
It is contract of assurance It is contract of insurance It is contract of insurance

Question 2.
I will give security against fraudulent practices of employees who deal with cash or other valuables

  1. Can you identify me?
  2. Explain my features. (6)

Answer:
1. Fidelity insurance.

2. This type of policy is taken by the employer of a fire to cover the risk arising out of fraud or dishonesty of his employees. Banks, Financial institutions and other business firms commonly use such insurance policies.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services

Question 3.
What are services? Explain their distinct characteristics? (6)
Answer:
Nature of Services
1. Intangibility:
Services are intangible, i.e., they cannot be touched. They are experiential in nature, e.g. Treatment by a doctor.

2. Inconsistency:
Since there is no tangible product, services have to be performed according to the demand and expectations of the different customers, e.g. Mobile services/Beauty parlour.

3. Inseparability:
Another important characteristic of services is the simultaneous activity of production and consumption being performed, i.e. They are inseparable, e.g. ATM may replace clerk but presence of customer is a must.

4. Less Inventory:
Services cannot be stored for future use.

5. Involvement:
Participation of the customer in the service delivery is a must.

Plus One Business Services Eight Mark Questions and Answers

Question 1.
Suppose you have visited a commercial bank in your locality. Then what information you have collected from there with regard to. (8)

  1. Deposit schemes
  2. Lending scheme
  3. Other services

Answer:
Functions of Commercial Banks:
a. Acceptance of deposits:
A commercial bank accepts following types of deposits. They are

1. Current deposit:
This type of account is usually opened by business men.lt carries no interest. There are no restrictions on number of deposits or withdrawals.

2. Savings deposit:
Savings accounts are for encouraging savings by individuals. Banks pay rate of interest as decided by RBI on these deposits. Withdrawal from these accounts has some restrictions.

3. Fixed deposit:
Fixed accounts are time deposits with higher rate of interest. They cannot be withdrawn before the maturity period.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services

b. Lending of funds:
The second important function of a commercial bank is to grant loans and advances out of the money received through deposits. These advances can be made in the form of overdrafts, cash credits, discounting trade bills, term loans, consumer credits and other miscellaneous advances.
1. Cash Credit:
Under this arrangement, the bank advances loan up to a specified limit against current assets. The bank opens an account in the name of the borrower and allows him to withdraw the money up to the sanctioned limit. Interest is charged on the amount actually withdraw.

2. Bank Overdraft:
Overdraft is an arrangement where customers are allowed to withdraw up to a level over and above their deposits in the current account. Interest is to be paid only on the actual amount of overdraft availed by him. OD is granted to businessmen against their current account.

3. Discounting of bill of exchange:
Under this, a bank gives money to its customers on the security of a bill of exchange before the expiry of the bill after deducting a small amount of discount.

4. Term Loans:
These loans are provided by the banks to their customers for a fixed period. Interest will be charged to the entire amount of loan.

c. Secondary Functions:
The secondary functions of commercial banks are as under:
1. Agency Functions:

  • Cheque facility: Bank collects customers cheque drawn on other bank. There are two types of cheques mainly
    Bearer cheques, which are encashable immediately at bank counters Crossed cheques which are to be deposited only in the payees account.
  • Payment of insurance premium, telephone charges, collection of dividend etc
  • Purchase and sale of shares and debentures
  • Acting as trustees or executors
  • Providing credit information

2. General utility services:

  • Providing lockers for safe custody of jewellery and others valuables of customers
  • Issuing letter of credit, pay orders, bank draft, credit cards, traveller’s cheques to customers.
  • Underwriting issues of shares and debentures
  • Providing foreign exchange to importers and traveller’s going abroad.

Bank Draft:
It is a financial instrument with the help of which money can be remitted from one place to another. The payee can present the draft on the drawee bank at his place and collect the money. Bank charges some commission for issuing a bank draft.

Pay Order:
It is almost like a bank draft. Banks issue pay order for local purpose and issue bank draft for outstations.

Question 2.
“For the smooth conducting of insurance the insurer and Insured should know the relevant factors underlying the insurance contracts.” In the light of this statement, can you state the principles of Insurance?
Answer:

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services 3

1. Utmost good faith:
The insured must disclose all material facts about the subject matter to the insured. Otherwise the insurer can cancel the contract. The insurer must disclose all the terms and conditions jn the insurance contract to the insured.

2. Insurable interest:
The insured must have an insurable interest in the subject matter of insurance. Insurable interest means the interest shown by the insured in the continued existence of the subject matter or the financial loss he is subjected to on the happening of an event against which it has been insured.

3. Indemnity:
All insurance contracts, except life insurance are contracts of indemnity. According to the principle of indemnity, in the event of occurrence of loss, the insured will be indemnified to the extent of the actual value of his loss or the sum insured which ever is less. The objective behind this principle is nobody should treat insurance contract as the source of profit.

4. Subrogation:
According to this principle, after the insured is compensated for the loss to the property insured by him the right of ownership of such property passes on to the insurer. This is because the insured should not be allowed to make any profit, by selling the damaged property.

5. Causa proxima:
When the loss is the result of two or more causes, the proximate cause for the loss alone will be considered by the insurance company for admitting the claim.

6. Contribution:
In certain cases, the same subject matter is insured with one or more insurer. In case there is a loss, the insured is eligible to receive a claim only up to the amount of actual loss suffered by him.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services

7. Mitigation of loss:
This principle states that it is the duty of the insured to take reasonable steps to minimize the loss or damage to the insured property. If reasonable care is not taken then the claim from the insurance company may be rejected.

Question 3.
Prepare a seminar paper on Types of warehouses. (8)
Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services 5
Answer:
Warehousing:
Warehousing means holding or preserving goods in huge quantities from the time of their purchase or production till their consumption. Warehousing is one of the important auxiliaries to trade. It creates time utility by bridging the time gap between production and consumption of goods.
Functions of Warehousing:

  • Warehouse helps in supplying the goods to the customers when it is needed.
  • By maintaining a balance of supply of goods warehousing leads to price stabilization.
  • By keeping the goods in the warehouse, the trader can relieve himself of the responsibility of keeping of goods.
  • The warehouse performs the function of dividing the bulk quantity of goods into smaller quantities.
  • Warehousing helps in the seasonal storage of goods to select businesses.
  • The functions of grading, branding and packing of goods can be done in warehouses.
  • The warehousing receipt can be used as a collateral security for obtaining loans.

Types of Warehouses:
Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4 Business Services 5
1. Private warehouses:
Private warehouses are owned by big business concerns or wholesalers for keeping their own products.

2. Public warehouses:
They are owned by some agencies, offer storage facilities to the public after charging certain fees. The working of public warehouses is subject to some govt, regulations. They are also known as Duty paid warehouses.

3. Bonded warehouses:
These warehouses are used to keep the imported goods before the payment of import duties. It offers many advantages to the importer, i.e. The importer can releases the goods in part by paying the proportionate amount of duty. The goods can be branded, blended and packed in the warehouse itself.

4. Government warehouses:
These warehouses are fully owned and managed by the government. For example, Food Corporation of India, State Trading Corporation, and Central Warehousing Corporation.

5. Co-operative warehouses:
Marketing co operative societies and agricultural oo operative societies have set up their own warehouses for members of their cooperative society.

Plus One Economics Chapter Wise Questions and Answers Chapter 2 Collection of Data

Students can Download Chapter 2 Collection of Data Questions and Answers, Plus One Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Economics Chapter Wise Questions and Answers Chapter 2 Collection of Data

Plus One Economics Collection of Data One Mark Questions and Answers

Question 1.
Which among the following is a method of primary data collection?
(i) Direct personal interview
(ii) Telephonic interview
(iii) Mailed questionnaire method
(iv) All the above
Answer:
(iv) All the above

Question 2.
Identify the random sampling method from the following.
(i) Judgment sampling
(ii) Stratified sampling
(iii) Convenience sampling
(iv) Quota sampling
Answer:
(ii) Stratified sampling

Question 3.
Which among the following is a source of secondary data?
(i) CSO
(ii) NSSO
(iii) RGI
(iv) All the above
Answer:
(iv) All the above

Plus One Economics Chapter Wise Questions and Answers Chapter 2 Collection of Data

Question 4.
Identify an advantage of sample method.
(i) Less time consuming
(ii) Less cost is needed
(iii) Gives more reliable information
(iv) All the above
Answer:
(iv) All the above

Question 5.
Name the different types of data.
Answer:
Primary data and secondary data

Plus One Economics Collection of Data Two Mark Questions and Answers

Question 1.
Differentiate between census survey and sample survey.
Answer:
Under census method, we collect information from each and every unit of population relating to the problem under investigation. On the other hand, under-sample method, rather than collecting information about all the units of population, we collect information from a few selected items from population.

Question 2.
Name the two important errors related to sampling?
Answer:

  1. Sampling error.
  2. Non – sampling error.

Question 3.
Give one example of primary data and one example of secondary data.
Answer:
The price of rice collected from a shop is primary data. The price of rice collected from a daily is secondary data.

Question 4.
Do samples provide better results than surveys? Give reasons for your answer.
Answer:
In sample method, we take part of the information. A small number of trained investigators can collect complete data. This information is reliable. In the case of doubt, enquiries can be undertaken for verification. Follow up is made easy in this case. In survey many difficulties are faced when the universe is very large. A large number of investigators are required in surveys. In case of doubts, it is difficult to undertake the enquiry again for verification.

Question 5.
List out the different steps involved in planning the survey.
Answer:

  1. Purpose of the survey
  2. Scope of the survey
  3. Unit of data collection
  4. Sources of data
  5. Technique of data collection
  6. Choice of the frame
  7. Degree of accuracy needed
  8. Miscellaneous considerations

Question 6.
Point out the advantages of sample method over census method?
Answer:
The advantages of sample method over census method are:

  1. Less time consuming
  2. Less cost is needed,
  3. Gives more reliable information
  4. Gives more detailed information.

Plus One Economics Chapter Wise Questions and Answers Chapter 2 Collection of Data

Question 7.
Suppose you want to select 50 households from a total of 5100 households by giving equal opportunity to all the units. Which method you would prefer for this purpose? Substantiate.
Answer:
Simple random sampling is appropriate for this purpose. It refers to that technique in which each and every unit of population has an equal opportunity of being selected in the sample. The selection depends purely on chance. Random samples may be selected by lottery method.

Plus One Economics Collection of Data Three Mark Questions and Answers

Question 1.
What are the advantages of mailed questionnaire method?
Answer:
Advantages of mailed questionnaire method are the following.

  • Least expensive
  • The only method to reach remote areas
  • No influence on respondents
  • Better for sensitive questions

Question 2.
How the non-sampling errors can be controlled?
Answer:
Sampling errors can be controlled by taking the following steps.

  1. Sample should be selected with care.
  2. Scientific methods of data collection should be used.
  3. The investigator should be well trained.
  4. A test check should be done.

Question 3.
List out the errors you would expect in the field survey by census method?
Answer:
Some of the errors expected in the field survey by census method are:

  1. Errors of measurement
  2. Errors due to mishandling of the questionnaire
  3. Recording mistakes
  4. Errors of non-response
  5. Arithmetic errors

Question 4.
Suppose the Wayanad district Panchayat wants to study the literacy of tribal population in the district.

  1. Which method would you suggest for data collection? Why?
  2. Give a popular instance Is this method suitable in all cases? If not, what are the drawbacks.

Answer:
1. I would suggest census method of data collection to the panchayat authorities in this situation. In this method, we get reliable information from all units of the population for which the study is conducted.

2. A popular instance where this method is used is population census conducted ever 10 years.

3. It is not suitable in all cases because there are some drawbacks – the effort, money and time required will be very high.

Question 5.
Give two examples each of sample, population and variable.
Answer:
1. Examples each of samples:

  • A representative group of few individuals taken form a particular locality to study the impact of rise in prices of petroleum products
  • A group of students from the school is selected to study average monthly expenditure

2. Examples each of population:

  • All individuals residing in a particular locality about . whom the information is to be collected
  • All students in a school for whom the information is to be collected.

3. Examples of each of variables:

  • Changing price of petrol
  • Increase or decrease in the production of rice.

Question 6.
List the demerits of telephone interviews.
Answer:
Demerits of telephone interview are:

  • Reactions cannot be watched
  • Limited use
  • Possibility of influencing respondents.

Plus One Economics Chapter Wise Questions and Answers Chapter 2 Collection of Data

Question 7.
Point out the precautions to be taken while using secondary data?
Answer:
While using secondary data it should be ensured that the following precautions are taken

  1. Ability of the collecting organisation
  2. Objective and scope of investigation
  3. Method of collection of secondary data
  4. Time and conditions of collection of data Definition of the unit, and
  5. Accuracy of the data

Question 8.
Point out the advantages of conducting pilot survey.
Answer:
Once the questionnaire is ready, it is advisable to conduct a try out with a small group which is known as Pilot Survey or PreTesting of the questionnaire. The pilot survey helps the investigator in the following ways.

  • The pilot survey helps in providing a preliminary idea about the survey.
  • It helps in pretesting of the questionnaire, so as to know the shortcomings and drawbacks of the questions.
  • Pilot survey also helps in assessing the suitability of questions, clarity of instructions, performance of enumerators and the cost and time involved in the actual survey

Question 9.
Explain the procedure of selecting a random sample of 3 students out of 10 in your class by using random number table.
Answer:
The procedures of selecting a random sample of 3 students out of 10 are as follows

  1. Prepare 10 slips of paper of identical size and shape
  2. Write the names of students on the slips
  3. Place the slips in a box and mix them properly
  4. Draw three slips without looking into the box
  5. The students, whose names appear on the slips drawn, constitute the required random sample.

Question 10.
Mention the important sources of collecting secondary data.
Answer:
The main sources of secondary data can be classified into two. They are:
A. Published sources

  • Government publications
  • Semi government publications
  • International publications
  • Private publications
  • Research publications

B. Unpublished sources

  • Studies made by scholars
  • Studies of various institutions

Question 11.
Prepare a note on NSSO?
Answer:
The National Sample Survey Organisation was established by the Government of India to conduct nationwide surveys on socioeconomic issues. The data collected by NSSO surveys on different socioeconomic subjects are released through reports and its quarterly journal Sarvekshana.

NSSO provides periodic estimates of literacy, school enrolment, utilisaiton of educational services, employment, unemployment, manufacturing and service sector enterprises, morbidity, maternity, child care, utilisation of the public distribution system, etc.

The NSSO also undertakes the fieldwork of annual surveys of industries, conducts crop estimation surveys, collects rural and urban retail prices for complication of consumer price index numbers.

Plus One Economics Chapter Wise Questions and Answers Chapter 2 Collection of Data

Question 12.
“Non-sampling errors are more serious than sampling errors”. Do you agree? Mention the possible nonsampling errors occur in sampling.
Answer:
Yes, I agree with the statement that non-sampling errors are more serious than sampling errors because a sampling error can be minimized by taking a larger sample. It is difficult to minimize non-sampling error, even by taking a large sample. Even a Census can contain nonsampling errors.
Some of the non-sampling errors are:
1. Errors in Data Acquisition:
This type of error arises from recording of incorrect responses. Suppose, the teacher asks the students to measure the length of the teacher’s table in the classroom. The measurement by the students may differ. The differences may occur due to differences in measuring tape, carelessness of the students, etc.

2. Non-Response Errors:
Non-response occurs if an interviewers unable to contact a person listed in the sample or a person from the sample refuses to respond. In this case, the sample observation may not be representative.

3. Sampling Bias:
Sampling bias occurs when the sampling plan is such that some members of the target population could riot possibly be included in the sample.

Plus One Economics Collection of Data Four Mark Questions and Answers

Question 1.
Prepare a chart showing different methods of sampling.
Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 2 Collection of Data img1

Question 2.
Give the advantages of direct personal interview.
Answer:
Advantages of direct personal interview.

  • Highest response rate
  • Allows use of all types of questions
  • Allow clarification of ambiguous questions
  • Better for using open ended questions

Question 3.
“Certain principles should be kept in mind while drafting a questionnaire” explain.
Answer:
The principles that are to be kept in mind while drafting a questionnaire are given below.

1. The questionnaire should not be too long. The number of questions should be as minimum as possible. Long questionnaires discourage people from completing them.

2. The series of questions should move from general to specific. The questionnaire should start from general questions and proceed to more specific ones.

3. The questions should be precise and clear.

4. The questions should not be ambiguous, to enable the respondents to answer quickly, correctly and clearly.

5. The question should not use double negatives. The questions starting with “Wouldn’t you” or “Don’t you” should be avoided, as they may lead to biased responses.

6. The question should not be a leading question, which gives a clue about how the respondent should answer.

7. The question should not indicate alternatives to the answer.

Plus One Economics Collection of Data Five Mark Questions and Answers

Question 1.
State true or false

  1. there are many sources of data
  2. telephone survey is the most suitable method of collecting data when their population is literate and spread over a large area
  3. data collecting by investigator is called secondary data
  4. non-sampling errors can be minimized by taking large samples
  5. there is a certain bias involved in the non-random selection of samples.

Answer:

  1. true
  2. true
  3. false
  4. false
  5. true

Plus One Economics Collection of Data Eight Mark Questions and Answers

Question 1.
Prepare a seminar report on the topic “methods of collecting primary data”
Answer:
“Methods of collecting primary data”
Introduction:
The plus one commerce batch of (Name of school) conducted a seminar on the topic “methods of collecting primary data” on 05/08/2017 at 11.0 a.m with the ample guidance of our economics teacher. Our class was divided into three groups to cover the entire area of the topic. Each group presented one objective each. They focused on the area given to them and the group leader presented the seminar paper after required preparation.

Content:
There are three basic ways of collecting data:

  1. Personal Interviews,
  2. Mailing (questionnaire) Surveys, and
  3. Telephone Interviews

1. Personal Interviews:
This method is used when the researcher has access to all the members. The researcher (or investigator) conducts face to face interviews with the respondents. Personal interviews are preferred due to various reasons. Personal contact is made between the respondent and the interviewer.

The interviewer has the opportunity of explaining the study and answering any query of the respondents. The interviewer can request the respondent to expand on answers that are particularly important. Misinterpretation and misunderstanding can be avoided.

Watching the reactions of the respondents can provide supplementary information. Personal interview has some demerits too. It is expensive, as it requires trained interviewers. It takes longer time to complete the survey.

Plus One Economics Chapter Wise Questions and Answers Chapter 2 Collection of Data

2. Mailing Questionnaire:
When the data in a survey are collected by mail, the questionnaire is sent to each individual by mail with a request to complete and return it by a given date. The advantages of this method are that it is less expensive. It allows the researcher to have access to people in remote areas too, who might be difficult to reach in person or by telephone.

It does not allow influencing of the respondents by the interviewer. It also permits the respondents to take sufficient time to give thoughtful answers to the questions. The disadvantages of mail survey are that there is less opportunity to provide assistance in clarifying instructions, so there is a possibility of misinterpretation of questions. Mailing is also likely to produce low response rates due to certain factors.

3. Telephone Interviews:
In a telephone interview, the investigator asks questions over the telephone. The advantages of telephone interviews are that they are cheaper than personal interviews and can be conducted in a shorter time. They allow the researcher to assist the respondent by clarifying the questions.

Telephone interview is better in the cases where the respondents are reluctant to answer certain questions in personal interviews. The disadvantage of this method is accessible to people, as many people may not own telephones. Telephone Interviews also obstruct visual reactions of the respondents.

Conclusion:
All the three groups presented their topics with necessary facts and figures. After the presentation, there was the question-answer session. The active participation of everyone made this session live and interesting. On the whole, the seminar was a big success.

Plus One Economics Chapter Wise Questions and Answers Chapter 3 Organisation of Data

Students can Download Chapter 3 Organisation of Data Questions and Answers, Plus One Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Economics Chapter Wise Questions and Answers Chapter 3 Organisation of Data

Plus One Economics Organisation of Data One Mark Questions and Answers

Question 1.
Under inclusive method ………….
(a) Upper-class limit is excluded
(b) Lower class limit is excluded
(c) Lower class limit is included
(d) None of the above
Answer:
(c) Lower class limit is included

Question 2.
Which of the following alternatives is true?
The class midpoint is equal to:
(a) The average bf the upper-class limit and the lower class limit.
(b) The product of the upper-class limit and the lower class limit.
(c) The ratio of the upper-class limit and the lower class limit.
(d) None of the above.
Answer:
(a) The average bf the upper-class limit and the lower class limit.

Question 3.
Which of the following alternatives is true?
The frequency distribution of two variables is known as ………
(a) Univariate Distribution
(b) Bivariate Distribution
(c) Multivariate Distribution
(d) None of the above
Answer:
(b) Bivariate Distribution

Question 4.
Which of the following alternatives is true?
Statistical calculations in classified data are based on …………
(a) the actual values of observations
(b) the upper-class limits
(c) the lower class limits
(d) the class midpoints
Answer:
(d) the class midpoints

Plus One Economics Chapter Wise Questions and Answers Chapter 3 Organisation of Data

Question 5.
Which of the following alternatives is true?
Under Exclusive method.
(a) the upper-class limit of a class is excluded in the class interval
(b) the upper-class limit of a class is included in the class interval
(c) the lower class limit of a class is excluded in the class interval
(d) the lower class limit of a class is included in the class interval
Answer:
(a) the upper-class limit of a class is excluded in the class interval

Question 6.
Which of the following alternatives is true?
Range is the ………….
(a) difference between the largest and the smallest observations
(b) difference between the smallest and the largest observations
(c) average of the largest and the smallest observations
(d) ratio of the largest to the smallest observation.
Answer:
(a) difference between the largest and the smallest observations

Plus One Economics Organisation of Data Two Mark Questions and Answers

Question 1.
What is classification?
Answer:
Classification of data is a technique with the help of which data is arranged into different groups or classes according to some common characteristics so as to facilitate the tabulation, analysis, and interpretation.

Question 2.
Distinguish between raw data and classified data.
Answer:
Raw data is the collected information in the form of numerical facts. It is the data collected as it is without any processing. The raw data is summurised and made comprehensible is known as classified data.

Question 3.
Distinguish between frequency table and frequency array.
Answer:
Frequency table is the classification of the data for a continuous variable. Frequency array is the classification of the data for a discrete variable.

Question 4.
Distinguish between univariate frequency distribution and bivariate frequency distribution.
Answer:
The frequency distribution of a single variable is called univariate distribution. The frequency distribution of two variables is known as bivariate frequency distribution.

Plus One Economics Chapter Wise Questions and Answers Chapter 3 Organisation of Data

Question 5.
In a city 45 families were surveyed for the number of domestic appliences they used. Prepare a frequency array based on their replies as recorded below.
1, 3, 2, 2, 2, 2, 1, 2, 2, 3, 3, 3, 3, 2, 4, 2, 7, 2, 0, 3, 1, 3, 3, 2, 3, 2, 2, 6, 1, 6, 2, 1, 5, 1, 5, 3, 4, 2, 4, 3, 4, 3
Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 3 Organisation of Data img1

Question 6.
Classification of data is a function similar to that of sorting letters in a post office. Do you agree? Explain.
Answer:
Yes, I do agree with the statement that the classification of data is a function similar to that of sorting letters in a post office.

The process of arranging data in groups or classes according to similarities is technically called classification. The classification is somewhat similar to that of sorting letters in a post office. Letters collected in a post office are sorted into different lots on geographical basis.

They are then put in separate bags. Thus by classification, we are trying to make different groups with similar characteristics. Units having a common characteristic place in one class and the whole data are thus divided into a number of classes.

Question 7.
List examples of variables and attributes in your daily life.
Answer:
1. Variables

  • Height
  • Weight
  • Marks obtained in final examination
  • Cricket score

2. Attributes

  • Caste religion
  • Sex
  • Marital status
  • Educational qualification

Plus One Economics Chapter Wise Questions and Answers Chapter 3 Organisation of Data

Question 8.
There are some problems in preparing frequency distribution on the basis of class interval’. Do you agree? Justify.
Answer:
Yes, I agree.
The following problems arise in preparing frequency distribution on the basis of class interval.

  1. Number of class intervals to be formed
  2. Size of class intervals
  3. Class limits.

Plus One Economics Organisation of Data Three Mark Questions and Answers

Question 1.
Answer the following

  1. To arrange things in appropriate order and to put them into homogenous groups means ………..
  2. a variable is continuous when it can be taken within the range ……….
  3. Population refers to ……………

Answer:

  1. classification
  2. any value
  3. both frequency and cumulative frequency

Question 2.
Name the three types of series.
Answer:

  1. Individual series
  2. Discrete series
  3. Continuous series

Plus One Economics Organisation of Data Four Mark Questions and Answers

Question 1.
Can there be any advantage in classifying things?
Answer:
The advantage of classification are the following.

  • It helps in the comparison of data.
  • It helps to understand relationship among variables.
  • It makes statistical analysis easier.
  • It highlights significant features of data at a glance.

Question 2.
State the meaning and example for the following.

  1. Chronological Classification.
  2. Spatial Classification
  3. Qualitative Classification
  4. Quantitative Classification.

Answer:
1. Chronological Classification:
In Chronological classification, data are classified either in ascending or in descending order with reference to time such as years, quarters, months, weeks, etc.
Example: population of India from 1951 to 2011

2. Spatial Classification:
In Spatial Classification the data are classified with reference to geographical locations such as countries, states, cities, districts, etc.
Example: yield of wheat in different states of India.

3. Qualitative Classification:
Attributes can be classified on the basis of either the presence or the absence of a qualitative characteristic. Such a classification of data on attributes is called a Qualitative Classification.
Example: grouping of people on the basis of gender

4. Quantitative Classification:
When the collected data of such characteristics are grouped into classes, the classification is a Quantitative Classification.
Example: marks obtained by students.

Plus One Economics Chapter Wise Questions and Answers Chapter 3 Organisation of Data

Question 3.
What is variable? Distinguish between a discrete and a continuous variable.
Answer:
A simple definition of variable does not tell you how it varies. Different variables vary differently and depending on the way they vary, they are broadly classified into two types:

  1. Continuous and
  2. Discrete.

A continuous variable can take any numerical value. It may take integral values (1, 2, 3, 4,…) or fractional values(1/2, 2/3, 3/4).

Unlike a continuous variable, a discrete variable can take only certain values. Its value changes only by finite “jumps”. It “jumps” from one value to another but does not take any intermediate value between them.

Question 4.
Differentiate between exclusive method and inclusive method.
Answer:
1. exclusive method, the upper-class limit is excluded but the lower class limit of a class is included in the interval.

2. On the other hand, under the inclusive method, the upper-class limit is included in the class interval.

Plus One Economics Organisation of Data Five Mark Questions and Answers

Question 1.
Differentiate between quantitative classification and qualitative classification.
Answer:
Classification done according to quantitative varieties like marks and wages etc. is termed as quantitative classification. On the other hand, classification according to attributes like honesty and beauty is known as qualitative classification.

Question 2.
Prepare a frequency distribution by inclusive and exclusive method for the following data.
28 27 6 16 15 2 14 5 19 25 19 20 28
32 37 13 15 11 32 4 6 9 3 36 12 8
4 1 8 3 18 12 7 17 15 22 29 21 23
2 9 4 10 5 20 20 33 27 21 27 18 9
31 18 9 7 1 26 24 20
Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 3 Organisation of Data img2

Plus One Economics Organisation of Data Eight Mark Questions and Answers

Question 1.
Define and illustrate the inclusive and exclusive method used in classification of data.
Answer:
1. Inclusive Method:
In comparison to the exclusive method, the Inclusive Method does not exclude the upper-class limit in a class interval. It includes the upper class in a class. Thus both class limits are parts of the class interval.
An example for inclusive method of frequency distribution is given below.
Plus One Economics Chapter Wise Questions and Answers Chapter 3 Organisation of Data img3

2. Exclusive Method:
The classes, by this method, are formed in such a way that the upper-class limit of one class equals the lower class limit of the next class. In this way, the continuity of the data is maintained. That is why this method of classification is most suitable in case of data of a continuous variable. Under the method, the upper-class limit is excluded but the lower class limit of a class is included in the interval.
An example for exclusive method of frequency distribution is given below.
Plus One Economics Chapter Wise Questions and Answers Chapter 3 Organisation of Data img4

Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion

Students can Download Chapter 6 Measures of Dispersion Questions and Answers, Plus One Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion

Plus One Economics Measures of Dispersion One Mark Questions and Answers

Question 1.
Find out the measure of dispersion from the following.
Answer:
(i) Range
(ii) Mean deviation
(iii) Standard deviation
(iv) All the above
Answer:
(iv) All the above

Question 2.
Variance is equal to:
(i) SD
(ii) SD²
(iii) √SD
(iv) SD/2
Answer:
(ii) SD²

Question 3.
Lorenz curve is used to measure
Answer:
(i) Literacy rate
(ii) Unemployment
(iii) Inequality
(iv) None of these
Answer:
(iii) Inequality

Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion

Question 4.
Which is the graphical method of determining dispersion?
(i) Range
(ii) Q.D
(iii) S.D
(iv) Lorenz curve
Answer:
(iv) Lorenz curve

Question 5.
Which among the following is not an absolute measure of dispersion?
(i) Range
(ii) Co-efficient of variation
(iii) Mean deviation
(iv) Standard deviation
Answer:
(ii) Co-efficient of variation

Question 6.
Which of the following is the most widely used measure of dispersion?
(i) Range
(ii) Standard deviation
(iii) Mean deviation
(iv) Quartile deviation
Answer:
(ii) Standard deviation

Question 7.
Which of the following is useful in comparing the variability of two or more distributions?
(i) Lorenz Curve
(ii) Standard deviation
(iii) Mean deviation
(iv) Quartile deviation
Answer:
(iii) Mean deviation

Question 8.
Which of the following is not based on all the values?
(i) Mean deviation from mean
(ii) Mean deviation from median.
(iii) Range
(iv) Standard deviation
Answer:
(iii) Range

Question 9.
Mean deviation
(i) is least when calculated from median and higher when calculated from mean.
(ii) is least when calculated from mean and higher when calculated from median.
(iii) same when calculated from median and mean.
(iv) cannot be calculated either from median or mean.
Answer:
(i) is least when calculated from median and higher when calculated from mean.

Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion

Question 10.
Mention the purpose for which Lorenz Curve is drawn.
Answer:
Lorenz curve is drawn for the presentation of inequality (or dispersion)

Plus One Economics Measures of Dispersion Two Mark Questions and Answers

Question 1.
Which measure of dispersion is the best and how?
Answer:
Standard deviation is the best measure of dispersion. Because it possesses most of the characteristics of an ideal measure of dispersion.

Question 2.
Range is not based on all the values. Do you agree? Substantiate.
Answer:
Yes, I agree with the statement that range is not based on all the values. It is simply the difference between the largest (L) and the smallest value (S) in a distribution. Therefore, range cannot be considered as a good measure of dispersion.

Question 3.
Give two limitations of range.
Answer:
Given below two limitations of range.

  1. range is unduly affected by extreme values
  2. range is not based on all values.

Question 4.
Differentiate between range and interquartile range.
Answer:
Range is the difference between the value of extreme while interquartile range is the difference between quarter extreme values of the series.

Question 5.
Some measures of dispersion depend upon the spread of values whereas some calculate the variation of values from a central value. Do you agree?
Answer:
Yes, I agree with this statement. Range and quartile deviation measure the dispersion by calculating the spread within which the values lie. That is they de-pend.on the spread of values. On the other hand, mean deviation and standard deviation calculate the variation of value from a central value.

Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion

Question 6.
Write the merits and demerits of mean deviation.
Answer:
1. Merits

  • It is based on all observations
  • It is less affected by extreme items
  • It is easy to calculate & simple to understand

2. Demerits

  • It ignores + signs
  • It is difficult to compute when
  • or median comes infraction.

Plus One Economics Measures of Dispersion Three Mark Questions and Answers

Question 1.
Calculate coefficient of standard deviation from the following information SD = 22.4, Mean = 159
Answer:
Coefficient of standard deviation = \(\frac{\mathrm{SD}}{\bar{x}} \times 100\)
\(=\frac{22.4}{159} \times 100=0.14 \times 100=14\)

Question 2.
Following are certain measures of dispersion. Classify them into absolute measures and relative measures.
{Range, Coefficient of variation, Mean Deviation, Coefficient of Quartile Deviation, coefficient of Range, Standard deviation}
Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion img1

Question 3.
Find odd ones out and justify your answer.

  1. range, quartile deviation, mode, standard deviation
  2. scatter, spread, variation, correlation
  3. Range, Mean Deviation, coefficient of variation, Standard deviation.

Answer:

  1. Mode. Others are methods of studying measures of dispersion.
  2. Correlation. Others are different expressions of dispersion.
  3. Coefficient of variation. Others are absolute measures of dispersion.

Question 4.
Calculate the coefficient of variation, if,
Number of Boys = 72
Average Height = 68
The variance of distribution = 9
Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion img2

Question 5.
Point out features of good measure of dispersion.
Answer:
The features of a good measure of dispersion are pointed out below:

  • It should be easy to understand
  • It should be simple to calculate
  • It should be rigidly defined
  • It should be based on all observations
  • It should not be affected by extreme values

Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion

Question 6.
The size of landholdings of 5 persons are given below:

Individuals Land in acres
A 1.5
B 3.0
C 0.5
D 6.0
E 2.0

 

  1. Analyse the data by measuring the dispersion using the easiest method.
  2. What are the drawbacks of that method?

Answer:
1. Range is the simplest method of studying dispersion.
Range = L – S
where L = Largest item
S = Smallest item
Range = 6 – 0.5 = 5.5 acres
2.

  • It is not based on each and every item of the distribution.
  • It is subject to fluctuations from sample to sample.
  • It cannot be computed in case of open-end classes.

Question 7.
If the sum of 10 values is 100 and the sum of their square is 1090, find the coefficient of variation.
Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion img3

Plus One Economics Measures of Dispersion Four Mark Questions and Answers

Question 1.
In a town, 25% of the persons earned more than Rs 45,000 whereas 75% earned more than 18,000. Calculate the absolute and relative values of dispersion.
Answer:
We get the following information from the above.
Q1 = Rs. 18000 and Q3 = Rs. 45000
Absolute measure of dispersion (quartile deviation)
= Q3 – Q1/2 = 45000 – 18000/2
= Rs. 13500
Relative measure of dispersion (coefficient of Q. D)
= Q3-Q1/Q3 + Q1
= 45000 -18000 / 45000 + 18000
= 27000/63000
= 3/7
= 0.43

Question 2.
What do you mean by dispersion? Name the important methods of measuring dispersion.
Answer:
Dispersion is the extent to which values in a distribution differ from the average of the distribution.
To quantify the extent of the variation, there are certain measures namely:

  1. Range
  2. Quartile Deviation
  3. Mean Deviation
  4. Standard Deviation
  5. Lorenz curve

Question 3.
Mention the merits of standard deviation.
Answer:
Merits of standard deviation are:

  1. It is rigidly defined
  2. It is based on all observations
  3. It takes algebraic signs in consideration
  4. It serves the basis of other measures like correlation.

Question 4.
The number of telephone calls received at an exchange in 245 successive one-minute intervals are shown in the following frequency distribution. Compute the mean deviation about the median.

No. of calls Frequency
0 14
1 21
2 25
3 43
4 51
5 40
6 39
7 12

Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion img4

Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion

Question 5.
The coefficient of variation of two series are 58% and 69% and their standard deviation are 21.2 and 15.6. Calculate their mean.
Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion img5

Question 6.
Calculate the standard deviation.

Marks Frequency
20 – 40 3
40-80 6
80-100 20
100-120 12
120-140 9

Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion img6

Plus One Economics Measures of Dispersion Five Mark Questions and Answers

Question 1.
Match the following.

A B
Measures of dispersion Karl Pearson
Standard deviation SD2
Graphical method SD/mean x 100
Variance Average of the second order
Quartile deviation Semi-interquartile range
Coefficient of variation Lorenz curve

Answer:

A B
Measures of dispersion Average of the second order
Standard deviation Karl Pearson
Graphical method Lorenz curve
Variance SD2
Quartile deviation Semi-inter quartile range
Coefficient of variation SD/mean x 100

Plus One Economics Measures of Dispersion Eight Mark Questions and Answers

Question 1.
What do you mean by Lorenz curve? Give the steps required in the Construction of the Lorenz Curve.
Answer:
The measures of dispersion discussed so far give a numerical value of dispersion. A graphical measure called Lorenz Curve is available for estimating dispersion. In constructing a Lorenz curve, the following steps are involved.

  1. Calculate class mid-points and find cumulative totals
  2. Calculate cumulative frequencies
  3. Express the grand totals and convert the cumulative totals in these columns into percentages,
  4. Now, on the graph paper, take the cumulative percentages of the variable on Y-axis and cumulative percentages of frequencies on X-axis. Thus each axis will have values from ‘0’to ‘100’.
  5. Draw a line joining Co-ordinate (0, 0) with (100,100). This is called the line of equal distribution
  6. Plot the cumulative percentages of the variable with corresponding cumulative percentages of frequency.
  7. Join these points to get the curve OAC.

Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion

Question 2.
Calculate

  1. standard deviation and
  2. mean deviation

from the following data.
X = 7, 9, 18, 11, 10, 8, 17, 13, 11, 16
Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion img7
1. Standard deviation
\(s=\sqrt{\frac{\sum d^{2}}{N}}\)
\(=\sqrt{\frac{134}{10}}=3.66\)

2. Median deviation
\(M D=\frac{\Sigma|d|}{N}=\frac{32}{10}=3.2\)

Question 3.
The following table gives the distribution of income of 100 families in a village. Calculate standard deviation.

Income No. of Families
0-1000 18
1000-2000 26
2000-3000 30
3000 – 4000 12
4000 – 5000 10
5000 – 6000 4

Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion img8

Question 4.
Write down the formula for finding out various measures of dispersion.
Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion img9
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion img10
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Measures of Dispersion img11