Oswaal CBSE Question Bank Class 9 Science, Chapterwise and Topicwise Solved Papers For 2025 Exams 9789359580777, 9359580775

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For 2024-25 Exam

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Chapterwise & Topicwise

SAMPLE QUESTION PAPERS

CLASS 9

SCIENCE

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20th EDITION

I SB N SYLLABUS COVERED

YEAR 2024-25 “9789359580777”

CENTRAL BOARD OF SECONDARY EDUCATION DELHI

PUBLISHED BY OSWAAL BOOKS & LEARNING PVT. LTD.

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DISC L AIM ER This book is published by Oswaal Books and Learning Pvt Ltd (“Publisher”) and is intended solely for educational use, to enable students to practice for examinations/tests and reference. The contents of this book primarily comprise a collection of questions that have been sourced from previous examination papers. Any practice questions and/or notes included by the Publisher are formulated by placing reliance on previous question papers and are in keeping with the format/pattern/guidelines applicable to such papers. The Publisher expressly disclaims any liability for the use of, or references to, any terms or terminology in the book, which may not be considered appropriate or may be considered offensive, in light of societal changes. Further, the contents of this book, including references to any persons, corporations, brands, political parties, incidents, historical events and/or terminology within the book, if any, are not intended to be offensive, and/or to hurt, insult or defame any person (whether living or dead), entity, gender, caste, religion, race, etc. and any interpretation to this effect is unintended and purely incidental. While we try to keep our publications as updated and accurate as possible, human error may creep in. We expressly disclaim liability for errors and/or omissions in the content, if any, and further disclaim any liability for any loss or damages in connection with the use of the book and reference to its contents”.

Kindle Edition

Contents l

Latest CBSE Syllabus

5 - 6

In each chapter, for better comprehension, questions have been categorized according to the typology issued by CBSE as follows : R - Remembering, U - Understanding,

A - Analyze,

AP - Application

C - Creating.

Unit I : Matter : Its Nature and Behaviour

1. 2. 3. 4.

Matter in Our Surroundings Nature of Matter Particle Nature and their Basic Units Structure of Atom

1 - 14 15 - 23 24 - 33 34 - 43 44 - 44 45 - 46

u Artificial Intelligence u Self Assessment Paper-1

Unit II : Organization in the Living World

5. Cell–Basic Unit of Life 6. Tissues

47 - 61 62 - 76 77 - 79 80 - 81

u Artificial Intelligence u Self Assessment Paper-2

Unit III : Motion, Force and Work









7. Motion

82 97 99 111 112 121 123 130 143 145 156 157

u Artificial Intelligence

8.

Force and Newton's Laws

u Artificial Intelligence

9. Gravitation

u Artificial Intelligence

10. Floatation 11. Work, Energy And Power u Artificial Intelligence

12. Sound

u Artificial Intelligence u Self Assessment Paper-3

Unit IV : Food Production

13. Improvement in Food Resources

- 96 - 98 - 110 - 111 - 120 - 122 - 129 - 142 - 144 - 155 - 156 - 159

160 - 174 175 - 176

u Self Assessment Paper-4

qq

LOOK OUT FOR 'HIGHLY LIKELY QUESTIONS'

These questions are selected by Oswaal Editorial Board. These are highly likely to be asked in the upcoming examinations.

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Preface Elevate Your Performance, Surpassing the Past Get ready for another epic journey through the academic wonders of the academic year 20242025 with your trustworthy companion—Oswaal Books! Remember last year’s triumphs? Well, buckle up because we are about to make this year even more awesome! As the legendary dancer Martha Graham once said, “Practice means to perform, repeatedly in the face of all obstacles, some act of vision, of faith, of desire.” We have taken this wisdom to heart and packed it into our brand-new Question Banks for 2024-2025. They are a magical mix of CBSE Board Updates, cool questions from the past, and specially crafted ones tailored to the Latest Typologies. Oh, and did we mention the fantastic Learning Resources that come with them?

What makes these Question Banks truly exceptional? • 100% Updated Syllabus & Question Typologies: We have got you covered with the latest and 100% updated curriculum along with the latest typologies of Questions. • Timed Revision with Topic-wise Revision Notes & Smart Mind Maps: Study smart, not hard! • Extensive Practice with 1000+ Questions & SAS Questions (Sri Aurobindo Society): To give you 1000+ chances to become a champ! • Concept Clarity with 500+ Concepts & Concept Videos: For you to learn the cool way— with videos and mind-blowing concepts • NEP 2020 Compliance with Competency-Based Questions & Artificial Intelligence: For you to be on the cutting edge of the coolest educational trends. If you are looking to conquer every study challenge, these Question Banks are your secret weapon. It is like having a superhero ally for your exams! So, let’s kick off this exciting journey, fill those learning gaps, and rock the year with ease and confidence. Big shoutout to our superhero team—the Oswaal Editorial Board! They’re the brains behind this incredible resource, working day and night just for you. And a massive thank you to you, our fellow Students, Parents & Teachers for your awesome inputs that make this book one-of-a-kind. Wishing you all the best, superheroes-in-the-making! Strive for greatness! Team Oswaal Books

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Syllabus SCIENCE CLASS IX (Code No. 086)



Marks : 80 Unit No.

Unit

Marks

I

Matter - Its Nature and Behaviour

25

II

Organisation in the Living World

22

III

Motion, Force and Work

27

IV

Food; Food Production

06

Total

80

Internal Assessment

20

Grand Total

100

Theme : Materials Unit I : Matter – Nature and Behaviour Definition of matter; solid, liquid and gas; characteristics - shape, volume, density; change of state- melting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation. Nature of matter : Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Physical and chemical changes (excluding separating the components of a mixture). Particle nature and their basic units : Atoms and molecules, Law of Chemical Combination, Chemical formula of common compounds, Atomic and molecular masses. Structure of atoms : Electrons, protons and neutrons, Valency, Atomic Number and Mass Number, Isotopes and Isobars.

Theme : The World of the Living Unit II : Organisation in the Living World Cell - Basic Unit of life : Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes - basic structure, number. Tissues, Organs, Organ System, Organism : Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).

Theme: Moving Things, Ideas Unit III : Motion, Force and Work

People

and

Motion : Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, elementary idea of uniform circular motion. Force and Newton's laws : Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration. Gravitation : Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall. Floatation : Thrust and pressure. Archimedes' Principle; Buoyancy. Work, energy and power : Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy (excluding commercial unit of Energy). Sound : Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Theme : Food Unit IV : Food Production Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.

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Syllabus Note for the Teachers: 1. The chapter Natural Resources (NCERT Chapter 14) will not be assessed in the year-end examination. However, learners may be assigned to read this chapter and encouraged to prepare a brief write up on any concept of this chapter in their Portfolio. This may be for Internal Assessment and credit may be given for Periodic Assessment/Portfolio. 2. The NCERT text books present information in boxes across the book. These help students to get conceptual clarity. However, the information in these boxes would not be assessed in the year-end examination. PRACTICALS Practicals should be conducted alongside the concepts taught in theory classes. (LIST OF EXPERIMENTS) 1. Preparation of: Unit-I (a) a true solution of common salt, sugar and alum (b) a suspension of soil, chalk powder and fine sand in water (c) a colloidal solution of starch in water and egg albumin/milk in water and distinguish between these on the basis of • transparency • filtration criterion • stability 2. Preparation of :  Unit-I (a) A mixture (b) A compound using iron filings and sulphur powder and distinguishing between these on the basis of: (i) appearance, i.e., homogeneity and heterogeneity (ii) behaviour towards a magnet (iii) behaviour towards carbon disulphide as a solvent (iv) effect of heat

3. Perform the following reactions and classify them as physical or chemical changes: Unit-I (a) Iron with copper sulphate solution in water (b) Burning of magnesium ribbon in air (c) Zinc with dilute sulphuric acid (d) Heating of copper sulphate crystals (e) Sodium sulphate with barium chloride in the form of their solutions in water 4. Preparation of stained temporary mounts of (a) onion peel, (b) human cheek cells & to record observations and draw their labeled diagrams. Unit-II 5. Identification of Parenchyma, Collenchyma and Sclerenchyma tissues in plants, striped, smooth and cardiac muscle fibers and nerve cells in animals, from prepared slides. Draw their labeled diagrams. Unit-II 6. Determination of the melting point of ice and the boiling point of water. Unit-I 7. Verification of the Laws of reflection of sound. Unit-III 8. Determination of the density of solid (denser than water) by using a spring balance and a measuring cylinder.  Unit-III 9. Establishing the relation between the loss in weight of a solid when fully immersed in Unit-III (a) Tap water (b) Strongly salty water, with the weight of water displaced by it by taking at least two different solids. 10. Determination of the speed of a pulse propagated through a stretched string/slinky (helical spring).  Unit-III 11. Verification of the law of conservation of mass in a chemical reaction. Unit-III

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UNIT-I

Matter: Its Nature and Behaviour

CHAPTER



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Syllabus

Study Time Reading Time: 3:00 Hr No. of Questions: 58

MATTER IN OUR SURROUNDINGS

Definition of matter; solid, liquid and gas; characteristics–shape, volume, density; change of state—melting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation.

Revision Notes Matter  Physical nature of Matter • Matter is made up of particles that vary in size, shape and nature. These small particles are called atoms. • The particles of matter are too small so they cannot be seen by naked eyes or simple microscope. • Characteristics of particles: Scan to know (i) Large number of particles constitute matter. more about this topic (ii) Particles of matter are very small in size.



(iii) Particles of matter have spaces between them. (iv) Particles of matter are continuously in motion. (v) Particles of matter attract each other. (vi) The force that exists between the particles is called interparticle force of attraction. (vii) The force of attraction between particles of solid, liquid and gas can be arranged in decreasing order as: Solid > Liquid > Gas. • Matter around us exists in three states: solid, liquid and gas.

Characteristics of Particles of Matter Scan to know more about this topic

 Solid state:

(i) All solids have definite shape, distinct boundaries and fixed volumes, that is, they have negligible compressibility. (ii) Solids have a tendency to maintain their shape when subjected to outside force.

States of Matter

(iii) Solids are rigid.

 Liquid state:

(i) Liquids have no fixed shape but have a fixed volume. They take up the shape of the container in which they are kept. (ii) Liquids flow and change shape, so they are not rigid and are called fluid.

Mnemonics Concept: Matter has three states – Solid, Liquid, Gas Mnemonics: Ma Sona Le Gayi Interpretations:

(iii) The rate of diffusion of liquid is higher than that M: Matter, S: Solid, L: Liquid, G: Gas of solids. This is due to the fact that in the liquid state, particles move freely and have greater space between each other as compared to particles in the solid state.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

MATTER IN OUR SURROUNDINGS

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 Gaseous state: (i) Gases are highly compressible as compared to solids and liquids. (ii) The liquefied petroleum gas (LPG) cylinder that we get in our home for cooking or the oxygen supplied to hospitals in cylinders is compressed gas. (iii) In the gaseous state, the particles move around randomly at high speed. Due to this random movement, the particles hit each other and also to the walls of the container. Pressure of gas is applied on the walls of the vessel by the irregular moving gas particles.  Interconversion of States of Matter The phenomenon of change of matter from one state to another and then back to the original state by altering the conditions like temperature and pressure is called the interconversion states of matter. Matter can change its state.  Water can exist in three states of matter:

(i) Solid, as ice,



(ii) Liquid, as water, and



(iii) Gas, as water vapour.

 The states of matter are inter-convertible. The state of matter can be changed by changing temperature or pressure.

Scan to know more about this topic

Interconversion of States of Matter

S.I. unit of temperature is Kelvin. T (K) = t (°C) + 273.  Melting Point: The temperature at which solid melts to form liquid at atmospheric pressure is called melting point. Melting point of ice is 273.16 K (0°C) The melting point of a solid is an indication of the strength of the force of attraction between its particles.  Boiling Point: The temperature at which a liquid starts boiling at the atmospheric pressure is known as boiling point. Boiling point of water is 373 K (100° C).  Latent heat of vaporisation is the heat energy required to change 1 kg of liquid to gas at atmospheric pressure at its boiling point. Boiling is a bulk phenomenon.  Latent heat of fusion is the amount of heat energy required to change 1 kg of solid into liquid at its melting point.  Freezing: The process in which liquid changes into solid is known as freezing. For example, freezing of water (liquid) into ice (solid).  Sublimation: Sublimation is the change of a solid directly into the gaseous state without passing through the liquid state upon heating and back to the solid state when the temperature is lowered. Evaporation • Evaporation takes place only at the surface of the liquid while boiling can take place in all parts of the liquid. • Evaporation is surface phenomenon. Particles from the surface gain enough energy to overcome the forces of attraction present in the liquid and change into the vapour state. Scan to know • Evaporation is a continuous or ongoing process. Evaporation causes cooling. more about this topic • The rate of evaporation is affected by the surface area exposed to atmosphere, temperature, humidity and wind speed. • Since evaporation is a surface phenomenon, therefore, it increases with an increase in surface area. • Evaporation increases with an increase in temperature. Evaporation • Evaporation decreases with an increase in humidity. • Evaporation increases with the increase in wind speed.  Plasma: The state consists of super energetic and super excited particles. These particles are in the form of ionised gases. The fluorescent tube and neon sign bulb consists of plasma stars. The sun and the stars glow because of the presence of plasma in them.

Key Words  Matter: Anything that has mass and occupies space is called matter.  Solid: Solid is defined as that form of matter which possesses rigidity, is incompressible and hence has a definite shape and a definite volume.  Liquid: Liquid is defined as that form of matter which possesses fluidity but is almost incompressible and hence has a definite volume but no definite shape.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

 Gas: Gas is defined as that form of matter which possesses fluidity but it is highly compressible and hence has neither definite shape nor definite volume.  Humidity: The amount of water vapour present in the air.  Density: It is the mass occupied by a solid per unit volume and is obtained by dividing the mass of a particular solid by the volume occupied.  Fusion: The process in which a solid changes to liquid state by absorbing heat at constant temperature.  Diffusion: The process in which particles of one substance occupy the vacant spaces present in the particles of the other substance, is called diffusion.  Condensation : The process in which a gas changes into liquid state by giving out heat at constant temperature.  Latent heat: The hidden heat which breaks the force of attraction between the molecules during change of state.  Latent heat of fusion: The amount of heat energy that is needed to convert one kg of a solid into the liquid state at atmospheric pressure at its melting point is termed as latent heat of fusion.  Boiling point: The temperature at which a liquid starts boiling at the atmospheric pressure is known as boiling point.  Freezing point: The temperature at which a liquid changes to solid by giving out heat at the atmospheric pressure.  Latent heat of vaporisation: The amount of heat energy that is needed to convert one kg of a liquid at its boiling point temperature into its vapour state without any rise in temperature, is termed as latent heat of vaporisation.  Melting point: The melting point of a solid may be defined as the temperature at which a solid melts to become a liquid at the atmospheric pressure.  Sublimation: Sublimation is the change of a solid directly into the gaseous state without passing through the liquid state upon heating and back to the solid state when the temperature is lowered.  Evaporation: The phenomenon of change of liquid to the vapour state at any temperature below the boiling point of the liquid is termed as evaporation.  Transpiration: The process of evaporation of water from the aerial parts of plants especially leaves is called transpiration.  Freezing: The process of conversion of liquid into solid is known as freezing.

Example

Write in brief, an activity to show the particulate nature of matter. List any two characteristics of particles of matter. Solution: Step I: Activity: (i) Take a 100 mL beaker. (ii) Fill half the beaker with water and mark the level of water. (iii) Dissolve some salt / sugar with the help of a glass rod. (iv) Observe any change in water level. Step II: Observation: Level of water remains same and salt/sugar, has now spread throughout water.

When we dissolve salt in water, the particles of salt get into the spaces between particles of water. This shows the particulate nature of matter.

Step III: Two characteristics of particles of matter are: (i) They are continuously moving. (ii) They attract each other.

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions 

(1 mark each)

Q. 1. The property to flow is unique to fluids. Which one of the following statements is correct?

(A) Only gases behave like fluids. (B) Gases and solids behave like fluids.

(C) Gases and liquids behave like fluids. (D) Only liquids are fluids. Ans. Option (C) is correct. Explanation: Gases and liquids tend to flow due to less force of attraction between their particles. Solids do not flow.

MATTER IN OUR SURROUNDINGS

Q. 2. A few substances are arranged in the increasing order of ‘forces of attraction’ between their particles. Which one of the following represents a correct arrangement? (A) Water, air, wind (B) Air, sugar, oil (C) Oxygen, water, sugar (D) Salt, juice, air Ans. Option (C) is correct. Explanation: It is because the force of attraction increases in the order: Gas < Liquid < Solid. Q.3. During summer, water kept in an earthen pot becomes cool because of the phenomenon of (A) Diffusion (B) Transpiration (C) Osmosis (D) Evaporation Ans. Option (D) is correct. Explanation: It is because of the phenomenon called evaporation. Earthen pot has a large number of tiny pores in its walls and some of the water molecules continuously keep seeping through these pores to outside the pot. This water evaporates continuously and takes the latent heat required for vaporisation from the remaining water. In this way, the remaining water loses heat and gets cooled. Q.4. Which one of the following sets of phenomena would increase on raising the temperature? (A) Diffusion, evaporation, compression of gases (B) Evaporation, compression of gases, solubility (C) Evaporation, diffusion, expansion of gases (D) Evaporation, solubility, diffusion, compression of gases Ans. Option (C) is correct. Explanation: Evaporation rate increases because on increasing temperature, kinetic energy of molecules increases, so the molecules present at the surface of the liquid leave the surface quickly and go into the vapour state. Diffusion and expansion of gases also increase as the molecules move more rapidly and try to occupy more space. Q.5. Seema visited a Natural Gas Compressing Unit and found that the gas can be liquefied under specific conditions of temperature and pressure. While sharing her experience with friends she got confused. Help her to identify the correct set of conditions. (A) Low temperature, low pressure (B) High temperature, low pressure (C) Low temperature, high pressure (D) High temperature, high pressure Ans. Option (C) is correct. Explanation: There is a lot of space between the particles of a gas. On applying high pressure, the particles of gas move so close that they start attracting each other sufficiently forming a

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liquid. When gas is compressed too much, heat is produced, so it is necessary to cool it. Cooling lowers the temperature of compressed gas and helps in liquifying it. Hence, a gas can be liquified by applying high pressure and lowering the temperature. Q. 6. Choose the correct statement of the following. (A) Conversion of solid into vapours without passing through the liquid state is called vaporisation. (B)  Conversion of solid into vapour without passing through the liquid state is called sublimation. (C) Conversion of vapours into solid without passing through the liquid state is called freezing. (D)  Conversion of solid into liquid is called sublimation. Ans. Option (B) is correct. Explanation: The conversion of liquid into gas (vapour) is called vaporisation. The conversion of liquid into solid is called freezing. The conversion of solid into liquid is called melting. Q. 7. Which condition out of the following will increase the evaporation of water? (A) Increase in temperature of water (B) Decrease in temperature of water (C) Less exposed surface area of water (D) Adding common salt to water Ans. Option (A) is correct. Explanation: On increasing the temperature, kinetic energy of water molecules increases and more particles get enough kinetic energy to go into the vapour state. This increases the rate of evaporation. On the other hand, decrease in temperature of water, less exposed surface area of water and addition of common salt to water decreases the rate of evaporation. Q. 8. On converting 25°C, 38°C and 66°C to Kelvin scale, the correct sequence of temperatures will be: (A) 298 K, 311 K and 339 K (B) 298 K, 300 K and 338 K (C) 273 K, 278 K and 543 K (D) 298 K, 310 K and 338 K Ans. Option (A) is correct. Explanation: On converting 25°C, 38°C and 66°C, to Kelvin scale, we get the following temperatures: 25°C + 273 = 298 K 38°C + 273 = 311 K 66°C + 273 = 339 K Q. 9. The boiling points of diethyl ether, acetone and n-butyl alcohol are 35°C, 56°C and 118°C, respectively. Which one of the following correctly represents their boiling points in Kelvin scale?

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX



(A) 306 K, 329 K, 391 K (B) 308 K, 329 K, 392 K (C) 308 K, 329 K, 391 K (D) 329 K, 392 K, 308 K

Ans. Option (C) is correct. Explanation: On applying the formula, T°C + 273 = K, Boiling point of diethyl ether = 35°C + 273 = 308 K Boiling point of acetone = 56°C + 273 = 329 K and Boiling point of n-butyl alcohol = 118°C + 273 = 391 K. Hence, the correct order of boiling points in Kelvin scale is 308 K, 329 K and 391 K. Q. 10. In which of the following conditions, the distance between the molecules of hydrogen gas would increase?

(i) Increasing pressure on hydrogen contained in a closed container.

(ii) Some hydrogen gas leaking out of the container. (iii) Increasing the volume of the container of hydrogen gas. (iv) Adding more hydrogen gas to the container without increasing the volume of the container.

(A) (i) and (ii) (B) (i) and (iv) (C) (ii) and (iii) (D) (ii) and (iv)

Ans. Option (C) is correct. Explanation: Some hydrogen gas leaking from the container leaves some vacant space inside the container. So, hydrogen gas molecules inside the container occupy all the space available and the distance between the molecules of hydrogen gas will be increased. Similarly, on increasing the volume of the container of hydrogen gas, more space will be available inside the container and hydrogen gas molecules will occupy all the space available. As a result, the distance between the molecules will be increased. So, option (ii) and (iii) will increase the distance between the molecules of hydrogen gas. On the other hand, on increasing pressure, hydrogen molecules will come closer and the distance between them will be decreased. Also, on adding more hydrogen gas molecules without increasing the volume of container will decrease the distance between molecules.

B Assertion and Reason (1 mark each) Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion: A particle of gas intermixes with each other. Reason: The intermixing of particles of two different types of matter on their own is called diffusion Ans. Option (B) is correct. Explanation: Particles of a gas are loosely packed. So, they move randomly due to space between them and intermix with other particles present there. Q. 2. Assertion: Heat energy when supplied to the solid, it starts melting. Reason: Solid particles take up the heat and helps in melting or fusion. Ans. Option (A) is correct. Explanation: When heat energy is supplied to the solid, the solid particle takes up the heat energy supplied to it and helps in melting. Q. 3. Assertion: A gas exerts pressure on the walls of the container. Reason: Rate of diffusion of gases is more than that of liquids. Ans. Option (B) is correct. Explanation: In the gaseous state, the particles move randomly at certain speed. Due to this movement, they hit each other and also on the walls of the container. This force exerts pressure on the walls. Q. 4. Assertion: Naphthalene ball disappears with time without leaving any solid residue. Reason: Naphthalene ball gets converted into vapours due to evaporation. [CBSE SQP 2020) Ans. Option (C) is correct. Explanation: Naphthalene ball disappears with time without leaving any solid residue because of sublimation. With time, these naphthalene balls sublime directly into vapour. It is a physical change and the process is known as sublimation.

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MATTER IN OUR SURROUNDINGS

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions Q. 1. Arrange the following substances in the increasing order of force of attraction between their particles: Oxygen, sugar, water.

Answering Tip

Ap [NCERT]

 Ans. Oxygen < water < sugar. 



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Q. 2. We can get the smell of perfume sitting several metres away. Give reason.[NCERT] Ans. Particles of perfume diffuse into the air and can be smelled even at a distance.  1 Q. 3. If the food is being cooked in the kitchen, name the process which brings smell. Ap [NCERT] Ans. Name of the process is diffusion.

1

Q. 4. A diver is able to cut through water in a swimming pool. Which property of matter does this observation show ? A [NCERT] Ans. Particles of water have space between them known as intermolecular space. 1

(1 mark each)

Read the question carefully and then write the

answer. The words used in the question should reciprocate in the answer so that it becomes easier for the examiner to correlate the question and answer.

Q. 5. Convert the following temperatures to the Celsius scale:

(i) 293 K

(ii) 470 K U (KVS 2019, NCERT)

Ans. (i) 293 K As 0°C = 273 K \ 293 K = (293 – 273)°C = 20°C (ii) 470 K As 0°C = 273 K \ 470 K = (470 – 273)°C = 197°C  [½ + ½]

Commonly Made Error

Q. 6. Convert the following temperatures to Kelvin scale:

Students write random answers.



(a) 25°C

(b) 373°C

Ans. (a) 25 + 273 = 298 K (b) 373 + 273 = 646 K

Short Answer Type Questions-I Q. 1. State four properties of liquids. Ans. Properties of liquid state are:

U

(i) Liquids have definite volume (ii) Liquids can flow. (iv) They have less intermolecular force of attraction as compared to solids but more intermolecular force of attraction as compared to gases. [½ ½ × 4] Q. 2. Give reasons:

U

(i) A gas completely fills the vessel in which it is kept.

(ii) A gas exerts pressure on the walls of the container. Ans. (i) In gas, the force of attraction between particles is negligible. So, they move freely in all the directions and acquire all the spaces in the vessel.

[½ + ½]

(2 marks each)

(ii) Due to freely moving particles and high kinetic energy, particles of gas move randomly in all the directions and hit the walls of the container exerting pressure. 

(iii) Liquids are compressible.

Ap

[1+1]

Q. 3. Why are solids incompressible? Ans. Solids are incompressible because the particles are closely packed and there is no space for their movement. [2] Q. 4. State one difference between gas and vapour.  U Ans. Gas: It is a stable state as compared to vapour, e.g., O2, H2. Vapour: It is an unstable state. On normal cooling, vapour changes into liquid state. [2] Q. 5. Convert the boiling point of water into Kelvin temperature. Ap Ans. Boiling point of water is 100 °C. 100 °C + 273 = 373 K [1 + 1]

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Q. 6. Illustrate three states of matter with a schematic A diagram. Ans. The three states of matter are solid, liquid and gas.

[2]

Short Answer Type Questions-II Q. 1. Prove that liquids have no fixed shape but have a fixed volume. Why the rate of diffusion of fluids is higher than that of solids? U+Ap [Board Term-I, 2016]



Ans. (i) Liquid take the shape of the container.

(ii) Particles are loosely packed, kinetic energy of particles is more than solids.



(iii) Rate of diffusion is directly proportional to the kinetic energy as kinetic energy is higher, rate of diffusion is also higher.



[CBSE Marking Scheme, 2016] [1 + 1 + 1]

Detailed Answer:

Experiment to prove that liquid has a fixed volume but no fixed shape:



Take a flask and a glass. Pour 50 mL of water in the flask. Then observe the shape of water in the flask. After then, pour the same water in a glass. The water will take the shape of the glass. However, you will find that 50 mL of water is 50 mL even after pouring the water into different beakers.



Rate of diffusion of liquids are higher than the solids because particles of liquid have more space between them compared to solid and so move freely than in solids. [2 + 1]

Q. 2. What are the three different states of matter? Which one of these has a definite shape, distinct boundaries and fixed volume? Compare the three on the basis of compressibility. U+Ap [Board Term-I, 2016]



Ans. Three states of matter are solid, liquid and gas. 

Solid has definite shape, distinct boundary and fixed volume. [1 + 1 + 1] [CBSE Marking Scheme, 2016]

Q. 3. What is the effect of change of pressure on physical state of matter ? Explain with an example of a gas. 

U [Board Term-I, 2015]

(3 marks each)

Ans. The physical state of matter can be changed by changing the pressure. By lowering temperature and increasing the pressure, gases can be changed into liquids and some solids can be changed into gases on decreasing the pressure. This happens with gases as there is lots of space between the particles of a gas and upon applying high pressure, particles come close to each other which upon cooling gets liquefied. [1 + 1 + 1]  [CBSE Marking Scheme, 2015] Q. 4. (i) After winters, people pack off their woollens by keeping naphthalene balls in them. With passage of time these balls become smaller in size. Why does this happen ? What type of change is involved during this process ? [NCERT] (ii) How can you convert a saturated solution into an unsaturated solution ?  Ap+U [Board Term-I, 2015] Ans. (i) Sublimation, naphthalene balls sublimate and become smaller in size. It is a physical change. (ii) By adding large quantities of the solvent into the solution.

[CBSE Marking Scheme, 2015] [1½ + 1½]

Detailed Answer: (i) With time, naphthalene balls sublime directly into vapour. It is a physical change and the process is known as sublimation. (ii) By adding large quantities of the solvent into the solution. Q. 5. List three characteristics of particulate nature of matter. U [NCERT] Ans. (i) Particles of matter have space between them. (ii) Particles of matter are continuously moving. (iii) Particles of matter attract each other.  [1 × 3] Q. 6. Give reasons: (i) A gas completely fills the vessel in which it is kept.

MATTER IN OUR SURROUNDINGS

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(ii) A gas exerts pressure on the walls of the container (iii) A wooden table should be called a solid.  R Ans. (i) In gas, the force of attraction between particles is negligible. So, they move freely in all the directions and acquire all the spaces in the vessel. (ii) Due to freely moving particles and high kinetic energy, particles of gas move randomly in all the directions and hit the walls of the container exerting pressure. (iii) Because it has a definite shape and fixed volume. [1 + 1 + 1] Q. 7. Design an activity to show that gases are highly compressible as compared to solids and liquids.  U [Board Term-I, 2012] Ans. Take three 100 mL syringes and remove their pistons. Also, close the nozzles of the syringes with rubber corks. Then, fill one syringe with chalk powder and the other with water. Now, insert the pistons back into the syringes and push them. You will observe that the pistons of the syringes (containing chalk pieces and water) require a large amount of force, while the piston of the third syringe is comparatively easier to push. Hence, it shows that gases are highly compressible. Piston



Fig: Set up to explain that particles of matter are very small. With every dilution, though the colour becomes light, it is still visible.



(iv) Keep on diluting the solution like this 5 to 8 times.

(v) This experiment shows that just a few crystals of potassium permanganate can change colour of a large volume of water (about 1000 mL). So, we conclude that there must be millions of tiny particles in just one crystal of potassium permanganate, which kept on dividing themselves into smaller and smaller particles. [3]  [CBSE Marking Scheme, 2012] Q. 9. The following triangle exhibits inter conversion of the three states of matter. Complete the triangle by labelling the arrows marked A, B, C and D.

Piston Piston

Chalk powder

Water

Cork

Cork

 Air Cork

[3] Q. 8. With the help of labelled diagram describe an activity to show that the particles of matter are very small. Use the following material that has been provided to you. 4 beakers, spatula, 4 test-tubes, distilled water and a few crystals of potassium permanganate.

U [Board Term-I, 2012]

U [KVS 2020, Board Term-I, 2016]

Ans. A → Melting, C → Condensation B → Evaporation, D → Sublimation[3]





U+Ap [Board Term-I, 2016]

Ans. Ice melts at 0°C. The temperature remains same when it changes

Ans. (i) Take 2-3 crystals of potassium permanganate and dissolve them in 100 mL of water. (ii) Take out approximately 10 mL of this solution and put it into 90 mL of clear water. (iii) Take out 10 mL of this solution and put it into another 90 mL of clear water.

[CBSE Marking Scheme, 2016]

Q. 10. At what temperature does the ice melt ? Why does the temperature remain same when it changes to liquid state ?





to liquid state because the heat supplied is continuously used up in changing the state from solid to liquid by overcoming the forces of attraction between the particles. This process is known as latent heat of fusion.  [1 + 2] [CBSE Marking Scheme, 2016]

Q. 11. While boiling the water, a student observed that temperature remains constant at 100° C till whole of water vaporises. Explain. 

Ap [Board Term-I, 2016]

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Ans. This is because the heat supplied is absorbed by the water particles and this heat increases their kinetic energy. Thus, because of an increase in kinetic energy, the bond between the water particles are cut down and they move more freely, compared to water.

Ans. There are pores in an earthen pot through which the liquid inside the pot evaporates. This evaporation makes the water inside the pot cool. In this way, water kept in an earthen pot becomes cool during summers.[3]  [CBSE Marking Scheme, 2015]



Q. 15. Distinguish among three states of matter with respect to property indicated: Density, Diffusion and Particle Motion.  U [Board Term-I, 2015]

Hence, they become gas. Thus, the temperature remains constant even though heat is supplied continuously to the water.[3]  [CBSE Marking Scheme, 2016] Q. 12. Give three reasons to justify that water is a liquid at room temperature. [NCERT]  U [Board Term-I, 2016]

Ans.

Ans. Fluidity, takes shape of the container, fixed volume.  [CBSE Marking Scheme, 2016] [3]

(ii) It takes the shape of the container in which it is kept. (iii) It can flow. Q. 13. Is the inter-conversion of three states of matter possible ? Illustrate with a schematic diagram.  Ans. Yes

R+U [Board Term-I 2016] [NCERT]

Liquid

Intermediate Lowest

(i)

Density

Highest

(ii)

Diffusion

Negligible Slower

(iii) Particle No Motion / movement

Detailed Answer: At room temperature, water is liquid because it has the following characteristics of liquid: (i) At room temperature, water has no shape but has fixed volume.

Solid





Yes, but confined

Gas Rapid Yes, free motion

[CBSE Marking Scheme, 2015] [1 +1 +1]

Q. 16. Why does the temperature remain constant during the change of state of matter ? Explain it on the basis of change of solid state into liquid state. 

U [Board Term-I, 2015]

Ans. Latent heat. Heat is used up in changing the state by overcoming the forces of attraction between the particles.  [CBSE Marking Scheme, 2015] [3] Detailed Answer: The temperature remains constant as the heat gets used up in changing the state by overcoming the forces of attraction between the particles. For example, a solid melts on heating. Its temperature does not rise until the entire solid is converted into liquid. This heat energy gets hidden into the content and is known as the latent heat.

[1 + 2]

Q. 17. Explain with the examples from your daily life where cooling is caused by evaporation. 

Commonly Made Error

Many students fail to draw the correct diagram.

Ans. When a liquid evaporates, it draws the latent heat of vaporisation from anything which it touches. Thus causes cooling. For examples: Sweating and Water in earthen pot. [CBSE Marking Scheme, 2014] [2 + 1] Q. 18. (i) Define latent heat of vaporisation. (ii) Give reasons for the following:

Answering Tip

Ap [Board Term-I, 2016, 2014]

Practice self-explanatory diagrams of interconversion of three states of matter with proper labelling, arrows and headings.

Q. 14. People of village use earthen pots to get cool water in summers. Explain the reason that why water remains cool in earthen pots. [NCERT] Ap [Board Term-I, 2015]

(a) You feel cold when you pour some nail polish remover on your palm. [NCERT] (b) During summer, sitting under a fan makes us comfortable. U+Ap Ans. (i) The amount of heat energy required to change 1 kg of a liquid to gas at atmospheric pressure at its boiling point is called latent heat of vaporisation.

MATTER IN OUR SURROUNDINGS

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(ii) (a) Particles gain heat energy from the palm and evaporate causing the palm to feel cool.

Ans.

(b) When we sit under a fan during summer, rate of evaporation of sweat increases due to increase in wind speed. Sweat takes heat from body to evaporate leaving us cool.  [1 + 1 + 1]

(c) Gases exerts pressure on the walls of the gas container due to the collision of molecules.

Q. 19.

(i) List any four characteristic properties of gases.

(ii) Steam produces more severe burns than boiling water. Why ? U+Ap [NCERT]

(i) (a) Gases neither have a definite shape nor a definite volume.

(b) Gases are compressible.

(d) Gases flow easily.



(ii) When water is converted into steam at 373 K, it absorbs energy equal to the latent heat of vaporisation. Thus, steam at 373 K has more heat energy than water at 373 K and hence steam produces more severe burns than boiling water. [½ × 4 + 1]

Long Answer Type Questions

(5 marks each)

Q. 1. Compare in tabular form, the properties of solids, liquids and gases with respect to: (i) shape, (ii) volume, (iii) compressibility, (iv) diffusion, (v) fluidity or rigidity. U [NCERT] Ans.

S. No.

Property

Solids

Shape

Definite shape

No definite shape, take the shape of the container

No definite shape, takes the shape of the container

(ii)

Volume

Definite volume

Definite volume

No definite volume

(iii)

Compressibility

Non-compressible

Slightly compressible

Highly compressible

(iv)

Diffusion

Can diffuse into liquids Diffusion is higher than solids.

(v)

Fluidity or rigidity

Rigid

Students sometimes get confused between different properties of states of matter.

Answering Tip

Learn the differences in tabular form for easy retention and understanding. Students should understand the nature of different states of matter and hence remember all the properties.

Q. 2.

[NCERT] (iii) What happens to the heat energy which is supplied to the solid once it starts melting?  Ap (i) (a) Gases and liquids do not have fixed shape.

Highly diffusible.

Less rigid and shows fluidity.

No rigidity and shows fluidity. [1 × 5] (b) Gases and liquids flow easily. (ii) The shape does not change when pressed i.e. it is hard and rigid. It has a definite shape and has high density. (iii) The heat energy supplied is taken up by solid particles and helps in melting or fusion. [2 + 2 + 1]

Commonly Made Error

Students sometimes write the same answer in different words.

Answering Tip

(i) List any two properties that liquids have in common with gases.

(ii) Give two reasons to justify that an iron almirah is a solid at room temperature.

Ans.

Gases

(i)

Commonly Made Error

Liquids



Avoid writing answers which are simply

repetition of the question. Do not overlook any part of a question and avoid being in a hurry to conclude the answer.

Q. 3. (a) Distinguish between solids and gases on the basis of following parameters: (i) Interparticle distance (ii) Interparticle forces of attraction,

12

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

(iii) Compressibility.[DDE 2014] (b) Give two factors that determine the rate of diffusion of a liquid in another liquid. 

U [Board Term-I, 2013]

Ans. (a) Differences between Solids and Gases: S. No. (i)

(ii)

Solids

Gases

Intermolecular space is small so the distance is less.

Intermolecular space is maximum so the distance is more.

Intermolecular force of attraction is maximum.

Intermolecular force of attraction is minimum.

(iii) Solids are rigid and incompressible.



Gases are non-rigid and are compressible.

(b) The two factors that determine the rate of diffusion of a liquid in another liquid are:



(i) Temperature



(ii) Pressure

[3 + 2] [CBSE Marking Scheme, 2013]

Q. 4. How the water changes into vapours at temperature below its boiling point ? List the factors affecting evaporation. Mention two examples from daily life where evaporation causes cooling.  [DDE 2017]  Ap [Board Term-I, 2016] Ans. Fractions of particles at the surface having higher kinetic energy, are able to break away from the forces of attraction of other particles and gets converted into vapour. Factors which affects rate of evaporation: (i) Surface area (ii) Temperature (iii) Humidity (iv)  Wind speed e.g., Sprinkling of water on the roof, water in earthen pots, etc. (or any other) [CBSE Marking Scheme, 2016] [1 + 2 + 2] Q. 5. When a solid melts the temperature of the system does not change after the melting point is reached even when we continue to supply heat. Give reason. Define latent heat of vaporisation. Which will cause more severe burns-boiling water or steam and why ? U [Board Term, I 2016] Ans. The heat gets used up in changing the state by overcoming the forces of attraction between the particles. Latent heat of vaporisation: The amount of heat energy that is required to change 1 L of a liquid into gas at atmospheric pressure at its boiling point. Steam, particles of steam absorb extra energy in the form of latent heat of vaporisation. [CBSE Marking Scheme, 2016] [5]

Q. 6. Explain with the examples from your daily life where cooling is caused by evaporation.  Ap [Board Term, I 2012, Board Term, I 2016]  [Board Term, I 2014] Ans.

(i) Rain water takes heat from road to get evaporated leaving road dry.

(ii) Cotton being good absorber of water, helps in absorbing the sweat and exposes it to the atmosphere for easy evaporation. (iii) In summer, trees absorb more water and minerals from the soil as the rate of transpiration increases. (iv) Acetone evaporates by taking heat from palm, leaving it cool. (v) Water evaporates from the roof taking heat from surroundings on hot summer day, leaving roof cool. [CBSE Marking Scheme, 2016] [1 × 5] Q. 7. (a) Explain the term diffusion. Illustrate with an activity that rate of diffusion increases with temperature. (b) Name two compressed gases which: (i) are used in our homes for cooking. (ii) are supplied to hospital in cylinders. U [Board Term, I 2015]

Ans. (a) When particles of one substance occupy the

vacant space present in the particles of the other substance, this is called diffusion. Activity: (i) Take 5 g of copper sulphate each in three beakers. (ii) Pour 100 mL of distilled water slowly in one of the beakers. (iii) Cover this beaker with a watch glass. (iv) Pour 100 mL of cold water in a second beaker slowly. (v) Place a third beaker containing 100 mL of water on a tripod stand for heating. (vi) Observe the diffusion process which begins in all the beakers. (vii) Record the time taken for the dissolution of copper sulphate in all the three cases. Conclusion: The rate of diffusion of copper sulphate in water is in the order: Beaker 3 > Beaker 2 > Beaker 1.



(b) (i) Liquified Petroleum Gas (LPG)



(ii) Oxygen.[1 + 2 + 1 + 1]

 [CBSE Marking Scheme, 2015] Q. 8. (a) Define melting point. Describe an activity with labelled diagram to find melting point of ice. (b) Explain why temperature remains same during melting of ice.  U+Ap [Board Term-I, 2015]

MATTER IN OUR SURROUNDINGS

Ans. (a) Melting point of a solid is defined as the temperature at which a solid melts to become a liquid at the atmospheric pressure. Activity: (i) Take about 150 g of ice in a beaker and suspend a laboratory thermometer so that its bulb is in contact with the ice. (ii) Start heating the beaker on a low flame. (iii) Note the temperature when the ice starts melting. (iv) Note the temperature when all the ice has converted into water. (v) Record your observations for this conversion of solid to liquid state.

13

Q. 9. Answer the following questions: (i) Out of boiling and evaporation which is a surface phenomenon ? Explain. In the absence of a refrigerator, butter is kept wrapped in a wet cloth during summer. Why ? (ii) Why do ice-cream appears colder than water at the same temperature? Ap [Board Term-I, 2015]; [NCERT]

Ans. (i) Evaporation is a surface phenomenon. Particles from the surface gain enough energy to overcome the forces of attraction present in the liquid and change into vapour state.

(vi) Now, put a glass rod in the beaker and heat while stirring till the water starts boiling.

Due to wet cloth, the temperature is comparatively lower than room temperature.

(vii) Keep a careful eye on the thermometer reading till most of the water has vaporised.



So, butter does not melt when remain wrapped in wet clothes.



(ii) Ice cream at 273 K, will take latent heat from the medium to convert itself into liquid at 273 K and then into liquid at higher temperature but in water such condition is not possible.



[CBSE Marking Scheme, 2015] 5

(viii) Record your observations. (b) The temperature remains constant during the melting of ice even though the heat is supplied regularly to increase the temperature for changing the state of matter. The heat of is consumed by the particles of ice and they vibrate faster breaking the forces of attraction and becomes liquid. Thermometer Iron stand

Glass stirrer

Q. 10. (a) Mention the physical state of water at: (i) 100°C, (ii)  0°C, (iii) 25°C [NCERT] (b) Convert the following temperatures into Celsius scale: (i) 298 K (ii) 300 K (iii) 280 K[NCERT]  Ans. (a) (i) Gas

U [Board Term-I, 2014]

(ii) Solid

Beaker

(b) (i) 298 – 273 = 25°C

Ice

(ii) 300 – 273 = 27°C (iii) 280 – 273 = 7°C Burner

(a)

(iii) Liquid

[1 + 1 + 1 + 1 + 1]

Commonly Made Error

Calculation error is commonly seen.

Answering Tips

The correct way to write is K and not °K. Remember the conversion formula

temperature, Kelvin to Celsius: T (K) = T (°C) + 273.

for

Q. 11. Account for the following: (i) When sugar crystals dissolve in water, the level of water does not rise appreciably.



(b) [3 + 1 + 1] [CBSE Marking Scheme, 2015]

(ii) Doctors advise to put strips of wet cloth on the forehead of a person having high fever. (iii) Naphthalene balls disappear with time without leaving any solid residue. [NCERT]

14

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

(iv) A wooden table should be called a solid. (v) Dogs generally hang out their tongue in summer.  Ap Ans. (i) Particles of sugar crystals occupy the space between the particles of water. (ii) The excess heat from the body is taken by high latent heat of vaporisation of water. As a result, temperature of body decreases.

(iii) By Sublimation process, naphthalene gets converted into vapours. (iv) Because a wooden table has a fixed shape as well as fixed volume and it is incompressible also. (v) Evaporation of saliva causes cooling. 

COMPETENCY AND CRITICAL THINKING BASED QUESTIONS

I. Read the passage and answer the following questions. Rahul added a crystal of CuSO4 in a glass of water and allowed it to settle at the bottom. After some time he observed that the blue colour appears just above the solid crystal and with the passage of time whole water in glass turns blue. (1) Name the phenomenon due to which this happens. (2) Identify the characteristic of particles of matter associated with this observation. (3) The state of matter in which particles just move around randomly because of very weak force of attraction is liquid. (True or False) (4) The arrangement of particles is less ordered in the ____________ state. However there is no order in the ___________ state. Ans. (1) Diffusion (2) Particles of matter are continuously moving. (3) False. The state of matter in which particles just move around randomly because of very weak force of attraction is gas. (4) Liquid, gaseous. II. Read the passage and answer any four following questions. In a laboratory, while doing an experiment, carbon dioxide was taken in an enclosed

[1 × 5]

(1 mark each)

cylinder and compressed by applying pressure and keeping low temperature. 1. In which state of matter, carbon dioxide occurs most commonly around us? (A) Solid (B) Liquid (C) Gas (D) Both (A) and (B) Ans. Option (C) is correct. 2. Which state of matter we will obtain after completion of the above given process? (A) Solid (B) Liquid (C) Gas (D) Both (A) and (B) Ans. Option (A) is correct. 3. Which process is responsible for conversion of gas to solid directly ? (A) Condensation (B) Sublimation (C) Diffusion (D) Melting Ans. Option (B) is correct. 4. The common name of the product obtained in the above process is __________. (A) Hot ice (B) Dry ice (C) Ice (D) Cold ice Ans. Option (B) is correct. 5. What is sublimation? (A) The process of conversion of liquid to gas. (B) The process of conversion of solid directly into vapours. (C) The process of conversion of solid to liquid. (D) The process of conversion of liquid to solid. Ans. Option (B) is correct. 

Study Time Reading Time: 2:30 Hr No. of Questions: 39

CHAPTER

2



Syllabus

NATURE OF MATTER

Elements, compounds and mixtures; Heterogeneous and homogeneous mixtures; colloids and suspensions. Physical and chemical changes (excluding separating the components of a mixture).

Revision Notes





 Matter can be classified as pure substances or mixtures.  A pure substance may either contain constituent particles of only one kind or of different kinds. A pure substance has a fixed composition. Element Scan to know  An element is a basic form of matter which cannot be broken down into simpler substances more about by any physical or chemical means. this topic  Elements can be broadly classified as metals, non-metals and metalloids.  Metals are one category of elements that have lustre. They conduct heat and electricity. They are sonorous. They are malleable and ductile.  Non-metals do not have lustre, are not sonorous and are bad conductors of heat and Elements and electricity. compounds  Metalloids are elements having properties intermediate between those of metals and nonmetals. Compound  A compound is a pure substance composed of two or more elements chemically combined in a fixed proportion. It can be broken down into simpler substances by chemical or electrochemical methods.  Properties of compounds are different from those of its constituent elements, whereas a mixture shows the properties of its constituent elements or compounds. Mixtures Scan to know more about  A mixture contains two or more elements or compounds which are mixed together in any this topic proportion. From a mixture, no new compound is formed. A mixture shows the properties of the constituent substances.  Mixtures are classified as homogeneous and heterogeneous mixtures.  Mixtures whose components mix completely with each other to make a uniform composition are called homogeneous mixtures. For example: Alloy. Mixtures  A heterogeneous mixture has a non-uniform composition.  Alloys are mixture of two or more metals or a metal and a non-metal and cannot be separated by physical methods.  The ability of a substance to dissolve in another substance is called solubility. Scan to know Solution more about  Homogeneous mixture of two or more substances is called a solution. this topic  Component of a solution present in small quantity is called a solute. Solute particles cannot be separated from the mixture by filtration.  Component of a solution present in large quantity is called a solvent.  Particles of a solution are smaller than 1 nm in diameter. They cannot be seen by naked eyes. Solutions

16

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

NATURE OF MATTER







17

 Particles of solution do not scatter beam of light.  Solution with high solute concentration is called concentrated solution and those with low concentration is called dilute solution.  The concentration of a solution is the amount of solute present in a given amount (mass or volume) of solvent Amount of solute or solution. Concentration of a solution = Amount of solution  Percentage by mass is one method of expressing concentration of solution.  There are two kinds of heterogeneous mixtures: colloids and suspensions. Colloids  Colloids are mixtures with particle sizes from 1 nm to 100 nm.  The component of colloid present in small amount is called dispersed phase.  The medium in which colloidal particles disperse or suspend themselves is called dispersion medium.  In a colloidal system, particles are always suspended and do not settle down. This constant colliding of the particles in continuous motion is called Brownian movement.  Scattering of a beam of light when light is passed through a colloidal solution is called the Tyndall effect.  Colloids are classified according to the state (solid, liquid or gas) of the dispersed medium or dispersing medium and the dispersed phase.  Colloid in which dispersed medium is a liquid and dispersed phase is solid is called as sol.  Colloid in which both dispersed phase and dispersed medium are in liquid state is called as an emulsion.  Colloid in which dispersed phase is either liquid or a solid and dispersed medium is a gas is called as aerosol. Suspension  A suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of medium. Particles of suspension are visible to naked eye. Suspensions are heterogeneous mixtures with particles that have a size greater than 1000 nanometers.  The change in which the shape, size, appearance or state of a substance may alter but its chemical composition remains the same is called a physical change. In a physical change, no new substance is formed.  Any change that involves the formation of a new substance and leads to a transformation of chemical identity is called chemical change.  Chemical changes are usually accompanied with heat exchange. Chemical changes are permanent changes which are usually irreversible.

Key Words  Matter: Anything that has mass and occupies space is called matter.  Pure substances: It consists of particles of only one kind of matter which are similar to one another and which cannot be separated into other kinds of matter by any physical process.  Element: It is defined as a basic form of matter which cannot be broken down into simpler substances by any chemical method.  Metals: They possess lustre. They are malleable and ductile, good conductors of heat and electricity and are sonorous.  Non-metals: They are neither malleable nor ductile. They are not lustrous and non-conductors of heat and electricity.  Metalloids or semi-metals: They have intermediate properties between those of metals and non-metals.  Compound: It is defined as a pure substance made up of two or more elements chemically combined in a fixed proportion by mass.  Mixtures: A mixture contains two or more substances (elements or compounds) which are physically mixed in any proportion but not chemically combined.  Solution: It is a homogeneous mixture of two or more substances. The major component of the solution is called the solvent and the minor component is called the solute.  Alloys: They are homogeneous mixtures. They may also be regarded as solid in solid solution.  Concentration of a solution: It is the amount of solute present per unit volume or per unit mass of the solution/ solvent.  Concentration of Solution: Mass of solute 1. Mass by mass percentage = × 100 Mass of solution

18

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

2. Mass by volume percentage =

Mass of solute × 100 Volume of solution

 Saturated solution: It is a solution which contains the maximum amount of the solute dissolved in a given quantity of the solvent at the given temperature and which cannot dissolve any more solute at that temperature.  Unsaturated solution: It is a solution which can dissolve more amount of solute in it at the given temperature.  Supersaturated solution: It is a solution which temporarily contains more solute than the saturation level.  Suspension: It is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium.  Colloids: They are heterogenous mixtures in which the size of the particles lies in between those of true solutions and suspensions.



Example

A solution contains 60 g of common salt in 240 g of water. Calculate the concentration in terms of mass by mass percentage of solution. Solution: Step I: Concentration of solution

 Mass of solute     100  Mass of solution 

Mass of common salt is 60 g. Mass of water is 240 g. Step II: Mass of solution = (60 + 240)g= 300 g. Step III: Concentration of solution  60    [2]   100  20%   300 

OBJECTIVE TYPE QUESTIONS Choice Questions A Multiple  (1 mark each) Q. 1. Which of the following statements are true for pure substances? (i) Pure substances contain only one kind of particles. (ii) Pure substances may be compounds or mixtures. (iii) Pure substances have the same composition throughout. (iv) Pure substances can be exemplified by all elements other than nickel. (A) (i) and (ii) (B) (i) and (iii) (C) (iii) and (iv) (D) (ii) and (iii) Ans. Option (B) is correct. Explanation: A pure substance is one which is made up of only one kind of atoms or molecules. They have the same composition throughout. Q. 2. A mixture of sulphur and carbon disulphide is (A) Heterogeneous and shows Tyndall effect (B) Homogeneous and shows Tyndall effect (C) Heterogeneous and does not show Tyndall effect (D) Homogeneous and does not show Tyndall effect Ans. Option (A) is correct. Explanation: A mixture of sulphur and carbon disulphide is a heterogeneous colloid and shows Tyndall effect. In a colloidal solution, the particles

are big enough to scatter light. This phenomenon of scattering of light by colloidal particles is known as Tyndall effect. Q. 3. Tincture of iodine has antiseptic properties. This solution is made by dissolving (A) iodine in potassium iodide (B) iodine in vaseline (C) iodine in water (D) iodine in alcohol Ans. Option (D) is correct. Explanation: Tincture of iodine is made by dissolving iodine in alcohol. Q. 4. Which of the following are homogeneous in nature? (i) Ice (ii) Wood (iii) Soil (iv) Air (A) (i) and (iii) (B) (ii) and (iv) (C) (i) and (iv) (D) (iii) and (iv) Ans. Option (C) is correct. Explanation: Ice and air are homogeneous in nature as their particles are not distinctly visible. Wood and soil are heterogeneous in nature. Q. 5. Two substances, A and B were made to react to form a third substance, A2B according to the following reaction 2A + B → A2B. Which of the following statements concerning this reaction are incorrect? (i) The product A2B shows the properties of substances A and B.

NATURE OF MATTER

19

(ii) The product will always have a fixed composition.

Q. 9. Which of the following are chemical changes? (i) Decaying of wood

(iii) The product so formed cannot be classified as a compound.

(ii) Burning of wood

(iv) The product so formed is an element. (A) (i), (ii) and (iii) (B) (ii), (iii) and (iv) (C) (i), (iii) and (iv) (D) (i), (ii) and (iv) Ans. Option (C) is correct. Explanation: A2B is a compound made up of two elements A and B in a fixed ratio. The properties of a compound are entirely different from those of its constituent elements. The composition of a compound is always fixed. Q. 6. Two chemical species X and Y combine together to form a product P which contains both X and Y, X + Y → P, X and Y cannot be broken down into simpler substances by simple chemical reactions. Which of the following concerning the species X, Y and P are correct? (i) P is a compound (ii) X and Y are compounds (iii) X and Y are elements (iv) P has a fixed composition (A) (i), (ii) and (iii) (B) (i), (ii) and (iv) (C) (ii), (iii) and (iv) (D) (i), (iii) and (iv) Ans. Option (D) is correct. Explanation: In this reaction, X and Y cannot be broken down into simpler substances by chemical reactions; therefore, X and Y are elements. A compound is a substance made up of two or more elements chemically combined in a fixed proportion by mass; therefore, P is a compound, having a fixed composition. Q. 7. Rusting of an article made up of iron is called (A) Corrosion and it is a physical as well as chemical change (B) Dissolution and it is a physical change (C) Corrosion and it is a chemical change (D) Dissolution and it is a chemical change Ans. Option (C) is correct. Explanation: Rusting of an article made up of iron is called corrosion. Corrosion is a chemical change because rust is a chemical compound (Hydrated iron oxide), which is totally different from element iron (Fe). Q. 8. Which of the following are physical changes? (i) Melting of iron metal (ii) Rusting of iron (iii) Bending of an iron rod (iv) Drawing a wire of iron metal (A) (i), (ii) and (iii) (B) (i), (ii) and (iv) (C) (i), (iii) and (iv) (D) (ii), (iii) and (iv) Ans. Option (C) is correct. Explanation: Rusting of iron is a chemical change. Rest three show physical changes, because in these three processes, iron changes its form, not the chemical composition.

(iii) Sawing of wood (iv) Hammering of a nail into a piece of wood (A) (i) and (ii) (B) (ii) and (iii) (C) (iii) and (iv) (D) (i) and (iv) Ans. Option (A) is correct. Explanation: Decaying of wood and burning of wood are chemical changes, because in these processes, the chemical composition of wood is changed and new substances are formed, which cannot be converted back into their original form. Sawing of wood and hammering of a nail into a piece of wood are physical changes.

B

Assertion and Reason 

(1 mark each)

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

(B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).



(C) Assertion (A) is true but reason (R) is false.

(D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion: Elements and compounds are pure substances. Reason: Properties of compounds are different from those of its constituent elements. Ans. Option (B) is correct. Explanation: A pure substance may either contain constituent particles of only one kind or of different kinds. A pure substance has a fixed composition. Thus, elements and compounds are example of pure substances. Properties of compounds are different from those of its constituent elements. Q. 2. Assertion: Alloys are homogenous mixture of metals. Reason: Alloys cannot be separated into their components by physical methods. Ans. Option (B) is correct. Explanation: An alloy is homogenous mixture because it shows the properties of its constituents. Q. 3. Assertion: Air is a mixture. Reason: Air can be separated into its constituents like oxygen, nitrogen, etc., by the physical process of fractional distillation. Ans. Option (A) is correct. Explanation: Air is a mixture, because it can be separated into its constituents like oxygen, nitrogen, etc., by the physical process of fractional

20

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

distillation. Also, air shows the properties of all the gases present in it. Q. 4. Assertion: A solution can scatter a beam of light passing through it. Reason: The particles of solution are smaller than 1 nm in diameter.

Ans. Option (D) is correct. Explanation: A solution does not scatter a beam of light passing through it as the size of the particles of solution is smaller.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions Q. 1. What is a substance ? U Ans. A substance is a kind of matter that cannot be separated into other kinds of matter by any physical process. [1]

Q. 2. What is a pure substance ?

U (NCERT) Ans. A substance which is made up of only a single type of particles is called a pure substance. [1]

Q. 3. What are mixtures ?

U

Q. 4. Define a solution.

U

Ans. Mixtures are constituted by more than one kind of pure form of matter. [1] Ans. A homogeneous mixture of two or more substances is called a solution. [1]

Q. 7. Give an example of a solid solution. K Ans. Brass (an alloy having 70% copper and 30% zinc). [1] Q. 8. Give an example of a gas in liquid solution. K Ans. Aerated drinks like soda water. [1] Q. 9. Define concentration of a solution. K Ans. It indicates the exact amount of solute dissolved in an exact amount of solvent or solution. [1] Q. 10. What is solubility ? U Ans. Solubility of a substance (solute) is the mass of the substance in grams that dissolves in 100 g of a solvent to form a saturated solution. [1]

Q. 5. Which of the following is/are pure substance(s)?

(NCERT) Ink, paper, water, milk, butter, ghee, sugar, blood, ice, iron, hydrochloric acid, calcium oxide, mercury, brick, wood, air. U Ans. Water, sugar, mercury, ice, iron, calcium oxide.  [1] Q. 6. Give an example of a liquid in liquid type solution. K Ans. Vinegar (acetic acid + water). [1]

(1 mark each)

Commonly Made Error

Students fail to write the correct definition. Some of them write it incomplete.

Answering Tip

Definitions of terms must be precise and complete.

Short Answer Type Questions-I

(2 marks each)

Q. 1. Can a homogeneous mixture have a variable

Q. 4. Write any two differences between physical and

composition ? Justify giving an example. U Ans. Yes, homogeneous mixtures can have a variable composition.

A chemical changes. Ans. Difference between physical and chemical changes:

For example, (i) Fe + H2SO4(aq) → FeSO4 + H2 (ii) FeS + H2SO4(aq) → FeSO4 + H2S Q. 2. List any two characteristics of colloid.

[2] K

S. No. Physical change

Chemical change

(i)

These are reversible changes and their chemical composition does not change.

These are irreversible changes and the chemical composition also changes.

(ii)

No new substance New substance is is formed, e.g., formed, e.g., Burning Tearing of paper. a matchstick.

Ans. (i) It is heterogeneous mixture. (ii) Particles of colloids scatter a beam of light. (Tyndall effect) [2] Q. 3. Identify colloid from the following mixtures: Muddy water, Sugar in water, Ink, Blood, Soda water. Ans. Ink, blood

U

[2]

[1 + 1]

NATURE OF MATTER

Q. 5. Classify the following as a chemical or physical change: (i) Water boils to form steam. (ii) Burning of paper. Ap

Ans. (i) Water boils to form steam is the change of water from liquid to gaseous state is a physical change. [1] (ii) Burning of paper is the chemical change of cellulose into carbon. [1]

Short Answer Type Questions-II Q. 1. Explain what is observed when a strong beam of light is focused on a colloidal solution of starch in water. Name the phenomenon.  A [KVS 2018, Board Term-I, 2016] Ans. When a strong beam of light is passed through colloidal solution of starch, the path of light becomes visible, which scatters through the solution. This phenomenon is known as Tyndall effect. [3] Q. 2. You are provided with a solution of substances 'X'. How will you test whether it is saturated or unsaturated with respect to 'X' at a given temperature? What happens when a hot saturated solution is allowed to cool?  A [Board Term-I, 2016] Ans. When no more quantity of 'X' substance can be dissolved in a solution at a given temperature, then the solution is saturated with respect to 'X'. When a hot saturated solution is allowed to cool, crystals of substance separate out from the solution. [3] Q. 3. (a) Why path of light is not visible in a solution when a beam of light is passed through it ? (b)  Classify each of following as solution, colloid or suspension: (i) A mixture whose particles are big enough to scatter a beam of light passing through it. (ii) A mixture whose particles settle down when it is left undisturbed. U [Board Term-I, 2016]

21

(3 marks each)

Q. 4. Rahul and Manav each were given a mixture of iron fillings and sulphur powder. Rahul heated the mixture strongly and a new substance was formed. Write three points of difference between the two. Ap [Board Term-I, 2016] Ans. Rahul has a compound, Manav has a mixture. S. No.

Compound

Mixture

(i)

Elements react to form a compound.

Elements or compounds get mixed together.

(ii)

Fixed composition.

Variable composition.

(iii)

Totally different properties.

Shows properties of constituent substances.

[CBSE Marking Scheme, 2016] [3] Q. 5. Name the only liquid metal and the only liquid non-metal. Mention two gaseous non-metals.  K [Board Term-I, 2015) Ans. Mercury is the only liquid metal and bromine is the only liquid non-metal. Two gaseous nonmetals are hydrogen and nitrogen. [1 + 1 + 1] Q. 6. (i) How tincture of iodine is prepared ? (ii) Define solubility. (iii) What would happen if you were to take a saturated solution at a certain temperature and cool it slowly ?  Ap+U [Board Term-I, 2016]

 Ans.

(i) By dissolving iodine in alcohol.

Ans. (a) Because of small particle size, they cannot scatter a beam of light. (b) (i) Colloids



(ii) Solubility: The amount of solute present in the saturated solution at a particular temperature is called its solubility.

(ii) Suspension (CBSE Marking Scheme, 2016] [1 + 1 + 1]



(iii) The solute particles will settle down. [CBSE Marking Scheme, 2016] [3]





Long Answer Type Questions

(5 marks each)

Q. 1. Classify different types of pure substances. Differentiate them on the basis of their chemical properties giving examples of each.  U [Board Term-I, 2016] Ans. Two types of pure substances are elements and compounds. Differences between elements and compounds are: S.No.

Compound

Element

(i)

A compound contains atoms of different elements chemically combined together in a fixed ratio. Compounds contain different elements in a fixed ratio arranged in a defined manner through chemical bonds.

An element is a pure chemical substance made of same type of atom. Elements are distinguished by their atomic number (number of protons in their nucleus).

(ii)

22

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

S.No. (iii) (iv) (v)

Compound

Element

A compound can be separated into simpler Elements cannot be broken down into simpler substances by chemical methods/reactions. substances by chemical reactions. The list of compounds is endless. There are about 117 elements that have been observed. and can be classified as metal, non-metal or metalloid. A compound is represented using a formula. An element is represented using symbols.

(vi)

e.g., Water (H2O), Sodium chloride (NaCl), e.g., Iron, copper, silver, gold and nickel, etc. Sodium bicarbonate (NaHCO3), etc.  (Any four) [1 + 4] Q. 2. (a) Can a homogeneous mixture have a variable Detailed Answer: composition ? Justify giving an example. (a) Homogeneous mixtures are the mixtures (b) What happens when: with uniform composition throughout like sugar solution or another. (i) Dilute sulphuric acid is added to a mixture of iron filings and sulphur (b) (i) Results in the formation of ferrous powder. sulphate and evolution of hydrogen. (ii) Dilute sulphuric acid is added to a Fe + H2SO4(aq) → FeSO4 + H2 mixture of iron filings and sulphur powder heated to red hot followed by (ii) Results in the formation of iron sulphide cooling. which reacts with sulphuric acid to form  Ap [Board Term-I, 2015] ferrous sulphate and release of hydrogen disulphide gas. Ans. (a) Homogenous mixtures are the mixtures with uniform composition throughout like Fe + S → FeS sugar solution or another. [3] FeS + H2SO4 → FeSO4 + H2S (b) (i) Fe + H SO (aq) → FeSO + H 2

4

4

2

(ii) FeS + H2SO4(aq) → FeSO4 + H2S [CBSE Marking Scheme, 2015] [1 + 1]

Commonly Made Error

Students often forget to write balanced chemical equation.

Q. 3. (a) State two ways by which you can change a saturated solution to unsaturated solution. (b) Distinguish between homogeneous and heterogeneous mixture by giving one example of each. U [NCERT] Ans. (a) (i) By increasing the heating the solution.

temperature/by

(ii) By increasing the amount of solvent. (b)

Answering Tip

Homogeneous Mixture

Heterogeneous Mixture

Uniform composition.

Non-uniform composition.

S. No.

Mention the name of the reactants or products

(i)

below the chemical formula in the chemical equation. Practise writing a balanced chemical equation.

(ii) No distinct boundaries Distinct boundaries of of separation. separation. e.g., sugar + water. e.g., sand + water. [2 + 3] Q. 4. Based on the following characteristics distinguish in tabular form the behaviour of true solution, suspension and colloidal solution. U [Board Term-I, 2012] [NCERT] Ans. S. No.

True Solution

Colloidal Solution

Suspension

(i)

A true solution is homogeneous mixture solute and solvent.

(ii)

It is transparent.

(iii)

The solute particles are very The solute particles are between The solute particles are quite large, small, i.e., less than 1 nm. 1-100 nm. i.e., more than 100 nm.

(iv)

The particles are not visible The particles are visible with the The particles are visible even with even with a powerful help of microscope. naked eye. microscope.

a A colloidal solution appears to be It is a heterogeneous mixture. of homogeneous but actually it is a heterogeneous mixture of solute and solvent. It is translucent.

It is opaque.

NATURE OF MATTER

S. No.

True Solution

Colloidal Solution

23

Suspension

(v)

The entire solution passes The particles can pass through The particles cannot pass through through filter paper as well as ordinary filter paper but not either a filter paper or through a semisemi-permeable membrane. through a semi-permeable permeable membrane. membrane.

(vi)

The solute particles do not The particles show Tyndall effect. show Tyndall effect.

(vii)

The particles do not settle due The particles do not settle due to The particles may settle due to gravity, to gravity, e.g., salt in water gravity, e.g., blood. e.g., chalk powder in water. solution.

The particles may or may not show Tyndall effect.

[Any five] [1 × 5]

COMPETENCY AND CRITICAL THINKING BASED QUESTIONS

(1 mark each)

I. Read the given passage and answer the following questions. The teacher instructed three students A, B and C respectively to prepare a 50 % (mass by volume) solution of sodium hydroxide. ‘A’ dissolved 50 g of NaOH in 100 mL of water, ‘B’ dissolved 50 g of NaOH in 100 g of water while ‘C’ dissolved 50 g of NaOH in water to make 100 mL of solution. (1) Which one of the following has made the desired solution? (2) Define concentration of a solution. (3) How will you prepare a 10% solution of sugar? (4) The two components of a solution are ___________ and ____________. Ans. (1)  Student C has made desired solution by dissolving 50 g NaOH in water to make the volume of the solution 100 mL. (2) It indicates the exact amount of solute dissolved in an exact amount of solvent or solution. (3) Dissolve 10 g of sugar in (100 – 10) = 90 g of water. (4) The two components of a solution are solute and solvent. II. Study the substances given in the box and answer any of the four questions given below:



(2) A student has written following statements about soil. Which of these is correct? (A) Soil is a homogeneous mixture. (B) Soil is a heterogeneous mixture. (C) Soil is a compound. (D) Soil is a suspension. Ans. Option (B) is correct. Soil is composed of small pieces of a variety of materials, so it is a heterogeneous mixture. (3) Give an example of gas – gas homogenous mixture. (A) Milk (B) Carbon dioxide (C) Water (D) Air. Ans. Option (D) is correct. (4) Sugar is a pure substance. It is because: (A) it cannot be separated. (B) it can be separated. (C) it contains carbon and hydrogen. (D) it is crystalline. Ans. Option (A) is correct. Sugar is a pure substance because it cannot be separated. (5) How can we say that milk is heterogeneous mixture? (A) As milk cannot be separated by physical process into its components such as water, fats etc.

Air, Soil, Filtered tea, Wood, Mercury, Milk, Calcium oxide, Ink (1) The homogeneous mixture is/are: (A) Only air (B) Air and filtered tea (C) Air and wood (D) Soil, calcium oxide and ink Ans. Option (B) is correct. Air and filtered tea are homogenous mixtures while wood and soil are heterogenous mixtures.

(B)  As milk can be separated by physical process into its components such as water, fats etc. (C) As milk is in liquid form (D) As milk is not crystalline Ans. Option (B) is correct. As milk can be separated by physical process into its components such as water, fats, etc.



Study Time Reading Time: 2:30 Hr No. of Questions: 38

CHAPTER

3



Syllabus

PARTICLE NATURE AND THEIR BASIC UNITS

Atoms and molecules, Law of chemical combination, Chemical formula of common compounds, Atomic and molecular masses.

Revision Notes  Laws of chemical combination : There are two laws of chemical combination : (i) Law of conservation of mass : Mass can neither be created nor be destroyed in a chemical reaction. (ii) Law of constant proportions or Law of definite proportions : In a chemical substance, the elements are always present in a definite proportion by mass.  Postulates of Dalton’s atomic theory : (i) Every matter is made up of very tiny particles called atoms. (ii) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction. (iii) Atoms of a given element are identical in mass and chemical properties.

Scan to know more about this topic

(iv) Atoms of different elements have different masses and chemical properties. (v) Atoms combine in the ratio of small whole numbers to form compounds. (vi) The relative number and kinds of atoms are constant in a given compound.

Atoms and Molecules

Atoms  Atoms are building blocks of all matters.  Atomic radius is measured in nanometers (1 m = 109 nm). Elements and their naming  Each element has a unique name and a unique symbol.  IUPAC (International Union of Pure and Applied Chemistry) approves names of the elements.  Rules for assigning symbols for atoms of various elements are as follows : (i) The abbreviation used to represent an element is generally the first letter of the element’s name in English.

English name of element Hydrogen Boron Oxygen Nitrogen Fluorine

Symbol H B O N F

PARTICLE NATURE AND THEIR BASIC UNITS

25

26

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

(ii) When the names of two or more elements begins with the same initial letter, the initial letter is followed by the letter appearing later in the name : Name of element Barium Bismuth Bromine Silicon Cadmium, Calcium

Symbol

Scan to know more about this topic

Ba Bi Br Si Cd, Ca

Laws of Chemical Combinations

(iii) Symbols of some elements are derived from their Latin / German or Greek names : Name of element



Latin/German/Greek name

Symbol

Sodium

Natrium

Na

Potassium

Kalium

K

Copper

Cuprum

Cu

Iron

Ferrum

Fe

Gold

Aurum

Au

Silver

Argentum

Ag

 One atomic mass unit is a mass unit exactly equal to 1/12th the mass of one C-12 atom.  Atoms of most elements are not able to exist independently. Atoms form molecules and ions. Molecules  Molecules of an element are formed by the atoms of the same type.  Atoms of same or different elements join together in definite proportions to form molecules of compounds.



 The number of atoms constituting a molecule is known as its atomicity.



Ions  An ion is a charged particle and can be negatively or positively charged.



 Ions may consist of a single charged atom or a group of atoms that have a net charge on them.



 Ionic compounds contain charged species called ions as their smallest unit.



 A group of atoms carrying a fixed charge on them are called polyatomic ions or radicals.



 The chemical formula of a compound is a symbolic representation of its composition. Valency  Valency is the combining capacity of an element.



 Valency can be used to find out how the atom(s) of an element will combine with the atom(s) of another element to form a chemical compound.



 Names and symbols of some ions : Valency

Name of ion

1

Sodium Potassium Silver Copper (I)*

Symbol Na+ K+ Ag+ Cu+

Non-metallic element Hydrogen Hydride Chloride Bromide Iodide

Symbol H+ H– Cl– Br– I–

Ammonium Hydroxide Nitrate Hydrogen Carbonate

NH4+ OH– NO3– HCO3–

2

Magnesium Calcium Zinc Iron (II)* Copper (II)*

Mg2+ Ca2+ Zn2+ Fe2+ Cu2+

Oxide Sulphide

O2– S2–

Carbonate Sulphite Sulphate

CO32– SO32– SO42–

3

Aluminium Iron (III)*

Al3+ Fe3+

Nitride

N3–

Phosphate

PO43–

* Some elements show variable valency which is represented by a roman numerical brackets.

Polyatomic ions

Symbol

PARTICLE NATURE AND THEIR BASIC UNITS

27

 Rules for writing the formula of a compound : (i) Formula of compound is given by writing side by side the symbols of constituent elements. (ii) Symbol of the more metallic element is written first in the formula. (iii) Number of atoms of each of the constituent element present in the molecule is indicated by subscript. (iv) When either of the ions or both the ions are polyatomic and their valency is more than one, we enclose the polyatomic ions in brackets. No brackets are necessary if the valency(ies) of polyatomic ion (s) is (are) 1. (v) While writing the formula of a compound if the valency numbers have a Highest Common Factor (H.C.F), divide the valency numbers by H.C.F so as to get the simplest ratio between the combining elements.



 The charges or valencies on the ion must be balanced.  Formula of a binary compound is written by criss-crossing the valencies of elements present in a molecule of the compound.  A chemical compound is always electrically neutral; hence the positive and negative valencies or charges of the ions in the compound must add upto zero.  Scientists use the relative mass scale to compare the masses of different atoms of elements.  Atoms of C-12 isotopes are assigned a relative atomic mass of 12 and the relative masses of all other atoms are obtained in comparison with the mass of a C-12 atom.  Relative mass of a molecule is expressed in atomic mass unit (u).



 Atoms of different elements are of different sizes and masses.



Key Words  Atom : Smallest particle of an element that shows all the properties of an element.  Atomic number : Number of protons in an atom of an element.  Molecule : Smallest particle of an element/compound that is capable of an independent existence and shows all the properties of that substance.  Anion : Negatively charged ion.  Cation : Positively charged ion.  Atomicity : Number of atoms present in one molecule of an element.  Radical : An atom or a group of atoms carrying positive or negative charge that behaves as a single unit in a chemical reaction.  Mole : Amount of substance that contains the same number of units as there are atoms in exactly 12 g of carbon-12 isotope.  Chemical formula : Expression of the composition of a substance by chemical symbols and numerical subscript.  Diatomic : A molecule which contains two atoms.  Triatomic : A molecule which contains three atoms.  Polyatomic : A molecule which contains more than two atoms.  Valency : Measure of combining capacity of an element with other atoms when it forms compounds or molecules.  Binary compound : Simplest compounds made up of two different elements. e.g., HCl, H2O.  Gram atomic mass : Atomic mass of an element expressed in terms of grams.  Molecular mass : Sum of the atomic masses of all the atoms in a molecule of the substance.  Formula unit mass : Sum of the atomic masses of all the atoms in a formula unit of a compound.



Example

(a) A sample of vitamin C is known to contain 2.58 × 1024 oxygen atoms. How many moles of oxygen atoms are present in the sample ? (b) Write one word for the following : (i)  In a balanced chemical equation, the sum of the masses of reactants and products remains unchanged.

(ii) A group of atoms carrying a fixed charge on them. (c)  Write chemical formulae of the following compounds : (i)  Sodium phosphate (ii)  Ammonium carbonate

28

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Solution: Step I: (a)  Number of moles

=



=

Given no. of particles Avogadro number 2.58 × 10 24 6.022 × 10 23

= 0.4284 × 10 = 4.284 moles Step II: (b)  (i)  Law of conservation of mass. (ii)  Polyatomic Ion Step III: (c)  (i) Na3 PO4 (ii) (NH4)2 CO3  [1 + 2 + 2 = 5]

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions 

(1 mark each)

Q. 1. Which of the following correctly represents 360 g of water? (i) 2 moles of H2O (ii) 20 moles of water (iii) 0.22 × 1023 molecules of water (iv) 1.2044 × 1025 molecules of water (A) (i) (B) (i) and (iv) (C) (ii) and (iii) (D) (ii) and (iv) Ans. Option (D) is correct. Explanation: (ii) and (iv) correctly represent 360 g of water. Q. 2. Which of the following statements is not true about an atom? (A) Atoms are not able to exist independently. (B)  Atoms are the basic units from which molecules and ions are formed. (C) Atoms are always neutral in nature. (D) Atoms aggregate in large numbers to form the matter that we can see, feel or touch. Ans. Option (D) is correct. Explanation: The molecules and ions aggregate together in large number to form the matter. We cannot see the individual molecules/ions with our eyes, but we can see the various substances which are a big collection of molecules/ions. Q. 3. The chemical symbol for nitrogen gas is (A) Ni (B) N2 (C) N+ (D) N Ans. Option (B) is correct. Explanation: Nitrogen molecule is diatomic molecule, therefore, it exists as N2 molecules. Q. 4. The chemical symbol for sodium is (A) So (B) Sd (C) NA (D) Na Ans. Option (D) is correct. Explanation: The chemical symbol for sodium is ‘Na’ derived from latin word natrium. Q. 5. The nucleus of an atom has positive charge because of:

(A) Neutrons (B) Protons (C) Both neutrons and protons D) Both electrons and protons Ans. Option (B) is correct. Q. 6. What is the valency of oxygen [Atomic number of oxygen = 8] ? (A) 2 (B) 4 (C) –2 (D) 1 Ans. Option (C) is correct. Explanation: Electronic configuration of oxygen is = 2, 6. Hence, valency = 8 – 6 = 2 Q. 7. Symbol of triatomic molecule is: (A) O3 (B) O2 (C) S8 (D) P4 Ans. Option (A) is correct. Explanation: Ozone has three oxygen atoms. Hence, it is triatomic. Q. 8. An element ‘X’ has a valency 3. Write the formula of its oxide. (A) X3O4 (B) X2O3 (C) XO3 (D) XO4 Ans. Option (B) is correct.

B

Assertion and Reason 

(1 mark each)

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion: Water molecules always contain hydrogen and oxygen in the ratio 1:8. Reason: Water obeys law of constant proportions irrespective of source and method of preparation. Ans. Option (A) is correct. Explanation: Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. This indicates the law of definite proportions.

PARTICLE NATURE AND THEIR BASIC UNITS

According to this law, the elements are always present in definite proportion by mass in a chemical substance. All pure samples of a compound contain the same elements combined together in the same proportion by mass. Q. 2. Assertion: Atoms can neither be sub-divided, created nor destroyed. Reason: This postulate of Dalton’s theory is the result of law of constant proportion. Ans. Option (C) is correct. Explanation: Atoms can neither be sub-divided, created nor destroyed. This postulate of Dalton’s atomic theory is the result of law of conservation of mass. Q. 3. Assertion: Carbonates are polyatomic ions. Reason : The carbonate ion consists of one carbon atom and three oxygen atoms and carries an overall charge of 2−.

29

Ans. Option (A) is correct. Explanation: The carbonate ion consists of one carbon atom and three oxygen atoms and carries an overall charge of 2−. The formula of the carbonate ion is CO32-. The atoms of a polyatomic ion are tightly bonded together and so the entire ion behaves as a single unit. Q. 4. Assertion: Relative atomic mass of the atom of element is the average masses of the atom as 1 compared to th the mass of one carbon-12 12 atom. Reason : Carbon-12 isotope is the standard reference for measuring atomic masses. Ans. Option (B) is correct. Explanation: Carbon-12 is taken as standard reference because no other nuclides have exactly whole number.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions Q. 1. State the Law of Conservation of Mass.  U [Board Term-II 2016] [DDE 2017] Ans. It states that mass can neither be created nor destroyed in a chemical reaction. [1]

Commonly Made Error

Students often write incorrect law or sometimes they miss the important operative words.

Answering Tip

Students should learn all the laws thoroughly. Understand and state law of conservation of mass in simple words, giving importance to operative words.

(1 mark each)

Q. 5. Write down the electron distribution of oxygen atom. How many valence electrons does it have ? (Atomic number of oxygen is 8)  U [Board Term-II 2015] Ans. O(8) = 2, 6 [½] K, L Number of valence electrons = 6 [½] [CBSE Marking Scheme, 2015] Q. 6. Mention the postulate of Dalton’s atomic theory which can successfully explain the law of definite proportions.  K [Board Term-II 2014] [NCERT] OR Which postulate of Dalton’s atomic theory can explain the law of definite proportions ?

Ans. Law of constant proportion states that in a chemical compound the elements are always present in a definite proportion by mass. [1] Q. 3. Write the symbols of : (i) Lead, (ii) Boron  U [Board Term-II 2016]

Ans. The Law of Definite Proportions is explained by the following postulate of Dalton : ‘‘The relative number and kinds of atoms are constant in a given compound.’’  [CBSE Marking Scheme, 2014] [1] Q. 7. If ‘K’ and ‘L’ shells of an atom are completely filled with electrons then : (i) what will be the total number of electrons in the atom ?

Ans. (i) Lead — Pb

(ii) find its valency.

(ii) Boron — B (½ + ½) Q. 4. Explain what do you understand by valence electrons ? U [Board Term-II 2015]

Ans. (i) Valence electrons = 2 (K), 8 (L) = 10

Q. 2. State the law of constant proportion. U [DDE 2017]

Ans. The electrons present in the outermost shell of an atom are called as valence electrons. [1] [CBSE Marking Scheme, 2015]

U [Board Term-II 2013]

(ii) Valency = 0

[½ + ½]

Commonly Made Error

Students often calculate incorrect values of

valence electrons and valency. It seems they are confused between these two terms.

30

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Q. 8. 10 g of silver nitrate solution is added to 10 g of sodium chloride solution. What change in mass do you expect after the reaction and why ? U+Ap [Board Term-II 2013]

Answering Tip

Lay stress on understanding the concept

Ans. No change in mass will take place because law of conservation of mass holds good. [1]

instead of rote learning.

Short Answer Type Questions-I Q. 1. Write the chemical formula of: (i) Sodium carbonate (ii) Ammonium chloride

C + O2  → CO2 U [NCERT]

Ans. (i) NaCO3 -2

+1

Na2CO3 (ii) NH4Cl −1

+1

NH4Cl  Q.2. Write the chemical formula for: (i) Zinc phosphate (ii) Lead carbonate

(2 marks each)

[1+1 = 2]

U

Ans. (i) Zn3(PO4)2

[1]

(ii) PbCO3

[1]

Q. 3. (i) Write down the name of compounds represented by the following formulae: (a)  Ca(OH)2 (b)  K2SO4 (ii) Give two examples of bivalent cations. U [NCERT] Ans. (a) (i) Calcium hydroxide [½] (ii) Potassium sulphate [½] (b) Bivalent cation = Mg2+, Ca2+, Zn2+, Fe2+, Cu2+ (Any two) [½ + ½] Q. 4. State the law of conservation of mass. If 12 g of carbon is burnt in the presence of 32 g of oxygen, how much carbon dioxide will be formed? U+Ap Ans. The law states that matter can neither be created nor destroyed or mass of reactants is always equal to that of product.

Carbon + Oxygen  →  Carbon dioxide Mass of reactants = 12 + 32 = 44g [½] Mass of product (CO2) = 44 g [½] (One mole of C reacts with one mole of oxygen to form one mole of CO2) Q. 5. The following data represents the distribution of electrons, protons and neutrons in atoms of four elements A, B, C, D. Element Protons Neutrons Electrons A 9 10 9 B 16 16 16 C 12 12 12 D 17 22 17 Answer the following questions: (i) Give the electronic distribution of element B. (ii) The valency of element A? (iii) The atomic number of element B? (iv) The mass number of element D?

(ii) Valency of A is 1.

[½]

(iii) Atomic number of element B is 16.

[½]

(iv) Mass number of element D is 39.

[½]

Q. 6. What are polyatomic ions ? List two examples. U+K (Board Term-II, 2012)



Ans. A group of atoms carrying a charge is known as polyatomic ion. 1+1 e.g., PO43–, SO42–, NH4+. [CBSE Marking Scheme, 2012]

(3 marks each) (ii)

Q. 1. Show the formation of chemical formulae of following compounds using their ions : (i) Ammonium sulphate (iii) Aluminium sulphide   U [Board Term-II, 2016] Ans. (i)



(iii)





Ap

Ans. (i) Electronic distribution of element B = 2, 8, 6  [½]

Short Answer Type Questions-II

(ii) Magnesium nitrate

[1]



[1 + 1 + 1 = 3]

PARTICLE NATURE AND THEIR BASIC UNITS

Q. 2. Define valency. Classify the following cations

on the basis of their valencies : Ca++, Cu++, NH4+, Al+++ (KVS 2018-19)  U+K [Board Term-II, 2016] Ans. The combining power (or capacity) of an element to displace or combine with number of hydrogen atoms is known as its valency. Ca++ → divalent NH4+ → monovalent ++ Cu → divalent Al+++ → trivalent  [1 + ½ + ½ + ½ + ½ = 3] Q. 3. (i) What are polyatomic ions ? Give one example. (KVS 2018-19) [NCERT] (ii) In an experiment, 14 g of sodium bicarbonate was allowed to react with 10 g of acetic acid. After the reaction was completed, it was found that only 16.67 g of the solution was left because a gas escaped from the container. What was the mass of the gas that escaped into the atmosphere ? Name the law applied.  Ap Ans. (i) Polyatomic ions are a group of atoms carrying charge. They are also called molecular ion. Example: Carbonate or phosphate or nitrate ion. (ii) Mass of the gas that escaped into the atmosphere = 24 – 16.67 = 7.33 g The law of conservation of mass is applied in this expression.  [½ + ½ + 1 + 1 = 3] [CBSE Marking Scheme, 2012] Q. 4. In a reaction 4.6 g of barium chloride reacted with 3.4 g of sodium sulphate. The products were 2.8 g of sodium chloride and 5.2 g of barium sulphate. The reaction takes place as follows : Barium chloride + Sodium sulphate → Sodium chloride + Barium sulphate Show that the above observation is in agreement with law of conservation of mass. State the law.  U [Board Term-II, 2016] [NCERT]



Ans. Barium chloride + Sodium sulphate → Sodium chloride + Barium sulphate



4.6 g + 3.4 g → 2.8 g + 5.2 g 8g→8g 8=8 This is in agreement with the ‘Law of Conservation of Mass’ that states matter can neither be created nor destroyed in a chemical reaction.  [CBSE Marking Scheme, 2016] [2 +1 = 3] Q. 5. Derive the molecular formulae for the following compounds : (i) Copper (II) bromide (ii) Ammonium carbonate (iii) Aluminium oxide. [NCERT Exemplar] R [Board Term-II, 2015]

Ans. (i) Cu

2+

Br

–

(ii) NH4+CO3–2



3+

(iii) Al O

CuBr2[1]

2–

[NH4]2CO3[1] Al2O3[1]

[CBSE Marking Scheme, 2015]

Q. 6. (a) From the symbol

32 16 S

state : (i) Atomic number of sulphur



(ii) Mass number of sulphur

(iii) Electronic configuration of sulphur (b) Which of the two elements given below would be chemically more reactive, ‘X’ of atomic number 18 or element ‘Z’ of atomic number 16 and why ? 

A [Board Term-II, 2015]

Ans. (a) (i) 16



(ii) 32

(iii) 16S = 2, 8, 6 [1 ½] (B) Z is more reactive than X. Electronic configuration of Z = 2, 8, 6 which may gain 2 electrons and thus is more reactive whereas X = 2, 8, 8 has complete octet and is inert. [1 ½] [CBSE Marking Scheme, 2015]

Long Answer Type Questions Q. 1.

31

(5 marks each)

(i) Name the international organization who approves names of elements.



(ii) Give an example with explanation to show that the law of conservation of mass applies to physical changes also.

It is found that there is no change in the weight i.e.,











This shows law of conservation of mass holds true for physical changes.[5]

U [Board Term-II, 2015]

Ans. (i) International Union of Pure and Applied Chemistry (IUPAC).

(ii) When ice melts into water it is a physical change. Take a piece of ice in a small flask, cork it and weigh it and denote it as Wice g.



Heat the flask gently and ice (solid) slowly melts into water (liquid). Then, weigh the flask again as Wwater g.



Wice = Wwater Heat ( ) Ice  Water

[CBSE Marking Scheme, 2015] Q. 2.

(i) If 18 g of pure water is electrolysed, 2 g of hydrogen and 16 g of oxygen is obtained. Which law of chemical combination is illustrated by this statement ?

(ii) State the law of constant proportion. Illustrate with the help of an example.

32

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

(iii) Which postulate of Dalton’s atomic theory is the result of law of conservation of mass? (iv) Which point of Dalton’s atomic theory came from law of constant proportions ?  [Board Term-II, 2013] Ans. (i) Law of constant proportion. (ii) A compound prepared by any method contains the same elements in the fixed ratio by mass. For example, H2O contains hydrogen and oxygen in the ratio 2 : 16, i.e., 1 : 8 by mass. (iii) Atoms can neither be sub-divided, created nor destroyed. (iv) Atoms of different elements combine in simple whole number ratios to form chemical compound. [1 + 2 + 1 + 1 = 5] Q. 3. When 3 g of carbon is burnt in 8 g of oxygen, 11 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3 g of carbon is burnt in 50 g of oxygen? Which law of chemical combination will govern your answer? State the law.  U [NCERT]

chemical substance. All pure samples of a compound contain the same elements combined together in the same proportion by mass. [2] [CBSE Marking Scheme, 2012] Q. 4.

(i) What do the following symbols / formulae stand for :

(a) 2O

(b) O2

(c) O3

(d) H2O

(ii) Give the chemical formula of the following compounds : (a) Potassium carbonate (b) Calcium chloride (iii) Calculate the formula unit mass of Al2(SO4)3 (Given : Atomic mass of Al = 27 u, S = 32 u, O = 16 u)  U+Ap (Board Term-II, 2012) Ans. (i) (a) Two atoms of oxygen : 2O[½]

(b) Diatomic oxygen : O2 molecule[½]



(c) Triatomic oxygen : O3 molecule[½]

(d) Two atoms of hydrogen and one atom of oxygen forming one molecule of water (H2O).[½]

Ans. When 3 g of carbon is burnt in 8 g oxygen, 11 g of carbon dioxide is produced. It means carbon and oxygen are combined in the ratio of 3 : 8 to form carbon dioxide. Thus, when there is 3 g carbon and 50 g oxygen, then also only 8 g of oxygen will be used and 11 g of carbon dioxide will be formed. The remaining oxygen is not used. [2] This indicates the law of definite proportions. [1] According to this law, the elements are always present in definite proportion by mass in a



(ii) (a) K2CO3 : Potassium carbonate[½]



(b) CaCl2 : Calcium chloride[½]

(iii) Al2(SO4)3

Al = 27 × 2 = 54 u



S = 32 × 3 = 96 u





O = 16 × 12 = 192 u

Formula unit mass = 54 + 96 + 192 = 342 u. [2] [CBSE Marking Scheme, 2012]

COMPETENCY AND CRITICAL THINKING BASED QUESTIONS

I. The following data represents the distribution of electrons, protons and neutrons in atoms of four elements A, B, C, D. Understand the data carefully and answer the following questions. Element



Protons

Neutrons

Electrons

A 9 10 9 B 16 16 16 C 12 12 12 D 17 22 17 (1) Select the correct electronic distribution of element B: (A) 2, 8, 1

(B) 2, 8, 2

(C) 2, 8, 4 (D) 2, 8, 6 Ans. Option (D) is correct. Explanation: The electronic configuration of element B = 2, 8, 6



(1 mark each)

(2) The Valency of element B: (A) 2 (B) –1

(C) – 2 (D) 3 Ans. Option (C) is correct. Explanation: The electronic configuration of element B is 2, 8, 6. There are six valence electrons in its outermost shell. So, valency = (No. of valence electrons – 8)



= 6 − 8 = – 2

(3) The atomic number of element D is: (A) 9

(B) 16

(C) 12 (D) 17 Ans. Option (D) is correct. Explanation: Atomic number of element = Number of protons in that element = Number of electrons.

PARTICLE NATURE AND THEIR BASIC UNITS

(4) The Mass number of element C is: (A) 17 (B) 22 (C) 24 (D) 14 Ans. Option (C) is correct. Explanation: Mass number = Number of Protons + Number of Neutrons = 12 + 12 = 24 II. The purpose of the item is to assess whether students understand and can interpret an electronic configuration and use one to calculate valence electron and then link it to calculate valency and atomic no.

33

Question(s) The electronic configuration of element X is 2. The electronic configuration of element Y is 2,8,2 (a) State the valence number of electrons of: Element X __________ Element Y __________ (b) State the valency of: Element X __________ Element Y __________ Ans. (a) X – 2; Y – 2; (b) X – 0; Y – 2; [CBSE – SAS] 

Study Time Reading Time: 3:00 Hr No. of Questions: 43

CHAPTER



STRUCTURE OF ATOM

4

Syllabus

Electrons, protons and neutrons; Valency, Atomic number and Mass Number, isotopes and isobars.

Revision Notes

 An atom is divisible and consists of charged particles.  Ionization of gases in the discharge tube proved that atoms have sub-atomic particles.  Summary of characteristics of electrons, protons and neutrons: Characteristics Symbol Relative charge Nature Discovered by Mass

Electron

Proton

Neutron

e

p

n

–1

+1

0

Negatively charged

Positively charged

Neutral

J. J. Thomson

E. Goldstein

James Chadwick

1/2000 times mass of hydrogen atom

1 unit

Mass is nearly equal to that of proton

 Thomson’s model of atom: Scan to know (i) An atom is a uniform sphere of positive charges (due to the presence of protons) as well more about this topic as negative charges (due to the presence of electrons) which are embedded in it. This model is often called the ‘Water Melon Model’. (ii) An atom, as a whole, is electrically neutral because the negative and positive charges are equal in magnitude.  Limitations of Thomson’s model of atom: The model failed to explain how protons and Structure of Atom electrons could be arranged in an atom so close to each other.  α-particles are charged particles having two units of positive charge and four units of mass, i.e., they are double-charged helium ions (He2+).  Observations predicted from a-particle scattering experiment by Rutherford based on Thomson’s model of atom are: (i) Rutherford expected that if the model proposed earlier by J. J. Thomson, according to which there is uniform distribution of positive and negative charge, was correct then α-particles striking the gold atoms would be uniformly deflected which was not the case. (ii) Since the α-particles were much heavier than the protons, he did not expect to see large deflections.  Selection of gold metal for Rutherford’s α-particle scattering experiment: Gold is easily malleable and can be beaten into very thin sheets.  Observations made by Rutherford from α-particle scattering experiment: (i) Most of the α-particles passed straight through gold foil without suffering any deflection from their original path. (ii) Some of the α-particles were deflected by the foil at small angles.

STRUCTURE OF ATOM

35

36

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

(iii) One out of every 12000 particles appeared to rebound.  Conclusions from Rutherford’s a-particle scattering experiment: (i) Most of the space inside the atom is empty. Hence, it allows the α-particles to pass straight through it without any deflection. (ii) Very few particles were deflected from their path, which suggests that the positive charge of the atom occupies very little space. (iii) The total volume occupied by a nucleus is very small compared to the total volume of the atom, as very few α-particles are reflected by 180o, and all the positive charge and mass of the gold atom were concentrated in a very small volume within the atom.  Rutherford’s nuclear model of an atom: (i) There is a positively charged centre in an atom called the nucleus and the entire mass of atom resides in the nucleus. (ii) Electrons revolve around the nucleus in well-defined circular orbits. (iii) Size of the nucleus is very small as compared to the size of an atom.  Defects in Rutherford’s model of atom: (i) Rutherford had proposed that electrons move around a positively charged nucleus at a very high speed in circular orbits. Electron would have to be accelerated centripetally (tending to move toward a center) to remain in a circular orbit, but according to electromagnetic theory, if charged body (electron) is accelerated around another charged body (nucleus) then there would be continuous radiation of the moving body (i.e., electron). This loss of energy would slow down the speed of electron and eventually electron would fall into nucleus. But Rutherford’s model could not explain such a collapse. (ii) Rutherford had proposed that electrons revolve around the nucleus in fixed orbits. He did not specify the number of electrons in each orbit.  Postulates put forward by Bohr regarding model of atom: (i) Electrons revolve around the nucleus in a limited number of orbits called discrete orbits of electrons that are also called as permissible orbits. (ii) While revolving in discrete orbits, the electrons do not radiate energy i.e., energy of an electron remains constant so long as it stays in a given orbit. Electrons present in different orbits have different energies. (iii) When an electron jumps from lower energy level to higher energy level, some energy is absorbed, while energy is released when electron jumps from higher energy level to lower one.  Orbits or shells are represented by the letters K, L, M, N... or the numbers, n = 1, 2, 3, 4....  Bohr-Bury scheme for distribution of electrons in different orbits: (i) Maximum number of electrons that can be accommodated in a shell is given by 2n2, where n is the shell number i.e., first shell can accommodate two electrons, second shell can accommodate eight electrons, third shell can accommodate 18 electrons and so on.

Fig. Distribution of electrons in different orbit. (ii) Outermost orbit of an atom can accommodate a maximum number of eight electrons. (iii) Electrons are not accommodated in a given shell, unless the inner shells are filled, i.e., the shells are filled in a step-wise manner.  Outermost shell of an atom is called valence shell.  Neutrons are situated in the nucleus of all the atoms, except hydrogen.  If the outermost shell of an atom is completely filled, its valency is 0.  Valency of elements having 1 to 4 electrons in the outermost shell are generally determined by the rule : Valency = Number of electrons in the outermost shell.  Valency of elements having number of electrons in outermost shell close to 8 is determined by the formula: Valency = 8 – Number of electrons in the outermost shell.  Significance of valence electrons: (i) Valence electrons are responsible for chemical changes.

STRUCTURE OF ATOM

37

(ii) Elements having same number of valence electrons in their atoms possess similar chemical properties because chemical properties of an element are determined by the number of valence electrons in an atom. (iii) Elements having different number of valence electrons in their atoms possess different chemical properties.  Protons and neutrons together are called nucleons.  All atoms of an element have the same atomic number.  Atomic number is denoted by ‘Z’ (Z = np).  For a neutral atom, number of protons and electrons are equal.  Mass number is denoted by ‘A’ (A = np + nN). np = No. of protons nN = No. of neutrons  Isotopes: (i) Isotopes are the atoms of same element having same atomic number but different mass number. (ii) Isotopes have similar chemical properties because they have same number of valence electrons. (iii) Isotopes have different physical properties such as boiling point and melting point because they have different mass numbers. (iv) Atomic masses of elements are fractional, due to the fact that all elements have isotopes. (v) Applications of isotopes: 1. An isotope of uranium is used in nuclear reaction. 2. An isotope of cobalt is used to remove brain tumour and their treatment. 3. Isotope of sodium has been used to diagnose restricted circulation of blood. (vi) Example: 3 isotopes of hydrogen–protium, deuterium and tritium.  Isobars: Isobars are the atoms of different elements with different atomic numbers, but same mass number. Example: 20Ca40, 18Ar40.

Key Words

 Canal rays: Positively charged radiations discovered by Goldstein in a gas discharge tube at low pressure and high voltage.



 Electron: Negatively charged particle.



 Proton: Positively charged particle.



 Neutron: Neutral particle.



 Energy level: Possible locations around an atom where electrons having specific energy values may be found.



 Octet: Shell which has eight electrons in the outermost shell.



 Valency: Combining capacity of an atom.



 Valence shell: Outermost shell of an atom.



 Valence electrons: Electrons present in the valence shell.



 Atomic number: Total number of protons present in the nucleus of an atom.



 Nucleons: A nucleon is one of the particles that make up the atomic nucleus.



 Mass number: Sum of the total number of protons and neutrons present in the nucleus of an atom.

Example Define the terms (a) isotope, (b) isobar giving one example of each. Name the element whose isotope is used in (i) nuclear reactor, (ii) treatment of cancer. Solution: Step I: (a)  The atoms of same element having same atomic number, but different mass numbers are called isotopes. e.g., 1H1, 1H2, 1H3.  (any other example) 1

(b)  The atoms of different elements with same mass number and different atomic numbers are called isobars. e.g., 20Ca40, 18As40. (any other example) Step II: Mass number is same (i.e, 40) [1] (i) Uranium

(ii) Cobalt 

[½ + ½]

38

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions

(1 mark each)



Q. 1. Which of the following correctly represent the electronic distribution in the Mg atom? (A) 3, 8, 1 (B) 2, 8, 2 (C) 1, 8, 3 (D) 8, 2, 2 Ans. Option (B) is correct. Explanation: Atomic number and the number of electrons in magnesium atom is 12. So, electronic configuration is 2, 8, 2 (because 12=2 + 8 + 2). Q. 2. Rutherford’s alpha particles scattering experiment resulted into discovery of (A) Electron (B) Proton (C) Nucleus in the atom (D) Atomic mass  (CBSE SQP, 2020) Ans. Option (C) is correct. Explanation: Rutherford discovered nucleus of the atom by his α-particle scattering experiment. Q. 3. The number of electrons in an element X is 15 and the number of neutrons is 16. Which of the following is the correct representation of the element? (A) (C)

31 15 X 16 15 X

31 (B) 16 X

15 (D) 16 X

Ans. Option (A) is correct. Explanation: Given that, number of electrons in element X = 15 and number of neutrons = 16. Atomic number = Number of protons = Number of electrons in neutral atom = 15 Mass number = number of protons + number of neutrons = 15+16=31 31 X. So, correct representation of element is 15 Q. 4. Dalton’s atomic theory successfully explained: (i) law of conservation of mass (ii) law of constant composition (iii) law of radioactivity (iv) law of multiple proportion (A) (i), (ii) and (iii) (B) (i), (iii) and (iv) (C) (ii), (iii) and (iv) (D) (i), (ii) and (iv) Ans. Option (D) is correct. Explanation: Dalton’s atomic theory successfully explained the laws of chemical combination but no point about radioactivity was mentioned by Dalton in his theory. Q. 5. Which of the following statements about Rutherford’s model of atom are correct? (i) Considered the nucleus as positively charged.

(ii) Established that the α-particles are four times as heavy as a hydrogen atom. (iii) Can be compared to solar system. (iv) Was in agreement with Thomson’s model. (A) (i) and (iii) (B) (ii) and (iii) (C) (i) and (iv) (D) Only (i) Ans. Option (A) is correct. Explanation: According to Rutherford model, a central positively charged nucleus is present in the atom and electrons revolve around it. Q. 6. Which of the following are true for an element? (i) Atomic number = number of protons +  number of electrons (ii) Mass number = number of protons +  number of neutrons (iii) Atomic number = number of protons =  number of neutrons (iv) Atomic number = number of protons =  number of electrons (A) (i) and (ii) (B) (i) and (iii) (C) (ii) and (iii) (D) (ii) and (iv) Ans. Option (D) is correct. Explanation: Atomic number = Number of protons = Number of electrons in neutral atom Mass number = Number of protons + Number of neutrons Q. 7. In the Thomson’s model of atom, which of the following statements are correct? (i) The mass of the atom is assumed to be uniformly distributed over the atom. (ii) The positive charge is assumed to be uniformly distributed over the atom. (iii) The electrons are uniformly distributed in the positively charged sphere. (iv) The electrons attract each other to stabilise the atom. (A) (i), (ii) and (iii) (B) (i) and (iii) (C) (i) and (iv) (D) (i), (iii) and (iv) Ans. Option (A) is correct. Explanation: Points (i), (ii) and (iii) are correct. According to Thomson’ model of the atom, an atom consists of a sphere of positively charge with negatively charged electrons embedded in it. These negative and positive charges in an atom are equal in magnitude, due to which an atom is electrically neutral.

STRUCTURE OF ATOM

Q. 8. Rutherford’s α-particle scattering experiment showed that (i) Electrons have negative charge. (ii) The mass and positive charge of the atom is concentrated in the nucleus. (iii) Neutron exists in the nucleus. (iv) Most of the space in atom is empty. Which of the above statements are correct? (A) (i) and (iii) (B) (ii) and (iv) (C) (i) and (iv) (D) (iii) and (iv) Ans. Option (B) is correct. Explanation: Points (ii) and (iv) are correct. An atom consists of a positively charged, dense and very small nucleus which have all the protons and neutrons. Positive charge is due to protons, as neutrons have no charge. Most of the space is empty because most of the alpha particles pass straight through the gold foil without any deflection. Electrons have negative charge. Q. 9. The ion of an element has 3 positive charges. Mass number of the atom is 27 and the number of neutrons is 14. What is the number of electrons in the ion? (A) 13 (B) 10 (D) 14 (D) 16 Ans. Option (B) is correct. Explanation: Given, Charge = + 3, mass number = 27 and number of neutrons = 14. Number of protons = Atomic number. Atomic number = Mass number – Number of neutrons = 27 – 14 = 13 This is the atomic number of aluminium. Therefore, this element is aluminium (Al). Number of electrons in the Al atom = 13 Number of electrons in the Al3 + ion = 13 – 3 = 10 as it is formed from the neutral atom by loss of 3 electrons. Q. 10. Identify the Mg2+ ion from the figure where, n and p represent the number of neutrons and protons respectively.



39

Ans. Option (D) is correct. Explanation: Electronic configuration of Mg atom = 2, 8,2 and that of Mg2+ ion = 2, 8 Number of protons in Mg atom = 12 Number of neutrons in Mg atom = 24 -12 = 12 [as mass number of Mg atom = 24 and number of neutrons = mass number – number of protons]. Therefore, option (D) is the correct answer.

B

Assertion and Reason 

(1 mark each)

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:

(A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).



(B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).



(C) Assertion (A) is true but reason (R) is false.

(D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion (A): All isotopes of a given element show the same type of chemical behaviour. Reason (R): The chemical properties of an atom are controlled by the number of electrons in the atom. Ans. Option (A) is correct. Explanation: All isotopes of an element have same number of electrons and same electronic configuration, hence their chemical properties are also same. Q. 2. Assertion: Atom is electrically neutral. Reason: Equal number of protons and electrons are present in an atom. Ans. Option (A) is correct. Explanation: Atom is electrically neutral because the total positive charge, i.e., protons is cancelled by the total negative charge, i.e., electrons. Q. 3. Assertion: Mass of an atom is the sum of its nucleons. Reason: Protons and neutrons are present in the nucleus of an atom. Ans. Option (A) is correct. Explanation: Mass of an atom is due to its nucleus which contains protons and neutrons. Collectively protons and neutrons are known as nucleons. Q. 4. Assertion: Inert elements show zero valency. Reason: Atoms of inert element have fully filled outermost orbit. Ans. Option (A) is correct. Explanation: Inert elements show zero valency. For example, Neon and Argon have completely filled electrons in their outermost shell with eight electrons. Thus, they cannot lose, gain or share electrons with atoms of other elements. Therefore, the valency of these gases is zero.

40

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions Q. 1. Define atomic number. 

U [Board Term-II, 2016]

Ans. It is the number of protons in the nucleus of an atom.

[CBSE Marking Scheme, 2016] [1]

Q. 2. What is an electron ? Who discovered it ? 

U [Board Term-II, 2016]

Ans. Electron is a negatively charged particle present outside the nucleus of an atom. It was discovered by J. J. Thomson in 1897. [½ ½ + ½]

Q. 3. Name the particles which determine the mass of an atom.

U [Board Term-II, 2016]

Ans. Protons and Neutrons.

[½ ½ + ½]

Q. 4. Study the composition of the two nuclei of two atomic species L and M. L = protons 20, neutrons 20 M = protons 18, neutrons 22 Give the relation between the two. Ap [Board Term-II, 2014]



Ans. L = nucleons = 20 + 20 = 40 M = nucleons = 18 + 22 = 40

[½ ½ + ½]

(1 mark each)

Both have same atomic mass but different atomic number. Relation between L and M: Isobars.[1] [CBSE Marking Scheme, 2014] Q. 5. There are 15 protons and 16 neutrons in the nucleus of an element. Calculate its atomic number and its atomic mass.  Ap (DDE 2017) Ans. Atomic number (Z) = Number of protons = 15 Atomic mass (A) = Number of protons + Number of neutrons = 15 + 16 = 31 [½ ½ + ½] Q. 6. Write any two observation which support the fact that atoms are divisible.  K [NCERT Exemplar] Ans. (i) Discovery of electrons and protons. (ii) Discovery that radioactivity is nothing but disintegration of atoms. [½ ½ + ½] Q. 7. Why did Rutherford select a gold foil in his a -ray scattering experiment ?  K [NCERT Exemplar] Ans. He selected a gold foil because it has high malleability. [1] 1] Q. 8. Write the notation of an atom ‘X‘, if the mass number is A and the atomic number is Z. U Ans. The notation of an atom is represented as: A X. [1] Z

Short Answer Type Questions-I Q. 1. Which atom, Na+ or He, has completely filled K and L shells ? Give reason to support your answer. A [Board Term-II, 2011, Set-A] OR Na+ has completely filled K and L shells. Explain. [NCERT] Ans. K L  M       K  L      K Na = 2, 8, 1      Na+ = 2, 8   He = 2 Na+ has completely filled K and L shells. Na atom gets converted into Na+ by losing one electron from its outermost shell. He atom has only K shell. [2] Q. 2. An atom of an element has three electrons in its 3rd orbit, which is the outermost shell. Write: (i) the electronic configuration (ii) atomic number (iii) number of protons (iv) valency

Ap [Board Term-II, 2011]

Ans.

(2 marks each) (i) Electronic configuration - 2, 8, 3



(ii) Atomic number - 13



(iii) Number of protons - 13



(iv) Valency - 3

[½ ½ × 4 = 2]

[CBSE Marking Scheme, 2011]



Q. 3. An element ‘Z’ forms the following compound when it reacts with hydrogen, chlorine, oxygen and phosphorus. ZH 3, ZCl 3, Z2 O3 and ZP (i) What is the valency of element Z ? (ii) Element ‘Z’ is metal or non metal? 

Ap [Board Term-II, 2011, Set-B]

Ans. (i) The valency of Z is 3.[1]

(ii) Z is a metal because it is electropositive and reacts with non-metals. [CBSE Marking Scheme, 2011] [1]

STRUCTURE OF ATOM

Q. 4. An atom of an element has one electron in the outer most M shell. State its: (i) Electronic configuration (ii) Number of protons (iii) Atomic number (iv) Valency of this element  Ap [Board Term-II 2011, Set-B] Ans. (i) Electronic configuration - 2, 8, 1 (ii) Number of protons - 11 (iii) Atomic number - 11 (iv) Valency of this element - 1 Q. 5. Give one word for the following: (i) Positively charged atom.

[½ ½ × 4 = 2]

(ii) A group of atoms carrying a charge.

U

Ans. (i) Positively charged atom:Cation

a revolution in the study of matter. Name the scientists who discovered these sub-atomic particles. U [Board Term-II, 2016] Ans. Electron: By J.J Thomson in 1897 Proton: By E. Goldstein in 1886 Neutron: By James Chadwick in 1932  [1 + 1 + 1] Q. 2. Illustrate that Na atom has completely filled K and L shells. U [Board Term-II, 2016] Ans. Electronic configuration of Na is: 2, 8, 1 K LM

K shell of Na has 2, L shell has 8 and M shell has one electron. [3] Q. 3. Write the electronic configuration and valency of the following: (i) Chlorine (ii) Sodium and (iii) Silicon U [Board Term-II, 2016] Ans. (i) E.C. of Cl = 2, 8, 7 Valency = – 1 (ii) E.C. of Na = 2, 8, 1 Valency = 1 (iii) E.C. of Si = 2, 8, 4 Valency = 4 [3] Q. 4. List the observations in α-particle scattering experiment which led Rutherford to make the following conclusions: (i) Most of the space in an atom is empty. (ii) Whole mass of an atom is concentrated in its centre. (iii) Centre is positively charged.  A [Board Term-II, 2016]

Ans. (i) Most of the alpha-particles passed through the gold foil without getting deflected. (ii) Very few particles were deflected from their path by 180°, indicating that whole mass of the atom is present at its centre.

[1]

(ii) Group of atoms carrying a charge: Polyatomic ion. [1] Q. 6. How many electrons, protons and neutrons will 19 be there in an element 9 X? What will be the valency of the element ? Ap Ans. No. of protons = Atomic number = 9 No. of protons + neutrons = Mass number = 19 Number of electrons = 9 Number of protons = 9 Number of neutrons = Mass number − Atomic number = 19 − 9 = 10 Electronic configuration of X = 2, 7 Valency of X = 1 (since, it requires one electron to complete its octet) [½ ½ × 4 = 2]

Short Answer Type Questions-II Q. 1. The discovery of subatomic particles led to

41

(3 marks each)

(iii) Few particles deflected at small and large angles from their path indicating that centre is positively charged.

[1 × 3 = 3]

26 26 Q. 5. There are two elements A13 and B14 . Find the number of sub-atomic particles in each of these elements. What is the relationship between the two ? Ap [Board Term-II, 2015] 26 Ans. A 13 electrons = 13 protons + 13 neutrons Atomic number = 26 – 13 = 13 26 B14 electrons = 14 protons + 12 neutrons Atomic number = 26 – 12 = 14[3] They are isobars.[CBSE Marking Scheme, 2015]

Q. 6.

(i) Why chemical properties of all the isotopes of an element are same ?

(ii) Name the isotopes used in the treatment of goitre and cancer. (iii) An element ‘X’ has 2 electrons in its M shell. What is its atomic number ? 

U [Board Term-II, 2015]

Ans. (i) This is because isotopes have same atomic number, so the number of valence electrons present in them is same and it is the valence electrons which take part in chemical reactions. So, the isotopes of an element have same chemical properties. (ii) Goitre – Isotope of iodine Cancer – Isotope of cobalt (iii) Atomic number of X = 12. [1 × 3 = 3] [CBSE Marking Scheme, 2015] Q. 7.

(i) What are canal rays ? State the nature of the constituents of canal rays.

(ii) Who discovered canal rays ? U [Board Term-II, 2014]

42

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Ans. (i) The positively charged radiations produced in the discharge tube at low pressure and high voltage are called canal rays. The canal rays have positively charged sub-atomic, particles known as protons (p).

(ii) E. Goldstein.

 Q. 8.

[CBSE Marking Scheme, 2014] [3] (i) Both Helium and Beryllium have two electrons in the valence shells. Helium is a noble gas whereas Beryllium is a metal, justify.

(ii) Hydrogen exists in three isotopic forms. Why are the isotopes of hydrogen chemically alike ?  Ans. (i) He2 = 2

composition of their nuclei and write their mass numbers. Also represent them in the form of U [Board Term-II, 2016] symbols. Ans. Isotopes of hydrogen are: Protium 1H1 (1 proton, 0 neutron), Mass no. 1 Deuterium 1H2 (1 proton, 1 neutron), Mass no. 2 Tritium 1H3 (1 proton, 2 neutrons), Mass no. 3  [3 + 1 +1 = 3] Q. 2. (i) Define valency. What conclusions can be drawn about the reactivity of an atom from its valency? (ii) Why does an atom of Argon have zero valency? Explain using the electronic configuration of Argon.  U [Board Term-II, 2016] Ans. (i) Valency: The combining capacity of an atom is known as its valency. Valency is the no. of electrons in outermost shell of an atom. The no. of electrons gained or lost or shared gives us the combining capacity of an atom and this decides whether an atom is reactive or not.[3] (ii) The atomic number of Argon = 18, i.e., it has 18 electrons. Hence, its electronic configuration will be 2, 8, 8 Since it has 8 electrons in its valence shell, So, its valency = 8 – 8 = (zero)[2] Q. 3. (a) Briefly explain Bohr-Bury scheme for the distribution of electrons in different shells. (b) An atom has 2 electrons in its outermost shell M. What is the atomic number of the element ? Also mention its name.  U (NCERT Exemplar) [Board Term-II, 2015] Ans. (a) The following rules are followed for writing the number of electrons in different energy levels or shells: (i) The maximum number of electrons present in a shell is given by the formula 2n2 where ‘n’ is the orbit number or energy level index 1, 2, 3 .... Hence, the maximum number of electrons in different shells are as follows:

Be4 = 2, 2

In case of He, its first shell is the last shell with 2 e– (stable). (ii) Hydrogen isotopes = H11, H12, H13, are chemically similar because they have same atomic number. [CBSE Marking Scheme, 2014] [3]

Long Answer Type Questions Q. 1. Hydrogen has three isotopes. State the

Ap [Board Term-II, 2014]

(5 marks each)

1st orbit or K-shell will be = 2 × 12 = 2, 2nd orbit or L-shell will be 2 × 22 = 8, 3rd orbit or M-shell will be 2 × 32 = 18, 4th orbit or N-shell will be 2 × 42 = 32 and so on. (ii) The maximum, number of electrons that can be accommodated in the outermost orbit is 8. (iii) Electrons are not accommodated in a given shell, unless the inner shells are filled. That is, the shells are filled in a step-wise manner:[3] (b) Atomic number of element = 12 K L M 2 8 2 Element is magnesium.[2] [CBSE Marking Scheme, 2015] Q. 4. Describe Rutherford’s a-particle scattering experiment and mention the important observations and conclusions drawn from this experiment.  U [Board Term-II, 2014] Ans. In this experiment, fast moving a-particles were made to fall on a thin gold foil. The following observations were made: (i) Most of the fast moving α-particles passed straight through the gold foil. (ii) Some of the α-particles were deflected by the foil by small angles. (iii) One out of every 12000 particles appeared to rebound. Conclusions: (i) Most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected. (ii) Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space. (iii) A very small fraction of α-particles was deflected by 180°, indicating that all the positive charge and mass of the gold atom were concentrated in a very small volume within the atom.[5] [CBSE Marking Scheme, 2014]

STRUCTURE OF ATOM

COMPETENCY AND CRITICAL THINKING BASED QUESTIONS

I. The electronic configuration of an element ‘X’ is 2, 8, 2: (1) The number of electrons present in the atom of element ‘X’ is __________. (A) 2 (B) 10 (C) 12 (D) 8

Ans. Option (C) is correct. Explanation: The electronic configuration of element X is 2, 8, 2. So its atomic number is 12 and the number of electrons is equal to the atomic number. (2) The element X is: (A) Non-metal (B) Metal (C) Metalloid (D) Noble gas Ans. Option (B) is correct. Explanation: Element X is a metal. (3) The Valency of element X is: (A) +2 (B) −2 (C) 0 (D) 1

43

(1 mark each)

Ans. Option (A) is correct. Explanation: The number of electrons present in the outermost shell is 2. So, the valency will be 2. (4) How many valence electrons are there in element X? (A) 2 (B) 3 (C) 8 (D) 10 Ans. Option (A) is correct. Explanation: The element present in the outermost shell are known as Valence electrons. In the outermost shell of element X, 2 valence electrons are present. II. Electrons revolve around the nucleus in a limited number of orbits called discrete orbits of electrons that are also called as permissible orbits. While revolving in discrete orbits, the electrons do not radiate energy i.e., energy of an electron remains constant so long as it stays in a given orbit. Electrons present in different orbits have different energies. [CBSE SAS] Q. 1. Name the Positively charged center in the nucleus. Ans Proton Q. 2. How the atomic number is represented? Ans. Atomic number is represented by 'Z'. Q. 3. What is the other name of orbits/shells? Ans. Energy level Q. 4. How the number of neutrons is calculated? Ans. Number of neutron= Atomic mass-Atomic number 

44

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Artificial Intelligence PARAMETERS

DESCRIPTION

Chapter Covered

Chapter 4: Structure of the Atom

Name of the Book

Science, Class 9, NCERT

Subject and Artificial Atomic structure integrated with interactive simulations. Intelligence Integrated

Learning Objectives

To familiarize students with the concept of electronic configuration so that they are able to place the subatomic particles correctly in an atom and distribute electrons in shells.

Time Required

40 min

Classroom Arrangement

Flexible

Material Required

Computer, Projector, laptop, markers. Chart papers, Coloured Post-its sketch pens etc Paper cutouts with name of elements and the number of protons neutrons and electrons in it. One computer per pair of student (Class can be taken in Computer lab).

AI CONCEPTS INTEGRATED

Scan to know more about this topic

Pre-Preparation Activities Divide the class in small groups and assigning them places in the computer lab. Providing them list of elements . Previous Knowledge

Methodology

Learning Outcomes

Follow up Activities

Reflections

Students know about the scientific theory of atoms(also known as atomic theory) by describing changes in the atomic model over time and why those changes were necessitated by experimental evidence. Students know that the max number of electrons in a shell is equal to 2n2 where n= shell number 1,2,3,.. Students will be seated in the computer lab in pairs with one computer per pair (ideally) The teacherrevises the pre knowledge using a small video after which the game can start Video link for revising the pre knowledge is given. (10 min) https://www.youtube.com/watch?v=Q3htqdKrHOk&t=260s Teacher (after the end of the video) can ask some probing questions to the students. ● What is the total charge of an atom? ● What is the function of neutrons in an atom? ● Can you name some conditions under which we can say that an atom is stable? ● What is your understanding of the term Electronic Configuration? ● Each pair writes their responses on a coloured post its and pastes it on a chart placed at the corner of the class. (Discussion about responses) (10MIN) Teacher then hands over the sheet of paper to each pair (on which the name of elements with their number of protons, neutrons and electrons are printed) Teacher asks each pair to make atoms using simulation (15 MIN). Teacher sums up the session announcing the follow up activity and HW (5 min). Students will be able to understand the meaning of electronic configuration and distribute electrons in various shells. Students would be able to virtually construct their own atom using simulation.

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Students would draw the structure of at least 2 atoms in their notebooks. (encourage them to use colouredpens/coding for representing protons, neutrons and electrons). Research Question: According to 2n2 rule the third shell should have 18 electrons but it actually has 8. What might be the reason for that? Feedback of simulation tool from students and where it can be used further? 

SELF ASSESSMENT PAPER - 01 MM: 30

Maximum Time: 1 hour

Q. 1. The number of electrons in an element X is 15 and the number of neutron is 16. Which of the following is the correct representation of the element? (A)

31 15

X

(B)

31 16

X

(C)

16 15

X

(D)

15 6

X

[1]

OR The property to flow is unique to fluids. Which one of the following statement is correct? (A) Only gases behave like fluids. (B) Gases and solids behave like fluids (C) Gases and liquids behave like fluids. (D) Only liquids are fluids [1] Q. 2. Rusting of an article made up of iron is called. (A) corrosion and it is a physical as well as chemical change. (B) dissolution and it is a physical change. (C) corrosion and it is a chemical change. (D) dissolution and it is a chemical change. [1] Q. 3. Which of the following are homogeneous in nature? (i) Ice (ii) Wood (iii) Soil (iv) Air (A) (i) and (iii) (B) (ii) and(iv) (C) (i) and(iv) (D) (iii) and(iv) [1] In the following questions a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as : (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true but Reason (R) is false. (D) Assertion (A) is false but Reason (R) is true. Q. 4. Assertion : Alloys are homogeneous mixture of metals. Reason : Alloys cannot be separated into their components by physical methods. [1] Q. 5. Assertion : Naphthalene ball disappears with time without leaving any solid residue. Reason : Naphthalene balls gets converted into vapours due to evaporation. [1] The following data represents the distribution of electrons, protons and neutrons in atom of four elements A,B,C,D. Understand the data carefully and the following question: [5] Element

Protons

Neutrons

Electrons

A

9

10

9

B

16

16

16

C

12

12

12

22

17

D 17 Q. 6. Select the correct electronic distribution of Element B (A) 2,8,1 (B) 2,8,2 (C) 2,8,4 (D) 2,8,6 Q. 7. The Valency of element A : (A) 4 (B) 1 (C) 2 (D) 3 Q. 8. The Atomic number of element B is : (A) 9 (B) 16 (C) 12 (D) 17

46

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Q. 9. The Mass number of elements D is : (A) 17 (B) 22 (C) 39 (D) 14 Q. 10. The number of valence electrons is element C : (A) 8 (B) 2 (C) 12 (D) 4 Q. 11. Arrange the following substance in the increasing order of force of attraction between their particle: Oxygen, Sugar, Water. [1] Q. 12. Define the term molecular mass. [1] Q. 13. Name the particles which determine the mass of an atom.[1] Q. 14. Calculate the molar mass of. (i) Ethyne (C2H2) (ii) Phosphorus molecule (P4) (Atomic mass of C = 12u, H = lu and P = 3lu) [2] Q. 15. Write any two differences between physical and chemical changes. [2] Q. 16. Give reasons: (i) Solids are incompressible (ii) A gas completely fills the vessel in which it is kept. [2] Q. 17. (i) Why chemical properties of all isotopes of an element are same? (ii) Name the isotopes used in the treatment of Goitre and Cancer. (iii) An element ‘X’ has 2 electron is its M-shell. What is its atomic number? [3] Q. 18. (i) How tincture of iodine is prepared? (ii) Define solubility (iii) What would happen if you to take a saturated solution at a certain temperature and cool it slowly? [3] Q. 19. Describe Rutherford’s a-particle scattering experiment and mention the important observations and conclusions drawn from this experiment. [5]



https://qr.page/g/2pw02IBpjvv

UNIT-II

Organisation in the Living World

CHAPTER



5

Syllabus

Study Time Reading Time: 4 Hr No. of Questions: 69

CELL— BASIC UNIT OF LIFE

Cell as a basic unit of life,, prokaryotic and eukaryotic cells; multicellular organisms; cell wall and cell membrane; cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus, nucleus, chromosomes-basic structure and number. TOPIC - 1 Cell-Prokaryotic Eukaryotic and multicellular organisms .... P. 47

Topic-1

Cell-Prokaryotic, Eukaryotic and multicellular organisms

Revision Notes

TOPIC - 2 Cellular components .... P. 54

Scan to know more about this topic

 In 1665, Robert Hooke first discovered and introduced the term 'cell'.  Cell is the structural and functional unit of all living organisms.  Organisms may be unicellular or multicellular. A single cell constitutes the unicellular organism whereas many cells coordinately function in case of multicellular organism. Microscope  The size, shape and volume of the cells are related to the specific function that they perform.  A cell generally shows plasma membrane, nucleus and cytoplasm. Mnemonics  In 1674, Leeuwenhoek observed the living cells in protists and bacteria. Concept: Cell Organelles present in cell are:  In 1831, Robert Brown discovered the nucleus in the Mnemonics: Cup Mein Garam Coffee Pina cell.  Purkinje coined the term “Protoplasm” for the fluid Roz Verna Egg Lena substance of the cell in 1839. Interpretation  The cell theory that all the plants and animals are C: Cytoplasm (Cup) composed of cells and the cell is the basic unit of life M: Mitochondria (Mein) was proposed by Schleiden and Schwann .  Virchow (1855) expanded the cell theory by suggesting G: Golgi bodies (Garam) that all cells arise from pre-existing cells. C: Centrosome (Coffee)  Plasma membrane is the outer covering of the cell P: Plastids (Pina) which separates the cellular components from the R: Ribosomes (Roz) external environment. V: Vacuoles (Varna)  In plants, cell wall is present external to the cell membrane which is rigid and made up of cellulose. E: Endoplasmic reticulum (Egg)  Prokaryotic cell: The primitive cells in which well L: Lysosomes (Lena) defined nuclear membrane is absent and genetic material lies as a single chromosome, i.e., nucleoid is known as a prokaryotic cell. The cell locks membrane - bound cell organelles. Nucleoid as hereditary material lies freely in the cytoplasm. For, e.g., Bacteria, Cyanobacteria.

cell (plasma) membrane

smooth ER (no ribosomes)

rough ER (endoplasmic reticulum)

lysosome golgi vesicles

pinocytotic vesicle

Gave cell theory and stated that all plants and animals are composed of cells

Discovered nucleus in cell in 1831.

In 1665, he had discovered cell in cork-slice with the help of a primitive microscope.

ribosome

cytoplasm

microtubules

centrioles (2) Each composed of 9 microtubule triplets

nucleus

nucleolus

golgi apparatus

mitochondrion

Cell - Basic Unit of Life

Mitochondria

Lysosomes

Cell Organelles

Golgi Apparatus

Endoplasmic reticulum

Vacuoles

Chloroplast

Animal Cell

Cell Theories

Types of cell

Eukaryotic cell

Plastids

Plant Cell

Plasma membrane or cell membrane

Cell wall

Nucleus

Nuclear Membrane

Chromatin network

Nucleolus

Second Level

Trace the Mind Map Third Level

Membrane that separates the cellular components from the outer environment.

It is present in only plant cells, external to the cell membrane.

First Level 

Prokaryotic cell



expended the cell theory in the year 1855 by suggesting that all the cells arise from pre-existing cell



48 Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

CELL — BASIC UNIT OF LIFE

49

 Eukaryotic cell: The cells that have well-defined nucleus and membrane bound organelles are celled eukaryotic cells. The hereditary material is covered by nuclear envelope. For, e.g., Plants, animals and protozoans.  Unicellular and Multicellular organisms: Single celled organisms are known as unicellular organisms for, e.g., protozoans, Chlamydomonas, bacteria etc., and they perform all the life processes like digestion, respiration, excretion and reproduction by a single cell. Multicellular organisms are made up of a number of cells specialized for performing different functions. They consists of tissues, organs and organ systems. For, e.g., Plants, Animals, Fungi etc. Difference between plant cell and Animal Cell:

Plant Cell

Animal Cell

1. Cellulose cell wall present external to cell 1. membrane

No cell wall. Outermost structure is cell membrane or plasma membrane.

2. Vacuoles are usually large

2.

Generally vacuoles are absent and if present they are usually small.

3. Plastids present

3.

Plastids absent

4. Centriole absent

4.

Centriole present

5. Golgi body present in the form of units known as 5. dictyosomes.

Golgi body well developed having 2 cisternae.

Key Words

 Cell: An autonomous self-replicating structure that forms the structural, functional and biological unit of all living organisms.  Prokaryotic cell: A cell characterised by the absence of a distinct, membrane-bound nucleus or membranebound organelles, and by DNA that is not organized into chromosomes. Scan to know  Nucleoid: An undefined nuclear region of the prokaryotic cell, containing the genetic more about material (nucleic acids). this topic  Eukaryotic cell: A cell containing a membrane-bound nucleus and membrane-bound organelles.  Unicellular organism: Organism having only one cell.  Multicellular organism: Organism consisting of more than one cell, where in the differentiated Preparation of Onion Peel slide cells perform specialised functions in the organism.

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions 

(1 mark each)

Q. 1. Which of the following can be made into a crystal? (A) A bacterium (B) An Amoeba (C) A virus (D) A sperm U Ans. Option (C) is correct. Explanation: A virus crystal is a collection of thousands of viruses.

Q. 2. The undefined nuclear region of prokaryotes are also known as: (A) Nucleus (B) Nucleolus (C) Nucleic acid (D) Nucleoid K Ans. Option (D) is correct. Explanation: The undefined nuclear region of a prokaryotic cell is called nucleoid. The prokaryotic cells lack true nucleus. A circular DNA lies naked in the cytoplasm.

50

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Q. 3. Amoeba acquires its food through a process, termed: (A) Exocytosis (B) Endocytosis (C) Plasmolysis (D) Both exocytosis and endocytosis U Ans. Option (B) is correct. Explanation: Amoeba acquires its food through a process, termed endocytosis. Endocytosis is the ingestion of material by the cells through their plasma membrane. Q. 4. Cell arises from pre-existing cell was stated by: (A) Haeckel (B) Virchow (C) Hooke (D) Schleiden K Ans. Option (B) is correct. Explanation: Schleiden (1836) and Schwann (1834) gave the cell theory, which was further refined by R. Virchow (1855). Virchow presented the idea that all cells arise from pre-existing cell. Q. 5. Cell theory was given by: (A) Schleiden and Schwann (B) Virchow (C) Hooke (D) Haeckel K Ans. Option (A) is correct. Explanation: Schleiden (1836) and Schwann (1834) gave the cell theory which states that all the plants and animals are composed of cells and cell is the basic unit of life. Q. 6. The only cell organelle seen in prokaryotic cell is: (A) Mitochondria (B) Ribosome (C) Plastids (D) Lysosomes A Ans. Option (B) is correct. Explanation: A prokaryotic cell lacks membranebound organelles like plastids, mitochondria and endoplasmic reticulum but smaller and randomly scattered ribosomes are seen. Q. 7. Living cells were discovered by: (A) Robert Hooke (B) Purkinje (C) Leeuwenhoek (D) Robert Brown K Ans. Option (C) is correct. Explanation: In 1674, Antonie Van Leeuwenhoek discovered the living cell.

B

Assertion and Reason 

(1 mark each)

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A)  Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion: Cells are the structural and functional unit of life. Reason: A single cell has the ability to form whole organism. Ans. Option (B) is correct. Explanation: Cells are the functional unit of life because they can perform all the activities required to sustain life. New cells are formed by division of pre-existing cells. A single cell has the ability to form whole organism for, e.g., Amoeba. Q. 2. Assertion: A single cell lives independently on its own. Reason: Single cell can perform all the life processes. U Ans. Option (A) is correct. Explanation: A single cell or unicellular organism like Amoeba and Paramecium can perform all the life processes and therefore can live independently on their own. Q. 3. Assertion: Multicellular organisms have higher survival rate than unicellular organisms. Reason: In Multicellular organism, dead cells are replaced by new cells. U Ans. Option (A) is correct. Explanation: Multicellular organisms are made up of many cells. Death or injury of any cells can be replaced by new cells. Thus, because of increased efficiency, organization and division of labour, multicellular organisms have higher survival rate. Q. 4. Assertion: Not all cells of human body look alike in terms of structure. Reason: Structure of cell is determined by the function it performs. U Ans. Option (A) is correct. Explanation: Yes, not all cells in our body look alike in terms of shape, size and structure. The shape, size and structure of cells are determined by the function they perform.

CELL — BASIC UNIT OF LIFE

51

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions Q. 1. Name the process by which Amoeba acquires its food.



Ans. Chlamydomonas and bacteria.[1]  [CBSE Marking Scheme, 2016]

U [Board Term-I, 2016]

OR Name the process through which an Amoeba acquires its food from the external environment. (Board Term-I 2013, 2016 )

Ans. Endocytosis. [CBSE Marking Scheme, 2016] [1] Q. 2. Name the process by which unicellular freshwater organisms and most plant cells tend to gain water. U [Board Term-I, 2016] Ans. Endosmosis. When cell is kept in hypotonic solution, it will gain water. [CBSE Marking Scheme, 2016] [1] Q. 3. Identify the single celled organisms from the following: Cockroach, Chlamydomonas, snake, mosquito, bacteria. A [Board Term-I, 2016]

(1 mark each)

Q. 4. List any two single celled (unicellular)

organisms. K Ans. Amoeba and Euglena. [½ + ½] Q. 5. Name: (i) the cells which have changing shape, (ii) the cells which have a typical shape. U

Ans. (i) Amoeba, (ii) Nerve cell.

[½ + ½] [CBSE Marking Scheme, 2012]

Q. 6. Mention the difference between Prokaryotes and Eukaryotes in terms of nuclear region present in them. K Ans. Prokaryotic cells do not have a well defined nuclear region. Genetic material is present as nucleoid whereas eukaryotic cells have a well defined nucleus. [½ + ½]

Short Answer Type Questions-I Q. 1. Cell size may range from a few micro metre to a metre. Support this statement with the help of examples. Ap [Board Term-I, 2014, DDE-2014) Ans. Many cells are visible only under a microscope e.g., Mycoplasma is the smallest cell and longest cell in human body is nerve cell or neuron.[2] [CBSE Marking Scheme, 2014] Q. 2. What is the fundamental unit of life ? Who discovered it ? How can they be observed ? U [NCERT] (DDE 2017) [Board Term-I, 2016]  Ans. Cell. Robert Hooke. Under a microscope. Some are big enough, can be seen with naked eye like egg.  [CBSE Marking Scheme, 2016] [1 + 1] Q. 3. Can a single cell live independently on its own? Explain by giving example. 

Ap [Board Term-I, 2016]

Ans. Yes, Unicellular organisms. Single cell can perform all the life processes. 

Amoeba, Paramecium etc. [CBSE Marking Scheme, 2016] [2]

(2 marks each)

Commonly Made Error

Students often forget to underline biological name of organism.

Answering Tip

Stress upon writing scientific names correctly

according to Binomial Nomenclature. Q. 4. Bacteria do not have chloroplast but some bacteria are photoautotrophic in nature and perform photosynthesis. Which part of bacterial cell performs this?  [NCERT Exemplar] Ans. Bacterial cell do not have chloroplast but yet some photoautotrophic bacteria perform photosynthesis due to the presence of chlorophyll incorporated in the membrane. In the cell membrane, there are reaction centers which specifically absorb light energy.  [2] Q. 5. Draw and label the parts of prokaryotic cell. R Ans.

Detailed Answer:  Yes, single-celled (unicellular) organisms can live independently. Unicellular organisms such as Amoeba, Paramecium, carry out digestion, respiration, excretion and reproduction on their own. 

Prokaryotic cell (Any 4 labelling) [1 + 1 = 2]

52

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Short Answer Type Questions-II Q. 1. What are living organisms made up of ? Why are they called structural and functional unit of life ? U [Board Term-I, 2016] [NCERT] Ans. Cell. Basic building blocks of all living organisms. Perform all the activities required to sustain life.  [CBSE Marking Scheme, 2016] [1 + 2 = 3] Q. 2. (a) Write two differences between nuclear region of a bacterial cell and nuclear region of an animal cell. (b) Which structures present in the nuclear region of a living cell bear genes ?  U [Board Term-I, 2012] Ans. (a) S. Nuclear region of Nuclear region No. bacterial cell of an animal cell (i)

Poorly defined and lacks nuclear membrane.

Well defined and membrane bound.

(ii)

Has single chromosome.

Has more than one chromosome.

(3 marks each)

Commonly Made Error

Students fail to draw a neat diagram. Many of

them label the parts incorrectly. Spelling error is commonly seen while labelling.

Answering Tip

Practice self-explanatory diagrams of plant and

animal cell with proper labelling, arrows and headings. Q. 4. Draw a diagram of a plant cell and label it’s any four parts. Ans.

(iii) Lacks true organelles.

Well defined membrane bound cell organelles present. (b) Chromosomes bear ‘genes’. [2 + 1 = 3]

Commonly Made Error

Students often get confused between nuclear region of bacterial cell and nuclear region of animal cell. They write opposite answers.

Ans. Differences between Eukaryotic cell:

Answering Tip

Lay stress on writing compatible differences and in tabular form.

Q. 3. Draw a neat diagram of an animal cell and label any four parts of it.

Ans.

(Any 4 labelling) [1 + ½ × 4] Q. 5. Write three differences between Prokaryotic and Eukaryotic cells. U OR How is prokaryotic cell different from a eukaryotic cell ? [NCERT] [KVS, 2018, 19]

[NCERT]

U [NCERT Exemplar]

Animal Cell (Any 4 labelling) [1 + ½ × 4]

S. Prokaryotic cell No. (i) Size: Generally small (1-10 µm) 1 µm = 10–6 m. (ii) Nuclear region: Contains only nucleic acid and is undefined due to the absence of nuclear membrane and is known as nucleoid. (iii) Membrane-bound cell-organelles absent.

Prokaryotic

and

Eurkaryotic cell Size: Generally large (5-100 µm) Nuclear region: Well defined and surrounded by a nuclear membrane.

Membrane bound cell organelles (e.g., chloroplasts, golgi bodies etc.) present.

[Any three] [1 × 3 = 3] [CBSE Marking Scheme, 2012]

CELL — BASIC UNIT OF LIFE

Commonly Made Error

Answering Tip

Students often fail to write proper differences.



Many of them forget to write the term nucleoid, while explaining nuclear region of prokaryotic cells.

Learn the differences between prokaryotic

and eukaryotic cell with diagram for proper understanding and retention.

Long Answer Type Questions Q. 1. On the basis of number of cells, living organisms are classified as unicellular and multicellular. (i) Name two unicellular organisms. (ii) What is meant by division of labour in multicellular organisms ? (iii) Name one prokaryotic and one eukaryotic unicellular organism. (iv) ‘Every multicellular organism has come from a single cell.’ Justify this statement. (v) Write one common feature between an Amoeba and white blood cells of humans. 

U+A [Board Term-I, 2016]

Ans. (i) Amoeba, yeast, Euglena (Any two) (ii) Cell → Tissue → Organ → Organ system → Organism (iii) Prokaryotic - Bacteria, Eukaryotic - Amoeba (iv) Yes, every cell of the multicellular organism has come from a single cell. After fertilisation, single cell is formed. The zygote is actually a single cell. The zygote gives rise to all the other cells of our body. (v) Do not have any shape.  [CBSE Marking Scheme, 2016] [2 + 1 + 2 = 5]

(5 marks each)

(iii) Prokaryotic - Bacteria, Eukaryotic - Amoeba. (iv) Yes, every cell of the multicellular organism has come from a single cell. After fertilisation, single cell is formed. The zygote is actually a single cell. The zygote gives rise to all the cells of our body. (v) Both, Amoeba and white blood cells of humans do not have any fixed shape. Q. 2. Do all cells of our body look alike in terms of shape, size and structure ? What similarities do they have? Illustrate by drawing diagrams of various cells present in human body. 

A [Board Term-I, 2016]

Ans. No, not all cells in our body look alike in terms of shape, size and structure. The shape, size and structure of cells are determined by the function which they performs.

ovum (egg-cell) muscle cells

red blood cells epithelial cell

Detailed Answer: (i) Amoeba and organisms.

Euglena

53

are

unicellular

(ii) Multicellular organisms are made up of millions and trillions of cells. All these cells perform specific functions. All the cells specialised for performing similar functions are grouped together as tissues in the body. Hence, a particular function is carried out by a group of cells at a definite place in the body. Similarly, different functions are carried out by different groups of cells in an organism. This is known as division of labour in multicellular organisms.

nerve cell

sperm

white blood cells

Different Types of Cell The eggs of many animals are spherical in shape, some are oval in shape, smooth muscle fibres are spindle shaped, nerve cells are elongated and RBCs are discoidal in shape. The size and number of cells also vary from organism to organism. Similarities: All cells have the organelles within them. [CBSE Marking Scheme, 2016] [2 + 1 + 2 = 5]

Q. 3. Make a comparison and write down ways in which plant cells are different from animal cells. U [NCERT] Ans. Differences between plant cells and animal cells: S. No.

Plant Cells

Animal Cells

(i)

Plant cells are larger in size.

(ii)

They contain cell wall made of cellulose, which Cell wall is absent. is present outside the plasma membrane.

Animal cells are comparatively smaller in size.

54

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

S. No.

Plant Cells

Animal Cells

(iii)

They contain plastids, i.e., chloroplast, leucoplast and chromoplast.

Plastids are absent.

(iv)

Centrosome is absent.

Centrosome is present.

(v)

Larger vacuoles are present.

Vacuoles either absent or very small in size.

(vi)

Food is stored in the form of starch.

Food is stored in the form of glycogen.

(vii)

Lysosomes either absent or very few in number.

More number of prominent lysosomes are present.



(Any five) [1 × 5 = 5]

Topic-2 Cellular components Revision Notes  Plasma membrane is a thin, selectively permeable membrane, covering the cell and is made up of lipids and proteins. Scan to know more about  Functions of plasma membrane: this topic (i) It separates the contents of a cell from its outside environment.

(ii) It regulates the flow of substances to and from the cell through diffusion, facilitated diffusion, active transport and endocytosis.  Osmosis is diffusion of water through a selectively permeable membrane.  If a cell is placed in different solutions: (i) Hypotonic solution: A cell placed in it will gain water. (ii) Hypertonic solution: A cell placed in it will lose water, also known as plasmolysis.

Plasma Membrane Scan to know more about this topic

(iii) Isotonic solution: A cell placed in it will neither gain nor lose water.  To know about the structure of chromosomes and their significance. in cell division. • Chromosomes are rod like or thread like structure which are formed by the condensation of chromatin fibers during cell division and reproduction. Fundamental Unit of Life • A chromosome consists of two similar threads called chromatids attached to each other at a point called centromere. • Chromosomes are made up of DNA molecules and proteins which carry and help to transfer information for inheritance of characters from parents to next generation. • Human has 46 pairs of chromosomes. Each parent contributes 23 chromosomes.  Cells of plants, fungi & bacteria: Contain both plasma membrane and cell wall. Cell wall is rigid, non-living and outer most covering, composed mainly of cellulose.  When placed in hypertonic solution, a living plant cell shows plasmolysis.  Cell wall provides mechanical strength to the cell. It permits the cell to withstand huge changes in the surrounding medium. Cell wall of plants, fungi consists of cellulose and chitin respectively. Bacterial cell is made up of peptidoglycan • Nucleus is an important, spherical, usually centrally located constituent of the cell and is bounded by double layered nuclear envelope. • The nucleus of a dividing cell shows rod-shaped chromosomes, made up of DNA and proteins. In a nondividing cell, the chromosomes elongate and take the form of thread-like chromatin. • DNA molecules are responsible for transmitting hereditary information from one generation to the next.  Nucleus controls all metabolic activities of the cell.  Depending on the presence or absence of nucleus, cells may be prokaryotic or eukaryotic. (i) Prokaryotic cells lack a well-defined nucleus and instead show nucleoid, an undefined nuclear region containing the genetic material. (ii) Eukaryotic cells possess a proper nucleus with nuclear membrane.

CELL — BASIC UNIT OF LIFE

55

 Cytoplasm is the fluid content of the cell, occurring between nucleus and plasma membrane. It stores several vital chemicals and is the site of certain important metabolic pathways. • Several specialised cell organelles are present in the cytoplasm. These organelles perform different kinds of metabolic activities and are kept separate from each other.  The various cell organelles include endoplasmic reticulum, golgi apparatus, lysosomes, mitochondria, plastids, vacuoles and centrosome.  Endoplasmic reticulum (ER) is an extensive, interconnected, membrane bound network of tubes and sheets.  Ribosomes are attached to the surface of Rough Endoplasmic Reticulum (RER) and are absent in Smooth Endoplasmic Reticulum (SER). • Functions of Endoplasmic Reticulum (ER): (i) It synthesises important proteins (RER) and lipids (SER). (ii) It provides a pathway for intracellular transport of materials. (iii) SER of liver cells is important for detoxification. • Ribosomes are the sites of protein synthesis. Golgi apparatus  Golgi apparatus is a network of stacked, flattened, membrane bound sacs and vesicles. • Golgi apparatus carries out the storage, modification and packaging of substances manufactured in the cell and is also involved in lysosome formation.  The spherical, sac-like lysosomes contain powerful digestive enzymes and form the waste disposal system of the cell. They are also known as ‘suicidal bags’. Mitochondria • Mitochondria and plastids are covered by two membranes and possess their own DNA and ribosomes. • Mitochondria are the ‘power house of the cell’, providing energy for various metabolic activities. Plastids • Chromoplasts and leucoplasts are the two types of plastids present in plant cells. • Chloroplasts are chromoplasts containing chlorophyll and carry out photosynthesis in plants. • Leucoplasts store starch, oil and protein granules. Vacuole  The large central vacuole of mature plant cells contains cell sap which provides turgidity to the cell and also stores important substances.  In unicellular organisms, vacuoles play an important role in nutrition and osmo-regulation.  Centrosome is found only in animal cells and consists of two centrioles. Centrosome helps in cell division.  The membrane-bound cell organelles are absent in prokaryotic cells.  The basic structural organisation of the cell helps it to perform important functions like respiration, nutrition, excretion and protein synthesis.

Key Words

 Diffusion: The spontaneous movement of a substance from a region of its higher concentration to a region of its lower concentration.  Osmosis: The movement of water through a semi-permeable membrane from a region of high water concentration to a region of low water concentration.  Hypertonic solution: A solution that has a higher solute concentration than the one to which it is compared.  Hypotonic solution: A solution that has a lower solute concentration than the one to which it is compared.  Isotonic solution: A solution that has the same tonicity as another solution with which it is compared.  Plasmolysis: Shrinkage or contraction of the protoplasm away from the wall of a living plant or bacterial cell, caused by loss of water through osmosis.  Cell organelle: A specialised sub-unit within a cell that has a specific function, and is usually enclosed within its own membrane.  Genes: A hereditary unit consisting of a sequence of DNA that occupies a specific location on the chromosomes and determines a particular characteristic in an organism.  Membrane biogenesis: The process of synthesising the biological membranes.  Plasma membrane: The thin, selectively permeable membrane composed of lipids and proteins which surrounds an entire cell and regulates the flow of substances to and from the cell.  Cell wall: The rigid, non-living, outer covering of certain cells (like plant and bacteria), composed mainly of cellulose. It provides the cell with structural support and protection.  Cytoplasm: The jelly like material of a cell that is enclosed within the plasma membrane, except the nucleus and contains the cell organelles.

56

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Example

Define osmosis. In what two ways it is different from diffusion ? Solution:  Osmosis is the movement of solvent (usually water) from a region of high water concentration to a region of low water concentration; it takes place through semi-permeable membrane whereas the diffusion does not require any membrane. In osmosis movement of solvent is involved whereas in diffusion movement of solid, liquid and gases are involved.  Osmosis is the process in which there is a movement of solvent (usually water) from a region



of high water concentration to a region of low water concentration. Differences between osmosis and diffusion: S.No.

Osmosis

Diffusion

(i)

It takes place through semipermeable membrane.

The diffusion does not require any membrane.

(ii)

Movement of solvent is involved.

Movement of solid, liquid and gases are involved.

OBJECTIVE TYPE QUESTIONS Choice Questions A Multiple  (1 mark each) Q. 1. A cell will swell up if: (A) The concentration of water molecules in the cell is higher than the concentration of water molecules in surrounding medium. (B)  The concentration of water molecules in surrounding medium is higher than water molecules concentration in the cell. (C)  The concentration of water molecules is same in the cell and in the surrounding medium. (D) Concentration of water molecules does not matter. U Ans. Option (B) is correct. Explanation: When concentration of water molecules in surrounding medium is higher than water molecules concentration in the cell, the cell will swell up due to endosmosis. Q. 2. Which of these options are not the functions of ribosomes? I. It helps in molecules.

manufacture

of

protein

II. It helps in manufacture of enzymes.

III. It helps in manufacture of hormones. IV. It helps in manufacture of starch molecules. (A) I and II (B) II and III (C) III and IV (D) IV and I K Ans. Option (C) is correct. Explanation: Ribosome plays an important role in synthesis of proteins and help in manufacture of Enzymes.

Q. 3. Which of these is not related to endoplasmic reticulum? (A) It behaves as a transport channel for proteins between nucleus and cytoplasm. (B)  It transport materials between various regions in cytoplasm. (C) It can be the site of energy generation. (D)  It can be the site for some biochemical activities of the cell. A Ans. Option (C) is correct. Explanation: Mitochondria are the site of energy generation. Q. 4. Following are a few definitions of osmosis. Read carefully and select the correct definition. (A) Movement of water molecules from a region of higher concentration to a region of lower concentration through a semi permeable membrane. (B)  Movement of solvent molecules from its higher concentration to lower concentration. (C) Movement of solvent molecules from higher concentration to lower concentration of solution through a permeable membrane. (D) Movement of solute molecules from lower concentration to higher concentration of solution through a semi permeable membrane. U Ans. Option (A) is correct. Explanation: The movement of water molecules through selectively permeable membrane along concentration gradient is called osmosis. Q. 5. Plasmolysis in a plant cell is defined as (A) Break down (lysis) of plasma membrane in hypotonic medium.

CELL — BASIC UNIT OF LIFE

(B)  Shrinkage of cytoplasm in hypertonic medium. (C) Shrinkage of nucleoplasm. (D) None of the above. U Ans. Option (B) is correct. Explanation: Plasmolysis in a plant cell is defined as shrinkage of cytoplasm in hypertonic medium. When a cell is immersed in hypertonic (very concentrated) solution, water diffuses out of the cell by the process of exosmosis. As a result, the cell shrinks. If exosmosis continues in a plant cell, the cytoplasm appears to be shrunken. This is called plasmolysis and the cell is said to be plasmolysed. Q. 6. Which of the following are covered by a single membrane? (A) Mitochondria (B) Vacuole (C) Lysosome (D) Both (B) and (C) K Ans. Option (D) is correct. Explanation: Vacuole and lysosome are covered by a single membrane while mitochondria, and plastid have double membrane. Q. 7. Which cell organelle plays a crucial role in detoxifying many poisons and drugs in a cell? (A) Golgi apparatus (B) Lysosomes (C) Smooth endoplasmic reticulum (D) Vacuoles K Ans. Option (C) is correct. Explanation: SER metabolises various toxic or poisonous substances such as drugs, aspirin, insecticides (DDT), petroleum products and pollutants. Q. 8. The proteins and lipids, essential for building the cell membrane, are manufactured by (A) Endoplasmic reticulum (B) Golgi apparatus (C) Plasma membrane (D) Mitochondria K Ans. Option (A) is correct. Explanation: Rough Endoplasmic Reticulum (RER) is associated with the synthesis of proteins while Smooth Endoplasmic Reticulum (SER) is associated with the synthesis of lipids. Q. 9. Select the odd one out. (A)  The movement of water across a semi permeable membrane is affected by the amount of substances dissolved in it. (B) Membranes are made of organic molecules like proteins and lipids. (C) Molecules soluble in organic solvents can easily pass through the membrane. (D) Plasma membranes contain chitin sugar in plants. U

57

Ans. Option (D) is correct. Explanation: Each cell is bounded by an extremely delicate, thin elastic living membrane, called the plasma membrane. The plasma membrane is made up of two layers of lipid (fat) molecules with protein molecules which is sandwiched between lipid layer.

B

Assertion and Reason 

(1 mark each)

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion: Human RBCs burst in hypotonic solution. Reason: Cells burst due to endosmosis. U Ans. Option (A) is correct. Explanation: Human red blood cells when they are placed in hypotonic solution, they burst due to endosmosis. Q. 2. Assertion: Mitochondria are semi-autonomous cell organelle. Reason: Mitochondria generate energy. K Ans. Option (B) is correct. Explanation: Mitochondria have their own DNA and ribosome to make proteins. It can replicate independent of nuclear DNA. Hence, it is known as semi-autonomous organelle. It is also associated with ATP production. Q. 3. Assertion: Chloroplast performs photosynthesis. Reason: Chloroplast comprises photosynthetic pigments. U Ans. Option (A) is correct. Explanation: Chloroplast is involved in the process of photosynthesis. It comprises a photosynthetic pigment called chlorophyll, which traps energy from the Sun. Q. 4. Assertion: Lysosomes are known as suicidal bag of cells. Reason: Lysosomes contain powerful enzymes capable of breaking down all organic material.  U Ans. Option (A) is correct. Explanation: During the disturbance in cellular metabolism lysosomes may burst and the enzymes digest their own cell. Therefore, lysosomes are also known as suicidal bags. Lysosomes are able to do this because they contain powerful enzymes capable of breaking down all organic material.

58

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions Q. 1. Describe the process of shrinking of a cell on being put in a strong salt solution.



U [Board Term-I, 2016]

Ans. Concentration of water inside the cell is greater.

By osmosis, water moves out and causes shrinkage of a cell called plasmolysis. [CBSE Marking Scheme, 2014, 2016] [1]

Q. 2. What is the primary function of leucoplasts ? 



K [Board Term-I, 2015] OR State the role of leucoplasts in plant cells. [Board Term-I, 2016]

Ans. They store starch, oils, protein granules.

[CBSE Marking Scheme, 2016] [1]

Q. 3. State any one function of Golgi apparatus. 

K [Board Term-I, 2016]

Ans. Golgi bodies stores proteins and enzymes. It helps in modification and packaging of products.  [1]

Q. 4. Name the plastid which gives red colour to tomato and purple colour to Brinjal .

K [Board Term-I 2015]

OR Name the plastid involved in conversion of a green brinjal to violet. Ans. Chromoplast. [1] Q. 5. Expand ATP. Where it is produced ?



Where is it prepared ? U [Board Term-I, 2013]

Ans. Ribosomes is a cell organelle that lacks membrane. It is prepared in the nucleolus.  [CBSE Marking Scheme, 2013] [1 + 1] Q. 2. What is endoplasmic reticulum ? Name the two types of endoplasmic reticulum. Write its main functions.  U [NCT 2014; 11] Ans. Endoplasmic reticulum is a membranous network enclosing a fluid-filled lumen. The two types of endoplasmic reticulum are Rough Endoplasmic Reticulum (RER) and Smooth Endoplasmic Reticulum (SER). RER has ribosomes attached to its surface. The ribosomes take part in protein synthesis. SER does not have any ribosomes on it and secretes lipids. Some proteins and lipids synthesised in ER are used for producing new cellular parts, specially the cell membrane, by biogenesis.  [CBSE Marking Scheme, 2014] [½ + 1 + ½]

U [NCT 2014]

Ans. Adenosine Triphosphate. It is produced in 

mitochondria. [CBSE Marking Scheme, 2014] [½ + ½]

Q. 6. State the significance of membrane biogenesis. U [DDE 2014]

Ans. Membrane biogenesis involves the synthesis of cell membrane with the help of proteins and lipids. The smooth endoplasmic reticulum helps in the manufacture of fats or lipids important for cell function and building cell membrane. [1] Q. 7. Identify the cell organelle which is known as the power house of the cell. State reason.   U [NCT-2014] Ans. Mitochondria; as energy is released from it.  [½ + ½] Q. 8. Name the cell organelle that is commonly termed as suicidal bags of the cell. [NCERT]  [Board Term-I 2013] [NCT-2014]  U [NCERT Exemplar] [DDE 2017] Ans. Lysosomes. [1]

Short Answer Type Questions-I Q. 1. Name a cell organelle which lacks membrane.

(1 mark each)

(2 marks each)

Q. 3. What is plasma membrane ? What are its functions ?  U Ans. Plasma membrane also called as cell membrane, is the outer covering of a cell that separates its contents from the surrounding medium. It is made up of lipids and proteins, and provides a mechanical barrier to protect the inner contents of the cell. It encloses the nucleus and cytoplasm of the cell.[1 + ½ × 2] Q. 4. What do you mean by a nucleoid ? U Ans. In prokaryotes and lower organisms like bacteria, the nuclear region of the cell may be poorly defined because of the absence of a nuclear membrane. Such an undefined and incipient nucleic region containing only naked nucleic acids without any membrane covering them is called a nucleoid. [2] U Q. 5. Write the function of chromatin material. Ans. The chromatin material mainly consists of deoxyribonucleic acid (DNA) which stores and transmit the hereditary information from one generation to another.[1 + 1]

CELL — BASIC UNIT OF LIFE

Q. 6. Write the main functions of cell wall. U Ans. (i) Cell wall provides shape as well as rigidity to the cell. (ii) It protects the protoplasm.

(iii) It is involved in the movement of materials in and out of the cell. (iv) Growth of cell wall determines the growth of cell. [½ × 4 = 2]

Short Answer Type Questions-II Q. 1. What is the energy currency of the cell ? Write it in expanded form. Which cell organelle is related to the currency ? U [Board Term-I, 2015] Ans. ATP; Adenosine Triphosphate Mitochondria.  [CBSE Marking Scheme, 2015] [3] Detailed Answer: ATP is the energy currency of the cell. Its expanded form is Adenosine Triphosphate. Mitochondria is organelle related to the currency. Q. 2. What happens when: (i) Methylene blue stain is added to human cheek cell. (ii) Rheo leaves are boiled in water and a drop of sugar is added to it. (iii) RBCs are kept in concentrated solution.  Ap [Board Term-I, 2015] Ans. (i) Because of its affinity for DNA and RNA, methylene blue will produce a darker stain leading the DNA in the nucleus to stand out so that nucleus can be clearly seen. (ii) On boiling, all the cells of Rheo leaves become dead. On adding sugar syrup nothing will happen as liquid cannot pass through dead cell membrane. (iii) On placing RBCs in concentrated solution, the water will come out and the cell will shrink as the concentration of solution outside is higher than inside the cell. As a result of osmosis, water comes out of the cell to maintain equilibrium. [1 + 1 + 1] Q. 3. (i) Where are chromosomes present in the cell? What is their chemical composition?

(3 marks each)

(ii) How many pairs of chromosomes are present in humans ?  U [Board Term-I, 2015] Ans. (i) Chromosomes are present in the nucleus of a cell. Their chemical composition is of DNA, proteins. (ii) Humans have 23 pairs of chromosomes.  [2 + 1] Q. 4. Discuss the role of: (i) Cellulose in cell wall (ii) Presence of deeply folded membrane in mitochondria (iii) Digestive enzymes in lysosomes.  U [Board Term-I, 2014] Ans. (i) Cellulose provides rigidity to the plant cell and helps it to withstand in dilute medium. (ii) Folds in mitochondria increase the surface area to help in ATP generating reactions. (iii) Digestive enzymes in lysosomes help in removal of viruses, worn out organelles, damaged cell. [1 + 1 + 1] Q. 5. How are the following related to each other ? (i) Chromatin network and chromosomes (ii) Chloroplast and chlorophyll (iii) Genes and DNA. 

A [Board Term-I, 2014]

Ans. (i) On cell division, chromatin network organise themselves into chromosomes. (ii) Chloroplast is a plastid which contains a green pigment called chlorophyll which is responsible for photosynthesis.



(iii) The segments of DNA are called genes. [CBSE Marking Scheme, 2014] [1 + 1 + 1]



Long Answer Type Questions Q. 1. How do respiratory gases move in and out of the cell ? Taking the example of photosynthesis explain how the exchange of gases takes place. What is the composition of plasma membrane?  A [Board Term-I, 2016]

(5 marks each)

Commonly Made Error



Ans. Diffusion. Conc. of CO2 is more in air as compared to cell so it moves in. O2 is produced in photosynthesis thus conc. of O2 is higher in cell so it moves out. Plasma membrane made up of proteins +lipids.  [CBSE Marking Scheme, 2016] [1 + 2 + 2]

59

Students often write osmosis, instead of diffusion.

Answering Tip

Students often write osmosis, instead of diffusion.

60

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Q. 2. (a) What are the consequences of the following conditions ? (i) A cell having higher water concentration than the surrounding medium.

Similarly, water moves out of the cell by exosmosis when a cell is placed in a hypertonic solution.  [1 + 1 + 1 + 2]

(ii) A cell having lower water concentration than the surrounding medium. (iii) A cell having equal water concentration to its surrounding medium. (b) Name the materials of which the cell membrane and cell wall are composed of.  A [Board Term-I, 2015] Ans. (a) (i)  When a cell possesses higher water concentration than the surrounding medium then exosmosis occurs in the cell due to the difference in concentration. As a result, the cell shrinks. (ii) When a cell has low water concentration than surrounding medium then endosmosis occurs that results in the swelling of the cell. (iii) A cell having equal water concentration to its surrounding medium will not show any change. (b) Cell wall is composed of cellulose and cell membrane is composed of lipids and proteins. [3 + 2] Q. 3. (i) Explain the terms:



(a) Endocytosis, (b) Plasmolysis.

(ii) What will happen if the organisation of a cell is damaged due to certain physical or chemical reasons ? (iii) How do substances like CO2 and water move in and out of the cell ?  Ans.

Commonly Made Errors

Students get confused between the technical





terms like plasmolysis, endocytosis etc. and write irrelevant stories. Sometimes they forget to answer the 2nd and 3rd part of the question.

Answering Tip

Comprehend what is being asked before

answering by reading the question carefully. Do not overlook any part of a question and avoid being in a hurry to conclude the answer.

Q. 4. (a) What is the name given to the thread shaped structures in the nucleus ? Why is it important? (b) Draw a diagram of the nucleus to show the given parts:  (i) nucleolus, (ii) nuclear pore, (iii) nuclear envelope. A Ans. (a) The thread shaped structures in the nucleus are known as chromosomes. These are important because they contain information for inheritance of features from parents to the next generation. (b) Nuclear envelope

A [NCERT]

(i) (a) Endocytosis: The flexibility of the cell membrane enables the cell to engulf food and other materials from its external environment. Such process is known as endocytosis.

(b) Plasmolysis: When a living plant cell loses water through osmosis, there is shrinkage or contraction of the contents of the cell away from the cell wall. This phenomenon is known as plasmolysis. (ii) When the organisation of a cell gets damaged, lysosomes will burst and their enzymes will eat up their own cell organelles. Therefore, lysosomes are also known as the “suicidal bags of the cell”. (iii) Gases like CO2 and O2 move in and out of the cell by diffusion from their higher concentration to lower concentration. Water enters the cell by endosmosis through semi-permeable plasma membrane from its higher concentration to lower concentration.

Nuclear pores Nucleolus

Chromosome

[2 + 3] Q. 5. In the given figure of an animal cell as observed under an electron microscope. A 

CELL — BASIC UNIT OF LIFE

(i) Name the parts labelled as 1 to 10.

3. Ribosome

(ii) Which parts are concerned with the following functions:

4. Smooth Endoplasmic reticulum

(a) Release of energy,

6. Nucleolus

(b) Protein synthesis, (c) Transmission of hereditary characters from parents to their offsprings.

(iii) Mention any two structures, found only in plant cell not in animal cell.

Ans. 

(i) 1. Mitochondria

2. Cytoplasm

5. Rough Endoplasmic reticulum 7. Nucleoplasm 8. Nuclear membrane 9. Centrosome 10. Golgi apparatus (ii) (a) Mitochondria, (b) Ribosome, (c) Nucleus (iii) (a) Cell wall and (b) Plastids 

COMPETENCY AND CRITICAL THINKING BASED QUESTIONS

I. All living Organisms are made up of cells and these cells perform all the functions essential for the survival of the Organism, e.g., respiration, digestion, excretion etc. In Unicellular organisms, a single cell carries out all these functions and in multicellular organisms different group of cells carry out different functions. (1) Cell was discovered by: (A) Robert Brown (B) Robert Hooke (C) A.V. Leeuwenhoek (D) Charles Darwin

 K Ans. Option (B) is correct. Explanation: In 1665, Robert Hooke First discovered and introduced the term cell. (2) The structural and functional unit of life is (A) Cell (B) Tissue (C) Mitochondria (D) Chloroplast. U Ans. Option (A) is correct. Explanation: Cell is the basic building block of all living organisms and perform all the activities required to sustain life. So, it is the structural and functional unit of life. (3) Select the Prokaryotic cell (A) Amoeba (B) Bacteria (C) Paramecuim (D) Hydra K Ans. Option (B) is correct. Explanation: Bacteria do not have a true nucleus or well developed nucleus. So, it is a Prokaryotic cell. (4) The Cell which do not have a fixed shape (A) Red Blood Cells (B) Paramecium (C) Amoeba (D) All of these U

61

[¼ × 10 + ½ × 5]

(1 mark each)

Ans. Option (C) is correct. Explanation: Both Amoeba and white blood cells of humans do not have any fixed shape. II. Study the diagram of a plant cell and answer the following Questions-

Q. 1. What is the composition of plant cell wall? Ans. Plant cell wall is made up of cellulose, hemicellulose and pectin. Q. 2. Which cell organelle is known as kitchen of the cell? Ans. Chloroplast is known as the kitchen of the cell because photosynthesis take place in the chloroplast. Q. 3. Which cell organelle is large in plant cell as compared to animal cell Ans. In a plant cell,vacuoles plays very important role as they store toxic metabolic by products and provide turgidity and provide turgidity and rigidity to plant cell Q. 4. Which cell organelle is called power house of the cell? Ans. Mitochondria is called power house of the cell. 

Study Time Reading Time: 4:00 Hr No. of Questions: 86

CHAPTER

6



Syllabus

TISSUES

Structure and functions of animal and plant tissue (Only four types of tissue in animals; meristematic and permanent tissues in plants).

Plant Tissues : Structure and

Topic-1 Functions Revision Notes

Plant Tissues:  Tissues ensure division of labour in multicellular organisms. • The tissues present in plants and animals are different owing to variations in their body organisation and mode of living. • Plants have two main types of tissues – meristematic tissues and permanent tissues. (a) Meristematic tissues may be apical, lateral or intercalary, depending on their location in the plant. (b) Permanent tissues are classified into simple and complex tissues.

TOPIC - 1 Plant Tissues : Structure and Functions.... P. 62 TOPIC - 2 Animal Tissues : Structure and Functions .... P. 69

Mnemonics Concept: Types of Meristematic tissue: Mnemonics: Accha Innam Laga Interpretation A: Apical meristem I: Intercalary meristem L: Lateral meristem

• Simple tissue shows only one type of cells whereas complex tissues consist of more than one type of cells, functioning as a unit.

Scan to know more about this topic

 Three types of simple permanent tissues are parenchyma, collenchyma and sclerenchyma. • Parenchyma is a supporting and storing tissue, composed of unspecialised, thin-walled cells with large intercellular spaces. • Collenchyma cells are elongated, with irregularly thickened cell walls. It provides mechanical support and elasticity to the plant. • Sclerenchyma is the main supporting tissue, consists of long and narrow cells with thick and lignified cell walls.

Complex Permanent Plant Tissues

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• Parenchyma and collenchyma are living tissues whereas sclerenchyma is a dead tissue.

 Epidermis is the outer protective covering of the plant and is usually layered by cuticle. • Stomatal pores, present in the epidermis, are essential for transpiration and gaseous exchange.

Parenchyma and Sclerenchyma Tissues

TISSUES

63

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

• In older plants, many layered cork is seen, made up of dead and compactly arranged cells.  Xylem and phloem are important types of complex tissues in plants. • Xylem is composed of tracheids, vessels, xylem parenchyma and xylem fibres. It conducts water and minerals from roots to aerial parts of the plant. • Phloem consists of sieve tubes, companion cells, phloem fibres and phloem parenchyma. It transports food from leaves and storage organs to all other parts of the plant.

Key words

 Tissues : A group of specialised cells with similar structure, working together to perform a common function.  Meristematic tissue : Tissue made up of actively dividing cells, present in the growing areas of the plant body.  Apical meristem : Meristem present at the growing tips of stem and root that cause the stem and root to increase in length.  Lateral meristem : Meristem located on the lateral portion of the plant and responsible for increasing the girth of its stem and root.  Intercalary meristem : Meristem found between already differentiated tissues, in locations such as the base of leaves or internode.  Permanent tissue : A well-differentiated plant tissue derived from meristematic tissue, which has lost its ability to divide.  Differentiation : The process by which a cell attains a permanent shape, size and function.  Simple permanent tissue : A permanent tissue composed of only one of cell type.  Complex permanent tissue : A permanent tissue composed of more than one type of cells which coordinate to perform a common function.  Chlorenchyma : Parenchyma whose cells contain chloroplasts and hence performs photosynthesis.  Aerenchyma : Parenchyma containing large air cavities, providing buoyancy to aquatic plants and allowing the circulation of gases.  Xylem : The complex tissue that conducts water and minerals in vascular plants and composed of tracheids, vessels, fibres, and parenchyma.  Phloem : The food-conducting tissue of vascular plants, consisting of sieve tubes, companion cells, fibres and parenchyma.  Epidermis : The outermost, protective layer of cells covering the surface of a plant.

OBJECTIVE TYPE QUESTIONS Choice Questions A Multiple  (1 mark each) Q.1. Which of the following tissues has dead cells? (A) Parenchyma (B) Sclerenchyma (C) Collenchyma (D) Epithelial tissue  K Ans. Option (B) is correct. Explanation: Sclerenchyma provides hardness and stiffness to the plant and is composed of dead cells. Q.2. Find out incorrect sentence. (A)  Parenchymatous tissues have intercellular spaces. (B)  Collenchymatous tissues are irregularly thickened at corners. (C)  Apical and intercalary meristems are permanent tissues. (D) Meristematic tissues, in its early stage, lack vacuoles. K

Ans. Option (C) is correct. Explanation: Apical and intercalary meristem bring primary growth (increase in height) and secondary growth (increase in diameter) respectively and are classified under meristematic ­tissues. Q.3. Girth of stem increases due to (A) Apical meristem (B) Lateral meristem (C) Vertical meristem (D) Intercalary meristem K Ans. Option (B) is correct. Explanation: Girth of the stem increases due to lateral meristematic tissue. They are found beneath the bark (called cork cambium) and in vascular bundles of dicot roots and stems (called vascular cambium) as thin layers. This increase in the diameter and girth of the plant is called secondary growth.

Q.4. Which cell does not have perforated cell wall?

(A) Tracheid

(B) Companion cells

(C) Sieve tubes

(D) Vessels

A

TISSUES

Ans. Option (B) is correct. Explanation: Companion cells possess numerous mitochondria and ribosomes and are supporting units of sieve tubes. They do not have perforated cell wall. Q.5. Meristematic tissues in plants are (A) Localised and permanent (B) Not limited to certain regions (C) localised and dividing cells (D) Growing in volume K Ans. Option (C) is correct. Explanation: Meristematic tissues consist of actively dividing cells and are present in the growing regions of plants, e.g., the tips of roots and stems. Q.6. Which is not a function of epidermis? (A) Protection from adverse condition (B) Preventing entry of pathogen (C) Conduction of water (D) Transpiration U Ans. Option (C) is correct. Explanation: The main function of epidermis is to protect the plant from desiccation and infection. In fact, cuticle of epidermis helps to reduce water loss by evaporation from the plant surface and also helps in preventing the entry of pathogen. Q.7. The dead element present in the phloem is (A) Companion cells (B) Phloem fibres (C) Phloem parenchyma (D) Sieve tubes K Ans. Option (B) is correct. Explanation: Phloem fibers are thick walled, elongated spindle shaped dead cells which possess narrow lumen. Q.8. Which of the following does not lose their nucleus at maturity? (A) Companion cells (B) Red blood cells (C) Vessel (D) Sieve tube cells K Ans. Option (A) is correct. Explanation: Companion cells do not lose nucleus at maturity. RBC, vessels and sieve tube cells lose their nucleus at maturity. Q.9. In desert plants, rate of water loss gets reduced due to the presence of (A) Cuticle (B) Stomata (C) Lignin (D) Suberin U Ans. Option (A) is correct. Explanation: Cuticle minimises the water loss through transpiration (with the help of stomata) and also reduces pathogen entry.

65

Q.10. A long tree has several branches. The tissue that helps in the sideways conduction of water in the branches is (A) Collenchymas (B) Xylem parenchyma (C) Parenchyma (D) Xylem vessels U Ans. Option (D) is correct. Explanation: Xylem vessels helps in the sideways conduction of water in the branches. They are very long tube-like structures formed by a row of cells placed end to end.

B

Assertion and Reason 

(1 mark each)

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion: Husk of coconut is made of sclerenchymatous tissue. Reason: Cells of sclerenchymatous tissue are dead with long and narrow walls thickened due to lignin. K Ans. Option (A) is correct. Explanation: Husk of coconut is made of sclerenchymatous tissue. The cells are dead with long and narrow walls thickened due to lignin. Q. 2. Assertion: Permanent tissue is composed of mature cells. Reason: Meristematic tissue is a group of actively dividing cells. K Ans. Option (B) is correct. Explanation: Meristematic tissues are made up of actively dividing cells, present in the growing areas of the plant body whereas Permanent tissue is a well-differentiated plant tissue derived from meristematic tissue, which has lost its ability to divide. Q. 3. Assertion: Guard cells are specialised epidermal cells. Reason: Stomata are found in the epidermis of leaves. U Ans. Option (B) is correct. Explanation: Stomata are present in the epidermis of leaves. They are essential for transpiration and gaseous exchange. Stomata are minute aperture bounded by two specialised epidermal cells called guard cells.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions Q. 1. Name the tissue which is responsible for increase in length of stem and root. U [Board Term-I, 2014] Ans. Apical meristem.  [CBSE Marking Scheme, 2014] [1] Q. 2. Name the four elements of phloem. K [DDE-2014] OR What are the constituents of phloem ?[NCERT] Ans. Sieve tubes, companion cells, phloem parenchyma and phloem fibres. [1]

(1 mark each)

Q. 3. What is meristematic tissue ? U Ans. Meristematic tissue is capable of dividing and is found in the developing regions of the plant. [1] Q. 4. What do you understand by differentiation ? K Ans. When meristematic tissues lose their ability to divide and become permanent in shape, size and function, the process is called differentiation. [1] Q. 5. What are stomata ? K Ans. Stomata are the small pores present in the epidermis of leaf. [1] Q. 6. What are vascular bundles ? K Ans. Vascular bundles consist of xylem and phloem. 

Short Answer Type Questions-I

[1]

(2 marks each)

Q. 1. What is apical meristem ? What is its function?  U [Board Term-I, 2013]

Q. 5. Explain the process of formation of cork. U [Board Term -I, 2016]

Ans. Apical meristems are the meristematic tissues which are found at the growing tips of stems and roots. It increases the length of the stems and roots and is responsible for the growth of plant.  [CBSE Marking Scheme, 2013] [1 + 1] Q. 2. Name the living component common to both the complex permanent tissues found in plants. What is its function ? A

Ans. Strip of secondary meristem replaces the

Ans. Living component common to xylem and phloem tissues is parenchyma. [1]

Its function is to store food and help in sideways conduction of water in xylem and food in phloem. [1]

Q. 3. Name the following tissues :

(i) The connective tissue found between the skin and muscles.

(ii) The epithelial tissue which forms the lining of the kidney tubules. (iii) The tissue which is present in the veins of leaves. U Ans. (i) Areolar, (ii) Cuboidal epithelium, (iii) Sclerenchyma.

[½ × 4]

Q. 4. Name the simple permanent tissue which :

(i) forms the basic packing tissue.

(ii) provides flexibility in plants.

U

Ans. (i) Parenchyma, (ii) Collenchyma. 

[1 + 1]



epidermis of the stem. Cells on the outside are cut off which forms the cork. [CBSE Marking Scheme, 2016] [2]

Q. 6. Write the location and function of collenchyma tissue. U Ans. Collenchyma is located in leaf stalks below the epidermis.

It provides flexibility in plants and easy bending and mechanical support. [1 + 1] Q. 7. Write any two characteristic features of parenchyma tissue. U Ans. (i) It consists of relatively unspecialised cells with thin cell walls. (ii) They are usually loosely packed so that large spaces are present between the cells. 

[Any two] [1 + 1]

Q. 8. List two points of differences between parenchyma and sclerenchyma. U (DDE 2017) Ans. Parenchyma tissue : In this, cells are found with thin cell walls and are usually loosely packed so that large intercellular spaces are found.

Sclerenchyma : Cells are dead and cell wall is thickened due to lignin. It provides strength to plants. [1 + 1] Q. 9. Given diagram is showing a longitudinal section of collenchyma tissue. Label the parts ‘M’, ‘N’, ‘O’ and ‘P’ in the given diagram.  U

TISSUES

67

Q. 10. With the help of labelled diagrams differentiate parenchyma and collenchyma.  U Ans.

Ans. M – Chloroplast, N – Nucleus, O – Cytoplasm, P – Intercellular space. [½ × 4 = 2]

thin primary cell wall

Parenchyma

Short Answer Type Questions-II Q. 1. (i) State one point of difference between xylem and phloem. (ii) Draw a neat diagram of xylem vessel and a tracheids.  U [NCERT Exemplar, DDE 2017] [Board Term-I, 2012] [DDE 2014] Ans. (i) Xylem conducts water in the plant body.

irregularly thickened primary cell wall

Collenchyma [1 + 1]

(3 marks each)

Q. 4. (a) Differentiate between epidermal and cork cells. (b) Why are they called protective tissues ?  U [Board Term-I, 2014] [NCERT Exemplar] Ans. (a) Epidermal Cells

Cork cells

Phloem transports food in the plant body.

Single layered

Multi-layered

(ii)

Living

Non-living

Secrete cutin

Secrete suberin

Present in younger Present plants. plants.

in

older

(b) They are called protective tissues because : (i) They protect from mechanical injury and infection. 

[1 + 1 + 1]

Q. 2. Classify meristematic tissues on the basis of the region they are present. Also, mention their functions. U [Board Term-I, 2016] Ans. Apical meristems : increase length of stem and roots. Lateral meristems : increases girth. Intercalary meristems : increase length of internodes. [CBSE Marking Scheme, 2016] [3] Q. 3. (i) Why do sclerenchyma cells have a narrow lumen ? (ii) Where are these tissues present and why ? A [Board Term-I, 2016] Ans. (i) Due to deposition of lignin. (ii) In stems around vascular bundles, in veins of leaves and in hard coverings of seeds and nuts. It makes the plant hard, stiff, provides strength.  [1 + 1 + 1] [CBSE Marking Scheme, 2016]





(ii) They prevent loss of water. [CBSE Marking Scheme, 2014] [2 + 1] Q. 5. List two characteristics of cork. How is it formed? Mention its role in trees.  U [Board Term-I, 2014] OR List the characteristics of cork. How are they formed ? Mention their role. [NCERT] Ans. (i) Non-living (ii) Compactly arranged (iii) No intercellular spaces (iv) Multilayered (v) Contains suberin [Any two] A strip of secondary meristem replaces the epidermis. Cells on the outside are cut forming cork. Protection, makes the plant impervious to gases, prevents loss of water, prevents mechanical injury or infection.  [CBSE Marking Scheme, 2014] [2 + 1] Q. 6. Write three distinguishing features between cells of meristematic and permanent plant tissues. U [Board Term-I, 2014]

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Ans.

Ans.

S.No. Meristematic tissue Permanent tissue (i) Cells possess dividing Cells generally do not ability. divide. (ii) Cells are living. Cells can be living or dead. (iii) Main function is to Performs various type of functions. bring about growth. [1 × 3 = 3] Q. 7. Label the following and give one function of each part labelled (i), (ii) and (iii).  A [NCERT]

(ii) Intercalary meristem : Helps in increasing internodes of plants like sugarcane. (iii) Apical meristem : Helps in growth of the stem and the root. [1 + 1 + 1] Q. 8. Explain in brief any three roles of epidermis in plants.  U OR Why is epidermis important for the plants ?  [NCERT] Ans.

(iii)

(i) Lateral meristem : Helps in growth and development of plant’s shoot girth.

(i) The epidermis protects all parts of the plants.

(ii) Epidermal cells on the aerial part of the plant often secrete a waxy, water resistant layer which helps in protection against water loss and mechanical injury.

(ii)

(iii) Protect against invasion of parasitic fungi. [1 × 3 = 3]

(i)

Long Answer Type Questions Q. 1. The growth of plant occurs only in specific regions: (i) Name the tissue which is responsible for this growth. (ii) State the different types of this tissue. (iii) Write one function of each of the above mentioned tissue. 

A [NCERT Exemplar]

Ans. (i) Meristematic tissue

[½]

(ii) The different types of meristematic tissue are : (A) Apical Meristem (B) Intercalary Meristem (C) Lateral Meristem.

[3 × ½ = 1 ½]

(iii) Refer SAQ-II/ Q. 7. [3] Q. 2. (a) Analyse the reason behind the following statements : (i) Epidermis is thicker in desert plants though it is usually single layered. (ii) Presence of waxy layer (secreted by epidermis) on the outer surface of plants. (b) Discuss the cell arrangement which supports the fact that epidermis is a protective tissue.  A [Board Term-I, 2013]



(b) Cells are elongated, flattened, closely packed. No intercellular spaces and form a continuous layer. [2 + 3]  [CBSE Marking Scheme, 2013] Detailed Answer : (a) (i) In desert habitat, protection against water loss is essential so, epidermis is thicker in desert plants. (ii) The waxy covering aids in protecting the plant against loss of water, mechanical injury and invasion by parasitic fungi. (b) Epidermis is the outermost covering of cells in plants. It is usually made up of a single layer of cells. On aerial parts of a plant epidermal cells often secrete a waxy, water resistant layer on their outer surface to prevent loss of water from plant. The cells of epidermis are present in a continuous layer without intercellular spaces. Small pores are present on the epidermis of leaf. These pores are called as stomata, which help in gaseous exchange and transpiration. As the plant grows older, a strip of secondary meristem replaces the epidermis of stem and forms a thick cork. [2 + 3]

Ans. (a) (i) In such habitat, protection against water loss is essential. (ii) Protect against water loss, mechanical injury.

(5 marks each)

Commonly Made Error

Students overlook parts of question in hurry.

TISSUES

Answering Tip

Read the question carefully and answer the questions part-by-part.

Q. 3. List any five characteristics of Cork. U Ans. The Characteristics of Cork are as follows (i) In older stems and roots Cork is found as the outer protective tissue.

69

(ii) The cells are arranged compactly without intercellular spaces. (iii) Matured Cork is dead filled with tannin, resins etc. (iv) It is formed by secondary lateral meristem called Cork Cambium. (v) The cells are multilayered thick and impermeable due to the deposition of suberin in their wall.[1+1+1+1+1]

Topic-2 Animal Tissues : Structure and Functions Revision Notes





 Animal tissues are grouped into 4 basic types – epithelial, connective, muscular and nervous tissue. Scan to know  Epithelial tissues are the covering or protective tissues which act as a barrier between the more about various systems of the body. It rests on a basement membrane and is composed of tightly this topic packed cells.  Connective tissues is the binding and supporting tissue of the animal body. Matrix forms the main bulk of this tissue, whereas the cells are loosely spaced and less in number.  Muscular tissues have the unique ability to contract and relax for doing mechanical or Animal tissues strenuous work. It help for the movement of organs.  Nervous tissues help to transmits signal to respond for the stimulus received by the sense organs. It is formed by the nerve cell called neurons.  Blood, bone, ligament, tendon, cartilage, areolar tissue and adipose tissue are important connective tissues present in our body. • Blood is a fluid connective tissue in which RBCs, WBCs and platelets are suspended and plays a significant role in the process of transportation. • Bone function is to protect, providing skeletal framework and anchoring are carried out by the strong and hard bone tissue. • Ligaments connect bones to bones whereas tendons connect bones to muscles. • Cartilage provides support and flexibility to the body parts. • Areolar tissue repairs the injured tissues and fills spaces within organs. These are found between the skin and muscles, around blood vessels and nerves and in bone marrow. • Adipose tissue serves as a fat reservoir and also carries out the function of insulator. It is found below the skin and between internal organs.  Muscular tissues is responsible for movements in our body through the contraction and relaxation of their contractile fibres of muscles.  Striated, unstriated and cardiac are three types of muscle tissues.  Nervous tissues help to transmits signal to respond for the stimulus received by the sense organs. It is formed by the nerve cell called neurons. • Nervous tissue is present in the brain, spinal cord and nerves. • Neuron is made up of cell body, dendrites and axon. • Neurons are specialised to receive and conduct impulses rapidly.

Key words

 Stratified epithelium : An epithelium composed of multiple layers of cells, with only the basal layer being in contact with the basement membrane.



 Ligament : A fibrous connective tissue that connects (or binds) bones to bones.



 Tendon : A fibrous connective tissue that connects bones to muscles.



 Voluntary muscles : Muscles which can be controlled according to our will.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Involuntary muscles : Muscles which are not under the control of our will. Multinucleate cell : Cell containing more than one nucleus. Uninucleate cell : Cell containing only one nucleus. Neuron : A cell of the nervous system specialised to conduct nerve impulses and made up of cell body, axon and dendrites.  Impulse : An electrical signal transmitted along a nerve fibre in response to a stimulus.    

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions 

(1 mark each)

Q. 1. Intestines absorb the digested food materials. What types of epithelial cells are responsible for that? (A) Stratified squamous epithelium (B) Columnar epithelium (C) Spindle fibres (D) Cuboidal epithelium K Ans. Option (B) is correct. Explanation: Columnar epithelium forms the lining of stomach, small intestine and colon, forming the mucous membrane. Their main function is absorption (e.g., stomach, intestine) and secretion (e.g., mucous by goblet cells). Q. 2. A person met with an accident in which two long bones of hand were dislocated. Which among the following may be the possible reason? (A) Tendon break (B) Break of skeletal muscle (C) Ligament break (D) Areolar tissue break U Ans. Option (C) is correct. Explanation: Dislocation of joint occurs when there is an abnormal separation in joint, which are held together by ligament. Therefore ligament break and result in dislocation of bone. Q. 3. While doing work and running, you move your organs like hands, legs, etc. Which among the following is correct ? (A)  Smooth muscles contract and pull the ligament to move the bones. (B)  Smooth muscles contract and pull the tendons to move the bones. (C)  Skeletal muscles contract and pull the ligament to move the bones. (D)  Skeletal muscles contract and pull the tendon to move the bones. U Ans. Option (D) is correct. Explanation: Skeletal muscles are striped, voluntary muscles, due to presence of alternate dark and light bands. They are called voluntary as they work according to our will. Q. 4. Which muscles act involuntarily? (i) Striated muscles (ii) Smooth muscles (iii) Cardiac muscles

(iv) Skeletal muscles (A) (i) and (ii) (B) (ii) and (iii) (C) (iii) and (iv) (D) (i) and (iv) K Ans. Option (B) is correct. Explanation: The working of both smooth and cardiac muscles are involuntary while skeletal or striated muscles move according to our will and are voluntary in action. Q. 5. Select the incorrect sentence. (A) Blood has matrix containing proteins, salts and hormones. (B) Two bones are connected with ligament. (C) Tendons are non-fibrous tissue and fragile. (D) Cartilage is a form of connective tissue. K Ans. Option (C) is correct. Explanation: A tendon is a white fibrous tissue which has great strength but limited flexibility. Tendon joins muscles to bones. Q. 6. Cartilage is not found in (A) Nose (B) Ear (C) Kidney (D) Larynx K Ans. Option (C) is correct. Explanation: Cartilage occurs at the joints of bones, in the nose, ear, trachea and larynx. It helps in smoothing the surface at the joints. It lends support and provides flexibility to the body parts. Q. 7. Fats are stored in human body as (A) Cuboidal epithelium (B) Adipose tissue (C) Bones (D) Cartilage K Ans. Option (B) is correct. Explanation: Adipose tissue stores fat and acts as an insulator. Q. 8. Bone matrix is rich in (A) Fluoride and calcium (B) Calcium and phosphorus (C) Calcium and potassium (D) Phosphorus and potassium K Ans. Option (B) is correct. Explanation: Bone cells are embedded in a hard matrix, made by calcium and phosphorus salts. Q. 9. Contractile proteins are found in (A) Bones (C) Muscles

(B) Blood (D) Cartilage

K

TISSUES

Ans. Option (C) is correct. Explanation: Contractile proteins are found in muscles, as they are associated with the movement of body or limbs. Q. 10. Voluntary muscles are found in (A) Alimentary canal (B) Limbs (C) Iris of the eye (D) Bronchi of lungs U Ans. Option (B) is correct. Explanation: Voluntary muscles are the muscles, which are under our complete control. Example includes the working and movement of limbs.

B

Assertion and Reason (1 mark each)



Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion: Specialisation in cells are advantageous for the organism. Reason: It increases the efficiency of an organism. U  Ans. Option (A) is correct. Explanation: Specialisation of cells into tissue, organ and organ systems is advantageous for organism as it increases the efficiency through division of labour. Q. 2. Assertion: Muscle fibers are contractile in nature. Reason: Cells of muscle tissue can shorten forcefully and again return to the relaxed state. U

71

Ans. Option (A) is correct. Explanation: Muscle fibres are said to be contractile in nature because cells of muscle tissue can shorten forcefully and then can again return to the relaxed state. This specialised property is known as contractility. Q. 3. Assertion: Thigh muscles can get tired but not heart muscles. Reason: Muscles of thigh are voluntary while muscles of heart are involuntary muscles. U Ans. Option (B) is correct. Explanation: Thigh muscles are striated muscles which soon get fatigued due to overwork. They are called voluntary muscles because they can be moved at the will of organism while the heart muscles are cardiac muscles which show rhythmic and automatic contractions. Heart muscles cannot be moved at one’s will and therefore are involuntary. Q. 4. Assertion: Epithelial tissues are covering or protective tissues. Reason: Materials are exchanged at the surface across the epithelial tissues. K Ans. Option (B) is correct. Explanation: Epithelial tissue forms a continuous layer over the free surfaces of many other tissues. Consequently, it covers the external surface of the animal body and the internal (luminal) surfaces of visceral organs, body cavities and blood vessels. Thus, it protects the underlying and overlying tissues. Another important function of epithelial tissue is exchange of materials. Q. 5. Assertion: Ciliated epithelium helps in movement of particles. A Reason: Cilia help in movement. Ans. Option (A) is correct. Explanation: The main function of the cilia is to move particles. It is present in inner surfaces of some hollow organs such as fallopian tubes, bronchioles and small bronchi and help in movement of the particles present there. Thus the function of ciliated epithelium (as it possesses cilia) is the movement of particles.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions Q. 1. Name the following tissues : (i) That forms the inner lining of our mouth. (ii) Present in the brain.

U [Board 2015]

Ans. (i) Epithelial tissue-squamous epithelium. (ii) Nervous tissue. Q. 2. Name the following tissues : (i) Found in the iris of the eye.

[½ + ½]

(ii) That connects two bones. U [Board 2014]

(1 mark each)

Ans. (i) Involuntary muscular tissue. (ii) Ligament 

[CBSE Marking Scheme, 2014] [½ + ½]

Q. 3. State two functions of the adipose tissues. 

U [DDE 2014]

Ans. (i) It helps in storage of fats. 

[½]

(ii) It act as an insulator.

[½]

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Q. 4. Which epithelial tissue is present on the tongue? U Ans. Stratified squamous epithelium. [1] Q. 5. Which tissue contains flat squamous cells arranged in many layers to prevent wear and tear of parts ? U Ans. Stratified squamous epithelium. [1] Q. 6. Name an animal whose skeleton is made up of cartilage. U

Ans. Shark fish. [1] Q. 7. What is the function of lymph ? U Ans. It helps in the exchange of materials between blood and body cells. [1] Q. 8. How are simple epithelial tissue and compound epithelial tissue different ? U Ans. Simple epithelial tissue is unilaminar while compound epithelial tissues is multilaminar. [1]

Short Answer Type Questions-I Q. 1. Name the type of tissue and its function present in the lining of the kidney tubules. A [Board Term-I, 2016]

Blood

Fluid matrix

Ans. Cuboidal epithelium. Function—Provides mechanical support. [CBSE Marking Scheme, 2016] [1 + 1] Q. 2. Name the tissue in animals which carries out similar function as the following tissues do in plants : (i) Epidermis (ii) Vascular Bundles Also write their functions. 

A [Board Term-I, 2016]

Ans. (i) Epithelial tissue, epidermal epithelium– protection. (ii) Blood. Function—Vascular

bundles

/

blood

transportation. [CBSE Marking Scheme, 2016] [1 + 1] Q. 3. Write the location of the following tissues :

(i) Unstriated muscle fibres

(ii) Cuboidal epithelium (iii) Adipose tissue (iv) Striated muscle fibres 

U [Board Term-I, 2016]

Ans. (i) Alimentary canal, iris of the eye, ureters, bronchi, etc. (any one)

(2 marks each)

(a)

Cells

(b)

(d)

(c)

 (Board 2015) Ans. (a) Plasma, (b) WBC, (c) Platelets, (d) RBC.  [½ × 4] Q. 5. Why are plants and animals made up of different tissues ? A [Board Term-I, 2014]



Ans. Plants are stationary thus their supportive tissue is made up of dead cells. Animals move, hence they possess living cells to provide energy for movement. Also, for the many more differences and functions in plants and animals, they are made up of different tissues. [CBSE Marking Scheme, 2014] [2] Q. 6. Write two locations of the following animal tissues: K (i) Simple Squamous Epithelial cells. (ii) Cuboidal Epithelium. [Board Term-I, 2012] Ans. (i) Oesophagus, lining of mouth. (ii) Lining of kidney tubules, ducts of salivary glands.  [1 + 1] Q. 7. (i) Name the connective tissue which connects two bones. (ii) Name the connective tissue present in external ear. U [Board Term-I, 2012]

(ii) Kidney tubules, ducts of salivary glands.



(iii) Below the skin, between internal organs.

 [CBSE Marking Scheme, 2012] Q. 8. Mention one region in the human body where adipose tissue is present and state one function of the tissue.  U [Board Term-I, 2012]

(iv) Limbs, tongue etc. [CBSE Marking Scheme, 2016] [2] Q. 4. Mention the different components of blood in the following diagram ? K

Ans.

(i) Ligament (ii) Cartilage. 

[1 + 1]

Ans. It is found below the skin. It acts as an insulator. [CBSE Marking Scheme, 2012] [1 + 1]

TISSUES

Short Answer Type Questions-II Q. 1.

(i) Name the animal tissue which is present in the larynx.

S.



Ans. (i) Cartilage.



(ii) Proteins and sugar.

Ligament

(i)

They join bone to They join bone to muscles. bone.

(ii)

They have limited They have elasticity. flexibility.

(iii) They have more They have strength. strength.

(iii) Smoothens body surfaces at joints, helps in easy bending. 

Tendon

No.

(iii) What functions does this tissue perform ? K [Board Term-I, 2016]

(3 marks each)

Ans. (a) Differences between tendon and ligament :

(ii) Write the chemical constituents of this tissue.



[1 + 1 + 1]

[CBSE Marking Scheme, 2016] Q. 2. Write two distinguishing features between the muscles present in the alimentary canal and limbs of man. Draw labelled diagrams of the two kinds of muscles.  A [Board Term-I, 2016] Ans. Alimentary canal : smooth muscle-spindle shaped, long pointed ends, uninucleate, involuntary etc. [Any two] Limbs : Striated-muscle, cylindrical, unbranched, voluntary, multinucleate. [Any two] Spindle shaped Nuclei muscle cell striations

73

less

[Any two] (b) Adipose tissue stores fat and provides insulation. [2 + 1] [CBSE Marking Scheme, 2015, 2014] Q. 5. Identify the type of tissues in the following : (i) Vascular bundle (ii) Inner lining of the intestine (iii) Lining of kidney tubule (iv) Iris of the eye (v) Muscles of the heart (vi) Bronchi of lungs. 

A [DDE 2014; Board Term-I, 2014]

Ans. (i) Complex tissues (ii) Columnar epithelium (iii) Cuboidal epithelium

Nucleus

(i) Smooth muscle

(ii) Striated muscle [2 + 1] Q. 3. Describe the structure, function and location of the nervous tissue. K [Board Term-I, 2015] Ans. Structure: The nervous tissue is made up neurons which consists of a cell body with a nucleus and cytoplasm, from which long thin hair-like parts arise. Dendrite

(iv) Involuntary muscular tissues (v) Cardiac muscles (vi) Ciliated cuboidal epithelium. [½ × 6 = 3] [CBSE Marking Scheme, 2014] Q. 6. Identify the two types of tissues given in the diagram. Write two distinguishing features of each of the two. A [Board Term-I, 2014] Haversian canal (contains blood vessels and nerve fibres)

Chondrocyte Hyaline matrix

Nerve ending Cell body

Nucleus



Axon



Function: On stimulation, the nerve cells transmit the stimulus very rapidly from one place to another within the body.



Location: Nervous tissues are located in the brain, spinal cord and nerves. [1 + 1 + 1] Q. 4. (a) State the differences between tendon and ligament. [Board Term-I, 2013] (b) Give the function of adipose tissues.  U [Board Term-I, 2015] [DDE 2014]

Canaliculus (contains slender process of bone cell or osteocyte)

(a)(a)

(b)

(b)

Ans. (a) Bone (Connective tissue): (i) It has a hard matrix. (ii) They are usually hollow. (b) Cartilage (Connective tissue): (i) This tissue is elastic and harder but softer than bone. (ii) The matrix of cartilage is solid but elastic.  [1 + 1 + 1]

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Long Answer Type Questions Q.1. Name a connective tissue which is the hardest. What makes it so hard ? List any three important functions of this tissue.  U [Board Term-I, 2016] Ans. Bone, Hard matrix is composed of Ca and P compounds.[2] Functions : (i) Forms framework of body. (ii) Anchors muscles. (iii) Supports the main organs of body and provides protection to them (e.g., brains, lungs). [3]

(5 marks each)

Q. 2. Identify the following tissues : (i) The epithelial tissue which has pillar like tall cells ? (ii) The cells of this tissue are filled with fat globules. (iii) The movement of this tissue pushes the mucus forward to clear respiratory tract. (iv) It gives buoyancy to lotus to help it afloat. (v) Tissue present in lung alveoli. 

A [Board Term-I, 2014]

Ans. (i) Columnar (ii) Adipose (iii) Ciliated columnar (iv) Aerenchyma (v) Squamous. 

[CBSE Marking Scheme, 2014] [1 × 5 = 5]

Q. 3. What are the differences between striated, unstriated and cardiac muscles ? Ans. Differences between striated, unstriated and cardiac muscles : S. No.

Striated muscles

U (DDE-2014; NCERT)

Unstriated (Smooth) muscles

Cardiac muscles

(i)

They are found in tongue, pharynx etc.

(ii)

Long, cylindrical with blunt Short, spindle shaped with Short, branched and cylindrical ends pointed ends with flat ends

(iii)

Multinucleate, peripheral

(iv)

They are voluntary in action.

(v)

Dark and light bands are No bands present. present.

limbs, They are present in the wall of They are present in the heart. visceral organs.

nuclei Uninucleate, nucleus in the One or two nuclei in the centre centre They are involuntary in action.

They are involuntary in action. Bands present. [1 × 5 = 5]

also supports the body. Its matrix is impregnated with phosphates and carbonates of calcium and magnesium which provides hardness to it. The matrix also contains ossein protein. The matrix is arranged in concentric rings which are called lamellae.

Commonly Made Error

Students often get confused with the terms and write opposite differences.

Answering Tip



Students

should learn the differences between striated, unstriated and cardiac muscles in tabular form for easy retention and understanding.

Bone cells lie between the lamellae in fluid-filled spaces called lacunae. Bone cells are also called osteocytes.

Q. 4. Draw a labelled diagram of neuron. 

A (NCT-2014)

Ans. For diagram : Refer SAQ-II/Q. 3. Q. 5. Describe the structure of bone and cartilage. 

U [DDE 2014, NCERT]

Ans. Bone : Bone is a solid, hard porous tissue. It is produced by osteocyte cells. It forms the natural skeleton and gives the body its basic structure and

Bone

TISSUES





The bone is surrounded by a connective tissue called periosteum by which muscles and tendons are attached to the bone. A long bone has a hollow cavity filled with bone marrow which is richly supplied with blood vessels. Cartilage : Cartilage has widely spaced cells. It is a solid but semi-rigid connective tissue. The solid matrix is composed of proteins and sugars. The cartilage cells are large and angular and they are called chondrocytes. They occur in clusters of 2 or 3 cells in the small spaces called lacunae scattered in the matrix. A sheath called perichondrium binds the cartilage.



Cartilage

75

[3 + 2]

Q. 6. Complete the following flowchart :

A [Board Term-I, 2014]

Ans. (a) Connective, (b) Plasma, (c) RBCs, (d) Platelets, (e) Neutrophil, (f) Eosinophil, (g) Lymphocyte, (h) Basophils, (i) Monocyte and (j) Bone or ligament or tendon. [CBSE Marking Scheme, 2014] [½ × 10]

COMPETENCY AND CRITICAL THINKING BASED QUESTIONS

I. Study the given diagram and answer the following questions. A B

C D

(1 mark each)

(1) Name the tissue shown in the diagram. (A) Xylem (B) Phloem (C) Epidermis (D) Cortex U Ans. Option (B) is correct. (2) What will happen if Phloem at the base of the branch is removed? (A) Plant will die (B) lower parts of the plants wilted. (C) No change occur (D) None of these U Ans. Option (B) is correct. Explanation: If the Phloem at the base of branch is removed, then lower area of the branch will not receive food from the leaves. But the plant

76



Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

will not die, as it will continue to receive food from other branches as food can move in phloem in both direction. (3) Identify the parts A,B,C and D.



(A) A - Companion cell, B - Sieve tube, C - Sieve plate, D - Phloem parenchyma



(B) A - Phloem Parenchyma , B - Companion cell, C - Sieve tube, D - Sieve plate



(C) A - Sieve plate, B - Sieve tube, C - Phloem parenchyma D - Companion cell



(D) A - Sieve tube, B - Phloem parenchyma, C - Sieve plate, D - Companion cell K

Ans. Option (C) is correct.

(4) Vascular bundles are :



(A) Only Xylem



(B) Only Phloem



(C) Only Cortex



(D) Both Xylem and Phloem

U

Ans. Option (D) is correct. Explanation: Vascular Bundles are Xylem and Phloem together .

II. Study the given diagram and answer the following questions.



(1) Identify the tissue :



(A) Parenchyma

(B) Collenchyma



(C) Sclerenchyma

(D) Aerenchyma

K

Ans. Option (B) is correct. Explanation: Collenchyma cells have thick deposits of cellulose in their cell walls and appear polygonal in cross section.

(2) Collenchymatous tissues are : (A) Temporary tissue (B) Complex Permanent tissue (C) Simple Permanent Tissue (D) Meristematic tissues K Ans. Option (C) is correct. (3) Collenchyma Tissues are found in : (A) Leaf stalks (B) Flower (C) Roots (D) None of these K Ans. Option (A) is correct. Explanation: Collenchymatous tissues form long flexible but strong strands in Leaf stalk. (4) The irregular thickening is due to the deposition of : (A) Suberin (B) Pectin (C) Lignin (D) Fats K Ans. Option (B) is correct. III. Sunil while playing football with his friends got injured suddenly. His friends took him to the hospital and the doctor told that he was suffering from sprain and advised bed rest. Every afternoon, his friends visited him to enquire about his health. [CBSE SAS] (1) During a sprain, which type of tissue are stresses? Ans. Ligaments Explanation: with severe sprain, ligaments tear completely or separate from the bone (2) Which tissue connects bones to muscles in humans? Ans. Tendons Explanation: A tendon is a fibrous connective tissue which attaches muscle to bones (3) Why dislocation of bones takes place? Ans. Due to breakage of bones Explanation: Ligaments are torned when dislocation occurs, as ligaments are flexible bands of fibrous tissue. (4) Which tissue connect one bone to another bone? Ans. Ligament Explanation: A Ligament is a fibrous connective tissue which attaches bone to bone and usually serves to hold structures together



TISSUES

77

Artificial Intelligence PARAMETERS

DESCRIPTION

Chapter Covered

Chapter 6: Tissues

Name of the Book

Science, Class 9, NCERT

Subject and Artificial Tissue - Plant Tissue and its types Intelligence Integrated Learning Objectives

● Identify the various types of plant tissues. ● To understand structural, functional and locational differences between various types of Simple permanent tissues -Parenchyma, Collenchyma and Sclerenchyma. ● Explain structural and functional differences of various types of complex permanent tissues- Xylem and Phloem.

Time Required

2 to 3 periods (35 mins each)

Classroom Arrangement

● Regular Classroom setup having laptop/ computer and internet.

Material Required

● Textbook ● Computer, projector, smart board and internet connectivity ● Pen, paper

AI CONCEPTS INTEGRATED

Unsupervised learning Natural Language processing

Pre-Preparation Activities ● Make four groups and give students salientfeatures of different types of plant tissues and ask them to guess the location of these in a plant. Previous Knowledge

● Students to know the difference between cell and tissue. ● Students to know the parts of a plant.

Methodology

● Students continue in the same groups and conduct an online search for various kinds of plant tissues and classify them. AI tool for showing ● Students will be able to define the term tissue and differentiate it from cell and tissue cell. ● Students to be guided to identify location of the same. Cell Cycle ● Student groups are asked to prepare a presentation on plant tissues and their functions. Scan to know For better understanding: https://youtu.be/lLnjo4Pf2JM more about this topic Growth of a plant http://platane.github.io/Procedural-Flower/examples/RampantFlower. html

Learning Outcomes

● Students will be able to define the term tissue and differentiate it from cell. ● Classify and compare different types of plant tissues as meristematic and permanent tissues. ● Classify meristematic tissue on the basis of its location.

Follow up Activities

● Make students solve a quiz on google form. ● Peer assessment- Let the students work in pairs, draw diagrams of different plant tissues and have it assessed by their partner. ● Observe and assess how much students have understood. ● Also assess if the lesson is better learnt by using AI tool

Reflections

● Know from students the role of AI application. 

78

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Artificial Intelligence PARAMETERS

DESCRIPTION

Chapter Covered

Chapter 6: Tissues

Name of the Book

Science, Class 9, NCERT

Subject and Artificial Epithelial Tissues : Types and Functions Intelligence Integrated

AI CONCEPTS INTEGRATED

Supervised learning, Data exploration Scan to know more about this topic

Learning Objectives

● To enable the students understand the body organisation and how tissues are formed. ● To enable the students define tissues. ● To familiarise the students with the four types of tissues found in animals. ● To enable the students describe the general characteristics and functions of epithelial tissues. ● To enable the students name the types of epithelial tissues found in our body and help the students identify an organ in which each is found. ● To enable the students explain how glands are formed from epithelium and classify the glands.

Time Required

4-5 periods of 35 minutes each

Classroom Arrangement

Regular classroom setup with projector or any other visual aid available to show the diagrams different kinds of epithelial tissues.

Material Required

● ● ● ●

Computers to display the powerpoint presentation. Permanent slides of epithelial tissues. Common classroom materials required for teaching. Computers with internet connection to search more about the epithelial tissue and make their own mind maps.

Pre-Preparation Activities ● Materials needed for self-guided learning of tissues. ● Preset microscope with permanent slides. ● Power point slides for supervised learning. Previous Knowledge

The students have the knowledge of cells which are the basic unit of life and how they are organised to form tissues.

Methodology

● The teacher introduces the concept of tissues and the four basic types Supervised learning of tissues found in the animals. ● The teacher divides the students into group of 4 students each and Data visualisation and each group are then instructed to study on their own a group of exploration tissues within a specified period of time through a microscope to promote hands on learning with microscopes. Scan to know ● The students are then instructed to observe the slides preset in the more about this topic microscope each showing a good field of view. ● The teacher then shows the histological images in the form of power point slides on a projector screen in the classroom and explains the primary and distinguishing anatomical features of each tissue , the etymologies of the names of tissues, the locations in the body and the roles each tissue plays in bodily functions. ● In the last the teacher instructs the students of each group to draw the tissues and label them correctly on the basis of their observation of the slides in microscope and images seen. ● The students are introduced with a link to make their own tree diagrams.

TISSUES

Learning Outcomes

● The students will be able to understand the body organisation of animals. ● The students will be able to familiarise themselves with the four types of tissues found in animals. ● The students will be able to describe the general characteristics and functions of epithelial tissues. ● The students will be able to name the types of epithelial tissues found in our body and help the students identify an organ in which each is found. ● The students will be able to explain how glands are formed from epithelium and classify the glands.

Follow up Activities

● Assessment worksheets ● Report on the use of the AI in the field of Histopathology using google search engine.

79



SELF ASSESSMENT PAPER - 02 MM: 30

Maximum Time: 1 hour

Q. 1. Amoeba acquires its food through a process, termed : [1] (A) Exocytosis (B) Endocytosis (C) Plasmolysis (D) Both exocytosis and endocytosis OR Chromosomes are made up of : [1] (A) DNA (B) Protein (C) DNA and Protein (D) RNA Q. 2. Choose the wrong statement: [1] (A) The nature of matrix differs according to the function of the tissue. (B) Fats are stored below the skin and in between the internal Organs. (C) Epithelial tissues have intercellular spaces between them. (D) Cells of striated muscles are multinucleated and unbranched. Q. 3. Girth of stem increases due to: [1] (A) Apical meristem (B) Lateral meristem (C) Vertical meristem (D) Intercalary meristem Direction: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion(A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 4. Assertion: Permanent tissue is composed of mature cells. Reason: Meristematic tissue is a group of actively dividing cells. [1] Q. 5. Assertion: Chloroplast perform photosynthesis. Reason: Chloroplast comprises photosynthesis pigments. Study the given diagram and answer the following questions. [5]

Q. 6. Identify the diagram: (A) Plant cell (B) Animal cell (C) Bacterial cell (D) None of the above Q. 7. Name the labelled part 1 which releases energy. (A) Chloroplast (B) Ribosome

SELF ASSESSMENT PAPER

81

(C) Centrosome (D) Mitochondria Q. 8. Which part in the cell help in protein synthesis: (A) Chloroplast (B) Ribosomes (C) Nucleus (D) Lysosomes Q. 9. The Organelle responsible for intracellular transport is: (A) Plastids (B) Cytosomes (C) Endoplasmic Reticulum (D) Centrioles Q. 10. Lysosome arises from: (A) Endoplasmic reticulum (B) Golgi apparatus (C) Nucleus (D) Mitochondria Q. 11. What are the types of complex permanent tissues? [1] Q. 12. State the role of leucoplasts in plant cells. [1] Q. 13. Name the following tissues: [1] (i) That forms the inner lining of our mouth. (ii) Present in the brain. Q. 14. Write two differences between cell wall and cell membrane. [2] Q. 15. List two characteristics of cork. Name the chemical present in them and mention its role. [2] Q. 16. Why are lysosomes Known as Suicidal bags? [2] Q. 17. (a) State the difference between tendon and ligament. [3] (b) Give the function of adipose tissues. Q. 18. (a) Which cell organelle would you associate with ATP production? How is this organelle able to make its own proteins? (b) A Student performed an experiment by placing the de-shelled egg in a concentrated salt solution for five minutes. what changes did he observe in the egg ? [3] Give reasons for the same. Q. 19. Mention the differences between striated , Unstriated and cardiac muscles? [5]



https://qr.page/g/4p3pfvOGaxL

UNIT-III

Motion, Force and Work

Study Time Reading Time: 3:00 Hr No. of Questions: 51

CHAPTER



7

Syllabus

Topic-1

MOTION Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, elementary idea of uniform circular motion.

Distance, displacement, velocity, Uniform and non-uniform motion, acceleration.

TOPIC - 1 Distance, displacement, velocity, Uniform and non-uniform motion, acceleration.  .... P. 82 TOPIC - 2 Uniformly Accelerated Motion .... P. 91

Revision Notes



Scan to know  State of Motion : more about • If the position of an object does not change with respect to time, it is said to be at rest. this topic • If the position of an object changes as time passes, it is said to be in motion. • Reference point is a fixed point with respect to which a body is at rest or in motion. • Rest and motion are relative terms.  Distance is the length of the actual path travelled by a body in a given time. Uniform and  Displacement is the shortest distance between the initial and final positions of the body in a Non-Uniform specified direction. Motion  A physical quantity which has both magnitude and direction is known as vector quantity. e.g., velocity, force.  A physical quantity which has only magnitude is known as scalar quantity. e.g., time, speed.  The S.I. unit of distance and displacement is metre.  A body is said to be in uniform motion, if it travels equal distances in equal intervals of time.  A body is said to have non-uniform motion if it travels unequal distances in equal intervals of time.  In non-uniform motion, speed of an object is not constant. The S.I. unit of speed is m/s.  Speed is the ratio of distance travelled to the time taken to cover that distance. Scan to know more about  Average speed of a body is the total distance travelled divided by the total time taken. this topic  Velocity is displacement per unit time. The S.I. unit of velocity is metre per second.  Average velocity is the total displacement divided by the total time taken.  Time is an independent variable, plotted along X-axis. Distance is a dependent variable, plotted along Y-axis in the distance-time graph. Distance,  Graphs are designed to make it easier for the reader to interpret and understand numerical Displacement, data. Speed and  The distance-time graph is a straight line parallel to time axis when the object is at rest. Velocity  The nature of distance-time graph is a straight line, with fixed slope, when the object is in the state of uniform motion.

MOTION

83

84

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

 Slope of the position-time graph gives the velocity of the object.  A more steeply inclined position-time graph indicates greater velocity. The nature of position-time graph is a curve having varying slope when the object has non-uniform motion.  If the velocity of a body remains constant, the velocity-time graph is a horizontal line parallel to the time axis.  If the velocity of the body changes uniformly at a constant rate, the velocity-time graph is a Scan to know straight line. more about  If the velocity of the object changes non-uniformly, the velocity-time graph is a curve having this topic increasing or decreasing slope.  The area enclosed by the velocity-time graph and the time axis represents the displacement.  The slope of the velocity-time graph gives the acceleration.  Acceleration is a vector quantity. Its SI unit is m/s2. What is  When a body travels along a circular path of constant radius with a constant speed then its Acceleration? motion is uniform circular motion.  In a uniform circular motion, velocity of a particle is not constant but its speed is constant, hence it is an accelerated motion.



Keywords  Distance : Distance is the actual length of the path followed by a moving object. Distance is a scalar quantity. SI unit of distance is metre .  Displacement : It is the shortest distance covered by a moving object from the point of reference (initial position of the body), in a specified direction. SI unit of displacement is metre.  Uniform speed : An object is said to be moving with uniform speed if it covers equal distances in equal intervals of time.  Non-uniform speed : An object is said to be moving with variable speed or non-uniform speed if it covers equal distances in unequal intervals of time or vice-versa.  Average speed : The average speed is the ratio of the total distance travelled by an object to the total time taken for the journey.  Instantaneous speed : The speed of a moving body at any particular instance of time is called instantaneous speed.  Velocity : It is defined as the distance covered by a moving object in a particular direction in unit time or speed in a particular direction.  Acceleration : It is defined as the rate of change of velocity of a moving body with time.  Uniform Acceleration : If the change in velocity in equal intervals of time is always the same, then the object is said to be moving with uniform acceleration.  Non-uniform or Variable Acceleration : If the change in velocity in equal intervals of time is not the same, then the object is said to be moving with variable acceleration.  Uniform velocity : A body is said to be moving with uniform velocity if it covers equal displacements in equal intervals of time in a specified direction.  Variable velocity : A body is said to be moving with variable velocity if it covers unequal displacements in equal intervals of time and vice-versa in a specified direction or if it changes the direction of motion.  Circular motion : Motion along circular track is called circular motion.



Example The following graph describes the motion of a girl going to meet her friend who stays 50 m from her house.



(i) How much time she takes to reach her friend’s house ? (ii) What is the distance travelled by the girl during the time interval 0 to 12 min ? (iii) During which time interval she is moving towards her house ? (iv) For how many minutes she was at rest, during the entire journey ? (v) Calculate the speed by which she returned home.

MOTION

Solution: (i) The friend’s house is 50 m away so the time taken to reach that distance from her initial position O to G is 14 min. (ii) Distance traversed in 0–2 m = 20m Distance traversed in 2–4 m = 0m Distance traversed in 4–6 m = 20m Distance traversed in 6–8 m = 0m Distance traversed in 8–10 m = 20m Distance traversed in 10–12 m = 0m

85

So, total distance traversed during 0 to 12 minute time interval = 20 + 0 + 20 + 0 + 20 + 0 = 60m (iii) During the time interval 8–10 min. (iv) Time when she was at rest = AB + CD + EF =2+2+2 = 6 min 50 (v) Speed with which she reached home = 2 × 60 = 0.417 m/s [1 × 5 = 5]

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions

(1 mark each)



Q. 1. A particle is moving in a circular path of radius r. The displacement after half a circle would be (A) Zero (B) π r (C) 2r (D) 2 π r Ans. Option (C) is correct. Explanation: Given, after half a circle, the particle will reach the diametrically opposite point i.e., from point A to point B. We know displacement is shortest path between initial and final point. So, displacement after half circle = AB=OA + OB [Given, OA and OB = r]

A

r

.

r B

O

= r + r= 2 r Hence, the displacement after half circle is 2r. Q. 2. The numerical ratio of displacement and distance for a moving object is : (A) Always less than 1 (B) Always equal to 1 (C) Always more than 1 (D) Equal or less than 1. Ans. Option (D) is correct. Explanation: Displacement of an object can be less than or equal to the distance covered by the object, It can never be more than the distance. Q. 3. Suppose a boy is enjoying a ride on a merry-goround which is moving with a constant speed of 10 ms–1. It implies that the boy is : (A) At rest (B) Moving with no acceleration

(C) In accelerated motion (D) Moving with uniform velocity. Ans. Option (C) is correct. Explanation: In merry-go-round, the speed is constant but velocity is not constant, because its direction goes on changing i.e., there is acceleration in the circular motion. So, we can say that the boy is in accelerated motion. Q. 4. From the given v-t graph (see figure), it can be inferred that the object is :





(A) In uniform motion (B) At rest (C) In non-uniform motion (D) Moving with uniform acceleration.

Ans. Option (A) is correct. Explanation: From the given v-t graph, it is clear that the velocity of the object is not changing with time i.e., the object is in uniform motion. Q. 5. Area under v-t graph represents a physical quantity which has the unit (A) m2 (B) m (C) m5 (D) ms−1 Ans. Option (B) is correct. Explanation: Area under v-t graph represents displacement whose unit is metre or (m). It is because, unit of velocity v = m/s and unit of time = s. Q. 6. Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs is shown in figure. Choose the correct statement.

86

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX C

Distance (m)

D A B

Time (s)

(A) Car A is faster than car D (B) Car B is the slowest (C) Car D is faster than car C (D) Car C is the slowest. Ans. Option (B) is correct. Explanation: The slope of distance-time graph represents the speed. From the graph, it is clear that the slope of distance-time graph for car B is the lowest compared to all other cars. Hence, car B is the slowest.

B

Assertion and Reason 

(1 mark each)

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A)  Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

(C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion : A boy riding on bicycle in a crowded street exhibit non uniform motion. Reason : The boy covers equal distances in equal intervals of time. Ans. Option (C) is correct. Explanation: Boy riding on the bicycle exhibit non-uniform motion. He covers unequal distances in equal intervals of time. Q. 2. Assertion : Velocity is the speed of an object in a particular direction. Reason : SI unit of velocity is same as speed. Ans. Option (B) is correct. Explanation: Velocity is the speed of an object moving in a direction, or the displacement of an object in unit time. Speed, on the other hand, is the distance travelled by an object in the given time. SI unit of velocity is same as speed i.e., m/s. Q. 3. Assertion : A stone tied with a piece of thread describing a circular path with constant velocity on being released moves in a straight line. Reason : Along the circular path direction of motion remains the same at every point. Ans. Option (C) is correct Explanation: In a circular path direction of motion changes at every point. The direction of motion at any point is the tangent to that point in circular path.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions

(1 mark each)

Q. 1. Mention the condition when the magnitude of the average velocity of an object is equal to its average speed.  U [Board Term-I, 2016]

Q. 4. Give an example of a motion in which acceleration is non-uniform.

Ans. If the body moves along a straight line in a particular direction. [CBSE Marking Scheme, 2016] [1] Q. 2. Mention the physical quantity shown by the slope of a velocity – time graph.  U [Board Term-I, 2016]

Ans. A car travelling in a straight road changes its speed by unequal amounts in equal intervals of time.[1]

Ans. Acceleration.  [CBSE Marking Scheme, 2016] [1] Q. 3. If the time interval between lightning and thunder is 2 seconds, what is the distance of the point of lightning? (Speed of sound in air is 346 ms–1)  A [Board Term-I, 2016] Ans. Speed of light is too high. So, the light reaches almost immediately. The speed of sound (v) is much lower than that of light. So, distance of lightning point, d = v×t = (346)×(2) = 692 m[1]



A [Board Term-I, 2015]

Q. 5. An object travels a distance of 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object ? A [Board Term-I, 2014]

Ans. Average speed = (16 + 16) / (4 + 2) =

32 6

= 5.33 m/s [CBSE Marking Scheme, 2014] [1]

Q. 6. An ant travels from P to Q and then moves from Q to R (as shown in the diagram). Show its resultant displacement in the diagram. A [Board Term-I, 2014]

MOTION

Q

87

R

Commonly Made Error

Many students are unable to answer this

question. It seems they are unaware of this concept.

P

Answering Tip

Ans. PR

Q

R



P

[CBSE Marking Scheme, 2014] [1]

Q. 7. Does the speedometer of a car measure its average speed ? A (NCERT, 2014] Ans. No. It measures its instantaneous speed.[1]

Students should thoroughly learn about the

speedometer, odometer, instantaneous speed and average speed. Many times students get confused while answering.

Q. 8. Can an object be at rest as well as in motion at the same time ? U Ans. Yes, an object may be at rest related to one reference point and at the same time it may be in motion related to another reference point.[1] Q. 9. If the displacement of a body is zero, is it necessary that the distance covered by it is also zero ?  A [NCERT Exemplar] [NCERT] Ans. No, When the body comes back to the starting position after travelling a distance, its displacement is zero though it has travelled some distance.[1]

Short Answer Type Questions-I Q. 1. Plot the velocity - time graphs showing +a and –a. How is distance calculated from velocity - time graph? A (NCT-2014; DDE-2014)

 Distance from velocity - time graph can be calculated by finding the area under the graph[2] Q. 2. Enumerate the differences between speed and velocity in a tabular form.  U [NCT 2014] OR Distinguish between speed and velocity. [NCERT] Ans. S. No. (i)

Speed

Velocity

Speed is defined as Velocity is the rate of change defined as the of distance. rate of change of displacement.

(2 marks each) (ii)

Speed is a scalar quantity.

Velocity is a vector quantity.

(iii)

Speed may or may not be equal to velocity.

A body may possess different velocities but the same speed.

(iv)

Speed can never be negative or zero.

Velocity can be negative, zero or positive. [½ × 4 ] Q. 3. What is the difference between uniform velocity U and non-uniform velocity ? Ans. Uniform Velocity : An object with uniform velocity covers equal distances in equal intervals of time in a specified direction e.g., an object moving with speed of 40 kmh–1 towards west has uniform velocity.

Non-uniform Velocity : When an object covers unequal distances in equal intervals of time in a specified direction, or if the direction of motion changes, it is said to be moving with a nonuniform or variable velocity e.g., revolving fan at a constant speed has variable velocity.[2] Q. 4. Make a velocity – time graph from the following displacement – time graph.  A

88

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Ans. From the graph

Velocity after 2s =













10 = 5 m/s 2 20 Velocity after 4s = = 5 m/s 4 30 Velocity after 6s = = 5 m/s 6 Velocity is constant, acceleration = 0 So, velocity - time graph will be drawn as : [1½]



[½]

Short Answer Type Questions-II Q. 1. Explain the following type of motion with one example for each.

(3 marks each)

Ans. (i) 0 to 10 sec. (ii)

(i) acceleration is positive

(ii) acceleration is negative (iii) acceleration is zero. 

A [Board Term-I, 2016]

Ans. (i) Here, the motion is accelerated motion, e.g., car moving on a road with increasing velocity. (ii) Here, the motion is retarded motion, e.g., brakes applied to a moving car. (iii) Here, the motion is uniform motion, e.g., car moving with a constant speed along a road. [1 + 1 + 1] Q. 2. The velocity-time graph shown represents the motion of a body.

below

Total displacement = area under the curve OABCF

= area of triangle OAD + area of trapezium DACF = 1/2 (OD × DA) + 1/2 (DA + FC)DF = 1/2(10 × 4) + 1/2(4 + 6)20 = 120 m Average velocity = 120/30 s = 4 m/s [CBSE Marking Scheme, 2016] [1½ + 1½] Q. 3. A particle moves over three quarters of a circle of radius r cm. Calculate the magnitude of its distance and displacement.  Ans.

(i) During which interval of time, the body is moving with maximum acceleration ?

(ii) Calculate the average velocity for the entire journey. U+A [Board Term-I, 2016]

A [Board Term-I, 2016]

89

MOTION





A particle moves over three quarters of a circle of radius r cm. 3 3 Distance travelled = 2 r  = pr cm 4 2





Displacement = AD





AD =

r2 + r2

= r 2 cm. [CBSE Marking Scheme, 2016] [3] Q. 4. On a 120 km long railway track, a train travels the first 80 km at a uniform speed of 160 km/h. Calculate the speed with which the train should move on the rest of the track so as to get the average speed of 80 km/h for the entire trip.  A [Board Term-I, 2016] Ans. Total length of the track = 120 km Total time taken to cover total distance = x Average speed = 80 km/hr Average speed = 120/x 80 = 120/x x = 120/80 = 1.5 hr

Q. 5. 

irst 80 km were covered with speed of 160 km/hr. F Time taken to cover this 80 km = Distance/speed = 80/160 = 0.5 hr As we know, total time taken is 1.5 hr, so when first 80 km is covered in 0.5 hr, then rest of the distance is covered in 1 hr. At the rest portion, Time taken to cover 40 km = 1 hr Speed of the train at the last portion, = Distance/time = 40/1 = 40 km/hr. [3] Name the physical quantities denoted by : (i) the slope of the distance–time graph (ii) the area under velocity–time graph (iii) the slope of velocity–time graph. A [Board Term-I, 2015] [DDE 2014) [Board Term-I, 2014]

Ans.



(i) Speed

(ii) Displacement (iii) Acceleration [1 + 1 + 1] [CBSE Marking Scheme, 2014]



Long Answer Type Questions

(5 marks each)

Q. 1. (a) Give two differences between distance and displacement. The position time graph for children ‘A’ and ‘B’ returning from their school ‘O’ to their homes ‘P’ and ‘Q’ returning are shown in fig. From the graph find : (i) Which of the two ‘A’ or ‘B’ lives closer to school? (ii) Which of the two ‘A’ or ‘B’ starts earlier from school ? (iii) Which of the two ‘A’ or ‘B’ walks faster ? [KVS, 2018-19] Q

A

O



t2 t1 time (b) The speed of a car increases from 18 km/h to 36 km/h in 10 seconds. Find its acceleration. U+A

 Ans.(a) S. No. 1.

Distance

Displacement Displacement is the shortest distance between the initial and final positions of an object in a given direction.

Distance covered Displacement may can never be be positive, negative negative. It is always or zero. positive or zero.

4.

Distance between two given points may be same or different for different paths chosen.

Displacement between two given points is always the same.

(Any two) (i) A (ii) A (iii) B. (b) Given, u = 18 km/h = 5 m/s v = 36 km/h = 10 m/s, t = 10 s a = (10 – 5)/10 = 0·5 m/s2 [2 + 1½ + 1½] Q. 2. (i) Give one similarity and one dissimilarity between the two graphs. y

y B

A

Distance is the length of the actual path covered by an object, irrespective of its direction of motion.

3.

B

Velocity (m/s)

position

B

Distance is a scalar Displacement is a quantity. vector quantity.

Velocity (m/s)

P

2.

time (s)

x

A

time (s)

x

(ii) What do you understand by the term acceleration? What is meant by its being positive or negative ? Explain with example. Write its SI unit.  U+A [Board Term-I, 2016]

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Ans.

(i) Similarity: Both have uniform acceleration.





Dissimilarity: In first graph body starts from rest. In second graph it has initial velocity.



(ii) Acceleration is the rate of change of velocity.



Positive: Positive acceleration increases 4s the velocity of an object. Example: Free falling body due to gravity.

Negative: Negative acceleration decreases the velocity. Example: Body thrown up with an initial velocity. SI Unit: ms–2[2+3] [CBSE Marking Scheme, 2016] Q. 3. (a) A train 100 m long is moving with a constant velocity of 60 kmh–1. Find the time it takes to cross the bridge 1 km long. (b) The slope of the line on a position-time graph reveals information about an object’s velocity. What conclusion can you draw regarding the motion of an object, if the position time graph is a : (i) horizontal line parallel to time axis. (ii) straight line originating from origin. (iii) curve.

A [Board Term-I, 2016]

Ans. (a) s = 1.1 km distance t= = 1.1 = 0.018 hours speed 60 (b) (i) the body is at rest. (ii) the body is at uniform motion. (iii) the body is at non-uniform motion.



Detailed Answer : (i) Impossible. Reason: Distance can never be zero when displacement is non zero. For example, in a circular motion distance is circumference of the circle . But the displacement is zero. (ii) Possible. Reason : Acceleration in opposite direction of motion at an instant makes the speed zero but the acceleration exists. Example: When an object is thrown upward, at the highest point its velocity becomes zero for an instant of time, but its acceleration (g) remains non-zero due to attraction of gravity. (iii) Impossible. Reason: Velocity and speed are related to displacement and distance, respectively, so the reason is the same as (i) (iv) Possible. Reason : Acceleration in opposite direction exists which retards the motion. Example: When a moving car applies the brake its velocity decreases. The acceleration is against the direction of motion of the car. (v) Possible. Reason : While speeding up, the final velocity is greater than the initial velocity so the acceleration is positive. Example: When a train starts from a station with zero initial velocity, it applies positive acceleration and gains speed.  [1 × 5] Q. 5. Study the velocity-time graph and calculate :

[CBSE Marking Scheme, 2016] [2 + 3]

Q. 4. State with reasons, if it is possible or impossible, for an object in motion to have : (i) Zero distance covered and may have nonzero displacement. (ii) Zero speed at an instant but non-zero acceleration at the same time. (iii) Zero speed and may have non-zero velocity. (iv) Acceleration opposite to the direction of motion. (v) Positive acceleration while speeding up.  A [Board, 2014]

(i) The acceleration from A to B

Ans. (i) Impossible, reason

(ii) The acceleration from B to C

(ii) Possible, reason

(iii) The distance covered in the region ABE

(iii) Impossible, reason

(iv) The average velocity from C to D

(iv) Possible, reason

(v) The distance covered in the region BCFE. U+A [NCERT] 

(v) Possible, reason [CBSE Marking Scheme, 2014]

MOTION

Ans.

(i) a =

(ii) a =

 25  0  3  0  20  25  43

(iv) v =

= 8.3 m/s2

 20  0  2

91

= 10 m/s

(v) Distance = Area of trapezium BCFE = – 5 m/s2

1 = (25 + 20) × (4 – 3) = 22.5 m 2 [1 × 5 = 5]

1 (iii) Distance = Area of triangle ABE = × 3 × 25 2 = 37.5 m

Topic-2 Uniformly Accelerated Motion Revision Notes  First Equation of Motion : Consider a particle moving along a straight line with uniform acceleration ‘a’. At t = 0, let the particle be at A and u be its initial velocity and when t = t, v be its final velocity.

Acceleration =

⇒  Second Equation of Motion :

change in velocity time

Equations of Motion

= v-u t

a = v - u t v – u = at v = u + at

(I equation of motion)

total distance travelled total time taken s Average velocity = t Average velocity =

...(i)

u+v 2 u+v s From equations (i) and (ii) = 2 t The first equation of motion is v = u + at. Substituting the value of v in equation (iii), we get u + u + at s = 2 t Average velocity can also be written as

or or

Scan to know more about this topic

s =

...(ii) ...(iii)

(u + u + at )t 2

s = ut +

 Third Equation of Motion : The first equation of motion is v = u + at v – u = at s Average velocity = t

1 2 at 2

Average velocity = u + v 2

(II equation of motion)

...(i) ...(ii) ...(iii)

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

From equation (ii) and equation (iii) we get,

u+v = s 2 t Multiplying equation (i) and equation (iv) we get, (v – u) (v + u) = at ×

...(iv) 2s t

Q. 1. A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is : (A) u/g

(B) u2/2g

(C) u2/g

(D) u/2g

Ans. Option (B) is correct. Explanation: Given, initial velocity = u, height = h and a = g (acceleration due to gravity). At the highest point, final velocity becomes zero i.e., v = 0. From, third equation of motion, v2= u2 − 2gh 0 = u2 − 2gh 2gh = u2 2 h= u 2g

Distance (m)

Distance (m)

Q. 2. Which of the following figure represents uniform motion of a moving object correctly?

Time (s) (b)

Distance (m)

Distance (m)

Time (s) (a)

Time (s) (b)

Distance (m)

Choice Questions A Multiple  (1 mark each)

Distance (m)

Time (s) OBJECTIVE TYPE QUESTIONS (a)

Distance (m)

Distance (m)

(v – u) (v + u) = 2as [a2 – b2 = (a + b) (a – b)] v2 – u2 = 2as (III equation of motion)  When a particle is moving upwards or downwards, the above equations of motion can be written as: (a) For upward motion (1) v = u – gt 1 (2) h = ut – gt2 2 (3) v2 – u2 = – 2gh (b) For downward motion (1) v = u + gt 1 (2) h = ut + gt2 2 (3) v2 – u2 = 2gh where, h = height covered, g = acceleration due to gravity, v = final velocity and u = initial velocity

Time (s) (c)

Time (s) (d)

Ans. Option (A) is correct. Explanation: For uniform motion, the distancetime graph is a straight line. This is because in uniform motion object covers equal distance in equal interval of time. Q. 3. If the displacement of an object is proportional to square of time, then the object moves with : (A) Uniform velocity (B) Uniform acceleration (C) Increasing acceleration (D) Decreasing acceleration. Ans. Option (B) is correct. Explanation: From second equation of motion, s= ut + 1 at2 2 If object start from rest i.e. its initial velocity (u) = 0 and has an acceleration (a) Then, s = 0 + 1 at2 2 s = 1 at2 2

MOTION

s ∝ t2, if a = constant So, the object moves with constant or uniform acceleration.

B

Assertion and Reason 

(1 mark each)

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A)  Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion : Motion of satellites around their planets is considered an accelerated motion.

93

Reason : During their motion, the speed remains constant, while the direction of motion changes continuously. Ans. Option (A) is correct. Explanation: Satellites revolve around their planets in almost circular orbits with constant speed. Thus, during their motion, the speed remains constant, while the direction of motion changes continuously. As a result, there is a change in their velocity. Therefore, the motion of satellites around their planets is considered as accelerated motion. Q. 2. Assertion : A tiger can accelerate from rest at the rate of 4 m/s2. Reason : The velocity attained by it in 10 s is 40 m/s. Ans. Option (A) is correct. Explanation: Initial velocity (u) = 0, acceleration (a) = 4 m/s2 v = u + at v = 0 + 4 × 10= 40 m/s.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions Q. 1. Give one example of a body whose average velocity is zero. A Ans. When the body returns back to initial point, e.g., when an athlete returns back to starting point, his average velocity is zero. [1]

Q. 2. Give one example of a body whose average speed is zero. A Ans. Average speed of a moving body can never be zero. [1]

Short Answer Type Questions-I Q. 1. Why is the motion of satellites around their planets considered an accelerated motion ? A Ans. Satellites revolve around their planets in almost circular orbits with constant speed. Thus, during their motion, the speed remains constant, while the direction of motion changes continuously. As a result, there is a change in their velocity. Therefore, the motion of satellites around their planets is considered as accelerated motion. [2]

(2 marks each)

Q. 2. A bridge is 500 m long. A 100 m long train crosses the bridge at a speed of 30 m/s. Find the time taken by train to cross it. A Ans. Total length of path covered by train = 500 m + 100 m = 600 m. Speed of train = 30 m/s. Time taken by train to cover the bridge distance = speed

=

Short Answer Type Questions-II Q. 1. A ball thrown up vertically returns to the thrower after 6 seconds. Find : (a) The velocity with which it was thrown up. (b) The maximum height it reaches. (c) Its position after 4 seconds. Ans. (a) Time of ascent = Time of descent So, to attain the maximum height ball takes 3s v = 0, g = – 9.8 m/s2 v = u + gt

(1 mark each)

600 = 20 seconds sec. 30

[2]

(3 marks each)

0 = u + (–9.8 × 3) u = 9.8 × 3 = 29.4 m/s = Velocity with which it was thrown (b) u = 29 m/s, v = 0, g = – 9.8 m/ s2, t = 3s h = ut + 1/2 gt2 = 29.4 × 3 + 1/2 × (–9.8) × (3)2 = 88.2 – 44.1 = 44.1 m = maximum height it reaches.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

(c) s = ut + 1/2 gt2 = 29.4 × 4 – 1/2 × 9.8 × (4)2 = 117.6 – 78.4 = 39.2 m Ball is 39.2 m above the ground after 4 seconds. [1+1+1] U+A [KVS, 2018-19] Q. 2. The brakes applied to a car produce an acceleration of 6 ms–2 in the opposite direction to the motion. If the car takes 2s to stop after the application of brakes, calculate the distance it travels during this time. A [Board Term-I, 2016]



Ans. Here From





a = – 6 m/s2, t = 2 s, v = 0, s = ? v = u + at 0=u–6×2 u = 12 m/s

From v2 − u2 = 2as

s= =

v 2 - u2 2a 0 - 12 × 12 2×6

= 12 m  [CBSE Marking Scheme, 2016] [3] Q. 3. A motor cycle moving with a speed of 5 m/s obtains an acceleration of 0.2 m/s2. Calculate the speed of the motor cycle after 10 seconds, and the distance travelled by it in this time. 

A [Board Term-I, 2014]

Ans. Here, u = 5 m/s, a = 0.2 m/s2, t = 10 s, s = ?, v = ? From, v = u + at v = 5 + 0.2 × 10 = 7 m/s From, v2 = u2 + 2as 72 = 52 + 2 × 0.2 × s 49 = 25 + 0.4 s



49 – 25 = 0.4 s



0.4 s = 24



s = 24 / 0.4



s = 60 m [CBSE Marking Scheme, 2014] [3]



Q. 4. From a station ‘X’ a train starts from rest and attains a speed of 54 km/h in 10 s. Then by applying brakes negative acceleration of 2.5 ms–2 is produced and it stops at station ‘Y’ in 6 s. Find the distance between station ‘X’and ‘Y’. A [Board Term-I, 2016]  Ans. Here, u = 0, v = 54 km/h = 15 m/s, t = 10 s v = u + at a=

v−u 15 = = 1.5 m/s2 t 10

The distance travelled, in first part 1 s1 = ut + at 2 2 = 0 +

1 × 1.5 × 102 2

= 75 m In the last part, u = 15 m/s, v = 0, t = 6s, a = – 2.5 m/s2 v2 = u2 + (– 2as2) or, s2 =

u2 2a

(15 )2 = 2 × 2.5 m / s2 = 45 m Distance between X and Y = s1 + s2 = (75 + 45) m = 120 m

Long Answer Type Question Q. 1. (a) When will you say a body is in: (i) Uniform acceleration. (ii) Non-uniform acceleration. (b) A train starts from rest and accelerates uniformly for 30 s to acquire a velocity of 108 km/h. It travels with this velocity for 20 min. The driver now applies brakes and the train retards uniformly to stop after 20 s. Find the total distance covered by the train.  R+Ap [Board Term-I, 2013] Ans. (a) (i) A body is in uniform acceleration if it travels in a straight path when its velocity increases equally or decreases by equal time intervals. (ii) A body is in non-uniform acceleration if it travels in a straight path when

[3]

(5 marks each)

its velocity increases or decreases by unequal amount in equal time intervals. (b) When train starts from rest, u = 0, t = 30 s, v = 108 km/h

= 108 ×

5 = 30 m/s. 18

v−u = 1 m/s2. t 1 2 Distance S1 = ut + at 2

a=

1 × 1 × 30 × 30 2 = 450 m. At uniform velocity of 30 m/s the train runs for 20 min = 1200 s,

=0+

MOTION

Distance S2 = 30 × 1200 = 36000 m. On applying brakes, t = 20 s, u = 30 m/s, v = 0

a=

v−u 0 − 30 = = – 1.5 m/s2 t 20

95

2 0   30  v 2 − u2 Distance S3 = = = 300 m 2   1.5  2a Total distance = S1 + S2 + S3 = 450 + 36000 + 300 = 36750 m.  [2 + 3]

COMPETENCY AND CRITICAL THINKING BASED QUESTIONS

(1 mark each)

I. The graph shows the distance-time graph of three objects A, B and C. Study the graph and answer any four questions.

Q. 1. Which of the three is travelling the fastest? (A) A (B) B (C) C (D) All of them are at rest. Ans. Option (B) is correct. Explanation: Speed = Distance/Time From the graph, we see that,

A travels from 6 to 12 km in 2 hours.



B travels from 0 to 12 km in 1.4 hours.



C travels from 2 to 12 km in 1.6 hours.

From graph distance travelled by A is 8 km. Q. 4. How far has B travelled by the time it passes C? (A) 12 km (B) 8 km (C) 16 km (D) 5 km Ans. Option (D) is correct. Explanation: The distance travelled by B is around 5 km.



Since, B travels the maximum distance, in the smallest time, it is travelling the fastest. Q. 2. All the three objects A,B & C __________ (A) once met at a single point. (B) were at the same point at the beginning. (C) will meet at the same point at the end. (D) were never at the same point on the road. Ans. Option (D) is correct. Explanation: All the three objects A, B and C never meet at a single point. Thus, all the three were never at the same point on the road. Q. 3. How far has C travelled when B passes A? (A) 12 km (B) 8 km (C) 10 km (D) 16 km Ans. Option (B) is correct. Explanation:

Q. 5. The distance-time graph is a ______________ when the object is at rest. (A) straight line passing through origin (B) straight line parallel to time axis (C) a curve (D) straight line intersecting y-axis Ans. Option (B) is correct.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Explanation: The slope is zero when straight line is parallel to time axis. And the slope of distance - time graph gives velocity. II. Read the following passage and answer following questions. The diagram shows a satellite travelling in uniform circular motion around the Earth.

Q. 1. Explain what is meant by the term uniform circular motion. Ans. Uniform circular notion is the motion of an object that travels along a circular path at a constant speed. In this type of motion, the direction of the object's velocity changes constantly as it moves around the circular path, but its speed remain constant. Q. 2. The satellite is kept in orbit by a force. What is the direction of this force? Ans. This force is always directed towards the centre of the celestial body, such as the Earth. This is known as centripetal force. Q. 3. State and explain whether this force does any work on the satellite. [CBSE – SAS] Ans. No, because force is perpendicular to the motion of the satellite OR there is no movement in the direction of the force.



MOTION

97

Artificial Intelligence PARAMETERS

DESCRIPTION

Chapter Covered

Chapter 8: Motion

Name of the Book

Science, Class 8, NCERT

Subject and Artificial Intelligence Integrated

Understanding Concept of Uniform and Non-Uniform motion using Google maps AI Applications and Data Acquisition and Data exploration.

Learning Objectives

● To understand the concept of data handling in motion. ● To identify steps of data handling - Sources of data - Collection of data - Organisation, representation and analysis of data. ● To differentiate between uniform and non-uniformmotion ● To understand application of Data Acquisition and Data exploration in real life situations.

Time Required

2 periods of 40 minutes each

Classroom Arrangement

Flexible

Material Required

Graph Paper, Colored Pen, paper, Black Board chalk, Laptops/ desktops and Internet connections.

Pre-Preparation Activities

● The students will be divided into groups for a discussion on types of motion they observe around them. ● Students will be asked to collect data on the types of motion identified by them, for eg. - Distance between their school and home and time taken to cover it. - Distance between their home and nearest shopping mall and time taken to cover it. - Time taken to travel to school in traffic jam.

Previous Knowledge

● Students must possess the knowledge of initial and final position for a given motion. ● Students should have knowledge of speed and distance problems. ● They should be aware about speed calculations in real life.

Methodology

● Ask students to collect data (distance and time) through Google Maps. Inform how Google maps help us to know about the real time needed to travel from oneplace to another on the basis of the speed of the vehicle. Ask students to calculate the time for the same distance if travelling by - Car - SchoolBus - On foot(walking). - Cycle ● After collecting data discuss analyse and represent data with the help of some graphical representation. Students through the graph will be able to differentiate between Uniform and Non-Uniform. Ask them to predict the type of motion before plotting graph (at the end compare their answers to see who wins) ● Ask the students to go on https://datavizcatalogue.com and explore various types of graphs and the way to use these. Ask them to select a representation which will suit their data best(line graph) Students will be able to recognise various patterns/trends out of their representations which can be used to represent this problem. Ask the students to explore the possibilities/applications of using AI in solving this problem.

AI CONCEPTS INTEGRATED

Data Acquisition Data exploration Google map Data Visualisation

Scan to know more about this topic

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Learning Outcomes

● Students will be able to understand and differentiate between Uniform and Non-Uniform motion. ● Students will be able to understand the dependence of speed/velocity on distance travelled and time taken.

Follow up Activities

● They will explore similar AI tools to deepen the learning of this concept. ● Students can use other AI tools for graphical representation of complete journey of train/flight.

Reflections

Discussion with Students on the role of AI application. 

Study Time Reading Time: 3:00 Hr No. of Questions: 50

CHAPTER

8



Syllabus

FORCE AND NEWTON'S LAWS

Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration.

TOPIC - 1 Force, Laws of Motion and Acceleration.... P. 99

Force, Laws of Motion and

Topic-1 Acceleration

TOPIC - 2 Inertia and conservation of momentum .... P. 106

Revision Notes

Scan to know Force is a push or pull acting upon an object. more about Balanced forces : The resultant of all the forces acting on a body is zero. this topic Unbalanced forces : The resultant of all the forces acting on a body is not zero. Newton’s first law of motion states that a body at rest will remain at rest and a body in motion will remain in uniform motion unless acted upon by an unbalanced force.  The net force acting on the object is zero, whenever balanced forces act on it. Force  The momentum of an object is the product of its mass and velocity and has the same direction as that of the velocity. Its SI unit is kg-m-s–1.  Newton’s second law of motion states that the rate of change of momentum of a body is directly proportional to the force and takes place in the same direction as the force.  Force is also defined as the product of mass and acceleration. Scan to know  The SI unit of force is kg-m-s–2. This is also known as Newton and represented by the symbol N. more about –2 this topic  A force of one Newton produces an acceleration of 1 m-s on an object of mass 1 kg.  Force of friction always opposes motion of objects.  Two forces resulting from the interaction between two objects are called action and reaction forces respectively.  Action and reaction forces act on two different bodies but they are equal in magnitude. Newton’s laws of Motion  Newton’s third law of motion : For every action there is an equal and opposite reaction; but action and reaction act on different bodies.

   

Keywords

 Force : A force is a physical quantity which, when unopposed, will change the motion, direction and shape of an object.



 Balanced Force : When two forces of equal magnitude act in opposite directions on an object simultaneously, then the object continues in its state of rest or a uniform motion in a straight line. Such forces acting on the object are known as balanced force.

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FORCE AND NEWTON'S LAWS



101

 Unbalanced Force : When two forces of unequal magnitudes act in opposite directions on an object simultaneously, then the object moves in the direction of the larger force. These forces acting on the object are known as unbalanced force.  Momentum : Momentum of a body is equal to the product of the mass (m) of the body   and the velocity v of the body. It is denoted by p .  Recoil Velocity : The velocity with which the gun moves backward after firing a bullet is known as recoil velocity.  Friction : Whenever a body slides or rolls over the surface of another body, a force comes into action which acts in the opposite direction of the motion of a body. This opposing force is called ‘friction’.  Resultant Forces : The resultant force or resultant of several forces acting simultaneously on a body is that single force which produces the same effect on a body as all these forces together produce.

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions 

(1 mark each)

Q. 4. A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is :

Q. 1. Which of the following statement is not correct for an object moving along a straight path in accelerated motion? (A) Its speed keeps changing (B) Its velocity always changes (C) It always goes away from the Earth (D) A force is always acting on it. Ans. Option (C) is correct. Explanation: For an object moving along a straight path in an accelerated motion, it is not necessary that it always goes away from the Earth. Q. 2. According to the third law of motion, action and reaction :

(A) Accelerated

(A) Always act on the same body.

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion: If a balanced force is applied on a wooden block it will move. Reason: Unbalanced force changes the state of motion or rest while balanced force does not. Ans. Option (D) is correct. Explanation: Force applied on both the sides of wooden block in equal magnitude counterbalance each other. Therefore do not bring about any change in the state of motion or rest of the object. Such types of forces are known as balanced forces. Unbalanced force changes the state of motion.

(B) Always act on different bodies in opposite directions. (C) Have same magnitude and direction. (D) Act on either body at normal to each other. Ans. Option (B) is correct. Explanation: According to third law of motion, action and reaction always act on different bodies in opposite directions. Q. 3. A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goalkeeper to : (A) Exert larger force on the ball (B)  Reduce the force exerted by the ball on hands (C) Increase the rate of change of momentum (D) Decrease the rate of change of momentum. Ans. Option (D) is correct. Explanation: The goalkeeper pulls his hands backwards after holding the ball to decrease the rate of change of momentum by increasing the time. By doing this, less force is exerted on his hands.

(B) Uniform (C) Retarded (D) Along circular tracks. Ans. Option (A) is correct. Explanation: If the coin falls behind the passenger that means the train is accelerated. When the coin is tossed it has same velocity as that of train but during the time it is in air its velocity becomes less than that of train (because the train is accelerated), so it falls behind the passenger.

B

Assertion and Reason 

(1 mark each)

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Q. 2. Assertion : When we stop pedalling a bicycle it slows down. Reason : Force of friction always acts in the direction of motion. Ans. Option (C) is correct. Explanation: When we stop pedalling a bicycle it slows down due to force of friction which acts in the opposite direction of motion. Q. 3. Assertion : If a spring is stretched from one side, the size and shape of the spring changes.

Reason : Unbalanced force acting on the spring changes the size and shape of the spring in the direction of application. Ans. Option (A) is correct. Explanation: If a spring is stretched from one side, the size and shape of the spring changes. It is because unbalanced force acting on the spring changes the size and shape of the spring in the direction of application.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions Q. 1. What is the momentum of a toy car of mass 200 g moving with a speed of 5 m/s ? A [Board Term-I, 2016] Ans. Momentum = Mass × velocity =

200 × 5 =1 kg ms–1. 1000

[CBSE Marking Scheme, 2016] [1] Q. 2. Shyam throws a heavy stone out of his boat. As a result, the boat moves in opposite direction. Why? A [Board Term-I, 2016] Ans. According to III Law of Motion for every action there is an equal and opposite reaction.  [CBSE Marking Scheme, 2016] [1] Q. 3. While getting down a moving bus, why should a person run in the same direction as that of the bus? A [Board Term-I, 2016] Ans. A person in a moving bus possesses inertia of motion. Thus, if he simply jumps out, his feet suddenly come to rest, but his body continues moving in the direction of bus. Thus, he may fall forward and may cause serious injury to himself. However, if the person starts running in the direction of the bus, his body will not come to rest and he will not fall in the forward direction.  [1] Q. 4. A runner presses the ground with his feet before he starts his run. Identify action and reaction in this situation. U [Board Term-I, 2016]

Ans. Action : Force exerted by the runner by his foot against the ground. Reaction : Ground exerts an equal and opposite force. [1] Q. 5. Calculate the net force acting on a bus boarded with passenger, of mass 2000 kg moving with a uniform velocity of 60 km/hr. 

A [Board Term-I, 2014]

Ans. Since velocity is uniform, acceleration will be zero therefore net force is zero. (F = ma) [CBSE Marking Scheme, 2014] [1] Q. 6. If a balloon filled with air and its mouth untied, the air is released from its mouth in the downward direction, and balloon moves upwards. Identify action and reaction in this case. U [Board Term-I, 2014]



Ans. Release of air in downward direction is action.

Movement of balloon in upward direction is reaction. [CBSE Marking Scheme, 2014] [1] Q. 7. In a tug of war, the rope does not move in any direction. Why ? A (Board Term-I 2013)



Ans. Because the forces applied by the two teams are equal and opposite.  [CBSE Marking Scheme, 2013] [1]

Short Answer Type Questions-I Q. 1. Define SI unit of force. A force of 2N acting on a body changes its velocity uniformly from 2 m/s to 5 m/s in 10s. Calculate the mass of the body. 

Ans. One Newton is the force that produces an acceleration of 1 ms−2 on a body of mass 1 kg.



Here, F = 2N, u = 2 m/s, v = 5 m/s, t = 10s, m = ?

(2 marks each) vu t 52  10  0.3 m/s2 20 F 2  6.67 kg m    10  a 3 3 a

U+A [Board Term-I, 2016]



(1 mark each)





[CBSE Marking Scheme, 2016] [2]

FORCE AND NEWTON'S LAWS

Q. 2. While driving vehicle how does the use of safety belts prevents accidents ? To show that a body remains at rest unless acted upon by an unbalanced force, mention one situation from every day life. A [Board Term-I, 2016]





Ans. During accident the speed of vehicle becomes suddenly zero and the driver has a tendency to lean forward and may get struck on the dashboard. Being ties with seat belt this tendency is prevented and thus the injury of the driver is prevented. Example : Pedalling of bicycle. When we stop pedalling, the bicycle begins to slow down. This is again because of the friction forces acting opposite to the direction of motion.



In order to keep the bicycle moving, we have to start pedalling again. It thus appears that an object maintains its motion under the continuous application of an unbalanced force. [CBSE Marking Scheme, 2016] [2]

Q. 3. Derive Newton’s first law of motion from the mathematical expression of the second law of motion. A [Board Term-I, 2015]

time, t. If u is zero then v will also be zero, i.e., object will remain at rest. [2] Q.4. Why is it easier to stop a tennis ball in comparison to a cricket ball moving with the same speed?  A [DDE 2014] Ans. Tennis ball is lighter (less mass) than a cricket ball. Tennis ball moving with same speed has less momentum (mass × velocity) than a cricket ball. It is easier to stop tennis ball having less momentum.  [2] Q. 5. Water sprinkler used for grass lawns begins to rotate as soon as the water is supplied. Explain the principle on which it works.  U [NCERT Exemplar] Ans. In works on third law of motion. As the water comes out of the nozzle of the sprinkler, an equal and opposite reaction force comes into play. So the sprinkler starts rotating. [2] Q. 6. Which is having higher value of momentum? A bullet of mass 10 g moving with a velocity of 400 m/s or a cricket ball of mass 400 g thrown with the speed of 90 km/hr. A





Momentum  P  = mass  m  velocity  v 

Ans.

Ans. Newton’s first law states that a body stays at rest if it is at rest and moves with a constant velocity until a net non-zero force is applied on it. Newton’s second law states that the net force applied on the body is equal to the rate of change in its momentum.



103

Mass of bullet  10g  10  10 3 kg  10 2 kg Velocity of bullet  4 00 m/s Momentum of bullet = 10 2 kg 4 00 m/s  4 kg m/s Mass of cricket ball  4 00 g  4 00 10 3 kg  0.4 00 kg Velocity of ball  9 0 km/hr

F  ma m( v  u) or F  t or Ft  mv  mu

90  1000 m  25 m/s 3600 s Momentum of ball  0.4 00  25  10 kg m/s =

That is, when F = 0, v = u for whatever time, t is taken. This means that the object will continue moving with uniform velocity, u throughout the

   

∴ The cricket ball has higher momentum.

Short Answer Type Questions-II

[2]

(3 marks each)

Q. 1. What type of force is acting in the graph given U+A [Board Term-I, 2016] above ? (iii) C

(i) A

v

v t

t

Ans. (i) Accelerating unbalanced force. (ii) No force. (iii) Retarding unbalanced force.  [CBSE Marking Scheme, 2016] [1 + 1 + 1 = 3]

B (ii) v



Q. 2. Total momentum of two bodies remains unchanged before and after the collision. Justify this statement. A [Board Term-I, 2016] t

104

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Ans. When two or more bodies act upon one another their total momentum remains constant, provided no external forces are acting.



Consider two particles A and B, which collide head on. The particles move in a straight line before and after collision. Let particle A have initial velocity u1 and particle B has initial velocity u2. The two particles will collide, if u1 > u2. Let after collision the final velocities of A and B becomes v1 and v2 respectively. The two particles will separate after collision, if v2 > v1. Let the two particles A and B have m1 and m2 respectively.



Initial momentum of particle A is m1u1



Initial momentum of particle B is m2u2



Final momentum of particle A is m1v1



Final momentum of particle B is m2v2



Rate of change of momentum of particle



A = (m1v1 – m1u1)/t = m1(v1 – u1)/t



⇒ m1u1 + m2u2 = m1v1 + m2v2 or initial momentum of the system = final momentum of the system, which is the law of conservation of momentum. [CBSE Marking Scheme, 2016] [3]

Q. 3. A force of 5N produces an acceleration of 8m/s2 in mass m1 and an acceleration of 24m/s2 in mass m2. What acceleration would it give if both the masses are tied together?  A [Board Term-I, 2016, 15)

Ans. Here, F = 5 N, a1 = 8 m/s2, a2 = 24 m/s2, a = ?





Mass m1 =

...(i)

5 Mass m2 = kg 24 When masses are tied together (m1 + m2)

Rate of change of momentum of particle



According to Newton’s second law of motion



FAB = (m1v1 – m1u1)/t = m1 (v1 – u1)/t and



FBA = (m2v2 – m2u2)/t = m2 (v2 – u2)/t



Then, by Newton’s third law of motion, we have FAB + FBA = 0 FAB = – FBA i.e., m1v1 – m1u1 = – (m2v2 – m2u2)



M=



=



(ii) their momentum. U+A [Board Term-I, 2016]

Ans. (a) The momentum, p of an object is defined as

the product of its mass, m and velocity, v. p = mv Unit of momentum (p) = unit of mass × unit of velocity = kg × ms–1 = kg ms–1

5 5 + 8 24 20 kg 24







F Now, a = M









=

[CBSE Marking Scheme, 2016] [3]







5 × 24 20

= 6 m/s2

Long Answer Type Questions Q. 1. (a) Define momentum. Derive its SI unit. How is force expressed in terms of momentum ? (b) Two balls A and B of masses ‘m’ and ‘2m’ are in motion with velocities 2v and 1v respectively. Compare : (i) their inertia.

F a1

5 = kg 8



B = (m2v2 – m2u2)/t = m2 (v2 – u2)/t ...(ii) Let FAB and FBA, be the force exerted by particle B on A and particle A on B respectively.





(5 marks each)

Force =

change in momentum time

(b) (i) Ball B has more inertia than A as its mass is 2 times that of A

(ii) Momentum A = m × 2v = 2 mv, while momentum of B = 2 m × v = 2mv

It is clear that both have same momentum.



Q. 2.

[CBSE Marking Scheme, 2016] [5]

(i) State Newton’s second Law of Motion.

Express it mathematically and find SI unit of force from it.

FORCE AND NEWTON'S LAWS

(ii)

Acceleration of hammer =

In the diagram given above, if the card is flicked away with a jerk, what will you observe ? Explain the reason for this observation. U+A [Board Term-I, 2015] Ans. (i) Newton’s second Law of Motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force. It is expressed mathematically as,

m  ( v  u) F ∝ t



F = km  ( v  u) = kma

105

( 0 - 5) 0.01

a = – 500 m/s2 Force applied by the nail on hammer F = ma = 0.5 × (– 500) = – 250 N. [1 + 2 + 1 + 1] Q. 5. (a) State the law that provides the formula for measuring force and the law which provides the definition of force. (b) Velocity time graph of a 50 g marble rolling on a floor is given below. Find : (i) time in which it stops. (ii) negative acceleration produced on it. (iii) positive force acting on the marble. 

A

[NCERT Exemplar]

t

Unit force applied to unit mass produces unit acceleration. Hence k = 1. ∴ F = ma (ii) We will observe that the card moves ahead allowing the coin to fall vertically into the glass. This is due to inertia of rest. The inertia of the coin tries to maintain its state of rest. [3 + 2] Q. 3. State Newton’s first law of motion. Show that Newton’s first law of motion is a special case of Newton’s second law. Determine the acceleration of a car of mass 800 kg, on application of a force of 200 N on it.  U+A [Board Term-I, 2015] Ans. Statement of Newton first law of motion : Refer Topic 1 ‘Revision Notes'

Try yourself. Refer Q.3 of SATQ-I. m = 800 kg, F = 200 N



F 200 1 a == = = 0.25 m/s2 m 800 4

Ans. (a) Formula for measuring force is given by Newton’s 2nd law. Second law of motion gives us a method to measure the force acting on an object as force is the product of its mass and acceleration. Definition of force is given by Newton’s first law. (b) From graph

(i) t = 25 s



(ii) a =

( 30 - 0 ) = 1.2 m/s2 25







(iii) F = ma = (50/1000) × 1.2 = 0.06 N



[½ + 1 + 1+ 1+ ½ + 1]

[2 + 3]

Q. 4. Give statement for Newton’s second law of motion. Deduce a mathematical formulation for it. Using above derived expression, calculate the force exerted by a nail on the hammer of mass 500 g moving at 5.0 m/s striking it. Consider that the nail stops the hammer in a short time of 0.01 s. A [DDE 2014] Ans. Refer Topic 1 'Revision Notes' Mathematical derivation : Try yourself. Refer Q. 2 (i) of LATQ. m = 500 g = 0.5 kg, u = 5 m/s, v = 0 t = 0.01 s

Commonly Made Error

Students often give incorrect explanation.

Answering Tips

Students should memorize the statements





of the laws of motion and their mathematical derivation. Lay stress on understanding the concept instead of rote learning.

106

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Topic-2 Inertia and conservation of momentum Revision Notes  Inertia • The property by the virtue of which an object tends to remain in the state of rest or of uniform motion unless acted upon by some force is called inertia. • The mass of a body is a measure of inertia. • Inertia is the inability of a body to change its state of rest or of uniform motion in a straight line by itself. • The inherent property of a body by virtue of which it cannot change its state of rest is called inertia of rest.  Effects of force are :





(i) It can produce motion in stationary bodies. (ii) It can stop moving bodies. (iii) It can change the speed and direction of motion of bodies. (iv) It can also bring about change in dimensions of a body.

Key words  Inertia : The tendency of a body to oppose or resist any change in its state of rest or uniform motion is called inertia of the body.  Inertia of Motion : The tendency of a body to oppose any change in its state of uniform motion is known as inertia of motion. e.g., : The passengers fall forward when a fast moving bus stops suddenly.  Inertia of Direction : The tendency of a body to oppose any change in its direction of motion is known as inertia of direction. e.g., : When a fast moving bus negotiates a curve on the road, passengers fall towards the centre of the curved road.



Example

(i) A ball is allowed to roll down from an inclined plane. It reaches the foot of the plane and continues to roll on the ground. It stops after travelling some distance. Is this the violation of law of inertia? Give reasons for your answer. (ii) A player lowers his hand while catching a ball. Explain reason behind his action.  Solution: (i) No. It is not the violation of law of inertia.

Law of inertia is obeyed only when no external force acts on a body. But in this case the friction due to the ground acts on the ball, so it comes to rest. (ii) Player lowers his hand because by doing so he increases the time in which velocity of ball comes to zero. This decreases the rate of change of momentum and so the impact of force is reduced. [1 + 1+ 1]

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions 

(1 mark each)

Q.1. The inertia of an object tends to cause the object: (A) To increase its speed (B) To decrease its speed

(C) To resist any change in its state of motion (D) To decelerate due to friction. Ans. Option (C) is correct. Explanation: The inertia of an object tends to cause the object to resist any change in its state of rest or motion.

FORCE AND NEWTON'S LAWS

Q.2.

Rocket works on the principle of conservation of: (A) Mass (B) Energy (C) Momentum (D) Velocity

Ans. Option (C) is correct. Explanation: Rocket works on the conservation of momentum. Q.3. An object of mass 2 kg is sliding with a constant velocity of 4 ms-1 on a frictionless horizontal table. The force required to keep the object moving with the same velocity is: (A) 32 N (B) 0 N (C) 2 N (D) 8 N Ans. Option (B) is correct. Explanation: Given, mass m = 2 kg, velocity v = 4 ms–1. As the object is moving with a constant velocity i.e., 4 ms–1, so the acceleration of the object is zero i.e., a = 0 and according to the property of inertia if there is no external force acting on the body, then body remains as it is i.e., if the body is at rest, it remains in rest and if it is in motion, it remains in motion. Q.4. A water tanker filled upto 2/3 of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would (A) Move backward (B) Move forward (C) Be unaffected (D) Rise upwards. Ans. Option (B) is correct. Explanation: On the sudden application of brake, the tanker will come in the state of rest but the water remains in the state of motion, so the water will move forward.

B

107

Assertion and Reason 

(1 mark each)

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q.1. Assertion: Sudden application of brakes in a moving car may cause injury. Reason: Inertia is the tendency of an object to keep moving or being at rest undisturbed. Ans. Option (B) is correct. Explanation: When we are in a moving car, our body is also moving with the velocity of car. Sudden brake applied stops the car as well as lower portion of our body while upper portion of our body is still in motion that causes injury. Q.2. Assertion: A five rupee coin has more inertia than one rupee coin. Reason: Inertia does not depend upon mass of the object. Ans. Option (C) is correct. Explanation: Inertia depends upon the mass of the object. Heavier objects offer more inertia than lighter objects. Q.3. Assertion: Momentum of an object which is the product of mass and velocity is a vector quantity. Reason: Momentum has both direction as well as magnitude. Ans. Option (A) is correct. Explanation: Momentum of an object which is the product of mass and velocity is a vector quantity. Momentum has both direction as well as magnitude.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions Q. 1. There are three solids made up of aluminium, steel and wood of the same shape and same volume. Which of them would have highest inertia ? (NCERT) Ans. Solid made of steel has the highest inertia because its density, hence mass is greater than aluminium and wooden solids. [1] Q. 2. Which has more inertia, a cricket ball or a rubber ball of the same size ? Give reason for your answer. A [Board Term-I, 2015]

(1 mark each)

Ans. Inertia of a body depends on its mass. A cricket ball has more mass than a rubber ball, thus it has greater inertia. [1] Q. 3. When a carpet is beaten with a stick it releases dust. Explain why ? A (NCT 2014) Ans. As dust possess inertia of rest, it resists the change and falls down. [1]

108

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Short Answer Type Questions-I Q. 1. State why Newton’s first law of motion is called law of inertia. A (DDE 2014) Ans. Inertia is a tendency of the object to resist change in its state. Newton’s first law of motion also states the same i.e., the object will remain in its present state unless an external force is applied. That’s why Newton’s first law is called Law of inertia. [2]

Q. 2. Why does the pillion rider fall forward when brakes are applied ? A Ans. During the ride, pillion rider and driver are in a state of motion. But when the driver applies brakes, the upper portion of the body of pillion rider continues moving forward on account of inertia of motion. Therefore, the pillion rider falls forward. [2]

Short Answer Type Questions-II Q. 1. Which accident will be more damaging, collision between two trucks moving with a speed of 50 km/hr or collision between two cars moving with a speed of 50 km/hr ? Explain.  A (Board Term-I 2014) Ans. Collision between trucks, because more will be the mass, more will be the inertia and therefore more will the momentum. Mass of the trucks being more than that of cars so collision of trucks will cause more damage. [3] Q. 2. (a) State the name of the object which has more inertia : (i) a rubber ball and a stone of the same size. (ii) an empty box and another similar box filled with clothes. (b) Give reasons for the following : (i) Luggage placed on the roof of a car or bus is tied with rope.  (NCERT) (ii) When a branch of a tree is shaken vigorously, some of the leaves drop down.  A Ans.

(a) (i) Stone.

(ii) Box filled with clothes.

[½] [½ ]

(b) (i) This is done to prevent luggage from falling when the vehicle is suddenly stopped or started the luggage develops a tendency to fall forward or backward due to inertia. [1]

(ii) On shaking, branch comes to state of motion and leaves which are in state

(3 marks each)

of rest experience a jerk due to which these get detached and fall down. [1] Q. 3. State the law of Inertia. Why do we fall in forward direction if a moving bus stops suddenly and fall in the backward direction if it suddenly accelerates from rest ? U+A (NCERT)  Ans. Law of Inertia : An object remains in its state of rest or of uniform motion in a straight line until an external unbalanced force acts on it.

When a moving bus stops suddenly, the bus slows down but our body tends to remain in the state of motion due to inertia of motion. Sudden start of bus brings motion to the bus as well as our feet but rest of the body still has inertia of rest due to which we fall backwards. [1 + 2] Q. 4. A body of mass 4 kg is dropped from a height of 20 m. Calculate the initial momentum and the momentum just before it strikes the ground. (g = 10 m/s2) A (Board Term-I 2014) Ans. Initial velocity u = 0 ∴ Initial momentum = 4 × 0 = 0 kg m/s Let the velocity just before striking the ground be u v2 = u2 + 2gh v2 = 0 + 2 × 10 × 20 v2 = 400 v = 20 m/s ∴ Final Momentum = 4 × 20 = 80 kg m/s [3]

Long Answer Type Questions Q. 1. What is meant by ‘inertia’ ? What are different types of inertia ? Give two examples in each case. U Ans. Inability of the body to change by itself its state of rest or state of uniform motion is called inertia. Types : Inertia of rest : e.g., : (i) When a card, placed on a tumbler and a coin on it, is flicked with a finger the coin placed over it falls in the tumbler.

(2 marks each)

(5 marks each)

(ii) Only the carom coin at the bottom of a pile is removed when a fast moving carom striker hits it. Inertia of motion : e.g., : (i) When a moving bus stops suddenly, the luggage might slide towards the front side of the bus and fall. (ii) We tend to fall forward when a bus suddenly stops. [1 + 2 + 2]

FORCE AND NEWTON'S LAWS

Q. 2.

(i) Define momentum. Write its S.I. unit. (ii) How much momentum will an object of mass 10 kg transfer to the floor, if it falls from a height of 5 m (g =10 m/s2).

(iii) Explain how a karate player can break a pile of tiles with a single blow of his hand.



U+A

Ans. (i) Try yourself. Refer Q.1 (a) of LATQ of Topic-1. (ii) v2 = u2 + 2gh v2 = (0)2 + 2(10) (5) v2 = 100 \ v = 10 m/s momentum = m × v = 10 × 10 = 100 kg m/s (iii) The karate player strikes the pile of tiles with his hand very fast. In doing so, the large momentum of fast moving hand is reduced to zero in a very short time. This exerts a very large force on the pile of tiles which is sufficient to break them. [1 + 1 + 2 + 1] Q. 3. (a) State the law of conservation of momentum. (b) Observe the following diagram and answer the questions given below :

(i) Which direction does the balloon move when the thread tied to its neck is removed and why ? (ii) State the conclusion drawn from this U+A activity. Ans. (a) When two or more bodies act upon one another their total momentum remains constant, provided no external forces are acting. (b) (i) Air from inside the balloon escapes from the mouth of the balloon. Balloon moves in opposite direction that is from left to right. (ii) Forces of action and reaction are equal and opposite. [2 + 1 + 1 + 1]

Commonly Made Error



Balloon

Students make error while writing law of conservation of momentum.

Answering Tip

Straw

Thread

Students should memorize the law of

conservation of momentum and its derivation as questions based on it are frequently asked in the exams.

COMPETENCY AND CRITICAL THINKING BASED QUESTIONS I. Study the following diagram and choose the correct options to answer any four questions given below: In the figure below the card is flicked with a push. It was observed that the card moves ahead while coin falls in glass.

(1) Give reason for the above observation. (A) The coin possesses inertia of rest, it resists the change and hence falls in the glass. (B) The coin possesses inertia of motion; it resists the change and hence falls in the glass. (C) The coin possesses inertia of rest, it accepts the change and hence falls in the glass. (D) The coin possesses inertia of motion, it accepts the change and hence falls in the glass. Ans. Option (A) is correct. (2) Name the law involved in this case. (A) Newton’s second law of motion. (B) Newton’s first law of motion.

109

(1 mark each)

(C) Newton’s third law of motion. (D) Law of conservation of energy Ans. Option (B) is correct. (3) If the above coin is replaced by a heavy five rupee coin, what will be your observation. Give reason. (A) Heavy coin will possess more inertia so it will not fall in tumbler. (B) Heavy coin will possess less inertia so it will fall in tumbler. (C) Heavy coin will possess more inertia so it will fall in tumbler. (D) Heavy coin will possess less inertia so it will not fall in tumbler. Ans. Option (C) is correct. (4) Name the law which provides the definition of force. (A) Law of conservation of mass (B) Newton’s third law. (C) Newton’s first law. (D) Newton’s second law. Ans. Option (C) is correct.

110

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

(5) State Newton’s first law of motion. (A) Energy can neither be created nor be destroyed, it can be converted from one form to another, total amount of energy always remains constant. (B) A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless it is acted upon by an external unbalanced force. (C) For every action in nature there is an equal and opposite reaction. (D) The acceleration in an object is directly related to the net force and inversely related to its mass. Ans. Option (B) is correct. II. Read the following paragraph and choose the correct options to answer any four questions given below : A large bus and a van, both moving with a velocity of magnitude v, have a head-oncollision and both the vehicles stop after the collision. The time of the collision is 1 sec. (1) The vehicle, which experiences smaller force of impact is ................................ . (A) Van (B) Bus (C) Van and Bus both (D) There will not be any effect on any of the vehicle Ans. Option (A) is correct. (2) The vehicle, which experiences the smaller momentum change is .................... . (A) Bus (B) Van (C) Bus and Van both (D) There will not be any effect on any of the vehicle Ans. Option (B) is correct. (3) The vehicle, which experiences the greater retardation? (A) Bus and Van both (B) Bus (C) Van (D) There will not be any effect on any of the vehicle

Ans. Option (A) is correct. (4) Which vehicle according to you will suffer the less damage? (A) Van (B) Bus and Van both will have similar impact (C) Bus (D) There will not be any effect on any of the vehicle Ans. Option (C) is correct. (5) Under the same conditions, if there was a heavy truck heavier than the bus, in place of van, which one will be more affected? (A) Bus (B) Truck (C) Bus and Truck both (D) There will not be any effect on any of the vehicle Ans. Option (A) is correct. III. Read the following passage and answer questions given below: Newton's first law of motion states that a body at rest will remain at rest and a body which is in motion continues to be in motion unless acted upon by an external force. In other words, objects that are undisturbed tend to remain in their current state of rest or motion, is called Inertia. All objects resist a change in their state of motion. 1. State Newton's first law of motion. Ans. Newton's first law state that if the object is in rest always be in rest or if it is in motion always be in motion unless untill a non-zero external force applied on it. 2. Why Newton's first law of motion is called the Law of Inertia ?  Ans. All objects resist a change in their motion. In other words, the tendency of undisturbed object to stay at rest or keep moving with the same velocity is called Inertia. 3. What will happen if no external force act on moving object ? Ans. It will continue to move with same speed in same direction. 4. Why people experience a jerk who are sitting in a moving bus when it stops ? Ans. This is due to Inertia of motion. 

FORCE AND NEWTON'S LAWS

111

Artificial Intelligence PARAMETERS

DESCRIPTION

Chapter Covered

Chapter 9: Force and Laws of Motion

Name of the Book

Science, Class 9, NCERT

Subject and Artificial ● Soft Buddy game to understand impact of force: Intelligence Integrated ● Introduction to the concept of push/pull through Pakku Pakia Game (by Rishabh) (Google Search). ● Application of force to be understood through sculpting game: ● Recap of the concept- Frictional Force through Hoverborad Game (AI Google Experiment). ● Concept of motion can be introduced through visualisation of following tool: ● Demo of gravitational force through simulation: https://youtu.be/ZTwrQSOHdX0

Learning Objectives

Students will be able to: ● Understand force and its effects. ● Understand meaning of balanced and unbalanced forces. ● Explain relationship between force and motion. ● Understand the role of friction in inhibiting motion.

Time Required

2-3 periods (40 min each)

Classroom Arrangement

Flexible

Material Required

Ball, Clay, Sponge, Wooden Block, Book Projector and Screen Computer Lab.

AI CONCEPTS INTEGRATED

Scan to know more about this topic

Scan to know more about this topic

Pre-Preparation Activities Materials through which different impacts of force can bediscussed. Situations to be developed for role play. Previous Knowledge

Teacher will show demo with different materials and students will have to identify what impact is caused by force in each case.

Methodology

With the help of role play and videos concept of friction alongwith Balanced/Unbalanced Forces will be discussed. https://youtu.be/DK81n-Fw9x0 https://youtu.be/d3QRNH_jNOI https://youtu.be/ZTwrQSOHdX0 Concept of Static, Sliding and Rolling Friction would be taken up with relevant examples. As an integrated approach, students will be made to play Tug of War in their sports class. Thereafter they will be made to discuss about muscular force, friction, unbalanced forces with a list of follow-up questions: What will happen if: • Team A is given smoother end of rope and Team B rough end. • Team A is provided with powder to rub their hands and Team B with sand. • Team A members are with shoes and Team B members are without shoes. ● With the help of so many real life examples the concept of Net Force/ Resultant Force will be clear to students and the students will be able to find logical answers to most of routine life incidents. ● Students will be able to identify force as a vector quantity and interpret the direction of motion in case of unbalanced forces.

Learning Outcomes

Follow up Activities

Picture based questions to check on the understanding about Types of Force and impact caused by force. Quiz on Quizizz.com (Let’s Reinforce the concept of Force) Assessment of chapter.

Reflections

Simple life activities would be interpreted by students in terms of physics.

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

Study Time Reading Time: 2:00 Hr No. of Questions: 42

CHAPTER

9



Syllabus

GRAVITATION

Gravitation; Universal law of gravitation; Force of Gravitation of the earth (gravity); Acceleration due to Gravity; Mass and Weight; Free fall.

Revision Notes



 According to the universal law of gravitation, the force of attraction between any two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. The law applies to objects anywhere in the universe. Such a law is said to be universal.  Universal gravitational constant G = 6.67 × 10-11 Nm2 kg–2.  Acceleration with which a body falls towards the centre of the Earth is called acceleration due to gravity (g).  The force of gravity decreases with increasing altitude. It also varies on the surface of the Earth, decreasing Scan to know from poles to the equator. more about  Mass is the quantity of matter contained in the body. this topic  Weight of the body is the force with which the Earth attracts the body.  The weight is equal to the product of mass and acceleration due to gravity.  Mass of a body does not change but weight of a body is different at different places.  Law of gravitation is a inverse square rule since F is inversely proportional to the square of d. Universal Law of Gravitation  Weight of an object on the Moon is one-sixth time of its weight on the Earth.

Key words

 Gravitation : It is the force of attraction between any two bodies in the universe.  Gravity : It is the force of attraction between the Earth and any object lying on or near its surface.  Newton’s universal law of gravitation : This law states that everybody in this universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematically, mm F = G 12 2 . r

where, G is Universal gravitational constant.  Universal gravitational constant : It is equal to the force of attraction between two bodies of unit mass each placed at a unit distance apart. It is denoted by G and its value is 6.67 × 10–11 Nm2/kg2.  Centripetal acceleration of the Moon : If the Moon is revolving with speed v in a circular orbit of radius r, then acceleration acting on it along the radius and towards the centre of its orbit is

ac =

v2 r

.

GRAVITATION

113

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

 Free fall : The motion of a body under the influence of force of gravity alone is called a ‘free fall’.  Acceleration due to gravity : The acceleration produced in the bodies due to Earth’s force of gravity is called acceleration due to gravity of earth. Its value on the Earth’s surface is 9.8 m/s2.  Centre of mass : The centre of mass of a body may be defined as the point at which whole mass of the body may be assumed to be concentrated.  Centre of gravity : The centre of gravity of a body is a point at which the resultant of all the parallel forces experienced by various particles of the body, due to attraction of Earth, passes irrespective of the orientation of the body.  Projectile : Any object thrown into space with some initial velocity and which moves thereafter under the influence of gravity alone is called a ‘projectile’. The path of a projectile is a parabola. Its horizontal range is maximum when the angle of projection is 45°, if the effect of frictional forces are neglected.  Weightlessness : The state when an object does not weight anything during free fall.







Example

(a) What is the relation between the mass m and the weight W of a body ? (b) What are the differences between mass and weight? Solution: (a) Weight of a body is the force of attraction of the Earth on that body. This force depends on the mass (m) of the body and the acceleration due to gravity (g). W=m×g

(b) Difference between mass and weight : S. No

Mass

Weight

(i)

The weight (W) of the body is directly proportional to the mass of the body.



Its value remains Its value changes constant at all from place to place places. due to change in the ‘g’. (ii) It is a scalar quantity. It is a vector quantity. (iii) It is never zero. It may be zero depending on the value of g. (iv) Its unit is kg. Its unit is N or kg- wt. [1 + 2]

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions 

(1 mark each)

Q. 1. Two objects of different masses falling freely near the surface of Moon would (A) Have same velocities at any instant (B) Have different accelerations (C) Experience forces of same magnitude (D) Undergo a change in their inertia. Ans. Option (A) is correct. Explanation: Objects of different masses falling freely near the surface of the Moon would have the same velocities at any instant because they will have same acceleration due to gravity. Q. 2. The value of acceleration due to gravity (A) is same on equator and poles (B) is least on poles (C) is least on equator (D) Increases from pole to equator

Ans. Option (C) is correct. Explanation: Earth is bulging out at equator and at poles it is a bit flattened. So the acceleration due to gravity experienced at the poles is slightly higher than that at the equator. Q. 3. The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become

(A)

F 4

(B)

F 2

(C) F (D) 2F Ans. Option (A) is correct. Explanation: We know that, according to force of gravitation, Gm1 m2 F= (G =Gravitational constant) 2 r where m1 and m2 are the masses of two objects respectively and r is the distance between the two masses.

GRAVITATION

Now, according to the question, if masses of both objects are halved. i.e.,

1 Thus, the new gravitational force will become 4 times of its original gravitational force. Q. 4. A boy is whirling a stone tied with a string in a horizontal circular path. If the string breaks, the stone (A) Will continue to move in the circular path. (B) Will move along a straight line towards the centre of the circular path. (C) Will move along a straight line tangential to the circular path. (D) Will move along a straight line perpendicular to the circular path away from the boy. Ans. Option (C)is correct. Explanation: In circular motion, the direction of velocity at a point is always along the tangent at that point. If string breaks, then the centripetal force acting on the stone becomes zero and it will move along a straight line tangential to the circular path. Q. 5. In the relation F = GM m/d2, the quantity G : (A) Depends on the value of g at the place of observation (B) Is used only when the Earth is one of the two masses (C) Is greatest at the surface of the Earth (D) Is universal constant of nature. Ans. Option (D) is correct. Explanation: The quantity G is universal constant of nature. It is applied to all the bodies present in universe. Q. 6. Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be :



1 (A) times 4

(B) 4 times



1 (C) times 2

(D) unchanged

Ans. Option (B) is correct. Explanation: We know that, according to gravitational force

Mm 2 r  F=G

Where, F = Force between two masses       M = First mass

…(Eq. 1)

115

      m = Second mass       G = Gravitational constant         r = Distance between two masses According to the question, F’ = New force when masses are doubled If mass of each particle is doubled i.e., M = 2M and m = 2m On putting these values in Eq.(1), we get

∴



Because,

(2 M )(2m) r2 Mm F=G 2 r F '= 4F

F' = G

So, Q. 7. The force of attraction between two unit point masses separated by a unit distance is called : (A) Gravitational potential (B) Acceleration due to gravity (C) Gravitational field (D) Universal gravitational constant Ans. Option (D) is correct. Explanation: The force of attraction between two unit point masses separated by a unit distance is called universal gravitational constant. Q. 8. An apple falls from a tree because of gravitational attraction between the Earth and apple. If F1 is the magnitude of force exerted by the Earth on the apple and F2 is the magnitude of force exerted by apple on Earth, then : (A) F1 is very much greater than F2 (B) F2 is very much greater than F1 (C) F1 is only a little greater than F2 (D) F1 and F2 are equal Ans. Option (D) is correct. Explanation: According to Newton’s universal law of gravitation, force exerted by the one body to other body is equal in magnitude and opposite in direction.

B

Assertion and Reason 

(1 mark each)

Directions : In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Q. 1. Assertion : On Moon, man feels lighter than Earth. Reason : Due to more gravitational force exerted by Moon on man. Ans. Option (C) is correct. Explanation: On Moon, man feels lighter than Earth. It is due to less gravitational force exerted by Moon on man. Q. 2. Assertion : If the distance between two bodies of mass m1 and m2 is increased by a factor of 5, 1 the gravitational force is reduced by . 25 Reason: The gravitational force is inversely proportional to the square of the distance between two bodies. Ans. Option (A) is correct.

Explanation: Gravitational force is given by GMm the formula F = . Gravitational force R2 is inversely proportional to the square of the distance between two bodies. Q. 3. Assertion : An object thrown vertically upward with certain velocity v, reaches maximum height and fall back with same velocity. Reason : Whenever an object falls towards the Earth, gravitational force of the Earth causes acceleration. Ans. Option (D) is correct. Explanation: When an object is thrown vertically upward with certain velocity, it will fall back freely. There will be a change in the magnitude of velocity due gravitational force of the Earth.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions Q. 1. Which force causes the things to fall towards the Earth ? U Ans. Force due to gravity. [1] Q. 2. Who found out the value of gravitational constant (G) ? U Ans. Henry Cavendish. [1] Q. 3. What is the unit of gravitational constant (G)?  U Ans. The unit of gravitational constant is Nm² kg–2.  [1] Q. 4. How is gravitation different from gravity ? U Ans. Gravitation is the force of attraction between any two bodies while gravity refers to attraction between any body and the Earth. [1] Q. 5. Is the value of ‘G’ dependent on the medium present between the two bodies ? U Ans. No. [1] Q. 6. What does a small value of G indicate ? U Ans. A small value of G indicates that the force of gravitational attraction between two ordinary sized objects is a very weak force. [1]

(1 mark each)

Q. 7. Give reason for the statement, “The value of g is greater at the poles than at the equator.” 

A [Board Term-I, 2015]

Ans. At poles the radius of the Earth is less than that at the equator.  [CBSE Marking Scheme, 2015] 1 Q. 8. What do you mean by weight of a body ? U Ans. Weight of a body is the force with which a body is attracted towards the centre of the Earth. [1] Q. 9. Write the SI unit of weight. U [Board Term I, 2015] Ans. SI unit of weight is Newton (N). [1] Q. 10. What kind of quantity is weight – a scalar or a vector ? U Ans. Weight is a vector quantity. [1] Q. 11. Give the relation between mass and weight of a body. U Ans. W = mg. [1]

Short Answer Type Questions-I

(2 marks each)

Q. 1. Name the positions on Earth where the value of ‘g’ is (i) maximum (ii) minimum? Justify your answer. U+A [Board Term-I, 2014] 

Q. 2. Explain what happens to the force between two objects if : (i) the mass of one object is doubled? (ii) the distance between the objects is tripled.  A [DDE 2014]

Ans. On Earth value of g is maximum at poles and minimum at the equator. At poles radius of Earth is less so value of g is more, at equator radius of Earth is more so value of g is less g ∝ 1/(R2)  [CBSE Marking Scheme, 2014] [1+1 = 2]

Ans. F = Gm1m d2 (i) If m1 = 2m, then F becomes double. 

(ii) If d1 = 3d, then F becomes one-ninth. [CBSE Marking Scheme, 2014] [1+1 = 2]

GRAVITATION

Q. 3. Explain an activity to show that, during a free fall heavier and lighter objects accelerate at the same rate. C [Board Term-I, 2014] Ans. Drop two balls with different mass from a tall building at the same time. They will reach the ground at the same time. Both the balls are at free fall and their initial velocity is same, that is zero. [Also it is known from equation of motion that, h = ½ gt2.] So the only way that they touch the ground at the same time is that acceleration is same for both the balls. This experiment proves that during free fall every object accelerate at the same rate, irrespective of its mass.  [CBSE Marking Scheme, 2014] [2] Q. 4. An astronaut carried a pot containing soil weighing 60 N from the Earth to the surface of Moon. He kept it there and just before return journey from Moon to Earth he weighed the soil there on the surface of Moon and found that it was only 10 N. Where did the rest of the soil go and how much mass of soil was lost? (gEarth = 10 ms–2) (gMoon= gEarth / 6) A [DDE-2014]



Ans. Given that, the weight of the soil on Earth is 60 N. gEarth = 10 ms–2



60 = 6 kg 10 Weight of the soil on Moon = 10 N







Mass on Moon, m2 =





Mass on the Earth, m1 =

10   = 6 kg 10 Since m1 = m2, hence there has been no loss in mass of the soil on the surface of Moon and decrease in weight was due to difference in the gravity. [CBSE Marking Scheme, 2014] [2]

Q. 5. Account for the following : (i) On Moon, man feels lighter than Earth . (ii) Mass is scalar while weight is a vector quantity. 

A [Board Term-I, 2016]

Ans. (i) D  ue to less gravitational force exerted by Moon on man. (ii) Mass do not have direction while weight has direction. 

[CBSE Marking Scheme, 2016) [1+1 = 2]

Q. 6. A stone and the Earth attract each other with an equal and opposite force. Why then we see only the stone falling towards the Earth but not the Earth rising towards the stone?  A [Board Term-1, 2016] Ans. Acceleration = Force / Mass. The mass of a stone is very small due to which the gravitational force produces a large acceleration in it. Due to very large mass of Earth, the same gravitational force produces negligible acceleration in the Earth.  [CBSE Marking Scheme, 2016] [2] Q. 7. What is the source of centripetal force that a planet requires to revolve around the Sun? On what factors does that force depend?  U [NCERT Exemplar] Ans. Gravitational force. This force depends on the product of the masses of the planet sun and the inverse of square of distance between them. [2] Q. 8. On the Earth, a stone is thrown from a height in a direction parallel to the Earth’s surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and why? E [NCERT Exemplar] Ans. Both stones will take the same time to reach the ground because the two stones fall from the same height and accelerate due to gravity alone. Both objects also have zero vertical velocity at the instant they are released. [2]

Short Answer Type Questions-II Q. 1. Explain : (i) Universal gravitational constant (ii) Free fall

U [Board Term-I, 2015]

Ans. (i) Universal gravitational constant is the constant ‘G’ appearing in Newton’s law of gravitation.

GMm

(3 marks each)

(ii) Free fall : Whenever objects fall towards the Earth under the gravitational force alone, we can say that the objects are in free fall. While falling there is no change in the direction of motion of the objects. But due to Earth’s attraction, there will be a change in the magnitude of the velocity.

, r2 where F is the force between two masses m and M at a distance r apart. The numerical value of G is equal to



6.673 × 10–11 Nm2 kg–2.

(ii) Gravitational force

F =

117

[1½ + 1½]

Q. 2. In which direction do the following forces act when an object is in motion ? Explain with the help of an example. (i) Frictional force 

U+A [Board Term-I, 2015]

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Ans.





when both are dropped simultaneously from roof.

(i) Frictional force : Backwards Example : If a book slides across the surface of a desk, then the desk exerts a frictional force in opposite (i.e., backwards) direction of its motion.

(ii) Gravitational force : Downwards

Example : When we throw a ball in the air, it returns to the ground. [CBSE Marking Scheme, 2015] [1½ + 1½] Q. 3. A man’s weight when taken at the poles is 600 N. Will his weight remain the same when measured at the equator ? Will there be an increase or decrease in his weight ? Explain.  A [Board Term-I, 2015] Ans. No, his weight will not remain same as that at the poles. There will be a decrease in his weight at the equator. As the radius of the Earth increases from the poles to the equator, the value of ‘g’ decreases from pole to equator.  [3] Q. 4. Give reasons : (i) A piece of paper takes much longer to fall than a stone through the same distance,

(ii) The mass is constant everywhere but the weight keeps changing. (iii) The value of ‘g’ keeps changing as we move away from the Earth whereas value of ‘G’ remains constant all over the universe. 

A [Board Term-I, 2016]

Ans. (i) This is because a piece of paper has larger surface area and therefore experiences more friction due to air than a stone which has less surface area.



(ii) Because acceleration due to gravity varies from place to place.



(iii) The value of g depends on altitude of the place and the mass of the Earth while G is called universal constant as its value remains constant at all the places in the universe.[3]



[CBSE Marking Scheme, 2014] [1 + 2]

Long Answer Type Questions Q. 1.

(i) A bar of metal has a mass 200 g and a certain weight. Mass remains the same when weighed at equator but weight decreases. Why ?

(ii) Differentiate between mass and weight. Write any three differences. 

A [Board Term-I, 2016]

Ans. (i) Weight is dependent on acceleration due to gravity. Since, on equator, acceleration due to gravity is less, so the weight of the bar of metal decreases. (ii) Difference between mass and weight Refer Example 1 (b)  [CBSE Marking Scheme, 2016] [2 + 3] Q. 2. Write three points of differences between mass and weight. How much would a 70 kg astronaut weigh on Moon? What would be his mass on the Earth and on the Moon? U+A [Board Term-I 2016, 15] Ans. Differences between mass and weight : Refer Example 1 (B) Mass of the astronaut on Moon =70 kg, g = 1.6 m/s2 on Moon W = m × g = 70 × 1.6 = 112N



This is the weight of astronaut on Moon. The mass of a body is constant everywhere in the universe. So, the mass of the astronaut would be same on the Earth as well as on the Moon i.e., 70 Kg. [CBSE Marking Scheme, 2016] [3 + 2]

(5 marks each)

Q. 3.

(i) Prove that if the Earth attracts two bodies placed at the same distance from the centre of Earth, with equal force; then their masses will be the same. (ii) Mathematically express the acceleration due to gravity that is expressed by a free falling object. (iii) Why is ‘G’ called a universal constant ?  U [Board Term-I, 2014]

Ans. (i) Let mass of first body be m1 Let mass of second body be m2 Force on 1st body = Force on 2nd body GMm1/R2 = GMm2/R2 So, m1 = m2 Hence proved. (ii) g = GM/R2 (iii) Its value is constant throughout the universe. [CBSE Marking Scheme, 2014] [3 + 1 + 1]

Q. 4. (a) Write the formula to find the magnitude of gravitational force between the Earth and an object on the Earths’ surface. (b) Derive how does the value of gravitational force ‘F’ change between two objects when the: (i) distance between them is reduced to half, and (ii) mass of one object is increased four U+A [Board Term-I, 2016] time.

GRAVITATION

119

related to the mass of the Earth and its radius? A+U (Board Term-I, 2015) Derive it.

Ans. (a) F = GMm

R2

(b) (i) According to the law of gravitation, the force of attraction acting between two bodies is given by,

1 Ans. (a) (i) g on Moon = gEarth× = 9.8 = 1.63 ms–2 6

F = GM1m2/R2



(i) When distance reduced to half

F´ =





between

them

is

Mm GM1m2 = G 1 22  R  R2    2

2 = 4F = 4G M1m R2 Thus, when the distance between the objects is reduced to half, the gravitational force increases by four times the original force.

(ii) When mass of one object is increased four time.

F’ = GM1 × 4m2/R2 = 4F So, as the mass of any one of the object

is increased four times, the force is also increased by four times. [CBSE Marking Scheme, 2016] [3 + 2] Q. 5. (a) A person weighs 110.84 N on Moon, whose acceleration due to gravity is 1/6 of that Earth. If the value of ‘g’ on Earth is 9.8ms–2 calculate

(i) ‘g’ on Moon.

(ii) Mass of person on Moon (iii) Weight of person on Earth (b) How does the value of g on the Earth is

(ii) Mass on Moon =

(iii) Weight on Earth = mg = 68 × 9.8 = 666.4 N. (b) Derivation of g = GM/R2 [CBSE Marking Scheme, 2015] [5] Detailed Answer : 1 1 = 9.8 × = 1.63 ms–2 6 6 (ii) Mass is constant everywhere and does not change from place to place. (a) (i) gmoon = gEarth×

Mass on Moon = F/gmoon = 110.84/1.63 = 68 kg (iii) Weight on Earth = 68 × 9.8 = 666.4 N. (b) According to Newton’s law of gravitation, the force of attraction between Earth and a body is given by GmM F=  ... (i) r2 Force produces an acceleration ‘g’. So, from Newton’s second law F = mg .... (ii) From equation (i) and (ii) we get



mg =



g=

COMPETENCY AND CRITICAL THINKING BASED QUESTIONS

I. Read the following paragraph and choose the correct options to answer any four questions given below: Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Q. 1. At what place on the Earth’s surface is the weight of a body minimum? (A) At the equator (B) At all places the weight will be equal (C) At the north pole (D) At the south pole Ans. Option (A) is correct. Q. 2. Which of the following statement is true? (A) The friend at equator will agree with the weight of gold bought at poles. (B) The friend at equator will not agree with the weight of gold bought at poles.

F 110.84 = = 68 kg g 1.63



GmM r2

GM r2

(1 mark each)

(C) The friends at the poles will not agree with the weight of gold bought at poles. (D) The friends at equator as well as at the poles will agree with the weight of gold bought at poles. Ans. Option (B) is correct. Q. 3. The value of g is greater at the poles than at the equator, so the weight of gold at the equator will be (A) more than the weight of gold at the poles. (B) less than the weight of gold at the poles. (C) same as the weight of gold at the poles. (D) zero Ans. Option (B) is correct. Q. 4. What is the relation between mass and weight? (A) Weight (W) of the body is always less than the mass of the body.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

(B) Weight (W) of the body is same as that of the mass of the body. (C) Weight (W) of the body is inversely proportional to the mass of the body. (D) Weight (W) of the body is directly proportional to the mass of the body. Ans. Option (D) is correct. Q. 5. What is the SI unit of Mass? (A) Watt (B) Newton (C) Kilogram (D) Pascal Ans. Option (C) is correct. II. Read the following paragraph and answer the questions given below: The question assesses the students' knowledge of universal law of gravitation, its variation by varying the mass, distance between the bodies, acceleration due to gravity and how universal law of gravitation is applicable to heavenly bodies.

The diagram shows the equation for universal gravitation where: F is a force, M is the mass of body A, m is the mass of body B and d is the distance between them. G is the universal gravitational constant Q. 1. Describe and explain the relationships between F, M, m and d?[2] Ans. F is attraction force between masses; F increases with mass; F decreases with the square of the distance; Q. 2. On Earth F is approximately 10 N/kg and is denoted by ‘g’. Describe two differences between g and G? [CBSE – SAS] [2] Ans. ‘g’ varies from place to place whereas G is constant; ‘g’ is a vector quantity whereas G is a scalar quantity; ‘g’ and G have different units

(Any two)



GRAVITATION

121

Artificial Intelligence PARAMETERS

DESCRIPTION

Chapter Covered

Chapter 10: Gravitation

Name of the Book

Science, Class 9, NCERT

Subject and Artificial Free fall Intelligence Integrated Acceleration due to gravity integrated with data visualisation and inklewriter. Learning Objectives

Students will be able to: ● Understand the concept of free fall. ● Analyse that while falling there is no change in the direction of motion of the objects. ● Explain that when objects fall towards the earth, it accelerates. ● List the factors on which acceleration due to gravity depends. ● Calculate the value of acceleration due to gravity on Earth. ● Explain why acceleration of a freely falling body is independent of its mass.

Time Required

2-3 classes of 40 minutes each.

Classroom Arrangement

Regular classroom setup, Science Laboratory, Computers with internet connection for each group of students.

Material Required

Textbooks, laptop/desktop with internet connection with the AI tools installed in it.

Pre-Preparation Activities ● Students should observe the objects falling vertically downwards to see if they fall uniformly or not. ● Some understanding of bar graphs/ histogram/ line graphs (from newspapers/ comparative charts). Previous Knowledge

● Students know the types of forces studied in class 8th. ● Understand the terms- velocity, acceleration. ● Objects fall towards the Earth because it pulls them.

Methodology

Activity 1: Teacher to perform the activity and ask the students to observe. Dropping any two objects (a heavy and a light body) from a certain height simultaneously. For example - a hollow and a solid body, a paper and a pen, a paper and a piece of stone. ● Students are asked to observe, which of the two bodies reach the ground faster? ● What if we increase the height from where the two bodies are released? ● Do the bodies fall with uniform speed? ● What difference will you observe if the activity is performed in the presence or absence of air? ● What are the forces acting on a freely falling body? ● After discussing the answers to all the above questions, the teacher explains the concept of free fall and acceleration due to gravity. ● Derive the factors on which acceleration due to gravity depends on. https://ncase.me/loopy/v1.1/ Task to be performed by students: After discussing the activity 1, students will be asked to open the above mentioned url on their desktop. The teacher will write all the factors which may or may not affect the motion of a freely falling body. Using the Loopy tool, students will categorize the given factors as dependent and independent factors. The dependent factors to be shown with a + sign and independent factors to be shown with a - sign.

AI CONCEPTS INTEGRATED

Datavizcatalogue Inklewriter/ my story on chatbot

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Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

Learning Outcomes

Students will be able to: ● Apply the concept of free fall in daily life. ● Explain why velocity of a freely falling body increases. ● Conclude that all objects hollow or solid, big or small, should fall at the same rate. ● State the factors on which acceleration due to gravity depends.

Follow up Activities

● In the last 5- 10 minutes, the teacher recaps the topics covered in each class.

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● Students are asked to read the story of Galileo, how he dropped different objects from the top of the Leaning Tower of Pisa in Italy to prove that all objects fall at the same rate. Ask them to use the AI tool My story on chatbot or https://www.inklewriter.com/to present their understanding in own words. After understanding the factors on which acceleration due to gravity depends on, students will be asked to collect data(from internet) for different values of acceleration due to gravity on all the planets and hence chose “comparison chart” from https://datavizcatalogue.com/search.html to show the difference in the value of g at different planets. Reflections

Discussion with Students on the role of AI application- dataviz catalogue. 

Study Time Reading Time: 2:00 Hr No. of Questions: 26

CHAPTER

10



Syllabus

FLOATATION

Thrust and pressure, Archimedes’ principle, buoyancy,

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Revision Notes

 All objects experience a force of buoyancy when they are immersed in a fluid.  Objects having density less than that of the liquid in which they are immersed, float on the surface of the liquid. If the density of the object is more than the density of the liquid in which it is immersed then it sinks in the liquid.  Archimedes’ principle : When a body is immersed partially or completely in a fluid it experiences an upthrust that is equal to the weight of the fluid displaced by the body.  Lactometers are used to determine the purity of a sample of milk.  Hydrometers are used for determining the density of liquids.  Density of different substances are different.

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Archimedes’ Principle

Key words

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Weight : Force by which an object is attracted towards the Earth. SI unit is Newton (N). Upthrust / buoyant force : The upward force exerted by a liquid on the body that is immersed in the liquid. Density : It is the mass of a unit volume. Its unit is kilogram per metre cube (kgm–3). Pressure : Force per unit area. SI unit is N/m2 or Nm–2 or Pascal. Gravity : Force of attraction due to Earth.

Example (i) Radius of an iron sphere is 0.21 cm. If density of iron is 7.8 g/cm3, calculate its mass. (ii) A pressure of 1000 Pa, acts on a surface of area 15 cm2 by a block of mass ‘m’. Calculate ‘m’. Calculate the new pressure exerted by the same block if the area of contact with the surface becomes 10 cm2.  Solution: Mass (i) Density = Volume M = d × V (a)

d = 7.8 g/cm3 Volume of sphere V =

4 3 pr 3

4 22 21 21 21 × × = × × 3 7 100 100 100 4 × 22 × 21 × 21 = cm3 100 × 100 × 100 Putting the value of V in eqn (a)

M=

7.8 × 4 × 22 × 21 × 21 100 × 100 × 100

(b)

124

Oswaal CBSE Question Bank Chapterwise & Topicwise, SCIENCE, Class-IX

FLOATATION



302702.4 100 × 100 × 100 = 0.3027024 g



= 0.30 g.



(ii)

=

P=

[3]

=

Now, P' =

125

1000 × 15 = 0.15 kg 10 × 100 × 100 mg 0.15  10  = 1,500 Pa[2] A 10  10( 4 )

PA F mg = ⇒m= g A A

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions 

(1 mark each)

Q.1. An object is put one by one in three liquids having different densities. The object floats with 1 , 2 and 3 parts of their volumes outside 9 11 7 the liquid surface in liquids of densities d1, d2 and d3 respectively. Which of the following statement is correct? (A) id1> d2> d3 (B) d1> d2< d3 (C) id1< d2> d3 (D) d1< d2< d3 Ans. Option (D) is correct. Explanation: Here, volumes of object outside the liquids of densities d1, d2, d3 respectively, are given as: 1 = 0.11, 2 = 0.18, and 3 =0.43. This 9 11 7 means buoyant face is maximum in liquid with density d3 and minimum in liquid with density d1. As buoyant force exerted by a liquid is directly proportional to its density, therefore, correct order of the densities of three liquids is: d1 < d2