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Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 7 Let’s Cultivate and Reap Goodness Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 7 Let’s Cultivate and Reap Goodness Question Answer Notes

Class 8 Basic Science Chapter 7 Notes Kerala Syllabus Let’s Cultivate and Reap Goodness Question Answer

Let’s Cultivate and Reap Goodness Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
Arrange the information given in boxes A and B in the table suitably.

Answer:

Space constraints	Synthetic fertiliser	Organic fertiliser
Vertical farming.	Urea	Manure
Terrace farming	Ammonium phosphate	Vermicompost
Sack farming	Superphosphate	Bone meal

Question 2.
Satheesh has 15 cents of homestead land and 30 cents of paddy field. He says that paddy cultivation is not profitable. What suggestions do you have to make farming profitable by making use of the homestead land and the paddy field?
Answer:
Integrated farming methods can be adopted. There is potential for vegetable cultivation, papaya cultivation, poultry farming, and cattle farming on the land. There is potential for duck farming and rearing of fishes along with paddy cultivation in the field.

Question 3.
Which is the odd one? What are the common features of the others?
a) Wick irrigation, Vertical farming, Drip irrigation, Mulching
b) Hydroponics, aquaponics, aeroponics, geographical indication
Answer:
a) Vertical farming, others are various irrigation methods.
b) Geographical indication, others are parts of smart farming.

Question 4.
‘Farming will be profitable only if all the pests are killed!’ What is your response to this comment from a farmer? How can effective pest control be implemented?
Answer:
I do not agree with this opinion of the farmer. Pest control methods should be selected taking into account the density and nature of the crops. The farmers’ need is not to kill the entire pests, but to control their growth in a way that does not damage the crops. Integrated pest control is a method that minimises the use of pesticides through automated pest control methods using various types of /lets and traps, friendly insects, and the cultivation of seeds that are resistant to pests.

Question 5.
Observe the illustration.

a) What are the benefits to farmers by producing such products?
b) Prepare a similar illustration of any other crop.
Answer:
a) The farmer’s income increases through the marketing of value-added products. New employment opportunities arise.

b) Pineapple jam, pineapple juice, pineapple squash, pineapple leaf fiber, pineapple wine, pineapple candy, dried pineapple pieces, pineapple vinegar, pineapple pickle.

Basic Science Class 8 Chapter 7 Question Answer Kerala Syllabus

Answers to the indicators on page no. 104
Question 1.
What are the ideas you have learned from the news reports?
Answer:

• A school student has caught the media’s attention by developing a simple machine for harvesting cassava.
• The little scientist has applied for a patent for his invention.
• A young farmer with a higher degree sells banana leaves when the price of banana leaves is low.
• He earns a good income by exploiting the market potential of nutritious leafy vegetables.
• He earns income by making several value-added products from turmeric
• He uniquely cultivates the rare Gandhakashala rice and markets it under a special brand, earning good sales.

Question 2.
What are the circumstances that have prompted farmers to choose new ways?
Answer:

• The use of light machinery for harvesting cassava has made harvesting easier.
• A unique idea that banana leaves can be sold when they are cheap.
• Utilising the market potential of nutritious leafy vegetables.
• Due to the fall in price of turmeric during the harvest, several value-added products have been created from turmeric and brought to the market.
• Gandhakshala rice, which is being grown elsewhere, is being cultivated in a unique way and brought to the market under a special brand.

Question 3.
What are the other possibilities to make farming profitable?
Answer:

• Construction of simple machines to save time and effort in farming.
• The sale potential of not only fruits but also other parts of plants, such as leaves.
• Value-added products from agricultural products.
• Agricultural products are branded and marketed in a special way.
• Growing crops that have high demand in the market.

Question 4.
Completed illustration 7.1

Answer:

Observe figure 7.3 (pg. no. 107) and prepare a note based on the indicators through discussion.

Question 5.
How does it help to overcome space constraints?
Answer:
Vertical farming uses shelves or stacked layers to grow plants upwards, allowing more crops to be grown in a smaller space.

Question 6.
How does it ensure availability of light?
Answer:
The plants are arranged in such a way that all layers receive sufficient sunlight.

Question 7.
How does it help to reduce the use of water?
Answer:
Since water drips from the top layer to the bottom, many plants reuse the same water, thereby conserving it.

Question 8.
How to Control Construction Costs?
Answer:
If you use waste materials like plastic bottles, old wood, and containers, the cost is lower. Otherwise, the cost will increase.

Answers of Indicators, Page No. 108 from the Textbook
Question 9.
Macronutrients and Micronutrients:
Answer:
Plants obtain the elements they need from the soil. Of these, nitrogen, potassium, phosphorus, calcium, magnesium, and sulfur are the elements that are required in large quantities. These are called macronutrients. However, elements like barium, boron, zinc, copper, manganese, iron, molybdenum, chlorine, and nickel are required in very small quantities and are known as micronutrients.

Question 10.
Need for the application of fertilisers:
Answer:
Fertiliser is applied to ensure the availability of all the elements necessary for crops to grow.

Analysis of Table 7.1, Pg.no. 110

Question 11.
Why is it necessary to include local vegetables in the diet?
Answer:
We can also get all the elements that plants get in various forms by consuming them. Native vegetables like Colocasia leaf, curry leaves, Drumstick leaves, and Sweet amaranth contain more protein, fiber, starch, calcium, iron, carotene, and vitamin C than other vegetables.

Question 12.
Observe the illustration 7.3 showing the methods of production of high-yield planting materials. Prepare notes on it.

(a) Grafting

• The process of joining the stem of one plant with the stem, including the roots, of another plant to form a single plant.
• Used in plants such as Mango tree and roses.
• Property: Combines the properties of both plants.

(b) Budding

• A method of attaching a bud from one plant to the stem of another plant.
• Used in roses, lemons, etc.
• Advantage: Faster production

(c) Layering

• A branch of a plant is bent to the ground and covered with soil to grow roots. Later, it is cut and planted.
• Used in jasmine, peach, strawberry, etc.
• Advantage: New plants are similar to the mother plant.

(d) Tissue culture

• Seedlings are produced by isolating tissues from suitable parts of the plant and growing them in a special nutrient medium.
• Used in bananas and orchids.
• Benefits: Rapid production of large numbers of healthy, disease-free plants.

(e) GM crops (Genetically Modified Crops)

• These are crops that have had their genes altered in a lab to improve their characteristics. Their production requires strict safety testing.
• Examples: Bt cotton, golden rice.
• Properties: Pest resistance, superior qualities, high yield.

(f) Hybridisation

• The process of crossing two plants with different traits to produce a new plant with superior traits.
• Used in wheat, rice, and tomatoes.
• Advantages: High yield, disease resistance.

Figure 7.10 Cartoon Analysis on page 113

Question 13.
What are the problems farmers face regarding water use?
Answer:
Heavy rain, floods, droughts, and water scarcity

Question 14.
What are the suggestions to solve the problems faced by farmers regarding water use?
Answer:
Water from heavy rainfall is stored in dams and reservoirs. The stored water is then transported to farmlands through systems such as canals during times of water scarcity.

Answers to the indicators on page 114
Question 15.
What are the main pests affecting crops in our region?
Answer:
Pod borer (Legume), Leaf-rolling Caterpillar (Okra), Stem Borer (Brinjal)

Question 16.
What are the various methods adopted by farmers in your area to control pests?
Answer:

• Use of chemical pesticides
• Use of natural enemies of pests (biological control)
• Use of biological pesticides.
• Farmers change the type of crop grown in a field each season. This prevents pests from multiplying because they do not get the same crop to attack each time.
• Use of pest-resistant varieties.
• Keeping the fields clean.
• Weeds and debris are removed from the fields regularly. This prevents pests from hiding in the fields and breeding.

Answers to the indicators on page no. 116, by analysing illustration 7.4

Question 17.
Integrated farming- advantages and possibilities:
Answer:
(a) Benefits of integrated farming:

• Better use of resources.
• No resources are wasted. Animal waste is used as fertiliser for plants. Water is reused.
• More income for farmers
• Farmers get income not only from crops, but also from selling milk, eggs, fish, and vegetables.
• Even if one crop (like a bad crop) fails, other crops (like fish or eggs) provide income.

(b) The possibilities of integrated farming:

• It can be used in villages, small farms, and even in urban areas where there is space.
• Helps in self-employment and sustainable agriculture.
• Promotes organic farming and protects the environment.

(c) Food security

• Integrated farming produces non-toxic food.
• There is less need for pesticides or artificial feed.
• It improves nutrition for families and consumers.

(d) Reducing production costs

• Money is saved on fertilisers and feed. Because the residue of one is used as fertiliser for the other.
• Waste is recycled efficiently.
• The need to buy products from outside will be reduced.

Answers to the indicators on page no. 119, based on illustration 7.6

Question 18.
What are the benefits of diversifying agriculture?
Answer:
Agriculture is not just about producing food. Farming for Banana leaves, horticulture, Poultry farming, medicinal plant cultivation, Betel cultivation, floriculture, ornamental fish farming, and raising pets are all ways of diversifying agriculture.

Benefits:

• More sources of income – Farmers can earn money from many activities, not just from one crop.
• Even if one crop fails, income can be generated from other crops.
• Efficient use of land and resources – Different activities can be done in the same area by using waste from one activity for another.
• Employment for more people
• Helps small families and women to get work done from home.
• Supports local markets and economies

Question 19.
Which of these can be done even by those with limited space?
Answer:

• Medicinal plant cultivation – on terraces, balconies or small gardens
• Rearing of chickens – in backyards or coops
• Ornamental fish rearing – in small tanks or containers
• Floriculture – in pots or small plots
• Raising of pets – at home
• Betel cultivation – in shaded areas or along fences

Class 8 Basic Science Chapter 7 Question Answer Extended Activities

Question 1.
Find out the common plant diseases in your area. Prepare a pictorial chart listing their pathogens, mode of transmission, symptoms and remedies and display it on the bulletin board.
Answer:

Diseases	Causative organism (Pathogen)	Symptoms	Mode of transmission	Remedial measures
Blight disease in Rice	Bacteria	The tips of the leaves turn yellow, and the leaves become dry.	Through infected seeds, rain, and wind	Avoid diseased seeds.
Wilt disease in Banana	Fungus	The leaves turn yellow and wilt.	Through soil, and through water	Destroy infected plants.
Mosaic disease in Tapioca	Virus	Mosaic pattern on leaves	Through insects	Use healthy seedlings and control pests

Question 2.
Collect information about the major agricultural research institutes in Kerala and their contributions to the agricultural sector, and prepare a list of them.
Answer:

• Kerala Agricultural University, Thrissur – Research on crops. Training for farmers
• Indian Institute of Spices Research, Kozhikode – Research on spices like ginger, pepper and turmeric
• Central Tuber Crops Research Institute, Thiruvananthapuram – Research on tuber crops
• Rubber Research Institute of India, Kottayam – Production of good varieties of rubber

Question 3.
You know that many machines are used in the agricultural sector to reduce human effort. The results of thinking about how to alleviate the difficulties of farmers led to the discovery of most of the machinery seen today. Design a model of an innovative device that will be useful to the farmers in your area.
Answer:
Hint: Think about common issues experienced by small-scale farmers in a tropical, monsoon-affected area, particularly related to labor and resource management.
(An example is given below for you…)
Farmers in Kerala, particularly those cultivating crops like coconuts, spices, and various vegetables, often face challenges such as:

• Labor Shortage & Cost
• Nutrient Management
• Post-Harvest Damage
• Accessibility
• Environmental Concerns

Device Name: Coco-Smart Harvester & Nutrient Analyzer.
The Coco-Smart Harvester & Nutrient Analyzer is a semi-autonomous, drone-based system designed to efficiently harvest tree-borne products (initially focusing on coconuts and later adaptable for other tall tree crops) while simultaneously assessing the nutrient needs of the tree/soil and delivering targeted solutions.

Operational Workflow:

1. Pre-Flight Planning: Farmer maps out the area using a user-friendly tablet app, identifying trees for harvesting or analysis.
2. Automated Scan & Analysis: The drone flies autonomously over the designated trees, conducting a hyperspectral scan and, if needed, deploying the soil sensor.
3. Harvesting (if applicable): Based on visual recognition (AI-powered to identify ripe products) and farmer input, the drone’s robotic arm engages the harvesting mechanism, gently collecting the product.
4. Nutrient Recommendation: The AI analyses the data and provides precise nutrient recommendations to the farmer’s app, indicating which trees need what specific nutrients and in what quantities.
5. Targeted Application: The drone can then be programmed to autonomously apply the recommended nutrients via its micro-dosing sprayer or granular dispenser.
6. Data Logging: All data (harvest quantity, nutrient status, application history) is logged for future analysis and improved farm management.

The Coco-Smart Harvester & Nutrient Analyzer aims to be a game-changer for farmers in Kerala, addressing critical challenges with a blend of robotics, AI, and precision agriculture.

Let’s Cultivate and Reap Goodness Class 8 Notes

Class 8 Basic Science Let’s Cultivate and Reap Goodness Notes Kerala Syllabus

• Innovative agricultural initiatives can strengthen rural economies, enhance food security, and inspire future generations to engage in farming.
• Various farming methods to make use of most of the space include sack farming, aquaponics, pet bottle farming, vertical farming, pot cultivation, and terrace farming.
• Elements that plants need in large quantities are called macronutrients and elements that plants need in small quantities are called micronutrients.
• Fertilisers are used to ensure the availability of all the elements that plants need to grow.
• Different types of fertilisers used in agriculture – organic fertilisers, artificial fertilisers, nano fertilisers, and biofertilisers.
• Grafting, budding, layering, and tissue culture are methods used to produce seedlings that have the same characteristics as the parent plant.
• GM crops are crops that can incorporate new characteristics into crops by changing the genetic structure through genetic engineering.
• Tissue culture is a technology that helps produce large numbers of plants that have the same characteristics as the parent plant.
• The greenhouse is made of sheets like plastic, nylon, and polyethene. It also helps in reducing the incidence of pests and diseases as it is covered on all sides.
• Drip irrigation is an irrigation method that uses pipes and valves to drop water into the root zone.
• Wick irrigation is a method of delivering water directly from a water source to the root zone of plants with the help of a cotton wick.
• Mulching is a traditional method of covering the soil in agricultural fields with leaves and straw to reduce water loss due to evaporation.
• Integrated pest control is a method that minimises the use of pesticides through automated pest control methods using various types of nets and traps, friendly insects, and the cultivation of seeds that are resistant to pests.
• Integrated farming is the management of diverse organisms together.
• Smart farming is the effective use of modern technologies in agriculture, such as hydroponics, aeroponics, etc.
• Depending on the characteristics of the land where the products are grown, there will be differences in the taste, colour, smell and nutritional value of the products. On the basis of this, agricultural products produced in some areas are given Geographical Indication (GI) status.
• There are apps that provide weather warnings, pest and disease warnings, expert advice on agriculture, market price levels and information about benefits for farmers.

INTRODUCTION

In a world that often feels overwhelmed by challenges and negativity, the concept of “cultivating and reaping goodness” offers a powerful and hopeful perspective. This idea isn’t just a feel-good platitude; it’s a profound call to action, urging us to intentionally nurture positive qualities, actions, and intentions within ourselves and our communities. Just as a farmer carefully tends to their crops, preparing the soil, planting seeds, and providing consistent care, we too can consciously foster an environment where goodness can flourish. This process involves recognising that every small act of kindness, every moment of empathy, and every effort to uplift others contributes to a larger harvest of positive outcomes. When we actively cultivate goodness, we not only transform our own lives but also create a ripple effect, inspiring and empowering those around us to do the same. Ultimately, this journey is about understanding that the positive impact we wish to see in the world begins with the seeds we choose to plant today. In this chapter, we will deal with various agricultural initiatives, careful utilisation of land, application of fertilisers, use of high-quality planting material for getting better yield, water utilisation and pest control measures and integrated farming in detail.

AGRICULTURAL INITIATIVES

• The ‘Karshaka Pratibha’ Puraskar is an award given by the state government to the best student farmer.
• Innovative agricultural initiatives can strengthen rural economies, enhance food security, and inspire future generations to engage in farming.

MAXIMISING LAND UTILISATION

• Various farming methods to make use of most of the space include sack fanning, aquaponics, pet bottle farming, vertical farming, pot cultivation, and terrace farming.
• Vertical farming is an innovative way to overcome space constraints.

APPLICATION OF FERTILISERS

• Elements that plants need in large quantities are called macronutrients.
• Elements that plants need in small quantities are called micronutrients.
• Fertilisers are used to ensure the availability of all the elements that plants need to grow.
• Different types of fertilisers used in agriculture – organic fertilisers, artificial fertilisers, nano fertilisers, and biofertilisers.

• All the elements that plants get are also available to us as food in various forms.
• Native varieties will grow in harmony with the local environment.

FOR BETTER YIELD, HIGH QUALITY PLANTING MATERIALS

• Grafting, budding, layering, and tissue culture are methods used to produce seedlings that have the same characteristics as the parent plant.
• GM crops are crops that can incorporate new characteristics into crops by changing the genetic structure through genetic engineering.
• Tissue culture is a technology that helps produce large numbers of plants that have the same characteristics as the parent plant.

UTILISATION OF WATER AND PEST CONTROL
UTILISATION OF WATER

• A greenhouse is a system that helps in cultivating crops both in the rainy season and in the summer.
• Drip irrigation is an irrigation method that uses pipes and valves to drop water into the root zone.
• Wick irrigation is a method of delivering water directly from a water source to the root zone of plants with the help of a cotton wick.
• Mulching is a traditional method of covering the soil in agricultural fields with leaves and straw to reduce water loss due to evaporation.

PEST CONTROL

• Although there are different methods for pest control, the population density of pests and the nature off the crops should be taken into account when choosing the pest control methods.
• Farmers do not want to kill all the pests, but rather control their growth in a way that does not damage the crops.
• Integrated pest control is a method that minimises the use of pesticides through automated pest control methods using various types of nets and traps, friendly insects, and the cultivation of seeds that are resistant to pests.

INTEGRATED FARMING

• Integrated farming is the management of diverse organisms together.
• Smart farming is the effective use of modern technologies in agriculture.
• Hydroponics – Plants are grown in nutrient solutions, and the amount of nutrients is detected with the help of sensors and provided as needed.
• Aeroponics – Water and nutrients are provided to the roots growing in the air in a timely manner with the help of sensors.
• If a system is set up for consumers to see the process from sowing to harvesting, the reliability and market value of the products can be increased.
• Depending on the characteristics of the land where the products are grown, there will be differences in the taste, colour, smell and nutritional value of the products. On the basis of this, agricultural products produced in some areas are given Geographical Indication (GI) status.
• Agriculture is not only about producing food.
• There are apps that provide weather warnings, pest and disease warnings, expert advice on agriculture, market price levels and information about benefits for farmers.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 8 Origin of Life, Origin of Living Things Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 8 Origin of Life, Origin of Living Things Question Answer Notes

Class 8 Basic Science Chapter 8 Notes Kerala Syllabus Origin of Life, Origin of Living Things Question Answer

Origin of Life, Origin of Living Things Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
Write which of the given statements are not related to eukaryotes.
a) A clear nuclear membrane is seen.
b) Cell organelles have no membranous covering.
c) Genetic material is found inside the nuclear membrane.
d) Organelles that perform various functions are seen.
Answer:
b) Cell organelles have no membranous covering.

Question 2.
Arrange the timeline properly.

Answer:

Question 3.
Match the following.

Answer:
a) Urey- Miller – Amino acids
b) Joan Oro – Nitrogenous bases
c) Sidney Fox – Proteinoid
d) Oparin-Haldane – Theory of Chemical evolution

Basic Science Class 8 Chapter 8 Question Answer Kerala Syllabus

Answers to the indicators on page no. 125
Question 1.
Characteristics of the primitive Earth’s atmosphere
Answer:

• High temperature.
• Hydrogen, methane, carbon dioxide, hydrogen sulfide, ammonia, water vapour.
• No free oxygen.

Question 2.
Formation of Ocean
Answer:
The condensation of steam, prolonged rainfall and the formation of oceans. The ocean is formed by dissolving varidus substances.

Question 3.
Energy sources that assist the formation of biomolecules.
Answer:
Sunlight, Lightning, Ultraviolet Rays, Volcanic Eruptions.

Question 4.
Formation of primitive cell
Answer:
The condensation of steam, prolonged rainfall and the formation of o’ceans. The ocean is formed by dissolving various substances. Simple organic particles were formed in the seawater. Then, complex organic particles were formed from simple organic particles. A complex molecule called nucleic acid and a fat layer were formed. Then, a primitive cell capable of self-replication was formed.

Question 5.
Completed table 8.1

Answer:

Indicators	Conditions in the primitive Earth	Experimental set-up
Gases	Hydrogen, Methane, Carbon dioxide, Hydrogen sulphide, Ammonia, Water vapour.	Methane, ammonia, hydrogen, Water vapour.
Energy source for chemical synthesis	Sunlight, Lightning, Ultraviolet Rays, and Volcanic Eruptions.	Energy from an electric spark through electrodes

Question 6.
Analyse illustration 8.5 based on the indicators and record your inferences in the Science Diary

Answer:
Prokaryotic cell: A few cell organelles, Cell organelles have no membranous covering, Membrane-bound nucleus absent.
Eukaryotic cell: Many cell organelles, Organelles with a membranous covering, Membrane-bound nucleus present.

Question 7.
Analyse illustration 8.6 from the textbook page number 128 and prepare a note on it.

Answer:
The eukaryotic cell engulfs the small aerobic bacteria. Instead of digesting the small cell, it protects it. These eventually become mitochondria. The eukaryotic cell engulfs the tiny photosynthetic bacteria. Instead of digesting the photosynthetic bacteria, it protects them and gradually transforms them into chloroplasts.

Question 8.
Completed Table 8.2, Pg.no. 129

Answer:

Indicators	Prokaryotic cell	Eukaryotic cell
Structure	Simple	Complex
Nucleus	Membrane-bound nucleus absent	Well-defined nucleus with a membrane covering
Cell organelles such as mitochondria and chloroplasts	Absent	Present

Question 9.
Analyse the timeline (Table 8.3) from Textbook Pg.no. 130, and prepare’ a note based on it.

Answer:
4 – 4.6 billion years ago: Origin of the Earth
3.5 – 2.5 billion years ago: Formation of the first life. Single-celled prokaryotes with simple structures
2.5 – 541 million years ago: Multicellular organisms called eukaryotes
541 – 252 million years ago: Plants and animals on land
252 – 66 million years ago: Dinosaurs
66 million years ago to the present: Emergence of mammals, evolution of humans

Question 10.
Conclusions drawn from the analysis of Illustration 8.7 on page 131 of the textbook

Answer:
Suppose we are preparing a calendar assuming the age of the universe to be just one year. One second in the said calendar is equivalent to approximately 438 years. Conclusions drawn from analyzing a model of the cosmic calendar:
The universe is born at exactly 00.00 on January 1st
August 1st Solar System, Earth
September 22nd Emergence of life
October 12th Prokaryotes
November 9th Eukaryotes
December 18th Vertebrate
December 20th Land Plants
December 26th Dinosaurs
December 27th Mammals
December 31st, 11.52 PM – Human
According to the cosmic calendar, man was born in the last moments of the last day of the year. This shows how recent the emergence of man is in the chronology of the universe.

Class 8 Basic Science Chapter 8 Question Answer Extended Activities

Question 1.
Make a model of the Urey-Miller experimental setup using the materials available from the surroundings and display it at the science corner.

Question 2.
Prepare a digital model/chart of the Cosmic calendar including more information and exhibit in the classroom.

Origin of Life, Origin of Living Things Class 8 Notes

Class 8 Basic Science Origin of Life, Origin of Living Things Notes Kerala Syllabus

• Scientists believe that life originated on Earth about 3.5 billion years ago.
• Theories related to the origin of life – Panspermia theory, Chemical evolution theory
• Harold Urey, Stanley Miller, Sydney Fox, and Joan Oro are some of the scientists who contributed to the field of the origin of Life.
• The simple structured cells, which were formed in the beginning, are called prokaryotes, and Eukaryotic cells with more complex structures are evolved from prokaryotic cells.
• As part of the evolutionary process, the formation of a nuclear membrane was the key feature of eukaryotic cells. Additionally, membrane-bound organelles were evolved, enabling them to perform specialised functions.
• Around 3.8 billion years ago, primitive living cells were formed from the molecules present in Earth’s oceans. By 3.5 billion years ago, prokaryotic cells had evolved.
• Approximately 2.5 billion years ago, the occurrence of photosynthesis led to the release of oxygen into the atmosphere. Later, eukaryotic, cells with organelles such as mitochondria and chloroplasts were evolved. Over time, simple multicellular organisms appeared around 800 million years ago, followed by more complex life forms.
• The Cosmic Calendar is a depiction that helps to easily understand the chronology from the creation of the universe to the origin of human beings.

INTRODUCTION

The quest to understand the origins of life is one of the most profound and challenging scientific endeavours. It delves into the fundamental question of how inanimate matter transitioned into the complex, self-replicating systems we recognise as living organisms. This field encompasses two closely related yet distinct concepts: the origin of life and the origin of living organisms.

The origin of life, often referred to as abiogenesis, explores the initial emergence of life from non-living chemical compounds. This involves understanding the conditions on early Earth—its atmosphere, oceans, and energy sources – that could have facilitated the formation of fundamental organic molecules like amino acids, nucleotides, and fatty acids. It further investigates how these simple building blocks could have self-assembled into more complex polymers (proteins, nucleic acids) and eventually organised into self-replicating systems enclosed by primitive membranes, forming the very first “protocells or primitive cells.”

The origin of living things, on the other hand, broadly refers to the subsequent evolution and diversification of these initial life forms into the vast array of organisms we see today, from the simplest bacteria to complex multicellular plants and animals. While the origin of life focuses on the first step, the origin of living things traces the evolutionary journey, driven by natural selection, adaptation, and genetic changes, that led to the incredible biodiversity on Earth. This includes the development of complex cellular structures (eukaryotes), the emergence of multicellularity, and the evolution of different modes of nutrition and reproduction.

Together, these two areas of inquiry seek to unravel the complete narrative of life’s emergence and evolution on our planet, providing insights into our own existence and the potential for life elsewhere in the universe. In this chapter, we will learn about these concepts.

ORIGIN OF LIFE

• Millions of years ago, life on Earth was very different from what we see today.
• Living things evolved in many ways, depending on changes in external factors such as habitat, climate, and food availability, as well as changes in the internal components of living cells, and the living things we see today were formed.
• Scientists believe that life originated on Earth about 3.5 billion years ago.
• Some of the simpler organisms released oxygen through photosynthesis, which led to the formation of more complex organisms.
• Complex plants and animals evolved. Over time, organisms evolved into the modern-day
• Theories related to the origin of life – Panspermia theory, Chemical evolution theory

Panspermia Theory Life originated elsewhere in the universe and was accidentally transported to Earth in the form of microorganisms or spores. These microscopic particles are referred to as Panspermia.

Chemical Evolution Theory The Chemical Evolution Theory explains that life originated as a result of changes in the combination of chemical substances in the ocean under the unique conditions of the primitive Earth.

• Although many theories explaining the origin of life have emerged over time,-the theory of chemical evolution is the one that is most supported by evidence and has received the most acceptance in the scientific world.
• Some of the scientists who contributed to this field:
  

  Harold Urey, Stanley Miller: They together proved that the fundamental units responsible for the origin of life can form from simple gases. Sydney Fox: Proved that molecules similar to proteins can be synthesised artificially. Joan Pro: Adenine, one of the key building blocks of nucleic acids, was artificially synthesised.

• Among the many experimental evidence supporting the theory of chemical evolution, the most notable is the Urey-MiUer experiment.
• The Urey-Miller experiment was conducted by artificially recreating the conditions of the primitive Earth in a laboratory.
• The Urey-Miller experiment demonstrated that organic molecules can form from inorganic components under suitable conditions.

FROM PRIMITIVE CELLS TO COMPLEX ORGANISMS

• The primitive cell is composed of nucleic acids capable of self-replication and a lipid layer covering it.
• Primitive forms like bacteria evolved from this primitive cell.
• These simple structured cells, which were formed in the beginning, tire called prokaryotes.
• Eukaryotic cells with more complex structures evolved from prokaryotic cells.

Prokaryotic cell • A few cell organelles. • Cell organelles have no membranous covering. • Membrane-bound nucleus absent.

Eukaryotic cell • Many cell organelles. • Organelles with membranous covering • Membrane-bound nucleus present.

• The eukaryotic cell engulfs the small aerobic bacteria. Instead of digesting the small cell, it protects it. These eventually become mitochondria.
• The eukaryotic cell engulfs the tiny photosynthetic bacteria. Instead, of digesting the photosynthetic bacteria, it protects them and gradually transforms them into chioroplasts.
• In prokaryotes, genetic material is scattered within the cytoplasm.
• As part of the evolutionary process, the formation of a nuclear membrane was the key feature of
  eukaryotic cells. Additionally, membrane-bound organelles were evolved, enabling them to perform
  specialised functions.
• Around 3.8 billion years ago, primitive living cells were formed from the molecules present in Earth’s oceans. By 3.5 billion years ago, prokaryotic cells had evolved.
• Approximately 2.5 billion years ago, the occurrence of photosynthesis led to the release of oxygen into the atmosphere. Later, eukaryotic cells with organelles such as mitochondria and chloroplasts were evolved. Over time, simple multicellular organisms appeared around 800 million years ago, followed by more complex life forms.

Cosmic calendar

• The Cosmic Calendar is a depiction that helps to easily understand the chronology from the creation of the universe to the origin of human beings.
• Suppose we were creating a calendar assuming the age of the universe to be just one year. One second in that calendar would be equivalent to approximately 438 years.
• According to the cosmic calendar, man was born in the final moments of the last day of the year.

The process of evolution is the cause of present biodiversity. It is difficult to predict the direction of evolution as it is an accidental phenomenon.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 4 Chemistry of Changes Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 4 Chemistry of Changes Question Answer Notes

Class 8 Basic Science Chapter 4 Notes Kerala Syllabus Chemistry of Changes Question Answer

Chemistry of Changes Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
What are the changes in the particle arrangement of substances in the activities given below?
a) Solid becomes liquid.
b) Liquid becomes gas.
c) Gas becomes liquid.
Answer:
a) Solid becomes liquid.

• Distance between the particles increases
• Attraction between the particles decreases
• Speed of movement of the particles increases
• Energy of the particle increases

b) Liquid becomes gas.

• Distance between the particles increases
• Attraction between the particles decreases
• Speed of movement of the particles increases
• Energy of the particle increases

c) Gas becomes liquid

• Distance between the particles decreases
• Attraction between the particles increases
• Speed of movement of the particles decreases
• Energy of the particle decreases

Question 2.
In the given chemical reactions, what is the main form of energy released/absorbed? Write what type of chemical reaction they are.
a) Ammonium chloride and Barium hydroxide react.
b) Copper plating on an iron bangle.
c) Glowing of firefly.
d) Decomposition of Potassium permanganate
e) Lighting an LED using lemons
Answer:

Given chemical reactions	Main form of energy released/absorbed	Type of chemical reaction
a) Ammonium chloride and Barium hydroxide reacts.	Heat energy is absorbed.	Thermochemical reaction
b) Copper plating on an iron bangle.	Electrical energy is absorbed.	Electrochemical reaction
c) Glowing of firefly	Light energy is released.	Photochemical reaction
d) Decomposition of Potassium permanganate	Heat is absorbed	Thermochemical reaction
e) Lighting an LED using lemon	Electrical energy is released.	Electrochemical reaction

Question 3.
Heat some crystals of potassium permanganate in a dry test tube. Bring a burning incense stick near the mouth of the test tube.
a) What do you observe?
b) Which is the gaseous product formed?
c) Which type of reaction is this?
Answer:
a) The incense stick flares up and burns.
b) Oxygen
c) Thermochemical reaction

Question 4.
A white cloth dipped in silver nitrate darkens when it is kept in sunlight.
a) Which form of energy is responsible for this chemical change?
b) What is the general name for this type of reaction?
Answer:
a) Light energy
b) Photochemical reactions

Question 5.
Sodium metal reacts with water to give substances.
a) Which are the reactants in this reaction?
b) Which products are formed?
Answer:
a) Sodium, Water
b) Sodium hydroxide, Hydrogen

Basic Science Class 8 Chapter 4 Question Answer Kerala Syllabus

GENERAL PROPERTIES OF MATTER

Question 1.
What do you see in the picture?
Answer:

• A candle burning
• Food is cooked
• A person sculpting or working with stone/metal
• A firecracker cone exploding
• The bunch of bananas is ripening

Substances undergo different types of changes here.

Question 2.
The stone is shaped into a sculpture, and the banana gets ripened. Are these changes of the same kind?
Answer:
No. The stone is shaped into a sculpture, is man-made, and the banana gets ripened is natural.
These changes may lead to the production of new substances.

All substances in the universe are made of matter.

GENERAL CHARACTERISTICS OF MATTER

• Has Mass
• Occupies Space (Has Volume)
• Made of Particles
• Particles have Space Between Them
• Particles Attract Each Other

EXPERIMENT:
Dip a stone hung on a thread into a beaker with a marked water level.

Question 3.
What happens to the water level?
Answer:
Water level increases

Question 4.
Why does the water level rise?
Answer:
Because the stone needs space to occupy. So, the stone displaces the water. The water level rises more.

Question 5.
What difference do you observe if a bigger stone is used?
Answer:
The water level rises more.

One of the properties of matter is that it occupies space. The space occupied by matter is its volume.

Question 6.
Weigh both the stones using a balance. What difference do you observe?
Answer:
Both stones have different masses
Matter has mass. This is another property of matter.

Mass of a substance is the measure of the quantity of matter contained in it.

Question 7.
Is air a form of matter?
Answer:
Yes, air is a form of matter.

Question 8.
Does air need space to occupy?
Answer:
Yes, air needs space to occupy.

EXPERIMENT:
Fix a towel inside a glass and immerse it upside down into the water taken in a beaker, as shown in the figure.

Question 9.
Does the towel get wet?
Answer:
The towel doesn’t get wet.

Question 10.
Why doesn’t water enter the glass?
Answer:
Because there is air inside the glass.

Question 11.
What happens to the water level in the beaker?
Answer:
The water level rises.

Question 12.
What does the difference in water level indicate?
Answer:
The volume of air

Question 13.
How can we find out whether air has mass?
Answer:
Find the mass of an uninflated football using a digital balance. After filling it with air, find the mass of the football again. You can see that the mass of the air-filled football is greater. The difference between the masses is the mass of air inside the football.
Now it is clear that the air. has mass.

Anything that occupies space and has mass is called matter.

Question 14.
What are the main states of matter?
Answer:
Matter primarily exists in three fundamental states: solid, liquid, and gas.

Question 15.
Complete the given table with respect to the three states of matter. Put the (✓) mark appropriately.

Answer:

Property	Solid	Liquid	Gas
Has definite mass	✓	✓	✓
Has definite volume	✓	✓	✗
Has a definite shape	✓	✗	✗

Question 16.
Heat some wax in a steel vessel. What happens?
Answer:
The wax melts and becomes liquid.

Question 17.
What happens when the liquid wax is cooled?
Answer:
It solidifies and becomes solid again.

Question 18.
But if a piece of paper is burnt to ash, can it be changed to paper again?
Answer:
No

Question 19.
Do all the changes occur at the same speed?
Answer:
No

Question 20.
Write more examples for slow and fast changes.

Answer:

Slow changes	Fast changes
Ice melts	Petrol bums
Milk turns into curd.	Hydrogen bums
Water turns into steam.	Acid dissolves in water.
The colour of metals fades.	Firecracker explodes
Rusting of iron	Alcohol mixes/dissolves in water.
Germination of the pea seed	Lighting a gas stove, Bursting of crackers

Question 21.
What is the difference between the burning of a piece of paper and the melting of wax?
Answer:
Burning paper is a chemical change, and the melting of wax is a physical change.

Question 22.
Classify the following changes into physical and chemical changes.
i) Curdling of milk
ii) Melting wax
iii) Burning a candle
iv) Formation of ice
v) Melting of ice
vi) Dissolution of salt in water
vii) Rusting of iron
viii) Burning of firewood
Answer:

Physical Change	Chemical Change
ii, iv, v, vi	i, iii, vii, viii

Question 23.
You have learnt about the arrangement of particles in different states of matter, i.e., solid, liquid and gas in the lower classes. Can you write about the distance and force of attraction between the particles, speed of movement and the energy of the particles in each state in the following table?

Answer:

	Solid	Liquid	Gas
Distance between the particles	very less	more than solid	very high
Attraction between the particles	very high	less than solid	very less
Speed of movement of the particles	very less	more than solid	very high
Energy of the particles	very less	higher than solid	very high

Question 24.
What happens to the distance between particles and the energy of particles when solids are heated?
Answer:
Distance between the particles: Increases
Energy of the particles: Increases

When a solid is heated, the distance between its particles and their energy increases and gradually it attains the particle arrangement of a liquid. When liquids are heated gradually, they become gases.

Question 25.
What happens when a gas is cooled?
Answer:
The distance between particles and the energy of particles decreases. Gradually, gas becomes liquid. If we further cool a liquid, it becomes solid.

SUBLIMATION
Solid substances like Camphor and Naphthalene, when heated, directly change to gases. This is known as Sublimation.

Question 26.
How does the water kept in a freezer change to ice? Discuss.
Answer:
At the low temperature inside the refrigerator, the speed of movement and energy of the molecules decrease, bringing them closer together. In this process, heat is released, and water turns into ice.
The change of substance from one physical state into another is known as a change of state.

Question 27.
Is a change of state a physical change or a chemical change?
Answer:
Change of state is a physical change

Question 28.
Which form of energy is absorbed or liberated during the change of state?
Answer:
Heat energy

Question 29.
What happens to the energy of particles when heat is absorbed or liberated?
Answer:
When heat is absorbed during a change of state, the energy of the particles increases.
When heat is liberated (released) during a change of state, the energy of the particles decreases.

Question 30.
Is the colour of potassium permanganate the same before and after the reaction?
Answer:
No

Question 31.
Take water in two beakers and add two or three crystals of fresh potassium permanganate into one of them. Add two or three crystals of the product obtained in the second beaker. Record the difference.

a. What is the change in colour you have observed?
Answer:
The pure permanganate solution is pink in colour. The solution of the substance after the reaction is green. This is due to a chemical change resulting in the formation of a new substance.

b. Is this a chemical change or a physical change?
Answer:
Chemical Change

The substances that take part in a chemical reaction are known as reactants, and the substances formed are known as products.

Question 32.
Find more examples of endothermic and exothermic reactions.

Answer:

Exothermic Reactions	Endothermic Reactions
Reaction between potassium permanganate and glycerine.	Decomposition of potassium permanganate
Reaction between magnesium and hydrochloric acid.	Reaction between ammonium chloride and barium hydroxide
Burning of Hydrogen in Air	Dissolution of Salts in Water
Reaction of Water with Quicklime	Process of heating Limestone to produce Quicklime

The reactions in which heat energy is liberated or absorbed are known as thermochemical reactions.

Question 33.
Which reaction is represented by the illustration

Answer:
Photosynthesis

a. What are the reactants and the products here?
Answer:
Reactant: Carbon dioxide and water
Product: Glucose and oxygen

b. Which form of energy is absorbed in photosynthesis
Answer:
Light energy

In photosynthesis, light energy is converted into chemical energy.

Question 34.
a. Do you know any reactions that liberate light energy?
Answer:
Combustion reactions: For example, burning of wood, candles, or fuels like natural gas, where heat and light are produced.
Bioluminescence: Chemical reactions occurring in living organisms that produce light (like fireflies).

b. How do fireflies produce light?
Answer:
The chemical luciferin (in fireflies) absorbs ultraviolet rays.

• This process is aided by the enzyme luciferase, which is also present in fireflies.
• As a result, visible light is emitted.
• This phenomenon is known as bioluminescence.
• Fireflies can control the amount of oxygen entering their body.
• This control allows them to regulate the intensity of light produced.
• Some species of marine organisms and worms also exhibit bioluminescence.

Question 35.
Some medicines are kept in brown-coloured bottles. What could be the reason?
Answer:
Some medicines decompose and change into other substances in the presence of light. Therefore, they are stored in brown bottles. Brown bottles do not allow light to pass through.

Question 36.
Silver nitrate is not stored in transparent bottles. Why?
Answer:
Silver nitrate is a compound with very low stability. It decomposes in the presence of light. Therefore, silver nitrate is not stored in transparent bottles.

Question 37.
Look at the figures given.

They are known as dry cells.

a. What is their use?
Answer:
Dry cells are used to convert chemical energy into electrical energy.

b. What are the parts of a dry cell?
Answer:

Electricity is produced because of chemical reactions that take place in a dry cell

In a dry cell chemical energy is conreated into electricl energy.

EXPERIMENT
Take some concentrated salt solution (sodium chloride) in a beaker. Add three or four drops of phenolphthalein into it. Dip the ends of copper wires connected to both the terminals of a battery into the solution.

a. What is your observation?
Answer:
When electricity is passed through, the solution turns pink

b. Which compound’s presence is indicated by pink colour?
Answer:
When sodium chloride undergoes dissociation, sodium hydroxide is formed as one of the products. This turns the solution pink.

c. Which form of energy is responsible for this reaction?
Answer:
Electrical energy

ELECTROLYSIS
The process of dissociation of a substance by absorbing electrical energy is known as electrolysis.
If electricity is passed through water containing a small amount of acid, it dissociates into hydrogen and oxygen.
Water → Hydrogen + Oxygen
Water dissociates by absorbing electrical energy.
Hence, this process is electrolysis.
Many such reactions take place by absorbing electricity.

ELECTROPLATING
Coating a metal with another metal using electricity is known as electroplating.

EXPERIMENT
Let’s coat copper on an iron bangle

a. Which solution is taken in the beaker?
Answer:
Copper sulphate solution

b. To which terminal of the battery is the copper plate connected?
Answer:
Positive

c. What about the bangle?
Answer:
Negative terminal of the battery

Question 38.
What all things should be taken care of during electroplating?
Answer:

• The metal that is to be plated should be connected to the positive terminal of the battery.
• The object on which the plating is to be done should be connected to the negative terminal of the battery.
• A salt solution containing ions of the metal being plated should be used as the electrolyte

SUITABLE SOLUTIONS FOR ELECTROPLATING EACH METAL:

Metal	Suitable solutions
Copper	Copper sulphate solution
Silver	A mixture of sodium cyanide and silver cyanide solutions
Gold	A mixture of sodium cyanide and gold cyanide solutions

Question 39.
Which metal plate should be connected to the positive terminal of the battery if silver is to be coated instead of copper?
Answer:
The plate of the metal to be plated should be connected to the positive terminal of the battery.

Question 40.
Which energy has caused the LED to glow?
Answer:
Electrical energy

Question 41.
Which is the source of this electrical energy?
Answer:
Here, the acid in the lemons reacts with metals to produce electricity.

If electrical energy is absorbed or produced during a chemical reaction, it is known as an electrochemical reaction.

Different forms of energy are absorbed or liberated in every chemical reaction. Reactions are known by the major form of energy involved in that reaction.

Chemical Reaction	Main Energy Change
Burning of substances	Liberates heat
Decomposition of substances on heating	Absorb heat
Bioluminescence	Liberates Light
A cell made of lemon.	Liberates electrical energy
Electrolysis of Sodium Chloride solution	Absorbs electrical energy

Question 42.
You have understood about the state of the matter and the changes they undergo. Among these, there are changes that are beneficial to human life and those that are not. Discuss in class how the changes in matter influence human life and prepare a note.
Answer:
Hints: –

• Living organisms produce the starch they need through photosynthesis. Photosynthesis also helps to maintain a stable level of oxygen in the atmosphere
• Through the combustion of fuels, thermal energy can be produced. This chemical change is very important for domestic needs and for the functioning of vehicles. The combustion of substances can also cause pollution.
• Electricity is generated through chemical reactions in electrochemical cells. The disposal of electrochemical cells causes environmental problems.
• The biodegradation of substances is a chemical reaction. It is helpful for environmental cleaning and waste disposal
• Chemical reactions in factories help in producing necessary products. By-products from factories can lead to pollution.

Class 8 Basic Science Chapter 4 Question Answer Extended Activities

Question 1.
Collect a little aluminium powder and iodine powder from the science lab. Mix them well. Add two or three drops of water to this mixture. Write the observations.
Answer:
Initially, nothing much happens when the powders are just mixed.

• Upon adding two or three drops of water, a vigorous reaction starts.
• Heat will be generated, and the mixture will become warm or even hot.
• Purple colored fumes (iodine vapour) will be seen rising from the mixture.
• A loud crackling or popping sound might be heard.
• The mixture might glow or sparkle briefly due to the intensity of the reaction.
• After the reaction subsides, a white or greyish solid (aluminium iodide) will be left behind.

Question 2.
Make a heap of ammonium dichromate powder on a tile. Make a small pit in it and insert the chemicals collected from matchsticks and ignite it. Record the changes you see.
Answer:
When the matchstick chemicals are ht in the pit, the orange powder starts to burn and glow.

• Suddenly, the heap of orange powder starts to grow bigger and pushes upwards, just like a volcano erupting!
• Bright orange sparks fly out from the top of the “volcano.”
• Instead of orange powder, a fluffy, dark green-black powder comes out and piles up around the base of the volcano.
• You can see smoke or steam rising from the reaction.
• The reaction continues for a while, making more and more of the green-black powder, and the “volcano” keeps growing taller.
• The entire process produces heat and light.

Chemistry of Changes Class 8 Notes

Class 8 Basic Science Chemistry of Changes Notes Kerala Syllabus

• Matter is anything that has mass and occupies space (volume).
• Properties of Matter:
  Mass: All matter has mass, (e.g., a big stone has more mass than a small one; air also has mass).
• Volume: All matter occupies space, (e.g., a stone displaces water; air inside a glass prevents water from entering).
• Matter mainly exists in three states:
  Solid:
  • Definite shape and definite volume.
  • Particles are very close and tightly packed.
  • Strong force of attraction between particles.
  • Particles vibrate in fixed positions (very low speed of movement).
  • Very low energy of particles.
    Liquid:
  • Definite volume but no definite shape (takes the shape of the container).
  • Particles are close but can move past each other.
  • Less attraction between particles than solids.
  • Higher speed of movement than solids.
  • Higher energy of particles than solids.
    Gas:
  • No definite shape and no definite volume (fills the entire container).
  • Particles are very far apart.
  • Very weak force of attraction between particles.
  • Particles move very fast and randomly (very high speed of movement).
  • Very high energy of particles.
• Changes in Matter
  • When matter changes from one state to another (e.g., solid to liquid, liquid to gas), it’s called a change of state
  • Role of Heat Energy:
    Heating (Absorption of Heat): Increases particle distance, speed, and energy.
    (Solid → Liquid → Gas).
    Cooling (Liberation/Release of Heat): Decreases particle distance, speed, and energy.
    (Gas → Liquid → Solid).
  • Sublimation: Some solids (like camphor, naphthalene) directly change into gas upon heating, without becoming liquid.
• Different Types of Changes
  Changes can be classified in various ways:
  • Physical Change: A change in appearance or form, but no new substance is formed. It’s often reversible. Examples: Melting wax, forming ice, dissolving salt in water, and melting ice.
  • Chemical Change: A process where new substances are formed with different properties. It’s usually irreversible. Examples: Burning paper/caridle/firewood, curdling of milk, rusting of iron,
  • Slow Changes and Fast Changes:
    • Slow: Rusting of iron, ripening of fruit, germination of seeds, milk turning to curd.
    • Fast: Burning petrol, explosion of firecrackers, lighting a gas stove.
• Natural Changes and Man-made Changes:
  • Natural: Ripening of bananas, rusting.
  • Man-made: Sculpting stone, burning fuel
• All chemical changes involve an energy change. Energy is either absorbed or released.
• Thermochemical Reactions: Reactions primarily involving heat energy.
  • Exothermic Reactions: Release heat energy (e.g., burning fuels, quicklime reacting With water, magnesium reacting with HCl). You’ll feel the container get warm/hot.
  • Endothermic Reactions: Absorb heat energy (e.g., decomposition of potassium permanga-nate by heating, ammonium chloride reacting with barium hydroxide). You’ll feel the container get cold.
• Photochemical Reactions: Reactions initiated or driven by light energy.
  • Absorb Light: Photosynthesis (plants convert CO2 and water using sunlight), decomposition of silver chloride (turns black in light).
  • Release Light: Bioluminescence (fireflies glowing), Chemiluminescence (glow sticks), combustion
    reactions (burning wood, candles).
  • Storing light-sensitive medicines/chemicals in brown bottles to prevent decomposition by light.
• Electrochemical Reactions: Reactions involving the conversion between chemical energy and electrical energy.
  • Release Electricity (Chemical to Electrical): Batteries/cells (e.g., dry cell, lemon battery).
  • Absorb Electricity (Electrical to Chemical): Electrolysis (breaking down substances using electricity, like water into hydrogen and oxygen).
  • Electroplating (coating one metal with another using electricity).
• Electroplating Setup:
  • The Object to be plated is connected to the negative terminal (cathode)
  • The plating metal is Connected to the positive terminal (anode)
  • The electrolyte contains ions of the plating metal.
• Changes in matter, both physical and chemical, significantly influence human life.
  Beneficial Changes:
  • Photosynthesis (produces food and oxygen).
  • Combustion of fuels (provides energy for homes, transport, industries).
  • Production of chemicals (fertilisers, medicines).
  • Biodegradation (waste disposal, environmental cleaning).
  • Electrochemical cells (batteries for devices).
    Harmful Changes
  • Pollution from fuel combustion.
  • Environmental problems from electro-chemical cell disposal.
  • Harmful effects of excessive/unscientific use of pesticides.
  • Pollution from industrial by-products.

INTRODUCTION

Many chemical reactions are happening around us. Changes from one form of matter to another can be categorised into various classifications, including physical and chemical changes, complex and relatively simple changes, natural and man-made changes, and permanent and temporary changes. Chemical changes are those that produce new substances. When chemical changes occur, energy is released or absorbed.

All substances around us exist in various forms of matter. Solid, liquid, and gas are the main states of matter. Matter can be changed from one state to another. This is called a phase transition. All changes occurring in nature have an impact on human life. This unit will cover these topics.

PHYSICAL CHANGE AND CHEMICAL CHANGE
Physical Change: A physical change is a process where the form or appearance of a substance changes, but its chemical composition remains the same. No new substances are formed.

Chemical Change: A chemical change is a process where a substance is converted into a different substance (or substances). This means new molecules are formed.

CHEMICAL CHANGES AND ENERGY CHANGE
EXPERIMENT:
Place some potassium permanganate crystals on a tile and pour glycerin in the middle. Record your observation.
Observation: An Intense reaction is taking place. Gases are being released. After a short while, the mixture ignites.

THERMOCHEMICAL REACTIONS
EXPERIMENT:
Take a piece of magnesium in a test tube and add diluted hydrochloric acid into it.
a. What do you observe?
Answer:
A vigorous reaction takes place with the release of a gas

b. How will you identify the gas formed here?
Answer:
If a burning matchstick is brought near the mouth of the test tube, the gas bums with a pop sound.

c. Which is this gas?
Answer:
When magnesium reacts with diluted hydrochloric acid, hydrogen gas is formed.

d. Touch the bottom of the test tube. What do you feel?
Answer:
As a result of the chemical reaction, heat is also produced along with hydrogen. Therefore, you will feel the bottom of the test tube become warm or hot.

e. What are the reactants of this reaction?
Answer:
Magnesium (Mg) and Hydrochloric Acid (HCl).

f. What are the products?
Answer:
Magnesium Chloride (MgCl₂) and Hydrogen gas (H₂).

Magnesium + Hydrochloric Acid (Dilute) → Magnesium Chloride + Hydrogen + Heat

EXPERIMENT:
Take quicklime in a steel cup. Add some water to it. Touch the cup after some time.
a. What do you feel?
Answer:
The steel cup becomes hot or warm.

b. What is the reason?
Answer:
The chemical reaction between quicklime (calcium oxide) and water releases a significant amount of heat. This type of reaction is called an exothermic reaction.
Quicklime + water → slaked lime + Heat

If heat is liberated as a result of a chemical reaction, such reactions are known as exothermic reactions.

EXPERIMENT:
Take some potassium permanganate in a test tube and heat it. Hold a burning incense stick near the mouth of this test tube.

a. What do you observe?
Answer:
An Incense stick will flare up or glow more brightly

b. Which gas helps flaring the incense stick?
Answer:
Oxygen

c. Which form of energy is used to decompose potassium permanganate?
Answer:
Heat energy (or Thermal energy)

EXPERIMENT:
Take some ammonium chloride in a watch glass, add some barium hydroxide into it and mix well with a glass rod. Touch the bottom of the watch glass.

a. What do you feel?
Answer:
The bottom of the watch glass becomes cold.

b. Is heat energy absorbed or liberated here?
Answer:
Heat energy is absorbed here.

If heat energy is absorbed during a chemical reaction, it is known as an endothermic reaction.

PHOTOCHEMICAL REACTIONS
The reactions in which light energy is absorbed or liberated are known as photochemical reactions.

EXPERIMENT
Take some silver nitrate solution in a watch glass and add sodium chloride solution to it. Dip two pieces of cotton in the product formed. Cover one of them with black paper and keep the other one open. Keep them aside for some time.

a. Record your observations.
Answer:
After some time, the piece of cotton dipped in the product (AgCl) and kept open to light will turn grey or blackish.
The piece of cotton dipped in the product (AgCl) and covered with black paper will remain white or show very little change in colour.

b. Which form of energy is responsible for the colour change
Answer:
Light energy.

c. Explain why a piece of cotton dipped in silver chloride turns black when exposed to light but remains white when covered.
Answer:
The reactants, silver nitrate and sodium chloride, react to form silver chloride. This silver chloride decomposes by absorbing light to form silver. This is the reason for the blackening of the cotton, which is kept open.
Silver nitrate + Sodium chloride → Silver chloride + Sodium nitrate
Silver chloride → Silver + chlorine

BATTERY MADE OF LEMONS
Arrange zinc and copper nails on lemons as shown in the figure. Connect them with copper wire. Then connect this arrangement to an LED.

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 5 Decimal Forms Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 5 Solutions Decimal Forms

Class 6 Kerala Syllabus Maths Solutions Chapter 5 Decimal Forms Questions and Answers

Decimal Forms Class 6 Questions and Answers Kerala Syllabus

Decimal Places (Page No. 69)

Question 1.
Split the numbers below according to place value:
(i) 4.5
(ii) 4.57
(iii) 4.572
(iv) 45.72
(v) 457.2
Answer:

Textbook Page No. 71

Question 1.
Now try to write 4 kilograms and 55 grams as kilograms in decimal form.
Answer:
4 kilograms 55 grams
55 grams means \(\frac {55}{1000}\) kilograms.
So, 4 kilograms 55 grams = 4\(\frac {55}{1000}\) kilograms
Splitting 4\(\frac {55}{1000}\) according to place value

So we can write the decimal form of 4\(\frac {55}{1000}\) as 4.055

Question 2.
Convert the measures below into the measures specified, using fractions and decimal forms.

Answer:

Decimals and Fractions (Page No. 74)

Question 1.
The decimal form of some numbers is given below. Write each of them as a fraction with a denominator of 10, 100, or 1000.
(i) 3.7
(ii) 3.07
(iii) 30.7
(iv) 3.72
(v) 37.2
(vi) 3.072
(vii) 30.72
Answer:
(i) \(\frac {37}{10}\)
(ii) \(\frac {307}{100}\)
(iii) \(\frac {307}{10}\)
(iv) \(\frac {372}{100}\)
(v) \(\frac {372}{10}\)
(vi) \(\frac {3072}{1000}\)
(vii) \(\frac {3072}{100}\)

Question 2.
Write the decimal form of the fractions given below.
(i) \(\frac {51}{100}\)
(ii) \(\frac {513}{10}\)
(iii) \(\frac {513}{100}\)
(iv) \(\frac {513}{1000}\)
(v) \(\frac {5130}{1000}\)
Answer:
(i) 5.1
(ii) 51.3
(iii) 5.13
(iv) 0.513
(v) 5.13

Addition and Subtraction (Page No. 79)

Question 1.
Anu made an 8.5 metre long festoon and Sarah made a 7.8 metre long one to decorate their classroom for the school anniversary. What is the total length of the festoon they made?
Answer:
Length of the festoon Anu made = 8.5 metres
Length of the Festoon Sarah made = 7.8 metres
Removing the measures and converting into a fraction
8.5 = \(\frac {85}{10}\)
7.8 = \(\frac {78}{10}\)
Adding the fraction
\(\frac{85}{10}+\frac{78}{10}=\frac{163}{10}\)
Converting to decimals
\(\frac {163}{10}\) = 16.3
Total length of the festoon = 16.3 metres

Question 2.
Amal needs 2.25 metres of cloth and Sagar, 1.85 metres for a school uniform. How many metres of cloth in all?
Answer:
Length of the cloth Amal needs for the school uniform = 2.25 metres
Length of the cloth Sagar needs for the school uniform = 1.85 metres
Removing measures and converting to fractions
2.25 = \(\frac {225}{100}\)
1.85 = \(\frac {185}{100}\)
Adding these fractions
\(\frac{225}{100}+\frac{185}{100}=\frac{200+100+25+85}{100}=\frac{410}{100}\)
Converting this to decimals \(\frac {410}{100}\) = 4.1
Total length of cloth in all is 4.1 metres.

Question 3.
A tin weighs 2.85 kilograms, and it is filled with 12.5 kilograms of rice. What is the total weight?
Answer:
Weight of the tin = 2.85 kilograms
Amount of rice filled in the tin = 12.5 kilograms
Removing measures and converting to fractions
2.85 = \(\frac {285}{100}\)
12.5 = \(\frac {125}{10}\)
Adding these fractions
\(\frac{285}{100}+\frac{125}{10}\)
To add these changing \(\frac {125}{10}\) to a form with denominator
\(\frac{125 \times 10}{10 \times 10}=\frac{1250}{100}\)
Adding these fractions,
\(\frac{1250}{100}+\frac{285}{100}\)
1000 + 200 + 250 + 85 = 1200 + 335 = 1535
\(\frac{1250}{100}+\frac{285}{100}=\frac{1535}{100}\)
Converting the fractions to decimals
\(\frac {1535}{100}\) = 15.35
Total weight = 15.35 kilograms

Question 4.
Bakul walks 2.25 kilometres in the morning and 1.5 kilometres in the evening every day. What is the total distance she walks each day?
Answer:
Distance Bakul walks in the morning = 2.25 kilometres
Distance Bakul walks in the evening = 1.5 kilometres
Total distance she walks each day = 2.25 kilometres + 1.5 kilometres
Removing measures and converting to fractions
2.25 = \(\frac {225}{100}\)
1.5 = \(\frac {15}{10}\)
To add these, change \(\frac {15}{10}\) to a form with denominator 100
\(\frac{15 \times 10}{10 \times 10}=\frac{150}{100}\)
Adding the fractions
\(\frac{225}{100}+\frac{150}{100}\)
200 + 100 + 25 + 50 = 375
\(\frac{225}{100}+\frac{150}{100}=\frac{375}{100}\)
Converting the fractions to decimals
\(\frac {375}{100}\) = 3.75
Total distance she walks each day = 3.75 kilometres

Question 5.
Two small bottles contain 0.850 litres and 0.375 litres of honey. If both the bottles are emptied into a large bottle, how much honey does it contain?
Answer:
Amount of honey in the first bottle = 0.850 litres
Amount of honey in the second bottle = 0.375 litres
Amount of honey in the large bottle = 0.850 litre + 0.375 litre
Removing measures and making into fractions
0.375 = \(\frac {375}{1000}\)
0.850 = \(\frac {850}{1000}\)
Adding these \(\frac{375}{1000}+\frac{850}{1000}\)
375 + 850 = 300 + 800 + 75 + 50 = 1225
\(\frac{375}{1000}+\frac{850}{1000}=\frac{1225}{1000}\)
Converting the fractions into decimals
\(\frac {1225}{1000}\) = 1.225
Amount of honey in the large bottle = 1.225 litres.

Textbook Page No. 82

Question 1.
From a rod 14.7 metres long, a piece 7.75 metres long is cut off. What is the length of the remaining piece?
Answer:
Length of the long rod = 14.7 metres
Length of the piece cut off from the long rod = 7.75 metres
Length of the remaining piece = 14.7 metre – 7.75 metres
Changing into fractions
14.7 = \(\frac {147}{10}\)
7.75 = \(\frac {775}{100}\)
To subtract change \(\frac {147}{10}\) to a form with denominator 100
\(\frac{147 \times 10}{10 \times 10}=\frac{1470}{100}\)
Subtracting \(\frac{1470}{100}-\frac{775}{100}=\frac{695}{100}\)
Changing back to decimals
\(\frac {695}{100}\) = 6.95
Length of the remaining piece = 6.95 metres.

Question 2.
There were 38.7 kilograms of rice in a sack, and 12.350 kilograms of this were used up. How much rice remains in the sack?
Answer:
Amount of rice in the sack = 38.7 kilograms
Amount of rice used up = 12.350 kilograms
Amount of rice remaining in the sack = 38.7 kilograms – 12.350 kilograms
38.7 = \(\frac {387}{10}\)
12.350 = \(\frac {12350}{1000}\)
\(\frac{387}{10}-\frac{12350}{1000}\)
\(\frac{387 \times 100}{10 \times 100}=\frac{38700}{1000}\)
Subtracting \(\frac{38700}{1000}-\frac{12350}{1000}=\frac{26350}{1000}\)
Changing back to decimals
\(\frac {26350}{1000}\) = 26.35
Amount of rice remaining in the sack = 26.35 kilograms

Question 3.
The perimeter of a rectangle is 24 centimetres and the length of one side is 6.4 centimetres. What is the length of the other side?
Answer:
Perimeter of rectangle = 2(length + breadth) = 24 centimetres
Length of one side = 6.4 centimetres
Length of the other side = 12 – 6.4 = 5.6 centimetres

Question 4.
There were 2.50 litres of oil in a bottle, and 0.475 litres of this were used for cooking. How much oil is left in the bottle?
Answer:
Amount of oil in the bottle = 2.50 litres
Amount of oil used for cooking = 0.475 litres
Amount of oil left in the bottle = 2.50 litres – 0.475 litres
2.50 = \(\frac {250}{100}\)
0.475 = \(\frac {475}{1000}\)
\(\frac{250}{100}-\frac{475}{1000}\)
\(\frac{250 \times 10}{100 \times 10}=\frac{2500}{1000}\)
Subtracting \(\frac{2500}{1000}-\frac{475}{1000}=\frac{2025}{1000}\)
Amount of oil remaining = 2.025 litres

Question 5.
What number must we add to 14.32 to get 16.43?
Answer:
Number to be added to 14.32 to get 16.43 = 16.43 – 14.32
16.43 = \(\frac {1643}{100}\)
14.32 = \(\frac {1432}{100}\)
Subtracting \(\frac{1643}{100}-\frac{1432}{100}=\frac{211}{100}\)
Converting back to decimals
\(\frac {211}{100}\) = 2.11

Class 6 Maths Chapter 5 Kerala Syllabus Decimal Forms Questions and Answers

Class 6 Maths Decimal Forms Questions and Answers

Question 1.
Split the numbers below according to place value.
(i) 3.6
(ii) 3.64
(iii) 3.641
(iv) 36.41
(v) 364.1
Answer:

Question 2.
Convert the following measures into the specified forms, using both fractions and decimal forms.

Answer:

Question 3.
The decimal form of some numbers is given below. Write each of them as a fraction with a denominator of 10, 100, or 1000.
(i) 4.2
(ii) 4.02
(iii) 40.2
(iv) 4.25
(v) 42.5
(vi) 4.025
(vii) 40.25
Answer:
(i) \(\frac {42}{10}\)
(ii) \(\frac {402}{100}\)
(iii) \(\frac {402}{10}\)
(iv) \(\frac {425}{100}\)
(v) \(\frac {4025}{1000}\)
(vi) \(\frac {4025}{100}\)

Question 4.
Write the decimal form of the fractions given below.
(i) \(\frac {9}{10}\)
(ii) \(\frac {47}{100}\)
(iii) \(\frac {381}{1000}\)
(iv) \(\frac {15}{10}\)
(v) \(\frac {245}{100}\)
(vi) \(\frac {7}{100}\)
(vii) \(\frac {82}{1000}\)
(viii) \(\frac {3456}{1000}\)
Answer:
(i) 0.9
(ii) 0.47
(iii) 0.381
(iv) 1.5
(v) 2.45
(vi) 0.07
(vii) 0.082
(viii) 3.456

Question 5.
John ran a distance of 4.25 kilometres and then walked another 2.5 kilometres. What is the total distance he covered?
Answer:
Distance John ran = 4.25 kilometres
Distance he walked = 2.5 kilometres
Total distance he covered = 4.25 kilometres + 2.5 kilometres
Converting into fractions
4.25 = \(\frac {425}{100}\)
25 = \(\frac{25}{10}=\frac{25 \times 10}{10 \times 10}=\frac{250}{100}\)
Adding \(\frac{425}{100}+\frac{250}{100}=\frac{675}{100}\)
Converting to decimals
\(\frac {675}{100}\) = 6.75
Total distance John covered = 6.75 kilometres

Question 6.
A baker has two bags of flour, one with 1.75 kilograms of flour and the other with 2.5 kilograms. If the baker combines all the flour into a single container, what is the total weight of the flour in the container?
Answer:
Total weight of the flour in the container = Amount of flour in Bag 1 + Amount of flour in Bag 2 = 1.75 kilograms + 2.5 kilograms
1.75 = \(\frac {175}{100}\)
2.5 = \(\frac{25}{10}=\frac{25 \times 10}{10 \times 10}=\frac{250}{100}\)
Adding \(\frac{175}{100}+\frac{250}{100}=\frac{425}{100}\)
Converting into decimals
\(\frac {425}{100}\) = 4.25
Total weight of the flour in the container = 4.25 kilograms

Question 7.
A ribbon was 15.8 metres long. If a piece measuring 4.25 metres was cut from it, how much ribbon is left?
Answer:
Length of the ribbon = 15.8 metres
Length of the piece cut off from the ribbon = 4.25 metres
Length of the ribbon left = 15.8 metres – 4.25 metres
15.8 = \(\frac{158}{10}=\frac{158 \times 10}{10 \times 10}=\frac{1580}{100}\)
4.25 = \(\frac {425}{100}\)
Subtracting \(\frac{1580}{100}-\frac{425}{100}=\frac{1155}{100}\)
Converting back to decimals
\(\frac {1155}{100}\) = 11.55
Length of the ribbon left = 11.55 metres

Question 8.
A water tank holds 50.5 litres of water. If 25.5 litres are used for gardening, how much water is left in the tank?
Answer:
Amount of water left in the tank = Amount of water that the water tank can hold – Amount of water used for gardening
50.5 = \(\frac {505}{10}\)
255 = \(\frac {255}{10}\)
Subtracting \(\frac{505}{10}-\frac{255}{10}=\frac{250}{10}\)
Converting to decimals
\(\frac {250}{10}\) = 2.5

Question 9.
A carpenter uses a 2.75-metre board and a 1.5-metre board for a project. What is the total length of the wood used?
Answer:
Total length of the wood used = 2.75 metre + 1.5 metre
2.75 = \(\frac {275}{100}\)
1.5 = \(\frac{15}{10}=\frac{15 \times 10}{10 \times 10}=\frac{150}{100}\)
Adding \(\frac{275}{100}+\frac{150}{100}=\frac{425}{100}\)
Converting to decimals
\(\frac {425}{100}\) = 4.25
Total length of the wood used = 4.25 metres

Question 10.
A plant is 15.6 centimetres tall. How many more centimetres must it grow to reach a height of 20.1 centimetres?
Answer:
Height of the plant = 15.6 centimetres
Target height = 20.1 centimetres
Height the plant must grow = 20.1 centimetres – 15.6 centimetres
20.1 = \(\frac {201}{10}\)
15.6 = \(\frac {156}{10}\)
Subtracting \(\frac{201}{10}-\frac{156}{10}=\frac{45}{10}\)
Converting to decimals
\(\frac {45}{10}\) = 4.5
The plant must grow 4.5 centimetres to reach the target height.

Class 6 Maths Chapter 5 Notes Kerala Syllabus Decimal Forms

→ In the decimal form of a number, the dot (decimal point) separates the whole number part and the fractional part.

→ Digits to the left of the decimal point represent ones, tens, hundreds, and so on;

→ The digits to the right represent tenths, hundredths, thousandths, and so on.

→ Decimals allow us to represent quantities that are not whole numbers with greater precision, making them essential for measurements, money, and other real-world applications.

→ To convert a measurement from centimetres to metres in decimal form, divide the number of centimetres by 100. This is equivalent to moving the decimal point two places to the left.

→ If you have a combination of metres and centimetres, first convert the centimetres to metres as a decimal and then add them to the whole number of metres.

→ To convert a measurement from centimetres to millimetres, multiply the number by 10.

→ To convert from millimetres to centimetres, divide the number by 10.

→ To add decimal numbers representing measurements (like 4.3 cm and 2.5 cm), align the decimal points and add them directly.

→ Alternatively, you can convert the measurements to a smaller unit (like millimetres) and complete the addition, then convert the result back to the original unit.

→ To subtract a decimal from another (like subtracting 3.2 cm from 8.5 cm), you can convert both decimals to fractions with a common denominator, subtract them, and then convert the result back into a decimal.

This chapter comprehensively covers the representation and manipulation of numbers beyond whole units. Key topics include understanding decimal places to denote fractional parts, establishing the crucial relationship and conversion techniques between decimals and fractions, and mastering the fundamental operations of addition and subtraction of decimal numbers to build foundational arithmetic skills.

Decimal Places
The length of a pencil can be said in different ways:

• 5 centimetres 7 millimetres
• 5\(\frac {7}{10}\) centimetres
• 5.7 centimetres

We can write other measures also like this:
5\(\frac {7}{10}\) litres = 5.7 litres
5\(\frac {7}{10}\) kilograms = 5.7 kilograms

We can drop all references to measures and simply say that 5.7 is the decimal form of the number 5\(\frac {7}{10}\).
5\(\frac {7}{10}\) = 5.7
Similarly, 4.29 is the decimal form of 4\(\frac {29}{100}\)
4\(\frac {29}{100}\) = 4.29

We write natural numbers using ones, tens, hundreds, and so on.
For example: 247 = 2 hundreds + 4 tens + 7 ones
Splitting 247.3
Split it as the sum of a whole number and a fraction.
247.3 = 247\(\frac {3}{10}\) = 247 + \(\frac {3}{10}\)
First, we split 247.3 as the sum of a whole number and a fraction, as
247.3 = 247\(\frac {3}{10}\) = 247 + \(\frac {3}{10}\)
The \(\frac {3}{10}\) here can be written as
\(\frac{3}{10}=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}\)
That is, 3 tenths. So, we can write 247.3 in terms of hundreds, tens, ones, and tenths:
247.3 = 2 hundreds + 4 tens + 7 ones + 3 tenths

Splitting 247.39
First, we write it as
247.39 = 247\(\frac {39}{100}\)= 247 + \(\frac {39}{100}\)
Then, we can split \(\frac {39}{100}\) as
\(\frac{39}{100}=\frac{30+9}{100}=\frac{30}{100}+\frac{9}{100}=\frac{3}{10}+\frac{9}{100}\)
The \(\frac {3}{10}\) here is 3 tenths: and \(\frac {9}{100}\) is 9 hundredths.
So 247.39 = 2 hundreds + 4 tens + 7 ones + 3 tenths + 9 hundredths

In the decimal form of a number, the dot separates the whole number part and the fractional part. Digits to the left of the dot show the multiples of ones, tens, hundreds, and so on; the digits to the right show the multiples of tenths, hundredths, thousandths, and so on.
For example, the two numbers used in the above examples can be split according to place value like this:

Question 1.
What is the decimal form of 23 metres 40 centimetres?
Answer:
Method 1
23 metres 40 centimetres = 23\(\frac {40}{100}\) metres = 23.40 metres
Taking only the numbers, we have
23\(\frac {40}{100}\) = 23.40
We can write the \(\frac {40}{100}\) here as
\(\frac{40}{100}=\frac{4}{10}\)
So, we get 23\(\frac {40}{100}\) = 23\(\frac {4}{10}\) = 23.4
This means 23.40 = 23.4

Method 2
Using place value

Thus, we can write 23 metres and 40 centimetres in two different ways:
23 metres 40 centimetres = 23.40 metres
23 metres 40 centimetres = 23.4 metres

Question 2.
What is the decimal form of 23 metres 4 centimetres?
Answer:
4 centimetres = \(\frac {4}{100}\) metre.
23 metres 4 centimetres =23\(\frac {4}{100}\) metres
Split 23\(\frac {4}{100}\) according to place value:
23\(\frac {4}{100}\) = 2 tens + 3 ones + 4 hundredths

The decimal form of 23\(\frac {4}{100}\) = 23.04

Question 3.
What is the decimal form of 23 metres and 4 millimetres?
Answer:
4 millimetres means \(\frac {4}{1000}\) metres.
So 23 metres 4 millimetres = 23\(\frac {4}{1000}\) metres
Split 23\(\frac {4}{1000}\) according to place value.

So, we can write the decimal form of 23\(\frac {4}{1000}\) as
23\(\frac {4}{1000}\) = 23.004

Converting Centimetres to a Decimal Form of Metres
To convert centimetres to a decimal form of metres, first, write the number of metres as the whole number part of your decimal.
Next, express the number of centimetres as a fraction of a metre. Since there are 100 centimetres in 1 metre, the denominator of your fraction will be 100. For example, 40 centimetres would be written as \(\frac {40}{100}\) metres.
Combine the whole number and the fraction to create a mixed number, such as 23\(\frac {40}{100}\).
To get the decimal form, divide the numerator of your fraction by 100. This is the same as moving the decimal point two places to the left. So, \(\frac {40}{100}\) becomes 0.40.
Finally, add the decimal to the whole number.
For example, 23 + 0.40 = 23.40.
The result is the length in meters.

Decimals and Fractions
Conversion of decimals into fractions
Start with the decimal number you want to convert.

Example: 7.3 centimetres
Write this in millimetres
7 centimetres = 70 millimetres
To convert \(\frac {3}{10}\) centimetres into millimetres:
\(\frac {3}{10}\) is three \(\frac {1}{10}\)
\(\frac {3}{10}\) centimetres = 3 millimetres
7.3 centimetres = 70 millimetres + 3 millimetres

Converting 73 millimetres into centimetres:
Divide it by 10
73 millimetres = \(\frac {73}{10}\) centimetres
Removing measure and writing as just numbers 7.3 = \(\frac {73}{10}\)

Question 4.
How do we write 7.31 metres as a fraction?
Answer:
Write it as a whole number and a fraction
7.31 metres = 7\(\frac {31}{100}\) metres
7 metres = 700 centimetres
\(\frac {31}{100}\) metres = 31 centimetres
7.31 metres = 700 centimetres + 31 centimetres = 731 centimeters
Converting this into metres
731 centimetres = \(\frac {731}{100}\) metres
7.31 metres = \(\frac {731}{100}\) metres

Question 5.
Convert 7.319 litres as a fraction.
Answer:
7.319 litres = 7\(\frac {319}{1000}\) litres
7 litres = 7000 millilitres
\(\frac {319}{1000}\) litres = 319 millilitres
7.319 litres = 7319 millilitres
Converting back to litres
7319 millilitres = \(\frac {7319}{1000}\) litres
7.319 litres = \(\frac {7319}{1000}\) litres

Converting 12.03 to a Fraction

• Step 1: Count the digits after the decimal point in 12.03. There are two digits (0 and 3).
• Step 2: The denominator is 100 because there are two digits after the decimal.
• Step 3: Remove the decimal point from 12.03 to get the numerator, which is 1203.
• Step 4: The final fraction is \(\frac {1203}{100}\).

Question 6.
What is the decimal form of \(\frac {1203}{1000}\)?
Answer:
Looking at the denominator, we can say there will be three digits after the decimal point.
\(\frac {1203}{1000}\) = 1.203

Addition and Subtraction

Addition of Decimal Numbers
A line 4.3 centimetres long was drawn and then extended by another 2.5 centimetres:

To find the total length of the line we have to add 4.3 centimetres and 2.5 centimetres

Method 1
Convert these to centimetres and millimetres.
4.3 centimetres = 4 centimetres 3 millimetres
2.5 centimetres = 2 centimetres 5 millimetres
And add the centimetres and millimetres separately.
4 centimetres + 2 centimetres = 6 centimetres
3 millimetres + 5 millimetres = 8 millimetres
The length of the line is 6 centimetres, 8 millimetres
Convert back to centimetres.
6 centimetres 8 millimetres = \(\frac {68}{10}\) centimetres = 6.8 centimetres

Method 2
Write the lengths in millimetres.
4.3 centimetres = 43 millimetres
2.5 centimetres = 25 millimetres
Add 43 and 25
43 + 25 = 40 + 20 + 3 + 5 = 68
Thus, the length of the line is 68 millimetres
Write as centimetres in decimal form.
68 millimetres = 6 centimetres 8 millimetres = 6.8 centimetres

Method 3
Remove the measures and write the numbers as fractions.
4.3 = \(\frac {43}{10}\)
2.5 = \(\frac {25}{10}\)
And these fractions we can add like this:
\(\frac{43}{10}+\frac{25}{10}=\frac{43+25}{10}=\frac{68}{10}\)
Write the fraction as a decimal number
\(\frac {68}{10}\) = 6.8
The length of the line is 6.8 centimetres.

Question 7.
Add 4.3 centimetres and 2.8 centimetres.
Answer:
Changing the lengths into millimetres
4.3 centimetres = 43 millimetres
2.8 centimetres = 28 millimetres
Adding 43 and 28
43 + 28 = 40 + 20 + 3 + 8 = 60 + 11 = 71
Thus, the length of this line is 71 millimetres.
Write in centimetres as a decimal:
71 millimetres = 7 centimetres 1 millimetre = 7.1 centimetres
Convert the numbers to fractions
4.3 = \(\frac {43}{10}\)
2.8 = \(\frac {28}{10}\)
Add the fractions:
\(\frac{43}{10}+\frac{28}{10}=\frac{43+28}{10}=\frac{71}{10}\)
Convert the fraction back to the decimal form.
\(\frac {71}{10}\) = 7.1
The length of the line is 7.1 centimetres.

Question 8.
A jar contains 3.5 litres of oil, and 6.25 litres more is poured into it. How much oil does the jar contain now?
Answer:
Convert just the numbers to fractions:
3.5 = \(\frac {35}{10}\)
6.25 = \(\frac {625}{100}\)
We can write \(\frac {35}{10}\) also as a fraction with denominator 100:
\(\frac{35}{10}=\frac{35 \times 10}{10 \times 10}\) = \(\frac {350}{100}\)
Now we can add like this:
\(\frac{35}{10}+\frac{625}{100}=\frac{350}{100}+\frac{625}{100}=\frac{300+600+50+25}{100}\)
= \(\frac{350+625}{100}\)
= \(\frac {975}{100}\)
Amount of oil the jar contains = 9.75 litres

Question 9.
A person bought 2.5 kilograms of rice and 3.125 kilograms of vegetables. What is the total weight?
Answer:
Converting into fractions
2.5 = \(\frac {25}{10}\)
3.125 = \(\frac {3125}{1000}\)
To add these, we change \(\frac {25}{10}\) to a form with a denominator of 1000.
\(\frac{25}{10}=\frac{25 \times 100}{10 \times 100}\) = \(\frac {2500}{1000}\)
Now, we add the fractions:
\(\frac{25}{10}+\frac{3125}{1000}=\frac{2500}{1000}+\frac{3125}{1000}\) = \(\frac{2500+3125}{1000}\)
One way of adding 2500 and 3125 is this:
2500 + 3125 = 2000 + 3000 + 500 + 125 = 5000 + 625 = 5625
So, we can continue our addition of fractions:
\(\frac{25}{10}+\frac{3125}{1000}=\frac{2500+3125}{1000}\) = \(\frac {5625}{1000}\)
Converting this to decimals:
\(\frac {5625}{1000}\) = 5.625
Thus, the total weight is 5.625 kilograms.

Subtraction of Decimal Numbers
Example: From an 8.5 centimetres long eerkkil, a 3.2 centimetres long piece is broken off. What is the length of the remaining piece?
Thinking in terms of numbers alone, what we need is to subtract 3.2 from 8.5.
We change the numbers to fractions.
8.5 = \(\frac {85}{10}\)
3.2 = \(\frac {32}{10}\)
Now we can subtract:
\(\frac{85}{10}-\frac{32}{10}=\frac{85-32}{10}\)
One way to subtract 32 from 85 is this:
85 – 32 = (80 – 30) + (5 – 2) = 50 + 3 = 53
So we get \(\frac{85}{10}-\frac{32}{10}=\frac{53}{10}\)
Finally, we switch back to decimals:
\(\frac {53}{10}\) = 5.3
Thus, the length of the remaining piece of eerkkil is 5.3 centimetres.

Question 10.
From an 8.5 centimetres long eerkkil, a 3.7 centimetres long piece is broken off. What is the length of the remaining piece?
Answer:
We start as before by converting the decimals to fractions:
8.5 = \(\frac {85}{10}\)
3.7 = \(\frac {37}{10}\)
And then subtract
\(\frac{85}{10}-\frac{37}{10}=\frac{85-37}{10}\)
We have seen in earlier classes that subtraction like 85 different ways.
For example,
85 – 37 = (85 – 35) – 2 = 50 – 2 = 48
85 – 37 = (87 – 37) – 2 = 50 – 2 = 48
85 – 37 = (85 – 40) + 3 = 45 + 3 = 48
Anyway, we find
\(\frac{85}{10}-\frac{37}{10}=\frac{48}{10}\)
Changing back to decimals,
\(\frac {48}{10}\) = 4.8
The remaining piece of eerkkil is 4.8 centimetres long.

Question 11.
There are 15 kilograms of rice in a sack. 4.25 kilograms from this are put in a bag. How much rice remains in the sack?
Answer:
Thinking just in terms of numbers, what we have to do is subtract 4.25 from 15.
Write 4.25 as a fraction:
4.25 = \(\frac {425}{100}\)
Write 15 as a fraction with a denominator of 100.
15 = \(\frac{15}{1}=\frac{15 \times 100}{1 \times 100}=\frac{1500}{100}\)
Subtracting \(\frac{1500}{100}-\frac{425}{100}=\frac{1500-425}{100}\)
We can do 1500 – 425 in several ways:
1500 – 425 = 1000 + 500 – 425 = 1000 + 75 = 1075
1500 – 425 – 1425 – 425 + 75 = 1000 + 75 = 1075
1500 – 425 = 1500 – 500 + 75 = 1000 + 75 = 1075
Thus, we have:
\(\frac{1500}{100}-\frac{425}{100}=\frac{1075}{100}\)
Changing back to decimals:
\(\frac {1075}{100}\) = 10.75
So, there are 10.75 kilograms of rice still in the sack.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 1 Measurement and Units Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 1 Measurement and Units Question Answer Notes

Class 8 Basic Science Chapter 1 Notes Kerala Syllabus Measurement and Units Question Answer

Measurement and Units Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
Identify the odd one out in each group and explain common features of the others.
I a) Kilogram b) Kilometre c) Second d) Mole
II a) Time b) Area c) Mass d) Electric current
III a) Metre b) Kilogram c) Second d) Degree Celsius
Answer:
I. b) Kilometre
Kilometre is a unit of length. Others are SI unit of mass, time and amount of substance respectively.

II. b)Area
Area is a derived quantity. Others are fundamental quantities.

III. d) Degree Celsius
This is a unit of temperature. Others are SI unit of length, mass and time respectively.

Question 2.
Different units of length are given below. Fill in the table below.

Answer:

Unit	Relationship with metre
Kilometre	1 km = 1000 metre
Millimetre	1 m = 1000 millimetre
Centimetre	100 cm = 1m

Question 3.
Convert the following measurements to SI units without changing their values.
a) 2000 g
b) 1 h
c) 1.5 km
d) 200 cm
Answer:
a) 2000 g = \(\frac{2000}{1000}\) = 2 kg
b) 1 h = 60 × 60 = 3600 s
c) 1.5 km= 1.5 × 1000 = 1500 m
d) 200cm = \(\frac{200}{100}\) = 2m

Question 4.
Different units of mass are given below. Arrange them in the ascending order of their values.
a) Kilogram
b) Milligram
c) Quinta
d) Gram
Answer:
b) Milligram < d) Gram < a) Kilogram < c) Quintal

Basic Science Class 8 Chapter 1 Question Answer Kerala Syllabus

In our daily life, it is necessary to measure and state the characteristic properties of objects and phenomena. Such measurable quantities are physical quantities.
Question 1.
Observe the following situations in our life. Find the physical quantities in each of them.


Record the quantities you identified, in the table.
Answer:

Situation	Physical quantity
1. Measuring the depth of a pit	Length
2. Measuring the weight of vegetables	Mass
3. Taking measurements by a tailor	Length
4. Using a stopwatch in a race	Time
5. Measuring blood pressure	Pressure
6. Measuring body heat	Temperature

Question 2.
Find and write more physical quantities that you are familiar with.
Answer:

• Electric current
• Amount of substance
• Luminous intensity

All the physical quantities cannot be measured directly. In situations where direct measurement is not possible, we can write them with reference to other physical quantities.

Question 3.
Find out the physical quantities mentioned in the table 1.1 and list them below.
Answer:

1. Length
2. Mass
3. Time
4. Temperature

Question 4.
Look at the pictures. What are the physical quantities in these situations?

Answer:
• Area
• Volume

Question 5.
Record how each of them is found out and complete the table appropriately.

Answer:

Situation	Physical Quantity	Method of Finding
For painting the wall	Area	Area = Length × Width
Measurement of medicine/liquid	Volume	Volume = Area of the measuring jar × Height

a) Which are the quantities used here to find area and volume?
Answer:
Length, width, area and height.

b) All of them are distances between two positions, aren’t they?
Answer:
Yes. All of them are distances between two positions.

The distance between two positions represents a physical quantity called length. We have used the fundamental quantity of length to find the quantities of area and volume. Such quantities that can be found out using fundamental quantities are called derived quantities.

Quantities that can be expressed in terms of fundamental quantities are derived quantities.

See, how the mass is marked on a gas cylinder.

Mass marked on the cylinder = 14.2 kilogram
Here, the physical quantity of mass is indicated using a numerical value i.e., 14.2 (magnitude) and a unit i.e., kilogram.

Question 6.
Similarly, complete the table with the physical quantities shown in the pictures below, along with their numerical values and units.

Answer:

Situation	Physical quantity	Numerical Value	Unit	Mode of marking measurements
Fig. 1.12	Temperature	37.8	Celsius	37.8° C
Fig. 1.13	Height	165	Centimetre	165 cm
Fig. 1.14	Mass	1	Kilogram	1kg

A physical quantity is expressed by a number indicating its value followed by its unit.

Question 7.
Tabulate the measurements from both the activities.

Answer:

Activity	Physical quantity	Reference object used for measurement	Recorded quantity
Measuring the height of the child	Length	Longer stick	2 stick
		Shorter stick	6 stick
Measuring the quantity of water	Volume	Larger glass	5 glass
		Smaller glass	10 glass

Question 8.
Analyze the table. Two different reference objects were used in each case to determine a physical quantity.
a) In both cases, are the measurements obtained the same?
Answer:
No, the measurements obtained are not the same.

b) Why are the measurements not equal?
Answer:
It is because the reference object used for each measurement is different.

c) When everyone uses the same reference object, isn’t measurement the same?
Answer:
When the same reference object is used, the measurement is the same.
When a physical quantity is measured anywhere in the world, the measurement should be the same. For this, everyone should adopt a fixed reference. This is called the unit of a physical quantity.

A unit is a standardised reference accepted universally to measure a physical quantity.

In the past, different units were used for measurement and recording in each region. For example, units like the foot, cubit and hand span were used locally to measure length.

Question 9.
Different units were also used in other countries. What would be the practical problems of using different units in different countries?
Answer:

• Low accuracy
• Difficulty for the people in other regions to analyse measurements.
• Lack of uniformity
• Difficulty with transaction

Today, the unit ’metre’ is used everywhere in the world to measure length.

There are internationally accepted units for all physical quantities. This is called the International System of Units, abbreviated as ‘SI’ units. Measurement using the SI units always has a universal result.

Now we can understand that with the help of SI units which always, has a universal result, parts of vehicles and equipment we use, even if manufactured in different countries can be perfectly assembled in any factory in the world.

Different units are required for the same physical quantity in various contexts. Larger units are used for larger quantities and smaller units for smaller quantities.

Question 10.
Now complete the following relationship given below.
Answer:
1 metre = 100 centimetre
1 centimetre = 10 millimetre
1 metre = 1000 millimetre
There are situations where we have to use smaller measurements.

Question 11.
Pay attention to the notice of a municipality.

Prohibited The sale of plastic bags below 30 micron is prohibited in shops within the limit of the municipality with effect from 30.10.2022.

What is the measurement mentioned in the notice?
Answer:
The measurement mentioned is micron. Micron is the abbreviation of micrometre.

Question 12.
How many micrometres would make one metre?
Answer:
1 metre = 1000000 micrometre

There are also situations where we need units larger than metre.
The abbreviation “km” on the traffic sign stands for kilometre.
1 kilometre = 1000 metre
Are there situations where we need even larger units? Read the following excerpt from a science article.

Scale of the Solar System Astronomical Unit (AU) is the average distance from the Earth to the Sun. It is approximately 150 million kilometre. A light year is the distance light travels in a year in vacuum. Light travels at a speed of approximately 300,000 km/s.

Question 13.
Discuss the situations where the units mentioned in the article are used.
Answer:
Astronomical unit is used to measure the distance between Earth and the Sun, distance from Sun to different planets and distance between planets in our solar system. Light year is used to measure the distance between Earth and stars, distance between galaxies or stars.

Question 14.
Look at the picture of weighing apples in a shop. The weight is measured by placing weight blocks on one side of the scale.

Here, why are weight blocks placed on one side?
Answer:
This is done to ensure that the apples taken have the same mass as weight of blocks.

The amount of matter contained in a substance is its mass.

Question 15.
Examine the picture of the weight blocks shown in figure below. What is written on them?

Answer:
The mass of the weight block is written on them

The unit of mass is the kilogram. Its symbol is ‘kg’.

We need units other than kilogram for mass.

Question 16.
You might have noticed the quantity of toothpaste and tablet printed on their packages. What does it mean?

Answer:
The quantity printed on them indicates their mass.

Milligram and gram are the smaller units of mass. 1 gram = 1000 milligrams

Question 17.
You might have seen trucks carrying load. Which are the larger units commonly used in such situations?
Answer:
Quintal and tonne are the larger units of mass commonly used in such situations.
Identify the relationship between the units of mass and kilogram from the table given below.

Unit	Relation to kilograms
Milligram	1 kilogram = 1000000 milligram
Gram	1 kilogram = 1000 gram
Quintal	1 quintal = 100 kilogram
Tonne	1 tonne = 1000 kilogram

Question 18.
Minute and hour are the other units used to denote time. Identify the relationship between these units and ‘second’.

Answer:

Unit	Relationship with second
Minute	1 minute = 60 second
Hour	1 hour = 3600 second

Question 19.
The figure shows thousand cubes each with sides of 1 cm arranged to form a large cube.

If volume of the large cube is 1 litre. Can you complete writing the relationship between various units based on the figure?
Answer:
1 litre = 1000 cm³
1 litre = 1000 millilitre

Question 20.
Take a cardboard box and calculate its volume. Fill it with sawdust and measure its mass. Then replace it with sand and find its mass. Tabulate the findings.

Answer:

Substance	Mass	Volume	\(\frac{\text { Mass }}{\text { Volume }}\)
Sawdust	30 g	200 cm3	\(\frac{30}{200}\) = 0.15 g/cm3
Sand	320 g	200 cm3	\(\frac{320}{200}\) =1.6 g/cm3

The mass of a substance per unit volume is called its density. Density = \(\frac{\text { Mass }}{\text { Volume }}\)

In the table given, even though the volume of sawdust and sand is the same, see how the mass per unit volume is calculated.

If volume is the same, objects with higher mass will have higher density. In the case of a particular substance, density is a fixed number.

Question 21.
Why is density displayed on the fuel dispenser in a petrol pump?
Answer:
Displaying density on the fuel dispenser in petrol pumps helps to check if the fuel is pure and not mixed with anything. If the fuel is adulterated with some impurities, then the density will change. To ensure that it is not adulterated, the density is displayed.

Question 22.
What are the characteristics of SI units?
Answer:

• They are standardised units.
• They are internationally accepted.
• Units of all other quantities can be expressed in terms of these units.

Question 23.
The table below shows common errors that may occur when writing units. Compare each of these with the correct version and suggest a general rule for each.

Answer:

Unit written incorrectly	General rules
1000 KG/M3 1.5 KG	Use lower case of the English alphabet.
1000 kgs/m3 1.5 kgs	Do not use the plural form for symbols.
1000kg/ m3 1.5kg	While writing units along with a numerical value, there must be a single space between them.
1000 kg/m/m/m	Do not use more than one slash in one derived unit.
1000 kg/ cubic metre 1000 kilogram per m3	Do not mix a symbol of a unit with the name of a unit.
1 kg 500 g	Do not use more than one unit to express a physical quantity.
273 Kelvin	Use only lowercase letters when writing the name of a unit instead of its symbol.

Now let us get familiar with some other rules.

Physical quantity	Correct method	Incorrect method	Rule
Force	N	n	The symbols of the units formed from the names of individuals should be written using uppercase of the English alphabet.
Length	60 cm is the length of the desk.	60 cm. is the length of the desk.	No full stop or comma should be used after the symbol. They can be used at the end of the sentence.
	The length of the desk is 60 cm.	The length of the desk is 60 cm
Energy	N.m Nm	Nm	A full stop/space should be used between the units formed as multiples of units.

Question 24.
What are the rules to be followed internationally when writing units and their symbols?
Answer:

1. Use lower case of the English alphabet to write the symbol of the units.
   e.g. 1000 KG/M³– wrong, 1000 kg/m³ – correct
2. Use only lowercase letters when writing the name of a unit.
   e.g. 273 Kelvin-wrong, 273 kelvin- correct
3. The symbols of the units formed from the names of individuals should be written using uppercase of the English alphabet.
   e.g. The unit of the physical quantity force is newton. This is named after sir Isaac Newton. The symbol is denoted as N.
4. While writing units along with a numerical value, there must be a single space between them, e.g. 1.5kg- wrong, 1.5 kg -correct
5. Do not use the plural form for symbols.
   e.g. 1.5 kgs- wrong, 1.5 kg -correct
6. Do not use more than one slash in one derived unit.
   e.g. 1000 kg/m/m/m -wrong, 1000 kg/m³– correct
7. Do not mix a symbol of a unit with the name of a unit.
   e.g. 1000 kg/ cubic metre-wrong, 1000 kg/m³– correct, 1000 kilogram per cubic metre (correct)
8. Do not use more than one unit to express a physical quantity.
   e.g. 1kg 500 g (wrong) 1.5 kg ( correct)
9. No full stop or comma should be used after the symbol. They can be used at the end of the sentence, e.g. 75 cm is the length of a table, (correct) 75 cm. is the length of a table, (wrong)
10. A full stop/space should be used between the units formed as multiples of units, e.g. Nm – wrong, N.m or N m – correct

Question 25.
What are the instruments used to measure length?
Answer:
Scale, tape.

Question 26.
Look at the picture given.

a) Measure the length of a pen using a scale and write it down.
b) Also, use a measuring tape to determine your height.
c) What is the unit on the scale/tape used?
d) What is the smallest measurement possible using the scale/measuring tape?
Answer:
a) Length of the pen = 14.7 cm
b) Your height = 138 cm
c) Centimetre
d) 0.1 cm

The smallest value that can be measured using an instrument is called its least count.

Question 27.
The least count of a commonly used scale is 0.1 cm . Are there instruments with a least count smaller than this? Find out and write.
Answer:
Vernier caliper is an instrument used to measure the length of rod, diameter of a cylinder or sphere shaped object etc. Least count: 0.01 cm (or 0.1 mm)
Screw gauge is an instrument used to measure the thickness of glass plates and diameter of thin wires. Least count – 0.001 cm (or 0.01 mm).

Question 28.
The figure shows some papers stacked together.

Measure the thickness of the paper stack and write it down.
Answer:
Number of papers in the paper stack = 50
Thickness of the paper stack = 5 cm
Thickness of one paper = \(\frac{\text { Thickness of the paper stack }}{\text { Number of papers }}\) = \(\frac{5}{500}\)
= 0.01 cm = 0.01 mm

Question 29.
What unit is used in the measuring jar?
Answer:
millilitre

Question 30.
What is the least count of the measuring jar?
Answer:
l ml
Initial water level before dipping the stone = 50 ml
Water level after dipping the stone = 78 ml
Volume of the stone = 78 – 50 = 28 ml

Class 8 Basic Science Chapter 1 Question Answer Extended Activities

Question 1.
Identify the different units used in our locality for measuring length and mass in the past.
Answer:
Some of the different units used in our locality for measuring length and mass in the past are tabulated.

Unit of length	Unit of mass
Vaara Muzham Feet Kol Furlong Mile	Chan Kizhi Edangazhi Nazhi Para Padi

Question 2.
Prepare a seminar paper on the rules to be followed when writing ‘units’.
Answer:
Title: Rules to be followed when writing units.

Introduction: In our day to day life and in science, we make use of different units to measure various physical quantities like length, mass, time etc. A physical quantity is expressed by a number indicating its value followed by its unit. It is very important to express units in the right way to avoid confusion and to maintain accuracy in measurements.

Rules to be followed when writing units
1. The symbols of units are normally written using small letters in the English alphabet.
e.g. m (metre), s (second)

2. The symbol of units named after persons should be expressed by capital letters of the English alphabet.
e.g. The unit of the physical quantity electric current is ampere. This is named after Andre-Marie Ampere. The symbol is A.

3. While writing the names of units never use capital letters.
e.g. kelvin (correct) Kelvin (wrong)

4. Never use the plural form for symbols.
e.g. 10 kg (correct) 10 kgs (wrong)

5. Never use full stop or comma after a symbol except at the end of a sentence.
e.g. 75 cm is the length of a table. (correct)
75 cm. is the length of a table, (wrong)

6. While writing derived units a slash (/) is used to denote division. But never use more than one slash in one derived unit.
e.g. m/s² (correct) m/s/s (wrong)

7. When a derived unit is expressed as the product of other units use a dot or a space between them, e.g. N.m or N m

8. Do not mix the name of a unit with the symbol.
e.g.kg/m³ (correct)
kilogram per cubic metre (correct)
kg/cubic metre (wrong)
kilogram per m³ (wrong)
kg per m³ (wrong)
kilogram/m³ (wrong)

9. While writing units along with a numerical value, there must be single space between them, e.g. 273 K (correct), 273K (wrong)

10. Never use more than one unit to express a physical quantity, e.g. 10.25 m (correct) 10 m 25 cm (wrong)

Conclusion
Following these rules will help us to write the units in its clear and correct manner so that it can be understood by all.

Measurement and Units Class 8 Notes

Class 8 Basic Science Measurement and Units Notes Kerala Syllabus

• Fundamental quantities are quantities that exist independently and cannot be expressed in terms of other quantities.
• Quantities that can be expressed in terms of fundamental quantities are derived quantities.
• A physical quantity is expressed by a number indicating its value followed by its unit.
• A unit is a standardised reference accepted universally to measure a physical quantity.
• There are internationally accepted units for all physical quantities. This is called the International System of Units, abbreviated as ‘SI’ units. Measurement using the SI units always has a universal result.
• The SI unit of length is metre. Its symbol is ‘m’. Centimetre, millimetre, kilometre, etc. are the other units of length.
• The amount of matter contained in a substance is its mass. The unit of mass is the kilogram. Its symbol is ‘kg’. Milligram and gram are the smaller units of mass.
• Quintal and tonne are the larger units of mass commonly used.
• The SI unit of time is the second. Its symbol is ‘s’.
• Minute and hour are the other units used to denote time.
• The volume of an object is the amount of space it occupies. The SI unit of volume is cubic metre.
  It’s symbol is m³.
• The mass of a substance per unit volume is called its density. Density = \(\frac{\text { Mass }}{\text { Volume }}\)
• Fundamental units are the units of fundamental quantities.
• Characteristics of SI units
  • They are standardised units.
  • They are internationally accepted.
  • Units of all other quantities can be expressed in terms of these units.
• Derived units are units that can be stated using fundamental units or that depend on fundamental units.
• The smallest value that can be measured using an instrument is called its least count.

INTRODUCTION

In our daily life, we often need to measure physical quantities like length, mass and time. In some situations, it is very important to be accurate in these measurements. Long ago, people had many problems because they did not have accurate ways to measure, and different places used different types of measuring scales. This caused confusion and mistakes. This chapter deals with fundamental quantities and derived quantities, units of physical quantities, fundamental and derived units, rules for writing the units and measuring instruments.

FUNDAMENTAL QUANTITIES AND DERIVED QUANTITIES

There are many physical quantities. Among them length, mass, time, electric current, temperature, amount of substance and luminous intensity are called fundamental quantities. All other quantities can be expressed in terms of these fundamental quantities.

Fundamental quantities are quantities that exist independently and cannot be expressed in terms of other quantities.

UNITS OF PHYSICAL QUANTITIES
Activity

• Mark the height of a child in your class on the wall using a pencil as shown in the figure 1.5. Each one in the class may measure the height using two sticks of different lengths.
• Fill a bucket with water. Measure the water in it with two glasses of different sizes (Figure 1.16).

DIFFERENT UNITS OF LENGTH

The SI unit of length is metre. Its symbol is ‘m’. Centimetre, millimetre, kilometre, etc. are the other units of length.

The picture shows part of a metre scale.

Take a metre scale from the science lab and examine it. You can see small and large lines on the metre scale. The distance between two consecutive large lines is one centimetre and the distance between small lines is one millimetre.

DIFFERENT UNITS OF TIME

The SI unit of time is the second. Its symbol is ‘s’.

VOLUME

The volume of an object is the amount of space it occupies. The SI unit of volume is cubic metre. It’s symbol is m3.

FUNDAMENTAL AND DERIVED UNITS

FUNDAMENTAL UNITS
In 1960, an international conference held in Paris approved the International System of Units or SI units as the universal system of units for measurements. Under this system, units were assigned to all the fundamental quantities.

Fundamental units are the units of fundamental quantities.

Note the fundamental units and their symbols given below.

DERIVED UNITS
We have learned about derived quantities. Such as volume and density, whose units are obtained from fundamental units.

We can write the derived units by relating fundamental units one another. Derived units are formed using fundamental units.

See how derived units are formulated in the table given.

Derived quantities	Equation	Unit
Area	Area = length × breadth	m × m = m2
Volume	Volume = length × breadth × height	m × m × m = m3
Density	Density = \(\frac{\text { Mass }}{\text { Volume }}\)	kg/m3

Derived units are units that can be stated using fundamental units or that depend on fundamental units.

RULES FOR WRITING THE UNITS
Observe the correct notation of units for two physical quantities.

Quantity	Unit
Mass of 1.5 litre of water	1.5 kg
Density of water	1000 kg/m3 1000 kilogram per cubic metre

MEASURING THE VOLUME USING A MEASURING JAR
Let’s try to find the volume of a stone. Pour some water into a measuring jar and mark its level. Tie the stone with a thread and dip into the water. Observe the rise in the water level. From this, we can calculate the volume of the stone which is equal to the volume of water displaced.

MEASURING TIME USING A STOPWATCH
A stopwatch is used to measure a time intervals. As shown in the figure, tie a metal ball using a thread and hang it. Pull the ball slightly and release it to oscillate. Observe the motion. Measure the time taken for 10 oscillations using a stopwatch. Record the measurement.

Time required for 10 oscillations = 10 s
A good understanding of physical quantities will help you in further studies and on the proper use of measurement and units in daily life.

Students can use Class 8 Malayalam Adisthana Padavali Notes Pdf and കുരുവിയും കാട്ടുതീയും Kuruviyum Kaattu Theeyum Summary in Malayalam to grasp the key points of a lengthy text.

Class 8 Malayalam Kuruviyum Kaattu Theeyum Summary

Kuruviyum Kaattu Theeyum Summary in Malayalam

കുരുവിയും കാട്ടുതീയും Summary in Malayalam

എഴുത്തുകാരനെ പരിചയപ്പെടാം

ആനന്ദിന്റെ കഥകൾ – സച്ചിദാനന്ദ്
തൂലികാനാമം – ആനന്ദ്

പ്രശസ്തനായ ഒരു മലയാള നോവലിസ്റ്റും എഴുത്തുകാരനുമാണ് ആനന്ദ് എന്നറിയപ്പെടുന്ന പി. സച്ചിദാനന്ദൻ. 1936 ൽ ഇരിങ്ങാലക്കുടയിലാണ് ജനിച്ചത്. തിരുവനന്തപുരം എൻജിനീയറിങ്ങ് കോളേജിൽ നിന്ന് സിവിൽ എൻജിനീ യറിങ്ങിൽ ബിരുദം. നാലുകൊല്ലത്തോളം പട്ടാളത്തിൽ സേവനമനുഷ്ഠി ച്ചിട്ടുണ്ട്. ന്യൂഡെൽഹിയിൽ സെൻട്രൽ വാട്ടർ കമ്മീഷനിൽ പ്ലാനിങ്ങ് ഡയറ ക്ടറായി വിരമിച്ചു. ശില്പ കലയിലും തത്പരനായ ആനന്ദിന്റെ പല നോവലു കളിലും മുഖച്ചിത്രമായി അദ്ദേഹം നിർമിച്ച ശില്പങ്ങളുടെ ഫോട്ടോയാണ് ഉപയോഗിച്ചിട്ടുള്ളത്. 2016 ലെ കൊച്ചിൻ മുസിരിസ് ബിനലയിൽ അദ്ദേഹം ശിൽപ്പങ്ങൾ പ്രദർശിപ്പിച്ചിരുന്നു.

വീടും തടവും, ജൈവമനുഷ്യൻ ഇവ കേരള സാഹിത്യ അക്കാദമി അവാർഡും മരുഭൂമികൾ ഉണ്ടാകുന്നത് വയലാർ അവാർഡും ഗോവർദ്ധനന്റെ യാത്രകൾ 1997ലെ കേന്ദ്ര സാഹിത്യ അക്കാദമി അവാർഡും നേടി. മഹാശ്വേതാദേവിയുടെ ‘കവി ബന്ദ്യഘടിഗായിയുടെ ജീവിതവും മരണവും’ എന്ന കൃതിയുടെ മലയാള വിവർത്തനത്തിന് 2012ൽ കേന്ദ്രസാഹിത്യ അക്കാദമി പുരസ്കാരം ലഭിച്ചു. 2019 ലെ എഴുത്തച്ചൻ പുരസ്കാരം ലഭിച്ചു.

വ്യവസ്ഥകളിലും ശീലങ്ങളിലും ക്രമപ്പെട്ടുപോയ മനുഷ്യരുടെ അകമേനിന്നു പുറപ്പെടുന്ന ഒച്ചയാണ് ആനന്ദിന്റെ കഥകൾ. കരച്ചിലോ, വിലാപങ്ങളോ അല്ല; സമൂഹത്തിന്റെയും ചരിത്രത്തിന്റെയും രാഷ്ട്രീയ ത്തിന്റെയും ജീർണ്ണസത്തകളിലേക്കുള്ള ശ്രദ്ധ ക്ഷണിക്കലാണ് ആ ശബ്ദം. ആ ഒച്ചകൾ നമുക്കു ചുറ്റും പ്രതിധ്വനിക്കുന്നു. സമകാലീനതയുടെ ആത്മകഥകളെന്നു വിശേഷിപ്പിക്കാ വുന്നവയാണ് ആനന്ദിന്റെ ശ്രദ്ധേയങ്ങളായ രചനകൾ.

പാഠസംഗ്രഹം

‘കുരുവിയും കാട്ടുതീയും’ ഏറെ സമകാലീന പ്രസക്തിയുള്ള ആഴമുള്ള സാമൂഹിക ചിന്ത കാഴ്ചവയ്ക്കുന്ന കഥയാണ് പ്രകൃതിയിലെ ദുരന്തമായി കാട്ടുതീ പടർന്നപ്പോൾ മൃഗങ്ങളും പക്ഷികളും ജീവനുവേണ്ടി ഓടി രക്ഷപ്പെടുന്നു. എന്നാൽ ഒരു കൊച്ചുകുരുവി മാത്രം, സ്വല്പമെങ്കിലും പ്രതിരോധിക്കാൻ ശ്രമിക്കുന്നു. തന്റെ ചെറുചുണ്ടിൽ വെള്ളം എടുത്ത് തീയിൽ തളിച്ച് തീ കെടുത്താൻ കുരുവി കഠിന ശ്രമം തുടരുന്നു. വലിയ കാട്ടുതീ അണയ്ക്കാൻ ചെറിയ ചുണ്ടിൽ വെള്ളം തേവുന്ന കുരുവിയെ കണ്ടു മഴദേവൻ പരിഹസിക്കുമ്പോൾ, കുരുവി മറുപടി പറയുന്നു: ‘ഞാൻ എനിക്കാവുന്നത് ചെയ്യുന്നു.’ മഴമേഘത്തിന് ‘തീയണയ്ക്കാൻ’ കഴിവുണ്ടായിട്ടും പ്രവർത്തിക്കാത്തിനെ വ്യംഗ്യമായി വിമർശിക്കുന്നു കുരുവി.

ഈ വാചകം കഥയുടെ ആത്മാവാണ്. മനുഷ്യൻ ശക്തരായും ശേഷിയുള്ളവരായും ഇരിക്കെ ഭൂമിയിലെ നശീകരണങ്ങളോടും ദുരന്തങ്ങളോടും ഉള്ള അവന്റെ മനോഭാവത്തെ ചോദ്യം ചെയ്യുകയാണ് ഈ കുരുവിയുടെ വാക്കുകളിലൂടെ കഥാകാരൻ. ഇവിടെ സാമാന്യേന ശേഷി കുറഞ്ഞ കൊച്ചുകുരുവി പോലുള്ള ജീവിയും സ്വന്തം കഴിവിനൊത്ത് വല്ലതും ചെയ്യുകയാണ്. അതിനുവേണ്ടിയാണ് കുരുവി ശ്രമിക്കുന്നത് പല നിർണായക സന്ദർഭങ്ങളിലും മനുഷ്യർ നമ്മൾ ഒരാൾ വിചാരിച്ചാൽ ഒന്നും സാധിക്കുകയില്ലെന്ന് പറഞ്ഞു മൗനദർശിയായി നിൽക്കാറുണ്ട്. ചെറുശേഷിയുള്ളതായിരുന്നാലും തന്റെ പ്രവർത്തി മറ്റുള്ളവരിൽ ഉത്തരവാദിത്ത ബോധം വളർത്തുമെന്നും മൃഗങ്ങൾ തന്നെപ്പോലെ ചിന്തിച്ചാൽ കാട്ടുതീ അണയ്ക്കാം എന്നുമുള്ള ബോധ്യമാകുരുവിക്ക് ഉണ്ടായിരിക്കണം. (നമ്മുടെ സ്വാതന്ത്ര്യ സമര ചരിത്രത്തിലെ ദണ്ഡിയാത്ര ഓർമ്മിക്കുക.) ഒരാളുടെ ചെറിയ പ്രവർത്തനം പോലും വലിയ പ്രചാരണങ്ങൾക്ക് വഴിവെക്കാം എന്നതിന്റെ മാതൃക ചിത്രമാണ് ഈ കുരുവി. അനാസ്ഥയുടെയും ഉപേക്ഷയുടെയും ആധുനിക സമൂഹ മനോഭാവത്തിന് എതിരായ പ്രതിരോധം തന്നെയാണിത്.

ഇന്നത്തെ കാലത്ത് പരിസ്ഥിതി പ്രശ്നങ്ങൾ, സാമൂഹിക അനീതികൾ, യുദ്ധങ്ങൾ, സാംസ്കാരിക മൂല്യച്ചുതി തുടങ്ങിയ പ്രശ്നങ്ങൾ മനുഷ്യന്റെ മുന്നിലുണ്ട്. അവയെ തടയാൻ പുതിയ തലമുറ മൗനപരമായ മനോഭാവം സ്വീകരിക്കുമ്പോൾ, അങ്ങേയറ്റത്തെ അനാസ്ഥയും നിരുത്തരവാദപരവും, പുരോഗമന കാഴ്ചപ്പാടുകളുടെ മറവിൽ നടത്തുന്ന ചൂഷണവും ഭൂമിയെ ആകെ ഒരു കാട്ടുതീ പോലെ പടർന്നുപിടിക്കുന്നത് കഥാകാരൻ ഈ കഥയിൽ ചേർത്ത് വച്ചിരിക്കുന്ന സത്യമാണ്.

‘കുരുവിയും കാട്ടുതീയും’ എന്ന കഥയിലൂടെ ആനന്ദ് സമകാലീന സമൂഹത്തിന് മുന്നിൽ ആധുനിക ചോദ്യങ്ങൾ ഉന്നയിക്കുന്നു: ‘നിനക്ക് എന്ത് ചെയ്യാനാകും,’, ‘നിനക്ക് അതിന് മനസ്സുണ്ടോ?’ അതിനുള്ള ഉത്തരം കഥാകൃത്ത് നല്കുന്നില്ല; മറിച്ച്, അത് വായനക്കാരനോട് തന്നെ ആവശ്യപ്പെടുകയാണ്. ചെറുതെങ്കിലും ഒരു കുരുവിയുടെ നീക്കം, വലിയൊരു മാറ്റത്തിന് തുടക്കമാകാമെന്നത് ഈ കഥയുടെ ഉദാത്തമായ സന്ദേശമാണ്.

പുതിയ പദങ്ങൾ

പ്രാണവായു = പ്രാണൻ നിലനിർത്തുന്ന വായു (ഓക്സിജൻ)
ജീവജലം = ജീവദായകമായ ജലം
അന്നം = ആഹാരം, ഭക്ഷ്യവസ്തു
മടയത്തം = വിഡ്ഢിത്തം / വിഫലമായ ശ്രമം
ആളിക്കത്തുന്ന = വളരെ വേഗത്തിൽ കത്തുന്ന വേഗത്തിൽ വ്യാപിക്കുന്ന

Practicing with Std 8 Malayalam Adisthana Padavali Notes Unit 1 Chapter 1 കുരുവിയും കാട്ടുതീയും Kuruviyum Kaattu Theeyum Notes Questions and Answers Pdf improves language skills.

കുരുവിയും കാട്ടുതീയും Question Answer Notes Std 8 Malayalam Adisthana Padavali Chapter 1

Class 8 Malayalam Adisthana Padavali Unit 1 Chapter 1 Notes Question Answer Kuruviyum Kaattu Theeyum

Class 8 Malayalam Kuruviyum Kaattu Theeyum Notes Questions and Answers

പാഠപുസ്തകത്തിലെ ചോദ്യങ്ങളും ഉത്തരങ്ങളും

കഥപറയാം

Question 1.
കഥ മൗനമായി വായിക്കൂ. സംഘമായി തിരിഞ്ഞ് ഒരാൾ ഒരുവാക്യം എന്ന ക്രമത്തിൽ ആശയം ചോർന്നുപോകാതെ പറയൂ.
Answer:
കൂട്ടുകാരെ മുകളിൽ പറഞ്ഞതുപോലെ ചെയ്യുമല്ലോ…

അഭിപ്രായ കുറിപ്പ്

Question 1.
കാട്ടുതീ ഒത്തിരി പക്ഷികളുടെയും മൃഗങ്ങളുടെയും ജീവനെടുക്കുന്നത് കാണുന്നില്ലേ? എന്ന് കുരുവി മഴ ദേവനോട് ചോദിക്കുന്നു എന്നാൽ മഴ ദേവൻ എന്താണ് ചെയ്തത് രണ്ടുപേരുടെയും മനോഭാവങ്ങളിൽ എന്ത് വ്യത്യാസമാണുള്ളത്? നിങ്ങൾ ആരുടെ ഭാഗത്ത് നിൽക്കുന്നു? നിങ്ങളുടെ അഭിപ്രായം എഴുതുക.
Answer:
കുരുവി തീ അണയ്ക്കാൻ ശ്രമിക്കുന്നപ്പോൾ മഴദേവൻ അവളെ പരിഹസിക്കുന്നു. ഇത് രണ്ടുവിധമുള്ള മനോഭാവങ്ങൾ പ്രതിപാദിക്കുന്നു:

മഴദേവൻ: പ്രതീക്ഷയില്ലാത്ത, സംഭവിച്ചുകൊണ്ടിരിക്കുന്ന സാഹചര്യങ്ങൾ അഥവാ സംഭവങ്ങൾ മാറ്റാൻ കഴിയില്ലെന്ന് കരുതുന്ന, മറ്റുള്ളവരുടെ ശ്രമത്തെ പരിഹസിക്കുന്ന, വിഫലം എന്ന് കരുതുന്ന, നോക്കിക്കൊണ്ടിരിക്കാൻ മാത്രം തയ്യാറുള്ള മനോഭാവത്തോട് കൂടിയവരെ പ്രതിനിധീകരിക്കുന്നു.

എന്നാൽ കുരുവി: ധൈര്യവും ഉത്തരവാദിത്വബോധവും നിറഞ്ഞവൾ ആണ്. വലിയത് ചെയ്തില്ലെങ്കിലും കഴിയുന്നതുമാത്രം ചെയ്യുന്നവൾ.സഹജീവികൾക്കായി കരുതുന്നവൾ. സ്വതന്ത്രമായി നീതിനിമിത്തം പ്രവർത്തിക്കുന്നവൾ. കുഞ്ഞിക്കിളിയുടെ പ്രവൃത്തി മാതൃകാപരമാണ്. അതുകൊണ്ടു തന്നെ ഞാൻ കുഞ്ഞിക്കിളിയുടെ ഭാഗത്തു നിൽക്കുന്നു. പ്രതീക്ഷയും കരുണയും ഉള്ള ഈ ചെറിയ പക്ഷിയുടെ മനോഭാവമാണ് യഥാർത്ഥത്തിൽ മനുഷ്യനാകാനുള്ള ആദ്യപടി. മാറ്റങ്ങൾ ചെറുതായിരുന്നാലും, അവ തുടങ്ങേണ്ടത് നമ്മിൽ കൂടെയാണ്. നോക്കിക്കൊണ്ടിരിക്കുന്നതിലല്ല, പ്രവർത്തനത്തിലായിരിക്കണം നമ്മുടെ പങ്ക്.

കഥയെഴുതൂ, പതിപ്പാക്കു

Question 1.
ഈ കഥയിലെ കുരുവിയും മഴ ദേവനും തമ്മിലുള്ള സംഭാഷണം ശ്രദ്ധിച് കുരുവിയും കാടും തമ്മിലുള്ള സംഭാഷണം ആയിരുന്നുവെങ്കിൽ കഥാഗതിയിൽ എന്തു മാറ്റമാണ് സംഭവിക്കുക സങ്കൽപ്പിച്ച് കഥയെഴുതി പതിപ്പാക്കും…
കുരുവിയും കാടും തമ്മിലുള്ള സംഭാഷനത്തിൽ സംഭവിക്കുന്ന മാറ്റം.
Answer:
സങ്കല്പ കഥ:
• കാട്ടുതീ വലിയ ശബ്ദത്തോടെ കാട്ടിലാകെ പടരുമ്പോൾ, കുരുവി തന്റെ കുഞ്ഞി ചുണ്ടുകളിൽ പുഴയിൽ നിന്ന് വെള്ളം കൊണ്ടുപോയി തീ കെടുത്താൻ ഒരു വിഫല ശ്രമം നടത്തുകയാണ് അതു കണ്ട് കാട് കുരുവിയോട് പറഞ്ഞു:

കാട്: കുരുവി തീ പടർന്നു പിടിക്കുന്നിടത്ത് നീ എന്താണ് ചെയ്യുന്നത്? നിന്റെ കുഞ്ഞി ചിറകുകൾക്ക് പൊള്ളലേൽക്കും മുമ്പ് ഇവിടെ നിന്നും ദൂരേക്ക് പറന്നു പോകൂ…

കുരുവി: ഞാൻ വസിക്കുന്ന കാട് ആണ് നീ. നിന്നെ ഈ അവസ്ഥയിൽ ഉപേക്ഷിച്ച് ഞാൻ എങ്ങനെ പോകും? (അതും പറഞ്ഞ് കുരുവി വീണ്ടും പുഴയിലേക്ക് ചെന്ന് വെള്ളവുമായി വന്നു)

കാട്: ‘നിന്റെ സ്നേഹം ഞാൻ അറിയുന്നു കുരുവി പക്ഷേ നിനക്ക് ഈ തീ അണയ്ക്കുവാൻ ആവുകയില്ല നീ നിന്റെ വലിപ്പവും കഴിവും മനസ്സിലാക്കണം. ഞാൻ നിന്റെ മാത്രമല്ല മറ്റു ജീവജാലങ്ങളുടെയും കാടാണ്. പക്ഷേ എല്ലാവരും സ്വന്തം ജീവന് വേണ്ടി പരക്കം പായുന്നത് നീ കാണുന്നില്ലേ…?
നിന്റെ കുഞ്ഞി ചുണ്ടിലെ രണ്ടിറ്റു വെള്ളം കൊണ്ട് ഈ തീ അണയുകയില്ല. സ്വന്തം ജീവനെങ്കിലും രക്ഷപ്പെടുത്തി കൊള്ളൂ ഇനിയും ഒരു കാട് ദൂരെയെങ്ങാനും കാണും.

കുരുവി: ‘നിന്റെ വേദന എനിക്ക് മനസ്സിലാകുന്നു. നിന്റെ സ്നേഹവും. തീ നിന്നെ തിന്നുമ്പോൾ ഞാനെന്തെങ്കിലും ചെയ്യേണ്ടതല്ലേ? എനിക്കാവുന്നതു ഞാൻ ചെയ്യുന്നു. നിന്റെ വിയോഗം നോക്കി എനിക്ക് മൗനമായി ഇരിക്കാൻ കഴിയില്ല. ഞാൻ പ്രത്യാശിക്കുകയാണ്. എന്റെ പ്രവർത്തി മറ്റുള്ളവരെ ചിന്തിപ്പിച്ചിരുന്നുവെങ്കിൽ അവരും എന്നോട് ചേർന്ന് നിന്നെ സംരക്ഷിക്കാൻ പ്രയത്നിച്ചിരുന്നു എങ്കിൽ… ഇവിടന്നങ്ങോട്ട് ഈ കാടിനെ സംരക്ഷിക്കാൻ, തീ പടരുന്നത് തടഞ്ഞു നിർത്തുവാൻ കഴിയുമായിരിക്കാം. എന്റെ ശ്രമം മറ്റുള്ളവർക്കും പ്രചോദനമാകുമെങ്കിൽ…

കാട്: ഞാൻ നിന്നിൽ അഭിമാനിക്കുന്നു…. നിനക്ക് വാസസ്ഥലമായതിലും നിന്നെ ഊട്ടിയതിലും.

കുരുവി : നീ നിന്റെ ഉത്തരവാദിത്വം നിർവഹിച്ചിരിക്കുന്നു. ഞാനും നിന്നെപ്പോലെ…… (മഴപെയ്തതായും തീ അണഞ്ഞതായും സങ്കൽപ്പിക്കാം. മറ്റു മൃഗങ്ങൾ ഒത്തുചേർന്ന് തീ അണച്ചതായും സങ്കൽപ്പിക്കാം കഥ കൂട്ടുകാരുടെ ആശയത്തിനൊത്ത് അവസാനിപ്പിക്കാം).

ആശയവ്യത്യാസം

Question 1.
ഇത് പലതവണ ആവർത്തിച്ചു
ഇത് പലതവണ ആവർത്തിക്കും
ഇത് പലതവണ ആവർത്തിക്കുന്നു
അടിവരയിട്ട് പദങ്ങളുടെ പ്രയോഗം വാക്യങ്ങളുടെ ആശയതലത്തിൽ ഉണ്ടാക്കുന്ന മാറ്റം എന്ത് അതിന്റെ കാരണം ചർച്ച ചെയ്യുക ഇത്തരം പദങ്ങൾ കൂടുതൽ കണ്ടെത്തി എഴുതുക.
Answer:
പദപ്രയോഗങ്ങളുടെ വ്യത്യാസവും അർത്ഥ വ്യത്യാസങ്ങളും

വാക്യങ്ങൾ:

1. ഇത് പലതവണ ആവർത്തിച്ചു → കഴിഞ്ഞിട്ടുള്ള പ്രവൃത്തി (ഭൂതകാലം)
2. ഇത് പലതവണ ആവർത്തിക്കും → വരാനിരിക്കുന്ന പ്രവൃത്തി (ഭാവികാലം)
3. ഇത് പലതവണ ആവർത്തിക്കുന്നു → ഇപ്പോൾ നടക്കുന്നതായ പ്രവൃത്തി (വർത്തമാനകാലം)

പദപ്രയോഗം	കാലം	ആശയം
ആവർത്തിച്ചു	ഭൂതകാലം (കഴിഞ്ഞത്)	ഒരു കാര്യം വളരെവട്ടം കഴിഞ്ഞുപോയിരിക്കുന്നു
ആവർത്തിക്കുന്നു	വർത്തമാനകാലം(ഇപ്പോൾ നടക്കുകയാണ്)	പ്രവർത്തനം ഇപ്പോഴും തുടരുന്നു.
ആവർത്തിക്കും	ഭാവികാലം നടക്കാൻ പോകുന്നു	ഇപ്പോഴും പ്രതീക്ഷയുണ്ട്.

ഈ വ്യത്യാസങ്ങൾ വാചകത്തിന്റെ തീവ്രതയും സമയബോധവും മാറ്റുന്നു. ഉദാഹരണത്തിന്, ‘അവൻ പഠിച്ചു’ എന്ന് പറയുമ്പോൾ പഠനം കഴിഞ്ഞതായി സൂചനയുണ്ട്, പക്ഷേ ‘പഠിക്കുന്നു’ എന്നത് ഇപ്പോഴും പ്രവർത്തനം നടക്കുകയാണെന്നത് വ്യക്തമാക്കുന്നു. ഈ ചെറിയ വ്യത്യാസങ്ങൾ കഥയിലെ സ്ഥിതിഗതികൾ വായനക്കാരന് കൂടുതൽ വ്യക്തമായി അനുഭവപ്പെടാൻ സഹായിക്കുന്നു.

ഉദാഹരണം

തുടർപ്രവർത്തനങ്ങൾ

Question 1.
തീയണക്കാൻ ശ്രമിച്ച കുരുവിയെ പരിഹസിച്ചതാര്?
Answer:
തീയണയ്ക്കാൻ ശ്രമിച്ച് കുരുവിയെ പരിഹസിച്ചത് മഴദേവൻ ആയിരുന്നു. കുരുവിയുടെ ചെറിയ ശ്രമം കാണുമ്പോൾ മഴ ദേവൻ അത് ഫലപ്രദമല്ലെന്ന് പരിഹസിക്കുന്നു. എന്നാൽ ചെറിയ സഹായം പോലും ചെയ്യാൻ മഴദേവൻ തയ്യാറാവുന്നുമില്ല.

Question 2.
“ഞാൻ എനിക്കാവുന്നത് ചെയ്യുന്നു”-
ഈ വാക്യം ആരോട് പറഞ്ഞു? ഈ വാക്യത്തിൽ പ്രതിഫലിക്കുന്ന മാനസികാവസ്ഥയെ വിശകലനം ചെയ്യുക. സന്ദർഭം വ്യക്തമാക്കുക.
Answer:
ഈ വാക്യം കുരുവിയാണ് പറയുന്നത്. കാട്ടുതീ പടർന്നപ്പോൾ കുരുവി തന്റെ കുഞ്ഞി ചുണ്ടിൽ വെള്ളം കൊണ്ടുവന്ന് തീയിൽ തളിച്ച് തീ അണയ്ക്കാൻ ശ്രമിക്കുന്നു. അതുകണ്ട് മഴദേവൻ പരിഹസിക്കുമ്പോൾ കുരുവി ഉത്തരമായി ഈ വാക്യം പറയുന്നു തന്റെ ചെറിയ ശ്രമം ചോദ്യം ചെയ്യുന്ന മഴദേവനോട് പറയുന്നു. ‘ഞാൻ എനിക്കാവുന്നത് ചെയ്യുന്നു.’
ഇവിടെ തെളിയുന്ന മാനസികാവസ്ഥ എന്തെന്നാൽ

ഈ വാക്യം പ്രതിനിധീകരിക്കുന്നത് ഒരു വ്യക്തിയുടെ ഉത്തരവാദിത്വബോധവും ധൈര്യവുമാണ്. കുരുവി തന്റെ കഴിവിനു എന്ത് ചെയ്യാനാവുമോ അതാണ് ചെയ്യുന്നത്. മറ്റുള്ളവരുടെ പരിഹാസം അതിന്റെ ചാരുത ഇല്ലാതാക്കുന്നില്ല.

ഇത് പാഠത്തിലെ മുഖ്യ സന്ദേശമായ ‘നമ്മിൽ ഒരാൾ ഒരു ചെറിയ ശ്രമം നടത്തിയാൽ പോലും വലിയ മാറ്റത്തിന് തുടക്കമാകാം’ എന്നതിനെ പ്രതിഫലിപ്പിക്കുന്നു.

Question 3.
ആനന്ദിന്റെ എഴുത്തിൽ പ്രകടമാകുന്ന സമകാലീനതയുടെ ആത്മാവിനെ’ കുറിച്ച് പ്രതിപാദിക്കുക.
Answer:
ആനന്ദിന്റെ എഴുത്തിൽ ‘സമകാലീനതയുടെ ആത്മാവ്’ എന്നത് ആധുനിക സാമൂഹിക പ്രശ്നങ്ങളെ നേരിട്ട് അഭിമുഖീകരിക്കുന്നു (പരിസ്ഥിതി നാശം, മനുഷ്യത്വമില്ലായ്മ, യുദ്ധ ഭീഷണി, അനാസ്ഥ, നിരുത്തരവാദിത്വം). ഏകാന്തമായെങ്കിലും പ്രതിരോധം നടത്താനുള്ള വ്യക്തിയുടെ നീക്കം (കുരുവി പ്രതിനിധാനം ചെയ്യുന്ന രീതിയിൽ). ഒരു വ്യക്തിയുടെ പ്രചോദനാത്മകമായ പ്രവൃത്തികൾ വലിയ മാറ്റത്തിന് തുടക്കമാകാമെന്ന സന്ദേശം. ആനന്ദിന്റെ കഥകൾ വായനക്കാരെ ചിന്തിപ്പിക്കുന്നു, മറുപടി നിർബന്ധിക്കുന്നില്ല – മറിച്ച് വായനക്കാരൻ സ്വയം ഉത്തരമെത്തേണ്ട അവസ്ഥയിലേക്കാണ് എഴുത്ത് നയിക്കുന്നത്.

Question 4.
കുരുവിയുടെ പ്രയത്നം കണ്ട് മറ്റു മൃഗങ്ങളും കുരുവിയോടൊപ്പം ചേർന്നു എന്നിരിക്കട്ടെ കഥയുടെ അവസാനം മാറ്റി എഴുതൂ.
Answer:
കുരുവിയുടെ കൃപയും പ്രതിബദ്ധതയും കണ്ട് കാട്ടിലുള്ള മറ്റു മൃഗങ്ങൾക്ക് അവരുടെ പ്രവർത്തിയിൽ ലജ്ജ തോന്നുന്നു. മൃഗങ്ങളെല്ലാവരും വെള്ളം കൊണ്ട് തീ അണയ്ക്കാൻ ശ്രമിക്കുന്നു. ഒടുവിൽ, എല്ലാ ജീവികളുടെയും ചേർന്ന പരിശ്രമം കാട്ടുതീ അണയ്ക്കാൻ കാരണമാകുന്നു. അവരുടെ കാടിന്റെ അവർ വീണ്ടെടുത്തു. ഒരുമയുടെ ബലം അവർക്ക് മനസിലായി.
മഴദേവനും ആ ശ്രമം കാണുമ്പോൾ നാണംകെട്ട് ക്ഷമ ചോദിക്കുന്നു.
സന്ദേശം: ‘നല്ലത് ചെയ്യുന്ന ഒരാളുടെ ശ്രമം മറ്റുള്ളവരെയും ഉണർത്തുന്നു.’

അധിക അറിവിലേക്ക്

സമാന ആശയം വരുന്ന കവികളുടെയും മഹത് വ്യക്തികളുടെയും വാക്കുകൾ
‘ചെറുതെങ്കിലും തനിക്കാവുന്നത് ചെയ്യുക’ എന്ന സന്ദേശം മലയാളം സാഹിത്യത്തിലും ലോകസാഹിത്യത്തിലും പലവിധരൂപങ്ങളിലായി പ്രതിഫലിച്ചിട്ടുണ്ട്. മനുഷ്യൻ തന്റെ പരിധിയിൽ നല്ലതൊരിക്കൽ ചെയ്താൽ തന്നെ അതിന് വലിയ പ്രഭാവമുണ്ടാകാമെന്ന ചിന്ത ഉൾക്കുന്ന വചനങ്ങളും കവിതാശകലങ്ങളും ചുവടെ പങ്കുവെക്കുന്നു;

മഹത് വചനങ്ങൾ (മഹത്വമുള്ള ഉദ്ധരണികൾ).
ഗാന്ധിജി
• “Be the change you wish to see in the world.”
(നിങ്ങൾ ലോകത്തിൽ കാണാൻ ആഗ്രഹിക്കുന്ന മാറ്റം തന്നെയാവുക.)
വലിയ വ്യതിയാനങ്ങൾ ചെറുതായെങ്കിലും വ്യക്തി തലത്തിൽ തുടങ്ങണം എന്ന സന്ദേശം.

മദർ തെരേസ:
• “Not all of us can do great things. But we can do small things with great love.”
(എല്ലാവർക്കും വലിയ കാര്യങ്ങൾ ചെയ്യാൻ കഴിയില്ല. എന്നാൽ വലിയ സ്നേഹത്തോടെ ചെറുകാര്യങ്ങൾ ചെയ്യാം.)
കുരുവിയുടെ ചിന്തയോടും പ്രവർത്തിയോടും ഏറെ സാമ്യമുണ്ട്.

സ്വാമി വിവേകാനന്ദൻ:
• “Each soul is potentially divine. Serve man, serve God.”
ഓരോ മനുഷ്യനും ദിവ്യതയുള്ളവൻ. സേവനത്തിലൂടെ ലോകം മാറാം.

മലയാള കവിതാശകലങ്ങൾ:

1. നന്മ ചെയ്യുവാൻ പുറപ്പെടുക,
നിന്റെ ഉള്ളമറിഞ്ഞു നീ ഭരിക്ക…’
♦ കുമാരനാശാൻ – ‘ചിന്താവിഷ്ടയായ സീത’:
ഭവനത്തിന്റെ അകത്തു നിന്ന് തന്നെ നന്മയുടെ വിത്തിടാൻ കൃത്യമായ ഒരു ആഹ്വാനം.

2. ‘ഞാൻ മനുഷ്യനെന്നതോർത്തു
ഞാൻ മനുഷ്യരെ സ്നേഹിച്ചു…’
♦ വള്ളത്തോൾ – ‘ജാതിവെറിക്കെതിരെ’:
ഒരാളുടെ മനസ്സിനുള്ളിലെ മാറ്റം സമൂഹത്തിന് മികച്ചതാവാൻ ഇടയാക്കുന്നു.

3. ‘ഞാൻ നിശ്ശബ്ദമായി നിന്നു
ചെറുതായെങ്കിലും തണൽ
നല്കാൻ ശ്രമിച്ചു…’
♦ സുഗതകുമാരി – ‘മൂലവൃക്ഷം’:

4. വലിയൊരു ലോകം
മുഴുവൻ നന്നാവാൻ
ചെറിയൊരു സൂത്രം
ചെവിയിലോതാം ഞാൻ
‘സ്വയം നന്നാവുക’
♦ കുഞ്ഞുണ്ണി മാഷ്
കുരുവിയുടെ നിലപാടിന്റെ കാവ്യപരമായ രൂപമാണ് ഇത്

5. ‘കർമണേവാധികാരസ്തേ മാ ഫലേഷു കദാചന’
ഭഗവത് ഗീത (ചാപ്റ്റർ 2, ശ്ലോകം 47):
(ഫലമിഛിക്കാതെ കർമ്മം ചെയ്യുക.)
കുരുവി ഫലത്തെക്കുറിച്ച് ചിന്തിക്കാതെ ദൈനംദിനം ചിന്താപരമായ കർമ്മം ചെയ്യുന്നു.

മറ്റു ഭാഷകളിൽ നിന്ന് കവിതാശകലങ്ങൾ
• “If I can stop one heart from breaking,
I shall not live in vain.”
♦ എമിലി ഡിക്കിൻസൺ:
ഒരാൾക്ക് പോലും ആശ്വാസം നൽകാനാകുന്നത് ജീവിതം ഫലപ്രദമാക്കിയതായി കാണുന്നു.

കിളിമൊഴി

(കിളിമൊഴി – പക്ഷികൾക്ക് വേണ്ടി 35 ഭാഷണങ്ങൾ) ഡോ. സലിം അലി.
പരിഭാഷ എസ്. ശാന്തി

നിരീക്ഷണക്കുറിപ്പ്

Question 1.
പക്ഷിനിരീക്ഷണത്തെക്കുറിച്ചുള്ള ഡോ. സാലിം അലിയുടെ നിർദേശങ്ങൾ ശ്രദ്ധിച്ചല്ലോ. അവ കൂടി പരിഗണിച്ചുകൊണ്ട് നിങ്ങൾക്കിഷ്ടപ്പെട്ട ഒരു പക്ഷിയെ നിരീക്ഷിച്ച് കുറിപ്പ് തയ്യാറാക്കുക.
Answer:
ഡോക്ടർ സലിം അലിയുടെ നിർദ്ദേശങ്ങൾ അനുസരിച്ച് ഒരു പക്ഷിയെ ഞാൻ നിരീക്ഷിച്ചെന്നു കണക്കാക്കി തയ്യാറാക്കിയ മാതൃക കുറിപ്പാണ്. നിങ്ങൾക്ക് ഇഷ്ടമുള്ള പക്ഷിയെ ഇവിടെ ചേർത്ത് ഈ മാതൃകയുടെ അടിസ്ഥാനത്തിൽ കൂട്ടുകാരും കുറുപ്പ് തയ്യാറാക്കണേ…

പക്ഷിനിരീക്ഷണ കുറിപ്പ്

പക്ഷിയുടെ പേര്: കുഞ്ചി തത്ത (Common Green Bee eater)

1. അവലോകനം ചെയ്ത സ്ഥലം
പാലക്കാട് ജില്ലയിലെ ഒരു കൃഷി ഭൂമിയിലായിരുന്നു ഈ പക്ഷിയെ കണ്ടത്. രാവിലെ ഏകദേശം 8 മണിയായിരുന്നു സമയം. വാനിൽ വെളിച്ചവും ചുറ്റും നിശബ്ദതയും ഉണ്ടായിരുന്നു.

2. ദൃശ്യവിവരണം
വലുപ്പം: ഏകദേശം ഒരു ചെറുതത്തയുടെ വലിപ്പം (16 – 18 സെ.മീ.).
നിറം: പച്ച നിറം ആധാരമായി കാണപ്പെടുന്നു. തലയിൽ ചെറിയ വെള്ള ചുവപ്പ് നിറമുള്ള പാലിളക്കം. കഴുത്ത് ചുറ്റി കറുത്ത വര.
ശരീരരൂപം: അതിസുന്ദരമായി നീളമുള്ള ചിറകും വാൽപ്പക്ഷ ത്തേയും കാണാം. രണ്ടു വാടികൾ കുറച്ച് നീളത്തിൽ നീളിയി രിക്കുന്നു.
കൊക്ക്: കുറച്ച് നീളമുള്ളതും വളഞ്ഞതും കറുത്ത നിറത്തി ലുള്ളതും.
കണ്ണുകൾ: ചെറുതായി ചുവപ്പു നിറം പതിച്ച കാഴ്ച.
കാൽ: ചെറുതും കറുത്തതും.
ഇരിപ്പിടം: വരമ്പത്ത് നാട്ടിയ ഒരു തൂണിന്മേൽ ഇരിക്കുകയായിരുന്നു.
കണ്ടത്: ഒറ്റയ്ക്കായിരുന്നു പക്ഷി.
നടപ്പുകൾ: വീണ്ടും വീണ്ടും വരമ്പത്ത് തൂണിൽ നിന്ന് പറന്നു ചെറു കീടങ്ങൾ പിടിച്ച് തിരികെ തൂണിലേയ്ക്ക് വരികയായിരുന്നു.
ശബ്ദം: ചെറുതായി ചിര് ചിര്’ പോലുള്ള ശബ്ദം ഉരിയുന്നു.

4. ആവാസ സ്ഥലം
ഈ
പക്ഷിയെ കൃഷിഭൂമികൾ, തുറന്ന ശുദ്ധവായുവുള്ള ഇടങ്ങൾ, വയൽ പാടങ്ങൾ, മരക്കൊമ്പുകൾ എന്നിവിടങ്ങളിൽ സാധാരണ യായി കാണാം.

5. നിരീക്ഷണ അഭിപ്രായം
കുഞ്ചി തത്തയെ നിരീക്ഷിക്കുന്നത് വളരെ രസകരമായ അനുഭവമാണ്. നാട്ടുവേലി തത്ത എന്നും ഇതിനെ പേരുണ്ട്. അതിന്റെ തെളിഞ്ഞ പച്ച നിറം പച്ചപ്പ് നിറഞ്ഞ പശ്ചാത്തലത്തിൽ ആകർഷകമായി നിറഞ്ഞുനിൽക്കുന്നു. തത്സമയം തന്നെ പറന്ന് കീടങ്ങൾ പിടിക്കുന്ന അതിന്റെ കൃത്യത ഉറ്റുനോക്കിയാൽ അതിന്റെ മനോഹാരിതയെ ഏറെ മികവോടെ കാണാം. ശാന്തമായ ഒരു രാവിലെ ഈ പക്ഷിയെ കാണാനായതിൽ ഞാൻ സന്തോഷവാനാണ്.

പക്ഷി നിരീക്ഷണം

വ്യവസ്ഥാധിഷ്ഠിതമായ പക്ഷിനിരീക്ഷണത്തിന് ഇന്ത്യയിൽ അടിസ്ഥാനമിട്ട ആളാണ് സാലിം അലി (സാലിം മുഇസുദ്ദീൻ അബ്ദുൾ അലി, നവംബർ 12, 1896 – ജൂലൈ 27, 1987) അദ്ദേഹത്തിന്റെ നിരീക്ഷണങ്ങൾ, ഭാരതത്തിലെ ജനങ്ങളിൽ പക്ഷിനിരീക്ഷണത്തിനും, പ്രകൃതി സ്നേഹത്തിനും അടിത്തറയിട്ടു. പക്ഷിനിരീക്ഷണ ശാസ്ത്രത്തെക്കുറിച്ചും പക്ഷികളെക്കുറിച്ചും സലിം എഴുതിയ ഗ്രന്ഥങ്ങൾ വിജ്ഞാനപ്രദവും പ്രസിദ്ധവുമാണ്. ഇവയിൽ കേരളത്തിലെ പക്ഷികളെ പറ്റിയെഴുതിയ ഗ്രന്ഥവും ഉൾപ്പെടും. ഒരു കുരുവിയുടെ പതനം’ അദ്ദേഹത്തിന്റെ ആത്മകഥയാണ്. പക്ഷിമനുഷ്യൻ എന്നും ഇദ്ദേഹം അറിയപ്പെടുന്നു

Class 8 Malayalam Adisthana Padavali Notes Unit 1 കനിവും കരുതലും

താഴെ കാണുന്ന ചിത്രങ്ങൾ ‘കനിവും കരുതലും’ എന്ന അധ്യായത്തിന്റെ ഭാഗമായി ഉൾപ്പെടുത്തിയതാണ്. ഇവ തമ്മിൽ ഉള്ള തിരിച്ചറിയലുകളും പ്രതിപാദ്യവും മനസ്സിലാക്കുന്ന പാഠഭാഗങ്ങളാണ് ഈ യൂണിറ്റിൽ പഠിക്കാൻ ഉള്ളത്. കുരുവിയും കാട്ടുതീയും, കൊച്ചു ദേവദാരു, പെരുമഴയത്ത് എന്നീ കഥകളുടെ ആകെ തുകയാണ് ഈ ചിത്രങ്ങൾ.

ചിത്രം 1: കാനനസൗന്ദര്യം – ചിത്രകാരൻ: മുരളി നാഗപ്പുഴ
(ഒ.എൻ.വി കുറുപ്പിന്റെ ഭൂമിക്ക് ഒരു ചരമഗീതം എന്ന പുസ്തകത്തിന്റെ കവർ ചിത്രം കൂടിയാണിത്)

മുരളി നാഗപ്പുഴയുടെ ചിത്രങ്ങൾ പച്ചപ്പും സാധാരണവും ഗ്രാമീണതയും അതിനോടൊപ്പം നിറങ്ങൾ ചാർത്തിയ കൗതുകങ്ങളും തിങ്ങി നിൽക്കുന്നുണ്ടായിരിക്കും അതുകൊണ്ടുതന്നെ അന്താരാഷ്ട്രതലത്തിൽ അവാർഡുകൾ നേടിയ ചിത്രങ്ങളാണ് മുരളി നാഗപ്പുഴയുടേത്.

ഈ ചിത്രം ശാന്തവും പച്ചപ്പും നിറഞ്ഞ ഒരു വനപ്രദേശം കാണിക്കുന്നു. കുട്ടികൾ വനത്തിൽ ശാസ്ത്രീയമായി പഠിക്കുകയും ആനന്ദിക്കുകയും ചെയ്യുന്നു.ആകാശത്ത് പറക്കുന്ന പക്ഷികൾ, തഴുകുന്ന മരങ്ങൾ, നിറഞ്ഞ ചെടികൾ എന്നിവ പ്രകൃതിയുടെ സമൃദ്ധിയും നിമിഷികതയും ചൂണ്ടിക്കാട്ടുന്നു. പ്രകൃതിയോട് സ്നേഹത്തോടെ സമീപിക്കുക, അതിൽ നിന്ന് പഠിക്കുക എന്നതാണ് ഈ ചിത്രത്തിന്റെ പ്രധാന താത്പര്യം.

ചിത്രം 2:- ചിത്രകാരൻ: സബീനബി
ആഗോളതാപനത്തെക്കുറിച്ചുള്ള ഒരു അന്താരാഷ്ട്ര കാർട്ടൂൺ (റഷ്യൻ എക്കോളജിക്കൽ മൂവ്മെന്റ് സംഘടിപ്പിച്ച) മത്സരത്തിൽ നാമനിർദ്ദേശം ചെയ്യപ്പെട്ട ആർട്ടിസ്റ്റ് സിബി ഷിബുവിന്റെ ചിത്രം.

ഈ ചിത്രത്തിൽ വരൾച്ച വിണ്ടുപോയ ഭൂമിക്ക് നടുവിലൂടെ രണ്ട് ആളുകൾ നടന്നു പോവുകയാണ്, അവരുടെ കൈയിൽ വിലയേറിയ ഒരു ജലത്തുള്ളിയുണ്ട്. തിളങ്ങുന്ന ആ ഒറ്റ തുള്ളി വെള്ളം, ഒരൊറ്റ വെള്ളതുള്ളിയെ ഏറ്റവും വിലപ്പെട്ടതായി കണക്കാക്കേണ്ടി വരുന്ന ആ നിമിഷം ഓർത്ത് നോക്കൂ. പ്രകൃതിയോട് അനാദരവ് കാട്ടുമ്പോൾ അതിന്റെ പ്രത്യാഘാതം എങ്ങനെയായിരിക്കും എന്നതിന്റെ മുന്നറിയിപ്പാണ് ഈ ചിത്രം.

ആദ്യ ചിത്രം പ്രകൃതിയോട് സ്നേഹപരമായ അനുഭാവവും സംരക്ഷണവും നൽകുമ്പോൾ രണ്ടാം ചിത്രം പ്രകൃതിദുരന്തത്തിന്റെ ദാരുണ ഫലങ്ങളെയും കാണിക്കുന്നു. ഈ രണ്ട് ചിത്രങ്ങളിലൂടെ മനുഷ്യന് പ്രകൃതിയോടുള്ള ഉത്തരവാദിത്വം വ്യക്തമാക്കുന്നു.

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 6 Multiples and Factors Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 6 Solutions Multiples and Factors

Class 6 Kerala Syllabus Maths Solutions Chapter 6 Multiples and Factors Questions and Answers

Multiples and Factors Class 6 Questions and Answers Kerala Syllabus

Multiples of Multiples (Page No. 85)

Question 1.
For each of the multiples given below, find the other numbers they are multiples of:
(i) Multiples of 8
(ii) Multiples of 10
(iii) Multiples of 12
Answer:
(i) The factors of 8 are 2 and 4
8 is a multiple of 2 and 4
Therefore, the multiples of 8 are also the multiples of 2 and 4

(ii) The factors of 10 are 2 and 5
10 is a multiple of 2 and 5
Therefore, the multiples of 10 are also the multiples of 2 and 5

(iii) The factors of 12 are 2, 3, 4, and 6
12 is a multiple of 2, 3, 4, and 6
Therefore, the multiples of 12 are also the multiples of 2, 3, 4, and 6.

Question 2.
Check whether each of the statements below is true or false. For true statements, explain why they are so. For the false statements, give an example in which it is not true.
(i) All multiples of 20 are multiples of 10
(ii) All multiples of 10 are multiples of 2
(iii) All multiples of 15 are multiples of 5
(iv) All multiples of 15 are multiples of 3
(v) All multiples of 5 are multiples of 15
(vi) All multiples of 3 are multiples of 15
Answer:
(i) True
Since 20 = 2 × 10,
So every multiple of 20 is a multiple of 10.

(ii) True
Since 10 = 2 × 5,
So every multiple of 10 is a multiple of 2.
Or
Since 10 is an even number, and every multiple of 10 ends in 0, it is divisible by 2
That is, every multiple of 10 is also a multiple of 2.

(iii) True
Since 15 = 3 × 5,
So every multiple of 15 is a multiple of 5.

(iv) True
Since 15 = 3 × 5,
So every multiple of 15 is a multiple of 3.

(v) False
Not all multiples of 5 are divisible by 15.
E.g.: 10 is a multiple of 5 but not of 15.

(vi) False
Not all multiples of 3 are divisible by 15.
E.g.: 6 is a multiple of 3, but not a multiple of 15.

Primary Factors (Page No. 88)

Question 1.
Can you write the numbers below as a product of primes?
(i) 24
(ii) 35
(iii) 36
(iv) 60
(v) 100
Answer:
(i) 24 = 2 × 12
= 2 × 2 × 6
= 2 × 2 × 2 × 3
Therefore 24 = 2 × 2 × 2 × 3

(ii) 35 = 5 × 7

(iii) 36 = 2 × 18
= 2 × 2 × 9
= 2 × 2 × 3 × 3
Therefore 36 = 2 × 2 × 3 × 3

(iv) 60 = 2 × 30
= 2 × 2 × 15
= 2 × 2 × 3 × 5
Therefore 60 = 2 × 2 × 3 × 5

(v) 100 = 2 × 50
= 2 × 2 × 25
= 2 × 2 × 5 × 5
Therefore 100 = 2 × 2 × 5 × 5

Textbook Page No. 89

Question 1.
Write each of the numbers below as a product of primes.
(i) 72
(ii) 105
(iii) 144
(iv) 330
(v) 900
Answer:
(i) 72 = 12 × 6
12 = 2 × 2 × 3
6 = 2 × 3
Therefore, 72 = 2 × 2 × 2 × 3 × 3

(ii) 105 = 21 × 5
21 = 3 × 7
5 = 5
Therefore 105 = 3 × 5 × 7

(iii) 144 = 12 × 12
12 = 2 × 2 × 3
Therefore 144 = 2 × 2 × 3 × 2 × 2 × 3

(iv) 330 = 10 × 33
10 = 2 × 5
33 = 3 × 11
Therefore, 330 = 2 × 3 × 5 × 11

(v) 900 = 30 × 30
30 = 2 × 3 × 5
Therefore, 900 = 2 × 3 × 5 × 2 × 3 × 5

All Factors (Page No. 89)

Question 1.
Find all the factors of the number below:
(i) 35
(ii) 77
(iii) 26
(iv) 51
(v) 95
Answer:
(i) 35 = 1 × 35 = 5 × 7
Factors of 35 are: 1, 5, 7, and 35

(ii) 77 = 1 × 77 = 11 × 7
Factors of 77 are: 1, 7, 11, and 77

(iii) 26 = 1 × 26 = 2 × 13
Factors of 26 are: 1, 2, 13, and 26

(iv) 51 = 1 × 51 = 3 × 17
Factors of 51 are: 1, 3, 17, and 51

(v) 95 = 1 × 95 = 5 × 19
Factors of 95 are: 1, 5, 19, and 95

Textbook Page No. 90

Question 1.
Write each of the numbers below as a product of three primes and find all its factors:
(i) 66
(ii) 70
(iii) 105
(iv) 110
(v) 130
Answer:
(i) 66 is the product of three prime numbers.
66 = 2 × 3 × 11
Factors of 66 are:
1
2, 3, 11
2 × 3 = 6
2 × 11 = 22
3 × 11 = 33
Therefore, the factors are: 1, 2, 3, 6, 11, 22, 33, and 66

(ii) 70 as the product of three prime numbers.
70 = 2 × 5 × 7
Factors of 70 are:
1
2, 5, 7
2 × 5 = 10
2 × 7 = 14
5 × 7 = 35
Therefore, the factors are: 1, 2, 5, 7, 10, 14, 35, and 70

(iii) 105 is the product of three prime numbers.
105 = 3 × 5 × 7
Factors of 105 are:
1
3, 5, 7
3 × 5 = 15
3 × 7 = 21
5 × 7 = 35
Therefore, the factors are: 1, 3, 5, 7, 15, 21, 35, and 105

(iv) 110 as the product of three prime numbers.
110 = 2 × 5 × 11
Factors of 110 are:
1
2, 5, 11
2 × 5 = 10
2 × 11 = 22
5 × 11 = 55
Therefore, the factors are: 1, 2, 5, 10, 11, 22, 55, and 110

(v) 130 is the product of three prime numbers.
130 = 2 × 5 × 13
Factors of 130 are:
1
2, 5, 13
2 × 5 = 10
2 × 13 = 26
5 × 13 = 65
Therefore, the factors are: 1, 2, 5, 10, 13, 26, 65, and 130

Prime Numbers (Page No. 92)

Question 1.
Find all primes less than 100. Find the primes that differ by 2 among these.
Answer:
Prime numbers less than 100 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Prime numbers that differ by 2 are:
(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73)

Question 2.
Can the product of two natural numbers be a prime?
Answer:
The product of two natural numbers can only be a prime number if one of the numbers is 1 and the other is a prime.
If both numbers are greater than one, their product will always have more than two factors, so the result cannot be a prime.

Question 3.
Can the sum of two prime numbers be prime?
Answer:
Yes, sometimes, but not always.
If the sum of two prime numbers is a prime only when one of the numbers is 2 (the only even prime).
If 2 is added to an odd prime, the result is strange and may be prime.
But if two odd primes are added, the result is even and will never be a prime (except 2 + 2 = 4, which is not a prime number).

Class 6 Maths Chapter 6 Kerala Syllabus Multiples and Factors Questions and Answers

Class 6 Maths Multiples and Factors Questions and Answers

Question 1.
For each of the multiples given below, find the other numbers they are multiples of:
(i) Multiples of 15
(ii) Multiples of 21
(iii) Multiples of 33
Answer:
(i) The factors of 15 are 3 and 5.
15 is a multiple of 3 and 5.
Therefore, the multiples of 15 are also the multiples of 3 and 5.

(ii) The factors of 21 are 3 and 7.
21 is a multiple of 3 and 7.
Therefore, the multiples of 21 are also the multiples of 3 and 7.

(iii) The factors of are 3 and 11.
33 is a multiple of 3 and 11.
Therefore, the multiples of 33 are also the multiples of 3 and 11.

Question 2.
Write the numbers below as a product of primes?
(i) 18
(ii) 40
(iii) 150
(iv) 210
(v) 300
Answer:
(i) 18 = 2 × 9 = 2 × 3 × 3
Therefore 18 = 2 × 3 × 3

(ii) 40 = 2 × 20
= 2 × 2 × 10
= 2 × 2 × 2 × 5
Therefore 40 = 2 × 2 × 2 × 5

(iii) 150 = 2 × 75
= 2 × 3 × 25
= 2 × 3 × 5 × 5
Therefore 150 = 2 × 3 × 5 × 5

(iv) 210 = 2 × 105
= 2 × 3 × 35
= 2 × 3 × 5 × 7
Therefore 210 = 2 × 3 × 5 × 7

(v) 300 = 2 × 150
= 2 × 2 × 75
= 2 × 2 × 3 × 25
= 2 × 2 × 3 × 5 × 5
Therefore 300 = 2 × 2 × 3 × 5 × 5

Question 3.
Write the following numbers as the product of three prime numbers, and find all the factors of them.
(i) 174
(ii) 385
(iii) 182
Answer:
(i) 174 is the product of three prime numbers.
174 = 2 × 3 × 29
Factors of 174 are:
1
2, 3, 29
2 × 3 = 6
2 × 29 = 39
3 × 29 = 78
Therefore, the factors are: 1, 2, 3, 6, 29, 58, 87, 174

(ii) 385 is the product of three prime numbers.
385 = 5 × 7 × 11
Factors of 385 are:
1
5, 7, 11
5 × 7 = 35
5 × 11 = 55
7 × 11 = 77
Therefore, the factors are: 1, 5, 7, 11, 35, 55, 77, 88, and 385.

(iii) 182 as the product of three prime numbers.
182 = 2 × 7 × 13
Factors of 182 are:
1
2, 7, 13
2 × 7 = 14
2 × 13 = 26
7 × 13 = 91
Therefore, the factors are: 1, 2, 7, 13, 14, 26, 91, and 182.

Class 6 Maths Chapter 6 Notes Kerala Syllabus Multiples and Factors

→ The multiples of a natural number are the product of that number with the natural numbers 1, 2, 3,…

→ All multiples of the multiple of a number are also multiples of that number.

→ All multiples of a number are also multiples of any of its factors.

→ A natural number greater than 1, which has no factors other than 1 and itself, is called a prime number.

→ Any composite number can be written as a product of primes.

→ The only even number among the prime numbers is 2.

Multiples and factors are fundamental concepts in mathematics that help us understand the relationships between numbers. A multiple of a number is the result of multiplying that number by any natural number, while a factor is a number that divides another number exactly, without leaving a remainder.

For example, 4 is a multiple of 2, and 2 is a factor of 4. Learning about multiples and factors is essential for solving problems involving divisibility, simplifying fractions, finding the greatest common factor (GCF), the least common multiple (LCM), and much more. In this chapter, we discuss multiples of multiples, primary factors, and prime numbers.

Multiples of Multiples
The multiples of a natural number are the product of that number with the natural numbers 1, 2, 3,…
For example:
The multiples of 2 are the numbers 2, 4, 6,… obtained by multiplying the natural number by 2.
The multiples of 4 are the numbers 4, 8, 12,… obtained by multiplying the natural number by 4.
Here, all the multiples of 4 can be written as multiples of 2 also:
1 × 4 = 4
2 × 4 = 8
3 × 4 = 12
4 × 4 = 16
5 × 4 = 20
……………

2 × 2 = 4
4 × 2 = 8
6 × 2 = 12
8 × 2 = 16
10 × 2 = 20
……………….

Similarly, if 6 is a multiple of 2 and 3.
So all multiples of 6 can be written as multiples of 2 and 3.

1 × 6 = 6
2 × 6 = 12
3 × 6 = 18
4 × 6 = 24
5 × 6 = 30
………………..

3 × 2 = 6
6 × 2 = 12
9 × 2 = 18
12 × 2 = 24
15 × 2 = 30
………………….

2 × 3 = 6
4 × 3 = 12
6 × 3 = 18
8 × 3 = 24
10 × 3 = 30
………………..

In general, all multiples of the multiple of a number are also multiples of that number.

The multiples of 15 can be written as the multiples of what numbers?
Answer:
15 is a multiple of 3 and 5.
So the multiples of 10 can be written as the multiples of 2 and 5.

We have seen that multiples can also be put in terms of factors.
For example:
4 is the multiple of 2 can also be written as 2 is a factor of 4.
6 is the multiple of 2 and 3 can also be written as 2 and 3 are two factors of 6.
12 is a multiple of 3, and 4 can also be written as 3 and 4 are factors of 12.

In general, we can say that all multiples of a number are also multiples of any of its factors.

14 is a multiple of 2 and 7. Express it in the form of factors.
Answer:
14 is a multiple of 2 and 7.
Since, 2 × 7 = 14
So 2 and 7 are two factors of 14.

70 is a multiple of 2, 5, and 7. Express it in the form of factors.
Answer:
70 is a multiple of 2, 5, and 7.
Since, 2 × 5 × 7 = 70
So 2, 5, and 7 are the factors of 70.

If 2, 3, and 7 are factors of a number. Then what is that number?
Answer:
2, 3, and 7 are factors of a number.
That means 2 × 3 × 7 = 42
Therefore, the number is 42, and multiples of 42 are 2, 3, and 7.

Primary Factors
Any number can be written as the product of its factors in different ways.
For example, consider the number 70,
1 × 70 = 70
2 × 35 = 70
5 × 14 = 70
10 × 7 = 70
We can write 70 as a product of three factors, without using 1:
That is 70 = 2 × 5 × 7
The only factors of each of the numbers 2, 5, and 7 are 1 and the number itself.
For any number 1, the number itself is are factor.

The numbers below 20 with factors 1 and itself are: 1, 2, 3, 5, 7, 11, 13, 17, 19.
Such numbers, excluding 1, are said to be prime numbers.

A natural number greater than 1, which has no factors other than 1 and itself, is called a prime number.

Numbers greater than 1, which are not primes, are called composite numbers.
For example, 4 is a composite number.
Since 4 = 2 × 2

A composite number can be written as the product of a prime number.

When a number is written as the product of two factors and any one of them is not a prime, then that factor can be written as the product of two factors. This can continue till all factors are prime.

Write 48 as the product of primes.
Answer:
48 = 2 × 24
= 2 × 2 × 12
= 2 × 2 × 2 × 6
= 2 × 2 × 2 × 2 × 3
Therefore 48 = 2 × 2 × 2 × 2 × 3

Write the following as the product of primes.
(i) 30
(ii) 45
(iii) 64
Answer:
(i) 30 = 2 × 15 = 2 × 3 × 5
Therefore 30 = 2 × 3 × 5

(ii) 45 = 5 × 9 = 5 × 3 × 3
Therefore 45 = 5 × 3 × 3

(iii) 64 = 2 × 32
= 2 × 2 × 16
= 2 × 2 × 2 × 8
= 2 × 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2 × 2
Therefore 64 = 2 × 2 × 2 × 2 × 2 × 2

Product of Primes:
Once we write two numbers as a product of primes, it is easy to write the product of these numbers also as a product of primes.
For example: 12 × 24 can be split like this:
12 = 2 × 2 × 3
24 = 2 × 2 × 2 × 3
Split 12 × 24 as shown below:
288 = 12 × 24
= (2 × 2 × 3) × (2 × 2 × 2 × 3)
= 2 × 2 × 3 × 2 × 2 × 2 × 3

To split a number into a product of primes, we first split it into the product of any two factors, then split each of these factors into a product of primes, and finally put these prime factors together.
For example: Split 140 into a product of primes,
140 = 14 × 10
Next, write 14 and 10 as products of primes
14 = 2 × 7
10 = 2 × 5
We can write 140 like this;
140 = 14 × 10
= (2 × 7) × (2 × 5)
= 2 × 2 × 5 × 7

Split the following numbers into a product of primes.
(i) 420
(ii) 180
(iii) 336
Answer:
(i) 420 = 15 × 28
15 = 3 × 5
28 = 2 × 2 × 7
Therefore, 420 = 2 × 2 × 3 × 5 × 7

(ii) 180 = 12 × 15
12 = 2 × 2 × 3
15 = 3 × 5
Therefore 180 = 2 × 2 × 3 × 3 × 5
(iii) 336 = 14 × 24
14 = 2 × 7
24 = 2 × 2 × 2 × 3
Therefore, 336 = 2 × 2 × 2 × 2 × 3 × 7

All Factors
If we know the prime factors of a number, we can find all its factors.
For example, the factors of 6 are: 1, 2, 3, and 6.

Write the prime factors of 15. And what are its other factors?
Answer:
Prime factors of 15 are: 3 and 5
Factors of 15 are: 1, 3, 5, and 15

Find all the factors of the numbers given below:
(i) 42
(ii) 54
(iii) 63
Answer:
(i) 42 = 1 × 42
= 2 × 21
= 3 × 14
= 6 × 7
Therefore, factors of 42 are: 1, 2, 3, 6, 7, 14, 21, and 42.

(ii) 54 = 1 × 54
= 2 × 27
= 3 × 18
= 6 × 9
Therefore, factors of 54 are: 1, 2, 3, 6, 9, 18, 27, and 54.

(iii) 63 = 1 × 63
= 3 × 21
= 7 × 9
Therefore, factors of 63 are: 1, 3, 7, 9, 21, and 63.

Product of Three Prime Numbers:
Now let’s look at the product of three different primes.

Write 42 as the product of three prime numbers and find all the factors?
Answer:
42 as the product of three prime numbers,
42 = 2 × 3 × 7
Factors of 42 are:
1
2, 3, 7
2 × 3 = 6
2 × 7 = 14
3 × 7 = 21
Therefore, the factors are: 1, 2, 3, 6, 7, 14, 21, and 42

Write the following numbers as the product of three prime numbers, and find all the factors of it?
(i) 102
(ii) 154
(iii) 195
Answer:
(i) 102 is the product of three prime numbers.
102 = 2 × 3 × 17
Factors of 102 are:
1
2, 3, 17
2 × 3 = 6
2 × 17 = 34
3 × 17 = 51
Therefore, the factors are: 1, 2, 3, 6, 17, 34, 51, and 102

(ii) 154 is the product of three prime numbers.
154 = 2 × 7 × 11
Factors of 154 are:
1
2, 7, 11
2 × 7 = 14
2 × 11 = 22
7 × 11 = 77
Therefore, the factors are: 1, 2, 7, 11, 14, 22, 77, and 154.

(iii) 195 as the product of three prime numbers.
195 = 3 × 5 × 13
Factors of 195 are:
1
3, 5, 13
1 × 5 = 15
3 × 13 = 39
5 × 13 = 65
Therefore, the factors are: 1, 3, 5, 13, 15, 39, 65, and 195

Prime Numbers
The only even number among the prime numbers, 2, 3, 5, 7, 11,… is 2.
All primes afterwards are odd numbers. But not all odd numbers are primes;
For example: 9 = 3 × 3, 15 = 3 × 5,……. They are not prime numbers.

There is no definite pattern for the odd primes.
For example, after 3, 5, 7 are consecutive primes that differ by 2, the next prime is not 9 (which is not a prime), but 11. Thus, the difference between 7 and 11 is 4. Similarly, after the prime 31, the next prime is 37, and their difference is 6; the prime after 89 is 97, with a difference of 8. But even as such consecutive primes drift further apart, there are consecutive primes like 41 and 43 or 71 and 73 in between, which are only 2 apart. There is a technique to list all primes less than a specified number. That is, first write all numbers up to 50 in rows and columns like this:

Strike off 1 from this. Then strike off all multiples of 2, except 2:

Keep 3 and strike off all multiples of 3:

Strike all multiples of 5, except 5 itself.
If we remove the multiples of 7 other than itself also, we can see that there are no multiples, except themselves, of the other numbers that remain:

Now the numbers not struck off are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
These are the prime numbers less than 50.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 3 Pressure Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 3 Pressure Question Answer Notes

Class 8 Basic Science Chapter 3 Notes Kerala Syllabus Pressure Question Answer

Pressure Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
Give reasons.
• Heavy vehicles have more tyres.
• Some passengers experience nosebleeds when airplanes fly at high altitudes.
• The size of the bubbles rising from the bottom of an aquarium increases gradually.
• The foundation of buildings is made wider.
Answer:
• Heavy vehicles are provided with more tyres to increase the surface area of contact with the surface of the road so that it reduce the pressure on each tyre and on the surface of the road and ensures smooth motion.
• As the altitude increases the atmospheric pressure decreases. This lower atmospheric pressure leads to nosebleeds in some passengers.
• At the bottom of the aquarium, the pressure is high. For a bubble coming up the pressure experienced is less. Hence the volume of bubble increases.
• As the foundation is made wider, the pressure exerted by the building on the ground decreases because surface area of contact between building and ground increases. Thus in turn makes the building stronger and safer.

Question 2.
Which of the bottles below shows the shape of rising bubbles correctly? Explain why.

Answer:
Bottle C shows the correct shape.

Liquid pressure is greater at the bottom. For a bubble coming up pressure decreases. Hence the size of the bubbles increases as it moves from bottom to top.

Question 3.
Observe the picture below. Which of the following jars containing the same liquid, will experience the highest pressure at the bottom? Explain why.

Answer:
All the three jars containing the same liquid will experience the same pressure. Liquid pressure depends on the depth and the density of the liquid. Since the liquids are same liquid pressure is not dependent on density here. Liquid pressure is same here as the depth of the liquid (or height of liquid column) is same in all the jars.

Question 4.
Cooking is faster in a pressure cooker than in an open vessel. Explain why.
Answer:
When we cook food in a pressure cooker, the pressure inside it increases as the temperature increases. The boiling point of water is raised by this increased pressure inside it. This allows food to be cooked at a higher temperature and makes cooking faster than in an open vessel.

Basic Science Class 8 Chapter 3 Question Answer Kerala Syllabus

Question 1.
Let’s now have a look at the figure given below, in which the same weight is applied in two different ways.

A brick is placed vertically and then horizontally on a sponge (Fig(a)). In both cases, does the weight of the brick remain the same? Are the compressions on the sponge equal?
Answer:
The weight of the brick remain the same in both cases. The compressions on the sponge is not equal.

Question 2.
The weight of the child remains the same whether they stand or lie on the mattress (Fig.(b)) as shown above. Yet, the mattress is compressed more while standing. What could be the reason for this?
Answer:
The mattress is compressed more while standing because the surface area in contact is less in this case.
Activity

Fix a single nail on the cardboard as shown in the figure and place a balloon on top of it. Place a small weight on the balloon. Observe what happens to the balloon. Now, nail multiple pins very close to each other as shown in the second figure. Place the same weight on top of the balloon. Observe the result.

Observation
When weight is placed on the balloon, placed on a single nail it bursts. When the same weight is placed on the top of the balloon, placed on the top of multiple pins it will not burst.

Question 3.
Record the observations from the above activity in the table below.

Answer:

Question 4.
Analyze the table.
Answer:
Even when the same force is applied in each case, the results vary depending on the change in surface area in contact.

Question 5.
What is pressure?
Answer:
Pressure is the force acting normally per unit area.

Question 6.

Who will win the race? Write your opinion.
Answer:
Duck will win the race. Hen cannot move at the same speed as that of duck in mud.

Question 7.
What could be the reason?
Answer:
Webbed feet of duck makes more surface area of contact with the mud. Thus the pressure experienced on its feet is less and it can walk over mud easily. But the hen’s feet make less surface area of contact with the mud and hence experience more pressure which makes it difficult for it to move over the mud.

Question 8.
See the figure of two concrete slabs placed in sand.

Each of them is made of cubes weighing 18 N fixed together. The area of one side of the cube is 0.36 m²


Referring to the figures, complete the following table.
Answer:

Measurements	Slab 1	Slab 2
Force experienced on the sand	18 N	18 N
Surface area of contact	0.36 × 4 = 1.44 m2	0.36 m2
Force experienced per unit area	18/1.44 = 12.5 N/m2	18/0.36 = 50 N/m2

The weight of the slab acts normally on the sand. This is called thrust.
The force acting normally on a surface is called thrust. The force acting normally per unit area is pressure.
If pressure is denoted by P, thrust by F and area by A, pressure = \(\frac{\text { Thrust }}{\text { Area }}\) P = \(\frac{F}{A}\)
Unit of pressure = \(\frac{\text { Unit of thrust }}{\text { Unit of area }}\)
The SI unit of pressure is pascal. 1 pascal = 1N/m²

Question 9.
Now, find out the pressure exerted by the above slabs, on the sand.
Answer:
Pressure exerted by slab 1 = 12.5 N/ m² or 12.5 Pa
Pressure exerted by slab 2 = 50 N/ m² or 50 Pa

Question 10.
A concrete block weighing 100 N is placed on sand in three different ways as shown in the figure below.

a) In which position will the highest pressure be exerted on the sand?
Answer:
Higher pressure will be exerted on the sand when the brick is placed vertically (position 3)

b) Which position will cause the concrete block to sink deeper into the sand? Why?
Answer:
The third position (vertical) will cause the concrete block to sink deeper into the sand. It is because the surface area of contact between the concrete block and sand is less here.

When thrust remains constant, the pressure is inversely proportional to the surface area. This means, when the surface area increases, the pressure decreases and vice versa.

Question 11.
Can you explain the reason why a lorry would not get stuck in the mud if a wooden plank is put.
Answer:
When a wooden plank is put over mud, the surface area of contact between the lorry and the mud increases, so the pressure experienced on the mud becomes less, so that the lorry will not get stuck in mud.

Question 12.
Try to lift your school bag by tying it with a thin twine. How do you feel?
Answer:
It is very difficult to lift a school bag by tying it with a thin twine because pressure felt on the hands is very high.

Question 13.
What if you lift it using a wide strap as shown in the figure? Can you explain the difference by relating its area to pressure?

Answer:
When the bag is lifted using a wide strap the surface area of contact between the bag and the hand is more, so the pressure felt on the hands is less.

But when thin strap is used, the surface area of contact between the bag and the hand is less, so the pressure felt on the hands is more. This makes it difficult to lift the bag.

Question 14.
Similarly, explain the following situations in the figure. Suggest more such situations.

Answer:

• The edge of a knife is made thinner – Thin edges have less surface area of contact which helps to exert more pressure on the objects to be cut.
• The wheels of bulldozers are connected by wide chains -Wide chains make more surface area of contact with the mud to exert less pressure over it. This helps to prevent the wheels getting stuck in the mud and ensures its smooth motion.
• A person lying on a bed of nails will not bleed- As the bed of nails make large surface area of contact with the person, the pressure exerted on the person is less. So person will not bleed.
• The basement of dams is made wider -Basements are made wider to increase the surface area of contact betw een walls and basement and thus to withstand the high pressure exerted by water stored in dams.

Question 15.
Do only solids exert pressure?
Answer:
No

Question 16.
Can’t liquids and gases exert pressure?
Answer:
Yes, Liquids and gases also exert pressure.
Activity
Tie a polythene cover tightly around your hand as shown in the figure. Dip your hand into a bucket of water.

Observation: The cover sticks to your hand.
Inference: This happens due to the pressure exerted by the water on the polythene cover.
Thus we can understand that, just like solids, liquids can also exert pressure.

The normal force exerted by a liquid is called thrust. The thrust that a liquid exerts per unit area is called liquid pressure. Liquids exert pressure on all sides of the container.

Question 17.
If the force of water jetting out is greater, wouldn’t the pressure also be.greater?
Answer:
Yes. When the force of water jetting out is greater the pressure also will be greater.

Question 18.
What happens to the water stream when the water level decreases?
Answer:
When the water level decreases the force of water jetting out becomes less and hence the pressure also will be lesser.

Question 19.
What can be inferred from this?
Answer:
As the force decreases due to decrease in water level pressure also decreases and pressure depends on the depth of the liquid.

Pressure increases as the depth in a liquid increases.

Activity
Attach a syringe on a board as shown in the figure below.

Attach one end of the I.V. set tube to the hub of the syringe. Attach the piston of the syringe to the bottom end of a pen barrel. Arrange it so that the pen barrel deflects when the piston moves. Now fix a thin stick to the top end of pen barrel as a pointer. Attach the other end of the I.V. set to the plastic bottle. Fill the plastic bottle with water and place it at a higher level than the syringe.

Question 20.
Raise the bottle. The force applied by the piston will change according to the pressure of the water. Now’ you can see the thin stick deflecting according to the force exerted in the piston, right?
Answer:
Yes. We can see the thin stick deflecting according to the force exerted in the piston.

Question 21.
Raise the bottle further. What do you observe? What is the reason?
Answer:
The thin stick deflects more. When the bottle is raised further, height of the liquid column increases and pressure of water increases and the force applied by the piston also increases.

Question 22.
Repeat the experiment using salt solution instead of water. Why does the thin stick deflect more?
Answer:
Water and saline water have different densities. Density of saline water is higher than that of water. When density of a liquid increases, pressure exerted by it increases. So the force exerted on the piston by saline water is more. That is why the piston moves more, causing greater deflection of the thin stick.

The density of a liquid influences its pressure. If the density of a liquid mercases, the pressure also increases.

The instrument used to measure liquid pressure is called a manometer.

Question 23.
Do you see any difference in the level of water in the tube?
Answer:
Yes . When slight pressure is applied the level of water in tube rises up.
Apply more pressure. You can see a difference in water levels also.

Question 24.
Now take different solutions in the beaker and measure the pressure at different levels. Write your observations in the table below.

Answer:

Position of the funnel	Difference in the water levels of U-tube/pressure
On the surface	In water	In saline water	In kerosene
Midway in the beaker	12 cm	12 cm	12 cm
At the bottom of the beaker	18 cm	20 cm	16 cm
Position of the funnel	24 cm	26 cm	20 cm

Question 25.
Analyse the table.
a) In which position is the pressure highest?
Answer:
Pressure is highest at the bottom of the beaker.

b) Which liquid exerts the highest pressure?
Answer:
Saline water exerts the highest pressure.

Question 26.
Fill a deep vessel with water. Dip a straw into the water and blow gently. Do bubbles come up?

Answer:
Yes, bubbles come up.

Question 27.
What happens to their size when they reach the surface?
Answer:
Size of the bubble increases.

Question 28.
Explain their change in size.
Answer:
The increase in size of bubble when it reach the surface is due to the decrease in pressure.

Question 29.
Explain why dams are built with a wider base.
Answer:
Dams are built with a wider base to withstand the pressure exerted by the water stored in it. When there is a wider base, the surface area of contact between water and the base of the dam is more and so pressure exerted decreases.

Question 30.
Can you think of more situations in daily life where liquid pressure is experienced?
Answer:

• Water tanks at home are placed at a height to exert liquid pressure to the pipelines in the building.
• Oil tankers are made of thick metals to withstand the pressure of huge volume of oil.
• When a bucket of water is carried on head pressure is felt on head.

Factors influencing gas pressure
Question 31.
Inflate a balloon at its maximum. What happens when the balloon is inflated further? Will it eventually burst?
Answer:
The balloon bursts as it fails to withstand the air pressure inside.

Question 32.
Find the change in the number of gas particles, in figure shown below, when more air is filled. What can be inferred from this?

Answer:
When more air is filled, the number of gas particles increases and the gas pressure also increases.

Gas pressure depends on the number of particles. As the number of particles increases, gas pressure also increases.

Question 33.
Fill the same quantity of air into two balloons of different sizes. Now the number of gas particles in both . balloons is equal, right? Which of the balloons has more pressure inside? What can be inferred from this?
Answer:
As the same quantity of air is filled into two balloons the number of gas particles in both balloons is equal. But the balloon with smaller size has more pressure as it has less volume. When the volume is less, the gas particles has less space to move around. This leads to the frequent collision with the walls of the balloon and hence more pressure is exerted.

Gas pressure depends on its volume.

Question 34.
Leave an inflated balloon in the sunlight. What do you observe? What is the change in the pressure when the air inside gets heated up?
Answer:
When the air inside gets heated up, the pressure inside the balloon increases and it bursts after some time.

Gas pressure depends on its temperature.

When we cook food in a pressure cooker, the pressure inside increases as the temperature increases.

Question 35.
List the factors influencing gas pressure.
Answer:

• Number of particles
• Volume
• Temperature

Question 36.
Is atmospheric pressure the same everywhere on the Earth? Observe the figure.

Which book will experience the highest pressure? Why?
Answer:
No, the atmospheric pressure is different at different places on earth. The book at the bottom experiences the highest pressure as it experiences the atmospheric pressure and the combined weight of all the books stacked above.

Question 37.
Similarly, we know that atmospheric pressure is the weight of air column experienced per unit area. So, what is the change that occurs in atmospheric pressure as altitude increases? Note your inferences.
Answer:
Atmospheric pressure decreases as height from the Earth’s surface increases.

Atmospheric pressure at sea level is known as standard atmospheric pressure. It is defined as the weight of a mercury column that is 0.76 m high, with a unit cross sectional area, exerting a pressure of 1 atm. The instrument used to measure atmospheric pressure is a barometer.

Question 38.
Take an aluminum can and fill it with hot water. Pour out the hot water and immediately seal the can. Immerse the can in cold water. What do you observe?

When the can cools down, the pressure inside the can decreases. The external atmospheric pressure is strong enough to crush the can.

Question 39.
Calculate the force exerted by atmospheric pressure on a table surface of area 1 m². Atmospheric pressure is 101325 Pascal.
Answer:
Force = Pressure × Area
= 1 Pa × 1 m²
= 101325 N
This is equivalent to the weight of an object with a mass of 10339 kg. If such a large force is exerted on the table, it doesn’t collapse. It is because the pressure is applied from all sides on it and so the forces get balanced.

Gas pressure is exerted in all directions, similar to that of liquids.

Activity
Fill a glass with water and cover it with a cardboard. Hold the cardboard with your hand and turn the glass upside down. Remove your hands slowly.

Question 40.
Does the water fall?
Answer:
The force due to atmospheric pressure can hold the entire weight of the water in the glass. So the water does not fall down.
The equilibrium of pressure between gases and liquids is a common phenomenon occurring in nature.

Question 41.
What happens to the candle flame?
Answer:
The candle flame goes out.

Question 42.
What happens to the air pressure inside the glass?
Answer:
Air pressure inside the glass decreases.

Question 43.
What is the change in the water level?
Answer:
Water level inside the glass rises.
The water level remains stable when the pressure inside the glass and the atmospheric pressure outside the glass are in equilibrium.

Question 44.
Gently pull back the piston of a syringe and seal the opening with your hand. Then pull the piston all the way back and release. What do you see?
Answer:
The piston will go back. When the syringe opening is sealed and the piston is pulled back and released, the piston will move into the syringe barrel. This movement is due to the difference in pressure between the outside and inside of the syringe. The higher outside pressure forces the piston back into the syringe barrel.

Question 45.
Why does the piston go back? Discuss the reason.
Answer:
The piston move back into the syringe barrel, because the pressure inside the syringe is less than the pressure outside. So the higher pressure outside forces the piston back into ehe syring barrel.

Question 46.
Read the following situations and explain them based on atmospheric pressure.
• Rubber suckers stick to smooth surfaces.
• Mountain climbers often experience nosebleeds at high altitudes.
• Holes are made on the injection bottle with a needle during a drip injection.
• Passengers travelling uphill in vehicles on a ghat road experience ear pain.
Answer:
• When a rubber sucker is pressed on a smooth surface, it pushes the air out. This creates low pressure inside. The higher air pressure outside pushes it onto the smooth surface making, it stick.
• Mountain climbers often experience nosebleeds at high altitudes because as altitude increases atmospheric pressure decreases.
• For the smooth flow of liquid there should be air flow. In the absence of a hole, air cannot enter the bottle and the flow stops. Making a hole lets air into the injection bottle and keeps the pressure balanced, ensuring continuous flow.
• As the altitude increases the atmospheric pressure decreases. But the pressure inside the ears remain the same for a while and this difference in pressure causes a feeling of pain in the ears.

Class 8 Basic Science Chapter 3 Question Answer Extended Activities

Question 1.
Conduct a study on the topic ‘Rain and Pressure’ and prepare a project report.
Answer:
An example of a project report is shown below.
Title: Rain and pressure

Introduction:
Rainfall is one of the most important phenomena. It plays a major role in agriculture, environment and many more day to day activities. One of the important factors that influence weather patterns is pressure. Here we explore the relationship between pressure and rain.

Main points to be included in the content

• The continuous movement of water through the phases of evaporation, condensation, precipitation, and collecting is known as the water cycle.
• Water vapour, which rises into the atmosphere, is created when the Sun heats water in lakes, rivers, and seas.
• In regions of low pressure, the air is generally less dense and rising.
• As warm air rises it cools down.
• When air cools, the vapour in it turns back into small liquid droplets or ice crystals to form clouds. This process is called condensation.
• Precipitation is the term for the water that returns to the earth as rain, snow, or hail when the clouds becomes heavy.
• The cycle then restarts when the water gets collected in lakes, rivers, or underground, with atmospheric pressure being a major factor in causing rain to fall and water vapour to rise.
• When the air pressure is high it means the air is generally more dense and sinking.
• Sinking air warms up.
• Warm air can hold more water vapour without condensing into droplets.
• So high pressure areas are usually associated with sunny weather and clear skies.
  (You can also refer to weather reports in newspapers and include points from it, add flowcharts and diagrams, various weather study reports)

Conclusion
Pressure and rainfall are interrelated.

Question 2.
Attach a pipe to each end of a plastic bottle, ensuring it airtight. One pipe should be longer and the other shorter. Place one end of the pipe in a bucket filled with water and placed at a highest level. Place the shorter pipe in another vessel. Squeeze and release the plastic bottle two or three times. Does the water from the bucket flow into the lower vessel? State the reason for this.
Answer:
Yes, the water from the bucket will flow into the lower vessel. When the bottle is squeezed, the pressure inside it increases. This higher pressure pushes air into the water through the longer pipe and forces water to rise up through it and then water is pushed down the longer pipe and into the shorter pipe, which then flows into the lower vessel due to gravity.

Pressure Class 8 Notes

Class 8 Basic Science Pressure Notes Kerala Syllabus

• The force acting normally on a surface is called thrust. The force acting normally per unit area is pressure.
• If pressure is denoted by P, thrust by F and area by A, Pressure = \(\frac{\text { Thrust }}{\text { Area }}\) P = \(\frac{\mathrm{F}}{\mathrm{~A}}\)
• Unit of pressure = \(\frac{\text { Unit of thrust }}{\text { Unit of area }}\). The SI unit of pressure is pascal. 1 pascal = 1N/m²
• When thrust remains constant, the pressure is inversely proportional to the surface area. This means, when the surface area increases, the pressure decreases and vice versa.
• The normal force exerted by a liquid is called thrust. The thrust that a liquid exerts per unit area is called liquid pressure.
• The factors affecting liquid pressure are
  • Height of the liquid column (h)-As height of the liquid column increases, liquid pressure increases.
  • Density of the liquid (d)-As density of liquid increases, liquid pressure increases.
• The instrument used to measure liquid pressure is called a manometer.
• The force exerted by a gas normally on a unit area is gas pressure.
• Factors influencing gas pressure.
  • Number of particles-As number of particles increases, gas pressure increases.
  • Volume-As volume increases, gas pressure decreases.
  • Temperature-As temperature increases, gas pressure increases.
• Atmospheric air exerts pressure on objects.
• The weight of air column per unit area on the earth’s surface is atmospheric pressure. The unit of atmospheric pressure is ‘bar’.
• Atmospheric pressure at sea level is known as standard atmospheric pressure. The instrument used to measure atmospheric pressure is a barometer.
• Gas pressure is exerted in all directions, similar to that of liquids.

INTRODUCTION

The concept of pressure is important in science and it is an inevitable part of our day to day life. It explains simple daily life situations like using a sharp knife to cut better, making use of syringes, air pumps to applications like making the base of dams wider and weather predictions. Not only solids, but liquids and gases also exert pressure. This chapter deals with topics like pressure and its measurement, liquid pressure, gas pressure and atmospheric pressure.

PRESSURE AND ITS MEASUREMENT
We have learnt about different types of forces. The earth exerts an attractive force on all objects. That is the weight of the object.

ACTIVITY TO FIND OUT THE FACTORS LIQUID PRESSURE DEPENDS ON
Make a hole at the bottom of a plastic bottle. Close the hole and fill the bottle with water. Open the hole and observe the water jetting out.

Repeat the experiment using bottles of different shapes and sizes.

MAKING A MANOMETER

Fix a plastic tube on a board in a U shape. Fill it with water. Connect a funnel to one end of the tube as shown in the figure. Make a diaphragm with a balloon on the mouth of the funnel. Attach a scale to the board. Now the manometer is ready.
Apply slight pressure with your finger on the balloon.

GAS PRESSURE
Just like solids and liquids, gases also have the ability to exert pressure. As we know, pressure is measured when air is filled into vehicle tyres.

Pressure is the force exerted by the particles of a gas on unit area of a surface. This is due to the collision of gas particles.

The force exerted by a gas normally on a unit area is gas pressure.

ATMOSPHERIC PRESSURE
The atmosphere is a blanket of air surrounding the Earth. The density of air decreases as we go higher.

ACTIVITY TO UNDERSTAND IF ATMOSPHERE CAN EXERT PRESSURE
Place a scale on the table as shown in the figure. Drop an ice cream ball on the scale from a small height. The scale falls down. Place an A4 size paper on the scale. Drop the ball again from the same height.

Now, the scale does not fall down. It is because of the weight of the air above the paper. The atmospheric pressure counteracts the force exerted by the ball.

Atmospheric air exerts pressure on objects.

From the above activity, we understood the influence of atmospheric pressure.

The weight of air column per unit area on the earth’s surface is atmospheric pressure. The unit of atmospheric pressure is ‘bar’.

PRESSURE BALANCE
Activity
Pour water in a flat container. Place a lit candle in it. Cover the candle with a glass.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 2 Motion and Force Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 2 Motion and Force Question Answer Notes

Class 8 Basic Science Chapter 2 Notes Kerala Syllabus Motion and Force Question Answer

Motion and Force Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
A car starts from A and reaches B, which is 75 m away. The uniform speed of the car is 25 m/s. Another car with a uniform speed of 30 m/s starts from A and reaches B through C. Which car will reach the destination first?

Answer:
Speed = \(\frac{\text { distance travelled }}{\text { time taken }}\)

Car 1
Distance travelled_from A to B = 75 m
Speed of car 1 = 25 m/s
Time taken bv Car 1 = \(\frac{\text { Distance }}{\text { Speed }}\) = 75 ÷ 25 = 3 S

Car 2
Distance travelled from A to B through C = AC + CB = 50 + 70 = 120 m
Speed of car 2 = 30 m/s
Time taken by Car 2 = \(\frac{\text { Distance }}{\text { Speed }}\) = 120 ÷ 30 = 4 s
So Car 1 will reach the destination first because it takes less time.

Question 2.
A bus starts from C and reaches D in 7 s. If the uniform speed of the bus is 50 m/s, find the distance from C to D.
Answer:
Speed of the bus = 50 m/s
Time taken = 7 s
Speed = \(\frac{\text { distance travelled }}{\text { time taken }}\)
Distance travelled from C to D = Speed × Time = 50 × 7 = 350 m

Question 3.
How long will it take to hear thunder from 12000 m away? (The speed of sound is 340 m/s).
Answer:
Speed of sound = 340 m/s
Distance travelled by thunder = 12000 m
Speed = \(\frac{\text { distance travelled }}{\text { time taken }}\)
Time taken to hear thunder \(\frac{\text { Distance }}{\text { Speed }}\) = 12000 ÷ 340 = 35.29 s

Question 4.
Complete the puzzle given below:

Answer:

• Starts moving a stationary object
• Changes the direction of motion
• Changes the speed of a moving object
• Changes the shape of an object
• Changes the size of an object

Basic Science Class 8 Chapter 2 Question Answer Kerala Syllabus

Question 1.
Analyze the picture given above. What are the situations shown?
Answer:

• Child sitting on a moving giant wheel
• Person riding a horse
• Log of wood in a moving lorry
• Twig in the beak of a flying bird
• Earth in the solar system
• A book on a table

Question 2.
Tabulate the above situations that the objects change their position relative to the surroundings and those do not.

Answer:

Objects changing position relative to surroundings	Objects not changing position relative to surroundings
• Child sitting on a moving giant wheel • Person riding a horse • Log of wood in a moving lorry • Twig in the beak of a flying bird • Earth in the solar system	• A book on a table

Objects that change the position with respect to their surroundings are considered to be in motion, while those that do not change position are considered to be stationary.

Question 3.
The above situations are given in the table below. Complete the table suitably by putting ✓ marks.

Answer:

From the table, we can infer that to determine whether an object is moving, we need to refer to another object. The object that is used as a reference is called the reference object.

The object taken to determine the state of motion or the state of rest of a body is called the reference object.

If an object changes its position relative to the reference object, it is said to be in motion, and if it does not change position, it is said to be stationary. Moving objects undergo a change in position.

Question 4.
The picture of a 400 m track is given below.

What will be the length of the path travelled by an athlete completing two rounds on this track?
Answer:
400 + 400 = 800 m

Distance is the length of path travelled by an object. The SI unit of distance is metre.

Odometre

An odometre records distance travelled by a vehicle in kilometre.

Question 5.
Since 1 km = 1000 m, calculate how far a vehicle has travelled so far, according to its odometer given above.
Answer:
1,44,969 km = 14,49,69,000 m

Question 6.
Figure 2.5 shows a railway track from A to F. Note down the distance traveled by the train upon reaching each place.

Answer:

Place	Distance traveled
At A	0 km
While reaching B	60 km
While reaching C	110 km
While reaching D	150 km
While reaching E	250 km
While reaching F	320 km

Question 7.
Bus A travelled 75 m in 5 seconds. Bus B travelled 112 m in 7 seconds. Which bus was faster?
Answer:
Speed of Bus A = \(\frac{\text { distance travelled }}{\text { time taken }}\) = \(\frac{75 m}{5 s}\) = 15 m/s
Speed of Bus B = \(\frac{\text { distance travelled }}{\text { time taken }}\) = \(\frac{112 m}{7 s}\) = 16 m/s
Bus B was faster.
Speedometer

Speedometre is the device that shows the speed of a vehicle.

Question 8.
The details of 3 children who participated in a 400 m running race in school sports meet are given in the table below. Can we find the fastest athlete among them.

Answer:

Child	Distance	Time	Speed
A	400 m	180 s	\(\frac{400}{180}\) = 2.22 m/s
B	400 m	120 s	\(\frac{400}{120}\) = 3.33 m/s
C	400 m	400 s	\(\frac{400}{400}\) = 1 m/s

Here, speed is the distance travelled in one second. The distance travelled in one second will not always be the same.

Question 9.
Look at the clock dial shown.
a) Complete the table based on the movement of the tip of the second hand (P in the figure). You can measure the distance between different points using a thread.


Answer:

Change in position	Distance (cm)	Time (s)
From A to C	10	10
From C to E	10	10
From E to G	10	10

Change in position	Distance (cm)	Time (s)
From A to D	15	15
From D to G	15	15
From G to J	15	15

b) What is the distance travelled in every 10 seconds?
Answer:
10 cm

c) What is the distance travelled in every 15 seconds?
Answer:
15 cm

If an object covers equal distances in equal time intervals, it is said to have uniform speed.

If an object travels equal distance in equal intervals of time, it is in uniform speed.

In our daily life, all movements are not in uniform speed.
Birds flying, humans walking, a ball rolling, etc., do not move with uniform speed. They move with nonuniform speed.

If an object travels unequal distance in equal intervals of time, it is in non-uniform speed.

Question 10.
We see vehicles ranging from slow-moving bicycles to high-speed cars on the roads. Road accidents due to excessive speed and carelessness are daily news. What can we do to avoid such road accidents? Discuss
Answer:
Road accidents caused by excessive speed and carelessness happen every day. To avoid such accidents, we : should always drive carefully and follow traffic rules. We must not rush or speed while driving. Wearing seat belts, paying attention while crossing the road, and avoiding distractions like using mobile phones while driving are also important. Being alert and responsible while on the road helps keep everyone safe.

Question 11.
Examine the figures.


Identify whether the force applied in each case is a contact or a non-contact force.

Answer:

Situation	Force	Through Contact / Non-Contact
Mango falling downward	Earth’s Gravitational Force	Through Non-Contact
Leaves swaying in the wind	Force applied by wind	Through Contact
Magnet attracting a pin	Magnetic force	Through Non-Contact
Hammering a nail	Muscular force	Through Contact
Swimming	Muscular force	Through Contact

The force experienced when objects come into contact with each other is called contact force. The force effected when there is no contact with the object is called non-contact force.

The standard unit of force is called the newton. It can be represented by the letter N. The capital letter is used because it is the first letter of the scientist’s name.

Question 12.
Where do we make use of friction in our life?
Answer:
Friction helps us walk, hold objects, drive, write, and stop vehicles. It provides grip and prevents slipping in everyday life.

Question 13.
Rub your palms together. Do you feel the warmth?
Answer:
When you rub your palms together, you can feel warmth. This happens because rubbing creates friction, which generates heat. Friction is the force that opposes the motion of two surfaces sliding over each other. So, the more you rub your palms, the more heat is produced, making your palms warm.

The same thing happens when striking a matchstick and lighting a lighter.

Advantages of friction

• Helps to hold objects firmly.
• Helps us in walking.
• Helps vehicles to move without slipping.
• Slows the fall and keeps the parachute open.

Question 14.
Is friction always beneficial?
Answer:
No, friction is not always beneficial. It has both advantages and disadvantages depending on the situation. Friction is helpful when we need to hold on or stop slipping. But in machines, too much friction is a problem because it wastes energy and makes parts wear out.

Disadvantages of friction

• Surfaces in contact wears out.
• Obstructs smooth movement of machine parts.
• Continuous friction between the bones during movement causes knee wear.

Question 15.
List some examples of lubricants.
Answer:

• Oil (such as motor oil, cooking oil)
• Grease (used in bicycles, machines)
• Wax (like bees wax)
• Soap solutions (used in some machinery)
• Graphite is a solid lubricant. It is commonly used as a lubricant between machine parts at high temperature.

Ball bearings seen in connection with tyres in vehicles and between hubs and the axles of ceiling fans and bicycles are used to reduce friction with the axle.

Airplanes and boats are made in special shapes to reduce friction. Smooth, streamlined designs help them move easily through air and water, making travel faster and easier by decreasing the resistance they face.

This method of reducing friction by changing the shape is called streamlining.

Question 16.
Write the various ways to reduce friction?
Answer:

• Use lubricants like oil, grease, or wax.
• Make objects streamlined.
• Use ball bearings in machines and vehicles.

Class 8 Basic Science Chapter 2 Question Answer Extended Activities

Question 1.
Prepare and present a seminar paper on the ways to reduce road accidents.
Answer:
Hints
Main Ways to Reduce Road Accidents
1. Follow Traffic Rules

• Obey speed limits
• Use signals and helmets

2. Improve Road Conditions

• Maintain roads properly
• Install clear signs

3. Raise Awareness

• Educate people about road safety
• Conduct awareness campaigns

4. Enforce Laws Strictly

• Penalties for Violations
• Regular checks for drunk driving

5. Use Technology

• Speed cameras
• Traffic lights and CCTV

6. Encourage Public Transport
• Less traffic, safer roads

7. Pedestrian Safety

• Crosswalks and footpaths
• Educate pedestrians

Question 2.
Prepare a science article on friction in daily life.
Answer:

• Friction is a force that opposes the motion of two surfaces sliding against each other. It is a common and important force we experience every day. Friction helps us in many activities, but sometimes it can also cause problems.
• Friction occurs when two surfaces come into contact and resist sliding. It acts in the opposite direction to movement. The amount of friction depends on the nature of the surfaces and how hard they press together.

Uses of Friction in Daily Life

• Walking and Running: Friction between shoes and the ground prevents slipping.
• Writing: Friction between pen and paper helps in writing.
• Driving: Friction between tyres and road helps in moving the vehicle without slipping.
• Climbing: Friction helps climbers grip rocks or walls.

Problems Caused by Friction

• Friction causes wear and tear of objects, like shoes or machine parts.
• It can slow down moving objects and waste energy.

How to Reduce Friction

• Use lubricants like oil or grease on machines.
• Smoothen surfaces to decrease friction.
• Use wheels and ball bearings.

Motion and Force Class 8 Notes

Class 8 Basic Science Motion and Force Notes Kerala Syllabus

• If an object changes its position relative to the reference object, it is said to be in motion, and if it does not change position, it is said to be stationary. Moving objects undergo a change in position.
• Distance is the length of path travelled by an object. The SI unit of distance is metre.
• An odometre records distance travelled by a vehicle in kilometre and a speedometre is the device that shows the speed of a vehicle in km/h.
• Speed is the distance travelled by an object in unit time. The SI unit of speed is m/s.
• If an object travels equal distance in equal intervals of time, it is in uniform speed.
• If an object travels unequal distance in equal intervals of time, it is in non-uniform speed.
• Force is a push or pull that changes the shape, size, volume, state of rest or state of motion of a body. The standard unit of force is called the newton.

INTRODUCTION

Understanding how objects move and interact is fundamental for the study of physics. This chapter introduces key concepts related to motion, including the difference between being at rest and in motion, and how distance and speed describe an object’s movement. We explore the forces that cause changes in motion, distinguishing between contact and non-contact forces, and examine the role of friction, a force that both opposes motion and its numerous practical applications in daily life. By grasping these fundamental ideas, we can better understand motion and force seen around us and learn how to control them to make our daily activities easier and safer.

STATE OF MOTION AND STATE OF REST
In our daily life we always engage in some activities. We move in various situations like running, playing and walking. A bird flying, a car running, leaves swaying in the wind – all these are different forms of motion.

SPEED

A photo finish picture of the 100 m race in the Olympics is given above. The athlete who completed 100 m in the least time would be the winner.

Experiment

Take a tall glass jar. Mark the top and bottom as A and B, respectively. Fill the glass jar with glycerin and drop a stone from the top. Start the stopwatch when the stone reaches A. Stop the watch when the stone reaches B.
Time taken for the stone to reach B = 5 s
Distance travelled = 30 cm
Distance travelled bv the stone in one second = \(\frac{30}{5}\) = 6 cm/s

Speed is the distance travelled by an object in unit time. The SI unit of speed is m/s. Speed = \(\frac{\text { distance travelled }}{\text { time taken }}\)

Although the SI unit of speed is m/s, the speed of vehicles is usually expressed in km/h.

Relation between km/h and m/s.
1 km/h = \(\frac{1000 m}{3600 s}\) = \(\frac{5}{18}\) m/s

FORCE

• Pulling a trolley
• Pushing a car
• Pushing a wall
• Pulling a table

We can see a push or a pull in all these situations. This is force. It is not only applied during pushing and pulling. Force can be applied in many ways, such as pressing, lifting, twisting, or even when objects attract each other (like gravity). Any action that causes a change in the motion or shape of an object involves applying force.

• Catching a ball
• Rolling dough to chapati
• Kicking a ball
• Hitting a cork

Here, we can see that force is applied to change the shape and direction of the object and to stop a moving object.

Force is a push or pull that changes the shape, size, volume, state of rest or state of motion of a body.

In all the situations shown above, force is applied through direct contact with the object. It is not necessary to have a direct contact in all the cases.

FRICTIONAL FORCE

You must have noticed that a ball rolling freely on the floor gradually comes to rest. The force responsible for this is called friction.

Let’s try an activity.
Take a wooden block and make one side smooth leaving the other rough. Slide its smooth surface down the inclined plane Then try to slide its rough side down the inclined plane.

Observation
More force is effected against the movement on a rough surface. This is frictional force. You can see that the speed decreases with the increase in friction.

When a surface moves or tries to move over another surface, a parallel force is produced between them against their relative motion. This is frictional force.

Activity

Slide a heavy box on a rough surface. Then move the same box on a trolley with tyres.

Observation

• The box moves faster on the trolley with tyres.
• The frictional force decreases in this case because rolling friction (or rolling resistance) is less than sliding friction.

Using tyres reduces the frictional force, making it easier and quicker to move the box.

More friction occurs when sliding. This is called sliding friction. When using a trolley, the tyre rolls. This is rolling friction. Sliding friction is greater than rolling friction.

WAYS TO REDUCE FRICTION
To understand the ways to reduce friction, we need to know the factors that influence it.
Activity

Take a wooden block, an ice block, and a rubber piece of the same mass. Allow them to slide down an inclined plane. The area in contact with the inclined surface should be equal in all cases.

Observations

• The rubber piece will slide slowly because of higher friction.
• The wooden block will slide faster than rubber but slower than ice.
• The ice block will slide the fastest because it has the least friction.

Nature of the surfaces in contact influence the frictional force
Experiment
Take some water in a glass jar and drop a round stone and a sharp-edged stone of equal mass into the water.

Observation
The round stone has a smooth shape, so it moves easily through water. It faces less friction and falls faster.
The sharp-edged stone has an uneven, rough shape. It faces more friction and falls slower.
From this, we can understand that the shape of the object affects friction.

Factors that affect friction

• Nature of surfaces in contact:
  Rough surfaces create more friction than smooth surfaces.
• Area of contact

Larger contact areas can increase friction.
We have seen that friction occurs when our palms are rubbed together. If oil is applied on the hands and rubbed together, they slip quickly. From this, it can be assumed thar friction is reduced when oil is applied.

Substances that help to reduce friction between contacting surfaces are lubricants.