ST. XAVIER’S SCHOOLS-CHANDIGARH, PANCHKULA,MOHALI,ZIRAKPUR

CLASS-VIII ENGLISH WORKSHEET 10

TOPIC: LETTER WRITING

Letters are of different types formal, informal, social letters and notes of invitation.

Formal letters are written to officials of firms, Editors of newspapers, Principal of schools or

colleges and people not known to us personally. The language used is courteous and simple.

FORMAT OF FORMAL LETTER

Sender’s address 20, Civil lines,

Chandigarh.

----------------- -----------------(space) Date 15th June,2020. ------------------ ------------------(space) The Editor, Receiver’s The Tribune,

address Chandigarh. -------------------- --------------------(space)

Salutation Sir/Dear Sir, -------------------- --------------------(space) Subject Request for------- --------------------- ---------------------(space)

_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________.

------------------ ------------------(space) Thanking you, ------------------- -------------------(space) Yours faithfully, Abhinav Rastogi. (full name) --------------------------------------------------------------------------------------------------------------------------------------

FORMAL LETTER-Write an application to the Principal of your school requesting him/her to

arrange an educational tour.

Sector-15, Panchkula, (your house address)

Haryana.

15th June,2020.

The Principal,

St. Xavier’s High School, (your school address)

Sector-20, Panchkula,

Haryana.

Dear Sir,

Subject: A request to organise an educational tour

With due respect, I wish to state that the students of class VIII, want to go on an educational

tour to the science city Kapurthala, which could enrich our knowledge for higher studies. It

will be an educational trip and will help us to gain a better understanding of the things we

study theoretically. This educational tour will enhance our knowledge about the scientific

facts and the real world.

I assure you Sir that there will be complete discipline throughout the trip. Hope that you will

give us such an opportunity with your kind consideration.

Thanking you,

Yours faithfully,

Suman Sharma. (your full name)

1. Write an application to the Principal of your school requesting him/her to issue a date

of birth certificate.

2. Write a letter to the Editor of a local newspaper making a case for public participation

in ensuring success of the ‘Clean India Mission’.

Answer key of Worksheet 9

Ex 1- 1. will have inaugurated 2. will have finished 3. will have arranged 4. will have gone 5.

will have prepared

Ex2- 1. will have been performing 2. will have been talking 3. will have been cycling 4. will

have been going 5. will have been sleeping

Ex3- 1. will find 2. will have prepared 3. will have been playing 4. will be announcing 5. will

become

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Letter for practise

CLASS – 8 S.ST WORKSHEET – 10(GEOGRAPHY)

CHAPTER – 4 (URBANIZATION)

URBANIZATION- Urbanization is the process of urban growth that leads to a

greater percentage of the population living in towns and cities. It is the

movement of people from rural areas to urban areas.

Though urbanization has a long history, the modern city worldwide has

developed only over the last 200 years.

THREE HISTORICAL processes have shaped modern cities in decisive ways -

The rise of INDUSTRIAL REVOLUTION

The establishment of COLONIAL RULE over large parts of the world

The development of DEMOCRATIC IDEALS

Industrialization changed the form of urbanization in the modern times.

The early Industrial cities of Britain such as Leeds and Manchester attracted

large number of migrants to textile mills set up in the late 18th century. In 1851,

more than three- quarters of the adults living in Manchester were migrants from

rural areas.

The AGRICULTURAL REVOLUTION, THE ENCLOSURE MOVEMENT

AND THE INDUSTRIAL REVOLUTION were the major factors which lead to

growth of MANCHESTER AND LONDON as the modern cities.

MUMBAI AND KOLKATA are the two oldest and important mega cities of

India. Both these cities developed during the colonial era.

ENCLOSURE MOVEMENT – Over the late eighteenth and early

nineteenth centuries the English countryside changed dramatically.

Before this time in large parts of England, the countryside was open. All

villagers had access to the commons. It was not partitioned into enclosed

lands privately owned by landlords. Enclosure was the legal process in

England during the 18th century of enclosing number of small

landholdings to create one large farm. Once enclosed, use of the land

became restricted to the owner, and it ceased to be common land for

communal use.

CAUSES OF URBANIZATION– People migrate from rural to urban

areas due to PUSH and PULL factors.

PUSH FACTORS – Push factors are those that push people from the

countryside to the major cities. This may include lack of economic

opportunities, division of land, lack of services (educational,

transportation health etc.), low pay and rural poverty.

PULL FACTORS – Pull factors are those that attract people to major

cities. This may include more economic opportunities, better jobs, better

quality of life, better educational, transportation and health services.

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EXERCISE

Answer the following questions –

Q1-What is urbanization?

Q2 –“Three historical processes have shaped modern cities in decisive

ways”. Mention the processes.

Q3 - Name any two colonial cities of India.

Q4 – State any two factors which lead to earlier industrial cities.

Q5 – Explain the push and pull factors responsible for urbanization.

Q6- Name any two earlier industrial cities.

Q7 – Name any two mega cities of India.

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Answer key of S.ST Worksheet – 9(Civics)

EXERCISE

1) Fill in the blanks:

a) The three organs of government are the legislature, the executive and the judiciary..

b) The lower house of India’s parliament is called the lok Sabha.

c) A bicameral state legislature consists of the Vidhan Sabha and the Vidhan Parishad

d) The vice president is the ex officio chairman of the Rajya Sabha.

2) Short Questions:

a) Explain the meanings of the terms ‘bicameral’ and ‘unicameral’.

Ans) A Bicameral Legislature has two houses- an upper house and a lower house. A

Unicameral Legislature has only one house

b) What is the meaning of Universal Adult Franchise?

Ans) All Indian citizens above the age of 18 can exercise franchise (the right to vote)

to elect their representatives. This is called universal adult franchise.

c) What do you understand by secret ballot?

Ans) Voting is done by a secret ballot. In this system, the ballot (vote) cast by each

voter remains secret.

d) What are the functions of the Election Commission?

Ans) Elections to the Indian legislature are conducted by the Election Commission.

The Election Commission fixes the dates of elections and marks out the

constituencies. It sets up polling booths, where people cast their votes. It also provides

ballot papers and electronic voting machine (EVMs), through which voters register

their choice. The commission then counts the votes and declares the results.

e) Who is a speaker? What are his functions?

Ans) The newly elected members elect from among themselves a presiding officer

called the speaker. The speaker conducts the proceedings of the house and maintains

order within the house. At the end of a debate, the speaker may take a vote on issue.

He can exercise his own casting vote (deciding vote) only in case of a tie.

e) The upper house of legislature is a permanent body. Justify this statement.

Ans) The Upper House is never dissolved. In other words it is not officially broken up

after a fixed term to be formed anew. One third of its members retire every two years.

Elections are then held for these vacancies. Each member is elected for a term of six

years.

3) Long Questions:

a) Who can contest the elections to a lower house of legislature?

Ans) To become a member of a lower house of legislature , a person

• Must be a citizen of India;

• Must be at least 25 years old;

• Must not be bankrupt or mentally unsound

• Must not hold any government job.

b) Describe the process of election to a lower house.

Ans) People choose their representatives through elections, Elections to a lower

house of legislature are usually held once every five years. For the purpose of

elections the territory of the country is divided into areas; each called a constituency.

Usually, several candidates contest elections from a constituency. A candidate may

represent a political party or be an independent (nonparty) candidate. The candidate

who gets the highest number of votes becomes the representative of the constituency

in the lower house of legislature.

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SUBJECT : PHYSICS CLASS : 8

TOPIC : PHYSICAL QUANTITIES AND MEASUREMENT

DENSITY: The density of a substance is its mass per unit volume. Density is the

measure of how compact the mass in a substance or object It is denoted by the

symbol ρ.

1. For a particular substance, density is independent of mass and volume. It is

dependent on a mass to volume ratio.

2.The density of a substance is a physical property.So, it does not changes with any

change in its shape or size.

3.Effect of temperature on density: Almost all substances expand on heating and

contract on cooling, but their mass does not change.So, the density of a substance

decreases with the increase in temperature and increases with the decrease in

temperature.

Exception : Water contracts on heating from 0oC – 4oC and expands on heating above

4oC. So, the density of water is maximum at 4oC. Density of water is 1g/cm3 or 1000

kg/m3 at 4oC.

4. For two different substances,

If mass is same, the density of the substance is inversely proportional to its

volume.

ρ α 1/V; If m is same

Equal masses of different substances have different volumes.

EXAMPLE – The volume of cotton is much larger than the volume of an equal mass

of iron. This is because the particles of iron are closely packed while those of cotton

are very loosely packed. In other words. Iron is denser than cotton

If volume is same, the density of the substance is proportional to its mass.

ρ α m; If v – same

Equal volumes of different substances have different masses.

EXAMPLE – The mass of iron is much more than the mass of an equal volume of

wood. This is because the particles of iron are closely packed while those of wood are

loosely packed. In other words. Iron is denser than wood.

5. Unit of density:

Unit of density = Unit of mass / Unit of volume

= Unit of mass / Unit of (length)3

Relationship between S.I. and C.G.S. units :

Vessels for measuring volume:

1. Measuring beaker: A measuring beaker is made up of glass, plastic or metal like

aluminium. It is used to take out a fixed volume of liquids. The capacity of a

measuring beaker is marked on it.

2. Eureka can: A Eureka can is a glass beaker with a side opening near its mouth

which is known as spout. Thus, the beaker can contain a volume of liquid up to the

spout. Any excess of liquid overflows through the spout.

3. Measuring cylinder: It is made up of glass or plastic and is graduated in millilitre

(mL) with its zero mark at the bottom.The graduations then increase upwards.The

capacity of a measuring cylinder is marked on it.

Determination of density of a regular solid:

1. First measure the mass M, of the given regular solid by using a beam balance.

2. Now, to find the volume V of the given regular solid, use the following formula:

3. Knowing the mass M and volume V of the given solid, calculate the density ρ of the

substance by using the formula.

Density of substance(ρ) = mass of the substance(M)/ volume of substance(V).

For example:- If mass of a cube of iron is M = 210 g

One side of cube = 3cm

Therefore, Volume of cube V = (one side)3 = (3)3 = 27 cm3

Density of iron (ρ) = M/V = 210 g /27 cm3= 7.78 g/cm3

Determination of density of an irregular solid:

1. First measure the mass of the given solid by using a beam balance.

2. To find the volume of the given solid, we use the displacement method.

3. Knowing the mass and volume of the given solid, calculate the density of a

substance by using the formula.

Density of substance(ρ) = mass of the substance(M)/ volume of substance(V).

For example:- If mass of solid is M = 78g

Therefore, Volume of solid V = 10 cm3

Density of solid (ρ) = M/V = 78/10 = 7.8 g/cm3

Determination of density of a liquid:

1. Measure the mass of the empty beaker by using a common beam balance.

2. Now take a measuring cylinder and pour liquid into it to a certain level say 60 mL.

3. Transfer the liquid into the empty beaker. Measure its mass again.

4. Find the difference, which gives the mass M of the liquid.

5. Calculate the density of liquid using the following formula:

Density of substance(ρ) = mass of the substance(M)/ volume of substance(V).

For example:- Mass of the liquid = mass ( liquid+ beaker) – mass (beaker) = 10 – 7

= 3g

Volume of the liquid = 60 mL = 60 cm3 (1mL = 1cm3)

Density of the liquid (ρ) = M/V = 3/60 = 0.05 g/cm3

Questions:

1. A piece of zinc of volume 86 cm3 has mass of 438.6 g. Find the density of iron.

2. The dimensions of a block of glass are 25 cm X 15cm X 4 cm. Find its density if its

mass is 4.5 kg.

3. Fill in the blanks:

a) Mass = density X _________.

b) 1 g/cm3 = __________kg/m3

c) 1 kg is the mass of ___________ mL of water at 4oC.

d) The C.G.S unit of density is _______.

4. A block of copper displaces 5 mL of water in a measuring cylinder. If the density of

silver is 8.8 kg/m3, find the mass of block.

5.State true or false:

a) Equal volumes of two different substances have equal masses.

b) The density of a piece of brass will change by changing its size or shape.

6. A given quantity of a liquid is heated. Which of the following quantity will vary

and how ?

a) mass, b) volume or c) density.

7. The density of alcohol is 500 kg/m3. Express it in g/cm3.

Answer Key of Science Worksheet-9 (Biology)

Q1 State whether the following statements are true or false. Page no-8

1) True 2) True 3) False 4) True 5) False 6) True

7) False 8) True 9) False 10 ) False 11 ) True 12) True

Quick check:-4 Page no- 10

A) State whether the following statements are true or false

1) True 2)True 3) False 4 ) False 5) False

B) Fill in the blanks

1) Sieve tubes 2)Palisade 3)Translocation 4 )Sieve tubes and companion cells

5) Macronutrient 6) Guttation

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SUBJECT – MATHS CLASS – VIII WORKSHEET– 10

(OPERATIONS ON SETS)

INTRODUCTION:

A well-defined collection of objects is called a set and it can be represented in three

ways:

(i) Description form

(ii) Roster form (Tabular form)

(iii) Set – Builder form (Rule Method)

CARDINAL NUMBER OF A SET: The number of elements in a set is called its

Cardinal number.

OPERATIONS ON SETS

UNION OF SETS: The union of two sets A and B is the set consisting of all those

elements which belong to either A or B or both and it is written as A U B (read as ‘A

union B’)

For Example: (i) If A = {a, b, c, d, e, f, g, h} and B = {a, e, i, o, u}

A U B = { a, b, c, d, e, f, g, h, i, o, u}

(ii) If A= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {1, 3, 5, 7, 9}

A U B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

INTERSECTION OF SETS: The intersection of two sets A and B is the set

consisting of all those elements which belong to both A and B and it is written as A ∩

B (read as ‘A intersection B’)

For Example: (i) If A = {2, 4, 6, 8, 10} and B = {4, 8, 12, 16, 20}

A ∩ B = { 4, 8}

(ii) If A= {a, b, c, d, e, f, g, h, i, j, k} and B = {a , b, c, d, e}

A ∩ B = { a , b, c, d, e}

DIFFERENCE OF TWO SETS: If A and B are two sets, then A – B is the set

consisting of all those elements which belong to A but do not belong to B

Thus, A – B = { x | x ∈ A and x ∉ B }

Similarly, B – A = { x | x ∈ B and x ∉ A }

**A – B is the set consisting of elements of A only and B – A is the set consisting of

elements of B only

For Example: If A = { a, b, c, d, e, f } and B = { a, e, i, o, u } , then

A – B = { b, c, d, f } and B – A = { i, o, u }

A – B ≠ B – A

COMPLEMENT OF A SET:

If ζ is the universal set and A is any set, then the complement of A is the set

consisting of those elements of ζ which do not belong to A and Complement of A is

denoted by A’ or or Ac

i.e. A’ = {x | x ∈ ζ and x ∉ A }

For Example: If A = {1, 3, 5, 7, 9} and ζ = {1, 2, 3, 4,……., 10}

A’ = {2, 4, 6, 8, 10}

OVERLAPPING (intersecting) SETS:

Two sets A and B are called Overlapping (intersecting or joint) sets if they have

atleast one element in common or we can say that two sets A and B are overlapping

sets if A ∩ B ≠

For Example: The sets A = {2, 3, 4, 5, 6} and B = {0, 3, 6, 9} are overlapping sets

because they have elements 3 and 6 in common. Here, A ∩ B ≠

DISJOINT SETS:

Two sets A and B are called Disjoint (or non-overlapping) sets if they have no

element in common or we can say that two sets A and B are disjoint sets if A ∩ B =

For Example: The sets A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8, 10} are disjoint sets

because they have no element in common. Here, A ∩ B =

Q (1) If ζ = {all digits in our number system} and A = {3, 4, 5, 8}, then find the

complement of A

Solution: Given ζ = {all digits in our number system}

ζ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and A = {3, 4, 5, 8}

A’ = {0, 1, 2, 6, 7, 9}

Q (2) If A = {factors of 24} and B = {factors of 36}, then find

(i) A U B (ii) A ∩ B (iii) A – B (iv) B – A

Also find Cardinal numbers of these sets.

Solution: The given sets in Roster form are:

A = {1, 2, 3, 4, 6, 8, 12, 24} and B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

(i) A U B = {1, 2, 3, 4, 6, 8, 12, 24, 9, 18, 36}

(ii) A ∩ B = {1, 2, 3, 4, 6, 12}

(iii) A – B = {8, 24}

(iv) B – A = {9, 18, 36}

Since A U B has 11 elements. Therefore, n (A U B) = 11

Similarly, n (A ∩ B) = 6 , n (A – B) = 2 and n (B – A) = 3

SOME FORMULAE RELATED TO CARDINAL NUMBER:

1. If A and B are finite sets, then:

(i) n (A U B) = n(A) + n(B) – n(A ∩ B)

(ii) n (A – B) = n(A U B) - n(B) or n(A) - n(A ∩ B)

(iii) n(B – A) = n(A U B) - n(A) or n(B) - n(A ∩ B)

(iv) n(A U B) = n (A – B) + n(B – A) + n(A ∩ B)

2. If ζ(universal set) is finite set and A is any set, then n(A) + n(A’) = n(ζ)

Q (3) If n (A) =16, n (B) = 13 and n ( A U B) = 22, then find n (A ∩ B)

Solution: We Know that:

n (A U B) = n(A) + n(B) – n (A ∩ B)

⇒ 22 = 16 + 13 – n (A ∩ B)

⇒ n (A ∩ B) = 16 + 13 - 22 = 29 – 22

⇒ n (A ∩ B) = 7

Q (4) If n (ζ) = 30, n ( A’) = 14, n (B) = 20 and n (A ∩ B) = 11, then find

(i) n (B’) (ii) n (A U B )

Solution: (i) We Know that: n (B’) = n(ζ) - n(B)

⇒ n(B’) = 30 – 20 = 10

(ii) We Know that: n (A’) = n(ζ) - n(A)

⇒ 14 = 30 – n (A)

⇒ n(A) = 30 – 14 = 16

Also we know that n (A U B) = n(A) + n(B) – n(A ∩ B)

⇒ n (A U B) = 16 + 20 – 11 = 25

Q (5) If n (ζ) = 40, n (A) =25, n (B) = 12 and n ((A U B )’) = 8, find

(i) n (A U B )

(ii) n (A ∩ B)

(iii) n (A – B)

Solution: (i) We Know that: n(A U B) + n((A U B)’) = n(ζ) [ Using n(A) + n(A’) =

n(ζ)]

⇒ n(A U B) + 8 = 40

⇒ n(A U B) = 40 – 8 = 32

(ii)We Know that: n (A U B) = n(A) + n(B) – n (A ∩ B)

⇒ 32 = 25 + 12 - n (A ∩ B)

⇒ n (A ∩ B) = 25 + 12 – 32 = 5

(iii) We Know that: n (A – B) = n(A) - n(A ∩ B)

⇒ n (A – B) = 25 – 5 = 20

Q (6) If n (A – B) = 15 , n (B – A) = 10 and n (A ∩ B) = 5, find

(i) n (A U B) (ii) n(A) (iii) n(B)

Solution: (i) We Know that: n(A U B) = n (A – B) + n(B – A) + n(A ∩ B)

⇒ n(A U B) = 15 + 10 + 5 = 30

(ii) We Know that: n (A – B) = n(A) - n(A ∩ B)

⇒ 15 = n(A) – 5

⇒ n(A) = 15 + 5 = 20

(iii) We Know that: n(B – A) = n(B) - n(A ∩ B)

⇒ 10 = n(B) – 5

⇒ n(B) = 10 + 5 = 15

SOME PRACTICE QUESTIONS

1. Find A’ when

(i) A = { 0, 1, 4, 7} and ζ = {0, 1, 2, 3, …….., 10}

(ii) A = {consonants} and ζ = {alphabets of English}

(iii) A = {odd natural numbers} and ζ = {whole numbers}

2. If A = {4, 5, 6} and B = {0, 1 , 2, 3, 4} then find:

(i) A U B

(ii) A ∩ B

(iii) A – B

(iv) B – A

Also find the Cardinal number of these sets.

3. If ζ = {1, 2, 3, ……9}, A = {1, 2, 3, 4, 6, 7, 8} and B = {4, 6, 8} then find

(i) A’ (ii) B’ (iii) A U B (iv) A ∩ B

(v) A – B (vi) B – A (vii) (A ∩ B)’ (viii) A’ U B’

4. If n (A) = 20, n (B) = 16 and n (A U B) = 30, find n (A ∩ B)

5. If n (ζ) = 40, n (A) = 20, n (B’) = 16 and n (A U B) = 32 then find n (B) and n (A

∩ B)

6. If n (ζ) = 40, n (A’) = 15, n (B) = 12 and n ((A ∩ B)’) = 32, find:

(i) n (A) (ii) n (B’) (iii) n (A ∩ B)

(iv) n (A U B) (v) n (A – B) (vi) n(B – A)

7. If n (A – B) = 12 , n(B – A) = 16 and n (A ∩ B) = 5, find:

(i) n (A) (ii) n (B) (iii) n (A U B)

ANSWER KEY OF MATHS WORKSHEET - 9

1. (i) 10 × 6 + 2 (ii) 100 × 8 + 10 × 0 + 3 (iii) 100 × 9 + 10 × 7 +

4

2. (i) 4 (ii) 9

3. 65

4. 4

5. Divisible by 2 and 4 : (i) and (ii) ; Divisible by 8: (iii)

6. Divisible by 3 and 9 : (i) and (iii)

7. Divisible by 6: (i)

8. Divisible by 11: (i) and (ii)

9. Divisible by 7: (i)

10. Divisible by 5: (i) and (ii) ; Divisible by 10: (i)

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CLASS-8 COMPUTER APPLICATIONS WORKSHEET-9

CHAPTER-9 CLOUD COMPUTING

Cloud is something that is present at a remote location. Cloud can provide services

over the network, i.e., either on the public networks, or on private networks, i.e.,

WAN, or LAN. Cloud is defined as a set of hardware, networks, storage, interfaces

and services, combined together to deliver different aspects of computing as a service

over a network or internet. All you need is an adequately featured device that allows

you to connect to the cloud.

Application, such as e-mail,

web conferencing, customer

relationship management

(CRM), all run in cloud.

Google Docs, Microsoft

OneDrive, Dropbox, Asus

Cloud Storage, Norton Online

backup etc., are some of the

common examples of free or

low-cost online storage for the

users. Thus, as an individual user, most of us are already using cloud services.

Let us take the example of Facebook. Most of you upload your photos, maintain

albums, update status etc., in real time, and these can be seen by your friends almost

instantaneously. Do you still follow the traditional way of attaching photos in an e-

mail and sending them to your friends? No! That is the power of a cloud. You do not

need a local storage or operating device. Once you upload a photo, it stays there

forever. It does not matter if you delete from your camera or phone; it simply stays in

the cloud.

Cloud computing is made up of two words, Cloud and Computing. Cloud is

ametaphor for internet, hence it means Computing on the internet. Thus,

Cloud Computing refers to manipulating, configuring, and accessing tht

applications online. It offers online data storage, infrastructure, and

applications.

CHARACTERISTICS OF CLOUD COMPUTING

ON DEMAN SELF-SERVICE In cloud computing, multiple clients can share

the resources and applications at the same time. These cloud services and resources

can be used on-demand and often bought on a subscription basis without human

interaction with the service providers.

BROAD NETWORK ACCESS Cloud Computing simply means that the

services can be accessed on the Internet anytime and anywhere in the world through

multiple devices such as mobile phones,

tablets, laptops, etc.

RESOURCE POOLING The

resources like virtual server space, network

connections, bandwidth etc., are pooled by

multiple users, simultaneously from any

location, without being interfered by the

other users.

RAPID ELASTICITY Sometimes, consumers require additional resources in

a small period of time. In such a scenario, Cloud Computing is the solution as it is

enabled with the elasticity feature. This allows consumers to transparently scale up

(increase) or scale down(decrease) the resources according to their computing needs.

MEASURED SERVICE Cloud Computing is based on a pay-for-what-you-

use model where resources usage is monitored, measured, and reported

transparently, based on utilization.

FILL IN THE BLANKS

HINT BOX [applications, network, internet, resources, CRM, remote)

1. Cloud is present at a ________ location.

2. Cloud can provide the service over the _______.

3. Cloud computing means computing on the ________.

4. In cloud computing, multiple clients can share the ________ and ________ at

the same time.

5. Applications like e-mail, web conferencing, ________ all run in cloud.

QUESTIONS:

1. Define cloud computing.

2. Write any 2 characteristics of Cloud Computing.

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Class 8 Worksheet-8 (For Mohali and Zirakpur Branches Only

Answer Key Class 8th Worksheet-7 (Mohali and Zirakpur)

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