Study Rankers

NCERT Solutions for Class 6th: पाठ 13 - मैं सबसे छोटी होऊँ हिंदी वसंत भाग-I

सुमित्रानंदन पंत

पृष्ठ संख्या: 118

प्रश्न अभ्यास

कविता से

1. कविता में सबसे छोटे होने की कल्पना क्यों की गई है?

उत्तर

कविता में सबसे छोटे होने की कल्पना इसलिए की गई है क्योंकि घर के सबसे छोटे सदस्य को सभी लोगों का प्यार और दुलार अधिक मिलता है और खासकर माँ के साथ तो उसका जुड़ाव कुछ ज्यादा ही होता है।

2. कविता में 'ऐसी बड़ी न होऊँ मैं' क्यों कहा गया है?

उत्तर

अपने माँ के स्नेह को हमेशा पाने के लिए, हमेशा उसके ममता के आँचल के साए में रहने के लिए कविता में 'ऐसी बड़ी न होऊँ मैं' कहा गया है।

3. कविता में किसके आँचल की छाया में छिपे रहने की बात कही गई है और क्यों?

उत्तर

कविता में माँ के आँचल की छाया में छिपे रहने की बात कही गई है क्योंकि माँ अपने बच्चों से सबसे अधिक प्यार करती है। उसके आँचल में बच्चा हमेशा अपने को निर्भय और सुरक्षित महसूस करता है।

4. आशय स्पष्ट करो -
हाथ पकड़ फिर सदा हमारे
साथ नहीं फिरती दिन-रात!

उत्तर

इन पंक्तियों का आशय है कि बड़े होने पर बचपन की तरह माँ हमारे साथ नहीं चलतीं। उनसे हमारा रिश्ता छूट जाता है।

कविता से आगे

5. कविता से पता करके लिखो कि माँ बच्चों के लिए क्या-क्या काम करती है? तुम स्वयं सोचकर यह भी लिखो कि बच्चों को माँ के लिए क्या-क्या करना चाहिए?

उत्तर

कविता के अनुसार माँ बच्चों को गोद में सुलाती हैं, अपने आँचल के साये में रखती हैं, हाथ पकड़कर चलना सिखाती हैं, खिलाती हैं, सजाती हैं तथा परियों की कहानियाँ सुनाने आदि का काम करती हैं।
बच्चों को भी अपनी माँ की बातों को मानना चाहिए, उन्हें परेशान नहीं करना चाहिए और उन्हें दुःख नहीं पहुँचाना चाहिए।

2. बच्चों को प्राय: सभी क्षेत्रों में बड़ा होने के लिए कहा जाता है। इस कविता में बालिका सबसे छोटी बनी रहनाक्यों चाहती है?

उत्तर

इस कविता में बालिका सबसे छोटी बनी रहना इसलिए चाहती हैं ताकि उसे हमेशा अपनी माँ का प्यार मिलता रहे। वो सदा अपनी माँ के आँचल में निर्भय और सुरक्षित रहे।

पृष्ठ संख्या: 119

भाषा की बात

1. 'पकड़-पकड़कर' की तरह नीचे लिखे शब्दों को पूरा करो और उनसे वाक्य भी बनाओ -
छोड़, बना, फिर, खिला, पोंछ, थमा, सुना, कह, दिखा, छिपा।

उत्तर

छोड़ (छोड़कर) - घर का सारा काम छोड़कर तुम बात करने में लगे हो।

बना (बनाकर) - रोहित ने चाय बनाकर पिलाया।

फिर (फिरकर) - तुम थोड़ी देर घूम-फिरकर आओ।

खिला (खिलाकर) - तुम्हें खिलाकर ही मैं खाऊँगा।

पोंछ (पोंछकर) - मोहित ने सामान को पोंछकर रखा था।

थमा (थमाकर) - इतना भारी सामान थमाकर वह भाग गया।

सुना (सुनाकर) - नानी मुझे कहानी सुनाकर सुलाया करती हैं।

कह (कहकर) - मैंने उसे तुम्हारी बात कहकर ही उसे पुस्तक दी।

दिखा (दिखाकर) - तुम्हें दिखाकर मैं हर काम नहीं करूँगा।

छिपा (छिपाकर) - हमें बड़ों से छिपाकर कोई काम नहीं करना चाहिए।

2. इन शब्दों के समान अर्थ वाले दो-दो शब्द लिखो -
हाथ, सदा, मुख, माता, स्नेह।

उत्तर

हाथ - कर, हस्त

सदा - हमेशा, सर्वदा

मुख - मुँह, आनन

माता - माँ, जननी

स्नेह - प्यार, प्रेम

3. कविता में 'दिन-रात' शब्द आया है। तुम भी ऐसे पाँच शब्द सोचकर लिखो जिनमें किसी शब्द का विलोम शब्द भी शामिल हो और उनके वाक्य बनाओ।

उत्तर

जीवन-मरण - जीवन-मरण तो इस जीवन का अहम हिस्सा है।

सुख-दुःख - जीवन में सुख-दुःख निरंतर चलते रहते हैं।

उल्टा-सीधा - तुम्हे कोई भी उल्टा-सीधा काम नहीं करना चाहिए।

लाभ-हानि - हर काम करने से पहले हमें उसके लाभ-हानि के बारे में सोचना चाहिए।

मित्र-शत्रु - मोहित को मित्र-शत्रु की पहचान नहीं है।

4. 'निर्भय' शब्दो में 'नि' उपसर्ग लगाकर शब्द बनाया गया है। तुम भी 'नि' उपसर्ग से पाँच शब्द बनाओ।

उत्तर

निश्छल

निकम्मा

निर्बल

निर्गुण

निराकार

5. कविता की किन्हीं चार पंक्तियों को गद्य में लिखो।

उत्तर

मैं सबसे छोटी होऊँ,

तेरी गोदी में सोऊँ,

तेरा अंचल पकड़-पकड़कर,

फिरूँ सदा माँ! तेरे साथ,

मैं सबसे छोटी होना चाहती हूँ ताकि मैं तेरी गोदी में सो पाऊँ। तेरा आँचल पकड़कर माँ मैं तुम्हारे साथ फिरना चाहती हूँ।

पाठ में वापिस जाएँ

NCERT Solutions for Class 6th: पाठ 14 - लोकगीत हिंदी वसंत भाग-I

भगवतशरण उपाध्याय

पृष्ठ संख्या: 125

प्रश्न अभ्यास

निबंध से

1. निबंध में लोकगीतों के किन पक्षों की चर्चा की गई है? बिंदुओं के रूप में उन्हें लिखो।

उत्तर

प्रस्तुत निबंध में लोकगीतों का इतिहास, उनकी रचनात्मकता, जनमानस में लोकप्रियता, स्त्रियों का लोकगीतों में योगदान, उनके विभिन्न प्रकार, उनके संगीत यंत्र, उनकी भाषा, नृत्य और लोकगीत जैसे अनेक बिन्दुओं पर चर्चा की गई है।

2. हमारे यहाँ स्त्रियों के खास गीत कौन-कौन से हैं?

उत्तर

हमारे यहाँ त्योहारों पर नहाते समय के, नहाने जाते हुए राह के, विवाह के, मटकोड़, ज्यौनार के, संबधियों के लिए प्रेमयुक्त गाली के, जन्म पर आदि अवसरों पर गाये जाने अलग-अलग गीत हैं, जो स्त्रियों के लिए खास गीत हैं।

3. निबंध के आधार पर और अपने अनुभव के आधार पर (यदि तुम्हें लोकगीत सुनने के मौके मिले हैं तो) तुम लोकगीतों की कौन-सी विशेषताएँ बता सकते हो?

उत्तर

लोकगीत की अनेक विशेषताएँ हैं-

• ये हमें गाँव के जन-जीवन से परिचित कराते हैं।

• इनके वाद्य यंत्र बहुत सरल होते हैं हैं जैसे ढोल, ढपली, थाल आदि।

• ये समूह में ऊँची आवाज़ में गाये जाते हैं जिस कारण हमारे अंदर उत्साह का संचार होता है।

• इन गीतों को गाने के लिए हमें किसी संगीत के ज्ञान की आवश्यकता नही होती है।

4. 'पर सारे देश के......अपने-अपने विद्यापति हैं' इस वाक्य का क्या अर्थ है? पाठ पढ़कर मालूम करो और लिखो।

उत्तर

पूरब की बोलियों में हमेशा मैथिल-कोकिल विद्यापति द्वारा लिखित गीत गाये जाते हैं परन्तु अगर वहाँ से निकलकर अन्य राज्यों या प्रदेशों में जाएँ तो उन लोगों के लोकगीतों की रचना करने वाले अपने-अपने विद्यापति मौजूद हैं।

पृष्ठ संख्या: 126

भाषा की बात

1. 'लोक' शब्द में कुछ जोड़कर जितने शब्द तुम्हें सूझें, उनकी सूची बनाओ। इन शब्दों को ध्यान से देखो और समझो कि उनमें अर्थ की दृष्टि से क्या समानता है। इन शब्दों से वाक्य भी बनाओ। जैसे - लोककला।

उत्तर

लोकतंत्र - दुनिया में अधिकतर देशों ने लोकतंत्र को अपना लिया है।

लोकमंच - लोकमंच आम जनमानस की परेशानियों को दिखाने का सबसे सरल तरीका है।

लोकहित - हमें वह कभी नहीं करना चाहिए जो लोकहित में ना हो।

लोकप्रिय - यह उत्पाद बाजार में बेहद लोकप्रिय है।

लोकमत - विपक्ष ने लोकमत का सम्मान किया।

पृष्ठ संख्या: 127

2. 'बारहमासा' गीत में साल के बारह महीनों का वर्णन होता है। नीचे विभिन्न अंकों से जुड़े कुछ शब्द दिए गए हैं। इन्हें पढ़ो और अनुमान लगाओ कि इनका क्या अर्थ है और वह अर्थ क्यों है। इस सूची में तुम अपने मन से सोचकर भी कुछ शब्द जोड़ सकते हो -
इकतारा, सरपंच, चारपाई, सप्तर्षि, अठन्नी, तिराहा, दोपहर, छमाही, नवरात्र।

उत्तर

• इकतारा - एक तार से बजने वाला वाद्ययंत्र

• सरपंच - पंचों का प्रमुख

• चारपाई - चार पैरों वाली

• सप्तर्षि - सात ऋषियों का समूह

• अठन्नी - पचास पैसे का सिक्का

• तिराहा - तीन रास्तों के मिलने की जगह

• दोपहर - जब दिन के दो पहर मिलते हों

• छमाही - छह महीने में होने वाला

• नवरात्र - नौ रातों का समूह

3. को, में, से आदि वाक्य में संज्ञा का दूसरे शब्दों के साथ संबंध दर्शाते हैं। पिछले पाठ (झाँसी की रानी) में तुमने का के बारे में जाना। नीचे 'मंजरी जोशी' की पुस्तक 'भारतीय संगीत की परंपरा' से भारत के एक लोकवाद्य का वर्णन दिया गया है। इसे पढ़ो और रिक्त स्थानों में उचित शब्द लिखो -

तुरही भारत के कई प्रांतों में प्रचलित है। यह दिखने .........अंग्रेज़ी के एस या सी अक्षर.........तरह होती है। भारत.........विभिन्न प्रांतों में पीतल या काँसे.........बना यह वाद्य अलग-अलग नामों.........जाना जाता है। धातु की नली.........घुमाकर एस.........आकार इस तरह दिया जाता है कि उसका एक सिरा संकरा रहे और दूसरा सिरा घंटीनुमा चौड़ा रहे। फ़ूँक मारने.........एक छोटी नली अलग.........जोड़ी जाती है। राजस्थान.........इसे बर्गू कहते हैं। उत्तर प्रदेश.........यह तूरी मध्य प्रदेश और गुजरात.........रणसिंघा और हिमाचल प्रदेश........ नरसिंघा.........नाम से जानी जाती है। राजस्थान और गुजरात में इसे काकड़सिंघी भी कहते हैं।

उत्तर

तुरही भारत के कई प्रांतों में प्रचलित है। यह दिखने मेंअंग्रेजी के एस या सी अक्षर कीतरह होती है। भारत केविभिन्न प्रांतों में पीतल या काँसे सेबना यह वाद्य अलग-अलग नामों सेजाना जाता है। धातु की नली कोघुमाकर एस काआकार इस तरह दिया जाता है कि उसका एक सिरा संकरा रहे और दूसरा सिरा घंटीनुमा चौड़ा रहे। फूँक मारने परएक छोटी नली अलग सेजोड़ी जाती है। राजस्थान मेंइसे बर्गू कहते हैं। उत्तर प्रदेश मेंयह तूरी मध्यप्रदेश और गुजरात मेंरणसिंघा और हिमाचलप्रदेश मेंनरसिंघा केनाम से जानी जाती है। राजस्थान और गुजरात में इसे काकड़सिंघी भी कहते हैं।

पाठ में वापिस जाएँ

NCERT Solutions for Class 6th: Ch 10 Motion and Measurement of Distance Science

Page No: 106

Exercises

1. Give two examples each, of modes of transport used on land, water and air.

Answer

Modes of transport used on land - Buses and Trains.
Modes of transport used on water - Ships and Boats.
Modes of transport used on air - Helicopters and Aeroplanes.

2. Fill in the blanks:
(i) One metre is ______________ cm.
(ii) Five kilometre is ______________ m.
(iii)Motion of a child on a swing is ______________.
(iv)Motion of the needle of a sewing machine is ______________.
(v) Motion of wheel of a bicycle is______________.

Answer

(i) One metre is 100 cm.
(ii) Five kilometre is 5000 m.
(iii)Motion of a child on a swing is periodic motion.
(iv)Motion of the needle of a sewing machine is periodic motion.
(v) Motion of wheel of a bicycle is circular motion.

3. Why can a pace or a footstep not be used as a standard unit of length?

Answer

The length of pace or a footstep varies from person to person. If the length of footstep will be used as standard unit for measurement of length then two measured quantity will not be same. Hence, footstep or pace is not a constant quantity and cannot be used as a standard unit of length.

4. Arrange the following lengths in their increasing magnitude:
1 metre, 1 centimetre, 1 kilometre,1 millimetre.

Answer

1 millimetre > 1 centimetre > 1 metre > 1 kilometre

5. The height of a person is 1.65 m. Express it into cm and mm.

Answer

1.65 m = 1.65 × 100 cm = 165 cm
1.65 m = 1.65 × 100 × 10 mm = 1650 mm

6. The distance between Radha's home and her school is 3250 m. Express this distance into km.

Answer

1km = 1000 m
3250 m = 3250/1000 km = 3.25 km

7. While measuring the length of a knitting needle, the reading of the scale at one end is 3.0 cm and at the other end is 33.1 cm. What is the length of the needle?

Answer

The actual count of the length of the needle start from 3.0 cm instead of 0 cm. Thus, the actual length of the needle is (33.1 - 3.0) cm = 30.1 cm

8. Write the similarities and differences between the motion of a bicycle and a ceiling fan that has been switched on.

Answer

Similarities:
(i) Both the ceiling fan and the wheels of a bicycle show circular motion.
(ii) Both are fixed at point.
Differences:
(i) A bicycle shows rectlinear motion while a ceiling fan does not.

9. Why could you not use an elastic measuring tape to measure distance? What would be some of the problems you would meet in telling someone about a distance you measured with an elastic tape?

Answer

An elastic tape is flexible and can be stretched. Thus, each time it will give different readings for the same length measured. Therefore, it is impossible to measure accurate distance from an elastic measuring tape.

10. Give two examples of periodic motion.

Answer

Two examples of periodic motion:
(i) Motion of a pendulum clock.
(ii) Motion of a swing.

Go To Chapters

NCERT Solutions for Class 12th: Ch 5 Principle and Inheritance of variation Biology

Exercises

Page No: 93

1. Mention the advantages of selecting pea plant for experiment by Mendel.

Answer

The advantages of selecting pea plant for experiment by Mendel:
• Peas have many visible contrasting characters such as smooth or wrinkled seeds, yellow or green seeds, smooth or inflated pods, green or yellow pods and tall or dwarf plants.
• It undergoes self-pollination easily producing offsprings with same traits generation after generation.
• Cross-pollination can be also done manually by emasculation in which the stamen of the flower is removed without affecting the pistil.
• Pea plants have a short life span and produce many seeds in one generation.

2. Differentiate between the following –

(a) Dominance and Recessive

Dominance	Recessive
Dominance allele is able to express itself even in the presence of its recessive allele.	Recessive allele or factor is unable to express its effect in the presence of dominant allele.
It does not require another similar allele to produce its effect on the phenotype, e.g., Tt is tall.	It produces its phenotypic effect only in the presence of a similar allele, e.g., tt is dwarf.

(b) Homozygous and Hetrozygous

Homozygous	Heterozygous
It is pure for a trait and breeds true, i.e., gives rise to similar homozygous individuals.	It is rarely pure and produces offspring with different genotypes on selling, e.g., TT. Tt and tt on selfing of Tt individuals.
Both the alleles of a character are similar. Example: TT, tt	It carries dissimilar alleles, Example: Tt
It produces only one type of gamete.	It produces two different kinds of gametes.

Monohybrid	Dihybrid
It is a cross between two pure organisms in order to study the inheritance of a single pair of alleles.	It is a cross between two pure organisms of a species in order to study the inheritance of two pairs of alleles.
It produces genotypic ratio of 1: 2: 1 in F₂ generation.	It produces a phenotypic dihybrid ratio of 9: 3: 3: 1 in F₂ generation.

3. A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?

Answer

By applying the formula 2ⁿ (where n = number of loci)

4 loci can produce = 2⁴ = 16 types of gametes.

4. Explain the Law of Dominance using a monohybrid cross.

Answer

Law of Dominance states that when two genes are present in an organism then dominant gene expresses itself and recessive gene does not show its effect and remain hidden.
When a cross pollination experiments is conducted between two organisms taking a single contrasting character at a time it is called monohybrid cross.
When two pea plants, one is pure tall (TT) and other is pure dwarf (tt) are crossed. All pants in F1 generation were found to be Tal l(TT). When these tall plants (Tt) were self fertilized, both the tall and dwarf seeds appeared in F2 generation in 3: 1 ratio. Hence, in F1generation, the dominant character (Tall) appeared and the recessive character (dwarf) got suppressed, which reappeared in F2 generation. Thus, this monohybrid cross explains Law of Dominance.

5. Define and design a test - cross?

Answer

A test cross is one in which an individual with an unknown dominant phenotype is crossed with a recessive individual for that trait. It is used to determine whether the individual is homozygous or heterozygous for a trait.

If the unknown is homozygous tall (TT), then crossing with dwarf recessive (tt) gives all tall offspring (Tt).

If the unknown is heterozygous tall (Tt), then crossing with dwarf results in 50% tall (Tt) and 50% dwarf (tt) progeny.

6. Using a Punnett square, work out the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.

Answer

7. When a cross in made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be

(a) tall and green.

(b) dwarf and green.

Answer

A cross between tall plant with yellow seeds and tall plant with green seeds will produce
(a) three tall and green plants
(b) one dwarf and green plant

Page No: 94

8. Two heterozygous parents are crossed. If the two loci are linked, then what would be the distribution of phenotypic features in the F1generation for a dihybrid cross?

Answer

If two loci are linked, they lie close on the same locus of a chromosome, they would separate. However, the chromosomes do segregate and end up in different gametes.

9. Briefly mention the contribution of T.H. Morgan in genetics.

Answer

TH Morgan's contribution in genetics are
• Morgan contributed to the understanding of sex-linked traits.
• He carried out several experiments on fruit flies and his dihybrid crosses revealed that genes for some traits did not segregate independently and the F2 ratio was a lot different from 9:3:3:1.
• He inferred that when two genes were situated on the same chromosome, the proportions of parental gene combinations were significantly higher than the non-parental type.
• He established the principle of linkage crossing over, sex-linked inheritance and discovered the relation.
• He established the technique of chromosome mapping.
• He observed and worked on mutation.

10. What is pedigree analysis? Suggest how such an analysis, can be useful.

Answer

Pedigree analysis is a record of occurrence of a trait in several generations of a family. It is based on the heritability of certain characteristic features in families.

Its usefulness are:

• It shows the origin of a trait and flow of that trait in a family.

• It is useful in knowing the possibility of expressive or recessive allele which may cause genetic disorder, such as colour blindness.

• It predicts the harmful effects of marriage between close relatives.

• It helps in genetic counselling to avoid disorders in children.

11. How is sex determined in human beings?

Answer

Human beings have 23 pairs of chromosomes in which 22 pairs of autosomes and one pair of sex chromosomes. The females have a pair of XX-chromosomes (homogametic) and the males XY (heterogametic) composition. The sex of the baby is determined by the type of male gamete that fuses with the female gamete. If the fertilizing sperm contains X chromosome, then the baby produced will be a girl and if the fertilizing sperm contains Y chromosome, then the baby produced will be a boy.

12. A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.

Answer

The child with blood group ‘O’ will have homozygous recessive alleles. Therefore, both the parents should be heterozygous, i.e., the genotype of the father will be IⁱA and of mother will be IⁱB.

Blood group: AB

The possible blood groups of other offsprings will be AB, A, B and O.

13. Explain the following terms with example
(a) Co-dominance
(b) Incomplete dominance

Answer

Co-dominance: Co-dominance is the phenomenon in which F1 generation resembles both the parents and both the parental characters are expressed simultaneously.
Example: If a plant with yellow coloured flower crosses with a plant of red coloured flower and in F 1 generation, all progeny show orange flowers because of co-dominance of both alleles for red and yellow colours.

Incomplete dominance: Incomplete dominance is a phenomenon in which the genes of allelomorphic pairs are not expressed as dominant and recessive but express themselves partially when present together in the hybrid.
Example: 4' O clock plant exhibit two types of flowers, red and white and hybrids are pink cloured flowers.

14. What is point mutation? Give one example.

Answer

Point mutation is a change in single base pair of DNA molecule. Example: Sickle cell anaemia. It involves mutation in a single base pair in the beta-globin chain of haemoglobin pigment of the blood. Glutamic acid in short arm of chromosome II gets replaced with valine at the sixth position.

15. Who had proposed the chromosomal theory of the inheritance?

Answer

Sutton and Boveri proposed the chromosomal theory of inheritance in 1902.

16. Mention any two autosomal genetic disorders with their symptoms.

Answer

Sickle cell anaemia: This is an autosome linked recessive trait that can be transmitted from parents to the offspring when both the partners are carrier for the gene or heterozygous.

Symptoms: Rapid heart rate, breathlessness, delayed growth and puberty, jaundice, weakness, fever, excessive thirst, chest pain, and decreased fertility are the major symptoms of sickle cell anaemia disease.

Down's syndrome: It is an autosomal disorder that is caused by the trisomy of chromosome 21.

Symptoms: The individual is short statured with round head, open mouth, protruding tongue, short neck, slanting eyes, and broad short hands. The individual also shows retarded mental and physical growth.

Go To Chapters

NCERT Solutions for Class 6th: पाठ - 15 नौकर हिंदी वसंत भाग-I

अनु बंधोपाध्याय

पृष्ठ संख्या: 112

प्रश्न अभ्यास

निबंध से

1. आश्रम में कॉलेज के छात्रों से गाँधी जी ने कौन-सा काम करवाया और क्यों?

उत्तर

आश्रम में कॉलेज के छात्रों से गाँधी जी ने गेंहूँ बीनने का काम करवाया। उन्हें अपने अंग्रेजी ज्ञान पर बड़ा गर्व था। बातचीत के अंत में उन्होंने गांधीजी से कोई कार्य माँगा चूँकि उन्हें लगा की वे उन्हें पढ़ने-लिखने सम्बंधित कार्य देंगे परन्तु गाँधी ने उनकी मंशा को भांपते हुए उन्हें गेहूँ बीनने का कार्य सौंप दिया।

2. 'आश्रम में गाँधी कई ऐसे काम भी करते थे, जिन्हें आमतौर पर नौकर-चाकर करते हैं'। पाठ से तीन ऐसे प्रसंगों को अपने अपने शब्दों में लिखो जो इस बात का प्रमाण हों।

उत्तर

• जब वे बैरिस्टरी से हज़ारो रुपये कमाते थे उस समय भी वे प्रतिदिन सुबह खुद चक्की पर आटा पीसा करते थे।

• आश्रम में वे सब्जियाँ छिलने का काम करते थे।

• आश्रम के नियमानुसार सभी लोगों को मिल-बांटकर बर्तन साफ़ करना पड़ता था। एक बार उन्होंने बर्तनों की सफाई खुद किया।

3. लंदन में भोज पर बुलाए जाने पर गाँधी जी ने क्या किया?

उत्तर

लंदन में भोज पर बुलाए जाने पर गाँधी जी ने वहाँ तश्तरियाँ धोने, सब्जियाँ साफ़ करने और अन्य छूट-पुट काम करने में छात्रों की मदद करने लगे।

4. गाँधी जी ने श्रीमती पोलक के बच्चे का दूध कैसे छुड़वाया?

उत्तर

गाँधी जी ने श्रीमती पोलक के बच्चे का दूध छुड़वाने के लिए वे बच्चे को माँ से दूर अपने बिस्तर पर सुलाते थे। वह चारपाई के पास एक बरतन में पानी भरकर रख लेते बच्चे को प्यास लगे तो उसे पिला दें। एक पखवाड़े तक माँ से अलग सुलाने के बाद बच्चे ने माँ का दूध छोड़ दिया।

5. आश्रम में काम करने या करवाने का कौन-सा तरीका गाँधी जी अपनाते थे? इसे पाठ पढ़कर लिखो।

उत्तर

गाँधीजी अपना काम स्वयं करते थे और दूसरों से काम करवाने में सख्ती भी बरतते थे। गाँधी जी को काम करता देख उनके अनुयायी भी उनका अनुकरण कर कार्य करने लगते थे। इस प्रकार गाँधी जी स्वयं के उदाहरण द्वारा लोगों को काम करने की प्रेरणा देते थे।

निबंध से आगे

6. गाँधी जी इतना पैदल क्यों चलते थे? पैदल चलने के क्या लाभ हैं? लिखो।

उत्तर

पैदल चलने से शरीर स्वस्थ रहता है। रोज पैदल चलने से शारीरिक फुर्ती बनी रहती है, शरीर में कमजोरी महसूस नहीं होती। व्यक्ति तरोताजा महसूस करता है।

पृष्ठ संख्या: 113

भाषा की बात

1. (क) 'पिसाई' संज्ञा है। पिसना शब्द से 'ना' निकाल देने पर 'पीस' धातु रह जाती है। पीस धातु में 'आई' प्रत्यय जोड़ने पर 'पिसाई' शब्द बनता है। किसी-किसी क्रिया में प्रत्यय जोड़कर उसे संज्ञा बनाने के बाद उसके रूप में बदलाव आ जाता है, जैसे ढोना से ढुलाई, बोना से बुलाई।
मूल शब्द के अंत में जुड़कर नया शब्द बनाने वाले शब्दांश को प्रत्यय कहते हैं।
नीचे कुछ संज्ञाएँ दी गई हैं। बताओ ये किन क्रियाओं से बनी हैं -
बुआई........................... कटाई...........................
सिंचाई........................... रोपाई...........................
कताई........................... रंगाई...........................

उत्तर

संज्ञा - क्रिया
बुआई - बोना
कटाई - काटना
सिंचाई - सींचना
रोपाई - रोपना
कताई - कातना
रंगाई - रंगना

पृष्ठ संख्या: 114

(ख) हर काम-धंधे के क्षेत्र की अपनी कुछ अलग भाषा और शब्द-भंडार होते हैं। ऊपर लिखे शब्दों का संबंध दो अलग-अलग कामों से है। पहचानो कि दिए गए शब्दों के संबंध किन-किन कामों से हैं।

उत्तर

दिए गए शब्द कृषि तथा कपड़े से संबंधित है।

2. (क) तुमने कपड़ो को सिलते हुए देखा होगा। नीचे इस काम से जुड़े कुछ शब्द दिए गए हैं। आस-पास के बड़ों से या दरजी से इन शब्दों के बारे में पूछो और इन शब्दों को कुछ वाक्यों में समझाओ।
तुरपाई बखिया कच्ची सिलाई चोर सिलाई

उत्तर

तुरपाई - कपडे में तुरपाई कर दो।

बखिया - रुमाल में बखिया लगा दो।

कच्ची सिलाई - पहले कपडे में कच्ची सिलाई कर दो।

चोर सिलाई - इस पैंट में चोर सिलाई करना।

3. नीचे लिखे गए शब्द पाठ से लिए गए हैं। इन्हें पाठ में खोजकर बताओ कि ये स्त्रीलिंग हैं या पुल्लिंग -
कालिख, भराई, चक्की, रोशनी, जेल, सेवा, पतीला

उत्तर

पुल्लिंग - पतीला

स्त्रीलिंग - कालिख, भराई, चक्की, रोशनी, जेल, सेवा

पाठ में वापिस जाएँ

NCERT Solutions for Class 9th: Ch 9 Areas of Parallelograms and Triangles Maths (Part-1)

Page No: 155

Exercise 9.1

1. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

Answer

(i) Trapezium ABCD and ΔPDC lie on the same DC and between the same parallel lines AB and DC.
(ii) Parallelogram PQRS and trapezium SMNR lie on the same base SR but not between the same parallel lines.
(iii) Parallelogram PQRS and ΔRTQ lie on the same base QR and between the same parallel lines QR and PS.
(iv) Parallelogram ABCD and ΔPQR do not lie on the same base but between the same parallel lines BC and AD.
(v) Quadrilateral ABQD and trapezium APCD lie on the same base AD and between the same parallel lines AD and BQ.
(vi) Parallelogram PQRS and parallelogram ABCD do not lie on the same base SR but between the same parallel lines SR and PQ.

Page No: 159

Exercise 9.2

1. In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Answer

Given,
AB = CD = 16 cm (Opposite sides of a parallelogram)
CF = 10 cm and AE = 8 cm
Now,
Area of parallelogram = Base × Altitude
= CD × AE = AD × CF
⇒ 16 × 8 = AD × 10
⇒ AD = 128/10 cm
⇒ AD = 12.8 cm

2. If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that
ar (EFGH) = 1/2 ar(ABCD).

Answer

Given,
E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD.
To Prove,
ar (EFGH) = 1/2 ar(ABCD)
Construction,
H and F are joined.
Proof,
AD || BC and AD = BC (Opposite sides of a parallelogram)
⇒ 1/2 AD = 1/2 BC
Also,
AH || BF and and DH || CF
⇒ AH = BF and DH = CF (H and F are mid points)
Thus, ABFH and HFCD are parallelograms.
Now,
ΔEFH and ||gm ABFH lie on the same base FH and between the same parallel lines AB and HF.
∴ area of EFH = 1/2 area of ABFH --- (i)
also, area of GHF = 1/2 area of HFCD --- (ii)
Adding (i) and (ii),
area of ΔEFH + area of ΔGHF = 1/2 area of ABFH + 1/2 area of HFCD
⇒ area of EFGH = area of ABFH
⇒ ar (EFGH) = 1/2 ar(ABCD)

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).

Answer

ΔAPB and ||gm ABCD are on the same base AB and between same parallel AB and DC.
Therefore,
ar(ΔAPB) = 1/2 ar(||gm ABCD) --- (i)
Similarly,
ar(ΔBQC) = 1/2 ar(||gm ABCD) --- (ii)
From (i) and (ii),
we have ar(ΔAPB) = ar(ΔBQC)

4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(APB) + ar(PCD) = 1/2 ar(ABCD)
(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
[Hint : Through P, draw a line parallel to AB.]

Answer

(i) A line GH is drawn parallel to AB passing through P.
In a parallelogram,
AB || GH (by construction) --- (i)
Thus,
AD || BC ⇒ AG || BH --- (ii)
From equations (i) and (ii),
ABHG is a parallelogram.
Now,
In ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH.
∴ ar(ΔAPB) = 1/2 ar(ABHG) --- (iii)
also,
In ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH.
∴ ar(ΔPCD) = 1/2 ar(CDGH) --- (iv)
Adding equations (iii) and (iv),
ar(ΔAPB) + ar(ΔPCD) = 1/2 {ar(ABHG) + ar(CDGH)}
⇒ ar(APB) + ar(PCD) = 1/2 ar(ABCD)

(ii) A line EF is drawn parallel to AD passing through P.
In a parallelogram,
AD || EF (by construction) --- (i)
Thus,
AB || CD ⇒ AE || DF --- (ii)
From equations (i) and (ii),
AEDF is a parallelogram.
Now,
In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF.
∴ ar(ΔAPD) = 1/2 ar(AEFD) --- (iii)
also,
In ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF.
∴ ar(ΔPBC) = 1/2 ar(BCFE) --- (iv)
Adding equations (iii) and (iv),
ar(ΔAPD) + ar(ΔPBC) = 1/2 {ar(AEFD) + ar(BCFE)}
⇒ ar(APD) + ar(PBC) = ar(APB) + ar(PCD)

5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 1/2 ar (PQRS)

Answer

(i) Parallelogram PQRS and ABRS lie on the same base SR and between the same parallel lines SR and PB.

∴ ar(PQRS) = ar(ABRS) --- (i)
(ii) In ΔAXS and parallelogram ABRS are lying on the same base AS and between the same parallel lines AS and BR.
∴ ar(ΔAXS) = 1/2 ar(ABRS) --- (ii)
From (i) and (ii),
ar(ΔAXS) = 1/2 ar(PQRS)

Page No: 106

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?

Answer

The field is divided into three parts. The three parts are in the shape of triangle. ΔPSA, ΔPAQ and ΔQAR.
Area of ΔPSA + ΔPAQ + ΔQAR = Area of PQRS --- (i)
Area of ΔPAQ = 1/2 area of PQRS --- (ii)
Triangle and parallelogram on the same base and between the same parallel lines.
From (i) and (ii),
Area of ΔPSA + Area of ΔQAR = 1/2 area of PQRS --- (iii)
Clearly from (ii) and (iii),
Farmer must sow wheat or pulses in ΔPAQ or either in both ΔPSA and ΔQAR.

Page No: 162

Exercise 9.3

1. In Fig.9.23, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar(ACE).

Answer

Given,

AD is median of ΔABC. Thus, it will divide ΔABC into two triangles of equal area.

∴ ar(ABD) = ar(ACD) --- (i)

also,

ED is the median of ΔABC.

∴ ar(EBD) = ar(ECD) --- (ii)

Subtracting (ii) from (i),

ar(ABD) - ar(EBD) = ar(ACD) - ar(ECD)

⇒ ar(ABE) = ar(ACE)

2. In a triangle ABC, E is the mid-point of median AD. Show that ar(BED) = 1/4 ar(ABC).

Answer

ar(BED) = (1/2) × BD × DE
As E is the mid-point of AD,

Thus, AE = DE
As AD is the median on side BC of triangle ABC,
Thus, BD = DC
Therefore,

DE = (1/2)AD --- (i)
BD = (1/2)BC --- (ii)

From (i) and (ii),

ar(BED) = (1/2)× (1/2) BC × (1/2)AD
⇒ ar(BED) = (1/2)× (1/2) ar(ABC)

⇒ ar(BED) = 1/4 ar(ABC)

3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Answer

O is the mid point of AC and BD. (diagonals of bisect each other)

In ΔABC, BO is the median.

∴ ar(AOB) = ar(BOC) --- (i)

also,

In ΔBCD, CO is the median.

∴ ar(BOC) = ar(COD) --- (ii)

In ΔACD, OD is the median.

∴ ar(AOD) = ar(COD) --- (iii)

In ΔABD, AO is the median.

∴ ar(AOD) = ar(AOB) --- (iv)

From equations (i), (ii), (iii) and (iv),

ar(BOC) = ar(COD) = ar(AOD) = ar(AOB)

So, the diagonals of a parallelogram divide it into four triangles of equal area.

4. In Fig. 9.24, ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that:

ar(ABC) = ar(ABD).

Answer

In ΔABC,

AO is the median. (CD is bisected by AB at O)

∴ ar(AOC) = ar(AOD) --- (i)

also,

In ΔBCD,

BO is the median. (CD is bisected by AB at O)

∴ ar(BOC) = ar(BOD) --- (ii)
Adding (i) and (ii) we get,
ar(AOC) + ar(BOC) = ar(AOD) + ar(BOD)
⇒ ar(ABC) = ar(ABD)

From Exercise 9.3 - Q. No. 5 to 16 (Page No. 163)

Go to Chapters

NCERT Solutions for Class 9th: Ch 9 Areas of Parallelograms and Triangles Maths (Part-2)

Page No: 163

5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC.
Show that
(i) BDEF is a parallelogram. (ii) ar(DEF) = 1/4 ar(ABC)
(iii) ar (BDEF) = 1/2 ar(ABC)

Answer

(i) In ΔABC,
EF || BC and EF = 1/2 BC (by mid point theorem)
also,
BD = 1/2 BC (D is the mid point)
So, BD = EF
also,
BF and DE will also parallel and equal to each other.
Thus, the pair opposite sides are equal in length and parallel to each other.
∴ BDEF is a parallelogram.

(ii) Proceeding from the result of (i),
BDEF, DCEF, AFDE are parallelograms.
Diagonal of a parallelogram divides it into two triangles of equal area.
∴ ar(ΔBFD) = ar(ΔDEF) (For parallelogram BDEF) --- (i)
also,
ar(ΔAFE) = ar(ΔDEF) (For parallelogram DCEF) --- (ii)
ar(ΔCDE) = ar(ΔDEF) (For parallelogram AFDE) --- (iii)
From (i), (ii) and (iii)
ar(ΔBFD) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF)
⇒ ar(ΔBFD) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = arar(ΔABC)
⇒ 4 ar(ΔDEF) = ar(ΔABC)
⇒ ar(DEF) = 1/4 ar(ABC)

(iii) Area (parallelogram BDEF) = ar(ΔDEF) + ar(ΔBDE)
⇒ ar(parallelogram BDEF) = ar(ΔDEF) + ar(ΔDEF)
⇒ ar(parallelogram BDEF) = 2× ar(ΔDEF) ⇒ ar(parallelogram BDEF) = 2× 1/4 ar(ΔABC) ⇒ ar(parallelogram BDEF) = 1/2 ar(ΔABC)

6. In Fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.
If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint : From D and B, draw perpendiculars to AC.]

Answer

Given,
OB = OD and AB = CD
Construction,
DE ⊥ AC and BF ⊥ AC are drawn.
Proof:
(i) In ΔDOE and ΔBOF,
∠DEO = ∠BFO (Perpendiculars)
∠DOE = ∠BOF (Vertically opposite angles)
OD = OB (Given)
Therefore, ΔDOE ≅ ΔBOF by AAS congruence condition.
Thus, DE = BF (By CPCT) --- (i)
also, ar(ΔDOE) = ar(ΔBOF) (Congruent triangles) --- (ii)
Now,
In ΔDEC and ΔBFA,
∠DEC = ∠BFA (Perpendiculars)
CD = AB (Given)
DE = BF (From i)
Therefore,ΔDEC ≅ ΔBFA by RHS congruence condition.
Thus, ar(ΔDEC) = ar(ΔBFA) (Congruent triangles) --- (iii)
Adding (ii) and (iii),
ar(ΔDOE) + ar(ΔDEC) = ar(ΔBOF) + ar(ΔBFA)
⇒ ar (DOC) = ar (AOB)

(ii) ar(ΔDOC) = ar(ΔAOB)
⇒ ar(ΔDOC) + ar(ΔOCB) = ar(ΔAOB) + ar(ΔOCB) (Adding ar(ΔOCB) to both sides) ⇒ ar(ΔDCB) = ar(ΔACB) (iii) ar(ΔDCB) = ar(ΔACB) If two triangles are having same base and equal areas, these will be between same parallels DA || BC --- (iv) For quadrilateral ABCD, one pair of opposite sides are equal (AB = CD) and other pair of opposite sides are parallel. Therefore, ABCD is parallelogram 7. D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC) = ar(EBC). Prove that DE || BC. Answer

ΔDBC and ΔEBC are on the same base BC and also having equal areas. Therefore, they will lie between the same parallel lines. Thus, DE || BC.

8. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that
ar(ΔABE) = ar(ΔAC)

Answer

Given,
XY || BC, BE || AC and CF || AB
To show,
ar(ΔABE) = ar(ΔAC)
Proof:
EY || BC (XY || BC) --- (i)
also,
BE∥ CY (BE || AC) --- (ii)
From (i) and (ii),
BEYC is a parallelogram. (Both the pairs of opposite sides are parallel.)
Similarly,
BXFC is a parallelogram.
Parallelograms on the same base BC and between the same parallels EF and BC.
⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) --- (iii)
Also,
△AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.
⇒ ar(△AEB) = 1/2ar(BEYC) --- (iv)
Similarly,
△ACF and parallelogram BXFC on the same base CF and between the same parallels CF and AB.
⇒ ar(△ ACF) = 1/2ar(BXFC) --- (v)
From (iii), (iv) and (v),
ar(△AEB) = ar(△ACF)

9. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. 9.26). Show that
ar(ABCD) = ar(PBQR).
[Hint : Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]

Answer

AC and PQ are joined.
ar(△ACQ) = ar(△APQ) (On the same base AQ and between the same parallel lines AQ and CP)
⇒ ar(△ACQ) - ar(△ABQ) = ar(△APQ) - ar(△ABQ)
⇒ ar(△ABC) = ar(△QBP) --- (i)
AC and QP are diagonals ABCD and PBQR. Thus,
ar(ABC) = 1/2 ar(ABCD) --- (ii)
ar(QBP) = 1/2 ar(PBQR) --- (iii)
From (ii) and (ii),
1/2 ar(ABCD) = 1/2 ar(PBQR)
⇒ ar(ABCD) = ar(PBQR)

10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

Answer

△DAC and △DBC lie on the same base DC and between the same parallels AB and CD.
∴ ar(△DAC) = ar(△DBC)
⇒ ar(△DAC) − ar(△DOC) = ar(△DBC) − ar(△DOC)
⇒ ar(△AOD) = ar(△BOC)

11. In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.
Show that
(i) ar(ACB) = ar(ACF)
(ii) ar(AEDF) = ar(ABCDE)

Answer

(i) △ACB and △ACF lie on the same base AC and between the same parallels AC and BF.

∴ ar(△ACB) = ar(△ ACF)

(ii) ar(△ACB) = ar(△ACF)

⇒ ar(△ACB) + ar(△ACDE) = ar(△ACF) + ar(△ACDE)

⇒ ar(ABCDE) = ar(△AEDF)

Page No: 164

12. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

Answer

Let ABCD be the plot of the land of the shape of a quadrilateral.

Construction,

Diagonal BD is joined. AE is drawn parallel BD. BE is joined which intersected AD at O. △BCE is the shape of the original field and △AOB is the area for constructing health centre. Also, △DEO land joined to the plot.
To prove:
ar(△DEO) = ar(△AOB)
Proof:
△DEB and △DAB lie on the same base BD and between the same parallel lines BD and AE.
ar(△DEB) = ar(△DAB)
⇒ ar(△DEB) - ar△DOB) = ar(△DAB) - ar(△DOB)

⇒ ar(△DEO) = ar(△AOB)

13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

[Hint : Join CX.]

Answer

Given,

ABCD is a trapezium with AB || DC.

XY || AC

Construction,

CX is joined.

To Prove,

ar(ADX) = ar(ACY)

Proof:
ar(△ADX) = ar(△AXC) --- (i) (On the same base AX and between the same parallels AB and CD)
also,

ar(△ AXC)=ar(△ ACY) --- (ii) (On the same base AC and between the same parallels XY and AC.)
From (i) and (ii),

ar(△ADX)=ar(△ACY)

14. In Fig.9.28, AP || BQ || CR. Prove that
ar(AQC) = ar(PBR).

Answer

Given,

AP || BQ || CR

To Prove,

ar(AQC) = ar(PBR)

Proof:

ar(△AQB) = ar(△PBQ) --- (i) (On the same base BQ and between the same parallels AP and BQ.)
also,

ar(△BQC) = ar(△BQR) --- (ii) (On the same base BQ and between the same parallels BQ and CR.)
Adding (i) and (ii),
ar(△AQB) + ar(△BQC) = ar(△PBQ) + ar(△BQR)

⇒ ar(△ AQC) = ar(△ PBR)

15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(AOD) = ar(BOC). Prove that ABCD is a trapezium.

Answer

Given,
ar(△AOD) = ar(△BOC)

To Prove,

ABCD is a trapezium.

Proof:

ar(△AOD) = ar(△BOC)

⇒ ar(△AOD) + ar(△AOB) = ar(△BOC) + ar(△AOB)

⇒ ar(△ADB) = ar(△ACB)
Areas of △ADB and △ACB are equal. Therefore, they must lying between the same parallel lines.
Thus, AB ∥ CD

Therefore, ABCD is a trapezium.

16. In Fig.9.29, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Answer

Given,

ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC)

To Prove,

ABCD and DCPR are trapeziums.

Proof:

ar(△BDP) = ar(△ARC)

⇒ ar(△BDP) -ar(△DPC) = ar(△DRC)

⇒ ar(△BDC) = ar(△ADC)

ar(△BDC) = ar(△ADC). Therefore, they must lying between the same parallel lines.
Thus, AB ∥ CD

Therefore, DCPR is a trapezium.

also,

ar(DRC) = ar(DPC). Therefore, they must lying between the same parallel lines.
Thus, DC ∥ PR

Therefore, DCPR is a trapezium.

From Exercise 9.1 to 9.3 (up to Q. No. 4)

Go to Chapters

NCERT Solutions for Class 12th: Ch 6 Molecular Basis of Inheritance Biology

Exercises

Page No: 125

1. Group the following as nitrogenous bases and nucleosides:

Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Answer

Nitrogenous bases are Adenine, Thymine and Uracil.
Nucleosides are Cytidine, Guanosine and Cytosine.

2. If a double-stranded DNA has 20 percent of cytosine, calculate the percentage of adenine in DNA.

Answer

The percent of cytosine = 20 (given)

% of C = % of G

Therefore, the percent of guanine = 20
The percent of thymine + adenine will be 100 - (20 + 20) = 60
Therefore, the percent of adenine will be 60/2 = 30

3. If the sequence of one strand of DNA is written as follows:
5′-ATGCATGCATGCATGCATGCATGC-3′
Write down the sequence of the complementary strand in the 3′ → 5′ direction.

Answer

The sequence of the complementary strand in the 3′→5′ direction will be 3'-TACGTACGTACGTACGTACGTACGTACG-5'.

4. If the sequence of a coding strand in a transcription unit is written as follows:
5'-ATGCATGCATGCATGCATGCATGCATGC-3'
Write down the sequence of m-RNA.

Answer

The sequence of a template strand will be
3'-TACGTACGTACGTACGTACGTACGTACG-5'
Thus, the sequence of mRNA will be
5'-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'

5. Which property of DNA double helix led Watson and Crick to hypothesise the semi-conservative mode of DNA replication? Explain.

Answer

The property of DNA double helix led Watson and Crick are:

• Two strands running opposite to each other, wherein bases will always pair with their counterpart-A with T and G with C (specific pairing).

• If H bonds break and bases of one strand lie exposed, unpaired, they will easily pair up with free nucleotides as well.

This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semi-conservative. where the two strands separate and act as a template for the synthesis of a new complementary strand.

6. Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.

Answer

• DNA-dependent DNA polymerase uses a DNA template to catalyse the polymerisation deoxynucleotides.
• DNA-dependent RNA polymerase catalyses the transcription of all types of RNA (in bacteria).
• DNA-dependent RNA polymerase-I transcribes rRNA.
• DNA-dependent RNA polymerase-II transcribes the precursor of mRNA (hnRNA).
• DNA-dependent RNA polymerase-III transcribes tRNA.

7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

Answer

According to Hershey and Chase experiment:
• They grew some bacteriophages on a medium that contained radioactive phosphorus and some in another medium that contained radioactive sulphur.

• Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein as phosphorus is present only in DNA.
• Viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.
• It was found that bacteria which were infected with bacteriophages that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.
• Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicated that proteins did not enter the bacteria from the viruses.
• This was a clear proof that DNA is the genetic material that is passed from virus to bacteria.

8. Differentiate between the followings:

(a) Repetitive DNA and Satellite DNA
(b) mRNA and tRNA
(c) Template strand and Coding strand

Answer

(a) Repetitive DNA and Satellite DNA

Repetitive DNA	Satelite DNA
It is the non-coding DNA with multiple copies of identical sequences which may lie in tandem or interspersed.	It refers to non-coding tandem repeat sequences.
These can be few base pairs to hundreds or thousands of base pairs.	These are generally short sequence repeats (up to 60 base pair long).
It appears as light bands.	It appears as small dark bands.

(b) mRNA and tRNA

mRNA	tRNA
It is called messenger RNA and carries the codes for amino acid sequence.	It is called transfer RNA as it carries amino acids to the site of protein synthesis.
It is a linear molecule.	It has clover leaf shape.
It is synthesised by RNA polymerase II	It is synthesised by RNA polymerase III.

Template strand	Coding strand
It is the strand which is transcribed into RNA.	It has the same sequence as mRNA.
It is called anti sense strand.	It is called sense or non-template strand.
It has 3'→ 5' polarity.	It has 5'→ 3' polarity.

9. List two essential roles of ribosome during translation.

Answer

Two essentials roles of ribosomes during translation are:
• The ribosome binds to mRNA and provides a platform for joining of amino acids.
• It also acts as a catalyst (23 SrRNA in bacteria is the enzyme ribozyme) for the formation of a peptide bond.

10. In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then why does lac operon shut down sometime after the addition of lactose to the medium?

Answer

Lactose regulates switching on and off of the lac operon.

If lactose is provided in the growth medium of the bacteria, it is transported into the cells through the action of permease.

The repressor of the operon is synthesized all the time from the i gene. It binds the repressor protein which binds to the operator region of the operon and prevents RNA polymerase from transcribing the

operon. In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase access to the promoter and transcription proceeds.

11. Explain (in one or two lines) the function of the followings :

(a) Promoter

(b) tRNA

Answer

(a) Promoter: It is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase.

(b) tRNA: tRNA or transfer RNA is a small RNA that reads the genetic code present on mRNA. It carries specific amino acid to mRNA on ribosome during translation of proteins.

12. Why is the human genome project called a mega project?

Answer

Human genome project is called a mega project because

• Its aim was to determine the nucleotide sequence of complete human genome which was a task of enormous magnitude.

• A total of 3 10 9 × base pairs were to be sequenced and the cost was about 9 billion US dollars.

• It required bioinformatics data base techniques and other contemporary devices for the analysis, storage and retrieval of information.

• Many countries worked jointly to complete this timed project.

13. What is DNA fingerprinting? Mention its application.

Answer

DNA fingerprinting is a technique used to identify and analyze the variations in various individuals at the level of DNA.

Its Application
• It is used in forensic science to identify potential crime suspects.
• It is used to settle parental disputes.
• It is used to identify and protect the commercial varieties of crops and livestock.
• It is used to find out the evolutionary history of an organism and trace out the linkages between groups of various organisms.

14. Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics

Answer

(a) Transcription: The process of copying genetic information from one strand of the DNA into RNA is known as transcription. RNA is assembled simply based on complementarity of the DNA strand, only uracil is substituted in place of thymine. Only a small segment of DNA that codes for a polypeptide is copied.

(b) Polymorphism: The variation in DNA arising through mutation at non-coding sequences is known as Polymorphism. Such variations are unique to specific sites of DNA and can occur due to deletion, insertion or substitution of bases. It can be observed by making fragments of DNA sample and separating them through electrophoresis.The polymorphism in a DNA sequence is the basis of genetic mapping of the human genome as well as DNA fingerprinting.

(c) Translation: It refers to the process of polymerisation of amino acids to form a polypeptide. The order and sequence of amino acids are defined by the sequence of bases in the mRNA. It occurs in cytoplasm in both prokaryotes and eukaryotes.

(d) Bioinformatics: It is the application of computer science and information technology which deals with handling, storing of huge information of genomics, processing information, analyzing data and creating new knowledge.

Go To Chapters

NCERT Solutions for Class 6th: Ch 11 Light, Shadows and Reflections Science

Page No: 113

Exercises

1. Rearrange the boxes given below to make a sentence that helps us understand opaque objects.

Answer

Opaque Objects Make Shadow

Page No: 114

2. Classify the objects or materials given below as opaque, transparent or translucent and luminous or non-luminous:
Air, water, a piece of rock, a sheet of aluminium, a mirror, a wooden board, a sheet of polythene, a CD, smoke, a sheet of plane glass, fog, a piece of red hot iron, an umbrella, a lighted fluorescent tube, a wall, a sheet of carbon paper, the flame of a gas burner, a sheet of cardboard, a lighted torch, a sheet of cellophane, a wire mesh, kerosene stove, sun, firefly, moon.

Answer

Opaque objects : a piece of rock, a sheet of aluminium, a mirror, a wooden board, a CD, an umbrella, a wall, a sheet of carbon paper, a sheet of cardboard

Transparent objects : Air, Water, a sheet of polythene, a sheet of plane glass, a sheet of cellophane, a wire mesh

Translucent objects : smoke, fog

Luminous objects : A piece of red hot iron, a lighted fluorescent bulb, the flame of a gas burner, a lighted torch, kerosene torch, sun, firefly

Non-luminous object : Air, water, a piece of rock, a sheet of aluminium, a mirror, a wooden board, a sheet of polythene, a CD, smoke, a sheet of plane glass, fog, an umbrella, a wall, a sheet of carbon paper, a sheet of cardboard, a sheet of cellophane, a wire mesh, moon.

3. Can you think of creating a shape that would give a circular shadow if held in one way and a rectangular shadow if held in another way?

Answer

A cylindrical shape can give a circular shadow when light is incident from its top or bottom. It will give a rectangular shadow when light is incident from its side.

4. In a completely dark room, if you hold up a mirror in front of you, will you see a reflection of yourself in the mirror?

Answer

No, we will not be able to see our reflection in the mirror because there is no source of light in the room.

Go To Chapters

NCERT Solutions for Class 6th: पाठ - 16 वन के मार्ग में हिंदी वसंत भाग-I

तुलसीदास

पृष्ठ संख्या: 145

प्रश्न अभ्यास

कविता से

1. प्रथम सवैया में कवि ने राम-सीता के किस प्रसंग का वर्णन किया है?

उत्तर

प्रथम सवैया में कवि ने राम-सीता के वन-गमन प्रसंग का वर्णन किया है।

2. वन के मार्ग में सीता को होने वाली कठिनाइयों के बारे में लिखो।

उत्तर

वन के मार्ग में जाते हुए सीता थक गईं। उनके माथे से पसीना बहने लगा और होंठ सूख गए। वन के मार्ग में चलते-चलते उनके पैरों में काँटें चुभने लगे।

3. सीता की आतुरता देखकर राम की क्या प्रतिक्रिया होती है?

उत्तर

सीता की आतुरता को देखकर श्रीराम व्याकुल हो उठते हैं। सीता को थका और प्यासा देख उनकी आँखों से आँसू बहने लगते हैं। वह इस बात से परेशान हो उठते हैं कि सीता को इतना कष्ट उनके वजह से झेलना पड़ रहा है।

4. राम बैठकर देर तक काँटे क्यों निकालते रहे?

उत्तर

राम से सीता की व्याकुलता देखी नहीं जा रही थी। लक्ष्मण पानी की तलाश में गए हुए थे इसलिए जब तक लक्ष्मण लौट कर आते हैं तब तक वे सीता की व्याकुलता और कष्ट को कम करना चाहते थे इसलिए राम बैठकर देर तक काँटे निकालते रहे क्योंकि अभी उन्हें और चलना था।

5. सवैया के आधार पर बताओ कि दो कदम चलने के बाद सीता का ऐसा हाल क्यों हुआ?

उत्तर

सीता राजा जनक की पुत्री थीं। उनका जीवन राजमहलों में बिता था। इस पकार की कठिनाइयों को उन्होंने ने कही नहीं देखा था। इसलिए अभ्यस्त न होने के कारण उनका ऐसा हाल हुआ।

6. 'धरि धीर दए' का आशय क्या है?

उत्तर

इस पंक्ति का आशय है धैर्य धारण करना। सीता वन मार्ग पर चलते हुए, राम का साथ देते हुए, तकलीफों को सहते हुए मन-ही-मन स्वयं को धीरज बँधा रहीं थीं।

अनुमान और कल्पना

• अपनी कल्पना से वन के मार्ग का वर्णन करो।

उत्तर

वन का मार्ग कठिनाइयों से भरा होता है। रास्ते में चारों तरफ लम्बे-लम्बे वृक्ष खड़े होते हैं और बीचों-बीच कँटीली झाड़ियाँ होती हैं। रास्ता उबड़-खाबड़, पत्थरीला और दलदली भी होता है। पेड़ों से घिरे होने के कारण सूरज की रोशनी भी काम पहुँचती है जिससे अँधेरा भी होता है साथ ही रास्ता जंगली जानवरों से भरा होता है।

भाषा की बात

• लखि - देखकर धरि - रखकर
पोछि - पोंछकर जानि - जानकर
ऊपर लिखे शब्दों और उनके अर्थ को ध्यान से देखो। हिंदी में जिस उद्देश्य के लिए हम क्रिया में 'कर' जोड़ते हैं, उसी के लिए अवधी में क्रिया में ि (इ) को जोड़ा जाता है, जैसे अवधी में बैठ + ि = बैठि और हिंदी में बैठ + कर = बैठकर। तुम्हारी भाषा या बोली में क्या होता है? अपनी भाषा के ऐसे छह शब्द लिखो। उन्हें ध्यान से देखो और कक्षा में बताओ।

उत्तर

छात्र अपनी मातृभाषा के छह शब्द लिखें।

पाठ में वापिस जाएँ

NCERT Solutions for Class 6th: Ch 12 Electricity and Circuits Science

Page No: 123

Exercises

1. Fill in the blanks :
(a) A device that is used to break an electric circuit is called _______________.
(b) An electric cell has _______________ terminals.

Answer

(a) A device that is used to break an electric circuit is called electric switch.
(b) An electric cell has two terminals.

2. Mark 'True' or 'False' for following statements:
(a) Electric current can flow through metals.
(b) Instead of metal wires, a jute string can be used to make a circuit.
(c) Electric current can pass through a sheet of thermo col.

Answer

(a) True
(b) False
(c) False

3. Explain why the bulb would not glow in the arrangement shown in Fig. 12.13.

Answer

Bulb will not glow in the arrangement because the holder of the tester used in the connection is made of plastic which is an insulator. Thus, current will not flow in the circuit.

Page No: 124

4. Complete the drawing shown in Fig 12.14 to indicate where the free ends of the two wires should be joined to make the bulb glow.

Answer

Earlier, the connection is not complete as one of the end of the cell is not connected to the switch and hence the bulb. Thus, there will be no flow of current in the circuit.
To complete the circuit, one of the end of the cell must be connected to the switch and then to the bulb. After this, bulb will start glowing as shown in the figure.

5. What is the purpose of using an electric switch? Name some electrical gadgets that have switches built into them.

Answer

An electric switch helps in making as well as breaking the circuit without removing the connection. When switch is ON, the current will flow and when switch is OFF, the current will not flow in the circuit. Thus, it also saves electricity and make the use of appliances easier.
Some electric gadgets are: Television, Iron, Washing Machine Refridgerator etc.

6. Would the bulb glow after completing the circuit shown in Fig. 12.14 if instead of safety pin we use an eraser?

Answer

No, the bulb will not glow when we will use an eraser instead of safety pin even after circuit is complete because rubber is an insulator and current will not flow through it.

7. Would the bulb glow in the circuit shown in Fig. 12.15?

Answer

No, the bulb will not glow because the wires from both terminals of the battery are connected to the one terminal of the bulb. In order to make the bulb glow, wires should be connected to the both terminals of the bulb.

8. Using the "conduction tester" on an object it was found that the bulb begins to glow. Is that object a conductor or an insulator? Explain.

Answer

That object must be an insulator because it allows the current to flow through it and thus helped in the glowing of bulb.

9. Why should an electrician use rubber gloves while repairing an electric switch at your home? Explain.

Answer

Rubber gloves are insulator which do not allow the current to flow through it. Therefore, an electrician use rubber gloves while repairing an electric switch at your home because it protect them from the shock of electricity and for the safety purpose.

10. The handles of the tools like screwdrivers and pliers used by electricians for repair work usually have plastic or rubber covers on them. Can you explain why?

Answer

The handles of the tools like screwdrivers and pliers used by electricians for repair work usually have plastic or rubber covers on them as plastic is an insulator which protect the electricians from severe electric shocks and accidents.

Go To Chapter

पाठ्य-पुस्तक 'स्पर्श-I' में प्रयुक्त वर्ण-विच्छेद शब्द - हिंदी व्याकरण Class 9th Course -'B'

पाठ 1- धूल

• धूलि = ध् + ऊ + ल + इ
• प्रेमी = प् + र् + ए + म् + ई
• कीमती = क् + ई + म् + अ + त् + ई
• श्रृंगार = श् + ऋ + न् + ग् + आ + र् + अ

• अभिजात = अ + भ् + इ + ज् + आ + त् + अ

• संसर्ग = स् + अं + स् + अ + र् + ग् + अ

• मिट्टी = म् + इ + ट् + ट् + ई

• स्मरण = स् + म् + अ + र् + अ + ण् + अ

पाठ 2 - दुःख का अधिकार

• विभिन्न = व् + इ + भ् + इ + न् + न् + अ

• घृणा = घ् + ऋ + ण् + आ

• व्यवधान = व् + य् + अ + व् + अ + ध् + आ + न् + अ
• ओझा = ओ + झ् + आ

• सूतक = स् + ऊ + त् + क् + अ

• विश्राम = व् + इ + श् + र् + आ + म् + अ

पाठ 3 - एवरेस्ट: मेरी शिखर यात्रा

• एवरेस्ट = ए + व् + अ + र् + ए + स् + ट् + अ
• दुर्गम = द् + उ + र् + अ + ग् + अ + म् + अ
• अग्रिम = अ + ग् + र् + इ + म् + अ
• सागरमाथा = स् + आ + ग् + अ + र् + अ + म् + आ + थ् + आ
• शेरपा =श् + ए + र् + अ + प् + आ
• अभियान = अ + भ् + इ + य् + आ + न् + अ
• नेतृत्व = न् + ए + त् + ऋ + त् + व् + अ
• ग्लेशियर = ग् + ल् + ए + श् + इ + य् + अ + र् + अ
• नौसिखिया = न् + औ + स् + इ + ख् + इ + य् + आ

पाठ 4 - तुम कब जाओगे अतिथि

• चतुर्थ = च् + त् + उ + र् + थ् + अ
• निस्संकोच = न् + स् + स् + अं + क् + ओ + च् + अ
• नम्रता = न् + अ + म् + र् + अ + त् + आ
• चट्टान = च् + अ + ट् + ट् + आ + न् + अ
• मध्यम = म् + अ + ध् + य् + अ + म् + अ
• सौहार्द = स् + औ + ह् + आ + र् + द् + अ
• देवता = द् + ए + व् + अ + त् + आ

पाठ 5 - वैज्ञानिक चेतना के वाहक चंद्रशेखर वेंकट रामन्

• समक्ष = स् + अ + म् + अ + क् + ष् + अ
• भौतिकी = भ् + औ + त् + इ + क् + इ
• प्रयोगशाला = प् + र् + अ + य् + ओ + ग् + अ + श् + आ + ल् + आ
• स्वीकार = स् + व् + ई + क् + आ + र् + अ
• तनख्वाह = त् + अ + न् + अ + ख् + व् + आ + ह् + अ
• तीव्रगामी = त् + ई + व् + व् + र् + अ + ग् + आ + म् + ई
• अनुसंधान = अ + न् + उ + स् + अं + ध् + आ + न् + अ

पाठ 6 - धर्म की आड़

• उत्पात = उ + त् + प् + आ + त् + अ
• उत्साह = उ + त् + स् + आ + ह् + अ
• अवस्था = अ + व् + स् + थ् + आ
• व्यापार = व् + य् + आ + प् + आ + र् + अ
• मुनष्यत्व = म् + उ + न् + अ + ष् + य् + अ + त् + व् + अ

पाठ 8 - शुक्रतारे के समान

• नक्षत्र = न् + अ + क् + ष् + अ + त् + र् + अ
• तेजस्वी = त् + ए + ज् + अ + स् + व् + ई
• स्याह = स् + य् + आ + ह् + अ
• व्यस्तता = व् + य् + अ + स् + त् + अ + त् + आ

व्याकरण सूची में वापस जाएँ

NCERT Solutions for Class 12th: Ch 7 Evolution Biology

Page No: 142

Exercises

1. Explain antibiotic resistance observed in bacteria in light of Darwinian selection theory.

Answer

In the presence of antibiotic, the bacteria that are sensitive to it will die. However, if there are any mutants in the population, that can somehow survive its effect, they will multiply and increase in numbers. After that, they will live as antibiotic resistant bacteria.

2. Find out from newspapers and popular science articles any new fossil discoveries or controversies about evolution.

Answer

Scientists have found the fossil of a 60-million-year-old creature in Morocco, which is the rabbit sized ancestor of the modern day elephant. Paleontologist Emmanuel Gheerbrant discovered the rabbit-size proto-elephant’s skull fragments in a basin 60 miles (100 kilometers) east of Casablanca, Morocco. The creature, called Eritherium azzouzorum, bolsters the case that whole new orders of mammals were already around less than 6 million years after global catastrophe ended the age of reptiles some 65.5 million years ago.

3. Attempt giving a clear definition of the term species.

Answer

Species can be defined as a group of organisms that can interbreed under natural conditions and produce fertile offsprings.

4. Try to trace the various components of human evolution (Hint : Brain size and function, skeletal structure, dietary preference, etc.).

Answer

Name	Features
Dryopithecus	Ape like, canines large, arms and legs are of equal size, ate soft fruits and leaves
Ramapithecus	More man-like, canines were small while molars were large, walked more erect, ate seeds and nuts
Australopithecus	Man-like, canines and incisors were small, walked upright, hunted with stone weapons, ate fruits, brain capacities were between 400-600cc.
Homo habilis	First human like being, canines were small, first tool makers, did not ate meats, brain capacities were between 650-800cc.
Homo erectus	Used stone and bone tools for hunting games, ate meat, brain capacity 900cc.
Homo neanderthalnsis	Cave dwellers, used hides to protect their bodies, and buried their dead, brain capacity 1400cc.
Homo sapiens (Modern human)	Modern man with high intelligence, developed art, culture, language etc., cultivated crops and domesticated animals.

5. Find out through internet and popular science articles whether animals other than man has self-consciousness.

Answer

Self-consciousness needs to be defined as the mental link or ones awareness of oneself as an individual or of one’s own being, actions, or thought. There are many other than humans, which have self-consciousness such as dolphins, crow, parrot, chimpanzee, gorilla, etc.

6. List 10 modern-day animals and using the internet resources link it to a corresponding ancient fossil. Name both.

Answer

Animals	Fossils
Man	Homo Sapiens
Dog	Leptocyon
Chimpanzee	Dryopithecus
Elephant	Moerithers
Horse	Eohippus
Gorilla	Dryopithecus
Camel	Protylopus
Whale	Protocetus
Fish	Arandaspis
Octopus	Belemnite

7. Practise drawing various animals and plants.

Answer

Draw various animals and plants from the chapter.

8. Describe one example of adaptive radiation.

Answer

Darwin finches of the Galapagos Islands is an example of adaptive radiation. They once had a common ancestor but as time passed they underwent evolution and adapted itself according to their food habitat.

9. Can we call human evolution as adaptive radiation?

Answer

No, human evolution cannot be called adaptive radiation because adaptive radiation is an evolutionary process that produces new species from a single, rapidly diversifying lineage, which is not the case with human evolution.

10. Using various resources such as your school Library or the internet and discussions with your teacher, trace the evolutionary stages of any one animal say horse.

Answer

The evolutionary stages of horse are:

• Eohippus: It appeared in the Eocene period about 52 million years ago. It was approximately the size of a fox (0.4 m), with a relatively short head and neck and a springy, arched back. It had four functional toes and a splint of 1 and 5 on each hind limb and a splint of 1 and 3 in each forelimb.

• Mesohippus: Approx, 40 million years ago in Oligocene period, Mesohippus which was slightly larger than Eohippus about 0.6 metre. It had three toes in each foot.

• Merychippus: In Miocene period the grazer Merychippus flourished. It had the size of approx 1m. It still had three toes in each foot, but it could run on one toe. The side toe did not touch the ground. The molars were adapted for chewing the grass.

• Pliohippus: Around 12 million years in Pilocene period, modern horse Pilohippus emerged. It had a single functional toe with splint of 2nd and 4th in each limb.

• Equus: Pliohippus gave rise to modern horse, Equus. It have one toe in each foot. They have incisors for cutting grass and molars for grinding food.

Go To Chapters

Extra Questions and Answer from Chapter 1 Nana Sahab ki putri devi Maina ko bahsm kar diya gya Kshitiz Bhaag I

निम्नलिखित गद्यांश को पढ़कर दिए गए प्रश्नों के उत्तर लिखिए -

1. कानपुर में भीषण हत्याकाण्ड करने के बाद अंग्रेज़ों का सैनिक दल बिठूर की ओर गया। बिठूर में नाना साहब का राजमहल लूट लिया गया; पर उसमें बहुत थोड़ी संपत्ति अंग्रेज़ों के हाथ लगी। इसके बाद अंग्रेज़ों ने तोप के गोलों से नाना साहब का महल भस्म कर देने का निश्चय किया। सैनिक दल ने जब वहाँ तोपें लगाईं उस समय महल के बरामदे में एक अत्यंत सुन्दर बालिका आकर खड़ी हो गई। उसे देखकर अंग्रेज सेनापति को बड़ा आश्चर्य हुआ, क्योंकि महल लूटने के समय वह बालिका वहाँ कहीं दिखाई न दी थी।

(क) नाना साहब का महल कहाँ था?
(ख) अंग्रेज़ों ने महल तोप के गोलों से भस्म करने का निश्चय क्यों किया?
(ग) अंग्रेज़ सेनापति को महल के बरामदे में लड़की दिखने से आश्चर्य क्यों हुआ?

उत्तर

(क) नाना साहब का महल बिठूर में था।
(ख) अंग्रेज़ों ने महल तोप के गोलों से भस्म करने का निश्चय इसलिए किया क्योंकि उन्हें अपनी सरकार से आदेश था की नाना के सभी निशानियों को मिटा दे।
(ग) अंग्रेज़ सेनापति को महल के बरामदे में लड़की दिखने से आश्चर्य इसलिए हुआ क्योंकि महल को लूटते वक़्त वह लड़की उन्हें कहीं नहीं दिखाई दी।

2. उस समय लण्डन के सुप्रिसद्ध 'टाइम्स' पत्र में छटी सितम्बर को एक लेख में लिखा गया 'बड़े दुःख का विषय है, कि भारत सरकार आज तक उस दुर्दान्त नाना साहब को नहीं पकड़ सकी, जिस पर समस्त अंग्रेज जाति का भीषण क्रोध है। जब तक हम लोगों के शरीर में रक्त रहेगा, तब तक कानपुर में अंग्रेजों के हत्याकाण्ड का बदला लेना हम लोग न भूलेंगे। उस दिन पार्लियामेंट की 'हाउस ऑफ़ लॉर्ड्स' सभा में सर टामस 'हे' की एक रिपोर्ट पर बड़ी हँसी हुई, जिसमे सर 'हे' ने नाना की कन्या पर दया दिखाने की बात लिखी थी।

(क) 'टाइम्स' पत्र के अनुसार दुःख का विषय क्या था?
(ख) 'टाइम्स' पत्र में छपे लेख का क्या मत था?
(ग) सर टॉमस 'हे' के रिपोर्ट पर हँसी क्यों हुई?

उत्तर

(क) भारत सरकार नाना साहब को पकड़ने में असफल रही जिसपर अंग्रज़ों के हत्याकाण्ड का आरोप था। यह 'टाइम्स' पत्र के अनुसार दुःख विषय था।
(ख) 'टाइम्स' पत्र में छपे लेख का मत था कि नाना के पुत्र, कन्या तथा अन्य कोई भी सम्बन्धी जहाँ भी मिले, मार डाला जाए।
(ग) सर टॉमस 'हे' ने अपने रिपोर्ट में नाना साहब, जिसपर समस्त अंग्रेज़ जाति का क्रोध था उसकी कन्या पर दया दिखाने की बात कर रहे थे।

निम्नलिखित प्रश्नों के उत्तर लिखिए -

1. नाना साहब अपनी पुत्री मैना को अपने साथ क्यों नहीं ले जा सके?

उत्तर

नाना साहब कानपुर में अंग्रेज़ों के सामने असफल रहे जिसके वजह से उन्हें जल्दी में भागना पड़ा इस कारण वे अपनी पुत्री मैना को अपने साथ नहीं ले जा सके।

2. मैना 'हे' को कैसे जानती थी? उसने 'हे' से क्या मदद माँगी? पाठ के आधार पर लिखिए।

उत्तर

'हे' की पुत्री मेरी मैना की सहेली थी। पहले 'हे' नाना साहब से मिलने उनके महल जाया करते थे और मैना को भी अपनी पुत्री समान प्यार करते थे। इस प्रकार मैना 'हे' को जानती थी। उसने 'हे' से राजमहल के रक्षा की माँग की।

3. सेनापति 'हे' ने मदद करने में असमर्थता के पीछे क्या वजह दिया?

उत्तर

सेनापति 'हे' ने मदद करने में असमर्थता के पीछे नाना साहब के प्रति अंग्रेज सरकार के कड़े रुख का वजह दिया। उन्होंने कहा कि वह जिस सरकार के नौकर हैं, उसकी आज्ञा वे नहीं टाल सकते।

4. सेनापति 'हे' ने जनरल आउटरम से क्या निवेदन किया?

उत्तर

सेनापति 'हे' ने जनरल आउटरम से नाना साहब के महल को किसी तरह बचाने का निवेदन किया।

5. पाठ के आधार पर जनरल आउटरम के चरित्र का विश्लेषण कीजिये।

उत्तर

जनरल आउटरम अंग्रेजी सेना का जनरल था। वह क्रूर, निर्दयी और पत्थर हृदय व्यक्ति था। उसने मैना को राजमहल के अवशेषों पर से रोते हुए देखा और हथकड़ी डालकर गिरफ्तार किया और कानपुर के किले में कैद कर दिया। उसने मैना की की आखिरी इच्छा भी पूरी नहीं की और उसी किले में मैना को आग जलाकर भस्म कर दिया गया।

6. मैना ने जनरल आउटरम से क्या प्रार्थना की?

उत्तर

मैना ने जनरल आउटरम से कुछ वक़्त और देने की प्रार्थना की ताकि वह राजमहल के अवशेषों पर जी भर कर रो ले।

7. मैना सैनिकों के प्रश्न का उत्तर क्यों नहीं दे रही थी?

उत्तर

मैना महल के टूट जाने से दुःख में डूबी थी। वह अपने ख्यालों में खोयी थी। उसे अपने हालात का आभास ना था और ना ही कोई डर था इस कारण वह सैनिकों के प्रश्न का उत्तर नहीं दे रही थी।

8. ब्रिटिश पार्लियामेंट में सेनापति 'हे' की किस बात पर हँसी हुई?

उत्तर

ब्रिटिश पार्लियामेंट में सेनापति 'हे के रिपोर्ट में लिखी हुई उस बात पर हँसी हुई जिसमे उन्होंने नाना की कन्या मैना पर दया दिखाने की बात कही थी।

9. अंग्रेज़ों का नाना साहब के प्रति इतना क्रूर रवैया क्यों था? पाठ के आधार पर लिखिए।

उत्तर

नाना साहब अंग्रेज़ों के कट्टर विरोधी थे। वे कानपुर में हुए अंग्रेज़ों के विरुद्ध आंदोलन के शीर्ष नेता थे। कानपुर में हुए अंग्रेज़ों के हत्याकाण्ड को उन्होंने करवाया था। इस कारण अंग्रेज़ों का नाना साहब के प्रति इतना क्रूर रवैया था।

10. महाराष्ट्रीय इतिहासवेत्ता महादेव चिटनवीस के पत्र में क्या छपा?

उत्तर

महाराष्ट्रीय इतिहासवेत्ता महादेव चिटनवीस के पत्र में छपा कि कल कानपुर के किले में एक भीषण हत्याकाण्ड हो गया। नाना साहब की एकमात्र कन्या मैना धधकती हुई आग में जलाकर भस्म कर दी गयी।

पठन सामग्री और सार - नाना साहब की पुत्री देवी मैना को भस्म कर दिया गया

NCERT Solutions of पाठ 5 - नाना साहब की पुत्री देवी मैना को भस्म कर दिया गया

CBSE has released Open Text Book Assessment (OTBA) Material for Class 9th 2016. The students can download the open text book material with sample questions. For Class 9th it is included in all subjects. This is only part of Summative Assessment II.

OTBA Matrial for Class IX

► Each text material includes sample questions and a suggestive marking scheme for use of teachers. It may also include the Value Based Questions.
► The textual material will be related to chosen concepts taken together from the syllabi.
► The Open Text based Assessment (OTBA) will have questions of higher order thinking skills and some of which may be subjective, creative and open ended.
► The question Papers in main subjects at Summative Assessment will be of 90 marks (in English of 70 marks + 20 marks for ASL) based on prescribed syllabus and question paper design.
► With the addition of OTBA of 10 marks, the total paper of Summative Assessment in a subject will be of 100 marks. However, weightage of SA will remain the same viz. 30%.

Subject & Code	Units/Themes	Theme for OTBA 2016	Marks Distribution (valid for one theme)
English (101 & 184)	Reading section (Based on the themes found in the course book )	1.Let's Welcome, Accept and Respect 2. Indigenous Games of India	5+5
Hindi (002 & 185)	Environment protection and Women Empowerment	1. Environment protection 2.Women Empowerment	5+5
Social Science	Unit 4: Economics (Food Security in India)	1. Together We Rise 2. Food Security in India	5+5
Science (086/090)	Unit 4: Our Environment	1. Handling Drought in our Country 2. Conservation of Water Bodies	2+3+5
Mathematics (041)	Unit 2: Linear Equations in two variables	1. Childhood Obesity in India 2. Energy Consumption	3+3+4

Download OTBA Material for Class for Different Subjects

Download OTBA Material for English

Download OTBA Material for Hindi

Download OTBA Material for Mathematics

Download OTBA Material for Science

Download OTBA Material for Social Science

CBSE has released Open Text Book Assessment (OTBA) Material for Class 11th 2016. OTBA is only applicable to three subjects in Class 11th i.e., Economics, Biology and Geography. This is only part of Summative Assessment II.

OTBA Material for Class XIth

► It is applicable to Economics, Geography and Biology only.
► A textual Material will be in form of an article, a case study, a diagram, a concept / mind map, a picture or a cartoon, problem / situation based on the concepts taught to students during the Term.
► Each text material includes sample questions and a suggestive marking scheme for use of teachers and may include the Value Based Questions.
► The textual material will be related to chosen concepts taken together from the syllabi.
► The Open Text based Assessment (OTBA) will have questions of higher order thinking skills and some of which may be subjective, creative and open ended.

Subject & Code	Units/Themes	Theme for OTBA 2016	Marks Distribution (valid for one theme)
Economics (030)	Unit 6: Development Experience of India	1. Special Economic Zones 2. Indo- Pak trade relations	5+5
Geography (029)	Unit 5:Water (Oceans) (Hydrosphere)	1. The Dynamic Ocean Current 2. Ocean Resources	5+5
Biology (044)	Unit 5: Human Physiology- (A)	1. Take care 2. The Ambient Air	5+5

Download OTBA Material for Different Subjects (in English)

Download OTBA Material for Economics

Download OTBA Material for Geography

Download OTBA Material for Biology

NCERT Solutions for Class 6th: Ch 13 Fun with Magnets Science

Page No: 134

Exercises

1. Fill in the blanks in the following
(i) Artificial magnets are made in different shapes such as __________, __________ and ____________.
(ii) The Materials which are attracted towards a magnet are called________.
(iii) Paper is not a ______ material.
(iv) In olden days, sailors used to find direction by suspending a piece of ___________.
(v) A magnet always has __________ poles.

Answer

(i) Artificial magnets are made in different shapes such as bar magnets, horse-shoe magnet and cylindrical or a ball-ended magnet.
(ii) The Materials which are attracted towards a magnet are called magnetic materials.
(iii) Paper is not a magnetic material.
(iv) In olden days, sailors used to find direction by suspending a piece of magnet.
(v) A magnet always has two poles.

2. State whether the following statements are true or false
(i) A cylindrical magnet has only one pole.
(ii) Artificial magnets were discovered in Greece.
(iii) Similar poles of a magnet repel each other.
(iv) Maximum iron filings stick in the middle of a bar magnet when it is brought near them.
(v) Bar magnets always point towards North-South direction.
(vi) A compass can be used to find East-West direction at any place.
(vii) Rubber is a magnetic material.

Answer

(i) False
(ii) False
(iii) True
(iv) False
(v) True
(vi) True
(vii) False

3. It was observed that a pencil sharpener gets attracted by both the poles of a magnet although its body is made of plastic. Name a material that might have been used to make some part of it.

Answer

The blade of the sharpener is made up of iron which is a magnetic substance which gets attracted by both the poles of a magnet although its body is made of plastic.

4. Column I shows different positions in which one pole of a magnet is placed near that of the other. Column II indicates the resulting Acids

Column I	Column II
N-N	_____________
N-___________	Attraction
S-N	____________
__________-S	Repulsion

Answer

Column I	Column II
N-N	Repulsion
N-S	Attraction
S-N	Attraction
S-S	Repulsion

5. Write any two properties of a magnet.

Answer

Two properties of a magnet:
(i) Magnet aligns in North-South direction when suspended freely.
(ii) A magnet has two magnetic poles.

6. Where are poles of a bar magnet located?

Answer

Poles of a bar magnet located at its two ends.

7. A bar magnet has no markings to indicate its poles. How would you find out near which end is its north pole located?

Answer

To locate its north pole, we would do the following steps:
(i) A bar is taken and suspended freely from the middle with the help of thread.
(ii) Allow the magnet to comes into rest.
(iii) The North pole of the magnet will face the north direction and South pole will face the south direction.
(iv) Mark the north pole of the magnet with the marker.

8. You are given an iron strip. How will you make it into a magnet?

Answer

Steps to make an iron strip into bar magnet:
(i) A flat strip of iron is taken.
(ii) On the iron strip, a bar magnet is placed and rubbed against it horizontally only in one direction.
(iii) When you reached the end of the strip, lift the magnet and again start rubbing from the initial position.
(iv) This process is repeated more than 40-50 times.
(v) After this, the iron strip will attained the property of magnet.

9. How is a compass used to find directions?

Answer

A compass has a magnetic needle attached to it which can rotates freely. The magnet always points to north-south direction which is marked on compass and thus helps in finding direction.

10. A magnet was brought from different directions towards a toy boat that has been floating in water in a tub. Affect observed in each case is stated in
Column I. Possible reasons for the observed affects are mentioned in
Column II. Match the statements given in Column I with those in Column II.

Column I	Column II
Boat gets attracted towards the magnet	Boat is fitted with a magnet with north pole towards its head
Boat is not affected by the magnet	Boat is fitted with a magnet with South pole towards its head.
Boat moves towards the magnet if North pole of the magnet is brought near its head.	Boat has a small magnet fixed along its length.
Boat moves away from the magnet when North pole is brought near its head.	Boat is made of magnetic material.
Boat floats without changing its direction.	Boat is made up of non-magnetic material.

Answer

Column I	Column II
Boat gets attracted towards the magnet	Boat is made of magnetic material.
Boat is not affected by the magnet	Boat is made up of non-magnetic material.
Boat moves towards the magnet if North pole of the magnet is brought near its head.	Boat is fitted with a magnet with South pole towards its head.
Boat moves away from the magnet when North pole is brought near its head.	Boat is fitted with a magnet with North pole towards its head.
Boat floats without changing its direction.	Boat has a small magnet fixed along its length.

Go To Chapter

NCERT Solutions for Class 6th: Ch 14 Water Science

Page No: 145

Exercises

1. Fill up the blanks in the following:
(a) The process of changing of water into its vapour is called _________________.
(b) The process of changing water vapour into water is called _________________.
(c) No rainfall for a year or more may lead to _________________ in that region.
(d) Excessive rains may cause _____________ .

Answer

(a) The process of changing of water into its vapour is called evaporation.
(b) The process of changing water vapour into water is called condensation.
(c) No rainfall for a year or more may lead to drought in that region.
(d) Excessive rains may cause floods.

2. State for each of the following whether it is due to evaporation or condensation:
(a) Water drops appear on the outer surface of a glass containing cold water.
(b) Steam rising from wet clothes while they are ironed.
(c) Fog appearing on a cold winter morning.
(d) Blackboard dries up after wiping it.
(e) Steam rising from a hot girdle when water is sprinkled on it.

Answer

(a)   Condensation
(b)   Evaporation
(c)   Condensation
(d)   Evaporation
(e)   Evaporation

3. Which of the following statements are “true” ?
(a) Water vapour is present in air only during the monsoon. ( )
(b) Water evaporates into air from oceans, rivers and lakes but not from the soil.( )
(c) The process of water changing into its vapour, is called evaporation.( )
(d) The evaporation of water takes place only in sunlight.( )
(e) Water vapour condenses to form tiny droplets of water in the upper layers of air where it is cooler.( )

Answer

(a) Not true
(b) Not true
(c) True
(d) Not true
(e) True

Page No: 146

4. Suppose you want to dry your school uniform quickly. Would spreading it near an anghiti or heater help? If yes, how?

Answer

Yes, spreading it near an anghiti or heater will surely help as heater and anghiti are source of heat which vaporize the water of the wet clothes and thus help in drying.

5. Take out a cooled bottle of water from refrigerator and keep it on a table. After some time you notice a puddle of water around it. Why?

Answer

The puddle of water seen around the cooled bottle of water is due to the condensation effect as the water vapour present in the air around the bottle get condensed after colliding with bottle.

6. To clean their spectacles, people often breathe out on glasses to make them wet. Explain why the glasses become wet.

Answer

Water vapour also gets released during exhalation process along with carbon dioxide. These water vapour gets attached with the glasses of the spectacles and then condensed in the presence of air surrounding it and thus making it wet.

7. How are clouds formed?

Answer

When the air moves up, it gets cooler and cooler and after reaching sufficient heights, the air
becomes so cool that the water vapour present in it condenses to form tiny drops of water called droplets which remain floating in air and thus clouds are formed.

8. When does a drought occur?

Answer

Drought occurs when an area does not receive rainfall for a period of year or more. Also, the soil continues to lose water by evaporation and transpiration and becomes dry and the ground water may also become scarce which may lead to drought.

Go To Chapter

'र' के अनेक रूप - हिंदी व्याकरण Class 9th Course -'B'

हिंदी वर्णमाला में 'र' वर्ण का विशेष स्थान है क्योंकि इसका प्रयोग लिखने में कई तरीकों से किया जाता है। हम 'र' वर्ण के अनेकों रूपों के बारे में उदाहरणों द्वार जानेंगे।

संयुक्त रूप से 'र' वर्ण का उच्चारण तीन तरीकों से होता है-

• रेफ (जैसे: पर्व)

• पाई वाले व्यंजनों के साथ (जैसे: क्रम )

• बिना पाई वाले व्यंजनों के साथ (जैसे: राष्ट्र)

रेफ

शब्द के बीच में पड़ते समय जब 'र' वर्ण के साथ कोई स्वर नहीं होता तब उसके ठीक बाद वाले वयंजन के ऊपर रेफ लगाया जाता है। जैसे:

• धर्म - इस शब्द में बिना स्वर वाले 'र' के बाद 'म' व्यंजन है इसलिए रेफ 'म' पर लगा है।

• कार्य - इस शब्द में बिना स्वर वाले 'र' के बाद 'य' व्यंजन है इसलिए रेफ 'य' पर लगा है।

जो अर्ध 'र' या रेफ़ शब्द के ऊपर लगता है, उसका उच्चारण हमेशा उस व्यंजन ध्वनि से पहले होता है।

पाई तथा बिना पाई वाले व्यंजनों के साथ 'र' का प्रयोग

जब 'र' से पहले प्रयुक्त व्यंजन बिना स्वर का होता है यानी अर्ध होता है और इसका उच्चारण प्रयुक्त वर्ण के बाद होता है तो 'र' से पहले पूरा वर्ण लिखा जाता है और 'र' के लिखने का रूप बदल जाता है।

ऐसी स्थिति में पाई तथा बिना पाई वाले व्यंजन के साथ इनका प्रयोग अलग होता है।

पाई वाले व्यंजनोंके बाद प्रयुक्त 'र' पाई के नीचे तिरछा होकर प्रयुक्त होता है। जैसे:

• क् + र = क्र, क्रोध, क्रम

• प् + र = प्र, प्रश्न, प्रधान

बिना पाई वाले व्यंजनोंके बाद प्रयुक्त 'र' व्यंजन के नीचे ( ्र ) के रूप में प्रयुक्त होता है। जैसे:

• ड् + र = ड्र, ड्रम, ड्रेस

• ट् + र = ट्र, ट्रैक, ट्रक

त् और श् के बाद 'र' का प्रयोग

त् के बाद जब 'र' आता है तो संयुक्ताक्षर 'त्र' बन जाता है। जैसे: त् + र = त्र, त्रिशूल, त्रिनेत्र
श् के बाद जब 'र' आता है तो संयुक्ताक्षर 'श्र' बन जाता है। जैसे: श् + र = श्र, श्रद्धा, श्रम

द् और ह् के बाद 'र' का प्रयोग

द् के बाद जब 'र' आता है तो यह 'द' के नीचे पाई के रूप में प्रयुक्त होता है। जैसे: द्रव्य, दरिद्र,
ह् के बाद जब 'र' आता है तो यह 'ह' में ही पाई के रूप में प्रयुक्त होता है। जैसे: ह्रस्व, ह्रास

र और ऋ की मात्राओं में अंतर

'र' और ऋ की मात्राओं में बहुत अंतर है। लेखन के साथ-साथ उच्चारण में भी इन दोनों की मात्राओं में साफ़ अंतर किन्तु फिर भी इनमें अक्सर गलतियाँ हो जाती हैं। नीचे दिए उदाहरणों से इस अंतर को समझें-

क् + ऋ = कृ, कृपा, कृष्ण
क् + र = क्र, क्रमांक, क्रिया

व्याकरण सूची में वापस जाएँ

Study Material of Packing (Summary, Character Sketch and Word Meanings)

Character Sketch

• Jerome: He is the narrator of the story. He is very proud of on his packing so he decide to do the packing for holiday but his real intention is different. He wants his friends would do the packing and he supervise thier work. He feel irritated when he is working and other people are relaxing. However, he does the packing task but forgotten many things outside so he have to do the packing many times. He enjoys the packing of foods by his friends and make them feel irritated.

• Montmorency: He is a pet dog. He had an ambition in his life to get in the way and get scolded. He felt that his day was not been wasted if he had made people mad and had things thrown at his head. His highest aim was to get somebody to stumble over him, and curse him steadily for an hour. He is disturbing George and Harris in packing work. He sat down on many things. He put his legs into the jam and scattered theteaspoons. He pretended that the lemons were rats.

Summary of the Chapter

Three friends George, Harris and Jerome, the narrator, decided to go on a Holiday. Jerome was very proud of his packing so he offered to do the packing task on which other two friends agreed. Both friends relaxed, but the narrator's intention was different. According to him, both would do packing task under his direction. The narrator hated other people relaxing while he was working. However, the narrator started packing and soon finished it.

Harris asked the narrator that he had put the boots in the package. The narrator searched and found he had forgotten. He opened the bag, put the boots and packed the boots. When he was going to close his bag a horrible idea came to his mind that had he packed his toothbrush! He turned everything. He found toothbrushes of George and Harris more than eighteen times but could not find out his toothbrush. He put the things back one by one. Then he found it inside a boot and packed the things once more.

After he finished packing, George asked if the soap was in. The narrator said that he didn't care about soap whether it was in bag or not. Again, he had to open the bag as he had packed his spectacles in it. The bag finally got finished at 10:05 p.m. Now, the hampers task remained to do which was taken by George and Harris on which narrator agreed.

Both started packing in cheerful nature showing narrator how to do. However, according to narrator they are worst packer in the world. The narrator could see the piles of plates and cups, and kettles, and bottles, and jars, and pies, and stoves, and cakes, and tomatoes, etc., and felt that the thing would soon become exciting. The narrator thinking was right as they started breaking the cup. Harris packed the strawberry jam on top of a tomato and squashed it. George stepped on the butter. The narrator came over and quietly sat on the edge of the table which irritated them as narrator felt. It made them nervous and excited. They couldn't able to find what they want. They packed the pies at the bottom and squashed them by putting heavy thing over them.

Both of them upset everything. George got the butter off his slipper and tried to put it in kettle, but it wouldn't go in and what was in wouldn't come out. At last, they dragged it out all and put it down the chair. Harris sat on it and butter stuck to him. The searched the whole room but couldn't able to find. Lastly, George got round at the back of Harris and saw it and they got it off, and packed it in the teapot.

Montmorency, pet dog had an ambition in his life to get in the way and get scolded. He felt that his day was not been wasted if he had made people mad and had things thrown at his head. His highest aim was to get somebody to stumble over him, and curse him steadily for an hour. He came and sat down on things, just when they were wanted to be packed. He put his leg into the jam, and he worried
the teaspoons, and he pretended that the lemons were rats, and got into the hamper and killed three of them before Harris could land him with the frying-pan. Harris said it was the narrator who encouraged the dog to do these things.

The packing was done at 12.50. Harris said that he hoped nothing would be found broken. They decided to go to sleep. George asked when he should wake them. After little arguments they decided to wake up at 6:30.

Terms and Meaning from the Chapter

Page No: 82

• Pride myself on: am proud of
• Uncanny: strange
• Intended: thought
• Potter about: do some unimportant things

Page No: 83

• Irritated: Annoyed
• Loll: sit
• Messing: creating confusion
• Superintend: supervise
• Strapped: tied with strap
• Horrible: shocking
• haunts: repeatedly give trouble
• Misery: difficult
• Perspiration: sweat
• Every mortal thing: every ordinary thing
• Rummaged: searched in a careless way

Page No: 84

• chaos: confusion
• Reigned: ruled
• Slammed: closed with force
• Hampers: large baskets for carrying food
• Light-hearted: carefree

Page No: 85

• Squashed: crushed
• Trod on: stepped on

Page No: 86

• Smashed: broke into pieces
• Scrape: drag
• Oath: pledge
• Indignantly: in anger
• Be sworn at: get scolded
• Squirm: enter
• Nuisance: problem

Page No: 87

• Stumble: fall over
• Conceit: his pride in himself
• Damp: wet

Page No: 88

• Reflection: thought
• Fellows: friend
• Split the difference: this means that they agreed on 6.30 because it was halfway between six and seven

View NCERT Solutions of Packing