Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam

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Kerala State Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness

Civic Consciousness Textbook Questions and Answers in Malayalam

Civic Consciousness Class 10 Notes Pdf Malayalam Medium

Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 2
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 3
Kerala Syllabus 10th Standard Social Science Notes

Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 5
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 6
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 7
Class 10 Kerala Syllabus Social Science Notes

Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 9
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 10
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 11
Kerala Syllabus 10th Standard Social Science Notes Malayalam Medium

Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 13
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 14
Civic Consciousness Class 10 Notes
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 16
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 17

Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 18
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 19
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 20
Social Science Textbook Class 10 State Syllabus
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 22

Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 23
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 24
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 25
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 26
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 27

Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 28
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 29
Kerala Syllabus 10th Standard Social Science Solutions Chapter 10 Civic Consciousness in Malayalam 30

Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा

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Kerala State Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा (यात्रावृत्त)

दिशाहीन दिशा Text Book Questions and Answers

दिशाहीन दिशा विश्लेषणात्मक प्रश्न

Hsslive Guru 10th Hindi Kerala Syllabus  प्रश्ना 1.
घर में चलते समय मन में यात्रा की कोई बनी हुई रूप-रेखा नहीं थी।” – लेखक के इस कथन के आधार पर बताएँ कि किसी यात्रा पर जान से पहले यात्रा की रूप-रेखा बनाना ज़रूरी है?
Hsslive Guru 10th Hindi Kerala Syllabus
उत्तर:
किसी यात्रा पर जाने से पहले यात्रा की रूपरेखा बनाना ज़रूरी है। यात्रा तो हम नई जगहों को पहचानने के लिए करते हैं। रूपरेखा बनाने से जगहों की सही जानकारी मिलती है. मसंदीदार जगह बड़ी चाव देख सकते हैं और अन्य जगह छोड भी सकते हैं। हम अपने समय का सही इस्तेमाल भी कर सकते हैं।

Hss Guru 10 Hindi Kerala Syllabus  प्रश्ना 2.
घने शहर की छोटी-सी तंग गली में पैदा हुए लेखक को कन्याकुमारी के विशाल समुद्र-तट के प्रति आत्मीयता का अनुभव होने का आधार क्या हो सकता है?
Hss Guru 10 Hindi Kerala Syllabus
उत्तर:
नुभवहीन बातों पर ज़्यादा रुचि रखना स्वाभाविक है। विपरीत के प्रति आकर्षण होना स्वाभाविक ही है। किसी तंग गली में जन्म लेने से कन्याकुमारी के विशाल तट में अपनापन का भाव जागृत होगा।

Hss Live Guru 10th Hindi Kerala Syllabus प्रश्ना 3.
‘मगर बात करने की जगह उसने मेरा बिस्तर लपेटकर खिड़की से बाहर फेंक दिया और खुद : मेरा सूटकेस लिए नीचे उतर गया।’ अविनाश के इस आचरण से मोहन राकेश और अविनाश के बीच की मित्रता का क्या अंदाज़ा मिल जाता है?
Hss Live Guru 10th Hindi Kerala Syllabus
उत्तर:
मोहन राकेश का एक दिली दोस्त है अविनाश। जब चाहे वह मोहन राकेश के साथ कोई भी आचरण कर सकता है। कुछ भी करने को उसके लिए स्वतंत्रता है। प्रस्तुत आचरण से उनके बीच का धनिष्ठ संबंध का परिचय मिलता है

10th Class Hindi Notes Kerala Syllabus प्रश्ना 4.
‘मगर आप चाहे तो चंद गज़लें तरन्नुम के साथ अर्ज कर सकता हूँ।’ इस कथन से आम जनता के साथ गज़लों के रिश्ते का क्या परिचय मिलता है?
10th Class Hindi Notes Kerala Syllabus
उत्तर:
आम जनता गज़लों से खूब परिचित थे। गज़ल आम जनता की ही कविता है। वे उसे गाते रहते हैं। क्योंकि उतनी मार्मिकता उसमें है। इसलिए बूढ़ा मल्लाह अब्दुल जब्बार भी शायर गालिब से परिचित थे। गायक न होते हुए भी मल्लाह कुछ गज़लें पेश करने को तैयार भी हुआ।

10th Hindi Notes Kerala Syllabus प्रश्ना 5.
‘उसके खामोश हो जाने से सारा वातावरण ही बदल गया।’ – इससे आपने क्या समझा?
10th Hindi Notes Kerala Syllabus
Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 6
उत्तर:
बूढ़ा मल्लाह झूम-झूमकर गज़ल गा रहा था। लेखक और मित्र भी उसके गायन में विलीन हो गए। उसका गला काफ़ी अच्छा था, सुनाने का अंदाज़ा भी शायराना था। गाते समय रात, सर्दी, नाव का हिलना इन सबका अनुभव नहीं हो रहा था। अब होने लगा। झील का विस्तार भी उतनी देर के लिए सिमट गया था, अब खुल गया।

दिशाहीन दिशा Text Book Activities & Answers

Hsslive Guru Hindi 10 Kerala Syllabus प्रश्ना 1.
संबंध पहचाने, सही मिलान करें।
Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 7

मोहन राकेश की बड़ी इच्छा थी  कि वहाँ जीवन बहुत सस्ता है।
समय और साधन की कमी से मोहन राकेश ने यात्रा करने का निश्चय किया।
 हाथ में पैसा आने पर  कि कन्याकुमारी चला जाऊँ
मोहन राकेश ने पहले सोचा था  कि समुद्र तट का सफ़र करें।
गोआ इसलिए हम जा सकते हैं मोहन राकेश समुद्र तट की यात्रा न कर सके।

उत्तर:

मोहन राकेश की बड़ी इच्छा थी  कि समुद्र तट का सफर करें।
समय और साधन की कमी से  मोहन राकेश समुद्र तट की यात्रा न कर सके।
 हाथ में पैसा आने पर  मोहन राकेश ने यात्रा करने का निश्चय किया।
मोहन राकेश ने पहले सोचा था कि कन्याकुमारी चला जाऊँ
गोआ इसलिए हम जा सकते हैं कि वहाँ जीवन बहुत सस्ता है।

Hsslive 10th Hindi Kerala Syllabus प्रश्ना 2.
पढ़ें, यात्रावृत्त के आधार पर उचित वाक्यों पर सही का निशान ✓ लगाएँ।
Hsslive Guru Hindi 10 Kerala Syllabus
Hsslive 10th Hindi Kerala Syllabus
उत्तर:
Kerala Syllabus 10th Standard Hindi Notes

Kerala Syllabus 10th Standard Hindi Notes प्रश्ना 3.
भोपाल ताल में अब्दुल जब्बार और अविनाश के साथ की सैर मोहन राकेश के लिए मज़ेदार थी। वे अपने अविस्मरणीय अनुभव दफ्तर के एक मित्र से बाँटना चाहते हैं। भोपाल ताल की सैर के अनुभवों का ज़िक्र करते हुए मित्र के नाम मोहन राकेश का पत्र लिखें।
Hss Live Guru Hindi 10 Kerala Syllabus
उत्तर:
भोपाल,
30-12-1952
प्रिय जयप्रकाशजी,
आप कैसे हैं? दफ़्तर में सब कुशल है न? कुछ बातें आपसे बाँटना चाहता हूँ। सोचा कि एक चिट्ठी लिखू। अब मैं भोपाल में हूँ। मुंबई के रास्ते में था। डिब्बे में सोने के लिए सीट भी मिली थी। लेकिन रात आई तो मैं भोपाल ताल की एक नाव में लेटा बुढ़े मल्लाह अब्दुल जब्बार से गज़लें सुन रहा था।

भोपाल स्टेशन पर मित्र अविनाश ने मुझसे मिलने आया था। बात करने की जगह उसने मेरा बिस्तर लपेटकर खिड़की से बाहर फेंक दिया और खुद मेरा सूटकेस लिए हुए नीचे उतरा। रात ग्यारह बजे के बाद हम लोग घूमने निकले। जब भोपाल ताल के पास आया तो मन लगा कि नाव लेकर कुछ देर तक झील की सैर करें। अचानक अविनाश ने कहा कि कितना अच्छा होता अगर इस वक्त हम में से कोई कुछ गा सकता। हमारी नाव का मल्लाह अब्दुल जब्बार गायक तो नहीं, मगर उसने कुछ गज़लें तरन्नम के साथ पेश किया। उसका गला अच्छा था। सुनाने का अंदाज़ भी शायराना था। एक के बाद दूसरी फिर तीसरी। हम दोनों उसके गायन में विलीन हो गए थे। जब वह खामोश हो गया तो वातावरण ही बदल गया। रात, सर्दी का नाव का हिलना सबका अनुभव पहले नहीं हो रहा था, अब होने लगा। फिर उससे गालिब की गज़लें सुनाया गया। भोपाल-ताल की सैर मज़ेदार था, दिल को छूनेवाली थी। दफ्तर में सबको मेरा नमस्कार कहना। बाकी सब अगले पत्र में। तुरंत ही जवाबी पत्र की प्रतीक्षा करते हुए।
Kerala Syllabus 10th Standard Hindi Guide

Hss Live Guru Hindi 10 Kerala Syllabus प्रश्ना 4.
पश्चिमी-तट की यात्रा निश्चय ही अवाच्य अनुभूति प्रदान करेगी। गोआ काफ़ी सुंदर जगह है। वहाँ की विशेषताओं को ध्यान में रखकर एक विवरणिका (ब्रॉशर) तैयार करें।
10th Standard Hindi Kerala Syllabus
उत्तर:
Hindi Notes 10th Class Kerala Syllabus

Kerala Syllabus 10th Standard Hindi Guide प्रश्ना 5.
चरित्र पर टिप्पणी लिखें।
बूढ़े मल्लाह ने एक गज़ल छेड़ दी। उसका गला काफ़ी अच्छा था और सुनाने का अंदाज़ | भी शायराना था। काफ़ी देर चप्पुओं को छोड़े वह झूम-झूमकर गज़लें सुनाता रहा।
‘दिशाहीन दिशाट के अब्दुल जब्बार का व्यक्तित्व बड़ा प्रभावशाली है। ये संकेत पढ़ें और अब्दुल जब्बार के चरित्र पर टिप्पणी लिखें।
1. गरीब
2. परिश्रमी .
3. खुशमिज़ाज .
4. सादा जीवन बितानेवाला
5. विनयशील .
6. गज़ल गायक
Standard 10 Hindi Kerala Syllabus
उत्तर:
गज़ल गायक – अब्दुलजब्बार
मोहनराकेश का यात्राविवरण ‘दिशाहीन दिशा’ के एक प्रभावशाली व्यक्ति है बूढ़ा मल्लाह श्री अब्दुल जब्बार । भोपाल-ताल की सैर में लेखक और मित्र का उनका परिचय होता है। हमेशा खुशमिज़ाज दिखाई पडनेवाला और सादा जीवन बितानेवाला था। दाढ़ी के ही नहीं छाति के भी बाल सफेद हो चुके थे। सर्दी के मौसम में भि सिर्फ तहमद लगाए आया था। भोपाल-ताल का नाविक अब्दुलजब्बार रात -दिन मेहनत करता रहता है। जब वह चप्पू चलाने लगता तो उसकी मांसपेशियाँ इस तरह हिलती जैसे उनमें फौलाद भरा हो । जब अविनाश गाने का आग्रह प्रकट किया तो बिना हिचक के तीन गज़लें छेड देता है। उसका गला काफ़ी अच्छा था। गज़लों से वह इतना परिचित था कि सुनाने का अंदाज़ भी शायराना था। कभी कभी नाव खोते समय चप्पुओं को छाड़े झूम-झूमकर गज़लें सुनाता था। असल में जब उसने गज़ले समाप्त की, वातावरण ही बदल गया था। अविनाश के अनुसार गालिब की चीज़ पेश करने को कहता है, तुरंत ही बिना एतराज के विनय के साथ गज़लें गाने लगा। मोटे तौर पर वह खुशमिज़ाज, विनयशील गरीब गज़लगायक अब्दुल जब्बार लेखक और मित्र के लिए उस रात अविस्मरणीय पात्र रहा।

10th Standard Hindi Kerala Syllabus प्रश्ना 6.
इंन शब्दों पर ध्यान दें :
Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 15

मुझे हमने
उसमें  उसकी
 इनका  मुझसे
उसने किसका

इनके मूल शब्दों को पहचानें और परिवर्तन के कारण पर चर्च करें।
Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 16
उत्तर:
मुझे — मैं
हमने — हम
उसमें — वह
उसकी — वह
इनका — ये
मुझसे — मैं
उसने — वह
किसका — कौन
मैं, हम, वह, ये, कौन आदि हिंदी के सर्वनाम है।
इन सर्वनाम के साथ कुछ प्रत्य लगाने से उपयुक्त शब्द मिलता है। इन्हें सर्वनाम का रूपांतरण कहते हैं। उदाहरण के लिए ‘मैं’ के साथ ‘को’ प्रत्यय लगाने से ‘मुझे’ या ‘मुझको’ शब्द मिलता है। मैं, हम, तू, तुम, आप, यह, ये, वह,वे, जो, कौन, कोई आदि हिंदी के सर्वनाम हैं।

Hindi Notes 10th Class Kerala Syllabus  प्रश्ना 7.
नमूने के अनुसार वाक्यों को बदलकर लिखें, अर्थ-भेद भी समझें।
Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 17
10th Standard Hindi Meaning Kerala Syllabus
उत्तर:
Hsslive Hindi 10th Kerala Syllabus

Standard 10 Hindi Kerala Syllabus प्रश्ना 8.
मान लें आप दिसंबर की छुट्टियों में दिल्ली जा रहे हैं। इसके लिए क्या-क्या पूर्व तैयारियाँ करेंगे। इस चार्ट की पूर्ति करें।
10th Standard Hindi Notes Kerala Syllabus
उत्तर:
Kerala Syllabus 10th Standard Hindi

दिशाहीन दिशा Additional Questions and Answers

10th Standard Hindi Meaning Kerala Syllabus प्रश्ना 1.
‘तीसरी गज़ल सुनकर वह खामोश हो गया। उसके खामोश हो जाने से सारा वातावरण ही बदल गया।’ बूढ़े मल्लाह अब्दुल जब्बार के साथ हुई भोपाल-ताल की सैर के बारे में लेखक अपनी डायरी में कुछ लिखते है। वह डायरी तैयार करें।
Hss Live Guru 10 Hindi Kerala Syllabus
26 दिसंबर 1952.
कल जो भोपाल- ताल की यात्रा करने का मन हुआ वह अविस्मरणीय रहा। रात साढ़े ग्यारह बजे मैं अविनाश के साथ भोपाल-ताल पर यात्रा की। हमारा मल्लाह अब्दुलजब्बार नामक एक बूढ़ा था। वह बहुत खुशमिज़ाज नज़र आया। अविनाश के आग्रह प्रकट करते ही उसने तीन गज़लें छेड़ दी। वाह ! हम उसपर विलीन हो गए थे। रात, सर्दी एवं नाव के हिलने का अंदाज़ा पहले नहीं हुआ था। झील का विस्तार भी गाते समय सिमट गया था। उसका गला काफ़ी अच्छा था। सुनाने का अंदाज़ भी शायराना था। नाव चलाने का बीच काफ़ी देर चप्पुओं को छोडे वह झूम -झूमकर गज़लें सुनाता रहा। तेज़ गर्मी में भी बेचारा सिर्फ एक तहमद लगाए बैठा था। जब वह चप्पू चलाने लगता तो उसकी मांसपेशियाँ इस तरह हिल्ती जैसे उनमें फौलाद भरा हो। मैं और अविनाश उसके गज़ल गायन में इतना विलीन हो गए थे कि लौट जाने की बात ही नहीं सोचा था। आगे उसने गालिब की गज़ल पैश की – “मुद्दत हुई है यार को मेहमाँ किए हुए……”! आहा! क्या बात है! यह दुनीया ही कुछ और है।

Hsslive Hindi 10th Kerala Syllabus प्रश्ना 2.
रात को ग्यारह के बाद हम घूमने निकले। भोपाल ताल के पास पहूँचे तो मन हो आया कि नाव लेकर कुछ देर झील की सैर की जाए। प्रस्तुत घटना को लेकर पटकथा का एक दृश्य लिखें।
दृश्य।:
Kerala Syllabus 10th Hindi Notes
दृश्य का विवरणः
(भोपाल ताल में नाव खोते हुए एक बूढ़ा मल्लाह नज़र आ रहा है। नाव में मोहनराकेश और मित्र अविनाश है। मोहनराकेश लेटे हुए है। मल्लाह सिर्फ तहमद पहना हुआ है। कडी सर्दी है। अविनाश गज़ल गाने का आग्रह प्रकट करता है। बूढ़ा मल्लाह गाने की धुन में है)

मोहनराकेश : बडी सुहानी रात है, कडी सर्द भी है।
अविनाश : हाँ, अगर हममें से कोई इस वक्त कोई गाना पेश करें तो कितना अच्छा होता।
अब्दुलजब्बार : मैं गा तो नहीं सकता, हुजूर ।
अविनाश : फिर भी कुछ प्रयास करें।
अब्दुलजब्बार : कुछ गज़लें तरन्नुम के साथ पेश करने का प्रयास करूँ?
मोहनराकेश
और अविनाश : (एकसाथ) ज़रूर, ज़रूर ।
(बूढा मल्लाह ताल- लय के साथ गज़लें गाने लगता है।)

दिशाहीन दिशा Summary in Malayalam and Translation

10th Class Hindi Notes Kerala Syllabus
Hindi Notes For Class 10 Kerala Syllabus
Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 26
Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 27
Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 28
Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 29
Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 30
Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 31

दिशाहीन दिशा शब्दार्थ

Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 32
Kerala Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 2 दिशाहीन दिशा 33

Kerala SSLC Class 10 Hindi Solutions

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 3 Reactivity Series and Electrochemistry

You can Download Reactivity Series and Electrochemistry Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solution Chapter 3 Reactivity Series and Electrochemistry

Reactivity Series and Electrochemistry Text Book Questions and Answers

Reactivity Series And Electrochemistry Kerala Syllabus Text Book Page No: 48

→ Which metal reacts vigorously?
Answer:
Sodium.

→ Which gas is formed as a result of this reaction?
Answer:
Hydrogen.

→ Write down its chemical equation.
Answer:
2Na + 2H2O → 2NaOH + H2

→ Complete the table (3.1) given below.
Reactivity Series And Electrochemistry Kerala Syllabus
Answer:
Sslc Chemistry Chapter 3 Kerala Syllabus

→ Based on your observation, arrange these metals in the decreasing order of their reactivity.
Answer:
Sodium > Magnesium > Copper
→ 2Mg + O2 →
Answer:
2Mg + O2 → 2MgO

Sslc Chemistry Chapter 3 Kerala Syllabus Text Book Page No: 49

→ Which metal among magnesium, copper, gold, sodium and aluminium, loses its lustre at a faster rate?
Answer:
Sodium

→ List the above metals in the decreasing order of their reactivity with air and thereby losing lustre
Answer:
Sodium > Magnesium > Aluminium > Copper > Gold.

Text Book Page No: 50

→ What happened to the Zn rod?
Answer:
Before the experiment the Zn rod was colourless. After the experiment Zn rod became blue due to the deposition of copper.

→ What is the reason for this?
Answer:
When the Zn rod is dipped in CuSO4 solution, the Cu2+ ions in the solution get deposited at the Zn rod as Cu atoms.

→ What is the reason for the change in intensity of the colour of CuSO4 solution?
Answer:
The blue colour of CuSO4 solution is due to the presence of Cu2+ ions. The change in intensity of the colour of CuSO4 solution because when the Zn rod is dipped in CuSO4 solution, the Cu2+ ions in the solution get, deposited at the Zn rod as Cu atoms.

→ Which is the metal that gets displaced here?
Answer:
Copper

→ Which is more reactive Zn or Cu?
Answer:
Zn

→ On the basis of the position of Zn and Cu in the reactivity series, can you explain why Cu had been displaced?
Answer:
Zn is placed above Cu in the reactivity series because Zn has a higher reactivity than Cu.

→ Isn’t it due to the higher reactivity of zinc (Zn) when compared to copper (Cu)?
Answer:
Yes.

Chemistry Class 10 Chapter 3 Kerala Syllabus Text Book Page No: 51

→ Is this reaction oxidation or reduction? Why?
Answer:
Oxidation. Because the losing of electrons is called oxidation.

→ The change that happened to Cu2+
Answer:
Cu–2+ + 2e → Cu

→ What is the name of this reaction? Why?
Answer:
Reduction. The gaining of electrons is called reduction.

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 3 Reactivity Series and Electrochemistry 29
Complete this chemical equation by assigning oxidation numbers.
Answer:
2 Ag+1 NO31–+ Cu°→ Cu2+ (NO3)–12 + 2Ag0

→ Which metal was oxidised in this case? Which metal was reduced?
Answer:
Metal which was oxidised: Cu
Metal ion which was reduced: Ag+

→ Write equations showing oxidation and reduction.
Answer:
Oxidation : Cu0 → Cu2+ + 2e
Reduction : Ag++ le → Ag0

Reactivity Series And Electrochemistry Sslc NotesText Book Page No: 52

→ Complete the table 3.3.
Sslc Chemistry Chapter 3 Notes Kerala Syllabus
Answer:
Chemistry Class 10 Chapter 3 Kerala Syllabus

Sslc Chemistry Chapter 3 Notes Pdf Kerala Syllabus Text Book Page No: 53

→ Which electrode has the ability to donate electrons in a cell constructed using these metals?
Answer:
Zn

→ Which one can gain electrons?
Answer:
Cu

→ Identify the chemical reaction that takes place at the Zn electrode. Tick ✓ the right one.
Answer:
Zn(s) → Zn2+ (aq) + 2e  (✓)
Zn2+(aq) + 2e → Zn(s)   (✘)

→ What is the reaction taking place here?
Answer:
Oxidation.

→ Write the chemical equation for the reaction taking place at the Cu electrode.
Answer:
Cu2+ (aq) + 2e → Cu(s)

→ Sketch the cell constructed.
Answer:
Reactivity Series And Electrochemistry Sslc Notes

→ Note down the reaction of the Galvanic cell.
Answer:
Cu(s) + 2Ag+(aq) → Cu2+ (aq) + 2Ag(s)
Cu(s) → Cu2+(aq) + 2e (Anode)
Ag+(aq) + le → Ag(s) (Cathode)

→ Direction of flow of electrons From Cu to Ag
Answer:
Mark the direction of electron how in the cell illustrated.

→ write the reactions taking place at cathode and anode.
Answer:
At cathode : Ag+ + le → Ag
At anode : Cu → Cu2+ + 2e

Hss Live Guru 10th Chemistry Kerala Syllabus Text Book Page No: 55

→ You have used three metals Zn, Cu and Ag. How many cells can be produced using these?
Answer:
Three.

→ Complete the Table 3.4 by writing anode and cathode in each.
Sslc Chemistry Chapter 3 Notes Pdf Kerala Syllabus
Answer:
Hss Live Guru 10th Chemistry Kerala Syllabus

→ What are the substances obtained when electricity is passed through acidified water?
Answer:
Hydrogen, Oxygen.

→ Do such type of chemical changes happen when electricity is passed through metals?
Answer:
Yes.

Sslc Chemistry Chapter 3 Questions And Answers Kerala Syllabus Text Book Page No: 56

→ To which electrodes are the positive ions attracted during electrolysis?
Answer:
Towards negative electrodes(Cathode)

→ To which electrodes are the negative ions attracted?
Answer:
Towards positive electrodes(Anode),

→ What changes happen to the ions which N are attracted to cathode?
Answer:
Reduction

→ What about the changes happening to the ions attracted to anode?
Answer:
Oxidation

→ Which ion is attracted to the positive electrode (anode)?
Answer:
Chloride ion (Cl)

→ What is the chemical reaction taking place there?
Answer:
2Cl (aq) → Cl2(g) + 2e

Text Book Page No: 57

→ Which is the gas liberated at the anode?
Answer:
Chlorine(Cl2)

→ Which is the ion attracted to the negative electrode (cathode)? Write the change happening to it?
Answer:
Na+ ions. These ions accept one electron and changes to sodium atom. That is sodium ions are reduced.

→ Which is the metal deposited at the cathode?
Answer:
Sodium (Na)

→ Which are the ions attracted to the positive electrode?
Ans.
Cl,OH

→ Which are the ions attracted to the negative electrode?
Answer:
Na+,
H3O+,
H2O.

Text Book Page No: 59

→ Which metal is connected to the negative terminal of the battery?
Answer:
Iron.

→ Which metal is connected to the positive terminal of the battery?
Answer:
Copper.

→ Which solution is used as the electrolyte?
Answer:
Copper sulphate solution.

→ What happens to Cu2+ ions at the cathode? Complete the equation.
Answer:
Cu2+ + 2e → Cu

→ What happened to the copper ions? Oxidation/Reduction?
Answer:
Reduction.

→ Complete the equation given below.
Answer:
Cu → Cu2+ + 2e

Text Book Page No: 60

→ Find out more examples and extend the list.
Answer:

  • Chromium plating is used in motor car etc.
  • To make metal coating easily corroding metals to prevent corrosion.
  • In ICs (Integrated Circuits) coating of gold /silver is made by electroplating.

Reactivity Series and Electrochemistry Let Us Assess

Kerala Genetics Question 1.
The solutions of ZnSO4, FeSO4, CuSO4 and AgNO3 are taken in four different test tubes. Suppose, an iron nail is kept immersed in each one
In which test tube the iron nail undergoes a colour change?
What is the reaction taking place here?
Justify your answer. (Refer reactivity series of metals).
Answer:
Iron nail immersed in solution of CuSO4 and AgNO3 undergoes a colour change.
i. Fe(s) + CuSO4(aq) →
FeSO4 (aq) + Cu (s).
ii. Fe(s) + 2AgNO3 (aq) →
Fe(NO3)2(aq) + 2Ag(s).
Iron displaces Cu from CuSO4 and Ag from AgNO3 because Fe has higher reactivity than Cu and Ag.

Genetic Engineering Syllabus Question 2.
Compare the electrolysis of molten potassium chloride and solution of potassium chloride. What are the processes taking place at the cathode and the anode?
Answer:
Molten KCl
KCl (s) → K+ + Cl
At the negative electrode:
K+ + le → K (reduction – cathode)
At the positive electrode:
2Cl → Cl2 + 2e (oxidation- anode)
Solution of potasium chloride.
At the negative electrode:
2H2O + 2e → H2 + 2OH (cathode).
At the positive electrode:
2Cl → Cl2 + 2e (anode).

Future Diary 10th Question 3.
You are given a solution of AgNO3, a solution of MgSO4, a Ag rod and a Mg ribbon. How can you arrange a Galvanic cell using these? Write down the reactions taking place at the cathode and the anode.
Answer:
At anode,
Mg (s) → Mg2+ (aq) + 2e
At cathode,
Ag (aq) + le → Ag (s)
Sslc Chemistry Chapter 3 Questions And Answers Kerala Syllabus

Reactivity Series and Electrochemistry Extended Activities

The Reactivity Question 1.
1. Keep two carbon rods immersed in copper sulphate solution. Then pass electricity through the solution.
i. At which electrode does colour change occur anode or cathode?
ii. Is there any change in the blue colour of the copper sulphate solution?
iii. Write down chemical equations for the changes occurring here.
Answer:
i. At cathode
ii. Colour fades
iii.At cathode: Cu2+ (aq) + 2e → Cu (s)
At anode: 2H2O → O2(g) + 4H+ (aq) + 2e

SSLC Chemistry Chapter 4 Question 2.
When acidified copper sulphate solution is electrolysed oxygen is obtained at the anode. What arrangements are to be made for this? Find the element deposited at the cathode.
Answer:
Kerala Syllabus 10th Standard Chemistry Chapter 3
Element deposited at the cathode: Copper.

Question 3.
a. When Galvanic cells are made using the metals like Mg, Cu, Zn and Ag, what will be the nature of reactions in each cell?
(Reactivity: Mg > Zn > Cu >Ag)
b. How many Galvanic cells can be made by using the metals Ag, Cu, Zn and Mg?
Chapter 7 Biology test Answers:
a. i. Cu-Ag cell
Anode: Cu (s) → Cu2+ (aq) + 2e
Cathode: Ag+ (aq) + le → Ag(s)
ii.Zn-Ag cell
Anode: Zn (s) → Zn2+ (aq) + 2e
Cathode: Ag+ (aq) + le → Ag (s)
(iii) Mg-Ag cell
Anode: Mg (S) → Mg2+ (aq) + 2e
Cathode: Ag+ (aq) + le → Ag (s)
(iv) Zn – Cu cell
Anode: Zn (s) → Zn2+ (aq) + 2e
Cathode: Cu2+ (aq)+ 2e → Cu(s)
(v) Mg-Cu cell
Anode: Mg (s) → Mg2+ (aq) + 2e
Cathode: Cu2+ (aq) + 2e → Cu (s)
(vi) Mg-Zn cell
Anode: Mg (s) → Mg2+ (aq) + 2e
Cathode: Zn2+(aq) + 2e → Zn(s)
b. 6 cells

Reactivity Series and Electrochemistry Orukkam Questions and Answers

Scope of Genetic Engineering Question 1.
1. Take cold water and Hot water in two test tubes, Add one or two drops of phenolphtha lein in it. Drop equally sized Mg ribbon in it.
a. In which test tube pink colour occured sharply?
b.Why did pink colour appear in that test tube so early?
c. Which gas evolved out from both test tubes?
d. Write balanced equation for the above mentioned reaction.
Answer:
a. Pink colour occurred sharply in the test tube with hot water.
b. Temperature is a factor that affects the rate of a reaction. When heated the kinetic energy of molecules increases and hence the rate of chemical reaction also increases which causes the pink colour to appear early.
c. Hydrogen
d. Mg(s) + 2H2O(l) → Mg(OH)2(aq) + H2(g)

SCERT Question Pool 2017 Question 2.
Cut a small sodium metal piece into two, watch it.
a. What change occurred on the surface of sodium metal?
b.Write one word for the process of this type of decomposition.
c. Write down the equations for this.
Answer:
a. After some time the cut piece of sodium will turn dull.
b.Corrosion.
c. 4Na(s) + O2(g) → 2Na2O
2Na(s) + 2H2O(l) → 2NaOH(s) + H2(g)
2NaOH (s) + CO2(g) → Na2CO3(s) + H2O(l)

Question 3.
Take equal quantities of dil HCl in five test tubes. Drop Mg, Zn, Fe, Cu in each test tube. Watch carefully.
a. Arrange metals in decreasing order of reactivity.
b.Write balanced equations for each reaction.
SCERT Question Pool Answer:
a. Mg > Zn > Fe > Cu
b. Zn + 2HCl → ZnCl2 + H2
Fe + 2HCl → FeCl2 + H2
Cu + HCl → No reaction
Mg + 2HCl → Mg Cl2 + H2

SCERT Question Pool Question 4.
Some metals and metallic compounds are given in the table. If the metal substitute the metal in the compound put a tick mark in the corresponding column and otherwise a cross mark in the column. Write down correct answer based on the table given below.

Metalsolution Mg Cu Zn Ag Fe
CuSO4 × ×
ZnSO4 × × × ×
AgNO4 ×
MgSO4 ×

a. Correct the table if necessary.
b. Is it possible to substitute lower positioned metals by top positioned metals in the reactivity series?
c. What type of reaction is this?
d. Write down balanced equations for all the true sign given in the table.
Answer:
a.

Metalsolution Mg Cu Zn Ag Fe
CuSO4 × ×
ZnSO4 × × × ×
AgNO4 ×
MgSO4 ×

b.Yes. It is possible.
c. Substitution reactions.
d.CuSO4 + Mg → Cu + MgSO4
CuSO4 + Zn → ZnSO4+ Cu
CuSO4 + Fe → FeSO4 + Cu
ZnSO4 + Mg → MgSO4 + Zn
AgNO3 + Mg → Ag + MgNO3
AgNO3 + Cu → CuNO3 + Ag
AgNO3 + Zn → ZnNO3 + Ag
AgNO3 + Fe → FeNO3 + Ag

Human Insulin Gene Question 5.
Draw maximum number of Galvanic cell using substances given in the table.
Salt bridge, Zinc rod, Copper rod, Voltmeter, Aluminium chloride, Copper sulphate, Zinc sulphate, Silver nitrate, Silver rod, Calcium chloride
a. Complete the table based on the figures drawn.

Galvanic Cell Electrode which Gives Electron Electrode which Gain Electron

B. Write down the general names used for an electrode which gives electrons,
c. Metals in that electrode in the reactivity series is (in the Top, Bottom)
d. General name of the Electrode which accepts electron.
e. Process of giving electron is
f. Process of Accepting electron is
g.Direction of the flow of Electron
h.Write down the balanced equation taking place in both electrodes.

Galvanic Cell

Electrode which Gives Electron

Electrode which accepts Electron

Answer:
Sslc Chemistry 3rd Chapter Notes Kerala Syllabus
Zn-Ag cell is also possible.
a.
Reactivity Series And Electrochemistry Questions And Answers
b. Anode.
c. In the top.
d. Cathode.
e. Oxidation.
f. Reduction.
g. From Anode to Cathode.
Hsslive 10th Chemistry Kerala Syllabus

Chemistry Reactivity Series Question 6.
Take Cupric chloride (CuCl2) solution in a beaker. Dip two graphic rod in it. Pass 5V electricity through it.
a. Why electricity passes through cupric chloride solution?
b.Which gas evolved out through positive electrode? How did you identify that gas?
c. Which product formed in negative electrode?
d. In which electrode oxidation and reduction take place?
e. Write one word for the process of chemical change happening in a Electrolyte while passing Electricity?
Answer:
a. Electrolytes are substances which conduct electricity in molten states or in aqueous solutions. In molten state ions of CuCl2 can more freely. These ions are responsible for the Conduction of electricity by electrolytes.
b.Chlorine.
c. Copper.
d.Oxidation takes place at the positive electrode and Reduction takes place at the negative electrode.
e. Electrolysis.

Reactivity Series Class 10 Question 7.
Take 25 ml water in a beaker and the pass electricity through it. Then add little sulphuric acid in it.
a. Why electricity didn’t pass through pure water?
b. What happens when dil H2SO4 is added?
c. Which type of ion formed more when sulphuric acid is added in water?
d. Complete the equation of the Ionization 0f H2SO4
H2SO4 → 2H+ +
Based on the equation given below write down the correct answers.
2H+ +  + 2H2O → 2H3O +  + SO42–
e. Complete the equation.
f. Write down the name of H3O+ ion?
g.Which ion is moving towards negative ion?
h.Complete the reaction taking place in the negative electrode.
2H3O+(aq) + 2e →  +
i. Whicfrion is having highest oxidation potential?
j. Complete the reaction taking place in positive electrode.
2H2O →  + 4H+
k. Ions remain in the beaker after the electrolysis are
I. What product form when these two combined together
Answer:
a. Since the number of ions is less, pure water does not allow the passage of electricity.
b. When dil H2SO4 is added the water becomes a good electrolyte. Hence electricity passes through it. When a little dilute sulphuric acid is mixed with water large quantity of hydronium ions are formed.
c. Hydronium ion (H3O+)
d. H2SO4 → 2H+ + SO42–
e. 2H+ + S042– + 2H2O → 2H3O+ + SO42–
f. Hydronium ion
g. H3O+
h. 2H3O+(aq) + 2e → H2(g) + 2H2O
i. H2O has the highest oxidation potential when compared to SO42–
j. 2H2O → O2(g) + 4H+ + 4e
k. 4H+, SO42–ions.
l. H2SO4 is formed when these 2 combine together.

Reactivity Series Chemistry Question 8.
a. Complete the table based on the Electrolysis of molten sodium chloride.

Electrodes

Reaction taking place

product

Anode
Cathode

b. Write down the reaction taking place in each electrodes and products formed in the electrolysis of sodium chloride solution.

Electrodes Reaction taking place product
Anode
Cathode

c. Why is hydrogen formed in the cathode instead of sodium?
d. Write one word for a solution undergoes chemical change when electricity passes through it.
e. Write the name of the above process.
f. Write down the uses of above type of reaction.
Answer:
Hsslive Chemistry 10th Kerala Syllabus
c. When Na+ ion and water are compared reduction occurs to water. Hence H2 is liberated at cathode.
H++ e → H,
H + H → H2.
d. Electrolyte.
e. Electrolysis.
f. Electroplating, Production of chemicals, Purification of metals.

Reactivity Series and Electrochemistry Evaluation Questions

Take little water in a test tube add two drops of phenolphthalein in it Same quantity of Kerosene is added to the mixture and a small piece of sodium is dipped in it.

Insulin Gene Question 1.
What kind of colour formed in the test tube? Why?
Answer:
Water become pink in colour. Because when phenolphthalein is added to water it becomes alkaline.

Question 2.
Which gas is bubbled on the surface of sodium metal?
Answer:
Hydrogen

Question 3.
Write balanced equation of the reaction between sodium and water.
Answer:
2Na(s) + 2H2O(l) → 2NaOH (aq) + H2(g).

Question 4.
What products occurs when Iron reacts with water vapour?
Answer:
Fe3O4 (Iron Oxide) and Hydrogen gas.

Question 5.
Lustre of magnesium disappeared fast when it placed in open space why?
Answer:
When Magnesium is kept in open space it reacts with atmosphere air and a light coat of magnesium oxide is formed. This is the reason for Magnesium to lose its lustre.

Question 6.
Verdigris formed on copper utensils disappeared after some days why?
Answer:
The copper in copper utensils react with atmospheric air and forms copper oxide. Due to this verdigris are formed on Copper utensils.

Question 7.
Lustre of aluminium utensils disappear after some days. Why?
Answer:
Aluminium reacts with atmospheric air and Aluminium oxide is formed. This process takes place slowly. Hence Aluminium utensils loose their lustre.

Question 8.
Write down the equation for the reaction between CuSO4 and iron nail? What type of reaction is this?
Answer:
CuSO4 + Fe → FeSO4 + Cu.
Redox reaction.

Reactivity Series and Electrochemistry SCERT Questions and Answers

Question 1.
5ml water is taken in 3 test tubes. Copper, sodium and magnesium of equal mass are dropped in different test tubes. Test tubes having copper and magnesium are heated.
a. Write the observations in the heated test tubes.
b. Write the equation for the reaction in the test tube in which sodium is dropped,
c. Arrange these metals in the decreasing order of their reactivity.
Answer:
a. Mg reacts with hot water liberating hydrogen, Copper does not react with hot water.
b. 2Na + 2H2O → 2NaOH + H2.
c. Na > Mg > Cu.

Question 2.
a. Which metal among copper, aluminium and gold loose its metallic lustre at a faster rate? Write the equation of the reaction.
b. Sodium is kept in Kerosene. Why?
Answer:
a. Aluminium, Al + 3O2 → 2Al2O3
b. Na reacts with air (oxygen) and water.

Question 3.
An experimental setup is made to compare the reactions of Mg, Zn and Cu with dilute hydrochloric acid.
a. Write the procedure and observation of the reaction.
b. Which is the gas evolved when zinc reacts with dilute hydrochloric acid?
Answer:
a. Take Mg, Zn and Cu in different test tubes and add dilute HCl to each.
Observation: Magnesium and Zinc reacts with dilute hydrochloric acid copper does not react with the acid.
b.Hydrogen.

Question 4.
Reactivity Series And Electrochemistry Notes Pdf Kerala Syllabus
a. What are the changes that can be observed with the iron rod and the colour of copper sulphate solution?
b. Write the equations of the oxidation and reduction reactions.
c. What will be the change if silver rod is used instead of iron rod? What is the reason?
Answer:
a. Copper is deposited on iron and the blue.colour of copper sulphate solution decreases.
b. Cathode – Cu2+(aq) + 2e → Cu (s) (Reduction)
Anode – Fe (s) → Fe2+(aq) + 2e (Oxidation)
c. No change occurs, Reactivity of silver is less than that of copper. In the reactivity series, the position of silver is below copper.

Question 5.
Sodium reacts with water.
a. Identify the gas evolved in the reaction h If two drops of phenol[ihthaloin is added to the water, what will be colour change of the resultant solution? Explain the reason?
Answer:
a. Hydrogen.
b. Colour changes to pink. Due to the presence of sodium hydroxide (alkaline nature).

Question 6.
Three Galvanic cells are given.
Hss Live Chemistry 10th Kerala Syllabus
a. Find out the most reactive metal and least reactive metal among them.
b. In cell, which electrode undergoes oxidation why?
c. Write the equation of the redox reaction occurring in cell 3 (Valency of A, B are 2.)
Answer:
a. Most reactive metal A, Least reactive metal B.
b. A Reactivity of A is higher than B.
c. A+ 2e → A2+,
C2+ + 2e → C ,
A + C2+ → A2+ C .

Question 7.
Some metals and salt solutions are given (Cu, Zn, Ag, ZnSO4, AgNO3, MgCl2)
a. Draw the diagram of a Galvanic cell that can be made using these substances.
b. Find out the anode and cathode of this cell and write the chemical equation for the reaction at cathode.
Answer:
Hsslive Chemistry Class 10 Kerala Syllabus
b. Anode – Zn,
Cathode – Ag
2Ag++ 2e→ 2Ag

Question 8.
Give reasons for the following.
a. Iron vessels are not used a boilers that are used to boil water.
b. Blue vitriol solution is not kept in iron vessels.
Answer:
a. Iron reacted with steam-heated to high temperature

Question 9.
Examine the given electrolytic cell.
Hss Live 10th Chemistry Kerala Syllabus
a. Which gas is evolved at the positive electrode?
b. Write the oxidation and reduction reactions of this cell.
c. What is the difference in the energy transformation of a Galvanic cell and an electrolytic cell?
Answer:
a. Chlorine / Cl2
b. 2Cl → Cl2 + 2e
Cu2+ + 2e → Cu
c. Galvanic cell – Chemical energy is converted to electrical energy.
Electrical Cell – Electrical energy is changed to chemical energy.

Question 10.
The solutions in the given table electrolyzed.
Hss Live Guru Class 10 Chemistry Kerala Syllabus
List any two areas in which electrolysis is made use of?
Answer:
a. i. Hydrogen.
ii. Chlorine.
iii.Chlorine.
iv. Hydrogen.
b.

  • Purification of metals.
  • Electroplating.
  • Production of chemicals.

Question 11.
The position of iron is below that of zinc in reactivity series. The cell formed by them is given. Correct the mistakes and redraw.
Sslc Chemistry Chapter 3 Notes English Medium Kerala Syllabus
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 3 Reactivity Series and Electrochemistry 20

Question 12.
Sodium chloride solution is electrolysed using platinum electrodes.
a. Write the chemical equation of the reaction at cathode.
b. What happens when phenolphthalein is added to the solution? State the reason?
Answer:
a. 2H2O + 2e → H2 + 2OH
b. Colour turns pink. Presence of sodium hydroxide in the solution.

Question 13.
The anode and cathode of two Galvanic cells are given.

Galvanic Cell

Anode

Cathode

Cell 1 Mg Zn
Cell 1 Zn Ag

A. Mg → Mg2+ + 2e
B. Zn2++2e → Zn
C. Ag+ +le → Ag
D. Zn → Zn2+ + 2e
E. Ag → Ag+ + le
F. Mg2+ +2e → Mg
a. Find out the reactions at the anode and cathode for each cell from the above.
b. Which metal can act only as cathode? Why?
Answer:
a. Cell 1: Anode Mg → Mg2+ + 2e
Cathode Zn2+ + 2e → Zn
Cell 2: Anode Zn → Zn2+ + 2e
Cathode Ag+ + le → Ag
b. Ag. Lesser tendency to give up electron that is it is a less reactive metal.

Question 14.
The chemical reactions of various Galvanic cells are given as incomplete in the table. Complete them.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 3 Reactivity Series and Electrochemistry 21
Answer:
a. Zn → Zn2++ 2e
b. Zn+Cu2+ → Zn2+ + Cu
c. Fe → Fe2+ + 2e
d. 2Ag+ + 2e- → 2Ag
e. Pb2++ 2e → Pb
f. Mg + Pb2+ → Mg2+ + Pb

Question 15.
The picture of a Galvanic Cell is given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 3 Reactivity Series and Electrochemistry 22
a. Identify A and B.
b. Give the direction of electron flow?
c. Write the chemical equation at the anode and cathode.
Answer:
a. A – Zn,
B – FeSO4Solution
b. From Zn to Fe
c. Anode Zn → Zn2+ + 2e
Cathode Fe2+ + 2e → Fe

Question 16.
An incomplete table about the electrolysis of different electrolytes are given below. Complete it.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 3 Reactivity Series and Electrochemistry 23
Answer:
a. Cu2+,
Cl,
H2O.
b. 2H2O → O2 + 4H+ + 4e
c. Na+,
Cl
d. Na+ + le → Na
e. 2Cl → Cl2 + 2e
f. 2H2O + 2e→ H2 + 2OH

Question 17.
5mI AgNO3 is taken in a test tube and a copper rod is dipped into
a. Identify the changes occurring with the copper rod and the solution?
b. Complete the equation of the reaction.
Cu + 2AgNO3 →  +
c. Write the equations of the oxidation and reduction reactions.
Answer:
a. Silver is deposited in copper rod. Colour of solution changes to blue.
b. Cu + 2 AgNO3 → Cu(NO3)2 + 2Ag
c. 2Ag+ + 2e → 2Ag (Reduction)
Cu → Cu2+ + 2e (Oxidation)

Reactivity Series and Electrochemistry Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
The symbols of certain metals are given below. Arrange them as they are given in the reactivity series.
Mg, Pb, Ag, Cu, Zn, Fe, Au, Sn.
Answer:
Mg, Zn, Fe, Sn, Pb, Cu, Ag, Au.

Question 2.
Analyse the table given below and answer the questions.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 3 Reactivity Series and Electrochemistry 24
a. Find out the metals which are likely to be A, B and C from the box given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 3 Reactivity Series and Electrochemistry 25
b. Write down the chemical equation between metal ‘B’ and water.
Answer:
a. A – Mg,
B – Cu,
C – Ca
b. Cu (s) + 2H2O (l) → Cu OH2 (aq) + H2 (g)

Very Short Answer Type Questions (Score 2)

Question 3.
Certain metals are given below:
Ag, Zn, Pb, Sn, Fe
a. When a galvanic cell is constructed using these metals, which one acts only as anode? Give the reason.
b. Draw Zn-Fe cell. Mark the direction of electron flow and write the chemical equation anode.
Answer:
a. Zn. Because Zn has a higher reactivity than the other 4 metals.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 3 Reactivity Series and Electrochemistry 26
Direction of electron flow from Zn to Fe Chemical reaction at anode:
Zn(s) → Zn2+(aq) + 2e

Question 4.
Zn (s) + 2AgNO3 (aq) →
Zn(NO3)2 (aq) + 2Ag
a. Write the oxidation state of each element in this displacement reaction.
b. Write the chemical equation for oxidation and reduction.
Answer:
a. Zn° (s) + 2Ag1+N5+ O32– (aq)
→ Zn2+(N5+O2–3)2 (aq) + 2Ag°

b.Oxidation:
Zn° (s) → Zn2+ (aq) + 2e
(Oxidation means losing of electrons)

Reduction:
Ag1+ (aq) + le → Ag (s)
(Reduction means gaining of electrons)

Question 5.
Based on the reactions given below, answer the following questions.
i. Aqueous solution of CuCl2 undergoes electrolysis using graphite rods.
ii. Molten KCl undergoes electrolysis.
iii. Aqueous solution of NaCI undergoes electrolysis.
a. In which all reactions Cl2 gas is formed? At which electrode is Cl2 gas formed?
b. In which reaction H2 gas is formed? Write the chemical equation of this reaction.
Answer:
a. Cl2 gas is formed in all the three reactions. Chlorine gas is formed at anode.

b. When aqueous solution of NaCI undergoes electrolysis, hydrogen gas is formed at cathode.
Equation: 2H2O + 2e → H2 + 2OH

Very Short Answer Type Questions (Score 3)

Question 6.
Take water in 4 different beakers and add a small piece of sodium, lead, iron and copper in each.
a. In which all solutions gas bubbles will be formed? Which gas is formed?
b. Which solution will turn pink on adding phenolphthalein? Why?
c. Write the chemical equation between this metal and water.
Answer:
a. Gas bubbles are formed in the beaker containing sodium. Hydrogen is tlie gas formed.

b. Beaker containing sodium turns pink because sodium reacts with water and forms an alkali, sodium hydroxide. Alkalies turn phenolphthalein pink.

c. 2Na (s) + 2H2O (l) → 2NaOH (aq) + H2 (g)

Question 7.
Analyse the picture given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 3 Reactivity Series and Electrochemistry 27
a. Identify ‘A’.
b. Write the chemical equation at ‘B’.
c. Add few drops of phenolphthalein to the remaining solution after electrolysis. What change will take place? Why?
Answer:
a. A – cathode.

b. Chemical equation at ‘B’:
2Cl → Cl2 + 2e

c. Solution turns pink. Because, after electrolysis K+ and OH ions are present in the remaining solution. That means KOH, an alkali is formed.

Question 8.
The flow of electron in certain galvanic cells are given below:
i. Cu → Ag
ii. Ag → Zn
iii.Na → Mg
iv.Fe → K
a. Choose the incorrect ones.
b. Explain your answer.
Answer:
a. ii. Ag → Zn,
iv. Fe → K are the incorrect ones.

b. Zn has a higher reactivity than Ag. Hence Zn undergoes oxidation (loses electrons). So the direction of electron flow is from Zn to Ag. Similarly, K has higher reactivity than Fe. Hence direction of electron flow is from K → Fe.

Question 9.
Which of the chemical reactions given below are wrong? Explain the reason.
a. Cu (s) + 2HCl (aq) →
CuCl2(aq) + H2 (g)
b. Mg(s) + Pb(NO3)2 (aq) →
Mg(N03)2 (aq) + Mg (s)
c. 3Fe (s) + 4H2O (l) →
Fe3O4(s) + 4H2(g)
Answer:
Equations (a) and (c) are wrong.
a. Copper cannot displace hydrogen from acids because it is placed below hydrogen in the reactivity series.

c. Fe reacts only with superheated steam. Fe does not react with water in liquid state.

Question 10.
Write the chemical equation of the electrolysis of water to which little sulphuric acid (H2SO4) is added.
Answer:
H2SO4 (l) + 2H2O (l) → 2H3O (aq) + SO22–(aq)
at cathode:
2H3O+ (aq) + 2e → H2(g) + 2H2O (l) (reduction)
at anode:
2H2O (l) → O2(g) + 4H (aq) + 4e (oxidation)

Long Answer Type Questions (Score 4)

Question 11.
Analyse the reactions and answer the following questions:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 3 Reactivity Series and Electrochemistry 28
a. Which among the following test tubes will undergo a chemical reaction?
b. What are these chemical reactions called?
c. Explain the oxidation and reduction reactions taking place here including the chemical equation?
Answer:
a. Chemical reaction takes place in test tube (ii) only.
b. These chemical reactions are called displacement reactions. .
c. Chemical equation: (including oxidation states)
Mg° (s) + Fe2+S6+O42– (aq) →
Mg2+S6+O2–4 (aq) + Fe° (s)
At Mg : Mg°(s) → Mg2+(aq) + 2e (oxidation)
At Fe2+, Fe2+ (aq) + 2e →  Fe°(s) (reductipn)

Question 12.
A students observation is given below:
i. When Zn is put in salt solution, Na gets deposited over Zn.
ii. Au reacts with water vapour and hydrogen gas is formed.
iii. Al reacts with acid and forms hydrogen gas.
iv. Mg reacts with hot water and forms hydrogen gas.
a. Which statements are incorrect?
b. Give reason for your answer.
Answer:
a. Statements (i) and (ii) are wrong.

b. i. Zn cannot displace sodium. Because sodium has a higher reactivity than Zn.
ii. Au does not react with water vapour.

Question 13.
“Sodium cannot be kept open in atmospheric air, and cannot be stored in water. So it is stored in kerosene.” Give explanation for the above statement with its chemical equation.
Answer:
Sodium which is placed above in the reactivity series reacts vigorously with water and oxygen.
4Na (s) + O2 (g) → 2Na2O (s)
2Na (s) + 2H2O (l) →
2NaOH (aq) + H2(g)
This NaOH reacts with CO2 in the atmospheric air and forms sodium carbonate.
2NaOH (s) + CO2 (g) → Na2CO3 + H2O (l)
To avoid the reaction with O2, H2O and CO2, sodium is stored in kerosene.

Question 14.
The direction of electron flow in certain galvanic cells are given below. (Symbols are not real)
i. B → A,
ii. E → C,
iii.D → E,
iv.A → D.
a. Arrange the metals A, B, C, D and E in the decreasing order of their reactivity.
b. Choose the reaction taking place at ‘C’ in cell (ii) E → C. Give the reason.
i. C+ (aq) + le → C(s)
ii. C (s) → C+ (aq) + le
iii.C (s) → C+ (aq) + le
Answer:
a. B,
A,
D,
E,
C.
b.C+ (aq) + le → C (s) is the correct equation. Because ‘E’ has higher reactivity than. So ‘C’undergoes reduction.

Question 15.
Give reason for the following.
a. CuSO4 solution is not stored in iron vessels.
b. Buttermilk is not stored in aluminium vessels.
Ans.
a. Fe displaces copper from copper sulphate solution and forms FeSO4.
b. Aluminium reacts with acid in the buttermilk.

Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis

You can Download Chemical Messages for Homeostasis Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Biology Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis

Chemical Messages for Homeostasis Text Book Questions and Answers

Chemical Messages For Homeostasis Kerala Syllabus 10th Question 1.
Which are the hormones you know? List them?
Answer:

  • Insulin
  • Thyroxine
  • Oestrogen

Hormones:
The endocrine glands play a vital role in coordinating and controlling life activities. Secretions of endocrine glands are called hormones. There secretions are chemical substances that belong to different categories such as proteins, peptides, steroids, fatty acids, etc. Endocrine glands do not have particular ducts carry hormones to various tissues. Hence they are known as ductless glands. Hormones are transported through blood. As these substances regulate cellular activities, they can be called chemical messages to cell.

Chemical Messages for Homeostasis Hormones In Target Cells

The cell which are acted upon by hormones are called target cells. Only cells having specific receptors can receive a particular hormone. A hormone-receptor complex is formed by the combination of each hormone molecule and its receptor. Following this, enzymes are activated within the cell. As a result, certain changes occur in cellular activities.
Chemical Messages For Homeostasis Kerala Syllabus 10th
After Digestion

Sslc Biology Chapter 3 Questions And Answers Kerala Syllabus
Pancreas helps in the digestive process. It functions as an endocrine gland too. It secretes two hormones namely insulin and glucagon.
The beta cells in the Islets of Langerhans glucagon.

Action of insulin and glucagon:
Sslc Biology Chapter 3 Notes Kerala Syllabus

Sslc Biology Chapter 3 Questions And Answers Kerala Syllabus Question 2.
Complete the illustration by including the production of hormones that regulate the level of glucose.
Answer:
Biology Class 10 Chapter 3 Notes Kerala Syllabus
Sslc Biology Chapter 3 Notes Kerala Syllabus Question 3.
How is the level of glucose in the blood maintained while fasting? Discuss
Answer:
When the level of glucose increases in blood the cells in the Islets of Langerhans produce insulin which converts the excess glucose into glycogen, protection and lipids.

Diabetes Mellitus

Diabetes is clinically referred to as a condition when the level of glucose before breakfast is above 126 mg/100ml of blood. It is caused either by decreased production of insulin or it$ malfunctioning. Symptoms: Increased appetite and thirst, Frequent urination, Traces glucose in urine Diabetes can be controlled through medicine, diet control and insulin injections.

Biology Class 10 Chapter 3 Notes Kerala Syllabus Question 4.
The increase of glucose in blood is said to be diabetes. Shouldn’t one be more energetic if the glucose level in his/her blood rises? What is your opinion? Write them down in your science diary.
Answer:
No. One be more should not energetic if the glucose level in his or her blood. Persons with diabetes experience loss of body weight, weakening of muscles and tiredness.

Regulation Of Metabolism

10th Class Biology Chapter 3 Kerala Syllabus

The anabolic and catabolic processes taking place in the body are metabolism. The thyroid gland is the endocrine gland contrails the metabolic process.

Functions of thyroxine:

Chemical Messages For Homeostasis Notes Pdf Kerala Syllabus

10th Class Biology Chapter 3 Kerala Syllabus Question 5.
How would be body activities be affected if sufficient amount of thyroxine is not produced?
Answer:

  • Low energy production
  • Bloating of body
  • Slowing down of heartbeat
  • Loss of appetite, lethargy
  • Dry skin

Undersecretion of thyroxine – Hypothyroidism:
The deficiency of thyroxine during the fetal stage or infancy leads to mental retardation and stunted growth. This condition is cretinism. Lack of thyroxine in adults leads to myxoedema.

Symptoms of Hypothyroidism:

  • Low metabolic rate
  • Sluggishness
  • Sleeplessness
  • Increase in body weight
  • Hypertension
  • Oedema

Oversecreation of thyroxine – Hyperthyroidism:
The condition in which all life activities controlled by thyroxine are accelerated due to the excessive production of thyroxine is referred to hyperthyroidism.

Symptoms of Hyperthyroidism:

  • High metabolic rate
  • Rise in body temperature
  • Excessive sweating
  • Increased heartbeat
  • Sleeplessness
  • Weight loss
  • Emotional imbalance

Goitre:
Iodine is essential for the production of thyroxine. The production of thyroxine is obstructed in the absence of iodine. In an attempt to produce more thyroxine; the thyroid gland enlarges. This condition is called goitre.

Chemical Messages For Homeostasis Notes Pdf Kerala Syllabus Question 6.
What is the importance of thyroxine in controlling life activities?
Answer:
Thyroxine is a hormone that influences metabolism in our body to a great extent.

Biology 3rd Chapter 10th Class Kerala Syllabus Question 7.
What are the problems caused by excessive production of thyroxine?
Answer:

  • Energy production increases and body weight decreases
  • Increased heartbeat
  • Increased appetite
  • Shivering of hands and profuse sweating
  • Persistent hyperthyroidism may lead to Graves disease, characterized by exophthalmic goiter.

Sslc Biology Chapter 3 Notes Pdf Kerala Syllabus Question 8.
What are the problems due to thyroxine deficiency?
Answer:
The deficiency of thyroxine retards mental and physical growth of children. This condition is called cretinism. In adults the deficiency of thyroxine results in a disease called myxoedema.

Sslc Biology Chapter 3 Solutions Kerala Syllabus Question 9.
How is iodine related to thyroid gland?
Answer:
Iodine is essential for the production of thyroxine. The production of thyroxine is obstructed in the absence of iodine.

Calcitonin:
It helps in maintaining the level of calcium in blood by depositing excess calcium in bones and by preventing the mixing of calcium with blood, from the bones.

Parathyroid Gland

Biology 3rd Chapter 10th Class Kerala Syllabus

The parathyroid gland is situated behind the thyroid gland. This gland secretes a hormone called parathormone. The function of this hormone is to raise the level of calcium in blood

Sslc Biology Chapter Wise Questions And Answers Kerala Syllabus Question 10.
Complete the illustration showing the maintenance of the level of calcium in blood by the action of calcitonin and parathormones
Answer:
Sslc Biology Chapter 3 Notes Pdf Kerala Syllabus

The hormone only upto youth:
Thymus gland:
The thymus gland is situated just below the sternum. The major function of thymus gland is to control, the activities and maturation of lymphocytes which help to impart immunity. This gland secretes thymosin, which is active during infancy. Hence it is known the ‘youth hormone’.

During Emergencies

These glands are situated above the kidneys. The outer part of the adrenal gland is known as the cortex and inner part is medulla. The adrenal cortex secretes aldosterone, cortisol and sex hormones. Adrenal medulla secretes epinephrine and norepinephrine.

Class 10 Biology Notes Chapter 3 Kerala Syllabus Question 11.
The structure of the adrenal glands and the hormones produced by them are illustrated below. On the basis of the indicators given, discuss and write down the notes in the science diary.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 9
i) Hormones secreted by the adrenal cortex
ii) The function of cortisol
iii) Maintenance of salt-water balance in the body
iv) The function of epinephrine and norepinephrine during emergencies.
Answer:
(i) Aldosterone and sex hormones are the hormones secreted by the adrenal cortex.
ii) The synthesis of glucose from protein and fat controls inflammation and allergy, slows down the action of defense cells.
iii) Aldosterone is the hormone that helps to maintains salt-water balance in the body by restating the lose of Na+ ions and by promoting the elimination of K+ ions through sweat, urine, etc.
iv) Epinephrine acts along with the sympathetic nervous system during emergencies. Thus we can resist or withdraw ourselves from such situations. Norepinephrine acts along with epinephrine.

Biological Clock
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 10
Pineal gland is seen in centre of the brain. It secretes the hormone, melatonin which helps in maintaining the rhythm of our daily activities. The production of melatonin is high at night and low during the day. When the level of melatonin increases we feel sleepy and when it decreases we wake up. This hormone also controls reproductive activities of organisms.

Behind Growth

Pituitary gland is a bibbed gland situated just below the hypothalamus in the brain. The anterior lobe produces tropic hormones which regulate the functions of other glands. The posterior lobe stores the hormones which are produced in the hypothalamus.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 11
Chemical Messages For Homeostasis Questions And Answers Question 12.
The hormones produced by the anterior lobe is listed in table. Analyse-it and complete the following worksheet in the science diary.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 12
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 13
Answer:
A – Stimulates the activity of thyroid giand
B – AdrenoCortico Tropic Hormone(ACTH)
C – Gonado Tropic Hormone(GTH)
D – Stimulates the activity of ovaries
E – Production of milk
F – Growth Hormone (GH) or Somatotropic Hormone (STH)

Hss Live Guru 10th Biology Kerala Syllabus Question 13.
How the variation in the production of somatotropin affects growth.
Answer:
Somatotropin promotes growth of the body during its growth phase. If the production of this hormone increases during the growth phase, it leads to the excessive growth of the body. This condition is gigantism. It causes another stage called dwarfism when its production decreases during the growth phase. Acromegaly is the condition caused by the excessive production of somatotropin after the growth phase. It is characterized by the growth of the bones on face, jaws and fingers.
The Posterior Lobe of pituitary gland – A storage center
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 14
The hormones oxytocin and vasopressin, which are secreted from the posterior lobe of the pituitary are actually produced in the neuro-secretory cells of the hypothalamus. The posterior lobe stores these two hormones and Releases them into blood when required.

Hss Live Guru 10 Biology Kerala Syllabus Question 14.
Observe the table and write down your inferences in the science diary.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 15
Answer:
Oxytocin and vasopressin are secreted from the hypothalamus and stored in the posterior lobe of pituitary gland. Through the connecting nerve fibers they are transported to pituitary gland. Oxytocin facilitates childbirth by stimulating the contraction of smooth muscles in the uterine wall and facilitates lactation. Vasopressin helps in the reabsorption of water in the kidneys.

Biology Chapter 3 Class 10 Kerala Syllabus Question 15.
Observe illustration given below which shows the action of vasopressin in kidneys. Based on the indicators given discuss and write a note in science diary.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 16
i) The functions of vasopressin in kidneys
ii) The reason for excessive production of urine during the rainy season
iii) The role of vasopressin in preventing loss of water from the body.
iv) Diabetes insipidus
Answer:
i) The hormone vasopressin regulator reabsorption of water and minerals from the glomerular filtrate.. When there is a reduction in the amount of water in blood, the production of vasopression increased. It accelerates the rate of reabsorption of water. When the quantity of water in blood increases the production of vasopressin decreases. As a result, the rate of reabsorption is also reduced.

ii) During rainy season, sweat production reduces. So the water loss through sweat is decreased. It wingcase the quantity of water in our body Comparatively high. Such situations demand elimination of excess water through urine.

iii) The production of vasopressin increases when there is a need to reduce loss of water through urine. As a result of this, more water is reabsorbed to the blood from kidneys, Thereby the loss of water through urine is reduced and regulate water loss in our body.

iv) The rate of resorption of water in the kidney is decreased when there is no sufficient amount of vasopressin. Hence excess amount of urine is excreted. This condition is called diabetes insipidus Symptoms include frequent urination.

Behind Sexual Characteristics

Testes and ovary, the male and female sex organs respectively, secrete different types of hormones.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 17
Prepare a table by including hormone, centre of production and function:
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 18

Prime Controller

Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 20
Oxytocin and vasopressin are secreted by the hypothalamus. In addition to this hypothalamus controls the pituitary gland by secreting a variety of releasing hormone are inhibitory hormones.

Biology Chapter 3 Class 10 Notes Kerala Syllabus  Question 16.
Observe the illustration given below on the functions of releasing hormones and inhibitory hormones. On the basis of indicatiors discuss and writes it down in the science diary.

Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 21
i) Action of releasing hormone
ii) Influence of tropic hormones in different glands.
iii) Action of inhibitory hormones
Answer:
i) Stimulates the anterior lobe of the pituitary and secretes tropic hormones.
ii) Tropic hormones stimulate the production of hormones of certain other important glands.
iii) Inhibits the production of tropic hormones in the anterior lobe of the pituitary gland.

Chemical Messages For Communication

Pheromones: Pheromones are chemical substances that are secreted in trace amounts to the surrounding in order to facilitate communication among organisms. Pheromones help in attracting mates, to inform the availability of food, to determine the path of travel and to inform about dangers.
Eg:- Musk in the nusk deer, civet on in civet cat, Bombycol in female silkworm.

Plant Hormones

There are certain chemical substances in plant cells to control and co-ordinate life activities. These are also called plant growth regulators.

Question 17.
Observe the illustration, which show plant hormones and their functions and complete the following table suitably.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 22
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 23
Answer:
a) Cell growth, cell elongation, fruit formation
b) Controls the dormancy of embryo in the seeds, dropping to leaves and fruits wilting of leaves, flowering, etc.
c) Gibberellins
d) Promotes cell division cell growth
e) Ethylene.

Artifical Plant Hormones

Auxins: NaphtheleneAceticAcid (NAA), Indol Butyric Acid (IBA), etc., are used for sprouting and the prevention of dropping of premature fruits. 2.4-D (2, 4-Dichloro phenoxy acetic acid) is used as a weedicide.

Gibberellins: It is used for increasing fruit size in grapes and apple and also for preventing ripening of fruits to assist in marketing.

Abscisic acid: As it accelerates the dropping of fruit, it is used for harvesting fruits at the same time.

Ethylene: Ethylene is used for the flowering of pineapple plants at a time and for the ripening of tomato, lemon, orange, etc. Ethyphon, a chemical which is available in liquid form gets transformed into ethylene when used in rubber trees, and it increases the production of latex.

Chemical Messages for Homeostasis Let Us Assess

Question 1.
Identifying the word-pair relationship and fill in the blank.
Thyroxine: Thyroid gland
Epinephrine:…………..
Answer:
Adrenal gland

Question 2.
Analyze the information given in the box and answer the questions.
X- The production of this hormone is more in night and less in day time.
Y – Hormones from the adrenal gland work along with the sympathetic system.
(a) Identify and name the hormone ‘X’ and its gland.
(b) Identify the hormones indicated as ‘Y’.
Answer:
a) Melatonin pineal gland
b) Epinephrine (Adrenaline)
c) Norepinephrine (Noradrenline)

Question 3.
Analyse the illustration and complete the table appropriately.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 24
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 25
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 26
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 27

Question 4.
The hormone that helps in the reabsorption of water in the kidneys.
a) TSH
b) ACTH
c) ADH
d) GTH
Answer:
ADH

Chemical Messages for Homeostasis Extended Activities

1. Conduct a seminar on the topic – The role of the Endocrine system in maintaining homeostasis’.
Main points:-

  • Situations which lead to change in homeostasis
  • How is homeostasis reinstated
  • Harmonious co-existence

2. Conduct a debate on ‘Use of artificial plant hormones – problems and possibilities’.

3. Collect information about novel laboratory tests related to diagnosis of diabetes and conduct an exhibition on World Diabetes Day.

Chemical Messages for Homeostasis More Questions and Answers

Question 1.
Correct the sentence if it is wrong
1. Endocrine glands are ductless glands
2. The alpha cells in the Islets of Langerhans secrete insulin.
3. Aldosterone slows down the action of defense cells.
4. Anti Diuretic hormone helps in the reabsorption of water in the kidneys.
5. Hypothalamus secretes inhibitory hormones which stimulate the anterior lobe of the pituitary gland.
Answer:
1. Endocrine glands are ductless glands
2. The beta cells in the islets of Langerhans secrete insulin.
3. Cortisol slows down the action of defense cells.
4. Anti Diuretic hormone helps in the reabsorption of water in the kidneys.
5. Hypothalamus secretes releasing hormones that stimulate the anterior lobe of the pituitary gland.

Question 2.
Endocrine glands are called ductless glands. Why?
Answer:
Endocrine glands do not have particular ducts to carry hormones to various tissues. Hence they are called ductless glands.

Question 3.
Name the hormone-producing centers situated in the brain?
Answer:
Hypothalamus, pituitary, Pineal

Question 4.
The gland which is active only during infancy?
Answer:
Thymus

Question 5.
Though hormones reach every part of the body through the blood, all hormones do not act upon all cell. Explain the reason.
Answer:|
The cell which are acted upon by hormones are called target cells. Only cells having specific receptors can receive a particular hormone. A hormone-receptor complex is formed by the combination of each hormone molecule and its receptor. Following this, enzymes are activated withfn the cell. As a result, certain changes occur in cellular activities.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 28

Question 6.
Name the digestive gland which is also functioning as an endocrine gland?
Answer:
Pancreas

Question 7.
What is the normal level of glucose in blood? How is this level maintained?
Answer:
The normal level of glucose is 70-110 mg/100 ml blood. The level of glucose in blood is maintained by the combined action of insulin and glucagon of the Islets of Langerhans tissues of the pancreas. Insulin, released from the beta cells of Islets of Langerhans, helps to reduce blood sugar by accelerating the process of cellular uptake of glucose and Conversion of glucose in to glycogen. When blood glucose level falls, glucagon, released from the alpha cells of Islets of Langerhans, converts glycogen to glucose and synthesizes glucose from amino acids.

Question 8.
Suppose a person is fasting in a day and takes heavy food on the very next day. How is the level of glucose in his body is maintained in these two days?
Answer:
While fasting glucagon converts glycogen or amino acids into glucose. When taking heavy food insulin enhances cellular uptakes of glucose and converts glucose into glycogen.

Question 9.
Diabetic patients frequently take insulin injections. Give reason?
Answer:
Insulin is helpful to reduce the excess glucose in the blood and to maintain its normal level

Question 10.
If the level of glucose increases one feels hunger, thirsty and fatigue instead of becoming energetic. Give reason?
Answer:
Increasing the level of glucose in blood adversely affects the normal functioning of the cells.

Question 11.
A doctor advised one of his patients to use iodized salt and to include more leafy vegetables and marine items in his diet. What should be reason for this recommendation?
Answer:
To prevent goitre. Deficiency of iodine may cause Goitre, a disorder affects on thyroid gland.

Question 12.
Under secretion of thyroxine: Hypothyroidism
Over secretion of thyroxine: ……………..?
Answer:
Hyperthyroidism

Question 13.
Overproduction, as well as underproduction of the hormone thyroxine, may lead to disorders’. Substantiate.
Answer:
Deficiency of thyroxine (Hypothyroidism) leads to cretinism in infants and myxoedema in adults. Excess production of thyroxine (Hyperthyroidism) leads to a condition, known as the Graves disease.

Question 14.
Persistent hyperthyroidism may leads to ……………. disease characterized by bulging of the eye balls.
Answer:
Graves disease

Question 15.
What is the normal level of calcium in the blood? How is this level maintained?
Answer:
9-11 mg/100 ml blood.
When the level of calcium in blood increases, thyroid gland secretes a hormone named calcitonin. It lowers the level of calcium in blood by depositing excess calcium in bones and by preventing the mixing of calcium with blood form the bones. When the level of calcium in blood decreases, parathyroid gland secretes parathormone. It increases blood calcium by reabsorbing it from the kidneys and also preventing the deposition of calcium in bones.

Question 16.
Complete the flow chart
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 66
Answer:
A. Medulla
B. Aldosterone
C. Cortisol
D. Norepinephrine

Question 17.
Overproduction of parathormone can weaken the bones. Why?
Answer:
The hormone, parathormone prevents the deposition of calcium in bones resulting its weakening.

Question 18.
The hormone which can be used to prevent allergy and inflammation? Can this hormone be given to diabetic patients? Why?
Answer:
Cortisol of adrenal gland. It cannot be given to diabetic patients as it increases the level of glucose in blood.

Question 19.
The pineal gland is known as the ‘biological clock’ in the body. Why?
Answer:
Melatonin, the secretion of the pineal gland helps to . maintain rhythm of our daily activities. Therefore pineal gland is called as the biological clock.

Qn. 20
What are the hormones of hypothalamus stored in the posterior lobe of pituitary gland? Mention its functions.
Answer:
Oxytocin – Facilitates childbirth by stimulating the contraction of smooth muscles in the uterine wall and also facilitates lactation Vasopressin (Anti Diuretic Hormone) – Helps in the reabsorption of water in the kidneys.

Question 21.
Give reasons.
Some times certain pregnant women need to take oxytocin injection.
Answer:
Oxytocin facilitates childbirth by stimulating the contraction of smooth muscles in the uterine wall. It also facilitates lactation.

Question 22.
Point out the functions of releasing hormones and inhibitory hormones.
Answer:
Releasing Hormones: Stimulate the anterior lobe of the pituitary to secretes tropic hormones and other hormones.
Inhibitory Hormones: Inhibit the production of tropic hormones and other hormones from the anterior lobe of the pituitary gland.

Question 23.
What is the reason behind the difference in the quantity of urine during summer and rainy season?
Answer:
The production of vasopressin is high during summer season where water loss is excessive through sweat. But its production is less during winter and rainy seasons and there is difference in the quantity of urine during summer and rainy seasons.

Question 24.
Why do Vasopressin is known as antidiuretic hormone (ADH)?
Answer:
Because vasopressin retains the quantity of water by inducing the kidneys to reabsorb it.

Question 25.
Complete the following table related with the hormonal functions of our sex organs.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 30
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 31

Question 26.
Identify the hormone defects concern with the following hints.
a) Insulin injection
b) Treatment using thyroxine
c) Food and medicine containing calcium
d) Seafood, vegetable and iodized salt.
Answer:
a) Diabetes
b) Myxoedema
c) Osteoporosis
d) Goiter

Question 27.
How is homeostasis of the body maintained?
Answer:
Homeostasis of the body maintained by the combined action of the quick nervous system and the slow endocrine system.

Question 28.
How are pheromones useful to animals?
Answer:
Pheromones help in attracting mates, to inform the availability of food, to determine the path of travel and to inform about dangers.

Question 29.
A farmer says pest control is made possible using pheromones. Can you say how?
Answer:
Artificial pheromones are used for pest control in agricultural field.

Question 30.
Identify the plant hormone that performs the following functions.
a) flowering and growth of leaves
b) ripening of fruits
c) dropping of leaves and fruits
d) growth of terminal bud.
Answer:
a) Gibberellin
b) Ethylene
c) Abscisic acid/ ethylene in excess amount.
d) Auxin

Question 31.
Site examples of situations where artificial plant hormones are applied widely.
Answer:
Ethylene is used for the flowering of pineapple plants at a time and for the ripening of tomato, lemon, orange, etc.
Ethyphon, a chemical which is available in liquid form gets transformed into ethylene when used in rubber trees, and it increases the production of latex.

Auxins: Naphthalene Acetic Acid (NAA), Indol Butyric Acid (IBA) etc. are used for sprouting and the prevention of dropping of premature fruits. 2,4- D (2, 4-Dichloro phenoxy acetic acid) is used as a weedicide.

Gibberellins: Used for increasing fruit size in grapes and apple and also for preventing ripening of fruits to assist in marketing.

Abscisic acid: As it accelerates the dropping of fruit, it is used for harvesting fruits at the same time.

Question 32.
The following figure shows the relationship of hypothalamus with an endocrine gland. (Model 2016)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 32
a) Write down the name of endocrine gland marked as X
b) Write down the name of hormone produced the A and B.
c) Mention the functions of hormones produced the B.
Answer:
Answer:
a) Pituitary gland
b) A – Tropic hormone
B – Oxytocin and vasopressin
c) Oxytocin helps to contraction of smooth muscles and vasopressin helps in the reabsorption of water in the kidneys.

Question 33.
Artificial hormones should be handled with care. What is your opinion?
Answer:
This statement is correct. Though artificial hormones are useful they should be handled with care as they are chemicals, which may cause health and environmental issues.

Question 34.
…………….. is used for increasing fruit size in grapes and apple.
Answer:
Gibberellins

Question 35.
………….. is a plant hormone, used for harvesting fruits in a field at the same time.
Answer:
Abscisic acid

Question 36.
The quantity of urine excreted by a person in different seasons is given below. Analyse-it and answer the following questions. (Model 2016)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 33
a) Write down the climate B and C
b) Analyse the difference shown in B and C and write down its reasons.
c) Which hormone is responsible for the excretion of excess water through urine.
Answer:
a) B – Rainy season or winter season, C – Summer season
b) In rainy season production of vasopressin is less it decreases the reabsorption of water in the kidneys. So raises the quantity of urine. In summer season production of vasopressin increases. It increases the reabsorption of water in kidneys and lowers the quantity of urine.
c) ADH or vasopressin

Question 37.
Given below is the blood test result of a person. Analyze the result and answer the following questions? (Model 2016)
Glucose – 200mg/100ml
Calcium -11 mg/100ml
a) Name the disease of the man mentioned in the test report.
b) Write down the name of hormone which related to this disease.
c) What is the cause of this disease?
Answer:
a) Diabetes mellitus
b) Insulin
c) It is caused either by the decreased production of insulin or its malfunctioning

Question 38.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 34
a) Complete the table based on the hormone somatotropin (Model 2014)
b) This hormone is not a tropic hormone. Why?
Answer:
a) (i) X-dwarfism 1) become dwarfs due to stunted growth of bones
ii) Y – gigantism 2) Growing tall with a heavy body
iii) Z-acromegaly 3) enlargement of internal organs and thickening of bones, especially in hands feet and face.

b) Somatotropin does not induce any other endocrine gland to release its hormone

Question 39.
The quantity of urine excreted by a person in different seasons is given below. Analyse-it and answer the following questions. (Model 2014)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 35
a) Which is the coldest season?
b) Which hormone is responsible for the variation in quantity of urine?
c) How this hormone regulates water level of the body.
Answer:
a) Season 3
b) ADH/Vasopressin
c) This hormone promotes reabsorption of water from renal tubules when normal level of water in blood decreases. The rate of reabsorption of water in the kidney is decreased when there is no sufficient amount of vasopressin.

Question 40.
Some hormones are given below. Make them into 4 pairs. Give reasons for pairing. (Model 2014)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 36
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 37
basis pairing:- Products of same gland

Question 41.
Observe the chart (March 2013)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 38
Write down the climate A and B
Answer:
A-Summer season
B – Rainy season or winter

Question 42.
Changes in the number of hormones produced will affect our bodily activities. Write down the changes occur in our body by the increase and decrease of the hormones given below.
a) Parathormone
b) Vasopressin
Answer:
a) Increase of parathormone – Bones fragile stones in urinary tract, high blood calcium
Decrease of parathormone – Blood calcium level decrease and it leads to tetany

b) Vasopressin
Production of vasopressin increases It accelerate the rate of reabsorption of water from kidney. So the loss of water through urine is reduced.
Production of vasopressin decreased The rate of reabsorption is reduced and more water discharged out through urine.

Question 43.
Rearrange B, C and D according to the data given A
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 39
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 40

Question 44.
“It is now that I understand why the cock crows early in the morning every day”. Anu said this during a classroom discussion on the rhythm of physiological activities.
a) Which is the hormone that regulates such activities?
b) Which gland secretes this hormone.
c) Write down more examples for such activities
Answer:
a) Melatonin
b) Pineal gland
c) It regulates the rhythm of life, reproductive activities of organisms with definite reproductive periods.

Question 45.
Fill up the blanks (Model 2012)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 41
Answer:
a) Thyroxine
b) Thyroid
c) Insulin
d) Diabetes
e) Pituitory gland
f) Dwarfism

Question 46.
Fill up the blanks
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 42
Answer:
a) – Gibberellin
b) – Helps in the ripening of fruits
c) – Abscisic acid

Chemical Messages for Homeostasis SCERT Questions and Answers

Question 1.
Observe the illustration given below and explain how hormones act in target cells.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 43
Answer:
The cell which are acted upon by hormones are called target cells. Only cells having specific receptors can receive a particular hormone. A hormone-receptor complex is formed by the . combination of each hormone molecule and its receptor. Following this, enzymes are activated within the cell. As a result, certain changes occur in cellular activities.

Question 2.
Some statements relate to endocrine system are given below. (Question Pool 2017)
A. Hormones are the secretions of endocrine glands.
B. Hormones are transported through lymph.
C. Hormones are transported through blood.
D. All the harmonies produced by the endocrine glands are proteins.
a) Choose the correct statement.
b) Imagine that particular hormone is not entering a particular cell. What may be the reason? Formulate two hypotheses.
Answer:
a) A, C
c) Receptors of that hormone in not in the cell

Question 3.
Examine the graph indicating the blood glucose level of different individuals before breakfast. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 44
a) Which individual is affected by diabetes mellitus?
b) Write two actions of insulin to prevent the rise in the level of glucose in blood.
c) Why do people having diabetes mellitus experience extreme fatigue?
Answer:
a) (B)
b) 1. Enhances the entry glucose into the cell.
2. Converts glucose to glycogen in liver and muscles.
c) Sufficient quantity of glucose i not reaching the cell. Energy production decreases. Excess amount of glucose is eliminated through urine.

Question 4.
Case sheets of two patients are given below. Analyze them and answer the questions. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 45
a) Which are the diseases whose symptoms are indicated above?’
b) Write the reasons for the diseases.
Answer:
a) Case -1 cretinism;
Case – 2 graves disease

b) Case -1 reasons
Deficiency of thyroxine during foetal stage and infancy.
Case-2 reasons
1. Persistent hyperthyroidism
2. Excessive production of thyroxine.

Question 5.
Analyse the table given below. Rearrange column Band C according to the indicators in Column A. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 46
Answer:
1 – (b) – (r)
2 – (c) – (p)
3 – (a) – (q)

Question 6.
Honey bees and termites live in colonies. (Question Pool 2017)
a) Name the chemical substance which helps them to live together.
b) Mention two uses of these chemical substances.
Answer:
a) Pheromones
b) 1. attracting mates
2. informing availability of food
3. determining the path of travel
4. informing the dangers

Question 7.
Observe the diagram and answer the questions. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 47
a) Which endocrine gland does ‘X’ indicate?
b) Which are the two hormones produced by the gland to control the physical activities with the sympathetic system?
Answer:
a) Adrenal gland
b) Epinephrine, Norepinephrine

Question 8.
Maintenance of the level of calcium in the blood is illustrated below. Analyse-it and answer the following questions. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 48
a) Name the hormone indicated as X’.
b) Which gland produces the hormone ‘Y’?
c) Write another activity performed by ‘X’ to raise the level of calcium in blood.
Answer:
a) Parathormone
b) Thyroid gland
c) Helps in the reabsorption of calcium from kidneys.

Question 9. (
Observe the diagram of the endocrine gland given below and answer the question. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 49
a) Name the part indicated as A and B.
b) Name the hormones synthesized by A. Explain their action.
Answer:
a) A Medulla
B Cortex
b) Epinephrine, Norepinephrine
Epinephrine – Helps to tide over emergency situations
Norepinephrine – acts along with epinephrine

Question 10.
An individual loses large quantities of water through urine (Question Pool 2017)
a) Which could be the disease?
b) Analyze the conditions that lead to this disease.
Answer:
a) Diabetes insipidus
b) ADH is synthesized by hypothalamus.
ADH increases the reabsorption of water into the kidney.
Synthesis of ADH decreases.

Question 11.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 50
a) Identify X and Y. (Question Pool 2017)
b) What is the function of ‘Y’?
Answer:
a) Portal vein – X
Posterior lobe of pituitary – Y
b) Stores the hormones vasopressin and oxytocin synthesized by hypothalamus and releases them into blood when required.

Question 12.
Given in the table below is to growth hormone. Complete the table suitably. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 51
Answer:
a) dwarfism
b) Excessive production of growth hormone during the growth phase.
c) Excessive production of somatotropin after the growth phase.
d) Growth of the bones on face, jaws and fingers.

Question 13.
Given below is a doctor’s comment at a seminar conducted as part of Diabetic day.
“In diabetic patients, the blood glucose level before breakfast is above 126mg/100ml.
Analyse the statement and enlist the reasons.
Answer:

  • Decreased production of insulin
  • Malfunctioning of insuline
  • Destruction of Beta Cells
  • Inactive insulin

Question 14.
Given below are a few statements related to hormones. Pick out the correct ones. (Question Pool 2017)
a) Estrogen helps to maintain embryo in the uterus.
b)Progesterone facilitates childbirth.
c) Prolactin helps in the production of milk.
d) Oxytocin n faci itates I a citation.
Answer:
c, d

Question 15.
Analyze the statements given below and write the reason. (Question Pool 2017)
a) Oxytocin is injected in pregnant women during childbirth, (delivery)
b) Feels sleepy during night, wakeup when day breaks.
Answer:
a) Facilitates childbirth by stimulating the contraction of smooth muscles in the uterine wall.
b) When the level of melatonin increases at night, we feel sleepy,
We wake up when the level of melatonin decreases during the day.

Question 16.
Analyse the table and identify the correct pair. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 52
Answer:
a) Somatotropin decreases during growth phase – dwarfism

Question 17.
Observe the table, re-arrange column Band C according to column A.
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 53
Answer:
1 – (b) – (S)
2 – (d) – (P)
3 – (a) – (Q)

Question 18.
A farmer named Balan cultivated oranges in his orchard. Now the trees are full of oranges. The price of oranges is Rs. 80/kg. (Question Pool 2017)
A) This farmer wants to harvest all fruits together.
B) Ripen them together.
a) Suggest two artificial plant hormones to satisfy the A, B needs of the farmer.
b) Uncontrolled use of plant hormones must be controlled. Evaluate this statement.
Answer:
A) a) A – Abscisic acid
B – Ethylene
b) Though artificial hormones are useful they should be handled with care as they are chemicals. Uncontrolled use of it may cause health and environmental issues.

Question 19.
Analyse the indicators and answer the question given below. (Question Pool 2017)
Indicators
Accelerates the growth and development of the brain in the foetal stage and infancy.
a) Which hormone are the indicators about?
b) Construct a flow chart relating the action of hypothalamus and pituitary in the synthesis of this hormone.
Answer:
a) Thyroxine
b)

Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 54

Question 20.
Artificial plant hormones are used extensively in the agricultural sector. (Question Pool 2017)
Write the name and function of two artificial plants. hormones belonging to the category, auxin.
Answer:
NAA- Sprouting, prevention of premature fall of fruits.
IBA – -do-
2, 4 – D – Weedicide

Question 21.
Artificial plant hormones are used extensively in the agricultural sector. Write a short note on the advantages and disadvantages of these (Question Pool 2017)
Answer:
Advantages:

  • Sprouting
  • Prevents premature fall of fruits
  • Medicinal action
  • Increases size of fruits
  • Ripening of fruits
  • Increases production of latex in rubber trees
  • Harvesting fruits at the same time.
  • Prevents early ripening of fruits

Disadvantages:

  • Environmental issues
  • Health issues

Question 22.
Choose the correct statement related to pheromones from those given below. (Question Pool 2017)
a) Pheromones are chemical substances secreted inside the body for communication.
b) This is the message to attract mates, determining the path of travel, etc.
c) Musk in the civet cat is a pheromone.
d) Bombycol is the pheromone secreted by the female silkworm.
Answer:
b, d

Question 23.
Analyze the box given below and complete the table suitably. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 67
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 56

Question 24.
Indicators related to the endocrine glands are given below. Analyze them and answer the questions. (Question Pool 2017)
1. Situated just below the sternum.
2. Active during infancy.
But constricts at puberty.
a) Name this endocrine gland?
b) Which is the hormone synthesized by this gland?
c) Write the function of this hormone.
Answer:
a) Thymus gland
b) Thymosin
c) Controls the activities and maturation of lymphocytes which help to impart immunity.

Question 25.
Given below is the illustration showing the hormones synthesized by the anterior lobe of the pituitary gland. Complete it Suitably. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 57
Answer:
a) Stimulates thyroid gland
b) ACTH
c) Production of milk
d) Enhances growth

Question 26.
Teacher: The TSH hormone synthesized by the pituitary gland acts on the thyroid gland. It is transported to the thyroid gland through blood. All hormones are transported like this through blood. (Question Pool 2017)
Amirtu: Can all the hormones synthesized by the pituitary gland reach the thyroid gland and act there? What is your answer for Ammu’s doubt?
Answer:
Receptors to receive other hormones synthesized by the pituitary gland are absent in the thyroid gland.

Question 27.
Plant hormones and their functions are given in two boxes below. Pair them suitably (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 58
Answer:
(a) – (iv)
(b) – (ii)
(c) – (i)
(d) – (iii)

Question 28.
The problems faced by two farmers are below. Suggest two artificial plant hormones to overcome this. (Question Pool 2017)
Satheesh: Excessive growth of weeds in the agricultural field.
Saneesh: Premature fall of fruit in the mango orchard.
Answer:
Satheesh: 2, 4- D
Saneesh: NAA /IBA

Question 29.
Observe the illustration given below and answer the questions. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 59
a) Write the names of the hormones ‘X’ and Y\
b) Mention two actions that take place in A and B.
c) Name the gland which synthesises X and Y.
Answer:
a) X-Insulin; Y-Glucagon
b) A-Converts glycogen to glucose
B – Converts glucose to glycogen
c) Pancreas

Question 30.
Identify the word pair relationship and fill in the blanks. (Question Pool 2017)
a) Civet cat:………………
Silkworm: Bombycol
b) Breaks up stored food: Gibberellins
helps in fruit ripening: ……………….
c) Vasopressin: Diabetes insipidus
Insulin: ………………
d) Dwarfism: somatotropin
Myxoedema: ………………..
Answer:
a) Civetone
b) Ethylene
c) Diabetes mellitus
d) Thyroxine

Question 31.
Pick the odd one out. Write the common features of the others. (Question Pool 2017)
a) Increases metabolic rate, increases energy production regulates growth in children, promotes production of milk.
b) Goitre, Acromegaly, Hypothyroidism, Hyperthyroidism.
c) Cortisol, Vasopressin, Epinephrine, Norepinephrine.
d) Ethylene, Cytokinin, Auxin, Pheromones.
Answer:
a) Increases the production of milk: all others are the activities of thyroxine.
b) Acromegaly: All others are disorders/diseases, related to thyroid gland
c) Vasopressin: All others are hormones of adrenal gland
d) Pheromones: All others plant hormones

Question 32.
Choose the correct statement. (Question Pool 2017)
a) Synthesis of vasopressin increases if the level of water in the blood increases.
b) Thyroid-stimulating hormone stimulates the activity of the thyroid gland.
c) Synthesis of insulin increases if the blood glucose level rises.
d) Deficiency of thyroxine causes cretinism in adults.
Answer:
b, c

Question 33.
Maintenance of the level of calcium in blood is illustrated below. Analyse-it and answer the questions. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 60
a) Which are the hormones indicated as ‘X’, ‘Y’?
b) Write the actions performed by ‘X’ in the bone and ‘Y’ in the kidney.
c) How does the deficiency of ‘Y’ affect the process of blood clotting?
Answer:
a) X – Calcitonin: Y – Parathormone
b) Action of X : Deposits excess calcium in bones.
Action of Y : Reabsorbs calcium into the blood in the kidney.
c) Deficiency of Y decreases the level of calcium in blood.
As calcium is required for blood clotting, the clotting process becomes slow.

Question 34.
Make suitable word pairs from the words given below. (Orukkam – 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 61
Answer:
Vasopressin- Diabetes insipidus
Dwarfism – Somatotropin
Cretinism – Thyroxin

Question 35.
Complete the illustration using the words given in the box. (Orukkam – 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 62
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 64

Question 36.
Observe the illustration and answer the following questions? (Orukkam – 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis - 65
a) Identify the parts marked as A , B and C ?
b) Name the hormones indicated as 1, 2, 3, 4 and 5?
c) What are the functions of the hormones Oxytocin and Prolactin?
d) What are the abnormalities caused by the difference in the production rate of the hormone marked as 1?
Answer:
a) A-Anterior lobe of pituitary
B – posterior lobe of pituitary
C – Hypothalamus

b) 1 – Somatotropin /growth hormone
2 – Vasopressin /ADH
3 – Tropic hormones
4, 5 – TSH/ACTH

c) Oxytocin facilitates childbirth by the contrac¬tion of smooth muscles in the uterine wall and also facilitates lactation. Vasopressin helps in the reabsorption of water in the kidney to prevent water loss through urine.

d) Dwarfism, Gigantism and Acromegaly

Question 37.
Identify the word pair relationship and complete the following. (Orukkam – 2017)
a) Alpha cells: Glucagon
Beta Cells: …………….
b) Prolactin: Production of milk
……………….: Facilitate lactation
c) Parathyroid: Parathormone
Thyroid
Answer:
a) Insulin
b) Oxytocin
c) Calcitonin

Question 38.
All hormones are being transported through the blood and reach all cells of the body, but all hormones are not functioning in all cells. Why? (Orukkam – 2017)
Answer:
Each hormone act only its target tissue, where spe¬cific receptors present to accept the same hormone.

Question 39.
The increased or decreased level of thyroxin may disrupt the homeostasis of the body. Explain? (Orukkam – 2017)
Answer;
Due to hypothyroidism (eg. cretinism) low metabolic rate, sluggishness, sleeplessness, increase in body weight, hypertension, oedema, etc.
Due to hyperthyroidism (eg. Graves disease) high metabolic rate, increased heartbeat, rise in body tem¬perature, sweating, sleeplessness, loss of weight, emotional imbalance.

Question 40.
Bees and termites are maintaining the colony life by using some chemical substances as chemical messages. (Orukkam – 2017)
a) What are these chemical substances?
b) Write the other uses of these chemical substances?
c) Give other examples for these chemical substances?
Answer:
a) Pheromones
b) To attract mates, to inform about food or dangers, to live in colonies, to follow one afterthe other.
c) Civetone in civet cat, Bombycol in female silkworm moth.

Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance

You can Download Mathematics of Chance Questions and Answers, Activity, Notes, Kerala Syllabus 10th Standard Maths Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance

Mathematics of Chance Text Book Questions and Answers

Textbook Page No. 71

Mathematics Of Chance Questions And Answers Kerala Syllabus 10th Standard Question 1.
A box contains 6 black and 4 white balls. If a ball is taken from it. What is the probability of it being black? And the probability of it being white?
Answer:
Total no. of balls = 10
No. of black balls = 6
No. of white balls = 4
Probability of it to be black = \(\frac { 6 }{ 10 }\) = \(\frac { 3 }{ 5 }\)
Probablitiy of to be white = \(\frac { 4 }{ 10 }\) = \(\frac { 2 }{ 5 }\)

Mathematics Of Chance Class 10 Kerala Syllabus Kerala Syllabus Question 2.
There are 3 red balls and 7 green balls in a bag, 8 red and 7 green balls in another.
i. What is the probability of getting a red ball from the first bag?
ii. From the second bag?
iii. If all the balls are put in a single bag, What is the probability of getting a red ball from it?
Answer:
i. Total no. of balls in the first bag = 10
No. of red balls = 3
P(red ball) = 3/10

ii. Total no. of balls in the second bag= 15 No. of red balls = 8
P(red ball) = 8/15

iii. Total no. of balls in both bags = 25
Total no. of red balls = 11
P(red ball) = 11/25

Mathematics Of Chance Questions And Answers Pdf Kerala Syllabus 10th Standard Question 3.
One is asked to say a two-digit number. What is the probability of it being a perfect square?
Answer:
Total no. of two-digit numbers = 90
Perfect squares from 10 to 99 are 16, 25, 36, 49, 64, 81.
There are 6 favourable numbers Probability = \(\frac { 6 }{ 90 }\) = \(\frac { 1 }{ 15 }\)

Mathematics Of Chance Extra Questions And Answers Kerala Syllabus 10th Standard Question 4.
Numbers from 1 to 50 are written on slips of paper and put in a box. A slip is to be drawn from it; but before doing so, one must make a guess about the number, either prime number or a multiple of five. Which is the better guess? Why?
Answer:
Probablity of getting prime numbers from 1 to 50 = \(\frac { 15 }{ 50 }\) = \(\frac { 3 }{ 10 }\)
Probability of getting a multiple of 5 from 1 to 50 = \(\frac { 10 }{ 50 }\) = \(\frac { 1 }{ 5 }\)
∴ Better guess would be prime numbers as probability of that is more.

Sslc Maths Mathematics Of Chance Kerala Syllabus 10th Standard Question 5.
A bag contains 3 red beads and 7 green beads. Another contains one red and one green more. The probability of getting a red from which bag is more?
Answer:
Total no. of beads in first bag = 10
No. of red beads in first bag = 3
P(red ball) in first bag = \(\frac { 3 }{ 10 }\)
Total no. of beads in second bag = 12
No. of red beads in second bag = 4
P(red ball) in second bag = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)
The probability of getting a red bead is more from the second bag since, \(\frac{1}{3}>\frac{3}{10}\left(\frac{10}{30}>\frac{9}{30}\right)\)

Textbook Page No. 72

In each picture below, the explanation of the green part is given. Calculate in each, the probability of a dot put without looking to be within the green part.

Mathematics Of Chance Pdf Kerala Syllabus 10th Standard Question 1.
A square got by joining the mid points of a bigger square.
Mathematics Of Chance Questions And Answers Kerala Syllabus 10th Standard
Answer:
∴ Probability = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)
Mathematics Of Chance Class 10 Kerala Syllabus Kerala Syllabus

Class 10 Maths Chapter 3 Mathematics Of Chance Kerala Syllabus Question 2.
A square with all vertices on a circle
Mathematics Of Chance Questions And Answers Pdf Kerala Syllabus 10th Standard
Answer:
Side of square = a = 2cm Radius
Mathematics Of Chance Extra Questions And Answers Kerala Syllabus 10th Standard
ie., The area of the square is 2/π the area of the circle.
Sslc Maths Mathematics Of Chance Kerala Syllabus 10th Standard
Probability = \(\frac{a^{2}}{\frac{\pi}{2} a^{2}}=\frac{4}{2 \pi}=\frac{2}{\pi}\)
Probablity ofthe dot falling on the square is \(\frac { 2 }{ π }\) = 0.64

Sslc Maths Chapter 3 Questions And Answers Kerala Syllabus Question 3.
Circle exactly fitting inside a square.
Mathematics Of Chance Pdf Kerala Syllabus 10th Standard
Answer:
If radius of circle is r
Side of the square = 2r
Area of the square = 4r2
Area of the circle= πr2
Class 10 Maths Chapter 3 Mathematics Of Chance Kerala Syllabus

Mathematics Of Chance Extra Questions Kerala Syllabus 10th Standard Question 4.
A triangle got by joining alternate vertices of a regular hexagon.
Sslc Maths Chapter 3 Questions And Answers Kerala Syllabus
Answer:
The area of regular hexagon having side a = \(\frac{3 \sqrt{3}}{2} a^{2}\)
Side of the triangle got by joining alternate vertices = √3a
Area of triangle = \(\frac{\sqrt{3}}{4}(\sqrt{3} a)^{2}=\frac{3 \sqrt{3}}{4} a^{2}\)
∴ Probability of the dot falling on the tri-angle = \(\frac{3 \sqrt{3}}{4} a^{2} \times \frac{2}{3 \sqrt{3} a^{2}}=\frac{2}{4}=\frac{1}{2}\)

Sslc Maths Chapter 3 Solutions Kerala Syllabus 10th Standard Question 5.
A regular hexagon formed by two overlapping equilateral triangles.
Mathematics Of Chance Extra Questions Kerala Syllabus 10th Standard
Answer:
Area of two equilateral triangles
Sslc Maths Chapter 3 Solutions Kerala Syllabus 10th Standard
Area of regular hexagon = \(\frac{3 \sqrt{3}}{2} a^{2}\)
Area of regular hexagon is half of the area of equilateral triangle.
∴ Probability of dot falling on the hexagon
Class 10 Maths Chapter 3 Kerala Syllabus

Textbook Page No. 75

Class 10 Maths Chapter 3 Kerala Syllabus Question 1.
Raj ani has three necklaces and free pairs of earrings, of green, blue and red stones. In what all different ways can she wear them? What is the probability of her wearing the necklace and earrings of the same color? Of different colors?
Answer:
Possible ways of wearing:

Necklace Earring
Green Green
Green Blue
Green Red
Blue Green
Blue Blue
Blue Red
Red Green
Red Blue
Red Red

Raj ani can wear the ornaments in 9 different ways.
Probability of wearing necklace and earring of same colour = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)
Probability of wearing necklace and earring of different colour = \(\frac { 6 }{ 9 }\) = \(\frac { 2 }{ 3 }\)

Sslc Maths Chapter 3 Notes Question 2.
A box contains four slips numbered 1, 2, 3, 4 and another box contains two slips numbered 1,2. If one slip is taken from each, what is the probability of the sum of numbers being odd? What is the probability of the sum being even?
Answer:

First box Second box Sum
1 1 2
1 2 3
2 1 3
2 2 4
3 1 4
3 2 5
4 1 5
4 2 6

Probability of sum being odd = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)
Probability of sum being even = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)

Kerala Syllabus 10th Standard Maths Chapter 3 Question 3.
A box contains four slips numbers 1, 2, 3, 4 and another contains three slips numbered 1, 2, 3. If one slip is taken from each, what is the probability of the product being odd? The probability of the product being even?
Answer:

First box Second box Product
1 1 1
1 2 2
1 3 3
2 1 2
2 2 4
2 3 6
3 1 3
3      . 2 6
3 3 9
4 1 4
4 2 8
4 3 12

Probability of sum being odd = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)
Probability of sum being even = \(\frac { 8 }{ 12 }\) = \(\frac { 2 }{ 3 }\)

10 Th Maths Text Book Questions And Answers Question 4.
From all two-digit numbers with either digit 1, 2 or 3 one number is chosen.
i. What is the probability of both digits being the same?
ii. What is the probability of the sum of the digits being 4?
Answer:
Two digit numbers
11, 12, 13, 21, 22, 23, 31, 32, 33.
i. P (both digits being same) = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)
ii. P(sum of digits being 4) = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)

10th Class Maths Textbook Answers Question 5.
A game for two players. First, each has to decide whether he wants odd number or even number. Then both raise some fingers of one hand. If the sum is odd, the one who chose odd at the beginning wins; if it is even, the one who chose even wins. In this game, which is the better choice at the beginning, odd or even?
Answer:
The results in the order when the first player raises one finger and second player raises one finger and so on.
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 12

Textbook Page No. 78

Maths Chapter 3 Class 10 Kerala Syllabus 10th Standard Question 1.
In class 10A, there are 30 boys and 20 girls. In 10B, there are 15 boys and 25 girls. One student is to be selected from each class.
i. What is the probability of both being girls?
ii. What is the probability of both being boys?
iii. What is the probability of one boy and one girl?
iv. What is the probability of at least one boy?
Answer:
i. Total no. of possible pairs = 50 × 40 = 2000
No. of pairs in which both are girls = 20 × 25 = 500
Probability of both being girls = \(\frac { 500 }{ 2000 }\) = \(\frac { 1 }{ 4 }\)

ii No. of pairs in which both are boys = 30 × 15 = 450
Probability of both being boys = \(\frac { 450 }{ 2000 }\) = \(\frac { 9 }{ 40 }\)

iii. No. of pairs in which one is boy and one is girl = 2000 – (500 + 450) = 2000 – 950 = 1050
Probability of one being boy and one girl = \(\frac { 1050 }{ 2000 }\) = \(\frac { 21 }{ 40 }\)

iv. No. of pairs in which atleast one is boy = 1050 + 450 = 1500
Probability in which atleast one is boy = \(\frac { 1050 }{ 2000 }\) = \(\frac { 3 }{ 4 }\)

Question 2.
One is asked to say a two-digit number.
i. What is the probability of both digits being the same?
ii. What is the probability of the first digit being larger?
iii. What is the probability of the first digit being smaller?
Answer:
i. Probability of two digits being same = \(\frac { 9 }{ 90 }\) = \(\frac { 1 }{ 10 }\)
(11, 22, 33, 44, 55, 66, 77, 88, 99)

ii. Numbers in which first digit is greater than second digit are 10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65, 70, 71, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 96, 97, 98,
45 outcomes, Required Probability = \(\frac { 45 }{ 90 }\) = \(\frac { 1 }{ 2 }\)

iii. Numbers in which first digit is smaller than second digit are
12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89,
36 total no.of favourable outcomes = 36
Probability =
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 13

Question 3.
Each two-digit number is written on a paper slip and these are all put in a box. What is the probability that the product of the digits of a number drawn is a prime number? What if three-digit numbers are used instead?
Answer:
Total two-digit numbers = 90
Product of the digits of a number drawn is a prime number 12, 13, 15, 17, 21, 31, 51, 71.
(1 is not a prime number)
Total number whose product of the digits drawn is a prime number = 8
Probability of product of the digits drawn is a prime number = \(\frac { 8 }{ 90 }\) = \(\frac { 4 }{ 45 }\)
Total three-digit numbers = 900
Product of the digits of a number drawn is a prime number 112, 113, 115, 117, 121, 131, 151, 171.
Total number whose product of the digits drawn is a prime number = 8
Probability of product of the digits drawn is a prime number = \(\frac { 8 }{ 900 }\) = \(\frac { 2 }{ 225 }\)

Question 4.
Two dice with faces numbered from 1 to 6 are rolled together. What are the possible sums? Which of these sums has the maximum probability?
Answer:
Each dice has the following numbers:
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 14

Mathematics of Chance Orukkam Questions & Answers

Worksheet 1

Question 1.
How many odd numbers are there below 25
Answer:
12 odd numbers

Question 2.
How many prime numbers are there below 30?
Answer:
10 prime numbers

Question 3.
Find the number of two-digit even numbers?
Answer:
90 two-digit even numbers

Question 4.
How many two digits perfect squares are there?
Answer:
6

Question 5.
Write all three-digit numbers that can be written using the digits 3, 6, 8 without repeating the digits.
Answer:
368, 386, 683, 638, 836, 863

Question 6.
How many multiples of 7 are there in between 100 and 300?
Answer:
First multiple of 7 = 105
Last multiple of 7 = 294
Number of multiples
= Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 15

Question 7.
There are 50 children in a class. Thirty of them are girls. There are 40 children in another class. 25 of them are boys. One student is taken from each class at random. What is the number of outcomes? How many outcomes contain both boys. How many outcomes contain both girls. How many outcomes have one boy and one girl?
Answer:
Total pairs = 50 × 40 = 2000
No. of pairs in which both are boys = 20 × 25 = 500
No. of pairs in which both are girls = 30 × 15 = 450
No. of pairs in which one is a boy and the other a girl = 30 × 25 + 20 × 15 = 750 + 300 = 1050

Worksheet 2

Question 8.
A fine dot is placed into the picture with-out looking into it. What is the probability of falling the dot in the small semicircle? What is the probability of falling the dot outside the small semicircle but inside the big semicircle?
Answer:
Let the radius of circle be r, then the radius of
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 16

Question 9.
P, Q, R are the midpoints of the sides of triangle ABC. Another triangle is drawn by joining these points. A fine dot is placed into the figure without looking into the picture. What is the probability of falling the dot in triangle PQR?
What is the probability of falling the dot outside the triangle?
Answer:
Area of each small triangle is 1/4th of area of large triangle ABC.
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 17
Probability of the dot falling on triangle PQR is = 1/4
Probability of the dot falling on small triangle is = 1/4
Probability of the dot falling inside the triangle PQR is \(\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\) less than that of outside the triangle.

Question 10.
What is the probability of occurring 53 Sundays in a leap year
Answer:
Leap year have 366 days.
That is 52 weeks and 2 days.
There two days are
Sunday – Monday, Monday – Tuesday, Tuesday – Wednesday, Wednesday – Thursday, Thursday – Friday, Friday – Saturday, Saturday – Sunday.
∴ Probability for to occur 53 Sunday is 2/7.

Question 11.
You can see a triangle inside a square. ABCD is a square. P, Q are the midpoints of C D and C B. A fine dot is placed into the figure without looking into the figure. What is the probability of falling the dot in triangle APQ?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 18

Question 12.
The value of 21, 22, 23… 250 are written in small papers and put it in the box. A paper is taken at random. What is the probability of getting a number having 4 in ones place? What is the probability of falling 8 in the one’s place?
Answer:
The 13 numbers having 4 in one’s place are 22, 26, 210 …, 250.
∴ Probability = \(\frac { 13 }{ 50 }\)
The 12 numbers having 8 in one’s place are 23, 27, 211 …, 247.
∴ Probability = \(\frac { 12 }{ 50 }\)

Worksheet 3

Question 13.
Numbers from 1 to 10 are written in small papers and placed in a box. One number is taken from the box at random. What is the probability of getting a prime number?
Answer:
Probability of getting a prime number = \(\frac { 4 }{ 10 }\) = \(\frac { 2 }{ 5 }\)

Question 14.
Two boxes contain tokens on which numbers 1, 2, 3, 4 are written One token is taken from each box. What is the probability of getting sum of the face numbers a prime number
Answer:
Pairs of numbers in tokens are
(1.1) , (1,2), (1,3), (1,4)
(2.1), (2, 2), (2, 3), (2, 4)
(3, 1), (3,2), (3; 3), (3,4)
(4, 1), (4, 2), (4, 3), (4,4).
Pairs getting sum as prime numbers
(1.1) , (1,2), (1,4), (2, 1), (2, 3), (3, 2), (3, 4), (4, 1), (4, 3)
Probability of getting sum of the face numbers a prime number = \(\frac { 9 }{ 16 }\)

Question 15.
One box contains 8 black balls and 12 white balls. Another box contains 9 black and 6 white balls. One ball is taken from each box at random. What is probability of getting both black? What is the probability of getting both white? What is the probability of getting one black and one white?
Answer:
Total pairs = 20 x 15 = 300
Number of pairs getting both black= 8 x 9 = 72
Probability of getting both black = \(\frac { 72 }{ 300 }\) = \(\frac { 6 }{ 25 }\)
Probability of getting one black and one white
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 19

Question 16.
In the figure, a triangle is drawn by joining the alternate vertices of a regular hexagon. A fine dot is placed into the figure at random. What is the probability of falling the dot in the triangle?
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 20
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 21
Probability of the dot falling on triangle
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 22

Question 17.
What is the probability of occurring four Wednesdays in 23 consecutive days in a month?
Answer:
23 days = 3 weeks + 2 days
Wednesday comes on Tuesday + Wednesday, Wednesday + Thursday when two days are taken.
Total probabilities (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
∴ Total probabilities = 7
∴ Probabilities = 2/7

Mathematics of Chance SCERT Question Pool Questions & Answers

Question 18.
One is asked to say a two digit number. What is the probability of being the number not a perfect square? [Score : 3, Time : 3 Minutes]
Answer:
Total number of two digit number : 90 (1)
Total number of two digit perfect squares: 6 (1)
Number of two digit numbers which are not perfeet squares : 90 – 6 = 84,
Probability = \(\frac { 84 }{ 90 }\) = \(\frac { 42 }{ 45 }\) (1)

Question 19.
A bag contains 10 blue balls and 12 yellow balls. Another contains 15 blue balls and 7 yellow balls.
a. What is the probability of getting a yellow ball from the first bag?
b. What is the probability of getting a yellow ball from the second bag?
C. If all the balls are put in a single bag, what is the probability of getting a yellow ball from it? [Score : 4, Time : 4 Minutes]
Answer:
a. Total number of balls in the first bag = 10 + 12 =22, Number of yellow balls = 12
Probability of getting a yellow ball = \(\frac { 12 }{ 22 }\) = \(\frac { 6 }{ 11 }\) (1)

b. Total number of balls in the second bag = 15 + 7 = 22, Number of a yellow ball = 7
Probability of getting a yellow ball = \(\frac { 7 }{ 22 }\) (1)

c. Total number of balls = 22 + 22 = 44
Number of yellow balls = 12 + 7 = 19 (1)
Probability of getting a yellow ball = \(\frac { 19 }{ 44 }\) (1)

Question 20.
A regular hexagon is drawn with its vertices on a circle. Without looking into the I picture, if one put dot in that picture, what is the probability of being the dot not in the regular hexagon? [Score: 4, Time: 3 Minutes] .
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 23
Answer:
Area of circle = πr2
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 24

Question 21.
In the figure, all the four shaded semicircles have same area. If we put a dot in the figure without looking into it, what is the probability of being the dot in the shaded semicircles? [Score : 3, Time: 5 Minutes]
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 25
Answer:
If the radius of the shaded semicircle is r,
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 26

Question 22.
What is the probability of getting 5 Sundays in December in a calendar year? [Score : 3, Time : 5 Minutes]
Answer:
There are 31 days in December. That means 4 full weeks and 3 days.
The probable three days are as shown below.
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 27
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 28

Question 23.
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 29
Two semicircles are drawn in a square as shown. If we put a dot in the figure, without looking into it, what is the probability of being the dot in the shaded region? [Score: 3, Time: 5 Minutes]
Answer:
If a is the side of a square,
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 30

Mathematics of Chance Exam Oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 24.
A bag contains 6 red balls, 8 green balls, and 8 white balls. One ball is drawn at random from the bag, find the probability of getting
i. A white or green ball
ii. Neither green ball nor a red ball.
Answer:
Red balls = 6
Green balls = 8
White balls = 8
Total number of balls = 6 + 8 + 8 = 22
a. The probability of getting a white or green ball = 16/20
b. The probability of getting neither green balls nor a red ball = 8/20

Question 25.
20 cards numbered 1, 2, 3, 4, ….19, 20 are put in a box. One boy draws a card from the box. Find the probability that the number on the card is:
i. Prime
ii. Divisible by 3
Answer:
Total number of outcomes = 20
a. Prime numbers from 1 to 17 are 2, 3, 5, 7, 11,13, 17, 19.
Number of outcomes = 8
The probability that the card drawn is prime number = 8/20

b. Numbers are divisible by 3 are 3, 6, 9, 12, 15, 18.
Number of outcomes = 6
The probability that the card drawn is divisible by 3 = 6/20

Question 26.
In a bag, there were 3 white balls and 5 black balls. From this one ball is taken, then
a. What is the probability of being blackball?
b. What is the probability of being white ball?
Answer:
Total no. of balls is 8 and in that 5 of then are black balls.
a. Probability of getting black balls = 5/8
b. 3 of the balls were white, so the probability of getting white balls = 3/8

Question 27.
a. How many two-digit natural numbers are there in all?
b. If we choose one number from the two-digit numbers, what is the probability that the sum of digits of that number will be 10?
Answer:
a. Number of two-digit natural numbers = 90

b. Numbers whose sum of digits will be 10 is 19, 28, 37, 46, 55, 64, 73, 82, 91
Probability that the sum of digits of that number will be 10 = \(\frac { 9 }{ 90 }\) = \(\frac { 1 }{ 10 }\)

Short Answer Type Questions (Score 3)

Question 28.
In selecting a two-digit number up to 50.
a. What is the probability of the digit in the ten’s place to be larger than the digit in the one’s place?
b. What is the probability of the digit in the tens place to be smaller than the digit in the one’s place?
Answer:
Total numbers of two-digit numbers up to 50 = 41
a. Number of numbers with the digit in the tens place to be larger than the digit in the units place = 11
Probability that the number with tens place digit is larger than the digit in the unit place = 11/41

b. Probability that the number with the tenth place digit is smaller than the digit in the unit place = 26/41

Question 29.
A black dice and a white dice are thrown at the same time,
a. Write all the possible outcomes.
b. What is the probability that the sum of the two numbers is to be 8.
c. What is the probability that being the same number to be on both dice?
Answer:
a. Total outcomes
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6) ‘
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6) .
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4.1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5.1) , (5,2), (5,3), (5,4), (5,5), (5,6)
(6.1), (6,2), (6,3), (6,4), (6,5), (6,6)
Total = 36

b. Sum of two numbers is to be 8 = 5
(2,6), (3,5), (4,4), (5,3), (6,2),
Probability = 5/36

c. Probability of being same number =(1,1),(2,2), (3,3), (4,4), (5,5), (6,6),
Total = 6
Probability = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

Question 30.
A box contains 400 electronic toy cars. Among them, 12 are defective. One toy is taken out at random. What is the probability that
a. It is a defective toy.
b. It is a non-defective toy.
Answer:
Total number of cars = 400
Number of defective toys = 12
Number of non-defective toys = 400 – 12 = 388
a. Probability of getting a defective toy = \(\frac { 12 }{ 400 }\) = \(\frac { 3 }{ 100 }\)

b. Probability of getting a non-defective toy = \(\frac { 388 }{ 400 }\) = \(\frac { 97 }{ 100 }\)

Long Answer Type Questions (Score 4)

Question 31.
Natural numbers from 1 to 30 are written on paper slips and kept in a box. If one slip is taken from the box,
a. What is the probability of this number to be even?
b What is the probability of this number to be a multiple of 3?
c. What is the probability of this number to be a multiple of 3 and 5?
d. What is the probability that this number be a natural number?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 31

Question 32.
There is one spot at one side of the cube, two on another side, three on the third side and so on. There are spots on all the six faces; in this order. Another cube which is marked in the same way is taken.
a. If both the cubes are thrown, what is the probability that the total number of spots on the upper faces is 6?
b. What is the probability that the sum of the spots on the upper faces is 9?
c. What is the probability that the sum of the spots be one?
d. What is the probability that the sum of the spots be a prime number?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 32

Question 33.
In a pack of 52 cards, half are red and the rest are black. There are 4 Suits of 13 cards each and having the signs ‘hearts’, ‘spade’, ‘clubs’ and ‘diamond’. If a card is picked from this pack.
a. What is the probability of it being black?
b. Whatistheprobabilityofitbeingaspade?
c. What is the probability of it being a spade or a diamond?
Answer:
a. Total number of cards = 52
Number of black cards = 26
Probability of the picked card being black = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

b. No. of spade cards = 13
Probability of a picked card being spade = \(\frac { 12 }{ 52 }\) = \(\frac { 1 }{ 4 }\)

c. No. of cards spade or diamond =13 + 3 = 26
Probability of a picked card being spade or diamond = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

Long Answer Type Questions (Score 5)

Question 34.
A man is asked to say a 3 digit number,
a. What is the probability that the first and last digits be equal?
b. What is the probability that the last two digits be ‘O’?
c. What is the probability that the last digits being greater than the first?
Answer:
Total 3 digit numbers = 900
a. Numbers with first and last digits are equal
101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393
There are 90 such numbers.
∴ Probability = \(\frac { 90 }{ 900 }\) = \(\frac { 1 }{ 10 }\)

b. The numbers with last two digits zero are 100, 200, 300,……… 900 total numbers 9.
∴ Probability = \(\frac { 9 }{ 900 }\) = \(\frac { 1 }{ 100 }\)

c. The numbers with the last digit greater than the first digit is 36
∴ probability = \(\frac { 36 }{ 90 }\) = \(\frac { 2 }{ 5 }\)

Question 35.
There are 10 black pearls and 5 white pearls in the box A. There are 8 black pearls and 7 white pearls in box B.
a. Which box has more probability of be ing the pearls black, when a pearl from each of the boxes is taken?
b. What is the probability to get a white pearl from the box A?
c. What is the probability to get a black pearl from box B?
d. If all the pearls in the box B is dropped in the box A, then what is the probability to get a black pearl from it?
Answer:
a. Box A because it has more black pearl
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 33

Question 36.
Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number of the card is
a. an even number
b. a number less than 16
c. a number which is a perfect square
d. a prime number less than 25
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 34

Mathematics of Chance Memory Map

When probabilities are explained in terms of numbers it is the ratio of number of favorable outcomes to the total number of outcomes.
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 35
The least probability will be 0 and the highest will be 1. The probability will be a number between 0 and 1.
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 36
If an event can be completed in ‘m’ ways and another event can be completed in ‘n’ ways then both the events can be completed one after the other in m x n ways.
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 37
Kerala Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance - 38

Kerala Syllabus 10th Standard Social Science Notes Chapter 2 World in the Twentieth Century

Kerala State Syllabus 10th Standard Social Science Notes Chapter 2 World in the Twentieth Century

Twentieth century is the period that influenced world history greatly. Imperialism was the developed form of capitalism which emerged in Europe after the Industrial Revolution. When capitalism developed into imperialism, it faced many crises. When the imperialist powers entered into mutual competitions in order to conquer the world, conflict became widespread. The international problems which surfaced during this period caused mutual mistrust and enmity. The conflicts among the imperialist powers ultimately led the entire world to a war. The growth of Fascism and Nazism, Second World War, international efforts for peace, cold war, non-alignment and globalisation are the other topics discussed in this unit.

→  Industrial Revolution: The changes which took place in England from the 18th century onwards came to be known as the Industrial Revolution. The basic feature of this was that human labour was substituted by machines.

→  Capitalism : The economic system in which production and distribution are controlled by capitalists with the aim to increase profit.

→  Imperialism : The practice of extending a nation’s political, economic and cultural dominance on another nation is imperialism.

→  Triple Alliance and Triple Entente : Military alliances that fought in the First World War.

→  Nationalism : A nation is defined as a people settling in a definite territory, speaking a common language, having a common culture and historical tradition. The ideology and programme of action based on this concept is called nationalism.

→  Aggressive nationalism : The policy of invading neighbouring countries, considering one’s nation as supreme and justifying whatever be the actions of the nation.

→  Pan-Slav Movement : The movement started under the leadership of Russia to unite the Slavic People of Serbia, Bulgaria, Greece, etc. in Eastern Europe.

→  Pan-German Movement : The movement started under the leadership of Germany to establish her dominance in Central Europe and Balkan provinces and to unite the Teutonic people.

→  Revenge Movement : The movement started under the leadership of France to regain her territories of Alsace-Lorraine which were captured by Germany in the Franco-Prussian war of 1871.

→  Treaty of Versailles : The treaty signed with Germany by the victorious powers after the First World War.

→  World Economic Depression : The economic crisis that started in 1929 and affected the whole world.

→  League of Nations : The international organisation formed after the first world war to maintain peace in the world.

→  Fascism, Nazism : The political ideology that supported dictatorship, racial superiority, aggressive nationalism and single party rule.

→  Munich Pact : The agreement that approved the claim of Germany over Sudetanland, a part of Czechoslovakia.

→  Policy of appeasement: Capitalist countries like Britain and France considered Soviet Union, being a socialist country, as their chief enemy and did not prevent fascist attacks. This policy which encouraged fascist attacks is known as policy of appeasement.

→  Non-Aggression Pact : The agreement signed between Germany and Soviet Union in 1939, by which they agreed not to attack each other and to partition Poland.

→  Teutonic People : The Germanic people are also called Teutonic peoples. Originally they belonged to Northern Europe. They spoke languages of the Germanic branch of the lndo-European language family.

→  Pearl Harbour Attack : The attack of Japan in 1941 on Pearl Harbour, the American naval base in the islands of Hawaii.

→  Hiroshima, Nagasaki : Japanese cities where atom bombs were dropped in 1945 by USA.

→  Hibakusha: The surviving victims of the atomic bombings of Hiroshima and Nagasaki.

→  United Nations Organisation : The international organisation formed after the Second World War to prevent war and maintain peace in the world.

→  Decolonization : The process of the colonies of Asia and Africa securing freedom from imperialist control.

→  Cold War : The enmity based on ideological conflict and diplomatic confrontations between US bloc and Soviet bloc.

→  Bipolar politics : USA led the Capitalist bloc and Soviet Union led the Socialist bloc after the Second World War. This ideological division between the power blocs is called bipolar politics.

→  Military Pacts : Military agreements formed among capitalist bloc and socialist bloc after the Second World War.

→  Non-alignment : The policy adopted by the newly independent countries of Asia and Africa not to join the power blocs and to follow an independent foreign policy.

→  Zionism: The movement that started to establ ish a homeland for the Jews.

→  PLO : Palestinian Liberation Organisation was a movement with the objective of establishing a nation for the Palestinians.

→  Oslo Pact: The pact signed between Israel and Palestine under the mediation of USA. By this, Israel agreed to recognise Palestine as a free nation.

→  Glasnost : The administrative reform started under Mikhail Gorbachev in Soviet Union to implement openness in political processes.

World In The Twentieth Century Notes

→  Perestroika: The administrative reform started under Mikhail Gorbachev to restructure the economic system of Soviet Union.

→  Unipolar world : USA emerged as a global power and centre of world politics following the disintegration of Soviet Union. The world order dominated by the USA is called unipolar world.

→  Neo imperialism: The multinational companies began to interfere in the economic, social and cultural sectors of the newly independent countries of Asia and Africa and Latin America for serving the interests of capitalist countries. This is known as neo imperialism.

→  Globalisation: The policy of transfer of products, ( services, raw materials, capital, latest technology and human resources across the borders of countries without any restriction.

→  Liberalisation: The policy of adoption of liberal regulations and taxation systems to facilitate the import of multinational products to domestic markets.

→  Privatisation : The policy of privatisation of public sector undertakings to promote private sector. (The process of reducing the role of public sector in the economy and increasing the role of private sector is known as privatisation).

World in the Twentieth Century – Famous Persons

→  Francis Ferdinand : The heir to the throne of Austria who was assassinated in June 1914 at Sarajevo. This was the immediate cause for the First World War.
→  Benito Mussolini : Leader of fascist reign in Italy.
→  Adolf Hitler: Leader ofNazi reign in Germany.
→  Matteotti: Eminent socialist thinker of Italy who opposed fascism. ‘
→  Woodrow Wilson : The US President who gave leadership to the formation of League of Nations.
→  Nelson Mandela ; Leader of anti – imperialist struggle in South Africa.
→  Quami Nkrumah : Leader of anti – imperialist struggle in Ghana.
→  Jomo Kenyatta : Leader of the freedom movement in Kenya.
→  Bernard Baruch : The American economist who first used the word ‘cold war’.
→  Architects of Non : Aligned movement:

  • Jawaharlal Nehru – India
  • Gamal Abdul Nasser – Egypt
  • Marshal Tito-Yugoslavia .
  • Ahmed Sukarno – Indonesia

→ Yasser Arafat: Founder President of Palestinian Liberation Organisation.

→  Mikhail Gorbachev : The last President of Soviet Union.

World in the Twentieth Century – Important Years and Events

  • 1871 – Franco Prussian War
  • 1904 – Moroccan Crisis
  • 1912 – Balkan Crisis
  • 1914 – Assassination of Francis Ferdinand
  • 1914-18 – First World War
  • 1919 – Paris Peace Treaty: Treaty of Versailles
  • 1924 – Mussolini in power
  • 1929 – World Economic Depression
  • 1933 – Hitlar as Chancellor of Germany
  • 1938 – Munich Pact
  • 1939-45 – Second World War
  • 1941 – Pearl Harbour Attack
  • 1945 – Atom bombs dropped at Hiroshima and Nagasaki in Japan by USA
  • 1945 – Formation ofUNO (October24,1945)
  • 1948 – Formation of Israel
  • 1955 – Emergence of Non-Aligned Movement
  • 1991 – Disintegration of Soviet Union
  • 1993 – Oslo Pact

Kerala Syllabus 10th Standard Social Science Notes

Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense

You can Download Soldiers of Defense Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Biology Solutions Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard BiologySolutions Chapter 5 Soldiers of Defense

Soldiers of Defense Text Book Questions and Answers

Sslc Biology Chapter 5 Kerala Syllabus Question 1.
Our surrounding are full of microorganisms. Most of them are pathogens too. Though we live in the midst of germs are we susceptible to diseases? What may be the reason?
Answer:
Numerous germs are present in our surroundings, that have the capacity to cause diseases. We are often in contact with them. There are several mechanisms in the human body which prevent the entry of germs. So we don’t get infected always.

Biology Chapter 5 Class 10 Kerala Syllabus Question 2.
What are the mechanisms in the body which prevent the entry of pathogens?
Answer:

  • A protein called keratin in skin, sebum, and acids.
  • Mucus in the trachea
  • Cilia in the bronchus
  • Hydrochloric acid in the stomach.
  • Cough and sneezing.
  • The wax in the ear.
  • The enzyme lysozyme in tears and saliva.
  • Blood, Lymph.

Sslc Biology Chapter 5 Notes Kerala Syllabus Body Coverings And Secretions

Sslc Biology Chapter 5 Kerala Syllabus

Biology Class 10 Chapter 5 Kerala Syllabus Question 3.
Skin is referred to as a “fort of resistance”, why?
Answer:
Keratin makes the skin a thick fort which prevents germs from entering it. So the skin is referred to as a fort of resistance.

10th Class Biology 5th Chapter Kerala Syllabus Question 4.
What is the function of cilia and mucus in the respiratory tract?
Answer:
Mucus in the trachea prevents the entry of germs into the lungs. The cilia in the bronchus wipe out dust that enters it.

HSSLive.Guru

Kerala Syllabus 10th Standard Biology Chapter 5 Question 5.
What are the methods in ears, eyes, and saliva to prevent germs?
Answer:
The enzyme lysozyme present in the tears and saliva are fight against germs. The wax in the ear prevents pathogens.

Class 10th Biology Chapter 5 Notes Kerala Syllabus Question 6.
What is the role of hydrochloric acid in the stomach to prevent germs that enter the body through food?
Answer:
Since hydrochloric acid is present stomach, the germs that enter through food and water are destroyed.

Class 10 Biology Chapter 5 Notes Kerala Syllabus Question 7.
Which are the secretions that help to defend pathogens? Analyze illustration and complete the table.
Biology Chapter 5 Class 10 Kerala Syllabus
Sslc Biology Chapter 5 Notes Kerala Syllabus
Answer:

Part of the body Secretion
Ear Ear wax
Mouth Lysozyme in saliva Lysozyme
Eye Lysozyme  in tears
Stomach HCI

Body Fluids And Defense

  • Body fluids like blood and lymph play an important role in defense mechanisms.
  • Controlling the entry of germs into the body.
  • Neutralizing germs and the toxic substances they produce, preventing their multiplication.

White blood cells and Defense actions

Biology Class 10 Chapter 5 Kerala Syllabus

Inflammatory Response

Lysozyme Antibodies Question 8.
Based on the indicators, analyze the following illustration. Write your inference in the science diary.
10th Class Biology 5th Chapter Kerala Syllabus
Answer:
The cells that get damaged by a wound or an infection produce certain chemical substances. These substances dilate the blood vessel thereby increasing the blood flow. Blood plasma and white blood cells reach the wound site. This is the reason for the swelling of the wound site. This defense mechanism is known as inflammatory response.

Question 9.
What is the advantage of the dilation of blood vessels at the wound site?
Answer:
The cells that get damaged by a wound or an infection produce certain chemical substances. These substances dilate the blood vessels thereby increasing the blood flow.

Label the Specializations of the Plasma Membrane Question 10.
Is inflammatory response a defense activity? Why?
Answer:
Inflammatory response is a defense activity. Inflammation formed in the body due to the changes in the wall of blood capillaries in a part of the body that affected a wound. When germs enter through – the wound, changes occur in the capillary wall of that part. It leads to inflammation. Flow of blood through these capillaries increases and as a result more leucocytes come out from the capillaries and destroy the germs by engulfing them. The affected parts swell and become red-colored due to the arrival of more blood at the affected part of the capillaries.

HSSLive.Guru

Question 11.
Prepare the flowchart which showing the stages of inflammatory response.
Answer:
Germs enter through wound → Produces chemical messages → Blood vessels dilate → White blood cells from the blood vessel reach the wound site → White blood cells destroy the germs.

Phagocytosis

Phagocytosis is the process of engulfing and destroying germs. The cells engaged in this process are called phagocytes. (Phago – to engulf, cyte – cell) Monocytes and neutrophils are phagocytes.

Question 12.
Stages of phagocytosis
Answer:
Kerala Syllabus 10th Standard Biology Chapter 5

Question 13.
Complete the flow chart by analyzing illustration showing the stages of phagocytosis.
Class 10th Biology Chapter 5 Notes Kerala Syllabus
Answer:
Class 10 Biology Chapter 5 Notes Kerala Syllabus

Blood Clotting

Blood clotting is a defense mechanism to prevent the loss of blood through wounds. In this process fibrin, the plasma protein forms a fibrous network. Blood cells get entangled in the network to form a blood clot.

Question 14.
Analyze the following illustration that details the stages of blood clotting
Biology Class 10 Chapter 5 Kerala Syllabus
Answer:
When a person gets a cut or wound the blood that flows out from the wound changes from a liquid to a gel, forming a clot which plays the wound and prevents further bleeding.

When the platelets come into contact with the atmospheric air, they burst and liberate thromboplastin. It converts prothrombin in the plasma to thrombin. This thrombin converts soluble fibrinogen molecules into insoluble fibrin. This fibrin filament form-fine network over the wound and trap blood corpuscles and platelets to form a clot. The clot seals the wounds and stops bleeding.

Healing Of Wounds

Healing of the wound is a stage after inflammatory response and blood clotting. When wound occurs new tissues are formed in place of the tissues damaged by the wound. In such situations, the wound scar does not remain. In cases when new tissues cannot be formed, the connective tissue heals the wound. In such situations, the wound scar remains.

Question 15.
In some situations, the wound scar remains. Why?
Answer:
When wound occurs, new tissues cannot be formed, then connective tissues heal the wound. In such situations, the wound scar remains.

Fever, A Defense Mechanism

Question 16.
The normal body temperature is 37°c (98.6°F). Body temperature rises during fever. Is it a disease or a symptom. Analyze the flowchart given and write your inferences in the science diary.
10th Class Biology Chapter 5 Kerala Syllabus
Answer:
Fever is not a disease. But it is a type of resistance activity. Though the body can control the multiplication of germs through mechanism like raising body temperature. The chemical substances produced by the white blood cells raises the body temperature. If the rise in body temperature persists for a long time, it may badly affect the internal organs including the brain. Hence it is necessary to seek medical assistance immediately.

Question 17.
Fever is the rise in the body temperature. Is it beneficial to the body?
Answer:
Our normal body temperature is 36.9°c. This temperature is suitable for the multiplication of germs. When infection occurs body rises the temperature through fever to reduce the capacity of multiplication of them.

Lymphocytes – The Warrior

Specific defense is the system which identifies and destroys pathogens. White blood cells known as lymphocytes are capable of destroying the pathogens in this way. Lymphocytes are of two types namely B lymphocytes and T lymphocytes. B lymphocytes mature in them bone marrow. T lymphocytes mature in the thymus gland.

B – Lymphocytes

B lymphocytes produce certain chemical substances to act against antigens. The chemical substances which act against antigens are called antibodies.
10th Class Biology 5th Lesson Kerala Syllabus

Antibodies destroy the pathogens in three different ways

  1. Destroy the bacteria by disintegrating their cell membrane.
  2. Neutralize the toxin of the antigens.
  3. Destroy the pathogens by stimulating other white blood cells.

T – Lymphocytes

10th Class Biology 5th Lesson Questions And Answers Kerala Syllabus

T lymphocytes stimulate other defense cells of the body. Moreover, these cells are capable of destroying cancer cells and cells affected by virus.

Question 18.
Complete illustration showing the defense mechanisms of blood.
Biology Class 10 Chapter 5 Notes Kerala Syllabus
Answer:

  • Inflammatory response
  • Phagocytosis
  • Blood clotting

Lymph And Defense

Biology Chapter 5 Class 10 Kerala Syllabus

The lymph formed from the blood and reabsorbed into blood has a prominent role in defense mechanisms, lymph contains plenty of lymphocytes. They destroy the disease-causing bacteria in lymph nodes and spleen.

Immunization

Defense mechanisms become slow when germs enter the body. This causes the spread and multiplication of germs. Immunization is the artificial method to make the defense cells alert against the attack of pathogens.

HSSLive.Guru

Question 19.
What are vaccines?
The substances used for synthesizing antibodies are called vaccines.

Question 20.
Which components of vaccine act as antigens?
Answer:
The components from alive or dead or neutralized germs neutralized toxins or cellular parts of the pathogens will be the component of each vaccine.

Question 21.
How do vaccines induce immunity?
Kerala Syllabus 10th Standard Biology Notes
Answer:
In induced immunity antibodies which can act against pathogens or toxins produced by them are synthesized in the body itself. The body prepares antibodies to act against these foreign bodies.
Hss Live Guru 10th Biology Kerala Syllabus

Treatment – Final Defense

Question 22.
Which are the different methods of treatment that we depend on?
Answer:

  • Ayurveda
  • Sidda
  • Unani
  • Naturopathy
  • Homeopathy
  • Allopathy (Modern medicine)

Question 23.
Observe the pictures given below and write the name and use of this equipment for diagnosis.
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 17
Answer:
A. Stethoscope – to measure heartbeat
B. Thermometer – To measure body temperature
C. Sphygmomanometer – To record blood pressure.

Question 24.
Given below is the table including a few other modern equipments and their uses.
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 18

Laboratory Tests

The report of a test showing the quantity of different factors in blood
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 19

Question 25.
Identify the specializations in medicine and the related areas and complete the table
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 20
Answer:

Specialization Related area
Cardiology Treatment of heart
Ophthalmology Treatment of eye
Neurology Treatment of brain/nerve diseases
Oncology Cancer treatment
ENT Treatment of diseases of ear, nose, throat.

Antibodies

The scientist Alexander Fleming, who first synthesized antibiotics in 1928. Antibiotics are used to resist bacterial diseases.

Question 26.
Antibiotics are very helpful but use of it should be with great care. Why?
Answer:

  • Regular use develops immunity in pathogens against antibiotics
  • Destroys useful bacteria in the body.
  • Reduces the quantity of some vitamins in the body.
  • Some of them cause allergy, problems to stomach, bones, and kidneys.

HSSLive.Guru

Question 27.
Is it proper to use antibiotics without recommendation by a doctor? Why? Discuss. Write your inferences in the science diary.
Answer:
No. Though antibiotics are effective medicines, their regular use brings many side effects. Therefore use medicines only by the instruction of the doctor. Doctors prescribe medicine by considering the dose, method of use, period of use, age of the patient, etc. indiscriminate use of them causes health problems. Regular use develops immunity in pathogens against antibiotics, destroys useful bacteria in the body and reduces the quantity of some vitamins. So self-treatment without the instruction of the doctor is not good.

First Aid

Question 28.
Observe figures A, B and C and identify the instance in which the following type of first aid is given.
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 21
Answer:

  • A – Breathing has stopped but heartbeat has not as in drowning electric shock, choking gas, suffocation, etc.
  • B – Bone and ligament injuries; and fractures
  • C – Choking occurs when an object swallowing

Blood Transfusion

The transfer of blood from one person to another is called blood transfusion. Certain instances such as blood is lost excessively in accidents, affected with diseases like blood cancer and surgical operations require blood transfusion.

Different Types of Blood Group

A, B, AB, O are the main blood groups. Carl Landsteiner proposed blood grouping on the basis of the presence or absence of A, B antigens seen on the surface of the red blood cells. The blood group in which Rh factor is present are positive blood groups and those without Rh factor are negative blood groups.

Question 29.
Can a patient receive blood from any person?
Answer:
No. Blood of certain persons cannot be received by others. The antigen present in the received blood and antibody in the recipient’s blood will react each other to form blood clot.

Question 30.
Observe the table and identify the various types of blood group, antigens, and antibodies present in them
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 22

Question 31.
Prepare posters on the greatness of donating blood
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 23

Defense Mechanisms In Plants

Question 32.
Complete the illustration by including different defense mechanisms in plants.
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 24
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 25

Let Us Assess

Question 1.
Which among the following is not included in non-specific body defense?
a) production of sebum
b) action of hydrochloric acid in the stomach
c) action of B lymphocytes
d) action of lysozyme in saliva
Answer:
c) action of B lymphocytes

Question 2.
Write the functions of the two types of lymphocytes in the defense mechanism of the body.
Answer:
B lymphocytes produce antibodies and it destroys the antigens, T- lymphocytes stimulate white blood cells and also destroys cancer cells.

HSSLive.Guru

Question 3.
What is the basis of grouping blood Into different types? Everybody cannot receive blood of all groups. Why?
Answer:
The basis of blood grouping is the presence of antigen seen on the surface of the red blood cells. When an antigen reaches one s blood, it stimulates defense activity to produce antibody. The antigen and antibody react each other and form a blood clot. Hence everyone cannot receive blood from aH blood groups.

Soldiers of Defense More Questions And Answers

Question 1.
Identify the diagram and mention the defense process taking place in A
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 26
Answer:
Skin.
A is the outermost Keratin layer. Keratin is a protein, it blocks the entry of germs.

Question 2.
Respiratory track is always free from germs. Why?
Answer:
Mucus in the trachea prevents the entry of germs into the lungs. The cilia in the bronchus wipe out dust that enters it. Cough and sneezing help to expel foreign bodies from the respiratory tract. So respiratory tract is always free from germs.

Question 3.
Complete the table suitably.
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 27
Answer:
a) Ear
b) Hydrochloric acid
c) eye / mouth
d) The outermost layer blocks the entry of germs/ Keratin / Sebum / Acids.

Question 4.
Is swelling of the wound site helpful or not? why?
Answer:
Yes, It is helpful. The wounds and cuts occur in the skin, that area swells and blood vessels dilate It increases the blood flow and more white blood cells can come out through the enlarged pores and destroy the germs.

Question 5.
Generally, bacteria are useful but some of them are pathogenic. How?
Answer:
After entering the body they multiply by binary fission and produce certain toxic substances, which either disrupt the cellular activities or destroy the cell itself.

Question 6.
Following are certain steps of a defense process identify the process.
1. Phagocytes reach near the pathogens.
2. Engulf pathogens in the membrane sac
3. Membrane sacs combine with lysosome.
4. The enzyme in the Iysosome destroys the pathogens.
5. Expels the remnants from phagocyte.
Answer:
Phagocytosis

Question 7.
Observe the illustration and identify the process.
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 28
Answer:
Phagocytosis

Question 8.
What are the factors needed for blood clotting?
Answer:
Prothrombin, Fibrinogen in plasma, Calcium ions, Vitamin K, Red blood cells, Platelets.

Question 9.
Blood clotting is a defense mechanism to prevent the loss of blood through wounds. Mention the different stages of this process.
Answer:

  • Tissues of the wounded part degenerate to form the enzyme thromboplastin.
  • Thromboplastin converts prothrombin in the plasma to thrombin.
  • Thrombin converts the fibrinogen in the plasma to fibrin.
  • Blood clot is formed by the entangling of platelets and red blood cells in the fibrin network.

Question 10.
Complete the flowchart showing the blood clotting.
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 29
Answer:
A) Thrombin
B) Fibrin

Question 11.
“Fever is not a disease, it is a defense mechanism.” Analyze the statement.
Answer:
Yes. The presence of toxin produced by the pathogens stimulate the white blood cells and hence the white blood cells produce chemical substance that raises the body temperature. The rise in body temperature reduces the rate of multiplication of pathogens and increases the rate of phagocytosis.

Question 12.
How is antibody destroy germs?
Answer:
Antibody destroys the bacteria by disintegrating their cell membrane and neutralize the toxin of the antigens by stimulating other white blood cells.

Question 13.
Define the following
1. Antigen
2. Antibody
3. Antibiotic
Answer:
Any foreign body that stimulates the defense mechanism is called an antigen. The chemical substance produced by the lymphocytes act against antigen is an antibody. Antibiotics are medicines used to resist bacterial diseases.

Question 14.
The graph shown below represents the difference in the number of two bacteria when taken a particular antibiotic by a patient.
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 30
Answer:
a) This antibiotic is effective and reduces the number of pathogens. But its antibiotics. It also destroys useful bacteria in the body.
b) Further use of these antibiotics is not effective because the harmful bacteria got resistance against it.

Question 15.
Complete the boxes according to the given hint.
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 31
Answer:
A – Regular use develops immunity in pathogens against antibiotics
B – Destroys useful bacteria in the body.
C – Reduces the quantity of some vitamins in the body.

Question 16.
Can a person with ‘A’ group blood receive blood from ‘B’ group person? Or it take place vice versa? Give reason for this.
Answer:
A person with ‘A’ group blood cannot receive or donate blood with ‘B’ group person. Because the antigen present in the received blood and antibody in the recipient’s blood will react each other and form a blood clot (Coagulation).

Question 17.
Give more examples of vaccine
Answer:

Vaccine Diseases
BCG Tuberculosis
OPV Polio
DPT Diphtheria, Petussis, Tetanus
MMR Mumps, Measles and Rubella
Hepatitis. B. Vaccine Hepatitis
TT Tetanus
Cowpox vaccine Small pox
Rabies vaccine Rabies

Question 18.
“Germs, both alive and dead are used to get immunity”. Substantiate the statement with vaccines used for rabies and tuberculosis. (March 2015)
Answer:
Germs, both alive and dead are used as vaccines. ‘ Dead germs are utilized in rabies vaccine which acts against rabies. Live, but inactivated vaccines are used in BCG vaccine against tuberculosis.

Question 19.
Observe the following figure and answer the given questions.
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 32
a) Label A and B.
b) How did they protect our body? (March 2015)
Answer:
a) A – small hair, B – Sebaceous gland
b) Since sebum is oily, water does not stick on to the skin. Covering of hair protects the body from cold and heat and also prevents the entry of foreign bodies.

Question 20.
Whichever be the type of germs infected, the initial symptom appear in human body will be the fever. Give reason. (March 2014)
Answer:
Bacterial infection produces toxic substances in body, body temperature is suitable for bacterial growth, in order to control the growth of bacteria body rises the temperature, fever is not a disease.

Question 21.
Constant use of antibiotics is not good for health. This is the opinion of Rahim.
a) Do you agree with his opinion? Why?
b) Give two specific examples for justifying your answer. (Model 2013)
Answer:
a) I agree with this. Constant use of antibiotics results a few side effects.
b) Constant use of antibiotics may destroy useful bacteria in the body, develop resistance in bacteria against antibiotics or reduces the level of certain vitamins in the body.

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Question 22.
How does the influence of the following action blocking germs.
a) Rise in body temperature.
b) Low oil content on skin.
c) Swelling occurs near wound.
d) Lymphocytes produce Antibodies. (March 2013)
Answer:
a) To resist the strengthening or increasing of causative organisms.
b) Waterproof and oily, germs cannot grow.
c) Flow of blood through the capillaries increases and more leucocytes comes out from the capillaries and destroy the germs by engulfing them.
d) Lymphocytes produce antibodies to destroy germs.

Question 23.
Whichever be the type of germs infected, the initial symptom appear in human body will be the fever Give reason. (March 2013)
Answer:
Bacterial infection produces toxic substances in body body temperature is suitable for bacterial growth, in order to control the growth of bacteria body rises the temperature, fever is not a disease.

Question 24.
The graph representing the difference in the number of two bacteria, while applying a particular antibiotic on a patient is shown below
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 33
A. Analyze the graph and record the findings.
B. Is this antibiotic effective against the bacteria? Why? (Model 2012)
Answer:
A. The harmful bacteria decrease in number in the first few weeks. Later they increase in number. Number of useful bacteria are decreasing gradually.
B. The antibiotic is not effective because the harmful bacteria got resistance against it. Moreover, number of useful bacteria decreases.

Soldiers of Defense Questions and Answers

Question 1.
Which among the following is the odd one? Why? Lymphocyte, Monocyte, Neutrophil, Basophil, Eosinophil (Question Pool-2017)
Answer:
Lymphocyte – Involved in specific defense

Question 2.
Skin is the largest sense organ of the body. It helps! us to sense heat, cold, touch, pressure, etc and it j acts as a soldier of defense of the body,
a) Does the skin have significance in defense as mentioned above? Justify. (Question Pool – 2017)
Answer:
Yes. The outermost Keratin, the protein layer blocks the entry of germs; sebum and some acids in the: skin-are disinfectants,

Question 3.
A table indicating primary level defense is given below. Arrange column B based on column A. (Question Pool – 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 34
Answer:
i) – c
ii) – d
iii) – a
iv) -b

Question 4.
Which among the following is the odd one and why? (Question Pool – 2017)
a) The Mucus of trachea destroys the pathogens.
b) The wax in the ear destroys pathogens.
c) Neutrophil destroys pathogens by engulfing them.
d) Lysozyme present in Saliva destroys pathogens
Answer:
C, Secondary defense

Question 5.
Nimisha’s hand got injured in an accident. After some time the wound area got swollen.
a) What is this type of activity known for?
b) Is it a defense mechanism? Why?
Answer:
a) Inflammatory response
b) Yes
Secondary level defense
Process to destroy pathogens in the body

Question 6.
Using the following statements,-prepare a flow chart of inflammatory response. (Question Pool-2017)
a) Production of chemical messages.
b) White blood cells destroy pathogens.
c) Blood vessels dilate.
d) Pathogens enter into the wound.
e) White blood cells come out from blood vessels.
f) Blood flow increases
Answer:
d
a
c
f
e
b

Question 7.
The given illustration includes white blood cells which act as a part of nonspecific defense. Fill up the blanks and complete the word web. (Question Pool -2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 35
Answer:
A – Neutrophil /Monocyte
B – Stimulates other white blood cells / dilates blood vessels
C – Eosinophil
D – Engulfs and destroys germs

Question 8.
When there is an injury or wound, the blood vessel of that part dilates. (Question Pool – 2017)
a) What is its benefit?
b) Which white blood cell dilates the blood vessel?
Answer:
a) The cells that get damaged by a wound or an infection produce certain chemical substances. These substances dilate the blood vessels thereby increasing the blood flow. Blood plasma and white blood cells reach the wound site and it destroys the germs,
b) Basophil

Question 9.
Observe the given illustration and answer the following questions. (Question Pool-2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 36
Answer:
a) Which is the process indicated in the illustration?
b) Which are the.white blood cells involved in the process?
c) Is it a specific defense mechanism? Justify
Answer:
a) Phagocytosis
b) Neutrophil, Monocyte
c) No
does not identify and destroy pathogens that enter to the body.

Question 10.
The flow chart given below indicates a type of defense mechanism occurring in the body. (Question Pool – 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 37
a) Complete the flow chart
b) Which process is it related to?
Answer:
a) i) Engulfs pathogen in the membrane sac
ii) The enzyme in the lysosome destroys the pathogens
iii) Expels the remnants
b) Phagocytosis

Question 11.
Blood clotting is a defense mechanism. Analyze the statement. (Question Pool – 2017)
Answer:

  • Prevents the entry of germs through wound
  • Prevents bleeding through wounds

Question 12.
Prepare the flow chart of the clotting of blood using the following statements. (Question Pool-2017)
a) Thromboplastin converts prothrombin to thrombin.
b) Blood flows from the wound.
c) Blood clot is formed.
d) Thrombin converts fibrinogen to fibrin.
e) Tissues degenerate to form the enzyme called thromboplastin.
f) The red blood cells and platelets entangle in the fibrin network.
Answer:
d
a
c
f
e
b

Question 13.
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 38
a) Identify A
b) B is a vitamin and C is an enzyme. Name them.
c) How does the lack of B or C affect the consequent chemical process? (Question Pool – 2017)
Answer:
A – Prothrombin
B – Vitamin K, C – Thromboplastin
C – Thrombin not formed
fibrinogen not converted to fibrin

Question 14.
Blood clot is formed by the entangling of red blood cells and platelets in the fibrin network.
White blood cells are not involved in this process. What explanation will you give for this? (Question Pool-2017)
Answer:

  • White blood cells do not have a definite shape
  • They come out through the fibrin network

Question 15.
One of the scars of the wound obtained by Binu while playing football remained even after 10 years. What explanation will you give for the scar remaining as such? (Question Pool-2017)
Answer:
When wound occurs, new tissues cannot be formed, then connective tissues heal the wound. In such situations, the wound scar remains

HSSLive.Guru

Question 16.
Fever is a defense mechanism. Is the statement correct? Justify your answer. (Question Pool-2017)
Answer:
Yes. The presence of toxin produced by the pathogens stimulate the white blood cell and hence the white blood cells produce chemical substance that raises the body temperature. The rise in body temperature reduces the rate of multiplication of pathogens and increases the rate of phagocytosis.

Question 17.
Complete the illustration (Question Pool-2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 39
Answer:
a) B – lymphocyte
b) T – Lymphocyte
i) Stimulates white blood cell and destroys pathogens
ii) Destroys the bacteria by disintegrating their cell membrane
iii) Stimulates defense cells
iv) Destroys the cell which is affected by virus.

Question 18.
After attending a class on immunity, Arun raised a question to his teacher. (Question Pool-2017)
“In spite of so many defense mechanisms in the body, why are we still affected by diseases?
a) What explanation will you give for Arun’s doubt?
Answer:

  • Bad habits
  • Unhealthy food habits
  • Unhygienic
  • Excess pathogens

Question 19.
The use of some modern equipment are given below. Identify the equipment. (Question Pool-2017)
a) To record electric waves in the brain.
b) To record electric waves in the heart muscle
c) To understand the structure of internal organs using ultrasonic sound waves.
Answer:
a) EEG
b) ECG
c) Ultrasound scanner

Question 20.
The doctor prescribed antibiotics to Sunil who is affected with cholera, but not to Anil who is affected with chickenpox. What is the reason? (Question Poo1 -2017)
Answer:
Antibiotics are used to prevent bacterial diseases. Chickenpox is a viral disease. Cholera is a bacterial disease. So the doctor prescribed antibiotics to Sunni

Question 21.
Enlist the demerits of antibiotics for Jose who is preparing for a seminar on the topic “The merits and demerits of Antibiotics. (Question Pool -2017)
Answer:

  • The frequent use of antibiotics produces disease defense in pathogens.
  • Destroys useful bacteria in the body.
  • Reduces the level of some vitamins in the body.

Question 22.
Ashiq who met with an accident was in need of blood. Antigen A and D and Antibody b was identified in his blood. (Question Pool -2017)
a) Name his blood group?
b) Whose blood, among the following can be accepted by ashiq?
(i) Venu = A+
(ii) Amal- AB+
(iii) Suhara – AB
(iv) Anoop – A
Answer:
a ) A+
b) (i)VenuA+
(ii) Anoop A

Question 23.
The table given below indicates blood groups.

Blood group Antigen Antibody
A (i) b
B B (ii)
(iii) A, B (iv)

Answer:
i) A
ii) a
iii) AB
iv) No
v) O
vi) No

Question 24.
Box A includes the major components of vaccines and box B includes the diseases against which they are used. Match them appropriately. (Question Pool -2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 40
Answer:
i) -d
ii) -c
iii) -b
iv) -a

Question 25.
Ravi prepared an illustration showing defense mechanisms in plants. Complete it. (Question Pool – 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 41
Answer:
a) Prevents the entry of germs which have crossed the cell wall, through cell membrane.
b) Bark
c) Cuticle in leaves
d) Cell wall

Question 26.
“This mode of treatment is a lifestyle in tune with nature rather than a mere method of treatment” This is a statement regarding a well-known mode of treatment.
a) Name the treatment.
b) Apart from this, name any two well-known modes of treatment. (Question Pool – 2017)
Answer:
a) Ayurveda
b) 1. Allopathy
2. Homeopathy/ etc.

Question 27.
Prepare two suitable placards to conduct an awareness rally in association with World Blood Donation day. (Question Pool-2017)
Answer:
2 placards contain appropriate concepts
Example: Donate blood Donate Life
Blood donation – Nothing to loose profits – Life

Question 28.
Match the following pairs (Question Pool – 2017)
a) T-lymphocyte: Thymus gland
B – lymphocyte:………………
b) EEG: to record electric waves in brain
………… to record electric waves in heart muscles
c) First Antibiotic: Alexander Flemming
First vaccine:……………..
d) Heartbeat: Stethoscope
Blood pressure:……………..
e) Antigen: Red blood cells
Antibody:…………………
Answer:
a) Bone marrow
b) ECG
c) Edward Jenner
d) Sphygmomanometer
e) Plasma

Question 29.
Given below is an equipment used for disease diagnosis. (Question Pool-2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 42
a) Identify the equipment
b) What is its use?
c) Name another equipment that works on the same principle.
Answer:
a) ECG
b) To record electric waves in heart muscles
c) EEG

Question 30.
“it is possible to build up a healthy society with hospitals, doctors, and medicines” This is Bashir’s opinion. Evaluate it (Question Pool -2017)
Answer:

  • The opinion of Bashir is wrong
  • Nutritious food
  • Healthy lifestyle
  • Hygiene, These are the factors which build up a healthy society.

Question 31.
“It is not necessary to detect blood groups if we can accept blood from anyone”. This was an argument put forward by Sivaprasad in a discussion on blood transfusion.
a) What is the base of blood group determination?
b) Can a person receive any blood from anyone? Why? (Question Pool – 2017)
Answer:
a) The presence of antigens A and B on the surface of Red blood cells,
b) 1. Not possible
2. When a foreign antigen reaches one’s blood, it stimulates the defense activity
3. The antigen present in the received blood and the antibody in the recipients, blood will react each other to forms a blood clot (coagulation)

Question 32.
Given below is the picture of white blood cells which are parts of specific defense? 32 (Question Pool – 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 43
a) Identify A and B
b) What is the role of A in specific defense?
c) Give anyone difference between A and B
Answer:
a) A – T- lymphocyte
B – B lymphocyte
b) 1. Stimulates other defense cells
2. Destroys cancer cells and virus affected cells.
c) B lymphocytes matured at bone marrow T lymphocytes matured at thymus

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Question 33.
Analyze the following statement and answer the following questions. 33 (Question Pool-2017)
When there is a wound, the body temperature rises.
a) What is the significance of white blood cells in this activity?
b) How does immunity become possible through a rise in temperature?
Answer:
a) The chemical substances produced by white blood cells rise body temperature
b) The presence of toxin produced by the pathogens stimulate the white blood cell and hence the white blood cells produce chemical substance that raises the body temperature. The rise in body tempera¬ture reduces the rate of multiplication of pathogens and increases the rate of phagocytosis.

Question 34.
Statements related to nonspecific defense and specific defense are given below. Identify the type of the defense and mark them using the letters N and S respectively.
(Orukkam – 2017)
a) The cilia in bronchus wipe out dust that enters it.
b) Destroy the bacteria by disintegrating their cell membrane.
c) The blood vessels near the wound diabetes.
d) The rise in body temperature reduces the rate of multiplication of pathogens.
e) B lymphocytes produce certain chemical substances against antigens.
f) Eosinophil produces chemical substances needed for inflammatory responses.
g) T lymphocytes destroys cancer cells.
h) The enzyme lysozyme present in tears destroys germs.
i) T lymphocytes destroy cancer cells.
j) Phagocytes engulf and destroy germs.
Answer:
a) N
b) S
c) N
d) N
e) S
f) N
g) S
h) N
i) S
j) N

Question 35.
Our body has the capacity to destroy germs those enter the body by breaking the first level defense Write your comment on this statement? (Orukkam – 2017)
( Hints: inflammation, different types WBCs and their functions, phagocytosis)
Answer:
The statement is true. When germs enter to the body the blood vessels diates (inflammation). This helps the white blood cells to act at the site A few white blood cells destroy germs by phagocytosis.

Question 36.
The basis of blood grouping is the presence of antigens in red blood cells. Complete the table given below based on this statement. (Orukkam-2017)
Blood groups
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 44
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 45
a) B lymphocytes produce antibodies against antigens.
b) We can use antigens as vaccines for the formation of antibodies in advance.
c) Neutralized toxins – Diphtheria
Alive but neutralized germs – Measles
Cellular parts of pathogens – Hepatitis B Killed germs – Cholera

Question 37.
Complete the illustration suitability related to antibiotics (Orukkam – 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 5 Soldiers of Defense - 46
Answer:
A) Prevent bacterial diseases
B) Prolonged use may develop immunity in germs
C) Destruction of useful bacteria
D) Deficiency of certain vitamins in the body

Question 38.
The wound scar does not remain always. Write reason? (Orukkam – 2017)
Answer:
If the wound is filled with same tissue, no wound scar occurs there.

Question 39.
Fill in the blanks by observing the relationship in the first pair. (Orukkam – 2017)
a) EEG: To record electric waves in the brain
ECG:……………………………..
b) Rabies: Killed germs
Typhoid:………………………..
Answer:
a – Records electric waves in the heart muscles
b – Alive but neutralized germs.

Question 40.
Name the first vaccine? Who developed this? Write the situation which leads to the development of vaccine? (Orukkam – 2017)
Answer:
Smallpox vaccine, developed by Edward Jenner. He observed no smallpox disease in people who had affected cowpox earlier.

Kerala SSLC Maths Model Question Paper 3 Malayalam Medium

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Kerala Syllabus 10th Standard Social Science Notes Chapter 6 Eyes in the Sky and Data Analysis

Kerala State Syllabus 10th Standard Social Science Notes Chapter 6 Eyes in the Sky and Data Analysis

Our ancestors made maps after collecting information about the surface of the earth and survey of earth’s surface which lasted for a long time. Aerial photography which developed later made map making quite easy. Satellite remote sensing and geographical information system which developed in the 1960’s as a result of the progress in science and technology paved the way for map making efficient and fast. The chapter helps you to understand how the launching of satellites and modern computer softwares help in the analysis of geophysical data.

→ Remote sensing : The method of collecting information about an object or phenomenon without actual physical contact.

→ Scanners : Sophisticated equipments that can detect electromagnetic radiation.

→ Sensors : The instruments used for data collection through remote sensing.

→ Active Remote Sensing : Remote sensing made with the help of artificial sources of energy is known as active remote sensing.

→ Passive Remote Sensing : Remote sensing carried out with the help of solar energy is known as passive remote sensing.

Platform : The carrier on which sensors are fixed is platform.

→ Terrestrial Photography : The method of obtaining earth’s photographs using cameras from the ground.

→ Aerial Remote Sensing : The method of obtaining photographs of the earth’s surface continuously from the sky using cameras mounted on aircrafts.

Social Science Short Notes For Class 10 Kerala Syllabus

→ Satellite Remote Sensing : The process of collecting information using sensors fitted on artificial satellites is called satellite remote sensing.

→ Stereoscope: The instrument used for obtaining three dimensional view from the stereo pairs is called stereoscope.

→ Stereo pair: Two aerial photographs of adjoining areas or two adjacent aerial photographs with overlap.

→ Overlap: Each aerial photograph contains about 60% of the area shown in the previous photograph is termed as overlap.

→ Stereoscopic vision : The three dimensional
view obtained while viewing a stereo pair through a stereoscope. ‘

→ Geostationary satellites : Artificial satellites that orbit the earth at a height of about 36000 km in accordance with the earth’s rotation.

→ Sun Synchronous satellites: Artificial satellites that revolve around the earth along poles at a height of 900 km from the earth’s surface.

→ Spectral Signature : The amount of reflected energy by each object.

→ Spatial Resolution : The size of the smallest object on the earth’s surface that a satellite sensor can distinguish.

→ Geographic Information System : Geographic Information System is a computer based information management system by which data collected from the sources of information like maps, aerial photographs, satellite imageries, tables and surveys are incorporated into the cqmputer using softwares, which are retrieved, analyzed and displayed in the form of maps, tables and graphs.

→ Spatial Data : Each feature on the earth’s surface has its own latitudinal and longitudinal location. Such information is known as spatial data.

→ Attributes: The additional information about the characteristics of each spatial data on the earth’s surface are called attributes.

→ Layers : The thematic maps prepared and stored in Geographic Information System for analytical purpose are called layers.

10th Geography Notes

→ Network Analysis : GIS analysis that deals with linear features on a map such as roads, railway and rivers.

→ Buffer Analysis : GIS analysis used for analyzing activities around a point feature or at a definite distance along a linear feature.

→ Overlay Analysis : GIS analysis used to identify the interrelationship of various surface features on earth and the changes they have undergone over a period of time.

→ Global Positioning System : A system using signals received from the artificial satellites to display the latitude, longitude, height and time of a place.

→ IRNSS : Indian Regional Navigation Satellite System is the satellite based navigation system developed by India as an alternative to GPS. It g is designed to provide accurate position information service.

Kerala Syllabus 10th Standard Social Science Notes

Kerala Syllabus 10th Standard English Solutions Unit 2 Chapter 2 My Sister’s Shoes

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Kerala State Syllabus 10th Standard English Solutions Unit 2 Chapter 2 My Sister’s Shoes (Screenplay)

My Sister’s Shoes (Screenplay) Textual Questions and Answers

My Sisters Shoes Summary Kerala Syllabus 10th Question 1.
The scene begins with a close-up shot. What frame would a filmmaker normally use when the characters (Cobbler &Ali) speak?
Answer:
Medium shot

My Sister’s Shoes Summary In English Kerala Syllabus 10th Question 2.
Apart from the cobbler’s shop what other details do you get about the space in which the actions in Scene 1 take place?
Answer:
A street. A door with curtain hanging on it which appears to open into a row house.

My Sisters Shoes Summary In English Kerala Syllabus 10th Question 3.
How many characters are introduced in scene 1? Who are they?
Answer:
Three, Cobbler, Ali and a man with a parcel.

My Sister’s Shoes Summary Kerala Syllabus 10th Question 4.
In scene 1 and 2 close up shots of the hands of the cobbler and Ali are shown. Do they serve any purpose? Explain.
Answer:
They are in close up because they emphasize the work they are doing. It will show what kinds of people are expected in the film. Such shots establish the ground work for the story to take off.

My Sister’s Shoes Notes Pdf Kerala Syllabus 10th Question 5.
Where is the location shifted to?
Answer:
The location is shifted to a footpath outside the bakery.

My Sister’s Shoes Summary In Malayalam 10th Question 6.
What could be the camera movement in this scene? Why is the camera moved in this manner?
Answer:
Boom/Crane shot. To show that the junk collector is * taking away the bundle of shoes.

Summary Of My Sisters Shoes Kerala Syllabus 10th Question 7.
What are the events taking place in scene 3?
Answer:
Ali goes to the vegetable shop. He takes potatoes and asks the shopkeeper to write it in the account. A junk collector comes and collects the junk. He thinks the bundle of shoes is also part of the junk, he throws it into his cart. In his eagerness to locate the bundle, Ali scatters vegetables on the ground. The shop keeper is angry and tells Ali to get lost.

My Sister’s Shoes Narration Kerala Syllabus 10th Question 8.
Which of the events is crucial in bringing about a twist tn the narrative?
Answer:
The taking away of the bundle of shoes by the junk collector.

My Sisters Shoes Questions And Answers Kerala Syllabus 10th Question 9.
What impression do you get about the financial status of Ali’s family? Support your answer with evidence from the text.
Answer:
Ali’s family is very poor. We see Ali buying things on credit. The shopkeeper tells Ali that the credit limit has reached and he should ask his mother to pay at least part of the payment due to him.

My Sister’s Shoes Questions And Answers Kerala Syllabus 10th Question 10.
What interesting strategy does the director make use of in presenting the ‘talks’ between Ali and his sister? Why do you think they communicate in such away?
Answer:
They communicate by writing in the notebook and then passing the notebook to the other. They do that because they don’t want to let the parents know about the loss of the shoes.

My Sisters Shoes Summary In Malayalam Kerala Syllabus 10th Question 11.
Do you think the sound of a hammer is used deliberately in this scene? What effect does it produce?
Answer:
Yes, it is used deliberately. It tells us that that father of Ali is a worker with wood (joiner or carpenter) as he is chopping wooden flints with a hammer on a block.

My Sisters Shoes Notes Character Sketch Kerala Syllabus 10th Question 12.
Apart from showing the characters, does the director make use of any visual image to add on to what he intends to communicate?
Answer:
Yes, he does. As an example we see the passing of the notebook between Ali and Zahra. They don’t want their parents to know about the loss of the shoes. Ali will be scolded or even beaten if his father comes to know that he was careless enough to lose the shoes of Zahra. Ali’s father is not rich enough to buy her another pair immediately.

Write a script for a short video on any one of the following themes. (Or, you may choose a theme other than those listed.) You may shoot it using a digital camera and upload it on YouTube.

My Sister’s Shoes Notes Kerala Syllabus 10th Question 13.
Leadership Quality, Unity of People, Incidents of Bravery, Child Labour, Evils of Smoking or Alcoholism, Wasting Food, Need for Social Change, Safe Driving, Preservation of Nature, Learning Disorders in Children, Values, Safe Earth, Save Tiger, Global Warming.
Answer:
The following points may help you.

1. Watch other people’s films You can teach yourself a lot about filmmaking by watching: short online videos, advertisements, feature films, etc. Look at what you like, and what you don’t like, and try and work out how and why the filmmaker made it that way.

2. Build your skills Learn how to use a camera and find out what it can do when you place it in different angles or use different lens settings. Try shooting different kinds of shots with your camera, recording good sound, and editing them. Watch a short scene from a film you like and see if you can copy it exactly.

3. Get organised Once you’ve got your idea, create a script and storyboards or shot lists. Use a digital still camera if you are not able to prepare the storyboard.

4. Keep it short and simple Have you got a strong idea? Write the idea down for your film in 50 to 75 words. If you can’t do that, it’s not clear enough. Keep the film short. People are more likely to watch an online video if they know it’s only 60  econds long.

5. Shoot separate shots Learn the different shot sizes. Use a variety of shots rather than just panning and zooming. Use plenty of close up shots to show the important things. Place your camera in different angles and not just from the front.

6. Get the sound right A good movie with a bad sound track will lose viewers faster than bad movies with good sound track. Use a microphone to record the sound right. If the sound cannot be recorded right, then fake it using sound effects, or edit our film to a recorded voiceover.

7. Edit it right Editing is an interesting job. It’s not about just getting rid of the bad stuff, it’s where your film will really come together. Get the pace right: make sure your film doesn’t drag, or that shots don’t flash past too quickly. Make sure your film makes sense. You can seek the help of technicians in this field if necessary.

My Sister’s Shoes About the author:

Kerala Syllabus 10th Standard English Solutions Unit 2 Chapter 2 My Sister’s Shoes 12
– Majid Majidi

From Children of Heaven (Bacheh-ye Aseman) directed by Majid Majidi Majid Majidi was born in 1959. He is an Iranian film director, producer and screenwriter. He has directed many feature films including The Colour of Paradise (1999), Baran (2001) and The Willow Tree (2005). Majidi directed the film “Children of Heaven” in 1998. It was nominated for the Academy Award for Best Foreign Language Film.

Kerala Syllabus 10th Standard English Solutions Unit 2 Chapter 2 My Sister’s Shoes 3

My Sister’s Shoes Summary in English

Page – 56
→ Scene 1

Cobbler’s shop
Close up of a cobbler stitching a girl’s shoe. Only the pink shoe and the cobbler’s hands are in the frame. The camera moves back to middle distance to show Ali. He is sitting on a low chair next to the cobbler and watching him work. The cobbler finishes the stitching of the shoe, picks up the other one of the pair and gives them to Ali.

Cobbler : That’ll be 30 Toumans.
Ali : Thankyou(giveSmoneytothecobbler)
Cobbler : Here is your change (picks up coins from the money box to give Ali).

Cut to the Street Outside

Page – 57
Seen from across the street. The cobbler’s shop can be seen. To the right of the shop is a door with a curtain. It looks the door opens into a row house. Ali is leaving the shop.

Ali’s voice: Thank you.
Cobbler’s voice : You are welcome. Goodbye.

A man with a parcel under his arm comes to the curtained door. He lifts the curtain and goes in. Ali comes out of the shop. He puts the shoes into a small black bag in his hands. He walks down the street and moves out of the frame on the left.

Kerala Syllabus 10th Standard English Solutions Unit 2 Chapter 2 My Sister’s Shoes 13

→ Scene 2

Bakery
Inside the bakery. Shots of bread being baked. Close up of a hand putting into the stove kneaded flour spread on a baking board and taking out the baked nan. The camera pans slightly to the right to show Ali picking up the nan dropped by the cook and stacking them on a cloth spread on a wooden plank. The camera moves back to showAli and three cooks sitting around the stove. They are kneading and putting it inside the stove. Ali finishes stacking the man and ties up the cloth into a bundle.

Page – 58
→ Scene 3

The footpath outside the bakery
A small group of men are waiting in a queue outside the bakery to buy nan. The outline of buildings in the street is seen in the distance. Ali comes from the bakery to the footpath. The parcel of nan is in his right hand and the parcel of shoes is in his left. He crosses the pavement and walks into a vegetable store. He keeps the bundle of nan on a pile of boxes of vegetables stacked in front of the shop. He places the bag of shoes in the small gap between two boxes.

On the sound track a hawker’s voice is heard : “Salt, salt for trade.” The shopkeeper is behind the counter.
Ali (to the shopkeeper) : Assalamu Alaikum, Aqbar Aqa, I need some potatoes.
Akbar, the shopkeeper, gives Ali a small black bag to put potatoes in. Ali starts picking the large, good potatoes in a box at the top of the heap.
Akbar : Not those, kid. Pick some down there.
Ali moves to his right, squats down, and fills the bag with smaller potatoes from a box on the floor. Close up of Ali picking the potatoes.

Cut to view of the shop from the pavement

A man pushing a handcart filled with junk enters the frame from the right and stops in front of the shop.
The junk collector (loudly), to the shop keeper : Assalamu Alaikum. With your permission.
The junk collector picks up the bundles of used polythene bags lying scattered on the floor near the boxes of vegetables and throws them into the cart. He sees Ali’s parcel of shoes. Thinking it is junk, he picks it up and puffin the cart.

The junk collector: Goodbye.
Akbar: Goodbye.
The junk collector leaves.

Cut to view from inside the shop

Page – 59
Akbar is on the left ledge of the frame counting money.
Cut to Ali filling his bag with potatoes.
Ali finishes filling the bag and hands it to Akbar for weighing. Akbar holds the bag in his hand to feel the weight for a moment and hands it back to Ali.
Akbar : Sixty five Toumans.
Ali : Mom said to put it on our tab.
Akbar : Tell her your account has reached its limit. She should pay at least part of it.
Ali : Alright.

Cut to view from outside the shop

Ali goes to the pile of boxes and picks up the bundle of nan. He then looks for the bag of shoes. Unable to find it, he places the bag of nan on top of the vegetable boxes and searches for the shoes underneath. He puts his hand and his head in the gap between the boxes of vegetables. It upsets them. The boxes fall and the vegetables scatter on the ground. Hearing the sound, Akbar comes. He sees the vegetables spread on the ground.

Kerala Syllabus 10th Standard English Solutions Unit 2 Chapter 2 My Sister’s Shoes 14

Akbar : (annoyed) What the hell are you doing?
Why did you spill these? Are you crazy?
Ali : (looking at him with guilt) My sister’s shows have disappeared.

Page – 60
Akbar : Get lost. Beat it.
Ali : I left my sister’s shows here.
Akbar : I said get lost! (bangs his fist on the pile of boxes.)
Ali runs away, scared.

Scene 4
Ali’s house
Middle distance shot of Ali and his sister Zahra reading their textbooks kneeling on the mattress. A baby’s cry is heard faintly on the soundtrack. The camera closes in on Zahra. She is writing in a notebook mumbling the words as she writes.

Kerala Syllabus 10th Standard English Solutions Unit 2 Chapter 2 My Sister’s Shoes 15

Ali: How am I going to school without shoes?
The camera moves back to middle distance. Now both Zahra and Ali are in the frame. Their parents are talking at the other end of the room and their talk can be heard on the soundtrack. We hear the banging of hammer in between.
Mother (off screed) : Go to your company store, tomorrow. We don’t have any formula left.

Zahra passes her notebook to Ali.
Father (offscreen): Don’t worry. Rahim Aqa’s wife had a slipped disc. Surgery made it worse.
Alternate close ups of Ali and Zahra listening to their parents’ talk and looking at each other.

Page – 61

Cut to

Middle distance shot of Mother lying on a bed, towards the left of the frame, leaning back against two propped up pillows. Father is sitting on a chair near the wall, chopping wooden flints with a hammer on a block. Behind him there are two ledges on which there are some vessels and clothes.

Father: You should learn to live with it.
Close up of Ali reading from the book Zahra has passed on to him.
Mother (off-screen): What do you want me to do? Do nothing all day?
Ali starts writing in the note book.
Father (off-screen): Well, the doctor has forbidden work.
Ali writes ‘you can go to school with slippers’ (mumbling the words as he writes) and passes the notebook to Zahra. Close up of the notebook.

Kerala Syllabus 10th Standard English Solutions Unit 2 Chapter 2 My Sister’s Shoes 16

Mother (off-screen) : Kokab Khanom’s sister had surgery, and she is fine.
Cut to close up Zahra taking the notebook and reading what Ali wrote on it.
Father (off-screen): Don’t everthink of surgery.
Alternate dose ups of Zahra and Ali looking at each other.
Cut to close up of Zahra writing in the notebook.

Page – 62
Zahra (mumbling as she writes): Ali, you have some nerve. You lost my shoes. I’ll tell dad.
Quick close up of Ali looking at Zahra. She passes the notebook to Ali. Close up of the notebook.
Two close-ups of Father working on the other end of the room looking at the camera. There are close-ups of Ali and Zahra and a middle-range shot showing them.
Ali writes in the notebook and passes it to Zahra.
Zahra writes ‘Then what shall Ido?’ in the notebook and passes it to Ali. Ali writes ‘you can wear my sneakers’ and passes the notebook back.
Zahra writes ‘I’ll wear them when you are back from school’.
During the passing of the notebook back and forth only the notebook is in close up and hands are visible in the frame.
Close up of Zahra’s writing. The stub of her pencil breaks. Ali puts his pencil on the notebook for her to write. Zahra does not take the pencil. Close up of pencil lying on the notebook.

Kerala Syllabus 10th Standard English Solutions Unit 2 Chapter 2 My Sister’s Shoes 17

My Sister’s Shoes Summary in Malayalam

Kerala Syllabus 10th Standard English Solutions Unit 2 Chapter 2 My Sister’s Shoes 4
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My Sister’s Shoes Glossary

Kerala Syllabus 10th Standard English Solutions Unit 2 Chapter 2 My Sister’s Shoes 1
Kerala Syllabus 10th Standard English Solutions Unit 2 Chapter 2 My Sister’s Shoes 2

Kerala Syllabus 10th Standard Social Science Solutions Chapter 8 Kerala towards Modernity in Malayalam

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Kerala Syllabus 10th Standard Hindi Solutions Unit 1 Chapter 4 बंटी

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टूटा पहिया Text Book Activities & Answers

टूटा पहिया अभ्यास के प्रश्न

Sslc Hindi Chapter 4 Notes Kerala Syllabus प्रश्ना 1.
“वहाँ उसके और ममी के बीच में बहुत सारी चीजें आ जाती हैं। वह बंटी की अपनी दुनिया थी।…..” इस हालत में बंटी की चिंताएँ क्याक्या हो सकती हैं? लिखें।
उत्तर:
बंटी की चिंताएँ
माँ मुझे यहाँ क्यों ले आती हैं? यह जगह मुझे बिलकुल अपरिचित है। यहाँ मेरा दम घुटता है। माँ तो यहाँ आकर माँ नहीं रह जाती हैं। उसका प्रिंसिपलवाला यह चेहरा मुझे बिलकुल पसंद नहीं आता। यहाँ माँ व्यस्त ही व्यस्त हैं। अपने लिए कुछ समय निकाल नहीं पातीं। मैं अपनी माँ को चाहता हूँ। किसी प्रिंसिपल को नहीं।

 

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Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away

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Keeping Diseases Away Text Book Questions and Answers

Keeping Diseases Away Kerala Syllabus 10th Question 1.
What are the reasons forthe spread of diseases and list out practical suggestions to avoid such situations?
Answer:
Keeping Diseases Away Kerala Syllabus 10th

Sslc Biology Chapter 4 Questions And Answers Question 2.
Expand the list by including pathogenic microorganisms?
Answer:

  • Bacteria
  • Virus
  • Fungus
  • Protozoa

Biology Class 10 Chapter 4 Notes Kerala Syllabus Question 3.
How are pathogens transmitted from one person to another?
Answer:
Sslc Biology Chapter 4 Questions And Answers
Communicable diseases:
Communicable diseases are caused by the invasion of pathogens into the body and are transmitted from one person to another.

Rat Fever (Leptospirosis)

Rat fever is a bacterial disease. Bacteria is a prokaryote. They enter the body and multiply through binary fission. The toxins produced by them destroy cells and cause disease.

Hss Live Guru 10th Biology Kerala Syllabus Question 4.
What are the measures to be taken to prevent the transmission of rat fever?
Answer:
For controlling diseases like rat fever, eliminate the vectors and take preventive medicines as per the directions of experts. Disease affected persons must undergo diagnostic tests to confirm their illness and should take proper medicines. Ensure personal hygiene and environmental hygiene.

Biology Class 10 Chapter 4 Kerala Syllabus Other Bacterial Diseases

Diphtheria:
Biology Class 10 Chapter 4 Notes Kerala Syllabus
Analyse the information and figure related to diphtheria and prepare a table

Pathogen Corynebacterium diphtheriae
Body parts affected Mucus membranes of the nose and the throat
Symptoms Fever, throat pain and inflammation the lymp glands of the throat
Transmission of Diseases Cough, sneezing or directly from the infected person to another person

Tuberculosis:

Pathogen Mycobacterium tuberculosis.
Major Symptoms Loss of body weight, fatigue, persistent cough
Transmission of Disease When the patient speaks, coughs or sneezes, the pathogens spread into the air and thereby to others.
Organs/Body parts Affected Tuberculosis mainly affects the lungs. But kidneys, bones, joints, brain etc. are also affected
Treatment Vaccine By administering antibiotics BCG is used as preventive vaccine against tuberculosis.

Hsslive Guru 10th Biology Kerala Syllabus Question 5.
Mode of transmission of other bacterial diseases
Answer:

Diseases Mode of infection of Pathogenic bacteria
Cholera, Typhoid Through contaminated water
Tetanus Through wounds
Tuberculosis Thorugh air
Anthrax Through contact with animals
Botulism Through stale food
Gonorrhoea, Syphilis Through sexual contact

Hsslive Guru Biology 10th Kerala Syllabus Viral Diseases

Virus has the simple structure with a DNA or RNA molecule within a protein coat. Virus has no cell organelles as seen in normal cells. Hence virus multiplies by taking control over the genetic mechanism of the host cells. Viruses infect not only human beings but also plants, other animals and even bacteria.
Hss Live Guru 10th Biology Kerala Syllabus

Nipah Virus And Transmission Of The Disease:
Biology Class 10 Chapter 4 Kerala Syllabus
Virus has the simple structure with a DNA or RNA molecule within a protein coat. Virus has no cell organelles as seen in normal cells. Hence virus multiplies by taking control over the genetic mechanism of host cells. Viruses infect human beings, plants, animals and bacteria

AIDS:
AIDS (Acquired Immuno Deficiency Syndrome) is a condition of gradual decrease of immunity by the destruction of lymphocytes by the infection of HIV. HIV enters the body and multiplies using the genetic mechanism of lymphocytes. Hence the number of lymphocytes decreases considerably and reduces the immunity of the body. Various other pathogens which enter the bod^ in such a situation make the condition of AIDS even more fatal
Hsslive Guru 10th Biology Kerala Syllabus

Class 10th Biology Chapter 4 Notes Kerala Syllabus Question 6.
What are the ways by which HIV spreads Write your inferences in science diary by analysing illustration?
Answer:
Hsslive Guru Biology 10th Kerala Syllabus

Sslc Biology Chapter 4 Notes Kerala Syllabus  Question 7.
How AIDS does not spread
Answer:

  • By touch, shaking hands, coughing sneezing etc
  • through insects like mosquitoes, house flies etc.
  • by staying together and sharing food.
  • Through the reception of blood and organs contaminated with HIV

Class 10 Biology Chapter 4 Notes Kerala Syllabus  Question 8.
Do we need to fear AIDS patients? What should be our attitude towards them Discuss?
Answer:
Unlike other diseases, AIDS can be completely prevented by taking right precautions. AIDS is not a communicable disease. Therefore AIDS virus will not be transferred through social contacts such as touching or sharing food, and water. It is transmitted by sexual contact transfusion of blood, using unsterilized needle and syringe, use of blade and the razor in the barbershops and mother to foetus. So show positive attitude towards AIDS patients. To provide adequate care and attention to AIDS patients and make their relatives and.the public aware of it.

Hepatitis:
Hepatitis is a liver diseases it is also caused by vims. Inflammation of the liver is its major symptom. When the flow of the bile secreted by the liver is blocked, an increase in the level of bile pigment, bilirubin in blood is noticed. This imparts dark yellow colour to the mucous membrane, white portion of the eyes and the nails. This is the external symptom of diseases.

Mode of transmission:
Disease gets transmitted through contaminated food and water blood components and excreta of the patient.

Biology Chapter 4 Class 10 Kerala Syllabus Question 9.
Complete the table of viral diseases and its modes of transmission
Class 10th Biology Chapter 4 Notes Kerala Syllabus
Answer:
a) Through body fluids
b) Chickenpox, Dengue fever
c) Chikungunya, Dengue fever.
d) Rabies

Fungal Diseases

Fungi are of various types. Some fungi are pathogenic. The toxins produced by the fungi cause diseases. Ringworm, Athletes’ foot are fungal diseases.
Given below is the table showing fungal diseases, symptoms and mode of infection

Diseases Caused By Protozoa

Protozoans are unicellular eukaryotes. Malaria is an. example for diseases caused by protozoa.

Sslc Biology Chapter Wise Questions And Answers Question 10.
Prepare a note by analysing illustration and information on malaria.
Sslc Biology Chapter 4 Notes Kerala Syllabus
Answer:
Malaria is caused by the protozoan, plasmodium. Malaria is spread by female Anopheles mosquito. High fever with shivering and profuse sweating are the major symptoms of malaria. Other symptoms include headache, vomiting diarrhoea, anaemia etc. Hoste is man. Symptoms repeat intermittently

Filariasis:
Filariasis is caused by filarial worms that are spread by Culex mosquito. The worms stay in the lymph ducts and obstruct the flow of lymph by blocking the ducts. This causes swelling in the lymph ducts in the legs.

Question 11.
What are the preventive measures to be taken against communicable diseases? What is the significance of observing “Dry Day” in schools and at home? Discuss.
Answer:
Keep the surroundings clean

  • Personal hygiene
  • Community hygiene
  • Eradication of mosquitoes
  • Awareness programme – Poster, Rally.

Our carelessness and unclean surroundings are the main reason for the spreading of these types of contagious diseases. Contaminated surroundings promote the multiplication of pathogens and vectors. So it is necessary to avoid situations that lead to communicable diseases. So observing dry day is most effective remedial measure.

Non Pathogenic Diseases

Question 12.
Are diseases caused only by pathogens?
Are they caused by other factors too?
Answer:
No, the diseases are not caused only by pathogens. Many diseases are caused by other factors too. The other factors are

  • Lifestyle
  • Deficiency of nutrients
  • Occupational
  • Genetic

Question 13.
Observe the illustration. Give example of diseases for each category.
Class 10 Biology Chapter 4 Notes Kerala Syllabus
Answer:

Lifestyle diseases Cancer, Diabetes, Stroke
Genetic diseases Haemophilia, Sickle cell, anaemia
Deficiency of nutrients Anaemia, Goiter, Marasmus
Occupational diseases Pneumoconiosis, Silicosis, Asbestosis.

Genetic Diseases

Genes regulate cellular activities. The defects in genes also bring about diseases. Such diseases are called genetic diseases. Haemophilia, sickle cell anaemia etc., are examples of genetic diseases.

Haemophilia:
Haemophilia is the condition that excess blood is lost even through minorwoundsdueto synthesis of protein fails when the genes that control protein synthesis become defective. As haemophilia is a genetic disease a complete cure is not possible. Temporary relief is brought in by injecting the deficient protein identified through clinical diagnosis.

Some social organisations are working for haemophlilia patients who need special care and attention. These organisations volunteerto provide adequate care to haemophilia patients and make their relatives and the public aware of it.

Sickle cell Anaemia:
The defects of genes may also cause deformities in the sequencing of amino acids which are the building blocks of haemoglobin. As a result of this, the structure of haemoglobin changes and this in turn decreases its oxygen carrying capacity.
Biology Chapter 4 Class 10 Kerala Syllabus

Question 14.
Why do haemophilia patients loose blood excessively, through minor wounds?
Answer:
Bloods clots with the help of some proteins present in blood plasma. The synthesis of proteins fails when the genes that control protein synthesis become defective. Hence excess blood is lost even through minor wounds.

Question 15.
How does the deformity of red blood cells in sickle cell anaemia patients affect their body?
Answer:
The changes occur in the structure of haemoglobin in red blood cells is cause of these diseases. The blood cells bend like sickle. Decreases the oxygen-carrying capacity of red blood cells. The RBCs in the shape of sickle aggregate and block the flow of blood through the blood vessels.

Cancer:
Cancer is caused by the uncontrolled division of cells and their spread to other tissues. The normal cell gets transformed into cancerous cells when the control system of cell division fails. Various reasons such as environmental factors, smoking, radiations, virus, hereditary facts and alternations in genetic material lead to the transformation of normal cells into cancerous cells.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 13

Question 16.
Early identification of cancer helps the treatment easy. What are the treatments for cancer patients?
Answer:
Surgery, chemotherapy, radiation therapy etc. are used in the treatment of cancer.

Question 17.
Many voluntary organisations are involved in our society to offer better care to cancer patients. What are the importance of such activities?
Answer:
These organisations offer better care to cancer patients. Give love and care to the patients. Give nutritious food, clothes, medicine and proper care. This helps make awareness about people.

Lifestyle Diseases

Lifestyle diseases are caused by unhealthy living style. The changes in food habits, lack of physical exercise, stress, bad habits like consumption of alcohol, drug abuse, smoking etc. lead to various lifestyle diseases

Question 18.
Prepare a table showing lifestyle disease and their causes.
Answer:

Diabetes Deficiency of insulin or its malfunctioning
Fatty Liver Deposition of excess fat in the liver
Stroke Rupture of blood vessels in brain, block of blood flow
Hypertension Decrease in the diameter of arteries due to deposition of fat
Heart Attack Block of blood flow due to deposition of fat in coronary arteries.

Question 19.
Observe the following illustration on some of the health problems associated with smoking
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 14
Prepare a poster showing slogans against the habit of smoking.
Answer:

  • Cigarette smoking is injurious to health.
  • Smoking is a social crime.
  • Quit smoking, it kill you

Question 20.
Collect more information on the bad effects of smoking and write it in your science diary.
Answer:
Cigarette smoke contains poisonous substances. They badly harm the respiratory system. Carcinogens present in the smoke cause cancer in respiratory system. Bronchitis and Emphysema are diseases prevalent in smokers. Tobacco smoke injures every part of the respiratory system. It injures the lining of the nose, throat, trachea and lungs. Tobacco smokers, therefore , face a higher risk of lung cancer that non-smokers.

Animal Diseases

Question 21.
Prepare table showing animal diseases and their pathogen.

Disease Pathogen
Anthrax, Inflammation of udder (Mastitis) Bacteria
Foot and mouth diseases Virus

Plant Diseases

Pathogen

Disease

Bacteria Blight disease in paddy, Wilt disease in brinjal
Virus Mosaic disease in peas and tapioca, Bunchy top of banana
Fungus Quick wilt in pepper, Bud rot of coconut

Keeping Diseases Away Let Us Assess

Question 1.
Which among the following is not a bacterial disease?
a) Tuberculosis
b) Nipah
c) Diphtheria
d) Anthrax
Answer:
b) Nipah

Question 2.
“Food safety will be adversely affected with the spread of plant diseases”.
a) Do you agree with this statement? Why?
b) Give two examples of plant diseases?
Answer:
a) Yes, I agree with this statement. We cannot make agriculture profitable by the maximum utilisation of farmland. The main reason for decreasing food production is only the plant diseases.
b) (1) Blight diseases
(2) Bunchy top of banana

Question 3.
Prepare a pamphlet including the major measures to be taken to prevent rat fever.
Answer:
For controlling diseases like rat fever, eliminate the vectors and take preventive medicines as per the directions of experts. Disease affected persons must undergo diagnostic tests to confirm their illness and should take proper medicines. Ensure personal hygiene and environmental hygiene.

Question 4.
What is the importance of vaccination in preventing diphtheria?
Answer:
Antitoxins which act against the toxins produced by Corynebacterium diphtheria are used to protect the uninfected cells. But if the disease becomes severe the patient cannot be recovered through medication. So vaccination is the best preventive method.

Question 5.
What health habits should be adopted to prevent lifestyle diseases?
Answer:

  • Avoid the use of food containing higher level of fat and salt.
  • Control diabetes and high blood pressure time to time
  • Take steps to minimize mental pressure.
  • Avoid smoking and alcohol consumption.
  • Have regular exercise.

Keeping Diseases Away Extended Activities

Question 1.
Prepare a checklist and collect information on various lifestyle diseases. Find out the lifestyle diseases that are seen commonly. (Hints: Preparing a questionnaire, data collection and analysis)

Question 2.
Prepare and exhibit posters highlighting the fact that social hygiene is as important as personal hygiene.

Keeping Diseases Away More Questions and Answers

Question 1.
Complete the table
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 15
Answer:

Disease Pathogens Vector of the pathogen
Dengue gever Dengue virus Aedes mosquitoes
Chikungunya CHIKV Aedes mosquitoes
Malaria Protozoa Anopheles mosquitoes
Filariasis Filarial worms Culex mosquitoes

Question 2.
List the diseases spread by the mosquitoes? Point out the preventive measure.
Answer:
Malaria, Dengue fever, Filariasis, Chikungunya, Yellow fever, Japanese encephalitis, Zika fever etc. are the diseases spread by mosquitoes.
Preventive measures

  • Avoid the circumstances leading to spread of mosquitoes
  • Observe “Dry Day” once in two weeks.
  • Keep our surroundings clean.
  • Practice use of measures like mosquito nets.
  • Avoid the habit of littering.
  • Do not throw garbage’s in the water bodies.

Question 3.
Suppose cholera is being spread in a particular locality. Suggest the precautionary measures to be adopted in the locality?
Answer:
The bacteria causing cholera spread through water or housefly in the dirty atmosphere. Therefore the first and foremost requirement is to keep the surroundings neat and clean Facilitates of clean atmosphere, personal hygiene and pure drinking water should be made available in the cholera affected locality.

Question 4.
Expansion of DOTS is …………..
Answer:
Directly observed treatment short course.

Question 5.
Complete the table given below.

Disease Symptom Mode of infection
a) ring Worm …………… …………..
b) ………….. ………….. …………….

Answer:

a) Ringworm round red blisters on the skin spread through contact
b) Athletes foot appearance of reddish scaly on the sole of the foot and between the toes Contact with contaminated water and soil

Question 6.
During rainy season Athletes’ foot is common’. Give reason?
Answer:
Athlete’s foot is a skin disease caused by certain types of fungi. Pathogens enterthroughthe toes when they come in contact with contaminated water and soil. That is why in rainy season it is very common

Keeping Away Question 7.
Given below is the graph showing the result of a health survey conducted by students in a locality. Analyse the graph and answerthe following questions.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 16
a) Identify the most common disease in that locality?
b) What are the symptoms of that disease?
c) Name the pathogen of each category of the disease.
Answer:
a) Tuberculosis
b) Loss of body weight, fatigue, persistence cough
c) Dengue fever, chikungunya-virus Tuberculosis, cholera – Bacteria

Kerala Syllabus 10th Standard Biology Question 8. Compare Haemophilia and Sickle cell anaemia?
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 17

10th Standard Biology Question 9.  What does the following picture indicate?
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 18
Answer:
Sickle cell Anaemia. Red blood cells bend like a sickle so the oxygen carrying capacity of red blood cells decreases.

Question 10.
Indications about a disease are given in the box. Identify the disease
1. Defects of genes cause deformities in the sequencing of amino acids which are the building blocks of Haemoglobin.
2. Red blood cells bend like a sickle.
3. Decreases the oxygen-carrying capacity of red blood cells.
Answer:
Sickle cell Anaemia.

Question 11.
Smokers and tobacco users are not only subjected to cancer but also many other defects. Evaluate the statement?
Answer:
The statement is correct. There is possibility of diseases like stroke, bronchitis, hypertension etc. apart from cancer in case of frequent users of tobacco products.

Question 12.
Point out some lifestyle habits to prevent heart diseases.
Answer:

  • Avoid the use of food containing higher level of fat and salt.
  • Control diabetes and high blood pressure time to time
  • Take steps to minimize mental pressure.
  • Avoid smoking and alcohol consumption.
  • Have regular exercise.

Question 13.
“In smokers the ability of blood to receive oxygen is low”.
a) What is the reason for this?
b) Name any two toxic substances present in tobacco? (Model 2015)
Answer:
a) A stable compound, ‘carboxyhaemoglobin’, forms in blood and therefore the transport of oxygen reduces.
b) Nicotine, CO, benzopyrin, Carcinogens (any two)

Question 14.
Analyse the graph showing the diseases that affected the crops in John’s-field and answer the following questions.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 19
a) Identify the most affected crop based on the inference on most prevalent disease.
b) Name the causative agents of the diseases A and C.
c) Which type of insects spread the disease B and D. (Model 2015)
Answer:
a) Paddy
b) Quickwilt- Fungus, Blight disease – Bacteria
c) Aphids

Question 15.
Match the column B and C with column A.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 19
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 21

Question 16.
Diseases are caused not only by micro organisms. Many of the diseases are caused by bad habits”.
a) Give two suitable examples for the statement given.
b) Substantiate the statement based on the topic habits and diseases. (March 2014)
Answer:
a) Smoking, use of drugs, use of alcohol (any two)
b) Side Effects (three each)

Question 17.
Observe the part of an awareness notice given below:

Precautions:
1. Avoid unprotected sexual relationship
2. Avoid unsterilized injection needles
a) Identify the disease.
b) How unsterilized injection needles transmit this disease? (March 2014)
Answer;
a) AIDS
b) Unsterilized needles used by an AIDS patient contain blood in which HIV present. Transmission through body fluid.

Question 18.
Given below is a graphical representation of the data of individuals who came for treatment on July 2013 in a Government Hospital, Kerala. Analyse the graph and answer the following questions.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 22
a) Which disease is common in this locality?
b) Write a major symptom of this disease.
c) Prepare two posters to be exhibited in an awareness programme against this disease. (March 2014)
Answer:
a) Dengue Fever
b) Blood platelet count decreases internal bleeding and red rashes on the chest or face.
c) Two posters

Question 19.
The blood test of a patient appeared the red blood cells as shown in the figure
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 23
a) Identify the affected disease.
b) What are the reasons for this disease?
c) Write a symptom of this disease.
d) How does this disease affect our body? (Model 2013)
Answer:
a) Sickle cell anaemia.
b) Structural deformity in the haemoglobin molecules of the red blood cells due to genetic disorder
c) RBC become sickle-shaped cells.
d) Anaemia due to decreased level of oxygen in the blood and therefore difficulty in doing hard work.

Question 20.
Observe carefully the graph illustrating the data of a survey conducted in certain towns by the Health Department. Analyse the graph and answer to the following questions.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 24
a) Which type of mosquito is dominated in town A?
b) Identify the disease which is likely to be spread in town B. Name the pathogen which causes that disease.
c) Write an important symptom of that disease. (Model 2013)
Answer:
a) Culex mosquitoes
b) Malaria, Protozoa.
c) Severe fever recurring at interval and severe headache.

Question 21.
Name any two bacterial diseases. (March 2012)
Answer:
Any two, For example- Cholera, Typhoid, Rat fever, Leprosy.

Question 22.
Given below are symptoms of a disease.
i) RBC undergo change in shape and transform into sickle-shaped cells.
ii) Oxygen transport to the tissue is reduced.
a) Identify the disease
b) What is the reason for this disease? (March 2012)
Answer:
Sickle cell anaemia, a genetic disorder which deforms the RBC and they become crescent shaped and content of haemoglobin in them is highly reduced.

Keeping Diseases Away Questions and Answers

Question 1.
What do you mean by disease?
Answer:
A disease is a particular abnormal condition, a deviation from the normal functioning of the body systems.

Question 2.
Qn. 2
The causal’Study of disease is called ……………
Answer:
Pathology

Question 3.
Name some pathogens?
Answer:
Virus, bacterim,fungus, protozoa and worms.

Question 4.
Name some communicable diseases?
Answer:
Dengue fever, chikungunya, chickenpox, cholera, Anthrax etc.

Question 5.
Write 2 slogans for the poster in association with mosquitoes day on August 20.
Answer:

  • Keep our surroundings clean.
  • Observe “Dry Day” once in two weeks

Question 6.
Prepare a flow chart showing the mode of infection of dengue fever.
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 25

Question 7.
Given belowthe picture of aedes mosquito name two diseases caused by this mosquito.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 26
Answer:
Denque fever, chikungunya

Question 8.
What is the structure of a virus?
Answer:
Virus has the simple structure with a DNA or RNA molecule within a protein coat.

Qn. 9
Name some viral diseases?
Answer:
Dengue fever, chikungunya, Ebola, chickenpox, SAARS, AIDS, Rabies.

Qn. 10
How does vims multiply?
Answer:
Virus multiplies by taking control over the genetic mechanism of host cells. During this process, the cell gets damaged leading to disease.

Qn. 11
“Virus requires a host cell to get multiplied” Give reason?
Answer:
Virus has no cell organelles as seen in normal cells It has the simple structure with a DNA or RNA molecule within a protein coat.

Question 12.
Identify the picture.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 27
Answer:
HIV virus (Human Immuno Deficiency Virus)

Question 13.
Expansion of AIDS?
Answer:
Acquired Immuno Deficiency Syndrome

Question 14.
Expansion of HIV?
Answer:
Human Immuno Deficiency vims

Question 15.
What is AIDS?
Answer:
AIDS (Acquired Immuno Deficiency Syndrome) is a * condition of gradual decrease of immunity by the destruction of lymphocytes by the infection of HIV (Human Immuno Deficiency Virus). HIV enters the body multiplies using the genetic mechanism of lymphocytes.

Question 16.
Read this newspaper cutting
“AIDS-affected children were denied admission in School”
What is your opinion about the decision of the school authorities? Is AIDS a communicable disease?
Answer:
The decision taken by the school authority to refuse admission for AIDS-affected children is totally wrong. This is because AIDS is not a communicable disease. Therefore AIDS vims will not be transferred through touching or sharing food, water, air etc.

Question 17.
Do you think that AIDS will spread through mosquitoes and houseflies?
Answer:
No

Question 18.
The following red ribbon is the international symbol to represent a disease identify the disease?
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 28

Question 19.
What are the precautions that can be taken to prevent the spread of virus causing AIDS?
Answer:?
Take precaution before receiving blood. Do not use syringe and needles already used. Have safety in sexual relationship.

Question 20.
Though bacteria are generally helpful, some of them are harmful. How?
Answer:
Bacteria multiply through binary fission immediately after entering the body. The toxins produced by these bacteria damage living cells and thereby bring
about diseases.

Question 21.
Name certain bacterial diseases?
Answer:
Cholera, Typhoid, Tetanus, Tuberculosis, Anthrax, Botulism, Gonorrhoea and Syphilis.

Question 22.
Complete the table of bacterial disease
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 28
Answer:
a) Cholera, Typhoid
b) Through wounds
c) Tuberculosis
d) Anthrax
e) Botulism
f) Through sexual contact

Question 23.
AIDS is a viral disease caused through sexual relations. Give examples of two bacterial diseases that are caused by similar manner?
Answer:
Gonorrhoea, Syphilis

Question 24.
Name a bacterial disease spread through air?
Answer:
Tuberculosis

Question 25.
……………… is a bacterial disease that affected the lungs.
Answer:
Tuberculosis.

Question 26.
Identify the bacteria which causes Tuberculosis?
Answer:
Mycobacterium tuberculosis.

Question 27.
………….. treatment is recommended for tuberculosis by World Health Organisation.
Answer:
DOTS

Question 28.
What are the major symptoms of Tuberculosis?
Answer:
Loss of body weight, fatigue, persistent cough are the major symptoms of tuberculosis.

Question 29.
Tuberculosis affects the lungs only’. Do you agree with this statement? Substantiate your answer?
Answer:
Tuberculosis mainly affects the lungs. It also affects kidneys, bones, joints, brain etc by this disease.

Question 30.
…………… is the vaccine used against tuberculosis across the world
Answer:
BCG vaccine

Question 31.
DOTS treatment is given to which category of patients?
Answer:
Patients suffering from Tuberculosis.

Question 32.
Write slogans to give awareness for controlling tuberculosis?
Answer:

  • Test your sputum if you have prolonged cough fora couple of weeks.
  • Take BCG vaccination immediately after the birth of children.

Question 33.
Name a bacterial disease which can be spread through castles?
Answer:
Anthrax

Question 34.
What is Botulism?
Answer:
Food poisoning caused by the growth of bacteria in improperly sterilized tinned meats and other preserved foods.

Question 35.
Give examples of fungal diseases?
Answer:
Ringworm. Athletes’ foot

Question 36.
How does fungal diseases spread?
Answer:
Fungal diseases spread through contact.

Question 37.
How does fungi cause diseases?
Answer:
The toxins produced by the pathogenic fungi cause diseases.

Question 38.
Malaria is caused by
Answer:
The protozoa plasmodium

Question 39.
Vector of malaria disease?
Answer:
Female Anopheles mosquitoes.

Question 40.
Mention the symptoms of malaria?
Answer:
High fiver with shivering and profuse sweating are the major symptoms of malaria. Other symptoms include headache, vomiting, diarrhoea, anaemia etc.

Question 41.
Filariasis is spread by mosquitoes.
Answer:
Culex mosquitoes

Question 42.
Filariasis is caused by worms.
Answer:
Filarial worms

Question 43.
Observe the figure and answer the following questions.
a) Name the disease?
b) Pathogen?
c) Vector?
d) How to prevent the disease?
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 30
Answer:
a) Filariasis
b) Filarial worms
c) Culux mosquitoes
d) Mosquito Control

Question 44.
Find the odd one out
a) Dengue fever, malaria, rabies, filariasis
b) Dengue fever, Rabies, chikungunya, Malaria,
Answer:
a) Rabis – rabies spread through animals, others spread through mosquitoes,
b) Malaria – Malaria is caused by protozoa, others are caused by virus.

Question 45.
Compare Malaria and filariasis with the help of the following indicators.
1. Pathogen
2. Vector
3. Symptoms
Answer:

Malaria Filariasis.
Pathogen Protozoa plasmodium Filarial worms
Vector Anopheles mosquito Culex mosquito
Symptoms High fever with shivering headache, vomiting diarrhoea sweating obstruct the flow of lymph by blocking the lymph ducts and that causes swelling in the lymph ducts in the legs

Question 46.
Give reasons “The legs of Filariasis patients’ swell?
Answer:
Because the filarial worms stay in the lymph ducts and obstruct the flow of lymph by blocking the ducts. This causes swelling in the lymph ducts in the legs.

Question 47.
Give example for a few diseases which are not communicable to person to person?
Answer:
Haemophilia, Sickle cell Anaemia, Cancer, Diabetes, Stroke.

Question 48.
Complete the word relation.
Diabetes: Lifestyle disease
Haemophilia: …………….
Answer:
Genetic disease

Question 49.
Differnciate Anaemia and Sickle cell Anaemia.
Answer:
Anaemia is a condition that develops when your blood lacks enough oxygen content due to the lack of iron content in the red blood cells. On the other hand, sickle cell anaemia is a serious inherited blood disorder where the red blood cells which carry oxygen around the body, develop abnormally. In cause of sickle cell anaemia changes occurs in the structure of haemoglobin in red blood cells. Red blood cell bend like a sickle and hence it is called sickle cell anaemia.

Question 50.
The condition that excess blood is lost even through minor wounds is known as ……………
Answer:
Haemophilia

Question 51.
World Haemophilia Day is observed in ……………….
Answer:
April 17

Question 52.
What do you mean by Haemophilia? What causes it? Is it curable by treatment?
Answer:
Haemophilia is the condition that excess blood is lost even through minor wounds due to synthesis of protein fails when the genes that control protein synthesis become defective. As haemophilia is a genetic disease a complete cure is not possible. Temporary relief is brought in by injecting the deficient protein identified through clinical diagnosis.

Question 53.
Why is sickle cell anaemia called a “Sickle disease”?
Answer:
In case of sickle cell, anaemia changes occur in the structure of haemoglobin in red blood cells. Red blood celsl bend like a sickle and hence it is called sickle cell anaemia.

Question 54.
Smoking causes ………………. cancer
Answer:
Lung cancer

Question 55.
Compare a patient of sickle cell anaemia with a healthy person?
Answer:
Compared to a healthy person, of a person having sickle cell anaemia decreases the oxygen carring capacity.

Question 56.
How does smoking affect the following organs? (Brian, Lungs and Heart)
Answer:
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 31

Question 57.
Identify the health issues caused by alcohol consumption?
Answer:
1) Causes kidney disease, ulcers, liver cancer, fatty liver etc.
2) Contributes to high blood pressure and stroke
3) Increases blood sugar levels which makes diabetes worse.

Question 58.
Give examples for diseases caused by unhealthy lifestyle?
Answer:
Diabetes, fatty liver, high blood pressure, stroke, heart attack.

Question 59.
‘India – world capital of Diabetes’ – Explain this news headline?
Answer:
Due to unhealthy food and lifestyle habits, the number of diabetic patients sharply increased in India. Apart from over consumption of food, absents of adequate exercise also contribute to diabetes.

Question 60.
‘Diabetes and high blood pressure are silent killers’ Why?
Answer:
Diabetes and hypertension can damage the kidneys. It increases the risk of blindness and dementia. Untreated hypertension increases the risk of heart disease and stroke. Due to these reasons, diabetes and hypertension are called silent killers.

Question 61.
Name some disease that affects the cattle?
Answer:
Anthrax, Foot and mouth disease, Inflammation of udder.

Question 62.
………………….. is the pathogen of Anthrax
Answer:
Bacillus Anthracis (Bacteria)

Question 63.
Complete the table of diseases that affect the domestic animals.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 32
Answer:
a) Virus – High fever, Blisters in the mouth and on feet, Drop in milk production, weight loss
b) Sudden fever, diarrhoea, mouth sores, nausea and vomiting
c) Inflammation of udder – Bacteria – swelling of udder, drop in milk production.

Question 64.
Complete the word relation?
Cattle : Antrax: Bacteria
Cattle: Foot and mouth diseases: ………………
Answer:
Virus

Question 65.
Prepare a table showing important plant diseases and their pathogen?
Answer:

Blight disease of paddy, Wilt disease of brinjal Bacteria
Mosaic disease in peas and tapioca, bunchy of banana Virus
Quickwilt in pepar Budrot of coconut Fungus

Question 66.
Complete the word relation
Paddy: Blight: Bacteria
Coconut : Budrot:………..
Answer:
Fungus

Question 67.
………………. is a viral disease which spreads from wild animals to human beings.
Answer:
Ebola

Question 68.
Where and when did Ebola was first idientified?
Answer:
Central Africa in March 2014.

Question 69.
Point out the symptoms of Ebola?
Answer:

  • Severe fever
  • Muscle pain
  • Vomiting
  • Diarrhoea
  • Internal and external bleeding

Keeping Diseases Away SCERT Questions and Answers

Question 1.
Polluted surroundings and stagnation of water lead to the multiplication of mosquitoes. Mosquitoes carry many diseases to man. One such disease results in the considerable decrease in the number of platelets.
a) Name the disease.
b) Which microbe causes this disease? (Question Pool 2017)
Answer:
a) Dengue
b) Dengue virus

Question 2.
Rearrange columns B & C suiting the pictures in column A. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 33
Answer:
i- b, R
ii – c, P
iii – a, Q

Question 3.
Choose the statements related to virus from those given below.
a) Multiplies by binary fission.
b) Has a simple structure with a DNA or RNA molecule within a protein coat.
c) Toxins produced by them damage living cells.
d) Multiplies by taking control over the genetic mechanism of host cells. (Question Pool 2017)
Answer:
b, d

Question 4.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 34
Shown above is one of the presentation slides prepared by Pradeep, as part of the International year of Microbes.
a) Which is the microbe mentioned in the slide?
b) Which disease is caused by this microbe?
c) How does this affect the immunity of the body?
Answer:
a)-HIV
b) – AIDS
c) 1. Multiplies using the genetic mechanism of lymphocytes.
2. Considerable decrease in the number of lymphocytes.
3. Disrupts immunity.

Question 5.
Analyse the statements related to the spread of AIDS and classify them suitably.
a) Through mosquitoes & houseflies.
b) Through body fluids.
c) Through extramarital sexual contact.
d) By touch, shaking hands, coughing etc.
e) From HIV infected mother to foetus.
f) When you sit near an HIV infected friend in the school.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 35
Answer:

Situations where  HIV spreads Situations where  HIV does not spread
(b) (a)
(c) (d)
(e) (f)

Question 6.
Complete the illustration of bacterial diseases given below. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 36
Answer:
A – Tuberculosis
B – Contaminated water/food
C – Tetanus

Question 7.
An interview with the doctor of the Primary Health Centre regarding tuberculosis, organised by the Health Club is given below. What is your explanation for the questions asked by the children?
a) Which bacteria causes this disease.
b) Write two symptoms of this disease.
c) Name the vaccine used to prevent tuberculosis. (Question Pool 2017)
Answer:
a) Mycobacterium tuberculosis
b) Loss of body weight, Fatigue, Persistent cough.
c) BCG

Question 8.
a) What is common between the diseases given in the box?
b) Pick out the odd one. Justify. (Question Pool 2017)
Answer:
a) Diseases spread through air
b) Tuberculosis is a bacterial disease and rest all virus diseases

Question 9.
The result of a survey conducted by the Health Department on mosquito-borne diseases is shown in the graph. Analyse the graph and answer the questions.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 37
a) Which disease affects more number of people?
b) Write symptoms of the disease B.
c) Suggest suitable measures to prevent the spread of the diseases shown in the graph. (Question Pool 2017)
Answer:
a) Filarial worm
b) High fever with shivering, profuse sweating, headache, vomiting, diarrhoea, anaemia etc.
c) Dengue fever, Malaria, Filariasis are the diseases spread by mosquitoes. So avoid the circumstances leading to spread of mosquitoes. Observe ‘Dry Day’ once in two weeks, keep our surroundings clean, practise use of measures like mosquito nets.

Question 10.
Make suitable pairs using the information given in the box.
Diabetes,
Goitre,
Lifestyle,
Haemophilia,
Deficiency of nutrients,
Genetic. (Question Pool 2017)
Answer:

  • Diabetes – lifestyle
  • Goitre – deficiency disease
  • Haemophilia – genetic

Question 11.
Anjanagets wounded on her foot while playing with her friends. Due to continuous bleeding, her parents take her to the hospital. The doctor’s diagnosis after thorough investigation is given below.
“This has happened as the blood is not clotting. This is a genetic disease.”
a) What is Anjana’s disease?
b) How can temporary relief be brought about for the disease? (Question Pool 2017)
Answer:
a) Haemophilia
b) Injecting the deficient protein helps in blood clotting.

Question 12.
Change in the shape of RBC due to genetic disease is shown in the figure.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 38
a) Name the disease shown in the figure.
b) How does the deformity of RBC affect the body?
Answer:
a) Sickle-cell anaemia
b) 1. Decreases the oxygen-carrying capacity of red blood cells.
2. Sickle-shaped RBC aggregate and block the blood flow through blood vessels.

Question 13.
Identify the given symptoms and tabulate them with the name of the disease as headings.
a) The deformity in the sequencing of amino acids of haemoglobin due to the defect of genes.
b) The defect in the production of protein for blood clotting.
c) Excess loss of blood even though a minor wound.
d) The oxygen-carrying capacity of red blood cells decreases. (Question Pool 2017)
Answer:

Haemophilia Sickle cell anaemia
b a
c d

Question 14.
You are invited to prepare a presentation slide for the Cancer awareness class, conducted by the Health Club. What explanation will you give to the ideas given below?
a) The disease cancer
b) Reasons for cancer
c) Treatment for cancer (Question Pool – 2017)
Answer:
a) Cells undergo uncontrolled division and spread to other tissues.
b) Environmental factors, smoking, radiations, virus, hereditary factors alteration of genetic materials, chemical substances.
c) Chemotherapy, surgery, radiation therapy

Question 15.
An early diagnosis of the disease is crucial in the treatment of cancer. Why? (Question Pool 2017)
Answer:
It is difficult for the patients to recover if the disease becomes severe.

Question 16.
Choose the ones related to lifestyle diseases from those given below.
a) Hereditary factors
b) Lack of exercise
c) Mental stress
d) Environmental factors ‘
e) Change in food habits
f) Alcoholism, smoking
Answer:
b, C, e, f

Question 17.
Arrange the indicators given in column B suitable to column A.

Disease Cause
A. Stroke i. deficiency of insulin or its malfunctioning.
B. Diabetes ii. deposition of excess fat in the liver.
C. Fatty liver iii. blockage of blood flow in the brain.
iv. decrease in the diameter of the artery due to deposition of fat.

Answer:
A – iii
B – i,
C – ii

Question 18.
Some health issues due to smoking are given below. Name the affected organ. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 39
Answer:
A – brain
B – lungs
C – any organ

Question 19.
Make a poster to provide awareness about the health issues due to smoking. (Question Pool 2017)
Answer:
Smoking is injurious to health

Question 20.
Classify the diseases given below into animal diseases and plant diseases. (Question Pool 2017)
Anthrax,
Blight disease,
Quick wilt,
Foot and
mouth disease,
Inflammation of udder
Answer:

Animal diseases Plant diseases
Anthrax Blight disease
Foot and mouth disease Quick wilt
Inflammation of the udder Bunchy top of banana

Question 21.
A study of the Agriculture Department on plant diseases in a panchayath is given below as a graph. Analyse this and answer the questions. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 40
a) Which is the mostly affected crop?
b) Name the disease that affects pepper.
c) What are the fungal diseases that affected the plants of that area?
Answer:
a) Paddy
b) Quick wilt
c) Bud rot; quick wilt

Question 22.
Analyse the slogan given in the placard and answer the questions
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 41
a) What is health?
b) What should be our attitude to patients? (Question Pool 2017)
Answer:
a) The complete physical, mental and social wellbeing of a person.
b) Compassion, mercy, sympathy, empathy, pity, service mindedness and helping mentality etc.

Question 23.
Given below is a doubt raised by Asna when she notices that lungs are mentioned in an advertisement against smoking.
“Does smoking affect only the lungs?”
As a science student, what explanation will you give for this question? (Question Pool 2017)
Answer:

  • No
  • Affects the Brain, Heart and Blood Vessels.

Question 24.
Nandu: Smoking causes cancer.
Mahesh: Smoking causes lifestyle diseases. Analyse the possibilities of these two statements and write your interpretation. (Question Pool 2017)
Answer:
The statements of both of them are correct. Smoking causes cancer and lifestyle diseases like stroke and hypertension.

Question 25.
Choose the correct statement.
a) Malaria, Filariasis, Cholera etc. are spread by mosquitoes.
b) Anthrax & Rabies are transmitted from animal to man.
c) Tuberculosis, SARS and chickenpox spread through air.
d) Syphilis, Gonorrhoea and Botulism spread (Question Pool 2017)
Answer:
b, c

Question 26.
Analyse the diseases given below and arrange them suitably in the boxes provided,
a. Blight
b. Botulism
c. Ebola
d. Inflammation
e. Quick wilt
f. Foot and of udder mouth disease

Plants Animals Man

Answer:
(a)
(d)
(b)
(e)
(f)
(c)

Question 27.
Classify the diseases given in the box suitably,
a. Anthrax
b.AIDS
c. Bud rot
d. Foot and mouth
e. Athletes foot
f. Tetanus disease
Answer:
Bacteria – (a), (f)
Virus – (b), (d)
Fungus – (c), (e)

Question 28.
Classify the diseases given below based on the mode of transmission. (Question Pool 2017)
1. SAARS,
2. Chikungunya,
3. AIDS,
4. Gonorrhoea,
5. Malaria,
6. Dengue,
7. Syphilis,
8. Chickenpox,
9. Tuberculosis
Answer:

Through air Through sexual contact Through Mosquitoes
1. SAARS 1. Syphilis 1. Dengue fever
2. Chickenpox 2. Gonorrhoea 2. Malaria
3. Tuberculosis 3. AIDS 3. Chikungunya

Question 29.

A B C
i. Blight a. Virus P. Pepper
ii. disease b. Bacteria Q.  Paddy
ii. Quick wilt c. Fungus R. Banana

Answer:
i) – (b) – (Q)
(ii) – (c) – (P)
(iii) – (a) – (R)

Question 30.
A few characteristic features of microorganisms are given below. Analyse them and complete the illustration. (Question Pool 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 42
Answer:
A – (iii)
B – (ii)
C – (iv)

Question 31.
Cause of all diseases are microorganisms. All microorganisms are pathogens. Evaluate this statement and justify your answer giving suitable examples. (Question Pool 2017)
Answer:

  • Do not agree with this statement.
  • Diseases occur without pathogens also, etc. diabetes/stroke/ cancer etc.
  • All micro organises are not pathogens. There are useful microorganisms also.

eg: Bacteria seen in the intestine and skin.

Question 32.
Read the statements and answer the questions given below.
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 43
a) Identify the disease?
b) Identify the pathogen?
c) Give scientific explanation against the tendency of isolating these persons from the society?
Answer:
a) AIDS
b) HIV (Human Immunodeficiency Virus)
c) This tendency is wrong. HIV spreads only through body fluids, but not by touching, biting, kissing, sneezing, cough, drinking or eating food.

Question 33.
“Unhealthy lifestyle invites diseases”. Justify the statement with examples. (Orukkam 2017)
Answer:
Unhealthy lifestyle may cause diseases like diabetes, fatty liver, hypertension, stroke and heart attack. Smoking may cause lung cancer, bronchitis, emphysema and decrease in the ability of heart.

Question 34.
The extent of the disease affected crops in Rajesh’s farm is represented graphically. Analyse the graph and answer the given questions. (Orukkam – 2017)
Kerala Syllabus 10th Standard Biology Solutions Chapter 4 Keeping Diseases Away - 44
a) Identify the most affected crop.
b) Identify the least affected crop.
c) Pair the pathogen and diseases of affected crops,
Answer:
a) Banana
b) Pepper
c) 1. Blight – Bacteria
2. Wilt – Bacteria
3. Bunch top – virus
4. Quick wilt – Fungus

Question 35.
a) What is cancer?
b) How normal cells get transformed into cancerous cells?
c) What are the methods adopted in cancer treatment? (Orukkam 2017)
Answer:
a) Cancer is the condition by which uncontrolled division of cells and their spread to other tissues
b) Environmental factors, smoking, radiations, viruses, hereditary factors and alteration in genetic materials.

Question 36.
The symptoms of a communicable disease are given below.
Loss of body weight, fatigue, persistent cough
a) Name the disease?
b) Identify the pathogen?
c) How this disease is transmitted? (Orukkam 2017)
Answer:
a) Tuberculosis
b) Mycobacterium tuberculosis (Bacteria)
c) Through air

Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles

You can Download Circles Questions and Answers, Activity, Notes, Kerala Syllabus 10th Standard Maths Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Maths Solutions Chapter 2 Circles

Circles Text Book Questions and Answers

Textbook Page No, 42

Circles Class 10 Kerala Syllabus Question 1.
Suppose we draw a circle with the bottom side of the triangles in the picture as diameter. Find out whether the top corner of each triangle is inside the circle, on the circle or outside the circle.
Circles Class 10 Kerala Syllabus
Answer:
Angle of the first triangle =110°
As 110° > 90° the top comer will be inside the circle
Angle of second triangle = 90°
∴ The top comer will be on the circle.
Angle of the third triangle = 70°
70° > 90°
∴ The top comer will be outside the circle

Sslc Maths Circles Questions And Answers Question 2.
For each diagonal of the quadrilateral shown, check whether the other two corners are inside, on or outside the circle with that diagonal as diameter
Sslc Maths Circles Questions And Answers
Answer:
The fourth angle in ABCD
= 360 – (110 + 105 + 55) = 360
Sslc Maths Chapter 2 Kerala Syllabus
Drawing diagonal AC and taking it as diameter of a circle As, ∠D = 90°
D will be on the circle. ∠B = 55° (90 > 55°)
∴ B will be outside the circle
Drawing diagonal BD and taking it as diameter of a circle.
∠A = 105°, ∠C = 110°
As both angles are greater than 90, they lie inside the circle.

Sslc Maths Chapter 2 Kerala Syllabus Question 3.
If circles are drawn with each side of a triangle of sides 5 centimetres, 12 centimetres and 13 centimetres, as diametres, then with respect to each circle, where would be the third vertex?
Answer:
As the sides are 5, 12, 13 cm and also
52 + 122 = 25 + 144 = 169 = 132
∴ ABC is a right triangle
Circles Class 10 Scert Solutions Kerala Syllabus
Taking BC as diameter and drawing a circle, ∠A (<90°), A will be outside the circle.
Taking AB as diameter and drawing a circle ∠C (<90°) C will be outside the circle.
Taking AC as diameter and drawing a circle, ∠B = 90°, B will be on the circle.

Circles Class 10 Scert Solutions Kerala Syllabus Question 4.
In the picture, a circle is drawn with a line as diameter and a smaller circle with half the line as diameter. Prove that any chord of the larger circle through the point where the circles meet is bisected by the small circle.
Sslc Maths Circles Questions And Answers Pdf
Answer:
∠ADO = ∠APB = 90°
(angle subtended by diameter is always 90°)
⇒ OD\\PB
Maths Chapter 2 Class 10 Kerala Syllabus
AO = OB (Radius of bigger circle)
(If in a triangle, the line drawn from midpoint of one side, is parallel to another side, then the line will bisect the third side)
Therefore AD = DP
(AB’s midpoint is ‘O’ and OD\\PB)

Sslc Maths Circles Questions And Answers Pdf Question 5.
Sslc Maths Circles Notes Kerala Syllabus
Use a calculator to determine up to two decimal places, the perimeter and the area of the circle in the picture.
Answer:
Sslc Maths Chapter 2 Circles Questions And Answers

Maths Chapter 2 Class 10 Kerala Syllabus Question 6.
The two circles in the picture cross each other at A and B. The points P and Q are the other ends of the diameters through A.
Sslc Maths Chapter 2 Circles Notes Kerala Syllabus
i. Prove that P, B, Q lie on a line.
ii. Prove that PQ is parallel to the line joining the centres of the circles and is twice as long as this line.
Answer:
Circles Class 10 Notes Pdf Kerala Syllabus
i. ∆ PAB (angle subtended by semicircle)
∠PBA = 90°
∆ ABQ be a triangle on the semi¬circle of centre D.
∴ ∠ABQ = 90°
∠PBA + ∠ABQ = 180 (Linear pair)
As AP, AQ are diameters of the circle. PQ be the line drawn through B per¬pendicularly to AB. Therefore P, B, Q lies on the same line.

ii.
Class 10 Maths Chapter 2 Circles Kerala Syllabus

Sslc Maths Circles Notes Kerala Syllabus Question 7.
Prove that the two circles drawn on the two equal sides of an isosceles triangle as diameters pass through the midpoint of the third side.
Answer:
10th Class Maths Chapter 2 Circles Kerala Syllabus
∠ADB= 90° (AABD angle subtended by semicircle)
∠CDA = 90°
∴ ∠ADB +∠CDA = 180° (linear pair)
∆ ABD ∆ ADC are right angled triangles.
In ∆ ABD
BD2 = AB2 – AD2 ( AB = AC )
= AC2 – AD2 = DC2
BD = CD

Sslc Maths Chapter 2 Circles Questions And Answers Question 8.
Prove that all four circles drawn with the sides of a rhombus as diameters pass through a common point.
Sslc Maths Chapter 2 Circles Kerala Syllabus
Prove that this is true for any quadrilat¬eral with adjacent sides equal, as in the picture.
Circles Class 10 Notes Kerala Syllabus
Answer:
Class 10 Maths Chapter 2 Kerala Syllabus
ABCD is a rhombus so diameter are perpendicular bisectors.
∠AOD = 90°
O be on the circle having diameter AD.
∠AOB = 90°, therefore
O be on the circle having diameter AB.
∠BOC= 90°, therefore
O be on the circle having diameter BC
∠DOC= 90°, therefore
O be on the circle having diameter DC
O be the common point on the circle.
∠A0D = ∠AOB and
∠COD = ∠BOC and,
AD = AB,
AO be the common side.
Circles Class 10 Notes State Syllabus
Δ AOD, Δ AOB are equal triangles.
OD = OB
Both the circles can passed through O.
A BCD is an isosceles triangle.
Those circles having diameters CD and BC are passing through midpoint of BD.
∴ O be common for the four circles. (Diameter)

Sslc Maths Chapter 2 Circles Notes Kerala Syllabus Question 9.
A triangle is drawn by joining a point on a semicircle to the ends of the diameter. Then semicircles are drawn with the other two sides as diameter.
Kerala Syllabus 10th Standard Maths Chapter 2
Prove that the sum of the areas of the blue and red crescents in the second picture is equal to the area of the triangle.
Answer:
Kerala Syllabus 10th Standard Maths Circles
Area of triangle = \(\frac { 1 }{ 2 }\) × 2r × h = rh
Area of the semicircle = \(\frac{\pi r^{2}}{2}\)
Area of the rest of the figure
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 19
The diameters of the red and blue semi-circles are the sides of the two triangles.
∴ Their areas
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 20
Area of red and blue crescents
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 21
= area of the triangle

Textbook Page No. 53

Circles Class 10 Notes Pdf Kerala Syllabus  Question 1.
In all the pictures given below, O is the centre of the circle and A, B, C are points on it. Calculate all angles of Δ ABC and Δ OBC in each.
Circles Class 10 State Syllabus Kerala Syllabus
Answer:
a.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 23
OA = OB (radii)
∴ ∠OAB = 20° ; (∠OAB = ∠OBA)
OC = OA (radii) ;
∠OAC = 30°
∠BAC = ∠OAB + ∠OAC
= 20 + 30 = 50°
∠BOC = 2x ∠BAC = 100°
OB = OC (radii)
∴ ∠OBC = ∠OCB = 40°
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 24
Angles of triangle ABC are ∠BAC = 50° ∠ABC = 60°, ∠ACB = 70°
Angles of ∆ BOC are
∠OBC = 40°, ∠OCB = 40°, ∠BOC = 100°

b.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 25
Angles of ∆ ABC are ∠ABC = 50°, ∠BAC = 60°, ∠BCA = 70°
Angles of ∆ AOC are
∠OAC = 40°, ∠AOC = 100°, ∠OCA = 40°

c.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 26
∠ACB = 180 – 55 = 125° ; OA = OC
∠CAO = ∠ACO = 70° ; ∠OBC = ∠BCO = 55° Angles of A OBC are
∠OBC = 55° ∠COB = 70° ∠BCO = 55°
Angles of ∆ ABC
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 27

Class 10 Maths Chapter 2 Circles Kerala Syllabus Question 2.
The numbers 1,4,8 on a clock’s face are joined to make a triangle.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 28
Calculate the angles of this triangle.
How many equilateral triangles can we make by joining numbers on the clock’s face?
Answer:
The angle subtended by two adjacent numbers at the centre of the clock is 30°
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 29
We can make 4 equilateral triangles by joining the numbers on the clock (1. 5, 9), (2, 6, 10), (3. 7, 11), (4, 8, 12)

10th Class Maths Chapter 2 Circles Kerala Syllabus Question 3.
In each problem below, draw a circle and a chord to divide it into two parts such that the parts are as specified.
i. All angles on one part 80°.
ii. All angles on one part 110°.
iii. All angles on one part half of all angles on the other.
iv. All angles on one part, one and a half times the angles on the other.
Answer:
i. ∠AOB = 160°
Therefore all angles in the are ACB arc 80°.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 30

ii. Draw angle as central angle 220° so angle on the small arc AB will be 110°.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 31

iii. Draw angle as central angle 120° or 240° All angles on one part will be 120°, and other part be 60°.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 32

iv. Draw a circle and draw central angle 144° All angles on the part APB will be 120° and All angles on the part AQB will be 108°.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 33

Sslc Maths Chapter 2 Circles Kerala Syllabus Question 4.
A rod bent into an angle is placed with its corner at the centre of a circle and it is found that \(\frac { 1 }{ 10 }\) of the circle lies within it. If
it is placed with its corner on another circle, what part of the circle would be within it?
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 34
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 35

Circles Class 10 Notes Kerala Syllabus Question 5.
In the picture, O is the centre of the circle and A, B, C are points on it. Prove that
∠OAC + ∠ABC = 90°
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 36
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 37

Class 10 Maths Chapter 2 Kerala Syllabus Question 6.
Draw a triangle of circumradius 3 centimetres and two of the angles 32\(\frac { 1° }{ 2 }\) and 37\(\frac { 1° }{ 2 }\)
Answer:
Draw a circle with radius 3 cm and central angle 65°.
Half of 65° is 32\(\frac { 1° }{ 2 }\) Similarly we can draw 75°
Join the points A, B and C Halfof75°is 373\(\frac { 1° }{ 2 }\) Complete the triangle.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 38

Circles Class 10 Notes State Syllabus Question 7.
In the picture, AB and CD are mutually perpendicular chords of the circle. Prove that the arcs APC and BQD joined together would make half the circle.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 39
Answer:
If ∠ADC = x
∠AOC =2x
(Angle subtended on the alternate arc is half of the central angle of arc )
If ∠BAD = y
∠BOD= 2y
∠AOC + ∠BOD = 2x + 2y
= 2 (x + y) = 2 × 90 = 180°
∴ The arcs APC and BQD joined together will make half the circle.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 40

Kerala Syllabus 10th Standard Maths Chapter 2  Question 8.
In the picture, A, B, C, D are points on a circle centred at O. The lines AC and BD are extended to meet at P. The line and BC intersect at Q. Prove that the angle which the small are AB makes at O is the sum of the angles it makes at P and Q.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 41
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 42

Textbook Page No. 59

Kerala Syllabus 10th Standard Maths Circles Question 1.
Calcula te the angles of the quadrilateral in the picture and also the angles between their diagonals:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 43
Answer:
Since
∠ACD = 30°
∠ABD = 30°
(Angle in the same segment of a circle)
Since ZCBD = 45°
∠CAD = 45°
Since ZBDC = 50°
∠BAC = 50°
∠ABC + ∠ADC = 180 (cyclic quadrilateral)
∠ABC = 75°
∴ ∠ADC = 180 – 75 = 105°
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 44
∠ADB = 105 – 50 = 55°
As, ∠BAD = 95°
∠DCB = 180 – 95 = 85°
∴ ∠ACB = 85 – 30 = 55°

Circles Class 10 State Syllabus Kerala Syllabus Question 2.
Prove that any outer angle of a cyclic quadrilateral is equal to the inner angle at the opposite vertex.
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 45
PQRS is cyclic Quadrilateral
∠PSR + ∠PQR = 180°
(sum of opposite angles)
∠PQR + ∠RQT = 180° (linear pair)
From this we get ∠PSR = ∠RQT

Maths Circles Class 10 State Syllabus Question 3.
Prove that a parallelogram which is not a rectangle is not cyclic.
Answer:
PQRS is cyclic quadrilateral.
∠P + ∠R = 180
Also in a parallelogram opposite angles will be equal.
∠P + ∠R = 180°
∠P = ∠R = 90°
This means that PQRS must be a rectangle, otherwise, it is not cyclic.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 46

Sslc Maths Chapter 2 Notes Kerala Syllabus Question 4.
Prove that any non-isosceles trapezium is not cyclic.
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 47
opposite angles are not supplementary.
∴ ABCD is not cyclic. Non-isosceles trapezium is not cyclic.

10th Class Maths Notes Kerala Syllabus Question 5.
In the picture, bisectors of adjacent angles of the quadrilateral ABCD intersect at P, Q, R, S.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 48
Prove that PQRS is a cyclic quadrilateral.
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 49
2x + 2y + 2z + 2w = 360°;
x + y + z + w = 180°
Δ DPC; ∠DPC = 180 – (w + z)
Δ ARB ; ∠ARB =180 – (x + y)
∠R + ∠P = 360 – (x + y + z + w) = 360 – 180 = 180°
In ΔBQC ZQ = m – (w + x)
In Δ ASD ∠S = 180 – (r + y)
Similarily ∠S +∠Q = 180.
PQRS is a cyclic quadrilateral.

Question 6.
i) The two circles below intersect at P, Q and lines through these points meet the circles at A, B, C, D. The lines AC and BD are not parallel. Prove that if these lines are of equal length, then ABDC is a cyclic quadrilateral.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 50
ii) In the picture, the circle on the left and right intersect the middle circle at P, Q, R, S; the lines joining them meet the left and right circles at A, B, C, D. Prove that ABDC is a cyclic quadrilateral.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 51
Answer:
i.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 52
∴ ABCD is a cyclic quadrilateral.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 53
ABCD is a cyclic quadrilateral.

Question 7.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 54
In the picture, points P, Q, R are marked on the sides BC, CA, AB of AABC and the circumcircles of ΔAQR and ΔBRP are drawn. M is a point where these circles intersect.

Prove that the circumcircle of ΔCPQ also passes through M.
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 55
Let M be a common point which three circles can passed.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 56
Therefore the circumcircle of ACPQ also passes through M

Textbook Page No. 67

Question 1.
In the picture, chords AB and CD of the circle are extended to meet at P.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 57
i. Prove that the angles of Δ APC and Δ PBD, formed by joining AC and BD, are the same.
ii. Prove that PA × PB = PC × PD.
iii. Prove that if PB = PD, then ABDC is an isosceles trapezium.
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 58
i. As ABCD is a cyclic quadrilateral.
If ∠BAC = x° then
∠BDC = 180 – x ∠BDP = x°
If ∠ACD = y° then ∠PBD = y°
As ∠APC is common angle.
Angles of Δ APC and Δ PBD are same
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 59
iii. If PB = PD ; AP= PC
In ABCD
ABDC is a cyclic quadrilateral, so their opposite angles are supplementary.
If AP = PC, in ∆ APC
∠A =∠C
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 60
As AB = CD
AC || BD
Adjacent angles are supplementary
∴ ABCD will be an isosceles trapezium

Question 2.
Draw a rectangle of width 5 centimetres and height 3 centimetres.
i. Draw a rectangle of the same area with width 6 centimetres.
ii. Draw a square of the same area.
Answer:
i. Draw a rectangle of length 5 cm and width 3cm.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 61
Extend AB up to 6cm.
(AE = 6cm) Draw an arc having radius as AE and A as centre. Extend DA and mark the point F.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 62
Extend BA towards left. Mark G as AD = AG
Draw Δ GFB.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 63
Circum circle of Δ GFB meets AD at D.
∴ AG × AB = AF × AH.
That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE and width AH.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 64

ii. Draw a rectangle of length 5cm and width 3cm. Area = 5 x 3 = 15 cm2.
Therefore side of a square will be √15.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 65
Draw a semicircle of diameter AH.
Extend BC, and mark the point F.
AB × BH = 5 × 3 = 15
AB × BH = BF2 ; BF = √5 cm
Area of BEGF = √15 × √15 = 15 cm2

Question 3.
Draw a square of area 15 square centimetres.
Answer:
Draw a rectangle of length 5cm and width 3cm. Area = 3 × 5 cm2 = 15 cm2. Side of the square is √15.Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 66
Draw a semicircle of diameter AH.
BC can touch the point F.
AB × BH = BF2 = √152 = 15 cm2.
Area of BEGF = 15 cm2.

Question 4.
Draw a square of area 5 square centimetres in three different ways. (Recall Pythagoras theorem)
Answer:
i. Draw a rectangle of length 5cm and width 1 cm.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 67
Draw a semicircle of diameter AE. Extend BC up to F. √5 is the side of the square BGHF.

ii. Draw a right-angled triangle of perpendicular sides 2 cm and 1cm.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 68
Hypotenuse will be y is cm. The area of the square ACDE is 5 cm2, because here we take the hypotenuse as sides of the square.

iii. Draw a right-angled triangle of hypotenuse 3 cm and one side 2cm.
Third side = \(\sqrt{3^{2}-2^{2}}=\sqrt{5} \mathrm{cm}\)
Draw a square having side BC, then area of the square BEDC will be 5cm2
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 69

Question 5.
In the picture, a line through the centre of a circle cuts a chord into two parts:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 70
What is the radius of the circle?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 71
The intersection may be with in the circle.
Chords AB & CD intersect at P i. e.,
AP × PB = CP × PD
(The intersection will be inside the circle)
AP × PB = CP × PD
4 × 6 = CP × (OP + OD)
24 = CP × (OP + OC)
24 = CP × (OP + OP + CP)
24 = CP × (5 + 5 + CP)
24 = CP × (10 + CP)
CP =2
Radius = CP + OP = 2 + 5 = 7cm

Question 6.
In the picture, a line through the centre of a circle meets a chord of the circle:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 72
What are the lengths of the two pieces of the chord?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 73
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 74
CP= 13 – PB
Therefore equation (1)
(13 – PB)PB = 40
PB = 5, 8
If PB = 5cm then PC = 8cm
If PB = 8cm then PC = 5cm
In figure PB > PC
PB = 8 cm, PC = 5 cm

Circles Orukkam Questions & Answers

Worksheet 1

Question 1.
In triangle ABC, AB = 8cm, BC = 6cm , AC = 10cm.
1. What kind of triangle is this?
2. What is the position of B based on the circle with AC as the diameter? Why?
3. What is the position of A based on the circle with BC as the diameter? Why?
4. What is the position of the point C based on the circle with diameter AB?
Answer:
In Δ ABC
AB2 + BC2 = 82 + 62 = 64 + 36 = 100 = AC2
Δ ABC is a right angled triangle.
If we draw a circle taken in a AC as diameter, ∠B = 90°, Therefore the point B on the circle.
Δ ABC is a right-angled triangle.
If we draw a circle taking BC as diameter, ∠A < 90°, Therefore the position of point A will be outside the circle.
If we draw a circle taking AB as diameter, ∠C < 90°, Therefore the position of point C will be outside the circle.

Question 2.
Three vertices of a parallelogram are on a circle and the fourth vertex is at the center. Find the angles of the parallelogram.

Mark a point P on the top of the figure on the circle, join AP and CP. If angle AP C = x then write? AOC
Write ABC?
Write ∠ABC + ∠APC ?
What is APC?
Find the angles
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 76
The angles of a parallelogram are 60, 120, 60 and 120.

Question 3.
In triangle ABC ,AB = AC, angle BAC = 30, BC = 5cm Find the radius of ABC
Draw the figure
Mark the center, BO and CO
Find the measure of angle BOC
Write the angles of triangle OBC
What kind of angle is triangle OBC
Write the radius of the circumcircle
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 77
∠OBC= 75 – 15 = 60 = OCB (∵OB = OC)
Angle of Δ OBC are: 60, 60, 60
ΔOBC is an equilateral triangle.
Radius of circum circle of Δ OBC = OB = OC = BC = 5 cm.

Question 4.
P QRS is cyclic.
∠P = 3x, ∠Q = y, ∠R = x, ∠ = 5 v
Find the angles
Draw circle , mark P, Q, R, S on it, complete PQRS Enter the given angles.
What is 3x + x? Find x
What is y + 5y? Find y
Find the angles
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 78
3x + x = 180° (Sum of the opposite angles of a cyclic quadrilateral is 180″)
=> 4x = 180° => 4x = 45°
y + 5y = 180 (Sum of the opposite angles of a cyclic quadrilateral is 180°)
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 79

Question 5.
In the figure ABC D is a trapezium. If the vertices are on a circle, prove that it is an isosceles trapezium draw figure
What is ∠A + ∠C?
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 80
What is ∠B + ∠C?
Write the relation between A and B. Write the conclusion.
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 81
∠A + ∠C = 180° (Sum of the opposite angles of a cyclic quadrilateral is 180°)
∠B + ∠C = 180° (AB || CD)
(Sum of the alternate angles of a cyclic quadrilateral is 180″)
∠A + ∠C = ∠B + ∠C
∴ ∠A = ∠B
∴ ABC’D is an isoceles trapezium

Question 6.
In ABC AB = AC . P is the midpoint of AB and Q is the midpoint of AC. Prove that BPQC is cyclic?
Draw figure. Mark PQ and complete BP QC?
Is PQ parallel to BC?
Note that ∠B = ∠C?
What is ∠C + ∠Q?
What is ∠B + ∠Q?
Write conclusion.
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 82
PQ || BC
(The line joining midpoints of two sides of a circle will be parallel to the third side).
∠B = ∠C (1)
(∵ AB = AC are given )
∠C + ∠Q = 180° (2)
(∵ PQ || BC, QC is the bisector, so the sum of alternate angles are 180°)
∠B + ∠Q = 180° .
From (1) and (2) we conclude that BPQC is an cyclic trapezium

Worksheet 2

Question 7.
Prove that ABCD given in the figure is cyclic
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 83
Draw figure and mark PQ
If ∠BAP =xthen
what is ∠BQP?
Find ∠PQD
Find ∠PDC? Why?
What is ∠A + ∠C?
Answer:
i. Quadrilateral ABPQ is cyclic
If ∠A = x, then ∠BQP = 1 80 – x
If ∠B = y, then ∠APQ = 180 – y
Quadrilateral PQCD is cyclic, SO
∠QCD = 180 – x (∠DPQ = x )
∠PDC = 180 – x (∠PQC = y)
∴ ABCD is a cyclic quadrilateral.

Worksheet 3

Question 8.
In the figure AB, C Dare extended and intersect at P. If AB = 5, BP = 3, P D = 2 then find CD? Draw the figure.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 84
Write the relation between PA, PB, PC, PD
Find C D
Answer:
PA x PB = PCxPD
If CD = x, then
⇒ 8 × 3 = (2 + x)2 ⇒ 24 = 4 + 2x
⇒ 2x = 20 ⇒ x = \(\frac { 20 }{ 2 }\) = 10
∴ CD= 10

Question 9.
In the figure AB is the diameter and CD is parallel to the diameter. AB = 8cm,BD = 2cm, find CD
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 85
Answer:
If we draw a perpendicular DP to AB from D. Then PAxPB = PD2.
Here PB = x, then PA = 8 – x.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 86
x(8 – x) = PD2, 22 = x2 + PD2
x(8 – x) = 4 – x2, 8x – x2 = 4 – x2
8x = 4, x = \(\frac { 1 }{ 2 }\)
Similarly, let us draw a perpendicular CQ to AB from C
AQ = ,PQ = 8– \(\left(\frac{1}{2}+\frac{1}{2}\right)\) = 7
CD = 7cm

Question 10.
Draw a rectangle of length 6cm and width 4cm. Draw another rectangle whose area equal to area of the first rectangle and one of the sides is 8cm.
Ans: Draw ABCD as in the given measurement. Mark E by extending AB 2cm more. AE = 8 will be 8cm. WithAas centre and AE radius draw an arc. This arc cut DA produced at F. Extend BA such that AD = AG and mark G. Draw triangle GF B and construct a circumcircle. The circle meets AD at H . Complete the rectangle AHIE
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 87

Worksheet 4

Question 11.
Draw an equilateral triangle of height 3cm. What is the length of a side?
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 88
Write the principle of construction. Students are advised to construct as in the steps given below.
Answer:
Draw a circle of radius 2cm and mark a diameter AB which is 4cm. Mark a point P from one end A is 3 cm apart on the diameter.
Draw a chord C D perpendicular to AB. Complete triangle C AD.
Using PA x PB = PD2, PD= √3.
Now we get AD = AC = C D = 2√3
Height AP = 3cm.

Worksheet 5

Question 12.
In the figure, PA is a tangent and O is the centre of the circle. P A = 17, ∠OPA = 30° then calculate the radius of the circle and distance from centre to the point P Triangle OAP with 30°, 60°, 90° is right triangle. Using the property of this special right triangle find the radius and the distance.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 89
Answer:
Δ OAP is a right-angled triangle having sides 30°, 60°and 90°.
Length of side which is opposite to the angle 90°, is twice the side which is opposite to the angle 30°.
Length of side which is opposite to the
angle 60°, is √3 times of the side which is opposite to the angle 30°
That is radius of the circle , OA = \(\frac { 17 }{ √3 }\)
Distance from centre to the point P
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 89a
Worksheet 7

Question 13.
Draw a circle and construct 30°, 150° angles on it.
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 90

Question 14.
Draw a circle and construct \(22 \frac{1}{2}^{0}\) on it.
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 91

Question 15.
In triangle ABC the radius of the circumcircle is 6 cm, ∠A = 70°, ∠B = 80°. Construct the triangle
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 92

Question 16.
Draw a rectangle of length 7cm, and width 5cm and construct a square whose area is same as the area of this rectangle.
Answer:
Draw a rectangle of length 7cm, and width 5cm. Area = 7 × 5 = 35 cm2.
Therefore length of one side of the square is √35.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 93
Draw a semicircle taking AH as diameter Extend BC, and mark a point F.
AB × BH = 7× 5 = 35
AB × BH = BF2 ;
BF = √35cm
Area of BEGF= √35 × √35 = 35 cm2

Question 17.
Draw a rectangle of one side 5cm, width 7cm. Construct another rectangle whose one side is 8cm and area equal to the area of the first rectangle.
Answer:
Draw a rectangle of length 7cm and width 5cm.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 94
Extend AB up to 8 cm (AE = 8cm) Draw an arc taking A as centre and AE as radius. Extend DA and mark the point F.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 95
Elongate BA towards left, and mark G such that AD = AG.
Draw Δ GFB
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 96
Circumcircle of A GFB will meet side AD on H.
AG × AB = AF × AH.
That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE and width AH.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 97

Question 18.
Draw a square of side 5cm and construct a rectangle having one side 7cm and area equal to area of the square.
Answer:
Draw a square of side 5cm.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 98
Extend AB to 7 cm
(AE = 7cm) Draw an arc taking A as centre and AE a radius. Extend DA and mark the point F.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 99
Elongate BA towards left, and mark G such that AD = AG.
Draw Δ GFB.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 100
Circumcircle of A GFB will meet side AD on H.
AG × AB = AF × AH.
That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE a width AH.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 101

Question 19.
What is the position of the vertex of an equilateral triangle with opposite side as the diameter?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 102
Angles of an equilateral triangle is 60° each. Position of the vertex of triangle with opposite side as the diameter is outside of the circle, because the angle is less than 90°.

Circles SCERT Questions & Answers

Question 20.
In the figure “ ABC is a right triangle
a. If a circle is drawn with AC as diameter find the position of B.
b. If a circle is drawn with BC as diameter, find the position of A. [ Score: 3 Time: 5 minutes]
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 103
Answer:
a. On the circle (1)
∠B = 90°

b. Outside the circle (1)
Position of the vertex of an triangle with opposite side as the diameter is outside the circle, because. ∠A <9o°.

Question 21.
A circle is drawn with AB as diameter. Find the positions of the points C, D, E related to the circle.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 104
[ Score: 3 Time: 5 minute]
Answer:
C inside the circle (1)
∠C > 90°.
D on the circle (1)
∠D = 90°.
E outside the circle (i)
∠E <90°.

Question 22.
In Δ ABC and Δ PQR, BC = QR, ∠ A = ∠P, ∠Q = 90°, QR = 5 cm, PQ = 12 cm.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 105
Find the diameter of the circumcircle of Δ ABC. [ Score: 4 Time: 6 minute]
Answer:
QR = BC (1)
∠A = ∠P (1)
PR Diameter of the circumcircle of Δ ABC (1)
Diameter = PR= \(\sqrt{12^{2}+5^{2}}=\sqrt{169}=13 \mathrm{cm}\) .(1)

Question 23.
PQ and RS are two mutually prependicular chords of a circle. < QPR=50° find< PQS. [ Score: 3, Time: 6 minute]
Answer:
∠PRS = 90 – 50 = 40° (1)
∠PQS = 40° (1)
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 106

Question 24.
O is the centre of the circle. If ∠BOC = 130° and ∠AOB = 110°. What is ∠AOC?
Find all angles of Δ ABC
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 107 [ Score: 3, Time: 3 minute]
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 108

Question 25.
Find all angles of the hexagon ABCDEF
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 109 [ Score: 4 Time:5 minute]
Answer:
∠ EFD = ∠EAD = 30°
∠ FE A = ∠FDA = 40°
∠FDE = ∠FAE = 35° (1)
∠BAC= ∠BDC = 45°
∠ABD= ∠ACD = 62°
∠ACB = ∠ADB= 35° (1)
∠A = 1480, ∠B = 100°
∠C = 97° ∠D= 155° (1)
∠E= 115° ∠F= 105° (1)

Question 26.
O is the centre of the circle and AB is a chord. AC is the bisector of ∠OAB. ∠OAB = 56°.
a. Prove that OC and AB are parallel,
b. Find ∠ABC and ∠OBE.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 110
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 111

Question 27.
O is the centre of the circle. AD and BC are perpendicular to XY. CB cuts the circle at E. Prove that –CE = AD.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 112
Answer:
∠AEB = 90° (Angle in a semi circle) ( 1)
∠AEC=90°
∴ AECD is a rectangle
∴ AD = CE (2)

Question 28.
ABCD is a parallelogram. A, B, E, F are the points on a circle. ∠DEF = 80° Find out the angles of the quadrilateral AEFB.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 113 [ Score: 4, Time: 4 minute]
Answer:
∠AEF = 180 – 80 = 100° (1)
∠ABF = 180 – 100 = 80° (1)
∠A = 180 – 80 = 100° (1)
∠EFB = 180 – 100 = 80° (1)

Question 29.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 114
O is the centre of the circle. ? ∠OCA = x °.
a. Find ∠OAC
b. Prove that ∠OCA + ∠ABC = 90°
c. Prove that ∠ADC – ∠OCA = 90° [Score: 4, Time: 4 minute]
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 115

Circles Exam Oriented Questions & Answers

Short Answer Type Questions (Score 2)

Question 30.
In the figure AB is the diameter. PC is perpendicular to AB. PC = 6cm, PB = 3cm. Find the radius of the semi-circle.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 116
Answer:
AP × PB = PC2
AP × 3 = 62
AP = 36/3 = 12 .
AB = 12 + 3 = 15
ie Radius = 15/2 = 7.5 cm

Question 31.
In Δ PQR, ZP = 60°, ∠R = 30° find whether the vertex Q on the circle with PR as diameter.
Answer:
∠P + ∠R = 60 + 30 = 90°.
∴ ∠Q = 180 – 90 = 90°.
So point Q on the circle.

Question 32.
In Δ ABC, ∠A = 60°, ∠B = 70°. Find whether the vertex C is inside or outside the circle with AB as diameter.
Answer:
The vertex C is outside the circle Since ∠C = 180 – (60 + 70) = 50° ∠90°

Question 33.
In the diagram, the central angle of arc ABC is 100° and ∠OAD is 30°. Find ∠OCD.
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 117
∠D = 50°
As ∠OAD
is an isosceles triangle.
∠ODA = 30°
∠ODC = 20°
∴∠OCD = 20°

Question 34.
In the figure, find ∠PQB, O is the centre.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 118
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 119

Short Answer Type Questions (Score 3)

Question 35.
The central angle of arc ABC is 60°, then find out the following,
i) ∠D
ii) Central angle of arc AEC,
iii) ∠B
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 120
Answer:
i. ∠D = 30°
ii. Then central angle of arc AEC = 300°.
iii. ∠B = 150°

Question 36.
Show that the arcs APC, BQD when joined make a semicircle?
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 121
Answer:
∠ADC is the half of the central angle of arc APC
The central angle of are APC = 2 ∠ADC
The central angle of arc BQD = 2 ∠DAB
In triangle AOD, CD ⊥ AB, ∠AOD = 90°,
∠DAO + ∠ADO = 180 – 90° =90°,
ie, ∠DAB + ∠ADC = 90°
2( ∠DAB + ∠ADC ) = 90° × 2 = 180°

Question 37.
The central angle of the complementary arc of a circle is 40° more than 3 times the central angle of the arc. Find out the central angles of each arc?
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 122
Answer:
x + 3x + 40 = 360
4x + 40 = 360
4x = 360 – 40 = 320
x = \(\frac { 320 }{ 4 }\) = 80
∴ central angle of arc
ABC = 80
central angle of arc ADC = 360 – 80 = 280°

Question 38.
ABCD in the diagram is a rectangle. Then find out the area of the circle.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 123
Answer:
ABCD is a rectangle ∠B = ∠D = 90°
AC in the diameter of the circle
AC = \(\sqrt{8^{2}+6^{2}}=\sqrt{64+36}=\sqrt{100}=10\)
∴ radius = 5cm
∴ Area of the circle = πr² = π × 5² = 25πcm²

Question 39.
In the figure, AD = 16cm, BD = 6cm, CD = 2cm. Find the length EF.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 124
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 125

Question 40.
In the given figure, O is the centre of the circle. If ∠OAP = 35° and ∠OBP = 40°, find the value of ∠x.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 126
Answer:
Join OP
Since OA = OP and
∠APO = ∠OAP=35°
Similarly, OB = OP and ∠OPB = ∠OBP = 40°
∠APB = 35°+ 40° = 75°
∠AOB = 2 × 750 = 150°

Long Answer Type Questions (Score 4)

Question 41.
In the figure find ∠APB,∠ABQ ; where O is the centre of the circle ∠OAP = 32° and  ∠OBP = 47° .
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 127
Answer:
JoinOP.
In OAP, OA = OP = Radius
∠OAP = ∠OPA = 32°
In OPR, OB = OP = radius
∠OBP = ∠OPB =47°
∠APB = 32°+ 47°= 79°
∠AQB = 180° – 79°= 10°

Question 42.
Draw a line of √7 cm.
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 128

Question 43.
Ois the centre of the circle as shown in the figure.
Find ∠CBD.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 129
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 130
Takeapoint E on the circle, join AE and CE.
∠AEC= \(\frac { 100 }{ 2 }\) = 50°
∠AEC + ∠ABC = 180° (Opposite angles of a cyclic quadrilaterals)
∠ABC = 130°
∠ABC + ∠CBD = 180° (linearpair)
130°+ ∠CBD = 180°
∠CBD = 50°

Long Answer Type Questions (Score 5)

Question 44.
In the figure O is the centre of the circle. Central angle of arc AXB is 60°, arc CYD is 80°. Then find all the angles of ΔAPD.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 131
Answer:
Central angle of arc AXB = 60°
i.e., ∠AOB = 60°
i.e., ∠ADP = 30°
Central angle of arc CYD = 80°
i.e., ∠COD = 80°
i e., ∠DPA = 40°
i e., ∠APD = 180° – (30 + 40) = 110°
Angles of = 30°, 40°, 110°

Question 45.
‘O’ is the centre of the circle ∠D = 80°, find the following measurements.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 132
a. ∠C
b. ∠ABC
c. ∠BAC
d. ∠F
Answer:
∠D = 80°
a. ∠C = 80° (∠D and ∠C are angled on a same arc So, both are equal)

b. ∠ABC = 90°
(AC is diameter, Angle of a hemisphere is right)

c. ∠BAC = 180 – (80 + 90)
= 180 – 170 = 100

d. ∠F= 180 – 80 = 100° (Opposite angles of a cyclic quadrilateral are equal)

Circles Memory Map

Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 133
Angle in a semicircle is right:
The angle made by any arc of a circle on the alternate arc is half the angle made at the centre.
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 134
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 135

All angles in an arc is equal:
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 136
If AB, CD are two chords, then
PA × PB = PC × PD

The area of the rectangle formed of parts into which a diameter of a circle is cut by a perpendicular chord is equal to the area of the square formed by half the chord.
PA × PB = PC2
Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles 137

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Kerala SSLC Malayalam Previous Year Question Paper March 2019 (Kerala Padavali)

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