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Students can Download Chapter 4 Motion in a Plane Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 4 Motion in a Plane

Summary
Motion In A Plane Class 11 Notes Pdf Introduction
In this chapter, we will study, about vector, its ’ addition, substraction and multiplication We then discuss motion of an object in a plane. We shall also discuss uniform circular motion in detail.

Plus One Physics Chapter 4 Notes Scalars And Vectors

a. Scalars:
A quantity which has only magnitude and no direction is called a scalar quantity.
Eg: length; volume, mass, time, work etc.

b. Vectors:
(i) The need for vectors:
In one dimensional motion, there are only two possible directions. But in two or three dimensional motion, infinite number of directions are possible. Hence quantities like displacement, velocity, force etc. cannot be represented by magnitude alone: Therefore in order to describe such quantities, not only magnitude but direction also is essential.

(ii) Vector:
A physical quantity which has both magnitude and direction is called a vector quantity.
Eg: Displacement, Velocity, Acceleration, Force, momentum.

1. Position and Displacement Vectors:
Position vector:
Consider the motion of an object in a plane. Let P be the position of object at time tw.r.t.origin given O.

A vector representing the position of an object P with respect to an origin O is called position vector \(\overrightarrow{\mathrm{OP}}\) of the object. This position vector may be represented
by an arrow with tail at O and head at P.

The length of the line gives the magnitude of the vector and arrow head (tip) indicates its direction in space. The magnitude of OP is represented by |\(\overrightarrow{\mathrm{OP}}\)|.

Displacement vector:

Consider the motion of an object in a plane. Let P be the position of a moving object at a time t and p¹ that at a later time t¹. \(\overrightarrow{\mathrm{OP}}\) and \(\overrightarrow{\mathrm{OP}^{1}}\) are the position vectors at time t and t¹ respectively. So the vector \(\overrightarrow{\mathrm{PP}^{1}}\) is called displacement vector corresponding to the motion in the time interval (t – t¹).

2. Equality of vectors:
Two vectors are said to be equal if they have the same magnitude and direction.

The above figure shows two vectors \(\vec{A}\) and \(\vec{B}\) having the same magnitude and direction.
∴ \(\vec{A}\) = \(\vec{B}\).

Motion In A Plane Class 11 Notes Hsslive Question 1.
Observe the following figures (a) and (b) and find which pair does represents equal vectors?

Answer:
Figure a represent that A and B are equal vectors. Two vectors A¹ and B¹ are unequal, because they were in different directions.

Multiplication Of Vectors By Real Numbers
Multiplying a vector \(\vec{A}\) with a positive number I gives a vector whose magnitude is changed by the factor λ.

The direction λ\(\vec{A}\) is the same as that of \(\vec{A}\).
Examples:

A vector \(\vec{A}\) and the resultant vector after multiplying \(\vec{A}\) by a positive number 2.

A vector A and resultant vector after multiplying it by a negative number-1 and -1.5.

Addition And Subtraction Of Vectors – Graphical Method
Vectors representing physical quantities of the same dimensions can be added or subtracted. The sum of two or more vectors is known as their resultant.

1. When two vectors are acting in the same direction:

2. When two vectors act in opposite direction:
In this case, the angle between the vectors is 180°.

The resultant of the two vectors is a new vector whose magnitude is the difference between the magnitudes of the two vectors and whose direction is the same as the direction of the bigger vector.

3. When two vectors are inclined to each other:
The sum of two vectors inclined at an angle q can be obtained either by

• the law of triangle of vectors
• the parallelogram law of vectors

(i) Triangle method:
This law states that if two vectors can be represented in magnitude and direction by the two sides of a triangle taken in the same order, then the resultant is represented in magnitude and direction by the third side of the triangle taken in the reverse order.

Explanation
Consider two vectors \(\vec{A}\) and \(\vec{B}\) as shown in figure.

(ii) Parallelogram law of vector addition:
This law states that if two vectors acting at a point can be represented in magnitude and direction by the two adjacent sides of a parallelogram, then the diagonal of the parallelogram through that point represents the resultant vector.
Explanation
Consider two vectors \(\vec{A}\) and \(\vec{B}\) as shown in figure.

To find \(\vec{A}\) + \(\vec{B}\), we bring theirtails to a common origin Q as shown below.


The diagonal of parallelogram OQSP, gives the resultantof (\(\vec{R}\) = \(\vec{A}\) + \(\vec{B}\)) of two vectors \(\vec{A}\) and \(\vec{B}\).
Note: Triangle and parallelogram law of vector addition gives the same result, ie. the two methods are equivalent.

4. Substraction of vectors:

To substract \(\vec{B}\) from \(\vec{A}\), reverse the direction of \(\vec{B}\).

Then add –\(\vec{B}\) with \(\vec{A}\) using parallelogram law or tri¬angle law.

The resultant of \(\vec{A}\) and \(\vec{B}\) is given by \(\vec{R}\).
Null vector or zero vector:
A vector having zero magnitude is called a zero vector or null vector. Null vector is represented by \(\vec{O}\). Since the magnitude is zero, we don’t have to specify its direction.
Properties of null vector:

Class 11 Physics Motion In A Plane Notes Pdf Question 2.
Explain a zero vector using an example.
Answer:
Suppose that an object which is at P at time t, moves to p¹ and then comes back to P. In this case displacement is a null vector.

Resolution Of Vectors Unit Vectors
A vector divided by its magnitude is called unit vector along the direction of that vector. A unit vector in the direction of \(\vec{A}\) is written as \(\hat{A}\).

Orthogonal unit vectors:

In the Cartesian coordinate system, the unit vectors along the X, Y and Z directions are represented by \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) respectively and are known as orthogonal unit vectors.
For unit vectors

Resolution of vector into rectangular components:
The components of a vector in two mutually perpendicular directions are called its rectangular components.
Explanation

Consider a vector \(\overrightarrow{\mathrm{A}}\) that lies in x-y plane as shown in figure. To resolve \(\overrightarrow{\mathrm{A}}\), draw lines from the head of \(\overrightarrow{\mathrm{A}}\) perpendicularto the coordinate axes as shown below.

The quantities A_x and A_y are called x and y components of the vector \(\overrightarrow{\mathrm{A}}\). Hence the vector \(\overrightarrow{\mathrm{A}}\) can be written in terms of rectangular components as

Magnitude of \(\overrightarrow{\mathrm{A}}\):

From the figure, the magnitude of \(\overrightarrow{\mathrm{A}}\) can be written as,

Vectors Physics Class 11 Notes Pdf Question 3.
A vector \(\overrightarrow{\mathrm{A}}\) in xyz plane is given below. A_x, A_y and A_z are the perpendicular components in x,y and z directions respectively.

1. Write \(\overrightarrow{\mathrm{A}}\) in terms of rectangular components.
2. Write the magnitude of \(\overrightarrow{\mathrm{A}}\).

Answer:

The magnitude of vector \(\overrightarrow{\mathrm{A}}\) is

Vector Addition – Analytical Method
The graphical method of adding vectors helps us in visualizing the vectors and the resultant vector. But this method has limited accuracy and sometimes tedious. Hence we use analytical method to add vectors.
Explanation

The vectors obey commutative and associative laws. Hence

Projectile Motion Class 11 Notes Question 4.
Find the magnitude and direction of the resultant of two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{A}}\) in terms of their magnitudes and angle between them.
Answer:

Consider two vectors \(\vec{A}(=\overrightarrow{O P}) \text { and } \vec{B}(=\overrightarrow{O Q})\) making an angle q. Using the parallelogram method of
vectors, the resultant vector \(\overrightarrow{\mathrm{R}}\) can be written as,

SN is normal to OP and PM is normal to OS. From the geometry of the figure
OS² = ON² + SN²
but ON = OP + PN
ie. OS² = (OP+PN)² + SN² ______(1)
From the triangle SPN, we get
PN = Bcosq and SN = Bsinq
Substituting these values in eq.(1), we get
OS² = (OP + Bcosq)² + (Bsinq)²
But OS = R and OP = A
R² = (A + Bcosq)² + B²sin²q
= A² + 2ABcosq + B²cos²q + B²sin²q
R² = A² + 2 ABcosq + B²

The resultant vector \(\overrightarrow{\mathrm{R}}\) make an angle a with \overrightarrow{\mathrm{A}}. From the right angled triangle OSN,

But SN = Bsinq PN = Bcosq

Motion In A Plane

1. Position vector and displacement vector Position vector:

Consider a small body located at P with reference to the origin O. The position vector of the point ‘P’

Displacement vector


where Dx = x¹ – x¹, Dy = y¹ – y
Velocity:
If Dt is the time taken to reach from P to P¹
The average velocity, \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}=\frac{\overrightarrow{\Delta r}}{\Delta \mathrm{t}}\) ____(3)
Substitute eq.(2) in eq.(3), we get

The direction of average velocity is the same as that of \(\overrightarrow{\Delta r}\).
The instantaneous velocity can be written as

Acceleration:
If the velocity of an object changes from \(\overrightarrow{\mathrm{v}} \text { to } \overrightarrow{\mathrm{v}^{1}}\) in time Dt, then its average acceleration is given by

Instantaneous acceleration:
The acceleration at any instant is called instantaneous acceleration. When Dt goes to zero, the average acceleration becomes instantaneous acceleration.
ie. Instantaneous acceleration

Motion In A Plane With Constant Acceleration
Consider an object moving in xy plane with constant acceleration ‘a’. Let \(\vec{u}\) be the initial velocity at t=0 and \(\vec{v}\) be the final velocity at time t.
Then by definition acceleration

In terms of components
v_x = u_x + a_xt
v_y = u_y + a_yt
Displacement in a plane
If \(\overrightarrow{\mathrm{r}_{0}}\) and \(\vec{r}\) be position vectors of particle at t = 0 and time t respectively, then
displacement = \(\vec{r}-\vec{r}_{0}\) _______(1)
For uniformly accelerated motion, displacement,

In terms of components
x = x₀ + u_xt + 1/2 a_xt²
y = y₀ + u_yt + 1/2 a_yt²
The eq.(2) shows that, the above motion in xy plane can be treated as two separate one dimensional motions along two perpendicular directions.

Relative Velocity In Two Dimensions
Consider two bodies A and B moving along a plane with velocities \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\). Then velocity of A relative to that of B is,

Similarly velocity of B relative to that of A

Projectile Motion
Projectile:
A body is projected into air and is allowed to move under the influence of gravity is called projectile.

Consider a body which is projected into air with a velocity u at an angle q. The initial velocity ‘u’ can be divided into two components ucosq along horizontal direction and using along vertical direction.

1. Time of flight:
The time taken by the projectile to cover the horizontal range is called the time of flight. Time of flight of projectile is decided by usinq. The time of flight can be found using the formula
s = ut + 1/2 at²
Taking vertical displacement s = 0, a = -g and initial vertical velocity = usinq, we get
0 = usinqt – 1/2gt²
1/2 gt² = usinqt

2. Vertical height:
Vertical height of body is decided by vertical component of velocity (usinq). The vertical displacement of projectile can be found using the formula v² = u² + 2as
When we substitute v=0, a = -g, s = H and u = usinq, we get
0 = (usinq)² + 2 – g × H
2gH = u²sin²q

3. Horizontal Range:
If we neglect the air resistance, the horizontal velocity (ucosq) of projectile will be a constant.
Hence the horizontal distance (R) can be found as
R = horizontal velocity × time of flight

The eq.(3) shows that, R is maximum when sin2q is maximum, ie. When q₀ = 45°.
The maximum horizontal range

Equation for path of projectile

Motion In A Plane Class 11 Pdf Question 5.
What is the shape of path followed by the projectile? Show that the path of projectile is parabola. The vertical displacement of projectile at any time t, can be found using the formula.
Answer:
S = ut+ 1/2at²
y = usinqt – 1/2gt²
But we know horizontal displacement, x = ucosq × t

In this equation g, q and u are constants. Hence eq.(4) can be written in the form
y = ax + bx²
where a and b are constants. This is the equation of parabola, ie. the path of the projectile is a parabola.

Uniform Circular Motion
The motion of an object along the circumference of a circle is called circular motion.
Uniform circular motion:
When an object follows a circular path at a constant speed, the motion is called uniform circular motion.
Period:
The time taken by the object to complete one full revolution is called the period.
Frequency:
The number of revolutions completed per second is called the frequency u of the circular motion.
If the period of a circular motion isT, its frequency

Angular Displacement (Dq):
The angle Dq in radians swept out by the radius vector in a given interval of time is called the angular displacement of the object.
Angular velocity:
The rate of change of angular displacement is called the angular velocity.

If T is the period of an object, then its radius vector sweeps out an angle of 2p radian.
Therefore in one second it sweeps out an angle \(\frac{2 \pi}{T}\).
∴ Angular velocity of the object

Expression for velocity and acceleration in uniform circular motion:

The direction of velocity is in the direction of tangent at that point. The change in velocity vectors \((\overrightarrow{\Delta v})\) is obtained by triangle law of vector as shown in figure (b).

a. Speed and angular speed in uniform circular motion:
Let the Dq be the angle constructed by the body during the time interval ∆t. The angular velocity can be written as

If the distance travelled by the object during the time Dt is Dr (ie. PP¹ = Ds) then speed

But Ds = RDq
where R = \(|\vec{r}|=|\overrightarrow{r^{\prime}}|\)
Substituting Dr = RDq in eq.(1)
we get

b. Acceleration in uniform circular motion:

The direction of this acceleration should be in the direction of \(\overrightarrow{\Delta V}\). The fig(b) shows that \(\overrightarrow{\Delta V}\) is towards the centre of the circular path. Hence the acceleration is directed towards the centre of the circle and is called centripetal acceleration.

The force which produces this centripetal acceleration is called centripetal force.
Centripetal force can be written as
F = ma_c

But ω = \(\frac{V}{R}\). Hence we get K

Students can Download Chapter 1 Introduction to Accounting 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 1 Introduction to Accounting

Plus One Accountancy Introduction to Accounting One Mark Questions and Answers

Plus One Business Studies Chapter 1 Notes Pdf Question 1.
Who among the following are not a user of accounting information?
(a) Management
(b) Investors
(c) Advertisers
(d) Lenders
Answer:
(c) Advertisers

Plus One Business Studies Chapter 1 Notes Question 2.
Spot the odd one out and state reason
(a) Loose Tools
(b) Copy Write
(c) Patent
(d) Goodwill
Answer:
(a) Loose Tools, is a Fixed assed, all others are intangible assets.

Plus One Business Studies Chapter 1 Question 3.
A Person who owes money to the business is a
(a) Debtor
(b) Investor
(c) Creditor
(d) Borrower
Answer:
(a) Debtor

Plus One Business Studies Notes Question 4.
Book-keeping is concerned with
(a) Analysis of transaction
(b) Recording of transaction
(c) Classification of transaction
Answer:
(b) Recording of transaction

Plus One Business Studies Notes In English Question 5.
Amount spent for purchasing fixed asset is a
(a) Revenue Expenditure
(b) Capital Expenditure
(c) Deferred Revenue Expenditure
Answer:
(b) Capital Expenditure.

Plus One Business Studies Chapter 1 In Malayalam Question 6.
Which quantitative characteristics of accounting in-formation is reflected when accounting information is clearly presented?
(a) Understandability
(b) Relevance
(c) Comparability
(d) Reliability
Answer:
(a) Understandability.

Plus One Business Studies Notes Pdf Question 7.
Which of the following is an example of a business transaction?
(a) Appointed Mr.Ram as the Manager of the business with a salary of Rs. 15,000.
(b) Obtain a loan of Rs. 1,00,000 to the business from Bank of India.
(c) Sent a quotation to Matha Traders worth Rs. 20,000.
Answer:
(b) Obtained a loan of Rs. 1,00,000 from Bank of India. Loan is taken meant for business. So it is a business transaction.

Hsslive Plus One Business Studies Notes Question 8.
Find the odd one out and state the reasons,
(a) share capital
(b) Debentures
(c) Sundry creditors
(d) Long-term loans.
Answer:
(c) Sundry creditors, all others are long term liabilities.

Business Studies Plus One Notes Question 9.
Value of goods remaining unsold at the end of an accounting period is termed as …………
Answer:
Closing Stocks

Plus One Business Studies Chapter Wise Notes Question 10.
Arun, a sole trader, draw Rs. 500 from the business for paying tuition fees to his child. This amount is termed as ……….
Answer:
Drawings.

Nature And Purpose Of Business Notes Pdf Question 11.
Assets minus liabilities are called ……….
Answer:
Capital.

Nature And Purpose Of Business Class 11 Notes Pdf Download Question 12.
………….. assets are those assets, which do not have any real value.
Answer:
Fictitious Assets.

Hss Live Business Studies Plus One Notes Question 13.
The amount earned by a business concern through sale of its products or providing services to customers is called ……
Answer:
Revenues.

Plus One Business Studies Notes In English Pdf Question 14.
The assets bought for long-term use in the business are termed as …………….. assets
Answer:
Fixed.

Hsslive Business Studies Plus One Notes Question 15.
Analysis of recorded data to bring entries of similar nature to one plane is called …………..
Answer:
Classifying.

Hsslive Plus One Business Studies Malayalam Notes Question 16.
A Person who is entitled to get money from the business is termed as …………
Answer:
Creditor

Question 17.
Information in financial reports is based on …………….. transaction.
Answer:
Economic.

Question 18.
All claims against the business are called ………………..
Answer:
Equity

Question 19.
The transaction is one wherein payment or receipt of money is postponed for a future date.
Answer:
Credit transaction.

Question 20.
Mr. Ismail, who is the owner of a Provision shop, took 50 kg. of rice worth Rs. 600 for his house-hold use. He should record this as
Answer:
Drawings.

Question 21.
Ravi, a trader purchased 100 notebooks from ‘Shyni stores’ on credit. How is Shyni stores related to Ravi?
Answer:
Shyni stores is the creditor of Ravi.

Question 22.
Who was the inventor of double-entry bookkeeping?
Answer:
Luca Pacioli

Question 23.
Expand AICPA
Answer:
The American Institute of Certified Public Accounts.

Question 24.
Identify the events not used to the accounting treatment.
(a) Commenced business with cash
(b) Bought Machinery for cash
(c) Cash Purchase of goods.
(d) The firm appointed an efficient Manger.
Answer:
(d) The firm appointed an efficient Manger.

Plus One Accountancy Introduction to Accounting Two Mark Questions and Answers

Question 1.
Define Accounting.
Answer:
According to American Institute of certified Public Accountants, “Accounting is the art of recording, classifying and summarizing in a significant manner and in terms of money, transactions and events which are in part, at least, of a financial character and interpreting the results thereof.”

Question 2.
‘Accounting is the language of the business’. Why?
Answer:
Accounting is the language of the business:
The performance of business in terms of profit or losses is conveyed to users of accounting information in a systematic manner. The financial position of the business concerned is revealed through accounting information.

Question 3.
“Raju sold goods to Rahim on credit”. What relation exists between them? What are the accounting terms involved in it?
Answer:

• Rahim – Debtor
• Raju – Creditor

Question 4.
You are the accountant of a firm. What are the functions to be performed by you?
Answer:
Accounting provides information regarding the financial status of a business and results of its operations. The following are the important functions of accountant of a firm.

1. Recording transactions by referring source documents.
2. Preparing journal, subdivision of journal.
3. Preparing ledger accounts.
4. Summarising.
5. Making statements of interpretation 0 Reporting to Management.

Question5.
Users of accounting are classified as under:

1. Internal users – Management, Investor, Creditor, Bank, Employees, Stock exchange.
2. External Users – Customers, Government, Researchers, Lenders.

Do you agree with this classification, if not correct it?
Answer:
No,
1. Internal Users:

• Management
• Employees

2. External Users:

• Investors
• Creditors
• Bank
• Stock Exchange
• Government
• Customers
• Lenders
• Researchers

Question 6.
All business transactions are events. But all events are not business transactions. Comment.
Answer:
Events can be anything, some events can be expressed in monetary terms while others are not. Only those events which can be expressed in money, terms are business transactions. Transaction is an event or economic activity of a businessman in his business having exchange of money or money’s worth. While events is part of the business transaction.

Question 7.
Classify the following expenses into capital expenditure and revenue expenditure.

• Machinery purchased
• Rent paid
• Interest paid
• Purchased building
• The amount for repair of the building

Answer:
1. Capital Expenditure:

• Machinery Purchase
• Purchased building

2. Revenue Expenditure:

• Rent paid
• Interest
• paid Amount for repair of building

Question 8.
How will you define Revenue and Expenses?
Answer:
1. Revenue:
These are the amounts earned by a business concern through sale of its products or providing services to customers. The common items of revenues are sales, commission received, rent received, interest received, etc.

2. Expenses:
The amount spent in the process of earning revenue is termed as expenses. Examples are Wages, Salaries, rent, Interest paid, electricity charges, etc.

Question 9.
What is a capital expenditure? Give some examples?
Answer:
Capital expenditure represents the amount spent for the acquisition of assets, the benefit from which is derived over a period that extends beyond the accounting year. It is long term in nature.
Examples: Furniture purchased, Land Purchased, Building purchased, etc.

Question 10.
Explain the meaning of Gain and Profit.
Answer:
1. Profit: The excess of revenues of a period over its related expenses during an accounting year is profit. Profit increases the investment of the owners.

2. Gain: A profit that arises from events or transactions which are incidental to business such as sale of fixed assets, winning a court case, receipt of interest and dividend, etc. Gain is irregular in nature. Gains are part of capital receipt. Gains are also known as “non-operating income.”

Question 11.
Match the following.

Answer:

Question 12.
Name the branches of accounting.
Answer:

1. Financial Accounting
2. Cost Accounting
3. Management Accounting

Plus One Accountancy Introduction to Accounting Three Mark Questions and Answers

Question 1.
Classify the following assets into suitable head Goodwill, Building, Land, Patent, Cash, Oilwell, Copy-write, Debtors, Stock, mines, Bill receivable, Preliminary expenses.
Answer:

Question 2.
Define assets, Liabilities, and capital.
Answer:
1. Assets:
Assets are properties and things of value owned by the business which can be expressed in monetary terms. Examples of Machinery, Buildings, Stock, Debtors, Furniture, etc.

2. Liabilities:
Liabilities are the obligations that an enterprise owes. These represent the amount payable by the business in the future. They represent the claim against the asset of business. Examples Loans, Creditors, Bills payable, etc.

3. Capital:
Capital is the investment made by the owners for use in the business. It is owner’s claim on the total assets of the business and is also called “owners equity”.

Question 3.
Distinguish between:

1. Goods and Assets
2. Expense and Loss

Answer:
1. Goods and Assets:

• Goods refer to things in which the trader deals. But assets refers to things with which the trader deals.
• Goods are meant for resale, while assets are kept in the business permanently with the help of which the business is carried on.

2. Expense and Loss:

• Costs incurred by a business in the process of earning revenue are known as expense.
• The excess of expenses of a period over its related revenues is termed as loss. It decreases in owner’s equity.

Plus One Accountancy Introduction to Accounting Four Mark Questions and Answers

Question 1.
Accounting has certain objectives to business Explain.
Answer:
The following are the important objectives of accounting.

1. Keeping of records of business transactions.
2. Ascertainment of Profit or Loss.
3. Ascertainment of financial position of business enterprises.
4. Providing meaningful information to different groups of people having interest in the business.

1. Keeping of records of business transaction:
The main purpose of accounting is to identify business transactions of financial nature and enter into appropriate books of accounts. The accounting records should be made properly and systematically, so that requisite information may be obtained at a glance.

2. Ascertainment of Profit or Loss:
The result of business (Profit or Loss) is available from the statement prepared for ascertaining it, called the Profit and Loss Account.

3. Ascertainment of financial position:
At the end of an‘ accounting year, a position statement known as the ‘Balance Sheet’ is prepared. The value of assets and liabilities are depicted in the Balance Sheet. The Balance sheet gives a true and fair view of the state of affairs of the concern.

4. Providing meaningful information to different groups of people having interest in the business:
Accounting records provide meaningful information to different groups of people having interest in the business.

Question 2.
Accounting information must possess certain qualitative characteristics. What are they?
Answer:
The following are the qualitative characteristic of accounting information.

1. Reliability: Accounting information will be reliable if it is free from error and faithfully represents what it seeks to represent.
2. Relevance: Information should be relevant and must be available in.time.
3. Understandability: Accounting information that is relevant must be capable of being understood by all its users.
4. Timeliness: Information must be available timely. If not, it loses its ability to influence decision.
5. Comparability: Accounting information should facilitate inter-firm comparison as well as interfirm comparison.
   Maximum Cputtishers

Question3.
Accounting and Book-keeping are viewed as distinct functions. Mention any four differences between Accounting and Book Keeping.
Answer:

Question 4.
“Accounting gives number of advantages to the business”. What are the important advantages of accounting?
Answer:
The following are the advantages of accounting
1. Provide Quantitative information:
Accounting helps in gathering quantitative information on profits earned by the business or loss sustained by them.

2. Helps in ascertaining financial position of the business that is, total assets owned and total liabilities owed.

3. Helps in making a systematic record of transactions, which can be used for future reference and appropriate retrieval.

4. Acts as an information system:
It provides adequate information to the interested users in a processed form.

5. Beneficial to different interested users of accounting information.

Question 5.
Accounting has certain ‘Limitations’. Explain.
The following are the limitations of accounting.
Answer:
1. It records only transactions which can be recorded in monetary terms:
Qualitative aspects like managerial skill, Services of experts, etc. are not recorded.

2. Accounting is a post mortem survey:
It records events as they have taken place. For example, expenses are recorded as incurred, assets are recorded at their cost of purchase. There is no scope for ascertaining what the appropriate expenditure or cost of acquisition should have been.

3. Effect of price level changes are not considered:
Transactions are always recorded in the books at cost price and not at market price.

4. Inexactness:
Accounting transactions are not exact. Different firms have their own different methods, so the results of the business will change in the practice.

Plus One Accountancy Introduction to Accounting Six Mark Questions and Answers

Question 1.
What are the different types of assets? Explain briefly.
Assets are things of value owned. They may be subdivided into the following.
Answer:
1. Fixed Assets:
Fixed Assets are assets held on long term basis, such as land, buildings, machinery, plant, furniture, etc. These assets are used for the normal operations of the business.

2. Current Assets:
These are assets held on a short-term basis such as debtors, bills receivables, stock, cash in hand, cash at bank, etc. It is also known as “Floating asset.”

3. Fictitious Assets:
These are those assets, which do not have a physical form. They do not have any real value. Actually, they are not the real assets but they are called assets on legal and technical grounds. Examples are preliminary expenses, discount on issue of shares or debentures, etc.

4. Tangible Assets:
Assets having physical existence which can be seen, touched are known as tangible assets. These assets are land, building, plant, equipment, etc.

5. Intangible Assets:
These assets have no physical existence which cannot be touched, seen or felt. Examples are Goodwill, trademark, patent, copyright.

6. Wasting Assets:
Assets, whose value goes on declining with the passage of time, are known as wasting assets. Mines, oilwells, quarries are its examples.

7. Liquid Assets:
Liquid assets are those assets, which can be converted into cash at short notice. The examples of liquid assets are cash in hand, cash at bank, debtors, bills receivable, etc. Liquid assets = Current Assets – (Stock + Prepaid Expenses)

Question 2.
“Accounting provides information to various users.” Discuss accounting as an information system.
Answer:
Accounting plays a significant role in society by providing information to management at all levels (internal users) and to those having a direct financial interest in the enterprise (external users), such as present and potential investors, creditors. Accounting information is also important to those having an indirect financial interest, such as regulatory agencies, tax authorities, customers, labour unions, stock exchange, and others.

Internal users, mainly management, need timely information on cost of sales, profitability, etc. for planning, controlling and decision making. External users who have limited authority, ability and resources to obtain the necessary information have to rely on financial statements. The external users are interested in the following.

1. Investors and Potential investors:
Information on the risks and returns on investments.

2. Suppliers and Creditors:
Information on whether amounts owed will be repaid when due and on the continued existence of the business.

3. Customers:
Information on the continued existence of the business and thus the profitability of a continued supply of products, parts, and after-sales services.

4. Employees:
They are interested in getting their salary, welfare measures, bonus, working conditions, etc. which are all related to financial performance of the business.

5. Lenders:
Information on the creditworthiness of the business and its ability to repay loans and pay interest.

6. Government and other regulators:
Information on the allocation of resources and the compliance to regulators.

Students can Download Chapter 1 Nature and Purpose of Business Notes, Plus One Business Studies Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Business Studies Notes Chapter 1 Nature and Purpose of Business

1st Standard Malayalam Textbooks Chapter 1 Contents

• Economic and non-economic activities
• Business and its characteristics
• Business, Profession and Employment
• Classification of business activities
• Industry and commerce
• Objectives of business
• Business risk
• Factors to be considered before starting a business.

Activities which human beings undertake are known as human activities. We can divide these activities into two categories.

1. Economic activities
2. Non-economic activities

1. Economic Activities:
The human activities that are undertaken with an objective to earn money or livelihood is known as economic activities. eg: A worker working in a factory, a doctor operating in his hospital, a manager working in the office, a teacher teaching in a school etc.

2. Non-economic activities:
Activities are undertaken to derive psychological satisfaction are known as non economic activities. eg: Mother preparing food for her children, Praying, listening to radio or watching television, playing football with friends, etc.

Types of economic activities:

• Business
• Employment
• Profession

1st Standard Malayalam Chapter 1 Business:
Business may be defined as an economic activity involving the production or purchase and sale of goods and services with the main object of earning profit by satisfying human needs in the society.
Characteristics of business

1. Business is an economic activity with the object of earning profit.
2. Business includes all the activities concerned with the production or procurement of goods and services.
3. There should be sale or exchange of goods and services for the satisfaction of human needs.
4. Business involves dealings in goods or services on a regular basis. Normally, one single transaction of sale or purchase is not treated as business.
5. One of the main objectives of business is to earn maximum profit.
6. Business involves risk and uncertainty of income. Risk means the possibility of loss due to change in consumer taste and fashion, strike, lockout competition, fire, theft etc.

Malayalam Text Book Std 1 Pdf Download Employment:
Employment refers to that type of economic activity in which people engage in some work for others regularly and get salary or wages in return of their services.
Characteristics of Employment

1. There must exist employer-employee relationship.
2. There must be a service contract between the employer and employee.
3. Employees get salary or wages for their services
4. Regularity in service.

First Standard Malayalam Textbooks Chapter 1 Profession:
Profession refers to an occupation which requires specialized knowledge, skip and training. Its objective is to provide service to the society.
Characteristics of Profession

1. A profession requires specialized knowledge, training and skill
2. The membership of a professional body is a must
3. Professionals have a code of conduct
4. They charge fee in return of their service.

Comparison of Business, Profession and Employment:

Classification of Business Activities: Business activities may be classified into two categories

• Industry
• Commerce.

Chart showing business activities

1 Standard Malayalam Textbooks Chapter 1 Industry:
Industry refers to economic activities, which are connected with conversion of resources into useful goods. Industries may be divided into 3 categories.
They are
1. Primary industries:
Primary industries are connected with the extraction and production of natural resources and reproduction and development of living organisms, plants, etc. Such industries are further divided into two.
(i) Extractive industries:
These industries extract products from natural resources. eg: mining, farrqing, hunting, fishing etc.

(ii) Genetic industries:
These industries are engaged in activities like rearing and breeding of animals, birds and plants. eg: diary faming, paultry farming, floriculture, pisciculture etc.

2. Secondary industries:
Secondary industries deal with materials extracted at the primary stage. Such goods may be used for consumption or for further production. Secondary industries are classified into two.
They are:
(i) Manufacturing industries:
Manufacturing industries engage in converting raw materials into finished goods. eg: Conversion of rubber into cotton, timber into furniture rubber into tyres etc. Manufacturing industries may be further divided into four categories. They are,

• Analytical industry which analyses and separates different elements from the same materials. eg: Oil refinery
• Synthetical industry which combines various ingredients into a new product. eg: cement
• Processing industry which involves successive stages for manufacturing finished products. eg: Sugar and paper industry.
• Assembling industry which assembles different component parts to make a new product. eg: television, car, computer, etc.

(ii) Construction industries:
These industries are involved in the construction of buildings, dams, bridges, roads etc.

3. Tertiary industries:
These are concerned with providing support services to primary and secondary industries. eg: Transport, banking, insurance, warehousing, communication, advertising etc.

1st Standard English Book Chapter 1 Commerce:
Commerce is defined as all activities involving the removal of hindrances in the process of exchange of goods. It includes all those activities, which are necessary for the free flow of goods and services from the producer to the consumer. Commerce includes trade and auxiliaries to trade.
Commerce = Trade + auxiliaries to trade
Functions of commerce:

Malayalam 1st Standard Chapter 1 Trade:
Trade refers to sale, transfer or exchange of goods. Trade may be classified into two broad categories
They are:

1. Internal trade
2. External trade

1. Internal, domestic ot dome trade:
is concerned with the buying and selling of goods and services within the geographical boundaries of a country. This may further be divided into two. They are:-
(a) Wholesale trade:
Under wholesale trade, the trader purchases goods in large quantities from the producers, and sells them in smaller quantities to the retailers.

(b) Retail trade:
Under the retail trade, the trader buys in comparatively smaller quantities from the wholesalers or producers and sells them to ultimate consumers.

2. External or Foreign trade:
Foreign trade consists of exchange of goods and services between two or more countries. Foreign trade may be divided in to three.

• Import trade: If goods are purchased from a foreign country, it is called import trade.
• Export trade: When goods are sold to a foreign country, it is known as export trade.
• Entrepot trade: When goods are imported for export to other countries, it is known as entrepot trade.

Auxiliaries to Trade (Aids to trade):
Activities which assist trade are called aids to trade or auxiliaries to trade.

1. Transport & Communication:
Transport facilitates the movement of raw material to the place of production and the finished products from factories to the place of consumption. Communication helps the producers, traders and consumers to exchange information with one another.

2. Banking & Finance:
Banking helps business activities to overcome the problem of finance. Commercial banks lend money in the form of overdraft, cash credit, loans and advances etc… and they also provide many services required for the business activity.

3. Insurance:
The goods may be destroyed while in production process or in transit due to accidents, or in storage due to fire or theft, etc. Insurance provides protection in all such cases.

4. Warehousing:
The goods should be stored carefully from the time they are produced till the time they are sold. This function is performed by warehouses.

5. Advertising:
Advertising helps in providing information about available goods and services and create in them a strong desire to buy the product.

Multiple Objectives of Business:
The main objectives of a business are:
1. Market standing:
A business firm can succeed only when it has a good market standing. Market standing refers to the position of an enterprise in relation to its competitors.

2. Innovation:
Innovation means developing new product or services orfinding new ideas and new methods of production and distribution. Innovation accelerates the growth of an enterprise.

3. Productivity:
Productivity is ascertained by comparing the value of output with the value of input. Every enterprise must aim at greater productivity through the best use of available resources.

4. Physical and financial resources:
The business must aim at maximum utilization of available physical and financial resources, i.e. men, material, money and machine in the best possible manner.

5. Earning Profit:
Earning maximum profit is the primary objective of every business. Profit is required for survival and growth of a business.

6. Manager performance and development:
Efficient managers are needed to conduct and co-ordinate business activities. So it is the objective of an enterprise to implement various programs for motivating the managers.

7. Worker performance and attitude:
Every enterprise must aim at improving its workers performance by providing fair salary, incentives, good working conditions, medical and housing facilities.

8. Social responsibility:
It refers to the obligation of business firm to contribute resources for solving social problems and work in a socially desirable manner.

Malayalam Text Book Std 1 Chapter 1 Business Risks:
The term ‘business risks’ refers to the possibility of inadequate profits or even losses due to uncertainties or unexpected events. Business enterprises may face two types of risk, i.e. speculative risk and pure risk.

Speculative risks involve both the possibility of gain as well as the possibility of loss. It arise due to change in demand, change in price etc. Pure risks involve only the possibility of loss or no loss. The chance of fire, theft or strike is examples of pure risks.
Nature of Business Risks:

1. Business risks arise due to uncertainties.
2. Risk can be minimized, but cannot be eliminated. It is an essential part of business.
3. Degree of risk depends mainly upon the nature and size of business.
4. Profit is the reward for risk taking.

Causes of Business Risks: Business risks arise due to a variety of causes.
They are:

1. Natural Causes: it includes natural calamities like flood, earthquakes, lightning, heavy rains, famine, etc.
2. Human Causes: Human causes include dishonesty, carelessness or negligence of employees, strikes, riots, management inefficiency, etc.
3. Economic causes: These include change in demand, change in price, competition, technological changes etc.
4. Political Causes: Change in Govt, policies, taxation, licensing policy etc.

Starting a Business – Basic Factors:
Factors to be considered for starting a business:
1. Selection of line of business:
The first thing to be decided by any entrepreneur of a new business is the nature and type of business to be undertaken.

2. Size of the firm:
If the market conditions are favorable, the entrepreneur can start the business at a large scale. If the market conditions are uncertain and risks are high, a small size business would be better choice.

3. Choice of form of ownership:
The selection of a suitable form of business enterprise i.e. Sole proprietorship, Partnership or a Joint stock company is an important management decision. It depends on factors like nature of business, capital requirements, liability of owners, legal formalities, continuity of business etc.

4. Location of business enterprise:
Availability of raw materials and labour, power supply and services like banking, transportation, communication, warehousing, etc., are important factors while making a choice of location.

5. Financing:
Proper financial planning must be done to determine (a) the requirement of capital, (b) source from which capital will be raised and (c) the best ways of utilizing the capital in the firm.

6. Physical facilities:
Availability of physical facilities including machines and equipment, building and supportive services is a very important factor to be considered at the start of the business.

7. Plant layout:
Layout means the physical arrangement of machines and equipment needed to manufacture a product.

8. Competent worked force:
Every enterprise needs competent and committed employees to perform various activities so that physical and financial resources are converted into desired outputs.

9. Tax planning:
The promoter must consider in advance the tax liability under various tax laws and its impact on business decision.

Students can Download Chapter 1 Biological Classification Notes, Plus One Botany Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Botany Notes Chapter 1 Biological Classification

Two kingdom classification:

• It was proposed by Linnaeus, include two Kingdoms-Plantae & Animalia.
• In two kingdom classification following things are not considered

Cell structure, nature of wall, mode of nutrition, habitat methods of reproduction and evolutionary relationship .

Five kingdom classification:

• It was first proposed by R H.Whittaker (1969).
• It includes Monera, Protista, Fungi, Plantae and Animalia
• Blue green algae are placed in kingdom Monera.
• Chlamydomonas, Chlorella with Paramoecium and Amoeba are placed in kingdom protista
• Chlorophyll less and non cellulosic (chitin)type plants are placed in kingdom Fungi
• All photosynthetic plants are placed in-kingdom plantae
• All animals with mode of nutrition(ingestion) placed in kingdom animalia.
• The characteristics of classification are

Kite Victers Plus One Botany Notes Pdf Kingdom monera:
Types of bacteria:
Based on shape, bacteria are of 4 types

1. Spherical – Coccus
2. Rod – shaped – Bacillus
3. Comma – shaped – Vibrium
4. Spiral – Spirillum.

Based on nutrition, bacteria are of 3 types:

1. Photosynthetic autotrophic: Bacteria can synthesise their own food by using chlorophyll in the presence of light.
2. Chemosynthetic autotrophic: Bacteria can synthesise their own food from inorganic substrates.
3. Heterotrophs: They are depend on other organisms for food.

Plus One Botany Notes Archaebacteria:

• These bacteria can live in extreme conditions.
• Its cell wall structure is different from other bacteria

Types of archaebacteria:

1. Halophiles: They are found in salty areas
2. Thermoacidophiles: They are found in hot springs.
3. Methanogens: They are found in marshy areas and guts of ruminant animals eg-cows, buffaloes etc.

Importance in industry:
They are responsible for the production of methane (biogas) from the dung.

Eubacteria Cyanobacteria (blue-green algae):

• They have chlorophyll a similar to green plants called as photosynthetic autotrophs.
• Some of them found in polluted water bodies.

Significance of cyanobacteria:
They can fix atmospheric nitrogen in specialised cells called heterocysts and increases fertility of soil eg: Nostoc and Anabaena

Chemosynthetic autotrophs:
They oxidise various inorganic substances such as nitrates, nitrites and ammonia and use the released energy for their ATP production.

Significance chemosynthetic autotrophs:
They play a great role in the recycling of nutrients like nitrogen, phosphorous, iron and sulphur.

Heterotrophic bacteria:

• They depend upon others for getting energy.
• Most of them are decomposers.
• They are helpful in making curd from milk, production of antibiotics, fixing nitrogen in legume roots, etc.

Disease caused by bacteria:
Some are pathogens causing disease to plants and animals, eg: Cholera, typhoid, tetanus, and citrus canker in plants.

Cell division in bacteria:
Reproduction:

• Bacteria reproduce mainly by fission.
• Some bacteria produce spores during unfavourable conditions.
• In sexual reproduction, transfer of DNA from one bacterium to the other takes place

Mycoplasmas:
Salient features:

• They are the smallest living cells can survive without oxygen
• They are pathogenic in animals and plants
• They lack a cell wall.

Plus One Botany Chapter Wise Notes Kingdom Protista:
Salient features:

• They are single-celled eukaryotes.
• Their cell body contains a well defined nucleus and other membrane-bound organelles.
• They are mainly aquatic.

Types of protist:
The kingdom include chrysophytes, Dianoflagellates, Euglenoids, Slime moulds and Protozoans.

Plus One Botany Chapters Chrysophytes:

• They are diatoms(chief ‘producers’ in the oceans) and golden algae (desmids).
• Most of them are photosynthetic.

Salient features:

• In diatoms the cell walls form two thin overlapping shells,which fit together as in a soap box.
• Their cell wall contain silica

Economic value:
Their cell wall deposited in ocean floor over billions of years in large amount called as ‘Diatomaceous earth’. It is used in polishing, filtration of oils and syrups.

Plus One Botany Notes Hsslive Dinoflagellates:
Salient features:

• They are marine and photosynthetic
• Most of them have two flagella one lies longitudinally and the other transversely.

Harmful effects:

• Red dianoflagellates -Gonyaulax undergoes rapid multiplication and sea appear red (red tides).
• Toxins released by them kill other marine animals such fishes.

Plus One Botany Victers Notes Euglenoids:
Salient features:

• They are fresh water organisms found in stagnant water.
• Pigments of Euglenoids are identical to those of higher plants.
• They have a protein rich layer called pellicle which makes their body flexible.
• They have two flagella, a short and a long one.

Mode of nutrition:
They are photosynthetic in the presence of sunlight (autotrophic) and predating on other smaller organisms in the absence of sunlight (heterotrophs). Hence nutrition is mixotrophic. Example: Euglena.

Plus One Botany Notes Victers Slime Moulds:
Salient features:

1. They are saprophytic protists.
2. They form an aggregation during favourable conditions called plasmodium which may grow and spread over several feet.
3. The plasmodium differentiates and forms fruiting bodies bearing spores at their tips during unfavourable conditions.
4. They are resistant and survive for many years under adverse conditions.

Plus One Botany Notes Pdf Protozoans:
They are heterotrophs and live as predators or parasites.
Type of protozoans:
1. Amoeboid protozoans:

• They live in freshwater or sea water
• They capture their prey by pseudopodia (false feet) as in Amoeba.
• Some are parasites as in Entamoeba

2. Flagellated protozoans:

• They possesss flagella.
• The parasitic forms cause diseases such as sleeping sickness. Example: Trypanosoma.

3. Ciliated protozoans:

• They are aquatic and have a cavity (gullet) that collect food from outside.
• They move with the help of cilia. Example: Paramoecium.

4 Sporozoans:
They are infectious due to the spore-like stage in their life cycle. eg: Plasmodium (malarial parasite) which causes malaria.

Hsslive Plus One Botany Notes Kingdom Fungi:
They are heteterotrophs, mainly 2 types

1. Saprophytes: They can absorb soluble organic matter from dead substrates
2. Parasites: They are depend on living plants and animals

Symbiotic associations:

1. Lichens-Fungi forms an association with algae
2. Mycorrhiza-Fungi forms association with roots of higher plants.

Disease caused bv fungi:
They cause diseases in plants and animals. eg: wheat rust disease by Puccinia.

Salient features:

• Yeast is unicellular fungus but others are multicellular.
• Mycelium: lt is body of fungi contains many hyphae
• Hyphae It is the long, slender thread-like structures
• Some hyphae with multinucleated cytoplasm are called coenocytic hyphae.
• Others have septae or cross walls in their hyphae. eg: Penicillium.
• The cell walls of fungi are composed of chitin and polysaccharides

Types of Reproduction:

1. Vegetative method: It takesplace by fragmentation, fission and budding.
2. Asexual method: It takesplace by spores called conidia orsporangiospores or zoospores.
3. Sexual reproduction: It takesplace by oospores, ascospores and basidiospores.
4. The spores are produced in special structures called fruiting bodies.

Steps of sexual cycle:

In this, two haploid hyphae come together and fuse results in diploid cells (2n).

Dikarvotic stage in funai:

1. In ascomycetes and basidiomycetes, after plasmogamy dikaryotic stage (n + n ) occurs for some time. Later these nuclei fuse and the cells become diploid.
2. The fungi form fruiting bodies in which reduction division occurs and forms haploid spores.

Hss Live Plus One Botany Notes Phvcomycetes:
Salient features:

• They are found in aquatic habitats and on decaying wood.
• Their mycelium is aseptate and coenocytic.

Types of reproduction:
1 Asexual reproduction:
It takes place by zoospores (motile) or by aplanospores (non-motile). These spores are produced in sporangium.

2 Sexual reproduction:
It is the fusion of gametes have similar structure (isogamous) or dissimilar structure (anisogamous or oogamous) and after fusion zygospore is formed. Examples are Mucor, Rhizopus (the bread mould) and Albugo (the parasitic fungi on mustard).

Hss Live Plus One Biology Notes Ascomycetes:
Salient features:

• They are commonly known as sac-fungi, eg-unicellular- yeast (Sacharomyces) or multicellular Penicillium.
• Some are coprophilous (growing on dung).
• Mycelium is branched and septate.
• The asexual spores are conidia produced on conidiophores.
• Sexual spores are called ascospores which are produced in sac like ascus.
• Spores are arranged in fruiting bodies called ascocarps.
• Examples are Aspergillus, Claviceps and Neurospora.

Economic value:

• Neurospora is used in genetic studies.
• Edible members are morels and buffles.

Plus One Botany Notes Chapter 1 Basidiomycetes:
Salient features:

• Their mycelium is branched and septate.
• Some members grow as parasites and disease causing organisms e.g. rusts and smuts
• Common basidiomycetes are mushrooms, bracket fungi or puffballs

Reproduction:

1. The asexual spores and sex organs are not found but vegetative reproduction by fragmentation.
2. In sexual reproduction, plasmogamy occur by fusion of two vegetative cells of different strains. It results dikaryotic mycelia which gives rise to basidium. Later, karyogamy and meiosis takeplace in the basidium and producing four basidiospores.
3. The basidia are arranged in fruiting bodies called basidiocarps.
4. Examples -Agaricus (mushroom) Ustilago (smut) and Puccinia (rust fungus).

Hsslive Botany Plus One Notes Deuteromvcetes:
Salient features:

• They are called as imperfect fungi because perfect stage or sexual reproduction is absent.
• Their asexual reproduction takes place with the help of conidia. ,
• The mycelium is septate and branched.

Economic value:

• Some members are decomposers play an important role in the mineral cycling.
• Examples are Alternaria, Colletotrichum and Trichoderma.

Hsslive Botany Notes Plus One kingdom Plantae

1. Majority members are eukaryotic, chlorophyll-containing organisms.
2. Few members are partially heterotrophic- insectivorous plants or parasite. eg: Bladder wort and Venus fly trap are examples of insectivorous plants and Cuscuta is a parasite.

Different types of plant group:
The kingdom Plantae includes algae, bryophytes, pteridophytes, gymnosperms and angiosperms.

Life cycle:
It has two distinct phases – the diploid sporophytic and the haploid gametophytic – that alternate with each other.

Plus One Botany Biological Classification Notes Kingdom Animalia:
Salient features:

• It includes heterotrophic eukaryotic organisms.
• They are multicellular and their cells lack cell wall.
• Their mode of nutrition is holozoic (ingestion of food).
• Most of them are capable of locomotion.

Plus One Botany Kite Victers Notes Pdf Viruses, Viroids And Lichens:
R.H. Whittaker not placed acellular organisms such as viruses, viroids and lichens in five kingdom classification.

VIRUSES:
Historical aspects and Discovery

1. Name virus that means venom or poisonous fluid was given by Pasteur D.J. Ivanowsky.
2. Extract of the infected plants of tobacco could cause infection in healthy plants and called the fluid as Contagium vivum fluidum (infectious living fluid).It was identified by M.W. Beijerinek (1898)
3. Viruses could be crystallised and crystals consist of proteins outside. It was identified by W.M. Stanley. (1935)

Salient features:

• Viruses are non living particle outside the living cell.
• It has an inert crystalline structure .
• They have living state inside the host and multiply by using host cell machinery-Ribosome. So they are called as obligate parasites.
• They are smaller than bacteria because they passed through bacteria-proof filters.

Structure of viruses:
Viruses contain proteins coat outside, either RNA or DNA inside, (i.e either single or double stranded RNA or double stranded DNA).

Symptoms and disease caused by viruses:
Disease:
Mumps, smallpox, herpes, influenza and AIDS
Symptoms in plants:

Viroids:
Strucure:
It has free RNA without protein coat.

Discovery:
T .O .Diener found that this infectious agent was smaller than viruses.

Disease:
It causes potato spindle tuber disease

Lichens:
They are symbiotic associations between algae and fungi.

Types of component:

1. The algal component is called phycobiont: (autotrophic) prepare food for fungi
2. The fungal component is called mycobiont: (heterotrophic) provide shelter and absorb mineral nutrients and water for algae.

Significance:
Lichens are very good pollution indicators i.e they do not grow in polluted areas.

Ncert Supplementary Syllabus
Six kingdom classification:
It was proposed by Carl Woese. It includes kingdoms like Archaebacteria, Eubacteria Protista, Mycota, Plantae and Animalia.

Students can Download Chapter 1 The Living World Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examination

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Plus One The Living World One Mark Questions and Answers

Plus One Zoology Exercise Answers Chapter 1 Question 1.
As we go from species to kingdom in a taxonomic hierarchy, the number of common characteristics
(a) Will decrease
(b) Will increase
(c) Remain same
(d) May increase or decrease
Answer:
(a) Will decrease

Plus One Zoology Chapter Wise Questions And Answers Pdf Question 2.
Which of the following ‘suffixes’ used for units of classification in plants indicates a taxonomic category of ‘family’.
(a) -Ales
(b) -Onae
(c) -Aceae
(d) none of the above
Answer:
(c) -Aceae

Zoology Chapter Wise Questions And Answers Question 3.
The term ‘systematics’ refers to:
(a) Identification and classification of plants and animals
(b) Nomenclature and identification of plants and animals
(c) Diversity of kinds of organisms and their relationship
(d) Different kinds of organisms and their classification
answer:
(c) Diversity of kinds of organisms and their relationship

Plus One Zoology Chapter Wise Questions And Answers Pdf Download Question 4.
Genus represents
(a) An individual plant or animal
(b) A collection of plants or animals
(c) Group of closely related species of plants or animals
(d) None of these
Answer:
(c) Group of closely related species of plants or animals

Plus One Zoology Previous Question Papers Chapter Wise  Question 5.
The taxonomic unit ‘Phylum’ in the classification of animals is equivalent to which hierarchical level in classfication of plants.
(a) Class
(b) Order
(c) Division
(d) Family
Answer:
(c) Division

Plus One Zoology Chapter Wise Questions And Answers Question 6.
Correct and rewrite the following if there is any mistake.
Sativa Oriza, Tigris Panthera
Answer:
Oriza sativa, Panthera tigris

Plus One Zoology Chapter Wise Previous Questions And Answers Question 7.
Kingdom – Carnivora
Phylum – Mammalia
Class – Felidsffe
Order – Chordata
Family – Tigris
Genus – Animalia
Sps – Panthera

Correct the mistakes of the flow chart to get the hierarchical arrangement of tiger in the ascending order.
Answer:
Kingdom – Animalia
Phylum – Chordata
Class – Mammalia
Order – Carnivora
Family – Felidae
Genus – Panthera
Sps – Tigris

Hsslive Plus One Zoology Chapter Wise Questions And Answers Question 8.
Match the following:

1. A Unit of classification – kingdom
2. Lowest taxonomic category – Genus
3. Panthera – Species
4. Highest Taxonomic category – Taxon

Answer:

1. A Unit of classification – Taxon
2. Lowest taxonomic category – Species
3. Panthera – Genus
4. Highest Taxonomic category – kingdom

Plus One Zoology Previous Question Papers Chapter Wise Hsslive Question 9.
Expand the term ICBN and ICZN.
Answer:
1. ICBN: International Code for Botanical Nomenclature.
2. ICZN: International Code for Zoological Nomenclature.

Plus One Zoology Chapter Wise Questions And Answers Pdf Hsslive Question 10.
Given below is the scientific name of Frog. Identify the correctly written name.
(a) Rana Tigrina
(b) Rana tigrina
Answer:
(b) Rana tigrina

Plus One Zoology Animal Kingdom Questions Question 11.
Rearrange the order of classification.
Genus, Family, Phylum, Species, Class, Kingdom, Order
Answer:
Species, Genus, Family, Order, Class, Phylum, Kingdom

The Living World Class 11 Questions And Answers Question 12.
Arrange the given terms in their taxonomic hierarchy.
Primata, Homosapien, Chordata, Mammalia, Hominidae.
Answer:
Homosapiens, Hominidae, Primata, Mammalia, Chordata.

Plus One The Living World Two Mark Questions and Answers

The Living World Class 11 Important Questions With Answers Question 1.
You are provided with a stuffed rabbit and a dried leaf. As a student of biology which taxonomic aid will you choose to store them.
Answer:
1. Stuffed rabbit – Museum
2. Dried leaf – Herbarium.

Question 2.
Find the odd one out and give reason.
Herbarium, botanical garden, museum, aquarium
Answer:
Aquarium – Aquarium is not a taxonomical aid.
All other are taxonomical aids.

Question 3.
Once you visited a museum. There are different kinds of animals and plants are preserved. How these plants and animals all preserved in a museum.
Answer:
Plants and animals are preserved in the containers or jars in preservative solutions. These may also be preserved as dry specimen. Insects are preserved in insect box.

Larger animals like birds and mammals are usually stuffed and preserved. Museums often have collections of skeleton of animals too.

Question 4.
Zoological parks are different from museums. Give reason?
Answer:
Zoological parks are the places where wild animals are kept in protected environments, that the conditions similar to their natural habitats. Museums have collection of preserved plants and animal specimens for study and reference.

Question 5.
Raju collected a skull of an animal and a living rare animal during a study tour. Select the suitable location for each from the list given in the brackets. (Botanical garden, Zoological park, Herbarium, Museum)
Answer:
Skull of an animal – museum
Living rare animal – Zoological park

Question 6.
Distinguish between taxonomic category and taxonomic hierarchy.
Answer:
Classification involves hierarchy of steps in which each step represents a rank or category. Since category is a part of overall taxonomic arrangement, it is called taxonomic category. All the categories together constitute taxonomic hierarchy.

Question 7.
Write the scientific name of following animals.

1. Lion
2. Frog
3. Housefly
4. Tiger

Answer:

1. Pantheraleo
2. Ranatigrina
3. Musca domestica
4. Panthera tigris

Question 8.
Fill in the blank spaces in the table given below.

Answer:

(a) Live
(b) Museum

Question 9.
Panthera jeo is the scientific name of Lion. List the rules you follow to write this scientific name.
Answer:

1. Biological names are generally in Latin and written in italics.
2. When handwritten, the words in a biological name are separately underlined.
3. The first word in a biological name represents the genus and second word represents the species name.
4. The first name (Genus) starts with capital letter and the second name (species) starts with small letter.

Question 10.
List the advantages of

1. Taxonomical key
2. Herbarium

Answer:
1. axonomical key is a taxonomical aid used for identification of plants and animals based on the similarities and dissimilarities. Key are generally analytical in nature.

2. Herbarium is a store house of collected plant specimens that are dried, pressed and preserved on sheets. These specimens, along with their descriptions on herbarium sheets, became a store house or repository for future use. Herbarium serves as quick reference system in taxonomical studies.

Question 11.
Give the terms.

1. The actual account of habital and distribution of plants of a given area.
2. Providing information for identification of names of species found in an area.
3. Contain information on any one taxon.
4. Identification of plants and animals based on the similarities and dissimilarities.

Answer:

1. Flora
2. Manuals
3. Monographs
4. Taxonomical key

Question 12.
Define a taxon. Give some examples of taxon at different hierarchical levels.
Answer:
Each category in the taxonomical hierarchy is considered as a taxonomic unit and is known as a taxon. The taxon used in the classification of animals are kingdom, phylum, class, order, family, genus and species.

Plus One The Living World NCERT Questions and Answers

Question 1.
What do we learn from identification of individuals and populations?
Answer:
In a diverse country like India can learn following things from identification of individuals and population:

1. Native place
2. MotherTongue
3. Costumes
4. Cuisine
5. Religion
6. Caste
7. Socio-economic Background

Question 2.
Can you identify the correct sequence of taxonomical categories?
(a) Species → Order → Phylum → Kingdom
(b) Genus → Species → Order → Kingdom
(c) Species → Genus → Order → Phylum
Answer:
As clear from the table in previous answer, (a) and (c) are showing the correct order.

Question 3.
Why are the classification system changing every now and then? ,
Answer:
In any branch of science nothing is written in concrete. Theories keep on changing as more relevant and correct theories are being discovered. In case of living beings certain species become extinct and some new species is being formed in every era.

This process of addition and deletion of species necessitates the continuous change of the classification system.

Question 4.
Define a taxon. Give some examples of taxa at different hierarchical levels.
Answer:
A taxon is a particular level of hierarchy in the system of classification of living beings. The following figure gives taxa at different hierarchical levels:

Plus One The Living World Multiple Choice Questions and Answers

Question 1.
Binomial nomenclature is described in the book
(a) Genera Plantarum
(b) Historia Plantarum
(c) Systema Naturae
(d) Flora Japonica
Answer:
(c) Systema Naturae

Question 2.
Which of the following is not the main criteria for five kingdom system of classification?
(a) Cell structure and thallus organization
(b) Mode of nutrition and reproduction
(c) Phylogenetic relationship.
(d) Gram staining
Answer:
(d) Gram staining

Question 3.
Two plants can be conclusively said to belong to the same species if they
(a) Can reproduce freely with each other and form seeds
(b) have more than 90 percent similar genes
(c) look similar and possess identical secondary metabolites
(d) have same number of chromosomes
Answer:
(a) Can reproduce freely with each other and form seeds

Question 4.
ICZN stands for
(a) International code of Botanical Nomenclature
(b) International code of Zoological Nomenclature
(c) international code of Viral Nomenclature
(d) International code of Zoo Nomenclature
Answer:
(b) International code of Zoological Nomenclature

Question 5.
In Mangifera indica L. Generic epithet is
(a) Indica
(b) Mangifera
(c) Linnaeus
(d) None of these
Answer:
(b) Mangifera

Question 6.
ICBN stands for
(a) Indian Congress of Biological Names
(b) International Code of Botanical Nomenclature
(c) International Congress of Biological Names
(d) Indian Code of Botanical Nomenclature
Answer:
(b) International Code of Botanical Nomenclature

Question 7.
Taxon is the
(a) taxonomic group of any rank
(b) procedure to assign a scientific name
(c) process of classification
(d) process by which anything is grouped into convenient categories based on characters
Answer:
(a) taxonomic group of any rank

Question 8.
Which of the following is famous for stating that “Population increases much faster than its food supply”?
(a) Fredrick Losch
(b) R Vircow
(c) T R Malthus
(d) Karl Von Baer
Answer:
(c) T R Malthus

Question 9.
Reproduction is the characteristic feature of living organisms. Which of the following can not reproduce?
(a) Amoeba and Paramecium
(b) Fungi and filamentous algae
(c) Humans and Ayes
(d) Mules and worker bees
Answer:
(d) Mules and worker bees

Question 10.
Nicotiana sylvestris flowers only during long days and N.tobacurh flowers only during short days. If raised in the laboratory under different photoperiods, they can be induced to flower at the same time and can be cross fertilized to produce self-fertile offspring. What is the best reason for considering N. sylvestris and N. tobocum to be separate species?
(a) They are physiologically distinct
(b) They are morphologically distinct
(c) They cannot interbreed in nature
(d) They are reproductively distinct
Answer:
(c) They cannot interbreed in nature

Question 11.
Which of the following book is not associated with Carolus Linnaeus, the father of Taxonomy and Nomenclature?
(a) Systema Naturae
(b) Genera Plantarum
(c) Species Plantarurn
(d) Historia Generalis Plantarum
Answer:
(d) Historia Generalis Plantarum

Question 12.
Philosophie Zoologique, a book written by Jean Baptiste de Lamarck is based on
(a) Survival of the fittest
(b) Natural Selection
(c) Inheritance of acquired characters
(d) Biogenetic law
Answer:
(c) Inheritance of acquired characters

Question 13.
The book “Philosophic Zoologique” was written by
(a) Lamarck
(b) Mendel
(c) Haeckel
(d) Hugo deVries
Answer:
(a) Lamarck

Question 14.
Binomial nomenclature was introduced by
(a) Linnaeus
(b) Darwin
(c) Aristotle
(d) deCandoile
Answer:
(a) Linnaeus

Students can Download Chapter 2 Animal Kingdom Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examination

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 2 Animal Kingdom

Plus One Animal Kingdom One Mark Questions and Answers

Plus One Zoology Animal Kingdom Questions Chapter 2 Question 1.
Fill in the blanks.

Answer:
i) B – Nereis
ii) B – Ctenoplanna
C – Ctenophora
iii) A – radula

Plus One Zoology Chapter Wise Questions And Answers Pdf Chapter 2 Question 2.
Write any two members of the Phylum Aschelminthes which are found parasitic on Human beings.
Answer:

1. Ascaris (Roundworm)
2. Wuchereria (Filaria worm)

Plus One Zoology Previous Question Papers Chapter Wise Chapter 2 Question 3.
In some animal groups, the body is found divided into compartments with at least some organs/ organ repeated. This characteristic feature is named
(a) Segmentation
(b) Metamerism
(c) Metagenesis
(d) Metamorphosis
Answer:
(b) Metamerism

Presence Of Column And Metamerism Are The Most Important Characters In Chapter 2 Question 4.
Given below are types of cells present in some animals. Each one is specialized to perform a single specific function except
(a) Choanocytes
(b) Interstitial cells
(c) Gastrodermal cells
(d) Nematocytes
Answer:
(b) Interstitial cells

Plus One Zoology Chapter Wise Questions And Answers Chapter 2 Question 5.
Which one of the following sets of animals share a four chambered heart?
(a) Amphibian, Reptiles, Birds
(b) Crocodiles, Birds, Mammals
(c) Crocodiles, Lizards, Turtles
(d) Lizards, Mammals, Birds
Answer:
(b) Crocodiles, Birds, Mammals

Hsslive Plus One Zoology Chapter Wise Questions And Answers Chapter 2 Question 6.
Which of the following pairs of animals has non glandular skin.
(a) Snake and Frog
(b) Chameleon and Turtle
(c) Frog and Pigeon
(d) Crocodile and Tiger.
Answer:
(c) Frog and Pigeon

Plus One Zoology Previous Question Papers Chapter Wise Hsslive Chapter 2  Question 7.
Birds and mammals share one of the following characteristics as a common feature.
(a) Pigmented skin
(b) Alimentary canal with some modification
(c) Viviparity
(d) Warm blooded nature
Answer:
(d) Warm blooded nature

Plus One Zoology Chapter Wise Questions And Answers Pdf Download Chapter 2 Question 8.
Note the relationship between the first two words and find a suitable word for the fourth place,

1. Coelenterata: radial symmetry, platyhelminthes, _______
2. Lizard: Poikilothermous, crow, _________

Answer:

1. bilaterally symmetrical
2. Homoiothermous

Plus One Zoology Chapter Wise Previous Questions And Answers Chapter 2 Question 9.

1. Annelida: Parapodia :: __________ : Comb plates
2. _________: Water vascular system :: Coelenterata : cnidoblast

Answer:

1. Ctenophora
2. Echinodermata

Plus One Zoology Chapter Wise Questions And Answers Pdf Hsslive Chapter 2 Question 10.
Malpighian tubule is the excretory organ of which phylum?
(a) Phylum porifera
(b) Phylum arthropoda
(c) Phylum Coelenterata
(d) Phylum mollusca
Answer:
(b) Phylum Arthropoda

Pick Out The Acoelomate Organisms From The Following Chapter 2 Question 11.
A chordate animal having flame cells as the excretory organ.
Answer:
Amphioxus

Animal Kingdom Class 11 Important Questions And Answers Chapter 2 Question 12.
From the pictures given below, find out the poikilothermic animals.

Answer:
B and D are poikilothermic animals.

Question 13.
Segmentation in the body is first observed in which of the following:
(a) Platyhelminthes
(b) Aschelminthes
(c) Annelida
(d) Arthropoda
Answer:
(c) Annelida

Question 14.
Shark has to swim continuously, otherwise, it will sink down. Give reason.
Answer:
Due to absence of air bladder.

Plus One Animal Kingdom Two Mark Questions and Answers

Question 1.
Arrange the phylum in order.
Arthropoda → Platyhelminthes → Porifera → Ctenophora → Cnidaria → Mollusca → Annelida → Echinodermata → Aschelminthesip
Answer:

• Presence of milk producing mammary gland.
• Presence of hair on skin.
• External ear or pinnae is present.
• Different types of teeth are present in the jaw.

Question 2.
Complete the blanks.

Answer:

Question 3.
Triploblastic animals are more complex than diploblastic animals. Do you agree with this statement? Justify.
Answer:
Triploblastic animals have more cell layers so they have the possibility of greater degree of cellular specialisation.

Question 4.
All vertebrates are chordates but all chordates are not vertebrates. Justify.
Answer:
Notochord is present in all vertebrates but vertebral column is present only in vertebrates and not in all chordates.

Question 5.
Copy and complete the table.

Answer:

Question 6.
Arrange the phylum in order.
Arthropoda → Platyhelminthes → Porifera → Ctenophora → Cnidaria → Mollusca → Annelida → Echinodermata → Aschelminthes
Answer:
Porifera → Cnidaria → Ctenophora→ Platyhelminthes → Aschelminthes → Annelida → Arthropoda → Mollusca → Echinodermata

Question 7.
Categorise and classify the following organisms and arrange them in a table with separate columns and provide appropriate headings.
Exocoetus, Physalia, Ascaris, Apis, Locusta, Corvus, Pita, Hydra, Sepia, Ancylostoma.
Answer:

Question 8.

1. Identify the phylum which exhibit metagenesis or alternation of generation.
2. What is meant by alternation of generation?

Answer:

1. Cnidaria
2. Alternation of sexual and asexual forms of organism: ie., Polyp asexually produce medusa, medusa sexually product polyp.

Question 9.
Name the following.

1. Phylum in which flatworms are included
2. Excretory organs of Annelids.
3. Largest phylum.
4. An oviparous mammal.

Answer:

1. Platyhelminthes
2. Nephridia
3. Arthropoda
4. Platypus

Question 10.
Apis, Prawn, Locust, Spider
Following animals have different habit and habitat. But they have many common characters.

1. Mention the common characters.
2. Identify their phylum.

Answer:

1. Joint footed animals
   • Metameric segmentation
   • Chitinous exoskeleton
2. Arthropoda

Question 11.
A list of animals are given below. Arrange them according to increase in complexity of organization. Scorpion, Earthworm, Liver fluke, Pigeon, Seaanemon, Sycon, Elephant, Anabas.
Answer:
Sycon, Seaanemon, Liver fluke, Earthworm, Scor¬pion, Anabas, Pegeon, Elephant.

Question 12.
Nithin is Studying in Std. XI. He collected some specimens during the field trip conducted by the Science Club of his School. Help Nithin to Classify the Animal in respective Phylum.
Prawn, Slug worm, Butterfly, Pila, Grass Hopper, Crab
Answer:

Question 13.
During a field trip Raju has collected some organisms with the following characters. Help him to identify the phyla of those organisms.

1. Metamerically segmented body.
2. Body covered with calcareous shell.
3. Dorso-ventrally flattened leaf like body.
4. Body divided into head, thorax and abdomen.

Answer:

1. Annelida
2. Mollusca
3. Platyhelminthes
4. Arthropoda

Question 14.
Categorise the following fishes into Osteichthyes and Chondrichthyes?

1. Exocoetus
2. Trygon

Answer:

1. Exocoetus – Oesteichthyes
2. Trygon – Chondrichthyes

Question 15.
Arrange the following terms in two columns correctly. Malpighian tubules, radula, metamerism, Bioluminescence, choanocytes, nematocytes, Phylum-coelenterate, phylum-Arthropoda, phylum- ctnophora, phylum-Mollusca, Phylum-Porifera.
Answer:

Question 16.
Match the following

Answer:

Question 17.
Observe the given organisms

1. Place these animals in proper phylum.
2. Segmentation in the body is first observed in which of the above phylum?

Answer:

1.  i) Arthropoda,
   ii) Porifera,
   ii) Annelida
2. Annelida

Question 18.
Observe the table given below and fill the blank columns A, B, C and from the animals given in brackets. (Ascaris, Starfish, Fasciola, Earthworm)

Answer:

Question 19.
1. Identify the animal given below.

2. Write one well marked property of the above animal.
Answer:

1. Pleurobrachia
2. Bioluminescence is well marked property of pleurobrachia

Question 20.
Representatives of some vertebrate classes are introducing themselves. Write down the name of the class in which they belong.

1. My gills are covered by operculum. I have bony endoskeleton.
2. I give birth to young ones. My body is covered by hair.
3. My skin is glandular? Have trilocular heart.
4. I live only in marine water. My endoskeleton is made up of cartilage.

Answer:

1. Osteichthyes
2. Mammalia
3. Amphibia
4. Chondrichthyes

Question 21.
During classroom discussion a student said that sponges are more complex than cnidarians. Do you agree with him. Justify.
Answer:
NO. Sponges are asymmetrical and body is formed of loose aggregate of cells. Cells are not organised to from tissues and organs. Cnidarians are radially symmetrical and tissue grade of organisation. So cnidarians are more complex than sponges.

Question 22.
Due to the absence of air bladder, fishes belonging to the class Chondrichthyes have to swim constantly. How important is the presence of air bladder in these fishes?
Answer:
Due to the absence of air bladder in chondrichthyes, they have to swim constantly to avoid sinking. If air bladder is present which regulates buoyancy.

Question 23.
Match the column A, B &C in the table given below:

Answer:

Question 24.
Identify the phylum whose larvae are bilaterally symmetrical, but adults are radially symmetrical.

1. Annelida
2. Arthropoda
3. Mollusca
4. Echinodermata

Mention two salient features of the phylum.
Answer:
4. Echinodermata
Salient features Presence of Echinodermata:

• Water vascular system
• Spiny bodies, Endoskeleton of Calcareous ossicles

Question 25.
Write the name of phylum.

1. Diploblastic, tissue grade of organisation, radially symmetrical, polymorphic animals.
2. Soft bodied, Unsegmented, Bilaterally symmetrical animals with open type circulation.
3. Triploblastic, bilaterally symmetrical, true coelomic animals with metamerism, closed circulation.
4. Triploblastic, chitinous exoskeleton and open circulation.

Answer:

1. Cnidaria
2. Mollusca
3. Annelida
4. Arthropoda

Question 26.
Observe the table given below and fill the blank columns a, b, c, and d from the animals given in brackets.

(Hydra, Shark, Spongilla, Obelia)
Answer:
(a) Hydra/Obelia
(b) Shark
(c) Spongilla
(d) Hydra/Obelia

Question 27.
Copy and complete the table.

Answer:

Question 28.
You are provided with two fishes Catla (Bony fish) and Shark (Cartilagenous fish). Prepare a table showing difference in:
(a) Position of mouth
(b) Air bladder
(c) Scales
(d) Fertilization
Answer:

Question 29.
Prepare a list of some animals that are found parasitic on human beings.
Answer:
Tapeworm (Taenia), Ascaris (Roundworm), Wuchereria (Filaria worm), Ancylostoma(Hookworm)

Question 30.

1. Identify the 2 forms of Cnidarians.
2. Mention any 2 difference between them.

Answer:

1. A – polyp
   B – medusa
2. Polyp: Asexual, sessile, mouth upwards Medusa: Sexual, Free swimming, Mouth downwards

Plus One Animal Kingdom Three Mark Questions and Answers

Question 1.
Classify the given organisms and arrange them in the order of their phylum.
Limulus, Corvus, Spongilla, Ascaris, Physalia, Nereis, Catla, Sepia, Echinus, Taenia, Pleurobrachia, Tiger, Viper, Toad.
Answer:

Question 2.
Match the following.

Answer:

Question 3.
Observe the diagram

1. Identify the phylum of this hypothetical organism.
2. List out the features that helps in identifying it.
3. Write about the fate of notochord in Urochordata, Cephalochordata and Chordata.

Answer:

1. Chordata
2. Notochord, Dorsal nerve cord, Pharyngeal gill slits, Post anal tail
3. In Urochordatarfiotochond is present only in larval tail. In Cephalochondata, notochord extends from head to tail region is persistent throughout their life. In vertebrata, the notochord is replaced by a cartilagenous or bony vertebral column in the adults.

Question 4.
Select the following items into their appropriate phylum.
Radula, Parapodia, Comb plate, Nephridia, Choanocytes, Flame cells.
Answer:

• Radula – Mollusca
• Parapodia – Annelida
• Comb plate – Ctenophora
• Nephridia – Annelida
• Choanocytes – Porifera
• Flame cells – Platyhelminthes

Question 5.

(a) Fill and complete the chart given below.

(b) Write any two fundamental characters of the phylum chordata.
(c) Classify Tetrapoda into classes:
Answer:
Male accessory ducts store and transport the sperms from testis to the outside through urethra. Male accessory glands secrete seminal Plasma, Which is rich in fructose, citrate, prostaglandins and certain enzymes. The secretion of cowper’s glands lubricate the penis.

Question 6.
Write down the functions of the following (any two) structures and assign their phyla.

1. Radula
2. Flame cells
3. Parapodia

Answer:

1. Radula: File like rasping organ for feeding.
   Phylum: Mollusca
2. Flame cells: Osmoregulation and excretion
   Phylum: Platyhelminthes
3. Parapodia: help in swimming
   Phylum: Annelida

Question 7.
1. Identify the organism A and B.

2. Which features of these organism enable you to identify them?
Answer:

1. A-Tapeworm
   B – Earthworm
2. Features of these organism:
   • Tapeworm: Scolex is present, suckers and hooks are present. No true segmentation.
   • Earthworm: True segmentation, Absence of Scolex, hook and suckers, Clitellum is present.

Question 8.

1. Which of the following animals exhibit metagenesis? (Ascaris Obelia Earthworm Crab)
2. To which phylum does it belong?
3. Write any two features of the phylum.

Answer:

1. Obelia
2. Cnidaria
3. Two features of the phylum:
   • Presence of Cnidoblasts
   • Cnidarians exhibit two basic forms called polyp and medusa.

Question 9.
Pick out the appropriate one from the term given within bracket and put against the corresponding phylum.

1. Porifera
2. Coelenterata
3. Platyhelminthes
4. Annelida
5. Arthropoda
6. Mollusca
7. Echinodermata
8. Chordata

(Hirudin, Flame cell, Choanocytes, Cnidoblast, Jointed legs, Radula, Notochord and Dermal Ossicles)
Answer:

1. Porifera – Choanocytes
2. Coelenterata – Cnidoblast
3. Platyhelminthes – Flame cells
4. Annelida – Hirudin
5. Arthropoda – Jointed legs
6. Mollusca – Radula
7. Echinodermata -Dermal ossicles
8. Chordata – Notochord

Question 10.

1. Identify the organism (A) and (B) and which class do they belong?
2. On which basis do you classify these animals?

(Hint: Write any 2 Identifying characters)
Answer:

1. Organism in (A) and (B)
   • A – Bony fish – Osteichthyes
   • B – Cartilagenous fish – Chondrichthyes

2.

Question 11.
From a fish market, you got a fish, on a close watching you friend says it is a cartilaginous fish.

1. Which characters helped him to identify it as a cartilaginous fish, (any four characters.)
2. Name the class it belongs.

Answer:

1. Characters:
   • a – Gillslits are separate and without operculum
   • b – Placoidscale
   • c – Mouth is located ventrally
   • d – heterocercal caudal fin
2. Chondrichthyes

Question 12.
Presence or absence of a cavity between the body wall and the gut wall is very important in classification.

1. Identify the different types of body cavities.
2. Give examples to each

Answer:

1. Different types of Body Cavities:
   • a – Coelomate
   • b – Pseudocoelomate
   • c – Acoelomate
2. Examples:
   • Coelomate – Chordates
   • Pseudocoelomate – Aschelminthes
   • Acoelomate – Platyhelminthes

Question 13.
From the following general characters find out corresponding/Class with an Example

1. Exclusively marine, triploblastic, spines on the skin, radially symmetrical in the adult and bilaterally symmetrical in the larval stage.
2. Marine, they migrate towards freshwater for spawning, then their larvae return to ocean after metamorphosis.
3. Triploblastic, bilaterally symmetrical, coelomate and metamerically segmented animals.

Answer:

1. Echinodermata
   • eg: Starfish
2. Class – Cyclostomata
   • eg: Petromyzon
3. Annelida
   • eg: Earthworm

Question 14.
Arrange the following points in a two-column table and give suitable heading for each column.

1. Notochord present
2. Post anal tail absent
3. Pharynx Perforated by gill slits
4. Notochord absent
5. Posts anal tail present
6. Gill slits are absent

Answer:

Question 15.
Observe the figure and answer the questions.

1. Identify the structure.
2. Name the phylum which possess this structure.
3. How this structure help the organism?

Answer:

1. Cnidoblastornematocyst
2. Cnidaria
3. Capture of Prey
   • Defense organ.

Question 16.
Note the relationship between first two words and suggest suitable words for the 4th place.

1. Planaria: Flartfe cells: Earthworm: _________
2. Jaw present: Gnathostomata:: Jaw absent: ___________

Answer:

1. Nephridia
2. Agnatha

Question 17.
Give reasons for the following.

1. Respiratory and circulatory system are absent in parasitic platyhelminthes and Aschelminthes.
2. Arthropods are the most successful invertebrate.
3. Body of endoparasites are covered with cuticle.

Answer:

1. Parasitic Plalyhelminthes and Aschelminthes lives in anaerobic condition. So respiratory and circulatory systems are absent in these parasitic forms.
2. Arthropods are most successful, because of the presence of unique chitinous cuticle.
3. In these parasites, the cuticle helps in escaping from the action of digestive enzymes.

Question 18.
Arthropodes are organisms with chitinous exoskeleton. Suppose exoskeleton is absent in arthropods. List the difficulties arthropodes has to face.
Answer:

1. Their body will be dried up due to evaporation
2. They couldn’t escape from predators.
3. They couldn’t live in all environments.

Question 19.
Suppose during your field visit for collection from a rocky seashore you have got some live specimens such as sea urchin, sea cucumber, sea anemone. Is it possible to keep them on an aquarium in your school. Give reasons for your answer.
Answer:
No. It is not possible.
Marine animals cannot live on freshwater because it leads to endosmosis and death occurs.

Question 20.

1. Why nematocysts are more concentrated on the oral end and tentacles of cnidarians?
2. What are the difficulties that coelenterate have to face if nematocysts were absent in body.

Answer:

1. Nematocysts are concerned with defence and offence. Tentacles are usually used for defence, offence and food collection. Hence nematocysts are more concentrated in the oral end.
2. The major difficulties cnidarians has to face in the absence of nematocysts are for food collection and escaping from enemies.

Question 21.

1. Which of the following show the body cavity of earthworm?
2. Identify the names of germlayers ‘a’ and ’b’.

Answer:

1. Figure C is the body cavity of earthworm. Because it is a true coelom
2. The names of germlayers
   • a-ectoderm
   • b-endQderm

Question 22.
Identify the characters listed below and put (✓) mark on appropriate places

Answer:

Question 23.

1. Identify the organisms.
2. Which feature of these organisms enable you to identify them?

Answer:

1. The organisms:
   • A-Tape worm
   • B – Ascaris
   • C – Earthworm
2. Features:
   • Tapeworm – Body is dorso-ventrally flattened and segmented.
   • Ascaris – Body is cylindrical and vermiform
   • Earthworm – Body is divided into similar segments and clitellum is present.

Question 24.
During a classroom discussion a student said that sponges are more complex than cnidarians. Do you agree with him. Jusftfy.
Answer:
1. Sponges: Cellular grade of organisation and cell aggregate body plan.

2. Cnidarians: Tissue level of organisation and blind sac body plan. So Cnidarians are more complex than sponges.

Question 25.
While comparing the digestive system of a roundworm and flatworm, a boy noted some differences. List out the differences.
Answer:
1. Flatworms: Digestive system incomplete, has only a single opening, ingestion and egestion occurs through the same opening.

2. Round worm: Complete digestive system, it has both mouth and anus. A muscular pharynx is also present in their digestive system.

Question 26.

1. Identify the invertebrate whose mouth is on ventral or lower side and anus is on dorsal or upper side.
2. Mention the phylum.
3. Comment on its General characters.

Answer:

1. Starfish
2. Echinodermata
3. Its General characters are:
   • Water vascular system
   • Tube feet
   • Coelomate
   • Spiny body

Question 27.
Match the following.

Answer:

• Arthropoda – Bombyx – Joint footed animals
• Annelida – Earthworm – Little rings
• Echinodermata – Antedon – Spiny bodies animals
• Mollusca – Pila – Soft bodies animals

Question 28.
Animals showing metameric segmentation are included under annelida. Body of tapeworm has numerous segments, but the animal is not included under annelida. How will you account for it?
Answer:
Tapeworm show false segmentation. In true segmentation number of segments is fixed and age of the segments are all same.

Question 29.
Complete the given flow chart showing the flow of water through canal system in sponges.

Answer:

Question 30.
Mention a single word for the following.

1. Sexes are not separate.
2. Body wall with three layers.
3. File like rasping organs in Mollusca.
4. Segmentation of animal body.
5. The property of a living organism to emit light.
6. Alternation of generation in Cnidarians.

Answer:

1. Hermaphrodite
2. Triploblastic
3. Radula
4. Metamerism
5. Bioluminescence
6. Metagenesis

Question 31.
The birds are well adapted for flying.

1. Write the general characters of Aves.
2. Give some of the flight adaptations seen in birds.

Answer:

1. General characters of Aves:
   • Presence of feathers
   • Presence of beak
   • Forelimbs are modified into wings.
   • Warm blooded
   • Respiration by lungs with air sacs.
   • Oviparous
2. Flight adaptations:
   • Forelimbs are modified into wings
   • Long bones are hollow with air cavities (Pneumatic)
   • Air sacs connected to lungs supplement respiration
   • Constant body temperature

Plus One Animal Kingdom NCERT Questions and Answers

Question 1.
What is the difference between direct and indirect development?
Answer:
1. Direct Development:
In direct development the young animal resembles an adult. There is no intermediate stage.

2. Indirect Development:
In indirect development there is intermediate stage, like larval stage. For example, frog before being developed into adult passthrough a tadpole stage. This is the case of indirect development.

Question 2.
What are the peculiar features that you find in parasitic platyhelminthes?
Answer:
In parasitic platyhelminthes hooks and suckers are present. Suckers help the parasite, in sucking the blood from the host.

Question 3.
What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?
Answer:
Arthropods are the first phylum to have well developed systems to carry out different activities. There is distinct system for respiration, locomotion and reproduction. Their survival capacity is great because of elaborate system. This has helped them survive in diverse conditions. They can live in water, on land and in air.

This can be one of the reasons why arthropods are the largest group among the animal kingdom. Another reason is their early development compared to animals of higher phylum.

Question 4.
Water vascular system is the characteristic of which group of the following:
(a) Porifera
(b) Ctenophora
(c) Echinodermata
(d) Chordata
Answer:
(c) Echinodermata

Question 5.
“All vertebrates are chordates but all chordates are not vertebrates”. Justify the statement.
Answer:
All chordates have notochord present in some stage of life.
The difference between vertebrates and nonvertebrates is as follows:
In vertebrates the notochord is present in the embryonic stage. This is replaced by a vertebral column during the adult stages.

Question 6.
How important is the presence of air bladder in Pisces?
Answer:
Presence of air-bladder in Pisces helps in buoyancy. This means that members of pisces don’t have to keep on swimming to remain floating.

Question 7.
What are the modifications that are observed in birds that help them fly?
Answer:
Following modification in birds help them fly:

1. Pneumatic or hollow bones make for a light weight skeleton.
2. Fore limbs are modified into wings to assist in flight.
3. Excertion of urine and faeces is through single opening facilitating weight reduction.
4. Aerodynamic body helps in flying.

Question 8.
Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?
Answer:
Usually number of eggs produced by oviparous mothers is greater than number of young ones produced by viviparous mothers. The main reason for this is the need of resources required for development of the embryo.
In oviparous the major part of development of the embryo takes place outside the uterus. This makes lesser burden on the mother.

On the other hand in viviparous animals the development takes place inside the uterus so lesser number of young ones can be successfully incubated. Moreover, once eggs are outside they are at risk of getting eaten by some predator because of their immobility, so need of more eggs is there to ensure continuity of progeny.

Plus One Animal Kingdom Multiple Choice Questions and Answers

Question 1.
Calcareous skeleton is found in
(a) echinoderms
(b) some sponges
(c) mollusca
(d) all the above
Answer:
(d) all the above

Question 2.
Which cannot be the character of cnidaria
(a) musculoepithelial cells
(b) gastrovascular cavity
(c) nerve cells and process
(d) organ grade organization
Answer:
(d) organ grade organization

Question 3.
A non-matching set in the following is
(a) sepia – cuttle fish
(b) octopus – devilfish
(c) limulus – king crab
(d) ancylostoma – pinworm
Answer:
(d) ancylostoma – pinworm

Question 4.
A character common to Echinoderms and chordates
(a) marine
(b) benthonic
(c) deuterostome
(d) none of the above
Answer:
(c) deuterostome

Question 5.
Largest animal in the world that feed on smallest plankton is the
(a) dolphin
(b) killer whale
(c) blue whale
(d) sea cow
Answer:
(c) blue whale

Question 6.
Which is common to amphibian, reptelia and fishes
(a) nucleated RBC
(b) dermal scales
(c) poikelothermic condition
(d) both a and d
Answer:
(d) both a and d

Question 7.
Which of the following has pseudocoelomate tube within a tube body plan
(a) hydra
(b) planaria
(c) ascaris
(d) pheretima
Answer:
(c) ascaris

Question 8.
Ink gland associated with alimentary canal is found in
(a) sepia
(b) earthworm
(c) starfish
(d) cockroach
Answer:
(a) sepia

Question 9.
Which is common to all tetrapods
(a) epidermal scales
(b) red coloured blood
(c) 12 pairs of cranial nerve
(d) ureotelism
Answer:
(b) red coloured blood

Question 10.
Which one of the following is not a characteristic of phylum annelida?
(a) closed circulatory system
(b) segmentation
(c) pseudocoelom
(d) ventral nerve cord
Answer:
(c) pseudocoelom

Question 11.
Respiratory pigment of mollusc is
(a) haemocyanin
(b) haemoglobin
(c) haemoerythrin
(d) both a and b
Answer:
(a) haemocyanin

Question 12.
Select the character that can be attributed to chondrithytes
(a) persistened notochord
(b) placoid scales
(c) poikelothermic body
(d) all the above
Answer:
(d) all the above

Question 13.
The central cavity of sponge is called
(a) spongocoel
(b) coelocentron
(c) canal system
(d) spongilla
Answer:
(c) canal system

Question 14.
Flame cell are excretory organ of
(a) hydra
(b) cockroach
(c) planaria
(d) frog
Answer:
(c) planaria

Question 15.
Pneumatic skeleton is a feature of
(a) amphibians
(b) reptiles
(c) fishes
(d) birds
Answer:
(d) birds

Question 16.
The number of gills present in osteichthyes is
(a) 2 pairs
(b) 6-15 pairs
(c) 5 pairs
(d) 4 pairs
Answer:
(d) 4 pairs

Students can Download Chapter 6 Work, Energy and Power 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 6 Work, Energy and Power

Plus One Physics Work, Energy and Power One Mark Questions and Answers

Plus One Physics Work Energy Power Questions Question 1.
Find the odd one out and find the relation connecting the remaining quantities. Joule, Calorie, Kilowatt, electron volt.
Answer:
kilowatt, unit of power
1 calorie = 4.2 joule
1 electron volt = 1.6 × 10^-19J.

Plus One Physics Chapter Wise Questions And Answers Question 2.
What is the work done by the tension in the string of simple pendulum?
Answer:
Zero

Plus One Physics Work Energy Power Question 3.
When is the exchange of energy is maximum during an elastic collision?
Answer:
When mass of two colliding bodies are same, there will be maximum exchange of energy.

Work Energy And Power Questions And Answers Question 4.
In atom, an electron is revolving around the nucleus. What is the work done?
Answer:
Work done is zero because work done by centripetal force is zero.

Plus One Physics Important Questions And Answers Pdf Question 5.
What is the type of collision when macroscopic particles collide?
Answer:
Perfectly inelastic collision.

Hsslive Plus One Physics Chapter Wise Questions And Answers Question 6.
Name the parameter which is a measure of degree of elasticity of a body.
Answer:
Coefficient of restitution.

Plus One Physics Chapter Wise Questions And Answers Pdf Hsslive Question 7.
What is the source of kinetic energy for falling rain drops?
Answer:
Gravitational potential energy.

Plus One Physics Work, Energy and Power Two Mark Questions and Answers

Work Energy And Power Class 11 Important Questions With Answers Question 1.
The law of conversation of energy states that energy can neither be created nor be destroyed but can only change from one form into another. A bus and a car, moving with the same kinetic energy are brought to rest by applying an equal retardation force by the breaking systems. Which one will come to rest at a shorter distance? Give the reason behind your answer.
Answer:
Change in K.E. = Force × Displacement
1/2 mv² = F × S
ie. KE α s
KE_car α S_car _____(1)
KE_bus α S_bus ______(2)

ie. S_car = S_bus
Both will travel equal distance.

Work Energy And Power Questions And Answers Pdf Question 2.
A body constrained to move alomg the Z-axis of a co-ordinate system is subjected to a constant force \(\bar{F}=(\hat{i}+2 \hat{\jmath}+3 \hat{k}) N\)

1. What is the magnitude of force along z direction.
2. What is the work done by this force in moving the body over a distance of 4m along z-axis.

Answer:

1. 3N
2. Work done = Force × Displacement = 3 × 4 = 12J.

Physics Questions On Work Energy And Power Question 3.
Match the following

Answer:
Collision of two balls – inelastic – TE and momentum Collision of two molecules – elastic – KE, TE, and momentum.

Plus One Physics Work, Energy and Power Three Mark Questions and Answers

Questions From Work Energy And Power Question 1.
A car of mass 1000kg moving with a speed 18mk/h on a horizontal road collides with a horizontally mounted spring of spring constant 6.25 × 10³N/m

1. What do you mean by Spring constant.
2. What is the maximum compression of the spring?

Answer:
1. Spring constant is the force required to stretch the spring by a unit distance.

2. \(\frac{1}{2}\)mv² = \(\frac{1}{2}\)kx², 18km/h = 5m/s.

x = 2m.

Question 2.
A man tries to lift a mass 200kg with a force 100N

1. Is he doing work? Explain.
2. If yes, find the amount of work done If No, find the force required to lift it.
3. If it is lifted to 2m in 10 seconds, find his power.

Answer:

1. No work is done, as there is no displacement, 100N force is insufficient to raise 200kg.
2. Force required to lift 200 kg = 200 × 9.8 = 1960N
3. Power = \(\frac{m g h}{t}=\frac{200 \times 9.8 \times 2}{10}\) = 392W.

Question 3.
Two cricket balls are colliding each other.

1. Name the collision
2. Say whether law of conservation of Kinetic Energy hold good in this case. Why?
3. State and prove the other conservation law applicable here.

Answer:

1. Inelastic collision
2. No, Total KE before collision is not equal to total KE after collision.
3. Proof and statement of law of conservation of momentum.

Plus One Physics Work, Energy and Power Four Mark Questions and Answers

Question 1.
Two cars A and B travelling with speeds 20m/s and 10m/s respectively applies breaks,.so that A comes to rest in 15 second and B in 7.5s

1. From the graph determine which of the two cars travelled further after brakes were applied and by how much distance it travelled?
2. Draw the velocity time graph of A and B in the same graph.
3. In the above process ,the wear and tear of which the car gets affected more ?

Answer:
1. The area of velocity time graph gives displacement distance travelled by the car A, S_A = 1/2 × 20 × 15 = 150 m
distance travelled by the car B, S_B = 1/2 × 10 × 7.5 = 37.5.

2.

3. Wear and tear gets affected more for the car A.

Question 2.
A sphere of mass m is moving with a velocity u and makes a head on collision with another identical mass which is at rest. It is observed that the stationary mass starts moving with a lesser velocity than u, after the collision.

1. Which physical quantity is conserved here?
2. Define coefficient of restitution.
3. Determine the ratio of the velocities of the two spheres after elastic collision if ‘e’ is the coefficient of restitution.

Answer:
1. conservation of linear momentum.

2. It is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision.

3. Coefficient of restitution,
e = \(\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\)
in this cos u₁ = u, u₂ = 0, v₁ = 0, v₂ = u
∴ e = \(\frac{u-0}{u-0}\)
e = 1.

Question 3.
From the table given below

1. Draw the force displacement curve
2. Analyse the graph & find the type of force involved
3. Estimate the workdone

Answer:
1.

2. Workdone by a variable force.

3. Workdone = Area of the graph
= \(\frac{1}{2}\)bh
= \(\frac{1}{2}\)5 × 10 = 25J.

Question 4.
Raju increased the speed of moving mass ‘50 kg’ from 2 m/s to 4m/s.

1. How much force will be required, if velocity change takes place with in 0.2 sec?
2. How much work is done by Raju?

Answer:
1. F = mass × acceleration
= 50 × \(\frac{(4-2)}{0.2}\)
= 500N.

2. w = \(\frac{1}{2}\)mv² – \(\frac{1}{2}\)mu².
=\(\frac{1}{2}\)50 (4² – 2²)
= 300 J.

Plus One Physics Work, Energy and Power Five Mark Questions and Answers

Question 1.
A car and a truck have the same kinetic energies at a certain instant while they are moving along two parallel roads. (Assume that the truck is heavier than the car)

1. Which one will have greater momentum?
2. Write the relationship between kinetic energy and linear momentum.
3. If the mass of truck is 100 times greater than that of the car, find the ratio between their velocities.

Answer:
1. Kinetic energy, of car, K.E_c = \(\frac{P_{c}^{2}}{2 m_{c}}\)
Kinetic energy of truck,

Since m_c < m_t
Hence P_t > P_c
∴ momentum of truck is greater than car.

2. KE = P²/2m.

3. KE_c = KE_t
1/2m_c V_c² = 1/2 x m_tV_t²
But m_t = 100m_c
\(V_{t}^{2}=\frac{V_{c}^{2}}{100}\)
Velocity of truck, Vt = \(\frac{V_{c}}{10}\)
ratio of velocity, 10V_t = V_c
10:1.

Question 2.
Raju dropped a rubber ball of mass m from a height h to the ground. He observed that the ball rebounds vertically and along the same line to a height h₁, which is less than h.

1. Is it an elastic or inelastic collision?
2. Find the velocity with which it strikes the ground?
3. If it is replaced by a solid aluminium ball, then what happens to the height of rebound?
4. If the rubber ball is allowed to fall on a spring placed on the ground then what change will Raju notice in the height of rebound?

Answer:
1. Inelastic collision.

2. The velocity with which ball strikes the ground,
v² = u² + 2as
v² = 0 + 2g × h
v = \(\sqrt{2 g h}\).

3. Height of rebound decreases.

4. Height of rebound depends on the state of potential energy stored in the spring. If ball falls on a compressed spring, the height of rebound increases due to potential energy given by the spring to bail.

Question 3.
A man tries to lift a mass 200kg with a force 100N.

1. Is he doing work? Explain.
2. If it is lifted to 2m in 10s, find the power.
3. Show that total mechanical energy is conserved fora freely falling body.

Answer:
1. No. Force required to lift the body is 2000N (w = mg = 200 × 10). But the applied force is 100N. Hence there is no displacement due to this applied force.

2. Power P = \(\frac{w}{t}=\frac{m g h}{t}=\frac{200 \times 10 \times 2}{10}\) = 400 watt.

3.

Consider a body of mass ‘m’ at a height h from the ground.
Total energy at the point A
Potential energy at A,
PE = mgh
Kinetic energy, KE = \(\frac{1}{2}\) mv² = 0
(since the body at rest, v = 0).
∴ Total mechanical energy = PE + KE
= mgh + 0 = mgh
Total energy at the point B
The body travels a distance x when it reaches B. The velocity at B, can be found using the formula.
v² = u² + 2as
v² = 0 + 2 gx

∴ KE at B, = \(\frac{1}{2}\) mv²
= \(\frac{1}{2}\) m2gx
= mgx
P.E. at B, = mg(h – x)
Total mechanical energy = PE + KE
= mg(h – x) + mgx
= mgh
Total energy at C
Velocity at C can be found using the formula
v² = u² + 2as
v² = 0 + 2 gh
∴ KE at C, = \(\frac{1}{2}\)mv²
= \(\frac{1}{2}\)m2gh
= mgh
P.E. at C = 0
Total energy = PE + KE = 0 + mgh = mgh.

Question 4.
An elevator of total mass 1800kg is moving up with a constant speed of 2m/s. A frictional force of 400N acts on this motion.

1. The direction of frictional force is______
   • Opposite to direction of motion
   • In the direction of motion
2. What is the work done by gravitational force.
3. What is the total work done by the elevator?

Answer:
1. Opposite to direction of motion

2. w = F × V
w = mg × 2
= 1800 × 10 × 2
w = 36000J

3. P = P_total × V
= (mg + F_fricti0n) × V
= (1800 × 10 + 4000) × 2
= (18000 + 4000) × 2
P = 44000w.

Question 5.
A stone of mass ‘m’ is to be thrown to a height h

1. What is the acceleration of the stone?
2. With what minimum velocity should it be thrown.
3. At what height does the KE and PE become equal?
4. Find the velocity at that height

Answer:
1. ^–g or ^–9.8m/s.

2. v = 0, a = -g, S = h
Substitute this values in
V² = u² + 2as we get
0 = u² – 2gh
u = \(\sqrt{2 g h}\)

3. at \(\frac{h}{2}\)., KE and PE are equal.

4. V² = U² + 2aS
= U² – 2g \(\frac{h}{2}\) (u² = 2gh) = U² – \(\frac{U^{2}}{2}\),
V² = \(\frac{U^{2}}{2}\),
V = \(\frac{U}{\sqrt{2}}\).

Question 6.
A toy gun, with a spring compresser 3cm is used to project a stone of mass 50gm to a height of 10m.

1. What is the potential energy of spring.
2. How much it should be compressed to throw the stone to a height 5m.
3. Find out the physical constant associated with the spring.

Answer:
1. PE of the spring = PE of the mass at the height h
= mgh = 0.050 × 9.8 × 10
= 5 × 9.8 = 4.9J

2. \(\frac{1}{2}\)kx₁² = mgh₁
\(\frac{1}{2}\)kx₂² = mgh₂

3. physical constant associated with the spring constant
\(\frac{1}{2}\)kx² = mgh

Question 7.
Find the odd one out and find the relation connecting the remaining quantities. Joule, Calorie, Kilowatt, electron volt.
Answer:
kilowatt, unit of power
1 calorie = 4.2 joule
1 electron volt = 1.6 × 10^-19J.

Question 8.
Atoy gun, with aspring compresser3cm is used to project a stone of mass 50gm to a height of 10m.

1. What is the potential energy of spring.
2. How much it should be compressed to throw the stone to a height 5m.
3. Find out the physical constant associated with the spring.

Answer:
1. PE of the spring = PE of the mass at the height h
= mgh = 0.050 × 9.8 × 10
= 5 × 9.8 = 4.9J

2. \(\frac{1}{2}\)kx₁² = mgh₁
\(\frac{1}{2}\)kx₂² = mgh₂

3. physical constant associated with the spring constant
\(\frac{1}{2}\)kx² = mgh

Question 9.

A graph paper is fitted on a board as shown in figure. Near to the graph paper a spring is placed. A pencil is attached to the end of the spring as shown in figure. The pencil is free to move on the graph paper. A stone of mass 50 gm is placed 1m above the spring. [Spring constant k = 98N/m]

1. The energy possessed by the stone due to its height is called_______
2. If this stone falls on the spring, find the length of mark that produced on the graph paper due to pencil [The change in P E of stone due to compression may be negleted]
3. What will happen to the length of mark, if spring having smaller spring constant is used? Justify.

Answer:
1. Potential energy.

2. \(\frac{1}{2}\)kx² = mgh
\(\frac{1}{2}\) × 98 × x² = 50 × 10^-3 × 9.8 × 1
x² = 100 × 10⁴
x = 10cm.

3. The length of mark will be decreased. Compression of spring depends on spring constant.

Plus One Physics Work, Energy and Power NCERT Questions and Answers

Question 1.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

1. work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
2. work done by gravitational force in the above case.
3. work done by friction on a body sliding down an inclined plane.
4. work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
5. work done by the resistive force of air on vibrating pendulum in bringing it to rest.

Answer:

1. +ve
2. -ve
3. -ve
4. +ve
5. -ve

Question 2.
The potential energy function fora particle executing linear simple harmonic motion is given by
V(x) = \(\frac{k x^{2}}{2}\), where k is the force constant of the oscillator, For k = 0.5N nm^-1, the graph of V(x) versus x is shown. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2m.
Answer:
We know that maximum potential energy = total energy
∴ (\(\frac{1}{2}\)kx²) max = 1 joule or \(\frac{1}{2}\) × 0.5 × (x²)_max = 1
or (x²)_max = 4 or (x)_max = ± 2m.

Question 3.
Choose the correct alternative:

1. When a conservative force does positive work on a body, the potential energy of the body increases/ decreases/remains unaltered.
2. Work done by a body against friction always results in a loss of its kinetic/potential energy.
3. The rate of change of total momentum of a many particle system is proportional to the external force/sum of the internal forces on the system.
4. In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/ total energy of the system of two bodies.

Answer:

1. decreases
2. kinetic energy
3. external force
4. total linear momentum and also total energy (if the system of two bodies is isolated).

Question 4.
State if each of the following statements is true or false.

1. In an elastic collision of two bodies, the momentum and energy of each body is conserved.
2. Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
3. Work done in the motion of a body over a closed loop is zero for every force in nature.
4. In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Answer:

1. False
2. False
3. False
4. False (true usually but not always).

Question 5.
A rain drop of radius 2mm falls from a height of 500m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) untill at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10ms^-1?
Answer:
r = 2 × 10^-3m,
volume = \(\frac{4}{3} \times \frac{22}{7}\) (2 × 10^-3)³ m³
p = 1000kgm^-3,
h = 250m
W= \(\frac{4}{3} \times \frac{22}{7}\) × 8 × 10^-9 × 1000 × 9.8 × 250J = 0.082J
Data reamains unchanged in the next half.

Question 6.
A bullet of mass 0.012kg and horizontal speed 70ms-1 strikes a block of wood of mass 0.4kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Answer:
If V be the velocity of the block after collision, the using law of conservation of momentum, we get
0.012 × 70 + 0 = (0.012 + 0.4)V
or V = \(\frac{0.012 \times 70}{0.412}\) ms^-1 = 2.04ms^-1
If h be the height through which block rises, then
(M + m) gh = \(\frac{1}{2}\) (M + m)V²
or h = \(\frac{v^{2}}{2 g}\) or
h = \(\frac{2.04 \times 2.04}{2 \times 9.8}\)m = 0.212m = 21.2 cm
Amount of heat produced in the block = loss of K.E.
= \(\frac{1}{2}\) × 0.012 × 70 × 70 – \(\frac{1}{2}\) × 0.412 × 2.04 × 2.04
= 29.4J – 0.857J = 28.543J.

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Kerala Plus One Computer Application Chapter Wise Questions Chapter 1 Fundamentals of Computer

Plus One Computer Application Fundamentals of Computer 1 Mark Questions and Answers

Plus One Computer Application Textbook Questions And Answers Question 1.
________ is a collection of unorganized fact.
Answer:
Data

Plus One Computer Application Chapter Wise Questions And Answers Question 2.
Data can be organized into useful ______.
Answer:
Information

Plus One Computer Application Textbook Solutions Question 3.
________ is used to help people to make decision.
Answer:
Information

Plus One Computer Application Chapter 1 Questions And Answers Question 4.
Processing is a series of actions or operations that convert inputs into _______.
Answer:
Output

Plus One Computer Application Textbook Answers Question 5.
The act of applying information in a particular con-text or situation is called ________.
Answer:
Knowledge

Plus One Computer Application Chapter Wise Questions And Answers Pdf Question 6.
What do you mean by data processing?
Answer:
Data processing is defined as a series of actions or operations that converts data into useful information.

Plus One Commerce Computer Application Textbook Answers Question 7.
Odd man out and justify your answer.
(a) Adeline
(b) 12
(3) 17
(d) Adeline aged 17 years is in class 12.
Answer:
d) This is information. The others are data.

Computer Application Exam Questions And Answers Question 8.
Raw facts and figures are known as ________.
Answer:
data

Plus One Computer Application Textbook Question 9.
Processed data is known as ______.
Answer:
Information

Plus One Computer Application Chapter Wise Notes Question 10.
Which of the following helps us to take decisions ?
(a) data
(b) information
(c) Knowledge
(d) intelligence
Answer:
(b) information

All In One Computer Application Class 10 Pdf Question 11.
Manipulation of data to get information is known as _________.
Answer:
Data processing

Class 7 Computer Chapter 1 Number System Question 12.
Arrange the following in proper order .
Process, Output, Storage, Distribution, Data Capture, Input.
Answer:
a) Data Capture
b) Input
c) Storage
d) Process
e) Output
f) Distribution

Question 13.
Pick the odd one out and give reason
a) Calculation
b) Storage
c) Comparison
d) Categorization
Answer:
b) Storage
It is one of the data processing stage the others are various operations in the stage Process.

Question 14.
Information may act as data. State true or False.
Answer:
False

Question 15.
Complete the Series.
a) 1012,1112,10012 ______,_______.
b) 10112,11102,100012 , ______,______.
Answer:
a) 1011, 1101
b) 10101,10111

Question 16.
What are the two basic types of data which are stored and processed by computers?
Answer:
Characters and number

Question 17.
The number of numerals or symbols used in a number system is its _____.
Answer:
Base

Question 18.
The base of decimal number system is _____.
Answer:
Base

Question 19.
MSD is _______.
Answer:
Most significant digit

Question 20.
LSD is ________.
Answer:
Least significant digit

Question 21.
Consider the number 627. Its MSD is _____.
Answer:
6

Question 22.
Consider the number 23.87. Its LSD is ______.
Answer:
7

Question 23.
The base of Binary number system is ________.
Answer:
2

Question 24.
What are the symbols used in Binary number system?
Answer:
0 and 1

Question 25.
Complete the following series.
(101)2, (111)2, (1001)2, ……..
Answer:
1011, 1101

Question 26.
State True or False.
In Binary, the unit bit changes either from 0 to 1 or 1 to 0 with each count.
Answer:
True

Question 27.
The base of octal number system is
Answer:
8

Question 28.
Consider the octal number given below and fill in the blanks.
0, 1,2, 3, 4, 5,6, 7, _
Answer:
10

Question 29.
The base of Hexadecimal number system is
Answer:
16

Question 30.
State True or False.
In Positional number system, each position has a weightage.
Answer:
True

Question 31.
In addition to digits what are the letters used in Hexa decimal number system.
Answer:
A(10), B(11), C(12), D(13), E(14), F(15)

Question 32.
Convert (1110.01011)2 to decimal.
Answer:
1110.01011 = 1 x 23 + 1 x 22 + 1 x 21 + 0 x 20 + 0 x 2- 1 + 1 x 2 – 2 + 0 x 2 – 3 + 1 x 2 – 4 + 1 x 2 – 5
= 8 + 4 + 2 + 0 + 0 + 0.25 + 0 + 0.0625 + 0.03125
= (14.34375)₁₀

Question 33.
1 KB is ______ bytes.
(a) 2⁵
b) 2¹⁰
c) 2¹⁵
d) 2²⁰
Answer:
d) 2²⁰

Question 34.
The base of hexadecimal number system is _______ Hexadecimal number system .
Answer:
16

Question 35.
A computer has no _______.
(a) Memory
(b) l/o device
(c) CPU
(d) IQ
Answer:
IQ

Question 36.
Real numbers can be represented in memory by using ______.
Answer:
Exponent and Mantissa

Question 37.
Consider the number 0.53421 x d) 10^-8 Write down the mantissa and exponent.
Answer:
Mantissa : 0.53421
Exponent:- 8

Question 38.
Characters can be represented in memory by using ______.
Answer:
ASCII Code

Question 39.
ASCII Code of ‘A’ is
Answer:
(100 0001)₂= 65 .

Question 40
ASCII Code of ‘a’ is
Answer:
(110 0001)₂= 97

Question 41.
Find MSD in the decimal number 7854.25
Answer:
Because it has the most weight

Question 42.
Which is the MSB of representation of-80 in SMR?
Answer:
It is 1 because In SMR if the number is negative then the MSB is 1.

Question 43.
Write 28.756 in Mantissa exponent form.
Answer:
28756 = .28756 x 100
= .28756 x10₂
= .28756 E + 2

Question 44.
ASCII stands for ______.
Answer:
American Standard Code for Information Interchange

Question 45.
List any two image file formats.
Answer:
BMP, GIF

Question 46.
Name the character representation coding scheme developed in India and approved by the Bureau of Indian Standards (BIS).
Answer:
lSCII(lndian Standard Code for Information Interchange)

Question 47.
Fill the series. Series
(151)₈, (153)₈, (155)₈ _____,_____.
Answer:
(157)₈, (161)₈

Question 48.
Meaningful and processed form of data is known as _______ Process
Answer:
Information

Question 49.
Choose the correct number system from the following to which the number 121 (one hundred and twenty one) belongs.
a) Octal and Decimal
b) Binary only
c) Binary, Octal, Decimal and Hexadecimal
d) Decimal only
Answer:
d) Decimal Only OR a) Octal and Decimal

Question 50.
Which one of the following is considered as brain of the computer?
a) Central Processing Unit
b) Control Unit
c) Arithmetic Logic Unit
d) Monitor
Answer:
Central Processing Unit

Question 51.
Which one of the following CPU resister helds address of next instruction to be executed by the processor?
a) Accumulator
b) Instruction Register (IR)
c) Memory address Register
d) Program Counter (PC)
Answer:
d) Program Counter (PC)

Question 52.
Processed data is known as ______.
a) facts
b) figures
c) information
d) raw material
Answer:
c) information

Plus One Computer Application Fundamentals of Computer 2 Marks Questions and Answers

Question 1.
Why do we store information?
Answer:
Normally large volume of data has to be given to the computer for processing so the data entry may be taken more days, hence we have to store the data. After processing these stored data, we will get Information as a result that must be stored in the computer for future references.

Question 2.
Which is the final stage in data processing?
Answer:
Distribution of information is the final stage in data processing

Question 3.
What is source document.
Answer:
Acquiring the required data from all the sources for the data processing and by using this data design a document, that contains all relevant data in proper order and format. This document is called source document.

Question 4.
Convert (106)₁₀ = ( )₂?
Answer:

Question 5.
Convert (106)₁₀ = ( )₈
Answer:

Question 6.
(106)₁₀ = ( )₁₆?
Answer:

Question 7.
Convert (55.625)₁₀ = ( )₂
Answer:

Question 8.
Convert (55.140625)₁₀ = ( )₈
Answer:

Question 9.
(55.515625)₁₀ = ( )₁₆
Answer:

Question 10.
Convert (101.101)₂ = ( )₁₀?
Answer:

Question 11.
Convert (71.24)₈ = ( )₁₀?
Answer:

Question 12.
Convert (AB.88)₁₆ = ( )₁₀
Answer:

Question 13.
Convert (1011)₂ = ( )₈?
Answer:

Question 14.
Convert (110100)₂ = ( )₁₆?
Answer:

Question 15.
(72)₈ = ( )₂?
Answer:

Question 16.
Convert (AO)₁₆ = ( )₂ ?
Answer:

Question 17.
Convert (67)₈ = ( )₁₆ ?
Answer:

Question 18.
Convert (A1)₁₆ = ( )₈ ?
Answer:

Question 19.
Write short notes about Unicode Unicode
Answer:
it is like ASCII Code. By using ASCII, we can represent limited, number of characters. But using Unicode we can represent all of the characters used in the written languages of the world.
Eg:- Malayalam, Hindi, Sanskrit …….

Question 20.
What is the use of the ASCII Code ?
Answer:
ASCII means American Standard Code for Information Interchange. It is a 7 bit code. Each and every character on the, keyboard is represented in memory by using ASCII Code.
Eg:- A’s ASCII Code is 65 (1000001).
a’s ASCII Code is 97 (1100001)

Question 21.
Define the term’bit’?
Answer:
A bit stands for Binary digit. That means either 0 or 1.

Question 22.
Convert the decimal number 31 to binary
Answer:

Question 23.
Find decimal equivalent of (10001 )₂
Answer:

Question 24.
If (X)₈ =(101011 )₂ then find X.
Answer:
Divide the binary number into groups of 3 bits and write down the corresponding octal equivalent.

Question 25.
Fill the blanks
(____)₂ = (AB)₁₆
Write down the 4 bit equivalent of each digit
Answer:

Question 26.
Represent-60 in 1’s complement form – 60 am 1’s complement form
Answer:
Change all 1 to 0 and all 0 to 1 to get the 1’s complement.
– 60 is in 1’s complement is 11000011

Question 27.
Define Unicode.
Answer:
The limitations to store more characters is solved by the introduction of Unicode.
It uses 16 bits so 2¹⁶ =65536 characters(i.e,world’s all written language characters) can store by using this.

Question 28.
Find the smallest number in the list.
a) (1101)₂
b) (A)₁₆
c) (13)₈
d) (15)₁₀
Answer:
Convert all the numbers into a common base i.e. to decimal

Question 29.
Represent – 83 in 1’s complement form.
Answer:
Divide the number 82 by 2 successively and write down the remainders from bottom to top + 83=01010011
To take 1 ‘s complement of a binary number change all 1 s to 0 and all 0’s to 1.
Hence – 83 is 10101100

Question 30.
a) Write the two’s complement form of the decimal number-119.
b) State the benefit of using two’s complement representation as compared to one’s complement form.
Answer:
Binary equivalent of 119 in 8 bit is (0111 0111)2.
To find the 2’s complement of -119. First find the 1’s complement and Odd 1 to it.

b) If a computer uses 8 bit word length, 1 ’s complement method can represent numbers from – 127 to + 127. That means only 127 + 128 = 255 numbers can represent. But 2’s complement method can represent numbers from – 128 to + 127. That means we can represent a total of 256 numbers. We can represent one number more in 2’s complement representation.

Question 31.
Write a short note on Unicode.
Answer:
It is like ASCII Code. By using ASCII, we- can represent limited number of characters. But using Unicode we can represent all of the characters used in the written languages of the world.
Eg:- Malayalam, Hindi, Sanskrit,….

Question 32.
“Central Processing Unit (CPU) is the brain of the computer”. What is the role of Control Unit (CU) in the CPU?
Answer:
All the activities of a computer is controlled by the control unit. That means the function of key board, mouse, monitor, memory etc. are controlled by the control unit.

Question 33.
Storage of data, capturing of data, processing of data, input of data, and output of data are the different stages in data processing. Write these stages in correct order?
Answer:
1) Capturing of data
2) Input of data
3) Storage of data
4) Processing of data
5) Output of data

Question 34.
There is a memory inside the CPU. What is its name? Write down its purpose In the computer.
Answer:
It stores data, intermediate results, Address, instructions etc for CPU to process temperarily.

Question 35.
Convert the hexadecimal (A2D)₁₆ into its octal equivalent.
Answer:
Step 1: First convert the number into binary for this do the following.

Plus One Computer Application Fundamentals of Computer 3 Marks Questions and Answers

Question 1.
Briefly explain data, information and processing with real life example.
Answer:
Consider the process of making coffee.
Here data is the ingredients – water, sugar,milk and coffee powder.
Information is the final product- i.e, Coffee Processing is the series of steps to convert the in-gradients into final product, Coffee. That is mix the water,sugar and milk and boil it. Finally pour the coffee powder.

Question 2.
Differentiate manual data processing and electronic data processing?
Answer:
In manual data processing human beings are the processors. Our eyes and ears are input devices. We get data either from a printed paper, that can be read using our eyes or heard with ears. Our brain is the processor and it can process the data, and reach in a conclusion known as result. Our mouth and hands are output devices.
In electronic data processing the data is processing with the help of a computer. In a super market, key board and hand held scanners are used to in. put data, the CPU process the data, monitor and printers (Bill) are output devices.

Question 3.
Complete the series.
(a) 324₈,327₈,332₈ ,______,______.
(b) 567₈,576₈,605₈ ,______,_______.
Answer:

Question 4.
Fill up the missing digits
(a) (4___)₈=(___110)₂
(b) ( __7___ )₈ = (100 ___ 110)₂
Consider the following:

Answer:
a) 4 ……..100
and 110 ……… 6
So (46)₈ = (100 110)₂
b) 100 ……… 4
7 ………. 111
110 …….. 6
So (476)₈ = (100 111 110)₂

Question 5.
Fill up the missing numbers.
(a) (A ___)₁₆ = ( ___ 1001)₂
(b) ( __ B ___ )₁₆ = (1000 ___ 1111)₂
consider the following:

Answer:

Question 6.
Complete the Series.
(a) 6ADD , 6ADF , 6AE1 ___,____.
(b) 14A9,14AF , 14B5 , ___,____.
Answer:
a) Consider the Seguence.-
6ADD, 6ADF, 6AE1,
Here the ’numbers’ are
0,1,2, 3, 4, 5,6, 7, 8, 9,A,,B, C, D, E, F, 10, 11, ………
The difference between 6ADD & 6ADF is 2

So Add 2 to 6AE1 we will ge 6AE3 Then add 2 to 6AE3 we will get 6AE5 Therefore the missing terms 6AE3, 6AE5
b) Consider the sequence.

So the missing terms are 14BB and 14C1

Question 7.
Find the octal numbers corresponding to the following numbers using shorthand method.
Short hand method
(a) (ADD)₁₆
(b)(DEAD)₁₆
Answer:
a) Step 1 : Write down the 4 bit binary equivalent of each digit.

Step 2 : Divide this number into groups of 3 bits starting from the right and write down the octal equivalent.

b) Step 1 : Write down the 4 bit binary equivalent of each digit.

Step 2 :Divide this number into groups of 3 bits starting from the right and write down the octal equivalent.

Question 8.
The numbers in column A have an equivalent number in another number system of column B.
Find the exact match.

Answer:

Question 9.
ASCII Is used to represent characters in memory. Is it sufficient to represent all characters used in the written languages of the world ? Propose a solution. Justify.
Answer:
No It is not sufficient to represent all characters used in the written languages of the world because , it is a 7 bit code so it can represent 2⁷ = 128 possible codes. To represent all the characters Unicode is used because it uses 4 bytes, so it can represent 232 possible codes.

Question 10.
If (126)x = (56)y , then find x and y.
Answer:

Question 11.
If (102)x = (42)y then (154)x = (___) y.
Answer:

Question 12.
a) Name various number systems commonly used in computers.
b) Include each of the following numbers into all possible number systems
Answer:
a) The number system are binary, octal, decimal and hexa decimal.
b) 123 Octal, decimal and hexa decimal
569 Decimal, hexa decimal
1101 Binary, Octal, Decimal, Hexa decimal

Question 13.
Fill up the missing digit. (Score 1)
(41)₈ = ( )₁₆
Answer:
Step 1 : Divide the number into one each and write doWn the 3 bits equivalent.
Step 2: Then divide the number into group of 4 bits starting from the right then write its equivalent hexa decimal.’

So the answer is 21.

Question 14.
Fill up the missing digit. (Score 2)
If (220)_a = (90)_b then (451 )_a = ( )₁₀
Answer:
It contains 2 & 9, so a and b 2, b 8. The values of a can be 8 or 19. The values of b can be 10 or 16, L.H.S > R.H.S. a The possible values of a and b are given below

Question 15.
Fill up the missing digit. . (Score.3)
If (121)_a = (441)_b then (121)_b = ()₁₀
Answer:

Question 16.
Fill up the missing digit. (Score 3)
If (128)_a = (450)_b then (16)_a = ()₁₀
Answer:

Question 17.
Fill up the missing digit.
(3A.6D)₁₆ = ( )₈
Answer:
Step I: Write down the 4 bits equivalent of each digit.

Step II : Divide this number into groups of 3 bit starting from the right side of the left side of the decimal point and starting from the left side of the right side of the decimal point.
So 00/111/010.011/011/010
Step III: Write the octal equivalent of each group: So we will get. (72.332)₈
(3A.6D)₁₆ = (72.332)₈

Question 18.
What are the various ways to represent integers in computer?
Answer:
There are three ways to represent integers in computer. They are as follows:
1) Sign Magnitude Representation (SMR)
2) 1’s Complement Representation
3) 2’s Complement Representation .
1) SMR : Normally a number has: two parts sign and magnitude, eg:- Consider a number +5. Here + is the sign and 5 is the magnitude. In SMR the most significant Bit (MSB) is used to represent the sign. If MSB is 0 sign is +ve and MSB is 1 sign is – ve.

Here MSB is used for sign then the remaining 7 bits are used to represent magnitude. So we can represent 27 = 128 numbers. But there are negative and positive numbers. So 128 + 128 = 256 number. The numbers are 0 to + 127 and 0 to – 127. Here zero is repeated. So we can represent 256 – 1 = 255 numbers.
2) 1’s Complement Representation : To get the 1’s complement of a binary number, just replace every 0 with 1 and every 1 with 0. Negative numbers are represented using 1’s complement but + ve number has no 1’s complement,
eg:- To find the 1 ‘s complement of 21 +21 = 00010101
To get the 1 ‘s complement change all 0 to 1 and all 1 to 0.
– 21 = 11101010
1’s complement of 21 is 11101010
3) 2’s Complement Representation : To get the 2’s complement of a binary number, just add 1 to its 1’s complement +ve number has no 2’s complement.
eg:- To find the 2’s complement of 21
+21 = 00010101
First lake the 1’s complement for this change all 1 to 0 and all 0 to 1

2’s complement of 21 is 1110 1011

Question 19.
Match the following.

Answer:
1-b
2-f
3-a
4-c
5-d
6-e

Question 20.
Pick invalid numbers from the following.
i) (10101 )₈
ii) (123)₄
iii) (768)₈
iv) (ABC)₁₆
Answer:
i) (10101 )₈ – Valid
ii) (123)₄ – Valid
iii) (768)₈ – Invalid. Octal number system does not contain the symbol 8
iv) (ABC)₁₆ – Valid

Question 21.
Find the largest number in the list
i) (1001 )₂
ii) (A)₁₆
iii) (10)₈
iv) (11)₁₀
Answer:
Convert all numbers into decimal

Question 22.
If (11011)₂ = (A)₈ = (B)₁₆ = (c)₁₀.
Find the value of A, B and C.
Answer:
i) To find the value of A, First divide the binary number (11011)₂ into groups of 3 bits (starting from the right) Then write down the corresponding octal number of each group for this insert a zero (0) in the left side of the binary number.

ii) To find the value of B, First divide the binary number (11011)₂ into groups of 4 bits (starting from the right). Then write down the corresponding.
Hexa decimal equivalent of each group. For this insert 3 zeroes in the left side of the binary number.

iii) To find the value of C find the decimal equivalent so as to do the following.

Question 23.
(i), Binary representation of +38 is 00100110. Which of the following is the 2’s compliment representation of -38?
a) 11011001
b) 00100111
c) 11011010
d) 11011011
ii) If the Octal representation of decimal number X is (64)8. Find the hexa decimal equivalent of X.
Answer:
i) (c) 11011010
ii) Step 1 : First convert the octal number (64)₈ into binary for this write down the 3 bit binary equivalent

Step II:- The obtained binary number is divided into groups of 4 bits, starting from the right. For this insert 2 zeroes in the left side. After that write down the corresponding Hexa decimal number .

Question 24.
Convert the decimal number 29 into binary. Using sign and magnitude form arid 1 ’s complement form represent +29 and -29 in memory in 8-bit word length?
Answer:

A computer with 8 bit word length, sign and magnitude representation of +29 is (00011101)₂ and -29 is (10011101)₂.
1 ’s complement form of (-29) is (11100010)₂.
Note : To find the 1 ’s complement of (+29), change all ones to zeroes and all zeroes to ones.

Question 25.
Data processing refers to the activities performed on dafa to generate information. List the stages of data processing.
Answer:
Data processing phases (6)
a) Capturing data – In this step acquire or collect data from the user to input into the computer.
b) Input- It is the next step. In this step appropriate data is extracted and feed into the computer.
c) Storage – The data entered into the computer must be stored before starting the processing.
d) Processing /Manipulating data – It is a laborious work. It consists of various steps like computations, classification, comparison, summarization etc. that converts input into output.
e) Output of information – In this stage we will get the results as information after processing the data.
f) Distribution of information – in this phase the information(result) will be given to the concerned persons / computers.

Question 26.
a) Convert (1010.11)₂ to decimal.
b) Find the missing terms in the following series.
Answer:

Plus One Computer Application Fundamentals of Computer 5 Marks Questions and Answers

Question 1.
Explain the components of Data processing
Answer:
Data processing consists of the techniques of sorting, relating, interpreting and computing items of data in order to convert meaningful information. The components of data processing are given below.
a) Capturing data – In this step acquire or collect data from the user to input into the computer.
b) Input – It is the next step. In this step appropriate data is extracted and feed into the computer.
c) Storage – The data entered into the computer must be stored before starting the processing.
d) Processing / Manipulating data – It is a laborious work. It consists of various steps like computations, classification, comparison, summarization, etc. that converts input into output.
e) Output of information – In this stage we will get the results as information after processing the data.
f) Distribution of information – In this phase the information(result) will be given to the concerned persons / computers.

Question 2.
Define computer. What are the characteristics?
Answer:
A computer is an electronic device used to perform operations at very high speed and accuracy. Following are the characteristics of the computer.
1) Speed : It can perform operations at a high speed.
2) Accuracy : It produces result at a high degree . of accuracy.
3) Diligence: Unlike human beings, a computer is
free from monotony, tiredness, lack of concentration etc. We know that it is an electronic machine. Hence it can work four hours without making any errors.
4) Versatility: it is capable of performing many tasks. It is useful in many fields.
5) Power of Remembering: A computer consists of huge amount of memory. So it can store and recall any amount of information. Unlike human beings it can store huge amount of data and can be retrieved when needed.

Disadvantages of computer

(1)No. IQ : It has no intelligent quotient. Hence they are slaves and human beings are the masters. It can’t take its own decisions.
(2) No feelings: Since they are machines they have no feelings and instincts. They can perform tasks based upon the instructions given by the humans (programmers)

Students can Download Chapter 2 Forms of Business Organisation Notes, Plus One Business Studies Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Business Studies Notes Chapter 2 Forms of Business Organisation

Plus One Business Studies Chapter 2 Notes Contets

• Sole Proprietorship – Meaning – Features Advantages & Disadvantages
• Joint Hindu Family Business (H.U.F) – Meaning – Features – Advantages & Disadvantages
• Partnership – Meaning – Features – Advantages & Disadvantages – Types of Partners – Types of Partnership – Partnership Deed – Registration
• Co operative Society – Meaning – Features Advantages & Disadvantages – Types of Co-operative Societies
• Joint Stock Company- Meaning – Features – Advantages & Disadvantages – Types of Companies-Choice of form of Business organisation

Various forms of business organisations are:

(a) Sole proprietorship,
(b) Joint Hindu family business,
(c) Partnership,
(d) Cooperative societies, and
(e) Joint stock company.

Plus One Business Studies Chapter 2 Sole proprietorship:
Sole proprietorship refers to a form of business organization which is owned, managed and controlled by an individual who is the recipient of all profits and bearer of all risks. It is the most common form of business organization.
Features:

1. The sole trader is the single owner and manager of the business.
2. The formation of a sole proprietorship is very easy. There are no legal formalities to form and close a sole proprietorship.
3. The liability of a sole trader is unlimited, i.e. in case of loss, his personal properties can be used to pay the business liabilities.
4. The entire profit of the sole trading business goes to the sole proprietor. If there is any loss it is also to be borne by the sole proprietor alone.
5. The sole trader has full control over the affairs of the business. So he can take quick decisions.
6. A sole trading concern has no legal existence separate from its owner.
7. The death, insolvency etc. of a sole trader causes discontinuity of business.

Merits:
1. Easy formation:
The formation of a sole proprietorship is very easy. There are no legal formalities to form and close a sole proprietorship.

2. Quick Decision:

The sole trader has full control over the affairs of the business. So he can take quick decisions and prompt actions in all business matters.

3. Motivation:
The entire profit of the sole trading business goes to the sole proprietor. It motivates him to work hard.

4. Secrecy:
A sole trader can keep all the information related to business operations and he is not bound to publish firm’s accounts.

5. Close Personal Relation:

The sole proprietor can maintain good personal contact with the customers and employees and thus, business runs smoothly.

Advantage And Disadvantage Of Cooperative Society Limitations

1. Limited capital: A sole trader can start business only on a small scale because of limited capital.
2. Lack of Continuity: Death, insolvency or illness of a proprietor affects the business and can lead to its closure.
3. Limited managerial ability: A sole proprietor may not be an expert in every aspect of management.
4. Unlimited liability: The liability of a sole trader is unlimited, i.e. in case of loss, his personal properties can be used to pay off the business liabilities.
5. Suitability: Sole proprietorship is suitable in the following cases.
   • Where the market is limited, localized and customers demand personalized services. Eg. tailoring, beauty parlour etc.
   • Where goods are unstandardized like artistic jewellery.
   • Where lower capital, limited risk & limited managerial skills are required as in case of retail store.

Joint Hindu Family Business (HUF):
It refers to a form of organisation where in the business is owned and carried on by the members of a joint Hindu family. It is also known as Hindu Undivided Family Business (H.U.F). It is governed by Hindu succession Act, 1956. It is found only in India.

The business is controlled by the head of the family who is the eldest member and is called karta. All members have equal ownership right over the property of an ancestor and they are known as co-parceners.

Features
1. Formation:
For a Joint Hindu family business there should be at least two members in the family and some ancestral property to be inherited by them.

2. Membership:

Membership by virtue of birth in the family.

3. Liability:
The Karta has unlimited liability. Every other coparcener has a limited liability up to his share in the HUF property.

4. Control:
The control of the family business lies with the karta. He takes all the decisions and is authorised to manage the business.

5. Continuity:

The business is not affected by the death of the Karta as in such cases the next senior male member becomes the Karta.

6. Minor Members:
The basis of membership in the business is birth in the family. Hence, minors can also be members of the business.

Merits
1. Effective control:
The karta has absolute decision making power. This avoids conflicts among members

2. Continuity of business:
The death of the karta will not affect the business as the next eldest member will then take up the position

3. Limited liability of members:
The liability of all the co-parceners except the karta is limited to their share in the business.

4. Increased loyalty:
Members are likely to work with dedication, loyalty and care, because the work involves the family name.

Limitation
1. Limited capital:
The capital of HUF is limited since the ancestral property only can be used for the business. This reduces the scope for business growth.

2. Unlimited liability:
The liability of Karta is unlimited. His personal property can be used to repay business debts.

3. Dominance of karta:
There is a possibility of differences of opinion among the members of the Joint Family. It may affect the stability of the business.

4. Limited managerial skills:
The karta may not be an expert in all areas of management. It may affect the profitability of the business.

Advantages And Disadvantages Of Cooperative Society Partnership:
The Indian Partnership Act, 1932 defines partnership as “the relation between persons who have agreed to share the profit of the business carried on by all or any one of them acting for all.”
Features

1. Formation: For the formation of a partnership, agreement between partners is essential.
2. Liability: The partners of a firm have unlimited liability. The partners are jointly and individually liable for payment of debts.
3. Risk bearing: The profit or loss shall be shared among the partners equally or in agreed ratio.
4. Decision making and control: The activities of a partnership firm are managed through the joint efforts of all the partners.
5. Lack of Continuity: The retirement, death, insolvency, insanity etc of any partner brings the firm to an end.
6. Membership: There must be at least two persons to form a partnership. The maximum number of persons is ten in banking business and twenty in non banking business.
7. Mutual agency: In partnership, every partner is both an agent and a principal.

Merits of Partnership:

1. Easy formation and closure:
A partnership firm can be formed and closed easily without any legal formalities.

2. Balanced decision making:
In partnership, decisions are taken by all partners. So they can take better decisions regarding their business.

3. Division of labour:
Division of labour is possible in partnership firm. Duties can be assigned to different partners according to their ability.

4. Large funds:
In a partnership, the capital is contributed by a number of partners. So they can start business on a large scale.

5. Sharing of risk:
The risks involved in running a partnership firm are shared by all the partners. This reduces the anxiety, burden and stress on individual partners..

6. Secrecy:
A partnership firm is not legally required to publish its accounts and submit its reports. Hence it can maintain confidentiality of information relating to its operations.

Limitations of Partnership:
1. Unlimited liability:
The partners of a firm have unlimited liability. The partners are jointly and individually liable for payment of debts.

2. Limited resources:
There is a restriction on the number of partners. Hence capital contributed by them is also limited.

3. Possibility of conflicts:
Lack of mutual understanding and co-operation among partners may affect the smooth working of the partnership business.

4. Lack of continuity:
The retirement, death, insolvency, insanity etc of any partner brings the firm to an end.

5. Lack of public confidence:
A partnership firm is not legally required to publish its financial reports. As a result, the confidence of the public in partnership firms is generally low.

Business Studies Class 12 Notes Chapter 2 Types of Partners:
1. Active partner:
A partner who contribute capital and takes active part in the business is called an active partner.

2. Sleeping (Dormant partner):
Partners who do not take part in the day to day activities of the business are called sleeping partners. He contributes capital, share profits and losses and has unlimited liability.

3. Secret partner:
A secret partner is one whose association with the firm is unknown to the general public. He contributes to the capital, takes part in the management, shares its profits and losses, and has unlimited liability.

4. Nominal partner (Quasi Partner):
A nominal partner neither contributes capital nor takes any active part in the management of the business. He simply lend his name to the firm. But, he is liable to third parties for all the debts of the firm.

5. Partner by estoppel:
If a partner by his talk or action leads others to believe that he is a partner in a firm, then he is known as partner by estoppel. However, he is liable to third parties.

6. Partner by holding out:
if a partner declares that a particular person is a partner of their firm, and such a person does not disclaim it, then he/she is known as ‘Partner by Holding out’. Such partners are not entitled to profits but are liable to third parties.

7. Minor Partner:
A minor can be admitted to the benefits of a partnership firm with the mutual consent of all other partners. In such cases, his liability will be limited to the extent of the capital contributed by him.

He will not be eligible to take an active part in the management of the firm. But, a minor can share only the profits and cannot be asked to bear the losses. However, he can inspect the accounts of the firm.

Class 11 Business Studies Chapter 2 Notes Types of Partnerships:
On the basis of duration, there are two types of partnerships:

1. Partnership at will
2. Particular partnership

1. Partnership at will:
This type of partnership exists at the will of the partners. It can continue as long as the partners want and is terminated when any partner gives a notice of withdrawal from partnership to the firm.

2. Particular partnership:
Partnership formed for the accomplishment of a particular project or for a specified time period is called particular partnership.

On the basis of liability, the two types of partnerships are:

1. General partnership
2. Limited partnership

1. General Partnership:
In general partnership, the liability of partners is joint and unlimited. Registration of the firm is optional. The existence of the firm is affected by the death, lunacy, insolvency or retirement of the partners.

2. Limited Partnership:
In limited partnership, the liability of at least one partner is unlimited whereas the rest may have limited liability. Registration of such partnership is compulsory. Such a partnership does not get terminated with the death, lunacy or insolvency of the limited partners. This form of partnership is permitted in India after the introduction of Small Enterprise Policy in 1991.

Class 11th Business Studies Chapter 2 Notes Partnership Deed:
The written agreement which specifies the terms and conditions that govern the partnership is called the partnership deed;
Contents

1. Name of firm
2. Nature of business and location of business
3. Duration of business
4. Investment made by each partner
5. Profit sharing ratio
6. Rights, duties and powers of the partners
7. Salaries and withdrawals of the partners
8. Terms governing admission, retirement and expulsion of a partner
9. Interest on capital and interest on drawings
10. Procedure for dissolution of the firm
11. Preparation of accounts and their auditing
12. Method of solving disputes

Registration of partnership:
According to Indian Partnership Act 1932, registration of a partnership is not compulsory, it is optional. However, they can register with the Registrar of firms of the state in which the firm is situated.
Procedure for Registration:

1. 1. Submission of application in the prescribed form to the Registrar of firms. The application should contain the following particulars:
   • Name of the firm
   • Location of the firm
   • Names of other places where the firm carries on business
   • The date when each partner joined the firm
   • Names and addresses of the partners
   • Duration of partnership. This application should be signed by all the partners.
2. Deposit of required fees with the Registrar of Firms.
3. The Registrar after approval will make an entry in the register of firms and will subsequently issue a certificate of registration. The consequences of non-registration of a firm are as follows:
   • A partner of an unregistered firm cannot file suit against the firm or other partner.
   • The firm cannot file a suit against third party.
   • The firm cannot file a case against its partner.

Business Studies Class 11 Notes Chapter 2 Co-operative Society:
The cooperative society is a voluntary association of persons, who join together with the motive of welfare of the members. The basis of co-operation is self help through mutual help, the motto is “each for all and all for each”.

The cooperative society is compulsorily required to be registered under the Cooperative Societies Act 1912. At least ten persons are required to form a society. The capital of a society is raised from its members through issue of shares.

Features:
The important features of a co-operative society are:

1. Voluntary membership: The membership of a cooperative society is voluntary. Membership is open to all, irrespective of their religion, caste, and gender.
2. Legal status: Registration of a cooperative society is compulsory.
3. Limited liability: The liability of the members of a cooperative society is limited to the extent of the amount contributed by them as capital.
4. Control: Management and control lies with the managing committee elected by the members.
5. Service motive: ‘Self help through mutual help’ or ‘each for all and’ all for each’ is the foundation of co-operative society.

Merits:

1. Equality in voting status: The principle of ‘one man one vote’governs the cooperative society.
2. Limited liability: The liability of members of a cooperative society is limited to the extent of their capital contribution.
3. Stable existence: Death, insolvency or insanity of the members do not affect continuity of a cooperative society.
4. Economy in operations: Co-operative society aims to eliminate middlemen. This helps in reducing cost.
5. Support from government: A co-operative society gets support from the government in the form of low taxes, subsidies and low interest rates on loans.
6. Easy formation: The cooperative society can be started with a minimum often members. Its registration procedure is simple involving a few legal formalities

Limitations:

1. Limited resources:
Resources of a cooperative society consists of limited capital contributions of the members.

2. Inefficiency in management:
Cooperative societies are unable to attract and employ expert managers because of their inability to pay them high salaries.

3. Lack of secrecy:
As a result of open discussions in the meetings of members it is difficult to maintain secrecy about the operations of a cooperative society.’

4. Government control:
cooperative societies have to comply with several rules and regulations related to auditing of accounts, submission of accounts, etc. It affects its freedom of operations.

5. Differences of opinion:
The different viewpoints of members in a co-operative society may lead to difficulties in decision making.

Class 11 Bst Chapter 2 Notes Types of co-operative society:
1. Consumer’s cooperative societies:
The consumer cooperative societies are formed to protect the interests of consumers. The society aims at eliminating middlemen to achieve economy in operations. It purchases goods in bulk directly from the wholesalers and sells goods to the members at the lowest price.

2. Producer’s cooperative societies:
These societies are set up to protect the interest of small producers. It supplies raw materials, equipment and other inputs to the members and also buys their output for sale.

3. Marketing cooperative societies:
Such societies are established to help small producers in selling their Products. It collects the output of individual members and sell them at the best possible price. Profits are distributed to members.

4. Farmer’s cooperative societies:
These societies . are established to protect the interests of farmers by providing better inputs at a reasonable cost. Such societies provide better quality seeds, fertilizers, machinery and other modern techniques for use in the cultivation of crops.

5. Credit cooperative societies:
Credit cooperative societies are established for providing easy credit on reasonable terms to the members. Such societies provide loans to members at low rates of interest.

Joint Stock company:
A company may be defined as a voluntary association of persons having a separate legal entity, with perpetual succession and a common seal. It is an artificial person created by law. The companies in India are governed by the Indian . Companies Act, 1956.

The capital of the company is divided into smaller parts called ‘shares’ which can be transferred freely, (except in a private company). The shareholders are the owners of the company. The company is managed by Board of Directors, elected by shareholders.
Features:
1. Incorporated association:
A company is an incorporated association, i.e. Registration of a company is compulsory under the Indian Companies Act, 1956.

2. Separate legal entity:
A company is an artificial person created by law. Company has a separate legal entity apart from its members. It can enter into contracts, own property, sue and be sued, borrow and lend money etc.

3. Formation:
The formation of a company is a time consuming, expensive and complicated process.

4. Perpetual succession:
A company has a continuous existence. Its existence not affected by death, insolvency or insanity of shareholders. Members may come and go, but the company continues to exist.

5. Control:
The management and control of the affairs of the company is in the hands of Board of directors who are elected the representatives of the shareholders.

6. Liability:
The liability of the shareholders is limited to the extent of the face value of shares held by them.

7. Common seal:
The Company being an artificial person acts through its Board of Directors. All documents issued by the company must be authenticated by the company seal.

8. Transferability of shares:
Shares of a joint stock company are freely transferable except in case of a private company.

Merits:

1. Limited liability:
The liability of the shareholders is limited to the extent of the face value of shares held by them. This reduces the degree of risk borne by an investor.

2. Transferability of shares:
Shares of a public company are freely transferable . It provides liquidity to the investor.z

3. Perpetual existence:
A company has a continuous existence. Its existence not affected by death, insolvency or insanity of shareholders.

4. Scope for expansion:
A company has large financial resources. So it can start business on a large scale.

5. Professional management:
A company can afford to pay higher salaries to specialists and professionals. This leads to greater efficiency in the company’s operations.

6. Public confidence:
A company must publish its audited annual accounts. So it enjoys public confidence.

Limitations:

1. Difficulty in formation:
The formation of a company is very difficult. It requires greater time, effort and extensive knowledge of legal requirements.

2. Lack of secrecy:
It is very difficult to maintain secrecy in case of public company, as company is required to publish its annual accounts and reports.

3. Impersonal work:
It is difficult for the owners and top management to maintain personal contact with the employees, customers and creditors.

4. Numerous regulations:
The functioning of a company is subject to many legal provisions and compulsions. This reduces the freedom of operations of a company.

5 Delay in decision making:
A company takes important decisions by holding company meetings. It requires a lot of time.

6. Oligarchic management:
Theoretically, a company is democratically managed but actually it is managed by few people, i.e board of directors. The Board of Directors enjoy considerable freedom in exercising their power which they sometimes ignore the interest of the shareholders.

7. Conflict in interests:
There may be conflict of interest amongst various stakeholders of a company. It affects the smooth functioning of the company.

8. Lack of motivation:
The company is managed by board of directors. They have little interest to protect the interest of the company.

Types of Companies:
A company can be either a private or a public company.
Private Company:
A private company means a company which:

1. restricts the right of members to transfer its shares
2. has a minimum of 2 and a maximum of 50 members
3. does not invite public to subscribe to its share capital
4. must have a minimum paid up capital of Rs.1 lakh

It is necessary for a private company to use the word private limited after its name.

Privileges of a private company:

1. A private company can be formed by only two members.
2. There is no need to issue a prospectus
3. Allotment of shares can be done without receiving the minimum subscription.
4. A private company can start business as soon . as it receives the certificate of incorporation.
5. A private company needs to have only two directors.
6. A private company is not required to keep an index of members.
7. There is no restriction on the amount of loans to directors in a private company.

Public Company:
A public company means a company which is not a private company. As perthe Indian Companies Act, a public company is one which:

1. has a minimum paid-up capital of Rs. 5 lakhs
2. has a minimum of 7 members and no limit on maximum members
3. can transfer its shares
4. can invite the public to subscribe to its shares.

A private company which is a subsidiary of a public company is also treated as a public company. A public company’must use the word limited after its name

Difference between a public company and private company:

A Comparative assessment of different forms of business organisation:

Choice of business organisation:
The important factors determining the choice of organization are:
1. Cost and Ease of formation:
From the point of view of cost, sole proprietorship is the preferred form as it involves least expenditure and the legal requirements are minimum. Company form of organisation, is more complex and involves greater costs.

2. Liability:
In case of sole proprietorship and partnership firms, the liability of the owners/ partners is unlimited. In cooperative societies and companies, the liability is limited. Hence, from the point of view of investors, the company form of organisation is more suitable as the risk involved is limited.

3. Continuity:
The continuity of sole proprietorship and partnership firms is affected by death, insolvency or insanity of the owners. However, such factors do not affect the continuity of cooperative societies and companies. In case the business needs a permanent structure, company form is more suitable.

4. Management ability:
If the organisation’s operations are complex in nature and require professionalized management, company form of organisation is a better alternative.

5. Capital:
If the scale of operations is large, company form may be suitable whereas for medium and small sized business one can opt for partnership or sole proprietorship.

6. Degree of control:
If direct control over business and decision making power is required, proprietorship may be preferred. But if the owners do not mind sharing control and decision making, partnership or company form of organisation can be adopted.

7. Nature of business:
If direct personal contact is needed with the customers, Sole proprietorship may be more suitable. Otherwise, the company form of organisation may be adopted.

Students can Download Chapter 4 Motion in a Plane 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 4 Motion in a Plane

Plus One Physics Motion in a Plane One Mark Questions and Answers

Motion In A Plane Previous Questions And Answers Chapter 4 Question 1.
A cricketer can throw a ball to a maximum horizontal distance of 100m. With the same speed how much high above the ground can the cricketer throw the same ball?
(a) 200m
(b) 150m
(c) 100m
(d) 50m
Answer:
(d) 50m
R_max = \(\frac{u^{2}}{g}\)
100 = \(\frac{u^{2}}{g}\) or u² = 100g
Using, v² = u² + 2as
0 = (100g) + 2(-g)h or h = 50m.

Motion In A Plane Previous Year Questions Chapter 4 Question 2.
If vectors \(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\) and \(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}-\mathrm{ak}\) are equal vectors, then the value of a is
(a) 5
(b) 2
(c) -3
(d) -5
Answer:
(d) Comparing vector, we get = \(+5 \hat{k}=-a \hat{k}\)
a = -5.

Motion In A Plane Questions And Answers Chapter 4 Question 3.
State which of the following algebraic operations are not meaningful
(a) Addition of a scalar to a vector
(b) Multiplication of any two scalars.
(c) Multiplication of vector by scalar
(d) Division of a vector by scalar Addition of a scalar to a vector
Answer:
(a) Addition of a scalar to a vector

Plus One Physics Chapter Wise Questions And Answers Chapter 4 Question 4.
What is the acceleration of train travelling at 40ms^-1 as it goes round a curve of 160m radius?
Answer:
a = \(\frac{v^{2}}{r}=\frac{40 \times 40}{160}\) = ms^-2.

Plus One Physics Important Questions And Answers Pdf Chapter 4 Question 5.
What provides centripetal force in the following cases.

1. Electron revolving around nucleus.
2. Earth revolving around sun

Answer:

1. Electrostatic force
2. Gravitational force

Hsslive Plus One Physics Chapter Wise Questions And Answers Chapter 4 Question 6.
Why a cyclist has to bend inwards while going on a circular track?
Answer:
The cyclist bends inwards to provide required centripetal force.

Plus One Physics Chapter Wise Questions And Answers Pdf Hsslive Chapter 4 Question 7.
A body executing uniform circular motion has constant
(i) velocity
(ii) acceleration
(iii) speed
(iv) angular velocity
Answer:
(iii) speed

Motion In A Plane Class 11 Question Bank Pdf Chapter 4 Question 8.
Name a quantity which remains unchanged during projectile motion.
Answer:
Horizontal component.

Motion In A Plane Questions And Answers Pdf Chapter 4 Question 9.
What is the effect of air resistance in time of flight and horizontal range?
Answer:
The effect of air resistance is to increase time of flight and decrease horizontal range.

Motion In A Plane Class 11 Important Questions Chapter 4 Question 10.
What is the angle between directions of velocity and acceleration at the highest point of trajectory of projectile?
Answer:
At the highest point velocity is horizontal and acceleration is vertical. So angle is 90°.

Motion In A Plane Class 11 Questions And Answers Pdf Chapter 4 Question 11.
Can a body have constant velocity and still have a varying speed?
Answer:
No.
If velocity is constant, speed also will be constant.

Important Questions Of Motion In A Plane Class 11 Chapter 4 Question 12.
Can a body have zero velocity, still accelerating?
Answer:
Yes.
When a body is at highest point of motion, its velocity is zero but acceleration is equal to acceleration due to gravity.

Motion In A Plane Class 11 Solved Problems Chapter 4 Question 13.
A quantity has both magnitude and direction. Is it necessarily a vector? Give an example.
Answer:
No. The given quantity will be a vector only if it obeys laws of vector addition.
Example: Current.

Motion In A Plane Class 11 Solved Problems Pdf Chapter 4 Question 14.
What is the angle between \(\vec{A} \times \vec{B}\) and \(\vec{B} \times \vec{A}\) ?
Answer:
These two vectors will be antiparallel. Hence θ = 180°.

Plus One Physics Motion in a Plane Two Mark Questions and Answers

Question 1.
A particle is projected with a velocity u so that its horizontal range is twice the greatest hieght attained. The horizontal range is
(a) \(\frac{u^{2}}{g}\)
(b) \(\frac{2 u^{2}}{g}\)
(c) \(\frac{4 u^{2}}{5g}\)
(d) None of these
Answer:
(c) Horizontal range, R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\);
Maximum height, H = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
As per question, R = 2H

or sin2θ = sin²θ
or 2sinθcosθ = sin²θ or tanθ = 2
Hence, sinθ = \(\left(\frac{2}{\sqrt{5}}\right)\) and cosθ = \(\left(\frac{1}{\sqrt{5}}\right)\)
Horizontal range, R = \(\frac{2 u^{2} \sin \theta \cos \theta}{g}\)

Question 2.

Answer:

Question 3.
Choose the correct alternative given below. A Particle executing uniform circular motion. Then its
(a) Velocity and acceleration are radial.
(b) Velocity and acceleration are tangential.
(c) Velocity is tangential, acceleration is radial.
(d) Velocity is radial, acceleration is tangential.
(e) In a circus, a rider rides in a circular track of radius Y in a vertical plane. The minimum velocity at the highest point of the track will be

• \(\sqrt{2 g r}\)
• \(\sqrt{ g r}\)
• \(\sqrt{3 g r}\)
• 0

Answer:
(c) Velocity is tangential, acceleration is radial.
(e) \(\sqrt{ g r}\)

Question 4.
Two non-zero vectors \(\bar{A}\) and \(\bar{B}\) are such that \(|\bar{A}+\bar{B}|=|\bar{A}-\bar{B}|\). Find the angle between them.
Answer:
\(|\bar{A}+\bar{B}|=|\bar{A}-\bar{B}|\)
|A² + B² + 2ABcosθ| = |A² + B² – 2ABcosθ|
|4ABcosθ| = 0
Since A and B are non-zero we get, cos θ = 0 or θ = 90°.

Question 5.
Consider a particle moving along the circumference of a circle of radius R with constant speed with a time period T.

1. During T, what is the distance coverd and displacement?
2. What is the direction of the velocity at each point?

Answer:

1. Distance = 2πR. Displacement = 0
2. Tangent to the circle at every point.

Question 6.
Classify into scalars and vectors. Frequency, velocity gradient, instantaneous velocity, Area.
Answer:

Question 7.
A body is projected so that it has maximum range R. What is the maximum height reached during the fight?
Answer:
At maximum range, θ = 45°

Plus One Physics Motion in a Plane Three Mark Questions and Answers

Question 1.
An electron of mass ‘m’ moves with a uniform speed v around the nucleus along a circular radius Y.

1. Derive an expression for the acceleration of the electron.
2. Explain why the speed of electron does not increase even though it is accelerated by the above acceleration.

Answer:
1.

The acceleration is directed towards the centre of the circle and is called centripetal acceleration.
a₀ = \(\frac{v^{2}}{R}\)
But V = Rω
Substituting we get
a_c = Rω²

2. The direction of centripetal force is towards the centre. The angle between force and displacement is 90°. Hence the work done by the centripetal force is zero. So speed does not increase.

Question 2.
A boy pulls his friend in a home made trolley by means of a rope inclined at 30° to the horizontal. If the tension in the rope is 400N.

1. Draw the vertical and horizontal components of tension in the rope.
2. Find the effective force pulling the trolley along the ground.
3. Find the force tending to lift the trolley off the ground.

Answer:
1.

2. Effective horizontal force = T Cos30°
= 400 × Cos 30°

3. Vertical force = T Sin 30°
= 400 × Sin 30°

Question 3.
A stone tied to the end of a string is whirled in a horizontal circle with constant speed.

1. Name the acceleration experienced by the stone.
2. Arrive at an equation for magnitude of acceleration experienced by the stone.

Answer:
1. Centripetal acceleration

2.

The acceleration is directed towards the centre of the circle and is called centripetal acceleration.
a₀ = \(\frac{v^{2}}{R}\)
But V = Rω
Substituting we get
a_c = Rω²

Plus One Physics Motion in a Plane Four Mark Questions and Answers

Question 1.
Two balls are released simultaneously from a certain height, one is allowed to fall freely and other thrown with some horizontal velocity.

1. Will they hit the ground together?
2. At any time during the fall will the velocities of the balls are same?
3. How does the path of the balls appear to a person standing on the ground?

Answer:

1. Both balls will reach at same time.
2. Total velocity of first body and second body is different. First ball has only downward velocity but second ball has both downward and horizontal velocity.
3. The path of first ball appears to be straight line and that of second ball appears to be parabola.

Question 2.
A ball is thrown straight up.

1. Obtain a mathematical expression for the height to which it travels.
2. What js its velocity and acceleration at the top?
3. Draw the velocity-time graph for the ball showing its motion up and down.

Answer:
1. u = u, v = o, a = -g, h = ?
We can find maximum height using the equation
v² = u² + 2as
0 = u² + 2 × – g × H
2gh = u²
H = \(\frac{u^{2}}{2 g}\)

2. Velocity is zero, but it has a acceleration and its value g = 9.8 m/s².

3.

Question 3.

1. Parallelogram law helps to find the magnitude and direction of the resultant of two forces. State the law.
2. For two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are acting at a point with an angle a between them, find the magnitude and direction of the resultant vector.
3. What will be the angle between two vectors of equal magnitude for their resultant to have the same magnitude as one of the vectors?

Answer:
1. Law of parallelogram of vectors:
If two vectors acting simultaneously at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram then the diagonal of the parallelogram passing through that point represents the resultant in magnitude and direction.

2.

Consider two vectors \(\vec{A}(=\overrightarrow{O P}) \text { and } \vec{B}(=\overrightarrow{O Q})\) making an angle θ. Using the parallelogram method of vectors, the resultant vector R can be written as,

SN is normal to OP and PM is normal to OS. From the geometry of the figure
OS² = ON² + SN²
but ON = OP + PN
ie. OS² = (OP+PN)² + SN² ______(1)
From the triangle SPN, we get
PN = Bcosθ and SN = Bsinθ
Substituting these values in eq.(1), we get
OS² = (OP + Bcosθ)² + (Bsinθ)²
But OS = R and OP = A
= A² + 2ABcosθ + B²cos²θ + B²sin²θ
² = A² +2 ABcosθ + B²
R = \(\sqrt{A^{2}+2 A B \cos \theta+B^{2}}\)
The resultant vector \(\overrightarrow{\mathrm{R}}\) make an angle a with \(\overrightarrow{\mathrm{R}}\). From the right angled triangle OSN,

But SN = Bsinθ and PN = Bcosθ

3. 120°

Question 4.
A ball of mass m is projected at an angle with the ground and it is found that its kinetic energy at the highest point is 75% of that at the point of projection.

1. Is it a one dimensional or a two-dimensional motion? Why?
2. Find the angle of projection
3. Determine another angle of projection which produces the same range.

Answer:
1. Two-dimensional motion. Projectile has two dimensions.

2. The K.E at highest point
E_k = E cos²θ,
where E = initial K.E
0.75 E = E cos²θ
cos²θ = 0.75 = 3/4
cosθ = \(\sqrt{3} / 2\)
θ = 60°

3. When an object is projected with velocity ‘u’ making an angle θ to with the horizontal direction, the horizontal range will be
R₁ = \(\frac{u^{2} \sin 2 \theta}{g}\) ____(1)
when an object is projected with velocity u making an angle (90° – θ) with the horizontal direction, then horizontal range will be

From eq (1) eq (2), we get R₁ = R₂, which means that we get same horizontal range for two angles θ and (90 – θ).

Question 5.
“The graphical method of adding vectors helps us in visualizing the vectors and the resultant vector. But, sometimes, it is tedious and has limited accuracy”.

1. Name the alternative method of vector addition.
2. Write a mathematical expression to find resultant of two vectors.
3. A particle is moving eastward with a velocity of 5m/s. |f in 10s, the velocity changes by 5 m/s northwards, what is the average acceleration in this time.

Answer:
1. Analytical method of vector addition.

2. R = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

3. Change in velocity

Average acceleration

Question 6.

Answer:

Question 7.
A projectile is an object projected into air with a velocity V so that it is moving under the influence of gravity.

1. What is the shape of the path of projectile?
2. As a projectile moves in its path, is there any point along the path where the velocity and acceleration vectors are perpendicular to each other
3. If E is energy with a projectile is projected.

(i) What is the Kinetic energy at the highest point.
(ii) What is P.E at highest point?

Answer:
1. Parabola

2. Yes, highest point

3. Energy with a projectile is projected:
(i) Kinetic energy E_k = 1/2mv²
Velocity at the highest point = Vcosθ
∴ K.E. at highest point = 1/2m(Vcosθ)²
= 1/2 mv²cos²θ
K.E. = E_k cos²θ

(ii) Potential energy at highest point,
P.E. = mgh
= mg\(\frac{v^{2} \sin ^{2} \theta}{2 g}\) = 1/2mv²sin²θ
P.E. = E_k sin²θ.

Plus One Physics Motion in a Plane Five Mark Questions and Answers

Question 1.
An object is projected with velocity U at an angle θ to the horizontal.

1. Obtain a mathematical expression for the range in the horizontal plane.
2. What are the conditions to obtain maximum horizontal range?
3. Find the maximum height of the object when its path makes an angle of 30° with the horizontal (velocity of projection = 8 ms^-1)

Answer:
1. If we neglect the air resistance, the horizontal velocity (ucosθ) of projectile will be a constant. Hence the horizontal distance (R) can be found as
R = horizontal velocity × time of flight

2. when θ = 45° we get maximum horizontal range

3. Height H

Question 2.
The path of projectile from A is shown in the figure. M is the mass of the particle.

When the particle moves from A to M.

1. a) What is the change in vertical velocity?
2. b) What is the change in speed?
3. What is the change in linear momentum?
4. The ceiling of a long hall is 25m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms^-1 can go without hitting the ceiling of the hall?

Answer:
1.

When the particle reaches at M, the vertical component of velocity becomes zero.
change in vertical velocity = u sinθ – 0
= u sinθ

2. change in speed = u cosθ – u

3. change in momentum = mu sinθ – 0
= mu sinθ

4.

θ = 33°48¹
The horizontal range of the ball is

Question 3.
A particle moving uniformly along a circle, experiences a force directed towards the centre and an equal and opposite force directed away from the centre.

1. Name the two forces directed towards and away from the centre.
2. Obtain an expression for the force directed towards the centre.
3. An aircraft executes a horizontal loop at a speed of 720 km hr^-1 with its wings banked at 15°. What is the radius of the loop?

Answer:
1. centripetal force and centrifugal force.

2.


force which produces this centripetal acceleration is called centripetal force.
Centripetal force can be written as F = ma.
F = m\(\frac{V^{2}}{R}\)

3. Speed of aircraft, 720 × \(\frac{5}{18}\) = 200 m/s
The velocity of aircraft, ν = \(\sqrt{r g \tan \theta}\)

Question 4.
A body is projected with a velocity ‘u’ in a direction making an angle θ with the horizontal.

1. Derive the mathematical equation of the path followed.
2. Draw the velocity-time graphs for the horizontal and vertical components of velocity of the projectile.
3. Obtain an expression for the time of flight of the projectile.

Answer:
1. The vertical displacement of projectile at any time t, can be found using the formula.
S = ut + 1/2at²
y = usinθt – 1/2gt²
But we know horizontal displacement,
x = ucosθxt

In this equation g, θ and u are constants. Hence eq.(4) can be written in the form
y = ax + bx²
where a and b are constants. This is the equation of parabola, ie. the path of the projectile is a parabola.

2.

3. The time taken by the projectile to cover the horizontal range is called the time of flight.
Time of flight of projectile is decided by usinθ. The time of flight can be found using the formula s = ut + 1/2 at²
Taking vertical displacement s = 0, a = -g and initial vertical velocity = usinθ, we get
0 = usinθt – 1/2gt²
1/2 gt² = usinθt
t = \(\frac{2 u \sin \theta}{g}\).

Question 5.
An object projected into air with a velocity is called a projectile.

1. What will be the range when the angle of projections are zero degrees and ninety degrees?
2. Show that fora projectile, the upward time of flight is equal to the downward time of flight.
3. At what angles will a projectile have the same range fora velocity?

Answer:
1. When θ = 0°
R = 0
When θ = 90°

2. Upward motion
u = u sinθ, a = -g, v = 0
V = u + at
0 = u sinθ + ^–gt
t_a = \(\frac{u \sin \theta}{g}\) ____(1)
Downward motion
u = 0, V= u sinθ, a = +g
V = u + at
u sinθ = 0 + gt
t_a = \(\frac{u \sin \theta}{g}\) ____(2)
eq (1) and (2), shows up ward time of flight is equal to downward time of flight.

3. θ, 90 – θ.

Question 6.
“An object that is in flight after being thrown or projected is called a projectile”.
1. Which of the following remains constant throughout the motion of the projectile?
(i) Vertical component of velocity
(ii) Horizontal component of velocity
2. Derive an expression for maximum range of a projectile.
3. Show that range of projection of a projectile for two angles of a projection a and (3 is same where α + β = 90°.

Answer:
1. (ii)

2. If we neglect the air resistance, the horizontal velocity (ucosθ) of projectile will be a constant. Hence the horizontal distance (R) can be found as
R = horizontal velocity × time of flight

3. Range of projectile R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\)
Case – 1
at angle α,
R_α = \(\frac{\mathrm{u}^{2} \sin 2 \alpha}{\mathrm{g}}\) ____(1)
Case – 2
at an angle β,

From (1) and (2), we get
R_α = R_β

Question 7.
A bullet is fired with a velocity ‘u’ at an angle ‘θ’ with the horizontal such that it moves under the effect of gravity.

1. What is the nature of its trajectory.
2. Arrive at an expression for time of flight of the bullet.
3. What is the relation between time of ascent and time of decent, when air resistance is neglected.
4. How the relation is affected when air resistance . is considered.

Answer:
1. Parabola

2. The time taken by the projectile to cover the horizontal range is called the time of flight. Time of flight of projectile is decided by usinθ. The time of flight can be found using the formula s = ut + 1/2 at²
Taking vertical displacement s = 0, a = -g and initial vertical velocity = usinθ, we get
0 = usinθt – 1/2gt²
1/2 gt² = usinθt
t = \(\frac{2 u \sin \theta}{g}\).

3. Time of ascent = time of descent

4. Time of descent > time of ascent

Plus One Physics Motion in a Plane NCERT Questions and Answers

Question 1.
State for each of the following physical quantities, if it is a scalar or a vector:
Volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Answer:

• Scalars: Volume, mass, speed, density, number of moles and angular frequency.
• Vectors: acceleration, velocity, displacement, and angular velocity.

Question 2.
Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Answer:
Work, current.

Question 3.
State, with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:

1. Adding any two scalars.
2. Adding a scalar to a vector of the same dimensions
3. Multiplying any vector by any scalar
4. Multiplying any two scalars
5. Adding any two vectors
6. Adding a component of a vector to the same vector.

Answer:

1. No. Scalars must represent same physical quantity.
2. No. Vector can be added only to another vector.
3. Yes. When a vector is multiplied by any scalar, we get a vector.
4. Yes. In multiplication, scalars may not represent the same physical quantity.
5. No. The two vectors must represent the same physical quantity.
6. Yes. The vector and its component must have the same dimensions.

Question 4.
Pick out the only vector quantity in the following list: temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
Answer:
Impulse.

Question 5.
Read each statement below carefully and state with reasons, if it is true or false:

1. The magnitude of a vector is always a scalar.
2. Each component of a vector is always a scalar.
3. The total path length is always equal to the magnitude of the displacement vector of a particle.
4. The average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time.
5. Three vectors not lying in a plane can never add up to give a null vector.

Answer:

1. True
2. False
3. False
4. True
5. True.

In order to give a null vector, the third vector should have the same magnitude and opposite direction to the resultant of two vectors.

Question 6.
A cyclist starts from the centre O of a circular park, then cycles along the circumference, and returns to the centre along QO as shown. If the round trip takes 10 minute, what is the

1. Net displacement.
2. Average velocity and
3. Average speed of the cyclist?

Answer:
1. Since both the initial and final positions are the same therefore the net displacement is zero.

2. Average velocity is the ratio of net displacement and total time taken. Since the net displacement is zero therefore the average velocity is also zero.

3. Average Speed

Question 7.
A passenger arriving in a new town wishes to go from the station to hotel located 10km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 minute, what is

1. The average speed of the taxi
2. The magnitude of average velocity? Are the two equal?

Answer:
Magnitude of displacement = 10km
Total path length = 23km
Time taken = 28 min

Average Speed = \(\frac{23 \mathrm{km}}{\frac{7}{15} \mathrm{h}}\)
= 49.3 kmh^-1
Magnitude of average velocity

Students can Download Chapter 1 Indian Economy on the Eve of Independence 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 1 Indian Economy on the Eve of Independence

Plus One Economics Indian Economy on the Eve of Independence One Mark Questions and Answers

Plus One Economics Chapter Wise Questions And Answers Chapter 1 Question 1.
Who among the following attempted to estimate India’s national income for the pre-independence period.
(i) Dadabhai Navoroji
(ii) V.K.R.V. Rao
(iii) R.C. Desai
(iv) All the above
Answer:
(iv) All the above

Plus One Economics Chapter Wise Questions And Answers Pdf Chapter 1 Question 2.
Zamindari system was introduced in India in:
(i) 1757
(ii) 1782
(iii) 1793
(iv) 1800
Answer:
(iii) 1793

Plus One Economics Chapter Wise Questions And Answers Pdf Download Chapter 1 Question 3.
The theory of demographic transition has ____ stages.
(i) 2
(ii) 3
(iii) 4
(iv) 5
Answer:
(iii) 4

Plus One Economics Chapter 1 In Malayalam  Question 4.
Choose the correct answer.
The first all India population census was conducted in the year.
(i) 1965
(ii) 1870
(iii) 1872
(iv) 1882
Answer:
(iii) 1872

Plus One Economics Textbook Answers Chapter 1 Question 5.
Occupational structure refers to:
(i) number of workers living in a country
(ii) size of working population in the industrial sector
(iii) distribution of working population among different occupations
(iv) nature of different occupations in the economy.
Answer:
(iii) distribution of working population among different occupations

Plus One Economics Important Questions And Answers Chapter 1 Question 6.
When was India’s first official census operation undertaken?
Answer:
India’s first official census took place in 1881.

Plus One Economics Chapter Wise Questions And Answers Pdf Malayalam Chapter 1 Question 7.
Per capita income means:
(i) National income / population
(ii) Population / national income
(iii) Total capital /population
(iv) None of these
Answer:
(i) National income / population

Plus One Economics Questions And Answers Chapter 1 Question 8.
At the time of independence, _____ % of population in India engaged in agricultural activities.
Answer:
72%

Plus One Economics Chapter Wise Questions And Answers Malayalam Chapter 1 Question 9.
How many stages are there in theory of demographic transition?
Answer:
Four Stages.

Plus One Economics Chapter Wise Question Papers Chapter 1 Question 10.
According to 2011 census, the total population of India is _____ crores?
Answer:
121 crores.

Plus One Economics Question And Answer Chapter 1 Question 11.
Which is regarded as the defining year to mark the demographic transition from its first to the second decision stage?
Answer:
1921 is considered as the dividing year. Before 1921, India was in the first stage of demographic transition. The second stage of transition started after 1921.

Hsslive Economics Plus One Chapter Wise Questions And Answers Chapter 1 Question 12.
Which of the following factors are true regarding the foreign trade of India under colonial period?
(i) Large export surplus
(ii) Import of commodities like wool, sugar, and indigo to India.
(iii) Light machinery produced in the factories of India.
(iv) Indian trade was dominated with China, Ceylon, and Persia.
Answer:
(i) Large export surplus

Plus One Economics Indian Economy on the Eve of Independence Two Mark Questions and Answers

Question 1.
Comment on Indian infant mortality rate and life expectancy.
Answer:
At the time of British rule, Indian infant mortality rate was very high. It was 218 per thousand. Life expectancy was very low. It was just 44 years.

Question 2.
Mention the growing regional variation in occupational structure of India under British rule.
Answer:
70 -75 percent of Indian workforce was engaged in agriculture 10 percent in manufacturing and 15-20 percent in services. But parts of the Madras presidency witnessed a fall in the dependence of the workforce on agricultural sector and an increase in manufacturing and service sector.

Question 3.
State whether true or false.

1. Rao’s estimate of national income during the colonial period is regarded very significant
2. The policy of British government made India an exporter of finished goods and importer of primary products.

Answer:

1. True
2. False. The British policy made India an importer of finished products and exporter of primary products.

Question 4.
Name some notable economists who estimated India’s per capita income during the colonial period?
Answer:
Dadabhai Naoroji, Finlay Shirras, William Digby, VKRV Rao, and RC Desai were the notable economists who estimated India’s per capita income during the colonial period.

Question 5.
Name some modern industries which were in operation in our country at the time of independence.
Answer:
The modern industries which were in operation in our country at the time of independence were:

• Cotton and jute textile mills
• The iron and steeUjompany
• Sugar industry
• Cement industry

Question 6.
What was the twofold motive behind the systematic de-industrialization effected by the British in pre-independence India?
Answer:
The primary motive of the colonial government behind the policy of systematic de-industrialization was two-fold:

1. To turn India into a market of finished products of those industries which were established in Britain.
2. To reduce India to a status of a mere exporter of important raw materials for the industries in Britain.

Question 7.
What do you mean by commercialisation of agriculture?
Answer:
Commercialisation of agriculture means the production of cash crops for the market rather than food crops for self-consumption.

Question 8.
Define economic development.
Answer:
Economic development refers to that process as a result of which, along with increase in real per capita income, there is reduction in inequality, poverty, illiteracy, and disease.

Question 9.
What do you mean by ‘tariff?
Answer:
Tariff is a tax or duty on imports which can be levied either on physical units.

Plus One Economics Indian Economy on the Eve of Independence Three Mark Questions and Answers

Question 1.
“The traditional handicrafts industries were ruined under the British rule”. Do you agree with this view? Give reasons in support of your answer.
Answer:
Yes. India’s traditional handicraft industries were ruined under British rule. There were many reasons behind this. The British adopted the strategy of systematically de-industrializing India. The main raw materials required by the handicraft industries such as cotton and silk were exported to Britain and instead imported to Indian markets the finished products which are low cost.

Thus, due to lack of raw materials, reduced markets and no government policies to protect the handicraft industry resulted in collapse of the Indian handicraft industries.

Question 2.
Highlight the salient features of India’s pre-independence occupational structure?
Answer:
The distribution of working persons across different industries and sectors is called occupational structure. The agricultural sector accounted for the largest share of workforce. About 72 percent people were engaged in agriculture, 10 percent in manufacturing and 18 percent in service sector.

Question 3.
Mention the objectives of British policies of infrastructure development in India?
Answer:
Basic infrastructure like railways, roads, telegraphs, etc. were developed during British period. The purpose behind this was to satisfy various colonial interests.

• Roads were built for the purpose of mobilising the army.
• The introduction of telegraph in India was basically to maintain law and order.
• Railways introduced to mobilise army and raw materials.

Question 4.
Name the states where the following iron and steel industries are located?

1. TISCO
2. IISCO
3. Bhila Steel Plant

Answer:

1. TISCO – (Tata Iron and Steel Co.) – Jharkhand
2. IISCO – (Indian Iron and Steel Co.) – West Bengal
3. Bhilai Steel Plant – Chattisgarh.

Question 5.
What are the major economic challenges of India today?
Answer:

• Low agricultural growth rate
• Rising prices
• Food inflation
• Rapid growth of population
• Poor infrastructure facilities
• Shortage of resources
• Poverty and unemployment

Question 6.
“During British rule land and labour were transformed into commodity”. Justify this statement.
Answer:
Land being the most important income yielding asset in the rural society became a much sought after commodity. The Zamindars, who got land in the British settlement and money lenders who acquires land against their debt, wanted to acquire more and more of it.

All artisans who lost craft and could not be accommodated on land came to town in search of employment. Some of them got employment in factories; some others became coolies and domestic servants. In their case also, labour power became a commodity Thus it can be concluded that during British rule land and labour were transformed into commodity.

Question 7.
What do you understand by the drain of Indian wealth during the colonial rule?
Answer:
The most important feature of India’s foreign trade throughout the colonial period was generation of large export surplus. But this caused scarcity in the domestic market. Further, this export surplus did not result in any flow of gold or silver into India. Rather, this was used to make payments for the expenses incurred by,

• expenses on war
• office setup of British government and
• Import of invisible items.

All these led to drain of Indian wealth during the colonial period.

Plus One Economics Indian Economy on the Eve of Independence Five Mark Questions and Answers

Question 1.
Match the columns

Answer:

Question 2.
The British rule made certain positive contributions. Do you agree? Substantiate your answer.
Answer:
Though the basic intention of the British government was not the development of Indian economy, it left some beneficial effects. The positive contribution of the British rule are noted below

1. Development of means of transport especially railways and communication
2. Political and economic unification of the country
3. Evolution of banking and monetary system
4. Stability, peace, and order
5. Emergence of market economy
6. Introduction of new system of education
7. Establishment of civil laws and courts

Question 3.
“India’s agriculture sector was stagnant during the colonial rule” Do you agree? Substantiate.
Answer:
Yes, India’s agriculture sector was stagnant during the colonial rule. The stagnation in agriculture sector was caused mainly because of the following.

1. The various systems of land settlement in India.
2. The British government did nothing to improve the condition of agriculture
3. There was immense misery and social tension among the cultivators because of the exploitation by the zamindars
4. Low level of technology, lack of irrigational facilities and use of negligible amount so fertilizers were also responsible for low production in agriculture.

Question 4.
“Our agricultural production received a set back due to the partition of the country”. Do you agree? Substantiate.
Answer:
Yes, I agree with the statement that our agricultural production received a set back due to the partition of the country India’s agricultural production received a further set back due to the country’s partition at the time of independence. A sizeable portion of the undivided country’s highly irrigated and fertile land went to Pakistan this had an Adverse impact upon India’s output from the agriculture sector.

Particularly affected was India’s jute industry since almost the whole of the jute producing area became part of East Pakistan (now Bangladesh). India’sjute goods industry (in which the country had enjoyed a world monopoly so far), thus, suffered heavily for lack of raw material.

Question 5.
Given below are certain items of export and import of India during British period. Classify them into a table. (Raw silk, Cotton, Silk, Raw cotton, Machinery, Wool, Chemicals, Edible oils, Medicines, Sugar, Woolen cloth, Indigo, Jute)
Answer:

Question 6.
Indicate the direction of trade at the time of independence.
Answer:
During the British period, the colonial government maintained monopoly over India’s trade. As a result, at the time of independence, more than half of India’s foreign trade was restricted to Britain while the rest was allowed with a few other countries like China, Srilanka, and Iran.

Question 7.
Give an account of the economic policies pursued by the colonial government in India?
Answer:
Since the Britishers had their vested interests in India, that could be seen in the economic policies implimented which meant only to promote their economy at the cost of India’s economy. The impact of the policies resulted into a major change in the structure of the Indian economy:

1. Transformed the country into a net supplier of raw materials and the consumer of the finished industrial products from Britain.

2. The agricultural sector continuously faced stagnation and enormous deterioration. Agricultural productivity become incrementally low.

3. No modern industrial base was allowed to come up. The policies meant to systematically deindustrialise the country.

4. India’s world-famous and centuries old handicraft industry perished due to the policies layed by colonial rule.

5. Due to all the above factors, the level of unemployment increased.

6. India’s foreign trade was mostly restricted to Britain.