X Swad Science

MAHARASHTRA STATE BOARD OF SECONDARY

AND HIGHER SECONDARY EDUCATION

SHIVAJINAGAR, PUNE-411 005

SECONDARY SCHOOL CERTIFICATE EXAMINATION

(STD-X)

SELF-STUDY BOOKLET FOR

PRIVATE CANDIDATES

SCIENCE AND TECHNOLOGY

PART - I AND II

MAHARASHTRA STATE BOARD OF SECONDARY

AND HIGHER SECONDARY EDUCATION

SHIVAJINAGAR, PUNE-411 005

SECONDARY SCHOOL CERTIFICATE EXAMINATION

(STD-X)

SELF-STUDY BOOKLET FOR

PRIVATE CANDIDATES

SCIENCE AND TECHNOLOGY

PART - I AND II

(ENGLISH MEDIUM)

N.B.

Exercise section - Page nos. 86 to 202 should be solved and is to be submitted at the centre as per instructions of the Head of the Centre.

Introductory

About the Booklet

This booklet is specially prepared for you by experienced teachers. The booklet has two sections.

Section I consists of summary of the chapter. It includes important points from

the chapters. It also includes some tips, instructions in between and some solved examples.

Section II consists of exercises for self study. They involve important questions

from your textbooks and also some additional exercises for practice.

How to use the booklet ?

1. The booklet consists of summary of each chapter from the textbook. But it includes only some important points to bring to your notice. So do not rely merely on the summary given. Read the complete chapter from the textbook before studying the summary.

2. Try to understand the concepts and principles given in the summary. If you do not understand, refer to the chapter in the textbook once again.

3. Study figures, diagrams, graphs, tables from the textbook. They are not included in the booklet.

4. Study the solved sample questions or examples. Try to concentrate on the format of answers sequence of the points included and steps of solving numerical examples carefully.

5. Once you become confident enough, be ready to solve the exercises given in section II of the booklet.

6. Try to write answers and solve examples on your own. In case of difficulty, refer to page numbers from the textbook given against each question. Re

-read the content, organise your answer and write it down. 7. Study the written answers till you master them.

Some tips for self study

-
Concentrate while studying. Sit erect. Avoid noise around as far as possible.

Try to understand the meaning of scientific terms used. This will help you to remember them.

aloud that part, if required.

Try to comprehend the definitions, formulae and equations. Write them again and again till you learn them by heart.

Study solved examples. Try solving them without seeing and check to get self feedback. Find similar examples from the textbook and solve them.

Seek guidance from peer group and teachers available in case of difficulty.

Practice drawing diagrams. Remember labels given to parts and structure.

You may find some chapters easy and some chapters difficult according to your interest. So start from easy chapters and slowly go to the difficult one. Remember the chapters are not difficult. Only thing is you probably do not understand them. Make effort to understand them. Do not skip any chapter.

Part- 1 Self Study Material

Chapter 1

Classification of Elements

(Marks - 5, Marks with option 8)

Introduction :

Man has discovered 116 elements uptil now. We use (Modern Periodic Table, (Long Form of Periodic Table) prepared by Moseley to study the elements.

Features of Modern Periodic Table :

(1) Atomic number of an element is equal to the number of protons or electrons in the atom of the element.

(Atom of any element contains equal number of protons and electrons)

(2) Moseley observed that properties of the elements are a periodic function of their atomic numbers. Therefore Moseley assumed atomic number (z) as the basis of modern periodic table.

(Definition of the periodic law-refer to page 2)

(3) There are 7 rows (known as periods) and 18 columns (known as Groups) (4) At the bottom of the periodic table there are 2 additional rows known as series. All

the elements in series belong to third group. Do not misunderstand series as periods.

(5) Place of element in the periodic table suggests -a) Serial No. of period

b) Whether the element is metal, non-metal or inert gas. c) reactivity of an element

d) Valency of an element

e) no. of electrons in the outermost orbit.

f) While studying periods, one has to refer the periodic table from left to right gradually. One can understand whether the element is metal, non-metal, metalloid or an inert gas.

g) Refer the group from top to bottom. No. of shells go on increasing from top to bottom gradually. All the elements from the same column contains equal number of electrons in the outermost shell.

h) The entire table is divided into four blocks based on electronic configuration of elements.

Block Group No. Name

s block 1,2 Normal elements

p block 13 to 17 Normal elements

p block 18 Inert elements

d block 3 to 12 Transition elements f block lanthenide, Inner transition

actinide series elements

The periodic table can be read like a cross-word puzzle.

Prepare a blank periodic table and solve the following.

Ex. 1 : The element carbon (C) is in second period and 14th _{group of the periodic table.}

Place element Carbon in the proper place.

Ex. 2 : Atomic no. of hydrogen is 1and that of lithium is 3. Place the elements in proper

place in the periodic table. (Note - Electronic configuration of hydrogen is 1, and lithium is 2,1) 2 3 4 5 6 7 1 periods 1/IA 2/IIA s-block Modern Periodic T able

Transition elements (Metals)

d-block 3 _IIIB 4 IVB 5 _VB 6 VIB 7 VIIB 89 VIII 10 11 IB 12 IIB ○ ○ ○○○○○○○ ○ 13 _IIIA 14 _IVA 15 VA 16 _VIA 17 _VIIA p-block (Normal elements) 5 B 6 C 14 S 32 Ge 33 As 51 Sb 52 Te 84 Po 85 At * * series - 1 * series - 2 * No. of Elements - period - first = 2, period second = 8 period third = 8, period fourth = 18, period fifth = 18 period sixth = 32,period seventh = 30

The bordering elements along the zig-zag line

are the Metalloids

f-block

Elements: z symbol (Electronic configuration) 18/ zero

Ex. 3 : Atomic no. of silicon is 14, which is greater than atomic no. of carbon by 8. Locate

the position of silicon in the periodic table, with the help of given information. Prepare a blank table (Table no. 1.2, page 4)

If we consider the table from left to right, we can observe that the chemical properties of elements change gradually. Table no. 1.2 consists of the molecular formulae of compounds formed by elements of the second period. The change in valency depends upon the change in electronic configuration of elements. Gradual change in the molecular fourmulae is synchronous with the change in properties of elements. While filling Table no. 1.2, first write symbol and valency of the element. After that write molecular formula of the corresponding compound.

e.g. LiCl

₆

Li

₆

1

Q. 1 Atomic no. of Mangnesium is 12. And molecular formula of Magnesium Chloride is MgCl₂. If the At. no. of aluminium is 13, Write the molecular formula of aluminium chloride.

Q. 2 Atomic no. of Carbon is 6, and molecular formula of Carbon chloride is CCl₄. Atomic no. of boron is 5. Find out and write the molecular formula of boron chloride.

Table 1.2 Compounds of Second Period Elements.

Element Chloride Oxide Hydride Valency

Table No. 1.3 Electronic Configuration of elements of first three Periods.

Molecular Formula of

Period (Shell) 1 K

1

I A/1 II A/2 III A/13 IV A/14 V A/15 VI A/16 VII A/17 2 (K,L) 3 (K,L,M) Valency Element : z symbol

Important points to remember :

Moseley's modern form of the Periodic Law, Periods, Groups, Series, Normal elements, Inert elements, Transition elements, Inner transition elements, Metalloid, Sub-atomic particles.

Solved Sample Questions

Give scientific reasons.

1) The last column on the right side of the modern periodic table is called zero group/18 or noble gases. (page no. 7)

Expected Answer

i) In this group elements have stable electronic configuration with complete duplet or complete octet.

ii) Because of this the valency of this group of element is zero.

iii) The atoms of this group of elements form neither ionic nor covalent bond with other atoms.

iv) The elements of the group have their last shell completely filled.

Write the difference between the following. Periods Groups

···

i) The horizontal rows of elements in the modern periodic table are called periods.

ii) The period number indicates the number of electronic shells present in an atom belonging to that period. iii) Elements in the same period

do not show similar Chemical properties.

iv) Seven rows are present.

i) The Vertical columns of elements in the modern periodic table are called groups.

ii) The group number indicates the number of electrons in outer shell of an atom belonging to that group. iii) The elements is the same

group show similar chemical properties.

iv) Eighteen columns are present.

Chapter 2

Electrolysis

(Marks - 4, Marks with option 7)

Introduction :

You must have read in the newspapers about mishaps and accidents caused due to lightening falling on the earth. Lightening occurs when the clouds collide with each other. When lightening falls on the earth its electric level is much greater than that of the earth which is zero. This level of electric energy is called potential. Difference between electric level is called potential difference. Difference between electric level causes electric current.

1) Conductivity of substance :

Every substance is made up of atoms and molecules. Substances having free electrons in large numbers in their atoms conduct electricity.

Electric conductors are three main types. We are going to study electrolytic conductors.

2) Comparison between Electrolytes and Non Electrolytes :

Electrolytes Non Electrolytes

3) Difference between anion and cation

Anion Cation

i) A substance whose aqueous solution can conduct electric current is called electrolyte. ii) Most of the electrolytes are

electrovalent compounds. iii) All acids, bases and salts

are electrolytes. Molten salts some covalent compounds (HCl, NH₃) undergo ionisation on dissolving in water. They are also electrolytes.

i) A substance whose aqueous solution can not conduct electric current is called non-electrolyte.

ii) Non-electrolytes are co-valent compounds.

iii) Organic compunds like glucose, urea, cane sugar alcohol are non electrolytes.

i) Anions are negatively charged ions.

ii) During electrolysis anion gives electrons and becomes neutral.

iii) During electrolysis anion attracts towards anode, hence the name anion.

i) Cations are positively charged ions.

ii) During electrolysis cation accepts electrons and becomes neutral.

iii) During electrolysis cation attracts towards cathode, hence the name cation.

Note - Positively charged electrode is known as anode and negatively charged electrode

is known as cathode.

4) For types of conductors and definitions of electrolyte and non-electrolytes refer to text

book page no. 13,14.

5) Electrolysis of Copper Chloride :

(Observe Fig 2.2 on 16 from the text-book) Both the electrodes are carbon rods. Open end of Carbon rod attached to positive terminal of the battery is called as anode. Open end of Carbon rod attached to the negative terminal of the battery is called as cathode.

The observation and inference of the experiment is as follows :

The chemical reaction during electrolysis of copper chloride solutions : a) Reaction at Cathode :

Positively charged copper ions attract towards negative electrode, accept electrons and convert into copper metal

b) Reaction at anode :

Negatively charged chloride ions attract towards positive electrode, give electrons and convert into chlorine gas.

Note : During electrolysis of copper chloride, chloride ion (Cl–_{) gives away electrons and}

copper ion (Cu2+_{) accepts the electrons and copper and chlorine form is elemental form.} 6) Electroplating :

Refer to Fig. no. 2.3 on page 18 of the text book.

Note the following points about the process

-a) An article to be electroplated should be used as negative terminal (cathode) b) A rod of the less active metal to be coated on the article is used as positive

terminal (anode)

c) The electrolyte used should be of the same metal, which is used as anode.

For example

-For electroplating of silver on copper article, Copper article should be treated as cathode and silver rod should be treated as anode.

Changes Observation Inference

1) At cathode 2) At anode 3) Colour of electrolyte 1) Reddish deposit is seen. 2) Bubbles of a gas are observed. 3) Blue colour starts

fading.

1) Copper is deposited at Cathode.

2) The gas given out is chlorine. 3) Due to decreasing concentration of copper chloride molecules colour fades. Cu2+ (aq) + 2e – →Cu (s) 2Cl– (aq)→ Cl2(g)2e –

The electrolyte should be a solution of silver nitrate. Go through and study Table no. 2.1 for the process.

7) Anodising Technique

-Get information about the process (described) on page no. 18,19 of the text book.) and practical applications.

8) Ion as charge Carriers in living system :

Every living being is made up of cells. Cell contains Cell-sap and inorganic ions. They control the functions of cells. (Read more information about the process on page nos. 19, 20 of the text book.)

Points to remember :

i) Electric potential is the level of electric energy.

ii) Electric current flows from higher electric potential to lower electric potential. iii) Electric conductors are of three types

1) Gaseous conductors

2) Metallic (Electronic) conductors 3) Electrolytic conductors

iv) Electrolytic conductors conduct electric current by movement of ions.

·

_{Electrolytes : These are the substances in the molten state or whose aqueous}

solutions conduct electric current.

·

_{Non-Electrolytes : These are the substances which do not conduct electricity either}

in the molten state or in the aqueous state.

·

_{Electrolytes : All acids, bases, salts.}

·

_{Non-Electrolytes : Carbon compounds like glucose, sugar, urea, alcohol etc.}

·

Electrolysis : It is the process of decomposition of an electrolyte by the passage of

electric current.

·

_{Electroplating and anodising are the applications of electrolysis.}

·

_{Electroplating is a process by which a metal or an alloy is coated with another less}

active but more attractive metal, using electrolysis.

·

_{Anodising is another application of electrolysis where the surface of the aluminium}

anode is covered by a thin film of aluminium oxide to make the aluminium surface resistant to corrosion and abrasion.

·

_{During electrolysis, cations move towards cathode and anions move towards the}

anode.

·

When aqueous copper chloride solution is electrolysed, copper is deposited at the cathode and chlorine is evolved at the anode. Due to ionisation of copper chloride, the no. molecules of copper chloride in the solution decreases and hence the colour fades.

·

_{Many phenomena in living organisms are controlled by ion transport across cell}

·

_{Healthy heart requires proper K}+_/Na+ _{balance in the body.} Solved Sample Questions :

1) Give scientific reason

-Distilled water is a bad conductor of electricity.

Ans : a) Pure distilled water being a covalent compound does not dissociate. b) As it does not dissociate, we cannot get cations and anions. As there is

no movement of ions, distilled water does not carry current. Hence it is a bad conductor of electricity.

Chapter 3

Strength of Solutions

(Marks - 4, Marks with option 6)

Introduction :

We use large number of solutions in our daily life. e.g. tea, coffee, sherbat,medicines, essences, preservatives. Solutions play vital role in our life. We are going to study solutions, types of solutions from a chemical point of view.

3.1 Study the definitions of - solution, solvent, solute, aqueous solution, non-aqueous

solution.

Types of solutions according to state of solute and solvent.

Note - Identify solvent, solute from the solutions used in daily life. 3.2 Ionisation and Dissociation

a) Ionisation

-Ionic solutions when dissolved in water turns into positive and negative ions. The process is called ionisation.

b) Dissociation

-The process of separation of positive ions and negative ions in the solution of ionic compound.

Solutions is called dissociation.

Note that in any solution, ionisation is the first stage and dissosiation in the second stage.

Acidity of Base Sr.

No.

Base No. of OH– _{ions in} Acidity the molecule

1. 2. 3. 4.

Sodium hydroxide (NaOH) Ammonium hydroxide (NH₄OH) Calcium hydroxide [Ca(OH)₂] Aluminium hydroxide [Al(OH)₃]

1 1 2 3 1 1 2 3 Sr. No.

Types of Solution Nature of Example Solvent Solute 1. 2. 3. 4. 5. Solid in liquid Gas in liquid Gas in gas Liquid in liquid Solid in solid liquid liquid gaseous liquid solid solid gaseous gaseous liquid solid salt solution soda water air water in milk alloys e.g. brass

Neutralisation :

The reaction in which H+ _{ions from acid and OH}– _{ions from base combine to form}

H₂O moecule, is known as neutralisation. The reaction between base and acid is NaOH + HCl

→

NaCl + H₂O

Note : ionic form of NaOH is Na+ _OH–

and that of HCl is H+ _Cl–

A) Points to remember (page nos. are given in the bracket) i) Solution, aqueous, solution non-aqueous solution (24)

ii) Acid, Base (25)

iii) Strond acid, weak acid, strong base, weak base (26) neutralisation reaction (26) iv) Equivalent weight : equi. w.t. of an acid

Equivalent weight of a base, basicity of an acid, (27)

acidity of base (28)

Normality (28)

3.3 Study Arrhenius theory from the text book (page no. 25 to 28). Also familiarize with

the terms like strong acid (HCl) strong base (NaOH) weak acid (CH₃COOH) weak base (NH₄OH)

Sr. No

Acid / Base Example Percentage of dissociation 1 2 3 4 Strong acid Weak acid Strong base Weak base Hydrochloric acid (HCl) Sulphuric acid (H₂SO₄) Nitric acid (HNO₃) Acetic acid Cytric acid Lactic acid

Sodium hydroxide (NaOH) Potassium hydroxide (KOH) ammonium hydroxide calcium hydroxide to a large extent to a lesser extent to a large extent to a lesser extent

3.4 Study the concept of equivalent weight (Page no. 27 of the text-book)

Go through the table of basicity of an acid and acidity of base.

Basicity of Acid :

B) Study the following formulae, Use appropriate formula to solve the numericals :

i) Equi. wt. of acid =

ii) Equi. wt. of base =

iii) Normality (N) =

iv) Molarity (M) =

v) No. of moles of solute (n) =

vi) from (iv) and (v) we get,

Molarity (M) =

M =

vii) Normality Equation :

N₁V₁ = N₂V₂

N₁ = Normality of an acid / N₂ = Normality of base V₁ = Volume of an acid / V₂ = Volume of base unit of Normality is N and that of volume is ml.

Sr. No.

Acid No. of H+ _{ions in} Basicity the molecule 1. 2. 3. 4. Hydrochloric acid (HCl) Nitric acid (HNO₃) Sulphuric acid (H₂SO₄) Phosphoric acid (H₃PO₄) 1 1 2 3 1 1 2 3

Molecular mass of base Acidity of base

Wt. of solute in gram

Gram equivalent wt. × Volume in litre Wt. of solute in gram

Mol. mass of solute × Volume in litre Mass of solute in gram

Mol. mass of solute

No. of moles of solute (n) Volume of solution in litre (V) n

V

Molecular mass of acid Basicity of acid

1) Relation between molecular mass and equivalent weight of an acid or a base.

Note : u is the unit.

2) In normality equation -N₁V₁ = N₂V₂

N₁ = Normality of an acid (unit N) V₁ = Volume of an acid (unit ml) N₂ = Normality of a base V₂ = Volume of a base

Points to Remember :

·

_{Solution - It is a homogeneous mixture of two or more different substnaces.}

·

_{Solution = Solute + Solvent}

Solute is the dissolved substance in the solution and is a minor component while solvent is the major component of the solution in which the solute is dissolved.

·

_{According to Arrhenius theory, when dissolved in water, an acid releases (H}+_{) ions and}

a base releases (OH–) ions.

·

_{Neutralisation - An acid and a base in their aqueous solutions react with each other}

to form salt and water

·

Basicity of an acid and acidity of a base depends on the number of replaceable H+

and OH– ions present in the acid and base respectively.

·

_{Standard Solution - A solution whose concentration is accurately known.}

·

_{Normality = Basicity × Molarity (For acid)}

·

_{Normality = Acidity × Molarity (For base)}

·

_{Normality equation : N}

1 × V1 = N2 × V2 .

·

Weight dissolved = Normality × Gram equivalent weight per litre of solution.

Acid / Base Molecular Mass Equivalent weight

1) HCl 2) H₂SO₄ 3) HNO3 4) NaOH 5) KOH 6) Ca(OH)₂ 36.5 u 98 u 63 u 40 u 56 u 74 u 36.5 u 1 = 36.5 u 98 u 2 = 49 u 63 u 1 = 63 u 40 u 1 = 40 u 56 u 1 = 56 u 74 u 2 = 37 u

1) Solved Sample Questions

2) Give Scientific reasons :

Due to excessive use of chemical fertilizers, fiels lose its fertility.

Ans : Land is basic in nature. Land becomes salty due to excessive use of chemical fertilizers. These basic substances in the soil react with the acids in the fertilizers and produce salts. In excessive salty soil, the plants cannot grow and therefore fields lose its fertility.

3) Solved examples

-i) Text book Q. 4 (35) ₁st _{example in the table.}

HCl - mass of solute in solution = 18.25 g. Volume of the solution = 1000 ml. Find normality. Normality of HCl =

=

=

= = 0.5 N.

ii) Q. 4 i) An acid is 0.01 N. 9 ml of this acid required 10 ml of basic solution for complete neutralisation. Find the normality of base and weight of base dissolved in 1000ml of solution.

(Equivalent weight of base = 56.)

Acids Bases

1) All acids contain (H+) ions

2) In aqueous solution acids release hydrogen ions (H+) 3) Acids are sour to

taste

4) Acids turn bule litmus to red.

1) All Bases contain (OH–) ions

2) In aqueous solution bases release hydroxyl ions (OH–)

3) Bases are corrosive, hence should not be tasted. 4) Bases turn red litmus to

blue. 1 2 18.25 36.5 1000 1000 ×

Weight of solute in gram

Gram equivalent weight × Volume in litre 18.25

Solution Given -Normality of acid N₁ = 0.01 N Volume of acid V₁ = 9 ml Volume of base V₂ = 10 ml Normality of base N₂ = ?

Weight of base dissolved in 1000ml of solution = ?

Acid Base N₁ × V₁ = N₂ × V₂ 0.01 × 9 = N₂ × 10

ˆ

= N2

ˆ

N2 = 0.009 N

ˆ

Normality of base = 0.009 N.

Weight of base dissolved in 1000 ml means in 1 litre of solution = Normality × Equivalent weight

= 0.009 × 56 = 0.504 g

···

0.01 × 9 10

Chapter 4

Current Electricity

(Marks - 3, Marks with option 6)

Introduction :

Now-a-days electricity is being used for daily activities of human being. So that when there is no power supply, life gets disturbed. For domestic use; electrical appliances like refrigerator, oven, hot plate, mixer etc. are used for daily activities in kitchen. For washing clothes; washing machine is used. So the demand for electricity is increasing, while the generation of electricity has not increased in the same rate that of the demand.

Knowledge : In the 9th std. text book previous study the structure of atom and the main

constituents of the atom.

A) Some important concepts, terms and definitions (Study carefully)

(In bracket page number is given from text book). 1) Static electricity (page 37)

2) Current electricity (page 37) 3) Conductors of electricity (page 38) 4) Insulators (page 38)

5) Semiconductors (page 38)

6) Simple Voltaic Cell (page 39 fig. 4.2)

7) Direction of conventional current in the wire (page 40 fig. 4.4) 8) Definitions and units of Coulomb, Volt and Ampere (42) 9) Ohm's law (page 44)

10) Super conductors (page 48) 11) Electromotive force (page 49)

12) The equivalent - resistance when connected in series (page 51) 13) The equivalent resistance when connected in parallel. (page 52)

B) Draw and lable the diagrams :

(Study carefully and practice)

1) Atoms of hydrogen and lithium (page 38) (fig. 4.2) 2) Simple Voltaic cell (page 39) (fig. 4,2)

3) Verification of Ohm's law [(page 46, fig. 4.7 (a)]

4) Connection diagram of resistances in series. (page 51, fig. 4.9) 5) Connection diagram of resistances in parallel. (page 52, fig. 4.10)

C) Observe the following table and classify domestic objects :

Conductors Insulators Semiconductors

Gold, Silver, Copper, Aluminium, Iron, Graphite, aqueous solution of salts, like NaCl Rubber, Plastic, glass, mica, porcelein, etc. Germenium, gallium, Silicon

D) Symbols commonly used in circuit diagrams study carefully and practice (page 43, table 4.1)

E) For details of concept of potential and potential difference (refer page 41 and 42) F) Resistance of a conductor and factors on which resistance of a conductor depends

(page 45)

G) What are Ohmic conductors and non Ohmic conductors, study V against I graphs for them. (page 47, 48, fig. 4.8)

H) Study the following formulae and make use of these for solving numericals.

1) Potential difference (P.D.) between two points

ˆ

V =

Here, V = Potential difference (Volts) W = Work done (Joules) Q = Electric charge (Coulomb)

ˆ

1 Volt =

2) Electric Current (I) Current =

ˆ

I =

Here, I = Electric Current (Ampere) Q = Electric charge (Coulomb) t = time (second)

ˆ

1 Ampere =

3) Ohm's law equation form : = Constant

Here V = potential difference (Volts) I = electric Current (Ampere)

The constant in the above relation is called resistance of the conductor and is denoted by R. Charge time Q t 1 Coulomb 1 Second V I V I

ˆ

= R W Q Work done (W) Electric charge (Q)

P.D. =

1 Joule 1 Coulomb

This is the symbolic form of Ohm's law. = 1 Ohm

(N. B. :- '

Ω

' This symbol is used for ohm.)

4) R = is the relation between electrical resistance (R), length of the conductor (L) and area of cross of the conductor (A).

ρ

is constant of proportion and is called resistivity of the material.

R = Electrical resistance (Ohm) L = Length of the conductor (meter) A = Area of cross section of conductor (m2 _{i.e. square meter)}

ρ

= Resistivity (Ohm-meter)

5) Resistances in series and equivalent resistances

R_s = R₁ + R₂ + R₃ + ... + R_n Here R_s = Equivalent resistance

R₁, R₂, R_{3 .}... R_n are various resistances connected in series combination. This is a general equation.

For three resistances connected in series combination the formula will be R_s = R₁ + R₂ + R₃

8) Resistances in parallel and equivalent resistance :

This will be a general form,

For three resistances connected in parallel combination use the following formula.

7) Learn the following conversions :

i) 1 Newton = 105 _dynes

ii) 1 Volt = 1000 millivolts = 103 _{millivolt (or mv)}

iii) 1 Ampere = 1000 milliampere = 103 _mA.

iv) 1 Meter = 100 cm. v) 1 sq. meter = 100 × 100 sq.cm. = 104 _cm2_. 1 Volt 1 Ampere

ρ

L A 1 R_p 1 R₁ 1 R₂ 1 R₃ = + + 1 R_p 1 R₁ 1 R₂ 1 R₃ = + + + ...+1 R_n

Learn the following symbols for electrical terms and make use of them.

8) Observe the table for difference between resistances in series and in parallel combition.

Some Illustrative Examples :

1) Calculate the potential difference V across a 10

Ω

resistor carrying a current 0.2A.

Ans. : Given Resistance (R) = 10

Ω

Electric current (I) = 0.2 A; V = ?

Formula : R = substituing the values,V_I

ˆ

10 =

ˆ

V = 10 × 0.2

ˆ

Potential difference (V) = 2 Volts

V 0.2

Term Symbol Unit (Mks) Unit symbol 1. Electric force 2. Electric charge 3. Potential difference 4. Electric current 5. Time 6. Electrical resistance F Q V I t R Newton Coulomb Volt Ampere Second Ohm N C V A s

Ω

Resistances in series Resistances in parallel

i) Many resistances are connected one after another

ii) Equivalent resistance will be given by R_s = R₁ + R₂ + R₃ + ... + R_n iii) R_s is greater than R₁,

R₂, ..., R_n.

iv) Series combination is used to increase the resistance of a circuit.

i) A number of resistances are connected between two common points. ii) Equivalent resistance R_p

will be given by

iii) R_p is less than R₁, R₂, ..., R_n.

iv) Parallel combination is used to reduce the resistance of a circuit. 1 R_p 1 R₁ 1 R₂ = + ...+1_R n

2) Calculate the resistance of the conductor if Potential difference is 1.5 Volt and 300 mA

current passes through it. Ans. : Given P.D. (V) = 1.5 Volt;

Current (I) = 300 mA = 0.3 A; R = ? Formula : R = Substituting the values

R = R = = 5

ˆ

Resistance of the conductor = 5

Ω

3) You are given two resistances 20 Ω and 5

Ω

. What will be their equivalent resistance (a) in series (b) in parallel combination.

Ans. : Given R₁ = 20

Ω

; R₂ = 5

Ω

. (a) In series combination,

R_s = R₁ + R₂ R_s = 20 + 5 = 25

Ω

.

ˆ

In series combination equivalent resistance = 25 ohms

(b) In parallel Combination,

ˆ

= +

ˆ

= =

ˆ

=

ˆ

Rp = 4 Ohm

4) Resistors of 16 ohms and 4 ohms are connected in parallel combination and 5 ohms

resistor is connected in series combination with them. Find the combined resistance, Ans. : Given : R₁ = 16

Ω

; R₂ = 4

Ω

,

these are connected in parallel, V I 1.5 0.3 15 3 1 R_p 1 R₁ 1 R₂ = + 1 R_p 1 20 1 5 1 R_p 1 + 4 20 5 20 1 R_p 1 4 1 R_p

ˆ

= +

ˆ

= +

ˆ

= 1 R₁ 1 R₂ 1 R_p 1 16 1 4 1 R_p 1 + 4 16

Now Rp will act as R₁ and R₂ will be 5

Ω

,

ˆ

In series combination,

R_s = R₁ + R₂

ˆ

Rs = +

ˆ

Rs = 3.2 + 5 = 8.2

Ω

.

5) 8 ohm and 4 ohm resistors are connected in parallel combination and the e.m.f. of the

cell is 3 volt and internal resistance ohm.Find the total current in the circuit. Ans. : Formula : Total voltage generated by the battary is given by the formula :

E = IR + Ir = I (R + r) ...I E = E.M.F. of the cell = 3 Volt,

Internal resistance of the cell (r) = ohm, First we have to find R = ?

In this case R₁ = 8 ohm and R₂ = 4 ohm resistors are connected in parallel combination,

Now substituting the values of E, R, r in eqn (I) we get, 3 = I ( + )

ˆ

3 = I ( )

ˆ

3 × = I

ˆ

I = Ampere = 0.75 A

ˆ

Total current in the circuit = 0.75 A.

1 R_p 1 R₁ 1 R₂

ˆ

= + 1 R_p 1 8 1 4

ˆ

= + 1 R_p 1 + 2 8

ˆ

= 8 3

ˆ

Rp = ohm 8 3 4 3 12 3 3 12 3 4 4 3

ˆ

=

ˆ

Rp =

Ω

, 1 R_p 5 16 16 5 16 5 5 1 4 3

6) Three resistance of each 12 ohm are connected 1st in series combination, then in

parallel combination. Determine their equivalent resistance in each case. Ans : Given R₁ = 12 ohm, R₂ = 12 ohm.

Also R₃ = 12 ohm,

First consider in series combination; R_s = R₁ + R₂ + R₃

ˆ

Rs = 12 + 12 + 12 = 36 ohm

Now in parallel combination,

Q. 2 Give scientific reasons :

1) Potential difference of source of current (i.e. cell) is less than e.m.f.

Ans : As e.m.f. of the source is the work done by the source in raising a unit positive charge from its lower potential end to higher potential end. The energy imported to the charge by the source is used to circulate the charge round the circuit against opposition to external circuit and internal resistance of the source. Therefore e.m.f. (E) is greater than the potential difference (V) between its terminals.

2) The plates of heating devices such as toaster, an electric iron are made of an alloy rather than a pure metal.

Ans. : A good conductor of a given size has a low resistance. Copper is a good conductor while some alloys like nichrome and constantan have a high resistance. So the heat will be produced more, when current is passed through nichrome coil or plates.The melting point of nichrome is also higher than pure metal. That is why the material of toaster electric iron are made of an alloy like nichrome rather than pure metal.

Instructions :

This lession is important from the point of examination as well as for scoring more marks. Different types of questions are asked on this lesson.

- Learn by heart the definitions and rules. - Practice the diagrams properly.

- Solve all types of sums for practice

Questions like 'Derive the equation' can be asked on this lesson. e.g. Derive an equation for equivalent resistance for three resistances in series / parallel.

1 R_p 1 R₁ 1 R₂

ˆ

= + + 1_R 3 1 R_p 1 12 1 12

ˆ

= + + 1₁₂ 1 R_p 1 + 1 + 1 12

ˆ

= = =

ˆ

=

ˆ

Rp = 4 ohm

ˆ

Resistance in series = 36 ohm and

resistance in parallel = 4 ohm 3 12 1 4 1 R_p 1 4

Chapter 5

Effects of Electric Current

(Marks - 3, Marks with option 6)

Introduction :

Today our most of the daily activities depend upon electric power, e.g. mixer, refrigerator, heater, oven, T.V., radios, even recharging of the cell phone requires electric power supply. When power supply breaks down, our many activities stop and life gets disturbed. In this chapter we are going to study the heating effect and magnetic effect of electric current and their practical applications.

Previous knowledge : In the previous chapter we have studied some important concepts

about electric current, like resistance of a conductor.

A) Some important conceptds, terms and definitions - study carefully them.

(Numbers in the bracket shows the page numbers from the text book) 1) Heating effect of an electric current (page 60)

2) Joule's law (page 63)

3) Magnetic effect of an electric current (page 66) 4) Right hand rule (page 69)

5) Direct current and Alternating current (page 71)

B) Draw and label the diagrams and practice

1) Verification of Joule's law (fig. 5.1) (page 61) 2) Oersted's experiment (fig. 5.3) (page 66)

3) Lines of magnetic field due to current carrying straight line conductor (fig. 5.5) (page 68)

4) Electric bell (fig. 5.9) (page 72)

5) Telephone ear piece (fig. 5.10) (page 73)

C) Study the following formulae carefully and make use of these formulae for solving numericals.

1) Joule's law formulae : H = I2_{Rt Joules}

Here, H = Quantity of heat produced I = Electric current (Ampere) R = Electrical resistance (Ohms) t = time (seconds)

To convert the above heat in calories, substitute 1 calorie = 4.18 Joules

By Ohm's law we substitute I = H = caloriesI

2 _{R t}

4.18 _V

we gets,

Again by same process we get,

ˆ

Various forms of Joule's formula will be as follows :

i) H = I2_{Rt Joules ii) H = calories}

iii) H = calories iv) H = calories * (Cal is the short form used for calories.)

2) Electric Power

P = VI

P = electric power (Watts)

V = Electric potential difference (Volts) I = Electric current (Ampere)

By Ohm's law we substitute, V = IR The above equation will be

-P = Here ,

V = Potential difference (Volts) R = Electrical resistance (ohms)

N.B. : 1) 1 Watt =

1 kilowatt = 1000 watts

2) If we observe the electric bulb,We see the following information

-(a) The power of the bulb (in Watts) (b) The potential difference (in Volts)

e.g. 40 Watts, 250 Volts. It means the power of the bulb is 40 Watts, and it works fully at 250 Volts P.D. Hence we observe dim light when the voltage is lowered.

D) Study carefully the following units.

1) The quantity of heat generated in a conductor depends upon the following factors ... (page 63)

2) Application of the heating effect of electric current (page 64, 65) 3) Magnetic effect of electric current (page 66, 67)

4) Solenoid and magnetic field produced by solenoid (page 70, 71) 5) Direct and alternating current information. (page 71)

H = caloriesV 2 _t 4.18 R H = caloriesV I t 4.18 V I t 4.18 V2 _t 4.18 R I2 _Rt 4.18 V2 R 1 Joule 1 second

6) Safety precautions while using electrical appliances:

a) Precautions against bad insulations (page 74) b) Precautions against improper earthing (page 75) c) General precautions for safety. (page 75)

E) Study the following information carefully :

1) The resistance of the conductor depends on the material used. Gold, silver, metals have very low resistance. But these are noble metals, hence for conduction of electric current wires of gold, silver are practically impossible. Copper and aluminium are also good conductors and have low resistance. So for electric wiring purpose copper or aluminium wires are used.

2) The resistance of the conductor also depends upon the area of cross-section. Smaller the area of cross-section, larger will be the resistance. Therefore in electric bulb very thin hairy wire of tungston metal in the form of coil is used. Due to coil formation length of the wire increases, which shows that resistance is directly proportional to the length, i.e. larger the length of wire, larger will be the resistance. So in heater, geyser the nichrome coil is used

3) In a safety fuse lead allloy wire is used. Because lead has very low melting point, hence when, high current flows through the circuit, the fuse wire gets heated and melted. The circuit is broken and current stops flowing, so the electric appliance will not be damaged. For different types of electric appliances fuses of suitable capacities are used.

4)

(N.B. - w.r.t. mean with respect to)

Solved Sample Questions Difference between A.C. and D.C.

Types of metal Uses w.r.t. electricity

1) Copper, aluminium 2) Nichrome

3) Tungston

Conduction wires, connection wires Heater, geysers, electric iron, etc. Electric bulb

Direct Current (D.C.) Alternating Current (A.C.)

1. The direct current always flows in one direction.

2. It is non-oscillating current generated by the source. 3. Generally D.C. is generated

from an electric cell/battery.

1. Alternating current reverses its direction periodically.

2. It is oscillating current generated by the source. 3. Alternating current is

Illustrative examples :

1) Calculate the heat generated in a coil of resistance 209 ohm and 0.5 amp. current is passed through it for 2 minutes.

Ans. : Given Resistance (R) = 209 ohm Current (I) = 0.5 amp.

time (t) = 2 min = 2 × 60 = 120 sec. Formula H = calorie.

ˆ

H = calories

ˆ

H = calories

ˆ

H = 1500 calories

ˆ

The heat generated in the coil = 1500 cals.

2) Calculate the heat generated in an electric iron, if P.D. is applied 240 Volts, and 418-mA current is passed for 1 minute.

Ans. : Given P.D. (V) = 240 Volts, Current (I) = 418 mA = 0.418 Amp. time (t) = 1 min = 60 sec.

Formula H = calorie.

ˆ

H = calories

ˆ

H = 1440 calories

ˆ

Heat generated in the electric iron = 1440 cals.

3) Find the resistance of 40 Watt, 240 Volt bulb. Ans. : Given : Power (P) = 40 Watt

P.D. (V) = 240 V R = ?

Formula P = Substituting the values.

ˆ

40 =

ˆ

R =

ˆ

R = 1440 ohms

ˆ

The resistance of the bulb = 1440 ohm.

I2 _{R t} 4.18 (0.5)2 _{× 209 × 120} 4.18 0.5 × 0.5 × 209 × 120 × 100 418 VIt 4.18 240 × 0.418 × 60 4.18 VIt 4.18 V2 R (240)2 R 240 × 240 40

V2 _t

R

4) Heat generated in a conductor of resistance 40

Ω

in 1 minute is 150 Joule. Calculate the P.D. applied across it to produce above heat.

Ans. : Given Resistance (R) = 40

Ω

time (t) = 1min = 60 sec. Heat generated (H) = 150 Joule

P.D. (V) = ?

Formula : H = Joules substituting the values.

ˆ

150 =

ˆ

150 × = V2

ˆ

= V2 V2= 100

ˆ

V= 100

ˆ

V= 10 volts.

P. D. between two points = 10 volts.

Q. 2 Give scientific reasons :

1) Filament of incandescent lamp is made of tungston :

Ans. :The resistance of tungston is very high, so if current is passed through tungston coil, due to its high resistance it becomes white hot which emits light. Also the melting point of tungston is very high. So it will not melt at that temperature. Therefore the filament of incandescent lamp is made of tungston.

2) Fuse is made of material having low melting point.

Ans.: The function of the fuse is to protect the electrical appliances from sudden rise of potential difference in the circuit. Therefore if such incidence happens, due to low melting point of the fuse it melts and the circuit breaks. Thus protecting the appliances connected in that circuit. Therefore generally fuse is made of lead alloy because the melting point of lead is low.

···

V2 _{× 60} 40 40 60 150 × 40 60

√

Chapter 6

Energy Sources

(Marks - 2, Marks with option 4)

Introduction :

In modern world, we need energy in every walk of life. Energy in various forms is used to cook food, for entertainment, for travel. Gas is used in kitchen, petrol or diesel is used in vehicles. These energy sources are depleting very fast; which has resulted in energy crisis.

Previous knowledge

-Students know the basic concepts of energy fuels. The important concepts / laws from the chapter. i) Classification of energy sources.

a) Non-renewable energy sources. b) Renewable energy sources.

(for definition of the above refer to the textbook page no. 81)

1) Energy sources and their classification :

2) Solar energy is the most promising energy source; but there are some limitations for

the use of solar energy. The point is discussed in detail on page no. 87,88 of the text book. Go through the details to understand the point.

3) Wind energy, tidal energy, geothermal energy are the sources of energy. But the

availability of these sources depends upon the environmental conditions. Try to collect more information about the hydroelectric projects in the country. Read about the micro and mini hydroelectric power plants from the text book (page no. 85, 86)

4) Observe the figures of solar cooker (fig. 6.9 page 88), solar water heater (fig, no. 6.10

page 89). Try to sketch and label the figures.

5) Solar Cells

It is one of the main device to convert solar energy directly into electrical energy. Solar cells are extensively used as main source of energy for artificial satellites.

6) Read more about nuclear energy and two ways to obtain the energy from the text - book

(page no. 90)

Non-renewable energy cources

Renewable energy cources

Firewood, cow dung, charcoal, coal, kerosene, cooking gas; petrol, diesel etc.

Wind energy, tidal energy, geothermal energy, energy from flowing water, solar energy, energy from biomass.

Difference between Nuclear fission and Nuclear fusion (Breeder reaction)

Note - Nuclear fission and breeder reaction are one and the same. Breeder reaction is

controlled chain reaction.

There is a slight difference between working of breeder reactor and burner reactor. Go through the text-book page nos. 90 and 91 for the same.

6) Bio-diesel :

This is a renewable energy source. Collect more information about the oil producing seed plants, cultivation of the plants, use of waste land to cultivate these plants. For merits of biodiesel fuel, see text-book page os. 93, 94.

7) Calorific value of fuel

-Get information from the text-book page no. 97. Know the merits of gaseous fuel.

8) Important figures :

Try to draw and label following diagrams i) Solar cooker (fig. 6.9 page 88)

ii) Solar Water heater (fig. 6.10, page 89) iii) Nuclear fission (fig. 6.11 page 91) iv) Biogas plant (fig. 6.12, page 93)

Points to remember

-1) Renewable and Non-renewable energy sources

2) Appliances based on the use of solar energy. 3) Ways to obtain nuclear energy

4) Types of fuels with proper examples 5) Calorific value of fuels - definition, unit

Solved sample questions Q. 1 Give scientific reasons.

1) We have to search for new renewable energy sources.

Ans. : i) Currently fossil fuels like petrol, diesel, kerosene, natural gas are mainly used as conventional energy sources.

ii) These sources are in great demand and are used in large quantities. But these sources are limited and getting exhausted at fast rate.

Nuclear fission (Breeder reaction)

Nuclear fusion

1) Radioactive material like Uranium - 235 converts into fragments releasing large amount of heat energy

2) Bombardment of neutrons is essential to carry out the reaction.

1) Hydrogen atoms

combine to form helium, releasing large amount of heat energy.

2) High temperature is required to carry out the reaction.

iii) Hence we have to search for new renewable energy sources.

2) Wind mills require specific locations.

Ans. : i) Location must be at some height and requires rigid support. ii) must have strong steady wind blowing for most of the year. iii) hence wind mills require specific locations.

Q. 2 Distinguish between.

···

Biomass Bio gas

i) Biomass refers to wood, biological waste products

ii) It burns with smoke

i) Biogas is the mixture of methane, carbon-dioxide and hydrogen-sulphide.

Chapter 7

Types of Energy

(Marks - 3, Marks with option 5)

Introduction :

We are familiar with different forms of energy — thermal energy, light energy, sound energy, magnetic energy. We are going to study mechanical energy in this chapter.

7.1 Points to remember –

i) What is energy? (page 101) ii) Types of energy

a) Potential energy (page 102) formula (page 104)

b) Kinetic energy (page 105) formula (page 106) iii) units - CGS, MKS (page106)

iv) Law of conservation of energy (page 107)

7.2 Study fig. no. 7.1 on page 103 and read more about the work done while lifting an object through the height 'h' from A to B

Note that, if

θ

is the angle between a constant force F acting on a body and a displacement 'r' of the body caused by this force, then the work done by the force is

W = F.r cos

θ

When F and r are in the same direction,

θ

becomes 0 and cos 0 = 1. Hence Work done = W = Fr.cos

θ

= F× r

Units of Potential energy and Kinetic energy:

Units of work and energy is same.

In CGS system, unit of energy is dyne-cm It is also called as 'erg'

In MKS system, unit of energy is Newton-meter It is also called as 'Joule'

1 Joule = 107 _erg. A) Potential Energy P.E. :

P.E. = mgh

P.E. = Potential Energy g = gravitational acceleration h = height of the object

It is clear from the above equation, that P.E. of a particular object will change according to the value of 'h'

When the object is at rest on the earth's surface 'h' becomes zero.

When angle between force and displacement is 900_{. Then work done becomes O}

because,

W = Fr cos

θ

= Fr cos 90

= Fr × 0 (

_ˆ

cos 900 _{= 0)}

= 0

When an object rolls on the ground h = 0, hence P.E. = 0

B) Kinetic Energy [K.E.]

K.E. = mv2

K.E. = Kinetic energy m = mass of an object v = velocity of the object

It is clear from the above equation, that K.E. of a particular object will change according to the value of 'v'.

When the object is at rest, its velocity is zero. Hence its K.E. becomes zero.

C) Law of conservation of Energy

-E = P.-E. + K.-E. (E = Total energy)

∴

E = mgh + mv2

i) When we throw the object upward or downward, the total energy of the object at any moment is always constant.

When we throw the object upward, its K.E. is maximum and P.E. is zero conversely, if we drop the object from a height, its PE is maximum and KE is minimum.

During transit, when one energy increases, the other decreases at the same rate. Hence total energy remains constant.

Note - While throwing the object upwards, K E changes from maximum to zero, at the

same time P.E. changes from zero to maximum.

ii) Study the various examples of transformation of energy from the text book. (page nos. 112, 113, 114).

Solved Sample questions Q.1 Give scientific reasons

i) The work done on an object by a conservative force is zero. If it has come back to the same point from where it started.

Ans. i) Work done on an object by conservative force depends only on initial and final positions and not on path followed.

ii) An object comes to the same point where it is started.

iii) Hence work done on an object by a conservative force is zero. 1

2

1 2

Q. 2 Solve the numericals.

1) Energy of 2J is used to lift a block of 0.5 kg. How high will it rise? (g = 10 m/s2₎

Ans. : Given E = Energy = 2J m = mass = 0.5 kg g = 10 m/s2 h = ? E = mgh

ˆ

2 J = 0.5 kg × 10 m/s2 × h

ˆ

h = = 0.4 m

(Height attained by the block = 0.4 m)

···

2 J 0.5 kg × 10m/s2

Chapter 8

Power

(Marks - 2, Marks with option 4)

Introduction :

In our daily life we use strength as power. But these two terms have different scientific meaning. If the same work is done for different time duration, then we say the power is different in scientific view. In this chapter we are going to study about power and its units and applications

Previous knowledge : In previous standard we might have been studied about work and

energy and their definitions and applications. We also know that the units of work and energy are same. viz. In CGS unit it is Erg, while in MKS unit it is Joule.

A) Some important concepts, definitions and units : (The number in the bracket is

page number from text-book)

1) Power : definition, unit (page 118)

2) Unit of work used in the industry. (page 119)

B) Study thoroughly the following information and practice for solving numericals. 1) Table No. 1 Work, power, Energy units

2) Other units of power :

1 kilowatt (KW) = 1000 Watt 1 Horse power (H.P.) = 746 Watt

3) In industry the unit of work, also energy is expressed as kilowatt-hour (K.W. hr.) For

consumption of electricity it is also KW-hr and is called as unit. We know that 1 Kilowatt = 1000 Watts

and 1 hour = 3600 seconds.

4) Study carefully the following.

1) Work = Force × displacement

ˆ

W = F × s Unit of force (F)

In CGS system the unit of force is dyne and in MKS system it is Newton. 1 Newton = 105 _dyne.

Units of displacement :

In CGS system unit of displacement is cm and in MKS system it is meter (m)

Physical quantity Symbol used CGS Unit MKS Unit Work Power Energy Force W P E F Erg. Erg/second Erg Dyne Joule

Joule/second i.e. watt Joule

If we lift any object through height 'h' (perpendicular) from ground level, in such case we use following formula for work.

W = mgh m = mass g = gravitational acceleration h = height 2) Power = taking w = Fs we get P = We know = V

ˆ

P = F × becomes P = F × V

N.B. Study carefully table No. 8.1 from the text book. Solved SampleQuesions

1) A man draws a bucket of water from a well 10 m deep; in 20 seconds. If the mass of water drawn is 20 kg. Find the power used by the man? (g = 10m/s2₎

Ans. Given m = 20 kg, g = 10m/s2_{; s = 10m, t = 20s, P = power =?}

Formula = Power (P) =

ˆ

P =

ˆ

P = 100 W

Power used by man = 100 W.

[N.B. - Here for simplicity of calculations g = 10 m/s2 _{is given]}

2) Calculate the power of the crane which lifts the load of 600 kg to height of 10m in 2 minutes. (g= 9.8 m/s2₎ Given : m = 600 kg, g = 9.8 m/s2_{, h = 10 m} t = 2 minutes = 120 seconds Power = ? Power = = = = 490 Watt power of crane = 490 Watt

Also note that in above formula P = and P = (i.e. h = s)

···

Work time

_ˆ

_{P =} W t Fs t s t s t mgs t 20 kg × 10 m/s2 _{× 10} 20 s W t mgh t 600 kg × 9.8 m/s2 × 10 m 120 seconds mgh t mgs t

Chapter 9

Sound

(Marks - 3, Marks with option 5)

9.0 Previous knowledge :

Sound is produced by a vibrating body. Vibrating tuning fork produces sound. Vibrating string, vibrating plates also produce sound. Harmonium and trumpet vibrate air and produce sound. These instruments are sources of sound.

9.1 Production of sound :

Sound travels in the form of longitudinal waves consisting of alternate compression and rarefaction. Due to vibrations of an object when adjacent layers of air are pressed, compression is formed. When adjacent layers get seperated, rarefactions are formed. [Observe fig 9.2 in the text book). When these longitudinal waves in the air reach the ear, the ear drum is set into vibrations. These vibrations are communicated to brain and we get sensation of hearing.

9.2 Medium is essential for propogation of sound Previous knowledge : solid, liquid and gas are the media

Points to study : i) Sound waves do not travel through vaccum. They need a medium for propogation. Refer to experiment on pg. 128 i.e. Sound waves need a medium.

ii) The velocity of sound in air at 00_{C is 332 m/s. Velocity of sound is different in different}

media.

Refer table no. (9.1) in the text book i.e. Velocity of sound in different media.

9.3 Propogation of sound in different media

Previous Knowledge : The number of cycles or vibrations completed in one second is

termed as the frequency of the vibrations. The MKS unit of frequency is Hertz. (i.e. Hz)

Points to study :

i) The sound which has frequency between 20 Hz to 20,000 Hz is audible sound. This frequency range is audible range for human beings.

ii) The sound having frequency below 20 Hz and above 20,000 Hz cannot be heard by human being.

iii) To understand how sound is propogated in air, refer experiment on page no. 129 in the text book.

iv) Study of table no. (9.1) on page no. 130, we can understand that the velocity of sound is maximum in the solids, less in the liquids and least in the gases. Velocity of light is greater than velocity of sound that is why we see flash of lighting from the clouds before the thunder is heard although both occur simultaneously.

9.4 Reflection of sound waves : Previous Knowledge :

i) Some hill-stations have echo points, which attract the tourists.

ii) When you shout into a well or inside an empty hall, you hear your own sound after some time.

i.e. reflection of sound.

Points to study : Echo effect means the reflection of same sound periodically. Sound is

incident on plane surface and it gets reflected. This sound is called as reflected sound. Echo can be heard only if the reflected sound reaches the ear at 1/10th _of

a second after the direct sound is heard.

Velocity of sound in air at ordinary temperatures is 340 m/s. The minimum distance covered within 1/10th _{second will be}

340 × = 34 m

In other words echo is heard only if the reflecting surface is at least at a distance of 17 meters

9.5 Effect of wind, Temperature and Humidity on velocity of sound :

Previous knowledge : Loudspeakers are used for announcements in the village fair and

advertisements.

Points to study :

i) Wind - When sound travels in the direction of the wind, the velocity of sound is greater. When sound travels in a direction opposite to that of the wind. The velocity of sound lowers.

ii) Temparature : Velocity of sound depends upon temperature. Increase in the temperature increases the velocity of sound in air.

iii) Humidity : Humidity depends upon the water vapour present in air. The velocity of the sound in moist air is greater than the velocity of the sound in dry air. Increases in humidity increases the velocity of sound in air.

9.6 Intensity of sound :

Intensity is the amount of sound energy received per second from the source of sound.

Unit of intensity of sound is decibel (dB). For sound intensity of different sources, refer to table (9.2) on page no. 136 in the text book.

9.7 Sound pollution or noise : Noise is sound dumped into the atmosphere.

1) Children making noise in the classroom 2) Noise of fire crakers in Diwali.

Unwanted sound is called noise. This unwanted sound makes sound pollution.

Causes of noise :

Noise is caused by various sources. 1) Internal sources.

2) External sources.

Noise pollution is a health hazard. 1

Refer to page 137 and 138 of textbook for effect of noise pollution and measures of noise control.

Solved Sample Questions Q. 1 What is Echo?

Ans. : Reflected sound is known as echo

Q. 2 How the principle of echo is used to measure the depth of sea?

Ans : The principle of echo is used in the (SONAR) system of ship to detect the depth of the sea. From the transmitter in the ship sharp pulses of sound are emitted. These pulses travel downword and get reflected from the sea bed. The reflected sound is detected by the receiver in the ship.

The time interval between the production of sound and its reflection is recorded. Knowing the velocity of sound in water and time, the depth of sea can be determined by following formula

Depth of sea = Velocity of sound in water ×

Q. 3 Whistle of a passing train is clearly heard on a quiet misty night.

Ans : At night when humidity tends to rise, the sound travels faster. Hence sound can be heard more clearly on a quiet misty night.

Hence whistle of a passing train is clearly heard on a quiet misty night (pg. no. 136)

Q. 4 Define - Intensity of sound

Ans Intensity is the amount of sound energy received per second from the source of sound.

Q. 5 What is sound pollution?

Ans. An unnecessary, unpleasant, intolerable or unwanted sound is called noise or sound pollution

Illustrative Examples :

Solve the following examples :

1) A person hears a thunder 6 seconds after a flash of lightning is seen, at what distance the lightning is struck neglecting speed of light?

(Speed of sound in air is 340m/s) Solution - Speed of sound = 340m/s Time = 6s

Distance = ?

Distance = speed of sound × time = 340 × 6

= 2040 m

(The lightning has struck at a distance 2040 m from the observer.)

2) A person observes a smoke from the cannon. After 3 seconds he hears the bang. The cannon is 1020 m away from the observer. Calculate the velocity of sound in air? Ans Solution

Distance = 1020 m time = 3 s

time 2

Speed of sound = ?

Distance = speed of sound × time 1020 = speed of sound × 3

ˆ

Speed of sound = = 340 m/s (Speed of sound = 340 m/s)

···

1020 3

Chapter 10

Heat

(Marks - 3, Marks with option 5)

Previous knowledge :

Heat is one of the forms of energy and it can be obtained by transforming any other form of the energy.

Sun, wood, charcoal, cooking gas are some of the sources of heat.

Heat can be produced by chemical reactions, flow of electric current and nuclear reactions.

10.1 Anomalous behaviour of water :

Previous knowledge : Any substance when heated expands and contracts on cooling.

Most of the liquids expand on heating and contract on cooling. But water shows a remarkable exceptional behaviour between OC0 _{to 4}0_C

When water is heated from 00_{C initially it contracts in volume upto 4}0_{C instead of}

expansion.

The behaviour of the water between 00_{C to 4}0_{C is called anomalous expansion of}

water.

The anomalous behaviour of water can be demonstrated with the help of Hope's apparatus.

Practice the diagram of Hope's Apparatus. Refer pg no. 144 in the text book

At 40_{C, the volume of water is minimum. Hence density of water is maximum at}

40_{C. Density of ice is less than water, therefore ice floats on water. If we take water in}

a glass and put some pieces of ice in it, we observe that ice floats on water. Following are some examples of anomalous expansion of water.

i) In cold countries; fishes, acquatic animals and plants remain alive.

ii) In cold countries sometimes water enters into the crevices of rocks. When temperature falls below 40_{C water expands and tremendous pressure is exerted}

on rocks. So these rocks crumble into pieces.

iii) In winter, the pipe lines carrying water burst when the temperature of atmosphere falls below 40_C

10.2 Humidity and dewpoint :

i) The presence of the water vapour in the atmosphere plays an important role in everyday life. The amount of the water vapour present in the atmosphere determines the nature of the weather.

Before raining we perspire a lot. Then we say that humidity in air is increased and soon it will rain.

Similarly the region nearby sea such as Bombay, Goa, Ratnagiri etc. percentage of humidity is more in these region. So we perspire more.

ii) In winter, we observe dewdrops on grass and on vehicles. If we keep pieces of ice in stainless steel glass then we observe dewdrops on the outer surface of the glass. Because the water vapour which is in air gets condensed on the outer surface of the glass. Study the definitions of absolute humidity, relative humidity and dew point from text book (pg. no. 147, 148)

10.3 Units of Heat :

Study the definitions of calorie, kilo calorie, specific heat capacity from text book (pg. no. 149, 150)

10.4 Specific heat capacity :

This property changes from material to material (Refer to the experiment on pg. no. 149 in the text book)

Specific heat capacity of water is highest (Study table no. 10.1 on pg no. 150. Study MKS and CGS unit of sp. heat capacity.

10.5 Principle of heat exchange :

We know that for daily bathing we mix cold water to hot water. Here hot water provides heat to cold water and cold water absorbs heat from the hot water. When hot body and cold body is kept together the temperature of hot body goes on decreasing while that of the cold body goes on increasing until both the bodies attain the same common temperature within a short while.

Formula : Heat lost by the hot body = Heat gained by the cold body.

Using this formula solve the numerical examples.

Solved Sample Questions Example :

1) Certain mass of water at 640_{C is mixed with an equal mass of water at 22}0_{C. What}

will be the resulting temperature of the mixture? Ans. Given, Let mass of water = m gm

Specific heat of water = 1 cal/g0_C

t₁ = 640_{C, t} 2 = 22

0_{C t}

3 =? (temp. of mixture)

Formula : Heat lost by hot body = Heat gained by cold body mc (t₁ – t₃) = mc (t₃ – t₂)

ˆ

t1 – t3 = t3 – t2

ˆ

64– t3 = t3 – 22

ˆ

m is same for hot and cold water

ˆ

2 t3 = (64 + 22) 0_C

ˆ

t3 =( ) 0_C

ˆ

t3 = 43 0_C

Ans. The temperature of the mixture will be 430_C.

86 2

* Specific heat of water is 1

It is the highest than other most known liquids and solids.

* Due to high specific heat, hot water is used in the hot water bags and used for heating purposes.

Chapter 11

Light

(Marks - 5, Marks with option 9)

Introduction :

Light is a form of energy. Light is necessary to see the objects. We can hardly see anything in dark. Light is electromagnetic waves.

A) Points to remember (text book page nos of the relevant topic in bracket)

-1) Visible light (158) 2) Refraction of light (159)

3) Refraction of light through a glass slab (161) 4) Refractive index (163)

5) Laws of refraction (165)

6) Critical angle, total internal reflection (166) 7) Types of lenses (167)

8) Some terms used in connection with lenses.

Centre of curvature, radii of curvature, optical axis, principal foci (167, 168) 9) Lens formula (176)

10) Focal length (f) of convex lens is positive and of concave lens is negative (fig. 11.13 (a) (b) page 175)

11) Cartesian sign conventions (174, 175) 12) Magnification and diopter (176, 177) 13) Passage of light through a prism (179,180) 14) Deviation and angle of minium deviation (181) 15) Defects of eye (185, 186, 187)

16) Persistence of vision (193)

B) Draw and practice the ray diagrams

i) Refraction of light through a glass slab (fig. 11.4, page 161)

ii) Path of light through a prism (fig. 11.16, page 181) iii) A simple microscope (fig. 11.25 page 190)

iv) A compound microscope (fig. 11.26, page 191) v) An astronimical telescope (fig. 11.27, page 193)

C) Note that to construct an image of a given object placed in front of a lens by using any two of the three specialised rays. Read carefully instructions on page no 170 of the text-book.

D) Go through the tables carefully -Table no. 11.2 and 11.3

E) Draw and practice the ray diagram