We are given the resistance R of the wire = 26 Ω, the diameter d = 0.3 mm = 3 × 10^-4 m, and the length l of the wire = 1 m.
Therefore, from Eq, the resistivity of the given metallic wire is
ρ = (RA/l) = (Rπd²/4l )
Substitution of values in this gives ρ = 1.84 × 10^–6 Ω m
The resistivity of the metal at 20°C is 1.84 × 10^–6 Ω m.
From Table, we see that this is the resistivity of manganese

We are given, potential difference V = 60 V, current I = 4 A.
According to Ohm's law, \(R=\frac{V}{I}=\frac{60 \mathrm{~V}}{4 \mathrm{~A}}=15 \Omega\).
When the potential difference is increased to 120 V the current is given by current \(=\frac{V}{R}=\frac{120 \mathrm{~V}}{15 \Omega}=8 \mathrm{~A}\)
The current through the heater becomes 8 A .

Justification: Women produce only one type of ovum (carrying X chromosome) and males produce two types of sperms (carrying either X or Y chromosome) in equal proportions. So the sex of a child is a matter of chance depending upon the type of sperm fertilizing the ovum.

Four methods: (i) Mechanical or barrier method OR Male or female condoms.
(ii) Use of hormonal preparations OR Oral Pills/i-pill/Saheli.
(iii) Use of loop or copper T OR IUCD.
(iv) Surgical method OR tubectomy/vasectomy.
Effect on health and prosperity:
(i) Health of women is maintained.
(ii) Parents can give more attention to children.
(iii) More resources can be made available.

The remains (or impressions) of dead animals or plants, that lived in the remote past, are known as fossils.
(a) Fossils provide evidence for evolution, e.g.r a fossil called Archaeopteryx had feathered wings like birds but teeth and tail like reptiles hence suggesting that birds and reptiles had a common ancestor.
(b) Fossils provide direct evidence of past life and thus provide convincing proof of organic evolution.

Differences between binary fission and multiple fission are given below

Variations allow organisms to exist in diverse habitats or niches. In its absence, a species may remain restricted to a particular area. If this area gets drastically altered due to various natural or man-made causes, the species may be wiped out. However, if some variations are present in few individuals, it would help them to colonise other habitats and survive. But, if variations are present in a single organism, there would be a very little chance for it to survive and species is lost forever.

(a) Metals are malleable and ductile i.e. it can be drawn into sheets and wires, it does not break due to high tensile strength and therefore it can be used for our needs.
(b) Nitric acid is strong oxidising agent, when metals react with nitric acid it does not release H₂ gas as oxidise it to form H₂ O.

Haematite (Fe₂O₃) is an important ore of iron.
In the upper part of the blast furnace, carbon monoxide reduces iron (III) oxide to iron metal.
\(\underset { Iron(III)oxide }{ { Fe }_{ 2 }{ O }_{ 3 } } +\underset { Carbonmonoxide }{ 3CO(g) } \rightarrow \underset { Ironmetal }{ 2Fe(I) } +\underset { Carbondioxide }{ { 3CO }_{ 2 }(g) } \)
Following reactions are involved in the blast furnace:
(i) Formation of carbon monoxide
\(\underset { Carbon }{ C(s) } +\underset { Oxygen }{ { O }_{ 2 }(g) } \rightarrow \underset { Carbondioxide }{ { CO }_{ 2 }(g) } +Heat\\ \underset { Carbondioxide }{ { CO }_{ 2 }(g) } +\underset { Coke }{ C(s) } \rightarrow \underset { Carbonmonoxide }{ 2CO(g) } \)
(ii) Reduction of iron (III) oxide or haematite to iron
\(\underset { Iron(III)oxide }{ { Fe }_{ 2 }{ O }_{ 3 } } +\underset { Carbonmonoxide }{ 3CO(g) } \rightarrow \underset { Ironmetal }{ 2Fe(I) } +\underset { Carbondioxide }{ { 3CO }_{ 2 }(g) } \)

(a) Resistance is defined as the property of a conductor due to which it opposes the flow of current through it. The S.I. unit of resistance is Ohm \(\left( \Omega \right) \).
(b) (i) The resistance (R) of a conductor is related to its length (l) and area of cross section (A) as
          \(R=\frac { p\times l }{ A } \)  Where \(\rho \) is the resistivity of the conducting material.
Let R₁= Resistance of the original wire
      R₂= Resistance of the wire when its length is doubled
I₁= Length of original wire
l₂= Length of original wire when doubled
A₁= Area of the original wire
A₂= Area when it is halved 
Therefore,
 \({ R }_{ 1 }=\frac { P\times { l }_{ 1 } }{ { A }_{ 1 } } \quad \quad ---\quad (1)\\ { R }_{ 2 }=\frac { P\times { l }_{ 2 } }{ { A }_{ 2 } } \quad \quad ---\quad (2)\)
Dividing equation (1) and (2) by equation (1), we obtain
\(\frac { { R }_{ 1 } }{ { R }_{ 2 } } =\frac { { l }_{ 1 } }{ { l }_{ 2 } } \times \frac { { A }_{ 1 } }{ { A }_{ 2 } } \quad ---\quad (3)\)
It is given that:
\({ l }_{ 1 }=2l_{ 1 }\quad and\quad { A }_{ 2 }=\frac { 1 }{ 2 } { A }_{ 1 }\\ \)
Putting these values in equation (3), we obtain
R₂=4R₁
Hence, the resistance of the wire will be 4 times of the original value.
(ii) The resistivity will not change on stretching the wire. Resistivity is the property of the conducting material. It does not depend on the dimension of the wire. As materials of both the wires are the same, the resistivity will not change.

(a) Short circuiting is the term used for defining the situation in which the neutral and live wires of an electric circuit come in direct contact. When too many electrical appliances are connected to a single socket, the appliances drawn much more current or power that the permissible limit from the main supply. This situation is termed as overloading.
(b) Resistivity of an alloy is much higher than its constituent metals Moreover, alloys do not oxidise as easily as the constituent metals at high temperature. Because of these reasons, coils of electric testers are made of an alloy rather than a pure meta.

(i) Sodium carbonate (washing soda) is used in glass, soap and paper industries.

(ii) It is used in the manufacture of sodium compounds such as borax.

(iii) Sodium carbonate can be used as a cleaning agent for domestic purposes.

(iv) It is used for removing permanent hardness of water.

Plaster of Paris is calcium sulphate hemihydrate \((CaSO_4.{1\over2}H_2O).\)

Gypsum, when heated at a temperature of about 380 K in a kiln. loses its three-fourth water of crystallisation and changes to plaster of Paris.

Following is the reaction involved:

\(\underset{Gypsum}{CaSO_4.2H_2O}\xrightarrow {Heat} \underset{Plaster\ of\ Paris}{CaSO_4.{1\over2}H_2O}+\underset{Water}{1{1\over2}H_2O}\)

Distilled water does not contain any ions and have no electrolytes. Hence, it does not conduct electricity. But in rain water, a small amount of electrolyte is present in the form of acids. These acids are produced due to dissolution of acidic oxides (e.g. SO₂, NO₂) in rain water and make it a better conductor of electricity.

CO₂ is the gas that extinguishes a burning candle.
CO₂ is produced when metal carbonate reacts with acid. Since, one of the products is calcium chloride.
Thus, the compound A should be calcium carbonate, The balanced chemical equation is
CaCO₃ (s) + 2HCl(aq) \(\rightarrow\) CaCl₂(aq) + CO₂(g) + H₂O(l)

(i) Gonorrhoea and syphilis are two common STD'S.
(ii) Prenatal sex-determination is prohibited by law. This is because of practice of female foeticide due to the desire of male child.

(i) Focal length, \(f(\mathrm{~m})=\frac{1}{P(D)}\)
\(f =\frac{1}{-2.5 D}=\frac{-10}{25} =-04 \mathrm{~m}=40 \mathrm{~cm} \)
\(f  =-40 \mathrm{~cm}, v=-10 \mathrm{~cm}, u=\text { ? (to find) }\)
Using lens formula, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(\frac{-1}{u}  =\frac{1}{40}+\frac{1}{-10}-\frac{1}{u}=\frac{1}{-40} =\frac{-1+4}{40}=\frac{3}{40}\)
\(\therefore \quad u  =\frac{-40}{3}=-13.3 \mathrm{~cm}\)
(ii) The lens is concave as the power is -ve.

OA = v=-10 cm, OB=-13.3 cm, OF = f=-40cm

 (i) (a) In Parallel  \(I=\overset { V }{ \underset { R }{ \_ \_ } } \) or 2 x 5 = V = 10 V
For I0  \(\Omega \) resistance I =  \(\overset { V }{ \underset { R }{ \_ \_ } } =\overset { 10 }{ \underset { 10 }{ \_ \_ } } \) =1A 
For 15 \(\Omega \) resistance I =  \(\overset { V }{ \underset { R }{ \_ \_ } } =\overset { 10 }{ \underset { 10 }{ \_ \_ } } \) = 0.7A 
(b)  \(I=\overset { V }{ \underset { R }{ \_ \_ } } \)  
or, 2 =\(\overset { V }{ \underset { 5 }{ \_ \_ } } \)
or V=10 V
(c) Resistance in Parallel
\(\overset { 1 }{ \underset { R }{ \_ \_ } } =\overset { 1 }{ \underset { 10 }{ \_ \_ } } +\overset { 1 }{ \underset { 15 }{ \_ \_ } } =\overset { 1 }{ \underset { 30 }{ \_ \_ } } =\overset { 1 }{ \underset { 6 }{ \_ \_ } } \)
So, R= 6 \(\Omega \)
So, Total resistance across AC
= 5 + 6 = 11\(\Omega \)
 (ii)
 
 If we connect all the resistance in parallel we will get the least total resistance.

In vegetative propagation, new plants are obtained from the parts of old plants like stems, roots and leaves, without the help of any reproductive organ.
There are two ways of vegetative propagation:
(a) Natural Vegetative Propagation, and
(b) Artificial Vegetative Propagation.
Natural vegetative propagation by leaves: The fleshy leaves of Bryophyllum bear adventitious buds in the notches along the leaf margin.
Grafting: In this method of reproduction, two plants of closely related varieties are joined together so that they live as one plant:
(a) The portion of a plant that is grafted on the other plant is called scion, and the plant in which grafting is performed is called the stock. .
(b) This method is applied to improve variety of fruits like mango, apple, peas, citrus and guava.
Advantages of vegetative propagation:
(i) Vegetative propagation is a cheaper, easier and more rapid method of propagation in plants than growing plants from their seeds.
(ii) Better quality of the plants can be maintained by this method.