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NCERT Solutions for Class 6th: Ch 15 Air Around us Science

Page No: 153

Exercises

1. What is the composition of air?

Answer

Air contains some gases mainly nitrogen, oxygen, small amount of carbon dioxide, water vapour and
dust particles.

2. Which gas in the atmosphere is essential for respiration?

Answer

Oxygen gas in the atmosphere is essential for respiration.

3. How will you prove that air supports burning?

Answer

Step I: Two shallow containers are taken and burning candles are fitted into it.
Step II: Now, water is filled in the container at two different levels.
Step III: Both the candles are then covered with glass.
Step IV: After sometimes, we observed that the candles having less water in its container lights off.

This happened because the container having more water contains more oxygen and therefore it burnt for more time. Thus, it is proved that air supports burning.

4. How will you show that air is dissolved in water?

Answer

Step I: Water in a pan is taken.
Step II: It is now boiled on stove.
Step III: We will observe tiny bubbles of water visible coming out from the bottom of the pan.

These tiny bubbles are coming due to the air dissolved in water. Thus, this proves that air is dissolved in water.

5. Why does a lump of cotton wool shrink in water?

Answer

The lump of cotton contains air in it. When it comes in contact with water, the air pushed out from the vacant space and dissolved in water and thus a lump of cotton wool shrink in water.

Page No: 154

6. The layer of air around the earth is known as ___________.

Answer

The layer of air around the earth is known as atmosphere.

7. The component of air used by green plants to make their food, is ___________.

Answer

The component of air used by green plants to make their food, is carbon dioxide.

8. List five activities that are possible due to the presence of air.

Answer

Five activities that are possible due to the presence of air:
(i) Helps in seed dispersal and pollination.
(ii) Sailing of ships.
(iii) Flying of the birds and kites.
(iv) Gives us oxygen to respire.
(v) Helps in the rotation of windmill.

9. How do plants and animals help each other in the exchange of gases in the atmosphere?

Answer

Plants take in carbon dioxide in during photosynthesis give out oxygen in the atmosphere while animals take in oxygen and breathe out carbon dioxide during respiration. In this way, plants and animals help each other in the exchange of gases in the atmosphere.

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Extra Questions and Answer from Chapter 5 Vaigyanik chetna ke vaahak Chandrashekhar Venkat Raman Sparsh Bhaag I

निम्नलिखित गद्यांश को पढ़कर दिए गए प्रश्नों के उत्तर लिखिए -

1. बात सन् 1921 की है, जब रामन् समुद्री यात्रा पर थे। जहाज के डेक पर खड़े होकर नीले समुद्र को निहारना, प्रकृत-प्रेमी रामन् को अच्छा लगता था। वे समुद्र की नीली आभा में घंटों खोये रहते। लेकिन रामन् केवल भावुक प्रकृति-प्रेमी ही नहीं थे। उनके अंदर एक वैज्ञानिक की जिज्ञासा भी उतनी ही सशक्त थी। यही जिज्ञासा उनसे सवाल कर बैठी- 'आखिर समुद्र का रंग नीला ही क्यों होता है? कुछ और क्यों नहीं?' रामन् सवाल का जवाब ढूँढने में लग गए। जवाब ढूँढ़ते ही वे विश्वविख्यात बन गए।

(क) रामन् समुद्री यात्रा पर कब गए थे? (1)
(ख) रामन् को जहाज के डेक से किसे निहारना अच्छा लगता था और क्यों? (2)
(ग) रामन् के अंदर की वैज्ञानिक जिज्ञासा क्या प्रश्न कर बैठी? (2)

उत्तर

(क) रामन् सन् 1921 में समुद्री यात्रा पर गए थे।
(ख) रामन् को जहाज के डेक से नील समुद्र को घंटों निहारना अच्छा लगता था क्योंकि वे भावुक प्रकृति प्रेमी थे।
(ग) समुद्र के पानी का रंग नीला ही क्यों होता है? कोई और क्यों नहीं होता है? रामन् के अंदर की वैज्ञानिक जिज्ञासा इन प्रश्नों को कर बैठी।

2. रामन् का मस्तिष्क विज्ञान के रहस्यों को सुलझाने के लिए बचपन से ही बैचैन रहता था। अपने कॉलेज के ज़माने से ही उन्होने शोधकार्यों में दिलचस्पी लेना शुरू कर दिया था। उनका पहला शोधपत्र फिलॉसॉफिकल मैगज़ीन में प्रकाशित हुआ था। उनकी दिली इच्छा तो यही थी कि वे अपना सारा जीवन शोधकार्यों को समर्पित कर दें, मगर उन दिनों शोधकार्यों को पूरे समय के कैरियर के रूप में अपनाने की कोई ख़ास व्यवस्था नहीं थी। प्रतिभावान छात्र सरकारी नौकरी की ओर आकर्षित होते थे। रामन् भी अपने समय के सुयोग्य छात्रों की भाँति भारत सरकार के वित्त विभाग में अफसर बन गए। उनकी तैनाती कलकत्ता में हुई।

(क) किन रहस्यों को सुलझाने में रामन् का मस्तिष्क बैचैन रहता था? (1)
(ख) रामन् की दिली इच्छा क्या था? (1)
(ग) वे सरकारी नौकरी की ओर क्यों आकर्षित हुए और उनकी तैनाती कहाँ हुई? (2)

उत्तर

(क) विज्ञान के रहस्यों को सुलझाने में रामन् का मस्तिष्क बैचैन रहता था।
(ख) रामन् की दिली इच्छा अपना सारा जीवन शोधकार्य को समर्पित कर देने की थी।
(ग) वे सरकारी नौकरी की ओर इसलिए आकर्षित हुए क्योंकि उन दिनों शोधकार्य को पूरे समय के कैरियर के रूप में अपनाने की कोई ख़ास व्यवस्था नहीं थी। अन्य प्रतिभावान छात्र सरकारी नौकरी की ओर आकर्षित होते थे। उनकी तैनाती कलकत्ता में हुई।

3. वाद्ययंत्रों पर किए जा रहे शोधकार्यों के दौरान उनके अध्यन के दायरे में जहाँ वायलिन, चैलो या पियानो जैसे विदेशी वाद्य आए, वहीँ वीना, तानपुरा और मृदंगम पर भी उन्होंने काम किया। उन्होंने वैज्ञानिक सिद्धांतों के आधार पर पश्चिमी देशों की इस भ्रान्ति को तोड़ने की कोशिश की कि भारतीय वाद्ययंत्र विदेशी वाद्ययंत्रों की तुलना में घटिया हैं। वाद्ययंत्रों के कंपन के पीछे छिपे गणित पर उन्होंने अच्छा-खासा काम किया और अनेक शोधपत्र भी प्रकाशित किए।

(क) रामन् ने किन भारतीय वाद्ययंत्रों पर काम किया? (1)
(ख) रामन् ने पश्चिमी देशों की किस भ्रान्ति को तोड़ा? (2)
(ग) उन्होंने वाद्ययंत्रों के किस विषय पर काम किया? (2)

उत्तर

(क) रामन् ने वीना, तानपुरा, मृदंगम् जैसे वाद्ययंत्रों पर काम किया।
(ख) रामन् ने पश्चिमी देशों की इस भ्रान्ति को तोड़ने की कोशिश की कि भारतीय वाद्ययंत्र विदेशी वाद्ययंत्रों की तुलना में घटिया हैं।
(ग) उन्होंने वाद्ययंत्रों के कंपन के पीछे छिपे गणित पर अच्छा-खासा काम किया और अनेक शोधपत्र भी प्रकाशित किए।

4. रामन् सरकारी नौकरी की सुख-सुविधाओं को छोड़ सन् 1917 में कलकत्ता विश्वविद्यालय की नौकरी में आ गए। उनके लिए सरस्वती की साधना सरकारी सुख-सुविधाओं से कहीं अधिक महत्वपूर्ण थी। कलकत्ता विश्विद्यालय की शैक्षणिक माहौल में वे अपना पूरा समय अध्यन, अध्यापन और शोध में बिताने लगे। चार साल बाद यानी सन् 1921 में समुद्र-यात्रा के दौरान जब रामन् के मस्तिष्क में समुद्र के नीले रंग की वजह का सवाल हिलोरें लेने लगा, तो उन्होंने आगे इस दिशा में प्रयोग किए, जिसकी परिणति रामन् प्रभाव के की खोज के रूप में हुई।

(क) रामन् कब कलकत्ता विश्वविद्यालय की नौकरी में आ गए? (1)
(ख) वे कलकत्ता विश्वविद्यालय में किस तरह समय बिताने लगे? (1)
(ग) कौन-सी बात रामन् प्रभाव के की खोज के रूप में सामने आई? (2)

उत्तर

(क) रामन् सन् 1917 में कलकत्ता विश्वविद्यालय की नौकरी में आ गए।
(ख) वे कलकत्ता विश्वविद्यालय में अपना पूरा समय अध्यन, अध्यापन और शोध में बिताने लगे।
(ग) सन् 1921 में जब रामन् समुद्री यात्रा पर थे उस समय समुद्र के नीले रंग होने की वजह का सवाल रामन् प्रभाव के की खोज के रूप में सामने आई।

निम्नलिखित प्रश्नों के उत्तर दीजिये-

1. रामन् ने किस रहस्य से पर्दा हटाया?

उत्तर

रामन् ने समुद्र की नील वर्णीय आभा से पर्दा हटाया।

2. किन प्रश्नों के उत्तर खोजने के बाद रामन् विश्वविख्यात बन गए?

उत्तर

'समुद्र के पानी का रंग नीला ही क्यों होता है? कोई और क्यों नहीं होता है?' इन प्रश्नों के उत्तर खोजने के बाद रामन् विश्वविख्यात बन गए।

3. रामन् के कॉलेज की पढाई कहाँ पूरी हुई?

उत्तर

रामन् ने कॉलेज की पढाई ए.बी.एन. कॉलेज तिरुचिरापल्ली से और फिर प्रेसीडेंसी कॉलेज मद्रास से की।

4. रामन् ने शोधकार्यों को छोड़कर सरकारी नौकरी क्यों अपनाया?

उत्तर

रामन् ने शोधकार्यों को छोड़कर सरकारी नौकरी इसलिए अपनाया क्योंकि उन दिनों शोधकार्य को पूरे समय के कैरियर के रूप में अपनाने की कोई ख़ास व्यवस्था नहीं थी। प्रतिभावान छात्र सरकारी नौकरी की ओर आकर्षित होते थे।

5. 'इंडियन एसोसिएशन फॉर द कल्टीवेशन ऑफ़ साइंस' एक अनूठी प्रयोगशाला क्यों थी?

उत्तर

इंडियन एसोसिएशन फॉर द कल्टीवेशन ऑफ़ साइंस' प्रयोगशाला का मुख्य उद्देश्य देश में वैज्ञानिक चेतना का विकास करना था। उस समय देश में ऐसे उद्देश्य के लिए समर्पित यह एकमात्र प्रयोगशाला थी इसलिए यह एक अनूठी प्रयोगशाला थी।

6. सर आशुतोष मुखर्जी ने रामन् के सामने क्या प्रस्ताव रखा?

उत्तर

सर आशुतोष मुख़र्जी ने रामन् के सामने सरकारी नौकरी छोड़कर कलकत्ता विश्वविद्यालय में प्रोफ़ेसर का पद स्वीकार करने का प्रस्ताव रखा।

7. रामन् की खोज से पहले अणुओं और परमाणुओं की आंतरिक संरचना के अध्यन के लिए किसका उपयोग किया जाता था?

उत्तर

रामन् की खोज से पहले अणुओं और परमाणुओं की आंतरिक संरचना के अध्यन के लिए इंफ्रा रेड स्पेक्ट्रोस्कोपी का उपयोग किया जाता था। यह तकनीक बेहद मुश्किल थी और गलतियों की संभावना भी अधिक रहती थी।

8. रामन् ने विदेशों में भी अपनी भारतीय पहचान को अक्षुण्ण रखा। स्पष्ट कीजिए।

उत्तर

अंतरराष्ट्रीय प्रसिद्धि के बाद भी रामन् ने अपने दक्षिण भारतीय पहनावे को नहीं छोड़ा। वे शुद्ध शाकाहारी थे और मदिरा से सख्त परहेज करते।

पठन सामग्री और सार - वैज्ञानिक चेतना के वाहक : चन्द्र शेखर वेंकट रामन्

NCERT Solutions for Class 9th: पाठ 5 - वैज्ञानिक चेतना के वाहक : चन्द्र शेखर वेंकट रामन्

Study Material and Notes of Ch 6 Population Ch 1 India - Size and Location Class 9th Geography

Topics in the Chapter

• Introduction
• Size and Distribution
→ India’s Population Distribution by Density
• Population Growth
→ Processes of Population Growth
• Age Composition
• Sex Ratio
• Literacy rates
• Occupational Structure
• Health
• Adolescent Population
• NPP 2000
• Relation between NPP 2000 and Adolescent Population

Introduction

• People make and use resources. They are also considered as resources having different quality.

• Population (total number of persons inhabiting a particular place like city, state, country etc.) is the central element in social studies. It is the point of reference from which all other elements are observed.

• Human beings are producers as well as consumers of the resources so information about population of a country such as their size, distribution are important.

• The census of India provides us with information regarding the population of our country. The data provided by the census cover below three major questions about the population:
→ Population size and distribution
→ Population growth and processes of population change
→ Characteristics or qualities of the population

Size and distribution

• As per March 2001, India’s population stood at 1,028 million, which account for 16.7 percent of the world’s population.

• According to the data, Uttar Pradesh is the most populous state of India with a population size of166 million, which account for 16 percent of total India's population.

• Sikkim's population - 0.5 million (5 lakhs) while Lakshadweep has 60 thousand people.

• The five states Uttar Pradesh, Maharashtra, Bihar, West Bengal, and Andhra Pradesh. Rajasthan holds almost half of Indian population (48.8%).

India’s Population Distribution by Density

• Population density is calculated as the number of persons per unit area.

• The population density of India in the year 2001 was 324 persons per sq km making it one of the most densely populated countries of the world.

• Densities vary from 904 persons per sq km in West Bengal to only 13 persons per sq km in Arunachal Pradesh.

• The reason for scarce (thinly distributed) population in some states such as Meghalaya, Orissa etc. are rugged terrain and unfavourable climatic conditions.

• Hilly, dissected and rocky nature of the terrain, moderate to low rainfall, shallow and less fertile soils have influenced population in Assam and most of the Peninsular states.

• The Northern Plains and Kerala in the south have high to very high population densities because of the flat plains with fertile soils and abundant rainfall.

Population Growth

• Population Growth refers to the change in the number of inhabitants of a country or territory during a specific period of time, say during the last ten years.

• The change can be expressed in two ways

→ in terms of absolute numbers

→ in terms of percentage change per year

• The absolute number is calculated by simply subtracting the earlier population (e.g. that of 1991) from the later population (e.g. that of 2001). It is referred to as the absolute increase.

• The rate of population is studied in per cent per annum, e.g. a rate of increase of 2 per cent per annum means that in a given year, there was an increase of two persons for every 100 persons in the base population. This is referred to as the annual growth rate.

• India’s population has been steadily increasing from 361 million in 1951 to 1028 million in 2001.

• Since 1981, however, the rate of growth started declining gradually as birth rates declined rapidly. But India has a very large population so when a low annual rate is applied to a very large population, it yields a large absolute increase.

• At this growth rate, India may overtake China in 2045 to become the most populous country in the world.

Processes of Population Change/Growth

• There are three main processes of change of population: birth rates, death rates and migration.

• Birth rate is the number of live births per thousand persons in a year. In India, birth rates have always been higher than death rates.

• Death rate is the number of deaths per thousand persons in a year. In India, there is rapid decline in death rates which is the main cause of growth of the Indian population.

• Till 1980, high birth rates and declining death rates resulted in higher rate of population growth but since 1981, birth rates have also started declining gradually, resulting in a gradual decline in the rate of population growth.

• Migration is the movement of people across regions and territories. This can be internal (within the country) or international (between the countries).

• Internal migration does not change the size of the population, but changes the distribution of population within the nation.

• In India, most migrations have been from rural to urban areas because of adverse conditions of poverty and unemployment in the rural areas and increased employment opportunities and better living conditions in city.

• Effects of Migrations:
→ It changes the population size.
→ It also changes the population composition of urban and rural populations in terms of age and sex composition.

• In India, the rural-urban migration has resulted in a steady increase in the percentage of population in cities and towns.

Age Composition

• The age composition of a population refers to the number of people in different age groups in a country.

• The number and percentage of a population found within the children, working age and aged groups are notable determinants of the population’s social and economic structure.

• Population of country can be grouped into three categories:

→ Children (below 15 years): economically unproductive, need to be provided with food, clothing, education and medical care. Comprises 34.4% of total India's population.

→ Working Age (15-59 years): economically productive and biologically reproductive. Considered as working population. Comprises 6.9% of total India's population.

→ Aged (Above 59 years): can be economically productive though they may have retired. May be working but they are not available for employment through recruitment. Comprises 58.7% of total India's population.

Sex Ratio

• Sex ratio is defined as the number of females per 1000 males in the population.

• Importance: measuring the extent of equality between males and females in a society at a given time. In India, sex ratio has always remained unfavourable to females.

• Census year with Sex ratio:

Census Year	Sex Ratio
1951	956
1961	951
1971	930
1981	934
1991	929
2001	933

Literacy rates

• According to the Census of 2001, a person aged 7 years. and above who can read and write with understanding in any language, is treated as literate. Low levels of literacy are a serious obstacle for economic improvement.

• The literacy rate in the country as per the Census of 2001 is 64.84 per cent; 75.26 per cent for males and 53.67 percent for females.

Occupational Structure

• The distribution of the population according to different types of occupation is referred to as the occupational structure.

• Occupations are generally classified into three categories:

→ Primary activities (related to land): It include agriculture, animal husbandry, forestry, fishing, mining and quarrying etc.

→ Secondary activities (related to industry): It include manufacturing industry, building and construction work etc.

→ Tertiary activities (related to services): include transport, communications, commerce, administration and other services.

• Developed nations have a high proportion of people in secondary, and tertiary activities while developing nations have a higher proportion of their workforce engaged in primary activities.

• In India, about 64 percent of the population is engaged only in agriculture. 13 percent are dependent on secondary and 20 percent are on tertiary sectors.

• In recent times, people are moving towards secondary and tertiary sectors because of growing industrialisation and urbanisation in recent times.

Health

• Health is an important component of population composition, which affects the process of development.

• There has been a significant improvement in health conditions in India. Death rates have declined from 25 per 1000 population in 1951 to 8.1 per 1000 in 2001 and life expectancy at birth has increased from 36.7 years in 1951 to 64.6 years in 2001.

• The improvement is due to:
→ Improvement in public health
→ Prevention of infectious diseases
→ Application of modern medical practices in diagnosis and treatment of ailments.

• Health is still a major concern for India because:
→ The per capita calorie consumption is much below the recommended levels and malnutrition afflicts a large percentage of our population.
→ Safe drinking water and basic sanitation amenities are available to only one- third of the rural population.

Adolescent Population

• The age-group of 10 to 19 years are considered as Adolescent Population. It constitutes one-fifth of the total population of India.

• They are most important future resources for any country. Nutrition requirements of adolescents are higher than those of a normal child or adult.

• In India, the diet available to adolescents is inadequate in all nutrients. A large number of adolescent girls suffer from anaemia (deficiency of red blood cells or haemoglobin).

• The awareness can be improved through the spread of literacy and education among adolescent girls.

National Population Policy

• The Government of India initiated the comprehensive Family Planning Programme in 1952 for improving individual health and welfare.

• It sought to promote responsible and planned parenthood on a voluntary basis.

• National Population Policy (NPP) 2000 is the peak of years of planned efforts.

What NPP aims at?

→ It provides a policy framework for imparting free and compulsory school education up to 14 years of age,
→ Reducing infant mortality rate to below 30 per 1000 live births
→ Achieving universal immunisation of children against all vaccine preventable diseases,
→ Promoting delayed marriage for girls
→ Making family welfare a people-centered programme.

Relation between NPP 2000 and Adolescents

• NPP 2000 identified adolescents as one of the major sections of the population that need greater attention.

• Besides nutritional requirements, the policy put greater emphasis on other important needs of adolescents including protection from unwanted pregnancies and sexually transmitted diseases (STD).

• The programmes started by NPP 2000 for adolescents aims at:
→ Encouraging delayed marriage and child-bearing.
→ Education of adolescents about the risks of unprotected sex.
→ Making contraceptive services accessible and affordable.
→ Providing food supplements, nutritional services.
→ Strengthening legal measures to prevent child marriage.

Do you know from chapter

• Only Bangladesh and Japan have higher average population densities than India.

• Kerala has a sex ratio of 1058 females per 1000 males, Pondicherry has 1001 females for every 1000 males, while Delhi has only 821 females per 1000 males and Haryana has just 861.

NCERT Solutions for Population

NCERT Solutions for Class 6th: Ch 16 Garbage in, Garbage out Science

Page No: 164

Exercises

1. (a) Which kind of garbage is not converted into compost by the redworms?
(b) Have you seen any other organism besides redworms, in your pit? If yes, try to find out their names. Draw pictures of these.
Answer

(a) The garbage which are non-biodegradable in nature like iron strips, plastics, poly bags etc. are not converted into compost by the redworms.
(b) Yes, these organisms include maggots, flies, cockroaches etc.

2. Discuss :
(a) Is garbage disposal the responsibility only of the government?
(b) Is it possible to reduce the problems relating to disposal of garbage?

Answer

(a) No, garbage disposal is not only the responsibility of the government. It is sole responsibility of both government and the people. Each and every citizen must be responsible for maintaining proper sanitation and disposable of waste materials produced. People should not litter at public places. They should also take care for the effective disposal of the waste produced at home, schools, hospitals etc.

(b) Yes, it is possible to reduce the problems relating to disposal of garbage by taking following measures:

(i) People should concern about generating less waste products and creating awareness among them.

(ii) Each and every product should be used efficiently.

(iii) Biodegradable and non-biodegradable wastes should be separated.

(iv) Waste products or garbage must be recycled and setting up centre for the treatment of them.

(v) We should minimize the use of plastic bags and use recyclable products.

3. (a) What do you do with the left over food at home?
(b) If you and your friends are given the choice of eating in a plastic plate or a banana leaf platter at a party, which one would you prefer and why?

Answer

(a) The left over food at home is given to animals on road, thrown in garbage or some times we use it to make compost.

(b) I will prefer to eat in banana leaf because it is a biodegradable product and environment friendly and can be decomposed.

4. (a) Collect pieces of different kinds of paper. Find out which of these can be recycled.
(b) With the help of a lens look at the pieces of paper you collected for the above question. Do you see any difference in the material of recycled paper and a new sheet of paper ?

Answer

(a) Papers without having plastic coating on it can be recycled.

(b) The recycled is slightly yellowish in color than the new sheet of paper. Recycled paper is also rough and of low quality than new sheet of paper.

5. (a) Collect different kinds of packaging material. What was the purpose for which each one was used? Discuss in groups.
(b) Give an example in which packaging could have been reduced?
(c) Write a story on how packaging increases the amount of garbage.

Answer

(a) Different kinds of packaging materials and its uses:
(i) Paper packaging materials: Mostly used for packing of light and durable and hard products like soaps, blades, chocolates etc.
(ii) Plastic packaging materials: Used for liquid items and medium weighted too like oils, drinking water bottles, shampoo etc.
(iii) Poly bags: Used for carrying groceries, vegetables etc.
(iv) Clothes and Jute packaging materials: Used for heavy weighted as well as medium weighted too like carrying clothes, grains, vegetables, fruits etc.

(b) If people started carrying their own carry bags for buying groceries and vegetables and fruits then there will be no need for packing those materials by using plastics and thus the packaging could have been reduced.

(c) Packaging increase the amount of garbage as the it is quite useless ones the product is delivered or been utilised. These useless packaging materials are thrown away and just add loads to the garbage. Many packaging materials can't even be reused. Some of them are made of plastics and thus a non-biodegradable item which add hazards to the environment.

6. Do you think it is better to use compost instead of chemical fertilisers? Why?

Answer

Yes, i think it is better to use compost instead of chemical fertilisers because:
(i) It is environment friendly and add natural fertility to the soil.
(ii) It doesn't have adverse effect on nature and creates no pollution.
(iii) The food items grown are also healthy and do not contains any chemicals.
(iv) It is cheaper than the chemical fertiliser.
(v) Soil will never loss its fertility if we use compost.

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नुक्ता - हिंदी व्याकरण Class 9th Course -'B'

किसी व्यंजन अक्षर के नीचे लगाये जाने वाले बिंदु को नुक्ता कहते हैं। उर्दू, अरबी, फ़ारसी भाषा से हिन्दी भाषा में आए क ख ग ज फ वर्ण को अलग से बताने के लिए इनका प्रयोग किया जाता है क्योंकि इन भाषाओँ से लिए गए शब्दों को हिन्दी में उच्चारित सही से नहीं किया जा सकता था। नुक्ता के प्रयोग से उस वर्ण के उच्चारण पर अधिक दबाव आ जाता है।

जैसे: ‘खुदा’ का अर्थ है हिंदी में ‘खुदी हुई ज़मीन’ और नुक्ता लग जाने से ‘ख़ुदा’ का अर्थ ‘भगवान’ हो जाता है।

कुछ नुक्ता वाले शब्द-

कमज़ोर, तूफ़ान, ज़रूर, इस्तीफ़ा, ज़ुल्म, फ़तवा, मज़दूर, ताज़ा, फ़कीर, फ़र्ज़, ज़ेवर, ज़ोर, फ़्रेंच, जि़ंदगी, इज़्ज़त, फ़रमान, रिलीज़, ब्लेज़र, ज़मानत, रफ़ू आदि।

क, ख, ग में नुक्ता का प्रयोग हिंदी भाषा में अनिवार्य नहीं है परन्तु 'ज़' और 'फ़' में नुक्ता लगाना आवश्यक है।

स्पर्श I में प्रयुक्त नुक्ता वाले शब्द-

साफ़, दर्ज़ा, ज़रा, बाज़ार, तरफ़, ज़माना, ख़रबूज़े, ज़िन्दा, बरफ़, तेज़, बर्फ़, काफ़ी, सब्ज़ियों, मेहमाननवाज़ी, ज़िक्र, शराफ़त, गुज़र, उफ़, अफ़सर, दफ़्तर, ज़ोर, प्रोफ़ेसर, गुज़रने, परहेज़, चीज़ें, पुर्ज़े, फ़ायदा, मज़हबी, ऐतराज़, नमाज़, आज़ाद, रोज़े, गिरफ़्तार, रोज़, फौजी, ज़ुल्मों, हफ़्ते, ज़रूरत, सफ़ेद, ताज़े, हज़ारों

NCERT Solutions for Class 12th: Ch 8 Human Health and Disease Biology

Page No: 164

Exercises

1. What are the various public health measures, which you would suggest as safeguard against infectious diseases?

Answer

Public health measures which should be taken to afeguard against infectious diseases are:
(i) Maintenance of personal and public hygiene:It is one of the most important methods for prevention of various infectious diseases. This measure consists of maintaining a hygienic body, taking of healthy and nutritious food, drinking clean water, etc. In public hygiene their is proper disposal of garbage, excreta, periodic cleaning of society, and cleaning of water reservoirs.
(ii) Isolation: To prevent the spread of air-borne infectious diseases like pneumonia, chicken pox, tuberculosis, etc., it is essential measure to keep the infected person in isolation with others to reduce the chances of spreading these infectious diseases.
(iii) Vaccination: Vaccination is the protection of the body from communicable diseases by injecting some agent that makes copy of the microbe inside the body. It helps in providing passive immunity to the body. Several vaccines are available against many diseases such as tetanus, polio, measles, mumps, etc.
(iv) Vector Eradication: Various diseases such as malaria, filariasis, dengue, and chikungunya spread through vectors. Thus, these diseases can be prevented by providing a clean environment and by the prevention of breeding of mosquitoes. This can be achieved by not allowing water to stagnate around public areas. Also, measures like periodic cleaning of coolers, use of mosquito nets and spreading of insecticides in drains, ponds, etc. can be undertaken to ensure a healthy environment. Introducing fish such as Gambusia in ponds also controls the breeding of mosquito larvae in still water against diseases.

2. In which way has the study of biology helped us to control infectious diseases?

Answer

→ Various advancements that have occurred in the field of biology have helped us gain a better understanding to fight against various infectious diseases.
→ Biology has developed as we have come to know about the life cycle of various parasites, pathogens, and vectors along with the modes of transmission of various diseases and the measures for controlling them.
→ Vaccination programmes against several infectious diseases such as small pox, chicken pox, tuberculosis, etc. have helps us to eradicate these diseases.
→ Biotechnology has helped in the preparation of developed and safe drugs and vaccines.
→ Antibiotics have also played a major role in the treatment of various infectious diseases.

3. How does the transmission of each of the following diseases take place?
(a) Amoebiasis          (b) Malaria       (c) Ascariasis     (d) Pneumonia

Answer

(a) Amoebiasis: It is a vector transmitted disease that spreads by the means of contaminated food and water. The vector involved in the transmission of this disease is the housefly. Its mode of transmission is Entamoeba histolytica.

(b) Malaria: It is a vector transmitted disease that spreads by the biting of the female Anopheles mosquito. Its mode of transmission is Ascaris lumbricoides.

(c) Ascariasis: It spreads through contaminated food and water. Its mode of transmission is Ascaris lumbricoides

(d) Pneumonia: It spreads by the sputum of a diseased person. Its mode of transmission is Streptococcus pneumoniae,

4. What measure would you take to prevent water-borne diseases?

Answer

Water-borne diseases such as cholera, typhoid, hepatitis B, etc. spread by drinking contaminated water.
→ These water-borne diseases can be prevented by proper disposal of garbage, excreta, regular cleaning.
→ Speading insecticide in community water reservoirs, boiling drinking water, etc

5. Discuss with your teacher what does ‘a suitable gene’ means, in the context of DNA vaccines.

Answer

A suitable gene of specific dna segment when inserted in the the body of host to produce specific type of protein which gives passive immunity to the organism and helps to fight with foreign organism.

6. Name the primary and secondary lymphoid organs.

Answer

(i) Primary lymphoid organs include the thymus and bone marr
(ii) Secondary lymphoid organs are the spleen, lymph nodes, tonsils, Peyer's patches of small intestine, and appendix.

7. The following are some well-known abbreviations, which have been used in this chapter. Expand each one to its full form:
(a) MALT    (b) CMI     (c) AIDS     (d) NACO     (e) HIV

Answer

(a) MALT- Mucosa-Associated Lymphoid Tissue
(b) CMI- Cell-Mediated Immunity
(c) AIDS- Acquired Immuno Deficiency Syndrome
(d) NACO- National AIDS Control Organization
(e) HIV- Human Immuno Deficiency virus

8. Differentiate the following and give examples of each:
(a) Innate and acquired immunity    (b) Active and passive immunity

Answer

(a) Innate and acquired immunity

Innate immunity	Acquired immunity
(i) It is a non-pathogen specific type of defense mechanism.	(i) It is a pathogen specific type of defense mechanism.
(ii) It is inherited from parents and protects the individual since birth.	(ii) It is acquired after the birth of an individual.
(iii) It operates by providing barriers against the entry of pathogenic agents.	(iii) It produces primary and secondary responses, which are mediated by B-lymphocytes and T-lymphocytes.
(iv) It does not specific memory.	(iv) It is observed by an immunological memory.

(b) Active and passive immunity

Active immunity	Passive immunity
(i) It is a type of acquired immunity in which the body produces its own antibodies against disease-causing antigens.	(i) It is a type of acquired immunity in which readymade antibodies are transferred from one person to another.
(ii) It shows long lasting effect.	(ii) It does not have long lasting effect.
(iii)It is slow. It takes time in producing antibodies and giving responses.	(iii) It is fast. It provides immediate respose.
(iv) Injecting microbes through vaccination inside the body is an example of active immunity.	(iv) Transfer of antibodies present in the mother's milk to the infant is an example of passive immunity.

9. Draw a well-labelled diagram of an antibody molecule.

Answer

10. What are the various routes by which transmission of human immuno-deficiency virus takes place?

Answer

AIDS (Acquired Immuno Deficiency Syndrome) is caused by the Human immunodeficiency virus (HIV).
This is transmitted by following modes -
(a) Unprotected sexual contact with a diseased person.
(b) Transfusion of blood from a healthy to a diseased person.
(c) Sharing infected needles or syringes.
(d) infected mother to a child through the placental connection.

11. What is the mechanism by which the AIDS virus causes deficiency of immune system of the infected person?

Answer

→ AIDS (Acquired Immuno Deficiency Syndrome) is caused by the Human immunodeficiency virus (HIV) via sexual or blood to blood contact.
→ After entering the human body, the HIV virus attacks and enters into the macrophages. Inside the macrophages, the RNA of the virus replicates with the help of enzyme reverse transcriptase and gives rise to viral DNA copy.
→ Then, this viral DNA incorporates into the host DNA and directs the synthesis of virus particles. → At the same time, HIV enters helper T- lymphocytes. It replicates and produces viral progeny.
→ These newly formed progeny viruses get released into the blood, attacking other healthy helper T-lymphocytes in the body.
→ As a result, the number of T-lymphocytes in the body of an infected person decreases in number, which causes decrease in immunity of person.

12. How is a cancerous cell different from a normal cell?

Answer

Normal cell	Cancerous cell
(i) Normal cells show the property of contact inhibition. Therefore, when these cells come into contact with other cells, they stop dividing.	(i) Cancerous cells lack the property of contact inhibition. Therefore, they continue to divide, thereby forming a mass of cells called tumor.
(ii) They undergo differentiation after attaining a specific growth.	(ii) They do not undergo differentiation.
(iii) These cells remain confined at a particular location.	(iii) These cells do not remain confined at a particular location. They move into neighboring tissues and disturbs the functioning.

13. Explain what is meant by metastasis.

Answer

The property of metastasis is perfomed by malignant tumors. These melingnent cells moves through different part of body by a pathological process. These cells divide uncontrollably, forming a mass of cells called tumor. From the tumor, some cells get shed off and enter into the blood stream. From the blood stream, these cells reach distant parts of the body and therefore, start the formation of new tumors by dividing actively.

14. List the harmful effects caused by alcohol/drug abuse.

Answer

Alcohol and drugs have several adverse effects on the individual, his family, and the society.
(i) Effects of alcohol:
→ Effects on the individual: Alcohol is injurious to the health of the individual. When an individual consumes excess alcohol, it causes damage to the liver and the central nervous system. As a result, other symptoms such as depression, fatigue, aggression, loss of weight and appetite may also be observed in the individual. Sometimes, extreme levels of alcohol consumption may also lead to heart failure, resulting coma and death. The immediate adverse effects of alcohol abuse are manifested in form of reckless bhehaviour, vandalism and voilence,Also, it is advisable for pregnant women to avoid alcohol as it may inhibit normal growth of the baby.
→ Effects on the family: Consumption of excess alcohol by any family member can have devastating effects on the family. It leads to several domestic problems such as quarrels, frustrations, insecurity, etc.
Effects of alcohol on society-
(a) Rash behaviour
(b) Malicious mischief and violence
(c) Disturbing social network
(d) Loss of interest in social activities, loss of interests in hobbies,change in eating and sleeping habites etc.

(ii) Effects of drugs: An individual who is addicted to drugs creates problems not only for himself but also for his family.
→ Effects on the addited individual: Drugs have an adverse effect on the central nervous system of an individual. This leads to the malfunctioning of several other organs of the body such as the kidney, liver, etc. The spread of HIV is most common in these individuals as they share common needles while injecting drugs in their body. Drugs have long-term side effects on both males and females. These side effects include increased aggressiveness, mood swings, and depression
→ Effects on the family and society: A person addicted to drugs creates problems for his family and society. A person dependant on drugs becomes frustrated, irritated, and anti-social.At the time of drug addicted becomes mental and financial distress.

15. Do you think that friends can influence one to take alcohol/drugs? If yes, how may one protect himself/herself from such an influence?

Answer

Yes, friends can influence one to take drugs and alcohol. A person can take the following steps for the prevention of themself against drug abuse:
→ by avoiding undue peer pressur as everyone has theire own field of interest which shouid be respected by theire teachers and family. One should not experiment with alcohol for curiosity and fun.
→ Avoid the company of friends who take drugs.
→ Seek help from parents and peers.a child should not pushed beyond his/her threshold limits.
→ Take proper knowledge and counseling about drug abuse. Devote your energy in other extra-curricular activities.
→ Seek immediate professional and medical help from psychologists and psychiatrists if symptoms of depression and frustration become apparent.
→ Get rid of the problems completely and lead a perfectly normal life by increasing their will power.

16. Why is that once a person starts taking alcohol or drugs, it is difficult to get rid of this habit? Discuss it with your teacher.

Answer

This question should be discussed with your subject teacher.

17. In your view what motivates youngsters to take to alcohol or drugs and how can this be avoided?

Answer

→ Various factors are responsible for motivating youngsters towards alcohol or drugs. Curiosity, need for adventure and excitement, experimentation for fun are the initial causes for motivating youngsters.
→ Some youngsters start consuming drugs and alcohol in order to overcome negative emotions (such as stress, pressure, depression, frustration) and to excel in various fields..such youngsters use to abuse alcohol due to their family pressure for academics .
→ Several mediums like television, internet, newspaper, movies promote various brend by the brand ambassadors like celebrities . Celebrities are also ideal of youngsters so the get much infiuenced by them.
→ Amongst these factors, reasons such as unstable and unsupportive family structures and peer pressure can also lead an individual to be dependant on drugs and alcohol.
Preventive measures against addiction of alcohol and drugs:
(i) Parents should motivate and try to increase the will power of their child.
(ii) Parents should educate their children about the ill-effects of alcohol. They should provide them with proper knowledge and counselling regarding the consequences of addiction to alcohol.
(iii) It is the responsibility of the parent to discourage a child from experimenting with alcohol. Youngsters should be kept away from the company of friends who consume drugs.
(iv) Children should be encouraged to devote their energy in other extra- curricular and recreational activities.
(v) Proper professional and medical help should be provided to a child if sudden symptoms of depression and frustration are observed.

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NCERT Solutions for Class 9th: Ch 13 Surface Areas and Volumes Maths

Page No: 213

Exercise 13.1

1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1m² costs Rs 20.

Answer

Length of plastic box = 1.5 m
Width of plastic box = 1.25 m
Depth of plastic box = 1.25 m
(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.
Surface area of the box = Lateral surface area + Area of the base
                                      = 2(l+b)×h + (l×b)
= 2[(1.5 + 1.25)×1.25] + (1.5 × 1.25) m²
= (3.575 + 1.875) m²
                                     = 5.45m²
The sheet required required to make the box is 5.45 m²
(ii) Cost of 1 m²of sheet= Rs 20
∴ Cost of 5.45m²of sheet = Rs (20 × 5.45) = Rs 109

2. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 per m².

Answer

length of the room = 5m
breadth of the room = 4m
height of the room = 3m
Area of four walls including the ceiling = 2(l+b)×h + (l×b)
                                                                = 2(5+4)×3 + (5×4) m²
= (54 + 20) m²
                                    = 74m²Cost of white washing = ₹7.50 per m²
Total cost = ₹(74×7.50) = ₹ 555

3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of₹10 per m² is₹15000, find the height of the hall.
[Hint : Area of the four walls = Lateral surface area.]

Answer

Perimeter of rectangular hall = 2(l + b) = 250 m
Total cost of painting = ₹15000
Rate per m²=₹10
Area of four walls = 2(l + b) h m²= (250×h) m²
A/q,
(250×h)×10 = ₹15000
⇒ 2500×h = ₹15000
⇒ h = 15000/2500 m
⇒ h = 6 m
Thus the height of the hall is 6 m.

4. The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?

Answer

Volume of paint = 9.375 m²=93750 cm²
Dimension of brick = 22.5 cm×10 cm×7.5 cm
Total surface area of a brick = 2(lb + bh + lh) cm²
                                             = 2(22.5×10 + 10×7.5 + 22.5×7.5) cm²
                                             = 2(225 + 75 + 168.75) cm²
                                             = 2×468.75 cm² = 937.5cm²
Number of bricks can be painted = 93750/937.5 = 100

5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?

Answer

(i) Lateral surface area of cubical box of edge 10cm = 4×10² cm² = 400 cm²
Lateral surface area of cuboid box = 2(l+b)×h
                                                       = 2×(12.5+10)×8 cm²
                                                       = 2×22.5×8 cm² = 360 cm²
Thus, lateral surface area of the cubical box is greater by (400 – 360) cm² = 40 cm²

(ii) Total surface area of cubical box of edge 10 cm =6×102cm2=600cm2
Total surface area of cuboidal box = 2(lb + bh + lh)
                                         = 2(12.5×10 + 10×8 + 8×12.5) cm²
                                                       = 2(125+80+100) cm²
                                                     = (2×305) cm²= 610 cm²
Thus, total surface area of cubical box is smaller by 10 cm²

6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?

Answer

(i) Dimensions of greenhouse:
l = 30 cm, b = 25 cm, h = 25 cm
Total surface area of green house = 2(lb + bh + lh)
                                                      = 2(30×25 + 25×25 + 25×30) cm²
                                                      = 2(750 + 625 + 750) cm²
                                                      = 4250 cm²
(ii) Length of the tape needed = 4(l + b + h)
                                                = 4(30 + 25 + 25) cm
                                                = 4×80 cm = 320 cm

7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹4 for 1000 cm² , find the cost of cardboard required for supplying 250 boxes of each kind.

Answer

Dimension of bigger box = 25 cm × 20 cm × 5 cm
Total surface area of bigger box = 2(lb + bh + lh)
                                                    = 2(25×20 + 20×5 + 25×5) cm²
                                                    = 2(500 + 100 + 125) cm²
                                                    = 1450 cm²
Dimension of smaller box = 15 cm × 12 cm × 5 cm
Total surface area of smaller box = 2(lb + bh + lh)
                                                   = 2(15×12 + 12×5 + 15×5) cm²
                                                     = 2(180 + 60 + 75) cm²
                                                   = 630 cm²
Total surface area of 250 boxes of each type = 250(1450 + 630) cm²
   = 250×2080 cm²= 520000cm²
Extra area required = 5/100(1450 + 630)× 250 cm²= 26000cm²

Total Cardboard required = 520000 + 26000 cm² = 546000cm²
Total cost of cardboard sheet = ₹ (546000 × 4)/1000 = ₹ 2184

8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4m × 3m?

Answer

Dimensions of the box- like structure = 4m × 3m × 2.5
Tarpaulin only required for all the four sides and top.
Thus, Tarpaulin required = 2(l+b)×h + lb
                                                           = [2(4+3)×2.5 + 4×3] m²
                                                           = (35×12) m²
                                                           = 47 m²
Page No: 216

Exercise 13.2

1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.

Answer

Let r be the radius of the base and h = 14 cm be the height of the cylinder.
Curved surface area of cylinder = 2πrh = 88 cm²
⇒ 2 × 22/7 × r × 14 = 88
⇒ r = 88/ (2 × 22/7 × 14)
⇒ r = 1 cm
Thus, the diameter of the base = 2r = 2×1 = 2cm

2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?

Answer

Let r be the radius of the base and h be the height of the cylinder.
Base diameter = 140 cm and Height (h) = 1m
Radius of base (r) = 140/2 = 70 cm = 0.7 m
Metal sheet required to make a closed cylindrical tank = 2πr(h + r)
                                              = (2 × 22/7 × 0.7) (1 + 0.7) m²
                                              = (2 × 22 × 0.1 × 1.7) m²
                                              =7.48 m²

3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

Answer

Let R be external radius and r be the internal radius h be the length of the pipe.

R = 4.4/2 cm = 2.2 cm

r = 4/2 cm = 2 cm
h = 77 cm
(i) Inner curved surface = 2πrh cm²
= 2 × 22/7 × 2 × 77cm²
= 968 cm²

(ii) Outer curved surface = 2πRh cm²

= 2 × 22/7 × 2.2 × 77 cm²
= 1064.8 cm²

(iii) Total surface area of a pipe = Inner curved surface area + outer curved surface area + areas of two bases

= 2πrh + 2πRh + 2π(R^{2 -} r²)
= [968 + 1064.8 + (2 × 22/7) (4.84 - 4)] cm²
= (2032.8 + 44/7 × 0.84) cm²
= (2032.8 + 5.28) cm²= 2038.08 cm²

Page No: 217

4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².

Answer

Length of the roller (h) = 120 cm = 1.2 m
Radius of the cylinder = 84/2 cm = 42 cm = 0.42 m
Total no. of revolutions = 500
Distance covered by roller in one revolution = Curved surface area = 2πrh
                                                                       = (2 × 22/7 × 0.42 × 1.2) m²
                                                                       = 3.168 m²
Area of the playground = (500 × 3.168) m²= 1584 m²

5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m².

Answer

Radius of the pillar (r) = 50/2 cm = 25 cm = 0.25 m
Height of the pillar (h) = 3.5 m.
Rate of painting = ₹12.50 per m²
Curved surface = 2πrh
                         = (2 × 22/7 × 0.25 × 3.5) m²
=5.5 m²
Total cost of painting = (5.5 × 12.5) = ₹68.75

6. Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Answer

Let r be the radius of the base and h be the height of the cylinder.
Curved surface area = 2πrh = 4.4 m²
⇒ 2 × 22/7 × 0.7 × h = 4.4
⇒ h = 4.4/(2 × 22/7 × 0.7) = 1m
⇒ h = 1m

7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of ₹40 per m².

Answer

Radius of circular well (r) = 3.5/2 m = 1.75 m
Depth of the well (h) = 10 m
Rate of plastering = ₹40 per m²
(i) Curved surface = 2πrh
                        = (2 × 22/7 × 1.75 × 10) m²
                              = 110 m²

(ii) Cost of plastering = ₹(110 × 40) = ₹4400

8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Answer

Radius of the pipe (r) = 5/2 cm = 2.5 cm = 0.025 m
Length of the pipe (h) = 28/2 m = 14 m
Total radiating surface = Curved surface area of the pipe = 2πrh
                                     = (2 × 22/7 × 0.025 × 28) m²
= 4.4 m²
9. Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank.

Answer

(i) Radius of the tank (r) = 4.2/2 m = 2.1 m
Height of the tank (h) = 4.5 m
Curved surface area = 2πrh m²
                                 = (2 × 22/7 × 2.1 × 4.5) m²
                                 = 59.4 m²
(ii) Total surface area of the tank = 2πr(r + h) m²
                    = [2 × 22/7 × 2.1 (2.1 + 4.5)] m²
                                = 87.12 m²

Let x be the actual steel used in making tank.
∴ (1 - 1/12) × x = 87.12
⇒ x = 87.12 × 12/11
⇒ x = 95.04 m²

10. In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Answer

Radius of the frame (r) = 20/2 cm = 10 cm
Height of the frame (h) = 30 cm + 2×2.5 cm = 35 cm
2.5 cm of margin will be added both side in the height.
Cloth required for covering the lampshade = curved surface area = 2πrh
                                                  = (2 × 22/7 × 10 × 35)cm²
                                    = 2200 cm²

11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Answer

Radius of the penholder (r) = 3cm
Height of the penholder (h) = 10.5cm
Cardboard required by 1 competitor = CSA of one penholder + area of the base
                                              = 2πrh + πr²
                                              = [(2 × 22/7 × 3 × 10.5) + 22/7 × 3²] cm²
                                              = (198 + 198/7) cm²
                                              = 1584/7 cm²
Cardboard required for 35 competitors = (35 × 1584/7) cm²
                                                               = 7920 cm²

Page No: 221

1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Answer

Radius (r) = 10.5/2 cm = 5.25 cm
Slant height (l) = 10 cm
Curved surface area of the cone = (πrl) cm²
                                                    =(22/7 × 5.25 × 10) cm²
                                                    =165 cm²
2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Answer

Radius (r) = 24/2 m = 12 m
Slant height (l) = 21 m
Total surface area of the cone = πr (l + r) m²
                                                = 22/7 × 12 × (21 + 12) m²
                                                = (22/7 × 12 × 33) m²
                                                = 1244.57 m²

3. Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface area of the cone.

Answer

(i) Curved surface of a cone = 308 cm²
Slant height (l) = 14cm
Let r be the radius of the base
∴ πrℓ = 308
⇒ 22/7 × r × 14 = 308
⇒ r =308/(22/7 × 14) = 7 cm

(ii) TSA of the cone = πr(l + r) cm²
                                 = 22/7 × 7 ×(14 + 7) cm²
                                 = (22 × 21) cm²
                                 = 462 cm²

4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹70

Answer

Radius of the base (r) = 24 m
Height of the conical tent (h) = 10 m
Let l be the slant height of the cone.
∴ l²= h²+ r²
⇒ l = √h²+ r²
⇒ l = √10²+ 24²
⇒ l = √100+ 576
⇒ l = 26 m
(ii) Canvas required to make the conical tent = Curved surface of the cone
Cost of 1 m² canvas = ₹70
∴ πrl = (22/7 × 24 × 26) m² = 13728/7 m²
∴ Cost of canvas = ₹ 13728/7 × 70 = ₹137280

5. What length of tarpaulin 3m wide will be required to make conical tent of height 8m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20cm (Use π = 3.14).

Answer

Radius of the base (r) = 6 m
Height of the conical tent (h) = 8 m
Let l be the slant height of the cone.
∴ l = √h²+ r²
⇒ l = √8²+ 6²
⇒ l = √100
⇒ l = 10 m
CSA of conical tent = πrl
        = (3.14 × 6 × 10) m² = 188.4 m²
Breadth of tarpaulin = 3 m
Let length of tarpaulin sheet required be x.
20 cm will be wasted in cutting.
So, the length will be (x - 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
[(x - 0.2 m) × 3] m = 188.4 m²
⇒ x - 0.2 m = 62.8 m
⇒ x = 63 m

6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹210 per 100 m².

Answer

Radius (r) = 14/2 m = 7 m
Slant height tomb (l) = 25 m
Curved surface area = πrl m²
                                =(227×25×7) m²
                              =550 m²
Rate of white- washing = ₹210 per 100 m²
Total cost of white-washing the tomb = ₹(550 × 210/100) = ₹1155

7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Answer

Radius of the cone (r) = 7 cm
Height of the cone (h) = 24 cm
Let l be the slant height
∴ l = √h²+ r²
⇒ l = √24²+ 7²
⇒ l = √625
⇒ l = 25 m
Sheet required for one cap = Curved surface of the cone
                                           = πrl cm²
                                           = (22/7 × 7 × 25) cm²
                                           = 550 cm²
Sheet required for 10 caps = 550 × 10 cm²= 5500 cm²

8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹12 per m² , what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)

Answer

Radius of the cone (r) = 40/2 cm = 20 cm = 0.2 m
Height of the cone (h) = 1 m
Let l be the slant height of a cone.
∴ l = √h²+ r²
⇒ l = √1²+ 0.2²
⇒ l = √1.04
⇒ l = 1.02 m
Rate of painting = ₹12 per m²

Curved surface of 1 cone = πrl m²
                                         = (3.14 × 0.2 × 1.02) m²
   = 0.64056 m²
Curved surface of such 50 cones = (50 × 0.64056) m²
= 32.028 m²
Cost of painting all these cones = ₹(32.028 × 12)
                                                   = ₹384.34
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Study Material and Notes of Ch 5 Natural Vegetation and Wildlife Class 9th Geography

Topics in the Chapter

• Introduction
• Three factors for biodiversity
→ Relief
→ Climate
→ Ecosystem
• Importance of Forests
• Types of vegetation
→ Tropical Evergreen Forests
→ Tropical Deciduous Forests
→ Tropical Thorn Forests and Scrubs
→ Montane Forests
→ Mangrove Forests
• Wildlife
• Conservation of flora and fauna
→ Governmental steps

Introduction

• India is one of the twelve mega bio-diversity countries of the world. India occupies 10th place in the world with about 47,000 plant species and 4th in Asia in plant diversity.

• 15,000 flowering plants in India constitutes 6 percent in the world’s total number of flowering plants. India also has approximately 90,000 species of animals as well as a rich variety of fish in its fresh and marine waters.

• A plant community which has grown naturally without human aid is called Natural vegetation.

• Natural vegetation that has been left undisturbed by humans for a long time is called virgin vegetation.

• Flora is used to denote plants of a particular region or period and fauna is species of animals of a particular region or period. Flora and fauna kingdom constitute biodiversity of a particular region or period.

• Relief, Climate and Ecosystem are three factors for biodiversity.

• Relief

→ Land: The nature of land influences the type of vegetation. Fertile lands supports agriculture while undulating and rough terrains have grassland and woodlands which give shelter to a variety of wild life.

→ Soil: Different types of soils provide basis for different types of vegetation. The sandy soils of the desert support cactus and thorny bushes while wet, marshy, deltaic soils support mangroves and deltaic vegetation. The hill slopes with some depth of soil have conical trees.

• Climate

→ Temperature: The character and extent of vegetation are mainly determined by temperature along with humidity in the air, precipitation and soil.

→ Sunlight: Due to differences in latitude, altitude, season and duration of the day there is variation in duration of sunlight.

→ Precipitation (Rains): Areas of heavy rainfall have more dense vegetation as compared to other areas of less rainfall. In India, almost the entire rainfall is brought in by the advancing southwest monsoon (June to September) and retreating northeast monsoons.

• Ecosystem: All the plants and animals in an area are interdependent and interrelated to each other in their physical environment, thus, forming an ecosystem.

• A very large ecosystem on land having distinct types of vegetation and animal life is called a biome.

• Importance of Forests:
→ These are renewable resources and play a major role in enhancing the quality of environment.
→ They modify local climate, control soil erosion, regulate stream flow, support a variety of industries, provide livelihood for many communities and offer panoramic or scenic view for recreation.
→ It controls wind force and temperature and causes rainfall. It provides humus to the soil and shelter to the wildlife.

• Natural vegetation in India has undergone many changes due to several factors such as the growing demand for cultivated land, development of industries and mining, urbanisation and over-grazing of pastures.

Types of Vegetation

• Types of vegetation may be identified in our country are:
(i) Tropical Evergreen Forests
(ii) Tropical Deciduous Forests
(iii) Tropical Thorn Forests and Scrubs
(iv) Montane Forests
(v) Mangrove Forests

• Tropical Evergreen Forests:

→ These forests are restricted to heavy rainfall areas of the Western Ghats and the island groups of Lakshadweep, Andaman and Nicobar, upper parts of Assam and Tamil Nadu coast.
→ They grow in an area of 200 cm of rainfall with a short dry season.
→ The trees reach great heights up to 60 metres or even above.
→ These forests have rich vegetation of all kinds – trees, shrubs, and creepers which give it a multilayered structure since the region is warm and wet throughout the year.
→ The forests appear green all the year round because there is no definite time for trees to shed their leaves.
→ Commercially important trees of this forest are ebony, mahogany, rosewood, rubber and cinchona.
→ Common animals found in these forests are elephants, monkey, lemur and deer. Plenty of birds, bats, sloth, scorpions and snails are also found.

• Tropical Deciduous Forests

→ Most widespread forest in India.
→ These are also called the monsoon forests and spread over the region receiving rainfall between 200 cm and 70 cm.
→ Trees of this forest-type shed their leaves for about six to eight weeks in dry summer.
→ These forests are further divided into two types: Moist and Dry deciduous.

→ Moist deciduous: Found in areas receiving rainfall between 200 and 100 cm. Present mostly in the
eastern part of the country– northeastern states, along the foothills of the Himalayas, Jharkhand, West Orissa and Chhattisgarh, and on the eastern slopes of the Western Ghats. Teak is the most dominant species of this forest. Commercially important species are Bamboos, sal, shisham, sandalwood, khair, kusum, arjun, mulberry.

→ Dry deciduous: Found in areas having rainfall between 100 cm and 70 cm. Present in the rainier parts of the peninsular plateau and the plains of Bihar and Uttar Pradesh. There are open stretches in which Teak, Sal, Peepal, Neem grow. A large part of this region has been cleared for cultivation and some parts are used for grazing. Common animals found are lion, tiger, pig, deer and elephant also huge variety of birds, lizards, snakes, and tortoises are found here.

• Tropical thorn forests and Scrubs

→ These are found in the region of with less than 70 cm of rainfall.
→ The natural vegetation consists of thorny trees and bushes. Acacias, palms, euphorbias and cacti are the main plant species.
→ Trees are scattered and have long roots penetrating deep into the soil to get moisture. The stems are succulent to conserve water. Leaves are mostly thick and small to minimize evaporation.
→ Common animals are rats, mice, rabbits, fox, wolf, tiger, lion, wild ass, horses and camels.

• Montane Forests

→ Montane forests have a succession of natural vegetation belts in the same order as we see from the tropical to the Tundra region.
→ Between a height of 1000 and 2000 metres, wet temperate type of forests containing evergreen broad-leaf trees such as oaks and chestnuts are predominate.
→ Between 1500 and 3000 metres, temperate forests containing coniferous trees like pine, deodar, silver fir, spruce and cedar, are found.
→ At higher elevations, temperate grasslands are common.
→ At high altitudes, generally more than 3,600 metres above sea-level, alpine vegetation found which have silver fir, junipers, pines and birches trees common.
→ Near snow line, shrubs and scrubs, they merge into the Alpine grasslands which are used extensively for grazing by nomadic tribes like the Gujjars and the Bakarwals.
→ At higher altitudes, mosses and lichens form part of tundra vegetation.
→ The common animals found in these forests are Kashmir stag, spotted dear, wild sheep, jack rabbit, Tibetan antelope, yak, snow leopard, squirrels, Shaggy horn wild ibex, bear and rare red panda, sheep and goats with thick hair.

• Mangrove Forests

→ These forests are found in the areas of coasts influenced by tides where mud and silt get accumulated.
→ Dense mangroves are the common varieties with roots of the plants submerged under water. These are deltas of the Ganga, the Mahanadi, the Krishna, the Godavari and the Kaveri
→ In the Ganga- Brahamaputra delta, sundari trees are found, which provide durable hard timber.
→ Palm, coconut, keora, agar, also grow in some parts of the delta.
→ Royal Bengal Tiger is the famous animal also turtles, crocodiles, gharials and snakes are found in these forests.

Wildlife

• India has about 2,000 species of birds which constitute 13% of the world’s total. There are 2,546 species of fish, which account for nearly 12% of the world’s stock. It also shares between 5 and 8 percent of the world’s amphibians, reptiles and mammals.

• Elephants are found in the hot wet forests of Assam, Karnataka and Kerala.

• One-horned rhinoceroses live in swampy and marshy lands of Assam and West Bengal.

• Rann of Kachchh is habitat of wild ass and camels are found in Thar desert.

• Indian bison, nilgai (blue bull), chousingha (four horned antelope), gazel and different species of deer are some other animals found in India.

• India is the only country in the world that has both tigers and lions. Gir forest in Gujrat is the natural habitat of lion whereas Tigers are found in the forests of Madhya Pradesh, the Sundarbans of West Bengal and the Himalayan region.

• The Himalayas harbour a hardy range of animals, which survive in extreme cold.

• Ladakh’s freezing high altitudes are a home to yak, the shaggy horned wild ox weighing
around one tonne, the Tibetan antelope, the bharal (blue sheep), wild sheep, and the kiang (Tibetan wild ass). The ibex, bear, snow-leopard and very rare red panda are found in certain parts.

• In the rivers, lakes and coastal areas, turtles, crocodiles and gharials are found.

• Birds like Peacocks, pheasants, ducks, parakeets, cranes and pigeons are some of the birds inhabiting the forests and wetlands of the country.

Conservation of Flora and Fauna

• The excessive exploitation of the plants and animal resources by human beings, disturbed the ecosystem. About 1,300 plant species are endangered and 20 species are extinct also few animals are endangered and some have become extinct.

• Causes of this threats are:
→ Hunting by greedy hunters for commercial purposes.
→ Pollution due to chemical and industrial waste and acid deposits
→ Introduction of alien species
→ Reckless cutting of the forests to bring land under cultivation and inhabitation.

• Governmental steps to protect flora and fauna of country are:
→ Fourteen biosphere reserves have been set up in the country to protect flora and fauna.
→ Financial and technical assistance is provided to many Botanical Gardens by the government since 1992.
→ Project Tiger, Project Rhino, Project Great Indian Bustard and many other eco- developmental projects have been introduced.
→ 89 National Parks, 490 Wildlife sanctuaries and Zoological gardens are set up to take care of Natural heritage.

• The Sunderbans in the West Bengal, Nanda Devi in Uttarakhand, the Gulf of Mannar in Tamil Nadu and the Nilgiris (Kerala, Karnataka and Tamil Nadu) have been included in the world network of Biosphese reserves.

List of fourteen bio-reserves in India are:
• Sunderbans (West Bengal)
• Simlipal (Odhisha)
• Gulf of Mannar (Tamil Nadu)
• Dihang-Dibang (Arunachal Pradesh)
• The Nilgiris in South India (in the states of Tamil Nadu, Karnataka and Kerala)
• Dibru Saikhowa (Arunachal Pradesh)
• Nanda Devi (Uttarakhand)
• Agasthyamalai (Kerala and Tamil Nadu)
• Nokrek (Meghalaya)
• Kanchenjunga (Sikkim)
• Great Nicobar (Bay of Bengal)
• Pachmari (Madhya Pradesh)
• Manas (Assam)
• Achanakmar-Amarkantak (Chhattisgarh)

Do you Know from the chapter

• The virgin vegetation, which are purely Indian are known as endemic or indigenous species but those which have come from outside India are termed as exotic plants.

• According to India State of Forest Report 2011, the forest cover in India is 21.05 per cent.

• Wildlife Protection Act was implemented in 1972 in India.

• The Gir Forest is the last remaining habitat of the Asiatic lion.

NCERT Solutions for Natural vegetation and Wildlife

NCERT Solutions for Class 12th: Ch 9 Strategies for Enhancement in Food Production Biology

Page No: 178

Exercises

1. Explain in brief the role of animal husbandry in human welfare.

Answer

(i) Animal husbandry deals with the scientific management of livestock which includes various aspects such as feeding, breeding, and control diseases to raise the population of animal livestock.
(ii) It usually deals with animals such as buffaloes cattle, pig, sheep, poultry, and fish which are useful for humans in various ways.
(iii) These animals are used for the production of commercially important products such as milk, meat, wool, egg, honey, silk, etc.
(iv) As increase in human population, the demand for the products from livestocks. So it is necessary to improve the management of livestock.

2. If your family owned a dairy farm, what measures would you undertake to improve the quality and quantity of milk production?

Answer

(i) Dairy farm management deals with processes which aim at improving the quality and quantity of milk production. Milk production is primarily dependent on choosing improved cattle breeds, provision of proper feed for cattle, maintaining proper shelter facilities, and regular cleaning of cattle.
(ii) Improved cattle breeds is an important factor of cattle management. Hybrid cattle breeds are introduced for improved productivity. Thus, it is important that hybrid cattle breeds should have a combination of various desirable genes such as high milk production and high resistance to diseases.
(iii) Cattle should also be given high quality and nutritious food including roughage, fibre concentrates, and high levels of proteins and other nutrients.
(iv) Cattle should be kept in well maintained housed in proper cattle-houses and should be kept in well ventilated roofs to prevent them from bad climatic conditions such as heat, cold, and rain.
(v) Regular baths and proper brushing should be ensured to control diseases. Also, regular check ups by a veterinary doctor for symptoms of various diseases.

3. What is meant by the term 'breed'? What are the objectives of animal breeding?

Answer

A breed is a improved variety of animals within a species. It is similar in most characters such as general appearance, size, configuration, and features with other members of the same species.
For example- Jersey and Brown Swiss are foreign breeds of cattle. These two varieties of cattle have the ability to produce abundant quantities of milk which is nutritious with high protein content.
Objectives of animal breeding:
(i) To improve the desirable qualities of the animal produce
(i) To increase the yield of animals
(iii) To produce disease-resistant varieties of animals.

4. Name the methods employed in animal breeding. According to you which of the methods is best? Why?

Answer

Animal breeding is the method of mating interrelated individuals. There are several methods involded in animals breeding, which can be classified into the following categories:

(i) Natural methods of breeding include inbreeding and out-breeding. Breeding between animals of the same breed is known as inbreeding, while breeding between animals of different breeds is known as out-breeding. Out-breeding of animals is of three types:
→ Out-crossing: In this type of out-breeding, the mating of animals occurs within the same breed.as they have no common ancestors up to the last 4-5 generations.
→ Cross-breeding: In this type of out-breeding, the mating occurs between different breeds of the same species, thereby producing a hybrid.
→ Interspecific hybridization: In this type of out-breeding, the mating occurs between different species.

(ii) Artificial methods of breeding include modern techniques of breeding. It involves controlled breeding experiments, which are of two types:-
→ Artificial insemination: It is a process of introducing the semen (collected from the male) into the oviduct or the uterus of the female body by the breeder. This method of breeding helps the breeder overcome certain problems faced in abnormal mating.
→ Multiple ovulation embryo technology (MOET): It is a technique for cattle improvement in which super-ovulation is induced by a hormone injection. Then, fertilization is achieved by artificial insemination and early embryos are collected. Each of these embryos are then transplanted into the surrogate mother for further development of the embryo.

The best method to carry out animal breeding is the artificial method of breeding, which includes artificial insemination and MOET technology. These technologies are scientific in nature. They help to minimize problems of normal mating and have a high success rate of crossing between mature males and females. Also, it ensures the production of hybrids with the desired qualities. This method is highly economical as a small amount of semen from the male can be used to inseminate several cattle as semen is not destroyed.

5. What is apiculture? How is it important in our lives?

Answer

(i) Apiculture is the practice of bee-keeping for the production of various products such as honey, bee's wax, etc. Honey is a highly nutritious food source and is used as an indigenous system of medicines.
(ii) It is useful in the treatment of many diseases such as cold, flu, and dysentery.
(iii) Other commercial products obtained from honey bees include bee's wax and bee pollen. Bee's wax is used for making cosmetics, polishes, and is even used in several medicinal preparations.
(iv) As demand of honey is increasing , people have started practicing bee-keeping on a large scale. It has become an income generating activity for farmers since it requires a low investment and is labour intensive.

6. Discuss the role of fishery in enhancement of food production.

Answer

(i) Fishery is an industry devoted with catching, processing, and marketing of fishes and other aquatic animals that have a high economic value.
(ii) Some commercially important aquatic animals are prawns crabs, oysters, lobsters, and octopus.
(iii) Fisheries play an important role in the Indian economy. This is because a large part of the Indian population is dependent on fishes as a source of food, which is both cheap and high in animal protein.
(iv) A Fishery is an employment generating industry especially for people staying in the coastal areas. Both fresh water fishes (such as Catla, Rohu, etc) and marine fishes (such as tuna, mackerel pomfret, etc.) are of high economic value.

7. Briefly describe various steps involved in plant breeding.

Answer

Plant breeding is purposeful manipulation of plants species in order to create desired plants that are better suited for cultivation, give better yield and are disease restistant. various steps involved in plant breeding are as follows:
(i) Collection of genetic variability: Genetic variability from various wild relatives of the cultivated species are collected to maintain the genetic diversity of a species. The entire collection of the diverse alleles of a gene in a crop is called the germplasm collection.
(ii) Evaluation of germplasm and selection of parents: The germplasm collected is then evaluated for the desirable genes. The selected plants with the desired genes are then used as parents in plant breeding experiments and are multiplied by the process of hybridization.
(iii) Cross-hybridization between selected parents: The next step in plant breeding is to combine the desirable characters present in two different parents to produce hybrids. It is a tedious job as one has to ensure that the pollen grains collected from the male parent reach the stigma of the female parent.
(iv) Selection of superior hybrids:the selection process is crucial to the success of breeding objective and requires careful scientific evaluation of the progeny. The progenies of the hybrids having the desired characteristics are selected through scientific evaluation. The selected progenies are then self-pollinated for several generations to ensure homozygosity.
(v) Testing, release, and commercialization of new cultivars: the newly selected lines are evaluated for theire yield and other agronomic traits of quality, disease resistance, by growing them in research fields for at least three growing seasons in different parts of the country. After thorough testing and evaluation, the selected varieties are given to the farmers for growing in fields for a large-scale production.

8. Explain what is meant by biofortification.

Answer

(i) Biofortification is a process of breeding crops with higher levels of vitamins, minerals, proteins, and fat content.
(ii) This method is employed to improve public health. Breeding of crops with improved nutritional quality is undertaken to improve the content of proteins, oil, vitamins, minerals, and micro-nutrients in crops.
(iii) It is also undertaken to upgrade the quality of oil and proteins. An example of this is a wheat variety known as Atlas 66, which has high protein content in comparison to the existing wheat. In addition, there are several other improved varieties of crop plants such as rice, carrots, spinach etc. which have more nutritious value and more nutrients than the existing varieties.

9. Which part of the plant is best suited for making virus-free plants and why?

Answer

Apical and axillary meristems of plants is used for making virus-free plants. In a diseased plant, only this region is not infected by the virus as compared to the rest of the plasnt region. Hence, the scientists remove axillary and apical meristems of the diseased plant and grow it in vitro to obtain a disease-free and healthy plant. Banana, sugarcane, and potato have been obtained using this method by scientist are virus free plants.

10. What is the major advantage of producing plants by micropropagation?

Answer

Micropropagation is a method of producing new plants in a short duration using plant tissue culture. Some major advantages of micropropagation are as follows:
(i) Micropropagation helps in the propagation of a large number of plants in a short span of time.
(ii) The plants produced are identical to the mother plant
(iii) It leads to the production of healthier plantlets, which exhibit better disease-resisting powers.

11. Find out what the various components of the medium used for propagation of an explant in vitro are?

Answer

The major components of medium used for propagation of explants in vitro are carbon sources such as sucrose, inorganic salts, vitamins, amino acids, water, agar-agar, and certain growth hormones such as auxins and gibberellins.

12. Name any five hybrid varieties of crop plants which have been developed in India.

Answer

The five hybrid varieties of crop plants which have been developed in India are:

Crop Plant	Hybrid Variety
Wheat	Sonalika and kalian sona
Rice	Jaya and Ratna
Cauliflower	Pusa shubra and Pusa snowball K-1
Cowpea	Pusa komal
Mustard	Pusa swarnim

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Extra Questions and Answer from Chapter 6 Kichad kaa kaavy Sparsh Bhaag I

निम्नलिखित गद्यांश को पढ़कर दिए गए प्रश्नों के उत्तर लिखिए -

1. हम आकाश का वर्णन करते हैं, पृथ्वी का वर्णन करते हैं, जलाशयों का वर्णन करते हैं, पर कीचड़ का वर्णन कभी किसी ने किया है? कीचड़ में पैर डालना कोई पसंद नहीं करता है, कीचड़ से शरीर गन्दा होता है, कपडे मैले हो जाते हैं। अपने शरीर पर कीचड़ उड़े यह किसी को अच्छा नहीं लगता और इसलिए कीचड़ किसी को सहानुभूति नहीं होती। यह सब यथार्थ है किन्तु तटस्थ्ता से सोचें तो हम देखेंगे कि कीचड़ में कुछ कम सौंदर्य नहीं है। पहले तो यह की कीचड़ का रंग बहुत सुन्दर है। पुस्तकों के गत्तों पर, घरों की दीवालों पर अथवा शरीर के कीमती कपड़ों के लिए हम सब कीचड़ के जैसे रंग पसंद करते हैं। कलाभिज्ञ लोगों की भट्टी में पकाये हुए मिट्टी के बर्तनों के लिए यही रंग बहुत पसंद है। फोटो लेते समय भी यदि उसमें कीचड़ का, एकाध ठीकरे का रंग आ जाए तो उसे वार्मटोन कहकर विज्ञ लोग खुश हो जाते हैं। पर लो, कीचड़ का नाम लेते ही सब बिगड़ जाता है।

(क) हम प्रकृति के किन रूपों का प्रायः वर्णन करते हैं? (1)
(ख) लोग कीचड़ को पसंद क्यों नहीं करते? (2)
(ग) कीचड़ का रंग कहाँ-कहाँ पसंद किया जाता है? (2)

उत्तर

(क) हम प्रकृति के सुंदर रूपों आकाश, पृथ्वी, जालशयों आदि का प्रायः वर्णन करते हैं।
(ख) लोग कीचड़ को पसंद इसलिए नहीं करते क्योंकि इससे शरीर गन्दा होता है, कपडे मैले हो जाते हैं।
(ग) कीचड़ का रंग पुस्तकों के गत्तों पर, घरों की दीवालों पर और शरीर के कीमती कपड़ों के लिए पसंद किया जाता है। कलाभिज्ञ लोगों की भट्टी में पकाये हुए मिट्टी के बर्तनों के लिए कीचड़ का रंग बहुत पसंद है।

2. कीचड़ देखना है तो गंगा किनारे या सिंधु के किनारे और इतने से तृप्ति ना हो तो सीधे खंभात पहुँचना चाहिए। वहाँ मही नदी के मुख से आगे जहाँ तक नज़र पहुँचे वहाँ तक सर्वत्र सनातन कीचड़ ही देखने को मिलेगा। इस कीचड़ में हाथी डूब जाएँगे ऐसा कहना, न शोभा दे ऐसी अलोपक्ति करने जैसा है। पहाड़ के पहाड़ उसमें लुप्त हो जाएँगे, ऐसा कहना चाहिए।
हमारा अन्न कीचड़ में से ही पैदा होता है इसका जाग्रत भान हर एक मनुष्य को होता तो वह कभी कीचड़ का तिरस्कार न करता।

(क) इस कीचड़ में हाथी भी डूब जाएँ -का क्या आशय है? इस कीचड़ में क्या-क्या डूब सकते हैं? (2)
(ख) मही नदी के मुख के आगे स्थित कीचड़ की क्या विशेषता है? (1)
(ग) मनुष्य कीचड़ का तिरस्कार करना कब छोड़ेगा? (1)

उत्तर

(क) इस कथन का आशय है कि कीचड़ बहुत जयादा तथा गहरा होता है। इस कीचड़ में पहाड़ के पहाड़ डूब सकते हैं।
(ख) मही नदी के मुख के आगे से जहाँ तक नजर पहुँचे वहाँ तक सर्वत्र सनातन कीचड़ ही देखने को मिलता है।
(ग) मनुष्य को जब यह जाग्रत भान हो जाएगा कि उसका अन्न कीचड़ से ही पैदा होता है तब वह इसका तिरस्कार करना छोड़ेगा।

3.नदी के किनारे जब कीचड़ सुखकर उसके टुकड़े हो जाते हैं, तब वे कितने सुन्दर दिखते हैं। ज्यादा गर्मी से जब उन्हीं टुकड़ों में दरारें पड़ती हैं और वे टेढ़े हो जाते हैं, तब वे सुखाये हुए खोपरे जैसे दीख पड़ते हैं। नदी किनारे मीलों तक जब समतल और चिकना कीचड़ एक-सा फैला हुआ होता है, तब वह दृश्य कुछ कम खूबसूरत नहीं होता। इस कीचड़ का पृष्ठ भाग कुछ सुख जाने पर उस पर बगुले और अन्य छोटे-बड़े पक्षी चलते हैं, तब तीन नाख़ून आगे और अँगूठा पीछे ऐसे अनेक पदचिह्न, मध्य एशिया के रास्ते की तरह दूर-दूर तक अंकित देख इसी रास्ते अपना कारवाँ ले जाने की इच्छा हमें होती है।

(क) नदी के किनारे कीचड़ कब सुखाये हुए खोपडे जैसे दिखाई देते हैं? (1)
(ख) नदी के किनारे कौन सा दृश्य खूबसूरत होता है? (1)
(ग) लेखक को क्या देख अपना कारवाँ उस रास्ते ले जाने की इच्छा होती है? (2)

उत्तर

(क) जब ज्यादा गर्मी पड़ती है तब नदी के किनारे सूखे पड़े कीचड़ कर टुकड़ों में दरारें पड़ जाती हैं तब वे सुखाये हुए खोपडे जैसे दिखाई देते हैं।
(ख) नदी के किनारे मीलों तक जब समतल और चिकना कीचड़ एक-सा फैला हुआ होता है, तब वह दृश्य खूबसूरत होता है।
(ग) कीचड़ का पृष्ठ भाग कुछ सूख जाने पर उस पर बगुले और अन्य छोटे-बड़े पक्षी चलते हैं, तब तीन नाख़ून आगे और अँगूठा पीछे ऐसे अनेक पदचिह्न, मध्य एशिया के रास्ते की तरह दूर-दूर तक अंकित देख लेखक को अपना कारवाँ इस रास्ते ले जाने की इच्छा होती है।

निम्नलिखित प्रश्नों के उत्तर दीजिये-

1. रंग की सारी शोभा कहाँ जमी थी?

उत्तर

रंग की सारी शोभा उत्तर में जमी थी।

2. कवि कीचड़ का वर्णन क्यों नहीं करते?

उत्तर

कवि बाहरी सौंदर्य को ज्यादा महत्व देते हैं, उनकी उपयोगिता को महत्व नहीं देते इसलिए कवि कीचड़ का वर्णन नहीं करते।

3. कीचड़ कब सुन्दर दिखते हैं?

उत्तर

नदी के किनारे जब कीचड़ सुखकर उसके टुकड़े हो जाते हैं तब कीचड़ सुन्दर दिखते हैं।

4. लेखक के अनुसार हमें कीचड़ देखने के लिए कहाँ जाना चाहिए?

उत्तर

लेखक के अनुसार हमें कीचड़ देखने के लिए गंगा के किनारे या सिंधु के किनारे जाना चाहिए। अगर इतने से भी संतुष्टि ना मिले तो खंभात जाना चाहिए।

5. आशय सपष्ट कीजिये "नदी के किनारे अंकित पद चिह्न और सींगों के चिह्नों से मानो महिषकुल के भारतीय युद्ध का पूरा इतिहास ही इस कर्दम लेख में लिखा हो भास होता है।"

उत्तर

इस कथन का आशय है कि नदी के किनारे कीचड़ सूखी कीचड़ में जब दो भैंस के बच्चे मस्त होकर अपने सींगों को उस कीचड़ में धँसाकर रौंदते और लड़ते हैं तो उनके लड़ने से कीचड़ में उनके पद चिह्न और सींगों के चिह्न अंकित हो जाते हैं जिसे देखकर ऐसा लगता है मानो महिषकुल के सभी भारतीय युद्ध का पूरा इतिहास कीचड़ में लेख के रूप लिख दिया गया हो।

कीचड़ का काव्य - पठन सामग्री और सार

NCERT Solutions for पाठ 6 - कीचड़ का काव्य

NCERT Solutions for Class 9th: Ch 13 Surface Areas and Volumes Maths

Page No: 225

Exercise 13.4

1. Find the surface area of a sphere of radius:
(i) 10.5 cm       (ii) 5.6 cm (iii) 14 cm

Answer

(i) Radius of the sphere (r) = 10.5 cm
Surface area = 4πr²
                    = (4 × 22/7 × 10.5 × 10.5) cm²
                    = 1386 cm²

(ii) Radius of the sphere (r) = 5.6 cm
Surface area = 4πr²
                     = (4 × 22/7 × 5.6 × 5.6) cm²
                     = 394.24 cm²

(iii) Radius of the sphere (r) = 14 cm
Surface area = 4πr²
                     = (4 × 22/7 × 14 × 14) cm²
                   = 2464 cm²

2. Find the surface area of a sphere of diameter:
(i) 14 cm         (ii) 21 cm        (iii) 3.5 m

Answer

(i) r = 14/2 cm = 7cm
Surface area = 4πr²
= (4 × 22/7 × 7 × 7)cm²
                                                =616cm²

(ii) r = 21/2 cm = 10.5 cm
Surface area = 4πr²
                     = (4 × 22/7 × 10.5 × 10.5) cm²
                     = 1386 cm²

(iii) r = 3.5/2 m = 1.75 m
Surface area = 4πr²
                    = (4 × 22/7 × 1.75 × 1.75) m²
                  = 38.5 m²

3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Answer

r = 10 cm
Total surface area of hemisphere = 3πr²
                                                     = (3 × 3.14 × 10 ×10) cm²
                                                     = 942 cm²

4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer

Let r be the initial radius and R be the increased radius of balloons.
r = 7cm and R = 14cm
Ratio of the surface area =4πr²/4πR²
                                        = r²/R²
                                        = (7×7)/(14×14) = 1/4
Thus, the ratio of surface areas = 1 : 4

5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm².

Answer

Radius of the bowl (r) = 10.5/2 cm = 5.25 cm
Curved surface area of the hemispherical bowl = 2πr²
= (2 × 22/7 × 5.25 × 5.25) cm²
                                                           = 173.25 cm²
Rate of tin - plating is = ₹16 per 100 cm²
Therefor, cost of 1 cm²=₹16/100
Total cost of tin-plating the hemisphere bowl = 173.25 × 16/100
                                                                     = ₹27.72

6. Find the radius of a sphere whose surface area is 154 cm².

Answer

Let r be the radius of the sphere.
Surface area = 154 cm²
⇒ 4πr²= 154
⇒ 4 × 22/7 × r²= 154
⇒ r²= 154/(4 × 22/7)
⇒ r²= 49/4
⇒ r = 7/2 = 3.5 cm

7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Answer

Let the diameter of earth be r and that of the moon will be r/4
Radius of the earth = r/2
Radius of the moon = r/8
Ratio of their surface area = 4π(r/8)²/4π(r/2)²
= (1/64)/(1/4)
                                          = 4/64 = 1/16
Thus, the ratio of their surface areas is 1:16

8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Answer

Inner radius of the bowl (r) = 5 cm
Thickness of the steel = 0.25 cm
∴ outer radius (R) = (r + 0.25) cm
                              = (5 + 0.25) cm = 5.25 cm
Outer curved surface = 2πR²
                                  = (2 × 22/7 × 5.25 × 5.25) cm²
                                  = 173.25 cm²

9. A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).

Answer

(i) The surface area of the sphere with raius r = 4πr²

(ii) The right circular cylinder just encloses a sphere of radius r.
∴ the radius of the cylinder = r and its height = 2r
∴ Curved surface of cylinder =2πrh
                                               = 2π × r × 2r
                                               = 4πr²
(iii) Ratio of the areas = 4πr²:4πr² = 1:1

Page No: 228

Exercise 13.5

1. A matchbox measures 4cm × 2.5cm × 1.5cm. What will be the volume of a packet containing 12 such boxes?

Answer

Dimension of matchbox = 4cm × 2.5cm × 1.5cm
l = 4 cm, b = 2.5 cm and h = 1.5 cm
Volume of one matchbox = (l × b × h)
                                         = (4 × 2.5 × 1.5) cm³= 15 cm³
Volume of a packet containing 12 such boxes = (12 × 15) cm³= 180 cm³

2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1m³ = 1000 l)

Answer

Dimensions of water tank = 6m × 5m × 4.5m
l = 6m , b = 5m and h = 4.5m
Therefore Volume of the tank =ℓbh m³
                                                   =(6×5×4.5)m3=135 m³
Therefore , the tank can hold = 135 × 1000 litres          [Since 1m3=1000litres]
                                                  = 135000 litres of water.

3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?

Answer

Length = 10 m , Breadth = 8 m and Volume = 380 m³
Volume of cuboid = Length × Breadth × Height
⇒ Height = Volume of cuboid/(Length × Breadth)
                = 380/(10×8) m
                = 4.75m

4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m³.

Answer

l = 8 m, b = 6 m and h = 3 m
Volume of the pit = lbh m³
                             = (8×6×3) m³
= 144 m³
Rate of digging = ₹30 per m³
Total cost of digging the pit = ₹(144 × 30) = ₹4320

5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

Answer

length = 2.5 m, depth = 10 m and volume = 50000 litres
1m³ = 1000 litres
∴ 50000 litres = 50000/1000 m³ = 50 m³
Breadth = Volume of cuboid/(Length×Depth)
             = 50/(2.5×10) m
             = 2 m

6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last?

Answer

Dimension of tank = 20m × 15m × 6m
l = 20 m , b = 15 m and h = 6 m
Capacity of the tank = lbh m³
                                 = (20×15×6) m³
                                 = 1800 m³
Water requirement per person per day =150 litres
Water required for 4000 person per day = (4000×150) l
                                                                = (4000×150)/1000
                                                                = 600 m³
Number of days the water will last = Capacity of tank Total water required per day
                                                        =(1800/600) = 3
The water will last for 3 days.

7. A godown measures 40m × 25m × 15m. Find the maximum number of wooden crates each measuring 1.5m × 1.25m × 0.5m that can be stored in the godown.

Answer

Dimension of godown = 40 m × 25 m × 15 m
Volume of the godown = (40 × 25 × 15) m³ = 10000 m³
Dimension of crates = 1.5m × 1.25m × 0.5m
Volume of 1 crates = (1.5 × 1.25 × 0.5) m³ = 0.9375 m³
Number of crates that can be stored =Volume of the godown/Volume of 1 crate
                                             = 10000/0.9375 = 10666.66 = 10666

8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Answer

Edge of the cube = 12 cm.
Volume of the cube = (edge)³ cm³
= (12 × 12 × 12) cm³
= 1728 cm³
Number of smaller cube = 8
Volume of the 1 smaller cube =1728/8 cm³ = 216 cm³
Side of the smaller cube = a
a³= 216
⇒ a = 6 cm
Surface area of the cube = 6 (side)²
Ratio of their surface area = (6 × 12 × 12)/(6 × 6 × 6)
                                           = 4/1 = 4:1

9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Answer

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Extra Questions and Answer from Chapter 14 Girgit Sparsh Bhaag II

निम्नलिखित गद्यांश को पढ़कर दिए गए प्रश्नों के उत्तर लिखिए -

1. "हुज़ूर! यह तो जनशांति भंग हो जाने जैसा कुछ दीख रहा है," सिपाही ने कहा।
ओचुमेलॉव मुड़ा और भीड़ की तरफ चल दिया। उसने काठगोदाम के पास बटन विहीन बास्केट धारण किये हुए उस आदमी को देखा, जो अपना दायाँ हाथ उठाये वहाँ मौजूद था तथा उपस्थित लोगों को अपना लुहलुहान ऊँगली दिखा रहा था। उसके नशीले-से हो आये चेहरे पर साफ लिखा था दिख रहा था - "शैतान की औलाद! मैं तुझे छोड़ने वाला नहीं! और उसकी ऊँगली भी जीत के झंडे की तरह गड़ी दिखाई दे रही थी। ओचुमेलॉव ने इस व्यक्ति को पहचान लिया। वह ख्यूक्रिन नामक सुनार था और इस भीड़ बीचोंबीच, अपनी अगली टाँगे पसारे, नुकीले मुँह और पीठ पर फैले पीले दागवाला, अपराधी-सा नज़र आता, सफ़ेद बारजोई पिल्ला, ऊपर से नीचे तक काँपता पसरा पड़ा था। उसकी आँसुओं से सनी आँखों में संकट आतंक की गहरी छाप थी।

(क) व्यक्ति की ऊँगली लहूलुहान कैसे हो गयी थी? (1)
(ख) घायल व्यक्ति के चेहरे पर कैसे भाव थे? (2)
(ग) कुत्ते की स्थिति कैसी थी? (2)

उत्तर

(क) कुत्ते द्वारा काटे जाने से व्यक्ति की ऊँगली लहूलुहान हो गयी थी।
(ख) घायल व्यक्ति के चेहरे पर कुत्ते के प्रति क्रोध के भाव थे। उसके चेहरे से लग रहा था कि वह उस कुत्ते को छोड़ने वाला नहीं है।
(ग) कुत्ता ऊपर से नीचे तक काँपता पसरा पड़ा था। उसकी आँसुओं से सनी आँखों में संकट और आतंक की गहरी छाप थी।

2. "हुज़ूर! मैं तो चुपचाप चला जा रहा था, "मुँह पर हाथ रखकर खाँसते हुए ख्यूक्रिन ने कहा -"मुझे मित्री मित्रिच से लकड़ी लेकर कुछ काम निपटाना था, तब अचानक इस कमबख्त ने अकारण मेरी ऊँगली काट खाई। माफ़ करें। आप तो जानते हैं मैं ठहरा कामकाजी आदमी... मेरा काम भी एकदम पेचीदा किस्म का है। मुझे लग रहा है एक हफ्ते तक मेरी यह ऊँगली अब काम करने लायक नहीं हो पाएगी। तो हुज़ूर! मेरी गुज़ारिश है कि इसके मालिकों से मुझे हरज़ाना तो दिलवाया जाए। यह तो किसी कानून में नहीं लिखा है हुज़ूर, कि आदमखोर जानवर हमें काट खाएँ और हम उन्हें बरदाश्त करते रहे। अगर हर कोई इसी तरह काट खाना शुरू कर दे तो हमारी ज़िंदगी तो नर्क हो जाए....

(क) उपर्युक्त कथन किसके हैं? (1)
(क) ख्यूक्रिन क्या काम करता था? उसकी ऊँगली कैसे कट गई? (2)
(ख) ख्यूक्रिन की क्या गुज़ारिश थी और क्यों? (2)

उत्तर

(क) उपर्युक्त कथन ख्यूक्रिन नामक व्यक्ति के हैं।
(ख) ख्यूक्रिन सुनार का काम करता था। कुत्ते द्वारा अकारण काटे जाने की वजह से उसकी ऊँगली कट गई।
(ग) ख्यूक्रिन की गुज़ारिश थी कि उसे कुत्ते के मालिक से मुआवज़ा मिलना चाहिए क्योंकि उसका काम पेचीदा है और ऊँगली काट जाने के कारण वह कई दिनों तक काम नहीं कर पाएगा।

3."मेरे ख्याल से यह जनरल झिगालॉव का है।" भीड़ में से एक आवाज़ उभरकर आई।
"जनरल झिगालॉव! हूँ येल्दीरीन मेरी कोट उतरवाने में मेरी मदद करो.... ओफ़्फ़ आज कितनी गरमी है। लग रहा है बारिश होकर रहेगी," वह ख्यूक्रिन की तरफ मुड़ा- "एक बात मेरी समझ में नहीं आती- आखिर इसने तुम्हें कैसे काट खाया? यह तुम्हारी ऊँगली तक पहुंचा कैसे? तू इतना लम्बा-तगड़ा आदमी और यह रत्ती भर का जानवर! जरूर ही तेरी ऊँगली पर कोई कील वगैरह गड़ गई होगी और तत्काल तूने सोचा होगा कि इसे कुत्ते के मत्थे मढ़कर कुछ हरज़ाना वगैरह ऐंठकर फायदा उठा लिया जाए। मैं तेरे जैसे शैतान लोगों को अच्छी तरह समझता हूँ।"

(क) किसे अचानक गर्मी लगने लगी? (1)
(ख) ख्यूक्रिन को कुत्ते ने नहीं काटा - इसे सिद्ध करने के लिए ओचुमेलॉव ने क्या तर्क दिए? (2)
(ग) ओचुमेलॉव के अनुसार ख्यूक्रिन ने ऊँगली कट जाने का दोष कुत्ते के मत्थे क्यों मढ़ा? (2)

उत्तर

(क) ओचुमेलॉव को अचानक गर्मी लगने लगी।
(ख) इसे सिद्ध करने के लिए ओचुमेलॉव ने ख्यूक्रिन से कहा कि तुम इतने लम्बे-तगड़े आदमी हो। यह छोटा जानवर तुम्हारी ऊँगली तक कैसे पहुँच सकता है।
(ग) ओचुमेलॉव के अनुसार ख्यूक्रिन ने ऊँगली कट जाने का दोष कुत्ते के मत्थे इसलिए मढ़ा क्योंकि वह कुत्ते के मालिक से हरजाना लेकर अपना फ़ायदा उठाना चाहता था।

4. "उधर देखो, जनरल साहब का बावर्ची आ रहा है। जरा उससे पता लगते हैं.... ओ प्रोखोर! इधर आना भाई। इस कुत्ते को तो पहचानो... क्या यह तुम्हारे यहाँ का है?"
"एक बार फिर से तो कहो! इस तरह का पिल्ला तो हमने कई जिंदगियों में नहीं देखा होगा।"
"अब अधिक जाँचने की जरूरत नहीं है," ओचुमेलॉव ने कहा- "यह आवारा कुत्ता है। इसके बारे में इधर खड़े होकर चर्चा करने की जरूरत नहीं है। मैं तुमसे पहले ही कह चुका हूँ कि यह आवारा है , तो है। इसे मार डालो और सारा किस्सा ख़त्म।"
"यह हमारा नहीं है," प्रोखोर ने आगे कहा- "यह तो जनरल साहब के भाई का है, जो थोड़ी देर पहले इधर पधारे हैं। अपने जनरल साहब को 'बारजोयस' नस्ल के कुत्तों में कोई दिलचस्पी नहीं है पर उनके भाई को यही नस्ल पसंद है।"

(क) ओचुमेलॉव ने किसे पुकारा? (1)
(ख) उसने कुत्ते की बारे में क्या बताया? (2)
(ग) ओचुमेलॉव ने कुत्ते को मार डालने को क्यों कहा? (2)

उत्तर

(क) ओचुमेलॉव ने जनरल साहब के बावर्ची प्रोखोर को पुकारा।
(ख) उसने बताया कि कुत्ता जनरल साहब का नहीं है। शायद कुत्ता जनरल साहब के भाई का है चूँकि उन्हें ही 'बारजोयस' नस्ल के कुत्ते पसंद हैं।

(ग) जब प्रोखोर ने बताया की कुत्ता जनरल साहब का नहीं है तब ओचुमेलॉव ने कुत्ते को आवारा बताते हुए मार डालने को कहा।

निम्नलिखित प्रश्नों के उत्तर दीजिये-

1. ख्यूक्रिन की बात पहली बार सुनने पर ओचुमेलॉव ने क्या कहा?

उत्तर

ख्यूक्रिन की बात पहली बार सुनने पर ओचुमेलॉव ने त्योरियाँ चढ़ाते हुए कहा कि वह कुत्ते के मालिक को छोड़ने नहीं वाला है। इस तरह कुत्ते को आवारा छोड़ कर कानून तोड़ने वाले मालिक को वह मजा चखाकर रहेगा। उस बदमाश आदमी को वह इतना जुर्माना ठोकेगा कि उसे पता चल जाए की जानवरों को आवारा छोड़ने का नतीजा क्या होता है।

3. येल्दीरीन ने ख्यूक्रिन को दोषी बताने के लिए क्या-क्या तर्क दिए?

उत्तर

येल्दीरीन ने ख्यूक्रिन को दोषी बताने के लिए कहा कि इसने ही अपनी जलती सिगरेट से इस कुत्ते की नाक यूँ ही जला डाली होगी वरना यह कुत्ता बेवकूफ है क्या जो इसे काट खाता। उसने ख्यूक्रिन को शैतान बताते हुए कहा कि वह हमेशा कोई ना कोई शरारत करता रहता है।

4. ओचुमेलॉव के अनुसार काटने वाला कुत्ता जनरल साहब का क्यों नहीं हो सकता था?

उत्तर

ओचुमेलॉव के अनुसार जनरल साहब के सभी कुत्ते महँगे और अच्छी नस्ल के हैं और काटने वाला कुत्ता मरियल-सा है इसलिए वह जनरल साहब का नहीं हो सकता था।

5. ख्यूक्रिन के अनुसार कानून किस बात की इजाजत नहीं देता?

उत्तर

ख्यूक्रिन के अनुसार कानून इस बात की इजाजत नहीं देता कि आदमखोर जानवर हमें काट खाएँ और हम उन्हें बरदाश्त करते रहें।

6. प्रोखोर ने कुत्ते के विषय में क्या बताया और ओचुमेलॉव पर इसका क्या प्रभाव पड़ा?

उत्तर

प्रोखोर ने कुत्ते के विषय में बताया कि वह कुत्ता जनरल साहब के भाई का है। इससे ओचेमेलॉव का व्यवहार पूरी तरह बदल गया। वह उस मरियल कुत्ते को प्यारा और मासूम डॉगी बताने लगा। उसने ख्यूक्रिन को भी धमकाया।

7. पाठ में गिरगिट के स्वभाव का कौन सा पात्र है और क्यों?

उत्तर

पाठ में गिरगिट के स्वभाव का इंस्पेक्टर ओचुमेलॉव है क्योंकि वह समय-समय पर खुद अपनी कही बातों से ही पलट जाता है।

8. 'गिरगिट' पाठ का मूल भाव स्पष्ट कीजिए।

उत्तर

'गिरगिट' पाठ में मूल रूप से शासन में फैले चापलूसी और भाई-भतीजावाद को दिखाया गया है। पुलिस लोगों की सुरक्षा छोड़कर अपने स्वार्थ के लिए अधिकारियों की चापलूसी करने में लगी है। ऐसे में आम आदमी न्याय से वंचित है।

9. पाठ का शीर्षक 'गिरगिट' की सार्थकता स्पष्ट कीजिए।

उत्तर

गिरगिट अपना रंग क्षण-क्षण बदलने के लिए मशहूर है। उसी तरह इस पाठ में भी इंस्पेक्टर ओचुमेलॉव अपनी बात, विचार और व्यवहार को पल-पल बदलते रहता है। जब वह मरियल कुत्ते को देखता है तब वह उसके मालिक को मजा चखाने की बात करता है परन्तु जब उसे पता लगता है कि कुत्ता जनरल साहब का है तब वह ख्यूक्रिन को ही दोषी ठहराने लगता है। फिर बाद में जब उसे लगता है कि कुत्ता जनरल साहब का नहीं है तब फिर वह मजा चखाने की बात करता है। इस तरह 'गिरगिट' शीर्षक इस पाठ के लिए उपयुक्त और सार्थक है।

गिरगिट - पठन सामग्री और सार

NCERT Solutions for Class 10th: पाठ 14- गिरगिट

NCERT Solutions for Class 9th: Ch 13 Surface Areas and Volumes Maths

Page No: 233

Exercise 13.7

1. Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm    (ii) radius 3.5 cm, height 12 cm

Answer

(i) Radius (r) = 6 cm
Height (h) = 7 cm
Volume of the cone = 1/3 πr²h
                                = (1/3 × 22/7 × 6 × 6 × 7) cm³
                                = 264 cm³
(ii) Radius (r) = 3.5 cm
Height (h) = 12 cm
Volume of the cone = 1/3 πr²h
                                = (1/3 × 22/7 × 3.5 × 3.5 × 12) cm³
                                = 154 cm³

2. Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm    (ii) height 12 cm, slant height 13 cm

Answer

(i) Radius (r) = 7 cm
Slant height (l) = 25 cm
Let h be the height of the conical vessel.
∴ h = √l²- r²
⇒ h = √25²- 7²
⇒ h = √576
⇒ h = 24 cm
Volume of the cone = 1/3 πr²h
= (1/3 × 22/7 × 7 × 7 × 24) cm³
                                = 1232 cm³
Capacity of the vessel = (1232/1000) l = 1.232 l

(i) Height (h) = 12 cm
Slant height (l) = 13 cm
Let r be the radius of the conical vessel.
∴ r = √l²- ^h²
⇒ r = √13²- 12²
⇒ r = √25
⇒ r = 5 cm
Volume of the cone = 1/3 πr²h
= (1/3 × 22/7 × 5 × 5 × 12) cm³
                                = (2200/7) cm³
Capacity of the vessel = (2200/7000) l = 11/35 l

3. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)

Answer

Height (h) = 15 cm
Volume = 1570 cm³
Let the radius of the base of cone be r cm
∴ Volume = 1570 cm³
⇒ 1/3 πr²h = 1570
⇒ 13 × 3.14 × r²× 15 = 1570
⇒ r²= 1570/(3.14×5) = 100
⇒ r = 10

4. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.

Answer

Height (h) = 9 cm
Volume = 48π cm³
Let the radius of the base of the cone be r cm
∴ Volume = 48π cm³
⇒ 1/3 πr²h = 48π
⇒ 13 × r²× 9 = 48
⇒ 3r²= 48
⇒ r²= 48/3 = 16
⇒ r = 4

5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Answer

Diameter of the top of the conical pit = 3.5 m
Radius (r) = (3.5/2) m = 1.75 m
Depth of the pit (h) = 12 m
Volume = 1/3 πr²h
             = (13 × 22/7 × 1.75 × 1.75 × 12) m³
             = 38.5 m³
1 m³= 1 kilolitre
Capacity of pit = 38.5 kilolitres.

6. The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find
(i) height of the cone      (ii) slant height of the cone
(iii) curved surface area of the cone

Answer

(i) Diameter of the base of the cone = 28 cm
Radius (r) = 28/2 cm = 14 cm
Let the height of the cone be h cm
Volume of the cone = 13πr²h = 9856 cm³
⇒ 1/3 πr²h = 9856
⇒ 1/3 × 22/7 × 14 × 14 × h = 9856
⇒ h = (9856×3)/(22/7 × 14 × 14)
⇒ h = 48 cm

(ii) Radius (r) = 14 m
Height (h) = 48 cm
Let l be the slant height of the cone
l² = h²+ r²
⇒ l² = 48²+ 14²
⇒ l² = 2304+196
⇒ l²= 2500
⇒ ℓ = √2500 = 50 cm

(iii) Radius (r) = 14 m
Slant height (l) = 50 cm
Curved surface area = πrl
                                 = (22/7 × 14 × 50) cm²
                                 = 2200 cm²

7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Answer

On revolving the ⃤ ABC along the side 12 cm, a right circular cone of height(h) 12 cm, radius(r) 5 cm and slant height(l) 13 cm will be formed.
Volume of solid so obtained = 1/3 πr²h
                                             = (1/3 × π × 5 × 5 × 12) cm³
                                             = 100π cm³

8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Answer

On revolving the ⃤ ABC along the side 12 cm, a cone of radius(r) 12 cm, height(h) 5 cm, and slant height(l) 13 cm will be formed.
Volume of solid so obtained =1/3 πr²h
                                              = (1/3 × π × 12 × 12 × 5) cm³
                                            = 240π cm³
Ratio of the volumes = 100π/240π = 5/12 = 5:12

9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Answer

Diameter of the base of the cone = 10.5 m
Radius (r) = 10.5/2 m = 5.25 m
Height of the cone = 3 m
Volume of the heap = 1/3 πr²h
                                = (1/3 × 22/7 × 5.25 × 5.25 × 3) m³
                                = 86.625 m³
Also,
l² = h²+ r²
⇒ l² = 3²+ (5.25)²
⇒ l² = 9 + 27.5625
⇒ l²= 36.5625
⇒ l = √36.5625 = 6.05 m
Area of canvas = Curve surface area
                         = πrl = (22/7 × 5.25 × 6.05) m²
                         = 99.825 m²(approx)

Page No: 236

Exercise 13.8

1. Find the volume of a sphere whose radius is
(i) 7 cm        (ii) 0.63 m

Answer

(i) Radius of the sphere(r) = 7 cm
Therefore, Volume of the sphere = 4/3 πr³
                                                     = (4/3 × 22/7 × 7 × 7 × 7) cm³
                                                     = 4312/3 cm³

(ii) Radius of the sphere(r) = 0.63 m
Volume of the sphere = 4/3 πr³
                                   = (4/3 × 22/7 × 0.63 × 0.63 × 0.63) m³
                                   = 1.05 m³

2. Find the amount of water displaced by a solid spherical ball of diameter.
(i) 28 cm                     (ii) 0.21 m

Answer

(i) Diameter of the spherical ball = 28 cm
Radius = 28/2 cm = 14 cm
Amount of water displaced by the spherical ball = Volume
                              = 4/3 πr³
                        = (4/3 × 22/7 × 14 × 14 × 14) cm³
                              = 34496/3 cm³

(ii) Diameter of the spherical ball = 0.21 m
Radius (r) = 0.21/2 m = 0.105 m
Amount of water displaced by the spherical ball = Volume
                              = 4/3 πr³
                              = (43×227×0.105×0.105×0.105) m³
                          = 0.004851 m³

3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?

Answer

Diameter of the ball = 4.2 cm
Radius = (4.2/2) cm = 2.1 cm
Volume of the ball = 4/3 πr³
                               = (4/3 × 22/7 × 2.1 × 2.1 × 2.1) cm³
                               = 38.808 cm³
Density of the metal is 8.9g per cm3
Mass of the ball = (38.808 × 8.9) g = 345.3912 g

4. The diameter of the moon is approximately one-fourth of the diameter of the earth.What fraction of the volume of the earth is the volume of the moon?

Answer

Let the diameter of the moon be r
Radius of the moon = r/2
A/q,
Diameter of the earth = 4r
Radius(r) = 4r/2 = 2r
Volume of the moon = v = 4/3 π(r/2)³
                                  = 4/3 πr³× 1/8
⇒ 8v = 4/3 πr³--- (i)
Volume of the earth = r³ = 4/3 π(2r)³
                            = 4/3 πr³× 8
⇒ V/8 = 4/3 πr³ --- (ii)
From (i) and (ii), we have
8v = V/8
⇒ v = 1/64 V
Thus, the volume of the moon is 1/64 of the volume of the earth.

5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Answer

Diameter of a hemispherical bowl = 10.5 cm
Radius(r) = (10.5/2) cm = 5.25cm
Volume of the bowl = 2/3 πr³
                                = (2/3 × 22/7 × 5.25 × 5.25 × 5.25) cm³
                                = 303.1875 cm³
Litres of milk bowl can hold = (303.1875/1000) litres
                                 = 0.3031875 litres (approx.)

6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Answer

Internal radius = r = 1m
External radius = R = (1 + 0.1) cm = 1.01 cm
Volume of iron used = External volume – Internal volume
                                  = 2/3 πR³- 2/3 πr³
                                  = 2/3 π(R³- r³)
                            = 2/3 × 22/7 × [(1.01)³−(1)³] m³
                                  = 44/21 × (1.030301 - 1) m³
                                  = (44/21 × 0.030301) m³
                                  = 0.06348 m³(approx)

7. Find the volume of a sphere whose surface area is 154 cm².

Answer

Let r cm be the radius of the sphere
So, surface area = 154cm²
⇒ 4πr²= 154
⇒ 4 × 22/7 × r²= 154
⇒ r²= (154×7)/(4×22) = 12.25
⇒ r = 3.5 cm
Volume = 4/3 πr³
             = (4/3 × 22/7 × 3.5 × 3.5 × 3.5) cm³
             = 539/3 cm³

8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹498.96. If the cost of white-washing is ₹2.00 per square metre, find the
(i) inside surface area of the dome,      (ii) volume of the air inside the dome.

Answer

(i) Inside surface area of the dome =Total cost of white washing/Rate of white washing
                                                        = (498.96/2.00) m²= 249.48 m²

(ii) Let r be the radius of the dome.
Surface area = 2πr²
⇒ 2 × 22/7 × r²= 249.48
⇒ r²= (249.48×7)/(2×22) = 39.69
⇒ r²= 39.69
⇒ r = 6.3m
Volume of the air inside the dome = Volume of the dome
                                                       = 2/3 πr³
= (2/3 × 22/7 × 6.3 × 6.3 × 6.3) m³
                                                       = 523.9 m³(approx.)

9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the
(i) radius r′ of the new sphere,    (ii) ratio of S and S′.

Answer

(i) Volume of 27 solid sphere of radius r = 27 × 4/3 πr³ --- (i)
Volume of the new sphere of radius r′ = 4/3 πr'³ --- (ii)
A/q,
4/3 πr'³= 27 × 4/3 πr³
⇒ r'³= 27r³
⇒ r'^³= (3r)³
⇒ r′ = 3r
(ii) Required ratio = S/S′ = 4 πr²/4πr′²= r²/(3r)²
                              = r²/9r² = 1/9 = 1:9

10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?

Answer

Diameter of the spherical capsule = 3.5 mm
Radius(r) = 3.52mm
              = 1.75mm
Medicine needed for its filling = Volume of spherical capsule
                                                  = 4/3 πr³
                                                  = (4/3 × 22/7 × 1.75 × 1.75 × 1.75) mm³
                                                  = 22.46 mm³(approx.)

Go to Part I (Exercise 3.1 to 3.3)
Go to Part II (Exercise 3.4 to 3.6)

Go To Chapters

NCERT Solutions for Class 10th: Ch 15 Probability Maths

Page No: 308

Exercise 15.1

1. Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = ___________ .
(ii) The probability of an event that cannot happen is __________. Such an event is called ________ .
(iii) The probability of an event that is certain to happen is _________ . Such an event is called _________ .
(iv) The sum of the probabilities of all the elementary events of an experiment is __________ .
(v) The probability of an event is greater than or equal to and less than or equal to __________ .

Answer

(i) Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) The probability of an event that cannot happen is 0. Such an event is called impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called sure or certain event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.

2. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.

Answer

(i) It does not have equally likely outcomes as it depends on various reasons like mechanical problems, fuels etc.
(ii) It does not have equally likely outcomes as it depends on the player how he/she shoots.
(iii) It has equally likely outcomes.
(iv)It has equally likely outcomes.

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Answer

Yes, tossing of a coin is a fair way of deciding which team should get the ball at the beginning of a football game because it has only two outcomes either head or tail. A coin is always unbiased.

4. Which of the following cannot be the probability of an event?
(A) 2/3 (B) -1.5 (C) 15% (D) 0.7

Answer

The probability of an event is always greater than or equal to 0 and less than or equal to 1.
Thus, (B) -1.5 cannot be the probability of an event.

5. If P(E) = 0.05, what is the probability of ‘not E’?

Answer

P(E) = 0.05
also, P(E) + P(not E) = 1
⇒ P(not E) = 1 - P(E)
⇒ P(not E) = 1 - 0.05
⇒ P(not E) = 0.95

6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?

Answer

(i) Since the bag contains only lemon flavoured.
Therefor, No. of orange flavoured candies = 0
Probability of taking out orange flavoured candies = 0/1 = 0

(ii) The bag only have lemon flavoured candies.
Probability of taking out lemon flavoured candies = 1/1 = 1

7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Answer

Let E be the event of having the same birthday.
P(E) = 0.992
⇒ P(E) + P(not E) = 1
⇒ P(not E) = 1 – P(E)
⇒ 1 - 0.992 = 0.008
The probability that the 2 students have the same birthday is 0.008

8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

Answer

No. of red balls = 3
No. of black balls = 5
Total no. of balls = 5+3 = 8
(i) Probability of drawing red balls = No. of red balls/Total no. of balls = 3/8

(ii) Probability of drawing black balls = No. of black balls/Total no. of balls = 5/8

9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ? (ii) white ? (iii) not green?

Answer

No. of red marbles = 5
No. of white marbles =8
No. of green marbles = 4
Total no. of balls = 5+8+4 = 17

(i) Favourable no. of elementary events = 5
Probability of taking out red marble = 5/17

(ii) Favourable no. of elementary events = 8
Probability of taking out red marble = 8/17

(iii) Favourable no. of elementary events = 4
Probability of taking out red marble = 4/17

Page No: 309

10. A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin ? (ii) will not be a ₹5 coin?

Answer

No. of 50p coins = 100
No. of ₹1 coins = 50
No. of ₹2 coins = 20
No. of ₹5 coins = 10
Total no. of coins = 100 + 50 + 20 + 10 = 180

(i) Favourable no. of elementary events = 100
Probability that it will be 50p coins = 100/180 = 5/9

(ii) Favourable no. of elementary events = 100+50+20 = 170
Probability that it will be 50p coins = 170/180 = 17/18

11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 15.4). What is the probability that the fish taken out is a male fish?

Answer

No. of male fish in the tank = 5
no. of female fish in the tank = 8
Total number of fish in the tank = 5 + 8 = 13
Favourable number events = 5
Probability of taking out a male fish = 5/13

12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at
(i) 8 ?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Answer

Possible no. of events = 8
(i) Favourable number of events = 1
Probability that it will point at 8 = 1/8

(ii) Odd numbers = 1, 3, 5 and 7
Favourable number of events = 4
Probability that it will be an odd number = 4/8 = 1/2

(iii) Numbers greater than 2 = 3, 4, 5, 6, 7 and 8
Favourable number of events = 6
Probability that a number greater than 4 = 6/8 = 3/4

(iv) Numbers less than 9 = 1,2,3,4,5,6,7,8
Favourable number of events = 8
Probability that a number less than 9 = 8/8 = 1

13. A die is thrown once. Find the probability of getting
(i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

Answer

Possible numbers of events on throwing a dice = 6
Numbers on dice = 1,2,3,4,5 and 6

(i) Prime numbers = 2, 3 and 5
Favourable number of events = 3
Probability that it will be a prime number = 3/6 = 1/2

(ii) Numbers lying between 2 and 6 = 3, 4 and 5
Favourable number of events = 3
Probability that a number between 2 and 6 = 3/6 = 1/2

(iii) Odd numbers = 1, 3 and 5
Favourable number of events = 3
Probability that it will be an odd number = 3/6 = 1/2

14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds

Answer

Possible numbers of events = 52

(i) Numbers of king of red colour = 2
Probability of getting a king of red colour = 2/52 = 1/26

(ii) Numbers of face cards = 12
Probability of getting a face card = 12/52 = 3/13

(iii) Numbers of red face cards = 6
Probability of getting a king of red colour = 6/52 = 3/26

(iv) Numbers of jack of hearts =1
Probability of getting a king of red colour = 1/526

(v) Numbers of king of spade = 13
Probability of getting a king of red colour = 13/52 = 1/4

(vi) Numbers of queen of diamonds = 1
Probability of getting a king of red colour = 1/52

15. Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Answer

Total numbers of cards = 5

(i) Numbers of queen = 1
Probability of picking a queen = 1/5

(ii) When queen is drawn and put aside then total numbers of cards left is 4
(a) Numbers of ace = 1
Probability of picking an ace = 1/4
(a) Numbers of queen = 0
Probability of picking a queen = 0/4 = 0

16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Answer

Numbers of defective pens = 12
Numbers of good pens = 132
Total numbers of pen = 132 + 12 = 144 pens
Favourable number of events = 132
Probability of getting a good pen = 132/144 = 11/12

17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer

(i) Total numbers of bulbs = 20
Numbers of defective bulbs = 4
Probability of getting a defective bulb = 4/20 = 1/5

(ii) One non defective bulb is drawn in (i) then the total numbers of bulb left is 19
Total numbers of events = 19
Favourable numbers of events = 19 - 4 = 15
Probability that the bulb is not defective = 15/19

18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.

Answer

Total numbers of discs = 50

(i) Total numbers of favourable events = 81
Probability that it bears a two-digit number = 81/90 = 9/10

(ii) Perfect square numbers = 1, 4, 9, 16, 25, 36, 49, 64 and 81
Favourable numbers of events = 9
Probability of getting a perfect square number = 9/90 = 1/10

(iii) Numbers which are divisible by 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90
Favourable numbers of events = 18
Probability of getting a number divisible by 5 = 18/90 = 1/5

Page No: 310

19. A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?

Answer

Total numbers of events = 6

(i) Total numbers of faces having A on it = 2

Probability of getting A = 2/6 = 1/3

(ii) Total numbers of faces having D on it = 1

Probability of getting A = 1/6

20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m?

Answer

Area of the rectangle = (3 × 2) m² = 6m²

Area of the circle = πr² = π(1/2)² m² = π/4 m²

Probability that die will land inside the circle = (π/4) × 1/6 = π/24

21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it?
(ii) She will not buy it?

Answer

Total numbers of pens = 144
Numbers of defective pens = 20
Numbers of non defective pens = 144 - 20 = 124

(i) Numbers of favourable events = 124
Probability that she will buy it = 124/144 = 31/36

(ii) Numbers of favourable events = 20
Probability that she will not buy it = 20/144 = 5/36

22. Refer to Example 13. (i) Complete the following table:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.

Answer

Events that can happen on throwing two dices are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total numbers of events : 6 × 6 = 36

(i) To get sum as 2, possible outcomes = (1,1)
To get sum as 3, possible outcomes = (1,2) and (2,1)
To get sum as 4, possible outcomes = (1,3); (3,1); and (2,2)
To get sum as 5, possible outcomes = (1,4); (4,1); (2,3); and (3,2)
To get sum as 6, possible outcomes = (1,5); (5,1); (2,4); (4,2); and (3,3)
To get sum as 7, possible outcomes = (1,6); (6,1); (5,2); (2,5); (4,3); and (3,4)
To get sum as 8, possible outcomes = (2,6); (6,2); (3,5); (5,3); and (4,4)
To get sum as 9, possible outcomes = (3,6); (6,3); (4,5); and (5,4)
To get sum as 10, possible outcomes = (4,6); (6,4) and (5,5)
To get sum as 11, possible outcomes = (5,6) and (6,5)
To get sum as 12, possible outcomes = (6,6)

Event: Sum on 2 dice	2	3	4	5	6	7	8	9	10	11	12
Probability	1/36	2/36	3/36	4/36	5/36	6/36	5/36	4/36	3/36	2/36	1/36

(ii) No, i don't agree with the argument. It is already justified in (i).

23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Answer

Events that can happen in tossing 3 coins = HHH, HHT, HTH, THH, TTH, HTT, THT, TTT
Total number of events = 8
Hinif will lose the game if he gets HHT, HTH, THH, TTH, HTT, THT
Favourable number of elementary events = 6
Probability of losing the game = 6/8 = 3/4

24. A die is thrown twice. What is the probability that
(i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Answer

(i) Consider the following events.
A = first throw shows 5,
B = second throw shows 5
P(A) = 6/36, P(B) = 6/36 and P(notB) = 5/6
⇒ P(notA) = 1– 6/36 = 30/36 = 5/6
Required probability = 5/6 × 5/6 = 25/36

(ii) Number of events when 5 comes at least once = 11
Probability = 11/36

Page No: 311

25. Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2

Answer

(i) The statement is incorrect
Possible events = (H,H); (H,T); (T,H) and (T,T)
Probability of getting two heads = 1/4
Probability of getting one of the each = 2/4 = 1/2

(ii) Correct. The two outcomes considered are equally likely.

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Extra Questions and Answer from Chapter 7 Dharm ki aad Sparsh Bhaag I

निम्नलिखित गद्यांश को पढ़कर दिए गए प्रश्नों के उत्तर लिखिए -

1. देश के सभी शहरों का यही हाल है। उबल पड़ने वाले साधारण आदमी का इसमें केवल इतना ही दोष है कि वह कुछ भी नहीं समझता-बुझता, और दूसरे लोग उसे जिधर जोत देते हैं, उधर जुट जाता है।यथार्थ दोष है, कुछ चलते-पुरज़े, पढ़े लिखे लोगों का, जो मूर्ख लोगों की शक्तियों का और उत्साह का दुरूपयोग इसलिए कर रहे हैं कि इस प्रकार, जाहिलों के आधार पर उनका नेतृत्व और बड़प्पन कायम रहे। इसके लिए धर्म और ईमान की बुराइयों से काम लेना उन्हें सबसे सुगम मालुम पड़ता है। सुगम है भी।

(क) कौन किसका दुरूपयोग कर रहा है और क्यों? (2)
(ख) साधारण आदमी का क्या दोष है? (1)
(ग) चलते-पुरज़े किन्हें कहा गया है और वे क्या करते हैं? (2)

उत्तर

(क) कुछ चालाक और पढ़े-लिखे लोग मूर्ख लोगों की शक्तियों और उत्साह का दुरूपयोग धर्म के नाम पर अपना नेतृत्व और बड़प्पन कायम रखने के लिए कर रहे हैं।
(ख) साधरण आदमी का दोष यह है की कुछ समझता-बुझता नहीं है, केवल उबल पड़ता है। दूसरे लोग जिधर जोत देते हैं उधर जुत जाता है।
(ग) चलते-पुरज़े कुछ पढ़े-लिखे लोगों को कहा गया है। वे लोग धर्म और ईमान की बुराइयों से लाभ उठाकर मूर्ख लोगों की शक्तियों का दुरुपयोग अपने फायदे के लिए करते हैं।

2. हमारे देश में धनपतियों का इतना ज़ोर नहीं है। यहाँ धर्म में नाम पर कुछ इन-गिने आदमी अपने हैं स्वार्थों की सिद्धि के लिए करोड़ों आदमियों की शक्ति का दुरूपयोग किया करते हैं। गरीबों का धनाढ्यों द्वारा चूसा इतना बुरा नहीं है, जितना बुरा यह है कि वहाँ है धन की मार, यहाँ पर है बुद्धि की मार। वहाँ धन दिखाकर करोड़ो को वश में किया जाता है और फिर मनमाना धन पैदा करने के लिए जोत दिया जाता है। यहाँ है बुद्धि पर परदा डालकर पहले ईश्वर और आत्मा का स्थान अपने लिए लेना और फिर धर्म, ईमान, ईश्वर और आत्मा के नाम पर अपनी स्वार्थ सिद्धि के लिए लोगों को लड़ाना-भिड़ाना।

(क) पाश्चात्य देशों और भारत में क्या अंतर है? (2)
(ख) आज धर्म के नाम पर क्या होता है? (1)
(ग) 'बुद्धि पर परदा डालना' का क्या अर्थ है? (1)

उत्तर

(क) पाश्चात्य देशों में धन दिखाकर लोगों को वश में करते हैं और भारत में धर्म के नाम पर लोगों की शक्ति को दुरूपयोग किया जा रहा है।
(ख) आज धर्म के नाम पर कुछ गिने-चुने लोग अपने स्वार्थ सिद्धि क लिए करोड़ों लोगों की शक्तियों करते हैं।
(ग) 'बुद्धि पर परदा डालना' का अर्थ है कुछ सोचने-समझने की ताकत को खत्म करना जैसा आज के समय में धर्म के नाम पर किया जा रहा है।

3. धर्म की उपासना के मार्ग में कोई रुकावट न हो। जिसका मन जिस प्रकार चाहे, उसी प्रकार धर्म की भावना को अपने मन में जगावे। धर्म और ईमान, मन का सौदा हो, ईश्वर और आत्मा के बीच का संबंध हो, आत्मा को शुद्ध करने और ऊँचा उठाने का साधन हो। वह किसी दशा में भी, किसी दूसरे व्यक्ति की स्वाधीनता को छीनने या कुचलने का साधन न बने। आपका मन चाहे, उस तरह का धर्म आप मानें और दूसरों का मन चाहे, उस प्रकार का धर्म वह माने। दोनों भिन्न धर्मों को मानने वालों को टकरा जाने के लिए कोई भी स्थान न हो। यदि किसी धर्म के मानने वाले कहीं जबरदस्ती टाँग अड़ाते हों, तो उनका इस प्रकार का कार्य देश की स्वाधीनता के विरुद्ध समझा जाए।

(क) धर्म किस बात का साधन है? (1)
(ख) विभिन्न धर्मों का संबंध कैसा होना चाहिए? (2)
(ग) कौन सा कार्य देश की स्वाधीनता के विरुद्ध समझा जाए? (2)

उत्तर

(क) धर्म आत्मा को शुद्ध करने और ऊँचा उठाने का साधन है।
(ख) विभिन्न धर्मों का संबंध ऐसा होना चाहिए कि उनके मानने वालों के टकरा जाने के लिए कोई भी स्थान ना हो।
(ग) यदि किसी धर्म के मानने वाले कहीं जबरदस्ती टाँग अड़ाते हों तो उनका यह कार्य देश की स्वाधीनता के विरुद्ध समझा जाए।

4. देश की स्वाधीनता के लिए जो उद्योग किया जा रहा था, उसका वह दिन निःसंदेह, बुरा था, जिस दिन स्वाधीनता के क्षेत्र में ख़िलाफ़त, मुल्ला, मौलवियों और धर्माचार्यों को स्थान दिया जाना आवश्यक समझा गया। एक प्रकार से उस दिन हमने स्वाधीनता के क्षेत्र में, एक कदम पीछे हटकर रखा था। अपने उसी पाप का फल आज हमें भोगना पड़ रहा है। देश को स्वाधीनता के संग्राम ही ने मौलना अब्दुल बारी और शंकराचार्य को देश के सामने दूसरे रूप में पेश किया, उन्हें अधिक शक्तिशाली बना दिया और हमारे इस काम का फल यह हुआ कि इस समय, हमारे हाथों से ही बढ़ाई इनकी और इनके से लोगों की शक्तियाँ हमारी जड़ उखाड़ने और देश में मज़हबी पागलपन, प्रपंच और उत्पात का राज्य स्थापित कर रही हैं।

(क) देश की स्वाधीनता का कौन सा दिन सबसे बुरा था? (1)
(ख) हमने कब स्वाधीनता के क्षेत्र में एक कदम पीछे हटकर रखा और क्यों? (2)
(ग) हमारे मज़हबी कार्य का फल क्या हुआ? (2)

उत्तर

(क) देश की स्वाधीनता का वह दिन सबसे बुरा था जिस दिन स्वाधीनता के क्षेत्र में ख़िलाफ़त, मुल्ला, मौलवियों और धर्माचार्यों को स्थान दिया गया।
(ख) स्वाधीनता के क्षेत्र में हमने धर्म को स्थान देकर एक कदम पीछे हटकर रखा क्योंकि हमने मज़हब को स्थान दिया जिससे टकराव की स्थिति और बढ़ गयी।
(ग) हमारे मज़हबी कार्यों का फल यह हुआ कि हमारे द्वारा बढ़ाई गई मुल्ला-मौलवियों और धर्माचार्यों की शक्ति हमारी जड़ें उखाड़ने, मज़हबी पागलपन, प्रपंच और उत्पात का राज्य स्थापित कर रही हैं।

5. ऐसे धार्मिक और दीनदार आदमियों से तो वे ला-मज़हब और नास्तिक आदमी कहीं अधिक अच्छे और ऊँचे हैं, जिनका आचरण अच्छा है, जो दूसरों के सुख-दुःख का ख्याल रखते हैं और जो मूर्खों को किसी स्वार्थ सिद्धि के लिए उकसाना बहुत बुरा समझते हैं। ईश्वर इन नास्तिक और ला-मज़हब लोगों को अधिक प्यार करेगा और वह अपने पवित्र नाम पर अपवित्र काम करने वालों से यही कहना पसंद करेगा, मुझे मानो या ना मानो, तुम्हारे मानने से ही मेरा ईश्वरत्व कायम नहीं रहेगा, दया करके मनुष्यत्व को मानो, पशु बनना छोडो और आदमी बनो।

(क) कौन लोग किससे अधिक अच्छे हैं? (2)
(ख) ईश्वर किन लोगों से प्यार करेगा? (1)
(ग) 'दया करके मनुष्यत्व को मानो, पशु बनना छोडो और आदमी बनो।' इस पंक्ति का क्या अर्थ है? (2)

उत्तर

(क) ला-मज़हबी और नास्तिक लोग जिनका आचरण अच्छा है, धार्मिक और ईमानदार लोगों से अच्छे हैं।
(ख) ईश्वर उनलोगो से अधिक प्यार करेगा जिनका आचरण अच्छा है, जो दूसरों लोगों के सुख-दुःख का ख्याल करते हैं और जो मूर्खों को किसी स्वार्थ सिद्धि के लिए उकसाना बहुत बुरा समझते हैं।
(ग) इस पंक्ति का अर्थ है कि हमें ईश्वर को मानने या ना मानने से पहले मनुष्यता को मानना चाहिए। अपने स्वार्थ के लिए धर्म के नाम पर उत्पात नहीं मचाना चाहिए। हिंसा रूपी पशु को त्यागकर दूसरों की भलाई का काम करना चाहिए।

निम्नलिखित प्रश्नों के उत्तर दीजिये-

1. आज धर्म और ईमान के नाम पर कौन-कौन से ढोंग किये जाते हैं?

उत्तर

आज धर्म और ईमान के नाम पर उत्पात, जिद और झगडे करवाये जाते हैं। अपने स्वार्थ को पूरा करने लिए धर्म को साधन बनाया जाता है और दंगे कराये जाते हैं। आम आदमी धर्म को जाने या ना जाने परन्तु धर्म के नाम पर जान देने और लेने के लिए तैयार हो जाता है।

2. पाश्चात्य देशों और हमारे देश में क्या अंतर है? पाठ के आधार पर लिखिए।

उत्तर

पाश्चात्य देशों में धन का बोलबाला है। वहाँ धनी लोग गरीब लोगों को धन दिखाकर उनका शोषण करते हैं। हमारे देश में धन का उतना ज़ोर नहीं है। यहाँ कुछ लोग बुद्धि पर पर्दा डाल धर्म के नाम पर स्वार्थ सिद्धि के लिए लोगों को आपस में भिड़ाते हैं।

3. लेखक के अनुसार धर्म की भावना कैसी होनी चाहिए?

उत्तर

लेखक के अनुसार धर्म का विषय व्यक्ति के मन के ऊपर हो। जिसका मन जिस प्रकार चाहे उसी प्रकार का धर्म माने। यह आत्मा को शुद्ध करने और ऊँचा उठाने का साधन है। यह किसी दूसरे व्यक्ति की स्वाधीनता को छीनने या कुचलने का साधन ना बने।

4. अजाँ देने, शंख बजाने, नाक दबाने और नमाज़ पढ़ने का नाम धर्म नहीं है। पाठ के आधार पर स्पष्ट कीजिए।

उत्तर

घंटों पूजा कर, शंख बजाकर और पंच-वक्ता नमाज़ अदा कर कोई सच्चा धार्मिक नहीं हो जाता। ऐसा करने के बाद अगर व्यक्ति बुरे काम में लिप्त है तो यह धर्म का पालन नहीं हुआ। शुद्धाचरण और सदाचरण ही सच्चा धर्म है। अगर आपका आचरण अच्छा नहीं है तो पूजा-पाठ और नमाज़ अदायगी व्यर्थ के कार्य हैं।

धर्म की आड़ - पठन सामग्री और सार

NCERT Solutions for Class 9th: पाठ 7- धर्म की आड़

Extra Questions and answers, study material and notes of Physical Education of Class 11th

The chapters, questions and answers and notes are full based on the latest syllabus as prescribed by CBSE. Students can take help from these study materials for the preparation of their Class 11th Physical Education Board Examination.

UNIT I - Changing Trends & Career in Physical Education
• Define Phy. Edu., Its Aims & Objectives
• Development of Phy. Edu. - Post Independence
• Concept & Principles of Integrated Phy. Edu.
• Concept & Principles of Adaptive Phy. Edu.
• Special Olympic Bharat
• Career Options In Phy. Edu.

UNIT II - Physical Fitness, Wellness & Lifestyle
• Meaning & Importance of Physical Fitness, Wellness & Lifestyle
• Components of physical fitness
• Components of wellness
• Preventing Health Threats Through Lifestyle Change
• Components of Positive Lifestyle

UNIT III - Olympic Movement
• Ancient & Modern Olympics
• Olympic Symbols, Ideals, Objectives & Values
• International Olympic Committee
• Indian Olympic Association
• Dronacharya Award, Arjuna Award & Rajiv Gandhi Khel Ratna Award
• Organisational set-up of CBSE Sports & Chacha Nehru Sports Award
• Paralympic Movement

UNIT IV - Yoga
• Meaning & Importance of Yoga
• Yoga as an Indian Heritage
• Elements of Yoga
• Introduction to - Asanas, Pranayam, Mediation & Yogic Kriyas
• Physiological benefits of Asana & Pranayam
• Prevention & Management of Common Lifestyle Diseases; Obesity, Asthma, Diabetes, Hyper-
Tension & Back-Pain

Unit V - Doping
• Concept & classification of doping
• Prohibited Substances & Methods
• Athletes Responsibilities
• Side Effects of Prohibited Substances
• Ergogenic aids & doping in sports
• Doping control procedure

Unit VI - Physical Activity Environment
• Introduction to physical activity
• Concept & need of sports environment
• Essential elements of positive sports environment
• Principles of physical activity environment
• Components of health related fitness
• Behaviour change technique for physical activity
• Exercise Guidelines at different stages of growth

UNIT VII - Test & Measurement in Sports
• Define Test & Measurement
• Importance of Test & Measurement In Sports
• Calculation of BMI & Waist - Hip Ratio
• Somato Types (Endomorphy, Mesomorphy & Ectomorphy)
• Procedures of Anthropromatric Measurement – Height, Weight, Arm & Leg Length And Skin
Fold

UNIT VIII - Fundamentals of Anatomy & Physiology
• Define Anatomy, Physiology & Its Importance
• Function of Skeleton System, Classification of Bones & Types of Joints
• Properties of Muscles
• Function & Structure of Muscles
• Function & Structure of Respiratory System, Mechanism of Respiration
• Structure of Heart & Introduction to Circulatory System
• Oxygen debt, second-wind

Unit IX - Biomechanics & Sports
• Meaning & Importance of Biomechanics in Phy. Edu. & Sports
• Newton‘s Law of Motion and its application in sports
• Levers & Its Types and its application in sports
• Equilibrium – Dynamic & Static and Centre of Gravity and its application in sports
• Force – Centrifugal & Centripetal and its application in sports

Unit X - Psychology & Sports
• Definition & Importance of Psychology in Phy. Edu. & Sports
• Define & Differentiate Between Growth & Development
• Developmental Characteristics at Different Stage of Development
• Adolescent Problems & Their Management
• Define Learning, Laws of Learning & Transfer of Learning
• Plateau & causes of plateau
• Emotion: Concept & controlling of emotion

Study Material and Notes of Define Physical Education, Its Aims & Objectives Class 11th Physical Education

What is Physical Education?

Literal meaning

Physical = related to body
education = knowledge

Thus,

Physical Education is knowledge about the body. In this subject, we read about various activities which are necessary for development and growth of the body.

There are so many definitions about Physical education given by various experts.

According to Oberteuffer, Physical Education is the sum of those experiences which comes to individual through movements.

According to J.B.Nash, Physical Education is that phase of the whole field of the education that deals with big muscle activities and their related responses.

According to H.C. Buck, Physical education is the part of general education programme, which is considered wit growth, development and education of children through the medium of big muscle activities. Physical activities are the tools. They are so selected and conducted as to influence every child's life physically, mentally, emotionally and morally.

According to Charles A Bucher, Physical education is an integral part of the total educational process and has as its aim the development physically, mentally, emotionally, and socially fit citizens through the medium of physical activities which have been selected with a view to realizing these outcomes.

According to C.C.Cowell, Physical education is the social process of change in the behavior of the human organism, originating primarily from the stimulus of social big-muscle play and related activities.

According to American Association for Health, Physical Education and Recreation (AAHPER),
Physical education is a way of education through physical activities which are selected and carried on with full regard to values in human growth, development, and behavior.

Aim of Physical education

The aim of physical education is the wholesome development of personality of an individual which means making an individual physical fit, mentally alert, emotionally balanced, socially well adjusted, morally true and spiritually uplifted.

About aim of Physical education:

According to J.F. Williams, Physical education should aim to provide skilled leadership, adequate facilities and ample time that will afford an opportunity for the individuals or groups to act in situations that are physically wholesome, mentally stimulating and satisfying, and socially sound.

According to National Planning of Physically Education and Recreation, The aim of physical education must be to make every child physically, mentally and emotionally fit and also to develop in him such personal and social qualities as will help him to live happily with others and build him up a good citizen.

Objectives of Physically education

Objectives are steps towards the attainment of aim. When an aim is achieved it becomes an objective.

There are different objectives of Physical education brought by the experts in this field:

According to J.B. Nash
• Organic Development
• Neuro-muscular development
• Interpretive development
• Emotional development

According to Irwin
• Physical
• Social Development
• Emotional Development
• Recreational development of skills
• Intellectual development

Therefore, the objectives of Physical education are:

• Organic Development: The primary objective is the development of our organic systems, such as respiratory system, circulatory system, digestive system, nervous system, muscular systems. Physical activities and exercises have various effects on our organic systems which increase efficiency, capacity, shape and size.

• Social Development: Various physical activities programs gives individual opportunity for social contact and group living which help them to adjust themselves in different situations and building relations. The qualities like cooperation, obedience, temperament, sacrifice, loyalty, sportsmanship, self-confidence develop. These developments help them to become good human being and forms a healthy society.

• Neuro-muscular co-ordination: The physical activities help in maintaining a better relationship between nervous system and muscular system. The devlopment neuro-muscular co-ordination develops control and balance of the body. Various games develops our ability of actvites such as running, bouncing, catching etc. It also helpful in proper use of energy.

• Emotional Development: The programs of physical education tell us how to control our various types of emotions such as anger, pleasure, jealousy, fear loneliness etc. It makes an individual emotinally balanced.

• Health Development: Physical education develops health through health education. It develops healthy habits of sleeping, exercises, food etc. It also reduces worries and anxieties through developing appropriate interests and habits of engaging in exercise and games.

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NCERT Solutions for Class 12th: Ch 11 Biotechnology: Principles and Processes Biology

Page No: 205

Exercises

1. Can you list 10 recombinant proteins which are used in medical practice? Find out where they are used as therapeutics (use the internet).

Answer

Recombinant proteins are obtained from the recombinant DNA technology. This technology involves the transfer of specific genes from an organism into another organism using vectors and restriction enzymes as molecular tools.

Ten recombinant proteins used in medical practice are:
(i) Insulin: used for the treatment of diabetes mellites
(ii) Interferon-α: Used for chronic hepatitis C
(iii) Interferon: Used for herpes and viral enteritis
(iv) Coagulation factor VII: Treatment of haemophilia A

(v) Coagulation factor IX: Treatment of haemophilia B

(vi) DNAase I: Treatment of cystic fibrosis
(vii) Anti-thrombin III: Prevention of blood clot
(viii) Interferon B: For treatment of multiple sclerosis
(ix) Human recombinant growth hormone: For promoting growth in an individual
(x) Tissue plasminogen activator: Treatment of acute myocardial infection

2. Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the product it produces.

Answer

3. From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?

Answer

Enzymes are smaller in size than DNA molecules. This is because DNA contains genetic information for the development and functioning of all living organisms. It contains instructions for the synthesis of proteins and DNA molecules.while, enzymes are proteins which are synthesized from a small strend of DNA known as 'genes', which are involved in the formatoin of the polypeptide chain.

4. What would be the molar concentration of human DNA in a human cell? Consult your teacher.

Answer

The molar concentration of human DNA in a human diploid cell is as follows:

⇒ Total number of chromosomes × 6.023× 10²³

⇒ 46 × 6.023× 10²³

⇒ 2.77 ×10²³ moles

Hence, the molar concentration of DNA in each diploid cell in humans is 2.77 ×10²³moles.

5. Do eukaryotic cells have restriction endonucleases? Justify your answer.

Answer

No, eukaryotic cells do not have restriction endonucleases. This is because the DNA of eukaryotes is highly methylated by a modified enzyme, called methylase. Methylation protects the DNA from the activity of restriction enzymes .These enzymes are present in prokaryotic cells where they help prevent the invasion of DNA by virus.

6. Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?

Answer

The shake flask method is used for a small-scale production of biotechnological products in a laboratory. whereas stirred tank bioreactors are used for a large-scale production of biotechnology products.
Stirred tank bioreactors have several advantages over shake flasks:
(i) Small volumes of culture can be taken out from the reactor for testing.
(ii) It has a foam breaker for regulating the foam.
(iii) It has a control system that regulates the temperature and pH.

7. Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.

Answer

The palindromic sequence is a certain sequence of the DNA that reads the same whether read from 5' → 3' direction or from 3' → 5' direction. They are the site for the action of restriction enzymes. Most restriction enzymes are palindromic sequences.
Five examples of palindromic sequences are-
(i) 5'-AGCT-3'
3'-TCGA-5'

(ii) 5'-GAATTC-3'
3'-CTTAAG-5'

(iii) 5'-AAGCTT-3'
3'-TTCGAA-5'

(iv) 5'-GTCGAC-3'
3'-CAGCTG-5'

(v) 5'-CTGCAG-3'
3'-GACGTC-5'

8. Can you recall meiosis and indicate at what stage a recombinant DNA is made?

Answer

Meiosis is a process that includes the reduction in the amount of genetic material. It is of two types, namely meiosis I and meiosis II. During the pachytene stage of prophase I, crossing over of chromosomes takes place where the exchange of segments between non-sister chromatids of homlogous chromosomes takes place. This results in the formation of recombinant DNA.

9. Can you think and answer how a reporter enzyme can be used to monitor transformation of host cells by foreign DNA in addition to a selectable marker?

Answer

A reporter gene can be used to monitor the transformation of host cells by foreign DNA.They act as a selectable marker to determine whether the host cell has taken up the foreign DNA or the foreign gene gets expressed in the cell. The researchers place the reporter gene and the foreign gene in the same DNA construct. Then, this combined DNA construct is inserted in the cell. then, the reporter gene is used as a selectable marker to find out the successful uptake of genes of interest .
Example of reporter genes - lac Z gene, which encodes a green fluorescent protein in a jelly fish.

Page No: 206

10. Describe briefly the following:
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing

Answer

(a) Origin of replication -Origin of replication is defined as the DNA sequence in a genome from where replication initiates. The initiation of replication can be either uni-directional or bi-directional. A protein complex recognizes the 'on' site, unwinds the two strands, and initiates the copying of the DNA.
(b) Bioreactors - Bioreactors are large vessels used for the large-scale production of biotechnology products from raw materials. They provide optimal conditions to obtain the desired product by providing the optimum temperature, pH, vitamin, oxygen, etc. Bioreactors have an oxygen delivery system, a foam control system, a PH, a temperature control system, and a sampling port to obtain a small volume of culture for sampling.
(c) Downstream processing - Downstream processing is a method of separation and purification of foreign gene products after the completion of the biosynthetic stage. The product is subjected to various processes in order to separate and purify the product. After downstream processing, the product is formulated and is passed through various clinical trials for quality control and other tests.

11. Explain briefly
(a) PCR
(b) Restriction enzymes and DNA
(c) Chitinase

Answer

(a) PCR: - Polymerase chain reaction (PCR) is a technique in molecular biology to amplify a gene or a piece of DNA to obtain its several copies. It is extensively used in the process of gene manipulation. The process involves in-vitro synthesis of sequences using a primer, a template strand, and a thermostable DNA polymerase enzyme obtained from a bacterium, called Thermus aquaticus. The enzyme utilizes building blocks dNTPs (deoxynucleotides) to extend the primer. In the first step, the double stranded DNA molecules are heated to a high temperature so that the two strands separate into a single stranded DNA molecule. This process is called denaturation. Then, this ssDNA molecule is used as a template strand for the synthesis of a new strand by the DNA polymerase enzyme and this process is called annealing, which results in the duplication of the original DNA molecule. This process is repeated over several cycles to obtain multiple copies of the rDNA fragment.

(b) Restriction enzymes are molecular scissors used in molecular biology for cutting DNA sequences from a specific site. It plays an important role in gene manipulation. The enzymes recognize a specific six-box pair sequence known as the recognition sequence and cut the sequence at a specific site. For example, the recognition site for enzyme ECORI is as follows
Restriction enzyme are categorized into two types:
(i) Exonuclease: It is a type of restriction enzyme that removes the nucleotide from either 5' or 3' ends of the DNA molecule.
(ii) Endonuclease: It is a type of restriction enzyme that makes a cut within the DNA at a specific site. This enzyme acts as an important tool in genetic engineering. It is commonly used to make a cut in the sequence to obtain DNA fragments with sticky ends, which are later joined by enzyme DNA ligase.

(c) Chitinase - Chitinase is a class of enzymes used for the degradation of chitin, which forms a major component of the fungal cell wall. Therefore, to isolate the DNA enclosed within the cell membrane of the fungus, enzyme chitinase is used to break the cell for releasing its genetic material.

12. Discuss with your teacher and find out how to distinguish between
(a) Plasmid DNA and Chromosomal DNA
(b) RNA and DNA
(c) Exonuclease and Endonuclease

Answer

(a) Plasmid DNA and Chromosomal DNA

Plasmid DNA is an extra-chromosomal DNA molecule in bacteria that is capable of replicating, independent of chromosomal DNA.

Chromosomal DNA is the entire DNA of an organism present inside chromosomes.

(b) RNA and DNA

RNA is a single stranded molecule.	DNA is a double stranded molecule.
It contains ribose sugar.	It contains deoxyribose sugar.
The pyrimidines in RNA are adenine and uracil.	The pyrimidines in DNA are adenine and thymine.
RNA cannot replicate itself.	DNA molecules have the ability to replicate.
It is a component of the ribosomes.	It is a component of the chromosomes.

It is a type of restriction enzyme that removes the nucleotide from 5' or 3' ends of the DNA molecule.

It is a type of restriction enzyme that makes a cut within the DNA at a specific site to generate sticky ends.

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NCERT Solutions for Class 8th: Ch 1 Rational Numbers Maths

Page No: 14

Exercise 1.1

1. Using appropriate properties find.
(i) -2/3 × 3/5 + 5/2 - 3/5 × 1/6   (ii) 2/5 × (-3/7) - 1/6 × 3/2 + 1/14 × 2/5

Answer

(i) -2/3 × 3/5 + 5/2 - 3/5 × 1/6
= -2/3 × 3/5 - 3/5 × 1/6 + 5/2    (by commutativity)
= 3/5(-2/3 - 1/6) + 5/2
= 3/5{(-4 - 1)/6} + 5/2
= 3/5(-5/6) + 5/2    (by distributivity)
= -15/30 + 5/2
= -1/2 + 5/2
= 4/2 = 2

(ii) 2/5 × (-3/7) - 1/6 × 3/2 + 1/14 × 2/5
= 2/5 × (-3/7) + 1/14 × 2/5 - (1/6 × 3/2)    (by commutativity)
= 2/5(-3/7 + 1/14) - 1/4
= 2/5{(-6 + 1)/14} - 1/4    (by distributivity)
= 2/5(-5/14) - 1/4
= -1/7 - 1/4
= (-4-7)/28
= -11/28

2. Write the additive inverse of each of the following.
(i) 2/8   (ii) -5/9   (iii) -6/-5   (iv) 2/-9   (v) 19/-6

Answer

(i) 2/8
Additive inverse = -2/8
(ii) -5/9
Additive inverse = 5/9
(iii) -6/-5 = 6/5
Additive inverse = -6/5
(iv) 2/-9 = -2/9
Additive inverse = 2/9
(v) 19/-6 = -19/6
Additive inverse = 19/6

3. Verify that : -(-x) = x for.
(i) x = 11/15 (ii) x = -13/17

Answer

(i) x = 11/15
The additive inverse of x = 11/15 is -x = -11/15 as 11/15 + (-11/15) = 0
The same equality 11/15 + (-11/15) = 0 , shows that the additive inverse of -11/15 is 11/15 or
-(-11/15) = 11/15 i.e. -(-x) = x

(ii) x = -13/17
The additive inverse of x = -13/17 is -x = 13/17 as (-13/17) + 13/17 = 0
The same equality 13/17 + (-13/17) = 0 , shows that the additive inverse of 13/17 is -13/17 or
-(13/17) = -13/17 i.e. -(-x) = x

4. Find the multiplicative inverse of the following.
(i) -13    (ii) -13/19    (iii) 1/5    (iv) -5/8 × -3/7    (v) -1 × -2/5    (vi) -1

Answer

The multiplicative inverse of a number is the reciprocal of that number.

(i) -13
Multiplicative inverse = -1/13
(ii) -13/19
Multiplicative inverse = -19/13
(iii) 1/5
Multiplicative inverse = 5
(iv) -5/8 × -3/7 = 15/56
Multiplicative inverse = 56/15
(v) -1 × -2/5 = 2/5
Multiplicative inverse = 5/2
(vi) -1
Multiplicative inverse = -1

5. Name the property under multiplication used in each of the following.
(i) -4/5 × 1 = 1 × -4/5 = -4/5
(ii) -13/17 × -2/7 = -2/7 × -13/17
(iii) -19/29 × 29/-19 = 1

Answer

(i) -4/5 × 1 = 1 × -4/5 = -4/5
Here 1 is the multiplicative identity.
(ii) -13/17 × -2/7 = -2/7 × -13/17
Commutavity
(iii) -19/29 × 29/-19 = 1
Multiplicative inverse

6. Multiply 6/13 by the reciprocal of -7/16.

Answer

Reciprocal of -7/16 = 16/-7
A/q,
6/13 × (Reciprocal of -7/16)
= 6/13 × 16/-7 = 96/-91 = -96/91

7. Tell what property allows you to compute 1/3 × (6 × 4/3) as (1/3 × 6) × 4/3.

Answer

By the property of associativity.
8. Is 8/9 the multiplicative inverse of

? Why or why not?

Answer

If it will be the multiplicative inverse then their product will be 1.

= -7/8
A/q,
8/9 × -7/8 = -7/9 ≠ 1
Hence, 8/9 is not the multiplicative inverse.

9. Is 0.3 the multiplicative inverse of

? Why or why not?

Answer

If it will be the multiplicative inverse then their product will be 1.

= 10/3

also, 0.3 = 3/10

A/q,
3/10 × 10/3 = 1
Hence, 0.3 is the multiplicative inverse.

Page No: 15

10. Write.
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.

Answer

(i) 0 is the rational number that does not have a reciprocal.

(ii) 1 and -1 are the rational numbers that are equal to their reciprocals.

(iii) 0 is the rational number that is equal to its negative.

11. Fill in the blanks.
(i) Zero has ________ reciprocal.
(ii) The numbers ________ and ________ are their own reciprocals
(iii) The reciprocal of – 5 is ________.
(iv) Reciprocal of 1/x, where x ≠ 0 is ________.
(v) The product of two rational numbers is always a _______.
(vi) The reciprocal of a positive rational number is ________.

Answer

(i) Zero has no reciprocal.

(ii) The numbers 1 and -1 are their own reciprocals
(iii) The reciprocal of -5 is -1/5.
(iv) Reciprocal of 1/x, where x ≠ 0 is x.
(v) The product of two rational numbers is always a rational numbers.
(vi) The reciprocal of a positive rational number is positive rational numbers.

Page No: 20

Exercise 1.2

1. Represent these numbers on the number line. (i) 7/4 (ii) -5/6

Answer

(i) 7/4 on the number line.

Divide line between two natural number in 4 parts. Thus, the rational number 7/4 lies at a distance of 7 points from 0 towards positive number line.

(ii) -5/6 on the number line.

Divide line between two natural number in 6 parts. Thus, the rational number -5/6 lies at a distance of 5 points from 0 towards negative number line.

2. Represent -2/11, -5/11, -9/11 on the number line.

Answer

-2/11, -5/11, -9/11 on the number line.

Divide line between two natural number in 11 parts. Thus, the rational number -2/11, -5/11, -9/11 lie at a distance of 2, 5, 9 points from 0 towards negative number line respectively.

3. Write five rational numbers which are smaller than 2.

Answer

2 can be written as 10/5.

Thus, 5 natural numbers smaller than 2 are:

9/5, 8/5, 7/5, 6/5 and 5/5

4. Find ten rational numbers between -2/5 and 1/2.

Answer

The numbers -2/5 and 1/2 can be written as -8/20 and 10/20

Thus, ten rational numbers between -2/5 and 1/2 are:

-7/20, -6/20, -5/20, -4/20, -3/20, -2/20, -1/20, 0, 1/20 and 2/20

5. Find five rational numbers between.

(i) 2/3 and 4/5 (ii) -3/2 and 5/3 (iii) 1/4 and 1/2

Answer

(i) Five rational numbers between 2/3 and 4/5

The numbers 2/3 and 4/5 can be written as 30/45 and 36/45

Thus, five rational numbers are:

31/45, 32/45, 33/45, 34/45 and 35/45

(ii) Five rational numbers between -3/2 and 5/3

The numbers -3/2 and 5/3 can be written as -9/6 and 10/6

Thus, five rational numbers are:

-8/6, -5/6, -2/6, 0 and 2/6

(iii) Five rational numbers between 1/4 and 1/2

The numbers 1/4 and 1/2 can be written as 7/28 and 14/28

Thus, five rational numbers are:

8/28, 9/28, 10/28, 11/28 and 12/28

6. Write five rational numbers greater than -2.

Answer

-2 can be written as -16/8.

Five rational numbers greater than -2 are:

-15/8, -14/8, -13/8, -12/8 and -11/8

7. Find ten rational numbers between 3/5 and 3/4.

Answer

The numbers 3/5 and 3/4 can be written as 48/80 and 60/80

Thus, ten rational numbers between 3/5 and 3/4 are:

49/80, 50/80, 51/80, 52/80, 53/80, 54/80, 55/80, 56/80, 57/80 and 58/80.

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NCERT Solutions for Class 12th: Ch 11 Biotechnology and its Applications Biology

Page No: 215

Exercises

1. Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because –
(a) bacteria are resistant to the toxin
(b) toxin is immature;
(c) toxin is inactive;
(d) bacteria encloses toxin in a special sac.

Answer

Toxin is inactive:
In bacteria, the toxin is present in an inactive form, called prototoxin, which gets converted into active form when it enters the body of an insect.

2. What are transgenic bacteria? Illustrate using any one example.

Answer

Transgenic bacteria contain foreign gene that is intentionally introduced into its genome. They are manipulated to express the desirable gene for the production of various commercially important products.
An example of transgenic bacteria is E.coli. In the plasmid of E.coli, the two DNA sequences corresponding to A and B chain of human insulin are inserted, so as to produce the respective human insulin chains. Hence, after the insertion of insulin gene into the bacterium, it becomes transgenic and starts producing chains of human insulin. after that, these chains are extracted from E.coli and combined to form human insulin.

3. Compare and contrast the advantages and disadvantages of production of genetically modified crops.

Answer

The production of genetically modified (GM) or transgenic plants has several advantages.

(i) Most of the genetically modified crops have been developed for pest resistance, which increases the crop productivity and reduces the reliance on chemical pesticides.
(ii) Many varieties of genetically modified food crops have been developed, which has good nutritional quality. For example, golden rice is a transgenic variety in rice, which is rich in vitamin A.
(iii) These plants prevent the loss of fertility of soil by increasing the efficiency of mineral usage.
(iv) They are highly resistant to unfavourable abiotic conditions.
(v) The use of genitically modified crops decreases the post harvesting loss of crops.

But, there are certain controversies regarding the use of genetically modified crops around the world. The use of these crops can affect the native biodiversity in an area. For example, the use of Bt toxin to decrease the amount of pesticide is posing a threat for beneficial insect pollinators such as honey bee. If the gene expressed for Bt toxin gets expressed in the pollen, then the honey bee might be affected. As a result, the process of pollination by honey bees would be affected. Also, genetically modified crops are affecting human health. They supply allergens and certain antibiotic resistance markers in the body. Also, they can cause genetic pollution in the wild relatives of the crop plants. Hence, it is affecting our natural environment.

Page No: 216

4. What are Cry proteins? Name an organism that produce it. How has man exploited this protein to his benefit?

Answer

Cry proteins are encoded by cry genes. These are toxins, which are produced by Bacillus thuringiensis bacteria. This bacterium contains these proteins in their inactive from. When the inactive toxin protein is ingested by the insect, it gets activated by the alkaline pH of the gut of the insects. This results in the lysis of epithelial cell and gradually the death of the insect. Thus, man has exploited this protein to develop certain transgenic crops with insect resistance.
For example- Bt cotton, Bt corn, etc.

5. What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency.

Answer

Gene therapy is a technique for correcting a defective gene through gene manipulation. It involves the insertion of a normal gene into the person to replace the affected gene.

For example, the introduction of gene for adenosine deaminase (ADA) in ADA deficient individual. The adenosine deaminase enzyme is important for the normal functioning of the immune system. The individual suffering from this disorder can be cured by transplantation of bone marrow cells. The first step involves the extraction of lymphocyte from the patient's bone marrow. Then, a functional gene for ADA is introduced into lymphocytes with the help of retrovirus. These treated lymphocytes containing ADA gene are then introduced into the patient's bone marrow. Thus, the gene gets activated producing functional T- lymphocytes and activating the patient's immune system.

6. Digrammatically represent the experimental steps in cloning and expressing an human gene (say the gene for growth hormone) into a bacterium like E. coli ?

Answer

DNA cloning is a method of producing multiple identical copies of specific template DNA. It involves the use of a vector to carry the specific foreign DNA fragment into the host cell. The mechanism of cloning and transfer of gene for growth hormone into E.coli is represented below.

7. Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and chemistry of oil?

Answer

Recombinant DNA technology (rDNA) is a technique used for manipulating the genetic material of an organism to obtain the desired result.
For example, this technology is used for removing oil from seeds. The costituentsf oil are glycerol and fatty acids. Using rDNA, one can obtain oilless seeds by stoping the synthesis of either glycerol or fatty acids. This is done by removing the specific gene responsible for the synthesis.

8. Find out from internet what is golden rice.

Answer

• Golden rice is a genetically modified variety of rice, Oryza sativa,which has been developed as a fortified food for areas where there is a shortage of dietary vitamin A.
• It contains a precursor of pro-vitamin A, called beta-carotene, which has been introduced into the rice through genetic engineering.
• The rice plant naturally produces beta-carotene pigment in its leaves. However, it is absent in the endosperm of the seed. This is because beta-carotene pigment helps in the process of photosynthesis while photosynthesis does not occur in endosperm.
• Since beta-carotene is a precursor of pro-vitamin A, it is introduced into the rice variety to fulfill the shortage of dietary vitamin A. It is simple and a less expensive alternative to vitamin supplements.
• However, this variety of rice has faced a significant opposition from environment activites. Therefore, they are still not available in market for human consumption.

9. Does our blood have proteases and nucleases?

Answer

No, human blood does not include the enzymes, nucleases and proteases. In human beings, blood serum contains different types of protease inhibitors, which protect the blood proteins from being broken down by the action of proteases. The enzyme, nucleases, catalyses the hydrolysis of nucleic acids that is absent in blood.

10. Consult internet and find out how to make orally active protein pharmaceutical. What is the major problem to be encountered?

Answer

Protein pharmaceutical cannot be taken orally because they can be degraded by the proteases of our alimentary canal.
Thus, major problem to be encountered is the action of digestive enzymes. It has to be made digestible for the digestive system and also protect it from the degradation of HCl present in stomach so it is coated by a film that is resistant to protein degrading enzymes.

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NCERT Solutions for Class 6th: Ch 15 Air Around us Science

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