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Balbharti Maharashtra State Board Class 10 Hindi Solutions Lokbharti Chapter 4 मन Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Hindi Lokbharti Chapter 4 मन

Hindi Lokbharti 10th Std Digest Chapter 4 मन Textbook Questions and Answers

कृति

सूचना के अनुसार कृतियाँ कीजिए:

प्रश्न 1.
लिखिए:

हाइकु द्वारा मिलने वाला संदेश
करते जाओ पाने की मत सोचो जीवन सारा।	भीतरी कुंठा नयनों के द्वार से आई बाहर।

उत्तर:

प्रश्न 2.
कृति पूर्ण कीजिए:

उत्तर:

प्रश्न 3.
उत्तर लिखिए:
a. मँझधार में डोले —-
b. छिपे हुए ——-
c. धुल गए ——-
d. अमर हुए ——-
उत्तर:
a. मँझधार में डोले – जीवन नैया।
b. छिपे हुए सितारे
c. घुल गए विषाद
d. अमर हुए गीतों के स्वर।

प्रश्न 4.
निम्नलिखित काव्य पंक्तियों का केंद्रीय भाव स्पष्ट कीजिए:
a. चलतीं साथ पटरियाँ रेल की फिर भी मौन।
b. काँटों के बीच खिलखिलाता फूल देता प्रेरणा।
उत्तर:
a. रेल की पटरियाँ अनंत काल से साथ चल रही हैं, परंतु वे सदा मौन रहती हैं। एक-दूसरे से कभी बात नहीं करती।
b. गुलाब का फूल काँटों के बीच भी हँसता है, खिलखिलाता है। वह हमें हर पल प्रेरणा देता है कि हमें परेशानियों से घबराए बिना अपना काम करते जाना है।

उपयोजित लेखन

वक्तृत्व प्रतियोगिता में प्रथम स्थान पाने के उपलक्ष्य में आपके मित्र/सहेली ने आपको बधाई पत्र भेजा है, उसे धन्यवाद देते हुए निम्न प्रारूप में पत्र लिखिए:
दिनांक: ……………………….
संबोधन: ……………………….
अभिवादन: ……………………….

प्रारंभ:
विषय विवेचन:
…………………………………………………………………………
…………………………………………………………………………
…………………………………………………………………………
…………………………………………………………………………
…………………………………………………………………………
…………………………………………………………………………

तुम्हारा/तुम्हारी,
……………………….
नाम: ……………………….
पता: ……………………….
ई-मेल आईडी: ……………………….
उत्तर:
दिनांक: 25/8/20
प्रिय अविनाश,
नमस्ते!
तुम्हारा पत्र अभी-अभी मिला। धन्यवाद।
अंतर विद्यालय वक्तृत्व प्रतियोगिता में प्रथम स्थान पाने के लिए तुम्हारा बधाई-पत्र मिला। पत्र पाकर दिल गदगद हो गया। वास्तव में मेरी इस सफलता में तुम जैसे मित्रों का मुझे सदा उत्साह दिलाते रहने का बड़ा हाथ है। तुम तो जानते हो, मंच पर बोलने में मुझे कितनी झिझक होती थी।

पर तुम जैसे मित्रों और हमारे कक्षा अध्यापक के निरंतर प्रोत्साहन से आज मुझे अंतर विद्यालय वक्तृत्व प्रतियोगिता में प्रथम स्थान प्राप्त करने का अवसर मिला है। में इसके लिए तुम जैसे अपने सभी मित्रों और अपने कक्षा अध्यापक नरेश कौशल जी का तहे दिल से आभारी हूँ।

मेरा, उत्साह बढ़ाने के लिए धन्यवाद!
तुम्हारा मित्र
राजेश शर्मा।
17, विमल मेंशन,
महात्मा गांधी रोड,
औरंगाबाद।
ई-मेल आईडी: rajesh@xyz.com

Hindi Lokbharti 10th Textbook Solutions Chapter 4 मन Additional Important Questions and Answers

कृतिपत्रिका के प्रश्न 3 (आ) के लिए)
पद्यांश क्र. 1

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

कृति 1: (आकलन)

(1) उत्तर लिखिए:
(i) खिले हुए ……………………….
उत्तर:
(i) खिले हुए – फूल।

कृति 2: (स्वमत अभिव्यक्ति)

प्रश्न.
फागुन के महीने में प्रकृति रंगों से रंग जाती है। इस विषय पर 25 से 30 शब्दों में अपने विचार लिखिए।
उत्तर:
फागुन का महीना बहुत महत्त्वपूर्ण होता है। इस महीने में प्रकृति में चारों ओर नवीनता दिखाई देती है। खेत सरसों के पीले-पीले फूलों से भर जाते हैं। इन्हें देखकर ऐसा लगता है जैसे जमीन पर पीले रंग की विशाल चादरें बिछाई दी गई हों। बीच-बीच में अलसी के नीले-नीले फूल पीले रंग पर छाप जैसे लगते हैं। पलाश के वन लाल रंग के बड़े-बड़े फूलों से लद जाते हैं।

दूर से इन वनों को देखकर ऐसा लगता है, मानो पेड़ों से आग की लपटें निकल रही हों। विभिन्न प्रकार के पेड़ों पर गुलाबी रंग की नई-नई कोंपलें आ जाती हैं। इन्हें देखकर लगता है जैसे ये पेड़ गुलाबी रंग के वस्त्रों से सज गए हैं। इनके अतिरिक्त फागुन के महीने में ही तो होली का त्योहार आता है जब चारों ओर तरह-तरह के रंगों और अबीर-गुलाल की बहार आ जाती है। लोग खुशी से एक-दूसरे को रंगों से सराबोर कर देते हैं। इस तरह फागुन के महीने में प्रकृति तरह-तरह के रंगों से रँग जाती है।

पद्यांश क्र. 2

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

कृति 1: (आकलन)

(1) उचित जोड़ियाँ मिलाइए:

‘अ’	‘आ’
(i) मछली	मौन
(ii) गीतों के स्वर	सूना
(iii) रेल की पटरियाँ	प्यासी
(iv) आकाश	अमर
	पीड़ा

उत्तर:

‘अ’	‘आ’
(i) मछली	प्यासी
(ii) गीतों के स्वर	अमर
(iii) रेल की पटरियाँ	मौन
(iv) आकाश	सूना।

(2) परिणाम लिखिए:
(i) सितारों का छिपना।
(ii) तुम्हारा गीतों को स्वर देना।
उत्तर
(i) सूना आकाश
(ii) गीतों का अमर होना।

(3) मन की ……………………. बरसी आँखें। इस हाइकु का सरल अर्थ लिखिए।
उत्तर
जब मन की पीड़ा बहुत गहरी हो जाती है, तो वह बादल बनकर आँसुओं के रूप में बरसने लगती है।

(4) तालिका पूर्ण कीजिए:

स्थिति	निवास	स्थान
मछली	प्यासी	सागर
सितारे	छिपे हुए	आकाश

कृति 2: (स्वमत अभिव्यक्ति)

प्रश्न.
‘आँखें देखने के अलावा और भी कई तरह के काम करती हैं’, इस विषय पर 25 से 30 शब्दों में अपने विचार स्पष्ट कीजिए।
उत्तर:
मनुष्य के शरीर में विभिन्न अंग होते हैं और वे अपनाअपना निर्धारित काम करते हैं। कुछ अंगों से निर्धारित कामों के अलावा और भी कई तरह के काम लिए जाते हैं। आँखें हमारे शरीर का महत्त्वपूर्ण अंग हैं। इनसे देखने का काम तो लिया ही जाता है, साथ ही साथ और भी कई काम लिए जाते हैं। आँखों से तरह-तरह के इशारे किए जाते हैं, जिन्हें सामनेवाला आदमी आसानी से समझ लेता है। आँखें तरेरकर क्रोध प्रकट किया जाता है।

आँखें झुकाकर शर्म प्रदर्शित की जाती है। मन में छुपी दुख देने वाली भावनाओं को आँखों में आँसू लाकर प्रकट किया जाता है। मन भारी होने पर लोग रोकर अपना मन हल्का करते हैं। कोई अचंभेवाली घटना होने पर वाणी के साथ-साथ आँखों से भी भाव प्रदर्शित होता है। आँखों का एक आवश्यक काम मनुष्य को निद्रावस्था में ले जाकर उसे आराम दिलाना है। इस तरह आँखें देखने के अलावा कई महत्त्वपूर्ण काम करती हैं।

मन Summary in Hindi

मन कविता का सरल अर्थ

1. घना अंधेरा ……………………………. आई बहार।

जब अँधेरा घना होता है, तब प्रकाश और अधिक चमकता है अर्थात जब प्रतिकूल परिस्थितियाँ घने अंधकार के रूप में हमें घेर लेती हैं, तब वहीं से एकाएक प्रकाश की किरणें फूट पड़ती हैं।

हमें पूरा जीवन काम करते रहना चाहिए। यह नहीं सोचना चाहिए कि हमें क्या प्राप्त होगा।

जीवन रूपी नैया यदि संसार रूपी सागर में डगमगा रही है, तो उसे कोई अन्य सँभालने के लिए नहीं आएगा। हमें स्वयं उसे पार लगाने के लिए प्रयास करना होगा।

फागुन का महीना अपने संग बसंत के विविध रंग लेकर आया है। यह समय उल्लास और उमंग का समय है। अतः हम सभी को कुछ समय के लिए चिंताओं और परेशानियों को भूलकर बसंत ऋतु का आनंद लेना चाहिए।

गुलाब का फूल काँटों के बीच भी हंसता है, खिलखिलाता है। वह हमें हर पल प्रेरणा देता है कि परेशानियों से घबराए बिना अपना काम करते जाना है।

जब नेत्रों से अश्रु बहते हैं, तो यह मानना चाहिए कि मन की कुंठा नयन रूपी द्वार से बाहर आ रही है।

2. खारे जल ……………………………. प्यासी ही रही। . . .

जब नेत्रों से अश्रु बहते हैं तो यह समझना चाहिए कि आँसुओं के खारे जल के साथ मन का संपूर्ण विषाद धुल गया है और मन पहले के समान पावन हो गया है।

प्रत्येक मनुष्य के जीवन में अनेक परेशानियाँ हैं, चिंताएँ हैं, और हैं अप्रिय प्रसंग। ऐसे में जीवन रूपी संग्राम में डटे रहना हमारी जिजीविषा का प्रमाण है।

जब आकाश में बादल बहुत घने होते हैं, तभी वर्षा होती है। उसी प्रकार जब मन की पीड़ा बहुत गहरी हो जाती है, तो वह बादल बनकर आँसुओं के रूप में बरसने लगती है।

रेल की पटरियाँ अनंत काल से साथ चल रही हैं, परंतु वे सदा मौन रहती हैं। एक-दूसरे से कभी बात नहीं करती।

सितारे आकाश का शृंगार हैं। वे आकाश की शोभा बढ़ाते हैं। जैसे ही सितारे बादलों की ओट में छिपे, आकाश सूना हो जाता है। ठीक इसी प्रकार कुछ लोग हमारे जीवन में अत्यंत महत्त्वपूर्ण होते हैं। उनके चले जाने पर या विमुख हो जाने पर मानो हमारा जीवन निरर्थक हो जाता है।

कवि के अंदर अनोखी सामर्थ्य होती है। वह जिन गीतों को स्वर देता है, वे अमर हो जाते हैं। इसी प्रकार कवि अपनी रचनाओं के द्वारा समाज में परिवर्तन ला सकता है।

सागर में अथाह जलराशि होती है, परंतु खारा होने के कारण अथाह होने पर भी वह जलराशि पीने योग्य नहीं होती। उसी प्रकार कोई व्यक्ति कितना भी बड़ा या धनवान क्यों न हो, यदि वह किसी जरूरतमंद के काम नहीं आ सकता तो उसका बड़प्पन व्यर्थ है।

मन विषय-प्रवेश:

प्रस्तुत कविता ‘मन’ जापान की लोकप्रिय विधा हाइकु’ पर आधारित है। यह विधा हिंदी साहित्य में स्वीकृति पा चुकी है। इस विधा को विश्व की सबसे छोटी कविता का स्थान प्राप्त है। इस कविता में कवि ने तीन-तीन छोटी पंक्तियों में अलग-अलग घटनाओं को सुंदर ढंग से पिरोया है। प्रस्तुत कविता की यह अपनी विशेषता है।

Balbharti Maharashtra State Board Class 10 Hindi Solutions Lokbharti Chapter 3 वाह रे! हमदर्द Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Hindi Lokbharti Chapter 3 वाह रे! हमदर्द

Hindi Lokbharti 10th Std Digest Chapter 3 वाह रे! हमदर्द Textbook Questions and Answers

कृति

सूचना के अनुसार कृतियाँ कीजिए:

प्रश्न 1.
संजाल पूर्ण कीजिए:

उत्तर:

प्रश्न 2.
अंतर स्पष्ट कीजिए:
प्राइवेट अस्पताल – सार्वजनिक अस्पताल
१. …………………….. – १. ……………………..
प्राइवेट वार्ड – जनरल वार्ड
१. …………………….. – १. ……………………..
उत्तर:

प्राइवेट अस्पताल	सार्वजनिक अस्पताल
प्राइवेट अस्पताल में अच्छी सुविधाएँ होती हैं।	सार्वजनिक अस्पताल में कई बार सुविधाओं का अभाव होता है।

प्राइवेट वॉर्ड	जनरल वॉर्ड
मिलने का कोई निश्चित समय नहीं होता।	मिलने का निश्चित समय होता है।

प्रश्न 3.
आकृति में लिखिए:

उत्तर:

प्रश्न 4.
कारण लिखिए
a. लेखक को अधिक गुस्सा अपनी पत्नी पर आया ……………………..
b. लेखक कहते हैं कि मेरी दूसरी टाँग उस जगह तोड़ना जहाँ कोई परिचित न हो ……………………..
उत्तर:
a. आगंतुक को रोते देखकर लेखक की पत्नी ने उसे कोई रिश्तेदार या करीबी मित्र समझकर टैक्सीवाले को किराये के पैसे दे दिए थे।
b. उस जगह लेखक के परिचित होंगे तो लेखक से समय-असमय मिलने आकर तंग करेंगे।

प्रश्न 5.
शब्दसमूह के लिए एक शब्द लिखिए:
a. वह स्थान जहाँ अनेक प्रकार के पशु-पक्षी रखे जाते हैं – ……………………..
b. जहाँ मुफ्त में भोजन मिलता है – ……………………..
उत्तर:
(i) चिड़ियाघर
(ii) लंगर (भंडारा)।

प्रश्न 6.
शब्द बनाइए:

उत्तर:

प्रश्न 7.
अभिव्यक्ति- मरीज से मिलने जाते समय कौन-कौन-सी सावधानियां बरतनी चाहिए, लिखिए।
उत्तर:
प्राय: सभी को कभी-न-कभी मरीजों से मिलने अस्पताल में जाना पड़ता है। मरीज से मिलने जाते समय कुछ सावधानियाँ बरतना अत्यंत आवश्यक है। मरीज से मिलने जाते समय हमें इस बात का ध्यान रखना चाहिए कि हमारी वजह से उसे कोई कष्ट न पहुँचे। बच्चे चुलबुले होते हैं। इसलिए मरीज के पास बच्चों को नहीं लेकर जाना चाहिए। बीमारी में दवा और पथ्य के साथ मरीज को आराम व अच्छी नींद आवश्यक है।

अत: मरीज के पास ज्यादा देर तक बैठना, जोर-जोर से बोलना, मरीज की बीमारी के बारे में नकारात्मक बातें करना आदि उचित नहीं है। जहाँ तक हो सके, मरीज का उत्साह बढ़ाना चाहिए। अस्पताल में डॉक्टर मरीज को उसकी आवश्यकता के अनुसार दवाएँ देते हैं। इसलिए मरीज से देसी नुस्खे आजमाने की बातें नहीं करनी चाहिए और न ही डॉक्टर की दवा के बारे में रोगी के मन में किसी तरह का भ्रम पैदा करना चाहिए।

भाषा बिंदु
प्रश्न 1.
निम्नलिखित वाक्यों में आए हुए संज्ञा शब्दों को रेखांकित करके उनके भेद लिखिए:
1. सोनाबाई अपने चार बच्चों के साथ आई। ……………………..
2. गाय बहुत दूध देती है। ……………………..
3. मैं रोज ईश्वर से प्रार्थना करता हैं। ……………………..
4. सैनिकों की टुकड़ी आगे बढ़ी। ……………………..
5. सोना-चाँदी और भी महँगे होते जा रहे हैं। ……………………..
6. गोवा देख मैं तरंगायित हो उठा। ……………………..
7. युवकों का दल बचाव कार्य में लगा था। ……………………..
8. आपने विदेश में भ्रमण तो कर लिया है। ……………………..
9. इस कहानी में भारतीय समाज का चित्रण मिलता है। ……………………..
10. सागर का जल खारा होता है। ……………………..
उत्तर:
1. सोनाबाई – व्यक्तिवाचक बच्चों – जातिवाचक।
2. गाय – जातिवाचक दूध – द्रव्यवाचक।
3. ईश्वर – जातिवाचक प्रार्थना- भाववाचक।
4. सैनिकों – जातिवाचक टुकड़ी – समूहवाचक।
5. सोना-चाँदी – द्रव्यवाचक।
6. गोवा – व्यक्तिवाचक।
7. युवकों – जातिवाचक दल – समूहवाचक। कार्य – भाववाचकी
8. विदेश – जातिवाचक भ्रमण – भाववाचका
9. कहानी – जातिवाचक समाज- समूहवाचक। चित्रण- भाववाचक।
10. सागर – जातिवाचक जल – द्रव्यवाचक।

प्रश्न 2.
पाठ में प्रयुक्त किन्हीं पाँच संज्ञाओं को ढूँढकर उनका वाक्यों में प्रयोग कीजिए।
उत्तर:

• साइकिल – मुझे साइकिल चलाना नहीं आता।
• जोश – कई लोग जोश में होश खो बैठते हैं।
• रेत – आन्या को सागर तट पर रेत का घर बनाना बहुत पसंद है।
• आत्मा – प्रत्येक आत्मा परमात्मा का अंश होती है।
• बंदर – बंदर और बच्चे एक जैसे शरारती होते हैं।

प्रश्न 3.
निम्नलिखित वाक्यों के रिक्त स्थानों में उचित सर्वनामों का प्रयोग कीजिए:
1. …………………….. सार्वजनिक अस्पताल के प्राइवेट वार्ड में हैं।
2. …………………….. बाजार जाओ।
3. …………………….. कारखाने में एक ही विभाग में काम करते थे।
4. इसे लेकर …………………….. क्या करोगे?
5. हृदय …………………….. है; …………………….. उदार हो।
6. लोग …………………….. कमरा स्वच्छ कर रहे हैं।
7. …………………….. रिसॉर्ट हमने पहले से बुक कर लिया है।
8. इसके बाद …………………….. लोग दिन भर पणजी देखते रहे।
9. …………………….. इसके पहले उसे मना करता।
10. काम करने के लिए कहा है …………………….. करो।
उत्तर:
1. वे सार्वजनिक अस्पताल के प्राइवेट वार्ड में हैं।
2. तुम बाजार जाओ।
3. हम कारखाने में एक ही विभाग में काम करते थे।
4. इसे लेकर तुम क्या करोगे।
5. हृदय वही है; तुम उदार हो।
6. लोग स्वयं कमरा साफ कर रहे हैं।
7. मैं रिसॉर्ट हमने पहले से बुक कर लिया है।
8. इसके बाद हम लोग दिन भर पणजी देखते रहे।
9. मैं इसके पहले उसे मना करता।
10. काम करने के लिए कहा है वही करो।

प्रश्न 4.
पाठ में प्रयुक्त सर्वनाम ढूँढ़कर उनका स्वतंत्र वाक्यों में प्रयोग कीजिए।
उत्तर:

• मैंने
  वाक्य: मैंने रेत का घर बनाया।
• तुझे
  वाक्य: शिक्षिका ने तुझे बुलाया है, मनन।
• वे
  वाक्य: वे मेरे चाचा हैं।
• कोई
  वाक्य: बाहर कोई है।
• आप
  वाक्य: कल आप कहाँ थे?
• मुझसे
  वाक्य: माँ ने गुस्से में कहा, मुझसे बात मत करो।
• उन्होंने
  वाक्य: उन्होंने मुझे घर तक पहुँचाया।
• मुझे।
  वाक्य: मुझे नींद आ रही है।

उपयोजित लेखन

प्रश्न.
निम्नलिखित मुद्दों के आधार पर किसी समारोह का वृत्तांत लेखन कीजिए:

• स्थान
• तिथि और समय
• प्रमुख अतिथि
• समारोह
• अतिथि संदेश
• समापन

उत्तर:
गांधी जयंती पर गांधी जी का स्मरण
अकोला, 3 अक्तूबर। अकोला के सरदार पटेल विद्यालय में कल 2 अक्तूबर को गांधी जयंती समारोह का आयोजन किया गया। विद्यालय में समारोह सुबह 10 बजे आयोजित किया जाना था। विद्यालय के विद्यार्थी 9 बजे से ही अपने-अपने स्थान पर बैठ गए थे।

विद्यालय के सभी अध्यापक मंच पर खादी का कुर्ता-पाजामा और खादी टोपी पहनकर विराजमान थे। प्रमुख अतिथि के रूप में शहर के वयोवृद्ध गांधीवादी जनार्दन पाटील उपस्थित थे। मंच पर गांधी जी की तस्वीर सुशोभित हो रही थी।

समारोह की शुरुआत ‘वंदे मातरम्’ गीत से हुई। विद्यालय के प्रधानाचार्य राम रतन जोशी ने उपस्थित लोगों का परिचय दिया और देश के लिए गांधी जी के योगदान की चर्चा की।

प्रमुख अतिथि जनार्दन पाटील ने गांधी जी के जीवन की कई घटनाओं के बारे में बताया। उन्होंने गांधी जी के हमेशा सत्य बोलने के आग्रह के बारे में बताया और कहा कि हमें सत्य के मार्ग पर चलना चाहिए। अपने लाभ के लिए कभी झूठ का सहारा नहीं लेना चाहिए।

विद्यालय के उपमुख्याध्यापक सुधीर देशपांडे ने प्रमुख अतिथि के प्रति आभार व्यक्त किया।

राष्ट्रगान के साथ समारोह का समापन हुआ।

Hindi Lokbharti 10th Textbook Solutions Chapter 3 वाह रे! हमदर्द Additional Important Questions and Answers

कृतिपत्रिका के प्रश्न 1 (अ) तथा 1(आ) के लिए

गद्यांश क्र.1

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

कृति 1: (आकलन)

प्रश्न 1.
प्रवाह तालिका पूर्ण कीजिए:

उत्तर:

प्रश्न 2.
आकृति पूर्ण कीजिए:

उत्तर:

प्रश्न 3.
संजाल पूर्ण कीजिए:

उत्तर:

कृति 2: (आकलन)

प्रश्न 1.
संजाल पूर्ण कीजिए:

उत्तर:

प्रश्न 2.
आकृति पूर्ण कीजिए:

उत्तर:

प्रश्न 3.
जोड़ियाँ मिलाइए:

‘अ’	‘आ’
(i) ऐक्सिडेंट	खुला निमंत्रण
(ii) टाँग	दुर्घटना
(iii) प्राइवेट वार्ड	रेत की थैली
(iv) सार्वजनिक अस्पताल में भरती होना	फ्रैक्चर

उत्तर:

‘अ’	‘आ’
(i) ऐक्सिडेंट	फ्रैक्चर
(ii) टाँग	रेत की थैली
(iii) प्राइवेट वॉर्ड	खुला निमंत्रण
(iv) सार्वजनिक अस्पताल में भरती होना	दुर्घटना

कृति 3: (शब्द संपंदा)

प्रश्न 1.
सूचना के अनुसार लिखिए:

उत्तर:

प्रश्न 2.
गद्यांश में प्रयुक्त उर्दू शब्द ढूँढकर लिखिए।
(i) ………………….
(ii) ………………….
(iii) ………………….
(iv) ………………….
उत्तर:
(i) जवाब
(ii) फिक्र
(ii) तकलीफ
(iv) मरीज।

प्रश्न 3.
गद्यांश में प्रयुक्त शब्द-युग्म ढूँढकर लिखिए।
(i) ………………….
(ii) ………………….
(ii) ………………….
(iv) ………………….
उत्तर:
(ii) मिलने-जुलने
(iii) सही-सलामत
(iv) परिचित-अपरिचित।

प्रश्न 4.
गद्यांश में प्रयुक्त उपसर्गयुक्त शब्द ढूँढ़कर उनके मूल शब्द और उपसर्ग अलग करके लिखिए।
(i) ………………….
(ii) ………………….
(iii) ………………….
उत्तर:
(i) अपरिचित = अ + परिचित।
(ii) दुर्घटना = दुर् + घटना।
(iii) हमदर्दी = हम + दर्दी।

कृति 4: (स्वमत अभिव्यक्ति)

प्रश्न.
सार्वजनिक अस्पतालों में मरीजों को होने वाली परेशानियों के विषय में अपने विचार लिखिए।
उत्तर:
देश में अनगिनत निजी अस्पताल हैं, परंतु देश की आधी से अधिक गरीब जनता सार्वजनिक अस्पतालों पर ही निर्भर है। इन अस्पतालों की हालत बहुत दयनीय है। इन अस्पतालों की एक्स-रे आदि मशीनों का कोई ठिकाना नहीं होता। गरीबों को वहाँ इलाज के स्थान पर तकलीफ ही मिलती है। सार्वजनिक अस्पतालों में समय पर डॉक्टर नहीं मिलते। डॉक्टर यदि मिल भी जाता है, तो दवाइयाँ नहीं मिलती।

इसलिए मरीजों को महँगे दामों पर बाहर से दवाएँ खरीदने को बाध्य होना पड़ता है। इसके अलावा डॉक्टर के साथ-साथ अस्पताल के कर्मचारियों का व्यवहार भी रोगियों के प्रति बहुत खराब होता है। ऐसे में इन अस्पतालों में मरीज का ढंग से इलाज नहीं हो पाता। इसलिए लोग इन अस्पतालों में जाने से कतराते हैं।

गद्यांश क्र.2
प्रश्न.
निम्नलिखित पठित गद्यांश पढ़कर दी गई सूचनाओं के अनुसार कृतियाँ कीजिए:

कृति 1: (आकलन)

प्रश्न 1.
वाक्य पूर्ण कीजिए:
(i) इनकी हमदर्दी में यह बात खास छिपी रहती है ………………………..।
(ii) उस दिन सोनाबाई अपने चार बच्चों के साथ आई तो ………………………..।
उत्तर:
(1) इनकी हमदर्दी में यह बात खास छिपी रहती है कि देख बेटा, वक्त सब पर आता है।
(ii) उस दिन सोनाबाई अपने चार बच्चों के साथ आई तो मुझे लगा कि आज फिर कोई दुर्घटना होगी।

प्रश्न 2.
आकृति पूर्ण कीजिए:

उत्तर:

कृति 2: (आकलन)

प्रश्न 1.
आकृति पूर्ण कीजिए:
(i) दर्द के मारे एक तो मरीज को वैसे ही यह नहीं आती – [ ]
(ii) कुछ लोग सिर्फ यह निभाने आते हैं – [ ]
(iii) इन लोगों को मरीज से यह नहीं होती – [ ]
(iv) कब मेरी टाँग टूटे, कब वे अपना यह चुकाएँ – [ ]
उत्तर:
(i) दर्द के मारे एक तो मरीज को वैसे ही यह नहीं आती [नींद]
(ii) कुछ लोग सिर्फ यह निभाने आते हैं – [औपचारिकता]
(iii) इन लोगों को मरीज से यह नहीं होती – [हमदर्दी]
(iv) कब मेरी टाँग टूटे, कब वे अपना यह चुकाएँ – [एहसान]

प्रश्न 2.
विधानों के सामने सत्य /असत्य लिखिए:
(i) मैंने तय किया कि आज मैं आँख ही नहीं खोलूँगा।
(ii) ऑफिस के बड़े साहब आए।
(iii) उन्होंने मेरी टाँग के टूटे हिस्से को जोर से दबाया।
(iv) कहिए, अब सिरदर्द कैसा है?
उत्तर:
(i) सत्य
(ii) असत्य
(iii) सत्य
(iv) असत्य।

प्रश्न 3.
आकृति पूर्ण कीजिए:

उत्तर:

कृति 3: (शब्द संपदा)

प्रश्न 1.
निम्नलिखित शब्दों का वचन बदलकर लिखिए:
(i) बेटा
(ii) टाँग
(iii) दुर्घटनाएँ
(iv) हिस्सा।
उत्तर:
(i) बेटा – बेटे
(ii) नींद – स्त्रीलिंग
(iii) दुर्घटनाएँ – दुर्घटना
(iv) वक्त – पुल्लिग।

प्रश्न 2.
निम्नलिखित शब्दों के लिंग पहचानकर लिखिए:
(i) दिन
(ii) नींद
(ii) फुरसत
(iv) वक्त।
उत्तर:
(i) दिन – पुल्लिग
(ii) आँख = नयन
(iii) फुरसत – स्त्रीलिंग
(iv) वक्त = समय।

प्रश्न 3.
निम्नलिखित शब्दों के समानार्थी शब्द लिखिए:
(i) नींद
(ii) आँख
(iii) दर्द
(iv) वक्त।
उत्तर:
(i) नींद = निद्रा
(iii) दर्द = पीड़ा
(ii) टाँग – टाँगें
(iv) हिस्सा – हिस्से।

गद्यांश क्र. 3

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

कृति 1: (आकलन)

प्रश्न 1.
कारण लिखिए:

(i) आगंतुक ने जब लेखक से आँख मिलाई तो एकदम चुप हो गया …………………………
उत्तर:
(i) आगंतुक किसी अन्य मरीज से मिलने आया था।

प्रश्न 2.
ऐसे दो प्रश्न बनाइए, जिनके उत्तर: निम्नलिखित हों:

(i) दवा की शीशी
(ii) औपचारिकता।
उत्तर:
(i) सोनाबाई की लड़की ने क्या पटक दी?
(ii) कुछ लोग क्या निभाने की हद कर देते हैं?

कृति 2: (आकलन)

प्रश्न 1.
आकृति पूर्ण कीजिए:

उत्तर:

प्रश्न 2.
आकृति पूर्ण कीजिए:

उत्तर:

प्रश्न 3.
गद्यांश में उल्लिखित शरीर के अंगों के नाम:

उत्तर:

कृति 3: (शब्द संपदा)

प्रश्न 1.
निम्नलिखित शब्दों के विरुद्धार्थी शब्द लिखिए:
(i) सिर
(ii) रोना
(iii) गलत
(iv) गुस्सा।
उत्तर:
(i) सिर x पैर
(ii) रोना x हँसना
(iii) गलत x सही
(iv) गुस्सा x प्यार।

प्रश्न 2.
गद्यांश में प्रयुक्त अंग्रेजी शब्द ढूँढकर लिखिए।
(i) …………………………
(ii) …………………………
(iii) …………………………
(iv) …………………………
उत्तर:
(i) टेबल
(ii) डांस
(iii) टैक्सी
(iv) प्रैक्टिस।

कृति 4: (स्वमत अभिव्यक्ति)

प्रश्न.
‘शकुन-अपशकुन’ के बारे में अपने विचार लिखिए।
उत्तर:
शकुन-अपशकुन समाज में प्रचलित एक अवधारणा है। इसमें यह माना जाता है कि कुछ विशेष प्रकार की परिघटनाएँ हमारे भविष्य का संकेत देती हैं। अनुकूल भविष्यवाणी करने वाले संकेतों को शुभ शकुन और प्रतिकूल भविष्यवाणी करने वाले संकेतों को अपशकुन कहा जाता है। हमारे देश में ही नहीं, अपितु संसार भर में लोग शकुन-अपशकुन पर विश्वास करते हैं। भारतीय संस्कृति में शकुन-अपशकुन का वर्णन वेदों, पुराणों और धार्मिक ग्रंथों में भी मिलता है।

काली बिल्ली द्वारा रास्ता काट जाना, किसी कार्य को आरंभ करते समय किसी का छींक देना, घर से बाहर जाते हुए व्यक्ति को किसी के द्वारा टोका जाना आदि समाज में बहुप्रचलित अपशकुन हैं। इन अपशकुनों को मानने वालों की संख्या कम नहीं है। इन अपशकुनों के चक्कर में आकर कभी-कभी लोगों को हानि भी उठानी पड़ती है, फिर भी वे इन्हें मानने से नहीं चूकते। ये मान्यताएँ मनुष्य को कमजोर बनाती हैं। वैज्ञानिक दृष्टि से इन शकुन-अपशकुनों को अंधविश्वास ही माना जाता है।

गद्यांश क्र.4

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

कृति 1: (आकलन)

प्रश्न 1.
आकृति पूर्ण कीजिए:

उत्तर:

प्रश्न 2.
कारण लिखिए:

(i) लेखक ने बड़ी मुश्किल से कवि लपकानंद को विदा किया …………………………
उत्तर:
(i) कवि लपकानंद जब कविता सुनाना शुरू करते, तो रुकने का नाम नहीं लेते थे।

कृति 2: (आकलन)

प्रश्न 1.
आकृति पूर्ण कीजिए:

उत्तर:

प्रश्न 2.
ऐसे दो प्रश्न बनाइए, जिनके उत्तर निम्नलिखित शब्द हों:
(i) डायरी
(ii) बड़े बेवफा।
उत्तर:
(i) कवि ने झोले से क्या निकाली?
(ii) हमदर्दी जताने वाले कैसे होते हैं?

प्रश्न 3.
आकृति पूर्ण कीजिए:

उत्तर:

कृति 3: (शब्द संपदा)

प्रश्न 1.
गद्यांश में प्रयुक्त शब्द-युग्म ढूंढकर उनको वाक्यों में प्रयोग कीजिए।
(i) …………………………
(ii) …………………………
उत्तर:
(i) दस-बीस – गोदाम में दस-बीस किलो गेहूँ पड़ा है।
(ii) चार-पाँच – चार-पाँच लड़कों को भेजो, कक्षा का फर्नीचर बाहर निकलवाना है।

कृति 4: (स्वमत अभिव्यक्ति)

प्रश्न.
कवियों की कविता सुनाने की आदत के बारे में अपने विचार लिखिए।
उत्तर:
कवि दो प्रकार के होते हैं। एक वे, जो सचमुच कवि होते हैं और अपने विचारों को मथकर उन्हें सुंदर और सुरुचिपूर्ण शब्दों के माध्यम से कागज पर उतारते हैं। उनकी कविता सुनकर श्रोता को आनंद के साथ-साथ एक दिशा भी मिलती है। दूसरे प्रकार के कवि वे होते हैं, जो अंत:करण से कवि नहीं होते। वे जबरन कवि बनकर कविता लिखना चाहते हैं। इनकी कविता कविता न होकर शब्दों का बेतरतीब समूह होती है।

जोड़-तोड़कर कविता तैयार करते ही ये श्रोता की तलाश करने लगते हैं और जो भी सामने मिल जाता है, उसे अपनी कविता सुनाए बिना नहीं छोड़ते। इनकी कविता सुनने के लिए कोई आसानी से तैयार नहीं होता। पर विद्वान कवि कभी अपनी कविता सुनाने की कोशिश नहीं करते। उनकी कविता सारगर्मित होती है और वे हर किसी को कविता सुनाते नहीं फिरते।

भाषा अध्ययन (व्याकरण)

प्रश्न.
सूचनाओं के अनुसार कृतियाँ कीजिए:

1. शब्द भेद:
निम्नलिखित वाक्यों में अधोरेखांकित शब्दों के शब्दभेद पहचानकर लिखिए:
(i) मैं अपनी टाँगों की ओर देखता हूँ।
(ii) मेरे दिमाग में एक नये मुहावरे का जन्म हुआ।
(iii) सोनाबाई के बच्चे खेलने लगे।
उत्तर:
(i) मैं – पुरुषवाचक सर्वनाम।
(ii) नये – गुणवाचक विशेषण।
(iii) सोनाबाई – व्यक्तिवाचक संज्ञा।

2. अव्यय:
निम्नलिखित अव्ययों का अपने वाक्यों में प्रयोग कीजिए:
(i) अकसर
(ii) इर्द-गिर्द
(iii) धीरे-धीरे।
उत्तर:
(i) मैं लपकानंद को देखकर अकसर भाग खड़ा होता हूँ।
(ii) मेरे इर्द-गिर्द अनेक लोग खड़े थे।
(iii) बड़े बाबू धीरे-धीरे मुझे हिलाने लगे।

3. संधि:
कृति पूर्ण कीजिए:।

संधि शब्द	संधि विच्छेद	संधि भेद
………………..	नै + इका	………………..
अथवा
दुर्बल	………………..	………………..

उत्तर:

संधि शब्द	संधि विच्छेद	संधि भेद
नायिका	नै + इका	स्वर संधि
अथवा
दुर्बल	दुः + बल	विसर्ग संधि

4. सहायक क्रिया:
निम्नलिखित वाक्यों में से सहायक क्रियाएँ पहचानकर उनका मूल रूप लिखिए:
(i) अस्पताल का खयाल आते ही में काँप उठा।
(ii) कोई भी आए मैं चुपचाप पड़ा रहूँगा।
(iii) बच्चे खेलने लगे।
उत्तर:
सहायक क्रिया – मूल रूप
(i) उठा – उठना
(ii) रहूँगा – रहना
(iii) लगे – लगना

5. प्रेरणार्थक क्रिया:
निम्नलिखित क्रियाओं के प्रथम प्रेरणार्थक और द्वितीय ‘ प्रेरणार्थक रूप लिखिए:

क्रिया	प्रथम प्रेरणार्थक रूप	द्वितीय प्रेरणार्थक रूप
(i) मानना
(ii) लिखना
(ii) जलना

उत्तर:

क्रिया	प्रथम प्रेरणार्थक रूप	द्वितीय प्रेरणार्थक रूप
(i) मानना	मनाना	मनवाना
(ii) लिखना	लिखाना	लिखवाना
(ii) जलना	जलाना	जलवाना

6. मुहावरे:
(1) निम्नलिखित कहावत का अर्थ लिखिए और वाक्य में प्रयोग कीजिए:
ढाक के तीन पात।
अर्थ: सदा एक-सी स्थिति।
वाक्य: छगनलाल ने सालभर में कई व्यवसाय बदले, पर हालत आज भी वही है ढाक के तीन पात।

(2) अधोरेखांकित वाक्यांश के लिए उचित मुहावरे का चयन कर वाक्य फिर से लिखिए:
सुमधुर गायन सुनकर श्रोताओं ने गायक की प्रशंसा की। (सराहना करना, बोलबाला होना)
उत्तर:
अर्थ: सराहना करना।
वाक्य: सुमधुर गायन सुनकर श्रोताओं ने गायक की सराहना की।

7. कारक:
निम्नलिखित वाक्यों में प्रयुक्त कारक पहचानकर उनका भेद लिखिए:
(i) मैंने उन्हें जल्दी से चाय पिलाई।
(ii) आप अस्पताल में हैं।
उत्तर:
(i) मैंने – कर्ता कारक
(ii) अस्पताल में – अधिकरण कारक।

8. विरामचिह्न:
निम्नलिखित वाक्यों में यथास्थान उचित विरामचिह्नों का प्रयोग करके वाक्य फिर से लिखिए:
(i) वे मुझे ऐसे देख रहे थे मानो उनकी एक आँख पूछ रही हो कहो कविता कैसी रही और दूसरी आँख पूछ रही हो बोल बेटा अब भी मुझसे भागेगा
(ii) सोनाबाई ने लड़की को घूरा फिर हँसते हुए बोली भैया पेड़े खिलाओ दवा गिरना शुभ होता है
(iii) मैंने कराहते हुए पूछा मैं कहाँ हूँ
उत्तर:
(i) वे मुझे ऐसे देख रहे थे, मानो उनकी एक आँख पूछ रही हो, ‘कहो, कविता कैसी रही?’ और दूसरी आँख पूछ रही हो, बोल, बेटा! अब भी मुझसे भागेगा?’
(ii) सोनाबाई ने लड़की को घूरा, फिर हँसते हुए बोली, “भैया, पेड़े खिलाओ, दवा गिरना शुभ होता है।”
(iii) मैंने कराहते हुए पूछा, “मैं कहाँ हूँ?”

9. काल परिवर्तन:
निम्नलिखित वाक्यों का सूचना के अनुसार काल परिवर्तन कीजिए:
(i) एक चेहरा बड़ी तेजी से जवाब देता है। (पूर्ण वर्तमानकाल)
(ii) मेरी आँख खुलते ही सबके चेहरों पर प्रसन्नता की लहर दौड़ जाती है। (सामान्य भूतकाल)
(iii) सोनाबाई फिर आती है। (सामान्य भविष्यकाल)
उत्तर:
(i) एक चेहरे ने बड़ी तेजी से जवाब दिया है।
(ii) मेरी आँख खुलते ही सबके चेहरों पर प्रसन्नता की लहर दौड़ गई।
(iii) सोनाबाई फिर आएगी।

10. वाक्य भेद:
(1) निम्नलिखित वाक्यों का रचना के आधार पर भेद पहचानकर लिखिए:
(i) जब आँख खुली तो मैंने स्वयं को बिस्तर पर पाया।
(ii) मैंने उसे जल्दी से चाय पिलाई और विदा किया।
उत्तर:
(i) मिश्र वाक्य
(ii) संयुक्त वाक्य।

(2) निम्नलिखित वाक्यों का अर्थ के आधार पर दी गई सूचना के अनुसार परिवर्तन कीजिए:
(i) मेरी टाँग टूटना एक दुर्घटना थी। (प्रश्नवाचक)
(ii) आज फिर कोई दुर्घटना होगी। (इच्छावाचक)
उत्तर:
(i) क्या मेरी टाँग टूटना एक दुर्घटना थी?
(ii) आज फिर कोई दुर्घटना न हो।

11. वाक्य शुद्धिकरण:
निम्नलिखित वाक्य शुद्ध करके लिखिए:
(i) अब मैं अपने टाँगों की ओर देखता है।
(ii) सोनाबाई से एक पल लड़की को घूरी।
(iii) गुप्ता जी की कमरा शायद बगल में हैं।
उत्तर:
(i) अब मैं अपनी टाँगों की ओर देखता हूँ।
(ii) सोनाबाई ने एक पल लड़की को घूरा।
(iii) गुप्ता जी का कमरा शायद बगल में है।

उपक्रम/कृति/परियोजना

किसी सार्वजनिक या ग्राम पंचायत की सभा में अंगदान’ के बारे में अपने विचार प्रस्तुत कीजिए।
उत्तर:
आदरणीय सरपंच महोदय, पंच परमेश्वर तथा अन्य सभी उपस्थित सज्जनो, आज मैं आप सभी के समक्ष अंगदान के विषय में अपने विचार प्रस्तुत करना चाहता हूँ। अंगदान वह प्रक्रिया है, जिसमें किसी व्यक्ति के शरीर का कोई अंग उसकी व उसके परिवार की सहमति से हटाकर किसी अन्य व्यक्ति को दे दिया जाता है। इस प्रक्रिया द्वारा एक व्यक्ति को नया जीवन मिल जाता है।

प्रत्यारोपण के लिए गुर्दे, लिवर, फेफड़े, हृदय, हड्डियाँ, अस्थि मज्जा, त्वचा, अग्न्याशय, कॉर्निया, आँत आदि का दान दिया जाता है। अंगदान की प्रक्रिया को दुनिया भर में प्रोत्साहित किया जाता है। भारत में यह कानूनन वैध है। अंगदान समाज के लिए एक चमत्कार साबित हुआ है। हालाँकि माँग की तुलना में आपूर्ति बहुत कम है।

वाह रे! हमदर्द Summary in Hindi

वाह रे! हमदर्द विषय-प्रवेश :

अस्पताल में इलाज के लिए भर्ती हुए मरीज को देखने जाने की परंपरा समाज में पुरानी है। इससे मरीज को खुशी होती है और कुछ समय के लिए उसका ध्यान अपने कष्ट से हट जाता है। पर कुछ मिलने वाले ऐसे होते हैं, जो मरीज के लिए परेशानी का कारण बन जाते हैं। प्रस्तुत हास्य-व्यंग्यात्मक निबंध में लेखक ने दुर्घटना के माध्यम से एक ऐसी ही स्थिति का चित्रण किया है। निबंध में जहाँ एक ओर समाज में विद्यमान परोपकार की भावना पर प्रकाश डाला गया है, वहीं दूसरी ओर बड़े ही रोचक ढंग से हमदर्द लोगों की मानसिकता को भी चित्रित किया गया है। कभी-कभी हमदर्दी भी रोगी की मानसिक पीड़ा का कारण बन जाती है।

वाह रे! हमदर्द मुहावरे – अर्थ

• ड़ाना – बाधा डालना।
• काँप उठना – भयभीत होना।

Balbharti Maharashtra State Board Class 10 Hindi Solutions Lokbharti Chapter 6 गिरिधर नागर Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Hindi Lokbharti Chapter 6 गिरिधर नागर

Hindi Lokbharti 10th Std Digest Chapter 6 गिरिधर नागर Textbook Questions and Answers

कृति

(कृतिपत्रिका के प्रश्न 2 (अ) तथा प्रश्न 2 (आ) के लिए)
सूचनानुसार कृतियाँ कीजिए:

प्रश्न 1.
संजाल पूर्ण कीजिए:

उत्तर:

प्रश्न 2.
प्रवाह तालिका पूर्ण कीजिए:

उत्तर:

प्रश्न 3.
इस अर्थ में आए शब्द लिखिए:

	अर्थ	शब्द
i.	दासी	………….
ii.	साजन	………….
iii.	बार-बार	………….
iv.	आकाश	………….

उत्तर:

	अर्थ	शब्द
i.	दासी	चेरी
ii.	साजन	पति
iii.	बार-बार	बेर-बेर
iv.	आकाश	अंबर

प्रश्न 4.
कन्हैया के नाम

उत्तर:

प्रश्न 5.
दूसरे पद का सरल अर्थ लिखिए।
उत्तर:
(i) निकट – ढिग
(ii) साजन – पति।

उपयोजित लेखन

निम्नलिखित शब्दों के आधार पर कहानी लेखन कीजिए तथा उचित शीर्षक दीजिए:
अलमारी, गिलहरी, चावल के पापड़, छोटा बच्चा
उत्तर:
जीव दया
एक गाँव में एक छोटा बच्चा रहता था। उसका नाम चिंटू था। एक दिन चिंटू अपने घर के बाहर खेल रहा था। उसने देखा कि सामने एक पेड़ के नीचे दो-तीन कौए किसी चीज पर चोंच मार रहे हैं और वहाँ से हल्की-हल्की चर्ची-चीं की आवाज आ रही है। चिंटू दौड़कर वहाँ पहुँचा और उसने उन कौओं को वहाँ से उड़ाया। उसने देखा कि एक छोटी-सी गिलहरी वहाँ ची-चीं कर रही थी। उसका शरीर कौओं की चोंच से घायल हो गया था। चिंटू ने अपनी जेब से रूमाल निकाला और डरे बिना धीरे से गिलहरी को उठा लिया। उसने घर के अंदर लाकर उसे पानी पिलाया, उसके घावों को साफ करके उन पर सोफामाइसिन लगाई और उसे मेज पर बैठा दिया।

गिलहरी कुछ देर बाद धीरे-धीरे मेज पर घूमने लगी। मेज पर एक प्लेट में चावल के पापड़ रखे थे। गिलहरी ने एक पापड़ उठाया और अपने अगले दोनों पंजों में पकड़कर धीरे-धीरे उसे खाने लगी। चिंटू को बहुत अच्छा लगा। उसने माँ से पूछा कि जब तक गिलहरी बिलकुल ठीक नहीं हो जाती क्या मैं उसे अपने पास रख सकता हूँ। अभी अगर वह बाहर जाएगी तो कौए उसे अपना आहार बना लेंगे। माँ को चिंटू की ऐसी सोच पर गर्व हुआ और उन्होंने खुशी-खुशी उसकी बात मान ली। चिंटू ने अपनी किताबों की खुली आलमारी के एक खाने में एक तौलिया बिछाकर गिलहरी को बैठा दिया। उसके पास चावल के कुछ पापड़ तथा अमरूद के कुछ टुकड़े रख दिए। तीन-चार दिन बाद जब गिलहरी अच्छी तरह दौड़ने लगी तो चिंटू ने उसे बाहर पेड़ पर छोड़ दिया।

सीख: हमें पशु-पक्षियों के प्रति दया भाव रखना चाहिए।

अपठित पद्यांश

सूचना के अनुसार कृतियाँ कीजिए:-

काम जरा लेकर देखो, सख्त बात से नहीं स्नेह से
अपने अंतर का नेह अरे, तुम उसे जरा देकर देखो।
कितने भी गहरे रहें गर्त, हर जगह प्यार जा सकता है,
कितना भी भ्रष्ट जमाना हो, हर समय प्यार भा सकता है।
जो गिरे हुए को उठा सके, इससे प्यारा कुछ जतन नहीं,
दे प्यार उठा पाए न जिसे, इतना गहरा कुछ पतन नहीं ।।

– (भवानी प्रसाद मिश्र)

प्रश्न 1.
उत्तर लिखिए:
a. किसी से काम करवाने के लिए उपयुक्त – ………….
b. हर समय अच्छी लगने वाली बात – ………….
उत्तर:

प्रश्न 2.
उत्तर लिखिए:
a. अच्छा प्रयत्न यही है – ………….
b. यही अधोगति है – ………….
उत्तर:

प्रश्न 3.
पद्यांश की तीसरी और चौथी पंक्ति का संदेश लिखिए।
उत्तर:

भाषा बिंदु

कोष्ठक में दिए गए प्रत्येक/कारक चिह्न से अलग-अलग वाक्य बनाइए और उनके कारक लिखिए:
[ने, को, से, का, की, के, में, पर, हे, अरे, के लिए]
………………………………………………………..
………………………………………………………..
………………………………………………………..
………………………………………………………..
………………………………………………………..
उत्तर:
(1) ने – ऋतु ने खाना बनाया।
(2) को – विपिन ने प्रगति को खाना खिलाया।
(3) से – हिमानी साइकिल से ऑफिस जाती है।
(4) का – शुभम हर्षित का भाई है।
(5) की – पूर्वी आयुष की बहन है।
(6) के – नीरज के तीन चाचा हैं।
(7) में – नीनू घर में है।
(8) पर – पेड़ पर बंदर कूद रहे हैं।
(9) हे – हे भगवान, कितना शोर है यहाँ।
(10) अरे – अरे! सलिल तुम कहाँ हो?
(11) के लिए – अंशु वारिजा के लिए फ्रॉक लाई।

Hindi Lokbharti 10th Textbook Solutions Chapter 6 गिरिधर नागर Additional Important Questions and Answers

पद्यांश क्र. 1

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

कृति 1: (आकलन)

प्रश्न 1.
आकृति पूर्ण कीजिए:
(i)

उत्तर:

(ii) a. श्रीकृष्ण के सिर पर है
b. मीरा इन्हें अपना पति मानती हैं –
उत्तर:
a. श्रीकृष्ण के सिर पर है – मोर मुकट
b. मीरा इन्हें अपना पति मानती हैं – गिरधर गोपाल

प्रश्न 2.
संजाल पूर्ण कीजिए:

उत्तर:

प्रश्न 3.
उचित जोड़ियाँ मिलाइए:

अ	आ
आँसुओं के जल से कुल की मर्यादा प्रेम बेल संत संगति के कारण	छोड़ दी आनंद फल लोक लाज खोई प्रेम की बेल सींची प्रेम से बिलोई।

उत्तर:

अ	आ
आँसुओं के जल से कुल की मर्यादा प्रेम बेल संत संगति के कारण	प्रेम की बेल सींचा। छोड़ दी। आनंद फल। लोक लाज खोई।

कृति 2: (शब्द संपदा)

प्रश्न 1.
निम्नलिखित शब्दों को शुद्ध रूप में लिखिए:
(i) भगत – …………………….
(ii) माखन – …………………….
(iii) आणंद – …………………….
(iv) जाके – …………………….
उत्तर:
(i) भगत – भक्त
(ii) माखन – मक्खन
(iii) आणंद – आनंद
(iv) जाके – जिसके।

प्रश्न 2.
निम्नालाखत शब्दा क समानाथा शब्दालाखए:
(i) मोर = …………………….
(ii) जगत = …………………….
(iii) दूध = …………………….
(iv) प्रेम = …………………….
उत्तर:
(i) मोर = मयूर
(ii) जगत = संसार
(iii) दूध = दुग्ध
(iv) प्रेम = प्यार।

प्रश्न 3.
निम्नलिखित अर्थवाले शब्द पद्यांश से चुनकर लिखिए:
(i) उद्घार करो – …………………….
(ii) मुझे – …………………….
उत्तर:
(i) उद्घार करो – तारो
(ii) मुझे – मोही।

कृति 3: (सरल अर्थ)

प्रश्न.
उपर्युक्त पद्यांश की अंतिम चार पंक्तियों का सरल अर्थ 25 से 30 शब्दों में लिखिए।
उत्तर:
मैंने दूध जमाने के पात्र में जमे दही को मथानी से बड़े प्रेम से बिलोया और उसमें से कृष्ण-प्रेम रूपी मक्खन को निकाल लिया। शेष छाछ रूपी निस्सार जगत को छोड़ दिया। कृष्ण-भक्तों को देखकर मैं प्रसन्न होती हूँ, परंतु संसार का व्यवहार देख मुझे दुख होता है और मैं रो पड़ती हूँ। हे गिरधरलाल, मीरा तो आपकी दासी है, उसे इस संसार रूप भव-सागर से पार लगाओ।

पद्यांश क्र. 2

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

कृति 1: (आकलन)

प्रश्न 1.
आकृति पूर्ण कीजिए:

(i) ये मीरा के प्रतिपालक हैं
(ii) कृष्ण के बिना इनको कहीं आश्रय नहीं है –
(iii) मीरा को प्रभु से मिलने की तीव्र यह है –
(iv) मीरा की यह संसार सागर में डूबने वाली है
उत्तर:
(i) ये मीरा के प्रतिपालक हैं – कृष्ण
(ii) कृष्ण के बिना इनको कहीं आश्रय नहीं है – मीरा
(iii) मीरा को प्रभु से मिलने की तीव्र यह है – लालसा
(iv) मीरा की यह संसार सागर में डूबने वाली है – नौका

प्रश्न 2.
पद्यांश में आए इस अर्थ के शब्द लिखिए:

(i) दासी
(ii) आश्रय
(iii) बार-बार
(iv) नौका।
उत्तर:
(i) दासी – चेरी
(ii) आश्रय – गती
(iii) बार-बार – बेर-बेर
(iv) नौका – बेरी।

प्रश्न 3.
विधानों के सामने सत्य / असत्य लिखिए:

(i) हरि बिना मीरा को कहीं आश्रय नहीं है।
(ii) कृष्ण मीरा के पति हैं और वे उनकी पत्नी।
(iii) मीरा बार-बार प्रभु की आरती करती हैं।
(iv) मीरा की जीवन नौका संसार सागर में डूबने वाली है।
उत्तर:
(i) सत्य
(i) असत्य
(iii) असत्य
(iv) सत्य।

प्रश्न 4.
आकृति पूर्ण कीजिए:

उत्तर:

कृति 2: (शब्द संपदा)

प्रश्न 1.
निम्नलिखित शब्दों के अर्थ लिखिए:

(i) कूण – …………………..
(ii) रावरी – …………………..
(iii) बेरी – …………………..
(iv) नेरी – …………………..
उत्तर:
(i) कूण – कहाँ
(ii) रावरी – आपकी
(iii) बेरी – बेड़ा
(iv) नेरी – पास, निकट।

प्रश्न 2.
निम्नलिखित शब्दों के लिंग पहचान कर लिखिए:
(i) पखावज …………………..
(ii) पिचकारी …………………..
(iii) गुलाल …………………..
(iv) चरण …………………..
उत्तर:
(i) पखावज – स्त्रीलिंग
(ii) पिचकारी – स्त्रीलिंग
(iii) गुलाल – पुल्लिंग
(iv) चरण – पुल्लिंग।

कृति 3: (सरल अर्थ)

प्रश्न.
उपयुक्त पद्यांश की ‘हरि बिन कूण आरति है तेरी।।’ पंक्तियों का सरल अर्थ 25 से 30 शब्दों में लिखिए।
उत्तर:
हे हरि, आपके बिना मेरा कौन है? अर्थात आपके सिवा मेरा कोई ठिकाना नहीं है। आप ही मेरा पालन करने वाले हैं और मैं आपकी दासी हूँ। मैं रात-दिन, हर समय आपका ही नाम जपती रहती हूँ। में बार-बार आपको पुकारती हूँ, क्योंकि मुझे आपके दर्शनों की तीव्र लालसा है।

पद्यांश क्र. 3

कृति 1: (आकलन)

प्रश्न 1.
पद्यांश में आए इस अर्थ के शब्द लिखिए:
बोर्ड की नमूना कृतिपत्रिका

(i) कमल
(ii) आकाश
(iii) श्रेष्ठ
(iv) संतोष।
उत्तर:
(i) कमल – कँवल
(ii) आकाश – अंबर
(iii) श्रेष्ठ – छतीयूँ
(iv) संतोष – संतोख।

प्रश्न 2.
जोड़ियाँ बनाइए:

उत्तर:
(i) सुर – राग
(ii) करताल – झनकार
(iii) घट – पट
(iv) प्रेम – पिचकार।

प्रश्न 3.
आकृति पूर्ण कीजिए:
(i) आकाश लाल हो गया है इससे –
(ii) गिरिधर नागर हैं मीरा के ये –
उत्तर:
(i) आकाश लाल हो गया है इससे – गुलाल से।
(ii) गिरिधर नागर हैं मीरा के ये – प्रभु।

प्रश्न 4.
उत्तर लिखिए: (बोर्ड की नमूना कृतिपत्रिका)

उत्तर:

कृति 2: (शब्द संपदा) (बोर्ड की नमूना कृतिपत्रिका)

पद्यांश से ढूँढ़कर लिखिए:
ऐसे दो शब्द जिनका वचन परिवर्तन से रूप नहीं बदलता –
(i) ……………………..
(ii) ……………………..
उत्तर:
(i) दिन
(ii) चरण।

कृति 3: (सरल अर्थ)

प्रश्न.
उपर्युक्त पद्यांश की प्रथम चार पंक्तियों का सरल अर्थ 25 से 30 शब्दों में लिखिए। (बोर्ड की नमूना कृतिपत्रिका)
उत्तर:
हे मेरे मन, फागुन मास में होली खेलने का समय अति अल्प होता है। अतः तू जी भरकर होली खेल। अर्थात मानव जीनव अस्थायी है, इसलिए भगवान कृष्ण से पूर्ण रूप से प्रेम कर ले। जिस प्रकार होली के उत्सव में नाच आदि का आयोजन होता है, उसी प्रकार कृष्ण-प्रेम में मुझे ऐसा प्रतीत होता है मानो करताल, पखावज आदि बाजे बज रहे हैं और अनहद नाद का स्वर सुनाई दे रहा है, जिससे मेरा हृदय बिना स्वर और राग के अनेक रागों का आलाप करता रहता है। मेरा रोम-रोम भगवान कृष्ण के प्रेम के रंग में डूबा रहता है। मैंने अपने प्रिय से होली खेलने के लिए शील और संतोष रूपी केसर का रंग घोला है। मेरा प्रिय-प्रेम ही होली खेलने की पिचकारी है।

भाषा अध्ययन (व्याकरण)

प्रश्न.
सूचनाओं के अनुसार कृतियाँ कीजिए:

1. शब्द भेद:
अधोरेखांकित शब्दों का शब्दभेद पहचानकर लिखिए:
(i) बाहर कोई नहीं है।
(ii) माँ को हंसी आ गई।
(iii) चतुर वैद्य विष से भी दवा का काम ले सकता है।
उत्तर:
(i) कोई – अनिश्चयवाचक सर्वनाम।
(ii) हँसी – भाववाचक संज्ञा।
(iii) चतुर – गुणवाचक विशेषण।

2. अव्यय:
निम्नलिखित अव्ययों का अपने वाक्यों में प्रयोग कीजिए:
(i) भी
(ii) इसलिए।
उत्तर:
(i) भी – हमारी बेटी गाय से भी अधिक सीधी है।
(ii) इसलिए – नीरज बीमार था, इसलिए दफ्तर नहीं गया।

3. संधि:
कृति पूर्ण कीजिए:
संधि शब्द – संधि विच्छेद – संधि भेद
…………. – उत् + लेख। – ……………
अथवा
तपोबल – …………… – ……………
उत्तर:
संधि शब्द – संधि विच्छेद – संधि भेद
उल्लेख – उत् + लेख – व्यंजन संधि
अथवा
तपोबल – तपः + बल – विसर्ग संधि

4. सहायक क्रिया:
निम्नलिखित वाक्यों में से सहायक क्रियाएँ पहचानकर उनका मूल रूप लिखिए:
(i) यह कुरसी दीवार की तरफ खिसका दो।
(ii) हम समय पर स्टेशन पहुंच गए।
उत्तर:
सहायक क्रिया – मूल रूप
(i) दो – देना
(ii) गए – जाना

5. प्रेरणार्थक क्रिया:
निम्नलिखित क्रियाओं के प्रथम प्रेरणार्थक और द्वितीय प्रेरणार्थक रूप लिखिए:
(i) पढ़ना
(ii) जीतना
(ii) करना।
उत्तर:
क्रिया – प्रथम प्रेरणार्थक रूप – द्वितीय प्रेरणार्थक रूप
(i) पढ़ना – पढ़ाना – पढ़वाना
(ii) जीतना – जिताना – जितवाना
(iii) करना – कराना – करवाना

6. मुहावरे:
(1) निम्नलिखित मुहावरों का अर्थ लिखकर वाक्य में प्रयोग कीजिए:
(i) चैन न मिल पाना
(ii) झेंप जाना।
उत्तर:
(i) चैन न मिल पाना।

अर्थ: बेचैनी अनुभव करना। वाक्य: मालिक की कड़ी बातें सुनकर मुनीम को चैन न मिल पाया।

(ii) झेंप जाना।
अर्थ: लज्जित होना, शरमाना। वाक्य: नकल करने पर मित्र की फटकार सुनकर अभिनव झेंप गया।

(2) अधोरेखांकित वाक्यांशों के लिए कोष्ठक में दिए गए उचित मुहावरे का चयन करके वाक्य फिर से लिखिए:
(नाक-भौं सिकोड़ना, गुदगुदा देना, सिर झुका देना)
(i) अप्रिय बात सुनकर सभी तिरस्कार प्रकट करेंगे।
(ii) जीवन में आनंददायी क्षण बहुत कम होते हैं।
उत्तर:
(i) अप्रिय बात सुनकर सभी नाक-भौं सिकोड़ेंगे।
(ii) जीवन में गुदगुदाने वाले क्षण बहुत कम होते हैं।

7. विरामचिह्न:
निम्नलिखित वाक्यों में यथास्थान उचित विरामचिह्नों का प्रयोग करके वाक्य फिर से लिखिए:
(i) मैं कहाँ से पैसे , पहले तो दूध की बिक्री के पैसे मेरे पास जमा रहते थे
(ii) सहसा कानों में आवाज आई काकी उठो मैं पूड़ियाँ लाई हूँ
उत्तर:
(i) “मैं कहाँ से पैसे दूँ? पहले तो दूध की बिक्री के पैसे मेरे पास जमा रहते थे।”
(ii) सहसा कानों में आवाज आई – “काकी, उठो मैं पूड़ियाँ लाई हूँ।”

8. काल परिवर्तन:
निम्नलिखित वाक्यों का सूचना के अनुसार काल परिवर्तन कीजिए:
(i) विदा का क्षण आ गया। (सामान्य भविष्यकाल)
(ii) आप छत पर क्या करते हैं? (अपूर्ण भूतकाल)
(iii) मेरी गाड़ी तो बिक जाती है। (पूर्ण वर्तमानकाल)
उत्तर:
(i) विदा का क्षण आ जाएगा।
(ii) आप छत पर क्या कर रहे थे?
(iii) मेरी गाड़ी तो बिक गई है।

9. वाक्य भेद:
(1) निम्नलिखित वाक्यों का रचना के आधार पर भेद पहचान कर लिखिए:
(i) संसार का व्यवहार देखकर मुझे दुःख होता है और मैं रो पड़ती हूँ।
(ii) मीराबाई कहती हैं कि अब उन्हें लोकलाज का कोई डर नहीं हैं।
उत्तर:
(i) संयुक्त वाक्य
(ii) मिश्र वाक्य।

(2) निम्नलिखित वाक्यों का अर्थ के आधार पर दी गई सूचना के अनुसार परिवर्तन कीजिए:
(i) संतो के साथ बैठकर मैंने लोकलाज त्याग दी है। (विस्मयादिबोधक वाक्य)
(ii) हे हरि, आप ही मेरा पालन करने वाले हैं। (आज्ञावाचक वाक्य)
उत्तर:
(i) हाय! संतों के साथ बैठकर मैंने लोकलाज त्याग दी है।
(ii) हे हरि, आप ही मेरा पालन करें।।

11. वाक्य शुद्धिकरण:
निम्नलिखित वाक्य शुद्ध करके लिखिए:
(i) लेखक ने अभिनेता की घमंड तोड़ी।
(ii) हमने क्रिकेट की मैच देखी।
उत्तर:
(i) लेखक ने अभिनेता का घमंड तोड़ा।
(ii) हमने क्रिकेट का मैच देखा।

गिरिधर नागर Summary in Hindi

गिरिधर नागर कविता का सरल अर्थ

1. मेरे तो गिरधर गोपाल …………………………….. तारो अब मोही।।

गिरि को धारण करने वाले, गायों के पालक कृष्ण के सिवा मेरा और कोई नहीं है। जिनके मस्तक पर मोर का मुकुट शोभित है, वे ही मेरे पति हैं। उनके लिए मैंने कुल की मर्यादा छोड़ दी है। चाहे कोई मुझे कुछ भी कहे। संतों के साथ बैठ-बैठकर मैंने लोकलाज त्याग दी है। मैंने अपने प्रेम रूपी बेल को अपने अश्रु रूपी जल से सींच-सींचकर बड़ा किया है। अब तो यह प्रेम-बेल फैल गई है और इसमें आनंद रूपी फल लगने लगा है। मैंने दूध जमाने के पात्र में जमे दही को मथानी से बड़े प्रेम से बिलोया और उसमें से कृष्ण-प्रेम रूपी मक्खन को निकाल लिया। शेष छाछ रूपी निस्सार जगत को छोड़ दिया। कृष्ण-भक्तों को देखकर मैं प्रसन्न होती हूँ, परंतु संसार का व्यवहार देख मुझे दुख होता है और मैं रो पड़ती हूँ। हे गिरधरलाल, मीरा तो आपकी दासी है, उसे इस संसार रूपी भव-सागर से पार लगाओ।

2. हरि बिन कूण गती मेरी …………………………….. मैं सरण हूँ तेरी।।

हे हरि, आपके बिना मेरा कौन है? अर्थात आपके सिवा मेरा कोई ठिकाना नहीं है। आप ही मेरा पालन करने वाले हैं और मैं आपकी दासी हूँ। मैं रात-दिन, हर समय आपका ही नाम जपती रहती हूँ। मैं बार-बार आपको पुकारती हूँ, क्योंकि मुझे आपके दर्शनों की तीव्र लालसा है। यह संसार विभिन्न प्रकार के दोषों और विकारों से भरा हुआ सागर है, जिसके बीच में घिर गई हैं। इस संसार रूपी सागर में मेरी नाव टूट गई है। हे प्रभु, आप शीघ्र इस नाव का पाल बाँधिए, अन्यथा यह जीवन-नौका इस संसार-सागर में डूब जाएगी। हे प्रियतम, आपकी यह विरहिणी निरंतर आपकी बाट जोहती रहती है। आपके आगमन की प्रतीक्षा करती रहती है। आपकी यह दासी मीरा सदा आपके नाम का स्मरण करती रहती है और आपकी शरण में आई है।

3. फागुन के दिन चार …………………………….. चरण कँवल बलिहार रे।।

हे मेरे मन, फागुन मास में होली खेलने का समय अति अल्प होता है। अतः तू जी भरकर होली खेल। अर्थात मानव जीनव अस्थायी है, : इसलिए भगवान कृष्ण से पूर्ण रूप से प्रेम कर ले। जिस प्रकार होली के : उत्सव में नाच आदि का आयोजन होता है, उसी प्रकार कृष्ण-प्रेम में मुझे ऐसा प्रतीत होता है मानो करताल, पखावज आदि बाजे बज रहे हैं और अनहद नाद का स्वर सुनाई दे रहा है, जिससे मेरा हृदय बिना स्वर और राग के अनेक रागों का आलाप करता रहता है। मेरा रोम-रोम भगवान कृष्ण के प्रेम के रंग में डूबा रहता है। मैंने अपने प्रिय से होली खेलने के लिए शील और संतोष रूपी केसर का रंग घोला है। मेरा प्रिय-प्रेम ही होली खेलने की पिचकारी है। उड़ते हुए गुलाल से सारा आकाश लाल हो गया है। अब मुझे लोक-लज्जा का कोई डर नहीं है, इसलिए मैंने हृदय रूपी घर के दरवाजे खोल दिए हैं। अंत में मीरा कहती हैं कि मेरे स्वामी गोवर्धन पर्वत को धारण करने वाले कृष्ण भगवान हैं। मैंने उनके चरण-कमलों में अपना सर्वस्व न्योछावर कर दिया है।

गिरिधर नागर विषय-प्रवेश :

प्रस्तुत काव्य में रसिक शिरोमणि श्रीकृष्ण की अनन्य उपासिका मीराबाई अपना प्रेम प्रकट कर रही हैं। सभी पदों में | श्रीकृष्ण के प्रति मीराबाई के प्रेम, उनकी भक्ति, उत्सुकता, प्रिय-मिलन की आशा, प्रतीक्षा आदि का मार्मिक चित्रण है।

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

Maharashtra State Board Class 10 Hindi Lokbharti Chapter 1 भारत महिमा

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

सूचना के अनुसार कृतियाँ कीजिए:

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

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

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

उत्तर:

b. भारतीय संस्कृति की दो विशेषताएँ:

उत्तर:

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

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

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

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

कृति 1: (आकलन)

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

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

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

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

कृति 2: (शब्द संपदा)

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

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

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

कृति 3: (सरल अर्थ)

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

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

पद्यांश क्र. 2

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

कृति 1: (आकलन)

प्रश्न 1.
आकृति पूर्ण कीजिए:

उत्तर:

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

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

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

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

कृति 2: (शब्द संपदा)

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

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

कृति 3: (सरल अर्थ)

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

पद्यांश क्र. 3

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

कृति 1: (आकलन)

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

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

प्रश्न 3.
संजाल पूर्ण कीजिए:

उत्तर:

प्रश्न 4.
आकृति पूर्ण कीजिए:

उत्तर:

कृति 2: (शब्द संपदा)

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

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

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

कृति 3: (सरल अर्थ)

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

भाषा अध्ययन (व्याकरण)

प्रश्न. सूचनाओं के अनुसार कृतियाँ कीजिए:

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

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

3. संधि:
कृति पूर्ण कीजिए:

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

उत्तर:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

भारत महिमा Summary in Hindi

भारत महिमा कविता का सरल अर्थ

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

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

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

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

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

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

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

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

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

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

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

भारत महिमा विषय-प्रवेश :

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

भारत महिमा मुहावरा –

• अर्थ निछावर करना – अर्पण करना, समर्पित करना।

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Class 9 Science Chapter 5 Acids, Bases and Salts Textbook Questions and Answers

1. Identify the odd one out and justify.

(a) Chloride, nitrate, hydride, ammonium
Answer:
Ammonium is the odd one out as it is a basic radical and rest all are acidic radicals. Generally, basic radicals are formed by the removal of electrons from the atom of metals such as Na⁺, Cu²⁺. But there are some exceptions, such as NH₄⁺.

(b) Hydrogen chloride, sodium hydroxide, calcium oxide, ammonia
Answer:
Hydrogen chloride is the odd one out. It is acidic and rest all are basic.

(c) Acetic acid, carbonic acid, hydrochloric acid, nitric acid
Answer:
Carbonic acid is the odd one out. It is a dibasic acid and rest are all monobasic acids.

(d) Ammonium chloride, sodium chloride, potassium nitrate, sodium sulphate
Answer:
Ammonium chloride is the odd one out, as it is made up of a strong acid and weak base and rest all are formed from strong acid and strong base.

(e) Sodium nitrate, sodium carbonate, sodium sulphate, sodium chloride
Answer:
Sodium carbonate is the odd one out, as it is made up of a weak acid and strong base, and rest all are formed from strong acid and strong base.

(f) Calcium oxide, magnesium oxide, zinc oxide, sodium oxide.
Answer:
Zinc oxide is the odd one out, as it is an amphoteric oxide, and rest all are basic oxides.

(g) Crystalline blue vitriol, crystalline common salt, crystalline ferrous sulphate, crystalline sodium carbonate.
Answer:
Crystalline common salt is the odd one out, as it does not contain water of crystallisation. It is an ionic compound and ionic compounds are crystalline in nature and rest all have their crystalline structure because of their water of crystallization.

(h) Sodium chloride, potassium hydroxide, acetic acid, sodium acetate.
Answer:
Acetic acid is the odd one out. It is an acid, the rest are all salts.

2. Write down the changes that will be seen in each instance and explain the reason behind it.

(a) 50ml water is added to 50ml solution of copper sulphate.

Answer:

• Copper sulphate solution is blue. It is a concentrated solution.
• When 50 ml of water is added to this concentrated solution, it becomes a diluted solution.
• The intensity of the blue colour is now different in this homogenous mixture.

(b) Two drops of the indicator phenolphthalein were added to 10ml solution of sodium hydroxide.

Answer:

• Sodiumhy droxide is a base and phenolphthalein is a synthetic indicator.
• Sodium hydroxide solution will turn pink if phenolphthalein is added to it.
• It is a test for identifying bases.

(c) Two or three filings of copper were added to 10ml dilute nitric acid and stirred.
Answer:
When copper metal reacts with dilute nitric acid, the metal does not displace hydrogen from the acid like reaction with other metals. Instead the reaction produces nitric oxide, (NO).

(d) A litmus paper was dropped into 2ml dilute HCl. Then 2ml concentrated NaOH was added to it and stirred.
Answer:
Blue litmus Paper:

• HCl is hydrochloric acid, so the blue litmus turns red.
• When equal amount of NaOH is added the colour again changes to blue and remains the same.

Red litmus paper:

• Red litmus paper shows no colour change in hydrochloric acid.
• When some amount of NaOH is added the colour changes to blue initially but when the amount of NaOH is sufficient the blue colour dissappears.
• Equal amounts of HC1 and NaOH results in the formation of NaCl, a salt, and the solvent water. This reaction is called the neutralization reaction.

(e) Magnesium oxide was added to dilute HCl and magnesium oxide was a added to dilute NaOH.
Answer:
(i) Magnesium oxide + dil HCl.
This is a neutralization reaction. Magnesium oxide is an insoluble base, it reacts with dilute HCl to produce a soluble salt MgCl₂ and water H₂O.
\(\mathrm{MgO}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{MgCl}_{2(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(n)}\)

(ii) Magnesium oxide + NaOH.
No chemical reaction takes place between magnesium oxide and sodium hydroxide.

(f) Zinc oxide was added to dilute HCl and zinc oxide was added to dilute NaOH.
Answer:

• Zinc oxide reacts with dilute hydrochloric acid to form zinc chloride and water. It is a neutralization reaction.
  \(\mathrm{ZnO}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{ZnCl}_{2(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
• Zinc oxide reacts with sodium hydroxide to form sodium zincate and water.
  \(\mathrm{ZnO}_{(\mathrm{s})}+2 \mathrm{NaOH}_{(\mathrm{aq})} \longrightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{t})}\)

(g) Dilute HCl was added to limestone.
Answer:

• When hydrochloric acid is added to limestone, carbon dioxide is liberated. Limestone is calcium carbonate.
  CaCO₃ + 2 HCl → CaCl₂ + CO₂ + H₂O
• Carbon dioxide is prepared in the laboratory using these chemicals.

(h) Pieces of blue vitriol were heated in a test tube. On cooling, water was added to it.
Answer:

• On heating, the crystalline structure of blue vitriol breaks down to form a colourless powder and water is released.
• This water is part of the crystal structure of blue vitriol.
• It is called water of crystallization.
• On adding water to the white powder, a solution was formed which has the same colour as the copper sulphate salt solution.

(i) Dilute H₂SO₄ was taken in an electrolytic cell and electric current was passed through it.
Answer:

• If pure water is used in the electrolytic cell, current does not flow even on putting on the switch.
• Pure water is a bad conductor of electricity. Dilute H₂SO₄ is acidulated water.
• The electrical conductivity of water increases on mixing with strong acid or base in it due to their dissociation and electrolysis of water takes place.
• H₂SO₄ is fully dissociated in aqueous solution. \(\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{H}^{+}+\mathrm{SO}_{4}^{2-}\)
• H₂O is a weak electrolyte and is only slightly dissociated \(\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}^{+}+\mathrm{OH}^{-}\)
• During electrolysis, the hydrogen ions migrate towards the cathode and are discharged there.

[H+ ions gains electrons and are converted to hydrogen gas]
\(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2(\mathrm{~g})}\)
Cathode reaction:
\(2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{i})}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2(\mathrm{~g})}+2 \mathrm{OH}_{(\mathrm{aq})}^{-}\)
Anode reaction:
\(2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \longrightarrow \mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{H}_{(\mathrm{aq})}^{+}+4 \mathrm{e}\)

• For every hydrogen ion discharged at the anode, another hydrogen ion is formed at the cathode.
• The net result is that the concentration of the sulphuric acid remains constant and electrolysis of water is overall reaction. \(2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{H}_{2}+\mathrm{O}_{2}\)
• The volume of the hydrogen gas formed near the cathode is double that of the oxygen gas formed near the anode.

3. Classify the following oxides into three types and name the types.
CaO, MgO, CO₂, SO₃, Na₂O, ZnO, Al₂O₃, Fe₂O₃
Answer:
There are three types of oxides : Basic oxides, Acidic oxides and Amphoteric oxides.

Basic oxides	Acidic oxides	Amphoteric oxides
CaO	CO2	ZnO
MgO	SO3	Al2O3
Na2O
Fe2O3

Generally metal oxides are basic in nature.
Exception : Al₂O₃ and ZnO are amphoteric oxides.
Generally non-metal oxides are acidic in nature.

4. Explain by drawing a figure of the electronic configuration.

a. Formation of sodium chloride from sodium and chlorine.
Answer:

b. Formation of a magnesium chloride from magnesium and chlorine.
Answer:

5. Show the dissociation of the following compounds on dissolving in water, with the help of chemical equation and write whether the proportion of dissociation is small or large.
Hydrochloric acid, Sodium chloride, Potassium hydroxide, Ammonia, Acetic acid, Magnesium chloride, Copper sulphate.
Answer:
(a) Hydrochloric acid (HCl)

• \(\mathrm{HCl}_{(g)} \frac{\text { Water }}{\text { dissociation }}>\mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)
• Hydrochloric acid is a strong acid, as on dissolving in water, it dissociates almost completely and the resulting aqueous solution contains mainly H ions and the concerned acidic radical.
• The proportion of dissociation is large.

(b) Sodium chloride (NaC1)

• \(\mathrm{NaCl}_{(s)} \frac{\text { Water }}{\text { dissociation }}>\mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{\text {(aq) }}^{-}\)
• When an ionic compound begins to dissolve in water, the water molecules push themselves in between the positive and negative ions of the compound and separate them from each other.
• The proportion of dissociation is large.

(c) Potassium hydroxide (KOH)

• \(\mathrm{KOH}_{(n)} \frac{\text { Water }}{\text { dissociation }} \mathrm{K}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}\)
• Potassium hydroxide is a strong base, as on dissolving in water, it dissociates almost completely and the resulting aqueous solution contains mainly OH⁺ ions and the concerned basic radical.
• The proportion of dissociation is large.

(d) Ammonia (NH₃)
(i) \(\mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{n}}-\frac{\text { Water }}{\text { uissociation }}>\mathrm{NH}_{4 \text { (aq) }}^{+}+\mathrm{OH}_{\text {(aq) }}^{-}\)
(ii) Ammonia dissolves in water to form NH₄OH (ammonium hydroxide). NH₄OH does not dissociate completely as it is a weak base. The aqueous solution contains a small proportion of OH^– ions and the concerned basic radical along with a large proportion of undissociated molecules of the base i.e. NH₄OH.
(iii) The proportion of dissociation is small.

(e) Acetic acid (CH₃COOH)

• \(\mathrm{CH}_{3} \mathrm{COOH}_{(l)} \frac{\text { Water }}{\text { dissociation }} \mathrm{CH}_{3} \mathrm{COO}_{(\text {aq) }}+\mathrm{H}_{(a z)}+{ }^{+}\)
• Acetic acid is a weak acid, on dissolving in water it does not dissociate completely, and the resulting aqueous solution contains H⁺ ion and the concerned acidic radical in small proportion along with large proportion of the undissociated molecules of the acid.
• The proportion of dissociation is small.

(f) Magnesium chloride (MgCl₂)

• \(\mathrm{MgCl}_{2(\mathrm{~s})} \frac{\text { Water }}{\text { dissociation }}>\mathrm{Mg}^{2+} \text { (aq) }+2 \mathrm{Cl}^{-} \text {(aq) }\)
• Magnesium chloride dissolves in water and forms magnesium ions and chloride ions. When an ionic compound begins to dissolve in water, the water molecules push themselves in between the ions of the compound and separate them from each other.
• The proportion of dissociation is large.

(g) Copper sulphate (CuSO₄)

• \(\mathrm{CuSO}_{4(s)} \frac{\text { Water }}{\text { dissociation }}>\mathrm{Cu}_{(\mathrm{aq})}^{2+}+\mathrm{SO}_{4}^{2-}\)
• When Copper sulphate is dissolved in water it forms copper ions and sulphate ions. When an ionic compound begins to dissolve in water, water molecules push themselves in between the ions of the compound and separate them from each other.
• The proportion of dissociation is large.

6. Write down the concentration of each of the following solutions in g/L and mol/L.

a. 7.3g HCl in 100ml solution
b. 2g NaOH in 50ml solution
c. 3g CH₃COOH in 100ml solution
d. 4.9g H₂SO₄ in 200ml solution
Answer:
To find : The concentration in g/L.
Solution:

7. Obtain a sample of rainwater. Add to it a few drops of universal indicator. Meausre its pH. Describe the nature of the sample of rainwater and explain the effect if it has on the living world.
Answer:

• pH of rain water is 6.5 that means rain water is slightly acidic.
• When we add universal indicator to rain water it turns orangish red, indicating pH value is between 0 to 7, which tells us that rain water is acidic in nature.
• Most of the plants grow best when pH of soil is close to 7. If the soil is too acidic or too basic, it affects plant growth.

8. Answer the following questions.

a. Classify the acids according to their basicity and give one example of each type.
Answer:

• Basicity of acids : The number of H⁺ ions obtainable by the dissociation of one molecule of an acid is called its basicity. The acids are classified as monobasic, dibasic and tribasic acids based on the number of H⁺ ions present.
• Examples of monobasic acid : HCl, HNO₃, CH₃COOH
• Examples of dibasic acid: H₂SO₄, H₂CO₃
• Examples of tribasic acid: H₃BO₃, H₃PO₄

b. What is meant by neutralization? Give two examples from everyday life of the neutralization reaction.
Answer:

• In neutralization reaction, an acid reacts with a base to form salt and water.
• In a neutralisation reaction the acid dissociates to form H⁺ ions and base dissociates to form OH ions.
• They combine to form H₂O molecules which mixes with the solvent.

Examples in daily life:

• When people suffer from acidity, they take some antacids to neutralise the acid in their stomach.
• If an ant stings us the pain is due to formic acid. It is neutralised by rubbing moist baking soda which is basic in nature.

c. Explain what is meant by electrolysis of water. Write the electrode reactions and explain them.
Answer:
Electrolysis of water:

• Electrolysis of water is the decomposition of water into oxygen and hydrogen gas due to an electric current being passed through acidified water.
• Cathode reaction:
  \(2 \mathrm{H}_{2} \mathrm{O}_{(l)}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2(\mathrm{~g})}+2 \mathrm{OH}_{(\mathrm{aq}}\)
• Anode reaction:
  \(2 \mathrm{H}_{2} \mathrm{O}_{(l)} \longrightarrow \mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{H}_{(\mathrm{aq})}^{+}+4 \mathrm{e}^{-}\)
• It is found that the volume of gas formed near the cathode is double that of the gas formed near the anode.
• Hydrogen gas is formed near the cathode and oxygen gas near the anode.
• From this, it is clear that electrolysis of water has taken place and its constituent element have been released.

9. Give a reason for the following.

a. Hydronium ions are always in the form H₃O⁺.
Answer:

• Acids in water gives H⁺ ions. These H⁺ ions do not exist freely in water.
• This is because H⁺ is a single proton, a hydrogen atom has only one proton and one electron.
• If the electron is removed to make H⁺, all that is left is an extremely tiny positively charged nucleus.
• This H ion will immediately combine with the surrounding water (H₂O) molecules to form (H₃O₄) hydronium ion.

b. Buttermilk spoils if kept in a copper or brass container.
Answer:

• Buttermilk contains an organic acid called as lactic acid.
• The lactic acid reacts with copper and brass and forms toxic compounds which are not fit for consumption.
• They are harmful and may cause food poisoning.
• So it is not advisable to keep buttermilk in brass or copper containers.

10. Write the chemical equations for the following activities.

(a) NaOH solution was added to HCl solution.
Answer:
When NaOH reacts with HCl, it gives NaCl and water.

(b) Zinc dust was added to dilute H₂SO₄.
Answer:
When zinc reacts with dilute sulphuric acid, it forms zinc sulphate and hydrogen gas is liberted.

(c) Dilute nitric acid was added to calcium oxide.
Answer:
When dilute nitric acid reacts with calcium oxide, it forms calcium nitrate and water.

(e) Carbon dioxide gas was passed through KOH solution.

(f) Dilute HCl was poured on baking soda.

11. State the differences.

a. Acids and bases
Answer:

Acids	Bases
(i) A substance which liberates H+ ions when dissolved in water is an acid (ii) Blue litmus turns red in an acid. (iii) The pH of an acid is less than 7. (iv) Acids are sour to taste (v) e.g. HCl, H2SO4	A substance which liberates OH– ions when dissolved in water is called a base. Red litmus turns blue in a base The pH of a base is greater than 7. Bases are bitter to taste, e.g. NaOH, KOH.

b. Cation and anion
Answer:

Cations	Anions
(i) Cations are ions with a net positive charge.	Anions are ions with a net negative charge.
(ii) Cations are generally formed by metals. When metals donate electrons, they have excess of protons, hence they form cations.	Anions are generally formed by non-metals. When non-metals accept electrons, they have excess of electrons, hence they form anions.
(iii) Cations are attracted towards the cathode which are negatively charged electrodes.	Anions are attracted towards the anode which are positively charged electrodes.
(iv) e.g.: Na+, Ca2+, Mg2+, K+ etc.	e.g.: O2 , S2-, Cl–, Br– etc.

c. the Negative electrode and the positive electrode.

Answer:

Negative Electrode	Positive Electrode
(i) Negatively charged electrodes are called as a cathode.	Positively charged electrodes are called as Anode.
(ii) Positively charged cations move towards the cathode or negative electrode.	Negatively charged anions move towards the anode or positive electrode.
(iii) Cathode accepts electrons from cations	Anode gives electrons to anions

12. Classify aqueous solutions of the following substances according to their pH into three groups : 7, more than 7, less than 7.

Common salt, sodium acetate, hydrochloric acid, carbon dioxide, potassium bromide, calcium hydroxide, ammonium chloride, vinegar, sodium carbonate, ammonia, sulphur dioxide.
Answer:

pH = 7	pH > 7	pH < 7
(a) common salt.	sodium acetate.	sulphur dioxide.
(b) potassium bromide.	sodium carbonate	hydrochloric acid.
(c)	ammonia.	carbon-dioxide.
(d)	calcium hydroxide.	ammonium chloride.
(e)		vinegar

Class 9 Science Chapter 5 Acids, Bases and Salts Intext Questions and Answers

Question 1.
How are the following substances classified into three groups with the help of litmus?
Lemon, tamarind, baking soda, buttermilk, vinegar, orange, milk, lime, tomato, milk of magnesia, water, alum.
Answer:

Basic Substance: Turns Red Litmus Blue	Acidic Substance: Turns Blue Litmus Red	Neutral Substance: No change in the colour of litmus
Baking Soda	Lemon	water
Lime	Tamarind
Milk of Magnesia	Buttermilk
	Vinegar
	Orange
	Milk
	Tomato
	Alum

Question 2.
Fill in the columns in the part of the following table:
Answer:

Question 3.
Complete the following table of the concentration of various aqueous solutions.
Answer:

Question 4.
Complete the following table of neutralization reactions and also write down the names of the acids, bases and salts in it.
Answer:

Question 5.
What are the names of the following compounds? NH₃, Na₂O, CaO.
Answer:
NH₃ : Ammonia
Na₂O : Sodium oxide
CaO : Calcium oxide

Question 6.
Into which type will you classify the above compounds, acid, base or salt?
Answer:
NH₃ : base
Na₂O : base
CaO : base

Question 7.
ive examples of monobasic, dibasic and tribasic acids.
Answer:
Monobasic acid examples: HNO₃, HCl, CH₃COOH
Dibasic acid examples: H₂SO₄, H₂CO₃
Tribasic acid examples: H₃BO₃, H₃PO₄

Question 8.
Give the three types of bases with their examples.
Answer:
Types of bases:
Monoacidic base examples : NaOH, KOH, NH₄OH
Diacidic base examples: Ca(OH)₂, Ba(OH)₂
Triacidic base examples: A(OH)₃, Fe(OH)₃

Question 9.
What are the colours of the following natural and synthetic indicators in acidic and basic solutions? Litmus, turmeric, jamun, methyl orange, phenolphthalein?
Answer:

Indicator	Colour in Acidic Solution	Colour in basic Solution
Litmus	Red	Blue
Turmeric	Yellow	Red
Jamun	Red	Blue-Green
Methyl orange	Red	Yellow
Phenolphthalein	Colourless	Pink

Question 10.
On mixing the substances as shown here, what are the resulting mixtures formed by mixing the following substances called?
(a) Water and salt
(b) Water and sugar
(c) Water and sand
(d) Water and sawdust
Answer:
(a) Water and salt – Homogeneous mixture
(b) Water and sugar – Homogeneous mixture
(c) Water and sand – Heterogeneous mixture
(d) Water and sawdust – Heterogeneous mixture

Question 11.
What would be the definition of an acid and a base with reference to the neutralization reaction?
Answer:

• Acid: An acid is a substance which neutralises a base to form salt and water.
• Base: A base is a substance which neutralises an acid to form salt and water.

Answer the following questions:

Question 1.
Take aqueous solution of sodium chloride, copper sulphate, glucose, urea, dil.H₂SO₄ and dil.NaOH in a beaker and test the electrical conductivity of the solutions. Answer the given below questions.

(a) With which solutions did the bulb glow?
Answer:
Solutions with which bulb glows: Aqueous solution of NaCl, CuSO₄, H₂SO₄ and NaOH.

(b) Which solutions are electrical conductors?
Answer:
Solutions which are electrical conductors:
NaCl, CuSO₄, H₂SO₄ and NaOH.

Question 2.
During electrolysis of copper sulphate, if electric current is passed through the electrolytic cell for a long time, what change would be seen at the anode?
Answer:

• When electricity is passed for a long time through copper sulphate solution, the following reaction is seen at the anode: Anode Reaction: Cu(s) → Cu²⁺(aq) + 2e^–
• All the copper atoms will get converted into copper ions and get deposited on the cathode. This process continues till the copper anode exists.

Question 3.
Would water be a good conductor of electricity?
Answer:
Pure water is not a good conductor of electricity.

Answer the following questions:

Question 1.
(i) With which solutions did the bulb glow?
(ii) Which solutions are electrical conductors?

Answer:
(i) The bulbs glows when the wire are immersed in NaCl solution.
(ii)

Solution	Results
1g Copper Sulphate Solution	Bulb glow
1g Glucose Solution	Bulb does not glow
1g Urea Solution	Bulb does not glow
5ml dil.H2SO4 Solution	Bulb glows
5ml dil. NaOH Solution	Bulb glows

Question 2.
What would be the definition of an acid and a base with reference to the neutralization reaction?
Answer:

• Acid: An acid is a substance which neutralises a base to form salt and water.
• Base: A base is a substance which neutralises an acid to form salt and water.

Question 3.
Dissolve 2g salt in 500ml pure water. Take 250 ml of this solution in a 500ml capacity beaker. Connect two electrical wires to the positive and negative terminals of a power supply. Remove the insulating cladding from about 2cm portions at the other ends of the wires. These are the two electrodes. Fill two test tubes upto the brim with the prepared dilute salt solution. Invert them on the electrodes without allowing any air to enter. Start the electric current under 6V potential difference from the power supply. Observe what happens in the test tubes after some time.
(a) Did you see the gas bubbles forming near the electrodes in the test tubes?
(b) Are these gases heavier or lighter than water?
(c) Are the volumes of the gases collected over the solution in the two test tubes the same or different?
(d) Test the solutions in the two test tubes with litmus paper, what do you see?
(e) Repeat the activity by using dilute H₂SO₄ as well as dilute NaOH as the electrolyte.
Answer:
(a) Yes, gas bubbles can be seen forming near the electrodes in the test tube.
(b) These gases are lighter than water.

(c) The volumes of gases collected over the solution in the two test tubes are different.
(d) The solution in the cathode turns red litmus blue while solution in anode turns blue litmus red.
(e) When the above experiment is repeated with dil H₂SO₄ and dil NaOH, Hydrogen gas is liberated at cathodes and oxygen gas is liberated at anode.

Answer the following questions:

Question 1.
Cut a lemon into two equal parts. Take the juice of each part into two separate beakers. Pour 10 ml of drinking water in one beaker and 20 ml in the second. Stir the solutions in both the beakers and taste them. Is there any difference in the tastes of the solutions in the two beakers? What is that difference?
Answer:

• In the above activity, the sour taste of the solutions is because of the solute, lemon juice in them.
• The quantity of the lemon juice is the same in both the solutions. Yet their tastes are different.
• The solution in the first beaker is more sour than the one in the second.
• Although the quantity of the solute is the same in both the solutions, the quantity of the solvent is different.
• Ratio of the quantity of the solute to the quantity of the resulting solution is different. This ratio is larger for the solution in the first beaker and therefore that solution tastes more sour.
• On the other hand, the proportion of the lemon juice in the total solution in the second beaker is smaller and taste is less sour.
• The taste of foodstuff depends upon the nature of the taste-giving ingredient and its proportions in the foodstuff.
• Similarly, all the properties of a solution depend on the nature of the solute and solvent and also on the proportion of the solute in a solution
• The proportions of a solute in a solution is called the concentration of the solute in the solution.

Question 2.
Take a big test tube. Choose a rubber stopper in which a gas tube can be fitted. Take a few pieces of magnesium ribbon in the test tube and add some dilute HCl to it. Take a lighted candle near the end of the gas tube and observe. What did you observe?

Answer:

• Magnesium metal reacts with dilute HCl and an inflammable gas, hydrogen, is formed.
• During this reaction, the reactive metal displaces hydrogen from the acid to release hydrogen gas.
• At the same time, the metal is converted into basic radical which combines with the acidic radical from the acid to form the salt.
  \(\mathrm{Mg}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq}) \longrightarrow} \underset{\rightarrow}{\mathrm{MgCl}_{2(\mathrm{aq})}}+\mathrm{H}_{2(\mathrm{~g})}\)

Question 3.
Take some water in a test tube and add a little red oxide (the primer used before painting iron articles) to it. Now add a small quantity of dilute HCl to it, shake the test tube and observe.
(a) Does the red oxide dissolve in water?
(b) What change takes place in the particles of red oxide on adding dilute HCl?
Answer:
(a) The chemical formula of red oxide is Fe₂O₃. It is insoluble in water.
(b) The water-insoluble red oxide reacts with HCl to produce a water-soluble salt FeCl₃.

• This gives a yellowish colour to the water.
• The following chemical equation can be written for this chemical change.
  \(\mathrm{Fe}_{2} \mathrm{O}_{3(\mathrm{~s})}+6 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow 2 \mathrm{FeCl}_{3(\mathrm{aq})}+3 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)

Question 4.
Fit a bent tube in a rubber cork. Take some lime water in a test tube and keep it handy. Take some baking soda in another test tube and add some lemon juice to it. Immediately fit the bent tube over it. Insert its other end into the lime water. Note down your observations of both the test tubes. Repeat the procedure using washing soda, vinegar and dilute HC1 properly. What do you see?
Answer:

• In this activity, when limewater comes in contact with the gas released in the form of an effervescence, it turns milky.
• This is a chemical test for carbon dioxide gas.
• When lime water turns milky, we infer that the effervescence is of CO₂.
• This gas is produced on the reaction of acids with carbonate and bicarbonate salts of metals.
• A precipitate of CaCO₃ is produced on its reaction with the lime water Ca(OH)₂.
• This reaction can be represented by the following chemical equation.
  \(\mathrm{Ca}(\mathrm{OH})_{2(\mathrm{aq})}+\mathrm{CO}_{2(\mathrm{~g})} \longrightarrow \mathrm{CaCO}_{3(\mathrm{~s})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
• Washing soda is sodium carbonate Na2CO₃. It will react same as baking soda (NaHCO₃).
• Vinegar and HCl are acids, they do not react chemically with lemon juice.

Class 9 Science Chapter 5 Acids, Bases and Salts Additional Important Questions and Answers

Select the correct option

Question 1.
…………………. acid is present in lemon.
(a) malic acid
(b) tartaric acid
(c) citric acid
(d) butyric acid
Answer:
(c) Citric acid

Question 2.
Tamarind contains …………………. acid.
(a) Lactic acid
(b) tartaric acid
(c) matlic acid
(d) butyric acid
Answer:
(b) tartaric

Question 3.
Butter milk contains …………………. acid.
(a) butyric acid
(b) Lactic acid
(c) matlic acid
(d) citric acid
Answer:
(a) Butyricacid

Question 4.
If the basic radical is H+ the type of compound is ………………… .
(a) neutral
(b) base
(c) acid
(d) alkali
Answer:
(c) Acid

Question 5.
The name of compound NH3 is ………………… .
(a) nitric acid
(b) ammonium
(c) nitride
(d) ammonia
Answer:
(b) ammonia

Question 6.
The bases which are soluble in water are called as ………………… .
(a) indicators
(b) acids
(c) alkalis
(d) salts
Answer:
(c) alkalis

Question 7.
H₃PO₄ is a …………………. acid.
(a) monobasic
(b) tribasic
(c) tetrabasic
(d) dibasic
Answer:
(b) tribasic

Question 8.
According to pH scale pure water has a pH of ………………… .
(a) 6
(b) 7
(c) 5
(d) 8
Answer:
(b) 7

Question 9.
With reference to neutralization, metallic oxides are ………………… in nature.
(a) basic
(b) neutral
(c) acidic
(d) saline
Answer:
(a) basic

Question 10.
Molecular formula of blue vitriol is ………………… .
(a) CuSO₃ 5H₂O
(b) CuSO₄ 4H₂O
(c) CUSO₃ 4H₂O
(d) CUSO₄ 5H₂O
Answer:
(d) CUSO₄ 5H₂O

Question 11.
Molecular formula of crystalline alum is ………………… .
(a) KSO₄, AISO₄, 24H₂O
(b) K₂SO₄, ALSO₄, 24H₂O
(c) K₂SO₄, AL₂(SO₄)₃, 24H₂O
(d) KSO₄, Al₂(SO₄)₃, 24H₂O
Answer:
(c) K₂SO₄, AL₂(SO₄)₃, 24H₂O

Question 12.
Molecular formula for sodium oxide is ………………… .
(a) Na₂O
(b) NaO₂
(c) NaO
(d) Na₂O₂
Answer:
(a) Na₂O

Question 13.
H₂CO₃ is …………………. acid.
(a) monobasic
(b) dibasic
(c) tribasic
(d) tetrabasic
Answer:
(b) dibasic

Question 14.
Molecular formula of Red oxide is ………………… .
(a) Fe₂O₃
(b) CuO
(c) Fe₃O₄
(d) Na₂O
Answer:
(a) Fe₂O₃

Question 15.
The positive terminal electrode is called as ………………… .
(a) anode
(b) cathode
(c) anion
(d) cation
Answer:
(a) anode

Question 16.
…………………. produced in stomach helps in digestion.
(a) Hydrochloric acid
(b) Oxalic acid
(c) Sulphuric acid
(d) Nitric acid?
Answer:
(a) Hydrochloric acid

Question 17.
The solution turns blue litmus red, its pH is likely to be ………………… .
(a) 7
(b) 4
(c) 14
(d) 9
Answer:
(b) 4

Question 18.
An ionic compound NaCl is formed by ………………… .
(a) Na⁺ and Cl^–
(b) Na⁺ and Cl⁺
(c) Na^– and Cl^–
(d) Na^– and Cl⁺
Answer:
(a) Na⁺ and Cl^–

Question 19.
pH of strong acid is ………………… .
(a) 0
(b) 7
(c) 8
(d) 14
Answer:
(a) 0

Question 20.
HCl + NaOH → NaCl + H₂O is a …………………. reaction.
(a) neutralization
(b) crystallisation
(c) electrolysis
(d) dissociation
Answer:
(a) neutralization

Question 21.
Adding water to acid is an …………………. reaction.
(a) endothermic
(b) exothermic
(c) neutralisation
(d) crystallisation
Answer:
(b) exothermic

Find the odd one out:

(a) Rose Petal, Turmeric, Phenolphthalein, Indigo.
Answer:
Phenolphthalein is odd one out as rest are natural indicators while phenolphthalein is a synthetic indicator.

(b) Lime water, Vinegar, Acetic acid, Tartaric acid.
Answer:
Lime water is odd one out as this is basic in nature while rest are acidic.

(c) NaHCO₃, HCl, H₂SO₄, HNO₃
Answer:
NaHCO₃ is odd one out as it is base while rest are acids.

(d) Oxalic acid, Nitric acid, Citric acid, acetic acid.
Answer:
Nitric acid is odd one out as others are weak acids while Nitric acid is a strong acid.

(e) Crystalline, Liquid, Gases, Solid.
Answer:
Crystalline is odd one out as this is nature of a compound while others are physical states of compounds.

(f) Ca(OH)₂, Mg (OH)₂, NaOH, NH₄OH.
Answer:
NaOH is odd one out even though all are bases but NaOH is highly soluble in water compared to others.

Name the following:

Question 1.
Name the three types of ionic compounds.
Answer:
The three types of ionic compounds are acids, bases and salts.

Question 2.
Name the two constituents of molecule of an ionic compound.
Answer:

• Cation (positive ion/ basic radical)
• Anion (negative ion/acidic radical).

Question 3.
Name any three acids with their molecular formula.
Answer:

• Hydrochloric acid – HC1
• Sulphuric acid – H₂SO₄
• Nitric acid – HNO₃

Question 4.
Name any three bases with their molecular formula.
Answer:

• Sodium hydroxide – NaOH
• Potassium hydroxide – KOH
• Calcium hydroxide – Ca(OH)₂

Question 5.
Name any three salts with their molecular formula.
Answer:

• Sodium chloride – NaCl
• Potassium sulphate – K₂ SO₄
• Calcium chloride – CaCl₂

Question 6.
Name any two strong acids.
Answer:

• Hydrochloric acid – HC1
• Sulphuric acid – H₂ SO₄

Question 7.
Name any two weak acids.
Answer:

• Acetic acid – CH₃ COOH
• Carbonic acid – H₂ CO₃

Question 8.
Name any two strong bases.
Answer:

• Sodium hydroxide – NaOH
• Potassium hydroxide – KOH

Question 9.
Name a weak base.
Answer:
Ammonium hydroxide – NH₄OH

Question 10.
Name any two alkalis.
Answer:

• Sodium hydroxide – NaOH
• Potassium hydroxide – KOH

Question 11.
Name any two acids with their basicity 1 (monobasic)
Answer:

• Hydrochloric acid – HC1
• Nitric acid – HNO₃

Question 12.
Name any two acids with their basicity 2 (dibasic)
Answer:

• Sulphuric acid – H₂ SO₄
• Carbonic acid – H₂CO₃

Question 13.
Name any two acids with their basicity 3 (tribasic)
Answer:

• Boric acid – H₃BO₃
• Phosphoric acid – H₃PO₄

Question 14.
Name any two bases with their acidity 1 (monoacidic)
Answer:

• Sodium hydroxide – NaOH
• Potassium hydroxide – KOH

Question 15.
Name any two bases with their acidity 2 (diacidic)
Answer:

• Calcium hydroxide – Ca(OH)₂
• Barium hydroxide – Ba(OH)₂

Question 16.
Name any two bases with their acidity 3 (triacidic)
Answer:

• Aluminium hydroxide – Al(OH)₃
• Ferric hydroxide – Fe(OH)₃

Question 17.
Name the two units to express the concentration of the solution.
Answer:

• mass of solute in grams dissolved in one litre of the solution grams per litre, (g/L).
• Number of moles of the solute dissolved in one litre of the solution. Molarity (M) of the solution.

Match the columns:

Question 1.

Column ‘A’	Column ‘B’
(1) HNO3 (2) H3PO4 (3) CH3COOH (4) H2CO3	(a) Acetic acid (b) Carbonic acid (c) Phosphoric acid (d) Nitric acid

Answer:
(1-d),
(2- c),
(3 – a),
(4 – b)

Question 2.

Column ‘A’	Column ‘B’
(1) NH4OH (2) Ca(OH)2 (3) Al(OH)3 (4) Ba(OH)2	(a) Aluminium Hydroxide (b) Barium Hydroxide (c) Calcium Hydroxide (d) Ammonium Hydroxide

Answer:
(1-d),
(2- c),
(3 – a),
(4 – b)

Question 3.

Column ‘A’ Solution		Column ‘B’ pH
(1)	Milk	(a)	0
(2)	Milk of Magnesia	(b)	14
(3)	1 M HCl	(c)	10.5
(4)	1 M NaOH	(d)	6.5

Answer:
(1-d),
(2- c),
(3 – a),
(4 – b)

Question 4.

Column ‘A’	Column’B’
(1) Urine (2) Apples (3) Orange (4) Butter	(a) Butyric acid (b) Uric acid (c) Malic acid (d) Citric acid

Answer:
(1 – b),
(2 – c),
(3 – d),
(4 – a)

Question 5.

Column ‘A’	Column ‘B’
(1) Crystalline blue vitriol (2) Crystalline green vitriol (3) Crystalline (4) washing soda Crystalline alum	(a) FeS04-7H20 (b) K2S04-A12(S04)3.24H20 (c) CuS04-5H20 (d) Na2CO310H2O

Answer:
(1-c),
(2- a),
(3 – d),
(4 – b)

State whether the following statements are true or false and if false. Correct the false statement:

(1) The separation of H and CI in HCI is in absence of water.
(2) \(\mathrm{NaOH}_{(\mathrm{s})} \frac{\text { Water }}{\text { dissociation }}>\mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
(3) H₂SO₄ is a strong acid.
(4) NaC1 is an ionic compound.
(5) Turmeric is synthetic indicator.
(6) Metal + Dilute acid forms salt and water.
(7) Copper oxide is called red primer.
(8) Oxide of non-metal + Acid → Salt + Water.
(9) Zinc oxide reacts with sodium hydroxide to form sodium zincate.
(10) Al₂O₃ is an amphoteric oxide.
(11) Blue vitriol ZnSO₄. 7H₂O.
(12) Molecular formula for crystalline Ferrous sulphate is Fe5O₄. 5H₂O.
(13) NaCl in water does not conduct electricity.
(14) Phenolphthalein is colourless in base.
Answer:
(1) False. The separation of H⁺ and Cl^– in HC1 is in presence of water.
(2) True
(3) True
(4) True
(5) False. Turmeric is a natural indicator.
(6) False. Metal + Dilute acid forms salt and hydrogen gas.
(7) False. Iron oxide is called red primer.
(8) False. Oxide of izan-metal + Base → Salt + Wafer.
(9) True
(10) True
(11) False. Blue Vitriol is CuSO₄. 5H₂O.
(12) False. Molecular formula for crystalline Ferrous sulphate is FeSO₄. 7H₂O.
(13) False. NaCl in water conducts electricity.
(14) False, Phenolphthalein is colourless in acid and pink in base.

Define the following:

Question 1.
Acid
Answer:
An acid is a substance which on dissolving in water gives rise to H ion as the only cation. For example, HC1, H₂ SO₄, H₂CO₃.

Question 2.
Base
Answer:
A base is a substance which on dissolving in water gives rise to the OW ion as the only anion. For, NaOH, Ca(OH)₂

Question 3.
Strong Acid
Answer:
On dissolving in water, a strong acid dissociates almost completely, and the resulting aqueous solution contains mainly H ions and the concerned acidic radical. e.g. HC1, HBr, HNO₃, H₂SO₄.

Question 4.
Weak Acid
Answer:
On dissolving in water a weak acid does not dissociate completely, and the resulting aqueous solution contains H+ ion and the concerned acidic radical in small proportion along with large proportion of the undissociated molecules of the acid, e.g., H₂CO₃ (Carbonic acid), CH₃COOH (Acetic acid)

Question 5.
Strong Base
Answer:
On dissolving in water, a strong base dissociates almost completely and the resulting aqueous solution contains mainly OH ions and the concerned basic radicals, e.g. NaOH, KOH, Ca(OH)₂, Na₂O.

Question 6.
Weak Base
Answer:
On dissolving in water, a weak base does not dissociate completely and the resulting aqueous solution contains a small proportion of OH ions and the concerned basic radical along with a large proportion of undissociated molecules of the base. e.g. NH₄ OH.

Question 7.
Alkali
Answer:
The bases which are highly soluble in water are called alkali, e.g. NaOH, KOH, NH3. Here, NaOH and KOH are strong alkalis while NH₃ is a weak alkali.

Question 8.
Basicity of acids
Answer:
The number of H⁺ ions obtainable by the dissociation of one molecule of an acid is called its basicity.

Question 9.
Acidity of bases
Answer:
The number of OH ions obtainable by the dissociation of one molecule of a base is called its acidity.

Question 10.
Concentration of solute in the solution.
Answer:
The proportion of a solute in a solution is called the concentration of the solute in the solution.

Question 11.
Concentrated solution.
Answer:
When the concentration of a solute in its solution is high, it is a concentrated solution.

Question 12.
Dilute solution
Answer:
When the concentration of a solute in its solution is low, it is a dilute solution.

Question 13.
Neutralization
Answer:
A reaction in which an acid reacts with a base to form salt and water is called a neutalization reaction.
NaOH + HC1 → NaCl + H₂O
Base + Acid → Salt + Water

Question 14.
Cathode
Answer:
The electrode connected to the negative terminal of a battery by means of a conducting wire is called a cathode.

Question 15.
Anode
Answer:
The electrode connected to the positive terminal of a battery by means of a conducting wire is called an anode.

Question 16.
Cations
Answer:
Cations are positively charged ions which are attracted towards negative electrode (cathode) when electricity is passed into a solution of an ionic compound.

Question 17.
Anions
Answer:
Anions are negatively charged ions which are attracted towards the positive electrode (anode) when electricity is passed into a solution of an ionic compound.

Question 18.
Electrolytic cell
Answer:
An assembly that consists of a container with an electrolyte and the electrodes dipped in it, is called an electrolytic cell.

Question 19.
Molarity of a solution
Answer:
The number of moles of the solute dissolved in one litre of the solution is called the molarity of that solution. The molarity of a solute is indicated by writing its molecular formula inside a square bracket for example [NaCl] = 1

Question 20.
Acid – base indicators
Answer:
Some natural and synthetic dyes show two different colours in acidic and basic solution, and such dyes are acid base indicators.

Explain the following chemical reactions with the help of balanced equations:

Question 1.
Magnesium reacts with dilute hydrochloric acid.
Answer:
When magnesium reacts with dilute hydrochloric acid, it forms magnesium chloride and hydrogen gas is liberated.

Question 2.
When copper reacts with nitric acid.
Answer:
When copper reacts with nitric acid, it forms copper nitrate and hydrogen gas is liberated.

Question 3.
When ferric oxide reacts with diluted hydrochloric acid.
Answer:
When Ferric oxide reacts with diluted hydrochloric acid, it forms ferric chloride and water.

Question 4.
When calcium oxide reacts with dilute hydrochloric acid.
Answer:
When calcium oxide reacts with dilute hydrochloric acid, it forms calcium chloride and water.

Question 5.
When Magnesium oxide reacts with dilute hydrochloric acid.
Answer:
When magnesium oxide reacts with dilute hydrochloric acid, it forms magnesium chloride and water.

Question 6.
When zinc oxide reacts with dilute hydrochloric acid.
Answer:
When zinc oxide reacts with dilute hydrochloric acid, it forms zinc chloride and water.

Question 7.
When aluminium oxide reacts with hydrogen fluoride
Answer:
When aluminium oxide reacts with hydrogen fluoride, it forms Aluminium fluoride arid water.

Question 8.
When carbon dioxide reacts with sodium hydroxide.
Answer:
When carbon dioxide reacts with Sodium hydroxide, it forms Sodium carbonate and water.

Question 9.
When carbon dioxide reacts with potassium hydroxide.
Answer:
When carbon dioxide reacts with potassium hydroxide, it forms potassium carbonate and water.

Question 10.
When sulphur trioxide reacts with sodium hydroxide.
Answer:
When sulphur trioxide reacts with sodium hydroxide, it forms sodium sulphate and water.

Question 11.
When calcium hydroxide reacts with carbon dioxide.
Answer:
When calcium hydroxide reacts with carbon dioxide, it forms calcium carbonate and water.

Question 12.
When sodium carbonate reacts with hydrochloric acid.
Answer:
When sodium carbonate reacts with hydrochloric acid, it forms sodium chloride, carbon dioxide and water.

Question 13.
When sodium carbonate reacts with sulphuric acid.
Answer:
When sodium carbonate reacts with sulphuric acid, it forms sodium sulphate, carbon dioxide and water.

Question 14.
When calcium carbonate reacts with nitric acid.
Answer:
When calcium carbonate reacts with nitric acid, it forms calcium nitrate, carbon dioxide and water.

Question 15.
When potassium carbonate reacts with sulphuric acid.
Answer:
When potassium carbonate reacts with sulphuric acid, it forms potassium sulphate, carbon dioxide and water.

Question 16.
When sodium bicarbonate reacts with hydrochloric acid. OR Dilute HC1 was poured on baking soda
Answer:
When sodium bicarbonate reacts with hydrochloric acid, it forms sodium chloride, carbon dioxide and water.

Question 17.
When potassium bicarbonate reacts with nitric acid.
Answer:
When potassium bicarbonate reacts with nitric acid, it forms potassium nitrate, carbon dioxide and water.

Question 18.
When sodium bicarbonate reacts with acetic acid.
Answer:
When sodium bicarbonate reacts with acetic acid, it forms sodium acetate, carbon dioxide and water.

Question 19.
When copper sulphate is heated.
Answer:
When copper sulphate is heated it loses it’s water of crystallization to form white anhydrous copper sulphate.

Question 20.
When Ferrous sulphate is heated.
Answer:
When ferrous sulphate is heated it loses its water of crystallization to form white anhydrous ferrous sulphate.

Q.2. (B)-3. Complete the following table.

Question 1.
Complete the following table.
Answer:

Complete the following reactions.

Question 1.
Answer:

Question 2.
Metal + Dilute acid → Salt + Hydrogen
Answer:

Question 3.
Metal oxide + Dilute acid → SaIt + Water
Answer:

Question 4.
Oxide of non-metal + base → salt + water
Answer:

Question 5.
Carbonate salt of metal + dilute acid → Another salt of metal + Carbon dioxide + water
Answer:

Question 6.
Bicarbonate salt of metal + dilute acid → Another salt of metal + carbon dioxide + water
Answer:

Give scientific reasons:

Question 1.
Ionic compound NaCl has very high stability.
Answer:

• The outermost shell of both Na⁺ and Cl^– ions is a complete octet.
• An electronic configuration with a complete octet indicates a stable state.
• A molecule of NaCl has Na⁺ and Cl^– ions. An ionic bond is formed between these ions.
• The force of attraction between them is very strong as it is formed between the oppositely charged Na⁺ and Cl^– ions.
• Therefore NaCl, an ionic compound has very high stability.

Question 2.
Ionic compound dissociates while forming an aqueous solution.
Answer:

• On dissolving in water, an ionic compound forms an aqueous solution.
• In the solid state, the oppositely charged ions in the ionic compound are sitting side by side.
• When an ionic compound being to dissolve in water, the water molecules push themselves in between the ions of the compound and it separates them from each other, that is to say, an ionic compound dissociates while forming an aqueous solution.

Question 3.
Blue coloured copper sulphate crystals become colourless on heating.
Answer:

• Copper sulphate crystals are blue in colour and crystalline in form due to presence of water of crystallisation.
• Each molecule of crystalline copper sulphate contains five molecules of water of crystallisation (CuSO₄.5H₂O).
• On heating, the copper sulphate crystals lose the water of crystallisation and turns into white amorphous powder called as anhydrous copper sulphate.
• Therefore, blue coloured copper sulphate crystals become colourless on heating.

Question 4.
During electrolysis of water, a few drops of sulphuric acid are added to it.
Answer:

• Pure water is a covalent compound and hence it is a non-electrolyte and does not conduct electricity.
• When a few drops of sulphuric acid (H₂SO₄) are added to water.
• Being a strong acid it dissociates almost completely in its solution forming H⁺ cations and \(\mathrm{SO}_{4}^{2-}\) anions.
• The movement of these ions in the solution towards the respective electrodes amount to the conduction of electricity through the solution.
• Therefore, during electrolysis of water, a few drops of sulphuric acid are added to it.

Question 5.
Glucose is a non-electrolyte.
Answer:

• Glucose is a covalent compound
• It does not form any ions in its aqueous solution.
• Due to this aqueous solution of glucose does not conduct electronic current.
• Hence, glucose is a non-electrolyte.

Question 6.
Pure water is a poor conductor of electricity but aqueous solution of sodium chloride conducts electricity.
Answer:

• Pure water does not contain any free ions.
• Sodium chloride (NaCl) is an ionic compound made up of sodium cation (Na⁺) and chloride anion (Cl)
• When sodium chloride is dissolved in water , these ions dissociates in its aqueous solution. ‘
• These ions are free to move in the solution and conduct electricity.
• Therefore, pure water is a poor conductor of electricity but aqueous solution of sodium Chloride conducts electricity.

Question 7.
When carbon dioxide gas is passed through freshly prepared lime water, the limewater turns milky.
Answer:

• Limewater traditionally means a weak solution of the alkali calcium hydroxide Ca(OH)₂.
• When CO₂ is passed through limewater, it reacts with calcium hydroxide to form insoluble particulates (precipitate of calcium carbonate (CaCO₃).
• Calcium carbonate is weak basic salt and this gives a milky white precipitate.
  
• Hence, lime water turns milky when CO₂ gas is passed through it.

Q.3.1. Answer the following:

Question 1.
Write down chemical equations for
(a) Zinc oxide reacts with sodium hydroxide
(b) Aluminium oxide reacts with sodium hydroxide.
Answer:
(a) When zinc oxide reacts with sodium hydroxide, it forms sodium zincate and water

(b) When Aluminium oxide reacts with sodium hydroxide, it forms sodium aluminate and water.

Question 2.
Can we call Al₂O₃ and ZnO acidic oxides on the basis of above reactions.
Answer:

• No, because they also react with acids to form their respective salts and water.
• So, they show the properties of basic oxides also.

Question 3.
Define ‘amphoteric oxides’ and give two examples.
Answer:

• Amphoteric oxides are those oxides which react with both adds as well as bases to form their respective salts and water.
• Amphoteric oxides show the properties of both acidic oxides as well as basic oxides. ZnO and Al₂ O₃ are amphoteric oxides.?

Question 4.
Take a solution of 1g copper sulphate in 50ml water in a 100 ml capacity beaker. Use a thick plate of copper as anode and a carbon rod as cathode. Arrange the apparatus as shown in the figure and pass an electric current through the circuit for some time. Do you see any changes?
Answer:

• A thin film of copper metal is deposited on cathode which is immersed in solution. There is no change in colour of solution
• The electrons from cathode combines with Cu²⁺ ion from the solution forming Cu atoms which were then deposited on the cathode.

Q.3.5. Answer in brief:

Question 1.
What are acids, bases and salts?
Answer:

• Compounds having H⁺ as the basic radical in their molecules are called Acids.
• Compounds having OH^– as the acidic radical in their molecule are called Bases.
• Ionic compounds which have a basic radical other than H⁺ and an acidic radical other than OH^– are called salts.

Question 2.
What is an ionic bond?
Answer:

• The molecule of an ionic compound has two constituents namely cation (positive ion / basic radical) and anion (negative ion / acidic radical).
• There is a force of attraction between these ions as they are oppositely charged, and that is called the ionic bond.
• The force of attraction between one positive charge on a cation and one negative charge on an anion makes one ionic bond.

Question 3.
Give examples to show that proportions of H⁺ and OH^– ions in aqueous solution determines the properties of those solutions.
Answer:
The examples to show that proportions of H⁺ and OH^– ions in aqueous solution determines the properties of those solutions are :

• The proportions of H⁺ and OH ions divides soil into the acidic, neutral and basic, types of soil.
• It is necessary for blood, cell sap etc to have H⁺ and OH^– ions in certain definite proportions for their proper functioning.
• Fermentation carried out with the help of micro-organisms, other biochemical processes and also many chemical processes require the proportion of H⁺ and OH^– ions to be maintained within certain limits.

Question 4.
What is pH scale?
Answer:

1. In 1909, the Danish scientist Sorensen introduced a convenient new scale of expressing H⁺ ion concentration which is found to be useful in chemical and biochemical processes.
2. It is the pH scale (pH: power of hydrogen). The pH scale extends from 0 to 14. According to this scale pure water has a pH of 7. pH 7 indicates a neutral solution. This pH is the midpoint of the scale.
3. The pH of an acidic solution is less than 7 and?

Question 5.
Give the pH of following solutions.
Answer:

Question 6.
What is universal indicator? Which is the most accurate method of measuring the pH of a solution?
Answer:

• In the pH scale, the pH of solutions varies from 0 to 14 in accordance with the strength of the acid or base.
• To show these variations in pH, a universal indicator is used.
• A universal indicator shows different colours at different values of pH. A universal indicator is made by mixing several synthetic indicators in specific proportions.
• The pH of a solution can be determined by means of a universal indicator solution or the pH paper made from it.
• However, the most accurate method of measuring the pH of a solution is to use an electrical instrument called pH meter.
• In this method, pH is measured by dipping electrodes into the solution.

Answer in detail:

Question 1.
Explain the Arrhenius theory of acids and bases.
Answer:
The Swedish scientist Arrhenius put forth a theory of acids and bases in the year 1887. This theory gives definitions of acids and bases as follows:
Acid : An acid is a substance which on dissolving in water gives rise to H ion as the only cation. For example, HCl, H₂SO₄, H₂CO₃

Base: Abase is a substance which on dissolving in water gives rise to the OH ion as the only anion, For example, NaOH, Ca(OH)₂

Question 2.
Write a short note on Neutralization.
Answer:

• Take 10 ml of dilute HCl in a beaker, go on adding dilute NaOH drop by drop and record the pH.
• Stop adding the NaOH when the green colour appears on the pH paper, that is when the pH of solution becomes 7.
• Both HCl and NaOH dissociate in their aqueous solutions.
• Addition of NaOH to HCl solution is like adding a large concentration of OH ions to a large concentration of H⁺ ions.
• However water dissociates into H⁺ and OH ions to a very small extent.
• Therefore on mixing the excess OH^– ions combines with excess H⁺ ions and forms H₂O molecules which mix with solvent water.
• This change can be represented by the ionic equation shown as follows. H⁺ + Cl^– + Na⁺ + OH^– → Na⁺ + Cl^– + H₂O
• It can be observed that Na⁺ and CT ions are there on both the sides. Therefore the net ionic reaction is H⁺ + OH^– → H₂O
• As NaOH solution is added drop by drop to the HCl solution, the concentration of ff goes on decreasing due to combination with added OH ions, and that is how the pH goes on increasing.
• When enough NaOH is added to HCl, the resulting aqueous solution contains only Na⁺ and Cl^– ions, that is, NaCl, a salt, and the solvent water. The only source of H⁺ and OH^– ions in this solution is a dissociation of water.
• Therefore, this reaction is called the Neutralization reaction. The Neutralization reaction is also represented by the following simple equation.
  

Question 3.
Explain the water of Crystallization.
Answer:

• Take some crystals of blue vitriol (CuSO₄. 5H₂O) in a test tube. Heat the test tube on a low flame of a burner.
• It was observed that on heating, the crystalline structure of blue vitriol broke down to form a colourless powder and water came out.
• This water was part of the crystal structure of blue vitriol. It is called water of crystallization.
• On adding water to the white powder a solution
  
  was formed which had the same colour as the solution in the first test tube.
• From this we come to know that no chemical change has occurred in the crystals of blue vitriol due to heating.
• Losing water on heating blue vitriol, breaking down of the crystal structure, losing blue colour and regaining blue colour on adding water are all physical changes.
  
• Similarly, ferrous Sulphate crystals also contain 7 molecules of water of crystallization which are lost on heating.
• The reaction is represented as
  
• Ionic compounds are crystalline in nature. These crystals are formed as a result of definite arrangement of ions.
• In the crystals of some compounds, water molecules are also included in this arrangement.
• That is the water of crystallization. The water of crystallization is present in a definite proportion of the chemical formula of the compound.

Question 4.
Explain the conduction of electricity through solutions of ionic compounds
Answer:

1. Electrons conduct electricity through electrical wires, and ions conduct electricity through a liquid or a solution.
2. Electrons leave the battery at the negative terminal, complete the electric circuit and enter the battery at the positive terminal.
3. When there is a liquid or a solution in the circuit, two rods, wires or plates are immersed in it. These are called electrodes.
4. Electrodes are made of conducting solid. The electrode connected to negative terminal of a battery by means of a conducting wire is called a cathode and the electrode connected to the positive terminal of a battery is called anode.
5. We have seen that salts, strong acids and strong bases dissociates almost completely in their aqueous solutions.
6. Therefore the aqueous solutions of all these three contain large number of cations and anions.
7. A characteristic of liquid state is the mobility of its particles. Due to its mobility the positive charged ions of the solution are attracted to the negative electrode or cathode.
8. On the other hand, the negative charged ions of the solution are attracted to the positive electrode or anode.
9. The movement of ions in the solution towards the respective electrodes accounts to the conduction of electricity through the solution.
10. From this we can understand that those liquids or solutions which contain a large number of dissociated ions conduct electricity.
    

 

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 6 Classification of Plants Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 6 Classification of Plants

Class 9 Science Chapter 6 Classification of Plants Textbook Questions and Answers

1. Match the proper terms from columns A and C with the description in column B.

2. Complete the sentences by filling in the blanks and explain those statements.
(angiosperms, gymnosperms, spore, Bryophyta, thallophyta, zygote)
a. ……………….. plants have soft and fiber-like body.

b. ……………….. is called the ‘amphibian’ of the plant kingdom.
Answer:
Bryophyta plant is called the ‘amphibian’ of the plant kingdom

c. In pteridophytes, asexual reproduction occurs by ……………….. formation and sexual reproduction occurs by ………………..formation.
Answer:
Spore, zygote: Pteridophyta plants show alteration of generation. One generation reproduces by spore-formation and the next generation reproduces sexually by zygote formation.

d. Male and female flowers of ……………….. are borne on different sporophylls of the same plant.
Answer:
Gymnosperms bear their male and female flowers on different sporophylls of the same plant

3. Answer the following questions in your own words.

a. Write the characteristics of subkingdom Phanerogams.
Answer:

• Plants which have special structures for reproduction and produce seeds are called Phanerogams.
• In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food.
• During the germination of the seed, the stored food is used for the initial growth of the embryo.
• Depending upon whether seeds are enclosed in a fruit or not phanerogams are classified into gymnosperms and angiosperms.

b. Distinguish between monocots and dicots.
Answer:

	Dicots	Monocots
Seed	Two cotyledons	Single cotyledon
Root	Well developed, primary root (Taproot)	Fibrous roots
Stem	Strong, hard. e.g. Banyan tree	Hollow, e.g. Bamboo False, e.g. Banana Disc-like, e.g. Onion.
Leaf	Reticulate venation	Parallel venation
Flower	Flowers with 4 or 5 parts or in their multiples (tetramerous or pentamerous)	Flowers with 3 parts or in multiples of three (trimerous).

c. Write a paragraph in your own words about the ornamental plants called ferns.
Answer:

• Ferns belong to the group of plants called Pteridophyta.
• They have well-developed roots, stem and leaves but do not bear flowers and fruits.
• They have separate tissues for the conduction of food and water.
• They reproduce with the help of spores formed along the back or posterior surface of their leaves.
• They reproduce asexually by spore formation and sexually by zygote formation.

d. Sketch, label and describe the Spirogyra.

e. Write the characteristics of the plants belonging to division Bryophyta.
Answer:

• Bryophyta group of plants are called the amphibians of the plant kingdom because they grow in moist soil but need water for reproduction.
• These plants are thalloid, multicellular and autotrophic.
• They reproduce by spore-formation.
• Their plant body structure is flat, ribbon-like, long, without true roots, stem and leaves.
• Instead, they have stem-like or leaf-like parts and root-like rhizoids.
• They do not have specific tissues for the conduction of food and water.
• Examples: Moss (Funaria), Anthoceros, Riccia etc.?

4. Sketch and label the figures of the following plants and explain them into brief.
Marchantia, Funaria, Fern, Spirogyra.

Question 1.
Spirogyra.
Answer:

• Spirogyra belongs to the division thallophyta. They are called as algae.
• It grows mainly in water.
• It does not have specific parts like root-stem- leaves-flowers but are autotrophic due to the presence of chlorophyll.
• The plant body of Spirogyra is soft and fibre-like.
• It has spirally arranged chloroplasts in its cell.

Question 2.
Funaria and Marchantia (Bryophyta)
Answer:

• These plants are called ‘amphibians’ of the plant kingdom because they grow mostly in soil and need water for reproduction.
• They do not have specific tissues for the conduction of food and water.
• The plant body is fiat, ribbon-like long, without true roots, stem and leaves
• Instead, they have stem-like or leaf-like parts and root like rhizoids.

Question 3.
Fern (Pteridophyta):
Answer:

• They have well-developed roots, stem and leaves for the conduction of food and water.
• They do not bear flowers and fruits.
• They reproduce with the help of spores present along the back or posterior surface of the leaves.

5. Collect a monocot and dicot plant available in your area. Observe the plants carefully and describe them in scientific language.

6. Which criteria are used for the classification of plants? Explain with reasons.
Answer:
Criteria for classification of plants:

• If plants do not bear flowers, fruits and seeds, they are non-seed bearing plants. If they bear flowers, fruit and seeds, they are seed-bearing plants.
• Presence or absence of conducting tissues- Plants such as pteridophytes, gymnosperms and angiosperms which possess conducting tissues are included in vascular plants whereas thallophytes and bryophytes which do not possess conducting tissues are included under non-vascular plants.
• Depending upon whether the seeds are enclosed in fruit or not, plants are classified as gymnosperms (naked-seeds) and angiosperms (seeds covered by fruit)
• Depending upon the number of cotyledons in seeds, plants are classified into dicotyledons and monocotyledons

Class 9 Science Chapter 6 Classification of Plants Intext Questions and Answers

Can you recall?

Question 1.
How have living organisms been classified?
Answer:
(i) Organisms have been classified based on the following:

• Cell structure
• Body Organisation
• Mode of nutrition
• Reproduction

(ii) Organisms are also classified at kingdom level and groups and subgroups.

Activity-based questions

Question 1.
You may have seen a lush green soft carpet on old walls, bricks and rocks in the rainy season. Scrape it gently with a small ruler, observe it under a magnifying lens and discuss.
Answer:

• It shows considerable tissue complexity and is differentiated into two main parts: a root and a shoot.
• They have a variety of specialized tissues within these two regions of the body.
• Same kind of cells are seen throughout the whole body except reproductive cells.

Question 2.
You may have seen ferns among the ornamental plants in a garden. Take a leaf of a fully grown fern and observe it carefully.
Answer:

• New leaves typically expand by the unrolling in a tight spiral manner.
• The anatomy of fern leaves can either be simple or highly divided.
• They show the presence of spores formed along the back or posterior surface of their leaves.

Question 3.
Observe all garden plants like Cycas, Christmas tree, Hibiscus, Lily, etc. and compare them. Note the similarities and differences among them. Which differences did you notice in gymnosperms and angiosperms?
Answer:
Cycas and Christmas tree are gymnosperms, whereas Hibiscus and lily are angiosperms.

1. Similarities: These plants have special structures for reproduction and produce seeds. During the germination of the seed, the stored food is used for the initial growth of the embryo.
2. Differences: In gymnosperms, reproductive organs have cones whereas in angiosperms reproductive organs have flowers.
3. In gymnosperms, seeds are without natural coverings whereas in angiosperms seeds are enclosed in natural coverings called fruits.

Question 4.
Soak the seeds of corns, beans, groundnut, tamarind, mango, wheat, etc. in water for 8 to 10 hrs. After they are soaked, check each seed to see whether it divides into two equal halves or not and categorize them accordingly.
Answer:
Monocots: com, wheat (it cannot be divided into equal halves)
Dicots: beans, groundnut, tamarind and mango (it can be divided into two equal halves)

Class 9 Science Chapter 6 Classification of Plants Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
The five-kingdom classification was proposed b7
(a) Robert Whittaker
(b) Robert Hooke
(c) Eichler
(d) Louis Pasteur
Answer:
(a) Robert Whittaker

Question 2.
In 1883, classified plants into two sub-kingdoms.
(a) Robert Whittaker
(b) Alexander Fleming
(c) Eichler
(d) Robert Hooke
Answer:
(c) Eichler

Question 3.
Ulothrix, ulva, sargassum belong to
(a) Bryophyta
(b) Thallophyta
(c) Pteridophyta
(d) Gymnosperms
Answer:
(b) Thallophyta

Question 4.
is a bryophyte.
(a) Ulva
(b) Nephrolepis
(c) Funaria
(d) Equisetum
Answer:
(c) Funaria

Question 5.
In the seeds are naked.
(a) Pteridophyta
(b) Angiosperms
(c) Gymnosperms
(d) Bryophyta
Answer:
(c) Gymnosperms

Question 6.
In the flowers are reproductive organs.
(a) Angiosperms
(b) Gymnosperms
(c) Pteridophyta
(d) Bryophyta
Answer:
(a) Angiosperms

Question 7.
In the flowers are tetramerous or pentamerous.
(a) Monocotyledons
(b) Dicotyledons
(c) Gymnosperms
(d) Pteridophyta
Answer:
(b) Dicotyledons

Question 8.
In monocotyledonous plants, the stem is
(a) hollow
(b) false
(c) disc-like
(d) all of these.
Answer:
(d) all of these

Question 9.
Lycopodium belongs to
(a) Thallophyta
(b) Bryophyta
(c) Gymnosperms
(d) Pteridophyta
Answer:
(d) Pteridophyta

Question 10.
Leaves of show reticulate venation.
(a) Bamboo
(b) Banana
(c) Onion
(d) Banyan
Answer:
(d) Banyan

Question 11.
Various types of fungi like yeasts and moulds are included in the group
(a) Thallophyta
(b) Halophyte
(c) Xenophyta
(d) Angiosperms
Answer:
(a) Thallophyta

Question 12.
Bryophytes have a root-like structure called
(a) Nodes
(b) Rhizoids
(c) Nodules
(d) Aerenchyma
Answer:
(b) Rhizoids

Question 13.
reproduce with the help of spores formed along the back or posterior surface of their leaves.
(a) Halophyta
(b) Pteridophyta
(c) Thallophyta
(d) Angiosperms
Answer:
(b) Pteridophyta

Question 14.
In ……………………….., the reproductive organs cannot be seen.
(a) Pteridophyta
(b) Cryptogams
(c) Thallophyta
(d) Angiosperms
Answer:
(b) Cryptogams

Question 15.
are mostly evergreen, perennial and woody.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(c) Gymnosperms

Question 16.
Gymnosperms bear male and female flowers on different of the same plant.
(a) Branches
(b) Roots
(c) Sporophylls
(d) Flowers
Answer:
(c) Sporophylls

Question 17.
In the seeds are not enclosed by fruit.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(c) Gymnosperms

Question 18.
In the seeds are enclosed by fruit.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(d) Angiosperms

Question 19.
The plants whose seeds cannot be divided into equal parts are called
(a) Algae
(b) Fungus
(c) Dicotyledons
(d) Monocotyledons
Answer:
(d) Monocotyledons

Question 20
The plants whose seeds can be divided into equal parts are called
(a) Algae
(b) Fungus
(c) Dicotyledons
(d) Monocotyledons
Answer:
(c) Dicotyledons

Find the odd one out:

Question 1.
Ulothrix, Ulva, Nephrolepis, Sargassum
Answer:
Nephrolepis: It belongs to division Pteridophyta whereas the others belong to division thallophyta.

Question 2.
Funaria, Marchantia, Anthoceros, Spirogyra
Answer:
Spirogyra:

Question 3.
Marsilea, Pteris, Lycopodium, Riccia
Answer:
Riccia:

Question 4.
Cycas, Mango, Apple, Banyan
Answer:
Cycas:

Question 5.
Onion, Papaya, Wheat, Green peas
Answer:
Green peas:

Complete the analogy:

(1) Spirogyra : Thallophyta : : Riccia :
(2) Moss : Bryophyta : : Selaginella :
(3) Nephrolepis : Pteridophyta :: Ulothrix :
(4) Pteridophyta : Roots :: Bryophyta :
(5) Gymnosperms : naked seeds : : Angiosperms :
(6) Dicotyledon : Reticulate venation : : Monocotyledon:
(7) Bamboo stem: Hollow:: Onion Stem:
(8) Monocotylendon : Tap root:: Dicotyledon :
Answer:
(1) Bryophyta
(2) Pteridophyta
(3) Thallophyta
(4) Rhizoids
(5) Covered seeds
(6) Parallel venation
(7) Disc like
(8) Fibrous roots

Difference between:

Question 1.
Thallophyta and Bryophyta
Answer:

Thallophyta	Bryophyta
These plants grow mainly in water	They grow in moist soil but need water for reproduction

Question 2.
Gymnosperms and Angiosperms
Answer:

Gymnosperms	Angiosperms
No natural covering on seeds	Seeds are formed in fruits

Question 3.
Algae and Moss
Answer:

Algae	Moss
These plants mainly grow in water.	These plants need water for reproduction.

State whether the following statements are true or false. Correct the false statements:

(1) Thallophyta are called as the amphibians of the plant kingdom.
(2) Fungi like yeasts and moulds are included in division bryophyta.
(3) Moss (Funaria) belongs to division bryophyta.
(4) Bryophyta have specific tissues for conduction of food and water.
(5) Plants belonging to Thallophyta group are only unicellular.
(6) Pteridophytes have well developed roots, stems and leaves.
(7) Pteridophytes reproduce with the help of spores formed along the back or posterior surface of their leaves.
(8) Nephrolepis belongs to division Pteridophyta.
(9) Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into monocots and dicots.
(10) Gymnosperms are mostly evergreen, perennial and woody.
(11) Gymnosperms bear male and female flowers on different sporophylls of different plants.
(12) In Angiosperms, the seeds are covered by fruits.
(13) Dicotyledonous plants show reticulate venation.
(14) Moncotyledonous plants have trimerous flowers.
(15) In dicotyledonous plants, the stem is strong and hard.
Answer:
(1) False. Thallophyta plants grow mainly in water.
(2) False. Fungi like yeasts and moulds are included in division thallophyta.
(3) True
(4) False. Bryophyta do not have specialised tissuesfor conduction of food and water.
(5) False. Plants belonging to thallophyta group may be unicellular or multicellular.
(6) True
(7) True
(8) True
(9) False. Depending whether seeds are enclosed in. a fruit or not, angiosperms are classified into monocots and dicots.
(10) True
(11) False. Gymnosperms bear male and female flowers on different sporophylls of the same plant.
(12) True
(13) False. Dicotyledonous plants show parallel venation.
(14) True
(15) True.

Give name

Question 1.
What are ornamental plants are called?
Answer:
Ferns

Question 2.
Plants with two cotyledons are called.
Answer:
Dicots

Question 3.
Plants with single cotyledon are called.
Answer:
Monocots

Question 4.
Type of venation showed by hibiscus plant leaves
Answer:
Reticulate venation

Question 5.
Type of venation showed by lily plant leaves
Answer:
Parallel venation

One line answers

Question 1.
Which plants are mostly evergreen, perennial and woody?
Answer:
Gymnosperms are mostly evergreen, perennial and woody.

Question 2.
Which type of venation showed by dicot plants?
Answer:
Leaves of dicot plants show reticulated venation.

Question 3.
Which type of venation showed by monocot plants?
Answer:
Leaves of monocot plants show parallel venation

Question 4.
How are angiosperms classified into monocot and dicot?
Answer:
Depending whether seeds and enclosed in fruit or not, angiosperms are classified into monocot and dicot

Question 5.
In which division are fungi like moulds and yeast classified?
Answer:
Fungi like moulds and yeast classified in division thallophyta.

Question 6.
Plants belonging to which group may be unicellular or multicellular?
Answer:
Plants belonging to thallophyta group may be unicellular or multicellular

Give scientific reason

Question 1.
Thallophyta plants have thin and fibre like body
Answer:
Thallophyta: These plants grow mainly in water i.e. fresh water as well as in saline water, therefore they usually have a soft and fibre-like (filamentous) body.

Question 2.
Bryophyta plants are called the amphibian plants.
Answer:
Bryophyta: They grow in moist soil but need water for reproduction. Therefore, they are called ‘amphibians of plant kingdom’.

Question 3.
Gymnosperms bear their male and female flowers on different sporophylls of the same plant
Answer:
Gymnosperms: As these plants do not take the assistance of pollinators i.e. vectors, the male and female flowers are present on the different sporophyll of the same plant for successful fertilisation.

Write note on

Question 1.
August W. Eichler
Answer:
In 1883, Eichler, a botanist, classified the Kingdom Plantae into two subkingdoms. As a result, two subkingdoms, cryptogams and phanerogams were considered for plant classification.

Question 2.
Thallophyta
Answer:
These plants grow mainly in water. This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll, is called algae. Algae show great diversity. They may be unicellular or multicellular, and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a soft and fibre-like body. Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Question 3.
Bryophyta
Answer:
This group of plants is called the amphibians’ of the plant kingdom because they grow in moist soil but need water for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by spore formation. The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves. Instead, they have stem-like or leaf-like parts and root-like rhizoids. They do not have specific tissues for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.

Question 4.
Pteridophyta
Answer:
Plants from this group have well developed roots, stem and leaves and separate tissues for conduction of food and water. But, they do not bear flowers and fruits. They reproduce with the help of spores formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce asexually by spore-formation and sexually by zygote formation. They have a well-developed conducting system.

Question 5.
Phanerogams
Answer:
Plants which have special structures for reproduction and produce seeds are called phanerogams. In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into gymnosperms and angiosperms.

Question 6.
Gymnosperms
Answer:
Gymnosperms are mostly evergreen, perennial and woody. Their stems are without branches. The leaves form a crown. These plants bear male and female flowers on different sporophylls of the same plant. Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.

Question 7.
Angiosperms
Answer:
The flowers these plants bear are their reproductive organs. Flowers develop into fruits and seeds are formed within fruits. Thus, these seeds are covered; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Complete the flow chart.

Question 1.
Living Organisms

Answer:

Question 2.
Kingdom: Plantae

Answer:

Distinguish between:

Question 1.
Bryophyta and Pteridophyta:
Answer:

Bryophyta	Pteridophyta
Bryophytes grow in soil but need water for reproduction.	Pteridophytes grow in soil.
Plant body is without specific parts like true roots, stem and leaves.	Plant body is differentiated into true roots, stem and leaves.
Conducting tissues for food and water absent.	Conducting tissues for food and water present.
Examples: Moss (Funaria), Marchantia, Anthoceros, etc.	Examples: Nephrolepis, Marsilea, Pteris, Adiantum, Lycopodium etc.

Question 2.
Angiosperms and Gymnosperms.
Answer:

Angiosperms	Gymnosperms
(i) In Angiosperms, the stems have branches.	(i) In Gymnosperms, the stems are without branches.
(ii) Reproductive organs are flowers.	(ii) Reprodcutive organs are cones.
(iii) Seeds are enclosed in natural coverings, i.e., fruits.	(iii) Seeds are not enclosed in natural coverings.
(iv) Examples: Mango, Bamboo, etc.	(iv) Examples: Cycas, Picea etc.

Question 3.
Cryptogams and Phanerogams.
Answer:

Cryptogams	Phanerogams
(iii) Their reproductive organs are hidden.	(iii) Their reproductive organs are exposed.
(iii) They reproduce by forming spores.	(iii) They reproduce by forming seeds.
(iii) They are less evolved plants.	(iii) They are highly evolved plants.
(iv) They are divided into Thallophyta,	(iv) They are divided into Gymnosperms and
Bryophyta, Pteridophyta.	Angiosperms.

Distinguish between:

Question 1.
Thallophyta
Answer:
Spirogvra, Ulothrix, Ulva, Sargassum

Question 2.
Bryophyta
Answer:
Moss (Funaria), Marchantia, Anthoceros, Riccia

Question 3.
Pteridophyta
Answer:
Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium

Question 4.
Gymnosperms
Answer:
Cycas, Picca (Christmas tree), Thuja (Morpankhi), Pinus (Deodar)

Question 5.
Angiosperms
Answer:
Tamarind, Mango, Apple, Lemon

Question 6.
Monocot plants
Answer:
Bamboo, bananas, corn, daffodils, garlic, ginger, grass, lilies, onions, orchids, rice, sugarcane, tulips, and wheat

Question 7.
Dicot plants
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

Observe the figure and answer the questions

1. Dicot Plants

Question 1.
What are the characteristics of the above plants in terms of root system?
Answer:
Well developed, primary root (Tap root)

Question 2.
What are the characteristics of the above plants in terms of flowers?
Answer:
Flowers with 4 or 5 parts or in their multiples (tetramerous or pentamerous)

Question 3.
What are the characteristics of the above plants in terms of leaf venations?
Answer:
Reticulate Venation

Question 4.
What are the characteristics of the above plants in terms of type of stem?
Answer:
Strong and hard

Question 5.
What are the characteristics of the above plants in terms of seed?
Answer:
Two cotyledons

Question 6.
Give example of the following types of plants
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

2. Monocot Plants

Question 1.
What are the characteristics of the above plants in terms of root system?
Answer:
Fibrous roots

Question 2.
What are the characteristics of the above plants in terms of flowers?
Answer:
Flowers with 3 parts or in multiples of three (trimerous).

Question 3.
What are the characteristics of the above plants in terms of leaf venations?
Answer:
Parallel Venation

Question 4.
What are the characteristics of the above plants in terms of type of stem?
Answer:
Hollow, False or Disc-like

Question 5.
What are the characteristics of the above plants in terms of seed?
Answer:
Single cotyledons

Question 6.
Give example of the following types of plants
Answer:
Bamboo, bananas, com, daffodils, garlic, ginger, grass, lilies, onions, orchids, rice, sugarcane, tulips, and wheat

3. Spirogyra

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division I Thallophyta.

Question 2.
Where does this plant grow?
Answer:
These plants grow mainly in water.

Question 3.
Are these types of plants unicellular or multicellular?
Answer:
They may be unicellular or multicellular and microscopic or large.

Question 4.
Are these types of plant autotropic?
Answer:
They are autotrophic due to the presence of chlorophyll but types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Question 5.
Do these plants have a root-stem-leaves-flowers system?
Answer:
They do not have specific parts like root-stem- leaves-flowers.

Question 6.
How is the body of these types of plants?
Answer:
These plants usually have a soft and fibre-like body.

4. Funaria

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division II Bryophyta.

Question 2.
Where does this plant grow?
Answer:
They grow in moist soil but need water for reproduction.

Question 3.
What are these group of plants called in the plant kingdom?
Answer:
This group of plants is called the ‘amphibians’ of the plant kingdom.

Question 4.
Are these types of plant autotropic?
Answer:
They reproduce by spore formation.

Question 5.
Do these plants have root-stem-leaves-flowers system?
Answer:
The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves.

Question 6.
What do these plants have instead of roots?
Answer:
They have root like rhizoids.

5. Fern

Question 1.
Which division of plants does this plant come? under?
Answer:
This plant come under Division III Pteridophy ta.

Question 2.
Where does this plant grow?
Answer:
They grow in soil.

Question 3.
How do these plants reproduce?
Answer:
These plants reproduce asexually by spore- formation and sexually by zygote formation.

Question 4.
Do these plants produce flowers and fruits?
Answer:
They do not bear flowers and fruits.

Question 5.
Do these plants have root-stem-leaves-flowers system?
Answer:
Plants from this group have well developed roots, stem and leaves and separate tissues for conduction of food and water.

Question 6.
Where are the spores formed in the plants body?
Answer:
The spores formed along the back or posterior surface of their leaves.

6. Cycas

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division III Phanerogams Division I Gymnosperms.

Question 2.
Explain structure of these types of plants?
Answer:
Gymnosperms are mostly evergreen perennial and woody.

Question 3.
How is stem and leaves of these types of plants?
Answer:
Their stems are without branches and the leaves form a crown.

Question 4.
Where are the male and female flowers located?
Answer:
These plants bear male and female flowers on different sporophylls of the same plant.?

Question 6.
Give some examples of these types of plants?
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

7. Monocot and Dicot plants

Question 1.
Which division of plants does this plant come under?
Answer:
ThisplantcomeunderDivisionlllPhanerogams Division II Angiosperms

Question 2.
How are the seeds of these types of planis?
Answer:
The seeds are formed within fruits thus these seeds are covered

Question 3.
How can we classify the plants according to their seeds in this division?
Answer:
The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Question 4.
How the venations are present on the leaves of these types of plants?
Answer:
These plants bear parallel or reticulated venations on the leaves.

Question 5.
How is the root system of these types of plants?
Answer:
The root systems of these types of plant are tap roots or fibrouš roots.

Complete the paragraph

Question 1.
Thallophyta plants grow mainly in …………….. . This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of …………….., is called algae. Algae show great diversity. They may be unicellular or …………….., and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a …………….. and fibre-like body. Various types of …………….. like yeasts and moulds which do not have …………….. are also included in this group.
Answer:
Thallophyta plants grow mainly in water. This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll, is called algae. Algae show great diversity. They may be unicellular or multicellular, and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a soft and fibre-like body. Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Question 2.
…………….. group of plants is called the amphibians’ of the plant kingdom because they grow in moist soil but need …………….. for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by …………….. formation. The structure of the plant body of bryophytes is flat, ribbon?like long, without true …………….., stem and leaves. Instead, they have stem-like or leaf?like parts and root-like ……………. . They do not have specific …………….. for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.
Answer:
Bryophyta group of plants is called the ‘amphibians’ of the plant kingdom because they grow in moist soil but need water for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by spore formation. The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves. Instead, they have stem-like or leaf-like parts and root-like rhizoids. They do not have specific tissues for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.

Question 3.
Plants from Pteridophyta group have well developed roots, stem and leaves and separate …………….. for conduction of food and water. But,
they do not bear …………….. and ……………… They reproduce with the help of …………….. formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce …………….. by spore-formation and sexually by …………….. formation. They have a well-developed conducting system.
Answer:
Plants from Pteridophyta group have well developed roots, stem and leaves and separate tissues for conduction of food and water. But, they do not bear flowers and fruits. They reproduce with the help of spores formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce asexually by spore-formation and sexually by zygote formation. They have a well-developed conducting system.

Question 4.
Phanerogams plants which have special structures for …………….. and produce …………….. In these plants, after the process of reproduction, seeds are formed which contain the …………….. and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in …………….. a or not, phanerogams are classified into …………….. and ……………. .
Answer:
Phanerogams plants which have special structures for reproduction and produce seeds. In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into gymnosperms and angiosperms.

Question 5.
Gymnosperms are mostly …………….., perennial and woody. Their stems are without …………….. The leaves form a …………….. These plants bear male and female flowers on different …………….. of the same plant …………….. of these plants do not have natural coverings, i.e. these plants do not form …………….. and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.
Answer:
Gymnosperms are mostly evergreen, perennial • and woody. Their stems are without branches. The leaves form a crown. These plants bear male and female flowers on different sporophylls of the same plant. Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.

Question 6.
The flowers of Angiosperms plants bear are their …………….. organs Flowers develop into …………….. and seeds are formed within …………….. . Thus, these seeds are ……………..; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called …………….. plants and those whose seeds cannot be divided into equal parts are called …………….. plants.
Answer:
The flowers of Angiosperms plants bear are their reproductive orgAnswer: Flowers develop into fruits and seeds are formed within fruits. Thus, these seeds are covered; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Answer the questions in detail:

Question 1.
Write the characteristics of Thallophyta.
Answer:

• Thallophyta plants grow mainly in water.
• The group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll are called algae.
• Algae show great diversity They may be unicellular or multicellular and microscopic or large.
• Some of these are found in freshwater while some are found in saline water.
• Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.
• Examples: Spirogyra, Ulothrix, Ulva, etc.

Question 2.
Write the characteristics of Gymnosperms.
Answer:

• Gymnosperms are mostly evergreen, perennial and woody.
• Their stems are without branches.
• The leaves form a crown.
• These plants bear male and female flowers on different sporophylls of the same plant.
• Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms (gmnos: naked, sperms: seeds)
• Examples: Cycas, Picea (christmas tree), Thuja, Pinus (deodar), etc.

Make concept diagram

Question 1.
Plant classification
Answer:

Question 2.
Taxonomy of carnivorous 1ant
Answer:

Question 3.
Taxonomy of mango plant
Answer:

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 3 Current Electricity Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 3 Current Electricity

Class 9 Science Chapter 3 Current Electricity Textbook Questions and Answers

1. The accompanying figure shows some electrical appliances connected in a circuit in a house. Answer the following questions.

A. By which method are the appliances connected?
Answer:
Appliances are connected in parallel.

B. What must be the potential difference across individual appliances?
Answer:
The potential difference across all appliances is same in parallel connection.

C. Will the current passing through each appliance be the same? Justify your answer.
Answer:
No, as every appliance has a different load (resistance), the current flowing through each appliance will be different.

D. Why are the domestic appliances connected in this way?
Answer:
The domestic appliances are connected in parallel as the potential difference remains same.

E. If the T.V. stops working, will the other appliances also stop working? Explain your answer.
Answer:
No, the other devices will not stop working as the current flowing through them is along different paths.

2. The following figure shows the symbols for components used in the accompanying electrical circuit.

Which law can you prove with the help of the above circuit?
Answer:

(b) This circuit can be used to prove Ohm’s law.
(c) V = 1R is the expression of Ohm’s law

3. Umesh has two bulbs having resistances of 15 W and 30 W. He wants to connect them in a circuit, but if he connects them one at a time the filament gets burnt. Answer the following.

A. Which method should he use to connect the bulbs?
B. What are the characteristics of this way of connecting the bulbs depending on the answer of A above?
C. What will be the effective resistance in the above circuit?

4. The following table shows current in Amperes and potential difference in Volts.

a. Find the average resistance.
b. What will be the nature of the graph between the current and potential difference? (Do not draw a graph.)
c. Which law will the graph prove? Explain the law.

5. Match the pairs

‘A’ Group – ‘B’ Group
1. Free electrons – a. V/ R
2. Current – b. Increases the resistance in the circuit
3. Resistivity – c. Weakly attached
4. Resistances in series – d. VA/LI

6. The resistance of a conductor of length x is r. If its area of crosssection is a, what is its resistivity? What is its unit?

7. Resistances R₁, R₂, R₃ and R₄ are connected as shown in the figure. S₁ and S₂ are two keys. Discuss the current flowing in the circuit in the following cases.

a. Both S₁ and S₂ are closed.
b. Both S₁ and S₂ are open.
c. S₁ is closed but S₂ is open.
Answer:
(a) When both S₁ and S₂ are dosed, the effective resistance of the circuit decreases and hence, current will increase.
(b) When both S₁ and S₂ are open, the effective resistance of the rircuit increases and hence, current will decrease.
(c) When S₂ is closed and S₂ is open, the effective resistance of the tircuit decreases and hence current will increase. [Current will be more than case (b) but less than in case (a)]

8. Three resistances x₁, x₂ and x₃ are connected in a circuit in different ways. x is the effective resistance. The properties observed for these different ways of connecting x₁, x₂ and x₃ are given below. Write the way in which they are connected in each case. (I-current, V-potential difference, x-effective resistance)

a. Current I flows through x₁, x₂ and x₃
b. x is larger than x₁, x₂ and x₃
c. x is smaller than x₁, x₂ and x₃
d. The potential difference across x₁, x₂and x₃ is the same
e. x = x₁ + x₂ + x₃
\(\text { f. } x=\frac{1}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}}\)

9. Solve the following problems.

A. The resistance of a 1m long nichrome wire is 6Ω. If we reduce the length of the wire to 70 cm. what will its resistance be? (Answer : 4.2Ω)
Answer:
The resistance of 70cm wire will be 4.2 Ω

B. When two resistors are connected in series, their effective resistance is 80Ω. When they are connected in parallel, their effective resistance is 20Ω. What are the values of the two resistances? (Answer : 40Ω, 40Ω)
Answer:
The values of the two resistances R₁ and R₂ are 40Ω and 40Ω.

C. If a charge of 420 C flows through a conducting wire in 5 minutes what is the value of the current? (Answer : 1.4 A)
Answer:
Given: Electric charge (Q) = 420 C
Time (t) = 5 min = 5 x 60
= 300 sec.
To find: Electric current (1) = ?
Formula:
\(I=\frac{Q}{t}\)
Solution:
\(I=\frac{Q}{t}\)
The current in the circuit is 1.4 A.

Class 9 Science Chapter 3 Current Electricity Intext Questions and Answers

Can you recall?

Question 1.
You must have seen a waterfall. Which way does the water flow?
Answer:
Water flows from a certain height of a mountain towards the ground.

Question 2.
Material: Copper and aluminium wires, glass rod, rubber.
Make connection as shown in figure 3.8. First connect a copper wire between points A and B and measure the current in the circuit. Then in place of the copper wire, connect the aluminium wire, glass rod, rubber, etc one at a time and measure the current each time. Compare the values of the current in different cases.

Also take different metal strips (Iron, Copper, Zinc, and Aluminium) and connect it in slot AB. Now observe the difference in the resistance using Ohm meter.
Answer:
When copper and aluminium wires are connected to the circuit, current flows through it, as both are good conductors of electricity. When glass rod or rubber was connected to the circuit, current does not flow through it, as both are bad conductors of electricity.

Copper displays lowest resistance while the resistance increases with aluminium, zinc and iron respectively.

Question 3.
Set up the experiment as shown in figure. Then remove the clamp from the rubber tube.

(a) What happens when the clamp is removed?
Answer:
When the clamp is removed, water flows from higher level to lower level.

(b) Does the water stop flowing? Why?
Answer:
Yes, the water stops flowing. This happens when the level of water becomes equal in both the bottles, i.e., there is no difference in the water levels.

(c) What will you do to keep the water flowing for a longer duration?
Answer:
The difference in the water level has to be maintained till that time. The difference must never be zero.

Question 4.
Point out the mistakes in the figure below:

Answer:
A: Wire is broken at the negative terminal. Bulb will not glow as the circuit is incomplete.
B: Wire is disconnected at the negative terminal. Bulb will not glow as the circuit is incomplete.
C: The circuit is complete. Therefore, bulb will glow.
D: Rubber is a bad conductor of electricity. Hence, it will not allow current to flow and the bulb will not glow.

Question 5.
Why are the bulbs in Figures B, C and D not lighting up?

Answer:

• In B, the blue wire is broken. Hence circuit is incomplete and current does not flow. Therefore, bulb will not light up.
• In C, the red wire is broken. Hence circuit is incomplete and current does not flow. Therefore, bulb will not light up.
• In D, both wires are connected to the same terminal. Hence, there is no potential difference and current does not flow. Therefore, bulb will not light up.

Class 9 Science Chapter 3 Current Electricity Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
1mA = …………… A.
(a) 10³
(b) 10^-3
(c) 10⁶
(d) 10^-6
Answer:
(a) 10³

Question 2.
To increase the effective resistance in a circuit the resistors are connected in ………….. .
(a) Series
(b) Parallel
(c) Both ways
(d) None of these
Answer:
(a) series

Question 3.
1 kilowatt hr = …………… joules.
(a) 4.6 x 10⁶
(b) 3.6 x 10⁶
(c) 30.6 x 10⁶
(d) 3.6 x 1O⁵
Answer:
(b) 3.6 x 10⁶

Question 4.
The voltage difference in India between the live and neutral wires is about ………….. .
(a) 110 V
(b) 220 V
(c) 440 V
(d) 60 V
Answer:
(b) 220 V

Question 5.
Resistivity is the specific property of a ………….. .
(a) Area of cross-section
(b) Temperature
(c) Length
(d) Material
Answer:
(d) material

Question 6.
If a P.D. of 12 V is applied across a 3Ω resistor then the current passing through it is ………….. .
(a) 36 A
(b) 4 A
(c) 0.25 A
(d) 15 A
Answer:
(b) 4 A.

Question 7.
In order to measure the electric current flowing through a circuit, we connect …………… with the circuit.
(a) a voltmeter in parallel
(b) a voltmeter in series
(c) an ammeter in parallel
(d) an ammeter in series
Answer:
(d) an ammeter in series

Question 8.
P and Q are two wires of same length and different cross-sectional areas and made of same material. Name the property which is same for both the wires.
(a) Resistivity
(b) Resistance
(c) Current
(d) Both (a) and (b)
Answer:
(a) Resistivity

Question 9.
The following is true for identical bulbs connected in parallel.
(a) All bulbs glow with unequal brightness.
(b) If one bulb is non-functional, all will stop working.
(c) All bulbs glow with equal brightness.
(d) Bulbs function for longer time.
Answer:
(c) All bulbs glow with equal brightness

Question 10.
The …………… wire is either yellow or green in colour.
(a) Live
(b) Neutral
(c) Earth
(d) Fuse
Answer:
(c) earth

Question 11.
A current flows through a circuit due to the difference in …………… between two points in the conductor.
(a) Gravity
(b) Potential
(c) Resistance
(d) Fuse
Answer:
(b) potential

Question 12.
…………… is the amount of charge flowing through a particular cross sectional area in unit time.
(a) Electric current
(b) Ampere
(c) Volt
(d) Force
Answer:
(a) Electric current

Question 13.
The flow of …………… constitutes the electric current in a wire.
(a) Protons
(b) Neutrons
(c) Electrons
(d) Gravitons
Answer:
(c) electrons

Question 14.
The conventional direction of flow of current is from …………… terminal to …………… terminal.
(a) Negative to positive
(b) Neutral to positive
(c) Positive to negative
(d) Positive to neutral
Answer:
(c) positive, negative

Question 15.
Current stops flowing when potential difference between two ends of a wire becomes ………….. .
(a) Zero
(b) Positive
(c) Negative
(d) Higher
Answer:
(a) zero

Question 16.
Resistances are connected in …………… so as to pass the same current through them.
(a) Series
(b) Parallel
(c) Reversed
(d) Disconnect
Answer:
(a) series

Question 17.
To decrease the effective resistance in a circuit, the resistances are connected in ………….. .
(a) Series
(b) Parallel
(c) Reversed
(d) Disconnect
Answer:
(b) parallel

Question 18.
1μV = …………… V
(a) 10²
(b) 10^-6
(c) 10⁶
(d) 10³
Answer:
(b) 10^-6

Question 19.
Good conductors contain a large number of ………….. .
(a) Protons
(b) Neutrons
(c) Electrons
(d) Gravitons
Answer:
(c) free electrons

Question 20.
Electrons flow from …………… terminal to …………… terminal in a conductor when a potential difference is applied.
(a) Negative to positive
(b) Neutral to positive
(c) Positive to negative
(d) Positive to neutral
Answer:
(a) negative, positive

Find the odd one out:

Question 1.
Voltmeter, Ammeter, Galvanometer, Thermometer
Answer:
Thermometer

Question 2.
Rubber, Silver, Copper, Gold
Answer:
Rubber

Question 3.
Wood, Glass, Steel, Rubber
Answer:
Steel

Question 4.
Graphite, Diamond, Fullerenes, Coal
Answer:
Fullerenes

Distinguish between:

Question 1.
Voltmeter and Ammeter
Answer:

Voltmeter	Ammeter
(i) It is an instrument used to measure the potential difference between two terminals of a cell.	(i) It is an instrument to measure the electric current flowing through a circuit.
(ii) It is connected in parallel with the cell.	(ii) It is connected in series with the cell.
(iii) It has a very high resistance.	(iii) It has a very low resistance.
(iv) Voltmeter has range of volts.	(iv) Ammeter has range of amps.

Question 2.
Ohmic conductors and Non-Ohmic conductors
Answer:

Conductors	Insulators
(i) Substances which have very low electrical resistances are called conductors.	(i) Substances which have extremely high electrical resistances are called Insulators.
(ii) They contain a large number of free electrons.	(ii) They contain practically no free electrons.
(iii) Conductors are mostly metals.	(iii) Insulators are mostly non metals.
(iv) Conductor example iron, copper.	(iv) Insulator example rubber, plastic.

Question 3.
Conductors and Insulators
Answer:

Resistance	Resistivity
(i) The hindrance to the flow of electrons is called resistance.	(i) Resistivity is the specific property of the material of a conductor.
(ii) The S.I. unit of resistance is ohm (Q).	(ii) The S.I. unit of resistivity is ohm-metre (Q – m).
(iii) It depends on temperature, area of cross-section, length of conductor and material of the conductor.	(iii) It depends on material of the conductor.
(iv) Resistance can be changed as it depends of external factor as well.	(iv) Resistivity cannot be changed as it depends of internal factors.

Question 4.
Resistance in Series and Resistance in Parallel
Answer:

Resistance in Series	Resistance in Parallel
(i) Effective resistance of the resistors is equal to the sum of their individual resistances.	(i) Inverse of the effective resistance is equal to the sum of the inverse of individual resistances.
(ii) The same current flows through each resistor.	(ii) The total current flowing through the circuit is the sum of the currents flowing through individual resistors.
(iii) The effective resistance is larger than each of the individual resistances.	(iii) The effective resistance of resistors connected in parallel is less than the least resistance of individual resistors.
(iv) This arrangement is used to increase the resistance in a circuit.	(iv) This arrangement is used to decrease the resistance in a circuit.

Question 5.
Answer:

Electric current	Potential difference
(i) The flow of electric charge per unit time is called electric current.	(i) The difference in potential between the positive and negative terminal of a cell is the potential difference of that cell.
(ii) The S.I. unit of electric current is ampere.	(ii) The S.I. unit of potential difference volt.
(iii) Ammeter is used to measure electric current.	(iii) Voltmeter is used to measure electric current.
(iv) Current is represented by: \(\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}\)	(iv) Potential difference is represented by: \(\mathrm{V}=\frac{\mathrm{W}}{\mathrm{Q}}\)

Make pair:

Question 1.
Copper : Conductor :: Rubber : ……………….
Answer:
Insulator

Question 2.
Aluminium : ………………. :: Indium oxide : Super Insulator
Answer:
Super conductor

Question 3.
Parallel Connection : \(\frac{1}{\mathrm{R}_{p}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\) :: Series Connection : ……………….
Answer:
R_s = R₁ + R₂

(4) Electric Current : ………………. :: Electric charge : Coulomb
Answer:
Ampere

(5) Electric resistance : Ohm :: Potential difference : ……………….
Answer:
Volt

State whether the following statements are true or false. Correct the false statements:

(1) The SI unit of charge is volt.
(2) Voltmeter is always connected in series with the device.
(3) The conventional direction of flow of current is from positive terminal to negative terminal.
(4) Silver and copper are good conductors.
(5) Resistivity of pure metals is more than alloys.
(6) Resistance in series arrangement is used to decrease resistance of circuit.
(7) A conducting wire offers less resistance to flow of electrons.
(8) Charges are measured in ampere.
(9) The unit of potential difference is ampere.
(10) Resistance of a conductor is inversely proportional to the length of the conductor.
(11) Ammeter is connected in parallel to the cell to measure current.
(12) Fuse is made of wire having high melting point.
Answer:
(1) False. The SI unit of charge is coulomb.
(2) False. Voltmeter is ahvays connected in parallel with the device.
(3) True
(4) True
(5) False. Resistivity of pure metals is less than alloys.
(6) False. Resistance in series arrangement is used to increase resistance of circuit.
(7) True
(8) False. Charges are measured in coulomb.
(9) False. The unit of potential difference is volt.
(10) False. Resistance of a conductor is directly proportional to the length of the conductor.
(11) False. Ammeter is connected in series to the cell to measure current.
(12) False. Fuse is made of wire having low melting point.

Answer the following in one sentence:

Question 1.
Which is the unit used to measure large voltages?
Answer:
Kilovolts and Megavolts are the units used to measure large voltages.

Question 2.
What is the SI unit of potential difference?
Answer:
The SI unit of potential difference is volt (V).

Question 3.
What is lightning?
Answer:
Lightning is the electric discharge travelling from clouds at high potential to earth’s surface which is at zero potential.

Question 4.
What is the unit of resistivity.
Answer:
The unit of resistivity is ohm metre (Qm).

Question 5.
Which substances are called conductors of electricity?
Answer:
Those substances which have very low electrical resistance are called conductors of electricity.

Question 6.
What is Earth wire?
Answer:
Earth wire is generally yellow or green colour, it is connected to a metal plate buried deep underground near the house and is for safety purpose.

Write formula:
(1) E1ectriccurrent \(=\frac{Q}{t}\)
(2) Electric charge = It
(3) Potential difference = IR
(4) Electric resistance \(=\frac{V}{I}\)
(5) Current \(=\frac{V}{R}\)
(6) Resistivity \(=\frac{RA}{L}\)

Give scientific reasons:

Question 1.
Free electrons are required for conduction of electricity.
Answer:

• Every atom of a metallic conductor has one or more outermost electrons which are very weakly bound to the nucleus.
• These are called free electrons. These electrons can easily move from one part of a conductor to its other parts. The negative charge of the electrons also gets transferred as a result of this motion.
• The free electrons in a conductor are the carriers of negative charge. Hence, free electrons are required for conduction of electricity.

Question 2.
Wood and glass are good insulators.
Answer:

• Those substances which have infinitely high electrical resistance are called insulators.
• Wood and glass have high resistance and negligible free electrons for conduction of electricity.
• Hence, wood and glass are good insulators.

Question 3.
Connecting wires in a circuit are made of copper and aluminium.
Answer:

• Copper and aluminum are good conductors of electricity.
• They have low electrical resistance and large number of free electrons.
• As they are malleable and ductile, they can be drawn into thin wires. Hence, connecting wires in a circuit are made of copper or aluminum.

Question 4.
A thick wire has a low resistance.
Answer:

• The resistance (R) of a wire is inversely proportional to the cross-sectional area (A) of a wire. i.e., R i
• Thus, greater is the cross-sectional area of a conductor (wire), lower is its resistance. Hence, a thick wire has a low resistance.

Question 5.
A series combination of resistances is used to increase the resistance of a circuit.
Answer:

• When resistances are connected in series, the effective resistance of the resistors is equal to the sum of their individual resistances. R_s = R₁ + R₂ ………….. R_n
• The effective resistance is larger than each of the individual resistances. Hence, This arrangement is used to increase the resistance in a circuit.

Question 6.
A parallel combination of resistances decreases the effective resistance of the circuit.
Answer:

• In a parallel combination, the inverse of the effective resistance is equal to the sum of the inverses of individual resistances. \(\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}} \ldots \ldots \cdot \frac{1}{\mathrm{R}_{\mathrm{n}}}\)
• The effective resistance of resistors connected in parallel is less than the individual resistors.
• Due to this, any addition of an individual resistance in parallel combination will decrease the overall resistance of the circuit. Hence, a parallel combination of resistance decreases the effective resistance of the circuit.

Question 7.
Lightning occurs from sky to earth.
Answer:

• Lightning is the electric discharge travelling from clouds at high potential to the earth’s surface, which is at zero potential.
• The earth is always at lower potential as compared to the clouds.
• Hence, lightning occurs from sky to earth.

Question 8.
In streetlights, bulbs are connected in parallel.
Answer:

1. Even if any one of the several bulbs connected in parallel becomes non-functional because of some damage to its filament, the circuit does not break as the current flows through the other paths, and the rest of the bulbs light up.
2. When several bulbs are connected in parallel, they emit the same amount of light as when they are connected individually in the circuit, while bulbs connected in series emit less light than when connected individually. Hence, streetlights are connected in parallel.

Numerical:
Numericals based on the formula: (1) Q= It (2) W= VQ

Question 1.
A current of 0.4 A flows through a conductor for 5 minutes. How much charge would have passed through the conductor?
Answer:
Given: Current (I) = 0.4 A
Time (t) = 5 min = 5 x 60 = 300 s
To find: Charge (Q) =?
Formula: Q = 1 x
Solution: Q = 0.4 x 300
Q= 120 C.
Charge passing through the conductor is 120

Question 2.
Find the amount of work done if 3 C of charge is moved through a potential difference of 9 V.
Answer:

The work done is 27 joule.

Question 3.
The resistance of the filament of a bulb is 1000Ω. It is drawing a current from a source of 230 V. How much current is flowing through it?
Answer:

The current flowing through the filament of bulb is 0.23 A.

Question 4.
The length of a conducting wire is 50 cm and its radius is 0.5 mm. If its resistance is 30Ω, what is the resistivity of its material?
Answer:


The resistivity of the wire is 4.71 x 10^-5 Qm.

Question 5.
A current of 0.24 A flows through a conductor when a potential difference of 24 V is applied between its two ends. What is its resistance?
Answer:

The resistance of a conductor is 100Ω.

Question 6.
If three resistors 15Ω, 3Ω and 4Ω each are connected in series, what is the effective resistance in the circuit?
Answer:
Given:
R₁ =15Ω
R₂ = 3Ω
R₃ = 4Ω
Effective resistance in series (R_s) = ?
R_s = R₁ + R₂ + R₃
R_s = 15 + 3 + 4
R_s = 22Ω
The effective resistance in the circuit is 22Ω.

Question 7.
Three resistances 15Ω, 20Ω and 10Ω are connected in parallel. Find the effective resistance of the circuit.
Answer:


The effective resistance of the circuit is 4.615 Ω. It is less than the least of the three i.e., 10Ω.

Write a note on the following:

Question 1.
Electric current
Answer:
An electric current is the flow of electrons through a conductor. Quantitatively, current
(I) is defined as the charge passing through a conductor in unit time.
\(T=\frac{Q}{t}\)

Question 2.
1 ampere
Answer:
One ampere current is said to flow in a conductor if one coulomb charge flows through it every second.
\(1 \mathrm{~A}=\frac{1 \mathrm{C}}{1 \mathrm{~s}}\)

Question 3.
1 volt
Answer:
The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to another.
\(1 \mathrm{~V}=\frac{1 \mathrm{~J}}{1 \mathrm{C}}\)

Question 4.
Potential Difference
Answer:
The amount of work done to carry a unit positive charge from point A to point B is called the electric potential difference between the two points.
\(V=\frac{W}{Q}\)

Question 5.
Conductor
Answer:
Those substances which have very low resistance are called conductors. Current can flow easily through such materials.

Question 6.
Insulators
Answer:
Those substances which have extremely high resistance and through which current cannot flow are called insulators.

Question 7.
1 ohm
Answer:
If one Ampere current flows through a conductor when one Volt potential difference is applied between its ends, then the resistance of the conductor is one Ohm.
\(\frac{1 \text { Volt }}{1 \text { Ampere }}=1 \mathrm{Ohm}\)

Question 8.
Potential
Answer:
The level of electric charge present is known as potential.

Question 9.
Ohm’s Law
Answer:
If the physical state of a conductor remains constant, the current (I) flowing through it is directly proportional to the potential difference (V) between its two ends.
V = IR

Question 10.
Superconductors
Answer:
The resistance of some conductors becomes nearly zero if their temperature is decreased up to a certain value close to 0 K. Such conductors are called superconductors.

Question 11.
Non-ohmic conductors
Answer:
Conductors which do not obey Ohm’s law are called non-ohmic conductors.

Complete the flow charts:

(1) Protection from Electricity

(2) Resistance

(3) Resistivity

Write properties/characteristics/advantages of the following:

Superconductors
Answer:
The resistance of these conductors becomes nearly zero if their temperature is decreased up to a certain value close to 0 K. Aluminium is an example of Super Conductor. Superconductors can be used in space missions to increase/ boost the signal strength. They are also used i in the data fibres to increase the speed of data transfer.

Give explanations of the given statements:

Question 1.
Safety precautions are to be taken while using electricity.
Answer:

• Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket.
• Before cleaning an electrical appliance it should be switched off and its plug removed from the socket.
• One’s hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles.
• As rubber is an insulator, it prevents the current from flowing’ through our body, thereby protecting it.
• If a person gets an electric shock, you should not touch that person. You should switch off the main switch or remove the plug from the socket if possible.
• If not, then you should use a wooden pole to push the person away from the electric wire.

Question 2.
In a domestic circuit colour code is followed while setting up electrical wiring.
Answer:

• The electricity in our homes is brought through the main conducting cable either from the electric pole or from underground cables.
• Usually, there are three wires in the cable.
  (a) Live wire which brings in the current. It has a red or brown insulation.
  (b) Neutral wire through which the current returns. It is blue or black.
  (c) Earth wire is of yellow or green colour. This is connected to a metal plate buried deep underground near the house and is for safety purposes.
• In India, the voltage difference between the live and neutral wires is about 220 V.
• Live and neutral wires are connected to the electric meter through a fuse.
• They are connected through a main switch, to all the conducting wires inside the home so as to provide electricity to every room.
• In each separate circuit, various electrical appliances are connected between the live and neutral wires.
• The different appliances are connected in parallel and the potential difference across every appliance is the same.

Question 3.
Fuse used in electrical circuit can save electrical objects from damage.
Answer:

• Fuse wire is used to protect domestic appliances.
• It is made of a mixture of substances and has a specific melting point.
• It is connected in series to the electric appliances. If for some reason, the current in the circuit increases excessively, the fuse wire gets heated and melts.
• The circuit gets broken and the flow of current stops, thus protecting the appliance.
• This wire is fitted in a groove in a body of porcelain-like non-conducting material. For domestic use, fuse wires with upper limits of 1 A, 2 A, 3 A, 4 A, 5 A and 10 A are used.

Question 4.
Bulbs arranged in parallel glow brighter than bulbs arranged in series.
Answer:

• The amount of light given out by bulbs in parallel combination will be more than that in series combination.
• In parallel combination the resistance of the overall circuit decreases whereas in series it increases, so the current flowing through the bulbs in parallel circuit is more.
• Due to this, intensity of light given out by bulbs in parallel combination is more than the bulbs in series combination.

Complete the following table:
Answer:

Solve the numerical:

Question 1.
The length of a conducting wire is 50 cm and its radius is 0.5 mm. If its resistance is 30Ω, what is the resistivity of its material?
Answer:

The resistivity of the wire is 4.71 x 10^-5 Ωm.

Question 2.
Determine the current that will flow when a potential difference of 33 V is applied between two ends of an appliance having a resistance of 110 Ω. If the same current is to flow through an appliance having a resistance of 500 Ω, how much potential difference should be applied across its two ends?
Answer:

The current is 0.3 A and potential difference to be applied is 150 V.

Question 3.
Determine the resistance of a copper wire having a length of 1 km and diameter of 0.5 mm.
Answer:
Given: Resistivity of copper (p)
= 1.7 x 10^-8 Ω m
Converting all measures into metres.
Length of wire (L) = 1 km
= 1000 m = 10³ m
Diameter of wire (d) = 0.5 mm
= 0.5 x 10^-3m
To find: Resistance of wire (R) = ?
Formula:
\(R=\rho \frac{L}{A}\)
Solution:
If d is the diameter of the wire then, its area of cross-section

The resistance of a copper wire is 85Ω and area of cross section is 0.2 x 10^-6 m²

Question 5.
Two resistors having resistance of 16 and 14 are connected in series. If a potential difference of 18 V Is applied across them, calculate the current flowing through the circuit and the potential difference across each individual resistor.
Answer:


The current in the circuit is 0.6 A and potential across 16 Ω retor is 9.6 volt and 14 Ω resistor is 8.4 voIt.

Question 6.
If the resistors 5 Ω, 10 Ω and 30 Ω are connected in parallel to battery of 12 V, find the effective resistance in the circuit. Calculate the total current and current in each resistor.
Answer:


(a) The total current is 4 A and current in each resistor is 2.4 A, 1.2 A and 0.4 A respectively.
(b) The effective resistance in the circuit is 3Ω.

Complete the diagram and answer the questions:

Question 1.
The following figure shows the symbols for components used in the accompanying electrical circuit.

(a) Place them at proper places and complete the circuit.
(b) Which law can you prove with the help of the above circuit?
(c) State expression for Ohm’s law
Answer:

(b) This circuit can be used to prove Ohm’s law.
(c) V = IR is the expression of Ohm’s law

Question 3.
Explain with the help of a diagram, what are free electrons and how they move through the conductor?

Answer:

• Every atom of a metallic conductor has one or more outermost electrons which are very weakly bound to nucleus.
• These are called free electrons.
• These electrons can easily move from one part of a conductor to its other parts.

Complete the paragraph:

Question 1.
If resistors are connected in series,
Answer:
The same current flows through each resistor. The effective resistance of the resistors is equal to the sum of their individual resistances. The potential difference between the two extremes of the arrangement is equal to the sum of the potential differences across individual resistors. The effective resistance is larger than each of the individual resistances. This arrangement is used to increase the resistance in a circuit. This type of connection is used in electrical heating equipment like geysers, iron, and hair dryers.

Question 2.
If a number of resistors are connected in parallel,
Answer:
The inverse of the effective resistance is equal to the sum of the inverses of individual resistances. The current flowing through an individual resistor is proportional to the r inverse of its resistance and the total current flowing through the circuit is the sum of the currents flowing through individual resistors. The potential difference across all r resistors is the same. The effective resistance of resistors connected in parallel is less than r the least resistance of individual resistors.

This arrangement is used to reduce the resistance in a circuit. Even if any one of the several bulbs connected in parallel becomes non-functional because of some damage to its filament, the circuit does not break as the current flows, through the other paths, and the rest of the bulbs light up. When several bulbs are connected in parallel, they emit the same amount of light as when they are connected individually in the circuit, while bulbs connected in series emit less light than when connected individually.

Read the paragraph and answer the questions.

Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket. Before cleaning an electrical appliance it should be switched off and its plug removed from the socket. One’s hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles. As rubber is an insulator, it prevents | the current from flowing through our body, thereby protecting it. If a person gets an electric shock, you should not touch that person. You should switch off the main switch and if the switch is too far or you do not know where it is located, then you should remove the plug from the socket if possible. If not, then you should use a wooden pole to push the person away from the electric wire.

(i) Why should the electrical sockets be fitted at a certain height?
Answer:
Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside.

(ii) Why plug wires should not be pulled out while removing any electrical device?
Answer:
Plug wires should not be pulled out while removing any electrical device as it may cause the wire to break causing short circuit which can lead to fire or death.

(iii) Why should a person wear footwear with rubber soles while handling electrical appliances. .
Answer:
As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it. Hence a person should wear footwear with rubber soles while handling electrical appliances.

(iv) Saee is touching an electrical button socket with wet hands what will you advise her and why?
Answer:
We will advise her to dry her hands before touching any electrical sockets or devices as water on the hands can cause an electrical short circuit producing shock to the person touching it.

(v) Sneha is getting an electrical shock what will you do the save her life?
Answer:
We should switch off the main switch and if the switch is too far or we do not know where it is located, then we should remove the plug from the socket if possible. If not, then we should use a wooden pole to push the person away from the electric wire.

(vi) Give a title to the above passage.
Answer:
Precautions to be taken while using electricity

Answer the questions in details:

Question 1.
Find the expression (i.e., derive the expression) for the resistors connected in series.
Answer:
Expression for the resistance connected in series:
(i) Let R₁, R₂ and R₃ be three resistances connected in series between C and D.
(ii) Let R_s be the effective resistance in circuit and V₁, V₂ and V₃ be the potential difference across R₁, R₂ and R₃ respectively.
(iii) Let the potential difference across CD be V.

(iv) In series combination.
v = v₁ + v₂ + v₃ ……………………(i)
By using Ohm’s law
V = IR_s
∴ V₁= IR₁, V₂ = IR₂ and V₃ = 1R₃
Substituting these values in equation (j) we get
IR_s = IR₁ + IR₂ + IR₃
∴ R_s = R₁ + R₂ + R₃
For ‘n’ number of resistors conneded in series we get
R_s = R₁ + R₂ + R₃ + …………………. + R_n

Question 2.
Find the expression (i.e., derive the expression) for the resistors connected in parallel.
Answer:
Expression for the resistance connected in parallel.
(i) Let R₁, R₂ and R₃ be the three resistances connected in parallel combination between points C and D and let R be their effective resistance.
(ii) Let I₁, I₂ and I₃ be the currents flowing through resistances R₁, R₂ and R₃ respectively. Let I be the current flowing through the circuit and V be the potential difference of the cell.

(iii) For parallel combination of resistances,

(iv) Substituting the values of (I, I₁, I₂ and I₃) in equation (i) we get

Question 3.
Find the expression for resistivity of a material.
Answer:
(i) At a given temperature, the resistance (R) of a conductor depends on its length (L), area of cross-section (A) and the material it is made of. If the resistance of a conductor is R, then

(ii) p is the constant of proportionality and is called the resistivity of the material.
(iii) The unit of resistivity in SI units is Ohm metre (Ω m).
(iv) Resistivity is a specific property of a material and different materials have different resistivity. ’

Make the concept diagram and explain:

Question 1.
Make the concept diagram of an electrical circuit and explain the working of a fuse.

Answer:

• Fuse wire is used to protect domestic appliances.
• It is made of a mixture of substances and has a specific melting point.
• It is connected in series to the electric appliances. If for some reason, the current in the circuit increases excessively, the fuse wire gets heated and melts. The circuit gets broken and the flow of current stops, thus protecting the appliance.
• This wire is fitted in a groove in a body of porcelain-like non-conducting material. For domestic use, fuse wires with upper limits of 1A, 2A, 3A, 4A, 5A, and lO Aareused.

(2) Show motion of electrons in an circuit and explain precautions while using an electrical device.
Answer:

• Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket.
• Before cleaning an electrical appliance it should be switched off and its plug removed from the socket.
• One’s hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles.
• As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it.
• If a person gets an electric shock, you should not touch that person. You should switch off the main switch or remove the plug from the socket if possible.
• 1f not, then you should use a wooden pole to push the person away from the electric wire.

Q.4.4.Complete the incomplete figure and give an explanation:

Answer:

If the physical state of a conductor remains constant, the current (I) flowing through it is directly proportional to the potential difference (V) between its two ends.

I α V
I = kV (k = constant of proportionality)

This is known as Ohm’s law.
We can obtain the SI unit of resistance from the above formula, Potential difference and current are measured in Volts and Amperes respectively. The unit o resistance is called Ohm. It is indicated by the symbol Ω.

The resistance of one Ohm : If one Ampere current flows through a conductor when one Volt potential difference is applied between its ends, then the resistance of the conductor is one Ohm.

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 2 Work and Energy Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 2 Work and Energy

Class 9 Science Chapter 2 Work and Energy Textbook Questions and Answers

1. Write detailed answers?

a. Explain the difference between potential energy and kinetic energy.
Answer:

Kinetic Energy	Potential Energy
(i) Kinetic energy is the energy possessed by the body due to its motion.	(i) Potential energy is the energy possessed by the body because of its shape or position.
(ii) K.E = 1/2 mv2	(ii) P.E = mgh
(iii) e.g., flowing water, such as when falling from a waterfall.	(iii) e.g., water at the top of a waterfall, before the drop.

b. Derive the formula for the kinetic energy of an object of mass m, moving with velocity v.
Answer:
Suppose a stationary object of mass ‘m’ moves because of an applied force. Let ‘u’ be its initial velocity (here u = 0). Let the applied force be ‘F’. This generates an acceleration a in the object, and after time T, the velocity of the object becomes equal to ‘v’. The displacement during this time is s. The work done on the object is
W = F x s ……………….. (1)
Using Newton’s 2nd law of motion,
F = ma ……………….. (2)
Using Newton’s 2nd equation of motion
\(s=u t+\frac{1}{2} a t^{2}\)
However, as initial velocity is zero, u = 0

c. Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.
Answer:
Let us look at the kinetic and potential energies of an object of mass (m), falling freely from height (h), when the object is at different heights.

As shown in the figure, the point A is at a height (h) from the ground. Let the point B be at a distance V, vertically below A. Let the point C be on the ground directly below A and B. Let us calculate the energies of the object at A, B and C.

(1) Let the velocity of the object be v_B when it reaches point B, having fallen through a distance x.

(2) When the object is stationary at A, its initial velocity is u = 0
∴ K.E = 1/2 mass x velocity2

(3) Let the velocity of the object be vc when it reaches the ground, near point C.

From equations (i) and (iii) we see that the total potential energy of the object at its initial position is the same as the kinetic energy at the ground.

d. Determine the amount of work done when an object is displaced at an angle of 300 with respect to the direction of the applied force.
Answer:
When an object is displaced by displacement ‘s’ and by applying force ‘F’ at an ’angle’ 30°. work done can be given as

e. If an object has 0 momenta, does it have kinetic energy? Explain your answer.
Answer:

• No, it does not have kinetic energy if it does not have momentum.
• Momentum is the product of mass and velocity. If it is zero, it implies that v = 0 (since mass can never be zero).
• Now K.E = ~ mv2, So if v = 0 then K.E also will be zero.
• Thus, if an object has no momentum then it cannot possess kinetic energy.

f. Why is the work done on an object moving with uniform circular motion zero?
Answer:

• In uniform circular motion, the force acting on an object is along the radius of the circle.
• Its displacement is along the tangent to the circle. Thus, they are perpendicular to each other.
  Hence θ = 90° and cos 90 = θ
  ∴ W = Fs cos θ = 0

2. Choose one or more correct alternatives.

a. For work to be performed, energy must be ….
(i) transferred from one place to another
(ii) concentrated
(iii) transformed from one type to another
(iv) destroyed

b. Joule is the unit of …
(i) force
(ii) work
(iii) power
(iv) energy

c. Which of the forces involved in dragging a heavy object on a smooth, horizontal surface, have the same magnitude?
(i) the horizontal applied force
(ii) gravitational force
(iii) reaction force in vertical direction
(iv) force of friction

d. Power is a measure of the …….
(i) the rapidity with which work is done
(ii) amount of energy required to perform the work
(iii) The slowness with which work is performed
(iv) length of time

e. While dragging or lifting an object, negative work is done by
(i) the applied force
(ii) gravitational force
(iii) frictional force
(iv) reaction force

3. Rewrite the following sentences using a proper alternative.

a. The potential energy of your body is least when you are …..
(i) sitting on a chair
(ii) sitting on the ground
(iii) sleeping on the ground
(iv) standing on the ground
Answer:
(iii) sleeping on the ground

b. The total energy of an object falling freely towards the ground …
(i) decreases
(ii) remains unchanged
(iii) increases
(iv) increases in the beginning and then decreases
Answer:
(iii) increases

c. If we increase the velocity of a car moving on a flat surface to four times its original speed, its potential energy ….
(i) will be twice its original energy
(ii) will not change
(iii) will be 4 times its original energy
(iv) will be 16 times its original energy.
Answer:
(ii) will not change

d. The work done on an object does not depend on ….
(i) displacement
(ii) applied force
(iii) initial velocity of the object
(iv) the angle between force and displacement.
Answer:
(iii) initial velocity of the object

4. Study the following activity and answer the questions.

1. Take two aluminium channels of different lengths.
2. Place the lower ends of the channels on the floor and hold their upper ends at the same height.
3. Now take two balls of the same size and weight and release them from the top end of the channels. They will roll down and cover the same distance.

Questions
1. At the moment of releasing the balls, which energy do the balls have?
2. As the balls roll down which energy is converted into which other form of energy?
3. Why do the balls cover the same distance on rolling down?
4. What is the form of the eventual total energy of the balls?
5. Which law related to energy does the above activity demonstrate? Explain.
Answer:
1. At the moment of releasing the ball they possess Potential energy as they are at a height above the ground.
2. As the balls roll down, the Potential energy is converted into Kinetic energy since they are now in motion.
3. Since they have been released from the same height, they will cover the same distance.
4. The eventual form of the total energy of the balls is “Mechanical Energy” i.e, a combination of Potential energy and Kinetic energy
5. The above activity demonstrates the “Law of Conservation of Energy”

5. Solve the following examples.

a. An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m? (Ans : 1224.5 kg)
Answer:
Given:
Power (P) = 2 kW = 2000 W
Height (h) = 10 m
Time (t) = 1 min = 60 s
Acceleration due to gravity (g) = 9.8 m/s²
To Find:
Mass of water (m)= ?
Formula:

Water lifted by the pump is 1224.5 kg

b. If the energy of a ball falling from a height of 10 metres is reduced by 40%, how high will it rebound? (Ans : 6 m)
Answer:
Given: Initial height (h₁) = 10m
Let Initial (P.E₁) = 100
Final (P.E₂) = 100 – 40
= 60

To Find:
Final height (h₂) = ?
Formula:
P.E. = mgh
Solution:

The ball will rebound by 6 m.

d. The velocity of a car increase from 54 km/hr to 72 km/hr. How much is the work done if the mass of the car is 1500 kg? (Ans. : 131250 J)
Answer:


Work done to increase the velocity = 131250 J

e. Ravi applied a force of 10 N and moved a book 30 cm in the direction of the force. How much was the work done by Ravi? (Ans: 3 J)
Answer:
Given:
Force (F) = 10 N
θ = 0°, (Since force and displacement are in same direction)
Displacement (s) = 30 cm = 30/100 m
To Find:
Work (W) = ?
Formula:
W = Fs cos θ
Solution:
W = Fs cos θ
Solution:
The work done by Ravi is 3J

Numericals For Practice

Class 9 Science Chapter 1 Laws of Motion Intext Questions and Answers

Question 1.
What are different types of forces? Give examples.
Answer:
Forces are of two types.

• Contact force e.g.: Mechanical force, frictional force, muscular force
• Non-contact force e.g.: gravitational force, magnetic force, electrostatic force

Question 2.
Monashee wants to displace a wooden block from point A to point B along the surface of a table as shown. She has used force F for the purpose.

(a) Has all the energy she spent been used to produce an acceleration in the block?
(b) Which forces have been overcome using that energy?
Answer:
(a) Only part of the energy applied by Minakshee is used in accelerating the block.
(b) Force of friction has been overcome using the energy.

Question 3.
Mention the type of energy used in the following examples.
(i) Stretched rubber string.
(ii) Fast-moving car.
(iii) The whistling of a cooker due to steam.
(iv) A fan running on electricity.
(v) Drawing out pieces of iron from garbage, using a magnet.
(vi) Breaking of a glass window pane because of a loud noise.
(vii) The drackers exploded in Diwali.
Answer:
(i) Potential energy
(ii) Kinetic energy
(iii) Sound energy
(iv) Electrical energy
(v) Magnetic energy
(vi) Sound energy
(vii) Sound energy, light energy and heat energy

Question 4.
Study the pictures given below and answer the questions:

(a) In which of the pictures above has work been done?
(b) From scientific point of view, when do we say that no work was done?
Answer:
(a) Girl studying : No work done
Boy playing with ball: Work is done
Girl watching T.V.: No work done Person lifting sack of grains : Work is done
(b) No work is said to be done when force is applied but there is no displacement.

Question 5.
Make two pendulums of the same length with the help of thread and two nuts. Tie another thread in the horizontal position.

Tie the two pendulums to the horizontal thread in such a way that they will not hit each other while swinging. Now swing one of the pendulums and observe. What do you see?

Answer:
You will see that as the speed of oscillation of the pendulum slowly decreases, the second pendulum which was initially stationary, begins to swing. Thus, one pendulum transfers its energy to the other.

Question 6.
Ajay and Atul have been asked to determine the potential energy of a ball of mass m kept on a table as shown in the figure. What answers will they get? Will they be different? What do you conclude from this?

Answer:

• According to Ajay P.E₁ = mgh₁ and according to Atul P.E₂ = mgh₂.
• Yes, the answer will be different as the two heights are different.
• Potential energy is relative.

Question 7.
Discuss the directions of force and of displacement in each of the following cases.
(i) Pushing a stalled vehicle.
(ii) Catching the ball which your friend has thrown towards you.
(iii) Tying a stone to one end of a string and swinging it round and round by the other end of the string.
(iv) Walking up and down a staircase; climbing a tree.
(v) Stopping a moving car by applying brakes.
Answer:
(i) Force and displacement are in the same direction.
(ii) Force and displacement are in the opposite direction.
(iii) Force and displacement are perpendicular to each other.
(iv) Force and displacement are in the opposite direction.
(v) Force and displacement are in the opposite direction.

Question 8.
(A) An arrow is released from a stretched bow.
(B) Water kept at a high flows through a pipe into the tap below.
(C) A compressed spring is released.
(a) Which words describe the state of the object in the above examples?
(b) Where did the energy required to cause the motion of the objects come from?
(c) If the obj ects were not brought in those states, would they have moved?
Answer:
(a) Words such as stretched bow, water kept at a height and compressed spring describe the state of the objects.
(b) The energy required for the objects came from its specific state or motion in the form of potential energy.
(c) No, if the objects were not brought in those states, they would have not moved.

Question 9.
Study the activity and answer the following questions.

(a) Figure A – Why does the cup get pulled?
(b) Figure B – What is the relation between the displacement of the cup and the force applied through the ruler?
(c) In Figure C-Why doesn’t the cup get displaced?
(d) What is the type of work done in figures A, B and C?
(e) In the three actions above, what is the relationship between the applied force and the displacement?
Answer:
(a) The cup gets pulled as the force of the nut and the displacement of the cup is in the same direction.
(b) The displacement of the cup and the force applied through the ruler is in the opposite direction.
(c) Tire cup does not get displaced as two equal forces are working in opposite directions.
(d) The work done in figure A is positive, figure B is negative and in figure C is zero.
(e) In figure A the applied force and the displacement is in the same direction, in figure B the applied force and the displacement is in the opposite direction and in figure C the applied force and displacement is perpendicular to each other.

Question 10.
From the following activities find out whether work is positive, negative or zero. Give reasons for your answers.
(a) A boy is swimming in a pond.
(b) A coolie is standing with a load on his head.
(c) Stopping a moving car by applying brakes.
(d) Catching the ball which you friend has thrown towards you.
Answer:
(a) A boy is swimming in a pond: The work done is positive because the direction of applied force and displacement are the same.
(b) A coolie is standing with a load on his head: The work done is zero because the applied force does not cause any displacement.
(c) Stopping a moving car by applying brakes: The work done is negative because the fore applied by the brakes acts in a direction opposite to the direction of motion of car.
(d) Catching the ball which you friend has thrown towards you : Negative work because the force required to stop the ball, acts opposite to the displacement of the ball.

Question 11.
(a) Can your father climb stairs as fast as you can?
(b) Will you fill the overhead water tank with the help of a bucket or an electrical motor?
(c) Suppose Raj ashree, Yash and Ranjeet have to reach the top of a small hill. Raj ashree went by car. Yash went cycling while Ranjeet went walking. If all of them choose the same path, who will reach first and who will reach last? (Think before you answer.
Answer:
(a) No, father takes more time to climb stairs.
(b) Overhead water tank can be filled with the help of one electric motor rather than filling it with bucket.
(c) Raj ashree will reach first, followed by Yash and Ranjeet will reach last because car moves faster than a cycle and a person walking.

Class 9 Science Chapter 1 Laws of Motion Additional Important Questions and Answers

1. Choose and write the correct option:

Question 1.
Forces are of …………………… types.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 2.
Example of Contact force is ………………….. .
(a) Gravitational Force
(b) Magnetic Force
(c) Electrostatic Force
(d) Muscular Force
Answer:
(d) Muscular Force

Question 3.
Example of Non-contact force is ………………….. .
(a) Mechanical Force
(b) Frictional Force
(c) Muscular Force
(d) Electrostatic Force
Answer:
(d) Electrostatic force

Question 4.
Work is said to be done on a body when a …………………… is applied on object causes displacement of the object.
(a) Direction
(b) Area
(c) Volume
(d) Force
Answer:
(d) force

Question 5.
W = ………………. .
(a) mgh
(b) mdh
(c) mv²
(d) mfe
Answer:
(a) mgh

Question 6.
The energy stored in the dry cell is in of ………………. energy.
(a) Light
(b) Chemical
(c) Solar
(d) Kinetic
Answer:
(b) chemical

Question 7.
The work done is zero if there is no ……………… .
(a) Direction
(b) Displacement
(c) Mass
(d) Angle
Answer:
(b) displacement

Question 8.
Flowing water has ………………. energy.
(a) Potential
(b) Chemical
(c) Solar
(d) Kinetic
Answer:
(d) kinetic

Question 9.
By stretching the rubber strings of a we store ………………. energy in it.
(a) Potential
(b) Chemical
(c) Electric
(d) Kinetic
Answer:
(a) potential

Question 10.
………………. is the unit of force.
(a) Both B and C
(b) Newton
(c) Dyne
(d) Volts
Answer:
(a) Both B and C

Question 11.
For a freely falling body, kinetic energy is ………………. at the ground level.
(a) Maximum
(b) Minimum
(c) Neutral
(d) Reversed
Answer:
(a) Maximum

Question 12.
Energy can neither be ………………. nor ……………… .
(a) Destroyed
(b) Created
(c) Saved
(d) Both A and B
Answer:
(d) Both A and B

Question 13.
Work and …………………… have the same unit.
(a) Energy
(b) Electricity
(c) Force
(d) Both B and C
Answer:
(a) Energy

Question 14.
S.I. unit of energy is ………………….. .
(a) Joule
(b) Ergs
(c) m/s²
(d) Both A and B
Answer:
(a) Joule

Question 15.
Work is the product of ………………….. .
(a) force and distance
(b) displacement and velocity
(c) kinetic and potential energy
(d) force and displacement
Answer:
(d) force and displacement

Question 16.
S.I. unit of work is ………………….. .
(a) dyne
(b) newton-meter or erg
(c) N/m2 or joule
(d) newton-meter or joule
Answer:
(d) newton-meter or joule

Question 17.
…………………… is the capacity to do work.
(a) Energy
(b) Force
(c) Power
(d) Momentum
Answer:
(a) Energy

Question 18.
Kinetic energy of a body (KE) = ………………….. .
(a) mv²
(b) ¹/₂ mv²
(c) mgh
(d) Fs
Answer:
(b) ¹/₂ mv²

Question 19.
Potential energy of a body is given by (P.E.) = ………………….. .
(a) Fs
(b) mgh
(c) ma
(d) mv²
Answer:
(b) mgh

Question 20.
1 hp = ………………….. .
(a) 476 watts
(b) 746 watts
(c) 674 watts
(d) 764 watts
Answer:
(b) 746 watts

Question 21.
…………………… is the commercial unit of power.
(a) kilowatt second
(b) dyne
(c) kilowatt
(d) erg
Answer:
(c) kilowatt

Question 22.
1 kWh = …………………… joules.
(a) 3.6 x 10³
(b) 3.6 x 10⁶
(c) 6.3 x 10⁶
(d) 6.3 x 10³
Answer:
(b) 3.6 x 10⁶

Based on Practicals

Question 23.
The work done by a force is said to be …………………… when the applied force does not produce displacement.
(a) positive
(b) negative
(c) zero
(d) none of these
Answer:
(c) zero

Question 24.
When some unstable atoms break up, they release a tremendous amount of …………………… energy.
(a) chemical
(b) potential
(c) nuclear
(d) mechanical
Answer:
(c) nuclear.

Name the following:

Question 1.
Unit of energy used for commercial purpose.
Answer:
Kilowatt-hour kW h is the unit of energy used for commercial purpose.

Question 2.
Unit used in industry to measure power.
Answer:
Horse power (hp) is the unit used in industry to express power.

Question 3.
SI unit of energy.
Answer:
SI unit of energy is Joule (J).

Question 4.
Two types of mechanical energy.
Answer:
Potential energy and kinetic energy are the two types of mechanical energy.

Question 5.
An example where force acting on an object does not do any work.
Answer:
In a simple pendulum, the gravitational force acting on the bob does not do any work as there is no displacement in the direction of force.

Question 6.
The relationship between 1 joule and 1 erg.
Answer:
1 joule = 107 erg.

Question 7.
Various forms of energy
Answer:
The various forms of energy are mechanical, heat, light, sound, electro-magnetic, chemical, nuclear and solar.

State whether the following statements are true or false:

(1) The potential energy of a body of mass 1 kg kept at height 1 m is 1 J.
(2) Water stored at some height has potential energy.
(3) Unit of power is joule.
(4) Mechanical energy can be converted into electrical energy.
(5) Work is a vector quantity.
(6) Power is a scalar quantity.
(7) The kilowatt hour is the unit of energy.
(8) The CGS unit of energy is dyne.
(9) The SI unit of work is newton.
(10) Kinetic energy has formula – mv²
Answer:
(1) False
(2) True
(3) False
(4) True
(5) False
(6) True
(7) True
(8) False
(9) False
(10) True

Find the odd man out.

Question 1.
Work, Energy, Power, Force.
Answer:
Force.

Question 2.
A stretched spring, A body placed in at some height, A bullet fired from gun.
Answer:
A bullet fired from gun.

Question 3.
A stretched spring, A rock rolling downhill, A bullet fired from gun.
Answer:
A stretched spring.

Write the formula of the following.

Question 1.
Kinetic energy
Answer:
\(\frac{1}{2}\)mv²

Question 2.
Potential energy
Answer:
mgh

Question 3.
Work
Answer:
Fs or Fs cosθ

Question 4.
Force
Answer:
ma

Question 5.
Power
Answer:
\(\frac{w}{1}\)

One line answer.

Question 1.
(i) When is work done said to be zero?
Answer:
Work done is zero when force acting on the body and its displacement are perpendicular to each other.

(ii) Which quantities are measured in ergs?
Answer:
Work and energy are measured in ergs.

(iii) What is the relationship between newton, meter and joule?
Answer:
1 joule = 1 newton x 1 meter

(iv) What is energy?
Answer:
The ability of a body to do work is called energy.

(v) Give 4 examples of energy
Answer:
Solar, wind, mechanical and heat.

(vi) Which device converts electrical energy into heat?
Answer:
Electric water heater (Geyser) converts electrical energy into heat.

(vii) What is the relationship between second, horsepower and joule?
Answer:
1 horse power = \(\frac{746 \text { joules }}{1 \text { second }\)

Question 2.
Find whether work is positive, negative or zero.

(a) Person moving along circle from A to B.
Answer:
Work done is positive as direction of applied force and displacement are the same.

(b) Person completing one circle and returns to position A.
Answer:
Work done is zero because there is no displacement for the person.

(c) Person pushing a car in the forward direction.
Ans,
Work done is positive as the motion of car is in the direction of the applied force.

(d) A car coming downhill even after pushing it in the opposite uphill direction.
Ans,
Work done is negative as the motion of car is in opposite direction of the applied force.

(e) Motion of the clock pendulum.
Answer:
work done is zero as there is no displacement of the pendulum and it comes back to its original position.

Give Scientific reasons:

Question 1.
A moving ball hits a stationary ball and displaces it.
Answer:

• The moving ball has certain energy.
• When it hits the stationary ball, the energy is transferred to the stationary ball, because of which it moves.
• Hence, a moving ball hits a stationary ball and displaces it.

Question 2.
Flowing water from some height can rotate turbine.
Answer:

• Flowing water has certain energy.
• When it hits the turbine, energy is transferred to the turbine, because of which it rotates.
• Hence, flowing water from some height can rotate a turbine.

Question 3.
A stretched rubber band when released regains its original length.
Answer:

• When we stretch a rubber band we give energy to it.
• This energy is stored in it.
• Hence, when we release it, it regains its original length.

Question 4.
Wind can move the blades of a windmill.
Answer:

• Wind has certain energy.
• When it hits the windmill energy is transferred to the windmill because of which it moves.
• Hence, wind can move the blades of a wind mill.

Question 5.
An exploding firecracker lights up as well as makes a sound.
Answer:

• The exploding firecracker converts the chemical energy stored in it into light and sound respectively.
• Here, energy is converted from one type to another.
• Hence, an exploding firecracker lights as well as makes a sound.

Question 6.
Work done on an artificial satellite by gravity is zero while moving around the earth.
Answer:

• When the artificial satellite moves around the earth in a circular orbit, gravitation force acts on it.
• The gravitational force acting on the satellite and its displacement are perpendicular to each other. i.e. 0 = 90°
• For 0 = 90°, work done is zero. [ v cos 90 = 0)
• Hence, work done on an artificial satellite by gravity is zero while moving around the earth.

Difference between :

Question 1.
Work and Power:
Answer:

Work	Power
(i) Work is the product of force and displacement. (ii) Work is given by the formula : W = Fs (iii) MKS unit – joule, CGS unit-erg	(i) Power is the rate of doing work. (ii) Power is given by the formula : \(\mathrm{P}=\frac{\mathrm{W}}{\mathrm{t}}\) (iii) MKS unit – joule/sec, CGS unit – erg/sec

Question 2.
Work and Energy:
Answer:

Work	Energy
(i) It is the product of the magnitude of the force acting on the body and the displacement of the body in the direction of the force. (ii) It is the effect of energy.	(i) It is the capacity to do work. (ii) It is the cause of work.

Solve the following:

Type – A

Formula:
W = Fs cosθ
If force and displacement are in same direction, then θ = 0°, and cos θ = 1
If force and displacement are in opposite direction, then θ = 180°, and cos θ = -1
If force and displacement are perpendiculars, then θ = 90°, and cos θ = 0

Question 1.
Pravin has applied a force of 100 N on an object, at an angle of 60° to the horizontal. The object gets displaced in the horizontal direction and 400 J work is done. What is the displacement of the object? (cos 600 =12)
To Find:
Displacement (s) = ?
Formula:
W = Fs cos θ
Solution:

The object will be displaced through 8 m.

Question 2.
A force of 50 N acts on an object and displaces it by 2 m. If the force acts at an angle of 60° to the direction of its displacement, find the work done.
Answer:
50 J

Question 3.
Raj applied a force of 20 N and moved a book 40 cm in the direction of the force. How much was the work done by Raj?
Answer:
8J

Type -B

Formula:
1) W = K.E = 1/2 mv2
2) W = P.E = mgh
• W = P.E, W = K.E
1 km/hr =
\(\frac{1000}{3600} \mathrm{~m} / \mathrm{s}=\frac{5}{18} \mathrm{~m} / \mathrm{s}\)

Question 4.
A stone having a mass of 250 gm is falling from a height. How much kinetic energy does it have at the moment when its velocity is 2 m/s?
Answer:


The kinetic energy of the stone is 0.5 J

Question 5.
500 kg water is stored in the overhead tank of a 10 m high building. Calculate the amount of potential energy stored in the water.
Answer:
Given:
Mass (m) = 500 kg
Height (h) = 10 m
Acceleration due to gravity (g) = 9.8 m/s²
To Find:
Potential energy (P.E) = ?
Formula:
P.E = mgh
Solution:
P.E = mgh
= 500 x 9.8 x 10
= 500 x 98
= 49000J
The P.E of the stored water is 49000 J

Question 6.
Calculate the work done to take an object of mass 20 kg to a height of 10 m. (g = 9.8 m/s²)
Answer:
Given:
Mass (m) = 20 kg
Acceleration due to gravity (g) = -9.8 m/s²
Displacement (s) = (h) = 10 m.
To Find:
Work done (W) = ?
Formula:
(i) W = P.E = mgh
Solution:
W = mgh
= 20 x (-9.8) x 10
= -1960J
The work done to take an object of mass 20 kg to a height of 10 m is -1960 J.

Question 7.
A body of 0.5 kg thrown upwards reaches a maximum height of 5 m. Calculate the work done by the force of gravity during this vertical displacement.
Answer:
Given:
Mass (m) = 0.5 kg
Acceleration due to gravity (g) = -9.8 m/s²
Displacement (s) = 5 m.
To Find:
Work done (W) = ?
Formula:
W = P.E = mgh
Solution:
W = mgh
= 0.5 x (-9.8) x 5
= -24.5 J
The work done by the force of gravity is -24.5 joule.

Question 8.
1 kg mass has a kinetic energy of 2 joule. Calculate its velocity.
Answer:
Given:
Mass (m) = 1 kg
Kinetic Energy (K.E) = 2 J
To Find:
Velocity (v) = ?
Formula:

The velocity is 2 m/s

Question 9.
A rocket of mass 100 tonnes is propelled with a vertical velocity 1 km/s. Calculate kinetic energy.
Answer:

The kinetic energy of the rocket is 5 x 10¹⁰ J

Type – C

Formula:
\(\text { 1) Power }=\frac{\text { work }}{\text { time }}=\frac{\text { mgh }}{t}\)

Power should be expressed in kW
Time should be expressed in hours
1 k Wh = 1 unit

Question 10.
Swaralee takes 20 s to carry a bag weighing 20 kg to a height of 5 m. How much power has she used?
Given:
Mass (m) = 20 kg
Height (h) = 5 m
Time (t) = 20s
Acceleration due to gravity (g) = 9.8 m/s²
To Find:
Power (P) = ?
Formula:
\(\mathrm{P}=\frac{\mathrm{mgh}}{\mathrm{t}}\)
Solution:
\(\begin{aligned}
P &=\frac{m g h}{t} \\
&=20 \times 9.8 \times \frac{5}{20} \\
&=9.8 \times 5
\end{aligned}\)
= 49 W
Power used by Swaralee is 49 W

Write notes on the following:

Question 1.
Derive the expression for potential energy.
Answer:
(i) To carry an object of mass ‘m’ to a height ‘h’ above the earth’s surface, a force equal to ‘mg’ has to be used against the direction of the gravitational force.

(ii) The amount of work done can be calculated as follows:
Work = force x displacement
∴ W = mg x h
∴ W = mgh

(iii) The amount of potential energy stored in the object because of its displacement.
PE = mgh (W = P.E)

(iv) Displacement to height h causes energy equal to mgh to be stored in the object.

Question 2.
When can you say that the work done is either positive, negative or zero?
Answer:

• When the force and the displacement are in the same direction, the work done by the force is positive.
• When the force and displacement are in the opposite directions, the work done by the force is negative.
• When the applied force does not cause any displacement or when the force and the displacement are perpendicular to each other, the work done by the force is zero.

Question 3.
Explain the relation between, the commercial and SI unit of energy.
Answer:
The commercial unit of energy is a kilowatt-hour (kWh) while the SI unit of energy is the joule. Their relation is
1 kWh = 1kW x 1hr
= 1000 Wx 3600 s
= 3600000J
(Watt x Sec = Joule)
1 kWh = 3.6 x 106 J.

Question 4.
How is work calculated if the direction of force and the displacement are inclined to each other?
Answer:
If the direction of force and the displacement are inclined to each other then, we must convert the applied force into the force acting along the direction of displacement.

If θ is angle between force and displacement, then force (F₁) in direction of displacement is

Complete the flow chart.

Question 1.
Transformation of energy

Answer:

Question 2.
Transformation of energy

Answer:

Write effects of the following with examples.

Question 1.
Force
Answer:

• A force can move a stationary object. The force of engine makes a stationery car to move.
• A force can stop a moving object. The force of brakes can stop a moving car.
• A force can change the speed of a moving object. When a hockey player hits a moving ball, the speed of ball increases.
• A force can change the direction of a moving object. In the game of carrom ,when we take a rebound then the direction of striker changes because the edge of the carrom board exerts a force on the strike.
• A force can change the shape and size of an object. The shape of kneaded wet clay changes when a potter converts it into pots of different shapes and sizes because the p otter applies force on the kneaded wet clay.

Give two examples in each of the following cases:

Question 1.
Potential energy
Answer:

• Water stored in a dam
• A compressed spring

Question 2.
Kinetic energy
Answer:

• Water flowing
• Bullet fired from a gun

Question 3.
Chemical energy
Answer:

• Chemical in cell
• Explosive mixture of a bomb

Question 4.
Zero work done
Answer:

• A stone tied to a string and whirled in a circular path
• Motion of the earth and other planets moving around the sun

Question 5.
Negative work done
Answer:

• A cyclist applies brakes to his bicycle, but the bicycle still covers some distance.
• When a body is made to slide on a rough surface, the work done by the frictional force.

Question 6.
Positive work done
Answer:
(i) A boy moving from the ground floor to the first floor.
(ii) A fruit falling down from the tree.
= 0.5 hr x 30 days
= 15 hrs
To Find:
Energy consumed = ?

The units of energy consumed in the month of April by the iron is 18 units.

Question 7.
A 25 W electric bulb is used for 10 hours every day. How much electricity does it consume each day?
Answer:
Given:
Power (P) = 25 W
25/1000 kW
Time (E) = 10 hrs
To Find:
Electric energy consumed = ?
Formula:
Electric energy consumed = power x time
Solutions:
Electric energy consumed = power x time
= 25/1000 x 10
= 0.25 kWh
The electric bulb consumes 0.25 kWh of electricity each day.

Question 8.
If a TV of rating 100W is operated for 6 hrs per day, find the amount of energy consumed in any leap year?
Answer:

= 2196 hrs.
To Find:
Electric energy consumed
Formula:
Electric energy consumed = power x time
Solution:
Electric energy consumed = power x time
= 0.1 x 2196
= 219.6 kWh
The amount of energy consumed is 219.6 kWh

Complete the paragraph.

Question 1.
………….. is the measure of energy transfer when a force (F) moves an object through a ………….. (d). So when ………….. is done, energy has been transferred from one energy store to another, and so: energy transferred = ………….. done. Energy transferred and work done are both measured in ………….. (J)
Answer:
Work is the measure of energy transfer when a force (F) moves an object through a distance (d). So when work is done, energy has been transferred from one energy store to another, and so: energy transferred = work done. Energy transferred and work done are both measured in joules (J).

Question 2.
………….. energy and ………….. done are the same thing as much as ………….. energy and work done are the same thing. Potential energy is a state of the system, a way of ………….. energy as of virtue of its configuration or motion, while ………….. done in most cases is a way of channeling this energy from one body to another.
Answer:
Potential energy and work done are the same thing as much as kinetic energy and work done are the same thing. Potential energy is a state of the system, a way of storing energy as of virtue of its configuration or motion, while work done in most cases is a way of channeling this energy from one body to another.

Question 3.
In physics, ………….. is the rate of doing work or, i.e., the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the ………….. equal to one ………….. per second.

Power is a ………….. quantity that requires both a change in the physical system and a specified time interval in which the change occurs. But more ………….. is needed when the work is done in a shorter amount of time.
Answer:
In physics, power is the rate of doing work or, i.e., the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt. equal to one joule per second.

Power is a scalar quantity that requires both a change in the physical system and a specified time interval in which the change occurs. But more power is needed when the work is done in a shorter amount of time.

Activity-based questions

Answer in detail:

Question 1.
State the expression for work done when displacement and force makes an angle θ OR State the expression for work done when force is applied making an angle θ with the horizontal force.

Answer:
Let ‘F’ be the applied force and Fj be its component in the direction of displacement. Let ’S’ be the displacement.

The amount of work done is given by W = F₁s ……………………………………… (1)
The force ‘F’ is applied in the direction of the string.

Let ‘θ’ be the angle that the string makes with the horizontal. We can determine the component ‘F₁‘, of this force F, which acts in the horizontal direction by means of trigonometry.
\(\begin{aligned}
\cos \theta=\frac{\text { base }}{\text { hypotenuse }} \\
\therefore \quad \cos \theta=\frac{\mathrm{F}_{1}}{\mathrm{~F}} \\
\therefore \quad \mathrm{F}_{1}=\mathrm{F} & \cos \theta
\end{aligned}\)
Substituting the value of F₁ in equation 1
Thus, the work done by F₁ is
W cos θ s
∴ W = Fscosθ

Question 2.
When a body is dropped on the ground from some height its P.E is converted into K.E but when it strikes the ground and it stops, what happens to the K.E?
Answer:
When a body is dropped on the ground, its K.E appears in the form of:

• Heat (collision between the body and the ground).
• Sound (collision of the body with the ground).
• The potential energy of change in state of the body and the ground.
• Kinetic energy is also utilized to do work i.e., the ball bounces to a certain height and moves to a certain distance vertically and horizontally till Kinetic energy becomes zero.
• The process in which the kinetic energy of a freely falling body is lost in an unproductive chain of energy is called the dissipation of energy.

Question 3.
Explain the statement “Potential Energy is relative”.
Answer:

• The potential energy of an object is determined and calculated according to a height of the object with respect to the observer.
• So, the person staying on 6th floor more potential energy than those staying on the 3rd floor.
• But, the person on the 6th floor will have lesser potential energy than on the 8th floor. Hence potential energy is relative.

 

Balbharti Maharashtra State Board Class 9 Geography Solutions Chapter 11 Transport and Communication Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Geography Solutions Chapter 11 Transport and Communication

Class 9 Geography Chapter 11 Transport and Communication Textbook Questions and Answers

1. Differentiate between:

(A) Railways and roadways
Answer:

Basis	Railways	Roadways
(1) Carrying Capacity	Carrying capacity is more	Carrying capacity is limited
(2) Distance	Suitable for long distances	Suitable for short distance
(3) Door to door service	Railway does not provide door to door service.	Roadways provide door to door service.
(4) Traffic	There is no problem of traffic jam on railways.	There is a problem of traffic jam on roadways.
(5) Pollution	Railways do not create a problem of air pollution.	Roadways create a problem of air pollution.

(B) Transportation and communication
Answer:

Basis	Transportation	Communication
(1) Meaning	Transportation is the movement of humans, animals and goods from one location to another.	Communication is the exchange of information, ideas and messages by speaking, writing or some other medium.
(2) Means	It is done through railways, roadways, waterways, airways and pipelines.	It is done through telephones, mobiles, video-conferencing, email and post etc.
(3) Threats	The threats like a traffic jams, accidents, noise pollution, air pollution are associated with transportation.	The threats like technical issues, cyber crimes, etc. are associated with the communication.

(C) Conventional and modern means of communication.
Answer:

Basis	Conventional means of Communication	Modem means of Communication
(1) Meaning	The means of communication – used since olden times – conventional means of communication.	The means of communication – used in modern times – modern means of communication.
(2) Examples	Letter, newspapers, radio, television.	Mobile phone, internet, etc.
(3) Interaction	May not facilitate the direct interaction between – sender and receivers of information.	Facilitates the direct interaction between – sender and the receivers of information.

2. Answer in detail:

Question 1.
‘Newspapers are used for communication’. Explain the statement.
Answer:

• The news related to economic events, politics, social issues, culture, education, etc. are published in newspapers.
• Newspapers arecheap means of communication. Through newspapers information gets spread to masses at a time.
• Newspapers are published in various languages.
• In this way, newspapers are used for communication.

Thus, newspapers are used for communication.

Question 2.
Explain how T.V. is a cheap means of communication.
Answer:

• Various programs, serials, etc. related to entertainment, social issues, culture, education, politics, economic events, sports, weather conditions, etc. are broadcasted on television.
• Through television the information is exchanged to masses at a time with high speed.
• Television can broadcast both audio and video for communication. In this way, television is a cheap means of communication.

Thus, TV is a cheap means of communication.

Question 3.
What types of communications can be done through mobiles?
Answer:

• Calling and SMS (Short message system) facilitates easy one-to-one communication, using mobiles anywhere and at any time.
• Video conferencing and applications like ‘Whatsapp’ allow one to communicate with many people simultaneously.
• Various functions like, money transfer, payment of bills, purchase of goods and services and online trading can be done using smart phones’ various apps like BHIM, SBI anywhere, etc.
• Internet and social media can also be accessed through mobile phones.

3. Name them on the basis of the given information:

Question 1.
Five cities with airways services in Maharashtra
Answer:
Mumbai, Pune, Nagpur, Kolhapur, Aurangabad, Nashik and Nanded.

Question 2.
Services available in post offices
Answer:

• Financial Services: Saving schemes, insurance services and mutual fund.
• Mail services: Speed post, postcard, parcel and courier.

Question 3.
National Highways near your area
Answer:

• Mumbai- Goa Highway (NH 66)
• Mumbai- Bangaluru highway (NH 04)
• Mumbai- Agra Highway (NH 08)
  Note: Answer may vary.

Question 4.
Ports along the coast of Maharashtra
Answer:

• Malvan
• Venture
• Vasai
• Dahanu
• Gharapuri

4. Identify the relation and match the columns making a chain

Group ‘A’	Group ‘B’	Group ‘C’
(1) Postal services	Roadways	Speed post
(2) Shivneri	World network of connected computers	Exchange of information
(3) Internet	Conventional means of communication	Comfortable journey
(4) RoRo transport	Railways	Energy, time and labour saving

Answer:

Group ‘A’	Group ‘B’	Group ‘C’
(1) Postal services	Conventional means of communication	Speed post.
(2) Shivneri	Roadways	Comfortable journey.
(3) Internet	World network of connected computers	Exchange of information.
(4) RoRo transport	Railways	Energy, time and labour saving.

5. Read the following maps and answer the questions:

Question 1.
In which region do you find a dense network of transport routes in the map?
Answer:
The central part of the district has dense network of transport routes.

Question 2.
How is the physiography of the region with dense network?
Answer:
The central part of the district has lower and medium elevation as compared to the Western part.

Question 3.
Which region has a sparse network of transport routes?
Answer:
The transport network is sparse in the Eastern part of the district.

Question 4.
How is the physiography of this region?
Answer:
The region with sparse network of transport . routes is comparatively of lower and medium elevations.

Question 5.
Look for the region lacking transport routes.
Answer:
The Western region lacks transport routes.

Question 6.
What kind of obstruction can you find there?
Answer:
Sahyadri Mountains and Shivsagar reservoir of Koyna dam are the obstructions found here.

Class 9 Geography Chapter 11 Transport and Communication Intext Questions and Answers

Let’s Recall

Complete the following table:

Transport	Means of	Used for
Route	Transport
Roadways	Rickshaw	Passengers
Roadways	Trucks
	Metro
Waterways
	Helicopter
Airways
	Submarine
Waterways		Freight
	Mules
Railways
Pipelines

Answer:

Transport	Means of Transport	Used for
Roadways	Rickshaw	Passengers
Roadways	Trucks	Goods
Railways	Metro	Passengers
Waterways	Cruise/Boats	Passengers
Airways	Helicopter	Passengers
Airways	Aeroplane	Passengers
Waterways	Submarine	Defence & Research
Waterways	Cargo-ship	Freight / Goods
Roadways	Mules	Goods
Railways	Cargo goods train	Goods
Pipelines	Pipes	Oil, Water and Gas

Can you tell?

We have given some specific conditions. In this context, tell with reasons which means of transport route will you take?

Question 1.
You have to reach Bhopal from Nagpur due to some emergency.
Answer:
Airways : As it is the fastest mode of transport.

Question 2.
You have to reach Kanyakumari carrying the message of cleanliness. There is no time limit for it.
Answer:
Roadways : Since roadways connect even the remotest places the message can reach to all persons.

Question 3.
Send the Alphonso mangoes from Konkan to Arab countries.
Answer:
Airways – Since mango is a perishable commodity, the fastest mode of transportation is used. ,

Question 4.
Indrayani variety of rice has to be exported from Pune to Cape Town of South Africa at low expenditure.
Answer:
In such conditions, we will choose waterways as a route and ship as a means of transport. Because rice is comparatively durable agricultural good in the given situation it has to be transported at low cost.

Question 5.
Large-scale production of vegetables in Nandurbar has taken place but is not fetching a good price. The Nagpur-Surat National Highway and the Surat-Bhusawal Railway line passes through the district.
Answer:
In a given situation, we will choose roadways and railways as a route and truck and goods’ ways/train respectively as a means of transport. Because the highway and railway line that pass through the district, connects the important towns and cities from the district.

Question 6.
You have to go to Singapore from your village/ town. You have 10 days to do the same.
Answer:
Roadway and Airways: I will first reach the nearest airport of a city by road and then take an Airway. It is the fastest mode of transport.

Can you tell?

Question 1.
Make a list of various means of communication you are aware of.
Answer:
Letters, radio, television, telephones, mobile phones, newspaper, internet, satellites etc. are the various means of communication.

Question 2.
How many of these do you actually use? Make a box around them.
Answer:

Question 3.
For what do you use them ?
Answer:
We use these means of communication to exchange important information, ideas, opinions etc. with friends, parents, relatives and teachers.

Question 4.
Who uses the remaining means ?
Answer:
The remaining means are used by parents, other relatives, businessmen from locality and government agencies.

Can you do it?

Observe the image and the instructions given on page 87 of the textbook and answer the following questions:

Question 1.
Which are the dates mentioned in the image?
Answer:
The dates mentioned are 15/5/2017 and 19/5/2017

Question 2.
What does the information in the image tell?
Answer:
The image informs that the person’s email account has been hacked. He cannot access his important files as they have been encrypted and to recover his files, he will have to pay a certain amount to the hacker.

Question 3.
What is the price asked for recovering the files and in what currency?
Answer:
The price asked for recovering the files is 300 US dollars in bitcoins.

Question 4.
What is the type of crime here?
Answer:
This is a cyber crime.

Give it a try
Think about the transport issues you come across during your journeys. Write the innovative changes you would suggest in the transport routes or means in your copy.

Question 1.
Congested city roads and Pollution
Answer:
Electric cars which are smaller and smarter.

Question 2.
Time-consuming travel
Answer:
Dedicated bus corridors, carpooling, more number of Expressways.

Give it a try

Question 1.
Look for the other uses of artificial satellites? Try to understand how they are related to your daily life?
Answer:
(a) The other uses of artificial satellites are as follows:

• Studying about other planets.
• Live broadcasting of a program/event from any region of the earth.
• Studying the resources on the earth’s surface.
• Regional planning.
• Planning defense strategies.
• Forecasting weather etc.

(b) The artificial satellites are directly or indirectly related to personal, social, educational, economic, cultural, political aspects in everyone’s daily life. For eg. through artificial satellites, one can enjoy a live program like award functions/cricket match, etc. on television.

Class 9 Geography Chapter 11 Transport and Communication Additional Important Questions and Answers

Complete the statement choosing the correct option from the bracket:

Question 1.
The price of the goods can be kept low if the transportation is ………….. .
(a) feasible
(b) expensive
(c) costly
(d) affordable
Answer:
(d) affordable

Question 2.
…………… growth gets a boost due to transportation.
(a) Physical
(b) Culture
(c) Economic
(d) Political
Answer:
(c) Economic

Question 3.
Freight transport through …………… is costlier than railways.
(a) trucks
(b) horses
(c) bullock-cart
(d) yak
Answer:
(a) trucks

Question 4.
The western part of Satara district is occupied by the …………… of its off shoots.
(a) Vindhyas
(b) Satpudas
(c) Sahyadris
(d) Aravallis
Answer:
(c) Sahyadris

Question 5.
The use of RORO (Roll on Roll off) services started in …………… railways in India.
(a) Goa
(b) Maharashtra
(c) Konkan
(d) Pune
Answer:
(c) Konkan

Question 6.
In the modem age, man-made …………… are an important and effective means of communication.
(a) planets
(b) asteroids
(c) satellites
(d) rockets
Answer:
(c) satellites

Question 7.
Satellite images obtained by …………… facilitate study of resources on Earth’s surface and help in regional planning.
(a) GPS
(b) radio
(c) remote sensing1
(d) drones
Answer:
(c) remote sensing

Question 8.
Communication is not just limited to talking on telephones or sending messages but …………… is also available now.
(a) tele-calling
(b) STD-Calling
(c) video-calling
(d) Local-calling
Answer:
(c) Video-calling

Question 9.
There is a correlation between transport routes and the …………… of region.
(a) soil type
(b) rainfall
(c) climate
(d) physiography2
Answer:
(d) physiography

Question 10.
Transport facilities can develop well in …………… region.
(a) mountainous
(b) plain
(c) forest
(d) plateau
Answer:
(b) plain

Question 11.
Shiv sagar reservoir of the …………… dam is located in the Satara district.
(a) Ram Krishna
(b) Bhakra-Nagal
(c) Koyna
(d) Tehri
Answer:
(c) Koyna

Question 12.
The …………… part of Satara district has a dense transport network.
(a) Western
(b) Central
(c) Eastern
(d) Southern
Answer:
(b) central.

Write answers in one sentence

Question 1.
What is transportation?
Answer:
The movement of goods and people from one place to another is called transportation.

Question 2.
Which are the different kinds of transport routes?
Answer:
Roadways, railways, waterways, airways and pipelines are the different kinds of transport routes.*

Question 3.
For what reasons does a region become devoid of any transport route?
Answer:
Due to mountains, valleys, rivers, reservoirs

1. remote sensing – obtaining information regarding any place or an object without actually establishing direct contact with it is called remote sensing.
2. physiography – nature and slope of land and undulating1 topography2 a region becomes devoid of any transport route.

Question 4.
Why does a dense network of transportation develop in some regions?
Answer:
A dense network of transportation develops in some regions due to lower and medium elevation, plains, flat and regular topography, etc.

Question 5.
What does Ro-Ro transport stand for?
Answer:
Ro-Ro transport is Roll-on, Roll-off transport.

Question 6.
Why was the Ro-Ro transport introduced?
Answer:
Freight transport by trucks is costlier than railways, so as a solution the Ro-Ro transport has been introduced.

Question 7.
What are Cyber Crimes?
Answer:
Crimes like website/email hacking, theft of information, economic frauds, wars, terrorism, etc. that are committed by using computers and internet are called ‘cyber crimes’.

Question 8.
How are BHIM app, SBI anywhere app useful?
Answer:
BHIM App, SBI Anywhere app, helps us to pay various bills, sell, buy and carry out various transactions through mobile phones.

Question 9.
Where was Ro-Ro service introduced for the first time in India?
Answer:
Ro-Ro service, was introduced for the first time in India by the Konkan Railway.

Give Reasons:

Question 1.
The development of transportation is an indicator of the development of that region.
Answer:

• With development of transportation there is an increase in the movement of freight and passengers of that region.
• Development of transportation develops industries and markets.
• Per capita Income (PCI) and Gross Domestic product (GDP) increases leading to economic growth.
• So it is said that development of transportation is an indicator of the development of that region.

Question 2.
Green Corridor saves many lives.
Answer:

1. Green Corridor is a route cleared of all traffic obstacles, so that a dead person’s (donor’s) organs can be speedily transported to the receiver
2. It is called ‘green’ corridor because the traffic lights are turned green for the speedy movement of the vehicle carrying the organ.
3. Thus, Green Corridor saves many lives.

Question 3.
Ro-Ro Transport helps to reduce cost of transport.
Answer:

• In Ro-Ro (Roll-on, Roll-off) transport, the trucks loaded with goods are transported to desired railway stations through a goods train.
• From there the trucks take the goods ahead to the desired locations.
• Ro-Ro transport helps to reduce the cost of transport as railways are used for the part of the distance.
• Ro-Ro transport also reduces cost of fuel and pollution caused by trucks.

Answer in details:

Question 1.
Give the Importance of transportation.
Answer:
Transportation is a basic infrastructure.
The development of transportation infrastructure is an indicator of the development of the particular region or a country.

The importance of transportation can be explained with the help of the following points :

• Extending trade and network.
• Rapid industrialisation.
• Availability of employment opportunities.
• Regional connectivity.
• Utility of the site.
• Overcoming scarcity (deficit).
• Decreasing regional imbalance1.
• Tourism development.

Question 2.
Give the Importance of a communication system.
Answer:

• Communication or exchange of information is an important process in today’s era. Communication is basic infrastructure.
• Man-made satellites are an important and effective means of communication.
• The exchange of messages through mobiles, watching progammes on television, getting updates of climatic condition etc. is possible simultaneously through man-made satellites.
• Satellite images obtained by remote sensing facilitate study of resources on earth’s surface and helps in regional planning.
• Many apps which can be used on mobile phones have been developed for the same.
• For e.g. BHIM app, SBI anywhere, etc. By using these communicational facilities, we can pay various bills, sell and buy goods and services and carry other such transactions.
• Nowadays, communication is not just limited to talking on telephone or sending messages but also video calling is available now.

Explain:

Question 1.
Factors to be kept in mind while selecting the route way and the means of transport.
Answer:
The following factors should be kept in mind while selecting the route way and means of transport:

• Distance
• Duration
• Cost
• Time
• Products
• Climate
• Market
• Routes and means
• Physiography

Question 2.
Importance of Transportation.
Answer:

1. The development of transportation infrastructure is an indicator of the development of the country or that region.
2. The reforms in the transport sector enhance the dynamicity of freight and passengers in a region.
3. Industries and markets develop. Economic growth gets a boost. Per Capita Income (PCI) and Gross Domestic Product (GDP) increases too.
4. Transportation leads to –
   • Extending trade and network
   • Rapid industrialisation
   • Availability of employment opportunities
   • Regional connectivity
   • Utility of the site
   • Overcoming scarcity (weakness)
   • The decrease in regional imbalance
   • Tourism development

Question 3.
Importance of man-made satellites.
Answer:

• In the modern age, man-made satellites are an important and effective means of communication.
• The exchange of messages through mobiles, watching programmes on TV and getting updated information regarding climatic conditions is possible simultaneously through man-made satellites.
• Satellite images obtained by remote sensing facilitate study of resources on earth’s surface and help in regional planning.

Thus, man-made satellites are important.

Question 4.
Green Corridor.
Answer:

• It happens that sometimes a dead person has donated his organs.
• In such cases, such organs need to be transported from the donor’s location to the receiver urgently.
• For this organ transfer, all types of routes are cleared of all obstacles. This is called Green Corridor.
• Consequently, this kind of rapid transport corridor can save the receiver’s life.

Question 5.
Threats associated with means of communication.
Answer:

• Besides facilities, means of communication have a few threats associated with them.
• Many crimes are happening through the internet like website/email hacking, fraud, theft, attack, wars and terrorism. Possibilities of threats like theft of information, economic frauds, attacking important websites etc. arise.
• Therefore, one should take precautions while using social networks.
• One should not reveal personal information before ensuring safety.
• One should not put any sensitive information or personal information on social networking sites, blogs, etc.