Heat will not flow. Because, the iron ball and water are at same temperature.

i) It results in fertilization which leads to the formation of fruits and seed.
ii) New varieties of plants are formed through new combination of genes in case of cross pollination.

Volume percentage = \(\frac{Volume \ of \ the \ solute}{Volume \ of \ the \ solution}\)
= \(\frac{7.5}{15}\) x 100 = 50%

Solubility is defined as the number of grams of a solute that is dissolved in 100 g of a solvent to form a saturated solution at a given temperature and pressure.

Characteristic studied	Dominant character	Recessive character
Seed shape	Round	Wrinkled
Seed colour	Yellow	Green
Seed-coat colour	Coloured	White

Gibberellin promote the production of male flowers in Cucurbits.

Hydra and yeast are the organisms which are reproduces through budding

(i) Aquatic animals live more in cold regions.
(ii) More amount of dissolved oxygen is present in the water of cold regions.
(iii) This shows that the solubility of oxygen in water is more at low pressure.

S. No	Ideal gas	Real gas
(i)	If the atoms or molecules of a gas do not interact with each other, then the gas is said to be an ideal gas or a perfect gas.	If the molecules or atoms of a gas interact with each other with a definite amount of intermolecular or interatomic force of attraction, then the gases are said to be a real gases.

(i) A spanner has a long handle to produce a larger moment of force by a small force applied normally at the end of its handle.
(ii) Moment of force = \(\overrightarrow{\mathbf{F}} \times \overrightarrow{\mathrm{d}}\)

DNA is a large molecule consisting of millions of nucleotides. Hence, it is also called a polynucleotide. Each nucleotide consists of three components.
(i) A sugar molecules - Deoxyribose sugar.
(ii) A nitrogenous base.
There are two types of nitrogenous bases in DNA.
They are
(a) Purines (Adenine and Guanine)
(b) Pyrimidines (Cytosine and Thymine)
(iii) A phosphate group
Nucleoside and Nucleotide:
Nucleoside = Nitrogen base + Sugar
Nucleotide = Nucleoside + Phosphate
The nucleotides are formed according to the purines and pyrimidines present in them.
Watson and Crick model of DNA:
(i) DNA molecule consists of two polynucleotide chains.
(ii) These chains form a double helix structure with two strands which run anti-parallel to one another.
(iii) Nitrogenous bases in the centre are linked to sugar-phosphate units which form the backbone of the DNA.
(iv) Pairing between the nitrogenous bases is very specific and is always between purine and pyrimidine linked by hydrogen bonds.
a) Adenine (A) links Thymine (T) with two hydrogen bonds (A = T)
b) Cytosine (C) links Guanine (G) with three hydrogen bonds( C ≡ G) This is called complementary base pairing.
(v) Hydrogen bonds between the nitrogenous bases make the DNA molecule stable.
(vi) Each turn of the double helix is 34 A^o (3.4 nm). There are ten base pairs in a complete turn.
(vii) The nucleotides in a helix are joined together by phosphodiester bonds.
Significance of DNA:
(i) DNA is responsible for the transmission of hereditary information from one generation to next generation.
(ii) It contains information required for the formation of proteins.
(iii) It controls the developmental process and life activities of an organism.

(i) The main part of the ovule is the nucellus which is enclosed by two integuments leaving an opening called as micropyle.
(ii) The ovule is attached to the ovary wall by a stalk known as funiculus.
(iii) Chalaza is the basal part.
(iv) The embryo sac contains seven cells and eight nuclei located within the nucellus.
(v) Three cells at the micropylar end form the egg apparatus and the three cells at the chalaza end are the antipodal cells.
(vi) The remaining two nuclei are called polar nuclei found in the centre.
(vii) In the egg apparatus one is the egg cell (female gamete) and the remaining two cells are the synergids.

Deliquescence substances	Hygroscopic substances
When exposed to the atmosphere at ordinary temperature, they absorb moisture and do not dissolve.	When exposed to the atmospheric air at ordinary temperature, they absorb moisture and dissolve.
Hygroscopic substances do not change its physical state on exposure to air.	Deliquescent substances change its physical state on exposure to air.
Hygroscopic substances may be amorphous solids or liquids.	Deliquescent substances are crystalline solids.
Hygroscopic substances are used as drying agents.	Deliquescent Substances are not used as drying agent.
Example : H2SO4.P2O5	Example : NaOH, KOH

There are three main factors which govern the solubility of a solute. They are:
a) Nature of the solute and solvent
b) Temperature
c) Pressure
a) Nature of the solute and solvent:
(i) The nature of the solute and solvent plays an important role in solubility.
(ii) Although water dissolves an enormous variety of substances, both ionic and covalent, it does not dissolve everything.
(iii) The phrase that scientists often use when predicting solubility is "like dissolves like".
(iv) The expression means that dissolving occurs when similarities exist between the solvent and the solute.
(v) For example: Common salt is a polar compound 4 and dissolves readily in polar solvent like water.
(vi) Non-polar compounds are soluble in non-polar solvents. For example: Fat dissolved in ether.
(vii) But non-polar compounds, do not dissolve in polar solvents; polar compounds do not dissolve in non-polar solvents.
b) Effect of Temperature:
Solubility of Solids in Liquid:
(i) Generally, solubility of a solid solute in a liquid solvent increases with increase in temperature.
(ii) For example, a greater amount of sugar will dissolve in warm water than in cold water.
(iii) In endothermic process, solubility increases with increase in temperature.
(iv) In exothermic process, solubility decreases with increase in temperature.
Solubility of Gases in liquid:
(i) Solubility of gases in liquid decrease with increase in temperature.
(ii) Generally, water contains dissolved oxygen.
(iii) When water is boiled, the solubility of oxygen in water decreases, so oxygen escapes in the form of bubbles.
(iv) Aquatic animals live more in cold regions because, more amount of dissolved oxygen is present in the water of cold regions.
(v) This shows that the solubility of oxygen in water is more at low temperatures.
c) Effect of Pressure:
(i) Effect of pressure is observed only in the case of solubility of a gas in a liquid.
(ii) When the pressure is increased, the solubility of a gas in liquid increases.
(iii) The common examples for solubility of gases in liquids are carbonated beverages, i.e. soft drinks, household cleaners containing aqueous solution of ammonia, formalin aqueous solution of formaldehyde, etc.

Statement:
i) There is no change in the linear momentum of a system of bodies as long as no net external force acts on them.
ii) Let us prove the law of conservation of linear momentum with the following illustration:

Proof:
(i) Let two bodies A and B having masses m₁ and m₂ move with initial velocity u₁ and u₂ in a straight line.
(ii) Let the velocity of the first body be higher than that of the second body. i.e., u₁ > u₂
(iii) During an interval of time t second, they tend to have a collision.
(iv) After the impact, both of them move along the same straight line with a velocity v₁ and v₂ respectively.
Force on body B due to A, F_A = m₂ (v₂ - u₂) / t
Force on body A due to B, F_B = m₁ (v₁ - u₁) / t
By Newton's III law of motion,
Action force = Reaction force
F_A = - F_B
m₁ (v₁ - u₁) / t = - m₂ (v₂ - u₂) / t
m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂
i) The above equation confirms in the absence of an external force, the algebraic sum of the momentum after collision is numerically equal to the algebraic sum of the momentum before collision.
ii) Hence the law of conservation of linear momentum is proved.

(i) According to Newton's second law, "the force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force".
(ii) This law helps us to measure the amount of force. So it is called as law of force'
(iii) Let, "m" be the mass of a moving body, moving along a straight line with an initial speed 'u'.
(iv) After a time interval of 't', the velocity of the body changes to 'v' due to the impact of an unbalanced external force F.
Initial momentum of the body, \( P_{i}=m u \)
Final momentum of the body, \( P_{f}=m v \)
Change in momentum, \( \Delta \mathrm{p}=\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}} \)
\(=m v-m u \)
By Newton's second law of motion,
\(\text {Force, } \ F \propto\) rate of change of momentum
\(\mathrm{F} \propto \) change in momentum / time
\(\mathrm{F} \propto \frac{\mathrm{mv}-\mathrm{mu}}{\mathrm{t}}\)
\(\mathrm{F}=\frac{\mathrm{km}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\)
Here, k is the proportionality constant.
k = 1 in all system of units. Hence,
\(F=\cfrac { m(v-u) }{ t } \)
Since acceleration = change in velocity / time, a = (v-u) / t. Hence, we have
F = m x a
Force = mass x acceleration

(i) Saturated solution:
1. A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution.
2. E.g. 36 g of sodium chloride in 100g of water at 25° C forms saturated solution.
3. Further addition of sodium chloride, leaves it undissolved.
(ii) Unsaturated solution:
Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature.
E.g. 109 or 20 g or 30 g of Sodium chloride in 100 g of water at 25° C forms an unsaturated solution.

Gears:
A gear is a circular wheel with teeth around its rim. It helps to change the speed of rotation of a wheel by changing the torque and helps to transmit power.
Seesaw:
(i) Most of you have played on the seesaw. Since there is a difference in the weight of the persons sitting on it, the heavier person lifts the lighter person.
(ii) When the heavier Person comes closer to the pivot point (fulcrum) the distance of the line of action of the force decreases.
(iii) It causes less amount of torque to act on it. This enables the lighter person to lift the heavier Person.
Steering Wheel:
A small steering wheel enables you to turn a car easily by transferring a torque to the wheels with less effort.

Auxins bring about a variety of physiological effects in different parts of the plant body. Some of them are discussed below:
(i) Auxins promote the elongation of stems and coleoptiles which makes them to grow.
(ii) Auxins induce root formation at low concentration and inhibit it at higher concentration.
(iii) The auxins produced by the apical buds suppress growth of lateral buds. This is called Apical Dominance.
(iv) Seedless fruits without fertilization are induced by the external application of auxins. (Parthenocarpy). Examples: Watermelon, Grapes, Lime etc.
(v) Auxins prevent the formation of abscission layer.

(i) Let M be the mass of the Earth and m be the mass of the object.
(ii) The entire mass of the earth is assumed to be concentrated at its centre. The radius of the earth is R (= 6378 km = 6400 km approximately).
(iii) By Newton's law of gravitation, the force acting on the object is given by
F = G M m / R²  ............. (A)
(iv) According to Newton's second law, the force acting on the object is given by the product of its mass and acceleration. Here acceleration of the is under action of gravity hence a =g.
F=ma =mg
F = weight = mg
Comparing equations (A) and (B), we get
g= GM/ R²
Acceleration due to gravity
g=\(\frac { GM }{ { R }^{ 2 } } \).

Structure of a Chromosome:
(i) The chromosomes are thin, long and thread like structures consisting of two identical strands called sister chromatids.
(ii) They are held together by the centromere.
(iii) Each chromatid is made up of spirally coiled thin structure called chromonema.
(iv) The chromonema has number of bead-like structures along its length which are called chromomeres.
(v) The chromosomes are made up of DNA, RNA, chromosomal proteins (histones and non-histones) and certain metallic ions.
(vi) These proteins provide structural support to the chromosome.
A chromosome consists of the following regions:
Primary constriction:
(i) The two arms of a chromosome meet at a point called primary constriction or centromere.
(ii) The centromere is the region where spindle fibres attach to the chromosomes during cell division.
Secondary constriction:
(i) Some chromosomes possess secondary constriction at any point of the chromosome.
(ii) They are known as the nuclear zone or nucleolar organizer (formation of nucleolus in the nucleus).
Telomere:
(i) The end of the chromosome is called telomere. Each extremity of the chromosome has a polarity and prevents it from joining the adjacent chromosome.
(ii) It maintains and provides stability to the chromosomes.
Satellite:
(i) Some of the chromosomes have an elongated knob-like appendage at one end of the chromosome known as satellite.
(ii) The chromosomes with satellites are called as the sat-chromosomes

(i) Pea plant is naturally self-pollinating and so is very easy to raise pure breeding individuals.
(ii) It has a short life span as it is an annual and so it was possible to follow several generations.
(iii) It is easy to cross-pollinate.
(iv) It has deeply defined contrasting characters.
(v) The flowers are bisexual.

A - Exine
B - Intine
C - Generative cell
D - Vegetative nucleus

Volume of the ethanol = 35 ml
Volume of the water = 65 ml
Volume percentage = \(\frac{\text { Volume of the solute }}{\text { Volume of the solution }} \times 100\)
Volume percentage =\(\frac{\text { Volume of the solute }}{\text { Volume of the solute + Volume of the solvent }} \times 100\) 
Volume percentage = \(\frac{35}{35+65}\times\)100
Volume percentage = \(\frac{35}{100 }\) x 100
= 35%

(i)  Mass percentage of the solute
Mass percentage of solute = \(\begin{aligned} & \frac{\text { Mass of the solute }}{\text { Mass of the solute +Mass of the solvent}} \times 100 \\ \end{aligned}\)
= \(\frac{16 × 100}{16 + 100}\)
\(=\frac{1600}{116}\)
Mass percentage of the solute = 13.79 %
(ii) Mass percentage of solvent = 100 - (Mass percentage of the solute)
= 100 – 13.79
= 86.21%

Area of copper rod,   A_o = 10 m²
Changes of Area of cross section,
Initial temperature \(\Delta \mathrm{A} =11 -10 =1 \mathrm{~m}^{2}  \)
\(\mathrm{~T}_{1} =90 \mathrm{~K}  \)
\(a_{\mathrm{A}} =0.0021 / \mathrm{K}  \)
\(\mathrm{T}_{2} =?  \)
\(\frac{\Delta A}{A_{0}} =a_{\mathrm{A}} \Delta \mathrm{T}  \)
\(\frac{1 }{10 } =0.0021\left[\mathrm{~T}_{2}-90\right]  \)
\(0.1 =0.0021\left[\mathrm{~T}_{2}-90\right]=\frac{0.1}{0.0021}+90=\mathrm{T}_{2}  \)
\(\mathrm{~T}_{2} =137.61 \mathrm{~K}\)
So the final temperature of a copper rod is 137.61 K