NCERT CBSE Standard 12 Solid State Chapter 1 Physical Chemistry

Solutions to Solid State Chapter 1 :

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Question 1.1 :

1.1 QA Solid State CBSE std 12 Solutions

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Gyan :

Ferromagnetism, Ferrimagnetism, Antiferromagnetism, Paramagnetism

4 Explain Ferromagnetism, ferrimagnetism

5 Explain Ferromagnetism, ferrimagnetism

6 Explain Ferromagnetism, ferrimagnetism

Question 1.2 :

1.2 QA Solid State CBSE std 12 Solutions

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Untitled

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14 bent face

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Question 1.3

1.3 QA Solid State CBSE std 12 Solutions

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Gyan Question :

1 How do crystalline solids differ from amorphous solids

2 How do crystalline solids differ from amorphous solids

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14 Fossil macher mundu

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Gyan Question :

1 Crystal gas lighter quartz watches

2 Crystal gas lighter quartz watches

3 Crystal gas lighter quartz watches

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Gyan Question

1 a crystal adopts BCC arrangement given edge length

Solution :

2 a crystal adopts BCC arrangement given edge length

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14 Girl catching her image from back

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Gyan Question :

3 categorize the crystalline solids

4 categorize the crystalline solids

5 categorize the crystalline solids

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Question 1.4 :

1.4 QA Solid State CBSE std 12 Solutions

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1 Solid has cubic structure x atoms located

2 Solid has cubic structure x atoms located

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Gyan Question :

6 space lattice unit cell monoclinic

7 space lattice unit cell monoclinic

8 space lattice unit cell monoclinic

9 space lattice unit cell monoclinic

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3 Compound formed by elements x and Y crystallizes

4 Compound formed by elements x and Y crystallizes

5 Compound formed by elements x and Y crystallizes

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Gyan Question :

3 solution contains 66 gm of acetone 46 gm water

Solution :

4 solution contains 66 gm of acetone 46 gm water

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Gyan Question :

10 Soild A adopts a orthorhombic unit

11 Soild A adopts a orthorhombic unit

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Question 1.5 :

1.5 QA Solid State CBSE std 12 Solutions

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Gyan Question :

6 Cubic solid is made up of 2 elements P and Q

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Question 1.6 :

1.6 QA Solid State CBSE std 12 Solutions

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Gyan Question :

7 An element crystallizes in bcc

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Question 1.7 :

1.7 QA Solid State CBSE std 12 Solutions

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Gyan Question :

1 element crystallizes adopting ABAB pattern

2 element crystallizes adopting ABAB pattern

3 element crystallizes adopting ABAB pattern

4 element crystallizes adopting ABAB pattern

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Gyan Question :

8 solid oxide ions adopt ccp arrangement

9 solid oxide ions adopt ccp arrangement

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Question 1.8 :

1.8 QA Solid State CBSE std 12 Solutions

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Gyan Question :

1 tetrahedral void closed packed array of N atoms

2 tetrahedral void closed packed array of N atoms

3 tetrahedral void closed packed array of N atoms

4 tetrahedral void closed packed array of N atoms

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Question 1.9 :

1.9 QA Solid State CBSE std 12 Solutions

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Gyan Question :

5 element A crystallizes adopting bcc

6 element A crystallizes adopting bcc

7 element A crystallizes adopting bcc

8 element A crystallizes adopting bcc

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Question 1.10 :

1.10a QA Solid State CBSE std 12 Solutions

1.10b QA Solid State CBSE std 12 Solutions

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Gyan Question :

9 calculate and show that percentage space occupied

10 calculate and show that percentage space occupied

11 calculate and show that percentage space occupied

12 calculate and show that percentage space occupied

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Question 1.11 :

1.11 QA Solid State CBSE std 12 Solutions

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13 calculate and show that percentage space occupied

14 calculate and show that percentage space occupied

15 calculate and show that percentage space occupied

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Question 1.12 :

1.12 QA Solid State CBSE std 12 Solutions

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Gyan Question :

16 In Chromiun III chloride CrCl3

17 In Chromiun III chloride CrCl3

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Question 1.13 :

1.13 QA Solid State CBSE std 12 Solutions

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Gyan Question :

1 Gold crystallizes cubic lattice adopting

2 Gold crystallizes cubic lattice adopting

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Question 1.14 :

1.14 QA Solid State CBSE std 12 Solutions

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Gyan Question :

3 diffraction of crystal with x rays

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Question 1.15 :

1.15 QA Solid State CBSE std 12 Solutions

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Gyan Question :

4 binary compound AB adopts fcc

5 binary compound AB adopts fcc

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Question 1.16 :

1.16 QA Solid State CBSE std 12 Solutions

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Gyan Question :

6 what is meant by radius ratio

7 what is meant by radius ratio

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Question 1.17 :

1.17 QA Solid State CBSE std 12 Solutions

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Gyan Question :

8 Draw unit cell of NACl structure

9 Draw unit cell of NACl structure

10 Draw unit cell of NACl structure

11 Draw unit cell of NaCl structure

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Question 1.18 :

1.18b QA Solid State CBSE std 12 Solutions

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Gyan Question :

12 unit cell of CsCl structure

13 unit cell of CsCl structure

14 unit cell of CsCl structure

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Gyan Question :

1 composition of a sample of Wustite

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Gyan Question :

1 sphalerite structure name compounds

2 sphalerite structure name compounds

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Question 1.19 :

1.18 QA Solid State CBSE std 12 Solutions

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Gyan Question :

3 element of density 6.8 gm per cc

4 element of density 6.8 gm per cc

5 element of density 6.8 gm per cc

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Gyan Question :

2 doping NaCl is doped with SrCl2

3 doping NaCl is doped with SrCl2

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Question 1.20 :

1.20 QA Solid State CBSE std 12 Solutions

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Gyan Question :

6 Potassium crystallises in bcc latice

7 Potassium crystallises in bcc latice

8 Potassium crystallises in bcc latice

9 Potassium crystallises in bcc latice

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14 Nude girl from side Nathan-De-Young

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Question 1.21 :

1.21 QA Solid State CBSE std 12 Solutions

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14 Painting

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Gyan Question :

10 copper crystallizes in fcc unit cell

11 copper crystallizes in fcc unit cell

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14 various ape skeletons

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Gyan Question :

7 Solid A+B- has NaCl type close packed

8 Solid A+B- has NaCl type close packed

9 Solid A+B- has NaCl type close packed

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15 distorted face

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Question 1.22 :

1.22 QA Solid State CBSE std 12 Solutions

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Gyan Question :

12 Radius of anion in ionic solid is 100 pm

13 Radius of anion in ionic solid is 100 pm

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15 Fossil leaf

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Question 1.23 :

1.23a QA Solid State CBSE std 12 Solutions

1.23b QA Solid State CBSE std 12 Solutions

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Gyan Question :

10 edge length of a cubic crystal

11 edge length of a cubic crystal

12 edge length of a cubic crystal

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15 Is it a garden pond

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Question 1.24 :

1.24 QA Solid State CBSE std 12 Solutions

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Gyan Question :

13 element having atomic mass 52 exists in bcc

14 element having atomic mass 52 exists in bcc

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15 painting

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Question 1.25

1.25 QA Solid State CBSE std 12 Solutions

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Gyan Question :

15 Iron crystallizes in 2 forms bcc structure

16 Iron crystallizes in 2 forms bcc structure

17 Iron crystallizes in 2 forms bcc structure

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16 bent face

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Question 1.26 :

1.26a QA Solid State CBSE std 12 Solutions

1.26b QA Solid State CBSE std 12 Solutions

1.26c QA Solid State CBSE std 12 Solutions

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Gyan Question :

18 crystal defects schottky effect

19 crystal defects schottky effect

20 crystal defects schottky effect

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16 fossil

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Gyan Question :

21 type of cubic lattice forrmed by Iron atoms

22 type of cubic lattice forrmed by Iron atoms

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Q.1 : Why are solids rigid?

ANS : The intermolecular forces of attraction that are present in solids are very strong. The constituent particles of solids cannot move from their positions i.e., they have fixed positions. However, they can oscillate about their mean positions. This is the reason solids are rigid.
Q.2 : Why do solids have a definite volume?
ANS : The intermolecular forces of attraction that are present in solids are very strong. The constituent particles of solids have fixed positions i.e., they are rigid. Hence, solids have a definite volume.
Q.3 : Classify the following as amorphous or crystalline solids:
Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.
ANS: Amorphous solids
Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass
Crystalline solids
Naphthalene, benzoic acid, potassium nitrate, copper
Q.4 : Why is glass considered a super cooled liquid?
ANS : Similar to liquids, glass has a tendency to flow, though very slowly. Therefore, glass is considered as a super cooled liquid. This is the reason that glass windows and doors are slightly thicker at the bottom than at the top.
Q.5 : Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?
ANS : An isotropic solid has the same value of physical properties when measured along different directions. Therefore, the given solid, having the same value of refractive index along all directions, is isotropic in nature. Hence, the solid is an amorphous solid.
When an amorphous solid is cut with a sharp edged tool, it cuts into two pieces with irregular surfaces.
Q.6 : Classify the following solids in different categories based on the nature of intermolecular forces operating in them:
Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.
ANS : Potassium sulphate → Ionic solid
Tin → Metallic solid
Benzene → Molecular (non-polar) solid
Urea → Polar molecular solid
Ammonia → Polar molecular solid
Water → Hydrogen bonded molecular solid
Zinc sulphide → Ionic solid
Graphite → Covalent or network solid
Rubidium → Metallic solid
Argon → Non-polar molecular solid
Silicon carbide → Covalent or network solid
Q.7 : Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it?
ANS : The given properties are the properties of a covalent or network solid. Therefore, the given solid is a covalent or network solid. Examples of such solids include diamond (C) and quartz (SiO2).
Q.8: Ionic solids conduct electricity in molten state but not in solid state. Explain.
ANS : In ionic compounds, electricity is conducted by ions. In solid state, ions are held together by strong electrostatic forces and are not free to move about within the solid. Hence, ionic solids do not conduct electricity in solid state. However, in molten state or in solution form, the ions are free to move and can conduct electricity.
Q.9 : What type of solids are electrical conductors, malleable and ductile?
ANS : Metallic solids are electrical conductors, malleable, and ductile.
Q.10 : Give the significance of a ‘lattice point’.
ANS: The significance of a lattice point is that each lattice point represents one constituent    particle of a solid which may be an atom, a molecule (group of atom), or an ion.
Q.11 : Name the parameters that characterize a unit cell.
ANS : The six parameters that characterise a unit cell are as follows.
(i) Its dimensions along the three edges, ab, and c
These edges may or may not be equal.
(ii) Angles between the edges
These are the angle ∝ (between edges and c), β (between edges a and c), and γ (between edges a and b).
Q.12 : Distinguish between
(i)Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centred unit cells.
ANS : (i) Hexagonal unit cell
For a hexagonal unit cell,
Monoclinic unit cell
For a monoclinic cell,
(ii) Face-centred unit cell
In a face-centred unit cell, the constituent particles are present at the corners and one at the centre of each face.
End-centred unit cell
An end-centred unit cell contains particles at the corners and one at the centre of any two opposite faces.
Q.13 : Explain how much portion of an atom located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell.
ANS : (i)An atom located at the corner of a cubic unit cell is shared by eight adjacent unit cells.
Therefore, portion of the atom is shared by one unit cell.
(ii)An atom located at the body centre of a cubic unit cell is not shared by its neighboNS uring unit cell. Therefore, the atom belongs only to the unit cell in which it is present i.e., its contribution to the unit cell is 1.
Q.14: What is the two dimensional coordination number of a molecule in square close packed layer?
ANS : In square close-packed layer, a molecule is in contact with four of its neighbours. Therefore, the two-dimensional coordination number of a molecule in square close-packed layer is 4.
Q.15 : A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
ANS: Number of close-packed particles = 0.5 × 6.022 × 1023 = 3.011 × 1023
Therefore, number of octahedral voids = 3.011 × 1023
And, number of tetrahedral voids = 2 × 3.011 × 1023 = 6.022 ×1023
Therefore, total number of voids = 3.011 × 1023 + 6.022 × 1023 = 9.033 × 1023
Q.16 : A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound?
ANS : The ccp lattice is formed by the atoms of the element N.
Here, the number of tetrahedral voids generated is equal to twice the number of atoms of the element N.
According to the question, the atoms of element M occupy of the tetrahedral voids.
Therefore, the number of atoms of M is equal to of the number
of atoms of N.
Therefore, ratio of the number of atoms of M to that of N is M: N 
Thus, the formula of the compound is M2 N3.
Q.17: Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centred cubic and (iii) hexagonal close-packed lattice?
ANS : Hexagonal close-packed lattice has the highest packing efficiency of 74%. The packing efficiencies of simple cubic and body-centred cubic lattices are 52.4% and 68% respectively.
Q.18: An element with molar mass 2.7 × 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m−3, what is the nature of the cubic unit cell?
ANS : It is given that density of the element, d = 2.7 × 103 kg m−3
Molar mass, M = 2.7 × 10−2 kg mol−1
Edge length, a = 405 pm = 405 × 10−12 m
= 4.05 × 10−10 m
It is known that, Avogadro’s number, NA = 6.022 × 1023 mol−1
Applying the relation,
This implies that four atoms of the element are present per unit cell. Hence, the unit cell is face-centred cubic (fcc) or cubic close-packed (ccp).
Q.19: What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
ANS : When a solid is heated, vacancy defect can arise. A solid crystal is said to have vacancy defect when some of the lattice sites are vacant.
Vacancy defect leads to a decrease in the density of the solid.
Q.20: What type of stoichiometric defect is shown by:
(i) ZnS (ii) AgBr
ANS : (i) ZnS shows Frenkel defect.
(ii) AgBr shows Frenkel defect as well as Schottky defect.
Q.21 : Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
ANS : When a cation of higher valence is added to an ionic solid as an impurity to it, the cation of higher valence replaces more than one cation of lower valence so as to keep the crystal electrically neutral. As a result, some sites become vacant. For example, when Sr2+ is added to NaCl, each Sr2+ ion replaces two Naions. However, one Sr2+ ion occupies the site of one Na+ ion and the other site remains vacant. Hence, vacancies are introduced.
Q.22 : Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
ANS : The colour develops because of the presence of electrons in the anionic sites. These electrons absorb energy from the visible part of radiation and get excited.
For example, when crystals of NaCl are heated in an atmosphere of sodium vapours, the sodium atoms get deposited on the surface of the crystal and the chloride ions from the crystal diffuse to the surface to form NaCl with the deposited Na atoms. During this process, the Na atoms on the surface lose electrons to form Na+ ions and the released electrons diffuse into the crystal to occupy the vacant anionic sites. These electrons get excited by absorbing energy from the visible light and impart yellow colour to the crystals.
Q.23 : A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?
ANS : An n-type semiconductor conducts because of the presence of extra electrons. Therefore, a group 14 element can be converted to n-type semiconductor by doping it with a group 15 element.
Q.24 : What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.
ANS : Ferromagnetic substances would make better permanent magnets.
In solid state, the metal ions of ferromagnetic substances are grouped together into small regions. These regions are called domains and each domain acts as a tiny magnet. In an unmagnetised piece of a ferromagnetic substance, the domains are randomly oriented. As a result, the magnetic moments of the domains get cancelled. However, when the substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field and a strong magnetic effect is produced.
The ordering of the domains persists even after the removal of the magnetic field. Thus, the ferromagnetic substance becomes a permanent magnet.
EXERCISE 
Q.1.1 : Define the term ‘amorphous’. Give a few examples of amorphous solids.
ANS: Amorphous solids are the solids whose constituent particles are of irregular shapes and have short range order. These solids are isotropic in nature and melt over a range of temperature. Therefore, amorphous solids are sometimes called pseudo solids or super cooled liquids. They do not have definite heat of fusion. When cut with a sharp-edged tool, they cut into two pieces with irregular surfaces. Examples of amorphous solids include glass, rubber, and plastic.
Q 1.2 : What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
ANS :  The arrangement of the constituent particles makes glass different from quartz. In glass, the constituent particles have short range order, but in quartz, the constituent particles have both long range and short range orders.
Quartz can be converted into glass by heating and then cooling it rapidly.
Q 1.3 : Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
(i) Tetra phosphorus decoxide (P4O10) (vii) Graphite
(ii) Ammonium phosphate (NH4)3PO4 (viii) Brass
(iii) SiC (ix) Rb
(iv) I2 (x) LiBr
(v) P4 (xi) Si
ANS : Ionic → (ii) Ammonium phosphate (NH4)3PO4(x) LiBr
Metallic → (viii) Brass, (ix) Rb
Molecular → (i) Tetra phosphorus decoxide (P4O10), (iv) I2, (v) P4.
Covalent (network) → (iii) SiC, (vii) Graphite, (xi) Si
Amorphous → (vi) Plastic
Q 1.4 : (i) What is meant by the term ‘coordination number’?
(ii) What is the coordination number of atoms:
(a) in a cubic close-packed structure?
(b) in a body-centred cubic structure?
ANS : (i) The number of nearest neighbours of any constituent particle present in the crystal lattice is called its coordination number.
(ii) The coordination number of atoms
(a) in a cubic close-packed structure is 12, and
(b) in a body-centred cubic structure is 8
Q 1.5 : How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
ANS : By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be determined.
Let ‘a’ be the edge length of a unit cell of a crystal, ‘d’ be the density of the metal, ‘m’ be the atomic mass of the metal and ‘z’ be the number of atoms in the unit cell.
Now, density of the unit cell 
[Since mass of the unit cell = Number of atoms in the unit cell × Atomic mass]
[Volume of the unit cell = (Edge length of the cubic unit cell)3]
From equation (i), we have:
Now, mass of the metal (m
Therefore, 
If the edge lengths are different (say ab and c), then equation (ii) becomes:
From equations (iii) and (iv), we can determine the atomic mass of the unknown metal.
Q 1.6 : ‘Stability of a crystal is reflected in the magnitude of its melting point’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?
ANS : Higher the melting point, greater is the intermolecular force of attraction and greater is the stability. A substance with higher melting point is more stable than a substance with lower melting point.
The melting points of the given substances are:
Solid water → 273 K
Ethyl alcohol → 158.8 K
Diethyl ether → 156.85 K
Methane → 89.34 K
Now, on observing the values of the melting points, it can be said that among the given substances, the intermolecular force in solid water is the strongest and that in methane is the weakest.
Q 1.7: How will you distinguish between the following pairs of terms:
(i) Hexagonal close-packing and cubic close-packing?
(ii) Crystal lattice and unit cell?
(iii) Tetrahedral void and octahedral void?
ANS : 
  1. A 2-D hexagonal close-packing contains two types of triangular voids (a and b) as shown in figure 1. Let us call this 2-D structure as layer A. Now, particles are kept in the voids present in layer A (it can be easily observed from figures 2 and 3 that only one of the voids will be occupied in the process, i.e., either a or b). Let us call the particles or spheres present in the voids of layer A as layer B. Now, two types of voids are present in layer B (c and d). Unlike the voids present in layer A, the two types of voids present in layer B are not similar. Void c is surrounded by 4 spheres and is called the tetrahedral void. Void d is surrounded by 6 spheres and is called the octahedral void.
Figure 1
Figure 2
Figure 3
Now, the next layer can be placed over layer B in 2 ways.
Case 1: When the third layer (layer C) is placed over the second one (layer B) in such a manner that the spheres of layer C occupy the tetrahedral voids c.
In this case we get hexagonal close-packing. This is shown in figure 4. In figure 4.1, layer B is present over the voids a and layer C is present over the voids c. In figure 4.2, layer B is present over the voids b and layer C is present over the voids c. It can be observed from the figure that in this arrangement, the spheres present in layer C are present directly above the spheres of layer A. Hence, we can say that the layers in hexagonal close-packing are arranged in an ABAB….. pattern.
Figure 4.1
Figure 4.2
Case 2: When the third layer (layer C) is placed over layer B in such a manner that the spheres of layer C occupy the octahedral voids d.
In this case we get cubic close-packing. In figure 5.1, layer B is present over the voids a and layer C is present over the voids d. In figure 5.2, layer B is present over the voids b and layer C is present over the voids d. It can be observed from the figure that the arrangement of particles in layer C is completely different from that in layers A or B. When the fourth layer is kept over the third layer, the arrangement of particles in this layer is similar to that in layer A. Hence, we can say that the layers in cubic close-packing are arranged in an ABCABC….. pattern.
Figure 5.1
Figure 5.2
The side views of hcp and ccp are given in figures 6.1 and 6.2 respectively.
Figure 6.1
Figure 6.2
(ii) The diagrammatic representation of the constituent particles (atoms, ions, or molecules) present in a crystal in a regular three-dimensional arrangement is called crystal lattice.
A unit cell is the smallest three-dimensional portion of a crystal lattice. When repeated again and again in different directions, it generates the entire crystal lattice.
(iii) A void surrounded by 4 spheres is called a tetrahedral void and a void surrounded by 6 spheres is called an octahedral void. Figure 1 represents a tetrahedral void and figure 2 represents an octahedral void.
Figure 1
Figure 2
 
Q 1.8 : How many lattice points are there in one unit cell of each of the following lattice?
(i) Face-centred cubic
(ii) Face-centred tetragonal
(iii) Body-centred
ANS : (i) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred cubic.
(ii) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred tetragonal.
(iii) There are 9 (1 from the centre + 8 from the corners) lattice points in body-centred cubic.
Q 1.9: Explain
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
ANS : (i) The basis of similarities between metallic and ionic crystals is that both these crystal types are held by the electrostatic force of attraction. In metallic crystals, the electrostatic force acts between the positive ions and the electrons. In ionic crystals, it acts between the oppositely-charged ions. Hence, both have high melting points.
The basis of differences between metallic and ionic crystals is that in metallic crystals, the electrons are free to move and so, metallic crystals can conduct electricity. However, in ionic crystals, the ions are not free to move. As a result, they cannot conduct electricity. However, in molten state or in aqueous solution, they do conduct electricity.
(ii) The constituent particles of ionic crystals are ions. These ions are held together in three-dimensional arrangements by the electrostatic force of attraction. Since the electrostatic force of attraction is very strong, the charged ions are held in fixed positions. This is the reason why ionic crystals are hard and brittle.
Q 1.10: Calculate the efficiency of packing in case of a metal crystal for
(i) simple cubic
(ii) body-centred cubic
(iii) face-centred cubic (with the assumptions that atoms are touching each other).
ANS : (i) Simple cubic
In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge.
Let the edge length of the cube be ‘a’ and the radius of each particle be r.
So, we can write:
a = 2r
Now, volume of the cubic unit cell = a3
= (2r)3
= 8r3
We know that the number of particles per unit cell is 1.
Therefore, volume of the occupied unit cell 
Hence, packing efficiency 
(ii) Body-centred cubic
It can be observed from the above figure that the atom at the centre is in contact with the other two atoms diagonally arranged.
From ΔFED, we have:
Again, from ΔAFD, we have:
Let the radius of the atom be r.
Length of the body diagonal, c = 4π
or, 
Volume of the cube, 
A body-centred cubic lattice contains 2 atoms.
So, volume of the occupied cubic lattice 
(iii) Face-centred cubic
Let the edge length of the unit cell be ‘a’ and the length of the face diagonal AC be b.
From ΔABC, we have:
Let r be the radius of the atom.
Now, from the figure, it can be observed that:
Now, volume of the cube, 
We know that the number of atoms per unit cell is 4.
So, volume of the occupied unit cell 
= 74%
Q 1.11 : Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10−8 cm and density is 10.5 g cm−3, calculate the atomic mass of silver.
ANS : It is given that the edge length, a = 4.077 × 10−8 cm
Density, d = 10.5 g cm−3
As the lattice is fcc type, the number of atoms per unit cell, z = 4
We also know that, NA = 6.022 × 1023 mol−1
Using the relation:
= 107.13 gmol−1
Therefore, atomic mass of silver = 107.13 u
Q 1.12: A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?
ANS : It is given that the atoms of Q are present at the corners of the cube.
Therefore, number of atoms of Q in one unit cell
It is also given that the atoms of P are present at the body-centre.
Therefore, number of atoms of P in one unit cell = 1
This means that the ratio of the number of P atoms to the number of Q atoms, P:Q = 1:1
Hence, the formula of the compound is PQ.
The coordination number of both P and Q is 8.
Q 1.13 : Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm−3, calculate atomic radius of niobium using its atomic mass 93 u.
ANS : It is given that the density of niobium, d = 8.55 g cm−3
Atomic mass, M = 93 gmol−1
As the lattice is bcc type, the number of atoms per unit cell, z = 2
We also know that, NA = 6.022 × 1023 mol−1
Applying the relation:
= 3.612 × 10−23 cm3
So, a = 3.306 × 10−8 cm
For body-centred cubic unit cell:
= 1.432 × 10−8 cm
= 14.32 × 10−9 cm
= 14.32 nm
Q 1.14: If the radius of the octachedral void is r and radius of the atoms in close packing is R, derive relation between r and R.
ANS : 
A sphere with centre O, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure that ΔPOQ is right-angled
∠POQ = 900
Now, applying Pythagoras theorem, we can write:
Q 1.15: Copper crystallises into a fcc lattice with edge length 3.61 × 10−8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm−3.
ANS : Edge length, a = 3.61 × 10−8 cm
As the lattice is fcc type, the number of atoms per unit cell, z = 4
Atomic mass, M = 63.5 g mol−1
We also know that, NA = 6.022 × 1023 mol−1
Applying the relation:
= 8.97 g cm−3
The measured value of density is given as 8.92 g cm−3. Hence, the calculated density 8.97 g cm−3 is in agreement with its measured value.
Q 1.16: Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?
ANS : The formula of nickel oxide is Ni0.98O1.00.
Therefore, the ratio of the number of Ni atoms to the number of O atoms,
Ni : O = 0.98 : 1.00 = 98 : 100
Now, total charge on 100 O2− ions = 100 × (−2)
= −200
Let the number of Ni2+ ions be x.
So, the number of Ni3+ ions is 98 − x.
Now, total charge on Ni2+ ions = x(+2)
= +2x
And, total charge on Ni3+ ions = (98 − x)(+3)
= 294 − 3x
Since, the compound is neutral, we can write:
2x + (294 − 3x) + (−200) = 0
⇒ −x + 94 = 0
⇒ x = 94
Therefore, number of Ni2+ ions = 94
And, number of Ni3+ ions = 98 − 94 = 4
Hence, fraction of nickel that exists as Ni2+
= 0.959
And, fraction of nickel that exists as 
= 0.041
Alternatively, fraction of nickel that exists as Ni3+ = 1 − 0.959
= 0.041
Q 1.17: What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
ANS : Semiconductors are substances having conductance in the intermediate range of 10-6 to 104 ohm−1m−1.
The two main types of semiconductors are:
(i) n-type semiconductor
(ii) p-type semiconductor
n-type semiconductor: The semiconductor whose increased conductivity is a result of negatively-charged electrons is called an n-type semiconductor. When the crystal of a group 14 element such as Si or Ge is doped with a group 15 element such as P or As, an n-type semiconductor is generated.
Si and Ge have four valence electrons each. In their crystals, each atom forms four covalent bonds. On the other hand, P and As contain five valence electrons each. When Si or Ge is doped with P or As, the latter occupies some of the lattice sites in the crystal. Four out of five electrons are used in the formation of four covalent bonds with four neighbouring Si or Ge atoms. The remaining fifth electron becomes delocalised and increases the conductivity of the doped Si or Ge.
p-type semiconductor: The semiconductor whose increased in conductivity is a result of electron hole is called a p-type semiconductor. When a crystal of group 14 elements such as Si or Ge is doped with a group 13 element such as B, Al, or Ga (which contains only three valence electrons), a p-type of semiconductor is generated.
When a crystal of Si is doped with B, the three electrons of B are used in the formation of three covalent bonds and an electron hole is created. An electron from the neighbouring atom can come and fill this electron hole, but in doing so, it would leave an electron hole at its original position. The process appears as if the electron hole has moved in the direction opposite to that of the electron that filled it. Therefore, when an electric field is applied, electrons will move toward the positively-charged plate through electron holes. However, it will appear as if the electron holes are positively-charged and are moving toward the negatively- charged plate.
Q 1.18: Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?
ANS : In the cuprous oxide (Cu2O) prepared in the laboratory, copper to oxygen ratio is slightly less than 2:1. This means that the number of Cu+ions is slightly less than twice the number of O2− ions. This is because some Cu+ ions have been replaced by Cu2+ ions. Every Cu2+ ion replaces two Cu+ ions, thereby creating holes. As a result, the substance conducts electricity with the help of these positive holes. Hence, the substance is a p-type semiconductor.
Q 1.19: Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
ANS : Let the number of oxide (O2−) ions be x.
So, number of octahedral voids = x
It is given that two out of every three octahedral holes are occupied by ferric ions.
So, number of ferric (Fe3+) ions
Therefore, ratio of the number of Fe3+ ions to the number of O2− ions,
Fe3+ : O2− 
= 2 : 3
Hence, the formula of the ferric oxide is Fe2O3.
Q 1.20: Classify each of the following as being either a p-type or an n-type semiconductor:
(i) Ge doped with In (ii) B doped with Si.
ANS : (i) Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be a p-type semiconductor.
(ii) B (a group 13 element) is doped with Si (a group 14 element). So, there will be an extra electron and the semiconductor generated will be an n-type semiconductor.
Q 1.21: Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?
ANS : For a face-centred unit cell:
It is given that the atomic radius, r = 0.144 nm
So, 
= 0.407 nm
Hence, length of a side of the cell = 0.407 nm
Q 1.22: In terms of band theory, what is the difference
(i) Between a conductor and an insulator
(ii) Between a conductor and a semiconductor
ANS : (i) The valence band of a conductor is partially-filled or it overlaps with a higher energy, unoccupied conduction band.
On the other hand, in the case of an insulator, the valence band is fully- filled and there is a large gap between the valence band and the conduction band.
(ii) In the case of a conductor, the valence band is partially-filled or it overlaps with a higher energy, unoccupied conduction band. So, the electrons can flow easily under an applied electric field.
On the other hand, the valence band of a semiconductor is filled and there is a small gap between the valence band and the next higher conduction band. Therefore, some electrons can jump from the valence band to the conduction band and conduct electricity.
Q 1.23: Explain the following terms with suitable examples:
(i) Schottky defect
(ii) Frenkel defect
(iii) Interstitials and
(iv) F-centres
ANS : (i) Schottky defect: Schottky defect is basically a vacancy defect shown by ionic solids. In this defect, an equal number of cations and anions are missing to maintain electrical neutrality. It decreases the density of a substance. Significant number of Schottky defects is present in ionic solids. For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature. Ionic substances containing similar-sized cations and anions show this type of defect. For example: NaCl, KCl, CsCl, AgBr, etc.
(ii) Frenkel defect: Ionic solids containing large differences in the sizes of ions show this type of defect. When the smaller ion (usually cation) is dislocated from its normal site to an interstitial site, Frenkel defect is created. It creates a vacancy defect as well as an interstitial defect. Frenkel defect is also known as dislocation defect. Ionic solids such as AgCl, AgBr, AgI, and ZnS show this type of defect.
(iii) Interstitials: Interstitial defect is shown by non-ionic solids. This type of defect is created when some constituent particles (atoms or molecules) occupy an interstitial site of the crystal. The density of a substance increases because of this defect.
(iv) F-centres: When the anionic sites of a crystal are occupied by unpaired electrons, the ionic sites are called F-centres. These unpaired electrons impart colour to the crystals. For example, when crystals of NaCl are heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl ions diffuse from the crystal to its surface and combine with Na atoms, forming NaCl. During this process, the Na atoms on the surface of the crystal lose electrons. These released electrons diffuse into the crystal and occupy the vacant anionic sites, creating F-centres.
                                                           
Q 1.24 : Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.
(i) What is the length of the side of the unit cell?
(ii) How many unit cells are there in 1.00 cm3 of aluminium?
ANS : (i) For cubic close-packed structure:
= 353.55 pm
= 354 pm (approximately)
(ii) Volume of one unit cell = (354 pm)3
= 4.4 × 107 pm3
= 4.4 × 107 × 10−30 cm3
= 4.4 × 10−23 cm3
Therefore, number of unit cells in 1.00 cm3 = 
= 2.27 × 1022

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Notes : Solid State

CBSE Solid State 1 Chapter 1 Concepts CBSE Solid State 2 Chapter 1 Concepts CBSE Solid State 3 Chapter 1 Concepts CBSE Solid State 4 Chapter 1 Concepts

CBSE Solid State 5 Chapter 1 Concepts

CBSE Solid State 6 Chapter 1 Concepts CBSE Solid State 7 Chapter 1 Concepts CBSE Solid State 8 Chapter 1 Concepts CBSE Solid State 9 Chapter 1 Concepts CBSE Solid State 10 Chapter 1 Concepts CBSE Solid State 11 Chapter 1 Concepts CBSE Solid State 12 Chapter 1 Concepts CBSE Solid State 13 Chapter 1 Concepts CBSE Solid State 14 Chapter 1 Concepts CBSE Solid State 15 Chapter 1 Concepts

CBSE Solid State 16 Chapter 1 Concepts

CBSE Solid State 17 Chapter 1 Concepts CBSE Solid State 18 Chapter 1 Concepts CBSE Solid State 19 Chapter 1 Concepts CBSE Solid State 20 Chapter 1 Concepts CBSE Solid State 21 Chapter 1 Concepts

CBSE Solid State 22 Chapter 1 Concepts

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http://zookeepersblog.wordpress.com/science-tuition-chemistry-physics-mathematics-for-iit-jee-aieee-std-11-12-pu-isc-cbse/

In youtube  search for Zookeeper Dezrina     you will get most videos. I say most because I do not upload all videos that I make. I have many more videos which are not in the net.

http://zookeepersblog.wordpress.com/iit-jee-capacitance/

Casimir,Polder,Davies,Unruh,BELL,Aspect,Galileo,Mosley,Chadwick,Feynman,Schrodinger
http://www.youtube.com/watch?v=672a2lLgxao

Synthesizing new life forms http://www.youtube.com/watch?v=AKxmqMH4w_A

http://zookeepersblog.wordpress.com/iit-jee-3d-geometry-solutions/

http://zookeepersblog.wordpress.com/iit-jee-algebra/

http://zookeepersblog.wordpress.com/iit-jee-area-problems/

http://zookeepersblog.wordpress.com/iit-jee-binomial-theorem/

http://zookeepersblog.wordpress.com/iit-jee-calculus/

http://zookeepersblog.wordpress.com/iit-jee-optics/

http://zookeepersblog.wordpress.com/iit-jee-co-ordinate-geometry/

http://zookeepersblog.wordpress.com/iit-jee-complex-number/

http://zookeepersblog.wordpress.com/iit-jee-current-electricity-circuits/

http://zookeepersblog.wordpress.com/iit-jee-determinant-and-matrices/

http://zookeepersblog.wordpress.com/iit-jee-differentiability/

http://zookeepersblog.wordpress.com/iit-jee-electromagnetics/

http://zookeepersblog.wordpress.com/iit-jee-electrostatics/

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The following Videos are available for you ( As of Now ).  These explain tricky Physics  and Mathematics Numericals.

Eventually I will try to give Videos for full course here for you.

These covers PU ( Pre University courses, school / college ) courses, IIT JEE, AIEEE ( All India Engineering Entrance Examination ) , CET ( Combined Engineering Test ), AIPMT ( All India Pre Medical Test ), ISc ( Intermediate Science / Indian School Certificate Exam ), CBSE ( Central Board Secondary Exam ), Roorkey Joint Entrance Test Questions ( Discontinued since 2002 ), APhO ( Asian Physics Olympiad ), IPhO ( International Physics Olympiad ), IMO ( International Mathematics Olympiad ) , NSEP ( National Standard Exam in Physics ), RMO ( Regional Math Olympiad , India ), INMO ( Indian National Maths Olympiad ), Irodov Solutions, Prof. H C Verma ( Concepts of Physics ) Solutions etc.

( You can see the history of Indian Participation in various Olympiads at -> http://zookeepersblog.wordpress.com/indian-participation-in-ipho-icho-ibo-and-astronomy-olympiad/ )

[ In each of these videos there is at-least 1 or more errors. Please tell me about those ]

In youtube if you search as ” Dezrina ” or ” Zookeeper Physics ” you should get to see all the Uploaded videos. Though we have many more study videos.

Thanks and Regards
Zookeeper ;-D    Subhashish Chattopadhyay

[ I suggest you see the videos starting with 1- first then starting with 2- ..... in that sequence. ]

[ Tell your friends about this link if you liked the videos ]

In case of doubts or suggestions, Please send me email at    mokshya@gmail.com

In youtube you can search for Dezrina as these have been uploaded with this login-id. or search for ” Zookeeper IIT JEE Physics “

Answers to -> Frequently Asked Questions ( FAQ ) [ commonly asked intelligent Questions :-)  ]

1 ) How do I prepare for IIT ?

Ans : – See the videos made by me ( in youtube search as Dezrina or “ Zookeeper Physics “ – will see all Uploaded ones. Though we have many more which have not been uploaded ). While watching the videos, take notes and try to solve the problems yourself by pausing the video. Tell me if any calculation is wrong. See the videos with 1- first then 2- and so on.  Write to IAPT Kothrud, Pune office to buy ( 150 Rs approx ) the book with previous papers of NSEP ( National Standard Exam in Physics – The 1st level ), INPhO ( Indian National Physics Olympiad – 2nd level ). Prepare with these and see how much you are scoring. You can guess your ALL INDIA rank easily from NSEP, and INPhO rank. Since 1998 the IIT JEE toppers have been mostly representing India in IPhO.

2 ) Some Videos have sound Problem … what do I do ?

Ans : – Only 4 videos have very slight sound echo problem. The same topic got covered again ( many times ) in other videos correctly. The room in which these 4 videos have been recorded had windows only on one side. That gave little bit echo problems. Also external noise of cars, auto, children shouting have randomly come in. You have to have good speakers with filters or good earphones with filters. We have checked mostly it is OK with these. ( If you are depending only on your embedded speakers of computer /screen / keyboard then there may be extra distortions. As these speakers are often not of good Quality. Also install latest KL Codecs ) In any case reduce the volume see the board, imagine sitting in the last bench and solving the problems of your own. See if your solution differs anywhere with the scribbles on the board.

3 ) Why are you giving these ( high Quality ) lecture for free ?

Ans : Well there are lot of good things free in this world. Linux, My-SQL, Open-Office ….. Go to sourceforge and get thousands of high quality software free along with source code. Yes all officially free …. Why do you think Richard Stallman, Zimmerman, ….. etc are considered Guru philosophers ? In Punjab and Gurudwaras worldwide there are so many Langars where you get better food than Restaurants. … why ? Why do you have Dharmasalas and subsidized rest rooms near hospitals / Famous Temples / various places ? in Iftar party anyone can eat for free …  why ?  I am teaching for 20 years now and observed most students can do much better if they have the self motivation to solve and practice. Cheap books are available in second hand bookstalls, where you get thousands of Numericals to solve ….. but most students will like to blow their time going and coming for tuition, travel time …. TV for hours and hours watching cricket / Tennis games, playing computer games …. My free lectures are not going to make much difference in spending of unnecessary money for coaching ….. I know very well , how much people enjoy .. , spending unnecessarily !!    Do you know that there are NO poor / needy students in Bangalore. Sometime back I had tried to teach for IIT JEE FREE. Discussed with a few NGOs and social service guys. Arranged rooms but got only 1 student. We had informed many people in many ways to inform students …. We did not get students who are ready to learn for free. So I am sure these lectures are NOT FREE. If anyone learns from these, s/he changes and that’s the gain / benefit. This change ( due to learning ) is very costly …. Most do not want to learn ………..

In youtube  search for Zookeeper Dezrina     you will get most videos. I say most because I do not upload all videos that I make. I have many more videos which are not in the net.

4 ) How can I get all your lectures ?

Ans : – Apart from my lectures there are approx 450GB of PCM ( Phy, Chem, Math ) lectures. It takes approx 3 years of continuous download from scattered sources. I have ( 20,000 )Thousands  of these. You can take ALL of them from me in an external 500 GB hard disk, instead of spending so much money and time again for downloading. These cover ( by Various  Professors )  everything of Chemistry, Physics, Maths… Lot of this is from outside India … as foreigners have much wider heart than Indians ( as most of GNU / open source software have been developed  by Non-Indians ). I observed the gaps in these videos, and thus I am solving IIT, APhO, Roorkey, IPhO Numericals. Videos made by me along with these videos gives a complete preparation.

Send me a mail at          mokshya@gmail.com          to contact me.

In youtube  search for Zookeeper Dezrina     you will get most videos. I say most because I do not upload all videos that I make. I have many more videos which are not in the net.

5 ) How do you get benefited out of this ?

Ans :- If anyone learns we all will have better people in this world. I will have better “ YOU “.

6 ) Why do you call yourself a Zookeeper ?

Ans :- This is very nicely explained at http://zookeepersblog.wordpress.com/z00keeper-why-do-i-call-myself-a-zoookeeper/
7 ) Where do you stay ?

Ans :- Presently I am in Bangalore.

8 ) If I need videos in a few topics can you make them for me ?

Ans :- Yes. You have to discuss the urgency with me. If I am convinced I will surely make these quickly for you and give you and ALL. I teach both Maths and Physics. So anything in these 2 subjects are welcome.

9 ) Why did you write an article saying there are No Poor students ?

Ans :- There are lots of NGOs and others working for rural / poor children education at lower classes. While very less effort is on for std 9 till 12. Also see the answer in question number ( 3 ) above. In last 20 years of teaching I never met a Poor child who was seriously interested in ( higher ) studies. As I have a mind / thinking of a ” Physicist “, I go by ” Experimental Observation “.

It is not about what is being said about poor in media / TV etc, or ” what it should be ” ( ? ) …. It is about what I see happening. Also to add ( confuse ? you more )…. You must be knowing that in several states over many years now girl students have better ( by marks as well as by pass percentage ) result in std 10 / Board Exams….. well but NEVER a girl student came FIRST in IIT JEE … why ? [ The best rank by a Girl student is mostly in 2 digits, very rarely in single digit ]  ????? So ????

10 ) How much do I have to study to make it to IIT ?

Ans :- My experience of  Teaching for IIT JEE since last 20 years, tells me, Total 200 hours per subject ( PCM ) is sufficient. If you see my Maths and Physics videos, each subject is more than 200 hours. So if someone sees all the videos deligently, takes notes and remembers, …… Done.

11 ) What is EAMCET ?

Ans :- Engineering Agriculture and Medicine Common Entrance Test is conducted by JNT University Hyderabad on behalf of APSCHE. This examination is the gateway for entry into various professional courses offered in Government/Private Colleges in Andhra Pradesh.

12 ) In your videos are you covering other Exams apart from IIT ?

Ans : – Yes. See many videos made by solving problems of MPPET, Rajasthan / J&K CET, UPSEAT ( UPES Engineering Aptitude Test ), MHCET, BCECE ( Bihar Combined Entrance Competitive Examination Board ), WB JEE etc

13 ) What is SCRA ?

Ans : – Special Class Railway Apprentice (SCRA) exam is conducted by Union Public Service Commission (UPSC) board, for about 10 seats.That translates into an astonishing ratio of 1 selection per 10,000 applicants. The SCRA scheme was started in 1927 by the British, to select a handful of most intelligent Indians to assist them in their Railway Operations, after training at their Railway’s largest workshop, i.e. Jamalpur Workshop, and for one year in United Kingdom. The selected candidates were required to appear in the Mechanical Engineering Degree Exmination held by Engineering Council (London).

Thanks for your time. To become my friend in google+  ( search me as  mokshya@gmail.com and send friend request )

Read http://edge.org/responses/what-scientific-concept-would-improve-everybodys-cognitive-toolkit

Temperature-Sea Levels-CO2-etc always have been fluctuating over ages-Global Warming
http://www.youtube.com/watch?v=DMH8O9v6YaY
The following video is a must see for full CO2 cycle, plates of Earth, Geological activities, stability of weather
http://www.youtube.com/watch?v=Qtv_JD_I5X8
The Great Global Warming swindle
http://www.youtube.com/watch?v=YtevF4B4RtQ
Article in Nature says CO2 increase is good for the trees http://thegwpf.org/science-news/6086-co2-is-greening-the-planet-savannahs-soon-to-be-covered-by-forests.html
Ice cap variations, Temperature and humidity fluctuations nicely explained
http://www.youtube.com/watch?v=pz-w4NWRObw
http://climaterealists.com/index.php?id=9752

BBC documentary Crescent and Cross shows the 1000 years of fight between Christians and Muslims. Millions have been killed in the name of Religion. To decided whose GOD is better, and which GOD to follow. The fight continues. http://www.youtube.com/watch?v=zqK-RuntywY
The Virus of Faith
http://www.youtube.com/watch?v=scarHc8RA0g
The God delusion
http://www.youtube.com/watch?v=LVr9bJ8Sctk
cassiopeia facts about evolution  http://www.youtube.com/watch?v=K7tQIB4UdiY
Intermediate Fossil records shown and explained nicely Fossils, Genes, and Embryos http://www.youtube.com/watch?v=fdpMrE7BdHQ

The Rise Of Narcissism In Women  http://www.youtube.com/watch?v=wZHKCbHGlS0
13 type of women whom you should never court http://timesofindia.indiatimes.com/life-style/relationships/man-woman/13-Women-you-should-never-court/articleshow/14637014.cms
Media teaching Misandry in India www.youtube.com/watch?v=-M2txSbOPIo

Summary of problems with women http://problemwithwomentoday.blogspot.in/2009/12/problem-with-women-today-what-in-hell.html
Eyeopener men ? women only exists www.youtube.com/watch?v=6ZAuqkqxk9A
Most unfortunate for men http://www.youtube.com/watch?v=73fGqUwmOPg

Miracles for Sale http://www.youtube.com/watch?v=iuP5uOI7Xwc
The Enemies of Reason http://www.youtube.com/watch?v=0CyMglakWoo
Each of you is an Activist in some way or other. You are trying to propagate those thoughts, ideas that you feel concerned / excited about. Did you analyze your effectiveness ? http://www.youtube.com/watch?v=61qn7S9NCOs Culturomics can help you :-D
Why some temples become ” FAMOUS ” ? How you can be manipulated ? Luck for others ? http://www.youtube.com/watch?v=O4mN33w5Ftw

see how biased women are. Experimental proof. Women are happy when they see another woman is beating a man ( see how women misbehave with men )
www.youtube.com/watch?v=LlFAd4YdQks
see detailed statistics at http://www.youtube.com/watch?v=5lHmCN3MBMI

An eye opener in Misandry http://www.youtube.com/watch?v=YiTaDS_X6CU

My sincere advice would be to be EXTREMELY careful ( and preferably away ) of girls. As girls age; statistically certain behavior in them has been observed. Most Male can NOT manage those behaviors… Domestic violence, divorce etc are rising very fast. Almost in all cases boys / males are HUGE loosers. Be extremely choosy ( and think from several angles ) before even talking to a girl. http://zookeepersblog.wordpress.com/save-the-male/
How women manipulate men http://www.angryharry.com/esWomenManipulateMen.htm

Gender Biased Laws in India http://zookeepersblog.wordpress.com/biased-laws/
Violence against Men http://www.youtube.com/watch?v=MLS2E-rRynE
Only men are victimised http://www.youtube.com/watch?v=4JA4EPRbWhQ

Men are BETTER than women http://www.menarebetterthanwomen.com/
see  http://www.youtube.com/watch?&v=T0xoKiH8JJM#!
Male Psychology http://www.youtube.com/watch?v=uwxgavf2xWE

Women are more violent than men http://www.independent.co.uk/news/uk/home-news/women-are-more-violent-says-study-622388.html

Misandry in Media http://www.youtube.com/watch?v=j7U0r7vIrgM
In the year 2010, 168 men ended their lives everyday ( on average ). More husbands committed suicide than wives. http://www.rediff.com/news/report/ncrb-stats-show-more-married-men-committing-suicide/20111028.htm

It is EXTREMELY unfortunate that media projects men as fools, women as superiors, Husbands as servants, and replaceable morons. In ad after ad worldwide from so many companies, similar msg to disintegrate the world is being bombarded. It is highly unacceptable misandry http://www.youtube.com/watch?v=oq14WHkFq30

It is NOT at all funny that media shows violence against MEN. Some advertisers are trying to create a new ” Socially acceptable culture ” of slapping Men ( by modern city women ). We ( all men ) take objection to these advertisements. We oppose this Misandry bad culture. Please share to increase awareness against Men bashing http://www.youtube.com/watch?v=D8ecN2rh0uU

Are you a nice person ? Just shout Wooooooooo , Eyye Eyye and enjoy to see someone in trouble …. Extension of Milgram Experiments – In a Mob also people become cruel step by step – http://www.youtube.com/watch?v=scOJqyiYVtk

Think what are you doing … why are you doing ? http://www.youtube.com/watch?v=qp0HIF3SfI4

Every Man must know this …  http://www.youtube.com/watch?v=cIFmQHJEG1M
Manginas, White Knights, & Other Chivalrous Dogs  http://www.youtube.com/watch?v=oXQDtBT70B8

!
!
: ****__********__***
…….. (””(`-“’´´-´)””)
……….)…..–…….–….(
………/…..(6…_…6)….\
………\……..(..0..)….;../
……__.`.-._..’=’…_.-.`.__
…./……’###.,.–.,.###.’…\
….\__))####’#’###(((__/
……##### u r #####
……..### SWEET. ###
……/….#########…\
..__\…..\..######/…../
(.(.(____)….`.#.´..(____).).)

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