# NCERT Solutions Class 9 Science

Science in 9th grade is a big leap from 8th grade. The content to learn is comparatively more vast. Some of the students of 9th grade and 10th grade come out with flying colours while some others find it an uphill task. The three main branches of science  – Physics, Chemistry, and Biology often appear challenging to many students. To overcome this issue, NCERT Solutions can be referred to help with this problem.

Students of Class 9 can refer to NCERT Solutions for Class 9 for quick learning. Class 9 Science NCERT Solutions by Extramarks make exam preparation easier as concepts are explained well to render in-depth knowledge to students. Understanding the concepts like Living life forms, Tissues, Force, Structure of atoms, etc., of physics, chemistry, and biology is crucial if students want to score better. So if it’s just doubt clearing or aiming to score better, CBSE Class 9 Science NCERT Solutions are something you can bank on.

The smooth but easy-to-understand language encourages self-learning among students.  These solutions are meant to strengthen the overall foundation of a student’s knowledge in Science so that their higher studies move ahead with a strong base.

Chapter 1: Matter in Our Surroundings

Chapter 2: Is Matter Around Us Pure

Chapter 3: Atoms and Molecules

Chapter 4: Structure of the Atoms

Chapter 5: The Fundamental Unit of Life

Chapter 6: Tissues

Chapter 7: Diversity in Living Organisms

Chapter 8: Motion

Chapter 9: Force and Laws of Motion

Chapter 10: Gravitation

Chapter 11: Work and Energy

Chapter 12: Sound

Chapter 13: Why Do We Fall Ill?

Chapter 14: Natural Resources

Chapter 15: Improvement in Food Resources

### NCERT Solutions of Class 9 Science Book Chapter  an overall view

Class 9 Science NCERT Solutions include all exercises, including end-of-chapter from the NCERT Class 9 Science textbook. NCERT Solutions assist students in a very simple language. It also includes solved numerical questions that will help students practice well for their exams. The chapters test your memory, logic, and numerical skills. Here is a  preview of the Class 9 CBSE Science Chapters;

### NCERT Solutions for Class 9 Science Chapter 1 Matter in Our Surroundings

Everything surrounding us that has mass and occupies space is made up of matter. “Matter in Our Surroundings” is a chapter that forms the basis of learning Chemistry. Due to this, it is crucial that this chapter is understood well. Physicists classify matter based on its physical and chemical properties. This chapter teaches students about the physical properties of matter.

Matter is made up of small particles that have space between them, are continuously moving, and are attracted to each other. The state of the matter can be changed and is interchangeable. This chapter will also make you realise how things around us change because of pressure and temperature.

### NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure

NCERT 9th Class Science Book Solutions have “Is Matter around Us Pure” as the second chapter. This chapter is a further continuation of the first one and discusses the different methods of separation, properties of solutions, classification of matter, compounds and their properties, difference between mixtures and compounds, and physical and chemical changes.

### NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules

Chapter 3 of NCERT Class 9 Science imparts knowledge about laws of chemical combination, molecules, atoms, how to write chemical formulas, etc. This chapter has a significant weightage in exams as one can expect at least 3 to 4 questions from this chapter. Mole concept of this chapter is an essential basis for studying chapter 11 as well. Therefore, build a strong  foundation because a thorough comprehension of the Mole concept is required to solve numerical problems

### NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atoms

This chapter of NCERT Science Class 9 discusses the varied atomic models different scientists have proposed and talks about the distribution of electrons in different orbits, calculation of atomic number, mass number, and valency. This chapter is l particularly useful for  students who plan to pursue science related studies in future..

### NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life

Chapter 5 is related to Biology. This chapter covers how the cell is the fundamental unit of life and further explains the cell structure which has parts like plasma membrane, nucleus, cell wall, etc. It has explanations and diagrams of animal and plant cells. These biological units define all their structural and functional abilities

### NCERT Solutions for Class 9 Science Chapter 6 Tissues

The chapter covers tissues and their types. Animal and plant tissue differences are explained in great detail with diagrams. This will help students form a strong foundational base in  Biology.

### NCERT Solutions for Class 9 Science Chapter 7 Diversity in Living Organisms

This chapter informs students about the classification of plants and animals. The hierarchy of the plant and animal kingdom is discussed in this chapter. It imparts knowledge about the living organisms being classified into mainly 5 kingdoms; Protista, Monera, Plantae, Fungi, and Animalia, and about how  they are  different  from one another.

### NCERT Solutions for Class 9 Science Chapter 8 Motion

NCERT Science Book  Chapter 8 is a mix of theory and numerical questions. It includes information about motion, rate of change of velocity, the speed with direction and the graphical representation of motion. Equations of motions and numerical problems related to the same are also a part of this chapter. This chapter will also  explain the fact that the earth is in constant motion and so are the things on it.

### NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion

The Force and Laws of Motion chapter enhances one’s conceptual understanding of force and motion. It discusses the 3 laws of motion, namely the first, second, and third law of motion, and exposes students to balanced and unbalanced forces.

### NCERT Solutions for Class 9 Science Chapter 10 Gravitation

Newton had wondered why the apple fell from a tree. This led to the existence of gravitational force being discovered and discussed by him. This chapter will discuss the gravitational force in detail (its importance, mass, weight, free fall, etc.) and includes numericals based on it. Good practice will be required to ace the numerical questions. So buckle up and practice well.

### NCERT Solutions for Class 9 Science Chapter 11 Work and Energy

Different examples, activities, and numericals are utilized to define the concept of work in this chapter. Energy and its forms are explained with examples for better understanding. .

### NCERT Solutions for Class 9 Science Chapter 12 Sound

This chapter encompasses the explanation of the production of sound and its journey from the source of production to one’s ears. Propagation of sound waves, echo, reverberation, etc. are a part of this chapter along with a few numericals as well.

### NCERT Solutions for Class 9 Science Chapter 13 Why Do We Fall Ill?

Chapter 13 deals with the significance of health, different types of diseases, be it acute or chronic  and their causes, etc. It imparts necessary information about diseases to students and how to stay fit and healthy.

### NCERT Solutions for Class 9 Science Chapter 14 Natural Resources

Natural resources are utilized by us on a daily basis. The importance of air, soil, water, ozone layer, and biochemical cycle,  as well as how humans are damaging them are some of the topics covered in this chapter.

### NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources

The contributions  of air, water, and soil are discussed in this last chapter of the NCERT Science Class 9 book. It also expands upon agricultural, farming, and dairy topics like crop production, storage of grains, production management, fertilizers, manure, nutritional value of food, fish production, animal husbandry, and practices like beekeeping.

## CBSE 2022-23 Class 9 Science Marking Scheme

 UNIT CHAPTERS FOR TERM- I MARKS I Matter-Its Nature and Behaviour – Chapter – 2 9 II Organisation in the Living World – Chapter 5 and 6 18 III Motion, Force and Work – Chapter 8 and 9 13 UNIT CHAPTERS FOR TERM- II MARKS I Matter: Its Nature and Behaviour – Chapter 3 and 4 18 II Organisation in the Living World – Chapter 13 8 III Motion, Force and Work – Chapter 10 and 11 14 Total theory (Term I + Term II) 80 Internal Assessment –  Term I 10 Internal Assessment –  Term II 10 Grand Total 100

### CBSE Class 9 Science Examination – Preparation Tips

• NCERT Solutions for Class 9 Science include many important questions that are a part of the Class 9 CBSE Exam Syllabus. Learn and understand all the concepts well. Science is something you can’t understand by rote learning. It’s the application of knowledge and understanding.
• Practice diagrams well. Do not draw diagrams with a pen. Do it with a pencil.
• Answer what you know first. Do not waste too much time on one question.
• Practise numerical questions well.
• Solve model question papers and previous years’ question papers to test yourself and to get an idea about the type of questions that will come in the exam.
• Physics questions will mostly be conceptual and numerical. Newton’s laws of motion and their application, diagrams like electric circuit diagram, lines of a magnetic field around a solenoid and bar magnet, image formation by lenses and mirrors, AC and DC generators, glass prism, the human eye, image formation for corrected and defected vision are some of the important  chapters you need to do it thoroughly. Numericals on lens and mirror and the series and parallel combination of resistances are frequently asked to test your skills.
• Chemistry  should be prepared well by understanding the modern periodic table with electronic configuration properly, memorising the elements placed in the Periodic table’s first and last two groups, and applications of acids, basis, and salt in day-to-day life. Learn  how to balance chemical equations and  take proper notes  on reactions involving conversion compounds which will act as revision notes during exams. It is advisable to focus on carbon compounds and the nomenclature containing functional groups.
• Biology can be aced by revising complex terminologies regularly, preparing Mendel’s experiments, understanding traits inheritance, functions of reproductive organs in humans, and the process of reproduction in plants. Diagrams  of different parts of a flower, respiratory system, representation of reflex action, human brain, male and female reproductive organs, etc. can be practiced for better results.

## Features of NCERT 9th Class Science Book Solutions

Here are the features of NCERT Science Book Solutions for Class 9;

• It includes all the questions and answers that are in the NCERT science textbook.
• It also includes MCQs, previous years’ question papers, short questions and answers, descriptive questions and answers, etc.
• The inclusion of diagrams to explain concepts makes understanding and learning the topics quicker and easier.
• Excellently curated answers help students score better.

Q.1 Compare the properties of electrons, protons and neutrons.

Ans

 Electron Proton Neutron They are present outside the nucleus. They are present in the nucleus. They are present in the nucleus. They are negatively charged. They are positively charged. They are neutral. They are represented as e–. They are represented as p+. They are represented as n0.

Q.2 What are the limitations of J.J. Thomson’s model of the atom?

Ans

According to J.J. Thomson’s model of an atom, an atom consists of a positively charged sphere and the electrons are embedded in it. However, later it was found that the positively charged particles are present at the centre of the atom (nucleus) and electrons revolve around the nucleus.

Q.3 What are the limitations of Rutherford’s model of the atom?

Ans

The limitations of Rutherford’s model of the atom are:

• According to the electromagnetic theory, if a charged particle moves around another charged particle, it accelerates and continuously loses energy in the form of the radiant energy. Loss of energy would slow down the speed of the electron and eventually, the electron would fall into the nucleus. However, such a collapse did not occur and Rutherford’s model was unable to explain the reason behind it.
• It does not say anything about the distribution of electrons around the nucleus and the energy of electrons.

Q.4 Describe Bohr’s model of the atom.

Ans

1. Electrons revolve around the nucleus in certain discrete orbits (energy levels) without losing any energy.
2. An electron present in a particular orbit possesses a definite amount of energy.
3. An electron does not radiate energy (lose energy) even though it accelerates the motion around the nucleus.

Q.5 Compare all the proposed models of an atom given in this chapter.

Ans

 Thomson’s Model Rutherford’s Model Bohr’s Model An atom consists of a positively charged sphere and the electrons are embedded in it. There is a positively charged centre in the atom, which is called the nucleus. Almost the whole mass of an atom resides in the nucleus. The electrons revolve around the nucleus in well-defined orbits. The size of a nucleus is very small as compared to the size of an atom. Electrons revolve around the nucleus in certain discrete orbits (energy levels) without losing any energy. An electron present in a particular orbit possesses a definite amount of energy. An electron does not radiate energy (lose energy) even though it accelerates the motion around the nucleus.

Q.6 Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.

Ans

The rules for distributing the electrons in various shells are given below.

1.The maximum number of electrons present in a shell (n) is given by the formula 2n2, where ‘n’ is an integer.

For example,

Maximum number of electrons in K shell is 2(1)2 = 2 electrons

Maximum number of electrons in L shell is 2(2)2 = 8 electrons

2.The maximum number of electrons in the outermost orbit of an atom is 8.

For example,

Total number of electrons 13

Expected distribution is 2,11 but actual distribution is 2,8,3

3.Electrons are not accommodated in a given shell, unless the inner shells are filled, i.e., shells are filled in a step-wise manner.

For example,

In order to accommodate 13 electrons in K, L and M shell, the correct distribution is 2,8,3 and not 2,7,4.

Q.7 Define valency by taking examples of silicon and oxygen.

Ans

Valency is the number of electrons gained, lost or shared by an atom so as to make the octet of the electrons in the outermost shell.

For example,

1. Atomic number of Silicon is 14.

Electronic configuration of Silicon is 2,8,4.

Since the number of valence electrons is 4 therefore, valency of Silicon is 4.

1. Atomic number of Oxygen is 8.

Electronic configuration of Oxygen is 2,6.

Since the number of valence electrons is more than 4 therefore, valency of Oxygen is 8 – 6 i.e., 2.

Q.8 Explain with examples
(i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. Give any two uses of isotopes.

Ans

(i) Atomic number:

• It represents the total number of protons present in the nucleus of an atom.
• It is denoted by Z.
• All the atoms of an element have the same atomic number.
• The number of electrons and protons in an atom are equal.

For example,

Atomic number of Boron is 5

Atomic number of Aluminium is 13

Atomic number of Argon is 18

(ii) Mass number:

• It represents the total number of protons and neutrons present in the nucleus of an atom.
• It is denoted by A.
• The mass of an atom is due to the protons and neutrons present in the nucleus of an atom.
• Protons and neutrons are collectively called the nucleons.

For example,

Mass number of Boron is 11

Mass number of Aluminium is 27

Mass number of Argon is 40

(iii) Isotopes:

• These are the atoms of the same element that have the same atomic number but different mass number.
• Isotopes of a particular element have the same chemical properties but different physical properties.

For example,

Carbon has three isotopes 12C, 13C, 14C

Boron has two isotopes 10B, 11B

(iv) Isobars:

• These are the atoms that have the same mass numbers but different atomic numbers.
• Isobars have different chemical as well as physical properties.

For example,

Ar40, 19K40, 20Ca40

C14, 7N14

Uses of Isotopes

1. An isotope of uranium is used as fuel in the nuclear reactor.
2. An isotope of cobalt is used in the treatment of cancer.

Q.9 Na+ has completely filled K and L shells. Explain.

Ans

Atomic number of Sodium (Na) is 11.

Electronic configuration of Sodium (Na) is 2,8,1.

In order to complete its octet it will lose its valence electron and form Sodium ion (Na+).

Electronic configuration of sodium ion (Na+) is 2,8.

In the electronic configuration of sodium ion both K and L shell are completely filled.

Q.10 If bromine atom is available in the form of, say, two isotopes 35Br79 (49.7%) and 35Br81 (50.3%), calculate the average atomic mass of bromine atom.

Ans

$\begin{array}{l}\text{Atomic mass of bromine =}\frac{\text{79u x 49}\text{.7 + 81u x 50}\text{.3}}{\text{49}\text{.7+50}\text{.3}}\\ \text{=}\frac{\text{8000}\text{.6}}{\text{100}}\text{u}\\ \text{= 80}\text{.0 u}\end{array}$

Q.11 The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 16/8X and 18/8 X in the sample?

Ans

$\begin{array}{l}\text{Let the percentage of the isotope}\text{X}_{\text{8}}^{\text{16}}\text{be y}_{\text{1}}\\ \text{Hence, the percentage of the isotope}\text{X}_{\text{8}}^{\text{18}}\text{is = 100 – y}\text{}_{\text{1}}\\ \text{Average atomic mass of X =}\frac{\text{(Percentage of}\text{X}_{\text{8}}^{\text{16}}\text{x 16u) + (100 – % of}\text{X}_{\text{8}}^{\text{16}}\text{) x 18u}}{\text{100}}\\ \text{or 16}\text{.2 u =}\frac{{\text{(y}}_{\text{1}}{\text{x 16 u) + (100 – y}}_{\text{1}}\text{) x 18 u}}{\text{100}}\\ {\text{or 1620 u = 16 y}}_{\text{1}}{\text{u + 1800 u –18 y}}_{\text{1}}\text{u}\\ {\text{or 18 y}}_{\text{1}}{\text{u – 16 y}}_{\text{1}}\text{u = 1800u – 1620 u}\\ {\text{or 2y}}_{\text{1}}\text{u = 180 u}\\ {\text{or y}}_{\text{1}}\text{=}\frac{\text{180}}{\text{2}}\text{= 90}\\ \text{Now, Percentage of}\text{X}_{\text{8}}^{\text{16}}\text{= 90 %}\\ \text{Percentage of}\text{X}_{\text{8}}^{\text{18}}\text{= 100 – 90 = 10 %}\end{array}$

Q.12 If Z = 3, what would be the valency of the element? Also, name the element.

Ans

Z represents the atomic number of an element. Atomic number 3 is of the element Lithium.

Electronic configuration of Lithium is 2,1. In order to complete its duplet it will lose its valence electron therefore, its valency is 1.

Q.13 Composition of the nuclei of two atomic species X and Y are given as under

 X Y Protons 6 6 Neutrons 6 8

Give the mass numbers of X and Y. What is the relation between the two species?

Ans

We know that

Mass number = Number of protons + Number of neutrons

Mass number of X = 6 + 6 = 12

Mass number of Y = 6 + 8 = 14

X and Y have the same atomic number but different mass numbers therefore; X and Y are the Isotopes.

Q.14 For the following statements, write T for True and F for False.

(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.

(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.

(c) The mass of an electron is about 1/2000 times that of proton.

(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.

Ans

(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons. (F)

(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral. (F)

(c) The mass of an electron is about 1/2000 times that of proton. (T)

(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine. (T)

Q.15 Which one of the following is a correct electronic configuration of sodium?

(a) 2,8 (b) 8,2,1 (c) 2,1,8 (d) 2,8,1.

Ans

2,8,1

Q.16 Complete the following table.

 Atomic number Mass number Number of neutrons Number of protons Number of electrons Name of the atomic species 9 – 10 – – – 16 32 – – – Sulphur – 24 – 12 – – – 2 – 1 – – – 1 0 1 0 –

Ans

 Atomic number Mass number Number of neutrons Number of protons Number of electrons Name of the atomic species 9 19 10 9 9 Fluorine 16 32 16 16 16 Sulphur 12 24 12 12 12 Magnesium 1 2 1 1 1 Deuterium 1 1 0 1 0 Protium

Q.17 Convert the following temperatures to the Celsius scale.
(a) 293 K (b) 470 K

Ans

$\begin{array}{l}\left(\text{a}\right)\text{Temperature in Celsius scale}=\text{Temperature in Kelvin scale}-\text{273}\\ \text{}=\text{293}-\text{2}73\\ \text{}=\text{20}{\text{\hspace{0.17em}}}^{\text{ο}}\text{C}\text{\hspace{0.17em}}\\ \left(\text{b}\right)\text{Temperature in Celsius scale}=\text{Temperature in Kelvin scale}-\text{273}\\ \text{}=\text{470}-\text{2}73\\ \text{}={\text{197}}^{\text{ο}}\text{C}\text{\hspace{0.17em}}\end{array}$

Q.18 Convert the following temperatures to the Kelvin scale.

(a) 25°C (b) 373°C

Ans

$\begin{array}{l}\left(\text{a}\right)\text{Temperature in Kelvin scale = Temperature in Celsius scale+273}\\ \text{= 25+273}\\ \text{= 298K}\text{\hspace{0.17em}}\\ \left(\text{b}\right)\text{Temperature in Kelvin scale = Temperature in Celsius scale + 273}\\ \text{= 373+273}\\ \text{= 646K}\end{array}$

Q.19 Give reason for the following observations.
(a) Naphthalene balls disappear with time without leaving any solid.
(b) We can get the smell of perfume sitting several metres away.

Ans

(a) Naphthalene is a sublimating substance. Therefore, naphthalene balls sublimate that is turns into vapour without changing into liquid. Therefore, they disappear with time without leaving any solid.
(b) We can get the smell of perfume sitting several meters away because of diffusion. The molecules of perfume spread out through the air and reach to great distance.

Q.20 Arrange the following substances in increasing order of forces of attraction between the particles— water, sugar, oxygen.

Ans

Oxygen is a gas; therefore, the force of attraction is negligible between its particles. Water is a liquid, so the force of attraction between its particles is more than liquid and less than solid. Since sugar is a solid, thus force of attraction between particles is greatest. Hence, the correct increasing order of forces of attraction between the particles is as follows

Oxygen < Water < Sugar

Q.21 What is the physical state of water at—
(a) 25°C (b) 0°C (c) 100°C ?

Ans

(a) At 25°C water exists as a liquid, with some water vapour above the water surface
(b) At 0°C water can exist as a solid as well as a liquid. Small amount of water vapour will be present in the air above the water/ice mixture.
(c) At 100°C water can exist in both liquid and gaseous states.

Q.22 Give two reasons to justify—
(a) Water at room temperature is a liquid.
(b) an iron almirah is a solid at room temperature.

Ans

(a) Water has no fixed shape but has a fixed volume that is,it occupies the shape of the container in which it is kept. Boiling point of water is 100°C and its freezing point is 0°C but the room temperature is between 0 and 100 0°C, therefore, it is a liquid at room temperature.
(b) All the atoms in iron are very tightly packed at the room temperature. Also its melting and boiling point is very high as it is a metal. So it is a solid at room temperature.

Q.23 Why is ice at 273 K more effective in cooling than water at the same temperature?

Ans

Ice is solid form of water which at its freezing point (273K) absorbs heat to get converted into liquid form at this constant temperature which is called the latent heat of fusion. When completely converted into water (liquid), it will absorb heat till its temperature becomes equal to that of the surrounding or the body to which it has contact.

But water at 273K will only absorb heat till its temperature becomes equal to that of the surrounding or the body to which it has contact.

Therefore, ice at 273 K more effective in cooling than water at the same temperature.

Q.24 What produces more severe burns, boiling water or steam?

Ans

Steam contains more heat, in the form of latent heat, as compared to boiling water. So when steam comes in contact with skin it will give more heat than boiling water, so steam causes more severe burns.

Burns caused by steam are more severe than those caused by the boiling water; though both are at the same temperature (100 °C).This is because, steam has 540 kcal/kg (latent heat) more energy than boiling water. When steam condenses on skin, it gives out 540 kcal/kg more energy than the boiling water.

Q.25 Name A, B, C, D, E and F in the following diagram showing change in its state

Ans

A. Fusion or melting
B. Vaporisation
C. Condensation
D. Solidification
E. Sublimation
F. Deposition

Q.26 A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Ans

Mass of the given compound = 0.24 g

Mass of boron = 0.096 g

Mass of oxygen = 0.144g

So,

$\begin{array}{l}\text{Percentage of boron by weight =}\frac{\text{Mass of boron}}{\text{Mass of the compound}}\text{x 100}\\ \text{=}\frac{\text{0}\text{.096g}}{\text{0}\text{.24g}}\text{x 100}\\ \text{= 40 %}\end{array}$ $\begin{array}{l}\text{Percentage of oxygen by weight =}\frac{\text{Mass of oxygen}}{\text{Mass of the compound}}\text{x 100}\\ \text{=}\frac{\text{0}\text{.144g}}{\text{0}\text{.24g}}\text{x 100}\\ \text{= 60 %}\end{array}$

Q.27 When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen?

Ans

Given that,

$\begin{array}{l}{\text{C + O}}_{2}\text{}\to {\text{CO}}_{2}\text{}\\ \text{3g 8g 11g}\end{array}$

The given data shows that in this reaction, 3.0 g of carbon and 8.0 g of oxygen react completely to produce 11.0 g of carbon dioxide.

In the second experiment, excess of oxygen (50 g) is present. When 3.00 g of carbon is burnt in 50.00 g of oxygen, it will use just 8.00 g of oxygen from the total of 50 g of oxygen to produce 11 g of carbon dioxide and the rest of 42 g (50 g – 8 g) of oxygen will be left behind without any change.

The above answer is governed by the following two laws of chemical combination:

1. Law of conservation of mass which states that mass can neither be created nor destroyed in a chemical reaction i.e., total mass of reactants remains same as the total mass of products.
Here, in terms of mass 3g of carbon and 50 g of oxygen combine to give 11 g of carbon dioxide and 42 g of unused oxygen keeping the total mass of combining chemicals by weight constant before and after the reaction.
Total mass of reactants = 3 + 50= 53 g

Total mass of products = 11 + 42 (un-used oxygen) = 53g
Total mass of reactants = total mass of products

Hence, the law of conservation of mass is proved.

2. Law of constant proportions states that in a chemical compound the elements are always present in definite proportions by mass. Here, irrespective of the excess quantity of oxygen available for burning of same weight of carbon, the amount of carbon dioxide will remain same. This means that (50 – 8) = 42 g of oxygen will remain unreacted.

Q.28 What are polyatomic ions? Give examples.

Ans

Polyatomic ions are the group of atoms carrying a net charge on them. This charge may be positive or negative.

Examples:

Ammonium ion (NH4+), Nitrate ion (NO3), Carbonate ion (CO32–), Sulphate ion (SO42–)

Q.29 Write the chemical formulae of the following.

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate

Ans

(a) Magnesium chloride: MgCl2

(b) Calcium oxide: CaO

(c) Copper nitrate: Cu(NO3)2

(d) Aluminium chloride: AlCl3

(e) Calcium carbonate: CaCO3

Q.30 Give the names of the elements present in the following compounds.

(a) Quick lime

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate.

Ans

 Compound Elements present (a) Quick lime (CaO) Carbon, Oxygen (b) Hydrogen bromide (HBr) Hydrogen, Bromine (c) Baking powder (NaHCO3) Sodium, Hydrogen, Carbon, Oxygen (d) Potassium sulphate (K2SO4) Potassium, Sulphur, Oxygen

Q.31 Calculate the molar mass of the following substances.

(a) Ethyne, C2H2

(b) Sulphur molecule, S8

(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)

(d) Hydrochloric acid, HCl

(e) Nitric acid, HNO3

Ans

(a) Gram atomic mass of carbon = 12 g

Gram atomic mass of hydrogen = 1g

Molar mass of ethyne (C2H2) = 2 x 12 + 2 x 1

= 26 g/mol

(b) Gram atomic mass of sulphur = 32 g

Molar mass of sulphur molecule (S8) = 8 x 32

= 256 g/mol

(c) Gram atomic mass of phosphorus = 31 g

Molar mass of phosphorus molecule (P4) = 4 x 31

= 124 g/mol

(d) Gram atomic mass of hydrogen = 1g

Gram atomic mass of chlorine = 35.5 g

Molar mass of hydrochloric acid (HCl) = 1 + 35.5

= 36.5 g/mol

(e) Gram atomic mass of hydrogen = 1g

Gram atomic mass of nitrogen = 14 g

Gram atomic mass of oxygen = 16 g

Molar mass of nitric acid (HNO3) = 1 + 14 + 3 x 16

= 63 g/mol

Q.32 What is the mass of:

(a) 1 mole of nitrogen atoms?

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?

(c) 10 moles of sodium sulphite (Na2SO3) ?

Ans

a. Mass of 1 mole of nitrogen atoms = Atomic mass of nitrogen expressed in grams = 14 g
b. Mass of 4 moles of aluminium atoms = 4 x Atomic mass of aluminium expressed in grams = 4 x 27 = 108 g
c.Molar mass of sodium sulphite (Na2SO3) = (2 x 23 + 32 + 3 x 16) g/mol

$\begin{array}{l}\\ \text{= (46 + 32 + 48) g/mol}\\ \text{= 126 g/mol}\\ \text{Mass of 10 moles of sodium sulphite = 10 mol x Molar mass of sodium sulphite}\\ \text{= 10 mol x 126 g/mol}\\ \text{= 1260 g}\end{array}$

Q.33 Convert into mole:

(a) 12 g of oxygen gas

(b) 20 g of water

(c) 22 g of carbon dioxide

Ans

(a)

$\begin{array}{l}\text{Given mass of oxygen gas = 12 g}\\ \text{Molar mass of oxygen gas (O2) = 2 x 16 g/mol}\\ \text{= 32 g/mol}\\ \text{The number of moles (n) =}\frac{\text{Given mass}}{\text{Molar mass}}\\ \text{=}\frac{\text{12 g}}{\text{32 g/mol}}\text{= 0}\text{.375 mol}\end{array}$

Hence, 12 g of oxygen gas = 0.375 mol of oxygen gas

(b)

$\begin{array}{l}\text{Given mass of water = 20 g}\\ \text{Molar mass of water (H2O) = (2 x 1 + 1 x 16) g/mol}\\ \text{= 18 g/mol}\\ \text{The number of moles (n) =}\frac{\text{Given mass}}{\text{Molar mass}}\\ \text{=}\frac{\text{20 g}}{\text{18 g/mol}}\text{= 1}\text{.11 mol}\end{array}$

Hence, 20 g of water = 1.11 mol of water

$\begin{array}{l}\text{(c)}\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\text{Given mass of carbon dioxide = 22 g}\\ \text{Molar mass of carbon dioxide (CO2) = (12 + 2 x 16) g/mol}\\ \text{= 44 g/mol}\\ \text{The number of moles (n) =}\frac{\text{Given mass}}{\text{Molar mass}}\\ \text{=}\frac{\text{22 g}}{\text{44 g/mol}}\text{= 0}\text{.5 mol}\end{array}$

Hence, 20 g of carbon dioxide = 0.5 mol of carbon dioxide

Q.34 What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Ans

$\begin{array}{l}\left(\text{a}\right)\\ \text{Mass of 0}\text{.2 mol atoms of oxygen = 0}\text{.2 mol x Molar mass of oxygen atom}\\ \text{= 0}\text{.2 mol x 16 g/mol}\\ \text{= 3}\text{.2 g}\\ \left(\text{b}\right)\\ \text{Molar mass of water}\left({\text{H}}_{2}\text{O}\right)\text{= (2 x 1 + 16) g/mol}\\ \text{= 18 g/mol}\\ \text{Mass of 0}\text{.5 mol molecules of water = 0}\text{.5 mol x Molar mass of water}\\ \text{= 0}\text{.5 mol x 18 g/mol}\\ \text{= 9 g}\end{array}$

Q.35 Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Ans

Mass of solid sulphur = 16 g

Molar mass of sulphur molecule (S8) = 8 × 32g/mol = 256 g/mol

Avogadro’s number = 6.022 × 1023

$\begin{array}{l}{\text{Number of moles of S}}_{8}\text{molecules =}\frac{\text{Given mass of sulphur}}{{\text{Molar mass of S}}_{8}\text{}}\\ \text{=}\frac{\text{16 g}}{\text{256 g/mol}}\text{=}\frac{\text{16}}{256}\text{mol}\\ \text{No}{\text{. of S}}_{8}\text{molecules in the sample = No}{\text{. of moles of S}}_{8}\text{x Avogadro’s number}\\ \text{=}\frac{16}{256}\text{x 6}{\text{.022 x 10}}^{\text{23}}\text{}\\ \text{= 3}{\text{.76 x 10}}^{\text{22}}\text{molecules}\end{array}$

Q.36 Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

Ans

$Mass of aluminium oxide = 0.051 g Molar mass of aluminium oxide ( Al 2 O 3 )=( 2×27+3×16 ) g/mol = 102 g/mol Number of moles of aluminium oxide molecules = Given mass Molar mass = 0.051 g 102 g/mol =5.0× 10 −4 molMathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@E1F9@$

From the stoichiometry,

Al2O3 ≡ 2 Al3+
∴ 1 mol of Al2O3 contains = 2 mol of Al3+ ions
∴ 5.0 × 10–4 mol Al2O3 contains = 2 × 5.0 × 10–4 mol of Al3+ ions = 10–3 mol of Al3+ ions

So,
Number of Al3+ ions present in 10–3 mol
= 10–3 mol × 6.022 × 1023 ions mol–1
= 6.022 × 1020 ions

Hence, 0.051 g of aluminium oxide (Al2O3) contains 6.022 × 1020 Al3+ ions.

Q.37 Which separation techniques will you apply for the separation of the following?

(a) Sodium chloride from its solution in water.

(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.

(c) Small pieces of metal in the engine oil of a car.

(d) Different pigments from an extract of flower petals.(e) Butter from curd.

(f) Oil from water.

(g) Tea leaves from tea.

(h) Iron pins from sand.

(i) Wheat grains from husk.

(j) Fine mud particles suspended in water.

Ans

(a) Sodium chloride from its solution in water: Evaporation
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride: Sublimation
(c) Small pieces of metal in the engine oil of a car: Filtration
(d) Different pigments from an extract of flower petals: Chromatography
(e) Butter from curd: Centrifugation
(f) Oil from water: Using separating funnel
(g) Tea leaves from tea: Filtration
(h) Iron pins from sand: Magnetic Separation
(i) Wheat grains from husk: Winnowing
(j) Fine mud particles suspended in water: Centrifugation

Q.38 Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.

Ans

Tea can be prepared by following steps:

1. Take some water as solvent in a kettle and boil it for few minutes.
2. Now add one tea spoon sugar, one tea spoon tea leaves and some milk as solute to water. They all together form a solution.
3. Now boil the solution again for few minutes so that sugar dissolves in solution as sugar is soluble in water.
4. Now filter the solution through a strainer. The insoluble tea leaves remain on the strainer as residue and the filtrate is collected in cup.

Q.39 Pragya tested the solubility of three different substances at different temperatures and collected the data as given below

(results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution)

(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?

(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.

(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?

(d) What is the effect of change of temperature on the Solubility of a salt?

Ans

$\begin{array}{l}\text{(a) Solubility of potassium nitrate at 313 K =}\frac{\text{62}}{\text{100}}\text{}\\ \text{100 g of water contains potassium nitrate = 62 g}\\ \therefore \text{50 g of water contains potassium nitrate=}\frac{\text{62}}{\text{100}}\text{x 50 = 31 g}\\ \text{Thus, 31 g potassium nitrate would be needed to}\\ \text{produce a saturated solution of potassium nitrate in}\\ \text{50 g of water at 313 K}\text{.}\end{array}$

(b) On cooling the solubility of salt decreases with the falling temperature. Hence, the crystals of potassium chloride will separate out on cooling.

(c) The maximum amount of the salt which is dissolved in 100 g of water to form a saturated solution at the given temperature is known as solubility. The solubilities of different salts at 293 K in 100g of water are given in column 2 in the table.

1. Solubility of potassium nitrate at 293K is 32g.
2. Solubility of sodium chloride at 293K is 36g.
3. Solubility of potassium chloride at 293K is 35g.
4. Solubility of ammonium chloride at 293K is 37g.

From the above data it is clear that ammonium chloride has the highest solubility at 293 K.

(d) Generally, the solubility of a salt increases with the increasing temperature. However, from the data given in the table, the increase is different for different salts. For example the solubility of potassium nitrate increases appreciably, that of ammonium chloride increases slightly, that of potassium chloride increases marginally while that of sodium chloride almost remains constant.

Q.40 Explain the following giving examples.
(a) saturated solution
(b) pure substance
(c) colloid
(d) suspension

Ans

(a) Saturated solution: A solution in which no more solute can be dissolved at a particular temperature is called as saturated solution. For example: Soft drinks are saturated with carbon dioxide, hence it gives off carbon dioxide through bubbles, the Earth’s soil is saturated with nitrogen.

(b) Pure Substance: A substance that shows same characteristics at a given temperature and pressure is called a pure substance. For example distilled water is a pure substance.

(c) Colloid: A colloid is a heterogeneous mixture that consists of particles bigger than that of a solution and smaller than that of suspension. For example: Smoke, butter, milk, etc. are colloids.

(d) Suspension: A suspension is a heterogeneous mixture in which solid is dispersed in liquid. The solute particles in suspension do not dissolve but remain suspended throughout the medium. The particles are large in size and visible to naked eye. For example: Chalk particles, muddy water etc.

Q.41 Classify each of the following as a homogeneous or heterogeneous mixture.

soda water, wood, air, soil, vinegar, filtered tea.

Ans

Homogeneous mixtures: Soda water, vinegar, filtered tea

Heterogeneous mixtures: Soil, wood

Air is a homogeneous mixture of different gases. However, if some dust or other particles are also present, then air becomes heterogeneous mixture.

Q.42 How would you confirm that a colourless liquid given to you is pure water?

Ans

Every liquid has a characteristic boiling point. Pure water has boiling point of 100 °C (373 K) at 1 atmospheric pressure. Hence, the purity of water can be confirmed by determining its boiling point.

Q.43 Which of the following materials fall in the category of a “pure substance”?

(a) Ice

(b) Milk

(c) Iron

(d) Hydrochloric acid

(e) Calcium oxide

(f) Mercury

(g) Brick

(h) Wood

(i) Air

Ans

Ice, iron, hydrochloric acid, calcium oxide and mercury fall in the category of pure substance.

Q.44 Identify the solutions among the following mixtures.

(a) Soil

(b) Sea water

(c) Air

(d) Coal

(e) Soda water

Ans

Sea water, air and soda water are solutions.

Q.45 Which of the following will show “Tyndall effect”?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution.

Ans

Milk and starch solution will show Tyndall effect because they are colloidal solutions.

Q.46 Classify the following into elements, compounds and mixtures.
(a) Sodium
(b) Soil
(c) Sugar solution
(d) Silver
(e) Calcium carbonate
(f) Tin
(g) Silicon
(h) Coal
(i) Air
(j) Soap
(k) Methane
(l) Carbon dioxide
(m) Blood

Ans

 Element Compound Mixture Sodium, Silver, Tin, Silicon Calcium carbonate, Methane, Soap, Carbon dioxide Soil, Sugar solution, Coal Air, Blood

Q.47 Which of the following are chemical changes?
(a) Growth of a plant
(b) Rusting of iron
(c) Mixing of iron filings and sand
(d) Cooking of food
(e) Digestion of food
(f) Freezing of water
(g) Burning of a candle

Ans

Growth of plant, rusting of iron, cooking of food, digestion of food and burning of a candle are chemical changes.

Q.48 Make a comparison and write down ways in which plant cells are different from animal cells.

Ans

 Features Animal Cell Plant Cell Size Small Comparatively larger than animal cell Cell wall Absent Present Plastids Generally absent except in Euglena Present Vacuoles Small Large

Q.49 How is a prokaryotic cell different from a eukaryotic cell?

Ans

 Features Prokaryotic cell Eukaryotic cell Size Generally small (1-10µm) Larger than prokaryotic cells (5-100 µm) Nuclear region Poorly defined, nuclear membrane absent, lacks a true nucleus Clearly defined, true nucleus is surrounded by a nuclear membrane Nucleolus Absent Present Chromosomes Single Many Membrane-bound cytoplasmic organelles Absent Present

Q.50 What would happen if the plasma membrane ruptures or breaks down?

Ans

Rupturing or breakdown of plasma membrane would not allow the cell to exchange substances from its surrounding through diffusion or osmosis. The selectively permeable nature of the plasma membrane will be disturbed as the cytoplasm of the cell would come in direct contact of the surrounding medium and the act of selection would stop. In case of severe rupturing of plasma membrane, the contents of the cell will leak out and would eventually cause the death of the cell.

Q.51 What would happen to the life of a cell if there was no Golgi apparatus?

Ans

In the absence of the Golgi apparatus, following activities will not take place in a cell:

1. Functions like storage, modification and packaging of products synthesised in the endoplasmic reticulum will not take place in a cell as all these functions are performed by the Golgi apparatus.
2. Formation of complex sugars from simple sugars will be hampered in the absence of Golgi apparatus.
3. Cleaning of the cell will be disturbed due to the absence of Golgi apparatus as it is involved in the formation of lysosomes that acts as a waste disposal system of a cell.

Q.52 Which organelle is known as the powerhouse of the cell? Why?

Ans

Mitochondria are known as the powerhouse of the cell. During cellular respiration, mitochondria produce energy in the form of ATP that is required to perform various activities in a cell. ATP (Adenosine triphosphate) is known as the energy currency of the cell. Due to this, mitochondria are known as the powerhouse of the cell.

Q.53 Where do the lipids and proteins constituting the cell membrane get synthesised?

Ans

Lipids and proteins constituting the cell membrane are synthesised in the endoplasmic reticulum. Lipids are synthesised in smooth endoplasmic reticulum. Ribosomes present on the rough endoplasmic reticulum serve as the site for protein synthesis.

Q.54 How does an Amoeba obtain its food?

Ans

Amoeba acquires its food by the process of endocytosis. Amoeba captures food from its surroundings through its plasma membrane. Being flexible in nature, plasma membrane enables the cell to engulf food particles from the surrounding medium by extending itself into the pseudopodia.

Q.55 What is osmosis?

Ans

Osmosis can be defined as the movement of water molecules from a region of high water concentration to a region of low water concentration through a selectively permeable membrane.

Q.56 Carry out the following osmosis experiment:
Take four peeled potato halves and scoop each one out to make potato cups. One of these potato cups should be made from a boiled potato. Put each potato cup in a trough containing water. Now,
(a) Keep cup A empty
(b) Put one teaspoon sugar in cup B
(c) Put one teaspoon salt in cup C
Put one teaspoon sugar in the boiled potato cup D
Keep these for two hours. Then observe the four potato cups and answer the following:
(i) Explain why water gathers in the hollowed portion of B and C.
(ii) Why is potato A necessary for this experiment?
(iii) Explain why water does not gather in the hollowed out portions of A and D.

Ans

(i) Water gathers in the hallowed portions of set-up B and C due to osmosis. In the cup of the potato B, sugar is kept and in the potato C, salt is kept. Inside the cups of the potatoes, the concentration of water is low as compared to the cells making the hallowed portion. Hence, water from its high concentration (from the potato cell) moves to the low concentration (out of the cell and in the hallowed portion) through the plasma membrane of the cells. Due to the movement of water from the potato cells to the hallowed portion by osmosis, water gathers in the hallowed portions of the potato cup.

(ii) In this experiment, potato A acts as control set-up. A control set-up is run in an experiment to study the differences that take place in or on the object or group being experiment on. It also increases the reliability of the final result by comparing the observations of the experimental set-up and control set-up.

(iii) Water does not gather in the hallowed portions of potato A and D because potato A is kept empty and is a control set-up in the experiment. Water does not enter the potato D as osmosis did not take place in it because the potato used here was boiled. Boiling denatured the proteins present in the plasma membrane of the cells of the potato and thus, disrupted the cell membrane. A selectively permeable membrane is required for osmosis to occur which is disrupted in this case. Therefore, osmosis does not occur. Hence, water does not enter the potato D.

Q.57 Which type of cell division is required for growth and repair of body and which type is involved in formation of gametes?

Ans

The type of cell division which is required for the growth and repair of the body is mitosis. Mitosis maintains the same chromosome number in parent and daughter cells. Thus the daughter cells retain the characters of the parent cell. This results in the growth of tissues and the overall development of the organism. With the help of mitosis, tissues are repaired as the worn out cells are replaced.

For the formation of gametes, meiotic division is required. It is also called reductional division. In this type of cell division, the chromosome number is reduced to half. Thus, this cell division occurs in diploid reproductive cells to form haploid cells that are gametes (sperm and egg).

Q.58 What are the advantages of classifying organisms?

Ans

Following are the advantages of classifying organisms:

1. Classifying organisms helps us in recognising and establishing the basic arrangement of a hierarchical structure among diverse species.
2. It helps us in establishing relationship and resemblances between various organisms and thus facilitates research of wide variety related with organisms.
3. It helps in understanding the evolution of organisms, as it gives us the information,allowing a reconstruction of phylogeny of life.

Q.59 How would you choose between two characteristics to be used for developing a hierarchy in classification?

Ans

We should choose the basic characteristics among several other characteristics for developing a hierarchy of classification. The basic characteristics that are chosen are free of any other characteristics in their effects on the form and function of the organism. The characteristics in the next level would be dependent on the previous one and would help in deciding the variety in the next level. In this way, the entire hierarchy of commonly related characteristics would be build to be used for classification. This helps in the establishment of hierarchy of characteristics.

Q.60 Explain the basis for grouping organisms into five kingdoms.

Ans

In 1959, Robert H Whittaker proposed a five kingdom classification of living organisms on the basis of cell structure, mode and source of nutrition and body organisation as main features. The five kingdoms proposed by Whittaker are Monera, Protista, Fungi, Plantae and Animalia.

The basis for grouping organisms into five kingdoms:

1. Complexity of cell structure- based on the presence of membrane bound cell organelles and well-defined nucleus the cells are broadly categorised into: prokaryotic and eukaryotic.
2. Cell structure: Presence or absence of cell wall is another important characteristic
3. Unicellular and multicellular organisms: Based on the number of cells present in the body of the organisms, they are categorised into unicellular and multicellular
4. Mode of nutrition: Organisms basically have two types of modes of nutrition autotrophic and heterotrophic and they are classified according to that.
5. Phylogenetic relationship: Phylogeny is the evolutionary history or ancestry of an organism. It is a belief that simple organisms are primitive and complex organisms are advanced. It is also believed that the advanced organisms are evolved from primitive ones. Thus, primitive and advance nature of organisms also helps in the classification of organisms into broad groups.

Q.61 What are the major divisions in Plantae? What is the basis for these divisions?

Ans

The kingdom Plantae is divided into five main divisions: Thallophyta, Bryophyta, Pteridophyta, Gymnosperms and Angiosperms.

The classification depends on the following criteria:

1. Differentiated or undifferentiated plant body
2. Presence or absence of vascular tissues
3. With or without seeds
4. Naked seeds or seeds inside fruits

1. The first level of classification depends on whether a plant body well differentiated or not. The plants that do not have a well differentiated body are grouped as thallophyta.
2. On the basis of the presence or absence of vascular tissues, plants are catogorised as bryophytes (plants without vascular tissues) and tracheophytes (plants with vascular tissues).
3. On the basis of the absence/presence of seed formation, tracheophyta is sub-divided into two divisions pteridophyta and phenerogams.
4. Pteridophytes do not produce seeds.
5. Phanerogams have well developed reproductive organs that produce seeds.
6. On the basis of the presence of naked seeds and seeds enclosed in fruits, phanerogams are further sub-divided gymnosperms and angiosperms respectively.
7. Gymnosperms are the seed bearing non-flowering plants, whereas angiosperms are flowering plants in which the seeds are enclosed within the fruit.

Q.62 How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?

Ans

Criteria for deciding divisions in plants are:

1. Differentiated or undifferentiated plant body
2. Presence/ absence of vascular tissues
3. With or without seeds
4. Naked seeds/seeds inside fruits

Criteria for deciding sub-groups in animals are:

Kingdom Animalia is divided into two major groups on the basis of the presence or absence of a notochord.

Non-chordates do not possess a notochord, while all members of a phylum chrodates possess a notochord.

Non-chordate is further divided into subgroups on the basis of the following features.

1. Presence or absence of tissues
2. Type of body symmetry (Radial, Bilateral)
3. Types of coelom (Acoelom, Pseudocoelom, eucoelom)
4. Types of true coelom (Enterocoelom and Schizocoelom)
5. Presence of number of germ layers during embryonic development (Diploblastic and Triploblastic)

On the basis of the above features, non-chordates are divided into the following sub-grpups.

Porifera, Coelentrata, Platyhelminthes, Nmatodes, Annilides, Molluscs, Arthropoda and Echinodermata

All members of the phylum chordate possess a non-chordate. However, some animals such as Balanoglossus, Amphioxus, Herdmanina, etc have a notochord which is either absent or does not run the entire length of the animal’s body. Therefore, these animals are kept in a separate subphylum called Protochordata.

The rest of the chordates are included in the subphylum vertebrata. The members of the sub-phylum vertebrata are advanced chordates. They are further categorised into five classes, Pisces, Amphibia, Reptilia, Aves and Mammalia.

Q.63 Explain how animals in vertebrata are classified into further Subgroups?

Ans

All the vertebrates are bilaterally symmetrical, triploblastic, coelomic, and segmented with complex differentiation of body tissues. Following features are common to all chordates:

1. notochord
2. dorsal nerve cord
3. triploblastic
4. paired gill-pouches
5. coelomate

Vertebrates are grouped into following five classes:

Q.64 Explain any one method of crop production which ensure high yield.

Ans

Crop rotation is one of the methods of crop production that ensures high yield. In this method, two or more varieties of crops are grown on the same land in sequential seasons. Every crop has some specific requirement of nutrients from the soil. If the same crop is grown in the subsequent seasons in a particular field, the field will be deficient in those nutrients that are specifically required by the growing crop. To avoid this, crops having different nutrient requirements are grown in subsequent seasons. For example, legumes which have nitrogen fixing bacteria in their root nodules supply the soil with nitrogen. Therefore, these legumes are rotated with nitrogen requiring cereals such as wheat and maize. This method reduces the need of fertilisers, thereby increasing the overall yield of crops.

Q.65 Why are manure and fertilisers used in fields?

Ans

Manure and fertilisers are used in the field to enrich the soil with the required nutrients. Manure helps in enriching the soil with organic matter and nutrients whereas fertilisers add chemical substances that are rich in nitrogen, phosphorus and potassium. Adding fertilisers and manures improve the fertility and the structure of the soil and ensures a healthy growth and development in plants. It is advised to use a balanced combination of manure and fertilisers in the soil to get a good yield.

Q.66 What are the advantages of intercropping and crop rotation?

Ans

Intercropping and crop rotation both play an important role in increasing the yield of crops.

1. Prevents the pests and diseases to spread throughout the field,
2. Maintains soil fertility,
3. Reduces the use of fertilisers and
4. Controls the growth of weeds

1. Prevents soil depletion
2. Increases soil fertility
3. Reduces soil erosion,
4. Reduces the use of fertilisers, and
5. Controls the growth of weeds

Q.67 What is genetic manipulation? How is it useful in agricultural practices?

Ans

Genetic manipulation is a process where a foreign gene from one cell (for a particular character) is introduced into the chromosome of another cell. When the gene for a particular character is introduced in a plant cell, a transgenic plant is produced. These transgenic plants exhibit characters directed by the newly introduced genes.

For example, let us assume there is a wild plant that produces small flowers. If the gene responsible for larger flower size is introduced in this wild plant, this plant would become transgenic, and would produce larger flowers. Similarly genes for higher yield, disease resistance, etc can be introduced in any desired plant. Gene manipulations play following important roles in agricultural practices.

1. Improves crop variety
2. Ensures food security and insect resistant crops
3. Improves the quality and yield of crops.

Q.68 How do storage grain losses occur?

Ans

Abiotic and biotic factors play a significant role in storage losses.

Abiotic factors such as moisture and temperature of the storage area and moisture content of the stored grain Biotic factors like insects, rodents, fungi, mites and bacteria. Both the factors cause degradation in quality, loss in weight and poor germinability of grains.

Q.69 How do good animal husbandry practices benefit farmers?

Ans

Good animal husbandry practices benefit farmers in the following ways:

1. The economic value of the animals increases when proper shelter and protection from pests and diseases is given.
2. Yield of animals produce such as meat, egg, milk, fur, etc increases when good animal husbandry practices are taken into consideration.
3. Improved breeds of the animal are obtained by good animal husbandry practices.

Q.70 What are the benefits of cattle farming?

Ans

Cattle farming is one of the methods of animal husbandry that is most beneficial to the farmers. Benefits of cattle farming are;

1. Quality and quantity of milk can be obtained
2. New varieties that are resistant to diseases and are very robust to be used for agricultural work can be produced by crossing two varieties with the desired traits

Q.71 For increasing production what is common in poultry, fisheries and bee keeping?

Ans

For increasing production common activities in poultry, fisheries and bee keeping that should be followed are;

1. Proper and regular cleaning of farms
2. Providing good quality of nutritious food
3. Proper care and protection from diseases and enemies (predators)

Q.72 How do you differentiate between capture fishing, mariculture and aquaculture?

Ans

 Capture fishing Mariculture Aquaculture In this process, fish are obtained from natural water bodies like, rivers, lakes, ponds etc. In this process, fish are marine fish are cultured in open seas. This culture involves the production of aquatic animals of high economic value like prawns, lobsters, fishes, crabs etc. In this process, fish are first located and then caught by using fishing net. In this process, echo-sounders and satellites are used to locate fish. After that fish are caught by using various fishing tools. In this process, animals are first located and then caught by using fishing net.

Q.73 Why is atmosphere essential for life?

Ans

Atmosphere plays vital role in the sustenance of life on earth. It is essential for life as:

1. It helps in maintaining the temperature of earth by acting as a blanket of gases.
2. It provides all the gases to earth that are necessary for the life on earth.
3. It contains a protective layers of gases i.e., ozone that prevents UV rays to reach earth.
4. Winds are generated and rain is regulated as a result of changes that take place in the atmosphere due to heating, and movement of air and water vapour present in it.
5. It also plays a vital role in water cycle.

Q.74 Why is water essential for life?

Ans

Water is essential for life due to following reasons:

1. Water helps in maintaining our body temperature.
2. All the metabolic activities in our body require water.
3. In both, plants and animals, water is required for transportation of substances.

Q.75 How are living organisms dependent on the soil? Are organisms that live in water totally independent of soil as a resource?

Ans

All the organisms are dependent on soil. Some organisms are directly dependent on soil whereas some are indirectly dependent on soil. For example, plants directly depend on soil, for their growth as they absorb minerals from the soil. Soil also acts as a template where roots anchor and provide support to plants. Herbivore and carnivore are indirectly dependent on soil. Soil provides home to many insects and burrowing animals.

The organisms that live in water are totally not independent on soil as a resource. The aquatic plants require minerals that are dissolved in water for their growth. The minerals are carried to the water bodies through rain water and get dissolved in them. Like land animals, aquatic animals are indirectly dependent on soil. Hence, it would be difficult to imagine an aquatic life without the supply of minerals from soil to water bodies.

Q.76 You have seen weather reports on television and in newspapers. How do you think we are able to predict weather?

Ans

Weather of a region can be predicted on the basis of the temperature of the region, speed and direction of the flowing wind, air pressure, rainfall, relative humidity by using various tools and instruments. Reports of all these elements are collected and analyzed by the meteorological department before making a weather forecast.

Q.77 We know that many human activities lead to increasing levels of pollution of the air, water bodies, and soil. Do you think that isolating these activities to specific and limited areas would help in reducing pollution?

Ans

Isolating the human activities to specific and limited areas would definitely help in reducing pollution of the air, water bodies, and soil. For example, a farmer is using manure and biological control agents for pest control to avoid soil pollution. On the other hand, a farmer is using fertiliser and insecticides in high amounts in his fields that are adjacent to the fields where manures and biological control agents are utilised. The fertilisers and insecticides would reach the adjacent fields through water channels and rain water and would cause soil pollution in the adjacent filed. But the level of pollution would be low as compared to the situation when both the farmers were using fertilisers and insecticides in their fields. Pollution in one area would easily spread in the adjacent area but to a lesser extent. Hence we can say that by limiting out activities to a specific are, we can reduce pollution level.

Q.78 Write a note on how forests influence the quality of air, soil and water resources.

Ans

Forests influence the quality of air, soil and water resources in the following ways:

1. Forests help in maintaining a balance between the level of oxygen and carbon dioxide in the air.
2. By controlling the speed of wind and water, forests prevent soil erosion.
3. Forests help in replenishing water resources by maintaining water cycle through transpiration.

Q.79 Define the term “tissue”.

Ans

Tissue is defined as a group of similar cells that are arranged and designed to perform a specific function

Q.80 How many types of elements together make up the xylem tissue? Name them.

Ans

Xylem tissue is a type of complex tissue. There are four types of elements that together make up the xylem tissue. The different types of elements are:

1. Trachieds: tubular structure
2. Vessels: tubular structures and along with trachieds they facilitate vertical conduction of water and minerals
3. Xylem parenchyma: stores food and facilitate sideways conduction of water.
4. Xylem fibres: supportive in function

Q.81 How are simple tissues different from complex tissues in plants?

Ans

 Character Simple tissues Complex Tissues Composition of cells Composed of only one type of cells Composed of different types of cells Origin and structure of cells Common origin and structure Different origin and structure Function Storage of food and supportive in nature Transport water, minerals, sugar and metabolites within plant body Examples Parenchyma, Collenchyma, Sclerenchyma Xylem and phloem

Q.82 Differentiate between parenchyma, collenchymas and sclerenchyma on the basis of their cell wall.

Ans

 Parenchyma Collenchyma Sclerenchyma Cell walls are thin. Cell wall is irregularly thickened at the corners. Cell wall is uniformly thickened. Cell wall is mainly composed of cellulose. Pectin and hemicelluloses are the main constituents of the cell wall in this tissue. The thick cell wall scelerenchyma are impregnated with lignin.

Q.83 What are the functions of the stomata?

Ans

Stomata perform the following two functions;

1. They allow the exchange of gases (oxygen and carbon dioxide) between plants and atmosphere.
2. They facilitate the release of water vapours from the leaves and hence regulate the process of transpiration.

Q.84 Diagrammatically show the difference between the three types of muscle fibres.

Ans

Q.85 What is the specific function of the cardiac muscle?

Ans

The specific function of cardiac muscle is to contract and relax rhythmically through the life and to facilitate the transportation of gases, nutrients and metabolites to various body parts through blood circulation.

Q.86 Differentiate between striated, unstriated and cardiac muscles on the basis of their structure and site/location in the body.

Ans

 Difference between striated, unstriated and cardiac muscles on the basis of their structure Character Striated muscles Unstriated muscles Cardiac muscles Size and shape Long and cylindrical Long, narrow and spindle shaped Short and cylindrical Branching Absent Absent present Number nucleus present Multinucleate Uninucleate Uninucleate Position of nucleus Peripheral Central Central Striations Present in the form of dark and light bands Absent Present in the form of faint bands Edges of the cell Blunt ends Tapered end Flat and wavy Difference between striated, unstriated and cardiac muscles on the basis of their location Striated muscles Unstriated muscles Cardiac muscles Location Tongue, hands, legs, etc. Iris of the eye, ureters, bronchi, etc. Heart

Q.87 Draw a lebelled diagram of a neuron.

Ans

Labelled diagram of a neuron:

Q.88 Name the following

(a) Tissue that forms the inner lining of our mouth.

(b) Tissue that connects muscle to bone in humans.

(c) Tissue that transports food in plants.

(d) Tissue that stores fat in our body.

(e) Connective tissue with a fluid matrix.

(f) Tissue present in the brain.

Ans

(a) Squamous epithelial tissue

(b) Tendon

(c) Phloem

(e) Blood

(f) Nervous tissue

Q.89 Identify the type of tissue in the following skin, bark of tree, bone, lining of kidney tubule, vascular bundle.

Ans

 Name of the organ Type of tissue present Skin Stratified squamous epithelial tissue Bark of tree Secondary meristem (cork) Bone Connective tissue Lining of kidney tubule Cuboidal epithelial tissue Vascular bundle Complex tissue: xylem and phloem

Q.90 Name the region in which parenchyma tissue is present.

Ans

Parenchyma tissue is present in the following plant parts:

Stem, leaves, roots, flowers and vascular tissues.

Q.91 What is the role of epidermis in plants?

Ans

Epidermis performs the following functions in plants:

1. It covers the entire plant and protects all the parts of the plant.
2. It secrets a waxy, water resistant layer on outer surface of the aerial parts of the plants and prevents water loss.
3. It protects plants against mechanical injury and invasion of parasitic fungi.
4. It possesses stomata that facilitate gaseous exchange between plant and the atmosphere and allows transpiration to take place.
5. In roots, it aids in water absorption through the root hair present in root cells.

Q.92 How does the cork act as a protective tissue?

Ans

Cork is a several layered thick outermost protective tissue present on mature plants. The cells of cork are compactly arranged without the intercellular spaces and are dead. A special chemical called suberin is present in the cell walls of the cork cells that make them impermeable to gases and water. Due to this, it protects plants against water loss. It also protects plants against mechanical injury and temperature extremes. It also protects plants against fungal diseases by not allowing the parasitic fungi to grow on the plants. .

Q.93 Complete the table:

Ans

Q.94 How many times did you fall ill in the last one year? What were the illnesses?

1. Think of one change you could make in your habits in order to avoid any of/most of the above illnesses.
2. Think of one change you would wish for in your surroundings in order to avoid any of /most of the above illnesses.

</ol>
Ans

Frequency of falling ill varies from person to person. It depends on personal factors such as the immunity of a person and environmental factors such as cleanliness in the surrounding in which he/she is living. Some of the common diseases from which generally children get affected are cold, cough, fever, stomach infection, dengue, malaria, typhoid, etc.

1. To avoid most of the above mentioned illnesses a person should have healthy eating habits. He/she should eat balanced diet that would help in his/her growth and development. A person should include lots of fruits in his diet as fruits boost up our immune system and provide minerals and vitamins that protect us from many diseases.
2. To avoid the spread of diseases like dengue, malaria, etc, one has to keep his surroundings clean so that the germs causing these diseases would not get a chance to grow and multiply. By doing so one can stop the diseases like dengue, malaria, typhoid, etc.

Q.95 A doctor/nurse/health-worker is exposed to more sick people than others in the community. Find out how she/he avoids getting sick herself/himself?

Ans

A doctor/nurse/health-workers, generally, take the following precautions:

1. They wear a mask while treating the infected person.
2. They keep themselves covered while moving around an infected area.
3. They eat healthy and nutritious food and drink lots of clean water.
4. They give importance to personal and community hygiene.

Q.96 Conduct a survey in your neighborhood to find out what the three most common diseases are. Suggest three steps that could be taken by your local authorities to bring down the incidence of these diseases.

Ans

The three most common diseases are:

1. Malaria
2. Jaundice
3. Typhoid

Steps that should be taken by the local authorities to bring down the incidence of these diseases:

1. Avoid collection of stagnant water in the surroundings to prevent mosquitoes from breeding
2. Ensuring the supply of clean drinking water
3. Managing the proper disposal of sewage and waste to check the breeding ground of germs or causative agents of various diseases

Q.97 A baby is not able to tell her/his caretakers that she/he is sick. What would help us to find out

1. that the baby is sick?
2. what is the sickness?

</ol>
Ans

1. Following behavioural changes in the baby would help us to find out that the baby is sick:
1. Constant crying of the baby,
2. Avoiding his/her meal
3. Mood swings in the baby
1. The sickness in a baby can be identified by the following symptoms or indications:
1. vomiting
2. fever
3. loose motions
4. dullness in activity

Q.98 Under which of the following conditions is a person most likely to fall sick?

1. when she is recovering from malaria.
2. when she has recovered from malaria and is taking care of someone suffering from chicken pox.
3. when she is on a four day fast after recovering from malaria and is taking care of someone suffering from chicken pox.

Why?

Ans

c. A person is most likely to fall sick when she is on a four day fast after recovering from malaria and is taking care of someone suffering from chicken pox.

Fasting just after the recovery from malaria would definitely weaken the immune system of the person and would not even help her to fight with any infection if she comes in contact with it. Under these conditions, if she takes care of someone who is suffering from chicken pox, she is more likely to get infected from chicken pox and will fall sick.

Q.99 Under which of the following conditions are you most likely to fall sick?

1. when you are taking examinations.
2. when you have travelled by bus and train for two days.
3. when your friend is suffering from measles.

Why?

Ans

c. You are most likely to fall sick when you will be in the company of your friend who is suffering from measles.

Measles is a highly contagious disease that spreads through droplets from the nose or mouth of the infected person. Therefore, if your friend is suffering from measles, you should stay away from him/her to protect yourself from the disease.

Q.100 What is sound and how is it produced?

Ans

Vibration in the particles causes sound. When a body vibrates, it forces the nearby particles of the medium to vibrate. Hence, a disturbance is created in the medium, which travels in the form of waves. This disturbance is called sound.

Q.101 Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.

Ans

When a body vibrates, it creates a region of high pressure and low pressure in its surrounding. These regions of high pressure and low pressure are called compressions and rarefactions. As the body continues to move forward and backward, it produces a series of compressions and rarefactions.

Q.102 Cite an experiment to show that sound needs a material medium for its propagation.

Ans

For this, take an electric bell and hang it inside an empty bell-jar fitted with a vacuum pump. Now, start removing the air from the bell-jar with the help of a vacuum pump. It can be observed that the sound of the ringing bell decreases. On further pumping off air, vacuum will be created inside the jar. At this moment, no sound can be heard from the ringing bell although one can see that the prong of the bell is still vibrating. This experiment shows that sound cannot travel through vacuum. Hence, sound requires a material medium for its propagation.

Q.103 Why is sound wave called a longitudinal wave?

Ans

The vibration of the medium that moves along the direction of the wave is known as longitudinal wave. In the case of sound wave, the particles of the medium vibrate in the direction parallel to the direction of the propagation of disturbance. Therefore, a sound wave is called a longitudinal wave.

Q.104 Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

Ans

Quality is the characteristic of the sound which helps us to recognise a particular person. Sound produced by two persons may have the same pitch and loudness, but the quality of the two sounds will be different.

Q.105 Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

Ans

The speed of sound is very less as compared to the speed of light. Hence, sound of thunder takes more time to reach the Earth as compared to light. Therefore, a flash is seen before than the thunder is heard.

Q.106 A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m s–1.

Ans

1. $Given, v 1 = 20 Hz, v 2 = 20 kHz = 20000 Hz Speed of sound in air, v = 344 ms -1 On using the relation, v = v× λ or, λ = v v Let, the corresponding wavelengths for frequencies 20 Hz and 20 kHz are λ 1 and λ 2 . ∴ λ 1 = v v 1 = 344 20 = 17.2 m λ 2 = v v 2 = 344 20000 = 0.0172 m MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@57CE@$

Q.107 Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.

Ans

1. $\begin{array}{l}\text{Consider the length of the aluminium rod = l}\text{.}\\ \text{Speed of sound wave in aluminium at 25 °C,}\\ {v}_{1}=6420{\text{ms}}^{\text{-1}}\\ Hence,\text{time taken by the sound wave to reach}\\ {\text{the other end, t}}_{1}=\frac{d}{{v}_{1}}=\frac{d}{6420}\text{}\dots \text{(i)}\\ \text{Speed of sound in air at 25 °C,}\\ {v}_{2}={\text{346 ms}}^{\text{-1}}\\ Hence,\text{the time taken by sound wave to reach the}\\ {\text{other end, t}}_{2}=\frac{d}{{v}_{2}}=\frac{d}{\text{346}}\text{}\dots \text{(ii)}\\ \text{From (i) and (ii),}\\ \text{}\frac{{\text{t}}_{\text{2}}}{{\text{t}}_{\text{1}}}=\frac{\frac{d}{346}}{\frac{d}{6420}}=\frac{6420}{346}=18.55\text{}\text{.}\end{array}$

Q.108 The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

Ans

1. $\begin{array}{l}\text{Given,}\text{\hspace{0.17em}}\text{Frequency}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{sound}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{100}\text{\hspace{0.17em}}\text{Hz}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Total}\text{\hspace{0.17em}}\text{time}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{1}\text{\hspace{0.17em}}\text{min}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{60}\text{\hspace{0.17em}}\text{s}\\ \text{On using the relation,}\\ \text{Frequency =}\frac{\text{Number}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{oscillations}}{\text{Total}\text{\hspace{0.17em}}\text{time}}\\ \text{Here,}\text{\hspace{0.17em}}\text{Number}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{oscillations}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{Frequency}\text{\hspace{0.17em}}\text{×}\text{\hspace{0.17em}}\text{Total}\text{\hspace{0.17em}}\text{time}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{100 Hz}\text{\hspace{0.17em}}\text{×}\text{\hspace{0.17em}}\text{60}\text{s\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{6000}\text{\hspace{0.17em}}\\ \text{Thus,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{source}\text{\hspace{0.17em}}\text{vibrate}\text{\hspace{0.17em}}\text{6000}\text{\hspace{0.17em}}\text{times}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{minute}\text{.}\end{array}$

Q.109 Does sound follow the same law of reflection as light does? Explain

Ans

Yes, the incident sound wave and the reflected sound wave make the same angle with the normal to the surface at the point of incidence. Moreover, the incident sound wave, the reflected sound wave, and the normal to the point of incidence all lie in the same plane.

Q.110 When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?

Ans

An echo can be heard when the time interval between the original sound and the reflected sound is at least 0.1 s. The speed of sound in a medium increases with an increase in temperature. Therefore, echo cannot be heard on a summer day as the time interval between the original sound and the reflected sound decreases.

Q.111 Give two practical applications of reflection of sound waves.

Ans

(i) Sound board: It is used to send the sound towards audience in a big hall or auditorium. This works on the basis of laws of reflection of sound waves.

(ii) Stethoscope: It is also based on reflection of sound. In a stethoscope, the sound of a patient’s heartbeat reaches to the doctor’s ear by multiple reflection of sound.

Q.112 A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m s–2 and speed of sound = 340 m s–1.

Ans

1. $\begin{array}{l}{\text{Given, h = 500 m, g = 10 ms}}^{\text{-2}}\text{,}\\ {\text{Speed of sound, v = 340 ms}}^{\text{-1}}\\ \text{The intial velocity (u) of the stone when it is dropped}\\ \text{is zero, i}\text{.e}\text{., u = 0}\\ \text{Let, it takes (t) time to reach on the surface of water}\\ \text{On using the relation, s = ut +}\frac{\text{1}}{\text{2}}{\text{gt}}^{\text{2}}\text{}\\ \therefore \text{500 = 0+ 0}{\text{.5 ×10×t}}^{\text{2}}\\ {\text{or, 100 = t}}^{\text{2}}\\ \text{or, t = 10 s}\\ \text{Now, let time taken by the sound to reach at the top the}\\ \text{tower be t’}\\ \text{then, h =v×t’}\\ \text{or, 500 = 340×t’}\\ \text{or, t’ =1}\text{.47 s}\\ \therefore \text{Total time = t+t’ = 10+1}\text{.47 = 11}\text{.47 s}\text{.}\\ \text{Hence, after 11}\text{.47 s later, the splash will be heard at}\\ \text{the top of the tower}\text{.}\end{array}$

Q.113 A sound wave travels at a speed of 339 m s–1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Ans

1. $\begin{array}{l}{\text{Given, Speed of sound, v = 339 ms}}^{\text{-1}}\\ \text{Wave length, λ = 1}\text{.5 cm = 0}\text{.015 m}\\ \text{On using the relation,}\\ \text{v = λ × v}\\ \therefore \text{v =}\frac{\text{v}}{\text{λ}}\text{=}\frac{\text{339}}{\text{0}\text{.015}}\text{= 22600 Hz}\\ \text{As the frequency range of audible sound for humans lies between 20 Hz to 20000 Hz}\text{.}\\ \text{Since, the frequency of the given sound is more than 20000 Hz, it is not audible}\text{.}\end{array}$

Q.114 What is reverberation? How can it be reduced?

Ans

The reverberation is the persistence of sound due to repeated reflection. When the source produces sound, it starts moving in all the directions. It is partly reflected back from the wall. This reflected sound reaches the other wall and again gets reflected partly. Thus, sound can be heard even after the source has stopped producing the sound.

To minimise reverberation, sound should be absorbed as it reaches the walls and the ceiling of a room. Sound absorbing materials like fibre board, heavy curtains, and cushioned seats can be used to reduce reverberation.

Q.115 What is loudness of sound? What factors does it depend on?

Ans

Loudness depends on the amplitude of the sound. The louder a sound, the more energy it has. Loudness is proportional to the square of the amplitude of vibrations.

Q.116 Explain how bats use ultrasound to catch a prey.

Ans

Bats generate high-pitched ultrasonic shrill. These high-pitched shrill are reflected by objects such as preys and comes back to the bat’s ear. Thus, a bat can easily determine the distance of a prey to catch it.

Q.117 How is ultrasound used for cleaning?

Ans

To clean an object when it is put in a cleaning solution, the ultrasonic sound waves are passed through the solution. The high frequency of these ultrasound waves detaches the dirt from the objects.

Q.118 Explain the working and application of a sonar.

Ans

It is a device that uses ultrasound propagation to navigate, communicate or detect underwater objects such as submarine, a sunken ship, an iceberg, etc. A SONAR apparatus consists of two parts:

1. A transmitter (for emitting ultrasonic waves).
2. A receiver (for detecting the reflected ultrasonic waves).

Both these parts are installed in a ship or a boat. The transmitter sends ultrasonic waves towards the ocean floor.

These waves when reflected back by an object or the ocean floor in form of ‘echo’ are detected by a detector.

Depth of an object can be calculated using time interval between generation of wave and reception of its echo and the speed of sound in water. This method is known as sonar.

Q.119 A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.

Ans

1. $\begin{array}{l}\text{Given, Time taken to hear to echo, t = 5 s}\\ \text{Distance of the object from the submarine, s = 3625 m}\\ \therefore \text{Total distance travelled = 2 s}\\ \therefore \text{Velocity, v =}\frac{2\text{s}}{t}\text{=}\frac{2\text{}×\text{3625 m}}{5}\\ {\text{= 1450 ms}}^{-1}\end{array}$

Q.120 Explain how defects in a metal block can be detected using ultrasound.

Ans

The ultrasounds cannot pass through a defective metal block but they are reflected back. This fact is used to detect defects in metal blocks. Ultrasound is passed through one end of a metal block and detectors are placed on the other end. Thus, defects in metal blocks can be detected by using ultrasound.

Q.121 Explain how the human ear works.

Ans

Pinna collects different sounds produced in our surroundings and sends these sounds to the ear drum through ear canal. At the end of ear canal, there is a thin membrane called the eardrum or tympanic membrane. When the compression of sound waves strikes the eardrum, it is pushed inward. When the rarefaction of sound waves strikes the ear drum, it moves outward. Thus, eardrum vibrates. Middle ear consists of three bones – hammer, anvil and stirrup. These bones amplify the vibrations from the eardrum and transmit these vibrations to the inner ear. Inner ear consists of cochlea that converts vibrations or pressure variations into electrical signals. These electrical signals sent to brain via auditory nerve. Brain interprets them as sound.

Q.122 Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.

• Suma is swimming in a pond.

• A donkey is carrying a load on its back.

• A wind-mill is lifting water from a well.

• A green plant is carrying out photosynthesis.

• An engine is pulling a train.

• Food grains are getting dried in the sun.

• A sailboat is moving due to wind energy.

Ans

• While swimming, Suma exerts a force on the water in backward direction. Hence, she swims in the forward direction caused by the forward reaction of water. Thus, the force causes a displacement. Hence, Suma does the work while swimming.

• When the donkey carries a load, it applies a force in the upward direction while displacement of the load is in the forward direction. In this case, displacement is perpendicular to the force. Hence, the work done is zero.

• A wind mill pulls water against the gravitational force. Thus, work is done by the wind mill to lift the water from the well.

• In photosynthesis process, there is no displacement of the leaves of the plant. Hence, the work done is zero.

• A force is applied by engine on the train. Therefore, the train moves in the direction of applied force. Here, displacement is in the direction of force applied. Hence, work is done by the engine on the train.

• Food grains do not move in the presence of solar energy. Hence, the work done is zero during the process of food grains getting dried in presence of the Sun.

• Wind exerts a force on the sailboat to push it in the forward direction. Thus, there is a displacement in the boat in the direction of force. Hence, work is done by wind on the boat.

Q.123 An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?

Ans

Work done by the force of gravity on an object depends only on vertical displacement. Vertical displacement can be calculated by the difference in the initial and final heights of the object. Here, vertical displacement is zero. Hence, work done by gravity W = mgh

Here, h = 0

Therefore, W = mg × 0 = 0

Q.124 A battery lights a bulb. Describe the energy changes involved in the process.

Ans

The chemical energy of battery is changed into electrical energy when a bulb is connected between the terminals of a battery. Later on, this electrical energy changes into light and heat energy. Hence, the transformation of energy in the given situation can be shown as:

1. $\text{Chemical}\text{\hspace{0.17em}}\text{energy}\text{\hspace{0.17em}}\to \text{\hspace{0.17em}}\text{Electrical}\text{\hspace{0.17em}}\text{Energy}\text{\hspace{0.17em}}\to \text{\hspace{0.17em}}\text{Light}\text{\hspace{0.17em}}\text{Energy}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\text{Heat}\text{\hspace{0.17em}}\text{Energy}$

Q.125 Certain force acting on 20 kg mass changes its velocity from 5 ms−1 to 2 ms−1. Calculate the work done by the force.

Ans

1. $\begin{array}{l}\text{Kinetic energy}\mathrm{E}\text{=}\frac{1}{2}{\mathrm{mv}}^{2}\\ \mathrm{Where},\text{mass, m = 20 kg, v = velocity}\\ {\text{(i) Kinetic energy E}}_{\text{1}}{\text{, when velocity is 5 ms}}^{\text{-1}}\\ {\text{K}}_{\text{1}}\text{=}\frac{1}{2}\left(20\mathrm{kg}\right){\left(5{\mathrm{ms}}^{-1}\right)}^{2}=250\text{J}\\ {\text{(ii) Kinetic energy (K}}_{2}{\text{) when velocity is 2 ms}}^{\text{-1}}\\ {\text{K}}_{2}=\frac{1}{2}\text{\hspace{0.17em}}\left(20\mathrm{kg}\text{\hspace{0.17em}}\right)\text{\hspace{0.17em}}{\left(2\text{\hspace{0.17em}}{\mathrm{ms}}^{-1}\right)}^{2}=\text{40 J}\\ \text{Work done is equal to the change in the kinetic energy}\\ {\text{Hence, W = K}}_{2}-{\text{K}}_{\text{1}}\text{}=40-250=\text{-210 J}\end{array}$

Q.126 A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.

Ans

Work done by gravity depends only on the vertical displacement of the body. The expression for work done by gravity is given by,

W = mgh

Here, vertical displacement (h) is zero.

Hence, W = mg × 0 = 0

Thus, work done by gravity on the body is zero.

Q.127 The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Ans

No. When the body falls freely from a height, then its potential energy converts into kinetic energy with time. The decrease in the potential energy is equal to the increase in the kinetic energy of the body. Thus, total mechanical energy of the body remains conserved.

Q.128 What are the various energy transformations that occur when you are riding a bicycle?

Ans

In the case of bicycle riding, the muscular energy of the rider changes into heat energy and kinetic energy of the bicycle. Heat energy gives heat to the rider’s body while kinetic energy provides a velocity to the bicycle. In this whole process, the total energy remains conserved.

Q.129 Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?

Ans

On pushing a huge rock, no muscular energy is transferred to the stationary rock. Here, muscular energy is transferred into heat energy which makes the body feel warm. Hence, there is no loss of energy.

Q.130 A certain household has consumed 250 units of energy during a month. How much energy is this in joules?

Ans

1. $Here, 1 unit = 1 kWh Also, 1 kWh=3.6×1 0 6 J ∴ 250 units of energy =250 ×3.6×1 0 6 J= 9×1 0 8 J MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqipz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@8C0D@$

Q.131 An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.

Ans

1. $Given, Height of object, h = 5 m mass of object, m = 40 kg acceleration due to gravity, g = 9 .8 ms -2 Gravitational potential energy is given by, W = mgh or, = 40 kg × 9 .8 ms -2 × 5 m= 1960 J Now, potential energy at half-way = 1960 J 2 = 980J. At this point, the object has an equal amount of potential energy and kinetic energy. Thus, half – way down, Kinetic energy = 980J.$

Q.132 What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

Ans

There are two conditions when the work is said to be done.

(i) A force acts on the body.

(ii) There is a displacement of the body due applied force in or opposite to the direction of force.

If the direction of force is perpendicular to displacement, then the work done is zero.

When a satellite revolves around the Earth, then the direction of force of gravity on the satellite is perpendicular to its displacement. Thus, the work done on the satellite by the Earth is zero.

Q.133 Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.

Ans

Yes. When an object is moving with constant velocity, the net force acting on it is zero. However, there is a displacement along the motion of the object. Therefore, there can be a displacement of an object without any force acting upon it.

Q.134 A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.

Ans

When the person holds a bundle of hay over his head, then bundle of hay is not displaced. However, force of gravity is still working on the bundle but the person is not applying any force on it. Thus, work done by the person on the bundle is zero.

Q.135 An electric heater is rated 1500 W. How much energy does it use in 10 hours?

Ans

1. $Given, Power, P = 1500 W = 1.5 kW Time, t = 10 h Now, Work done = energy consumed by the heater and Energy consumed = Power × time Hence, = 1.5 kW × 10 h = 15 kWh Thus, energy consumed by the heater in 10 h is 15 kWh. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaacaqGhbGaaeyAaiaabAhacaqGLbGaaeOBaiaacYcacaqGGaGaaeiuaiaab+gacaqG3bGaaeyzaiaabkhacaqGSaGaaeiiaiaabcfacaqGGaGaaeypaiaabccacaqGXaGaaeynaiaabcdacaqGWaGaaeiiaiaabEfacaqGGaGaaeypaiaabccacaqGXaGaaeOlaiaabwdacaqGGaGaae4AaiaabEfaaeaacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGubGaaeyAaiaab2gacaqGLbGaaeilaiaabccacaqG0bGaaeiiaiaab2dacaqGGaGaaeymaiaabcdacaqGGaGaaeiAaaqaaiaab6eacaqGVbGaae4DaiaabYcacaqGGaGaae4vaiaab+gacaqGYbGaae4AaiaabccacaqGKbGaae4Baiaab6gacaqGLbGaaeiiaiaab2dacaqGGaGaaeyzaiaab6gacaqGLbGaaeOCaiaabEgacaqG5bGaaeiiaiaabogacaqGVbGaaeOBaiaabohacaqG1bGaaeyBaiaabwgacaqGKbGaaeiiaiaabkgacaqG5bGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGObGaaeyzaiaabggacaqG0bGaaeyzaiaabkhaaeaacaqGHbGaaeOBaiaabsgacaqGGaGaaeyraiaab6gacaqGLbGaaeOCaiaabEgacaqG5bGaaeiiaiaabogacaqGVbGaaeOBaiaabohacaqG1bGaaeyBaiaabwgacaqGKbGaaeiiaiaabccacaqG9aGaaeiiaiaabcfacaqGVbGaae4DaiaabwgacaqGYbGaaeiiaiabgEna0kaabccacaqG0bGaaeyAaiaab2gacaqGLbaabaGaaeisaiaabwgacaqGUbGaae4yaiaabwgacaqGSaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeypaiaabccacaqGXaGaaeOlaiaabwdacaqGGaGaae4AaiaabEfacaqGGaGaey41aqRaaeiiaiaabgdacaqGWaGaaeiiaiaabIgaaeaacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaab2dacaqGGaGaaeymaiaabwdacaqGGaGaae4AaiaabEfacaqGObGaaeiiaaqaaiaabsfacaqGObGaaeyDaiaabohacaqGSaGaaeiiaiaabwgacaqGUbGaaeyzaiaabkhacaqGNbGaaeyEaiaabccacaqGJbGaae4Baiaab6gacaqGZbGaaeyDaiaab2gacaqGLbGaaeizaiaabccacaqGIbGaaeyEaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiAaiaabwgacaqGHbGaaeiDaiaabwgacaqGYbGaaeiiaiaabMgacaqGUbGaaeiiaiaabgdacaqGWaGaaeiiaiaabIgacaqGGaGaaeyAaiaabohacaqGGaGaaeymaiaabwdacaqGGaGaae4AaiaabEfacaqGObGaaeOlaaaaaa@1A74@$

Q.136 Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

Ans

According to law of conservation of energy, energy can be neither created nor destroyed. It can only be converted from one form to another.

When a pendulum oscillates from its mean position O to either of its extreme positions P or Q, it rises through a height h above the mean level P. At this point, the kinetic energy of the bob changes completely into potential energy. The kinetic energy becomes zero, and the bob possesses only potential energy. As it goes towards point O, its potential energy decreases progressively. Accordingly, the kinetic energy increases. As the bob reaches point O, its potential energy becomes zero and the bob possesses only kinetic energy. This process is repeated as long as the pendulum oscillates.

After some time, bob comes to rest because air resists its motion. The pendulum loses its kinetic energy to overcome this friction and stops after some time.

Here, the law of conservation of energy is not violated because the energy lost by the pendulum to overcome friction is gained by its surroundings. Thus, the total energy of the pendulum and the surrounding system remain conserved.

Q.137 An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?

Ans

1. $\begin{array}{l}\text{When an object of mass}\left(\text{m}\right)\text{is moving with a velocity}\left(\text{v}\right)\text{then, its kinetic energy}\left(\text{K}\right)\text{is given as,}\\ \text{K =}\frac{1}{2}{\text{mv}}^{\text{2}}\\ \text{In order to bring the object to rest, an amount of work is need to be done that which is equal to the kinetic energy}\\ \text{of the object}\text{.}\\ \text{Hence, work done}\left(\text{W}\right)\text{on the object to bring it to rest}\\ \text{would be, W =}\frac{1}{2}{\text{mv}}^{\text{2}}.\end{array}$

Q.138 Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

Ans

1. $\begin{array}{l}\text{Given, Mass, m = 1500 kg}\\ {\text{Velocity, v = 60 kmh}}^{\text{-1}}\text{= 60 ×}\frac{5}{18}{\text{ms}}^{\text{-1}}\\ \text{The work done to stop the car would be equal to the kinetic energy acquired by it}\text{.}\\ {\text{Hence, kinetic energy, E}}_{\text{k}}\text{=}\frac{1}{2}{\text{mv}}^{\text{2}}\\ \therefore {\text{E}}_{\text{k}}\text{=}\frac{1}{2}×1500×{\left(60×\frac{5}{18}\right)}^{2}=208333.3{\text{}}^{\text{}}\text{J}\\ \text{Hence, 20}{\text{8333.3}}^{\text{}}\text{J of work is required to stop the car}\text{.}\end{array}$

Q.139 In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

Ans

1. In first case, the direction of force acting on the block is perpendicular to the displacement. Thus, work done by force on the block will be zero.

2. In the second case, the direction of force acting on the block is in the direction of displacement. Hence, work done by force on the block will be positive.

3. In the third case, the direction of force acting on the block is opposite to the direction of displacement. Hence, work done by force on the block will be negative.

Q.140 Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?

Ans

The statement of Soni is correct. Acceleration in an object may be zero even when several forces are acting on it. This condition occurs when all the forces cancel out each other. For a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero.

Q.141 Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.

Ans

1. $\begin{array}{l}\text{Given, Power of each device, P = 500 W = 0}\text{.50 kW}\\ \text{Time, t = 10 h}\\ \text{Energy consumed = Power × Time}\\ \text{= 0}\text{.50 kW × 10 h = 5 kWh}\\ \text{Hence, the energy consumed by four equal rating}\\ \text{devices in 10 h = 4 × 5 kWh = 20 units}\text{.}\end{array}$

Q.142 A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Ans

In free fall, potential energy of an object decreases and kinetic energy increases. When the object comes in contact with ground, potential energy converts into kinetic energy. As the object hits the hard ground, all its kinetic energy gets converted into heat energy and sound energy. The object can also deform the ground depending upon the nature of the ground and the amount of kinetic energy possessed by the object.

Q.143 How does the force of gravitation between two objects change when the distance between them is reduced to half?

Ans

1. $\begin{array}{l}\text{Gravitational force}\left(\text{F}\right)\text{acting between two objects is inversely proportional to the square of the distance}\\ \left(\text{r}\right)\text{between them}\text{.}\\ \text{Mathematically,}\\ \text{F}\propto \frac{1}{{\text{r}}^{\text{2}}}\\ \text{When distance them becomes}\frac{r}{2}\text{then,}\\ \text{F}\propto \text{}\frac{1}{{\left(\frac{\text{r}}{2}\right)}^{2}}\text{}\propto \text{}\frac{4}{{r}^{2}}\\ \text{Thus, if the distance is reduced to half, the gravitational force weill become four times larger}\text{.}\end{array}$

Q.144 Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?

Ans

All the objects fall towards the ground under the acceleration due to gravity and this acceleration is constant and does not depend upon the mass of the falling object. Thus, a heavy object does not fall faster than a light object.

Q.145 What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106 m.)

Ans

1. $\begin{array}{l}\text{Given, Mass of the object, m = 1 kg}\\ {\text{Mass of the Earth, M = 6 × 10}}^{\text{24}}\text{kg}\\ \text{According to universal law of gravitation,}\\ \text{Force, F =}\frac{\text{GMm}}{{\text{r}}^{\text{2}}}\text{}...\left(\text{i}\right)\\ \text{Where, G = universal gravitational constant}\\ {\text{= 6.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\\ \text{Since, the object is on the earth surface hence,}\\ {\text{r = radius of earth = 6.4×10}}^{\text{6}}\text{m}\\ \text{On putting these values in}\left(\text{i}\right)\text{, we have}\\ {\text{= 6.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\text{×}\frac{\left({\text{6×10}}^{\text{24}}\text{kg}\right)\left(\text{1 kg}\right)}{{\left({\text{6.4 ×10}}^{\text{6}}\text{m}\right)}^{\text{2}}}\\ \text{= 9.8 N}\end{array}$

Q.146 The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

Ans

The universal law of gravitation states that two objects attract each other with equal forces, but in opposite directions. Hence, the Earth attracts the moon with an equal force with which the moon attracts the earth.

Q.147 If the moon attracts the earth, why does the earth not move towards the moon?

Ans

The Earth and the moon exert equal gravitational forces to each other. Due to heavier mass of Earth, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. Therefore, the Earth does not move towards the moon.

Q.148 What happens to the force between two objects, if

(i) the mass of one object is doubled?

(ii) the distance between the objects is doubled and tripled?

(iii) the masses of both objects are doubled?

Ans

(i) According to universal law of gravitation, force is directly proportional to the product of the masses. If the mass of one object is doubled, then the gravitational force will also get doubled.

(ii) According to universal law of gravitation, force is inversely proportional to the square of the distance between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value. Also, if the distance gets tripled, then the gravitational force becomes one-ninth of its original value.

(iii) Force is directly proportional to the product of the masses. When masses of both the objects are doubled, then the gravitational force becomes four times.

Q.149 What is the importance of universal law of gravitation?

Ans

Importance of the universal law of gravitation

(i) It binds us to the earth.

(ii) It explains the motion of moon around the earth.

(iii) It explains the motion of planets around the sun.

Q.150 What is the acceleration of free fall?

Ans

An object falling towards the earth under the effect of gravitation is said to be in free fall. Acceleration of free fall also called acceleration due to gravity is 9.8 ms−2 near the surface of earth. It is constant for all the objects.

Q.151 What do we call the gravitational force between the Earth and an object?

Ans

Gravitational force acting between the earth and an object is called the weight of the object.

Q.152 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. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.]

Ans

As the gold purchased at the equator weighs less than at the poles so the value of acceleration due to gravity (g) for an object is greater at poles than at the equator. Therefore, Amit’s friend will not agree with the weight of the gold bought.

Q.153 Why will a sheet of paper fall slower than one that is crumpled into a ball?

Ans

When the paper is crumbled in the shape of a ball, its surface area decreases. Thus, air does not exert too much resistance to its motion and it falls faster than the sheet of paper.

Q.154 Gravitational force on the surface of the moon is only

1. $\frac{\text{1}}{\text{6}}$

as strong as gravitational force on the Earth. What is the weight in Newtons of a 10 kg object on the moon and on the Earth?

Ans

1. $\begin{array}{l}\text{Let,}\text{\hspace{0.17em}}{\text{Weight of an object on the moon = W}}_{\text{m}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Weight}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}{\text{an object on the Earth = W}}_{\text{E}}\\ \text{Hence,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{W}}_{\text{m}}\text{=}\frac{\text{1}}{\text{6}}{\text{× W}}_{\text{E}}\\ \text{Also, Weight = Mass × Acceleration}\\ \text{Acceleration due to gravity, g = 9}{\text{.8 ms}}^{\text{-2}}\\ \text{Hence, weight of a 10 kg object on the Earth}\\ \text{= 10 kg × 9}{\text{.8 ms}}^{\text{-2}}\text{= 98}\text{\hspace{0.17em}}\text{N}\\ \text{And,}\text{\hspace{0.17em}}\text{weight}\text{\hspace{0.17em}}\text{of the same object on the moon}\\ \text{=}\frac{\text{1}}{\text{6}}\text{× 98}\text{\hspace{0.17em}}\text{N =16}\text{.3}\text{\hspace{0.17em}}\text{N}\end{array}$

Q.155 A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

(i) the maximum height to which it rises.

(ii) the total time it takes to return to the surface of the earth.

Ans

1. $Given, u = 49 ms -1 , Here, v = final velocity of the ball = 0 Let, s = maximum height achieved by the ball On using third equation of motion, V 2 =u 2 + 2 as Where, a = g = acceleration due to gravity It would be negative, as the ball has thrown upward On putting these values, we have, 0 = ( 49 ms -1 )×( 49 ms -2 ) 2×9 .8 ms -2 = 122.5 m Let, t = time taken by the ball to reach at height of 122.5 m. On using the first equation of motion under gravity , v = u + at On putting the values, 0 = 49 ms -1 + t × ( -9.8 ) t = 49 ms -1 9 .8 ms -2 = 5 s Here, Time of ascent = Time of descent Hence, total time taken by the ball of return = 5 s + 5 s = 10 s$

Q.156 A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Ans

1. $Given, inital velocity, u = 0 height of the stone, h = 19.6 m acceleration due to gravity, g = 9.8 ms -2 Let, final velocity of the stone, v = ? On using the third equation of motion under gravity, v 2 = u 2 + 2gh On putting the values, v 2 = 0 + 2 (9.8 ms -2 )(19.6 m) v 2 = ( 19 .6ms -1 ) 2 v = 19.6 ms -1$

Q.157 A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Ans

1. $Given, inital velocity, u = 40 ms -1 final velocity, v = 0 ms -1 gravitational acceleration for upward motion, g = -10 ms -2 Let h be the height attained by the stone Using third equation of motion under gravity, v 2 = u 2 +2gh On putting the values, 0 = ( 40 m s -1 ) 2 +2( -10 m s -2 )h ∴ h = 1600 m 2 s -2 20 m s -2 =80 m Now, total distance covered by the stone after reaching on the earth = 80 m + 80 m = 160 m. The net displacement of the stone when it reaches to the ground = 80 m – 80 m = 0.$

Q.158 Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 1024 kg and of the Sun = 2 × 1030 kg. The average distance between the two is 1.5 × 1011 m.

Ans

1. $\begin{array}{l}{\text{Given, Mass of the Sun, m}}_{\text{1}}{\text{= 2 × 10}}^{\text{30}}\text{kg}\\ {\text{Mass of the Earth, m}}_{\text{2}}{\text{= 6 × 10}}^{\text{24}}\text{kg}\\ \text{Average distance between the Earth and the Sun, r = 1}{\text{.5 × 10}}^{\text{11}}\text{m}\\ \text{Universal gravitational constant, G = 6}\text{.67}×{\text{10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\\ \text{As per universal law of gravitation,}\\ \text{Force, F = G}\frac{{\text{m}}_{\text{1}}{\text{m}}_{\text{2}}}{{\text{r}}_{\text{2}}}\\ \text{On putting the given values,}\\ \text{=}\left(\text{6}\text{.67}×{\text{10}}^{\text{11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\right)\frac{\left(2×{10}^{30}\text{kg}\right)×\left(6×{10}^{24}\text{kg}\right)}{{\left(1.5×{10}^{11}\text{m}\right)}^{2}}\\ \text{= 3}{\text{.57 × 10}}^{\text{22}}\text{N}\end{array}$

Q.159 A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

Ans

1. $\begin{array}{l}\text{}\end{array}$ $Let, two stones meet at point O in time t. For the stone dropped from the tower. Initial velocity, u = 0 Let the displacement of the stone in time t from the top of the tower = s g = 10 ms -2 On using the second equation of motion under gravity, s = ut + 1 2 gt 2 or, = 0 + 1 2 ( 10 ms -2 ) t 2 or, s = 5t 2 …(i) For the stone thrown upward Initial velocity, u = 25 ms -1 Let the displacement of the stone form the ground in time t = s’ Acceleration due to gravity, g = -10 ms -2 On using the second equation of motion under gravity, s = ut + 1 2 gt 2 or, = 25t + 1 2 ( -10 ms -2 ) t 2 or, s’ = ( 25 ms -1 ) t – 5t 2 ….( ii ) Now, from the figure, s+s’ = 100 m ..( iii ) ∴ 5t 2 +25t – 5t 2 =100 m ⇒ t = 100 m 25 ms -1 = 4 s Hence, Both the stones will meet after 4 s. Now, from ( 1 ), distance covered by first stone in 4 s s = 5t 2 = 5 ( 4 s ) 2 = 80 m From ( iii ), distance covered by second stone in 4 s, s’ = 100 m – 80 m = 20 m$

Q.160 A ball thrown up vertically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s

Ans

1. $\begin{array}{l}\text{(a)\hspace{0.17em}Time of ascent of the ball = Time of descent of the ball}\\ \text{As total time of journey = 6 s}\\ \therefore \text{\hspace{0.17em}Time of ascent = 3 s}\\ {\text{For upward motion, v = 0, g = -9.8 ms}}^{\text{-2}}\\ \text{Now, by using the relation, v = u + gt}\\ {\text{or, 0 = u + (-9.8 ms}}^{\text{-2}}\text{) × 3\hspace{0.17em}s}\\ {\text{or,\hspace{0.17em}u = 29.4 ms}}^{\text{-1}}\\ {\text{Therefore,\hspace{0.17em}the ball was thrown upwards with a velocity of 29.4 ms}}^{\text{-1}}.\\ \left(\text{b}\right)\\ {\text{Here,\hspace{0.17em}Initial velocity, u = 29.4 ms}}^{\text{-1}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Final velocity, v = 0}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Acceleration due to gravity, g = -9.8 ms}}^{\text{-2}}\\ \text{Let, the maximum height = h}\\ \text{on\hspace{0.17em}using the relation}\\ \text{h = ut +}\frac{\text{1}}{\text{2}}{\text{\hspace{0.17em}gt}}^{\text{2}}\\ \text{On putting the given values,}\\ {\text{h = 29.4 ms}}^{\text{-1}}\text{× 3 s +}\frac{\text{1}}{\text{2}}\left({\text{-9.8\hspace{0.17em}m\hspace{0.17em}s}}^{\text{-2}}\right){\left(\text{3\hspace{0.17em}s}\right)}^{\text{2}}\text{= 44.1\hspace{0.17em}m}\\ \text{(c)}\\ \text{After 3 s,\hspace{0.17em}the ball attains the maximum height}\\ {\text{At maximum height, initial velocity, u = 0 m s}}^{\text{-1}}\\ \text{Position of the ball after 4 s of the throw is given by the}\\ \text{distance travelled by it during its downward}\\ \text{journey in (4s – 3s) = 1 s}\\ \text{On\hspace{0.17em}using the relation for downward motion,}\\ \text{s\hspace{0.17em}=\hspace{0.17em}ut +}\frac{\text{1}}{\text{2}}{\text{at}}^{\text{2}}\\ \text{s\hspace{0.17em}=\hspace{0.17em}0\hspace{0.17em}×\hspace{0.17em}t\hspace{0.17em}+}\frac{\text{1}}{\text{2}}\text{×\hspace{0.17em}}\left({\text{9.8\hspace{0.17em}ms}}^{\text{-2}}\right){\left(\text{1s}\right)}^{\text{2}}\\ \text{=\hspace{0.17em}4.9\hspace{0.17em}m}\\ \text{As total height = 44.1 m}\\ \text{Therefore, the\hspace{0.17em}}\mathrm{h}\text{eight\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}ball\hspace{0.17em}from\hspace{0.17em}the\hspace{0.17em}ground\hspace{0.17em}after â€‹4\hspace{0.17em}s would\hspace{0.17em}be\hspace{0.17em}}\\ \text{=\hspace{0.17em}}\left(\text{44.1\hspace{0.17em}m – 4.9â€‹\hspace{0.17em}m}\right)\text{= 39.2\hspace{0.17em}m.}\end{array}$

Q.161 In what direction does the buoyant force on an object immersed in a liquid act?

Ans

The buoyant force acts in the upward direction for an object immersed in a liquid.

Q.162 Why does a block of plastic released under water come up to the surface of water?

Ans

A plastic block under the water experiences two forces. One is the gravitational force, which pulls the block downwards and the other is the buoyant force which pushes the block upwards. If the upward buoyant force is greater than the downward gravitational force, then the block comes up to the surface of the water as soon as it is released within water.

Q.163 The volume of 50 g of a substance is 20 cm3. If the density of water is 1 g cm−3, will the substance float or sink?

Ans

1. $\begin{array}{l}\text{Density}\text{of}\text{the}\text{substance}=\frac{\text{Mass}\text{of}\text{the}\text{substance}}{\text{Volume}\text{of}\text{the}\text{substance}}\\ =\frac{50 g}{20{\text{cm}}^{3}}=\text{2}.\text{5}{\text{gcm}}^{-3}\end{array}$

if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid. Conversely, the density of an object is more than the density of a liquid, then it sinks in the liquid.

Here, the density of the substance is more than the density of water i.e., 1 g cm−3. Therefore, the substance will sink into the water.

Q.164 The volume of a 50 g sealed packet is 350 cm3. Will the packet float or sink in water if the density of water is 1 g cm−3? What will be the mass of the water displaced by this packet?

Ans

1. $\begin{array}{l}\text{Density of packet =}\frac{\text{Mass}}{\text{Volume}}\text{=}\frac{\text{500 g}}{{\text{350 cm}}^{\text{3}}}\text{=1}{\text{.428 gcm}}^{\text{-3}}\\ \text{Here, the density of the packet is more than the density of water}\text{.}\\ \text{Therefore, it will sink into the water}\text{.}\\ \text{The mass of water displaced}\\ \text{= volume of packet × density of water}\\ {\text{= 350 cm}}^{\text{3}}{\text{× 1 gcm}}^{\text{-3}}\text{=350g}\end{array}$

Q.165 An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Ans

Yes. It is possible that an object is moving with a non-zero velocity even when the object experiences a net zero external unbalanced force. It is possible only when the object has been moving with a constant velocity in a certain direction. In order to change the state of motion, a net non-zero external unbalanced force must be applied on the object.

Q.166 When a carpet is beaten with a stick, dust comes out of it. Explain.

Ans

Due to inertia, an object opposes any change in its state of rest or state of motion. As soon as the carpet is beaten with a stick, the carpet comes into motion while the dust particles oppose their state of rest. As per Newton’s first law of motion, the dust particles remain stationary, while the carpet moves. Therefore, the dust particles come out of the carpet.

Q.167 Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Ans

When the bus starts accelerating in forward direction, it comes in the state of motion but the luggage kept on the roof remains in the state of rest. Therefore, with the forward movement of the bus, the luggage remains at its original position and finally falls from the roof of the bus. Hence, it is advised to tie any luggage kept on the roof of a bus.

Q.168 A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.

Ans

The correct option is (c)

Explanation: Here, the ball comes to rest due to frictional force which opposes its motion. Frictional force always acts in the direction opposite to the direction of motion. Therefore, this force tries to stop the ball.

Q.169 Rutherford’s alpha-particle scattering experiment was responsible for the discovery of

(a) Atomic Nucleus (b) Electron

(c) Proton (d) Neutron

Ans

Atomic nucleus

Q.170 Isotopes of an element have
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons
(d) different atomic numbers.

Ans

Isotopes of an element have different number of neutrons.

Q.171 Number of valence electrons in Cl ion are:
(a) 16
(b) 8
(c) 17
(d) 18

Ans

8

Q.172 An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Ans

1. $\begin{array}{l}\text{Given, diameter of circular track, d = 200 m}\\ \text{time required to cover the distance of circular}\\ \text{track = 40 s}\\ \text{Now, radius of the track, r =}\frac{d}{2}\text{=}\frac{200 m}{2}\text{= 100 m}\\ \text{Circumference = 2}\pi \text{r = 2}\pi \text{}\left(100 m\right)=200\pi \text{m}\\ âˆµ\text{distance covered by athlete in 40 s =}200\pi \text{m}\\ \therefore \text{in 1 s, the distance covered by the athlete}\\ \text{=}\frac{200\pi \text{m}}{40}\text{=5}\pi \text{m}\\ \text{So, the distance covered in 2 minutes and 20 seconds,}\\ \text{i}\text{.e}\text{., 140 s is given by = (140}×\text{5}\pi \right)\text{m =700}\pi \text{m}\\ \text{= 700}×\frac{22}{7}\text{m =2200 m}\end{array}$ $\begin{array}{l}\text{The athlete completes one round in 4}0\text{s}\\ \text{So, the time taken in 3 complete rounds = 3}×\text{40 s}=\text{12}0\text{s}\\ \text{The displacement after 3 complete rounds would be zero}\text{.}\\ Now,\text{the time left = (140 s -120 s) = 20 s}\\ In\text{20 s},\text{he moves at the opposite end of the}\\ \text{initial position hence, the displacement will be equal to the}\\ \text{diameter of the circular track i}\text{.e}\text{., 200 m}\text{.}\end{array}$

Q.173 Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Ans

(a)

1. $\begin{array}{l}\text{Given, distance covered from A to B = 300 m}\\ \text{time taken = 2 min 30 seconds = 150 s}\\ \text{Average speed =}\frac{\text{Total distance covered}}{\text{Total time taken}}\\ \text{=}\frac{\text{300 m}}{\text{150 s}}\text{= 2}{\text{ms}}^{\text{-1}}\text{}\\ \text{displacement between A to B = 300 m}\\ \text{time taken = 2 min 30 seconds = 150 s}\\ \text{Average velocity =}\frac{\text{Net displacement}}{\text{Total time taken}}\text{}\\ \text{=}\frac{\text{300 m}}{\text{150 s}}{\text{= 2 ms}}^{\text{-1}}\text{}\end{array}$

(b)

1. $\begin{array}{l}\text{Total distance covered from A to C = AB +BC = 300 m + 100 m = 400}\\ \text{Total time taken = time taken from A to B + time taken from B to C =}\end{array}$ $\begin{array}{l}\text{150 s + 60 s = 210 s}\\ \text{Average speed =}\frac{\text{Total distance covered}}{\text{Total time taken}}\\ \text{=}\frac{\text{400 m}}{\text{210 s}}\text{= 1}{\text{.90 ms}}^{\text{-1}}\text{}\\ \text{Net displacement = AB – BC = 300 m – 100 m = 200 m}\\ \text{time taken = 210 s}\\ \text{Average speed =}\frac{\text{Total distance covered}}{\text{Total time taken}}\text{}\\ \text{=}\frac{\text{200 m}}{\text{210 s}}{\text{= 0.952 ms}}^{\text{-1}}\text{}\end{array}$

Q.174 Abdul, while driving to school, computes the average speed for his trip to be 20 kmh−1. On his return trip along the same route, there is less traffic and the average speed is 30 km h−1. What is the average speed for Abdul’s trip?

Ans

1. $\begin{array}{l}\text{While going to school:}\\ {\text{Given, Average speed = 20 kmh}}^{\text{-1}}\\ \text{Let, the distance travelled to reach school = d}\\ {\text{time taken = t}}_{\text{1}}\\ \text{Average speed =}\frac{\text{total distance}}{\text{total time taken}}\\ \therefore \text{20 =}\frac{\text{d}}{{\text{t}}_{\text{1}}}\text{}\\ {\text{or, t}}_{\text{1}}\text{=}\frac{\text{d}}{\text{20}}\\ \text{While returning from school:}\\ {\text{Given, average speed = 40 kmh}}^{\text{-1}}\text{}\\ \text{Let, the distance travelled while returning from school = d}\\ {\text{time taken = t}}_{\text{2}}\\ \therefore \text{40 =}\frac{\text{d}}{{\text{t}}_{\text{2}}}\\ {\text{or, t}}_{\text{2}}\text{=}\frac{\text{d}}{\text{40}}\\ \\ \text{Now, Average speed =}\frac{\text{Total distance travelled}}{\text{total time taken}}\\ \text{=}\frac{\text{2d}}{{\text{t}}_{\text{1}}{\text{+t}}_{\text{2}}}\\ \text{=}\frac{\text{2d}}{\frac{\text{d}}{\text{20}}\text{+}\frac{\text{d}}{\text{40}}}\text{=}\frac{\text{80}}{\text{3}}\text{=26}{\text{.67 ms}}^{\text{-1}}\end{array}$

Q.175 A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms−2 for 8.0 s. How far does the boat travel during this time?

Ans

1. $\begin{array}{l}{\text{Given, u= 0, a= 3 ms}}^{\text{–}}{}^{\text{2}}\text{, t = 8 s}\\ \\ \text{On using second equation of motion,}\\ \text{s = ut +}\frac{\text{1}}{\text{2}}{\text{at}}^{\text{2}}\\ On\text{putting given values,}\\ \text{s = 0 +}\frac{\text{1}}{\text{2}}{\text{×3 ms}}^{-2}\text{×}{\left(\text{8 s}\right)}^{\text{2}}\text{= 96 m}\\ \text{Hence, the boat travels a distance of 96 m}\text{.}\end{array}$

Q.176 A driver of a car travelling at 52 kmh−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 kmh−1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Ans

1. $\begin{array}{l}{1}^{st}\text{Case:}\\ {\text{Given, u = 52 kmh}}^{\text{-1}}\text{= 14}{\text{.4 ms}}^{\text{-1}}\\ Time\text{taken to stop the car, t = 5 s}\\ \text{v = 0}\\ Dis\mathrm{tan}ce{\text{covered, s}}_{\text{1}}\text{= area OAC}\\ \text{=}\frac{1}{2}×\text{14}{\text{.4 ms}}^{-1}×\text{5 s = 36 m}\\ {2}^{nd}\text{Case:}\\ {\text{Given, u = 3 kmh}}^{\text{-1}}\text{= 0}{\text{.833 ms}}^{\text{-1}}\\ Time\text{taken to stop the car, t = 10 s}\\ \text{v = 0}\\ Dis\mathrm{tan}ce{\text{covered, s}}_{\text{2}}\text{= area ODB}\\ \text{=}\frac{1}{2}×0.833\text{\hspace{0.17em}}m{s}^{-1}×10\text{\hspace{0.17em}}s\text{= 4}\text{.15 m}\\ Therefore,\text{the distance covered in first case is greater}\\ \text{than the distance covered in second case}\text{.}\end{array}$

Q.177 The given figure shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?

Ans

(a)

1. $\begin{array}{l}\text{speed =}\frac{\text{distance}}{\text{time}}\\ \mathrm{slope}\text{of graph =}\frac{y-\text{axis}}{x-\text{axis}}=\frac{\text{distance}}{\text{time}}\\ \therefore \text{Speed = Slope of the graph}\\ \text{Slop of B is greater than A and C}\text{. Hence,}\\ \text{it is travelling the fastest}\text{.}\end{array}$

(b) As, A, B and C will never meet at a single point at same time therefore, they will never be at the same point on the road.

(c)

1. $\begin{array}{l}\text{From graph, 7 small boxes = 4 km}\\ \therefore \text{1 box =}\frac{4}{7}\text{km}\\ \text{Initially, C is 4 blocks away from the}\\ \text{origin}\text{.}\\ \therefore \text{Intial distance of object C from origin =}\frac{16}{7}\text{km}\end{array}$ $\begin{array}{l}\text{B passes A at point P. Object C is at point Q when B passes A.}\\ \text{Distance of point Q from origin}\\ \text{= 8 km}\\ \text{Distance covered = 8 km –}\frac{16 km}{7}\text{=}\frac{56 km-16 km}{7}\text{=}\frac{40 km}{7}\\ \text{= 5}\text{.714 km}\\ \text{Therefore, C has travelled a distance of 5}\text{.714 km}\\ \text{when B passes A}\text{.}\end{array}$ $\begin{array}{l}\text{(d)}\end{array}$ $\begin{array}{l}\text{B passes C at point R.}\end{array}$ $\begin{array}{l}\text{Distance travelled by B at the time it passes C}\\ \text{= 9 boxes =}\frac{4}{7}×9.25=\frac{36}{7}=5.28\text{km}\\ \text{Therefore, B has travelled a distance of 5}\text{.28 km when}\\ \text{B passes A}\text{.}\end{array}$

Q.178 A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms−2, with what velocity will it strike the ground? After what time will it strike the ground?

Ans

1. $\begin{array}{l}Given,{\text{s = 20 m, a = 10 ms}}^{-2},\text{u = 0}\\ \text{v = ?}\\ \text{On using third equation of motion,}\\ {\text{v}}^{2}={\text{u}}^{2}+2as\\ \text{= 0 + 2}×\text{10}×20=400\\ or\text{v =}\sqrt{400}=20{\text{ms}}^{-1}\\ Now,\text{using first equation of motion}\\ \text{v = u + at}\\ \therefore \text{20 = 0 + 10}×\text{t}\end{array}$

⇒ t = 2 s.

Q.179 The speed-time graph for a car is shown here.

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?

Ans

(a)

Distance travelled by the car the first 4s = shaded area = 12 × 4s × 6ms-1 = 12cm

(b)

The part of the graph from 6 s to 10 s represents uniform motion of the car.

Q.180

State which of the following situations are possible and give an example for each of these:(a) an object with a constant acceleration but with zero velocity.

(b) an object moving with an acceleration but with uniform speed.

(c) an object moving in a certain direction with an acceleration in the perpendicular direction.

</div>
Ans

(a) It is possible.

Example: An apple has zero velocity at maximum height when it is thrown upward. Although, it will have constant acceleration due to gravity.

(b) It is possible.

Example: When a truck is moving in a circular track, it is moving with an acceleration but with uniform speed.

(c) It is possible.

Example: When a truck is moving in a circular track, its acceleration is perpendicular to its direction of motion.

Q.181 An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Ans

$Given, radius of circular path, r = 42250 km = 42250 km × 1000 m 1 km = 4 .225×10 7 m time period, T = 24 hours = 24 hr × 60 min 1 hr × 60 s 1 min = 86400 s orbital velocity of satellite, v = ? On using the relation, v = 2πr T On putting the given values, = 2×3.14×( 4 .225×10 7 m ) 86400 s = 3070 .94 ms -1$

Q.182 A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)

Ans

$Given, intial velocity of the truck, u = 0 distance travelled, s = 400 m time taken, t = 20 s acceleration, a = ? On using the relation s = ut + 1 2 at 2 400 m = 0(20 s) + 1 2 a(20 s) 2 or, a = 2 ms -2 Now, 1 metric tonne = 1000 kg ∴ 7 metric tonnes = 7×1000 kg = 7000 kg By using Newton’s second law, F = ma = (7000 kg) (2 ms -2 ) = 14000 N Thus, the acceleration of the truck is 2 ms -2 and the force acting on it is 14000 N.$

Q.183 A stone of 1 kg is thrown with a velocity of 20 ms−1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

Ans

$Given, u = 20 ms -1 v = 0 s = 50 m mass of the stone, m = 1 kg On using the third equation of motion, v 2 = u 2 + 2as (a = acceleration of the stone) ∴ 0 2 = (20 ms -1 ) 2 +2×a×50 m or, a = – 400 m 2 s −2 100 m = – 4 ms -2 Here, negative sign shows that acceleration is acting against the motion of the stone. By using Newton’s second law of motion, Force, F = ma or, = 1 kg ×(-4 ms -2 ) = -4 N Therefore, the force of friction between the stone and the ice is – 4 N.$

Q.184 A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:

(a) the net accelerating force and

(b) the acceleration of the train.

Ans

$\begin{array}{l}\left(a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}Given,}\text{force exerted by the engine, F = 40000 N}\\ \text{frictional force due to tracks, F’ = 5000 N}\\ \text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{â€Š}\text{\hspace{0.17em}}\therefore {\text{Net force, F}}_{\text{net}}=40000-5000\text{= 35000 N}\text{}\\ \text{(b)}\text{\hspace{0.17em}}\text{Net force on the wagons},\text{}{F}_{a}=\text{35}000\text{N}\\ \text{Mass of the 5 wagons},\text{}m=\text{2}000×\text{5}=\text{1}0000\text{kg}\\ \text{Mass of the engine},\text{}m\prime =\text{8}000\text{kg}\\ \therefore \text{Total mass},\text{M}=m+m\prime =\text{18}000\text{kg}\\ \text{Using Newton}’\text{s second law of motion,}\\ {\text{F}}_{a}=Ma\\ \text{or, a}\text{=}\frac{{\text{F}}_{a}}{M}\text{=}\frac{35000}{18000}=\text{1}{\text{.944 ms}}^{\text{-2}}\\ \text{Hence, the acceleration of the wagons and the train is}\\ 1.944{\text{ms}}^{\text{-2}}.\\ \end{array}$

Q.185 An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms−2?

Ans

$Given, m = 1500 kg v = 0 a = -1 .7 ms -2 On using Newton’s second law of motion, Force, F = m × a =1500 kg × ( -1 .7 ms -2 )= -2550 N Therefore, the force between the automobile and the road is 2550 N, in the direction opposite to the motion of the automobile.$

Q.186 What is the momentum of an object of mass m, moving with a velocity v?

1. (mv)2
2. mv2
3. ½ mv2
4. mv

</ol>
Ans

The correct option is (d).
Explanation:
1. $\begin{array}{l}\text{Given, mass of the object = m}\\ \text{velocity of the object = v}\\ \text{Momentum, P = mass}×\text{velocity}\\ \text{= mv}\end{array}$

Q.187 Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Ans

When the force of 200 N is applied on wooden cabinet, an equal and opposite force acts on it. This opposite force is the frictional force. Therefore, a frictional force of 200 N is exerted on the cabinet.

Q.188 Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

Ans

1. $Given, masses of the objects, m 1 =m 2 = 1.5 kg velocities of the objects, v 1 = 2 .5 ms -1 v 2 =- 2 .5 ms -1 Here, velocity of m 2 is considered negative as it is moving in opposite direction before collision. Both the objects stick together after collision. Hence, total mass M = m 1 +m 2 = 1.5 kg + 1.5 kg = 3 kg Let, velocity of the combined mass after collision = V As per conservation law of linear momentum, m 1 v 1 + m 2 v 2 = MV ∴ (1.5 kg × 2 .5 ms -1 )+(1.5 kg × -2 .5 ms -1 ) = 3V or, V = 0 Thus, the velocity of the combined object after collision will be zero.$

Q.189 According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

Ans

The static friction acting between the truck and the road is very large as the truck is very massive. Thus, truck does not move when someone pushes it. Therefore, it can be concluded that the applied force in one direction is cancelled out by the frictional force of equal amount acting in the opposite direction.

Q.190 A hockey ball of mass 200 g travelling at 10 ms−1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms−1. Calculate the magnitude of change of
momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Ans

1. $Given, mass of the hockey ball, m = 200 g = 0.2 kg velocity of the ball, v 1 = 10 ms -1 Initial momentum, P 1 = mv 1 Now, hockey ball travels in the opposite direction with velocity, v 2 = – 5 ms -1 Final momentum, P 2 = mv 2 ∴ Change in momentum, ΔP = P 1 -P 2 = m[ v 1 – v 2 ] = 0.2 kg[ 10 ms -1 – (-5 ms -1 ) ] = 3 kg ms -1 . The change in momentum comes out to be 3 kgms -1 and the negative sign indicates that the ball started moving in opposite direction.$

Q.191 A bullet of mass 10 g travelling horizontally with a velocity of 150 ms−1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Ans

1. $Given, inital velocity of the bullet, u = 150 ms -1 final velocity of the bullet, v = 0 (as the bullet comes to rest) time taken to come to rest for bullet, t = 0.03 s mass of the bullet, m = 0.01 kg On using the first equation of motion, v = u +at or, 0 = 150 + a×0.03 or, a = -5000 ms -2 (Here, negative sign shows that the velocity of the bullet is decreasing) On using the third equation of motion, v 2 = u 2 +2as or, 0 2 = (150) 2 +2×(−5000)×s or, s = 22500 10000 =2.25 m Hence, the distance of penentration of the bullet into the block is 2.25 m. On using Newton’s second law of motion, Force exerted by the wooden block on the bullet, F = ma = 0.01×5000=50 N. 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Q.192 An object of mass 1 kg travelling in a straight line with a velocity of 10 ms-1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Ans

1. $\begin{array}{l}\mathrm{Given},\text{mass of the object, m = 1 kg}\\ \text{mass of the wooden block, M = 5 kg}\\ \mathrm{velocity}{\text{of the object before collision, u}}_{\text{1}}{\text{= 10 ms}}^{\text{-1}}\\ \mathrm{velocity}{\text{of the wooden block befor collision, u}}_{\text{2}}\text{= 0}\\ \\ \mathrm{Total}\text{momentum before collision,}\\ {\text{P}}_{\text{inital}}={\text{mu}}_{\text{1}}{\text{+Mu}}_{\text{2}}\text{= 1}×\text{10}+\text{5}×0=10{\text{kgms}}^{\text{-1}}\\ \mathrm{After}\text{collision, both the blocks stick together, hence}\\ \mathrm{the}\text{the combined mass = M+m = 5+1 = 6 kg}\\ \text{Let, the velocity of the combined mass is V}\\ \text{Then, as per conservation law of momentum,}\\ {\text{P}}_{\text{inital}}{\text{= P}}_{\text{final}}\\ \text{10 = 6}×\text{V}\\ \text{or V =}\frac{10}{6}{\text{= 1.67 ms}}^{\text{-1}}\\ \text{Hence, the velocity of the combined mass after}\\ {\text{collision would be 1.67 ms}}^{\text{-1}}\text{.}\\ {\text{Now, final momentum, P}}_{\text{final}}\text{= (M+m)V =6}×\text{1.67}\\ {\text{=10 kgms}}^{\text{-1}}\text{.}\end{array}$

Q.193 An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms-1 to 8 ms-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Ans

1. $\begin{array}{l}\mathrm{Given},{\text{inital velocity, u = 5 ms}}^{\text{-1}}\\ {\text{final velocity, v = 8 ms}}^{\text{-1}}\\ \text{mass , m = 100 kg}\\ \text{time taken by the object to accelerate, t = 6 s}\\ \\ \therefore \text{Intial momentum = mu = 100}×{\text{5 = 500 kgms}}^{\text{-1}}\\ \text{Final momentum = mv = 100}×8{\text{= 800 kgms}}^{\text{-1}}\\ \mathrm{Force}\text{exerted on the object, F =}\frac{\mathrm{mv}-\mathrm{mu}}{\mathrm{t}}\\ \text{=}\frac{800-500}{6}\text{=}\frac{300}{6}\text{= 50 N.}\end{array}$

Q.194 Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.

Ans

As per the conservation of momentum,

(momentum of motorcar + momentum of insect)before

= (momentum of motorcar + momentum of insect)after

∴ Change in in momentum = 0

The direction of the insect gets reversed when it stuck on the windscreen and the velocity of the insect changes to a great extend. The motorcar continues moving with a constant velocity in the forward direction.

As, Kiran suggested that the insect suffers a greater change in momentum as compared to the motorcar hence, it is correct. The momentum of the insect after collision becomes very high as the motorcar is moving at a very high speed. Therefore, the momentum gained by the insect is equal to the momentum lost by the motorcar.

Secondly, Akhtar is also correct as the mass of the motorcar is very large as compared to the mass of the insect and the speed of the motorcar is also very high.

Now, Rahul’s explanation is correct as both the car and the insect experienced equal forces caused by the Newton’s action-reaction law. But Rahul’s statement becomes incorrect when he said that the system suffers a change in momentum because the momentum before the collision is equal to the momentum after the collision.

Q.195 How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s−2.

Ans

$Given, mass of the dumbell, m = 10 kg distance covered by the dumbell, s = 80 cm = 0.8 m acceleration in the downward direction, a = 10 ms -2 initial velocity of the dumbell, u = 0 final velocity of the dumbell, v = ? On using the third equation of motion, v 2 = u 2 +2as On putting the given values, v 2 = 0 2 + 2(10 ms -2 )(0.8 m) = 16 m 2 s −2 or, v = 16 m 2 s −2 = 4 ms -1 Therefore, the momentum with which the dumbell hits the floor is = mv= 10 kg × 4 ms -1 = 40 kgms -1 .$