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
(ENGLISH MEDIUM)

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.
Pay more attention to highlighted words / statements in the textbook. Read
(iii)

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.
The success is yours !

(iv)

Pc-2/d/Delight/D1 to D5/ Science & Tecg, P1/Std.X science & tech. p1-D1 ok A

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

Self Study Booklet (English Medium) (1) Science and Technology Part I & II

s-block Modern Periodic Table p-block
(Normal elements)
Elements: z symbol

periods
1/IA (Electronic configuration)
13 14 15 16 17 18/

lithium is 2,1)
1 2/IIA IIIA IVA VA VIA VIIA zero
Transition elements (Metals)
5 6
2 d-block B C
3 4 5 6 7 8 9 10 11 12
○ ○ ○ ○ ○ ○ ○ ○
14

○
3 IIIB IVB VB VIB VIIB VIII IB IIB S

○
32 33
4 Ge As
51 52
5 Sb Te
* 84 85
6 Po At
Place element Carbon in the proper place.

*
7

The bordering elements along the zig-zag line
No. of Elements - period - first = 2, period second = 8 are the Metalloids
period third = 8, period fourth = 18, period fifth = 18
The periodic table can be read like a cross-word puzzle.

period sixth = 32,period seventh = 30
Prepare a blank periodic table and solve the following.

f-block

series - 1 *
series - 2 *

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

place in the periodic table. (Note - Electronic configuration of hydrogen is 1, and
Ex. 2 : Atomic no. of hydrogen is 1and that of lithium is 3. Place the elements in proper

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
Formula of

Chloride
Molecular

Oxide
Hydride

Valency

Table No. 1.3 Electronic Configuration of elements of first three Periods.
Period 1 Element : z symbol Zero / 18
(Shell) 1 (Electronic configuration)
K

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


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 i) The Vertical columns of
elements in the modern elements in the modern
periodic table are called periodic table are called
periods. groups.
ii) The period number indicates ii) The group number indicates
the number of electronic the number of electrons in
shells present in an atom outer shell of an atom
belonging to that period. belonging to that group.
iii) Elements in the same period iii) The elements is the same
do not show similar Chemical group show similar chemical
properties. properties.
iv) Seven rows are present. iv) Eighteen columns are
present.

···


Chapter 2
Electrolysis
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
i) A substance whose aqueous i) A substance whose aqueous
solution can conduct electric solution can not conduct
current is called electrolyte. electric current is called non-
electrolyte.
ii) Most of the electrolytes are ii) Non-electrolytes are co-
electrovalent compounds. valent compounds.
iii) All acids, bases and salts iii) Organic compunds like
are electrolytes. Molten salts glucose, urea, cane sugar
some covalent compounds alcohol are non electrolytes.
(HCl, NH3) undergo ionisation
on dissolving in water. They
are also electrolytes.

3) Difference between anion and cation
Anion Cation
i) Anions are negatively i) Cations are positively charged
charged ions. ions.
ii) During electrolysis anion ii) During electrolysis cation
gives electrons and accepts electrons and
becomes neutral. becomes neutral.
iii) During electrolysis anion iii) During electrolysis cation
attracts towards anode, attracts towards cathode, hence
hence the name anion. 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 :

Changes Observation Inference
1) At cathode 1) Reddish deposit is 1) Copper is deposited at
seen. Cathode.
2) At anode 2) Bubbles of a gas 2) The gas given out is
are observed. chlorine.
3) Colour of 3) Blue colour starts 3) Due to decreasing
electrolyte fading. concentration of copper
chloride molecules
colour fades.

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
Cu2+(aq) + 2e– →Cu(s)
b) Reaction at anode :
Negatively charged chloride ions attract towards positive electrode, give electrons and
convert into chlorine gas.
2Cl– (aq)→ Cl2(g)2e–

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.


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
membrane.


· 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

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.
Sr. Types of Solution Nature of Example
No. Solvent Solute
1. Solid in liquid liquid solid salt solution
2. Gas in liquid liquid gaseous soda water
3. Gas in gas gaseous gaseous air
4. Liquid in liquid liquid liquid water in milk
5. Solid in solid solid solid alloys e.g. brass

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. Base No. of OH– ions in Acidity
No. the molecule
1. Sodium hydroxide (NaOH) 1 1
2. Ammonium hydroxide (NH4OH) 1 1
3. Calcium hydroxide [Ca(OH)2] 2 2
4. Aluminium hydroxide [Al(OH)3] 3 3


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. Acid / Base Example Percentage of
No dissociation

1 Strong acid Hydrochloric acid (HCl) to a large
Sulphuric acid (H2SO4) extent
Nitric acid (HNO3)
2 Weak acid Acetic acid to a lesser
Cytric acid extent
Lactic acid
3 Strong base Sodium hydroxide (NaOH) to a large
Potassium hydroxide (KOH) extent
4 Weak base ammonium hydroxide
calcium hydroxide 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 :

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

B) Study the following formulae, Use appropriate formula to solve the numericals :
Molecular mass of acid
i) Equi. wt. of acid =
Basicity of acid

Molecular mass of base
ii) Equi. wt. of base =
Acidity of base

Wt. of solute in gram
iii) Normality (N) =
Gram equivalent wt. × Volume in litre

Wt. of solute in gram
iv) Molarity (M) =
Mol. mass of solute × Volume in litre

Mass of solute in gram
v) No. of moles of solute (n) =
Mol. mass of solute

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

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

n
M = V
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.


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

Acid / Base Molecular Mass Equivalent weight
1) HCl 36.5 u 36.5 u
= 36.5 u
1
2) H2SO4 98 u 98 u
= 49 u
2
3) HNO3 63 u 63 u
= 63 u
1
4) NaOH 40 u 40 u
= 40 u
1
5) KOH 56 u 56 u
= 56 u
1
74 u
6) Ca(OH)2 74 u = 37 u
2

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 : N1 × V1 = N2 × V2 .
· Weight dissolved = Normality × Gram equivalent weight per litre of solution.


1) Solved Sample Questions
Acids Bases
–
+ 1) All Bases contain (OH )
1) All acids contain (H )
ions ions
2) In aqueous solution 2) In aqueous solution
acids release bases release hydroxyl
–
hydrogen ions (H ) + ions (OH )
3) Acids are sour to 3) Bases are corrosive, hence
taste should not be tasted.
4) Acids turn bule litmus 4) Bases turn red litmus to
to red. blue.
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.
Weight of solute in gram
Normality of HCl = Gram equivalent weight × Volume in litre

18.25
=
36.5 × 1000
1000

18.25
=
36.5

= 1 = 0.5 N.
2
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.)


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 0.01 × 9 N2
10 =

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

···


Chapter 4
Current Electricity

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


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
Work done (W)
P.D. = Electric charge (Q)
W
ˆ V= Q

Here, V = Potential difference (Volts)
W = Work done (Joules)
Q = Electric charge (Coulomb)
1 Joule
ˆ 1 Volt = 1 Coulomb

2) Electric Current (I)
Charge
Current = time

ˆI = Q
t
Here, I = Electric Current (Ampere)
Q = Electric charge (Coulomb)
t = time (second)
1 Coulomb
ˆ 1 Ampere =
1 Second
3) Ohm's law equation form :
V
= Constant
I
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.
V
ˆ = R
I


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

(N. B. :- ' Ω' This symbol is used for ohm.)
L
4) R = ρ A 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
Rs = R1 + R2 + R3 + ................. + Rn
Here Rs = Equivalent resistance
R1, R2, R3 .............. Rn are various resistances connected in series combination.
This is a general equation.
For three resistances connected in series combination the formula will be
Rs = R1 + R2 + R3
8) Resistances in parallel and equivalent resistance :
1 1 1 1 1
Rp = R1 + R2 + R3 + .............+
Rn
This will be a general form,
For three resistances connected in parallel combination use the following formula.
1 1 1 1
= + +
Rp R1 R2 R3

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.


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

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

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

i) Many resistances are i) A number of resistances
connected one after are connected between
another two common points.
ii) Equivalent resistance ii) Equivalent resistance Rp
will be given by will be given by
Rs = R1 + R2 + R3 + 1 1 1 1
................. + Rn Rp = R1 + R2 .....+ Rn
iii) Rs is greater than R1,
R2, ......., Rn. iii) Rp is less than R1, R2,
iv) Series combination is ......., Rn.
used to increase the iv) Parallel combination is
resistance of a circuit. used to reduce the
resistance of a circuit.

Some Illustrative Examples :
1) Calculate the potential difference V across a 10Ω resistor
carrying a current 0.2A.
Ans. : Given Resistance (R) = 10Ω
Electric current (I) = 0.2 A; V = ?
V
Formula : R = I substituing the values,
V
ˆ 10 = 0.2 ˆ V = 10 × 0.2
ˆ Potential difference (V) = 2 Volts


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 = ?
V
Formula : R = I Substituting the values
1.5 15
R = R = = 5
0.3 3
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 = 20Ω; R2 = 5Ω.
(a) In series combination,
Rs = R1 + R2 Rs = 20 + 5 = 25Ω.
ˆ In series combination equivalent resistance = 25 ohms
(b) In parallel Combination,
1 1 1
Rp = R1 + R2

ˆ 1 = 1 + 1
Rp 20 5

5
ˆ 1 = 1 + 4 =
20
Rp 20

1
ˆ 1 =
Rp 4
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 = 16Ω ; R2 = 4Ω,
these are connected in parallel,
1 1 1
ˆ R = +
p
R1 R2

1 1 1
ˆ R = 16 + 4
p

1 1 + 4
ˆ R = 16
p


ˆ 1 =
5
Rp 16
16
ˆ Rp = 5 Ω,

Now Rp will act as R1 and R2 will be 5 Ω,
ˆ In series combination,
Rs = R1 + R2
16 5
ˆ Rs = 5 + 1

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
4
cell is 3 volt and internal resistance ohm.Find the total current in the circuit.
3
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,
4
Internal resistance of the cell (r) = ohm,
3
First we have to find R = ?
In this case R1 = 8 ohm and R2 = 4 ohm resistors are connected in parallel
combination,
1 1 1
ˆ = +
Rp R1 R2
1 1 1
ˆ = +4
Rp 8
1 1 + 2
ˆ =
Rp 8
ˆ Rp = 8 ohm
3
Now substituting the values of E, R, r in eqn (I) we get,
8 4
3 = I ( 3 + 3 )
12
ˆ3=I( 3 )
3 3
ˆ 3 × 12 = I ˆ I = 4 Ampere = 0.75 A
ˆ Total current in the circuit = 0.75 A.


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;
Rs = R1 + R2 + R3
ˆRs = 12 + 12 + 12 = 36 ohm
Now in parallel combination,
1 1 1 1
ˆ R = R + R + R
p 1 2 3

1 1 1 1
ˆ = + +
Rp 12 12 12

1 1 + 1 + 1 3 1
ˆ = = =
Rp 12 12 4

ˆ 1 = 1 ˆ Rp = 4 ohm
Rp 4
ˆ Resistance in series = 36 ohm and
resistance in parallel = 4 ohm
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.
···


Chapter 5
Effects of Electric Current

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 = I2Rt 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
I2 R t
H = calories
4.18 V
By Ohm's law we substitute I =
R


we gets,
V2 t
H = calories
4.18 R
Again by same process we get,

V I t
H = calories
4.18
ˆ Various forms of Joule's formula will be as follows :
I2 Rt
i) H = I2Rt Joules ii) H = calories
4.18
V2 t V I t
iii) H = calories iv) H = calories
4.18 R 4.18
* (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 -
V2
P = Here ,
R
V = Potential difference (Volts)
R = Electrical resistance (ohms)

N.B. : 1) 1 Watt = 1 Joule
1 second
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)


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)
Types of metal Uses w.r.t. electricity
1) Copper, aluminium Conduction wires, connection wires
2) Nichrome Heater, geysers, electric iron, etc.
3) Tungston Electric bulb

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

Difference between A.C. and D.C.
Direct Current (D.C.) Alternating Current (A.C.)
1. The direct current always flows 1. Alternating current reverses its
in one direction. direction periodically.
2. It is non-oscillating current 2. It is oscillating current
generated by the source. generated by the source.
3. Generally D.C. is generated 3. Alternating current is
from an electric cell/battery. generated by A. C. generator.


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.
I2 R t
Formula H = calorie.
4.18

(0.5)2 × 209 × 120
ˆH= 4.18 calories

0.5 × 0.5 × 209 × 120 × 100
ˆH= 418 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.
VIt
Formula H = calorie.
4.18

240 × 0.418 × 60
ˆH= 4.18
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 = ?
V2
Formula P = Substituting the values.
R
2 240 × 240
ˆ 40 = (240) ˆR= 40
R

ˆ R = 1440 ohms
ˆ The resistance of the bulb = 1440 ohm.


4) Heat generated in a conductor of resistance 40Ω 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) = ?
V2 t
Formula : H = Joules substituting the values.
R

V2 × 60
ˆ 150 = 40

ˆ 150 × 40 = V2
60

ˆ 150 × 40 = V2
60
2
V = 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.

···


Chapter 6
Energy Sources

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 :
Non-renewable Renewable
energy cources energy cources
Firewood, cow dung, Wind energy, tidal
charcoal, coal, energy, geothermal
kerosene, cooking energy, energy from
gas; petrol, diesel flowing water, solar
etc. energy, energy from
biomass.
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)


Difference between Nuclear fission and Nuclear fusion (Breeder reaction)
Nuclear fission Nuclear fusion
(Breeder reaction)
1) Radioactive material like 1) Hydrogen atoms
Uranium - 235 converts combine to form helium,
into fragments releasing releasing large amount
large amount of heat of heat energy.
energy 2) High temperature is
2) Bombardment of required to carry out the
neutrons is essential to reaction.
carry out the 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.


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 i) Biogas is the mixture
wood, biological waste of methane, carbon-
products dioxide and hydrogen-
sulphide.
ii) It burns with smoke ii) It burns without smoke

···


Chapter 7
Types of Energy

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 cosθ
When F and r are in the same direction,
θ becomes 0 and cos 0 = 1. Hence
Work done = W = Fr.cosθ = 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 (ˆcos 900 = 0)
= 0
When an object rolls on the ground h = 0, hence P.E. = 0
B) Kinetic Energy [K.E.]
1
K.E. = 2 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. 1 (E = Total energy)
∴ E = mgh + 2 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.


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
2J
ˆ h = 0.5 kg × 10m/s2 = 0.4 m

(Height attained by the block = 0.4 m)

···


Chapter 8
Power

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


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
Work
2) Power = time W
ˆP= t
taking w = Fs we get
Fs s
P= We know =V
t t
s
ˆP=F× t
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 =?
mgs
Formula = Power (P) = t
20 kg × 10 m/s2 × 10
ˆP=
20 s

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 = ?
W mgh
Power = = t
t
600 kg × 9.8 m/s2 × 10 m
= 120 seconds
= 490 Watt
power of crane = 490 Watt mgh mgs
Also note that in above formula P = t and P = t (i.e. h = s)
···


Chapter 9
Sound

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 00C 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
1
340 × = 34 m
10
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.


Refer to page 137 and 138 of textbook for effect of noise pollution and measures
of noise control.
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
time
Depth of sea = Velocity of sound in water ×
2
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


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

ˆ Speed of sound = 1020
3
= 340 m/s
(Speed of sound = 340 m/s)

···


Chapter 10
Heat

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 40C
When water is heated from 00C initially it contracts in volume upto 40C instead of
expansion.
The behaviour of the water between 00C to 40C 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 40C, the volume of water is minimum. Hence density of water is maximum at
0
4 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 40C 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 40C
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.

Example :
1) Certain mass of water at 640C is mixed with an equal mass of water at 220C. What
will be the resulting temperature of the mixture?
Ans. Given, Let mass of water = m gm
Specific heat of water = 1 cal/g0C
t1 = 640C, t2 = 220C t3 =? (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)0C

ˆ t3 =( 86 )0C
2
ˆ t3 = 430C
Ans. The temperature of the mixture will be 430C.


* 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

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
fig. 11.10(b) (page 173)


Note : The image of an object placed anywhere in front of concave lens is always virtual
erect and diminished.
G) Differenciate

Convex lens Concave lens
1) Convex lens is also called 1) Concave lens is also called
as converging lens as diverging lens
2) 2)

3) Convex lens is thicker at 3) Concave lens is thinner at
the centre than at the the centre than at the
edges. edges.
4) Focal length is positive 4) Focal length is negative.
5) If any object is placed at 5) Image is always virtual
F1 and beyond the image erect, diminished.
is real and inverted. If it is
placed between F1 and O,
the image is virtual and
erect.

A simple microscope A compound microscope

1) Consists of a convex lens 1) Consists of two convex
lenses.
(Objective and eye piece)
2) An object is placed within 2) An object is placed just
the focal length of convex beyond the focal length of
lens the objective lens.
3) We get magnified image 3) We get highly magnified
image.
4) It is used by watch makers, 4) It is used to observe plant
jewellers and while reading cell, animal cells, bacteria
small font words. etc.


Short sightedness Long sightedness
1) Defect is due to slight 1) Defect is due to slight
elongation of the eye ball flattening of the eye ball in
or bulging of the eye lens. the horizontal direction.
2) A person suffereing from 2) A person suffering from this
this defect can see nearby defect can see distant
objects clearly but unable objects clearly but unable
to see objects at long to see nearby objects.
distances.
3) The image of the distant 3) The image of the nearby
object is in front of retina. object is formed behind the
retina.
4) It can be corrected by 4) It can be corrected by
using concave lens of using convex lens of
suitable focal length. suitable focal length.

Remember the formulae :
Real depth sin i
i) Reflactive index of glass = Apparent depth =
sin r
∠ i = angle of incidence, ∠ r = angle of refraction
Real depth
ii) Refractive index of water in a tank = Apparent depth

iii) Lens formula :
1 1 1
– =
v u f
v = image distance
u = object distance
f = focal length
iv) Magnification (m) :
q v
m = p = u

m = magnification
q = height of the image
p = height of the object


v) Power of a lens

P = 1
f
P = Power of the lens in diopters
f = focal length in meters
vi) If two lenses of focal lengths f1 and f2 are kept in contact with each other, their
combined focal length is given by,
1 1 1 1 1
= + , P1 = , P2 =
f f1 f2 f1 f2
P = P1 + P2 P = Power

Note : Go through the topics 'Persistence of vision' and 'Perception of colour'
(page nos. 193, 194, 195)

Q. 1 Give scientific reason
1) Rapid movement of burning incense stick appears as a complete circle of red
light.
Ans. Principle : The impression of image formed on retina lasts for 1/16 of a second evern
after the object is removed. This effect is called Persistence of vision.
(1 mark)
Explanation : While the successive positions of the burning tip are being seen by
you, the images of the previous positions are still there on the retina and you get
the illusion of the complete circle of red light.
(1 mark)
Solve the numerical.
An image of a object placed in front of a convex lens at a distance 20 cm is formed
at a distance of 60 cm from the lens. The image is real.
Then find :
1) focal length of the lens
2) size of object is 6 cms, find the size of the image
i) u = – 20 cm v = 60 cm (Image is real) Find f = ?

By using
1 1 1 1
– = mark
v u f 2
ˆ 1 – 1 = 1
60 (– 20) f
1 1 1
ˆ + =
60 20 f


ˆ1+3 = 1
60 f
4 1
ˆ = f
60
60
ˆ f = = 15 cm
4

ˆ The focal length of lens is 15 cms. 1/2 mark
ii) size of object (P) = 6 cm
size of image (q) = .........?
q v
ˆ = 1/2 mark
P u
q 60
ˆ =
6 – 20
60 × 6
ˆ q = = – 18 cm
– 20
– sign shows the image is inverted image.
ˆ Size of the image is 18 cms.
4) What is power of accomodation ? 2 marks
Ans We see objects when its image is formed on retina. In normal human eye retina to
lens distance does not change. So for different object distance (u) the focal length
(f) is adjusted such that the image distance (v) remains constant. This ability of the
eye lens of adjusting the focal length is known as Power of accomodation. For
observing distant objects the lens becomes thin and while observing nearer objects
it becomes slightly bulging. This is brought about by the ciliary muscles.

···


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