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Kerala State Syllabus 8th Standard Maths Solutions Chapter 10 Statistics

Statistics Text Book Questions and Answers

Textbook Page No 184

Class 8 Maths Chapter Statistics Kerala Syllabus Question 1.
The number of members in 50 households of a village are listed below.

Make a frequency table and answer these questions:
i. How many households have just two members?
ii. How many households have four or less?
iii. How many households have ten or more?
iv. Households of what size occurs the most?
Solution:

i. 5
ii. 5 + 11 + 9 = 25
iii. 1 + 1 = 2
iv. 3

8th Standard Maths Statistics Kerala Syllabus Question 2.
There are 44 children in class 8B. The list shows how far they come from, in kilometres.

Make a frequency table and answer these questions:
i. How many children are from exactly 1 kilometre away?
ii. How many are from more than 5 kilometres?
iii. How many are from between 5 and 10 kilometres?
iv. How many are from more than 10 kilometres?
Solution:

i. 3
ii. 21
iii. 23
iv. 4

8th Standard Maths Guide Kerala Syllabus  Question 3.
The scores of 35 children in a test are given below:

Make a frequency table and answers these questions:
i. How many children scored 20?
ii. How many children got scores between 10 and 20?
iii. How many scored less than 10?
iv. What is the score most number of children got?
Solution:
Score Tally Number of children

i. 1
ii. 32
iii. 0
iv. 15

Textbook Page No 188

Statistics Class 8 Notes Kerala Syllabus Question 4.
Given below are the highest temperatures (in degree Celsius) one day in 40 towns. Make a frequency table.

Solution:

8th Class Maths Statistics Kerala Syllabus Question 5.
The heights (in centimeters) of 45 people who look part in a physical fitness test are given below? Make a frequency table.

Solution:

Textbook Page No 190

Hss Live Guru Maths 8th Kerala Syllabus Question 6.
The table shows the times 30 children took to complete a long distance race. Draw a histogram of this.

Solution:

Hsslive Guru Maths 8th Standard Kerala Syllabus Question 7.
The table shows the daily incomes of 6o households in a locality.

Draw a histogram.
Solution:

Hss Live Std 8 Maths Kerala Syllabus Question 8.
Detail of rainfall in June and July are given in the table below. Draw a histogram.

Solution:

Hsslive Guru 8th Class Maths Kerala Syllabus Question 9.
The time taken by 25 women and 23 men to complete a race are given in the table below. Draw separate histograms for men and women.

Solution:

Kerala Syllabus 8th Standard Maths Notes Question 10.
The weights of 45 children in a class are listed below.


Make a frequency table and draw a histogram.
Solution:

Additional Questions And Answers

8th Standard Maths Kerala Syllabus Question 1.
Complete the table below on the basis of the histogram.

Solution:

8th Class Maths Notes Kerala Syllabus Question 2.
Teacher conducted a test in her class of 45 students. Their score out of a total of 10 are given below.

a. Construct a frequency table representing these details?
b. Construct a bar graph?
Solution:

Hss Live Guru Class 8 Maths Kerala Syllabus Question 3.
The list below gives the daily wages earned by 30 laboures. Prepare a frequency table of these.

a. What is the daily wages of most labourers?
b. How many get 250 rupees a day?
c. How many get the least amount of wages?
Solution:

a. 225
b. 6
c. 5

Class 8 Maths Chapter 10 Kerala Syllabus Question 4.
The table shows daily expenditure of 60 household in a locality. Draw histrogram.

Solution:

8th Standard Maths Notes Kerala Syllabus Question 5.
Given below are the amount of rainfall (in mm) one day in 61 towns. Make a frequency table.


Solution:

8th Std Maths Solution Guide Kerala Syllabus Question 6.
The runs that a batsman got in 40 one – day cricket matches are given below. Make a frequency table and answer these questions.

a. How many centuries did he get?
b. How many half centimes?
c. In how many games did he score less than 50?
Solution:

Question 7.
The height of 30 childrens in a class are listed below. Mark a frequency table and draw a histogram.

Solution:

You can Download मेरी ममतामई माँ Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Hindi Solutions Unit 5 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Hindi Solutions Unit 5 Chapter 1 मेरी ममतामई माँ (संस्मरण)

मेरी ममतामई माँ Textual Questions and Answers

मेरी ममतामई माँ विश्लेषणात्मक प्रश्न

Kerala Syllabus 9th Standard Hindi Solutions Unit 5 Chapter 1 प्रश्ना 1.
एकल परिवार और संयुक्त परिवार की अपनी – अपनी विशेषताएँ हैं। चर्च करें।

उत्तर:
कल परिवार का अर्थ है कि एक ऐसा परिवार जिसमें पति, पत्नी और उनके बच्चे ही रहते हैं। एक ऐसा परिवार जहाँ माता – पिता, बेटे – बहू, पोते – पोती, चाचा-चाची और ताऊ- ताई आदि एकसाथ रहते हैं उसे हम संयुक्त परिवार कहते हैं।
विशेषताएँ:
1. एकल परिवार छोटे होते हैं जबकि संयुक्त परिवार बड़े होते हैं।
2. एकल परिवार में खर्च कम होते हैं जबकि संयुक्त में अधिक खर्चे होते हैं।
3. एकल परिवार में लोगों को एकांत अधिक मिल पाता है जबकि संयुक्त परिवार में इस चीज़ की कमी होती है।
4. एकल परिवार में बच्चे अपने दादा-दादी, नाना-नानी के प्यार से वंचित रह जाते हैं जबकि संयुक्त परिवार में उन्हें अपने दादा – दादी, नाना – नानी आदि का भरपूर प्यार और संस्कार मिल पाते हैं ।
5. एकल परिवार में अच्छा-बुरा सलाह के लिए किसी बड़े अनुभवी का साथ नहीं मिल पाता जबकि संयुक्त में अनुभवी व्यक्ति की सलाह से कई समस्याएँ चुटकियो में सुलझ जाती है।
6. एकल परिवार में बच्चे संस्कृति और संस्कारो में पीछे रह जाते हैं जबकि संयुक्त परिवार में बच्चों को अच्छे संस्कार और संस्कृति को जानने का मौका मिल पाता है।
7. एकल परिवार में यदि कोई एक व्यक्ति बीमार हो जाता है तो देखभाल के लिए कोई भी नहीं होता है जबकि संयुक्त परिवार में यदि कोई बीमार हो जाता है तो सहायता और देखभाल के लिए कई लोग मिल जाते हैं और बड़ों का अनुभव भी मिलता है।

Kerala Syllabus 9th Standard Hindi Notes प्रश्ना 2.
‘शहर में सबसे पहले मैं ही लोगों तक समाचार-पत्र पहूँचाता था।’- इस प्रस्ताव से बालक कलाम का कौन-सा मनोभाव प्रकट होता है?


उत्तर:
कलाम एक ईमानदार बालक है। वह अपना दायित्व ठीक तरह से निभानेवाला है। समय की पाबंदी लगाने में वह सफल रहा। वह हमेशा मेहनती नज़र आता है।

9th Hindi Notes Kerala Syllabus प्रश्ना 3.
“उन्होंने अपने हिस्से की भी सारी रोटियाँ तुम्हें दे दी”-भाई की इस बात पर बालक कलाम की प्रतिक्रिया क्या होगी?

उत्तर:
बालक कलाम को अपनी माँ के प्रति ममता के मारे सिहरन आ गई। वह अपने आपको रोक नहीं पाया। दौड़कर माँ के पास गया और भावावेश में उनसे लिपट गया।

Std 9 Hindi Notes Kerala Syllabus प्रश्ना 4.
‘नारी ईश्वर की सुंदर रचना है।’ कलाम को ऐसा क्यों लगा?

उत्तर:
कलाम नारी को आदर करनेवाले है। कलाम अपनी माँ को ईश्वर समान मानते हैं। माँ ही उनके लिए सबकुछ है। माँ की ममता कलाम को सदा लिपटती थी। उनकी हर उन्नति पर माँ का हाथ रहा था। वे अपनी माँ के नाश्ते के बारे में अग्नि की उडान’ पर याद करते हैं। शायद इसीलिए उन्हें ऐसा लगा होगा।

मेरी ममतामई माँ Text Book Activities

Class 9 Hindi Notes Kerala Syllabus प्रश्ना 1.
मिलान करें:

उत्तर:

9th Class Hindi Notes Kerala Syllabus प्रश्ना 2.
बातचीत लिखें।
‘उस दिन पहली बार मुझे सिहरन की अनुभूति हुई। मैं अपने आपको रोक नहीं सका। दौड़कर अपनी माँ के पास गया और भावावेश में उनले लिपट गया।’ इस प्रसंग पर बालक कलाम और माँ के बीच क्या-क्या बातें हुई होंगी?

उत्तर:
कलाम : (दौडकर आता हुआ) माँ ………… ओ माँ ……..
माँ . : (लिपटती हुई) क्या हुआ बेटा?
कलाम : यह आपने क्या कर दिया? मुझे सिहरन हुई।
माँ : क्या बात है?
कलाम : आज आपने अपने हिस्से की रोटियाँ भी मुझे दे दी!
माँ : हाँ, तुम तो भूखा था न, इसलिए।
कलाम : तो आप स्वयं भूखी रहकर मुझे दे दिया?
माँ : तुम तो पढ़ाई और कमाई एकसाथ करते है न।
कलाम : इतनी कुरबानी मेरे लिए क्यों माँ?
माँ : क्योंकि मैं तेरी माँ हूँ और तुम मेरे लाडले हो।
कलाम : हमारे घर की हालत उतनी अच्छी नहीं है माँ।
माँ : मैं संभालूँगी बेटा। तुम फिकर मत कर।
कलाम : आप भूखों मत मरेगी। यह मेरी वादा है।
माँ : (प्यार से सहलाती हुई) जा जाकर कुछ पढ़ाई कर।

Kerala Syllabus 9th Standard Hindi प्रश्ना 3.
नमूने के अनुसार बदलकर लिखें।

उत्तर:

मेरी ममतामई माँ विधात्मक प्रश्न

Hindi Kerala Syllabus 9th Standard प्रश्ना 1.
“उन्होंने अपने हिस्से की भी सारी रोटियाँ तुम्हें दे दी। एक ज़िम्मेदार बेटा बनो और अपनी माँ को भूखों मत मारों।” बड़े भाई ने कलाम को डाँटा तो उसे क्या लगा होगा? उसकी डायरी कल्पना करके लिखें।

उत्तर:
25 जून 2019
बडे भाई का डाँट सुनकर आज पहली बार मुझे सिहरन की अनुभूति हुई। मैं अपने आपको रोक नहीं सका। मैं दोडकर अपनी माँ के पास गया। फिर भावावेश में उनसे लिपट गया। माँ की कुरबानी इतनी क्यों? घर की हालत उतनी अच्छी नहीं थी। युद्ध : के कारण सभी वस्तुओं की किल्लत हो गई थी। विशाल संयुक्त परिवार में रोटियों की भी कमी महसूस हुई थी। फिर भी माँ स्वयं भूखी रहकर उनकी सारी रोटियाँ मुझे दे दी। सिर्फ माँ ऐसा कर सकती है। वे कितने दयालू, स्नेहशील और धार्मिक प्रवृत्ति की महिला थीं। सदा मैं उनसे प्रेरित थी। मेरी माँ ही सबकुछ है। आगे मेरी दृष्टि उनके स्वास्थ्य पर भी होगा। अगले दिन की प्रतीक्षा में….

मेरी ममतामई माँ Additional Questions and Answers

9th Standard Hindi Notes Kerala Syllabus प्रश्ना 1.
‘कलाम के घर में खुशी और गम का आना जाना लगा रहता था’- इसका कारण क्या होगा?

उत्तर:
कलाम का परिवार संयुक्त परिवार है। संयुक्त परिवार में सदस्यों की संख्य ज़्यादा है। तो हमेशा कुछ खुशी या कुछ गम आते जाते रहेंगे।

9th Std Hindi Notes Kerala Syllabus प्रश्ना 2.
कलाम के गणित के शिक्षक की शिक्षण रीति कैसी थी?

उत्तर:
गणित के शिक्षक श्री स्वामियार निःशुल्क ट्यूशन पढ़ाते थे। वे एक साल में पाँच ही छात्रों को पढ़ाते थे। उनका एक ही शर्त था कि बच्चे स्नान करके पाँच बजे कक्षा में उपस्थित हो जाएँ।

Kerala Syllabus 9th Standard Notes Hindi प्रश्ना 3.
कलाम की चरित्रगत विशेषताओं पर टिप्पणी तैयार करें।

उत्तर:
सुबह पाँच बजे उठकर श्री स्वामियार के पास गणित पढ़ने जाते थे। साढ़े पाँच बजे घर वापस लौटता। बाद में नमाज़ उदा करने को और कुरान शरीफ़ सीखने के लिए पिता के साथ जाता। उसके बाद तीन किलोमीटर दूरी पर रामेश्वरम रोड़ रेलवेस्टेशन पैदल जाता था। प्रस्तुत पंक्तियों से बालक कलाम बहुत ही परिश्रमी जान पडता है और ईमानदार भी धनुष्कोडी मेल से समाचार पत्रों का बंडल लेता और तेज़ी से शहर के लोगों तक पहुँचाता था। कलाम की समय पर पाबंदी यहाँ बहुत ही स्पष्ट है। वह अपनी ज़िम्मेदारी ठीक तरह से निमाने वाला भी है। वह अपनी माँ को बहुत प्यार करता है। जब उसका भाई उसे डाँटा तो तुरंत ही दौडकर जाते हुए माँ को लिपट लेता है। वह नारी को ईश्वर की सुंदर रचना मानता है। स्त्रियों को आदर करने में वह सदा आगे है। इसप्रकार स्वाभिमानी, परिश्रमी, सेवाव्रती, ज़िम्मेदार एवं मानवता के व्यक्तित्व है बालक कलाम।

मेरी ममतामई माँ Summary in Malayalam and Translation

मेरी ममतामई माँ शब्दार्थ

You can Download दोहे Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 3 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Hindi Solutions Unit 3 Chapter 3 दोहे (कविता)

दोहे Summary in Malayalam and Translation

बड़ा हुआ तो क्या हुआ, जैसे पेड़ खजूर।
पंथी को छाया नाहिं, फल लागै अतिदूर।।

Kabir Ke Dohe Hindi Poem 8th Standard भावार्थ :
इस दोहे में कबीरदास कहते हैं कि खजूर के पड़ के समान बडा होने से कोई फायदा नहीं है। बहुत लंबा होने के कारण पथिक को छाया नहीं मिलती है। फल भी बहुत ऊँचाई पर लगते हैं। वह भी लोगों केलिए अप्राप्य है। मतलब है कि हम बड़े होकर भी दूसरों को कोई उपकार नहीं करते तो उस बडप्पन से किसी को कोई लाभ नहीं। दूसरों की भलाई करनेवाले ही सच्चे अर्थ में बडे होते हैं।

रहीम वे नर धन्य हैं, पर उपकारी अंग।
बाँटन पारे को लगे, ज्यों मेहंदी को रंग।।

8th Class Hindi Dohe Kerala Syllabus भावार्थ :
रहीस कहते हैं – वे नर धन्य है जिनका शरीर सदा सबका उपकार है। जिसप्रकार मेहंदी बाँटनेवाले के अंग पर भी मेहंदी का रंग लगा जाता है उसी प्रकार परोपकारी का शरीर भी सुशोभित होता है। .

Rahim Ke Dohe Question Answer Kerala Syllabus दोहे शब्दार्थ Word meanings

You can Download जल-बैंक Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 4 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Hindi Solutions Unit 4 Chapter 3 जल-बैंक

जल-बैंक पाठ्यपुस्तक के प्रश्न और उत्तर

Jal Bank Hindi Story Kerala Syllabus 8th प्रश्ना 1.
‘चप्पे-चप्पे पर जल बैंक खुल गए हैं।’ लेखक की इस कल्पना के पीछे भविष्य का कौन-सा संकेत है?

उत्तर:
पानी की कमी एक गंभीर समस्या बनती जा रही है। अब भी हम इस दिशा पर ध्यान न रखेंगे तो भविष्य बड़ी आपत्ति हो जाएगी। पानी के लिए हमें किसी जल बैंक जैसी व्यवस्थाओं का आश्रय लेना पड़ेगा।

Hss Live Guru 8 Hindi Kerala Syllabus प्रश्ना 2.
‘पानी-पानी होना’, ‘पानीदार होना’ आदि प्रयोगों का मतलब क्या है?

उत्तर:
‘पानी-पानी होना’ का मतलब है- अत्यंत लज्जित होना और ‘पानीदार होना’ का मतलब होना- धनी होना।

Hss Live Guru Hindi 8th Kerala Syllabus प्रश्ना 3.
‘घर में पीने के वास्ते भी एक बूंद नहीं है’ लोग ऐसा झूठ क्यों कहते हैं?

उत्तर:
पानी की कमी बहुत है। लेकिन कुछ लोग उसके उपयोग में सावधानी नहीं बरते। कुछ लोग पानी के बिना तरसते हैं। उनके सामने लोग झूठ बोलते हैं कि पीने के लिए भी एक बूंद पानी नहीं है। यहाँ लोगों की संकुचित भाव की ओर संकेत है।

Hss Live Guru 8th Hindi Kerala Syllabus प्रश्ना 4.
‘चारों उँगलियाँ पानी में’ इस प्रयोग से आप क्या समझते हैं?

उत्तर:
‘चारों उँगलियाँ पानी में’ का मतलब है- जीवन में बड़ा नुकसान होना।

जल-बैंक Textbook Activities

Hsslive Guru Hindi Class 8 Kerala Syllabus प्रश्ना 1.
चर्चा करें।
जल बैंक की संकल्पना कैसी लगी?
वर्तमान समाज में इसकी प्रासंगिकता क्या है?

उत्तर:
जल दुर्लभता आज की विकट समस्या है। ऐसी स्थिति में जलबैंक की संकल्पना उचित है।

Hsslive Guru 8th Hindi Kerala Syllabus प्रश्ना 2.
जल-संरक्षण की आवश्यकता पर नारे बनाएँ।

जल है तो कल है।
जल जीवन का आधार।
जल नहीं तो हम नहीं।
जल ही जीवन है।
बूंद-बूंद का संरक्षण, जीवन का संरक्षण।
आज संजोएँ हरेक बूंद, कल को बनाएँ खूब सुहाना।

Hsslive Guru 8th Class Hindi Kerala Syllabus प्रश्ना 3.
उपर्युक्त नारों की मदद से पोस्टर तैयार करें।

जल-बैंक Summary in Malayalam and Translation

 जल-बैंक शब्दार्थ Word meanings

You can Download Sound Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 19 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 19 Sound

Sound is a form of energy which is familiar to us and necessary for communication. Three components are essential to experience the sound. Source of sound, medium, and the ear.

Source of sound

The sources that produce sound is called sources of sound. We can classify the sources of sound into two. Natural sources of sound and manmade sources of sound. Sound is produced by the vibration of object. The object that produces sound is called the source of sound.

Hss Live Guru 8th Physics Kerala Syllabus Natural frequency

When a body is set into vibration it vibrates with particular frequency of its own. This frequency is called its natural frequency. The unit of frequency is hertz (Hz). The constituents those affect the natural frequency of an object are the nature of the object, length of the object, surface area, tension of the object, etc.
Frequency (f) = Number of vibration(n)/ time(f)

Pitch and loudness

The sharpness of the sound heard is called the pitch. It depends on its frequency of sound. Cuckoo’s cry, female voice etc have high pitch and male voice, duck’s sound, lion’s roar, etc have low pitch. Loudness is the measure of audibility of a person. This depends mainly on frequency of sound and the sensory ability of the ear. Its unit is decibel (dB).

Kerala Syllabus 8th Standard Physics Notes Propagation of sound

A medium is necessary for sound to propagate. Sound is propagated not only through air but also through other substances. Loss of hearing is a disability of the ear. The people having damage to ear by birth effect many difficulties for commu¬nication, for ability to speak, vulnerability to danger, etc.

Class 8 Physics Notes Kerala Syllabus Limit of audibility

We cannot here sound of all frequencies. We can hear the sound of frequency in between 20 Hz and 20000Hz. Sounds with frequency less than 20 Hz are called infrasonic and that greater than 20000Hz are called ultrasonic. Ultrasonic sounds are used in the instrument sonar and in medical field.

Hss Live Guru 8th Basic Science Kerala Syllabus Noise pollution

Kerala is one of the places which is highest noise pollution in the world. Noise effects not only ear but mental, emotional level, and physical problems. Reduce the use of air horns, use silencers in vehicles, planting trees, etc are some ways to reduce noise pollution.

Sound Textbook Questions & Answers

Hsslive Guru Physics 8th Standard Kerala Syllabus Question 1.
If a tuning fork vibrates 480 times in one second, what would be its natural frequency?
Answer:
Natural frequency = 480Hz

Kerala Syllabus 8th Standard Chemistry Notes Question 2.
If a simple pendulum oscillates 10 times in 10 seconds, what would be its frequency?
Answer:
n = 10
t = -10 s
\(\mathrm{f}=\frac{n}{t}=\frac{10}{10 s}=1 \mathrm{Hz}\)

Hsslive Guru 8th Class Physics Kerala Syllabus Question 3.
What are the factors influencing the natural frequency of a body?
Answer:
i. Nature of the object
ii. Length
iii. Tention
iv. Surface area
v. Area of cross-section

Hss Live Physics Class 8 Kerala Syllabus Question 4.
The frequencies of certain tuning forks are given below. Find out which among these have the highest and the smallest pitches.
(256 Hz, 512 Hz, 480 Hz, 288 Hz)
Answer:
High pitch = 512 Hz
Low pitch = 256 Hz

Kerala Syllabus 8th Standard Basic Science Notes Question 5.
In the sources of sound given below, vibration in which main part produces sound?
a. Chenda
b. Flute
c. Vocal cord
Answer:
a. Chenda – Diaphragm
b. Flute-Air
c. Vocal box – vocal cord

Kerala Syllabus 8th Standard Chemistry Notes Malayalam Medium Question 6.
Design an activity to prove that sound can he propagated even through solid substances.
Answer:
Answer:
Press the ear on one end of a iron rod and beat on the other end with another iron rod or, toy telephone

8th Standard Physics Notes Kerala Syllabus Question 7.
Say whether the following statements given below are true or false.
If false, rewrite it by making necessary changes.
a. Sound cannot travel through vacuum.
b. When frequency of sound increases, pitch decreases.
Answer:
a. correct
b. false, when frequency decreases pitch decreases

Kerala Syllabus 8th Standard Physics Question 8.
‘Bats can catch prey even in the dark’. Do you agree with this statement? Explain your inference.
Answer:
Agree. Bats can produce and hear ultrasonic waves. The sound they produced reflects by hitting on objects. The bats can analyze the returning sounds.

8th Class Biology Notes Pdf Kerala Syllabus Question 9.
How do human beings contribute to noise pollution?
Answer:

• Air horn
• Loudspeaker
• The sounds of vehicles etc

8th Standard Chemistry Notes Pdf Kerala Syllabus Question 10.
Which unit represents loudness? (Hz, m/s, dB, W)
Answer:
dB

Sound Additional Questions & Answers

8th Standard Chemistry Notes Kerala Syllabus Question 1.
Complete the table

Answer:

Basic Science Class 8 Sound Kerala Syllabus Question 2.
Fill suitably
Frequency – hertz; Loudness- ………….
Answer:
decibel (dB)

Class 8 Science Notes Kerala Syllabus Question 3.
Table the following as those having high pitch and low pitch separately.
Cuckoo’s cry, lion’s roar, female voice, duck’s sound, male voice, air horn
Answer:

Class 8 Physics Kerala Syllabus Question 4.
Table the natural source and manmade source from the following. Lips, chenda, flute, tabla, violin, vocal cord, sound of birds
Answer:

Vibration Solutions Question 5.
What is the relation between the length of the pendulum and frequency?
Answer:
When the length of the pendulum increases, the frequency decreases.

Question 6.
Will the intensity of sound increase when the instrument like chenda and maddalam are beaten strongly? Justify your answer.
Answer:
Yes. When beaten strongly the diagram vibrates with greater amplitude and increases frequency and loudness

Question 7.
What are the demerits of sound pollution?
Answer:

• Causes mental stress.
• Causes emotional strain
• Causes deafness
• Increase the blood pressure

Question 8.
Write the ways to reduce sound pollution.
Answer:

• Ban the air horn
• Control loudspeakers
• Use silencers in vehicles
• Control the sound having more than 50 dB near the hospitals

Question 9.
Write two uses of ultrasonic sound.
Answer:
Use in SONAR
To find out the disease and treatment in medical field

Question 10.
Can we hear the sound from Galton whistle having a frequency of 30000Hz? Why?
Answer:
No. We cannot hear the sound having a frequency greater than 20000Hz

Question 11.
What are the problem faced by deaf people?
Answer:
The people having damage to hereby birth effect many difficulties for com¬munication, for ability to speak, vulnerability to danger, etc.

Question 12.
Complete the table.

Answer:
a. 5
b. 5
c. 20

Question 13.
An object vibrates 200 times in one second. What is its frequency?
Answer:
frequency = \(\frac{n}{t}=\frac{200}{1}=200 \mathrm{Hz}\)

Question 14.
How many times vibrate a body of frequency 290 Hz in 12 seconds.
Answer:
n = f × t
f = 290 Hz,
t = 12 s
n = 290 × 12 = 3480 times

Question 15.
Name the sounds of frequency below 20 Hz and above 20000 Hz.
Answer:
below 20 Hz = Infrasonic
Above 20000 Hz = ultrasonic

You can Download सफेद गुड़ Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 5 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Hindi Solutions Unit 5 Chapter 1 सफेद गुड़

सफेद गुड़ पाठ्यपुस्तक के प्रश्न और उत्तर

Safed Gud Story In Hindi Kerala Syllabus 8th प्रश्ना 1.
माँ के आसमान की ओर देखने का क्या कारण होगा?

उत्तर:
माँ चाहती होगी कि बेटे को पैसा दें। लेकिन वह विवश है। वह अपनी विवशता के कारण आकाश की ओर ताकती है। पैसे के अभाव में वह हमेशा ऐसा करती है। शायद वह ईश्वर से विनती करती होगी।

Safed Gud Kahani Ka Saransh Likho Kerala Syllabus 8th प्रश्ना 2.
वह अपने को धिक्कारने लगा और इस बुरे ख्याल के लिए ईश्वर से क्षमा माँगने लगा’ -यहाँ लड़के का कौन-सा मनोभाव प्रकट है?

उत्तर:
इससे बच्चे का विवेकशीलता प्रकट होता है। उसकी सच्चाई और गलती पर पछताने का भाव यहाँ स्पष्ट है।

Gud In Hindi Kerala Syllabus 8th प्रश्ना 3.
वह एक अठन्नी ही नहीं थी, उस गरीब पर ईश्वर की कृपा थी -ऐसा क्यों कहा है?

उत्तर:
वह पैसे के लिए बहुत तरसता था। बहुत प्रार्थना भी किया था। इसी अवसर पर उसे यह अठन्नी मिली। इसलिए उसे लगा कि वह ईश्वर की कृपा है।

सफेद गुड़ Textbook Activities

सफेद गुड कहानी का सारांश Kerala Syllabus 8th प्रश्ना 1.
निम्न्लिखित वाक्य पर घ्यान दें —

माँ बैठी फटे कपड़े सिल रही थी।

बताएँ, रेखांकित शब्दों में क्या संबंध है?

उत्तर:

Gud Meaning In Hindi Kerala Syllabus 8th प्रश्ना 4.
इस प्रकार आपसी संबंध रखनेवाले शब्द पाठ से चुनकर लिखें।

उत्तर:

Prarthana Poem In Hindi 8th Class Summary Kerala Syllabus 8th प्रश्ना 5.
लड़के की गुड़ खाने की इच्छा सफल नहीं हुई। उसके विचारों को डायरी के रूप में लिखें।

उत्तर:
25 जनवरी 2016
बहुत दिनों से गुड़ खाने की इच्छा थी। आज सोचा कि वह सफल हो गया। लेकिन क्या कहूँ, वह सपना सपना ही रह गया। नमक खरीदने बाज़ार जाते समय एक पैसे के लिए प्रार्थना करता रहा। ठीक उसी समय ज़मीन से एक अठन्नी पड़ी मिली। खुशी का ठिकाना न था। अफ़सोस है कि दुकानदार को देते समय वह हाथ से खिसक गया, धनिया के डिब्बे में। ढूँढ़ने से चिकना-सा पत्थर मिला। दुकानदार ने कहा कि मुफ्त में ले लो। लेकिन मन नहीं हुआ। नमक खरीद कर दुखी मन से वापस चला आया।

सफेद गुड़ Summary in Malayalam and Translation

सफेद गुड़ शब्दार्थ Word meanings

Students can Download Kerala SSLC Maths Previous Year Question Paper March 2019 English Medium, Kerala SSLC Maths Model Question Papers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala SSLC Maths Previous Year Question Paper March 2019 English Medium

Time: 2½ Hours
Total Score: 80 Marks

Instructions

• Read each question carefully before writing the answer.
• Give explanations wherever necessary.
• First 15 minutes is Cool-off time. You may use the time to read the questions and plan your answers.
• No need to simplify irrationals like √2, √3, π etc., using approximations unless you are asked to do so.

Answer any three questions from 1 to 4. Each question carries 2 scores. (3 × 2 = 6)

Question 1.
In the figure, O is the centre of the circle. ∠AOC = 80°
a) What is the measure of ∠ABC?
b) What is the measure of ∠ADC?

Answer:
a) \(\angle \mathrm{ABC}=\frac{1}{2} \times \angle \mathrm{AOC}=\frac{1}{2} \times 80=40^{\circ}\)
b) ∠ADC = 180 – ∠ABC = 180 – 40 = 140°

Question 2.
a) Write the first integer term of the arithmetic sequence \(\frac{1}{7}, \frac{2}{7}, \frac{3}{7} \ldots\)
b) What is the sum of the first 7 terms of this se-quence?
Answer:
a) First integer term is \(\left(\frac{7}{7}\right)=1\)
b) Sum of first 7 terms = \(\frac { 1 }{ 7 }\) (1 + 2 + 3 + ….. + 7)
= \(=\frac{1}{7} \times \frac{7 \times 8}{2}=4\)

Question 3.
a) If C(-1, k) is a point on the line passing through the points A(2, 4) and B(4, 8). Which number is k?
b) What is the relation between the x coordinate and the y coordinate of any point on this line?
Answer:
a) Since C is on the line AC,
slope of AB = slope of BC
\(\begin{aligned}
\frac{8-4}{4-2} &=\frac{8-k}{4–1} \\
\frac{4}{2} &=\frac{8-k}{5}
\end{aligned}\)
⇒ 4 × 5 = 2 (8 – k) = 20
⇒ 20 = 16 – 2k
⇒ 2k = 16 – 20 = -4
⇒ k = -2
b) Let (x, y) be a point on the line
\(\begin{aligned}
&\frac{y-4}{x-2}=\frac{8-4}{4-2}\\
&\frac{y-4}{x-2}=\frac{4}{2}
\end{aligned}\)
⇒ 2 (y – 4) = 4(x – 2)
⇒ 2y – 8 = 4x – 8
⇒ 4x – 2y – 8 + 8 = 0
⇒ 4x – 2y = 0
⇒ 2x – y = 0
⇒ y = 2x

Question 4.
a) Find P(1) if P(x) = x² + 2x + 5
b) If (x – 1) is a factor of x² + 2x + k, What number is k?
Answer:
a) P(1) = 1² + 2 × 1 + 5 = 1 + 2 + 5 = 8
b) Since x – 1 is a factor
P(1) = 0
⇒ 1² + 2 × 1 + k = 0
⇒ 1 + 2 + k = 0
⇒ k = -3

Answer any five questions from 5 to 11. Each question carries 3 scores. (5 × 3 = 15)

Question 5.
a) What is the remainder on dividing the terms of the arithmetic sequence 100, 107, 114.. by 7?
b) Write the sequence of all three digit numbers. Which leaves remainder 3 on division by 7? Which is the last term of this sequence?
Answer:
a) 100 = 7 × 14 + 2
107 = 7 × 15 + 2
Remainder on dividing the terms by 7 is 2
b) x₁ = 101, x₂ = 108, ……
Sequence : 101, 108, 115 ….
x_n = d_n + (f – d) = 7n + 94
To find largest 3 digit terms
7n + 94 < 1000
7n < 906
n < \(\frac { 906 }{ 7 }\) = 129.4 = 129
x₁₂₉ = 7 × 129 + 94 = 997

Question 6.
AB is the diameter of the. circle. D is a point on the circle. ∠ACB + ∠ADB + ∠AEB = 270°. The measure of one among ∠ACB, ∠ADB, ∠AEB is 110°. Write the measures of ∠ADB, ∠ACB, and ∠AEB.

Answer:
Since AB is the diameter, ∠ADC = 90°
∠ACB is obtuse.
∴ ∠ACB = 110°
∴ ∠AEB = 270 – (90 + 110) = 270 – 200 = 70°
∴ ∠ADB = 90°, ∠ACB = 110°, ∠AEB = 70°

Question 7.
If x is a natural number
a) What number is to be added to x² + 6x to get a perfect square?
b) If x² + ax + 16 is a perfect square which number is ‘a’?
c) If x² + ax + b is a perfect square prove that a² = 4b.
Answer:
a) x² + 6x = x² + 2 × x × 3
Add 3² to it we get
x² + 2x × 3 + 9 = (x + 3)²
b) x² + ax + 16 = x² + ax + 4²
ax = 2 × x × 4 = 8x
∴ a = 8 or a = -8
c) x² + ax + b is a perfect square
\(x^{2}+2 \times x \times \frac{a}{2}+\left(\frac{a}{2}\right)^{2}\) is a perfect square
\(\begin{aligned}
&b=\left(\frac{a}{2}\right)^{2}\\
&b=\frac{a^{2}}{4}, a^{2}=4 b
\end{aligned}\)

Question 8.
In the figure ∠B = 90°, ∠C = 44°

a) What is the measure of ∠A?
b) Which among the following is tan 44°?
\(\left(\frac{\mathrm{AB}}{\mathrm{BC}}, \frac{\mathrm{AB}}{\mathrm{AC}}, \frac{\mathrm{BC}}{\mathrm{AB}}, \frac{\mathrm{BC}}{\mathrm{AC}}\right)\)
c) Prove that tan 44° × tan 46° = 1.
Answer:
a) ∠A = 90 – 44 = 46°
b) \(\frac{A B}{B C}\)
c) \(\tan 44^{\circ} \times \tan 46^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}} \times \frac{\mathrm{BC}}{\mathrm{AB}}=1\)

Question 9.
Draw a circle of radius 3 centimetres. Mark a point P at a distance 6 centimetres from the centre of the circle. Draw tangents from P to the circle.
Answer:

• Draw the circle having radius 3 cm, mark the centre C, and point P at the distance 6 cm from the centre.
• Draw perpendicular bisector of CP and mark midpoint of CP as O.
• Draw a circle with centre O and radius OP. The circle cut the first circle at A and B
• PA and PB are the tangents
  

Question 10.
a) Find the coordinates of the point on the x-axis, which is at a distance 4 units from (3, 4).
b) Find the coordinates of the points on the x-axis at a distance 5 units from (3, 4).
Answer:
a) The point on x axis at the distance 4 cut from (3, 4) is (3, 0)
b) Let P (x, 0) be the point
(x – 3)² + (0 – 4)² = 5²
⇒ (x – 3)² = 25 – 16
⇒ (x – 3)² = 9
⇒ x – 3 = 3, -3
If x – 3 = 3, x = 6
If x – 3 = -3, x = 0
Points : (0, 0), (6, 0)

Question 11.
The given figure is the lateral face of a square pyramid. AB = AC = 25 centimetres and BD = DC = 15 centimetres.
a) What is the length of its base edge?
b) Find the lateral surface area of the pyramid.

Answer:
a) a = 30 cm
b) AD² + CD² = AC²
AD² + 15² = 25²
AD² = 625 – 225 = 400
AD = 20
Latral face area = 2al = 2 × 30 × 20 = 1200 cm²

Answer any 7 questions from 12 to 21. Each question carries 4 scores. (7 × 4 = 28)

Question 12.
In triangle ABC, ∠A = 30°, ∠B = 80°, circumradius of the triangle is 4 centimetres. draw the triangle. Measure and write the length of its smallest side.
Answer:

• Draw the circle of radius 4 cm and a radius OA.
• Mark a point Bon the circle such that ∠COB = 60° and draw OB.
• Mark a point C on the circle such that ∠AOC = 160°
• Join AB, BC and AC, and complete ∆ABC.

Question 13.
Find the following sums:
a) 1 + 2 + 3 + ……………. + 100
b) 1 + 3 + 5 + ………….. + 99
c) 2 + 4 + 6 + ………….. + 100
d) 3 + 7 + 11 + ……………. + 199
Answer:

Question 14.
A box contains some green and blue balls. 7 red balls are put into it. Now the probability of getting a
red ball from the box is \(\frac{7}{24}\) and that of a blue ball is \(\frac{1}{6}\)
a) How many balls are there in the box?
b) How many of them are blue?
c) What is the probability of getting a green ball from the box?
Answer:
a) Total number of balls = 24
b) Let x be the number of blue balls
\(\frac{x}{24}=\frac{1}{3}\)
⇒ 3x = 24
⇒ x = 8
c) Number of green balls = 24 – (7 + 8) = 24 – 15 = 9
Probability of. getting green ball \(=\frac{9}{24}=\frac{3}{8}\)

Question 15.
The land is acquired for road widening from a square ground, as shown in the figure. The width of the acquired land is 2 metres. Area of the remaining, ground is 440 square metres.

a) What is the shape of the remaining ground?
b) What is the length of the remaining ground?
Answer:
a) Rectangle
b) x (x – 2) = 440
⇒ x² – 2x = 440
⇒ x² – 2x + 1 = 441
⇒ (x – 1)² = 21²
⇒ x – 1 = 21
⇒ x = 1 + 21 = 22
Length 22m, width 20m

Question 16.
In the figure, P is the centre of the circle. A, B and D are points on the circle. ∠P = 90°, AD = 5 centimetres.

a) What is the measure of ∠A?
b) What is the area of triangle APD?
c) Find the area of the parallelogram ABCD
Answer:
a) AP = PD, ∠A = 45°
b) Area of APD = \(\frac { 1 }{ 2 }\) × AP × PD
\(\begin{aligned}
&=\frac{1}{2} \times \frac{5}{\sqrt{2}} \times \frac{5}{\sqrt{2}}\\
&=\frac{25}{4}
\end{aligned}\)
= 6.25 cm²
c) Area of ABCD = AB × PD
\(=2 \times \frac{5}{\sqrt{2}} \times \frac{5}{\sqrt{2}}\)
= 25 cm²

Question 17.
a) Draw the coordinate axes and mark the points A(1, 1), B(7, 1)
b) Draw an isosceles right triangle ABC with AB as hypotenuse.
c) Write the coordinates of C.
Answer:
a) A(1, 1) and B(7, 1)
b) The midpoint of AB is E (4, 1)
AE = 3 unit
Move 3 unit up from E and mark C (4, 4).
ABC will be an isosceles right triangle.

c) Coordinates of C = (4, 1 + 3) = (4, 4)

Question 18.
In the figure chord, BC is extended to P. Tangent from P to the circle is PA. AQ is the bisector of ∠BAC

a) Write one pair of equal angles from the figure.
b) If ∠PAC = x and ∠PCA = y prove that ∠BAC = y – x
c) Prove that ∠PAQ = \(\frac{y+x}{2}\)
Answer:
a) ∠ABC = ∠CAP
b) ∠B = ∠PAC = x
∠B = ∠BAC = ∠ACP = y
x + ∠BAC = y
∠BAC = y – x
c) ∠PAQ = ∠PAC + ∠CAQ

Question 19.
If x – 1 is a factor of the second degree polynomial P(x) = ax² + bx + c and P(0) = -5.
a) What is the value of c?
b) Prove that a + b = 5.
c) Write a second degree polynomial whose one factor is x = 1.
Answer:
a) P(0) = -5 , c = -5
b) P(1) = 0, a + b + c = 0
a + b – 5 = 0
a + b = 5
c) If x – 1 is a factor, sum of the coefficients willbe zero.
P(x) = 3x² + 2x – 5 = 0

Question 20.
A circular sheet of paper is divided into two sectors. The central angle of one of them is 160°.
a) What is the centre of the remaining sector?
b) These sectors are bent into cones of maximum volume. If the radius of the small cone is 8 centimetres, what is the radius of the other?
c) What is the slant height of the cones?
Answer:
a) Central angle of second sectoral plate = 360 – 160 = 200°
l × 160 = 360 × 8
\(\ell=\frac{360 \times 8}{160}\) = 18 cm
Radii of sectors are equal .
\(\begin{aligned}
&\frac{r}{18}=\frac{200}{360}\\
&\begin{aligned}
r &=18 \times \frac{200}{360} \\
&=10 \mathrm{cm}
\end{aligned}
\end{aligned}\)
Radius of the big cone = 10 cm
c) 18 cm (Equal to radius of the circle)

Question 21.
Equation of the line AB is 3x – 2y = 6. P is a point on the line. The line intersects the y-axis at A and the x-axis at B.
a) What is the x coordinate of A?
b) What is the length of OA?
c) What is the length of OB?
d) The x coordinate and the y coordinate of P are the same. Find the coordinates of P.

Answer:
a) x coordinate of A = 0
b) When x = 0,
3x – 2y = 6, y = -3
Coordinates of A (0, -3)
OA = 3 unit
c) y coordinate of B is 0
3x – 2 × 0 = 6, x = 2
Coordinates of B (2, 0)
OB = 2 unit
d) Since coordinates fo P are (x, x)
3x – 2x = 6
x = 6
P (6, 6)

Answer any five questions from 22 to 28. Each question carries 5 score. (5 × 5 = 25)

Question 22.
If the terms of the arithmetic sequence \(\frac{2}{9}, \frac{3}{9}, \frac{4}{9}, \frac{5}{9} .\) are represented as x1, x2, x3, ….. then
a) x₁ + x₂ + x₃ = ______
b) x₄ + x₅ + x₆ = _______
c) Find the sum of first 9 terms.
d) What is the sum of first 200 terms?
Answer:

d) Sum of 300 terms = 1 + 2 + 3 + ….. + 100
\(=\frac{100 \times 101}{2}\)
= 5050

Question 23.
Draw a rectangle of area 12 square centimetres. Draw a square having the same area.
Answer:

1. Draw rectangle ABCD of side AB = 6 cm, BC = 2 cm.
2. Produce AB to E such that BC = BE
3. Mark the midpoint of AB as O
4. Draw a semicircle with O as the centre and OE as the radius.
5. BC, when produced meet the semicircle at G.
6. Draw a square wit

Question 24.
A boy standing at one bank of a river sees the top of a tree on the other bank directly opposite to the boy pt an elevation of 60°. Stepping 40 metres back, he sees the top at an elevation of 30°.
a) Draw a rough figure and find the height of the tree
b) What is the width of the river?
Answer:
a) PC = 40 m
∠BPC = 30°
∠BCA = 60°

In ΔPCB, ∠PCB = 120°
∠PBC = 30°
ΔPCB is an isoscles triangle
BC = PC = 40 cm
Consider ΔCBA
This is a 30° – 60° – 90° triangle side opposite to 30° in ΔCAB is = 20 m
AB = 2 × \frac{40}{\sqrt{3}} = 20√3 m
b) Width of the river is 20 m

Question 25.
Circle with centre O touches the sides of the triangle at P, Q and R, AB = AC, AQ = 4 centimetres and CQ = 6 centimetres.

a) What is the length of CP?
b) Find the perimeter and the area of the triangle.
c) What is the radius of the circle?
Answer:
a) CP = CQ = 6 cm
b) AC = AQ + QC = 4 + 6 = 10 cm
AB = AC = 10 cm
AR = AQ = 4 cm
BR = AB – AR = 10 – 4 = 6 cm
BP = BR = 6 cm
BC = BP + PC = 6 + 6 = 12 cm
Perimeter of ∆ABC = AB + BC + AC = 10 + 12 + 10 = 32 cm
Area of ∆ABC = \(\frac { 1 }{ 2 }\) × BC × AP
= \(\frac { 1 }{ 2 }\) × 12 × 8
= 48 cm² \([\mathrm{AP}=\sqrt{\mathrm{AB}^{2}-\mathrm{BP}^{2}}]\)
c) \(r=\frac{A}{S}=\frac{48}{16}=3 \mathrm{cm}\)

Question 26.
The radius of a cylinder is equal to its height. If the radius is taken as r, the volume of the cylinder is πr² × r = πr³. Like this find the volumes of the solids, with the following measures.

a) What is the ratio of the volumes of the cone, hemisphere, cylinder and the sphere?
b) A solid metal sphere of radius 6 centimetres is melted and recast into solid cones of radius 6 centimetres and height 6 centimetres. Find the number of cones.
Answer:

Question 27.
C is the centre of the circle passing through the origin. Circle cuts the y-axis at A(0, 4) and the x-axis at B(4, 0)

a) Write coordinates of C
b) Write the equation of the circle.
c) (0, 0) is the point on the circle. There is one more point on the circle with x and y coordinates equal. Which is that point?
Answer:
a) \(C\left(\frac{0+4}{2}, \frac{4+0}{2}\right)=C(2,2)\)
b) Radius \(=\frac{A B}{2}=\frac{4 \sqrt{2}}{2}=2 \sqrt{2}\) Unit [1 : 1 : √2 ]
Equation = (x – 2)² + (y – 2)² = (2√2)²
= x² – 4x + 4 + y² – 4y + 4 = 8
= x² + y² – 4x – 4y = 0
c) If x = y
x² + x² – 4x – 4x = 0
2(x² – 4x) = 0
x² – 4x = 0
x(x – 4) = 0
x = 0, x = 4
The required point is (4, 4)

Question 28.
The table below shows the number of children in a class, sorted according to their heights.

If the students are directed to stand in a line according to the order of their heights starting from the smallest, then
a) The height of the child at what position is taken as the median?
b) What is the assumed height of the child in the 17th position?
c) Find the median height.
Answer:
n = 7 + 9 + 10 + 10 + 9 = 45 (odd)

a) \(\frac{45+1}{2}\) = 23rd chil’s height comes in the middle.
the median when 10 cm height in divided into 10 children, each one’s share is \(\frac{10}{10}=1\)
Height of 17th child = 150 + \(\frac { 1 }{ 2 }\) = 150.5
b) Take 150.5 as the fist term and 1 as the common difference 7th term is the height of 23rd child.
c) Median = 150.5 + 6 × 1 = 150.5 + 6 = 156.5 cm.

Read the following. Understand the Mathemati¬cal concepts in it and answer the questions. (6 × 1 = 6)

Question 29.
The remainders obtained on dividing the powers of two by 7 have an interesting property. We can understand it from the table given below.

If the powers are 1, 4, 7, … the remainder is 2
If the powers are 3, 6, 9,.. the remainder is 1
a) What is the remainder on dividing 28 by 7?
b) Write the sequence of powers of 2 leaving remainder 1 on division by 7.
c) Check whether 2019 is a term f the arithmetic sequence 3, 6, 9,…
d) What is the remainder on dividing 22019 by 7?
e) Write the algebraic form of the arithmetic sequence 1, 4, 7 …
f) Write the algebraic form of the sequence 21, 24, 27, ……. (powers of two leaving remainder 2 on division by 7).
Answer:
a) 4
b) 2⁰, 2³, 2⁶, 2⁹, ….. (1, 8, 64, 512)
c) Since 2019 is a multiple of 3, it is a term of the sequence 3, 6, 9, …..
d) Remainder = 1
e) x_n = 2n – 2
f) x_n = 2^3n-2

You can Download Genetics for the Future Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Biology Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Biology Solutions Chapter 7 Genetics for the Future

Genetics for the Future Text Book Questions and Answers

Sslc Biology Chapter 7 Kerala Syllabus Question 1.
Observe illustration on the various stages in the production of bacteria that are capable of producing insulin. Analyze it based on the indicators and write down the inferences.

a) How insulin-producing bacteria are created?
b) What is the change that occurred in the genetic constitution of the bacteria that can produce insulin?
c) Will the future generation of this bacteria have the ability to produce insulin? Why?
Answer:
a) Insulin-producing bacteria are created through Genetic Engineering. Cutting the gene responsible for the production of insulin and joining it with bacterial DNA (plasmid).
b) The gene responsible for the production of insulin become part of the bacterial DNA.
c) Yes, the bacteria containing genes that the ability which controlling insulin production They replicates and forms more in number.

10th Class Biology 7th Lesson Kerala Syllabus Question 2.
Observe the collage given below analysis and prepare notes about it.

Answer:
It is criticized that genetically modified varieties are threat to indigenous varieties and may cause health issues to human. There are possibilities to use the genetically modified organisms are bioweapons that might be applied any country to their enemies is called Bioware. This becomes a threat to the existence of human beings.

Genetic Engineering

The use of microorganisms and biological processes for various human requisites is called biotechnology. The ability of fungi and bacteria to convert sugar into alcohol was utilized to make wine, appam and cake. These can be considered as traditional methods of biotechnology. Genetic engineering is the modern form of biotechnology. Genetic engineering is the technology of controlling traits of Organisms by bringing about desirable changes in the genetic constitution of organisms.

Biology Class 10 Kerala Syllabus Question 3.
What is the basis of genetic engineering?
Answer:
The basis of genetic engineering is the discovery of the fact that genes can be cut and joined.

• Restriction Endonuclease: The enzyme used to cut DNA at specific sites. This enzyme is known as genetic scissors.
• Liqase: The enzyme ligase is used for joining DNA at specific sites, this enzyme is called as genetic glue.
• Vectors: Plasmids in bacteria are generally used as vectors. A gene from one cell is transferred to another cell by using suitable vectors.

Kerala Syllabus 10th Standard Biology Notes Pdf Question 4.
How is the new genes become a part of the genetic constitution of target cells?
Answer:
DNA with ligated genes enter the target cell. Thus the new genes become a part of the genetic constitution of target cells.

10th Standard Biology Malayalam Medium Question 5.
Scope of genetic engineering
Answer:
a) Gene therapy
b) Genetically modified animals and crops
c) Forensic test

Gene Therapy

Kerala Syllabus 10th Standard Biology Pdf Question 6.
Why is gene therapy essential?
Gene therapy is the method of curing genetic diseases by removing disease-causing genes from the genome and inserting normal functional genes. Gene therapy is beneficial for the sustenance of humankind.

Human Genome Project

Hss Live 10th Biology Kerala Syllabus Question 7.
What is the significance of the human genome project?
Answer:
The human genome includes about 30000 genes present in his 46 chromosomes. The secret of human genome is revealed through a project, known as the Human Genome Project started in 1990 and ended in 2003 in various laboratories of the world. The Gene mapping technology helped us to identify the location of a gene in the DNA.

10th Biology Notes Malayalam Medium Kerala Syllabus Question 8.
What is the benefit of gene mapping ?
Answer:
Gene mapping is a technology by which we can locate a specific gene in the DNA responsible for a particular trait.

10th Class Biology Notes Kerala Syllabus Question 9.
What is Junk genes ?
Answer:
In human DNA, majority of genes, except the genes that code for protein are non-functional. They are called junk genes.
The relevance of the Human Genome Project:

• Human genome has about 24000 functional genes.
• Major share of human DNA includes junk genes
• There is only 0.2 percent difference in DNA among humans.
• About 200 genes in human genome are identical to those in bacteria.

Genetically Modified Animals& Crops

10th Biology Guide Kerala Syllabus Question 10.
Proteins that can be used for the treatment of diseases in humans are produced through genetic engineering.
Answer:

Sslc Biology Notes Pdf Kerala Syllabus Question 11.
One of the future promises of genetic engineering is pharm animals. What do you mean by pharm animals?
Answer:
Genes responsible for the production of human insulin and growth hormones etc. are identified and inserted in animals like cows or pigs to transform them into ‘pharm animals’ (animals providing pharmaceuticals or medicines). Medicines thus produced can be extracted from the blood or milk of such animals.

Sslc Biology Textbook Solutions Kerala Syllabus Question 12.
Instead of bacteria, animals like cows or pigs are used as medicinal animals or pharm animals. Why?
Answer:
It is easy to rear animals like cows or pigs than the culturing of bacteria. Moreover, medicines can be extracted from their blood or milk.

Dna Finger Printing

The arrangement of nucleotides in the DNA of each person differs. This finding leads to the DNA testing. The technology of testing the arrangement of nucleotides in each person also differs. Hence this technology is also called DNA fingerprinting. Alec Jeffrey in 1984 paved the way for DNA testing.

Sslc Biology Notes Malayalam Medium Pdf Question 13.
How are persons identified through DNA testing?
Answer:
The arrangement of nucleotides in the DNA of each person differs.

Question 14.
What is the basis of DNA testing?
Answer:
The arrangement of nucleotides in the DNA of each person differs. This finding lead to the DNA testing.

Question 15.
How is it possible to identify the person’s blood relatives?
Answer:
The arrangement of nucleotides among close relatives have many similarities. So DNA fingerprinting is helpful to find out hereditary characteristics, to identify real parents in cases of parental dispute.

Question 16.
What is the scope of DNA testing?
Answer:
DNA fingerprinting helpful to find out hereditary characteristics to identify real parents in cases of parental is missing due to natural calamities or wars. DNA of the skin hair, nail blood and other body fluids obtained from the place of murder, robbery, etc. is compared with the DNA of suspected persons. Thus the real culprit can be identified from among the suspected persons through this method.

Question 17.
Observe the collage given below analysis and prepare notes about it.

Answer:
It is criticized that genetically modified varieties are threat to indigenous varieties and may cause health issues to humans. There are possibilities to use the genetically modified organisms are bioweapons that might be applied any country to their enemies is called Bioware. This becomes a threat to the existence of human beings.

Question 18.
We should utilize science and technologies for the well being of man and other living beings. On the basis of the above statement, should we favor gene technology, which has a few demerits? Share your thought.
Answer:
Though there are certain possibilities of misuse, gene technology has many merits. Hence, we should utilize science and technologies for the well being of man and other living beings.

Question 19.
Is it right to misuse technologies that are used for human progress? As such possibilities prevail, can we promote genetic engineering, organize a debate in the class on this topic.

Let Us Assess

Question 1.
Which of the following is not a part of modern genetic engineering?
a) DNA profiling
b) Gene mapping,
c) DNA fingerprinting.
d) X-ray diffraction.
Answer:
d) X-ray diffraction.

Question 2.
Gene therapy is an example of the benefits of science for human existence.
a) What is gene therapy?
b) What was the discovery that led to gene therapy?
c) How does gene therapy become useful to human beings?
Answer:
a) Gene therapy is the method of curing genetic diseases by removing disease-causing genes from the genome and inserting normal functional genes.
b) Gene mapping
c) We can cure genetic diseases and disorders by gene therapy.

Question 3.
‘Since genetic engineering has many harmful effects, it can’t be promoted’. Do you agree to this statement? Why?
Answer:
No. Though there are certain possibilities of misuse, gene technology has many merits (like medicines, vaccines, treatment of genetic diseases, production of high yield and resistant varieties of food crops). Hence, we should utilize science and technologies for the well being of man and other living beings.

Extended Activities

Question 1.
Prepare a slide presentation including the stages of production of insulin through genetic engineering.
Answer:

Question 2.
Prepare a science excerpt collecting pictures and news related to genetic engineering.

Genetics for the Future More Questions And Answers

Question 1.
Traditionally, human beings adopted and utilized various methods of biotechnology. Substantiate this statement with suitable examples.
Answer:

• Yeast (a fungus) was used to prepare food items like bread.
• Bacteria and fungi were utilized to convert sugar into alcohol or acids.
• Practiced the method of selecting and rearing of cattle or crops of superior hybrid variety

Question 2.
Give example for modern biotechnological practices.
Answer:

• Development of human insulin-producing bacteria.
• Production of ‘pharm animals’, that yielding medicines or vaccines.

Question 3.
The scope of modem biotechnology is endless. Substantiate this statement providing apt examples.
Answer:
The statement is true. Organisms that can withstand adverse conditions, beautiful flowers, amazing animals, effective vaccines, food crops, etc. can be developed through biotechnology.

Question 4.
Describe the stages in the production of human insulin bacteria through the process of genetic engineering.
Answer:

• From human DNA, cut the gene responsible for the production of insulin.
• This gene is joined with cutting of bacterial DNA (plasmid)
• Insert the joined DNA in the bacterial cell.

Question 5.
Both the genetic scissors and genetic glue are used in the process of genetic engineering. What do you mean by these?
Answer:
The enzymes like Restriction endonuclease, used to cut DNA at specific sites, are generally called as ‘genetic scissors’. The enzymes like Ligase, used for joining DNA at specific sites, are generally called as ‘genetic glue’.

Question 6.
Observe the given illustration and answer to the following questions

a) Name the technology indicated here.
b) Name the general term for enzymes that use to cut genes.
c) Which is the vector used in this process?
Answer:
a) Genetic engineering
b) Genetic scissor
c) Bacterial DNA (plasmid)

Question 7.
Define ‘vectors’ in genetic engineering.
Answer:
Vectors are other DNA (usually bacterial DNA), by which genes can be transferred from one cell to another.

Question 8.
Paravur Fire Tragedy: The remnants of body parts sent for DNA test to identify missed persons.
What is the test indicating in this news? How is it possible to identify any person from minute remnants of their body parts?
Answer:
DNA fingerprinting (DNA profiling or DNA Testing). DNA of the skin, hair, nail, blood and other body fluid obtained from the place is compared through DNA profiling with the DNA of suspected person’s blood relatives.

Question 9.
Identify this person. What is the technology that he put forwarded in 1984? Mention its importance.

Answer:
Alec Jeffrey.
He paved the way for DNA testing, by which we can find out hereditary characteristics, identify real parents in the case of parental dispute and also can identify persons found after long periods of missing.

Question 10.
Genetic engineering caused tremendous changes in medicinal field, how? Answer with examples.
Answer:

• Human insulin-producing bacteria.
• Pharm animals, which produce human insulin and growth hormones
• Medicine producing plants

Question 11.
Instead of bacteria, animals like cows or pigs are used as medicinal animals or pharm animals. Why?
Answer:
It is easy to rear animals like cows or pigs than the culturing of bacteria. Moreover, medicines can be extracted from their blood or milk.

Question 12.
Explain briefly about the merits of genetic engineering in the fields of food and agriculture?
Answer:
Genetic engineering influenced in the production of genetically modified disease resistant and high yield varieties of animals, food crops and cash crops.

Question 13.
Bioweapons are crucial threat to human beings. What are bioweapons? Which is the technology behind biowar?
Answer:
Bioweapons are genetically modified pathogens that might be applied any country to their enemies. Genetic engineering is the technology behind this kind of BioWare.

Question 14.
Make a few logo sentences that can be used for the awareness programme against the misuse of science and technology.
Answer:

• Genetic modification can be allowed only for the benefit of mankind.
• Avoid all weapons including bioweapons, save life.
• Science and technologies are meant for protection, not for destruction.

Question 15.
“Genetic engineering is the branch of Science that transforms the living world”.
a) What is your opinion on the above statement?
b) Give reason to substantiate your opinion.
Answer:
a) I agree with this statement
b) In agriculture, medicine, use of superbugs, DNA fingerprinting, misuses

Question 16.
BT Brinjal is less subject to pest attacks. (March 2013)
a) How is it possible?
b) What are the advantages of genetic modifications?
c) What are the harmful effects of genetic modifications?
Answer:
a) The gene which introduced in these plants causes the production of a protein since this protein can destroy pests.
b) Today insulin without any side effects is being manufactured through genetic engineering.

• Superbugs are one of the products of genetic engineering
• Bt cotton and Bt Brinjal are produced.
• The DNA fingerprinting which is used to prove disputed parentage and criminal offenses.

c) A possibility to the pathogens which will not yield to any medicine.

• Superbugs introduced in oil fields destroy oil fields

Genetics for the Future Questions & Answers

Question 1.
Analyze the word pair relationship and fill in the blanks: (Question Pool-2017)
a) Restriction endonuclease: genetic scissors
……………………………………..: genetic glue
b) DNA profiling: Tests the arrangement of nucleotides
………………………………………: Identifies the loca tionofageneinthe DNA
Answer:
a) Ligase
b) Gene mapping

Question 2.
Choose the right statement from those given below: (Question Pool – 2017)
i) Gene mapping is a technology that identifies the location of a gene in the DNA.
ii) The sum of genetic material presents in an organism is called its DNA.
iii) Enzyme Ligase is used to join the genes.
iv) Gene therapy is the technology that tests the arrangement of nucleotides.
Answer:
i) Gene mapping is a technology that identifies the location of a gene in the DNA.
iii) Enzyme Ligase is used to join the genes.

Question 3.
‘Pharm animals’ is one of the promises of genetic engineering. What is the significance of this concept? (Question Pool-2017)
Answer:

• Genes responsible for the production of insulin and growth hormones are inserted into animals, transforming
• them into pharm animals.
• These animals are easy to be reared and cared when compared to bacteria.
• Medicines can be extracted from their blood or milk.

Question 4.
Observe the logo given below. What does it indicate? (Question Pool-2017)

Answer:
Human Genome Project

Question 5.
“Gene therapy becomes the remedy for genetic diseases.” (Question Pool-2017)’
This is a note in Sethu’s science diary.
Do you agree to this note? Justify your opinion. (2)
Answer:

• Yes
• Gene therapy is the treatment for curing genetic diseases by removing disease-causing genes and inserting functional genes in the genome.

Question 6.
Suma murder case – trace of hair obtained from the site of incidence enabled to identify killer. (Question Pool-2017)
a) Read the above news. Name the technology that helped to find the killer?
b) Cite two other uses of this technology
Answer:
a) DNA fingerprinting
b) 1. to solve parental dispute
2. to identify culprits
3. to identify persons

Question 7.
“Insulin-producing bacteria created” – news report Santhosh raises the following doubts about the news. What explanations would you give as a student of Genetics? ‘
a) Which is the technology that helped to create insulin-producing bacteria?
b) Will the next generation of this bacteria be able to produce insulin? Give reason.
Answer:
a) Genetic Engineering
b) 1. Yes
2. Because the gene responsible for the production of insulin is there in the next generations

Question 8.
Given below are the various steps involved in the production of insulin through genetic engineering, Arrange them appropriately. (Question Pool – 2017)
a) Producing active insulin from this
b) Cutting the gene responsible for the production of insulin from human DNA.
c) Bacteria produce inactive form of insulin.
d) Isolating bacterial DNA.
e) Joining the gene with bacterial DNA and inserting it into the bacterial cell.
f) Providing a favorable medium for the multiplication of bacterial
Answer:
b → d → c → f → c → a

Question 9.
A debate has been organized in the topic. ‘Genetic Engineering – scope and challenges’. (Question Pool – 2017)
List out 3 scopes encountered in the field of Genetic Engineering for Anoop and 3 challenges for Safa respectively.
Answer:
Scopes: In the field of medicine, food crops, cash crops, cattle management, nature conservation, gene therapy, etc.
Challenges: Genetic modifications – violation of rights, bioweapons, BioWare, threat to indigenous varieties, health problems in man, superbugs, etc.

Question 10.
Identify the odd one and write the common feature of others: (1) (Question Pool – 2017)
DNA profiling, Electrocardiogram, gene mapping, gene therapy
Answer:
a) Electrocardiogram
Others related to genetic engineering

Question 11.
Given below is a word tree prepared by Appu for classroom presentation. Help him to complete the tree by choosing the words given in the box: (2) (Question Pool-2017)
Junk genes, Ligase, Gene therapy DNA profiling, Restriction endonuclease, Gene mapping, Plasmid, Genetic engineering.

Answer:
a) Restriction endonuclease
b) Ligase
c) DNA profiling
d) junk genes
e) Gene mapping
f) Gene therapy

Question 12.
Observe the table and form matching pairs. (2) (Question Pool-2017)

Answer:
a) ii
b) iv,
c) i,
d) ii

Question 13.

Didn’t you read the news report? (Question Pool -2017)
a) What is the basis of DNA test?
b) How is it possible to identify relations through DNA test?
Answer:
a) The arrangement of nucleotides in the DNA differs in different individuals
b) The arrangement of nucleotides among close relatives have many similarities.

Question 14.
Analyse the table given below and answer the following questions. (Orukkam – 2017)

a) What are the desirable characters that you like from hybridization between crop A and B?
b) Is there any chance for getting plants with undesirable characters in the same hybridization? Explain the reason for this chance in the light of Mendel’s experiment in pea plant?
c) Can you suggest a remedy for this problem?
Answer:
a) More productivity, and resistance to the disease.
b) Yes. Four different types of plants may forms in the ratio of. 9 : 3 : 3 : 1, as a result of the self-pollination of plants A and B.
c) Genetically modified plants is the remedy for this.

Question 15.
Read the statement given below and answer the following questions. (Orukkam – 2017)
Gene mapping is the method to identify the location of gene in the DNA responsible for a particular trait.
a) How does gene mapping help in insulin production?
b) What is the significance of pharm animals?
c) What is meant by gene therapy?
Answer:
a) We can locate the correct position of gene responsible for insulin production.
b) Pharmammals produce medicines, growth ‘ hormones, human insulin, etc.
c) The process in which new active gene is added in the place of diseased or inactive genes to rectify genetic diseases is known as gene therapy.

Question 16.
Complete the illustration which represents the scope and misuses of genetic engineering. (Orukkam – 2017)

Answer:
A) Food crops and cash crops with high productivity and disease resistance
B) Medicines: From pharm animals, plants and microorganism
C) DNA fingerprinting: To identify person in disputes.
D) Possibility of BioWare: Using genetically modified organisms as bioweapons
F) Violation of rights: Genetic modification is intrusion upon the freedom of organisms.

Question 17.
Explain the difference between”traditional biotechnology and modern biotechnology with suitable examples (Orukkam – 2017)
Answer:
In traditional biotechnology process, we select and use organisms having desirable qualities, eg. Yeasts were used in bread making.

Question 18.
What are the scope of DNA fingerprinting and gene mapping? (Orukkam – 2017)
Answer:
DNA fingerprinting:

• To identify. real parents
• To find out hereditary characteristics
• To identify apt persons found after a long period of missing.

Gene mapping:

• To identify the correct position of genes in DNA responsible for each characteristic.
• To rectify genetic disorders through gene therapy.
• To produce new varieties of organisms with desirable qualities.

Question 19.
Write down any two arguments that evolved during the debate about the topic “Is genetic engineering for human progress?” From support and against group. (Orukkam – 2017)
Answer:
Environment: Neutralizing substances that cause pollution to nature, Gene therapy, New desirable varieties.
Misuses: Threat to indigenous varieties: Genetically modified varieties cause harm to indigenous varieties.

You can Download Fibres and Plastics Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 17 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 17 Fibres and Plastics

We have variety of substances around us for improving our lifestyle. The natural resources are utilised to produce a variety of modern materials.

Fibres And Plastics Class 8 Kerala Syllabus Polymers

Cotton. Silk, jute, wool, rubber etc are the molecules belonging the group of polymers. Polymers are macromolecules formed by the combination of large number of simple molecules called monomers. Fibres are the polymers suitable for the manufacture of strong threads. Plastics are the polymers which can be moulded into different shapes. Rubber is an elastic polymer.

Manmade polymers

In order to overcome the demerits like less availability, less durability etc several synthetic polymers have been prepared through chemical methods. Synthetic threads have demerits too. Some of them are low aeration, low ability to absorb water, high inflammability etc

Basic Science Class 8 Chapter 17 Kerala Syllabus Plastic

These are the substances that changed the very face of human life which are having different properties.

Thermoplastics and thermosetting plastics

The plastic that gets softened on heating and hardened on cooling is thermoplastics. The plastic which remains soft when heated during its manufacture, and gets hardened permanently on cooling is thermosetting plastics.

Plastics and Pollution

Even though plastics are very useful substance, the careless use and misuse lead to environmental pollution. Plastics give many benefits to mankind. Forest conservation, household utility etc are some of the benefits. There are many ways to reduce pollution due to plastics. Some of them reduce the use of dispo¬sable plastics, reduce; the overuse of plastic materials, use other material like glass, ceramic materials instead of plastic etc.

Fibres and Plastics Textbook Questions and Answers

Basic Science For Class 8 Chapter 17 Kerala Syllabus Questions 1.
Polymers are macromolecules formed by the combination of many monomers.
a. How are polymers classified?
b. Classify the following:
Cotton, Wool, Nylon, Silk, Terylene, Jute, Polyester
Answer:
a. i. On the basis of its formation as natural and man-made.
ii. On the basis of structure as linear branches chain, cross-linked chain
iii. On the basis of process as addition polymer and condensation polymer.
iv. On the basis of molecular strength as fibres and plastics
b. Natural: wool, silk, jute, cotton Man-made: Nylon, Terylene, Polyester

Kerala Syllabus 8th Standard Physics Notes Questions 2.
Some monomers and polymers are given in the following table:

a. What is meant by the terms ‘monomer’ and ‘polymer’?
b. What is the common system of nomenclature of polymers?
Analyse the table and find out.
Answer:
a. Monomer: A Simple molecule having a particular structure
Polymer: Macromolecules formed by combination of large number of monomers.
b. Add the word ‘poly’ before the name of monomer.

Kerala Syllabus 8th Standard Chemistry Notes Question 3.
Natural fibres and synthetic fibres are used in the field of textile manufacturing.
a. Compare their merits and demerits and tabulate.
b. Which of these clothes is most suited for the following situations? Give reason.
i. While cooking in the kitchen
ii. To wear during summer
Answer:
a. Merits:

• Comfortable to wear
• Not inflammable
• Ability to absorb water
• High aeration

Demerits:

• Less available
• Less durability
• Wrinkle easily
• Cannot dry easily on getting wet

b. Natural fibres:- Because they are not inflammable, have ability to absorb sweat, high aeration

8th Class Biology Notes Pdf Kerala Syllabus Question 4.
You know what thermoplastics and thermosetting plastics are.
a. Which of these plastics cannot be recycled?
b.You might have noticed that those who collect old plastics do not accept certain type of plastic articles. What are they? What may be the reason for this?
Answer:
a. Thermoplastic
b. Thermosetting plastic. Because they can’t be recycled.

Kerala Syllabus 8th Standard Biology Notes Question 5.
Some argue that plastics are to be completely banned as they cause environmental pollution. What is your view?
Answer:
No. Without plastic, we cannot manage daily life. Control the use of plastic. avoid disposable plastic products and use thermoplastic materials.

8th Class Biology Notes Pdf Malayalam Medium Question 6.
The school science club has N decided to conduct a poster propaganda for creating awareness about pollution due to plastics. Prepare some posters for this.
Answer:
Avoid disposable plastic products use glass, ceramic utensils or natural substances
Use paper or natural materials fo decorations
Don’t dump plastic materials in soil.

8th Standard Basic Science Textbook Kerala Syllabus Question 7.
What suggestions can you propose to realise the concept of ‘plastic waste-free school’? List your findings.
Answer:
Reduce: Buy only what you need because a better way to reduce waste is by not creating it.
Reuse: Instead of throwing out plastic product you no longer think you need, try repurposing them because plastics exist for long time. Eg: use refill in ball pens, use thick plastics which can be reused, reuse the plastic bottle. The banning of plastics below the thickness of 30 micron aims at this. When thickness increases a tendency will arise to use it again.

Recycling: Is the process by which the plastics which is rendered useless is heated and subjected to certain processes to produces new materials. Recycling the rate of environmental pollution

Refuse: We can avoid the use of plastic when it is not necessary. Avoid thin plastic covers when things are bought. Use cloth bag, paper bag or thick plastic covers which can be used for long time

Fibres and Plastics Additional Questions and Answers

8th Biology Notes Malayalam Medium Kerala Syllabus Question 1.
Differentiate between thermoplastic and thermosetting plastic.
Answer:
The plastic that gets softened on heating and hardened on cooling is thermoplastic. When heating physical change occurs.
The plastic which remains soft when heated and gets hardened permanently on cooling is thermosetting plastics. Chemical change occurs when heating.

Kerala Syllabus 8th Standard Biology Notes Malayalam Medium Question 2.
Given some occasions of using plastic. Find the peculiarity of plastic used and fill the table.
Answer:

Answer:

Kerala Syllabus 8th Standard Chemistry Notes Malayalam Medium Question 3.
Write 4 occasions which plastic is harmful to daily life.
Answer:

1. Environmental pollution when it is thrown without any control.
2. When burning it cause air pollution.
3. Hindrance in drainages
4. The water absorption property of soil decreases when plastics are dumped in soil.

Kerala Syllabus 8th Standard Notes Question 4.
What are the uses of plastic in the field of human health?
Answer:
To produce IV tubes, bottles To produce heart valves To produce packets

Synthetic Fibres And Plastics Class 8 Notes Kerala Syllabus Question 5.
What are the uses of plastic in the field of production of house building?
Answer:
To produce roof materials, doors, plumbing and wiring materials.

Class 8 Chemistry Notes Kerala Syllabus  Question 6.
List the peculiarities of plastic.
Answer:
Can mould in any shape, longlasting, insulator, do not conduct heats, not reacting with chemicals and water, will burn.

Question 7.
Separate natural and artificial polymer from the list
Rubber, wool, Pvc, Bakelite, nylon, rayon cellulose, silk, polythene, polyester
Answer:

Question 8.
List the merits of natural and artificial polymer
Answer:

Question 9.
If we heat polyethene cover can we convert into earlier stage? Justify.
Answer:
No. Undergoes chemical change because it is thermosetting plastic.

Question 10.
Filling in the blanks.
1. Natural polymer is ……………
2. Insulin is a ………….
3. is a monomer of polythene. …………..
4. Nylon is type of plastic. …………….
5. Bakelite is an example for plastic. …………..
6. Natural polymer which has elastic nature is ……………
7. Inflammable tendency is higher in …………….
8. Thermoplastic is a ……………. type polymer.
9. The constituent unit of polymer is called
10. Polymers of plant origin are made up of …………..
Answer:
1. Starch
2. Protein
3. Ethylene
4. thermoplastic
5. thermosetting
6. Rubber
7. Synthetic fibres.
8. linear
9. monomer
10. cellulose

Question 11.
Give examples for natural fibres.
Answer:
Coconut husk, cotton, hemp, silk etc.

Question 12.
What is vulcanization? What is the use of it?
Answer:
’The process of heating rubber with sulphur is called vulcanization. By vulcanization rubber retains its form and to increase the hardness, Moreover, tensile strength, elasticity keeping stability at high temperature etc. are increased.

Question 13.
Give four difference between natural rubber and synthetic rubber.
Answer:
1. Synthetic rubber is harder than natural rubber.
2. Natural rubber is easily flammable.
3. The elasticity of natural rubber is lower than that of synthetic rubber.
4. Synthetic rubber keeps stability at higher temperature.

Question 14.
Classify the following into thermoplastic and thermosetting plastics.
Bakelite, polyvinyl chloride, nylon, celluloid, urea-formaldehyde, Teflon, polyester, polythene.
Answer:

Question 15.
Find out odd one and give reason. Bakelite, polyester, polythene, melamine formaldehyde.
Answer:
Polythene: All the others are thermosetting plastics.

You can Download The Paths Traversed by Life Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Biology Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Biology Solutions Chapter 8 The Paths Traversed by Life

The Paths Traversed by Life Text Book Questions and Answers

The Paths Traversed By Life Kerala Syllabus 10th Question 1.
Analyze the following illustration and prepare a note on the theory of chemical evolution.

Answer:
Gases like hydrogen, ammonia, water vapor and methane which present in the atmosphere of primitive earth reacted together to form simple organic molecules like amino acids. Thunder and lightning ultraviolet radiations, volcanic eruptions are energy sources. Condensation of water vapor present in the atmosphere and the resulting incessant rain led to the formation of oceans. Simple organic molecules are formed first in the primitive ocean. By further reactions, complex molecules were found including genetic material to evolve the first primitive cell.

Urey – Miller Experiment

The Path Traversed By Life Kerala Syllabus 10th Question 2.
Which were the conditions of the primitive earth, recreated by Stanley Miller and Harold Urey?
Answer:
Stanley Miller and Harold Urey re-created the experimental setup, in which the glass flask considered as the primitive atmosphere that contained methane, ammonia, hydrogen and water vapor. Instead of lightning or other energy sources, they passed high voltage electricity through the gaseous mixture. The condensed water from this gaseous mixture was considered as the primitive ocean.

Sslc Biology Chapter Wise Questions And Answers Question 3.
The organic substances synthesized through Urey- Miller experiment?
a) Amino acids
b) Nucleic acids
c) Nucleotides
d) DNA
Answer:
a) Amino acids.

Kerala Syllabus 10th Standard Biology Notes Pdf Question 4.
Observe the illustration and answer the questions that follow:

a) Name the scientists who set up this experiment.
b) What was the aim of their experiment?
c) Name the gases which are filled within the ‘A’ of the figure.
d) Which type of organic molecules did they synthesize through this experiment?
Answer:
a) Stanley Miller and Harold Urey.
b) To prove the theory of chemical evolution (Oparin- Haldane hypotheses) scientifically.
c) Methane, ammonia, hydrogen and water vapor.
d) Amino acids

Hsslive Guru 10th Biology Kerala Syllabus Question 5.
Analyze the major events related to the origin of life?

Answer:
Eukaryotes are originated from primitive prokaryotic cell. Gradually colonies of membrane-bound eukaryotic cells are formed. This led to the emergence of multicellular organisms.

Evolution – Through Theories

A) Lamarckism:

Biology Class 10 Kerala Syllabus Question 6.
Explain the ideas of J.B. Lamarck about organic evolution.
Answer:
Theory of Inheritance of Acquired Characters. Continuous use or disuse of an organ results variations to develop changes in the structure of that organ (Acquired characters). These will be transmitted to the next generation to form new species.

Hss Live Guru 10th Biology Kerala Syllabus Question 7.
Why did scientists criticize Lamarck’s view?
Answer:
The changes in the body (Acquired characters) that occur in the lifetime of an organism do not affect its genetic constitution and hence not possible to transmit to the next generation.

B) Darwinism:

Hss Live Guru Biology 8 Kerala Syllabus Question 8.
The items given in the box indicates a scientist.
a) Identify the scientist.
b) What idea did he put forward in the field of evolution?
1. The origin of species
2. HMS Beagle
3. Galapagos Islands
4. Survival of the favorable ones.
Answer:
a) Charles Darwin.
b) Theory of Natural Selection

Kerala Syllabus 10th Standard Biology Pdf Question 9.
Given below is the illustration of the ‘Darwin’s finches’ on the basis of indicators, analyze illustration.

a) Which peculiarity of the finches attracted Darwin.
b) How do these peculiarities help finches in their survival?
Answer:
The differences in the beaks of finches attracted Darwin.
The finches of Darwin’s had beaks adapted to their feeding habits insectivore finches have small beaks, cactus feeding finches have long and sharp beaks, woodpecker finches feed on worms in tree trunks have sharp beaks and ground finches feed on seeds have large beaks.

10th Standard Biology Malayalam Medium Question 10.
Complete the table of Darwin finches which showed differences in food and food habits.
Answer:

Sslc Biology Chapter Wise Questions Kerala Syllabus Question 11.
Compare the ideas of Thomas Robert Malthus and main concepts of Theory of Natural Selection’ put forward by Darwin.
Answer:

The Theory Of Natural Selection

Hss Live Guru Biology 10 Kerala Syllabus Question. 12
Describe the theory of Natural Selection proposed by Charles Darwin.
Answer:
Variations develop in each species. Only those variations, which are favorable to that nature, survive and those which are unfavorable get eliminated, According to Darwin, organisms of one kind, when produced in large numbers (Over Production), compete for food, space, mate, and other limited resources (Struggle for Existence). In this struggle, only organisms with favorable variations survive in that nature (Survival of the Fittest). Over a long period, the favorable variations accumulate, resulting the formation of new species.

Hss Live Guru 10 Biology Kerala Syllabus  Question 13.
What was the limitation in Darwin’s theory? Who gave sufficient explanations to this?
Answer:
Darwin could not explain the reasons for continuous variations in organisms. However, Hugo deVries explained that one of the reasons for variations in organisms is mutation (sudden changes that occur in genes).

Scert Class 10 Biology Notes Kerala Syllabus Question 14.
Describe Neo Darwinism.
Answer:
Neo Darwinism is the modified version of Darwin’s theory in the light of new information from the branches of genetics, cytology, geology, and paleontology about the reasons of variations occurred in organisms. Hugo deVries first supported Darwin by his theory of mutation.

Sslc Biology Notes Malayalam Medium Pdf Mutation

Sudden changes that occur in genes are called mutation.

Mutation Theory:
Mutation theory explains that new species are formed by the inheritances of sudden changes that occur in genes. This theory formulated by a Dutch scientist Hugo de Vries.

Evidence Of Evolution

Sslc Biology Notes Pdf Kerala Syllabus Question 15.
Mention the branches of science which provide evidence to organic evolution.
Answer:

• Paleontology (fossil study)
• Comparative morphological studies
• Biochemistry-physiology
• Molecular biology

Kerala Syllabus 10th Standard Biology Question 16.
Define fossils.
Answer:
Fossils are remnants of primitive organisms preserved in earth crust.

Scert Biology Class 10 Kerala Syllabus Question 17.
Analyze the following illustration.

Answer:
The study of fossils reveals that complex structured organisms are evolved from primitive simple organisms.
Certain linking fossils reveal the evolution of one form of organisms from another form.
Extinction of some species as well as the emergence of new species.

Comparative Morphological Studies

Question 18.
‘Comparative study of structure gives evidence to evolution’. Evaluate this statement.
Answer:
Though there are differences in the external structure (morphology) among different organisms, there are certain similarities in their internal structure (anatomy). The evidence from the comparative morphological studies justifies the inferences that all organisms were evolved from a common ancestor.

Question 19.
The morphological and anatomical structure of forelimbs in lizard, bat and sea cow are shown here. Observe the illustration and answer to the following questions.

a) Are these forelimbs differ morphologically or anatomically or both? What is the reason behind this difference?
b) What is the term used for such organs that are similar in structure but different in function?
Answer:
a) Only morphological differences are there among these organs. Reason for these differences are their adaptations to live their own habitats.
b) Homologous organs.

Question 20.
Homologous organs are seen among reptiles, birds, and mammals. What do you mean by homologous organs?
Answer:
Organs that are similar in structure but perform different functions are called homologous organs.

Biochemistry And Physiology

Question 21.
Observe the illustration and write the inference.

Answer:
All organisms are made up of cells with protoplasm. There are similarities among the cell organelles and cellular activities. Enzymes control chemical reactions and energy is stored in ATP molecules in all organisms. Hereditary factors are gene , seen in DNA and the structure of DNA is alike in all. Carbohydrates, proteins, and fats are the basic substances. There are similarities in growth, excretion, etc.

Question 22.
What evolutionary interference can be arrived from the evidence from the comparative morphological, Biochemical and physiological studies?
Answer:
The evidence from the comparative morphological, Biochemical and physiological studies justify the inferences that all organisms were evolved from a common ancestor.

Molecular Biology

Question 23.
What evidences of organic evolution do the study of molecular biology provide us?
Answer:
Through a comparative study of protein molecules in different species, the evolutionary relationship (similarity/difference) among organisms can be identified. For instance, we can analyze the similarities or differences in the sequence of amino acids in the beta chain of hemoglobin molecules of different mammals and thereby we can understand about the evolutionary relationship among them.

Question 24.
The differences of the sequential arrangement of amino acids in the beta chain of hemoglobin of man with other animals are given below.

Which animal is so close to human beings? What is the reason for this?
Answer:
Chimpanzee.
There is no difference in the sequential arrangement of amino acids in the beta chain of hemoglobin in man and chimpanzee.

Question 25.
How molecular studies can infer the period of separation of different groups from their ancestors?
Answer:
Mutations are the main reason for evolutionary changes. Through the molecular studies, we can find out how mutation occurs in the genes that determine amino acid sequences in protein molecules. From this, we can infer the period of separation of different group of organisms from their ancestors.

Evolution Of Human Beings

Question 26.
An evolutionary tree relating to certain organisms including humans is given below. Analyze and prepare a note.
Answer:

Question 27.
Which organisms is the closest to humans in specific characters?
Answer:
Chimpanzee

Question 28.
Do you agree with the statement that man is evolved from monkeys? What is your opinion?
Answer:
This statement is wrong. Man come under the group Hominoidea while monkeys are included in Cercopithecoidea. It is believed that both the ancestors of man and monkeys are evolved from a common ancestor.

Question 29.
Name of a few animals are given in the box. Out of these which one is more similar to man? Mention any two peculiarities of the other animals.
Baboons, Chimpanzees, Monkeys

Question 30.
Analyze the following illustration which shows the evolutionary history of modern man.

Answer:

Question 31.
The primitive man might be lived in African continent Do you agree this statement? What is your opinion?
Answer:
This statement might be right, because, fossils of Ardipithecus ramidus, Australopithecus afarensis, Homo habilis, and Homo erectus were discovered ” from African continent.

Question 32.
How do modern men differ from the other groups of human beings?
Answer:
Modern man have developed brains and equipped with advanced technologies.

Question 33.
Do the interventions of modern man cause any change to natural evolutionary process? How?
Answer:
Yes. Biodiversity is on a dangerous decline due to the interference of human beings in nature and natural resources. By human interventions, climatic changes brought in as well as the extinction of many organisms.

Question 34.
Here is an incomplete illustration of human evolutionary tree. Find out the missing links.

Answer:
A: Australopithecus
B: Homo habilis
C: Homo neanderthalensis.

Let Us Assess

Question 1.
Which concept is put forward by the theory of natural selection?
a) Origin of life
b) Origin of species
c) Origin of eukaryotes
d) Chemical evolution of life
Answer:
b) Origin of species

Question 2.
List the main concepts that indicate how the biodiversity seen today has been developed from prokaryotes.
Answer:
Origin of prokaryotes 3500 million years ago
Origin of eukaryotes 1500 million years ago
Origin of multicellular organisms 1000 million years ago

Question 3.
How does the interference of human beings in nature influence the process of evolution? How do these affect the existence of other organisms?
Answer:
Biodiversity is on a dangerous decline due to the interference of human beings in nature and natural resources. By human interventions, climatic changes brought in as well as the extinction of many organisms.

Extended Activities

Question 1.
Prepare and exhibit a model of the experimental set up constructed by Urey-Millerto scientifically prove the theory of chemical evolution.

Question 2.
Prepare a chart illustrating the evolutionary tree of man. (See Question 70)

The Paths Traversed by Life More Questions And Answers

Question 1.
The hypothesis which explain how do the evolution occur in living beings is illustrated below:

a) Write the name of scientist who proposed this hypothesis.
b) This hypothesis was not fully accepted. Give reason. (Model 2014)
Answer:
a) Lamarck
b) variations affect genetic constitution only transferred to next generation

Question 2.
Rearrange the following based on the relationship between different Primates. Chimpanzee – Gorilla – Gibbon – Orangutan (March 2014)
Answer:
Gibbon → Orange → Orangutan → Gorilla → Chimpanzee

Question 3.
DDT was sprayed on some insects as an experiment. Among them some died and some 3 survived A second generation was created by mating the survived ones. Again DDT sprayed. This process was repeated on five generations. The result is given in the table. Analyze it arid answer the following questions.

a) Interpret the result given in the table.
b) What scientific explanation can you give for this result?
c) What may be the result if the experiment is continued? (March 2014)
Answer:
a) After each generation’s resistance to DDT increases, ie. variation increases.
b) It is an example for Darwin’s theory of Natural selection. Due to struggle existence variations will arises and transmitted into next generations.
c) The total population becomes resistant to DDT / Evolution of DDT resistant species.

Question 4.
Name of certain animals belong to primates is given below.
a) Loris,
b) Gibbon,
c) Monkey
i) Which among them is closest to human beings according to the theory of evolution.
ii) Write down any two characteristics of the others. (March 2013)
Answer:
i) b) Gibbon
ii) Lorin
1. Noctural
2. Solitary
Monkey
1. Diurnal
2. Colonial life

Question 5.
Even though chemicals are used continuously, mosquitoes can’t be destructed completely. Write down scientific explanations for this statement on the basis of the theory of evolution. (March 2013)
Answer:
Mosquitoes develop resistant power in them against the chemicals. This is the survival of the fittest. Those mosquitoes which have the ability to resist the action of the chemicals survive, the others die.

Question 6.
Rearrange the following animals according to the evolutionary series.

b) List out any two evidences of organic evolution.
c) Man is involved in the evolution of other living beings. Substantiate your answer. (Model 2012)
Answer:
a) Gibbon, Orang-utan, Gorilla, Chimpanzee, Man.
b) Morphological similarities among animals and similar biochemical reactions indicate about a common ancestor/Different type of fossils are available / Scientific classification or taxonomy and molecular biology indicate that organisms are evolved from common ancestor and how they are interrelated or different, (any 2 points)
c) Yes. The involvement of man in his environment altered the natural evolutionary processes. Human influence may lead to mutation and other natural calamities which lead to evolution or extinction of certain species. Genetic engineering processes caused the production of new species, (any 2 points)

The Paths Traversed by Life Questions & Answers

Question 1.
Identify the odd one from those given below, and write the feature common to others. (Question Pool – 2017)
a) Monkey, gibbon, orangutan, gorilla.
b) acquired characters, overproduction, struggle for existence, favorable variations.
Answer:
a) 1. Monkey
2. Others belong to Hominoidea
b) 1. Acquired characters
2. Others are related to the theory of natural section

Question 2.
Analyze the word pair relationship and fill in the blanks.
a) Monkey: Cercopithecoidea
Chimpanzee:………………….
b) Theory of Natural Selection: Charles Darwin
Mutation theory:………………
Answer:
a) Hominoidea
b) Hugo deVries

Question 3.
The stages related to the origin of life are given below, b Analyse and arrange them correctly. (Question Pool – 2017)
a) Organic compounds
b) Prokaryotic cells
c) Chemical evolution
d) eukaryotic cells
e) Multicellular organism
f) Colonies of eukaryotic cells.
Answer:
(c) → (a) → (b) → (d) → (f) → (e)

Question 4.
Match the following

Answer:
a) (iii)
b) (i)
c) (ii)

Question 5.
An illustration related to chemical evolution is given below. Complete the illustration using the information given in the box. (Question Pool – 2017)
i) RNA, DNA
ii) polysaccharides, peptides, fats
iii) Presence of H₂, N₂, CO₂
iv) Monosaccharides, aminoacids

Answer:
A – (iii)
B – (i)
C – (iv)
D – (ii)

Question 6.
Identify the statements that are related to chemical evolution:
i) Life originated in some other planets in the universe and accidentally reached the earth.
ii) Life originated as a result of the changes that occurred in the chemical substances in water, under specific conditions of primitive earth.
iii) The theory is supported by the organic substances found in the meteors that fell on earth.
iv) A.I.Oparin and J.B.S.Haldane are the proponents of the theory.
Answer:
ii) Life originated as a result of the changes that occurred in the chemical substances in water, under specific conditions of primitive earth.
iv) A.I.Oparin and J.B.S.Haldane are the proponents of the theory.

Question 7.
Tabulate the data appropriately in the box given below.
i) Chemical evolution
ii) Natural selection
iii) Panspermia theory
iv) Mutation theory

Answer:
Origin of life i) Chemical evolution and iii) Panspermia theory
Evolution – ii) Natural selection and iv) Mutation theory

Question 8.
Complete the illustration related to the evolution of human beings appropriately. (Question Pool – 2017)

Answer:
a) Cercopithecidae
b) Hominoidea
c) Small brain

Question 9.
Complete the table using the data given in the box. (Question Pool – 2017)
Most primitive members of the human race, cranial capacity is 460cm³, made weapons from stones and bone pieces, cranial capacity is 1000 cm³, thick chin, and large teeth, cranial capacity is 1700cm³, slender body, cranial capacity 610cm³, cranial capacity 325cm³, contemporary of modem man.

Answer:

Question 10.
Given below is an illustration of the differences observed by Darwin in the beaks of the finches in the Galapagos island. (Question Pool-2017)

Finches with different beaks emerged from the ancestor finch. Substantiate the statement.
Answer:
Though the finches were similar in sound and nesting habits, only they showed differences in food and food habits. (Insectivore finches have small beaks, cactus feeding finches have long and sharp beaks, woodpecker finches feed on worms in tree trunks have sharp beaks and ground finches feed on seeds have latg beaks, etc.) So, Darwin thought that they were evolved from a common ancestor.

Question 11.
The links in the evolutionary history of modern man are given in the box. Complete the illustration choosing the appropriate ones from the box. (Question Pool – 2017)
Ardipithecus ramidus, Homo neanderthalensis, Homohabilis, Homo erectus, Homo sapiens, Australopithecus afarensis

Answer:
a) Australopithecus afarensis
b) Homohabilis
c) Homoneanderthalensis

Question 12.
‘The constant use of antibiotics develops resistance in bacteria”. (Question Pool-2017)
Substantiate the above statement on the basis of the theory of natural selection.
Answer:
The constant use of antibiotics develops resistance in bacteria. It is an example for Darwin’s theory of natural selection. Due to struggle for existence variations will arise and transmitted to next generation. So bacteria become resistant to antibiotic.

Question 13.
Complete the illustration given below, related to the evidences that support the evolution of new species. (Question Pool – 2017)

Answer:
B – Comparative morphology
D – Molecular biology

Question 14.
Scientific study of the remnants, body parts and imprints of primitive organisms are evidences on evolution.
a) What inferences do we arrive at, through such scientific studies?
b) How will you explain these inferences as evidences on evolution?
Answer:
a) 1. Primitive fossils have simple structure
2. Recently formed fossils have complex structure
3. Certain fossils are connecting links between different species.
b) 1. Organisms with complex structures are formed from those with simple structures.
2. Certain fossils indicate the evolution of one species from another species.

Question 15.
The forelimbs of the organisms shown in the picture below, do not show any similarity. Hence they do not have any evolutionary relationship. (Question Pool-2017)

How will you respond to this statement? Substantiate
Answer:
Do not agree

• The similarity in blood vessels, nerves, muscles and bones.
• Such anatomical resemblances indicate that they evolved from a common ancestor
• Differences in their external appearance are their adaptations to like in different habitats

Question 16.
Arrange the following links in human evolution in the ascending order of their cranial capacity. (Question Pool-2017)
Homo sapiens, Ardipithecus,
Homo erectus. Homo habilis
Answer:
Ardipithecus → Homo habilis → Homo erectus → Homo sapiens

Question 17.
Arrange the links in human evolution appropriately: (Question Pool – 2017)

Answer:
A -Ardipithecus
B-Australopithecus
C- Homo erectus
D-Homo sapiens

Question 18.
There is a common ancestor for all the different species that exist today.
Explain how Biochemical and Physiological studies substantiate the above statement?
Answer:
All organisms are made up of cells with protoplasm, There are similarities among the cell organelles and cellular activities. Enzymes control chemical reactions and energy is stored in ATP molecules in all organisms. Hereditary factors are gene, seen in DNA and the structure of DNA is alike in all. Carbohydrates, proteins, and fats are the basic substances. There are similarities in growth, excretion, etc.

Question 19.
An excerpt from the article “Darwin view of evolution” is given below. (Question Pool – 2017)
Variations often occur in organisms. New species arise when these variations are subjected to natural selection. But Darwin could not explain the reason for these variations.
a) Explain the reason for variations on the basis of Genetics.
b) How was Darwinism revised later?
Answer:
a) Mutations taking place in gene, chromosome; This brings about variations,
b) Darwinism was revised as Neodarwinism in the light of new information form the branches of genetics, cytology, geology, and paleontology.

Question 20.
The table given below shows the difference in amino acids obtained from a comparative study of the (3 chains of hemoglobin of different organisms Analyse the table and answer the questions. (Qn. Pool-2017)

a) Which organism is more closely related to man on the basis of evolution? Substantiate your observation.
b) Explain the reason for the difference in amino acids of hemoglobin of the organisms listed in the table on biochemical basis.
Answer:
a) 1. Chimpanzee
2. No difference in the amino acid sequence in the Beta-chain of hemoglobin
b) 1. Mutations may occur in the genes that determine amino acid sequence
2. This causes changes in amino acid sequence.

Question 21.
A few concepts of scientists like Darwin and Malthus are given below. Classify them in the table given below. (Question Pool – 2017)
a) Selection by nature leads to the diversity of species.
b) Rate of food production does not increase proportionately to the increase in population.
c) Those organisms that overcome the unfavourable situations will survive.
d) Scarcity of food and starvation leads to struggle for existence.

Answer:
Concepts of Darwin – a, c
Concepts of Malthus – b, d

Question 22.
An excerpt from the science article. ‘Man and Evolution’ is given below. Analyze the excerpt and answer the questions. (Question Pool -2017)
Certain evolutionary features make man different from other animals included in evolutionary history. This helped him in his dominance over nature and other organisms. His interference had created a negative impact on the existence of other organisms.
a) What are the features that make man different from other animals?
b) Has man’s interference led to Biodiversity deterioration as mentioned in the excerpt? Evaluate.
Answer:
a) High cranial capacity, ability to stand erect, ability to walk in two legs, ability to make and use machines and tools, cultural development
b) Yes, Climate changes, deteriorating habitats, extinction.

Question 23.
Observe the illustration and answer the questions given below. (Orukkam – 2017)

a) What are the characteristic features of Cercopithecidae group?
b) Name the group which includes man and gorilla. What are the characteristics features of this group?
c) Which organism is close to man from the evolutionary point of view?
Give explanation for this on the basis of molecular biology?
Answer:
a) Small brain, Longtail
b) Hominoidea
c) Developed brain, free moving hands
d) Chimpanzee
The amino acid sequencing in the beta chain of hemoglobin in both chimpanzees and man is almost same.

Question 24.
There exist certain scientific proofs about the formation of different species by evolution. Justify this statement. (Orukkam – 2017)
(Hints – Fossils, Comparative morphological studies, Molecular biology)
Answer:
a) Evidence from fossil studies-Agradual change from simple structure to complex structure, Linking between two groups of organisms.
b) Comparative study of homologous organs Reveals the existence of a common ancestor.
c) Molecular biology proves the evolutionary relationship among different groups of organism.

Question 25.
The different views regarding the evolution of species are given below. (Orukkam – 2017)

a) Name the scientists who proposed those views.
b) Name the view which was not accepted by scientific world. Why?
Answer:
a) A – Gene Lamarck,
B – Hugo devices
C – Charles Darwin
b) Theory of Lamarck was not accepted. The acquired characters described by him, make no change in genes, to affect evolutionary change.

Question 26.
Fill in the blanks by observing the relationship in the first pair. (Orukkam – 2017)
a) Cranial capacity 610 cu.cm: Homo habilis
Cranial capacity 1430 cu.cm : ……………………
b) Gibbon: Hominoidea
Monkey:………………….
Answer:
a) Homoneanderthalansis
b) Cercopithecidae

Question 27.
What do you mean by homologous organs? What evidences do they give for evolution? (Orukkam – 2017)
Answer:
Organs which are different in external structure and function, but similar in internal structure are called homologous organs. The comparative study of homologous organs reveals the possibility of a common ancestor to such animals.

You can Download For the Continuity of Generations Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 14 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 14 For the Continuity of Generations

For the Continuity of Generations Textbook Questions and Answers

For The Continuity Of Generations Biology Kerala Syllabus 8th Reproduction

In nature, there are various methods of reproduction to produce new generations.

For The Continuity Of Generations Kerala Syllabus 8th Budding

This is a method of asexual reproduction seen in Hydra, Yeast etc. The buds formed from the parent body detaches from it when mature and develop into a new organism.

Basic Science Class 8 Chapter 14 Kerala Syllabus Binary fission

Binary fission is seen in prokaryotes. Existing cell divides to form 2 new cells. In favorable conditions, bacteria-like organisms reproduce by binary fission

Kerala Syllabus 8th Standard Chemistry Notes Spore formation

This type of asexual reproduction is seen mainly in fungus. Spores are microscopic cells that can survive unfavourable conditions and develop into new organism on the return of favourable season.

Basic Science For Class 8 Chapter 14 Kerala Syllabus Pollination, Fertilization

Flowers are the sex organs in plants. Pollination is the process of transfer of pollen grains to the stigma of the flower. After pollination pollen tube grows towards the ovary. Simultaneously generative nucleus in the pollen grain divides at the pollen tube and two sperms are formed.
One of the sperms fuses with the ovum and forms the zygote. This process is called fertilization. The second sperm fuses with the polar nucleus in the ovary and forms the endosperm. Zygote develops into embryo and endosperm becomes the stored food for the growth of embryo.

Indicators (Text Book Page No:200)

Kerala Syllabus 8th Standard Chemistry Notes Malayalam Medium Questions 1.
Formation of male gametes
Answer:
After the pollination pollen tube grows towards the ovary. Simultaneously generative nucleus in the pollen grain divides at the pollen tube and two sperms are formed.

8th Class Biology Notes Pdf Malayalam Medium Questions 2.
Formation of embryo
Answer:
One of the sperms fuses with the ovum and forms the zygote. This process is called fertilization.

8th Class Biology Notes Pdf Kerala Syllabus Questions 3.
Formation of endosperm and its function
Answer:
The second sperm fuses with the polar nucleus in the ovary and forms the endosperm. Zygote develops into embryo and endosperm becomes the stored food for the growth of embryo.

8th Standard Chemistry Textbook Kerala Syllabus Male Reproductive system

The main parts of male reproductive system are Vas deferens, prostate gland, penis, testis etc.

Basic Science Class 8 Chapter 14 Solution Kerala Syllabus Vas deferens

Carries sperms from testis to urinary tract.

8th Standard Chemistry Textbook Kerala Syllabus Prostate gland

Secretes a fluid that contains factors required for the movement and nourishment of sperms.

Biology Class 8 Malayalam Medium Kerala Syllabus Penis

Deposits sperms into vagina

Basic Science Question Answer Chapter Wise Class 8 Testis

Produces sperms and male hormones.

Class 8 Science Notes Pdf State Syllabus  Sperm

Sperm is motile. It has a head, middle piece and tail. Mitochondria present in the middle piece provides energy for movement. In the head, nucleus containing paternal chromosomes are present.

Indicators (Text Book Page No:201)

Questions 4.
Characteristics of sperms
Answer:
Sperms:- Sperms are motile. It has a head, middle piece and tail. Mitochondria present in the middle piece provide energy for movement. In the head, nucleus containing paternal chromosomes is present.

Questions 5.
Location of testes and the production of sperms
Answer:
Testis are found in the scrotal sac outside abdominal cavity. Sperms are produced in the testis.

Questions 6.
Importance of glands
Answer:
Sperms reach the penis along with the fluid produced by accessory glands and they are secreted to outside. Secretions provide a medium for the movement of sperms and nourishment

Female Reproductive System

Ovaries, Oviduct, Uterus, Endometrium, Vagina, etc., are the parts of female reproductive system.
Ovary – Produces Ovum and female hormones.
Oviduct – Fertilization takes place in oviduct.
Uterus – Foetus completes its development in uterus.
Endometrium – Inner layer of Uterus. Foetus attach the endometrial wall.
Vagina – Uterus opens out through Vagina. Sperms are deposited here.

Ovum

Ovum is larger than sperms. They are non-motile. There are specified membranes outside the cell membrane of ovum.

Indicators (Text Book Page No:202)

Questions 7.
Characteristics of ovum
Answer:
Ovum is larger than sperms. They are non-motile. There are specified membranes outside the cell membrane of ovum.

Questions 8.
Function of ovary
Answer:
Produces ovum and female hormones.

Questions 9.
Completing the table (Text Book Page No:202)

Answer:

Menstruation

With the onset of ovulation, there are certain preparations in the uterus to facilitate embryonic development. Endometrium becomes thickened, and more capillary are formed. But if fertilization does not take place the newly formed tissues disintegrate and peel off from the uterine wall. These get eliminated to outside along with blood and mu¬cus through vagina. This process is called menstruation.

Fertilization

Fertilization is the process of fusion of ovum, that releases from the ovary with the sperm in the oviduct.

Placenta

It is the part that connects embryo with the endometrium. Oxygen and nutrients are supplied to the foetus through the umbilical cord formed from the placenta. Waste materials from the foetus are also eliminated through placenta.

Questions 10.
Completing the table (Text Book Page NO:204)

Answer:

Indicators (Text Book Page No:206)

Question 11.
What is adolescence?
Answer:
Adolescence is the period between puberty and adulthood.

Question 12.
How does adolescence influence the physical and mental development of an individual?
Answer:
The characteristics of adolescence like brain development, rapid increase in the height and weight, growth of reproduction organs, increased efficiency of glands, etc., influence the physical and natural development of individuals.

Question 13.
Why is the rate of adolescence growth higher in girls than in boys?
Answer:
In girls, the parts of brain that control physical and mental changes develop fast.

Question 14.
Is there any need to be anxious about the physical changes during adolescence? Why?
Answer:
No Adolescence is only a stage in growth. Bodily changes are part of development into a fully mature organism.

Adolescence and Food

Indicators (Text Book Page No:206)

Question 15.
What is the circumstance that led to the supply of iron-folic acid tablets to students?
Answer:
In adolescence severe anemia, due to deficiency of Iron was reported.

Question 16.
What is the role of food habits to overcome this situation?
Answer:
Including leafy vegetables, eggs, amaranthus, liver, etc., in the diet will help to check diseases like anemia.

Question 17.
How should the food habits be regularised so as to ensure the availability of nutrients for the rapid growth of body in adolescence?
Answer:
Vegetables, leafy vegetables, milk, egg, cereals, pulses, etc., should be included in the diet.

Need for assertiveness

Question 18.
What is your response towards this statement?
Answer:
Agree

Question 19.
Can you cite such instances?
Answer:
Invitation to practice alcohol, drugs, sex abuse, etc., temptation to do the dont’s, instances of crimes, etc.

Question 20.
How will you respond if such instances occur in your life?
Answer:
Say ‘No’.

Let us assess (Text Book Page No:211)

Question 21.
Which of the following activities takes place after fertilization in plants?
A. Pollen tube grows
B. Egg is formed in the ovary
C. Ovule becomes the seed
D. Male gametes are formed
Answer:
C. Ovule becomes the seed

Question 22.
Which part helps in the transportation of materials without mixing maternal and foetal blood?
A. Endometrium
B. Uterus
C. Placenta
D. Amni
Answer:
C. Placenta

Question 23.
Sequentially arrange the process that takes place after pollination in plants.
1. Embryo is formed
2. Pollen tube grows
3. Fertilization takes place
4. Male gametes are formed
5. Zygote is formed
6. Generative nucleus divides
Answer:
1. Pollen tube grows
2. Generative nucleus divides
3. Sperms are formed
4. Fertilization takes place
5. Zygote is formed
6. Foetus is developed

Question 24.
Substantiate the statement: “Excessive likes and dislikes of food materials adversely affects the health”.
Answer:
Avoiding food in order to become slim causes many diseases like anorexia. Overeating causes obesity and other chronic diseases.

Question 25.
Home hygiene and social hygiene are as important as personal hygiene for health. Do you agree with this opinion of the doctor who led an awareness class on health? Why?
Answer:
Definitely, unhygienic environment helps in the proliferation of pathogens. The dangers caused by pollution is much dreaded. Therefore home hygienic and social hygiene are very important as personal hygiene.

Question 26.
“Adolescence is full of challenges and possibilities”.
a. What are the challenges faced by adolescents?
b. What are your suggestions to overcome these challenges?
Answer:

• Causes mental and emotional disturbances.
• Confusion about his/her role in the society.
• Poor understanding about hygiene, immaturity.
• Possibility to addiction to bad habits.
• To survive these challenges
• Practice to say ‘No’ to wrong ways
• Parents should take care to make the environment in home pleasant
• Be careful while choosing friends.
• Take part in social services.
• Participate in co-curricular activities in the school.

Question 27.
It is easy to be addicted to drugs. But to escape from it is not that easy.
a. What should be our approach towards drugs?
b. What are the harmful effects of drugs?
Answer:
Never use drugs in your life. A single-use may push us to its addiction.

Harmful Effects

• Disease that leads to cancer are caused.
• Loss of social recognition
• Breaks interpersonal relationships. Loss of peace in the family and distortion of family relationship.
• Financial crisis arise
• Possibility to indulge in crimes

For the Continuity of Generations Additional Questions & Answers

Question 28.
Prepare a short note on the mode of reproduction in the organisms mentioned below. (Hydra, Bacteria, Fungi)
Answer:
Hydra: reproduces by budding. Buds are formed on the parent body. When it grows it gets detached from the parent and develops into new organisms.
Bacteria: reproduces by binary fission. An existing cell divides to two new cells.
Fungi: reproduces by spore formation.

Question 29.
Night-blooming flowers are while and having intense fragrance why?
Answer:
The smell and color of flowers is to attract insects.

Question 30.
“The disappearance of certain plants caused the extinction of some insects”- substantiate this statement.
Answer:
Certain insects depend only specific plants for their food and reproductive purposes; especially butterflies. Hence their destruction way leads to the extinction of the dependant species.

Question 31.
What is the function of tube nucleus and generative nucleus?
Answer:
Generative nucleus divides to form sperms. Tube nucleus disintegrates.

Question 32.
How does endosperm form?
Answer:
The second sperm fuses with the polar nucleus in the ovary and forms the endosperm.

Question 33.
What is the fate of Zygote, endo spermete?
Answer:
Zygote develops into embryo and endosperm forms the stored food for the development of embryo.

Question 34.
Find out the parts of the male reproductive system that performs the following functions?
1. Secrets the fluid containing nutrients for sperms
2. Secretes male hormones.
3. Carries sperms from the testis to ureter.
Answer:
1. Prostate gland
2. Testis
3. Vas deferens

Question 35.
Arrange the following state¬ments under the headings sperm Ovum.
1. Nonmotile
2. Parts like head, middle piece and tail are present.
3. Nucleus is present in the head,
4. Mitochondria is in middle piece.
Answer:

Question 36.
Are Ovulation and menstruation the same process?
Answer:
No. Ovulation is the process of release of mature ovum from the ovary. If fertilization does not take place the newly formed tissues in the uterine wall peel off and come, out along with blood and mucus through vagina. This process is called menstruation.

Question 37.
Find the odd one. Write the reason.
Endometrium, androgen, Oes-trogen, progesterone.
Answer:
Endometrium – Inner layer of uterus. Others are hormones.

Question 38.
How does placenta form? What is its function?
Answer:
Placenta is formed of fetal tissues and uterine tissues. Oxygen and nutrients are supplied to the fetus and waste materials are eliminated through the umbilical cord formed from placenta.

Question 39.
Fill in the blanks
1. Mushrooms are plants that reproduce by means of
2. During favorable conditions amoeba reproduces by
3. In tapeworm, both the male and female sex organs are found in the same organism. So it is said to be a
4. Each in man contains about eight hundred or more long coiled, minute tubules called seminiferous tubules.
5. The intimate mechanical and
physiological connection between fetal and maternal tissue is provided by
6. A cord containing blood vessels which connect the placenta with the fetus is called.
7.0n an average the length of menstrual cycle in females is completed in days.
8. A person infected with AIDS loses his
9. Male gamete fuse with ovule to form
10. The period of sexual maturity in human beings called
Answer:
1. spores
2. binary fission
3. hermaphrodite
4. testis
5. placenta
6. umbilical cord
7. 28
8. immunity
9. zygote
10. puberty

Question 40.
What is ‘after-birth’?
Answer:
About 15 minutes after the delivery of the baby, the placenta detaches from the uterine wall and is expelled out as ’after-birth’.

Question 41.
The lungs of the foetus are filled with fluid so it cannot breathe. But it doesn’t feel breathlessness. Why?
Answer:
A special tissue develops between uterine wall and the embryo called placenta. The growing foetus gets nutrients and oxygen from the mother’s blood through the placenta and umbilical cord. So it doesn’t feel breathlessness.

Question 42.
How is the structure of sperms adapted for fertilization?
Answer:
The vibrating tail of sperm helps to reach up to to ovum. The mitochondria present in the sperm provides energy for this movement. The enzyme produced by the acrosome present on the head of the sperm helps the sperm nucleus in penetrating the sheath of the ovum and entering the. ovum.

You can Download Equal Triangles Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 9 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers

Negative Numbers Text Book Questions and Answers

Textbook Page No 165

Negative Numbers Class 8 Kerala Syllabus Question 1.
Using the principles above, compute the following:
i. 5 – 10
ii. -10 + 5
iii. -5 – 10
iv. -5 – 5
vi. \(-\frac{1}{2}+1 \frac{1}{2}\)
vii. \(-\frac{1}{2}-1 \frac{1}{2}\)
viii. \(-\frac{1}{2}+\frac{1}{2}\)
Solution:
i. 5 – 10 = – (10 – 5) = -5
(since for x – y = -(y – x ))

ii. – 10 + 5 = 5 – 10 = – (10 – 5) = -5
(since for -x + y = y – x and x – y = – (y – x))

iii. -5 – 10 = – (5 + 10 ) = -15
(since for – x – y = -(x + y))

iv. -5 – 5 = – (5 + 5) = -10
(since for -x – y = -(x + y))

v. -5 + 5 = 5 – 5 = 0
(since for -x + y = y – x)

Kerala Syllabus 8th Standard Maths Notes Pdf Question 2.
Take as x different positive numbers, negative numbers and zero, and compute x + 1, x – 1, 1 – x. Check whether the equations below hold for all numbers.
i. (1 + x) + (1 – x) = 2
ii. x – (x – 1) = 1
iii. 1 – x = -(x – 1)
Solution:
If x = 1
x + 1 = 1 + 1 = 2
x – 1 = 1 – 1 = 0
1 – x = 1 – 1 = 0
If x = 2
x + 1 = 2 + 1 = 3
x – 1 = 2 – 1 = 1
1 – x = 1 – 2 – 1
If x = o
x + 1 = 0 + 1
x – 1 = 0 – 1 = -1
1 – x = 1 – 0 = 1
If x =-1
x + 1 = -1 + 1 = 1 – 1 = 0
x – 1 = -1 – 1 = -2
1 – x = 1 – (-1) = 1 + 1 = 2
If x = -2
x + 1 = -2 + 1 = -1
x – 1 = -2 – 1 = -3
1 – x = 1 – (-2) = 1 + 2 = 3

i. (1 + x) + (1 – x)
In x = 1, (1 + x) + (1 – x) = 2 + 0 = 2
In x = 2, (1 + x) + (1 – x) = 3 + (-1) = 3 – 1 = 2
In x = o, (1 + x) + (1 – x) = 1 + 1 = 2
In x = -1, (1 + x) + (1 – x) = 0 + 2 = 2
In x – 2, (1 + x) + (1 – x) – 1 + 3 = 3 – 1 = 2
(1 + x) + (1 – x) = 2 , for all values of x

ii. x – (x – 1)
In x = 1, x – (x – 1) = 1 – 0 = 1
In x = 2, x – (x – 1) = 2 – 1 = 1
In x = o, x – (x – 1) = 0 – (-1) = 1
In x = -1, x – (x – 1) = -1 – (-2) = -1 + 2 = 1
In x =-2, x – (x – 1) = -2 – (-3) = -2 + 3 = 1
x – (x – 1) = 1, for all values of x

iii. 1 – x
In x = 1, 1 – x = o = -(x – 1)
In x = 2, 1 – x = -1 = -(1) = -(x – 1)
In x = o, 1 – x = 1 = -(-1) = -(x – 1)
In x = -1, 1 – x = 2 = -(-2) = -(x – 1)
In x = -2, 1 – x = 3 = -(-3) = -(x – 1)
1 – x = -(x – 1), for all values of x

Hsslive Guru Maths 8th Standard Kerala Syllabus Question 3.
Taking different numbers as x, y and compute x + y, x – y. Check whether the following hold for all kinds of numbers.
i. (x + y) – x = y
ii. (x + y) – y = x
iii. (x – y) + y = x
Solution:
If we take x = 6 and y = 2, x + y = 6 + 2 = 8 and x – y = 6 – 2 = 4
If x = -6 and y = 2, x + y = -6 + 2 = -4 and x – y = -6 – 2 = -8
If x = 6 and y = -2, x + y = 6 + (-2) = 4
and x – y = 6 – (-2) = 8
If x = -6 and y = -2, x + y = (-6) + (-2)
= -8 and x – y = (-6) – (-2 ) = -4

i. (x + y) – x
If x = 6 and y = -2, (x + y) – x
= 8 – 6 = 2 = y
If x = -6 and y = 2, (x + y) – x
= -4 – (-6) = -4 + 6 = 2 = y
If x = 6 and y = -2, (x + y) – x
= 4 – 6 = – 2 = y
If x = -6 and y = -2, (x + y) – x
= -8 – (-6) = -8 + 6 = -2 = y
(x + y) – x = y, for all values of x and y

ii. (x + y) – y
If x = 6 and y = 2, (x + y) – y
= 8 – 2 = 6 = x
If x = -6 and y = 2, (x + y) – y
= -4 – 2 = -6 = x
If x = 6 and y = -2, (x + y) – y
= 4 – (-2) = 4 + 2 = 6 = x
If x = -6 and y = -2, (x + y) – y
= -8 – (-2) = -8 + 2 = -6 = x
(x + y) – y = x, for all values of x and y

iii. (x – y) + y
If x = 6 and y = 2, (x – y) + y
= 4 + 2 = 6 = x
If x = -6 and y = 2, (x – y) + y
= -8 + 2 = -6 = x
If x = 6 and y = -2, (x – y) + y
= 8 + (-2) = 8 – 2 = 6 = x
If x = -6 and y =- 2, (x – y) + y
= -4 + (-2) = -6 = x
(x – y) + y = x, for all values of x and y

8th Standard Maths Notes Kerala Syllabus Question 4.
Check whether the equation are identities. Write the patterns got from each, on taking x = 1, 2, 3, 4, 5 and x = -1, -2, -3, -4, -5.
i. -x + (x + 1) = 1
ii. -x + (x + 1) + (x + 2) – (x + 3) = o
iii. -x – (x + 1) + (x + 2) + (x + 3) = 4
Solution:
i) -x + (x + 1) = 1
If x = 1, -x + (x + 1)
= -1 + (1 + 1) = -1 + 2 = 1
If x = 2, -x + (x + 1)
= -2 + (2 + 1) = -2 + 3 = 1
If x = 3, -x + (x + 1)
= -3 + (3 + 1) = -3 + 4 = 1
If x = 4 -x + (x + 1)
= 4 + (4 + 1) = -4 + 5 = 1
If x = 5, -x + (x + 1)
= -5 + (5 + 1) = -5 + 6 = 1
If x = -1, -x + (x + 1)
= -(-1) + (-1 + 1) = 1 +0 = 1
If x = -2, -x + (x + 1)
= -(-2) + (-2 + 1) = 2 + (-1) = 1
If x = -3, -x + (x + 1)
= -(-3) + (-3 + 1) = 3 + (-2) = 1
If x = -4, -x + (x + 1)
= -(-4) + (-4 + 1) = 4 + (-3) = 1
If x = -5, -x + (x + 1)
= -(-5) + (-5 + 1) = 5 + (-4) = 1
It is an identity.

ii. -x + (x + 1) + (x + 2) – (x + 3) = 0
If x = 1, -x + (x + 1) + (x + 2) – (x + 3)
= -1 + (1 + 1) + (1 + 2) – (1 + 3)
= -1 + 2 + 3 – 4 = 0
If x = 2, -x + (x + 1) + (x + 2) – (x + 3)
= -2 + (2 + 1) + (2 + 2) – (2 + 3)
= -2 + 3 + 4 – 5 = 0
If x = 3, -x + (x + 1) + (x + 2) – (x + 3)
= -3 + (3 + 1) + (3 + 2) – (3 + 3)
= -3 + 4 + 5 – 6 = o
If x = 4 -x + (x + 1) + (x + 2) – (x + 3)
= -4 + (4 + 1) + (4 + 2) – (4 + 3)
= -4 + 5 + 6 – 7 = 0
If x = 5, -x + (x + 1) + (x + 2) – (x + 3)
= -5 + (5 + 1) + (5 + 2) – (5 + 3)
= -5 +6 +7 -8= 0
If x = -1, -x + (x + 1) + (x + 2) – (x + 3)
= -(-1) + (-1 + 1) + (-1 + 2) – (-1 + 3)
= 1 + 0 + 1 – 2 = 0
If x = -2, -x + (x + 1) + (x + 2) – (x + 3)
= 2 + (-2 + 1) + (-2 + 2) – (-2 + 3)
= 2 + -1 + 0 – 1 = 0
If x = -3, -x + (x + 1) + (x + 2) – (x + 3)
= 3 + (-3 + 1) + (-3 + 2) – (-3 + 3)
= 3 + -2 + -1 – 0 = 0
If x = -4, -x + (x + 1) + (x + 2) – (x + 3)
= 4 + (-4 + 1) + (-4 + 2) – (-4 + 3)
= 4 + -3 + -2 – (-1) = 0
If x = -5, -x + (x + 1) + (x + 2) – (x + 3)
= 5 + (-5 + 1) +(-5 + 2) – (-5 + 3)
= 5 + -4 + -3 – (-2) = o
It is an identity.

iii. -x – (x + 1) + (x + 2) + (x + 3) = 4
If x = 1, -x – (x +1) + (x + 2) + (x + 3)
= -1 – (1 + 1) + (1 + 2) + (1 + 3)
= -1 – 2 + 3 + 4 = 4
If x = 2, -x – (x + 1) + (x + 2) + (x + 3)
= -2 – (2 + 1) +(2 + 2) + (2 + 3)
= -2 – 3 + 4 + 5 = 4
If x = 3, -x – (x + 1) + (x + 2) + (x + 3)
= -3 – (3 + 1) + (3 + 2) + (2 + 3)
= -3 – 4 + 5 + 6 = 4
If x = 4 -x – (x + 1) + (x + 2) + (x + 3)
= -4 – (4 + 1) + (4 + 2) + (4 + 3)
= -4 – 5 + 6 + 7 = 4
If x = 5, -x – (x + 1) + (x + 2) + (x + 3)
= -5 – (5 + 1) + (5 + 2) +(5 + 3)
= -5 – 6 + 7 + 8 = 4
If x = -1, -x – (x + 1) + (x + 2) + (x + 3)
= -(-1) – (-1 + 1) + (-1 + 2) + (-1 + 3)
= 1 – 0 + 1 + 2 = 4
If x = -2, -x – (x + 1) + (x + 2) + (x + 3)
= 2 – (-2 + 1) + (-2 + 2) + (-2 + 3)
= 2 – (-1) + 0 + 1 = 4
If x = -3, -x – (x + 1) + (x + 2) + (x + 3)
= 3 – (-3 + 1) + (-3 + 2) + (-3 + 3)
= 3 – (-2) + – 1 + 0 = 4
If x = -4, -x – (x + 1) + (x + 2) + (x + 3)
= 4 – (-4 + 1) + (-4 + 2) + (-4 + 3)
= 4 – (-3) + – 2 + (-1) = 4
If x = -5, -x – (x + 1) + (x + 2) + (x + 3)
= 5 – (-5 + 1) + (-5 + 2) + (-5 + 3)
= 5 – (-4) + -3 + (-2) = 4
It is an identity.

Kerala Syllabus 8th Standard Maths Notes Question 5.
Taking different numbers, positive, negative and zero, as x, y, z and compute x + (y + z) and (x + y) + z. Check whether the equation, x + (y + z) = (x + y) + z holds for all these numbers.
Solution:

Textbook Page No 178

Class 8 Maths Kerala Syllabus Question 6.
Take various positive and negative numbers as x, y, z and compute (x + y) z and xz + yz. Check whether the equation (x + y) z = xz + yz holds for all these.
Solution:

Kerala Syllabus 8th Standard Maths Guide Question 7.
In each of the following equations, take x as the given numbers and compute the numbers y.
i. y = x², x = -5, x = 5
ii. y = x² + 3x + 2, x = -2
iii. y =x² + 5x + 4, x = -2, x = -3
iv. y = x³ + 1, x = -1
v. y = x³ + x² + x + 1
Solution:
i. y = x² , x = -5 , x = 5
If x =-5, y = x² = (-5 )² = -5 × -5 = 25
If x = 5, y = x² = (5)² = 5 × 5 = 25

ii. y = x² + 3x + 2, x = -2
If x = -2, y = x² + 3x + 2 = (-2)² + 3x(-2) + 2
= 4 – 6 + 2 = -2 + 2 = 0

iii. y = x² + 5x + 4, x = -2 , x = -3
If x = -2, y = x² + 5x + 4 = (-2 )² + 5x(-2) + 4
= 4 – 10 + 4 = -6 + 4 = -2
If x = -3, y = x² + 5x + 4 = (-3)² + 5x(-3) + 4
= 9 – 15 + 4 = -6 + 4 = -2

iv. y = x³ + 1, x = -1
If x = -1, y = x³ + 1 = (-1 )³ + 1
= (-1)(-1)(-1) + 1 = 1 × (-1) + 1 = 0

v. y = x³ + x² + x + 1, x = -1
If x = -1, y = x³ + x² + x + 1 = (-1)³ +(-1)² + (-1) + 1 = -1 + 1 – 1 + 1 = o

Kerala Syllabus 8th Standard Maths Question 8.
For a point starting at a point P and travelling along a straight line, time of travel is taken as t and the distance from P as s. The relation between s and t is found to be s = 12t – 2t², where distances to the right are taken as positive numbers and to the left as negative numbers.
i. Is the position of the point to the right or left of P, till 6 seconds?
ii. Where is the position at 6 seconds?
iii. After 6 seconds?
(Here it is convenient to write 12t – 2t² = 2t(6 – t).
Solution:
i. S = 12t – 2t²
Distance to the point when time is 1 second = 12t – 2t² = 12 × 1 – 2 × 1² = 12 – 2 = 10 m
Distance to the point when time is 5 second = 12t – 2t² = 12 × 5 – 2 × 5² = 60 – 50 = 10 m
Since the distance to the point till 6 seconds is positive. So the position of the point is on the right of P.
ii. Distance to the point when time is 6 second = 12t – 2t² = 12 × 6 – 2 × 62 = 72 – 72 = o
At the 6^th second, the point is at P.
iii. Distance to the point when time is 7 second (after 6 sec)= 12t – 2t² = 12 × 7 – 2 × 7²
= 84 – 2 × 49 = 84 – 98 = -14 metres
This is a negative number, So the position of the point is on the left of P.

Textbook Page No 179

8th Std Maths Guide Kerala Syllabus Question 9.
Natural numbers, their negatives and zero are together called integers. How many pair of integers are there, satisfying the equation. x² + y² = 25?
Solution:
It is convenient to write it as a table

Textbook Page No 180

Kerala Syllabus 8th Standard Maths Notes Malayalam Medium Question 10.

Solution:

Class 8 Kerala Syllabus Maths Question 11.

Solution:

Class 8 Maths Scert Solutions Kerala Syllabus Question 12.
In the equation \(z=\frac{x}{y}-\frac{y}{x}\), take x as the numbers given below and calculate the number z.
i. x = 10, y = -5
ii. x = -10, y = 5
iii. x = -10, y = -5
iv. x = 5, y = -10
v. x = -5, y = 10
Solution:

Additional Questions and Answers

Maths Class 8 Kerala Syllabus  Question 1.
Match the following

Solution:

8th Class Maths Notes Kerala Syllabus Question 2.

Solution:

Question 3.
Which of the following number is the largest (-1)⁶, (-1)¹⁰, (-1)², (-1)⁵⁰
Solution:
If the power of (-1) is even then answer will be 1
If the power of (-1) is odd then answer will be -1
(-1)⁶ = 1
(-1)¹⁰ = 1
(1)² = 1
(-1)⁵⁰ = 1
All the given numbers are equal.

Question 4.

Solution:

Question 5.
Complete the following table

Solution:

Question 6.
Write whether the answer got on doing the following operations are positive number or negative number.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
Calculate (-1)¹⁰ + (-1)¹⁷ + (-1)²¹ + (-1)²⁶ + (-1)⁷⁷
Solution:
(-1)¹⁰ + (-1)¹⁷ + (-1)²¹ + (-1)²⁶ + (-1)⁷⁷
= 1 + (-1) + (-1) + 1 + (-1)
= 1 – 1 – 1 + 1 – 1 = 2 – 3 = -1

Question 10.
Simplify [(-4) × (-5)] + [-16 × \(\frac{-1}{2}\) ]
Solution:
-4 × -5 = 20

Question 11.
If x = 8 and y = -3; find the values of x + y, y + x, x – y, y – x, -x – y and – y – x
Solution:
x + y = 8 + -3 = 5
y + x = -3 + 8 = 5
x – y = 8 – (-3) = 8 + 3 = 11
y – x = -3 – 8 = -11
– x – y = -8 – (-3) = -8 + 3 = -5
-y – x = -(-3) – 8 = 3 – 8 = -5

Question 12.
If x = 7, y = -6 and z = -2, find the value of
i. (x + y) + z
ii. x + (y + z)
iii. xyz
iv. (x + y)z
v. xy + xz
Solution:
i. (x + y) + z = (7 + -6) + -2 = 1 – 2 = -1
ii. x + (y + z) = 7 + (-6 + -2)
= 7 – 8 = -1
iii. xyz = 7 × -6 × -2 = 84
iv. (x + y)z = (7 + -6) × -2
= 1 × -2 = -2
v. xy + xz = (7 × -6) + (7 × -2)
= -42 – 14
= -56

Question 13.
Compute y = x² + 9x – 5, for take x as the given number,
i. x = 5
ii. x = -2
iii. x = o
iv. x = -3
Solution:
i. x = 5
y = x² + 9x – 5
= 5² + 9 × 5 – 5 = 25 + 45 – 5
= 20 + 45 = 65

ii. x = -2
y = x² + 9x – 5
=(2)² + 9 × (-2) -5
= 4 – 18 – 5 = 4 – 23 = – 19

iii. x = 0
y = x² + 9x – 5
= 0 + 9 × 0 – 5 = -5

iv. x = -3
y = x² + 9x – 5
= (-3)² + 9 × (-3) – 5
= 9 – 27 – 5 = 9 – 32 = -23

Question 14.
Find y = x⁴ + x³ + x² + x + 1, If x = -1
Solution:
y = x⁴ + x³ + x² + x + 1
= (-1)⁴ + (-1)³ + (-1)² + (-1) + 1
= 1 + (-1) + 1 + (-1) + 1
= 1 – 1 + 1 – 1 + 1
= 0 + 0 + 1 = 1

Question 15.
Complete the table

Solution:

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Kerala State Syllabus 8th Standard Hindi Solutions Unit 4 Chapter 2 इस बारिश में

इस बारिश में पाठ्यपुस्तक के प्रश्न और उत्तर

इस बारिश में कविता Kerala Syllabus 8th Chapter 2 प्रश्ना 1.
‘उसी के पास अब मेरी / बारिश भी चली गई’ से आपने क्या समझा?

उत्तर:
यह एक किसान का रोदन है। यह रोदन किसान की हालत की ओर संकेत करता – है। आजकल किसानों की ज़मीन छीन ली जाती है। यहाँ किसान यह व्यक्त करता है कि ज़मीन के साथ बारिश भी चली गई है। यहाँ खेती और बारिश के घने संबंध स्पष्ट होते हैं।

Is Barish Mein Kerala Syllabus 8th Chapter 2 प्रश्ना 2.
जिसकी नहीं कोई ज़मीन/उसका नहीं कोई आसमान’ इन पंक्तियों का क्या तात्पर्य है?

उत्तर:
इन पंक्तियों का मतलब है कि किसानों का ज़मीन से अटूट संबंध है। ज़मीन नष्ट होने पर किसान का अस्तित्व नष्ट हो जाता है। यह उस से जीने के सब सपने नष्ट हो जाते हैं।

इस बारिश में Textbook Activities

इस बारिश में कविता Meaning In Malayalam Chapter 2 प्रश्ना 1.
कविता में ‘उसी के लिए’ दोहराया गया है। यह प्रयोग किन-किन की ओर संकेत करता है?

उत्तर:
किसानों का शोषण करनेवालों की ओर यहाँ संकेत है।

इस बारिश में कविता का सारांश Kerala Syllabus 8th Chapter 2 प्रश्ना 2.
‘हल नहीं / बैल नहीं’ -इसमें ‘हल’ और ‘बैल’ किन-किनके प्रतीक हैं?

उत्तर:
हल’ और ‘बैल’ खेती और किसानी ज़िंदगी के प्रतीक हैं।

Hss Live Guru 8 Hindi Kerala Syllabus Chapter 2  प्रश्ना 3.
निम्नलिखित आशयवाली पंक्तियाँ चुनकर लिखें।

फसल होने पर कर्ज चुकाने की किसान की झूठी प्रतीक्षा भी नहीं रह गई।

उत्तर:
अगली फसल होते ही सब चुकता कर दूंगा/अब तो मेरी झूठी/ये गुज़ारिश भी चली गई।

Hss Live Guru 8th Hindi Kerala Syllabus Chapter 2 प्रश्ना 4.
आस्वादन टिप्पणी तैयार करें।

उत्तर:
नरेश सक्सेना समकालीन हिंदी कविता के समर्पित कार्यकर्ताओं की अग्रिम पंक्ति में हैं। ‘इस बारिश में’ नामक कविता में किसान की ज़मीन छीन जाने की कथा है। यह एक किसान का बारिश के मौसम का शोकगीत है। आकाश में कई दूर छा जानेवाले बादलों को देखकर किसान आह भरता है। अपनी ज़मीन छिन जाने पर किसान खेती न कर सकता। भूमंडलीकरण के दौर के किसानों की सिसकियाँ यहाँ हम देख सकते हैं। इस कविता क द्वारा कवि किसान लोगों की त्रासदी की ओर हमारा ध्यान आकर्षित करते हैं। कवि का कहना है कि अब बारिश भी ज़मीन के पीछे चली गई है। धरती की छाती से सौंधी सुगंध भी छिन गई मिट्टी के लिए उठती है।

अब किसान के लिए हल और बैल नहीं, खेतों के बीच का रास्ता नहीं, कहीं हरियाली की बूंद भी दिखाई न देता। किसान के जीवन का ताल, प्रतीक्षा का नक्षत्र सब नष्ट हो चुकी है। फसल होने पर कर्ज चुकाने की किसान की प्रतीक्षा भी नहीं रह गई। किसान की अपनी खेत – खलिहानों से दूर रहने की विवशत इस कविता में हम देख सकते हैं। ज़मीन छिन जाने पर किसान भयानक शोषण का शिकार बन जाता है। कवि . इस कविता दवारा यही कहना चाहता है। एक कविता तभी समसामयिक मानी जाती है जब वह तत्कालीन समस्याओं का
संबोधन करती है। इस बारिश में’ नामक यह कविता इस कसौटी पर खरा उतरता है।

इस बारिश में Summary in Malayalam and Translation

इस बारिश में शब्दार्थ Word meanings