INDEX

S. NO. SUBJECT NAME PAGE NO.

SPECIMEN COPY

. -

. -

1. PHYSICS 1-6

2 CHEMISTRY 7 27

3 MATHEMATICS 28 36

4. BIOLOGY 37- 42

5. ANSWER KEY 43

CLASS - VIII (IJSO-FOUNDATION)

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1PAGE # 1

MOTION

EQUATIONS OF UNIFORMLY ACCELERATED MOTION

(a) 1st Equation of Motion :

Consider a body having initial velocity u. Suppose

it is subjected to a uniform acceleration a so that

after time t its final velocity becomes v. Now we

know,

Acceleration =Time

velocityinchange

a =t

u�v

or v = u + at .......(i)

(b) 2nd Equation of Motion :

Suppose a body has an initial velocity u and uniform

acceleration �a� for time t so that its final velocitybecomes v. The distance travelled by moving bodyin time t is s then the average velocity = (v+u) /2 .Distance travelled = Average velocity × time

s = t2

vu

s = t

2atuu

(as v=u+at)

s = t2

at2u

s =

2atut2 2

s = 2at

2

1ut .......(ii)

(c) 3rd Equation of Motion :

Distance travelled = Average velocity × time

s = t2

vu

.......(iii)

from equation (i) t = a

u�v

Substituting the value of t in equation (iii), we get

s =

2uv

au�v

a2uv

s22

2as = v2 � u2 or v2 = u2 + 2as ......(iv)

(d) Distance covered in nth second :

Distance covered by a body in t second is

S = ut + 21

at2 .

Distance covered by a body along a straight line in

n second is

Sn = un +

21

an2 .......(v)

Distance covered by a body along a straight line in

( n�1) second is

Sn�1

= u (n �1) + 21

a (n�1)2 .......(vi)

The distance covered by the body in nth second

will be- :

snth

= sn � s

n �1

snth

= un +21

an2 � { u (n�1) + 21

a (n�1)2 }

Snth

= un +21

an2 � {nu �u + 21

a (n2+1 �2n)}

Snth

= un + 21

an2 � {un � u + 2a

2an2

� an}

Snth

= un + 21

an2 � un + u � 2a

�2

an2

+ an

Snth

= u + a

21

�n

Snth

= u + a

21�n2

Snth= u + 2

a (2n � 1 ) .......(vii)

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2PAGE # 2

TO SOLVE NUMERICAL PROBLEMS

(i) If a body is dropped from a height then its initial

velocity u = 0 but has acceleration (acting). If a

body starts from rest its initial velocity u = 0 .

(ii) If a body comes to rest its final velocity v = 0 or,

if a body reaches the highest point after being thrown

upwards its final velocity v = 0 but has acceleration

( acting).

(iii) If a body moves with uniform velocity, its

acceleration is zero i.e. a = 0.

(iv) Motion of a body is called free fall if only force

acting on it is gravity (i.e. earth�s attraction).

MOTION UNDER GRAVITY (UNIFORMLY

ACCELERATED MOTION)

The acceleration with which a body travels under

gravity is called acceleration due to gravity g. Its

value is 9.8 m/s2 ( or 10 m /s 2 ). If you have to

take g = 10 m/s2 then it must be mentioned in the

question otherwise take g = 9.8 m/s2.

(i) If a body moves upwards (or thrown up ) g is

taken negative (i.e. motion is against gravitation of

earth).

So we can form the equation of motion like ,

v = u � gt, h = u t �21

gt2 , v2 = u2 � 2 gh.

(Here �g replaces a)

(ii) If a body travels downwards (towards earth) then

g is taken + ve. So equations of motion becomes

v = u + gt , s = ut + 21

gt2 , v2 = u2 + 2gh.

(Here g replaces a)

(iii) f a body is projected vertically upwards with

certain velocity then it returns to the same point of

projection with the same velocity in the opposite

direction.

(iv) The time for upward motion is the same as for

the downward motion.

ILLUSTRATION

1. A car is moving at a speed of 50 km/h. Two seconds

there after it is moving at 60 km/h. Calculate the

acceleration of the car.

Sol. Here u = 50 km/h = 185

50 m/s = 18250

m/s

and v = 60 km/h = 185

60 = 18300

m/s

Since a = t

u�v = 2

18250

�18300

= 2

1850

= 3650

= 1.39 m/s2

2. A car attains 54km/h in 20 s after it starts. Find the

acceleration of the car.

Sol. u = 0 (as car starts from rest)

v = 54 km/h =185

54 = 15 m/s

a =t

u�v a =

200�15

= 0.75 m/s2

3. A ball is thrown vertically upwards with a velocity of 20

m/s. How high did the ball go ? (Take g = 9.8 m/s2).

Sol. u = 20 m/s , a = � g = � 9.8 m/s2 (moving

against gravity)

v = 0 ( at highest point), s = ?

v2 � u2 = 2as

(0)2 � (20)2 = 2 (�g) s

� 400 = 2 ( � 9.8) s

� 400 = �19.6 s

CIRCULAR MOTION

When a particle moves in a plane such that its distance

from a fixed (or moving) point remains constant, then

its motion is known as circular motion with respect to

that fixed (or moving) point. The fixed point is called

centre, and the distance of particle from it is called

radius.

3PAGE # 3

KINEMATICS OF CIRCULAR MOTION

(a) Angular Position :

To decide the angular position of a point in space weneed to specify (i) origin and (ii) reference line. Theangle made by the position vector w.r.t. origin, with

the reference line is called angular position. Clearlyangular position depends on the choice of the origin

as well as the reference line. Circular motion is atwo dimensional motion or motion in a plane.

P'P

Y

O r X

Suppose a particle P is moving in a circle of radius rand centre O.The angular position of the particle P at a given instantmay be described by the angle between OP and OX.This angle is called the angular position of theparticle.

(b) Angular Displacement and Angular

Velocity :

In a circular motion, the angular displacement of a body isthe angle subtended by the body at the centre in a giveninterval of time. t is represented by the symbol (theta).The angular displacement per unit time is called theangular velocity. t is represented by the symbol (omega).

Let a body moves along a circle of radius r and performa uniform circular motion. Let the body be at point P tostart with and reach point Q after time t.

Then, angular displacement = PCQ = and

angular velocity = t

(i.e. = t)

If the time period of the body is T, the angulardisplacement = 2c

Hence = T2

r

Q

PC

x

But T1

= N (frequency)

There = 2N

For an arc of length, Linear distance =

Angular displacement, = r

Hence, = rFor a time interval t,

Linear velocity, v = t

Angular velocity, = t

= tr

= rv

Hence, v = r.

b) Centripetal Force :

If there is no force acting on a body it will move in astraight line (with constant speed). Hence if a body ismoving in a circular path or any curved path, there mustbe some force acting on the body.

If speed of body is constant, the net force acting on thebody is along the inside normal to the path of the bodyand it is called centripetal force.

Centripetal force (Fc) = mac = r

mv2

= m 2 r

SIMPLE HARMONIC MOTION

(a) Periodic Motion :

When a body or a moving particle repeats its motionalong a definite path after regular intervals of time, itsmotion is said to be Periodic Motion and interval oftime is called time period or harmonic motion period(T). The path of periodic motion may be linear, circular,elliptical or any other curve.Eg.: Rotation of earth about the sun.

(b) Oscillatory Motion :

�To and Fro� type of motion is called an Oscillatory

Motion. It need not be periodic and need not have

fixed extreme positions.

Eg. : Motion of pendulum of a wall clock.

The oscillatory motions in which energy is conserved

are also periodic.

The force/torque (directed towards equilibrium point)

acting in oscillatory motion is called restoring force/

torque.

Damped oscillations are those in which energy is

consumed due to some resistive forces and hence

total mechanical energy decreases.

(c) Definition of Simple Harmonic Motion :

If the restoring force/torque acting on the body in

oscillatory motion is directly proportional to the

displacement of body/particle and is always directed

towards equilibrium position then the motion is called

Simple Harmonic Motion (SHM). It is the simplest (easy

to analysis) form of oscillatory motion.

TYPES OF SHM

(a) Linear SHM :

When a particle moves to and fro about an equilibrium

point, along a straight line, then its motion is known as

linear SHM. A and B are extreme positions and M is

mean position.

AM = MB = Amplitude

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

MBA

4PAGE # 4

(b) Angular SHM :

When body/particle is free to rotate about a given axis

executing angular oscillations, then its motion is known

as angular SHM.

Equation of Simple Harmonic Motion :

The necessary and sufficient condition for SHM is F = � kx

Where k = Force constant or spring constant

x = displacement from mean position.

or ma = � kx

a = �mk

x

Here negative sign show that F will always be towards

mean position or F and x are in opposite direction.

a = �2x ( Where = m

k)

It is the equation of SHM.

CHARACTERISTICS OF SHM

(a) Displacement :

It is defined as the distance of the particle from the

mean position at that instant. Displacement in SHM at

time t is given by x = A sin (t +), here is initial

phase.

extreme postion

equilibrium postion

extreme postion

AmplitudeA

AmplitudeB

NOTE : In the figure shown, path of the particle is a

straight line.

(b) Amplitude :

It is the maximum value of displacement of the particle

from the equilibrium position.

Amplitude = 21

[distance between two extreme

positions]

It depends on the energy of the system.

(c) Angular Frequency () :

= f2T

2ð

ð and its units is rad/s.

(d) Frequency (f) :

Number of oscillations completed in unit time interval

is called frequency of oscillations.

f = ð2T

1 ù , its units is Hz or s�1 .

(e) Time period (T) :

The smallest time interval after which the oscillation

repeats itself is called the time period.

T = k

m2

2ð

ð

ù

4. For a particle performing SHM, equation of motion is

given as a + 4x = 0. Find the time period ?

(Here a is acceleration of particle)

Sol. a = �4x, 2 = 4, = 2

T = ù

ð2 = .

RELATIVE MOTION

Motion is a combined property of the object under studyas well as the observer. It is always relative, there is nosuch thing like absolute motion or absolute rest. Motionis always defined with respect to an observer orreference frame.

Relative Motion In One Dimension :

Relative Position : It is the position of a particle w.r.t.observer.In general if position of A w.r.t. origin is xA and that of B w.r.t.origin is xB then �Position of A w.r.t. B� � xAB is

AB A Bx x x

Origin B A

xAB

xA

xB

Relative Velocity :

Definition : Relative velocity of a particle A with respectto B is defined as the velocity with which A appears tomove if B is considered to be at rest. In other words, itis the velocity with which A appears to move as seen byB considering itself to be at rest.

Relative velocity in one dimension :If xA is the position of A w.r.t. ground, xB is position of Bw.r.t. ground and xAB is position of A w.r.t. B then we cansay vA = velocity of A w.r.t. groundvB = velocity of B w.r.t. groundand vAB = velocity of A w.r.t. BThus

AB A Bv v v

5PAGE # 5

5. An object A is moving with 5m/s and B is moving with20m/s in the same direction. (Positive x-axis)(i) Find velocity of B with respect to A.(ii) Find velocity of A with respect to B

Sol. (i) vB = +20m/s vA = +5m/svBA = vB � vA = +15m/s

(ii) vB = +20m/s, vA = +15m/svAB = vA � vB = � 15m/s

Note : vBA= �vAB

EXERCISE

1. A body covers half the distance with a speed of20 m/s and the other half with a speed of 30 m/s.The average speed of the body during the wholejourney is :(A) Zero (B) 24 m/s(C) 25 m/s (D) None of these

2. A body is thrown vertically upwards and risesto a height of 10 m. The velocity with which thebody was thrown upwards is (g = 9.8 m/s2) :(A) 10 m/s (B) 20 m/s(C) 14 m/s (D) None of these

3. A body strikes the floor vertically with a speedu and rebounds at the same speed. The changein speed would be :(A) u (B) 3u(C) 2u (D) Zero

4. If a trolley starts from rest with an accelerationof 2 m/s2, the velocity of the body after 4s wouldbe :(A) 2 m/s (B) 8 m/s(C) 6 m/s (D) 4 m/s

5. The speed of a body describing its motion is :(A) direction (B) state(C) type (D) rapidity

6. When the distance of an object travels is directlyproportional to the length of time, it is said totravel with(A) zero velocity(B) constant speed(C) constant acceleration(D) uniform velocity

7. A body has uniform acceleration if its :(A) speed changes at a uniform rate(B) velocity changes at a uniform rate(C) speed changes at non-uniform rate(D) velocity remains constant

8. A girl swims in a swimming pool of length 100 m.She swims from one end to another end andreaches the starting point again in2 minutes. Then the average speed of theswimmer is :(A) 100 ms�1 (B) 0.83 ms�1

(C) 1.67 ms�1 (D) zero

9. The acceleration of car that comes to stop froma velocity of 10 m/s in distance of 25 m is :(A) �2 m/s2 (B) �4 m/s2

(C) �8 m/s2 (D) �16 m/s2

10. A stone is thrown in vertically upward direction

with a velocity of 5m/s. If the acceleration of

the stone during its motion be 10m/s2 in

downward direction, what will be the height

attained by the stone ?

(A) 1.25 m (B) 1.50 m

(C) 2 m (D) 3.5 m

11. A body with initial velocity 8 m/s moves along a

straight line with constant acceleration and

travels 640 m in 40s. Find the average velocity

during this interval.

(A) 8 m/s (B) 16 m/s

(C) 24 m/s (D) 32 m/s

12. A balloon starts rising from the ground with an

acceleration of 1.25 m/s2. After 8s a stone is

released from the balloon. How much time stone

will take to reach the ground ? (g = 10 m/s2)

(A) 4 s (B) 2s

(C) 22 s (B) 24 s

13. A ball is dropped from a height of 5m onto asandy floor and penetrates the sand upto 10cmbefore coming to rest find the retardation insand, assuming it to be uniform.(A) 9.8 m/s2 (B) 10 m/s2

(C) 100 m/s2 (D) 500 m/s2

14. A body moving with a constant retardation instraight line travels 5.7m and 3.9 m in 6 th and9th second respectively. When will the bodycome momentarily to rest ?(A) 10 s (B) 15 s(C) 20 s (D) 25 s

6PAGE # 6

15. Two simple pendulums of length and 4 aresuspended from same point and brought asidetogether and released at the same time. If thetime period of smaller pendulum is T there afterhow much time will they be together again andmoving in same direction.

4

(A) T/2 (B) T

(C) 2T (D) None of these

16. At moon the weight of things become 1/6th

of weight of earth . What is the ratio of time

period of simple pendulum at earth to that on

the moon.

(A) 6 : 1 (B) 6 : 1

(C) 1 : 6 (D) 1 : 6

7PAGE # 7

MATTER

INTRODUCTION

There are a large number of things around us whichwe see and feel. For example, we can see a book infront of us. A book occupies some space. The spaceoccupied by the book is called its volume. If we pickup the book, we can also feel its weight. So, weconclude that the book has some mass. We cannotsee the air around us, yet if we fill a balloon with airand then weigh it carefully, we will find that not onlydoes air occupy space (bounded by the balloon), butit also has mass.

Things like a book and air are examples of matter.Other examples of matter are wood, cloth, paper, ice,steel, water, oil etc. Further, that matter offersresistance is borne out by the fact that we cannotdisplace an object from one place to another withoutapplying some force. We have to apply force to pickup a stone from the ground. Thus , matter can bedefined as follows -

Anything that occupies space, has mass and offersresistance is called matter.

PHYSICAL NATURE OF MATTER

(a) Matter is Made up of Particles :

(i) Everything around us is made up of many tinypieces or particles.

(ii) Particles which make up the matter are constantlymoving.

(iii) Particles which make up matter are atoms ormolecules.

(i) Evidences for the presence of particles in matter :Most of the evidences for the existence of particles inmatter and their motion come from the experimentson diffusion and Brownian motion.

Evidence - 1Dissolving a solid in a liquid : Take a beaker. Fill halfof it with water. Mark the level of water in the beaker.Add some sugar to the water and dissolve it with thehelp of a glass rod. You will see that the sugar hasdisappeared, but there is no change in the level ofwater.

Conclusion : This can be explained by assuming thatmatter is not continuous, rather it is made up ofparticles. Sugar contains a large number of separateparticles. These particles when dissolved in wateroccupy the vacant spaces between the particles ofwater. That is why, the water level in the beaker didnot rise. Had sugar been continuous, like a block ofwood, the water level in the beaker would have risen.

Experiment to show that matter is made of particles

Evidence - 2Movement of pollen grains in water : The bestevidence for the existence and movement of particlesin liquids was given by Robert Brown in 1827. RobertBrown suspended extremely small pollen grains inwater. On looking through the microscope, it wasfound that the pollen grains were moving rapidlythroughout water in a very irregular way (or zig-zagway).Conclusion : Water is made up of tiny particles whichare moving very fast (The water moleculesthemselves are invisible under the microscopebecause they are very, very small). The pollen grainsmove on the surface of water because they areconstantly being hit by the fast moving particles ofwater. So, though the water particles (or watermolecules) are too small to be seen, but their effecton the pollen grains can be seen clearly. The randommotion of visible particles (pollen grains) caused bythe much smaller invisible particles of water is anexample of Brownian motion (after the name of thescientist Robert Brown who first observed thisphenomenon.)

Brownian motion : Zig-zag motion (in a very irregularway) of particles is known as brownian motion.Brownian motion can also be observed in gases.Sometimes, when a beam of light enters in a room,we can see tiny dust particles suspended in air whichare moving rapidly in a very random way. This is anexample of Brownian motion in gases. The tiny dustparticles move here and there because they areconstantly hit by the fast moving particles of air.The existence of Brownian motion gives twoconclusions.� Matter is made up of tiny particles.

� Particles of matter are constantly moving.

Note :Brownian motion increases on increasing thetemperature.

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8PAGE # 8

(b) Characteristics of Particles of Matter :

The important characteristics of particles of matterare the following :

(i) The particles of matter are very, very small

(ii) The particles of matter have spaces between them

(iii) The particles of matter are constantly moving :This property can be explained by diffusion.

(A) Diffusion :�Intermixing of particles of two different

types of matter on their own is called diffusion.�t isthe phenomenon in which the movement ofmolecules or particles occur from their higherconcentration towards their lower concentration.

e.g. : When a perfume bottle is opened in one cornerof a room, its fragrance spreads in the whole roomquickly. This happens because the particles ofperfume move rapidly in all directions and mix withthe moving particles of air in the room.

(A) Experiment : We take a gas jar full of brominevapours and invert another gas jar containinig air overit, then after some time, the red-brown vapours ofbromine sperad out into the upper gas jar containingair.

(B) Conclusion : In this way, the upper gas jar whichcontains colourless air in it, also turns red-brown.The mixing is due to the diffusion of bromine vapours(or bromine gas) into air.

Diffusion of bromine vapour (or bromine gas) into air

Note :The particles of matter possess kinetic energy andso are constantly moving. As the temperature rises,particles move faster.

(iv) Particles of matter attract each other : Thereare some forces of attraction between the particles ofmatter which bind them together.

(A) Cohesive Force : The force of attraction betweenthe particles of same substances is called cohesiveforce.

(B) Adhesive Force : The force of attraction betweenthe particles of different substances is calledadhesive force.

e.g. : If we take a piece of chalk, a cube of ice and aniron nail and beat them with a hammer, chalk willeasily break into smaller pieces, but more force willbe required to break a cube of ice and iron nail willnot break.

Reason : The reason for this is, that the force ofattraction is quite weak in between the chalk particles,but force of attraction in between the particles of icecube is a bit stronger, while force of attraction inbetween the particles of iron is very-very strong.

RIGID AND FLUID

(i) Rigid : Rigid means �unbending� or inflexible. A

solid is a rigid form of matter so that it maintains itsshape when subjected to outside force.

(ii) Fluids : Fluids are the substances which havetendency to flow. A liquid is a fluid form of matterwhich occupies the space of the container. Liquidshave a well defined surface. A gas is a fluid form ofmatter which fills the whole container in which it iskept.

Note :Liquids and gases are known as fluids.

CLASSIFICATION OF MATTER

On the basis of physical states, all matter can beclassified into three groups:-(a) Solids (b) Liquids (c) Gases

COMPARISON OF THE CHARACTERISTICS OF THREE STATES OF MATTER

Property Solid state Liquid state Gaseous state

Interparticle spaces Very small spacesComparatively large spaces than solids

Very large spaces

Interparticle forces Very strong Weak Very weak

Nature Very hard and rigid Fluid Highly fluid

Compressibility Negligible Very small Highly compressible.

Shape and volumeDefinite shape and volume

Indefinite shape, but definite volume

Indefinite shape as well as volume

Density High Less than solid state Very low density

Kinetic energy

LowComparatively high than solids

Very high

Diffusion Negligible Slow Very fast

9PAGE # 9

Gases are Highly Compressible therefore :

(i) LPG (Liquefied Petroleum Gas) is used in our home

for cooking.

(ii) Oxygen cylinders supplied to hospitals contain

liquid oxygen.

(iii) These days C.N.G. (Compressed Natural Gas) is

used as fuel in vehicles.

Note :

Gaseous particles move randomly at high speed and

hit each other and also walls of the container, so exert

pressure.

INTERCONVERSION OF STATES OF MATTER

The phenomenon of change of matter from one state

to another state and back to original state, by altering

the conditions of temperature and pressure, is called

interconversion of states of matter.

The various states of matter can be interchanged into

one another by altering the conditions of -

(a) Temperature (b) Pressure.

(a) Altering the Temperature of Matter :

(i) Interconversion of solid into liquid and vice versa :

Solids can be converted into liquids by heating them.

Similarly liquids can be cooled to form solids.

e.g. :ce at 00C changes into water at 00C, when heat

energy is supplied to it. The water at 00C changes

into ice at 00C on freezing.

Activity -

To study the change of state from ice to water.

Materials required -

A 100 cc beaker, a thermometer (Celsius), a glass

stirrer, a wire gauze, a tripod stand, a Bunsen burner,

an iron stand, ice cubes.

Method -

Half fill the beaker with ice cubes and place it over a

wire gauze and tripod stand. Suspend a Celsius

thermometer from the iron stand, such that its bulb is

touching the water level. Place a glass stirrer in the

ice.

Record the temperature of ice. You will find it is 00 C

(273 K). Now heat the beaker on a low bunsen flame

and continuously stir the contents of beaker. Record

the temperature five to six times, till all the ice melts.

You will observe that temperature through out

remains 00C (273 K), till all the ice melts.

stand

Change of state from ice to water

(A) Melting or Fusion: The process due to which asolid changes into liquid state by absorbing heatenergy is called melting or fusion.

(B) Freezing or Solidification: The process due towhich a liquid changes into solid state by giving outheat energy is called freezing or solidification.

(C) Melting Point: The constant temperature at whicha solid changes into liquid state by absorbing heatenergy at 1 atm pressure is called its melting point.

(D) Freezing Point: The constant temperature at whicha liquid changes into solid state by giving out heatenergy at 1 atm pressure is called freezing point.

Note :The numerical value of freezing point and meltingpoint is same.Melting point of ice = Freezing point of water = 0ºC

(273.16 K).

Explanation: On increasing the temperature of solids,the kinetic energy (K.E.) of particles increases. Dueto increase in K.E., the particles start vibrating withgreater speed. The energy supplied by heatovercomes the force of attraction between theparticles. Then, the particles leave their fixed positionsand start moving freely and thus solid melts.

Latent Heat of Fusion : The amount of heat energythat is required to change 1 kg of solid into liquid atatmospheric pressure and its melting point is knownas the latent heat of fusion. (In Greek Latent meansHidden) Latent heat of fusion of ice = 3.34 × 105 J/kg.

Note :Particles of water at 00C (273 K) have more energy ascompared to particles in ice at the same temperature.

(ii) Interconversion of liquid into gaseous state andvice versa: Liquids can be converted into gases byheating them. Similarly, gases can be converted intoliquids by cooling them.

e.g. : Water at 1 atm pressure changes into gas(steam) at 1000C by absorbing heat. Steam at 1000Cchanges into water by giving out energy.

10PAGE # 10

Activity -To study the change of state from water to steam.

Materials required -A 100 cc beaker, a thermometer (Celsius), a glassstirrer, a wire gauze, a tripod stand, a Bunsen burner,an iron stand, tap water.

Method -Half fill the beaker with water and place it over a wiregauze and tripod stand. Suspend a Celsiusthermometer from the iron stand, such that its bulb istouching the water level. Place a glass stirrer in thewater.Record the temperature of water. Heat the beaker ona low Bunsen flame and continuously stir the waterwith glass stirrer. Go on recording the temperature tillwater starts boiling. Allow the water to boil for fewminutes and record its temperature.

You will notice that temperature of water rises till itstarts boiling. The temperature of boiling water is 1000C(373 K). If we continue heating the water it changesinto steam, but the temperature remains constant, i.e.,1000C (373 K).

Change of state from water to steam

(A) Boiling or Vaporisation: The process due to whicha liquid changes into gaseous state by absorbingheat energy is called boiling.

(B) Condensation or Liquefaction: The process dueto which a gas changes into liquid state by giving outheat energy is called condensation.

(C) Boiling Point: The constant temperature at whicha liquid rapidly changes into gaseous state byabsorbing heat energy at atmospheric pressure iscalled boiling point.

(D) Condensation Point:- The constant temperatureat which a gas changes into liquid state by giving outheat energy at atmospheric pressure is calledcondensation point.

Note :The numerical value of condensation point andboiling point is same.Condensation point of vapour (water) = Boiling pointof water = 100ºC (373.16 K).

Explanation : When heat is supplied to water, particlesstart moving faster. At a certain temperature, a pointis reached when the particles have enough energy tobreak the forces of attraction between the particles. Atthis temperature the liquid starts changing into gas.

Latent heat of vaporisation: The amount of heat whichis required to convert 1 kg of the liquid (at its boilingpoint) to vapour or gas without any change intemperature. Latent heat of vaporisation of water= 22.5 × 105 J/kg.

Note :Particles in steam, that is water vapour at 373 K havemore energy than water at the same temperature.Because steam has absorbed extra energy in theform of latent heat of vaporisation.

CURVE [TEMPERATURE-TIME GRAPH]

We can show the change of temperature with time inthe form of a temperature-time graph drawn by usingthe readings obtained in the above experiment. Sucha time-temperature graph is shown in figure.

C

wat

er(li

quid

)Time of heating (in minutes)

Tem

pera

ture

(in

°C

)100

0A B

(solid

liquid)

D

(liquid gas)

Temperature Time Graph

In this graph at point A, we have all ice. As we heat it,the ice starts melting to form water but the tempera-ture of ice and water mixture does not rise. It remainsconstant at 0°C during the melting of ice. At point B,

all the ice has melted to form water. Thus, we haveonly water at point B. Now, on heating beyond point B,the temperature of water (formed from ice) starts ris-ing as shown by the sloping line BC in the graph.

(iii) Direct interconversion of solid into gaseous stateand vice versa: The changing of solid directly intovapours on heating and of vapours directly into solidon cooling is known as sublimation.

� The solid which undergoes sublimation to form

vapour is called �sublime�.

� The solid obtained by cooling the vapours of a solidis called �sublimate�.

e.g. : Ammonium Chloride (NH4Cl), iodine, camphor,

naphthalene (moth balls) and anthracene.

Vapo

risat

ion

Conde

nsat

ionFreezing

Liquid

SublimationSolidGas

Melting

Sublimation

Interconversion of states of matter

11PAGE # 11

Specific HeatThe specific heat of a substance is the amount ofheat which is required to raise the temperature of aunit mass of the substance by 1º C. Now, if we

measure the heat in �joules� and mass in �kilograms�,

then the definition of specific heat becomes.The specific heat of a substance is the amount ofheat in joules required to raise the temperature of1 kilogram of the substance by 1ºC.

The specif ic heat of a substance is usuallyrepresented by the symbol C (Sometimes, however,the specific heat of a substance is also representedby the letter's). The specific heat of a substance variesslightly with temperature. The change in the specificheat of a substance with temperature is due to thechanges which occur in the structure and organizationof the molecules in a substance with change intemperature.

Units of Specific HeatThe unit of specific heat depends on the units in which�heat� and �mass� are measured. Now, the S.I. unit of

heat is �joule� and that of mass is �kilogram�, so, the

S.I. unit of specific heat is �joules per kilogram per

degree celcius�, which is written in short form as :

J/kg°C or J kg�1 °C�1.

The difference in various states of matter is due tothe different intermolecular spaces between theirparticles. So when a gas is compressed theintermolecular space between its particles decreasesand ultimately it will be converted into liquid.Pressure and temperature determine the state of asubstance. So, high pressure and low temperaturecan liquefy gases.

e.g. : Carbon dioxide (CO2) is a gas under normal

conditions of temperature and pressure. It can beliquefied by compressing it to a pressure 70 timesmore than atmospheric pressure.Solid CO

2 is known as �Dry ice�. Solid CO

2 is extremely

cold and used to �deep freeze� food and to keep ice-

cream cold.

Unit of pressure :Atmosphere (atm) is a unit for measuring pressureexerted by a gas.The S. unit of pressure is Pascal (Pa.)1 atm = 1.01 × 105 Pa.

Note :When pressure is lowered the boiling point of liquid islowered. This helps in rapid change of liquid into gas.

EVAPORATION

The phenomenon of change of a liquid into vapoursat any temperature below its boiling point is calledevaporation.

Water changes into vapours below 1000C. Theparticles of matter are always moving and are neverat rest. At a given temperature in any gas, liquid orsolid, there are particles with different K.E.

In case of liquids, a small fraction of particles at thesurface, having higher K.E., is able to break the forcesof attraction of other particles and gets converted intovapour.

Note :The atmospheric pressure at sea level is 1 atm.

(a) Factors Affecting Evaporation:

(i) Temperature: With the increase in temperature therate of evaporation increases.Rate of evaporation T

Reason : On increasing temperature more numberof particles get enough K.E. to go into the vapour state.

(ii) Surface Area : Rate of evaporation Surface areaSince evaporation is a surface phenomena, if thesurface area is increased, the rate of evaporationincreases. So, while putting clothes for drying up wespread them out.

(iii) Humidity of Air : Rate of evaporation Humidity1

Humidity is the amount of water vapour present in air.When humidity of air is low, the rate of evaporation ishigh and water evaporates more readily. Whenhumidity of air is high, the rate of evaporation is lowand water evaporates very slowly.

(iv) Wind Speed : Rate of evaporation Wind speedWith the increase in wind speed, the particles of watervapour move away with the wind. So the amount ofwater vapour decreases in the surroundings.

(v) Nature of substance : Substances with highboiling points will evaporate slowly, while substanceswith low boiling points will evaporate quickly.

Differences between evaporation and boiling

Evaporation Boiling It is a surface phenomenon.

It is a bulk phenomenon.

It occurs at all temperatures below B.P.

It occurs at B.P. only.

The rate of evaporation depends upon the surface area of the liquid, humidity temperature & wind speed

The rate of boiling does not depend upon the

surface area, wind speed, and humidity.

(b) Cooling Caused by Evaporation:

The cooling caused by evaporation is based on thefact that when a liquid evaporates, it draws (or takes)the latent heat of vaporisation from �anything� which

it touches.

For example :� f we put a little of spirit, ether or petrol on the palm ofour hand then our hand feels very cold.

� Perspiration (or sweating) is our body�s method of

maintaining a constant temperature.

(c) We Wear Cotton Clothes in Summer :

During summer, we perspire more because of themechanism of our body which keeps us cool. Duringevaporation, the particles at the surface of liquid gainenergy from the surroundings or body surface.The heat energy equal to latent heat of vaporisation,is absorbed from the body, leaving the body cool.Cotton, being a good absorber of water helps inabsorbing the sweat.

12PAGE # 12

(d) Water droplets on the outer surface of

a glass containing ice cold water :

If we take some ice cold water in a glass then we willobserve water droplets on the outer surface of glass.

Reason : The water vapour present in air on comingin contact with glass of cold water, loses energy. Sowater vapour gets converted to liquid state, which wesee as water droplets.

PLASMA

This state consist of super energetic and super excitedparticles. These particles are in the form of ionisedgases.For eg: Neon sign bulb and fluorescent tubeNeon sign bulb � Neon gas

Fluorescent tube � Helium gas

When electrical energy flows through gas, it getsionised and hence plasma is created.Plasma glows with a special colour depending onnature of gas. Sun and the stars glow because of thepresence of plasma.

BOSE-EINSTEIN CONDENSATE (B.E.C.)

The B.E.C. is formed by cooling a gas of extremelylow density, about one-hundred-thousandth thedensity of normal air, to super low temperature.

Substance : A substance is a kind of matter that cannot

be separated into other kinds of matter by any physical

process. For example, sugar dissolved in water can

be separated from water by simply evaporating the

water but it cannot be broken into its components by

any physical process so here sugar is a substance.

PURE SUBSTANCE

A homogeneous material which contains particles ofonly one kind and has a definite set of properties iscalled a pure substance.Examples : Iron, silver, oxygen, sulphur, carbon dioxideetc., are pure substances because each of them hasonly one kind of particles.

(a) Characteristics of A Pure Substance :

(i) A pure substance is homogeneous in nature.

(ii) A pure substance has a definite set of properties.These properties are different from the properties ofother substances.

(iii) The composition of a pure substance cannot bealtered by any physical means.

Matter

Pure substances

Elements

Mixtures

Compounds

Only one type of particlesare present ( no impurities)

More than one

type ofparticles are present

Homogeneousmixtures(true solutions)

(b) Elements :

A pure substance, which cannot be subdivided intotwo or more simpler substances by any physical orchemical means is called an element.(i) Examples : Hydrogen, oxygen, nitrogen, copper,zinc, tin, lead, mercury, etc. are all elements as theycannot be subdivided into simpler parts by anyphysical or chemical means. A substance made upof the atoms with same atomic number is called anelement.

(ii) Classification of elements :

(A) On the basis of physical states, all elements canbe classified into three groups:-(1) Solids (2) Liquids (3) Gases

It has been found that :

� Two elements exist as liquids at room temperature.

They are mercury and bromine.

� Eleven elements exist as gases at room

temperature. They are hydrogen, nitrogen, oxygen,fluorine, chlorine, helium, neon, argon, krypton, xenonand radon.

� Remaining 98 elements are solids at room

temperature.

(B) Elements can be classified as metals andnon-metals. There are 22 non-metals and 89 metals.

� Amongst the metals, only mercury is a liquid metal.

All other metals are solids.� Amongst the 22 non-metals : 10 non-metals are

solids. They are boron, carbon, silicon, phosphorus,sulphur, selenium, arsenic, tellurium, iodine andastatine. 1 non-metal, bromine, is a liquid. Five non-metals, hydrogen, nitrogen, oxygen, fluorine andchlorine are chemically active gases. Six non-metals,helium, neon, argon, krypton, xenon and radon arechemically inactive gases. These are also callednoble gases, inert gases or rare gases.

METALLOIDS :There are a few elements which show someproperties of metals and other properties of non-metals. For example they look like metals but theyare brittle like non-metals. They are neitherconductors of electricity like metals nor insulators likenon-metals, they are semiconductors. The elementswhich show some properties of metals and someother properties of non-metals are called metalloids.Their properties are intermediate between theproperties of metals and non-metals. Metalloids arealso sometimes called semi-metals. The importantexamples of metalloids are : Boron (B), Silicon (Si),Germanium (Ge), Arsenic (As), Antimony (Sb),Tellurium (Te) Polonium (Po) and Astatine (At).

Note :Hydrogen is the lightest element.

13PAGE # 13

(C) Elements can be classified as normal elements and radioactive elements. The elements which do not give outharmful radiations are called normal elements. Elements from atomic number 1 to atomic number 82 are normalelements. The elements which give out harmful radiations are called radioactive elements. Elements from atomicnumber 83 to atomic number 112 and 114, 116 and 118 are radioactive in nature.

14PAGE # 14

(c) Compounds :

A pure substance, which is composed of two or moredifferent elements, combined chemically in a definiteratio, such that it can be broken into elements only bychemical means is called compound.

The two or more elements present in a compoundare called constituents or components of thecompound. For example, water is a compound ofhydrogen and oxygen, combined together in the ratioof 1 : 8 by weight. The water can be broken into itsconstituents only by electro-chemical method, i.e., bypassing electric current through it.

TYPES OF COMPOUNDS

(A) On the basis of constitutents elements :

(i) Inorganic compoundsThese compounds have been mostly obtained fromnon-living sources such as rocks and minerals. Afew examples of inorganic compounds are : commonsalt, marble, washing soda, baking soda, carbondioxide, ammonia, sulphuric acid etc.

(ii) Organic compoundsThe word �organ� relates to different organs of living

beings. Therefore, organic compounds are thecompounds which are obtained from living beingsi.e., plants and animals. It has been found that all theorganic compounds contain carbon as their essentialconstituent. Therefore, the organic compounds arequite often known as �carbon compounds�. A fewcommon examples of organic compounds are :methane, ethane, propane (all constituents of cookinggas), alcohol, acetic acid, sugar, proteins, oils, fatsetc.

(B) On the basis of their properties :(i) Acids : Compounds which give hydronium ion inaqueous solution for e.g. hydrochloric acid, sulphuricacid, nitric acid, formic acid etc.

(ii) Bases : Compounds which give hydroxide ion inaqueous solution for e.g.Sodium hydroxide,Potassium hydroxide

(iii) Salts : It is formed by the chemical reactionbetween acids and bases for e.g. ammoniumchloride, zinc sulphate etc.

MIXTURES

Most of the materials around us are not puresubstances, but contain more than one substances,elements or compounds. Such materials are calledmixtures.

(a) Definition :

When two or more substances (elements,compounds or both) are mixed together in anyproportion, such that they do not undergo anychemical change, but retain their individualcharacteristics, the resulting product is called amixture.

(b) Types of Mixture :

(i) Homogeneous Mixture : A mixture in which differentconstituents are mixed uniformly is called ahomogeneous mixture.

Examples : All solutions, such as solutions ofcommon salt, copper sulphate, sugar etc. areexamples of homogeneous mixtures. Similarly, alloyssuch as brass, bronze etc. are homogeneous solidsolutions of metals and air is homogenous mixtureof gases.

(ii) Heterogeneous Mixture : A mixture in whichdifferent constituents are not mixed uniformly is calleda heterogeneous mixture.

Examples : A mixture of sand and salt, iron powderand sulphur powder, soil etc. are examples ofheterogeneous mixtures.

15PAGE # 15

TRUE SOLUTIONS

A homogeneous mixture of two or more substancesis called a solution. Usually we think of a solution asa liquid that contains either a solid or a liquid or a gasdissolved in it. However, this is not true. We can alsohave a solid solution and gaseous solution as in thecase of alloys and air respectively.

(a) Components of a Solution :

The substances present in a homogeneous solutionare called components of the solution. A solutionbasically has two components, i.e., a solvent and asolute.

(i) Solvent : The component of a solution which ispresent in large proportion , is called solvent.

Note :

Usually, a solvent is the LARGER component of thesolution.

For example : In the solution of copper sulphate inwater, water is the solvent. Similarly, in paints,turpentine oil is the solvent.

(ii) Solute : The component of the solution which ispresent in small proportion is called solute.

For example: In the solution of common salt in water,the common salt is solute. Similarly, in carbonateddrinks (soda water), carbon dioxide gas is the solute.

Note :Usually, solute is the SMALLER component of thesolution.

(b) Characteristics of a True Solution :

(i) A true solution is always clear and transparent, i.e.,light can easily pass through it without scattering.

(ii) The particles of a solute break down to almostmolecular size and their diameter is of the order of 1nm (10�9 m) or less.

(iii) A true solution can completely pass through afilter paper as particle size of solute is far smallerthan the size of pores of filter paper.

(iv) A true solution is homogeneous in nature.

(v) In a true solution, the particles of solute do notsettle down, provided temperature is constant.

(vi) From a true solution, the solute can easily berecovered by evaporation or crystallisation.

(c) Types of Solution:

(A) On the basis of concentration :(i) Saturated solution : A solution, which at a giventemperature dissolves as much solute as it is capableof dissolving, is said to be a saturated solution.(ii) Unsaturated solution : When the amount of solutecontained in a solution is less than the saturation level,the solution is said to be an unsaturated solution.

(iii) Super saturated solution : A solution, whichcontains more of the solute than required to make asaturated solution, is called a super saturated solution.

(B) On the basis of solvent :(i) Aqueous Solutions : The solutions obtained by

dissolving various substances in water are called

aqueous solutions.

The common examples are :

(i) Common salt dissolved in water.

(ii) Sugar dissolved in water.

(iii) Acetic acid disssolved in water etc.

(ii) Non-Aqueous Solutions : The solutions obtained

by dissolving the substances in liquids other than

water are called non-aqueous solutions. The

common non-aqueous solvents are alcohol, carbon

disulphide, carbon tetrachloride, acetone, benzene

etc. Examples of non-aqueous solutions are :

(i) Iodine dissolved in carbon tetrachloride.

(ii) Sulphur dissolved in carbon disulphide.

(iii) Sugar dissolved in alcohol etc.

(C) On the bases of physical state of solute andsolvent :

(i) Solid-Solid solutions : All alloys are solid solutionsof metals. Brass is a solid solution of approximately30% of zinc and 70% of copper. In this solid solution,copper (larger component) is solvent and zinc(smaller component) is solute. Similarly, Bell Metalis a solid solution of 80% of copper and 20% of tin, inwhich copper is the solvent and tin is the solute.

(ii) Solid-Liquid solutions : Sugar solution is anexample, in which sugar is the solute and water isthe solvent. Similarly, common salt solution is anexample, in which common salt is the solute andwater is the solvent. In case of tincture of iodine,iodine is the solute and ethyl alcohol is the solvent.

(iii) Liquid-Liquid solutions : In case of an alcoholicdrink, ethyl alcohol is solute and water is solvent.Similarly, in case of vinegar, acetic acid is solute andwater is solvent.

(iv) Liquid-Gas solutions : In case of aerated drinks(soda water), carbon dioxide is the solute and wateris the solvent.

(v) Gas-Gas solutions : Air is a homogeneous mixtureof two main gases, i.e., 78% of nitrogen and 21% ofoxygen. In this mixture, nitrogen is solvent and oxygenis solute. Similarly, the petrol fed into the engines ofautomobiles is a mixture of petrol vapour and air.

(d) Concentration of a Solution :

It is defined as the amount of solute present in agiven quantity of the solution. The most commonmethod for expressing the concentration of a solutionis called percentage method. The concentration ofsolution refers to the percentage of solute present inthe solution. Furthermore, the percentage of solutecan be expressed in terms of :

16PAGE # 16

(i) mass of the solute (ii) volume of the solute.

(i) Concentration of a solution in terms of mass

percentage of solute : If a solution is formed by

dissolving a solid solute in a liquid solvent then the

concentration of solution is expressed in terms of

mass percentage of solute and is defined as under :

The concentration of solution is the mass of the solute

in grams, which is present in 100 g of a solution.

Note :

It is very important to keep in mind that the percentage

concentration of a solution refers to mass of solute in

100 g of solution and not 100 g of solvent, i.e., water.

The concentration of a solution in terms of mass

percentage of solute is calculated by the formula given

below :

Concentration of solution

= 100grams)(insolutionofMass

grams)(insoluteofMass

100grams)](insolventofMasssoluteof[Mass

grams)(insoluteofMass

(ii) Concentration of a solution in terms of volume

percentage of solute : If a solution is formed by

dissolving a liquid solute in a liquid solvent, then the

concentration of the solution is expressed in terms of

volume percentage of solute. The concentration of a

solution is the volume of the solute in milliliters, which

is present in 100 milliliters of a solution.

Note :

It is very important to keep in mind that the percentage

concentration of solution refers to volume of solute in

100 ml of solution and not 100 ml of solvent, i.e.,

water.

The concentration of a solution in terms of volume

percentage of the solute is calculated by the formula

given below :

Concentration of solution =

100ml)(insolutionofVolume

ml)(insoluteofVolume

= 100ml)(insolvent]ofVolumesoluteof[Volume

ml)(insoluteofVolume

Note :

The concentration of a solution is a pure percentage

number and has NO UNITS.

(c) Examples :

1. What is the meaning of 15% solution of NaCl ?Sol. 15% solution of NaCl is a solution 100 g of which

contains 15 g of NaCl and 85 g of water.

2. Calculate the amount of glucose required to prepare250 g of 5% solution of glucose by mass.

Sol. % of solute = solutionofMass

soluteofMass× 100

5 = 250

soluteofMass× 100

Mass of solute = 100

2505 =

10125

= 12.5 g

3. A solution contains 50 mL of alcohol mixed with 150mL of water. Calculate concentration of this solution.

Sol. This solution contains a liquid solute (alcohol) mixedwith a liquid solvent (water), so we have to calculatethe concentration of this solution in terms of volumepercentage of solute (alcohol). Now, we know that :

Concentration of solution = solutionofVolume

soluteofVolume× 100

Here, Volume of solute (alcohol) = 50 mLAnd. Volume of solvent (water) = 150 mLSo, Volume of solution = Volume of solute + Volumeof solvent= 50 + 150 = 200 mLNow, putting these values of �volume of solute� and

�volume of solution� in the above formula we get :

Concentration of solution = 20050

× 100 = 2

50

= 25 percent (by volume)Thus, the concentration of this alcohol solution is 25percent.

4. How much water should be added to 16 ml acetoneto make its concentration 48% ?

Sol. Concentration of solution = solutionofVol.

soluteofVol.× 100

x16

× 100 = 48 x = 4816

× 100 = 33.33 ml

Volume of solvent =33.33 � 16 = 17.33 ml.

SUSPENSIONS

A heterogeneous mixture of insoluble particles ofsolute, spread throughout a solvent, is called asuspension. The particle size (diameter) in asuspension is more than 10�5 cm. The particles havea tendency to settle down at the bottom of the vesseland can be filtered out, because their size is biggerthan the size of the pores of the filter paper.

(a) Examples :

(i) Muddy water, in which particles of sand and clayare suspended in water.

(ii) Slaked lime suspension used for white-washinghas particles of slaked lime suspended in water.

(iii) Paints in which the particles of dyes are suspendedin turpentine oil.

17PAGE # 17

(b) Characteristics of Suspensions :

(i) The size of particles is more than 10�5 cm indiameter.

(ii) The particles of suspension can be separatedfrom solvent by the process of filtration.

(iii) The particles of suspension settle down, whenthe suspension is kept undisturbed.

(iv) A suspension is heterogeneous in nature.

(v) More scattering takes place in suspensions,because of bigger size of particles.

Note :The process of settling of suspended particles underthe action of gravity is called sedimentation.

A heterogeneous solution in which the particle size isin between 10�7 cm to 10�5 cm, such that the soluteparticles neither dissolve nor settle down in a solventis called colloidal solution.In a colloidal solution, relatively large suspendedparticles are called dispersed phase and the solventin which the colloidal particles are suspended iscalled continuous phase or dispersing medium.

(a) Examples of Colloidal Solutions :

Few examples of colloidal solutions are as follows :� blood � Milk � Writing ink

� Jelly � Starch solution � Gum solution

� Tooth paste� Soap solution � Liquid detergents

� Mist and fog.

(b) Characteristics of Colloidal Solutions :

(i) The size of colloidal particles is in between 10�7

cm to 10�5 cm.

(ii) The particles of a colloidal solution are visibleunder a powerful microscope.

(iii) The particles of a colloidal solution do not settledown with the passage of time.

(iv) The particles of a colloidal solution can easilypass through filter paper.

(v) The colloidal solutions are heterogeneous innature.

(vi) Colloidal solutions are not transparent, buttranslucent in nature.

(vii) The particles of a colloidal solution scatter light,i.e., when strong beam of light is passed through thecolloidal solution, the path of beam becomes visible.

Scattering of Light (Tyndall Effect)

If a beam of light is passed through pure water or asalt solution, the path of light is visible but when astrong beam of light is passed through a colloidalsolution and viewed at right angles with the help of amicroscope, the path of light shows up a bright coneof bluish light.

Tyndall effect shown by colloid in a beaker

This luminosity of path of beam is known as Tyndalleffect and the illuminated path is known as Tyndallcone. (Tyndall being the name of the scientist whostudied this phenomenon first).

Tyndall effect is caused due to the scattering of lightby the colloidal particles. The true solutions do notscatter light and hence do not show Tyndall effect.

Tyndall effect can be seen when a fine beam of lightenter in a room through a small hole. This happensdue to scattering of light particles of dust and smokein the air of the room.Tyndall effect can be observed when sunlight passesthrough a dense forest. In the forest, fog contains tinydroplets of water which act as particles of colloiddispersed in air.

(viii) The particles of a colloidal solution are electricallycharged.ElectrophoresisThe collodial solutions contain either positively ornegatively charged particles and, therefore, when anelectric current is passed through them, the particlesmove towards either of the oppositely chargedelectrodes. Subsequently, they get discharged on theelectrodes and precipitate out. For example, when anegatively charged As2S3 solution is taken in a U-tube into which Platinum electrodes, connected to asource of E.M.F. are dipped, the colloidal particlesmove towards the positive electrode .The migration of colloidal particles under the influenceof an electric field is known as electrophoresis.

Electrophoresis showing migration of colloidalparticles

(c) Classification of Colloids :

The colloids are classified according to the state ofdispersed phase (solid, liquid or gas) and the stateof dispersing medium. A few common examples areshown in the table :

18PAGE # 18

Note :

Colloidal solutions can be separated by the process of CENTRIFUGATION.

S.No. Property True solution Colloidal solution Suspension

1 Nature Homogeneous Heterogeneous Heterogeneous

2 Particle size Diameter less than 1 nm

(or 10Å) or 10-7 cm

Diameter between 1-100 nm (or 10-1000Å) or

10-7 to 10-5 cm

Diameter more than100 nm (or 1000Å)

or 10-5 cm.3 Filtrability Passes through an

ordinary filter paperas well as animal orvegetable membranes

Passes throughordinary filter paperbut not through animalor vegetable membranes

Do not pass throughfilter paper or animalor vegetablemembranes

4 Visibility Particles are completelyinvisible

Particles themselves areinvisible but their presence can be detected by ultramicroscope since they scatter light.

Particles visible tothe naked eye orunder a microscope

5 Diffusion Diffuse rapidly Diffuse slowly Do not diffuse

6 Tyndall effect Not shown Shown May be shown

7 Appearance of solution

Clear and transparent Generally clear and transparent Opaque

SEPARATION OF

HETEROGENEOUS MIXTURES

Heterogeneous mixtures can be separated into theirrespective components by simple physical methodssuch as handpicking, sieving, filtration.

Generally following physical properties are consideredin the separation of the constituents of a mixture.

(i) Densities of the constituents of the mixture.

(ii) Melting points and boiling points of the constituentsof the mixture.

(iii) Property of volatility of one or more constituents ofthe mixture.

(iv) Solubility of the constituents of the mixture indifferent solvents.

(v) Ability of the constituents of the mixture to sublime.

(vi) Ability of the constituents of the mixture to diffuse.

Note :However, for separating homogeneous mixturesspecial techniques are employed depending uponthe difference in one or more physical properties ofthe constituents of the mixture.

TECHNIQUES USED FOR SEPARATING

THE COMPONENTS OF A MIXTURE

(A) Separation of mixture of two solids :

(a) By Sublimation:

The changing of solid directly into vapours on heating

and of vapours into directly solid on cooling is known

as sublimation.

(i) Separation of a mixture of common salt and

ammonium chloride :This method is used in the

separation of such solid-solid mixtures where one of

the components sublimes on heating. However, it is

useful only if the components of the mixture do not

react chemically on heating. The table shows the list

of mixtures which can be separated by the process

of sublimation.

19PAGE # 19

Solid-Solid Mixture Sublimable Solid

Common salt and Ammonium chlorideammonium chloride

Sand and iodine Iodine

Common salt and iodine Iodine

Sodium sulphate and Benzoic acidbenzoic acid

Iron filings and naphthalene Naphthalene

(ii) Method :

� Place the mixture of common salt and ammonium

chloride in a china dish and heat it over a low Bunsen

flame.

� Place a clean glass funnel in an inverted position in

the china dish and close the mouth of its stem with

cotton wool.

� The ammonium chloride in the mixture sublimes to

form dense white fumes. These fumes condense on

the cooler sides of the funnel in the form of fine white

powder.

� When the mixture gives off no more white fumes, lift

the funnel, scrap the fine white powder from its sides

on a piece of paper. This is pure ammonium chloride.

The residue left behind in the funnel is sodium

chloride.

Separation by sublimation Note :

Dry ice (solid CO2), Naphthalene, Anthracene, Iodine

etc. are sublimable solids.

(b) By Using a Suitable SolventIn some cases, one constituent of a mixture is solublein a particular liquid solvent whereas the otherconstituent is insoluble in it. This difference in thesolubilities of the constituents of a mixture can beused to separate them. For example, sugar is solublein water whereas sand is insoluble in it, so a mixtureof sugar and sand can be separated by using wateras solvent. This will become more clear from thefollowing discussion.

To Separate a Mixture of Sugar and SandSugar is soluble in water whereas sand is insolublein water. This difference in the solubilities of sugarand sand in water is used to separate them. This isdone as follows. The mixture of sugar and sand istaken in a beaker and water is added to it. The mixtureis stirred to dissolve the sugar . The sand remainsundissolved.

Separation of sugar and sand mixture The sugarsolution containing sand is filtered by pouring over afilter paper kept in a funnel. Sand remains as a residueon the filter paper and sugar solution is obtained asa filtrate in the beaker kept below the funnel. The sugarsolution is evaporated carefully to get the crystals ofsugar. In this way, a mixture of sugar and sand hasbeen separated by using water as the solvent.

(B) Separation of mixture of a solid and a liquid :

(a) By Evaporation :

(i) Separation of coloured component (dye) from

blue ink : The process of evaporation is suitable for

the separation of non-volatile soluble solid (dye) from

its liquid solvent (water).

20PAGE # 20

(ii) Method :

� Heat sand in an iron vessel by placing it over a tripodstand. This arrangement is called sand bath.

� Place a china dish on the sand bath. Pour about 5 ccof the ink into the china dish.

� Heat gently evaporates water from the ink such that itdoes not boil. In a few minutes the water evaporatesleaving behind dry blue ink. Method of evaporation issuitable for the following solid-liquid mixtures.

(b) By Centrifugation

The method of separating finely suspended orcolloidal particles in a liquid, by whirling the liquid ata very high speed is called centrifugation.

(i) Principle of centrifugation : It is based on theprinciple that when a very fine suspension or acolloidal solution is whirled rapidly, then the heavierparticles are forced towards the bottom of liquid andthe lighter stay at the top.(ii) Separation of cream from milk : The process ofcentrifuging is employed in separating cream frommilk. This process is generally employed inseparating colloidal solutions which easily passthrough the filter paper.

spin

CENTRIFUGEMethod :

� Pour full cream milk in the test tube with a pivot inyour laboratory centrifuge.

� Shut the lid of the centrifuge and switch on the current.When the centrifuge starts working, the tubecontaining milk swings out in the horizontal positionand whirls around its axis at a high speed.

� The centrifugal force (in the outward direction) pushesthe heavier particles outward, i.e., towards the bottomof the mixture. Thus, the heavier particles of theproteins, carbohydrates, etc. are pushed towards thebottom of the tube, but the lighter particles of the fatstay near the top of the tube and hence separate.

(iii) Applications of centrifugation :

� It is employed in milk dairies to separate cream fromthe milk.

� It is employed in diagnostic laboratories in testingurine samples.

� It is employed in blood banks to separate differentconstituents of blood.

� It is used in drying machines to squeeze out waterfrom the wet clothes.

(c) By Chromatography :

The process of separation of different dissolvedconstituents of a mixture by adsorbing them over anappropriate adsorbent material is calledchromatography.The adsorbent medium is generally magnesiumoxide, alumina or filter paper. The solvent generallyused for dissolving a mixture of two or moreconstituents is water or alcohol.The different constituents of a mixture get adsorbeddifferently on the same adsorbent material, becausethey have different rates of movement. The rate ofmovement of each adsorbed material depends upon :

� The relative solubility of the constituents of mixture ina given solvent.

� The relative affinity of the constituents of mixture forthe adsorbent medium.If a filter paper is used as an adsorbent material forthe separation of various constituents of a mixture,then this method of separation of mixture is calledpaper chromatography.Paper chromatography is very useful in separatingvarious constituents of coloured solutes present in amixture of lime, ink, dyes etc.

Note :Kroma means colour in Greek language andtechnique of chromatography was first applied for theseparation of colours, so this name was given.

(i) Separation of coloured constituents present in amixture of ink and water.

(ii) Method :

� Take a filter paper 22 cm long, 5 cm broad and stickits smaller end to a glass rod with the help of gum.On the other end, measure a distance of 2 cm fromlower end and mark a small point. On this point pourone or two drop of the ink.

21PAGE # 21

� Suspend this filter paper in a wide and tall cylinder asshown in Figure. Gradually, pour water into thecylinder till the lower end of filter paper slightly dips inthe water. Cover the cylinder with a glass lid to preventany evaporation and leave the apparatus undisturbedfor an hour. The water rises up the filter paper andreaches the ink mark. This water then dissolvesvarious constituents of the ink, gets adsorbed by thefilter paper in different amounts. More the constituentgets adsorbed, the lesser it moves upward and viceversa.

� When the solvent (water) reaches near the top of filterpaper, the filter paper is removed from water and dried.On the filter paper will be seen a band of colours, ofvarious constituents.

� A filter paper with separated bands of variousconstituents of a coloured substance is calledchromatogram.

(iii) Advantages :

� It can be carried out with a very small amount ofmaterial.

� The substances under investigation do not get wastedin chromatographic separation.(iv) Applications :

� It is used to separate colours from dye.� It is used in the separation of amino acids.� It is used in the separation of sugar from urine.� It is used in the separation of drugs from the samples

of blood.

(C) Separation of mixture of two liquids :

(a) By Distillation:

Distillation is the process of heating a liquid to formvapour and then cooling the vapour to get the backliquid.Distillation can be represented as :

Liquid Heating

CoolingVapour (or Gas)

Note :

The liquid obtained by condensing the vapour in theprocess of distillation is called DISTILLATE .

(i) Liebig condenser : Liebig condenser is a watercondenser. It is a long glass tube surrounded by awider glass tube (called water jacket) having an inletand outlet for water. During distillation, cold water fromtap is circulated through the outer tube of condenser.This water takes away heat from the hot vapourpassing through the inner tube of condenser andcauses its condensation.Process of simple distillation is used to recover bothsalt as well as water , from a salt-water mixture (orsalt solution) and to separate of components of amixture containing two miscible liquids that boilwithout decomposition and have sufficient differencein their boiling points.

SIMPLE DISTILLATION

(ii) Fractional distillation : Separation of mixture oftwo miscible liquids for which the difference in theboiling points is less : In case of two liquids whichhave very close boiling points, both the liquids tend todistil over in different proportions. It means lesserthe boiling point of a liquid, more is the proportion ofit distilling over.The above problem can be avoided by using afractionating column. It gives the effect of repeateddistillation by offering resistance to the passage ofvapour.The process of separation of two miscible liquids bythe process of distillation, making use of theirdifference in boiling points, is called fractionaldistillation.

Note :The process of fractional distillation is useful only, ifthe difference in the boiling points of the two miscibleliquids is less than 25ºC.

22PAGE # 22

(A) Method :

� The process of fractional distillation is similar to the

process of distillation, except that a fractionating

column is attached.

� The design of a fractionating column is such that the

vapours of one liquid (with a higher boiling point) are

preferentially condensed as compared to the vapours

of the other liquid (with lower boiling point).

DIFFERENT TYPES OF FRACTIONATING COLUMNS

� Thus, the vapours of the liquid with low boiling point,pass on to the Liebig�s condenser where they

condense. The liquid so formed is collected inreceiver.

� The thermometer shows a constant reading as longas the vapour of one liquid are passing to Liebig�scondenser. As soon as the temperature starts rising,the receiver is replaced by another receiver to collectsecond liquid.

(b) By Separating Funnel :

(i) Separation of a mixture of two immiscible liquids :The separation of two immiscible liquids is basedon the difference in their densities. The apparatusused for separation is separating funnel. It is a longglass tube provided with a tap at its bottom. The tablebelow shows different immiscible liquids which canbe separated by separating funnel.

Benzene and water Water Benzene

Kerosene oil and water Water Kerosene oil

Turpentine oil and water Water Turpentine oil

Chloroform and water Chloroform

Mustard oil and water Water Mustard oil

ImmiscibleLiquid-liquid Mixture

HeavierLiquid

LighterLiquid

Water

(ii) Method :

� Close the tap of separating funnel and clamp it in avertical position in an iron stand.

� Pour the immiscible liquid mixture (say benzene-watermixture) in the separating funnel. Allow the mixture tostand for half an hour or more.

� The immiscible components of the mixture, i.e.,benzene and water separate out into two distinctlayers. The benzene forms the lighter layer on the topand the water forms the heavier layer at the bottom.

� Place a conical flask or a beaker under the nozzle ofthe separating funnel. Turn the tap gently so that thewater trickles in the flask or the beaker drop by drop.Once the water is drained out, close the tap.

� Now place another conical flask or a beaker underthe nozzle of separating funnel. Open the tap to drainout benzene.

Separation by separating funnel

(iii) Applications :

� This method is used for separating any twoimmiscible liquids.

� This method is used in separation of slag (a wastematerial) from the molten metals during theirextraction. For example, during the extraction of ironfrom its ore, the molten iron and slag collect at thebase of blast furnace. The slag being less densefloats up the surface of molten iron. They are drainedout from two different outlets.

SEPARATION OF GASES FROM AIR

In order to separate the major components of air, it isfirst purified, then liquefied and finally fractionallydistilled. The steps involved in the process are asfollows -

(a) Purification of Air :

(i) Air generally contains carbon dioxide gas, hydrogensulphide gas and sulphur dioxide gas as impurities.In addition to it there are dust particles also .(ii) First of all air is washed by passing it throughwater, where the dust particles are removed.

(iii) The washed air is passed through dilute causticsoda solution, where the gases like carbon dioxide,sulphur dioxide and hydrogen sulphide are removed.(iv) The purified air, however, contains moisture. Themoist air is passed through pipes, maintained at atemperature below � 20º C, where water vapour

present in it freezes and hence, air becomes dry.

(v) The air leaving the cooling pipes is free from allimpurities.

23PAGE # 23

(b) Liquefaction of Air :

(i) The cool air, free from all impurities is compressed

to a pressure 200 times more than the atmospheric

pressure. The compression raises the temperature

of the air.

(ii) The hot compressed air is then passed through

cooling tank in which cold water enters from one end

and warm water leaves from the other end.

(iii) The compressed and cooled air is passed through

a spiral pipe, placed in a vacuum flask. The end of

spiral pipe is provided with a fine jet.

(iv) When compressed air suddenly escapes fromthe jet, its pressure suddenly falls. Thus, its moleculesmove wide apart. When the molecules move wideapart, they need energy. This energy is taken by themolecules from themselves and hence, theirtemperature drops.

(v) The air so cooled, is now at a pressure equal tothat of atmosphere. This cooled air rises up and inthe process further cools the incoming compressedair in spiral tube. The air is then sucked again by thecompression pump and the cycle is repeated. Withevery cycle, the temperature of air drops, till it liquefies.

(c) Fractional Distillation of Air :

(i) The liquid air mainly consists of nitrogen and oxygen,and is at a temperature of � 200º C.

(ii) The boiling point of liquid nitrogen is � 195º C and

that of liquid oxygen is � 183º C.

(iii) The liquid is gradually warmed to � 195º C, when

nitrogen starts boiling off from the liquid air. Thenitrogen gas so formed, is compressed and filled insteel cylinders.

(iv) The liquefied oxygen left behind, is also changedto gas and then filled in compressed state in steelcylinders.

PHYSICAL AND CHEMICAL CHANGES

Some kind of change always takes place in the matterwhen it is subjected to energy changes. Almost allthe changes (except nuclear changes) taking placein the matter can be classified under two headings,these are as follows -

(a) Physical Changes :

Definition : A change which alters some specificphysical property of the matter, like its state, texture,magnetic or electrical conditions or its colour, withoutcausing any change in the composition of itsmolecules, is called physical change, provided it getsreversed, if the cause producing the change isremoved.

Following points need special consideration :

(i) No new or different product is formed : Thecomposition of molecules of the substance remainsunaltered.

Example : Ice melts to form water. In this exampleonly the appearance (state) of matter has changedfrom solid to liquid. However, the composition of themolecules of ice or water remains same, i.e., for every1 g of hydrogen there is 8 g of oxygen required . Thus,only a physical change has occurred.

(ii) The change is temporary and reversible : Itmeans the change can be reversed by altering thecauses which produce the change.Example : The water formed from ice can be changedback to ice by placing it in a freezing mixture (a mixtureof ice and common salt).

Note :On altering the experimental conditions, the changewhich gets reversed, is a physical change.

(iii) There is no net gain or loss of energy : Theamount of energy required to bring about a physicalchange is generally equal to the amount of energyrequired to reverse the change. Thus, there is no netenergy change involved.

Example : If 1 g of water at 100º C on changing into

steam at 100ºC needs 2260 J of heat energy, then 1

g of steam at 100º C on changing into water at 100º

C, gives out 2260 J of heat energy. Thus, the net energychange is zero.

(iv) There is no change in the weight of substance :During a physical change it is only the energy whichis added or removed. No matter is added during aphysical change. Similarly, no matter is removedduring a physical change. Therefore, mass of thesubstance remains same.

24PAGE # 24

SOME EXAMPLES INVOLVING PHYSICAL CHANGES :

1. Switching on an electric bulb

2. Rubbing a permanent magnet on a steel rod.

3. Action of heat on iodine

4. Dissolving of common salt in water.

The bulb glows and givesout heat and light energy.

The steel rod gets magnetised.If it is brought near iron nails,they get attracted.

The brownish grey crystals of iodine change to form violet vapours. On cooling the vapourscondense on cooler parts of thetest tube to form crystals.

The white crystalline salt disappears in water. However, the water tastes exactly like commonsalt. Moreover, common salt can be recovered by evaporation.

The physical appearance of the bulb changes.

The steel rod acquires the property of attracting piecesof iron.

Change in state and colour.

Change of state.

Physical Change Observation Change in Physical Property

Some Common Examples of Physical Changes :

� Formation of dew.

� Evaporation of water.

� Crystallisation of sugar from its solution.

� Ringing of an electric bell.

� Breaking of a glass pane.

� Freezing of ice cream.

� A rock rolling down a hill.

� Bending of a glass tube by heating.

� Melting of wax.

� Sublimation of camphor.

(b) Chemical Change :

Definition : A change which alters the specific

properties of a material by bringing about a change

in its molecular composition, followed by a change in

state, is called a chemical change.

Following points need special consideration :

(i) A chemical change results in the formation of

one or more new products : The products formed

have different properties than the original substance.

Thus, the composition of the molecules of products

is different from the original substance.

Example : Heating of sugar

When sugar is gently heated in a test tube, it melts. It

gradually changes to brown colour, giving a large

amount of steamy fumes. In the end a black mass is

left which consists of carbon. Thus, new substances,

viz. carbon and water (steam), are formed. In this

change, the arrangement between the molecules of

carbon, hydrogen and oxygen breaks. The hydrogen

and oxygen atoms separate from carbon atoms and

join together to form water. The carbon atoms are set

free and are left as black residue.

SteamCarbonSugar heat

(ii) The weight of the substance undergoing

chemical change usually changes :

Example : During the heating of sugar, the weight of

the black residue is far less than the actual weight of

the sugar. However, this is an apparent change in

weight. If we take the weight of steam into account

and add to it the weight of carbon, then total weight

will be equal to the weight of sugar crystals. Thus,

strictly speaking, total weight of substances taking

part in a chemical change remains constant.

(iii) The chemical change is permanent and

irreversible : It means the change will not reverse by

altering the experimental conditions.

Example : The sugar, which has decomposed on

heating to form carbon and steam will not change to

sugar on cooling.

(iv) During chemical change energy is either

absorbed or given out : The various atoms in a

chemical compound are joined by attractive forces

commonly called bonds. The making or breaking of

the bonds always requires exchange of energy. Thus,

some amount of heat is either absorbed or given out

during a chemical change.

25PAGE # 25

SOME EXAMPLES INVOLVING CHEMICAL CHANGES :

Chemical Change Observation Equation

1. Burning of magnesium in air

When a magnesium ribbon is heated in a flame of Bunsen burner, it catches fire and burnswith a dazzling white flame to formwhite ash.

When iron (silver grey) is leftexposed to moist air for a few days, reddish brown powdery mass (rust) is found on its surface

When LPG (Liquefied Petroleum Gas) is burnt, it burns with a pale blue flame and liberates colourlessgas carbon dioxide along with steam.

Magnesium + Oxygen Magnesium oxide

Iron + Oxygen+ Water vapour Rust

Butane (LPG) + Oxygen Carbon dioxide + Water

2. Rusting of iron

3. Burning of LPG

SOME COMMON EXAMPLES OF CHEMICAL CHANGES

:� Burning of wood or charcoal � Burning of candle � Digestion of food � Curdling of milk

� Formation of biogas (Gobar gas) � Burning of petrol or diesel

� Smoking of cigarette � Drying of paint � Rusting of iron

� Ripening of fruit �Clotting of blood � Fading of the colour of a dyed cloth

� Baking of cake � Photosynthesis � Formation of wine � Butter turning rancid

� Electrolysis of water into hydrogen and oxygen � Formation of water from hydrogen and oxygen

(C) DIFFERENCE BETWEEN PHYSICAL AND CHEMICAL CHANGES

26PAGE # 26

EXERCISE

1. The quantity of matter present in an object is called

its -

(A) weight (B) volume

(C) mass (D) density

2. Which of the following statements is/are correct ?

(A) Interparticle spaces are maximum in the gaseous

state of a substance .

(B) Particles which constitute gas follow a zig-zag

path.

(C) Solid state is the most compact state of substance.

(D) All are correct

3. In sublimation process -

(A) solid changes into liquid.

(B) liquid changes into gas.

(C) solid changes directly into gas.

(D) None of these

4. During evaporation of liquid -

(A) the temperature of the liquid falls.

(B) the temperature of the liquid rises.

(C) the temperature of the liquid remains unchanged.

(D) all statements are wrong.

5. Which of the following statement is not true about

colloidal solution ?

(A) These are visible under powerful microscope.

(B) Their particles do not settle down with passage

of time.

(C) Their particles are electrically charged.

(D) These are homogeneous in nature.

6. Which of the following method is used for separation

of different components of petroleum?

(A) Fractional distillation (B) Sublimation

(C) Chromatography (D) Simple distillation

7. When common salt is added is ice -

(A) its melting point decreases.

(B) its melting point increases.

(C) its melting point does not change from 0ºC

(D) ice becomes harder.

8. Which of the following statements is false ?

(A) Melting and freezing point of a substance are the

same.

(B) Evaporation of liquid takes place only at its boiling

point.

(C) Pure water has no taste

(D) Water allows sunlight to pass through it.

9. Some matter and their groups are given in column P

and Q respectively. (P) (Q)(a) Air (i) Element(b) O

2(ii) Mixture

(c) Copper sulphate (iii) Base(d) Sodium hydroxide (iv) SaltThe correct option is -(A) a (ii), b (iv), c (i), d(iii)(B) a (iv), b (iii), c (ii), d (i)(C) A (i), b (ii), c (iii), d (iv)(D) a (ii) , b (i), c (iv), d (iii)

10. The water boils when :(A) Saturated vapour pressure of water becomesequal to the atmospheric pressure(B) Boiling point of water becomes more thanatmospheric pressure(C) Saturated vapour pressure of water is less thanatmospheric pressure(D) Vapour pressure of water becomes more thanatmospheric pressure

11. Which of the following is a chemical change ?(A) Melting of Wax(B) Dissolving sugar in water(C) Beating aluminium to make aluminium foil(D) Burning of Coal

12. The gas you use in kitchen is called liquefiedpetroleum gas (LPG). In the cylinder, it exists as aliquid. When it comes out of the cylinder, it becomesa gas (process A), then it burns (process B). Choosethe correct statement.(A) Process A is a chemical change.(B)Process B is a chemical change(C) Both processes A and B are chemical changes.(D) None of these processes is a chemical change.

13. The temperature remain same during melting, whileall the ice changes into water due to the :(A) latent heat of fusion.(B) latent heat of vapourisation.(C) latent heat of evaporation.(D) latent heat of sublimation.

14. Crystallization is considered better than evaporationfor obtaining pure crystal of sugar because in evapo-ration on heating -(A) sugar sublimes.(B) Sugar particles will evaporate.(C) Sugar particles will decompose.(D) Sugar particles will melt.

15. The principle behind fractional distillation techniquein separation of two liquids is -(A) difference in Melting point(B) difference in Boiling point(C) difference in Concentration(D) difference in Solubility

27PAGE # 27

16. Solubility of a gas in a liquid increases on -(A) increasing temperature.(B) decreasing pressure.(C) increasing pressure.(D) increasing temperature and pressure.

17. Fusion is the process of conversion of -(A) liquid into gas. (B) solid into gas.(C) solid into liquid. (D) liquid into solid.

18. Carbon burns in oxygen to form carbon dioxide. Theproperties of carbon dioxide are -(A) similar to carbon(B) similar to oxygen(C) totally different from both carbon and oxygen(D) much similar to both carbon and oxygen

19. A thermometer is inserted into a beaker filled with iceat 0ºC. The beaker is heated slowly. The temperature

does not rise for some time. This is because -(A) ice is very cold(B) heat was used for changing ice at 0ºC to water at

0ºC

(C) the density of water is more than ice(D) the density of water is less than the ice

20. Separation of cream from milk is done by :(A) filtration(B) centrifugation method(C) evaporation(D) boiling

21. What sublimate will be obtained when a mixture ofsand, sulphur, common salt and iodine is sublimed ?(A) Sand (B) Iodine(C) Sulphur (D) Common salt

22. Purity of organic liquid can be checked by itscharacteristic -(A) boiling point(B) volume(C) solubility in water(D) solubility in alcohol

23. Carbon tetra chloride and benzene are -(A) immiscible liquid (B) miscible liquid(C) both ( and ) (D) None of these

24. A pure substance can only be -(A) a compound(B) an element(C) an element or a compound(D) a heterogeneous mixture

25. Which of the following statements is not true aboutsuspension ?(A) The particles of suspension can be separatedfrom solvent by the process of filtration.(B) When the suspension is kept undisturbed thenthe particles of suspension settle down.(C) A suspension is homogeneous in nature.(D) Scattering of particles take place in suspension.

26. In which of the following, dispersed phase is a liquidand dispersion medium is a gas ?(A) Cloud (B) Smoke(C) Gel (D) Soap bubble

27. Which of the following statements is/are correct ?(A) Intermolecular forces of attraction in solids aremaximum.(B) Intermolecular forces of attraction in gases areminimum.(C) Intermolecular spaces in solids are minimum.(D) All of the above

28. A liquid disturbed by stirring comes to rest aftersometime due to its property of -(A) Compressibility (B) Diffusion(C) Viscosity (D) All of these

29. Which of the following statements regardingmelting point and freezing point of a substance istrue ?(A) Melting point of a substance is more than its

freezing point.(B) Melting point of a substance is less than its freez-ing point.

(C) Melting point and freezing point of a substanceare same.(D) None of these.

30. Which of the following conditions is most favourablefor converting a gas into liquid ?(A) High pressure, low temperature(B) Low pressure, low temperature(C) Low pressure, high temperature(D) High pressure, high temperature

28PAGE # 28

there exists

: such that

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> is greater than

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belongs to

angle

|| parallel to

is congruent to

therefore

perpendicular to

~ difference

= is equal to

is not equal to

less than or equal to

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SYMBOLS & IT�S MEANING

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29PAGE # 29

INTRODUCTION

Number System is a method of writing numerals torepresent numbers.

Ten symbols 0,1, 2, 3, 4, 5, 6, 7, 8, 9 are used torepresent any number (however large it may be) in ournumber system.

Each of the symbols 0,1, 2, 3, 4, 5, 6, 7, 8, 9 is called adigit or a figure.

INTEGERS

The set of integers is the set of natural numbers, zeroand negative of natural numbers simultaneously. Theset of integers is denoted by or Z.

Z = { ........� 4, � 3, � 2, � 1, 0, 1, 2, 3, 4 ...}

NATURAL NUMBER

Counting numbers 1, 2, 3, 4, 5, ....are called naturalnumbers.

The set of natural numbers is denoted by Ni.e., N = {1, 2, 3, 4, 5, .........}.

WHOLE NUMBER

All natural numbers together with zero are called wholenumbers, as 0, 1, 2, 3, 4, ... are whole numbers.

The set of whole numbers is denoted by W, i.e.W = { 0, 1, 2, 3, 4, 5 .....}. So, W = N {0}, where N is theset of natural numbers.

0 is the smallest whole number, there is no largestwhole number i.e. the number of the elements in theset of whole numbers is infinite.

Every natural number is a whole number.

i.e. N W i.e. N is a subset of W..

0 is a whole number, but not a natural number,i.e. 0 W but 0 N

EVEN NUMBERS

Whole numbers which are exactly divisible by 2 arecalled even numbers.

The set of even numbers is denoted by 'E', such thatE = {0, 2, 4, 6, 8, .....}.

NUMBER SYSTEM

ODD NUMBERS

Natural numbers which are not exactly divisible by 2are called odd numbers. O = {1, 3, 5, 7, 9.....}

PRIME NUMBERS

Natural numbers having exactly two distinct factors i.e.1 and the number itself are called prime numbers.2, 3, 5, 7, 11, 13, 17, 19,... are prime numbers.

2 is the smallest and only even prime number.

IDENTIFICATION OF PRIME NUMBER

Step (i) Find approximate square root of given number.

Step (ii) Divide the given number by prime numbers lessthan approximately square root of number. If givennumber is not divisible by any of these prime numberthen the number is prime otherwise not.

Ex.1 Is 131 a prime number ?Sol. Approximate square root = 12

Prime number < 12 are 2, 3, 5, 7, 11. But 131 is notdivisible by any of these prime number. So, 131 is aprime number.

COMPOSITE NUMBERS

Natural numbers having more than two factors arecalled composite numbers.

4, 6, 8, 9, 10, 12, 14, 15, 16, 18... are compositenumbers.

Number 1 is neither prime nor composite number.

All even numbers except 2 are composite numbers.

Every natural number except 1 is either prime orcomposite number.

There are infinite prime numbers and infinite compositenumbers.

CO-PRIME NUMBER OR RELATIVELY

PRIME NUMBERS

Two natural numbers are said to be co-primenumbers or relatively prime numbers if they have only1 as common factor. For ex. 8, 9 ; 15, 16 ; 26, 33 etc. areco-prime numbers.

Co-prime numbers may not themselves be primenumbers. As 8 and 9 are co-prime numbers, but neither8 nor 9 is a prime number.

Every two consecutive natural numbers are co - primes.

30PAGE # 30

TWIN PRIMES

Pairs of prime numbers which have only onecomposite number between them are called twinprimes.For example : 3, 5 ; 5, 7 ; 11,13 ; 17, 19 ; 29, 31 ; 41, 43;59, 61 and 71, 73 etc. are twin primes.

RATIONAL NUMBERS

These are real numbers which can be expressed in the

form of p/q, where p and q are integers and 0q .

For example : 32

, 1537

, �1917

.

All natural numbers, whole numbers and integers arerational.

Rational numbers include all Integers (without anydecimal part to it), terminating fractions (fractions inwhich the decimal parts terminating e.g. 0.75, � 0.02etc.) and also non-terminating but recurring decimals.e.g. 0.666....., � 2.333...., etc.

FRACTIONS

(a) Common fraction : Fractions whose denominator isnot 10.

(b) Decimal fraction : Fractions whose denominator is 10or any power of 10.

(c) Proper fraction : Numerator < Denominator i.e.5

3.

(d) Improper fraction : Numerator > Denominator i.e.35 .

(e) Mixed fraction : Consists of integral as well as fractional

part i.e.72

3 .

(f) Compound fraction : Fraction whose numerator and

denominator themselves are fractions. i.e.5/72/3

.

Improper fraction can be written in the form of mixedfraction.

IRRATIONAL NUMBERS

All real numbers which are not rational are irrationalnumbers. These are non-recurring as well asnon-terminating type of decimal numbers.

For example : 2 , 3 4 , 32 , 32 , 4 7 3 etc.

PURE RECURRING DECIMAL

A decimal is said to be a pure recurring decimal if thea digit or set of digits after the decimals are repeated.

Thus, 3

1 = 0.333...... = 3.0 ,

7

22= 3.142857142857..... = 3.142857 .

Conversion of decimal numbers into rationalnumbers of the form plq.

Short cut method for pure recurring decimal : Writethe repeated digit or digits only once in the numeratorand take as many nines in the denominator as thereare repeating digits in the given number.

For example : (i) 0. 3 = 3/9 or 1/3 (ii) 0. 387 = 387/999

MIXED RECURRING DECIMAL

A decimal is said to be a mixed recurring decimal ifthere is at least one digit after the decimal point, whichis not repeated.

Short cut method for mixed recurring decimal : Forma fraction in which numerator is the difference betweenthe number formed by all the digits (take the digitsonce which are repeating after decimal) and that formedby the digits which are not repeated and thedenominator is the number formed by as many ninesas there are repeated digits followed by as many zerosas the number of non-repeated digits.

Ex.2 Change 3574.0 in the form of p/q.

Sol. 3574.0 = 9900

747435 =

9900

7361.

Ex.3 Change 12 453. in the form of p/q.

Sol. 990

123�12345

=990

12222.

COMPARISON OF FRACTIONS

Suppose some fractions are to be arranged inascending or descending order of magnitude. Then,convert each one of the given fractions in the decimalform, and arrange them accordingly.

Now, 53

= 0.6, 76

= 0.857, 97

= 0.777.....

Since 0.857 > 0.777....> 0.6, so76

>97

>53

.

REAL NUMBERS

The set of rational numbers and irrational numberstaken together is known as a set of real numbers.

The absolute value of a real number I x I is defined as

| x | =

0xif ,x

0xif ,x

For example :| 2 |= 2 ; 2 > 0 and | � 2 | = � (� 2 ) = 2; � 2 < 0.

31PAGE # 31

BODMAS RULE

This rule depicts the correct sequence in which theoperations are to be executed so as to find out thevalue of a given expression.Here,�B� stands for �Bracket�, �O� for �of�, �D� for Division,�M� for Multiplication�, �A� for �Addition� and �S� forsubtraction�.Thus, in simplifying an expression, first of all thebrackets must be removed, strictly in the order ( ), { }and [ ].After removing the brackets, we must use the followingoperations strictly in the order :(i) Of (ii) Division (iii) Multiplication(iv) Addition (v) Subtraction.

Vinculum (or Bar) : When an expression containsVinculum, before applying the �BODMAS� rule, we

simplify the expression under the Vinculum.

Ex.4 Simplify :

61

41

21

221

41

141

3

Sol. Given exp.

12

23

2

5

2

1

4

5

4

13

=

12

1

2

5

2

1

4

5

4

13

=

12

130

2

1

4

5

4

13

=

24

29

4

5

4

13

=

24

2930

4

13

=

24

1

4

13=

24

4

13= 78.

SQUARE AND SQUARE ROOTS

Squares : When a number is multiplied by itself thenthe product is called the square of that number.

Perfect Square : A natural number is called a perfectsquare if it is the square of any other natural numbere.g. 1, 4, 9,... are the squares of 1, 2, 3,... respectively.

Ex. 5 Find the least perfect square which is exactly divisibleby each of the numbers 6, 9, 15 and 20.

Sol.

1,1,1,1

5,5,1,15

5,5,3,13

5,15,9,33

10,15,9,32

20,15,9,62

LCM = 3 × 5 × 2 × 3 × 2 = 180.

The least multiple of 180 which is a perfect square is180 × 5 = 900.

Square roots : The square root of a number x is thatnumber which when multiplied by itself gives x as theproduct. As we say square of 3 is 9, then we can alsosay that square root of 9 is 3.The symbol used to indicate the square root of a

number is � � , i.e. 81 = 9, 225 = 15 ...etc.

We can calculate the square root of positive numbers

only. e.g. 25 = 5.

Properties of Square Roots :

(i) If the unit digit of a number is 2, 3, 7 or 8, then it doesnot have a square root in N.

(ii) If a number ends in an odd number of zeros, then itdoes not have a square root in N.

(iii) The square root of an even number is even andsquare root of an odd number is odd.

e.g. 81 = 9, 256 = 16, 324 = 18 ...etc.

(iv) Negative numbers have no square root in set ofreal numbers.

Ex.6 If 18225 = 135, then find the value of

( 25.182 + 8225.1 + 018225.0 + 00018225.0 ) .

Sol. Given exp.

( 25.182 + 8225.1 + 018225.0 + 00018225.0 )

= 210

18225 + 410

18225 + 610

18225 + 810

18225

= 10

18225 + 210

18225 + 310

18225 + 410

18225

= 10

135 +

100

135 +

1000

135 +

10000

135

= 13.5 + 1.35 + 0.135 + 0.0135 = 14.9985.

CUBE AND CUBE ROOTS

Cube : If any number is multiplied by itself three timesthen the result is called the cube of that number.

Perfect cube : A natural number is said to be a perfectcube if it is the cube of any other natural number.

Cube roots : The cube root of a number x is thatnumber whose cube gives x.

The cube root of x is denoted by the symbol 3 x . Thus,

3 8 = 2, 3 27 = 3, 3 64 = 4, 3 125 = 5 and so on.

Ex.7 By what least number 675 be multiplied to obtain anumber which is a perfect cube ?

Sol. 675 = 5 × 5 × 3 × 3 × 3.

To make it a perfect cube, it must be multiplied by 5.

Ex.8 Find the cube root of .000216.

Sol. (0.000216)1/3 = 3/1

610

216

=

3/1

222 101010

666

= 210

6 =

100

6 = 0.06

32PAGE # 32

FACTORS AND MULTIPLES

Factors : �a� is a factor of �b� if there exists a relation

such that a × n = b, where �n� is any natural number.

Number of factors : For any composite number C,

which can be expressed as C = ap × bq × cr ×....., where

a, b, c ..... are all prime factors and p, q, r are positive

integers, then the number of factors is equal to (p + 1)

× (q + 1) × (r + 1)....

For example : 36 = 22 × 32.

So the factors of 36 = (2 +1) × (2 + 1) = 3 × 3 = 9.

Ex.9 Find the total number of factors in the expression

(4)11 × (7)5 × (11)2.

Sol. (4)11 × (7)5 × (11)2 = (2 × 2)11 × (7)5 × (11)2

= 222 × 75 × 112.

Total number of factors = (22 + 1)(5 + 1)(2 + 1) = 414.

DIVISIBILITY

Division Algorithm : General representation of result is,

DivisormainderRe

QuotientDivisor

Dividend

Dividend = (Divisor × Quotient ) + Remainder

Ex.10 On dividing 4150 by certain number, the quotient is

55 and the remainder is 25. Find the divisor.

Sol. 4150 = 55 × x + 25

55x = 4125

x = 55

4125 = 75.

NOTE :

(i) (xn � an) is divisible by (x � a) for all the values of n.

(ii) (xn � an) is divisible by (x + a) and (x � a) for all the

even values of n.

(iii) (xn + an) is divisible by (x + a) for all the odd values of n.

Test of Divisibility :

No. Divisiblity Test

2 Unit digit s hould be 0 or even

3 The s um of digits of no. s hould be divis ible by 3

4 The no form ed by las t 2 digits of given no. s hould be divis ible by 4.

5 Unit digit s hould be 0 or 5.

6 No s hould be divis ible by 2 & 3 both

8 The num ber form ed by las t 3 digits of given no. s hould be divis ible by 8.

9 Sum of digits of given no. s hould be divis ible by 9

11The difference between s um s of the digits at even & at odd placesshould be zero or m ultiple of 11.

25 Las t 2 digits of the num ber s hould be 00, 25, 50 or 75.

Ex.11 Ajay multiplied 484 by a certain number to get the

result 3823a. Find the value of �a�.

Sol. 3823a is divisible by 484, and 484 is a factor of 3823a.

4 is a factor of 484 and 11 is also a factor of 484.

Hence, 3823a is divisible by both 4 and 11.

To be divisible by 4, the last two digits have to be

divisible by 4.

�a� can take two values 2 and 6.

38232 is not divisible by 11, but 38236 is divisible by

11.

Hence, 6 is the correct choice.

Ex.12 Find the smallest number of 6 digit which is exactlydivisible by 111.

Sol. Smallest number of 6 digits is 100000.On dividing 100000 by 111, we get 100 as remainder.Number to be added = (111, � 100) = 11.

Hence, required number = 100011.

REMAINDERS

The method of finding the remainder without actually

performing the process of division is termed as

remainder theorem.

Remainder should always be positive. For example if

we divide �22 by 7, generally we get �3 as quotient

and �1 as remainder. But this is wrong because

remainder is never be negative hence the quotient

should be �4 and remainder is + 6. We can also get

remainder 6 by adding �1 to divisor 7 (7 �1 = 6).

Ex.13 A number when divided by 296 gives a remainder 75.

When the same number is divided by 37, then find the

remainder.

Sol. Number = (296 × Q) + 75

= (37 × 8Q) + (37 × 2) + 1 = 37 × (8Q + 2) + 1.

Required remainder = 1.

Ex.14 A number being successively divided by 3, 5 and 8

leaves remainders 1, 4 and 7 respectively. Find the

respective remainders if the order of divisors be

reversed.

Sol.

714z81y5

x3

z = (8 × 1 + 7) = 15 ; y = (5z + 4) = (5 × 15 + 4) = 79 ;

x = (3y + 1) = (3 × 79 + 1) = 238.

Now,

214536295

2388

Respective remainders are 6, 4, 2.

33PAGE # 33

We are having 10 digits in our decimal number system

and some of them shows special characterstics like

they repeat their unit digit after a cycle, for example 1

repeat its unit digit after every consecutive power. So, its

cyclicity is 1, on the other hand digit 2 repeat its unit digit

after every four power, hence the cyclicity of 2 is four.

The cyclicity of digits are as follows :

Digit Cyclicity

0, 1, 5 and 6 1

4 and 9 2

2, 3, 7 and 8 4

So, if we want to find the last digit of 245, divide 45 by 4.

The remainder is 1 so the last digit of 245 would be

same as the last digit of 21 which is 2.

Ex.15 Find the unit digit in the product (771 × 659 × 365).

Sol. Unit digit in 74 is 1.

Unit digit in 768 is 1.

Unit digit in 771 is 3.

[1 × 7 × 7 × 7 given unit digit 3]

Again, every power of 6 will give unit digit 6.

Unit digit in 659 is 6.

Unit digit in 34 is 1.

Unit digit in 364 is 1 . Unit digit in 365 is 3.

Unit digit in (771 × 659 × 365)

Unit digit in (3 × 6 × 3) = 4.

Ex.16 Find unit�s digit in y = 717 + 734

Sol. 717 + 734 = 71 + 72 = 56, Hence the unit digit is 6.

HIGHEST POWER DIVIDING A FACTORIAL

Factorial n : Product of n consecutive natural numbers

is known as �factorial n� it is denoted by �n!�.

So, n! = n(n � 1)(n � 2)...321. e.g. 5! = 5 × 4 × 3 × 2 × 1 =

120.

The value of factorial zero is equal to the value of

factorial one. Hence 0! = 1 = 1!

The approach to finding the highest power of x dividing

y! is

32 x

y

x

yxy

......., where [ ] represents just

the integral part of the answer and ignoring the

fractional part.

Ex.17 What is the highest power of 2 that divides 20!

completely?

Sol. 20! = 1 × 2 × 3 × 4 ×....× 18 × 19 × 20 = 1 × (21) × 3 ×

(22) × 5 × (21 × 31) × 7 × (23) × ..... so on. In order to find

the highest power of 2 that divides the above product,

we need to find the sum of the powers of all 2 in this

expansion. All numbers that are divisible by 21 will

contribute 1 to the exponent of 2 in the product

12

20 = 10. Hence, 10 numbers contribute 21 to the

product. Similarly, all numbers that are divisible by

22 will contribute an extra 1 to the exponent of 2 in the

product, i.e 22

20 = 5. Hence, 5 numbers contribute an

extra 1 to exponents. Similarly, there are 2 numbers

that are divisible by 23 and 1 number that is divisible

by 24. Hence, the total 1s contributed to the exponent

of 2 in 20! is the sum of ( 10 + 5 +2 +1) = 18. Hence,

group of all 2s in 20! gives 218 x (N), where N is not

divisible by 2.

If 20! is divided by 2x then maximum value of x is 18.

Ex.18 What is the highest power of 5 that divides of

x = 100! = 100 × 99 × 98 × ...... × 3 × 2 × 1.

Sol. Calculating contributions of the different powers of 5,

we have 15

100= 20, 25

100 = 4.

Hence, the total contributions to the power of 5 is 24,

or the number 100! is divisible by 524.

Ex.19 What is the highest power of 6 that divides 9!

Sol. By the normal method. 69

= 1 and 26

9 = 0. Thus

answers we get is 1 which is wrong. True there is just

one multiple of 6 from 1 to 9 but the product 2 × 3 = 6

and also 4 × 9 = 36, can further be divided by 6. Thus,

when the divisor is a composite number find the

highest power of its prime factors and then proceed.

In this case, 9! can be divided by 27 and 34 and thus by

64 (In this case we need not have checked power of 2

as it would definitely be greater than that of 3).

, BASE SYSTEM

The number system that we work in is called the

�decimal system�. This is because there are 10 digits

in the system 0-9. There can be alternative system

that can be used for arithmetic operations. Some of

the most commonly used systems are : binary, octal

and hexadecimal.

These systems find applications in computing.

Binary system has 2 digits : 0, 1.

Octal system has 8 digits : 0, 1,..., 7.

Hexadecimal system has 16 digits : 0, 1, 2,..., 9, A , B,

C, D, E, F.

After 9, we use the letters to indicate digits. For instance,

A has a value 10, B has a value 11, C has a value 12,...

so on in all base systems.

The counting sequences in each of the systems would

be different though they follow the same principle.

34PAGE # 34

Conversion : Conversion of numbers from (i) decimal

system to other base system. (ii) other base system to

decimal system.

(i) Conversion from base 10 to any other base :

Ex.20 Convert (122)10

to base 8 system.

Sol.

10

718

2158

1228

The number in decimal is consecutively divided by the

number of the base to which we are converting the

decimal number. Then list down all the remainders in

the reverse sequence to get the number in that base.

So, here (122)10

= (172)8.

Ex.21 Convert (169)10

in base 7.

Sol. 03241697

77

133

Remainder

(169)10 =(331)7

(ii) Conversion from any other base to decimal

system :

Ex.22 Convert (231)8 into decimal system.

Sol. (231)8 , the value of the position of each of the numbers

( as in decimal system) is :

1 = 80 × 1

3 = 81 × 3

2 = 82 × 2

Hence, (231)8 = (80 × 1 + 81 × 3 + 82 × 2)

10

(231)8

= (1 + 24 + 128)10

(231)8

= (153)10

ALPHA NUMERICS NUMBERS

Ex.23 If a � b = 2, and

0cc

bb

aa

then find the value of a, b and c.

Sol. These problems involve basic number

(i) aa + bb = 11(a + b) (ii) aa, bb are two-digit numbers.

Hence, their sum cannot exceed 198. So, c must be 1.

(iii) Hence, cc0 = 110. This implies a + b = 10 or a = 6

and b = 4.

Such problems are part of a category of problems

called alpha numerics.

Ex.24 If

1 a 4x 3 b

8 c 87 2

5 d 8

s

t

then, find the value of b + a, where all

the digits are different.Sol. Let us consider 1 a 4 × 3 = s72.

a × 3 results in a number ending in 6.

As 16 and 26 is ruled out, a is 2.Thus, s = 3, t, 4Now 1 a 4 × b = 8c8 ; b = 2 or 7

Again 2 is ruled out because in that case, product wouldbe much less than 800. b = 7.Hence, a = 2, b = 7, c = 6, d = 8, s = 3 and t = 4.

1. In numbers from 1 to 100 digit �0� appears_______

times.

(A) 9 (B) 10

(C) 11 (D) 12

2. How many numbers are there containing

2 digits.

(A) 90 (B) 99

(C) 100 (D) 89

3. Which of the following statement is true?

(A) Every whole number is a natural number

(B) Every natural number is a whole number

(C) �1� is the least whole number

(D) None of these

4. What least number should be added to 1330 to get a

number exactly divisible by 43?

(A) 46 (B) 1

(C) 3 (D) 7

5. The last digit of the number (373)333 is :

(A) 1 (B) 2

(C) 3 (D) 9

6. The two missing numbers shown with asterisk in the

equation 21

**3

5 = 19 are :

(A) 6, 3 (B) 7, 3

(C) 8,3 (D) 11, 3

7. Given 5 = 2.236 the value of 45 + 605 � 245

correct to 3 decimal places is :

(A) 15.652 (B) 11.180

(C) 18.652 (D) 16.652

35PAGE # 35

8. A student was asked to multiply a number by 23

.

Instead he divided the number by 23

and obtained a

number smaller by 32

, the number is :

(A) 54

(B) 53

(C) 32

(D) 21

9. Which of the following statements is true ?

(A) 187

125

94

32

(B) 32

94

125

187

(C) 32

125

187

94

(D) 187

94

125

32

10. Which of the following rational numbers lie between

73

and 89

?

(A) 21

(B) 0

(C) 1512

(D) None of these

11. 018.0 can be expressed in the rational form as :

(A) 1000

18(B)

99018

(C) 990018

(D) 99918

12. The least number which must be subtracted from 2509to make it a perfect square is :(A) 6 (B) 9(C) 12 (D) 14

13. If x y = 22 yx , the value of )221( )221( is :

(A) �7 (B) 0(C) 2 (D) 9

14. The value of

41

2

13

11

54

is :

(A) 3140

(B) 94

(C) 81

(D) 4031

15. If A and B are real numbers and A2 + B2 = 0, then :(A) A > 0, B < 0 (B) A < 0, B > 0(C) A = 0 = B (D) A = � B

16. The sum of three consecutive odd numbers is alwaysdivisible by :I. 2 II. 3 III. 5 IV. 6(A) only I (B) only II(C) only I and III (D) only II and IV

17. 461 + 462 + 463 + 464 is divisible by :(A) 3 (B) 10(C) 11 (D) 13

18. Which of the following fractions is less than 8

7 and

greater than 3

1 ?

(A) 4

1(B)

24

23

(C) 12

11(D)

24

17

19. Simplify : 18 � [5 � {6 + 2(7 � 58 )}].

(A) 13 (B) 15(C) 27 (D) 32

20. 5 �

71

61

5.021

243

is equal to :

(A) 284

23(B) 3

6

1

(C) 310

3(D) 5

10

1

21. If 2805 2.55 = 1100, then 280.5 25.5 =(A) 1.1 (B) 1.01(C) 0.11 (D) 11

22. The least number by which 294 must be multiplied tomake it a perfect square, is :(A) 2 (B) 3(C) 6 (D) 24

23. The number of prime factors of (3 × 5)12 (2 × 7)10 (10)25

is :(A) 47 (B) 60(C) 72 (D) 94

24. On dividing a number by 999, the quotient is 366 andthe remainder is 103. The number is :(A) 364724 (B) 365387(C) 365737 (D) 366757

25. Evaluate : 38535

2)]23(5[8

.

(A) 2 (B) 3(C) 4 (D) 5

26. The value of 254488130214 :

(A) 14 (B) 15(C) 16 (D) 17

36PAGE # 36

27. 3)22( × 89

is :

(A) a rational number (B) an irrational number(C) undefined (D) none of these

28. The sum of the digit of a number 10n � 1 is 3798. The

value of n is :(A) 422 (B) 431(C) 501 (D) 673

29. The value of ])41

121

121

1(21

121

1[21

1 is :

(A) 21

(B) 41

(C) 161

(D) 51

1

30. If 21.1 = 1.1, then 000121. is equal to :

(A) 0.0011 (B) 0.011(C) 0.11 (D) 11.0

31. Between two positive integers, there are K integers,then K is :(A) Finite (B) Infinite(C) Finite under some conditions(D) Infinite under some conditions

32. The largest natural number by which the product ofthree consecutive even natural number is alwaysdivisible is :(A) 16 (B) 24(C) 48 (D) 96

33. If a number x is divided by 95, then remainder is 30. Ifthe same number x is divided by 5, then what isremainder ?(A) 2 (B) 3(C) 4 (D) 0

34. A gardener plants tree in rows and finds that each rowcontains twice as many tree as these are rows. If thenumber of trees be 5408. Then the number of tree ineach row is :(A) 100 (B) 104(C) 108 (D) 112

35. x and x + y are the square of two consecutive naturalnumber. What is the square of the next natural number ?(A) x + 2y (B) x + 2y + 2(C) x + 3y (D) x + y2

36. If (12 +22 +32 +42 +52)2 =P1

, then P is :

(A) 3125

1(B) 3125

(C) 3025

1(D) 3025

37. In the equation 243

× 132

+ 91211

= 165

+ x, x is equals

to :

(A) 1232

(B) 10127

(C) 1032

(D) 9121

38. Which of the following is an irrational number :(A) 3.65789 (B) 3.65789125634.....(C) 3.65786578..... (D) 3.666......

39. The number of such squares having area less then 8times their sides (numerically), will be :(A) 1 (B) 5(C) more than 6 (D) 0

40. If the fraction nm

is negative, which of the following

cannot be true ?

(A) mn

> nm

(B) mn < 0

(C) (n � m) < 0 (D) mn3 > 0

41. Physical Instructor wants to arrange boys in rows toform a perfect square. He finds that in doing so, 25boys are left out. If the total number of boys is 1250then find the number of boys in each row is :(A) 25 (B) 125(C) 45 (D) 35

42. Find out (A + B + C + D) such that AB x CB = DDD,where AB and CB are two-digit numbers and DDD is athree-digit number.(A) 21 (B) 19(C) 17 (D) 18

43. S is a six digit number beginning with 1. If the digit 1 ismoved from the leftmost place to the rightmost placethe number obtained is three times of S, What is thesum of the digits of S ?(A) 12 (B) 15(C) 18 (D) 27

44. If ABC x CBA = 65125, where A, B and C are singledigits, then A + B + C = ?(A) 18 (B) 15(C) 8 (D) 7

45. If 27 = 123 and 31 = 133,Than 15 = ?(A) 13 (B) 31(C) 11 (D) 33

46. What is the largest power of 12 that would divide 49! ?(A) 22 (B) 23(C) 24 (D) 20

47. The highest power of 3 which is a factor of the product

of all the integers from 1 to 200 is-

(A) 100 (B) 97

(C) 102 (D) None of these

48. How many zero�s are there in the end of the multipli-

cation !16!8!4 !16!8!4

(A) 8! + 16! (B) 8! + 2.16!(C) 8! + 3.16! (D) 8!.3.16!

49. The digit in the unit place in the expansion of 427 is :(A) 2 (B) 4(C) 6 (D) 8

50. The unit digit in the expression (36234) (33512) � (5429)(25123) will be :(A) 6 (B) 8(C) 0 (D) 5

37PAGE # 37

CELL DIVISION

STRUCTURE OF CHROMOSOME

Chromosomes are the vehicles of heredity whichpossess DNA and are enclosed inside the nucleus.They are capable of self reproduction and maintainingmorphological and physiological properties throughsuccessive generations.Each chromosome consistsof two strands which are called as chromatids.Thetwo chromatids of a chromosome are joined togetherat a point called as centromere.

(a) Size and Shape of Chromosomes :

(i) Size : Size of chromosomes is variable in differentorganisms, different tissues and at different stagesof the cell cycle.

(ii) Shape : It is usually determined by the position ofits centromere. On this basis chromosomes can beof following types :

(A) Metacentric : They are V � shaped. These havecentromere in the middle of the chromosomes sothat the two arms are almost equal.

(B) Submetacentric : They are L shaped. In themcentromeres are slightly away from the midpoints, sothat the two arms are unequal.

(C) Acrocentric : They are J-shaped with centromereat subterminal position.

(D) Telocentric : They are I-shaped, having terminalcentromere.

� NOTE : These shapes can be observed durig anaphsestage of cell division.

(b) Number of Chromosomes :

Each species has a fixed number of chromosomes

in it�s cells.In an ordinary human cell 23 pairs ofchromosomes are present.So, there are twochromosomes, of each kind. These twochromosomes of each kind are called as homologouschromosomes.

� A cell which has the complete set of chromosomeswith two of each kind is called as diploid cell. In otherwords a diploid cell has two sets having twochromosomes of each type.

� The gametes (or sex cells) of human beings aredifferent from their other body cells because theycontain only half the number of chromosomes. A cellwhich has half the number of chromosomes, is calledas haploid cell. In other words a haploid cell has onlyone set of each type of chromosomes.

� Human gametes called sperm and egg have only 23chromosomes which is half the number ofchromosomes of other body cells. So, a gamete is ahaploid cell.

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� Females consist of two similar gametes and thereforecalled as homogametic and males consist ofdissimilar gametes and therefore called asheterogametic. During spermatogenesis two typesof sperm cells will be produced one which containsX-chromosome and the other which contains Ychromosome.During oogenesis each egg cell containone X-chromosomes.

� If X � chromosome of male fuses with X-chromosomeof female it will produce a female child.

� If Y � chromosome of male fuses with X-chromosomeof female it will produce a male child.

(c) Properties of Chromosomes :

The chromosomes must possess five importantproperties :

(i) Replication : Synthesis of new DNA molecule whichis identical to the parent DNA molecule.

(ii) Transcription : Synthesis of RNA molecule whichis identical to the DNA molecule.

(iii) Change in appearance.

(iv) Repair : It means correction of damaged parts byDNA.

(v) Mutation : Development of genetic changes.

(d) Functions of Chromosomes :

(i) They carry hereditary characters from parents tooffsprings.

(ii) They help the cells to grow, divide and maintainitself by synthesis of proteins.

(iii) They undergo mutation and thus contribute to theevolution of animals.

(iv) They guide cell differentiation during development.

(v) They also help in metabolic process.

(vi) They bring about continuity of life.

CELL DIVISION

Cell division was first observed by Nageli in plant cell

(1842).

(a) Cell Cycle :

It is a series of programmed cyclic changes by whichthe cell duplicates its contents and divides into twoparts. It is divided into two phases :

Mitotic division phase

D.N.A.Synthesis

(3-4 hr)G phase2

(1 hr)

M phase

(12 hr)

G phase1S-phase

(6-8 hr)

Cell cycle

(i) Long non dividing (I � phase) or interphase.

(ii) Short dividing M � phase or mitotic phase

(i) Long non dividing (I � phase) or interphase : t is acomplex of changes that occurs in a newly formed cellbefore it is able to divide. It lasts throughout the life. Itinvolves replication of DNA and synthesis of nuclearproteins and duplication of centriole. Synthesis of energyrich components also takes place.This occurs in threestages i.e. G1 (First growth phase), S (Synthesis phase),G2 (Second growth phase)

(ii) Short dividing M � phase : t is the phase of celldivision. It consists of karyokinesis (nuclear division)and cytokinesis (cytoplasmic division).It is of two types :

(b) Mitosis :

� Term mitosis was given by Flemming.

� It is also called as somatic division as it occurs duringformation of body cells.

� It is studied in plants, in meristems and in animals inbone marrow, skin and base of nails.

� It is an equational division in which a parent cell dividesinto two identical daughter cells, each of them containsthe same number and kind of chromosomes as arepresent in parent cell.

� It occurs in two steps :(i) Karyokinesis (ii) Cytokinesis

(i) Karyokinesis : Division of nucleus. It is divided infour steps :

(A) Prophase : Chromatin fibres condense to formchromosomes.They shorten and become distinct witheach having two chromatids attached to centromere.Centrosomes reach the poles and form spindlefibres.Nucleolus disappears, nuclear membranedisappears.

(B) Metaphase : Chromosomes attach to spindlefibres arise from each pole and lie at the equator,forming a metaphase plate.Chromosomes areshortest and thickest in this stage.

(C) Anaphase : Shortest phase, In this phasecentromere of each chromosome divides to form twodaughter chromosomes.They remain attached topoles through spindle fibres and start moving towardspole and become shortened. They appear in differentshapes.

� V � Shaped (Metacentric)

� L � Shaped (Submetacentric)

� J � Shaped (Acrocentric)

� I � Shaped (Telocentric)

(D) Telophase : Nucleus is reconstituted,chromosomes uncoil, elongate and form chromatinfibre.Nucleolus & nuclear envelope reappears formingtwo daughter nuclei

39PAGE # 39

(ii) Cytokinesis : It is referred to the division of

cytoplasm.It begins towards the middle of anaphase

and completes with the completion of telophase.By

this the complete cytoplasm including matrix as well

as organelles divides equally.In animals it occurs by

formation of cleavage furrow in the middle by

constriction in plasma membrane.In plants it occurs

by cell plate formation.

40PAGE # 40

(c) Meiosis :

� It occurs only once in the life cycle of gametes.� It is a double division in which a diploid cell divides

twice to form four haploid cells.� It can be studied in anthers of unopened flowers in

plants and in testis of grasshopper in animals.Itconsists of two phases :

(i) Interphase : Size of nucleus increases to threetimes. It also involves G

1 � S � G

2 phase.

(ii) M � phase : It occurs in two steps

(A) Meiosis � I, (B) Meiosis � II

(A) Meiosis � I : Also called as reduction division.

Diploid state changes to haploid state.It occurs in four

steps :

� Prophase � I : It is the longest phase of meiosis. Ithas following stages :

� Leptotene : Chromatin fibres condense to formchromosomes. There are two chromosomes of eachtype which are diploid and are called as �homologouschromosomes�.

� Zygotene : Homologous chromosomes join bysynapsis and form bivalents which are actually tetradswith half the number of individual chromosomes,pairing proceeds in zipper like fashion formingsynaptonemal complex.

� Pachytene : There occurs exchange of segmentsbetween non sister chromatids of bivalents and iscalled as crossing over.

� Diplotene : Synaptonemal complex is dissolved,tetrads are formed. At some places nonsisterchromatids of two homologous chromosomes remainattached forming, chiasmata.

� Diakinesis : Chiasmata shifts towards ends,nucleolus degenerates.

� Metaphase � I : Spindles are formed and bivalentsform a double whorl or double metaphase plate.

� Anaphase � I : Chiasmata disappears, homologouschromosomes separate by disjunction formingdyads.They move towards poles and form two groupsof haploid chromosomes.

� Telophase � I : Chromosomes elongate,nucleoplasm & nuclear envelope reappears.

(B) Meiosis � II : It is also called as equational divisionand maintains the haploid number of chromosomes.No replication of DNA occurs in this stage.

� Prophase � II : Chromatin fibres shorten and formchromosomes.Nuclear envelope and nucleolus startdisappearing.

� Metaphase � II : Chromosomes form singlemetaphasic plate by arranging themselves onequator.

� Anaphase � II : Centromere divides into two andseparates two chromatids of chromosome into twoindependent daughter chromosomes or chromatids.

� Telophase � II : The four groups of chromosomesorganize themselves into 4 haploid nuclei.Chromatinfibres are formed, nucleolus and nuclear envelopeare reappeared.

Fig. Various Stages of Meiosis

� Significance of mitosis : It is essential for growth,repair, differentiation, maintenance of chromosomenumber etc.

� Significance of meiosis : It produces variations andessential for sexual reproduction. It maintains thechromosome number in each generation of livingorganisms.

41PAGE # 41

Differences between mitoitc and meiotic cell division

S.No. Mitosis Meiosis1. It occurs in all som atic cells . It occurs in reproductive cells (germ cells )

2. In the resultant daughter cells , the num ber of chrom osom es rem ains the sam e (i.e., diploid), hence, called equational divis ion.

In resultant daughter cells , the num ber of chrom osom es reduces to half (i.e., haploid), hence, called reductional divis ion.

3. By m itos is two daughter cells are produced By m eios is four daughter cells are produced.

4. During m itos is no cross ing over takes place During m eios is cross ing over takes place.

5. Daughter cells have identical chrom osom es which are also identical to that of parent cell (i.e. rem ains cons tant)

Chrom osom es of the daughter cells are with com bined com ponents (genes ) of both parents (i.e., genetic variability occurs )

(c) Amitosis :

� It is also known as Direct or Incipient cell division.� First described by Remak (1841).� It is a very simple cell division. It occurs without spindle

formation and appearance of chromosomes, also thenuclear membrane remains intact. Both cell and its

nucleus elongate, constrict in middle and break offinto nearly equal halves.

� It occurs in abnormal case. It occurs in prokaryotes(E.g. Bacteria, cyanobacteria etc.) and eukaryotes (E.g.Amoeba, Yeast, Foetal membrane cells, Endospermcells of seed, Diseased cell and Old tissues).

EXERCISE

1. Crossing over in diploid organism is responsible for :(A) dominance of genes(B) recombination of linked genes(C) linkage between genes(D) sagregation of genes

2. Which one of following structures will not be commonto meiotic cells of higher plants ?(A) Cell plate (B) Centriole(C) Centromere (D) Spindle fibres

3. How many mitotic divisions are needed for a singlecell to make 128 cells ?(A) 7 (B) 14 (C) 28 (D) 64

4. Nuclear membrane reappears in(A) anaphase (B) metaphase(C) telophase (D) none of the above

5. The stage of the meiosis in which nucleolus andnuclear membrane disappear and chromosomesbecome distinct is(A) prophase (B) metaphase(C) anaphase (D) telophase

6. In which of the following stages chromosomes arethin and long thread -like ?(A) Leptotene (B) Zygotene(C) Pachytene (D) Diplotene

7. In which phase of mitosis, the chromosomes arearranged around the equator of the spindle ?(A) Prophase (B) Metaphase(C) Anaphase (D) Telophase

8. Chromosomes are distinctly visible in(A) anaphase (B) metaphase(C) prophase (D) telophase

9. Series of cell division is(A) prophase, metaphase, anaphse, telophase(B) prophase, anaphase, metaphase, telophase(C) prophase, metaphase, telophase, anaphase(D) anaphase, metaphase, telophase, prophase

42PAGE # 42

10. The actual shape of chromosomes can be seen in(A) metaphase - I of meiosis(B) anaphase - I of meiosis(C) metaphase of mitosis(D) anaphase of mitosis

11. Duplication of chromosomes takes place in(A) S-phase (B) G

1 - phase

(C) G2-phase (D) M-phase

12. The chromosome number is reduced to half in(A) mitosis (B) meiosis(C) binary fission (D) parthenogenesis

13. Chiasmata represents the sites of

(A) synapsis (B) disjunction

(C) crossing over (D) terminalization

14. Chromosomes other than sex chromosomes are

called as

(A) allosomes (B) autosomes

(C) microsomes (D) none of the above

15. In humans the number of chromosomes in a diploid

cell is

(A) 23 (B) 46 (C) 44 (D) 30

43PAGE # 43

ANSWER KEY

MOTION (PHYSICS)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Ans. B C D B D B B C A A B A D B C C

MATTER (CHEMISTRY)

Ques 1 2 3 4 5 6 7 8 9 10 11Ans C D C A D A A B D A D

Ques 12 13 14 15 16 17 18 19 20 21 22Ans B A C B A C C B B B A

Ques 23 24 25 26 27 28 29 30Ans B C C A D C C A

NUMBER SYSTEM(MATHEMATICS)

Ques. 1 2 3 4 5 6 7 8 9 10

Ans. C A B C C B A A B A

Ques. 11 12 13 14 15 16 17 18 19 20

Ans. D B D C C B B D C A

Ques. 21 22 23 24 25 26 27 28 29 30

Ans. D C D C D B A A A BQues. 31 32 33 34 35 36 37 38 39 40Ans. A C D B B C A B C D

Ques. 41 42 43 44 45 46 47 48 49 50Ans. D A D C D A B C B A

CELL DIVISION (BIOLOGY)

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15A. B B A C A A A B A C A B C B B

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