Tuesday, 29 December 2015

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

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.

(c) Exonuclease and Endonuclease
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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Saturday, 26 December 2015

Notes of Define Phy. Edu., Its Aims & Objectives| Class 11th Physical Education

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.

Study Material and Notes of Physical Education - Class 11th

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

Tuesday, 22 December 2015

पाठ - 7 धर्म की आड़ अन्य परीक्षापयोगी प्रश्न और उत्तर। स्पर्श भाग - I

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) साधारण आदमी का दोष यह है की वह 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

Monday, 21 December 2015

NCERT Solutions for Class 10th: Ch 15 Probability Maths

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
23456789101112
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.


Friday, 18 December 2015

NCERT Solutions for Class 9th: Ch 13 Surface Areas and Volumes Maths (Part-3)

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 πr--- (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)

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