# NCERT Solutions Class 10 Science

Science is a subject that can be challenging to grasp in terms of topics and concepts after you reach class ten. In such instances, having NCERT solutions for class 10 science can be quite beneficial, especially if students desire to do well in their exams. Extramarks offers carefully designed solutions for NCERT class 10th science solutions that will assist you in looking deeper into the chapters and gaining a better understanding of the concepts covered in each one. We recognize that topics like Biology, Chemistry, and Physics, especially in Class 10, can be difficult to handle.

## NCERT Solutions for Class 10 Science 2022-23 Free PDF

Don’t worry because guidance is straight here when you need it. All you have to do is go to the Extramarks website and make sure you get the best solutions for each chapter of the class 10 science textbook. The solutions are always up-to-date with the most recent editions of the books and include all of the answers to all of the textbook’s questions. Extramarks class 10 science NCERT solutions have been properly created to help you understand the concepts and learn how to respond correctly in board exams. The NCERT solutions for class 10 science comprise 16 chapters organised into three parts, each of which focuses on a different topic of science.

NCERT solutions for class 10 science are essential information that aids students in understanding difficult topics and preparing for the Class 10 board exams. Examining the answers to the textbook’s questions can help you assess your understanding  of a topic and identify your strengths and limitations. Our subject specialists have created these NCERT Solutions for Class 10 Science in such a way that students can grasp all of the concepts contained in the CBSE 10 science syllabus. Students who are looking for better solutions can download NCERT 10th Science Solutions to assist them in revising the entire syllabus and achieving higher grades in their exams.

### NCERT Solutions Class 10 Science Book All Chapters Brief:

#### Chapter 1 – Chemical Reactions and Equations – Term I

The foundation of chemistry is Chapter 1 Chemical Reactions and Equations for Class 10 Science NCERT Solutions. Different forms of reactions, chemical equations are covered in this topic. The NCERT Solution is available as a free PDF download from Extramarks official website. These solutions were created by our topic experts and are extremely useful for exam preparation. Our knowledgeable instructors will walk you through the topics and answer all of your questions.

#### Chapter 2 – Acids, Bases and Salts – Term I

If you want to do well on your Class 10 Science Board exam as well as lay a strong basis for JEE, NEET, and other professional options, NCERT Solutions for Class 10 Science Chapter 2 can help. Students will learn about diverse chemical substances and their characteristics in this chapter, which is divided into acids, bases, and salts. Important topics such as the reaction of acids with hydrogen, metal oxides, and carbonates get in-depth explanations.

#### Chapter 3 – Metals and Non-metals – Term I

Here are all of the NCERT Solutions for Class 10 Science Chapter 3. The students will learn about numerous metals and non-metals in this chapter. This includes their physical and chemical properties, as well as how they interact with other metallic salts in various settings. There are questions based on the reactivity series, which is a crucial topic that must be grasped. Students must also understand how different metals and their ores are extracted.

#### Chapter 4 – Carbon and its Compounds – Term II

NCERT Solutions for Class 10 Science, Carbon and its Compounds are available on Extramarks. Students will learn about all of the properties of carbon, as well as the qualities of its bond, in this chapter. This section is crucial in developing solid organic chemistry fundamental principles. You can also get Carbon and its Compounds Class 10 science NCERT Solutions from Extramarks website to help you go through the entire curriculum and get the best grades possible in your exams.

#### Chapter 5 – Periodic Classification of Elements – Term II

The introduction of NCERT Solutions for Class 10 Science Chapter 5 aims to aid students in understanding the periodic table’s classification of different elements which is one  of the most essential subjects in this chapter. It explains the layout, the placement of various metals, non-metals, and the many trends that are seen. This covers characteristics such as metallicity, electronegativity, and atomic radius, among others. Students should pay close attention to this issue because numerous questions can be asked in a variety of ways.

Chapter 6 – Life Processes – Term I

Class 10th is one of the most important years in a student’s life; this class can help you learn all of the fundamental principles you’ll need in the years ahead. Students learn about metabolism in Class 10 Science chapter 6, which is a chemical process that occurs in the human body and is responsible for controlling the human state of the cells in an organism. Catabolism and anabolism are the two types of metabolism. Extramarks is a website that offers students free NCERT Solutions for class 10 science and other study tools.

#### Chapter 7 – Control and Coordination

One of the most captivating chapters in Class 10 Science is Control and Coordination. Through the numerous subjects discussed in this chapter, students can pursue knowledge on a wide range of topics. NCERT Solutions for Class 10 Science Chapter 7 Control and Coordination presents students with the most simple strategies for achieving satisfactory results by covering the nervous system in animals and the way of coordination in plants.

Chapter 8 – How do Organisms Reproduce – Term II

The reproduction of organisms is the focus of Chapter 8 in Science Class 10. Chapter 8, Science Class 10, teaches you everything you need to know about reproduction. This chapter is extremely significant in Class 10 in terms of human and plant life, and you should thoroughly understand it. NCERT Solutions Class 10 Science Chapter 8 is written by highly qualified biological science instructors. NCERT Solutions offer the highest-quality notes for Chapter 8 of Class 10 Science in a format that is easy to use. Students can also look out for this chapter in NCERT solutions class 9.

#### Chapter 9 – Heredity And Evolution – Term II

‘Heredity and Evolution’ is the topic of NCERT solutions for Class 10 Science, Chapter 9. It is a very important and high-scoring chapter in the Science curriculum. This chapter includes numerous fundamental principles and key notions that have a lot of weight and significance, especially in light of future research. Evolution theory and classifications, speciation, and human evolution all of these topics are well covered in the NCERT solutions for Class 10. There are diagrammatic representations and flow charts to help in better understanding.

#### Chapter 10 – Light Reflection and Refraction – Term I

NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction are crucial for achieving a good grade on the Class 10 Science board exam. Because this is the first chapter of Physics, it is critical that students become interested as they read it. Concepts and definitions of reflection, refraction, and light behavior against spherical mirrors are all key subjects in this chapter. Students do not need to be concerned about the credibility of NCERT Solutions for Class 10 Chapter 10 Science because they were made and reviewed by some of the most experienced and knowledgeable faculty. The NCERT Solutions for Class 10 Science assist students in grasping specific concepts that are likely to appear in the test.

#### Chapter 11 -The Human Eye and Colorful World – Term I

NCERT solutions for chapter 11 The Human Eye and Colorful World have clear concepts and problems. This chapter covers topics such as the function and structure of the human eye, as well as typical eye disorders and abnormalities and how they are treated. Students will learn with the help of well-labeled graphics, that these three flaws are explained. Students will learn about natural phenomena such as rainbow production, air refraction, star twinkling, and more in this chapter.

#### Chapter 12 – Electricity – Term II

Chapter 12 of Science for Class 10 is on electricity. Basic definitions of electric current, circuit, potential, and potential difference are covered in Chapter 12 of the Science part. The chapter exposes students to one of the most important rules, Ohm’s law, on which practically all inquiries are founded. NCERT Science Solutions for Class 10 Electricity can be quite beneficial to students. Students will be able to grasp even the most difficult ideas in electricity because of the clear language and diagrammatic description.

#### Chapter 13 – Magnetic Effects of Electric Current – Term II

This chapter, like the previous one, contains a lot to study and comprehend. Many of today’s gadgets are based on the principles of electricity and magnetism, which are intricately intertwined. Not just for the examination, but also for future studies in classes 11 and 12, it is critical for the student to grasp these principles.

We are pleased to offer the NCERT Solutions for Class 10 Science, namely Physical Science Chapter 13 – Magnetic Effects of Electric Current Class 10, as a free PDF. Our site provides students with free access to the magnetic effect of electric current Class 10 NCERT Solutions PDF.

#### Chapter 14 – Sources of Energy

The NCERT solutions for Class 10 feature all of the questions as well as detailed solutions of chapter 14 Sources of Energy contains a wealth of information on energy and its different applications.

Extramarks is a website that offers students free NCERT Solutions from class 1 and other study tools. Students who wish to pursue science in class 11 and 12 will benefit from having a thorough understanding of the NCERT class 10 science chapter 14 solutions.

#### Chapter 15 – Our Environment – Term II

Class 10 Science Chapter 15 Solutions for Our Environment will assist students in understanding all aspects of environmental science, from the fundamentals to the more complicated issues. This chapter is quickly becoming a necessity; information is the only thing that can rescue the earth. This is what makes this chapter so crucial and is a foundation for NCERT solutions class 11 & 12. Furthermore, if one attempts to learn it in a systematic manner, it is a very high-scoring chapter.

Chapter 16 – Sustainable Management of Natural Resources – Only for Internal Assessment

Here are the complete NCERT Solutions for Class 10 Science Chapter 16 Sustainable Management of Natural Resources and the Topics covered include wildfires and forests, the resources they provide, water, and non-renewable resources such as petroleum and coal. NCERT solutions for Class 10 Science from Extramarks help students learn these topics in a systematic way, emphasising the importance of natural resource management. With these solutions, we ensure that students have a fundamental knowledge of ideas and that their doubts are cleared.

Download NCERT Solutions for Class 10 Science that have been prepared by experienced tutors. CBSE Class 10 Science Evaluation Scheme (Theory)

 Units Term – I Marks I Chemical Substances – Nature and Behavior: Chapter 1, 2 and 3 16 II World of Living: Chapter 6 10 III Natural Phenomena: Chapter 10 and 11 14 Units Term – II Marks I Chemical Substances – Nature and Behavior: Chapter 4 and 5 10 II World of Living: Chapter 8 and 9 13 IV Effects of Current: Chapter 12 and 13 12 V Natural Resources: Chapter 15 05 Total Theory (Term I + II) 80 Internal Assessment: Term I 10 Internal Assessment: Term II 10 Grand Total 100

Benefits of Referring to Class 10 Science Solutions

These NCERT Solutions for class 10 science chapters are necessary for students who are intending to take the CBSE examination. These solutions will assist you in learning more about the marking scheme, question structure, difficulty level and nature of questions, section-wise marking, and much more in the lead up to your exams. Here are a few advantages of using these solutions.

• Prepare for your exam by memorising questions.

Students can easily ensure that their preparation is on point by understanding the pattern of distinct questions that appear in the examination. As a result, they are more confident and certain about taking the board exam, resulting in higher grades.

• More time management practice is required.

It cannot be denied that with more practice, time management improves, as it aids in increasing the pace with which you answer questions throughout your tests. As a result, you can effectively manage time if you have complete awareness of the answers and solutions supplied by us.

One of the best aspects of the NCERT solutions for class 10th is that they are written by professors and masters who have extensive expertise in teaching the topic. As a result, the correctness and intricacies of the chapters are undeniable.

• Give A Stress-Free Examination

While students may feel concerned during their board examinations, they can easily practise more and more with the help of these NCERT solutions, reducing their tension and ensuring that they perform well in the tests.

#### How NCERT Solutions Class 10 Science is helpful while preparing for the exam?

Science is an important subject in Class 10 for students interested in pursuing further studies in this domain. Students should follow the class 10 science NCERT solutions and be thorough with them in order to get good grades in this subject. NCERT solutions for class 10 science will assist them in reviewing the topics and determining how well they understand the subject and can use the NCERT Solutions for Class 10 Science to double-check whether they answered all of the questions correctly after completing the exercise question.

By referring to the solutions, students will be able to determine which topic they should concentrate on, allowing them to learn more quickly. To solve all of the problems in the NCERT solutions for Class 10 Science, students must know the proper technique. As a result, we have offered chapter-by-chapter NCERT Solutions to assist them in studying for their exams. For additional preparation, students can use the NCERT Workbook Solutions Class 10 Science.

As there aren’t many options on the question paper, students taking the science test in class 10 should know that they must answer all of the questions in a clear and concise manner. The paper comprises 80 questions, each of which has different length and marks. There are also some practical concerns. A, B, and C are the three sections of the paper. The first segment contains one-mark questions, while the second section contains a few three-mark short-answer questions, and the third section contains long-answer questions that are typically worth five points apiece.

The NCERT Solutions for Class 10 science will give students a better grasp and knowledge of the topics. They will be able to understand all of the major themes and, as a result, will be able to better answer the questions posed in the tests if they have a clear concept. Not to mention that it helps a great deal in terms of improving overall grades in board exams. In that instance, our NCERT Solutions can be really beneficial. All of these solutions and answers have been carefully answered by masters and academics. These answers will provide students with all of the information they need to prepare for their class 10 board exams.

Other NCERT Resources for Class 10 Science

In addition to learning from NCERT Class 10 Science Solutions, students should consult other study tools such as the NCERT Class 10 Science Exemplar, the NCERT Class 10 Science textbook, and the NCERT Class 10 Science Syllabus. These study tools are likewise based on the CBSE Class 10 science syllabus and aid students in their board exam preparation. After students have completed the whole curriculum, they can turn to these study materials for quick revision before the exam.

Q.1 Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test tube A, while acetic acid (CH3COOH) is added to test tube B. Amount and concentration taken for both the acids are same. In which test tube will the fizzing occur more vigorously and why?

Ans

In both the test tubes hydrogen gas is formed. The hydrochloric acid (HCl) is stronger acid than acetic acid (CH3COOH) so more hydrogen gas is formed in test tube A which results in vigorous fizzing.

Q.2 Fresh milk has a pH of 6. How do you think the pH will change as it turns into curd? Explain your answer.

Ans

When milk is turned into curd then its pH value will decrease due to the production of lactic acid in curd which is acidic in nature.

Q.3 A milkman adds a very small amount of baking soda to fresh milk.

(a)Why does he shift the pH of the fresh milk from 6 to slightly alkaline?
(b)Why does this milk take a long time to set as curd?

Ans

1. By adding small amount of baking soda, the milkman shifts the pH of the fresh milk from 6 to slightly alkaline so that in basic form it will not spoil easily and will take a longer time for setting into curd.
2. This milk takes a long time to set as curd because the lactic acid produced in the milk due to bacterial action is neutralised by the baking soda. Hence, this milk takes longer time to set as curd.

Q.4 Plaster of Paris should be stored in a moisture-proof container. Explain why?

Ans

The Plaster of Paris should be stored in a moisture-proof container as it absorbs water from moisture and turn into hard mass called Gypsum as shown in following reaction:

$\underset{\text{PlasterofParis}}{{\text{CaSO}}_{4}.\frac{1}{2}{\text{H}}_{2}\text{O}}\text{}+\text{}1\frac{1}{2}{\text{H}}_{2}\text{O}\to \text{}\underset{\text{Gypsum}}{{\text{CaSO}}_{4}\cdot 2{\text{H}}_{2}\text{O}}$

Q.5 What is a neutralisation reaction? Give two examples.

Ans

The reaction between an acid and a base to give salt and water is known as neutralisation reaction. The effect of a base is nullified by an acid and vice-versa. In general, a neutralisation reaction can be written as:

$\begin{array}{l}\text{Base}+\text{Acid}\to \text{Salt}+\text{water}\\ \text{NaOH}+\text{HCl}\to \text{NaCl}+{\text{H}}_{2}\text{O}\\ {\text{Ca(OH)}}_{\text{2}}+{\text{H}}_{2}{\text{SO}}_{4}\text{}\to {\text{CaSO}}_{4}+{\text{H}}_{2}\text{O}\end{array}$

Q.6 Give two important uses of washing soda and baking soda.

Ans

(a) Two important uses of washing soda are:
(i) It is used in the manufacturing of soap and glass.
(ii) It is used to remove the permanent hardness of water.

(b) Two important uses of baking soda are:
(i) It is used for making baking powder.
(ii) It is used in soda- fire extinguishers.

Q.7 Ethane, with the molecular formula C2H6 has

(a) 6 covalent bonds.
(b) 7 covalent bonds.
(c) 8 covalent bonds.
(d) 9 covalent bonds.

Ans

7 covalent bonds

Q.8 Butanone is a four-carbon compound with the functional group

(a) carboxylic acid.
(b) aldehyde.
(c) ketone.
(d) alcohol.

Ans

(c) ketone

Q.9 While cooking, if the bottom of the vessel is getting blackened on the outside, it means that

(a) the food is not cooked completely.
(b) the fuel is not burning completely.
(c) the fuel is wet.
(d) the fuel is burning completely.

Ans

(b) the fuel is not burning completely.

Q.10 Explain the nature of the covalent bond using the bond formation in CH3Cl.

Ans

Carbon has 4 valence electrons. It can neither lose four of its electrons nor gain four electrons as both the processes require large amount of energy. In order to complete the octet, it shares each of the four electrons with each of the three hydrogen atoms and one chlorine atom. Since, bonds are formed because of sharing of electrons, hence these are covalent bonds.

Q.11 Draw the electron dot structures for

(a) ethanoic acid.
(b) H2S.
(c) propanone.
(d) F2.

Ans

(a) Ethanoic acid

(b) H2S

(c) Propanone

(d) F2

Q.12 What is a homologous series? Explain with an example.

Ans

A homologous series is a series of organic compounds with similar general formula, possessing similar chemical properties due to the presence of the same functional group, and show a gradation in physical properties as a result of increase in molecular size and mass.

The members of homologous series are called homologue.

Compounds of same homologous series differ from their consecutive members by one carbon atom and two hydrogen atoms, i.e. by CH2.

Example:

Alkanes; such as, Methane, Ethane, Propane, Butane, etc. belong to the same homologous series.

CH4 (Methane) and C2H6 (Ethane) differ by CH2

C2H6 (Ethane) and C3H6 (Propane) differ by CH2 and so on.

Q.13 How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?

Ans

Difference in Physical Properties

Ethanol is a liquid at room temperature with pleasant odour while ethanoic acid has vinegar like smell. The melting point of ethanoic acid is 17 °C. This is below room temperature and hence, it freezes during winters.

Difference in Chemical Properties

Ethanoic acid reacts with metal carbonates and metal hydrogencarbonates to form salt, water and carbon dioxide gas while ethanol does not react with them.

Example:

$\begin{array}{l}\text{Carboxylic acid + Meta carbonate}\to \text{Salt + Water + Carbon-dioxide}\\ 2{\text{CH}}_{3}\text{COOH}+{\text{Na}}_{2}{\text{CO}}_{3}\text{}\to \text{}2{\text{CH}}_{3}\text{COONa}+{\text{H}}_{2}\text{O}+{\text{CO}}_{2}\end{array}$

Ethanol does not react with sodium hydroxide (NaOH) while ethanoic acid reacts with sodium hydroxide to form sodium ethanoate and water

${\text{CH}}_{3}\text{COOH}+\text{NaOH}\to {\text{CH}}_{3}\text{COONa}+{\text{H}}_{2}\text{O}$

Ethanol is oxidised to give ethanoic acid in presence of acidified KMnO4 while no such reaction takes place with ethanoic acid in presence of acidified KMnO4.

Q.14 Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?

Ans

Soap is a sodium or potassium salt of long-chain fatty acid which has cleansing properties in water.

Soap molecule consists of two parts hydrophobic and hydrophilic. The hydrophilic part is the ionic end of the soap molecule which is soluble in water. The hydrophobic part is the organic end and is insoluble in water. Since the dirt present in clothes is organic in nature, the hydrophobic part entraps dirt and hydrophilic part remains suspended in water. Thus, many more molecules of soap are attached to dirt having their one end suspended in water to form clusters. These clusters with entrapped dirt are known as micelle.

No, micelle will not be formed in other solvents like ethanol.

Q.15 Why are carbon and its compounds used as fuels for most applications?

Ans

The carbon and its compounds are used as fuels for most applications because

1. Most of the carbon compounds produce a lot of heat and light when burnt in air.
2. The amount of heat released can be handled and used easily.
3. Once ignited carbon and its compounds keep on burning without any additional requirement of heat energy. Saturated hydrocarbons burn with a clean flame and no smoke is produced.
4. The carbon compounds, used as fuel, have high calorific values.

Q.16 Explain the formation of scum when hard water is treated with soap.

Ans

Hard water contains Mg2+ and Ca2+ ions. Soap is sodium or potassium salt of long-chain fatty acid. When hard water is treated with soap, Mg2+ and Ca2+ ions present in hard water react with soap to form Magnesium and Calcium salts of fatty acid which are insoluble in water and are known as scum.

Q.17 What change will you observe if you test soap with litmus paper (red and blue)?

Ans

Soap is basic in nature; therefore, it will turn red litmus blue. On the other hand, the colour of blue litmus will remain blue.

Q.18 What is hydrogenation? What is its industrial application?

Ans

Hydrogenation is the process of addition of hydrogen. For example: Ethene reacts with hydrogen when heated in the presence of nickel catalyst to form ethane.

$\underset{\text{Ethene}}{{\text{CH}}_{2}={\text{CH}}_{2}}\text{}\stackrel{\text{NiHeat}}{\to }\text{}\underset{\text{Ethane}}{{\text{CH}}_{3}-{\text{CH}}_{3}}$

It is used for making vanaspatee ghee from vegetable oil in industries.

Q.19 Which of the following hydrocarbons undergo addition reactions:

C2H6, C3H8, C3H6, C2H2 and CH4

Ans

Unsaturated hydrocarbons(alkenes and alkynes) undergo addition reactions. In the given list, C3H6 and C2H2 are unsaturated hydrocarbons and therefore, undergo addition reactions.

Q.20 Give a test that can be used to differentiate chemically between butter and cooking oil.

Ans

Butter contains saturated fats while cooking oil contains unsaturated fats. If we add bromine water to them, the reddish brown colour of bromine will disappear in the sample of cooking oil while it will remain same in the sample of butter.

Q.21 Explain the mechanism of the cleaning action of soaps.

Ans

A soap molecule is a sodium or potassium salt of long chain carboxylic acid. It consists of two parts:

1. A non-polar part which consists of long chain hydrocarbons called tail. It is hydrophobic.
2. An ionic part consisting of carboxylate ion called polar head. It is hydrophilic.

When a dirty cloth is dipped into a soap solution, its molecules arrange themselves around the dirt particles in such a way that the hydrophobic hydrocarbon chain gets embedded inside the dirt particles and the hydrophilic head projects outside the dirt particles like bristles.

As the hydrophilic part is polar, it interacts with the water molecules present around the dirt particles. This result into the formation of spherical clusters called Micelle.

The soap molecules form aggregates around the dirt particles which get pulled away from the surface of the cloth when the dirty cloth is agitated in soap solution.

The dirt particles get suspended in water due to which the soap water becomes dirty and cloth is cleaned.

Q.22 A solar water heater cannot be used to get hot water on

(a) a sunny day.
(b) a cloudy day.
(c) a hot day.
(d) a windy day.

Ans

The correct option is (b).
Explanation: A solar water heater requires bright and intense sunlight to function. Due to absence of sunlight on a cloudy day, solar energy is not available for the solar heater to work properly.

Q.23 Which of the following is not an example of a bio-mass energy source?

(a) Wood
(b) Gobar-gas
(c) Nuclear energy
(d) Coal

Ans

The correct option is (c).

Explanation: Bio-mass Energy Source: The energy source which can be obtained from plants materials and animal wastes are called bio-mass energy source.
Wood is a plant material, gobar gas is formed from animal dung and coal is a fossil fuel obtained from the buried remains of plants and animals. Therefore, these are bio mass products.
Nuclear energy is released during nuclear fission and fusion and produces tremendous amount of energy. Hence, nuclear energy is not an example of bio-mass energy source.

Q.24 Most of the sources of energy we use represent stored solar energy. Which of the following is not ultimately derived from the Sun’s energy?

(a) Geothermal energy
(b) Wind energy
(c) Nuclear energy
(d) Bio-mass

Ans

The correct option is (c).
Explanation: Nuclear energy is produced during nuclear fission and fusion. In nuclear fission, the nucleus of heavy atoms (such as uranium, plutonium) is bombarded with low-energy neutrons. Therefore, uranium atom breaks into two lighter nuclei. This reaction produces huge amount of energy. In nuclear fusion, two lighter nuclei are fused together to form a relatively heavier nuclei. This reaction produces tremendous amount of energy. These reactions can be carried out in the absence or presence of sunlight.

Q.25 Compare and contrast fossil fuels and the Sun as direct sources of energy.

Ans

Fossil Fuels: The fuels such as coal, petroleum, natural gas etc. are the energy sources those are found under the Earth’s crust. Nowadays, these energy sources are being used in a large scale. These energy sources are limited in the nature. These are non-renewable sources of energy and require millions of years for their formation.

Solar Energy: It is a renewable and direct source of energy produced by the Sun. The Sun has been shining from several years. Solar energy is free of cost to all in a huge amount.

Q.26 Compare and contrast bio-mass and hydro electricity as sources of energy.

Ans

Bio-mass is a renewable source of energy which is obtained from dead plants and animal wastes. It is the result of natural process. Therefore, it can be naturally refilled.

Examples: Gobar-gas, wood, etc.

Hydro-electricity is also a renewable source of energy. It is obtained from the potential energy stored in water at a height.

Example: Electricity generated at dams.

Q.27 What are the limitations of extracting energy from-

(a) the wind?
(b) waves?
(c) tides?

Ans

(a) Wind mills are used to harness wind energy. A windmill cannot generate electricity if the speed of wind is less than 15 km/h. Moreover, a large number of windmills are required, which covers a large area.

(b) To extract energy from waves, very strong ocean waves are required.

(c) It requires high tides to extract energy from tides. Moreover, the occurrence of tides depends on the relative positions of Sun, Moon, and Earth.

Q.28 On what basis would you classify energy sources as

(a) renewable and non-renewable?
(b) exhaustible and inexhaustible?

Are the options given in (a) and (b) the same?

Ans

(a) The source of energy that cannot be exhausted in nature is called renewable source of energy.
Examples: Sun, wind, moving water, bio-mass, etc.

The source of energy that can be exhausted in nature is called non-renewable source of energy.
Examples: Coal, petroleum, natural gas, etc.

(b) Exhaustible sources are those sources of energy, which will deplete after a few hundred years later.
Examples: Coal, petroleum, etc.

Inexhaustible resources of energy are those sources which will not deplete in future.
Example: Bio-mass.

Yes. The options given in (a) and (b) are same.

Q.29 What are the qualities of an ideal source of energy?

Ans

A source of energy which is considered as an ideal source of energy must be economical, easily accessible, pollution free, easy to store and transport and able to produce huge amount of heat and energy on burning.

Q.30 What are the advantages and disadvantages of using a solar cooker? Are there places where solar cookers would have limited utility?

Ans

Advantages: Solar cooker requires energy of Sun to operate. Solar energy is inexhaustible renewable source of energy and free for all and available in nature in huge amount.

Disadvantages: Solar cookers are very expensive. They do not work in absence of sunlight. Therefore, on a cloudy day, it becomes useless.

The regions where days are very short or places with cloud covers round the year, have limited utility for solar cooker.

Q.31 What are the environmental consequences of the increasing demand for energy? What steps would you suggest to reduce energy consumption?

Ans

There is a huge demand of fossil fuels due to industrialisation in all over the world. Burning of fossil fuels releases many harmful gases in the atmosphere and increases the level of green house gas content in the atmosphere. It has increased the problem of global warming.

The consumption of fossil fuels cannot be completely reduced. However, some measures can be taken such as using electrical appliances wisely and not wasting electricity. In place of using our own vehicles, public transport system with mass transit must be adopted on a large scale.

Q.32 Which of the following correctly describes the magnetic field near a long straight wire?

(a) The field consists of straight lines perpendicular to the wire.
(b) The field consists of straight lines parallel to the wire.
(c) The field consists of radial lines originating from the wire.
(d) The field consists of concentric circles centred on the wire.

Ans

The correct option is (d).
Explanation: The magnetic field lines around a straight current-carrying conductor are concentric circles and the centres of these circles lie on the wire.

Q.33 The phenomenon of electromagnetic induction is

(a) the process of charging a body.
(b) the process of generating magnetic field due to a current passing through a coil.
(c) producing induced current in a coil due to relative motion between a magnet and the coil.
(d) the process of rotating a coil of an electric motor.

Ans

The correct option is (c).
Explanation: A current is induced in the coil when a coil and a magnet are moved relative to each other. This phenomenon is called electromagnetic induction.

Q.34 The device used for producing electric current is called a

(a) generator.
(b) galvanometer.
(c) ammeter.
(d) motor.

Ans

The correct option is (a).
Explanation: Electric current is generated by an electric generator. An electric generator converts mechanical energy into electric energy.

Q.35 The essential difference between an AC generator and a DC generator is that

(a) AC generator has an electromagnet while a DC generator has permanent magnet.
(b) DC generator will generate a higher voltage.
(c) AC generator will generate a higher voltage.
(d) AC generator has slip rings while the DC generator has a commutator.

Ans

The correct option is (d).
Explanation: The main difference between AC generator and DC generator is that an AC generator has two rings known as slip rings while a DC generator has two half rings known as commutator.

Q.36 At the time of short circuit, the current in the circuit
(a) reduces substantially.
(b) does not change.
(c) increases heavily.
(d) vary continuously.

Ans

The correct option is (c).

Explanation: In the case of short-circuit, the resistance of the circuit becomes zero. Hence, the magnitude of the current flowing through the circuit increases quickly.

Q.37 State whether the following statements are true or false.

(a) An electric motor converts mechanical energy into electrical energy.
(b) An electric generator works on the principle of electromagnetic induction.
(c) The field at the centre of a long circular coil carrying current will be parallel straight lines.
(d) A wire with a green insulation is usually the live wire of an electric supply.

Ans

(a) False
Explanation: An electric motor converts electrical energy into mechanical energy.

(b) True
Explanation: An electric generator produces electricity by rotating a coil in presence of a magnetic field. It works on the principle of electromagnetic induction.

(c) True
Explanation: A long circular coil behaves like a long solenoid. The magnetic field lines inside the solenoid are parallel to each other.

(d) False
Explanation: Earth wire has green insulation colour while live wire has red insulation cover.

Q.38 List two methods of producing magnetic fields.

Ans

Two methods for producing magnetic field are as follows:

(a) By using permanent magnets.
(b) By using a current carrying conductor.

Q.39 How does a solenoid behave like a magnet? Can you determine the north and south poles of a current–carrying solenoid with the help of a bar magnet? Explain.

Ans

Solenoid: It is a long coil of circular loops of insulated copper wire. When electric current is passed through it, the magnetic field lines are produced around it. The magnetic field produced by a solenoid is similar to the magnetic field of a bar magnet.

When the north pole of a bar magnet is brought near the end connected to the negative terminal of the battery, the solenoid repels the bar magnet. Since like poles repel each other, the end connected to the negative terminal of the battery behaves like the north pole of the solenoid and the other end behaves like a south pole. Therefore, one end of the solenoid behaves as a north pole and the other end behaves as a south pole.

Q.40 When is the force experienced by a current–carrying conductor placed in a magnetic field largest?

Ans

When the direction of electric current is perpendicular to the direction of the magnetic field, the force experienced by a current-carrying conductor is maximum.

Q.41 Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?

Ans

The direction of magnetic field can be found with the help of Fleming’s left hand rule. The direction of magnetic field inside the chamber would be perpendicular to the direction of current and direction of force and it would be either upward or downward. As the negatively charged electrons are moving from back wall to the front wall hence, the direction of current would be from the front wall to the back wall. The direction of magnetic force is rightward. Therefore, according to Fleming’s left hand rule, the direction of magnetic field inside the chamber would be downward.

Q.42 Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in an electric motor?

Ans

An electric motor is a device which converts electrical energy into mechanical energy. It is based on the principle that current carrying conductor experiences a force, when placed in a magnetic field.

When a current is allowed to flow through the coil ABCD by closing the key, the coil starts rotating anti-clockwise. It is because a downward force acts on length AB and at the same time, an upward force acts on length CD. Thus, the coil rotates anti-clockwise. Current in the length AB flows from A to B and the magnetic field acts from left to right, normal to length AB. Hence, according to Fleming’s left hand rule, a downward force acts on the length AB. Similarly, current in the length CD flows from C to D and the magnetic field acts from left to right, normal to the flow of current. Therefore, an upward force acts on the length CD. These two forces cause the coil to rotate anti-clockwise. After half a rotation, the position of AB and CD interchanges. The half-ring P comes in contact with brush X and half-ring Q comes in contact with brush Y. Hence, the direction of current in the coil ABCD gets reversed. The current flows through the coil in the direction DCBA. The reversal of current through the coil ABCD repeats after each half rotation. As a result, the coil rotates unidirectional. The split rings help to reverse the direction of current in the circuit. These rings are called as commutator.

Q.43 Name some devices in which electric motors are used.

Ans

The devices in which electric motors are used are washing machines, water pumps, electric fans and electric mixers.

Q.44 A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is (i) pushed into the coil, (ii) withdrawn from inside the coil, (iii) held stationary inside the coil?

Ans

A current is induced in a solenoid when a bar magnet is moved relative to it. This is the principle of electromagnetic induction.

(i) When a bar magnet is pushed into a coil, a current is induced for a moment. Therefore, the needle of the galvanometer deflects for a short time in a particular direction.

(ii) When the bar magnet is moved away from the coil, a current is again induced for a moment in opposite direction. Thus, the needle of the galvanometer deflects for a short time in the opposite direction.

(iii) When a bar magnet remains stationary inside the coil, no current will be induced in the coil. Therefore, galvanometer will show no deflection.

Q.45 Two circular coils A and B are placed closed to each other. If the current in the coil A is changed, will some current be induced in the coil B? Give reason.

Ans

On changing the magnitude of electric current in coil A, the magnetic field linked with it also changes and hence, the magnetic field around coil B also changes. Due to this, an electric current induces in coil B. This effect is called electromagnetic induction.

Q.46 State the rule to determine the direction of a (i) magnetic field produced around a straight conductor carrying current, (ii) force experienced by a current carrying straight conductor placed in a magnetic field which is perpendicular to it, and (iii) current induced in a coil due to its rotation in a magnetic field.

Ans

(i) Maxwell’s right hand thumb rule is used to find the direction of magnetic field produced around a straight conductor carrying current.

(ii) Fleming’s left hand rule is used to find the direction of force experienced by a current carrying straight conductor placed in a magnetic field which is perpendicular to it

(iii) Fleming’s right hand rule is used to find the direction of current induced in a coil due to its rotation in a magnetic field.

Q.47 Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of brushes?

Ans

It is an electrical device which converts mechanical energy into electrical energy.
Principle: When a loop is moved in a magnetic field, an electric current is induced in the coil. An electric generator produces electricity by rotating a coil in a magnetic field.

Working: It consists of a coil ABCD which is mounted on a rotor shaft. The coil’s axis of rotation is placed perpendicular to the magnetic field (N-S). The two other ends of the armature are connected separately to two split rings P and Q. The split rings are then connected to two conducting carbon brushes X and Y. A galvanometer G is connected with external circuit.

If the coil ABCD is rotated clockwise, then the length AB moves upwards while length CD moves downwards. Since the lengths AB and CD are moving in a magnetic field, a current will be induced in both of them due to electromagnetic induction. Length AB is moving upwards and the magnetic field acts from left to right. Hence, according to Fleming’s right hand rule, the direction of induced current will be from A to B. Similarly, the direction of induced current in the length CD will be from C to D.

The direction of current in the coil is ABCD. Thus, the galvanometer shows a deflection in a particular direction. After half a rotation, length AB starts moving down whereas length CD starts moving upward. The direction of the induced current in the coil gets reversed as DCBA. As the direction of current gets reversed after each half rotation, the produced current is called an alternating current.

In the case of DC generator, instead of slip rings, two split rings are used. In this arrangement, brush X always remains in contact with the length of the coil that is moving up whereas brush Y always remains in contact with the length that is moving down. The split rings P and Q act as a commutator.

The direction of current induced in the coil will be ABCD for the first rotation and DCBA in the second half of the rotation. Hence, a unidirectional current is produced in a DC generator.

Q.48 When does an electric short circuit occur?

Ans

When the resistance of an electric circuit becomes very low, the magnitude of electric current flowing through the circuit becomes very high. This happens when many appliances are connected to a single socket and thus, results short-circuiting. Moreover, when a live wire touches neutral wire directly, the current flowing in the circuit increases suddenly. Thus, a short circuit occurs.

Q.49 What is the function of an earth wire? Why is it necessary to earth metallic appliances?

Ans

An earth wire joins a metallic body of an electric appliance to the earth so that any leakage of electric current is transferred to the ground. This avoids any electric shock to the user. Thus, earthing of the electrical appliances is very important.

Q.50 The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to

(a) presbyopia.
(b) accommodation.
(c) near-sightedness.
(d) far-sightedness.

Ans

Answer: The correct option is (b).
Explanation: Power of accommodation is the ability of eye lens to change its focal length so that the image of an object can be focused on the retina.

Q.51 The human eye forms the image of an object at its

(a) cornea.
(b) iris.
(c) pupil.
(d) retina.

Ans

The correct option is (d).
Explanation: In the case of human eye, image is formed at retina.

Q.52 The least distance of distinct vision for a young adult with normal vision is about

(a) 25 m.
(b) 2.5 cm.
(c) 25 cm.
(d) 2.5 m.

Ans

The correct option is (c).
Explanation: Least distance of distinct vision is the smallest distance at which the human eye can see the objects clearly without any strain. For a normal eye, it is 25 cm.

Q.53 The change in focal length of an eye lens is caused by the action of the

(a) pupil.
(b) retina.
(c) ciliary muscles.
(d) iris.

Ans

The correct option is (c).
Explanation: The curvature of the eye lens can be changed by the relaxation or contraction of ciliary muscles. The change in curvature of the eye lens changes the focal length of the eyes. Thus, the change in focal length of an eye lens is caused by the action of ciliary muscles.

Q.54 A person needs a lens of power – 5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting
(i) distant vision, and
(ii) near vision?

Ans

$\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.17em}}\text{Power}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the lens used for correcting distance vision= -5.5}\\ \text{Focal length}\left(\text{f}\right)\text{\hspace{0.17em}}\text{of the required lens}\\ \text{f =}\frac{\text{1}}{\text{P}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{1}}{\text{-5}\text{.5}}\text{=-0}\text{.181}\text{\hspace{0.17em}}\text{m}\\ \text{Hence, the focal length of the lens for correcting distant vision = – 0.181 m.}\\ \left(\text{ii}\right)\text{â€‹}\text{\hspace{0.17em}}\text{Power}\text{â€‹}\text{\hspace{0.17em}}\text{of the lends used for correcting near vision = +1}\text{.5}\text{\hspace{0.17em}}\text{D}\\ \text{Focal length (f) of the required lens,}\\ \text{f =}\frac{\text{1}}{\text{P}}\\ \text{f}\text{\hspace{0.17em}}\text{=}\frac{\text{1}}{\text{1}\text{.5 D}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\text{0}\text{.667}\text{\hspace{0.17em}}\text{m}\\ \text{Hence, the focal length of the lens for correcting near vision is 0}\text{.667}\text{\hspace{0.17em}}\text{m}\text{.}\end{array}$

Q.55 The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

Ans

$\begin{array}{l}\text{Given,}\text{\hspace{0.17em}}\text{u}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{¥}\\ \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{v}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{-80}\text{\hspace{0.17em}}\text{cm}\\ \text{On}\text{\hspace{0.17em}}\text{using lens formula,}\\ \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}}\frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \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{–}\frac{\text{1}}{\text{80}}\text{–}\frac{\text{1}}{\text{¥}}\text{=}\frac{\text{1}}{\text{f}}\\ \text{or,}\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{f}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{-80}\text{\hspace{0.17em}}\text{cm}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{-0}\text{.8}\text{\hspace{0.17em}}\text{m}\\ \text{Now,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{P}\text{\hspace{0.17em}}\text{=}\frac{\text{1}}{\text{f}\left(\text{metres}\right)}\\ \text{}\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{P}\text{\hspace{0.17em}}\text{=}\frac{\text{1}}{\text{-0}\text{.8}}\text{=-1}\text{.25}\text{\hspace{0.17em}}\text{D}\text{.}\\ \text{A}\text{\hspace{0.17em}}\text{concave lens of power -1}\text{.25 D is required by the person to correct the defect}\text{.}\end{array}$

Q.56 Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Ans

It is a visual defect in which an eye cannot see the nearby objects clearly. The image of the object, in this case, is formed beyond the retina and hence person experiences difficulty in understanding the object.

This defect of vision can be corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina. The convex lens actually creates a virtual image of a nearby object (o’) at the near point of vision (o) of the person suffering from hypermetropia.
A person can clearly see the object kept at 25 cm, if the image of the object is formed at his near point, which is given as 1 m.

$\begin{array}{l}\text{Given,}\text{\hspace{0.17em}}\text{u}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{-25}\text{\hspace{0.17em}}\text{cm}\\ \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{v}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{-1}\text{\hspace{0.17em}}\text{m=-10}\text{\hspace{0.17em}}\text{cm}\\ \text{On}\text{\hspace{0.17em}}\text{usig}\text{\hspace{0.17em}}\text{lens}\text{\hspace{0.17em}}\text{formula,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{1}}{\text{-100}}\text{–}\frac{\text{1}}{\text{-25}}\text{=}\frac{\text{1}}{\text{f}}\\ \text{or,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f=}\frac{\text{100}}{\text{3}}\text{\hspace{0.17em}}\text{cm}\text{\hspace{0.17em}}\text{=33}\text{.3}\text{\hspace{0.17em}}\text{cm}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{0}\text{.33}\text{\hspace{0.17em}}\text{m}\\ \text{Now,}\text{\hspace{0.17em}}\text{P}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{1}}{\text{f}\left(\text{metres}\right)}\\ \therefore \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{P=}\frac{\text{1}}{\text{0}\text{.33m}}\text{=+3D}\text{.}\\ \text{Hence,}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{convex}\text{\hspace{0.17em}}\text{lens}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{power}\text{\hspace{0.17em}}\text{+3D}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{required}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{correct}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{defect}\text{.}\end{array}$

Q.57 Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Ans

As the ciliary muscles of eyes are incapable to contract beyond a certain limit hence, a normal eye cannot see clearly the objects kept closer than 25 cm. If an object is kept at a distance less than 25 cm from the eye, then the object seems blurred and produces strain in the eyes.

Q.58

argin-left: 0.5in; text-indent: -0.5in; vertical-align: baseline;”>What happens to the image distance in the eye when we increase the distance of an object from the eye?

Ans

For an eye, the image distance always remains constant as the size of eyes cannot increase or decrease. The increase in the object distance is balanced by the change in the focal length of the eye lens. The eye changes its focal length focal length in such a way that the image is always formed at retina.

Q.59 Why do stars twinkle?

Ans

Twinkling of stars is due to the atmospheric refraction of light. As stars are too far from the earth therefore, they can be considered as point sources of light. When the light coming from stars passes through the earth’s atmosphere, it gets refracted at different levels because of the variation in the air density at different levels of the atmosphere. When the star light refracted by the atmosphere, it comes more towards us. It appears brighter when it comes less towards us.

Q.60 Explain why the planets do not twinkle.

Ans

Planets do not twinkle like stars because planets are much closer to the earth and are thus seen as extended objects. If we consider planet as a collection of large number of point-sized sources of light, sources of light the variation in amount of light entering our eye from all the individual point sized sources will average out to zero, thereby nullifying the twinkling effect.

Q.61 Why does the Sun appear reddish early in the morning?

Ans

During sunrise, the light coming from the Sun has to travel a larger distance in the earth’s atmosphere before entering to our eyes. During this, the shorter wavelengths of lights are scattered out and only longer wavelengths are able to reach our eyes. As blue colour has a shorter wavelength and red colour has a longer wavelength, the red colour reaches to our eyes after the atmospheric scattering of light. Hence, the Sun seems reddish while sunrise.

Q.62

argin-left: 0.5in; text-indent: -0.5in; vertical-align: baseline;”>Why does the sky appear dark instead of blue to an astronaut?

Ans

There is no atmosphere in the outer space. Due to this, the sunlight does not scatter and hence, no scattered light reaches to the eyes of astronauts in the space and hence, the sky appears black to them.

Q.63 Which of the statements about the reaction below are incorrect?
2PbO(s) + C(s) →2Pb(s) + CO2(g)
(b) Carbon dioxide is getting oxidised.
(c) Carbon is getting oxidised.
(d) Lead oxide is getting reduced.
(i) (a) and (b)
(ii) (a) and (c)
(iii) (a), (b) and (c)
(iv) all

Ans

(i) (a) and (b)
Reason: In this reaction lead oxide is reduced and carbon is oxidised.

Q.64 Fe2O3+ 2Al →Al2O3+ 2Fe

The above reaction is an example of a
(a) combination reaction.
(b) double displacement reaction
(c) decomposition reaction.
(d) displacement reaction.

Ans

(d) displacement reaction.
Reason: Aluminium has replaced iron in iron oxide.

Q.65 What happens when dilute hydrochloric acid is added to iron fillings? Tick the correct answer.

(a) Hydrogen gas and iron chloride are produced.
(b) Chlorine gas and iron hydroxide are produced.
(c) No reaction takes place.
(d) Iron salt and water are produced.

Ans

(a) Hydrogen gas and iron chloride are produced. (âœ“)

Q.66 What is a balanced chemical equation? Why should chemical equations be balanced?

Ans

A chemical reaction is balanced when the number of atoms of each element on both sides of arrow is equal.

$\text{Mg}\left(\text{s}\right)+{\text{H}}_{2}{\text{SO}}_{4}\left(\text{aq}\right)\text{}\to {\text{MgSO}}_{4}\left(\text{aq}\right)+{\text{H}}_{2}\left(\text{g}\right)$

A balanced chemical equation tells us about:

1. The formula, symbols, names and physical states of the reactants and products.
2. The relative number of atoms and molecules of the reactants and products participating in the reaction.
3. The ratio of moles and masses of the reactants and products.
4. The ratio of volumes of gaseous reactants and products.

A chemical reaction should be balanced because according to law of conservation of mass, matter can neither be created nor be destroyed.

Q.67 Translate the following statements into chemical equations and then balance them.

(a) Hydrogen gas combines with nitrogen to form ammonia.
(b) Hydrogen sulphide gas burns in air to give water and sulpur dioxide.
(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.

Ans

(a) N2(g) + 3H2 (g) → 2NH3 (g)
(b) 2H2 S(g) + 3O2 (g) → 2H2O(l)+ 2SO2(g)
(c) 3BaCl2(aq) + Al2(SO4)3(aq) → 2 AlCl3 (aq) + 3BaSO4(s)
(d) 2K (s)+ 2H2O(l)+ → 2KOH(aq) + H2(g)

Q.68 Balance the following chemical equations.

(a) HNO3 + Ca(OH)2 → Ca(NO3)2+ H2O
(b) NaOH + H2SO4 → Na2SO4+ H2O
(c) NaCl + AgNO3 → AgCl + NaNO3
(d) BaCl2+ H2SO4 → BaSO4+ HCl

Ans

(a) 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O
(b) 2NaOH + H2SO4 → Na2SO4 + 2H2O
(c) NaCl + AgNO3 → AgCl + NaNO3
(d) BaCl2+ H2SO4 → BaSO4 + 2HCl

Q.69 Write the balanced chemical equations for the following reactions.

(a) Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water
(b) Zinc + Silver nitrate → Zinc nitrate + Silver
(c) Aluminium + Copper chloride → Aluminium chloride + Copper
(d) Barium chloride + Potassium sulphate → Barium sulphate + Potassium chloride

Ans

(a) Ca(OH)2 + CO2 → CaCO3+ H2O
(b) Zn + 2AgNO3 → Zn(NO3)2 + 2Ag
(c) 2Al + 3CuCl2 → 2AlCl3 + 3Cu
(d) BaCl2+ K2SO4 → BaSO4+ 2KCl

Q.70 Write the balanced chemical equation for the following and identify the type of reaction in each case.

(a) Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s)
(b) Zinc carbonate(s) → Zinc oxide(s) + Carbon dioxide(g)
(c) Hydrogen(g) + Chlorine(g) → Hydrogen chloride(g)
(d) Magnesium(s) + Hydrochloric acid(aq) → Magnesium chloride(aq) + Hydrogen(g)

Ans

(a) Double displacement reaction
2KBr(aq) + BaI2 (aq) → 2 KI(aq) + BaBr2(s)

(b) Decomposition reaction
ZnCO3(s) → ZnO(s) + CO2(g)

(c) Combination reaction
H2(g) + Cl2 (g) → 2HCl(g)

(d) Displacement reaction
Mg(s) +2HCl(aq) → MgCl2 (aq) + H2(g)

Q.71 What does one mean by exothermic and endothermic reactions? Give examples.

Ans

The reactions in which heat is released along with the formation of products are called exothermic reactions. For example,

C + O2 → CO2
CaO + H2O → Ca(OH)2

The reactions that require energy to occur are known as endothermic reactions.
Example – When silver chloride is left in the sunlight, it absorbs heat and turns grey because of formation of silver metal.

$2\text{AgCl}\left(\text{s}\right)\text{}\stackrel{\text{Sunlight}}{\to }\text{}2\text{Ag}\left(\text{s}\right)+{\text{Cl}}_{2}\left(\text{g}\right)$

Q.72 Why is respiration considered an exothermic reaction? Explain.

Ans

During respiration glucose is broken into carbon dioxide and water and energy is released. Therefore,it is an exothermic reaction.

Q.73 Why are decomposition reactions called the opposite of combination reactions? Write equations for these reactions.

Ans

In a decomposition reaction a single substance decomposes to form two or more substances. It is exactly opposite of combination reaction in which two or more reactants combine to form a single product. Therefore, decomposition reactions are called the opposite of combination reactions. For examples

Decomposition reactions:

$\begin{array}{l}2{\text{FeSO}}_{4}\left(\text{s}\right)\text{}\stackrel{\Delta }{\to }{\text{Fe}}_{2}{\text{O}}_{3}\left(\text{s}\right)+{\text{SO}}_{2}\left(\text{g}\right)+{\text{SO}}_{3}\left(\text{g}\right)\\ 2\text{Pb}{\left({\text{NO}}_{3}\right)}_{2}\left(\text{s}\right)\text{}\stackrel{\Delta }{\to }\text{}2\text{PbO}\left(\text{s}\right)+4{\text{NO}}_{2}\left(\text{g}\right)+{\text{O}}_{2}\left(\text{g}\right)\end{array}$

Combination Reactions:

$\begin{array}{l}{\text{N}}_{2}\left(\text{g}\right)+3{\text{H}}_{2}\left(\text{g}\right)\text{}\to \text{}2{\text{NH}}_{3}\left(\text{g}\right)\\ \text{Mg}\left(\text{s}\right)+{\text{Cl}}_{2}\left(\text{g}\right)\text{}\to {\text{MgCl}}_{2}\left(\text{s}\right)\end{array}$

Q.74 Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.

Ans

$\begin{array}{l}{\text{CaCO}}_{3}\left(\text{s}\right)\text{}\stackrel{\text{Heat}}{\to }\text{CaO}\left(\text{s}\right)+{\text{CO}}_{2}\left(\text{g}\right)\\ 2\text{AgCl}\left(\text{s}\right)\text{}\stackrel{\text{Sunlight}}{\to }\text{}2\text{Ag}\left(\text{s}\right)+{\text{Cl}}_{2}\left(\text{g}\right)\\ 2{\text{H}}_{2}\text{O}\left(1\right)\text{}\stackrel{\text{Eteceicity}}{\to }\text{}2{\text{H}}_{2}\left(\text{g}\right)+{\text{O}}_{2}\left(\text{g}\right)\end{array}$

Q.75 What is the difference between displacement and double displacement reactions? Write equations for these reactions.

Ans

In a displacement reaction an atom or a group of atoms present in a molecule is displaced by another atom while in a double displacement reaction two compounds exchange their ions to form two new compounds.

Displacement reaction
CuSO4(aq) + Fe(s) → Cu(s) + FeSO4(aq)

Double displacement reaction
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)

Q.76 In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reaction involved.

Ans

2 AgNO3(aq) + Cu (s) → Cu (NO3)2(aq) + 2Ag (s)

Q.77 What do you mean by a precipitation reaction? Explain by giving examples.

Ans

Chemical reactions in which precipitate is formed are called precipitation reactions. In these reactions, usually an insoluble salt is formed which settles down as precipitate.

Na2SO4 (aq) + BaCl2 (aq) → BaSO4 (s) + 2NaCl (aq)

In this reaction, when sodium sulphate reacts with barium chloride, the white precipitate of barium sulphate is formed along with the sodium chloride.

Q.78 Explain the following in terms of gain or loss of oxygen with two examples each.

(a) Oxidation
(b) Reduction

Ans

(a) Oxidation: The reaction that involves addition of oxygen to a substance is called an oxidation reaction.
For example: CuO (s) + H2 (g) → Cu (s) + H2O (g); H2 is oxidised to form H2O

(b) Reduction: The reaction that involves removal of oxygen from a substance is called a reduction reaction.
For example: ZnO (s) + C (s) → Zn (s) + CO (g); ZnO is reduced to form Zn

Q.79 A shiny brown coloured element ‘X’ on heating in air becomes black in colour.

Name the element ‘X’ and the black coloured compound formed.

Ans

The shiny brown coloured element is ‘Copper’. When it is heated in air, it becomes black in colour due to the formation of black coloured compound, copper oxide.

Q.80 Why do we apply paint on iron articles?

Ans

The paint prevents iron from coming in contact with air and moisture. Thus, painting prevents rusting of iron articles.

Q.81 Oil and fat containing food items are flushed with nitrogen. Why?

Ans

Nitrogen acts as an antioxidant and prevents oil and fat containing food items from being oxidised. Therefore, food items are flushed with nitrogen.

Q.82 Explain the following terms with one example each.

(a) Corrosion
(b) Rancidity

Ans

(a) Corrosion is a process in which the surface of metals are attacked by substances around it such as moisture, acids, chemicals, etc. The examples of corrosion are

1. Rusting of iron
2. Green coating on copper

(b) Rancidity is the deterioration of fatty and oily food by the oxidation leading to unpleasant smell and taste. Example: When packets of oily snacks are kept in open for a long time, their taste and smell become bad. Such oily food is no longer safe to eat.

Q.83 Which of the following pairs will give displacement reactions?

(a) NaCl solution and copper metal
(b) MgCl2 solution and aluminium metal
(c) FeSO4 solution and silver metal
(d) AgNO3 solution and copper metal.

Ans

(d) AgNO3 solution and copper metal

Q.84 Which of the following methods is suitable for preventing an iron frying pan from rusting?

(a) Applying grease
(b) Applying paint
(c) Applying a coating of zinc
(d) All of the above.

Ans

(c) Applying a coating of zinc

Q.85 An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be:

(a) calcium
(b) carbon
(c) silicon
(d) iron.

Ans

(a) Calcium

Q.86 Food cans are coated with tin and not with zinc because

(a) Zinc is costlier than tin.
(b) Zinc has a higher melting point than tin.
(c) Zinc is more reactive than tin.
(d) Zinc is less reactive than tin.

Ans

(c) Zinc is more reactive than tin which may react with food items and make it unfit for health.

Q.87 You are given a hammer, a battery, a bulb, wires and a switch.

(a) How could you use them to distinguish between samples of metals and non-metals?
(b) Assess the usefulness of these tests in distinguishing between metals and non-metals.

Ans

1. If a substance can be beaten into thin sheets with the help of a hammer then it is a metal, whereas if it gets broken into pieces then it is non-metal. We can use the battery, bulb, wires, and a switch to set up a circuit with the sample. If the sample conducts electricity and bulbs starts to glow, then it is a metal otherwise it is a non-metal.

2. When a substance fulfills both the criteria then it can be confirmed as a metal. We know that there are some exceptions also for example sodium is metal which is not malleable in fact it is brittle. Graphite, a non-metal (allotrope of carbon) is a good conductor of electricity. Hence, either of the tests cannot confirm a metal or non-metal; when the test is done in isolation.

Q.88 What are amphoteric oxides? Give two examples of amphoteric oxides.

Ans

Amphoteric oxides are the oxides, which react with both acids and bases to form salt and water. Examples: Zinc oxide (ZnO) and Aluminium oxide (Al2O3)

Q.89 Name two metals which will displace hydrogen from dilute acids, and two metals which will not.

Ans

Metals that are more reactive than hydrogen displace it from dilute acids. For example, sodium and potassium displace hydrogen from dilute acids. On the other hand less reactive metals like copper, silver do not displace hydrogen from dilute acids.

Q.90 In the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte?

Ans

In the electrolytic refining of a metal M:
Anode is impure, thick block of metal M
Cathode is thin strip or wire of pure metal M
Electrolyte is salt solution of metal M to be refined

Q.91 Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it, as shown in figure below.

(a) What will be the action of gas on:
(i) dry litmus paper?
(ii) moist litmus paper?

(b) Write a balanced chemical equation for the reaction taking place.

Ans

(a) When sulphur is burnt in air then sulphur dioxide gas is formed.
(i) Sulphur dioxide gas has no action on dry litmus paper.
(ii) Sulphur dioxide gas turns moist blue litmus paper red because sulphur dioxide reacts with moisture to form sulphurous acid.

$\text{(b)S}\left(\text{s}\right)+{\text{O}}_{2}\left(\text{g}\right)\text{}\to \text{}\underset{\text{sulphurdioxide}}{{\text{SO}}_{2}\left(\text{g}\right)}$

Q.92 State two ways to prevent the rusting of iron.

Ans

The two ways by which rusting of iron can be prevented are:

1. By oiling, greasing or painting the surface becomes waterproof and the moisture and oxygen present in the air cannot come into direct contact with iron. Hence, rusting is prevented.
2. By Galvanization: In this method an iron article is coated with a layer of zinc metal, which prevents the iron from coming in contact with oxygen and moisture. Hence, rusting is prevented.

Q.93 What type of oxides are formed when non-metals combine with oxygen?

Ans

Non-metals combine with oxygen to form acidic oxides or neutral oxides. Examples of acidic oxides are SO2, CO2 etc. and examples of neutral oxides are NO, CO etc.

Q.94 Give reasons:

(a) Platinum, gold and silver are used to make jewellery.
(b) Sodium, potassium and lithium are stored under oil.
(c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking.
(d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction.

Ans

1. Platinum, gold, and silver are used to make jewellery because they are very less reactive metals. Also they are lustrous and do not corrode easily.
2. Sodium, potassium, and lithium are very reactive metals. They react vigorously with air as well as water; therefore, they are kept immersed in kerosene oil in order to prevent their contact with air and moisture.
3. Aluminium is a highly reactive metal and is resistant to corrosion. This is because aluminium reacts with oxygen present in air to form a thin layer of aluminium oxide. This oxide layer is very stable and prevents further reaction of aluminium with oxygen. Also, it is light in weight and a good conductor of heat. Hence, it is used to make cooking utensils.
4. Carbonate and sulphide ores are usually converted into oxides during the process of extraction because it is easier to obtain metals from their oxides as compared to their carbonates and sulphides.

Q.95 You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.

Ans

Copper reacts with moist carbon dioxide in air to form copper carbonate and as a result, copper vessel loses its shiny brown surface forming a green layer of copper carbonate. The sour substances like lemon or tamarind contain citric acid that neutralises the basis copper carbonate and dissolves the layer. That is why; tarnished copper vessels are cleaned with lemon or tamarind juice to give the surface of the copper vessel its characteristic lustre.

Q.96 Differentiate between metal and non-metal on the basis of their chemical properties.

Ans

 Metals Non-metals Metals are electropositive. They lose electron readily to form a cation. Non metals are electronegative. They gain electron readily to form anion. Metals are lustrous. Non-metals are non-lustrous except graphite. Metals are good conductors of heat and electricity. Non-metals are non-conductors of heat and electricity except graphite. Metals react with oxygen to form basic oxide. $4\text{Na}+\text{O}\to \text{}2{\text{Na}}_{2}\text{O}$ These have ionic bond. Non-metals react with oxygen to form acidic or neutral oxide oxides. $\begin{array}{l}2\text{C}+{\text{O}}_{2}\text{}\to \text{}2\text{CO(neutral oxide)}\\ \text{C}+{\text{O}}_{2}\text{}\to {\text{CO}}_{2}\text{(acidic oxide)}\end{array}$ These have covalent bond. Metals react with water to form oxides and hydroxides. Some metals react with cold water, some with hot water, and some with steam. $2\text{Na}+2{\text{H}}_{2}\text{O}\to \text{}2\text{NaOH}+{\text{H}}_{2}↑$ They do not react with water. Metals react with dilute acids to form a salt and evolve hydrogen gas. However, Cu, Ag, Au, Pt, Hg do not react. $2\text{Na}+2\text{HCl}\to \text{}2\text{NaCl}+{\text{H}}_{2}↑$ Non-metals do not react with dilute acids. These are not capable of replacing hydrogen. Metals act as reducing agents as they can easily lose electrons. $\text{Na}\to {\text{Na}}^{+}+{\text{e}}^{-}$ Non-metals act as oxidizing agents as they can gain electrons. ${\text{Cl}}_{2}+2{\text{e}}^{-}\text{}\to \text{}2{\text{Cl}}^{-}$

Q.97 A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used?

Ans

The solution he had used was Aqua regia which is the mixture of concentrated hydrochloric acid and concentrated nitric acid in the ratio of 3:1. It is a fuming, highly corrosive liquid that is capable of dissolving metals like Gold and Platinum. Since the outer layer of the gold bangles is dissolved in aqua regia so their weight was reduced drastically.

Q.98 Give reasons why copper is used to make hot water tanks and not steel (an alloy of iron).

Ans

Copper is used to make hot water tanks and not steel because copper does not react with cold water, hot water or steam. However, iron reacts with steam. If the hot water tanks are made of steel (an alloy of iron), then iron would react vigorously with the steam formed from hot water.

$3\text{Fe}+4{\text{H}}_{2}\text{O}\to {\text{Fe}}_{3}{\text{O}}_{4}+4{\text{H}}_{2}$

That is why copper is used to make hot water tanks, and not steel.

Q.99 Which of the following statements is not a correct statement about the trends when going from left to right across the periods of periodic table.

(a) The elements become less metallic in nature.
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.

Ans

(c) The atoms lose their electrons more easily.
Reason : When we move from left to right across the periods of the periodic table, the non-metallic character increases. Therefore, the tendency to lose electrons decreases.

Q.100 Element X forms a chloride with the formula XCl2, which is a solid with a high melting point. X would most likely be in the same group of the Periodic Table as

(a) Na (b) Mg (c) AI (d) Si

Ans

(b) Mg

Reason: X would most likely be in the same group as of the Magnesium (Mg) in the Periodic Table because the formula of chloride of X (XCl2) suggests that the valency of X is +2.

Q.101 Which element has
(a) two shells, both of which are completely filled with electrons?
(b) the electronic configuration 2, 8, 2?
(c) a total of three shells, with four electrons in its valence shell?
(d) a total of two shells, with three electrons in its valence shell?
(e) twice as many electrons in its second shell as in its first shell?

Ans

(a) Neon. It has two shells, both of which are completely filled with electrons (2 electrons in K shell and 8 electrons in L shell).

(b) Magnesium. Its electronic configuration is 2, 8, 2.

(c) Silicon. It has a total of three shells, with four electrons in its valence shell (2 electrons in K shell, 8 electrons in L shell and 4 electrons in M shell).

(d) Boron. It has a total of two shells, with three electrons in its valence shell (2 electrons in K shell and 3 electrons in L shell).

(e) Carbon. It has twice as many electrons in its second shell as in its first shell (2 electrons in K shell and 4 electrons in L shell).

Q.102 (a) What property do all elements in the same column of the Periodic Table as boron have in common?
(b) What property do all elements in the same column of the Periodic Table as fluorine have in common?

Ans

(a) All the elements in the same column as boron have the same number of valence electrons (3).

(b) All the elements in the same column as fluorine have the same number of valence electrons (7). All of them will have valency equal to -1.

Q.103 An atom has electronic configuration 2, 8, 7.

(a) What is the atomic number of this element?
(b) To which of the following elements would it be chemically similar?

(Atomic numbers are given in parentheses.)
N(7) F(9) P(15) Ar(18)

Ans

(a) The atomic number of this element is 17.
(b) It would be chemically similar to F (atomic number 9) having configuration as 2, 7 which means same number of valence electrons as the given element has.

Q.104 The position of three elements A, B and C in the Periodic Table are shown below –

 Group 16 Group 17 — — — A — — B C

(a) State whether A is a metal or non-metal.
(b) State whether C is more reactive or less reactive than A.
(c) Will C be larger or smaller in size than B?
(d) Which type of ion, cation or anion, will be formed by element A?

Ans

(a) A is a non-metal.
(b) C is less reactive than A because reactivity decreases down the group in halogens.
(c) C will be smaller in size than B because on moving from left to right in a period, the nuclear charge increases and therefore, electrons come closer to the nucleus.
(d) Element A will form an anion because it accepts an electron to complete its octet.

Q.105 Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the Periodic Table. Write the electronic configuration of these two elements. Which of these will be more electronegative? Why?

Ans

Electronic configuration of Nitrogen is 2,5 and that of Phosphorus is 2, 8, 5.

Nitrogen will be more electronegative than phosphorus because electronegativity decreases on moving from top to bottom in a group.

Q.106 How does the electronic configuration of an atom relate to its position in the Modern Periodic Table?

Ans

Electronic configuration of an element gives the information about number of shells and the number of valence electrons present in the element.

We can find the group number with the help of number of valence electrons.

Number of shells present in an element is equal to period in which this element is placed.

Thus, by knowing electronic configuration we know the group number and period number of an element, which indicates the position of an element in the periodic table.

Q.107 In the Modern Periodic Table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12, 19, 21 and 38. Which of these have physical and chemical properties resembling calcium?

Ans

The element with atomic number 12 (electronic configuration 2,8,2) and 38 (electronic configuration 2,8,2) have same chemical properties as that of calcium because both of them have same number of valence electrons (2) like calcium.

Q.108 Compare and contrast the arrangement of elements in Mendeléev’s Periodic Table and the Modern Periodic Table.

Ans

 S.No. Mendeleev’s periodic table Modern periodic table 1. Elements are arranged in the increasing order of their atomic masses. Elements are arranged in the increasing order of their atomic numbers. 2. No place for isotopes. No such problem occurs as elements are arranged on the basis of their atomic numbers. 3. Groups were divided in subgroups A and B. Groups are not subdivided. 4. It has 8 groups. It has 18 groups.

Q.109 Which one of the following materials cannot be used to make a lens?

(a) Water
(b) Glass
(c) Plastic
(d) Clay

Ans

The correct option is (d).
Explanation: The light can pass through a lens. As clay does not allow light to pass through it hence, it cannot be used to make a lens.

Q.110 The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?

(a) Between the principle focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principle focus.

Ans

The correct option is (d).
Explanation: Concave mirror forms a virtual, erect and enlarged image of an object when the object is placed between the pole and focus of the mirror.

Q.111 Where should an object be placed in front of a convex lens to get a real image of the size of the object?

(a) At the principle focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principle focus.

Ans

The correct option is (b).

Explanation: On placing the object at a distance twice of the focal length (at centre of curvature) in front of a convex lens, the image is formed at the centre of curvature on the other side of the lens. This image is real, inverted, and of the same size as the object.

Q.112 A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be

(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex and the lens is concave.

Ans

The correct option is (a).
Explanation: As per sign convention, the focal length of a concave mirror and a concave lens are considered as negative.

Q.113 No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be-

(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.

Ans

The correct option is (d).
Explanation: A plane mirror always forms a virtual and erect image of same size as that of the object. In the same way, a convex mirror forms a virtual and erect image of smaller size of the object placed in front of it.

Q.114 Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

(a) A convex lens of focal length 50 cm
(b) A concave lens of focal length of 50 cm
(c) A convex lens of focal length of 5 cm
(d) A concave lens of focal length 5 cm

Ans

The correct option is (c).
Explanation: When a convex lens is placed between the radius of curvature and focal length, it forms magnified image of the given object. Moreover, magnification is more for convex lenses having shorter focal length.

Q.115 We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Ans

A concave mirror always forms an erect image when object is placed between pole (P) and the principal focus (F). Therefore, the object should be placed anywhere between the pole and the focus of the concave mirror. The image formed in this case will be virtual, erect, and magnified in nature.

Q.116 Name the type of mirror used in the following situations.

(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace

Ans

(a) Concave mirror
(b) Convex mirror
(c) Concave mirror.

Reasons:
(a) In the headlights of cars, concave mirror is used as it produces parallel beam of light when the light source (bulb) is at the principal focus of the concave mirror.
(b) A convex mirror forms a virtual, erect, and diminished image of the objects placed in front of. Hence, a driver can see most of the traffic behind him.
(c) A concave mirror can converge the light incident on it at a single point known as principal focus. Therefore, it can be used to produce a large amount of heat at that point.

Q.117 One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your
observations.

Ans

The convex lens will produce a compete image of the object kept in front of it even half of the lens is covered with black paper.

i. When the upper half of the lens is covered:

The rays of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object.

ii. When the lower half of the lens is covered:

The rays of light coming from the object are refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object.

Q.118 An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Ans

$\begin{array}{l}\text{Given},\text{object distance (u) = -25 cm}\\ {\text{object height (h}}_{\text{o}}\text{)=5 cm}\\ \text{focal length (f) = +10 cm}\\ \text{As per lens formula,}\\ \text{}\frac{\text{1}}{\text{v}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\frac{\text{1}}{\text{u}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{\text{1}}{\text{f}}\\ ⇒\text{}\frac{\text{1}}{\text{v}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{1}}{\text{f}}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\frac{\text{1}}{\text{u}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{1}}{\text{10}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\frac{\text{1}}{\text{25}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{15}}{\text{250}}\\ ⇒\text{v}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{250}}{\text{15}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{16}\text{.66 cm}\\ \text{As image distance is positive hence, the image is}\\ \text{formed at the other side of the lens}\text{.}\\ \text{Magnification (m) =}\frac{image\text{distance}}{object\text{distance}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}-\frac{\text{v}}{\text{u}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}-\frac{\text{16}\text{.66}}{\text{25}}\\ \text{=}\text{\hspace{0.17em}}-\text{0}\text{.66}\\ \text{Here, negative sign shows that the image is real and}\\ \text{formed behind the lens}\text{.}\\ \text{Again, Magnification (m)}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{image height}}{\text{object height}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{{\text{h}}_{\text{i}}}{{\text{h}}_{\text{o}}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{{\text{h}}_{\text{i}}}{\text{5}}\\ {\text{or, h}}_{\text{i}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{m}\text{\hspace{0.17em}}\text{×}\text{\hspace{0.17em}}{\text{h}}_{\text{o}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}-\text{0}\text{.66}\text{\hspace{0.17em}}\text{×}\text{\hspace{0.17em}}\text{5}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}-\text{3}\text{.3 cm}\\ \text{Hence, the image formed is inverted}\text{.}\end{array}$

Q.119 A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Ans

$Given, focal length of concave lens ( f )= −15 cm image distance ( v )= −10 cm On using lens formula, 1 v – 1 u = 1 f ⇒ 1 v – 1 u = 1 f =− 1 10 cm − 1 ( -15 cm ) =− 5 150 cm ⇒ u =−30 cm Here, negative object distance shows that the object is placed 30 cm in front of the lens. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@3563@$

Q.120 An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Ans

$Given, focal length ( f ) = +15 cm object distance ( u ) = −10 cm As per mirror formula, 1 v + 1 u = 1 f ⇒ 1 v = 1 f − 1 u = 1 15 cm − 1 ( -10 cm ) = 25 150 cm v = 6 cm As ( v ) is positive hence, image is formed behind the mirror. Magnification ( m )= image distance object distance =− v 4 =− 6 ( −10 ) =+0.6 Here, positive sign shows that the image formed is virtual and erect. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6A3C@$

Q.121 The magnification produced by a plane mirror is +1. What does this mean?

Ans

$\begin{array}{l}For\text{a mirror, magnification (m) is given as,}\\ \text{m =}\frac{{h}_{i}}{{h}_{o}}\\ Where,{\text{h}}_{\text{i}}={\text{image height , h}}_{\text{o}}=\text{object height}\\ \text{When magnification produced by a plane mirror is +1,}\\ \text{the image formed would be of the same size as that}\\ \text{of object}\text{. Also, the positive sign indicates that the}\\ \text{image formed is virtual and erect}\text{.}\end{array}$

Q.122 An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature
and size.

Ans

$\begin{array}{l}\text{Given,}\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{u}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{-20}\text{\hspace{0.17em}}\text{cm,}\text{\hspace{0.17em}}{\text{h}}_{\text{o}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{5}\text{\hspace{0.17em}}\text{cm,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{R}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{30}\text{\hspace{0.17em}}\text{cm}\\ \text{focal}\text{\hspace{0.17em}}\text{length}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{(f)}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{R}}{\text{2}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{30 cm}}{\text{2 cm}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{15}\text{\hspace{0.17em}}\text{cm}\\ \text{As}\text{\hspace{0.17em}}\text{per}\text{\hspace{0.17em}}\text{mirror}\text{\hspace{0.17em}}\text{formula,}\\ \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}}\frac{\text{1}}{\text{v}}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\frac{\text{1}}{\text{u}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{1}}{\text{f}}\\ ⇒\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}}\frac{\text{1}}{\text{v}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{1}}{\text{f}}\text{\hspace{0.17em}}\text{–}\text{\hspace{0.17em}}\frac{\text{1}}{\text{u}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{1}}{\text{15 cm}}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\frac{\text{1}}{\text{20 cm}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{7}}{\text{60 cm}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{v =}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{8}\text{.57}\text{\hspace{0.17em}}\text{cm}\\ \text{As}\text{\hspace{0.17em}}\text{v}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{positive}\text{\hspace{0.17em}}\text{hence,}\text{\hspace{0.17em}}\text{image}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{formed}\text{\hspace{0.17em}}\text{behind}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{mirror}\text{.}\\ \text{Magnification}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{(m)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{–}\frac{\text{v}}{\text{u}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{–}\text{\hspace{0.17em}}\frac{\text{8}\text{.57 cm}}{\text{(-20 cm)}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{0}\text{.428}\\ \text{The}\text{\hspace{0.17em}}\text{positive}\text{\hspace{0.17em}}\text{value}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{magnification}\text{\hspace{0.17em}}\text{shows}\text{\hspace{0.17em}}\text{that}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{image}\text{\hspace{0.17em}}\text{formed}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{virtual}\text{.}\\ \text{Again,}\text{\hspace{0.17em}}\text{Magnification}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{(m)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{\text{h}}_{\text{i}}}{{\text{h}}_{\text{o}}}\\ \text{or,}\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{h}}_{\text{i}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{m × h}}_{\text{o}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{.428}\text{\hspace{0.17em}}\text{×5}\text{cm}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\text{.14}\text{\hspace{0.17em}}\text{cm}\\ \text{The}\text{\hspace{0.17em}}\text{positive}\text{\hspace{0.17em}}\text{value}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{image}\text{\hspace{0.17em}}\text{height}\text{\hspace{0.17em}}\text{indicates}\text{\hspace{0.17em}}\text{that}\text{\hspace{0.17em}}\text{the image}\text{\hspace{0.17em}}\text{formed}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{erect}\text{.}\\ \text{Therefore,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{image}\text{\hspace{0.17em}}\text{formed}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{virtual,}\text{\hspace{0.17em}}\text{erect}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{smaller in}\text{\hspace{0.17em}}\text{size}\text{.}\end{array}$

Q.123 An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.

Ans

$\begin{array}{l}\mathrm{Given},\text{u = -27 cm}\\ {\text{h}}_{\text{o}}=\text{7 cm}\\ \text{f = -18 cm}\\ \text{As per mirror formula,}\\ \text{}\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\\ ⇒\text{}\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}}=\frac{1}{-18}-\frac{1}{\left(-27\right)}=-\frac{1}{54}\\ \text{v= -54 cm}\\ \mathrm{Hence},\text{the screen should be placed at a distance of}\\ \text{54 cm in front of the given mirror.}\\ \text{Magnification (m) =}-\frac{\mathrm{v}}{\mathrm{u}}=-\frac{\left(-54\right)}{\left(-27\right)}=-2\\ \mathrm{Here},\text{negative sign shows that the image formed is real.}\\ \mathrm{Again},\text{m =}\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}}\\ \mathrm{or},{\text{h}}_{\text{i}}=\text{m}×{\mathrm{h}}_{\mathrm{o}}=7×\left(-2\right)=-14\text{}\mathrm{cm}\\ \mathrm{The}\text{negative value of image height shows that the image}\\ \text{formed is inverted.}\end{array}$

Q.124 Find the focal length of a lens of power – 2.0 D. What type of lens is this?

Ans

$\begin{array}{l}\text{Given, power of lens, P = -2 D}\\ \text{P =}\frac{1}{\text{f}\left(\text{metres}\right)}\\ \therefore \text{f =}\frac{1}{2D}=–0.5\text{m}\\ \text{As a concave lens has a negative focal length.}\\ \text{Hence, this lens is a concave.}\end{array}$

Q.125 A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Ans

$\begin{array}{l}\text{Given, power of lens P = 1}\text{.5 D}\\ \text{On using the formula for power}\\ \text{P =}\frac{1}{\text{f}\left(\text{metres}\right)}\text{}\\ \therefore \text{f =}\frac{1}{1.5 D}=0.66\text{m}\\ \text{As a convex lens has a positive focal length}\text{. Hence it is convex lens}\text{.}\end{array}$

Q.126 A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw

Ans

(c) TtWW

Explanation: Since all the progeny bore violet flowers, it means that the tall plant having violet flowers has WW genotype for violet flower colour. Since the progeny is both tall and short, the parent plant was not a pure tall plant; its genotype must be Tt. Therefore, the genetic makeup of the tall parent can be depicted as TtWW.

Gametes of TtWW (Tall, violet flower) will be TW, TW, tW and tW.

Gametes of ttww (Short, white flower) will be tw, tw, tw and tw.

On crossing-

Therefore, half the progeny will be tall and half will be short but all of them have violet flowers.

Q.127 An example of homologous organs is

(a) our arm and a dog’s fore-leg.
(b) our teeth and an elephant’s tusks.
(c) potato and runners of grass.
(d) all of the above.

Ans

(d) all of the above
Explanation: Homologous organs are the organs having the same structure, origin and constituting parts but different functions.

Q.128 In evolutionary terms, we have more in common with

(a) a Chinese school-boy.
(b) a chimpanzee.
(c) a spider.
(d) a bacterium.

Ans

(a) A Chinese school-boy

Explanation: All the human beings present on the earth belong to a species Homo sapiens.

Q.129 A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?

Ans

Let us assume that children with light coloured eyes can either have LL or Ll or ll genotype. If the children have LL genotype, then their parents will also be of LL genotype

LL X LL

LL

If the children with light-colouered eyes have ll genotype, then their parents will also have ll genotype.

ll X ll

ll

Therefore, from the above given information, it cannot be concluded whether light eye colour is dominant or recessive.

Q.130 How are the areas of study – evolution and classification – interlinked?

Ans

Genetic variations form the basis for evolution and play an important role in the origin of new species. Because of similarity in their inherited body designs organisms appear to be the same. The two species are said to be closely related if they have characteristics in common. And they are likely to have a common ancestor more recently. Based on the similarities and differences of characteristic, classification of organisms necessarily involves, organising them in different groups. Among diverse species, classification of organism helps us in recognising the fundamental arrangement of a hierarchical structure. Classification helps us in establishing the resemblances and relationships between various organisms and also facilitates studies or research of wide variety related with organisms. In fact, classification of species is an expression of their evolutionary relationship. Thus, we can say the areas of study – evolution and classification are interlinked.

Q.131 Explain the terms analogous and homologous organs with examples.

Ans

Analogous organs: The organs having the same functions but different in structure, origin and constituting parts are called analogous organs. For example, the wings of bats and the wings of birds are analogous organs as their origins, designs, structure and components are different but they have a common use for flying.

Homologous organs: The organs having the same structure, origin and constituting parts but different functions are called homologous organs. For example: the forearms of a horse and the hands of human.

Q.132 Outline a project which aims to find the dominant coat colour in dogs.

Ans

Let us assume that a male dog having genotype BB (black skin coat) is mating with a female dog having genotype bb (white skin coat).

Offspring produced by this mating are all black and have a genetic make-up Bb.

Hence, it can be concluded now that black skin coat is dominant over the white skin coat.

Q.133 Explain the importance of fossils in deciding evolutionary relationships.

Ans

Fossils are the remains or preserved traces of organisms, which died millions of years ago. The study of fossils has helped us in establishing evolutionary link between the two species. The study of fossils also helps in identifying the origin of new species from the existing one. So, fossils have an importance in deciding evolutionary relationship.

Q.134 What evidence do we have for the origin of life from inanimate matter?

Ans

An experiment conducted in 1953, by Stanley L Miller and Harold C Urey serves the evidence for the origin of life from inanimate matter. In the experiment, they created an atmosphere containing molecules like ammonia, methane and hydrogen sulphide, but no oxygen. This was similar to the atmosphere that was thought to exist on the early earth. The gases were maintained at a temperature just below 100°C and to stimulate lightening sparks were passed through the mixture of gases. At the end of the week, almost 15 % of the carbon from methane gas had been transformed into simple compounds of carbon including amino acids which make up protein molecules and support life in its basic form. Thus, suggesting that life arose anew on earth.

Q.135 Explain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually?

Ans

Sexual reproduction gives rise to more viable variations than asexual reproduction. It is because of the inbuilt tendency to variation during reproduction, both because of errors in DNA copying, and modes of sexual reproduction. The genes of the traits or characteristics are transmitted from one generation to the next generation and dominant characters are expressed while asexual reproduction not only helps in survival of species but also support diversity in long run. Drift in genetic traits get accumulated spanning across generations, this gives rise to formation of new species.

Q.136 How is the equal genetic contribution of male and female parents ensured in the progeny?

Ans

Equal genetic contribution of male and female parents is ensured in the progeny through inheritance of equal number of chromosomes from both parents. For example, in humans, there are 23 pairs of chromosomes. Out of these 23 pairs, the 22 pairs are known as autosomes and the remaining one pair is known as sex chromosomes represented as X and Y. Females have a pair of two X sex chromosomes whereas males have a pair of one X and one Y chromosome.

The male parent produces gametes having 22 autosomes and one X or Y chromosome where as the female parent produces gamete having 22 autosomes and one X chromosome. During fertilisation, the male gamete fuses with the female gamete resulting in the formation of zygote. The zygote in the progeny receives an equal contribution of genetic material from both the parents.

Q.137 Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?

Ans

We agree with the statement that only variations that confer an advantage to an individual organism will survive in a population. All the variations do not survive in the environment. The chances of their survival depend on the acceptance or rejection by the nature. Different individual have different kinds of variations. A polar bear can withstand extreme cold, will survive better in cold regions. Selection of variants by nature establishes the basis for evolution.

Q.138 The kidneys in human beings are a part of the system for

1. nutrition.
2. respiration.
3. excretion.
4. transportation.

Ans

c) Excretion

Explanation: Kidney is a bean-shaped organ. It helps in the removal of metabolic wastes.

Q.139 The xylem in plants are responsible for

1. transport of water.
2. transport of food.
3. transport of amino acids.
4. transport of oxygen.

Ans

a) transport of water

Explanation: Xylem is the vascular tissue that facilitates the movement of water in plants. Movement of food and amino acids takes place through the vascular tissue phloem. Transport of oxygen in plants occurs through stomata.

Q.140 The autotrophic mode of nutrition requires

1. carbon dioxide and water.
2. chlorophyll.
3. sunlight.
4. all of the above.

Ans

d) all of the above

Explanation: Autotrophic nutrition takes place by the process of photosynthesis. Carbon dioxide, water, chlorophyll and sunlight are required to carry out the process of photosynthesis.

Q.141 The breakdown of pyruvate to give carbon dioxide, water and energy takes place in

1. cytoplasm.
2. mitochondria.
3. chloroplast.
4. nucleus.

Ans

a) mitochondria.

Explanation: Breakdown of glucose into pyruvate takes place in cytoplasm. Then further breakdown of pyruvate into CO2, water and energy takes place in mitochondria.

Q.142 How are fats digested in our bodies? Where does this process take place?

Ans

From the stomach, partially digested food, that is acidic, enters the small intestine. Bile juice is released into the small intestine through the gall bladder. The acidity of the partially digested food is nutralised by the bile salts present in the bile juices. Bile salts also emulsify fat molecules by breaking their large globules into many small globules so that the pancreatic enzymes can act effectively on them. Pancreatic enzyme like lipase present in pancreatic juice digests fats and converts them into fatty acid and glycerol.

This entire process takes place in the small intestine.

Q.143 What is the role of saliva in the digestion of the food?

Ans

Saliva, a secretion of salivary glands, helps in digestion of food in the following two ways:

1. The saliva moistens the food and helps in its easy swallowing and movement in oesophagus.
2. The saliva contains an enzyme called salivary amylase that initiates the breakdown of carbohydrates such as starch into sugar.

Q.144 What are the necessary conditions for autotrophic nutrition and what are its by products?

Ans

The necessary conditions for autotrophic nutrition are the availability of:
1. Carbon dioxide
2. Sunlight (source of energy)
3. Chlorophyll pigment (that captures energy)
4. Water
The products formed are carbohydrate (glucose) and oxygen.

$6{\text{CO}}_{2}\text{}+\text{}6{\text{H}}_{2}\text{O}\underset{\text{chlorophyll}}{\overset{\text{light energy}}{\to }}{\text{C}}_{6}{\text{H}}_{12}{\text{O}}_{6}+6{\text{O}}_{2}$

Q.145 What are the differences between aerobic and anaerobic respiration. Name some organisms that use the anaerobic mode of respiration.

Ans

Differences between aerobic respiration and anaerobic respiration are:

 Aerobic respiration Anaerobic respiration It occurs in the presence of oxygen. It occurs in the absence of oxygen. The final products are carbon dioxide, energy and water The final products are alcohol, carbon dioxide and energy The amount of energy released is high The amount of energy released is lower than aerobic respiration. It takes place in cytoplasm and mitochondria of a cell. It takes place in cytoplasm only.

Some of the organisms that use anaerobic mode of respiration are bacteria and fungi like yeast.

Q.146 How are the alveoli designed to maximise the exchange of gases?

Ans

The alveoli are small balloon-like structures present in the lungs. The walls of alveoli consist of extensive network of blood capillaries. Each lung contains 300-350million alveoli. The alveolar surface when spread out covers the area about 80m2. This large surface area makes the gaseous exchange efficient.

Q.147 What would be the consequences of a deficiency of haemoglobin in our bodies?

Ans

Haemoglobin is the respiratory pigment in humans. It facilitates the transportation of oxygen in the body cells to carry out cellular respiration. Thus, deficiency of haemoglobin can directly affect the oxygen supplying capacity of the blood which can further lead to deficiency of oxygen in our body cells. The deficiency of haemoglobin also causes disease such as anaemia.

Q.148 Describe double circulation in human beings. Why is it necessary?

Ans

Circulation in human beings comprises two phases: pulmonary circulation and systemic circulation. During each cycle of the circulation, the blood goes through the heart twice.
Flow of blood in the heart
The human heart comprises four chambers: the right atrium, the right ventricle, the left atrium and the left ventricle.

The superior and inferior vena cava collects deoxygenated blood from the upper and lower regions of the body respectively and transfers this de-oxygenated blood to the right atrium of the heart.

The right atrium then contracts and transfers the deoxygenated blood to the right ventricle.

From the right ventricle, the deoxygenated blood is then transferred into the two pulmonary arteries, which pumps the blood to the lungs for oxygenation.

From the lungs, oxygenated blood is transferred to the left atrium of the heart through the pulmonary veins. This complete process is called pulmonary circulation.

Then the left atrium contracts and pushes the oxygenated blood into the left ventricle.

The left ventricle then contracts and forces the blood into the aorta which gives rise to many arteries that distribute the oxygenated blood to all the regions of the body.

This completes the systemic circulation.

Importance of Double blood circulation:

The separation of oxygenated and deoxygenated blood in the separate chambers of heart and blood vessels allows a more efficient supply of oxygen to the body cells.

The sufficient system of oxygen supply is very useful in animals that have high energy requirements to maintain their body temperature such as warm-blooded animals like birds and mammals.

Q.149 What are the differences between the transport of materials in xylem and phloem?

Ans

Differences between the transport of materials in xylem and phloem are:

 Transport of materials in xylem Transport of materials in phloem Xylem facilitates the transport of water in plants. Phloem facilitates the transport of food in plants. Transport of water is in upward direction from roots to the other parts of the plants Transport of food occurs in both the directions i.e., upward and downward. Simple physical forces like transpiration pull play an important role in the transport of water in xylem. Energy in the form of ATP is required in the transport of food in phloem.

Q.150 Compare the functioning of alveoli in lungs and nephrons in the kidneys with respect to their structure and functioning.

Ans

 Alveoli Nephron Structure They are tiny balloon-like structure present inside the lungs. Nephron is the structural and functional unit of kidney. They are present in the form of tubular structures inside the kidneys. The walls of alveoli are one cell thick and posses an extensive network of blood capillaries that help in exchange of gases. It is composed of glomerulus, Bowman’s capsule and a long renal tubule and is surrounded by the cluster of thin walled blood capillaries. Function The alveoli function as the sites for gaseous exchange. The exchange of O2 and CO2 takes place between the blood of the capillaries that surround the alveoli and the gases present in the alveoli. Being the basic filtration unit of kidney, nephron removes the nitrogenous wastes such as urea or uric acid from the blood and forms urine. Renal artery divides into many capillaries in glomerulus of the nephron. The blood gets filtered in the glomerulus and then the filtrate passes through the tubules of the nephron and gets collected in a collecting duct. As the filtrate flows in the tube, some substances such as glucose, amino acids and water are selectively re- absorbed and get converted into urine. Urine from many nephrons is collected into the collecting duct that passes the urine into the ureter.

Q.151 Asexual reproduction takes place through budding in

1. amoeba.
2. yeast.
3. plasmodium.
4. leishmania.

Ans

b. Yeast

Explanation: Out of the four organisms mentioned, only in yeast asexual reproduction takes place though budding. Asexually Amoeba and Leishmania reproduce through binary fission and Plasmodium reproduces by multiple fission method.

Q.152 Which of the following is not a part of the female reproductive system in human beings?

1. Ovary
2. Uterus
3. Vas deferens
4. Fallopian tube

Ans

Vas deferens

Explanation: Ovary, uterus and fallopian tube are the parts of the female reproductive system in human beings whereas vas deferens is the part of male reproductive system in human beings.

Q.153 The anther contains

1. sepals.
2. ovules.
3. carpel.
4. pollen grains.

Ans

d) Pollen grains

Explanation: The anther contains pollen grains. Sepals are generally green in colour and protect flower in bud stage. Ovules are present in ovary of the carpel (female reproductive organ of plant).

Q.154 What are the advantages of sexual reproduction over asexual reproduction?

Ans

Advantages of sexual reproduction over asexual reproduction are:

1. Sexual reproduction is responsible for variations in a population.
2. Variations in individuals are responsible for survival of a species in an ecosystem.
3. Sexual reproduction plays an important role in origin of new species and is a beginning for evolution.
4. Sexual reproduction helps in providing stability to populations of species.

Q.155 What are the functions performed by the testis in human beings?

Ans

Testes are the male reproductive organs in human beings. Functions of testis are:

1. It produces male gametes i.e., sperms.
2. It also produces male hormone called testosterone that regulates the formation of sperms and is responsible for secondary sexual characters in human males at the time of puberty.

Q.156 Why does menstruation occur?

Ans

Menstruation is a process in which blood and mucosal tissues are discharged periodically from the inner lining of uterus through the vagina. In human female, every month one egg is released from one ovary. Simultaneously the uterus prepares itself for pregnancy. During this phase, the inner lining of the uterus gets thickened and new blood vessels are formed that will nourish the future embryo. If pregnancy does not occur, then the lining of the uterus breaks down slowly and the content is released in the form of blood and mucus through the vagina and the process is known as menstruation.

Q.157 Draw a lebelled diagram of the longitudinal section of a flower.

Ans

Q.158 What are the different methods of contraception?

Ans

The different methods of contraception are:

1. Barrier method: In this method, the meeting of ovum with sperm is checked with the help of a mechanical barrier. Mechanical barrier such as condoms are made of thin and flexible rubber. The mechanical barrier is available for both males and females. These are used to cover penis in males and vagina in females and prevent sperms from entering the uterus during sexual intercourse to avoid pregnancy.
2. Oral contraceptive: This is method in which hormones are introduced in the female partner to prevent the release of egg to avoid pregnancy. The hormones are taken orally in the form of tablets or pills.
3. Intrauterine devices (IUDs): Contraceptive devices such as the loop or Copper T are placed in the uterus that prevent sperms from reaching the egg and prevent pregnancy to occur.
4. Surgical methods: Transfer of gametes can be blocked by using surgical methods. Surgical method is an irreversible method of contraception. Surgically, in males, vas deferens is blocked to stop the sperm transfer. This process is known as vasectomy. In females, surgically fallopian tubes are blocked to carry the eggs to the uterus. This surgical procedure in females is called tubectomy.

Q.159 How are the modes for reproduction different in unicellular and multicellular organisms?

Ans

Unicellular organisms have single cell. Generally, unicellular organisms reproduce by asexual methods like binary fission, multiple fission, budding, etc. For example, Amoeba reproduces by binary fission, yeast reproduces by budding, and Plasmodium reproduces by multiple fission method. Multicellular organisms are made up of either simple cells or complex tissues and organs. Some of their cells are specialised to perform the function of reproduction. Multicellular organisms can employ the asexual methods of reproduction as well as sexual methods of reproduction. Asexual methods include budding, fragmentation, spore formation and vegetative propagation. Complex multicellular organisms possess a well developed reproductive system having specialised reproductive organs.

Q.160 How does reproduction help in providing stability to populations of species?

Ans

In an ecosystem, organisms of different species use their ability of reproduction to survive and maintain their populations in their niches.

In an ecosystem, every individual of a species struggle for food, shelter and protection for their survival and existence. Factors like death at the end of natural life cycle, competition, predation, migration and sudden environmental changes keep on reducing population of species on a regular basis. The process of reproduction is the only way to overcome this continuous depletion of individuals from the populations of diverse species. Therefore, reproduction helps in providing stability to populations of species.

Q.161 What could be the reason for adopting contraceptive methods?

Ans

Contraceptive methods are mainly adopted because of the following reasons:

1. These devices prevent unwanted pregnancies to occur.
2. These devices not only help in restricting the number of children but also help in maintaining age gap between children. The age gap between children help parents in good bringing up of their children and less number of children ensures the improved living standards of a family.
3. These devices also prevent the spread of sexually transmitted diseases (STDs).
4. These devices play a role in controlling increase in human population and help in reducing pressure on life supporting natural resources such as food, air, water, fuel, etc.

Q.162 Which of the following is a plant hormone?
(a) Insulin
(b) Thyroxin
(c) Estrogen
(d) Cytokinin

Ans

d) Cytokinin

Explanation: Insulin, thyroxin and oestrogen are animal hormones.

Q.163 The gap between two neurons is called a
(a) dendrite.
(b) synapse.
(c) axon.
(d) impulse.

Ans

b) Synapse

Explanation: Dendrite and axon are the parts of a neuron. Impulse is the electrical signal that travels through a neuron.

Q.164 The brain is responsible of
(a) thinking
(b) regulating the heart beat
(c) balancing of the body
(d)all of the above

Ans

d) all of the above

Explanation: Our brain is a part of the nervous system that plays an important role in the control and coordination of higher animals. Thinking, regulating heart beat and balancing of the body are some of the functions that are controlled by the brain.

Q.165 What is the function of receptors in our body? Think of situations where receptors do not work properly? What problems are likely to arise?

Ans

The receptors are the sensory structures that on receiving environmental stimuli generate a nerve impulse. For example, our skin possesses many receptors that respond to the environmental stimuli such as increase or decrease in temperature, pain, etc. Taste receptors present on tongue help in identifying different tastes.

Functions of the receptors are:

1. They sense the external stimuli.
2. They also trigger impulse after perceiving stimuli.

When the receptors are damaged, the external stimuli, transferring signals to the brain are not felt. For example, when we drink hot coffee or tea, taste receptors on the tongue gets damaged. These damaged taste buds do not allow us to enjoy our food for a little while as they are not able to perceive the external stimuli i.e., taste of food.

Q.166 Draw the structure of a neuron and explain its function.

Ans

Neuron is the structural and functional unit of nervous system. The parts of a neuron are cyton (cell body), axon, and dendrites.

Functions of the parts of a neuron:

Dendrite: It receives information from the axon of another neuron and conducts the message in the form of an impulse in the cyton (cell body).

Cyton (cell body): It consists of nucleus, mitochondria, and other cell organelles and concerned with the growth and maintenance of the neuron.

Axon: It helps in conducting messages in the form of impulse away from the cyton (cell body).

Q.167 How does phototropism occur in plants?

Ans

Phototropism refers to the orientation of organisms in response to light. In plants, phototropism occurs due to the presence of a plant hormone called auxin. Auxin is synthesised at the shoot tips of the plants and help the plant cells to grow longer. In a case, when light is coming from one side of a plant, the plant hormone auxin diffuses towards the side of the plant shoot where no light is coming. The cells of the shoot which is away from the light start growing longer under the influence of plant hormone auxin and this causes bending of the plant towards light (phototropism).

Q.168 Which signals will get disrupted in case of a spinal cord injury?

Ans

In case of a spinal cord injury, the signals coming from the nerves and the receptors and the signals reaching the effector organ will be disrupted.

Q.169 How does chemical coordination occur in plants?

Ans

Plants do not possess a nervous system. Plants respond to the external and internal stimuli by showing different types of movements. In plants, growth, development and response to the stimuli is controlled and coordinated by certain chemical substances produced in their cells called hormones. A plant hormone is an organic compound which is synthesised at plant parts away from where they act and when required diffuses to the area of action. The five major types of plant hormones are auxins, gibberellins, cytokinins, abscisic acid and ethylene. These plant hormones either act as growth promoters or growth inhibitors. Auxins, gibberellins, cytokinins are growth promoters and abscisic acid and ethylene act as growth inhibitor.

Q.170 What is the need for a system of control and coordination in an organism?

Ans

An organism comprises millions of cells, tissues, organs and organ systems. In an organism, all the organ systems work in coordination with each other and are responsible for the sustenance and survival of an organism in its environment. The more complex is the organism, the more coordination is required. It is very much necessary that various organ systems of an organism should work together in controlled and coordinated manner so that different body parts can function as a single unit and respond appropriately to the stimuli. In the absence of this system of control and coordination, our body will not be able to function properly. In animals, the control and coordination are provided by the nervous system and endocrine system whereas in plants nervous system is absent.

Q.171 How are involuntary actions and reflex actions different from each other?

Ans

Involuntary actions cannot be consciously controlled. For example, we cannot consciously control the contraction and relaxation of our heart muscles. Involuntary actions are directly under the control of the brain. On the other hand, the reflex actions such as removal of hand just after getting pricked by a thorn shows sudden response and do not involve any thinking process. This clearly shows that the reflex actions are not under the control of brain as the involuntary actions are.

Q.172 Compare and contrast nervous and hormonal mechanisms for control and coordination in animals.

Ans

 Character Nervous Control and Coordination Hormonal Control and Coordination Form of Information Nerve impulse Chemical messenger Mode of transmission of information Through axons and dendrites Through blood Flow of information Rapid Slow Response Quick Slow Duration of effect Short-lived Prolonged effects

Q.173 What is the difference between the manner in which movement takes place in a sensitive plant and the movement in our legs?

Ans

The movement that takes place in a sensitive plant such as Mimosa pudica is an example of the response to touch stimulus. In response to the stimulus of touch and for this movement to occur, the plant cells change their shape by changing the amount of water in them. Movement in our legs is a voluntary action. In a voluntary action, the signals are passed to the brain and hence are consciously controlled. In animal muscle cells, when a nerve impulse reaches the muscle cells, the muscle cells move by changing their shape. Muscle cells possess special proteins that on receiving the nerve impulse change their shape and arrangement which further cause the movement of the organ.

Q.174 Which of the following groups contain only biodegradable items?

1. Grass flower and leather
2. Grass, wood and plastic
3. Fruit-peels, cake and lime juice
4. Cake, wood and grass

Ans

c) Fruit-peels, cake and lime juice and d)cake, wood and grass

Q.175 Which of the following constitute a food chain?

1. Grass, wheat and mango
2. Grass, goat and human
3. Goat, cow and elephant
4. Grass, fish and goat

Ans

b) Grass, goat and human

Q.176 Which of the following are environment friendly practices?

1. Carrying cloth bags to put purchases in while shopping
2. Switching off unnecessary lights and fans
3. Walking to school instead of getting your mother to drop you on her scooter
4. All of the above

Ans

d) All of the above

Q.177 What will happen if we kill all the organism in one trophic level?

Ans

Death of all the organisms of one trophic level would affect the organisms of the previous trophic level and the next trophic level. If all the organisms of one trophic level are killed then the number of organisms of the next trophic level would decrease as they feed on the organisms of the trophic level. The number of organisms of the previous trophic level would increase because the organisms that feed on them are killed.

For example in the food chain comprising of, Grass —> Goat —> Lion, if all the goats are killed then the number of lions would decreases due to lack of food and grass population would increase due to under grazing.

Q.178 Will the impact of removing all the organisms in a trophic level be different for different trophic levels? Can the organisms of any trophic level be removed without causing any damage to ecosystem?

Ans

Yes, the impact of removing all the organisms in a trophic level will be different for different trophic levels. If the organisms of first trophic level i.e., producers are removed, then the entire food chain will be disturbed and finally all the organisms might die. If the organisms of the topmost trophic level are removed then a disturbance in the ecosystem would occur as there would arose a competition in the organisms of the second topmost trophic level for food and space due to their overproduction in the absence of the organisms of the topmost trophic level.

Removal of organisms of any trophic level would definitely create a disturbance in the ecosystem. Hence, in an ecosystem, organisms of any trophic level cannot be removed.

Q.179 What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?

Ans

Biological magnification is a phenomenon of accumulation or increase in the concentration of some toxic substance in the bodies of organisms of each trophic level.

Pesticides like DDT etc are generally sprayed on plants to eradicate pests. These pesticides are then washed down into soil and water bodies. Plants absorb these substances with water whereas aquatic animals such as fishes take up these substances from the water body. On eating these plants and aquatic animals, these toxic substances enter our body.

Yes, the magnification is different at different trophic levels of the ecosystem. The animals at the higher trophic levels will receive more of these toxic substances along with their food than animals at the lower trophic level. Thus, at each tropic level the magnification will be high as compared to the previous trophic level.

Q.180 What are the problems caused by the non-biodegradable wastes that we generate?

Ans

The waste that is not degraded in the nature is called non-biodegradable waste. The non-biodegradable wastes cause the following problems:

1. They increase soil and water pollution.
2. Non-biodegradable waste like polyethene bags blocks the drains.
3. Disposal of these non-biodegradable waste releases poisonous gases that pollute the air also.

Q.181 If all the waste we generate is biodegradable, will this have no impact on the environment?

Ans

Though the biodegradable waste do not stay for long in the environment but their excess pose various threats to the environment. Accumulated biodegradable waste like kitchen waste serves as a breeding ground for various flies, mosquitoes, microorganisms etc that may cause various diseases and epidemics in their surroundings.

Q.182 Why is damage to ozone layer a cause for concern? What steps are being taken to limit this damage?

Ans

Hole in ozone layer will allow ultraviolet rays to reach the earth’s surface. These ultraviolet (UV) rays are harmful for living organisms in the following ways:

1. These rays may cause skin disease, such as skin cancer.
2. These rays cause retarded growth and destruction of pigments in plants.
3. UV rays may disturb the ecological balance by killing the microorganisms, decomposers and other useful microbes

Following steps should be taken to prevent damage of ozone layer:

1. According to Montreal Protocol (1989) the compounds like chlorofluorocarbons which cause depletion or hole in ozone layer should be used judiciously
2. Use of chemical pesticides should be minimised
3. Use of public transport should be encouraged

Q.183 A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio

$\frac{\mathrm{R}}{\mathrm{R}‘}$

is

$\begin{array}{l}\left(\mathrm{a}\right)\text{}\frac{1}{25}\\ \left(\mathrm{b}\right)\text{}\frac{1}{5}\\ \left(\mathrm{c}\right)\text{5}\\ \text{(d) 25}\end{array}$

Ans

The correct option is (d).

$\begin{array}{l}\text{The resistance of a piece of wire is}\mathrm{R}\text{and it is proportional}\\ \text{to the length of wire. As the wire is cut into five equal parts}\\ \text{h}\mathrm{ence},\text{resistance of each part =}\frac{\mathrm{R}}{5}\\ \text{Let the equivalent resistance of these piece of wire in}\\ \text{parallel is R’ then,}\\ \frac{1}{\mathrm{R}‘}=\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}=\frac{25}{\mathrm{R}}\\ \mathrm{or},\text{}\frac{\mathrm{R}}{\mathrm{R}‘}\text{= 25}\end{array}$

Q.184 Which of the following terms does not represent electrical power in a circuit?

(a) I2R
(b) IR2
(c) VI

$\left(\mathrm{d}\right)\frac{{\mathrm{V}}^{2}}{\mathrm{R}}$

Ans

$\begin{array}{l}\text{Electrical power is given by,}\\ \text{P= VI}\dots \text{(1)}\\ \text{According to ohm’s law,}\\ \text{V = IR}\dots \text{(2)}\\ \text{Where, V = potential difference}\\ \text{I = current}\\ \text{R = resistance}\\ \therefore \text{P = VI}\\ \text{From equation (1), P = (IR)×I}\\ \therefore {\text{P = I}}^{\text{2}}\text{R}\\ \text{From equation (2),}\\ \text{I =}\frac{\text{V}}{\text{R}}\\ \therefore \text{P =}\frac{{\text{V}}^{\text{2}}}{\text{R}}\text{}\\ {\text{Hence, P = VI = I}}^{\text{2}}\text{R =}\frac{{\text{V}}^{\text{2}}}{\text{R}}\\ {\text{Thus, IR}}^{\text{2}}\text{does not represent electrical power}\text{.}\end{array}$

Q.185 An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be

(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W

Ans

$\begin{array}{l}{\text{Given, V}}_{\text{1}}{\text{= 220 V, V}}_{\text{2}}=110\text{V}\\ {\text{P}}_{\text{1}}=\text{100 W}\\ \mathrm{On}\text{using the relation,}\\ {\text{P}}_{\text{1}}\text{=}\frac{{\text{V}}_{1}^{2}}{\mathrm{R}}\\ \text{or, R =}\frac{{\mathrm{V}}_{1}^{2}}{{\mathrm{P}}_{1}}\text{=}\frac{{\left(220\text{V}\right)}^{2}}{\left(100\text{W}\right)}=484\text{}\mathrm{\Omega }\\ \text{If the bulb is operated on 110 V, then the energy}\\ \text{consumed,}\\ {\text{P}}_{\text{2}}=\frac{{\text{V}}_{2}^{2}}{\mathrm{R}}=\text{}\frac{{\left(110\text{V}\right)}^{2}}{484\text{\hspace{0.17em}}\mathrm{\Omega }}=\text{25 W}\\ \text{Therefore, the power consumed will be 25 W.}\end{array}$

Q.186 Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –

(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1

Ans

$\begin{array}{l}\text{The correct option is}\left(\text{c}\right).\\ \text{Suppose, R = resistance of each wire}\\ \therefore \text{The equivalent resistance in series}\\ {\text{R}}_{\text{s}}\text{= R+R = 2R}\\ \text{If V is the applied potential difference, then it is the}\\ \text{voltage across the equivalent resistance.}\\ {\text{V = I}}_{\text{s}}×{\text{2R (I}}_{\text{s}}\text{= current in series combination)}\\ \\ ⇒{\text{I}}_{\text{s}}\text{=}\frac{\mathrm{V}}{2\mathrm{R}}\\ \mathrm{The}\text{heat dissipated in time t is,}\\ {\text{H}}_{\text{s}}={\text{I}}_{\mathrm{s}}^{2}×2\mathrm{R}×\mathrm{t}={\left(\frac{\mathrm{V}}{2\mathrm{R}}\right)}^{2}×2\mathrm{R}×\mathrm{t}\\ ⇒{\text{H}}_{\text{s}}=\frac{{\mathrm{V}}^{2}\mathrm{t}}{2\mathrm{R}}\text{}...\text{(i)}\\ \mathrm{The}\text{equivalent resistance of the parallel connection is}\\ {\text{R}}_{\text{p}}=\frac{1}{\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}}=\frac{\mathrm{R}}{2}\\ \mathrm{V}{\text{is the applied potential difference across R}}_{\text{p}}.\\ \therefore \text{}\mathrm{V}={\text{I}}_{\mathrm{p}}×\frac{\mathrm{R}}{2}\end{array}$

Q.187 How is a voltmeter connected in the circuit to measure the potential difference between two points?

Ans

A voltmeter need to be connected in parallel in a circuit to measure the potential difference between two points.

Q.188 A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10−8 Ωm. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Ans

$\begin{array}{l}\text{Given,}\rho =1.6×{10}^{-8}\text{}\Omega \text{m}\\ \text{A =}\pi {\text{r}}^{\text{2}}\\ r=\frac{0.5}{2}mm=\frac{0.0005}{2}m\\ \text{R = 10}\Omega \\ Hence,\text{length of the wire}\\ \text{l =}\frac{RA}{\rho }=\frac{10×3.14{\left(\frac{0.0005}{2}\right)}^{2}}{1.6×{10}^{-8}}=122.72\text{m}\\ \text{If the diameter of the wire is doubled then, new diameter}\\ \text{= 2}×\text{0}\text{.5 mm =1 mm =0}\text{.001m}\\ \text{Hence, Resistance R’ =}\rho \frac{l}{A}=\frac{1.6×{10}^{-8}×122.72}{\pi ×{\left(\frac{1}{2}×{10}^{-3}\right)}^{2}}\\ \text{= 2}\text{.5}\Omega \end{array}$

Q.189 The value of current (I) flowing in a given resistor for the corresponding values of potential difference (V) across the resistor are given below-

 I (amperes) 0.5 1 2 3 4 V (volts) 1.6 3.4 6.7 10.2 13.2

Plot a graph between V and I and calculate the resistance of that resistor.

Ans

When the graph is plotted between voltage and current then, it is called V-I characteristics. Here, the voltage is plotted on X axis and current is plotted on Y axis.

$\begin{array}{l}\mathrm{The}\text{slop of the line gives the value of resistance (R)}\\ \text{as,}\\ \text{Slope, =}\frac{1}{\mathrm{R}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{2}{6.8}\\ \therefore \mathrm{R}=\text{}\frac{6.8}{2}=3.4\text{}\mathrm{\Omega }\text{}.\end{array}$

Q.190 When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Ans

$\begin{array}{l}\mathrm{According}\text{to ohm’s law,}\\ \text{V = IR}\\ \text{or, R =}\frac{\mathrm{V}}{\mathrm{I}}\\ \text{Here, potential difference, V = 12 volts}\\ \text{i = 2.5 mA}\\ \text{=2.5}×{\text{10}}^{\text{-3}}\text{A}\\ \therefore \text{R =}\frac{12}{2.5×{\text{10}}^{\text{-3}}}\text{=4.8}×{\text{10}}^{\text{-3}}\mathrm{\Omega }.\text{}\end{array}$

Q.191 A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 resistors?

Ans

$\begin{array}{l}\text{The current would be same throughout the series circuit.}\\ \text{Equivalent resistance of the circuit,}\\ {\text{R}}_{\text{eq}}=\left(0.2+0.3+0.4+0.5+12\right)\text{}\mathrm{\Omega }=13.4\text{}\mathrm{\Omega }\text{}\\ \mathrm{V}=\text{9 V}\\ \text{By using ohm’s law,}\\ \text{V =IR}\\ \mathrm{or},\text{I =}\frac{\mathrm{V}}{\mathrm{R}}\\ \therefore \text{I =}\frac{9}{13.4\text{}\mathrm{\Omega }}\text{= 0.671 A}\\ \text{Therefore, the current that would flow through}\\ \text{12}\mathrm{\Omega }\text{resistor is 0.671 A.}\end{array}$

Q.192 How many 176 resistors (in parallel) are required to carry 5 A on a 220 V line?

Ans

$\begin{array}{l}Given,\text{V = 220 volt}\\ \text{I = 5 A}\\ \text{Equivalent resistance (R) of the combination is}\\ \text{given as}\\ \text{}\frac{1}{R}\text{= x}×\frac{1}{176}\\ \text{or, R =}\frac{176}{x}\\ \text{From ohm’s law,}\\ \text{}\frac{V}{I}=\frac{176}{x}\\ or,\text{x =}\frac{176×5}{220}=4\\ Hence,\text{4 resistors of 176}\Omega \text{are required to draw the}\\ \text{given amount of current}\text{.}\end{array}$

Q.193 Show how would you connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω (ii) 4 Ω.

Ans

(i)

$\begin{array}{l}\text{When two resistors are connected in parallel and}\\ \text{the third one is connected in series with these}\\ \text{two then, Two 6}\mathrm{\Omega }\text{resistors are connected in parallel.}\\ \text{Their equivalent resistance (R’) will be}\\ \text{}\frac{1}{\mathrm{R}‘}=\text{}\frac{1}{6\text{}\mathrm{\Omega }}+\frac{1}{6\text{}\mathrm{\Omega }}=\frac{2}{6\text{}\mathrm{\Omega }}=\frac{1}{3\text{}\mathrm{\Omega }}\\ \\ \text{or, R’ = 3}\mathrm{\Omega }\\ \text{The third 6}\mathrm{\Omega }\text{resistor is connected in series with 3}\mathrm{\Omega }.\\ {\text{Hence, the equivalent resistance R}}_{\text{1}}\text{= (6+3)}\mathrm{\Omega }\text{= 9}\mathrm{\Omega }\end{array}$

(ii)

$\begin{array}{l}\text{Two resistors of 6 Ω are in series. Their equivalent}\\ \text{resistance will be = (6}+6\text{) Ω = 12 Ω}\\ \text{The third 6 Ω resistor is connected in parallel with 12 Ω.}\\ {\text{Hence, equivalent resistance (R}}_{\text{2}}\text{)}\\ \frac{\text{1}}{{\text{R}}_{\text{2}}}\text{=}\frac{\text{1}}{\text{12\hspace{0.17em}Ω}}\text{+}\frac{\text{1}}{\text{6Ω}}\text{=}\frac{\text{3}}{\text{12Ω}}\text{=}\frac{\text{1}}{\text{4Ω}}\\ {\text{or, R}}_{\text{2}}\text{= 4 Ω}\end{array}$

Q.194 Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Ans

$\begin{array}{l}\mathrm{Given},\text{V = 220 V}\\ \text{i = 5 A}\\ {\text{P}}_{\text{1}}\text{= 10 W}\\ \text{On using the relation,}\\ \text{P =}\frac{{\mathrm{V}}^{2}}{{\mathrm{R}}_{1}}\\ {\text{or, R}}_{\text{1}}\text{=}\frac{{\mathrm{V}}^{2}}{{\mathrm{P}}_{1}}\\ \therefore \text{=}\frac{{\left(220\right)}^{2}}{10}\text{= 4840}\mathrm{\Omega }\\ \text{As per ohm’s law,}\\ \text{V= IR}\\ \therefore \text{R =}\frac{\mathrm{V}}{\mathrm{I}}\text{=}\frac{220}{5}=44\text{}\mathrm{\Omega }\\ \mathrm{Resistance}\text{of each electric bulb,}\\ {\text{R}}_{\text{1}}=4840\text{}\mathrm{\Omega }\\ \therefore \text{}\frac{1}{\mathrm{R}}=\frac{1}{{\mathrm{R}}_{1}}+\frac{1}{{\mathrm{R}}_{1}}+.\dots +\mathrm{upto}\text{x times}\\ \text{or, x =}\frac{{\mathrm{R}}_{1}}{\mathrm{R}}=\frac{4840}{44}=110\\ \mathrm{Thus},\text{110 electric bulbs are connected in parallel.}\end{array}$

Q.195 A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 resistances which may be used separately, in series or in parallel. What are the currents in the three cases?

Ans

$\begin{array}{l}\mathrm{Given},\text{Supply voltage, V = 220 V}\\ \text{Resistance, R = 24}\mathrm{\Omega }\text{}\\ \text{(i) When coils are used separately:}\\ \text{By using the relation,}\\ {\text{V = I}}_{\text{1}}{\text{R}}_{\text{1}}\\ {\text{Where, I}}_{\text{1}}\text{= current flowing through the coil}\\ \text{}\therefore {\text{I}}_{\text{1}}\text{=}\frac{\text{V}}{{\text{R}}_{\text{1}}}\text{=}\frac{\text{220}}{\text{24}}\text{= 9.166 A.}\\ \\ \text{(ii) When coils are connceted in series:}\\ \text{total resistance of the two coils}\\ {\text{R}}_{\text{2}}\text{= (24+24)}\mathrm{\Omega }\text{= 48}\mathrm{\Omega }\text{}\\ \text{By using the relation,}\\ {\text{V = I}}_{2}{\text{R}}_{2}\\ \therefore {\text{I}}_{\text{2}}\text{=}\frac{\text{V}}{{\text{R}}_{\text{2}}}\text{=}\frac{\text{220}}{\text{48}}\text{= 4.58 A.}\\ \text{(iii) When coils are connceted in parallel:}\\ {\text{Total resistance (R}}_{\text{3}}\right)\text{would be}\\ \text{}\frac{1}{{\text{R}}_{\text{3}}}=\frac{1}{24}+\frac{1}{24}=\frac{2}{24}\\ \mathrm{or},{\text{R}}_{\text{3}}=12\mathrm{\Omega }\\ \text{On using the relation}\\ {\text{V = I}}_{3}{\text{R}}_{3}\\ \therefore {\text{I}}_{3}=\text{}\frac{\mathrm{V}}{{\text{R}}_{3}}=\text{}\frac{220}{12}=18.33\text{A.}\end{array}$

Q.196 Compare the power used in the 2 Ω in each of the following circuits:
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{Given, Potential difference, V = 6 V}\\ {\text{R}}_{\text{1}}{\text{= 1 Ω, R}}_{\text{2}}\text{= 2 Ω}\\ \text{}\therefore {\text{Equivalent resistance in series, R = R}}_{\text{1}}{\text{+ R}}_{\text{2}}\\ \text{}=\text{(1+2) Ω = 3 Ω}\\ \text{On using the relation,}\\ \text{V = IR (I = current through the circuit)}\\ \\ \text{}\therefore \text{I =}\frac{\text{V}}{\text{R}}\text{=}\frac{\text{6}}{\text{3}}\text{= 2 A}\\ \text{Now, power in the circuit}\\ {\text{P = I}}^{\text{2}}{\text{R = (2)}}^{\text{2}}\text{×2=8 W.}\\ \text{(ii) Given, potential difference, V = 4 V}\\ {\text{R}}_{\text{1}}=\text{12}\mathrm{\Omega },{\text{R}}_{2}=2\text{}\mathrm{\Omega }\\ \text{The voltage across each component of a}\\ \text{parallel circuit remains the same. Hence, voltage across}\\ \text{}2\text{}\mathrm{\Omega }\text{resistor will be 4 V.}\\ \text{Power consumed by 2}\mathrm{\Omega }\text{is given by}\\ \text{P =}\frac{{\mathrm{V}}^{2}}{\mathrm{R}}=\frac{{4}^{2}}{2}=\text{8 W.}\end{array}$

Q.197 Two lamps, one rated 100 W and 220 V, and the other 60 W at 220 W, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{potential}\text{difference across each bulb will be 220}\\ \text{V because no division of voltage occurs in a parallel}\\ \text{circuit.}\\ \text{Current drawn by the bulb of rating 100 W is given by,}\\ {\text{P = VI}}_{\text{1}}\\ \therefore {\text{I}}_{\text{1}}\text{=}\frac{\mathrm{P}}{\mathrm{V}}=\frac{100W}{220V}\\ \mathrm{Similarly},\text{current drawn by the bulb of rating 60 W is}\\ \text{given by,}\\ {\text{I}}_{\text{2}}\text{=}\frac{\mathrm{P}}{\mathrm{V}}=\frac{60W}{220V}\\ \therefore \text{Current drawn from the line =}\frac{100W}{220V}+\frac{60W}{220V}=0.727\text{}\mathrm{A}\end{array}$

Q.198 Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Ans

$\begin{array}{l}\text{Energy consumed by a TV set of power 250 W in 1 h}\\ \text{is given by the expression}\\ \text{H = Pt = 250 W}×\text{3600 s = 9}×{\text{10}}^{\text{5}}\text{}\mathrm{J}\\ \mathrm{Similarly},\text{}\mathrm{energy}\text{consumed by a toaster of power 1200 W}\\ \text{in 10 minutes = 1200}×\text{600 = 7.2}×{\text{10}}^{\text{5}}\text{}\mathrm{J}\\ \text{Hence, the energy consumed by a 250 W TV in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.}\end{array}$

Q.199 An electric heater of resistance 8 Ω  draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Ans

$\begin{array}{l}\mathrm{Given},\text{R = 8}\mathrm{\Omega }\\ \text{I = 15 A}\\ \mathrm{Rate}\text{of heat produced can be expressed as power,}\\ {\text{P = I}}^{\text{2}}\mathrm{R}\\ \mathrm{On}\text{putting the values,}\\ \text{P =}{\left(\text{15 A}\right)}^{2}\text{\hspace{0.17em}}×8=1800\text{W or}1800\text{}\frac{\mathrm{J}}{\mathrm{s}}.\end{array}$

Q.200 Explain the following.

(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?

Ans

(a) Due to high melting point and high resistivity of tungsten, it does not burn at a high temperature. As the electric lamps glow at very high temperatures hence, the tungsten is used as heating element of electric bulbs.
(b) Bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of pure metals. Due to high resistivity, they produce large amount of heat.
(c) In a series circuit, there is a voltage distribution. Each element of a series circuit receives a small voltage for a large supply voltage. Thus, the amount of current decreases and the device becomes hot. Therefore, series arrangement is not used in domestic circuits.
(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A).
(e) The wires made of copper or aluminium have low resistivity and they are good conductors of electricity. Therefore, they are frequently used for electricity transmission.

1. How many chapters are present in the NCERT solutions for Class 10 Science?

The NCERT Solutions for Class 10 Science are separated into five parts and contain 16 chapters.

Unit I – Chemical Substances – Nature & Behavior (5 chapters)

Unit II – World of Living (4 chapters)

Unit III – Natural Phenomenon (2 chapters)

Unit IV – Effects of Current (2 chapters)

Unit V – Natural Resources (3 chapters).

2. Which are the most essential chapters in the NCERT Solutions for Class 10 Science?

The NCERT Solutions for Class 10 Science contain approximately 16 chapters, each of which is equally important. As a result, neglecting any of these will result in a lower score on the board exams. If students have any questions, they can ask their lecturers for clarification or use free internet study materials. Students can use Extramarks answers to acquire a better understanding of the chapters based on the CBSE exam pattern.

3. Is the NCERT Solutions for Class 10 Science PDFs enough to score good in the board exam?

The NCERT Solutions for Class 10 Science cover all of the textbook’s exercise questions. The solutions are sufficient to clear students’ doubts and notions, which are crucial for board exams. Aside from textbooks, students can use additional study tools to learn shortcut tips and tricks for rapidly answering difficult questions. By consulting the study resources, students will be able to gain a clear sense of the areas that are crucial from an exam standpoint.

4. Will the questions in the board exam appear from NCERT Solutions for Class 10 Science?

The NCERT Solutions for Class 10 Science contain the highest number of questions in the board exam. Students can refer to the solutions framed by the professors at Extramarks while solving textbook questions to get a clear understanding of the subject. The solutions are written in such a way that students may grasp even the most difficult concepts with ease. Students can use the solutions both online and offline without having to worry about running out of time.