below its boiling point  

particles from their solution cannot be separated by centrifugation

\(\frac { 1 }{ 12 } \) of C-12 isotopes of carbon

10^-15m

Golgi apparatus

adipose tissue

cell structure, body organisation and mode of nutrition

zero

it always falls towards the earth

98 N

4:1

frequency

Typhoid

biosphere

Italian bee

Air is matter because it has mass and occupies space.

(a) Non-metals do not conduct heat and electricity but carbon in the form of graphite is a good conductor of electricity.
(b) Non-metals are brittle but carbon in the form of diamond is very shining substances

Electronic configuration is 2, 8, 5
Number of electrons=15
Number of protons=15
Atomic number=15

Phloem consists of following four elements:
(i) Sieve tubes
(ii) Companion cells
(iii) Phloem parenchyma
(iv) Phloem fibres.
The most important element of phloem is sieve tubes because they translocate food materials from leaves to other parts of plant.

(a) It would have contributed in disturbing the ecological balance of the pond/lake or any such water bodies.
(b) He is environment friendly because he is conserving biodiversity and not destroying species unnecessarily.
(c) By spreading awareness in the community members about the importance of diversity, by developing empathy and love for all living organisms.

Vector quantities The Physical quantities which need both magnitude and direction for their complete description are called 'vectors' or 'vector quantities' Displacement,velocity,force,etc are all vector quantities.

Here, m = 1000 kg,  u = 45 km/h \(=\frac { 45\times 1000 }{ 3600 } =\frac { 25 }{ 2 } m/s,\)   v = 0,   t = 5s
\(\therefore\)  Force, \(F=m\left( \frac { v-u }{ t } \right) =1000\left( \frac { 0-25/2 }{ 5 } \right) =-2500\quad N.\) 
Negative sign shows the retarding nature of the force.

Density of wood = R.D x Density of wate
= 0.6 x 10³ kg m^-1 = 600 kg m^-3
Volume of wooden block = 2 m x 0.25 m x 0.10 m =0.05 m³
Mass of wooden block    = Volume x Density
= 0.05 m³ x 600 kg m^-3 =30 kg.

Mechanical waves. The waves which require a material medium for their propagation are called mechanical waves. They are also called elastic waves because their propagation depends on the elastic properties of the medium.
Examples of mechanical waves:
(i) Sound waves in air.
(ii) Waves over water surface.
(iii) Waves produced during earthquake. These are known as seismic waves.

(i) Communicable diseases - Malaria, Flu, Jaundice, etc
(ii) Non-Communicable disease - Diabates, cancer, heart disease, etcv

Water pollutants:
(i) Microorganism e.g., protozoans, bacteria and viruses
(ii) Eggs or larvae of disease causing vectors
(iii) Oils
(iv) Heavy metals
(v) Radioactive wastes
(vi) Detergents
(vii) Sewage disposals and domestic wastes.

Manure helps in enriching soil with mainly organic matter and small quantities of nutrients. The Bulk of organic matter in the form of manure helps in increasing water holding capacity in sandy soil. In clayey soils the large quantities of organic matter help in drainage and avoiding water logging.

Particles of matter are always in a state of motion.They move to interect with other particles and distribute themselves equally in all available space.This intermixing of particles of two substances on their own is called diffusion.Particles of a gas diffuse faster than particles of a liquid.
Example: Light an agarbatti in one corner of the room, and stand in the other corner. Very soon, you will feel smell of agarbatti.The particles of perfume in the agarbatti stick mix with particles of air and spread out eventually and reach to us even at a distance.this is due to diffusion of agarbatti particles into particles of air.

Characteristics of compounds
(i) Compounds are the substance formed by chemical combination of two or more elements.
(ii) The constituents elements are present in a fixed ratio.
(iii) A chemical reaction takes place using the formation of a compound.
(iv) Properties of a compound are different to those of its elements.
(v) Constituent elements cannot be separated by physical processes.

Number of atomes in 1 mole or 24 g Mg = 6.022 x 10²³
Number of atoms in 6 g Mg = \(\frac { 6 }{ 24 } \times 6.022\times { 10 }^{ 23 }\)
= 2.5 x 10²³ atoms
Now 6.022 x 10²³ number of atoms in grams of sodium = 23 g
2.5 x 10²³ number of atoms in grames of sodium = \(\frac { 23\times { 2.5\times 10 }^{ 23 } }{ 6.022\times { 10 }^{ 23 } } \)
= 9.548 x 10^-1 g.

Isotopes have the following uses:
(i)Treatment of diseases-Radioactive isotopes are used for the treatment of dreadful diseases like cancer.Cobalt-60 is used ti kill malignant cells in patients suffering from cancer.
(ii)Radioactive isotopes are used in chemical analysis.
(iii)Diseases in plants are investigated by using radioactive isotopes.
(iv)Production of energy: Uranium-235 can be subjected to fission process and thus production of electricity.
(v)Dating of plants, animals/humans beings obtained from ancient times after excavation by using carbon-14
(vi)I-131 is used in the treatment of thyroid disorders.
(vii)P-32 is used in the treatment of leukemia.

The second equation of motion. Suppose a body starts with initial velocity u and due to uniform acceleration a, its final velocity becomes v after t. Then
Average velocity = \(\frac { Intial\quad velocity+Final\quad velocity }{ 2 } =\frac { u+v }{ 2 } \)
So, the distance covered by the body in time t is
s = Average velocity x Time
=\(\frac { u+v }{ 2 } \) x t=\(=\frac { u+(u+at) }{ 2 } Xt\)   \([\therefore v=u+at]\)
\(=\frac { 2ut+at^{ 2 } }{ 2 } \)
or \(s=ut+\frac { 1 }{ 2 } at^{ 2 }\) 

Given that, mass of the boat, m = 50 kg
As we know that, when a body floats in river. Its apparent weight is zero.
Therefore, buoyant force = weight of body = mg = 50 x 10
i.e. Buoyant force = 500 N

(i) Exposure to air borne diseases can be prevented by providing living conditions that are not overcrowded.
(ii) Exposure to water borne microbes can be prevented by providing safe drinking water i.e., water treated to kill any microbial contamination.
(iii) The person suffering from an infectious disease should be kept in isolation
(iv) One should take balanced diet to make the immune system strong.This will also help in preventing infectious diseases.

There are two types of sources of water pollution. These are:
(i) The sources which are located to close to water source and discharge pollutants directly into the particular water source. These may be factories, power plants, underground coal mines and oil wells situated near to water sources. It is possible to treat the pollutants before their release into the water body.
(ii) Some sources are scattered and do not have specific situation for discharging pollutants into a particular water body. These are run-off water from agricultural fields, gardens, lawns, roads, logging sites, construction sites etc. It is very difficult to monitor, regulate and treat non-point sources.

(i) This is because solid carbon dioxide on decreasing pressure gets converted into gaseous carbon dioxide.
(ii) The water from the ground gets evaporated taking the latent heat of vaporisation from the ground and surrounding air which then becomes cool.

Homogeneous mixtures
(a) Solid solution. Alloys, e.g., brass, steel, bronze.
(b) liquid solution. Sugar solution in water, seawater, alcohol in water.
(c) Gaseous solution. Air, natural gas.
Heterogeneous mixtures
(a) Solid + Solid Sugar and sand, iron filings and sulphur, copper sulphate and potassium permanganate
(b) Solid +  Liquid Pastes, suspensions, etc.
(c) Solid + gas Smok
(d) Liquid + Liquid Milk, emulsions.
(e) liquid + Gas Soap bubbles.

In the begining, the names of elements were derived form the name of the place they were found for the first time or form some specific colours.For example, the name of copper was taken from Cyprus, a place where it was found.Glod was taken from the English word meaning yellow.

From table
Number of proton oxygen atom=8
Number of neutron oxygen atom=8
Mass number of oxygen = 8+8=16U
Number of protons in sulphur atom=16
Number of neutrons in sulphur atom=16
Mass number of sulphur atom=16+16=32U

(i) We will give ORS (Oral Rehydration Solution) from time to time till we reach to doctor or doctor comes to attend the patient. ORS is available in powder form packed in sachets.
(ii) We can make ORS in home by the following way:
Ingredients
1. One level teaspoon of salt
2. Eight level teaspoon of sugar
3. One litre of clean drinking or boiled water and then cooled (5 cupfuls-each cup about 200 .ml.),
Preparation Method
Stir the mixture till the salt and sugar dissolve.
(a) Arrange community lectures, debates to sensitise harmful effects of dehydration of the body.
(b) Conveying importation of ORS and its preparation in home

The practice of naming is one of the greatest inventions of man. Millions of animals and plants are named differently in different languages in different parts of the world. It means the names used for one organism differs from place to place. For example, Argemone a common weed has as many as ten to twele Hindi names alone. Some call it 'Firangi Dhatura', Sial Kanta', 'Pila Dhatura', 'Bharband', 'Kandhari', 'Satya nasi buti', 'Katela', 'Jahriber', 'Ugar Kanta' etc. Same thing is applicable to animals and other plants. Sometimes the common names are misleading. Same name may be used for different organisms in different parts.
To avoid this confusion Linnaeus gave a system in which an organism is given a name with two components-the first part is the generic name and the second part is the specific name. Thus the scientific name of common cat is Felis domestica, and that of tiger is Felis tigris. This system of giving a name with two components to an organism is called 'Binomial system of nomenclature'. The scientific names thus given to the organism are recognized all over the world. Hence, confusion is avoided on this aspect.

Non-uniform motion .If an object covers unequal distances in equal intervals of time,it is said to be in non-uniform motion. Most of the motions seen in our daily life are non-uniform. For example,if we drop a ball  from the roof of a building we will note that the ball covers 4.9 m in the 1st second 14.7 m in the 2nd second 24.5 m in the 3rd second , and so on That is the ball covers increasingly larger distances in successive seconds as it falls down. Thus, the motion of a freely falling body is non-uniform
Other examples of non-uniform motion:
(i) A stone dropped from the top of a building
(ii) A ball thrown vertically upwards
(iii) The motion of a train as it leaves the station
(iv) The motion of a bus as it approaches a bus-stop
(v) The motion of a ball rolling down an inclined plane.

Newton's first law of motion. According to this law, a body at rest or in uniform motion will remain at rest or in uniform motion unless an unbalanced force acts upon it. This law consists of three parts:
(i) First part, says that a body at rest continues in its state of rest. For example, a person standing in a bus falls backward when the bus suddenly starts moving forward. When the bus moves, the lower part of his body begins to move along with the bus while the upper part of his body continues to remain at rest due to inertia. That is why a person falls backward when the bus starts.
(ii) Second part, says that a body uniform motion continues moving in straight line path will a uniform speed. for example, when a moving bus suddenly stops, a person sittng in it falls forward. As the bus stops, the lower part of his body comes to rest along with the bus while upper part of his body continues to remain in motion due to inertia and so he fall forward.
(iii) Third part, says that a body moving with a uniform speed in a straight line cannot change itself its direction of motion. For example, when a bus takes a sharp turn, a person sitting in the bus gets a force acting away from the center of the curved path due to his tendency to move in the original direction. He has to hold on a support to prevent himself from swaying away in the turning bus.

Difference between mass and weight:

No the bundle of hay remains stationary i.e., displacement is zero.So, the work done is also zero.

Time period. The time taken by an oscillating body to complete one oscillation is called its time period. It is denoted by T. Its SI unit is second (s).
Frequency. The number of oscillations or vibrations completed by an oscillating body in one second is called its frequency. It is denoted by v (Greek letter nu).
SI unit of frequency=per second (s^-1) = cycles per second (cps) = hertz (Hz).
Relation between time period and frequency:
Let T=time period of an oscillating body. Then number of oscillations completed in T second = 1
Number of oscillations completed in 1 second =\(\frac { 1 }{ T } \)
But number of oscillations completed in 1 second = frequency (v)
∴ \(v=\frac { 1 }{ T } \)
Hence frequency is equal to the reciprocal of time period.

It is very necessary to replenish forests for the following reasons:
(i) Rainfall: Trees give out enormous amount of water during transpiration. This water vapour helps in the formation of rain clouds. If trees are destroyed and not replenished the rainfall in the area will reduce. This may lead to less number of trees to grow.
(ii) Soil erosion: If trees are cut at a large rate the soil becomes barren. The top soil which is rich in organic matter will be washed away by water or carried away by wind. Thus reduction in rainfall and soil erosion may lead to the formation of desert.
(iii) Carbon dioxide-oxygen balance: Forests have very large number of trees. They give out \({ O }_{ 2 }\) and take in \({ CO }_{ 2 }\) in day time by the process of photosynthesis. Thus they help in maintaining carbon dioxide-oxygen balance in the atmosphere.

Crop Rotation: The growing of different crops on the same field or a piece of land in a pre-planned succession is called crop rotation.
Selection of crops for crop rotation: Selection of crops depends on the -
(i) Moisture conditions i.e., availability water through rain or irrigation.
(ii) Length of rainy season.
(iii) Type of soil and nutrient status.
(iv) Availability of inputs-fertilisers, pesticides, manpower and machine power.
(v) Duration of crop-short or long.
(vi) Risk involved i.e., marketing and processing facilities.
For example, rabi crops of mustard, chickpea, linseed, barley, sunflower, and wheat are grown on conserved moisture.
In areas with assured soil moisture throughout the year, depending other inputs, rice, wheat, sugarcane, potato, and barseem are some of the major components of crop rotation.
Farmers near cities and towns prefer to grow vegetables and flowers for higher income.

Mass of silver (Ag) ornament = mg
Mass of gold used for polishing = \(\frac{1}{100}\times\) mg = 0.01 mg
Atomic mass of Ag = 108 u
\(\therefore\) 1 mole of Ag = 108 g = 6.022 x 10²³ atoms
Thus, 108 g of Ag have atoms = 6.022 x 10²³
\(\therefore\) mg of Ag have atoms = \(\frac{6.022\times 10^{23}}{108}\)
Similarly, atomic mass of gold (Au) = 197 u
1 mole of Au = 197 g = 6.022 x 10²³ atoms
Thus, 197 g of Au have atoms = 6.022 x 10²³
\(\therefore\) 0.01 mg of Au will have atoms = \(\frac{6.022\times 10^{23}\times 0.01\ m}{197}\)
\(\therefore\) Ratio of the number of atoms of gold and silver = \(\frac{6.022\times 10^{23}}{197}\times 0.01\ m; \frac{6.022\times 10^{23}}{108}m\)
= \(\frac{1}{19700}:\frac{1}{108}\) 
= 108: 19700

(i) 1 Mole of Carbon atom = 6.022 x 10²³ atoms
6.022 x 10²³ atoms of carbon weigh = 12 g
1 atom of carbon weighs = \(\frac { 12 }{ 6.022\times { 10 }^{ 23 } } \\ \)  
= 1.99 x 10^-23
(ii) (a) n = 0.5 mol; M = 14 x 2 = 28 g; m = ?
m = n x M = 0.5 x 28 = 14.0 g
(b) n = 0.2 mol; M = 16 g; m = ?
m = n x M = 0.2 x 16 = 3.2 g
(c) n = 4 mol; M = 27 g; m = ?
m = n x M = 4 x 27 = 108 g

(i) Lysosomes are membrane-bound sacs filled with digestive enzymes.These enzymes are made by rough endoplasmic reticulum.
(ii) Lysosomes are a kind of waste disposal system of the cell.
(iii) During the disturbance in cellular metabolism, e.g. when a cell gets damaged, lysosomes present in the cell may burst and the enzymes digest the damaged cell. Hence, lysosomes are called as 'suicidal bags' of a cell.
(iv) Lysosomes break up the foreign materials entering into the cell, such as bacteria or food into smal piece

((t) Formula to find the magnitude of gravitational
Force, \(F=\cfrac { GMm }{ { R }^{ 2 } } \)
where, M = mass of the earth
m = mass of the object
R = radius of the earth
and universal gravitational constant,
G = 6.67 X 10^-11 N-m²/kg²
(ii) (a) Let gravitational force be F when the distance between them is R
\(F=\cfrac { GMm }{ { R }^{ 2 } } \)
Now, when the distance reduces to half
\({ F }^{ ' }=\cfrac { GMm }{ \left( \cfrac { R }{ 2 } \right) ^{ 2 } } =\cfrac { 4GMm }{ { R }^{ 2 } } \)
On dividing Eq. (i) by Eq. (ii), we get
\(\cfrac { F }{ { F }^{ ' } } =\cfrac { GMm }{ { R }^{ 2 } } \times \cfrac { { R }^{ 2 } }{ 4GMm } \)
F' = 4F
(b) When the mass becomes 4 times
\(\cfrac { F }{ { F }^{ ' } } =\cfrac { GMm }{ { R }^{ 2 } } \times \cfrac { { R }^{ 2 } }{ 4GMm } \)
\(\Rightarrow\) F' = 4F