INDEX CLASS - VIII (VISTAAR-EXPERT) S. NO. SUBJECT NAME
PAGE NO.
SPECIMEN COPY 1.
PHYSICS
1-6
2.
CHEMISTRY
7-27
3.
MATHEMATICS
28-36
4.
BIOLOGY
37-42
6.
MENTAL ABILITY
43-46
7.
ANSWER KEY
47
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MOTION EQUATIONS OF UNIFORMLY ACCELERATED MOTION
(a) 1 st Equation of Motion :
v 2 u2 s 2a 2as = v2 – u2
or v2 = u2 + 2as ......(iv)
Consider a body having initial velocity u. Suppose it is subjected to a uniform acceleration a so that
(d) Distance covered in n th second :
after time t its final velocity becomes v. Now we
Distance covered by a body in t second is
know, change in velocity Acceleration = Time v–u a= t or v = u + at
1 2 at . 2
S = ut +
Distance covered by a body along a straight line in n second is
.......(i)
(b) 2 nd Equation of Motion :
1 2 an .......(v) 2 Distance covered by a body along a straight line in Sn = un +
( n–1) second is
Suppose a body has an initial velocity u and uniform acceleration ‘a’ for time t so that its final velocity becomes v. The distance travelled by moving body in time t is s then the average velocity = (v+u) /2 . Distance travelled = Average velocity × time u v t s= 2
u u at t s= 2
Sn–1 = u (n –1) +
1 a (n–1)2 2
.......(vi)
The distance covered by the body in nth second will be- : snth = sn – sn –1 snth = un +
1 2 1 an – { u (n–1) + a (n–1)2 } 2 2
(as v=u+at)
2ut at 2 s= 2
2u at t s= 2
s = ut
1 2 at 2
Snth = un +
1 2 1 an – {nu –u + a (n2+1 –2n)} 2 2
Snth = un +
1 2 an 2 a – an} an – {un – u + 2 2 2
Snth = un +
1 2 an2 a – + an an – un + u – 2 2 2
Snth=
1 u + a n – 2
Snth=
2n – 1 u+a 2
.......(ii)
(c) 3 rd Equation of Motion : Distance travelled = Average velocity × time u v t s= 2
from equation (i)
.......(iii)
t=
v–u a
Substituting the value of t in equation (iii), we get
Snth=
u+
a (2n – 1 ) 2
.......(vii)
v – u v u a 2
s=
PAGE # 11
TO SOLVE NUMERICAL PROBLEMS
ILLUSTRATION
(i) If a body is dropped from a height then its initial
1. A car is moving at a speed of 50 km/h. Two seconds
velocity u = 0 but has acceleration (acting). If a
there after it is moving at 60 km/h. Calculate the
body starts from rest its initial velocity u = 0 .
acceleration of the car.
(ii) If a body comes to rest its final velocity v = 0 or, if a body reaches the highest point after being thrown
Here u = 50 km/h = 50
Sol.
5 250 m/s = m/s 18 18
upwards its final velocity v = 0 but has acceleration ( acting).
and v = 60 km/h = 60
5 300 = m/s 18 18
(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).
300 250 50 – v–u 18 18 18 Since a = = = 2 t 2
= 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
50 = 1.39 m/s2 36
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)
question otherwise take g = 9.8 m/s2. v = 54 km/h = 54 (i) If a body moves upwards (or thrown up ) g is taken negative (i.e. motion is against gravitation of earth).
a=
5 = 15 m/s 18
v–u 15 – 0 a= = 0.75 m/s2 t 20
So we can form the equation of motion like , v = u – gt, h = u t –
1 2 2 gt , v = u2 – 2 gh. 2
(Here –g replaces a)
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
(ii) If a body travels downwards (towards earth) then
against gravity)
g is taken + ve. So equations of motion becomes
v = 0 ( at highest point), s = ? v2 – u2 = 2as
v = u + gt , s = ut +
1 2 2 gt , v = u2 + 2gh. 2
(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.
(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. PAGE # 22
b) Centripetal Force :
KINEMATICS OF CIRCULAR MOTION (a) Angular Position : To decide the angular position of a point in space we need to specify (i) origin and (ii) reference line. The angle made by the position vector w.r.t. origin, with the reference line is called angular position. Clearly angular position depends on the choice of the origin as well as the reference line. Circular motion is a two dimensional motion or motion in a plane.
If there is no force acting on a body it will move in a straight line (with constant speed). Hence if a body is moving in a circular path or any curved path, there must be some force acting on the body. If speed of body is constant, the net force acting on the body is along the inside normal to the path of the body and it is called centripetal force. Centripetal force (Fc) = mac =
Y
mv 2 = m 2 r r
SIMPLE HARMONIC MOTION P' P O
(a) Periodic Motion : X
r
Suppose a particle P is moving in a circle of radius r and centre O. The angular position of the particle P at a given instant may be described by the angle between OP and OX. This angle is called the angular position of the particle. (b) Angular Velocity :
Displacement
and
Angular
In a circular motion, the angular displacement of a body is the angle subtended by the body at the centre in a given interval of time. t is represented by the symbol (theta). The angular displacement per unit time is called the angular velocity. t is represented by the symbol (omega). Let a body moves along a circle of radius r and perform a uniform circular motion. Let the body be at point P to start 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 angular displacement = 2c Q Hence = But
2 T
C
1 = N (frequency) T
r
When a body or a moving particle repeats its motion along a definite path after regular intervals of time, its motion is said to be Periodic Motion and interval of time 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 :
x
If the restoring force/torque acting on the body in
P
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
There = 2N
to analysis) form of oscillatory motion.
For an arc of length, Linear distance = Angular displacement, =
r
TYPES OF SHM
Hence, = r For a time interval t,
(a) Linear SHM :
Linear velocity, v = t
point, along a straight line, then its motion is known as linear SHM. A and B are extreme positions and M is
Angular velocity, = Hence, v = r.
When a particle moves to and fro about an equilibrium
v = rt = t r
mean position. AM = MB = Amplitude
M A
B PAGE # 33
(b) Angular SHM :
(d) Frequency (f) :
When body/particle is free to rotate about a given axis executing angular oscillations, then its motion is known
Number of oscillations completed in unit time interval is called frequency of oscillations.
as angular SHM.
1 ù , its units is Hz or s–1 . T 2ð
Equation of Simple Harmonic Motion :
f=
The necessary and sufficient condition for SHM is F = – kx
(e) Time period (T) :
Where k = Force constant or spring constant
The smallest time interval after which the oscillation
x = displacement from mean position. or ma = – kx
repeats itself is called the time period.
k a=– x m
T=
Here negative sign show that F will always be towards mean position or F and x are in opposite direction. a = –2x
( Where =
k ) m
4.
2ð m 2ð ù k
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
It is the equation of SHM.
CHARACTERISTICS OF SHM
T=
2ð = . ù
(a) Displacement :
RELATIVE MOTION 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
Motion is a combined property of the object under study as well as the observer. It is always relative, there is no such thing like absolute motion or absolute rest. Motion is always defined with respect to an observer or reference 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
x AB x A xB B
Amplitude
A
Amplitude
xB
xAB xA
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 =
1 [distance between two extreme 2
positions] It depends on the energy of the system. (c) Angular Frequency () : =
2ð 2 ðf and its units is rad/s. T
Origin
B
A
Relative Velocity : Definition : Relative velocity of a particle A with respect to B is defined as the velocity with which A appears to move if B is considered to be at rest. In other words, it is the velocity with which A appears to move as seen by B 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 B w.r.t. ground and xAB is position of A w.r.t. B then we can say vA = velocity of A w.r.t. ground vB = velocity of B w.r.t. ground and vAB = velocity of A w.r.t. B Thus
v AB v A vB PAGE # 44
5.
An object A is moving with 5m/s and B is moving with 20m/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
8.
A girl swims in a swimming pool of length 100 m. She swims from one end to another end and reaches the starting point again in 2 minutes. Then the average speed of the swimmer is : (A) 100 ms–1 (B) 0.83 ms–1 –1 (C) 1.67 ms (D) zero
9.
The acceleration of car that comes to stop from a 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
Sol. (i) vB = +20m/s vA = +5m/s vBA = vB – vA = +15m/s (ii) vB = +20m/s, vA = +15m/s vAB = vA – vB = – 15m/s Note :
vBA= –vAB
EXERCISE 1.
2.
3.
4.
5.
6.
7.
A body covers half the distance with a speed of 20 m/s and the other half with a speed of 30 m/s. The average speed of the body during the whole journey is : (A) Zero (B) 24 m/s (C) 25 m/s (D) None of these A body is thrown vertically upwards and rises to a height of 10 m. The velocity with which the body 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 A body strikes the floor vertically with a speed u and rebounds at the same speed. The change in speed would be : (A) u (B) 3u (C) 2u (D) Zero
with a velocity of 5m/s. If the acceleration of the stone during its motion be 10m/s 2 in downward direction, what will be the height attained by the stone ?
11.
When the distance of an object travels is directly proportional to the length of time, it is said to travel with (A) zero velocity (B) constant speed (C) constant acceleration (D) uniform velocity 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
(B) 1.50 m
(C) 2 m
(D) 3.5 m
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.
12.
(A) 8 m/s
(B) 16 m/s
(C) 24 m/s
(D) 32 m/s
A balloon starts rising from the ground with an acceleration of 1.25 m/s2. After 8s a stone is
If a trolley starts from rest with an acceleration of 2 m/s2, the velocity of the body after 4s would be : (A) 2 m/s (B) 8 m/s (C) 6 m/s (D) 4 m/s The speed of a body describing its motion is : (A) direction (B) state (C) type (D) rapidity
(A) 1.25 m
released from the balloon. How much time stone will take to reach the ground ? (g = 10 m/s2) (A) 4 s
(B) 2s
(C) 2 2 s
(B) 4 2 s
13.
A ball is dropped from a height of 5m onto a sandy floor and penetrates the sand upto 10cm before coming to rest find the retardation in sand, assuming it to be uniform. (A) 9.8 m/s2 (B) 10 m/s2 2 (C) 100 m/s (D) 500 m/s2
14.
A body moving with a constant retardation in straight line travels 5.7m and 3.9 m in 6 th and 9th second respectively. When will the body come momentarily to rest ? (A) 10 s (B) 15 s (C) 20 s (D) 25 s
PAGE # 55
15.
Two simple pendulums of length and 4 are suspended from same point and brought aside together and released at the same time. If the time period of smaller pendulum is T there after how much time will they be together again and moving in same direction.
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
(C) 1 :
6
(B) 6 : 1 (D) 1 : 6
4
(A) T/2
(B) T
(C) 2T
(D) None of these
PAGE # 66
MATTER INTRODUCTION There are a large number of things around us which we see and feel. For example, we can see a book in front of us. A book occupies some space. The space occupied by the book is called its volume. If we pick up the book, we can also feel its weight. So, we conclude that the book has some mass. We cannot see the air around us, yet if we fill a balloon with air and then weigh it carefully, we will find that not only does air occupy space (bounded by the balloon), but it also has mass.
Experiment to show that matter is made of particles Evidence - 2 Movement of pollen grains in water : The best evidence for the existence and movement of particles in liquids was given by Robert Brown in 1827. Robert Brown suspended extremely small pollen grains in water. On looking through the microscope, it was found that the pollen grains were moving rapidly throughout water in a very irregular way (or zig-zag way). Conclusion : Water is made up of tiny particles which are moving very fast (The water molecules themselves are invisible under the microscope because they are very, very small). The pollen grains move on the surface of water because they are constantly being hit by the fast moving particles of water. So, though the water particles (or water molecules) are too small to be seen, but their effect on the pollen grains can be seen clearly. The random motion of visible particles (pollen grains) caused by the much smaller invisible particles of water is an example of Brownian motion (after the name of the scientist Robert Brown who first observed this phenomenon.)
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 offers resistance is borne out by the fact that we cannot displace an object from one place to another without applying some force. We have to apply force to pick up a stone from the ground. Thus , matter can be defined as follows Anything that occupies space, has mass and offers resistance is called matter.
PHYSICAL NATURE OF MATTER (a) Matter is Made up of Particles : (i) Everything around us is made up of many tiny pieces or particles. (ii) Particles which make up the matter are constantly moving. (iii) Particles which make up matter are atoms or molecules. (i) Evidences for the presence of particles in matter : Most of the evidences for the existence of particles in matter and their motion come from the experiments on diffusion and Brownian motion. Evidence - 1 Dissolving a solid in a liquid : Take a beaker. Fill half of it with water. Mark the level of water in the beaker. Add some sugar to the water and dissolve it with the help of a glass rod. You will see that the sugar has disappeared, but there is no change in the level of water. Conclusion : This can be explained by assuming that matter is not continuous, rather it is made up of particles. Sugar contains a large number of separate particles. These particles when dissolved in water occupy the vacant spaces between the particles of water. That is why, the water level in the beaker did not rise. Had sugar been continuous, like a block of wood, the water level in the beaker would have risen.
Brownian motion : Zig-zag motion (in a very irregular way) 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 which are moving rapidly in a very random way. This is an example of Brownian motion in gases. The tiny dust particles move here and there because they are constantly hit by the fast moving particles of air. The existence of Brownian motion gives two conclusions. • Matter is made up of tiny particles. • Particles of matter are constantly moving.
Note : Brownian motion increases on increasing the temperature. PAGE # 7
77 7
(b) Characteristics of Particles of Matter : The important characteristics of particles of matter are the following :
(i) The particles of matter are very, very small
Note : The particles of matter possess kinetic energy and so are constantly moving. As the temperature rises, particles move faster. (iv) Particles of matter attract each other : There are some forces of attraction between the particles of matter which bind them together.
(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) Cohesive Force : The force of attraction between the particles of same substances is called cohesive force.
(A) Diffusion :“Intermixing of particles of two different types of matter on their own is called diffusion.”t is the phenomenon in which the movement of molecules or particles occur from their higher concentration towards their lower concentration.
(B) Adhesive Force : The force of attraction between the particles of different substances is called adhesive force.
e.g. : When a perfume bottle is opened in one corner of a room, its fragrance spreads in the whole room quickly. This happens because the particles of perfume move rapidly in all directions and mix with the moving particles of air in the room.
e.g. : If we take a piece of chalk, a cube of ice and an iron nail and beat them with a hammer, chalk will easily break into smaller pieces, but more force will be required to break a cube of ice and iron nail will not break.
(A) Experiment : We take a gas jar full of bromine vapours and invert another gas jar containinig air over it, then after some time, the red-brown vapours of bromine sperad out into the upper gas jar containing air.
Reason : The reason for this is, that the force of attraction is quite weak in between the chalk particles, but force of attraction in between the particles of ice cube is a bit stronger, while force of attraction in between the particles of iron is very-very strong.
(B) Conclusion : In this way, the upper gas jar which contains colourless air in it, also turns red-brown. The mixing is due to the diffusion of bromine vapours (or bromine gas) into air.
RIGID AND FLUID (i) Rigid : Rigid means ‘unbending’ or inflexible. A solid is a rigid form of matter so that it maintains its shape when subjected to outside force. (ii) Fluids : Fluids are the substances which have tendency to flow. A liquid is a fluid form of matter which occupies the space of the container. Liquids have a well defined surface. A gas is a fluid form of matter which fills the whole container in which it is kept.
Note : Liquids and gases are known as fluids.
CLASSIFICATION OF MATTER On the basis of physical states, all matter can be classified into three groups:(a) Solids (b) Liquids (c) Gases
Diffusion of bromine vapour (or bromine gas) into air
COMPARISON OF THE CHARACTERISTICS OF THREE STATES OF MATTER Property
Solid state
Liquid state
Gaseous state
Interparticle spaces
Very small spaces
Comparatively 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 volume
Definite 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
Low
Comparatively high than solids
Very high
Diffusion
Negligible
Slow
Very fast
PAGE # 8
88 8
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.
stand
Note : Gaseous particles move randomly at high speed and hit each other and also walls of the container, so exert pressure.
Change of state from ice to water (A) Melting or Fusion: The process due to which a solid changes into liquid state by absorbing heat energy is called melting or fusion.
INTERCONVERSION OF STATES OF MATTER The phenomenon of change of matter from one state
(B) Freezing or Solidification: The process due to which a liquid changes into solid state by giving out heat energy is called freezing or solidification.
to another state and back to original state, by altering the conditions of temperature and pressure, is called interconversion of states of matter.
(C) Melting Point: The constant temperature at which a solid changes into liquid state by absorbing heat energy at 1 atm pressure is called its melting point.
The various states of matter can be interchanged into one another by altering the conditions of (a) Temperature
(b) Pressure.
(D) Freezing Point: The constant temperature at which a liquid changes into solid state by giving out heat energy at 1 atm pressure is called freezing point.
(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.
Note : The numerical value of freezing point and melting point 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. Due to increase in K.E., the particles start vibrating with greater speed. The energy supplied by heat overcomes the force of attraction between the particles. Then, the particles leave their fixed positions and start moving freely and thus solid melts.
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.
Latent Heat of Fusion : The amount of heat energy that is required to change 1 kg of solid into liquid at atmospheric pressure and its melting point is known as the latent heat of fusion. (In Greek Latent means Hidden) Latent heat of fusion of ice = 3.34 × 105 J/kg.
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.
Note : Particles of water at 00C (273 K) have more energy as compared to particles in ice at the same temperature. (ii) Interconversion of liquid into gaseous state and vice versa: Liquids can be converted into gases by heating them. Similarly, gases can be converted into liquids by cooling them. e.g. : Water at 1 atm pressure changes into gas (steam) at 1000C by absorbing heat. Steam at 1000C changes into water by giving out energy.
PAGE # 9
99 9
Explanation : When heat is supplied to water, particles start moving faster. At a certain temperature, a point is reached when the particles have enough energy to break the forces of attraction between the particles. At this temperature the liquid starts changing into gas.
Activity To study the change of state from water to steam. Materials required A 100 cc beaker, a thermometer (Celsius), a glass stirrer, 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 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 water. Record the temperature of water. Heat the beaker on a low Bunsen flame and continuously stir the water with glass stirrer. Go on recording the temperature till water starts boiling. Allow the water to boil for few minutes and record its temperature.
Note : Particles in steam, that is water vapour at 373 K have more energy than water at the same temperature. Because steam has absorbed extra energy in the form of latent heat of vaporisation.
CURVE [TEMPERATURE-TIME GRAPH] We can show the change of temperature with time in the form of a temperature-time graph drawn by using the readings obtained in the above experiment. Such a time-temperature graph is shown in figure.
Temperature (in °C)
You will notice that temperature of water rises till it starts boiling. The temperature of boiling water is 1000C (373 K). If we continue heating the water it changes into steam, but the temperature remains constant, i.e., 1000C (373 K).
Latent heat of vaporisation: The amount of heat which is required to convert 1 kg of the liquid (at its boiling point) to vapour or gas without any change in temperature. Latent heat of vaporisation of water = 22.5 × 105 J/kg.
C (liquid
D gas)
wa (liq ter uid )
100
(solid liquid)
0
A
B
Time of heating (in minutes) Temperature Time Graph
(A) Boiling or Vaporisation: The process due to which a liquid changes into gaseous state by absorbing heat energy is called boiling. (B) Condensation or Liquefaction: The process due to which a gas changes into liquid state by giving out heat energy is called condensation. (C) Boiling Point: The constant temperature at which a liquid rapidly changes into gaseous state by absorbing heat energy at atmospheric pressure is called boiling point.
Note : The numerical value of condensation point and boiling point is same. Condensation point of vapour (water) = Boiling point of water = 100ºC (373.16 K).
• The solid which undergoes sublimation to form vapour is called ‘sublime’. • The solid obtained by cooling the vapours of a solid is called ‘sublimate’. e.g. : Ammonium Chloride (NH4Cl), iodine, camphor, naphthalene (moth balls) and anthracene. Liquid
Gas
g tin ng el zi M ree F
(D) Condensation Point:- The constant temperature at which a gas changes into liquid state by giving out heat energy at atmospheric pressure is called condensation point.
(iii) Direct interconversion of solid into gaseous state and vice versa: The changing of solid directly into vapours on heating and of vapours directly into solid on cooling is known as sublimation.
V C ap on o de risa ns tio at n io n
Change of state from water to steam
In this graph at point A, we have all ice. As we heat it, the ice starts melting to form water but the temperature of ice and water mixture does not rise. It remains constant at 0°C during the melting of ice. At point B, all the ice has melted to form water. Thus, we have only water at point B. Now, on heating beyond point B, the temperature of water (formed from ice) starts rising as shown by the sloping line BC in the graph.
Sublimation Sublimation
Solid
Interconversion of states of matter
10 10 10
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Specific Heat The specific heat of a substance is the amount of heat which is required to raise the temperature of a unit 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 of heat in joules required to raise the temperature of 1 kilogram of the substance by 1ºC. The specific heat of a substance is usually represented by the symbol C (Sometimes, however, the specific heat of a substance is also represented by the letter's). The specific heat of a substance varies slightly with temperature. The change in the specific heat of a substance with temperature is due to the changes which occur in the structure and organization of the molecules in a substance with change in temperature. Units of Specific Heat The 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 to the different intermolecular spaces between their particles. So when a gas is compressed the intermolecular space between its particles decreases and ultimately it will be converted into liquid. Pressure and temperature determine the state of a substance. So, high pressure and low temperature can liquefy gases. e.g. : Carbon dioxide (CO2) is a gas under normal conditions of temperature and pressure. It can be liquefied by compressing it to a pressure 70 times more than atmospheric pressure. Solid CO2 is known as ‘Dry ice’. Solid CO2 is extremely cold and used to ‘deep freeze’ food and to keep icecream cold.
In case of liquids, a small fraction of particles at the surface, having higher K.E., is able to break the forces of attraction of other particles and gets converted into vapour.
Note : The atmospheric pressure at sea level is 1 atm. (a) Factors Affecting Evaporation: (i) Temperature: With the increase in temperature the rate of evaporation increases. Rate of evaporation T Reason : On increasing temperature more number of particles get enough K.E. to go into the vapour state. (ii) Surface Area : Rate of evaporation Surface area Since evaporation is a surface phenomena, if the surface area is increased, the rate of evaporation increases. So, while putting clothes for drying up we spread them out. (iii) Humidity of Air : Rate of evaporation
1 Humidity
Humidity is the amount of water vapour present in air. When humidity of air is low, the rate of evaporation is high and water evaporates more readily. When humidity of air is high, the rate of evaporation is low and water evaporates very slowly. (iv) Wind Speed : Rate of evaporation Wind speed With the increase in wind speed, the particles of water vapour move away with the wind. So the amount of water vapour decreases in the surroundings. (v) Nature of substance : Substances with high boiling points will evaporate slowly, while substances with low boiling points will evaporate quickly. Differences between evaporation and boiling Evaporation It is a surface phenomenon. It occurs at all temperatures below B.P.
Boiling It is a bulk phenomenon. It occurs at B.P. only.
The rate of evaporation The rate of boiling does depends upon the surface not depend upon the area of the liquid, humidity surface area, wind speed, temperature & wind speed and humidity.
(b) Cooling Caused by Evaporation: Unit of pressure : Atmosphere (atm) is a unit for measuring pressure exerted 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 is lowered. This helps in rapid change of liquid into gas.
The cooling caused by evaporation is based on the fact 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 of our hand then our hand feels very cold. • Perspiration (or sweating) is our body’s method of maintaining a constant temperature.
EVAPORATION The phenomenon of change of a liquid into vapours at any temperature below its boiling point is called evaporation. Water changes into vapours below 100 0C. The particles of matter are always moving and are never at rest. At a given temperature in any gas, liquid or solid, there are particles with different K.E.
(c) We Wear Cotton Clothes in Summer : During summer, we perspire more because of the mechanism of our body which keeps us cool. During evaporation, the particles at the surface of liquid gain energy 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 in absorbing the sweat.
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(d) Water droplets on the outer surface of a glass containing ice cold water :
Substance : A substance is a kind of matter that cannot
If we take some ice cold water in a glass then we will observe water droplets on the outer surface of glass.
process. For example, sugar dissolved in water can
be separated into other kinds of matter by any physical be separated from water by simply evaporating the
Reason : The water vapour present in air on coming in contact with glass of cold water, loses energy. So water vapour gets converted to liquid state, which we see as water droplets.
water but it cannot be broken into its components by any physical process so here sugar is a substance.
PURE SUBSTANCE
PLASMA This state consist of super energetic and super excited particles. These particles are in the form of ionised gases. For eg: Neon sign bulb and fluorescent tube Neon sign bulb – Neon gas Fluorescent tube – Helium gas When electrical energy flows through gas, it gets ionised and hence plasma is created. Plasma glows with a special colour depending on nature of gas. Sun and the stars glow because of the presence of plasma.
A homogeneous material which contains particles of only one kind and has a definite set of properties is called a pure substance. Examples : Iron, silver, oxygen, sulphur, carbon dioxide etc., are pure substances because each of them has only 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 of other substances.
BOSE-EINSTEIN CONDENSATE (B.E.C.) The B.E.C. is formed by cooling a gas of extremely low density, about one-hundred-thousandth the density of normal air, to super low temperature.
(iii) The composition of a pure substance cannot be altered by any physical means.
Matter
Mixtures More than one type of particles are present
Pure substances Only one type of particles are present ( no impurities)
Elements
Compounds
Homogeneous mixtures (true solutions)
(b) Elements : A pure substance, which cannot be subdivided into two or more simpler substances by any physical or chemical means is called an element. (i) Examples : Hydrogen, oxygen, nitrogen, copper, zinc, tin, lead, mercury, etc. are all elements as they cannot be subdivided into simpler parts by any physical or chemical means. A substance made up of the atoms with same atomic number is called an element.
• 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 and astatine. 1 non-metal, bromine, is a liquid. Five nonmetals, hydrogen, nitrogen, oxygen, fluorine and chlorine are chemically active gases. Six non-metals, helium, neon, argon, krypton, xenon and radon are chemically inactive gases. These are also called noble gases, inert gases or rare gases.
(ii) Classification of elements :
METALLOIDS : There are a few elements which show some properties of metals and other properties of nonmetals. For example they look like metals but they are brittle like non-metals. They are neither conductors of electricity like metals nor insulators like non-metals, they are semiconductors. The elements which show some properties of metals and some other properties of non-metals are called metalloids. Their properties are intermediate between the properties of metals and non-metals. Metalloids are also sometimes called semi-metals. The important examples of metalloids are : Boron (B), Silicon (Si), Germanium (Ge), Arsenic (As), Antimony (Sb), Tellurium (Te) Polonium (Po) and Astatine (At).
(A) On the basis of physical states, all elements can be 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, xenon and radon. • Remaining 98 elements are solids at room temperature. (B) Elements can be classified as metals and non-metals. There are 22 non-metals and 89 metals.
Note : Hydrogen is the lightest element.
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(C) Elements can be classified as normal elements and radioactive elements. The elements which do not give out harmful radiations are called normal elements. Elements from atomic number 1 to atomic number 82 are normal elements. The elements which give out harmful radiations are called radioactive elements. Elements from atomic number 83 to atomic number 112 and 114, 116 and 118 are radioactive in nature.
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(c) Compounds : A pure substance, which is composed of two or more different elements, combined chemically in a definite ratio, such that it can be broken into elements only by chemical means is called compound. The two or more elements present in a compound are called constituents or components of the compound. For example, water is a compound of hydrogen and oxygen, combined together in the ratio of 1 : 8 by weight. The water can be broken into its constituents only by electro-chemical method, i.e., by passing electric current through it. TYPES OF COMPOUNDS (A) On the basis of constitutents elements : (i) Inorganic compounds These compounds have been mostly obtained from non-living sources such as rocks and minerals. A few examples of inorganic compounds are : common salt, marble, washing soda, baking soda, carbon dioxide, ammonia, sulphuric acid etc. (ii) Organic compounds The word ‘organ’ relates to different organs of living beings. Therefore, organic compounds are the compounds which are obtained from living beings i.e., plants and animals. It has been found that all the organic compounds contain carbon as their essential constituent. Therefore, the organic compounds are quite often known as ‘carbon compounds’. A few common examples of organic compounds are : methane, ethane, propane (all constituents of cooking gas), alcohol, acetic acid, sugar, proteins, oils, fats etc. (B) On the basis of their properties : (i) Acids : Compounds which give hydronium ion in aqueous solution for e.g. hydrochloric acid, sulphuric acid, nitric acid, formic acid etc.
(ii) Bases : Compounds which give hydroxide ion in aqueous solution for e.g.Sodium hydroxide, Potassium hydroxide (iii) Salts : It is formed by the chemical reaction between acids and bases for e.g. ammonium chloride, zinc sulphate etc.
MIXTURES Most of the materials around us are not pure substances, but contain more than one substances, elements or compounds. Such materials are called mixtures. (a) Definition : W hen two or more substances (elements, compounds or both) are mixed together in any proportion, such that they do not undergo any chemical change, but retain their individual characteristics, the resulting product is called a mixture. (b) Types of Mixture : (i) Homogeneous Mixture : A mixture in which different constituents are mixed uniformly is called a homogeneous mixture. Examples : All solutions, such as solutions of common salt, copper sulphate, sugar etc. are examples of homogeneous mixtures. Similarly, alloys such as brass, bronze etc. are homogeneous solid solutions of metals and air is homogenous mixture of gases. (ii) Heterogeneous Mixture : A mixture in which different constituents are not mixed uniformly is called a heterogeneous mixture. Examples : A mixture of sand and salt, iron powder and sulphur powder, soil etc. are examples of heterogeneous mixtures.
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TRUE SOLUTIONS A homogeneous mixture of two or more substances is called a solution. Usually we think of a solution as a liquid that contains either a solid or a liquid or a gas dissolved in it. However, this is not true. We can also have a solid solution and gaseous solution as in the case of alloys and air respectively. (a) Components of a Solution : The substances present in a homogeneous solution are called components of the solution. A solution basically has two components, i.e., a solvent and a solute. (i) Solvent : The component of a solution which is present in large proportion , is called solvent.
Note : Usually, a solvent is the LARGER component of the solution. For example : In the solution of copper sulphate in water, water is the solvent. Similarly, in paints, turpentine oil is the solvent. (ii) Solute : The component of the solution which is present in small proportion is called solute. For example: In the solution of common salt in water, the common salt is solute. Similarly, in carbonated drinks (soda water), carbon dioxide gas is the solute.
Note : Usually, solute is the SMALLER component of the solution. (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 almost molecular size and their diameter is of the order of 1 nm (10–9 m) or less. (iii) A true solution can completely pass through a filter paper as particle size of solute is far smaller than 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 not settle down, provided temperature is constant. (vi) From a true solution, the solute can easily be recovered by evaporation or crystallisation. (c) Types of Solution: (A) On the basis of concentration : (i) Saturated solution : A solution, which at a given temperature dissolves as much solute as it is capable of dissolving, is said to be a saturated solution. (ii) Unsaturated solution : When the amount of solute contained in a solution is less than the saturation level, the solution is said to be an unsaturated solution. (iii) Super saturated solution : A solution, which contains more of the solute than required to make a saturated 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 and solvent : (i) Solid-Solid solutions : All alloys are solid solutions of metals. Brass is a solid solution of approximately 30% of zinc and 70% of copper. In this solid solution, copper (larger component) is solvent and zinc (smaller component) is solute. Similarly, Bell Metal is a solid solution of 80% of copper and 20% of tin, in which copper is the solvent and tin is the solute. (ii) Solid-Liquid solutions : Sugar solution is an example, in which sugar is the solute and water is the solvent. Similarly, common salt solution is an example, in which common salt is the solute and water 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 alcoholic drink, ethyl alcohol is solute and water is solvent. Similarly, in case of vinegar, acetic acid is solute and water is solvent. (iv) Liquid-Gas solutions : In case of aerated drinks (soda water), carbon dioxide is the solute and water is the solvent. (v) Gas-Gas solutions : Air is a homogeneous mixture of two main gases, i.e., 78% of nitrogen and 21% of oxygen. In this mixture, nitrogen is solvent and oxygen is solute. Similarly, the petrol fed into the engines of automobiles is a mixture of petrol vapour and air. (d) Concentration of a Solution : It is defined as the amount of solute present in a given quantity of the solution. The most common method for expressing the concentration of a solution is called percentage method. The concentration of solution refers to the percentage of solute present in the solution. Furthermore, the percentage of solute can be expressed in terms of :
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(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
(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.
mass percentage of solute and is defined as under :
Calculate the amount of glucose required to prepare 250 g of 5% solution of glucose by mass.
The concentration of solution is the mass of the solute in grams, which is present in 100 g of a solution.
Mass of solute Sol. % of solute =
Note : 5 =
It is very important to keep in mind that the percentage concentration of a solution refers to mass of solute in
percentage of solute is calculated by the formula given below : Concentration of solution =
Mass of solute (in grams) Mass of solution (in grams)
250
× 100 5 250 125 = = 12.5 g 100 10
3.
A solution contains 50 mL of alcohol mixed with 150 mL of water. Calculate concentration of this solution. Sol. This solution contains a liquid solute (alcohol) mixed with a liquid solvent (water), so we have to calculate the concentration of this solution in terms of volume percentage of solute (alcohol). Now, we know that : Volume of solute
Concentration of solution =
100
Volume of solution
× 100
(ii) Concentration of a solution in terms of volume
Here, Volume of solute (alcohol) = 50 mL And. Volume of solvent (water) = 150 mL So, Volume of solution = Volume of solute + Volume of solvent = 50 + 150 = 200 mL Now, putting these values of ‘volume of solute’ and ‘volume of solution’ in the above formula we get :
percentage of solute : If a solution is formed by
Concentration of solution =
Mass of solute (in grams) [Mass of solute Mass of solvent ](in grams)
100
dissolving a liquid solute in a liquid solvent, then the 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.,
50 50 × 100 = 200 2
= 25 percent (by volume) Thus, the concentration of this alcohol solution is 25 percent.
concentration of the solution is expressed in terms of
× 100
Mass of solute
Mass of solute =
100 g of solution and not 100 g of solvent, i.e., water. The concentration of a solution in terms of mass
Mass of solution
4.
How much water should be added to 16 ml acetone to make its concentration 48% ? Vol. of solute Sol. Concentration of solution = × 100 Vol. of solution 16 16 × 100 = 48 x = × 100 = 33.33 ml 48 x
Volume of solvent =33.33 – 16 = 17.33 ml.
water. The concentration of a solution in terms of volume percentage of the solute is calculated by the formula given below : Concentration of solution = Volume of solute (in ml) Volume of solution (in ml)
100
SUSPENSIONS A heterogeneous mixture of insoluble particles of solute, spread throughout a solvent, is called a suspension. The particle size (diameter) in a suspension is more than 10–5 cm. The particles have a tendency to settle down at the bottom of the vessel and can be filtered out, because their size is bigger than the size of the pores of the filter paper. (a) Examples :
=
Volume of solute (in ml) [Volume of solute Volume of solvent] (in ml)
100
Note : The concentration of a solution is a pure percentage number and has NO UNITS.
(i) Muddy water, in which particles of sand and clay are suspended in water. (ii) Slaked lime suspension used for white-washing has particles of slaked lime suspended in water. (iii) Paints in which the particles of dyes are suspended in turpentine oil.
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(b) Characteristics of Suspensions : (i) The size of particles is more than 10 –5 cm in diameter. (ii) The particles of suspension can be separated from solvent by the process of filtration. (iii) The particles of suspension settle down, when the 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 under the action of gravity is called sedimentation.
Tyndall effect shown by colloid in a beaker This luminosity of path of beam is known as Tyndall effect and the illuminated path is known as Tyndall cone. (Tyndall being the name of the scientist who studied this phenomenon first). Tyndall effect is caused due to the scattering of light by the colloidal particles. The true solutions do not scatter light and hence do not show Tyndall effect.
A heterogeneous solution in which the particle size is in between 10–7 cm to 10–5 cm, such that the solute particles neither dissolve nor settle down in a solvent is called colloidal solution. In a colloidal solution, relatively large suspended particles are called dispersed phase and the solvent in which the colloidal particles are suspended is called 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 visible under a powerful microscope. (iii) The particles of a colloidal solution do not settle down with the passage of time. (iv) The particles of a colloidal solution can easily pass through filter paper.
Tyndall effect can be seen when a fine beam of light enter in a room through a small hole. This happens due to scattering of light particles of dust and smoke in the air of the room. Tyndall effect can be observed when sunlight passes through a dense forest. In the forest, fog contains tiny droplets of water which act as particles of colloid dispersed in air. (viii) The particles of a colloidal solution are electrically charged. Electrophoresis The collodial solutions contain either positively or negatively charged particles and, therefore, when an electric current is passed through them, the particles move towards either of the oppositely charged electrodes. Subsequently, they get discharged on the electrodes and precipitate out. For example, when a negatively charged As2S3 solution is taken in a Utube into which Platinum electrodes, connected to a source of E.M.F. are dipped, the colloidal particles move towards the positive electrode . The migration of colloidal particles under the influence of an electric field is known as electrophoresis.
(v) The colloidal solutions are heterogeneous in nature. (vi) Colloidal solutions are not transparent, but translucent in nature. (vii) The particles of a colloidal solution scatter light, i.e., when strong beam of light is passed through the colloidal solution, the path of beam becomes visible. Scattering of Light (Tyndall Effect) If a beam of light is passed through pure water or a salt solution, the path of light is visible but when a strong beam of light is passed through a colloidal solution and viewed at right angles with the help of a microscope, the path of light shows up a bright cone of bluish light.
Electrophoresis showing migration of colloidal particles (c) Classification of Colloids : The colloids are classified according to the state of dispersed phase (solid, liquid or gas) and the state of dispersing medium. A few common examples are shown in the table :
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Note : Colloidal solutions can be separated by the process of CENTRIFUGATION.
S.No.
Property
1 2
Nature Particle size
3
Filtrability
4
Visibility
5 6 7
True solution
Colloidal solution
Heterogeneous Diameter more than 100 nm (or 1000Å) or 10 -5 cm. Do not pass through filter paper or animal or vegetable membranes Particles visible to the naked eye or under a microscope
Diffusion
Homogeneous Heterogeneous Diameter less than 1 nm Diameter between 1-100 nm (or 10-1000Å) or (or 10Å) or 10 -7 cm 10-7 to 10-5 cm Passes through an Passes through ordinary filter paper ordinary filter paper as well as animal or but not through animal vegetable membranes or vegetable membranes Particles are completely Particles themselves are invisible invisible but their presence can be detected by ultramicroscope since they scatter light. Diffuse rapidly Diffuse slowly
Tyndall effect
Not shown
Shown
May be shown
Appearance of solution
Clear and transparent
Generally clear and transparent
Opaque
SEPARATION OF HETEROGENEOUS MIXTURES Heterogeneous mixtures can be separated into their respective components by simple physical methods such as handpicking, sieving, filtration. Generally following physical properties are considered in the separation of the constituents of a mixture. (i) Densities of the constituents of the mixture. (ii) Melting points and boiling points of the constituents of the mixture. (iii) Property of volatility of one or more constituents of the mixture. (iv) Solubility of the constituents of the mixture in different solvents.
Suspension
Do not diffuse
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
(v) Ability of the constituents of the mixture to sublime.
the components sublimes on heating. However, it is
(vi) Ability of the constituents of the mixture to diffuse.
useful only if the components of the mixture do not
Note : However, for separating homogeneous mixtures special techniques are employed depending upon the difference in one or more physical properties of the constituents of the mixture.
react chemically on heating. The table shows the list of mixtures which can be separated by the process of sublimation.
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Solid-Solid Mixture
Sublimable Solid
Common salt and ammonium chloride
Ammonium chloride
Sand and iodine
Iodine
Common salt and iodine
Iodine
Sodium sulphate and benzoic acid
Benzoic 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 of sugar and sand mixture The sugar solution containing sand is filtered by pouring over a filter paper kept in a funnel. Sand remains as a residue on the filter paper and sugar solution is obtained as a filtrate in the beaker kept below the funnel. The sugar solution is evaporated carefully to get the crystals of sugar. In this way, a mixture of sugar and sand has been separated by using water as the solvent.
Separation by sublimation Note : Dry ice (solid CO2), Naphthalene, Anthracene, Iodine etc. are sublimable solids. (b) By Using a Suitable Solvent In some cases, one constituent of a mixture is soluble in a particular liquid solvent whereas the other constituent is insoluble in it. This difference in the solubilities of the constituents of a mixture can be used to separate them. For example, sugar is soluble in water whereas sand is insoluble in it, so a mixture of sugar and sand can be separated by using water as solvent. This will become more clear from the following discussion. To Separate a Mixture of Sugar and Sand Sugar is soluble in water whereas sand is insoluble in water. This difference in the solubilities of sugar and sand in water is used to separate them. This is done as follows. The mixture of sugar and sand is taken in a beaker and water is added to it. The mixture is stirred to dissolve the sugar . The sand remains undissolved.
(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).
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(ii) Method :
•
•
Heat sand in an iron vessel by placing it over a tripod stand. This arrangement is called sand bath.
•
Place a china dish on the sand bath. Pour about 5 cc of the ink into the china dish.
•
Heat gently evaporates water from the ink such that it does not boil. In a few minutes the water evaporates leaving behind dry blue ink. Method of evaporation is suitable for the following solid-liquid mixtures.
The centrifugal force (in the outward direction) pushes the heavier particles outward, i.e., towards the bottom of the mixture. Thus, the heavier particles of the proteins, carbohydrates, etc. are pushed towards the bottom of the tube, but the lighter particles of the fat stay near the top of the tube and hence separate. (iii) Applications of centrifugation :
•
It is employed in milk dairies to separate cream from the milk.
•
It is employed in diagnostic laboratories in testing urine samples.
•
It is employed in blood banks to separate different constituents of blood.
•
It is used in drying machines to squeeze out water from the wet clothes. (c) By Chromatography : The process of separation of different dissolved constituents of a mixture by adsorbing them over an appropriate adsorbent material is called chromatography. The adsorbent medium is generally magnesium oxide, alumina or filter paper. The solvent generally used for dissolving a mixture of two or more constituents is water or alcohol. The different constituents of a mixture get adsorbed differently on the same adsorbent material, because they have different rates of movement. The rate of movement of each adsorbed material depends upon :
(b) By Centrifugation The method of separating finely suspended or colloidal particles in a liquid, by whirling the liquid at a very high speed is called centrifugation. (i) Principle of centrifugation : It is based on the principle that when a very fine suspension or a colloidal solution is whirled rapidly, then the heavier particles are forced towards the bottom of liquid and the lighter stay at the top. (ii) Separation of cream from milk : The process of centrifuging is employed in separating cream from milk. This process is generally employed in separating colloidal solutions which easily pass through the filter paper.
•
The relative solubility of the constituents of mixture in a given solvent.
•
The relative affinity of the constituents of mixture for the adsorbent medium. If a filter paper is used as an adsorbent material for the separation of various constituents of a mixture, then this method of separation of mixture is called paper chromatography. Paper chromatography is very useful in separating various constituents of coloured solutes present in a mixture of lime, ink, dyes etc.
Note : Kroma means colour in Greek language and technique of chromatography was first applied for the separation of colours, so this name was given.
spin
(i) Separation of coloured constituents present in a mixture of ink and water.
CENTRIFUGE Method : •
Pour full cream milk in the test tube with a pivot in your laboratory centrifuge.
•
Shut the lid of the centrifuge and switch on the current. W hen the centrifuge starts working, the tube containing milk swings out in the horizontal position and whirls around its axis at a high speed.
(ii) Method : •
Take a filter paper 22 cm long, 5 cm broad and stick its smaller end to a glass rod with the help of gum. On the other end, measure a distance of 2 cm from lower end and mark a small point. On this point pour one or two drop of the ink.
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•
Suspend this filter paper in a wide and tall cylinder as shown in Figure. Gradually, pour water into the cylinder till the lower end of filter paper slightly dips in the water. Cover the cylinder with a glass lid to prevent any evaporation and leave the apparatus undisturbed for an hour. The water rises up the filter paper and reaches the ink mark. This water then dissolves various constituents of the ink, gets adsorbed by the filter paper in different amounts. More the constituent gets adsorbed, the lesser it moves upward and vice versa.
•
When the solvent (water) reaches near the top of filter paper, the filter paper is removed from water and dried. On the filter paper will be seen a band of colours, of various constituents.
•
A filter paper with separated bands of various constituents of a coloured substance is called chromatogram.
(i) Liebig condenser : Liebig condenser is a water condenser. It is a long glass tube surrounded by a wider glass tube (called water jacket) having an inlet and outlet for water. During distillation, cold water from tap is circulated through the outer tube of condenser. This water takes away heat from the hot vapour passing through the inner tube of condenser and causes its condensation. Process of simple distillation is used to recover both salt as well as water , from a salt-water mixture (or salt solution) and to separate of components of a mixture containing two miscible liquids that boil without decomposition and have sufficient difference in their boiling points.
SIMPLE DISTILLATION (ii) Fractional distillation : Separation of mixture of two miscible liquids for which the difference in the boiling points is less : In case of two liquids which have very close boiling points, both the liquids tend to distil over in different proportions. It means lesser the boiling point of a liquid, more is the proportion of it distilling over. The above problem can be avoided by using a fractionating column. It gives the effect of repeated distillation by offering resistance to the passage of vapour. The process of separation of two miscible liquids by the process of distillation, making use of their difference in boiling points, is called fractional distillation.
(iii) Advantages : •
It can be carried out with a very small amount of material.
•
The substances under investigation do not get wasted in 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 form vapour and then cooling the vapour to get the back liquid. Distillation can be represented as :
Liquid
Heating Cooling
Vapour (or Gas)
Note : The liquid obtained by condensing the vapour in the process of distillation is called DISTILLATE .
Note : The process of fractional distillation is useful only, if the difference in the boiling points of the two miscible liquids is less than 25ºC.
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PAGE # 21
(ii) Method :
(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).
•
Close the tap of separating funnel and clamp it in a vertical position in an iron stand.
•
Pour the immiscible liquid mixture (say benzene-water mixture) in the separating funnel. Allow the mixture to stand for half an hour or more.
•
The immiscible components of the mixture, i.e., benzene and water separate out into two distinct layers. The benzene forms the lighter layer on the top and the water forms the heavier layer at the bottom.
•
Place a conical flask or a beaker under the nozzle of the separating funnel. Turn the tap gently so that the water 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 under the nozzle of separating funnel. Open the tap to drain out benzene.
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 in receiver.
•
The thermometer shows a constant reading as long as the vapour of one liquid are passing to Liebig’s condenser. As soon as the temperature starts rising, the receiver is replaced by another receiver to collect second liquid.
Separation by separating funnel (iii) Applications : •
This method is used for separating any two immiscible liquids.
•
This method is used in separation of slag (a waste material) from the molten metals during their extraction. For example, during the extraction of iron from its ore, the molten iron and slag collect at the base of blast furnace. The slag being less dense floats up the surface of molten iron. They are drained out from two different outlets.
SEPARATION OF GASES FROM AIR (b) By Separating Funnel : (i) Separation of a mixture of two immiscible liquids : The separation of two immiscible liquids is based on the difference in their densities. The apparatus used for separation is separating funnel. It is a long glass tube provided with a tap at its bottom. The table below shows different immiscible liquids which can be separated by separating funnel. Lighter Liquid
Immiscible Liquid-liquid Mixture
Heavier Liquid
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
Water Mustard oil
In order to separate the major components of air, it is first purified, then liquefied and finally fractionally distilled. The steps involved in the process are as follows (a) Purification of Air : (i) Air generally contains carbon dioxide gas, hydrogen sulphide 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 through water, where the dust particles are removed. (iii) The washed air is passed through dilute caustic soda solution, where the gases like carbon dioxide, sulphur dioxide and hydrogen sulphide are removed. (iv) The purified air, however, contains moisture. The moist air is passed through pipes, maintained at a temperature 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 all impurities.
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PAGE # 22
(iv) When compressed air suddenly escapes from the jet, its pressure suddenly falls. Thus, its molecules move wide apart. When the molecules move wide apart, they need energy. This energy is taken by the molecules from themselves and hence, their temperature drops.
(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.
(v) The air so cooled, is now at a pressure equal to that of atmosphere. This cooled air rises up and in the process further cools the incoming compressed air in spiral tube. The air is then sucked again by the compression pump and the cycle is repeated. With every cycle, the temperature of air drops, till it liquefies.
(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.
(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.
Example : Ice melts to form water. In this example only the appearance (state) of matter has changed from solid to liquid. However, the composition of the molecules of ice or water remains same, i.e., for every 1 g of hydrogen there is 8 g of oxygen required . Thus, only a physical change has occurred.
(iii) The liquid is gradually warmed to – 195º C, when nitrogen starts boiling off from the liquid air. The nitrogen gas so formed, is compressed and filled in steel cylinders. (iv) The liquefied oxygen left behind, is also changed to gas and then filled in compressed state in steel cylinders.
(ii) The change is temporary and reversible : It means the change can be reversed by altering the causes which produce the change. Example : The water formed from ice can be changed back to ice by placing it in a freezing mixture (a mixture of ice and common salt).
PHYSICAL AND CHEMICAL CHANGES Some kind of change always takes place in the matter when it is subjected to energy changes. Almost all the changes (except nuclear changes) taking place in the matter can be classified under two headings, these are as follows (a) Physical Changes : Definition : A change which alters some specific physical property of the matter, like its state, texture, magnetic or electrical conditions or its colour, without causing any change in the composition of its molecules, is called physical change, provided it gets reversed, if the cause producing the change is removed. Following points need special consideration : (i) No new or different product is formed : The composition of molecules of the substance remains unaltered.
Note : On altering the experimental conditions, the change which gets reversed, is a physical change. (iii) There is no net gain or loss of energy : The amount of energy required to bring about a physical change is generally equal to the amount of energy required to reverse the change. Thus, there is no net energy 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 energy change is zero. (iv) There is no change in the weight of substance : During a physical change it is only the energy which is added or removed. No matter is added during a physical change. Similarly, no matter is removed during a physical change. Therefore, mass of the substance remains same.
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SOME EXAMPLES INVOLVING PHYSICAL CHANGES : Physical Change
Observation
Change in Physical Property
1. Switching on an electric bulb
The bulb glows and gives out heat and light energy.
The physical appearance of the bulb changes.
2. Rubbing a permanent magnet on a steel rod.
The steel rod gets magnetised. If it is brought near iron nails, they get attracted.
The steel rod acquires the property of attracting pieces of iron.
3. Action of heat on iodine
The brownish grey crystals of iodine change to form violet vapours. On cooling the vapours condense on cooler parts of the test tube to form crystals.
Change in state and colour.
4. Dissolving of common salt in water.
The white crystalline salt Change of state. disappears in water. However, the water tastes exactly like common salt. Moreover, common salt can be recovered by evaporation.
Some Common Examples of Physical Changes :
viz. carbon and water (steam), are formed. In this change, the arrangement between the molecules of
• Formation of dew.
carbon, hydrogen and oxygen breaks. The hydrogen
• Evaporation of water.
and oxygen atoms separate from carbon atoms and join together to form water. The carbon atoms are set
• Crystallisation of sugar from its solution. • Ringing of an electric bell.
free and are left as black residue.
Sugar heat Carbon Steam
• Breaking of a glass pane. (ii) The weight of the substance undergoing • Freezing of ice cream.
chemical change usually changes :
• A rock rolling down a hill.
Example : During the heating of sugar, the weight of
• Bending of a glass tube by heating.
the black residue is far less than the actual weight of the sugar. However, this is an apparent change in
• Melting of wax.
weight. If we take the weight of steam into account
• Sublimation of camphor.
and add to it the weight of carbon, then total weight
(b) Chemical Change :
will be equal to the weight of sugar crystals. Thus, strictly speaking, total weight of substances taking
Definition : A change which alters the specific
part in a chemical change remains constant.
properties of a material by bringing about a change in its molecular composition, followed by a change in
(iii) The chemical change is permanent and
state, is called a chemical change.
irreversible : It means the change will not reverse by
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.
altering the experimental conditions. Example : The sugar, which has decomposed on heating to form carbon and steam will not change to sugar on cooling.
Thus, the composition of the molecules of products
(iv) During chemical change energy is either
is different from the original substance.
absorbed or given out : The various atoms in a
Example : Heating of sugar
chemical compound are joined by attractive forces
When sugar is gently heated in a test tube, it melts. It
commonly called bonds. The making or breaking of
gradually changes to brown colour, giving a large
the bonds always requires exchange of energy. Thus,
amount of steamy fumes. In the end a black mass is
some amount of heat is either absorbed or given out
left which consists of carbon. Thus, new substances,
during a chemical change.
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PAGE # 24
SOME EXAMPLES INVOLVING CHEMICAL CHANGES :
Chemical Change
Equation
Observation
1. Burning of magnesium in air
When a magnesium ribbon is heated in a flame of Bunsen burner, it catches fire and burns with a dazzling white flame to form white ash.
2. Rusting of iron
When iron (silver grey) is left exposed to moist air for a few days, reddish brown powdery mass (rust) is found on its surface
3. Burning of LPG
When LPG (Liquefied Petroleum Gas) is burnt, it burns with a pale blue flame and liberates colourless gas carbon dioxide along with steam.
Magnesium + Oxygen Magnesium oxide
Iron + Oxygen+ Water vapour Rust
Butane (LPG) + Oxygen Carbon dioxide + Water
SOME COMMON EXAMPLES OF CHEMICAL CHANGES :• Burning of wood or charcoal • Burning of candle
• Digestion of food
• Formation of biogas (Gobar gas)
• Burning of petrol or diesel
• Smoking of cigarette
• Drying of paint
• Curdling of milk
• Rusting of iron
• Ripening of fruit
•Clotting of blood
• Fading of the colour of a dyed cloth
• Baking of cake
• Photosynthesis
• Formation of wine
• Electrolysis of water into hydrogen and oxygen
• Butter turning rancid
• Formation of water from hydrogen and oxygen
(C) DIFFERENCE BETWEEN PHYSICAL AND CHEMICAL CHANGES
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PAGE # 25
9.
EXERCISE 1.
The quantity of matter present in an object is called its -
2.
(A) weight
(B) volume
(C) mass
(D) density
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?
7.
(A) Fractional distillation
(B) Sublimation
(C) Chromatography
(D) Simple distillation
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.
Some matter and their groups and Q respectively. (P) (a) Air (b) O2 (c) Copper sulphate (d) Sodium hydroxide The 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)
are given in column P (Q) (i) Element (ii) Mixture (iii) Base (iv) Salt
10. The water boils when : (A) Saturated vapour pressure of water becomes equal to the atmospheric pressure (B) Boiling point of water becomes more than atmospheric pressure (C) Saturated vapour pressure of water is less than atmospheric pressure (D) Vapour pressure of water becomes more than atmospheric 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 liquefied petroleum gas (LPG). In the cylinder, it exists as a liquid. When it comes out of the cylinder, it becomes a gas (process A), then it burns (process B). Choose the 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, while all 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 evaporation for obtaining pure crystal of sugar because in evaporation 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 technique in separation of two liquids is (A) difference in Melting point (B) difference in Boiling point (C) difference in Concentration (D) difference in Solubility
26 26 26
PAGE # 26
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. The properties 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 ice at 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 of sand, 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 its characteristic (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 about suspension ? (A) The particles of suspension can be separated from solvent by the process of filtration. (B) When the suspension is kept undisturbed then the 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 liquid and 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 are maximum. (B) Intermolecular forces of attraction in gases are minimum. (C) Intermolecular spaces in solids are minimum. (D) All of the above 28. A liquid disturbed by stirring comes to rest after sometime due to its property of (A) Compressibility (B) Diffusion (C) Viscosity (D) All of these 29. W hich of the following statements regarding melting point and freezing point of a substance is true ? (A) Melting point of a substance is more than its freezing point. (B) Melting point of a substance is less than its freezing point. (C) Melting point and freezing point of a substance are same. (D) None of these. 30. Which of the following conditions is most favourable for 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
27 27 27
PAGE # 27
SYMBOLS & IT’S MEANING
there exists
:
such that
<
is less than
>
is greater than
infinity
belongs to
angle
||
parallel to
is congruent to
therefore
perpendicular to
~
difference
=
is equal to
is not equal to
less than or equal to
greater than or equal to
intersection
union
contains (is a super set of)
is contained in (is a sub set of)
is equivalent to
is implied by
implies
PAGE # 28
NUMBER SYSTEM
INTRODUCTION Number System is a method of writing numerals to represent numbers.
Ten symbols 0,1, 2, 3, 4, 5, 6, 7, 8, 9 are used to represent any number (however large it may be) in our number system.
ODD NUMBERS
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.
Each of the symbols 0,1, 2, 3, 4, 5, 6, 7, 8, 9 is called a digit or a figure.
INTEGERS
The set of integers is the set of natural numbers, zero and negative of natural numbers simultaneously. The set 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 natural numbers.
The set of natural numbers is denoted by N i.e., N = {1, 2, 3, 4, 5, .........}.
WHOLE NUMBER
All natural numbers together with zero are called whole numbers, 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 the set of natural numbers.
0 is the smallest whole number, there is no largest whole number i.e. the number of the elements in the set 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
Whole numbers which are exactly divisible by 2 are called even numbers.
The set of even numbers is denoted by 'E', such that E = {0, 2, 4, 6, 8, .....}.
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 less than approximately square root of number. If given number is not divisible by any of these prime number then 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 not divisible by any of these prime number. So, 131 is a prime number.
COMPOSITE NUMBERS Natural numbers having more than two factors are called composite numbers.
4, 6, 8, 9, 10, 12, 14, 15, 16, 18... are composite numbers.
Number 1 is neither prime nor composite number.
All even numbers except 2 are composite numbers.
Every natural number except 1 is either prime or composite number.
There are infinite prime numbers and infinite composite numbers.
CO-PRIME NUMBER OR RELATIVELY PRIME NUMBERS
Two natural numbers are said to be co-prime numbers or relatively prime numbers if they have only 1 as common factor. For ex. 8, 9 ; 15, 16 ; 26, 33 etc. are co-prime numbers.
Co-prime numbers may not themselves be prime numbers. As 8 and 9 are co-prime numbers, but neither 8 nor 9 is a prime number.
Every two consecutive natural numbers are co - primes.
EVEN NUMBERS
Natural numbers which are not exactly divisible by 2 are called odd numbers. O = {1, 3, 5, 7, 9.....}
PAGE # 29
Conversion of decimal numbers into rational numbers of the form plq.
TWIN PRIMES
Short cut method for pure recurring decimal : Write the repeated digit or digits only once in the numerator and take as many nines in the denominator as there are repeating digits in the given number.
Pairs of prime numbers which have only one composite number between them are called twin primes. For example : 3, 5 ; 5, 7 ; 11,13 ; 17, 19 ; 29, 31 ; 41, 43; 59, 61 and 71, 73 etc. are twin primes.
For example : (i) 0. 3 = 3/9 or 1/3 (ii) 0. 387 = 387/999
RATIONAL NUMBERS These are real numbers which can be expressed in the
MIXED RECURRING DECIMAL
form of p/q, where p and q are integers and q 0 . For example :
A decimal is said to be a mixed recurring decimal if there is at least one digit after the decimal point, which is not repeated.
2 37 17 , ,– . 3 15 19
All natural numbers, whole numbers and integers are rational.
Rational numbers include all Integers (without any decimal part to it), terminating fractions (fractions in which the decimal parts terminating e.g. 0.75, – 0.02 etc.) and also non-terminating but recurring decimals. e.g. 0.666....., – 2.333...., etc.
Short cut method for mixed recurring decimal : Form a fraction in which numerator is the difference between the number formed by all the digits (take the digits once which are repeating after decimal) and that formed by the digits which are not repeated and the denominator is the number formed by as many nines as there are repeated digits followed by as many zeros as the number of non-repeated digits. Ex.2 Change 0.7435 in the form of p/q.
FRACTIONS (a) Common fraction : Fractions whose denominator is not 10. (b) Decimal fraction : Fractions whose denominator is 10 or any power of 10. (c) Proper fraction : Numerator < Denominator i.e.
3 5
Sol. 0.7435 =
Ex.3 Change 12 .3 45 in the form of p/q. Sol.
=
(d) Improper fraction : Numerator > Denominator i.e.
part i.e. 3 (f)
12222 . 990
COMPARISON OF FRACTIONS
2 . 7
Suppose some fractions are to be arranged in ascending or descending order of magnitude. Then, convert each one of the given fractions in the decimal form, and arrange them accordingly.
Compound fraction : Fraction whose numerator and denominator themselves are fractions. i.e.
12345 – 123 990
.
5 . 3 (e) Mixed fraction : Consists of integral as well as fractional
7435 74 7361 = . 9900 9900
2/3 . 5/7
Now,
Improper fraction can be written in the form of mixed fraction.
3 6 7 = 0.6, = 0.857, = 0.777..... 5 7 9
Since 0.857 > 0.777....> 0.6, so
6 7 3 > > . 7 9 5
IRRATIONAL NUMBERS REAL NUMBERS All real numbers which are not rational are irrational numbers. These are non-recurring as well as non-terminating type of decimal numbers. For example :
2,
3
4 , 2 3 , 2 3 ,
47
A decimal is said to be a pure recurring decimal if the a digit or set of digits after the decimals are repeated. 1 = 0.333...... = 0.3 , 3
22 = 3.142857142857..... 7
= 3. 142857 .
The set of rational numbers and irrational numbers taken together is known as a set of real numbers.
3 etc.
PURE RECURRING DECIMAL
Thus,
The absolute value of a real number I x I is defined as
x, if x 0 |x|= x, if x 0 For example : | 2 |= 2 ; 2 > 0 and | – 2 | = – (– 2 ) = 2; – 2 < 0.
PAGE # 30
BODMAS RULE This rule depicts the correct sequence in which the operations are to be executed so as to find out the value of a given expression. Here,’B’ stands for ‘Bracket’, ‘O’ for ‘of’, ‘D’ for Division, ‘M’ for Multiplication’, ‘A’ for ‘Addition’ and ‘S’ for subtraction’. Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order ( ), { } and [ ]. After removing the brackets, we must use the following operations strictly in the order : (i) Of (ii) Division (iii) Multiplication (iv) Addition (v) Subtraction. Vinculum (or Bar) : When an expression contains Vinculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Vinculum.
1 1 1 1 1 1 Ex.4 Simplify : 3 4 1 4 2 2 2 4 6
13 5 1 5 3 2 Sol. Given exp. 4 4 2 2 12
=
13 5 1 5 1 4 4 2 2 12
Square roots : The square root of a number x is that number which when multiplied by itself gives x as the product. As we say square of 3 is 9, then we can also say that square root of 9 is 3. The symbol used to indicate the square root of a number is ‘
only. e.g.
(ii) If a number ends in an odd number of zeros, then it does not have a square root in N. (iii) The square root of an even number is even and square root of an odd number is odd. e.g. 81 = 9,
( 182 .25 + 1.8225 +
0.018225 +
0.00018225 ) .
Sol. Given exp. ( 182 .25 + 1.8225 +
=
=
=
=
13 1 13 4 24 = 4 24 = 78.
2 3, 9, 15, 10
324 = 18 ...etc.
Ex.6 If 18225 = 135, then find the value of
13 30 29 4 24
2 6, 9, 15, 20
256 = 16,
(iv) Negative numbers have no square root in set of real numbers.
=
Ex. 5 Find the least perfect square which is exactly divisible by each of the numbers 6, 9, 15 and 20. Sol.
25 = 5.
(i) If the unit digit of a number is 2, 3, 7 or 8, then it does not have a square root in N.
13 5 29 4 4 24
Perfect Square : A natural number is called a perfect square if it is the square of any other natural number e.g. 1, 4, 9,... are the squares of 1, 2, 3,... respectively.
225 = 15 ...etc.
Properties of Square Roots :
13 5 1 30 1 4 4 2 12
Squares : When a number is multiplied by itself then the product is called the square of that number.
81 = 9,
We can calculate the square root of positive numbers
=
SQUARE AND SQUARE ROOTS
’ , i.e.
=
18225 102
+
18225 + 10
18225 104
18225 102
0.018225 +
18225
+
+
106
18225
+
+
103
0.00018225 )
18225 108
18225 104
135 135 135 135 + + + 10 100 1000 10000
= 13.5 + 1.35 + 0.135 + 0.0135 = 14.9985.
CUBE AND CUBE ROOTS Cube : If any number is multiplied by itself three times then the result is called the cube of that number. Perfect cube : A natural number is said to be a perfect cube if it is the cube of any other natural number. Cube roots : The cube root of a number x is that number whose cube gives x. The cube root of x is denoted by the symbol 3
8 = 2,
3
27 = 3,
3
64 = 4,
3
3
x . Thus,
125 = 5 and so on.
3 3, 9, 15,
5
3 1, 3,
5,
5
5 1,
1,
5,
5
Ex.7 By what least number 675 be multiplied to obtain a number which is a perfect cube ? Sol. 675 = 5 × 5 × 3 × 3 × 3. To make it a perfect cube, it must be multiplied by 5.
1,
1,
1,
1
Ex.8 Find the cube root of .000216. 1/ 3
LCM = 3 × 5 × 2 × 3 × 2 = 180. The least multiple of 180 which is a perfect square is 180 × 5 = 900.
216 106
Sol. (0.000216)1/3 = =
6 10 2
=
666
1/ 3
2 2 2 10 10 10
=
6 = 0.06 100
PAGE # 31
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.
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 exactly divisible 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 DIVISIBILITY The method of finding the remainder without actually performing the process of division is termed as
Division Algorithm : General representation of result is, Dividend Re mainder Quotient Divisor Divisor
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
Dividend = (Divisor × Quotient ) + Remainder Ex.10 On dividing 4150 by certain number, the quotient is 55 and the remainder is 25. Find the divisor.
remainder is never be negative hence the quotient should be –4 and remainder is + 6. We can also get
Sol.
remainder 6 by adding –1 to divisor 7 (7 –1 = 6).
4150 = 55 × x + 25 55x = 4125
Ex.13 A number when divided by 296 gives a remainder 75. 4125 x= = 75. 55
When the same number is divided by 37, then find the remainder.
NOTE : (i) (xn – an) is divisible by (x – a) for all the values of n.
Sol. Number = (296 × Q) + 75 = (37 × 8Q) + (37 × 2) + 1 = 37 × (8Q + 2) + 1. Required remainder = 1.
(ii) (xn – an) is divisible by (x + a) and (x – a) for all the
Ex.14 A number being successively divided by 3, 5 and 8
even values of n. n
leaves remainders 1, 4 and 7 respectively. Find the
n
(iii) (x + a ) is divisible by (x + a) for all the odd values of n.
respective remainders if the order of divisors be Test of Divisibility : No.
reversed. Divisiblity Te st
Sol. 3 x 5 y 1 8 z 4 1 7
2
U nit 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
U nit digit s hould be 0 or 5.
z = (8 × 1 + 7) = 15 ; y = (5z + 4) = (5 × 15 + 4) = 79 ;
6
N o s hould be divis ible by 2 & 3 both
x = (3y + 1) = (3 × 79 + 1) = 238.
8
The num ber form ed by las t 3 digits of given no. s hould be divis ible by 8.
Now,
9
Sum of digits of given no. s hould be divis ible by 9
8 238 5 29 6 3 5 4 1 2
The difference betw een s um s of the digits at even & at odd places 11 s hould be zero or m ultiple of 11. 25 Las t 2 digits of the num ber s hould be 00, 25, 50 or 75.
Respective remainders are 6, 4, 2.
PAGE # 32
expansion. All numbers that are divisible by 21 will contribute 1 to the exponent of 2 in the product
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
20
= 10. Hence, 10 numbers contribute 21 to the 21 product. Similarly, all numbers that are divisible by
repeat its unit digit after every consecutive power. So, its cyclicity is 1, on the other hand digit 2 repeat its unit digit
22 will contribute an extra 1 to the exponent of 2 in the
after every four power, hence the cyclicity of 2 is four.
product, i.e
= 5. Hence, 5 numbers contribute an 22 extra 1 to exponents. Similarly, there are 2 numbers
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
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
So, if we want to find the last digit of 245, divide 45 by 4.
divisible by 2. If 20! is divided by 2x then maximum value of x is 18.
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.
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,
100 = 20, = 4. 51 52 Hence, the total contributions to the power of 5 is 24,
Unit digit in 771 is 3. [1 × 7 × 7 × 7 given unit digit 3]
we have
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.
20
100
or the number 100! is divisible by 524. Ex.19 What is the highest power of 6 that divides 9!
9 9 = 1 and 2 = 0. Thus 6 6 answers we get is 1 which is wrong. True there is just
Sol. By the normal method.
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,
Ex.16 Find unit’s digit in y = 717 + 734 Sol. 717 + 734 = 71 + 72 = 56, Hence the unit digit is 6.
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
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 =
as it would definitely be greater than that of 3). ,
BASE SYSTEM
120.
The value of factorial zero is equal to the value of
The number system that we work in is called the ‘decimal system’. This is because there are 10 digits
factorial one. Hence 0! = 1 = 1! The approach to finding the highest power of x dividing
in the system 0-9. There can be alternative system that can be used for arithmetic operations. Some of
y y y y! is 2 3 ......., where [ ] represents just x x x
the integral part of the answer and ignoring the fractional part. Ex.17 What is the highest power of 2 that divides 20! completely? 1
Sol. 20! = 1 × 2 × 3 × 4 ×....× 18 × 19 × 20 = 1 × (2 ) × 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
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. PAGE # 33
Conversion : Conversion of numbers from (i) decimal
1a4 x3b
system to other base system. (ii) other base system to Ex.24 If
decimal system.
then, find the value of b + a, where all
8c8 s 72
(i) Conversion from base 10 to any other base : Ex.20 Convert (122)10 to base 8 system. Sol.
8 122 8 8
15 1 0
2 7 1
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.
t 5d8 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, 4 Now 1 a 4 × b = 8c8 ; b = 2 or 7 Again 2 is ruled out because in that case, product would be much less than 800. b = 7. Hence, a = 2, b = 7, c = 6, d = 8, s = 3 and t = 4.
So, here (122) 10 = (172)8. Ex.21 Convert (169)10 in base 7.
7 169 7 24 7 3 Sol. 0
1 3 3
1.
In numbers from 1 to 100 digit “0” appears_______ times. (A) 9 (C) 11
Remainder 2.
(169)10 =(331)7 (ii) Conversion from any other base to decimal
How many 2 digits.
numbers
are
(A) 90 (C) 100
system : Ex.22 Convert (231)8 into decimal system.
(B) 10 (D) 12
3.
Sol. (231)8 , the value of the position of each of the numbers
there
containing
(B) 99 (D) 89
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
( as in decimal system) is : 1 = 80 × 1
(D) None of these
3 = 81 × 3 4.
2 = 82 × 2 0
1
Hence, (231)8 = (8 × 1 + 8 × 3 + 8 × 2)10 (231)8
= (1 + 24 + 128)10
(231)8
= (153)10
(C) 3 5.
6.
aa Ex.23 If a – b = 2, and
then find the value of a, b and c.
Sol. These problems involve basic number 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.
The last digit of the number (373)333 is : (B) 2 (D) 9
The two missing numbers shown with asterisk in the 3 1 equation 5 * = 19 are : * 2
cc 0
(i) aa + bb = 11(a + b) (ii) aa, bb are two-digit numbers.
(D) 7
(A) 1 (C) 3
ALPHA NUMERICS NUMBERS
b b
What least number should be added to 1330 to get a number exactly divisible by 43? (A) 46 (B) 1
2
7.
(A) 6, 3
(B) 7, 3
(C) 8,3
(D) 11, 3
Given
5 = 2.236 the value of
45 +
605 –
correct to 3 decimal places is : (A) 15.652 (C) 18.652
(B) 11.180 (D) 16.652
Such problems are part of a category of problems called alpha numerics.
PAGE # 34
245
8.
A student was asked to multiply a number by Instead he divided the number by
number smaller by
(A)
3 and obtained a 2
2 , the number is : 3
4 5
(B)
2 (C) 3
9.
3 . 2
3 5
1 (D) 2
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 always divisible 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
Which of the following statements is true ?
7 and 8
18. Which of the following fractions is less than (A)
2 5 4 7 3 9 12 18
greater than
5 2 7 4 (B) 18 12 9 3
(C)
(D)
5 2 4 7 9 18 12 3
(B) 0 (D) None of these
11. 0.018 can be expressed in the rational form as : 18 18 (A) (B) 1000 990 18 18 (C) (D) 9900 999 12. The least number which must be subtracted from 2509 to make it a perfect square is : (A) 6 (B) 9 (C) 12 (D) 14 x 2 y 2 , the value of (1 2 2 ) (1 2 2 ) is :
(A) –7 (C) 2
(B) 0 (D) 9
1
1 3
40 31 1 (C) 8
(A)
(B)
23 24
(C)
11 12
(D)
17 24
(B) 15 (D) 32
3 1 1 1 20. 5 – 4 2 2 0.5 6 7 is equal to :
9 3 and ? 8 7 1 (A) 2 12 (C) 15
5
1 4
(A) 13 (C) 27
10. Which of the following rational numbers lie between
14. The value of 4
(A)
19. Simplify : 18 – [5 – {6 + 2(7 – 8 5 )}].
2 5 4 7 3 12 9 18
13. If x y =
1 ? 3
(A) 2
23 84
(B) 3
1 6
(C) 3
3 10
(D) 5
1 10
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 to make 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 and the remainder is 103. The number is : (A) 364724 (B) 365387 (C) 365737 (D) 366757
is : 25. Evaluate :
1 1 2 4
(A) 2 (C) 4 4 9 31 (D) 40
(B)
8 [5 ( 3 2)] 2 . 53 58 3
(B) 3 (D) 5
26. The value of 214 130 88 44 25 (A) 14 (C) 16
:
(B) 15 (D) 17 PAGE # 35
3
27. (2 2 ) ×
38. Which of the following is an irrational number : (A) 3.65789 (B) 3.65789125634..... (C) 3.65786578..... (D) 3.666......
9 is : 8
(A) a rational number (C) undefined
(B) an irrational number (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 1 1 1 1 1 1 29. The value of 1 [1 1 (1 1 1 )] is : 2 2 2 2 2 4 1 (A) 2 1 (C) 16
30. If
1 (B) 4 1 (D) 1 5
1 .21 = 1.1, then
(A) 0.0011 (C) 0.11
.000121 is equal to : (B) 0.011 (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
39. The number of such squares having area less then 8 times their sides (numerically), will be : (A) 1 (B) 5 (C) more than 6 (D) 0 m 40. If the fraction is negative, which of the following n cannot be true ? n m > m n (C) (n – m) < 0
(A)
(B) mn < 0 (D) mn3 > 0
41. Physical Instructor wants to arrange boys in rows to form a perfect square. He finds that in doing so, 25 boys are left out. If the total number of boys is 1250 then 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 a three-digit number. (A) 21 (B) 19 (C) 17 (D) 18
32. The largest natural number by which the product of three consecutive even natural number is always divisible is : (A) 16 (B) 24 (C) 48 (D) 96
43. S is a six digit number beginning with 1. If the digit 1 is moved from the leftmost place to the rightmost place the number obtained is three times of S, What is the sum of the digits of S ? (A) 12 (B) 15 (C) 18 (D) 27
33. If a number x is divided by 95, then remainder is 30. If the same number x is divided by 5, then what is remainder ? (A) 2 (B) 3 (C) 4 (D) 0
44. If ABC x CBA = 65125, where A, B and C are single digits, then A + B + C = ? (A) 18 (B) 15 (C) 8 (D) 7
34. A gardener plants tree in rows and finds that each row contains twice as many tree as these are rows. If the number of trees be 5408. Then the number of tree in each row is : (A) 100 (B) 104 (C) 108 (D) 112 35. x and x + y are the square of two consecutive natural number. 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 = 1 3125 1 (C) 3025
(A)
37. In the equation 2
1 , then P is : P
(B) 3125 (D) 3025 2 5 3 11 ×1 +9 = 1 + x, x is equals 3 6 4 12
to : 2 3 2 (C) 10 3
(A) 12
7 12 1 (D) 9 12
(B) 10
45. If 27 = 123 and 31 = 133, Than 15 = ? (A) 13 (C) 11
(B) 31 (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 multiplication 4! 4! 8!8! 16!16! (A) 8! + 16! (C) 8! + 3.16!
(B) 8! + 2.16! (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 PAGE # 36
CELL DIVISION (ii) Shape : It is usually determined by the position of its centromere. On this basis chromosomes can be of following types :
STRUCTURE OF CHROMOSOME Chromosomes are the vehicles of heredity which possess DNA and are enclosed inside the nucleus. They are capable of self reproduction and maintaining morphological and physiological properties through successive generations.Each chromosome consists of two strands which are called as chromatids.The two chromatids of a chromosome are joined together at a point called as centromere.
(A) Metacentric : They are V – shaped. These have centromere in the middle of the chromosomes so that the two arms are almost equal. (B) Submetacentric : They are L shaped. In them centromeres are slightly away from the midpoints, so that the two arms are unequal. (C) Acrocentric : They are J-shaped with centromere at subterminal position. (D) Telocentric : They are I-shaped, having terminal centromere.
•
NOTE : These shapes can be observed durig anaphse stage 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 of chromosomes are present.So, there are two chromosomes, of each kind. These two chromosomes of each kind are called as homologous chromosomes.
(a) Size and Shape of Chromosomes :
•
(i) Size : Size of chromosomes is variable in different organisms, different tissues and at different stages of the cell cycle.
•
•
A cell which has the complete set of chromosomes with two of each kind is called as diploid cell. In other words a diploid cell has two sets having two chromosomes of each type. The gametes (or sex cells) of human beings are different from their other body cells because they contain only half the number of chromosomes. A cell which has half the number of chromosomes, is called as haploid cell. In other words a haploid cell has only one set of each type of chromosomes. Human gametes called sperm and egg have only 23 chromosomes which is half the number of chromosomes of other body cells. So, a gamete is a haploid cell.
PAGE # 37
•
• •
Females consist of two similar gametes and therefore called as homogametic and males consist of dissimilar gametes and therefore called as heterogametic. During spermatogenesis two types of sperm cells will be produced one which contains X-chromosome and the other which contains Y chromosome.During oogenesis each egg cell contain one X-chromosomes. If X – chromosome of male fuses with X-chromosome of female it will produce a female child. If Y – chromosome of male fuses with X-chromosome of female it will produce a male child.
(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 a complex of changes that occurs in a newly formed cell before it is able to divide. It lasts throughout the life. It involves replication of DNA and synthesis of nuclear proteins and duplication of centriole. Synthesis of energy rich components also takes place.This occurs in three stages i.e. G1 (First growth phase), S (Synthesis phase), G2 (Second growth phase) (ii) Short dividing M – phase : t is the phase of cell division. It consists of karyokinesis (nuclear division) and cytokinesis (cytoplasmic division).It is of two types :
(c) Properties of Chromosomes : The chromosomes must possess five important properties : (i) Replication : Synthesis of new DNA molecule which is identical to the parent DNA molecule. (ii) Transcription : Synthesis of RNA molecule which is identical to the DNA molecule.
(b) Mitosis :
• •
(iii) Change in appearance.
•
(iv) Repair : It means correction of damaged parts by DNA.
•
(v) Mutation : Development of genetic changes. (d) Functions of Chromosomes : (i) They carry hereditary characters from parents to offsprings.
•
(A) Prophase : Chromatin fibres condense to form chromosomes.They shorten and become distinct with each having two chromatids attached to centromere. Centrosomes reach the poles and form spindle fibres.Nucleolus disappears, nuclear membrane disappears.
(iii) They undergo mutation and thus contribute to the evolution of animals. (iv) They guide cell differentiation during development. (v) They also help in metabolic process. (vi) They bring about continuity of life.
(B) Metaphase : Chromosomes attach to spindle fibres arise from each pole and lie at the equator, forming a metaphase plate.Chromosomes are shortest and thickest in this stage.
CELL DIVISION Cell division was first observed by Nageli in plant cell (1842).
(C) Anaphase : Shortest phase, In this phase centromere of each chromosome divides to form two daughter chromosomes.They remain attached to poles through spindle fibres and start moving towards pole and become shortened. They appear in different shapes.
(a) Cell Cycle : It is a series of programmed cyclic changes by which the cell duplicates its contents and divides into two parts. It is divided into two phases : Mitotic division phase r) (1 h hase Mp
S-phase
(12 hr) G1 phase
(6-8 hr) D.N.A. Synthesis Cell cycle
It is also called as somatic division as it occurs during formation of body cells. It is studied in plants, in meristems and in animals in bone marrow, skin and base of nails. It is an equational division in which a parent cell divides into two identical daughter cells, each of them contains the same number and kind of chromosomes as are present in parent cell. It occurs in two steps : (i) Karyokinesis (ii) Cytokinesis (i) Karyokinesis : Division of nucleus. It is divided in four steps :
(ii) They help the cells to grow, divide and maintain itself by synthesis of proteins.
(3-4 hr) G2 phase
Term mitosis was given by Flemming.
•
V – Shaped (Metacentric)
•
L – Shaped (Submetacentric)
•
J – Shaped (Acrocentric)
•
I – Shaped (Telocentric) (D) Telophase : Nucleus is reconstituted, chromosomes uncoil, elongate and form chromatin fibre.Nucleolus & nuclear envelope reappears forming two daughter nuclei
PAGE # 38
(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.
PAGE # 39
(B) Meiosis – II : It is also called as equational division and maintains the haploid number of chromosomes. No replication of DNA occurs in this stage.
(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.It consists of two phases : (i) Interphase : Size of nucleus increases to three times. It also involves G1 – S – G2 phase.
•
Prophase – II : Chromatin fibres shorten and form chromosomes.Nuclear envelope and nucleolus start disappearing.
•
Metaphase – II : Chromosomes form single metaphasic plate by arranging themselves on equator.
•
Anaphase – II : Centromere divides into two and separates two chromatids of chromosome into two independent daughter chromosomes or chromatids.
•
Telophase – II : The four groups of chromosomes organize themselves into 4 haploid nuclei.Chromatin fibres are formed, nucleolus and nuclear envelope are reappeared.
(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. It has following stages :
•
Leptotene : Chromatin fibres condense to form chromosomes. There are two chromosomes of each type which are diploid and are called as “homologous chromosomes”.
•
Zygotene : Homologous chromosomes join by synapsis and form bivalents which are actually tetrads with half the number of individual chromosomes, pairing proceeds in zipper like fashion forming synaptonemal complex.
•
Pachytene : There occurs exchange of segments between non sister chromatids of bivalents and is called as crossing over.
•
Diplotene : Synaptonemal complex is dissolved, tetrads are formed. At some places nonsister chromatids of two homologous chromosomes remain attached forming, chiasmata.
•
Diakinesis : Chiasmata shifts towards ends, nucleolus degenerates.
•
Metaphase – I : Spindles are formed and bivalents form a double whorl or double metaphase plate.
•
Anaphase – I : Chiasmata disappears, homologous chromosomes separate by disjunction forming dyads.They move towards poles and form two groups of haploid chromosomes.
•
Telophase – I : Chromosomes elongate, nucleoplasm & nuclear envelope reappears.
Fig. Various Stages of Meiosis
•
•
Significance of mitosis : It is essential for growth, repair, differentiation, maintenance of chromosome number etc. Significance of meiosis : It produces variations and essential for sexual reproduction. It maintains the chromosome number in each generation of living organisms.
PAGE # 40
Differences between mitoitc and meiotic cell division S.No. 1. 2.
Mitosis It occurs in all s om atic cells . In the res ultant daughter cells , the num ber of chrom os om es rem ains the s am e (i.e., diploid), hence, called equational divis ion.
Meiosis It occurs in reproductive cells (germ cells ) In res ultant daughter cells , the num ber of chrom os om 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 cros s ing over takes place
During m eios is cros s ing over takes place.
5.
Daughter cells have identical chrom os om es which are als o identical to that of parent cell (i.e. rem ains cons tant)
Chrom os om 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 the nuclear membrane remains intact. Both cell and its
•
5.
The stage of the meiosis in which nucleolus and nuclear membrane disappear and chromosomes become distinct is (A) prophase (B) metaphase (C) anaphase (D) telophase
6.
In which of the following stages chromosomes are thin and long thread -like ? (A) Leptotene (B) Zygotene (C) Pachytene (D) Diplotene
7.
In which phase of mitosis, the chromosomes are arranged 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
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 common to meiotic cells of higher plants ? (A) Cell plate (B) Centriole (C) Centromere (D) Spindle fibres
3.
How many mitotic divisions are needed for a single cell 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
nucleus elongate, constrict in middle and break off into 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, Endosperm cells of seed, Diseased cell and Old tissues).
PAGE # 41
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) G1 - 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
PAGE # 42
NUMBER SERIES Number series problems deal with numbers. While attempting to solve the question, you have to check the pattern of the series. Series moves with certain mathematical operations. You have to check the pattern. Type of questions asked in the examination : (i) Find the missing term(s). (ii) Find the wrong term(s).
Ex 6.
Sol.
Number Series In this type of series, the set of given numbers in a series are related to one another in a particular pattern or manner. The relationship between the numbers may be • Consecutive odd/even numbers,
Ex 7.
Sol.
• Consecutive prime / composite numbers, • Squares/cubes of some numbers with/without variation of addition or substraction of some number,
Ex 8.
• Sum/product/difference of preceding number(s), • Addition/subtraction/multiplication/division by some number, and
Sol.
• Many more combinations of the relationship given above. Directions : (1 to 13) Find the missing numbers : Ex 1.
Sol. Ex 2.
Sol. Ex 3.
Sol.
Ex 4.
Sol.
Ex 5.
Sol.
3, 5, 7, 9, 11, 13, 15, 17, ? (A) 14 (B) 19 (C) 15 (D) 21 (B) Each term has a common difference = + 2. Hence, next term = 17 + 2 = 19. 2, 3, 5, 7, 11, ?, 17 (A) 14 (B) 13 (C) 10 (D) 12 (B) The series is made up of consecutive prime numbers. Therefore, the missing term is 13. 1, 4, 9, 16, 25, ? (A) 35 (B) 36 (C) 37 (D) 49 (B) Each term is a square of 1, 2, 3, 4 and so on 12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25. Hence, next term = 62 = 36. 2, 5, 10, 17, ? (A) 24 (B) 25 (C) 26 (D) 27 (C) Each term is a square of 1, 2, 3, 4 and so on and 1 is added to it, i.e. 12 + 1, (2)2 +1, (3)2 +1,..........= 2, 5, 10, 17.... Hence, next term = (5)2 + 1 = 26. 2, 3, 10, 15, 26, 35, ? (A) 48 (B) 51 (C) 49 (D) 50 (D) The series exhibits the pattern of n2 + 1, n2 – 1, alternately, n taking values 1, 2, ......
Ex 9.
Sol. Ex 10.
Sol.
Ex 11.
Sol.
Ex 12.
1, 8, 9, 64, 25, 216, ?, ? (A) 49, 64 (B) 343, 64 (C) 49, 512 (D) 343, 512 (C) Odd positioned digits are squares of 1, 3, 5 and so on, i.e. 12 = 1, 32 = 9, 52 = 25 and so on. Similarly, even positioned digits are cubes of 2, 4, 6, etc., i.e. 23 = 8, 43 = 64, 63 = 216. Therefore, the next term would be 72 i.e. 49 and 83 = 512 respectively. 0, 7, 26, ?, 124, 215 (A) 51 (B) 37 (C) 63 (D) 16 (C) Each term is a cube of 1, 2, 3, 4 and so on and 1 subtracted from it, i.e. 13 – 1, 23 – 1, 33 – 1, 43 – 1, 53 – 1, 63 – 1. Therefore, the term replacing the question mark would be 43 – 1 = 64 – 1 = 63. 3, 4, 10, 33, 136, ? (A) 240 (B) 430 (C) 685 (D) 820 (C) The terms of the series are, previous term × 1 + 1, previous term × 2 + 2, previous term × 3 + 3 and so on. Hence, the next term will be 136 × 5 + 5 = 680 + 5 = 685. 11, 15, 21, 29, ? (A) 40 (B) 41 (C) 37 (D) 39 (D) This series consists of increasing numbers. The pattern is +4, +6, +8,........ 3, 6, 18, 72, 360, ? (A) 720 (B) 1080 (C) 1600 (D) 2160 (D) The sequence in the given series is × 2, × 3, × 4, × 5, × 6. Hence, the missing number is 360× 6 =2160. 6, 12, 7, 11, 8, 10, 9, ? (A) 8 (C) 11 (B) Alternate series (i) 6, 7, 8, 9 (ii) 12, 11, 10, ? 0, 5, 22, 57, 116, ? (A) 205 (C) 192 0
Sol.
Difference
(B) 216 (D) 207
5
(A) Difference 5
22 17
12
(B) 9 (D)10 [Difference series]
57 35
18
116 59
24
205 89
30
Hence, the next term = 205
PAGE # 43
Ex 13.
Sol.
151, 158, 172, 182, ? (A) 210 (B) 193 (C) 197 (D) 203 (B) 1 + 5 + 1 = 7, The difference between 151 & 158 is seven (7) 1 + 5 + 8 = 14, The difference between 158 & 172 is (14). 1 + 7 + 2 = 10 .......... and so on, Missing term = 182 + 11 = 193.
Direction : (14 to16) Find the wrong term : 14. 2, 5, 9, 11, 14 (A) 2 (B) 5 (C) 9 (D) 11 Sol. (C) Series : + 3, + 3, + 3, .......... The next term is got by adding 3 in preceeding term. 2 + 3 = 5, 5 + 3 = 8 9 is wrong term. 15.
Sol.
10, 100, 1100, 11000, 111000, 1210000. (A) 1210000 (B) 11000 (C) 100 (D) 111000 (D) Given series is :
111000 is wrong. The correct term is 121000.
16.
Sol.
2, 6, 11, 17, 23, 32, 41 (A) 6 (C) 23 (C) Given series is :
6.
7.
8.
9
10.
11.
+5 +6 +7
+8
EXERCISE-1
2.
3.
4.
5.
2, 3, 5, 7, ? (A) 9 (C) 11
(B) 10 (D) 14
0, 6, 20, 42, ? (A) 64 (C) 80
(B) 72 (D) 84
3, 8, 35, 48, ?, 120 (A) 72 (C) 80
(B) 64 (D) 99
4, 25, 64, 121, 196, ? (A) 384 (C) 225
(B) 256 (D) 289
210, 120, ?, 24, 6, 0 (A) 64 (C) 35
(B) 48 (D) 60
(B) 66 (D) 94
8, 15, 28, 53, ? (A) 120 (C) 104
(B) 106 (D) 102
4, 8, 12, 24, 36, 72, ? A) 98 (C) 144
(B) 100 (D) 108
12, 15, 18, 21, ? (A) 24 (C) 22 3, 6, 12, 24, ?, 96 (A) 84 (C) 52
(B) 23 (D) 25 (B) 50 (D) 48
13.
4, 7, 3, 6, 2, 5, ? (A) 6 (C) 3
(B) 5 (D) 1
14.
4, 7, 10, 11, 22, 17, 46, 25, ? (A) 58 (B) 69 (C) 86 (D) 94
15.
2, 2, 4, 4, 6, 8, 8, ? (A) 10 (C) 14
(B) 12 (D) 16
16.
2, 3, 10, 15, 26, ? (A) 34 (C) 36
17.
1, 4, 27, 16, 125, 36, ? (NTSE Stage-I / Raj./2007) (A) 216 (B) 343 (C3) 64 (D) 49
18..
336, 210, 120, ?, 24, 6, 0 (NTSE Stage-I / Raj./2007) (A) 40 (B) 50 (C) 60 (D) 70
19.
3, 4, 8, 17, 33, ? (A) 58 (C) 49
20.
8, 13, 21, 34, 55, ? (NTSE Stage-I / Raj./ 2007) (A) 60 (B) 68 (C) 89 (D) 76
21.
480, 480, 240, 80, 20, ?
Directions : (1 to 50) Find the missing numbers : 1.
4, 10, 22, 46, ? (A) 56 (C) 76
2, 10, 19, 29, 40, 52, 65, 79, 94, ? (A) 110 (B) 109 (C) 108 (D) None of these
(B) 17 (D) 32
+9
(B) 210 (D) 258
12.
24 2, 6, 11, 17, 23, 32, 41 +4
2, 12, 36, 80, 150, ? (A) 194 (C) 252
(A) 4 (C) 5
(NTSE Stage-I / Raj./2007) (B) 35 (D) 37
(NTSE Stage-I / Raj./ 2007) (B) 69 (D) 98
(NTSE Stage-II, 2007) (B) 1 (D) 10 PAGE # 44
22.
1, 1, 2, 2, 3, 4, 4, 8, 5, 16, ? (NTSE Stage-II, 2007) (A) 6 (B) 32 (C) 8 (D) 7
23.
2, 5, 11, 23, 47, ? (A) 92 (C) 95
24.
12, 21, 23, 32, 34, 43, 45, ? (NTSE Stage-II, 2007) (A) 54 (B) 48 (C) 77 (D) 9
25.
27.
28.
(NTSE Stage-II, 2007) (B) 16, 35 (D) 16, 36
5, 6, 13, 26, 45, ? (A) 68 (C) 70
(NTSE Stage-I / Raj./ 2008) (B) 74 (D) 82
190, 94, 46, 22, 10, 4, ? (NTSE Stage-I / Raj./ 2008) (A) 3 (B) 2 (C) 1 (D) 0 128, 110, 90, 68, ? (A) 36 (C) 44
(NTSE Stage-I / Raj./ 2008) (B) 42 (D) 48
1, 2, 4, 7, ?, 16 (A) 9 (C) 12
30.
6, 8, 9, 12, 14, 18, ? (NTSE Stage-I / Raj./ 2008) (A) 21 (B) 19 (C) 23 (D) 20
31.
4, 9, 19, 34, 54, ? (A) 66 (C) 79
(NTSE Stage-II, 2008) (B) 75 (D) 84
31, 29, 24, 22, 17, ?, ? (A) 15, 13 (C) 14, 12
NTSE Stage-II, 2008) (B) 10, 8 (D) 15, 10
33.
3, 6, 11, 18, ? (A) 19 (C) 30
(NTSE Stage-II, 2008) (B) 27 (D) 37
34.
3, 8, 15, 24, ? (A) 30 (C) 36
(NTSE Stage-II, 2008) (B) 35 (D) 49
35.
36.
38.
7, 12, 22, 37, ?, 82, 112 (NTSE Stage-I / Raj./ 2009) (A) 62 (B) 57 (C) 52 (D) 42
39.
11, 13, 17, 19, ?, 25 (NTSE Stage-I / Raj./ 2009) (A) 20 (B) 21 (C) 23 (D) 22
40.
5, 9, 17, 33, ?, 129 (A) 72 (C) 65
41.
2, 5, 4, 10, 7, 15, 11, 20, ?, ? (NTSE Stage-II, 2009) (A) 12, 21 (B) 16, 25 (C) 13, 25 (D) 17, 30
42.
0, 6, 24, 60, 120, ? (A) 180 (C) 196
43.
57, 54, 58, 55, 59, 56, 60, ? (NTSE Stage-II,2011) (A) 64 (B) 63 (C) 58 (D) 57
44.
27, 31, 40, 56, 81, 117, ?
14, 1, 21, 4, 28, 9, ?,?
29.
32.
1, 4, 9, ?, 25, 36 (A) 11 (C) 21
(NTSE Stage-II, 2007) (B) 90 (D) 91
(A) 9, 42 (C) 35, 16 26.
37.
(NTSE Stage-I / Raj./ 2008) (B) 11 (D) 13
(A) 156 (C) 166
(NTSE Stage-I / Raj./ 2009) (B) 19 (D) 16
(NTSE Stage-II, 2009) (B) 67 (D) 58
(NTSE Stage-II, 2009) (B) 224 (D) 210
(NTSE Stage-II,2011) (B) 165 (D) 169
45.
55, 168, 57, 120, 60, 80, 62, 48, 65, 24, ?, ? (NTSE Stage-II,2011) (A) 69, 11 (B) 67, 8 (C) 8, 71 (D) 6, 72
46.
8, 7, 16, 5, 32, 3, 64, 1, 128, (?) (NTSE Stage-I / Raj./ 2012) (A) 18 (B) 13 (C) –1 (D) 3
47.
16, 33, 65, 131, (?), 523 (NTSE Stage-I / Raj./ 2012) (A) 261 (B) 521 (C) 613 (D) 721
48.
5, 2, 17, 4, (?) , 6, 47, 8, 65 (NTSE Stage-I / Raj./ 2012) (A) 29 (B) 30 (C) 31 (D) 32
4, 10, 23, 50, 105, ? (NTSE Stage-I / Raj./ 2009) (A) 215 (B) 210 (C) 216 (D) 439
49.
912, 303, 102, 33, ?, 3, 2 (NTSE Stage-I / Raj./ 2009) (A) 12 (B) 10 (C) 8 (D) 6
1, 2, 4, 8, (?), 32 (NTSE Stage-I / Raj./ 2012) (A) 10 (B) 12 (C) 14 (D) 16
50.
2, 3, 10, 15, 26, (?)(NTSE Stage-I / Raj./ 2012) (A) 36 (B) 35 (C) 39 (D) 48
PAGE # 45
12.
EXERCISE-2 Directions : (1 to 21) Find the wrong term of the series : 1.
3, 7, 9, 21, 27, 66, 81, 189, 243
13.
3, 9, 27, 82, 243 (A) 27
(B) 54
(C) 82
(D) 162
5, 9, 17, 35, 65, 129
(NTSE Stage-I / Raj./ 2007)
2.
(A ) 27
(B) 66
(C) 243
(D) 21
27, 34, 40, 45, 49, 53, 54 , 55 (NTSE Stage-I / Raj./ 2007)
3.
(A) 53
(B) 45
(C) 56
(D) 34
4.
(NTSE Stage-I / Raj./ 2009)
14.
(A) 3
(B) 6
(C) 20
(D) 54
15.
(B) 36
(C) 68
(D) 130
(B) 35
(C) 17
(D) 9
1, 5, 6, 11, 17, 27, 45, 73
(A) 27
(B) 45
(C) 17
(D) 11
3, 6, 11, 18, 28, 38, 51, 66 (NTSE Stage-I / Raj./ 2009)
0, 2, 10, 36, 68, 130 (NTSE Stage-I / Raj./ 2007) (A) 10
(A) 65
(NTSE Stage-I / Raj./ 2009)
0, 2, 3, 6, 6, 20, 9, 54, 12 (NTSE Stage-I / Raj./ 2007)
(NTSE Stage-I / Raj./ 2009)
16.
(A) 18
(B) 28
(C) 38
(D) 51
320, 254, 200, 155, 122, 100, 89 (NTSE Stage-I / Raj./ 2009)
5.
9, 54, 44, 264, 254, 1520, 1514 (NTSE Stage-I / Raj./ 2007)
6.
(A) 1514
(B) 1520
(C) 264
(D) 44
17.
8.
(B) 320
(C) 254
(D) 200
6, 8, 9, 12, 14, 18, 22, 26, 30 (NTSE Stage-I / Raj./ 2012)
10, 15, 26, 35, 48, 63, 82 (NTSE Stage-I / Raj./2008)
7.
(A) 155
(A) 48
(B) 26
(C) 63
(D) 82
3, 10, 30, 66, 127, 218 (NTSE Stage-I / Raj./2008) (A) 3
(B) 66
(C) 30
(D) 218
18.
(A) 12
(B) 22
(C) 26
(D) 30
3, 7, 9, 28, 27, 84, 81, 448, 243 (NTSE Stage-I / Raj./ 2012)
7, 9, 17, 42, 91, 172, 293
(A) 84
(B) 81
(C) 28
(D) 7
(NTSE Stage-I / Raj./2008)
9.
(A) 91
(B) 42
(C) 17
(D) 9
19.
(NTSE Stage-I / Raj./ 2012)
2, 12, 24, 34, 68, 78, 158, 166 (NTSE Stage-I / Raj./2008)
10.
(A) 68
(B) 78
(C) 158
(D) 166
20.
(A) 94
(B) 46
(C) 22
(D) 3
0, 5, 15, 50, 128 (NTSE Stage-I / Raj./ 2012)
2, 6, 10, 20, 30, 42, 56 (NTSE Stage-I / Raj./2008)
11.
190, 94, 46, 22, 10, 3
(A) 6
(B) 10
(C) 20
(D) 30
7, 9, 16, 25, 41, 68, 107, 173
21.
(A) 5
(B) 17
(C) 35
(D) 128
9, 63, 5, 35, 1, 8 (NTSE Stage-I / Raj./ 2012)
(NTSE Stage-II, 2008) (A) 16
(B) 41
(A) 63
(B) 5
(C) 68
(D) 107
(C) 35
(D) 8
PAGE # 46
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 Ans Ques Ans Ques Ans
1 C 12 B 23 B
2 D 13 A 24 C
3 C 14 C 25 C
4 A 15 B 26 A
5 D 16 A 27 D
6 A 17 C 28 C
7 8 A B 18 19 C B 29 30 C A
9 D 20 B
10 A 21 B
11 D 22 A
(MATHEMATICS)
NUMBER SYSTEM 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. Ques. Ans. Ques. Ans.
D 31 A 41 D
C 32 C 42 A
D 33 D 43 D
C 34 B 44 C
D 35 B 45 D
B 36 C 46 A
A 37 A 47 B
A 38 B 48 C
A 39 C 49 B
B 40 D 50 A
CELL DIVISION (BIOLOGY) Q. A.
1 B
2 B
3 A
4 C
5 A
6 A
7 A
8 B
9 A
10 C
11 A
12 B
13 C
14 B
15 B
NUMBER-SERIES(MENTAL ABILITY) Que.
1
Ans.
2
3
4
5
6
7
8
9
10
D
C
A
C
B
B
D
D
C
C
Que. 11
12
13
14
15
16
17
18
19
20
Ans.
B
B
B
A
D
B
A
C
A
B
Que. 21
22
Ans.
B
C
PAGE # 4747
