Class 9

A thorough understanding of Kerala Syllabus 9th Standard Biology Textbook Solutions Chapter 3 Respiration and Excretion Notes Questions and Answers English Medium can improve academic performance.

SCERT Class 9 Biology Chapter 3 Notes Questions and Answers Respiration and Excretion

Std 9 Biology Chapter 3 Notes Pdf Kerala Syllabus English Medium Solutions Questions and Answers

Class 9 Biology Chapter 3 Let Us Assess Answers Respiration and Excretion

Question 1.
Which among the following given below is not the characteristic feature of an efficient gaseous exchange surface?
a) Thick cell wall
b) Proximity to blood capillaries
c) Moist membrane
d) Large surface area
Answer:
a. Thick cell wall

Question 2.
Redraw the diagram and label the parts.

Answer:
a) Trachea
b) Lungs
c) Diaphragm

Question 3.
Write the role of each of the following in the exchange of gases.
a) Plasma
b) RBC
c) Haemoglobin
d) Tissue fluid
Answer:
a) Plasma: The cells need oxygen for metabolism, which creates carbon dioxide as a waste product. The carbon dioxide is absorbed from the cells by the blood plasma (some of it binds to hemoglobin too) and is transported back to the lungs in the bloodstream.

b) RBC: Oxygen molecules attach to red blood cells, which travel back to the heart. At the same time, the carbon dioxide molecules in the alveoli are blown out of the body the next time a person exhales.

c) Haemoglobion: Haemoglobin is a molecule that is responsible for carrying almost all of the oxygen in the blood. It is composed of four subunits, each with a heme group plus a globin chain. A single haemoglobin molecule can carry four O₂ molecules along with the blood.

d) Tissue fluid: It is also called lymph or interstitial fluid helps to bring oxygen and nutrients to cells and to remove waste products from them.

Question 4.
Redraw the diagram, label the parts, and write their functions.

Answer:
a) Kidney: They remove waste products from the blood and produce urine. Kidneys control the levels of many substances in the blood. Kidneys help to control our blood pressure.
b) Ureters: The ureters are the part of the urinary system, The ureters’ role in the process is to carry urine from the kidneys to the bladder. Contractions in the ureter force urine away from the kidneys and into the bladder.
c) Urinary bladder: A hollow organ that stores urine from the kidneys before disposal by urination.
d) Urethra: The urethra acts as a passage to expel urine from the urinary bladder to the outside of the body.

Question 5.
Analyse the figure and answer the questions.

a) Write the names of the circulations X and Y.
b) Write the names of the blood vessels i, ii, iii, iv, y.
c) What is the role of these circulations in the exchange and transport of gases?
d) Explain the role of these circulations in the process of excretion.

a) X- Alveolar exchange of gases (Pulmonary circulation), Y – Systemic exchange of gases (Systemic circulation).

b) (i) Inferior vena cava, (ii) Superior vena cava, (iii) Pulmonary artery, (iv) Pulmonary vein, (v) Aorta

c) Pulmonary circulation moves blood between the heart and the lungs. It transports deoxygenated blood to the lungs to absorb oxygen and release carbon dioxide. The oxygenated blood then flows back to the heart. Systemic circulation moves blood between the heart and the rest of the body.

d) The pulmonary veins by pulmonary circulation return oxygenated blood to the heart, which releases CO₂ from the deoxygenated blood to the lungs for gaseous exchange. The release of CO₂ is a kind of excretion. In systemic circulation, blood with oxygen, nutrients, and hormones travels from the heart to the rest of the body. In the veins, the blood picks up waste products as the body uses up oxygen, nutrients, and hormones.

Extended Activities

Question 1.
Visit a primary health centre and conduct an interview with a doctor on diseases affecting the lungs and kidneys.
Answer:
(Hints: Some question answers regarding this topic are given below for your reference.)

1. Can you explain some common lung diseases and their symptoms?
Answer:
Pulmonary fibrosis and sarcoidosis are examples of lung tissue disease.

2. How can lifestyle choices, such as smoking or pollution exposure, impact lung health?
Answer:
Smoking is one of the biggest contributors to lung-related health problems. It damages the airways and structures of the lungs, leading to chronic conditions like chronic bronchitis and emphysema, as well as various types of lung cancer. Quitting smoking is the best way to improve lung health.

3. Can you discuss the relationship between lung health and kidney function?
Answer:
The function of the lungs is also closely linked to the kidneys, both in health and in disease. In fact, regulation of acid-base balance, control of blood pressure, and fluid homeostasis are closely linked to the interaction of the kidneys and lungs.

Question 2.
Construct models of the respiratory system, kidney, and related parts and display them in the class.
Answer:
Materials:
Respiratory System:

• Cardboard boxes (different sizes for various organs)
• Balloons (represent lungs)
• Straws (represent trachea and bronchi)
• Pipe cleaners (represent smaller airways)
• Tape
• Markers/ paints
  Labels

Kidneys:

• Play dough or modeling clay (in different colors)
• Beans or pebbles (represent nephrons)
• Pipe cleaners or string (represent ureters)
• Cardboard or plastic sheet (to create a base)
• Markers/ paints
• Labels

Hints for Construction:

Research: Before diving in, have students research the basic structure and function of the respiratory system and kidneys. This will guide their model building process.

Divide and Conquer: Break the class into groups, each focusing on either the respiratory system or the kidneys. This allows for focused work and better model representation.

Creative Representation: Emphasize creativity! Cardboard boxes can become lungs, balloons can represent their inflation, and straws can depict the airways. Playdough can be molded into kidney shapes, beans as nephrons, and pipe cleaners as ureters.

Labeling is Key: Labels are crucial for understanding. Encourage students to label the different parts of their models (lungs, trachea, bronchi, nephrons, ureters, etc.) with clear descriptions of their function.

Showcase and Explain: Once the models are built, each group can present their work to the class. This allows students to explain the structure and function of the respiratory system or kidneys using their model as a visual aid.

Connect the Systems: Discuss the connection between these two systems. Explain how the respiratory system provides oxygen needed by the kidneys for proper function, and how the kidneys filter waste products produced by the body during respiration.

Question 3.
Organize an awareness class on organ donation.
Answer:
(Hints: important points to be included in the awareness class are given below)

• The Critical Need for Organ Donation
• Types of Organs and Tissues Donated
• The Organ Donation and Transplantation Process
• Dispelling Myths and Misconceptions
• Legal Aspects of Organ Donation

By including these key points, you can create a comprehensive and informative organ donation awareness class that empowers individuals to make informed decisions and potentially save lives.

Question 4.
Organize and implement programs to make your home and school waste-free.
Answer:
Here are some programs you can implement to reduce waste at home and school:

At Home:
Reduce:

• Plan meals: Reduce impulse buys and food waste by planning meals and creating a grocery list.
• Buy in bulk (with caution): Purchase staples like grains or nuts in bulk to minimize packaging, but ensure you’ll use everything before it expires.
• Choose reusable products: Opt for reusable shopping bags, water bottles, and coffee mugs instead of disposable alternatives.
• Fix instead of toss: Try to repair broken items instead of throwing them away.
• Borrow or rent: Instead of buying something you’ll rarely use, consider borrowing it from a friend or renting it.

Reuse:

• Get creative with containers: Repurpose old jars, containers, and boxes for storage or crafting projects.
• Donate or sell unwanted items: Give pre-loved clothes, toys, or furniture a second life through donations or online marketplaces.
• Compost food scraps: Start a compost bin to turn food scraps into nutrient-rich fertilizer for your plants (check local regulations for composting restrictions).

Recycle:

• Learn your local recycling guidelines: Understand what materials your local program accepts and ensure proper sorting.
• Invest in recycling bins: Have designated bins for recyclables (paper, plastic, glass, etc.) to make sorting easier.
• Recycle electronics responsibly: Don’t throw away old electronics; many electronics stores or specialized facilities accept them for proper recycling.

At School:
Reduce:

• Paperless communication: Encourage online communication (school websites, email) for announcements and assignments whenever possible.
• Double-sided printing: Set printers to default to double-sided printing to reduce paper usage.
• Reusable lunchware: Promote reusable lunch containers and water bottles among students and staff.
• Reusable alternatives: Replace single-use items like paper towels with cloth towels in classrooms and kitchens.

Reuse:

• Refill stations: Install water bottle refill stations to discourage the use of disposable water bottles.
• Donate or sell old textbooks: Organize textbook donation drives or sales to give them a second life.
• Creative project materials: Encourage the use of recycled or repurposed materials for art projects.

Recycle:

• Clearly labeled recycling bins: Place clearly labeled recycling bins throughout the school for easy sorting of paper, plastic, and other recyclables.
• Composting programs: If feasible, explore starting a school composting program for food scraps (check local regulations).
• Educational campaigns: Organize workshops or presentations to educate students and staff about the importance of waste reduction and recycling.

Respiration and Excretion Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
What is the role of oxygen in respiration and burning?
Answer:
Oxygen is essential for the respiration of living organisms, and it supports combustion because it is essential for burning components. Burning of candles involves the addition of oxygen to carbon hydrogen which forms carbon dioxide, water vapour and energy in the form of heat and light. Respiration utilises oxygen from the atmosphere to break down glucose in food and release carbon dioxide.

Question 2.
What are the products of respiration and burning?
Answer:
Both processes require oxygen for the production of energy. They produce carbon dioxide as a byproduct.

Question 3.
What is the respiratory surface in human beings? How is it arranged in our body?
Answer:
The respiratory surface in humans is the alveoli. They are tiny ballon-shaped sacs found in the lungs which exchange carbon dioxide and oxygen. The human respiratory system is composed of a pair of external nostrils, nasal chamber, pharynx, larynx, and trachea (the windpipe), which is bifurcated when enters into two lungs, bronchi again divides into fine branches called bronchioles inside the lungs and at the ends of the bronchioles we can find some sac-like structures called the alveoli. Alveoli are covered by thin blood capillaries through this exchange of gases takes place.

Question 4.
Complete the by analyse the description and prepare a short note on the respiratory surface in human beings.
Answer:

• Nostrils
• Nasal Cavity
• Trachea
• Bronchi

The respiratory surface is the area or space where the gas exchanges take place, It should have larger surface-to-volume ratio, moist inner surface, and thin wall in contact with blood capillaries Gaseous exchange in the lungs takes place between alveoli and alveolar blood capillaries. The respiration surface in humans is the alveoli.

Question 5.
Don’t you involved in sports and physical exercises? Is there any change in the rate o ventilation (breathing) during such activities? Do the activity given below?

• Form a group of two children each.
• Take rest for 5 minutes. Both of them record the number of inspirations that happen within a minute during this time.
• Record the time using a stop watch.
• In an interval of one minute, record the number of inspirations two more times.
• Then record the number of inspirations of the two children after running for three minutes as mentioned above.
• Continue to record the number of inspirations in every minute till they reach the normal condition.
• Complete the table given below using the recorded results, draw a line graph and compare the rate of ventilation of both the children.

Answer:

Question 6.
Analyse the figure and illustration and complete the table.

Indicators	Inspiration	Expiration
Action of intercostal muscles
Movement of ribs
Change that occurs to the diaphragm
The volume of the thoracic cavity	Increases	Decreases
The pressure of air in the lungs
Movement of air

Answer:

Indicators	Inspiration	Expiration
Action of intercostal muscles	Contracts	Relaxes
Movement of ribs	Rises	Lowers
Change that occurs to the diaphragm	Contracts	Relaxes
The volume of the thoracic cavity	Increases	Decreases
The pressure of air in the lungs	Decreases	Increases
Movement of air	From outside to the lungs	From lungs to the outside

Question 7.
What are the characteristic features of the walls of the alveolus and blood capillaries?
Answer:
The walls of the alveoli are very thin, which makes it easier for oxygen and CO₂ to pass between the alveoli and very small blood vessels (capillaries). Both are made up of a single layer of cells.

Question 8.
What is the importance of moisture in the wall of the alveolus?
Answer:
Moisture is important for gas exchange because, in order for oxygen and carbon dioxide to effectively cross the alveoli membrane, they must dissolve in an aqueous-like solution. With the right amount of moisture, gas exchange occurs properly.

Question 9.
Mention the concentration level of O₂ and CO₂ in the alveolus and in the blood capillaries.
Answer:
During inspiration, the concentration of oxygen inside the alveoli is more than that in blood capillaries whereas the concentration of carbon dioxide in blood capillaries is more than that in alveoli.

Question 10.
How does the exchange of O₂ and CO₂ between the alveolus and the blood capillaries happen?
Answer:
The exchange of gases (oxygen and carbon dioxide) occurs between the alveoli and blood capillaries by simple diffusion.

Indicators	Glycolysis	Krebs cycle
Site of activity
Substances that take part in the chemical process
Products
Requirement of oxygen

Answer:

Indicators	Glycolysis	Krebs cycle
Site of activity	Cytoplasm	Mitochondria
Substances that take part in the chemical process	Glucose converted into pyruvic avid	Pyruvic acid is converted into carbon dioxide and water
Products	Pyruvic acid and 2 ATP molecules	Carbon dioxide, water and 28 ATPs
Requirement of oxygen	Oxygen is not required	Oxygen is required

Question 11.
Complete the chemical process illustrated below by including the missing reactants required for cellular respiration and the products formed.

Answer:

Question 12.
Photosynthesis and respiration are metabolic processes that take place in the living world. Compare these two processes and revise Table replacing glycolysis and Krebs cycle with photosynthesis and respiration respectively.
Answer:

Indicators	Photosynthesis	Respiration
Site of activity	Chloroplast	Mitochondria
Substances that take part in the chemical process	CO2 , Water, Sunlight, Chlorophyll	Glucose
Products	Cilucose, Water, Energy	CO2, Water, Energy
Requirement of oxygen	Oxygen not required	Oxygen is required

Question 13.
What are the conditions that lead to the decrease in the level of haemoglobin? What are the different types of anaemia? Find out.
Answer:
Conditions that lead to the decrease in the level of haemoglobin:

• Blood loss, which can occur because of: Bleeding in your digestive tract, such as from ulcers, cancers, etc.
• Frequent blood donation
• Heavy menstrual bleeding

Different types of anaemia:

• Haemolytic anaemia
• Vitamin deficiency anaemia
• Iron deficiency anaemia
• Sickle cell anaemia, etc.

Question 14.
What are the healthy habits that can be followed to prevent anaemia? Discuss.
Answer:

• Eat foods rich in iron and B vitamins.
• Eat fruits and vegetables high in vitamin C, which helps your body absorb iron.
• Ask your doctor about iron supplements if you don’t get enough iron in your diet.
• Get checked every year or 2 if you’re a woman of childbearing age who has heavy menstrual periods or a previous diagnosis of anaemia.

Question 15.
Analyse illustration and description. Now prepare a short note on respiration in plants.

Answer:
Plants do not have a respiratory system or separate organs for the transport of gases. But they have special features in leaf, stem, and root for the exchange of gases.

The small pores on the surface of leaves and young stems are called stomata. Stomata remain open in the daytime and close at night. During the day photosynthesis and respiration take place simultaneously in the mesophyll cells of leaves. Since the rate of photosynthesis is high during the day, oxygen formed is used for respiration and the excess oxygen is expelled. CO₂ formed as a result of respiration is utilized for photosynthesis and insufficient CO₂ is received from the atmosphere. At night when stomata close, respiratory gas exchange occurs through diffusion.

Lenticels are the lens-shaped openings found on the surface of mature stems and roots. They also help in gaseous exchange in plants. Plant roots have special cells that perform gaseous exchange with the air that is present in the soil particles.

Question 16.
Does anaerobic respiration take place in human beings? Find out.
Answer:
Yes. In humans, aerobic respiration takes place in all cells as part of energy production in mitochondria. But in certain conditions, our cells lack oxygen to produce enough energy. For example, when we are doing strenuous exercises or something like that, our muscle cells lack oxygen. In such situations, muscle cells respire without oxygen which results in the production of lactic acid in cells.

Question 17.
Mention the differences in the cellular respiratory processes in bacteria and yeast.
Answer:
Both bacteria and yeast respire anaerobically within the cytoplasm of the cell. In both cases, the respiratory substances are the same, ie, glucose. But the end products are different. In lactobacillus, glucose is anaerobically broken down with the production of 2 molecules of ATP.

This breaking produces lactic acid through an intermediate production of pyruvic acid. In the case of yeast cells, alcohol and CO₂ are the end products of anaerobic respiration. Here also glucose is first converted into pyruvic acid with the production of two ATP molecules and then to alcohol and CO₂.

Question 18.
Some situations in daily life are given below. Discuss how anaerobic respiration is beneficial in such situations and prepare short notes on the same.
a) Yeast is added to leaven the dough.
b) Curd is added to milk to prepare curd.
Answer:
a) Yeast is added to leaven the dough:
Yeasts ferment the sugar and produce carbon dioxide in the process. Fermentation is otherwise called anaerobic respiration. Carbon dioxide results in the leavening or rising of the dough to three times its original size.

b) Curd is added to milk to prepare curd:
A teaspoon of curd is added to milk contains the bacteria Lactobacillus. As the bacteria gets more lactose, it will produce lactic acid in the milk converting the entire milk into curd.

Question 19.
Which are the major excretory substances in our body?
Answer:
The main excretory product generated by the human body is urea. The urea is excreted out of the body via urine. Sweat is another excretory substance from skin.

Question 20.
If CO₂ is not eliminated on time, how does it adversely affect the body? Find out.
Answer:
CO₂ transported through our body in three different ways, such as bicarbonate ions, as Carbonic acid when combine with water, then as CO₂ as such. Whern excess CO₂ accumulated in the body it will combines with water molecules which are abaundant in our body results in the formation of excess amount of carbonic acid. Presence of carbonic acid increases the acid content in our body.

Question 21.
How does the formation of ammonia happens?
Answer:
Amino acids are formed by the breakdown of proteins in the cells. As a result of the metabolic activities of amino acids, several nitrogenous by-products are formed. The most harmful among these is ammonia.

Question 22.
Name the organ where urea is synthesised?
Answer:
Liver

Question 23.
How does Urea synthesis happens?
Answer:
The ammonia formed in tissues diffuses into blood through tissue fluids and blood transports it to the liver. In liver with the help of certain enzymes, ammonia combines with carbon dioxide and water to form urea.

Question 24.
Explain the elimination of urea.
Answer:
Kidneys play a major role in the excretion of urine which contains waste materials including urea. Urea is made when foods containing protein (such as meat, poultry, and certain vegetables) are broken down in the body. Urea is carried in the blood to the kidneys. This is where it is removed, along with water and other wastes in the form of urine

Question 25.
Which are the excretory organs in our body?
Answer:

• Kidney – Excretes water and salts through urine.
• Liver – Synthesises urea
• Skin – Secretes sweat
• Lungs – Eliminates Carbon dioxide

Question 26.
Analyze illustrations based on the hints and find out how the structure of kidneys and nephrons are suitable to remove waste materials and complete the given table.

Hints	Parts
Blood vessel which carries blood to the kidneys
Blood vessel which carries blood away from the kidneys
Ultra filters present in the kidneys
The double walled cup-shaped structure present at one end of the nephron
Network of minute capillaries present in the Bowman’s capsule
Blood vessel which carries bkxxl to the capillary network
Blood vessel which carries blood away from the capillary- network
The long tubule which connects the Bowman’s capsule and the collecting duct
The part where renal tubules enter, collects urine and carries it to the pelvis.
The initial part of ureter
Part that carry urine to urinary bladder.

Answer:

Hints	Parts
Blood vessel which carries blood to the kidneys	Renal artery
Blood vessel which carries blood away from the kidneys	Renal vein
Ultra filters present in the kidneys	Nephron
The double-walled cup-shaped structure present at one end of the nephron	Bowman’s capsule
A network of minute capillaries present in the Bowman’s capsule	Glomerulus
Blood vessel which carries blood to the capillary network	Afferent vessel
Blood vessel which carries blood away from the capillary network	Efferent vessel
The long tubule that connects the Bowman’s capsule and the collecting duct	Renal tubules
The part where renal tubules enter, collects urine and carries it to the pelvis.	Collecting duct
The initial part of the ureter	Pelvis
The part that carry urine to the urinary’ bladder	Ureters

Question 27.
Analyze the illustration and complete the given worksheet using the hints.


Answer:

Question 28.
Why all components in the glomerular filtrate are not present in urine? Find out the components that are reabsorbed and the components that are secreted.
Answer:
Because some of the components in the glomerular filtrate are reabsorbed into the blood and some are secreted into the renal tubules. so that all compoments in the glomerular fillrate are not present in urine.
Components that are reabsorbed: Glucose, amino acids, NaCl , and other essential salts
Components which are secreted: Hydrogen ions, potassium ions, etc

Question 29.
Is diagnosis of diseases possible through urine tests?
Answer:
Yes. Urine as a Sample to Diagnose Sexually Transmitted Diseases. Urine is extremely useful to diagnose infections of the reproductive tract.

Question 30.
Kidney stone, nephritis, uraemia etc. are some diseases affecting the kidneys. Prepare a presentation including the causes, symptoms, etc of these diseases and present it in the class.
Answer:

Diseases	Causes	Symptoms
Kidney stone	Not drinking enough water	Severe pain in your back or side, Blood in urine, A burning feeling when urinating.
Nephritis	Diabetes, genetic disorder link that affects the kidneys.	Decreased urine output, High blood pressure, urine appears dark, tea-coloured, or cloudy
Uraemia	Diabetes, High blood pressure	Nausea, Vomiting, Loss of appetite

Question 31.
Analyze the illustration and prepare a short note on how hemodialysis is done.

Answer:
Hemodialysis is a process of filtering the blood of a person whose kidneys are not working normally. Hemodialysis helps control blood pressure and balance important minerals, such as potassium, sodium, and calcium, in your blood. In this process, blood with a high quantity of waste materials is passed to the dialysis unit after adding heparin to prevent coagulation.

When blood flows through the dialysis unit the waste materials in the blood diffuse into the dialysis fluid. This fluid is removed in due course. Antiheparin is added to the purified blood and is returned to the body.

Question 32.
Why is the dialysis fluid removed from the dialysis unit in due course of time?
Answer:
If the dialysis fluid is not removed properly, it can cause swelling and increase your blood pressure, which makes your heart work harder. Too much fluid can build up in the lungs, making breathing difficult. Hemodialysis removes fluid as the blood is filtered through the dialysis machine.

Question 33.
When is kidney transplantation required?
Answer:
Kidney transplantation is done when a person whose kidneys no longer function properly.

Question 34.
Complete the given table by collecting information about the main excretory products and excretory organs of the organisms.

Organism	Main excretory product	Main excretory organ/mechanism
Amoeba	Ammonia	Contractile vacuole
Earthworm
Insects
Frog
Reptiles
Birds

Answer:

• The kidney is the main excretory organ in human beings, but it is different in various organisms.
• Plants do not have special excretory systems like animals.
• The Hydathodes are also known as the water stomata

The excretory material is different in each organism depending on its life processes. The kidney is the main excretory organ in human beings, but it is different in various organisms.

Organism	Main excretory product	Main excretory organ/Mechanism
Amoeba Earthworm Insects Fishes Frogs Reptiles Birds	Ammonia Urea Uric acid Ammonia Urea Uric acid Uric acid	Contractile vacuole Nephridia Malpighian tubules Kidney Kidney Kidney Kidney

Question 35.
Analyze illustration and prepare a short note on excretion in plants.

Answer:
Animals have extensive systems for excretion. But plants do not have special excretory systems like animals. The different plant structures that eliminate different forms of waste are leaves, stems, ageing tissues, xylem, roots, bark, fruits, and flowers.

The process of excretion in plants occurs in the following ways: Through the stomata of leaves and lenticels of stems, oxygen, carbon dioxide, and water vapour are expelled as gaseous wastes. Some waste materials are gathered in tree bark and leaves. The wastes are removed when the leaves and bark are shed.The Hydathodes are also known as the water stomata; they are found at the tips of the veins of the leaves and help in the removal of water and salts from plants.

Question 36.
Analyze illustration and prepare a short note on how the excretory organs help in maintaining homeostasis.

Answer:
The excretory organs includes of the kidneys, liver, skin, large intestine, and the lungs. The excretory system maintains homeostasis through the elimination of excess water and wastes from the body.
Refer to topic 9 for more information.

Maintenance of homeostasis is the sign of life. The excretory system maintains homeostasis through the elimination of excess water and wastes from the body. The excretory structures consists of the kidneys, liver, skin, large intestine, and the lungs.

Kidneys:

• Regulate Salt and water balance.
• Regulation of blood pressure.
• Regulation of pH
• Expels waste

Liver:

• Regulates metabolism.
• Neutralises toxins.

Skin:

• Regulation of temperature, salt, and water.

Lungs:

• Elimination of CO₂
• Regulation of O₂ level.
• Regulation of pH

Any change in the internal environment disrupts homeostasis. The major factors that may cause a change in the internal environment are:

• Improper lifestyles
• Wrong food habits
• Over-nutrition and under-nutrition,
• Lack of exercise
• Mental stress
• Alcoholism
• Smoking and the use of addictive drugs
• Pollution
• Lack of hygiene
• The abundance of pathogens
• Improper use of medicines, and contact with toxic substances.

The external environment is also as important as the internal environment. The External environment should be kept waste-free for the well-being of living organisms. Every individual should develop the right attitude towards this. Prepare a master plan for making the school and its surroundings waste-free on behalf of the health club and implement it as part of the school master plan.

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 5 Malayalam Medium അഭിന്നകഗുണനം can save valuable time.

Kerala SCERT Class 9 Maths Chapter 5 Solutions Malayalam Medium അഭിന്നകഗുണനം

Class 9 Maths Chapter 5 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 5 Malayalam Medium Textual Questions and Answers

Question 1.
ഒരേ വലുപ്പമുള്ള നാലു സമഭുജത്രികോണങ്ങളിൽ രണ്ടെണ്ണം നെടുകെ മുറിച്ചതും, രണ്ടെണ്ണം മുഴുവനായും ചേർത്തുവച്ച് ഒരു ചതുരമുണ്ടാക്കാം:

സമഭുജത്രികോണങ്ങളുടെയെല്ലാം വശങ്ങളുടെ നീളം 2 സെന്റിമീറ്റർ ആണെങ്കിൽ, ചതുരത്തിന്റെ ചുറ്റളവും പരപ്പളവും എത്രയാണ്?
Answer:
ത്രികോണങ്ങളുടെ വശങ്ങളുടെ അളവുകൾ പരിഗണിക്കുകയാണെങ്കിൽ;

എന്ന് കിട്ടും.
അപ്പോൾ ചതുരത്തിന്റെ വശങ്ങളുടെ നീളം;

വീതി = 2 സെ.മീ
നീളം = √3 + √3 = 2√3 സെ.മീ

ചതുരത്തിന്റെ ചുറ്റളവ് = 2(വീതി + നീളം)
= 2(2 + 2√3)
= 4 + 4√3
≈ 4+4 × 1.732
≈ 4 + 6.928
≈ 10.928 സെ.മീ

ചതുരത്തിന്റെ പരപ്പളവ് = വീതി × നീളം
= 2 × 2√3
= 4√3
≈ 4 × 1.732
≈ 6.928 സെ.മീ

Question 2.
ഒരു സമചതുരവും, അതിന്റെ വശങ്ങളുടെ രണ്ടു മടങ്ങ് നീളമുള്ള വശങ്ങളോടുകൂടിയ ഒരു സമഭുജത്രികോണവും ചുവടെ കാണുന്നതുപോലെ മുറിച്ചു മാറ്റിയടുക്കി ഒരു ലംബകമുണ്ടാക്കാം.

സമചതുരത്തിന്റെ വശങ്ങളുടെ നീളം 2 സെന്റിമീറ്റർ എന്നെടുത്താൽ, ലംബകത്തിന്റെ ചുറ്റളവും, പരപ്പളവും എത്രയായിരിക്കും?
Answer:
സമചതുരങ്ങൾ രണ്ടായി മുറിച്ചതിൽ ഒരു കഷ്ണത്തിന്റെ അളവുകൾ;

സമഭുജത്രികോണം രണ്ടായി മുറിച്ചതിൽ ഒരു കഷണത്തിന്റെ അളവുകൾ;

ലംബകത്തിന്റെ വശങ്ങളുടെ അളവുകൾ;

ലംബകത്തിന്റെ ചുറ്റളവ് = 2 + 2√3 + 2 + 2√2 + 2√3 + 2√2
= 4 + 4√3 + 4√2
= 4 (1 + √3 + √2)
≈ 4(1 +1.732 + 1.414)
≈ 4 × 4.146
≈ 16.584 m.

ലംബകത്തിന്റെ പരപ്പളവ് = സമചതുരത്തിന്റെ പരപ്പളവ് + സമഭുജത്രികോണത്തിന്റെ പരപ്പളവ്
= 22 + \(\frac{4 \times 2 \sqrt{3}}{2}\)
= 4 + 4√3
= 4 (1 + √3)
≈ 4 (1 + 1.732)
≈ 4 × 2.732
≈ 10.928 ചതു. സെ.മീ

Question 3.
രണ്ടു സമചതുരങ്ങൾ ചേർത്തുവച്ച രൂപമാണ് ചിത്രത്തിൽ. ഈ രൂപത്തിന്റെ താഴത്തെ വശത്തിന്റെ നീളം, സെന്റിമീറ്റർ വരെ കൃത്യമായി കണക്കാക്കുക.

Answer:
വലിയ സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം = \(\sqrt{27}\)
= \(\sqrt{9 \times 3}\)
= √9 × √3
= 3 × √3
= 3 × 1,73
= 5.19 6m.

ചെറിയ സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം = \(\sqrt{12}\)
= √4 × √3
= 2 × √3
= 2 × 1.73
= 3.46 സെ.മീ.

താഴത്തെ വശത്തിന്റെ നീളം
= വലിയ സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം + ചെറിയ സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം
= 5.19 + 3.46
= 8.65 സെ.മീ.

Question 4.
രണ്ടു സമചതുരങ്ങൾ ഒരു മൂലയിൽ ചേർത്തുവച്ച രൂപമാണ് ചിത്രത്തിൽ. ചരിഞ്ഞ വരയുടെ നീളം കണ്ടുപിടിക്കുക.

Answer:
വശത്തിന്റെ നീളം 2 ആയിട്ടുള്ള ഒരു സമചതുരത്തിന്റെ വികർണ്ണത്തിന്റെ നീളം 2 ആണെന്ന് നമുക്കറിയാം. അതുകൊണ്ട്,
വലിയ സമചതുരത്തിന്റെ വികർണ്ണത്തിന്റെ നീളം = 3/2 സെ.മീ
ചെറിയ സമചതുരത്തിന്റെ വികർണ്ണത്തിന്റെ നീളം = 22 സെ.മീ
ചരിഞ്ഞ വരയുടെ നീളം = 3√2 + 2√2 = 5√2 സെ.മീ

Question 5.
ചിത്രത്തിലെ മട്ടത്രികോണത്തിന്റെ മൂന്നാമത്തെ വശത്തിന്റെ നീളവും ചുറ്റളവും കണക്കാക്കുക.

Answer:
പൈഥാഗറസ് സിദ്ധാന്തം അനുസരിച്ച് മട്ടത്രികോണത്തിന്റെ മൂന്നാമത്തെ വശത്തിന്റെ നീളം
= \(\sqrt{50-18}\)
= \(\sqrt{32}\)
= \(\sqrt{16×2}\)
= √16 × √2
= 4 × √2
= 4√2 സെ.മീ .

വലിയ സമചതുരത്തിന്റെ വശത്തിന്റെ നീളം
= \(\sqrt{50}\)
= \(\sqrt{25 \times 2}\)
= \(\sqrt{25}\) × √2
= 5 √2

ചെറിയ സമചതുരത്തിന്റെ വശത്തിന്റെ നീളം = \(\sqrt{18}\)
= \(\sqrt{9×2}\)
= √9 × √2
= 3 × √2
= 3√2 സെ.മീ .
മട്ടത്രികോണത്തിന്റെ ചുറ്റളവ് = 4√2 + 5√2 + 3√2
= 12√2 സെ.മീ

Question 6.
ചുവടെയുള്ള സംഖ്യാ ജോടികളിൽ ചിലതിന്റെ ഗുണനഫലം എണ്ണൽ സംഖ്യയോ ഭിന്നസംഖ്യയോ ആണ് അവ കണ്ടുപിടിക്കുക.
(i) √3, √12
Answer:
√3 × √12 = √3 × \(\sqrt{3 \times 4}\)
= √3 × √3 × √4
= 3 × 2
= 6
ഗുണനഫലം എണ്ണൽസംഖ്യ.

(ii) √3, √1.2
Answer:
\(\sqrt{3} \times \sqrt{1.2}=\sqrt{3 \times 1.2}\)
= \(\sqrt{3.6}\)

(iii) √5, √8
Answer:
\(\sqrt{5} \times \sqrt{8}=\sqrt{5 \times 8}\)
= \(\sqrt{40}\)
= \(\sqrt{4 \times 10}\)
= √4 × √10
= 2 × √10
= 2√10

(vi) √0.5, √8
Answer:
√0.5 × √8 = \(\sqrt{0.5 \times 8}\)
= √4
= 2
ഗുണനഫലം എണ്ണൽസംഖ്യ.

(v) \(\sqrt{7 \frac{1}{2}}, \sqrt{3 \frac{1}{3}}\)
Answer:
\(\sqrt{7 \frac{1}{2}} \times \sqrt{3 \frac{1}{3}}=\sqrt{\frac{15}{2}} \times \sqrt{\frac{10}{3}}\)
= \(\sqrt{\frac{15}{2} \times \frac{10}{3}}\)
= \(\sqrt{5 \times 5}\)
= \(\sqrt{25}\)
= 5
ഗുണനഫലം എണ്ണൽസംഖ്യ.

(vi) \(\sqrt{\frac{2}{5}}, \sqrt{\frac{1}{10}}\)
Answer:
\(\sqrt{\frac{2}{5}} \times \sqrt{\frac{1}{10}}=\sqrt{\frac{2}{5} \times \frac{1}{10}}\)
= \(\sqrt{\frac{1}{25}}\)
= \(\frac{1}{5}\)
ഗുണനഫലം ഭിന്നസംഖ്യ.

Question 7.
(√2 + 1) (√-2 – 1) = 1 എന്ന് തെളിയിക്കുക. ഇത് ഉപയോഗിച്ച്:
i. \(\frac{1}{\sqrt{2}-1}\) രണ്ടു ദശാംശസ്ഥാനം വരെ കണക്കാക്കുക.
ii. \(\frac{1}{\sqrt{2}+1}\) രണ്ടു ദശാംശസ്ഥാനം വരെ കണക്കാക്കുക.
Answer:
(√2 + 1)(√2 – 1) = (√2)² – 1²
= 2 – 1
= 1

i. \(\frac{1}{\sqrt{2}-1}=\frac{(\sqrt{2}+1)(\sqrt{2}-1)}{\sqrt{2}-1}\)
= √2 + 1
= 1.41 + 1
= 2.41

ii. \(\frac{1}{\sqrt{2}+1}=\frac{(\sqrt{2}+1)(\sqrt{2}-1)}{\sqrt{2}+1}\)
= √2 – 1
= 1.41 – 1
= 0.41

Question 8.
ചിത്രത്തിൽ സമഭുജത്രികോണത്തിന്റെ വശങ്ങളുടെ നീളം മില്ലിമീറ്റർ വരെ കൃത്യമായി കണക്കാക്കുക.

Answer:
താഴെ തന്നിരിക്കുന്ന ത്രികോണം പരിഗണിക്കുക.

പൈഥാഗറസ് സിദ്ധാന്തം ഉപയോഗിച്ച്;
(2x)² – x² = 4²
4x² – x² = 16
3x² = 16
x² = \(\frac{16}{3}\)
x = \(\sqrt{\frac{16}{3}}\)
= \(\frac{\sqrt{16}}{\sqrt{3}}\)
= \(\frac{4}{1.732}\)
= 2.30 സെ.മീ

വശങ്ങളുടെ നീളം = 2 × 2.30
= 4.60 സെ.മീ

Question 9.
ചിത്രത്തിലെ ചുവന്ന ത്രികോണങ്ങളെല്ലാം സമഭുജമാണ്. പുറത്തെ സമചതുരത്തിന്റെയും, അകത്തെ സമചതുരത്തിന്റെയും വശങ്ങൾ തമ്മിലുള്ള അംശബന്ധം എന്താണ്?

Answer:
സമഭുജത്രികോണത്തിന്റെ ഒരു വശത്തിന്റെ നീളം ‘a’ എന്നെടുത്താൽ
പുറത്തെ സമചതുരത്തിന്റെ വശത്തിന്റെ നീളം = a + a√3 = a(1 + √3) = a (√3 + 1)
അകത്തെ സമചതുരത്തിന്റെ വശത്തിന്റെ നീളം = a + a√3 – (a + a)
= a + a√3 – 2a
= a√3 – a
= a (√3 − 1)
പുറത്തെ സമചതുരത്തിന്റെയും, അകത്തെ സമചതുരത്തിന്റെയും വശങ്ങൾ തമ്മിലുള്ള അംശബന്ധം

= \(\frac{\mathrm{a}(1+\sqrt{3})}{\mathrm{a}(\sqrt{3}-1)}\)
= \(\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
= (√3 + 1) (√3 – 1)

Question 10.
\(\sqrt{2 \frac{2}{3}}\) = 2\(\sqrt{\frac{2}{3}}\) എന്നും \(\sqrt{3 \frac{3}{8}}=3 \sqrt{\frac{3}{8}}\) എന്നും തെളിയിക്കുക. ഇതുപോലുള്ള മറ്റു സംഖ്യകൾ കണ്ടുപിടി ക്കാമോ?
Answer:

ഇതുപോലുള്ള മറ്റു സംഖ്യകളുടെ പൊതുരൂപം ഇങ്ങനെ വിശദീകരിക്കാം;

Question 11.
ചുവടെയുള്ള സംഖ്യാജോടികളിൽ ആദ്യസംഖ്യയെ രണ്ടാം സംഖ്യകൊണ്ട് ഹരിച്ചാൽ ചിലത് എണ്ണൽസംഖ്യയോ ഭിന്നസംഖ്യയോ കിട്ടും. അവ കണ്ടുപിടിക്കുക.
(i) √72, √2
(ii) √27, √3
(iii) √125, √50
(iv) √10, √2
(v) √20, √5
(vi) √18, √8
Answer:
(i) \(\frac{\sqrt{72}}{\sqrt{2}}=\frac{\sqrt{36 \times 2}}{\sqrt{2}}=\frac{\sqrt{36} \times \sqrt{2}}{\sqrt{2}}\)
= \(\sqrt{36}\)
= 6
ആദ്യസംഖ്യയെ രണ്ടാം സംഖ്യകൊണ്ട് ഹരിച്ചാൽ എണ്ണൽസംഖ്യ കിട്ടും.

(ii) \(\frac{\sqrt{27}}{\sqrt{3}}=\frac{\sqrt{9 \times 3}}{\sqrt{3}}=\frac{\sqrt{9} \times \sqrt{3}}{\sqrt{3}}\)
= √9
= 3
ആദ്യസംഖ്യയെ രണ്ടാം സംഖ്യകൊണ്ട് ഹരിച്ചാൽ എണ്ണൽസംഖ്യ കിട്ടും.

(iii) \(\frac{\sqrt{125}}{\sqrt{50}}=\frac{\sqrt{25 \times 5}}{\sqrt{25 \times 2}}\)
= \(\frac{\sqrt{25} \times \sqrt{5}}{\sqrt{25} \times \sqrt{2}}\)
= \(\frac{\sqrt{5}}{\sqrt{2}}\)

(iv) \(\frac{\sqrt{10}}{\sqrt{2}}=\frac{\sqrt{2 \times 5}}{\sqrt{2}}\)
= \(\frac{\sqrt{2} \times \sqrt{5}}{\sqrt{2}}\)
= √5

(v) \(\frac{\sqrt{20}}{\sqrt{5}}=\frac{\sqrt{4 \times 5}}{\sqrt{5}}\)
= \(\frac{\sqrt{4} \times \sqrt{5}}{\sqrt{5}}\)
= √4
= 2
ആദ്യസംഖ്യയെ രണ്ടാം സംഖ്യകൊണ്ട് ഹരിച്ചാൽ എണ്ണൽസംഖ്യ കിട്ടും.

(vi) \(\frac{\sqrt{18}}{\sqrt{8}}=\frac{\sqrt{9 \times 2}}{\sqrt{4 \times 2}}=\frac{\sqrt{9} \times \sqrt{2}}{\sqrt{4} \times \sqrt{2}}\)
= \(\frac{\sqrt{9}}{\sqrt{4}}=\frac{3}{2}\)
ആദ്യസംഖ്യയെ രണ്ടാം സംഖ്യകൊണ്ട് ഹരിച്ചാൽ ഭിന്നസംഖ്യ കിട്ടും.

Question 12.
ചില സമഭുജത്രികോണങ്ങളുടെ ഒരു വശത്തിന്റെ നീളം ചുവടെ കൊടുത്തിരിക്കുന്നു. പരപ്പളവ് കണക്കാക്കുക.
i) 10 സെ.മീ
ii) 5 സെ.മീ
iii) √3 സെ.മീ
Answer:
സമഭുജത്രികോണങ്ങളുടെ പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) (വശO)²
i. പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) × a
= \(\frac{\sqrt{3}}{4}\) × 10²
= \(\frac{\sqrt{3}}{4}\) × 100
= √3 × 25
~ 1.732 × 25
~ 43.3 സെ.മീ

ii. പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) × a²
= \(\frac{\sqrt{3}}{4}\) × 5²
= \(\frac{\sqrt{3}}{4}\) × 25
= √3 × 6.25
~ 6.25 × 1.732
~ 10.825 ചതു.സെ.മീ

iii. പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) × a²
= \(\frac{\sqrt{3}}{4}\) × (√3)²
= \(\frac{\sqrt{3}}{4}\) × 3
= √3 × 0.75
~ 1.732 × 0.75
~ 1.299 ചതു.സെ.മീ

Question 13.
വശങ്ങളുടെയെല്ലാം നീളം 6 സെന്റിമീറ്റർ ആയ സമഷഡ്ഭുജത്തിന്റെ പരപ്പളവ് കണക്കാക്കുക.
Answer:
ഒരു സമഷഡ്ഭുജത്തിനെ അതിന്റെ വശങ്ങളുടെ അതേ നീളമുള്ള വശങ്ങളുള്ള ആറ് സമഭുജത്രികോ ണങ്ങളായി മുറിക്കാൻ പറ്റും.

സമഷഡ്ഭുജത്തിന്റെ പരപ്പളവ്
= ആറ് സമഭുജത്രികോണങ്ങളുടെയും പരപ്പളവിന്റെ തുക
= 6 × ഒരു സമഭുജത്രികോണത്തിന്റെ പരപ്പളവ്
= 6 × \(\frac{\sqrt{3}}{4}\) × 6²
= 6 × √3 × 9
= 54 × √3
= 54 × 1.732
= 93.528 ചതു.സെ.മീ

Question 14.
ഉയരം 12 സെന്റിമീറ്റർ ആയ സമഭുജത്രികോണത്തിന്റെ ചുറ്റളവും പരപ്പളവും കണക്കാക്കുക.
Answer:
ഉയരം = 12 സെ.മീ
\(\frac{\sqrt{3}}{4}\) × വശം = 12
വശം = \(\frac{12 \times 2}{\sqrt{3}}\)
= \(\frac{3 \times 4 \times 2}{\sqrt{3}}\)
= \(\frac{4 \times \sqrt{3} \times \sqrt{3} \times 2}{\sqrt{3}}\)
= 8√3

പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) × (വശം)
= \(\frac{\sqrt{3}}{4}\) × (8√3)²
= \(\frac{\sqrt{3}}{4}\) × 64 × 3
= √3 × 16 × 3
≈ 1.732 × 48
≈ 83.136 ചതു.സെ.മീ

ചുറ്റളവ് = 3 × വശം
= 3 × 8√3
= 24√3 സെ.മീ
≈ 24 × 1.732
≈ 41.568 സെ.മീ

Question 15.
സമാന്തരവശങ്ങൾ തമ്മിലുള്ള അകലം 6 സെന്റിമീറ്റർ ആയ സമഷഡ്ഭുജത്തിന്റെ ചുറ്റളവും പരപ്പളവും കണക്കാക്കുക.
Answer:
ഒരു സമഷഡ്ഭുജത്തിനെ അതിന്റെ വശങ്ങളുടെ അതേ നീളമുള്ള വശങ്ങളുള്ള ആറ് സമഭുജ ത്രികോണങ്ങളായി മുറിക്കാൻ പറ്റും.

അങ്ങനെനോക്കിയാൽ,
സമാന്തരവശങ്ങൾ തമ്മിലുള്ള അകലം = 2 × ഒരു സമഭുജത്രികോണത്തിന്റെ ഉയരം
6 = 2 × \(\frac{\sqrt{3}}{2}\) × സമഭുജത്രികോണത്തിന്റെ വശത്തിന്റെ നീളം
= 2 × \(\frac{\sqrt{3}}{2}\) × സമഷഡ്ഭുജത്തിന്റെ വശത്തിന്റെ നീളം
\(\frac{6}{2} \times \frac{2}{\sqrt{3}}\) = സമഷഡ്ഭുജത്തിന്റെ വശത്തിന്റെ നീളം

സമഷഡ്ഭുജത്തിന്റെ വശത്തിന്റെ നീളം = \(\frac{6}{\sqrt{3}}\)
= \(\frac{3 \times 2}{\sqrt{3}}\)
= \(\frac{\sqrt{3} \times \sqrt{3} \times 2}{\sqrt{3}}\)
= √3 × 2
= 2√3 സെ.മീ

സമഷഡ്ഭുജത്തിന്റെ ചുറ്റളവ് = 6 × സമഷഡ്ഭുജത്തിന്റെ വശത്തിന്റെ നീളം
= 6 × 2√3
= 12√3
≈ 12 × 1.732
≈ 20.784 സെ.മീ

സമഷഡ്ഭുജത്തിന്റെ പരപ്പളവ് = ആറ് സമഭുജത്രികോണങ്ങളുടെയും പരപ്പളവിന്റെ തുക
= 6 × \(\frac{\sqrt{3}}{4}\) × (2√3)²
= 6 × \(\frac{\sqrt{3}}{4}\) × 4 × 3
= 18 × √3
≈ 18 × 1.732
≈ 31.176 ചതു.സെ.മീ

Question 16.
വശങ്ങളുടെ നീളം 8 സെന്റിമീറ്റർ, 6 സെന്റിമീറ്റർ, 6 സെന്റിമീറ്റർ ആയ ത്രികോണത്തിന്റെ ഉയരവും പരപ്പളവും കണക്കാക്കുക.
Answer:

ഉയരം = \(\sqrt{6^2-4^2}\)
= \(\sqrt{36-16}\)
= \(\sqrt{20}\)
= \(\sqrt{4 \times 5}\)
= √4 × √5
= 2 × √5
= 2√5 സെ.മീ

പരപ്പളവ് = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 8 × 2√5
= 8√5 ചതു.സെ.മീ

Question 17.
ചില ത്രികോണങ്ങളുടെ വശങ്ങളുടെ നീളം ചുവടെ പറഞ്ഞിരിക്കുന്നു. ഓരോന്നിന്റെയും പരപ്പളവ് കണക്കാക്കുക.
i. 4 സെന്റിമീറ്റർ, 5 സെന്റിമീറ്റർ, 7 സെന്റിമീറ്റർ
ii. 4 സെന്റിമീറ്റർ, 13 സെന്റിമീറ്റർ, 15 സെന്റിമീറ്റർ
iii. 5 സെന്റിമീറ്റർ, 12 സെന്റിമീറ്റർ, 13 സെന്റിമീറ്റർ
Answer:
i. a = 4 സെ.മീ
b = 56 സെ.മീ
c = 7 സെ.മീ

ii. a = 4 സെ.മീ
b = 13 സെ.മീ
c = 15 സെ.മീ

iii. a = 5 സെ.മീ
b = 12 സെ.മീ
c = 13 സെ.മീ

Irrational Multiplication Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
തന്നിരിക്കുന്നവ ചെറുതിൽനിന്ന് വലുതിലേക്ക് എന്ന രീതിയിൽ എഴുതുക.
i. 3√5 and 4√3
ii. 2√5, 5√2 and 3√7
Answer:
(i) \(\sqrt{53}=\sqrt{3 \times 3 \times 5}=\sqrt{45}\)
4√3 = \(\sqrt{4 \times 4 \times 3}=\sqrt{48}\)
48 > 45 ⇒ \(\sqrt{48}>\sqrt{45}\)
⇒ 4√3 > 3√5
ചെറുതിൽനിന്ന് വലുതിലേക്ക് എന്ന രീതിയിൽ എഴുതിയാൽ; 3√5, 4√3

(ii) 2√5 = √2 × 2 × 5 = √20
5√2 = √5 × 5 × 2 = √50
3√7 = √3 × 3 × 7 = √63
20 < 50 < 63 ⇒ √20 <√50 < √63
⇒ 2√5 < 5√2 < 3√7
ചെറുതിൽനിന്ന് വലുതിലേക്ക് എന്ന രീതിയിൽ എഴുതിയാൽ; 2√5, 5√2, 3√7

Question 2.
x = √0.5, y = √32, z = √128
a) xy, yz, xz എന്നിവ കണ്ടെത്തുക.
Answer:

b) xy + yz + xz കണ്ടുപിടിക്കുക.
Answer:
xy + yz + xz = 4 + 64 + 8
= 76

c) y = 8x എന്ന് തെളിയിക്കുക
Answer:
8x = 8 × \(\sqrt{0.5}\)
= \(\sqrt{64 \times 0.5}\)
= \(\sqrt{32}\)
= y

Question 3.
√8 നെ \(\sqrt{4 \times 2}\) = 2√2 എന്ന് എഴുതാം
a) ഒരു പൂർണ്ണസംഖ്യയുടെയും √2 വിന്റെയും ഗുണനമായി √18 നെയും √32 നെയും എഴുതുക.
Answer:
\(\sqrt{18}=\sqrt{9 \times 2}\) = 3√2
\(\sqrt{32}=\sqrt{16 \times 2}\) = 4√2

b) √2 + √8 + √18 – √32 നെ ചെറുതാക്കി എഴുതുക.
Answer:
\(\sqrt{2}+\sqrt{8}+\sqrt{18}-\sqrt{32}\)
= √2 + 2√2 + 3√2 – 4√2
= 6√2 – 4√2
= 2√2

Question 4.
ചുറ്റളവ് 18 സെ.മീ ആയ സമഭുജത്രികോണത്തിന്റെ ഏരിയ എത്ര?
Answer:
ചുറ്റളവ് = 18
3 × വശം = 18

വശം = \(\frac{18}{3}\)
= 6 സെ.മീ

പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) × (വശം)²
= \(\frac{\sqrt{3}}{4}\) × 6²
= √3 × 9
= 9√3 ചതു.സെ.മീ

Question 5.
38 ഉം 7/8 ഉം തമ്മിൽ ഗുണിക്കുക.
Answer:
3√8 × 7√8 = 3 × √8 × 7 × √8
= 3 × 7 × √8 × √8
= 21 × 8
= 168

A thorough understanding of Kerala Syllabus 9th Standard Biology Textbook Solutions Chapter 2 Digestion and Transport of Nutrients Notes Questions and Answers English Medium can improve academic performance.

SCERT Class 9 Biology Chapter 2 Notes Questions and Answers Digestion and Transport of Nutrients

Std 9 Biology Chapter 2 Notes Pdf Kerala Syllabus English Medium Solutions Questions and Answers

Class 9 Biology Chapter 2 Let Us Assess Answers Digestion and Transport of Nutrients

Question 1.
Which one of the following is related to the digestion of Fat?
a) Protease
b) Lipase
c) Amylase
d) Carbohydrase
Answer:
b) Lipase

Question 2.
An illustration related to the circulatory system of human beings is given below. Analyse it and answer the questions.

a) Which letter indicates Pulmonary artery?
Answer: A

b) Which blood vessel is indicated by the letter D?
Answer: Aorta

c) Does the blood that has entered the ventricles return to the atria? Why?
Answer:
No, the blood do not flow back to the atria. Because, tricuspid valve and bicuspid valve prevents the back flow of blood from ventricles to atria.

d) What is the importance of double circulation in human?
Answer:
The same amount of blood passes through the heart twice. This type of circulation is called double circulation. It includes pulmonary circulation and systemic circulation. Double circulation helps to maintain the level of oxygen.

Question 3.
A flowchart on the path of nutrients is given below. Observe it and answer the questions.

a) Name the blood vessels indicated by the letters A, B and C. A, B,C
Answer:
A – Portal vein
B – Hepatic vein
C – Venacava

b) Do all the nutrients absorbed from the small intestine have the same path? Explain.
Answer:
Nutrients like amino acids and glucose are absorbed into the blood capillaries of the villus and are transported to liver through portal vein. Then they are carried to venacava through hepatic vein. Substances like fatty acid and glycerol are absorbed by lacteal of villus. They are carried to venacava by lymph vessel. Venacava carries all these nutrients to the heart.

Question 4.
Which among the following processes takes place by utilising energy?
a) Entry of water into the root cells.
b) Entry of sucrose into the sieve tube.
c) Loss of water from leaves through transpiration.
d) Conduction of water molecules through xylem tubes.
Answer:
b) Entry of sucrose into the sieve tube

Question 5.
Redraw the figure given below, name and label the parts based on the indicators.

a) Part where odontoblast cells are seen
Answer: Pulp cavity

b) Tissue that holds the tooth in the gum.
Answer: Cementum

c) The living tissue by which tooth is made.
Answer: Dentine

Extended Activities

Question 1.
Conduct an awareness class on the topic ‘Bad habits and health of the heart’.
Answer:
Main Points

• Heart disease is the leading cause of death globally.
• While some risk factors are uncontrollable (age, genetics), many lifestyle habits significantly influence heart health:
• Excessive saturated and trans fats raise bad cholesterol, clogging arteries.
• High sodium intake increases blood pressure, putting strain on the heart.
• A sedentary lifestyle increases the risk of obesity, diabetes, and heart disease
• Smoking increases the risk of blood clots and heart attacks.
• Limit saturated and trans fats, processed foods, and added sugars
• Regular exercise strengthens your heart muscle and improves blood flow.
• Ensure adequate sleep for better overall health and stress management.

Question 2.
Organize a nutritious food fest under the auspices of the school health club using locally available food items.
Answer:

Digestion and Transport of Nutrients Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Which are the different stages of nutrition?
Answer:

• Ingestion
• Mechanical Digestion
• Chemical Digestion
• Absorption
• Transport
• Egestion of digestive waste

Question 2.
Compare the nutritional processes of amoeba, a unicellular organism and hydra, a multicellular organism and complete the table.

Hints	Amoeba	Hydra
Body structure	Unicellular
Means to help ingestion
Part where digestion takes place	Inside the cell
Egestion of digestive wastes

Answer:

Hints	Amoeba	Hydra
Body structure	Unicellular	Multicellular
Means to help ingestion	Pseudopodia	Mouth
Part where digestion takes place	Inside the cell	Outside the cell
Egestion of digestive wastes	Through the cell surface	Through the mouth

Question 3.
Which are the parts where mechanical digestion takes place?
Answer:

• Mouth
• Stomach
• Small intestine

Question 4.
How does mechanical digestion take place in these parts? prepare a note based on the indicators.
Indicators:

• Structure of tooth.
• Mechanical digestion in mouth, stomach and small intestine.

Answer:
Teeth facilitate the process of breaking down food into smaller pieces and grinding it during chewing. Enamel is the hardest substance in the human body, serving as the non-living outer covering of the tooth. Beneath the enamel is dentine, a living tissue that forms the bulk of the tooth. The innermost part of the tooth is the pulp cavity, which contains soft connective tissue called pulp, along with blood vessels, nerves, and odontoblast cells. The tooth is anchored in the gum socket by cementum, a calcium-containing connective tissue.

The strong peristalsis occurring in the stomach transforms food into a paste form. The circular muscles of the stomach maintain food within the stomach for a sufficient duration. The small intestine involves mechanical processes, namely peristalsis and segmentation, to help in the movement of food and the blending of food with digestive fluids.

Question 5.
What are the things to be taken care for proper dental care? Conduct an interview with a doctor, prepare a poster and exhibit it.
Answer:
Interview
Interviewer: What are the most important things people can do to maintain a good oral health?
Doctor:

• Brush your teeth twice a day
• Flossing once a day removes plaque and food particles from between the teeth, where a brush can’t reach
• Schedule dental checkups and cleanings at least once a year.

Interviewer: Are there any additional tips you recommend for optimal dental health?
Doctor:

• Limit sugary drinks and foods
• Don’t use tobacco products
• Drink plenty of water
• Consider using mouthwash

Poster

Question 6.
Why doesn’t food enter trachea while swallowing it?
Answer:
Tongue compresses the food into balls with the help of palate. Uvula closes the nasal cavity that opens to the pharynx. Posterior part of the tongue allows food to move over the epiglottis into the oesophagus. Trachea rises up and is closed by the epiglottis.

Question 7.
What is the reason for saying that one should not talk while eating food? Find out.
Answer:
Talking while eating can cause food to enter the trachea instead of oesophagus. This will trigger coughing or choking.

Question 8.
How far is the structure of villus suitable for the absorption process? Prepare note according to the indicators.
Indicators:

• Villus and surface area of absorption.
• Lacteal and absorption.
• Blood capillaries and absorption.

Answer:
Villi increase the surface area of absorption of nutrients to a great extent within the small intestine. Absorption of simple nutrients and 90 percent of water takes place through villi. Villi are lined with single-layered epithelial cells, which are the primary surface for nutrient absorption.

Blood capillaries within the villus, formed from an arterial branch, absorb glucose, fructose, galactose, and amino acids, and then unite to form veins that exit the villus. Additionally, the lacteal, a branch of the lymph vessel, absorbs fatty acids and glycerol into the lymph.

Question 9.
Complete the illustration related to the blood components.

Answer:

Question 10.
Find out the functions of different proteins in plasma.
Answer:
Albumin – Regulates blood pressure
Globulin – Helps in defence
Fibrinogen – Plays a major role in the coagulation of blood

Question 11.
How is the quantity of tissue fluid regulated in the space between cells? Find out.
Answer:
Along with the formation of tissue fluid, it absorbs into the lymph capillaries. So there is no increase of pressure in the intercellular space.

Question 12.
Complete the flowchart using the hints.
Hints:

• Portal vein
• Lacteal of villus
• Amino acids
• Glycerol
• Hepatic vein
• Lymph vessel

Answer:

Question 13.
Which are the phases included in a cardiac cycle?
Answer:

• Atrial systole
• Ventricular systole
• Joint diastole

Question 14.
How the function of heart is maintained in people with a functionless pacemaker. Find out.
Answer:
In people with functionless pacemaker, artificial pacemaker is implanted for the functioning.

Question 15.
Apart from the wrist, which are the other parts of the body where we can feel pulse? Find out.
Answer:
Neck, Groin, Foot

Question 16.
Complete the chart

SI . No	Name of Children	Rate of heart beat		Pulse rate
		At rest	After doing exercise	At rest	After doing exercise
1
2

Answer:

SI No	Name of Children	Rate of heart heat		Pulse rate
		At rest	After doing exercise	At rest	After doing exercise
1.	Megha	72/m in	120/min	72/min	120/min
2.	Nikhil	74/min	125/min	74/min	125/min

Question 17.
What is the reason for variations of blood pressure? How does it affect the body? Find out.
Answer:
The normal rate of blood pressure is 120/80 mmHg. The disease condition in which the blood pressure increases above the normal rate is called hypertension. This happens due to many reasons. Unhealthy habits such as excess use of salt and fat, smoking, lack of exercise etc. The condition in which the blood pressure rate goes below the prescribed rate is called hypotension. These hypotension and hypertension may lead to stroke or heart attack.

Question 18.
How does the body utilise each nutrient? Find out.
Answer:
Carbohydrates are broken down into simple sugars in the digestive system. Glucose enters the bloodstream and is transported to cells, where it’s used for immediate energy production or stored for later use as glycogen in the liver and muscles.

Proteins are broken down into amino acids during digestion. Amino acids are then absorbed into the bloodstream and used to build, repair, and maintain various body structures and functions.

Fats are broken down into fatty acids and glycerol during digestion. Fatty acids are then transported to cells for energy storage or used to build cell membranes and signaling molecules.

Vitamins are absorbed in small amounts from food and transported throughout the body. They work in conjunction with enzymes to regulate numerous metabolic processes.

Students rely on Kerala Syllabus 9th Standard Chemistry Notes Pdf Download Chapter 7 Non Metals Extra Questions and Answers to help self-study at home.

Kerala Syllabus Std 9 Chemistry Chapter 7 Non Metals Extra Questions and Answers

Question 1.
a) What are the chemicals required for the laboratory preparation of Hydrogen?
b) Hydrogen-filled balloons rise up in the air. Why?
Answer:
a) Zinc and Hydrochloric acid
b) The density of hydrogen is less than air.

Question 2.
Hydrogen is used as a fuel because of its high calorific value.
a) What is calorific value?
b) Write any other advantage of hydrogen fuel.
c) Hydrogen is not commonly used as domestic fuel. Give reason
Answer:
a) The calorific value of a fuel is the heat energy released from one unit mass of that fuel on complete combustion.
b) Availability is plenty / High calorific value/pollution free.
c) Bum with an explosion, it is very difficult to store and transport.

Question 3.
a) What are the chemicals required to prepare hydrogen in the laboratory?
b) Write the balanced chemical equation of this reaction?
c) Which type of reaction is this?
Answer:
a) Dilute hydrochloric acid and Zinc
b) Zn + 2HCl → ZnCl₂ + H₂
c) Displacement reaction

Question 4.
List the uses of hydrogen
Answer:

• For the industrial production of ammonia and methanol.
• To saturate unsaturated oils.
• As a fuel.

Question 5.
a) Which fuel has the highest calorific value?
b) What will be the product formed when hydrogen burns in the air?
Answer:
a) Hydrogen
b) Water (H₂O)

Question 6.
Write the reaction between Hydrogen and chlorine.
Answer:
Hydrogen combines with chlorine in the presence of sunlight to form hydrogen chloride.
H₂2 + Cl₂ 2HCl

Question 7.
Write any three properties of Hydrogen.
Answer:
Insoluble in water, Hydrogen gas bums with a pop sound, Hydrogen is a colourless and odourless gas

Question 8.
Carbon is an element that shows allotropy.
a) What is allotropy?
b) Non-crystalline allotropes of carbon are commonly called __________
Answer:
a) Different forms of the same element having different physical properties but with the same chemical properties are known as Allotropes, and this phenomenon is called Allotropy.

b) Amorphous carbon.
The structure of two crystalline allotropes of carbon is given below,

Question 9.
The structure of two crystalline allotropes of carbon is given below.
a) Which among these is the structure of a diamond?


b) Which of these crystalline structures has no free electron?
c) Graphite is used as a lubricant. Why?
Answer:
a) Figure B
b) Figure B (Diamond)
c) Layers are held together by weak van der Waals forces. Hence, one layer can slide over another.

Question 10.
Carbon monoxide and carbon dioxide are two compounds formed by the combination of carbon and oxygen.
a) Write the balanced chemical equation showing the formation of carbon monoxide
b) Which are the gases mixed with carbon monoxide to prepare water gas and producer gas respectively?
c) Inhaling of excess carbon monoxide leads even to death. Why?
Answer:
a) 2C + O₂→ 2CO
b) Water gas → CO + H₂, Producer gas → CO + N₂
c) It reacts with the haemoglobin in the blood and forms carboxyhaemoglobin. As a result, the oxygen-carrying capacity of blood decreases, leading even to death.

Question 11.
When eggshell is added to Hydrochloric acid, a gas is produced.
a) Identify this gas.
b) Write the chemical equation of this reaction.
Answer:
a) Carbon dioxide
b) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Question 12.
Carbon dioxide (CO₂) is an important oxide of carbon.
a) Name the process by which CO₂ is utilised by plants.
b) Write an environmental problem due to the tremendous increase in the amount of CO₂ in the atmosphere.
c) What is dry ice? Write any one use.
Answer:
a) Photosynthesis
b) Climate change/ Greenhouse effect
c) The solid form of carbon dioxide.
Used as refrigerant / To create foggy effects in stage shows.

Question 13.
Writ any three uses of carbon dioxide.
Answer:
As gaseous fuel, To produce industrial fuels like water gas and producer gas, As a reducing agent in metallurgy.

Question 14.
a) Why does graphite conduct electricity?
b) Write any two properties of graphene
c) What are Buckyballs and Buckytubes?
Answer:
a) Graphite conducts electricity due to delocalised electrons and layered structure.

b) High electrical conductivity
High thermal conductivity

c) Fullerene has a hollow spherical structure consisting of pentagons and hexagons. They are
known as Bucky balls.
Fullerene with a cylindrical structure is used as carbon nanotubes. These are called Buckytubes.

Question 15.
What are the different ways in which nitrogen is obtained by plants?
Answer:

• By decomposition
• Through fertilisers
• By nitrogen fixing by bacteria

Question 16.
“It is said that lightning is a boon to plants”. Justify the statement.
Answer:
During lightning, the triple bond in nitrogen breaks, and nitrogen combines with the atmospheric oxygen to form nitric oxide (NO)
N₂ + O₂ → NO
Nitric oxide thus combines with oxygen to form nitrogen dioxide (NO₂)
2NO + O₂ → 2NO₂
Nitrogen dioxide dissolves in rainwater in the presence of oxygen and reaches the soil as nitric acid (HNO₃)
4NO₂ + 2H₂O + O₂ → 4HNO₃
Nitric acid reacts with the minerals in the soil to form nitrate salts, which the plants absorb. So, it is said that lightning is a boon to plants.

Question 17.
Can you list out which other means are there for getting a greater amount of elements for plants?
Answer:

• Use of organic fertilisers
• Use chemical fertilisers
• Bio decomposition
• Through nitrogen fixation

Question 18.
Write any two advantages of using organic fertilizers.
Answer:
Eco-friendly, Preserve the nature of the soil.

Question 19.
Nitrogen is inert in nature. Because it contains ………………. in the molecule.
(a) single bond (b) double bond (c) triple bond (d) ionic bond
Answer:
Triple bond

Question 20.
Classify the following gases into molecules having single bond, double bond and triple bond (a) Nitrogen (b) Chlorine (c) oxygen
Answer:
Single bond – Chlorine, Double bond – Oxygen, Triple Bond – Nitrogen

Question 21.
List the merits and limitations of the application of organic fertilisers.
Answer:
Merits:

• Eco friendly
• Preserve the nature of the soil

Limitations:

• Biodegradation needs time, so it cannot easily be absorbed by plants in time.
• The presence of microorganisms is necessary.
• Storage and transportation are not easy

Question 22.
Heat potassium permanganate in a dry boiling tube. When a glowing match stick is shown into
this boiling point tube, it flares up.
a) The presence of which gas is indicated by the flaring up of the glowing match stick?
b) Write down the chemical equation of the reaction taking place in the boiling tube.
c) To which category does this reaction belong?
Answer:
a) Oxygen
b) 2KMnO₄ + Heat → K₂MnO₄ + MnO₂ + O₂
c) Decomposition

Question 23.
Suggest a method to prepare oxygen in the laboratory
Answer:
Oxygen can be produced in the laboratory by the strong heating of potassium permanganate. 2KMnO₄ + heat → K₂MnO₄ + MnO₂ + O₂
Oxygen can also be prepared in the laboratory, by adding a small pinch of manganese dioxide in hydrogen peroxide (H₂O₂)
2H₂O₂ → 2H₂O + O₂

Question 24.
a) What is ozone?
b) Where is ozone found in the atmosphere?
c) What are chlorofluorocarbons (CFCs)?
Answer:
a) Ozone is an allotrope of oxygen. Three oxygen atoms combine to form an ozone molecule, O3.
b) In the stratosphere.
c) Chlorofluorocarbons are a class of compounds containing chlorine fluorine and carbon.

Question 25.
How CFC causes ozone layer depletion.
Answer:
Chlorofluorocarbons released into the atmosphere reach the stratosphere and break down by the action of ultraviolet radiations and release chlorine. This chlorine decomposes ozone molecules into oxygen. This disturbs the equilibrium in the ozone-oxygen cycle process.

Question 26.
What is the importance of ozone in the atmosphere?
Answer:
Ozone acts like a shield, protecting us from the sun’s harmful rays. It forms and breaks down in a continuous cycle to maintain a balance.

• The sun emits harmful ultraviolet (UV) radiation.
• UV radiation breaks apart oxygen molecules (O₂) into oxygen atoms (O).
• These oxygen atoms combine with other oxygen molecules to form ozone (O₃)
• Ozone absorbs more harmful UV radiation, protecting Earth’s surface.
• Some ozone molecules break down back into oxygen, but the formation of new ozone balances this process.
• This constant cycle helps maintain a stable level of ozone in the atmosphere, protecting us from harmful UV rays.

Question 27.
What are the uses of oxygen?
Answer:

• It’s the breath of life for all living organisms.
• For combustion, oxygen is necessary.
• As an oxidant in rocket fuels.
• For artificial respiration.
• For decomposition of wastes.

Question 28.
How can we reduce the rate of depletion of the ozone layer?
Answer:
Today, the use of CFC is being controlled in most countries. Harmful CFCs are replaced nowadays with safer substances. This has helped in reducing the rate of depletion of the ozone layer.

Question 29.
The picture given below shows the laboratory preparation of chlorine. Analyse the picture and answer the following questions.

a) What are the chemicals required for the preparation of chlorine gas?
b) Why is chlorine gas passed through water?
c) Which substance is used to remove moisture from chlorine?
d) How is bleaching powder prepared?
Answer:
a) KMnO₄ & HCl
b) To remove the traces of hydrogen chloride vapours that come out along with chlorine.
c) Sulphuric acid (H₂SO₄)
d) By passing dry chlorine gas over dry slaked lime.

Question 30.
List out the chlorine compounds you are familiar with.
Answer:
Hydrogen chloride (HCl), Sodium chloride (NaCl), Potassium chloride (KCl)

Question 31.
In the laboratory preparation of chlorine, why is chlorine gas passed through water?
Answer:
Hydrogen chloride vapours coming along with chlorine is removed by passing it through water.

Question 32.
Write the balanced chemical equation for the preparation of chlorine. Write the name of the reactant involved in the preparation of chlorine.
Answer:
2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂
Potassium permanganate and Con. HCl are the reactants.

Question 33.
Write the uses of chlorine.
Answer:
For bleaching.
For the preparation of insecticides.
For removing stains in the fabrics.
For purification of water.
For the preparation of bleaching powder.

Question 34.
How is bleaching powder produced, and what is the chemical process involved in its bleaching action?
Answer:
Bleaching powder is prepared by passing dry chlorine gas over dry slaked lime. Chlorine gas reacts
with the water to give hydrochloric acid and hypochlorous acid (HClO).
Cl₂ + H₂O → HCl + HOCl
Hypochlorous acid decomposes and liberates atomic oxygen
HOCl → HCl + [O] This atomic oxygen oxidises coloured substances

Question 35.
Which of the following is used to purify water?
(a) Hydrogen
(b) Oxygen
(c) Nitrogen
(d) Chlorine
Answer:
(d) Chlorine

Question 36.
Chlorine gas is passed through concentrated sulphuric acid. Why?
Answer:
Sulphuric acid can absorb the water vapour present in chlorine gas when it passes through the concentrated sulphuric acid (as a drying agent).

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 7 Malayalam Medium ന്യൂനസംഖ്യകൾ can save valuable time.

Kerala SCERT Class 9 Maths Chapter 7 Solutions Malayalam Medium ന്യൂനസംഖ്യകൾ

Class 9 Maths Chapter 7 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 7 Malayalam Medium Textual Questions and Answers

Question 1.
ചുവടെക്കൊടുത്തിരിക്കുന്ന പട്ടിക പൂർത്തിയാക്കുക.

Answer:
6 + (-10) = 6 – 10
= -(10 – 6)
= -4

-6 + 10 = 10 – 6
= 4

-6 + (-10) = -6 – 10
= -(6 + 10)
= -16

-6 + 6 = 6 – 6
= 0

6 + (-6) = 6 – 6
= 0

x	y	x + y
6	-10	-4
-6	10	4
-6	-10	-16
-6	6	0
6	-6	0

Question 2.
ചുവടെപ്പറയുന്ന ഓരോ കണക്കിലും പറഞ്ഞിരിക്കുന്ന തരത്തിലുള്ള രണ്ടു ജോടി x, y കണ്ടുപിടി ക്കുക.
(i) x അധിസംഖ്യ, y ന്യൂനസംഖ്യ, x + y = 1
(ii) x ന്യൂനസംഖ്യ, y അധിസംഖ്യ, x + y = 1
(iii) x അധിസംഖ്യ, y ന്യൂനസംഖ്യ, x + y = 1
(iv) x ന്യൂനസംഖ്യ, y അധിസംഖ്യ, x + y = 1
Answer:
(i) x = 2, y = -1 ഉം x = 5, y = -4 ഉം
(ii) x = -3, y = 4 ഉം x = -6, y = 7 20
(iii) x = 1, y = -2 ഉം x = 3, y = -4 ഉം
(iv) x = -3, y = 2 ഉം x = -5, y = 4 ഉം

Question 3.
ചുവടെയുള്ള പട്ടിക പൂർത്തിയാക്കുക.

Answer:
ആദ്യത്തെ വരി
(x + y) + z = (2 + 4) + (-5)
= 6 – 5
= 1

x + (y + z) = 2 + (4 + (-5))
= 2 + (4 – 5)
= 2 + (-1)
= 2 – 1
= 1

രണ്ടാമത്തെ വരി
(x + y) + z = ( (2 + (-4)) + 5
= (2 – 4)+ 5 = -2 + 5
= 5 – 2 = 3

x + (y + z)
= 2 + (-4 + 5)
= 2 + (5 – 4)
= 2 + 1
= 3

മൂന്നാമത്തെ വരി
(x + y) + z = (-2 + ·4) + (-5)
= (4 – 2) – 5 = 2 – 5
= -(5 – 2)
= -3

x + (y + z) = -2+(4 + (-5))
2 + (4 – 5) )
= -2 + – (5 – 4)
= -2 + (-1)
= -2 – 1
= -(2 + 1)
= -3

നാലാമത്തെ വരി
(x + y) + z = (2 + (-4)) + (-5)
= (2 – 4) + (-5)
= -2 – 5
= -(2 + 5)
= -7

x + (y + z) = 2 + (-4 + (-5))
= 2 + (-4 – 5)
= 2 + -(4 + 5)
= 2 + (-9)
= 2 – 9
= -(9 – 2)
= -7.

അഞ്ചാമത്തെ വരി
(x + y) + z = (-2 + 4) + 5
= (4 – 2) + 5
= 2 + 5
= 7

x + (y + z)
= -2 + (4 + 5)
= -2 + 9
= 9 – 2
= 7

ആറാമത്തെ വരി
(x + y) + z = (-2 + (-4)) + 5
= -(2 + 4) + 5
= -6 + 5
= 5 – 6
= -1

x + (y + z) = -2 + (-4 + 5)
= -2 + (5 – 4)
= -2 + 1
= 1 – 2
= -1

ഏഴാമത്തെ വരി
(x + y) + z = (-2 + (-4)) + (-5)
= (-2 – 4) – 5
= -(2 + 4) – 5
= -6 – 5
= -(6 + 5)
= -11

x + (y + z) = -2 + (-4 + (-5))
= -2 + (-4 – 5)
= -2 – 9
= -(2 + 9)
= -11

x	y	z	(x + y) + z	x +(y + z)
2	4	-5	1	1
2	-4	5	3	3
-2	4	-5	-3	-3
2	-4	-5	-7	-7
-2	4	5	7	7
-2	-4	5	-1	-1
-2	-4	-5	-11	-11

Question 4.
x, y, z ഇവ പല അധിസംഖ്യകളായും ന്യൂനസംഖ്യകളായും എടുത്ത്, x – (y – z) ഉം (x – y) + z ഉം കണക്കാക്കുക. രണ്ടും ഒരേ സംഖ്യയാണോ?
Answer:
x = 1, y = -1, z = 2 എന്നെടുക്കുക.
x – (y – z) = 1 -(-1 – 2)
= 1 – (-3)
= 1 + 3 = 4

(x – y) + z = (1 – (-1)) + 2
= (1 + 1) + 2
= 2 + 2
= 4
രണ്ടും ഒരേ സംഖ്യയാണ്.

Question 5.
ചുവടെപ്പറയുന്ന ഓരോ കണക്കിലും, പറഞ്ഞിരിക്കുന്ന തരത്തിലുള്ള രണ്ടു ജോടി x, y കണ്ടുപിടി ക്കുക.
(i) x അധിസംഖ്യ, y ന്യൂനസംഖ്യ, x – y = 1
(ii) x ന്യൂനസംഖ്യ, y അധിസംഖ്യ, x – y = 1
Answer:

Question 6.
(i) a, b, c, d ഇവ അടുത്തടുത്തുള്ള നാലു എണ്ണൽ സംഖ്യകളോ, അല്ലെങ്കിൽ അവയുടെ ന്യൂനങ്ങ ളായോ എടുത്ത് b- – c + d കണക്കാക്കുക.
(ii) സംഖ്യകൾ ഏതായാലും ഇതു പൂജ്യമാകുന്നത് എന്തുകൊണ്ടാണെന്ന് ബീജഗണിതമുപയോഗിച്ച് വിശദീകരിക്കുക.
(iii) a – b = c + d എന്നതിനുപകരം a + b – c – d ആയാലോ?
(iv) a – b + c – d ആയാലോ?
Answer:
i) a = 1, b = 2, c = 3, d = 4 എന്നെടുക്കുക.
a – b – c + d = 1 – 2 – 3 + 4
= -1 – 3 + 4
= -4 + 4
= 0

ii) Let a = x, b = x + 1, c = x + 2, d = x + 3 എന്നെടുക്കുക.
a – b – c + d = x – (x + 1) – (x + 2) + (x + 3)
= x – x – 1 – x – 2 + x + 3

iii) a + b – c – d = 1 + 2 – 3 – 4
= 3 – 3 – 4
= 0 – 4
= -4

iv) a – b + c – d = 1 – 2 + 3 – 4
= -1 + 3 – 4
= 2 – 4
= -2

Question 7.
x, y, 2 ആയി പല അധിസംഖ്യകളും ന്യൂനസംഖ്യകളും എടുത്ത് (x + y) ഉം xz + yz ഉം കണക്കാക്കുക. എല്ലാറ്റിലും (x + yz = xz + yz എന്ന സമവാക്യം ശരിയാകുന്നുണ്ടോ എന്ന് പരിശോധിക്കുക.
Answer:
x = 1, y = 2, z = -1 എന്നെടുത്താൽ,
(x + y)z = (1 + 2)(-1) = 3 × (-1) = -3.
xz + yz = 1 × (-1) + 2 × (-1) = -1 + (-2) = -1 − 2 = -3.
ഇവിടെ, (x + y)z = xz + yz.

x = -1, y = 3, z = -4 എന്നെടുത്താൽ,
(x + y)z = (-1+3)(-4) = 2 × (-4) = -8.
xz + yz = -1 × (-4) + 3 × (-4) = 4 + (-12) = 4 – 12 = -8.
ഇവിടെ, (x + y)z = xz + yz.

Question 8.
(x + y)(u + v) = xu + xv + yu +v എന്ന സമവാക്യം x, y, u, v ഇവയ്ക്കു പകരം -x, -y, -u, -v എടുത്താലും ശരിയാകുമെന്ന് തെളിയിക്കുക.
Answer:
(-x + (-y))(-u + (-v)) = (-x- y)(-u – v)
= -(x + y) × -(u + v)
= (x + y)(u + v)
= xu + xv + yu + yv

Question 9.
ചുവടെയുള്ള സമവാക്യങ്ങളിലെല്ലാം x ആയി പറഞ്ഞിട്ടുള്ള സംഖ്യകൾ എടുക്കുമ്പോൾ, y ആയി കിട്ടുന്ന സംഖ്യ കണ്ടുപിടിക്കുക.
Answer:
(i) y = x², x = -1
(ii) y = x² + 3x + 2, x =’-1
(iii) y = x² + 3x + 2, x = -2
(iv) y = (x + 1)(x + 2), x = -1
(v) y = (x + 1)(x + 2), x = -2
Answer:
i) y = (-1)²
= -1 × -1
= 1

ii) y = (-1)² + 3 × (-1) + 2
= 1 + (-3) + 2
= 1 – 3 + 2
= -2 + 2
= 0

iii) y = (-2)² + 3 × (-2) + 2
= 4 + (-6) + 2
= 4 – 6 + 2
= -2 + 2
= 0

iv) y = (-1 + 1)(-1 + 2)
= 0 × 1
= 0

v) y = (-2 + 1)(-2 + 2)
= -1 × 0
= 0

Question 10.
y = x² + 4x + 4 എന്ന സമവാക്യത്തിൽ x ആയി പല അധിസംഖ്യകളും, ന്യൂനസംഖ്യകളും എടുത്ത് y കണക്കാക്കുക. എന്തുകൊണ്ടാണ് x എന്തായാലും y അധിസംഖ്യയോ പൂജ്യമോ തന്നെ ആകുന്നത്?
Answer:
x = -1, എന്നെടുത്താൽ, y = (-1)² + 4 × (-1) + 4 = 1 – 4 + 4 = -3 + 4 = 1.
x = 1, എന്നെടുത്താൽ, y = 1² + 4 × 1 + 4 = 1 + 4 + 4 = 9.
x = 2, എന്നെടുത്താൽ, y = 2² + 4 × 2 + 4 = 4 + 8 + 4 = 16.
x = -2, എന്നെടുത്താൽ, y = (-2)² + 4 × (-2) + 4 = 4 – 8 + 4 = -4 + 4 = 0.
ഇവിടെ, y = x² + 4x + 4 = (x + 2)²
പൂജ്യമല്ലാത്ത ഏതൊരു സംഖ്യയുടെയും വർഗ്ഗം അധിസംഖ്യ ആയിരിക്കും. പൂജ്യത്തിന്റെ വർഗ്ഗം പൂജ്യം തന്നെയാണ്. ചുരുക്കിപറഞ്ഞാൽ ഏതൊരു സംഖ്യയുടെയും വർഗ്ഗം പൂജ്യമോ അധിസംഖ്യയോ ആയിരിക്കും. അതുകൊണ്ടാണ് x എന്തായാലും y അധിസംഖ്യയോ പൂജ്യമോ തന്നെ ആകുന്നത്.

Question 11.
എണ്ണൽ സംഖ്യകൾ, അവയുടെ ന്യൂനങ്ങൾ, പൂജ്യം ഇവയെ എല്ലാം പൊതുവായി പൂർണ്ണസംഖ്യകൾ എന്ന് പറയാം.
(i) x + y = 25 എന്ന സമവാക്യം ശരിയാകുന്ന എത്ര ജോടി പൂർണ്ണസംഖ്യകൾ കണ്ടുപിടിക്കാം?
(ii) x – y = 25 എന്ന സമവാക്യം ശരിയാകുന്ന എത്ര ജോടി പൂർണ്ണസംഖ്യകൾ കണ്ടുപിടിക്കാം?
Answer:
i) 0² + 5² = 25 → (0, 5)
5² + 0² = 25 → (5,0)
0² + (-5)² = 25 → (0, -5) .
(-5)² + 0² = 25 → (-5, 0)
3² + 4² = 25 → (3, 4)
42 + 3² = 25 → (4, 3)
(-3)² + (-4)² = 25 → (-3,-4)
(-4)² + (-3)² = 25 → (-4,-3) ‘
(-3)² + 4² = 25 → (-3, 4)
4² + (-3)² = 25 → (4, -3)
(-4)² + 3² = 25 → (-4, 3)
3² + (4)² = 25 → (3,-4)
പന്ത്രണ്ടു ജോടി പൂർണ്ണസംഖ്യകൾ.

ii) 5² – 0² = 25
(-5)² – o² = 25
13² – 12² = 25
(-13)² – 12² = 25
13² – (-12)² = 25
(-13)² – (-12)² = 25
ആറു ജോടി പൂർണ്ണസംഖ്യകൾ : (5, 0), (-5, 0), (13, 12), (-13, 12), (13, -12), (-13, -12)

Question 12.
1 × 2 × 3 × 4 × 5 = 120. ആണല്ലോ. (-1) × (-2) × (-3) × (-4) × (-5) എത്രയാണ്?
Answer:
(-1) × (-2) × (-3) × (-4) × (-5) = 2 × (-3) × (-4) × (-5)
= -6 × (-4) × (-5)
= 24 × (-5)
= -120

Question 13.
(1 × 2 × 3 × 4 × 5) + [(-1) × (-2) × (-3) × (-4) × (-5)] എത്രയാണ്?
Answer:
1 × 2 × 3 × 4 × 5 = 120
(-1) × (-2) × (-3) × (-4) × (-5) = -120
(1 × 2 × 3 × 4 × 5) + [(-1) × (-2) × (-3) × (-4) × (-5)]
= 120 + (-120)
= 120 – 120
= 0

Question 14.
y = \(\frac{1}{x}\) എന്ന സമവാക്യത്തിൽ, x ആയി 2, -2, \(\frac{1}{2}\), –\(\frac{1}{2}\) എന്നീ സംഖ്യകളെടുത്താൽ y ആയി കിട്ടുന്ന സംഖ്യകൾ കണക്കാക്കുക.
Answer:
x = 2, എന്നെടുത്താൽ, y = \(\frac{1}{2}\)
x = -2, എന്നെടുത്താൽ, y = \(\frac{1}{-2}\) = -(\(\frac{1}{2}\))
x = \(\frac{1}{2}\), എന്നെടുത്താൽ, y = \(\frac{1}{\frac{1}{2}} = 2\) = 2
x = –\(\frac{1}{2}\), എന്നെടുത്താൽ, y = \(\frac{1}{\frac{-1}{2}} = 2\) = 2

Question 15.
y = \(\frac{1}{x-1}+\frac{1}{x+1}\) എന്ന സമവാക്യത്തിൽ x = -2 എന്നെടുക്കുമ്പോഴും x = –\(\frac{1}{2}\) എന്നെടുക്കുമ്പോഴും y ആയി കിട്ടുന്ന സംഖ്യകൾ കണക്കാക്കുക.
Answer:
x = -2 എന്നെടുത്താൽ,
y = \(\frac{1}{-2-1}+\frac{1}{-2+1}\)
= \(\frac{1}{-3}+\frac{1}{-1}\)
= \(-\left(\frac{1}{3}+\frac{3}{3}\right)\)
= \(-\left(\frac{4}{3}\right)\)

x = –\(\frac{1}{2}\) എന്നെടുത്താൽ,
y = \(\frac{1}{-\left(\frac{1}{2}\right)-1}+\frac{1}{-\left(\frac{1}{2}\right)+1}\)
= \(\frac{1}{-\frac{3}{2}}+\frac{1}{\frac{1}{2}}\)
= \(-\frac{2}{3}+\frac{2}{1}=\frac{2}{1}-\frac{2}{3}\)
= \(\frac{4}{3}\)

Question 16.
z = \(\frac{x}{y}-\frac{y}{x}\) എന്ന സമവാക്യത്തിൽ, x, y ആയി ചുവടെപ്പറയുന്ന സംഖ്യകളെടുക്കുമ്പോൾ 2 ആയി കിട്ടുന്ന സംഖ്യകൾ കണക്കാക്കുക.
(i) x = 10, y = -5
Answer:
Z = \(\frac{10}{-5}-\frac{-5}{10}\)
= -2 – \(\frac{-1}{2}\)
= -(2 –\(\frac{1}{2}\))
= \(\frac{-3}{2}\)

(ii) x = -10, y = 5
Answer:
Z = \(\frac{-10}{5}-\frac{5}{-10}\)
= – 2 – \(\frac{1}{-2}\)
= -(2 – \(\frac{1}{2}\))
= \(\frac{-3}{2}\)

(iii) x = -10, y = -5
Answer:
Z = \(\frac{-10}{-5}-\frac{-5}{-10}\)
= 2 – \(\frac{1}{2}\)
= \(\frac{3}{2}\)

Class 9 Maths Chapter 7 Malayalam Medium Intext Questions and Answers

Question 1.
ചുവടെ പറഞ്ഞിരിക്കുന്ന കണക്കുകൾ ചെയ്യുക.
(i) 6 – 8
Answer:
6 – 8 = -(8 – 6)
= -2

(ii) -6 + 8
Answer:
-6 + 8 = 8 – 6
= 2

(iii) -6 – 8
Answer:
-6 – 8 = -(6 + 8)
= -14

(iv) 2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
Answer:
2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
= \(\frac{5}{2}-\frac{7}{2}=-\left(\frac{7}{2}-\frac{5}{2}\right)\)
= \(-\left(\frac{7-5}{2}\right)=-\left(\frac{2}{2}\right)\)
= -1

(v) -2\(\frac{1}{2}\) + 3\(\frac{1}{2}\)
Answer:
-2\(\frac{1}{2}\) + 3\(\frac{1}{2}\)
= \(-\frac{5}{2}+\frac{7}{2}\)
= \(\frac{7}{2}-\frac{5}{2}=\frac{2}{2}\)
= 1

(vi) -2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
Answer:
-2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
= \(-\frac{5}{2}-\frac{7}{2}\)
= \(-\left(\frac{5}{2}+\frac{7}{2}\right)=-\left(\frac{12}{2}\right)\)
= -6

Question 2.
സ്ഥാനങ്ങൾ അടയാളപ്പെടുത്തിയ വരയിൽ, ആദ്യസ്ഥാനവും സ്ഥാനമാറ്റവും വേറെ സംഖ്യകളാക്കിയ ഒരു പട്ടികയാണ് ചുവടെ.

ചിത്രം വരച്ച് അവസാനസ്ഥാനങ്ങൾ കണക്കാക്കി പട്ടികയിൽ എഴുതുക.
Answer:
ആദ്യത്തെ വരി;

രണ്ടാമത്തെ വരി

മൂന്നാമത്തെ വരി

നാലാമത്തെ വരി

അഞ്ചാമത്തെ വരി

ഏഴാമത്തെ വരി

എട്ടാമത്തെ വരി;

ന്യൂനസംഖ്യകൾ

സ്ഥാനമാറ്റം	ആദ്യസ്ഥാനം	അവസാനസ്ഥാനം
7	3 വലത്തേയ്ക്ക്	10
3	7 വലത്തേയ്ക്ക്	10
-7	3 വലത്തേയ്ക്ക്	-4
-3	7 വലത്തേയ്ക്ക്	4
7	3 ഇടത്തേയ്ക്ക്	4
3	7 ഇടത്തേയ്ക്ക്	-4
-7	3 ഇടത്തേയ്ക്ക്	-10
-3	7 ഇടത്തേയ്ക്ക്	-10

Negative Numbers Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
തന്നിരിക്കുന്ന നമ്പറുകൾ ഉപയോഗിച്ച് z = x – y എന്ന സമവാക്യത്തിലെ 2 ന്റെ മൂല്യം കണക്കാക്കുക.
) x = 7, y = 2
b) x = -3, y = -6
c) x = -8, y = 3
d) x = -4, y = 9
Answer:
a) x = 7, y = 2
z = x – y
= 7 – 2
= 5

b) x = -3 y = -6
z = x- y = -3- (-6)
= -3 + 6
= 3

c) x = -8 y = 3
z = x – y
= -8 – 3
= -11

d) x = -4 y = 9
z = x – y
= -4 – 9
= -13

Question 2.
x = 4, y = -3, z = 8 ആയാൽ തന്നിരിക്കുന്നവ കണക്കാക്കുക.
a) (x + y) + z
b) x + (y + z)
c) xyz
d) (x + y)z
e) xy + xz
Answer:
a) (x + y) + z
= (4 – 3) + 8
= (4 – 3) + 8
= 1 + 8
= 9

b) x + (y + z)
= 4 + (-3 + 8)
=4 + 5
= 9

c) yz = 4 × (-3) × 8
= -12 × 8
= -96

d) (x + y)z = (4 – 3) × 8
= 1 × 8
= 8

e) xy + xz = (4 × (-3)) + (4 × 8)
= -12 + 32
= 20

Question 3.
x ന് തന്നിരിക്കുന്ന മൂല്യങ്ങൾ എടുത്ത് y = x² + 9x – 5 കണക്കാക്കുക.
a) x = 1
b) x = -3
c) x = 0
Answer:
a) x = 1
y = x² + 9x – 5
= 1² + (9 × 1) – 5
= 1 + 9 – 5
= 10 – 5
= 5

b) x = -3
y = x² + 9x – 5
= (-3)² + (9 × (-3)) – 5
= 9 – 27 – 5
= -23

c) x = 0
y = x² + 9x – 5
= 0² + (9 × 0) – 5
= -5

Question 4.
x = -1 ആയാൽ y = x⁴ + x³ + x² + x + 2 കണക്കാക്കുക.
Answer:
y = (-1)⁴ + (-1)³ + (-1)² + (-1) + 2
= 1 + (-1) + 1 + (-1) + 2
= 1 – 1 + 1 – 1 + 2
= 2.

Question 5.
തന്നിരിക്കുന്ന സമവാക്യങ്ങൾ സർവ്വസമവാക്യങ്ങൾ ആണോ എന്ന് പരിശോധിക്കുക. x = 1, 2, 3 എന്നും x = -1, -2, -3 എന്നും എടുത്ത് ഓരോന്നിലും നിന്നും കിട്ടുന്ന പാറ്റേൺ എഴുതുക.
a) -x + (x + 3)
b) (x + 2) – (x + 3)
c) -x – (x + 1) + 2x + 1
Answer:
x = 1 ആയാൽ,
a) -x + (x + 3) = -1 + (1 + 3)
= -1 + 4
= 3

b) (x + 2) – (x + 3) = (1 + 2) – (1 + 3)
= 3 – 4
= -1

c) -x – (x + 1) + 2x + 1 = -1 – (1 + 1) + 2 + 1
= -1 – 2 + 2 + 1
= 0

ഇതുപോലെ, x = 2, 3 എടുക്കുമ്പോഴും ഇവ സർവ്വസമവാക്യങ്ങൾ ആണെന്ന് കാണാം. ഇനി, if x = -1 ആയാൽ,
a) -x + (x + 3) = -(-1) + (-1 + 3)
= 1 + 2
= 3

b) (x + 2) – (x + 3) = (-1 + 2) – (-1 + 3)
= 1 – 2
= -1

c) -x-(x + 1) + 2x + 1 = -(-1) – (-1 + 1) – 2 + 1
= 1 – 0 – 1
= 0
ഇതുപോലെ, x = -2, -3 എടുക്കുമ്പോഴും ഇവ സർവ്വസമവാക്യങ്ങൾ ആണെന്ന് കാണാം.

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 8 Malayalam Medium വൃത്തങ്ങളുടെ അളവുകൾ can save valuable time.

Kerala SCERT Class 9 Maths Chapter 8 Solutions Malayalam Medium വൃത്തങ്ങളുടെ അളവുകൾ

Class 9 Maths Chapter 8 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 8 Malayalam Medium Textual Questions and Answers

Question 1.
വ്യാസം 2 മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവ് ഏകദേശം 6.28 മീറ്ററാണെന്നു അളന്നു കണ്ടു പിടിച്ചു.
i) വ്യാസം 4 മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവെത്രയാണെന്ന് അളക്കാതെ എങ്ങനെ കണക്കാക്കും?
ii) വ്യാസം 1 മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവോ?
iii) വ്യാസം 3 മീറ്ററായാലോ?
Answer:
i) വ്യാസം 2 മീറ്ററായുള്ള വൃത്തത്തിന്റെ ചുറ്റളവ് 6.28 ആയതിനാൽ വ്യാസം 4 മീറ്ററായുള്ള വൃത്ത ത്തിന്റെ ചുറ്റളവ് 6.28 ന്റെ ഇരട്ടി ആയിരിക്കും. അതായതു 2 × 6.28 = 12.56

ii) വ്യാസം 1 മീറ്ററായാൽ വൃത്തത്തിന്റെ ചുറ്റളവ് = \(\frac{6.28}{2}\) = 3.14 മീ

iii) വ്യാസം 1 മീറ്റർ ആയാൽ വൃത്തത്തിന്റെ വ്യാസം 3.14
∴ വ്യാസം 3 മീറ്റർ ആയാൽ വൃത്തത്തിന്റെ വ്യാസം = 3 × 3.14 = 9.42 മീ

Question 2.
ഒരു കമ്പി വളച്ച് 4 സെന്റിമീറ്റർ വ്യാസമുള്ള വൃത്തമുണ്ടാക്കി. ഇതിന്റെ പകുതി നീളമുള്ള കമ്പി വളച്ചുണ്ടാക്കുന്ന വൃത്തത്തിന്റെ വ്യാസമെന്തായിരിക്കും ?
Answer:
വൃത്തങ്ങളുടെ ചുറ്റളവുകൾ മാറുന്നത് വ്യാസങ്ങൾ മാറുന്ന അതെ തോതിലാണ് .ആയതിനാൽ കമ്പിയുടെ നീളം പകുതിയാക്കി അത് വളച്ചുണ്ടാക്കുന്ന വൃത്തത്തിന്റെ വ്യാസവും പകുതിയായി കുറയുന്നു .
അതായതു വൃത്തത്തിന്റെ വ്യാസം 2 സെന്റിമീറ്റർ ആയിരിക്കും.

Question 3.
ചുവടെ വരച്ചിരിക്കുന്ന വൃത്തങ്ങളുടെയെല്ലാം ചുറ്റളവ് കണക്കാക്കുക.

Answer:
ചിത്രം: 1

ചിത്രത്തിൽ കാണിച്ചിരിക്കുന്നതുപോലെ സമഷഡ്ഭുജത്തെ ആറു തുല്ല്യ സമഭുജത്രികോണങ്ങളാക്കി ഭാഗിക്കുന്നു
ഇവിടെ ത്രികോണം OAB പരിഗണിച്ചാൽ,
OA എന്ന വശം വൃത്തത്തിന്റെ ആരത്തിനു തുല്ല്യമാണ്
OA = r = 2 സെമീ
ആയതിനാൽ,
വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π
= 2 × π × 2
= 47 സെമീ

ചിത്രം: 2

ചിത്രത്തിലെ സമചതുരം ABCD എന്ന് പരിഗണിച്ചാൽ, വികർണ്ണം AC വൃത്തത്തിന്റെ വ്യാസത്തിനു തുല്യമാണ്
ആയതിനാൽ,
AB = BC = 2 സെമീ, കോൽ B = 90 °
AC = \(\sqrt{2^2+2^2}=\sqrt{8}=2 \sqrt{2}\) സെമീ
വൃത്തത്തിന്റെ ആരം = \(\frac{1}{2}\) × 2√2 = 2 സെമീ
വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π × √2 = 2√2 π സെമീ

ചിത്രം: 3

ചിത്രത്തിൽനിന്നും,
PR = \(\sqrt{2^2+(1.5)^2}=\sqrt{6.25}\) = 2.5 സെമീ
വൃത്തത്തിന്റെ ആരം = \(\frac{1}{2}\) × 2.5 = 1.25 സെമീ
വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π × 1.25 = 2.5π സെമീ

Question 4.
ഒരു വൃത്തത്തിന്റെ കേന്ദ്രത്തിൽ നിന്നും 4 സെന്റിമീറ്റർ അകലെയുള്ള ഞാണിന്റെ നീളം 6 സെന്റി മീറ്ററാണ്. വൃത്തത്തിന്റെ ചുറ്റളവ് കണക്കാക്കുക.
Answer:
വൃത്തത്തിന്റെ കേന്ദ്രത്തിൽ നിന്നും ഞാണിലേക്കുള്ള ദൂരം = 4 സെമീ
ഞാണിന്റെ നീളം = 6 സെമീ

മുകളിൽ പറഞ്ഞ ചിത്രത്തിൽ നിന്നും AC യുടെ നീളം AB യുടെ പകുതിയാണ്
അതായത്,
AC = \(\) = 3 സെമീ
ത്രികോണം AOC പരിഗണിച്ചാൽ,
(OA)² = (AC)² + (OC)²
r² = 4²+ 3²
r² = 16 + 9 = 25
r = √25 = 5 സെമീ
വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π × r = 2π × 5 = 10 സെമീ

Question 5.
ചുവടെയുള്ള ചിത്രത്തിൽ, വൃത്തത്തിലെ മൂന്നു ബിന്ദുക്കൾ മൂലകളായി പാദവും ഉയരവും 4 സെന്റിമീറ്ററായ സമപാർശ്വത്രികോണം വരച്ചിരിക്കുന്നു.

വൃത്തത്തിന്റെ ചുറ്റളവ് കണക്കാക്കുക.
Answer:

മുകളിൽ പറഞ്ഞിരിക്കുന്ന ചിത്രത്തിൽ O വൃത്തകേന്ദ്രമാണ്
OA = OB = OC = r
CD = 4 സെമീ ആയതിനാൽ,
OD = CD – OC = 4 – r സെമീ
AD = 2 സെമീ

ത്രികോണം AOD യിൽ
(AO)² = (AD)² + (OD)²
r² = 2² + (4 – r)²
r² = 4 + 16
8r = 20
r = \(\frac{20}{8}=\frac{5}{2}\) = 2.5 സെമീ

വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π × r
= 2π × 2.5
= 5π സെമീ

Question 6.
ചുവടെയുള്ള ചിത്രങ്ങളിലെല്ലാം, വൃത്തങ്ങളുടെ കേന്ദ്രങ്ങൾ ഒരേ വരയിലാണ്. ആദ്യത്തെ രണ്ടു ചിത്രങ്ങളിൽ, ചെറിയ വൃത്തങ്ങൾക്ക് ഒരേ വ്യാസമാണ്.

എല്ലാ ചിത്രങ്ങളിലും, ചെറിയ വൃത്തങ്ങളുടെ ചുറ്റളവുകളുടെ തുകയാണ് വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ് എന്നു തെളിയിക്കുക.
Answer:
ചിത്രം: 1

ചെറിയ വൃത്തങ്ങളുടെ വ്യാസം തുല്ല്യമാണ്
വ്യാസം d എന്ന് എടുത്താൽ
ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = π × വ്യാസം = πd
ചെറിയ രണ്ടു വൃത്തങ്ങളുടെ ചുറ്റളവ് = 2πd
വലിയ വൃത്തത്തിന്റെ വ്യാസം = d + d = 2d
വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = π × 2d = 2πd
അതായത്, ചെറിയ വൃത്തങ്ങളുടെ ചുറ്റളവുകളുടെ തുകയാണ് വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ്

ചിത്രം: 2

ചെറിയ വൃത്തങ്ങളുടെ വ്യാസം d എന്ന് കരുതിയാൽ
ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = πd
ചെറിയ മൂന്നു വൃത്തങ്ങളുടെ ചുറ്റളവ് = 3πd
വലിയ വൃത്തത്തിന്റെ വ്യാസം = d + d + d = 3d
വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = π × 3d = 3rd
അതായത്, ചെറിയ വൃത്തങ്ങളുടെ ചുറ്റളവുകളുടെ തുകയാണ് വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ്

ചിത്രം: 3

ചിത്രത്തിൽ നിന്നും ചെറിയ വൃത്തങ്ങളുടെ വ്യാസം p, q, r എന്നു കരുതാം
അങ്ങനെയെങ്കിൽ,
ആദ്യത്തെ ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = πp
രണ്ടാമത്തെ ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = πq
മൂന്നാമത്തെ ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = πr
മൂന്നു ചെറിയ വൃത്തങ്ങളുടെ ചുറ്റളവ് = πp + πq + πr = π(p + q + r)
വലിയ വൃത്തത്തിന്റെ വ്യാസം = (p + q + r)
വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = π (р + q + r)
അതായത്, ചെറിയ വൃത്തങ്ങളുടെ ചുറ്റളവുകളുടെ തുകയാണ് വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ്

Question 7.
ചുവടെയുള്ള ചിത്രത്തിൽ, ഒരേ കേന്ദ്രമായ രണ്ടു വൃത്തങ്ങൾ വരച്ചിരിക്കുന്നു.

വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ്, ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവിനേക്കാൾ എത്ര കൂടുതലാണ്?
Answer:
ചെറിയ വൃത്തത്തിന്റെ ആരം എന്ന് കരുതാം
വലിയ വൃത്തത്തിന്റെ ആരം = r + 1 സെമീ
ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = 2πr സെമീ
വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π (r + 1) = 2πr + 2π സെമീ
അതായതു, വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ്, ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവിനേക്കാൾ 2π കൂടുതൽ ആണ്

Question 8.
ഒരു വൃത്തത്തിന്റെ കേന്ദ്രത്തിൽ നിന്നും 3 സെന്റിമീറ്റർ അകലെയുള്ള ഞാണിന്റെ നീളം 4 സെന്റിമീറ്ററാണ്. വൃത്തത്തിന്റെ പരപ്പളവ് കണക്കാക്കുക.
Answer:

cerardo = \(\sqrt{3^2+2^2}=\sqrt{9+4}=\sqrt{13}\)
പരപ്പളവ് = πr² = π(√13)² = 13π m

Question 9.
ചുവടെയുള്ള ചിത്രങ്ങളിൽ, വൃത്തത്തിന്റെയും സമബഹുഭുജത്തിന്റെയും പരപ്പളവുകൾ തമ്മിലുള്ള വ്യത്യാസം രണ്ടു ദശാംശസ്ഥാനങ്ങൾ വരെ കണക്കാക്കുക?

Answer:
ചിത്രം 1
സമചതുരത്തിന്റെ വികർണ്ണത്തിന്റെ നീളം = 2.5 സെമീ

സമചതുരത്തിന്റെ പരപ്പളവ് = വികർണ്ണത്തിന്റ വർഗ്ഗത്തിന്റെ പകുതിയാണ്
= \(\frac{2.5^2}{2}=\frac{6.25}{2}\)
= 3.125 ച.സെമീ

വൃത്തത്തിന്റെ വ്യാസം = 2.5 സെമീ
വൃത്തത്തിന്റെ ആരം = 1.25 സെമീ

വൃത്തത്തിന്റെ പരപ്പളവ് = πr²
= π(1.25)²
= 4.91 ച.സെമീ

പരപ്പളവുകൾ തമ്മിലുളള വ്യത്യാസം = 4.91 – 3.125
= 1.79 ച.സെമീ

ചിത്രം 2
സമഷഡ്ഭുജം 6 സമഭുജത്രികോണങ്ങൾ ചേർന്നതാണ്.
സമഭുജത്രികോണത്തിന്റെ ഒരുവശം = വൃത്തത്തിന്റെ ആരം = 2 സെമീ
ഒരു സമഭുജത്രികോണത്തിന്റെ പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) × (വശം)²
= \(\frac{\sqrt{3}}{4}\) × (2)²
= √3
= 1.73 ച. സെമീ

സമഷഡ്ഭുജത്തിന്റെ പരപ്പളവ് = 6 × 1.73
= 10.38 ച.സെമീ

വൃത്തത്തിന്റെ പരപ്പളവ് = πr² = 1(2)
= 12.56 ച.സെമീ

പരപ്പളവുകൾ തമ്മിലുളള വ്യത്യാസം = 12.56 – 10.38
= 2.18 ച.സെമീ

Question 10.
ഒരു സമചതുരത്തിന്റെ നാലു മൂലകളിൽക്കൂടിയും, ഒരു ചതുരത്തിന്റെ നാലു മൂലകളിൽക്കൂടിയും വൃത്തങ്ങൾ വരച്ചത് ചുവടെ കാണിച്ചിരിക്കുന്നു.

രണ്ടു വൃത്തങ്ങളുടെയും പരപ്പളവ് കണക്കാക്കുക.
Answer:
ചിത്രം 1
വ്യാസം = സമചതുരത്തിന്റെ വികർണം
വ്യാസം = \(\sqrt{(3)^2+(3)^2}\)
= \(\sqrt{2 \times 9}\)
= 3√2 സെമീ

വൃത്തത്തിന്റെ പരപ്പളവ് = πr²
= π × \(\left(\frac{3 \sqrt{2}}{2}\right)^2\)
= 4.5 π ച.സെമീ

ചിത്രം 2
= \(\sqrt{(4)^2+(2)^2}\)
= \(\sqrt{20}\) സെമീ

ആരം = \(\frac{\sqrt{20}}{2}\) സെമീ

പരപ്പളവ്= π × (\(\frac{\sqrt{20}}{2}\))²
= 5 ച.സെ.മീ

Question 11.
ഒരു സമചതുരം വരച്ച്, അതിന്റെ നാലു മൂലകൾ കേന്ദ്രമായും, വശത്തിന്റെ പകുതി ആരമായും വൃത്തങ്ങൾ വരയ്ക്കുക. ആദ്യത്തെ സമചതുരത്തിന്റെ അതേ വലുപ്പമുള്ള നാലു സമച തുരങ്ങൾ ചേർന്ന സമചതുരം വരച്ച്, അതിനുള്ളിൽ കൃത്യമായി ചേർന്നിരിക്കുന്ന വൃത്തം വരയ്ക്കുക.

വലിയ വൃത്തത്തിന്റെ പരപ്പളവ്, നാലു ചെറുവൃത്തങ്ങളുടെയും പരപ്പളവുകളുടെ തുക യാണെന്ന് തെളിയിക്കുക.
Answer:
ചിത്രം 1
ചെറിയ വൃത്തത്തിന്റെ ആരം = r സെമീ
ഒരു ചെറിയ വൃത്തത്തിന്റെ പരപ്പളവ് = πr² ച.സെമീ
4 ചെറിയ വൃത്തത്തിന്റെ പരപ്പളവ് = 4 × πr² ച.സെമീ
= 4πr² ച.സെമീ

ചിത്രം 2
വൃത്തത്തിന്റെ ആരം = രണ്ടു ചെറിയ വ്യത്തങ്ങളുടെ ആരങ്ങളുടെ തുകയാണ് = 2r
വൃത്തത്തിന്റെ പരപ്പളവ് = (2r)²
= 4πr² ച.സെമീ
ചെറിയ നാല് വ്യത്തങ്ങളുടെ പരപ്പളവുകളുടെ തുകയാണ് വലിയ വ്യത്തത്തിന്റെ പരപ്പളവ്

Question 12.
ചുവടെയുള്ള രണ്ട് ചിത്രങ്ങളിലെയും സമചതുരങ്ങൾക്ക് ഒരേ വലുപ്പമാണ്. ചിത്രങ്ങളിലെ പച്ച ഭാഗങ്ങൾക്ക് (ഷെയ്ഡ് ചെയ്ത ഭാഗം) ഒരേ പരപ്പളവാണെന്നു തെളിയിക്കുക.

Answer:
ചിത്രം 1
സമചതുരത്തിന്റെ വശം = 2a സെമീ
സമചതുരത്തിന്റെ പരപ്പളവ് = (2a)² = 4 a² ച.സെമീ
ഒരു വൃത്തത്തെ നാലു തുല്യഭാഗങ്ങളാക്കി സമചതുരത്തിന്റെ നാലു മൂലകളിൽ വെച്ചിരിക്കുന്നു അതിനാൽ, വൃത്തഭാഗത്തിന്റെ ആരം = a സെമീ
വൃത്തഭാഗത്തിന്റെ പരപ്പളവ് = \(\frac{\pi a^2}{4}\) ച.സെമീ
സമചതുരത്തിന്റെ നാലു മൂലകളിൽ വെച്ചിരിക്കുന്നു വൃത്തഭാഗത്തിന്റെ പരപ്പളവ്
= 4 × \(\frac{\pi a^2}{4}\) = πa² ച.സെ.മീ
പച്ചഭാഗത്തിന്റെ (ഷെയ്ഡ് ചെയ്ത ഭാഗം) പരപ്പളവ് = (4 a² – πa²) ച.സെമീ

ചിത്രം 2
സമചതുരത്തിന്റെ വശം = 2a സെമീ
സമചതുരത്തിന്റെ പരപ്പളവ് = (2a)² = 4 a² ച.സെമീ
വൃത്തത്തിന്റെ ആരം = a സെമീ
വൃത്തത്തിന്റെ പരപ്പളവ് = πa² ച.മീ
പച്ചഭാഗത്തിന്റെ (ഷെയ്ഡ് ചെയ്ത ഭാഗം) പരപ്പളവ് = (4a² – πa²) ച.മീ

Question 13.
ഒരു സമഷഡ്ഭുജത്തിന്റെ മൂലകൾ കേന്ദ്രമായി വൃത്തഭാഗങ്ങൾ വരച്ച്, ചുവടെക്കാണുന്ന രൂപം വെട്ടിയെടുക്കുന്നു. മുറിച്ചെടുത്ത രൂപത്തിന്റെ പരപ്പളവ് കണക്കാക്കുക.

Answer:
ചിത്രത്തിൽ സമഷഡ്ഭുജത്തിന്റെ ആറു മൂലകളിലും ഓരോ വ്യത്തഭാഗങ്ങൾ കാണാൻ കഴിയും. സമഷഡ്ഭുജത്തിന്റെ ഓരോ മൂലയുടെയും കോൺ 120° ആണ്
ഇതിൽ നിന്നും മൂന്നു വൃത്തഭാഗങ്ങൾ ചേർന്നാൽ ഒരു പൂർണ്ണ വൃത്തം ലഭിക്കും
ചിത്രത്തിൽ ഇത്തരത്തിൽ ആറു വൃത്തഭാഗങ്ങളാണ് ഉള്ളത്
അതിനാൽ ഇതിൽനിന്നും രണ്ടു പൂർണ്ണ വൃത്തങ്ങൾ ലഭിക്കും
അതായത്, ആറു വൃത്തഭാഗങ്ങൾ രണ്ടു പൂർണ്ണ വൃത്തങ്ങൾക്ക് തുല്യമായി കാണാൻ കഴിയും വൃത്തഭാഗത്തിന്റെ ആരം = 1 സെമീ
വൃത്തത്തിന്റെ പരപ്പളവ് = πr²
= π ച.സെമീ

രണ്ടു വൃത്തത്തിന്റെ പരപ്പളവ് = 2 × πr²
= 2π ച.സെമീ

സമഷഡ്ഭുജത്തിന്റെ പരപ്പളവ് = \(\frac{1}{2}\) × 2 × √3 × 6
= 6√3 ച.സെമീ

മുറിച്ചെടുത്ത രൂപത്തിന്റെ പരപ്പളവ് = സമഷഡ്ഭുജത്തിന്റെ പരപ്പളവ് – രണ്ടു പൂർണ്ണ
വൃത്തത്തിന്റെ പരപ്പളവ്
= (6√3 – 2π) ച.മീ

Question 14.
ഒരു സമചതുരത്തിനുള്ളിൽ, ചുവടെക്കാണുന്നതുപോലെ വൃത്തഭാഗങ്ങൾ വരയ്ക്കുന്നു.

ചിത്രത്തിൽ നീലനിറം കൊടുത്തിരിക്കുന്ന (ഷെയ്ഡ് ചെയ്ത ഭാഗത്തിന്റെ പരപ്പളവ്, സമചതുരത്തിന്റെ പരപ്പളവിന്റെ പകുതിയാണെന്നു തെളിയിക്കുക.
Answer:
സമചതുരത്തിന്റെ വശം = a സമീ
സമചതുരത്തിന്റെ പരപ്പളവ് = a² ച.സെമീ
ഷെയ്ഡ് ചെയ്ത ഭാഗത്തിന്റെ പരപ്പളവ് കാണാൻ സമചതുരത്തിന്റെ പകുതി ഭാഗത്തിൽ
വരുന്ന അർദ്ധവ്യത്തത്തിന്റെ പരപ്പളവും കൂടാതെ ബാക്കി വരുന്ന ഭാഗത്തിലുള്ള രണ്ടു കാൽ വ്യത്തങ്ങൾക്കിടയിലുള്ള ഭാഗത്തിന്റെ പരപ്പളവു കണ്ടു പിടിക്കുകയാണ് വേണ്ടത്,
അർദ്ധവ്യത്തത്തിന്റെ പരപ്പളവ് = \(\frac{\pi a^2}{8}\) ച.സെമീ
രണ്ടു കാൽ വൃത്തങ്ങൾക്കിടയിലുള്ള ഭാഗത്തിന്റെ പരപ്പളവ് = [\(\frac{a^2}{2}\) – 2 × \(\frac{1}{4}\) × π × \(\left(\frac{a}{2}\right)^2\)] ച.സെമീ
ഷെയ്ഡ് ചെയ്ത ഭാഗത്തിന്റെ പരപ്പളവ് = \(\frac{\pi a^2}{8}+\frac{a^2}{2}-\frac{\pi a^2}{8}=\frac{a^2}{2}\) ച.മീ

Circle Measures Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
ഒരു കമ്പി വളച്ച് 6 സെന്റിമീറ്റർ വ്യാസമുള്ള വൃത്തമുണ്ടാക്കി. ഇതിൻറെ ഇരട്ടി നീളമുള്ള കമ്പി വളച്ചുണ്ടാക്കുന്ന വൃത്തത്തിന്റെ വ്യാസമെന്തായിരിക്കും ?
Answer:
വൃത്തങ്ങളുടെ ചുറ്റളവുകൾ മാറുന്നത് വ്യാസങ്ങൾ മാറുന്ന അതെ തോതിലാണ്. ആയതിനാൽ കമ്പിയുടെ നീളം ഇരട്ടിയാക്കി അത് വളച്ചുണ്ടാക്കുന്ന വൃത്തത്തിന്റെ വ്യാസവും ഇരട്ടിയാക്കുന്നു. അതായതു വൃത്തത്തിന്റെ വ്യാസം 12 സെന്റിമീറ്റർ ആയിരിക്കും.

Question 2.
വ്യാസം 3 മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവ് ഏകദേശം 9.42 മീറ്ററാണെന്നു അളന്നു കണ്ടു പിടിച്ചു.
i) വ്യാസം 6 മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവെത്രയാണെന്ന് അളക്കാതെ എങ്ങനെ കണക്കാക്കും?
ii) വ്യാസം 1\(\frac{1}{2}\) മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവോ?
Answer:
i) വ്യാസം 2 മീറ്ററായുള്ള വൃത്തത്തിന്റെ ചുറ്റളവ് 9.42 ആയതിനാൽ വ്യാസം 6 മീറ്ററായുള്ള വൃത്തത്തിന്റെ ചുറ്റളവ് 9.42 ന്റെ ഇരട്ടി ആയിരിക്കും. അതായതു 2 × 9.42 = 18.84 മീ
ii) വ്യാസം 1\(\frac{1}{2}\) മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവ് = \(\frac{9.42}{2}\) = 4.7 മീ

Question 3.
20 സെമീ നീളമുള്ള ഒരു കമ്പി വളച്ചൊരു വൃത്തം ഉണ്ടാക്കിയിരിക്കുന്നു. അതെ കമ്പിയുടെ പകുതി വെട്ടി മറ്റൊരു വൃത്തമായി വളച്ചാൽ പുതിയ വൃത്തത്തിന്റെ വ്യാസം എത്രയായിരിക്കും.
Answer:
കമ്പിയുടെ ആകെ നീളം = 20 സെമീ
കമ്പിയുടെ പകുതി നീളം = \(\frac{20}{2}\) = 10 സെമീ

കമ്പിയുടെ നീളം വൃത്തത്തിന്റ ചുറ്റളവിനു തുല്യമായിരിക്കും
അതായത്,
10 = π × വ്യാസം
വ്യാസം = \(\frac{10}{\pi} /latex] സെമീ

Question 4.
ചുറ്റളവ് 81 സെമീ ആയ വൃത്തത്തിന്റെ കേന്ദ്രത്തിൽ നിന്നും 2 സെമീ അകലെയുള്ള ഞാണിന്റെ നീളം കണ്ടെത്തുക.

Answer:
വൃത്തത്തിന്റെ ചുറ്റളവ് = 2πг
8π = 2πr
r = [latex]\frac{8}{2}\) = 4 സെമീ

ത്രികോണം POR പരിഗണിച്ചാൽ,
(OP)² = (OR)² + (PR)²
4² = 2² + (PR)²
PR² = 16 – 4 = 12
PR = √12 = 2√3 സെമീ

Question 5.
ചിത്രത്തിലെ സമഷഡ്ഭുജത്തിന്റെ ചുറ്റളവ് 24 സെമീ ആണ്

i) വൃത്തത്തിന്റെ ആരമെത്ര?
ii) വൃത്തത്തിന്റെ ചുറ്റളവെത്ര?
Answer:
സമഷഡ്ഭുജത്തിന്റ ചുറ്റളവ് = 6 × ഒരു വശത്തിന്റെ നീളം
24 = 6 × ഒരു വശത്തിന്റെ നീളം
ഒരു വശത്തിന്റെ നീളം = \(\frac{24}{6}\) = 4 സെമീ

ചിത്രത്തിൽ കാണിച്ചിരിക്കുന്നതുപോലെ സമഷഡ്ഭുജത്തെ ആറുതുല്ല്യ സമഭുജത്രികോണങ്ങളാക്കി
മാറ്റുന്നു
വശത്തിന്റെ നീളം = വൃത്തത്തിന്റെ ആരം
വൃത്തത്തിന്റെ ആരം = 4 സെമീ
വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π
= 2π × 4
= 8 സെമീ

Question 6.
ആരം 8 സെന്റിമീറ്റർ ആയ വൃത്തത്തിൻറെ ചുറ്റളവും പരപ്പളവും കണക്കാക്കുക.
Answer:
വൃത്തത്തിന്റെ ആരം = 8 സെമീ
ചുറ്റളവ് = 2 πr = 2 × π × 8 = 16 π സെമീ
പരപ്പളവ് = πr² = π × 8² = 64 ച.സമീ

Question 7.
ഒരു വൃത്തത്തിന്റെ വ്യാസം 10 സെന്റീമീറ്ററാണ്. വൃത്തത്തിന്റെ പരപ്പളവ് കണ്ടുപിടിക്കുക.
Answer:
വൃത്തത്തിന്റെ വ്യാസം = 10 സെമീ
ആരം = 5 സെമീ
പരപ്പളവ് = πr² = π × 5² = 25 ച.മീ

Question 8.
i) ആരം 15 സെന്റിമീറ്ററായ ഒരു ചക്രത്തിന്റെ ചുറ്റളവ് എന്ത്? ഈ ചക്രം 5 തവണ കറങ്ങിയാൽ എത്ര ദൂരം സഞ്ചരിക്കും?
ii) ഈ ചക്രത്തിന്റെ രണ്ട് മടങ്ങ് ആരമുള്ള ചക്രം 5 തവണ കറങ്ങുകയാണെങ്കിൽ എത്ര ദൂരം സഞ്ചരിക്കും
Answer:
i) ചക്രത്തിന്റെ ആരം = 15 സെമീ
ചുറ്റളവ് = 2πr = 2 × π × 15 = 30 π സെമീ
ചക്രം 5 തവണ കറങ്ങിയാൽ സഞ്ചരിക്കുന്ന ദൂരം = 5 × 30 π = 150 π സെമീ

ii) ചക്രത്തിന്റെ ആരം = 2 × 15 = 30 സെമീ
ചുറ്റളവ് = 2πr = 2 × π × 30 = 60 7 സെമീ
ചക്രം 5 തവണ കറങ്ങിയാൽ സഞ്ചരിക്കുന്ന ദൂരം = 5 × 60 π = 300 π സെമീ

Question 9.
ചിത്രത്തിന്റെ AB വലിയ അർധവ്യത്തത്തിന്റെ വ്യാസമാണ്. AC, CD, DB ഇവ വ്യാസങ്ങളായ അർധവൃത്തങ്ങൾ വരച്ചിരിക്കുന്നു AC = 4 സെന്റിമീറ്റർ, CD = 4, DB = ഭാഗത്തിന്റെ പരപ്പളവ് എത്രയാണ്?

Answer:
AB വ്യാസമായാ അർധവ്യത്തത്തിന്റെ പരപ്പളവ് = πr² = π(3.5)² = 12.25π ച.സെ.മീ
AC വ്യാസമായാ അർധവ്യത്തത്തിന്റെ പരപ്പളവ് = πr² = π(2)² = 4π ച.സെ.മീ
CD വ്യാസമായാ അർധവ്യത്തത്തിന്റെ പരപ്പളവ് = πr² = π(1) = π ച.സെ.മീ
DB വ്യാസമായാ അർധവ്യത്തത്തിന്റെ പരപ്പളവ് = πr² – π(0.5)² = 0.25π ച.സെ.മീ
ഷെയ്ഡ് ചെയ്ത ഭാഗത്തിന്റെ പരപ്പളവ് = AB വ്യാസമായാ അർധവ്യത്തത്തിന്റെ പരപ്പളവ് – (AC വ്യാസമായ അർധവൃത്തത്തിന്റെ പരപ്പളവ് + CD വ്യാസമായ അർധവ്യത്തത്തിന്റ പരപ്പളവ്+ DB വ്യാസമായ അർധവ്യത്തത്തിന്റെ പരപ്പളവ്)
= 12.25π – (4π + π + 0.25π) = 7π ച.സെ.മീ

Students rely on Kerala Syllabus 9th Standard Chemistry Notes Pdf Download Chapter 5 Chemical Kinetics Extra Questions and Answers to help self-study at home.

Kerala Syllabus Std 9 Chemistry Chapter 5 Chemical Kinetics Extra Questions and Answers

Question 1.
Reactions can be classified as Direct combination, decomposition, simple displacement, double decomposition, and Redox reactions. State which of the following types takes place in the reactions given below.
(i) Cl₂ + 2KI → 2KCl + I₂
(ii) 2Mg + O₂ → 2MgO
(iii) AgNO₃ + HCl → AgCl + HNO₃
(iv) 4HNO₃ → 4NO₂ + 2H₂O + O₂
Answer:
(i) Cl₂+ 2KI → 2KCl + I₂: Simple displacement
(ii) 2Mg + O₂ → 2MgO: Direct combination
(iii) AgNO₃ + HCl → AgCl + HNO₃: Double decomposition
(iv) 4HNO₃ → 4NO₂ + 2H₂O + O₂: Decomposition reaction

Question 2.
Give balanced equations for reactions involving the evolution of a gas in addition of dilute acid to:
(a) sodium sulphite
(b) calcium carbonate
Answer:
(a)Na₂SO₃ + H₂SO₄ [dil.] → Na₂SO₄ + H₂O + SO₂ [g]
Sulphur dioxide gas has evolved.

(b) CaCO₃ + 2HCl [dil.] → CaCl₂ + H₂O + CO₂ [g]
Carbon dioxide gas has evolved.

Question 3.
Define the following types of chemical changes or reactions with a suitable example of each.
(a) Direct combination reaction or synthesis
(b) Decomposition reaction
(c) Displacement reaction or substitution reaction
(d) Double decomposition reaction
Answer:
(a) A chemical reaction in which two or more elements or compounds react to form one new compound is called a Direct combination reaction or synthesis.
For example,
Hydrogen bums in the air to give water:
2H₂ [g] + O₂ [g] → 2H₂O [1]

(b) A chemical reaction in which a chemical compound decomposes into two or more simpler
substances (elements and/or compounds) is called a Decomposition reaction.
For example,
Calcium carbonate, on heating, decomposes to Calcium Oxide and Carbon dioxide:
CaCO₃ \(\xrightarrow{\Delta}\) CaO + CO₂

(c) A chemical reaction in which an element or radical has replaced another element in a compound is known as a Displacement reaction or Substitution reaction.
For example,
Magnesium displaces Copper from Copper [II] sulphate solution:
Mg + CuSO₄ → MgSO₄ + Cu

(d) A chemical reaction in which both reactants [compounds] are decomposed to give two new compounds by exchanging their radicals is known as a Double decomposition reaction.
For example,
Silver nitrate + Potassium chloride → Silver chloride + Potassium nitrate
AgNO₃ + KCl → AgCl + KNO₃

Question 4.
Define a displacement reaction with a suitable example. State how it is represented. Give a reason why zinc displaces hydrogen from dilute sulphuric acid, but copper does not.
Answer:
A chemical reaction that takes place when an element [or radical] has replaced another element in a compound is known as a Displacement reaction It is represented as :
X + YZ → Y + XZ
Eg: Magnesium + Copper [II] sulphate → Magnesium sulphate + Copper
Mg + CuS0₄ → MgS0₄ + Cu
As a more electropositive metal displaces a less electropositive metal from its aq. soln. therefore, zinc, being more electropositive, is placed above [H] in the activity series and displaces hydrogen from dilute sulphuric acid, whereas copper, being less electropositive, is placed below [H] in the electrochemical series and cannot displace hydrogen from sulphuric acid.

Question 5.
Differentiate between a decomposition reaction and a double decomposition reaction.
Answer:

Decomposition reaction	Double decomposition reaction
A chemical reaction in which a compound decomposes to give two new elements / a new compound & an element / two new compounds is called a Decomposition reaction.	A chemical reaction in which both reactants [compounds] are decomposed to give two new compounds by exchanging their radicals is called a Double decomposition reaction. It is represented as XY + AB → XB + AY
For example: 2HgO → 2Hg + O2	For example: CaCl2 + Na2CO3 → 2NaCl + CaCO3↓

Question 6.
Match the chemical reactions in List I with the appropriate answer in List II.

List I	List II
AB → A + B	Endothermic reaction
X+Y– + A+B– → X+B– + A+Y–	Decomposition reaction
X + YZ → XZ + Y	Double displacement reaction
X + Y → XY – Δ (on heating)	Displacement reaction

Answer:

List I	List II
AB → A + B	Decomposition reaction
X+Y– + A+B– → X+B– + A+Y–	Double displacement reaction
X + YZ → XZ + Y	Displacement reaction
X + Y → XY – Δ (on heating)	(on heating)

Question 7.
Analyse the chemical equations and answer the following questions.
NaOH + HCl → NaCl + H₂O
(a) What is the salt formed in this reaction?
(b) What is the name of this type of reaction?
Answer:
(a) Sodium chloride
(b) Double displacement reaction

Question 8.
(a) What is the chemical, required to prepare hydrogen in the laboratory
(b) Write the balanced chemical equation of this reaction
(c) Which type of reaction is this
(Decomposition, Combination, Displacement, Double decomposition)
Answer:
(a) Dilute hydrochloric acid and Zinc
(b) Zn + 2HCl → ZnCl₂ + H₂
(c) Displacement reaction

Question 9.
A small quantity of sodium chloride solution is taken in a test tube. One drop of silver nitrate solution is added to this.
(a) What do you observe?
(b) NaCl + AgNO₃ → X + NaNO₃
Answer:
(a) A white curdy precipitate of silver chloride (AgCl) is formed
(b) Silver chloride (AgCl)

Question 10.
How does a physical change differ from a chemical change?
Answer:
Both physical and chemical changes involve a change in matter, but they differ in the way the matter is altered.

Physical change	Chemical change
The composition remains the same	Composition Changes
No new substances are formed	New substances are formed with different properties
Reversible	Irreversible
It may involve energy transfer	Involves energy transfer
Example: Melting of ice into water	Example: Burning wood

Question 11.
What are the methods usually adopted to make firewood burn faster?
Answer:

• Provide more air
• Split up into small pieces
• Make firewood dry

Question 12.
Describe an experiment to prove that the nature of the reactants affects the rate of a chemical reaction.
Answer:
Materials required for the experiment. Zn, Mg, dil. HCl and test tubes.
Procedure
Take an equal volume of dil. HCl in two test tubes. Add Zn to one and Mg of the same mass to the other. Hydrogen gas is produced in both test tubes. The rate of reaction is faster in the test tube containing Mg. This is because Mg is more reactive than Zn.

Question 13.
Why does the rate of reaction increase when concentration increases?
Answer:
As the concentration of reactants increases, the number of molecules per unit volume and the number of effective collisions increase. Consequently, the rate of reaction increases.

Question 14.
What is the relation between the rate of reaction and surface area? Write an experiment to prove it.
Answer:
Take an equal volume of dil. HC1 in two beakers. Add a small piece of marble into one and marble powder of equal mass into the other. The reaction rate is greater when marble powder is used. The rate of reaction increases when surface area increases.

Question 15.
Why does the rate of reaction increase when surface area increases?
Answer:
When solids are made into small pieces or powder, their surface area increases. As a result, the number of molecules undergoing effective collision also increases. Hence, the rate of reaction increases.

Question 16.
Write an experiment to prove the relation between temperature and rate of reaction.
Answer:
Materials required: Sodium thiosulphate, hydrochloric acid, water, boiling tube, spirit lamp – Procedure: Prepare a dilute solution of sodium thiosulphate in a beaker. Take equal volumes of this solution in two boiling tubes. Heat one boiling tube for some time. Add dilute hydrochloric acid in equal amounts in both boiling tubes.
Observation: Reaction is faster in a heated test tube.

Question 17.
What are catalysts?
Answer:
Catalysts are substances that alter the rate of chemical reactions without undergoing anypermanent chemical change.

Question 18.
The chemical reaction between marble and dilute HC1 is given.
CaCO₃ + 2HCl → CaCl₂ + H₂O + X
(a) Which gas is formed here? How can you identify this gas?
(b) Suggest any two ways you would choose to increase the rate of this chemical reaction.
Answer:
(a) Carbon dioxide (CO₂). Show a burning splinter in the gas. It will get extinguished.
OR
Show a glass rod dipped in clear lime water into the gas. The lime water will turn milky

(b) 1) Use powdered marble
2) Use concentrated HCl
3) Heat the mixture.

Question 19.
Sulphur pieces do not react with cold, concentrated nitric acid. But sulphur powder reacts.
(a) Explain the reason why the rate of chemical reaction is increased here.
(b) Suppose you want to increase the rate of reaction again. Which way would you choose? Give reason
Answer:
(a) When surface area increases, the rate of effective collision increases. So, the rate of reaction increases.

(b) Heat the mixture.
When temperature increases, more molecules will get threshold energy. So rate of effective collision and rate of reaction increases.

Question 20.
Small amounts of phosphoric acid are usually added to hydrogen peroxide to prevent its decomposition.
a. What is the function of phosphoric acid here?
b. By which name are these types of substances known?
c. Which substance would you add to increase the rate of decomposition of hydrogen peroxide?
Answer:
(a) Slow down the rate of decomposition of hydrogen peroxide
(b) Negative catalyst
(c) Manganese dioxide

Question 21.
The experiments conducted by two students are given below.
Experiment 1:
2 mL of sodium thiosulphate solution is taken in a test tube and heated, and 2 mL of diluted HC1 solution is added.
Experiment 2:
2 mL of sodium thiosulphate solution is taken in a test tube, and to it, 2 mL of diluted HC1 solution is added.
(a) In which experiment is the precipitate formed quickly? Justify your answer.
(b) Write the balanced equation for the reaction.
Answer:
(a) In a heated test tube. (Experiment 1) When temperature increases, more molecules get threshold
energy. So, the rate of effective collision increases, and the rate of reaction increases.

(b) Na₂S₂O₃ + 2HCl → 2NaCl + SO₃ + S

Question 22.
Some materials available in the laboratory are given below. Magnesium ribbon, marble powder, marble pieces, dilute HCl, concentrated HCl.
a. Which materials will you choose for the preparation of more CO₂ in less time?
b. Write the balanced chemical equation of the reaction.
Answer:
(a) Marble powder and concentrated HCl
(b) CaCO₂ + 2HCl → CaCl₂ + H₂O + CO₂

Question 23.
What are the factors that influence the speed of chemical reactions?
Answer:

• Nature of the reactants
• Concentration of the reactants
• Total surface area
• Temperature
• Presence of catalyst

Question 24.
Find a reason for the following.
a. Splitted-up logs burn fast.
b. It is said that drinking water along with food is not good.
c. Certain medicines are kept in refrigerators.
Answer:
(a) When logs are split up, the total surface area increases. So, the number of collisions per unit area increases, and the speed of chemical reaction (burning) increases.

(b) Intake of water will dilute the digestive enzymes and hydrochloric acid produced in the stomach which help the digestive process. So it is better to avoid drinking water half an hour before and half an hour after food.

(c) A rise in temperature may cause the decomposition of certain chemicals. Such substances are kept in refrigerators.

Question 25.
Match the following suitably.

Activity	Catalyst
To reduce the decomposition of hydrogen peroxide	Manganese dioxide
Industrial production of sulphuric acid	Phosphoric acid
Industrial production of ammonia	Vanadium pentoxide
To speed up the decomposition of potassium permanganate	Iron

Answer:

Activity	Catalyst
To reduce the decomposition of hydrogen peroxide	Phosphoric acid
Industrial production of sulphuric acid	Vanadium pentoxide
Industrial production of ammonia	Iron
To speed up the decomposition of potassium permanganate	Manganese dioxide

Question 26.
When, eggshell is added to Hydrochloric acid, a gas is produced
(a) Identify this gas
(b) Write the chemical equation of this reaction.
(c) If we add powdered eggshell, what happens to the rate of the above chemical reaction? Why?
Answer:
(a) Carbon dioxide (CO₂)

(b) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

(c) The reaction becomes faster. When the solid reactants are made into small pieces or-powders, their surface area increases. More reactant molecules come in contact and take part in collision, resulting in an increase in the number of effective collisions. Thus, the speed of the reaction increases.

The comprehensive approach in Kerala Syllabus 9th Standard Physics Textbook Solutions Chapter 1 Refraction of Light Notes Questions and Answers English Medium ensures conceptual clarity.

Std 9 Physics Chapter 1 Notes Solutions Refraction of Light

SCERT Class 9 Physics Chapter 1 Notes Solutions Kerala Syllabus Refraction of Light Questions and Answers

Class 9 Physics Chapter 1 Let Us Assess Answers Refraction of Light

Question 1.
Ray diagram showing the path of light through mediums A and B is given.

a) In which medium will the speed of light be less A or B ?
b) Which will be the optically denser medium? Justify your answer.
Answer:
a) Speed of light will be less in medium B.
b) Bis the optically denser medium. Here, when light enters from A to B, the light ray deviates towards the normal. It happens when light enters from an optically rarer medium to a denser medium. So B will be the optically denser medium.

Question 2.
Complete the given diagram. Mark the angle of incidence and angle of refraction.

Answer:

Question 3.
Light passes from medium X to Y. Here, the angle of refraction is greater than the angle of incidence.
a) In which medium is the speed of light more?
b) Which is the medium of larger refractive index?
c) Draw the path of light.
Answer:
a) Y
b) X
c)

Question 4.
Refractive index of different mediums are given in the table.

Medium	Refractive index
Crown glass	1.52
Glycerine	1.47
Sunflower oil	1.47
Water	1.33
Flint glass	1.62

a) In which medium does light travel with maximum speed?
b) Will a ray of light entering obliquely from glycerine to sunflower oil deviate? Explain the reason.
c) Light is transmitted from glass to each medium listed in table above If light is incident at an angle of 30°, which medium will have the largest angle of refraction? Why?
Answer:
a) In water
b) No, As speed of light is same in both these mediums, (as refractive index of both mediums are same) light does not undergo refraction.
c) In water. Among these, water has least refractive index.

Question 5.
Observe the figure. Ray of light incident on two mediums is depicted.

a) Which is the medium of higher optical density? Why?
b) Which is the medium of greater refractive index?
Answer:
a) Medium 1 is the medium of higher optical density.
The angle of refraction in medium 1 is less than that in medium 2. That is when light enters from air to medium 1, the refracted ray bends more towards the normal.
b) Medium 1 has greater refractive index. Observe the figure.

Question 6.
Observe the figure.

a) Which figure indicates total internal reflection?
b) Which figures indicate refraction?
Answer:
a) Fig (i)
b) Fig(ii) and Fig (iii)

Question 7.
The critical angle of glass is 42o. Choose the angle of incidence for which total internal reflection takes place.
a) 40°
b) 49°
c) 38°
d) 42°
Answer:
49°

Question 8.
We can see many small prisms in the reflectors used in motorcycles. Describe the benefits of using them.
Answer:
Many small prisms are used in reflectors used in motor cycles. Light falling on the prisms undergoes total internal reflection and reflects back without loosing the intensity of light. The image can be seen with greater visual clarity.

Question 9.
Observe the table.

Medium	Refractive index
Air	1.0003
Kerosene	1.33
Turpentine oil	1.44
Crown glass	1.47
Diamond	1.52

a) Choose the medium in which light has the least speed.
b) The speed of light in air is 3 × 10⁸ m/s. What is the speed of light in kerosene?
c) When a ray of light enters obliquely from air to diamond, will the refracted ray deviate towards the normal, or away from the normal? Justify your answer.
Answer:
a) Diamond

c) The refracted ray deviates towards the normal. Diamond has greatest optical density. As light enters obliquely from air to diamond, it deviates towards the normal.

Question 10.
The path of light through mediums A, B, C and D are given. Choose the correct figures. (The optical density of the mediums are in the order A<B<C<D.)

Answer:
Fig(b), (d), (e)

Question 11.
Speed of light in ethanol is lesser than that in methanol. Which medium has lower refractive index? Why?
Answer:
Methanol has lower refractive index. Speed of light in methanol is higher than that in ethanol.

Class 9 Physics Chapter 1 Extended Activities Answers Refraction of Light

Question 1.
Make periscopes using prisms instead of mirrors. Exhibit them.
Answer:
Hint:
Periscopes are devices used to see over or around objects by using a series of mirrors or prisms. Usually, periscopes use mirrors angled at 45 degrees, but prisms can also be used.

Materials required

• 2 right-angled prisms
• 2 PVC pipes or cardboard tubes (30-40 cm each)
• Strong glue
• Cutter
• Sandpaper

Steps:
1. Prepare Tubes

• Cut two PVC pipes or cardboard tubes to desired length.
• Smooth edges with sandpaper.

2. Fix Prisms

• Attach a prism inside each tube end with the hypotenuse facing the tube opening.
• Secure with hot glue or adhesive.

3. Join Tubes

• Align tubes at a right angle.
• Securely join them.
• Optionally, paint inside black to reduce reflections.
• Ensure all parts are securely fixed.
• For demonstration,position the periscope to look over obstacles and explain how prisms helps in providing a clear image in periscopes.

Question 2.
Find out the critical angle in different mediums like glycerine, water, coconut oil, glass etc., and compare them. Prepare a report including different stages like objective, materials, method of study, results etc.
Answer:
Short Report on Critical Angle in Various
Media
Objective
To determine the critical angle of various substances, such as glass, coconut oil, glycerine and water.

Material	Refractive index
Glass	1.52
Coconut oil	1.45
Glycerine	1.47
Water	1.33

Method of study
Formula to find critical angle
Snell’s law helps us to find the critical angle (θ_c). The critical angle is the angle of incidence above which total internal reflection occurs for light travelling from a denser to a rarer medium.
Snell’s law states that n₁sin (θ₁) = n₂sin(θ₂)
When θ₂ = 90° (at critical angle), sin(θ₂) = 1
So we have sin (θ_c) = \(\frac{n_2}{n_1}\)
θ_c = sin¹\(\left(\frac{n_2}{n_1}\right)\)

where n₁ is the refractive index of the denser medium and n₂ is the refractive index of the rarer medium (air). Refractive index of air is taken as n = 1
Calculating critical angle
Critical angle for glass

Substances	Critical angle
Glass	41.14°
Coconut oil	43.60°
Glycerine	42.86°
Water	48.75°

It can be observed that water has the largest critical angle, indicating that it has the lowest refractive index among the mediums studied and glass has the smallest critical angle, indicating it has the highest refractive index. Because of their same refractive indices, glycerine and coconut oil have relatively close critical angles.

Conclusion
The critical angle is one important property in optics that varies depending on the refractive index of the medium. The critical angles for glycerine, water, coconut oil, and glass were computed and compared in this study. Applications requiring light transmission and total internal reflection, such as fibre optics and different optical devices, require an understanding of these angles.

Question 3.
Can the angle of refraction be 90° when light enters from an optically rarer medium to a denser medium? Do an activity and write it down in the science diary.
Answer:
Objective
To find out if angle of refraction can be 90° when it moves from air (less dense) to water or glass (denser).
Assumption
Angle of refraction cannot be 90° when it goes from a less dense to a denser medium.

Materials required

• Laser pointer or flashlight
• Protractor
• Clear container (like a glass tank)
• Water
• Piece of glass or plastic
• Pen or pencil

Procedure

• Fill the clear container with water and place it on a table.
• Put the piece of glass or plastic next to the container.
• Shine the laser or flashlight at the water surface at different angles (0°, 30°, 45°, 60°, etc.)
• Measure the angle of incidence and the angle of refraction.
• Record Observations

Results

• When light goes from air to water, the angle of refraction is always less than the angle of incidence.
• The same happens with the glass or plastic.

Snell’s law states that n₁sin (θ₁) = n₂sin(θ₂) Here n, and n, are the refractive indices of the rarer and denser media, respectively, and θ₁, and are the θ₂, angles of incidence and refraction.
When light passes from a rarer to a denser medium, the refractive index of the denser medium (n₂) is greater than that of the rarer medium (n₁). Thus, the sine of the angle of refraction sin(θ₂) (cannot be greater than 1, limiting θ₂ to less than 90°.

Conclusion
Light cannot bend at a 90° angle when it goes from a less dense medium (like air) to a denser medium (like water or glass). This is because the refraction law shows it is impossible.

Refraction of Light Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Have you ever observed the straw in a glass of lemonade? Does it appear bent? What could be the reason?

Answer:
The straw in a glass of lemonade appears bent because of the phenomenon of light called refraction.

Question 2.
Don’t you think the light rays reflected from the straw dipped in lemonade travel through different mediums before they reach our eyes?
Answer:
Yes

Activity
Fill three fourth of a trough with water. Add two or three drops of milk into it. Cover the trough with a transparent sheet of paper. Fill the remaining space in the trough with a smoke. Then flash the light from a laser torch such that light should fall obliquely on the surface of water as shown in figure below.

Observe the path of light and record the observations in the science diary. Ray diagram showing the path of light travelling from air to water

Question 3.
Which are the mediums through which light travels within the trough?
Answer:
Air, water
Light is travelling from air to water

Question 4.
How will the path of light ray be, when it passes through only one medium?
Answer:
The path of light will be in a straight line.

Question 5.
Is there any deviation in the direction of the ray of light when it enters obliquely from one medium to another?
Answer:
Yes, there is deviation.

Question 6.
Where does the ray of light undergo a change in the direction?
Answer:
At the surface of water.

Activity
Allow light from a laser torch to fall perpendicular to the surface of water taken in a trough.

Question 7.
Is there a deviation in the path of the ray of light? Record your observation.
Answer:
No, there is no deviation in the path of the ray of light because light is incident normally.

When light travels through a single medium the path of light will be a straight line. A ray of light undergoes a deviation at the surface of separation when it enters obliquely from one medium to another. There will be no deviation in the path of the ray of light incident normally.

Activity
Observe the speed and a direction of a toy car moving from a smooth surface to a rough surface.

Question 8.
Is there a change in the direction of motion of the car?
Answer:
Yes, the direction of motion of the car changes.

Question 9.
Where does this change occur?
Answer:
This change occurs at the surface of separation of the two media.

Question 10.
Did the car move with the same speed on both surfaces?
Answer:
No, the car moves with different speed on both surfaces.
Note:The change in direction is caused by the change in the speed of the car as it moves from one surface to the other.

Reason for the change in the direction of light when it passes from one medium to another
The speed of light is different in different mediums. The change in the speed of light causes the change in the direction of the ray of light, when it passes from one medium to another. Examine the table showing the speed of light through various mediums.

Medium	Speed of Light (approximate)
Air	3 × 108 m/s
Water	2. 25 × 108 m/s
Glass	2 × 108 m/s
Diamond	1.25 × 108 m/s

By examining the table we can understand that the speed of light differs in various mediums due to the difference in their optical densities.

The ability of a medium to influence the speed of light through the medium is its optical density.

The speed of light will be lower in a medium of higher optical density (optically denser medium). The speed of light will be higher in a medium of lower optical density (optically rarer medium). Note that optical density has no relation with material density.

Question 11.
Arrange the mediums in the table shown below in the increasing order of their optical densities.

Medium	Speed of Light (approximate)
Air	3 × 108 m/s
Water	2. 25 × 108 m/s
Glass	2 × 108 m/s
Diamond	1.25 × 108 m/s

Answer:
Air <Water<Glass<Diamond

Activity to understand about refraction
A straw is kept obliquely in a glass as shown in figure below. Then fill the glass with water.

Question 12.
What difference do you observe?
Answer:
The straw appears bent when water is poured.

Question 13.
Why does the straw appear bent?
Answer:
It is due to refraction of light that the straw appears bent.

Activity
A straw is kept in a trough. Observe the path of the rays of light from the straw falling on the eyes before and after filling the trough with water.

Question 14.
Is there a deviation for the ray of light falling on the eye from point B of the straw before pouring water?
Answer:
After the water is poured, the ray of light from the immersed part of the straw entering air from water is deviated at the surface of separation.

Question 15.
Doesn’t the ray of light appear to come from C, though it is actually coming from B?
Answer:
The straw appears bent, as the actual position B seems to have risen to the position C.

Refraction

When a ray of light enters obliquely from one medium to another of different optical densities, it undergoes a deviation at the surface of separation of the mediums. This phenomenon is refraction. The ability of light to undergo refraction in a medium depends on the optical density of the medium. The role of refractive index is very important in designing various optical devices.

Refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in the medium.

Refractive index (n) = \(\frac{\text { The speed of light in vacuum (c) }}{\text { The speed of light in the medium (v)}}\) n = \(\frac{c}{v}\) The speed of light in vacuum (c) = 3 × 108 m/s

Question 16.
The speed of light in different mediums is given in the table below. Find the refractive index of each medium and complete the table.

Medium	Speed of light (v)	Refractive index (n)
Air	3 × 108 m/s	n = \(\frac{c}{v}\) = \(\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{3 \times 10^8 \mathrm{~m} / \mathrm{s}}\) = 1
Glass	2 × 108 m/s
Water	2.25 × 108 m/s

Answer:

Medium	Speed of light (v)	Refractive index (n)
Air	3 × 108 m/s	n = \(\frac{c}{v}\) = \(\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{3 \times 10^8 \mathrm{~m} / \mathrm{s}}\) = 1
Glass	2 × 108 m/s	n = \(\frac{c}{v}\) = \(\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{2 \times 10^8 \mathrm{~m} / \mathrm{s}}\) = 1.5
Water	2.25 × 108 m/s	n = \(\frac{c}{v}\) = \(\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{2.25 \times 10^8 \mathrm{~m} / \mathrm{s}}\) = 1.33

Question 17.
If the refractive index of diamond is 2.4, what is the speed of light that passes through it?
Answer:
Refractive index of diamond, n = \(\frac{c}{v}\) = 2.4
V = \(\frac{c}{v}\) = \(\frac{3 \times 10^8}{2.4}\)
v = 1.25 × 10⁸ m/s

Question 18.
How is the speed of light related to refractive index?
Answer:
As refractive index increases, speed of light decreases. The speed of light is less in a medium of higher refractive index. The speed of light is higher in a medium of less refractive index.

Question 19.
Write down how optical density is related to refractive index?
Answer:
Optical density will be higher in a medium having higher refractive index.
Note: A ray of light entering obliquely from one medium to another with different optical densities undergoes deviation at the surface of separation of the two mediums.

Question 20.
Is the deviation of light the same when it enters from air to water and from water to air when the angle of incidence is the same?
Answer:
No, the deviation of light is not the same when it enters from air to water and from water to air when the angle of incidence is the same.
Ray diagrams showing refraction
Observe the figures.

Incident ray:The ray falling on the surface of separation of the two mediums is the incident ray. Refracted ray:The ray which undergoes refraction is the refracted ray.
Angle of incidence: The angle between incident ray and the normal(NN’) .
Angle of refraction: The angle between refracted ray and the normal is the angle of refraction.

Question 21.
Find the incident ray, refracted ray, angle of incidence and angle of refraction from each figure and complete the table.

Answer:

	Fig.(1)	Fig (2)
Incident ray	AB	PQ
Refracted ray	BC	QR
Angle of incidence (i)	∠ABN	∠PQN
Angle of refraction (r)	∠CBN	∠NQR
The angle of refraction is greater than/ less than the angle of incidence	The angle of refraction is less than the angle of incidence	The angle of refraction is greater than the angle of incidence

Question 22.
How does the direction of light change when a ray of light enters obliquely from air to water? (deviates towards the normal/deviates away from the normal)
Answer:
deviates towards the normal

Question 23.
How does the direction of light change when a ray of light enters obliquely from water to air? (deviates towards the normal / deviates away from the normal)
Answer:
deviates away from the normal

Question 24.
Observe the figures that show light passing through different pairs of mediums.

Find answers to the following questions based on the optical densities of mediums.
a) Which are the figures in which the ray of light enters obliquely from an optically denser medium to a rarer medium?
Answer:
Fig.(4), Fig.(6)

b) In this case, to which direction does the refracted ray deviate? (towards the normal / away from the normal)
Answer:
away from the normal

c) Choose the figures in which the refracted ray deviates towards the normal.
Answer:
Fig.(3), Fig.(8)

d) In which situation will a ray of light deviate towards the normal as it passes from one medium to another?(from an optically denser medium to a rarer medium/from an optically rarer medium to a denser medium)
Answer:
from an optically rarer medium to a denser medium.

e) There is no refraction of light in Fig.(5) and Fig.(7). What may be the reason?
Answer:
When a ray incident normally at the surface of separation of mediums, it does not undergo refraction.

Note: When light enters from one medium to another, the incident ray, the refracted ray and the normal at the point of incidence, will be on the same plane. When light enters from an optically rarer medium to a denser medium, it deviates towards the normal. When light enters from an optically denser medium to a rarer medium, it deviates away from the normal. A ray incident normally at the surface of separation of mediums does not undergo refraction.

Allow light from a laser torch to fall on a glass slab.

Observe the path of light.

Question 25.
At which points do refraction occur for the ray of light?
Answer:
Refraction occurs at the points Q and R.

Question 26.
Draw the ray diagram showing the refraction of a ray of light in a glass slab
Answer:

Note: When light passes from air to glass it moves towards the normal and when it moves from glass to air it bends away from the normal. In a glass slab, the incident ray (PQ) and emergent ray (RS) are parallel to each other. The emergent ray is just shifted to the side, but it remains parallel to the original path.

Question 27.
Allow light to incident normally on the glass slab. Does refraction occur?
Answer:
No, the ray incident normally on the glass slab does not undergo refraction.

Put a coin in a vessel. Walk backwards looking at the coin. When the coin disappears, ask one of your classmates to pour water into the vessel.

Question 28.
What do you observe? Analyse the figure given above and find out the reason.
Answer:
On pouring water, the ray of light coming from the coin undergoes refraction and reaches the eye. Hence, the coin becomes visible again.

Question 29.
Place a glass slab on the letters in a textbook. The letters appear raised. What may be the reason? Find out.
Answer:
The ray of light coming from the letters undergoes refraction through the glass slab and reaches the eye. So the letters appear raised.

Question 30.
Observe the figure given below.

You can see a coin lying under the water in a trough. Try to take the coin out by looking through any one side of the trough.
Can you do it with ease? What may be the reason?
Answer:
We cannot do it with ease. When light coming from the coin enters from an optically denser medium to a rarer medium, it deviates away from the normal. So by looking through any one side of the trough we cannot locate the actual position
of the coin.

Question 31.
The bottom of a pond appears elevated when viewed from a distance than from a nearer point. Why?
Answer:
When light coming from the bottom of the pond enters from an optically denser medium to a rarer medium, it deviates away from the normal. As the distance from the pond and the observer increases, the angle of refraction made by the incident ray increases. So the bottom of a pond appears elevated.

Question 32.
People who engage in bow fishing aim at a point slightly below the perceived position of the fish. Why?
Answer:
When the reflected light coming from the fish enters from an optically denser medium to a rarer medium, it is deviated away from the normal at the surface of separation. This refracted light appears to come from a position just above the actual position. So people who engage in bow fishing aim at a point slightly below the perceived position of the fish.

Observe the figures.

Question 33.
Does the ray of light from the star reach our eyes by travelling in a straight line?
Answer:
No, the ray of light from the star does not reach our eyes by travelling in a straight line.

Question 34.
An illustration of the path of a ray of light from a distant star through the Earth’s atmosphere is given in Fig.(9) .Doesn’t the ray of light undergo irregular deviations? What may be the reason?
Answer:
Yes. The ray undergoes irregular deviations.
Stars appear as point sources of light because they are at a greater distance from the Earth compared to the planets. The light coming from the stars reaches our eyes by traversing through the atmosphere. The optical density of the medium through which the light travels goes on changing as the physical conditions (pressure, temperature etc.) of the layers of atmosphere change continuously. Hence, the light undergoes an irregular refraction. Therefore, when the light rays from the stars reach the eyes after refracted several times, the star cannot be seen continuously at the same position. This is the reason for the twinkling of stars.

Question 35.
Even after the Sun has passed the western horizon, the Sun is visible for some more time. Similarly, the Sun can be seen a few seconds before it reaches the eastern horizon in the morning. What is the reason? Analyse the figure given below and record the explanation in the science diary.

Answer:
This is due to atmospheric refraction. The temperature difference causes density difference in the layers of atmosphere. Due to this, optical density of the medium through which the light travels goes on changing. Light enters to the atmosphere undergoes continuous refraction. We can see the apparent position of the Sun just before and just after the actual Sunrise and actual Sunset.

Question 36.
Will refraction occur whenever light enters obliquely from one medium to another?
Answer:
No

Question 37.
When light passes through the fibres of the decorative lamps as shown in figure, it emerges only through the fibre tips. What may be the reason?
Answer:
It is due to total internal reflection of light.
Activities to understand total internal reflection

Activity 1
Allow light from a laser torch to fall at different angles on the surface of water taken in a trough.

Question 38.
Which are the mediums through which light passes inside the trough?
Answer:
water, air

Question 39.
Does the light enter from an optically denser medium to a rarer medium or from an optically rarer medium to a denser medium?
Answer:
Here, light enters from an optically denser medium to a rarer medium (That is from water to air).

Question 40.
Increase the angle of incidence gradually. What do you observe?
Answer:
Here the light enters from an optically denser medium to rarer medium. The angle of refraction goes on increasing with the increase in the angle of incidence. The incident ray reflects back to the water completely when the angle of incidence is above a specific value.

Question 41.
When does the light reflect completely to the same medium, without refraction?
Answer:
The incident ray reflects completely to the same medium when the angle of incidence is above a specific value.

Activity 2

Draw a circle on a chart paper. Mark angles on the paper as if two protractors are placed together. Keeping this on the surface of a table, place a thick semicircular glass slab on the circle drawn on the chart paper as shown in figure given above. Allow the light from a laser torch to fall on this slab at different angles.

Question 42.
Observe the angle of incidence and angle of refraction in each case. Draw the ray diagram.
Answer:

Question 43.
From which medium to which does the ray of light enter?
(from an optically denser medium to a rarer medium/from an optically rarer medium to a denser medium)
Answer:
From an optically denser medium to a rarer medium.

Question 44.
What is the change in the angle of refraction when the angle of incidence increases?
Answer:
The angle of refraction increases, when the angle of incidence increases.

Question 45.
What is the angle of incidence when the angle of refraction is 90o ?
Answer:
42°.

Question 46.
What peculiarity do you notice when the light is incident at an angle greater than this angle of incidence?
Answer:
The incident ray is reflected back completely to the same medium without undergoing refraction.

Critical Angle When a ray of light enters from an optically denser medium to a rarer medium, the angle of incidence at which the angle of refraction becomes 90° is the critical angle. In the glass-air pair, the critical angle is 42°.Total Internal Reflection When a ray of light enters from an optically denser medium to a rarer medium, at an angle of incidence greater than the critical angle, the ray is reflected back completely to the same medium without undergoing refraction. This phenomenon is total internal reflection.

Question 47.
The path of light through different mediums is given. Analyse the figures and answer the following questions.

Air
(a) Which of the above figures represent total internal reflection?
Answer:
Fig. (a), (e)

b) What is the critical angle of glass in this case?
Answer:
42°

c) Will total internal reflection occur for a ray of light entering from water to air at an angle of 500? Why? What is the critical angle for the water-air pair?
Answer:
Total internal reflection will occur for a ray of light entering from water to air at an angle of 50°. Because the angle of incidence is greater than the critical angle of water. The critical angle for the water-air pair is 48.6°

d) What are the two conditions required for total internal reflection to occur?
Answer:
The conditions required for total internal reflection to occur are

• The ray should enter from an optically denser medium to a rarermedium.
• The angle of incidence should be greater than the critical angle.

Note: If only one medium is mentioned while stating the critical angle, the second medium will be either air or vacuum.

Question 48.
Observe the figure.

The bottom of the aquarium is seen above the surface of water. What may be the reason?
Answer: The light coming from the bottom of the aquarium undergoes internal reflection at the surface of water. Hence, the bottom is seen above the surface of water. The distance from the surface of water to the bottom of the aquarium is the same as the distance from the surface to the image caused by total internal reflection.

Fibres in decorative lamp

One end of each optical fibre in the decorative lamp is connected to a suitable source of light. Light rays from this source travel through the fibre. While travelling through the fibre, it makes an angle of incidence greater than the critical angle with the walls of the fibre. Hence The ray should enter from an optically denser the light undergoes successive total internal reflection and emerges through the other end.

Optical illusions caused by total internal reflection in our surroundings.

Question 49.
During summer season there seems to be water logging on roads when viewed from a distance. What may be the reason?

Answer: During summer season the layers of air closer to the road have a low optical density as it is warmer than the upper layers. The optical density of the air increases gradually as we go higher. When light rays coming from the surrounding objects pass through different layers of air with different optical densities, they undergo refraction and then total internal reflection. Such deviated light rays fall on our eyes. Hence their image appears to have formed on roads. This phenomenon is known as mirage. Such familiar images are usually seen on the surface of water. That is why water seems to be logged on roads when viewed from a distance

Reflector

A large number of small prisms are fixed inside the reflectors in the tail lamps of vehicles. The rays of light incident on a prism get reflected back due to total internal reflection. It can be seen even in dim light.

Question 50.
How does a ray of light incident on a prism get reflected? Explain on the basis of the critical angle of glass.

Answer:
The light ray is incident normally on the side PQ. Hence, there is no refraction. The critical angle of glass is 42o. The angle of incidence at A and B is 45°. Hence the light falling on A undergoes total internal reflection and reaches B. There it undergoes total internal reflection again and comes out of the reflector. The same process happen in other prisms in the reflector as well.

Two equilateral prisms are used in periscope. Due to total internal reflection, the image can be seen with greater visual clarity.

The outer coating of optical fibre with higher refractive index is known as cladding and inner coating is known as the core. Core has a lower refractive index. Light rays passing along the core at an angle greater than the critical angle undergoes total internal reflection. Thousands of optical signals can be sent simultaneously through single cable without the loss of intensity. Such signals can be sent to distant places with the speed of light. This is the reason for using optical fibre cables in communication.

Uses of optical fibre cables
Optical fibres are used in the medical field and communication field.

Question 51.
The use of optical fibres in medical field is increasing day by day. Collect information increasing day by day. Collect information regarding this and present in your class.
Answer:
Hint:
These are some of the applications in medical field which make use of optical fibres
Endoscopy uses optical fibres to obtain the image of the inside of the body.

An optical fibre bundle can be inserted into the body by a physician. Some transport light into the body, while others return light that has been reflected off interior body surfaces. This enables the physician to look clearly into the body to diagnose conditions like cancer

Picture showing endoscope used to view inside of stomach.
Optical fibres helps to observe what is happening during keyhole surgery.
Laser Surgery
Optical fibres help deliver laser light for precise surgical procedures. The fibres guide laser light to the target area for cutting or repairing tissues.
Uses: Common in eye surgeries (like LASIK), skin treatments, and removing tumors.
Comparison of total internal reflection and reflection from a plane mirror

Question 52.
Statements regarding total internal reflection and reflection from plane mirror are given. Tabulate them suitably.

• Occurs only when a ray of light enters from an optically denser medium to a rarer medium at an angle greater than the critical angle.
• The ray of light is not completely reflected.
• Reflection occurs on a surface at any angle of incidence.
• The ray of light is completely reflected.

Answer:

Reflection from a plane mirror	Total internal reflection
The ray of light is not completely reflected	The ray of light is completely reflected.
Reflection occurs on a surface at any angle of incidence.	Occurs only when a ray of light enters from an optically denser medium to rarer medium at an angle greater than the critical angle.

Students rely on Kerala Syllabus 9th Standard Chemistry Textbook Solutions Chapter 6 Notes Questions and Answers English Medium to help self-study at home.

Kerala SCERT Class 9 Chemistry Chapter 6 Solutions

Kerala Syllabus Std 9 Chemistry Chapter 6 Notes Solutions Questions and Answers

Class 9 Chemistry Chapter 6 Let Us Assess Answers Solutions

Question 1.
Define the following terms and answer the related questions given below.
a. Solute – What is the solute in soda water?
b. Solvent – Name the solvent that dissolves common salt.
c. Solution – Differentiate between dilute solution and concentrated solution.
d. Solubility – Explain what is meant by the solubility of copper sulphate at 25°C.
Answer:
a. The component present in a small quantity or the component that is dissolved in the solvent is called the solute.
The solute in soda water is carbon dioxide (CO₂) gas.

b. The component that is present in large quantity or that dissolves the other components in it is called the solvent.
The solvent in common salt is water.

c. A solution is a homogeneous mixture of two or more substances.
A solution containing a small amount of solute is known as a dilute solution and the one which contains a large amount of solute is known as a concentrated solution.

d. The amount of solute in grams required to saturate 100 g of a solvent at a given temperature is the solubility of the solute in that solvent.
The amount of copper sulphate present in 100 g of water at 25°C is its solubility.

Question 2.
How much water in mL is needed to dissolve 200 g potassium nitrate for preparing a solution having a concentration of 20% ?
Ans:
Mass percentage = \(\frac{\text { mass of the solute }}{\text { mass of the solution }}\) × 100
Let x be the mass of the solvent.
20 = \(\frac{200}{200+x}\) × 100
200 + \(\frac{2000}{200+x}\)
20(200 + x) = 20000
4000 + 20x = 20000
20x = 20000 – 4000
= 16000
x = \(\frac{16000}{20}\) = 800
∴ 800 mL of water is required to dissolve 200g potassium nitrate.
OR
20% = \(\frac{20}{100}\) = \(\frac{200}{1000}\)
Mass of solute = 20Qg
Mass of solution = 1000 g
∴ Mass of solvent = Mass of solution – Mass of solute = 1000 – 200 = 800 g

Question 3.
You must have seen that if sodium metal or sodium chloride salt is placed in water, it disappears. Do the two changes occur in the same way? Justify your answer.
Answer:
The changes that occur are not the same. When sodium metal is placed in water, it undergoes a chemical reaction and disappears, whereas sodium chloride does not undergo a chemical reaction; instead, it gets dissolved in water.

Question 4.
A solution of sodium chloride is given. Suggest a simple method to prove whether it is saturated or unsaturated.
Answer:
By adding more sodium chloride to the solution, we can test whether the solution is saturated or unsaturated.

• If the added sodium chloride dissolves, the solution is unsaturated.
• If the added sodium chloride does not dissolve, the solution is saturated.

Question 5.
Suppose we want to change a saturated solution to an unsaturated solution. Suggest any two methods.
Answer:
A saturated solution can be converted into an unsaturated solution either by heating or by adding more solute.

Question 6.
The solubilities of certain salts in a saturated solution prepared in 100 g water at different temperatures are tabulated below.

a. Which salt shows the maximum solubility at low temperatures?
b. How does solubility change when temperature increases?
c. What is the amount of solute required to prepare a saturated solution of potassium nitrate in 50 g water at 40°C?
d. Which salt given in the table does not show much difference in solubility with varying temperature?
Answer:
a. Sodium Chloride

b. Solubility increases with an increase in temperature.

c. From the table, the amount of solute required to prepare a saturated solution of potassium nitrate in 100 g water at 40°C = 62 g
∴ The amount of solute required to prepare a saturated solution of potassium nitrate in 50 g water at40°C = \(\frac{62}{2}\) = 31 g

d. Sodium Chloride

Question 7.
Haven’t you noticed the instruction ‘Shake well before use’ on bottles containing certain medicines?
a. To which of the following categories do these medicines belong?
(Colloid, suspension, solution)
b. What is the reason for the instruction on the label?
Answer:
a. Suspension
b. Particles of a suspension settle down over time, so shaking is required to mix them again properly.

Question 8.
Classify and tabulate the mixtures given below.
(Gold ornaments, dilute acid, muddy water, rice starch water, smoke, ink, amalgam, mayonnaise, blood, lime water, ayurvedic decoction).

True Solution	Suspension	Colloid

Answer:

True Solution	Suspension	Colloid
True Solution Gold ornaments Dilute acid Amalgam	Suspension Muddy water Lime water Ayurvedic decoction	Rice starch water Smoke Ink Mayonnaise

Question 9.
The solubility of three atmospheric gases in water is given in the graph below.

a. Which gas shows maximum solubility in water?
b. What is the solubility of oxygen at 20°C?
c. What is the change in the solubility of gases as temperature increases?
d. Suggest a method to increase the solubility of CO₂ in water.
e. Prepare a note on the solubility of gases in water compared with that of solids.
f. What may beithe changes that occur in nature if oxygen shows more solubility in water? Prepare a note.
Answer:
a. Carbon dioxide

b. 0.004g/100g

c. As temperature increases, the solubility of gases decreases.

d. Decrease temperature

e. The solubility of solids and gases in water varies based on temperature. The solubility of solids in water increases with temperature, whereas the solubility of gases in water decreases with temperature.

f. The changes that occur in nature if oxygen shows more solubility in water are:

• Atmospheric oxygen decreases. Affects living organisms.
• Amount of gases like carbon dioxide increases.
• Slow down or even prevent combustion.
• Leading to slower biological decomposition and an increase in pollutants.

Question 10.
A small crystal of copper sulphate is suspended in a supersaturated solution and ’ an unsaturated solution of copper sulphate. What will be the observation on the next day?
Answer:
When a crystal of copper sulphate is added to an unsaturated solution of copper sulphate, the crystal will dissolve, increasing the concentration of the solution.
In the supersaturated solution, copper sulphate crystal will grow larger as the excess solute in the solution begins to crystallise around it.

Question 11.
Why are stabilisers used in soft drinks?
Answer:
Stabilisers are used in order to prevent the settling of particles in soft drinks.

Extended Activities

Question 1.
Prepare a supersaturated solution of sodium silicate (Na₂SiO₃) in a trough. It dissolves easily in hot water. After it reaches room temperature, add seed crystals of various chemicals to it. Copper sulphate, ferrous sulphate, ferric chloride, cobalt chloride, nickel sulphate or other available ones can be added. Keep the trough still. After a few days, it can be seen that chemicals have grown upwards from the crystals like spikes.
Answer:
Hints:
Here are the approximate colours of the crystals you might expect to form:

• Copper Sulphate: Blue
• Ferrous Sulphate: Green
• Ferric Chloride: Yellow-brown
• Cobalt Chloride: Pink or red
• Nickel Sulphate: Green

Question 2.
You might have seen the process of making tea in the kitchen. What are the ingredients used to make tea? Present the process of making tea using a note. The presentation should include terms such as solute, solvent, solution, dissolution, insoluble, filtration, precipitation, etc.
Answer:
We need water, tea, milk, and sugar to prepare tea.

• Place a quantity of water (solvent) in a pan and heat it.
• Add a certain amount of sugar (solute) and heat it until the sugar is entirely dissolved. The resulting homogeneous mixture is referred to as a solution.
• Add tea leaves (or tea) into the solution and boil the mixture to its boiling point.
• Next, add milk and then proceed to boil once more.
• Filter the mixture through the tea strainer to collect the filtrate, which consists of the soluble substances, specifically tea, in a cup.
• The tea leaves are insoluble and are left as a precipitate in the filter.

Question 3.
You are familiar with several situations in which mixtures are used in everyday life. Make a table showing such situations and the mixtures used and present in the class.
Answer:

Situation	Mixture
Baking a cake	Flour, sugar, eggs, milk, etc.
Cleaning with a detergent	Water and detergent
Cooking a meal	Various ingredients (e.g., vegetables, spices, oil)
Drinking a soda	Water, carbon dioxide, flavourings, and sweeteners
Using toothpaste	Water, abrasives, flavourings, and fluoride
Air we breathe	Nitrogen, oxygen, carbon dioxide, and other gases
Ocean water	Water, salts, minerals, and other substances
Soil	Minerals, organic matter, water, and air

Question 4.
Diverse dietary styles and the desire to discover and try out new and delicious dishes are widespread among us. We also hear news about food poisoning nowadays. Prepare a script that calls for fostering a healthy food culture and present it in the class in the form of a play. A collage can also be made by collecting newspaper articles and other notes related to food safety.
Answer:
Hints:
Key points to include in the script:

• Food hygiene practices: Proper handwashing, cleaning surfaces and utensils.
• Safe food handling: Separating raw and cooked foods, avoiding cross-contamination.
• Proper food storage: Refrigerating perishable foods, using expiration dates.
• Beware of food allergies and intolerances.

Collage ideas:

• Newspaper articles about food poisoning outbreaks
• Food safety tips and infographics
• Images of different cuisines and food cultures
• Quotes about food and health

Solutions Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
You are familiar with many substances which are soluble in water. Can you list them?
Answer:
Common Salt, Sugar, Alcohol, Acids, Salts. Alkalies. Some of the gases.

Question 2.
Now, can you name some substances which are insoluble?
Answer:
Enamel paint, Sand, Oils, Clay, Wooden powder, Coal, Limestone

The substance that dissolves is called solute, and the substance in which the solute dissolves is called solvent. When a solute dissolves in a solvent, a solution is formed.

We have learned that sugar solution is a homogeneous mixture. Solutions are homogeneous mixtures of two or more substances.

A mixture which shows the same property throughout is a homogeneous mixture.

Another important fact is that the components in the mixture do not react chemically.

Question 3.
Some solutions are given in the table. Find out the solute and the solvent, identify their physical states.
Answer:

In most of the cases, the physical state of the solvent itself is the state of the solution.
Why is pure air a gaseous solution? Notice the diagram showing the components of atmospheric air.

We have seen that pure air is a homogeneous mixture of various gases.
How do we find out the solute and solvent in a solution?

Question 4.
Brass, which is an alloy, contains 34% zinc (Zn) and 66% copper (Cu). Find out the solute and the solvent in brass from the table and record it.
Answer:
Solute – Zinc
Solvent – Copper
The component present in lesser quantity is the solute and the component present in greater quantity is the solvent.

Question 5.
Which component in atmospheric air is the solvent? Why?
Answer:
Nitrogen. The component present in larger quantity in atmospheric air is Nitrogen.So Nitrogen is the solvent.
In aqueous solutions, water is always the solvent, irrespective of its quantity.

Question 6.
In the beginning of this lesson, we saw that there are substances which are insoluble in water and soluble in some other solvents. Which is the solvent that dissolved the enamel paint?
Answer:
Turpentine

Alcohol, carbon disulphide (CS₂), benzene (C₆H₆), carbon tetrachloride (CCl₄) etc, are some organic solvents. All liquids may not mix together to form solutions.

Question 7.
Try to dissolve kerosene, petrol etc., separately in water. Do they dissolve? What may be the reason?
Answer:
Water is a polar compound while kerosene, petrol etc are non polar compounds. A solute dissolves in a solvent based on the general principle “Like dissolves like”. That is why table salt which is an ionic compound dissolves in water, whereas others like kerosene and petrol do not. Covalent compounds generally do not dissolve in water.

How table salt dissolve in water?
Table salt (NaCl) is an ionic compound made of sodium and chlorine. Its solid-state contains organised sodium and chloride ions. Table salt crystals dissolve into an ionic solution when the attraction force between hydrated ions decreases. The Na⁺/Cl^– ions in table salt crystal dissolve in water as tiny hydrated ions. Polar solvents dissolve polar solutes. Solutes dissolve in solvents because ‘Like dissolves like’

Question 8.
Take an equal quantity of water in two beakers. Add one or two crystals of potassium permanganate (KMnO₄) to one beaker. Add five or six crystals to the second beaker.
Observe the difference in the colour of the solutions in the two beakers.
Beaker 1: Light purple colour
Beaker 2: Dark purple colour
What is the reason for the difference in the colour of the two solutions?
Answer:
It is due to the difference in the amount of solute dissolved in the solution.
The solution containing a greater amount of solute is said to be more concentrated.
From this, we can understand that the concentration of a solution depends on the amount of solute.

The concentration of a solution refers to the amount of solute dissolved in a fixed amount of solvent. A solution containing a small amount of solute is known as a dilute solution and the one which contains a large amount of solute is known as a concentrated solution.

There are several units to express the concentration of a solution.

Question 9.
For example, if 10 g common salt is dissolved in 90 g water how is the concentration calculated
in terms of mass percentage?
Answer:
Mass of solute = 10 g
Mass of solvent = 90 g
Mass of solution = 10 g + 90 g = 100 g
Mass percentage of solution = \(\frac{10}{100}\) × 100 = 10%
What does it mean if the concentration of the common salt is 15 %?
It can be understood that 15 g of common salt is present in 100 g of the solution.

Question 10.
A solution is prepared by dissolving 2 g of substance A in 18 g water. Calculate the mass percentage of the solution.
Answer:
Mass of solute = 2 g
Mass of solvent = 18 g
Mass of solution = 2g + 18g = 20g
Mass percentage of solution = \(\frac{2}{20}\) × 100 = 10%
The concentration of solvents in terms of mass percentage is important in the production of many products in the industry.
For example, the bleaching solution used for industrial purposes is the aqueous solution of 3.62 mass percentage of sodium hypochlorite. This shows the importance of accuracy in measurement.

Question 11.
You saw that when a saturated solution of ammonium chloride was prepared, a little solute remained undissolved. Is it possible to dissolve it?
Try heating it. Add some more ammonium chloride and continue heating. What happens? Does the solute dissolve?
Answer:
Heating a saturated solution of ammonium chloride increases its solubility, allowing more solute to dissolve. However, if you keep adding solute to the heated solution, it will eventually become saturated again and no more solute will dissolve.

Question 12.
Now, leave the solution undisturbed and allow it to cool gradually to room temperature. Does the solute get precipitated? Record your observations.
Answer:
Yes, the solute will get precipitated when the solution cools down to room temperature. As the temperature decreases, the solubility of ammonium chloride in water also decreases. This means that the solution can no longer hold as much dissolved ammonium chloride as it could at a higher temperature. As a result, the excess ammonium chloride will precipitate out of the solution, forming crystals.

A solution which contains more amount of solute than required to saturate it at a particular temperature is called supersaturated solution.

Question 13.
Prepare supersaturated solutions of copper sulphate (CuSO₄) and potassium nitrate (KNO₃) in separate beakers. Into these solutions, hang one small crystal each (seed crystal) of copper sulphate and potassium nitrate using a thread. What change do you observe after some time? Observe it again after a day. Record your observations.
Answer:
We can see that the crystals are growing. This is due to more amount of solutes attaching to the crystals. This process will continue until the solution becomes saturated.

The process by which crystals of the solute are formed when a supersaturated solution is cooled slowly is called crystallization.

Question 14.
Take 50 mL water each in two beakers. Take 100 g each of powdered common salt (NaCl) and sodium bicarbonate (NaHCO₃) in separate dishes. Prepare saturated solutions by adding common salt to the first beaker and sodium bicarbonate to the second beaker in small portions, stirring slowly. By calculating the amounts of the two salts remaining in each dish from 100 g solute, we can find out the amount of solute required to prepare the saturated solution. Did the two solutes dissolve to the same extent?
Record your observations.
Answer:
You will likely find that sodium chloride is more soluble in water than sodium bicarbonate. This means that more sodium chloride can dissolve in a given amount of water compared to sodium bicarbonate. Therefore, the saturated solution of sodium chloride will contain a higher concentration of dissolved salt than the saturated solution of sodium bicarbonate.

The amount of solute in grams required to saturate 100 g of a solvent at a given temperature is the solubility of the solute in that solvent.

Here, we can see that it is the nature of the substance that affects solubility. You have learnt from the previous experiments that a change in temperature affects solubility.

Question 15.
Does solubility increase with an increase of temperature?

Let us examine the graph given below. This is called a solubility curve.
This is a graph that shows the solubility of certain salts in with respect to temperature.

a. What is the change that generally occurs in solubility with a rise in temperature? Increases/Decreases
Answer:
Increases

b. Which is the substance whose solubility decreases with a rise in temperature?
Answer:
Calcium Sulphate

c. Which substance shows maximum solubility at 30°C?
Answer:
Potassium Nitrate

d. What is the peculiarity in the solubility of sodium chloride compared to other salts?
Answer:
Its solubility is the same in almost all temperatures.

e. Prepare a note on the relation between temperature and solubility.
Answer:
The solubility of almost all substances increases with rise in temperature.
Example: Potassium Chloride and Potassium Chlorate
The solubility of a few substances, like calcium sulphate, decreases with a rise in temperature. The temperature has no influence on the solubility of certain substances like sodium chloride.

Are there any other factors which influence solubility?
We can see that sodium chloride is highly soluble in water. But it is only slightly soluble in alcohol, and insoluble in kerosene. This may be due to the difference in the nature of the solvents. Normally carbon dioxide (CO₂) gas is only slightly soluble in water. You must have seen that when a bottle of soda is opened, CO₂ gas comes out from the bottle with a brisk effervescence. Soda water is prepared by dissolving carbon dioxide in water under high pressure. When the bottle is opened, the pressure decreases, causing the carbon dioxide to come out as bubbles.

Question 16.
Which is the factor that affects solubility here?
Answer:
Pressure

Question 17.
On the basis of the observations you have made so far, try to list the factors which affect solubility. Answer:
1. Nature of the solute ‘
2. Temperature
3. Pressure
4. Nature of the solvent

Question 18.
Note the given solubility curve of Glauber’s salt (Na₂SO₄.10H₂O).

Find out the characteristics of the solubility of Glauber’s salt that differentiate it from other salts.
Answer:
The solubility of Glauber’s salt at 0°C is 47.6 g. On heating, the solubility increases to a certain point and reaches the maximum solubility of 49.7 g at 32.4°C. It can be seen that its solubility decreases on further heating above 32.4°C. Above 32.4°C, the kinetic energy of the solute molecules increases and the molecular structure Na2S04 is attained. Due to this, heat is liberated and the solubility tends to decrease. Glauber’s salt is cited as an example of anomalous solubility.

Question 19.
Are solutions formed when substances are mixed together? Let us mix it up.
Beaker 1: Water + sugar
Beaker 2: Water + chalk powder

a. Stir the contents in both beakers well, and observe.
In which of these beakers are the particles visible?
Answer:
Beaker 2

b. Let us examine the beakers after keeping them undisturbed for some time.
In which of these beakers did the particles settle down?
Ans:
Beaker 2

c. Stir the mixtures in both the beakers well, and pass an intense beam of light through the sides of the beakers.
In which of these beakers is the path of the light beam visible?

Answer:
Beaker 2

d. Filter each mixture using filter papers. Particles of which beaker could be separated?
Answer:
Beaker 2

The mixture of sugar and water is a true solution. Since it is a homogeneous mixture, it has a uniform composition throughout the mixture. The mixture of chalk powder and water is known as a suspension.

If the components in a mixture are not distributed uniformly, it is known as a heterogeneous mixture.

Question 20.
Find out more examples of heterogeneous mixtures.
Answer:
Mixture of water and chalk powder
Muddy Water
Paint
Mixture of oil and water
Ayurvedic decoction

Question 21.
Based on which aspects do we consider muddy water as a suspension?
Answer:
Muddy water is considered a suspension because:

• The particles are visible.
• The particles can be separated.
• The particles do not dissolve.

Question 22.
Now, you can tabulate the characteristics of a true solution and a suspension, from what you have observed. Fill up the blank spaces.
Answer:

Activity	True solution	Suspension
Filtering using a filter paper	Particles cannot be separated by filtration.	Particles can be separated by filtration.
Passing an intense beam of light	Path of light beam is not visible	Path of light beam is visible.
Keeping it undisturbed	Particles settle down	Particles do not settle down

The reason for the difference in observations is due to the difference in the size of the solute particles. The size, of the solute particles is very small in a true solution. They cannot be seen with naked eye Qr even a microscope. As the particles are very minute, they cannot scatter a beam of light. Hence, the path of light beam is not visible. In the case of a suspension, the solute particles are large enough to be seen with the naked eye. They scatter light. They gradually settle down due to gravity. We can filter these out also.

Question 23.
Take a mixture of water and milk in a beaker and repeat the previously done activity. Tick (√) your observations.
Answer:

We cannot see the particles in it with the naked eye as in a solution. They also do not settle down as in a suspension. But on passing an intense beam of light, the path of the beam is visible.

Colloidal mixtures are intermediate between solutions and suspensions. Hence, we can understand that milk is a colloid.

Question 24.
In cinema theatres and smart classrooms where visuals are shown using a projector, have you noticed that the path of light beam can be clearly seen if there are dust particles in air? What is the reason for this?
Answer:
The size of the dust particles is equal to the particle size of colloids. The reason why the path of the
light beam is visible in a dusty room is because the dust particles scatter the light in all directions. This scattered light enters our eyes, making the path of the beam appear visible.

Question 25.
By analysing the above activities, list the characteristics of true solution, suspension and colloid.
Answer:

True solutions	Suspensions	Colloids
• When kept undisturbed, particles do not settle down • Particles cannot be separated by filtration • On passing an intense beam of light, the path of the light beam is not visible. • Particle size (in nm) is 0.1 to 1 nm	• When kept undisturbed, particles settle down • Particles can be separated by filtration • On passing an intense beam of light, the path of the light beam is visible. • Particle size (in nm) is more than 1000 nm	• When kept undisturbed, particles do not settle down • Particles cannot be separated by filtration • On passing an intense beam of light, its path is visible • Particle size (in nm) is 1 to 1000 nm

Let us do an experiment.

Question 26.
Prepare a solution by adding 2g sodium thiosulphate (Na₂S₃O₃.5H₂O) to a beaker containing 50 mL of water. Place the beaker in the path of a beam of light, add a few drops of dilute hydrochloric acid, and stir well. Observe for a while. Record your observation.
Answer:
As a result of the chemical reaction, sulphur gets precipitated in colloidal form. A path of the light beam is formed. With the passage of time, more and more sulphur particles get separated, and the size of the particles increases. Finally, sulphur settles down in the form of suspension.

The phenomenon of scattering of a beam of light by colloidal particles is known as Tyndall effect.

Question 27.
Try to find out more examples of this phenomenon from daily life.
Answer:

• Sunlight filtering through trees: The dust particles in the air scatter the sunlight, making the beams visible.
• Foggy mornings: The water droplets in the fog scatter sunlight, creating a hazy appearance.
• Sunbeams through clouds: The light scatters off tiny water droplets in the clouds, creating a visible path.
• Car headlights in fog: The light scatters off the fog particles, making the beam visible.
• Smoke from a campfire: The smoke particles scatter sunlight, making the smoke appear visible.

Question 28.
Colloids have several applications in the modern world. Food production, treatment of wastewater, production of cosmetics, medicines, oil/water emulsions, mayonnaise, gels, and dairy products are all related to colloids. Identify the ingredients used in these and prepare a note. Present it in the class.
Answer:
Colloidal mixtures are intermediate between solutions and suspensions. We cannot see the particles in it with the naked eye as in a solution. They also do not settle down as in a suspension. But on passing an intense beam of light, the path of the beam is visible. Colloids play a vital role in various industries and in everyday life.

Food Production

• Milk: A colloid of fat droplets dispersed in water.
• Mayonnaise: An emulsion of oil in water, stabilised by egg yolk.
• Ice cream: A colloid of air bubbles dispersed in a mixture of milk, cream, sugar, and flavourings.
• Whipped cream: A colloid of air bubbles dispersed in cream.

Cosmetics

• Lotions and creams: Emulsions of oil in water or water in oil.
• Sunscreens: Colloids of zinc oxide or titanium dioxide dispersed in a liquid.
• Mascara: A suspension of pigments in a liquid.

Medicines

• Drug delivery systems: Colloids can be used to deliver drugs to specific tissues or organs.
• Vaccines: Often contain colloids of antigens or pathogens.
• Nanoparticles: Used in drug delivery and imaging.

Gels

• Hair gel: A gel of polymers in water.
• Contact lens solution: A gel of polymers in water.

All colloids are made up of two important components. They are dispersion medium and dispersed phase. Dispersed phase will be dispersed in dispersion medium Thau are eight major sub classes of colloids. They differ according to the physical state of the above two components.

Question 29.
How long are they fit for consumption?
Answer:
These natural drinks are best consumed fresh to retain their taste, nutrients, and health benefits. Chemical substances can be added to preserve them for a long time. The chemicals added to beverages in order to prevent the settling of particles are called stabilisers.

Some Stabilisers are

• Brominated vegetable oil
• Sucrose acetate isobutyrate
• Glyceryl ester of rosin

Substances used to prevent the spoilage of food are commonly known as preservatives. Examples are vinegar and salt.
Preservatives maintain the quality or enhance the taste of food. Certain chemicals are added to artificial drinks to give them attractive colours.
Some of such chemicals are given in the table.

Soft drinks are also known as carbonated drinks, fizzy drinks and pop drinks. Many of these are made by passing carbon dioxide through water, adding artificial colours, preservatives, sucrose or artificial sugar, caffeine etc. Artificial sweeteners can stimulate our taste buds and make us crave more sweets. Aspartame, saccharin, fructose com syrup, etc., which are many times sweeter than sugar and relatively cheaper, are the main villains in sweets.

Question 30.
Do soft drinks have any nutritive value?
Answer:
Soft drinks may provide temporary refreshment, but they offer virtually no nutritional benefits and can have negative health effects when consumed frequently.

Question 31.
How do such chemical substances which are added to food and drinks affect the body?
Answer:
Although these soft drinks have fewer calories than sugar, they raise blood triglycerides, produce unhealthy fats, and cause other health issues. Soft drinks can cause non-alcoholic liver cirrhosis and excess hunger. Such sweeteners raise uric acid in the blood and cause various disorders. Soft drink use can also induce obesity and tooth problems, according to studies.

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 10 Malayalam Medium ബഹുപദങ്ങൾ can save valuable time.

Kerala SCERT Class 9 Maths Chapter 10 Solutions Malayalam Medium ബഹുപദങ്ങൾ

Class 9 Maths Chapter 10 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 10 Malayalam Medium Textual Questions and Answers

Question 1.
ഒരു വശത്തിന്റെ നീളം മറ്റേ വശത്തിന്റെ നീളത്തേക്കാൾ 1 സെന്റിമീറ്റർ കുറവായ ചതുരങ്ങളിൽ, ചെറിയ വശത്തിന്റെ നീളം x സെന്റിമീറ്റർ എന്നെടുക്കുക.
i) ഇവയുടെ ചുറ്റളവുകൾ p(x) സെന്റിമീറ്റർ എന്നെടുത്ത്, x, p(x) ഇവ തമ്മിലുള്ള ബന്ധം സമവാക്യമായി എഴുതുക.
ii) ഇവയുടെ പരപ്പളവുകൾ a(x) ചതുരശ്രസെന്റിമീറ്റർ എന്നെടുത്ത്, x, a(x) ഇവ തമ്മിലുള്ള ബന്ധം സമവാക്യമായി എഴുതുക.
iii) p(1), p(2), p(3), p(4), p(5) എന്തെങ്കിലും ക്രമം കാണുന്നുണ്ടോ?
iv) a(1), a(2), a(3), a(4), a(5) എന്നിവ കണക്കാക്കുക. എന്തെങ്കിലും ക്രമം കാണുന്നുണ്ടോ?
Answer:
i) ചെറിയ വശത്തിന്റെ നീളം = x സെമീ
വലിയ വശത്തിന്റെ നീളം = x + 1 സെമീ
ചുറ്റളവ് = 2 (x + x + 1) = 4x + 2
സെമീ p(x) = 4x + 2

ii) പരപ്പളവ്, a(x) = x (x + 1) = x + x² ച.സെമി

iii) p(x) = 4x + 2
p(1) = 4 × 1 + 2 = 6
p(2) = 4 × 2 + 2 = 10
p(3) = 4 × 3 + 2 = 14
p(4) = 4 × 4 + 2 = 18
p(5) = 4 × 5 + 2 = 22
ഇവ നാല് വർധിച്ചു വരുന്ന ശ്രേണിയാണ്.

iv) a(x) = x² + x
a(1) = 1² + 1 = 2.
a(2) = 2² + 2 = 6
a(3) = 3² + 3 = 12
a(4) = 4² + 4 = 20
a(5) = 5² + 5 = 30
ഇവ തുടർച്ചയായ ഇരട്ടസംഖ്യകൾ കൂട്ടിവരുന്ന സംഖ്യാശ്രേണിയാണ്

Question 2.
ചിത്രത്തിൽക്കാണിച്ചിരിക്കുന്നതുപോലെ, ഒരു ചതുരത്തിന്റെ നാലു മൂലകളിൽ നിന്നും ഒരേ വലുപ്പമുള്ള ചെറു സമചതുരങ്ങൾ വെട്ടിമാറ്റി, മേലോട്ട് മടക്കി, ഒരു പെട്ടി ഉണ്ടാക്കുന്നു.

Answer:
i) വെട്ടിയെടുക്കുന്ന സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം x സെന്റിമീറ്റർ എന്നെടുത്ത്, പെട്ടിയുടെ മൂന്നളവുകളും എഴുതുക.
ii) പെട്ടിയുടെ വ്യാപ്തം v(x) ഘനസെന്റിമീറ്റർ എന്നെടുത്ത്, x ഉം v(x) ഉം തമ്മിലുള്ള ബന്ധം സമവാക്യമായി എഴുതുക.
iii) v (\(\frac{1}{2}\)), v(1), v (1\(\frac{1}{2}\)) ഇവ കണക്കാക്കുക.
Answer:
i) പെട്ടിയുടെ നീളം = 7 – 2x
പെട്ടിയുടെ വീതി = 5 – 2x
പെട്ടിയുടെ ഉയരം = x

ii) v(x) = (7 – 2x)(5 – 2x)(x)
= (35 – 24x + 4x²)(x)
= 4x³ – 24x² + 35x

ii)) v (1\(\frac{1}{2}\)) = (7 – 2 × \(\frac{1}{2}\))(5 – 2 × \(\frac{1}{2}\))(\(\frac{1}{2}\))
= 6 × 4 × \(\frac{1}{2}\)
=12 ഘ.സെമീ

v(1) = (7 – 2)(5 – 2)(1)
= 5 × 3 × 1 = 15 ഘ.സെമീ

v(1\(\frac{1}{2}\)) = (7 – 2 × (1\(\frac{1}{2}\))) (5 – 2 × (1\(\frac{1}{2}\)))(1\(\frac{1}{2}\))
= 4 × 2 × \(\frac{3}{2}\)
= 12 ഘ.സെമീ

Question 3.
ഒരു മീറ്റർ നീളമുള്ള കയറുകൊണ്ട് ഉണ്ടാക്കാവുന്ന ചതുരങ്ങളുടെ ഒരു വശത്തിന്റെ നീളം x സെന്റിമീറ്റർ എന്നും, ചതുരത്തിനകത്തെ പരപ്പളവ് 3(x) ചതുരശ്രസെന്റിമീറ്റർ എന്നുമെടുക്കുക.
i) x ഉം a(x) ഉം തമ്മിലുള്ള ബന്ധം സമവാക്യമായി എഴുതുക.
ii) a(10), 3(40) ഇവ ഒരേ സംഖ്യ ആകുന്നത് എന്തുകൊണ്ടാണ് ?
iii) x ആയി രണ്ടു വ്യത്യസ്ത സംഖ്യകളെടുക്കുമ്പോൾ a(x) ആയി ഒരേ സംഖ്യതന്നെ കിട്ടാൻ, സംഖ്യകൾ തമ്മിലുള്ള ബന്ധം എന്തായിരിക്കണം?
Answer:
i) ചതുരത്തിന്റെ ഒരു വശം = x
ചതുരത്തിന്റെ മറ്റേ വശം = 50 – x
ആയതിനാൽ, ചതുരത്തിന്റെ പരപ്പളവ്, a(x) = x(50 – x) = 50x – x²

ii) a(10) = 10(50 – 10) = 10 × 40 = 400
a(40) = 40(50 – 40) = 10 × 40 = 400
ഇവ രണ്ടും ഒന്നാണ്. അതായത്, ചതുരത്തിന്റെ നീളവും വീതിയും കൂട്ടിയാൽ 50 ആണല്ലോ. അപ്പോൾ ഒരു വശം 10 ആയാൽ മറ്റേ വശം 40 ആണ്

iii) കൂട്ടിയാൽ 50 കിട്ടുന്ന ഏത് സംഖ്യകൾക്കും ഇത് ശരിയാണെന്നു കിട്ടും. ഉദാഹരണമായി
x = 20 ഉം x = 30 എടുത്താൽ
a(20) = 20(50-20) = 20 × 30 = 600
a(30) = 30(50 – 30) = 30 × 20 = 600

Question 4.
ചുവടെയുള്ള കണക്കുകളിലെല്ലാം, പറഞ്ഞിരിക്കുന്ന അളവുകൾ തമ്മിലുള്ള ബന്ധം ബഹുപദമാണോ എന്നു പരിശോധിക്കുക. തീരുമാനത്തിന്റെ കാരണവും എഴുതുക.
i) സമചതുരാകൃതിയായ ഒരു മൈതാനത്തിനു ചുറ്റും 1 മീറ്റർ വീതിയിലൊരു പാത യുണ്ട്. മൈതാനത്തിന്റെ ഒരു വശത്തിന്റെ നീളവും, പാതയുടെ പരപ്പളവും തമ്മിലുള്ള ബന്ധം.
ii) – ലിറ്റർ വെള്ളവും, 3 ലിറ്റർ ആസിഡും ചേർന്ന ദ്രാവകത്തിൽ, വീണ്ടും ഒഴിക്കുന്ന ആസിഡിന്റെ അളവും, ദ്രാവകത്തിലെ ആസിഡിന്റെ ശതമാനത്തിലുണ്ടാകുന്ന മാറ്റവും.
iii) 3 മീറ്ററും, 4 മീറ്ററും ഉയരമുള്ള രണ്ടു കമ്പുകൾ 5 മീറ്റർ അകലത്തിൽ നിലത്തു കുത്തനെ നാട്ടിയിരിക്കുന്നു. ഒരു കമ്പിന്റെ മുകളിൽ നിന്ന് ഒരു കയറു വലിച്ചു നിലത്തുറപ്പിച്ച്, അവിടെനിന്ന് രണ്ടാമത്തെ കമ്പിലേക്ക് വലിച്ചു കെട്ടണം:

ഒരു കമ്പിന്റെ ചുവട്ടിൽനിന്ന് നിലത്തു കയർ ഉറപ്പിച്ച സ്ഥാനത്തേക്കുള്ള അകലവും, കയറിന്റെ നീളവും തമ്മിലുള്ള ബന്ധം
Answer:
i)

പാതയുടെ പരപ്പളവ് = വലിയ സമചതുരത്തിന്റെ പരപ്പളവ് – ചെറിയ സമചതുരത്തിന്റെ പരപ്പളവ് = (x + 2)² = x²
= 4x + 4
= 4(x + 1)
ഇതൊരു ബഹുപദമാണ് കാരണം ഇവിടെ കൃത്യങ്കം ഒരു എണ്ണൽ സംഖ്യയാണ്.

ii) ദ്രാവകത്തിൽ \(\frac{3}{10}\) ഭാഗം ആസിഡാണ്. അതിൽ x ലിറ്റർ ആസിഡ് ചേർക്കുമ്പോൾ ദ്രാവകത്തിന്റെ അളവും x ലിറ്റർ വർധിക്കും. അപ്പോൾ ദ്രാവക ത്തിന്റെയും ആസിഡിന്റെയും ശതമാനം \(\left(\frac{3+x}{10+x}\right)\) × 100 ഇത് ബഹുപദമല്ല, കാരണം ഇവിടെ കൃത്യങ്കം ഒരു എണ്ണൽ സംഖ്യയല്ല.

iii)

കയറിൻറെ തീളം. = \(\sqrt{x^2+9}+\sqrt{(5-x)^2+16}\)
ഇത് ബഹുപദം അല്ല.

Question 5.
ചുവടെപ്പറഞ്ഞിരിക്കുന്ന ക്രിയകളോരോന്നും ബീജഗണിതവാചകമായി എഴുതുക. ഏതെല്ലാമാണ് ബഹുപദമെന്ന് വിശദീകരിക്കുക.
i) സംഖ്യയുടെയും അതിന്റെ വ്യുൽക്രമത്തിന്റെയും തുക.
ii) സംഖ്യയുടെയും അതിന്റെ വർഗമൂലത്തിന്റെയും തുക.
iii) സംഖ്യയോട് അതിന്റെ വർഗമൂലം കൂട്ടിയതും, സംഖ്യയിൽ നിന്ന് വർഗമൂലം കുറച്ചതും തമ്മിലുള്ള ഗുണനഫലം.
Answer:
സംഖ്യ x ആയി എടുത്താൽ
i) x + \(\frac{1}{x}\), ബഹുപദമല്ല
ii) x + √x, ബഹുപദമല്ല
iii) (x – √x)(x + √x) = x² – x, ഇത് ബഹുപദമാണ്

Question 6.
ചുവടെയുള്ള ബഹുപദങ്ങളിൽ, p(1) ഉം p(10) ഉം കണക്കക്കുക.
i) p(x) = 2x + 5
ii) p(x) = 3x² + 6x + 1
iii) p(x) = 4×3 + 2x² + 3x + 7
Answer:
i) p(1) = 2 × 1 + 5 = 7
p(10) = 2 × 10 + 5 = 25

ii) p(1) = 3 × 1² + 6 × 1 + 1 = 10
p(10) = 3 × 10² + 6 × 10 + 1 = 361

iii) p(1) = 4 × 13 + 2 × 1² + 3 × 1 + 7 = 16
p(10) = 4 × 103 + 2 × 10² + 3 × 10 + 7 = 4237

Question 7.
ചുവടെയുള്ള ബഹുപദങ്ങളിൽ, p(0), p(1), p(-1) ഇവ കണക്കാക്കുക:
i) p(x) = 3x + 5
ii) p(x) = 5x – 8
iii) p(x) = 3x² + 6x + 1
iv) p(x) = 2x² – 5x + 3
v) p(x) = 4x³ + 2x² + 3x + 7
vi) p(x) = ax³ + bx² + cx + d
Answer:
i) p(0) = 3 × 0 + 5 = 5
p(1) = 3 × 1 + 5 = 8
p(-1) = 3 × -1 + 5 = 2

ii) p(0) = 5 × 0 – 8 = -8
p(1) = 5 × 1 – 8 = -3
p(-1)= 5 × -1 – 8 = -13

iii) p(0) = 3 × 0 + 6 × 0 + 1 = 1
p(1) = 3 × 1² + 6 × 1 + 1 = 10
p(-1)= 3 × (-1)² + 6 × – 1 + 1 = -2

iv) p(0) = 2 × 0 – 5 × 0 + 3 = 3
p(1) = 2 × 1² – 5 × 1 + 3 = 0
p(-1)= 2 × (-1)² – 5 × -1 + 3 = 10

v) p(0) = 4 × O + 2 × O + 3 × O + 7 = 7
p(1) = 4 × 1³ + 2 × 1² + 3 × 1 +7 = 16
p(-1)= 4 × (-1)³ + 2 × (-1)² + 3 × -1 + 7 = 2

vi) p(0) = a × 0 + b × 0 + c × 0 + d = d
p(1) = a × 1³ + b × 1² + c × 1 + d = a + b + c + d
p(-1)= a × (-1)³ + b × (-1)² + c × -1 + d = -a + b- c + d

Class 9 Maths Chapter 10 Malayalam Medium Intext Questions and Answers

Question 1.
ഒരു ചോദ്യത്തിൽ നിന്നും തുടങ്ങാം

i) ചിത്രത്തിലെ ചതുരത്തിന്റെ ചുറ്റളവെത്ര?
ii) ചതുരത്തിന്റെ നീളവും വീതിയും 1 സെന്റിമീറ്റർ വർധിപ്പിച്ചാൽ ചുറ്റളവ് എന്തായിരിക്കും? ചതുരത്തിന്റെ നീളവും വീതിയും 2 സെന്റിമീറ്റർ വർധിപ്പിച്ചാൽ ചുറ്റളവ് എത്ര?
iii) നീളവും വീതിയും 3 സെന്റിമീറ്റർ കൂട്ടിയാലോ?
iv) നീളവും വീതിയും x സെന്റിമീറ്റർ കൂട്ടിയാൽ ചുറ്റളവ് എന്താകും?
Answer:
i) ചുറ്റളവ് = 2(3 + 2) = 10 സെമീ

ii) നീളം 1 സെമീ കൂടിയാൽ = 3 + 1 = 4 സെമീ
വീതി 1 സെമീ കൂടിയാൽ = 2 + 1 = 3 സെമീ
അപ്പോൾ , പുതിയ ചുറ്റളവ് = 2(4 + 3) = 2 × 7
= 14 സെമീ

ഇത് മറ്റൊരു രീതിയിൽ കണ്ടെത്താം . ഇവിടെ നാലു വശവും 1 സെന്റീമീറ്റർ വീതം കൂടിയാൽ ആകെ 4 സെന്റീമീറ്റർ കൂടും.
അപ്പോൾ , പുതിയ ചുറ്റളവ് = 10 + 4
= 14 സെമീ

അതുപോലെ, നാലു വശവും 2 സെന്റീമീറ്റർ വീതം കൂടിയാൽ ആകെ 8 സെന്റീമീറ്റർ കൂടും.
അപ്പോൾ , പുതിയ ചുറ്റളവ് = 10 + 4 × 2
= 18 സെമീ

iii) നാലു വശവും 3 സെന്റീമീറ്റർ വീതം കൂടിയാൽ ആകെ 12 സെന്റീമീറ്റർ കൂടും. അപ്പോൾ , പുതിയ ചുറ്റളവ്
= 10 + 4 × 3
= 22 സെമീ

iv) നാലു വശവും x സെന്റീമീറ്റർ വീതം കൂടിയാൽ ആകെ 4x സെന്റീമീറ്റർ കൂടും.
അപ്പോൾ , പുതിയ ചുറ്റളവ് = 10 + 4 x

Polynomials Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
p(x) = x² + x + 1 എന്നെടുത്ത് (x + 1)p(x) – (x – 1)p(x) എന്ന ബഹുപദം കണക്കാക്കുക
Answer:
(x + 1)p(x) – (x – 1)p(x) = p(x)[(x + 1) – (x – 1)]
= p(x)[x + 1 − x + 1]
= p(x) × 2
= (x² + x + 1) × 2
= 2x² + 2x + 2

Question 2.
p(x) = 2x³ – 5x² + 6x – 3 എന്നെടുത്താൽ p(0) എന്താണ് ?
Answer:
p(0) = 2 × 0³ − 5 × 0² + 6 × 0 – 3 = -3

Question 3.
p(x) = 4x – 4, p(1), p(2) ആയാൽ p(1), p(2) എന്നീ സംഖ്യകൾ കണക്കാക്കുക.
Answer:
p(1) = 4 × 1 – 4 = 0
p(2) = 4 × 2 – 4 = 4

Question 4.
q(1) = 1 ഉം, q(2) = 5 ഉം ആയ ഒരു ഒന്നാം കൃതി ബഹുപദം എഴുതുക.
Answer:
q(x) = ax + b
q(1) = a + b
എന്നാൽ q(1) = 1 ആയതിനാൽ
a + b = 1 …(i)
q(2) = ax² + b = 2a + b
എന്നാൽ q(2) = 5 ആയതിനാൽ
2a + b = 5…(ii)
(ii) – (i) ⇒ a = 4
a = 4 (i) ആരോപിച്ചാൽ
4+ b = 1
b = 1 – 4 = -3
ഇതിൽ നിന്നും, q(x) = 4x – 3

Question 5.
p(x) = 2x² + 3x + 5
a) p(1), p(0) ഇവ കാണുക.
b) p(0) = 2, p(1) = 5 ആയ ഒരു രണ്ടാംകൃതി ബഹുപദം എഴുതുക.
Answer:
a) p(1) = 2 × 1² + 3 × 1 + 5
= 2 + 3 + 5 = 10
p(0) = 2 × 0² + 3 × 0 + 5 = 5

b) p(x) = ax² + bx + c എന്നെടുത്താൽ
p(0) = a × 0² + b × 0 + c = c
എന്നാൽ p(0) = 2 ആയതിനാൽ
c = 2
p(1) = a × 1² + b × 1 + c = a + b + c

എന്നാൽ p(1) = 5 ആയതിനാൽ
a + b + c = 5
a + b = 3
ഇതിൽ നിന്നും, p(x) = 2x² + x + 2
(a + b = 3 ആകുന്ന ഏതു സംഖ്യകളും എടുക്കാം. a ≠ 0)