Category: Class 10

Balbharti Maharashtra State Board Class 10 Hindi Solutions Lokbharti Chapter 1 भारत महिमा Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Hindi Lokbharti Chapter 1 भारत महिमा

Hindi Lokbharti 10th Std Digest Chapter 1 भारत महिमा Textbook Questions and Answers

सूचना के अनुसार कृतियाँ कीजिए:

प्रश्न 1.
निम्नलिखित पंक्तियों का तात्पर्य लिखिए:
a. कहीं से हम आए थे नहीं → …………………….
b. वही हम दिव्य आर्य संतान → …………………….
उत्तर:
(a) हम भारतवासी किसी अन्य देश से आकर यहाँ नहीं बसे। हम यहीं के निवासी हैं। सभ्यता के प्रारंभ से हम यहीं रहते आए हैं।
(b) भारतवासी आर्य थे और हम उन्हीं आर्यों की दिव्य संतानें हैं।

प्रश्न 2.
उचित जोड़ियाँ मिलाइए:
संचय
सत्य
अतिथि
रत्न
वचन
दान
हृदय
तेज
देव
उत्तर:
(i) संचय – दान
(ii) सत्य – वचन
(iii) अतिथि – देव
(iv) रत्न – तेज।

प्रश्न 3.
लिखिए.
a. कविता में प्रयुक्त दो धातुओं के नाम:

उत्तर:

b. भारतीय संस्कृति की दो विशेषताएँ:

उत्तर:

प्रश्न 4.
प्रस्तुत कविता की अपनी पसंदीदा किन्हीं दो पंक्तियों का भावार्थ लिखिए।
उत्तर:
हमारे संचय में था दान, अतिथि थे सदा हमारे देव वचन में सत्य, हृदय में तेज, प्रतिज्ञा में रहती थी टेव। हम भारतीय दीन-दुखियों की सेवा करने के लिए सदैव तत्पर रहते हैं। हम यदि धन और संपत्ति का संग्रह करते भी थे तो दान के लिए करते थे। दानवीरता भारतीयों का गुण रहा है। महर्षि दधीचि और कर्ण जैसे दानवीर इसी भूमि पर हुए हैं। हमारे देश में अतिथियों को देवता के समान माना जाता था। भारतीय सत्यवादी हरिश्चंद्र की संतानें हैं। हमारे हृदय में तेज था, गौरव था। हम सदा अपनी प्रतिज्ञा पर अटल रहते थे। भारतीयों का मानना था- प्राण जाएँ,: पर वचन न जाएँ।

प्रश्न 5.
निम्नलिखित मुद्दों के आधार पर पद्य विश्लेषण कीजिए:
a. रचनाकार का नाम
b. रचना का प्रकार
c. पसंदीदा पंक्ति
d. पसंदीदा होने का कारण
e. रचना से प्राप्त संदेश
उत्तर:
a. रचनाकार का नाम → जयशंकर प्रसाद।
b. रचना की विधा → कविता।
c. पसंद की पंक्तियाँ → व्योमतम पुंज हुआ तब नष्ट, अखिल संसृति हो उठी अशोक। (सूचना: विद्यार्थी अपनी पसंद की पंक्ति लिखेंगे।)
d. पंक्तियाँ पसंद होने का कारण → हम भारतीयों ने पूरे विश्व में ज्ञान का प्रसार किया, जिसके कारण समग्र संसार आलोकित हो गया। अज्ञान रूपी अंधकार का विनाश हुआ और संपूर्ण सृष्टि के सभी दुख-शोक दूर हो गए।
e. रचना से प्राप्त संदेश/प्रेरणा → हमें सदैव अपने देश और इसकी संस्कृति पर गर्व करना चाहिए। जब भी आवश्यकता पड़े, देश के लिए अपना सर्वस्व न्योछावर कर देने के लिए तत्पर रहना चाहिए।

Hindi Lokbharti 10th Textbook Solutions Chapter 1 भारत महिमा Additional Important Questions and Answers

पद्यांश क्र. 1
प्रश्न.
निम्नलिखित पठित पद्यांश पढ़कर दी गई सूचनाओं के अनुसार कृतियाँ कीजिए:

कृति 1: (आकलन)

प्रश्न 1.
पद्यांश से ऐसे दो प्रश्न तैयार कीजिए, जिनके उत्तर निम्नलिखित शब्द हों:
(i) अभिनंदन
(ii) आलोक।
उत्तर:
(i) उषा ने हँसकर क्या किया?
(ii) जब भारतीयों ने ज्ञान का प्रचार किया तो संसार में क्या फैला?

प्रश्न 2.
पद्यांश में प्रयुक्त इन शब्दों से सहसंबंध दर्शाने वाले शब्द लिखिए:
(i) हिमालय – ……………………..
(ii) किरण – ……………………..
(iii) विमल – ……………………..
(iv) कोमल – ……………………..
उत्तर:
(i) हिमालय -आँगन
(ii) किरण – उपहार
(iii) विमल -वाणी
(iv) कोमल -कर

प्रश्न 3.
विधानों के सामने सत्य/असत्य लिखिए:
(i) जब पूरा विश्व जगा तो भारतवासी भी जग गए।
(ii) वीणापाणि ने अपने हाथ में वीणा ली।
(iii) हिमालय के आँगन में किरणों का उपहास मिला।
(iv) सप्तसिंधु में सातों स्वर गूंजने लगे।
उत्तर:
(i) असत्य
(ii) सत्य
(iii) असत्य
(iv) सत्य।

प्रश्न 4.
उचित जोड़ियाँ मिलाइए:
(i) उषा – आलोक
(ii) हीरक – संगीत
(iii) विश्व – अभिनंदन
(iv) वीणा – हार
उत्तर:
(i) उषा – अभिनंदन।
(ii) हीरक – हार
(iii) विश्व – आलोक
(iv) वीणा -संगीत।

कृति 2: (शब्द संपदा)

प्रश्न 1.
पद्यांश में से ढूँढ़कर उपसर्गयुक्त शब्द लिखिए:
(i) ………………. (ii) ……………….
उत्तर:
(i) अभिनंदन
(ii) उपहार।

प्रश्न 2.
अनेक शब्दों के लिए एक-एक शब्द लिखिए:
(i) गले में पहनने की मूल्यवान माला
(ii) सितार जैसा वह वाद्य जो सब वाद्यों में श्रेष्ठ माना जाता है, ……………….
उत्तर:
(i) हार
(ii) वीणा।

प्रश्न 3.
निम्नलिखित शब्दों के लिए पद्यांश में प्रयुक्त शब्द ढूँढ़कर लिखिए:
(i) संपूर्ण
(ii) शोकरहित
(iii) संसार
(iv) आकाश।
उत्तर:
(i) संपूर्ण – अखिल
(ii) शोकरहित – अशोक
(iii) संसार – संसृति
(iv) आकाश – व्योम।

कृति 3: (सरल अर्थ)

प्रश्न.
उपर्युक्त पद्यांश की प्रथम चार पंक्तियों का सरल अर्थ 25 से 30 शब्दों में लिखिए।
उत्तर:
भारत देश हिमालय के आँगन के समान है। प्रतिदिन उषा भारत को सूर्य की किरणों का उपहार देती है, मानो हँसकर भारत-भूमि का अभिनंदन कर रही हो। ओस की बूंदों पर जब प्रातःकालीन सूर्य की रश्मियाँ पड़ती हैं, तो ओस की बूंदें चमकने लगती हैं और ऐसा लगता है मानो, उषा ने भारत को हीरों का हार पहना दिया हो।

सबसे पहले ज्ञान का उदय भारत में ही हुआ। अर्थात सबसे पहले हम जाग्रत हुए। फिर हमने पूरे विश्व में ज्ञान का प्रसार किया। इसके कारण समग्र संसार आलोकित हो गया। अज्ञानरूपी अंधकार का विनाश हुआ और संपूर्ण सृष्टि के सभी दुख-शोक दूर हो गए।

पद्यांश क्र. 2

प्रश्न.
निम्नलिखित पठित पद्यांश पढ़कर दी गई सूचनाओं के अनुसार कृतियाँ कीजिए:

कृति 1: (आकलन)

प्रश्न 1.
आकृति पूर्ण कीजिए:

उत्तर:

प्रश्न 2.
सही विकल्प चुनकर वाक्य फिर से लिखिए:
(i) भारत में केवल …………………………. की ही विजय नहीं रही। (चाँदी/लोहे/सोने)
(ii) यहाँ …………………………. भिक्षु की तरह रहते थे। (लोग/लड़के/सम्राट)
(iii) हमसे चीन को …………………………. की दृष्टि मिली। (धर्म/कर्म/धन)
(iv) हमारा देश सदा प्रकृति का …………………………. रहा। (खिलौना/आँगन /पालना)
उत्तर:
(i) भारत में केवल लोहे की ही विजय नहीं रही।
(ii) यहाँ सम्राट भिक्षु की तरह रहते थे।
(iii) हमसे चीन को धर्म की दृष्टि मिली।
(iv) हमारा देश सदा प्रकृति का पालना रहा।

प्रश्न 3.
उपर्युक्त पद्यांश पर आधारित ऐसे दो प्रश्न तैयार कीजिए, जिनके उत्तर निम्नलिखित शब्द हों:
(i) सम्राट
(ii) धर्म।
उत्तर:
(i) कौन भिक्षु होकर रहते?
(ii) चीन को कौन-सी दृष्टि मिली?

प्रश्न 4.
निम्नलिखित पंक्तियों का तात्पर्य लिखिए:
(ii) प्रकृति का रहा पालना यहीं।
उत्तर:
(ii) हमें प्रकृति ने प्रत्येक वस्तु मुक्तहस्त से प्रदान की। यहाँ की शस्य श्यामला भूमि, हिमाच्छादित गिरि शिखर, घाटियाँ, वादियाँ, सदानीरा नदियाँ, झरने, फल-फूल, संसाधनों से भरपूर जंगल सभी अनुपम हैं।

प्रश्न 5.
आकृति पूर्ण कीजिए:
(i) हमने गोरी को इसका दान दिया – [ ]
(ii) भारत की धरती पर इसकी धूम रही – [ ]
उत्तर:
(i) हमने गोरी को इसका दान दिया – [दया का]
(ii) भारत की धरती पर इसकी धूम रही – [धर्म की]

कृति 2: (शब्द संपदा)

प्रश्न 1.
निम्नलिखित शब्द-समूहों के लिए शब्द लिखिए:
(i) बहुमूल्य चमकीले प्रसिद्ध खनिज पदार्थ, जो आभूषणों आदि में जड़े जाते हैं –
(ii) छोटे बच्चों के लिए एक प्रकार का झूला या हिंडोला –
(iii) वह स्थान जहाँ किसी का जन्म हुआ हो –
(iv) बौद्ध संन्यासियों के लिए प्रयोग किया जाने वाला शब्द –
उत्तर:
(i) रत्न
(ii) पालना
(iii) जन्मस्थान
(iv) भिक्षु।

प्रश्न 2.
निम्नलिखित शब्दों के विलोम शब्द लिखिए:
(i) विजय x ………………….
(ii) धर्म x ………………….
(iii) भूमि x ………………….
(iv) जन्म x ………………….
उत्तर:
(i) विजय x पराजय
(ii) धर्म x अधर्म
(iii) भूमि x आकाश
(iv) जन्म – मरण।

कृति 3: (सरल अर्थ)

प्रश्न.
उपर्युक्त पद्यांश की अपनी पसंदीदा किन्हीं दो पंक्तियों का सरल अर्थ 25 से 30 शब्दों में लिखिए।
उत्तर:
विजय केवल लोहे की नहीं, धर्म की रही धरा पर धूम भिक्षु होकर रहते सम्राट, दया दिखलाते घर-घर घूम। भारतीयों ने शस्त्रों के बल पर दूसरे देशों को नहीं जीता, बल्कि उन्होंने प्रेमभाव से लोगों के हृदय जीते हैं। भारत में प्राचीन काल से ही लोगों के मन में धर्म की भावना रही है। यहाँ वर्धमान महावीर और गौतम बुद्ध जैसे त्यागी धर्मपुरुष हुए हैं, जिन्होंने अपना विशाल साम्राज्य छोड़कर भिक्षु का स्वरूप धारण किया और घर-घर घूमकर लोगों का कष्ट दूर करने का प्रयास किया, धर्म का प्रचार किया।

पद्यांश क्र. 3

प्रश्न.
निम्नलिखित पठित पद्यांश पढ़कर दी गई सूचनाओं के अनुसार कृतियाँ कीजिए:

कृति 1: (आकलन)

प्रश्न 1.
निम्नलिखित पंक्तियों का तात्पर्य लिखिए:
(i) किसी को देख न सके विपन्न।
उत्तर:
(i) भारतीय कभी किसी को दुखी नहीं देख सके। दीन-दुखियों की सेवा करने के लिए हम भारतीय सदैव तत्पर रहते हैं।

प्रश्न 2.
आकृति पूर्ण कीजिए:
(i) हम चरित्र के ऐसे थे – [ ]
(ii) हम दान के लिए यह करते थे – [ ]
(iii) हमारे लिए ये देवता के समान थे – [ ]
(iv) हमें अपने गौरव पर यह था – [ ]
उत्तर:
(i) हम चरित्र के ऐसे थे [पवित्र]
(ii) हम दान के लिए यह करते थे [संचय]
(iii) हमारे लिए ये देवता के समान थे [अतिथि]
(iv) हमें अपने गौरव पर यह था [गर्व]

प्रश्न 3.
संजाल पूर्ण कीजिए:

उत्तर:

प्रश्न 4.
आकृति पूर्ण कीजिए:

उत्तर:

कृति 2: (शब्द संपदा)

प्रश्न 1.
पद्यांश से उपसर्ग वाले दो शब्द ढूँढकर लिखिए:
(i) ……………………. (ii) …………………….
उत्तर:
(i) अतिथि
(ii) अभिमान।

प्रश्न 2.
निम्नलिखित शब्दों के समानार्थी शब्द लिखिए:
(i) पूत = …………………….
(ii) गर्व = …………………….
(iii) प्रतिज्ञा = …………………….
(iv) प्यारा = …………………….
उत्तर:
(i) पूत – पावन
(ii) गर्व = घमंड
(iii) प्रतिज्ञा = प्रण
(iv) प्रिय = प्यारा।

प्रश्न 3.
पद्यांश से शब्द ढूँढकर लिखिए:
(i) पवित्र शब्द के लिए प्रयुक्त शब्द – …………………….
(ii) गरीब शब्द के लिए प्रयुक्त शब्द – …………………….
उत्तर:
(i) पवित्र शब्द के लिए प्रयुक्त शब्द – पूत
(ii) गरीब शब्द के लिए प्रयुक्त शब्द – विपन्न।

कृति 3: (सरल अर्थ)

पदय विश्लेषण
सूचना: यह प्रश्नप्रकार कृतिपत्रिका के प्रारूप से हटा दिया गया है। लेकिन यह प्रश्न पाठ्यपुस्तक में होने के कारण विद्यार्थियों के अधिक अभ्यास के लिए इसे उत्तर-सहित यहाँ समाविष्ट किया गया है।

भाषा अध्ययन (व्याकरण)

प्रश्न. सूचनाओं के अनुसार कृतियाँ कीजिए:

1. शब्द भेद:
अधोरेखांकित शब्दों का शब्दभेद पहचानकर लिखिए:
(i) राजा दशरथ वृद्ध दंपति के सामने बैठ गए।
(ii) सड़क कदाचित कच्ची थी।
उत्तर:
(i) दशरथ – व्यक्तिवाचक संज्ञा।
(ii) सड़क – जातिवाचक संज्ञा।

2. अव्यय:
निम्नलिखित अव्ययों का अपने वाक्यों में प्रयोग कीजिए:
(i) बहुत
(ii) सामने
(iii) किंतु।
उत्तर:
(i) प्रयाग बहुत थक गया था।
(ii) स्कूल के सामने एक बगीचा है।
(iii) घर में दीपक तो था, किंतु उसमें तेल न था।

3. संधि:
कृति पूर्ण कीजिए:

संधि शब्द	संधि विच्छेद	संधि भेद
उज्ज्व	………………	………………
अथवा
प्रश्न + उत्तर	………………	………………

उत्तर:

संधि शब्द	संधि विच्छेद	संधि भेद
उज्ज्वल	उत् + ज्वल	व्यंजन संधि
अथवा
प्रश्नोत्तर	प्रश्न + उत्तर	स्वर संधि

4. सहायक क्रिया:
निम्नलिखित वाक्यों में से सहायक क्रियाएँ पहचानकर उनका ‘मूल रूप लिखिए:
(i) इस पद ने मोहिनी मंत्र का जाल बिछा दिया।
(ii) बालक भूमि पर लेट गया।
उत्तर:

सहायक क्रिया	मूल रूप
(i) दिया	देना
(ii) गया।	जाना

5. प्रेरणार्थक क्रिया:
निम्नलिखित क्रियाओं के प्रथम प्रेरणार्थक और द्वितीय प्रेरणार्थक रूप लिखिए:
(i) दौड़ना
(ii) बोलना
(iii) रोना।
उत्तर:

क्रिया	प्रथम प्रेरणार्थक रूप	द्वितीय प्रेरणार्थक रूप
(i) दौड़ना।	दौड़ाना	दौड़वाना
(ii) बोलना	बुलाना	बुलवाना
(iii) रोना	रुलाना	रुलवाना

6. मुहावरे:
(1) निम्नलिखित मुहावरों का अर्थ लिखकर वाक्य में प्रयोग कीजिए:
(i) दृष्टि फेरना
(ii) राह देखना।
उत्तर:
(i) दृष्टि फेरना।
अर्थ: नजर डालना।
वाक्य: नेताजी ने श्रोताओं पर दृष्टि फेरी।

(ii) राह देखना।
अर्थ: प्रतीक्षा करना।
वाक्य: विद्यार्थी कई दिनों से छुट्टियों की राह देख रहे थे।

(2) अधोरेखांकित वाक्यांश के लिए कोष्ठक में दिए गए उचित मुहावरे का चयन करके वाक्य फिर से लिखिए: (सपने की संपत्ति होना, चल बसना, भनक पड़ना)
(i) हफ्ते भर की बीमारी में मरीज चला गया।
(ii) दारोगाजी ने उड़ती हुई खबर सुनी कि कल दंगा होने वाला है।
(ii) ऐसा भूकंप आया कि क्षण भर में सारी चहल-पहल विलुप्त हो गई।
उत्तर:
(i) हफ्ते भर की बीमारी में मरीज चल बसा।
(ii) दारोगाजी के कान में भनक पड़ी कि कल दंगा होने वाला है।
(iii) ऐसा भूकंप आया कि क्षण भर में सारी चहल-पहल सपने की संपत्ति हो गई।

7. कारक:
निम्नलिखित वाक्यों में प्रयुक्त कारक पहचानकर उसका भेद लिखिए:
(i) नारी महान है।
(ii) वह किसी को किसी प्रकार की कमी नहीं होने देती।
(iii) प्रेरणा का सूक्ष्म प्रभाव होता है।
उत्तर:
(i) नारी – कर्ता कारक
(ii) किसी को – कर्म कारक
(iii) प्रेरणा का – संबंध कारक।

8. विरामचिह्न:
निम्नलिखित वाक्यों में यथास्थान उचित विरामचिह्नों का प्रयोग करके वाक्य फिर से लिखिए:
(i) क्या बताऊँ गाय ने दूध देना बंद कर दिया है बूढ़ी हो गई है इस जमाने में गाय भैंस पालने का खर्चा
(ii) हे मेरे मित्रो परिचितो आओ अपने सारे बदले लेने का यही वक्त है
उत्तर:
(i) “क्या बताऊँ। गाय ने दूध देना बंद कर दिया है, बूढ़ी हो गई है। इस जमाने में गाय-भैंस पालने का खर्चा …।”
(ii) “हे मेरे मित्रो, परिचितो! आओ, अपने सारे बदले लेने का यही वक्त है।”

9. काल परिवर्तन:
निम्नलिखित वाक्यों का सूचना के अनुसार काल परिवर्तन कीजिए:
(i) मनु पीछे की ओर मुड़ता है। (सामान्य भूतकाल)
(ii) तुम्हारा मुख लाल होता है। (अपूर्ण वर्तमानकाल)
(iii) रोगी की अवस्था बदल जाती है। (पूर्ण भूतकाल)
उत्तर:
(i) मनु पीछे की ओर मुड़ा।
(ii) तुम्हारा मुख लाल हो रहा है।
(iii) रोगी की अवस्था बदल गई थी।

10. वाक्य भेद:
(1) निम्नलिखित वाक्यों का रचना के आधार पर भेद पहचानकर लिखिए:
(i) भारतीय चरित्र के पवित्र होते हैं।
(ii) बादल आए किंतु पानी नहीं बरसा।
उत्तर:
(i) सरल वाक्य
(ii) संयुक्त वाक्य।

(2) निम्नलिखित वाक्यों का अर्थ के आधार पर दी गई सूचना के अनुसार वाक्य परिवर्तन कीजिए:
(i) तुम्हें अपना ख्याल रखना चाहिए। (आज्ञावाचक)
(ii) मास्टर जी ने पुस्तकें लाने के लिए पैसे दिए। (प्रश्नवाचक)
उत्तर:
(i) तुम अपना ख्याल रखो।
(ii) क्या मास्टर जी ने पुस्तकें लाने के लिए पैसे दिए?

11. वाक्य शुद्धिकरण:
निम्नलिखित वाक्य शुद्ध करके लिखिए:
(i) क्रोध से उसकी नेत्र लाल हो गए।
(ii) राम ने हिरण का शिकार की।
(iii) मैं मेरा काम करता है।
उत्तर:
(i) क्रोध से उसके नेत्र लाल हो गए।
(ii) राम ने हिरन का शिकार किया।
(iii) में अपना काम करता हूँ।

भारत महिमा Summary in Hindi

भारत महिमा कविता का सरल अर्थ

1. हिमालय के आँगन …………………………………… मधुर साम संगीत।. . .

हमारा यह प्यारा भारत देश हिमालय के आँगन के समान है। प्रतिदिन उषा भारत को सूर्य की किरणों का उपहार देती है। तब ऐसा लगता है मानो हँसकर वह भारत-भूमि का अभिनंदन कर रही हो। ओस की बूंदों पर जब प्रातःकालीन सूर्य की रश्मियाँ पड़ती हैं तो ऐसा लगता है जैसे उषा ने भारत को हीरों का हार पहना दिया हो।

सबसे पहले ज्ञान का उदय भारत में ही हुआ अर्थात सबसे पहले हम जाग्रत हुए। फिर हमने पूरे विश्व में ज्ञान का प्रसार किया। इसके कारण समग्र संसार आलोकित हो गया। अज्ञान रूपी अंधकार का विनाश हुआ और संपूर्ण सृष्टि के सभी दुख-शोक दूर हो गए।

वाणी की देवी वीणापाणि (सरस्वती) ने इसी पवित्र भूमि पर प्रेम के साथ अपने कमल के समान कोमल करों में वीणा उठाई, उसकी झंकार से सप्तसिंधुओं में सातों स्वरों का मोहक सरगम गूंजने लगा, मधुर संगीत का जन्म हुआ। इसी महान देश में संगीत के वेद सामवेद की रचना हुई।

2. विजय केवल …………………………………… आए थे नहीं।. . .

भारत के लोगों ने शस्त्रों के बल पर देशों को नहीं जीता। यहाँ प्राचीन काल से ही लोगों के मन में धर्म की प्रखर भावना रही है और उन्होंने संसार में धर्म का प्रचार किया। यहाँ गौतम बुद्ध और वर्धमान महावीर जैसे धर्मपुरुष हुए हैं, जिन्होंने विशाल साम्राज्य छोड़कर भिक्षु का स्वरूप धारण किया और घर-घर घूमकर लोगों का कष्ट दूर करने का प्रयास किया, धर्म का प्रचार किया। हमने मोहम्मद गोरी को पराजित करने के बाद भी दयापूर्वक क्षमा कर दिया। हमारे देश से ही चीन को धर्म की दृष्टि मिली। (भारत के महान सम्राट अशोक ने अपने पुत्र महेंद्र और पुत्री संघमित्रा को बौद्ध धर्म के प्रचार के लिए चीन, स्वर्ण भूमि अर्थात जावा और श्रीलंका भेजा) जावा और श्रीलंका के लोगों को पंचशील (अहिंसा, अस्तेय, अपरिग्रह, सत्य, ब्रह्मचर्य आदि) के सिद्धांत मिले।

भारतवासियों ने कभी किसी की संपत्ति या किसी का राज्य छीनने का प्रयास नहीं किया। हमें प्रकृति ने प्रत्येक वस्तु मुक्तहस्त से प्रदान की। प्रकृति की हमारे देश पर महान कृपा रही है। (यहाँ की शस्य श्यामला भूमि, हिमाच्छादित गिरि शिखर, घाटियाँ, वादियाँ, सदानीरा नदियाँ, झरने, फल-फूल, संसाधनों से भरपूर जंगल सभी अनुपम हैं) भारत सदा से हमारी जन्मभूमि है। हम इसी देश की संतानें हैं। हम बाहर के किसी स्थान से आकर यहाँ नहीं बसे हैं। (जैसा कि कुछ विदेशियों का कहना है।)

3. चरित थे पूत …………………………………… प्यारा भारतवर्ष।. . .

भारत के लोग सदा से चरित्रवान रहे हैं। हमारी भुजाओं में भरपूर शक्ति रही है। भारतीयों में वीरता की कभी कमी नहीं रही। साथ ही नम्रता सदा हमारा गुण रहा है। हमने कभी अपनी उपलब्धियों पर घमंड नहीं किया। हमें अपनी सभ्यता और संस्कृति पर गर्व रहा है। हम कभी किसी को दुखी नहीं देख सके। दीन-दुखियों की सेवा करने के लिए हम भारतीय सदैव तत्पर रहते हैं। ‘हम यदि धन और संपत्ति का संग्रह करते भी थे, तो दान के लिए करते थे। दानवीरता भारतीयों का गुण रहा है। हमारे देश में अतिथियों को सदा देवता के समान माना जाता था। भारत के लोग सत्य बोलना अपना धर्म मानते थे। (भारतीय सत्यवादी हरिश्चंद्र की संतानें हैं।) हमारे हृदय में तेज था, गौरव था। हम सदा अपनी प्रतिज्ञा पर अटल रहते थे। प्राण जाए, पर वचन न जाए हमारा जीवनमूल्य रहा है।

आज भी हम भारतीयों की धमनियों में उन्हीं पूर्वजों का रक्त प्रवाहित हो रहा है। आज भी हमारा देश वैसा ही है। आज भी भारतीयों में वैसा ही साहस है। भारतीय आज भी ज्ञान के क्षेत्र में सबसे आगे हैं। आज भी हम पहले के समान शांति के पुजारी हैं। देशवासियों में वैसी ही शक्ति है। हम उन्हीं आर्यों की दिव्य संतानें हैं।

हम जब तक जिएँ, इसी देश के लिए जिएँ। हमें इसकी सभ्यता और संस्कृति पर अभिमान है और हर्ष है कि हमने इस भूमि पर जन्म लिया है। यह हमारा प्यारा भारतवर्ष है। यदि कभी आवश्यकता पड़े, तो इसके लिए अपना सर्वस्व भी न्योछावर कर दें।

भारत महिमा विषय-प्रवेश :

प्रकृति ने हमारे देश भारत की रचना बड़े प्यार से की है। हमारा देश हिमालय की गोद में बसा हुआ है। हमारा देश सबसे पहले जाग्रत हुआ था और इसकी संस्कृति सबसे पुरानी है। प्रस्तुत कविता में छायावाद के प्रवर्तक जयशंकर प्रसाद जी ने हमारे प्यारे देश भारत के इसी महिमामंडित अतीत का मनोरम चित्रण किया है। कवि की आकांक्षा है कि हमें सदैव अपने देश पर, इसकी सभ्यता और संस्कृति पर गर्व करना चाहिए। आवश्यकता पड़ने पर, हमें देश के लिए अपना सर्वस्व न्योछावर कर देने के लिए तत्पर रहना चाहिए।

भारत महिमा मुहावरा –

• अर्थ निछावर करना – अर्पण करना, समर्पित करना।

Balbharti Maharashtra State Board Class 10 Marathi Solutions Kumarbharti Chapter 4 उत्तमलक्षण (संतकाव्य) Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Marathi Kumarbharti Chapter 4 उत्तमलक्षण (संतकाव्य)

Marathi Kumarbharti Std 10 Digest Chapter 4 उत्तमलक्षण Textbook Questions and Answers

प्रश्न. पुढील कवितेच्या आधारे दिलेल्या सूचनांनुसार कृती करा :

कृती १ : (आकलन)

प्रश्न 1.
आकृत्या पूर्ण करा :
(i)

उत्तर :

(ii)

उत्तर :

प्रश्न 2.
चौकटी पूर्ण करा :

उत्तर :

प्रश्न 3.
पुढील गोष्टींबाबत संत रामदास कोणती दक्षता घ्यायला सांगतात. (मार्च ‘१९)

उत्तर :

कृती २ : (आकलन)

प्रश्न 1.
शब्दजाल पूर्ण करा :
(i)

उत्तर :

(ii)

उत्तर :

(iii)

उत्तर :

प्रश्न 2.
पुढील व्यक्तींशी कसे वागावे, असे संत रामदास म्हणतात :

उत्तर :

प्रश्न 3.
पुढील गोष्टींबाबत कोणती दक्षता घ्यावी, ते लिहा :

उत्तर :
(i) तोंडाळासी भांडू नये.
(ii) संतसंग खंडू नये.
(iii) सत्यमार्ग सोडू नये.

प्रश्न 4.
असत्य विधान ओळखा :
(i) संतसंग सोडू नये.
(ii) अपकार घेऊ नये.
(iii) व्यापकपण सांडू नये.
(iv) खोटेपणाच्या पंथाला जाऊ नये.
उत्तर :
असत्य विधान – अपकार घेऊ नये.

प्रश्न 5.
अचूक विधान ओळखा : (मार्च ‘१९)
(i) पैज, होड लावावी.
(ii) सत्याची वाट धरावी.
(iii) पापद्रव्य सहज जोडावे.
(iv) नेहमी अभिमानाने वागावे.
उत्तर :
अचूक विधान – सत्याची वाट धरावी.

प्रश्न 6.
तुमच्यातील प्रत्येकी तीन गुण व तीन दोष शोधून लिहा.
उत्तर :
नमुना उत्तर :
गुण : (i) मी रोज व्यायाम करतो.
(ii) मी खोटे बोलत नाही.
(iii) मी आईला कामात मदत करतो.

दोष : (i) मला चटकन राग येतो.
(ii) मी ताटात अन्न टाकतो.
(iii) माझे अक्षर चांगले नाही.

कृती ३ : (काव्यसौंदर्य)

प्रश्न 1.
‘सभेमध्ये लाजों नये। बाष्कळपणे बोलों नये।’ या ओळीतील विचार स्पष्ट करा.
उत्तर :
‘उत्तमलक्षण’ या कवितेमध्ये संत रामदास यांनी आदर्श गुणसंपन्न व्यक्तीची लक्षणे समजावून सांगितली आहेत. त्यांपैकी एक लक्षण उपरोक्त चरणात सूचित केले आहे.

मनुष्य हा समाजप्रिय प्राणी आहे. माणसांमध्ये तो नित्य वावरत असतो. समूहामध्ये आदर्श व्यक्तीचे वर्तन कसे असावे, हे सांगताना संत रामदास म्हणतात – सभेमध्ये वावरताना, आपले मत मांडताना कधीही लाजू नये. स्पष्टपणे आपले म्हणणे मांडावे; परंतु त्याच वेळी बालिशपणे बोलून आपले हसे करून घेऊ नये. निरर्थक असे वक्तव्य करू नये. बाष्कळपणे बोलू नये. उत्तम पुरुषाचे एक मर्मग्राही लक्षण या ओवीतून मांडले आहे.

प्रश्न 2.
‘आळसें सुख मानूं नये,’ या ओळीचा तुम्हांला समजलेला अर्थ स्पष्ट करा.
उत्तर :
‘उत्तमलक्षण’ या ओव्यांमध्ये संत रामदास यांनी उत्तम व्यक्तीची लक्षणे विशद करताना आळस हा माणसाचा शत्रू आहे, असे ठासून प्रतिपादिले आहे.

‘आळस’ हा माणसाच्या अंगी असलेला दुर्गुण आहे. आळसामुळे कार्य करायला उत्साह राहत नाही व त्यामुळे बरीच कामे खोळंबून राहतात. ‘आळसे कार्यभाग नासतो!’ या समर्थ रामदासांच्या उक्तीमध्ये हेच तत्त्व सांगितले आहे. माणसाच्या मनाला जे षड्विकार जडतात, त्यात ‘आळस’ हा एक विकार आहे. दैनंदिन कामांमध्ये आळसाला स्थान देऊ नये. आळसामुळे प्रगती खुंटते, भविष्य अंधारते. आळसामुळे मनाला जडत्व प्राप्त होते व माणूस नाकर्ता होतो. आळशी माणसाला समाजात मान मिळत नाही. म्हणून आळसात सुख मानू नये, समाधान मानू नये, असे समर्थ रामदास सांगतात.

Marathi Kumarbharti Class 10 Textbook Solutions Chapter 4 उत्तमलक्षण Additional Important Questions and Answers

प्रश्न. पुढील कवितेसंबंधी त्याखाली दिलेल्या मुद्द्यांच्या आधारे कृती सोडवा :

प्रश्न 1.
कविता-उत्तमलक्षण.

उत्तर : उत्तमलक्षण.
(१) प्रस्तुत कवितेचे कवी : संत रामदास.
(२) कवितेचा रचनाप्रकार : ओवी.
(३) कवितेचा काव्यसंग्रह : श्रीदासबोध.
(४) कवितेचा विषय : उत्तम माणसाची लक्षणे.
(५) कवितेतून व्यक्त होणारा (स्थायी) भाव : आदर्श माणसे घडवण्याचा ध्यास.

(६) कवितेच्या कवींची लेखनवैशिष्ट्ये : प्रस्तुत कवितेची रचना ओवी या छंदात केलेली आहे. ओवी हा उच्चारणाला सुलभ असा अत्यंत लवचीक रचनाप्रकार आहे. त्यामुळे या कवितेतील भाषा ओवी या छंदाला साजेशी सुबोध व सर्वसामान्यांना सहज समजेल अशी आहे. साहजिक कवितेची आवाहन शक्ती वाढली आहे. चुकीचे वर्तन व चांगले वर्तन या दोन्ही बाबी समर्थांनी स्पष्ट शब्दांत कोणतीही संदिग्धता न ठेवता सांगितल्या आहेत. समर्थांच्या या रचनेत प्रासादिकता आढळते.

(७) कवितेची मध्यवर्ती कल्पना : प्रस्तुत कवितेत संत रामदासांनी उत्तम, गुणसंपन्न, आदर्श व्यक्तीची लक्षणे सांगितली आहेत. समाजात वावरताना कसे वागावे, काय करावे व काय टाळावे यांचे सुस्पष्ट शब्दांत निवेदन केले आहे. समाज नीतिमान व कर्तबगार व्हावा, ही तळमळ या पदयपाठातून स्पष्टपणे जाणवते.

(८) कवितेतून व्यक्त होणारा विचार : प्रस्तुत कवितेत संत रामदासांनी, व्यक्तीने समाजात कसे वागावे याचे मार्गदर्शन केले आहे. नेहमीच सावध मनाने वागावे. इतरांना सहकार्य करावे. कोणाशीही कपटाने वागू नये. तोंडाळ, वाचाळ माणसांना टाळावे. आळस झटकून टाकावा. पूर्ण विचार करून वागावे. उपकाराची परतफेड करावी. नेहमी उदारपणाने वागावे. मनाचा मोठेपणा बाळगावा. परावलंबी होऊ नये. नेहमी सत्याने वागावे. कुप्रसिद्धी टाळावी इत्यादी अनेक गुणांचे आचरण करण्यास या कवितेत संत रामदासांनी सांगितले आहे.

(९) कवितेतील आवडलेलो ओळ :
अपकीर्ति ते सांडावी। सत्कीर्ति वाडवावी।।
विवेके दृढ धरावी । वाट सत्याची ।।

(१०) कविता आवडण्याची वा न आवडण्याची कारणे : मला ही कविता खूप आवडली. एकतर हे विचार पटकन पटण्यासारखे आहेत. ही कविता उच्चारताना, वाचताना आनंद होतो. उच्च, उदात्त विचार मनात घोळवल्याने मनही आनंदित होते. सर्व गुण-अवगुण रामदासांनी अत्यंत स्पष्टपणे, परखडपणे सांगितले आहेत. कुठेही शब्दांचा बोजडपणा नाही. प्रत्येक शब्दागणीक अर्थ स्पष्ट होत जातो. मुख्य म्हणजे दैनंदिन जीवनात वागताना आवश्यक असलेल्या गुणांचे मार्गदर्शन घडत असल्याने कविता आपली, स्वतःची, स्वतःसाठी असलेली वाटत राहते.

(११) कवितेतून मिलणाग संदेश : प्रत्येक व्यक्तीने चांगले वागण्याचाच प्रयत्न केला पाहिजे. त्यासाठी वाईट गुण कोणते व चांगले गुण कोणते हे स्पष्टपणे समजून घ्यावे. प्रत्येकाने उत्तम माणूस बनण्याचा प्रयत्न करावा. यातूनच चांगला व समर्थ समाज निर्माण होतो.

कृतिपत्रिकेतील प्रश्न २ (इ) साठी…

प्रश्न. पुढील ओळींचे रसग्रहण तुमच्या शब्दांत लिहा :

प्रश्न 1.
‘जनी आर्जव तोडूं नये । पापद्रव्य जोडूं नये ।
पुण्यमार्ग सोडूं नये । कदाकाळीं ।।’ (मार्च ‘१९)
उत्तर :
आशयसौंदर्य : ‘उत्तमलक्षण’ या श्रीदासबोधातील एका समासामध्ये समर्थ रामदासांनी गुणसंपन्न आदर्श व्यक्तिमत्त्वाची महत्त्वाची लक्षणे सांगितली आहेत. त्यांपैकी उपरोक्त ओवीमध्ये तीन लक्षणांचा ऊहापोह केला आहे.

काव्यसौंदर्य : समाजात वावरताना व्यक्तीने कोणते आचरण करावे हे सांगताना संत रामदास म्हणतात – लोकांचे मन मोडू नये. लोकांनी केलेली विनंती धुडकावू नये. उलट जनभावनांचा आदर . करावा. तसेच वाईट मार्गाने संपत्ती साठवू नये. अशी संपत्ती हे पापाचे धन असते. म्हणून सत्शील मार्गाने जीवन व्यतीत करावे. पुण्यमार्ग आचारावा. कधीही पुण्यमार्गाने जाण्याचे सोडू नये.

भाषिक वैशिष्ट्ये : वरील ओवीमध्ये जनांसाठी खूप सुगम निरूपण केले आहे. ‘तोडू नये, जोडू नये, सोडू नये’ अशा सोप्या यमकांद्वारे संदेशामध्ये आवाहकता आली आहे. ओवीछंदाला साजेशी सुबोध भाषा वापरल्यामुळे जनमानसावर तत्त्व ठसवणे सुलभ झाले आहे. पापद्रव्य व पुण्यमार्ग यांतील विरोधाभास ठळकपणे उठून दिसतो. ओवीमध्ये प्रासादिकता हा गुण आढळतो.

प्रश्न 2.
‘अपकीर्ति ते सांडावी। सत्कीर्ति वाडवावी।
विवेकें दृढ धरावी। वाट सत्याची।।’ (मार्च ‘१९)
उत्तर :
आशयसौंदर्य : संत रामदासांनी ‘उत्तमलक्षण’ या कवितेत आदर्श गुणवान व्यक्तीची वैशिष्ट्ये सांगताना या ओळींमधून सद्वर्तन कशा प्रकारे करावे, याची शिकवण दिली आहे.

काव्यसौंदर्य : संत रामदास म्हणतात – लोक आपल्याला दूषणे देतील व निंदा करतील असे वर्तन कदापिही करू नये. ज्या वागण्याने आपली अपकीर्ती होईल, असे वागणे टाळावे. उलट आपल्या व्यक्तिमत्त्वाची कीर्ती पसरेल, अशी वागणूक करायला हवी. स्वतः चांगले वागून सत्कीर्ती वाढवायला हवी. त्यासाठी बुद्धीचा विवेक महत्त्वाचा ठरतो. सद्विचाराने, विवेकाने सत्याचा मार्ग ठामपणे आचरावा. विवेकबुद्धी ठोस असणे गरजेचे आहे.

भाषिक वैशिष्ट्ये : सन्मार्गाचे लक्षण सांगताना जनसामान्यांना समजतील असे तीन मुद्दे या ओळीत सहजपणे सांगितले आहेत. अपकीर्ती व सत्कीर्ती तसेच सांडावी व वाढवावी या विरोधी शब्दांमुळे ओवीची खुमारी वाढली आहे. दृढ धरणे हा वाक्प्रचार चपखलपणे उपयोगात आणला आहे. जनमानसावर तत्त्व ठसवण्याची समर्थांची हातोटी समर्थपणे व्यक्त झाली आहे.

व्याकरण व भाषाभ्यास

कृतिपत्रिकेतील प्रश्न ४ (अ) आणि (आ) यांसाठी…

अ. व्याकरण घटकांवर आधारित कृती:

१. समास :

प्रश्न 1.
पुढील तक्ता पूर्ण करा : (ठळक अक्षरांत उत्तरे दिली आहेत.)

२. अलंकार :
पुढील ओळी वाचून तक्ता पूर्ण करा :
प्रश्न 1.
‘ऊठ पुरुषोत्तमा । वाट पाही रमा ।
दावि मुखचंद्रमा । सकळिकांसी।।’

प्रश्न 2.
‘नयनकमल’ हे उघडित हलके जागी हो जानकी.

उत्तर :

३. वृत्त :
पुढील ओळींचे लगक्रम लिहा :

प्रश्न 1.
ऐकू न ये तुज पिकस्वर मंजुळे का?
उत्तर :

प्रश्न 2.
मना सज्जना भक्तिपंथेचि जावे.
उत्तर :

उत्तमलक्षण Summary in Marathi

कवितेचा भावार्थ :
उत्तम पुरुषाची (आदर्श व्यक्तीची) लक्षणे सांगताना संत रामदास श्रोत्यांना म्हणतात – श्रोतेहो, तुम्हांला आता मी उत्तम, गुणवान व्यक्तीची लक्षणे सांगतो, ती सावध मनाने ऐकावीत. ही लक्षणे ऐकून तुम्ही सर्व गोष्टी जाणण्याची खूण अंगी बाणवावी. ।।१।।

पूर्ण माहिती असल्याशिवाय कुठल्याही रस्त्याने जाऊ नये. वाटेमधील अडथळे जाणून घेतल्याशिवाय प्रवास करू नये. फळाचे अंतरंग (गुण) ओळखल्याशिवाय ते खाऊ नये. रस्त्यात पडलेली वस्तू अचानक पटकन उचलू नये. ।।२।।

लोकांनी केलेली विनंती लक्षात घ्यावी. तिला अमान्य करू नये. पाप किंवा कपट करून मिळवलेल्या संपत्तीचा साठा करू नये. कपट करून पैसा मिळवू नये. पुण्याचा मार्ग म्हणजे सच्छील मार्ग, सद्वर्तन कधीही सोडू नये. ।।३।।

जी तोंडाळ, भांडकुदळ व्यक्ती असेल, त्या व्यक्तीशी कधीही भांडू नये. व्यर्थ बडबड करणाऱ्या वाचाळ माणसाशी तंटा करू नये. सज्जनाची संगत कधी मध्येच सोडू नये, (हे कायम मनात ठेवा.) मनापासून झालेली संतांची मैत्री सोडू नये. ।।४।।

काम न करता ऐदीपणे, आळस करून आनंद घेऊ नये. आळसात सुख नसते. कुणाबद्दल उखाळ्यापाखाळ्या करून चहाडी करू नये. दुसऱ्याबद्दल खोटेनाटे बोलणे मनातही आणू नये. संपूर्ण शोध घेतल्याशिवाय कोणतेही काम करू नये. (काम अर्धवट सोडू नये.) ।।५।।

सभेमध्ये, समूहाच्या बैठकीमध्ये लाजू नये, मुखदुर्बळ राहू नये. मोकळ्या मनाने बोलावे परंतु बालिशपणे बोलू नये. बाष्कळ बडबड करू नये, कोणत्याही प्रकारची पैज, शर्यत लावू नये. कुणाशीही स्पर्धा करू नये. ।।६।।

कोणाचे उपकार सहसा घेऊ नयेत आणि घेतलेच तर त्यांची लगेच परतफेड करावी, उपकारातून लवकर उतराई व्हावे. दुसऱ्यांना दुःख देऊ नये, त्यांना व्यथित करू नये. कुणाचा विश्वासघात किंवा बेइमानी मुळीच करू नये. ।।७।।

स्वत:च्या मनाचा मोठेपणा कधी सोडून देऊ नये. मन कोते करू नये, ते व्यापक ठेवावे. परावलंबी होऊ नये. कुणावरी आपल्या आयुष्याचे ओझे लादू नये. ।।८।।

सत्याचा मार्ग कधी सोडू नये. खरेपणाने वागावे, असत्याच्या, खोटेपणाच्या मार्गाने जाऊ नये. खोटारडेपणा करून वागू नये. खोटेपणाचा कधीही वृथा अभिमान, व्यर्थ गर्व करू नये. ।।९।।

अपकीर्तीला बळी पडू नये. कुप्रसिद्धी टाळावी. चांगली कीर्ती वाढवावी. चांगल्या प्रकारे प्रसिद्ध पावावे. सारासार विचाराने, विवेकाने वर्तन करून सत्यमार्ग पत्करावा. ।।१०।।

उत्तमलक्षण शब्दार्थ

• उत्तम – आदर्श, गुणसंपन्न,
• लक्षण – गुणवैशिष्ट्य.
• श्रोतीं – ऐकणाऱ्या लोकांनो.
• सावधान – सावध होणे, सजग होणे.
• उत्तम – उत्कृष्ट, जेणें – ज्याने.
• बाणे – बाणणे, अंगिकारणे, सवय लागणे,
• सर्वज्ञ – सारे जाणणारा.
• पुसल्याविण – विचारल्याशिवाय,
• येकायेकी – एकदम, पटकन.
• जनीं – लोकांचे. आर्जव – विनंती.
• पापद्रव्य – पापाने (कपटाने) मिळवलेली संपत्ती.
• जोडू नये – साठवू नये.
• पुण्यमार्ग – चांगला रस्ता, सद्वर्तन,
• कदाकाळी – कोणत्याही वेळी.
• तोंडाळ – वाटेल ते बोलणारा, भांडकुदळ.
• वाचाळ – व्यर्थ बडबड करणारा.
• तंडो नये – तंटा (भांडण) करू नये.
• संतसंग – सज्जन माणसाची संगत.
• खंडू नये – तोडू नये.
• अंतर्यामी – मनातून, हृदयातून.
• आळस – काम न करणे.
• चाहाडी – एखादयाबद्दल वाईट सांगणे, आगलावेपणा.
• कार्य – काम,
• सभा – समूहाची बैठक.
• बाष्कळपणा – बालिशपणा.
• पैज – स्पर्धा, शर्यत.
• होड – पैज,
• परपीडा – दुसऱ्याला छळणे, दुःख देणे.
• विश्वासघात – बेइमानी.
• व्यापकपण – (मनाचा) मोठेपणा.
• पराधेन – परावलंबी, दुसऱ्यावर विसंबणे.
• वोझें – (स्वत:चा) भार,
• कोणीयेकासी – कोणावरही.
• सत्यमार्ग – खऱ्याचा मार्ग, सद्वर्तन.
• असत्य – खोटेपणा.
• पंथे – मार्गाने, वाटेने.
• कदा – कधीही.
• अभिमान – व्यर्थ गवं.
• अपकीर्ति – बेअब्रू, वाईट प्रसिद्धी,
• सत्कीर्ति – चांगली प्रसिद्धी.
• वाडवावी – वाढवावी.
• विवेके – चांगल्या विचाराने.
• दृढ – ठाम, मजबूत, ठोस.

Maharashtra State Board Class 10 Social Science Book Solutions

Maharashtra State Board Class 10 History Solutions Answers

• Chapter 1 Historiography Development in the West
• Chapter 2 Historiography Indian Tradition
• Chapter 3 Applied History
• Chapter 4 History of Indian Arts
• Chapter 5 Mass Media and History
• Chapter 6 Entertainment and History
• Chapter 7 Sports and History
• Chapter 8 Tourism and History
• Chapter 9 Heritage Management

Maharashtra State Board Class 10 Political Science Solutions Answers

• Chapter 1 Working of the Constitution
• Chapter 2 The Electoral Process
• Chapter 3 Political Parties
• Chapter 4 Social and Political Movements
• Chapter 5 Challenges faced by Indian Democracy

Maharashtra State Board Class 10 Geography Solutions Answers

• Chapter 1 Field Visit
• Chapter 2 Location and Extent
• Chapter 3 Physiography and Drainage
• Chapter 4 Climate
• Chapter 5 Natural Vegetation and Wildlife
• Chapter 6 Population
• Chapter 7 Human Settlements
• Chapter 8 Economy and Occupations
• Chapter 9 Tourism, Transport and Communication

Balbharti Maharashtra State Board Class 10 English Solutions Unit 2.3 Connecting the Dots Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 English Kumarbharati Textbook Solutions Unit 2.3 Connecting the Dots

Maharashtra Board Class 10 English Solutions Unit 2.3 Warming Up Questions and Answers

Question 1.
(A) Connect the dots to get what means a lot to you.

The word is ……………………………… .
Answer:

The word is SUN

(B) With your benchmark, use the letters given above to make a word register of ‘computers’. Set a time -limit of 5 minutes and compare your list with that of other classmates.
Answer:
AI, adobe, android, arithmetic, binary, browser, byte, bit, download, document, data, database, disk, format, http, hardware, homepage, java, keyboard, key, Microsoft, malware, memory, network, netscape, program, reboot, spam, spreadsheet, software, virus, web, windows, update, zip.

Question 2.
You are quite familiar with computers, especially the personal computer. Form pairs and make a list of famous computer manufacturing companies. One is given to you.
(a) Apple
(b) ……………………….
(c) ……………………….
(d) ……………………….
(e) ……………………….
(f) ……………………….
Answer:
(a) Apple
(b) Samsung
(c) IBM
(d) Lenovo
(e) Foxconn
(f) HP Inc.

Question 3.
Complete the web by filling the various benefits of computers.

Answer:

Question 4.
(A) Expand the following into their full forms.
(a) that’s
(b) didn’t
(c) here’s
(d) can’t
(e) I’ve
Answer:
(a) that’s – that is
(b) didn’t – did not
(c) here’s – here is
(d) can’t – cannot
(e) I’ve – I have

(B) Write the shortened forms of the following.
(a) You have ……………………….
(b) I would ……………………….
(c) It is ……………………….
(d) You are ……………………….
(e) He will ……………………….
(f) I had ……………………….
(g) will not ……………………….
(h) shall not ……………………….
(i) are not ……………………….
(j) need not ……………………….
(k) must not ……………………….
(l) ought not ……………………….
Answer:
(a) you have – you’ve
(b) I would – I’d
(c) it is – it’s
(d) you are – you’re
(e) he will – he’ll
(f) I had – rd
(g) will not – won’t
(h) shall not – shan’t
(i) arc not – aren’t
(j) need not – needn’t
(k) must not – mustn’t
(l) ought not – oughtn’t

Connecting the Dots Class 10 English Workshop Questions and Answers Maharashtra Board

Question 1.
(A) Rearrange the incidents in the life of Steve Jobs in chronological order.
(a) Steve Jobs started Next.
(b) Jobs underwent a surgery.
(c) Jobs learned about serif and sans serif type faces.
(d) Jobs returned to Apple Inc.
(e) Jobs married Laurene.
(f) Jobs was diagnosed with cancer.
(g) Jobs dropped out of Reed College.
Answer:
(a) Jobs started Next.
(b) Jobs underwent surgery.
(c) Jobs learned about serif and san serif typefaces.
(d) Jobs returned to Apple Inc.
(e) Jobs married Laurene.
(f) Jobs was diagnosed with cancer.
(g) Jobs dropped out of Reed College.

(B) Read the third story again. Complete the flow-chart given below.

Answer:

Question 2.
Read the lesson. Refer to a dictionary and match the words in column ‘A’ with their meanings in column ‘B’.

No	‘A’		‘B’
(a)	diagnosis	(i)	the power believed to control events
(b)	devastating	(ii)	complete list of items especially in a special order and description.
(c)	intuition	(iii)	act of identifying the nature of a problem or illness.
(d)	calligraphy	(iv)	power of understanding situations or people’s feelings before hand.
(e)	destiny	(v)	causing great destruction
(f)	catalogue	(vi)	beautiful handwriting done with a special pen or brush.

Answer:

No	‘A’		‘B’
(a)	diagnosis	(iii)	act of identifying the nature of a problem or illness.
(b)	devastating	(v)	causing great destruction
(c)	intuition	(iv)	power of understanding situations or people’s feelings before hand.
(d)	calligraphy	(vi)	beautiful handwriting done with a special pen or brush.
(e)	destiny	(a)	the power believed to control events
(f)	catalogue	(ii)	complete list of items especially in a special order and description.

Question 3.
Go through all the three stories. Identify some qualities of Steve Jobs and complete the web chart.

Answer:

Question 4.
Complete the following table.
‘The Three Stories in the Life of Steve Jobs’

	About	Setbacks	Reactions	Achievements and benefits
First story
Second story
Third story

Answer:

	About	Setbacks	Reac­tions	Achieve­ments and benefits
First story	College days	no boarding and lodging	managed with friends and at a temple	learnt calligraphy
Second story	starting ’Apple’	fired from the company	started a new company ‘Next’	most creative period; renaissance’ of ‘Apple’
Third story	death	diagnosed with cancer	surgery, cure	learnt not to waste time living someone else’s life

Question 5.
Say HOW?
→ the calligraphy classes helped Steve Jobs after 10 years.
Answer:
10 years later when Steve Jobs was designing the first Macintosh computer, he recalled what he had learned in the calligraphy classes about serif and san serif typefaces as well as other elements that go into great typography. He designed all this into the Macintosh computer.

→ You can connect dots.
Answer:
We can’t connect dots by looking forward; we can only connect them looking backwards. We have to trust that the dots will somehow connect in our destiny.

→ Jobs reacted later on, after the shock of being fired from Apple.
Answer:
After he overcame the shock of being fired from Apple, Jobs felt the lightness of being a beginner again. He felt free to enter one of the most creative periods of his life. He started two companies – Next and Pixar.

→ Jobs was cured of a rare cancer.
Answer:
Jobs was diagnosed with a very rare form of pancreatic cancer. It was curable by surgery. Jobs had the surgery and he was cured.

→ Jobs acquired the famous words ‘Stay Hungry. Stay Foolish’
Answer:
When Jobs was young, he used to read an amazing publication called The Whole Earth Catalogue’. In the final issue was a photograph of an early morning country road beneath which were the words ‘Stay Hungry, Stay Foolish’. Steve acquired these words from the magazine.

Question 6.
Besides those given at the end of the talk by Steve Jobs, pick out other pieces of advice that Jobs gives in his speech.
Answer:
(a) Learn to connect the dots.
(b) Learn whatever you can whenever you can.
(c) We can learn even from negative experiences.
(d) Have trust that somehow the dots will connect in your future.
(e) Have trust in your gut. destiny, life, karma, whatever …
(f) Love what you do.
(g) Keep looking for what you love and don’t ever settle for less.
(h) Even under the most devastating circumstances, you can start all over again.
(i) Don’t lose faith.
(j) Your time is limited, so don’t waste it lining someone else’s life.

Question 7.
(A) Use the following idioms/phrases in sentences of your own.
(a) drop in …………………..
(b) drop out …………………..
(c) stumble on …………………..
(d) look backwards …………………..
(e) look forward …………………..
(f) let (someone) down …………………..
(g) sign off …………………..
(h) begin anew …………………..
Answer:
(a) Though I am not a member of the club, I often play tennis there as a drop in.
(b) I decided to drop out of karate classes as it was taking too much of my time.
(c) While surfing the net, he stumbled into a portal that showed only horror movies.
(d) After his terrible experience in his native place, he went to the city and never looked backwards.
(e) He tried to forget his terrible experience in his native place and looked forward.
(f) He placed great trust in his secretary, and she never let him down.
(g) I would like to sign off by telling you a short story with a beautiful message.
(h) It is never too late to forget past mistakes and begin anew.

(B) Fill in the gaps in choosing the appropriate idioms.
(drown out, hits in the head with a brick, get one’s affairs in order, connect the dots, begin to dawn, stay hungry)
(a) The wealthy landlord made a will before he could die.
(b) Those who aspire for success should always to learn more.
(c) If you you will realise that crime ultimately leads to poverty.
(d) The siren of the ambulance all other traffic noise.
(e) When the father learned about his son’s misdeeds, it
(f) On reading exactly same essays in both answer sheets, it the examiner, that the students had cheated during exams.
Answer:
(a) Getting his affairs in order
(b) Stay hungry
(c) Connect the dots
(d) Drowned out
(e) It hit him on the head with a brick.
(f) Began to dawn on

Question 8.
(A) Name the Tense of the Verbs underlined to include Time (Past/Present/Future) and Aspect (Simple/Continuous/Perfect/Perfect Continuous)
(1) I slept on the floor
(2) We were designing the first Macintosh computer
(3) It had made all the difference
(4) I am fine, now
(5) I have been facing death
(6) I shall be telling you three stories
Answer:
(1) Time – Past; Aspect – Simple.
(2) Time – Past Tense; Aspect – Progressive (continuous).
(3) Time – Past tense Aspect – Perfect
(4) Time – Present; Aspect – Simple.
(5) Time – Present; Aspect – Perfect.
(6) Time – Future; Aspect – Progressive (continuous).

(B) Change the Tense as instructed.
(1) I got fired. (Future Perfect)
(2) Life hits you in the head. (Present Perfect Continuous)
(3) The dots will somehow connect. (Past Perfect)
(4) I started a company. (Present Continuous)
(5) My doctor advised me. (Past Perfect Continuous)
Answer:
(1) Jobs got fired.
(2) Life hits you in the head.
(3) The dots had somehow connected.
(4) I started a company.
(5) My doctor had been advising me.

Question 9.
Read the News item and write an application for a suitable job in the same company. Attach a seperate CV/Resume.
May 19, 2016
Apple Opens Development Office in Hyderabad
(A) The new office in Hyderabad will focus on development of maps, Apple products, like iPhone, iPad, Mac. etc. This will create upto 4000 jobs
Answer:

Answer:
Ratan Shah
11 Salsa Apts.
Hafeczpeth
Hyderabad
Telangana – 500 049.
27th May, 2020

The HR Manager
Apple Development Office
18-23, Rd Number 2
Financial District
Nanakram Guda
Hydcrabad
Telangana — 500 032.

Subject : Application for post of Systems Analyst ‘
Sir,
I read the news item in the Times of India’ dated May 20, which stated that Apple has opened a Development office in Hyderabad, which is likely to create up to 4000 jobs. I am interested in applying for the post of Systems Analyst.

I have a bachelor’s degree In Computer Information Systems (CIS) and 6 months experience in Cornputronic& Ltd. as Systems Analyst. I am well-versed in analyzing, designing and implementing Information Systems. I wish to further my prospects and hence am applying to your company.

I do hope that you will give me the opportunity to prove my mettle.

Yours truly,
Ratan Shah

Attachments: CV, photocopies of Academic and Professional Certificates
Resume:
Name: Ratan Shah
Age: 24 years
Educational Qualification: B.Com (1st class); B.Computer Science (CIS)
Experience: 6 months experience in Computronics Ltd. as Systems Analyst
Marital Status: Single
Special Interests: Chess, Football, Cycling
Contact information: Address as above
Mobile no.: 097

(B) Imagine you are already working as an Engineer in Apple Development Office, Hyderabad. Write an application for 2 weeks leave to the HR Manager as you have to undergo an urgent surgery.
Answer:
Ms. Sara Kanchwala
11 Salsa Apts.
Hafeezpeth
Hyderabad
Telangana 500 049.
3 September, 2020

The HR Manager,
Apple Development Office
18-23, Rd. Number 2
Financial District
Nanakram Guda
Hyderabad
Telangana – 500 032.

Subject : ApplicatIon’ for leave of absence due to Imminent surgery (4th September to 17th September)

Dear Sir,
I had a bad fall a couple of days back and tore a ligament ‘In my knee. The pain is Intense and my doctor has advised me that immediate surgery is necessary.

Please grant me 2 weeks medical leave as the doctor has advised a fortnight’s complete bed rest post-op. Thanking you in advance,

I remain,
Yours truly,
Sara Kanchwala

Question 10.
Prepare a speech on the title “The Will to Win” to be delivered before the class during a competition.
Hints –

• Title
• Introduction
• Objective and Illustrations
• Specific examples
• Purpose of the title.
• Sources/Resourses for implementation.
• Usefulness/Benefits
• Conclusion.

Answer:
The Will to Win

Friends,
The topic before us today is The Will to Win’. This title brings to my mind the story of the hare and the tortoise. When the hare challenged the tortoise to a race, the tortoise knew very well that its speed was in no way comparable to that of the hare. Yet this little animal agreed to the race. And why, may I ask you? Only because it had the will to win. Come what may, the tortoise had to prove to the hare that it could win if it wanted to. And finally it won!

This, in fact, is the driving force behind all great ventures and achievements. The Will to Win!

I remember the time when I went to Std. V. My father had been transferred from Agra to Bhusaval. A subject that I had trouble mastering was the language Marathi. Needless to say in the first term I flunked very badly. But I was a student who usually scored high marks in all subjects. This failure was devastating.

However, I did not let it faze me. I took great interest in class lessons. Read my Marathi textbook over and over. Made friends who spoke fluently in Marathi and ventured speaking to them. At first they poked fun at me. But within a month I had picked up the basic structure of the language and began writing answers to questions on my own. Believe it or not, at the end of the term I topped the class in Marathi. ; Today I can speak in the language as if it is my mother tongue.

Friends, my message to you today is that you can do anything. All you require is the will. If you have the will, the skill will follow. So set your goals, define your objectives and I wish all of you the very best in ; whatever you decide to do in life.

Be sure of one thing: If you have the will, You Will Win!

Thank you.

Question 11.
Project :
Collect more information about Steve Jobs with the help of Internet. Complete it with images into a file.

Question 12.
Do you remember doing some activities in your childhood that you didn’t like it. Form pairs and make a list of all those activities. Do you think, any one of these activities have helped you in solving your problems? Share your experience with the class.

• chopping vegetables:
• cleaning the home; sweeping and dusting
• hanging out clothes to dry
• looking after my little brother
• helping dad to repair the bicycle

Answer:
Once while driving home from school, the chain of my bicycle got dislodged and I almost fell off. There was no bicycle repair shop nearby and no passers-by were able to help me. Since I had experience helping dad to repair his bicycle, somehow I managed to get the chain fixed and wobbled back home. My hands were black and dirty with grease and oil, but at least I didn’t have to walk back home, pushing my bicycle along.

Question 13.
Write True or False for these statements: (The answers are given directly and underlined.)
Answer:
(1) Steve Jobs slept in his dorm room. False
(2) Steve took his required courses as a registered student of Reed College. False
(3) During Steve’s College days, one had to pay 5 cents deposit for a Coke bdttle. True
(4) Steve ha’d comfortable college experiences. False

Question 14.
What basic course in Reed College helped while designing the Mac?
Answer:
The basic course of calligraphy in Reed College helped while designing the Mac.

Question 15.
Arrange the following incidents in Steve Jobs’ life in proper sequence based on this passage:
(a) Jobs stayed as a drop-in for 18 months.
(b) Jobs used to sleep on the floor in his friend’s house.
Answer:
(a) Jobs stayed as a drop-in for 18 months.
(b) Jobs used to sleep on the floor in his friend’s house.

Question 16.
Complete the following: (The answers are given directly and underlined.)
(1) Steve’s first story is about connecting the dots.
(2) Steve got one good meal every Sunday night at the Hare Krishna temple.
Answer:
Connecting the dots.
Hare Krishna temple.

Question 17.
What did Steve Jobs do for two years after he joined Reed College?
Answer:
After joining Reed College, for two years Steve would stop taking the required classes that didn’t interest him. Instead he began dropping in on the ones that looked interesting.

Question 18.
List the hardships that Steve faced.
Answer:

• Steve didn’t have a dorm room and so had to sleep on the floor in friends’ rooms.
• Steve returned Coke bottles for the 5 cent deposits to buy food with.
• Steve would walk seven miles across town every Sunday night to get one good meal a week at the Hare Krishna temple.

Question 19.
Write from the passage a phrase that means ‘Apart from this, I have nothing more to tell.’
Answer:
That’s it.

Question 20.
Choose the correct contextual meaning of the phrase: connecting the dots.
(a) making a pattern with the help of dots
(b) associating one previous idea with other ideas that follow
(c) joining dots in a puzzle to get the correct picture
(d) understanding a procedure
Answer:
(b) associating one previous idea with other ideas that follow

Question 21.
Match the words in column A with their meanings in column B:

A		B
(i)	intuition	(a)	a set of characters like letters, symbols, etc. in one design
(ii)	calligraphy	(b)	the style and appearance of printed matter
(iii)	typography	(c)	power of understanding the feelings of people
(iv)	typeface	(d)	beautiful handwriting done with a special pen or brush

Answer:

(i)	intuition	(c)	power of understanding the feelings of people
(ii)	calligraphy	(d)	beautiful handwriting done with a special pen or brush
(iii)	ypography	(b)	the style and appearance of printed matter
(iv)	typeface	(a)	a set of characters like letters, symbols, etc. in one design

Question 22.
It was one of the best decisions I ever made. (Rewrite using ‘better than’.)
Answer:
It was better than most other decisions I ever made.

Question 23.
Personal Response: What impression of Steve Jobs do you get from this passage?
Answer:
In this passage, it appears that Steve Jobs is a student who goes by his impulses. He has a thirst for knowledge, but prefers subjects that he finds interesting and avoids those that he finds uninteresting, even though he has enrolled for them. He is ready to face all kinds of hardships in order to study what he wants to. Steve Jobs had a natural curiosity and intuition. He also had an instinct about what makes something really great and the habit of storing it away in his mind for future use.

Question 24.
Fill in the blanks: (The answers are given directly and underlined.)
(1) Steve designed the Macintosh computer.
(2) Windows copied the Macintosh computer.
(3) It was impossible to connect the dots looking forward when Steve was in college.
(4) You can only connect the dots looking backward.
Answer:
(1) Macintosh
(2) Windows
(3) forward
(4) backward

Question 25.
Complete the following: (The answers are given directly and underlined.)
(1) The Mac ‘computer which he designed would not have had multiple typefaces or proportionally spaced fonts.
(2) Since Windows just copied Mac, it was likely no personal computer would have had them.
Answer:
(1) not have had multiple typefaces or proportionally spaced fonts.
(2) no personal computer would have had them.

Question 26.
Complete the following: (The answers are given directly and underlined.)
Words connected with typography from the passage are: typeface and font.
Answer:
typeface and font.

Question 27.
Match the words /phrases in column A with their meanings in column B:

(A)		(B)
(i)	gut	(a)	Macintosh computer.
(ii)	destiny	(b)	having several parts.
(iii)	Mac	(c)	the power believed to control events.
(iv)	multiple	(d)	courage and determination.

Answer:

A		B
(i)	gut	(d)	courage and determination
(ii)	destiny	(c)	the power believed to control events
(iii)	Mac	(a)	Macintosh computer
(iv)	multiple	(b)	having several parts

Question 28.
It was impossible to connect the dots looking forward. (Pick out the verbs and say if they are finite or non-finite.)
Answer:
was – finite; to connect, looking – non-finites

Question 29.
It was very clear. (Rewrite as an exclamatory sentence.)
Answer:
How clear it was!

Question 30.
Mac would never have had multiple typefaces. (Rewrite as an interrogative sentence.)
Answer:
Would Mac ever have had multiple typefaces?

Question 31.
Personal Response: Write about something which you learned in the past and which has helped you in the present.
Answer:
When I was in Std. V, mother enrol led me for dancing classes, I had to stop when I reached Std. IX. However, now whenever I need a break or I am feeling stressed, I put on some music and dance. That gives me relief from stress.

Question 32.
Name the following: (The answers are given directly and underlined.)
Answer:
(1) The world’s most successful animation studio. Pixar
(2) The company that Steve Jobs took five years to establish. Next
(3) The company that bought Next. Apple
(4) Steve Jobs’ wife. Laurene

Question 33.
What setback did Jobs suffer when he was thirty?
Answer:
When Jobs was thirty, he was fired from the company which he himself had started. This was devastating and a major setback in his life.

Question 34.
Match the words/phrases in column A with their meanings in column B:

A		B
(i)	renaissance	(a)	causing great destruction
(ii)	to start over	(b)	lost one’s job
(iii)	fired	(c)	revival
(iv)	devastating	(d)	to begin again

Answer:

A		B
(i)	renaissance	(c)	revival
(ii)	to start over	(d)	to begin again
(iii)	fired	(b)	lost one’s job
(iv)	devastating	(a)	causing great destruction

Question 35.
Fill in the blanks choosing the appropriate idioms: (at the heart of, hit him on the head with brick, begin to dawn, lose faith)
(1) Even if you don’t succeed at first, don’t lose faith in yourself.
(2) It is corruption in high places that lies at the heart of the non-development of this locality.
Answer:
(1) Lose faith
(2) At the heart of

Question 36.
Personal Response: What does this second story of Jobs convey to you?
Answer:
The second story of Jobs – about love and loss – conveys to us that even if our efforts result in complete disaster, we should continue believing in ourselves. It tells us that we should love our work. If we haven’t yet found it, we should keep on searching. We should never accept less than what we aim for in life.

Question 37.
Which quality of Steve Jobs impresses you the most? How would you apply it in your life?
Answer:
I admire Steve’s quality of not giving up even after facing a terribly shocking loss. It inspires me never to be disheartened by failure, but to always keep trying. Even if one loses everything, one has to have the courage to start all over again.

Question 38.
Arrange the following incidents in Steve Jobs’ life in proper sequence based on this passage:
(a) Jobs was advised to get his affairs in order.
(b) Jobs was cured of a rare form of cancer.
Answer:
(a) Jobs was advised to get his affairs in order.
(b) Jobs was cured of a rare form of cancer.

Question 39.
Fill in the blanks: (The answers are given directly and underlined.)
(1) Steve’s third story is about death.
(2) The doctor describes the last stage of cancer as the ‘prepare to die’ stage.
(3) Steve ‘says that you should follow your heart and intuition.
(4) We should not waste time.
Answer:
(1) death
(2) ‘prepare to die’
(3) heart and intuition.
(4) time.

Question 40.
What does Jobs warn you about life and dogma?
Answer:
Jobs warns us not be trapped by dogma. Dogma is living with the results of other people’s thinking. The overpowering influence of other’s opinions should not drown out our own inner voice.

Question 41.
How does Jobs close his address to the graduate students?
Answer:
Steve Jobs closes his address to the graduate students by telling them about an amazing publication called The Whole Earth Catalogue’. In the final issue on the back cover there was a photograph of an early morning country road. Beneath it were the words ‘Stay Hungry, Stay Foolish’. Jobs wishes the students the same words as his farewell message to them.

Question 42.
Write from the passage four medical words / phrases / terms.
Answer:
pancreatic cancer, diagnosis, biopsy, surgery

Question 43.
Match the words in column ‘A’ with their meanings in column ‘B’:

A		B
(a)	diagnosis	(1)	a set of beliefs held by an organisation which others are expected to accept without argument.
(b)	catalogue	(2)	act of identifying the nature of a problem or illness.
(c)	dogma	(3)	in a new or different way.
(d)	anew	(4)	complete list of items especially in a special order and description.

Answer:

A		B
(a)	diagnosis	(2)	act of identifying the nature of a problem or illness.
(b)	catalogue	(4)	complete list, of items especially in a special order and description.
(c)	dogma	(1)	a set of beliefs held by an organisation which others are expected to accept without argument
(d)	anew	(3)	in a new or different way.

Question 44.
Choose correct question tags for the sentences and rewrite: wasn’t it? isn’t it? aren’t I? will you?
(a) Don’t be trapped by dogma.
(b) It was their farewell message.
(c) Your time is limited.
(d) I’m fine now.
Answer:
(a) Don’t be trapped by dogma, will you?
(b) It was their farewell message, wasn’t it?
(c) Your time is limited, isn’t it?
(d) I’m fine now, aren’t I?

Question 45.
Don’t waste it. (Rewrite without ‘don’t’.)
Answer:
Avoid wasting it.

Question 46.
Personal Response: “Your time is limited. So don’t waste it living someone else’s life.” What are your views about these statements?
Answer:
These are the conclusions that Steve Jobs came to when he discovered that he was suffering from pancreatic cancer. Coming face to face with death, he realized how very little time we have on earth to realize our true potential. Most of our lives are spent following the dictates of others. By this statement, Steve Jobs means that we should spend every moment of our lives trying to discover our own potential and determine our own purpose and path in life.

Question 47.
(1) Pick out the compound words from the given words: dropping, calligraphy, backwards, graduate, photograph
(2) Pick out the gerund and use it in your own sentence: I could begin dropping in on the ones that looked interesting.
(3) Punctuate the sentence: heres one ekample reed college offered perhaps the best calligraphy instruction in the country
(4) Spot the error and rewrite the correct sentence: At the calligraphy class, I learn about what make great typography great.
(5) Identify the type of sentence: Don’t settle.
(6) Find out two hidden words of at least 4 letters each from the word ‘devastating’.
(7) Form present participles in which the last letter is doubled: drop, quit
(8) Arrange these words in alphabetical order: friend, follow, found, freed.
Answer:
(1) backwards = back + wards; photograph = photo + graph
(2) Gerund: dropping Sentence: Stop dropping in here without any warning.
(3) Here’s one example: Reed College offered perhaps the best calligraphy instruction in the country.
(4) At the calligraphy class, I learned about what makes great typography great.
(5) Imperative Sentence in the Negative.
(6) devastating-vast, taste (gate, stain.)
(7) dropping, quitting
(8) follow, found, freed, friend

Question 48.
(1) Write the homophone for the word ‘story’ and frame a sentence.
(2) Rewrite the sentence using the past perfect continuous tense: We were designing the first Macintosh computer.
(3) Rewrite beginning with underlined part: The heaviness of being successful was replaced by the lightness of being a beginner again.
(4) Write a word register for the word: ‘doctor’ (4 words).
Answer:
(1) Homophones: story – storey Sentence: I live in a seven-storys building.
(2) We had been designing the first Macintosh computer.
(3) The lightness of being a beginner again replaced the heaviness of being successful.
(4) physician, surgeon, medico, healer, practitioner. (MD, medic, scientist, specialist)

Question 49.
(1) Use the word,‘trust’ in sentences as a noun as well as a verb.
(2) Underline the modal auxiliary and state its function: I had to take a calligraphy class.
Answer:
(1) (a) I have great trust in my teacher, (noun)
(b) Trust in God. (verb)
(2) I had to take a calligraphy class, (necessity/ obligation/compulsion)

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 1 Heredity and Evolution Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 1 Heredity and Evolution

Question 1.
Complete the following diagram.

Answer:

Question 2.
Read the following statements and justify same in your own words with the help of suituble examples.
a. Human evolution began approximately 7 crore years ago.
Answer:

• Approximately around 7 crore years back the ice age began on the earth. In such conditions, dinosaurs became extinct. The evolution and diversity of mammals started during this time. Due to change in climate the forest cover also declined rapidly.
• Ancestors of monkey-like animals were Lemur like animals which evolved during this time period.
• The tails of these monkey-like creatures started vanishing very gradually around 4 crore years ago.
• The body and brain both increased in volume forming first ape like animals. The monkey like ancestors gave rise to two evolutionary links to apes and human like animals.
• Later, the human evolution took place by changes in the brain volume, the ability to walk upright, excessive use of hand for manipulations.
• This journey of human evolution began 7 crore years ago. But the true wise and intelligent man arose around 50,000 years ago.

b. Geographical and reproductive isolation of organisms gradually leads to speciation.
Answer:

• Every species survives in specific geographical conditions. The requirements of food and habitat, is specific for each species. Their reproductive ability and period is also different.
• Therefore, the individuals from one species cannot reproduce with individuals from other species.
• When they are separated by a distance or geographical barriers they are said to be isolated geographically.
• When they cannot reproduce with each other, they are said to be isolated reproductively.
• The ancestor species of both these subspecies may be the same but due to isolation over a very long-time duration, there is genetic variation between the two. Therefore, the isolation leads to speciation.

c. Study of fossils is an important aspect of study of evolution.
Answer:
Answer:

• Fossils offer palaeontological evidence for the evolutionary process.
• Due to some natural calamities the organisms get buried during ancient times.
• The impressions and remnants of such organisms remain preserved underground. The hot lava also traps some organisms or their impressions. All such formations form fossils.
• Study of fossils help the researcher to understand the characteristics of the organisms that existed in the past.
• Carbon dating method also helps in finding out exact age of the fossil. According to the structure of earth’s crust the fossils are obtained at specific depths.
• The oldest ones are obtained at the depth while the relatively recent ones occupy the upper surface. Thus fossils of invertebrates were seen in very old Palaeozoic era. Later were seen fossils of Pisces, Amphibia and Reptilia. The Mesozoic era was dominated by reptiles while Coenozoic era showed presence of mammals.
• In this way, study of fossils unfold the evolutionary secrets.

d. There is evidences of fatal Science among chordates.
[Please read the above question as: Among different chordates there are embryological evidences.]
Answer:

• Very young embryos of fish, amphibians, reptiles, birds and mammals show quite similar structure in the early stages.
• As the further growth takes place, they acquire different patterns.
• The initial similarity between the vertebrate embryos is an evidence that during evolution, there was a common ancestor for all the vertebrate classes.
• This is called embryological evidence for vertebrate evolution.

Question 3.
complete the statements by choosing correct options from bracket.
(Genes, Mutation, Translocation, Transcription, Gradual development, Appendix)
a. The causality behind the sudden changes was understood due to ………… principle of Hugo de Vries.
Answer:
Mutation

b. The proof for the fact that protein synthesis occurs through ……….. was given by George Beadle and Edward Tatum.
Answer:
Genes

c. Transfer of information from molecule of DNA to mRNA is called as …………… process.
Answer:
Transcription

d. Evolution means ………….
Answer:
Gradual development

e. Vestigial organ ……….. present in human body is proof of evolution.
Answer:
Appendix

Question 4.
Write short notes based upon the information known to you.
a. Lamarckism.
Answer:
(1) Lamarckism consists of two theories which were proposed by Jean Baptiste Lamarck. These are as follows: (a) Use and disuse of the organs (b) Inheritance of acquired characters.
(2) In theory of use and disuse of organs, Lamarck says : The characters of organs develop because specific activities that the organisms perform. If such organ is not used it gets degenerated. Thus the morphological changes take place due to activities or inactivity of a particular organism.
(3) To emphasise this theory, he quoted following examples. Due to constant extension of neck to eat foliage from the top of the trees, giraffe’s neck became long. Similarly blacksmith has strong arms due to constant work. Flightless ostrich and emu did not fly and hence their wings became useless. Aquatic birds like swan and duck made their feet suitable for swimming by living in water. Snake lost limbs as it tried burrowing mode.
(4) Such acquired characters are passed from one parental generation to the offspring. This is called inheritance of acquired characters.
(5) The theory of inheritance of acquired characters is not accepted as such transmission of acquired character does not take place. Only genetic characters are transmitted.

b. Darwin’s theory of natural selection.
Answer:

• Charles Darwin proposed the theory of natural selection after making many observations on different specimens. He published a concept ‘Survival of the fittest’.
• Darwin explains this concept as follows: All the organisms reproduce prolifically. Therefore, there is always a competition for food, mate, etc. Only adaptations for sustaining this struggle.
• Natural selection plays important role by selecting only those organisms which are fit to live. Those that do not have better adaptations, perish. Selected sustaining organisms then perform reproduction and form new species in a very long period of time.
• Darwin published his views in the book titled ‘Origin of Species’.

c. Embryology.
Answer:

• Embryology is the study of developing embryos.
• These embryos in their initial stages are very similar to each other.
• These similarities decrease later in the development.
• This similarity in initial stages indicate that these vertebrates have originated from a common ancestor.
• In evolutionary science, comparative study of embryos of various vertebrates provide evidence for evolution.

d. Evolution.
Answer:

• The sequential changes in the groups of living organisms that take place very gradually is called evolution.
• Evolution is also described as the formation of new species due to natural selection.
• The process of evolution takes millions of years for development and speciation of different organisms.
• Changes in stars and planets in space and the changes in biosphere occurring on the Earth are all included under study of evolution.
• Due to evolution organisms become fit, biodiversity is increased, and new species are created.
• Different scientists have put forth theories to explain the process of evolution. Among these Charles Darwin’s theory of natural selection and speciation is accepted worldwide.

e. Connecting link.
Answer:
Some living organisms possess some characters in them which are the distinctive features of different groups or phyla. Such individuals connect these two groups by sharing the characters of both and hence they are known as connective links.

Examples: (1) Peripatus: Peripatus is the connecting link between Annelida and Arthropoda. It shows characters of both animal phyla. Like annelid worm, it shows segmented body, thin cuticle and parapodia. Like an arthropod, it shows open circulatory system and tracheal system for respiration.
(2) Duck Billed platypus: This is a connecting link between reptiles and mammals. Like reptiles it lays eggs but like mammals it has mammary glands and hairy skin.
(3) Lung fish: Lung fish is a connecting link between fishes and amphibians. Though a fish, it shows lungs for respiration as in amphibian animals.
(4) Connecting links indicate the direction and hierarchy of evolution.

Question 5.
Define heredity. Explain the mechanism of hereditary changes.
Answer:
(1) Heredity: Heredity is the process by which the biological characters from parental generation are transmitted to the next generation through genes.

(2) The mechanism of hereditary changes:

• Mutation: Sudden change in the parental DNA can cause mutations. This results into changes in the hereditary characters.
• At the time of meiosis, the crossing over takes place. This creates new recombination of the genetic information. Therefore, the haploid gametes produced carry changed hereditary characters.

Question 6.
Define vestigial organs. Write names of some vestigial organs in human body and write the names of those animals in whom same organs are functional.
Answer:

• Vestigial organs are degenerated or underdeveloped organs of organisms which do not perform any function.
• According to the principle of natural selection, such organs are on the verge of disappearance. But it takes many millions of years for its complete vanishing.
• The vestigial organs in one animal may be of use but to other kind of the animal as they still perform regular functions.
• Appendix is vestigial for humans, it does not perform any function but in ruminant animals it is concerned with digestion.
• Ear muscles are vestigial for us but in monkeys and cattle they are functional.
• Names of vestigial organs in human body-Appendix, tail-bone or coccyx, wisdom teeth and body hair.

Question 7.
Answer the following questions.
a. How are the hereditary changes responsible for evolution?
Answer:
Hereditary characters are transmitted from parental generation to the offspring. These characters are maintained through inheritance. But the genes which are beneficial for the organisms in helping them to adapt to the environment are transmitted to the next generations in a greater proportion. This happens due to natural selection.

The process of evolution happens at a very slow pace. The favourable genes are preserved in the species as they bring about better survival of the individuals. Such individual reproduces more efficiently and evolve. The individuals with unfavourable genes are not selected by nature and are thus removed from the population through natural death. The fuel for evolution is thus truly supplied by the hereditary changes.

b. Explain the process of formation of complex proteins.
Answer:
The proteins are synthesised in following steps, viz. transcription, translation and translocation. Protein synthesis takes place according to the sequence of nucleotides present on the DNA molecule with the help of RNA molecules. This is known as central dogma of protein synthesis.

1. Transcription: In the process of transcription, mRNA is produced as per the nucleotide sequence on the DNA. For this the two strands DNA are separated. Only one strand participates in the formation of mRNA. The sequence of nucleotides which is complementary to that of present on DNA is copied on mRNA. Instead of thymine present in DNA, uracil is added on the mRNA. Transcription takes place in nucleus but the mRNA leaves nucleus, carries the genetic code and enters the cytoplasm. This genetic code is always in triplet form arid hence is known as triplet codon. The code for each amino acid always consists of three nucleotides.

2. Translation: Each mRNA may carry thousands of codons. But each codon is specific for only one amino acid. The tRNA molecule brings the required amino acid as per the code present on mRNA. There is anticodon on each tRNA which is complementary to the codon on mRNA. This process is known as translation.

3. Translocation: In translocation, the ribosome keeps on moving from one end of mRNA molecule to other end by distance of one triplet codon. While this process is taking place, rRNA, helps in joining the amino acids together by peptide bonds. The peptide chains later come together to form complex protein molecules.

c. Explain the theory of evolution and mention the proof supporting it.
Answer:
1. Theory of evolution:

• According to the theory of evolution, first living material was in the form of protoplasm which was formed in ocean.
• Gradually, it gave rise to unicellular organisms. Changes took place in these unicellular organisms which made them evolve into larger and more complex organisms.
• All evolutionary changes were very slow and gradual taking about 300 crore years to happen.
• Different types of organisms were developed as the changes and development that occurred in living organisms wefts all round and multi-dimensional.
• Hence, this overall process of evolution is called organizational and progressive.
• Variety of plants and animals developed from the ancestors having different structural and functional organization during the process of evolution.

2. Proof here means evidences of evolution.
These evidences are as follows:

• Morphological evidences
• Anatomical evidences
• Vestigial organs
• Palaentological evidences
• Connecting links
• Embryological evidences.

d. Explain with suitable examples importance of anatomical evidences in evolution. (July 2019)
Answer:

• There are similarities in the structure and anatomy of different animal groups. E.g. human hand, forelimb of bull, patagium of bat and flipper of whale are all similar in their internal anatomy. There is similarity in the bones and joints of all these specimens.
• External morphology does not show any similarity. Use of each of the organ is also different in different animals. Structurally, they may not be related.
• However, the similarities in the anatomy is an evidence that they may have a common ancestor.
• In this way, the anatomical evidence throws light on the process of evolution.

e. Define fossil. Explain importance of fossils as proof of evolution.
Answer:

• Fossils offer palaeontological evidence for the evolutionary process.
• Due to some natural calamities the organisms get buried during ancient times.
• The impressions and remnants of such organisms remain preserved underground. The hot lava also traps some organisms or their impressions. All such formations form fossils.
• Study of fossils help the researcher to understand the characteristics of the organisms that existed in the.past.
• Carbon dating method also helps in finding out exact age of the fossil. According to the structure of earth’s crust the fossils are obtained at specific depths.
• The oldest ones are obtained at the depth while the relatively recent ones occupy the upper surface. Thus fossils of invertebrates were seen in very old Palaeozoic era. Later were seen fossils of Pisces, Amphibia and Reptilia. The Mesozoic era was dominated by reptiles while Coenozoic era showed presence of mammals.
• In this way, study of fossils unfold the evolutionary secrets.

f. Write evolutionary history of modern man.
Answer:
(1) Ancestors of humans developed from animals which resembled lemur like animals.
(2) Around seven crore years ago, monkey-like animals evolved from some of these lemur like animals.
(3) Then after about 4 crore years ago, in Africa the tails of these monkey like creatures very gradually disappeared.
(4) Simultaneously, there was enlargement in their body and brain volume too. The hands also improved and were provided with opposable thumb. In this way, ape-like animals were evolved.
(5) These ape-like animals independently gave rise to two lines of evolution, one giving rise to apes like gibbon and orangutan in the South and North-East Asia and gorilla and chimpanzee which stayed in Africa around 2.5 crores of years ago.
(6) The other line of evolution gave rise to human like animals around 2 crore years ago.
(7) The climate became dry and this resulted into reduction of forest cover. This made arboreal apes to descend on the land and start terrestrial mode.
(8) Due to this, there were changes in the pefvic
girdle and vertebral column. The hands were also freed from locomotion and thus they became more manipulative.
(9) Later, journey of hominoid species started from around 2 crores years ago. The first record of human like animal is ‘Ramapithecus’ ape from East Africa.
(10) Ramapithecus → Australopithecus → Neanderthal man → Cro-Magnon are the important steps in human evolution.
(11) Neanderthal man was said to be the first wise man. The increasing growth of brain made man more and more intelligent and thinking animal.
(12) Later, more than biological evolution, it was cultural evolution, when man started agriculture, animal , rearing. There was development of civilizations, arts and science etc. About 200 years ago there were industrial inventions and thus man now rules the earth.

Project:

Project 1.
Make a presentation on human evolution using various computer softwares and arrange a group disscussion over it in the class room.

Project 2.
Read the book – ‘Pruthvivur Manus Uparcich’ written by Late Dr. Sureshchandra Nadkarni and note your opinion on evolution.

Can you recall? (Text Book Page No. 1)

Question 1.
Which component of the cellular nucleus of living organisms carries hereditary characters?
Answer:
The chromosomes made up of nucleic acids and proteins, present in the nucleus of the cell are the components that carry hereditary characters in living organisms.

Question 2.
What do we call the process of transfer of physical and mental characters from parents to the progeny?
Answer:
The process of transfer of physical and mental characters from parents to the progeny is called inheritance or heredity.

Question 3.
Which are the components of the DNA molecule?
Answer:
DNA molecule is made up of two helical strands consisting of deoxyribose sugar, phosphoric acid and pairs of nitrogenous bases. These three together is called a nucleotide.

Choose the correct alternative and write its alphabet against the sub-question number:

Question 1.
Darwin has published a book titled …………..
(a) Natural selection
(b) Mutation
(c) Fall of a sparrow
(d) Origin of species
Answer:
(d) Origin of species

Question 2.
The …………. man evolved about 50 thousand years ago.
(a) Cro Magnon
(b) Neanderthal
(c) Java man
(d) Ramapithecus
Answer:
(a) Cro Magnon

Question 3.
About 10 thousand years ago, ………….. started to practise agriculture.
(a) Gorilla
(b) wise man
(c) Ramapithecus
(d) Australopithecus
Answer:
(b) wise man

Question 4.
………………. can be considered as the first example of wise-man.
(a) Australopithecus
(b) Ramapithecus
(c) Cro Magnon
(d) Neanderthal man
Answer:
(d) Neanderthal man

Question 5.
………. is a connecting link between Annelida and Arthropoda. (March 2019)
(a) Duck-billed platypus
(b) Peripatus
(c) Lung fish
(d) Whale
Answer:
(b) Peripatus

Question 6.
………… years ago human brain was sufficiently evolved to call him wise man.
(a) 50,000
(b) 30,000
(c) 20,000
(d) 10,000
Answer:
(a) 50,000

Question 7.
The process by which the gene in the nucleotide suddenly changes its position is called ………. (Board’s Model Activity Sheet)
(a) translation
(b) translocation
(c) mutation
(d) transcription
Answer:
(c) mutation

Question 8.
…………. is not the vestigial organ in the human body. (Board’s Model Activity Sheet)
(a) appendix
(b) Coccyx
(c) Canine
(d) Wisdom teeth
Answer:
(c) Canine

Write whether the following statements are true or false with proper justification for your answer:

Question 1.
It takes thousands of years for a useful structure to disappear.
Answer:
False. (The useful structures of the body do not disappear. The functioning of the body is easier due to such organs. It takes thousands of years for a functionless organ to disappear.)

Question 2.
Dr. Har Govind Khorana was awarded Nobel prize for his invention and publication in the journal Radio carbon.
Answer:
False. (Willard Libby was awarded Nobel prize for his invention and publication in the journal Radio carbon.)

Question 3.
Mesozoic era was dominated by variety of mammals.
Answer:
False. (Mesozoic era dominated by variety of reptiles.)

Question 4.
It seems that invertebrates have been slowly originated from vertebrates.
Answer:
False. (Vertebrates have been slowly originated from invertebrates in course of evolution. The primitive type of organisms always give rise to complex life forms. The invertebrates from Palaeozoic era gradually gave rise to vertebrates.)

Question 5.
The decaying process of C-12 occurs continuously from the dead remains of living organisms.
Answer:
False. (The decaying process of C-14 occurs continuously from the dead remains of living organisms. C-12 is not radioactive and hence it does not show decaying process.)

Question 6.
The theory of natural selection which mentions ‘Survival of fittest’ is given by Lamarck.
Answer:
False. (The theory of natural selection which mentions ‘Survival of fittest’ is given by Darwin.)

Question 7.
Changes acquired during life time are transferred to next generation.
Answer:
False. (Changes acquired during life time are not heritable. They are not transferred to next generation. Only the genes are transferred to the next generation.)

Question 8.
Each species grows in specific geographical conditions and has specific food, habitat, reproductive ability and period.
Answer:
True. (Each species has specifically evolved characters due to evolution and speciation.)

Question 9.
Humans walking with upright posture were confined to Africa only during prehistoric period.
Answer:
False. (Humans walking upright existed in Africa and China, Indonesia of Asian continent too.)

Question 10.
Industrial society was established about 200 years ago.
Answer:
True. (After the development and specialization of human brain, he started indulging in science and technology. Before; this period the idea of industrialization was not existing.)

Match the columns:

Question 1.

Scientist	Discovery
(1) Johann Gregor Mendel	(a) Chromosomes of grasshopper
(2) Hugo de Vries	(b) DNA is genetic material
	(c) Pioneer of the modern genetics
	(d) Mutational theory

Answer:
(1) Johann Gregor Mendel – Pioneer of the modern genetics.
(2) Hugo de Vries – Mutational theory.

Question 2.

Scientist	Discovery
(1) Walter, Sutton	(a) Chromosomes of grasshopper
(2) Mclyn McCarthy	(b) DNA is genetic material
	(c) Pioneer of the modern genetics
	(d) Mutational theory

Answer:
(1) Walter, Sutton – Chromosomes of grasshopper.
(2) Mclyn McCarthy – DNA is genetic material.

Question 3.

Evidences of evolution	Examples
(1) Morphological evidences	(a) Duck billed Platypus and Peripatus
(2) Anatomical evidences	(b) Remnants and impressions
	(c) Human hand and fore limb of bull
	(d) Shape and venation of leaf

Answer:
(1) Morphological evidences – Shape and venation of leaf.
(2) Anatomical evidences – Human hand and fore limb of bull.

Question 4.

Evidences of evolution	Examples
(1) Palaeontological evidences	(a) Duck billed Platypus and Peripatus
(2) Connecting links	(b) Remnants and impressions
	(c) Coccyx and wisdom tooth
	(d) Human hand and fore limb of cat

Answer:
(1) Palaeontological evidences – Remnants and impressions.
(2) Connecting links – Duck billed Platypus and Peripatus.

Find the odd one out:

Question 1.
Transcription, Translation, Translocation, Mutation
Answer:
Mutation. (All others are stages of protein synthesis.)

Question 2.
Bones of the hands, structure of nostrils, position of eyes, structure of ear pinnae
Answer:
Bones of the hands. (All the others are morphological evidences.)

Question 3.
Venation, Shape of seeds, Leaf petiole, Leaf shape
Answer:
Shape of seeds. (All the others are morphological evidences in plants.)

Question 4.
Human hand, wing of cockroach, forelimb of bull, flipper of whale
Answer:
Wing of cockroach. (All others are anatomical evidences, they are homologous organs.)

Identify the correlation between the first two words and suggest the suitable words in the fourth place:

Question 1.
mRNA : Transcription :: tRNA :…………
Answer:
Translation

Question 2.
Peripatus : Connecting link :: Appendix :……….
Answer:
Vestigial organs

Question 3.
Open circulatory system : Arthropods :: Thin cuticle and parapodia :………..
Answer:
Annelida

Question 4.
Between Annelida and Arthropoda : Peripatus ::……….: Lungfish
Answer:
Pisces/Fish and Amphibia

Question 5.
Theory of natural selection : Charles Robert Darwin :: Theory of inheritance of acquired characters :…………
Answer:
Jean Baptiste Lamarck

Question 6.
Survival of fittest : Darwin :: Acquired characters :……….
Answer:
Lamark

Question 7.
Wisdom teeth : Vestigial organs :: Lungfish :………..
Answer:
Connecting link.

Define the following:

Question 1.
Heredity.
Answer:
The transfer of biological characters from one generation to another through genes is called heredity.

Question 2.
Transcription.
Answer:

Question 3.
Translation.
Answer:
The process of bringing tRNA possessing anticodon that is complementary to the codon on mRNA for protein synthesis is called translation.

Question 4.
Translocation.
Answer:
The process of movement of the ribosome from one end of mRNA to other end by the distance of one triplet codon is called translocation.

Question 5.
Mutation.
Answer:
Sudden and drastic change that occurs in the genetic material is called mutation.

Question 6.
Species.
Answer:
The group of organisms that cap produce fertile individuals through natural reproduction is called a species.

Name the following:

Question 1.
Three Scientists who proved that except viruses, all living organisms have DNA as genetic material.
Answer:
Oswald Avery, Mclyn McCarthy and Colin MacLeod.

Question 2.
Genetic disorder caused due to mutation:
Answer:
Sickle cell anaemia.

Question 3.
Fish that can breathe with help of lungs:
Answer:
Lung fish.

Question 4.
Vestigial organs in human beings:
Answer:
Appendix, tail-bone or coccyx, wisdom teeth and body hair.

Question 5.
Important stages in the journey of human evolution:
Answer:

• Animals like Lemur
• Egyptopithecus
• Dryopithecus
• Ramapithecus
• Australopithecus
• Skilled Human
• Homo erectus i.e. Man with erect posture
• Neanderthal man
• Cro-Magnon man.

Distinguish between the following:

Question 1.
Transcription and Translation.
Answer:
Transcription:

1. In the process of transcription, the sequence of nucleotides present on the DNA molecule is copied
   and carried to the cytoplasm by mRNA.
2. The process of transcription takes place in nucleus.
3. During transcription, RNA is produced from DNA.
4. Only mRNA takes part in transcription.

Translation:

1. In the process of translation, the specific amino acids are picked up according to the codons brought by mRNA.
2. The process of translation takes place in ribosomes located in cytoplasm.
3. During translation, proteins are produced with the help of RNA.
4. mRNA, tRNA and rRNA take part in translation.

Question 2.
Ape and Human.
Answer:
Ape:

1. Brain of the apes is smaller in size.
2. Ape cannot walk upright.
3. Ape is less intelligent as compared to human.
4. Apes are arboreal in their habitat and they spend more time on the trees.
5. The forelimbs of ape are longer than the hind limbs.

Human:

1. Brain of humans is larger in size.
2. Humans can walk upright.
3. Human is considered to be the most intelligent animal.
4. Humans are terrestrial in their habitat. They cannot stay on the trees.
5. The forelimbs of humans are shorter than the hind limbs.

Give scientific reasons:

Question 1.
Some of the characters of parents are seen in their offspring.
Answer:

• The parental genes are transferred to their progeny through male and female gametes.
• These genes carry hereditary characters.
• Since they are transmitted from the parents to their offspring, one can see the parental characters in their offspring.

Question 2.
Darwin’s work on evolution has been a milestone.
Answer:
(1) Darwin has proposed two very important theories of evolution, viz. Theory of natural selection and Theory of origin of species.
(2) The evolution has taken place on the earth for last many crores of years.
(3) The exact nature and process of these evolutionary changes become clear after studying Darwinism. (4) The observations made by Darwin at that time are now tested according to the modern development in science and are found to be correct. Thus, his work is said to be a milestone.

Question 3.
Peripatus is said to be a connecting link between Annelida and Arthropoda.
Answer:

• Peripatus shows segmented body, thin cuticle, and parapodia-like organs.
• These characters are typical of Annelids.
• Similarly, it also shows tracheal respiration and open circulatory system which is a characteristic feature of Arthropods.
• Since Peripatus shares both these characters, it is said to be a connecting link between j Annelida and Arthropoda.

Question 4.
Vertebrates have been slowly originated from invertebrates.
Answer:

• When the carbon dating method was used to assess the age of fossils, it was understood that invertebrates were present on the earth much before the vertebrates.
• The fossils of invertebrates are present in lower layers of earth’s strata.
• They were seen in Palaeozoic era of geological time period. Vertebrates dominated during Coenozoic era.
• Their fossils are seen in the upper strata of the earth’s crust.
• The structural complexity also increased in vertebrates. All these facts indicate that Vertebrates have slowly originated from invertebrates.

Question 5.
During human evolution the hands became available for use.
Answer:

• During human evolution, the climate of earth started becoming dry.
• This resulted in loss of forest cover.
• The apes which were arboreal on the trees thus descended and started walking on land.
• The lumbar bones underwent change and the apes started walking upright on the grasslands.
• The vertebral column also underwent change. Due to upright posture the forelimbs were freed from locomotion.
• The legs started bearing the weight of the body and the hands became available for use.

Read the following statements and justify the same in your own words with the help of suitable examples:

Question 1.
Geographical and reproductive isolation of organisms gradually leads to speciation.
Answer:

• Every species survives in specific geographical conditions. The requirements of food and habitat, is specific for each species. Their reproductive ability and period is also different.
• Therefore, the individuals from one species cannot reproduce with individuals from other species.
• When they are separated by a distance or geographical barriers they are said to be isolated geographically.
• When they cannot reproduce with each other, they are said to be isolated reproductively.
• The ancestor species of both these subspecies may be the same but due to isolation over a very long-time duration, there is genetic variation between the two. Therefore, the isolation leads to speciation.

Question 2.
Study of fossils is an important aspect of study of evolution.
Answer:
(1) Fossils offer palaeontological evidence for the evolutionary process.
(2) Due to some natural calamities the organisms get buried during ancient times.
(3) The impressions and remnants of such organisms remain preserved underground. The hot lava also traps some organisms or their impressions. All such formations form fossils.
(4) Study of fossils help the researcher to understand the characteristics of the organisms that existed in the.past.
(5) Carbon dating method also helps in finding out exact age of the fossil. According to the structure of earth’s crust the fossils are obtained at specific depths.
(6) The oldest ones are obtained at the depth while the relatively recent ones occupy the upper surface. Thus fossils of invertebrates were seen in very old Palaeozoic era. Later were seen fossils of Pisces, Amphibia and Reptilia. The Mesozoic era was dominated by reptiles while Coenozoic era showed presence of mammals.
(7) In this way, study of fossils unfold the evolutionary secrets.

Question 3.
There is evidences of fatal Science among chordates.
[Please read the above question as: Among different chordates there are embryological evidences.]
Answer:

• Very young embryos of fish, amphibians, reptiles, birds and mammals show quite similar structure in the early stages.
• As the further growth takes place, they acquire different patterns.
• The initial similarity between the vertebrate embryos is an evidence that during evolution, there was a common ancestor for all the vertebrate classes.
• This is called embryological evidence for vertebrate evolution.

Question 4.
Human evolution began approximately 7 crore years ago.
Answer:

• Approximately around 7 crore years back the ice age began on the earth. In such conditions, dinosaurs became extinct. The evolution and diversity of mammals started during this time. Due to change in climate the forest cover also declined rapidly.
• Ancestors of monkey-like animals were Lemur like animals which evolved during this time period.
• The tails of these monkey-like creatures started vanishing very gradually around 4 crore years ago.
• The body and brain both increased in volume forming first ape like animals. The monkey like ancestors gave rise to two evolutionary links to apes and human like animals.
• Later, the human evolution took place by changes in the brain volume, the ability to walk upright, excessive use of hand for manipulations.
• This journey of human evolution began 7 crore years ago. But the true wise and intelligent man arose around 50,000 years ago.

Answer the following questions:

Question 1.
Answer the following questions: (March 2019)
(a) What do you mean by central dogma?
Answer:
Information about protein synthesis is present in DNA. As per this information, proteins are produced by DNA through RNA molecules. This is called central dogma.

(b) What is transcription?
Answer:
The process of synthesis of mRNA as per the nucleotide sequence present in DNA is called transcription. The nucleotide sequence on mRNA is complimentary to that of the single DNA strand used in synthesis. Instead of thymine, mRNA possesses uracil.

(c) What is meant by triplet codon?
Answer:
The code for each amino acids always consists of three nucleotides which is known as triplet codon.

Question 2.
Which animal is called a connecting link between Reptiles and Mammals? (Board’s Model Activity Sheet)
Answer:
Duck billed platypus is called a connecting link between Reptiles and Mammals.

Question 3.
In which way is science of heredity useful these days?
Answer:
The science of heredity is useful in the following ways:

• For diagnosis of hereditary disorders.
• For treatment of hereditary disorders
• For prevention of hereditary disorders
• For production of hybrid varieties of animals and plants
• For using microbes in the industrial processes.

Question 4.
What is meant by carbon dating method?
Answer:
(1) Carbon dating method is technique used for determining the age of fossils.
(2) After the death of the organisms, their consumption of carbon stops. But right from that moment the decaying process of C-14 occurs continuously.
(3) This results in change in the ratio between C-14 and C-12. C-12 is not radioactive as C-14.
(4) Thus the time passed since the death of a plant or animal is calculated by measuring the radioactivity of C-14 and ratio of C-14 to C-12 present in their body.
(5) The points noted during carbon dating are:

• The period after the organism has been dead.
• The activity of C-14 in the dead organism.
• Ratio between C-14 and C-12.

Question 5.
Answer the following questions:
(a) Describe briefly the Darwin’s theory of natural selection.
Answer:
Charles Darwin (1809-1882) proposed the theory of natural selection.
Theory of natural selection: ‘The survival of fittest’, i.e., organisms which are fit for survival, evolve while those that are not, perish. The natural selection thus acts to produce new species.

(b) What were the objections raised against Darwinism?
Answer:
Objections raised against Darwinism:

1. There are other factors too for evolution and just not the Natural Selection.
2. Arrival of useful and useless modifications were not explained by Darwin, though he said about the survival of the fittest.
3. He has not given any explanation about slow changes and abrupt changes occurring during evolution.

(c) Which book was published by Darwin to explain this theory? (Board’s Model Activity Sheet)
Answer:
Charles Darwin wrote the book ‘Origin of Species’.

Question 6.
What were the objections raised against Darwinism?
Answer:
Some of the main objections raised against Darwinism are as follows:

• There are other factors too for evolution and just not the Natural Selection.
• Arrival of useful and useless modifications were not explained by Darwin, though he said about the survival of the fittest.
• He has not given any explanation about slow changes and abrupt changes occurring during evolution.

Question 7.
Answer the following questions:
(a) Explain in brief-Lamarck’s principle of ‘use or disuse of organs’.
Answer:
The theory of use and disuse of organs says that the morphological characters of organism develop because of specific activities that the organisms perform. If some organ is not used it gets degenerated. If excessively, used, it develops. Thus, the morphological changes take place due to activities or non-working of a particular body parts in an organism.

(b) Give two examples.
Answer:
Due to constant extension of neck to eat foliage from the top of the trees, giraffe’s neck became long. Similarly, blacksmith has strong arms due to constant work. The flightless ostrich and emu did not fly and hence their wings became useless. Aquatic birds like swan and duck made their feet suitable for swimming by living in water. Snake lost limbs as it tried burrowing mode.

(c) What are acquired characters?
Answer:
Acquired characters are those characters which are obtained during the life time by any organism and passed on to next generations.

Write short notes:
(OR)
Write short notes based upon the information known to you:

Question 1.
Theory of evolution.
Answer:

• According to the theory of evolution, first living material was in the form of protoplasm which was formed in ocean.
• Gradually, it gave rise to unicellular organisms. Changes took place in these unicellular organisms which made them evolve into larger and more complex organisms.
• All evolutionary changes were very slow and gradual taking about 300 crore years to happen.
• Different types of organisms were developed as the changes and development that occurred in living organisms was all round and multi-dimensional.
• Hence, this overall process of evolution is called organizational and progressive.
• Variety of plants and animals developed from the ancestors having different structural and functional organization during the process of evolution.

By choosing appropriate words given in the bracket, complete the paragraph:

Question 1.
(translation, anticodon, tRNA, mRNA, amino acids, triplet codon, transcription, DNA)
The …….. formed in nucleus comes in cytoplasm. It brings in the coded message from DNA. The message contains the codes for amino acids. The code for each amino acid consists of three nucleotides. It is called as ‘………..’. Each mRNA is made up of thousands of triplet codons. As per the message on mRNA, ……… are supplied by the ………. For this purpose, tRNA has ‘…………’ having complementary sequence to the codon on mRNA. This is called ‘………..’.
Answer:
The mRNA formed in nucleus comes in cytoplasm. It brings in the coded message from DNA. The message contains the codes for amino acids. The code for each amino acid consists of three nucleotides. It is called as ‘triplet codon’. Each mRNA is made up of thousands of triplet codons. As per the message on mRNA, amino acids are supplied by the tRNA. For this purpose, tRNA has ‘anticodon’ having complementary sequence to the codon on mRNA. This is called ‘translation’.

Question 2.
(Cultural, agriculture, fire, brain, Cro-Magnon, Homo sapiens, Neanderthal)
Evolution of upright man continued in the direction of developing its ………. for the period of about 1 lakh years and meanwhile he discovered the ………. Brain of man, 50 thousand years ago had been sufficiently evolved to the extent that it could be considered as member of the species ………… Neanderthal man can be considered as the first example of wise-man. The ……….. man evolved about 50 thousand years ago and afterwards, this evolution had been faster than the earlier. About 10 thousand years ago, wise-man started to practise the ………. It started to rear the cattle-herds and established the cities. ………..development took place later.
Answer:
Evolution of upright man continued in the direction of developing its brain for the period of about 1 lakh years and meanwhile he discovered the fire. Brain of man 50 thousand years ago had been sufficiently evolved to the extent that it could be considered as member of the species Homo sapiens. Neanderthal man can be considered as the first example of wise-man. The Cro-Magnon man eyolved about 50 thousand years ago and afterwards, this evolution had been faster than the earlier. About 10 thousand years ago, wise-man started-to practise the agriculture. It started to rear the cattle-herds and established the cities. Cultural development took place later.

Read the paragraph and answer the questions given below:

With the help of RNA, the genes present in the form of DNA participate in the functioning of cell and thereby control the structure and functioning of the body. Information about protein synthesis is stored in the DNA and synthesis of appropriate proteins as per requirement is necessary for body. These proteins are synthesized by DNA through the RNA. This is called ‘Central Dogma’. mRNA is produced as per the sequence of nucleotides on DNA. Only one of the two strands of DNA is used in this process. The sequence of nucleotides in mRNA being produced is always complementary to the DNA strand used for synthesis. Besides, there is uracil in RNA instead of thymine of DNA. This process of RNA synthesis is called ‘transcription’.

Questions and Answers:

Question 1.
Which part of the cell control the structure and functioning of the body?
Answer:
Genes present in the form of DNA along with RNA control the structure and functioning of the body.

Question 2.
How is a specific protein synthesised in the cell?
Answer:
The information of protein synthesis is stored in the DNA which is utilised as per the requirement of the body. Later the proteins are synthesised by DNA through the RNA.

Question3.
What is the similarity between mRNA and DNA?
Answer:
The sequence of nucleotides on DNA is copied on mRNA. The nucleotide sequence on mRNA is thus complementary to DNA.

Question 4.
Give one difference between RNA and DNA.
Answer:
RNA has uracil instead of thymine which is present in DNA.

Question 5.
Define central dogma.
Answer:
Central dogma is the concept that proteins are synthesised by DNA through the RNA.

Diagram-based questions:

Question 1.
Observe the figure 1.3 of transcription given on page 9 in this chapter and answer the following questions:

(1) What is the sequence of nucleotides present on one strand of the DNA?
Answer:
A T G C A A T T

(2) According to the above sequence on DNA, what will be the transcribed sequence on the mRNA molecule?
Answer:
U A C G U U A A

(3) Which enzyme is taking part in the above process of transcription?
Answer:
RNA polymerase takes part in the process of transcription.

Question 2.
Observe the figure 1.5 of translation and translocation, given on page 9 this chapter and answer the following questions:

(1) Which is the initiation codon? Where is it present?
Answer:
AUG is the initiation codon, which is present on the mRNA.

(2) What are the types of RNA present inside the ribosome? Which triplet codon is present on it?
Answer:
There are two molecules of tRNA present inside the ribosome. The triplet codons present on them are UAC and AAG respectively.

(3) Which genetic code is present on mRNA that is leaving the nucleus? What must be the sequence on the DNA to have such code on mRNA?
Answer:
The mRNA that leaves the nucleus has genetic code: A U G U U C A A A
The genetic code on DNA therefore should be as follows: T A C A A G T T T

Question 3.
Observe the figure 1.6 given on page 10 from this chapter. Answer the following question based on your observations:

What is the significance of this figure from the viewpoint of evolution? Explain in brief.
Answer:
In the figure, the process of mutation is shown. The original nucleotide sequence of TGC is replaced by new mutated sequence GAT. The change in the nucleotide sequence will change the DNA.

This will result in the change in genes and then changing the hereditary characters. Due to such change in genes, the evolution proceeds. The mutation so formed can be minor or major. The greater the impact of the change, the evolution takes place rapidly. The mutation thereby produce recombinations leading to diversity.

Question 4.
Observe the picture and answer the following questions:

(1) Which evidence of evolution is shown in the picture?
Answer:
Embryological evidences of evolution are shown in this picture.

(2) What can be proven with this proof?
Answer:
The similarities in the initial embryonic stages of different vertebrates shows that there was a common origin of all of them. Thus embryological evidences prove that there was common vertebrate ancestor.

(3) Give one more example of evidence of evolution.
Answer:
Palaeontological evidences such as vestigial organs and connecting links are another examples of evolutionary evidences.

Question 5.
Which concept/theory do you remember after seeing this picture of Giraffes? Describe it in brief.

Answer:

• The picture is based on the Lamarck’s principle of ‘use and disuse of organs’.
• The morphological characters of organism develop because of specific activities that the organisms perform.
• If some organ is not used it gets degenerated. If excessively used, it develops further.
• Thus, the morphological changes take place due to activities or non-working of a particular body parts in an organism. Due to constant extension of neck to eat foliage from the top of the trees, giraffe’s neck became long.

Activity-based Questions:

Try this: (Text Book Page No. 4)

Observe the above images and note the similarities between given animal images and plant images.
Answer:
The above pictures of the animals show similarities such as structure of mouth, position of eyes, structure of nostrils and ear pinnae and body fur. In pictures of plants there are similarities in characters like leaf shape, leaf venation, leaf petiole, etc.
These above morphological evidences show that there may be a common ancestor for all of the species shown.

Observe and Discuss:

Question 1.
Observe the pictures given below. (Text Book Page No. 5)

Answer:
(1) Fossils offer palaeontological evidence for the evolutionary process.
(2) Due to some natural calamities the organisms get buried during ancient times.
(3) The impressions and remnants of such organisms remain preserved underground. The hot lava also traps some organisms or their impressions. All such formations form fossils.
(4) Study of fossils help the researcher to understand the characteristics of the organisms that existed in the.past.
(5) Carbon dating method also helps in finding out exact age of the fossil. According to the structure of earth’s crust the fossils are obtained at specific depths.
(6) The oldest ones are obtained at the depth while the relatively recent ones occupy the upper surface. Thus fossils of invertebrates were seen in very old Palaeozoic era. Later were seen fossils of Pisces, Amphibia and Reptilia. The Mesozoic era was dominated by reptiles while Coenozoic era showed presence of mammals.
(7) In this way, study of fossils unfold the evolutionary secrets.

Question 2.
Observe the pictures given and discuss the characters observed. (Text Book Page No. 6)

Answer:
Some living organisms possess some characters in them which are the distinctive features of different groups or phyla. Such individuals connect these two groups by sharing the characters of both and hence they are known as connective links.

Examples: (1) Peripatus: Peripatus is the connecting link between Annelida and Arthropoda. It shows characters of both animal phyla. Like annelid worm, it shows segmented body, thin cuticle and parapodia. Like an arthropod, it shows open circulatory system and tracheal system for respiration.
(2) Duck Billed platypus: This is a connecting link between reptiles and mammals. Like reptiles it lays eggs but like mammals it has mammary glands and hairy skin.
(3) Lung fish: Lung fish is a connecting link between fishes and amphibians. Though a fish, it shows lungs for respiration as in amphibian animals.
(4) Connecting links indicate the direction and hierarchy of evolution.

Project: (Do it your self)

Project 1.
Internet is my friend: (Text Book Page No. 3)
Collect the information from the internet about Big-Bang theory related with the formation of stars and planets and present it in your class.

Project 2.
Use of ICT: (Text Book Page No. 4)
Collect the information of geological dating and present it in the classroom.

Project 3.
Use of ICT: (Text Book Page No. 5)
Find how the vestigial organs in certain animals are functional in others. Present the information in your class and send it to others.

Project 4.
Internet is my friend: (Text Book Page No. 8)
Collect the pictures and information of various species of monkeys from internet.

Balbharti Maharashtra State Board Class 10 History Solutions Chapter 7 Sports and History Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 History Solutions Chapter 7 Sports and History

Question 1.
(A) Choose the correct option from the given options and complete the statement.
(1) The ancient event of Olympic competitions used to be held at ………………………… .
(a) Olympia, Greece
(b) Rome
(c) India
(d) China
Answer:
(a) Greece

(2) The wooden dolls made in Maharashtra are known as ………………………… .
(a) Thaki
(b) Kalichandika
(c) Gangavati
(d) Champavati
Answer:
(a) Thaki

(B) Identify and write the wrong pair in the following set.
(1) Mallakhamb – Outdoor game based on physical skills
(2) Water polo – Water sport
(3) Skating – Adventurous ice sport
(4) Chess – Outdoor game
Answer:
(4) Chess – Outdoor game

Question 2.
Write notes :
(1) Toys and Festivals
Answer:

• Toys and festivals are inter-related since ancient times.
• Toys are used for decoration in different cultures and religions during festivals.
• In some cultures toys are distributed as gifts. Santa Claus gifts children toys during Christmas.
• As part of Diwali celebration in Maharashtra, model forts are made displaying images of Chhatrapati Shivaji Maharaj, his soldiers and animals which are toys.
• Clay images of snakes and bullocks are- sold during festivals like Bail pola and Nagpanchami.

(2) Sports and movies
Answer:

• The presence of sports was limited to a scene in the movies made earlier.
• In recent times, biographical movies are made on sportspersons and on sports.
• Movies like Lagaan and Dangal are made related to cricket and wrestling respectively.
• Biographical movies are made on Mary Kom, and the Phogat sisters.
• Movies are made on careers of famous sprinter Milkha Singh.
• Bharat Ratna Sachin Tendulkar and Cricketer Mahendra Singh Dhoni.
• Overall, movies and sports are related from the silent era till date.

Question 3.
Explain the following statements with reasons.
(1) Currently the structure of sports economy has been significantly affected.
Answer:

• The process of globalisation has influenced the field of sports in the 20th-21st century.
• International matches of various sports like Cricket, Football, etc. are*telecast in every corner of the world.
• Fans watch these matches for entertainment,r and aspiring players to learn more.
• The citizens of the non-participating countries also watch these matches.
• Retired players get a chance on television channels as commentators.
• Matches garner a large audience, hence the commercial companies look at it as an opportunity to advertise and sell their products.
• All these factors have led to change in the structure of sports economy.

(2) Toys can tell us about cultural history.
Answer:
A tradition of making different type of toys for entertainment is going on since ancient time.

• Toys give us an idea about the cultural and religious development of that period.
• Clay models of forts and the images of Chhatrapati Shivaji Maharaj placed on fort gives us an idea about the structures of forts during that period.
• An ivory doll found at Pompeii, an ancient city in Italy, sheds light on Indo-Roman trade and cultural relations.
• The mention of games, toys and flying and dancing dolls in Kathasaritsagara give us an idea about cultural history of toys. In this way, we come to know about cultural history from toys.

Question 4.
Write detailed answers to the following questions.
(1) Write about the history of sports equipment and toys in ancient India.
Answer:

1. The history of sports equipment and toys dates back to epic age. The ancient Indian literature and epics mention various games such as games of dice, wrestling, horse and chariot race.
2. Sports are of two types ‘Indoor Games’ and ‘Outdoor Games’. Indoor games such as chess, card games, dice, carrom, kachkavadya or Indian ludo, Bhatukli were very popular. It can be noted that all of them required equipment to play.
3. Cards to play card games, dice to play game of dice, a board and pieces to play chess, bunch of seeds or stones to play sagargote; (playing house) to play Bhatukli.
4. Likewise, Outdoor games like marbles, lagori (seven stones), vitti-dandu, bhavare (tops) all require material like marbles, stones, tops, a small and large stick to play.
5. A Sanskrit play by Shudraka is named as Mrichchhakatika. It means a clay cart. A clay cart was a toy used to play during Harappan period.
6. Kathasaritsagara has very interesting descriptions of games and toys. There are descriptions of flying dolls. There is a mention that on pressing a key some dolls used to fly, some used to dance and some used to make sounds.

(2) Explain the close tie between sports and history.
Answer:
Sports and history are closely related with each other.

1. It is a must for a sports writer to know the history of the game he chooses to write on.
2. In order to write a review on any sport competition, the critic should have knowledge of competitions held in the past.
3. A comparative study of the skills, techniques and strategies used in the past and developments or improvements in the present makes the review comprehensive.
4. The writer has to resort to history while writing columns or articles on sports events like Olympics or Asiad or any national or international matches.
5. While commentating on Akashvani and Doordarshan, an expert commentator needs to have good knowledge of the history of the game, previous records of illustrious and eminent players, statistical analysis and historical anecdotes related to the game and players.
6. Coaches, special experts, selection committee should have information of the players, their strength and weakness and also history of the players in the opposing team. Even players should know history of their competitors.
   In short, it is essential to know the history of all the aspects related to sports.

(3) Explain the difference between indoor and outdoor games.
Answer:

Indoor games	Outdoor games
1. Most of the indoor games are played by sitting at one place. They are played in a closed environment.	1. Outdoor games are played on a field.
2. Indoor games require skills but physical exercise is negligible.	2. Outdoor games need more physical exercise and skill.
3. As there is no exertion in indoor games, so it is not essential to develop stamina.	3. Outdoor games require stamina and strength.
4. Indoor games do not involve adventure.	4. Outdoor games involve adventures at times, e.g.. Auto racing
5. Indoor games includes Chess, Indian Ludo (Playing house) Bhatukli and many more.	5. Outdoor games involve national and international games like Kabaddi, Kho-kho, Hockey, Cricket, etc.
6. With the exception of chess and carrom no competitions are held for rest of the Indoor games.	6. National and international competitions are held of almost all outdoor games.

Project
(1) Collect information about your favorite sports and its players.
(2) Discuss the hardships the sportspersons have to face while training for the sport with the help of information gathered through movies and literature.

Question 5.
Complete the sentences by choosing the correct option:
(a) The activity which combines physical exercise and entertainment is ……………………….. .
(a) Show
(b) Attitude
(c) Sports
(d) Competition
Answer:
(c) Sports

(b) ……………………….. was looked upon as a game and entertainment by ancient people.
(a) Dancing
(b) Playing
(c) Singing
(d) Hunting
Answer:
(d) Hunting

(c) ……………………….. and its various tactics were devised by Balambhat Deodhar, the physical trainer of Peshwa Bajirao II.
(a) Kabaddi
(b) Atyapatya
(c) Khokho
(d) Mallakhamb
Answer:
(d) Mallakhamb.

(d) The Indian Government has honoured Sachin, Tendulkar with ……………………… for his illustrious achievements in the field of cricket.
(a) Padma Shri
(b) Khel Ratna
(c) Arjuna Award
(d) Bharat Ratna
Answer:
(d) Bharat Ratna

(e) ……………………… is the national game of India.
(a) Hockey
(b) Cricket
(c) Football
(d) Kabaddi
Answer:
(a) Hockey

(f) The National Sports Day of India is celebrated on 29th August which is the birth date of …………………….. .
(a) Khashaba Jadhav
(b) Sachin Tendulkar
(c) Major Dhyan Chand
(d) Bal J. Pandit
Answer:
(c) Major Dhyan Chand

(g) Vishnubhat Godse in his book Maza Pravas wrote that sports and physical activity had great importance in the daily schedule of ………………………
(a) Tatya Tope
(b) Queen of Jhansi Lakshmibai
(c) Bahadur Shah Jaffar
(d) Nanasaheb Peshwa
Answer:
(b) Queen of Jhansi Lakshmibai

(g) Maruti Mane is known for …………………….. .
(a) Hockey
(b) Kabaddi
(c) Marathon
(d) Wrestling
Answer:
(d) Wrestling.

(h) …………………….. wrote the play Mrichchhakatika which means a clay cart.
(a) Harshvardhan
(b) Shudraka
(c) Bhavbhuti
(4) Kalidas
Answer:
(b) Shudraka

(i) The findings in the excavations of …………………….., an ancient city in Italy includes an ivory doll made by Indian craftsmen.
(a) Rome
(b) Athens
(c) Sparta
(d) Pompeii
Answer:
(d) Pompeii

(j) An interesting description of games and toys is found in ……………………. .
(a) Shakuntal
(b) Panchatantra
(c) Mrichchhakatika
(d) Kathasaritsagara
Answer:
(d) Kathasaritsagara

(k) Major Dhyan Chand was honoured in 1956 with …………………….. for his marvellous achievements in hockey.
(a) Padma Shri
(b) Padma Bhushan
(c) Padma Vibhushan
(d) Bharat Ratna
Answer:
(c) Padma Bhushan

(l) …………………….. was the first Indian female boxer to win a bronze medal in the Olympics.
(a) P 7 Sindhu
(b) Mary Kom
(c) Geeta Phoghat
(d) Saina Nehwal
Answer:
(b) Mary Kom.

Question 6.
Identify the wrong pair in the following and write it:
(1)

(1) Mallakhamb	Outdoor game based on physical skills
(2) Water Polo	Water sport
(3) Skating	Adventure ice sports
(4) Chess	Outdoor game

Answer:
Wrong pair: Chess – Outdoor game

(2)

(1) Mallakhamb trainer	Balambhat Deodhar
(2) Wizard of Hockey	Milkha Singh
(3) First Indian female boxer	Mary Kom
(4) First Indian female wrestlers	Phogat sisters

Answer:
Wrong pair: Wizard of Hockey – Milkha Singh

Question 7.
Do as directed
(A) Complete the graphical description:
(1)

Answer:

Answer:

Answer:

Answer:

(B) Prepare a Tree-Diagram on typs of games:
Answer:

Question 8.
Write short notes:

(a) Indigenous Games:
Answer:

1. The games which have their origin in India and are an important part of Indian culture are indigenous games.
2. They are of two types – ‘Indoor Games’ and ‘Outdoor Games’. Indoor Games are played within a closed environment and a number of them are played by sitting at one place. Chess, card games, dice, carrom, etc. are indigenous indoor games.
3. An open space or preferably a playground is required to play outdoor games. Kabaddi, Atyapatya, Kho-kho etc. are indigenous outdoor games.
4. The special feature of indigenous games is that they do not require high cost material and hence are less expensive. Phugadi, Zimma, Bhatukali are some of the indigenous games played by girls. In modern times, all national and international games are played by both girls and boys.

Question 9.
Explain the following statements with reasons:

(a) Major Dhyan Chand is called the Wizard of Hockey.
Answer:

• Major Dhyan Chand was part of hockey teams as a player in 1928 and 1932 which won gold medal at Olympics.
• He was also captain of the Indian Hockey team which won a Gold Medal at the Berlin Olympics in 1936.
• He shot 25 goals against America and Japan in the 1932 Olympics.
• He shot more than 400 goals in his entire career which include national and international matches. Owing to his brilliant achievement he is called the ‘Wizard of Hockey’.

(b) Globalisation has influenced sports.
Answer:

1. No sport is limited to any one country. Television and other media channels telecast matches widening the reach of sports in all comejcs, of the world.
2. International competition Asiad, Paralympics, Cricket Wc watched by people irrespective r
   part of the world.
3. World Cup matches of cricket, hockey and football are held.
4. No country has a monopoly on any sport which means that globalisation has influenced sports.

(c) A commentator should know the history of the game.
Answer:

• The mere description of a live match is not enough for commentators.
• A commentator should have good knowledge of the history of the game, previous records (who made or who broke) and eminent players in the past as well as events related to different competitions.
• Along with the history and information of the playground, commentator should narrate records made by the players in different matches.
• It will make his commentary interesting. Therefore, it is essential for the commentator to know the history of the game.

Question 10.
Answer the following questions in 25-30 words:
(a) Explain the importance of sports.
Answer:
Sports has gained great importance for the following reasons:

• Sports helps us to overcome our pains, worries and sufferings. We feel relaxed and refreshed by playing games.
• Games which involve a lot of physical activity not only provide good exercise but also help in building a tenacious and strong body.
• One. can develop courage; determination and sportsmanship playing games. A sense of cooperation and teamspirit develops when we participate in games which require collective participation.
• Team games also help in developing a leadership! quality.

(b) Write about the history of sports.
Answer:

• It is a natural instinct in human beings to play. From beginning of civilisation till date man has played different types of games for his entertainment.
• Hunting was a way of obtaining food for the ancient people as well as considered a game.
• Horse and chariot races, wrestling, game of dice (dyut) are mentioned in ancient Indian literary texts and epics.
• Dolls, whistles, toy carts were discovered in the excavations at Harappa. So, it can be said that the history of sports is as old as the history of man.

(c) Write about the importance of sports in education field.
Answer:

• Sports are an integral part „ of education. The making of a player begins at school level.
• Many types of sports events are held at the international level. To make the players competent they are given opportunity to play at district, state and national level.
• They are promoted and sponsored by the government and private sectors. Talented and ranking players get State scholarship or National scholarship.
• Seats are reserved for them in colleges and Universities. It has been observed that the foundation of successful players is laid in school life.

(d) What do we need to know while making movies on sports?
Answer:
While making a movie on sports the makers should have complete information of the. sport as well as its history.

1. Nowadays special research teams are appointed by production houses which do thorough research on the subject of the movie.
2. In order to gather information on the sportsperson or the sport it is essential to study books, articles, columns written by eminent sports writers.
3. If the movie is on a sportsperson, all interviews published in national and international magazines and newspapers should be read.
4. Factors such as period, type of equipment used, sports wear, dressing style, social life needs to be studied.
5. General understanding of the people about the game, practises and famous sportspersons is required.

Question 11.
Read the following passage and answer the following questions:

(a) Complete the Concept Map:

Answer:

(b) How are the experts in history helpful regarding international sports competition?
Answer:
Experts in history are helpful to write and critically analyse the game.

(c) How are professional opportunities available in field of sports?
Answer:
There are many professional opportunities available in the field of sports.

• Writers are in demand who can write on sports and critics to write reviews are in demand.
• Commentators are in demand on Television, radio and various other private channels. Experts and assistants are needed to provide information regularly.
• Coaches train the players, playground staff to maintain the field, umpires, etc.
• Cameramen, computer experts and team of assistants to have uninterrupted transmission. Trained and qualified referees are required to work at district, national and international levels.
• Overall, a great number of job opportunities are available in the field of sports.

Question 12.
Write a detailed answer to the following:

(a) Write about the history of sports literature and toys in ancient India.
Answer:
A new enterprise is developing in publishing related to sports in India. There is extensive written work on various sports.

• Many books related to sports and biographies are published. Encylopaedias are being written on sports.
• An independent encyclopaedia is written on exercise. The History of Mallakhamb is recently published.
• Sports magazines are published fortnightly and monthly.
• Many newspapers have allotted a separate section or last pages for news related to sports.
• ‘Shatkar’ was a sports magazine published some years ago. There is ample of literature available on sports.

(b) Trace the development of toys and their importance.
Answer:

• Toys and games have been essentially part of entertainment from ancient times. Every I developing society has made toys for the 8 entertainment and education of their children.
• Toys were found at archaeological sites at various places. The toys were made of clay, baked clay, terracotta and ivory.
• Either a mould was used to make the toy or it was fashioned by hand.
• Toys and the material used to make them were indicators of the development and advancement of civilisations.
• An interesting description of flying dolls is found in Kathasaritsagara. The dolls used to fly, made some sound and some danced when a key was pressed.
• Toys give us information about the period it was made, how they were made, religious and 8 cultural practices and technical know-how of the 8 people. ’

Question 13.
Observe the picture and write information about the event it is related to:

Answer:

1. This picture is the logo of the modern Olympic Games. The five interlocked rings represent the five continents of the world.
2. The rings coloured blue, yellow, black, green and red on a white field are known as the ‘Olympic rings’. The symbol was originally designed in 1912 by Pierre de Coubertin.
3. Olympic rings are the symbol of games which were first played in the ancient city of Olympia. They were held after every four years.
4. The ancient Olympics had fewer events than the modern games and only Greek men were allowed to participate. Events such as Horse and Chariot race, Footrace, Wrestling, Boxing, Discus Throw, Pentathlon were held.
5. The Greeks standardised rules of the sports were laid which was helpful to organise the games systematically.
6. The modern Olympic games are also held every four years. It is a great honour for sportspersons to participate and win the Olympic medals.

Memory Map

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 7 Lenses Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

Question 1.
Match the columns in the following table and explain them:

Column 1	Column 2	Column 3
Farsightedness	Nearby object can be seen clearly	Bifocal  lens
Presbyopia	Faraway object can be seen clearly	Concave  lens
Nearsightedness	Problem of old age	Convex  lens

Answer:

Column 1	Column 2	Column 3
Farsightedness	Faraway object can be seen clearly	Convex  lens
Presbyopia	Problem of old age	Bifocal  lens
Nearsightedness	Nearby object can be seen clearly	Concave  lens

1. Farsightedness:
Hypermetropia or farsightedness is the defect of vision in which a human eye can see distant objects clearly but is unable to see nearby objects clearly.
In this case the image of a nearby object would fall behind the retina instead of on the retina.

Possible reasons for hypermetropia:
(1) Curvature of the cornea and the eye lens decreases. Hence, the converging power of the eye lens becomes less.
(2) The distance between the eye lens and retina decreases (relative to the normal eye) and the focal length of the eye lens becomes very large due to the flattening of the eyeball.

2. Presbyopia:
Presbyopia is the defect of vision in which aged people find it difficult to see nearby objects
comfortably and clearly without spectacles.

Reason of presbyopia: The power of accommodation of eye usually decreases with ageing. The muscles near the eye lens lose their ability to change the focal length of the lens.

Therefore, the near point of the eye lens shifts rarther from the eye, This defect is corrected using a. convex lens of appropriate power. The lens converges light rays before they fall on the eye lens such that the action of the eye lens forms the image on the retina.

3. Nearsightedness:
Myopia or nearsightedness is the defect of vision in which a human eye can see nearby objects distinctly but is unable to see distant objects clearly as they appear indistinct.
In this case the image of a distant object is formed in front of the retina instead of on the retina.
[Figs. 7.29 (a), 7.29 (b)]

Possible reasons for myopia:
(1) The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large. (2) The distance between the eye lens and the retina increases as the eyeball elongates.

Myopia is corrected using a suitable concave lens. Light rays are diverged by the concave lens before they strike the eye lens. A concave lens of proper focal length is chosen to produce the required divergence. Hence, after the converging action of the eye lens, the image is formed on the retina. [Fig. 7.29 (c)]

Question 2.
Draw a figure explaining various terms related to a lens.
Answer:
(1) Centre of curvature (C): The centres of the spheres whose parts form the surfaces of a lens are called the centres of curvature of the lens. A lens has two centres of curvature C₁, and C₂ for its two spherical surfaces.

(2) Radii of curvature (R₁, R₂): The radii of the spheres whose parts form surfaces of a lens are called the radii of curvature of the lens.

(3) Principal axis: The imaginary straight line passing through the two centres of curvature of a lens is called the principal axis of the lens.

(4) Optical centre (O): The point inside a lens on the principal axis, through which light rays pass without changing their path is called the optical centre (O) of the lens.

(5) Principal focus (F): When light rays parallel to the principal- axis are incident on a convex lens, they converge at a point on the principal axis. This point is called the principal focus (F) of the convex lens. Light rays travelling parallel to the principal axis of a concave lens diverge after refraction in such a way that they appear to be coming out of a point on the principal axis. This point is called the principal focus of the concave lens. A lens has two principal foci F₁ and F₂.
[Note: In this chapter, the terms focus and the principal focus are used in the same sense.]

(6) Focal length (f): The distance between the optical centre and the principal focus of a lens is called the focal length (f) of the lens.

C₁, C₂: Centres of curvature, R₁, R₂: Radii of curvature, O: Optical centre.
The cross sections of convex and concave lenses are shown in parts (a) and (b) of Fig. 7.4. The surface marked as 1 is part of sphere S₁ while the surface marked as 2 is part of sphere S₂.

P₁, P₂, P₃: Incident rays of light,
Q₁, Q₂, Q₃: Refracted rays of light, O: Optical centre

F₁, F₂: Principal foci of the lens, f: Focal length of the lens

Question 3.
At which position will you keep an object in front of a convex lens so as to get a real image of the same size as the object? Draw a figure.
Answer:
At 2F₁.

Question 4.
Give scientific reasons:
a. A simple microscope is used for watch repairs.
Answer:
(1) when an object is placed within the focal length of a magnifying glass or simple microscope (convex lens), its larger and erect image is obtained on the same side of the lens as that of the object.

(2) By adjusting the distance between the object and the lens, the image can be obtained at the minimum distance of distinct vision. Thus, a watch repairer can see the minute parts of a watch more clearly with the aid of a magnifying glass (a simple microscope) than with the naked eye, without any stress on the eye. Hence, watch repairers use a magnifying glass (a simple microscope) while repairing the watches.

b. One can sense colours only in bright light.
Answer:
(1) The retina in the eye is made of many light sensitive cells. The rod-shaped cells respond to the intensity of light while the cone-shaped cells j respond to various colours.
(2) The cone-shaped cells do not respond to faint light. They function only in bright light. Hence, one can sense colours only in bright, light.

c. We cannot clearly see an object kept at a distance less than 25 cm from the eye.
Answer:
(1) When we try to see a nearby object, the eye lens becomes more rounded and its focal length decreases. Then a clear image of the object is formed on the retina of the eye.
(2) The focal length of the eye lens cannot be decreased beyond some limit. Therefore we cannot clearly see an object kept at a distance less than 25 cm from the eye.

Question 5.
Explain the working of an astronomical telescope using refraction of light.
Answer:
Construction of a refracting telescope: It consists of two convex lenses called the objective lens (directed towards the object) and the eyepiece (directed towards the eye). The focal length and diameter of the objective lens are respectively greater than the focal length and diameter of the eyepiece. The objective lens is fitted at one end of a long metal tube.

A metal tube of smaller diameter is fitted in this metal tube and the eyepiece is fitted at the outer end of the smaller tube. With the help of a screw it is possible to change the distance between the eyepiece and the objective lens by sliding the tube fitted with the eyepiece. The principal axes of the objective lens and the eyepiece are along the same line. A telescope is usually mounted on a stand.

Working: When the objective lens is pointed towards the distant object to be observed, the rays of light from the distant object, which are almost parallel to each other, pass through the objective lens. The objective lens collects maximum amount of light as it is large in size. It forms a real, inverted and diminished image in the focal plane of the objective lens. Now, the position of the eyepiece is adjusted such that this image falls just within the focal length of the eyepiece and serves as the object for the eyepiece which works as a simple microscope.

The final image is highly magnified, virtual, on the same side as that of the object and inverted with respect to the original object. The final image can be observed by keeping the eye close to the eyepiece. If the image formed by the objective lens lies in the focal plane of the eyepiece, the final image is formed at infinity.

Question 6.
Distinguish between the following:
a. Farsightedness (Hypermetropia) and Nearsightedness (Myopia).
Answer:
Farsightedness:

1. In hypermetropia, a human eye can see distant distinctly but is unable to see nearby objects clearly.
2. In this case, the image of a nearby object would be formed behind the retina.
3. This defect can be corrected using a convex lens of appropriate power.

Nearsightedness:

1. In myopia, a human eye can see near objects distinctly, but is unable to see distant objects clearly.
2. In this case, the image of a distant object is formed in front of the retina.
3. This defect can be corrected using a concave lens of appropriate power.

b. Concave lens and Convex lens.
Answer:
Concave lens:

1. A concave lens has its surfaces curved inwards.
2. It is thicker at the edges than in the middle.
3. It can form only a virtual image.
4. It can form only a diminished image.

Convex lens:

1. A convex lens has its surfaces puffed up outwards.
2. It is thicker in the middle than at the edges.
3. It can form a real image as well as a % virtual image.
4. It can form a magnified, diminished or the same sized image (relative to the object) depending on the position of the object.

Question 7.
What is the function of the iris and the muscles connected to the lens in the human eye?
Answer:
When the incident light is very bright, the muscles of the iris stretch to reduce the size of the pupil. When the incident light is dim, the muscles of the iris relax to increase the size of the pupil. Thus, the iris controls the size of the pupil and thereby regulates the amount of light entering the eye. (Fig. 7.26)

When a distant object is to be observed, the ciliary muscles relax so that the eye lens becomes flat. This increases the focal length of the lens. Therefore, a sharp image of the distant object is formed on the retina.
Thus, we can see a distant object clearly. (Fig. 7.27)

when an object closer to the eye is to be observed. the ciliary muscles contract increasing the curvature. of the eye lens. The eye lens, therefore, becomes rounded. This decreases the focal length of the lens. Therefore, a sharp image of the nearby object is formed on the retina. Thus, we can see a nearby object clearly. (Fig. 7.27)

Question 8.
Solve the following examples:
i. Doctor has prescribed a lens having I power + 1.5 D. What will be the focal length of the lens? What is the type of the lens and what must be the defect of vision?
Solution:
Data: P = + 1.5 D, f = ?
Focal length of the lens, f = \(\frac{1}{P}=\frac{1}{1.5 \mathrm{D}}\)
= \(\frac{10}{15}\) m = 0.6667 m = 0.67 m
P is positive. This shows that the lens is convex. The defect of vision is farsightedness (hypermetropia).

ii. 5 cm high object is placed at a distance of 25 cm from a converging lens of focal length of 10 cm. Determine the position, size and type of the image.
Solution:
Data: Converging lens, f = 10 cm
u = – 25 cm, h₁ = 5 cm, v = ?, h₂ = ?


The height of the image = -3.3 cm (inverted image ∴ minus sign).
(iii) The image is real, inverted and smaller than the object.

iii. Three lenses having powers 2, 2.5 and 1.7 D are kept touching in a row. What is the total power of the lens combination?
Solution:
Data: P₁ = 2 D, P₂ = 2.5 D, P₃ = 1.7 D, P = ?
Total power of the lens combination,
P = P₁ +P₂ + P₃
= 2 D + 2.5 D + 1.7 D
= 6.2 D.

iv. An object kept 60 cm from a lens gives a virtual image 20 cm in front of the lens. What is the focal length of the lens? Is it a converging lens or diverging lens?
Solution:
Data: u = -60 cm, v = -20 cm, f = ?

∴ The focal length of the lens, f = – 30 cm. As f is negative, it is a diverging lens.

Project:

Question 1.
Make a Powerpoint presentation about the construction and use of binoculars. (Do it your self)

Can you recall? (Text Book Page No. 80)

Question 1.
Indicate the following terms related to spherical mirrors in figure 7.1: pole, centre of curvature, radius of curvature, principal focus.

Answer:

Question 2.
How are concave and convex mirrors constructed?
Answer:
The given part of a hollow spherical glass can be converted into a concave mirror by (i) polishing (silvering) its inner side (inner surface or concave surface) to make it reflecting or (ii) coating its outer side with a thin layer of silver and painting it with red colour to protect the silver coating.
[Note: Case (i) corresponds to the front surface silvered concave mirror.]

The given part of a hollow spherical glass can be converted into a convex mirror by (i) polishing (silvering) its outer side (outer surface or convex surface) to make it reflecting or (ii) coating its inner side with a thin layer of silver and painting it with red colour to protect the silver coating.
[Note: Case (i) corresponds to the front surface silvered convex mirror. ]

Use your brain power! (Text Book Page No. 85)

Question 1.
From equations (1) and (2) what is the relation between h₁, h₂, u and v?
Answer:
M = \(\frac{h_{2}}{h_{1}}\) …….(1)
Also, M = \(\frac{v(\text { image distance })}{u \text { (object distance) }}\)……(2)
\(\frac{h_{2}}{h_{1}}\) = \(\frac{v}{u}\)

Try this (Text Book Page No. 88)

(1) Try to read a book keeping it very far from your eyes.
(2) Try to read a book keeping it very close to your eyes.
(3) Try to read a book keeping it at a distance of 25 cm from your eyes.
At which time do you see the alphabets clearly? Why?
Answer:
Minimum distance of distinct vision: Though the focal length of the eye lens is adjustable, it cannot be decreased below a certain limit. Hence, if an object is very close to the eye, it cannot be seen clearly. For a normal human eye, the minimum distance from the eye at which an object is clearly visible without stress on the eye, is called the minimum distance of distinct vision. For the normal human eye, it is 25 cm.

Use your brain power! (Text Book Page No. 89)

Question.
(1) Why do we have to bring a small object near the eyes in order to see it clearly?
(2) If we bring an object closer than 25 cm from the eyes, when can we not see it clearly even though it subtends a bigger angle at the eye?
Answer:
(1) when a small object is brought near the eyes, its apparent size increases. Therefore, it is
seen clearly.

(2) Minimum distance of distinct vision: Though the focal length of the eye lens is adjustable, it cannot be decreased below a certain limit. Hence, if an object is very close to the eye, it cannot be seen clearly. For a normal human eye, the minimum distance from the eye at which an object is clearly visible without stress on the eye, is called the minimum distance of distinct vision. For the normal human eye, it is 25 cm.

Try this (Text Book Page No. 91)

Question 1.
Take a burning incense stick in your hand and rotate it fast along a circle.
Answer:
A circle of red light is seen.

Question 2.
Draw a cage on one side of a cardboard and a bird on the other side. Hang the cardboard with the help of a thread. Twist the thread and leave It. What do you see and why?
Answer:
The bird appears to be inside the cage. This happens due to persistence of vision.
Persistence of vision: We see an object when its image is formed on the retina. The image disappears when the object is removed from our sight. But this is not instantaneous and the image remains imprinted on the retina for about \(\frac{1}{16}\) th of a second after the removal of the object. The sensation on the retina persists for a while. This effect is known as the persistence of vision.

It is due to persistence of vision that we continue to see the object in its position for about \(\frac{1}{16}\) th of a second after it is removed.
Example: When a burning stick of incense is moved fast in a circle, a circle of red light is seen.

Can you tell? (Text Book Page No. 91)

Question 1.
How do we perceive different colours?
Answer:
(1) In nature we field objects of various colours. Perception of colour means to be able to respond to colour.
(2) We can distinguish between various colours due to perception of colour.
(3) The cone-shaped cells on the retina of the eye respond to the various colours when light is bright and communicate to the brain about the colours of the image formed on the retina. This gives us the proper idea about the colours of the object.
(4) If, in the retina of a person, the cone-shaped cells responding to certain specific colours are absent, the person is unable to distinguish between the colours. As a result, he lacks perception of colour.

Fill in the blanks and rewrite the statements:

Question 1.
The focal length of a………..lens is positive.
Answer:
The focal length of a convex lens is positive.

Question 2.
The focal length of a………..lens is negative.
Answer:
The focal length of a concave lens is negative.

Question 3.
The magnification produced by a………..lens is always positive.
Answer:
The magnification produced by a concave lens is always positive.

Question 4.
The power of a………..lens is positive.
Answer:
The power of a convex lens is positive.

Question 5.
The power of a………..lens is negative.
Answer:
The power of a concave lens is negative.

Question 6.
The focal length of a lens with power 2.5 D is………..
Answer:
The focal length of a lens with power 2.5 D is 40 cm (0.4 m).

Question 7.
The power of a lens with focal length 20 cm is………..
Answer:
The power of a lens with focal length 20 cm is 5D.

Question 8.
The minimum distance of distinct vision for a normal human eye is………..
Answer:
The minimum distance of distinct vision for a normal human eye is 25 cm.

Question 9.
If two lenses with focal lengths 10 cm and 20 cm respectively are kept in contact with each other, the effective power of the combination is………..
Answer:
If two lenses with focal lengths 10 cm and 20 cm respectively are kept in contact with each other, the effective power of the combination is 15 D.

Question 10.
A………..lens is used as a simple microscope.
Answer:
A convex lens is used as a simple microscope.

Rewrite the following statements by selecting the correct options:

Question 1.
Inside water, an air bubble behaves………..
(a) like a flat plate
(b) like a concave lens
(c) like a convex lens
(d) like a concave mirror
Answer:
Inside water, an air bubble behaves like a concave lens.

Question 2.
…………represents the lens formula.

Answer:
(b) \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) represents the lens formula.

Question 3.
The power of a convex lens of focal length 25 cm is………..
(a) +4.0 D
(b) 0.25 D
(c) -4.0 D
(d) -0.4D
Answer:
The power of a convex lens of focal length 25 cm is +4.0 D

Question 4.
A lens does not produce any deviation of a ray of light passing through………..
(a) it’s centre of curvature
(b) it’s optical centre
(c) it’s principal focus
(d) an axial point at a distance 2F from its centre
Answer:
A lens does not produce any deviation of a ray of light passing through its optical centre.

Question 5.
The image formed by a concave lens is always………..
(a) virtual and erect
(b) real and erect
(c) virtual and inverted
(d) real and inverted
Answer:
The image formed by a concave lens is always virtual and erect.

Question 6.
A convex lens forms a virtual image of an object placed………..
(a) at infinity
(b) at a distance 2F from the lens
(c) at a distance F from the lens
(d) between the principal focus and the optical centre of the lens.
Answer:
A convex lens forms a virtual image of an object placed between the principal focus and the optical centre of the lens.

Question 7.
When an object is placed at 2F₁ of a convex lens, its image is formed………..
(a) at F₁
(b) at 2F₂
(c) beyond 2F₂
(d) on the same side as the object
Answer:
When an object is placed at 2F₁ of a convex lens, its image is formed at 2F₂

Question 8.
To obtain an image of the same size as that of an object with the help of a convex lens, the object should be placed………..
(a) at infinity
(b) beyond F₁
(c) between F₁ and 2F₁
(d) at 2F₁
Answer:
To obtain an image of the same size as that of an object with the help of a convex lens, the object should be placed at 2F₁

Question 9.
When an object is placed between O and F₁ in front of a convex lens, the image formed is ………..
(a) enlarged and erect
(b) diminished and erect
(c) real and enlarged
(d) diminished and inverted
Answer:
When an object is placed between O and F₁ in front of a convex lens, the image formed is enlarged and erect.

Question 10.
When an object is placed at any finite distance from a concave lens, the image is formed ………..
(a) between F₁ and 2F₁
(b) beyond 2F₁
(c) at F₁
(d) between F₁ and O on the same side as the object.
Answer:
When an object is placed at any finite distance from a concave lens, the image is formed between F₁ and O on the same side as the object.

Question 11.
A student obtained a clear image of window grills on the screen. But the teacher told him to get the image of a tree far away, instead of the window. To get a clear image, the lens must be ……….. (Practice Activity Sheet – 2)
(a) moved towards the screen
(b) moved away from the screen
(c) moved behind the screen
(d) moved far away from the screen
Answer:
A student obtained a clear image of window grills on the screen. But the teacher told him to get the image of a tree far away, instead of the window. To get a clear image, the lens must be moved towards the screen.

Question 12.
The image obtained while finding the focal length of a convex lens is ……….. (Practice Activity Sheet – 3)
(a) real and erect
(b) virtual and erect
(c) real and inverted
(d) virtual and inverted
Answer:
The image obtained while finding the focal length of a convex lens is real and inverted.

Question 13.
Yash found out f₁ and f₂ of a symmetric convex lens experimentally. Then which of the following conclusions is true? (Practice Activity Sheet – 4)
(a) f₁ = f₂
(b) f₁ > f₂
(c) f₁ < f₂
(d) f₁ ≠ f₂
Answer:
(a) f₁ = f₂

State whether the following statements are true or false: (If a statement is false, correct it and rewrite it.)

Group (A)

Question 1.
Power of a lens, P = \(\frac{1}{f}\).
Answer:
True.

Question 2.
If the power of a lens is 2 D, its focal length = 0.5 m.
Answer:
True.

Question 3.
A concave lens is a converging lens. (March 2019)
Answer:
False. (A concave lens is a diverging lens.)

Question 4.
A convex lens is a diverging lens.
Answer:
False. (A convex lens is a converging lens.)

Question 5.
A concave lens always forms a virtual image.
Answer:
True.

Question 6.
A convex lens always forms a virtual image.
Answer:
False. (A convex lens forms a real image or a virtual image depending on the object distance.)

Question 7.
Due to the light sensitive cells in the eye, we get information about the brightness or dimness of the object and the colour of the object.
Answer:
True.

Question 8.
The focal length of a concave lens is negative.
Answer:
True.

Question 9.
The magnification produced by a concave lens is positive or negative depending on the object distance.
Answer:
False. (The magnification produced by a concave lens is always positive.)

Question 10.
The magnification produced by a convex lens is positive or negative depending on the object distance.
Answer:
True.

Question 11.
A concave lens is used as a magnifying glass.
Answer:
False. (A convex lens is used as a magnifying glass.)

Question 12.
A convex lens is used as a simple microscope.
Answer:
True.

Question 13.
A concave lens is used to correct myopia.
Answer:
True.

Question 14.
A convex lens is used to correct hypermetropia.
Answer:
True.

Group (B)

Question 1.
When red light falls on the eyes, the cells responding to red light get excited more than those responding to other colours and we get the sensation of red colour.
Answer:
True.

Question 2.
When an object is placed in front of a concave lens, its image is obtained on the opposite side of the object.
Answer:
False. (When an object is kept in front of a concave lens, its image is obtained on the same side of the lens as the object.)

Question 3.
The image formed by a concave lens is always virtual.
Answer:
True.

Question 4.
The principal focus of a convex lens is virtual.
Answer:
False. (The principal focus of a convex lens is real.)

Question 5.
An object of height 2 cm forms an image of height 3 cm when placed in front of a concave lens.
Answer:
False. (An object of height 2 cm forms an image of height less than 2 cm when placed in front of a concave lens.)

Question 6.
Absence of rod like cells results in colour blindness.
Answer:
False. (Absence of conical cells results in colour-blindness.)

Question 7.
Nearsightedness can be corrected using spectacles having convex lenses.
Answer:
False. (Nearsightedness can be corrected using spectacles having concave lenses.)

Question 8.
Farsightedness can be corrected using spectacles having convex lenses of suitable focal length.
Answer:
True.

Question 9.
As one grows old, ciliary muscles become weak.
Answer:
True.

Question 10.
In a simple microscope, the object is placed within the focal length of the convex lens.
Answer:
True.

Question 11.
A compound microscope forms an erect and real image of a small object.
Answer:
False. (A compound microscope forms an inverted and virtual image of a small object.)

Question 12.
In a compound microscope, a real image acts as an object for the eyepiece.
Answer:
True.

Question 13.
In television, we see a continuous picture due to persistence of vision.
Answer:
True.

Question 14.
The conical cells can respond differently to red, green and blue colours.
Answer:
True.

Question 15.
The rod like cells respond to colours and communicate the presence of colours in the retinal image of the brain.
Answer:
False. (The rod like cells respond to the intensity of light and communicate the degree of brightness and darkness, to the brain.)

Question 16.
The conical cells respond to the intensity of light and communicate the degree of brightness and darkness to the brain.
Answer:
False. (The conical cells respond to colours and communicate the presence of colours in the retinal image to the brain.)

Question 17.
Generally, using the same objective lens, but different eyepieces, different magnification can be obtained.
Answer:
True.

Find the odd one out and give the reason:

Question 1.
Simple microscope, Compound microscope, Telescope, Myopia.
Answer:
Myopia. It is a defect of vision; others are instruments.

Question 2.
Myopia, Presbyopia, Hypermetropia, Spectrometer.
Answer:
Spectrometer. It is an instrument; others are defects of vision.

Question 3.
Presbyopia, Retina, Nearsightedness, Farsightedness.
Answer:
Retina. It is a part of the eye; others are defects of vision.

Question 4.
Compound microscope, Kaleidoscope, Simple microscope, Astronomical telescope.
Answer:
Kaleidoscope. Others are optical instruments.

Question 5.
TV, Motion picture, Complete circle formed by a revolving burning incense stick, Colour blindness.
Answer:
Colour-blindness. Others are examples of persistence of vision.

Question 6.
Planets, Stars, Satellites, Rainbow.
Answer:
Rainbow. Others are celestial bodies.

Considering the correlation between the words of the first pair, pair the third word accordingly with proper answer:

Question 1.
Nearsightedness: Elongated eyeball :: Farsightedness:………
Answer:
Flattened eyeball

Question 2.
Convex lens : Converging :: Concave lens :………..
Answer:
Diverging

Question 3.
Object at 2F₁ of a convex lens : Image at 2F₂ :: Object at F₁ :………..
Answer:
Image on the opposite side at infinity

Question 4.
Magnification positive : Erect image :: Magnification negative :………..
Answer:
Inverted image

Question 5.
Convex lens : Positive power of the lens :: Concave lens:
Answer:
Negative power of the lens.

Question 6.
\(\frac{1}{f(\text { in metre })}\) : Power of the lens (in dioptre) :: \(\frac{\text { Image distance }}{\text { Object distance }}\)
Answer:
Magnification.

Question 7.
Focal length : Metre :: Power of a lens :………..
Answer:
Dioptre.

Question 8.
Iris : Pupil :: Ciliary muscles :……….
Answer:
Eye lens

Question 9.
Nearsightedness : Concave lens :: Farsightedness :………..
Answer:
Convex lens

Question 10.
Nearsightedness : Image in front of the retina :: Farsightedness :………..
Answer:
Image behind the retina

Question 11.
Observation of stars and planets : Telescope :: Repairing a watch :……….
Answer:
Simple microscope

Question 12.
Cinema : Persistence of vision :: Rainbow :………
Answer:
Refraction, dispersion and internal reflection of light.

Match the following:

Question 1.

Column A	Column B
(1) Conical cells	(a) Intensity of light
(2) Rod like cells	(b) Colour of an image
(3) Pupil	(c) Iris
(4) Cornea	(d) Aperture
	(e) Transparent

Answer:
(1) Conical cells – Colour of an image
(2) Rod like cells – Intensity of light
(3) Pupil – Aperture
(4) Cornea – Transparent.

Question 2.

Column A	Column B
(1) Magnification	(a) \(\frac{1}{f}\)
(2) Power of a lens	(b) \(\frac{h_{2}}{h_{1}}\)
(3) Focal length	(c) f
(4) Distance of an object from a lens	(d) u
	(e) \(\frac{h_{1}}{h_{2}}\)

Answer:
(1) Magnification: \(\frac{h_{2}}{h_{1}}\)
(2) Power of a lens: \(\frac{1}{f}\)
(3) Focal length: f
(4) Distance of an object from a lens: u.

Question 3.

Answer:
(1) Lens: \(\frac{1}{f}: \frac{1}{v}-\frac{1}{u}\)
(2) Magnification: \(\frac{h_{2}}{h_{1}}\)
(3) Refractive index: \(\frac{\sin i}{\sin r}\)

Question 4.

Column A	Column B
(Convex lens)	(a) Image virtual, erect and enlarged
(1) Object at 2F1	(b) Image real, inverted and of the same size
(2) Object between F1 and 2F1	(c) Image real, inverted and highly diminished
(3) Object between O and F1	(d) Image real, inverted and highly enlarged
(4) Object at infinity	(e) Image real, inverted and enlarged

Answer:
(1) Object at 2F₁ – Image real, inverted and of the same size
(2) Object between F₁ and 2F₁ – Image real, inverted and enlarged
(3) Object between O and F₁ – Image virtual, erect and enlarged
(4) Object at infinity – Image real, inverted and highly diminished

Question 5.

Column A	Column B
(1) Nearsightedness	(a) Ciliary muscles become weak
(2) Farsightedness	(b) Image in front of the retina
(3) Presbyopia	(c) Colour-blindness
	(d) Image behind the retina

Answer:
(1) Nearsightedness – Image in front of the retina
(2) Farsightedness – Image behind the retina
(3) Presbyopia – Ciliary muscles become weak

Question 6.

Column A	Column B
(1) Convex lens	(a) To see small objects clearly
(2) Astronomical telescope	(b) To observe minute objects
(3) Compound microscope	(c) To observe astronomical objects such as stars, planets, etc.
(4) Simple microscope	(d) Presbyopia
	(e) Power of a lens

Answer:
(1) Convex lens – Presbyopia
(2) Astronomical telescope – To observe astronomical objects such as stars, planets, etc.
(3) Compound microscope – To observe minute objects
(4) Simple microscope – To see small objects clearly.

Question 7.

Column A	Column B
(1) Persistence of vision	(a) Lenses and mirrors are used
(2) Reflecting telescope	(b) To see objects far away from us
(3) Telescope	(c) Motion picture
(4) Compound microscope	(d) To observe blood
	(e) Convex lens

Answer:
(1) Persistence of vision – Motion picture
(2) Reflecting telescope – Lenses and mirrors are used
(3) Telescope – To see objects far away from us
(4) Compound microscope – To observe blood corpuscles.

Name the following:

Question 1.
Name the lens which forms a real image or a virtual image depending on the position of the object.
Answer:
A convex lens.

Question 2.
Name the lens which produces magnification always less than 1.
Answer:
A concave lens.

Question 3.
Name the lens which always forms an image virtual and smaller than the object.
Answer:
A concave lens.

Question 4.
Name the lens used to obtain the image on a screen.
Answer:
A convex lens.

Question 5.
Name the lens for which the image always lies between the object and the lens.
Answer:
A concave lens.

Question 6.
Name the instrument used to observe bacteria.
Answer:
A compound microscope.

Question 7.
Name the instrument used to observe planets.
Answer:
An astronomical telescope.

Answer the following questions in one sentence each:

Question 1.
An object is placed at 60 cm from a convex lens of focal length 20 cm. State the nature and size of the image relative to that of the object.
Answer:
The image is real, inverted and smaller than the object.

Question 2.
If an object is placed at 50 cm from a convex lens of focal length 25 cm, what will be the image distance?
Answer:
The image distance will be.50 cm.

Question 3.
An object is placed at 40 cm from a convex lens of focal length 20 cm. State the nature and the size of the image relative to that of the object.
Answer:
The image is real, inverted and of the same size as that or the object.

Question 4.
An object is placed at 30 cm from a convex lens of focal length 20 cm. State the nature and the size of the image relative to that of the object.
Answer:
The image is real, inverted and larger than the object.

Question 5.
An object is placed at 15 cm from a convex lens of focal length 25 cm. State the nature and size of the image relative to that of the object.
Answer:
The image is virtual, erect and larger than the object.

Question 6.
State the type of lens that can be used to burn paper in sunlight at noon.
Answer:
A convex lens can be used to burn paper in sunlight at noon.

Question 7.
State the type of lens used to correct myopia.
Answer:
A concave lens is used to correct myopia.

Question 8.
State the type of lens used to correct hypermetropia.
Answer:
A convex lens is used to correct hypermetropia.

Question 9.
If two lenses with focal lengths 10 cm and – 20 cm respectively are kept in contact with each other, what will be the effective power of the combination of the lenses?
Answer:
The effective power of the combination of the lenses will be + 5 D.

Question 10.
If two lenses with focal lengths – 10 cm and 40 cm respectively are kept in contact with each other, what can you say about the behaviour of the combination of the lenses?
Answer:
The combination of the lenses will behave as a concave lens.

Answer the following questions:

Question 1.
What is a lens?
Answer:
A lens is a transparent material bound by two surfaces, out of which at least one surface is spherical.
[Note: A lens is normally made of glass or plastic.]

Question 2.
In which instruments have you seen a lens?
Answer:
We have seen a lens in a microscope and a telescope.

Question 3.
How is a lens different from a mirror?
Answer:
A mirror has one reflecting surface. By reflection of light, it forms an image of the object placed in front of it. A mirror is not transparent. A lens has two surfaces that form an image by refraction of light. A lens is transparent.

Question 4.
Make a list of optical devices you know.
Answer:
Microscope, telescope, binoculars, camera, projector.

Question 5.
Do you know which is the natural optical device?
Answer:
Yes. The eye is the natural optical device.

Question 6.
What is a convex lens?
Answer:
A lens having both spherical surfaces puffed up outwards is called a convex lens or double convex lens or biconvex lens. It is thicker in the middle than at the edges.
[Note: A convex lens is also called a converging lens.]

Question 7.
What is a concave lens?
Answer:
A lens having both spherical surfaces curved inwards is called a concave lens or double concave lens or biconcave lens. It is thicker at the edges than in the middle.
[Note: A concave lens is also called a diverging lens.]

Question 8.
Draw neat labelled diagrams: Types of lenses.
Answer:

[Note: Positive meniscus behaves as a convex lens as it is thicker in the middle than at the edges. Negative meniscus behaves as a concave lens as it is thicker at the edges than in the middle.]

Question 9.
In general, when a ray of light passes through a lens, there occurs a change in its direction of propagation. Why?
Answer:
The working of a lens is similar to that of a triangular prism. When a ray of light passes through a lens, it is refracted twice: When entering the lens and when emerging from the lens. There is a change 5 in its direction of propagation every time and as both the changes occur in the same sense, the direction of propagation of the emergent ray is different from that of the incident ray.

Question 10.
State the rules used for drawing ray diagrams for the formation of an image by a convex lens.
Answer:
Rules used for drawing ray diagrams for the formation of an image by a convex lens:
(1) When the incident ray is parallel to the principal axis, the refracted ray passes through the principal focus.

(2) When the incident ray passes through the principal focus, the refracted ray is parallel to the principal axis.

(3) When the incident ray passes through the optical centre of the lens, it passes without changing its direction.

Question 11.
In the case of a convex lens, show the path of the refracted ray when the incident ray of light (1) is parallel to the principal axis of the lens (2) passes through the focus of the lens (3) passes through the optical centre of the lens.
Answer:
1.

2.

3.

Question 12.
Draw neat and well labelled ray diagrams for image formation by convex lens when an object is (1) at infinity (2) beyond 2F₁ (3) at 2F₁ (4) between F₁ and 2F₁ (5) at focus F₁ (6) between focus F₁ and optical centre O. Also, in each case, state the position, nature and size of the image relative to that of the object.
Answer:
(1) Object at infinity:

In this case, the image is formed at focus F₂ of the convex lens. It is real, inverted and highly diminished (point-sized).

(2) Object beyond 2F₁:

In this case, the image is formed between F₂ and 2F₂. It is real, inverted and diminished.

(3) Object at 2F₁:

In this case, the image is formed at 2F₂. It is real, inverted and of the same size as that of the object.

(4) Object between F₁ and 2F₁:

In this case, the image is formed beyond 2F₂. It is real, inverted and magnified (enlarged).

(5) Object at focus F₁:

In this case, the image is formed at infinity. It is real, inverted and infinitely large (highly magnified).

(6) Object between focus F₁ and optical centre O:

In this case, the image is formed on the same side of the lens as the object. It is virtual, erect and larger than the object.
[Note: Here, the image is virtual. Hence, it is shown by a dotted line.]

Question 13.
Observe the following figure and complete the table: (March 2019)

Answer:

Point	Answer
(i) Position of the object	Between F1 and O
(ii) Position of the image	On the same side of the lens as the object
(iii) Size of the image	Very large
(iv) Nature of the image	Virtual and erect

Question 14.
At which position will you keep an object in front of a convex lens to get a real image smaller than the object? Draw a figure.
Answer:
The object should be placed beyond 2F₁.

Question 15.
State the rules used for drawing ray diagrams for the formation of an image by a concave lens.
Answer:

1. When the incident ray is parallel to the principal axis, the refracted ray, when extended backwards, passes through the principal focus.
2. When the incident ray is directed towards the principal focus F₂, the refracted ray is parallel to the principal axis.
3. When the incident ray passes through the optical centre of the lens, it passes without changing its direction.

Question 16.
State the characteristics of an image formed by a concave lens.
Answer:
The image formed by a concave lens is always virtual, erect and smaller than the object. It is on the same side of the lens as the object. Generally, it is formed between the optical centre of the lens and the principal focus F₁. If the object is at infinity, the image is a point image formed at F₁.

Question 17.
In the case of image formation by a concave lens, what can you say about the position, nature and size of the image relative to the size of the object?
Answer:
Image formation by a concave lens :
(1) If the object is at infinity, the image is formed at the focus of the lens, on the same side of the lens as the object. It is virtual, erect and much smaller than the object (point image).

(2) If the object is at any finite distance from the lens, the image is formed on the same side of the lens as the object and between the focus and the optical centre of the lens. It is virtual, erect and smaller than the object. The image distance is less than the object distance.

Question 18.
Draw a ray diagram to show image formation by a concave lens.
Answer:

PQ : Object
P’Q’ : Image (virtual, therefore shown by a dotted line),
O : Optical centre,
F₁ : Principal focus,
f : Focal length of the lens
[Note: If in a Board examination, incomplete diagram (as shown below) is given, students should complete it and label its parts as shown in Figure.]

Question 19.
State the Cartesian sign convention for refraction of light (image formation) by a lens.
Answer:
Cartesian sign convention for refraction of light (image formation) by a lens:
In this case, the optical centre (O) of the lens is taken as the origin and the principal axis of the lens is taken as X-axis of the coordinate system.

(1) The object is always placed at the left of the lens.
All distances parallel to the principal axis are measured from the optical centre of the lens.
(2) All distances measured to the right of the origin are taken as positive while distances measured to the left of the origin are taken as negative.
(3) Distances measured perpendicular to and above the principal axis are taken as positive.
(4) Distances measured perpendicular to and below the principal axis are taken as negative.
(5) The focal length of a convex lens is positive and that of a concave lens is negative. (Fig. 7.21)

Question 20.
(i) What is a lens formula? (ii) State it.
Answer:
(i) The relationship between the object distance (u), image distance (v) and focal length (J) of a lens is called the lens formula.
(ii) It is \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
[Note: The lens formula holds good for all values of u and v and is applicable to a convex lens as well as a concave lens. The sign convention for u, v and f must tie used in solving numerical examples.]

Question 21.
What is meant by the magnification produced by a lens? State the formulae for it.
Answer:
The magnification (M) produced by a lens is the ratio of the height of the image (h₂) to the height of the object (h₁).
M = \(\frac{h_{2}}{h_{1}}\)……….(1)
Also M = \(\frac{v(\text { image distance })}{u \text { (object distance) }}\)……………(2)

Question 22.
When is the magnification produced by a lens (1) positive (2) negative?
Answer:
The magnification produced by a lens is
(1) Positive when the image is virtual (as it is erect)
(2) Negative when the image is real (as it is inverted).

Question 23.
Express the magnification produced by a lens in terms of the focal length of the lens and (1) the object distance (2) the image distance.
Answer:
Magnification (M) produced by a lens

where f is the focal length of the lens:

(2) From eq. (2), we have

Question 24.
An object is kept in front of a lens of focal length + 10 cm. Describe the nature of the image in the following cases: (1) The object” distance is 25 cm. (2) The object distance is 5 cm.
Answer:
Since, the focal length of the lens ( +10 cm) is positive, it is a convex lens.
(1) If an object is kept at 25 cm from the lens, the image will be real, inverted and smaller than the object.
(2) If an object is kept at 5 cm from the lens, the image will be virtual, erect and larger than the object.

Question 25.
Anu and Anand have concave and convex lenses respectively. They took lenses in sunlight and tried to burn two pieces of paper of equal areas and temperature. State which lens will burn the paper. Give the reason. Explain with the help of a diagram, why the other paper did not burn.
Answer:
(1) The convex lens will burn the paper. See Fig. 7.22 for reference. The ray of sunlight will converge at the principal focus of the lens. Hence, if the paper is held at the focus, it will burn due to concentration of heat energy.

(2) The paper held in front of the concave lens, will not burn. For reference, see Fig. 7.23. The concave lens will diverge the rays of sunlight falling on it. Hence, the paper will not burn.

Question 26.
To obtain a magnified real image of a small film strip, which type of lens is used? Where is the film strip placed to obtain the image on the screen?
Answer:
To obtain a magnified real image of a small film strip, a convex lens is used. The film strip is placed between F₁ and 2F₁ and the screen is placed on the other side of the lens.

Question 27.
When an object of height 2 cm is placed in front of a convex lens, the height of the image is found to be 3 cm. State the nature and position of the image giving reason.
Answer:
When an object is placed between the optical centre and the principal focus of a convex lens, the image formed by the lens is virtual and larger than the object. When an object is placed between F₁ and 2F₁ the image formed by the lens is real and larger than the object.

In the above case, if the image is virtual, it will be erect and on the same side of the lens as that of the object. If the image is real, it will be inverted and beyond 2F₂ on the other side of the lens with respect to the object.

Question 28.
You are given a lens which gives a virtual, erect and enlarged image. What type of lens is it?
Answer:
Since the lens gives a virtual, erect and enlarged image, it must be a convex lens.

Question 29.
When an object of height 3 cm is placed in front of a concave lens, the height of the image is found to be 6 cm. State, giving the reason, whether the given statement is true or false.
Answer:
When an object is placed in front of a concave lens, the image formed by the lens is always smaller than the object. In the statement given in the question, the height of the image is reported as greater than that of the object. Hence, the statement given in the question is false.

Question 30.
State two uses of a concave lens.
Answer:

1. A concave lens is used to correct myopia (nearsightedness).
2. In some optical instruments, a combination of a concave lens and a convex lens is used.

Question 31.
State two uses of a convex lens.
Answer:
A convex lens is used (1) to read words in small print (2) to correct hypermetropia (Far-sightedness).

Question 32.
An object is kept in front of a lens of j focal length – 20 cm. Describe the nature of the image when the object distance is 25 cm.
Answer:
Since the focal length of the lens (- 20 cm) is negative, it is a concave lens.
If an object is kept at 25 cm from the lens, the image will be virtual, erect and smaller than the object.
[Note: The nature of the image is independent of the object distance as it is a concave lens.]

Question 33.
An object is placed in front of a convex lens of focal length 20 cm. If the object distance is changed from 60 cm to 40 cm, what can you say about the size of the image relative to that of the object?
Answer:
In this case, the focal length (f) of the lens is 20 cm.
∴ 2f = 40 cm.
When the object distance is 60 cm (which is greater than 2f), the image will be smaller than the object. When the object distance becomes 40 cm (which is equal to 2f), the image will be of the same size as that of the object.

Question 34.
What is the power of a lens?
Answer:
The capacity of a lens to converge or diverge incident rays is called its power. The power (P) of a lens is the inverse of the focal length (f) of the lens.
P = \(\frac{1}{f}\)

Question 35.
What is the unit of power of a lens? Define it.
Answer:
The unit of power of a lens is the dioptre (D).
One dioptre is the power of a lens whose focal length is one metre.
1 dioptre (D) = \(\frac{1}{1 \text { metre }(\mathrm{m})}\)
[Note: The dioptre, the SI unit of power of a lens, is denoted by D.]

Question 36.
What is the sign of the power of (i) a convex lens (ii) a concave lens?
Answer:
The power of a convex lens is positive while that of a concave lens is negative.

Question 37.
If there is an increase or decrease in the focal length of a lens, what will be the effect on the power of the lens?
Answer:
The power of a lens is the inverse of its focal length. Hence, if there is an increase in the focal length of a lens, the power of the lens will decrease accordingly. Similarly, if there is a decrease in the focal length of a lens, the power of the lens will increase accordingly.

Question 38.
If two lenses of focal lengths f₁ and f₂ are kept in contact with each other, state the formula for the focal length of the combination. If P₁ and P₂ are the powers of these lenses, state the formula for the power of the combination.
Answer:
If two lenses of focal lengths f₁ and f₂ are kept in contact with each other, the focal length (f) of the combination is given by \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

[Two lenses kept in contact with each other]
If P₁ and P₂ are the powers of these lenses, the power (P) of the combination is given by P = P₁ + P₂.
[Note: The figures are given only for reference.]

Question 39.
Draw a neat labelled diagram to show the structure of the human eye.
Answer:

Question 40.
What is cornea?
Answer:
The cornea is a thin and transparent cover (membrane) on the human eye through which light enters the eye. Maximum refraction of light rays entering the eye occurs at the cornea.

Question 41.
What is iris?
Answer:
The iris is a dark fleshy screen (muscular diaphragm) behind the cornea in the human eye. Its colours are different for different people.

Question 42.
What is pupil?
Answer:
The pupil is a small circular opening of changing diameter at the centre of the iris in the human eye.

Question 43.
What is the use of the pupil in the human eye?
Answer:
The pupil in the human eye is useful for controlling and regulating the amount of light entering the eye. The pupil contracts in the presence of too much light and dilates when light is insufficient, thus changing the amount of light entering the eye.

Question 44.
With reference to the functioning of the pupil in the human eye, what is adaptation?
Answer:
The tendency of the pupil in the human eye to adjust the opening for light, depending on the intensity of incident light, to control and regulate the amount of light entering the eye is called adaptation.

Question 45.
What is the shape and the size of the human eyeball?
Answer:
The human eyeball is approximately spherical in shape with a diameter of about 2.4 cm.

Question 46.
Name the part of the human eye that forms a transparent bulge on the surface of the eyeball.
Answer:
The cornea forms a transparent bulge on the surface of the eyeball.

Question 47.
Which part of the human eye is located just behind the pupil?
Answer:
A transparent biconvex crystalline lens is located just behind the pupil in the human eye.

Question 48.
What is retina?
Answer:
The retina is a light sensitive screen consisting of a delicate membrane with a large number of light sensitive cells.

Question 49.
What is the nature of the eye lens and what does the eye lens do?
Answer:
The eye lens is a double convex transparent crystalline lens, just behind the pupil. The eye lens provides small adjustment of focal length to form a real and inverted sharp image on the retina.

Question 50.
What happens when light falls on the retina?
Answer:
When light falls on the retina, light sensitive cells of the retina are activated. They generate electrical signals which are passed by optic nerves to the brain. The brain interprets the signals and processes the information such that we perceive the object as it is.

Question 51.
What are ciliary muscles?
Answer:
The muscles which hold the eye lens in its position, and bring about changes in the shape (curvature) of the eye lens, and hence of focal length are known as ciliary muscles.

Question 52.
What is the focal length of the eye lens of a normal eye in relaxed position of eye muscles?
Answer:
The focal length of the eye lens of a normal eye in relaxed position of eye muscles is about 2 cm.

Question 53.
Where does the second focal point of the eye lens of a normal eye in relaxed position of eye muscles lie?
Answer:
The second focal point of the eye lens of a normal eye in relaxed position of eye muscles lies on the retina.

Question 54.
What is meant by power of accommodation of the eye?
Answer:
The ability of the eye lens to adjust its focal length is called the power of accommodation of the eye.

Question 55.
Explain the term power of accommoda¬tion of the eye.
(OR)
Write a short note on the power of accommodation of the eye.
Answer:
Power of accommodation of the eye: The eye lens is held in its position by the ciliary muscles. When we look at a nearby object, the ciliary muscles compress the eye lens so that it becomes rounded. Hence, the focal length of the eye lens decreases. Therefore, the image is formed on the retina of the eye and hence the nearby object is seen clearly.

When we look at a distant object, the ciliary muscles relax so that the eye lens becomes flat. Hence, the focal length of the eye lens increases. Therefore, the image is formed on the retina of the eye and hence the distant object is seen clearly. This ability of the eye lens to adjust its focal length is called the power of accommodation of the eye.

Question 56.
What is meant by accommodation? How is it brought about?
Answer:
The process of focusing the eye on objects at different distances is called accommodation. It is brought about by changing the curvature of the f elastic eye lens making it thinner or thicker.

Question 57.
The human eye is very similar to a photographic camera. The figure given shows the main parts of a photographic camera. Now answer the following questions:

(1) Name the parts of the human eye similar to the following parts of the photographic camera :
(a) Photographic film (b) Aperture.
(2) State one difference between the human eye lens and camera lens.
(3) Name the muscles which adjust the curvature of the eye lens.
(4) Which phenomenon of light is responsible for the working of the eye?
Answer:
(1) (a) The retina in the human eye is similar to the photographic film in a camera.
(b) The pupil in the human eye is similar to the aperture in a camera.
(2) In a photographic camera, the focal length of the lens changes when the position of the lens is changed. In the human eye, the focal length of the eye lens is changed by the ciliary muscles and the distance of the image from the eye lens is fixed.
(3) The ciliary muscles adjust the curvature of the eye lens.
(4) The refraction of light is responsible for the working of the eye.

Question 58.
Have you seen a photographic camera in which a film is used? Compare the human eye with it. State similarities between them. State the points of difference between them.
Answer:
Yes, we have seen a photographic camera in which a film is used.
Cameras, in general, have various shapes and sizes. Some cameras are much bigger than the human eye while some are smaller than the human eye. Here we shall consider a simple camera.

Similarities: In the case of a camera as well as the human eye, it is possible to control the amount of incoming light with the help of a diaphragm and an aperture. Both use a convex lens for focusing. The photographic film in a camera is coated with a photosensitive material. The retina in the eye consists of a large number of light sensitive cells. The photographic film in a camera is processed using chemicals and then prints (photographs) can be obtained using the appropriate paper.

In the human eye, the electrical signals generated by light sensitive cells are passed by optic nerves to the brain which interprets them.

Differences: Cameras come in a variety of sizes and shapes unlike the human eye. Unlike the human eye, a wide variation in exposure time is possible in the case of cameras. The human eye is sensitive in the visible region (red to violet) of the electromagnetic spectrum, while a much wider range of the electromagnetic spectrum can be covered with cameras designed for specific J purposes. In comparison with the human eye, a wider view and range can be covered by a camera.

In comparison with the human eye, a wider intensity (of light) range can be covered with a camera. The retina is indispensable in the human eye, while cameras without a photographic film have been designed with the help of photosensitive materials and are in current use.

[Note: With advances in technology, improved cameras are designed all the time, and the list of differences between the human eye and a camera in general would be practically endless.]

Question 59.
What is meant by the minimum distance of distinct vision?
Answer:
The minimum distance from the normal eye, at which an object is clearly visible without stress on the eye is called the minimum distance of distinct vision.

Question 60.
Explain the term minimum distance of distinct vision.
(OR) Write a short note on distance of distinct vision.
Answer:
Minimum distance of distinct vision; Though the focal length of the eye lens is adjustable, it cannot be decreased below a certain limit. Hence, if an object is very close to the eye, it cannot be seen clearly. For a normal human eye, the minimum distance from the eye at which an object is clearly visible without stress on the eye, is called the minimum distance of distinct vision. For the normal human eye, it is 25 cm.

Question 61.
State four reasons related to problems of vision.
Answer:
Problems of vision are related to (i) weakening of ciliary muscles (ii) change in the size of the eyeball (iii) irregularities on the surface of cornea (iv) formation of a membrane over the eye lens.

Question 62.
What is myopia or nearsightedness? What are the possible reasons of myopia? How is myopia corrected? Explain with diagrams.
Answer:
Myopia or nearsightedness is the defect of vision in which a human eye can see nearby objects distinctly but is unable to see distant objects clearly as they appear indistinct.
In this case the image of a distant object is formed in front of the retina instead of on the retina. [Figs. 7.29 (a), 7.29 (b)]

Possible reasons of myopia: (1) The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large. (2) The distance between the eye lens and the retina increases as the eyeball elongates.

Myopia is corrected using a suitable concave lens. Light rays are diverged by the concave lens before they strike the eye lens. A concave lens of proper focal length is chosen to produce the required divergence. Hence, after the converging action of the eye lens, the image is formed on the retina. [Fig. 7.29 (c)]

Question 63.
Observe the following diagram and answer the questions.
(a) Which eye defect is shown in this diagram?
(b) What are the possible reasons for this eye defect?
(c) How is this defect corrected? Write it in brief.

Answer:
(a) Myopia or Nearsightedness

(b) Possible reasons for the defect:
(i) The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large.
(ii) The eyeball elongates so that the distance between the lens and the retina increases.

(c) Correction of the defect: This defect can be corrected using spectacles with concave lenses.
A concave lens diverges the incident rays and these diverged rays can be converged by the lens in the eye to form an image on the retina.

Question 64.
What is the sign of the power of the lens used to correct myopia?
Answer:
The power of the lens used to correct myopia is negative.
[Note: It is a concave lens. Negative focal length Negative power.]

Question 65.
In a Std. X class, out of 40 students, 10 students use spectacles, 2 students have ( positive power and 8 students have negative power of lenses in their spectacles.
Answer the following questions:
(1) What does the negative power indicate?
(2) What does the positive power indicate?
(3) Generally which type of spectacles do most of the students use?
(4) What defect of eyesight do most of the students suffer from?
(5) Give two possible reasons for the above defect.
Answer:
(1) The negative power indicates a concave lens or myopia.
(2) The positive power indicates a convex lens or hypermetropia.
(3) Generally, most of the students use spectacles with concave lenses.
(4) Most of the students suffer from myopia.
(5) Two possible reasons for myopia:

1. The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large.
2. The distance between the eye lens and the retina increases as the eyeball elongates.

Question 66.
What is hypermetropia or farsightedness? What are the possible reasons of hypermetropia? How is hypermetropia corrected? Explain with figures.
Answer:
Hypermetropia or farsightedness is the defect of vision in which a human eye can see distant objects clearly but is unable to see nearby objects clearly.
In this case the image of a nearby object would fall behind the retina instead of on the retina.
[Figs. 7.31 (a), 7.31 (b)]

Possible reasons of hypermetropia:
(1) Curvature of the cornea and the eye lens decreases. Hence, the converging power of the eye lens becomes less. (2) The distance between the eye lens and retina decreases (relative to the normal eye) and the focal length of the eye lens becomes very large due to the flattening of the eyeball.

Hypermetropia is corrected using a suitable convex lens. Light rays are converged by the convex lens before they strike the eye lens. A convex lens of proper focal length is chosen to produce the required convergence. Hence, after the converging action of the eye lens, the image is formed on the retina. [Fig. 7.31 (c)]

Question 67.
What is the sign of the power of the lens used to correct hypermetropia?
Answer:
The power of the lens used to correct hypermetropia is positive.
[Note: It is a convex lens. Positive focal length ∴ Positive power.]

Question 68.
Given below is a diagram showing a defect of human eye.

Study it and answer the following questions:
(1) Name the defect shown in the figure. (Practice Activity Sheet – 3)
(2) Give two possible reasons for this defect of eye in human beings.
(3) Name the type of lens used to correct the eye defect.
(4) Draw a labelled diagram to show how the defect is rectified by using the lens.
Answer:
A lens having both spherical surfaces puffed up outwards is called a convex lens or double convex lens or biconvex lens. It is thicker in the middle than at the edges.
[Note: A convex lens is also called a converging lens.]

Question 69.
Observe the following figures and complete the table. (Practice Activity Sheet – 1)
Answer:

Question 70.
What is presbyopia? State the reason for this defect. How is presbyopia corrected?
Answer:
Presbyopia is the defect of vision in which aged people find it difficult to see nearby objects comfortably and clearly without spectacles.

Reason of presbyopia: The power of accommodation of eye usually decreases with ageing. The muscles near the eye lens lose their ability to change the focal length of the lens.
Therefore, the near point of the eye lens shifts farther from the eye.

This defect is corrected using a convex lens of appropriate power. The lens converges light rays before they fall on the eye lens such that the action of the eye lens forms the image on the retina.

Question 71.
What is a bifocal lens?
Answer:
A bifocal lens is a lens of which the upper part is a concave lens to correct myopia and the lower part is a convex lens to correct hypermetropia.

[Note: A person suffering from myopia as well as hypermetropia, uses a bifocal lens. Nowadays, the defects of vision such as myopia and hypermetropia can be corrected using contact lenses or by laser surgery.]

Question 72.
(A) Anil cannot see the blackboard writing clearly, but he can see nearby objects clearly.
(i) What is the eye defect he is suffering from?
(ii) How is it corrected?
(B) Anil’s uncle cannot see nearby objects clearly, but he can see distant objects clearly.
(i) What is the eye defect he is suffering from?
(ii) How is it corrected?
Answer:
(A) Anil cannot see the blackboard writing clearly, but he can see nearby objects clearly.
(i) This defect is called myopia (nearsightedness).
(ii) It is corrected using spectacles having concave lenses of appropriate power.

(B) Anil’s uncle cannot see nearby objects clearly, but he can see distant objects clearly.
(i) This defect is called hypermetropia (farsightedness).
(ii) It is corrected using, spectacles having convex lenses of appropriate power.

Question 73.
When are bifocal lenses used in spectacles?
Answer:
When a person cannot see nearby objects as well as distant objects clearly, bifocal lenses are used in spectacles.

Question 74.
Aniket from Std. X uses spectacles. The power of the lenses in his spectacles is -0.5 D. Answer the following questions:
(1) State the type of’ lenses used in his spectacles.
(2) Name the defect of vision Aniket is suffering from.
(3) Find the focal length of the lenses used in his spectacles.
Answer:
(1) Concave lenses are used in the spectacles used by Aniket.
(2) Aniket is suffering from myopia (near-sightedness).
(3) Focal length of the lenses used in his spectacles

= -2 m (Concave lens ∴ Minus sign)

Question 75.
Sunita from Std. X uses spectacles. Her spectacle number is -1.5 D. Answer the following questions:
(1) Name the defect of eye from which she is suffering.
(2) What type of lens is she using?
(3) Find the focal length of the lens.
Answer:
(1) Myopia.
(2) Concave.

(Concave lens ∴ minus sign)
This is the focal length of the lens.

Question 76.
Surabhi from Std. X uses spectacles. The power of the lenses in her spectacles is 0.5 D. Answer the following questions from the given information: (March 2019)
(i) Identify the type of lenses used in her spectacles.
(ii) Identify the defect of vision Surabhi is suffering from.
(iii) Find the focal length of the lenses used in her spectacles.
Answer:
(i) Convex lenses are used in the spectacles used by Surabhi.
(ii) Surabhi is suffering from hypermetropia (farsightedness).
(iii) Focal length of the lenses used in her spectacles

Question 77.
My grandfather uses a bifocal lens in his spectacles. Explain why.
Answer:
In old age, people usually suffer from both myopia and hypermetropia. Therefore, they need spectacles having bifocal lenses.

The upper part of a bifocal lens is a concave lens to correct myopia. The lower part of a bifocal lens is a convex lens to correct hypermetropia.

Question 78.
State uses of concave lens.
Answer:

1. Concave lenses are used for proper working of medical equipment, scanner, CD player – the instruments that employ laser rays.
2. One or more concave lenses are used in a small safety device, fitted in the peep hole in a door, due to which we can see a large area outside the door.
3. Concave lenses are used in spectacles to correct nearsightedness (myopia).
4. A concave lens is used to spread light emitted by the small bulb in a torch over a wide area.
5. A concave lens is used in front of the eyepiece or inside the eyepiece fitted in a camera, telescope and microscope – the instruments employing convex lenses.

Question 79.
State uses of a convex lens.
Answer:
Convex lenses are used in a simple microscope, compound microscope, refracting telescope, camera, projector, spectroscope, spectacles for correcting farsightedness (hypermetropia) and binoculars.

Question 80.
What is meant by the apparent size of an object? With a neat and labelled diagram, explain the relation between the apparent size of an object and the angle subtended by the object at the eye.
Answer:
An object appears small or big depending upon the size of its image formed on the retina of the eye. The size of an object as perceived by the eye is called the apparent size of the object. Consider two objects of the same size, one held near the eye and the other away from the eye as shown in the following figure (Fig. 7.34). The nearby object (PQ) appears larger than the distant object (P₁Q₁).

Also, the angle a subtended by the nearby object at the eye is larger than the angle β subtended by the distant object at the eye. This shows that the apparent size of an object depends upon the angle subtended by the object at the eye. The greater the angle subtended by the object at the eye, the greater is the apparent size of the object. Similarly, the smaller the angle subtended by the object at the eye, the smaller is the apparent size of the object.

Question 81.
With a neat labelled diagram, explain the working of a simple microscope. State uses of a simple microscope.
(OR)
What does a simple microscope consist of? What is the order of magnification obtained by a simple microscope? What is a simple microscope used for?
Answer:
A simple microscope consists of a convex lens of short focal length, usually fixed in a suitable frame with a handle or mounted on a stand.

The object is placed in front of the convex lens of short focal length such that the object distance is less than the focal length. The image is virtual and larger than the object. It is formed on the same side of the lens as the object.

A maximum magnification of about 20 can be obtained by a simple microscope. A simple microscope is used by watch repairers to observe small parts of a watch and by jewellers to examine ornaments. A simple microscope (also called a magnifying glass) is also used to read words in small print.

Question 82.
With a neat labelled diagram, explain the construction and working of a compound microscope.
Answer:
Construction of a compound microscope:

(1) A compound microscope consists of a metal tube fitted with two convex lenses at the two ends. These lenses are called the objective lens (the lens directed towards the object) and the eyepiece (the lens directed towards the eye). Both the lenses are small in size, but the cross section of the objective lens is less than that of the eyepiece. The objective lens has a short focal length. The focal length of the eyepiece is more than that of the objective lens.

(2) The metal tube is mounted on a stand. The principal axes of the objective lens and the eyepiece are along the same line. The distance between the object and the objective lens can be changed with a screw. It is possible to change the distance between the objective lens and the eyepiece.

Working :
(1) The object to be observed is illuminated and placed in front of the objective lens, slightly beyond the focal length of the objective lens. Its real, inverted and enlarged image is formed by the objective lens on the other side.

(2) This intermediate image lies within the focal length of the eyepiece. It serves as an object for the eyepiece. The eyepiece works as a simple microscope. The final image is virtual, highly enlarged and inverted with respect to the original object. It can be formed at the minimum distance of distinct vision from the eyepiece. The final image is observed by keeping the eye close to the eyepiece.

Question 83.
State two uses of a compound microscope.
Answer:
Uses of a compound microscope:

1. It is used to observe blood corpuscles, plant and animals cells, microorganisms like bacteria, etc.
2. It is used in a pathological laboratory to observe blood, urine, etc.
3. It is a part of a travelling microscope used for measurement of very small distance.

Question 84.
What will happen if in a compound microscope, the objective lens is large in size and has a focal length?
Answer:
If the objective lens of a compound microscope is large in size, in addition to the light coming from an object, other unwanted light will be incident on the objective lens. Hence, the image will not be seen clearly. If the objective lens has a large focal length, the magnification produced by it will be less.

Question 85.
(a) In which type of microscope do you find the lens arrangement as shown in the following diagram? (Practice Activity Sheet – 1)

(b) Write in brief, the working of this microscope.
(c) Where is this microscope used?
Answer:
(a) Compound microscope.

(b)
An object appears small or big depending upon the size of its image formed on the retina of the eye. The size of an object as perceived by the eye is called the apparent size of the object. Consider two objects of the same size, one held near the eye and the other away from the eye as shown in the following figure (Fig. 7.34). The nearby object (PQ) appears larger than the distant object (P₁Q₁).

Also, the angle a subtended by the nearby object at the eye is larger than the angle β subtended by the distant object at the eye. This shows that the apparent size of an object depends upon the angle subtended by the object at the eye. The greater the angle subtended by the object at the eye, the greater is the apparent size of the object. Similarly, the smaller the angle subtended by the object at the eye, the smaller is the apparent size of the object.

(c) A simple microscope is used by watch repairers to observe small parts of a watch and by jewellers to
examine ornaments. A simple microscope (also called a magnifying glass) is also used to read words in small print.

Question 86.
(i) Which type of microscope has the arrangement of lenses shown in the following figure?

(ii) Label the figure correctly.
(iii) Write the working of this microscope.
(iv) Where is this microscope used?
(v) Suggest a way to increase the efficiency of this microscope. (Practice Activity Sheet – 2)
Answer:
(i) Compound microscope.

(ii)

(iii) Working:
(1) The object to be observed is illuminated and placed in front of the objective lens, slightly beyond the focal length of the objective lens. Its real, inverted and enlarged image is formed by the objective lens on the other side.

(2) This intermediate image lies within the focal length of the eyepiece. It serves as an object for the eyepiece. The eyepiece works as a simple microscope. The final image is virtual, highly enlarged and inverted with respect to the original object. It can be formed at the minimum distance of distinct vision from the eyepiece. The final image is observed by keeping the eye close to the eyepiece.

(iv)

1. It is used to observe blood corpuscles, plant and animals cells, microorganisms like bacteria. etc.
2. It is used in a pathological laboratory to observe blood, urine, etc.
3. It is a part of a travelling microscope used for measurement of very small distance.

(v) Lenses with appropriate focal lengths should be selected.

Question 87.
State the use of a telescope.
Answer:
A telescope is used to observe a distant object such as mountain, moon, planet, star in the magnified form.

Question 88.
Observe the following figure and answer the questions. (Practice Activity Sheet – 3)
(a) Which optical instrument shows arrangement of lenses as shown in the figure?
(b) Write in brief the working of this optical instrument.

(c) How can we get different magnifications in this optical instrument?
(d) Draw the figure again and labelled it properly.
Answer:
(a) Refracting telescope.

(b) working: When the objective lens is pointed towards the distant object to be observed, the rays
of light from the distant object, which are almost parallel to each other. pass through the objective lens. The objective lens collects maximum amount of light as it is large in size. It forms a real, inverted and diminished image in the focal plane of the objective lens. Now, the position of the eyepiece is adjusted such that this image falls just within the focal length of the eyepiece and serves as the object for the eyepiece which works as a simple microscope.

The final image is highly magnified, virtual, on the same side as that of the object and inverted with respect to the original object. The final image can be observed by keeping the eye close to the eyepiece. If the image formed by the objective lens lies in the focal plane of the eyepiece, the final image is formed at infinity.

(c) We can get different magnifications by using the eyepiece with different focal lengths.

(d)

Question 89.
What is persistence of vision? Give one example of persistence of vision.
Answer:
Persistence of vision: We see an object when its image is formed on the retina. The image disappears when the object is removed from our sight. But this is not instantaneous and the image
remains imprinted on the retina for about \(\frac{1}{16}\) th of a second after the removal of the object.

The sensation on the retina persists for a while. This effect is known as the persistence of vision. It is due to persistence of vision that we continue to see the object in its position for about \(\frac{1}{16}\) th of a second after it is removed.
Example: When a burning stick of incense is moved fast in a circle, a circle of red light is seen.

Question 90.
Name two devices whose working is based on the phenomenon of persistence of vision.
(OR)
Name any two applications based on persistence of vision.
Answer:
The working of a television set and motion picture is based on the phenomenon of persistence of vision.
[Note: These are the examples of persistence of vision in daily life.

Question 91.
How is the phenomenon of persistence of vision used in motion pictures?
Answer:
In motion pictures, photographs of a moving object are taken at the rate of more than sixteen pictures per second. These photographs are projected on the screen at the same rate.
Each picture is slightly different from the other. As a result of persistence of vision, we get the impression of observing the object in continuous motion.

Question 92.
Name the two types of light sensitive cells present in the retina of the human eye. What are their functions?
Answer:
(1) The retina of the human eye contains a large number of light sensitive cells. These cells are of two shapes : (i) rods and (ii) cones.
(2) The rod-like cells respond to the intensity of light.
(3) The conical cells respond to various colours of light. They respond differently to red, green and blue colours. They do not respond to faint light.

Question 93.
When do you say that a person is colour blind?
Answer:
When a person is unable to distinguish between certain colours, he is said to be colour blind.
[Note: (i) Except for being colour-blind, their eyesight is normal, (ii) Rod-shaped cell ≡ rod-like cell, cone-shaped cell = Conical cell.]

Question 94.
Explain the perception of colour in the human eye.
(OR)
Explain in short perception of colour.
(OR)
Write a note on perception of colour.
Answer:
(1) In nature we firld objects of various colours. Perception of colour means to be able to respond to colour.
(2) We can distinguish between various colours due to perception of colour.
(3) The cone-shaped cells on the retina of the eye respond to the various colours when light is bright and communicate to the brain about the colours of the image formed on the retina. This gives us the proper idea about the colours of the object.
(4) If, in the retina of a person, the cone-shaped cells responding to certain specific colours are absent, the person is unable to distinguish between the colours. As a result, he lacks perception of colour.

Question 95.
What is colour-blindness?
Answer:
(1) The retina of the human eye contains a large number of light sensitive cells. These cells are of two shapes : (i) rods and (ii) cones.
(2) The cone-shaped cells respond to various colours of light when light is bright.
(3) Thus, the perception of colour is due to the presence of the cone-shaped cells in the retina.
(4) In the retina of some persons, cone-shaped cells responding to certain specific colours are absent. Hence, these persons are unable to distinguish between certain colours, i.e., they are colour-blind. This defect is known as colour¬blindness.

Question 96.
Why are some persons colour-blind? What is the cause of this defect?
Answer:
In the retina of some persons, cone-shaped cells responding to certain specific colours are absent. Hence, these persons are unable to distinguish between certain colours, i.e., they are colour-blind.

Question 97.
What are the difficulties faced by a colour-blind person?
Answer:
(1) A colour-blind person cannot distinguish between different colours. For example, he cannot distinguish between red and green colours. Also he cannot distinguish between blue and green colours. Red and green, both appear grey. Since a colour¬blind person cannot distinguish between red and green colours, it is difficult for him to cross a road. There is a possibility of an accident while crossing a road.

(2) A colour-blind person cannot distinguish between two objects of different colours, which are otherwise identical, e.g., clothes.

(3) A colour-blind person may have an inferiority complex and hence may find it difficult to mix with other persons.

Give scientific reasons:

Question 1.
A convex lens is known as a converging lens.
Answer:
When rays of light parallel to the principal axis of a convex lens pass through the lens, they converge to a point on the principal axis. Hence, a convex lens is known as a converging lens.

Question 2.
A concave lens is called a diverging lens.
Answer:
When rays of light parallel to the principal axis of a concave lens pass through the lens, they appear to diverge from a point on the principal axis. Hence, a concave lens is called a diverging lens.

Question 3.
In old age, a bifocal lens is necessary for some persons.
Answer:
(1) Some people, in old age, suffer from myopia (nearsightedness) as well as hypermetropia (farsightedness).
(2) Myopia is corrected using a concave lens of appropriate power. Hypermetropia is corrected using a convex lens of appropriate power. Therefore, they need a bifocal lens.

Question 4.
A person suffering from myopia (nearsightedness) uses spectacles of concave lenses.
Answer:
(1) A person suffering from myopia can see nearby objects clearly as the image of a nearby object is formed on the retina, but cannot see distant objects clearly as the image of a distant object is formed in front of the retina instead of on the retina.

(2) A concave lens diverges the rays of light passing through it. When spectacles of concave lenses of appropriate power are used, the parallel rays coming from a distant object are diverged to proper extent before they are incident on the eye lens. Therefore, after the converging action of the eye lens, the image of a distant object is formed on the retina of the eye and hence the distant object can be seen clearly.

Question 5.
A person suffering from hypermetropia (farsightedness) uses spectacles of convex lenses.
Answer:
(1) A person suffering from hypermetropia can see distant objects clearly as the image of a distant object is formed on the retina, but cannot see nearby objects clearly as the image of a nearby object would be formed behind the retina instead of on the retina.

(2) A convex lens converges the rays of light passing through it. When spectacles of convex lenses of appropriate power are used, the rays of light coming from a nearby object are converged to proper extent before they are incident on the eye lens. Therefore after the converging action of the j eye lens, the image of a nearby object is formed on j the retina of the eye and hence the nearby object | can be seen clearly.

Question 6.
You cannot enjoy watching a movie from a very short distance from the screen in a cinema hall.
Answer:
(1) The less the distance between the screen in a cinema hall and the person watching the movie, the more is the intensity of light falling on the eye.
(2) This results in great contraction of the pupil of the eye causing a strain. Hence, you cannot enjoy watching a movie from a very short distance from the screen in a cinema hall.

Question 7.
The rays of light travelling through the optical centre of a lens pass without changing their path.
Answer:
The portion of a lens near the optical centre is like a very thin slab of glass. Hence, the rays of light travelling through the optical centre of a lens pass without changing their path.

Question 8.
A convex lens converges the rays of light falling on it.
Answer:

• A convex lens can be regarded as made of a very large number of portions of triangular prisms. The bases of these prisms are towards the central thicker portion of the lens.
• A ray of light passing through a prism bends towards its base. Hence, a convex lens converges the rays falling on it.

Question 9.
A concave lens diverges the rays of light falling on it.
Answer:

• A concave lens can be regarded as made of a very large number of portions of triangular prisms. The bases of these prisms are towards the edges of the less, i.e, away from the central thinner portion of the lens.
• A ray of light passing through a prism bends towards its base. Hence, a concave lens diverges the rays of light falling on it.

Question 10.
When a burning stick of incense is moved fast in a circle, a circle of red light is seen.
Answer:
The impression of the image on the retina lasts for about \(\frac{1}{16}\) th of a second after the removal of the object. If a burning stick of incense is moved at a rate of more than sixteen revolutions per second, we see a circle of red light due to persistence of vision.

Question 11.
Colour-blind persons are unable to distinguish between different colours.
Answer:
(1) The cone-shaped cells in the retina of a person respond to colours. This makes the perception of colours possible.
(2) In the retina of colour-blind persons, cone-shaped cells responding to certain specific colours are absent. Hence, they are unable to distinguish between different colours.

Question 12.
It is risky to issue a driving license to a person suffering from colour-blindness.
Answer:
A colour-blind person cannot distinguish between different colours. If a driver is colour-blind, he will not be able to distinguish between the colours of the signal and the colours on different sign boards. This will lead to an accident. Hence, it is risky to issue a driving license to a person suffering from colour-blindness.

Distinguish the following:

Question 1.
Real image and Virtual image.
Answer:
Real image:

1. A real image is formed when the light rays starting from an object meet after reflection or refraction.
2. It can be projected on a screen.
3. It is inverted with respect to the object.

Virtual image:

1. A virtual image is formed when the light rays starting from an object (when extended backward) appear to meet after reflection or refraction.
2. It cannot be projected on a screen.
3. It is erect with respect to the object.

Question 2.
Simple microscope and Compound microscope.
Answer:
Simple microscope:

1. In a simple microscope, only one convex lens is used.
2. In this case, the object is placed within the focal length of the convex lens.
3. Its magnifying power is much less than that of a compound microscope.
4. It is used to observe minute parts of a watch, to read words in small print, etc.

Compound microscope:

1. In a compound microscope, two convex lenses, objective and eyepiece, are used.
2. In this case, the object is placed beyond the focal length of the objective lens.
3. Its magnifying power is much greater than that of a simple microscope.
4. It is used to observe blood corpuscles, plant and animal cells, etc.

Question 3.
Compound microscope and Astronomical refracting telescope.
Answer:
Compound microscope:

1. In a compound microscope, the focal length and cross section of the objective lens are respectively smaller than the focal length and cross section of the eyepiece.
2. In this case, to observe the object, the distance between the object and the objective lens is adjusted.
3. It forms a magnified image of a small object.
4. It is used to observe blood corpuscles, plant and animal cells, etc.

Astronomical refracting telescope:

1. In an astronomical refracting telescope, the focal length and cross section of the objective lens are respectively greater than the focal length and cross section of the eyepiece.
2. In this case, to observe the object, the distance between the objective lens and eyepiece is adjusted.
3. It forms a near image of a distant object.
4. It is used to observe sateulites, planets, stars, etc.

Question 4.
Simple microscope and Astronomical refracting telescope.
Answer:
Simple microscope:

1. In a simple microscope, only one convex lens is used.
2. In this case, the object is placed within the focal length of the convex lens.
3. In this case, the image is erect.
4. It is used to observe minute parts of a watch, to read words in small print, etc.

Astronomical refracting telescope:

1. In an astronomical refracting telescope, two convex lenses, objective lens and eyepiece are used.
2. In this case, the object is far away from the objective lens.
3. In this case, the image is inverted.
4. It is used to observe satellites, planets, stars, etc.

Read the following paragraph and answer the questions given below it:

Construction of a compound microscope:
(1) A compound microscope consists of a metal tube fitted with two convex lenses at the two ends. These lenses are called the objective lens (the lens directed towards the object) and the eyepiece (the lens directed towards the eye). Both the lenses are small in size, but the cross section of the objective lens is less than that of the eyepiece. The objective lens has a short focal length. The focal length of the eyepiece is more than that of the objective lens.(2) The metal tube is mounted on a stand. The principal axes of the objective lens and the eyepiece are along the same line. The distance between the object and the objective lens can be changed with a screw. It is possible to change the distance between the objective lens and the eyepiece.
Working:
(1) The object to be observed is illuminated and placed in front of the objective lens, slightly beyond the focal length of the objective lens. Its real, inverted and enlarged image is formed by the, objective lens on the other side.
(2) This intermediate image lies within the focal length of the eyepiece. It serves as an object for the eyepiece. The eyepiece works as a simple microscope. The final image is virtual, highly enlarged and inverted with respect to the original object. It can be formed at the minimum distance of distinct vision from the eyepiece. The final image is observed by keeping the eye close to the eyepiece.
Use: This microscope is used to observe blood cells, microorganisms, etc.

Question 1.
In a compound microscope, which lens has greater focal length?
Answer:
In a compound microscope, the eyepiece has greater focal length.

Question 2.
Where do you place the object to be observed with a compound microscope?
Answer:
In a compound microscope, the object to be observed is placed in front of the objective lens, slightly beyond the focus of the objective lens.

Question 3.
State which distance is adjusted to observe the object with a compound microscope.
Answer:
To observe the object with a compound microscope, the distance between the object and objective lens is adjusted.

Question 4.
State the nature of the final image in compound microscope relative to the object.
Answer:
In a compound microscope, the final image is highly enlarged, inverted and virtual relative to the object.

Question 5.
State the use of a compound microscope.
Answer:
A compound microscope is used to observe blood cells, microorganisms, etc.

Fill in the blanks for a convex lens:

Question 1.

f (m)	0.2	—————–	0.1
P (D)	—————	2	——————

Answer:
[P (D) = \(\frac{1}{f(\mathrm{m})}\)]

f (m)	0.2	0.5	0.1
P (D)	5	2	10

Question 2.

h1 (cm)	—————-	5	10
h2 (cm)	-30	-20	—————-
M	-2	—————–	-0.5

Answer:
[M = \(\frac{h_{2}}{h_{1}}\)]

h1 (cm)	15	5	10
h2 (cm)	-30	-20	-5
M	-2	-4	-0.5

Solve the following numerical problems:

Problem 1.
An object Is kept at 60 cm in front of a convex lens. Its real image is formed at 20 cm from the lens. Find the focal length or the lens.
Solution:
Data: Convex lens, u = -60 cm,

The focal length of the lens = 15 cm.

Problem 2.
The focal length of a convex lens is 20 cm. If an object of height 2 cm is placed at 30 cm from the lens, find (i) the position and nature of the Image (ii) the height of the image (iii) the magnification produced by the lens.
Solution:
Data: Convex lens, f = 20 cm,
u = -30 cm, h₁ = 2 cm, v = ?, h₂ = ?, M = ?

The image will be formed at 60 cm from the lens and on the other side of the lens with respect to the object. It is a real image.

h₂ is negative. This shows that the image is inverted.
The height of the image -4 cm.
(iii) M = \(\frac{h_{2}}{h_{1}}=\frac{-4 \mathrm{cm}}{2 \mathrm{cm}}\) = -2
M is negative, indicating that the image is inverted.
The magnification produced by the lens = -2.

Problem 3.
When a pm of height 3 cm is fixed at 10 cm from a convex lens, the height or the virtual image formed is 12 cm. Find the focal length of the lens.
Solution:
Data: Convex lens, h₁ =3 cm,
h₂ = 12 cm (virtual image), u = -10 cm, f = ?

The focal length of the lens = 13.33 cm [approximately]

Problem 4.
At what distance from a convex lens of focal length 2.5 m should a boy stand so that his image is half his height?
Solution:
Data: Convex lens, f= 2.5 m,
M= –\(\frac{1}{2}\), u=?

∴ u = -3f = -3 × 2.5 m = -7.5 m
This is the object distance.
The boy should stand at 7.5 m from the convex lens so that his image is half his height.

Problem 5.
A convex lens forms a real image or a pencil at a distance of 40 cm from the lens. The image formed is of the same size as the object. Find the focal length and power of the lens. At what distance is the pencil placed from the lens?
Solution:
Data: Convex lens, v = 40 cm,
M = -1, f = ?, h₁ = ?, u = ?
M = = -1 = \(\frac{v}{u}\)
∴ u = -v = -40 cm (object distance)
The pencil is placed at 40 cm from the convex lens.

The focal length of the lens 20 cm.

The power of the lens = 5 D.

Problem 6.
A spherical lens is used to obtain an image on a screen. The size of the image is four times the size of the object. What is the type of lens and at what distance is the screen placed from the lens?
Solution:
Data: M = -4, type of lens? v = ?
As the image formed by the lens is obtained on a screen, it is a real image. The lens is, therefore, a convex 1ens.

The distance of the screen from the lens = 5f.

Problem 7.
An object of height 5 cm Is held 20 cm away from a converging lens of focal length 10 cm. Find the position, nature and size of the image formed.
Solution:
Data: Converging lens (convex lens),
f = 10 cm, h₁ = 5 cm, u = -20 cm, v = ?, h₂ = ?

The image is real and inverted. it is formed at 20 cm from the lens and on the other side of the lens relative to the object.

The height of the image, h₂ = -5 cm
Thus, it is numerically the same as the height of the object.

Problem 8.
An object is placed at 10 cm from a convex lens of focal length 12 m. Find the position and nature of the image.
Solution:
Data: Convex lens, u = -10 cm,
f = 12 cm, v = ?


∴ v = -60 cm
It is negative.
The image is formed at 60 cm from the lens and on the same side of the lens relative to the object. It is virtual, erect, and enlarged.

Problem 9.
An object of height 4 cm is placed in front of a concave lens of focal length 40 cm. If the object distance is 60 cm, find the position and height of the image.
Solution:
Data: f = -40 cm (concave lens),
u = -60 cm, h₁ = 4 cm, v = ?, h₂ = ?

The image Is formed at 24 cm from the lens.
It is on the same side as the object.

The height of the image is 1.6 cm.

Problem 10.
What is the power of a convex lens having focal length 0.5 m?
Solution:
Data: Convex lens, f= 0.5 m, P = ?
P = \(\frac{1}{f}=\frac{1}{0.5 \mathrm{m}}\) = 2D
The power of the lens = 2D.

Problem 11.
The power of a convex lens is 2.5 dioptres. Find its focal length.
(OR)
Calculate the focal length of a corrective lens having power +2.5 D.
Solution:
Data: Convex lens, P = +2.5 D, f = ?
P = \(\frac{1}{f}\)
∴ 2.5 D = \(\frac{1}{f}\)
∴ f = \(\frac{1}{2.5 \mathrm{D}}\) = 0.4 cm = 40 cm
The focal length of the lens = 40 cm.

Problem 12.
Two convex lenses of focal length 20 cm each are kept in contact with each other. Find the power of their combination.
Solution:
Data: f₁ = 20 cm = 0.2 m,
f₂ =20 cm = 0.2 m, P (combination) = ?

∴ Focal length of the combination or the lenses, f = 0.1 m.
P = \(\frac{1}{f}=\frac{1}{0.1 \mathrm{m}}\) = 10 D
The power of the combination of the lenses, P = 10 D.

Problem 13.
Two convex lenses of equal focal lengths are kept in contact with each other. If the power of their combination is 20 D, find the focal length of each convex lens.
Solution:
Data: Convex lens, P = 20 D, f₁ = f₂ = ?
The focal length (f) of the combination of the lenses is given by

This gives the focal length or each convex lens.

Problem 14.
If a convex lens of focal length 10 cm and a concave lens of focal length 50 cm are kept in contact with each other, (i) what will be the focal length of the combination? (ii) what wiil be the power of the combination? (iii) what will be the behaviour of the combination (behaviour as a convex lens/concave lens)?
Solution:
Data: f₁ = +10 cm = +0.1 m (convex lens),
f₂ = -50 cm = -0.5 m (concave lens),
f (combination) = ?, P (combination) = ?

The focal length or the combination of the lenses = 0.125 m = 12.5 cm.

The power of the combination of the lenses 8.D.
(iii) The focal length of the combination of the lenses is positive. This shows that the combination will behave as a convex lens.

Numerical problems for practice:

Problem 1.
Find the focal length of a convex lens which produces a real image at 60 cm from the lens when an object is placed at 40 cm in front of the lens.
Answer:
24 cm

Problem 2.
Find the focal length of a convex lens which produces a virtual image at 10 cm from the lens when an object is placed at 5 cm from the lens.
Answer:
10 cm

Problem 3.
A real image is obtained at 30 cm from a convex lens of focal length 7.5 cm. Find the distance of the object from the lens.
Answer:
u = -10 cm

Problem 4.
An object is kept at 20 cm in front of a convex lens and its real image is formed at 60 cm from the lens. Find (1) the focal length of the lens (2) the height or the image if the height of the object is 6 cm.
Answer:
(1) 15 cm
(2) h₂ = -18 cm]

Problem 5.
An object is kept at 10 cm in front of a convex lens. Its image is formed on the screen at 15 cm from the lens. Calculate (1) the focal length of the lens (2) the magnification produced by the lens.
Answer:
(1) 6 cm
(2)M = -1.5

Problem 6.
An object is kept at 60 cm in front of a convex lens of focal length 15 cm. Find the image distance and the nature of the image. Also find the magnification produced by the lens.
Answer:
v = 20 cm. The image is real, inverted and smaller than the object. M = –\(\frac{1}{3}\)]

Problem 7.
An object of height 2 cm is kept at 30 cm from a convex lens. Its real image is formed at 60 cm from the lens. Find the focal length and power of the lens.
Answer:
f = 20 cm, P = 5 D

Problem 8.
If the power of a lens is 4 dioptres, find its focal length.
Answer:
25 cm

Problem 9.
Find the power of a convex lens of focal length 40 cm.
Answer:
2.5 D

Problem 10.
Find the power of a convex lens of focal length 12.5 cm.
Answer:
8 D

Problem 11.
If for a lens, f = – 20 cm, what is the power of the lens?
Answer:
-5 D

Problem 12.
An object of height 4 cm is kept in front of a concave lens of focal length 20 cm. If the object distance is 30 cm, find the position and the height of the image.
Answer:
v = -12 cm, h₂ = 1.6 cm

Problem 13.
If two convex lenses of focal lengths 10 cm and 5 cm are kept in contact with each other, what is their combined focal length?
Answer:
\(\frac{10}{3}\) cm [approximately 3.33 cm]

Problem 14.
If a convex lens of focal length 20 cm and a concave lens of focal length 30 cm are kept in contact with each other, (i) What will be the focal length of the combination? (ii) What will be the power of the combination? (iii) What will be the behaviour of the combination?
Answer:
(i) f = 60 cm
(ii) P = \(\frac{5}{3}\) D = 1.6667 D (approximately)
(iii) The combination will behave as a convex lens.

Problem 15.
A concave lens of focal length 12 cm and a convex lens of focal length 20 cm are kept in contact with each other, (i) Find the focal length of the combination, (ii) What will be the behaviour of the combination?
Answer:
(i) f = -30 cm
(ii) The combination will behave as a concave lens.

Balbharti Maharashtra State Board Class 10 Political Science Solutions Chapter 2 The Electoral Process Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Political Science Solutions Chapter 2 The Electoral Process

Question 1.
Choose the correct option from the given options and complete the sentences.
(1) The Election Commissioner is appointed by the …………………………. .
(a) President
(b) Prime Minister
(c) Speaker of Loksabha
(d) Vice President
Answer:
(a) President

(2) …………………………. was appointed as the first Chief Election Commissioner of independent India.
(a) Dr. Rajendra Prasad
(b) T.N. Sheshan
(c) Sukumar Sen
(d) Neela Satyanarayan
Answer:
(c) Sukumar Sen

(3) Constituencies are created by …………………………. committee of the Election Commission.
(a) Selection
(b) Delimitation
(c) Voting
(d) Timetable
Answer:
(b) Delimitation

Question 2.
State whether the following statements are true or false. Give reasons for your answer.
(1) The Elections Commission lays down the code of conduct during elections.
Answer:
The above statement is True. Reasons:

• It ensures free and fair elections.
• Maipractices during the election come under control.
• Due to the strict observance of the code of conduct in the last few’ elections, the common voters have become confident.

(2) Under special circumstances the Election Commission holds re-elections in a particular constituency for a second time.
Answer:
The above statement is True. Reasons :

• Sometimes, the representative of Lok Sabha, Vidhan Sabha or the local self governmènt resigns from his/her constituençy.
• In some cases, death of the representative occurs.
• In such special situations, the Election Commission has to conduct an election for a second time. It is called By-elections.

(3) The state government decides as to when and in how many stages the elections would be held in a particular State.
Answer:
The above statement is False. Reasons :

• The entire process of conducting elections is entrusted upon and managed by the Election Commission.
• If this responsibility is given to the state government it may adopt a biased approach.
• Hence, the Constitution has formed the Election Commission an independent body to carry out the responsibility.

Therefore, it is decided by the Election Commission as to when and in how many stages it will conduct elections.

Question 3.
Explain the concept.
(1) Reorganising the constituencies
Answer:
(1) The Election Commission of India formed constituencies for Lok Sabha and Legislative Assembly.
(2) The Election Commission had decided upon the constituencies before the first election. As the years passed, there was a lot of migration of the people for business and other activities from the villages to cities.
(3) This changed the demography to large extent. Number of voters in some constituencies reduced while in some it increased to a very great extent. This disturbed the ratio of- seats allotted as compared to population in those constituencies.
(4) Hence, the need to readjust the constituencies arose. The Delimitation Commission of the election commission does the work of reorganising or restructuring of constituencies.

(2) Midterm Elections

Question 4.
Complete the following picture.

Answer:


Answer:

Answer:

Answer:

Question 5.
Answer in brief.
(1) Explain the functions of the Election Commission.
Answer:
The functions of the Election Commissipn are:
(1) Prepare the voters’ list.
(2) Decide election timetable and decide the entire process of holding elections.
(3) Scrutinize the applications of the candidates.
(4) Conduct free and fair elections and do all the work related to it.
(5) Give recognition and also de-recognize political parties.
(6) Resolve all the disputes and complaints regarding elections.

(2) Write some additional information about post of the Election Commissioner.
Answer:
(1) The Election Commission in India has one Chief Election Commissioner and two other Chief Commissioners.
(2) All the commissioners are appointed by the President.
(3) The Chief Election Commissioner of India is usually a member of the Indian Civil Service or. Indian Administrative Service.
(4) The responsibility of conducting free and fair elections to the Parliament and State Legislatures lies with the Election Commissioner.
(5) In order do safeguard the independence of the Election Commissioner, he cannot be easily removed from the post for any political reasons.

(3) Explain the meaning of Code of Conduct.
Answer:
(1) After the announcement of elections till the declaration of results, the Election Commission enforces the Code of Conduct.
(2) It explains the rules to be followed by the government, political parties candidates and voters before and during elections.
(3) Code of conduct is adopted to control malpractices during elections. It ensure free and fair ecections.

Project
Organise a mock poll in the school to understand the process of voting.
Answer:

Memory Map

Question 6.
Choose the correct option from the given options and complete the sentences:
(a) Article of Indian Constitution created the independent body of Election Commisšion.
(a) 351
(b) 370
(c) 324
(d) 301
Answer:
(c) 324

(b) system exists in India.
(a) Single-party
(b) Two-party
(c) Multi-party
(d) No-party
Answer:
(c) Multi-party.

(c) The right to give recognition or de-recognize a political party lies with ……………….. .
(a) President
(b) Election Commission
(c) Parliament
(d) Vice-President
Answer:
(b) Election Commission

(d) There are constituencies of Lok Sabha at present.
(a) 288
(b) 350
(c) 500
(d) 543
Answer:
(d) 543

(e) from the present state of Himachal Pradesh was the first voter.
(a) Sukumar Sen
(b) Sham Sharan Negi
(c) Prem Kumar Ghumal
(d) P N. Chadda
Answer:
(b) Sham Sharan Negi

(f) Due to EVM, people can also vote easily.
(a) elder
(b) salaried
(c) Divyanga
(d) Transgender
Answer:
(c) Divyanga

(g) The first elections in India were held in
(a) 1948-49
(b) 1949-50
(c) 1950-51
(d) 1951-52
Answer:
(d) 1951-52.

Question 7.
State whether following statements are True or False. Give reasons for your answer :
(a) There should be secrecy in Election process.
Answer:
The above statement is False. Reasons :

• Election should be conducted in a free and fair environment.
• If the elections are not held in free environment then there are chances of malpractices and corruption.
• Then, it will be impossible to elect the honest and efficient candidates.

(b) The Election Commission has started awareness campaign for registration of voters.
Answer:
The above statement is True. Reasons :

• The responsibility of preparing and updating electoral roll lies with the Election Commission.
• The Election Commission starts an awareness campaign to create awareness among new eligible voters so that they register themselves in the voter’s list.
• The Indian voter is not -much aware about the election process.
• Special voter’s awareness campaign is run for voter’s registration.
• For their awareness National Voter’s Day is celebrated every year.

(c) Every candidate who fills the nomination form can contest election.
Answer:
The above statement is False. Reasons :

• Every candidate of a party or independent candidate has to be personally present to fill the nomination form.
• It is necessary for him or her to give complete information in the nomination form as decided by the Election Commission.
• The nomination forms are then scrutinized. If there are irregularities in a nomination paper and if the information is found to be false the nomination forms are rejected.

Therefore, it is not possible for every candidate who fills the nomination form to contest election.

(d) Sometimes, the Election Commission has to conduct mid-term elections. OR Explain the concept : Mid-Term Elections.
Answer:
The above statement is True. Reasons :

• If the elected government in power loses its majority before completing its term.
• If no party gets complete majority, then two or more parties come together and form a coalition government.
• Such coalition government collapses if any party withdraws the given support.
• In such situations, the government is left with no option other than resigning.
• If there is no alternative available to form government then the Parliament or Vidhan Sabha is dissolved before completing its term. In such a scenario, the Election Commission has to conduct mid-term elections.

Question 8.
Explain the following concepts :

(a) What is representation?
Answer:
Modern democracy is a representative democracy. In a democracy it is not possible to involve the entire population in the ecision-making process. This resulted in the starting of the practice of electing some people on behalf of entire population
as representatives who would run the government. The representatives who form the government are expected to be responsible to the people and give preference to the welfare of the people.

• Direct and Indirect or representative democracy rire two types of democracy.
• In modem nation-states; the population has increased to a great èxtent.
• So it is impossible to involve all the people in decision-making process.
• Thus, th practice of electing some people on behalf of entire population as representatives started.
• The elected representatives form government and work for the welfare of the people.

(b) Election Commission :
Answer:
In India, the Election Commission is central to the process of elections. Art. 324 of the Indian Constitution has established this autonomous body which consists of one Chief Election Commissioner and two other commissioners.

1. One of the most important features of a democratic nation is elections at regular intervals. Holding free and fair elections at regular intervals is essential for a democratic system.
2. Under the Article 324 of the Constitution, Election Commission was formed in 1950. The President appoints one Chief Election Commissioner and two additional commissioners. It is an autonomous body.
3. The rank and powers of all the three commissioners are the same. The declaration of dates of the elections to the announcement of the results the entire procedure is monitored by the Election Commission.
4. The Election Commission does not have its own staff to carry out this procedure. So they carry out the work with help of government employees and teachers. Special provisions are made for all finances incurred by the Election Commission.

Question 9.
Write short notes :
(a) Journey from Ballot box to EVM machine :
Answer:

1. From the first election in 1951-52 till 1999, elections were held using ballot box. Twenty lakh ballot boxes were used in the first election. Voters used to cast his or her vote by stamping in front of the candidate’s name and put them in the metal boxes.
2. Electronic Voting Machines (EVMs) were first used for 5 seats in Rajasthan, 5 seats in Madhya Pradesh and 6 seats in New Delhi 1998 in Legislative Assembly.
3. EVM machines were used at all polling booths in the general elections held in 2004. It proved to be a very useful device.
4. It has been improvised since its first use. Due to the use of EVMs the results are declared early and at a very fast rate.

(b) Recognition to Political Parties :
Answer:

• India has a multi-party system with recognition accorded to national, state and regional level parties by the Election Commission.
• Their recognition depends on the voting percentage received by them in the assembly elections and number of elected representatives of their party.
• If any party does not fulfill these criteria, its recognition is cancelled.
• The Election Commission allots appropriate symbols to parties and independent candidates. All political parties should have recognition of the Election Commission.

Question 10.
Complete the concept map :

(a) Prepare a flow chart on the process of election.
Answer:

(b) Which two conditions among following is the violation of code of conduct?
(1) The candidate distributes items of household use. –
(2) Promises made to resolve the water problem if elected.
(3) To go from door to door to meet voters and request them to vote.
(4) To appeal on the basis of caste and religion to get support.
Answer:
(1) The candidate distributes items of household use.
(2) To appeal on the basis of caste and religion to get support.

Question 11.
Answer in brief :
(a) Why is it important to conduct elections?
Answer:
It is important to conduct elections because of the following reasons :

• The existence and working of democracy depends on elections.
• All political parties get a chance to rule.
• Elections help to bring a change in power through peaceful meAnswer:
• It not only changes government policies but also society.

(b) What are the conditions for voting?
Answer:
The following are the conditions for voting:

• The person should be a citizen of India.
• He should have completed 18 years of age.
• His name should appear in voters’ list.
• The person should have photo identity card issued by the Election Commission of India.

(c) What action is taken by the Election Commission if disputes arise regarding elections?
Answer:

• If any disputes arise regarding the elections, the Election Commission is empowered to take final decisions.
• The Election Commission conducts a thorough inquiry about the said dispute.
• If there is evidence of any malpractices during elections, in any constituency, it declares the elections invalid and announces re-polls.
• If any candidate breaks the code of conduct and contests elections, he/she is barred by the Election Commission from contesting elections.

(d) What challenges are faced by the Election Commission to conduct free and fair elections?
Answer:
The following challenges are faced by the Election Commission tcx conduct free and fair elections :

• Managing the large geographical landscape and huge electoral population.
• To stop misuse of money and muscle power during elections.
• Barring candidates with criminal background from contesting elections.
• Conducting elections successfully in politically criminalised environment.
• Conducting elections in spite of increasing instances of violence and making them a success.

(e) What are the advantages of EVM machines?
Answer:
The battery operated Electronic Voting Machine (EVM) has more advantages than the ballot box. They are as follows :

• It saves tonnes of paper used to make ballot paper.
• So, it conserves the environment as it stops the reckless cutting of trees required to make paper.
• If the voter does not wish to cast his vote in favour of any candidate contesting, he can make use of NOTA (None Of The Above).
• It makes counting of the votes much faster which enables the election officer to declare result in a short time.
• It is helpful for disabled (Divyanga) people to cast vote.

(f) Explain the features of procedures of voting during the first Lok Sabha Election.
Answer:

• It was a challenge to prepare voters’ list at the time of the first election. Illiteracy rate was very high in our country. Therefore, the procedure to vote and making the voter list was a challenge.
• 20 lac steel boxes were made and election symbols of political parties were stuck on it.
• Blank ballot papers were given to the voters and they were supposed to drop in the box having the election symbols of the party they decide to vote for.
• Even the illiterate people could vote because of this system.

Question 12.
Give your opinion :
(a) When candidates have only the condition of age, why should they give other information to Election Commission? Answer:

• While filling the form candidates should reveal information about his property assets and if there are any criminal charges against him.
• When candidates have only the condition of age as eligibility, why should they give other information to election commission?
• Why are the candidates required to give the information of their property to Election Commission?
• Such candidates if elected can misuse power and amass wealth with corrupt practices.
• With criminal background they can even threaten voters to vote for him.
• His nomination could get cancelled based on the information.

(b) Why is it so?
(A) Some constituencies are reserved for scheduled castes and scheduled tribes.
Answer:

• It is difficult for the people of scheduled castes and scheduled tribes to get representation as they are scattered in different parts.
• Without a representative it is difficult to discuss their problems in Parliament.
• Lack of representative will hinder their progress. Hence some constituencies are reserved for scheduled castes and scheduled tribes.
• Some constituencies are kept reserved for Scheduled caste and Scheduled tribes.
• Every political party has an election symbol.
• At the time of voting and counting of votes, the official representatives of political parties remain present.
• Recognised parties have equal opportunity to present their side before media such as television and radio.

(B) Why every political party has an election symbol?
Answer:

• After independence, the literacy rate was quite low in India.
• It was not possible for the voters to read the name of the candidate and vote.
• Therefore, the Election Commission gave symbols to political parties and independent candidates which helped the voters to identify and decide whom to vote for.

(C) At the time of voting and counting of votes, the official representatives of political parties remain present.
Answer:

• There are incidences of duplicate voters who register in multiple constituencies.
• There are cases of rigging of EVM or booth capturing.
• Such incidences are brought to light by representatives who are present at polling centres.
• During the counting process, if the EVM machine looks tampered, the representative can raise an objection.

(D) All recognised parties should get an equal opportunity to express their opinion on media such as television and radio.
Answer:

• All political parties should get a fair chance to express their agenda.
• Their ideas and philosophy should reach the people.
• Television and radio are owned by the government.
• Political parties have equal right on both.
• Hence, all the recognised parties can express their opinion on Doordarshan and Radio.

(c) Think!
(A) How political parties suffer due to family monopoly in the party? OR What are the disadvantages of dynasty rule?
Answer:

1. If only one family has domination on the political party because of dynasty rule then others are not given leadership opportunity.
2. It is impossible to have all the members of the family efficient. An inefficient heir can cause damage to the party.
3. The growth and expansion of party comes to a halt because of such heir. His faults seep into the party making it weak in the long term.
4. The nature of such a party become dictatorial. Opposing views are suppressed and the internal democracy in the party vanishes.
5. If the heir does not have progressive thoughts then the party becomes regressive and of obsolete ideology.
6. How political parties suffer due to family monopoly in the party?
7. What do you understand by the system of ‘one vote one value’?

(B) What do you understand by the system ‘One Vote One Value’?
Answer:

• There is great importance in political and social equality in democracy.
• According to this ideology, ‘One Vote One Value’ is very important.
• In a democracy, each vote has the same value. The value of the vote of a Prime Minister and a common man is same.
• Under military rule or dictatorship or during monarchy the value of a vote for privileged classes was more. There was no importance given to the vote of the common man.
• ‘One Man One Vote’ indicates all the people in the country have same status. This is the gift of democracy.

(d) Voting is our duty as well as responsibility to vote.
Answer:

• It is enshrined in the fundamental principles of our Constitution to vote.
• It is not only our duty but also responsibility.
• Democracy exists because of elections. People should elect honest and efficient representatives through election.
• If voters show no interest in voting then the government will ignore people’s welfare.
• Hence I feel it is not only the duty of every citizen to vote but also his responsibility.
• Government has to observe the code of conduct declared by the Election commission.

(e) What measures should be taken to increase the credibility of elections?
Answer:
To increase the credibility of elections the following measures should be taken :

• 50% seats should be reserved for women candidates by every party.
• Candidates with a criminal background should be permanently barred from contesting any elections.
• The misuse of money should be stopped during elections. The government should incur the expenditure.
• Candidates who resort to malpractices should be immediately booked. A strict inquiry and action should be taken against them by the court.
• Laws and regulations should be followed strictly by the political parties before giving election tickets.
• If the political parties do not co-operate with the above terms, the Election Commission should cancel their recognition.

(f) Which rules would you include in Code of Conduct for voters?
Answer:
The following rules should be included in Code of Conduct for voters :

• The voters who abstain from voting should be fined and government should suspend all the facilities given to them.
• If it is proved that the voter has accepted money or any kind of gifts, he should be punished.
• The action of voters should not instigate common people.
• They should not involve in bogus voting.
• They should not resort to illegal means for voting.
• The candidate distributes items of household use.
• Promise made to resolve the water problem if elected.
• To go from door to door to meet voters and request them to vote.
• To appeal on the basis of caste and religion to get support.

 

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light

Question 1.
Fill in the blanks and explain the completed statements:
a. Refractive index depends on the………….of light.
Answer:
Refractive index depends on the velocity of light.
It is an experimental fact. (There is no question of explanation.)

b. The change in…………of light rays while going from one medium to another is called refraction.
Answer:
The change in the direction of propagation of light rays while going from one medium to another is called refraction. This is definition of refraction. It is assumed that the ray of light passes obliquely from one medium to another. (There is no question of explanation.)

Question 2.
Prove the following statements:
a. If the angle of incidence and angle of emergence of a light ray falling on a glass slab are i and e respectively, prove that i = e. (Practice Activity Sheet – 4)
Answer:
In the following figure, SR || PQ and NM is the refracted ray. Hence, r = i₁.
Now _gn_a = sin i/sin r and _an_g = sin i₁/ sin e.

Also gna = \(\frac{1}{{ }_{\mathrm{a}} n_{\mathrm{g}}}\)
∴ \(\frac{\sin i}{\sin r}=\frac{\sin e}{\sin i_{1}}\)
As r = i₁, it follows that sin i = sin e
∴ i = e.

b. A rainbow is the combined effect (an exhibition) of the refraction, dispersion, and total internal reflection of light (taken together). (Practice Activity Sheet – 1)
(OR)
With a neat labelled diagram, explain how the formation of rainbow occurs.
Answer:
(1) The formation of a rainbow in the sky is a combined result of refraction, dispersion, internal reflection and again refraction of sunlight by water droplets present in the atmosphere after it has rained.

Here, for simplicity only violet and red colours are shown. The remaining five colours lie between these two.

(2) The sunlight is a mixture of seven colours: violet, indigo, blue, green, yellow, orange and red. After it has stopped raining, the atmosphere contains a large number of water droplets. When sunlight is incident on a water droplet, there is (i) refraction and dispersion of light as it passes from air to water (ii) internal reflection of light inside the droplet and (iii) refraction of light as it passes from water to air.

(3) The refractive index of water is different for different colours, being maximum for violet and minimum for red. Hence, there is dispersion of light (separation into different colours) as it passes from air to water. [ See above Figure for reference.]

(4) The combined action of different water droplets, acting like tiny prisms, is to produce a rainbow with red colour at the outer side and violet colour at the inner side. The remaining five colours lie between these two.
The rainbow is seen when the sun is behind the observer and water droplets in the front.

Question 3.
Mark the correct answer in the following questions :
A. What is the reason for the twinkling of stars?
(i) Explosions occurring in stars from time to time
(ii) Absorption of light in the earth’s atmosphere
(iii) Motion of stars
(iv) Changing refractive index of the atmospheric gases
Answer:
Changing refractive index of the atmospheric gases.

B. We can see the Sun even when it is little below the horizon because of
(i) reflection of light
(ii) refraction of light
(iii) dispersion of light
(iv) absorption of light
Answer:
refraction of light

C. If the refractive index of glass with respect to air is 3/2, what is the refractive index of air with respect to glass?
(i) \(\frac{1}{2}\)
(ii) 3
(iii) \(\frac{1}{3}\)
(iv) \(\frac{2}{3}\)
Answer:
(iv) \(\frac{2}{3}\)

Question 4.
Solve the following examples:
a. If the speed of light in a medium is 1.5 × 10⁸ m/s, what is the absolute refractive index of the medium? (Practice Activity Sheet – 1 and 4)
Solution:
Data: v = 1.5 × 10⁸ m/s,
c = 3 × 10⁸ m/s, n = ?
n = \(\frac{c}{v}=\frac{3 \times 10^{8} \mathrm{m} / \mathrm{s}}{1.5 \times 10^{8} \mathrm{m} / \mathrm{s}}\) = 2
This is the absolute refractive index of the medium.

b. If the absolute refractive indices of glass and water are \(\frac{3}{2}\) and \(\frac{4}{3}\) respectively, what is the refractive index of glass with respect to water?
Solution:

This is the refractive index of glass with respect to water.

Project:

Question 1.
Using a laser and soap water. study the refraction of light under the guidance of your teacher. (Do it your self)

Can you recall? (Text Book Page No. 73)

Question 1.
What is meant by reflection of light?
Answer:
Reflection of light: When light is incident on the surface of an object, in general, it is deflected in different directions. This process is called reflection of light.

Question 2.
What are the laws of reflection?
Answer:
Laws of reflection of light:

1. The incident ray and the reflected ray of light are on the opposite sides of the normal to the reflecting surface at the point of incidence and all the three are in the same plane.
2. The angle of incidence j and the angle of reflection are equal in measure.

Can you recall? (Text Book page No. 75)

Question 1.
If the refractive index of the second medium with respect to the first medium is ₂n₁ and that of the third medium with respect to the second medium is ₃n₂, what and how much is ₃n₁.
Answer:
₃n₁ is the refractive index of the third medium with respect to the first medium.

∴ ₃n₁ = ₂n₁ × ₃n₂.

[Suppose medium 1 = air, medium 2 ≡ ice and medium 3 ≡ diamond. Then, ₂n₁ ÷ 1.31, ₃n₂ = 1.847
∴ ₃n₁ = ₂n₁ × ₃n₂ = 1.31 × 1.847 = 2.42 which is the refractive index of diamond with respect to air.]

Can you tell? (Textbook page No. 76)

Question 1.
Have you seen a mirage which is an illusion of the appearance of water on a hot road or in a desert?
Answer:
Due to the changes in refraction of light, the light rays coming from a distant object appear to be coming from the image of the object inside the ground. This is called a mirage. When the earth’s surface is heated by the sun, the temperature of air increases. This produces a layer of hot air of lower density (mass per unit volume) and lower refractive index at the surface.

Hot air works as an optically rarer medium relative to cool air. When the temperature changes rapidly in the vertical direction, as refraction of light takes place, the angle of refraction changes continuously. The rays of light from the top of an object such as a car or tree cross the rays from the bottom of the object on their way to the observer’s eye.

Hence, an inverted image is formed below the object’s true position and downward towards the surface in the direction of air at higher temperature. In this case, some rays of light bend back up into the denser air above figure. Mirage produces an impression of water near the hot ground.

Question 2.
Have you seen that objects beyond and above a holy fire appear to be shaking? Why does this happen?
Answer:
The temperature of air beyond and above a holy fire changes all the time. Hence, the density of air also changes constantly. Hence, the direction of propagation of the rays of light approaching us from the objects beyond and above the holy fire changes constantly. Therefore, those objects appear to be shaking.

Use your brain power! (Text Book Page No. 77)

Question 1.
From incident white light how will you obtain white emergent light by making use of two prisms?
Answer:

• Take a prism. Allow white light to fall on it.
• Obtain a spectrum.
• Take a second identical prism. Place it parallel to the first prism in an upside down position with the first prism [as shown in Figure]
• Allow the colours of the spectrum to pass through the second prism.
• Obtain the beam of light emerging from the other side of the second prism.

The beam of light emerging from the other side of the second prism is a beam of white light.

Explanation: White light is made up of seven colours. The first prism produces dispersion of white light while the second prism combines light of different colours to produce white light again. The net deviation of a ray of light is zero.
[Note: This experiment is due to sir Isaac Newton. It proved that it was not the prism which added colours to the white light but a property of the white light itself.]

Question 2.
You must have seen chandeliers having glass prisms. The light from a tungsten bulb gets dispersed while passing through these prisms and we see coloured spectrum. If we use an LED light instead of a tungsten bulb, will we be able to see the same effect?
Answer:
Light emitted by LED (light-emitting-diode) does not have all wavelengths in the region 400 nm to 700 nm. Hence, its spectrum is not the same as that of light from a tungsten bulb or as that of sunlight.

Fill in the blanks and rewrite the statements:

Question 1.
The phenomenon of change in the………..of light when it passes obliquely from one transparent medium to another is called refraction.
Answer:
The phenomenon of change in the direction of propagation of light when it passes obliquely from one transparent medium to another is called refraction.

Question 2.
The refractive index depends upon the…………of propagation of light in different media.
Answer:
The refractive index depends upon the velocity of propagation of light in different media.

Question 3.
The process of separation of light into its component colours while passing through a medium is called………..
Answer:
The process of separation of light into its component colours while passing through a medium is called dispersion of light.

Question 4.
When a light ray travels obliquely from air to water, it bends………the normal at the point of incidence.
Answer:
When a light ray travels obliquely from air to water, it bends towards the normal at the point of incidence.

Question 5.
When a light ray travels obliquely from benzene to air, it bends…………the normal at the point of incidence.
Answer:
When a light ray travels obliquely from benzene to air, it bends away from the normal at the point of incidence.

Question 6.
In glass, the speed of red ray is……violet ray.
Answer:
In glass, the speed of red ray is greater than that of violet ray.

Question 7.
The speed of light in glass is………in water.
Answer:
The speed of light in glass is less than that in water.

Question 8.
The speed of light in water is…………in benzene.
Answer:
The speed of light in water is greater than that in benzene.

Question 9.
Rainbow occurs due to refraction, dispersion,……….and again refraction of sunlight by water droplets.
Answer:
Rainbow occurs due to refraction, dispersion, internal reflection and again refraction of sunlight by water droplets.

Question 10.
In dispersion of sunlight by a glass prism,………..ray is deviated the least.
Answer:
In dispersion of sunlight by a glass prism, red ray is deviated the least.

Rewrite the following statements by selecting the correct options:

Question 1.
The change in the direction of propagation of light when it passes obliquely from one transparent medium to another is called………
(a) dispersion
(b) scattering
(c) refraction
(d) reflection
Answer:
(c) refraction

Question 2.
When a ray of light travels from air to glass slab and strikes the surface of separation at 90°, then it…………
(a) bends towards the normal
(b) bends away from the normal
(c) passes unbent
(d) passes in zigzag way
Answer:
(c) passes unbent

Question 3.
If a ray of light passes from a denser medium to a rarer medium in a straight line, the angle of incidence must be…………
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Answer:
(a) 0°

Question 4.
A ray of light strikes a glass slab at an angle of 50° with the normal to the surface of the slab. What is the angle of incidence?
(a) 50°
(b) 25°
(c) 40°
(d) 100°
Answer:
(a) 50°

Question 5.
If a ray of light propagating in air strikes a glass slab at an angle of 60° with the surface of the slab, the angle of refraction is…………
(a) more than 30 °
(b) less than 30 °
(c) 60°
(d) 30°
Answer:
(b) less than 30 °

Question 6.
A ray of light gets deviated When it passes obliquely from one medium to another medium because………..
(a) the colour of light changes
(b) the frequency of light changes
(c) the speed of light changes
(d) the intensity of light changes
Answer:
(c) the speed of light changes

Question 7.
The speed of light in turpentine oil is 2 × 10⁸ m/s. The absolute refractive index of turpentine oil is about……..[Speed of light in vacuum ≈ 3 × 10⁸ m/s]
(a) 1.5
(b) 2
(c) 1.3
(d) 0.67
Answer:
(a) 1.5

Question 8.
LASER stands for………..
(a) light amplification by stimulated emission of radiation
(b) light and sound energy radiation
(c) light and simulated energy radiation
(d) light amplification by sound energy radiation
Answer:
(a) light amplification by stimulated emission of radiation

Question 9.
Out of the following……….has the highest absolute refractive index.
(a) fused quartz
(b) diamond
(c) crown glass
(d) ruby
Answer:
(b) diamond

Question 10.
The absolute refractive index…………
(a) is expressed in dioptre
(b) is expressed in m/s
(c) of air is about \(\frac{4}{3}\)
(d) has no unit
Answer:
(d) has no unit

Question 11.
The speed of light in a medium of refractive index n is………., where c is the speed of light
in vacuum.
(a) \(\frac{c}{n}\)
(b) nc
(c) \(\frac{n}{c}\)
(d) \(\sqrt{\frac{c}{n}}\)
Answer:
(a) \(\frac{c}{n}\)

Question 12.
The speed of light in a transparent medium having absolute refractive index 1.25 is……….[Speed of light in vacuum ≈ 3 × 10⁸ m/s]
(a) 1.25 × 10⁸ m/s
(b) 2.4 × 10⁸ m/s
(c) 3.0 × 10⁸ m/s
(d) 1.5 × 10⁸ m/s
Answer:
(b) 2.4 × 10⁸ m/s

Question 13.
…………light is deviated the maximum in the spectrpm of white light obtained with a glass prism.
(a) Red
(b) Yellow
(c) Violet
(d) Blue
Answer:
(c) Violet

Question 14.
………..light is deviated the least in the spectrum of white light obtained with a glass prism.
(a) Red
(b) Yellow
(c) Violet
(d) Blue
Answer:
(a) Red

Question 15.
A ray of light makes an angle of 50° with the surface S1 of the glass slab. Its angle of incidence will be………….(March 2019)
(a) 50°
(b) 40°
(c) 140°
(d) 0°
Answer:
(a) 50°

Question 16.
A glass slab is placed in the path of convergent light. The point of convergence of light:
(a) moves away from the slab
(b) moves towards the slab
(c) remains at the same point
(d) undergoes a lateral shift
Answer:
(a) moves away from the slab

Question 17.
In refraction of light through a glass slab, the directions of the incident ray and the refracted ray are………… (Practice Activity Sheet – 1)
(a) perpendicular to each other
(b) non-parallel to each other
(c) parallel to each other
(d) intersecting each other
Answer:
(c) parallel to each other

Question 18.
If we gradually increase the angle of incidence of a ray of light passing through a prism, then………….. (Practice Activity Sheet – 4)
(a) the angle of deviation goes on decreasing
(b) the angle of deviation decreases but after certain value of incident angle, deviation angle increases
(c) the angle of deviation goes on increasing
(d) the angle of deviation increases but after certain value of incident angle, deviation angle decreases
Answer:
(b) the angle of deviation decreases but after certain value of incident angle, deviation angle increases

State whether the following statements are True or False. (If a statement is false, correct it and rewrite it.):

Question 1.
The incident ray and the refracted ray of light are on the opposite sides of the normal at the point of incidence.
Answer:
True.

Question 2.
The refractive index of a medium (such as glass) does not depend on the wavelength of light.
Answer:
False. (The refractive index of a medium depends on the wavelength of light.)

Question 3.
When a light ray travels obliquely from an optically rarer medium to an optically denser medium, it bends away from the normal.
Answer:
False. (When a light ray travels obliquely from an optically rarer medium to an optically denser medium, it bends towards the normal.)

Question 4.
When a light ray travels obliquely from glass to air, it bends towards the normal.
Answer:
False. (When a light ray travels obliquely from glass to air, it bends away from the normal.)

Question 5.
If the angle of incidence is 0°, the angle of refraction is 90°.
Answer:
False. (If the angle of incidence is 0°, the angle of refraction is also 0°.)

Question 6.
In dispersion of white light by a glass prism, yellow colour is deviated the least.
Answer:
False. (In dispersion of white light by a glass prism, red colour is deviated the least.)

Question 7.
In vacuum, the speed of light does not depend upon the frequency of light.
Answer:
True.

Question 8.
In glass, the speed of violet ray is less than that of red ray.
Answer:
True.

Question 9.
In a material medium, the speed of light depends on the frequency of light.
Answer:
True.

Question 10.
The velocity of light is different in different media.
Answer:
True.

Question 11.
Wavelength of red light is close to 700 nm.
Answer:
True.

Question 12.
Wavelength of orange light is greater than that of blue light.
Answer:
True.

Find the odd one out and give the reason:

Question 1.
Reflection, Neutralization, Refraction, Dispersion.
Answer:
Neutralization. It is associated with a chemical reaction between an acid and an alkali; others are phenomena associated with light.

Answer the following questions in one sentence each:

Question 1.
Mention any two phenomena in nature where refraction of light takes place.
Answer:
Mirage and twinkling of a star.

Question 2.
What is the angle of refraction when the angle of incidence is 0°?
Answer:
When the angle of incidence is 0°, the angle of refraction is also 0°.

Question 3.
In refraction of light, \(\frac{\sin i}{\sin r}\) = constant in sin a particular case. What is this constant called?
Answer:
The constant \(\frac{\sin i}{\sin r}\) (in a particular case) is called the refractive index of the second medium with respect to the first medium.

Question 4.
If the refractive index of medium 2 with respect to medium 1 is 5/3, what is the refractive index of medium 1 with respect to medium 2?
Answer:
The refractive index of medium 1 with respect to medium 2 is 0.6.

Question 5.
In dispersion of sunlight by a glass prism, which colour is deviated the least?
Answer:
In dispersion of sunlight by a glass prism, red colour is deviated the least.

Question 6.
In dispersion of sunlight by a glass prism, which colour is deviated the most?
Answer:
In dispersion of sunlight by a glass prism, violet colour is deviated the most.

Question 7.
What is the wavelength of violet light?
Answer:
The wavelength of violet light is (about) 400 nm.

Question 8.
State the relation between ₂n₁ and critical angle.
Answer:
₂n₁ = sin i, where i is the critical angle.

Answer the following questions:

Question 1.
What is meant by refraction of light?
Answer:
The change in the direction of propagation of light when it passes obliquely from one transparent medium to another is called refraction of light.

Question 2.
Why is there a change in the direction of propagation of light when it passes obliquely from one transparent medium to another?
Answer:
The velocity of light is different in different media. Hence, there is a change in the direction of propagation of light when it passes obliquely from one transparent medium to another.

Question 3.
In the case of refraction of light through a glass slab, the emergent ray is parallel to the incident ray, but it is displaced sideways. Why does this happen?
Answer:

The first refraction takes place as light passes obliquely from air to glass. In this case, the ray of light bends towards the normal at point N. The second refraction takes place as light passes obliquely from glass to air. In this case, the ray of light bends away from the normal at point M. The faces PQ and SR of the glass slab are parallel. Hence, the extent of bending of light at SR is equal in magnitude but opposite in sense relative to the bending of light at PQ. Hence, the emergent ray of light (MD) is parallel to the incident ray of light (AN), but it is displaced sideways as shown in Figure.

Question 4.
Define angle of incidence and angle of refraction.
Answer:
(1) The angle made by the incident ray of light with the normal to the surface at the point of incidence is called the angle of incidence.

(2) The angle made by the refracted ray of light with the normal to the surface at the point of incidence is called the angle of refraction.

[Note: The angle e in Fig. 6.3 is also called the angle of emergence as it is the angle made by the emergent ray with the normal to the surface at the point of emergence. ]

Question 5.
Repeat the activity “Refraction of light passing through a glass sl^b” by replacing the glass slab by a transparent plastic slab.
(i) What similarity do you observe?
(ii) What difference do you notice?
Answer:
(i) Similarity: The emergent ray is parallel to the incident ray, but it is displaced sideways.
(ii) Difference: For a given angle of incidence, the extent of refraction (bending) is different (in general, less) for a transparent plastic slab relative to the glass slab.

Question 6.
State the laws of refraction of light.
Answer:
Laws of refraction of light:
(1) The incident ray and the refracted ray are on the opposite sides of the normal to the surface at the point of incidence and all the three, i.e., the incident ray, the refracted ray and the normal are in the same plane.

(2) For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant (Snell’s law). This constant is called the refractive index of the second medium with respect to the first medium.
[Note: Here, a ray means a ray of light.]

Question 7.
How is refraction of light related to refractive index?
Answer:
When a ray of light travels obliquely from an optically rarer medium (lower refractive index) to an optically denser medium (higher refractive index), the ray bends towards the normal. When a ray of light travels obliquely from an optically denser medium to an optically rarer medium, the ray bends away from the normal. For a given angle of incidence (i ≠ 0), the extent of refraction (bending) of light is different in different media.

If the refractive index of the second medium with respect to the first medium is greater than 1, the greater the refractive index, the greater is the bending of the ray of light towards the normal. If the refractive index of the second medium with respect to the first medium is less than 1, the greater the refractive index, the lesser is the bending of the ray of light away from the normal.

Question 8.
Define the refractive index of the second medium with respect to the first medium.
(OR)
What is meant by refractive index?
Answer:
The refractive index of the second medium with respect to the first medium is defined as the ratio of the sine of the angle of incidence to the sine of the angle of refraction when the ray of light is obliquely incident at the boundary separating the

two media and travels from the first medium to the second medium. (See Fig. 6.4.)
(OR)
The refractive index of the second medium with respect to the first medium is defined as the ratio of (the magnitude of) the velocity of light in the first medium to (the magnitude of) the velocity of light in the second medium.

[Note: Velocity is a vector, i.e., it has magnitude and direction. In definition of refractive index, we consider only the magnitude of velocity of light (speed of light). Velocity of light in a medium depends on the physical condition of the medium as well as the frequency of light. Velocity of light is different in different media. For a given medium, the refractive index depends on the colour of light (frequency of light.)]

Question 9.
State the formulae for the refractive index of the second medium with respect to the first medium.
Answer:
The refractive index of the second medium with respect to the first medium,
₂n₁ = \(\frac{\sin i}{\sin r}=\frac{v_{1}}{v_{2}}\)
where i is the angle of incidence, r is the angle of refraction (as the ray of light passes obliquely from the first medium to the second medium), v₁ is the magnitude of the velocity (speed) of light in the first medium and v₂ is the magnitude of the velocity of light in the second medium.

Question 10.
Define absolute refractive index.
Answer:
The absolute refractive index of a medium is defined as the ratio of the magnitude of the velocity of light in vacuum to the magnitude of the velocity of light in the medium.

[Note: The speed of light is maximum in vacuum, about 3 × 10⁸ m/s. When light travels from one medium to another, there occurs a change in its speed and wavelength (A). But its frequency (v) remain the same.]

Question 11.
Obtain the relation between the refractive index of the second medium with respect to the first medium and the refractive index of the first medium with respect to the second medium.
Answer:
Let v₁ = speed of light in the first medium, v₂ = speed of light in the second medium, ₂n₁ = refractive index of the second medium With respect to the first medium and ₁n₂ = refractive index of the first medium with respect to the second medium.
By definition, ₂n₁ = \(\frac{v_{1}}{v_{2}}\) and ₁n₂ = \(\frac{v_{2}}{v_{1}}\)
Hence,

(OR)
₁n₂ × ₂n₁ = 1.

Question 12.
If the refractive index of a certain material with respect to air is 1.5, what is the refractive index of air with respect to that material?
Answer:
As the refractive index of the given material with respect to air is 1.5, the refractive index of air with respect to the material is
\(\frac{1}{1.5}=\frac{1}{3 / 2}=\frac{2}{3}\) = 0.6667 (approximately)

Question 13.
Explain the terms optically rarer medium and optically denser medium with examples.
Answer:
When we consider two media (such as air and glass), the medium with lower refractive index is called the optically rarer medium (in the present case, air) and the medium with higher refractive index is called the optically denser medium (glass, in the present case).

The higher density does not necessarily mean higher refractive index. For example, the density of water is greater than that of kerosene, but the absolute refractive index of water is less than that of kerosine. Thus, when we consider water and kerosine, water is an optically rarer medium while kerosine is an optically denser medium.

If we consider kerosene and benzene, kerosine is an optically rarer medium while benzene is an optically denser medium.

Question 14.
A ray of light is incident obliquely at a boundary separating two media. What is its behaviour if (1) the refractive index of the second medium is greater than that of the first medium (2) the refractive index of the first medium is greater than that of the second medium? Draw the corresponding neat and labelled diagrams.
Answer:
Consider a ray of light incident obliquely at a boundary separating two media.
(1) If the refractive index of the second medium is greater than that of the first medium, the ray bends towards the normal at the point of incidence as it travels from the first medium (optically rarer medium) to the second medium (optically denser medium). The angle of refraction (r) is less than the angle of incidence (i). (Fig. 6.6)

Fig. 6.6: A ray of light travelling from a rarer medium to a denser medium (Schematic diagram)

Fig. 6.7: A ray of light travelling from a denser medium to a rarer medium (Schematic diagram)

(2) If the refractive index of the first medium is greater than that of the second medium, the ray bends away from the normal at the point of incidence as it travels from the first medium (optically denser medium), to the second medium (optically rarer medium). The angle of refraction (r) is greater than the angle of incidence (i). (Fig. 6.7)

[Note In this chapter, a rarer medium means an optically rarer medium and a denser medium means optically denser medium unless stated otherwise.]

Question 15.
Observe the following figure and write accurate conclusion regarding refraction of light. (Practice Activity Sheet – 2)

Answer:
When a light ray passes obliquely from a rarer medium to a denser medium, it bends towards the normal.

Question 16.
What happens when a ray of light is incident normal to the interface between two media? Draw the corresponding neat and labelled diagram.
Answer:
When a ray of light is incident normal to the interface between two media, the ray propagates undeviated as it travels from the first medium to the second medium irrespective of the refractive indices of the two media. In this case, the angle of incidence (i) is zero and so also the angle of refraction (r).

Fig. 6.9: A ray of light incident normal to the interface between two media propagates without any change in its direction of propagation

Question 17.
Draw a neat and labelled diagram to show the path of a ray of light in air and glass when the ray is incident obliquely on a glass slab. Show the (i) incident ray (ii) refracted ray (iii) emergent ray (iv) angle of incidence (v) angle of refraction (vi) angle of emergence in the diagram.
(OR)
Draw a neat and labelled diagram to show refraction of light through a glass slab.
Answer:

Fig. 6.10: The path of the ray of light in air and glass when the ray is incident obliquely on a glass slab
In Fig. 6.10, i = angle of incidence, r = angle of refraction and e = angle of emergence.

Question 18.
Observe the given figure and name the following rays:

(i) ray AB
(ii) Ray BC
(iii) ray CD
Answer:
(i) The ray AB is the incident ray.
(ii) The ray BC is the refracted ray.
(iii) The ray CD is the emergent ray.

Question 19.
A plane mirror is kept at the bottom of a trough with water in it as shown in the following figure (Fig. 6.12). The ray of light emerging from a source at the point S outside the trough reaches the point A on the surface of water. Draw a neat ray diagram to show the subsequent path of light and complete the ray diagram.

Answer:

Question 20.
Give two examples of the effect of atmospheric refraction on a small scale in local environment.
Answer:

1. The occurrence of a mirage
2. Flickering of an object seen through a turbulent stream of hot air rising above the Holi fire are examples of the effect of atmospheric refraction on a small scale in local environment.

Question 21.
What is a mirage? With a neat labelled diagram, explain the conditions under which it is seen.
Answer:
Due to the changes in refraction of light, the light rays coming from a distant object appear to be coming from the image of the object inside the ground. This is called a mirage.

When the earth’s surface is heated by the sun, the temperature of air increases. This produces a layer of hot air of lower density (mass per unit volume) and lower refractive index at the surface. Hot air works as an optically rarer medium relative to cool air. When the temperature changes rapidly in the vertical direction, as refraction of light takes place, the angle of refraction changes continuously.

The rays of light from the top of an object such as a car or tree cross the rays from the bottom of the object on their way to the observer’s eye. Hence, an inverted image is formed below the object’s true position and downward towards the surface in the direction of air at higher temperature. In this case, some rays of light bend back up into the denser air above figure. Mirage produces an impression of water near the hot ground.

Question 22.
Explain in brief the flickering of an object seen through a turbulent stream of hot air rising above the Holi fire.
Answer:
During the Holi fire, the temperature of the air just above the fire becomes much greater than that of the air further up. The hot air has lower density (mass per unit volume) and lower refractive index. It becomes an optically rarer medium. The cool air has higher density and higher refractive index. It is an optically denser medium relative to hot air. Hence, in refraction of light, the angle of refraction changes continuously due to a continuous variation in refractive index.

As the physical conditions of air change rapidly, the apparent position of an object fluctuates rapidly. This gives rise to the flickering of an object seen through a turbulent stream of hot air rising above the Holi fire.

Question 23.
With a neat labelled diagram, explain twinkling of a star. Also explain why a planet does not twinkle.
Answer:
(1) As a star is far away from the earth, it appears as a point source of light. The density of air decreases with height above the earth’s surface. Hence, the refractive index of air also decreases with height. When starlight enters the earth’s atmosphere, it undergoes refraction continuously in the medium with gradually changing refractive index. The bending of starlight occurs towards the normal as it passes from the optically rarer part of the medium to the optically denser part.

(2) Hence, when a star is observed near the horizon, its apparent position is slightly higher than the actual position (See below figure).

(3) Further, the apparent position varies with time as the medium is not stationary due to mobility of air and change in temperature. When more light is refracted towards the observer the star appears bright. When less light is refracted towards the observer, the star appears dim.

(4) Thus there is fluctuation in the brightness of a star when observed from the earth. This is called twinkling of a star.

(5) Compared to stars, planets are relatively closer to the earth. Hence, a planet appears as a collection of a large number of point sources. Due to the changes in the refractive index of air, there is a change in the position and brightness of these point sources.

There is an increase in intensity of light coming from some point sources while there is a decrease in intensity of light coming from equal number of other point sources, on an average. The average brightness of a planet remain the same. Also, there is no change in the average position of a star. Hence, a planet does not twinkle.

Question 24.
What is the correct reason for blinking/flickering of stars? Explain it.
(a) The blasts in the stars.
(b) Absorption of star light by the atmosphere.
(c) Motion of the stars.
(d) Changing refractive index of gases in the atmosphere. (Practice Activity Sheet – 2)
Answer:
(d) Changing refractive index of the gases in the atmosphere results in blinking/flickering of stars.

Explanation:
(1) As a star is far away from the earth, it appears as a point source of light. The density of air decreases with height above the earth’s surface. Hence, the refractive index of air also decreases with height. When starlight enters the earth’s atmosphere, it undergoes refraction continuously in the medium with gradually changing refractive index. The bending of starlight occurs towards the normal as it passes from the optically rarer part of the medium to the optically denser part.

(2) Hence, when a star is observed near the horizon, its apparent position is slightly higher than the actual position (See below figure).

(3) Further, the apparent position varies with time as the medium is not stationary due to mobility of air and change in temperature. When more light is refracted towards the observer the star appears bright. When less light is refracted towards the observer, the star appears dim.

(4) Thus there is fluctuation in the brightness of a star when observed from the earth. This is called twinkling of a star.

(5) Compared to stars, planets are relatively closer to the earth. Hence, a planet appears as a collection of a large number of point sources. Due to the changes in the refractive index of air, there is a change in the position and brightness of these point sources.

There is an increase in intensity of light coming from some point sources while there is a decrease in intensity of light coming from equal number of other point sources, on an average. The average brightness of a planet remain the same. Also, there is no change in the average position of a star. Hence, a planet does not twinkle.

Question 25.
With a neat labelled diagram, explain advanced sunrise and delayed sunset.
Answer:
(1) The sunrise (the appearance of the sun above the horizon) is advanced due to atmospheric refraction of sunlight. An observer on the earth sees the sun two minutes before the sun reaches the horizon. A ray of sunlight entering the earth’s atmosphere follows a curved path due to atmospheric refraction before reaching the earth. This happens due to a gradual variation in the refractive index of the atmosphere.

For the observer on the earth, the apparent position of the sun is slightly higher than the actual position. Hence, the sun is seen before the sun reaches the horizon.

(2) Increased atmospheric refraction of sunlight occurs also at the sunset (the sun disappearing below the horizon). In this case, the observer on the earth continues to see the setting sun for two minutes after the sun has dipped below the horizon, thus delaying the sunset.

The advanced sunrise and delayed sunset increases the duration of day by four minutes.

Question 26.
Water in a swimming pool or water tank appears shallower than its depth. Why?
Answer:
When light rays travel obliquely from an optically denser medium (water, in this case) to an optically rarer medium (air, in this case), they bend away from the normal at the point of incidence. As a result, the bottom of a swimming pool or water tank appears raised to an observer standing near the edge of the pool or the tank. Therefore, the swimming pool or water tank appears shallower than its depth.

Question 27.
Place a coin at the bottom of a glass jar containing water. Now tilt the jar suitably. When viewed at a suitable angle, the coin appears to be floating. Why?
Answer:
When light rays travel obliquely from an optically denser medium (water, in this case) to an optically rarer medium (air, in this case), they bend away from the normal at the point of incidence. As a result, the coin appears raised. Therefore, when the jar is tilted suitably and observed at a suitable angle, the coin appears to be floating.

Question 28.
State the wavelength range of electromagnetic radiation to which our eyes are sensitive.
Answer:
Our eyes are sensitive to light (electromagnetic radiation). Its wavelength range is 400 nm to 700 nm.
[Note: Wavelength (λ) goes on decreasing and frequency (ν) goes on increasing from red (λ ≃ 700 nm) → orange → yellow → green → blue → indigo → violet (A ≃ 400 nm). c = vλ, where c is the speed of light in vacuum.]

Question 29.
What do you mean by dispersion of light? What is a spectrum of light? Name the different colours of light in the proper sequence in the spectrum of white light.
(OR)
What do you mean by dispersion? Name the different colours of light in the proper sequence in the spectrum of white light.
Answer:
The process of separation of light into its component colours while passing through a medium is called dispersion of light. The band of coloured components of a light beam is called spectrum.
The different colours of light in the spectrum of white light are violet, indigo, blue, green, yellow, orange and red.

Question 30.
What is a prism?
Answer:
A prism is a transparent medium bound by two plane surfaces inclined at an angle. Normally it is made of glass and has triangular cross section.

Question 31.
With a neat labelled diagram, describe the experiment to demonstrate dispersion of sunlight (white light) by a prism.
Answer:
Experiment:
(1) Procedure: Keep a glass prism on a table in a dark room. Hold a plane mirror outside the room so that it reflects a beam of sunlight into the room. Allow this beam to pass through a narrow slit made in cardboard and then fall on the prism. Place a white screen on the other side of the prism as shown in the following figure. [Fig. 6.17]

(2) Observations:

1. A pattern of various colours is observed on the screen. This pattern is called the spectrum.
2. It is found that in dispersion, the ray corresponding to violet colour deviates the most.
3. The ray corresponding to red colour deviates the least.
4. The deviation of rays corresponding to other colours is intermediate.

(3) Conclusion: When sunlight (white light) is incident on a prism, dispersion of light takes place, forming a spectrum.

[Notes: (1) This experiment is due to Sir Isaac Newton (1642 – 1727), English physicist and mathematician. (2) If in a Board examination, incomplete diagram (as shown in Fig. 6.18) is given, students should complete it and label its parts as shown in Fig. 6.17.]

Question 32.
How does the dispersion of white light take place when it passes through a glass prism?
Answer:
When rays of light are incident on a prism, they are refracted twice, while travelling from air to glass and then from glass to air. Even when the incident rays are directed away from the base of the prism, the emergent rays bend towards the base of the prism, as the prism is triangular. Thus, the rays are deviated as they pass through the prism.

The refractive index of glass is different for different colours. Therefore, the rays corresponding to different colours are deviated to different extents. White light is a mixture of seven colours : violet, indigo, blue, green, yellow, orange and red. Hence, when white light is incident on a prism, a spectrum of seven colours is obtained.

The refractive index of glass is maximum for violet light and minimum for red light. Hence, violet light is deviated the most and red light is deviated the least. The deviation of rays corresponding to other colours is intermediate. In this manner, the dispersion of light takes place when it passes through a glass prism. [For reference, see Fig. 6.17.]

Question 33.
What is a spectrum? Why do we get a spectrum of seven colours when while light is dispersed by a prism?
(OR)
Explain how a spectrum is formed.
Answer:
A band of coloured components of a light beam is called a spectrum. When white light is incident on a prism, the rays corresponding to different colours bend through different angles on refraction.

Of the various colours in the visible region, red light bends the least and violet light bends the most. Each colour emerges through the prism along a different path and becomes, distinct. Hence, we get a spectrum of seven colours.

Question 34.
What is partial reflection of light?
Answer:
When light travels from a denser medium to a rarer medium, it is partially reflected, i.e., part of light comes back into the denser medium as per the laws of reflection. This is called partial reflection of light.

[Note: Partial reflection of light occurs even when light travels from a rarer medium to a denser medium. The rest of light is refracted.]

Question 35.
Explain the terms total internal reflection and critical angle.
Answer:
Figure 6.20 shows passage of light from water (denser medium) to air (rarer medium).

The ray of light incident at the boundary separating the two media bends away from the normal on refraction. Here, the angle of refraction r, is greater than the angle of incidence i.
Now _an_w = \(\frac{\sin i}{\sin r}\) < 1. Here, _an_w is the refractive index of sin r air with respect to water.

As _an_w is constant, r increases as i increases. For r = 90°, the ray travels along the boundary. If i is increased further, as r cannot be greater than 90°, light does not enter air. There is no refraction of light and all the light enters water on reflection. This is called total internal reflection.
For r = 90°, _an_w = \(\frac{\sin i}{\sin 90^{\circ}}\) = sin i. This angle i is sin 90° called the critical angle.

Question 36.
Swarali has got the following observations while doing an experiment. Answer her questions with the help of observations. (Practice Activity Sheet – 2)
Swarali observed that the light bent away from the normal, while travelling from a denser medium to a rarer medium. When Swarali increased the values of the angle of incidence (i). the values of the angle of refraction (r) went on increasing. But at a certain angle of incidence, the light rays returned into the denser medium.
So, Swarali has some questions. Answer them.
(a) Name this certain value of i. What is the value of r at that time?
(b) Name this process of returning light in the denser medium. Explain the process.
Answer:
(a) Critical angle r = 90°
(b) Total internal reflection.

As light goes from a denser to rarer medium, if the value of the angle of incidence increases, then the value of the angle of refraction also increases. But after a specific angle of incidence called the critical angle, the light gets reflected back into the denser medium.

The ray of light incident at the boundary separating the two media bends away from the normal on refraction. Here, the angle of refraction r, is greater than the angle of incidence i.
Now _an_w = \(\frac{\sin i}{\sin r}\) < 1. Here, _an_w is the refractive index of sin r air with respect to water. As _an_w is constant, r increases as i increases. For r = 90°, the ray travels along the boundary. If i is increased further, as r cannot be greater than 90°, light does not enter air. There is no refraction of light and all the light enters water on reflection. This is called total internal reflection.

For r = 90°, anw = \(\frac{\sin i}{\sin 90^{\circ}}\) = sin i. This angle i is sin 90° called the critical angle.

Question 37.
The observations made by Swarali while doing the experiment are given below. Based on these write answers to the questions:
Swarali found that the light ray travelling from the denser medium to a rarer medium goes away from the normal. If the angle of incidence (i) is raised by Swarali, the angle of refraction (r) went on increasing. However, after certain value of the angle of incidence, the light ray is seen to return back into the denser medium. (March 2019)
(i) What is the specific value of ∠i called?
(ii) What is the process of reflection of incident ray into a denser medium called?
(iii) Draw the diagrams of three observations made by Swarali.
Answer:
(i) Critical angle
(ii) Total internal reflection
(iii)

Question 38.
Define total internal reflection of light.
Answer:
When light travels from a denser medium to a rarer medium, if the angle of incidence is greater than the critical angle, there is no refraction of light and all the light is reflected in the denser medium. This is called total internal reflection of light.

Question 39.
Define critical angle.
Answer:
When light travels from a denser medium to a rarer medium, the angle of incidence for which the angle of refraction becomes 90°, is called the critical angle.

Question 40.
If the refractive index of a rarer medium with respect to a denser medium is 0.5, what is the critical angle?
Answer:
₂n₁ = 0.5 = sin i
∴ Critical angle i = 30°.

Question 41.
Name the devices in which total internal reflection of light is used.
Answer:

1. Total internal reflecting prisms are used in a camera, binoculars, periscope.
2. Total internal reflection of light is used in optical fibres.

[Note: Total internal reflection of light plays an important role in sparkling brilliance of a diamond.]

Question 42.
Explain why an empty test tube held obliquely in water appears shiny to an observer looking down.
Answer:
When an empty test tube is held obliquely in Water in a beaker, some light rays passing from water to air are incident at an angle greater than the critical angle. They are, thus, totally internally reflected as shown, and the surface of the test tube has a silvery shine.

Question 43.
Observe the given figure and answer the following questions. (Practice Activity Sheet – 3)

(a) Identify and write the natural process shown in the figure.
(b) List the phenomena which are observed in this process.
(c) Redraw the diagram and show the above phenomena in it.
Answer:
(a) The natural process shown in the figure is formation of rainbow.
(b) The phenomena observed in this process are refraction, internal reflection and dispersion of light.
(c)

Write a short note on the following:

Question 1.
Refraction observed in the atmosphere.
Answer:
When a ray of light passes obliquely from an optically rarer medium to an optically denser medium, it bends towards the normal at the point of incidence. If opposite is the case, the ray bends away from the normal.

Atmosphere is never static. Air is mobile and its density and temperature are not uniform. As a result, in general, the path of a ray of light through atmosphere of varying refractive index is a curve. The refractive index of cool air is greater than that of hot air.

Atmospheric refraction of light results in many interesting optical phenomena such as twinkling of a star, advanced sunrise and delayed sunset, mirage and flickering of an object seen through a turbulent stream of hot air rising from a fire.

Question 2.
Dispersion of light.
Answer:
The process of separation of light into its component colours while passing through a medium is called dispersion of light. When white light passes through a glass prism, it spreads out into a band of different colours (components) called the spectrum of light. The colours in the spectrum of white light are violet, indigo, blue, green, yellow, orange and red.

Formation of a rainbow is an example of dispersion of light in nature. In this case, raindrops are responsible for dispersion of sunlight.

Dispersion takes place because the refractive index of a material such as glass or water, is different for different colours. It is maximum for violet colour and minimum for red colour. Hence, in the spectrum of white light (sunlight) obtained with a prism, violet light is deviated the most while red light is deviated the least. The deviation of light corresponding to other colours lies in between.

Give scientific reasons:

Question 1.
A coin kept in a bowl is not visible when seen from one side. But, when water is poured in the bowl, the coin becomes visible.
Answer:
(1) When the bowl is empty, the rays of light coming from the coin are obstructed by the side of the bowl, and hence the coin is not visible when seen from one side of the bowl.

(2) When water is poured in the bowl, the rays of light coming from the coin travel from water (denser medium) to air (rarer medium). Hence, they bend away from the normal on refraction. Therefore, the coin appears to be raised and becomes visible when observed from one side of the bowl.

Question 2.
A pencil dipped in water obliquely appears bent at the surface of water.
(OR)
When a pencil is partly immersed in water and held in a slanting position, it appears to be bent at the boundary separating water and air.
Answer:
(1) When a pencil is partly immersed in water and held in a slanting position, the rays of light coming from the immersed part of the pencil emerge from water (a denser medium) and enter air (a rarer medium). During this propagation, they bend away from the normal on refraction.

The pencil appearing bent at the boundary of water and air (schematic diagram)

(2) As a result, the immersed part of the pencil does not appear straight with respect to the part outside the water, but appears to be raised. Hence, a pencil dipped obliquely in water appears bent at the surface of the water.

Question 3.
The shadow of the edge of an empty vessel is formed due to the slanting rays of the sun. When water is poured in the vessel, the shadow is shifted.
Answer:
(1) When the slanting rays of the sun are obstructed by the edge of the empty vessel, the shadow of the edge is formed.

(2) When water is poured in the vessel, the slanting rays of the sun travel from air (rarer medium) to water (denser medium). During this propagation, they bend towards the normal on refraction. Hence, some part in the region of the shadow is now illuminated and the shadow appears to have shifted.

Question 4.
The bottom of a pond appears raised.
Answer:
(1) The rays of light coming from the bottom of a pond bend away from the normal as they travel from water (denser medium) to air (rarer medium).

(2) Hence, they appear to come from a point above the actual point from which they come.
Therefore, the bottom of the pond appears raised.

Question 5.
While shooting a fish in a lake, the gun is aimed below the apparent position of the fish.
Answer:
(1) The rays of light coming from the fish bend away from the normal as they travel from water (denser medium) to air (rarer medium).
(2) Hence, the position of the fish in water appears to be above Its real position. Therefore, while shooting a fish in a lake, the gun is aimed below the apparent position of the fish.

Question 6.
The sun is seen on the horizon a little before sunrise.
(OR)
The sun is seen on the horizon for sometime even after sunset.
Answer:
(1) The earth is surrounded by an atmosphere which is denser near the surface of the earth. When the rays of light from the sun enter the earth’s atmosphere from outer space, they travel from a rarer medium to a denser medium. Hence, they bend towards the normal on refraction.

(2) Hence, even when the sun is below the horizon while rising or setting, its rays reach us due to refraction and it appears to be on the horizon. Therefore, the sun is seen on the horizon a little before sunrise as well as for some time even after sunset.

Distinguish between:

Question 1.
Reflection of light and Refraction of light:
Answer:

Reflection of light	Refraction of light
1. The rays of light, before and after reflection, travel in the same medium.	1. In refraction of light, the rays travel from one medium to another medium.
2. In reflection, the angle of incidence and the angle of reflection are equal.	2. In refraction, when the rays travel obliquely from one medium to another medium, the angle of incidence and the angle of refraction are not equal.
3. In reflection, there is no change in the speed and wavelength of light.	3. In refraction, there occurs a change in the speed and wavelength of light.
4. In reflection, there is no dispersion of light.	4. Generally, in refraction, there occurs dispersion of light.

[Note: The frequency of light remains the same in reflection and refraction.]

Complete the following or Solve and fill in the blanks :

Question 1.

Speed of light in the first medium (v1)	Speed of light in the second medium (v2)	Refractive index 2n1	Refractive index 2n1
3 × 108 m/s	1.2 × 108 m/s	————————–	————————–
————————–	2.25 × 108 m/s	4/3	————————–
2 × 108 m/s	————————–	————————–	1.5

Answer:

Speed of light in the first medium (v1)	Speed of light in the second medium (v2)	Refractive index 2n1	Refractive index 2n1
3 × 108 m/s	1.2 × 108 m/s	2.5	0.4
3 × 108 m/s	2.25 × 108 m/s	4/3	0.75
2 × 108 m/s	3 × 108 m/s	2/3	1.5

Formulae:
₂n₁ = v₁/v₂, ₁n₂ = v₂/v₁

Solve the following examples/numerical problems:
c = 3 × 10⁸ m/s

Problem 1.
The speed of light in a transparent medium is 2.4 × 10⁸ m/s. Calculate the absolute refractive index of the medium.
Solution:
Data: c = 3 × 10⁸ m/s,
v = 2.4 × 10⁸ m/s, n = ?

The absolute refractive index of the medium = 1.25.

Problem 2.
The velocity of light in a medium is 2 × 10⁸ m/s. What is the refractive index of the medium with respect to air, if the velocity of light in air is 3 × 10⁸ m/s?
Solution:
Data: v₁ = 3 × 10⁸ m/s,
v₂ = 2 × 10⁸ m/s, ₂n₁ = ?
₂n₁ = \(\frac{v_{1}}{v_{2}}\)
\(=\frac{3 \times 10^{8}}{2 \times 10^{8}}\)
= 1.5
The refractive index of the medium with respect to air is 1.5.

Problem 3.
Light travels with a velocity 1.5 × 10⁸ m/s in a medium. On entering second medium its velocity becomes 0.75 × 10⁸ m/s. What is the refractive index of the second medium with respect to the first medium? (Practice Activity Sheet – 3)
Solution:
Given: Velocity of light in the first medium = v₁ = 1.5 × 10⁸ m/s,
velocity of light in the second medium = v₂ = 0.75 × 10⁸ m/s,
refractive index of the second medium with respect to the first medium = ₂n₁ = ?
₂n₁ = \(\frac{v_{1}}{v_{2}}\)
₂n₁ = \(\frac{1.5 \times 10^{8}}{0.75 \times 10^{8}}\) = 2
Hence, the refractive index of the second medium with respect to the first medium is 2.
[Note : The absolute refractive index of the second medium = \(\frac{3 \times 10^{8} \mathrm{m} / \mathrm{s}}{0.75 \times 10^{8} \mathrm{m} / \mathrm{s}}\) = 4 (greater than that of diamond, not likely).]

Problem 4.
The refractive index of water is 4/3 and the speed of light in air is 3 × 10⁸ m/s. Find the speed of light in water.
Solution:
Data: ₂n₁ = 4/3, v₁ = 3 × 10⁸ m/s, v₂ = ?

The speed of light in water = 2.25 × 10⁸ m/s.

Problem 5.
The speed of light in water and glass is 2.2 × 10⁸ m/s and 2 × 10⁸ m/s respectively. What is the refractive index of (i) water with respect to glass (ii) glass with respect to water?
Solution:
Data: u_w = 2.2 × 10⁸ m/s,
v_g= 2 × 10⁸ m/s, _wn_g = ?, _gn_w = ?

The refractive index of water with respect to glass = 0.909 (approximately).

The refractive index of glass with respect to glass = 1.1 (approximately).

Numerical Problems For Practice:
(Given: C = 3 × 10⁸m/s)

Problem 1.
The speed of light in a transparent medium is 2 × 10⁸ m/s. Find the absolute refractive index of the medium.
Solution:
1.5

Problem 2
The absolute refractive index of a transparent medium is 5/3. Find the speed of light in the medium.
Solution:
1.8 × 10⁸ m/s

Problem 3.
The absolute refractive index of a transparent medium is 2.4 and the speed of light in that medium is 1.25 × 10⁸ m/s. Find the speed of light in air.
Solution:
3 × 10⁸ m/s

Problem 4.
The speed of light in water is 2.25 × 10⁸ m/s and that in glass is 2 × 10⁸ m/s. Find the refractive index of (i) the glass with respect to water (ii) water with respect to the glass.
Solution:
(i) 1.125
(ii) 0.889 (approximately)

Problem 5.
If the refractive index of a certain glass with respect to water is 1.25, find the refractive index of water with respect to the glass.
Solution:
0.8

Problem 6.
If the absolute refractive index of glass is 1.5 and that of water is \(\frac{4}{3}\), find the refractive index of water with respect to glass.
Solution:
\(\frac{8}{9}\)

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 5 Heat Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 5 Heat

Question 1.
Fill in the blanks and rewrite the sentences:
a. The amount of water vapour in air is determined in terms of its………..
Answer:
The amount of water vapour in air is determined in terms of its absolute humidity.

b. If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their………….
Answer:
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacities.

c. When a liquid is getting converted into solid, the latent heat is……….  (Practical Activity Sheet – 1 and 2)
Answer:
When a liquid is getting converted into solid, the latent heat is released.

Question 2.
Observe the following graph. Considering the change in volume of water as its temperature is raised from 0 °C, discuss the difference in the behaviour of water and other substances. What is this behaviour of water called?

Answer:
If the temperature of water is raised from 0 °C to 10 °C, its volume goes on decreasing in the range 0 °C to 4 °C. It is minimum at 4 °C. The volume of water goes on increasing in the range 4 °C to 10 °C.

In general, when a substance is heated, its volume goes on increasing with temperature. Thus, in the range 0 °C to 4 °C, behaviour of water is different from other substances. It is called anomalous behaviour of water.

Question 3.
What is meant by specific heat capacity?
How will you prove experimentally that different substances have different specific heat capacities?
Answer:
The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C is called the specific heat capacity of that object.

Question 4.
While deciding the unit for heat, which temperatures interval is chosen? why?
Answer:
While deciding the unit for heat, the temperature interval chosen is 14.5 °C to 15.5 °C. For the reason, see the information given in the following box.

Question 5.
Explain the following temperature vs time graph:

(Practice Activity Sheet – 1 and 4; March 2019)
Answer:
The graph shows what happens when a mixture of ice and water is heated continuously. The temperature of the mixture remains constant (0 °C) till all the ice melts as shown by the line AB. This temperature is the melting point of ice. On further heating, the temperature rises steadily from 0 °C to 100 °C as shown by the line BC, At 100 °C water starts converting into steam. This temperature is the boiling point of water. Further heating does not change the temperature and the conversion waters steam continues as shown by the line CD.

Question 6.
Explain the following:
a. the role of anomalous behaviour of water in preserving aquatic life in regions of cold climate?
Answer:

In cold regions, during winter, the temperature of the atmosphere falls well below 0 °C. As the temperature decreases, the water at the surfaces of lakes and ponds starts contracting. Hence, its density increases and it sinks to the bottom. This process continues till the temperature of all the water in a lake falls to 4 °C. As the water at the surface cools further, i.e., its temperature falls below 4 °C, it starts expanding instead of contracting. Therefore, its density decreases and it remains at the surface.

The temperature of the water at the surface continues to fall to 0 °C. Finally, the water at the surface is converted into ice, but the water below the layer of ice is at 4 °C. Ice is a bad conductor of heat. Hence, the layer of the ice at the surface does not allow transfer of heat from the water to the atmosphere. As the water below the layer of ice remains at 4 °C, fish and other aquatic animals and plants can survive in it.

b. How can you relate the formation of water droplets on the outer surface of a bottle taken out of a refrigerator with formation of dew?
Answer:
At a given temperature, there is a limit on how much water vapour the given volume of air can hold. The lower the temperature, the lower is the capacity of air to hold water vapour.

The temperature of a bottle kept in a refrigerator is lower than room temperature. Hence, when the bottle is taken out of the refrigerator, the temperature of the air surrounding the bottle is lowered. Therefore, the capacity of the air to hold water vapour becomes less. Hence, the excess water vapour condenses to form water droplets (like dew) on the outer surface of the bottle.

c. In cold regions in winter, the rocks crack due to anomalous expansion of water.
Answer:
Sometimes water enters into crevices of the rocks. When the temperature of the atmosphere falls below 4 °C, water expands. Even when water freezes to form ice, there is increase in its volume. As there is no room for expansion, it exerts a tremendous pressure on the rocks which crack and break up into small pieces.

Question 7.
a. What is meant by latent heat? How will the state of matter transform if latent heat is given off?
Answer:
When a solid is heated, initially, its temperature increases. Here, the heat absorbed by the body (substance) is used in increasing the kinetic energy of the particles (atomic, molecules, etc.) of the body as well as for doing work against the forces of attraction between them. As the heating is continued, at a certain temperature (melting point), solid is converted into liquid. In this case, the temperature remains constant and the heat absorbed is used for weakening the bonds and conversion into liquid phase (liquid state). This heat is called the latent heat of fusion.

When a liquid is converted into the gaseous phase (gaseous state), at the boiling point, the heat absorbed is used for breaking the bonds between the atoms or molecules. This heat is called the latent heat of vaporization. Some solids, under certain conditions, are directly transformed into the gaseous phase. Here the heat is absorbed but the temperature remains constant. The absorbed heat is used for breaking the bonds between atoms or molecules. This heat is called the latent heat of sublimation.

In general, latent heat is the heat absorbed or given out by a substance during a change of state at constant temperature.
In transformations from liquid to solid, gas to liquid and gas to solid, latent heat is given out by the body (substance).
(Note: change of state = change of phase)

b. Which principle is used to measure the specific heat capacity of a substance?
Answer:
The principle of heat exchange is used to measure the specific heat capacity of a substance. This principle is as follows: If a system of two objects is isolated from the environment by keeping it inside a heat resistant box, then no energy can leave the box or enter the box. In this situation, heat energy lost by the hot object = heat energy gained by the cold object.

c. Explain the role of latent heat in the change of state of a substance.
Answer:
When a solid is heated, initially, its temperature increases. Here, the heat absorbed by the body (substance) is used in increasing the kinetic energy of the particles (atomic, molecules, etc.) of the body as well as for doing work against the forces of attraction between them. As the heating is continued, at a certain temperature (melting point), solid is converted into liquid. In this case, the temperature remains constant and the heat absorbed is used for weakening the bonds and conversion into liquid phase (liquid state). This heat is called the latent heat of fusion.

When a liquid is converted into the gaseous phase (gaseous state), at the boiling point, the heat absorbed is used for breaking the bonds between the atoms or molecules. This heat is called the latent heat of vaporization. Some solids, under certain conditions, are directly transformed into the gaseous phase. Here the heat is absorbed but the temperature remains constant. The absorbed heat is used for breaking the bonds between atoms or molecules. This heat is called the latent heat of sublimation.

In general, latent heat is the heat absorbed or given out by a substance during a change of state at constant temperature.
In transformations from liquid to solid, gas to liquid and gas to solid, latent heat is given out by the body (substance).
(Note: change of state = change of phase)

d. what basis and how will you determine whether air is saturated with vapour or not?
Answer:
Whether the air is saturated with water vapour or not is determined on the basis of the extent of water vapour present in the air. If the relative humidity is 100%, the air is saturated with water vapour. In that case, we can see the formation of water droplets on the leaves of plants/grass.
If the relative humidity is less than 100%, the air is not saturated with water vapour.

Question 8.
Read the following paragraph and answer the questions:
If heat is exchanged between a hot and cold object, the temperature of the cold object goes on increasing due to gain of energy and the temperature of the hot object goes on decreasing due to loss of energy.

The change in temperature continues till the temperatures of both the objects attain the same value. In this process, the cold object gains heat energy and the hot object loses’ heat energy. If the system of both the objects is isolated from the environment by keeping it inside a heat resistant box (meaning that the energy exchange takes place between the two objects only), then no energy can flow from inside the box or come into the box.
(1) Heat is transferred from where to where?
(2) Which principle do we learn about from this process?
(3) How will you state the principle briefly?
(4) Which property of the substance is measured using this principle?
Answer:
(1) Heat is transferred from a hot object to a cold object.
(2) This process shows the principle of heat exchange.
(3) In this process, the cold object gains heat energy and the hot object loses energy. If a system of two objects is isolated from the surroundings, heat energy lost by the hot object = heat energy gained by the cold object.
(4) This principle is used to measure the specific heat capacity of a substance.

Question 9.
Solve the following problems:
a. Equal heat is given to two objects A and B of mass 1 g. The temperature of A increases by 3 °C and B by 5°C. Which object has more specific heat? And by what factor?
Solution:
Data: m = 1 g, Δ T₁ = 3 °C, Δ T₂ = 5 °C,
Q same
Here, Q = mc₁ ΔT₁ = mc₂ ΔT₂

Thus, c₁ > c₂
The specific heat of A is more than that of B and

b. Liquid ammonia is used in ice factory for making ice from water. If water at 20 °C is to be converted into 2 kg, ice at 0 °C, how many grams of ammonia is to be evaporated?
(Given: The latent heat of vaporization of 1 ammonia = 341 cal/g)
Solution:
Data : m₁ = 2kg, ΔT₁=20 °C – 0 °C
= 20 °C, c₁ = 1 kcal/kg·°C, L₁ (ice) = 80 kcal/kg,
L₂ (vaporization of ammonia) = 341 cal/g = 341 kcal/kg, m₂ =?
Q₁ (heat lost by water) = m₁c₁ ΔT₁ + m₁L₁
= 2kg × 1 kcal/kg·°C × 20 °C + 2 kg × 80 kcal/kg
=40 kcal + 160 kcal = 200 kcal
Q₂ (heat absorbed by ammonia) = m₂L₂
= m₂ × 34l kcal/kg
According to the principle of heat exchange, Q₁ = Q₂
∴ 200 kcal = m₂ × 341 kcal/kg
∴ m₂ = \(\frac{200}{341}\) kg = 0.5864 kg = 586.4 g
586.4 g of ammonia are to be evaporated.

c. A thermally insulated pot has 150 g ice at temperature 0 °C. How much steam of 100 °C has to he mixed to it, so that water of temperature 50 °C will be obtained?
(Given: Latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g °C)
Solution:
Data: m₁ = 150 g, ΔT₁ = 50 °C – 0 °C
= 50 °C, c_w = 1 cal/g.°C, L₁ = 80 cal/g, L₂ = 540 cal/g,
Δ T₂ = 100°C – 50 °C = 50 °C, m₂ = ?
Q₁ (heat absorbed by ice) = m₁L₁
= 150 g × 80 cal/g = 12000 cal
Q₂ (heat absorbed by water formed on melting of ice) =m₁ c_w ΔT₁
= 150 g × 1 cal/g·°C × 50 °C = 7500 cal
Q₃ (heat given out by steam) = m₂L₂
= m₂ × 540 cal/g
Q₄ (heat given out by water formed on condensation of steam)
= m₂ c_w ΔT₂ = m₂ × 1 cal/g·°C × 50 °C
According to the principle of heat exchange,
Q₁ + Q₂ = Q₃ + Q₄
∴ 12000 cal + 7500 cal = m₂ × 540 cal/g + m₂ × 50 cal/g
∴ 19500 cal = m₂ (540 + 50) cal/g
∴ m₂ = \(\frac{19500}{590}\) g
33.5 g of steam is to be mixed.

d. A calorimeter has mass 100 g and specific heat 0.1 kcal/kg ·°C. It contains 250 g of liquid at 30 °C having specific heat of 0.4 kcal/kg·°C. If we drop a piece of ice of mass 10 g at 0 °C into the liquid, what will be the temperature of the mixture?
Solution:
Data: m₁ = 100 g, c₁ = 0.1 kcal/kg·°C,
= 0.1 cal/g·°C, T₁ = 30 °C, m₂ = 250 g,
c₂ = 0.4 kcal/kg·°C = 0.4 cal/g·°C, T₂ = 30 °C,
m₃ = 10 g, T₃ = 0 °C, L = 80 cal/g,
c (water) = 1 cal/g·°C, T = ?
Q₁ (heat lost by calorimeter) = m₁c₁ (T- T₁),
Q₂ (heat lost by liquid) = m₂c₂ (T – T₂),
Q₃ (heat absorbed by ice) = m₃ L,
Q₄ (heat absorbed by water formed on melting of ice) = m₃c (T – 0 °C)
According to the principle of heat exchange,
Q₁ + Q₂ = Q₃ + Q₄
∴ m₁c₁ (T₁ – T) + m₂c₂ (T₂ – T) = m₃L + m₃c (T – 0 °C)
∴ m₁c₁T₁ – m₁c₁T + m₂c₂T₂ – m₂c₂T = m₃L + m₃c (T – 0°C)
∴ m₁c₁T₁ + m₂c₂T₂ = m₃L + (m₁c₁ + m₂c₂ + m₃c)T
∴ 100g × 0.1 cal/g°C × 30 °C + 250g × 0.4 cal/g.°C × 30 °C J
= 10 g x× 80 cal/g + (100 g × 0.1 cal/g.°C + 250 g × 0.4 cal/g.°C + 10 g × 1 cal/g.°C) T
∴ (10 + 100 + 10) T = (300 + 3000 – 800)°C
∴ 120 T = 2500 °C
∴ T = \(\frac{2500}{120}\) °C = \(\frac{125}{6}\) °C = 20.83 °C
This is the temperature of the mixture.

Project:
Take help of your teachers to make a working model of Hope’s apparatus and perform the experiment. Verify the results you obtain. [Do it your self]

Can you recall? (Text Book Page No. 62)

Question 1.
What is the difference between heat and temperature?
Answer:
Heat is a form of energy. Particles of matter (atoms, molecules, etc.) possess potential energy and kinetic energy. Total energy (potential energy + kinetic energy) of all particles of matter in a given sample is called it’s thermal energy. When two bodies at different temperatures are in thermal contact with each other, there is transfer of thermal energy from a body at higher temperature to a body at lower temperature. This energy in transfer is called heat. It is expressed in joule, calorie and erg.

Temperature is a quantitative measure of degree of hotness or coldness of a body. It is expressed in °C, °F or K (kelvin). Temperature determines the direction of energy transfer.

Question 2.
What are the different ways of heat transfer?
Answer:
Ways of heat transfer: conduction, convection and radiation.
[Note: heat ≡ heat energy. In the textbook, both the terms are used.]

Use your brain power! (Text Book Page No. 63)

Question 1.
Is the concept of latent heat applicable during transformation of gaseous phase to liquid phase and from liquid phase to solid phase?
Answer:
Yes.

Question 2.
Where does the latent heat go during these transformations?
Answer:
During these transformations, the latent heat is given out by the substance to the surroundings.

Use your brain power! (Text Book Page No. 64)

Question 1.
In the above experiment, the wire moves through the ice slab. However, the ice slab does not break. Why?
Answer:
When the thin wire with two equal weights attached to its ends is hung over the block of ice, it exerts pressure on the ice below it. Due to this, the melting point of the ice below the wire is lowered and some ice melts. The wire passes through the water so formed.

The water above the wire is no longer under pressure and, therefore, refreezes. Once again the ice below the wire melts, and the wire passes through it, and the process continues. In this way, due to alternate melting of ice and refreezing of water, the wire cuts right through the block of ice leaving the block intact.

Question 2.
Is there any relationship of latent heat with regelation?
Answer:
Yes. when the ice melts, heat is absorbed, but the temperature does not change. Also, when water refreezes, heat is given out, but the temperature does not change. This heat absorbed or given out is the latent heat.

Question 3.
You know that as we go higher than the sea level, the boiling point of water decreases. What would be the effect on the melting point of a solid?
Answer:
As we go higher than the sea level, the melting point of solids (i) that expand on melting is lowered due to a decrease in pressure (ii) that contract on melting is raised due to a decrease in pressure.

[The wire used in the experiment is made of a metal (usually copper). Metals are good conductors of heat. Hence, exchange of heat between the portion of the ice above the wire and that below the wire takes place readily.]

Can you tell? (Text Book Page No. 64)

Question 1.
We feel that some objects are cold, and some are hot. Is this feeling related in some way to our body temperature?
Answer:
Yes. If the temperature of the object is lower than our body temperature, e.g., ice, we feel the object is cold. If the temperature of the object is higher than our body temperature, e.g., hot water, we feel the object is hot.

Use your brain power! (Text Book Page No. 66)

Question 1.
How will you explain the following statements with the help of the anomalous behaviour of water?
(1) In regions with cold climate, the aquatic plants and animals can survive even when the atmospheric temperature goes below 0 °C.
(2) In cold regions in winter the pipes for water supply break and even rocks crack.
Answer:

In cold regions, during winter, the temperature of the atmosphere falls well below 0 °C. As the temperature decreases, the water at the surfaces of lakes and ponds starts contraeting. Hence, its density increases and it sinks to the bottom. This process continues till the temperature of all the water in a lake falls to 4 °C. As the water at the surface cools further, i.e., its temperature falls below 4 °C, it starts expanding instead of contracting. Therefore, its density decreases and it remains at the surface.

The temperature or the water at the surface continues to fall to 0 °c. Finally, the water at the surface is converted Into ice, but the water below the layer of ice is at 4 °C. Ice is a bad conductor of heat. Hence, the layer of the ice at the surface does not allow transfer of heat from the water to the atmosphere. As the water below the layer of ice remains at 4 °C, fish and other aquatic animals and plants can survive in it.

(2) Sometimes water enters into crevices of the rocks. when the temperature of the atmosphere falls below 4 °C, water expands. Even when water freezes to form ice, there is increase in its volume. As there is no room for expansion, it exerts tremendous pressure on the rocks which crack and break up Into small pieces.

In cold countries, in winter, the temperature of the atmosphere falls below 0 °C. when the temperature of water falls below 4 °C, it expands. Hence, the water in pipes expands. Even if ice is rormed, there is an increase in the volume.

As there is no room for expansion, water (or ice) exerts a large pressure on the pipes. Hence, the pipelines carrying water burst.

Fill in the blanks and rewrite the sentences:

Question 1.
The amount of water vapour in air is determined in terms of its…………..
Answer:
The amount of water vapour in air is determined in terms of its absolute humidity.

Question 2.
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their……………
Answer:
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacities.

Question 3.
When a liquid is getting converted into solid, the latent heat is…………. (Practical Activity Sheet – 1 and 2)
Answer:
When a liquid is getting converted into solid, the latent heat is released.

Rewrite the following statements by selecting the correct options:

Question 1.
……….is used to study the anomalous behaviour of water.
(a) Calorimeter
(b) Joule’s apparatus
(c) Hope’s apparatus
(d) Thermos flask
Answer:
(c) Hope’s apparatus

Question 2.
When water boils and is converted into steam, then………..
(a) heat is taken in and temperature remains constant
(b) heat is taken in and temperatures rises
(c) heat is given out and temperature lowers
(d) heat is given out and temperature remains constant
Answer:
(a) heat is taken in and temperature remains constant

Question 3.
When steam condenses to form water,………..
(a) heat is absorbed and temperature increases
(b) heat is absorbed and temperature remains the same
(c) heat is given out and temperature decreases
(d) heat is given out and temperature remains the same
Answer:
(d) heat is given out and temperature remains the same

Question 4.
The temperature of ice can be decreased below 0 °C by mixing………..in it. (Practice Activity Sheet – 3)
(a) saw dust
(b) sand
(c) salt
(d) coal
Answer:
(c) salt

Question 5.
Ice/water is a substance that………..
(a) expands on melting and contracts on freezing
(b) contracts on melting and does not undergo change in volume on freezing
(c) contracts on melting and expands on freezing
(d) does not undergo any change in volume on melting or freezing
Answer:
(c) contracts on melting and expands on freezing

Question 6.
Heat absorbed when 1 g of ice melts at 0 °C to form 1 g of water at the same temperature is………..cal.
(a) 80
(b) 800
(c) 540
(d) 54
Answer:
(a) 80

Question 7.
The latent heat of vaporization of water is………..
(a) 540 cal/g
(b) 800 cal/g
(c) 80 cal/g
(d) 54 cal/g
Answer:
(a) 540 cal/g

Question 8.
The latent heat of fusion of ice is………..
(a) 540 cal/g
(b) 80 cal/g
(c) 800 cal/g
(d) 4cal/g
Answer:
(b) 80 cal/g

Question 9.
If the temperature of water is decreased from 4 °C to 10 °C, then its………..
(a) volume decreases and density increases
(b) volume increases and density decreases
(c) volume decreases and density decreases
(d) volume increases and density increases
Answer:
(b) volume increases and density decreases

Question 10.
At 4 °C, the density of water is………..
(a) 10 g/cm³
(b) 4g/cm³
(c) 4 × 10³ kg/m³
(d) 1 × 10³ kg/m³
Answer:
(d) 1 × 10³ kg/m³

Question 11.
The density of water is maximum at………..
(a) 0 °C
(b) – 4 °C
(c) 100 °C
(d) 4 °C
Answer:
(d) 4 °C

Question 12.
………..heat is needed to raise the temperature of 1 kg of water from 14.5 °C to 15.5 °C.
(a) 4180 J
(b) 103 J
(c) 1 cal
(d) 4180 cal
Answer:
(a) 4180 J

Question 13.
………..heat is needed to convert 1 g of water at 0 °C and at a pressure of one atmosphere into 1 g of steam under the same conditions.
(a) 80 cal
(b) 540 cal
(c) 89 J
(d) 540 J
Answer:
(b) 540 cal

Question 14.
Water expands on reducing its temperature below………..°C. (March 2019)
(a) 0
(b) 4
(c) 8
(d) 12
Answer:
(b) 4

State whether the following statements are true or false. (If a statement is false, correct it and rewrite it.):

Question 1.
Specific latent heat of fusion is expressed in g/cal.
Answer:
False. (Specific latent heat of fusion is expressed in cal/g.)

Question 2.
If the temperature of water is raised from 0 °C to 10 °C, its volume goes on increasing.
Answer:
False. (If the temperature of water is raised from 0 °C to 10 °C, its volume goes on decreasing in the range 0 °C to 4 °C and then goes on increasing in the range 4 °C to 10 °C.)

Question 3.
At dew point relative humidity is 100%.
Answer:
True.

Question 4.
1 kcal = 4.18 joules.
Answer:
False. (1 kcal = 4180 joules.)

Question 5.
Specific heat capacity is expressed in cal/g·°C
Answer:
True.

Question 6.
Latent heat of fusion, Q = mL.
Answer:
True.

Question 7.
If the relative humidity is more than 60%, we feel that the air is humid.
Answer:
True.

Question 8.
If the relative humidity is less than 60%, we feel that the air is dry.
Answer:
True.

Question 9.
Relative humidity has no unit.
Answer:
True.

Question 10.
Absolute humidity is expressed in kg/m3.
Answer:
True.

Identify the odd one and give the reason:

Question 1.
Temperature, conduction, convection, radiation.
Answer:
Temperature. It is a physical quantity. Others are modes of transfer of heat.

Question 2.
The joule, The erg, The calorie, The newton.
Answer:
The newton. It is a unit of force. Others are units of energy (as well as work.)

Question 3.
cal/g, cal/g·°C, k cal/kg·°C, erg/g·°C.
Answer:
cal/g. It is a unit of specific latent heat. Others are units of specific heat capacity.

Match the columns:

Column A	Column B
1. Latent heat	a. Q = mc ΔT
2. Specific heat capacity	b. Q = mL
3. Heat absorbed or given out by a body when its temperature changes.	c. kcal
	d. cal/g·°C

Answer:
(1) Latent heat – Q = mL
(2) Specific heat capacity – cal/g·°C
(3) Heat absorbed or given out by a body when its temperature changes – Q = mc ΔT.

Answer the following questions in one sentence each:

Question 1.
State units of temperature.
Answer:
Units of temperature: °C, °F and K (kelvin).

Question 2.
State units of energy.
Answer:
Units of energy: the erg, the joule, the calorie.

Question 3.
State the relation between the joule and the calorie.
Answer:
1 calorie = 4.18 joules.

Question 4.
State the relation between the erg and the joule.
Answer:
1 joule = 10⁷ ergs.

Question 5.
State the relation between the erg and the kilocalorie.
Answer:
1 kilocalorie = 4.18 × 10¹⁰ ergs.

Question 6.
State the relation between the joule and the kilocalorie.
Answer:
1 kilocalorie = 4.18 × 10³ joules.

Question 7.
When heat energy is absorbed by an object, ΔT represents the rise in temperature. What would ΔT represent if the object loses heat energy? (Practice Activity Sheet – 4)
Answer:
If the object loses heat energy, ΔT would represent the decrease in temperature.

Answer the following questions:

Question 1.
Define latent heat of fusion.
(OR)
What is latent heat of fusion? State its units.
Answer:
When a solid is converted into liquid at constant temperature (melting point of the substance) the amount of heat absorbed by it is called the latent heat of fusion.
Heat is a form of energy. Hence, latent heat is expressed in units joule, erg, calorie or kilocalorie.

Question 2.
Define specific latent heat of fusion.
(OR)
What is specific latent heat of fusion? State its units.
Answer:
The amount of heat energy absorbed at constant temperature by unit mass of a solid to convert into liquid phase is called the specific latent heat of fusion.
It is expressed in units J/kg, erg/g, cal/g, kJ/ kg and kcal/kg.

[Note: Specific latent heat (L) = \(\frac{\text { latent heat }(Q)}{\text { mass of the substance }(m)}\)
:. SI unit of specific latent heat = SI unit of energy / SI unit of mass = J/kg]

Question 3.
Explain the term latent heat of vaporization.
Answer:
When a liquid is heated continuously, initially, its temperature increases. Later, at a certain stage, its temperature does not increase even when heat is supplied to it. At this temperature, heat absorbed by the liquid is used for breaking the bonds between its atoms or molecules, i.e., for doing work against the forces of attraction between the atoms or molecules and conversion into gaseous phase.

This heat is called the latent heat of vaporization and the constant temperature at which this change of state occurs is called the boiling point of the liquid.

Question 4.
Define boiling point of a liquid.
(OR)
What is boiling point of a liquid?
Answer:
The constant temperature at which a liquid transforms into gaseous state is called the boiling point of the liquid.
[Note: On application of pressure, the boiling point of a liquid is raised. On reducting the pressure, the boiling point is lowered.]

Question 5.
Define specific latent heat of vaporization.
OR
What is specific latent heat of vaporization?
Answer:
The amount of heat energy absorbed at constant temperature by unit mass of a liquid to convert into gaseous phase is called the specific latent heat of vaporization.

Question 6.
The specific latent heat of fusion of ice is 80 cal/g. Explain this statement.
Answer:
When 1 g of ice at a pressure of one atmosphere and at a temperature 0 °C is converted into 1 g of water, heat absorbed by the ice is 80 cal.

Question 7.
The specific latent heat of fusion of silver is 88.2 kJ/kg. Explain this statement.
Answer:
When 1 kg of silver at a pressure of one atmosphere and at a temperature of 962 °C (melting point of silver) is converted into 1 kg of silver in liquid phase, heat absorbed by the silver is 88.2 kJ.

Question 8.
The specific latent heat of vaporization of water is 540 cal/g. Explain this statement.
Answer:
When 1 g of water at a pressure of one atmosphere and at a temperature of 100 °C is converted into 1 g of steam, heat absorbed by the water is 540 cal.

Question 9.
Define regelation.
(OR)
What is regelation?
Answer:
The phenomenon in which the ice converts to liquid due to applied pressure and then re-converts to ice once the pressure is removed is called regelation.

Question 10.
The terms hot and cold are used in relative context. Explain.
Answer:

(1) Take three large bowls, P, Q and R. Fill bowl P with cold water, bowl Q with lukewarm water, and bowl R with hot water.
(2) Immerse your right hand in bowl P, and left hand in bowl R for about five seconds.
(3) Now, immerse both the hands in bowl Q at the same time.
(4) You will find that the water in bowl Q appears warm to your right hand, and cold to your left hand. Thus, the hand immersed in cold water for some time finds the lukewarm water hot while the one immersed in hot water finds the same lukewarm water cold. This experiment shows that the terms hot and cold are relative.

Question 11.
Draw a neat labelled diagram of Hope’s apparatus. Explain how this apparatus can be used to demonstrate anomalous behaviour of water. Draw a graph of temperature of water against time.
Answer:

The figure shows Hope’s apparatus. Initially, the cylindrical container in Hope’s apparatus is filled with water at about 12 °C and the flat bowl is filled with a freezing mixture of ice and salt.

The temperature of water in the upper part of the container (T₂) is recorded by thermometer T₂ and that of water in the lower part of the container (T₁) is recorded by thermometer T₁. Figure shows variation of temperature of water with time.

Initially, both the thermometers show the same temperature (say, 12 °C). In a short time, the temperature shown by the lower thermometer starts decreasing, while the temperature shown by the upper thermometer does not change very much.

This process continues till the temperature shown by the lower thermometer falls to 4 °C and remains constant thereafter. This shows that in the temperature range 12 °C to 4 °C, the density of the water in the central part of the container goes on increasing and hence the water sinks to the bottom. It means that water contracts, i.e., its volume decreases as its temperature falls from 12 °C to 4 °C.

As the temperature of the water in the central part of the container becomes less than 4 °C, the temperature shown by the upper thermometer begins to fall rapidly to 0 °C. But the temperature shown by the lower thermometer remains constant (4 °C). Later, the heading shown by the lower thermometer decreases to 0 °C.

In the temperature range 4 °C to 0 °C, the water moves upward. This shows that the density of water goes on decreasing in this range. It means that water expands, i.e. its volume increases as its temperature falls from 4 °C to 0 °C.

Thus, the volume of a given mass of water is minimum at 4 °C, i.e., the density of water is maximum at 4 °C.

In the above figure, the point of intersection of the two curves shows the temperature at which the density of water is maximum. This temperature is 4 °C.

Question 12.
A mountaineer climbing on the Everest, experienced the following facts. Explain each fact with the scientific reason : (1) He found j fishes alive below the ice (2) Time required for cooking was more as he went higher (3) He saw many times cliffs falling suddenly (4) He saw tubes carrying water broken.
Answer:
Explanation:
(1) Water expands as its temperature decreases from 4 °C to 0 °C. Water is converted into ice at 0 °C. The density of water is more than that of ice. Fishes can remain alive in the water (at 4 °C) below the ice.
(2) At high altitudes, atmospheric pressure is low and hence water boils at a temperature lower than its normal boiling point. Therefore, the time required for cooking food is more at higher altitudes.
(3) Water expands while freezing. Hence, the water present in the crevices of the rocks exerts a tremendous pressure on the rocks, while freezing. Therefore, the cliffs fall.
(4) Water expands while freezing. Hence, the water in the tube exerts a large pressure on the tube, while freezing. Therefore, the tube carrying water breaks.

Question 13.
What is humidity?
Answer:
The moisture, i.e., the presence of water vapour, in the atmosphere is called humidity.

Question 14.
When is air said to be saturated with water vapour?
Answer:
When air contains maximum possible water vapour, it is said to be saturated with water vapour at that temperature.

Question 15.
What does the amount of water vapour needed to saturate air depend on?
Answer:
The extent of water vapour needed to saturate air depends on the temperature. The greater the temperature, the greater is the amount of water vapour needed to saturate air.

Question 16.
When is air said to be unsaturated with water vapour?
Answer:
When air contains water vapour less than its capacity to hold water vapour at that temperature, it is said to be unsaturated with water’vapour.

Question 17.
What is dew point temperature?
(OR)
Define dew point temperature.
Answer:
If the temperature of unsaturated air is decreased, a temperature is reached at which the air becomes saturated with water vapour. This temperature is called the dew point temperature.

Question 18.
Name the physical quantity used to express the amount of water vapour present in air.
Answer:
Absolute humidity.

Question 19.
Define absolute humidity.
(OR)
What is absolute humidity? State its unit.
Answer:
The mass of water vapour present in a unit volume of air is called absolute humidity. Generally it is expressed in kg/m³.

Question 20.
Define relative humidity.
(OR)
What is relative humidity? Write the formula for % relative humidity.
Answer:
The ratio of the actual mass of water vapour content in the air for a given volume and temperature to that required to make the same volume of air saturated with water vapour at the same temperature is called the relative humidity.

% Relative humidity = [the actual mass of water vapour content in the air for a given volume and temperature ÷ the mass of water vapour required to make the same volume of air saturated with water vapour at the same temperature] × 100%.

Question 21.
What is the value of relative humidity at the dew point temperature?
Answer:
At the dew point temperature, relative humidity is 100%.

Question 22.
The mass of water vapour in air enclosed in a certain space is 60 g and the mass of water vapour needed to saturate the same air with water vapour under the same conditions is 100 g. What is the corresponding % relative humidity?
Answer:
Here, % relative humidity = (\(\frac{60 \mathrm{g}}{100 \mathrm{g}}\)) × 100% = 60%

Question 23.
During winter, sometimes we see a white trail at the back of a flying aeroplane in a clear sky. Explain why.
Answer:
In winter, air temperature is low. Hence, when an aeroplane flies, the vapour released by its engine condenses and forms white clouds. If the relative humidity of the air surrounding the plane is high, we see this white trail at the back of the plane for a long time before it disappears. If.the relative humidity is low, the white trail is short and disappears quickly. If the relative humidity is very low, there is no formation of the white trail.

Question 24.
State two effects of humidity present in atmosphere.
Answer:
Effects of humidity present in atmosphere: When the temperature of air falls below the dew point, dew and fog are formed.

Question 25.
Explain how dew and fog are formed.
(OR)
Write a short note on formation of dew and fog.
Answer:
At a particular temperature, a given volume of air can contain a certain maximum amount of water vapour. Normally, the temperature of air during the day is such that air is not saturated with water vapour present in it.

As the temperature falls, the capacity of air to hold water vapour becomes less. During a cold night, the temperature of air may fall to the dew point, or even below the dew point. If the temperature falls below the dew point, the excess of water vapour in air condenses on the surfaces of cold bodies and dew is formed. If the water vapour condenses on the fine dust particles present in the atmosphere, mist or fog is formed.

Question 26.
State the units of heat.
Answer:
Units of heat: joule, erg, calorie, kilocalorie.

Question 27.
Define the kilocalorie.
Answer:
The amount of heat necessary to raise the temperature of 1 kg of water by 1 °C from 14.5 °C to 15.5 °C is called one kilocalorie.

Question 28.
Define the calorie.
Answer:
The amount of heat necessary to raise the temperature of 1 g of water by 1 °C from 14.5 °C to 15.5 °C is called one calorie.

Question 29.
State the relation between the kilocalorie and the calorie.
Answer:
1 kilocalorie = 10³ calories.

Question 30.
Study the following procedure and answer the questions below:  (Practice Activity Sheet – 2)
1. Take 3 spheres of iron, copper and lead of equal mass.
2. Put all the 3 spheres in boiling water in a beaker for some time.
3. Take 3 spheres out of the water. Put them immediately on a thick slab of wax.
4. Note the depth that each sphere goes into the wax.
(i) Which property of a substance can be studied with this procedure?
(ii) Describe that property in minimum words.
(iii) Explain the rule of heat exchange with this property.
Answer:
(i) Specific heat.
(ii) Specific heat: The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C.
(iii) According to the rule/principle of heat exchange, heat energy lost by the hot object = heat energy gained by the cold object.

In this activity, heat absorbed by the iron sphere is transmitted more in the wax, hence the sphere goes deepest into the wax, while the lead sphere absorbs less heat, resulting in less transmission of heat in the wax, hence, the sphere goes the least depth into the wax.

Question 31.
Write the symbol for specific heat capacity. State the units of specific heat capacity.
Answer:
Symbol for specific heat capacity: c.
Units of specific heat capacity: J/kg·°C,
erg/g·°C, cal/g·°C, kcal/kg·°C.
[ Notes: (1) Specific heat capacity

In SI, heat is expressed in joule (J), mass in kg and temperature in kelvin(K).
∴ SI unit of specific heat capacity = \(\frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}\). (2) The specific heat capacity of a substance depends upon its constituent particles (atoms, molecules, etc.), interaction between them, structure of the substance (atomic/molecular arrangement), temperature of the substance, etc.]

Question 32.
Explain the principle of heat exchange. Ans. Suppose two objects A and B at different temperature T₁ and T₂ respectively are enclosed in a box of heat resistant material as shown in figure.

Let m₁ = mass of A, m₂ = mass of B, c₁ = specific heat capacity of A, c₂ = specific heat capacity of B and T = common temperature attained by A and B after the heat exchange between A and B. Here, no heat leaves the box or enters the box from outside. Hence, if T₁ > T₂, heat energy lost by A (Q₁) = heat energy gained by B (Q₂).
∴ m₁c₁ (T₁ – T) = m₂c₂ (T – T₂)
[Note: If m₁, c₁, T₁, T, m₂ and T₂ are known, c₂ can be determined.]

Question 33.
The specific heat capacity of silver is 0.056 kcal/kg·°C. Explain this statement.
Answer:
The amount of heat needed to raise the temperature of 1 kg of silver by 1 °C is 0.056 kcal.

Question 34.
Explain how the specific heat capacity of a solid can be determined (measured) by the method of mixture.
Answer:
A hot solid is put in water in a calorimeter. The mixture is stirred continuously and the maximum temperature of the mixture is measured with a thermometer. Heat exchange between the hot solid, water and calorimeter results in sill bodies attaining the same temperature after some time. Hence, according to the principle of heat exchange, heat lost by the solid = heat gained by the water in the calorimeter + heat gained by the calorimeter.

Now, heat lost by the solid (Q) = mass of the solid × its specific heat capacity × decrease in its temperature, heat gained by the water (Q₁) = mass of the water × its specific heat capacity × increase in its temperature and heat gained by the calorimeter (Q₂) = mass of the calorimeter × its specific heat capacity × increase in its temperature.

Heat lost by the hot object = heat gained by the calorimeter + heat gained by the water. Q = Q₂ + Q₁
Using this equation, the specific heat capacity of the solid can be determined (measured) when the other quantities are known.

Give scientific reasons:

Question 1.
Even though heat is supplied to boiling water, there is no increase in its temperature.
Answer:
Once water starts boiling, all the heat supplied to it is used in conversion of water into steam at the boiling point of water. Hence, there is no rise in its temperature.

Question 2.
Burns from steam are worse those from boiling water at the same temperature.
Answer:
1. A given quantity of steam contains more heat than the same quantity of boiling water at the same temperature.
2. When steam comes in contact with one’s body, it releases extra heat of 540 calories per gram and causes a more serious burn than that caused by boiling water.

Question 3.
In winter, the pipelines carrying water burst in cold countries.
Answer:
1. In cold countries, in winter, the temperature of the atmosphere falls below 0 °C. When the temperature of water falls below 4 °C, it expands. Hence, the water in pipes expands. Even if ice is formed, there is an increase in the volume.

2. As there is no room for expansion, water (or ice) exerts a large pressure on the pipes. Hence, the pipelines carrying water burst.

Question 4.
If crushed ice is pressed and then the pressure is released, a lump of ice is formed.
Answer:
1. When crushed ice is pressed, its melting point is lowered and some ice melts to form water.
2. When pressure is released, the melting point becomes normal and the water freezes to form ice forming a lump.

Question 5.
In cold countries, in winter, even when the water of lakes freezes, aquatic animals and plants can survive.
Answer:
1. In cold countries, in winter, a layer of ice is formed on the surface of lakes when the atmospheric temperature falls below 0 °C. However, below this layer, there is water at 4 °C.
2. Ice, being a bad conductor of heat, does not allow transfer of heat from this water to the atmosphere. Hence, aquatic animals and plants can survive in this water.

Question 6.
Water droplets are seen on’ the outer surface of a cold drink bottle.
Answer:
1. The temperature of the outer surface of a cold drink bottle is less than that of the atmosphere.
2. Therefore, the excess of water vapour from the air condenses to form droplets on the outer surface of the cold drink bottle.

Question 7.
During cold nights, sometimes dew is formed.
Answer:
1. During a cold night, the temperature of air may fall to the dew point, or even below the dew point. 2. If the temperature falls below the dew point, the excess of water vapour in air condenses on the surfaces of cold bodies and dew is formed.

Question 8.
When you enter a warm room after being outside on a frosty early morning, your spectacles ‘steam up’.
Answer:
1. On a frosty early morning, the temperature of air outside a warm room is lower than the dew point.
2. Hence, when you enter the room from outside, some water vapour in the room condenses on the glass of your spectacles, i.e., the spectacles ‘steam up’.

Question 9.
A plastic bottle, completely filled with water, when kept in a freezer, is likely to break.
Answer:
The temperature of air in the freezer (deep freeze) compartment of a refrigerator is less than 0 °C. 2. When a plastic bottle, completely filled with water, is kept in this compartment, the temperature of water falls below 4 °C and the water expands. Even when water freezes and ice is formed, there is an increase in the volume. It exerts a large pressure on the sides of the bottle and hence the bottle is likely to break.

Question 10.
The outer surface of a beaker containing ice cubes becomes wet in a short while.
Answer:
1. When ice cubes are placed in a beaker, ice starts melting. The heat required for melting is absorbed from the surrounding air and also from the beaker to some extent.
Hence, the temperature of the air and beaker falls.

2. The capacity of air to hold water vapour depends upon the temperature of the air, and this capacity decreases as the temperature decreases. At a certain low temperature, the surrounding air becomes saturated with water vapour present in it. As the temperature falls further, the air is unable to hold all the water vapour.

Hence, the extra water vapour starts condensing on the cold outer surface of the beaker in the form of minute drops. Therefore, the outer surface of the beaker containing ice cubes becomes wet in a short while.

Distinguish between the following:

Question 1.
Absolute humidity and Relative humidity.
Answer:
Absolute humidity:

1. Absolute humidity is the mass of water vapour present in a unit volume of air.
2. It is commonly expressed in kg/m³.

Relative humidity:

1. Relative humidity is the ratio of the mass of water vapour in a given volume of air at a given temperature to the mass of water vapour required to saturate the same volume of air at the same temperature.
2. It does not have unit.

Solve the following examples/Numerical problems:
[Use the data given in the Tables on pages 130 and 131.]

Problem 1.
Calculate the amount of heat required to convert 5 g of ice of 0 °C into water at 0 °C. (Specific latent heat of fusion of ice = 80 cal/g)
Solution: Here, m = 5 g, L = 80 cal/g; Q = ?
Amount of heat required, Q = mL
= 5 g × 80 cal/g
= 400 calories.

Problem 2.
Find the amount of heat required to convert 10 g of water at 100 °C into steam. (Specific latent heat of vaporization of water = 540 cal/g)
Solution: Here, m = 10 g, L = 540 cal/g; Q = ?
Amount of heat required, Q = mL
= 10 g × 540 cal/g
= 5400 calories.

Problem 3.
Calculate the amount of heat required to convert 15 g of water at 100 °C into steam. (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
m = 15 g, L = 540 cal/g; Q = ?
Amount of heat required, Q = mL
= 15 g × 540 cal/g
= 8100 calories.

Problem 4.
How many calories of heat will be absorbed when 3 kg of ice at 0 °C melts?
Solution:
m = 3 kg = 3000 g; L = 80 cal/g; Q = ?
Quantity of heat absorbed, Q = mL
= 3000 g × 80 cal/g
∴ Q = 240000 calories.

Problem 5.
Calculate the amount of heat required to convert 10 g of water at 30 °C into steam at 100 °C. (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
Here, m = 10 g; c = 1 cal/g·°C
T₂ – T₁ = 100 °C – 30 °C = 70 °C; L = 540 cal/g; Q = ?
Amount of heat required, Q = mc (T₂ – T₁) + mL
= 10 g × 1 cal/g·°C × 70 °C + 10 g × 540 cal/g
= 700 cal + 5400 cal
∴ Q = 6100 calories.

Problem 6.
If water of mass 80 g and temperature 45 °C is mixed with water of mass 20 g and temperature 30 °C, what will be the maximum temperature of the mixture?
Solution:
Data : m₁ = 80 g, T₁ = 45 °C, m₂ = 20 g,
T₂ = 30 °C, T = ?
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
∴ m₁c (T1 – T) = m₂c (T – T₂)
∴ m₁T₁ – m₁T = m₂T – m₂T₂
∴ m₁T₁ + m₂T₂ = (m1 + m₂)T
∴ Maximum temperature of the mixture,

= (36 + 6) °C
= 42°C.

Problem 7.
When water of mass 70 g and temperature 50 °C is added to water of mass 30 g, the maximum temperature of the mixture is found to be 41 °C. Find the temperature of water of mass 30 g before hot water was added to it.
Solution:
Data : m₁ = 70 g, T₁ = 50 °C, m₂ = 30 g, T = 41 °C, T₂ = ?
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
∴ m₁c (T₁ – T) = m₂c (T – T₂)
∴ m₁T₁ – m₁T = m₂T – m₂T₂
∴ m₂T₂ = (m₁ + m₂) T – m₁T₁

This is the required temperature.

Problem 8.
Find the heat needed to raise the temperature of a silver container of mass 100 g by 10 °C. (c = 0.056 cal/g.°C)
Solution:
Data: m = 100 g, ΔT = 10 °C, c = 0.056 cal/g·°C
Heat needed to raise the temperature of the container = mc ΔT
= 100 g × 0.056 cal/g·°C × 10 °C .
= 56 calories.

Problem 9.
If steam of mass 100 g and temperature 100 °C is released on an ice slab of temperature 0 °C, how much ice will melt?
Solution:
Data: m₁ = 100 g, L₁ = 540 cal/g,
T₁ = 100 °C, mass of ice, m = ?, L₂ = 80 cal/g, c (water) = 1 cal/g·°C
According to the principle of heat exchange, heat lost by hot body = heat gained by cold body. Conversion of steam into water:
Q₁ = m₁L₁ = 100 g × 540 cal/g = 54000 cal
Decrease in the temperature of this water to 0 °C:
Q₂ = m₁c × (T₁ – 0 °C) = 100 g × 1 cal/g·°C × (100 °C – 0 °C) = 10000 Cal
Melting of ice: Q₃ = mL₂
= m × 80 cal/g
Now, Q₁ + Q₂ = Q₃
∴ (54000 + 10000) cal = m × 80 cal/g
∴ m = \(\frac{64000}{80}\) = 800 g
800 g of ice will melt.

Numerical problems for practice:

Problem 1.
Calculate the amount of heat required to convert 80 g of ice at 0 °C into water at the same temperature. (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
6400 cal

Problem 2.
Find the heat required to convert 20 g of ice at 0 °C into water at the same temperature. (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
1600 cal

Problem 3.
Calculate the quantity of heat released during the conversion of 10 g of ice cold water (temperature 0 °C) into ice at the same temperature. (Specific latent heat of freezing of water = 80 cal/g)
Solution:
800 cal

Problem 4.
How many calories of heat will be absorbed when 2 kg of ice at 0 °C melts? (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
160000 cal

Problem 5.
How much heat will be required to convert 20 g of water at 100 °C into steam at 100 °C? (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
10800 cal

Problem 6.
Find the heat absorbed by 25 g of water at 100 °C to convert into steam at the same temperature. (Specific latent heat of vaporization of water = 540 cal/g.)
Solution:
13500 cal

Problem 7.
If water of mass 60 g and temperature 50 °C is mixed with water of mass 40 g and temperature 30 °C, what will be the maximum temperature of the mixture?
Solution:
42 °C

Problem 8.
If water of mass 60 g and temperature 60 °C is mixed with water of mass 60 g and temperature 40 °C, what will be the maximum temperature of the mixture?
Solution:
50 °C

Problem 9.
Find the heat needed to raise the temperature of a piece of iron of mass 500 g by 20 °C. (c = 0.110 cal/g·°C)
Solution:
1100 cal

Problem 10.
Water of mass 200 g and temperature 30 °C is taken in a copper calorimeter of mass 50 g and temperature 30 °C. A copper sphere of mass 100 g and temperature 100 °C is released into it. What will be the maximum temperature of the mixture? [c (water) = 1 cal/g·°C, c (copper) =0.1 cal/g·°C]
Solution:
33.26 °C

Problem 11.
A copper calorimeter of mass 100 g and temperature 30 °C contains water of mass 200 g and temperature 30 °C. If a piece of ice of mass 40 g and temperature 0 °C is added to it, what will be the maximum temperature of the mixture? [c (copper) = 0.1 cal/g·°C, c (water) = 1 cal/g·°C, L = 80 cal/g]
Solution:
12.4 °C