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MP Board Class 8th Social Science Solutions Chapter 6 स्थलमण्डल-स्थल एवं स्थलाकृतियाँ

MP Board Class 8th Social Science Chapter 6 अभ्यास प्रश्न

प्रश्न 1.
निम्नलिखित प्रश्नों के सही विकल्प चुनकर लिखिए –
(1) पृथ्वी का लगभग कितने प्रतिशत भाग जलमण्डल से घिरा हुआ है?
(क) 61 प्रतिशत
(ख) 71 प्रतिशत
(ग) 81 प्रतिशत
(घ) 51 प्रतिशत
उत्तर:
(ख) 71 प्रतिशत

(2) किस देश में सबसे अधिक भूकम्प आते हैं ?
(क) भारत
(ख) फ्रांस
(ग) जापान
(घ) श्रीलंका
उत्तर:
(ग) जापान

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिए –
(1) पृथ्वी का लगभग ……………… प्रतिशत भाग स्थल द्वारा घिरा हुआ है।
(2) पृथ्वी की सबसे. बाहरी पर्त को भूपर्पटी या ………… कहते हैं।
(3) सियाल में सिलिका तथा ……………….. धातुओं की प्रधानता होती है।
(4) प्रति 32 मीटर गहराई पर ………………. सेल्सियस तापमान बढ़ जाता है।
उत्तर:

1. 29
2. बाहरी परत
3. एल्युमिनियम
4. 12

MP Board Class 8th Social Science Chapter 6 अति लघु उत्तरीय प्रश्न

प्रश्न 3.
(1)भू-पटल की प्राथमिक शैलें कौन-सी हैं ?
उत्तर:
आग्नेय शैलें भू-पटल की प्राथमिक शैलें हैं।

(2) धरातल की तीन प्रमुख स्थलाकृतियों के नाम लिखिए।
उत्तर:

• पर्वत
• पठार
• मैदान

धरातल की तीन प्रमुख स्थलाकृतियाँ हैं।

(3)शैलों में जब लहरनुमा मोड़ पड़ जाते हैं तो उन्हें क्या कहते हैं ?
उत्तर:
शैलों में जब लहरनुमा मोड़ पड़ जाते हैं तो उन्हें ‘वलन’ कहते हैं।

(4)संसार में सबसे अधिक ज्वालामुखी कहाँ हैं ?
उत्तर:
संसार में सबसे अधिक ज्वालामुखी प्रशान्त महासागर के चारों ओर तटीय भागों तथा महाद्वीपीय क्षेत्रों में हैं।

MP Board Class 8th Social Science Chapter 6 लघु उत्तरीय प्रश्न

प्रश्न 4.
(1) वलन तथा भ्रंशन में क्या अन्तर है ?
उत्तर:
पृथ्वी के अन्दर क्षैतिज भू-संचलन द्वारा जब शैलों में लहरनुमा मोड़ पड़ जाते हैं तो इन्हें ‘वलन’ कहते हैं जबकि क्षैतिज भू-संचलन से उत्पन्न दबाव तथा तनाव के कारण शैलों के टूटकर अलग होने की प्रक्रिया को ‘भ्रंशन’ कहते हैं।

(2) अवसादी शैलों का निर्माण कैसे होता है ?
उत्तर:
जल, वायु एवं हिम द्वारा बहाकर लाये कंकड़, पत्थरों के छोटे-छोटे कण, जीवाश्म आदि (अवसाद) भू-भाग या समुद्र तल में परतों के रूप में जमा हो जाते हैं और ये अवसाद की परतें गर्मी तथा दबाव के कारण कठोर हो जाती हैं। यही कठोर पदार्थ अवसादी शैल कहलाते हैं।

(3) ज्वालामुखी किसे कहते हैं ? उद्भेदन के दो कारण दीजिए।
उत्तर:
ज्वालामुखी’ भू-पटल पर एक गोल छेद या दरार वाला खुला भाग होता है। इससे होकर पृथ्वी के अत्यन्त तप्त | भू-गर्भ से गैसें, तरल लावा, ऊष्ण जल, चट्टानों के टुकड़े, राख व धुआँ आदि निकलता है। प्लेटों का खिसकना व भूकम्प ज्वालामुखी उद्भेदन के दो प्रमुख कारण हैं।

(4) भूकम्प से लाभ तथा हानियाँ लिखिए।
उत्तर:
भूकम्प से लाभ –

• इससे कभी-कभी उपजाऊ भूमि उभर आती है।
• नवीन भू-आकारों का निर्माण होता है।
• इनसे बहुमूल्य खनिज पदार्थ धरातल पर आ जाते हैं।
• इनसे नीचे हो जाने वाले भू-भाग पर झीलों का निर्माण होता है। भूकम्प से हानियाँ
• इससे जन-धन की हानि होती है। मनुष्य, पशु आदि मर जाते हैं। इमारतें गिर जाती हैं। रेलें, सड़कें टूट जाती हैं, कारखाने नष्ट हो जाते हैं।
• नदियों के मार्ग रुकने से भयंकर बाढ़ आ जाती है। समुद्र में बहुत ऊँची विनाशकारी लहरें उठती हैं।
• भूखण्डों में दरारें पड़ जाती हैं तथा कुछ भाग नीचे स जाता है।

MP Board Class 8th Social Science Chapter 6 दीर्घ उत्तरीय प्रश्न

प्रश्न 5.
(1) शैल किसे कहते हैं ? शैलों के विभिन्न प्रकार बताइए।
उत्तर:
धरातल की रचना करने वाले सभी पदार्थ शैल कहलाते हैं। अर्थात् जिन पदार्थों से भूपृष्ठ का निर्माण हुआ है, उन्हें शैल कहते हैं। शैलों के तीन प्रकार हैं –

(1) आग्नेय शैलें – ये शैलें भूपृष्ठ की प्रारम्भिक शैलें हैं। इन्हें प्राथमिक शैलें भी कहते हैं। ये शैलें पृथ्वी के आन्तरिक भाग में पिघले पदार्थों के ठण्डे होने से बनी हैं। भूपृष्ठ के नीचे अति गर्म पिघला पदार्थ भू-पर्पटी में अथवा उसके ऊपर ठण्डा होकर कठोर हो जाता है, उसे आग्नेय शैल कहते हैं।

(2) अवसादी शैल – जल, वायु एवं हिम द्वारा बहाकर लाये गये कंकड़, पत्थरों के छोटे-छोटे कण, जीवाश्म आदि भू-भाग या समुद्र तल में परतों के रूप में जमा होते जाते हैं। इस प्रकार जमे हुए पदार्थ को ‘अवसाद’ कहते हैं। यही अवसाद की पर्ते गर्मी तथा दबाव के कारण कठोर हो जाती हैं तो उन्हें अवसादी या परतदार शैल कहा जाता है।

(3) कायान्तरित शैलें – जब आग्नेय तथा अवसादी शैलों के रूप, रंग और गुण में आन्तरिक ताप तथा दबाव के कारण पूर्ण रूप से परिवर्तन हो जाता है तो उन्हें कायान्तरित या परिवर्तित शैल कहा जाता है।

(2) पृथ्वी की संरचना को रेखाचित्र द्वारा समझाइए।
उत्तर:
पृथ्वी की सबसे बाहरी व ऊपरी पर्त को भू-पर्पटी या स्थलमण्डल कहते हैं। इसी पर प्राणी जगत निवास करता है। यह मण्डल हल्की जलज शैलों से बना हआ है और इसकी मोटाई लगभग 10 से 70 किमी है। स्थलमण्डल की ऊपरी परत को सियाल (Sial) भी कहते हैं। इसमें सिलिका तथा ऐल्युमीनियम –

पृथ्वी की आन्तरिक संरचना

दो धातुओं की प्रधानता है। सियाल के नीचे की परत को सीमा (Sima) कहते हैं। इसमें सिलिका और मैग्नीशियम धातुओं की प्रधानता है। इसके नीचे मैंटल है। इसमें ओलिवाइन और पाइरॉक्सिन खनिजों की प्रधानता है। इसके नीचे पृथ्वी का क्रोड है जिसे ‘नीफे’ (Nife) कहते हैं।

(3) ज्वालामुखी के मानव जीवन पर होने वाले प्रभाव बताइए।
उत्तर:
ज्वालामुखी के मानव जीवन पर होने वाले प्रमुख प्रभाव इस प्रकार हैं –
ज्वालामुखी से विभिन्न स्थलाकृतियों की रचना होती है, जैसे मैदान, पठार, पर्वत आदि। इनसे हमें बहुमूल्य खनिज पदार्थों की प्राप्ति होती है। ज्वालामुखी से निकला लावा चारों ओर फैलकर कालान्तर में उपजाऊ मिट्टी का निर्माण करता है। शान्त ज्वालामुखी के मुख में वर्षा का जल भरने से झीलों का निर्माण होता है। ज्वालामुखी से कई लाभ के साथ-साथ हानियाँ भी होती हैं। ज्वालामुखी उद्गार से मानव, जीव-जन्तु, वनस्पति, कृषि क्षेत्र, मानव आवास एवं बड़े-बड़े नगर, गाँव जलकर नष्ट हो जाते हैं अथवा दबकर ध्वस्त हो जाते हैं।

(4) भूकम्प किसे कहते हैं ? भूकम्प आने के कारण लिखिए
उत्तर:
भूकम्प – भूकम्प शब्द दो शब्दों से बना है – भू तथा कम्प, जिसका सामान्य अर्थ ‘पृथ्वी का कम्पन’ है। जिस तरह किसी शान्त जल में पत्थर का टुकड़ा फेंकने पर गोलाकार लहरें, केन्द्र से चारों ओर प्रवाहित होती हैं उसी तरह भूगर्भ उद्गम केन्द्र (गड़बड़ी वाले स्थान) से भूकम्प लहरें चारों ओर फैलती हैं। भूकम्प की उत्पत्ति जिस स्थान पर होती है, उसे ‘भूकम्प केन्द्र (फोकस)’ कहते हैं।
भूकम्प आने के कारण –

• भूगर्भ में यदाकदा अचानक हलचल हो जाने के कारण पृथ्वी की सतह पर. भूकम्प आ जाते हैं।
• तीव्र ज्वालामुखी उद्भेदन होने पर भी भूकम्प आते हैं।
• कभी – कभी भूगर्भ में बनने वाली गैसों एवं जल वाष्प भी कमजोर भू-पटल को हिला देती है जिससे भूकम्प आते हैं।

(5) निम्नलिखित शैलों को दिये गये शैलों के प्रारूप में अंकित कीजिए
(1) संगमरमर
(2) कोयला
(3) ग्रेनाइट
(4) चूने का पत्थर
(5) बेसाल्ट
(6) हीरा
उत्तर:
शैलों का प्रारूप:

MP Board Class 8th Social Science Solutions

MP Board Class 12th Chemistry Solutions Chapter 1 The Solid State

The Solid State NCERT Intext Exercises

Question 1.
Why are solids rigid?
Answer:
Solids are rigid due to the presence of strong intermolecular forces between the constituent particles.

Question 2.
Why do solids have a definite volume
Answer:
In solids, the constituent particles are bonded together by strong attractive forces between them. By the increase or decrease of pressure, the intermolecular space remains unaffected. Thus, the volume of solids is definite.

Question 3.
Classify the following as amorphous or crystalline solids:
Polyurethane, naphthalene, benzoic acid, Teflon, potassium nitrate, cellophane, polyvinyl chloride, fiberglass, copper.
Answer:
Amorphous solids: Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass.
Crystalline solids : Benzoic acid, potassium nitrate, copper.

Question 4.
Why is glass considered a supercooled liquid?
Answer:
Like liquids, glass has tendency to flow but very slowly. Therefore, it is called supercooled liquid.

Question 5.
Refractive index of a solid is observed to have the same value along with all directions. Comment on the nature of this solid. Would it show cleavage property?
Answer:
Since refractive index of a solid is observed to have the same value along with all directions, the solid must be amorphous. On cutting this, solid will undergo irregular cleavage.

Question 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.
Answer:
Potassium sulphate : Ionic, Tin : Metallic, Benzene : Molecular (non-polar), Urea: Molecular (polar), Ammonia: Molecular (H-bonded), Water: Molecular (H-bonded), Zinc sulphide: Ionic, Graphite: Covalent or network, Rubidium: Metallic, Argon: Molecular (non-polar), Silicon carbide : Covalent or network.

Question 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 ?
Answer:
Covalent solid.

Question 8.
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Answer:
In solid-state, the ions are fixed and the ions cannot move in an applied electric field. But in a molten state, the ions are free to move and hence conduct electricity.

Question 9.
What type of solids are electrical conductors, malleable and ductile?
Answer:
Metallic solids, they are conductor due to the presence of free electron in them.

Question 10.
Give the significance of a ‘lattice point’.
Answer:
The lattice point represents the position of the constituent particles in the crystal lattice. When lattice points are joined by straight lines, they bring out the geometry of the lattice.

Question 11.
Name the parameters that characterize a unit cell.
Answer:
A unit cell is characterized by :

1. The dimensions along the three edges. These are represented by a, b, and c.
2. The angle between the edges. These are represented by α,β, and γ. The angle α is between b and c,β is between a and c and γ is between a and b (For diagram refer book).

Question 12.
Distinguish between
1. Hexagonal and monoclinic unit cells
2. Face centred and end centred unit cells.
Answer:
1. Hexagonal unit cell:
a = b ≠ c and α = β = 90°, γ = 120°
Monoclinic unit cell:
a ≠ b ≠ c and α = γ = 90°, β ≠ 90°

2. Face-centred unit cell:
Particles at the comers as well as at the centre of each face.
Number of particles per unit cell = 8 × \(\frac{1}{8}\) + 6 × \(\frac{1}{2}\) = 4

End – centred unit cell:
Particles at the corners and at the centres of two end faces.
Number of particles per unit cell = 8 × \(\frac{1}{8}\) + 2 × \(\frac{1}{2}\) = 2

Question 13.
Explain, how many portions of an atom located at (i) corner and (ii) body centre of a cubic unit cell is part of its neighbouring unit cell?
Answer:

1. In a crystal, each comer of a unit cell is common to 8 unit cells and hence each comer atom is shared by 8 unit cells. Thus, \(\frac{1}{8}\) atom is part of its neighbouring unit cell.
2. The atom located at the body centre of a cubic unit cell belongs to only one unit cell. Thus it is not shared with the neighbouring unit cell.

Question 14.
What is the two-dimensional coordination number of a molecule in a square close-packed layer?
Answer:
4.

Question 15.
A compound forms a hexagonal close-packed structure. What is the total number of voids in 0-5 mol of it? How many of these are tetrahedral voids?
Answer:
The number of octahedral voids = N
The number of tetrahedral voids = 2N
∴ Total number of voids = 3N
The number of close packed atoms = 0.5 mol = 0.5 × 6.022 × 10²³
∴ Total number of voids = 3 × 0.5 × 6.022 × 10²³ = 9.03 × 10²³
The number of tetrahedral voids = 2 × 0.5 × 6.022 × 10²³ = 6.022 × 10²³

Question 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?
Answer:
Since N forms ccp arrangement, it will have 4 atoms in a unit cell.
Number of N atoms in unit cell = 4
For each atom, there are two tetrahedral voids so that there are 8 tetrahedral voids per unit cell.
No. of M atoms = \(\frac { 1 }{ 3 } \) × 8 = \(\frac { 8 }{ 3 } \)
Formula = M_8/3 N₄ or M₂N₃.

Question 17.
Which of the following lattices has the highest packing efficiency
(i) simple cubic
(ii) body-centered cubic and
(iii) hexagonal close-packed lattice?
Answer:
The packing efficiencies are:
Simple cubic = 52.4%
Body centred cubic = 68%
Hexagonal close packed = 74%
∴ Hexagonal close-packed lattice has the highest packing efficiency.

Question 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?
Answer:
We know that Z = \(\frac{a^{3} \times \mathrm{N}_{\mathrm{A}} \times d}{\mathrm{M}}\)
Where a = 405pm = 405 x 10^-10
d = 2.7 × 10³ kg m^-3
M = 2.7 × 10^-2 kg m^-1
= 2.7g mol^-1
N_A = 6.203 x 10²³

Z =  4
∴ The element has fee (cep) unit cell.

Question 19.
What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
Answer:
Vacancy defect. It results in a decrease in the density of the substance since some of the lattice sites are occupied.

Question 20.
What type of stoichiometric defect is shown by : (i) ZnS, (ii) AgBr.
Answer:
(i) ZnS, shows Frenkel defect due to large difference in size of ions.
(ii) AgBr, shows both Frenkel defect and Schottky defect.

Question 21.
Explain, how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it ?
Answer:
When a cation of higher valence is added as an impurity, some of the sites of the original cations are occupied by the cations of higher valence and produce vacant sites.

Question 22.
Ionic solids, which have anionic vacancies due to metal excess defect, develop colour ? Explain with the help of a suitable example.
Answer:
The anionic vacancies due to metal excess defect in ionic solids are occupied by free electrons to maintain electrical neutrality. These impart colour by excitation of these electrons when they absorb energy from the visible light falling on the crystals. For example :
When NaCl is heated in presence of sodium vapours, Na⁺ ions are in excess, Cl^– ions leave their normal site and come to the surface. The vacant site of anion is occupied by electron forming F-centre. They absorb light from visible region and radiate complementary colour.

Question 23.
A group-14 element is to be converted into an n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?
Answer:
n-type semiconductors are obtained by doping of a higher group impurity. Hence, to convert group 14 element to n-type semiconductor, it should be doped with a Group 15 element.

Question 24.
What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic? Justify your answer.
Answer:
Ferromagnetic materials would make better permanent magnets than ferrimagnetic materials because, in ferromagnetic solids, the magnetic moments of unpaired electrons spontaneously align themselves in the same direction. However, in ferrimagnetic solids the magnetic moments of the domains are aligned in the parallel and anti-parallel direction in unequal numbers.

The Solid State NCERT Textbook Exercises

Question 1.
Define the term ‘amorphous’. Give a few examples of amorphous solids.
Answer:
The term ‘amorphous’ (Greek) means ‘no forms’ and in amorphous solids, the constituent particles have only short-range order, e.g. Glass, plastics, rubber.

Question 2.
What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Answer:
Glass is an amorphous solid in which the constituent particles (SiO₄ tetrahedra) have only short-range order and there is no long-range order. On melting quartz and then cooling it rapidly, it is converted into glass.

Question 3.
Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
(i) Tetra phosphorus decoxide (P₄O₁₀)
(ii) Ammonium phosphate (NH₄)₃PO₄
(iii) SiC
(iv) I₂
(v) P₄
(vi) Plastic
(vii) Graphite
(viii) Brass
(ix) Rb
(x) LiBr
(xi) Si.
Answers:
Ionic: Ammonium phosphate, LiBr
Metallic: Brass, Pb
Molecular: I₂, P₄O₁₀, white phosphorus
Covalent: Graphite, SiC, Si, red phosphorus
Amorphous: Plastic

Question 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-centered cubic structure?
Answer:
(i) The coordination number of constituent particles (atom, ion, or molecule) in a crystal is the number of constituent particles which are the immediate neighbor of the particle in the crystal. In ionic crystals, the coordination number of an ion in the crystal is the number of oppositely charged ions surrounding that particular ion.
(ii) a] 12 b] 8.

Question 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.
Answer:
Atomic mass (M)

Question 6.
‘Stability of a crystal is reflected in the magnitude of its melting points’, 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?
Answer:
The stability of a crystal depends on the force of attraction so: Higher the melting point of the crystal, the stronger the intermolecular force of attraction, hence greater is the stability of a crystal.

Melting point of the H₂O, C₂H₅OH, diethyl ether and methane given as below :

On the basis of melting point, the strength of the intermolecular forces between these molecules follows the order :
Water > Diethyl ether > Ethyl alcohol > Methane.

Question 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?
Answer:
(i) During the formation of crystals, the constituent particles are closely packed. In close packing the maximum available space is occupied. The stability of the crystals depends on the extent of close packing. In close packing the solid particles are considered as identical spheres.

(ii) Crystal lattice :
The regular arrangement of the constituent particle (atoms, molecules, or ions) of a crystal in three-dimensional space.

Unit cell :
It is the smallest repeating unit in three-dimensional space, which is repeated again and again to give the complete lattice.

(iii)Tetrahedral void :

1. It is the open space between four touching spheres of two layers of atoms.
2. The radius of tetrahedral void relative to radius of sphere is 0-225.

Octahedral void :

1. It is the open space between six touching spheres of two layers of atoms.
2. The radius of octahedral void relative to radius of sphere is 0-414.

Question 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.
Answer:
Lattice points in face centred cubic or face centred tetragonal = 8 (at corners) + 6 (at face centres) = 14
However, particles per unit cell = 8 × \(\frac{1}{8}+\) + 6 × \(\frac{1}{2}+\) =4
lattice points in body centred cube = 8 (at comers) + 1 (at body centre) = 9.
However, particles per unit cell = 8 × \(\frac{1}{8}+\) + 1 = 2

Question 9.
Explain:
1. The basis of similarities and differences between metallic and ionic crystals.
2. Ionic solids are hard and brittle.
Answer:

1. Ionic solids:
In ionic solids, the constituent particles are positive and negative ions. The positive ions are surrounded by negative ions and vice versa. The ions are held together by a strong electrostatic force of attraction.

Since the ions are firmly fixed in the crystal lattice, they are hard and brittle with high melting point, high heat of fusion and lattice energy. These are insulators in the solid-state but are good conductors in solution or molten state. e.g. NaCl, MgSO₄, CaF₂ etc.

2. Molecular solids:
Molecular solids are made up of small discrete molecules. They are subdivided into the following categories.

(a) Non-polar molecular solids:
In these solids, the non-polar molecules like H₂, Cl₂, I₂ etc are held together by weak dispersion forces or London forces. They are soft, non-conductors of electricity and have low melting points.

(b) Polar molecular solids:
Molecules having polar covalent bonds are held together by dipole-dipole interactions. Solid SO₂, NH₃ etc are examples of this type. They are soft and non-conductors of electricity.

(c) Hydrogen bonded molecular solids:
The molecules are held together by strong hydrogen bonds.
e.g. Ice.

3. Ionic solids have strong interionic attractive forces and hence are hard. When an ionic solid is subjected to distortion, the ions of the same charge are brought close together and the repulsive forces between them cause the crystal to fracture. Hence are brittle.

Question 10.
Calculate the efficiency of packing in the case of a metal crystal for:
(i) Simple cubic
(ii) Body centered cubic
Solution:
(i) Packing efficiency of simple cube

(ii) Packing efficiency in body centred cubic structure

(iii) Packing efficiency of face centred cubic structure

Question 11.
Silver crystallizes in fee lattice. If edge length of the cell is 4.077 × 10^-8 cm and density is 10-5 gcm-3, calculate the atomic mass of silver.
Solution:

Z = 4 (fcc lattice), d = 10.5 gcm, N = 6.022 x 10, (a = 4.077 × 10^-8 cm)

Question 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?
Answer:
As atoms Q are present at the 8 corners of the cube, therefore, number of atoms of in the unit cell = \(\frac { 1 }{ 8 }\) × 8 = 1
As atoms P are present at the body centre, therefore, number of atoms of x in the unit cell = 1
∴Ratio of atoms x:y =1:1
Hence the formula of the compound is PQ coordination number of each of P and Q = 8.

Question 13.
Niobium crystallizes in a body-centered cubic structure. If density is 8.55 gem-3, calculate atomic radius of niobium using its atomic mass 93u.
Solution:
Density = 8.55 g cm^-3
Let, length of the edge = a cm
Number of atoms per unit cell, Z = 2 (bcc)
Atomic mass, M = 93 g mol^-1
density, d = \(\frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}\)

= 1.431 x 10^-10m = 0.143 nm

Question 14.
If the radius of the octahedral void is r and the radius of the atoms in close packing is R, derive a relation between r and R.
Solution:
Atoms covering the octahedral void from the top and below are not shown in the figure. Centre of the octahedral void is C and its radius is equal to r. Atoms surrounding the voids are of radius R.

According to the figure,
Radius of atom surrounding the void BA = R
BC = Radius of void + Radius of outer atom = R + r
∠ABC = 45°
in triangle ABC \(\frac {AB}{BC} \) = cos 45° = \(\frac{1}{\sqrt{2}}\) = 0.707
or \(\frac {AB}{BC} \) = 0.707
or 0.707 R + 0.707 r = R
or o.293 = 0.707r
\(\frac {r}{R} \) = \(\frac {0.293}{0.707} \) = 0.414

Question 15.
Copper crystallizes into a fee lattice with edge length 3.61 × 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 gcm^-3.
Solution:
We know that, d = \(\frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}\)
For fcc Z = 4, Atomic mass of copper = 63.5, a = 3.16 x 10^-8

= 8.96 g cm^-3
This value is close to measured value.

Question 16.
The analysis shows that nickel oxide has the formula NiO_0.98 O_1.00. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions?
Solution:
The formula Ni₀₉₈ O_1.00 shows that Ni : O = 0.98 : 1.00 = 98 : 100.
Thus, if there are 100 O-atoms, then Ni atoms = 98
charge on 100 O2′ ions = 100 x (-2) = – 200
suppose Ni atoms as Ni²⁺ = x
Then Ni atoms as Ni³⁺ = 98 – x
The total charge on Ni2+ and Ni³⁺ = (+2) x + (+3) (98 – x) = 294 – x
As metal oxide is neutral, total charge on
cations = total charge on anions
294 – x = 200
x = 94
∴ % of Ni as Ni²⁺ = 96%
% of Ni as Ni³⁺ = 100 – 96 = 4.

Question 17.
What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
Answer:
The solids which have intermediate conductivities generally between 10^-6 to 10⁴ Ω^-1m^-1 are called semiconductors, e.g. Germanium and silicon. The two main types of semiconductors are as follows.

(i) n-type semiconductor:
When a silicon crystal is doped with atoms of group-15 elements, such as P, As, Sb or Bi, then only four of the five valence electrons of each impurity atom participate in forming covalent bonds and fifth electron is almost free to conduct electricity. Silicon that has been doped with a group-15 element is called an n-type semiconductor.

(ii) p-type semiconductor:
When a silicon crystal is doped with atoms of group-13 elements, such as B, Al, Ga or In, each impurity atom forms only three covalent bonds with the host atoms. The place where the fourth electron is missing is called a hole which moves through the crystal-like positive charge and hence increases its conductivity. Silicon that has been doped with group-13 elements is called a p-type semiconductor.

Question 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?
Answer:
The ratio less than 2:1 in Cu₂O shows that some cuprous (Cu⁺) ions have been replaced by cupric (Cu²⁺) ions. To maintain electrical neutrality, every two Cu⁺ ions will be replaced by one Cu²⁺ ion thereby creating a hole. As conduction will be due to presence of these positive holes, hence it is a p-type semiconductor.

Question 19.
Ferric oxide crystallizes 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.
Solution:
O^2- ions form cubic close-packed lattice.
∴ Number of Fe³⁺ = \(\frac{2}{3}\) of octahedral voids = \(\frac{2}{3}\)
∴ Formula of the compound = \(\mathrm{Fe}_{3, \mathrm{O}}^{3+} \mathrm{O}^{2-}\) = Fe₂O₃

Question 20.
Classify each of the following as being either a p-type or an-type semiconductor :
(i) Ge doped with In
(ii) B doped with Si.
Answer:
(i) Ge is group 14 element and In is group 13 element. Hence, an electron deficit hole is created, and therefore, it is p-type.
(ii) B is group 13 element and Si is group 14 element, there will be the free electron. Hence it is p-type.

Question 21.
Gold (atomic radius = 0.144 nm) crystallizes in a face-centered unit cell. What is the length of a side of the cell?
Solution:
For fee lattice, edge length, a = 2\(\sqrt{2}\)r = 2\(\sqrt{2}\) × 0.144 nm = 0.407 nm

Question 22.
In terms of band theory, what is the difference:
(i) between a conductor and an insulator?
(ii) between a conductor and a semiconductor?
Answer:
(i) The energy gap between the valence band and the conduction band in an insulator is very large whereas in a conductor the energy gap is either very small or there is overlapping between the valence band and conduction band.

(ii) In a conductor, the energy gap between the valence band and conduction band is very small or there is overlapping between valence band and conduction band. But in a semiconductor, there is always a small energy gap between them.

Question 23.
Explain the following terms with suitable examples :
(i) Schottky defect
(ii) Frenkel defect
(iii) Interstitials and
(iv) F-centres.
Answer:
(i) Schottky defect: This defect arises in the crystal when one cation and one anion is missing from their normal lattice site and as a result vacancies are created. Since, the number of missing positive ions is equal to the missing negative ions, the crystal as a whole is electrically neutral. Due to this defect, the density of the crystal decreases.

This defect generally occurs in strongly ionic compounds with high coordination number and where positive and negative ions are almost of similar sizes, e.g., NaCl and CsCl.

(ii) Frenkel defect: This defect is due to vacancy at a cation site. The cation leaves its correct lattice site and moves to another position between the two. Frenkel defects are common in ionic compound which possesses low co-ordination number in which there is large difference between the size of positive and negative ions. As there is no absence of ions from the lattice the density remains the same, e.g. ZnS, AgCl.

(iii) Interstitial defect: When some constituent particles (atoms or molecules) occupy an interstitial site the crystal is said to have interstitial defect. This defect increases the density of the substance. Nonionic solids are examples of this defect.

(iv) F-centres: Alkalihalides like NaCl and KCl show this type of defect when crystals of NaCl are heated in an atmosphere of sodium vapour the sodium atoms are deposited normal on the surface of the crystal. The Cl^– ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl. This happens by the loss of electrons by sodium atoms to form Na⁺ ions. The released electron diffuse into the crystal and occupy anionic sites.

As a result crystal has now excess of sodium. The anionic sites occupied by unpaired electrons are called F-centres. They impart yellow colour to the crystal of NaCl. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals. Other examples are LiCl, KCl etc.

Question 24.
Aluminium crystallizes 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 aluminum?
Solution:
(i) For a cubic close-packed structure, the length of the side of the unit cell is related to the radius.

Question 25.
If NaCI is doped with 10-3 mol % of SrCl₂, what is the concentration of cation vacancies?
Solution:
Introduction of one Sr²⁺ introduces one cation vacancy because Sr²⁺ replaces two Na⁺ ions.
∴ Introduction of 10^-3 moles of SrCl₂ per 100 moles of NaCl would introduce 10^-3 mole cation vacancies.
∴ Number of vacancies per mole of NaCl = \(\frac{10^{-3}}{100}\) = 10^-5 mole = 10^-5 × 6.022 × 10²³ = 6.02 × 10¹⁸ vacancies.

Question 26.
Explain the following with suitable examples :
(i) Ferromagnetism
(ii) Paramagnetism
(iii) Ferrimagnetism
(iv) Anti-ferromagnetism
(v) 12-16 and 13-15 group compounds.
Answer:
On the basis of magnetic properties solids are classified into the following categories :
(i) Ferromagnetic: A substance that shows unusually large paramagnetism and shows permanent magnetism even in absence of a magnetic field is called ferromagnetic.
Examples : Fe, Co, Ni, CrO₂, Fe₃O₄, alnico (alloy of Al, Ni, Co, Fe and Cu)
A ferromagnetic substance if once magnetised remains magnetised permanently. Ferromagnetism arises due to the spontaneous alignment of magnetic moments (due to unpaired electrons) in the same direction.

(ii) Paramagnetic: A substance which is attracted by a magnetic field is called para-magnetic. Paramagnetism arises due to the presence of permanent dipoles due to unpaired electrons in atoms, ions, or molecules.

Example: Cu⁺², Fe³⁺, TiO, CuO, O₂ etc.
(iii) Ferrimagnetic: A substance which shows a fairly good paramagnetic character is called ferrimagnetic.
In ferrimagnetic substances, the alignment of magnetic moments in opposite directions are not equal. As a result, the substance attains net magnetic moments, e.g. Fe₃O₄ is a ferrimagnetic substance which on heating upto 850 K becomes paramagnetic.

(iv) Antiferromagnetic: A substance which shows much-reduced paramagnetism than expected is called antiferromagnetic.
In antiferromagnetic substances, the alignment of magnetic moments is equal and in opposite directions. Hence, the net magnetic moment is zero.
Examples: V₂O₃,Cr₂O₃,MnO,Mn₂O₃,MnO₂,FeO,Fe₂O₃,CoO, Co₃O₄,NiO

(v) 12-16 and 13-15 group compounds: Various types of compounds are formed by the mixture of elements of group 13 and 15 and group 12 and 16 whose average valency is like Ge and Si is 4. Of these specific compounds of group 13-15 are InSb, AlP and GaAs. Gallium arsenious are accelerated sensitive semiconductors. Semiconductors brought a revolutionary change in the manufacture of devices. ZnS, CdS, CdSe, and HgTe are examples of compounds of group 12-16. Bonds of these compounds are not totally covalent and their ionic properties depend on the electronegativity of both the elements present.

The Solid State Other Important Questions and Answers

The Solid State Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Due to Frenkel defect, the density of ionic solids :
(a) Decreases
(b) Increases
(c) Does not change
(d) It changes.

Question 2.
In CsCl each Cl is surrounded by how many Cs :
(a) 8
(b) 6
(c) 4
(d) 2.

Question 3.
Frenkel defect is not shown by :
(a) AgBr
(b)AgCl
(c) KBr
(d) ZnS.

Question 4.
In NaCl crystal number of oppositely charged ions situated at an equal distance are:
(a) 8
(b) 6
(c) 4
(d) 2.

Question 5.
The best conductor of electricity is:
(a) Diamond
(b) Graphite
(c) Silicon
(d) Carbon (Non-crystalline).

Question 6.
Which type of point defect is found in NaCI crystal or KCl crystal:
(a) Frenkel defect
(b) Schottky defect
(c) Lattice defect
(d) Impurity defect.

Question 7.
How many space lattices (Bravais lattice) can be obtained from various crystal systems:
(a) 7
(b) 14
(c) 32
(d) 230.

Question 8.
Diamond is a:
(a) H-bond solid
(b) Ionic solid
(c) Covalent solid
(d) Glass.

Question 9.
The Co-ordination number of Ca²⁺ ions in fluoride structure is :
(a) 4
(b) 6
(c) 8
(d) 3.

Question 10.
8 : 8 Co-ordination number is found in which compound :
(a) MgO
(b) Al₂O₃
(C) CsCl
(d) All of these.

Question 11.
The co-ordination number of body-centered cubic cell is:
(a) 8
(b) 12
(c) 6
(d)4.

Question 12.
The density of unit cell is :

Question 13.
The number of tetrahedral voids in the unit cell of cubic close packing :
(a) 4
(b) 8
(c) 6
(d) 2.

Question 14.
Intra-ionic distance of CsCl will be :
(a) a
(b) \(\frac { a }{ 2 } \)
(c) \(\frac{\sqrt{3}}{2} a\)
(d) \(\frac{2 a}{\sqrt{3}}\)

Question 15.
The number of atoms in the body-centered cubic unit cell is:
(a) 1
(b) 2
(c) 3
(d) 4

Question 16.
Which of the following is Bragg equation :
(a) nλ = 2Φ sin θ
(b) nλ = 2d sin θ
(c) nλ = sin θ
(d) \(n \frac{\theta}{2}=\frac{d}{2} \sin \theta\)

Question 17.
Constituents of the covalent crystal are:
(a) Atom
(b) Molecule
(c) Ion
(d) All of these.

Question 18.
Number of Na atom present in the unit cell of NaCl crystal is :
(a) 1
(b) 2
(c) 3
(d) 4.

Question 19.
What type of magnetic substance are Fe, Co, Ni:
(a) Paramagnetic
(b) Ferromagnetic
(c) Diamagnetic
(d) Antiferromagnetic.

Question 20.
The correct example of Frenkel defect is :
(a) NaCI
(b) CsCl
(c) KCl
(d) AgCI.

Question 21.
Dry ice (solid CO₂) is a/an :
(a) Ionic crystal
(b) Covalent crystal
(c) Molecular crystal
(d) Metallic crystal.

Question 22.
Co-ordination number of Cs in CsCl:
(a) Like Cl i.e., 6
(b) Like Cl i.e., 8
(c) Unlike Cl i.e., 8
(d) Unlike Cl i.e., 6.

Question 23.
Structure of NaCl crystal:
(a) Tetragonal
(b) Cubic
(c) Orthorhombic
(d) Monoclinic.

Question 24.
Each Na⁺ ion in NaCl crystal is surrounded by :
(a) Three Cl^– ions
(b) Eight Cl^– ions
(c) Four Cl^– ions
(d) Six Cl^– ions.

Question 25.
For increasing of electro-conductivity in a solid crystal, mixing of impurities is known as:
(a) Schottky defect
(b) Frenkel defect
(c) Doping
(d) Electronic defect

Question 26.
Which type of lattice is found in KCl crystal:
(a) Face centred cubic
(b) Body centred cubic
(c) Simple cubic
(d) Simple tetragonal.

Question 27.
Number of atoms in a body-centered cubic unit cell of a monoatomic substance is:
(a) 1
(b) 2
(c) 3
(d) 4.

Question 28.
Radius ratio limit for tetrahedral symmetry is :
(a) 0.155
(b) 0.414
(c) 0.732
(d) 0.225.

Question 29.
The defect produced due to a cation and an anion vacancy in a crystal lattice is known as:
(a) Schottky defect
(b) Frenkel defect
(c) Crystal defect
(d) Ionic defect.

Question 30.
If co-ordination number of Cs⁺ is 8 in CsCl then co-ordination number of Cl^– ion is :
(a) 8
(b) 4
(c) 6
(d) 12.

Answers:
1. (c), 2. (a), 3. (c), 4. (b), 5. (b), 6. (b), 7. (b), 8. (c), 9. (c), 10. (c), 11. (a), 12. (a), 13. (b), 14. (c), 15. (b), 16. (b), 17. (a), 18. (d),’ 19. (b), 20. (d), 21. (c), 22. (b), 23. (b), 24. (d), 25. (c), 26. (a), 27. (b), 28. (d), 29 (a), 30. (a).

Question 2.
Fill in the blanks :

1. The defect produced due to the removal of a cation and an anion from a crystal lattice is called ………………..
2. If in a crystal lattice a cation leaves its lattice site and occupies a space in the interstitial site then the defect is called ……………….
3. The cause of electric conduction of NaCl in its molten state is its ………………..
4. Due to ……………….. defect the density of crystal decreases.
5. Total ……………….. types of crystal system are there.
6. ……………….. proposed the concept of the atom for the first time.
7. The ratio of the cation and anion present in a crystal is known as ………………..
8. The process of adding a small amount of impurities in an element or compound is called ………………..
9. Total 14 types of unit cells are there which are known as ………………..
10.  In NaCl crystal structure, the coordination number of both Na⁺ and Cl^– ion is ………………..
11. ……………….. defect is found in ZnS and AgCl crystal.
12. Due to the Schottky defect, the density of crystal ………………..
13. In metallic solids, conductivity is due to the presence of …………..
14. Point defects are found in ………….. crystals. Substances which are attracted in the magnetic field are called …………..
15. For a unit cell, if r = \(\frac{a}{\sqrt{8}}\) then it will be ………….. type of unit cell.
16. The conductivity of semiconductor ………….. on increasing temperature.

Answers:

1. Schottky defect
2. Frenkel defect
3. Free ions
4. Schottky
5. Seven
6. Kannad
7. Radius ratio
8. Doping
9. Bravais lattice
10. Six
11. Frenkel
12. Decreases
13. Free electron
14. Ionic
15. Paramagnetic substance
16. fcc
17. Increases.

Question 3.
Match the following

Answers:
1. (b)
2. (d)
3. (c)
4. (a)

Answers:
1. (c)
2. (d)
3. (a)
4. (b).

Answer:
1. (d)
2. (c)
3. (b)
4. (a).

Question 4.
Answer in one word/sentence:

1. Give two examples of metallic crystals.
2. Give two examples of covalent crystals.
3. Give two examples of ionic crystal.
4. What is the co-ordination number of F^– in CaF₂?
5. What type of crystal is SiC ?
6. What is the value of the co-ordination number of hexagonal close packing structures?
7. Write the formula of radius ratio.
8. What is the type of structure of NaCl crystal?
9. Give an example of body centered cubic cell.
10. Give an example of a compound which has both Schottky and Frenkel type of defect.
11. Give two examples of amorphous or non-crystalline solid.
12. Write Bragg equation.
13. What is effect on the density of a substance or crystal due to the Schottky defect?
14. State the co-ordination number of CsCl and NaCl.
15. Write the formulae of two superconductors’ substances.
16. Give an example of Frenkel defect.
17. Give an example of a superconductor.
18. The radius ratio of the tetrahedral void is.

Answers:

1. Copper, Nickel
2. Diamond, Graphite
3. NaCl, NaNO₃
4. 4
5. Covalent
6. 12
7. 
8. Cubic
9. CsCl
10. AgBr
11. Glass,plastic,
12. nλ = 2d sin θ
13. Due to schottky defect, density of substance decreases
14. Co-ordination number of CsCl = 8 : 8, Co-ordination number of NaCl = 6:6
15. (i) λBa3- Cu₂O₇, (ii) Bi₂Ca₂Sr₂Cu₃O₁₀
16. AgCl
17. Ba_0.7K_0.3BIO₃
18. 0.225.

The Solid State Very Short Answer Type Questions

Question 1.
What are crystalline solids? Crystalline solids are of how many types?
Answer:
Solids in which the constituent particles like atoms, molecules, or ions are in a definite order, with a definite geometry are known as crystalline solids.

Crystalline solids are of four types:

1. Ionic crystal
2. Covalent crystal
3. Molecular crystal
4. Metallic crystal.

Question 2.
Write the formula of density of unit cell.
Answer:

Question 3.
What is crystal lattice?
Answer:
Geometry of a crystal in which unit cells are arranged in a definite order and form a crystal-like shape of the unit cell is known as crystal lattice.

Question 4.
What is a unit cell?
Answer:
The smallest unit formed by the arrangement of constituent particles atoms, ions, or molecules of a crystal in an ordered form is known as the unit cell of the crystal.

Question 5.
Give two-two examples of each of the following :

1. Diamagnetic substance
2. Paramagnetic substance
3. Ferromagnetic substance
4. Antiferromagnetic substance
5. Ferrimagnetism.

Answer:

1. Diamagnetic substance – TiO₂, NaCl
2. Paramagnetic substance – Cu⁺², Fe⁺³
3. Ferromagnetic substance – Fe, Co
4. Antiferromagnetic substance – MnO₂, MnO
5. Ferrimagnetism – Fe₃O₄, Ferrite.

Question 6.
Write the structure and coordination number of the following:
(1) CsCl
(2) NaCl
(3) Zn.
Answer:
(1) CsCl – Structure : Cubic, Co-ordination number : 8.
(2) NaCl – Structure: Octahedral, Co-ordination number : 6.
(3) Zn – Structure : Tetrahedral, Co-ordination number : 4.

Question 7.
What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Answer:
Glass is an amorphous solid in which constituent particles are orderly arranged in a short-range order. Quartz is a crystalline form of silica in which SiO₄ units are orderly arranged in a long-range order.
Quartz can be converted into glass by melting the glass and then cooling it rapidly.

Question 8.
Which are the seven fundamental crystal systems on the basis of crystal geometry?
Answer:
Seven types of crystal system are following:

1. Cubic
2. Tetragonal
3. Orthorhombic
4. Monoclinic
5. Hexagonal
6. Rhombohedral
7. Triclinic.

Question 9.
Write the names of different type of cubic system,
Answer:
Cubic system is of three types :

1. Simple cubic (see)
2. Body centred cubic (bcc)
3. Face centred cubic (fcc).

Question 10.
What is the co-ordination number of Na⁺ and Cl^– in the structure of NaCl?
Answer:
In NaCl structure each Na⁺ is surrounded by 6 Cl^– and each Cl^– is surrounded by 6 Na⁺ ions.
Thus, Co-ordination number of Na⁺ = 6
Co-ordination number of Cl^– = 6.

Question 11.
What is co-ordination number ? What is the effect of temperature and pressure on co-ordination numbers?
Answer:
The number of neighbouring ions around the constituent particles of a crystal lattice is known as its co-ordination number. At high-pressure co-ordination number increases and at low pressure it decreases.

Question 12.
What information is obtained by X-ray diffraction study of crystals?
Answer:
By X-ray diffraction study of crystals, spacing between crystal planes of constituents particles of crystals is known.

Question 13.
What type of solids are electrical conductors : metallic or ductile ?
Answer:
Metallic solids.

Question 14.
Give the significance of a ‘lattice point’.
Answer:
Each lattice point represents one constituent particle of the solid. The constituent particle may be an atom, a molecule (group of atom) or an ion.

Question 15.
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?
Answer:
As the solid has same value of refractive index along all directions, this means that it is isotropic and hence amorphous. Being an amorphous solid, it would not show a clean cleavage when cut with a knife. Instead it would break into pieces with irregular surfaces.

The Solid State Short Answer Type Questions

Question 1.
What is Schottky defect ?
Answer:
This type of defect is found in crystals in which both the cation and anion leave their normal lattice site and make their place vacant. In this defect, density decreases but r electrical neutrality is maintained. Like NaCl, CsCl etc.

Question 2.
What is radius ratio of ions ?
Answer:
Ratio of radii of cation and anion in any crystal is called radius ratio.

For example, in NaCl, ionic radii of Na⁺ and Cl- ions are 95 pm and 181 pm respectively.
Radius ratio in NaCl = \(\frac { 95 }{ 181 } \) = 0.52
Due to presence of other forces, in the crystal the observed value of radius ratio is less as in NaCl crystal it is 0414.

Question 3.
Describe briefly the structure of CsCl.
Answer:
It is AB type ionic crystal with body centred cubic structure. In this Cs⁺ ion in centre of cube and Cl^– ions at comers of cube (or vice versa). Co-ordination number of caesium chloride is 8 : 8 and radius ratio of Cs⁺ and Cl^– ions is 0.732.

In the unit cell of CsCl one Cs⁺ ion and one Cl^– ion are present.
Cs⁺ = 8 (at Centre) × 1 = 1
Cl^– = 8 (at corners) × \(\frac { 1 }{ 8 } \) = 1

Question 4.
Calculate the density of unit cell.
Answer:

If number of particles (atom, molecule, ions) of unit cell is Z and mass of each particle is m, then
Mass of unit cell = m × Z …….(1)
If molar mass of the substance is M, then mass of each particle :

Question 5.
Why is glass considered as super cooled liquid ?
Answer:
Glass is an amorphous solid. Like liquids it has tendency to flow, though very slowly. The proof of this fact is that the glass panes in the windows or doors of old buildings are invariably found to be slightly thicker at the bottom that at the top.

Question 6.
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Answer:
In the molten state, ionic solids dissociate to give free ions and hence can conduct electricity. However in the solid state, as the ions are not free but remain held together by strong electrostatic forces of attraction, they cannot conduct electricity in the solid state.

Question 7.
What type of defect can arise when a solid is heated ? Which physical property is affected by it and in what way ?
Answer:
When a solid is heated, vacancy defect is produced in the crystal. This is because on heating some atoms or ions leave the lattice site completely, some lattice sites become vacant. As a result of this defect the density of the substance decreases because some atoms ions leave the crystal completely.

Question 8.
A compound consists of A and B exhibits cubic structure. Atoms are arranged at corners of cube while B are at the centre of faces. What will be the formula of the compound?
Answer:
Atoms situated at corners are 8 A atoms which are shared by 8 cubes. So in the unit cell:
Number of A atom = 8 × \(\frac { 1 }{ 8 } \) = 1
B atoms are present in centre of 6 faces and each face is shared by two cubes
So number of B atoms = 6 × \(\frac { 1 }{ 2 } \) = 3
Formula of compound will be AB₃.

Question 9.
Prove that in face centred cubic (fee) structure, there are four atoms in a unit cell.
Answer:
Face Centred Cubic Structure : In this structure, one atom is also situated in each face along with the corners of cube. The atom in each face is shared by two faces.
Thus, number of atom in each unit cell = 8 × \(\frac { 1 }{ 8 } \) (at 8 comers) + 6 × \(\frac { 1 }{ 2 } \) (at 6 faces)
= 1 + 3 = 4.

Question 10.
Write Bragg equation.
Answer:
Bragg equation is as follows : 2d sin θ = nλ.
Where, d = Distance between two consecutive planes in a crystal,
θ = Incident angle of X-rays
n = Simple whole number
λ. = Wavelength of X-rays.
By this distance d between the planes of the crystal is determined.

Question 11.
Name the parameters that characterise a unit cell.
Answer:
A unit cell is characterised by (i) Its dimensions along the three edges a, b and c. These edges may or may not be mutually perpendicular
(ii) Angles between the edges, a, (between b and c), β (between a and c) and γ (between a and b). Thus, a unit cell is characterised by six parameters a, b, c, α, β and γ.

Question 12.
Window glass of old buildings appear milky. Why?
Answer:
In day time, glass becomes hot and cools down at night. This way, the process of annealing takes place. Due to annealing in many years, glass develops crystalline property and window glass appear milky.

Question 13.
Common salt sometimes appear yellow instead of being colourless. Why?
Answer:
In common salt, due to metal excess defect the anion Cl^– disappears from its lattice site but leaves an electron thereby which the crystal remains electrically neutral. A hole is formed at the vacant space of anion. This hole is known as F-centre. Due to this reason NaCl appear yellow.

Question 14.
With the increase in temperature, electrical conductivity of semiconductors increases. Why?
Answer:
The energy gap between valence band and conduction band is less. Thus, with the increase in temperature, some electrons from the valence band jump to the conduction band due to which electrical conductivity increases with the increase in temperature.

Question 15.
Which of the following lattices has the highest packing efficiency :
(i) Simple cubic
(ii) Body centered cubic
(iii) Hexagonal close-packed lattice?
Answer:
Packing efficiency for simple cubic = 52-4%
Body centred cubic = 68%
Hexagonal close packed = 74%
Hence, Hexagonal close-packed (hcp) has the highest packing efficiency.

Question 16.
What type of stoichiometric defect is shown by (i) ZnS (ii) AgBr?
Answer:
(i) ZnS shows Frenkel defect because its ions have a large difference in size.
(ii) AgBr shows both Frenkel and Schottky defects.

Question 17.
Explain how many portions of an atom located at (i) Corner and (ii) the Body center of a cubic unit cell is part of its neighbouring unit cell?
Answer:
(i) An atom at the corner is standing by eight adjacent unit cells. Hence, portion of the atom at the corner not belongs to one unit cell = \(\frac { 1 }{ 8 } \)
(ii) The atom at the body centre of a cubic unit cell is not stand by another unit cell. Hence, it belongs fully to the unit cell.

Question 18.
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
Answer:
Let d – Density of the unit cell and volume of the unit cell = a3 (in case of the cubic crystal)
Mass of an atom present in the unit cell

Question 19.
The stability of a crystal is reflected in the magnitude of its melting point. Comment. Collect melting point of solid water, ethyl alcohol, diethyl ether, and methane from a data book. What can you say about the intermolecular forces between these molecules?
Answer:
Higher the melting point, the greater are the forces holding the constituent particles together and hence greater is the stability.
Melting points of the substance are given below:

Question 20.
A group 14 element is to be converted into an n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?
Answer:
n-type semiconductor means conduction due to the presence of excess negatively charged electrons. Hence, to convert group 14 element into the n-type semiconductor, it should be doped with group 15 element.

Long Answer Type Questions

Question 1.
Give the difference between crystalline solid and amorphous solid.
Answer:
Differences between Crystalline solid and Amorphous solid:

Crystalline solid:

1. Structure of crystalline solid are of definite geometrical shape.
2. In the internal structure, particle are arranged systematically.
3. Crystalline solids are solids in real sense.
4. Melting points of these compounds are sharp and definite.
5. These exhibit anisotropy.
6. Cooling curve are not continuous.
7. Crystalline solids possess less energy.

Amorphous solid

1. No any definite geometrical shape in the structure.
2. No any definite arrangement of particles in the internal structure. Amorphous solids are supercooled liquids.
3. Melting point of these compounds are not definite and sharp.
4. These exhibit isotropy.
5. Cooling curves are continuous.
6. Amorphous solids possess higher energy.

Question 2.
Distinguish between :
(i) Hexagonal and monoclinic unit cell
(ii) Face centred and end centred unit cell.
Answer:
(i) For hexagonal unit cell a = b ≠ c, α = β = 90°, γ = 120°
For monoclinic unit cell a ≠ b ≠ c, α = γ= 90°, β = 90°
(ii) A face-centered unit cell has one constituent particle present at the centre of each face in addition to the particles present at the comers.
An end centred unit cell has one constituent particle each at the centre of any two opposite faces in addition to the particles present at the corners.

Question 3.
Give differences between Schottky and Frenkel defect.
Answer:
Differences between Schottky and Frenkel defect:

Schottky Defect:

1. Cation and anion completely leave their lattice sites.
2. Density of the crystal decreases by this defect.
3. This defect generally occur in ionic compounds with high co-ordination number where size cation and anion are nearly same.
4. No effect on dielectric constant.

Frenkel Defect:

1. Cation leave its normal lattice site and occupies the intestitial site.
2. No effect on the density of the crystal.
3. It is found in ionic compounds with low co-ordination number and where size of cation and anion differ largely.
4. Magnitude of dielectric constant increase.

Question 4.
What do you understand by imperfections in crystals? What are the rea¬sons for imperfections?
Answer:
It is generally supposed that arrangement of constituent particles in crystal struc¬ture is completely regular, but actually it is very hard to get a such complete ideal crystal. Crystal structures have many imperfections or defects. Reasons for these imperfections are following:
(i) Temperature: Crystal which have no imperfections or defects are known as ideal crystal. However such crystals exist only at absolute zero temperature because the energy of crystals at 0 K is minimum. At any temperature above 0 K, there are crystals which have some departure from complete order arrangement.

(ii) Presence of impurities: Sometimes the presence of impurities causes disorder in regular arrangement of crystals which is responsible for imperfections and defects.

Question 5.
State the importance of radius ratio in crystal structure.
Answer:
Importance of radius ratio in crystal structure: Cations have tendency to get surrounded by maximum number of anions, hence larger be the size of cation greater will be its co-ordination number. We can understand this by taking examples of NaCl and CsCl. In NaCl, small sized Na⁺ has co-ordination number 6, while in CsCl large-sized Cs⁺ ion has co-ordination number 8. Hence, radius ratio is closely related to co-ordination number.

Radius ratio, Co-ordination number and Structural arrangement

Question 6.
Ionic solids, which have anionic vacancies due to metal excess defect develop colour. Explain with the help of suitable example.
Answer:
Due to anion vacancies basic halides like NaCl, KCl etc., show this type of metal excess defect. Taking the example of NaCl, when its crystals are heated in presence of sodium vapour some chloride ion leave their lattice sites to combine with sodium to form NaCl. For this reaction to occur Na atoms lose electrons to form Na⁺ ions. The electron has released diffuse into the crystal to occupy the anion vacancies created by Cl^– ions. The crystal now has excess of sodium. The sites occupied by unpaired electrons are called F- centres. They impart yellow colour to the crystal because they absorb energy from the visible light and get excited.

Question 7.
What type of substances would make better permanent magnets: ferromagnetic or ferrimagnetic? Justify your answer.
Answer:
Ferromagnetic substances make better permanent magnets. This is because the metal ions of ferromagnetic substances are grouped into small regions called ‘Domains’. Each domain acts as a tiny magnet. These domains are randomly oriented. 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 field is produced. This ordering of domains persists even when the external magnetic field is removed. Hence, the ferromagnetic substance becomes a permanent magnet.

Question 8.
How many lattice points are there in one unit cell of each of the following lattice:
(i) Face centered cubic
(ii) Face centered tetragonal
(iii) Body centered.
Solution:
(i) Lattice points in face-centred cubic lattice = 4.
(ii) Face centred tetragonal = 8 (at corners) + 6 (at the face centre) = 14.
However, particles per unit cell
= 8 × \(\frac { 1 }{ 8 } \) + 6 × \(\frac { 1 }{ 2 } \)
= 1 + 3
= 4.
(iii) Lattice points in body centred cube
= 8 (at comers) + 1 (at the body centre)
= 9
However, particles per unit cell = 8 × \(\frac { 1 }{ 8 } \) + 1 = 2.

The Solid State Numerical Questions

Question 1.
A compound forms a hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
Solution:
No. of atoms in the close packing = 0.5 mol
= 0.5 × 6.022 × 10²³
= 3.011 × 10²³
No. of octahedral voids = No. of atoms in the packing .
= 3.011 × 10²³
No. of tetrahedral voids = 2 × No. of atoms in the packing
= 2 × 3.011 × 10²³
= 6.022 × 10²³
Total No. of voids = 3.011 × 10²³ + 6.022 × 10²³
= 9.033 × 10²³.

Question 2.
A compound is formed by two elements M and N. The elements N forms ccp and atoms of M occupy 1/3rd of the tetrahedral voids. What is the formula of the compound?
Solution:
Suppose the atoms N in the ccp = n
∴ No. of tetrahedral voids = 2n
As 1/3rd of the tetrahedral voids are occupied by atoms M, therefore,
No of atoms M = \(\frac { 2n }{ 3 } \)
∴ Ratio of M : N = \(\frac { 2n }{ 3 } \) : n
Hence, the formula is M₂N₃

Question 3.
The structure of a solid AB is like NaCl. Radii of cation A is 100 pm, find out the radii of the anion.
Solution:
For NaCl structure value of radius ratio (\(\frac{r^{+}}{r^{-}}\)) is in between 0.414 – 0.732.
Radius of cation is 100 pm.

Question 4.
The radii of A⁺ and B⁺ ions are 0.95 Å and 1.81 Å respectively. Find out the co-ordination number of A⁺.
Or,
The ionic radii of Na⁺ and Cl^– are 95 pm and 181 pm respectively.
What will be the co-ordination number of Na⁺?
Solution:
Radii of Na⁺ = 95 pm
Radii of Cl^– =181 pm
Radius ratio = \(\frac{r^{+}}{r^{-}}\) = \(\frac { 95 }{ 181 } \) = 0.524
Radius ratio is between 0.414 and 0.732. Thus, the co-ordination number of Na⁺ or A⁺ will be 6.

Question 5.
Core length of a face centered cubic crystal is 400 pm calculate the density of element Atomic mass of element is 60.
Solution:
Density \((d)=\frac{Z \times M}{N_{0} \times a^{3}}\)
Where, Z = Number of atoms = 4, in face centred cubic structure.
M = Atomic mass = 60, N₀ = Avogadro number (6.023 × 10²³)
a = Core length (400 p.m.)

Question 6.
Silver crystallises in fee 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.
Solution:

Where d = Density of the material
a = Length of the edge of the cell
N_A = Avogadro number
Z = No. of atoms

∴ Atomic mass of silver = 107.08 g mol^-1.

Question 7.
Atomic mass of a face centred cubic (fee) is 60 g mol^-1 and the edge of its face is 400 pm. Determine the density of the element
Solution:
Volume of unit cell (a³) = (Length of edge)³
= (400 × 10^-12 m)³
= 64 × 10^-30 m³
= 64 × 10^-30 (10² cm)³
= 64 × 10^-24 cm³
Number of atoms in fee unit cell = 4

Question 8.
Structure of CuCl like ZnS is cubic. If density of CuCl is 3-4 g cm^-3, then determine the length of edge of the unit cell.
Solution:
ZnS has a fee structure, thus the same structure will be of CuCl. If length of edge of unit cell is a, number of atoms in fee is Z, molecular mass M and Avogadro number is N_A, then,

MP Board Class 12th Chemistry Solutions

MP Board Class 8th Maths Solutions Chapter 4 प्रायोगिक ज्यामिती Ex 4.1

प्रश्न 1.
निम्नलिखित चतुर्भुजों की रचना कीजिए –
(i) चतुर्भुज ABCD जिसमें –

• AB = 4.5 cm
• BC = 55 cm
• CD = 4 cm
• AD = 6 cm
• AC = 7 cm है।

(ii) चतुर्भुज JUMP जिसमें –

• JU = 3.5 cm
• UM = 4 cm
• MP =5cm
• PJ =4.5 cm
• PU = 6.5 cm है।

(iii) समान्तर चतुर्भुज MORE जिसमें –

• OR = 6 cm
• EO = 7.5 cm
• MO = 7.5 cm है।

(iv) समचतुर्भुज BEST जिसमें

• BE = 4.5 cm
• ET= 6 cm

हल:
(i) रचना के चरण –

1. 1. सर्वप्रथम रेखाखण्ड AC = 7 cm बनाया।
   2. A को केन्द्र मानकर AB = 4.5 cm त्रिज्या लेकर एक चाप लगाया तथा C को केन्द्र मानकर CB = 5.5 cm त्रिज्या लेकर चाप लगाया जो पहले चाप को बिन्दु B पर काटता है।
      

1. पुनः A व C को केन्द्र मानकर AD = 6 cm तथा CD = 4cm के चाप लगाए जो AC के विपरीत D बिन्दु पर काटते हैं।
2. AB, BC, CD तथा AD को मिलाया।

इस प्रकार बनी आकृति ABCD अभीष्ट चतुर्भुज है।

(ii) रचना के चरण –

1. सर्वप्रथम रेखाखण्ड PU = 6.5 cm बनाया।
2. P व U को केन्द्र मानकर क्रमशः PM = 5 cm तथा UM = 4 cm त्रिज्या लेकर चाप लगाए जो M बिन्दु पर काटते हैं।
   
3. पुनः P और U को केन्द्र मानकर PJ = 4.5 cm तथा UJ = 3.5 cm त्रिज्याएँ लेकर PU के विपरीत चाप लगाए जो J बिन्दु पर काटते हैं।
4. PM, UM, PJ तथा UJ को मिलाया।
5. इस प्रकार बनी आकृति JUMP अभीष्ट चतुर्भुज है।

(iii) रचना के चरण –

1. सर्वप्रथम रेखाखण्ड OR = 6 cm बनाया।
2. 0 व R को केन्द्र मानकर OE = RE = 7.5 सेमी. त्रिज्याएँ लेकर चाप लगाए जो E बिन्दु पर काटते हैं।
3. E व 0 को केन्द्र मानकर क्रमश: EM = 6 cm तथा OM = 7.5 cm के चाप लगाए जो M बिन्दु पर काटते हैं।
4. OE, RE, EM तथा OM को मिलाया।
5. इस प्रकार बनी आकृति MORE अभीष्ट समान्तर चतुर्भुज है।

(iv) रचना के चरण –

1. सर्वप्रथम रेखाखण्ड BE = 4.5 cm बनाया।
2. B व E बिन्दु को केन्द्र मानकर क्रमश: BT = 4.5 cm तथा ET = 6 cm त्रिज्याएँ लेकर चाप लगाए जो T बिन्दु पर काटते हैं।
   
3. पुनः T व E को केन्द्र मानकर TS = ES = 4.5 cm त्रिज्याएँ लेकर चाप लगाए जो S बिन्दु पर काटते हैं।
4. BT, ET, TS तथा ES को मिलाया।
5. इस प्रकार बनी आकृति BEST अभीष्ट समचतुर्भुज है।

पाठ्य-पुस्तक पृष्ठ संख्या # 68

सोचिए, चर्चा कीजिए और लिखिए (क्रमांक 4.3)

प्रश्न 1.
उपर्युक्त उदाहरण 2 में क्या हम पहले AABD खींचकर चतुर्थ बिन्दु C को ज्ञात करके चतुर्भुज की रचना कर सकते हैं?
उत्तर:
हम जानते हैं कि किसी त्रिभुज की रचना के लिए कोई भी तीन मापें होनी चाहिए। लेकिन ∆ABD के लिए केवल दो मापें AD और BD दी हुई हैं। अतः चतुर्भुज की रचना नहीं की जा सकती है।

प्रश्न 2.
क्या आप एक चतुर्भुज PQRS की रचना कर सकते हैं जिसमें PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm और SQ = 4 cm है? अपने उत्तर की पुष्टि कीजिए।
उत्तर:
हम चतुर्भुज PQRS की रचना नहीं कर सकते हैं। क्योंकि ∆QSP की रचना नहीं की जा सकती है।

यहाँ दो भुजाओं SQ + QP ≯  SP अर्थात् 4 cm + 3 cm ≯  7.5 अतः चतुर्भुज PQRS की रचना सम्भव नहीं है। SQ = 4 cm की रेखा चतुर्भुज के लिए सम्भव नहीं है।

MP Board Class 8th Maths Solutions

MP Board Class 7th Science Solutions Chapter 16 Water: A Precious Resource

Activities

Activity – 1
Given here is the rainfall map of India (Fig.) It gives the average annual rainfall in different regions of our country?

Question 1.
Are you blessed with sufficient rainfall?
Answer:
No.

Question 2.
Is there sufficient water available in your area throughout the year?
Answer:
Yes.

Question 3.
Can we attribute this to mismanagement of water resources?
Answer:
Yes.

Water: A Precious Resource Text Book Exercises

Question 1.
Mark T if the statement is true and F if it is false:

1. The freshwater stored in the ground is much more than that present in the rivers and lakes of the world.
2. Water shortage is a problem faced only by people living in rural areas.
3. Water from rivers is the only source for irrigation in the fields.
4. Rain is the ultimate source of water.

Answer:

1. True
2. False
3. False
4. True

Question 2.
Explain how groundwater is recharged?
Answer:
The ground water get recharged through the process of infiltration. The infiltration means seeping in water from rain, rivers and lakes into the empty spaces and crack deep below the ground.

Question 3.
There are ten tubewells in a lane of fifty houses. What could be the long term impact on the water table?
Answer:
The effect on the water table depends on the replenishment of the underground water. As only five families will share a tubewell, the water used for daily domestic purpose will not effect the water table as such. But if there is acute shortage of rains the water used by the families will not replenished and water table will fall down.

Question 4.
You have been asked to maintain a garden. How will you minimise the use of water?
Answer:
By collecting rain water. Also to minimise the wastage of water we will use the new technique of drip irrigation, which directly throws water at the base of plants.

Question 5.
Explain the factors responsible for the depletion of water table?
Answer:
Increasing population:
Increasing population creates demand for construction of houses, shops, offices, roads and pavements. This decreases the open areas like parks, and playgrounds. This, in turn, decreases the seepage of rainwater into the ground. What could be the consequence? Recall that a pukka floor does not allow water to seep in easily, while in a grass lawn water seeps through in no time.

Moreover a huge amount of water is required for construction work. Often groundwater is used for this purpose. So, on the one hand we are consuming more groundwater, and on the other we are allowing lesser water to seep into the ground. This results in depletion of water table. In fact, the water table in some parts of many cities has gone down to alarmingly low levels.

Increasing industries:
Water is used by all the industries. Almost everything that we use needs water somewhere in its production process. The number of industries is increasing continuously. Water used by most of the industries is drawn from the ground.

Agricultural activities:
A majority of farmers in India depend upon rains for irrigating their crops. Irrigation systems such as canals are there only in a few places. Even these systems may suffer from lack of water due to erratic rainfall. Therefore, farmers have to use groundwater for irrigation. Population pressure on agriculture forces increasing use of groundwater day by day. This results in depletion of water table.

Question 6.
Fill in the blanks with the appropriate answers:

1. The water bearing layer of the earth is …………………………
2. The process of water seepage into the ground is called ………………………
3. The water bearing layer of the earth is ………………………
4. The process of water seepage into the ground is called ………………………….

Answer:

1. Well and handpumps
2. Solid (ice), liquid (water), gas (water vapour)
3. Aquifer
4. Infiltration.

Question 7.
Which one of the following is not responsible for water shortage?

1. Rapid growth of industries
2. Increasing population
3. Heavy rainfall
4. Mismanagement of water resources.

Answer:
3. Heavy rainfall.

Question 8.
Choose the correct option. The total water –

1. In the lakes and rivers of the world remains constant.
2. Under the ground remains constant.
3. In the seas and oceans of the world remains constant.
4. Of the world remains constant.

Answer:
3. In the seas and oceans of the world remains constant.

Question 9.
Make a sketch showing groundwater and water table? Lab T it?
Answer:

Extended Learning Activities and Projects

Question 1.
Role play?
You are a water detective in your school? You have a term of six members? Survey the campus and make a note of the following:

1. Total number of taps
2. Number of taps leaking
3. Amount of water wasted due to leakage
4. Reasons of leakage
5. Corrective measures taken.

Answer:
Do yourself.

Question 2.
Groundwater pumped out?
Try to find out if there are any hand pumps in your neighborhood. Go to the owner or the users of a few of these and find out the depth at which they struck water? If there are any differences/think of the probable reason? Write a brief report and discuss it in your class. If possible, visit a place water boring is going on to install a hand pump? Watch the process carefully and find out the depth of the water table at that place?
Answer:
Do yourself.

Question 3.
Catching rainwater – Traditional methods? Form groups of 4 to 5 students in the class and prepare a report on the various traditional ways of water harvesting. If possible, use the following web link: www.rainwaterharvesting.org.?
Answer:
Do yourself.

Question 4.
Conservation of water? Carry out a campaign to conserve water at home and in the school. Design posters to remind others of the importance of water resources?
Answer:
Do yourself.

Question 5.
Create a logo? Hold a competition to create a logo or a symbol depicting water scarcity?
Answer:
Do yourself.

Water: A Precious Resource Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Question (a)
…………………………. is celebrated as the “world water day”?
(a) 22 March
(b) 20 March
(c) 22 April
(d) 20 April.
Answer:
(a) 22 March

Question (b)
Water is a Unique substance because?
(a) It is consumable
(b) It is colourless
(c) It exits in all three states
(d) Both (a) and (b)
Answer:
(c) It exits in all three states

Question (c)
Year …………………….. was observed as the International Year of fresh water to make people aware of this dwindling natural resource?
(a) 2002
(b) 2003
(c) 2004
(d) 2005.
Answer:
(b) 2003

Question (d)
The earth’s surface is covered with water about?
(a) 70%
(b) 71%
(c) 72%
(d) 73%.
Answer:
(b) 71%

Question (e)
Main source of water on earth is –
(a) rain
(b) snow
(c) glaciers
(d) all the above.
Answer:
(a) rain

Question (f)
One of these is not a source of surface water –
(a) lake
(b) rain
(c) river
(d) spring.
Answer:
(d) spring.

Question (g)
This is level …………………. of ground water?
(a) sea level
(b) rain
(c) aquifer
(d) (b) and (c).
Answer:
(d) (b) and (c).

Question (h)
Water has maximum density at –
(a) 0°C
(b) 4°C
(c) 98.4°C
(d) 100°C.
Answer:
(b) 4°C

Question (i)
Water is –
(a) a mixture
(b) a compound
(c) an element
(d) none of these.
Answer:
(b) a compound

Question (j)
The purest form of natural water is –
(a) river
(b) sea
(c) rain
(d) spring.
Answer:
(c) rain

Question (k)
Common salt content of sea water is –
(a) 1.5%
(b) 2.5%
(c) 3.5%
(d) 10%.
Answer:
(b) 2.5%

Question 2.
Fill in the blanks:

1. The amount of water recommended by the United Nations for drinking, cooking, washing and maintaining proper hygiene is a minimum of ……………………… liters per person per day.
2. The water that is fit for use is ………………………..
3. Solid form of …………………. water, and …………… , is present as ice caps at the poles of the earth.
4. ……………………. water is present in lakes, oceans, rivers and even underground.
5. The process of seeping of water into ground is called ………………………..
6. Water in the aquifiers can be usually pumped out with the help of …………………………. or ………………………..
7. Increase in population, industrial and agricultural activities are some common factors affecting ……………………….. table.
8. The rain water can be used to recharge the …………………………
9. Bhujpur in the Kutch area of ………………….. has a very erratic rainfall.
10. ……………………… irrigation is a technique of watering plants.

Answer:

1. 50
2. Fresh water
3. Snow, ice
4. Liquid
5. Infiltration
6. Tubewells, handpumps
7. Water
8. Groundwater
9. Gujarat
10. Drip.

Question 3.
Which of the following statements are true (T) or false(F):

1. Water is essential for all living beings.
2. Water exists in three forms.
3. There is an uneven distribution of water.
4. Water shortage has become a matter of concern throughout the world.
5. The gaseous form is the water vapour present in the air around us.
6. Many villages do not have water supply system.
7. The water – table does not varies from place to place.
8. Water drawn from under the ground gets replenished by seepage of rain water.
9. Rainwater harvesting is not necessary.
10. Rajasthan is a hot and dry place.
11. Water allows sunlight to pass through it.
12. Pure water has no taste.

Answer:

1. True
2. True
3. True
4. True
5. True
6. True
7. False
8. True
9. False
10. True
11. True
12. True.

Water: A Precious Resource Very short Answer Type Questions

Question 1.
What is the most important property of water?
Answer:
The most important property of water is its ability to dissolve various substances. Hence, it is called a universal solvent.

Question 2.
What are the sources of water?
Answer:
The two main sources of natural water are surface water and underground water or subsoil water.

Question 3.
What are the different states of water?
Answer:
The different states of water are solid, liquid and gas.

Question 4.
What are the characteristics of pure water?
Answer:
The pure water is colourless, odourless, tasteless and transparent.

Question 5.
What do you mean by a resource?
Answer:
Materials that we get from the environment to meet our needs are called resources.

Question 6.
A part from water, what are the other natural resources?
Answer:
A part from water, air, forests, minerals and fossilfuels are natural resources.

Question 7.
What is the percentage of water that is available for human beings?
Answer:
Only a tiny fraction about 0.01% of total water is used by human beings.

Question 8.
List the sources of water for your daily use?
Answer:
Sources of water for our daily use are well, lakes, rivers and ponds.

Question 9.
Define saturated solution?
Answer:
A saturated solution of a solute at a given temperature is a solution which contains as much of the solute as it can dissolve at that temperature.

Question 10.
Define unsaturated solution?
Answer:
If a solution contains less of the solute than what it can dissolve at that temperature, then it is called all unsituated  solution.

Question 11.
Name three forms of water?
Answer:
Three forms of water are solid, gaseous and liquid.

Question 12.
Which is universal solvent?
Answer:
Water.

Question 13.
What is the percentage of water (by weight) that an average elephant body has?
Answer:
An average elephant body has 80% water by weight.

Question 14.
If you have sample of tap water, well water and sea water, which do you think has the highest amount of salts?
Answer:
Sea water has highest amount of salts.

Question 15.
List the salts that can be obtained from the sea?
Answer:

1. Common salt (NaCl)
2. Sodium bromide (NaBr)
3. Potassium iodide (Kl)
4. Calcium salts.

Question 16.
Which sea has the highest salinity?
Answer:
Dead sea has the highest salinity.

Question 17.
Name two gaseous fuels which are prepared from water?
Answer:
Water gas and hydrogen gas.

Question 18.
Define water table?
Answer:
The depth at which water is found at a particular place is called water table.

Question 19.
What is Bawri?
Answer:
Bawri is a traditional way of collecting water.

Question 20.
What is potable water?
Answer:
The water which is fit for drinking is called potable water.

Question 21.
Define salinity of water?
Answer:
Water which has salts dissolved in it and is salty in taste is called salinity of water.

Question 22.
Define water cycle?
Answer:
The water from the earth reaches the atmosphere and from there it ultimately comes back to land, this is known as water cycle.

Question 23.
How does water cycle help in maintaining global climate?
Answer:
Liquid water on heating turns into water vapours. Water vapour on cooling again forms liquid water. This change of state of water over and over again makes the cycle in nature.

Question 24.
How are clouds formed?
Answer:
Water vapours being lighter and rise up in the atmosphere. At the upper layer of the atmosphere, where the temperature is lower. The vapours get condensed into tiny water droplets and clouds are formed.

Question 25.
Which unwanted material could be present in the water that you get from your local water supply?
Answer:
The unwanted materials in our water supply are dust particles, germs, bacteria and many other materials as impurities.

Question 26.
Define water harvesting?
Answer:
Most of the water that we get as rainfall just flows aways. This is a waste of precious natural resource. The rainwater can be used to recharge the groundwater. This is known as water harvesting or rainwater harvesting.

Question 27.
Define drip irrigation?
Answer:
Drip irrigation is a technique of watering plants by making use of narrow tubings which deliver water directly at the base of the plant.

Water: A Precious Resource Short Answer Type Questions

Question 1.
Wrote a short note on source of water?
Answer:
Water is available to us through various sources. Rain water, lake Water, river water, sea water are some of the sources. Natural water from all these sources contain some impurities these are the dissolved minerals and salts. Apart from that, they contain some insoluble impurities too.

Question 2.
List the sources of water on the earth?
Answer:
There are two main sources of natural water:

1. Underground water
2. Surface water.
3. Underground Water: There are mainly two types of underground water –
   • (a) Well water
   • (b) Spring water.

2. Surface water:
It is of three types:

• Rain water
• River and lake water
• Sea water.

Question 3.
Write a short rote an river water?
Answer:
The river water is mainly due to rain or due to melting of snow on the mountains. The river water usually carries some suspended impurities and some soluble impurities. The soluble impurities are the dissolved salts and minerals. Apart from that they may carry certain micro – organisms and organic matter due to contamination.

Question 4.
Write a short note on sea water?
Answer:
Sea water is the largest source of natural water. But it is not fit either for drinking or for irrigation directly. This is because it contains a large number of dissolved salts of chlorides, bromides, iodides, sulphates and carbonates. The sea water is a rich source for common salt. It is recovered from sea water by evaporation and then is purified to remove the other impurities present.

Question 5.
Write a short note an underground water?
Answer:
Water present underneath the earth is tapped as it is an excellent source of pure drinking water. We usually dig wells. Depending on the geographical conditions of the place, the depth in which water is available is estimated. Well water contains some dissolved impurities.

Again the nature of the impurities depends on the geographical conditions. We also get water from springs under the earth. Such springs usually contain some dissolved mineral salts. However the spring water does not contain suspended impurities as it comes from deep under the earth.

Question 6.
Write a short note on rain water?
Answer:
It is the purest form of natural water available to us. Though the first few droplets of rain bring along with them some dust particles and micro-organisms, the later drops are free from impurities.

Question 7.
What are the properties of water?
Answer:
Properties of Water:

1. Pure water is colourless, odourless and tasteless.
2. Pure water freezes at 0 degree Celsius and boils at 100 degree celsius.
3. The density of water is 1 g/cc. It means that 1 cc of water at 1 degree Celsius weighs 1 gram.

Question 8.
What are the different ways by which water vapour is put into the atmosphere?
Answer:
The different ways by which water vapours put into atmosphere are as follows:

1. Factories and thermal stations produce a lot of steam and put is into the atmosphere.
2. Water in ponds, rivers and lakes, ocean gets evaporated due to atmospheric heat.
3. Plants throughout water vapours during respiration.

Question 9.
What makes the sea water so saline?
Answer:
Rivers flowing from different places bring water containing dissolved salts and minerals into the sea. Sea water is continuously evaporated by the heat of the sun. This evaporated water forms clouds and again fall on the earth as rain. This rain water again dissolves salts into it and goes into the sea. Thus the amount of salts in sea water goes on increasing.

Question 10.
How will you show that saline water is not fit for agriculture?
Answer:
Take two healthy plants of same type. Water both the plants separately one with ffesh water and other with saline water. Continue to water similarly for seven days and observe. The plant which was watered with saline water wilts. This shows that saline water is not fit for plants or agriculture.

Water: A Precious Resource Long Answer Type Questions

Question 1.
Draw a diagram to show water cycle?
Answer:

Question 2.
What are the causes of water scarcity?
Answer:
Causes of Water Scarcity:

1. Overpopulation
2. Over – irrigation for agriculture
3. Water pollution
4. Climatic change and variability
5. Indiscriminate cutting of forest
6. Increase in demand for water
7. Misuse of water
8. Poor water resource management
9. Disputes between the states for sharing water
10. Destruction of natural water reservoirs.

Question 3.
What are the methods to prevent water scarcity? Ans. Methods to prevent water scarcity:

1. Control population by adopting small family norms
2. Prevent misuse and wastage of water
3. Prevent water pollution
4. Replace large flush tanks with smaller ones.
5. Build dams, bunds and tanks to store rainwater.
6. Replace large flush tanks with smaller ones.
7. Recharge underground water
8. Recycle waste water for gardening
9. Follow drip irrigation or low water sprinklers for agriculture.

Question 4.
What are the effect of water scarcity on the life of people?
Answer:
Effect of water scarcity on the life of the people:

1. Long walk to fetch water.
2. Have to pay high price for buying water.
3. Low scope for generating employment.
4. Uncertainty over availability of water for agriculture.
5. Area will remain backward and neglected.
6. Increase incidents of crime.
7. More incidence of diseases due to poor sanitary conditions.
8. Fall in the productivity of animals.

MP Board Class 7th Science Solutions

MP Board Class 6th General English Solutions Practice Test Paper 2

Question 1.
Recite any poem from your textbook.
(अपनी पाठ्य-पुस्तक से कोई कविता का पाठ करें।)
Answer:
Students may recite any poem given in the text book.

Question 2.
Read a paragraph from your textbook.
(अपनी पाठ्थ पुस्तक से एक परिच्छेद पढ़ें)
Answer:
Students may read a paragraph from any lesson from the textbook.

Question 3.
Fill in the blanks and complete the poem.
(खाली स्थानों को भरकर कविता पूरी कीजिए।)
Answer:
(1) When the blazing sun is gone.
When there nothing shines upon.
Then you show your little light.
Twinkle twinkle all the night.
(2) Here comes the elephant. Swaying along with his cargo of children. All singing a song.

Question 4.
(1) Match the animals with their young ones.
(जानवरों को उनके शिशुओं से मिलान करें।)

Answer:
1. → (d)
2. → (c)
3. → (e)
4. → (b)
5. → (a)

(2) Rearrange the letters and spell the names of the games.
(वर्गों को पुनः व्यवस्थित कर खेलों के सही नाम लिखें।)
Answer:
dabinmont – badminton
toolfalb – football
oldu – ludo
keycho – hockey
iddabka – kabaddi

Question 5.
(1) Fill in the blanks with ‘and’ or ‘but’.
(‘and’ और ‘but’ से रिक्त स्थान भरिए।)
Answer:
Seema and Meena are friends. Meena is beautiful but Seema is not so beautiful. Seema and Meena both like to sing. Seema sings well but Meena does not sing so well. Meena and Seema both like to go to school daily.

(2) Rewrite with proper punctuation marks and capitals.
(उचित विराम चिह्न और बड़े अक्षर लगाकर लिखें।)
once mr giles came for inspection he gave five words to write as a spelling exercise one of the words was kettle I misspelt it.
Answer:
Once Mr. Giles came for inspection. He gave five words to write as a spelling exercise. One of the words was ‘kettle’. I misspelt it.

(3) Make sentences using the past tense of the verbs:
(क्रियाओं (verbs) के Past tense का प्रयोग कर वाक्य बनाइए!:)
(think, run, read, make, cry.)
Answer:

1. Ram thought about a plan.
2. Meena ran too fast.
3. He read loudly.
4. She made a mess.
5. The child cried loudly.

Question 6.
1. Rewrite the following sentences removing the inverted commas |“……”| and change the speech.
(इन्वटैड कोमा |“……”| हटाकर निम्न वाक्यों को पुनः लिखें और कथन बदलें।)

Question 1.
Raman says, “I want to be a teacher.”
Answer:
Raman says that he wants to be a teacher.

Question 2.
Mona says, “I want to be a pilot.”
Answer:
Mona says that she wants to be a pilot.

Question 3.
I says, “I want to be a lawyer.”
Answer:
I say that I want to be lawyer.

Question 4.
Radha says, “I want to be a doctor”.
Answer:
Radha says that she wants to be a doctor.

Question 5.
Mohan says, “I want to be a judge.”
Answer:
Mohan says that he wants to be a judge.

Question 7.
(1) Say whether the statements are ‘true’ or ‘false’:
(सही या गलत कथन बताइए।)

1. Meena is tall, strong and well built.
2. Yudhisthir is a Kaurav prince.
3. The elephant has wide ears.
4. The traveller does not like the stars.
5. The twinkle of the starts is wonderful.

Answer:

1. False
2. False
3. True
4. False
5. True.

(2) Answer the following questions:
(निम्न प्रश्नों के उत्तर दीजिए।)

Question 1.
What happens when the sun goes away?
Answer:
When the sun goes away and nothing shines upon. Then the little star shows its light and twinkles whole night.

Question 2.
What is the cargo of the elephant?
Answer:
The cargo of the elephant is the children on it.

Question 3.
Who gives the signal to start the race?
Answer:
The games teacher gives the signal to start the race.

Question 4.
What was the test?
Answer:
The test was to shoot the right eye of the bird sitting on the tree.

Question 8.
Write a message to your friend that you cannot come to his birthday party as you are going to your grandma’s home.
(अपने मित्र को एक सन्देश लिखें कि आप अपनी दादी माँ के घर जा रहे हैं इसलिए उसके जन्मदिन की पार्टी पर नहीं आ सकते हैं।)
Answer:
Dear Anshul,

Wishing you a very happy birthday. I wanted to inform you that I would be unable to attend your birthday party today as I have to go urgently to my grandma’s home because she is ill.

Yours
Rajan.

MP Board Class 6 English Solutions

MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.2

Question 1.
Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centers.
Solution:
We have a circles having its centers at O and O’ two equal chords AB and CD such that they subtend ∠AOB and ∠COD respectively at their centers, i.e. at O and O’.

We have to prove that
∠AOB = ∠CO’D
Now, in ∆AOB and ∆CO’D, we have
AO = CO’ [Radii of the same circle]
BO = DO’ [Radii of the same circle]
AB = CD [Given]
∆AOB = ∆CO’D [SSS criterion]
Their corresponding parts are equal.
∴ ∠AOB = ∠CO’D

Question 2.
Prove that if chords of congruent circles subtend equal angles at their centers, then the chords are equal.
Solution:
We have circles having their centers at O and O’, and its two chords AB and CD such that
∠AOB = ∠CO’D
we have to prove that
AB = CD
In ∆AOB and ∆COO’D, we have:
AO = CO’ [Radii of the same circle]
BO = DO’ [Radii of the same circle]
∠AOB = ∠CO’D [Given]
∆AOB = ∆CO’D [SAS criterion]
Their corresponding parts are equal, i.e.,
AB = CD.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.1

Question 1.
Fill in the blanks:

1. The center of a circle lies in …………of the circle.
2. A point, whose distance from the center of a circle is greater then its radius lies in …………..of the circle.
3. The longest chord of a circle is a ………… of the circle.
4. An arc is a ………….. when its ends are the ends of a diameter.
5. Segment of a circle is the region between an arc and ……….. of the circle.
6. A circle divides the plane, on which it lies, in ……….. parts.

Solution:

1. interior
2. exterior
3. diameter
4. semicircle
5. the chord
6. three.

Question 2.
Write True or False. Give reasons for your answers.

1. Line segment joining the center to any point on the circle is a radius of the circle.
2. A circle has only finite number of equal chords.
3. If a circle is divided into three equal arcs, each is a major arc.
4. A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
5. Sector is the region between the chord and its corresponding arc.
6. A circle is a plane figure.

Solution:

1. True [∵ All points on the circle are equidistant from the center]
2. False [∵ A circle can have an infinite number of equal chords.]
3. False [∵ Each part will be less than a semicircle.]
4. True [∵ Diameter = 2 x Radius]
5. False [∵ The region between the chord and its corresponding arc is a segment.]
6. True [∵ A circle is drawn on a plane.]

MP Board Class 9th Maths Solutions

MP Board Class 7th General English Essay Writing

1. Holi

India is the country where different festivals are celebrated every month. Most important festivals of India are Holi, Raksha Bandhan, Dusshera, Deepawali, Id and Christmas.

Holi is the festival of colours. People express their happiness by throwing colours on their friends, relatives and neighbours. They make fun. They use dry and wet colours.

People are lost in singing songs relating to the festival and season. They feel happy by welcoming their friends with sweetmeats and tea or coffee or cold drink. Sugarcane fields look full of cheerfulness of the farmers who work there. Grain fields look very beautiful.

Some people throw mud on one another and they shout. Everyone, small or big bears a smile on his face.

कठिन शब्द – different = अनेक; festivals = त्यौहार; celebrated = मनाये जाते हैं; express = प्रकट करना; fun = मजाक; dry and wet = सूखे और गीले; relating to = सम्बन्धित; season = मौसम; are lost = लीन हो जाते हैं; welcoming = स्वागत करके; sweetmeats = मिठाईयाँ; sugarcane fields=ईख के खेत; cheerfulness = प्रसन्नता; look = प्रतीत होते हैं; grain = अनाज; one another = एक-दूसरे पर; shout = शोर मचाते हैं; bear a smile = मुस्कुराहट लिए हुए; face = चेहरा।

2. My Hobby

“All work and no play makes Jack a dull boy.” There should be some amusement after work. A hobby is a thing which we do in our free time. There are several hobbies like singing, dancing, gardening etc. My hobby is gardening.

I have made a small garden. I grow vegetables, fruits and flowers in it. I work in my garden in the morning and in the evening. It gives me pleasure to work in the garden. It also saves money as I don’t have to buy vegetables from the market. It has also increased my knowledge of science. Gardening is a useful hobby as it controls pollution.

कठिन शब्द – amusement = मनोरंजन; pleasure = खुशी; saves = बचाना; increased = बढ़ाना; knowledge – ज्ञान; pollution = प्रदूषण; useful = उपयोगी।

3. My Pet Animal

I am fond of pet animals. Moti is my pet dog. It watches our house. It barks aloud if some stranger comes to our house. I take Moti for a morning walk daily. I throw a ball and Moti brings it back.

I feed it with bread and milk. I wash it daily. One night it started barking loudly. When we got up we saw some thieves trying to enter our house. Seeing us and Moti they ran away. Thus Moti saved us. I love Moti a lot.

कठिन शब्द – pet = पालतू; watch = निगरानी करना; feed = खिलाना; bark = भौंकना; loudly = तेज आवाज में; enter = घुसना; thieves = चोर; saved = बचाया।

4. Annual Day of My School

Every year in the month of February, ‘Annual day’ in our school in celebrated. Many students from different classes and sections participate in different games and activities. My friend Reeta from Class VII A, took part in 100 metre race. Other participants did their best. They all ran very fast but Reeta stood first in this race.

Reeta is the thinnest girl of her class but she is the strongest. She started from the very line and finished the race on it. Everyone praised her performance.

कठिन शब्द – annual day = वार्षिकोत्सव; celebrated = मनाया जाता है; different classes = भिन्न-भिन्न कक्षाओं के; sections = वर्गों के; participate = भाग लेते हैं; thinnest = सबसे पतली; strongest = सबसे अधिक ताकतवर; praised = प्रशंसा की; performance = कार्य प्रणाली के।

5. My Favourite Teacher

I study in Modern Public School, Raipur. There are many teachers in our school. But Mr. Gupta is my favourite teacher. He teaches us English. He is also our class teacher. He is smart and handsome. He works hard with his students. He helps weak students.

He teaches well. His examination results are always very good. He is both strict and kind. We obey him. He is also good at games. He plays cricket well. I like him very much.

कठिन शब्द – favourite = प्रिय; smart = चुस्त; handsome = सुन्दर; work hard = मेहनत करना; weak = कमजोर; well = अच्छा; strict = कठोर; obey = आज्ञा पालन करना।

6. A Journey by Train

In the last winter vacations my uncle invited us to spend a week in Delhi. My sister and I decided to leave for Delhi by the Taj Express.

We My sister and I reached the railway station at 4 p.m. I bought two tickets. We got into the train and occupied two seats. At 4:15 p.m. the train left Gwalior. After an hour it reached Morena. Then it started passing over the Chambal bridge very slowly. We saw the famous Chambal river and its ravines.

At 10:30 p.m. the train finally reached Nizamuddin railway station. Our uncle and aunt were waiting for us. They were very happy to see us. They took us home in their car.

कठिन शब्द – winter vacations = शीत अवकाश; invited = निमन्त्रण दिया; spend = बिताना; decide = निश्चय; reached = पहुँचे; bought = खरीदे; occupied = घेर लेना; ravines = बीहड़; wait = इन्तजार।

7. A Visit to A Market

I went to my uncle’s house in these summer vacations, in Agra. There I visited the Rajamandi Bazar with my cousin sister..

There was a large crowd in the market. All the shops were decorated with lights. There were shops of almost all articles. There were shops of gifts, garments, shoes, books, etc. Street vendors were also there selling various articles like purses, cosmetics etc. at cheaper rates.

I bought a pair of shoes and a nice dress for my sister. I really enjoyed there.

कठिन शब्द – crowd = भीड़; decorated = सजी हुई; articles = वस्तुएँ; cosmetics = सौन्दर्य प्रसाधन; enjoyed = लुल्फ उठाना

8. A Visit to a Circus

Last week my parents and I went to see the Apollo Circus. There was a huge crowd inside the tent. But we had special passes so we got a seat in front.

In the beginning there was a show of dwarfs.

They were so small in height and did such actions that everyone laughed. After that there were several small shows of different animals like monkey, elephant, lion, tiger. The ring master controlled them. There was also a belle dance. The dancers had a flexible body. In the end came the joker for whom I was waiting. He just made us laugh and laugh. I greatly enjoyed in the circus.

कठिन शब्द – special = ख़ास; dwarfs = बैन आदमी; beginning = शुरुआत में; flexible = लचीली।

9. My Best Friend

I have several friends but Raja is my best friend. He is my neighbour. He also studies in my class.

He is a very good boy. He is very intelligent. He always stands first in class. He is not proud. He helps his classmates. He is a ‘friend in need’. His father is a bank manager. His mother is a house wife.

He is good at sports and games too. He is cheerful and polite. I love him for his qualities and nature.

कठिन शब्द – several = कई; neighbour = पड़ोसी; studies = पढ़ता है; intelligent = बुद्धिमान; classmates = सहपाठी; house wife= गृहिणी; proud = घमण्डी; cheerful = प्रसन्नचित; qualities = गुण; nature = व्यवहार।

10. The Independence Day

15th August is an important day in the history of India. On this day in 1947 our country became free. It is our national festival. It is celebrated all over the country with great joy. We also celebrated it in our school this year.

All the students were in neat and clean uniforms. Our principal unfurled the national flag. We, then sang National Anthem. Some students presented different types of programmes. The principal gave speech in the end on our freedom fighters. The function was over after the distribution of sweets.

कठिन शब्द – important = महत्त्वपूर्ण; history = sfera; celebrated = 45MI; unfurled = 46<F1; freedom fighters = देशभक्त; distribution = वितरण।

11. My Mother

My mother is an ideal one. She is kind and affectionate. She loves her children very much. Whenever anyone of us falls ill,she is the one who tends and nutses the sick. Being of polished manners, she teaches us lessons on truthfulness, thrift and health. She also helps us in completing our home work. But she is strict also, and never tolerates any mischief in the house.

Her love towards her family is very great. She wishes to make her family ideal and for this she is always busy. She never neglects her duty. Out of immense love for the family she manages the kitchen, sews and washes clothes. She spares no pains in giving us comfort.

I am proud of my mother.

कठिन शब्द – affectionate = स्नेही; tends = सेवा करना; thrift = मितव्ययिता; tolerate = सहन करना; immense = अत्यधिक।

12. A policeman

A policeman looks very smart in his uniform. He is generally dressed in Khaki but policemen on traffic duty are clad in white. A policeman wears a cap with a badge of his department on his head. He wear’s a leather belt round his waist.

The duty of a policeman is very hard. He has to control traffic in the burning head of the sun and heavy rains. He has to look after the life and property of the public. He detects wrongdoers and brings them to book. He keeps law and order in the country.

The policeman is respected by public for his honesty and hard work. He, sometimes becomes impolite due to frustration caused by his hard duty. Certainly, our life and property are safe only by the presence of police force in our country.

कठिन शब्द – uniform = वेशभूषा; waist = कमर; wrong-doers = गलती करने वाले; impolite = अशिष्ट,अभद्र; frustration = निराशा कुण्ठा।

MP Board Class 7th English Solutions

MP Board Class 9th Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.2

Question 1.
In fig. given below, ABCD is a parallelogram, AE ⊥ DCand CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Solution:
ar (ABCD) = AE x DC ….(i)
ar (ABCD) = AD x CF ….(ii)
From (i) and (ii), we get
AE x DC = AD x CF
⇒ AE x AB = AD x CF (∵ AB = DC)

8 x 16 = AD x 10
\(\frac{8×16}{10}\) = AD
\(\frac{128}{10}\) = AD
AD = 12.8 cm.

Question 2.
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) \(\frac{1}{2}\) = ar (ABCD)
Solution:
Given
ABCD is a ∥^gm E, F, G, H, are the mid-points of AB, BC, CD and DA respectively.

To prove:
ar (EFGH) = \(\frac{1}{2}\) ar (ABCD)
Construction: Join GE.
Proof:
ABCD is a ∥gm
∴ AB ∥ DC and AB = DC
⇒ \(\frac{1}{2}\)AB ∥ \(\frac{1}{2}\)DC and \(\frac{1}{2}\) AB = \(\frac{1}{2}\) DC
AE ∥ DG and AE = DG
∴ AEGD is a parallelogram.
Similarly BCGE is also a parallelogram.
∥gm AEGD and AEGH both lie on the same base EG and are between the same parallels AD and EG.
ar (EGH) = \(\frac{1}{2}\) ar {AEGD) …(1)
Similarly ar {GEF) = \(\frac{1}{2}\) ar {BCGE) …(2)
Adding (1) and (2), we get
ar (EGH) + ar (GEF) = ar (AEGD) + \(\frac{1}{2}\) ar {BCGE)
ar (EFGH) ar {EFGH) = \(\frac{1}{2}\) [ar (AEGD) + ar (BCGE)]

Question 3.
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
Solution:
Given:
ABCD is a ∥^gm
To prove:
ar (APB) = ar (BQC)..
Proof:
∆APB and ∥^gm ABCD lie on the same base BC and are between the same parallels AB and DC.

ar(APB) = \(\frac{1}{2}\) ar (ABCD)
∆BQC and ∥^gm ABCD lie on the same base BC and are between the same parallels BC and AD.
ar (BQC) = \(\frac{1}{2}\) ar (ABCD) …(2)
From (1) and (2), we get
ar (APB) = ar (BQC)

Question 4.
In Fig. given below, P is a point in the interior of a parallelogram ABCD, Show that

1. ar (APB) + ar (PCD) = \(\frac{1}{2}\) ar (ABCD)
2. ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

[Hint: Thorough P, draw a line parallel to AB.]
Solution:
Given:
P is any point in the interior of ∥^gm ABCD.
To prove:

1. ar (APB) + ar (PCD) \(\frac{1}{2}\) ar (ABCD)
2. ar (APD) + ar (PBC) ar (APB) + ar (PCD)

1. Construction:
Draw a line EF passing through point P parallel to AB.
Proof:
AB ∥EF (By construction) …(1)
AB ∥ DC (∵ ABCD is a ∥gm) …(2)
EF ∥ DC [From (1) and (2)]

In quadrilateral ABFE, AB ∥ EF and AE ∥ BF (∵ AD ∥ BC)
∴ ABFE is a ∥gm
∆ APB and ∥gm ABEF lie on same base base AB and are between the same parallels AB and EF.
ar (APB) = \(\frac{1}{2}\) ar (ABFE) …. (A)
Similarly, ar (PCD) = \(\frac{1}{2}\) ar (DCFE) …(B)
Adding (A) and (B), we get
ar (APB) + ar (PCD) = \(\frac{1}{2}\) ar (ABFE) + \(\frac{1}{2}\) (DCFE)
= \(\frac{1}{2}\) [ar (ABFE) + ar (DCFE)]
= \(\frac{1}{2}\) ar (ABCD) …(3)

2. Construction:
Draw a line GH passing through point P parallel to AD.
Proof:
AD ∥ GH (by construction) …(4)
AD ∥ BC (∵ ABCD is a ∥^gm) …(5)
GH ∥ BC [From (4) and (5)]
In quadrilateral ADHG,
AD ∥ GH (by construction)
and AG ∥ DH (∵ AB\\DC)
∴ ADHG is a parallelogram.
∆APD and ∥gm ADHG lie on the same base AD and lie betweeiwthe same parallels AD and GH.
ar(APD) = \(\frac{1}{2}\) ar (ADHG) …(6)
Similarly, ar (BCP) = \(\frac{1}{2}\) ar (BCHG) …(7)
Adding (6) and (7), we get
ar (APD) + ar (BCP) = \(\frac{1}{2}\) ar (ADHG) + \(\frac{1}{2}\) ar (BCHG)
= \(\frac{1}{2}\) ar (ADHG) + ar (BCHG)
= \(\frac{1}{2}\) [ar (ADHG) + ar (BCHG)]
= \(\frac{1}{2}\) ar (ABCD) …(8)
From (3) and (8), we get
ar (APB) + ar (PCD) = ar (APD) + ar (BCP)

Question 5.
In Fig. given below PQRS and ABRS are parallelograms and X is any point on side BR. Show that

1. ar (PQRS) = ar (ABRS)
2. ar (AZS) = \(\frac{1}{2}\) ar (PQRS)

Solution:
Given
PQRS and ABRS are H801 and X is any point on side BR.
To prove:

1. ar (PQRS) = ar (ABRS)
2. ar (AXS) \(\frac{1}{2}\) ar (PQRS)

Proof:
1. ∥^gm PQRS and ABRS lie on the same base SR and are on the same parallels SR and PB.
∴ ar (PQRS) = ar (ABRS) …(i)

2. D AXS and ∥^gm ABRS lie on the same base AS and are on the same parallels AS and BR.
ar (∆ AXS) = \(\frac{1}{2}\) ar (ABRS) …(ii)

From (i) and (ii), we get
ar (AXS) = \(\frac{1}{2}\) ar (PQRS)

Question 6.
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
Solution:
The field will be divided in three parts. The shapes of three parts are triangles.
Given:
PQRS is a ∥^gm
To prove:
ar (APQ) = ar (APS) + ar (AQR)
Construction:
Draw a line AB ∥ SP.
Proof:
∆APQ and ∥^gm PQRS are on the same base PQ and are between the same parallels PQ and SR.

ar (PQRS) = \(\frac{1}{2}\) ar (PQRS)
⇒ 2ar (APQ) = ar (PQRS)
2ar (APQ) = ar (APS) + ar (APQ) + ar (AQR)
2ar (APQ) – ar (APQ) = ar (APQ) ar (AQR)
ar (APQ) = ar (APS) + ar (AQR)

MP Board Class 9th Maths Solutions

MP Board Class 7th Science Solutions Chapter 18 Wastewater Story

Activities

Activity – 1
We have given one example of the use of clean water. You can add many more?

Answer:

Activity – 2
Locates an open drain near your home, school or on the roadside and inspect water flowing through it? Record color, odour and any other observation. Discuss with your friends and your teacher and fill up the following Table?
Table: Contaminant survey

Wastewater Story Text Book Exercises

Question 1.
Fill in the blanks:

1. Cleaning of water is a process of removing ……………………….
2. Waste – water released by houses is called ………………………
3. Dried ……………………… is used as manure.
4. Drains get blocked by …………………….. and ………………………..

Answer:

1. Contaminants
2. Sewage
3. Dung
4. Plastic, sludge.

Question 2.
What is sewage? Explain why it is harmful to discharge untreated sewage into rivers or seas?
Answer:
Sewage is waste – water released by homes, hospitals, offices, industries and other users. It also includes rainwater that has run down the street during a heavy rain or storm. The water that washes off roads and roof tops carries harmful substances with it. Basically sewage is a liquid waste. Most of it is water, which has dissolved and suspended impurities which are called contaminants. That is why it is harmful to discharge untreated sewage in to rivers or seas.

Question 3.
Why should oils and fats be not released in the drain? Explain.
Answer:
Oils and fats should not be released in the drains because they harden the soil in the pipes and block them. Fats get clogged in the holes of the soil in the drain and block it. It does not allow the waste – water to flow and thus the whole sewer system is blocked.

Question 4.
Describe the steps involved in getting clarified water from waste – water?
Answer:
Treatment of waste – water involves physical, chemical, and biological processes, which remove physical, chemical and biological matter that contaminates the waste – water.

1. Waste – water is passed through bar screens. Large objects like rags, sticks, cans, plastic packets, napkins are removed.

2. Water goes to a grit and sand removal tank. The speed of the incoming waste-water is decreased to allow sand, grit and pebbles to settle down.

3. The water is then allowed to settle in a large tank which is sloped towards the middle. Solids like faces settle at the bottom and are removed with a scraper. This is the sludge. A skimmer removes the float able solids like oil and grease. Water so cleared is called clarified water.

4. The sludge is transferred to a separate tank where it is decomposed by the anaerobic bacteria. The bio gas produced in the process can be used as fuel or can be used to produce electricity.

5. Air is pumped into the clarified water to help aerobic bacteria to grow. Bacteria consume human waste, food waste, soaps and other unwanted matter still remaining in clarified water. After several hours, the suspended microbes settle at the bottom of the tank as activated sludge. The water is then removed from the top.

Question 5.
What is sludge? Explain how it is treated?
Answer:
Sludge is the collected solid waste from the waste – water during the treatment in water treatment plant. Sludge is decomposed in a separate tank by the anaerobic bacteria. The activated sludge is about 97% water. The water is removed by sand drying beds or machines. Dried sludge is used as manure, returning organic matter and nutrients to the soil.

The treated water has a very low level of organic material and suspended matter. It is discharged into a sea, a river or into the ground. Nature cleans it up further. Sometimes it may be necessary to disinfect water with chemicals like chlorine and ozone before releasing it into the distribution system.

Question 6.
Untreated human excreta is a health hazard? Explain?
Answer:
Untreated human excreta is a health hazard. It may cause water pollution and soil pollution. Both the surface water and groundwater get polluted. Groundwater is a source of water for wells, tubewells, springs and many rivers. Thus, it becomes the most common route for water borne diseases. They include chlorea, typhoid, polio, meaningities, hepatitis and dysentery.

Question 7.
Name two chemicals used to disinfect water?
Answer:
Ozone, chlorine.

Question 8.
Explain the function of bar screens in a waste – water treatment plant?
Answer:
Bar screens clear the waste – water of all the physical impurities. Large waste objects like napkins, plastics, can sticks, rags etc. are removed from the waste – water through the bar screens.

Question 9.
Explain the relationship between sanitation and disease?
Answer:
Untreated human excreta is a health hazard. It may cause water pollution and soil pollution. Both the surface water and groundwater get polluted. Groundwater is a source of water for walls, tubewells, springs and many rivers. Thus, it becomes the most common route for water borne diseases. They include cholera, typhoid, polio, meningitis, hepatitis and dysentery. That is proper sanitation is must to avoid some of the deadliest diseases.

Question 10.
Outline your role as an active citizen in relation to sanitation?
Answer:
As active citizen we should take care of our personal and environmental sanitation. We should make people around us, aware of the benefits of sanitation we should help the municipal corporations and gram panchayats to cover all the open drains and remove the unhygienic and disease causing substances thrown in open.

Question 11.
Here is a crossword puzzle: Good luck!

Across

1. Liquid waste products
2. Solid waste extracted in sewage treatment
3. A word related to hygiene
4. Waste matter discharged from human body.

Down

1. Used water
2. A pipe carrying sewage.
3. Micro – organisms which causes cholera.
4. A chemical to disinfect water.

Answer:

Question 12.
Study the following statements about ozone:

1. It is essential for breathing of living organisms.
2. It is used to disinfect water.
3. It absorbs ultraviolet rays.
4. Its proportion in air is about 3%.

Which of these statements are correct?

1. (a), (b) and (c)
2. (b) and (c)
3. (a) and (d)
4. All four.

Answer:

2. (b) and (c).

Extended Learning – Activities and Projects

Question 1.
Construct a crossword puzzle of your using the keywords?
Answer:

Across:

1. Mixing with air.
2. Decomposed product of leaves.
3. Does not read oxygen
4. Needs oxygen.

Down:

1. Necessary for hygiene
2. Solid waste
3. Pipes to carry sewage

Question 2.
Then and now: Talk to your grand parents and other elderly people in the neighbourhood? Find out the sewage disposal systems available to them. You can also write letters to people living in far off places to get more information. Prepare a brief report on the information you collected?
Answer:
Do yourself.

Question 3.
Visit a sewage treatment plant?
It could be as exciting and enriching as a visit to a zoo, a museum, or a park? To guide your observation here are a few suggestions?

Record in your notepad:
Place ……………….. Date …………………… Time
Name of the official at the plant ………………………. Guide/Teacher …………………………

1. The location of the sewage plant.
2. Treatment capacity.
3. The purpose of screening as the initial process?
4. How is air bubbled through the aeration tank?
5. How safe is the water at the end of the treatment? How is it tested?
6. Where is the water discharged after treatment?
7. What happens to the plant during heavy rains?
8. Is bio gas consumed within the plant or sold to other consumers?
9. What happens to the treated sludge?
10. Is there any special effort to protect nearby houses from the plant?
11. Other observations.

Answer:
Do with the help of your subject teacher.

Wastewater Story Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative :

Question (a)
This is not the cause of water pollution?
(a) Floods
(b) Rains
(c) Chemicals
(d) Open defecation.
Answer:
(b) Rains

Question (b)
To improve sanitation following new technique is being used?
(a) Vermi – processing toilets
(b) Sewer system
(c) Onsite sewage disposal
(d) Both (a) and (b).
Answer:
(d) Both (a) and (b).

Question (c)
Following should not be disposed off in the drains?
(a) Tissue Papers
(b) Excreta
(c) Oils and Fats
(d) Waste water.
Answer:
(c) Oils and Fats

Question (d)
Following process is not a part of waste – water treatment?
(a) Decomposition
(b) Grit and Sand Removal
(c) Evaporation
(d) Chlorination.
Answer:
(c) Evaporation

Question (e)
Polluted water causes disease likes –
(a) Hepatitis
(b) Typhopid
(c) Cholera
(d) Diarrhoea.
Answer:
(c) Cholera

Question (f)
In a public sewerage system, the largest sewers are –
(a) Filters
(b) Interceptors
(c) Waterways
(d) None of these.
Answer:
(b) Interceptors

Question 2.
Fill in the blanks:

1. …………………… bacteria is used to treat sludge.
2. Sewage is a liquid waste which causes water and soil
3. Waste – water is treated in a sewage treatment ………………………
4. …………………… and …………………….. are the products of water clarification.
5. By – products of waste – water treatment are sludge and ………………………
6. Addition of disease causing organisms in water is called water …………………….
7. …………………………… in open cause health hazards.
8. ……………………….. is used as manure.
9. To improve sanitation, low cost ……………………….. sewage disposal systems are being encouraged.
10. Adopting good sanitation practices should be our way of ……………………………….

Answer:

1. Anaerobic
2. Pollution
3. Plant
4. Sludge and bio gas
5. Bio – gas
6. Contamination
7. Defecation
8. Activated sludge
9. Onsite
10. Life.

Question 3.
Which of the following statements are true (T) or false(F):

1. Sewage contains pure water for drinking.
2. Used water is waste – water.
3. Waste – water could not be reused.
4. Open drain system is a breeding place for flies.
5. Manholes are located at every 50 m to 60 m in the sewerage.
6. Waste – water is passed through for screens.
7. Eucalyptus trees absorb all surplus waste – water rapidly and release pure water vapour into the atmosphere.

Answer:

1. False
2. True
3. False
4. True
5. True
6. True
7. True.

Wastewater Story Very Short Answer Type Questions

Question 1.
Define sewage?
Answer:
Sewage is water that contains waste products produced by human beings. It is also called waste water.

Question 2.
Which is the world water day?
Answer:
22nd March.

Question 3.
Which is proclaimed as the International Decade for action on water for life?
Answer:
United Nations proclaimed the period 2005 – 2015 as the international Decade for action on “Water for life”.

Question 4.
What do you mean by cleaning of water?
Answer:
Cleaning of water is a process of removing pollutants before it enters a water body or is reused.

Question 5.
What is sewage treatment?
Answer:
This process of wastewater treatment is commonly known as “Sewage Treatment”.

Question 6.
In how many steps sewage treatment divided?
Answer:
The sewage treatment in most cities involve two main steps primary and secondary treatment. Some cities also require an additional step called tertiary treatment.

Question 7.
What do you mean by primary treatment?
Answer:
The primary treatment removes the heaviest solid material from sewage. This process removes about half the suspended solids and bacteria in sewage. Sometimes chlorine gas is added to kill most of the remaining bacteria.

Question 8.
What do you mean by secondary treatment?
Answer:
The secondary treatment removes from 85% to 90% of the solids and oxygen consuming wastes remaining in sewage after it has undergone primary treatment. The most common methods of secondary treatment are the activated sludge process and the trickling filtration process.

Question 9.
Why should we plant eucalyptus along sewage ponds?
Answer:
These trees absorb all surplus waste – water rapidly and release pure water vapour into the atmosphere.

Question 10.
Write certain inorganic impurities in the waste – water?
Answer:
Metals, phosphates and nitrates.

Question 11.
Name certain disease causing micro – organism?
Answer:
Bacterias, Viruses etc.

Question 12.
Which process removes the solids like faces and other substances from the waste – water?
Answer:
Grit and sand removal tank.

Question 13.
How is sand, grit on pebbles settled down?
Answer:
Water goes to a grit and sand removal tank. The speed of the incoming waste-water is decreased to allow sand, grit and pebbles to settle down.

Question 14.
How is dry sludge used?
Answer:
Dried sludge is used as manure, returning organic matter and nutrients to the soil.

Question 15.
Who decomposes the sludge?
Answer:
Anaerobic bacteria decompose the sludge.

Question 16.
Which instrument is used to remove floatable solids from the waste – water?
Answer:
A skimmer is used to remove floatable impurities.

Question 17.
What helps to clean the clarified water?
Answer:
Aerobic bacteria helps to clean the clarified water.

Question 18.
Why is ozone and chlorine used?
Answer:
Ozone and chlorine is used to kill the bacteria etc. present in the clarified water.

Question 19.
Why is air pumped to clarified water?
Answer:
Air is pumped into the clarified water to help aerobic bacteria to grow. Bacteria consume human waste, food waste, soaps and other unwanted matter still remaining in clarified water.

Wastewater Story Short Answer Type Questions

Question 1.
Explain sewage?
Answer:
Waste water including human excreta which flows from our homes into the drains is called domestic sewage. This contains microbes which cause water – borne diseases. This waste water is often dumped into water bodies.

Question 2.
How is water polluted?
Answer:
Water is used various purposes in homes, industries and agriculture. When water is used for cleaning, bathing, washing, dying etc. it pollutes the water. Unwanted waste materials and chemicals etc get added in the water and this wastes the water.

Question 3.
What is done to improve sanitation?
Answer:
To improve sanitation, low cost onsite sewage disposal stem are being encouraged. Examples are septic tanks, chemical toilets, composting pits. Septic tanks are suitable for places where there is no sewerage system, for hospitals isolated buildings or a Luster of 4 to 5 houses.

Question 4.
What is vermi process toilet?
Answer:
A design of a toilet in which humans excreta is treated by earthworms has been tested in India. It has been found to be a novel, j low water – use toilet for safe processing of human waste. The operation of the toilet is very simple and hygienic. The human excreta is completely converted to vermi cakes resource much needed for soil.

Wastewater Story Long Answer Type Questions

Question 1.
How defection in open cause health hazards?
Answer:
Due to lack of proper sewage disposal system a large amount of people in India defecates in open. They use riverbeds, railway lines, fields and drains for this purpose. These excreta dries down and percolate in soil with rain water. It pollutes the ground water.

Excreate along river bed pollutes the river water. In this way water on the ground and under the ground get polluted. This polluted water contains the micro – organisms of various communicable diseases like cholera, typhoid, hepatitis and meaning it is dysentery etc.

Question 2.
Suggest some better house keeping practices?
Answer:

1. Cooking oil and fats should not be thrown down the drain. They can harden and block the pipes. In an open drain the fats dog the soil pores redunt. Its effectiveness in filtering water. Oil and fats should he thrown in the dustbin.

2. Chemicals like paints, vents, insecticides, motor oil, medicines may kill microbes it help purity water. So they should not be thrown in the drain.

3. Used ten – leaves, solid food remains, soft toys, cotton, sanitary towels, etc. should also be thrown in the dustbin. These waste choke the drains. They do not allow free flow of oxygen. This hampers the degradation process.

Question 3.
What is the composition of sewage?
Answer:
Sewage is a complex mixture containing suspended solids, organic and inorganic impurities. nutrients, saprotrophic and disease causing bacteria and other microbes.

1. Organic impurities:
Human faces, animal waste, oil, urea (urine), pesticides, herbicides, fruit and vegetable waste, etc.

2. Inorganic impurities:
Nitrates, phosphates, metals. Nutrients Phosphorus and nitrogen.

3. Bacteria:
Such as which cause cholera and typhoid.

4. Other microbes:
Such as which cause dysentery.

Question 4.
How is sewage treated?
Answer:
Domestic sewage should be treated before being discharged into the river. Sewage is treated by first separating the solid material by sedimentation and filtration. Compressed air is then passed through the liquid which is then chlorinated to kill micro – organisms. The solid matter separated from sewage can be used to generated bio gas which can be used as fuel. The sludge that is left can be used as manure in the fields to grow organic foods.

Question 5.
What are the different ways in which solid waste can be disposed off?
Answer:
The different ways are as:

1. Domestic wastes like fruit and vegetable waste, leftover food, leaves of potted plants can be converted into compost and used as manure.
2. Most of the solid waste is buried in low lying areas to level uneven land. This is called landfill.
3. Wastes coming from industries such as metals can be recycled and used again.
4. Broken plastic articles like plastic bags, buckets, bowls, cups, plates, etc. can be melted and remolded to make new articles.
5. The waste disposal on a large scale is done by the municipality of a city using incinerators. The solid waste is burnt at high temperature. Ash is removed from time to time.

Question 6.
Explain sewage treatment plant.
Answer:
Sewage is water that contains waste products by human beings. It is also known as waste water. In fact sewage comes from the sinks and toilets of homes, restaurants, factories and office buildings. The sewage mainly consists of dissolved material that cannot be seen and bits of such solid matter as human waste and ground up rubbish.

It also contains harmful chemical and disease producting bacteria. Most sewage ultimately goes into lakes, rivers and oceans. However, in many western nations, the sewage is treated in some way before it goes into the waterways as a semi – clear liquid called effluent. Most methods used to treat sewage convert organic sewage into inorganic compounds viz. nitratres, sulphates and phosphates. Some of these compounds serve as food for algae. As the algae decay using excess of oxygen from water, the fish and plants in water will ultimately die.

Question 7.
Explain rural sewerage system?
Answer:
Many rural areas not served by public sewers. In such areas, most home owners use septic tanks to treat their sewage. These tanks are concrete or steel containers buried underground at home and buildings. Sewage flows into a septic tank through a pipe connecting the tank with a building.

Solids in the sewage sink to the bottom of the tank as sludge or float to the surface as scum. Effluent then flows from the tank into a system of pipes with open joints that allow sewage effluent to be gradually distributed into the soil. The soil bacteria then destroy the remaining organic material in the influent.

In a septic tank, bacteria in the sewage attack and digest the sludge and scum. The digestion process changes most to the wastes into gas and a harmless substance called humus. The gas then escapes into the air. The humus in the tank should be pumped our periodically and taken to a sewage treatment plant.

Question 8.
Explaining urban sewerage system?
Answer:
In a public sewerage system, the largest sewers, known as interceptors, carry the sewage to a wastewater treatment plant. The sewage treatment in most cities involves two main steps, primary treatment and secondary treatment. Some cities also require an additional step called tertiary treatment. At a treatment plant, sewage first passes through a screen that traps the largest pieces of matter. It then flows through a grit chamber, where heavy inorganic matter, such as sand, settles down.

The liquid next flows into a large primary sedimentations tank. Many suspended solids sink to the bottom of this tank and form a muddy material called sludge. Grease floats to the surface, where it is removed by a process called skimming. The effluent is then released into waterways.

Primary treatment removes bout half of the suspended solids and bacteria in sewage. Sometimes chlorine gas is added after primary or secondary treatment to kill most of the remaining bacteria. The secondary treatment removes about 85 to 90 percent of the solids and oxygen consuming wastes remaining in sewage after it has undergone primary treatment.

Question 9.
Explain the most common methods of secondary treatment in urban sewerage system?
Answer:
The most common methods of secondary treatment are:

1. The activated sludge process.
2. The trickling filtration process.

In case of activated sludge process, the influent from the primary sedimentation tank flows into a second tank called an aeration tank. The useful bacteria move through the liquid and change the organic matter into less harmful substances. The liquid then flows into a final sedimentation tank, where the sludge settles down to the bottom.

The influent is then discharged into waterways. In case of trickling filtration process, the filters are filled with crushed rocks. As sewage is distributed over the rocks, it reacts with slime that develops on the rocks. The slime contains useful bacteria that change organic material in the sewage into less harmful substances. These substances are removed in a final sedimentation tank, where they fall to the bottom as sludge.

Sometimes tertiary treatment is also used after primary and secondary treatment to produce purer effluent. The tertiary treatment methods include chemical treatment, microscopic screening, radiation treatment, etc. Tertiary treatment makes effluent safer to discharge into waterways.

MP Board Class 7th Science Solutions