Formula: The moment of a force M = F × d
Given: F = 40 N and d = 90 cm = 0.9 m.
Hence, moment of the force = 40 × 0.9 = 36 N m.

Compound microscope:
A Compound microscope is used to see the tiny objects has better magnification power than simple microscope.
Construction:
(i) A compound microscope consists of two convex lenses.
(ii) The lens with the shorter focal length is placed near the object, and is called as 'objective lens' or 'objective piece'.
(iii) The lens with larger focal length and larger aperture placed near the observer's eye is called as 'eye lens' or 'eye piece'.
(iv) Both the lenses are fixed in a narrow tube with adjustable provision.
Working:
(i) The object (AB) is placed at a distance slightly greater than the focal length of objective lens \(\left(u>f_{0}\right)\).
(ii) A real, inverted and magnified image \(\left(\mathrm{A}^{\prime} \mathrm{B}^{\prime}\right)\) is formed at the other side of the objective lens.
(iii) This image behaves as the object for the eye lens.
(iv) The position of the eye lens is adjusted in such a way, that the image (A' B') falls within the principal focus of the eye piece.
(v) This eye piece forms a virtual, enlarged and erect image (A" B") on the same side of the object.
(vi) Compound microscope has 50 to 200 times more magnification power than simple microscope.

(i) Human beings have 23 pairs of chromosomes out of which 22 pairs are autosomes and one pair (23^rd pair) is the sex chromosome.
(ii) The female gametes or the eggs formed are similar in their chromosome type (22+XX).
(ii) Therefore, human females are homogametic.

(iv) The male gametes or sperms produced are of two types.
(v) They are produced in equal proportions.
(vi) The sperm bearing (22+X) chromosomes and the sperm bearing (22+Y) chromosomes.
(vii) The human males are called heterogametic.
(v) It is a chance of probability as to which category of sperm fuses with the egg.
(vi) If the egg (X) is fused by the X-bearing sperm an XX individual (female) is produced.
(vii) If the egg (X) is fused by the Y-bearing sperm an XY individual (male) is produced.
(viii) The sperm, produced by the father, determines the sex of the child. The mother is not responsible in determining the sex of the child.
(ix) Now let's see how the chromosomes take part in this formation.
(x) Fertilization of the egg (22+X) with a sperm (22+X) will produce a female child (44+XX) while fertilization of the egg (22+X) with a sperm (22+Y) will give rise to a male child (44+XY).

(i) Dihybrid cross involves the inheritance of two pairs of contrasting characteristics (or contrasting traits) at the same time.
(ii) The two pairs of contrasting characteristics chosen by Mendel were shape and colour of seeds: round-yellow seeds and wrinkled-green seeds.
(iii) Mendel crossed pea plants having round - yellow seeds with pea plants having wrinkled green seeds.
Mendel made the following observations:
(i) Mendel first crossed pure breeding pea plants having round-yellow seeds with pure breeding pea plants having wrinkled green seeds and found that only round yellow seeds were produced in the first generation (F₁).
(ii) No wrinkled-green seeds were obtained in the F₁ generation.
(iii) From this it was concluded that round shape and yellow colour of the seeds were dominant traits over the wrinkled shape and green color of the seeds.
(iv) When the hybrids of F₁ generation pea plants having round-yellow seeds were cross-breed by self pollination, then four types of seeds having different combinations of shape and colour were obtained in second generation or F₂ generation.
(v) They were round-yellow, round-green, wrinkled yellow and wrinkled-green seeds.
(vi) The ratio of each phenotype (or appearance) of seeds in the F₂ generation is 9:3:3:1. This is known as the Dihybrid ratio.
(vii) From the above results it can be concluded that the factors for each character or trait remain independent and maintain their identity in the gametes.
(viii) The factors are independent to each other and pass to the offsprings (through gametes).

Results of a Dihybrid Cross:
(i) Mendel got the following results from his dihybrid cross.
Four Types of Plants:
(i) A dihybrid cross produced four types of F₂ offsprings in the ratio of 9 with two dominant traits, 3 with one dominant trait and one recessive trait, 3 with another dominant trait and
another recessive trait and 1 with two recessive traits.
New Combination:
(i) Two new combinations of traits with round green and wrinkled yellow had appeared in the dihybrid cross (F₂ generation).

i) The top layers of soil contain humus and mineral salts, which are vital for the growth of plants.
ii) Removal of upper layer of soil by wind and water is called soil erosion.
iii) Soil erosion causes a significant loss of humus, nutrients and decrease the fertility of soil.
iv) The direct and primary effect of soil erosion is soil loss and nutrient leaching resulting in reduction of land productivity.
v) Annual floods causes damages to crops, property and lives.
vi) Deforested rain forest soil becomes dry and nutrient-deficient as there is no longer vegetation to hold water and nutrients in place.
vii) Heavy rains further erode soil and saturate waterways with excess nutrients, disrupting the food chains of tropical ecosystems.
viii) Eroded sediments can even change the course of rivers, which suffer from huge deposits of silt from deforestation.
ix) Desertification is another possible consequence of erosion.

Sources of solid waste:
a) Municipal wastes
b) Hospital wastes
c) Industrial wastes
d) e - wastes
Solid waste management:
Solid-waste management involves the collection, treatment and proper disposing of solid material that is discarded from the household and industrial activities.
Methods of solid wastes disposal:
i) Segregation:
It is the separation of different type of waste materials like biodegradable and non biodegradable wastes.
ii) Sanitary landfill:
a) Solid wastes are dumped into Iow lying areas.
b) The layers are compacted by trucks to allow settlement.
c) The waste materials get stabilized in about 2 - 12 months.
d) The organic matter undergoes decomposition.
iii) Incineration:
It is the burning of non-biodegradable solid wastes (medical wastes) in properly constructed furnace at high temperature.
iv) Composting:
Biodegradable matter of solid wastes is digested by microbial action or earthworms and converted into humus.

(i) Application of gibberellins on plants stimulate extraordinary elongation of internode. E.g: Corn and Pea.
(ii) Treatment of rosette plants with gibberellin induces sudden shoot elongation followed by flowering. This is called bolting.
(iii) Gibberellins promote the production of male flowers in monoecious plants (Cucurbits).
(iv) Gibberellins break dormancy of Potato tubers.
(v) Gibberellins are efficient than auxins in inducing the formation of seedless fruit - Parthenocarpic fruits (Development of fruits without fertilization) E.g: Tomato.

(i) Insulin used in the treatment of diabetes.
(ii) Human growth hormone used for treating children with growth deficiencies.
(iii) Blood clotting factors are developed to treat haemophilia.
(iv) Tissue plasminogen activator is used to dissolve blood clots and prevent heart attack.
(v) Development of vaccines against various diseases like Hepatitis B and rabies.

The reproductive period is marked by characteristic events repeated almost every month in physiologically normal woman (28 days with minor variation) in the form of a menstrual flow
The menstrual cycle consists of 4 phases:
(i) Menstrual or Destructive phase 
(ii) Follicular or Proliferative Phase
(iii) Ovulatory Phase
(iv) Luteal or Secretory Phase 

The neurons may be of different types based on their structure and functions.

Structurally the neurons may be of the following types:
(i) Unipolar neurons
(ii) Bipolar neurons
(iii) Multipolar neurons
(i) Unipolar neurons:
i) Only one nerve process arises from the cyton which acts as both axon and dendron. 
ii) Unipolar neurons found in early embryos but not in adults.
(ii) Bipolar neurons:
i) The cyton gives rise to two nerve processes of which one acts as an axon while another as a dendron. 
ii) Bipolar neurons found in retina of eye and olfactory epithelium of nasal chambers.
(iii) Multipolar neurons:
i) The cyton gives rise to many dendrons and an axon. 
ii) Multipolar neurons found in cerebral cortex of brain.

A human brain is formed of three main parts:
(a) forebrain
(b) midbrain and
(c) hindbrain.
(1) Forebrain:
A. Cerebrum(cortex)
B. Diencephalon
i) Thalamus ii) Hypothalamus
(2) Midbrain
(3) Hindbrain:
A. Cerebellum
B. Pons
C. Medulla Oblongata

WBC's are colourless and nucleated cells. Granulocytes contain granules in their cytoplasm. Agranulocytes do not contain granules in their cytoplasm, So they are classified as Granulocytes and Agranulocytes.
Granulocytes: Granulocytes contain granules in their cytoplasm. They are of three types.
(i) Neutrophils:
i) They are large in size and have a 2-7 lobed nucleus.
ii) These corpuscles form 60% - 65% of the total leucocytes.
iii) Their number are increased during infection and inflammation.
(ii) Eosinophils:
i) Eosinophils has a bilobed nucleus and constitute 2% - 3% of the total leucocytes.
ii) Their number increases during conditions of allergy and parasitic infections.
iii) It brings about detoxification of toxins.
(iii) Basophils:
i) Basophils have lobed nucleus.
ii) They form 0.5% - 1.0 % of the total leucocytes. 
iii) They release chemicals during the process of inflammation.
(iv) Agranulocytes:
i) Granules are found in the cytoplasm of these cells.
ii) The Agranulocytes are of two types:
(i) Lymphocytes:
i) These are about 20 - 25% of the total leucocytes.
ii) They produce antibodies during bacterial and viral infections.
(ii) Monocytes:
i) They are the largest of the leucocytes and are amoeboid in shape.
ii) These cells form 5 - 6% of the total leucocytes.
iii) They are phagocytic and can engulf bacteria.

Yes. Lifestyle can be modified so as the risk factors of cardio vascular diseases can be avoided. Thereby we can avoid heart diseases.
The following measures can be advocated to change life style:
Diet management:
(i) Reduction in the intake of calories, low saturated fat and cholesterol rich food, low carbohydrates and common salt are some of the dietary modifications.
(ii) Diet rich in polyunsaturated fatty acids (PUFA) is essential.
(iii) Increase in the intake of fiber diet, fruits and vegetables, protein, minerals and vitamin are required.
Physical activity:
(i) Regular exercise, walking and yoga are essential for body weight maintenance.
(ii) There should be some physical activity in between long sitting hours.
Addictive substance avoidance:
Alcohol consumption and smoking are to be avoided.

A Nuclear reactor is a device in which the nuclear fission reaction takes place in a self-sustained and controlled manner to produce electricity.
(i) Fuel:
a) A fissile material is used as fuel.
b) The commonly used fuel material is uranium
(ii) Moderator:
a) A moderator is used to slow down the high-energy neutrons to get slow neutrons.
b) Graphite and heavy water are the commonly used moderators.
(iii) Control rod:
a) Control rods are used to control the number of neutrons in order to have a sustained chain reaction.
b) Mostly boron or cadmium rods are used as control rods.
(iv) Coolant:
a)A coolant is used to remove the heat produced in the reactor core to produce steam.
b) This steam is used to run a turbine to produce electricity.
c) Water, air and helium are some of the coolants.
(v) Protection wall:
 A thick concrete lead wall is built around the nuclear reactor in order to prevent the harmful radiations from escaping into the environment.

1)  0.3 mole of aluminium (Atomic mass of Al = 27)
\(\text { Number of moles }=\frac{\text { Mass of } \mathrm{Al}}{\text { Atomic mass of } \mathrm{Al}}\)
Mass = No. of moles × atomic mass
So, mass of Al = 0.3 × 27
= 8.1 g
2) 2.24 litre of SO₂ gas at S.T.P
 Molecular mass of SO₂ = 32 + (16 × 2)
= 32 + 32 = 64
\(\text { Number of moles of } \mathrm{SO}_2=\frac{\begin{array}{c} \text { Given volume of } \mathrm{SO}_2 \text { at S.T.P } \end{array}}{\begin{array}{c} \text { Molar volume } \mathrm{SO}_2 \text { at S.T.P } \end{array}}\)
Number of moles of SO₂ = \(\frac{2.24}{22.4}\)
= 0.1 mole
\(\text { Number of moles }=\frac{\text { Mass }}{\text { Molecular mass }}\)
Mass = No. of moles × molecular mass
Mass = 0.1 × 64
Mass of SO₂ = 6.4 g
iii) 1.51 × 10²³ molecules of water
Molecular mass of H₂O = 18
\(\text { Number of moles }=\frac{\begin{array}{c} \text { Number of molecules of }\text { water } \end{array}}{\text { Avogadro's number }}\)
= 1.51 × 10²³ / 6.023 × 10²³
= 1 / 4
= 0.25 mole
\(\text { Number of moles }=\frac{\text { Mass }}{\text { Molecular mass }}\)
0.25 = mass / 18
Mass = 0.25 × 18
Mass = 4.5 g
iv) 5 × 10²³ molecules of glucose
Molecular mass of glucose = 180
\(\text { Mass of glucose }=\frac{\begin{array}{c} \text { Molecular mass } \times \text { number of particles } \end{array}}{\text { Avogadro's number }}\)
= (180 × 5 × 10²³) / 6.023 × 10²³
= 149.43 g

Echo:
(i) An echo is the sound reproduced due to the reflection of the original sound from various rigid surfaces such as walls, ceilings, surfaces of mountains, etc.
a) Conditions necessary for hearing echo:
(i) The persistence of hearing for human ears is 0.1 second.
(ii) This means that you can hear two sound waves clearly, if the time interval between the two sounds is at least 0.1 s.
(iii) Thus, the minimum time gap between the original sound and an echo must be 0.1 s.
(iv) The above criterion can be satisfied only when the distance between the source of sound and the reflecting surface would satisfy the following equation:
\(\text { Velocity } =\frac{\text { distance travelled by sound }}{\text { time taken }}  \)
\(v =\frac{2 d}{t}  \)
\(d =\frac{v t}{2}\)
since, t = 0.1 second, then \(\mathrm{d}=\frac{331}{0.20}=\frac{\mathrm{v}}{20}\)
(v) Thus the minimum distance required to hear an echo is 1 / 20^th part of the magnitude of the velocity of sound in air.
(vi) If you consider the velocity of sound as 344 ms^-1, the minimum distance required to hear an echo is 17.2 m.
b) Applications of echo:
(i) The principle of echo is used in obstetric ultrasonography, which is used to create real-time visual images of the developing embryo or fetus in the mother's uterus.
(ii) This is a safe testing tool, as it does not use any harmful radiations.
c) Calculation of speed of sound:
(i) The sound pulse emitted by the source travels a total distance of 2d while travelling from the source to the wall and then back to the receiver.
(ii) The time taken for this has been observed to be 't'. Hence, the speed of sound wave is given by
\(\text { Speed of Sound }=\frac{\text { distance travelled }}{\text { time taken }}=\frac{2 \mathrm{d}}{\mathrm{t}}\)

i) The electricity produced in power stations is distributed to all the domestic and industrial consumers through overhead and underground cables.
ii) In our homes, electricity is distributed through the domestic electric circuits wired by the electricians.
iii) The first stage of the domestic circuit is to bring the power supply to the main-box from a distribution panel, such as a transformer.
Main Box Contains:
a) Fuse Box:
i) The fuse box contains either a fuse wire or a miniature circuit breaker (MCB).
ii) The function of the fuse wire or a MCB is to protect the house hold electrical appliances from overloading due to excess current.
iii) An MCB is a switching device, which can be activated automatically as well as manually. It has a spring attached to the switch, which is attracted by an electromagnet when an excess current passes through the circuit.
iv) It has a spring attached to the switch, which is attracted by an electromagnet when an excess current passes through the circuit.
v) Hence, the circuit is broken and the protection of the appliance is ensured.
b) Meter:
i) The meter is used to record the consumption of electrical energy.
Insulated Wire:
i) The electricity is brought to houses by two insulated wires.
ii) Out of these two wires, one wire has a red insulation and is called the "live wire'.
iii) The other wire has a black insulation and is called the 'neutral wire'.
iv) Both, the live wire and the neutral wire enter into a box where the main fuse is connected with the live wire.
v) After the electricity meter, these wires enter into the main switch, which is used to discontinue the electricity supply whenever required.
vi) After the main switch, these wires are connected to live wires of two separate circuits.
5A rating circuit:
i) Out of these two circuits, one circuit is of a 5 A rating, which is used to run the electric appliances with a lower power rating, such as tube lights, bulbs and fans.
15 A rating circuit:
i) The other circuit is of a 15 A rating, which is used to run electric appliances with a high power rating, such as air-conditioners, refrigerators, electric iron and heaters.
ii) It should be noted that all the circuits in a house are connected in parallel, so that the disconnection of one circuit does not affect the other circuit.
iii) One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.
iv) The electricity supplied to your house is actually an alternating current having an electric potential of 220 V.

a) Joule's law of heating:
a) Joule's law of heating states that the heat produced in any resistor is:
(i) Directly proportional to the square of the current passing through the resistor.
(ii) Directly proportional to the resistance of the resistor.
(iii) Directly proportional to the time for which the current is passing through the resistor.
b) Alloy of nickel and chromium have the following properties:
(i) It has high resistivity,
(ii) It has a high melting point,
(iii) It is not easily oxidized.
c) The fuse wire is connected in series, in an electric circuit. When a large current passes through the circuit, the fuse wire melts due to Joule's heating effect and hence the circuit gets disconnected. Therefore, the circuit and the electric appliances are saved from any damage. The fuse wire is made up of a material whose melting point is relatively low.

(i) Propulsion of rockets is based on law of conservation of linear momentum as well as Newton's III law of motion.
(ii) Rockets are filled with a fuel (either liquid or solid) in the propellant tank.
(iii) When the rocket is fired, this fuel is burnt and a hot gas is ejected with high speed from the back nozzle producing a huge momentum.
(iv) To balance this momentum, an equal and opposite reaction force is produced combustion chamber which makes the rocket project forward.
(v) While in motion, the mass of the rocket gradually decreases, until the fuel is completely burnt out.
(vi) Since there is no net external force acting on it, the linear momentum of the system is conserved.
(vii) The mass of the rocket decreases with altitude, which results in the gradual increase in velocity of the rocket.
(viii) At one stage, it reaches a velocity, which is sufficient to just escape from the gravitational pull of the Earth. This velocity is called escape velocity.

Here, although the solution is extremely dilute, the concentration given is not of an acid or a base but that of H⁺ ions. Hence, the pH can be calculated from the relation:
pH = –log₁₀[H⁺]
given [H⁺] = 1.0 × 10^–8 mol litre^–1
pH = –log₁₀10^–8 = –(–8 × log₁₀10)
= –(–8 × 1) = 8

NaOH is a strong base and dissociates in its solution as:
NaOH_(aq) → Na⁺_(aq) + OH^–_(aq)
One mole of NaOH would give one mole of OH^– ions. Therefore,
[OH^–] = 1 × 10^–4 mol litre^–1
pOH = –log₁₀[OH^–] = –log₁₀ × [10^–4]
= –(–4 × log₁₀10)= –(–4) = 4
Since, pH + pOH = 14
pH = 14 – pOH = 14 – 4
= 10

(i) ₁₃AI²⁷ + ₂He⁴  \(\longrightarrow \) ₁₅P³⁰ + ₀n¹
(ii) ₁₂Mg²⁴ + ₀n¹  \(\longrightarrow \) ₁₁Na²⁴ + ₁H¹
(iii) ₉₂U²³⁸ + ₀n¹ \(\longrightarrow \) ₉₃Np²³⁹ + _-1e⁰
A is alpha particle, B is neutron, C is proton, and D is electron.

At 30°C, 36 g of sodium chloride is dissolved in 100 g of water.
\(\therefore\) Mass of sodium chloride that would be need for 100 g of water = 36 g
\(\therefore\) Mass of sodium chloride dissolved in 50 g of water = \(\frac {36 \times 50}{100}\)
=18 g

Mass of the solute = 1.5 g
Mass of the solvent = 15 g
Solubility of the solute = \(\frac{\text { Mass of the solute }}{\text { Mass of the solvent }} \times 100\) 
Solubility of the solute = \(\frac{1.5}{15}\times 100\) 
= 10 g

A) Since the resistors are connected in parallel, the potential difference across each resistor is same (i.e. V=10V)
Therefore, the current through R₁ is,
\({ I }_{ 1 }=\frac { V }{ { R }_{ 1 } } =\frac { 10 }{ 5 } =2A\)
Current through \({ R }_{ 2 }={ I }_{ 2 }=\frac { V }{ { R }_{ 2 } } =\frac { 10 }{ 10 } =1A\)
Current through \({ R }_{ 3 }={ I }_{ 3 }=\frac { V }{ { R }_{ 3 } } =\frac { 10 }{ 20 } =0.5A\)
B) Total current in the circuit, I = I₁ + I₂ + I₃
= 2 + 1 + 0.5 = 3.5 A
C) Total resistance in the circuit  \(\frac { 1 }{ { R }_{ P } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \)
\(=\frac { 1 }{ 5 } +\frac { 1 }{ 10 } +\frac { 1 }{ 20 } \)
\(=\frac { 4+2+1 }{ 20 } \)
\(\frac { 1 }{ { R }_{ P } } =\frac { 7 }{ 20 } \)
Hence, \({ R }_{ P }=\frac { 20 }{ 7 } =2.857\Omega \) 

Given that, d = 1.5m; D = 25cm = 0.25m (For a normal eye).
From equation (2.8), the focal length of the correction lens is
\(f=\cfrac { d\times D }{ d-D } =\cfrac { 1.5\times 0.25 }{ 1.5-0.25 } =\cfrac { 0.375 }{ 1.25 } =0.3m\)

When the source is moving towards the stationary listener, the expression for apparent frequency is
\(n'=\left( \frac { v }{ v-{ v }_{ s } } \right) n\)
= \(\left( \frac { v }{ v-\left( \frac { 1 }{ 10 } \right) v } \right) n=\left( \frac { 10 }{ 9 } \right) n\)
= \(\left( \frac { 10 }{ 9 } \right) \times 90=100\) Hz
n'=100 Hz

Given:
\(\mathrm{f}=10 \mathrm{~cm}, \mathrm{u}=-20 \mathrm{~cm}, \mathrm{v}=?  \)
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}  \Rightarrow \frac{1}{v} =\frac{1}{f}+\frac{1}{u} \)
\(\frac{1}{v} =\frac{1}{10}+\frac{1}{-20}=\frac{1}{10}-\frac{1}{20}  \)
\(\frac{1}{v} =\frac{2-1}{20}=\frac{1}{20}  \)
\(\mathbf{v} =20 \mathrm{~cm}  \)
Image distance = 20 cm
Nature of image:
Nature of the image is real, enlarged and inverted image.

Data:
Initial pressure (P₁)= P
Final Pressure (P₂) = 4P
Initial volume (V₁) = 20cc = 20cm³
Final volume (V₂) = ?
Using Boyle's Law, PV = constant
P₁V₁ = P₂V₂
\({ V }_{ 2 }=\frac { { P }_{ 1 } }{ { P }_{ 2 } } \times { v }_{ 1 }\)
\(=\frac { P }{ 4P } \times 20{ cm }^{ 3 }\)
V₂ = 5 cm³

(i) The ideal gas equation is an equation, which relates to all the properties of an ideal gas.
(ii) An ideal gas obeys Boyle's law and Charles's law and Avogadro's law.
(iii) According to Boyle's law,
PV = constant ... (1)
(iv) According to Charles's law,
V/T = constant ... (2)
(v) According to Avogadro's law,
V/n = constant ... (3)
(vi) After combining equations (1), (2), and (3), we can get the following equation.
PV/nT = constant ... (4)
The above relation is called the combined gas law
(vii) If you consider a gas which contains μ moles of the gas, the number of atoms contained will be equal to μ times the Avogadro number, N_A. i.e.,
\(n=\mu N_A\) ...... (5)
(viii) Using equation (5), equation (4) can be written as
\(PV/\mu N_AT\) = constant
(ix) The value of constant in the above equation is taken to be k_B, which is called as Boltzmann's constant. Its value is (1.381 x10^-23JK^-1). Hence, we have the following equation:
\(PV/\mu N_AT=K_B\)
\(PV=\mu N_AK_BT\)
(xi) Here, \(\mu N_AK_B=R\), which is termed as universal gas constant whose value is 8.31 J mol^-1 K^-1.
PV = RT ...... (6)
Ideal gas equation is also called as equation of state because it gives the relation between the state variables and it is used to describe the state of any gas.

(i) Polyuria occurs in people diagnosed with Diabetes mellitus, if blood glucose levels have risen too high.
(ii) Regular exercise, along with a good diet, can reduce the risk of diabetes.

Polyploldys are characterised by gigantism
An organism having more than two sets of chromosomes is called polyploid. It causes increase in size. Polys = Many + aploos = One fold + eidos = Form (E.g) Watermelon

(i) During transpiration, water is lost from leaves.
(ii) Still transpiration is essential for the movement of water and minerals from the root to the healthy parts of the plant.
(iii) But excess transpiration may result in drying up of the leaves or wilting and loss of soil water.
(iv) Hence it is termed as a necessary evil.

(i) Vinu will get faster dissolution of sugar.
(ii) Temperature is one of the factors.
(iii) It will affect the solubility of a solute in a liquid.
(iv) Solvent increases with increase in temperature.

(i) Octopus, cockroach and frog all have eyes.
(ii) Octopus-belongs to mollusc which have simple eye without lens.
(iii) Cockroach are invertebrate have compound eyes.
(iv) Frog which is vertebrate, have highly specialised lens.
(v) However, all of them perform the same function that is vision.
(vi) Thus a common evolutionary origin can be established on the basis of eye.

(i) Fossils are remains or impressions of organisms that lived in the remote past.
(ii) Fossil provide the evidence that the present animal have originated from previously existing ones through the process of continuous evolution.

(i) Canines are absent.
(ii) Hence, a gap is seen between the incisors and premolars.
(iii) This is called diastema.
(iv) It helps in mastication and chewing of food in herbivorous animals.

(i) On passing ' C ' through water it becomes acidic.
(ii) Therefore the gas ' C ' must be a non-metal oxide \(\left(\mathrm{CO}_{2}\right)\).
(iii) So a solid compound must be a calcium carbonate.
(iv) It decomposes into calcium oxide and carbon dioxide. (C)
\(\mathrm{CaCO}_{3(\mathrm{~g})} \rightarrow \mathrm{CaO}_{(\mathrm{S})}+\mathrm{CO}_{2(\mathrm{~g})} \uparrow\\ \quad \mathrm{A} \quad \quad \quad \quad \mathrm{B} \quad \quad \quad \quad \mathrm{C}\)

(a) Ionic bond.
(b) Electronegativity property.
(c) (i)  Along the period, from left to right in the periodic table, the electronegativity increases, because of the increase in the nuclear charge which in turn attracts the electrons more strongly.
(ii) On moving down a group, the electronegativity of the element decreases because of the increased number of valence shells.

The light-dependent reactions occur in the thylakoids (grana) and the light independent reactions (Calvin cycle) occur in the stroma.

In alpha decay atomic number decreases by two and mass number decreases by four. So
232 - 208 = 24
Mass number decreased by 24. So 6 alpha particle would come out.
\({ }_{90} \mathrm{X}^{232} \rightarrow{ }_{78} \mathrm{Y}^{208}+6{ }_{2} \mathrm{He}^{4}\)
In beta decay atomic number increases by one
78 + 4 = 82
So, 4 beta particle would come out.
\({ }_{78} \mathrm{Y}^{208} \rightarrow{ }_{82} \mathrm{Z}^{208}+4_{-1} \mathrm{e}^{\mathrm{o}}\)
So, six alpha particle and four beta particles would come out.

(i) 
(ii) IUPAC name - Butane-2-ol (or) 2-Butanol.
(iii) The compound is saturated, because these compounds are do not decolourize with bromine water.

b) Light
The light wave has a longer wave length. Because, it has much greater speed.

i) 1 mole of calcium carbonate are involved
ii) Gram molecular mass of \(\mathrm{CaCO}_{3} \rightarrow \mathrm{Ca} \times 1+\mathrm{C} \times 1+\mathrm{O} \times 3\)
40 x 1 + 12 x 1 + 16 x 3 
40 + 12 + 48 = 100 g
Gram molecular mass of CaCO₃ is 100 g / mol.
iii) 1 mole of CO₂ are there in above equation.