Class 10

Punjab State Board PSEB 10th Class Hindi Book Solutions Chapter 3 नीति के दोहे Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Hindi Chapter 3 नीति के दोहे

Hindi Guide for Class 10 PSEB नीति के दोहे Textbook Questions and Answers

(क) विषय-बोध

I. निम्नलिखित प्रश्नों के एक-दो पंक्तियों में उत्तर दीजिए

प्रश्न 1.
रहीम जी के अनुसार सच्चे मित्र की क्या पहचान है?
उत्तर:
रहीम जी के अनुसार जो व्यक्ति मुसीबत में काम आता है, वही सच्चा मित्र होता है।

प्रश्न 2.
ज्ञानी व्यक्ति संपत्ति का संचय किस लिए करते हैं?
उत्तर:
ज्ञानी व्यक्ति संपत्ति का संचय परोपकार अथवा दूसरों की भलाई के लिए करते हैं।

प्रश्न 3.
बिहारी जी के अनुसार किस का साथ शोभा देता है?
उत्तर:
बिहारी जी के अनुसार एक जैसे स्वभाव अथवा प्रकृति वालों का साथ शोभा देता है।

प्रश्न 4.
बिहारी जी ने मानव को आशावादी होने का क्या संदेश दिया है?
उत्तर:
बिहारी जी ने संदेश दिया है कि मनुष्य को निराश न हो कर आने वाले अच्छे दिनों के लिए आशावादी होना चाहिए।

प्रश्न 5.
छल और कपट का व्यवहार बार-बार नहीं चल सकता-इस के लिए वृन्द जी ने क्या उदाहरण दिया है?
उत्तर:
वृन्द जी के अनुसार जैसे काठ की हांडी बार-बार नहीं चढ़ती उसी प्रकार छल और कपट का व्यवहार भी बार-बार नहीं चल सकता।

प्रश्न 6.
निरंतर अभ्यास से व्यक्ति कैसे योग्य बन जाता है? वृन्द जी ने इस के लिए क्या उदाहरण दिया है?
उत्तर:
जैसे बार-बार रस्सी घिसने से पत्थर पर भी निशान पड़ जाते हैं, वैसे ही निरंतर अभ्यास से अयोग्य व्यक्ति भी योग्य बन जाता है।

प्रश्न 7.
शत्रु को कमज़ोर या छोटा क्यों नहीं समझना चाहिए?
उत्तर:
शत्रु को कमज़ोर या छोटा नहीं समझना चाहिए क्योंकि शत्रु के प्रति लापरवाही बहुत हानि पहुँचा सकती है जैसे तिनकों के बड़े ढेर को आग का छोटा-सा अंगारा क्षण भर में जला कर राख कर देता है।

II. निम्नलिखित पद्यांशों की सप्रसंग व्याख्या करें

(1) रहिमन देखि बड़ेन को, लघु न दीजिये डारि।
जहां काम आवे सुई, का करे तरवारि।।
उत्तर:
रहीम जी कहते हैं कि किसी बड़े व्यक्ति अथवा वस्तु को देखकर हमें छोटे व्यक्ति अथवा वस्तु को छोड़ नहीं देना चाहिए अथवा उसका तिरस्कार नहीं करना चाहिए क्योंकि आवश्यकता के समय जहाँ सुई काम आती है, वहाँ तलवार किसी काम नहीं आती।

(2) कनक-कनक ते सौ गुनी, मादकता अधिकाय।
वह खाये बौराय है, यह पाये बौराय।
उत्तर:
महाकवि बिहारी कहते हैं कि धतूरे की तुलना में स्वर्ण में सौगुना अधिक नशा होता है क्योंकि धतूरे को खाने पर नशा होता है जबकि सोने के प्राप्त होने पर ही नशा हो जाता है। जो नशा धतूरा खाने पर होता है, उससे कहीं अधिक नशा धन-वैभव के प्राप्त होने पर होता है।

(3) मधुर वचन ते जात मिट, उत्तम जन अभिमान।
तनिक सीत जल सो मिटे, जैसे दूध उफान।।
उत्तर:
कवि कहता है कि मधुर वचनों अथवा मीठी वाणी के बोलों से किसी भी अभिमानी व्यक्ति के गर्व को उसी प्रकार से शांत किया जा सकता है जैसे थोड़े से ठंडे पानी के छींटों से उबलते हुए दूध के उफ़ान को कम कर लिया जाता है।

(ख) भाषा-बोध

निम्नलिखित शब्दों के विपरीत शब्द लिखें:

सम्पत्ति = ———–
उत्तम = ———–
हित = ———–
आशा = ———–
बैर। = ———–
उत्तर:
शब्द – विपरीत शब्द
संपति – विपत्ति
उत्तम – अधम
हित – अहित।
आशा – निराशा
बैर – मिलाप।

निम्नलिखित शब्दों के विशेषण शब्द बनाएं:

प्रकृति = ———–
विष = ———–
बल = ———–
मूल = ———–
हित = ———–
व्यापार। = ———–
उत्तर:
शब्द विशेषण
शब्द विशेषण शब्द विशेषण
प्रकृति – प्राकृतिक
विष – विषैला
बल – बलवान
मूल – मूलभूत
हित – हितैषी
व्यापार। – व्यापारिक।

निम्नलिखित शब्दों की भाववाचक संज्ञा बनाएं:

लघु = ———–
मादक = ———–
एक = ———–
मधुर। = ———–
उत्तर:
शब्द – भाववाचक संज्ञा
लघु – लघुता
मादक – मादकता
एक – एकता
मधुर – मधुरता।

(ग) पाठ्येतर सक्रियता

प्रश्न 1.
अध्यापक महोदय उपर्युक्त नीति के दोहों पर आधारित शिक्षाप्रद कहानियाँ छात्र/छात्राओं को सुनाएं और उनसे भी इस प्रकार की कोई सच्ची घटना अथवा कहानी सुनाने के लिए कहें।
उत्तर:
(विद्यार्थी स्वयं करें।)

प्रश्न 2.
छात्र-छात्राएँ इस प्रकार के अन्य दोहों का संकलन कर विद्यालय की भित्ति पत्रिका पर लगाएँ।
उत्तर:
(विद्यार्थी स्वयं करें।)

प्रश्न 3.
कक्षा में ‘दोहा गायन प्रतियोगिता’ में सक्रिय रूप से भाग लें।
उत्तर:
(विद्यार्थी स्वयं करें)

प्रश्न 4.
रहीम अथवा अन्य कवियों के द्वारा रचित दोहों की ऑडियो/वीडियो सी०डी० लेकर अथवा इंटरनेट के माध्यम से सुनें/देखें। उत्तर:
(विद्यार्थी स्वयं करें)

(घ) ज्ञान विस्तार

रहीम, बिहारी तथा वृंद ने अपने काव्य की रचना दोहा छंद में की है, जिसके पहले तथा तीसरे चरण में 13-13 तथा दूसरे तथा चौथे चरण में 11-11 मात्राएँ होती हैं, जैसे-

यहाँ पहली तथा तीसरी में 13-13 और दूसरी तथा चौथी में 11-11 मात्राएँ हैं, इसलिए दोहा छंद हुआ। (I = एक मात्रा, s = दो मात्राएँ)

PSEB 10th Class Hindi Guide नीति के दोहे Important Questions and Answers

प्रश्न 1.
रहीम ने सब को साधने का क्या उपाय बताया है?
उत्तर:
रहीम जी मानते हैं कि एक की साधना पूरी तरह करने से सब सध जाते हैं तथा मनुष्य को अपना लक्ष्य भी प्राप्त हो जाता है।

प्रश्न 2.
सुई के महत्त्व से रहीम जी क्या कहना चाहते हैं?
उत्तर:
कवि यह कहना चाहते हैं कि संसार में कोई भी वस्तु महत्त्वहीन नहीं होती है। छोटी-से छोटी वस्तु का भी अपना महत्त्व होता है। इसलिए कुछ बड़ा पा कर छोटे को त्यागना नहीं चाहिए क्योंकि जो काम छोटी-सी सुई कर सकती है वह काम तलवार नहीं कर सकती।

प्रश्न 3.
बिहारी के अनुसार किस का नशा नशीले पदार्थ के सेवन से भी अधिक होता है?
उत्तर:
बिहारी जी के अनुसार सोने अर्थात् धन-संपत्ति का नशा नशीले पदार्थ से भी सौ गुणा अधिक होता है क्योंकि जिस के पास धन-संपत्ति आ जाती है वह उसी के नशे में अहंकारी बन जाता है।

प्रश्न 4.
बिहारी के अनुसार गुणवान कौन होता है?
उत्तर:
बिहारी जी का मानना है कि किसी गुणहीन व्यक्ति को बार-बार गुणी-गुणी कहते रहने से वह गुणवान नहीं बन जाता क्योंकि सच्चा गुणी तो वही होता है जिसमें स्वाभाविक रूप से सद्गुण होते हैं।

प्रश्न 5.
वृन्द जी के अनुसार मधुर वाणी के क्या लाभ हैं?
उत्तर:
वृन्द जी के अनुसार मीठे वचनों का प्रयोग करने से क्रोधी तथा अभिमानी व्यक्ति के क्रोध एवं अहंकार को उसी प्रकार शांत कर सकते हैं जैसे उबलते हुए दूध के उफान को ठंडे जल के छींटे मारने से शांत किया जाता है।

एक पंक्ति में उत्तरात्मक प्रश्न

प्रश्न 1.
रहीम के अनुसार लोग कब हमारे सगे बन जाते हैं?
उत्तर:
जब हमारे पास सम्पत्ति होती है।

प्रश्न 2.
बिहारी ने किस में धतूरे से भी अधिक मादकता बताई है?
उत्तर:
बिहारी ने सोने में धतूरे से भी अधिक मादकता बताई है।

प्रश्न 3.
काजल की शोभा कहाँ होती है?
उत्तर:
काजल आँखों में सुशोभित होता है।

प्रश्न 4.
मधुर वचनों से क्या लाभ हैं?
उत्तर:
मधुर वचनों से क्रोधी के क्रोध को शांत तथा अभिमानी के गर्व को शांत किया जा सकता है।

बहुवैकल्पिक प्रश्नोत्तरनिम्नलिखित प्रश्नों के उत्तर एक सही विकल्प चुनकर लिखें

प्रश्न 1.
सज्जन किसके लिए धन एकत्र करते हैं?
(क) परहित
(ख) स्वहित
(ग) राजहित
(घ) परलोक हित।
उत्तर:
(क) परहित

प्रश्न 2.
गुलाब के मूल से कौन अटका रहता है?
(क) कलि
(ख) अलि
(ग) कली
(घ) आली।
उत्तर:
(ख) अलि

प्रश्न 3.
अरक का अर्थ है
(क) चंद्रमा
(ख) तारे
(ग) दीपक
(घ) सूर्य।
उत्तर:
(घ) सूर्य

एक शब्द/हाँ-नहीं/सही-गलत/रिक्त स्थानों की पूर्ति के प्रश्न

प्रश्न 1.
ठंडे जल के छींटे से किसका उफान मिट जाता है? (एक शब्द में उत्तर दें)
उत्तर:
दूध का

प्रश्न 2.
सोने को पाकर ही मनुष्य बौरा जाता है। (सही या गलत में उत्तर लिखें।)
उत्तर:
सही

प्रश्न 3.
जहाँ तलवार काम आती है वहाँ सुई भी काम आती है। (सही या गलत में उत्तर लिखें)
उत्तर:
गलत

प्रश्न 4.
वृक्ष अपने फल स्वयं खाते हैं। (हाँ या नहीं में उत्तर दें)
उत्तर:
नहीं

प्रश्न 5.
शत्रु को कम समझने से हानि होती है। (हाँ या नहीं में उत्तर दें)
उत्तर:
हाँ

प्रश्न 6.
जैसे ……….. काठ की, चढ़ें न ………. बार।
उत्तर:
हाँडी, दूजी

प्रश्न 7.
सोहतु संगु ……… सों, यहै कहै ……… लोग।
उत्तर:
समानु, सब

प्रश्न 8.
रहिमन देखि ………… को, लघु न ………… डारि।
उत्तर:
बड़ेन, दीजिये।

नीति के दोहे दोहों की सप्रसंग व्याख्या

1. कहि रहीम सम्पति सगे, बनत बहुत बहु रीत।
विपत कसौटी जे कसे, सोई साँचै मीत।।

शब्दार्थ:
कहि = कहते हैं। सम्पति = धन-दौलत। सगे = सगे-संबंधी। बनत = बनते हैं। बहु = अनेक। रीति = प्रकार। विपत = मुसीबत। जे = जो। कसौटी कसे = कसौटी पर खरा उतरना। सोई = वही। साँचे = सच्चा। मीत = मित्र।

प्रसंग:
प्रस्तुत दोहा रहीम द्वारा रचित ‘नीति के दोहे’ पाठ में से लिया गया है। इस दोहे में कवि ने सच्चे मित्र के लक्षण बताये हैं।

व्याख्या:
रहीम जी कहते हैं कि जब पास में धन-दौलत होती है तो अनेक प्रकार से लोग हमारे सगे-संबंधी-मित्र आदि बन जाते हैं परंतु जो मुसीबत रूपी कसौटी पर कसे जाने के समय साथ देता है वही सच्चा मित्र होता है।

विशेष:

1. कठिनाई में काम आने वाले ही सच्चा मित्र होता है।
2. भाषा सरल, सरस, सहज और दोहा छंद है। अनुप्रास अलंकार है।

2. एकै साधे सब सधै, सब साधै सब जाय।
रहिमन सींचे मूल को, फूलै फलै अधाय।।

शब्दार्थ:
मूल = जड़। फूलै = फूल आना। फलै = फल आना। अधाय = तृप्त होना।

प्रसंग:
प्रस्तुत दोहा रहीम जी द्वारा रचित ‘नीति के दोहे’ में से लिया गया है। इस दोहे में कवि ने एकनिष्ठ भाव से एक की आराधना करने पर बल दिया है।

व्याख्या:
कवि कहता है कि एकनिष्ठ भाव से एक ईश्वर की आराधना करने से सब को साध लिया जाता है जबकि सब की साधना करने से सभी हाथ नहीं आते अथवा सब कुछ नष्ट हो जाता है। रहीम जी उदाहरण देकर समझाते हैं कि जैसे किसी वृक्ष की जड़ को सींचने से वह फलता-फूलता है तथा उसके फलों को खा कर सब तृप्त हो जाते हैं।

विशेष:

1. अपना ध्यान एक लक्ष्य की ओर केंद्रित करने से ही सफलता की प्राप्ति होती है।
2. भाषा सहज, सरल, भावानुकूल तथा दोहा छंद है। अनुप्रास अलंकार है।

3. तरुवर फल नहिं खात हैं, सरवर पियहिं न पान।
कहि रहीम पर काज हित, सम्पति संचहिं सुजान।

शब्दार्थ:
तरूवर = वृक्ष, पेड़। खात = खाना। सरवर = तालाब। पान = पानी। परकाज = परोपकार, दूसरे की भलाई। हित = के लिए। संचहिं = एकत्र करना। सुजान = अच्छे लोग, सज्जन।

प्रसंग:
प्रस्तुत दोहा रहीम जी द्वारा रचित ‘नीति के दोहे’ से लिया गया है। इस दोहे में कवि ने परोपकार की महत्ता पर प्रकाश डाला है।

व्याख्या:
कवि कहता है कि वृक्ष कभी भी अपने फल नहीं खाता और तालाब भी अपना जल कभी नहीं पीता। रहीम जी कहते हैं कि दूसरों की भलाई के लिए ही सज्जन धन-दौलत एकत्र करते हैं।

विशेष:

1. मनुष्य को अपनी धन-संपत्ति का सदुपयोग दूसरों की भलाई के लिए करना चाहिए।
2. भाषा सरल, सरस, दोहा छंद तथा अनुप्रास अलंकार है।

4. रहिमन देखि बड़ेन को, लघु न दीजिये डारि।
जहाँ काम आवे सुई, का करे तरवारि।।

शब्दार्थ:
लघु = छोटा, तुच्छ। तरवारि = तलवार।

प्रसंग:
प्रस्तुत दोहा रहीम जी द्वारा रचित ‘नीति के दोहे’ से लिया गया है। इस दोहे में कवि ने कहा है कि हमें छोटी वस्तु के महत्त्व को भी अनदेखा नहीं करना चाहिए।

व्याख्या:
रहीम जी कहते हैं कि किसी बड़े व्यक्ति अथवा वस्तु को देखकर हमें छोटे व्यक्ति अथवा वस्तु को छोड़ नहीं देना चाहिए अथवा उसका तिरस्कार नहीं करना चाहिए क्योंकि आवश्यकता के समय जहाँ सुई काम आती है, वहाँ तलवार किसी काम नहीं आती।

विशेष:

1. हमें बड़ी वस्तु अथवा संपन्न व्यक्ति को देखकर छोटी वस्तु अथवा निर्धन व्यक्ति की उपयोगिता को भूलना नहीं चाहिए। दोनों का सम्मान करना चाहिए।
2. भाषा सरल, भावानुरूप, दोहा छंद तथा अनुप्रास अलंकार है।

5. कनक-कनक ते सौगुनी, मादकता अधिकाय।
बह खाये बौरात है, इहिं पाये बौराय।।

शब्दार्थ:
कनक = धतूरा, सोना। मादकता = नशा। बौराय = पागल हो जाता है।

प्रसंग:
प्रस्तुत दोहा बिहारी द्वारा रचित ‘नीति के दोहे’ से लिया गया है जिसमें कवि ने यह बताया है धन और वैभव का नशा मनुष्य को पागल बना देता है। धनी व्यक्ति नशा करने वाले व्यक्ति की तरह व्यवहार करने लगता है।

व्याख्या:
महाकवि बिहारी कहते हैं कि धतूरे की तुलना में स्वर्ण में सौगुना अधिक नशा होता है क्योंकि धतूरे को खाने पर नशा होता है जबकि सोने के प्राप्त होने पर ही नशा हो जाता है। जो नशा धतूरा खाने पर होता है, उससे कहीं अधिक नशा धन-वैभव के प्राप्त होने पर होता है।

विशेष:

1. धन-संपत्ति पाकर मनुष्य अहंकारी बन जाता है।
2. दोहा छंद, ब्रज भाषा तथा यमक अलंकार है।

6. इहि आशा अटक्यो रहै, अलि गुलाब के मूल।
हो इहै बहुरि बसन्त ऋतु, इन डारनि पै फूल॥

शब्दार्थ:
इहि = इस। अलि = भँवरा। मूल = जड़। होइहै = हो जाएगी। बहरि = फिर। _प्रसंग-प्रस्तुत दोहा बिहारी द्वारा रचित ‘नीति के दोहे’ से लिया गया है। इस दोहे में कवि ने मनुष्य को सदा आशावादी रहने का संदेश दिया है।

व्याख्या:
कवि कहता है कि फूल न होने पर भी इस उम्मीद से भँवरा गुलाब की जड़ के पास रहता है कि फिर से बसंत ऋतु आएगी और इन डालियों पर फूल खिल जाएँगे। भाव यह है कि मनुष्य को कभी निराश नहीं होना चाहिए क्योंकि दुःख के बाद सुख आता ही है।

विशेष:

1. कवि ने मनुष्य को अपना कर्म करते हुए निरन्तर आशावादी बने रहने का संदेश दिया है।
2. ब्रज भाषा, दोहा छंद तथा अनुप्रास अलंकार है।

7. सोहतु संगु समानु सो, यहै कहै सब लोग।
पान पीक ओठनु बनैं, नैननु काजर जोग।।

शब्दार्थ:
सोहतु = शोभामान होता है। संगु = साथ। यहै = यही। काजर = काजल।

प्रसंग:
प्रस्तुत दोहा बिहारी द्वारा रचित ‘नीति के दोहे’ से लिया गया है। इस दोहे में कवि बताता है कि एक जैसे स्वभाव वालों का साथ सदा बना रहता है।

व्याख्या:
कवि कहता है कि एक जैसे स्वभाव वाले लोगों का साथ ही सदा रहता है ऐसा ही सब लोग भी कहते हैं क्योंकि पानी की पीक की लालिमा सदा ओंठों पर तथा काजल आँखों में सुशोभित होता है। पान की पीक की लालिमा ओंठों के लिए तथा काजल आँखों के लिए बना है।

विशेष:

1. समान प्रकृति तथा स्वभाव के व्यक्तियों का साथ सदा बना रहता है।
2. ब्रज भाषा, दोहा छंद तथा अनुप्रास अलंकार है।

8. गुनी गुनी सबकै कहैं, निगुनी गुनी न होतु।
सुन्यौ कहूँ तरू अरक तें, अरक-समान उदोतु॥

शब्दार्थ:
गुनी = गुणवान। कहैं = कहने से। निगुनी = गुणहीन। तरू = वृक्ष । अरकतें = आक के। अरक-समान = सूर्य के समान। उदोतु = प्रकाशवान।

प्रसंग:
प्रस्तुत दोहा बिहारी द्वारा रचित ‘नीति के दोहे’ से लिया गया है। इस दोहे में कवि ने स्पष्ट किया है कि कहने मात्र से ही गुणहीन व्यक्ति गुणवान नहीं हो सकता।

व्याख्या:
कवि कहता है कि सब लोगों के द्वारा किसी गुणहीन व्यक्ति को बार-बार गुणवान कहने से वह गुणहीन व्यक्ति गुणवान नहीं बन सकता क्योंकि कहीं यह नहीं सुना कि आक के वृक्ष में भी सूर्य के समान तेज तथा उजाला है। जैसे आक कहने से आक का वृक्ष अरक अर्थात् सूर्य नहीं हो सकता वैसे ही गुणहीन को गुणी-गुणी कहते रहने से वह गुणवान नहीं हो सकता है।

विशेष:

1. गुणहीन को गुणी कहते रहने से उसे गुणवान नहीं बनाया जा सकता।
2. ब्रज भाषा, दोहा छंद, अनुप्रास तथा पुनरुक्ति प्रकाश अलंकार है।

9. करत करत अभ्यास के, जड़मति होत सुजान।
रसरी आवत जात ते, सिल पर परत निसान।।

शब्दार्थ:
जड़मति = मूर्ख। सुजान = विद्वान्, बुद्धिमान। रसरी = रस्सी। सिल = पत्थर, चट्टान।

प्रसंग:
प्रस्तुत दोहा वृन्द द्वारा रचित ‘नीति के दोहे’ से लिया गया है। इस दोहे में कवि ने यह बताया है कि अभ्यास करने से मूर्ख भी विद्वान् बन सकता है।

व्याख्या:
कवि कहता है कि बार-बार अभ्यास करने से मूर्ख व्यक्ति भी विद्वान् बन सकता है जैसे बार-बार रस्सी के घिसने से पत्थर पर भी निशान पड़ जाते हैं।

विशेष:

1. परिश्रम करने से व्यक्ति अपने लक्ष्य को अवश्य प्राप्त कर लेता है।
2. भाषा अत्यंत सरल, भावपूर्ण, दोहा छंद और पुनरुक्ति प्रकाश अलंकार है।

10. फेर न ह्वै है कपट सों, जो कीजै व्यापार।
जैसे हाँडी काठ की, चढ़े न दूजी बार।।

शब्दार्थ:
फेर = दुबारा। ह्वै = फिर से होना। कपट = छल। काठ = लकड़ी। दूजी = दूसरी।

प्रसंग:
प्रस्तुत दोहा वृन्द द्वारा रचित ‘नीति के दोहे’ से लिया गया है। इस दोहे में कवि ने यह स्पष्ट किया है कि छल-कपट का व्यवहार बार-बार नहीं चलता है।

व्याख्या:
कवि कहता है कि यदि कोई व्यक्ति छल-कपट और चालाकी से अपना कार्य करता है तो उसकी चालाकी एक बार तो चल जाती है परंतु बार-बार नहीं चलती जैसे काठ की हांडी एक बार तो चढ़ जाती है परंतु दोबारा नहीं चढ़ सकती।

विशेष:

1. छल-कपट का व्यवहार बार-बार नहीं चलता है।
2. भाषा सरल, भावपूर्ण, दोहा छंद है।

11. मधुर वचन ते जात मिट, उत्तम जन अभिमान।
तनिक सीत जल सों मिटे, जैसे दूध उफान।।

शब्दार्थ:
मधुर = मीठे। वचन = शब्द, वाणी। अभिमान = घमंड, अहंकार। तनिक = थोड़े से। सीत = ठंडा। उफान = उबाल।

प्रसंग:
प्रस्तुत दोहा वृन्द द्वारा रचित ‘नीति के दोहे’ से लिया गया है। इस दोहे में कवि ने मधुर वाणी के प्रभाव का वर्णन किया है।

व्याख्या:
कवि कहता है कि मधुर वचनों अथवा मीठी वाणी के बोलों से किसी भी अभिमानी व्यक्ति के गर्व को उसी प्रकार से शांत किया जा सकता है जैसे थोड़े से ठंडे पानी के छींटों से उबलते हुए दूध के उफ़ान को कम कर लिया जाता है।

विशेष:

1. मधुर वाणी के प्रयोग से क्रोधी व्यक्ति के क्रोध तथा अभिमानी के गर्व को भी शांत किया जा सकता है।
2. भाषा सरल, सहज, भावपूर्ण तथा दोहा छंद है।

12. अरि छोटो गनिये नहीं, जाते होत बिगार।
तृण समूह को तनिक में, जारत तनिक अंगार।

शब्दार्थ:
अरि = शत्रु, दुश्मन। गनिये = मानना, गिनना। बिगार = बिगड़ना। तृण = तिनका। तनिक = क्षण भर में। जारत = जलाना। तनिक = छोटा-सा। अंगार = आग का अंगारा।

प्रसंग:
प्रस्तुत दोहा वृन्द द्वारा रचित ‘नीति के दोहे’ से लिया गया है। इस दोहे में कवि ने संदेश दिया है कि अपने छोटे-से-छोटे शत्रु को भी कभी कमजोर नहीं समझना चाहिए।

व्याख्या:
कवि कहता है कि अपने शत्रु को कभी भी छोटा, तुच्छ या अपने से कमजोर नहीं समझना चाहिए क्योंकि इस से बहुत हानि हो जाती है। जिस प्रकार तिनकों के ढेर को आग का केवल एक अंगारा क्षणभर में जला कर नष्ट कर देता है वैसे ही शत्रु को कम समझने से हानि होती है।

विशेष:

1. कभी भी किसी कार्य अथवा शत्रु को अपने से कम नहीं समझना चाहिए।
2. भाषा सहज, सरल, भावपूर्ण, दोहा छंद और अनुप्रास अलंकार है।

नीति के दोहे Summary

नीति के दोहे रहीम कवि परिचय

रहीम का पूरा नाम अब्दुर्रहीम खानखाना था। इनका जन्म सन् 1553 ई० में हुआ था। इनके पिता अकबर बादशाह के अभिभावक मुग़ल सरदार बैरम खाँ खानखाना थे। अकबर के राज्यकाल में ये अकबर के दरबार के नवरत्नों में से एक थे। ये संस्कृत, अरबी, फारसी, ब्रज, अवधी आदि भाषाओं के विद्वान् तथा हिंदी काव्य के रचयिता थे। ये अत्यंत पानी तथा परोपकारी व्यक्ति थे। गोस्वामी तुलसीदास इनके प्रिय मित्र थे। इन्होंने मुग़ल साम्राज्य के विस्तार के लिए अनेक युद्ध भी लड़े थे। जहाँगीर के शासनकाल में इन्हें राजद्रोह के अपराध में बंदी बना लिया गया था तथा इनकी सारी जागीर भी छीन ली गई थी। इनकी वृद्धावस्था बहुत ग़रीबी में बीती थी। सन् 1625 ई० में इनका देहांत हो गया था।

रचनाएँ-इनकी प्रमुख रचनाएँ रहीम सतसई, बरवै नायिका भेद, श्रृंगार सोरठ, मदनाष्टक और रास पंचाध्यायी हैं। इनकी भाषा अत्यंत सरल, सहज तथा शैली भावानुरूप है। इन्होंने दोहा, सोरठा, सवैया, कवित्त, बरवै छंदों का प्रयोग किया है।

नीति के दोहे बिहारी कवि परिचय

रीतिकाल के सप्रसिद्ध कवि बिहारी का जन्म सन् 1603 ई० में बसुआ गोबिंदपुर गाँव (ग्वालियर) में हुआ था। इनके पिता का नाम केशवराय था। बिहारी ने अपनी शिक्षा महात्मा नरहरिदास के आश्रम में रह कर ग्रहण की थी। इनका विवाह मथुरा में हुआ था तथा इनकी पत्नी भी विदुषी एवं कवयित्री थी। बिहारी को मुग़ल सम्राट शाहजहाँ ने अपने दरबार में आमंत्रित किया था। जयपुर नरेश महाराजा जयसिंह के ये दरबारी कवि थे। एक बार जयपुर के राजा जयसिंह अपनी नव विवाहिता पत्नी के राग-रंग में इतने अधिक तल्लीन हो गए थे कि अपना समस्त राज-काज भी भुला बैठे थे तब बिहारी ने उन्हें निम्नलिखित दोहा लिख कर भिजवाया था-
“नाहिं पराग नहिं मधुर मधु, नहिं विकास इहि काल।
अलि कली ही सौ बंध्यौ, आगे कौन हवाल।”

इस दोहे ने महाराज को सचेत कर दिया था तथा वे राज-काज में रुचि लेने लगे थे। बिहारी रीति काल के प्रतिनिधि कवि माने जाते हैं। सन् 1664 ई० में वृंदावन में इनका स्वर्गवास हो गया था।

रचनाएँ: बिहारी द्वारा रचित केवल ‘सतसई’ ही मिलती है। इसमें सात सौ तेरह दोहे हैं। कुछ विद्वानों ने दोहों की संख्या सात सौ छब्बीस भी मानी है। इस रचना में मुख्य रूप से श्रृंगार रस की प्रधानता है। कवि ने श्रृंगार रस के अतिरिक्त नीति, वैराग्य, भक्ति आदि से संबंधित कुछ दोहों की रचना की है। बिहारी सतसई की लोकप्रियता से प्रभावित होकर इसकी अनेक टीकाएँ भी की गई हैं। संस्कृत, उर्दू, फारसी, खड़ी बोली, अंग्रेज़ी, मराठी आदि भाषाओं में भी इसके अनुवाद किए गए हैं।

नीति के दोहे वृन्द कवि परिचय

वृन्द रीतिकाल के ‘सूक्तिकार’ कवियों में प्रमुख माने जाते हैं। इनका जन्म सन् 1685 ई० में मेड़ता (मेवाड़) जोधपुर में हुआ था। ये कृष्णगढ़ नरेश महाराज राजसिंह के गुरु थे। इनके संबंध में प्रसिद्ध है कि ये कृष्णगढ़ नरेश के साथ औरंगज़ेब की सेना में ढाका तक गए थे। इनके वंशजों के बारे में कहा जाता है कि वे अब भी कृष्णगढ़ में रहते हैं। इनका देहावसान सन् 1765 ई० में हुआ था।

रचनाएँ-वृन्द की प्रमुख रचना ‘वृन्द सतसई’ है। इसमें नीति से संबंधित सात सौ दोहे हैं। इनकी अन्य रचनाएँ ‘श्रृंगार शिक्षा’, ‘पवन पचीसी’, ‘हितोपदेश संधि’, ‘वचनिका’ तथा ‘भावपंचाशिका’ हैं। इनकी काव्य भाषा अत्यंत सरल, सहज, सरस तथा भावपूर्ण है। अपने काव्य में इन्होंने सूक्तियों का बहुत सुंदर तथा सटीक प्रयोग किया है।

नीति के दोहे रहीम दोहों का सार

पाठ्य-पुस्तक में रहीम जी के चार दोहे संकलित हैं। पहले दोहे में कवि ने कहा है कि अच्छे समय में तो सभी मित्र बन जाते हैं परंतु सच्चा मित्र वही होता है जो मुसीबत के समय साथ देता है। दूसरे दोहे में कवि एकनिष्ठ भाव से एक की ही साधना करने पर बल देते हैं। तीसरे दोहे में स्वार्थ की अपेक्षा परमार्थ करने का लाभ बताया गया है तथा चौथे दोहे में बड़ी वस्तु देखकर छोटी वस्तु की उपयोगिता को नहीं भूलने के लिए कहा गया है।

नीति के दोहे बिहारी दोहों का सार

पाठ्यपुस्तक में बिहारी के चार दोहे संकलित हैं। पहले दोहे में धन-संपत्ति के नशे, दूसरे दोहे में मनुष्य के आशावादी होने का, तीसरे दोहे में एक जैसे स्वभाव वालों के परस्पर मेल-जोल से रहने तथा चौथे दोहे में कवि ने गुणों के महत्त्व का वर्णन किया है।

नीति के दोहे बिहारी दोहों का सार

पाठ्यपुस्तक में वृन्द के चार दोहे संकलित हैं। पहले दोहे में कवि ने परिश्रम का महत्त्व बताया है, जिससे हम अपना लक्ष्य प्राप्त कर सकते हैं। दूसरे दोहे में कवि ने बताया है कि जैसे काठ की हांडी बार-बार नहीं चढ़ती वैसे ही चालाकी भी बार-बार नहीं की जा सकती। तीसरा दोहा मधुर वचन का प्रभाव स्पष्ट करता है। चौथे दोहे में अपने शत्रु को भी कमज़ोर नहीं समझने का संदेश दिया गया है।

Punjab State Board PSEB 10th Class Hindi Book Solutions Chapter 2 पदावली Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Hindi Chapter 2 पदावली

Hindi Guide for Class 10 PSEB पदावली Textbook Questions and Answers

(क) विषय बोध

I. निम्नलिखित प्रश्नों के उत्तर एक या दो पंक्तियों में दीजिए

प्रश्न 1.
श्री कृष्ण ने कौन-सा पर्वत धारण किया था?
उत्तर:
श्री कृष्ण ने गोवर्धन पर्वत को धारण किया था।

प्रश्न 2.
मीरा किसे अपने नयनों में बसाना चाहती है?
उत्तर:
मीरा नंद के पुत्र श्रीकृष्ण को अपने नयनों में बसाना चाहती है।

प्रश्न 3.
श्रीकृष्ण ने किस प्रकार का मुकुट और कुंडल धारण किए हैं ?
उत्तर:
श्रीकृष्ण ने मोर पंखों का मुकुट और मकर की आकृति के कुंडल धारण किए हुए हैं।

प्रश्न 4.
मीरा किसे देखकर प्रसन्न हुई और किसे देखकर दुःखी हुई ?
उत्तर:
मीरा भक्तों को देखकर प्रसन्न हुई और सांसारिक माया-मोह में फंसे हुए लोगों को देख कर दुःखी हुई।

प्रश्न 5.
संतों की संगति में रहकर मीरा ने क्या छोड़ दिया ?
उत्तर:
संतों की संगति में रहकर मीरा ने कुल की मर्यादा तथा लोकलाज को छोड़ दिया था।

प्रश्न 6.
मीरा अपने आँसुओं के जल से किस बेल को सींच रही है?
उत्तर:
मीरा अपने आँसुओं के जल से श्रीकृष्ण के प्रति अपने हृदय में विद्यमान प्रेम रूपी बेल को सींच रही है।

प्रश्न 7.
पदावली के दूसरे पद में मीराबाई गिरिधर से क्या चाहती है ?
उत्तर:
मीराबाई ने गिरिधर को अपने पति के रूप में स्वीकार कर उनसे चाहा है कि केवल वे ही उसका अब उद्धार कर सकते हैं। वह उनकी शरण को सदा के लिए पाना चाहती है।

II. निम्नलिखित पद्यांशों की सप्रसंग व्याख्या कीजिए

(क) बसौ मेरे नैनन में नंद लाल।
मोहनि मूरति साँवरी सूरति नैना बनै विसाल।
मोर मुकुट मकराकृत कुंडल अरुण तिलक दिये भाल।
अधर सुधारस मुरली राजति उर वैजन्ती माल।
छुद्र घंटिका कटि तट सोभित नुपूर शब्द रसाल।
मीरा प्रभु सन्तन सुखदाई भक्त बछल गोपाल॥
उत्तर:
कवयित्री अपनी कामना व्यक्त करते हुए कहती है कि हे नंद के पुत्र श्री कृष्ण, आप मेरे नेत्रों में निवास करने की कृपा करो। आपकी मोहित करने वाली आकृति तथा सांवले रंग की सूरत है। आपके नेत्र बहुत बड़े-बड़े हैं। आप ने मोर के पंखों का मुकुट सिर पर और कानों में मकर की आकृति के कुंडल धारण किए हुए हैं। आपके माथे पर लाल रंग का तिलक सुशोभित हो रहा है। आप के होठों पर अमृत समान मधुर स्वर रस की वर्षा करने वाली बाँसुरी तथा हृदय पर वैजंती माला विराजमान है। छोटी-छोटी घंटियाँ आप की कमर पर बंधी हुई हैं तथा पैरों में घुघरू बंधे हैं जिनकी मधुर गुंजार सुनाई दे रही है। मीरा के प्रभु संतों को सुख प्रदान करते हैं तथा अपन भक्तों की सदा रक्षा करते हैं।

(ख) मेरे तो गिरिधर गोपाल, दूसरो न कोई।
जाके सिर मोर मुकुट, मेरो पति सोई।
तात मात भ्रात बंधु, आपनो न कोई।
छोड़ि दई कुल की कानि, कहा करै कोई।
संतन ढिग बैठि बैठि, लोक लाज खोई।
अँसुअन जल सींचि सींचि, प्रेम बेलि बोई।
अब तो बेलि फैल गई, आनंद फल होई।
भगत देखि राजी भई, जगत देखि रोई।
दासी मीरा लाल गिरधर, तारौ अब मोही।
उत्तर:
मीरा जी कहती हैं कि मेरा तो सर्वस्व गोवर्धन पर्वत धारी श्रीकृष्ण हैं। उनके अतिरिक्त मेरा किसी से कोई संबंध नहीं। मोर-मुकुट धारण करने वाले श्रीकृष्ण ही मेरे पति हैं। पिता, माता, भाई, सगा-संबंधी इनमें अब मेरा अपना कोई नहीं है। मैंने अपने परिवार की मान-मर्यादा को छोड़ कर उन्हें अपना लिया है। इसलिए मेरा अब कोई क्या कर सकता है अर्थात् मुझे किसी की परवाह नहीं है। मैं लोक-लाज की चिंता छोड़कर संतों के पास बैठती हूँ। मैंने आँसुओं के जल से सींच-सींच कर कृष्ण प्रेम की बेल को बोया है। अभिप्राय यह है कि श्रीकृष्ण के प्रति मीरा के मन की प्रेम रूपी बेल का विकास हो चुका है। अतः अब उसे किसी प्रकार से भी नष्ट नहीं किया जा सकता। मीरा जी कहती हैं कि वे प्रभु भक्त को देख कर तो प्रसन्न होती हैं पर संसार को देखकर रो पड़ती हैं। भाव यह है कि माया-मोह में लिप्त प्राणियों के दयनीय अंत की कल्पना मात्र से मीरा का हृदय दुःख से भर जाता है। मीरा स्वयं को श्रीकृष्ण की दासी मानती हुई उनसे अपने उद्धार की प्रार्थना करती है।

(ख) भाषा-बोध

निम्नलिखित शब्दों के दो-दो पर्यायवाची शब्द लिखें

भाल = ———–
प्रभु = ———–
जगत = ———–
वन। = ———–
उत्तर:
शब्द पर्यायवाची शब्द
भाल – माथा, मस्तक, ललाट, कपाल।
प्रभु – ईश्वर, परमेश्वर, जगदीश, भगवान्।
जगत – संसार, विश्व, भुवन, दुनिया।
वन – विपिन, जंगल, कानन, अटवी।

निम्नलिखित भिन्नार्थक शब्दों के अर्थ लिखकर वाक्यों में प्रयोग कीजिए

कुल = ———–
कूल = ———–
कटि = ———–
कटी। = ———–
उत्तर:
कुल = वंश-श्री राम का जन्म रघुकुल में हुआ था।
कूल = किनारा-यमुना नदी के कूल पर स्नानार्थियों की भीड़ थी।
कटि = कमर-बाल कृष्ण की कटि पर छोटे-छोटे घुघरू बंधे हुए थे।
कटी = फटना-उसकी कमीज़ नीचे से कटी हुई है।

(ग) पाठ्येतर सक्रियता

प्रश्न 1.
मीराबाई द्वारा वर्णित श्रीकृष्ण के बाल रूप का एक चित्र लेकर अपने विद्यालय की भित्ति पत्रिका पर लगाएँ। (विद्यार्थी स्वयं करें।)
उत्तर:
किलकत कान्ह घुटरुवन आवत।
मनिमय कनक नंद के आँगन, मुख प्रतिबिंब पकरिबे धावत।
कबहुँ निरखि हरि आप छाँह को, पकरन को चित चाहत।
किलकि हँसत राजत द्वै द॑तियाँ, पुनि-पुनि तिहि अवगाहत।
कनक-भूमि पर कर-पग-छाया, यह उपमा एक राजत।
प्रति-कर प्रति-पद प्रतिमनि वसुधा, कमल बैठकी साजत।
बाल-दसा-सुख निरखि जसोदा, पुनि-पुनि नंद बुलावति।
अँचरा तर लै ढाँकि ‘सूर’ के प्रभु कौं दूध पियावति। (सूरदास)

प्रश्न 2.
मीराबाई की तरह अन्य कृष्ण भक्त कवियों जैसे सूरदास, रसखान आदि ने श्रीकष्ण के बाल सौंदर्य का वर्णन अपनी रचनाओं में किया है। अपने स्कूल की पुस्तकालय से ऐसे कुछ पद संकलित करें।
उत्तर:
सूरदास और रसखान का एक-एक पद यहाँ दिया जा रहा है। अन्य पद विद्यार्थी स्वयं संकलित करें।
धूर भरे अति शोभित स्याम जू तैसी बनी सिर सुंदर चोटी।
डोलन खात फिरै अँगना पग पैंजनी बाजती, पीरी कछोटी॥
वा छवि को रसखानि बिलोकत बारत काम-कला निज कोटी।
काग के भाग बड़े सजनी हरि हाथ सों लै गयो माखन रोटी॥ (रसखान)

प्रश्न 3.
मीराबाई के पदों की सी०डी०/डी०वी०डी० देखें या इंटरनेट पर इन पदों को सुन कर आनंद लें।
उत्तर:
(विद्यार्थी स्वयं करें।)

प्रश्न 4.
कृष्ण जन्माष्टमी के अवसर पर ‘मीरा पद गायन’ प्रतियोगिता का आयोजन करें।
उत्तर:
(विद्यार्थी स्वयं करें।)

(घ) ज्ञान-विस्तार

मीरा श्रीकृष्ण के प्रेम में दीवानी भक्तिन कही जाती है। वह उन्हें ‘गिरिधर,’ ‘गोपाल’ आदि नामों से पुकारती थी। श्री कृष्ण को गिरिधर इसलिए कहते हैं क्योंकि उन्होंने इंद्र के प्रकोप से वृंदावन के लोगों की रक्षा करने के लिए गोवर्धन पर्वत को छतरी के समान अपनी उंगली पर धारण कर लिया था और वृंदावनवासियों को घनघोर वर्षा से बचाया था। वे ‘गोपाल’ इसलिए कहलाते हैं क्योंकि वे गौओं को चराया करते थे।

PSEB 10th Class Hindi Guide पदावली Important Questions and Answers

प्रश्न 1.
श्रीकृष्ण की आकृति तथा वर्ण कैसा है?
उत्तर:
श्रीकृष्ण की मन को मोहित करने वाली आकृति तथा सांवले वर्ण की शारीरिक छवि है।

प्रश्न 2.
श्रीकृष्ण के माथे की शोभा में वृद्धि किस से हो रही है?
उत्तर:
श्रीकृष्ण के माथे की शोभा में वृद्धि उनके माथे पर लगे हुए लाल रंग के तिलक से हो रही है।

प्रश्न 3.
श्रीकृष्ण की कौन-सी ध्वनि मन को आकर्षित करती है?
उत्तर:
श्रीकृष्ण की कमर में छोटी-छोटी घंटियाँ बँधी हुई हैं तथा पाँवों में छोटे-छोटे घुघरू बंधे हैं, जिनसे निकलने वाली ध्वनि मन को आकर्षित करती है।

प्रश्न 4.
मीरा ने किसे अपना पति कहा है/
उत्तर:
मीरा ने मोर पंखों का मुकुट धारण करने वाले श्रीकृष्ण को अपना पति कहा है।

प्रश्न 5.
‘तात मात भ्रात बंधु, आपनो न कोई ‘ से मीरा का क्या आशय है?
उत्तर:
मीरा का मानना है कि अब इस संसार में श्रीकृष्ण के अतिरिक्त उसका अपना कोई पिता, माता, भाई-बंधु नहीं

प्रश्न 6.
मीरा की प्रेम बेलि पर कैसा फल लगा है?
उत्तर:
मीरा की प्रेम-बेलि पर श्रीकृष्ण से मिलन रूपी आनंद का फल लगा है।

प्रश्न 7.
मीरा ने श्रीकृष्ण की कैसी छवि को अपने मन में कल्पित किया था ?
उत्तर:
मीरा ने अपने मन में कल्पित किया था कि श्रीकृष्ण अपार शोभावान थे। उनकी छवि शोभा से युक्त थी। उनका सांवला रंग था। उनकी आँखें अति संदर और बड़ी-बड़ी थीं। उनके सिर पर मोर-मुकुट शोभा देता था। कानों में कुंडल और माथे पर लाल रंग का तिलक अपार शोभा देता था। उनके होठों पर बाँसुरी अति सुंदर लगती है। वैजन्ती माला उनकी छाती की शोभा बढ़ाती है।

प्रश्न 8.
मीराबाई ने श्रीकृष्ण के प्रति अपने कैसे भावों को अभिव्यक्त किया था ?
उत्तर:
मीरा ने स्पष्ट रूप से स्वीकार किया था कि श्रीकृष्ण ही उसके स्वामी थे। उनके अतिरिक्त कोई भी दूसरा उसका नहीं था। वे ही उसके पति थे। उन्हीं के कारण उसने अपने सारे सांसारिक रिश्ते-नाते त्याग दिए थे। उन्हीं के लिए उसने सभी सामाजिक और दुनियादारी के बंधन सदा के लिए छोड़ दिए थे। उसने अपने आँसुओं से ही अपने प्रेम को पोषित किया था।

एक पंक्ति में उत्तरात्मक प्रश्न

प्रश्न 1.
मीरा ने किसे अपना पति माना है ?
उत्तर:
मीरा ने सिर पर मोर के पंख का मुकुट धारण करने वाले श्रीकृष्ण को अपना पति माना है।

प्रश्न 2.
श्रीकृष्ण की कमर में क्या बंधा हुआ है?
उत्तर:
श्रीकृष्ण की कमर में छोटी-छोटी घंटियाँ बंधी हुई हैं।

प्रश्न 3.
मीरा ने किनके साथ बैठकर लोक लाज गँवा दी थी?
उत्तर:
मीरा ने संतों के साथ बैठकर लोक लाज गँवा दी थी।

प्रश्न 4.
मीरा श्रीकृष्ण को कहां बसाना चाहती है?
उत्तर:
मीरा श्रीकृष्ण को अपने नेत्रों में बसाना चाहती है।

नलिखित प्रश्नों के उत्तर एक सही विकल्प चुनकर लिखें

प्रश्न 1.
‘श्रीकृष्ण का वर्ण कैसा है
(क) गौर
(ख) सांवला
(ग) सुनहरा
(घ) आसमानी।
उत्तर:
(ख) साँवला

प्रश्न 2.
श्रीकृष्ण ने कैसे कुंडल पहने हुए हैं
(क) मत्स्याकृत
(ख) सर्पाकृत
(ग) मकराकृत
(घ) वृत्ताकृत।
उत्तर:
(ग) मकराकृत

प्रश्न 3.
मीरा ने किस जल से प्रेम बेल बोई है
(क) गंगाजल
(ख) आँसुओं का जल
(ग) यमुना जल
(घ) घट जल।
उत्तर:
(ख) आँसुओं का जल

एक शब्द/हाँ-नहीं/सही-गलत/रिक्त स्थानों की पूर्ति के प्रश्न

प्रश्न 1.
अधर ……….. मुरली राजति।
उत्तर:
सुधा रस

प्रश्न 2.
मेरे तो ………. गोपाल।
उत्तर:
गिरिधर

प्रश्न 3.
भगत देखि ………… भई, जगत ……….. रोई।
उत्तर:
राजी, देखि

प्रश्न 4.
मीरा ने किसे छोड़ दिया था ? (एक शब्द में उत्तर दें।)
उत्तर:
कुल की कानि को

प्रश्न 5.
मीरा ने कहा मेरे पिता-माता-भाई-सगे-संबंधी सब अपने हैं। (हाँ या नहीं में उत्तर दें)
उत्तर:
नहीं

प्रश्न 6.
श्रीकृष्ण के नेत्र विशाल हैं। (हाँ या नहीं में उत्तर दें)
उत्तर:
हाँ

प्रश्न 7.
मीरा की प्रेम बेल अब फैल गई है। (सही या गलत में उत्तर दें)
उत्तर:
सही

प्रश्न 8.
मीरा गिरधर को तारने के लिए कह रही है। (सही या गलत में उत्तर दें)
उत्तर:
सही।

पदावली पदों की सप्रसंग व्याख्या

1. बसौ मेरे नैनन में नन्द लाल।
मोहनि मूरति साँवरी सूरति नैना बनै विसाल।
मोर मुकुट मकराकृत कुंडल अरुण तिलक दिये भाल।
अधर सुधारस मुरली राजति उर वैजन्ती माल।
छुद्र घंटिका कटि तट सोभित नुपूर शब्द रसाल।
मीरा प्रभु सन्तन सुखदाई भक्त बछल गोपाल॥

शब्दार्थ:
बसौ = निवास करो। नैनन = नेत्र। नन्दलाल = नंद के पुत्र श्रीकृष्ण। मोहनि = मोहित करने वाली, आकर्षित करने वाली। मूरति = आकृति। साँवरी = सांवले रंग की। विसाल = बड़े-बड़े, विशाल। मकराकृत = मकर की आकृति के। अरुण = लाल भाल = माथा। अधर = होंठ। सुधारस = अमृत रस। मुरली = बांसुरी। राजति = सुशोभित होना। उर = हृदय। छुद्र = छोटी। कटि = कमर। नुपूर =घुघरू। रसाल = मीठा, मोहक, मधुर। बछल = रक्षक, वत्सल।

प्रसंग:
प्रस्तुत पद मीराबाई द्वारा रचित पदावली से लिया गया है। इस पद में कवयित्री ने बाल कृष्ण की मनोहारी छवि का वर्णन करते हुए उसे अपने नेत्रों में बसाने का वर्णन किया है।

व्याख्या:
कवयित्री अपनी कामना व्यक्त करते हुए कहती है कि हे नंद के पुत्र श्री कृष्ण, आप मेरे नेत्रों में निवास करने की कृपा करो। आपकी मोहित करने वाली आकृति तथा सांवले रंग की सूरत है। आपके नेत्र बहुत बड़े-बड़े हैं। आप ने मोर के पंखों का मुकुट सिर पर और कानों में मकर की आकृति के कुंडल धारण किए हुए हैं। आपके माथे पर लाल रंग का तिलक सुशोभित हो रहा है। आप के होठों पर अमृत समान मधुर स्वर रस की वर्षा करने वाली बाँसुरी तथा हृदय पर वैजंती माला विराजमान है। छोटी-छोटी घंटियाँ आप की कमर पर बंधी हुई हैं तथा पैरों में घुघरू बंधे हैं जिनकी मधुर गुंजार सुनाई दे रही है। मीरा के प्रभु संतों को सुख प्रदान करते हैं तथा अपन भक्तों की सदा रक्षा करते हैं।

विशेष:

1. कवयित्री ने श्रीकृष्ण के बालरूप को संतों के लिए सुखदायी तथा भक्तों की रक्षा करने वाला मानते हुए उन्हें इसी रूप में अपने नेत्रों में बसने की कामना व्यक्त की है।
2. भाषा सहज, सरस, भावपूर्ण राजस्थानी शब्दों से युक्त है। अनुप्रास अलंकार तथा गेयता का गुण विद्यमान है।

2. मेरे तो गिरिधर गोपाल, दूसरो न कोई।
जाके सिर मोर मुकुट, मेरो पति सोई।
तात मात भ्रात बंधु, आपनो न कोई।
छोड़ि दई कुल की कानि, कहा करै कोई।
संतन ढिग बैठि बैठि, लोक लाज खोई।
अँसुअन जल सींचि सींचि, प्रेम बेलि बोई।
अब तो बेलि फैल गई, आनंद फल होई।
भगत देखि राजी भई, जगत देखि रोई।
दासी मीरा लाल गिरधर, तारौ अब मोही।

शब्दार्थ:
गिरिधर = गोवर्धन पर्वत को धारण करने वाले श्रीकृष्ण। दूसरो = अन्य। जाके = जिसके। पति = स्वामी, पति। छांडि = त्यागना, छोड़ना। दई = दी। कानि = मर्यादा। कहा = क्या। करिहै = करेगा। ढिग = पास, निकट। सींचि = सींचना। आणंद = आनंद। राजी = प्रसन्न। जगति = संसार। तारो = उद्धार करना, मुक्ति देना।

प्रसंग:
प्रस्तुत पद मीराबाई की पदावली से अवतरित किया गया है। इसमें कवयित्री ने भगवान् श्रीकृष्ण को अपने पति रूप में मानकर उनके प्रति अपनी अनन्य भक्ति भावना का परिचय दिया है।

व्याख्या:
मीरा जी कहती हैं कि मेरा तो सर्वस्व गोवर्धन पर्वत धारी श्रीकृष्ण हैं। उनके अतिरिक्त मेरा किसी से कोई संबंध नहीं। मोर-मुकुट धारण करने वाले श्रीकृष्ण ही मेरे पति हैं। पिता, माता, भाई, सगा-संबंधी इनमें अब मेरा अपना कोई नहीं है। मैंने अपने परिवार की मान-मर्यादा को छोड़ कर उन्हें अपना लिया है। इसलिए मेरा अब कोई क्या कर सकता है अर्थात् मुझे किसी की परवाह नहीं है। मैं लोक-लाज की चिंता छोड़कर संतों के पास बैठती हूँ। मैंने आँसुओं के जल से सींच-सींच कर कृष्ण प्रेम की बेल को बोया है। अभिप्राय यह है कि श्रीकृष्ण के प्रति मीरा के मन की प्रेम रूपी बेल का विकास हो चुका है। अतः अब उसे किसी प्रकार से भी नष्ट नहीं किया जा सकता। मीरा जी कहती हैं कि वे प्रभु भक्त को देख कर तो प्रसन्न होती हैं पर संसार को देखकर रो पड़ती हैं। भाव यह है कि माया-मोह में लिप्त प्राणियों के दयनीय अंत की कल्पना मात्र से मीरा का हृदय दुःख से भर जाता है। मीरा स्वयं को श्रीकृष्ण की दासी मानती हुई उनसे अपने उद्धार की प्रार्थना करती है।

विशेष:

1. मीरा श्रीकृष्ण के प्रति समर्पित है। उन्हें वह अपने पति रूप में मानती है। वह उनसे अपने उद्धार की प्रार्थना करती है।
2. ब्रज मिश्रित राजस्थानी भाषा का प्रयोग है। अनुप्रास अलंकार है।

पदावली Summary

पदावली कवयित्री परिचय

मीराबाई कृष्ण भक्तिकालीन कवियों में प्रमुख स्थान रखती हैं। उनका काव्य हृदय का काव्य है जिसमें कोमल भावधारा का स्वाभाविक प्रवाह है। उनका जन्म गाँव कुड़की, जोधपुर (राजस्थान) के राव रत्न सिंह राठौर के घर सन् 1498 ई० में हुआ था। शैशवावस्था में ही माता की मृत्यु हो जाने के कारण उनका पालन-पोषण दादा ने किया। दादा की वैष्णव भक्ति ने मीरा को भी प्रभावित किया। मीरा का विवाह मेवाड़ के राणा सांगा के पुत्र भोजराज के साथ हुआ था। विवाह के कुछ समय पश्चात् ही उनके पति की मृत्यु हो गई। मीरा को संसार से विरक्ति हो गई। वे कृष्ण की उपासना में मग्न रहने लगीं। वे मंदिरों में कृष्ण की मूर्ति के आगे नाचतीं और साधु-संगति में अपना समय व्यतीत करतीं। उनके इस प्रकार के आचरणों से रुष्ट होकर उनके देवर ने उन्हें मारने के अनेक प्रयत्न किए, परंतु प्रभु-भक्ति में लीन रहने वाली मीरा का बाल भी बांका न हुआ। अपने परिवार के बुरे व्यवहार से तंग आकर मीरा वृंदावन और फिर द्वारिका चली गई थी। वहीं उन्होंने अपनी देह त्याग दी थी। उनकी मृत्यु सन् 1573 ई० में मानी जाती है।

रचनाएँ-मीरा की निम्नलिखित रचनाएँ हैं-
नरसी जी का मायरा, गीत गोबिंद की टीका, राग गोविंद, राग सोरठा के पद आदि। विद्वानों के अनुसार, ‘पदावली’ ही उनकी प्रामाणिक रचना है।

मीरा प्रमुख रूप से भक्त गीतकार हैं। उनके पदों में भक्त हृदय की पुकार है। सरसता, मधुरता एवं मौलिकता की दृष्टि से उनके पद बेजोड़ हैं। माधुर्य भाव की भक्ति के कारण इनके काव्य में श्रृंगार रस के दोनों पक्षों संयोग एवं वियोग का सुंदर चित्रण हुआ है। इनके काव्य में प्रकृति-चित्रण के सुंदर चित्र भी मिलते हैं। वेदना-भाव के मिश्रण ने इन चित्रों को बहुत मार्मिक बना दिया है।

मीरा की भाषा में ब्रज और राजस्थानी शब्दों की प्रमुखता है लेकिन उसमें पंजाबी, गुजराती आदि के शब्दों का भी प्रयोग किया गया है। इनके काव्य में भावों को अधिक महत्त्व दिया गया है। इन्होंने भावों के अनुकूल सरल, सरस और भावपूर्ण शब्दों का प्रयोग किया है। राजस्थानी के शब्दों की प्रमुखता दिखाई देती है। राजस्थान के भक्तों में तो इनका स्थान सर्वोपरि है।

पदावली पदावली का सार

पाठ्य-पुस्तक में मीराबाई के दो पद संकलित हैं। पहले पद में कवयित्री ने नंदलाल, बालक कृष्ण की मनमोहक छवि को अपने नेत्रों में बसाने की कामना करते हुए उनकी सांवली सूरत, बड़ी-बड़ी आँखों, मोर के पंखों के मुकुट, मकर की आकृति के कुंडलों तथा माथे पर लगे तिलक की प्रशंसा की है। उनके होंठों पर मधुर स्वर उत्पन्न करने वाली मुरली तथा हृदय पर वैजंती माला विराजमान है। उनकी कमर पर छोटी-छोटी घंटियाँ तथा पैरों में घुघरू बंधे हैं, जिनका मधुर स्वर वातावरण में गूंज रहा है। मीरा के प्रभु का यह रूप संतों को सुख देने वाला तथा भक्तों की रक्षा करने वाला है।

दूसरे पद में कवयित्री श्रीकृष्ण को अपना सर्वस्व मानते हुए कहती हैं कि उसका उनके अतिरिक्त किसी अन्य से कोई संबंध नहीं है। मोर मुकुट धारी ही उसके स्वामी हैं। उसने कुल की मर्यादा का त्याग कर दिया है, इसलिए अब उसे किसी की कोई चिंता नहीं है। वह लोकलाज छोड़कर संतों के साथ विचरण कर रही है। उसने अपने आँसुओं से कृष्ण प्रेम की बेल को बोया है, जो अब खूब फैल गई है, जिस पर आनंद रूपी फल लग गए हैं। वह प्रभ भक्ति से प्रसन्न है तथा सांसारिक माया-मोह के बंधनों को देखकर रोती है। वह अपने आराध्य से अपनी मुक्ति की प्रार्थना करती है।

Punjab State Board PSEB 10th Class Science Important Questions Chapter 9 Heredity and Evolution Important Questions and Answers.

PSEB 10th Class Science Important Questions Chapter 9 Heredity and Evolution

Long Answer Type Questions

Question 1.
Describe briefly Mendel’s experiment.
Answer:
Mendel’s Experiment. Gregor Mendel (1822-1884) was an Austrian monk. He conducted experiments with garden pea (Pisum sativum). The results thus formulated the laws of inheritance.

He studied inheritance of each character separately.

• He selected two pure varieties of pea (Pisum sativum) which differed in size. One of them was tall and the other dwarf.
• He cross-pollinated them. He placed the pollen of tall one on the stigmas of dwarf and vice versa.
• The hybrid seeds obtained in both the cases were sown. Whichever way the cross was made, on germination the seeds grew into plants which were all tall. This first hybrid generation, is called the first filial generation and is usually writte. as F₁.
• The hybrids of F₁ generation were all similar to the tall parent. The resu t of this generation surprised Mendel. He expected the hybrids to be intermediate in size.
• Accordingly the character which appeared in the F₁ generation (tallne. s in this case) he called dominant and the other which did not appear he called recessive.
• Mendel’s next step was to allow the F₁ hybrids to self-pollinate and produce seeds. He collected the seeds, planted them and observed the results. He found that three-fourths of the plant of F₂ generation were tall like the original tall parent and one- fourth dwarf like the original dwarf parent. The result of F₂ generation was all the more surprising to Mendel.

Question 2.
Describe present-day concept of evolution.
Answer:
1. Modern Concept of Evolution: The modern concept of evolution is a modified form of Darwin’s theory of natural selection and is often called Neo-Darwinism. It comprises genetic variation, natural selection and isolation.

• Mutations: These have been recognized as the ultimate source of biological changes and hence the raw material of evolution. The mutation in chromosomes may be due to changes in structure, number or gene.
• Gene Recombination takes place during crossing over in meiosis. New combinations of genes produce new phenotypes.
• Hybridisation is the intermingling of the genes of the members of closely related species.
• Genetic drift is the elimination of the genes of some original characteristics of a species by extreme reduction due to different regions.
• In monoparental reproduction only chromosomal and gene mutation are sources of genetic variation.

2. Natural Selection: If differential reproduction i.e. some individuals produce abundant offspring, some only a few and some organisms none) continues for many generations, genes of the individuals which produce more offspring will become predominant of the gene pool of the population. Thus natural selection occurs through differential reproduction in successive generations.

Migration of individuals from one to other population is an accessory factor for speciation (origin of new species).

3. Isolation: By selecting the most suitable genotypes, natural selection guides different populations into different adaptive channels. The reproductive isolation between the populations due to certain physical barriers or others leads to the formation of new species. The isolation plays a significant role in evolution.

Question 3.
Why is Mendel known as father of genetics?
Or
Write contribution of Mendel.
Answer:
Gregor Johann Mendel in 1866 demonstrated the wray in which characters are transmitted from one generation to another and suggested that each cell of an organism contains two factors for each character, both of which separate and are passed on to different progeny through different gametes. Thus Mendel laid the foundation of genetics, the science of heredity and variation, hence it is proper to call him father of Genetics.

Question 4.
Explain Monohybrid cross.
Or
Explain Mendel’s Law of segregation with an example of monohybrid cross.
Answer:
Monohybrid cross. It is a cross in which only one character is considered at a time. In a cross between tall and dwarf plant, the size of stem is considered.

Mendel made a cross between pure tall (TT) and a pure dwarf (tt) pea plant as follows :

• Mendel selected a tall pea plant represented by the genes (TT) and a dwarf pea plant represented by the genes (tt). Pea plant is self pollinating.
• He removed the anthers of a tall plant and stigma of a dwarf plant, deposited pollen grains of dwarf plant on the stigma of tall plant and obtained seeds.
• When the seeds were placed in the soil and allowed to grow he obtained only tall plant (Tt), although he was expecting the plants of an intermediate size. These plants were labelled as plants of First filial generation (F₁ generation).
• In F₁ generation all plants were tall because the character tall is dominant over the dwarf.
• He allowed the plants of F₁ generation to self-pollinate. Gametes formed by meiosis contained only one gene of a character because genes separate at the time of gamete formation.
• The seeds thus obtained were placed in the soil and allowed to grow.
• The plants formed were tall and dwarf in the ratio of (3:1).
• Genotypically pure tall, hybrid tall and dwarf were in the ratio of 1 : 2 : 1. These were labelled as plants of F₂ generation.

Diagrammatically monohybrid cross can be represented as shown below :

Monohybrid Cross in a Pea Plant

Question 5.
Explain the law of independent assortment with a dihybrid cross.
Answer:
Law of independent assortment. According to this law, the factors of different pairs of contrasting characters, do not influence each other. They are independent of one another in their assortment to form new combination during gamete formation.

Dihybrid cross: A cross in which two characters are taken into consideration during experimentation, such a cross is called dihybrid cross. A cross between a pea plant with yellow smooth and a pea plant with green wrinkled seeds are considered.

Explanation: A cross is made between pea plant having yellow smooth seeds (YYSS) and a pea plant with green wrinkled seeds (yyss). At the time of cross pollination, yellow smooth (YYSS) produce gametes with genes (YS) and green wrinkled will produce gametes with gene (ys). Gametes unite at random. The seeds obtained when placed in soil will grow to form plants and produce seeds which are yellow smooth (YySs) because yellow and smooth characters are dominant over green and wrinkled. These are called as plants of Fi generation.

When plants of Fx generation are allowed to self pollinate gametes formed YS, Ys, yS and ys by meiosis, they unite at random forming seeds. The plants thus obtained were called as F₂ generation. They are yellow smooth (YYSS, YySS, YsSS, YYSs); yellow wrinkled (YYss, Yyss), green smooth (yySS, yySs) and green wrinkled (yyss) in the ratio of 9 : 3 : 3 : 1. The result of dihybrid cross can be shown below in the chequer board.

From the above dihybrid cross, it can be derived that each gene is assorted independently of the other during its passage from one generation to the other or Law of Independent Assortment is justified.

Result of dihybrid cross

Question 6.
State the hypothesis of Oparin and Haldane about the primeval Earth condition. What do you understand by Haldane’s hot dilute soup? State its significance.
Answer:
Alexander I. Oparin (1894-1980), a Russian biochemist and J.B.S. Haldane (1892-1964), a British scientist have put forward the concept of abiogenesis. According to Oparin and Haldane primeval earth had reducing condition and the atmosphere was free from oxygen. Oxygen remained bound in H₂0 and metallic oxides on the surface of rocks and its particles. The early gas cloud was rich in Hydrogen, in form of methane (CH₄ and ammonia (NH₃) and water (H₂O). The organic molecules formed due to the atmospheric reaction accumulated slowly and gradually in the sea and constitute what is called “hot dilute soup”.

Significance of hot dilute soup. Thus conditions of reducing nature are unable to oxidize these organic compounds which form the basis of life.

Question 7.
Summarise Miller’s simulation experiment for organic synthesis. Comment on its efficacy.
Answer:
Miller’s experiment. Miller (1953) made the first successful simulation experiment to assess the validity of the claim for origin of organic molecules.

Stanley Miller’s Experiment for the artificial synthesis of organic compounds

Miller sealed in a spark chamber a mixture of water, methane, ammonia, hydrogen gas. He made arrangement for boiling water. The trap in turn, was connected with the flask for boiling water. After 18 days, a significant amount of simple major organic compounds such as amino acids like glycine, alanine and aspartic acid and peptide chains began to appear. Simple sugars, urea and short chain fatty acids were also formed.

Question 8.
How human evolution take place over the years?
Answer:
Human evolution.
The study of human evolution indicates that all of human heings belong to a single species Homo sapiens that evolved in Africa.

• DNA sequences have been used for studying human evolution.
• Due to the diversified human forms and features, skin colour is the common way for identifying the races.
• Few thousand years ago some ancestors left Africa while others stayed back.
• The residents spread across Africa and the migrant spread across the planet from Africa to West Asia, Central Asia, Eurasia, South Asia, East Asia, Indonesia, Philippines, Australia and America.
• They went forward and backward with groups separating from each other, or sometime coming together.
• Like all other species, they were also living their lives to the best of their ability.In atmosphere, this spark is provided by U.V. light or other energy source.

Short Answer Type Questions

Question 1.
What is meant by heredity?
Answer:
Heredity. It is defined as the transmission of characters from parents to offspring or from one generation to the successive generations of living beings.

We observe in our daily life that similarities tend to be greatest between members of a family-between the offspring of parents. Children tend to resemble parents, even grandparents and persons of earlier generations. The similarities are not due to coincidence but rather due to inheritance or heredity.

Question 2.
What is genetics?
Answer:
Genetics: It is that branch of science which deals with study of heredity (inheritance of characters) and variations. It deals with inborn characteristics of the organisms. Genetics also deals with inborn differences between offsprings of family and related organisms.
Genes are carriers of characters and present on chromosomes. Mendel is considered as “Father of genetics.”

Question 3.
What are the causes of variations in clones?
Answer:
Causes of variations in clones :
Clones have the same genetic make up but variations appear in clones due to following reasons :

• Inaccuracies during DNA copying.
• Effect of environment termed acquired variations.
• Mutations: These are sudden stable abrupt changes and they are discontinuous inheritable as produced due to changes in genetic make up.

Question 4.
Explain the term variation.
Answer:
Variation: No living organisms are alike and they vary appreciably in many structural and functional aspects. These differences between individual organisms are called variations.

Question 5.
(a) Write one difference between continuous and discontinuous variations.
(b) Give differences between germinal and somatic variations.
Or
What do you mean by discontinuous variation?
Answer:
(a) Differences between continuous and discontinuous variations

Continuous variations	Discontinuous variations
They are small indistinct differences from the normal conditions and called fluctuations.	They are large distinct differences from the parents and termed as mutations or sport.

(b) Differences between germinal variations and somatic variations

Germinal variations	Somatic variations
1. They are caused due to changes in germ cells.	1. They are caused due to changes in the somatic cells.
2. These variations are heritable.	2. They are not heritable.

Question 6.
What is the importance of variations?
Answer:
Importance of variations :

• They enable the organism to adapt themselves to changing environment.
• They form raw material for evolution and development of new species.

Question 7.
Mention the information source of making protein in the cell. What is the basic event in reproduction.
Answer:
DNA (Deoxyribose Nucleic Acid) directs the synthesis of proteins through mRNA (messenger RNA).

DNA copying is essential part of the process of reproduction.

• DNA copying provides cellular apparatus in the daughter cells.
• DNA in daughter cells will be able to control the functioning of daughter cells.
• DNA copies will retain the traits.

Question 8.
Why did Mendel choose garden pea for his experiments?
Answer:
Mendel selected pea plant (Pisum sativum) because :

• Many varieties were available with observable alternative forms for a trait or characteristic.
• Peas are normally self pollinated; as their corolla completely enclose the reproductive organs until pollination is completed.
• It was easily available.
• It has pure lines for experimental purpose, i.e., they always breed true.
• It has contrasting characters. The traits were seed colour, pod colour, pod shape, flower shape, position of flower, seed shape and plant height.
• Its life cycle was short and produced large number of offsprings.
• The plant is grown easily and does not require care except at the time of pollination.

Question 9.
Make a table showing characters of pea selected by Mendel.
Answer:
Characters of garden pea (Pisum sativum) selected by Mendel

Heritable variations	Dominant	Recessive
1. Plant height	Tall (T)	Dwarf (t)
2. Flower and pod position	Axial (A)	Terminal (a)
3. Pod colour	Green (G)	Yellow (g)
4. Pod shape	Inflated (I)	Constricted (i)
5. Seed coat	Coloured (C)	White (c)
6. Seed shape	Round (R)	Wrinkled (r)
7. Seed (cotyledon) colour	Yellow (Y)	Green (y)

Question 10.
What is monohybrid cross?
Answer:
Monohybrid cross. It is a cross in which only one character is considered at a time, e.g., in a cross between tall and dwarf plant, the size of stem is considered. Mendel made a cross between pure tall (TT) and a pure dwarf (tt) pea plant.

He obtained all tall (hybrid) plants in FL generation. On selfing these plants produced tall and dwarf in the ratio of 3:1. The genotypic ratio of 1 : 2 : 1 and phenotypic ratio of 3 : 1 is termed monohybrid ratio.

Question 11.
State and explain principle of dominance.
Answer:
Law of Dominanc J: According to this law, when two factors of a character are unlike, one of them will manifest in the body and is called dominant while the other remains hidden and is termed recessive factor.

The law can be well explained by the monohybrid cross by studying the following crosses :
1. Pure tall = TT, Hybrid tall = Tt

Results of cross between pure tall and pure dwarf

Gametes of TT parent = \(\frac{1}{2}\) T + \(\frac{1}{2}\) T
Gametes of Tt parent = \(\frac{1}{2}\) T+ \(\frac{1}{2}\) t
The 50% are pure tall and 50% hybrid tall. Then pure tall plants will produce 100% tall in F₂ generation and hybrid plants will produce in the ratio of 1 : 2 : 1 in the F₂ generation.

2. When the cross is made between pure tall and pure dwarf, we get results as follows (Figure)

Question 13.
What is dihybrid ratio?
Answer:
Dihybrid ratio: The ratio obtained in a dihybrid cross is called dihybrid ratio. It is 9 : 3 : 3 : 1.

Question 14.
What is a gene? What is the nature of gene?
Answer:
Gene: Term gene was coined by Johansen (1900). A hereditary determiner specifying a biological function; a unit of inheritance (DNA) located in a fixed place on the chromosome is called gene.

Mendel considered every character as unit which is controlled by a factor presently called gene. Chemically gene is a segment of DNA which controls one character and physically is a part of chromosome.

Question 15.
Where are genes located? What is the chemical nature of gene?
Answer:
Genes are located on chromosomes. Chemically gene is a segment of DNA (Deoxyribose nucleic acid).

Question 16.
In man four types of blood groups A, B, AB and O are controlled by three alleles of a gene. What is the mechanism of inheritance of the blood groups?
Answer:
More than two forms exists for certain genes. It is an example of multiple alleles. A well known example is ABO blood types in human. The four human blood groups, A, B, AB and O are phenotypes of the trait.

Three different alleles I^A, I^B and i of gene determine the phenotypes of the four blood groups. The six types of genotypes are as follows :

Phenotype (Blood Group)	Genotype
O	ii
A	IAIA or IAi
B	IBIB or IBi
AB	IAIB

Both I^A and I^B are dominant over i. Since a person with genotype IAIB has AB blood groups. It is an example of codominance.

Thus ABO blood groups exhibit three genetic aspects :

1. Dominant-recessive mechanism.
2. Multiple alleles
3. Co-dominance.

Question 17.
A man with type A blood has a wife with type B. They have a child with type O blood. Give the genotype of all the three. What other blood groups can be expected in the future offspring of this couple?
Answer:
Genotypes. Man (I^A I^O), Mother I^B I^O and child I^OI^O.
Blood group of the future offspring. A type, B type, o type and AB type. It is based on the following cross :

Question 18.
Define genetic engineering. Write applications of genetic engineering.
Answer:
Genetic Engineering: The method of artificial synthesis of new genes and their subsequent “transplatation” in the genome of an organism or methods of correcting the defective genes is called genetic engineering.

Applications of genetic engineering

• Genetic engineering has introduced a new form of medicine called gene therapy which may be used in treating, crippling, hereditary diseases like haemophilia, phenylketonuria.
• With the help of genetic engineering it may be possible to produce new plants and animals having a new design and specific character according to will.
• Gene coding for vitamins, antibiotics or hormones from higher animals to bacteria is also possible which will help to produce chemicals which are impossible to get or to synthesise.

Question 19.
Give a graphic representation of mechanism of gene expression.
Answer:
Mechanism of gene expression

Question 20.
What do you understand by origin of life? Explain.
Answer:
Origin of Life. The oldest surviving terrestrial rocks, about 4.3 billion years old, contain no definite trace of life, at least not recognisable,as yet. Some rocks, about 3.9 billion years old, contain carbonates. Geologists interpret that these carbonates have resulted from life processes. Therefore, life was present on Earth about 3.9 billion years ago. However, the oldest microfossils discovered so far are that of photosynthetic cyanobacteria.

Question 21.
Write the contribution of Urey and Miller.
Answer:
Urey and Miller conducted experiment which supported that life originated by chemosynthesis. The chemosynthetic theory (Oparin-Haldane) states that life originated from non-living matter is based on the presence of methane and ammonia in the atmosphere. It required a high temperature, high energy radiations and electric discharges.

Question 22.
What are homologous organs? Give examples.
Answer:
Homologous organs. The organs of different classes have different forms because they have to perform different functions but their structures basically remains similar. Such organs are called homologous organs.

Examples of homologous organs. 1. The wings of bird and bat, flipper (fin) of whale, structure of human forearm are different in forms because these have to perform

Homologous organs

different functions. Studies of the bones forming the skeleton of these organs, would reveal similarity in construction. In fact, these are the forms of fore-arms which have originated from pentadactyl forms and due to the different functions they are performing hence transformed into different forms.

2. In plants, the homologous organs may be a thorn of Bougainvillea or a tendril of cucurbita both arising in axillary position.

Question 23.
What are analogous organs? Give examples.
Answer:
Analogous organs. The organs which are similar in appearance and perform the same function but differ in their fundamental structure and origin are called analogous organs.

Examples:

• Wings of birds and insects.
• Leaves of a plant and cladodes of Ruscus are also analogous organs.

Wings of insect and bird

Question 24.
Differentiate homologous organs and analogous organs.
Answer:
Differences between homologous organs and analogous organs

Homologous Organs	Analogous Organs
1. Some organs of different organisms resemble in structure and bear the same relation to the body.	1. These organs which are functionally similar but morphologically different are called analogous organs.
2. They have same fundamental plan of structure e.g. leaves of all higher plants arise from the nodes and bear an axillary bud in their axils.	2. Their basic structure is different e.g. wings of vertebrates and insects perform the same function of flying.

Question 25.
Are the fossils being formed at present time?
Answer:
Fossil formation. Fossilization of a dead organism or its parts usually begins when it is buried before it has a chance to decay. The organism sinks into a bog or a marsh or to the bottom of a lake, sea or river. In some cases it is buried by wind-driven sands. Even after burial decay can occur so that soft body parts decompose, a fact which emphasizes again that the hard parts are the ones that mou commonly persist as fossil. The buried parts that do not decay are preserved and mud or sand hardens to rock, the fossil becomes entombed.

Fossilization is a hit or miss effect. Only those organisms become fossilized that happen to die in a spot where they can be buried by natural process before their carcasses are destroyed by scavengers. Thus under such conditions fossils are being formed at the present time.

Question 26.
What is the physical method of determining the age of fossils?
Answer:
Determination of age of fossils (dating of fossils). Once the fossils are unearthed palaeontologists try to determine their position in the historical sequence of life. The absolute age of fossil is difficult to calculate because older the fossil, less precise is tne calculation.

Physical method. Before we begin to determine the age of fossil, we have to gain some perspective about the age of earth.
The age of the earth is estimated to be near about 300 crore years. This life span of earth has been divided into six principal eras. Out of these, three eras are subdivided into smaller span known as periods or epochs.

Question 27.
Discuss the importance of artificial selection in the derivation of the concept of natural selection.
Answer:
Importance of artificial selection. From his enquiries on breeding domesticating plants and animals, Darwin obtained clear evidence for selection, in this case of artificial selection. The breeders selected and perpetuated those variant types that interested them or seemed to be useful to them. Similar to artificial selection, natural selection also controls the speciation. But natural selection is too slow to observe.

Question 28.
Distinguish between microevolution and macroevolution. Narrate the significance of population genetics in evolution.
Answer:
Evolution on the grand scale of geological time, is called macroevolution while evolution at genetic level is microevolution. Microevolution is actually operative at genetic level change.

Significance of Population Genetics. The gene frequency of a population is called population genetics. Evolution occurs within populations as the relative frequencies of different variations of DNA change over time. If genes change, then enzymes automatically change and represent two different forms of individuals and definitely result in evolution.

Question 29.
What is variation? Name the basic processes that cause variations among organisms. Discuss the role of migration in evolution.
Answer:
Variations: The features which differ among the individuals are called variations.
Causes of variation: Mutation, recombination, gene migration, genetic drift and natural selection.
Role of Migration: Few populations are isolated from the other populations of same species, usually some migration takes place if the migrating individuals breed within the new population then immigrant will add new alleles to the local gene pool of host population.

Question 30.
Define variations in relation to species. Why is variation beneficial to the species?
Answer:
Variation. No living organisms are alike and they vary appreciably in many structural and functional aspects. These differences between individual organisms are called variations.

The useful variation in individuals of a species will enable them to adapt according to the changes and new needs. Thus they will enable the survival of the species.

Question 31.
Explain Genetic Drift.
Answer:
Genetic Drift: The term genetic drift refers to the chance elimination of the genes of certain traits when a section of population migrates or dies of natural calamity. It dramatically alters the gene frequency of the remaining population. It eliminates certain alleles and fixes the other alleles, thereby reducing the genetic variability of the population. For example, in case of snowstorm, the individuals having alleles (characters) that provide resistance to cold survive, whereas others die.

Question 32.
What is reproductive isolation?
Answer:
Reproductive isolation. Speciation is not likely to occur simply by changes in the genotype of a population. The populations with different genotypes appearing in them must be isolated so that differences may accumulate to the level of speciation. Else interbreeding of emerging populations will result in mixing of their genotypes and disappearance of differences between them. Isolation preserves the integrity of a species by checking hybridisation.

Question 33.
Give a brief account of present day concept of evolution.
Answer:
Present day concept of evolution

• appearance of genetic variations in certain individuals of a population by
  1. migration
  2. non-random mating
  3. genetic drift
  4. chromosomal changes
  5. gene mutation
  6. recombination of genes, and
  7. hybridization.
• spreading of genetic variations in a sub-group of a population by natural selection through differential reproduction in successive generations.
• some sort of reproductive isolation of a subgroup of population having the genotypes selected by nature from other subgroups, and
• accumulation of genetic variations to sufficiently alter the individuals of the subgroup to become a new species.

Question 34.
How is the equal genetic contribution of male and female parent ensured in the progeny?
Answer:
In sexual reproduction, both the parents contribute equal amount of genetic material (genes) to the offspring. This means that for each trait there will be two alternatives in the sexually reproducing organisms. Out of these two alternatives, one is called dominant trait and the other is called recessive trait. There will be some progeny with new combination. DNA controls the traits and are copied from one generation to the next generation. Inaccuracies do occur during DNA copying which is more prominent in sexual reproduction. These variation in DNA copying gets inherited. Accumulation of variation generation after generation altogether leads to evolution of a new species.

Question 35.
How does the creation of variations in a species ensure survival?
Answer:
Genetic variations arise in sexually producing organisms as a result of following mechanism.

• Crossing over during gamete formation.
• Random segregation of chromosome during meiosis at the time of gamete formation.
• Random rejoining of gametes having different genetic set up in the chromosomes during fertilisation.
• These variations form the raw materials of evolution.

Only variations that confer an advantage to an individual organism will survive in a population.
The organism with useful variations will adapt and survive. Moreover they leave behind more offsprings and populations with such genetic variations will survive.

Very Short Answer Type Questions

Question 1.
How are new organisms formed?
Answer:
New organisms are produced as a result of reproduction.

Question 2.
Do you find more variation amongst human or in sugarcane fields.
Answer:
Human beings show more variations.

Question 3.
What do the progeny get from parents?
Answer:
Inherited traits and variations.

Question 4.
What special features you find in second generation?
Answer:
The members of second generation inherit some characters from their parents and new variations are also produced.

Question 5.
Organisms reproducing by asexual reproduction show a few variation.
Answer:
These organisms carry out mitosis during reproduction. As there is no meiosis no new combinations of characters are formed.

Question 6.
What is the basis of evolution?
Answer:
Variations are raw materials of evolution.

Question 7.
What are functional unit of hereditary material?
Answer:
Genes.

Question 8.
Coin the term for transfer of characters from parents to offspring.
Answer:
Heredity/Inheritance.

Question 9.
Who is the Father of Genetics?
Answer:
Gregor Mendel.

Question 10.
Mendel worked on which plant having alternate traits?
Answer:
Pisum sativum (garden pea).

Question 11.
Mention any two of the seven contrasting traits of garden pea selected by Mendel.
Answer:

1. Height-Tall/Dwarf
2. Seed shape-Round/Wrinkled.

Question 12.
What is dominance?
Answer:
The expression of heritable trait present is heterozygous condition.

Question 13.
What is recessive?
Answer:
An allele that is not expressed phenotypically when present in heterozygous conditions.

Question 14.
Information for synthesis of proteins is stored in which part of cell.
Answer:
DNA.

Question 15.
What is gene for protein?
Answer:
Segment of DNA which directs synthesis of protein is called gene for protein.

Question 16.
Name the structure or mole-cules which control the traits.
Answer:
Gene.

Question 17.
Name the carriers of genes.
Answer:
Chromosomes.

Question 18.
What is gene according to molecular structure?
Answer:
A segment of DNA that provides information for the synthesis of gene for protein is called gene.

Question 19.
Name the alternative form of gene.
Answer:
Allele.

Question 20.
How many pairs of chromo-somes are present in male/female human?
Answer:
23 pairs.

Question 21.
Write sex-chromosome of female.
Answer:
XX.

Question 22.
Write sex-chromosome of male.
Answer:
XY.

Question 23.
Which chromosome determine the sex of child.
Answer:
XX-chromosome in female and XY-chromosome in male.

Question 24.
If in vegetation of green plants, what will happen if green beetles are not distinguished as compared to red beetles.
Answer:
The population of green beetles will increase and subsequently red population will decrease.

Question 25.
What is evolution?
Answer:
Descend with modifications is termed evolution.

Question 26.
What is the basis of evolution?
Answer:
Variations are the basis of evolution.

Question 27.
What is the effect of starvation on DNA in beetles?
Answer:
No effect.

Question 28.
What are the causes of inbuilt tendency of variations?
Answer:

1. Sexual reproduction
2. Errors in DNA copying

Question 29.
Why crows could not eat coloured beetles?
Answer:
Crows could not see green coloured beetles as they get matched with green leaves and bushes.

Question 30.
What is meant by the term extinction?
Answer:
Elimination of a species is termed extinction.

Question 31.
What is the basis of Darwin’s theory of evolution?
Answer:
Natural selection/Survival of the fittest.

Question 32.
Who proposed the theory of origin of life from abiotic chemicals?
Answer:
Haldane.

Question 33.
Name the scientist who conducted experiments to prove abiotic origin of life?
Answer:
Stanley Miller and Urey.

Question 34.
What is genetic drift?
Answer:
It refers to chance elimination of genes of certain traits when a section of a population dies or migrates.

Question 35.
Define macroevolution.
Answer:
Macroevolution involves large scale changes among group of species

Question 36.
Genetic drift occur in small or large population.
Answer:
Small in number.

Question 37.
What is the structural unit of life?
Answer:
Cell.

Question 38.
Name a cell without nucleus.
Answer:
Micro-organism (Bacteria).

Question 39.
What is fossil?
Answer:
Fossils are preserved remains, traces of organisms that lived in the past.

Question 40.
The species sharing more common characteristics will be close or distant related?
Answer:
Closely related.

Question 41.
Give the three key factors of the modern concept of evolution.
Answer:
Genetic variations, natural selection and isolation.

Question 42.
What is the cause of sickle-cell anaemia?
Answer:
Gene mutation which changes the composition of haemoglobin and shapes of RBCs.

Question 43.
Name any fossil animal which serves as connecting link. Which two groups does it connect?
Answer:

1. Archaeopteryx
2. It is a connecting link between reptiles and birds.

Question 44.
What are factors of Mendel?
Answer:
Genes.

Question 45.
What is the modern term given to the factor of Mendel?
Answer:
Gene is the fector of Mendel; It is chemically a segment of DNA.

Question 46.
Coin one word for a class of individuals which are morphologically similar.
Answer:
Phenotype.

Question 47.
Coin the term for the character which does not allow the expression of a contrasting character in a hybrid.
Answer:
Dominant.

Question 48.
Coin the term for individual which breed true for its character.
Answer:
Homozygous.

Question 49.
What are inherited traits?
Answer:
Inherited traits. The distinguish-able feature of a character is called a trait. The traits which are passed from parents to offsprings are called inherited traits.

Question 50.
What is Mendel’s monohybrid ratio?
Answer:
3 :1.

Question 51.
Write down Mendel’s dihybrid ratio for phenotypes.
Answer:
9 : 3 : 3: 1

Question 52.
Write the genotype of man with blood group ‘A’.
Answer:
IAIA, IAI°.

Question 53.
What are two major functions of DNA?
Answer:
Replication and expression of genetic information in the form of poly-peptide (Protein).

Question 54.
Define speciation.
Answer:
Origin of species is termed speciation.

Question 55.
What is paleontology?
Answer:
Study of fossils is termed paleontology.

Question 56.
What are the kinds of organs shown in the figure?

Answer:
Homologous organs.

Question 57.
What are the kind of organs shown in the figure?

Answer:
Analogous Organ.

Question 58.
Write the blood group of progeny P and Q.

Answer:Blood group of P = A
Blood Group of Q = 0

Question 59.
What are vestigial organs?
Answer:
These are organs of the body which are non-functional in the possessor but were functional in the ancestralss and related organism.

Question 60.
Write any four vestigial organs of human.
Answer:
Muscles of ear lobes, appendix, wisdom tooth, hair of chest.

Question 61.
State ‘Biogenetic law’.
Answer:
Ontogeny repeats phylogeny.

Question 62.
Name the gases used by Urey and Miller for their experimnt.
Answer:
CH₄, NH₃, H₂ and H₂O.

Question 63.
What is biogenesis?
Answer:
Life is always formed from pre-existing life.

Question 64.
What is locus?
Answer:
Site of gene on a chromosome is called locus.

Question 65.
In the diagram what is the sex of (A) and (B)?

Answer:
The sex of (A) is female (XX) The sex of (B) is male (XY)

Multiple Choice Questions

Question 1.
Branch of biology deals with heredity and variation is called
(A) Paleontology
(B) Evolution
(C) Genetics
(D) Ecology.
Answer:
(C) Genetics

Question 2.
The factors which represent the contrasting pairs of characters are called
(A) Dominant
(B) Recessive
(C) Determinants
(D) Alleles.
Answer:
(D) Alleles

Question 3.
Two allelic genes are located on
(A) the same chromosome
(B) two homologous chromosomes
(C) two non-homologous chromosomes
(D) any two chromosomes.
Answer:
(B) two homologous chromosomes

Question 4.
Mendel’s law of segregation is based on separation of alleles during
(A) gamete formation
(B) seed formation
(C) pollination
(D) embryonic development.
Answer:
(A) gamete formation

Question 5.
The ratio of phenotype in F₂ generation of a dihybrid cross is:
(A) 3:1
(B) 1:2:1
(C) 2:1
(D)9:3:3:1.
Answer:
(D) 9: 3 : 3: 1

Question 6.
The composition male sex chromosomes is :
(A) XX
(B) XYX
(C) YXY
(D) XY.
Answer:
(D) XY

Question 7.
How many pairs of chromosomes are present in male and female?
(A) 33
(B) 43
(C) 23
(D) 46.
Answer:
(C) 23

Question 8.
Who postulated ‘the natural selection’ as the basis of evolution ?
(A) Darwin
(B) Haldane
(C) Lamark
(D) Newton.
Answer:
(A) Darwin

Question 9.
How many years back human has started to grow the wild cabbage as food?
(A) 20
(B) 200
(C) 2000
(D) 20000.
Answer:
(C) 2000

Question 10.
How the age of fossils is determined?
(A) fossil dating
(B) DNA
(C) gene
(D) biological evolution.
Answer:
(C) gene

Fill in the blanks:

Question 1.
_________ are units of heredity.
Answer:
Genes.

Question 2.
There are _________ pairs of chromosomes in human.
Answer:
23.

Question 3.
Site of gene on a chromosome is called _________
Answer:
Locus.

Question 4.
Continuity of life is maintained through _________ and _________
Answer:
Genetics and Evolution.

Question 5.
Mendel formulated Law of purety of gametes on the basis of _________
Answer:
Dihybrid cross.

Question 6.
The composition of female sex-chromosomes is _________
Answer:
XX.

Question 7.
Wing of butterfly and wing of birds are examples of organs.
Answer:
Analogous.

Punjab State Board PSEB 10th Class Science Book Solutions Chapter 9 Heredity and Evolution Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 9 Heredity and Evolution

PSEB 10th Class Science Guide Heredity and Evolution Textbook Questions and Answers

Question 1.
A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as :
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw.
Answer:
(c) TtWW.

Question 2.
An example of homologous organs is :
(а) our arm and a dog’s fore-leg
(б) our teeth and an elephant’s tusks
(c) potato and runners of grass
(d) All of the above.
Answer:
(d) All of the above.

Question 3.
In evolutionary terms, we have more in common with :
(а) a Chinese school-boy
(b) a chimpanzee
(c) a spider
(d) a bacterium.
Answer:
(a) a Chinese school-boy.

Question 4.
A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:
On this basis we cannot say that light eye colour is dominant or recessive until a cross is made between parent having light eye colour and another with dark eye colour. Only then it will be possible to predict the dominant or recessive nature of the gene.

Question 5.
How are the areas of study of evolution and classification interlinked?
Answer:
Evolution and classification are interlinked as evident from following points :

• Characteristics are shared by most of the organisms. The characteristic in the next level of classification will be shared by most and not by all.
• Cell designs also indicate this relationship.
• Groups formed during classification are related to their similarities.

Question 6.
Explain the terms homologous and analogous organs with example.
Answer:
Homologous organs: The organs of different classes have different forms because they have to perform different functions but their structures basically remain similar. Such organs are called homologous organs.

Example:

• Fore limbs of amphibians, birds and mammals have same fundamental structural plans but perform different functions.
• In plants, the homologous organs may be a thorn of Bougainvillea or a tendril of cucurbita both arising in axillary position.

Analogous organs: The organs are quite different in their structure and origin but similar in function. Such organs are known as analogous organs. The presence of analogous organs proves that different structures can be modified to perform a similar function. Analogy indicates convergent evolution.
Examples. The wings of insects and vertebrates perform the same function.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
Make a chart or thermocol sheet showing the following monohybrid cross

Dominance of black coat colour in dogs

Question 8.
Explain the importance of fossils in deciding evolutionary relationship.
Answer:

Question 9.
What evidence do we have for the origin of life from inanimate matter?
Answer:
Urey and Miller provided experimental evidence regarding origin of life from inanimate matter. They assembled an atmosphere similar to that, thought to exist on early earth.

In a spark flask they collected ammonia, methane and hydrogen sulphide, but no free oxygen over water at a temperature just below 100°C and sparks were passed through the mixture of gases to stimulate lightning. At the end they obtained organic molecules such as amino acid, urea, sugars. Amino acids which make up protein molecules. Thus they showed life originated from inanimate matter.

Question 10.
Explain how sexual reproduction gives rise to more viable Variation than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:
Genetic variations arise in nature as a result of following mechanism during sexual reproduction are more viable and raw materials of evolution.

• Crossing over during gamete formation.
• Random segregation of chromosome during meiosis at the time of gamete formation has decreased.
• Random rejoining of gametes having different genetic set up in the chromosomes during fertilisation.

Question 11.
How is equal genetic contribution of male and female parents ensured in the progeny?
Answer:
During sexual reproduction fusion of gametes having haploid set of chromosomes belonging to male and female parents ensure equal contribution.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Yes. The organism with useful variations will adapt and survive. Moreover they leave behind more offsprings and populations with such genetic variations will survive.

Science Guide for Class 10 PSEB Heredity and Evolution InText Questions and Answers

Question 1.
If a trait A exists in 10% of population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer:
In asexually reproducing organism trait B originated earlier. The variations in a population are only due to inaccuracies of DNA copying.

Question 2.
How does the creation of variations in a species promote survival?
Answer:
The useful variation in individuals of a species will enable them to adapt according to the changes and new needs. Thus they will enable the survival of species.

Question 3.
How do Mendel’s experiments show that gene may be dominant or recessive?
Answer:
Mendel conducted experiments on garden pea plant selecting seven visible contrasting characters. He selected and crossed homozygous tall pea plant having the genotype TT with a homozygous dwarf pea plant having the genotype tt. Fx generation consists only of tall plants, having genotype Tt. Since they have an allele for dwarfness also, they are all hybrids. The expressed allele T for tallness is dominant over the unexpressed allele t for dwarfness. The fact that the allele for dwarfness is present in the F1 plants can be verified by interbreeding them when F2 progeny will consist of both tall and dwarf plants in the ratio of 3 : 1. On this basis he proposed “Law of Dominance.”

Question 4.
How do Mendel’s experiments proved that traits are inherited indepen dently?
Answer:
Mendel proposed a law on the basis of a dihybrid cross between two homozygous parents. He selected a dominant plant with round and yellow seeds and a recessive plant with wrinkled and green seeds, yields Fx offspring showing the dominant form of both traits, viz. round and yellow. Fx plants, on selling, produce F2 progeny with two parental and two new recombinant phenotypes, that is round yellow: round green: wrinkled yellow: wrinkled green in the ratio of 9 : 3 : 3: 1. This ratio is called Mendel’s dihybrid phenotypic ratio. The factors (genes) of different traits are independent of each other in their distribution into the gametes and in the progeny. This is Mendel’s law of independent assortment.

Question 5.
A man with blood group A married a person with blood group O. Their daughter has blood group O. Is this information enough to tell you which of the blood group trait A or O is dominant. Why or why not?
Answer:
As blood groups is hereditary character, the knowledge of blood groups of parents can give information about the possible blood groups of children and vice-versa.

In this case illustration is as follow :

In the above cross, 50 per cent of progeny will have A blood group and 50 per cent O blood group.
At the same time this data is insufficient. It is not mentioned father has homozygous or heterozygous A blood group. If it is homozygous A then 100 per cent of progeny will have A blood group as Gene IA is dominant over Gene I°.

Question 6.
How is the sex of child determined in human beings?
Answer:
Determination of the sex of child. Sex chromosomes determine sex in human beings. In males, there are 44 +
XY chromosomes, whereas, in female there are 44 + XX chromosomes. Here,
X and Y chromosomes determine sex in the human beings.

Sex determination in man (Note that all the eggs carry X-chromosome but one-half of the sperm carry an X-chromosome and one half carry a Y-chromosome)

Two types of gametes are formed in male, one type is having 50%
X-chromosome, whereas the other type is having Y-chromosome. In female, gametes are of one type and contain X-chromosome.

The females are homogametic. If male gamete having Y-chromosome (androsperm) undergoes fusion with female gamete having X-chromosome the zygote will have XY chromosomes and this gives rise to male child.

If the male gamete having Fig. 9.1. Sex determination in man (Note X-chromosome undergoes fusion with that all the eggs carry X-chromosome but female gamete having X-chromosome, one-half of the sperm carry an the zygote will be having XX-chromosome X-chromosome and one half carry a and this gives rise to female child. Y-chromosome)

Question 7.
What are different ways in which individuals with a particular trait may increase in a population?
Answer:
A particular population with specific traits will increase due to following reasons :

• Sexual reproduction which results into variations.
• Inheritance of variations.
• Natural Selection. The individuals with special traits survive the attack of their predators and multiply while the others will perish.
• Genetic drift provides diversity without any adaptation.

Question 8.
Why are traits acquired during life-time of an individual not inherited?
Answer:
Change in non-reproductive tissue (somatic cells) cannot be passed on to the DNA of germ cells. Thus the acquired trait will die with the death of individual. It is non- heritable and cannot be passed on to its progeny. Changes that occur in DNA of germ cells are inherited.

Question 9.
Why are the small number of surviving tigers is a cause of worry from the point of view of genetics?
Answer:
As the population of tigers is decreasing, there is loss of genes from the gene pool. There cannot be recombinations and variations. Hence no evolution. If number falls suddenly they may become extinct.

Question 10.
What factors could lead to the rise of new species?
Answer:
Factors leading to rise of new species

• Genetic variations
• Mutations
• Natural selection
• Reproductive isolation
• Origin of new species.

Question 11.
Will geographical isolation be a major factor in the speciation by a self- pollinating plant species? Why or why not?
Answer:
No, m self-pollinating species, geographical isolation will not play any role for speciation because the self-pollination is occurring on the same plant.

Question 12.
Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?
Answer:
No, as there is neither genetic drift nor gene flow play any role during speciation. Moreover asexual reproduction involves single parent and natural geographical barrier can occur between different organisms.

Question 13.
Give an example of characteristic being used to determine how close two species are in evolutionary terms.
Answer:
Homologous organs helps to identify the relationship between organisms. These characteristics in different organisms would be similar because they are inherited from a common ancestor. Example. Fore limbs of mammals having same basic structural plans in birds, reptiles and mammals however the functions get modified in different species.

Question 14.
Can the wing of butterfly and wing of a bat be considered homologous organs? Why or why not?
Answer:
Wings of insects and wings of birds have different basic structural plan and origin. They perform the same function. Thus they are analogous organs and not homologous organs.

Question 15.
What are fossils? What do they tell us about the process of evolution?
Answer:
Fossils are preserved remains, tracks or traces of organisms that lived in the past. Fossils have been found linking all major groups of vertebrates.

Significance of fossils

• Fossils are direct evidence in support of evolution.
• Living forms with simple organization appeared earlier than the complex forms. We can conclude this because fossils of lower layers of the earth are simple as compared to fossils of the upper layers.
• Several forms bearing intermediate characters indicate the transition from an earlier simple to a later complex.
• Fossils of Archaeopteryx serve as a missing link between reptiles and birds. This bird has wings and unlike birds, it had teeth and a long tail.
• On the basis of the fossil records, the complete evolutionary history of present-day horse has been studied.

Question 16.
Why are human beings which look so different from each other in terms of size, colour, and looks are said to be belonging to the same species?
Answer:

• DNA studies have shown that human beings belong to the same species.
• The number of chromosomes is the same.
• All have originated from a common ancestor.
• They interbreed among themselves to produce fertile young ones of their own kind.

Question 17.
In evolutionary terms can we say that which among bacteria, spiders, fish, and chimpanzees have a ‘better’ body design? Why or why not?
Answer:
Chimpanzees have a better body design as compared to the other three mentioned. They are better adapted for locomotion, communication, and thinking.

Punjab State Board PSEB 10th Class Hindi Book Solutions Chapter 1 दोहावली Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Hindi Chapter 1 दोहावली

Hindi Guide for Class 10 PSEB दोहावली Textbook Questions and Answers

(क) विषय-बोध

I. निम्नलिखित प्रश्नों के उत्तर एक या दो पंक्तियों में दीजिए

प्रश्न 1.
तुलसीदास जी के अनुसार राम जी के निर्मल यश का गान करने से कौन-से चार फल मिलते हैं ?
उत्तर:
तुलसीदास जी के अनुसार राम जी के निर्मल यश का गान करने से धर्म, अर्थ, काम और मोक्ष नामक चार फल मिलते हैं।

प्रश्न 2.
मन के भीतर और बाहर उजाला करने के लिए तुलसी कौन-सा दीपक हृदय में रखने की बात करते हैं?
उत्तर:
तुलसी हृदय में श्रीराम नाम रूपी मणियों के दीपक को रखने की बात करते हैं।

प्रश्न 3.
संत किस की भाँति नीर-क्षीर विवेक करते हैं?
उत्तर:
संत हंस की भाँति नीर-क्षीर विवेक करते हैं।

प्रश्न 4.
तुलसीदास के अनुसार भव सागर को कैसे पार किया जा सकता है?
उत्तर:
तुलसीदास के अनुसार श्रीराम से स्नेह, सांसारिक प्राणियों से समता तथा राग, रोष, दोष, दुःख आदि का त्याग करने से भव सागर को पार किया जा सकता है।

प्रश्न 5.
जो व्यक्ति दूसरों के सुख और समृद्धि को देखकर ईर्ष्या से जलता है, उसे भाग्य में क्या मिलता है?
उत्तर:
जो व्यक्ति दूसरों की सुख-समृद्धि से ईर्ष्या में जलता है उसे अपने जीवन में कभी सुख की प्राप्ति नहीं होती।

प्रश्न 6.
रामभक्ति के लिए गोस्वामी तुलसीदास किसकी आवश्यकता बतलाते हैं?
उत्तर:
गोस्वामी तुलसीदास रामभक्त के लिए ईश्वर की भक्ति, नीर-क्षीर विवेकी होने और परम आस्तिकता के भावों की आवश्यकता बतलाते हैं। रामभक्ति के लिए संत समागम और हृदय की पवित्रता आवश्यक है।

II. निम्नलिखित पद्यांशों की सप्रसंग व्याख्या कीजिए

(1) प्रभु तरुतर कपि डार पर, ते किए आपु समान।
तुलसी कहुँ न राम से, साहिब सील निधान।
(2) सचिव, वैद, गुरु तीनि जो, प्रिय बोलहिं भयु आस।
राज, धर्म, तन तीनि कर, होइ बेगिही नास।
उत्तर:
(1) कवि कहता है कि प्रभु श्री राम जी तो वृक्षों के नीचे और बंदर वृक्षों की डालियों पर रहते थे, परंतु ऐसे बंदरों को भी उन्होंने अपने समान बना लिया। तुलसीदास जी कहते हैं कि श्रीरामजी जैसे शीलनिधान स्वामी अन्य किसी स्थान पर कहीं भी नहीं हैं।

(2) कवि कहता है कि स्वामी से तो वह सेवक बड़ा होता है जो अपने धर्म के पालन को करने में निपुण होता है। इसीलिए स्वामी श्रीराम तो सागर पर पुल बंधने के बाद ही समुद्र पार कर सके परंतु उनका सेवक हनुमान तो बिना पुल के ही समुद्र लांघ गया था।

(ख) भाषा-बोध

निम्नलिखित शब्दों के विपरीत शब्द बनाएं-

शब्द = विपरीत शब्द
संपत्ति = ——–
सेवक = ——–
भलाई = ——–
लाभ = ———-
उत्तर:
शब्द = विपरीत शब्द
संपत्ति = विपत्ति।
सेवक = स्वामी।
भलाई = बुराई।
लाभ = हानि।

2. निम्नलिखित शब्दों की भाववाचक संज्ञा बनाएं-

दास = ——–
निज = ——–
गुरु = ——–
जड़। = ———
उत्तर:
शब्द . विपरीत शब्द
दास = दासता।
निज = निजता।
गुरु = गुरुत्व।
जड़ = जड़ता।

3. निम्नलिखित के विशेषण बनाएँ-

धर्म = ——–
मन = ———
भय = ———
दोष। = ———
उत्तर:
शब्द – विशेषण शब्द
धर्म = धार्मिक।
मन = मानसिक।
भय = भयनक।
दोष = दोषी।

(ग) पाठ्येतर सक्रियता

प्रश्न 1.
अपने विद्यालय के पुस्तकालय से गोस्वामी तुलसीदास से संबंधित पुस्तकों से उनके जीवन की अन्य घटनाओं के बारे में जानकारी प्राप्त करें।
उत्तर:
(विद्यार्थी स्वयं करें।)

प्रश्न 2.
तुलसीदास द्वारा रचित दोहों की ऑडियो या वीडियो सी०डी० लेकर अथवा इंटरनेट से इन दोहों को सुनकर आनंद लें और स्वयं भी इन को याद कर लय में गाने का अभ्यास करें।
उत्तर:
(विद्यार्थी स्वयं करें।)

प्रश्न 3.
इंटरनेट के माध्यम से राष्ट्रीय दूरदर्शन पर दिखाए ‘तुलसीदास’ के जीवन पर आधारित सीरियल को ग्रीष्म अवकाश में देखें। उत्तर:
(विद्यार्थी स्वयं करें।)

(घ) ज्ञान विस्तार

रामभक्त गोस्वामी तुलसीदास जी द्वारा रचित हनुमान चालीसा का प्रथम दोहा ‘श्री गुरु चरन सरोज रज’ कवि की हनुमान जी के प्रति श्रद्धा-भक्ति का प्रतीक है। हनुमान जी को केसरी नंदन तथा अंजनि पुत्र भी कहते हैं। केसरी इनके पिता तथा अंजना माता थी। लंका में सीता का समाचार लाते समय ये एक ही छलांग में समुद्र लांघ गए थे और जब इनकी पूंछ में आग लगा दी गई थी तब इन्होंने रावण की सारी लंका ही जला दी थी। लक्ष्मण मूर्छा के समय ये ही लंका से सुषेण वैद्य को तथा उसके कहने पर संजीवनी बूटी को लाए थे। इन्होंने ही सुग्रीव की श्रीराम जी के साथ मित्रता करवा कर उसे उसका राज्य दिलवाया था। इस प्रकार हनुमान महाबली तथा श्री राम जी के अनन्य सेवक माने जाते हैं।

PSEB 10th Class Hindi Guide दोहावली Important Questions and Answers

प्रश्न 1.
कवि के अनुसार श्रीराम जी जैसा शीलनिधान अन्य कोई क्यों नहीं है?
उत्तर:
श्रीराम जैसा शील निधान अन्य कोई भी नहीं था क्योंकि श्रीराम ने वानरों को भी पूरा सम्मान दिया था।

प्रश्न 2.
बिना हरि कृपा के क्या प्राप्त नहीं हो सकता?
उत्तर:
हरि कृपा के बिना संत-समागम नहीं प्राप्त हो सकता।

प्रश्न 3.
ईर्ष्यालु व्यक्ति की क्या दशा होती है?
उत्तर:
ईर्ष्यालु व्यक्ति सदा ईर्ष्या की आग में जलता रहता है तथा उसका कभी भी भला नहीं होता है।

प्रश्न 4.
भगवान् से भक्त को बड़ा बताने के लिए तुलसीदास ने कौन-सा उदाहरण दिया है?
उत्तर:
श्रीराम को लंका जाने के लिए समुद्र पर पुल बंधवा कर जाना पड़ा जबकि उनके भक्त हनुमान छलांग लगा कर ही समुद्र पार कर लंका पहुँच गए थे।

प्रश्न 5.
चापलूस दरबारियों से क्या हानि होती है?
उत्तर:
चापलूस दरबारी सदा राजा की हाँ में हाँ मिलाते हैं, उसकी आलोचना कभी भूलकर भी नहीं करते जिससे राजा अपने राज्य, धर्म और अपने शरीर का भी नाश झूठे अभिमान के कारण कर देता है।

प्रश्न 6.
कैसी भक्ति के बिना श्रीराम अपने भक्तों पर कृपा नहीं करते?
उत्तर:
जिस भक्ति में श्रीराम के प्रति विश्वास नहीं होता उन भक्तों पर श्रीराम कृपा नहीं करते।

(क) एक पंक्ति में उत्तरात्मक प्रश्न

प्रश्न 1.
तुलसी ने हृदय में कैसे दीपक को रखने की बात कही है?
उत्तर:
तुलसी ने हृदय में श्रीराम नाम रूपी मणियों के दीपक के रखने की बात कही है।

प्रश्न 2.
राम की कृपा के बिना स्वप्न में भी क्या नहीं मिलता ?
उत्तर:
राम की कृपा के बिना स्वप्न में भी जीवन में चैन नहीं मिलता।

प्रश्न 3.
श्रीराम ने सागर कैसे पार किया था ?
उत्तर:
श्रीराम ने सागर पर बने पुल से सागर पार किया था।

प्रश्न 4.
हंस की भांति नीर-क्षीर विवेक कौन करते हैं?
उत्तर:
संत हंस की भांति नीर-क्षीर विवेक करते हैं।

(ख) निम्नलिखित प्रश्नों के उत्तर एक सही विकल्प चुनकर लिखें

प्रश्न 1.
‘श्री गुरु चरन सरोज रज’ में सरोज है
(क) सरोवर
(ख) कमल
(ग) सरिता
(घ) धूल।
उत्तर:
(ख) कमल

प्रश्न 2.
‘निज मन मुकुरु सुधारि’ में मुकुरु है
(क) मुकरना
(ख) मौसम
(ग) शीशा
(घ) सुधरना।
उत्तर:
(ग) शीशा

प्रश्न 3.
भगवान् पर क्या किए बिना भक्ति नहीं हो सकती
(क) दया
(ख) विश्वास
(ग) प्रार्थना
(घ) समझौता।
उत्तर:
(ख) विश्वास

प्रश्न 4.
‘दोहावली’ किसकी रचना है
(क) तुलसीदास
(ख) वृन्द
(ग) बिहारी
(घ) रहीम।
उत्तर:
(क) तुलसीदास

(ग) एक शब्द/हाँ या नहीं/सही-गलत/रिक्त स्थानों की पूर्ति के प्रश्न

प्रश्न 1.
संत किस की भाँति नीर-क्षीर विवेक करते हैं? (एक शब्द में उत्तर दीजिए)
उत्तर:
हँस

प्रश्न 2.
राम जी का निर्मल यशगान करने से चार फल मिलते हैं। (हाँ या नहीं में उत्तर लिखें)
उत्तर:
हाँ

प्रश्न 3.
अपने धर्म में निपुण सेवक साहब से बड़ा नहीं होता। (सही या गलत लिख कर उत्तर दें)
उत्तर:
गलत

प्रश्न 4.
तुलसीदास राम के भक्त थे। (सही या गलत लिख कर उत्तर दें)
उत्तर:
सही

प्रश्न 5.
बरनऊं रघुबर ……. जसु।
उत्तर:
बिमल

प्रश्न 6.
गिरिजा संत ……….. सम।
उत्तर:
समागम

प्रश्न 7.
राज, धर्म, तन तीनि कर, होइ ……. नास।
उत्तर:
बेगिही।

दोहावली दोहों की सप्रसंग व्याख्या

1. श्री गुरु चरन सरोज रज, निज मन मुकुरु सुधारि।
बरनऊँ रघुबर विमल जसु, जो दायकु फल चारि॥

शब्दार्थ:
चरन = पैर। सरोज = कमल। रज = धूल। मकरु = दर्पण। बरनऊँ = वर्णन करूँ। विमल = निर्मल, उज्ज्वल। जसु = यश। दायकु = देने वाले। फल चारि = चार फल-धर्म, अर्थ, काम और मोक्ष।

प्रसंग:
प्रस्तुत दोहा तुलसीदास द्वारा रचित ‘दोहावली’ से लिया गया है। यह दोहा हनुमान चालीसा का प्रारंभिक दोहा है। इस दोहे में गुरु वंदना के बाद श्रीराम के पवित्र चरित्र के गुणगान करने की कवि ने कामना की है।

व्याख्या:
कवि कहता है कि मैं श्री गुरु महाराज के चरण कमलों की धूल से अपने मनरूपी दर्पण को स्वच्छ और पवित्र करके श्री रघुवीर रामजी के निर्मल पवित्र यश का वर्णन करता हूँ, जो चारों फलों धर्म, अर्थ, काम और मोक्ष को देने वाला है।

विशेष:

1. कवि अपने सद्गुरु की कृपा प्राप्त कर श्रीराम के उज्ज्वल चरित्र का गुणगान करने की कामना कर रहा है।
2. भाषा ब्रज मिश्रित अवधी, दोहा छंद, अनुप्रास, उपमा तथा रूपक अलंकार हैं।

2. राम नाम मनी दीप धरु, जीह देहरी द्वार।
तुलसी भीतर बाहरु हुँ, जौ चाहसि उजियार॥

शब्दार्थ:
मनी = मणियाँ। दीप = दीपक। जीह = जीभ । देहरी = दहलीज। चाहसि = चाहता है। उजियार = उजाला।

प्रसंग:
प्रस्तुत दोहा तुलसीदास द्वारा रचित ‘दोहावली’ से लिया गया है, जिसमें कवि ने श्रीराम के नाम स्मरण की महिमा का वर्णन किया है।

व्याख्या:
तुलसीदास जी कहते हैं कि हे मानव, तू राम नाम रूपी मणि-दीपक को मुखरूपी द्वार की जीभरूपी दहलीज पर रख। यदि तू भीतर और बाहर दोनों ओर उजाला चाहता है। भाव यह है कि राम नाम रूपी मणियों से बने दीपक को हृदय में धारण करने से अज्ञान रूपी अंधेरा नष्ट हो जाता है तथा ज्ञान रूपी उजाला हो जाता है।

विशेष:

1. राम नाम के स्मरण से समस्त दोषों का नाश हो जाता है तथा हृदय के बाहर और भीतर उजाला हो जाता है।
2. ब्रज भाषा में अवधी के शब्दों का मिश्रण है। दोहा छंद, रूपक अलंकार है।

3. जड़ चेतन गुन दोषभय, बिस्व कीन्ह करतार।
संत हंस गुन गहहिं पय, परिहरि बारि विकार॥

शब्दार्थ:
जड़ = निर्जीव। चेतन = सजीव। बिस्व = संसार। करतार = ईश्वर, परमात्मा। गहहिं = लेकर। पय = दूध। परिहरि = छोड़कर। बारि = पानी। विकार = बुराई।

प्रसंग:
प्रस्तुत दोहा तुलसीदास द्वारा रचित ‘दोहावली’ में से लिया गया है, जिसमें कवि ने संतों की विशेषता का वर्णन किया है।

व्याख्या:
तुलसीदास कहते हैं कि परमात्मा ने इस जड़-चेतन संसार को गुण और दोष से युक्त बनाया है, परंतु संत हँसों के समान नीर-क्षीर विवेकी होने के कारण दोष रूपी जल को त्याग कर गुण रूपी दूध को ग्रहण करते हैं।

विशेष:

1. संत सदा सद्गुणों से युक्त होते हैं। वे विकारों से रहित होते हैं।
2. अवधी भाषा, दोहा छंद, अनुप्रास तथा रूपक अलंकार का प्रयोग सराहनीय है।

4. प्रभु तरुतर कपि डार पर, ते किए आपु समान।
तुलसी कहुँ न राम से, साहिब सील निधान। 

शब्दार्थ:
तरुतर = वृक्ष के नीचे। कपि = वानर। आयु = अपने।

प्रसंग:
प्रस्तुत दोहा तुलसीदास के द्वारा रचित ‘दोहावली’ से लिया गया है। इस दोहे में कवि ने श्रीराम द्वारा वानरों को दिए गए सम्मान का वर्णन किया है।

व्याख्या:
कवि कहता है कि प्रभु श्री राम जी तो वृक्षों के नीचे और बंदर वृक्षों की डालियों पर रहते थे, परंतु ऐसे बंदरों को भी उन्होंने अपने समान बना लिया। तुलसीदास जी कहते हैं कि श्रीरामजी जैसे शीलनिधान स्वामी अन्य किसी स्थान पर कहीं भी नहीं हैं।

विशेष:

1. श्रीराम की सबके प्रति समतावादी दृष्टि का उल्लेख किया गया है। वे कोई भी भेदभाव न करते हुए सबको अपने समान बना लेते थे।
2. अवधी भाषा, दोहा छंद, अनुप्रास अलंकार है।

5. तुलसी ममता राम सो, समता सब संसार।
राग न रोष न दोष दुःख, दास भए भव पार॥

शब्दार्थ:
ममता = स्नेह, प्रेम। समता = बराबर। राग = प्रेम। रोष = क्रोध। भव = संसार।

प्रसंग:
प्रस्तुत दोहा तुलसीदास द्वारा रचित ‘दोहावली’ से लिया गया है। इस दोहे में कवि ने श्रीराम के प्रति आस्था रखने तथा सांसारिक प्राणियों से समभाव रखने की प्रेरणा दी है।

व्याख्या:
तुलसीदास जी कहते हैं कि श्रीराम से ममता रखो तथा संसार के प्राणियों के प्रति समभाव रखो, इससे मनुष्य राग, रोष, दोष, दुःख से मुक्त हो जाता है तथा श्रीराम का दास होने के कारण इस संसार रूपी सागर से पार हो जाता है।

विशेष:

1. सांसारिक माया-मोह के बंधनों से मुक्त होकर भव-सागर पार करने के लिए श्रीराम के प्रति ममता होनी आवश्यक है।
2. अवधी भाषा, दोहा छंद, अनुप्रास तथा रूपक अलंकार हैं।

6. गिरिजा संत समागम सम, न लाभ कछु आन।
बिनु हरि कृपा न होइ सो, गावहिं वेद पुरान॥

शब्दार्थ:
गिरिजा = पार्वती। सम = समान। आन = अन्य, दूसरा।

प्रसंग:
प्रस्तुत दोहा तुलसीदास द्वारा रचित ‘दोहावली’ से लिया गया है। इस दोहे में कवि ने संत-समागम की महिमा का वर्णन किया है।_

व्याख्या:
कवि कहता है कि शिव जी पार्वती को संतों के सम्मेलन की महिमा का वर्णन करते हुए कहते हैं कि हे पार्वती, संतों के साथ मिल बैठकर उनके विचार सुनने के समान संसार में अन्य कुछ भी लाभकारी नहीं है तथा संत-समागम भी श्रीराम की कृपा के बिना नहीं मिलता। ऐसा वेद-पुराणों में भी कहा गया है।

विशेष:

1. संत-समागम प्रभु कृपा से प्राप्त होता है तथा इसके समान लाभदायक संसार में और कुछ भी नहीं है।
2. भाषा अवधी, दोहा छंद तथा अनुप्रास अलंकार हैं।

7. पर सुख संपत्ति देखि सुनि, जरहिं जे जड़ बिनु आगि।
तुलसी तिन के भाग ते, चलै भलाई भागि॥

शब्दार्थ:
पर = पराया, दूसरे का। जरहिं = जलना, ईर्ष्या करना। जड़ = मूर्ख भाग = भाग्य। भागि = भाग जाना, चले जाना।

प्रसंग:
प्रस्तुत दोहा तुलसीदास द्वारा रचित ‘दोहावली’ से लिया गया है, जिसमें कवि ने ईर्ष्यालु व्यक्ति की दुर्दशा का वर्णन किया है।

व्याख्या:
गोस्वामी तुलसी दास कहते हैं कि दूसरे के सुख और संपत्ति को देख-सुन कर जो मूर्ख बिना आग के ही जलते रहते हैं, उन लोगों के भाग्य से भलाई स्वयं ही भाग जाती है। भाव यह है कि दूसरों की उन्नति को देखकर ईर्ष्या करने वाले व्यक्ति का कभी भी भला नहीं होता है।

विशेष:

1. ईर्ष्यालु व्यक्ति का कभी भी भला नहीं होता है और न ही सुख की प्राप्ति होती है।
2. भाषा अवधी-ब्रज का मिश्रण, दोहा छंद, अनुप्रास अलंकार हैं।

8. सचिव वैद गुरु तीनि जो, प्रिय बोलहिं भयु आस।
राज, धर्म, तन तीनि कर, होइ बेगिही नास॥

शब्दार्थ:
सचिव = मंत्री। वैद = वैद्य। प्रिय बोलहिं = मीठा बोलना, मुँह देखी बोलना। भयु = डर से। तीनि = तीनों। बेगिही = शीघ्र ही। नास = नाश, विनाश।

प्रसंग:
प्रस्तुत दोहा तुलसीदास द्वारा रचित ‘दोहावली’ से लिया गया है, जिसमें कवि ने राजा के भय से उसकी हाँ में हाँ मिलाने वाले चापलूसों के परिणाम का वर्णन किया है।

व्याख्या:
कवि कहता है कि यदि किसी राजा का मंत्री, वैद्य और गुरु–ये तीनों राज-भय से अथवा किसी लोभलालच से उसकी बात जैसी है वैसी ही मान लेते हैं अथवा उसकी हाँ में हाँ मिलाते हैं तो उसके राज्य, धर्म और शरीर तीनों का शीघ्र ही विनाश हो जाता है।

विशेष:

1. तुलसी दास का मानना है कि रावण के भय से उसके सभासद सदा उसकी हाँ में हाँ मिलाते थे, इसलिए उसका विनाश हो गया था। इसलिए ऐसे चापलूसों से बचना चाहिए।
2. भाषा अवधी और ब्रज है। दोहा छंद तथा अनुप्रास अलंकार है।

9. साहब ते सेवक बड़ो, जो निज धरम सुजान।
राम बाँध उतरै उद्धि, लांघि गए हनुमान॥

शब्दार्थ:
साहब = स्वामी। उद्धि = समुद्र। बाँध = पुल। सुजान = निपुण।

प्रसंग:
प्रस्तुत दोहा तुलसीदास द्वारा रचित दोहावली से लिया गया है, जिसमें कवि ने भगवान् से अधिक उनके भक्त की प्रशंसा की है।

व्याख्या:
कवि कहता है कि स्वामी से तो वह सेवक बड़ा होता है जो अपने धर्म के पालन को करने में निपुण होता है। इसीलिए स्वामी श्रीराम तो सागर पर पुल बंधने के बाद ही समुद्र पार कर सके परंतु उनका सेवक हनुमान तो बिना पुल के ही समुद्र लांघ गया था।

विशेष:

1. स्वामी की कृपा से सेवक स्वामी से भी बड़ा काम कर सकता है।
2. भाषा अवधी, ब्रज, दोहा छंद, अनुप्रास अलंकार हैं।

10. बिनु बिस्वास भगति नहिं, तेहि बिनु द्रवहिं न राम।
राम कृपा बिनु सपनेहुँ, जीवन नहिं विश्राम।

शब्दार्थ:
द्रवहिं = द्रवित होना, पिघलना, दया करना। लह = प्राप्त करना, लेना।

प्रसंग:
प्रस्तुत दोहा तुलसीदास द्वारा रचित ‘दोहावली’ से लिया गया है। इस दोहे में कवि ने अपने आराध्य पर पूर्ण विश्वास रखते हुए भक्ति करने का संदेश दिया है।

व्याख्या:
कवि कहता है कि बिना भगवान् पर विश्वास किए उनकी भक्ति नहीं हो सकती। विश्वास से रहित भक्ति से श्रीराम अपने भक्त पर दया नहीं करते और राम की कृपा के बिना स्वप्न में भी जीवन में चैन नहीं मिलता है। भाव यह है कि विश्वासपूर्वक भक्ति करने से ही परमात्मा की प्राप्ति होती है।

विशेष:

1. ईश्वर की भक्ति उस पर पूर्ण विश्वास रख कर ही करनी चाहिए तभी ईश्वर की कृपा होती है।
2. भाषा अवधी, ब्रज हैं। दोहा छंद और अनुप्रास अलंकार है।

दोहावली Summary

दोहावली  कवि परिचय।

राम-भक्त कवियों में तुलसीदास का नाम विशेष आदर से लिया जाता है। उनका जन्म सन् 1532 ई० में राजापुर, ज़िला बाँदा में हुआ था। कुछ विद्वान् उत्तर प्रदेश के एटा जिले में सोरों नामक ग्राम को उनका जन्म स्थान मानते हैं। इनके पिता का नाम आत्मा राम तथा माता का नाम हुलसी था। मूल नक्षत्र में उत्पन्न होने के कारण माता-पिता ने इन्हें त्याग दिया था। इनका बचपन दर-दर की ठोकरें खाते हुए अनेक कष्टों में बीता। बाद में बाबा नरहरिदास ने उन्हें सहारा दिया। उनके पास रहकर उन्होंने शिक्षा प्राप्त की थी। बाद में काशी के महान् विद्वान शेष सनातन से उन्होंने वेद-शास्त्रों और इतिहास-पुराण का ज्ञान प्राप्त किया। विद्वान बनकर वे वापस राजपुर लौटे थे। तब दीनबंधु पाठक ने अपनी पुत्री रत्नावली से इनका विवाह करवा दिया था। वे अपनी पत्नी रत्नावली से बहुत प्यार करते थे। एक दिन वह इनको बताए बिना अपने मायके चली गई। तुलसीदास जी को जब पता चला तो अन्धेरी रात तथा मूसलाधार वर्षा में रत्नावली के पास पहुँच गए। इस पर रत्नावली ने उन्हें फटकार सुनाई और राम के चरणों में स्नेह लगाने की प्रेरणा दी।

लाज न लागत आपको दौरे आयह साथ,
धिक धिक ऐसे प्रेम को कहां कहौ हौं, नाथ।

इस घटना से तुलसीदास का हृदय ग्लानि से भर गया। उन्होंने संसार को त्याग कर अपना सारा जीवन भगवान राम की आराधना तथा भक्तिपरक साहित्य लिखने में लगा दिया। सन् 1623 ई० में तुलसीदास जी का स्वर्गवास हुआ।

रचनाएँ-तुलसीदास के नाम से 37 पुस्तकें स्वीकार की जाती हैं। लेकिन इनमें से तुलसी के प्रामाणिक ग्रन्थ बारह ही हैं, वे हैं–रामचरितमानस, वैराग्य संदीपनी, रामललानहछू, बरवै रामायण, पार्वती मंगल, जानकी मंगल, रामाज्ञा-प्रश्न, दोहावली, कवितावली, गीतावली, कृष्ण गीतावली और विनय पत्रिका।

तुलसीदास ने अपने काव्य में राम को विष्णु का अवतार मान कर उन्हें ईश्वर पद प्रदान किया है-‘सोई दशरथ सुत हित, कौसलपति भगवान्।’ उन्होंने माना है कि राम ही धर्म का उद्धार करने वाले हैं तथा उनमें शील, शक्ति और सौंदर्य के गुण विद्यमान हैं। राम के माध्यम से कवि ने अपने काव्य में आदर्श समाज की कल्पना की है। उन्होंने राम, सीता, भरत, लक्ष्मण, कौशल्या, हनुमान आदि के द्वारा आदर्श गृहस्थ, आदर्श समाज और आदर्श राज्य की कल्पना को साकार रूप दिया है। इनकी रचनाओं का मूल रस शांत है। लेकिन स्थान-स्थान पर अन्य सभी रसों का सुंदर प्रयोग दिखाई दे जाता है। कवितावली के बालकांड में वात्सल्य रस के सुंदर उदाहरण दिए गए हैं। तुलसीदास एक श्रेष्ठ कवि और सच्चे लोकनायक हैं।

इन्होंने अपने काव्य में जीवन के विविध रूपों को प्रस्तुत किया है। इन्होंने अवधी और ब्रज भाषाओं का संदर प्रयोग किया है। इनकी अधिकांश रचनाएँ अवधी में हैं। लेकिन ‘विनय पत्रिका’ में ब्रज भाषा का प्रयोग किया गया है। इनके प्रबंध काव्य में दोहाचौपाई छंदों का प्रयोग अधिक है तो मुक्तक काव्यों में गीति शैली’ की प्रधानता है। इन्होंने आवश्यकतानुसार उर्दू, फ़ारसी, बुंदेली, भोजपुरी आदि शब्दों का प्रयोग किया है। तुलसीदास वास्तव में ही उत्कृष्ट कोटि के भक्त कवि हैं।

दोहावली दोहों का सार

पाठ्यपुस्तक में तुलसीदास द्वारा रचित दस दोहे संकलित हैं, जिनमें कवि की भक्ति एवं नीति से संबंधित भावनाएँ व्यक्त हुई हैं। पहले दोहे में कवि अपने गुरु की वंदना कर अपने पवित्र मन से चारों फलों को देने वाला श्रीराम के पावन चरित्र के गुणगान करने की कामना करता है। दूसरे दोहे में श्रीराम रूपी मणियों के दीपक के प्रकाश से मन के अंधकार को दूर करने तथा तीसरे दोहे में संतों को नीर-क्षीर विवेकी हँसों के समान बताया गया है जो गुणों को अपना कर समस्त विकार त्याग देते हैं। चौथे दोहे में श्री राम की वानरों को सम्मान देने, पांचवें दोहे में श्रीराम के प्रति ममता तथा संसार के सभी लोगों से समता का व्यवहार रखने और छठे दोहे में संतों के समागम के लाभ का वर्णन किया गया है।

सातवें दोहे में दूसरों की संपत्ति को देखकर ईर्ष्या करने वालों की दुर्दशा का वर्णन है। आठवें दोहे में कवि ने राम भक्त हनुमान की प्रशंसा की है। नौवें दोहे में चापलूस सभासदों से राजा को सावधान रहने के लिए कहा गया है क्योंकि जी हजूरी करने वालों से धर्म, शरीर और राज्य का नाश हो जाता है। दसवें दोहे में कवि ने स्पष्ट किया है कि पूर्ण आस्था से भक्ति करने पर ही श्रीराम अपने भक्तों पर कृपा करते हैं तथा श्रीराम की कृपा के बिना स्वप्न में भी शांति नहीं मिलती है।

Punjab State Board PSEB 10th Class Science Book Solutions Chapter 8 How do Organisms Reproduce? Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce?

PSEB 10th Class Science Guide How do Organisms Reproduce? Textbook Questions and Answers

Question 1.
Asexual reproduction takes place through budding in :
(a) Amoeba
(b) Yeast
(c) Plasmodium
(d) Leishmania.
Answer:
(b) Yeast.

Question 2.
Which of the following is not a part of the female reproductive system in human beings?
(a) Ovary
(b) Uterus
(c) Vas deferens
(d) Fallopian tube
Answer:
(c) Vas deferens.

Question 3.
The anther contains :
(a) Sepals
(b) Ovules
(c) Carpel
(d) Pollen Grains
Answer:
(d) Pollen grains.

Question 4.
What are advantages of sexual reproduction over asexual reproduction?
Answer:
Advantages of sexual reproduction:
Sexual reproduction has a dual significance for the species.

• It results in multiplication and perpetuation of the species.
• It contributes to evolution of the species by introducing variation in a population much more rapidly than asexual reproduction.

Question 5.
What are the functions performed by the testis in human beings?
Answer:
Functions of testis

• Produce male gametes, called sperms.
• Leydig’s cells secrete male sex hormone namely testosterone into blood.

Question 6.
Why does menstruation occur?
Answer:
During menstruation, broken down endometrium is passed out as menstrual flow along with unfertilized egg (ovum).

Question 7.
Draw a labelled diagram of Vertical section of flower.
Answer:
Vertical section of flower

V.S of Folwer

Question 8.
Wliat are different methods of contraception?
Answer:
The various methods of birth control are :

Flowchart: Brief account on birth control measures

Question 9.
How are modes of reproduction different in unicellular and multicellular organisms?
Answer:
Unicellular organisms mostly reproduce by asexual methods except for a few such as Paramecium, Eimeria, these organisms reproduce by sexual methods also. The multicellular organisms have more complex body. They reproduce both by asexual and sexual methods. But sexual reproduction is the more common method.

Question 10.
How does reproduction help in providing stability to populations of species?
Answer:
Reproduction is the process by which organisms increase their populations. The rate of births and deaths in a given population determine its size.

Question 11.
What could be the reasons for adopting contraceptive methods?
Answer:

• To control population.
• For family planning.
• For good health of women and children.

Science Guide for Class 10 PSEB How do Organisms Reproduce? InText Questions and Answers

Question 1.
What is the importance of DNA copying in reproduction?
Answer:

• DNA copying is called DNA replication. In this process one copy each of replicated DNA will be passed on two daughter cells. Each cell formed will have its own cellular apparatus to control the activities of the daughter cells.
• It maintains the body designs and features in different generations of the species.
• Variations may be introduced during copying of DNA. This inbuilt tendency for variation during reproduction forms the basis of evolution.

Question 2.
Why is variation beneficial to the species but not necessarily for the individuals?
Answer:
Variations present in some populations would enable them to survive. In case of changing environmental condition such as global warming. Thus it is useful for survival of species over time.

Question 3.
How does binary fission differ from multiple fission?
Answer:
Differences between binary fission and multiple fission

Binary fission	Multiple fission
1. Parent cell divides into two daughter cell.	1. Parent cell divides into many daughter cells.
2. It takes place during favourable conditions.	2. It takes place during unfavourable conditions.
3. Residual cytoplasm is left.	3. Nothing is left with parent.
4. Cytoplasm divides after every nuclear division.	4. Cytoplasm does not divide after every nuclear division.

Question 4.
How will an organism be benefited if it reproduces by spores?
Answer:

• Spores are covered by thick walls, that protect them until they come into contact with another substratum on which they can grow.
• They are produced in large numbers.

Question 5.
Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?
Answer:

• Regeneration is not reproduction because most organism would not depend on being cut up.
• It is not certain that cut up portion will be able to survive and give rise to a new individual.

Question 6.
Why is vegetative propagation practised for growing some types of plants?
Answer:
Advantages of vegetative propagation

• It is usually a means of propagating such plants which do not produce viable seeds.
• It is rapid method of producing young ones.
• It helps in retaining useful characters from generation to generation.

Question 7.
Why is DNA copying an essential part of the process of reproduction?
Answer:

• DNA copying provides cellular apparatus in the daughter cells.
• DNA in daughter cells will be able to control the functioning of daughter cells.
• DNA copies will retain the traits.

Question 8.
How is the process of pollination different from fertilisation?
Answer:
Differences between pollination and fertilization

Pollination	Fertilization
1. It is the transference of pollen grains from anther to stigma.	1. It is fusion of a male and a female gamete.
2. An external agent is required like wind, water and animal.	2. It does not require any external agent.

Question 9.
What is the role of seminal vesicles and prostate gland?
Answer:
Role of seminal vesicles. These glands secrete viscous secretions, which contain fructose and prostaglandins. The fructose is the source of energy for sperms and prostaglandins stimulate uterine contraction and thus may help the sperm to move towards the female’s oviduct.

Role of prostate gland. It secretes an alkaline milky fluid that aid in sperm motality. The fluid contains small amount of citric acid, some lipids and a few enzymes. It also contains bicarbonate ions which give the semen its alkaline pH.

Question 10.
What are the changes seen in girls at the time of puberty?
Answer:
Changes at the time of puberty: These changes occur under the influence of hormones FSH (Follicle stimulating hormone) and estrogen.

• Growth of breast and external genitalia.
• Darkening of nipple skin.
• Broadening of pelvis.
• Growth of pubic and axillary hair.
• Increase in subcutaneous fat.
• Initiation of menstruation and ovulation.

Question 11.
How does embryo get nourishment inside the mother’s body?
Answer:
Embryo gets nourishment through placenta. Placenta is a disc embedded in the wall of uterus. It contains villi on the embryo side of tissue. On the mother side are blood spaces which surround villi. Placenta serves to bring the foetal and maternal blood close enough to permit exchange of materials between the two. Placenta also acts as endocrine gland.

Question 12.
If a woman is using a copper-T, will it help her in protecting sexually transmitted diseases?
Answer:
Copper-T prevents fertilization but the chances of infection persist. Thus it will not help her in protecting from sexually transmitted diseases.

Punjab State Board PSEB 10th Class Science Book Solutions Chapter 7 Control and Coordination Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 7 Control and Coordination

PSEB 10th Class Science Guide Control and Coordination Textbook Questions and Answers

Question 1.
Which of the following is plant hormone?
(A) Insulin
(B) Thyroxin
(C) Oestrogen
(D) Cytokinin.
Answer:
(D) Cytokinin.

Question 2.
The gap between two neurons is called :
(A) Dendrite
(B) Synapse
(C) Axon
(D) Impulse.
Answer:
(B) Synapse.

Question 3.
The brain is responsible for :
(A) Thinking
(B) Regulating the heart beat
(C) Balancing
(D) All of above.
Answer:
(D) All of above.

Question 4.
What is the function of receptors in the body? Think of situations where receptors do not work properly. What problems are likely to arise?
Answer:
Receptors. They are meant for receiving and detecting the information from environment. They are located in sense organs. They receive the information detected by tips of dendrites and convey them as electrical impulses.
If receptors do not detect the information, there will not be any co-ordination. It may lead to accidents. Body response will not be there.

Question 5.
Draw a labelled diagram of neuron and explain its function.
Or
Draw the structure of neuron and label the following on it.
Nucleus, Dendrite, Cell body and Axon.
Answer:
Functions of Neuron (Nerve cell).

• Nerve cells are specialised for conducting information via electrical impulses from one part of the body to another part.
• Dendrites acquire the information.
• Axon conducts information as electrical impulse.
• Terminal arborization pass the information as chemical stimulus at synapse for onward transmission.

A Nerve Cell

Question 6.
How does phototropism Occur in planes?
Answer:
Phototropism.
It is an established fact that plants bend towards light when they are exposed to it from one side of long axis. The aerial parts are positively phototropic and the roots and other underground parts bend away from light. These movements are due to interaction of light and auxins. The unilateral growth causes bending of stem as tip grows more rapidly.

Question 7.
Which signals will get disrupted in case of spinal cord injury?
Answer:

• Spinal cord mainly controls reflex actions in the body. Spinal cord is made up of nerves which supply information to think about. Thus these actions will get disrupted in case of injury to spinal cord.
• Sensation and movement are restricted.

Question 8.
How does chemical co-ordination occur in plants?
Answer:
Chemical coordination in plants

• Plants do’not have well organised nervous and muscular system, still they respond to stimuli and co-ordinate in the best possible way.
• Different kinds of stimuli trigger the release of chemicals called plant hormones or phytohormones.
• These phytohormones help to coordinate growth development and responses to the environment.
• Auxins stimulate the cells to grow longer on one side of shoot in response to light thus it bends way.
• Gibberellins help in growth of stem.
• Cytokinins promote cell division.
• Abscisic acid inhibits growth. Its effects include wilting of leaves.

Question 9.
What is the need for a system of control and co-ordination in an organism?
Answer:
Control and Co-ordination in the body is of two types i.e. nervous control and hormonal control. Nervous control is rapid. It takes place through electrical signals called nerve impulses. The hormonal control is through chemical messengers called hormones secreted by endocrine (ductless) glands and carried by blood to the target organs.

Question 10.
How are involuntary actions different from reflex action?
Answer:
Reflex action is sudden and quick response to input (stimulus). It saves from many accidents. It is controlled by spiral cord. The involuntary actions are controlled by midbrain and hind brain. These are the actions in which thinking does not have any control. These actions are blood pressure, salivation and cycling.

Question 11.
Compare and contrast the nervous and hormonal mechanism for control and coordination in animals.
Answer:
Differences between endocrine system and nervous system

Endocrine System	Nervous System
1. The action of endocrine system is often very diverse, affecting many cells and sometimes several organs found in different parts of the body.	1. The action of nervous system is limited to a few muscle fibres or gland cells of an organ or organ system.
2. The system is not directly connected to organs or tissues under its control.	2. The system is directly connected to tissues or organs under its control.
3. It exerts its control through hormones or chemical regulators poured into circulatory system.	3. Nervous system exerts its control through chemical stimulants poured directly over the tissues or organs.
4. The information is transmitted slowly.	4. The information is transmitted almost instantaneously.
5. The system takes time to produce response. It, therefore, regulates those processes where the response is not immediately required.	5. It controls process where an immediate response is required.
6. The effect is long lasting.	6. The effect is short lived.

Question 12.
What is the difference between the manner is which movement takes place in a sensitive plant and the movement in our leg.
Answer:
Movement of human leg is under the control of nervous system. It is under the
direction of motor area which, controls voluntary muscles of leg. The sensitive plant “Touch me not,” moves its leaves in response to touch but without any regulation by nervous tissue.

Science Guide for Class 10 PSEB Control and Coordination InText Questions and Answers

Question 1.
What is the difference between reflex action and walking?
Answer:
Differences between reflex action and walking:

Reflex Action	Walking
1. It is an inborn action and present in an individual right from birth.	1. Walking is acquired through learning.
2. It is controlled by spinal cord.	2. It is controlled by hind brain.
3. It cannot be changed.	3. It can be changed.
4. It is involuntary action.	4. To start with it is voluntary and later on it becomes involuntary.
5. Response is given by muscles and glands.	5. Response is given by muscles.

Question 2.
What happens at the synapse between two neurons?
Answer:
Synapse is a junction between terminal ends of one axon and dendrites of adjacent neuron. There is a gap between the two called synaptic cleft. As the electrical impulse reaches the terminal knobs, they release chemicals called neurotransmitters. These chemicals cross the gap and start a similar electrical impulse in the dendrite of next neuron. Synapse ensures that nerve impulse travels in one direction only.

Question 3.
Which part of the brain maintains posture and equilibrium of the body?
Answer:
Cerebellum part of hind brain maintains posture and equilibrium.

Question 4.
How do we detect the smell of an agarbatti (Incense stick)?
Answer:
Sensory information regarding smell is received by olfactory lobes of brain. As the air passes through the nasal chambers, the olfactory epithelial cells get stimulated and convey the information as electrical impulses to brain which have the power of interpretation.

Question 5.
What is the role of the brain in reflex action?
Answer:
Thinking centres are located in brain. Brain is the co-ordinating centre. Brain and spinal cord in coordination with each other control all voluntary and involuntary actions. The spinal reflexes are produced in the spinal cord but the message of reflex action taken also goes to brain where the thinking process occurs. Some reflex arcs involve the brain rather than the spinal cord only.

Question 6.
WTiat are plant hormones? Name any two.
Answer:
Plant hormones. These are chemical compounds secreted by plants which diffuse all around the other cells and regulate the activities. Plant hormones help to co-ordinate growth, development and responses to the environment.
Examples : Auxins and Cytokinin.

Question 7.
How is the movement of leaves of the sensitive plant different from the movement of a shoot towards light?
Answer:
Sensitive plant show seismonastic movements. It is due to turgidity of cells. The movement of a shoot is a tropic movement.

Movement of the leaves of sensitive plant	Movement of shoot towards light
1. It is a nastic movement which does not depend on the direction of stimulus.	1. It is a tropic movement.
2. The stimulus is touch.	2. The stimulus is light.
3. It is caused by sudden loss of water from the swelling present at the base of the leaves.	3. It is caused by the unequal growth on the two sides of the shoot.
4. It is not a growth movement.	4. It is a growth movement.

Question 8.
Give an example of a plant hormone that promotes growth.
Answer:
IAA = Indole-3-Acetic Acid.

Question 9.
How do auxins promote the growth of a tendril around a support?
Answer:
Auxins synthesised in the tip helps the cells to grow longer. Some plants like the pea plant climb up other plants or support by means of tendrils. These tendrils are sensitive to touch. When they come in contact with any support, the part of the tendril in contact with the object does not grow as rapidly as the part of the tendril away from the object. This causes the tendril to circle around the object and thus cling to it. It is due to accumulation of auxins.

Question 10.
Design an experiment to demonstrate hydrotropism.
Answer:
Experiment to demonstrate hydrotropism

Demonstration of hydrotropism

• Take a porous pot and fill it with water.
• Keep a few freshly germinated pea seedlings in a dried sand.
• As the water is not available in sand the root growing will bend towards porous pot filled with water.
• You wall observe a hydrotropic curvature of the root as it grows towards water.
• This bending of root show the movement as a response towards water.

Question 11.
How does chemical coordination take place in animals?
Answer:
Chemical coordination. It is brought about by chemical messengers called hormones. They are secreted by endocrine glands (ductless glands). The hormones are carried by the blood to the site of action (target organs). The hormones are consumed during their action. Hormones provide wide ranging changes.

Question 12.
Why is the use of iodised salt advisable?
Answer:
Iodine is necessary for the thyroid gland to make thyroxine hormone. Thyroxine regulates carbohydrate, protein and fat metabolism. It is required for growth. In case of deficiency of thyroxine, a disorder called goitre is caused. Thus, the use of iodised salt is advisable .to prevent iodine deficiency disorders in the body.

Question 13.
How does our body respond when adrenaline is secreted into blood?
Answer:
Adrenaline is secreted by the adrenal gland during emergencies. It prepares the body to respond effectively. Following actions occur :

• Heartbeats faster so as to pump the blood to muscles that need more energy.
• The blood supply to the digestive system and skin is reduced due to the contraction of muscles around these organs. This helps in diverting blood supply to muscles.
• Breathing becomes fast.

Question 14.
Why are some patients with diabetes treated by giving injections of insulin?
Answer:
Insulin hormone is secreted by Islets of Langerhans of the pancreas. This hormone helps in regulating sugar levels in the blood. Its deficiency results in high sugar levels and causes many harmful effects. To bring the sugar level to normal in such diabetic patients, insulin injections are given.

Punjab State Board PSEB 10th Class Science Important Questions Chapter 6 Life Processes Important Questions and Answers.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Long Answer Type Questions

Question 1.
What are the kinds of heterotrophic organisms on the basis of nutrition?
Answer:
Heterotrophic nutrition is a mode of nutrition in which organisms obtain readymade organic food from outside source. The organisms that depend upon outside sources for obtaining organic nutrients are called heterotrophs. It is a characteristic feature of all animals and non-green plants, that are unable to utilize carbon and synthesi s organic compounds necessary for life, but depends upon organic sources of carbon. They are thus dependent upon autotrophic organisms (Plants) and are called as heterotrophs.

It is of the following types :

• Saprophytic nutrition: In this type of nutrition, an organism lives upon dead organic sources such as dead plants and dead animals. These usually secrete dissolving and digesting enzymes and absorb the liquefied molecules so formed e.g. yeast, bread moulds and dung moulds etc.
• Parasitic nutrition: In this type of nutrition, an organism lives totally at the expense of others and derives its food material and shelter from the other .These organisms which derive food material are called parasites and the organism from which food is derived is called as host. This type of nutrition is termed as parasitic or holozoic nutrition: It is also known as parasite-host relationship e.g. Cuscuta, Ascaris etc.
• Holozoic nutrition. It is a mode of heterotrophic nutrition which involves intake of solid pieces of food. Since solid food is taken in, Holozoic nutrition is also called ingestive nutrition. Holozoic nutrition (GK. “Holo”-Whole; “Zoon”-Animal) is found in animals and protists. The food may consist of another animal, plant or its parts.

Depending upon the source of food, holozoic organisms are of three types :

1. Herbivores: These organisms obtain their food from plants e.g. cow, rat, deer and goat etc.
2. Carnivores: These organisms take the flesh of other organisms as their food e.g. tiger, cheetah, snake, eagle etc.
3. Omnivores: These can feed on plants and flesh of other organisms e.g. human, cockroach, crow etc.

Question 2.
What is photosynthesis? Describe the significance of photo-synthesis.
Answer:
Photosynthesis (Photos-Light, Synthesis-putting together) may be defined as an anabolic process in which green plants manufacture complex organic food substances (carbohydrate) from simple inorganic compounds like carbon dioxide and wrater in presence of sunlight with the aid of chlorophyll and evolve out oxygen as a byproduct of the process. Thus photosynthesis is a process in which radiant energy is converted into chemical energy

In other words photosynthesis is a series of oxidation- reduction reaction in which CO₂ is reduced and H₂O is oxidized to produce carbohydrates and oxygen.

• Chloroplasts are the actual sites for photosynthesis.
• All green parts of a plant are involved in photosynthesis.
• Leaves are the most important organs of photosynthesis.

Section of Leaf showing site of photosynthesis

Significance of photosynthesis:

• Photosynthesis is a source of all our food and fuel. It is the only biological process that acts as the driving vital force for the whole animal kingdom and for the non-photosynthetic organisms.
• It drives all other processes of biological and abio- logical world. It is responsible for the growth and sustenance of our biosphere.
• It provides organic substances, which are used in the production of fats, proteins, nucleoproteins, pigments, enzymes, vitamins, cellulose, organic acids, etc. Some of them become structural parts of the organisms.
• It makes use of simple raw materials such as CO₂, H₂O and inexhaustible light energy for the synthesis of energetic organic compounds.
• It is significant because it provides energy in terms of fossil fuels like coal and petrol obtained from plants, which lived millions and millions of years ago.
• Plants, from great trees to microscopic algae, are engaged in converting light energy into chemical energy, while man with all his knowledge in chemistry and physics cannot imitate them.

Question 3.
Which organelle is responsible for photosynthesis? Describe the role of chlorophyll.
Answer:
Chloroplast is the organelle responsible for photosynthesis. Chloroplasts contain green pigment called as chlorophyll. Photosynthetic pigments occur in the granum. They constitute the pigment system called photosystem. About 250 to 400 pigment molecules are present in a photosystem.

The primary function of photosystems is to trap light energy and converts it to chemical energy.

• Chloroplast was discovered by Schimper.
• Number of chloroplasts is variable in different species of plants.
• In lower plants like algae they are 1 or 2 number.
• In higher plants their number varies from 40 -100 per palisade cell or more.
• Chloroplasts also have variable shapes, for example cup-shaped, ribbon-shaped etc. in algae while it is discoidal in higher plants.

A typical structure of chloroplast is a double membranous structure having two parts.

• Grana: It is a lamellar system consisting of stacks of granum lamella each bounded by a membranous box called as thylakoid. They are 40 – 60 per cell. Number of thylakoids per grana is 50 or more. Chlorophyll molecules are found inside the thylakoid membrane where they trap solar energy in the form of small energy packets called ‘photon’ or ‘quanta’. Grana are interconnected to each other by a channel called as stroma lamellae or Fret’s channel.
• Stroma: It is a non-pigmented proteinaceous matrix in which grana remain embedded. It contains enzymes for dark reaction.
• Mechanism of Photosynthesis: Photosynthesis is formation of organic food from carbon dioxide and water with the help of sunlight inside chlorophyll containing cells. Oxygen is produced as by-products.

Oxygen comes from water. Hydrogen of water is used to reduce carbon dioxide to form carbohydrate.

Photosynthesis occurs in two main steps:

• The first step is dependent upon light and the second step is not dependent upon light. Hence, the former is called light reaction or photochemical phase while the latter, the dark reaction or biosynthetic phase of photosynthesis.
• Water is split up during photosynthesis by the process called photolysis. It provides reductant for carbon dioxide. Oxygen is liberated. All the liberated oxygen, therefore, comes from water.
• The photolysis of water in photosynthesis was discovered by Hill and hence it is also known as Hill reaction.
• The light reaction of photosynthesis is followed by the dark reaction. In this, CO₂ is first fixed by ribulose diphosphate and from this fixed CO₂ phosphoglyceric acid is formed. The phosphoglyceric acid thus formed ultimately forms carbohydrates.
• The basic organic compound formed in photosynthesis is often considered to be glucose. The storage product of plants is commonly starch.
• A chemical equation of photosynthesis is

In plants and most algae it occurs in the chloroplasts i.e., light reaction and dark reaction.

Question 4.
Discuss the steps of light reaction and dark reaction.
Answer:
Steps of light reaction.
Two main steps of light reaction are :
(a) Photolysis of water
(b) Conversion of fight energy into chemical energy.

• Light reaction takes places in the thylakoid membranes and the intergranal lamellae of the chloroplast in the presence of fight.
• Two photosystems (PSI and PSII) work in a coordinated manner.
• H₂O splits into H⁺ and OH^–.
• H+ is used to reduce NADP to NADPH₂ which is used in dark reaction.
• Photophosphorylation takes place in two ways—cyclic and non-cyclic.
• Light energy is converted into chemical energy.
• The end-products are ATP and NADPH₂.
• As a result of photolysis of water, oxygen is evolved as a by-product.
  4H₂O → 4 OH^– + 4 H⁺
  4 OH^– + 4e^– → 4 OH^–
  OH → 2H₂O + O₂

Summary of dark reaction :

• Dark reaction takes place in the stroma of the living chloroplast.
• Atmospheric CO₂ is absorbed.
• The end products of fight reaction (ATP and NADPH₂) are also used.
• All green plants operate C₃ photosynthetic pathway. Some monocot plants like maize, sugarcane operate both C₃ and C₄ photosynthetic pathways.
• First end product of photosynthesis is sugar.

Question 5.
Explain process of nutrition in Amoeba.
Answer:
Nutrition in Amoeba. Amoeba is omnivorous, i.e. it feeds on smaller animals, plants, micro-organisms and fragments of larger organisms. Nutrition is holozoic. Ingestion can occur at any place on the surface since a regular mouth is absent. Ingestion occurs through phagocytosis or engulfing the food particle in an invagination of the body. The engulfed food particle comes to fie inside a food vacuole. The latter is surrounded by a membrane.

Digestion of food within food vacuole

Digestion: The digestion is intracellular and the food vacuoles act as temporary stomach for digestion.
Absorption: It occurs by diffusion and distribution takes place by cyclosis.
Assimilation: Assimilation of digested material occurs in a single cell.
Egestion: The undigested food is eliminated through the surface of the cell, where the food vacuole containing the undigested food bursts and discharges its contents to the outside.

Question 6.
Difference between the followings :
(i) Autotrophic and Heterotrophic nutrition
(ii) Herbivore and Carnivore.
Answer:
(i) Autotrophic and Heterotrophic nutrition :

Autotrophic Nutrition	Heterotrophic Nutrition
1. Food is self-manufactured.	1. Food is obtained readymade from outside.
2. An external source of energy is required for synthesis of food.	2. An external source of energy is not required. The required energy is present in the food obtained from outside.
3. Inorganic substances constitute the raw materials for manufacturing food.	3. Inorganic substances are not much required.
4. Chlorophyll is present for trapping light energy.	4. Chlorophyll is absent
5. Digestion is absent.	5. An external or internal digestion is required for conversion of complex organic materials into simpler and soluble ones.
6. Organisms performing autotrophic nutrition function as producers. Examples: Green plants, some bacteria, some protists.	6. Organisms performing heterotrophic nutrition function as consumers. Animals, many protists and monerans.

(ii) Herbivore and Carnivore

Herbivore	Carnivore
Animals winch cat only plants. e.g. Cow, goat etc.	They feed on flesh of other animals, e.g. Lion, vulture etc.

Question 7.
Describe human alimentary canal. Draw a labelled diagram of human alimentary canal.
Or
Describe the human alimentary canal with the help of a suitable diagram.
Answer:

• Mammalian (human) alimentary canal comprises mouth, buccal cavity, pharynx, oesophagus, stomach, small intestine (duodenum, jejunum and ileum), large intestine (caecum, colon and rectum) and anus.
• The mouth, bounded by two lips open into oral cavity.
• Buccal cavity contains teeth and tongue, and receives saliva from 3 pairs of salivary glands. Teeth are meant for cutting and mastication of food.
• Pharynx: It is a vertical tube. It is a cross passage for food and air. It has uvula and epiglottis which closes the internal nares and glottis respectively during swallowing of food to ensure the passage of food into oesophagus (food pipe).
• Oesophagus: It is a 25 cm long narrow muscular straight tube. It opens into stomach. Oesophagus propels the swallowed food into stomach.
• Stomach: It is a sac-like structure situated in the upper part of abdominal cavity below the diaphragm. Large part of this sac is situated left of the median line.
• Small Intestine: It is the longest part of alimentary canal. It is thin-walled and highly coiled tubular structure. It is about 3-3.5 metres long and occupies most part of abdominal cavity. It is coiled upon itself. Its inner lining is thrown into numerous villi.

Human digestive system.

Large intestine: The large intestine is about 1.5 metres long. It is divided into following parts, i.e. the vermiform appendix, the colon and the rectum. Caecum is a blind tube and represented by vermiform appendix (5-8 cm) and is present below the junction of small and large intestine. Rectum is the last part and opens to the outside by anus guarded by anal sphincter.

Question 8.
Describe the process of digestion of food in man.
Answer:
Digestion: The process of conversion of non-diffusible form of food into the simple and diffusible form by chemical and mechanical processes in the alimentary canal is called digestion.

• The process of digestion starts in the mouth cavity and is completed in the intestine.
• In the mouth, food gets mixed up with saliva secreted by salivary glands.
• Saliva contains an enzyme ptyalin (salivary amylase) which breaks polysaccharide starch into disaccharide maltose.

• The food from the mouth cavity called bolus passes into the stomach through the oesophagus.
• The gastric glands of the stomach secrete gastric juice which contains hydrochloric acid, protein digesting
• enzyme-pepsin Rennin, mucus and small amount of gastric lipase are also components of gastric juice.
• Pepsin breaks down proteins into peptones and proteoses in acidic medium of stomach.
• Rennin undies the milk.
• Muscles present on the wall of stomach churn and propel the food called chyme forward into duodenum.
• The digested food moves from stomach to duodenum of the small intestine.
• Duodenum receives bile from juices from liver and pancreatic juice from pancreas.
• The pancreatic juice contains trypsin, amylase and lipase.
• The proteins, fats and carbohydrates are further digested into diffusible form amino acids, glycerol and fatty acids, glucose and fructose.
• The intetinal juice consists of amylolytic, proteolytic and lipolytic enzymes.
• Finally, the digestion is completed in the ileum with the secretion of the intestinal juice by intestinal glands.
• Emulsion form of food called chyle is ready for absorption.

Question 9.
Name the constituents of blood and state the functions of each.
Answer:

Constituents	Functions
1. Plasma, (i) Serum	It contains proteins as well as organic and inorganic substances in solution.
(ii) Fibrinogen	It serves to carry the nutritive and waste materials, antibodies, enzymes and hormones.
2. Red blood corpuscles (R.B.Cs.)	Clotting of blood.
(Erythrocytes)	They help to transport oxygen.
3. White blood corpuscles (W.B.Cs)	They help to defend our body against bacteria, as well as the toxins which these organisms may produce. They also help to remove useless dead tissues from the blood.
4. Blood Platelets or Thrombocytes	They play a vey important role in bringing about the coagulation of blood.

Question 10.
State the functions of blood.
Answer:
Blood performs a number of functions in the body, the most important of which are as follows :

• Blood supplies nutrients and oxygen to various organs and cells of the body.
• It carries the waste matter formed in the cells to the excretory organs.
• It regulates the temperature of the body.
• It supplies hormones to different parts of the body.
• It prevents the body from various diseases by destroying the pathogenic germs.
• It prevents excessive loss of nutrients from cuts and wound by forming a clot.

Question 11.
Describe the structure of human heart.
Answer:
Heart.

• It is a highly efficient, pumping organ of body. Human heart consists of 4 chambers: upper smaller right and left atria (auricles) with thinner wall and lower larger right and left ventricles with thicker walls.
• Atria open into the respective ventricles by atrioventricular apertures guarded by atrioventricular valves.
• The two atria are separated from each other by interatrial septum, and the two ventricles are separated from each other by interventricular septum.
• The sinoatrial node (SAN) or the pacemaker is located in the upper wall of right atrium.

Valves of the heart.
Valves are muscular flaps which prevent the blood to flow back through it. Two types of heart valves are distinguished :

1. The Atrioventicular valves: These valves separate the atria from the ventricles. The right side of the heart possesses the tricuspid valve or right atrioventricular valve and the left side of the heart possesses the bicuspid or mitral valve.
2. Semilunar valves: These are located in the arteries leaving the heart. The pocket-shaped pulmonary semilunar valves lies in the opening where the pulmonary trunk leaves the right ventricle and aortic semilunar valve lies at the opening between the left ventricle and aorta.

Question 12.
Draw Human heart and label its parts.
Or
Draw a sectional view of the human heart and label on it, Aorta, Right Ventricle and Pulmonary veins.
Answer:
Human Heart

V.S. of Human heart.

Question 13.
Explain respiratory system of human.
Answer:

Human respiratory system

Question 14.
Describe the mechanism of breathing in human beings.
Answer:
Mechanism of breathing
Breathing. It is simply inhaling fresh air rich in oxygen and exhaling foul air rich in carbon dioxide. Respiratory system of man consists of :

1. Respiratory tract.
2. Respiratory organs (lungs).

Respiratory tract. It is the tract or the path through which fresh air enters the body reaches the lungs and foul air (rich in CO₂) leaves the lungs to come out. It consists of Nasal cavity → Pharynx → Larynx → Trachea → Bronchi → Bronchioles → Alveoli (of Lungs).
Breathing is accomplished through changes in size and air pressure of chest cavity. It involves Inspiration and Expiration.

Inspiration:
Intake of fresh air is called inspiration. It occurs when the chest cavity is increased in size and therefore decrease in pressure.

Expiration:
The expulsion of foul air (rich in CO₂) from lungs is called expiration. It results when the chest cavity is reduced in size.

Question 15.
Draw a well labelled diagram of human excretory system.
Answer:
Human Excretory System.

Human excretory system.

Question 16.
Describe the structure of human urinary system.
Answer:
The urinary system consists of the following :

1. Kidneys
2. Ureters
3. Urinary bladder
4. Urethra.

1. Kidneys: The kidneys are a pair of bean-shaped delicate organs. They are situated one on each side of the mid-dorsal line of the abdominal cavity, just below the level of the stomach.

2. Ureters: They are two tubes about 30 cm long, emerging from each kidney with the pelvis of which they are continuous. The ureters run downwards and inwards and open into the urinary bladder.

3. Urinary bladder: It serves as a reservoir for the urine. It is a hollow muscular organ lined by stratified epithelium. Its average capacity for storage is about 500 mm. It is situated in the cavity of the pelvis just behind the pubic symphysis.

4. Urethra: The urethra in two sexes differ. The male urethra is about 20 cm in length. The female urethra is a short duct of about 4 cm long and it extends from the urinary bladder to the external urethra orifice which is in the vestibule just above and anterior to the vaginal orifice.

Short Answer Type Questions

Question 1.
What are life processes?
Answer:
Life processes: Every living organisms takes food, derives energy, removes wagte materials from their bodies and responds to changes in their environment. These activities are called life processes. In all living organisms there occur the basic life processes such as nutrition, respiration, transportation, excretion and reproduction, which are necessary for survival.

Question 2.
What is nutrition? Briefly explain the two major kinds of nutrition.
Or
Define autotrophic nutrition.
Answer:
Nutrition: All living organisms need matter to build up the body and energy to operate the metabolic reaction that sustains life. The materials which provide these two primary requirements of life are called nutrients or foods. The sum total of processes by which organisms obtain matter and energy is termed nutrition.

Modes of Nutrition:
1. Autotrophic or Holophytic nutrition: All green plants and certain protozoans (Euglena) have evolved a mechanism to directly use the energy7 of sunlight for preparing organic food in their own body from simple raw materials i.e. C02 and H20. These simple inorganic materials are transformed into glucose and oxygen is evolved.

2. Heterotrophic Nutrition: Animal, fungi, (Amoeba) and many bacteria cannot utilize solar energy. They use chemical bond-energy of organic molecules synthesized by other organisms in building their own organic molecules. Such a mode of feeding is termed as heterotrophic nutrition, and the organisms having it are called heterotrophs.

It is of three types :

1. Saprophytic
2. Parasitic
3. Holozoic.

Question 3.
Write a note on saprophytic nutrition.
Answer:
Saprophytic nutrition (Sapros = rotten; phyton = plant) It this organism releases some juices to soften or digest the food and then absorbs the nutrient. Thus they decompose the dead organic matter into simpler substance. Fungi (yeast, moulds, mushrooms) and many bacteria are saprophytic in nutrition.

The saprophytic mode of nutrition can best be exemplified by the common bread mould, Rhizopus. It converts the complex organic food materials of bread i.e. starch into soluble sugars with the help of starch digesting enzymes. These soluble sugars are then absorbed by the fungus.

Question 4.
Explain the two main steps of photosynthesis.
Answer:
Photosynthesis occurs in two main steps.

• The first step is dependent upon light and the second step is not dependent upon light. Hence, the former is called light reaction or photochemical phase while the latter, the dark reaction or biosynthetic phase of photosynthesis.
• Water is split up during photosynthesis by the process called photolysis. It provides reductant for carbon dioxide. Oxygen is liberated. All the liberated oxygen, therefore, comes from water.
• The photolysis of water in photosynthesis was discovered by Hill and hence it is also known as Hill reaction.
• The light reaction of photosynthesis is followed by the dark reaction. In this, CO₂ is first fixed by ribulose diphosphate and from this fixed CO₂ phosphoglyceric acid is formed. The phosphoglyceric acid thus formed ultimately forms carbohydrates.
• The basic organic compound formed in photosynthesis is often considered to be glucose. The storage product of plants is commonly starch.
• A chemical equation of photosynthesis is

In plants and most algae it occurs in the chloroplasts i.e., light reaction and dark reaction.

Question 5.
Give a brief account of factors affecting the process of photosynthesis.
Answer:
Factors affecting the process of photosynthesis.

• Temperature: The rate of photosynthesis increases with increase in temperature upto a maximum of 35°C. However, the rate starts decreasing if the temperature rises beyond 30°C.
• Water: The rate of photosynthesis is slow in water-deficient conditions.
• Carbon dioxide: The rate of photosynthesis increases with an increase in carbon dioxide concentration upto a certain level, beyond which there is no effect on the rate of photosynthesis.
• Anatomy of leaf
• Chlorophyll contents.

Question 6.
What are stomata? Explain role in respiration.
Or
How respiration takes place in roots and stems in plants?
Answer:
Respiration in plants. Plants, during the process of photosynthesis, give off oxygen which is utilized during respiration.

Plant respiration occurs at slower rate.

• In the leaves Stomata are the openings located on the surface of the leaves which are guarded by two kidney-shaped guard cells. Through stomatal opening, air can pass into or out of leaves.
• In woody stems. Lenticelo are the breathing pores located on the surface of the woody stems through which air can pass. Plants do not have any specialised ventilation mechanism.
• In roots. Roots take up oxygen present in between particles through root hairs by diffusion. Root hairs are simple extensions of epidermal cells of root in contact with soil.

In the older parts of roots: Older portions of root hairs lacks root hairs. They are covered with protective layer of dead cells having very small openings called lenticels through which gaseous exchange take place between soil and inner living cell.

Question 7.
Demonstrate with experiment that O₂ is evolved during photosynthesis.
Answer:
Take a beaker filled with water. Add a pinch of baking soda (NaHCO₃) to it and put a Hydrilla plant (Aquatic plant) in it. Cover the plant with a funnel. Invert a test tube containing water over the stem of the funnel. Keep this apparatus in the bright sunlight. After some time bubbles start emerging out from the plant, which gets collected in the upper part of the test tube. Remove the test tube and test the gas with a lighted splinter, it keeps on glowing showing that the gas is a supporter of combustion. Thus, the experiment clearly shows that O₂ is evolved during photosynthesis.

Question 8.
To demonstrate that photosynthesis take place in the presence of chlorophyll
Procedure :

• Take a potted plant with variegated leaves-for example, money plant or crotons.
• Keep the plant in a dark room for three days so that all the starch gets used up.
• Now keep the plant in the sunlight for about six hours.
• Pluck a leaf from the plant. Mark the green areas in it and trace them on a sheet of paper.
• Dip the leaf in boiling water for a few minutes.
• After this immerse it in a beaker containing alcohol.
• Carefully place the above beaker in a water-bath and heat till the alcohol begins to boil.
• What happens to the colour of the leaf? What is the colour of the solution?
• Now dip the leaf in a dilute solution of iodine for a few minutes.
• Take out the leaf and rinse off the iodine solution.
• Observe the colour of the leaf and compare this with the tracing of the leaf done in the beginning (Fig.).

What can you conclude about the presence of starch in various areas of the leaf?

• Observation and Conclusion: Only green patches of variegated leaf take on blue colour. Other parts remain unchanged.
• Photosynthesis take place only in chlorophyll-containing patches of leaf.

Question 9.
To demonstrate that CO₂ necessary for photosynthesis.
Answer:
Procedure:

• Take two healthy potted plants which are nearly of the same size.
• Keep them in a dark room for three days.
• Now place each plant on separate glass plates. Place a watch-glass containing potassium hydroxide by the side of one of the plants.

The potassium hydroxide is used to absorb carbon dioxide.

Experimetal set-up (a) with potassium hydroxide (b) without potassium hydroxide

• Cover both the plants with separate bell-jars as shown in Figure.
• Use vaseline to seal the bottom of the jars to the glass plates so that the set-up is air-tight.
• Keep the plants in sunlight for about two hours.
• Pluck a leaf from each plant and check for the presence of starch as in the above activity.
• Do both the leaves show the presence of the same amount of starch?

What can you conclude from this activity?
Observation and Conclusion. The plant containing CO₂ in the surrounding carry out photosynthesis. Other plant does not. It proves that CO₂ is essential for photosynthesis.

Question 10.
Demonstrate with experiment that light is essential for photosynthesis.
Answer:
Take a de-starched potted plant, which has been kept in dark for 3 to 4 days. Cover one of its leaves completely with a carbon paper so that no light falls on it. Keep the plant in light for 4 to 6 hours. Test the covered leaf and uncovered leaf for starch with iodine test. The covered leaf will show negligible amount of starch, while the uncovered leaf will give positive test for starch. The process clearly shows that light is necessary for photosynthesis.

Question 11.
How is respiration different from breathing?
Answer:
Differences between Breathing and Respiration

Breathing	Respiration
1. It is ventilation or bringing in of oxygenated air and giving out deoxygenated air.	1. Respiration of animals includes breathing, gaseous exchange and catabolic breakdown of food.
2. It is a physical process.	2. Respiration is both a physical and physiological process.
3. Breathing does not liberate energy.	3. It liberates energy.
4. It is restricted to organs where gaseous exchange occurs between blood and atmospheric air.	4. Respiration involves every living ceil of the body.

Question 12.
List three differences between respiration in plants and respiration in animals.
Answer:
Differences in respiration in plants and animals

Respiration in plants	Respiration in animals
1. All the cells of plant parts (root, stem, leaves) perform the respiration individually.	1. It is performed by specific respiratory organs for all the cells of body.
2. There is little transport of gases from one part to the other.	2. Transport of gases is maximum.
3. Rate of respiration is low.	3. Rate of respiration is high.

Question 13.
What is the function of epiglottis?
Answer:
Epiglottis is a flap like structure present at the top of glottis. It closes glottis during swallowing of food thus checks the entry of food into respiratory passage.

Question 14.
Why is food necessary for living organisms?
Answer:
Food provide energy to raw materials for growth and maintenance.

Utility of components of food.

• Carbohydrates are mainly used for producing energy.
• Fats serve as stored concentrated fuel for energy production.
• Proteins are mainly used to build up tissues.
• Mineral salts and vitamins regulate metabolic processes and growth.
• Water is essential for all biological activities.

Question 15.
Explain the mechanism of breathing in human.
Answer:
Inhalation or Inspiration

• The entry of air from outside into alveoli of lungs through respiratory tract is called inhalation.
• The air enters when thoracic cavity expands due to contraction of intercostal muscles attached to ribs and peripheral muscles of the diaphragm.
• Thus the thorax moves upward, outward and forward.
• It increases the volume of thoracic cavity and the pressure decreases.
• Thus air from outside rushes into alveoli of lungs through nostrils, nasal chambers, trachea, bronchi and bronchioles.
• The alveolar sac gets filled with oxygen rich air.

Exhalation or Expiration is concerned with the expelling of carbon dioxide from lungs.

• It takes place when the volume of the thoracic cavity decreases and the pressure of the contained air in the thoracic cavity increases.
• Air passes out through the respiratory tract from the lungs.

Question 16.
What is the effect of sternuos exercise on rate of breathing and why?
Answer:
Normally man breathes about 15-18 times per minute but during hard exercise the breathing rate increases to 20 to 25 times per minute. It is due to the fact that body needs more of energy thus requires more of oxygen.

Question 17.
Give an outline of Calvin-Benson Cycle.
Answer:
Melvin Calvin and Andy Benson discovered this cycle, hence it is called Calvin cycle.

Calvin-Benson Cycle

Question 18.
Label the parts in the figure.
Answer:

Answer:

1. Pseudopodium
2. Food particle
3. Ingestion
4. Food Vacuole
5. Waste
6. Digestion of food.

Question 19.
Describe the role of intestinal juice.
Answer:
Intestinal juice too is alkaline (pH 8.3).

It has many enzymes :

• Intestinal amylase hydrolyses the remaining starch and glycogen to maltose.
• Maltase changes maltose to glucose.
• Sucrase converts sucrose into glucose and fructose.
• Lactase hydrolyses lactose to glucose and fructose.
• Dipeptidases hydrolyse dipeptides to amino acids.
• Intestinal lipase splits emulsified fats into fatty acids and glycerol.
• Alkaline emulsion of digestion products formed in the small intestine is called chyle.

Question 20.
Explain absorption of food in the small intestine.
Answer:
Absorption. The process of diffusion of digested food into blood present in the blood capillaries of smail intestine is called absorption. Inner lining is thrown in fold called villi. They increase surface area for absorption of food.
Glucose, amino acid, vitamins, mineral salts and water diffuses into blood present in blood capillaries of numerous villi of the small intestine. The fatty acids and glycerol diffuses into lymph present in lymph vessels called lacteals. The digested food is carried to the liver by the hepatic portal vein. The fatty acid and glycerol unite in lymph to form fat. Most of the fat passes as a milky emulsion. After absorption, the undigested- food passes into large intestine.

Question 21.
State the functions of stomach and large intestine.
Functions of Stomach
Answer:

• Storage of food.
• Mechanical breakdown of food.
• Partial digestion of food.

Functions of large intestine
Colon:

• Its wall absorbs the water from undigested food.
• Absorption of digested food also takes place in this region which has not been absorbed by ileum.

Question 22.
Differentiate the following :
Respiration and Photosynthesis
Answer:
Differences between Respiration and Photosynthesis

Respiration	Photosynthesis
1. It is a catabolic process in which food substrates are broken down.	1. It is an anabolic process in which food substrates are synthesized.
2. It takes place in all living cells.	2. It is carried out only by the chlorophyll containing cells of plants.
3. CO2 and H2O are produced.	3. CO2 and H2O are used.
4. CO2 is given out.	4. O2 is released as a byproduct.
5. Chemical energy is converted into ATP and some energy is lost as heat.	5. Radiant energy’of light is converted into 1 chemical energy.

Question 23.
What is fermentation? How is it important?
Answer:
The slow decomposition of organic matter into simpler substances in the presence of enzymes is known as fermentation. It is a type of anaerobic respiration. Fermentation literally means a chemical change accompanied by effervescence. The anaerobic breakdown of glucose to carbon dioxide and ethanol is a form of respiration referred to fermentation. It is normally carried by yeast cells and accounts for the production of alcohol in alcoholic bevefages. In fermentation process, if glucose is converted into ethanol then it is called ethanolic fermentation. When glucose is converted into organic acids such as lactic acid, then this type of fermentation is known as lactic acid fermentation. It is carried out by the bacterium Bacillus acidilacti.

Question 24.
Briefly explain the human respiratory system and also label the parts 1, 2, 3, 4, 3 and 6.

Answer:
Human Respiratory System. Respiratory system of human beings and other mammals consists of air passage or respiratory tract, a pair of lungs.

Respiratory tract is made up of nostrils nasal cavity, pharynx, larynx, trachea and bronchi.

1. Left Lung
2. Heart
3. Diaphragm
4. Trachea
5. Larynx
6. Nasal Cavity

The lungs are a pair of brownish grey coloured spongy structures situated in the thoracic cavity. The left lung consists of two lobes while the right lung consists of three lobes.

Each lobule of a lung consists of bronchioles which terminate into a bunch of spherical thin walled air sacs, called alveoli.

Each alveolus or air sac has a diameter of 75 to 300 microns and has a very thin wall. The walls of the alveoli are elastic and are supplied with capillaries. Gases are exchanged between the capillaries and the air sacs through these thin walls.

Question 25.
What are the different modes of respiration in animals?
Answer:
In animals such as earthworm, respiration is by skin.

• The insects have an elaborate tracheal system of respiration.
• Fishes respire through gills.
• Respiratory system of human beings and other mammals consists of air passage or Respiratory tract, and a pair of lungs.
• Respiratory tract is made up of nasal cavity, pharynx, larynx, trachea and bronchi. The lungs are a pair of brownish grey coloured spongy structures situated in the thoracic cavity.
• The left lung consists of two lobes while the right lung consists of three lobes.
• Each lobule of a lung consists of bronchioles which terminate into a bunch of spherical thin-walled air sacs, called alveoli.

Question 26.
Differentiate between respiration and combustion.
Answer:
Difference between respiration and combustion.

Respiration	Combustion
1. It occurs in living cells.	1. It occurs in non living and dead cells.
2. It is a complex biochemical process controlled by several enzymes.	2. It is a chemical process.
3. Heat produced is much less.	3. A large amount of heat is produced.
4. There is no flame or light produced.	4. This process is usually accompanied with flame and light.

Question 27.
What are parasitic nutrition?
Answer:
Parasitic nutrition (Para = besides ; sitos = food). In this an organism (parasite) depends upon the organism (host) for its nutritional requirements. Many bacteria, viruses, fungi, some non-green plants and many animals have this mode of nutrition.

For example, a fungus Puccinia is a parasite on wheat and barberry plants; Cuscuta or dodder plant grows as a parasite on many plants; tapeworms and round worms are parasites in the body of man etc.

Parasites are of two kinds :

1. Ectoparasites and
2. Endoparasites

Question 28.
Differentiate between Light reaction and Dark reaction in photosynthesis.
Answer:
Difference between light reaction and Dark reaction

Light reaction	Dark reaction
1. Light reaction is light induced chemical reaction.	1. Dark reaction requires no light and is purely enzyme controlled reaction.
2. Energy rich compounds like ATP and NADPH2 are synthesized.	2. The energy rich compounds are used to produce the organic compounds.
3. Oxygen is liberated.	3. No liberation of oxygen.
4. It takes place in the grana of the chloroplasts.	4. It takes place in the stroma of chloroplasts.

Question 29.
What are the functions of liver?
Answer:
Functions of liver

1. Role in digestion. Bile produced by liver helps in the digestion of food as follows.
   • It emulsifies the fats with its salts.
   • It prevents decomposition of food by checking the growth of bacteria.
   • It neutralizes the acid coming from the stomach along with food and provide alkaline medium in the intestine required for action of enzymes of pancreas and intestinal glands.
2. Regulation of Blood Sugar: The liver separates the excess of sugar from the blood and stores it in its cells as glycogen (animals starch).
3. Formation of Glycogen from non-carbohydrates Sources.
4. Deamination: In the liver, the amino acids coming from the alimentary canal are sorted out, ammonia is formed. Ammonia is converted to less toxic urea.
5. Excretion: Liver collects haemoglobin of the worn-out red blood corpuscles and changes it into bile pigments.

Question 30.
Differentiate between Saprophytic and Parasitic nutrition.
Answer:

Saprotrophic Nutrition

Parasitic nutrition

Many organisms absorb fluid food through the body surface. This is called saprotrophic nutrition. Bacteria and fungi flourish on dead, decaying organic matter of both plant and animal origin. They secrete digestive enzymes onto this matter. The enzymes hydrolyze the organic matter into simple soluble products that are then absorbed. This method of taking up organic food is known as saprophytic nutrition.

The organisms obtain nutrients from a living host without helping it any way. Examples. Liver fluke lives in the bile duct of sheep and absorbs nutrient. Other examples include several fungi, bacteria and a few higher non-green plants such a cuscuta.

Question 31.
Discuss the fate of food in the oral cavity of man.
Answer:
Fate of food in mouth cavity. The partial digestion of food occurs due to the action of Ptyalin enzyme. It acts on starches and forms maltose.
The sufficiently masticated, partially digested food forms bolus. It is swallowed into oesophagus through gullet by raising the throat aided by the muscles of pharynx.

Question 32.
What is a digestive gland? Name the various digestive glands of man and their secretions.
Answer:
Digestive gland. A gland that secretes digestive juice which is helpful in the digestion of food is called a digestive gland.

Digestive glands of Man

Name of digestive gland	Name of digestive juice/Secretion
1. Salivary gland	Saliva
2. Gastric glands	Gastric juice
3. Pancreas	Pancreatic juice
4. Liver	Bile
5. Intestinal glands	Intestinal juice

Question 33.
Write the enzymes of the pancreatic juice, the substrates they digest and the products of their digestive action.
Answer:
Enzymes of the pancreatic juice

Name of Enzyme	Substrate	Name of end products
1. Amylase	Starch, glycogen	Maltose and Isomaltose
2. Trypsin	Proteoses, Peptones and Proteins	Peptides and amino acid
3. Lipase	Emulsified lipids	Glycerol and fatty acids

Question 34.
What is meant by photosynthesis? What are the basic requirements for the process of photosynthesis?
Answer:
The manufacture of organic compounds from carbon dioxide and water in the presence of sunlight inside the chlorophyll containing cells of the plant is called photosynthesis. The overall reaction of photosynthesis is

Basic requirements for photosynthesis. CO2, water, chlorophyll and solar energy. Photosynthesis is a photo-biochemical process in which energy rich compounds such as carbohydrates (glucose) are synthesized from simple inorganic compounds like CO2 and water in the presence of sunlight and chlorophyll. Oxygen is a by-product.

Question 35.
What are the different ways in which glucose is oxidized to provide energy in various organisms?
Answer:
Breaking down of glucose involves two step process. In the first step, it is broken into three carbon molecule called pyruvate. The pyruvate is further broken down into energy in following different ways in various organisms :

• Aerobic Respiration: In this case, pyruvate is broken down into water and carbon dioxide along with release of energy. It commonly occurs in mitochondria of cells.
• Anaerobic Respiration in Yeast: In yeast cells during fermentation pyruvate is converted into ethanol and carbon dioxide in the absence of oxygen.
• Anaerobic Respiration in Muscles: Due to lack of oxygen, e.g. during vigorous running or exercise, in human muscles, pyruvate is converted into lactic acid.

Question 36.
Explain how exchange of materials takes place between Blood and Tissues.
Answer:
Arteries supply fresh blood with 02 and food materials to different body organs. Inside the body organ the artery divides into smaller branches called arterioles. The arterioles further divide into extremely thin walled blood capillaries. The blood capillaries form an extensive network inside the body organ. They make their way through the tissue cells. Blood plasma along with the dissolved materials comes out of the thin walls of the blood capillaries and collects into the tissue. It is then called tissue fluid, which acts as an intermediate medium between blood and tissue cells.

The tissue fluid contains different materials such as oxygen, amino acids, glucose, mineral ions and proteins etc. which are needed by the body cells. The body cells take up the required materials from the tissue fluid and release their wastes such as C02 and nitrogenous wastes. These enter through the blood capillary wall and dissolve into the blood plasma or enter into the red blood cells and are carried away.

Question 37.
Outline inhalation-exhalation cycle.
Answer:

• Inhalation: Lowering of diaphragm → Rising of rib cage → Gas (O₂) passes to Alveoli
• Exhalation: Air is forced out → Rising of diaphragm → Lowering of ribcage

Question 38.
Leaves of a healthy potted plant are coated with vaseline to block the stomata. Will this plant remain healthy for long? State three reasons to support your answer.
Answer:
No, the plant will not remain healthy because no exchange of gases will take place. It will lead to :

• low respiration
• no photosynthesis occurs
• no transpiration.

Hence plant will not remain healthy and may die eventually.

Question 39.
What are
(i) stomata and
Answer:
Stomata are tiny apertures found on the surface of the leaf, which regulate the exchange of respiratory gases and transpiration.

(ii) lenticels?
Answer:
Lenticels are the raised pores in the woody plants that allow the exchange of gases between the atmosphere and the internal tissues.

Question 40.
Briefly explain breathing, external resperation exchange of gases and tissue resperation.
Answer:

• Breathing involves inhaling of oxygen rich fresh air and exhaling of carbon dioxide rich foul air. The respiratory surface is richly supplied with blood for this purpose. Oxygen of the inhaled air is taken up by blood while carbon dioxide of the blood passes into the air for exhalation.
• The exchange of gases between the blood and the air at the respiratory surface is. known as external respiration.
• Oxygen absorbed by the blood at the respiratory surface is taken to various parts of the body through arteries. Blood loses the oxygen contained in it to tissue fluid, from where it picks up carbon dioxide. The latter is brought to the respiratory surface by blood.
• Tissue respiration, also called internal respiration, is the exchange of gases between the tissue cells and the blood involving uptake of oxygen by tissue cells, oxidation of respiratory substrate and elimination of carbon dioxide by the cells.

Question 41.
Why is mitochondria termed ‘power house’ of the cell?
Answer:

• Most of the aerobic respiration occurs inside the mitochondria and therefore, the latter are also called power houses of the cells.
• Mitochondria are site of synthesis, storage and transport of ATP.
• ATP (Adenosine triphosphate) is the energy currency of the living organisms.
• Adenosine triphosphate. The energy released during cellular respiration is immediately used to synthesise a molecule called ATP from ADP and inorganic phosphate

ATP is used to fuel all other activities in the cell. Therefore, it is said to be the energy currency for . most cellular processes.

Question 42.
The figure shows different ways in which glucose is oxidised to provide energy in various organisms?
Fill up the blanks : A _________ B _________ C _________, D. _________ E _________ F _________

Answer:
Different pathways to provide energy from glucose
A. glucose
B. Pyruvate
C. Absence of Oxygen
D. Presence of oxygen
E. Lactic Acid F. Carbondioxide

Question 43.
Fill in the blanks 1, 2, 3, 4, 5, 6 in the figure.

Double circulation of blood in birds and mammals.
Answer:

1. Right vetricle
2. Lungs
3. Left Auricle
4. Left Ventricle
5. Body parts (except Lungs)
6. Right Auricle

Question 44.
State differences between artery and vein.
Answer:
Differences between Artery and Vein

Artery	Vein
1. An artery carries blood from the heart to different organs of the body.	1. A vein collects blood from different organs of the body and brings it back to the heart.
2. Blood flows under great pressure.	2. Blood flows under less pressure.
3. It has a thick muscular wall.	3. Wall is thin
4. It is non collapsible.	4. It is collapsible.
5. It contains oxygenated blood (Exception pulmonary artery).	5. It contains deoxygenated blood (Exception pulmonary vein)
6. Valves are absent.	6. Valves are present.
7. Mostly deep seated.	7. Mostly superficial.

Question 45.
Why are WBCs called ‘soldiers of the body’?
Answer:
WBC (White Blood Corpuscles) or leucocytes engulf and destroy the foreign particles in the body. Hence they are called ‘soldiers of the body.’

Question 46.
What are hypertension and hypotension?
Answer:
Hypertension. It is the high blood pressure which is caused due to emotions such as worry, excitement, fear etc.
Hypotension. It is the low blood pressure when it falls below the normal level.

Question 47.
Name the three major types of blood vessels. Explain briefly.
Answer:
The three main types of blood vessels are :

• Arteries have thick elastic walls and their diameter may be even 1 cm. These blood vessels carry the blood from the heart to the various parts of the body.
• Capillaries. Arteries divide into thin arterioles and arterioles further ramify into capillaries (1 micron diameter). The wall of a capillary is made up of a single layer of cells. The muscles and elastic fibres are absent in the capillaries. The walls of these capillaries are so thin that the exchange of food materials, waste materials and gases takes place between the blood and protoplasm of cells (liver, lung etc.) through them.
• Veins. The capillaries again reunite to form venules and venules unite to form veins. These venules and veins return the blood to the heart.

Diagram showing the relationship among blood vessels

Question 48.
Write a note on lymphatic system in human beings.
Answer:
Lymphatic system: The lymphatic system comprises colourless fluid, the lymph; a network of fine channels, the lymphatic capillaries; tubes of varied sizes, the lymphatic vessels; and the lymph nodes. Tissue or interstitial fluid present in the spaces between tissue cells is formed by filtration of protein-free fluid from the blood. Tissue fluid passes into lymphatic capillaries to form lymph. The latter is carried by lymphatic vessels to the veins. Lymphatic vessels have lymph nodes which filter lymph, removing microorganisms and cellular debris and adding lymphocytes.

Question 49.
Write functions of lymph.
Answer:
Functions of lymph

• It drains excess tissue fluid from the extracellular spaces back into the blood.
• Some of the fluid from the digestive tract is absorbed in the lymph. The lymphatic vessels store this fluid temporarily, and release it gradually so that the kidneys do not face a sudden pressure of urine excretion.
• It carries carbon dioxide and nitrogenous waste materials that diffuse into the tissue fluid to the blood.
• It takes lymphocytes and antibodies from the lymph nodes to the blood.

Question 50.
Give differences between blood and lymph.
Answer:
Differences between Blood and Lymph

Blood	Lymph
1. It consists of plasma, erythrocytes, leucocytes and platelets.	1. It consists of plasma and leucocytes (lymphocytes most abundant).
2. It is red in colour due to the presence of haemoglobin in erythrocytes.	2. It is colourless as haemoglobin is absent,
3. Its plasma has more proteins, calcium and phosphorus.	3. Its plasma has fewer proteins and less calcium and phosphorus.
4. It transports materials in the body.	4. It acts as middle man between blood and body tissue.

Question 51.
How are xylem and phloem well suited for transport of materials in plants? Explain.
Or
How are water and mineral transported in plants?
Answer:
Xylem and phloem are well suited to carry water, minerals and food in plants. Vessels in the xylem are cylindrical in shape with their ends open and are placed one above the other so as to form a continuous column stretching from roots to leaves. So, the water and minerals absorbed by the roots are carried upwards to the leaves. This is known as transportation.

Similarly, phloem has sieve tubes that are also cylindrical but the ends are not open instead covered with sieve (perforated) plate. These tubes are also placed one above the other, forming a continuous column from leaves to other parts of the plant body. The food synthesized in the leaves is carried to other paxts of the plant body through phloem. Sucrose is the main form in which carbohydrates are translocated in plants.

Xylem vessels that transport water and mineral sahs

Phioem tubes that conduct prepared food.

Question 52.
Mention any three methods adopted by plants to minimise the transpiration rate.
Answer:
Three methods adopted by plants to minimise the rate of transpiration are :

1. In some cases leaves are rolled to cover stomata (e.g. some grasses)
2. The stomata may be sunken (e.g. Nerium)
3. In some cases, leaves may be dropped or absent as in most cacti.

Question 53.
Write a short note on root pressure.
Answer:
Root pressure: During absorption, water is forced into the xylem vessels by the surrounding cortical cells with a certain force. This induces a pressure which is responsible for ascent of sap to many feet in xylem. This pressure which is developed due to the activity of root is called as root pressure.
Root pressure is a vital phenomenon and depends upon the activity of living root cells. The magnitude of root pressure varies from 2-8 atm.

Question 54.
How is transpiration useful to plants?
Answer:
Advantages of transpiration

• It has cooling effect on the plants as excess of sun’s energy is dissipated.
• It helps in the removal of excess of water from the plant.
• It causes ascent of sap.
• It helps to maintain water cycle.
• It increases the amount of sugar and mineral content in the fruit.
• It is needed to permit photosynthesis to take place.

Question 55.
What are the disadvantages of transpiration?
Answer:

• More plants die from excessive water loss by transpiration.
• Due to high rate of transpiration plants suffer from loss of turgidity.

Question 56.
What is translocation?
Answer:
Translocation: The long distance transport of the organic food from a source to a sink is known as translocation.

Question 57.
(a) What is blood pressure?
Answer:
Blood pressure is the force that blood exerts against the wall of a vessel. This pressure is much greater in arteries than in veins. The pressure of blood inside the artery during ventricular systole is called systolic pressure and pressure in artery during ventricular diastole is called diastolic pressure.

(b) What are normal value of systolic and diastolic blood pressure
Answer:
The normal systolic pressure is about 120 mm of Hg and diastolic pressure is 80 mm of Hg.

(c) How is it measured?
Answer:

Measurement of blood pressure

(d) Name the instrument used to measure blood pressure.
Answer:
Blood pressure is measured with an instrument called sphygmomanometer.
High blood pressure is also called hypertension and is caused by the constriction of arterioles, which results in increased resistance to blood flow. It can lead to the rupture of an artery and internal bleeding.

Question 58.
Write a note on mechanism of blood clotting.
Answer:
Blood Clotting: At the site of injury, blood platelets disintegrate and release enzyme thromboplastin. Thromboplastin in the presence of calcium ions, transforms the inactive prothrombin into active thrombin. Thrombin converts soluble fibrinogen into insoluble fibrin. This forms a meshwork of fibrils which entangle blood corpuscle and transforms the liquid blood into a gel or clot.

Mechanism of Blood Clotting
Unless the blood clots at the site of the injury, there will be loss of blood (blood haemorrhage).
Such a loss of blood will lead to death of the person.

Question 59.
Briefly describe excretory system.
Answer:

• Kidneys are a pair of bean-shaped reddish brown organs which lie in the lumbar part of abdomen along the dorsal wall, one on either side of the vertebral column.
• Each kidney receives a renal artery from dorsal aorta and sends a renal vein to inferior vena cava.
• The excretory waste products ,are filtered out in the kidney.
• Each kidney contains about 1.2 million excretory units called uriniferous tubules or nephrons.
• Uriniferous tubule is a long, twisted, narrow, tubular structure which consists of ‘ Bowman’s capsule, neck, proximal convoluted tubule, loop of Henle and distal convoluted tubule.
• Bowman’s capsule is a blind cup-shaped end of uriniferous tubule with a tuft of blood capillaries called glomerulus.
• Ureters are two distensible tubes which connect the kidneys with the urinary bladder.
• Urinary bladder is a median pear-shaped bag-like structure that occurs in she pelvic region of abdominal cavity.
• Urinary bladder can hold 300-800 ml of urine.
• Urethra is a tubular connection between the urinary bladder and the external Opening of urinary tract.

Question 60.
Label the parts of human excretory system.
Answer:

1. Kidney
2. Ureter
3. Urinary Bladder.

Question 61.
How does the excretion take place in Amoeba?
Answer:
The excretory product of Amoeba is ammonia. Special excretory organelle is lacking in Amoeba. C02 and ammonia are excreted by diffusing in solution through plasma membrane. The concentration of ammonia is always higher in Amoeba than in the surrounding water. The water enters through plasma membrane by “endosmosis”. Ammonia is formed in cytoplasm by metabolism.

Surplus water enters contractile vacuole. This surplus water can rupture the animal’s body. Thus size of contractile vacuole increases, when the contractile vacuole is fully expanded with water, it moves towards the periphery. As the contractile vacuole comes in close contact with the plasma membrane, it bursts. Thus excess of water (surplus water) is discharged in the surrounding water. This phenomenon of controlling the amount of water in the body is called as “osmoregulation”.

Excretion in Amoeba

Question 62.
State one main function of the following :
(i) Glomerulus
Answer:
The filtration of blood in the nephron takes place in the glomerulus.

(ii) Malpighian capsule
Answer:
Malpighian capsule is concerned with ultrafiltration.

(iii) Sweat gland
Answer:
Sweat glands produce sweat containing urea, uric acid and salts. Sweat evaporates to bring down the body temperature to normal.

(iv) Nephron of kidney tubule
Answer:
Through the kidney tubule or nephrons, filtration of urea, uric acid, water and some salts occur from the blood.

(v) Loop of Henle.
Answer:
Loop of Henle is useful in the absorption of water and secretion of urea.

Question 63.
Name the chief organs of excretion in man. Mention the waste products that they excrete.
Answer:
The chief excretory organs and the waste products removed by them are :

1. Kidneys – Urea in the form of urine.
2. Lungs – Carbon dioxide.
3. Skin – Water and salts as sweat.

Question 64.
Name the following :
(i) A process by which the unwanted nitrogenous wastes are eliminated from the body.
Answer:
Excretion

(ii) Major excretory organs of man.
Answer:
Kidneys

(iii) The structural and functional units of kidney.
Answer:
Nephrons

(iv) A tuft of blood capillaries found in the Bowman’s capsule of nephron.
Answer:
Glomerulus

Question 65.
Name the following :
(i) The structure that brings urine from the kidney to the urinary bladder.
Answer:
Ureter

(ii) Thin membranous sac serving as the reservoir of urine.
Answer:
Urinary bladder

(iii) Any two organic constituents of normal human urine.
Answer:
Urea, creatinine

(iv) The chief nitrogenous waste product in the human urine and the organ which produces it.
Answer:
Urea, liver

(v) Name two excretory products formed by the liver.
Answer:
Bile pigments (Bilirubin, Biliverdin), urea.

Question 66.
What do you understand by artificial kidney? Name the principle on which it works.
Or
Write a note on haemodialysis.
Answer:
Artificial Kidney or Haemodialysis. In case of acute kidney failure, the poisonous materials may accumulate in the body fluids which will cause oedema and finally lead to death of the patient. In such cases, artificial kidney is used. It works on the principle of dialysis and separates wastes from the blood. The process is called haemodialysis.

Artificial kidney contains a number of tubes with a semipermeable lining.
Its functioning is similar to kidney, but it is different since there is no reabsorption involved.

In this, blood from an artery is diverted through a cellophane tube, having pores equal to those of glomerular capillaries, placed in a circulating bath. Concentration of bath-fluid is kept equal to that of normal plasma. The pores in the cellophane tube allow small sized wastes like urea, ammonium salts etc. to pass through but do not allow the passage of blood cells, proteins, fats etc. Diffusion of small and useful substances like glucose, amino acids etc is prevented by keeping their concentration in the dialysis fluid equal to the normal plasma. Blood from dialyser is returned to the body through a vein.

Flow of blood through an artificial kidney for haemodialysis.

Question 67.
What are the functions of tongue?
Answer:
Functions of tongue.

• It helps in mastication of food.
• It bears taste buds and helps in the sensation of taste of food.
• It takes part in the modification of sound production.
• It acts as a brush and cleans the teeth.
• It aids in deglutition of food.

Question 68.
(i) Label the part A-E
Answer:

A. Interventricular System B. Right Ventricle C. Right Atrium D. Left Ventricle E. Aorta

(ii) Write Function of E.
Answer:
Function of E-Aorta carries blood to all parts of body.

Very Short Answer Type Questions

Question 1.
Why are maintenance processes required?
Answer:
They are required to prevent breakdown.

Question 2.
List three characteristics of living organisms.
Answer:

1. Growth
2. Movements
3. Repair and maintenance of their structures.

Question 3.
What are life processes?
Answer:
The processes which together perform the maintenance jobs are collectively termed as life processes.

Question 4.
What is the basic requirement of maintenance?
Answer:
Energy is needed by living organisms for maintenance.

Question 5.
What are sources of energy for living organisms?
Answer:
Carbon based molecules i.e. food obtained from environment.

Question 6.
List the common reactions required to obtain energy from carbon based molecules.
Answer:
Oxidising-reducing reactions.

Question 7.
Name any four life processes required for maintenance.
Answer:

1. Nutrition
2. Respiration
3. Transportation
4. Excretion.

Question 8.
What are nutrients?
Answer:
The substances which provide materials for growth, energy and maintenance are called nutrients.

Question 9.
What is nutrition? Why is it necessary?
Answer:
Nutrition. The sum total of processes by which living organisms obtain food materials and prepare them for use in the growth, repair and providing energy is termed nutrition.

Question 10.
What is food?
Answer:
Food provides energy. It Provides raw materials for growth and maintenance.

Question 11.
What is holozoic nutrition?
Answer:
Holozoic nutrition. When the nutrients are ingested as solid organic food matter, it is called holozoic nutrition.

Question 12.
What is the source of food for heterotrophs?
Answer:
All heterotrophs obtain food from autotrophs.

Question 13.
Name the process which prepares food is autotrophs.
Answer:
Photosynthesis.

Question 14.
Why are green plants called producers?
Answer:
Green plants prepare their food from CO₂ and H₂O in the presence of sunlight and chlorophyll. All other organisms obtain food from green plants, thus called as producers.

Question 16.
In which spectrum of light maximum photosynthesis occurs?
Answer:
Red light.

Question 17.
Where is chlorophyll present in cells of leaves?
Answer:
Chloroplast.

Question 18.
How does oxygen produced during photosynthesis enter the atmosphere?
Answer:
Oxygen passes out of green leaves through stomata and diffuses into the atmosphere.

Question 19.
Where does photolysis occur in plant?
Answer:
Photolysis occurs in chloroplasts present in the cell.

Question 20.
Write chemical reaction of photosynthesis.
Answer:

Question 21.
What are the advantages of cooked food?
Answer:
Human beings consume cooked food. Cooking makes it soft, palatable, tasty and easier to digest.

Question 22.
What is the role of CO₂ during photosynthesis?
Answer:
CO₂ provides carbon for synthesis of glucose (C₆H₁₂O₈) during photosynthesis.

Question 23.
What is the role of stomata in green leaves?
Answer:
Stomata are minute pores through which exchange of gas occurs.

Question 24.
Name the minerals obtained from soil by plants.
Answer:
Nitrogen, phosphorus, potassium, iron, magnesium and other minerals.

Question 25.
What is the role of nitrogen plants?
Answer:
Nitrogen is constituent of proteins and nitrogen bases of nucleic acids.

Question 26.
What is the food of Amoeba?
Answer:
Food of Amoeba consists of small protozoans, algae, rotifers, bacteria, and diatoms. It also feeds upon bits of organic matter.

Question 27.
Is food vacuole of Amoeba, temporary structure or a permanent one?
Answer:
Food vacuole is a temporary structure.

Question 28.
What is saliva?
Answer:
Saliva is a digestive juice secreted by salivary glands present in oral cavity.

Question 29.
Name the enzyme present in saliva.
Answer:
Salivary amylase.

Question 30.
Name the organ through which blood passes into stomach from oral cavity.
Answer:
Pharynx and oesophagus.

Question 31.
Name the largest part of alimentary canal.
Answer:
Small intestine.

Question 32.
Which of the animals need long large intestine?
Answer:
Animals eating grasses need large intestine for digestion of cellulose.

Question 33.
What is the nature of food entering small intestine from stomach?
Answer:
Acidic in nature.

Question 34.
Name the juices which convert acidic food into alkaline in small intestine.
Answer:
Bile and pancreatic juice.

Question 35.
Name the digestive juice secreted by liver.
Answer:
Bile juice.

Question 36.
Name the digestive juice secreted by pancreas.
Answer:
Pancreatic juice.

Question 37.
List the enzymes which help in the digestion of proteins.
Answer:

• Pepsin
• Trypsin
• Chymotryposin
• Peptidases.

Question 38.
What is approximate length of human alimentary canal?
Answer:
9 to 10 metres.

Question 39.
Name the largest gland present in human body.
Answer:
Liver.

Question 40.
Name the gland which is exocrine as well as endocrine in nature.
Answer:
Pancreas.

Question 41.
What are three parts of small intestine?
Answer:

1. Duodenum
2. Jejunum
3. Ileum.

Question 42.
Differentiate between chyme and chyle.
Answer:

• Chyme is a semisolid, semidigested acidic food which passes from stomach into duodenum.
• Chyle is emulsion form, completely digested, alkaline food present is small intestine ready for absorption.

Question 43.
Name the four enzymes present in pancreatic juice.
Answer:

1. Pancreatic amylase
2. Pancreatic lipase
3. Trypsin
4. Chymotrypsin.

Question 44.
Maximum water is absorbed in which part of alimentary canal.
Answer:
Large intestine.

Question 45.
What is digestion?
Answer:
Digestion. Chemical and mechanical break down of complex, non-diffusible form of food into simple diffusible form of food by action of enzyme.

Question 46.
What is assimilation?
Answer:
The absorption and digestion of food or nutrients by the body or any biological system.

Question 47.
What is egestion?
Answer:
The act or process of discharging undigested food as faeces from a cell in case of unicellular organisms.

Question 48.
What are villi present is small intestine and not in stomach?
Answer:
Small intestine is site of absorption of digested food, villi increase the surface area for absorption of food. No absorption of food occurs in stomach thus villi are absent.

Question 49.
What happens to pyrnvate produced during anaerobic respiration?
Answer:
Pyruvate produced at the end of glycolysis is converted to C02 and ethanol.

Question 50.
What is aerobic respiration?
Answer:
Aerobic respiration is the process of producing cellular energy involving oxygen cells break down food in the mitochondria in a long, multistep process that produces roughly 36 ATP.

Question 51.
What is anaerobic res-piration?
Answer:
A form of incomplete intracellular breakdown of sugar or other organic compounds in the absence of oxygen that releases energy.

Question 52.
What is ATP?
Answer:
ATP is Adenosine triphosphate. It is the energy currency of life. ATP is a high- energy molecule found in every cell. Its job is to store and supply the cell with needed energy.

Question 53.
Expand NADP.
Answer:
NADP. Nicotinamide Adenine Dinucleotide.

Question 54.
Name three animals which respire through skin.
Answer:

1. Earthworm
2. Leech
3. Frog.

Question 55.
Where does Krebs cycle occur in the body?
Answer:
Krebs cycle is completed in mitochondria of cells.

Question 56.
What is hypoxia?
Answer:
Hypoxia is a condition of shortage of oxygen in the body due to strangulation on cyanide poisoning.

Question 57.
Name the respiratory sub-strate.
Answer:
Glucose.

Question 58.
What is the end product of aerobic respiration?
Answer:
Carbon dioxide and water.

Question 59.
What is fermentation?
Answer:
Incomplete breakdown of glucose in the absence of oxygen in microbes such as bacteria and yeast. It forms CO2₂, ethanol and energy is released in small amount.

Question 60.
Man breaths how many times per minute?
Answer:
12-15 times.

Question 61.
What are the functions of ATP?
Answer:
ATP provide energy for all metabolic reactions in the body such as movements, synthesis, cell division.

Question 62.
What is bark?
Answer:
The tissue ontside the outer most covering of the stem or root is called bark.

Question 63.
Name the tissue which transport food in plants.
Answer:
Phloem.

Question 64.
What is xylem?
Answer:
Xylem is a complex tissue which transport water and minerals from roots to other parts of the plant.

Question 65.
Name two substances which enter the root through the root hairs.
Answer:

1. Water
2. Soluble minerals from soil.

Question 66.
What is ascent of sap?
Answer:
The process by which water absorbed by the root is carried to aerial parts of plant is called ascent of sap.

Question 67.
Name the transport tissue of body.
Answer:
Blood and lymph.

Question 68.
Name the three types of blood vessels.
Answer:

1. Arteries
2. Veins
3. Capillaries.

Question 69.
List types of circulation in human body.
Answer:
Double circulation involving pulmonary circulation and systemic circulation.

Question 70.
Name the fluid medium of blood.
Answer:
Plasma.

Question 71.
Why is the S-A node called pace-maker of the heart?
Answer:
S-A node, being self-excitatory, initiates a wave of contraction in the heart.

Question 72.
Name the organs which play role in circulation of blood.
Answer:
Heart. It is a muscular, pumping organ. It pumps the blood in the body.

Question 73.
Name the types of cells which destroy harmful bacteria in the body.
Answer:
White blood corpuscles (W.B.C.)

Question 74.
Name the instrument used to measure blood pressure.
Answer:
Sphigmomanometer.

Question 75.
Name the artery which carry impure (deoxygenated) blood from heart to lungs.
Answer:
Pulmonary artery.

Question 76.
Expand ECG.
Answer:
Electrocardiogram.

Question 77.
Write normal blood pressure in human body.
Answer:
120/80 → Systolic = 120 and Diastolic = 80

Question 78.
Name the largest artery.
Answer:
Aorta.

Question 79.
What is excretion?
Answer:
Excretion. Elimination of nitrogenous waste materials from body is called excretion.

Question 80.
How many nephrons are present in each kidney?
Answer:
About 10 lakhs.

Question 81.
What are the excretory structures of amoeba?
Answer:
Contractile vacuole.

Question 82.
What is a malpighian body (renal corpuscle)?
Answer:
Bowman’s capsule and glomerulus.

Question 83.
Name the reservoir of urine in the body.
Answer:
Urinary bladder.

Question 84.
What is micturition?
Answer:
Act of passing out of urine from urinary bladder is called micturition.

Question 85.
Name the structures which store wastes in plants.
Answer:
Central vacuoles.

Question 86.
What are resins and gums?
Answer:
These are storage wastes of plants.

Question 87.
Define autotrophic nutrition.
Answer:
The organisms prepare their own food from raw materials like C02 and H20 in the presence of sunlight. It takes place in green plants containing chlorophyll.

Question 88.
Define heterotrophic nutrition.
Answer:
The mode of taking readymade organic food material is called heterotrophic nutrition. It may be holozoic (ingestive) or saprophytic and parasitic (absorptive).

Multiple Choice Questions

Question 1.
Among the following which is a parasitic plant?
(A) Plasmodium
(B) Cuscuta
(C) Amoeba
(D) Rhizobium.
Answer:
(B) Cuscuta

Question 2.
Dark reaction and light reaction of photosynthesis takes place is:
(A) stroma and grana of chioroplast respectively
(B) grana and stroma of chioroplast respectively
(C) grana only
(D) stroma only.
Answer:
(A)stroma and grana of chioroplast respectively

Question 3.
Chemical reaction takes place during dark reaction of photosynthesis is:
(A) photo1ysi
(B) hydrolysis
(C) carbon dioxide is bonded with RUBP
(D) nitrogen fixation.
Answer:
(C) carbon dioxide is bonded with RUBP

Question 4.
Plants are green in colour because:
(A) they absorb green light only
(B) they reflect green light
(C) they absorb green light but reflect all other lights
(D) none of the above are correct.
Answer:
(B) they reflect green light.

Question 5.
The nutrition in Mucor is:
(A) parasitic
(B) autotrophic
(C) saprophytic
(D) holozoic.
Answer:
(C) saprophytic

Question 6.
In amoeba the digestion is intracellular because :
(A) amoeba is unicellular
(B) amoeba is multicellular ‘
(C) amoeba is found in pond
(D) amoeba is microscopic animal.
Answer:
(A) amoeba is unicellular

Question 7.
Which of the following has no digestive enzyme?
(A) Saliva
(B) Bile
(C) Gastric juice
(D) Intestinal juice.
Answer:
(B) Bile

Question 8.
C02 acceptor during dark reaction of photosynthesis is :
(A) RUBP
(B) PEP
(C) NADPH
(D) ATP.
Answer:
(A) RUBP

Fill in the blanks:

Question 1.
Viruses show _________ movements.
Answer:
Molecular.

Question 2.
Growth, _________ and repair and _________ are characteristics of life.
Answer:
movements, maintenance.

Question 3.
_________ is needed by living organisms for movements and maintenance.
Answer:
Energy.

Question 4.
_________ is the largest gland of body.
Answer:
Liver

Question 5.
_________ and are the raw materials for photosynthesis.
Answer:
C02 and H20.

Question 6.
RBC transport _________ in the body.
Answer:
Oxygen.

Question 7.
Xylem and _________ are the main conducting tissues in plants.
Answer:
Phloem.

Question 8.
Translocation of food takes place through _________ of phloem.
Answer:
Sieve tubes.

Punjab State Board PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements Important Questions and Answers.

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Long Answer Type Questions

Question 1.
What do you understand by Dobereiner’s triads? Give some examples to support it.
Answer:
In 1817, a German Chemist, Dobereiner, gave a classification in which the similar elements were arranged in groups of three called Triads. The arrangement was such that the atomic mass of the middle element was almost the average of the atomic masses of the first and third elements. For example, if A, B and C are the elements present in the triad, then

Atomic mass of B =\(\frac{\text { Atomic mass of } \mathrm{A}+\text { Atomic mass of } \mathrm{C}}{2}\)

Examples of triads.
A few examples of triads are listed as given ahead :
1. Lithium, Sodium. Potassium
Atomic mass of Lithium (Li) = 7
Atomic mass of Potassium (K) = 39
Atomic mass of Sodium (Na) = \(\frac{39+7}{2}\) = 23
Actual atomic mass of sodium = 23

This group of triads is called Alkali Metal Group. All the elements present are metals, have valency equal to one (1) and dissolve in water to form soluble hydroxides called alkalis.

2. Calcium, Strontium, Barium
Atomic mass of Calcium (Ca) = 40
Atomic mass of Barium (Ba) = 137
Atomic mass of Strontium (Sr) = \(\frac{137+40}{2}=\frac{177}{2}\) = 88.5
Actual atomic mass of strontium = 88

This group of triads is called Alkaline Earth Metals Group. All the elements present are metals, have valency equal to two (2) and their oxides dissolve in water to form hydroxides which are alkaline in nature.

Question 2.
(a) What was Dobereiner’s basis of classifying elements?
Answer:
Dobereiner gave a classification in which the elements were arranged in a group of three elements called triads. The arrangement was such that the atomic masses of the middle elements were almost the average of the atomic masses of the first and third elements.

(b) What is the basis of classification of elements according to Mendeleev?
Answer:
The major contribution in the classification of the elements is by Mendeleev, a Russian chemist. He studied in detail the properties of the elements and made a very important observation. He stated that when elements are arranged in order of increasing atomic masses, the elements with similar properties recur after a definite gap. He based his classification of law called Mendeleev’s Periodic Law. The law may be stated as:

Physical and chemical properties of the elements are periodic function of their atomic weights or atomic masses.

Question 3.
Give a brief discussion of the Mendeleev’s classification of the elements.
Answer:
Mendeleev’s classification of the elements is based upon the Mendeleev’s periodic law. The law helped him to develop a table called Mendeleev’s Periodic Table. The table has been divided into vertical columns which are called Groups and horizontal rows which are known as Periods.

These are briefly discussed as follows :
1. Groups: These are the vertical rows. There are in all eight groups. The elements present in first seven groups are called Normal Elements. The elements present in group VIII are called the Transition Elements. Each group (I to VII) has been further divided into sub-groups which are called A and B. The inert gas or noble gas elements (He, Ne, Ar, Kr, Xe) were not known at that time. Therefore, they were not shown in the table. All the elements placed in a group have the same valency. Ail the elements present in a sub-group have the similar properties. For example, group I-B includes element Cu (Copper), Ag (Silver) and Au (Gold). They have similar properties.

2. Periods: In the periodic table horizontal rows are called periods. There were in all six periods in the original periodic table. The seventh period was added later on and this is not shown in the periodic table. The properties of the elements present in a period change systematically. For example, in every period, the first element is a typical metal. As we move from left to right, the metallic character gradually decreases and non- metallic character increases. For example, in period 2, the first element Li (Lithium) is a metal while the last element F (Fluorine) is a non-metal. The Mendeleev’s Periodic Table is shown below :

Mendeleev’S Periodic Table

Question 4.
(a) Why do we classify elements?
Answer:
To simplify and systematize the study of known elements.

(b) What were the two criteria used by Mendeleev in creating his Periodic Table?
Answer:

1. Mendeleev arranged the elements in order of increasing atomic masses.
2. Mendeleev considered the compounds formed by the elements oxygen and hydrogen.

(c) Why did Mendeleev leave some gaps in his Periodic Table?
Answer:
For the discovery of new elements.

(d) In Mendeleev’s Periodic Table, why was there no mention of Noble gases like7 Helium, Neon and Argon?
Answer:
Because noble gases were not known at that time.

(e) Would you place the two isotopes of chlorine, Cl-35 and Cl-37 in different slots because of their different atomic masses or in the same slot because their chemical properties are the same? Justify your answer.
Answer:
In the same slot because Cl-35 and Cl-37 have same chemical properties.

Question 5.
How did Mendeleev’s Periodic Table help in the discovery of new elements?
Answer:
When Mendeleev gave the periodic table, only 63 elements were known. The classification was based on two major properties :

• Elements are arranged in order of increasing atomic masses.
• Elements present in a group have similar properties.

Many elements were not known at the time the periodic table was given. Therefore, Mendeleev left gaps for these elements in the periodic table. But the properties of these elements could be predicted. For example, let us suppose that Rb (Rubidium) a member of group IA was not known when Mendeleev framed the periodic table. But its properties could be predicted. It was expected to be a metal with valency equal to 1. It was expected to be soluble in water to form a soluble hydroxide RbOH which is an alkali. This helped in the discovery of the element. In this manner, a number of elements could be discovered.

Question 6.
Discuss some major merits of the Mendeleev’s Periodic Table.
Answer:
Merits of Mendeleev’s Periodic Table
Mendeleev’s periodic table was the first proper systematic classification of the elements. The important merits of the table are listed as follows :
1. Systematic study of elements. With the classification of elements into groups and periods, their study became quite systematic. For example, if the properties of one particular element in a group are known, those of the other elements could be predicted. Actually, elements placed in a group are expected to show similar characteristics.

2. Correction of wrong atomic masses. The periodic table helped in correcting the atomic masses of some of the elements because the elements were arranged in order of their increasing atomic masses.

3. Prediction of new elements. At the time Mendeleev gave the periodic table, only 63 elements were known. While arranging these elements in groups and periods, certain gaps were left. These gaps represented some undiscovered elements. But the properties of these unknown elements could be predicted from their positions in the respective groups. This helped, later on, to discover these elements.

Question 7.
On the basis of Mendeleev’s Periodic Table given following, answer the questions that follow the table :

(a) Name the element which is in
(i) 1st group and 3rd period
Answer:
Sodium

(ii) VII group and 2nd period.
Answer:
Fluorine

(b) Suggest the formula for the following :
(i) oxide of nitrogen
Answer:
N₂O₅

(ii) hydride of oxygen.
Answer:
H₂O.

(c) In group VIII of the Periodic Table, why does cobalt with atomic mass 58.93 appear before nickel having atomic mass 58.71?
Answer:
Because the elements with similar properties could be grouped together.

(d) Besides gallium, which two other elements have since been discovered for which Mendeleev had left gaps in his Periodic Table?
Answer:
Scandium and Germanium.

(e) Using atomic masses of Li, Na and K, find the average atomic mass of Li and K and compare it with the atomic mass of Na. State the conclusion drawn from this activity.
Answer:
Average atomic mass of Li and K = \(\frac{6.939+39.102}{2}=\frac{46.04}{2}\) = 23.02
Atomic mass of Na = 22.99
Hence Atomic mass of Na = Average of atomic masses of Li and K = \(\frac{46.04}{2}\) = 23.02

Question 8.
Point out the major defects in Mendeleev’s Periodic Table.
Answer:
Defects in Mendeleev’s Periodic Table:
Mendeleev’s periodic table was quite helpful in the classification of the elements.

But it had certain defects also. These are discussed as follows :
1. Position of hydrogen. Hydrogen was placed at the top of group LA. It is a non-metal where all other elements included in the group are metals.

2. Position of isotopes. The periodic table is based on the basis of the atomic masses of the elements and the elements with different atomic masses must be given separate places in the table. If this is correct, all the isotopes of an element must be allotted separate positions. For example, there are three isotopes for hydrogen and they must be given three separate places in the table. But only one position for hydrogen has been given.

3. Wrong order of atomic masses of some elements. In the table, the elements are arranged in order of increasing atomic masses. This means that the element with higher atomic mass must be placed after the element with the lower atomic mass. But in the table, there are some anomalies. For example, Co (Cobalt) with atomic mass 58-9 should be placed after Ni (Nickel) with atomic mass 58-7. But it has been placed before nickel.

4. Elements with similar properties placed in different groups. In the periodic table, it has been found that the elements with similar properties are placed in different groups. For example, copper and mercury have many common properties. But copper has been placed in group I B and mercury in group II B.

5. No similarity in the elements placed in sub-groups. The elements present in different sub-groups of the same group are expected to have common properties. But these are quite different. For example, elements in group I A are very soft and reactive metals but elements in group IB are hard and less reactive in nature.

6. No explanation for the cause of periodicity. Mendeleev was not in a position to explain why the elements included in a group show similar properties.

Question 9.
Give a brief description of Long Form of Periodic Table.
Answer:
The Long form of periodic table has been formed by arranging the elements in order of increasing atomic numbers. It is based upon Modern periodic law which states that the properties of the elements are the periodic function of their atomic numbers. Just as in case of the Mendeleev’s table, this periodic table has also been divided into Periods and Groups.
(A) Groups
These are the vertical columns. In all, there are eighteen groups in the table. The details of the groups are as follows :

Long Form of Periodic Table of Modern Periodic Table

• Group 1: The elements present in group I or 1 are called Alkali Metals.
• Group 2: The elements which are present in group IIA or 2 are called Alkaline Earth Metals.
• Groups 13 to Group 18: There are in all six groups. The Groups 13 to 16 are named after the first element present in the family. For example, Group IIIA or 13 is called Boron Family because first member is boron.
• Group 17 consists of a family called Halogen Family.
• The group 18 is also called zero group because the elements have zero valency. These elements are ail gases. They have very little tendency to take part in chemical combination. These are also called Noble Gases.

In the latest Long Form of Periodic Table, groups of Alkali metals and Alkaline erttH metals are given numbers 1 and 2. The transition of elements are numbered from its,2 The non-metals are included in groups 13 to 18.

Group No.	Name of Family
Group 13	Boron Family
Group 14	Carbon Family
Group 15	Nitrogen Family
Group 16	Oxygen Family
Group 17	Halogens
Group 18 or zero group	Noble Gases

• Group 3 to Group 12: There are in all ten groups. These are all metals and are called Transition elements. When we go down each group, the metallic character further increases.
• Group 3: Group 3 also includes fourteen elements belonging to Lanthanide family. These are called Lanthanides because they start after Lanthanum (La) with Z = 57. These are present in the 6th period as shown in the table.
• It also includes another fourteen elements called Actinides. These are present in 7th period. These are so called as they come after Actinium (Ac) with Z = 89. These are placed at the bottom of the table for convenience.

(B) Periods
Periods are the horizontal rows which are present in the Long Form of Periodic Table. Different periods have different number of elements and their atomic numbers are continuous. There are in all seven periods. The seventh period is still incomplete. The number of elements which are included in each period are given below :

Period	No. of elements	Name of the Period
1	2	Shortest Period
2	8	Short Period
3	8	Short Period
4	18	Long Period
5	18	Long Period
6	32	Longest Period
7	20	Incomplete Period

Question 10.
What is periodicity? What is the cause of periodicity?
Answer:
Periodicity may be defined as the repetition of the similar properties of the elements placed in a group and separated by definite gaps of atomic numbers (8, 8, 18, 18, 32).

Cause of Periodicity. The properties of the elements, particularly the chemical properties, are linked with number of electrons present in the outermost shell of their atoms which is also called Valence shell. Elements with similar valence shell electronic configurations are expected to have similar properties.

It may be noted that all the elements which are present in a group have the same number of electrons in the valence shells of their atoms. In other words, the same valence shell electronic arrangement gets repeated after definite gaps of atomic numbers (8, 8, 18, 18, 32). Therefore, the elements placed in a group show similar properties.

Example, Let us write the electronic distribution of the first four members of the alkali metals present in group I.

All the four elements have one electron each in the valence shell of their atoms. They have, therefore, similar properties.

Question 11.
(a) What were the two major shortcomings of Mendeleev’s periodic table? How have these been removed in the modern periodic table?
Answer:
The shortcomings of Mendeleev’s periodic table are :

• Isotopes of an element find different positions in periodic table.
• Some chemically similar elements have been separated and some dissimilar elements are placed together.

In these Modern Periodic Table
Modern Table

• Isotopes of an element occupy the same position because they have same atomic number.
• The similar elements are grouped together and dissimilar elements are separated.

(b) Two elements X and Y have atomic numbers 12 and 16 respectively. Write the electronic configuration for these elements. To which period of the modern periodic table do these two elements belong? What type of bond will be formed between them and why?
Answer:

Element	Electronic Configuration	Period
X12	2, 8, 2	3rd
Y16	2, 8, 6	3rd

They will form ionic bonds because two electrons are transferred from X to Y so that they get their octets complete :

Question 12.
Why is Long Form of Periodic Table regarded better than Mendeleev’s Periodic Table?
Or
How could Modern Periodic Table remove various anomalies of Mendeleev’s Periodic Table?
Answer:
Long Form of Periodic Table is regarded better than the Mendeleev’s periodic table due to the following reasons:

• It is based upon atomic number which is considered better than the atomic mass because the properties of the elements are related to the atomic number.
• It explains why the elements placed in a group show similar properties but Mendeleev’s Periodic Table gives no explanation for the same.
• All groups in the Periodic Table are independent groups and there are no sub¬groups as in Mendeleev’s Periodic Table.
• Many defects in the Mendeleev’s Periodic Table have been removed.
• There is no confusion regarding the position of isotopes because all the isotopes of an element have the same atomic number.
• The periodic table is more systematic than the Mendeleev’s table and is easy to remember.

Question 13.
The following tables shows the position of six elements A, B, C, D, E and F in the periodic table.

Using the above table answer the following questions :
(a) Which element will form only covalent compounds?
Answer:
E

(b) Which element is a metal with valency 2?
Answer:
D

(c) Which element is non-metal with valency of 3?
Answer:
B

(d) Out of D and E, which one has a bigger atomic radius and why?
Answer:
D

(e) Write a common name for the family of elements C and F,
Answer:
The noble gases.

Question 14.
The question refers to the elements of the periodic table with atomic number from 3 to 18.

(a) Which of these :
(i) is/are noble gas?
Answer:
H, P

(ii) is it a halogen?
Answer:
G, O

(iii) is an alkali metal?
Answer:
A, I

(iv) is it a metal with valency 2?
Answer:
A, I

(b) Write the electronic arrangement of G.
Answer:
G has the electronic configuration = 2, 7

(c) If A combines with F, what would be the formula of resulting compound?
Answer:

Short Answer Type Questions

Question 1.
Atomic number is considered to be a more appropriate parameter than atomic mass classification of elements in a periodic table. Why?
How does atomic size of elements vary on moving from
(i) Left to right in a period.
(ii) from top to bottom in a group.
Give reasons for your answers.
Answer:

• This is because atomic number is a more fundamental property of an atom.
• As we move from left to right along a period atomic radius decreases due to increase in effective nuclear charge.
• As we move from top to bottom in a group, atomic radius increases. This is due to addition of new electronic shells.

Question 2.
Define periodic law. Why was it necessary to change the basis of classification from atomic masses to atomic numbers?
Answer:
Periodic Law: The properties of elements are a periodic function of their atomic numbers. It was necessary to change the basis of classification from atomic masses to atomic numbers because atomic number and not atomic mass is the fundamental property of an element.

Question 3.
What do you understand by the term periodicity? Do the properties of two elements placed in a group the same? Illustrate.
Answer:
The repetition of the similar properties of the elements placed in a group and separated by definite gaps of atomic numbers (8, 8, 18, 18, 32) is called periodicity. The elements placed in a group show similar properties, e.g. consider group 1 elements.

Element	Symbol	Electronic configuration
Lithium	(3)	Li – 2, 1
Sodium	(11)	Na – 2, 8, 1
Potassium	(19)	K – 2, 8, 8, 1
Rubidium	(37)	Rb – 2, 8, 18, 8, 1
Cesium	(55)	Cs – 2, 8, 18, 18, 1
Francium	(87)	Fr – 2, 8, 18, 32, 18, 8, 1

These elements show similar properties because they have similar outer electronic configurations.

Question 4.
What was wrong with Dobereiner’s classification of elements?
Answer:
Dobereiner classified the elements in group of three in such a way that the atomic mass of the middle element was the mean of the first and the third elements. But he could not find many triads of elements. Therefore, the classification was rejected.

Question 5.
What properties do ail elements in the same column of the periodic table as boron have in common?
Answer:
The elements of Boron family in the periodic table show

• Tricovalency
• Form trihalides
• Form trioxides
• React with halogens to form halides.

Question 6.
Indicate the atomic number of elements of period 3 of Modern periodic table :
(a) non-metals
Answer:
14, 15, 16, 17

(b) elements forming negative ions.
Answer:
15, 16, 17

(c) elements with high melting points.
Answer:
11, 12, 13, 14

(d) elements forming positive ions.
Mention the atomic number only.
Answer:
11, 12, 13

Question 7.
Define atomic radius. Give its units.
Answer:
Atomic radius. It may be defined as the distance between the centre of nucleus and the outermost shell of an isolated atom.

Also the atomic radius of a non-metallic element is defined as half the distance between the nuclei of two atoms bound by a single covalent bond.
Units = Å or pm (picometre)
e.g. atomic radius of hydrogen atom = 37 pm.

Question 8.
How does atomic radius vary down a group and along a period?
Answer:
Variation in a group. The atomic radius generally increases from top to bottom in a group due to the addition of a new shell.
Variation along a period. The atomic radius decreases on moving from left to right due to the increase in nuclear charge.

Question 9.
Write down the electronic configuration of elements with atomic numbers 2, 14, 17, 19. Indicate the group of the periodic table to which they belong.
Answer:
The information is being given in a tabular form.

Atomic number	Electronic configuration K L M N	Group
2	2	18
14	2, 8, 4	14
17	2, 8, 7	17
19	2, 8, 8, 1	1

Question 10.
Locate the following group in the periodic table :
(a) Alkali metals
Answer:
Alkali metals: Group 1 or IA

(b) Halogens
Answer:
Halogens: Group 17 or VIIA

(c) Alkaline earth metals
Answer:
Alkaline earth metals: Group 2 or IIA

(d) Noble gases.
Answer:
Noble gases: Group 18 or zero

Question 11.
What properties do the elements in the same vertical column of the periodic table as fluorine have in common?
Answer:
These are :

• They form diatomic molecules F₂, Cl₂, Br₂ I₂
• They are non-metals.
• They show a valency of one.

Question 12.
Write the chemical electronic configuration of nitrogen (N = 7) and phosphorous (P = 15).
Answer:

Question 13.
(i) Name the members of the alkaline earth family.
Answer:
The members of the alkaline earth family are :
Be, Mg, Ca, Sr, Ba, Ra.

(ii) To which group do they belong?
Answer:
They belong to group 2.

(iii) Which member is radioactive in nature?
Answer:
The radioactive member in them is Radium (Ra).

(iv) Which member is the least reactive?
Answer:
The least reactive member of the family is Beryllium (Be).

Question 14.
Why are the members of group 1 called alkali metals?
Answer:
The members of group 1 are called alkali metals because all of them are water soluble. They react with water to form soluble hydroxides. The soluble hydroxides of the metals are called alkalies.

Question 15.
An atom has the electronic configuration 2, 8, 7.
(a) What is the atomic number of this element?
Answer:
The atomic number of the element = Total number of electrons
= 2 + 8 + 7 = 17

(b) To which of the following elements would it be chemically similar? (atomic numbers are given) N (7), F (9), P (15), Ar (18).
Answer:
The electronic configurations of given atom and N, F, P and Ar are.

Atom	Electronic configuration
	K	L	M
	2	8	7
N	2	5
F	2	7
P	2	8	5
Ar	2	8	8

Since F has same number of electrons in the outermost shell as the given atom. Hence, the given atom is chemically similar to F.

Question 16.
What physical and chemical properties of elements were used by Mendeleev in creating his periodic table? List two observations which posed a challenge to Mendeleev’s Periodic Law.
Answer:
He selected the compounds of the elements with oxygen and hydrogen. He gave a table based upon atomic weights of the elements.

Limitations of Mendeleev’s Classification

• The position of hydrogen was uncertain.
• The isotopes of elements were not given proper positions in the periodic table.

Question 17.
(a) What are amphoteric oxides? Choose the amphoteric oxides from amongst the following oxides :
Na₂O, ZnO, Al₂O₃, CO₂, HaO
Answer:
The oxides which can react both with acids as well as bases to produce salts and water.

(b) Why is it that non-metals do not displace hydrogen from dilute acids?
Answer:
This is because non-metals cannot lose electrons and cannot reduce H⁺ ions from acid to H₂.

Question 18.
What are noble gas elements? Why are they so called?
Answer:
Noble gas elements are the. elemehts present in group 18 of the periodic table which is also called zero group, It means that the valency of the elements is zero. Actually, whereas the first member helium has two electrons in its only shell, the atoms of the remaining elements (Neon, Argon, Krypton, Xenon and Radon) have eight electrons in their outermost shells. They do not have any tendency to combine with atoms of other elements. Hence, they show zero valency. These are also called noble gases because they do not take part in chemical combination.

Question 19.
How is metallic character of an element defined? How does the metallic character of the elements change in a group?
Answer:
The metallic character of an element may be expressed in terms of its tendency to lose electrons and to form positive ion.
M (Element) → M⁺ + e^–

In a group the metallic character increases’ downwards. For example, among the elements of group 2, Beryllium (Be) is the least metallic. At the same time, radium (Ra) which is the last element is maximum metallic in nature.

Question 20.
Why do the elements present in a group show similar chemical properties?
Answer:
The properties of the elements, particularly the chemical properties are related to valence shell electronic distribution. The elements with the same valence shell electronic distribution have the similar chemical properties. For example, the members, of alkaline earth metal family (Group 2) have two electrons in the valence shell of their atoms. They therefore, show similar chemical properties.

Question 21.
How does the reactivity of the metals vary in a group?
Answer:
In a group, containing metals, the reactivity increases down the group. For example, in the metals of group 1 (Alkali metals), Lithium reacts with water very slowly. Sodium is more reactive and potassium is. still more reactive than sodium.

Question 22.
Name the elements present in the second period. Give their electronic configuration.
Answer:
The second period of the Long Form of Periodic Table has eight elements. The first element is Lithium and the last element is Neon. The electronic configuration of the elements are given below:

Question 23.
Why do not the elements present in a period show same valency?
Answer:
The valency of the element is related to the number of electrons in the outermost energy shell of its atom. Since the elements present in a period have different number of valence electrons or outermost electrons, they show different valencies.

Question 24.
The metallic character of the elements in a period decreases from left to the right. Justify.
Answer:
In every period, as we move from left to right, the metallic character of the elements decreases gradually. This is shown with the elements present in the third period.

Question 25.
Give symbols for :
(a) a metal belonging to second group of the periodic table.
Answer:
The metal belonging to second group is calcium (Ca).

(b) a metal belonging to the third group of the periodic table.
Answer:
The metal belonging to third group is aluminium (Al).

(c) two non-metals belonging to the halogen family.
Answer:
The two non-metals of halogen family are fluorine (F) and chlorine (Cl).

Question 26.
Write electronic structures of:
(i) Potassium
(ii) Lithium
(iii) Fluorine.
Answer:
The electronic configurations of the atoms are given below :

Question 27.
Name two other elements which are in the same family as
(i) carbon
Answer:
Carbon belongs to the group 16. Two other elements are silicon (Si) and germanium (Ge).

(ii) fluorine
Answer:
Fluorine belongs to group 17. Two other elements are chlorine (Cl) and bromine (Br).

(iii) sodium.
Answer:
Sodium belongs to group 1. Two other elements are lithium (Li) and potassium (K).

Question 28.
Carbon (atomic number 6) and silicon (atomic number 14) are elements in the same group of the periodic table. Give the electronic arrangements of the carbon and silicon atoms and state the groups in which these elements occur.
Answer:
The required information may be given in a tabular form as follows :

Element	Atomic no.	Electronic arrangement	Group
Carbon (C)	6	2, 4	14
Silicon (Si)	14	2, 8, 4	14

Question 29.
Sodium and aluminium have atomic numbers of 11 and 13 respectively. They are separated by one element in the periodic table and have valencies of 1 and 3 respectively. Chlorine and potassium are also separated by one element in the periodic table (their atomic numbers are 17 and 19 respectively) and yet both have valency of one. Explain your answer.
Answer:
Sodium and aluminium: The electronic configurations of the elements are given below :
Sodium (Na) = 2, 8, 1
Aluminium (Al) = 2, 8, 3

The valency in this case is given by the number of valence electrons. Therefore, the valency of sodium is 1 and that of aluminium is 3.

Chlorine and potassium: The electronic configurations of the elements are given below :
Chlorine (Cl) = 2, 8, 7
Potassium (K) = 2, 8, 1

Question 30.
Give the atomic number and electronic distribution of:
(i) The third alkali metal
Answer:
Potassium (19) 2, 8, 8, 1

(ii) The second alkaline earth metal
Answer:
Magnesium (12) 2, 8, 2

(iii) The first halogen
Answer:
Fluorine (9) 2, 7

(iv) The second noble gas.
Answer:
Neon (10) 2, 8.

Question 31.
Observe the following elements in the Modern Periodic Table.

Name the elements A, B, C and D. Also indicate noble gas
Answer:
(A) Lithium
(B) Chlorine
(C) Neon
(D) Potassium C is a noble gas

Question 32.
Match the following :

(a) Fluorine	(i) Metalloid
(b) Neon	(ii) Halogen
(c) Sodium	(iii) Noble gas
(d) Arsenic	(iv) Alkali metal

Answer:

(a) Fluorine	(ii) Halogen
(b) Neon	(iii) Noble gas
(c) Sodium	(iv) Alkali metal
(d) Arsenic	(i) Metalloid

Question 33.
How many electrons can be present in the valence shells of metal atoms and non-metal atoms?
Answer:
Metal atoms have 1, 2 or 3 electrons in their valence shells whereas non-metal atoms have 4 to 7 electrons in their valence shells.

Question 34.
How are the various groups of the Modern Periodic Table designated according to the IUPAC system and old system?
Answer:
The designations of various groups of the Modern periodic table are :

Question 35.
What are the uses of Modern Periodic Table?
Answer:

• Systematic study of the elements. In the periodic table, the elements with similar properties are placed together in the same group. If we know the properties of one element of the group, the properties of other elements belonging to the same group can be predicted. Thus, there is no need to study the properties of all the elements.
• Properties of an element can be predicted from the position of the element in the periodic table. For example, if the element belongs to group IA or IIA it is likely to be a reactive metal, and if it belongs to group VII A it is likely to be a reactive non-metal.
• It has led to the discovery of many new elements.

Very Short Answer Type Questions

Question 1.
How many elements have been discovered so far?
Answer:
114.

Question 2.
How are elements classified?
Answer:
The elements have been classified on the basis of their properties.

Question 3.
Why are group IA elements called alkali metals?
Answer:
This is because all these elements are metals and their oxides and hydroxides give alkaline solutions in water.

Question 4.
What is the basis of Modern Periodic table?
Answer:
It is based upon Modern Periodic law.

Question 5.
Name the family to which halogens belongs?
Answer:
Halogen family.

Question 6.
Name the second elements of group 14.
Answer:
Silicon.

Question 7.
How many valence electrons and present in halogen elements?
Answer:
Seven.

Question 8.
How many elements are present in 4th period?
Answer:
18.

Question 9.
How many electrons are present in Mg²⁺ ion?
Answer:
Ten.

Question 10.
Out of Na and Mg which was larger size?
Answer:
Na

Question 11.
What is the valency of nitrogen?
Answer:
Three.

Question 12.
Out of Na and K which is more reactive?
Answer:
Sodium (Na).

Question 13.
Name the group number of halogen family.
Answer:
Group 17 (or VII-A).

Question 14.
Name the last element of third period?
Answer:
Argon.

Question 15.
What is Dobereiner’s Triad?
Answer:
A group of three elements having similar properties is called Dobereiner’s Triad.

Question 16.
A, B and constitute the Dobereiner s Traid. Atomic mass of A and C are 7 and 23 respectively. Calculate atomic mass of B.
Answer:
\(\frac{7+23}{2}=\frac{30}{2}\) = 15

Question 17.
Name the elements discovered after Mendeleev’s Periodic Table?
Answer:
Scandium (Se), Gallium (Ga), Germanium (Ge) etc.

Question 18.
Name the groups and periods in Mendeleev’s Periodic Table?
Answer:
8 groups and 7 periods.

Question 19.
How does atomic radii as we move from left along a period in the Periodic Table?
Answer:
It decreases.

Question 20.
An element has the electronic configuration 2, 8, 3. What is its group number in Modern Periodic Table?
Answer:
Group 13.

Question 21.
Give the basis of Dobereiner’s classification.
Answer:
Doberenier Triads.

Question 22.
Give the characteristics of Dobereiner’s Triads.
Answer:
The atomic mass of the central element is the average of masses of the other two elements.

Question 23.
What is the drawback of Doberenier’s Triads?
Answer:
All the known elements could be grouped into Triads.

Question 24.
There are three alements A, B and C. The atomic masses of A and C are 7 and 39. What is the atomic mass of B on the bais of Dobereiner’s Traids?
Answer:
Atomic mass of B = \(\frac{7+39}{2}\) = 23.

Question 25.
There are three elements X, Y, Z. Atomic masses of X and Z are 35.5 and 127. What will be atomic mass of Y on the basis of Dobereiner’s Traids?
Answer:
Atomic mass of Y
\(\frac{35.5+127}{2}=\frac{162.5}{2}\) = 81.25

Question 26.
Write Newland’s Law of Octaves for classification of elements.
Answer:
When the elements are arranged in order of increasing atomic masses, eighth elements has properties similar to the first element.

Question 27.
How many element were classified by Newland.
Answer:
Upto mass number 40.

Question 28.
Indicate the group number and period number of P in the modern Periodic Table.
Answer:
Group-15, Period-3.

Question 29.
An element has the electronic configuration 2, 8, 8, 2. Indicate its group and Period in the Modern Periodic Table.
Answer:
Group-12, Period-4.

Question 30.
An element M is in the group 13 of Modern Periodic Table write the formula of its oxide.
Answer:
M₂O₃.

Question 31.
Give the groups and periods in the Modern Periodic Table.
Answer:
Group-18, Periods-7.

Question 32.
Give the electronic configuration of ₁₇Cl³⁵. Also indicate its position in the Periodic Table.
Answer:
Electronic configuation of ₁₇Cl³⁵ = 2, 8, 7
Period number = 3
Group number =17. (VII A)

Question 33.
Give the name and electronic configuation of element with atomic number 9.
Answer:
Fluorine and it has the electronic configuation = 2, 7.

Question 34.
What is Modern Periodic Law?
Answer:
It states that the properties of the elements are the periodic functions of their atomic numbers.

Question 35.
Who gave Newland’s Law of Octaves.
Answer:
Newland.

Question 36.
Define Mendeleev’s periodic law.
Answer:
It states that the properties of the elements are the periodic functions of their atomic masses.

Question 37.
What is the basis of Mendeleev’s Modern Table?
Answer:
It is based upon Mendeleev periodic law and in this table the elements are arranged in order of increasing atomic masses.

Question 38.
How many groups are present in Mendeleev’s Period Table.
Answer:
Eight.

Question 39.
Name the next elements after P in Modern Periodic Table.
Answer:
Sulphur (S).

Question 40.
Give the group number of nitrogen and phosphorus.
Answer:
They belong to group number 15.

Question 41.
Out of Mg and Al which is more metallic?
Answer:
Mg.

Question 42.
Out of Be, Mg, Ca, Al which does not belong to same group?
Answer:
Al.

Question 43.
In which group noble gases are present?
Answer:
Group 18 or group zero.

Question 44.
Na and S are present in the third period of Modern Periodic Table. Which is more metallic and why?
Answer:
Na is more metallic due to larger size than S.

Question 45.
What is metallic character?
Answer:
It is the tendency of an atom of the element to form positive ions by losing electrons.

Question 46.
What is the trend in metallic character on moving from left to right along a period?
Answer:
It decreases.

Multiple Choice Questions:

Question 1.
Who gave Law of octaves?
(A) Newland
(B) Dobereiner
(C) Mendeleef
(D) Lother Mayer.
Answer:
(A) Newland

Question 2.
In Mendeleef’s periodic table which element was discovered in the gap between Boron and Aluminium ?
(A) Na
(B) Ca
(C) Ga
(D) Ba.
Answer:
(C) Ga

Question 3.
According to Mendeleef’s periodic law, the elements are arranged in order of A increasing:
(A) Atomic numbers
(B) Decreasing atomic number
(C) Increasing atomic masses
(D) Decreasing atomic masses.
Answer:
(C) Increasing atomic masses

Question 4.
Which element occupied gap left in Mendeleers periodic table?
(A) Germanium
(B) chlorine
(C) Oxygen
(D) Silicon.
Answer:
(A) Germanium

Question 5.
An element has the electronic configuration 2, 8, 2. It is present in group:
(A) 2
(B) is
(C) 8
(D) 10.
Answer:
(A) 2

Question 6.
Which element shows metallic character?
(A) 2, 8, 2
(B) 2, 8, 4
(C) 2, 8, 8
(D) 2, 7.
Answer:
(A) 2, 8, 2

Question 7.
Which shell is largest shell?
(A) K
(B)L
(C) M
(D) N.
Answer:
(D) N

Fill in the blanks :

Question 1.
Out of Na and Mg, ______ has bigger size.
Answer:
Na.

Question 2.
Number of elements known in Mendeleefs periodic table were ______
Answer:
63.

Question 3.
Oxygen and sulphur belong to same ______
Answer:
group.

Question 4.
The elements of group 17 are called ______
Answer:
halogens.

Question 5.
The valency of the members of noble gas family is ______
Answer:
zero.

Question 6.
The halogens belong to group ______
Answer:
17.

Punjab State Board PSEB 10th Class Science Book Solutions Chapter 5 Periodic Classification of Elements Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

PSEB 10th Class Science Guide Periodic Classification of Elements Textbook Questions and Answers

Question 1.
Which of the following statements is not a correct statement about the trends when going from left to right across the periods of periodic table.
(а) The elements become less metallic in nature.
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.
Answer:
(c) The atoms lose their electrons more easily.

Question 2.
Element X forms a chloride with the formula XCl₂, which is a solid with a high melting point. X would most likely be in the same group of the periodic table as :
(a) Na
(b) Mg
(c) Al
(d) Si
Answer:
(b) Mg

Question 3.
Which element has :
(а) two shells, both of which are completely filled with electrons?
Answer:
Neon

(b) the electronic configuration 2, 8, 2?
Answer:
Magnesium

(c) a total of three shells, with four electrons in its valence shell?
Answer:
Silicon

(d) a total of two shells, with three electrons in its valence shell?
Answer:
Boron

(e) twice as many electrons in its second shell as in its first shell?
Answer:
Carbon.

Question 4.
(a) What property do all elements in the same column of the periodic table as Boron have in common?
Answer:
All elements of this column have 3 electrons in their valence shell like Boron.

(b) What property do all elements in the same column of the periodic table as Fluorine have in common?
Answer:
All elements of this column have 7 electrons in their valence shell like fluorine.

Question 5.
An atom has electronic configuration 2, 8, 7.
(а) What is the atomic number of this element?
Answer:
17

(b) To which of the following element would it be chemically similar? (atomic numbers are given in parenthesis).
N (7), F (9), P (15), Ar (18).
Answer:
F (9) (2, 7)

Question 6.
The position of three elements A, B and C in the periodic table are as shown below :

(а) State whether A is a metal or non-metal.
Answer:
A is non-metal.

(b) State whether C is more reactive or less reactive than A.
Answer:
C is less reactive than A.

(c) Will C be larger or smaller in size than B.
Answer:
C has smaller size than B.

(d) Which type of ion, cation or anion, will be formed by element A?
Answer:
Anion, A-.

Question 7.
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the periodic table. Write the electronic configurations of these two elements. Which of these will be more electronegative? Why?
Answer:

	K	L	M
N7 has electronic configuration	2	5
P15 has electronic configuration	2	8	7

Nitrogen is more electronegative than Phosphorus due to smaller size.

Question 8.
How does the electronic configuration of an atom relate to its position in the Modern Periodic Table?
Answer:
By knowing the electronic configuration of an element, we can know its period number from the number of shells present in its atom and from number of electrons in the valence shell of its atoms we can know its group number, e.g. let us consider the case of sodium atom.

Atomic number of sodium = 11
Its electronic configuration = 2, 8, 1 (K) (L) (M)
∴ Number of shells = 3
∴ Sodium belongs to 3rd period.

Also sodium atom has one electron in its valence shell.
∴ It is present in first group.
∴ Sodium lies in the first group and third period of Modern Periodic Table.

Question 9.
In the Modern Periodic Table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12,19, 21 and 38. Which of these have physical and chemical properties resembling calcium?
Answer:
Element with atomic numbers 12 and 38 because they have two electrons in their valence shells like calcium (2, 8, 8, 2).

Question 10.
Compare and contrast the arrangement of elements in Mendeleev’s Periodic table and the Modern Periodic table.
Answer:
Similarities :

• In both the elements are arranged in groups and periods.
• In both similar elements are placed in same group.
• Both the classification make the study of elements simple and systematic.

Differences :

Mendeleev’s Periodic Table	Modern Periodic Table
1. The elements are arranged in order of increasing mass numbers.	1. The elements are arranged in order of increasing atomic numbers.
2. It has 8 vertical columns called groups.	2. It contains eighteen vertical columns called groups.
3. Groups like group VIII ‘have been divided into sub groups A and B.	3. Each group is an independent group.
4. Inert gases are not included in this table.	4. Inert gases are included in this periodic table.

Science Guide for Class 10 PSEB Periodic Classification of Elements InText Questions and Answers

Question 1.
Did Dobereiner’s triads also exist in the columns of Newlands’ octaves? Compare and find out.
Answer:
Yes, Dobereiner’s triads also existed in columns of Newlands’ octaves. These are:

H	Li	Be
F	Na	Mg
Cl	K	Ca

Question 2.
What were the limitations of Dobereiner’s classification?
Answer:

• Dobereiner could find only three triads from the elements known at that time.
• It is applicable only to a few elements.

Question 3.
What were the limitations of Newlands’ law of octaves?
Answer:

• It is applicable upto calcium only.
• Sometimes two elements were put in the same slot.
• After the discovery of noble gas, law of octave is not valid.

Question 4.
Use Mendeleev’s periodic table to predict the formulae for the oxides of following elements :
K, C, Al, Si, Ba.
Answer:
K₂O, CO₂, Al₂O₃, SiO₂, BaO.

Question 5.
Besides Gallium, which other elements have since been discovered to fill the gaps left by Mendeleev in his periodic table? (any two)
Answer:
Scandium and Germanium.

Question 6.
What were the criteria used by Mendeleev in creating his periodic table?
Answer:

• The formulae of hydrides and oxides formed by an element wrere treated as one of the basic properties of an element for its classification.
• It is based upon Mendeleev’s periodic law which predicts that the properties of the elements are the periodic function of their atomic masses.

Question 7.
Why do you think the noble gases are placed in a separate group?
Answer:
These gases were discovered very late because they are very inert and placing them in a separate group, does not disturb the existing order put forward by Mendeleev.

Question 8.
How could Modern Periodic Table remove various anomalies of Mendeleev Periodic Table?
Answer:

• Isotopes of an elements occupy same position in the periodic table due to same atomic number.
• There is a logical separation of elements into subgroups.
• It is based upon the fundamental property of an element i.e. atomic number.

Question 9.
Name two elements you would expect to show same kind of chemical reactivity as magnesium. What is the basis for your choice?
Answer:
Calcium and Strontium because they have same number of valence electrons.

Question 10.
Name :
(а) three elements that have only a single electron in their outermost shells.
Answer:
Lithium, Sodium and Potassium.

(b) two elements that have two electrons in their outermost shells.
Answer:
Magnesium, Calcium.

(c) three elements with filled outermost shells.
Answer:
Neon, Argon, Krypton.

Question 11.
(a) Lithium, Sodium, Potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?
Answer:
One electron in their valence shells and are metals.

(b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common?
Answer:
Eight electrons in their valence shells.

Question 12.
In the modern periodic table, of the first ten elements, which are metals?
Answer:
Lithium and Beryllium.

Question 13.
By considering their position in the periodic table, which one of the following elements would you expect to have the most metallic characteristics?
Ga Ge As Se Be.
Answer:
Gallium.