1.The ground water is constantly used for irrigation and for human use sufficient rainwater is not avail to recharge the ground water.
This situation has occured due to deforestation and axing the trees in urban and rural areas Trees facilitate to seep rainwater into sail that reaches to ground water reserviour .Trees check the speed of flowing rainwater .Thus,they provide sufficient time to perculate in to the soil.
2. We can reground water bycharge  
(a) afforestion planting more trees in urban and rural areas and (b) resitriction  on cutting trees.
(b) Building small chevk-dams which stop the rainwater from flowingaway and lead to an increase in ground water levels.
3. (i) We can organise campaigns for crediting  awarness among masses about the value of ground water level, methods that can be used to recharge ground water level,methods(ii) that can be used to recharge ground water and stop depletion of ground water.
(ii) With the help of teaches we can make soak pits to recharge ground water with the rainwater.

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.

(i) Vaccination against small pox.
(ii) BCG vaccination against tuberculosis.
(iii) Polio drops against polio disease
(iv) Vaccination against chicken pox.
(v) Vaccination against Hepatitis
(vi) DPT vaccination against dipheria, pertusis
(viii) Immunisation against measles.
Major health problems are:
(i) Hepatites
(ii) Chicken pox
(iii) Tuberculosis
(iv) Tetanus.

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.

(i) \(u=0,\quad g=-9.8\quad m/s^{ 2 },\quad s=-4.9m\) 
\(s=ut+\frac { 1 }{ 2 } { gt }^{ 2 }\) 
\(\therefore -4.9=0-\frac { 1 }{ 2 } \times 9.8\times { t }^{ 2 }\quad or\quad t=\sqrt { \frac { 2\times 4.9 }{ 9.8 } } =1\ s.\) 
(ii) \(u=0,\quad s=-4.9m,\quad g=-9.8\quad m/s^{ 2 },\) 
\({ v }^{ 2 }-{ u }^{ 2 }=2gs\) 
\(\therefore \ { v }^{ 2 }-0=2\times (-9.8)\times (-4.9)\quad or\quad v=\sqrt { 2\times 9.8\times 4.9 } =9.8\ m/s.\) 
(iii) \(u=0,\quad g=-9.8\quad m/s^{ 2 },\quad s--7.9m\)
\(\because { v }^{ 2 }-{ u }^{ 2 }=2gs\) 
\(\\ \therefore { \ v }^{ 2 }-0=2\times (-9.8)\times (-7.9)\) 
\(\\ v=\sqrt { 2\times 9.8\times 7.9 } m/s=10.46\quad m/s\)
(iv) Acceleration after 1 s and after 2 s will be 9.8 m/s² , because acceleration due to gravity remains almost same for small heights.

(a)  Measurement of force from Newton's second law. Suppose a force F acts on a body of mass m and changes its velocity from u to v in t seconds. Then
Initial momentum of the body, \({p}_{1}=mu\)
Final momentum of the body,  \({p}_{2}=mu\)
Change of momentum  \(= {p}_{2}-{p}_{1}=mv-mu=m\left(v-u\right)\)
Time taken \(= t\)
\(\therefore\) Rate of change of momentum  \(= \frac {Change \ of\ momentum} {time \ taken} = \frac {m\left(v-u\right)}{t}=ma\) 
Where a  is the acceleration of the body.
According to Newton's second law, the rate of change of momentum is directly proportional to the applied force, so
 \(F \infty \ ma.\)             or F = kma
Where k is constant. The unit of force is so chosen that k is equal to one. If m = 1, a = 1 and F = 1, then
1 = k. 1. 1    or  k = 1
F = ma     or   \(Force = Mass \times Acceleration.\)  
So a unit force is that force which produces a unit acceleration in a body of unit mass.
(b) The S.I. unit of force is Newton. One Newton is that force which produces an acceleration of \(1 m/ {s}^{2}\) in a body of mass 1
1 newton \(= 1 kg \times 1 m/s^2\) or \(1N = 1 kg m/s^2\)  
Thus, the second law of motion gives us a method to measure force. If the mass and acceleration of a body are known, we can determine the force acting on it.

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.

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.

(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

(i)Cathode rays travel in straight lines and thus cast shadows of objects placed in their path.
(ii)Cathode rays possess material particles because they can rotate a light paddle wheel placed in their path.
(iii)They are deflected towards positive plate thus showing that theses are negatively charged particles knowns as electrons
(iv)They ionise  gas through which they pass
(v)They are deflected by magnetic fields.
(vi)The nature of cathode rays is independent of the material of cathode.Hence they are common constituents of all matter
(vii)They can penetrate through thin metallic sheet
(viii)They can produce X-rays
(ix)The mass of a cathode ray particle is very-very small as compared to the mass of the atom from which it is formed

(a) Water at room temperature is a liquid because it has fluidity and assumes the shape of the containing vessel.
(b) An almirah is a solid because it is rigid and has a fixed shape.

Given, initial velocity,u = 20 m/s
Maximum height that the ball will reach, h = ?
From the third equation of motion, v² = u² + 2gh
[\(\therefore\) at maximum height, v = 0]
\(\Rightarrow \quad h=\cfrac { -u^{ 2 } }{ 2g } =\cfrac { -\left( 20 \right) ^{ 2 } }{ 2\left( -10 \right) } =\cfrac { 400 }{ 20 } =20m\)
This means that ball will reach the height of20 m and comes back. It will pass the window rwo times. (1/2)
Now, to calculate the time that ball will take to reach 15 m height.
From the second equation of upward motion,
\(h=ut-\cfrac { 1 }{ 2 } gt^{ 2 }\) [.:g is in the downward direction]
15 = 20t \(\cfrac { 1 }{ 2 } \)(10)t²
\(\Rightarrow\) 5t²- 20t +15 =0 2
t² -  4t + 3 = 0
t² - 3t -t + 3 = 0
t (t -3) -1 (t -3)  = 0
(t - l) (t - 3) = 0
\(\Rightarrow\)   t = I,3
Thus, ball will pass the window at 1 s and 3 s, respectively

The steps used for the genetic improvement of plants are:
(i) Introduction: It is the transportation of crop plants from their native place of cultivation to the new place where they never grown earlier (introduction of new crop plants).
(ii) Selection of crop plants: This involves selection of some desirable characters for plants are high yield, resistance to diseases, pests, drought ect.
(iii) Hybridization: This is the process of crossing between genetically dissimilar plants to obtain improved varieties. The crossing may be
(a) intervarietal i.e., crossing between two varieties of a same species.
(b) interspecific i.e., cross between two different species of the same genus.
(c) intergeneric i.e., cross between plants of different genera.
In crop improvement, intervarietal hybridization (plant breeding is extremely useful).

A. Affect of addition of undesirable substances
(i) This may cause poisoning of water such as addition of mercury salt, lead, etc. cause the death of aquatic life such fishes.
(ii) Addition of estables, sewage, fertiliser, etc. cause growth of enormous algae and microbes that cause oxygen deficiency and death of aquatic life.
(iii) The polluted water may cause serious diseases in human beings, such as diahorrea, minamata (due to mercury salts) and many other diseases.
B. Change in temperature
(i) Affects the balance between various organism and may affect their breeding.
(ii) The eggs and larvae of various animals susceptible to temperature changes. For example, trout eggs fail to hatch at higher temperature.

(i) Immediate cause of disease is Infectious organisms like virus, bacteria, fungi, protozoan etc.For example, the cause of loose motions can be an infection with a virus, Here viruses is the immediate cause of disease.
(ii) Contributory cause of disease:
(a) poor health due to lack of good nourishment.
(b) poor economic condition due to which one does not get proper and sufficient food and lives in unhygienic conditions.
(c) genetic defects which cause some people more prone to disease than others.
(d) Lack of public services.

Characteristics of wave motion:
(i) It is the disturbance which travels forward through the medium and not the particles of the medium, the particles of the medium merely vibrate about their mean positions.
(ii) Each particle receives vibrations a little later than its preceding particle.
(iii) The velocity with which wave travels is different from the velocity of the particles with which they vibrate about their mean positions.
(iv) The wave velocity remains constant in a given medium while the particle velocity changes continuously during its vibration about mean position.

(a) The above table shows that the motion is accelerated and acceleration is increasing uniformly with time.
(b) As the acceleration is increasing uniformly, the force acting on the body is also increasing uniformly with time.

Let the one way distance =X km
Time taken in the forward journey at a speed of 50 km/h = \(\frac { Distance }{ Speed } =\frac { X }{ 50 } h\)
Time taken in the return journey at a speed of 40 km/h = \(\frac { X }{ 40 } h\)
Total time for the whole journey = \(\frac { x }{ 50 } +\frac { x }{ 40 } =\frac { 4x+5x }{ 200 } =\frac { 9x }{ 200 } h\)
Total distance covered = x + x = 2 x  km
\(\therefore \) Average speed = \(\frac { Total\quad distance }{ Total\quad time } \)
\(=\frac { 2x }{ 9x } =\frac { 2xX200 }{ 9x } =\frac { 400 }{ 9 } =\) 44.44 km/h.

(i) Difference between Meristematic cells and Permanent cells

(ii) Difference between Parenchyma and Collenchyma

J.J.Thomson. Since the discharge tube experiment suggested the presence of negatively charged particles in an atom that is neutral, J.J.Thomson suggested that electrons are embedded in a sphere of positive charge.
E.Rutherford- \(\alpha\)-ray scattering experiments on gold foil suggested that all the positive charge is located in a very small space which is 10^-5 times the radius of an atom. Therefore, Rutherford gave a model in which electrons are revolving around the nucleus.
Neils Bohr. Since charged bodies moving in circular motion emit radiations. This will lead to loss of energy of the moving electron and ultimately giving an unstable model of an atom. To explain the stability of atom and atomic spectra, Bohr suggested that electrons are moving around the nucleus in orbits that have fixed energy shells. There is a loss or gain in energy of an electron only when it moves from one orbit to the other.

The solubility of potassium nitrate is 62 g at 313 K. it means it is the max. amount soluble at 313 K in 100 g of solution.
Amount of salt = 62 g
Amount of water = 100 - 62 = 38 g
Amount of KNO₃ in 50 g water = \({62\over38}\times {50\over1}={81.58 g}\)
The solubility of a salt increase with rise in temperature.

Desirable agronomic characteristics in crop plants help give higher yield (i.e., higher productivity). It varies from crop to crop. For example, desirable characters for fodder crops are tallness and profuse branching, For cereal crops dwarfness is desired character so that lessnutrients are consumed by these crops.

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.

Different
Reason: (i) When we think about disease, we think about individual sufferers.But when we think about health, we think about societies and communities.
(ii) It is possible to be in poor health without suffering from a particular disease.This is particularly true of social and mental health. 

Relationship between frequency, wavelength and wave velocity. Since wavelength is the distance traveled by the wave during the time a particle of the medium completes one vibration, therefore if  \(\lambda \) be the wavelength and T the time-period, then the wave travels a distance \(\lambda \) in time T. Hence \(Wave\quad velocity=\frac { Distance }{ Time } \) or \(v=\frac { \lambda }{ T } \)
or  \(\upsilon =v\lambda \) \(\left[ \because \quad \frac { 1 }{ T } =frequency\quad \left( v \right) \right] \)
∴ Wave velocity=Frequency x Wavelength.
The wave velocity in a number remains constant under the same physical conditions.

(i) The reaction force of the ground on the feet of the horse.
(ii) The force which the horse exerts on the cart.

Here, v = 90 km/h = \(\frac { 90X1000m }{ 3600s } =\) 25 m/s
s = 100 m
Minimum time, \(t=\frac { s }{ v } =\frac { 100 }{ 25 } =4s\)

(a) Snake has scales on the skin, has bones and crawls making lateral loops. The body of a snake is not segmented like earthworm.
(b) They loosen the soil and supplement the work of the farmer's ploughing. They increase the fertility of the soil. Its burrow make the way for air to plant roots. So, earthworms are regarded a great friend of the agriculturists.
(c) Because he prevented cruelty toward animals and also helped conserving biodiversity.

Meristematic cells have:
(i) Thin cell walls.
(ii) Abundant or dense cytoplasm and single large nucleus.
(iii) Spherical, oval, polygonal or rectangular shape.
(iv) No intercellular spaces between them.
(v) Either no vacuoles at all or a few vacuoles.

Mass of one electron = 9.1x10^-31kg
No. of electrons in 1 kg \(={1\over 9.1}\times10^{31}\)
No. of electrons in 30 kg = \({30\over 9.1}\times10^{31}=3.3\times10^{31}\)
Number of electrons in 30 kg mass is much larger than the population of India (10⁹).

Elements:
(i) Sodium,
(ii) hydrogen 
(viii) Chlorine 
(xii) iron fillings,
(xiii) oxygen
Compounds:
(iv) Carbon dioxide 
(vii) Water,
(x) Common salt 
(xvi) Ammonia
(xviii) pure marble,
(xxii) Sugar
Mixtures:
(iii) brass,
(v) Gunpowder,
(vi) Soda water,
(ix) Air, 
(xi) Stainless steel
(xiv) Milk
(xv) blood,
(xvii) Glass
(xix) emulsion
(xx) Wood.

The specific function of cardiac muscle is to contract and relax rhythmically throughout life.

(a) Gravitational force acting on the 50 kg,
mg= 50x 9.8 = 490N 
(b) Gravitational force acting on the 50 kg mass due to jupiter,
\({ F }_{ jupiter }=\cfrac { G\times { M }_{ jupiter }\times { { M }_{ person } } }{ \left( distance\quad of\quad jupiter\quad from\quad the\quad earth \right) ^{ 2 } } \)
\({ F }_{ jupiter }=\cfrac { 6.67\times { 10 }^{ -11 }\times 2\times { 10 }^{ 27 }\times 50 }{ 6.3\times { 10 }^{ 11 }\times 6.3\times { 10 }^{ 11 } } \)
\({ F }_{ jupiter }=\cfrac { 6.67\times 2\times 50\times { 10 }^{ -11+27-22 } }{ 6.3\times 6.3 } \)
F_Jupiter = 1.68 X 10^-5 N
\({ F }_{ saturn }=\cfrac { G\times M_{ saturn }\times { M }_{ person } }{ \left( distance\quad of\quad saturn\quad from\quad the\quad earth \right) ^{ 2 } } \)
\({ F }_{ saturn }=\cfrac { 6.67\times { 1 }0^{ -11 }\times 6\times { 10 }^{ 26 }\times 50 }{ 1.28\times 10^{ 12 }\times 1.28\times 10^{ 12 } } \)
\({ F }_{ saturn }=\cfrac { 6.67\times 6\times 50 }{ 1.28\times 1.28 } \times { 10 }^{ -11+26-24 }\)
F_saturn = 0.12x 10^-5N
\(\therefore\) Total gravitational force due to the jupiter and the saturn = (1.68x 10^-5+0.12x 10^-5)N
= 1.8x 10^-5 N
Thus, the combined force due to the planets jupiter and saturn (1.8 x10^-5) N is negligible as compared to the gravitational force due to the earth.
(ii) We know that, g at equator is less than g at poles (Antarctica). Thus, weight at equator is less than weight at pole (Antarctica). A bag of sugar weighs w at a certain place on the equator. If this bag is taken to Antarctica, then it will weigh more due to greater value of g.

a = 8 \(ms^{ -2 }\),v = 32 \(ms^{ -1 }\) 
(ii) t = \(2\sqrt { 2\quad s } \)