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Shemushi Sanskrit Class 9 Solutions Chapter 12 वाङ्मनःप्राणस्वरूपम्

Class 9 Sanskrit Shemushi Chapter 12 वाङ्मनःप्राणस्वरूपम् Textbook Questions and Answers

अभ्यासः

प्रश्न 1.
एकपदेन उत्तरं लिखत –

(क) अन्नस्य कीदृशः भागः मनः?
(ख) मध्यमानस्य दनः अणिष्ठः भागः किं भवति?
(ग) मनः कीदृशं भवति?
(घ) तेजोमयी का भवति?
(ङ) पाठेऽस्मिन् आरुणिः कम् उपदिशति?
(च) “वत्स! चिरञ्जीव”-इति कः वदति?
(छ) अयं वाहः कस्मात् उपनिषद: संग्रहीत?
उत्तर:
(क) अणिष्ठः
(ख) सर्पिः
(ग) आशितान्न-अणिष्ठः
(घ) वाक्
(ङ) श्वेतकेतुम्
(च) आरुणिः
(छ) छान्दोग्योपनिषदः

प्रश्न 2.
अधोलिखितानां प्रश्नानामुत्तराणि संस्कृतभाषया लिखत –

(क) श्वेतकेतुः सर्वप्रथमम् आरुणिं कस्य स्वरूपस्य विषये पृच्छति?
(ख) आरुणिः प्राणस्वरूपं कथं निरूपयति?
(ग) मानवानां चेतांसि कीदृशानि भवन्ति?
(घ) सर्पिः किं भवति?
(ङ) आरुणेः मतानुसारं मनः कीदृशं भवति?
उत्तर:
(क) श्वेतकेतुः सर्वप्रथमम् आकर्णि मनसः स्वरूपस्य विषये पृच्छति।
(ख) आरुणिः प्राणस्वरूपविषये कथयति ‘पीतानाम् अपां योऽणिष्ठः सः प्राणः’ इति।
(ग) मानवः यादृशम् अन्नादिकं गृह्णाति तादृशम् एव तेषाम् चेतांसि भवन्ति।
(घ) मध्यमानस्य दध्नः योऽणिमा ऊर्ध्वः समदीषति स तत्सर्पिः भवति।
(ङ) आरुणे: मतानुसारं मनः अन्नमयं भवति।

प्रश्न 3.
(अ) ‘अ’ स्तम्भस्य पदानि ‘ब’ स्तम्भेन दत्तैः पदैः सह यथायोग्यं योजयत –

अ – ब

1. मनः – अन्नमयम्
2. प्राण: – तेजोमयी
3. वाक् – आपोमयः

उत्तर:
अ – ब

1. मनः – अन्नमयम्
2. प्राणः – आपोमयः
3. वाक् – तेजोमयी

(आ) अधोलिखितानां पदानां विलोमपदं पाठात् चित्वा लिखत –

(क) गरिष्ठः
(ख) अधः
(ग) एकवारम्
(घ) अनवधीतम्
(ङ) किञ्चित्
उत्तर:
(क) अणिष्ठः
(ख) ऊर्ध्वम्,
(ग) भूयः,
(घ) अवधीतम्,
(ङ) सर्वम्।

प्रश्न 4.
उदाहरणमनुसृत्य निम्नलिखितेषु क्रियापदेषु ‘तुमुन्’ प्रत्ययं योजयित्वा पदनिर्माणं कुरुत –
यथा- प्रच्छ् + तुमुन् – प्रष्टुम्
(क) श्रु + तुमुन् = ………………..
(ख) वन्द् + तुमुन् = ………………..
(ग) पत् + तुमुन् = ………………..
(घ) कृ + तुमुन्। = ………………..
(ङ) वि + ज्ञा + तुमुन् = ………………..
(च) वि + आ + ख्या + तुमुन् = ………………..
उत्तर:
(क) श्रोतुम्,
(ख) वन्दितुम्.
(ग) पठितुम्.
(घ) कर्तुम्,
(छ) विज्ञातुम्,
(च) व्याख्यातुम्।

प्रश्न 5.
(अ) निर्देशानुसारं रिक्तस्थानानि पूरयत –

(क) अहं किञ्चित् प्रष्टुम् ………..। (इच्छ – लट्लकारे)
(ख) मनः अन्नमयं ……… (भू – लट्लकारे)
(ग) सावधानं …………..। (श्रु – लोट्लकारे)
(घ) तेजस्वि नौ अधीतम् ……… (अस् – लोट्लकारे)
(ङ) श्वेतकेतुः आरुणे: शिष्यः ………..। (अस् – लङ्लकारे)
उत्तर:
(क) इच्छामि,
(ख) भवति,
(ग) श्रण.
(घ) अस्तु,
(ङ) आसीत्।

(आ) उदाहरणमनुसृत्य वाक्यानि रचयत –
यथा- अहं स्वदेशं सेवितुम् इच्छामि।

(क) …………. उपदिशामि।
(ख) ……….. प्रणमामि।
(ग) …………. आज्ञापयामि।
(घ) ………….. पृच्छामि।
(ङ) …………. अवगच्छामि।
उत्तर:
(क) अहं शिष्यम् उपदिशामि।।
(ख) अहम् गुरुम् प्रणमामि।
(ग) अहम् शिष्यं पुस्तकम् आनेतुम् आज्ञापयामि।
(घ) अहम् गुरुं प्रश्नं पृच्छामि।
(ङ) अहम् भवतः सङ्क्तम् अवगच्छामि।

प्रश्न 6.
(अ) सन्धिं कुरुत –

(क) अशितल्य + अनस्य = …………….
(ख) इति + अपि + अवधार्यम् = …………….
(ग) का + इयम् = …………….
(घ) नौ + अधीतम् = …………….
(ङ) भवति + इति = …………….
उत्तर:
(क) अशितस्यान्नस्य,
(ख) इत्यप्यवधार्यम्.
(ग) केयम्,
(घ) नावधीतम्,
(ङ) भवतीति।

(आ) स्थूलपदान्यधिकृत्य प्रश्ननिर्माणं कुरुत –

(क) मथ्यमानस्य दनः अणिमा ऊर्ध्व समुदीषति।
(ख) भवता घृतोत्पत्तिरहस्यं व्याख्यातम्।
(ग) आरुणिम् उपगम्य श्वेतकेतुः अभिवादयति।
(घ) श्वेतकेतुः वाग्विषये पृच्छति।
उत्तर:
(क) कस्य दध्नः अणिमा ऊवं समुदीषति?
(ख) केन घृतोत्पत्तिरहस्यं व्याख्यातम्?
(ग) आरुणिम् उपगम्य कः अभिवादयति?
(घ) श्वेतकेतुः कस्यविषये पृच्छति?

प्रश्न 7.
पाठस्य सारांशं पञ्चवाक्यैः लिखत।
उत्तर:
(क) अन्नमयं मनः भवति।।
(ख) आपोमयः प्राणः भवति एवं जलमेव जीवनं भवति।
(ग) तेजोमयी वाक् भवति।।
(घ) अश्यमानस्य तेजसः यः अणिमा, स ऊर्ध्वः समुदीपति, सा खलु वाग्भवति।
(ङ) यादृशमन्नादिकं मानवः ग्रह्णाति तादृशमेव तस्य चित्तादिकं भवति।

Class 9 Sanskrit Shemushi Chapter 12 वाङ्मनःप्राणस्वरूपम् Additional Important Questions and Answers

अतिरिक्त कार्यम्

प्रश्न 1.
निम्नलिखितम् संवाद पठित्वा तदाधारिताना प्रश्नानाम् उत्तराणि लिखत –

1. श्वेतकेतुः – भगवन्! श्वेतकेतुरहं वन्दे।
आरुणिः – वत्स! चिरञ्जीव।।
श्वेतकेतुः – भगवन्! किञ्चित्प्रष्टुमिच्छामि।
आरुणिः – वत्स! किमद्य त्वया प्रष्टव्यमस्ति?
श्वेतकेतुः – भगवन्! ज्ञातुम् इच्छामि यत् किमिदं मनः?
आरुणिः – वत्स! अशितस्यान्नस्य योऽणिष्ठः तन्मनः।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. कः प्रश्न प्रष्टुम् इच्छति?
2. श्वेतकेतुः कस्मात् प्रश्नं पृच्छति?
3. अन्नस्य कीदृशः भागः मनः?

उत्तर:

1. श्वेतकेतुः
2. आरुणे
3. अणिष्ठ:

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. श्वेतकेतुः कम् प्रणामम् करोति?
2. मनः किम् अस्ति?

उत्तर:

1. श्वेतकेतुः आरुणिम् प्रणामम् करोति।
2. अशितस्यान्नस्य योऽणिष्ठः तन्मनः।

(iii) भाषिककार्यम (केवलं प्रश्नत्रयमेव)

1. ‘तत् मनः’ अत्र विशेषणपदं किम्?
2. अहम् वन्दे’ अत्र ‘अहम्’ पदं कस्मै प्रयुक्तम्?
3. “किञ्चित् प्रष्टुम् इच्छामि’ अत्र ‘इच्छामि’ क्रियापदस्य कर्ता कः?
4. भक्षितस्य’ इति पदस्य अर्थे गद्यांशे किं पदं प्रयुक्तम्?

उत्तर:

1. तत्,
2. श्वेतकेतवे,
3. श्वेकेतुः (अहम्),
4. अशितस्य।

2. श्वेतकेतुः – कश्च प्राणः?
आरुणिः – पीतानाम् अपां योऽणिष्ठः स प्राणः।
श्वेतकेतुः – भगवन्! का इयं वाक्?
आरुणिः – वत्स! अशितस्य तेजसा योऽणिष्ठः सा वाक्। सौम्य! मनः अन्नमय, प्राणः आपोमयः वाक् च तेजोमयी भवति इत्यप्यवधार्यम्।
श्वेतकेतुः – भगवन्! भूय एव मां विज्ञापयतु।।
आरुणिः – सौम्य! सावधानं शृणु! मध्यमानस्य दध्नः योऽणिमा, स ऊर्ध्वः समुदीपति। तत्सर्पिः भवति।
श्वेतकेतुः – भगवन्! भवता घृतोत्पत्तिरहस्यम् व्याख्यातम्। भूयोऽपि श्रोतुमिच्छामि।
आरुणिः – एवमेव सौम्य! अश्यमानस्य अन्नस्य योऽणिमा, स ऊर्ध्वः समुदीपति। तम्मनो भवति। अवगतं न वा?

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. घृतोत्पत्तिरहस्यम् कः वदति?
2. अपाम् अरिष्टः भागः कः भवतिः
3. श्वेतकेतुः कस्य शिष्यः आसीत्?

उत्तर:

1. आरुणिः,
2. प्राणः
3. आरुणे:

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. सर्पिः किम् भवति?
2. वाक्, प्राण: मनश्च कीदृशानि भवन्ति?

उत्तर:

1. मध्यमानस्य दनः योऽणिमा, स ऊर्ध्वः समुदीषति। तत्सर्पिः भवति।
2. वाक् तेजोमयी, प्राणः आपोमयः मनः अन्नमयं च भवति।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. उपरि’ इति पदस्य अर्थे किम् पदम् संवाद प्रयुक्तम्?
2. वक्तुम्’ इति पदस्य संवादे विपर्ययपदं किं प्रयुक्तम्?
3. ‘सौम्य’ इति पदम् कस्मै प्रयुक्तम्?
4. ‘केयं वाक्’ अत्र विशेषणपदं किम् प्रयुक्तम्?

उत्तर:

1. ऊर्ध्वः,
2. श्रोतुम्,
3. श्वेतकेतवे,
4. इयम्

3.श्वेतकेतुः – सम्यगवगतं भगवन्!
आरुणिः – वत्स! पीयमानानाम् अपां योऽणिमा स ऊर्ध्वः समुदीपति स एव प्राणो भवति।
श्वेतकेतुः – भगवन्! वाचमपि विज्ञापयतु।
आरुणिः – सौम्य! अश्यमानस्य तेजसो योऽणिमा, स ऊर्ध्वः समुदीपति। सा खलु वाग्भवति। वत्स! प त्वां विज्ञापयितुमिच्छामि यत् अन्नमयं भवति मनः, आपोमयो भवति प्राणा: तेजोमयी च भवति वागिति। किञ्च यादृशमन्नादिकं गृहाति मानवस्तादृशमेव तस्य चित्तादिकं भवतीति मदुपदेशसारः। वत्स! एतत्सर्व हवयेन अवधारय।
श्वेतकेतुः – यदाज्ञापयति भगवन्। एष प्रणमामि।
आरुणिः – वत्स! चिरञ्जीव। तेजस्वि नौ अधीतम् अस्तु (आवयोः अधीतम् तेजस्वि अस्तु )।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. क: सम्यक् अवगच्छति?
2. चिरञ्जीव! इति कः कथयति?
3. श्वेतकेतः कं प्रणमति?

उत्तर:

1. श्वेतकेतुः,
2. आरुणिः
3. आरुणिम्

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकर्मव)

1. आरुणेः उपदेशस्य सारः किम् अस्ति?
2. मनः कीदृशं भवति?

उत्तर:

1. आरुणे: उपदेशस्य सारः अस्ति यत् मनुष्यः यादृशम् अन्नादिकं गृह्णाति तादृशमेव तस्य चित्तादिकं भवति।
2. अन्नमयं मनः भवति।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. अत्र ‘प्रणमामि’ इति क्रियापदस्य कर्ता कः?
2. ‘पठितम्’ इति पदस्य पर्यायपदं संवादे किम्?
3. ‘भवतीति मदपदेशसार:’ अत्र ‘मत’ पदम् कस्मै प्रयुक्तम्?
4. अनुच्छेदे ‘अधः’ इत्यस्य पदस्य कः विपर्ययः आगतः?

उत्तर:

1. श्वेतकेतुः (अहम्)
2. अधीतम्,
3. आरुणये,
4. ऊर्ध्वः

प्रश्न 2.
निम्नवाक्येषु रेखाङ्कित पदानाम् स्थानेषु प्रश्नवाचक पदं लिखत –

प्रश्न 1.
मथ्यमानस्य दनः अणिमा ऊर्ध्य समुदीषति।
(क) कस्य
(ख) किम्
(ग) कम्
(घ) कुत्र
उत्तर:
(घ) कुत्र

प्रश्न 2.
भवता घृतोत्पत्तिरहस्यम् व्याख्यातम्।
(क) कम्
(ख) किम्
(ग) केन
(घ) कीदृशम्
उत्तर:
(ख) किम्

प्रश्न 3.
आरुणिम् उपगम्य श्वेतकेतुः अभिवादयति।
(क) किम्
(ख) कम्
(ग) कः
(घ) के
उत्तर:
(ख) कम्

प्रश्न 4.
श्वेतकेतुः वाग्विषये पृच्छति।
(क) कस्य
(ख) किम्
(ग) कः
(घ) कीदृशः
उत्तर:
(ग) कः

प्रश्न 5.
अशितस्यान्नस्य यः अणिष्ठः तत् मनः।
(क) किम्
(ख) क;
(ग) कस्य
(घ) कम्
उत्तर:
(ग) कस्य

प्रश्न 6.
पीतानाम् अपां यः अणिष्ठः सः प्राणः।
(क) किम्
(ख) कीदृशाणाम्
(ग) कम्
(घ) कस्य
उत्तर:
(ख) कीदृशाणाम्

प्रश्न 7.
अशितस्य तेजसा योऽणिष्ठः सा वाक्।
(क) कीदृशः
(ख) कः
(ग) कस्य
(घ) किम्
उत्तर:
(क) कीदृशः

प्रश्न 8.
अहम् भूयोऽपि श्रोतुमिच्छामि।
(क) कथम्
(ख) कतिवारम्
(ग) कदा
(घ) किम्
उत्तर:
(ख) कतिवारम्

प्रश्न 9.
एतत् सर्वम् हृदयेन अवधारय।
(क) कथम
(ख) किम्
(ग) केन
(घ) कम्
उत्तर:
(ग) केन

प्रश्न 10.
मानवः यादृशं अन्नं गृह्णाति तादृशं तस्य चित्तादिकं भवति।
(क) कः
(ख) कम्
(ग) कथम्
(घ) किम्
उत्तर:
(क) कः

प्रश्न 11.
मानवः यादृशं अन्नं गृह्णाति तादृशं तस्य चित्तादिकं भवति।
(क) कीदृशम्
(ख) कम्
(ग) किम्
(घ) कः
उत्तर:
(क) कीदृशम्

प्रश्न 12.
मनः अन्नमयं भवति।
(क) कस्य
(ख) कः
(ग) किम्
(घ) केषाम्
उत्तर:
(ग) किम्

प्रश्न 13.
प्राणः अपोमयः भवति।
(क) का
(ख) कीदृशः
(ग) कस्य
(घ) किम्
उत्तर:
(ख) कीदृशः

प्रश्न 14.
वाक् तेजोमयी भवति।
(क) का
(ख) कः
(ग) कथम्
(घ) कदा
उत्तर:
(क) का

प्रश्न 15.
मनः अन्नमयं भवति।
(क) केषाम्
(ख) कस्य
(ग) केन
(घ) कीदृशम्
उत्तर:
(घ) कीदृशम्

प्रश्न 16.
प्राणः अपोमयः भवति।
(क) कथम्
(ख) कदा
(ग) का
(घ) केन
उत्तर:
(ग) का

प्रश्न 17.
वाक् तेजोमयी भवति।
(क) कस्य
(ख) कीदृशी
(ग) कस्य
(घ) किम्
उत्तर:
(ख) कीदृशी

‘अ’ वर्गस्य पर्यायपदानि ‘ब’ वर्गण दत्तः पदैः सह यथायोग्यं योजयत –

‘अ’ वर्ग: – ‘ब’ वर्गः
1. अपाम् – भक्ष्यमाणस्य
2. अपोमयः – उत्तमरीत्या
3. अवधार्यम् – घृतम्
4. भूयोऽपि – प्रणमामि
5. वाक् – लधिष्ठः
6. अवगतम् – जलानाम्
7. ऊर्ध्व – पुनरपि
8. अस्यमानस्य – अवागच्छम्
9. सम्यक् – जलमयः
10. सर्पिः – उपरि
11. वन्दे – अवगन्तव्यम्
12. प्रष्टव्यम् – वाणी
13. अणिष्ठः – प्रष्टुम् योग्यम्
14. तेजोमयः – समुत्तिष्ठति
15. विज्ञापयतु – तेजोयुक्तम्
16. उपदेशान्ते – आवयोः
17. श्रोतुम् – अग्निमय:
18. समुदीषति – प्रबोधयतु
19. तेजस्वि – प्रवचनान्ते
20. नौ – आकर्णयितुम्
उत्तर:
1. – जलानाम्,
2. – जलमयः,
3. – अवगन्तव्यम्,
4. – पुनरपि,
5. – वाणी,
6. – अवागच्छम्,
7. – उपरि,
8. – भक्ष्यमाणस्य,
9. – उत्तमरीत्या,
10. – घृतम्,
11. – प्रणमामि,
12. – प्रष्टुम् योग्यम्,
13. – लधिष्ठः,
14. – अग्निमयः,
15. – प्रबोधयतु,
16. – प्रवचनान्ते,
17. – आकर्णयितुम्,
18. – समुत्तिष्ठति,
19. – तेजोयुक्तम्,
20. – आवयोः।

प्रश्न 4.
निम्नवाक्यानि घटनाक्रमानुसार पुनर्लिखत –

प्रश्न 1.

1. भगवन् मनः किं भवति?
2. अशितस्य तेजसः यः अणिष्ठः भागः भवति सा वाक् अस्ति।
3. अनेन शिष्यः सन्तुष्टः भूत्वा तं प्रणमति।
4. श्वेतकेतुः नाम शिष्यः गुरुम् आरुणिं प्रश्न पृच्छति।
5. भगवन् प्राणः कः?
6. गुरुः वदति यत् अशितस्य अन्नस्य यः अणिष्ठः भागः तत् मनः भवति।
7. गुरु: अकथयत्-पीतानाम् अपां य: अणिष्ठः भागः सः प्राणः भवति।
8. पुनः श्वेतकेतुः वदति यत् वाक् काऽस्ति?

उत्तर:

1. श्वेतकेतुः नाम शिष्यः गुरुम् आरुणिं प्रश्नं पृच्छति।
2. भगवन् मनः किं भवति?
3. गुरुः वदति यत् अशितस्य अन्नस्य यः अणिष्ठः भागः तत् मनः भवति।
4. भगवन् प्राणः कः?
5. गुरुः अकथयत्-पीतानाम् अपां यः अणिष्ठः भागः सः प्राणः भवति।
6. पुनः श्वतकतुः वदति यत् वाक् काऽस्ति?
7. अशितस्य तेजस: य: अणिष्ठः भागः भवति सा वाक् अस्ति।
8. अनेन शिष्यः सन्तष्टः भत्वा तं प्रणमति।

प्रश्न 2.

1. गुरुः आरुणिः वदति-पीयमानानाम् अपाम् यः अणिमा ऊर्ध्वः समुदीपति स एव प्राणः भवति।
2. वत्स! अश्यमानस्य तेजसः च अणिमा भागः ऊर्ध्वः उद्गच्छति सा एव वाक् भवति।
3. गुरुः कथयति-सौम्य! आश्यमानस्य अन्नस्य यः अणिमा अंशः सः एव मनः भवति।
4. गुरोः आरुणे: प्रियः शिष्यः श्वेतकेतुः आसीत्।
5. पुनः शिष्यः पृच्छति-गुरुवर! प्राणश्च कः?
6. भगवन् मनः किमस्ति?
7. भगवन्! कथयतु भवान् यत् वाक् च कः कथ्यते?
8. स: गुरुम् उपगम्य मनसः, प्राणस्य वाचः च विषये प्रश्नानि पृच्छति।

उत्तर:

1. गुरोः आरुणे: प्रियः शिष्यः श्वेतकेतुः आसीत्।
2. स: गुरुम् उपगम्य मनसः, प्राणस्य वाचः च विषये प्रश्नानि पृच्छति।
3. भगवन् मनः किमस्ति?
4. गुरुः कथयति-सौम्य! आश्यमानस्य अन्नस्य यः अणिमा अंशः सः एव मनः भवति।
5. पुनः शिष्यः पृच्छति-गुरुवर! प्राणश्च कः?
6. गुरुः आरुणिः वदति-पीयमानानाम् अपाम् यः अणिमा ऊर्ध्वः समुदीपति स एव प्राणः भवति।
7. भगवन्! कथयतु भवान् यत् वाक् च क; कथ्यते?
8. वत्स! अश्यमानस्य तेजसः च अणिमा भागः ऊर्ध्वः उद्गच्छति सा एव वाक् भवति।

प्रश्न 5.
‘क’ स्तम्भे विशेषणानि ‘ख’ स्तम्भे विशेष्याणि दत्तानि। तानि समुचित योजयत –

‘क’ स्तम्भः – ‘ख’ स्तम्भः
1. रहस्यम् – इदम्
2. यः – वाक्
3. व्याख्यातम् – मनः
4. सा – प्राण:
5. अन्नमयम् – वाणी
6. अपोमयः – मनः
7. तेजोमयी – अणिष्ठः
8. अन्नमयम् – मनः
उत्तर:
1. – मन:
2. – अणिष्ठः
3. – रहस्यम्
4. – वाक्
5. – मनः
6. – प्राण:
7. – वाणी
8. – मनः

प्रश्न 6.
‘अ’ वर्गस्य विपर्ययपदानि ‘ब’ वर्गे दत्तः पदैः सह मेलयत –

‘अ’ वर्ग: – ‘ब’ वर्गः
1. श्रोतुम् – असम्यक
2. रहस्यम् – अवधीतम्
3. अन्ते – सर्वम्
4. सौम्यः – अधः
5. सम्यक् – ‘वक्तुम्
6. भूयः – लधिष्ठः
7. ऊर्ध्वम् – प्राकटम्
8. अनवधीतम् – आरम्भ
9. किञ्चित् – चञ्चल:
10. गरिष्ठः – एकवारम्
उत्तर:
1. – वक्तुम्
2. – प्राकटम्
3. – आरम्भे
4. – चञ्चलः
5. – असम्यक्
6. – एकबारम्
7. – अध:
8. – अवधीतम्
9. – सर्वम्
10. – लधिष्ठः

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Shemushi Sanskrit Class 9 Solutions Chapter 5 सूक्तिमौक्तिकम्

Class 9 Sanskrit Shemushi Chapter 5 सूक्तिमौक्तिकम् Textbook Questions and Answers

अभ्यासः

प्रश्न 1.
एकपवेन उत्तरं लिखत –

(क) वित्ततः क्षीणः कीदृशः भवति?
(ख) कस्य प्रतिकूलानि कार्याणि परेषां न समाचरेत्?
(ग) कुत्र दरिद्रता न भवेत्?
(घ) वृक्षाः स्वयं न खादन्ति?
(ङ) का पुरा लघ्वी भवति?
उत्तर:
(क) अक्षीणः
(ख) आत्मनः
(ग) प्रियवाक्यप्रदाने
(घ) फलानि
(ङ) सज्जनमैत्री

प्रश्न 2.
अधोलिखितप्रश्नानाम् उत्तराणि संस्कृतभाषया लिखत –

(क) यत्नेन किं रक्षेत् वित्तं वृत्तं वा?
(ख) अस्माभिः (किं न समाचरेत्) कीदृशम् आचरणं न कर्त्तव्यम्?
(ग) जन्तवः केन तुष्यन्ति?
(घ) सज्जनानां मैत्री कीदृशी भवति?
(ङ) सरोवराणां हानिः कदा भवति?
उत्तर:
(क) यत्नेन वृत्तं रक्षेत्।
(ख) अस्माभिः आत्मनः प्रतिकूलम् आचरणं न कर्त्तव्यम्।
(ग) जन्तवः प्रियवाक्यप्रदानेन तुष्यन्ति।
(घ), सज्जनानां मैत्री पुरा लघ्वी पश्चात् च वृद्धिमती भवति।
(ड) सरोवराणां हानिः मरालैः सह विप्रयोगेन भवति।

प्रश्न 3.
‘क’ स्तम्भे विशेषणानि ‘ख’ स्तम्भे च विशेष्याणि वत्तानि, तानि यथोचित योजयत –

‘क’ स्तम्भः – ‘ख’ स्तम्भः
(क) आस्वाद्यतोयाः – 1. खलानां मैत्री
(ख) गुणयुक्तः – 2. सज्जनानां मैत्री
(ग) दिनस्य पूर्वाद्धभिन्ना – 3. नद्यः
(घ) दिनस्य परार्द्धभिन्ना – 4. दरिद्रः
उत्तर:
‘क’ स्तम्भः – ‘ख’ स्तम्भः
विशेषणम् – विशेष्यः
(क) आस्वाद्यतोयाः – 3. नद्यः
(ख) गुणयुक्तः – 4. दरिद्धः
(ग) दिनस्य पूर्वार्द्धभिन्ना – 1. खलानां मैत्री
(घ) दिनस्य पराद्धभिन्ना 2. सज्जनानां मैत्री

प्रश्न 4.
अधोलिखितयोः श्लोकद्वयोः आशयं हिन्दीभाषया आङ्ग्लभाषया वा लिखत –

(क) आरम्भगुर्वी क्षयिणी क्रमेण
लघ्वी पुरा वृद्धिमती च पश्चात्।
दिनस्य पूर्वार्द्धपरार्द्धभिन्ना
छायेव मैत्री खलसज्जनानाम् ।।

(ख) प्रियवाक्यप्रदानेन सर्वे तुष्यन्ति जन्तवः।
तस्मात्तदेव वक्तव्यं वचने का दरिद्रता ।।
उत्तर:
भाव हिन्दी में –
(क) दुष्टों और सज्जनों की मित्रता में अंतर स्पष्ट करते हुए आचार्य भर्तृहरि कहते हैं कि जिस प्रकार छाया दिन में आरंभ में बड़ी होती है तथा धीरे-धीरे छोटी होती जाती है। उसी प्रकार दुष्टों की मित्रता पहले गहरी होती है और धीरे-धीरे कम होती जाती है। इसके विपरीत जिस प्रकार दोपहर में छाया छोटी होती है और धीरे-धीरे बढ़ती है, इसी प्रकार सज्जनों की मित्रता पहले कम तथा धीरे-धीरे दूसरों के गुण स्वभाव आदि समझकर बढ़ती है।

भाव अंग्रेजी में –
While explaining the difference between the wicked and the gentlemen, scholar Bharathari says that the shadow at the beginning of the day is small, and increases gradually. In a very manner, the wicked people have deep friendship in the initial stage but lose their friends slowly. On the contrary, like the shadow in the noon is smaller and develops gradually, in the same manner, the gentlemen have fewer friends in the beginning and increase their friends, comprehending the quality and nature of others.

भाव हिन्दी में –
(ख) मीठे वचन बोलने से सारे प्राणी प्रसन्न होते हैं, इसलिए हमें हमेशा मीठा ही बोलना चाहिए। मीठे वचन बोलने में कंजूसी नहीं करनी चाहिए, क्योंकि मीठे वचनों की कोई कमी नहीं है।

भाव अंग्रेजी में –
All human beings become happy hearing the sweet words. Therefore, we should always speak sweet words, he should not restrain ourselves from speaking sweet words, because there is no dearth of sweet words.

प्रश्न 5.
अधोलिखितपदेभ्यः भिन्नप्राकृतिकं पदं चित्वा लिखत –

(क), वक्तव्यम्, कर्त्तव्यम्, सर्वस्वम्, हन्तव्यम्।
(ख) यत्नेन, वचने, प्रियवाक्यप्रदानेन, मरालेन।
(ग) श्रूयताम्, अवधार्यताम्, धनवताम्, क्षम्यताम्।
(घ) जन्तवः, नद्यः, विभूतयः, परितः।
उत्तर:
(क) सर्वस्वम्
(ख) वचने
(ग) धनवताम्
(घ) परितः

प्रश्न 6.
स्थूलपदान्यधिकृत्य प्रश्नवाक्यनिर्माणं कुरुत –

(क) वृत्ततः क्षीणः हतः भवति।
(ख) धर्मसर्वस्वं श्रुत्वा अवधार्यताम्।
(ग) वृक्षाः फलं न खादन्ति।
(घ) खलानाम् मैत्री आरम्भगुर्वी भवति।
उत्तर:
(क) कस्मात् क्षीणः हतः भवति?
(ख) किं श्रुत्वा अवधार्यताम्?
(ग) के फलं न खादन्ति?
(घ) केषाम् मैत्री आरम्भगुर्वी भवति?

प्रश्न 7.
अधोलिखितानि वाक्यानि लोट्लकारे परिवर्तयतयथा –

सः पाठं पठति। – सः पाठं पठतु। – ……………….
(क) नद्यः आस्वाधतोयाः सन्ति। – ……………….
(ख) सः सदैव प्रियवाक्यं वदति। – ……………….
(ग) त्वं परेषां प्रतिकूलानि न समाचरसि। – ……………….
(घ) ते वृत्तं यत्नेन संरक्षन्ति। – ……………….
(ङ) अहम् परोपकाराय कार्य करोमि। – ……………….
उत्तर:
(क) नद्यः आस्वाद्यतोयाः सन्ति। – नद्यः आस्वाद्यतोयाः सन्तु।
(ख) सः सदैव प्रियवाक्यं वदति। – सः सदैव प्रियवाक्यं वदतु।
(ग) त्वं परेषां प्रतिकूलानि न समाचरसि। – त्वं परेषां प्रतिकूलानि न समाचर।
(घ) ते वृत्तं यत्नेन संरक्षन्ति। – ते वृत्तं यत्नेन संरक्षन्तु।
(ङ) अहम् परोपकाराय कार्य करोमि। – अहं परोपकाराय कार्य करवाणि।

परियोजनाकार्यम्

(क) परोपकारविषयकं श्लोकद्वयम् अन्विष्य स्मृत्वा च कक्षायां सस्वर पठ।
(ख) नद्याः एक सुन्दरं चित्र निर्माय संकलय्य वा वर्णयत यत् तस्याः तीरे मनुष्याः पशवः खागाश्च निर्विघ्नं जलं पिबन्ति।
उत्तर:
(क) 1. परोपकाराय वहन्ति नद्यः परोपकाराय दुहन्ति गावः। परोपकाराय फलन्ति वृक्षाः परोपकारार्थमिदं शरीरम्।।
2. श्रोत्रं श्रुतेनैव न कुण्डलेन, दानेन पाणिनं तु कङ्कणेन। विभाति कायः करुणापराणा, परोपकारेण न चन्दनेन। (छात्र इन श्लोकों को याद करें तथा अध्यापक के सहयोग से इनका कक्षा में सस्वर पाठ करें।)
(ख) छात्र अध्यापक के सहयोग से नदी का चित्र बनाएँ तथा वर्णन करें कि उसके तट पर मनुष्य, पशु-पक्षी सब बिना कष्ट पानी पीते हैं।

Class 9 Sanskrit Shemushi Chapter 5 सूक्तिमौक्तिकम् Additional Important Questions and Answers

अतिरिक्त कार्यम्

प्रश्न 1.
निम्नलिखितम् श्लोक पठित्वा तदाधारिताना प्रश्नानाम् उत्तराणि लिखत –

1. वृत्तं यत्नेन संरक्षेद् वित्तमेति च याति च।
अक्षीणो वित्ततः क्षीणो वृत्ततस्तु हतो हतः॥

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. जीवने किम् आगच्छति गच्छति च?
2. वित्ततः क्षीणः कीदृशः भवति?
3. वृत्ततः क्षीणः कीदृशः भवति?

उत्तर:

1. वित्तम् (धनम्)
2. अक्षीणः
3. हतः

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. मनुष्यः कदा अक्षीणः एव तिष्ठति?
2. वृत्तं केन संरक्षेत्?

उत्तर:

1. वित्ततः क्षीणः तु मनुष्यः अक्षीणः एव तिष्ठति।
2. वृत्तः यत्नेन संरक्षेत्।

(iii) भाषिककार्यम् (केवल प्रश्नत्रयमेव)

1. ‘संरक्षेत्’ इति क्रियापदस्य कर्मपदं किम्?
2. ‘आयाति’ इति क्रियापदस्य विपर्ययपदं पद्यांशे किम्?
3. ‘आचरणम्’ इति पदस्य समानार्थकम् पदं श्लोकात् चित्वा लिखत
4. ‘धनम्’ इत्यर्थे किं पदं प्रयुक्तम्?

उत्तर:

1. वित्तम्
2. याति
3. वृत्तम्
4. वित्तम्

2. श्रूयतां धर्मसर्वस्वं श्रुत्वा चैवावधार्यताम्।
आत्मनः प्रतिकूलानि परेषां न समाचरेत्।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. किं श्रुत्वा अवधार्यताम्?
2. श्रुत्वा किं कर्त्तव्यम्?
3. कस्य प्रतिकूलानि न समाचरेत्?

उत्तर:

1. धर्मसर्वस्वं
2. अवधार्यताम्
3. आत्मनः

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. अस्माभिः किं न समाचरेत्।
2. आत्मनः प्रतिकूलानि केषाम् न समाचरेत्?

उत्तर:

1. अस्माभिः आत्मनः प्रतिकूलानि न समाचरेत्।
2. आत्मनः प्रतिकूलानि परेषां न समाचरेत्।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. ‘आचरणं कर्तव्यम्’ इति पदस्य पर्यायपदं श्लोके किं प्रयुक्तम्?
2. ‘अवधार्यताम्’ इति क्रियापदस्य कर्म पदं किम्?
3. ‘अनुकुलानि’ इति पदस्य विपर्यय पदं श्लोकात् चित्वा लिखत

उत्तर:

1. समाचरेत्
2. धर्मसर्वस्वम्।
3. प्रतिकूलानि

3. प्रियवाक्यप्रवानेन सर्वे तुष्यन्ति जन्तवः।
तस्मात् तवेव वक्तव्यं वचने का वरिद्रता ।।

(i) एकपवेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. जन्तवः केन प्रसीदन्ति?
2. सर्वदा कीदृशं वाक्यं वक्तव्यम्?
3. प्रियवाक्यप्रदानेन के प्रसीदन्ति?

उत्तर:

1. प्रियवाक्यप्रदानेन
2. प्रियं
3. जन्तवः

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. कुत्र दरिद्रता न भवेत्?
2. सर्वे जन्तवः केन तुष्यन्ति?

उत्तर:

1. प्रियवचने दरिद्रता न भवेत्।
2. सर्वे जन्वतः प्रियवाक्य प्रदानेन तुष्यन्ति।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. ‘प्राणिनः’ इति पदस्य पर्यायपदं किम्?
2. ‘तदेव’ इति पदं कस्मै आगतम्?
3. ‘अप्रियवाक्य’ इति पदस्य विपर्यायपदं श्लोकात् चित्वा लिखत।
4. ‘प्रसन्नाः भवन्ति’ इत्यर्थे किं पदं प्रयुक्तम्?

उत्तर:

1. जन्तवः
2. प्रियवाक्याय
3. प्रियवाक्य
4. तुष्यन्ति

4. पिबन्ति नद्यः स्वयमेव नाम्भः। स्वयं न खावन्ति फलानि वृक्षाः।
नादन्ति सस्यं खलु वारिवाहाः परोपकाराय सतां विभूतयः।।

(i) एकपवेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. स्वयमेव अम्भः काः न पिबन्ति?
2. वृक्षाः किं न खादन्ति?
3. वारिवाहाः किं न अदन्ति?

उत्तर:

1. नद्यः
2. फलानि
3. सस्यम्

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. सज्जनानाम् विभूतयः किमर्थ भवन्ति?
2. नद्यः किं न पिबन्ति?

उत्तर:

1. सज्जनानाम् विभूतयः परोपकाराय भवन्ति।
2. नद्यः स्वयमेव अम्भः न पिबन्ति।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. अन्नम्’ इति अर्थे किं पदं श्लोके प्रयुक्तम्?
2. अत्र ‘नद्यः’ इति कर्तृपदस्य क्रियापदं कि?
3. श्लोके ‘जलम्’ पदस्य कः पर्यायः आगतः?
4. श्लोके मेघाः (जलदाः) इत्यर्थे किं पदं प्रयुक्तम्?

उत्तर:

1. सस्यम्
2. पिबन्ति
3. अम्भः
4. वारिवाहाः

5. गुणेष्वेव हि कर्तव्यः प्रयत्नः पुरुषः सवा।
गुणयुक्तो वरिवोऽपि नेश्वरैरगुणैः समः।

(i) एकपवेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. पुरुषैः केषु प्रयत्न कर्तव्यः?
2. पुरुषैः कदा प्रयत्नः कर्त्तव्यः?
3. सर्वदा कैः प्रयत्नः कर्तव्यम्?

उत्तर:

1. गुणेषु
2. सदा
3. पुरुषैः

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. गुणयुक्तः दरिद्रः अपि कैः समः न भवति।
2. ईश्वरैः गुणैः समः कः न भवति?

उत्तर:

1. गुणयुक्तः दरिद्रः अपि गुणहीनैः अगुणैः ईश्वरैः समः न भवति।
2. गुणयुक्त दरिद्रः अपि गुणहीनैः ईश्वरैः गुणैः समः न भवति।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. कर्तव्यः’ इति क्रियापदस्य श्लोके कर्ता कः?
2. श्लोके ‘धनिकः’ इति पदस्य विलोम पदं किम प्रयुक्तम?
3. ‘गुणसम्पन्नः’ इति पदस्य पर्यायपदं श्लोकात् अवचित्य लिखत।
4. श्लोकात् एकं अव्ययपदं चित्वा लिखत।

उत्तर:

1. पुरुषैः
2. दरिद्रः
3. गुणयुक्तः
4. ‘हि’/ ‘सम’

6. आरम्भगुर्वी क्षयिणी क्रमेण लध्वी पुरा वृद्धिमती च पश्चात्।
दिनस्य पूर्वाद्धंपरार्द्धभिन्ना छायेव मैत्री खलसज्जनानाम्॥

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. केषाम् मैत्री आरम्भगुर्वी भवति?
2. केषाम् मैत्री पश्चात् वृद्धिमती भवति?
3. खलसज्जनानाम् मैत्री कीदृशी भवति?

उत्तर:

1. खलानाम्
2. सज्जनानाम्
3. छायेव

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. दिनस्य पूर्वार्द्धछायेव केषाम् मैत्री भवति?
2. दिनस्य अपराईछायेव केषाम् मैत्री भवति?

उत्तर:

1. दिनस्य पूर्वार्द्धछायेव खलानाम् मैत्री भवति।
2. दिनस्य अपरार्द्धछायेव सज्जानानाम् मैत्री भवति?

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. श्लोके ‘लध्वी’ इति पदस्य विपर्ययपदं किं प्रयुक्तम्?
2. अत्र ‘भवति’ इति क्रिया पदस्य कर्ता कः?
3. ‘आदौ दीर्घा’ इत्यर्थे श्लोके किं पदं प्रयुक्तम्?
4. ‘सज्जनानाम्’ इति पदस्य विपर्ययपदं किम्?

उत्तर:

1. गुर्वी
2. मैत्री
3. आरम्भगुर्वी
4. खलानाम्

7. यत्रापि कुत्रापि गता भवेयु
हंसा महीमण्डलमण्डनाय।
हानिस्तु तेषां हि सरोवराणां
येषां मरालैः सह विप्रयोगः।।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. हानिः केषाम् भवति?
2. हंसाः किमर्थम् गच्छेयुः?
3. हंसानां कैः सह विप्रयोगः सरोवराणां हानि करोति?

उत्तर:

1. सरोवराणाम्
2. महीमण्डलमण्डनाय
3. मराले:

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. सरोवराणां हानिः कथं भवति?
2. हंसाः सर्वदा किमर्थम् भवन्ति?

उत्तर:

1. सरोवराणां हानिः मरालैः सह विप्रयोगेन भवति।
2. हंसाः सर्वदा महीमण्डलमण्डनाय भवन्ति।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. ‘मरालः’ इति पदस्य पर्यायपदं पद्यांशे किम् प्रयुक्तम्?
2. हंसाः’ इति कर्तृपदस्य क्रियापदं किम् अत्र अस्ति
3. लाभः’ इतिपदस्य विलोमपदं किं? ।
4. अत्र ‘वियोगः’ इत्यर्थे किं पदं श्लोके प्रयुक्तम्?

उत्तर:

1. हंसः
2. गताः
3. हानिः
4. विप्रयोगः

8. गुणा गुणज्ञेषु गुणा भवन्ति
ते निर्गुणं प्राप्य भवन्ति दोषाः।
आस्वाद्यतोयाः प्रवहन्ति नद्यः
समुद्रमासाद्य भवन्त्यपेयाः।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. गुणाः निर्गुणं प्राप्य किं भवन्ति?
2. गुणाः कुत्र/केषु गुणाः भवन्ति?
3. आस्वाद्यतोयाः काः प्रवहन्ति

उत्तर:

1. दोषाः
2. गुणज्ञेषु
3. नद्यः

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. कीदृश्य: नद्यः प्रवहन्ति?
2. गुणाः कदा दोषाः भवन्ति?

उत्तर:

1. आस्वाद्यतोया: नद्यः प्रवहन्ति।
2. गुणाः निर्गुणं प्राप्य दोषाः भवन्ति।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. अत्र श्लोके ‘नद्यः’ इति कर्तृपदस्य क्रियापदं किम्?
2. आसाद्य’ इति पर्यायपदं पद्यांशे किम् अत्र अस्ति?
3. श्लोके नद्यः’ इति पदस्य विशेषणपदं कि?
4. ‘जलम्’ इत्यर्थे अत्र किं पदं प्रयुक्तम्?

उत्तर:

1. प्रवहन्ति
2. प्राप्य
3. आस्वाद्यतोयाः
4. तोयम्

प्रश्न 2.
निम्नवाक्येषु रेखाङ्कित पदानां स्थानेषु प्रश्नवाचकपदं लिखत –

प्रश्न 1.
वृत्तं यत्नेन संरक्षेत्।
(क) केन
(ख) केभ्यः
(ग) कस्य
(घ) कस्मात्
उत्तर:
(क) केन

प्रश्न 2.
वित्ततः क्षीणः अक्षीणः भवति।
(क) कस्य
(ख) केन
(ग) कस्मात्
(घ) कः
उत्तर:
(ग) कस्मात्

प्रश्न 3.
वृत्ततः तु हतो हतः।
(क) कस्य
(ख) क:
(ग) कस्मात्
(घ) केन
उत्तर:
(ग) कस्मात्

प्रश्न 4.
वित्ततः क्षीणः अक्षीणः भवति।
(क) का
(ख) कीदृशः
(ग) कस्मात्
(घ) काम्
उत्तर:
(ख) कीदृशः

प्रश्न 5.
श्रूयतां धर्मसर्वस्वम्।
(क) किम्
(ख) का
(ग) कस्मात्
(घ) काम्
उत्तर:
(क) किम्

प्रश्न 6.
आत्मनः प्रतिकूलानि परेषां न समाचरेत्।
(क) कः
(ख) कम्
(ग) कस्य
(घ) काम्
उत्तर:
(ग) कस्य

प्रश्न 7.
आत्मनः प्रतिकूलानि परेषां न समाचरेत्।
(क) केषाम्
(ख) काम्
(ग) कस्मात्
(घ) काम्
उत्तर:
(क) केषाम्

प्रश्न 8.
प्रियवाक्यप्रदानेन सर्वे तुष्यन्ति जन्तवः।
(क) केभ्यः
(ख) केन
(ग) के
(घ) काम्
उत्तर:
(ग) के

प्रश्न 9.
प्रियवाक्यप्रदानेन सर्वे जन्तवः तुष्यन्ति।
(क) का
(ख) कम्
(ग) केन
(घ) काम्
उत्तर:
(ग) केन

प्रश्न 10.
तस्मात् तदेव वक्तव्यम्।
(क) के
(ख) किमेव
(ग) कीदृशम्
(घ) काम्
उत्तर:
(ख) किमेव

प्रश्न 11.
मधुरवचने दरिद्रता न भवति।
(क) के
(ख) कीदृशे
(ग) कस्मिन्
(घ) केषु
उत्तर:
(ग) कस्मिन्

प्रश्न 12.
मधुरवचने दरिद्रता न भवेत्।
(क) कस्याः
(ख) का
(ग) कः
(घ) केषु
उत्तर:
(ख) का

प्रश्न 13.
नद्यः स्वयमेव जल न पिबन्ति।
(क) काः
(ख) का
(ग) कः
(घ) केषु
उत्तर:
(क) काः

प्रश्न 14.
वृक्षाः फलानि न खादन्ति।
(क) काः
(ख) के
(ग) कः
(घ) केषु
उत्तर:
(ख) के

प्रश्न 15.
बारिवाहाः सस्य न अदन्ति।
(क) काः
(ख) काम्
(ग) कस्य
(घ) के
उत्तर:
(घ) के

प्रश्न 16.
सतां विभूतयः परोपकाराय।।
(क) काम्
(ख) केषाम्
(ग) के
(घ) केषु
उत्तर:
(ख) केषाम्

प्रश्न 17.
सतां विभूतयः परोपकाराय।
(क) केषाम्
(ख) किमर्थम् /कस्मै
(ग) कस्य
(घ) केषु
उत्तर:
(ख) किमर्थम् /कस्मै

प्रश्न 18.
पुरुषैः सदा गुणेष्वेव हि प्रयत्न कर्त्तव्यः।
(क) कै:
(ख) के
(ग) केन
(घ) केषु
उत्तर:
(क) कै:

प्रश्न 19.
आरम्भगुर्वी मैत्री क्रमेण क्षयिणी भवति।
(क) कीदृशः
(ख) कीदृशी
(ग) केन
(घ) केषु
उत्तर:
(ख) कीदृशी

प्रश्न 20.
सज्जनानाम् मैत्री धायेव भवति।
(क) का
(ख) केषाम्
(ग) केन
(घ) केषु
उत्तर:
(क) का

प्रश्न 21.
खलानाम् मैत्री छायेव भवति।।
(क) केषाम्
(ख) काम्
(ग) का
(घ) केषु
उत्तर:
(क) केषाम्

प्रश्न 22.
हंसा: महीमण्डलमण्डलनाय कुत्रापि गच्छेयुः।
(क) काय
(ख) कस्मै किमर्थम्
(ग) के
उत्तर:
(ख) कस्मै किमर्थम्

प्रश्न 23.
सरोवराणां हि हानिः भवति।
(क) केषाम्
(ख) के
(ग) कस्मात्
(घ) कस्य
उत्तर:
(क) केषाम्

प्रश्न 24.
मरालैः सह विप्रयोगः सरोवराणां हानिः भवति।।
(क) काः
(ख) केभ्यः
(ग) काय
(घ) कैः
उत्तर:
(घ) कैः

प्रश्न 25.
नद्यः आस्वाद्यतोयाः प्रवहन्ति।
(क) कीदृशे
(ख) कीदृशी
(ग) कीदृश्यः
(घ) कैः
उत्तर:
(ग) कीदृश्यः

प्रश्न 26.
समुद्रम् आसाद्य जलं अपेयम् भवति।
(क) कीदृशम्
(ख) काम्
(ग) केषु
(घ) कस्य
उत्तर:
(क) कीदृशम्

प्रश्न 27.
निर्गुणं प्राप्य गुणाः अपि दोषाः भवन्ति।
(क) कम्
(ख) किम्
(घ) काम्
(ग) कैः
उत्तर:
(ख) किम्

प्रश्न 28.
गुणज्ञेष गुणाः गुणाः भवन्ति।
(क) कम्
(ख) केषु
(ग) कैः
(घ) काम्
उत्तर:
(ख) केषु

प्रश्न 3.
अन्वय लेखनम्

उचितैः पदैः अन्वयं सम्पूरयत

1. वृतं यत्नेन संरक्षेत् वित्तमेति च याति च।
अक्षीणो वित्ततः क्षीणो वृत्ततस्तु हतो हतः।
अन्वयः – वृत्तं

1. ……… संरक्षेत् वित्तं तु जीवने
2. ……… याति च। (य:) वित्ततः
3. ……… भवति (तस्य किमपि नष्टं न भवति। किन्तु यदि कोऽपि)
4. ……… वृत्ततः क्षीणः भवति तदा सः हतो हतः एव भवति।

मञ्जूषा – क्षीणः, एति, वृत्ततः, यत्नेन
उत्तर:

1. यत्नेन
2. एति
3. क्षीणः
   वृत्ततः

2. श्रूयतां धर्मसर्वस्वं श्रुत्वा चैवावधार्यताम्।
आत्मनः प्रतिकूलानि परेषां न समाचरेत्।।
अन्वयः – धर्मसर्वस्वं श्रूयतां

1. ……… च
2. ……… अवधार्यताम्
3. ……… प्रतिकूलान
4. ……… न समाचरेत्।

मञ्जूषा – परेषां, श्रुत्वा, एव, आत्मनः
उत्तर:

1. श्रुत्वा
2. एव
3. आत्मनः
4. परेषां

3. प्रियवाक्यप्रदानेन सर्वे तुष्यन्ति जन्तवः।
तस्माद् तदेव वक्तव्यं वचने का दरिद्रता।।
अन्वयः – सर्वे जन्तवः

1. ……… प्रदानेन
2. ……… तस्मात्
3. ……… एव
4. ……… वचने का दरिद्रता।

मञ्जूषा – वक्तव्यं, प्रियवाक्य, तुष्यन्ति, तत्
उत्तर:

1. प्रियवाक्य
2. तुष्यन्ति
3. तत्
4. वक्तव्यं

4. पिबन्ति नद्यः स्वयमेव नाम्भ: स्वयं न खादन्ति फलानि वृक्षाः।
नादन्ति सस्यं खलु वारिवाहाः परोपकाराय सतां विभूतयः॥
अन्वयः – नद्यः

1. ……… एव अम्भः न
2. ……… वृक्षाः स्वयं फलानि न
3. ……… वारिवाहाः खलु
4. ……… न अदन्ति सतां विभूतयः परोपकाराय (एव भवति)।

मञ्जूषा – खादन्ति, सस्यं, पिबन्ति, स्वयम् |
उत्तर:

1. स्वयम्
2. पिबन्ति
3. खादन्ति
4. सस्य

5. गुणेष्वेव हि कर्तव्यः प्रयत्नः पुरुषैः सदा।
गुणयुक्तो दरिद्रोऽपि नेश्वरैरगुणैः समः।।
अन्वयः – पुरुषैः

1. ……… गुणेषु
2. ……… हि प्रयत्नः
3. ……… दरिद्रः
4. ……… गुणयुक्तः (नरः) ईश्वरैः गणैः समः (न भवति)

मञ्जूषा – अपि, एव, कर्तव्यः, सदा।
उत्तर:

1. सदा
2. एवं
3. कर्तव्यः
4. अपि

6. आरम्भगुर्वी क्षयिणी क्रमेण
लघ्वी पुरा वृद्धिमती च पश्चात्।
दिनस्य पूर्वार्द्धपरार्द्धभिन्ना
छायेव मैत्री खलसज्जनानाम्।।
अन्वयः – आरम्भगुर्वी

1. ……… क्षयिणी पुरा
2. ……… पश्चात् वृद्धिमती पूर्वाद्ध
3. ……… च
4. ……… छाया इव भिन्ना मैत्री भवति।।

मञ्जूषा – अपरार्द्ध, दिनस्य, लम्वी, क्रमेण
उत्तर:

1. क्रमेण
2. लघ्वी
3. अपरार्द्ध
4. दिनस्य

7. यत्रापि कुत्रापि गता भवेयु
हंसा महीमण्डलमण्डनाया
हानिस्तु तेषां हि सरोवराणां
येषां मरालैः सह विप्रयोगः॥
अन्वयः – महीमण्डल मण्डनाय

1. ……… यत्र अपि
2. ……… अपि
3. ……… भवेयुः, हानिः तु
4. …… सरोवराणां हि भवति येषां मरालैः सह विप्रयोगः।

मञ्जूषा – हंसाः, तेषा, गताः, कुत्र
उत्तर:

1. हंसाः
2. कुत्र
3. गताः
4. तेषां

8. गुणाः गुणज्ञेषु गुणा भवन्ति ते निर्गुणं प्राप्य भवन्ति दोषाः।
आस्वाद्यतोयाः प्रवहन्ति नद्यः – समुद्रमासाद्य भवन्त्यपेयाः॥
अन्वयः – गुणा

1. ……… गुणाः भवन्ति निर्गुणं प्राप्यते
2. ……… भवन्ति। नद्यः
3. ……… प्रवहन्ति (ता:) समुद्रम्
4. ……… अपेयाः भवन्ति। ।

मञ्जूषा – अस्वाद्यतोयाः, असाद्य, गुणज्ञेषु, दोषाः
उत्तर:

1. गुणज्ञेषु
2. दोषा:
3. आस्वाद्यतोया:
4. आसाद्य

प्रश्न 4.
निम्न श्लोकानि पठित्वा भावलेखनम् कुरुत –

1. वृत्तं यलेन संरक्षेद् वित्तमेति च याति च।
अक्षीणो वित्ततः क्षीणो वृत्ततस्तु हतो हतः॥
अस्य भावोऽस्ति – मानवः स्वजीवने सदैव

1. ………… रक्षणं प्रयत्नेन कुर्यात्, धन तु तस्य
2. ……… आयाति गच्छति च। यदि मनुष्यः धनेन
3. …………. भवति तर्हि किमपि न हीयते परन्तु यदि
4. …… हीनः भवति तदा तु तदा जीवनमेव निरथकं भवति, सः तु जीवितः एव मृत इव भवति।

मञ्जूषा – आचरणेन, आचरणस्य, जीवने, हीनः
उत्तर:

1. आचरणस्य
2. जीवने
3. हीन:
4. आचरणेन

2. श्रूयतां धर्मसर्वस्वं श्रुत्वा चैवावधार्यताम्।
आत्मनः प्रतिकूलानि परेषां न समाचरेत्।
अस्य भावोऽस्ति – महात्मा विदुरोऽकथयत् यत् हे जनाः! धर्मस्य

1. ……… प्रथम शृणुत पश्चात् तद्जीवनेषु धारयत।
2. ……… स्वरोऽस्ति यत्
3. ……… विपरीतम् आचरणं कदापि अन्येभ्यः
4. ………. न कुरुत। अर्थात् यदाचरणं भवतां कृते उचितं हितकर वा न स्यात् तदन्येषां कृते न कुरुत।

मञ्जूषा – जनेभ्यः, धर्मस्य, सारम्, आत्मनः |
उत्तर:

1. सारम्
2. धर्मस्य
3. आत्मनः
4. जनेभ्यः

3. प्रियवाक्यप्रदानेन सर्वे तुष्यन्ति जन्तवः।
तस्माद् तदेव वक्तव्यं वचने का दरिद्रता॥
अस्य भावोऽस्ति – आचार्य चाणक्यो वदति यत् अस्मिन्

1. ……… सर्वे देहधारिण प्रिय
2. ……… श्रुत्वा एव प्रसीदन्ति। अतः सर्वैः जनैः प्रयत्नेन सदैव तदेव प्रियं
3. ……… वक्तव्यम्। यतः वचन कथने कापि दरिद्रता न आगच्छति अथवा कापि
4. ……… न भवति।

मञ्जूषा – वचनं, संसारे, दरिद्रता, वाक्यम् (वचनम्) |
उत्तर:

1. संसारे
2. वचनं
3. वाक्यम्
4. दरिद्रता

4. पिबन्ति नद्यः स्वयमेव नाम्भः।
स्वयं न खादन्ति फलानि वृक्षाः।
नादन्ति सस्यं खलु वारिवाहाः
परोपकाराय सतां विभूतयः।।
अस्य भावोऽस्ति-अस्मिन् जगति यथा

1. …… स्वजल स्वयमेव न पिबन्ति अर्थात् अन्येभ्यः प्राणिभ्यः । जलं यच्छन्ति। वृक्षाः स्वफलानि कदापि स्वयं न
2. ……… अन्यान् जीवान् खादयन्ति मेघाश्च स्वसामर्थ्यण । विवर्धितान्
3. ……… स्वयं न खादित्वा सर्वेभ्यः प्राणिभ्यः यच्छन्ति, तथैव सज्जनानां
4. ……… स्वार्थाय न भूत्वा परोपकारार्थमेव भवन्ति।

मञ्जूषा – सम्पत्तयः, सस्यान्, नद्यः, खादित्वा
उत्तर:

1. नद्यः
2. खदित्वा
3. सस्थान्
4. सम्पत्तय

5. गुणेष्वेव हि कर्तव्यः प्रयत्नः पुरुषैः सदा।
गुणयुक्तो दरिद्रोऽपि नेश्वरैरगुणैः समः।
अस्य भावोऽस्ति- यत् अस्मिन् संसारे जनाः सदैव

1. ……… ग्रहणे एव
2. ……… कुयुः। यतः जगति गुणवन्तः
3. ……… अपि गुणहीनेषु
4. ……… श्रेष्ठतमाः मन्यन्ते। अतः संसारे गुणानाम् अतीव महत्त्वम् अस्ति ।

मञ्जूषा – प्रयत्नान्. धनवत्सु. दरिद्धाः, गुणानाम् ।
उत्तर:

1. गुणानाम्
2. प्रयत्नान्
3. दरिद्धाः
4. धनवत्सु

6. आरम्भगुर्वी क्षविणी क्रमेण
लध्वी पुरा वृद्धिमती च पश्चात्।।
दिनस्य पूर्वार्द्धपरार्द्धभिन्ना
छायेव मैत्री खलसज्जनानाम्।।
अस्य भावोऽस्ति – कविः भर्तृहरिः नीतिशतके कथयति यत् यथा दिनस्य पूर्वार्द्ध प्राणिनाम् छाया

1. ……… दीर्घा पश्चात् शनैः शनै:
2. ……… समायाति तथैव खलानां दुष्यणां वा मैत्री अपि प्रारम्भे गुर्वी भवति शनैः शनैश्च क्षयिणी भवति। एवमेव यथा परार्द्ध समये
3. ……… छाया पूर्वं लघ्वी भवति पुनश्च शनैः शनैः वृद्धिम् प्राप्तनोति तथैव सज्जनाया
4. ……… अपि आरम्भे लघ्वी भवति एवं शनैः शनैः वृद्धिमायाति।

मञ्जूषा – प्राणिनाम्, आरम्भे, मैत्री, क्षयम् ।
उत्तर:

1. आरम्भे
2. क्षयम्
3. प्राणिनाम्
4. मैत्री

7. यत्रापि कुत्रापि गता भवेयु
हंसा महीमण्डलमण्डनाय।
हानिस्तु तेषां हि सरोवराणां
येषां मरालैः सह विप्रयोगः।।
अस्य भावोऽस्ति-हंसाः अर्थात् श्रेष्ठा:

1. ……… यत्र कुत्रापि स्थानेषु गच्छेयुः तेषां
2. ……… तेषां स्थानाना मण्डनं भवति परन्तु
3. ……… तु तेषां सरोवराणां स्थानानां वा भवति यैः सह तेषां हंसानां सज्जनानां वा
4. ……… भवति। अतः सदैव सौभाग्यार्थं सज्जनानां सङगति कुर्यात्।

मञ्जूषा – गमनेन, हानिः (दुर्भाग्य), वियोगः, जनाः ।
उत्तर:

1. जनाः
2. गमनेन
3. हानिः (दुर्भाग्य)
4. वियोगः

8. गुणाः गुणज्ञेषु गुणा भवन्ति
ते निर्गुणं प्राप्य भवन्ति दोषाः।
आस्वाद्यतोयाः प्रवहन्ति नद्यः
समुद्रमासाद्य भवन्त्यपेयाः।।
भावार्थ- गुणाः सदैव गुणीनां

1. ……… प्राप्य सद्गुणाः भवन्ति परन्तु गुणहीनान् प्राप्य द्रोषरूपे
2. ……… भवन्ति यथा नद्याः जल
3. ……… शुद्ध पेयं च भूत्वा अपि समुद्रे मिलित्वा
4. ……… भवति। अत: गुणीजनैः कदापि दुर्जनस्य संगतिः न कर्तव्या।

मञ्जूषा – स्वादिष्ट, संगतिं, अपेयं, परिवर्तिताः
उत्तर:

1. संगति
2. परिवर्तिताः
3. स्वादिष्टं,
4. अपेयं

प्रश्न 5.
निम्न ‘क’ वर्गीय पदानाम् ‘ख’ वर्गीय पदेषु पर्यायचयनम् कुरुत –

‘क’ वर्ग: – ‘ख’ वर्ग:
1. गुणज्ञेषु – गुणरहितैः
2. मरालैः – तरवः महीरुहाः
3. पुरा – कथनीयम्
4. वृद्धिमती – न पेयाः
5. अगुणैः – गुण-सम्पन्न:
6. दरिद्रः – गुणज्ञातृषु जनेषु
7. गुणयुक्तः – समृद्धयः
8. वारिवाहा: – हंसै:
9. अम्भ: – प्रसन्नाः भवन्ति
10. वृक्षाः – आचरणं कर्त्तव्यम्
11. तुष्यन्ति – जलम्
12. समाचरेत् – मेघाः
13. वक्तव्यम् – निर्धनः
14. विभूतयः – वृद्धिम् उपगता
15. अपेयाः – प्राचीनकाले
उत्तर:
1. – गुणज्ञातृषु जनेषु
2. – हंसैः
3. – प्राचीनकाले
4. – वृद्धिम् उपगता
5. – गुणरहितैः
6. – निर्धनः
7. – गुण सम्पन्न
8. – मेघाः
9. – जलम्
10. – तरवः महीरुहाः
11. – प्रसन्नाः भवन्ति
12. – आचरणं कर्त्तव्यम्
13. – कथनीयम्
14. – समृद्धयः
15. – न पेयाः

प्रश्न 6.
‘क’ स्तम्भे विशेषणानि ‘ख’ स्तम्भे विशेष्याणि दत्तानि। तानि समुचित योजयत

‘क’ स्तम्भः – ‘ख’ स्तम्भः

1. गताः – सरोवराणाम्
2. तेषाम् – जन्तवः
3. सर्वे – मैत्री
4. तत्- नद्यः
5. गुर्वी – दरिद्रता
6. अपेयाः – पुरुषः
7. का – कथा
8. गुणयुक्तः – वक्तव्य
9. लब्धी – नारी
10. श्रेष्ठमा – हंसाः

उत्तर:

1. – हंसाः
2. – सरोवराणाम्
3.  – जन्तवः
4.  – वक्तव्य
5.  – मैत्री
6.  – नद्यः
7. – दरिद्रता
8.  – पुरुषः
9.  – कथा
10.  – नारी

प्रश्न 7.
निम्न ‘क’ वर्गीय पदानाम् ‘ख’ वर्गीय पदेषु विपर्ययचयनम् कुरुत-

‘क’ वर्गः ‘ख’ वर्ग:
1. प्रतिकूलानि – कटुः
2. अक्षीणः – श्रवणीयम्
3. एति – गुणहीनः
4. वक्तव्यम् – गुर्वी
5. प्रियं – वृद्धिमती
6. सतां/सज्जनानाम् – अनुकूलानि
7. दरिद्रः – क्षीणः
8. गुणयुक्तः – याति
9. आरम्भः – दुर्जनानाम्
10. लष्वी – धनिकः
11. पुरा – अन्तः
12. क्षयिणी – पश्चात्
उत्तर:
1. – अनुकूलानि
2. – क्षीणः
3. – याति
4. – श्रवणीयम्
5. – कदुः
6. – दुर्जनानाम्
7. – धनिकः
8. – गुणहीनः
9. – अन्तः
10. – गुर्वी
11. – पश्चात्
12. – वृद्धिमती

We have given detailed NCERT Solutions for Class 9 Sanskrit Shemushi Chapter 8 लौहतुला Questions and Answers come in handy for quickly completing your homework.

Shemushi Sanskrit Class 9 Solutions Chapter 8 लौहतुला

Class 9 Sanskrit Shemushi Chapter 8 लौहतुला Textbook Questions and Answers

अभ्यासः

प्रश्न 1.
एकपदेन उत्तरं लिखत –

(क) वाणिक्यपुत्रस्य किं नाम आसीत्?
(ख) तुला कैः भक्षिता आसीत्?
(ग) तुला कीदृशी आसीत्?
(घ) पुत्रः केन हतः इति जीर्णधनः वदति?
(ङ) विवदमानौ तौ द्वावपि कुत्र गतौ?
उत्तर:
(क) जीर्णधनः
(ख) मूषकैः
(ग) लौहघटिता
(घ) श्येनेन
(ङ) राजकुलम्

प्रश्न 2.
अधोलिखितानां प्रश्नानाम् उत्तराणि संस्कृतभाषया लिखत –

(क) देशान्तरं गन्तुमिच्छन् वणिक्पुत्रः किं व्यचिन्तयत्?
(ख) स्वतुला याचमानं जीर्णधनं श्रेष्ठी किम् अकथयत्?
(ग) जीर्णधनः गिरिगुहाद्वार कया आच्छाद्य गृहमागतः?
(घ) स्नानानन्तरं पुत्रविषये पृष्टः वणिक्पुत्रः श्रेष्ठिनं किम् अवदत्?
(ङ) धर्माधिकारिण: जीर्णधनश्रेष्ठिनौ कथं तोषितवन्तः?
उत्तर:
(क) देशान्तरं गन्तुमिच्छन् वणिक्पुत्रः व्यचिन्तयत् “यत्र स्ववीर्यतः भोगाः भुक्ताः वसन्ति तत्र यः विभवहीनः वसेत् सः पुरुषाधमः अस्ति।”
(ख) स्वतुला याचमानं जीर्णधनं श्रेष्ठी अकथयत्-“भोः! त्वदीया तुला मूषकैः भक्षिता” इति।
(ग) जीर्णधनः गिरिगुहाद्वार बृहच्छिलया (बृहत् शिलया) आच्छाद्य गृहमागतः।
(घ) स्नानानन्तरं पुत्रविषये पृष्टः वणिक्पुत्रः श्रेष्ठिनम् अवदत्- “नदीतटात् सः बाल: श्येनेन हृतः” इति।
(ङ) धर्माधिकारिणः जीर्णधनश्रेष्ठिनौ परस्परं संबोध्य तुला-शिशु-प्रदानेन तोषितवन्तः।

प्रश्न 3.
स्थूलपदान्यधिकृत्य प्रश्ननिर्माणं कुरुत –

(क) जीर्णधनः विभवक्षयात् देशान्तरं गन्तुमिच्छन् व्यचिन्तयत्।
(ख) श्रेष्ठिनः शिशुः स्नानोपकरणमादाय अभ्यागतेन सह प्रस्थितः।
(ग) वणिक् गिरिगुहा बृहच्छिलया आच्छादितवान्।
(घ) सभ्यैः तौ परस्पर संबोध्य तुला-शिशु-प्रदानेन सन्तोषितौ।
उत्तर:
(क) कः विभवक्षयात् देशान्तरं गन्तुमिच्छन् व्यचिन्तयत्?
(ख) श्रेष्ठिनः शिशुः स्नानोपकरणमादाय केन सह प्रस्थितः?
(ग) वणिक् गिरिगुहां कथम् कया आच्छादितवान्?
(घ) सभ्यः तौ परस्परं संबोध्य कथं सन्तोषितौ?

प्रश्न 4.
अधोलिखितानां श्लोकानाम् अपूर्णोऽन्वयः प्रदत्तः पाठमाधृत्य तम् पूरयत –

(क) यत्र देशे अथवा स्थाने ……….. भोगाः भुक्ताः ……….. विभवहीनः यः ……….. पुरुषाधमः।
(ख) राजन्! यत्र लौहसहस्त्रस्य ……….. मुषका: ……….. तत्र श्येनः ……….. हरेत् अत्र संशयः न।
उत्तर:
(क) यत्र देशे अथवा स्थाने स्ववीर्यतः भोगाः भुक्ताः तस्मिन् विभवहीनः यः वसेत् स पुरुषाधमः।
(ख) राजन्! यत्र लौहसहस्रस्य तुलाम् मूषकाः खादन्ति तत्र श्येनः बालकम् हरेत् अत्र संशयः न।

प्रश्न 5.
तत्पदं रेखाङ्कितं कुरुत यत्र –

(क) ल्यप् प्रत्ययः नास्ति।
विहस्य, लौहसहस्रस्य, संबोध्य, आदाय।
(ख) यत्र द्वितीया विभक्तिः नास्ति।
श्रेष्ठिनम्, स्नानोपकरणम्, सत्त्वरम्, कार्यकारणम्।
(ग) यत्र षष्ठी विभक्तिः नास्ति।
पश्यतः, स्ववीर्यतः, श्रेष्ठिनः, सभ्यानाम्।
उत्तर:
(क) लौहसहस्रस्य
(ख) सत्वरम्
(ग) स्ववीर्यतः

प्रश्न 6.
सन्धिना सन्धिविच्छेदेन वा रिक्तस्थानानि पूरबत –

(क) श्रेष्ठ्याह = …………….. + आह
(ख) …………. = द्वौ + अपि
(ग) पुरुषोपार्जिता = पुरुष + …………..
(घ) ……………… = यथा + इच्छया
(ङ) स्नानोपकरणम् = ………….. + उपकरणम
(च) ………………. = स्नान + अर्थम्
उत्तर:
(क) श्रेष्ठी
(ख) द्वावपि
(ग) उपार्जिता
(घ) यथेच्छया
(ङ) स्नान
(च) स्नानार्थम्

प्रश्न 7.
समस्तपदं विग्रहं वा लिखत –
विग्रहः – समस्तपदम्

(क) स्नानस्य + उपकरणम् = ………………..
(ख) …………… + …………… = गिरिगुहायाम्
(ग) धर्मस्य + अधिकारी = ………………..
(घ) …………… + …………… = विभवहीनाः
उत्तर:
(क) स्नानोपकरणम्
(ख) गिरेः + गुहायाम्
(ग) धर्माधिकारी
(घ) विभवेन + हीना:

प्रश्न 7.
(अ) यथापेक्षम् अधोलिखितानां शब्दानां सहायतया “लौहतुला” इति कथायाः सारांश संस्कृतभाषया लिखत –

वणिक्पुत्रः – स्नानार्थम् – लौहतुला – अयाचत् – वृत्तान्तं – ज्ञात्वा – श्रेष्ठिन – प्रत्यागतः – गतः – प्रदानम्
उत्तर:
एकदा जीर्णधनः नाम वणिक्पुत्रः धनोपार्जनाम देशान्तरं गन्तुम् अचिन्तयत्। तस्य गृहे एका ‘लौहतुला’ आसीत्। वणिक्पत्रः लौहतलाम् श्रेष्ठिनः गृहे रक्षित्वा सः देशान्तरं अस्थितः। देशान्तरं भ्रान्त्वा पुन: स्वपुरम् प्रत्यागत्य सः तुलामयाचत्। सः श्रेष्ठी उवाच-“तुला तु मूषके: भक्षिता।” ततः जीर्णधनः श्रेष्ठिनः पुत्रेण सह स्नानार्थ गतः। स्नात्वा सः श्रेष्ठिपुत्रं गिरिगुहायां प्रक्षिप्य, तद्द्वारं च बृहच्छिलया आच्छाद्य गृहम् आगतः। ततः सः वणिक् श्रेष्ठिनं स्वपुत्रविषये अपृच्छत्। वणिक् उवाच-“नदीतटात् सः श्येनेन हृतः” इति। सः शीघ्रमाह-“श्येन: बालं हर्तुं न शक्नोति अत: समर्पय में सुतम्। जीर्णधनः अवदत् यत्र लौहसहस्रस्य तुला मूषकाः खादन्ति तत्र बालकः श्येनेन हरेत्।” एवं विवदमानौ तौ राजकुलं गतौ। सर्व वृत्तान्तं ज्ञात्वा धर्माधिकारिभिः तुला-शिशु-प्रदानेन तौ द्वौ सन्तोषितौ।

Class 9 Sanskrit Shemushi Chapter 8 लौहतुला Additional Important Questions and Answers

अतिरिक्त कार्यम् 

प्रश्न 1.
गद्यांशम् पठित्वा प्रश्नानाम् उत्तराणि लिखत –

1. आसीत् कस्मिंश्चिद् अधिष्ठाने जीर्णधनो नाम वणिक्पुत्रः। स च विभवक्षयात् देशान्तरं गन्तुमिच्छन् व्यचिन्तयत् यत्र देशेऽथवा स्थाने भोगा भुक्ताः स्ववीर्यतः।
तस्मिन् विभवहीनो यो वसेत् स पुरुषाधमः॥ तस्य च गृहे लौहघटिता पूर्वपुरुषोपार्जिता तुला आसीत्। तां च कस्यचित् श्रेष्ठिनो गृहे निक्षेपभूतां कृत्वा देशान्तर प्रस्थितः। ततः सुचिरं कालं देशान्तरं यथेच्छया भ्रान्त्वा पुनः स्वपुरम् आगत्य तं श्रेष्ठिनम् अवदत्-“भोः श्रेष्ठिन्! दीयतां मे सा निक्षेपतुला।” सोऽवदत्-“भोः! नास्ति सा, त्वदीया तुला मूषकैः भक्षिता” इति।
जीर्णधनः अवदत्-“भोः श्रेष्ठिन्! नास्ति दोषस्ते, यदि मूषकैः भक्षिता। ईदशः एव अयं संसारः। न किञ्चिदत्र शाश्वतमस्ति। परमहं नद्यां स्नानार्थं गमिष्यामि। तत् त्वम आत्मीयं एनं शिशु धनदेवनामानं मया सह
स्नानोपकरणहस्तं प्रेषय” इति।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. वणिक्पुत्रस्य नाम किम् आसीत्?
2. लौहघटिता तुला कीदृशी आसीत्।
3. जीर्णधनः कुत्र प्रस्थित:?

उत्तर:

1. जीर्णधनः,
2. पूर्वपुरुषोपार्जिता
3. देशान्तरम्।

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव

1. वाणिवपुत्रः स्वपुरमागत्य श्रेष्ठिनम् किम् अवदत्?
2. श्रेष्ठी किम् अवदत्?

उत्तर:

1. वणिक्पुत्रः स्वपुरमागत्य श्रेष्ठिनम् अवदत्- “भोः श्रेष्ठिन्! दीयतां में सा निक्षेपतुला।”
2. श्रेष्ठी अवदत् “भोः। नास्ति सा, त्वदीया तुला मूषक: भक्षिता” इति।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. अत्र “जीर्णधनः” इति कर्तृपदस्य क्रियापदं किम्?
2. अस्मिन् अनुच्छेदे ‘काल’ इति पदस्य विशेषणपदं किम् अत्र?
3. ‘अशाश्वतं’ इति पदस्य विपर्यय पदम् किम् अत्र?
4. अनुच्छेदे ‘अस्ति’ इति क्रियापदस्य विपर्ययपदं किम् प्रयुक्तम्?

उत्तर:

1. आसीत्
2. सुचिरं
3. शाश्वतं
4. आसीत्

2. स श्रेष्ठी स्वपुत्रम् अवदत्-“वत्स! पितृव्योऽयं तव, स्नानार्थं यास्यति, तद् अनेन साकं गच्छ” इति। अथासौ श्रेष्ठिपुत्रः धनदेवः स्नानोपकरणमादाय प्रहष्टमनाः तेन अभ्यागतेन सह प्रस्थितः। तथानुष्ठिते स वणिक् स्नात्वा तं शिशुं गिरिगुहायां प्रक्षिप्य, तद्वारं बृहत् शिलया आच्छाद्य सत्त्वर गृहमागतः। सः श्रेष्ठी पृष्टवान्-“भोः! अभ्यागत! कथ्यतां कुत्र मे शिशुः यः त्वया सह नदीं गतः”? इति। स अवदत्-“तव पुत्रः नदीतटात् श्येनेन हृतः” इति। श्रेष्ठी अवदत्- “मिथ्यावाविन्! कि क्वचित् श्येनो बाल हर्तुं शक्नोति? तत् समर्पय मे सुतम् अन्यथा राजकुले निवेदयिष्यामि।” इति। सोऽकथयत्-“भोः सत्यवादिन्! यथा श्येनो बालं न नयति, तथा मूषका अपि लौहघटिता तुला न भक्षयन्ति।
तदर्पय मे तुलाम्, यदि वारकेण प्रयोजनम्।” इति।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. वणिक्शिशुः कीदृशं मनः तेन सह प्रस्थितः?
2. लौहघटितातुला केन खादिता?
3. तेन सह कः प्रस्थितः?

उत्तर:

1. (प्रसन्नमनः) प्रहृष्टमनाः
2. मूषकैः
3. वणिक्शिशुः

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. वणिक्पुत्रः किम् कृत्वा गृहमागतः?
2. श्रेष्ठी स्वपुत्रम् किम् उवाच?

उत्तर:

1. वणिक्पुत्रः स्नात्वा तं शिशु (वणिक्शिशं) गिरिगुहायां प्रक्षिप्य, तद्द्वार बृहत् शिलया आच्छाद्य सत्वर गृहम् आगतः।
2. श्रेष्ठी स्वपुत्रमुवाच-“वत्स! पितृव्योऽयं तव, स्नानार्थ यास्यति, तद् गम्यताम् अनेन साकम्।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. अत्र ‘मिथ्यावादिन्’ इति पदस्य विपर्ययपदं किम् प्रयुक्तम्?
2. ‘भक्षयन्ति’ इति क्रियापदस्य कर्तृपदं किम् अस्मिन् अनुच्छेदे?
3. अत्र गद्यांशे ‘शीघ्रम्’ इति पदस्य पर्यायपदं किम् प्रयुक्तम्?
4. ‘पुत्रम्’ इत्यर्थे किं पदं प्रयुक्तम् अत्र?

उत्तर:

1. सत्यवादिन्,
2. मूषकाः
3. सत्वरम्
4. सुतम्

3. एवं विवदमानौ तौ द्वावपि राजकुलं गतौ। तत्र श्रेष्ठी तारस्वरेण अवदत्- “भोः। वञ्चितोऽहम्! वञ्चितोऽहम्! अब्रह्मण्यम्! अनेन चौरेण मम शिशुः अपहृतः” इति। अथ धर्माधिकारिणः तम् अवदन्-“भोः! समयंता श्रेष्ठिसुतः । सोऽवदत्- “किं करोमि? पश्यतो मे नदीतटात् श्येनेन शिशुः अपहृतः ”। इति। तच्छ्रुत्वा ते अवबन्-भोः। भवता सत्यं नाभिहितम्-कि श्येनः शिशु हतु समर्थों भवति? सोऽवदत्-भोः भोः! श्रूयतां मद्वचः
तुला लौहसहस्रस्य यत्र खादन्ति मूषकाः। राजन्तत्र हरेच्छ्येनो बालकं, नात्र संशयः॥ ते अपृच्छन्- “कथमेतत्”। ततः स श्रेष्ठी सध्यानामग्रे आदितः सर्व वृत्तान्तं न्यवेदयत्। ततः न्यायाधिकारिणः विहस्य तौ द्वावपि सम्बोध्य
तुला-शिशुप्रदानेन तोधितवन्तः।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. राजकुलं को गती?
2. श्रेष्ठी कीदृशेन स्वरेण प्रोवाच?
3. किं कुर्वाणौ तौ द्वावपि राजकुलं गतौ?

उत्तर:

1. द्वावपि
2. तारस्वरेण
3. विवदमानौ।

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. श्रेष्ठी तारस्वरेण किम् उवाच?
2. अत्र किं संशयः न अस्ति?

उत्तर:

1. श्रेष्ठी तारस्वरेण उवाच- “भोः! अब्रह्मण्यम्! अब्रह्मण्यम्! अनेन चौरेण मम शिशुिः अपहृतः” इति।
2. यत्र लौहसहस्रस्य तुलाम् मूषकाः खादन्ति तत्र श्येनः अपि बालकम् हरेत् अत्र संशयः न अस्ति।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. ‘सर्व’ इति पदस्य विशेष्य पदं किम् अत्र?
2. गद्यांशे अत्र “सन्तोषितौ” इति क्रियापदस्य कर्तृपदं किम्?
3. “प्रारम्भतः” इति पदस्य पर्यायपदं किम अस्मिन् अनुच्छेदे?
4. अत्र “उच्चस्वरेण” इति अर्थ किम् पदं प्रयुक्तम्?

उत्तर:

1. वृत्तान्तम्
2. तौ
3. आदितः
4. तारस्वरेण।

प्रश्न 2.
निम्नवाक्येषु रेखाङ्कित पदानां स्थानेषु प्रश्नवाचक पद लिखत –

प्रश्न 1.
लौहघटिता पूर्वपुरुषोपार्जिता तुलासीत्।
(क) कीदृशा
(ख) कीदृशी
(ग) कीदृशम्
(घ) कम्
उत्तर:
(ख) कीदृशी

प्रश्न 2.
सः श्रेष्ठिनो गृहे रक्षित्वा विदेशं अगच्छत्।
(क) केन
(ख) कस्य
(ग) कैः
(घ) कम्
उत्तर:
(ख) कस्य

प्रश्न 3.
त्वदीया तुला मूषकैः भक्षिता।
(क) केन
(ख) कया
(ग) कैः
(घ) कम्
उत्तर:
(ग) कैः

प्रश्न 4.
संसारे किञ्चिदपि शाश्वतं नास्ति।
(क) किम्
(ख) कीदृशम्
(ग) कस्य
(घ) कम्
उत्तर:
(ख) कीदृशम्

प्रश्न 5.
संसारे किञ्चिदपि शाश्वतं नास्ति।
(क) कुत्र
(ख) कीदृशम्
(ग) कम्
(घ) किम्
उत्तर:
(क) कुत्र

प्रश्न 6.
वणिक्शिशुः अभ्यागतेन सह प्रस्थितः।
(क) केन
(ख) कः
(ग) कैः
(घ) कस्य
उत्तर:
(क) केन

प्रश्न 7.
तत् द्वार बृहत् शिलया आच्छाद्य आगतः।
(क) केन
(ख) कया
(ग) कस्या
(घ) किम्
उत्तर:
(ख) कया

प्रश्न 8.
सः शिशुं गिरिगुहायां प्रक्षिप्य आगतः।
(क) कया
(ख) कस्याः
(ग) कस्याम्
(घ) किम्
उत्तर:
(ग) कस्याम्

प्रश्न 9.
वणिक्पुत्रः सत्वरं गृहं आगतः।
(क) कम्
(ख) किम्
(ग) कुत्र
(घ) कस्या
उत्तर:
(ग) कुत्र

प्रश्न 10.
मम शिशुः त्वया सह नीं गतः।
(क) कम्
(ख) कस्या
(ग) केन
(घ) किम्
उत्तर:
(ग) केन

प्रश्न 11.
क्वचित् श्येनो बालं हर्तुम् समर्थोऽस्ति।
(क) को
(ख) कः
(ग) कोहशः
(घ) कान्
उत्तर:
(ख) कः

प्रश्न 12.
श्रेष्ठी तारस्वरेण उवाच।
(क) की
(ख) का
(ग) क:
(घ) कोहशः
उत्तर:
(ग) क:

प्रश्न 13.
तौ विवदमानौ राजकुलं गतौ।
(क) कम्
(ख) को
(ग) के
(घ) क:
उत्तर:
(ख) को

प्रश्न 3.
अन्वय लेखनम् –

1. यत्र देशेऽथवा स्थाने भोगा भुक्ताः स्ववीर्यतः।
तस्मिन् विभवहीनो यो वसेत् स पुरुषाधमः॥
अन्वय- यत्र

1. ……….. अथवा
2. ……….. स्ववीर्यतः भोगा:
3. ……….. तस्मिन्
4. ……….. य: वसेत् स पुरुषाधमः।

मञ्जूषा – विभवहीनः, स्थाने, देशे, भुक्ता
उत्तर:

1. देशे
2. स्थाने
3. भुक्ता:
4. विभवहीनः।

2. तुलां लोहसहस्रस्य यत्र खादन्ति मूषकाः।
राजन्तत्र हरेच्छ्येनो बालकं, नात्र संशयः॥
अन्वय- राजन्!

1. ……….. लोहसहस्रस्य
2. …….. मुषका:
3. ……. तत्र
4. ……….. बालकम् हरेत् अत्र संशयः ना

मञ्जूषा – तुलाम्, यत्र, श्येनः, खादन्ति |
उत्तर:

1. यत्र
2. तुलाम्
3. खादन्ति
4. श्येनः।

प्रश्न 4.
निम्न श्लोकनि पठित्वा भावलेखनम् –

1. यत्र देशेऽथवा स्थाने भोगा भुक्ताः स्ववीय॑तः ।
तस्मिन् विभवहीनो यो वसेत स पुरुषाधमः ॥
अस्य भावोऽस्ति- यत्र देशे

1. ….. स्थाने
2. ……… भोगाः भुक्ता
3. ………….. यः विभवहीन:
4. ……….. सः पुरुषः अधमः।।

मञ्जूषा –  वसेत्, स्ववीर्यतः, अथवा, तस्मिन् ।
उत्तर:

1. अथवा
2. स्ववीर्यतः
3. तस्मिन्
4. वसेत्।

2. तुला लौहसहस्रस्य यत्र खादति मूषकाः।
राजन्तत्र हरेच्छ्येनो बालकं नात्र संशयः।।
अस्य भावोऽस्ति-

1. ………… यत्र
2. …………… तुला
3. …….. खादन्ति तत्र श्वेनः
4. ……….. हरेत् अत्र संशयः न।।

मञ्जूषा – बालकम्, मूषकाः, लौहसहस्रस्य, राजन् |
उत्तर:

1. राजन्
2. लौहसहस्रस्य
3. मूषकाः
4. बालकम्।

प्रश्न 5.
निम्नवाक्यानि घटनाक्रमानुसार पुनर्लिखत –

प्रश्न 1.

1. भोः! नास्ति सा, त्वदीया तुला मूषकैक्षिता इति।
2. स: एकदा विदेशं गच्छन् अचिन्तयत्।
3. कस्मिंश्चिद् अधिष्ठाने जीर्णधनः नामक: वाणिक्पुत्रः आसीत्।
4. धनिकः नृपस्य समीपं न्यायार्थम् अगच्छत्।
5. भोः श्रेष्ठिन्! दीयता में सा निक्षेपतुला।
6. इमाम् लौहतुलां श्रेष्ठिनः गृहे निक्षिप्य गच्छामि।
7. कथं श्येनः पुत्रं नेतुं समर्थोऽस्ति।
8. अस्य पुत्रस्य अपहरणं श्येनेन कृतम्।

उत्तर:

1. करिमश्चिद् अधिष्ठाने जीर्णधनः नामकः वाणिक्पुत्रः आसीत्।
2. स: एकदा विदेशं गच्छन् अचिन्तयत्।
3. इमाम् लौहतुलां श्रेष्ठिनः गृहे निक्षिप्य गच्छामि।
4. भोः श्रेष्ठिन्! दीयतां में सा निक्षेपतुला।
5. भो: नास्ति सा, त्वदीया तुला मूषकैक्षिता इति।
6. कथं श्येनः पुत्रं नेतुं समर्थोऽस्ति।
7. धनिकः नृपस्य समीपं न्यायार्थम् अगच्छत्।
8. अस्य पुत्रस्य अपहरणं श्येनेन कृतम्।

प्रश्न 2.

1. तव तुला मूषकाः खादिताः।
2. न्यायाधीशस्य आज्ञया धनिकः तस्य तुलाम् अयच्छत्।
3. न्यायाधीशः जीर्णधनस्य वार्ता श्रुत्वा अवदत्-‘कथमेतत् भवितुं शक्नोति’?
4. तव पुत्रं श्येन: नीतवान्।
5. जीर्णधनः सर्वां वार्ता कथयति।
6. एकदा एकः जीर्णधन: नाम वाणिवपुत्रः न्वयसत्।
7. सः स्वलौहतुलां धनिकस्य समीप निक्षिप्य विदेशम् अगच्छत्।
8. जीर्णधनः स्नानार्थं धनिकस्य पुत्रं नीत्वा नदीम् अगच्छत्।

उत्तर:

1. एकदा एकः जीर्णधनः नाम वाणिवपुत्रः न्वयसत्।
2. सः स्वलौहतुलां धनिकस्य समीपं निक्षिप्य विदेशम् अगच्छत्।
3. तव तुलां मूषकाः खादिताः।
4. जीर्णधनः स्नानार्थ धनिकस्य पुत्रं नीत्वा नदीम् अगच्छत्।
5. तव पुत्रं श्येन: नीतवान्।
6. न्यायाधीश: जीर्णधनस्य वार्ता श्रुत्वा अवदत्-‘कथमेतत् भवितुं शक्नोति’?
7. जीर्णधनः सर्वां वार्ता कथयति।
8. न्यायाधीशस्य आज्ञया धनिकः तस्य तुलाम् अयच्छत्।

प्रश्न 6.
‘क’ स्तम्भे लिखिताना पदाना पर्यायाः ‘ख’ स्तम्भे लिखिताः। तान् यथासमक्ष योजयत् –

‘क’ स्तम्भः – ‘ख’ स्तम्भः

1. अधिष्ठाने – एतादृशः
2. निक्षेपः – धनाभावात्
3. ईदृशः – देहि
4. विभवक्षयात् – प्रारम्भतः
5. समर्पय – बोधयित्वा
6. आदितः – हसित्वा
7. संबोध्य – न्यासः
8. विहस्य – स्थाने

उत्तर:

1. – स्थाने
2. – न्यासः
3. – एतादृशः
4. – धनाभावात्
5. – देहि
6. – प्रारम्भत :
7. – बोधयित्वा
8. – हसित्वा

प्रश्न 7.
‘क’ स्तम्भे विशेषणानि ‘ख’ स्तम्भे विशेष्याणि दत्तानि। तानि समुचित योजयत –

‘क’ स्तम्भः – ‘ख’ स्तम्भः

1. सुचिरे – तुलाम्
2. कस्मिंश्चित् – वृत्तान्तम्
3. लौहघटिताम् – पुरुषः
4. सर्व – तौ
5. अधमः – अधिष्ठाने
6. विवादमानौ – काले

उत्तर:

1. – काले
2. – अधिष्ठाने
3. – तुलाम्
4. – वृत्तान्तम्
5. – पुरुषः
6. – तौ

प्रश्न 8.
निम्न पदानाम् दत्तेषु विपर्ययपदेषु शुद्धं विपर्यय सह मेलनं कुरुत’ –

पदानि – विपर्ययाः

1. – गन्तुम् – मदीया
2. – सत्यवादिन्! – पुरुषाधमः
3. – अस्ति – अपर्य
4. – गृहाण – श्रेष्ठिन्!
5. – त्वदीया – आदितः
6. – सत्वरम् – आसीत्
7. – दुःखितमनाः – मिध्यावादिन!
8. – पुरुष श्रेष्ठ – चिरम्
9. – अन्ततः – प्रहृष्टमनाः
10. – भिक्षुक! – आगन्तुम्

उत्तर:

1. – आगन्तुम्,
2. – मिथ्यावादिन्!
3. – आसीत्
4. – अपर्य
5. – मदीया
6. – चिरम्
7. – प्रहष्टमनाः
8. – पुरुषाधमः
9. – आदितः
10. – श्रेष्ठिन्!।

We have given detailed NCERT Solutions for Class 6 Sanskrit Ruchira Chapter 3 शब्द परिचयः 3 Textbook Questions and Answers come in handy for quickly completing your homework.

NCERT Solutions for Class 6 Sanskrit Ruchira Chapter 3 शब्द परिचयः 3

Class 6th Sanskrit Chapter 3 शब्द परिचयः 3 Textbook Questions and Answers

अभ्यासः

प्रश्न 1.
(क) उच्चारणं कुरुत- (उच्चारण कीजिए- Read it out.)

(ख) चित्राणि दृष्ट्वा पदानि उच्चारयत- (चित्रों को देखकर शब्दों का उच्चारण कीजिए Look at the pictures and pronounce these words.)

उत्तर:
(क) छात्र स्वयं उच्चारण करें। ये सभी शब्द नपुंसकलिङ्ग हैं।
(ख) छात्र चित्रों को देखकर पदों का उच्चारण करें। उच्चारण करते समय चित्र की ओर ध्यान दें ताकि पद का अर्थ याद रहे। उच्चारण शुद्ध हो। प्रत्येक पद का उच्चारण यदि दो बार
करें, तो लाभ होगा। सभी नपुंसकलिङ्ग पद हैं।

प्रश्न 2.
(क) वर्णसंयोजनं कृत्वा पदं कोष्ठके लिखत- (वर्णों को जोड़कर शब्द कोष्ठक में लिखिए Combine the letters and write down the word in the box.)

(ख) अधोलिखितानां पदानां वर्णविच्छेदं कुरुत- (निम्नलिखित पदों का वर्ण-विच्छेद कीजिए Separate the letters in the words given below.)

उत्तर:
(क) खनित्रम्, पुराणानि, पोषकाणि, कङ्कतम्।

प्रश्न 3.
चित्राणि दृष्ट्वा संस्कृतपदानि लिखत- (चित्र देखकर संस्कृत शब्द लिखिए- Look at the pictures and write the -words in Sanskrit.)

उत्तर:
(क) आम्रम्
(ख) पर्णम्
(ग) व्यजनम्
(घ) करवस्त्रम्
(ङ) सूत्रम्
(च) छत्रम्

प्रश्न 4.
चित्रं दृष्ट्वा उत्तरं लिखत- (चित्र देखकर उत्तर लिखिए- Look at the pictures and write the answer.)

उत्तर:
(क) मयूरौ नृत्यतः।
(ख) एते क्रीडनके स्तः।
(ग) बालिकाः पठन्ति।
(घ) पुष्पाणि विकसन्ति।

प्रश्न 5.
निर्देशानुसारं वाक्यानि रचयत- (निर्देशानुसार वाक्य बनाएँ- Transform the sentences as directed.)

यथा- एतत् पतति। (बहुवचने) — ………………… एतानि पतन्ति।
(क) एते पणे स्तः। (बहुवचने) — …………………
(ख) मयूरः नृत्यति। (बहुवचने) — …………………
(ग) एतानि यानानि। (द्विवचने) — …………………
(घ) छात्रे लिखतः। (बहुवचने) — …………………
(ङ) नारिकेलं पतति। (द्विवचने) — …………………
उत्तर:
(क) एतानि पर्णानि सन्ति।
(ख) मयूराः नृत्यन्ति।
(ग) एते याने।
(घ) छात्राः लिखन्ति।
(ङ) नारिकेले पततः।

प्रश्न 6.
उचितपदानि संयोज्य वाक्यानि रचयत- (उचित पदों का मेल करके वाक्य बनाइए- Match the appropriate words and frame sentences.)

कोकिले — विकसति
पवनः — नृत्यन्ति
पुष्पम् — उत्पतति
‘खगः — वहति
मयूराः — गर्जन्ति
सिंहाः — कूजतः
उत्तर:
(क) कोकिले कूजतः।
(ख) पवनः वहति।
(ग) पुष्पम् विकसति।
(घ) खगः उत्पतति।
(ङ) मयूराः नृत्यन्ति।
(च) सिंहाः गर्जन्ति।

Class 6th Sanskrit Chapter 3 शब्द परिचयः 3 Additional Important Questions and Answers

प्रश्न 1.
चित्रं दृष्ट्वा संस्कृतपदं लिखत। (प्रत्येक चित्र देखकर संस्कृत पद लिखिए। Look at each picture and write down the word in Sanskrit.)

उत्तर:
(क) उपनेत्रम् – कमलम् – सङ्गणकम्
(ख) पर्णम् – क्रीडनकम् – नारिकेलम्

प्रश्न 2.
वर्णसंयोजनम् कुरुत। (वर्णसंयोग कीजिए- Join the alphabets.)

उत्तर:
(i) सङ्गणकम्
(ii) सूत्रम्
(iii) उद्यानम्
(iv) क्रीडनकम्

(ख) वर्णविच्छेदं कुरुत। (वर्ण-विच्छेद कीजिए- Separate the alphabets.)

पर्णम् = ………………….
वातायनम् = ………………….
कङ्कतम् = ………………….
कदलीफलम् = ………………….
उत्तरम्-

प्रश्न 3.
अधोदत्ताम् तालिकां पूरयत। (निम्नलिखित तालिका को पूरा कीजिए- Complete the table given below.)

उत्तर:


उत्तर:

प्रश्न 4.
संस्कृतेन लिखत- (संस्कृत में लिखिए– Write down the word in Sanskrit.)

यथा — खेत – क्षेत्रम्
(क) छाता …………….
(ख) नारियल …………….
(ग) छुरी …………….
(घ) कुञ्जी …………….
(ङ) कंघा …………….
उत्तर:
(क) छत्रम्
(ख) नारिकेलम्
(ग) छुरिका
(घ) कुञ्चिका
(ङ) कङ्कतम्/ककतम्

प्रश्न 5.
प्रत्येकं स्तम्भात् उचितं पदम् आदाय वाक्यानि रचयत। (प्रत्येक स्तम्भ से उचित पद लेकर वाक्य बनाइए- Frame sentences by picking out the appropriate word from each column.)

उत्तर:

प्रश्न 6.
अधोदत्तायां तालिकायाम् रिक्तस्थानानि पूरयत। (तालिका में रिक्त स्थान भरिए- Fill in the blanks in the table given below.)

उत्तर:

बहुविकल्पीयप्रश्नाः

प्रश्न 1.
उचित विकल्पं चित्वा वाक्यपूर्तिं कुरुत। (उचित विकल्प चुनकर वाक्य पूरे कीजिए। Complete the sentences by picking out the correct option.)

(क)
(i) ……….. आम्राणि मधुराणि। (एतत्, एते, एतानि)
(ii) ………. उद्यानम् सुंदरम्। (सः, सा, तत्)
(ii) ………. बसयाने कुत्र गच्छतः। (ते, सा, तानि)
(iv) ………. कोकिला कूजति। (एषः, एषा, एतत्)
(v) ………. जनाः गच्छन्ति। (सः, तौ, ते)
उत्तर:
(i) एतानि
(ii) तत्
(iii) ते
(iv) एषा
(v) ते

(ख)
(i) अत्र ………. अस्ति। (विश्रामगृह, विश्रामगृहम्, विश्रामगृहः)
(ii) …………. विकसति। (कमलम्, कमल, कमलः)
(iii) ………. मधुराणि सन्ति। (कदलीफलम्, कदलीफले, कदलीफलानि)
(iv) ………. कुत्र स्त? (पुस्तकम्, पुस्तके, पुस्तकानि)
(v) ………. चलति। (व्यञ्जनम्, व्यजनः, व्यजनम्)
उत्तर:
(i) विश्रामगृहम्
(ii) कमलम्
(iii) कदलीफलानि
(iv) पुस्तके
(v) व्यजनम्

प्रश्न 2.
उचितं विकल्पं चित्वा चित्रस्य समक्षं लिखत। (उचित विकल्प चुनकर चित्र के सामने लिखिए।Pick out the correct option and write down in front of the picture.)


उत्तर:
(क) करवस्त्रम्
(ख) सोपानम्
(ग) वातायनम्
(घ) आम्रम्
(ङ) व्यजनम्

प्रश्न 3.
उचित-विकल्पेन रिक्तस्थानानि पूरयत। (उचित विकल्प द्वारा रिक्त स्थान भरिए। Fill in the blanks with the correct option.)

(क)
(i) पुष्पाणि …………….. (विकसति, विकसतः, विकसन्ति)
(ii) बालिके कुत्र …………….. ? (गच्छति, गच्छतः, गच्छन्ति)
(iii) ………………. पतन्ति । (पर्णाः, पर्णम्, पर्णानि)
(iv) ……………….. कुत्र अस्ति? (उद्यान, उद्यानम्, उद्यानानि)
(v) …………….. अत्र अस्ति। (वातायन, वातायनः, वातायनम्)
उत्तर:
(i) विकसन्ति
(ii) गच्छतः
(iii) पर्णानि
(iv) उद्यानम्
(v) वातायनम्।

(ख)
(i) पर्णम् ………….। (पतति, उत्पतति)
(ii) सुवर्णकारः अंगुलीयकम् ………….। (चालयति, रचयति)
(iii) सौचिकः ………….। (सीव्यति, सूचयति)
(iv) अजाः ………….। (चरन्ति, कूजन्ति)
(v) जवनिका …………… (चलति, दोलति)
उत्तर:
(i) पतति
(ii) रचयति
(iii) सीव्यति
(iv) चरन्ति
(v) दोलति।

We have given detailed NCERT Solutions for Class 10 Sanskrit Shemushi Chapter 10 भूकंपविभीषिका Questions and Answers come in handy for quickly completing your homework.

Shemushi Sanskrit Class 10 Solutions Chapter 10 भूकंपविभीषिका

Class 10 Sanskrit Shemushi Chapter 10 भूकंपविभीषिका Textbook Questions and Answers

अभ्यासः

प्रश्न 1.
एकपदेन उत्तरं लिखत-

(क) कस्य दारुण-विभीषिका गुर्जरक्षेत्र ध्वंसावशेषेषु परिवर्तितवती?
उत्तराणि:
भूकम्पस्य

(ख) कीदृशानि भवनानि धाराशायीनि जातानि?
उत्तराणि:
बहुभूमिकानि

(ग) दुर्वार-जलधाराभिः किम् उपस्थितम्?
उत्तराणि:
महाप्लावनदृश्यम्

(घ) कस्य उपशमनस्य स्थिरोपायः नास्ति?
उत्तराणि:
दैवीप्रकोपो/भूकम्पो

(ङ) कीदृशाः प्राणिनः भूकम्पेन निहन्यन्ते?
उत्तराणि:
विविशाः

प्रश्न 2.
अधोलिखितानां प्रश्नानाम् उत्तराणि संस्कृतभाषया लिखत-

(क) समस्तराष्ट्र कीदृशे उल्लासे मग्नम् आसीत्?
उत्तराणि:
समस्त राष्ट्र नृत्य-गीतवादित्राणाम् उल्लासे मग्नम् आसीत्।

(ख) भूकम्पस्य केन्द्रबिन्दुः क: जनपद: आसीत्?
उत्तराणि:
भूकम्पस्य केन्द्रबिन्दुः कच्छजनपदः आसीत्।

(ग) पृथिव्याः स्खलनात् कि जायते?
उत्तराणि:
पृथिव्याः स्खलतात् महाविनाशदृश्य जायते।

(घ) समग्रं विश्वं कै: आतंकित: दृश्यते?
उत्तराणि:
समग्र विश्वः प्राकृतिक (दैवीय) आपदाभिः आतंकित: दृश्यते।

(ङ) केषां विस्फोटैरपि भूकम्पो जायते?
उत्तराणि:
ज्वालामुखपर्वतानां विस्फोटैरपि भूकम्पो जायते।

प्रश्न 3.
स्थूलपदानि आधृत्य प्रश्ननिर्माणं कुरुत-

(क) भूकम्पविभीषिका विशेषेण कच्छजनपदं ध्वंसावशेषेषु परिवर्तितवती।
उत्तराणि:
केषु

(ख) वैज्ञानिकाः कथयन्ति यत् पृथिव्याः अन्तर्गर्भ, पाषाणशिलानां संघर्षणेन कम्पनं जायते।
उत्तराणि:
के

(ग) विवशाः प्राणिनः आकाशे पिपीलिकाः इव निहन्यन्ते।
उत्तराणि:
कुत्र/कस्मिन्

(घ) एतादृशी भयावहघटना गढ़वालक्षेत्रे घटिता।
उत्तराणि:
कीदृशी

(ङ) तदिदानीम् भूकम्पकारणं विचारणीय तिष्ठति।
उत्तराणि:
किम्

प्रश्न 4.
‘भूकम्पविषये’ पञ्चवाक्यमितम् अनुच्छेद लिखत।
उत्तराणि:

1. भूकम्पधरायाः विनाशलौली भवति।
2. यदा धरायाः अन्तः स्थितासु शिलासु संघर्षणं भवति तदा भूकम्प: जायते।
3. भूकस्पेन अपरिमित लावाराः धरातलात् निर्गत्य नदीवेगेन प्रवहन्तः ग्रामेषु नगरेषु वा महाविनाशं कुर्वन्ति।
4. अतः अस्माभिः प्रकृतेः विरुद्धानि कार्याणि न करणीयानि।
5. बहुभूमिक भवननिर्माणं नदीतटं बन्धान्, वृक्षाणां कर्तनम् का न करणीयम्।

प्रश्न 5.
कोष्ठकेषु दत्तेषु धातुषु निर्देशानुसारं परिवर्तनं विधाय रिक्तस्थानानि पूरयत

(क) समग्रं भारतं उल्लासे मग्नः ___________ (अस् + लट् लकारे)
(ख) भूकम्पविभीषिका कच्छनपदं विनष्टं ___________ (कृ + क्तवतु + ङीप)
(ग) क्षणेनैव प्राणिनः गृहविहीनाः ___________ (भू + लङ, प्रथमः पुरुषः बहुवचनम्)
(घ) शान्तानि पञ्चतत्त्वानि भूतलस्य योगक्षेमाभ्यां ___________ (भू + लट्, प्रथम प्रथम-पुरुषः बहुवचनम्)
(ङ) मानवाः ___________ यत् बहुभूमिकभवननिर्माणं करणीयम् न वा? (पृच्छ् + लट्, प्रथम-पुरुषः बहुवचनम्)
(च) नदीवेगेन ग्रामाः तदुदरे ___________ (सम् + आ + विश् + विधिलिङ, प्रथम पुरुषः एकवचनम्)
उत्तराणि:
(क) अस्ति
(ख) कृतवती
(ग) अभवन्
(घ) भवन्ति
(ङ) पृच्छन्ति
(च) समाविशेयुः

प्रश्न 6.
सन्धि / सन्धिविच्छेदं च कुरुत-

(अ) परसवर्णसन्धिनियमानुसारम्
(क) किञ्च = ___________ + च
(ख) ___________ = नगरम् + तु
(ग) विपन्नञ्च = ___________ + ___________
(घ) ___________ = किम + नु
(ङ) भुजनगरन्तु = ___________ + ___________
(च) ___________ = सम् + चयः
उत्तराणि:
(क) किम्
(ख) नगरन्तु
(ग) विपन्नम् + च
(घ) किन्नु
(ङ) भुजनगरम् + तु
(च) सञ्चयः

(आ) विसर्गसन्धिनियमानुसारम्

(क) शिशवस्तु = ___________ + ___________
(ख) ___________ = विस्फोटैः + अपि
(ग) सहस्रशोऽन्ये = _________ + अन्ये
(घ) विचित्रोऽयम् = विचित्र: + _________
(ड) _________ = भूकम्पः + जायत
(च) वामनकल्प एव = ___________ + ___________
उत्तराणि:
(क) शिशवः + तु
(ख) विस्फोटैरपि
(ग) सहस्रशः
(घ) अयम्
(ङ) भूकम्पो जायते
(च) वामनकल्पः + एव

प्रश्न 7(अ).
‘क’ स्तम्भे पदानि दत्तानि ‘ख’ स्तम्भे विलोमपदानि, तयोः संयोगं कुरुत-

उत्तराणि:

प्रश्न 7(आ).
‘क’ स्तम्भे पदानि दत्तानि ‘ख’ स्तम्भे समानार्थकपदानि, तयोः संयोगं कुरुत-

उत्तराणि:

प्रश्न 8(अ).
उदाहरणमनुसृत्य प्रकृति-प्रत्यययोः विभागं कुरुत-
यथा- परिवर्तितवती – परि + वृत् + क्तवतु + डीप् (स्त्री)
धृतवान् – _________ + _________
हसन् – _________ + _________
विशीर्णा – वि + शृ + क्त + _________
प्रचलन्ती – _________ + _________ + शतृ + डीप् (स्त्री)
हतः – _________ + _________
उत्तराणि:
धृतवान् – धृ + क्तवतु
हसन् – हस् + शतृ
विशीर्णा – वि + शृ + क्त + टाप् (स्त्री)
प्रचलन्ती – प्र + चल् + शतृ + डीप् (स्त्री)
हतः – हन् + क्त

प्रश्न 8(आ).
पाठात् विचित्य समस्तपदानि लिखत-
महत् च तत् कम्पन – _________
दारुणा च सा विभीषिका – _________
ध्वस्तेषु च तेषु भवनेषु – _________
प्राक्तने च तस्मिन् युगे – _________
महत् च तत् राष्ट्र तस्मिन् – _________
उत्तराणि:
महत् च तत् कम्पन – महत्कम्पनम्
दारुणा च सा विभीषिका – दारुणविभीषिका
ध्वस्तेषु च तेषु भवनेषु – ध्वस्तभवनेषु
प्राक्तने च तस्मिन् युगे – प्राग्युगे
महत् च तत् राष्ट्र तस्मिन् – महद्राष्ट्र

योग्यताविस्तारः
हमारे वातावरण में भौतिक सुख-साधनों के साथ-साथ अनेक आपदाएँ भी लगी रहती हैं। प्राकृतिक आपदाएँ जीवन को अस्त-व्यस्त कर देती हैं। कभी किसी महामारी की आपदा, बाढ़ तथा सूखे की आपदा या तूफान के रूप में भयंकर प्रलय-ये सब हम अपने जीवन में देखते तथा सुनते रहते हैं। भूकम्प भी ऐसी ही आपदा है, जिस पर यहाँ दुष्टिपात किया गया है। इस पाठ के माध्यम से यह बताया गया है कि किसी भी आपदा में बिना किसी घबराहट के, हिम्मत के साथ किस प्रकार हम अपनी सुरक्षा स्वयं कर सकते हैं।

भूकम्प परिचय- भूमि का कम्पन भूकम्प कहलाता है। वह बिन्दु भूकम्प का उद्गम केन्द्र कहा जाता है, जिस बिन्दु पर कम्पन की उत्पत्ति होती है। कम्पन तरंग के रूप में विविध दिशाओं में आगे चलता है। ये तरंगें सभी दिशाओं में उसी प्रकार फैलती हैं जैसे किसी शान्त तालाब में पत्थर के टुकडों को फेंकने से तरंगें उत्पन्न होती हैं।

धरातल पर कुछ स्थान ऐसे हैं जहाँ, भूकम्प प्रायः आते ही रहते हैं, उदाहरण के अनुसार-प्रशान्त महासागर के चारों ओर के प्रदेश, हिमाचल प्रदेश, गड़ा एवं ब्रह्मपुत्र का तटीय भाग, इन क्षेत्रों में अनेक भूकम्प आए जिनमें से कुछ तो अत्यधिक भयावह और विनाशकारी थे। सुनामी भी एक प्रकार का भूकम्पन ही है जिसमें भूमि के भीतर अत्यन्त गहराई से तीव्र कम्पन उत्पन्न होता है। यही कम्पन समुद्र के जल को काफी ऊँचाई तक तीव्रता प्रदान करता है। फलस्वरूप तटीय क्षेत्र सर्वाधिक प्रभावित होते हैं। सुनामी का भीषण प्रकोप 20 सितम्बर, 2004 को हुआ। जिसकी चपेट में भारतीय प्रायद्वीप सहित अनेक देश आ गये। क्षिति, जल, पावक, गगन और समीर इन पञ्चतत्वों में सन्तुलन बनाए रखकर प्राकृतिक आपदाओं से बचा जा सकता है। इसके विपरीत असन्तुलित पञ्चतत्वों से सृष्टि विनष्ट हो सकती है।

भूकम्पविषये प्राचीनतम्
प्राचीनैः, ऋषिभिः अपि स्वस्वग्रन्थेषु भूकम्पोल्लेखः कृतः येन स्पष्टं भवति यत् भूकम्पाः प्राचीनकालेऽपि आयान्ति स्म।
यथा-
वराहसंहितायाम्-
क्षितिकम्पमाहुरेके मह्यन्तर्जलनिवासिसत्त्वकृतम्
भूभारखिन्नदिग्गजनिः श्वास समुद्भवं चान्ये।
अनिलोऽनिलेन निहितः क्षिती पतन सस्वनं करोत्यन्ये
केचित् त्वदृष्टकारितमिदमन्ये प्राहुराचार्याः॥
मयूरचित्रे
कदाचित् भूकम्पः श्रेयसेऽपि कल्पते। एतादृशाः अपि उल्लेखाः अस्माकं साहित्ये समुपलभ्यन्ते यथा वारुणमण्डलमौशनसे-
प्रतीच्यां यदि कम्पेत वारुणे सप्तके गणे
द्वितीययामे रात्रौ तु तृतीये वारुणं स्मृतम्।
अत्र वृष्टिश्च महती शस्यवृद्धिस्तथैव च,
प्रज्ञा धर्मरताश्चैव भयरोगविवर्जिताः॥
उल्काभूकम्पदिग्दाहसम्भवः शस्यवृद्धये।
क्षेमारोग्यसुभिक्षार्थ वृष्टये च सुखाय च।
भूकम्पसमा एव अग्निकम्पः वायुकम्पः, अम्बुकम्पः इत्येवमन्येऽपि भवन्ति।

Class 10 Sanskrit Shemushi Chapter 10 भूकंपविभीषिका Additional Important Questions and Answers

अतिरिक्त प्रश्नाः

1. गद्यांशान् पठित्वा प्रश्नानाम् उत्तराणि लिखत-

(क) एकोत्तर द्विसहस्रनीष्टाब्ने (2001 ईस्वीये वर्षे) गणतन्त्र-दिवस-पर्वणि यदा समग्रमपि भारतराष्ट्रं नृत्य-गीतवादित्राणाम् उल्लासे मग्नमासीत् तदाकस्मादेव गुर्जर- राज्यं पर्याकुलं, विपर्यस्तम्, क्रन्दनविकलं, विपन्नञ्च जातम्। भूकम्पस्य दारुण- विभीषिका समस्तमपि गुर्जरक्षेत्र विशेषेण च कच्छजनपदं ध्वंसावशेषु परिवर्तितवती। भूकम्पस्य केन्द्रभूतं भुजनगरं तु मृत्तिकाक्रीडनकमिव खण्डखण्डम् जातम्। बहुभूमिकानि भवनानि क्षणेनैव धराशायीनि जातानि। उत्खाता विद्युद्दीपस्तम्भाः। विशीर्णाः गृहसोपान-मार्गाः। फालद्वये विभक्ता भूमिः। भूमिग दुपरि निस्सरन्तीभिः दुर्वार-जलधाराभिः महाप्लावनदृश्यम् उपस्थितम्। सहस्त्रमिताः प्राणिनस्तु क्षणेनैव मृताः। ध्वस्तभवनेषु सम्पीडिता सहस्रशोऽन्ये सहायतार्थं करुणकरुणं क्रन्दन्ति स्म। हा दैव! क्षुत्क्षामकण्ठाः मृतप्रायाः केचन शिशवस्तु ईश्वरकृपया एव द्वित्राणि दिनानि जीवन धारितवन्तः।

प्रश्न 1.
एकपदेन उत्तरत-
(i) भूकम्पः कुत्र अभवत्?
(ii) कस्याः क्रीडनकमिव भुजनगरं खण्डम् जातम्?
उत्तराणि:
(i) गुर्जरराज्ये
(ii) मृत्तिकायाः

प्रश्न 2.
पूर्णवाक्येन उत्तरत-
(i) भूकम्पस्य केन्द्रभूतं भुजनगरे किम् किम् अभवत्?
उत्तराणि:
(i) भूकम्पस्य केन्द्रभूतं भुजनगरे तु मृतिकाक्रीडनकमिव खण्डखण्डम् जातम्। बहुभूमिकानि भवनानि क्षणेनैव धराशायीनि जातानि। उत्खाता विद्युद्दीपस्तम्भाः।

प्रश्न 3.
भाषिककार्यम्-
(i) ‘भवनानि’ इति विशेष्यपदस्य विशेषणपदं किम्?
(ii) ‘आसीत्’ इति क्रियापदस्य कर्तृपदं किम्?
(iii) अस्मिन् गद्यांशे सम्बोधन पदं किम् प्रयुक्तम्।
उत्तराणि:
(i) बहुभूमिकानि
(ii) भारत
(iii) हा दैव!

(ख) इयमासीत् भैरवविभीषिका कच्छ-भूकम्पस्या पञ्चोत्तर-द्विसहस्रख्रीष्टाब्दे (2005 ईस्वीये वर्षे) अपि कश्मीर-प्रान्ते पाकिस्तान देशे च धरायाः महम्कम्पनं जातम्। यस्मात्कारणात् लक्षपरिमिताः जनाः अकालकालकवलिताः। पृथ्वी कस्मात्प्रकम्पते वैज्ञानिकाः इति विषये कथयन्ति यत् पृथिव्या अन्तर्गर्भे विद्यमानाः बृहत्यः पाषाण-शिला यदा संघर्षणवशात् त्रुट्यन्ति तदा जायते भीषणं संस्खलनम्, संस्खलनजन्य कम्पनञ्च। तदैव भयावहकम्पनं धरायाः
उपरितलमप्यागत्य महाकम्पनं जनयति येन महाविनाशदृश्यं समुत्पद्यते।

प्रश्न 1.
एकपदेन उत्तरत-
(i) कच्छ-भूकम्पस्य विभीषिका कीदृशी आसीत्?
(ii) धरायाः महत्कम्पनं कुत्र जातम्?
उत्तराणि:
(i) भैरव
(ii) कश्मीरप्रान्ते

प्रश्न 2.
पूर्णवाक्येन उत्तरत-
(i) पृथ्वी कस्मात् प्रकम्पते इति विषये वैज्ञानिकाः किम् कथयन्ति?
उत्तराणि:
(i) पृथ्वी कस्मात् प्रकम्पते इति विषये वैज्ञानिकाः कथयन्ति यत् पृथिव्या अन्तर्गर्भ विद्यमानाः बृहत्यः पाषाणशिला यदा संघर्षणवशात् त्रुट्यन्ति तदा भीषणं संस्खलनम् जायते, संस्खलनजन्य कम्पनञ्च।

प्रश्न 3.
भाषिककार्यम्-
(i) ‘वैज्ञानिकाः’ कर्तृपदस्य क्रियापदं किम्?
(क) पृथिव्या
(ख) कथयन्ति
(ग) अन्तर्गर्भ
(घ) इति
उत्तराणि:
(ख) कथयन्ति

(ii) ‘मृताः’ इत्यर्थे किं पदं प्रयुक्तम्?
(क) जनाः
(ख) कालकवलिताः
(ग) परिमिताः
(घ) पृथ्वी
उत्तराणि:
(ख) कालकवलिताः

(iii) ‘जनाः’ इति विशेष्यपदस्य विशेषणपदं किम्?
(क) कालकवलिताः
(ख) लक्षपरिमिताः
(ग) विद्यमानाः
(घ) वैज्ञानिका
उत्तराणि:
(ख) लक्षपरिमिताः

(ग) ज्वालामुखपर्वतानां विस्फोटैरपि भूकम्पो जायत इति कथयन्ति भूकम्पविशेषज्ञाः। पृथिव्याः गर्भे विद्यमानोऽग्निर्यदा खनिजमृत्तिकाशिलादिसञ्चयं क्वथयति तदा तत्सर्वमेव लावारसताम् उपेत्य दुर्वारगत्या धरा पर्वतं वा विदार्य बहिर्निष्क्रामति। धूमभस्मावतं जायते तदा गगनम्। सेल्सियश-ताप-मात्राया अष्टशतातामुपगतोऽयं लावारसो यदा नदीवेगेन प्रवहति तदा पार्श्वस्थग्रामा नगराणि वा तदुदरे क्षणेनैव समाविशन्ति।
निहन्यन्ते च विवशाः प्राणिनः। ज्वालामुगिरन्त एते पर्वता अपि भीषणं भूकम्पं जनयन्ति।

प्रश्न 1.
एकपदेन उत्तरत-
(i) केषाम् विस्फोटैः भूकम्पो जायते?
(ii) ज्वालामुखपर्वतानां कृते के कथयन्ति?
उत्तराणि:
(i) ज्वालामुखपर्वतानां
(ii) भूकम्पविशेषज्ञाः

प्रश्न 2.
पूर्णवाक्येन उत्तरत-
(i) ज्वालामुखपर्वतानां कृते भूकम्प विशेषज्ञाः किम् कथयन्ति?
उत्तराणि:
(i) ज्वालामुखपर्वतानां कृते भूकम्पविशेषज्ञाः कथयन्ति यत् पृथिव्याः गर्भ विद्यमानोऽग्निर्यद। खनिजमृत्तिकाशिलादि सञ्चयं क्वथयति तदा तत्सर्वमेव लावारसताम् उपेत्यर्वारगत्या धरा पर्वतं वा विदार्य बहिनिष्क्रामति।

प्रश्न 3.
भाषिककार्यम्-
(i) ‘भूकम्पविशेषज्ञाः’ इति कर्तृपदस्य क्रियापदं किम्?
(ii) ‘अन्तः’ इतिपदस्य विपर्ययपदं किं प्रयुक्तम्?
(iii) ‘विवशाः’ इति विशेषणपदस्य विशेष्यपदं किम्?
उत्तराणि:
(i) कथयन्ति
(ii) बहिः
(iii) प्राणिनः

(घ) यद्यपि दैवः प्रकोपो भूकम्पो नाम, तस्योपशमनस्य न कोऽपि स्थिरोपायो दृश्यते। प्रकृति समक्षमद्यापि विज्ञानगर्वितो मानवः वामनकल्प एव तथापि भूकम्परहस्यज्ञाः कश्चयन्ति यत् बहुभूमिकभवननिर्माणं न करणीयम्। तटबन्धं निर्माय बृहन्मात्र नदीजलमपि नैकस्मिन् स्थले पुञ्जीकरणीयम् अन्यथा असन्तुलनवशाद् भूकम्पस्सम्भवति। वस्तुतः शान्तानि एव पञ्चतत्त्वानि क्षितिजलपावकसमीरगगनानि भूतलस्य योगक्षेमाभ्यां कल्पन्ते। अशान्तानि खलु तान्येव महाविनाशम् उपस्थापयन्ति।

प्रश्न 1.
एकपदेन उत्तरत-
(i) कस्य प्रकोपस्य नाम भूकम्पः जायते?
(ii) कस्य स्थिरोपायो न दृश्यते?
उत्तराणि:
(i) दैवस्य
(ii) भूकम्पस्य

प्रश्न 2.
पूर्णवाक्येन उत्तरत-
(i) भूकम्परहस्यज्ञाः किम् कथयन्ति?
उत्तराणि:
(i) भूकम्परहस्यज्ञाः कथयन्ति यत् बहुभूमिकभवन-निर्माण न करणीयम्। तटबन्ध निर्माय बृहन्मात्र नदीजलमपि न एकस्मिन् स्थले पुजीकरणीयम् अन्यथा असन्तुलनवशात् भूकम्पः सम्भवति।

प्रश्न 3.
भाषिककार्यम्-
(i) ‘करणीयम्’ इति क्रियापदस्य कर्तृपदं किम्?
(क) भवननिर्माणं
(ख) निर्माय
(ग) तटबन्धं
(घ) मानव
उत्तराणि:
(क) भवननिर्माणं

(ii) ‘विज्ञानगर्वितः’ इति विशेषणपदस्य विशेष्यपदं किम्?
(क) वामनकल्पः
(ख) मानवः
(ग) कोऽपि
(घ) प्रकृतिः
उत्तराणि:
(ख) मानवः

(iii) भूकम्परहस्यज्ञाः इति कर्तृपदस्य क्रियापदं किम्?
(क) वस्तुतः
(ख) कथयन्ति
(ग) करणीयम्
(घ) निर्माय
उत्तराणि:
(ख) कथयन्ति

2. रेखातिपदम् प्रश्ननिर्माणं कुरुत-

(क) भारतराष्ट्रं नृत्य-गीतवादित्राणाम् उल्लासे मग्नमासीत्।
(ख) गुर्जर-राज्यं पर्याकुलं विपन्नञ्च जातम्।
(ग) भुजनगरं तु मृत्तिकाक्रीडनकमिव खण्डरखण्डम् जातम्।
(घ) बहुभूमिकानि भवनानि धराशायीनि जातानि।
(ङ) भूमिग दुपरि निस्सरन्तीभिः दुर्वार-जलधाराभिः महाप्लावनदृय॑म् उपस्थितम्।
(च) ज्वालामुखपर्वतानां विस्फोटैरपि भूकम्पो जायते।
(छ) भूकम्पविशेषज्ञाः इति कथयन्ति।
(ज) प्रकृति समक्षमद्यापि विज्ञानगर्वितो मानवः वामनकल्प अभवत्।
(झ) पाकिस्तानदेशे धरायाः महत्कम्पनं जातम्।
(ञ) पाषाणशिला संघर्षणवशात् त्रुट्यन्ति।
(ट) निहन्यन्ते च विवशाः प्राणिनः।
(ठ) अनेके भूकम्परहस्यज्ञाः कथयन्ति।
(ड) लक्षपरिमिताः जनाः अकालकालकवलिताः अभवन्।
(ढ) सम्पीडिताः सहायतार्थम् करुणकरुणं क्रन्दन्ति स्म।
उत्तराणि:
(क) कस्मिन्
(ख) कीदृशम्
(ग) कीदृशमिव
(घ) कीदृशानि
(ङ) काभिः
(च) केषाम्
(छ) के
(ज) कीदृशः
(झ) कस्याः
(ञ) कस्मात्
(ट) कीदृशाः
(ठ) के
(ड) के
(ढ) किमर्थम्

3. उचितं उत्तराणि चित्वा प्रश्ननिर्माणं कुरुत-

(क) ज्वालामुगिरन्तः एते पर्वता अपि भीषणं भूकम्पं जनयन्ति।
(i) किम्
(ii) कम्
(iii) कीदृशम्
(iv) काम
उत्तराणि:
(iii) कीदृशम्

(ख) नद्याः जलम् एकस्मिन् स्थले न पुजी कारणीयम्।
(i) कस्याः
(ii) का
(iii) किम्
(iv) कम्
उत्तराणि:
(i) कस्याः

(ग) पृथिव्याः गर्भे विद्यमानोऽग्निः क्वथयति।
(i) का
(ii) कस्याम्
(iii) कस्याः
(iv) किम्
उत्तराणि:
(ii) कस्याः

(घ) शिशवः तु द्वित्राणि दिनानि जीवनं धारितवन्तः।
(i) कति
(ii) कानि
(iii) कीदृशानि
(iv) कान्
उत्तराणि:
(i) कति

(ङ) सहस्रमिता: प्राणिनः तु क्षणनैव मृताः।
(i) काः
(ii) के
(iii) कति
(iv) कीदृशाः
उत्तराणि:
(iii) कति

(च) कश्मीरप्रान्ते धरायाः महत्कम्पनं जातम्।
(i) कुत्र
(ii) के
(iii) काः
(iv) कीदृशे
उत्तराणि:
(i) कुत्र

(छ) उत्खाता विद्युद्दीपस्तम्भाः।
(i) का
(ii) के
(iii) कीदृशाः
(iv) कथम्
उत्तराणि:
(ii) के

(ज) भूमिः फालडये विभक्ता आसीत्।
(i) क:
(ii) काः
(iii) का
(iv) की
उत्तराणि:
(iii) का

(झ) लक्षपरिमिताः जनाः अकालकालकवालिताः।
(i) कति
(ii) काः
(iii) के
(iv) कीदृशाः
उत्तराणि:
(i) कति

(ज) असन्तुलनवशात् भूकम्पः सम्भवति।
(i) कस्मात्
(ii) कात्
(iii) कस्मै
(iv) के
उत्तराणि:
(i) कस्मात्

4. अधोलिखिपदानां तेषाम् पर्यायपदानि सह मेलनम् कुरुत-


उत्तराणि:

5. अधोलिखितपदानां तेषाम् विपर्ययपदानि सह मेलनं कुरुत-

उत्तराणि:

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 14 Important Extra Questions Semiconductor Electronics: Materials, Devices and Simple Circuits

Semiconductor Electronics Important Extra Questions Very Short Answer Type

Question 1.
Draw the energy band diagram for a p-type semiconductor.
Answer:
The energy level diagram is shown below.

Question 2.
Draw the voltage-current characteristic of a p-n junction diode in forwarding bias and reverse bias.
Answer:
The characteristics are as shown.

Question 3.
Draw the voltage-current characteristic for a Zener diode.
Answer:
The V-l characteristic of the Zener diode is as shown.

Question 4.
Draw the energy band diagram for n-type semiconductor.
Answer:
The diagram is as shown.

Question 5.
An ac input signal of frequency 60 Hz is rectified by an
(i) Half wave and an
Answer:
The output frequency remains the same in a half-wave rectifier, i.e. 60 Hz.

(ii) Full-wave rectifier. Write the output frequency in each case.
Answer:
The output frequency becomes twice the input frequency in the case of the full-wave rectifier, i.e. 120 Hz.

Question 6.
Give the ratio of the number of holes and the number of conduction electrons in an intrinsic semiconductor.
Answer:
The ratio is one.

Question 7.
What is the depletion region in a p-n junction?
Answer:
It is a thin layer between p and n sections of the p-n junction which is devoid of free electrons and holes.

Question 8.
Name an impurity which when added to pure silicon makes it a
(i) p-type semiconductor
Answer:
Boron, aluminum, etc.

(ii) n-type semiconductor.
Answer:
Phosphorous, antimony, etc.

Question 9.
Which type of biasing gives a semiconductor diode very high resistance?
Answer:
Reverse biasing.

Question 10.
Identify the biasing in the figure given below.

Answer:
Forward biasing.

Question 11.
Draw the circuit symbol of (a) photodiode, and (b) light-emitting diode.
Answer:
The circuit symbols are as shown below.

Question 12.
What is the function of a photodiode? (CBSE AI 2013C)
Answer:
It functions as a detector of optical signals.

Question 13.
When a p-n junction diode is forward biased, how will its barrier potential be affected? (CBSEAI 2019)
Answer:
Potential barrier decreases in forwarding bias.

Question 14.
Name the junction diode whose l-V characteristics are drawn below: (CBSE Delhi 2017)

Answer:
Solar cell.

Question 15.
How does the width of the depletion region of a p-n junction vary if the reverse bias applied to it decreases?
Answer:
With the increase in the reverse bias, the depletion layer increases.

Question 16.
How does the width of the depletion region of a p-n junction vary if the reverse bias applied to it decreases?
Answer:
If the reverse bias decreases, the width of the depletion layer also decreases.

Question 17.
Why is the conductivity of n-type semiconductors greater than that of p-type semiconductors even when both of these have the same level of doping?
Answer:
It is because in n-type the majority carriers are electrons, whereas in p-type they are holes. Electrons have greater mobility than holes.

Question 18.
How does the conductance of a semiconducting material change with rising in temperature?
Answer:
Increases with an increase in temperature.

Question 19.
How is a sample of an n-type semiconductor electrically neutral though it has an excess of negative charge carriers?
Answer:
It is because it contains an equal number of electrons and protons and is made by doping with a neutral impurity.

Question 20.
How is the bandgap, Eg, of a photodiode related to the maximum wavelength, λ_m, that can be detected by it?
Answer:
E_g = \(\frac{h c}{\lambda_{m}}\)

Question 21.
Zener diodes have higher dopant densities as compared to ordinary p-n junction diodes. How does it affect the
(i) Width of the depletion layer
Answer:
Junction width will be small and

(ii) Junction field?
Answer:
The junction field will be high.

Question 22.
Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction? (NCERT Exemplar)
Answer:
No, because the voltmeter must have a resistance very high compared to the junction resistance, the latter being nearly infinite.

Semiconductor Electronics Important Extra Questions Short Answer Type

Question 1.
Draw a labeled circuit diagram of a full-wave rectifier using a p-n junction.
Answer:
The diagram is as shown.

Question 2.
What is a solar cell? How does it work? Give one of its uses.
Answer:
It is a p-n junction used to convert light into electrical energy. In such a diode, one region either the p-type or the n-type is made so thin that light falling on the diode is not absorbed appreciably before reaching the junction. The thin region in the solar cell is called the emitter and the other is called the base. The magnitude of current depends upon the intensity of light reaching the junction. A solar cell can be used to charge storage batteries during the daytime, which can be used during the night.

These are used as power supplies for satellites and space vehicles.

Question 3.
Draw the output signal in a p-n junction diode when a square input signal of 10 V as shown in the figure is applied across it. (CBSE AI 2019)

Answer:
The diode will conduct only when it is forward biased. Therefore, till the input voltage is + 5 V, we will get an output across R, accordingly the output waveform shown in the figure.

Question 4.
The following diagrams, indicate which of the diodes are forward biased and which are reverse biased.

Answer:
(a) Forward biased.
(b) Reverse biased.
(c) Forward biased,
(d) Reverse biased.

Question 5.
Mention the important considerations required while fabricating a p-n junction diode to be used as Light-Emitting Diode (LED). What should be the order of bandgap of an LED if it is required to emit light in the visible range? (CBSE Delhi 2013)
Answer:
The important considerations are

• It should be heavily doped.
• The diode should be encapsulated with a transparent cover so that emitted light can come out.

The semiconductor used for the fabrication of visible LEDs must at least have a bandgap of 1.8 eV.

Question 6.
In the given circuit diagram shown below, two p-n junction diodes D₁ and D₂ are connected with a resistance R and a dc battery E as shown. Redraw the diagram and indicate the direction of flow of appreciable current in the circuit. Justify your answer.

Answer:
The redrawn diagram showing the flow of appreciable current is shown below.

Here diode D₂ is forward biased, hence it conducts. Therefore appreciable current will pass through it. However, diode 0, is reverse biased, hence negligible current will flow through it.

Question 7.
The diagram below shows a piece of pure semiconductor S in series with a variable resistor R and a source of constant voltage V. Would you increase or decrease the value of R to keep the reading of ammeter (A) constant when semiconductor S is heated? Give reason.

Answer:
When a semiconductor is heated, its resistance decreases. As a result, the total resistance of the circuit will decrease. In order to maintain constant current flow, the total resistance of the circuit must remain constant. Hence, the external resistance has to be increased to compensate for the decrease of resistance of the semiconductor.

Question 8.
Two semiconductor materials X and Y showed in the figure are made by doping germanium crystal with indium and arsenic respectively. The two are joined end to end and connected to a battery as shown,
(i) Will the junction be forward or reverse biased?
(ii) Sketch a V-l graph for this arrangement.

Answer:
Material X is p-type and material Y is n-type.
(i) The junction is reverse biased.
(ii) For the V-l graph
The characteristics are as shown.

Question 9.
Draw the output waveform across the resistor (figure). (NCERT)

Answer:
It is a half-wave rectifier, therefore only the positive cycle will be rectified. Thus the output waveform is as shown.

Question 10.
(i) Name the type of a diode whose characteristics are shown in figure
(a) and figure (b).
(ii) What does point P in figure (a) represent?
(iii) What do the points P and Q in figure (b) represent? (NCERT Exemplar)

Answer:
(i) Zener junction diode and solar cell.
(ii) Zener breakdown voltage
(iii) P-open circuit voltage.
Q-short circuit current

Question 11.
A Zener of power rating 1 W is to be used as a voltage regulator. If Zener has a breakdown of 5 V and it has to regulate voltage which fluctuated between 3 V and 7 V, what should be the value of R, for safe operation (figure)? (NCERT Exemplar)

Answer:
Given P = 1 W, Vz = 5V, Vs = 7 V, Rs = ?
We know that

Question 12.
Two material bars A and B of the equal area of the cross-section are connected in series to a dc supply. A is made of usual resistance wire and B of an n-type semiconductor.
(i) In which bar is the drift speed of free electrons greater?
(ii) If the same constant current continues to flow for a long time, how will the voltage drop across A and B be affected? Justify each answer. (CBSE Sample Paper 2018-19)
Answer:
(i) Drift speed in B (n-type semiconductor) is higher.
Reason: Since the two bars A and B are connected in series, the current through each is the same.
Now l = neAv_d
Or
v_d = \(\frac{1}{n e A}\) ⇒ v_d ∝ \(\frac{1}{n}\) (As l and A are same).

As n is much lower in semiconductors, drift velocity will be more.

Question 13.
Explain how the width of the depletion layer in a p-n junction diode changes when the junction is (i) forward biased and (ii) reverse biased. (CBSE Delhi 2018C)
Answer:
The width of the depletion region in a p-n junction diode decreases when it is forward biased because the majority of charge carriers flow towards the junction. While it increases when the junction diode is reverse biased because the majority of charge carriers move away from the junction.

Semiconductor Electronics Important Extra Questions Long Answer Type

Question 1.
Define the terms ‘potential barrier’ and ‘depletion region’ for a p – n junction diode. State how the thickness of the depletion region will change when the p-n junction diode is (i) forward biased and (ii) reverse biased.
Answer:
Potential barrier: The potential barrier is the fictitious battery, which seems to be connected across the p-n junction with its positive terminal in the n-region and the negative terminal in the p-region.

Depletion region: The region around the junction, which is devoid of any mobile charge carriers, is called the depletion layer or region.

1. When the p-n junction is forward biased, there is a decrease in the depletion region.
2. When the p-n junction is reverse biased, there is an increase in the depletion region.

Question 2.
Explain (i) forward biasing and (ii) reverse biasing of a p-n junction diode.
Answer:
(i) A p-n junction is said to be forward-biased if its p-type is connected to the positive terminal and its n-type is connected to the negative terminal of a battery.
(ii) A p-n junction is said to be reverse-biased if its n-type is connected to the positive terminal and its p-type is connected to the negative terminal of a battery. The diagrams are as shown.

Question 3.
Draw V-l characteristics of a p-n junction diode. Answer the following questions, giving reasons:
(i) Why is the current under reverse bias almost independent of the applied potential up to a critical voltage?
(ii) Why does the reverse current show a sudden increase at the critical voltage? Name any semiconductor device which operates under the reverse bias in the breakdown region. (CBSEAI 2013)
Answer:
The characteristics are as shown.

(i) This is because even a small voltage is sufficient to sweep the minority carriers from one side of the junction to the other side of the junction.

(ii) As the reverse bias voltage is increased, the electric field at the junction becomes significant. When the reverse bias voltage V = V_z critical voltage, then the electric field strength is high enough to pull valence electrons from the host atoms on the p-side which are accelerated to the n-side. These electrons account for the high current observed at the breakdown.

Zener diode operates under the reverse bias in the breakdown region.

Question 4.
Draw the energy band diagrams of (i) n-type and (ii) p-type semiconductor at temperature T > 0 K.
In the case of n-type Si semiconductors, the donor energy level is slightly below the bottom of the conduction band, whereas in p-type semiconductors, the acceptor energy level is slightly above the top of the valence band. Explain what role do these energy levels play in conduction and valence bands. (CBSE AI 2015 C)
Answer:
For energy bands
(i) The energy level diagram is shown below.

(ii) The diagram is shown as

In the energy band diagram of n-type Si semiconductor, the donor energy level E_A is slightly below the bottom Ec of the conduction band and electrons from this level move into the conduction band with a very small supply of energy. At room temperature, most of the donor atoms get ionized but very few (-10-12) atoms of Si get ionized. So the conduction band will have most electrons coming from the donor impurities.

Similarly, for p-type semiconductors, the acceptor energy level E_A is slightly above the top E_v of the valence band. With the very small supply of energy, an electron from the valence band can jump to the level E_A and ionize the acceptor negatively. Alternately, we can also say that with a very small supply of energy, the hole from level EA sinks down into the valence band. Electrons rise up and holes fall down when they gain external energy.

Question 5.
Give reasons for the following:
(i) High reverse voltage does not appear across an LED.
Answer:
It is because the reverse breakdown voltage of LED is very low, i.e. nearly 5 V.

(ii) Sunlight is not always required for the working of a solar cell.
Answer:
Because solar cells can work with any light whose photon energy is more than the bandgap energy.

(iii) The electric field, of the junction of a Zener diode, is very high even for a small reverse bias voltage of about 5 V. (CBSE Delhi 2016C)
Answer:
The heavy doping of p and n sides of the p-n junction makes the depletion region very thin, hence for a small reverse bias voltage, the electric field is very high.

Question 6.
State the reason why the photodiode is always operated under reverse bias. Write the working principle of operation of a photodiode. The semiconducting material used to fabricate a photodiode has an energy gap of 1.2 eV. Using calculations, show whether it can detect light of a wavelength of 400 nm incident on it. (CBSE Al 2017C)
Answer:
It is easier to observe the change in the current with the change in the light intensity if a reverse bias is applied. Thus photodiode is used in the reverse bias mode even when the current in the forward bias is more the energy gap (E_g) of the semiconductor, then electron-hole pairs are generated due to the absorption of photons.

Due to the electric field of the junction, electrons and holes are separated before they recombine. The direction of the electric field is such that electrons reach the n-side and holes reach the p-side. Electrons are collected on the n-side and holes are collected on the p-side giving rise to an emf. When an external load is connected, current flows.

Given λ = 400 nm,
Energy of photon
E = \(\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{400 \times 10^{-9} \times 1.6 \times 10^{-19}}\) = 3.105 eV

Since the bandgap is lesser than this energy, therefore it will be able to detect the wavelength.

Question 7.
Explain the two processes involved in the formation of a p-n junction diode. Hence define the term ‘barrier potential’. (CBSE Delhi 2017C)
Answer:
The two processes are
(i) Diffusion and
(ii) Drift

• Diffusion: The holes diffuse from the p-side to the n-side and electrons diffuse from the n-side to the p-side.
• Drift: The motion of charge carriers due to the applied electric field which results in the drifting of holes along the electric field and of electrons opposite to the electric field.

The potential barrier is the fictitious battery that seems to be connected across the junction with its positive end on the n-type and the negative end on the p-type.

Question 8.
Explain briefly how a photodiode operates.
Answer:
A Photodiode is again a special purpose p-n junction diode fabricated with a transparent window to allow light to fall on the diode. It is operated under reverse bias. When the photodiode is illuminated with light (photons) with energy (hv) greater than the energy gap (E_g) of the semiconductor, then electron-hole pairs are generated due to the absorption of photons. The diode is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode.

Question 9.
Name the p-n junction diode which emits spontaneous radiation when forward biased. How do we choose the semiconductor, to be used in these diodes, if the emitted radiation is to be in the visible region?
Answer:
The p-n junction diode, which emits spontaneous radiation when forward biased, is the “light-emitting diode” or LED.

The visible tight is from 400 nm to 700 nm and the corresponding energy is between 2.8 eV to 1.8 eV. Therefore, the energy gap of the semiconductor to be used in LED, in order to have the emitted radiation be in the visible region, should be 1.8 eV. Phosphorous doped gallium arsenide and gallium phosphide are two such suitable semiconductors.

Question 10.
The figure shows the V-l characteristic of a semiconductor diode designed to operate under reverse bias. (CBSE Al 2019)

(a) Identify the semiconductor diode used.
Answer:
The diode used is Zener diode

(b) Draw the circuit diagram to obtain the given characteristics of this device.
Answer:
The circuit diagram is as shown.

(c) Briefly explain one use of this device.
Answer:
The Zener diode can be used as a voltage regulator in its breakdown region. The Zener voltage remains constant even when the current through the Zener diode changes.

Question 11.
With the help of a diagram, show the biasing of a light-emitting diode (LED). Give its two advantages over conventional incandescent lamps.
Answer:
The biasing of a light-emitting diode (LED), has been shown below.

Two main advantages of LED over conventional incandescent lamps are as follows:

1. Low operational voltage and less power consumption.
2. Fast action and no warm-up time required.
3. The bandwidth of emitted light is 100 A to 500 A or in other words, it is nearly (but not exactly) monochromatic.
4. Long life and ruggedness.
5. Fast on-off switching capability.

Question 12.
(a) Writetheprincipleofasemiconductor device which is used as a voltage regulator.
Answer:
(a) Zener diode is used as a voltage regulator
Principle: It is based on the Principle that when breakdown voltage V₂ takes place, there is a large change in the reverse current even with the insignificant change in the reverse bias voltage.

(b) With the help of a circuit diagram explain its working.
Answer:

Working: If the reverse voltage across a Zener diode is increased beyond the breakdown voltage V_z, the current increases sharply and large current l_z flows through the Zener diode and the voltage drop across Rs increases maintaining the voltage drop across RL at constant value V_o = V_z.

On the other hand, if we keep the input voltage constant and decrease the load resistance RL, the current across the load will increase. The extra current cannot come from the source because drop-in Rs will not change as the Zener is within its regulating range. The additional load current will pass through the Zener diode and is known as Zener current l_z so that the total current (l_L + l_z) remains constant.

(c) Draw its l-V characteristics. (CBSE 2019C)
Answer:
l-V Characteristics of Zener diode

Question 13.
With what considerations in view, a photodiode is fabricated? State its working with the help of a suitable diagram.
Even though the current in the forward bias is known to be more than in the reverse bias, yet the photodiode works in reverse bias. What is the reason? (CBSE Delhi 2015)
Answer:
It is fabricated with a transparent window to allow light to fall on the diode. It is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode.

When the photodiode is illuminated with light (photons) with energy (h_w) greater than the energy gap (E_g) of the semiconductor, then electron-hole pairs are generated due to the absorption of photons. The diode is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode. Due to the electric field of the junction, electrons and holes are separated before they recombine.

The direction of the electric field is such that electrons reach the n-side and holes reach the p-side. Electrons are collected on the n-side and holes are collected on the p-side giving rise to an emf. When an external load is connected, current flows. The magnitude of the photocurrent depends on the intensity of incident light (photocurrent is proportional to incident light intensity). The diagram is as shown.

It is easier to observe the change in the current with a change in the light intensity if a reverse bias is applied. Thus photodiode is used in the reverse bias mode even when the current in the forward bias is more.

Question 14.
Draw the circuit diagram of a full-wave rectifier and explain its working. Also, give the input and output waveforms. (CBSE Delhi 2019)
Answer:
The circuit diagram is as shown.

The two ends S₁ and S₂  of a center-tapped secondary of a transformer are connected to the P sections of the two diodes D₁ and D₂ respectively. The n-sections of the two diodes are joined together and their com¬mon junction is connected to the central tap C of the secondary winding through a load resistance R_L. The input is applied across the primary and the output is ob¬tained across the load resistance R_L. The arrows show the direction of the current.

Assume that the end A of the secondary is positive during the first half cycle of the supply voltage. This makes diode D₁ forward biased and diode D₂ reverse biased. Thus diode D₁ conducts and an output is obtained across the load R_L.

During the second half cycle of the supply voltage, the polarities of the secondary windings reverse. A becomes negative and B becomes positive with respect to the central terminal C. This makes diode D₂ forward biased. Hence it conducts and an output is obtained across R_L.

The input-output waveforms are as shown.

Question 14.
Draw the circuit diagram to show the use of a p-n junction diode as a half-wave rectifier. Also show the input and the output voltages, graphically. Explain its working.
Answer:
The diagram is as shown.

The input and output waveforms are as shown.

A p-n junction diode is used as a half-wave rectifier. Its work is based on the fact that the resistance of the p-n junction becomes low when forward biased and becomes high when reversing biased. These characteristics of a diode are used in rectification.

Question 15.
Distinguish between conductors, insulators, and semiconductors on the basis of the band theory of solids.
Answer:
The diagrams are as shown.

Metals: A distinguishing character of all conductors, including metals, is that the valence band is partially filled or the conduction and the valence band overlap. Electrons in states near the top of the filled portion of the band have many adjacent unoccupied states available, and they can easily gain or lose small amounts of energy in response to an applied electric field. Therefore these electrons are mobile and can contribute to electrical and thermal conductivity. Metallic crystals always have partially filled bands figure (ii).

Insulators: In the case of insulators, there is a large energy gap of approximately 6 eV depending upon the nature of the crystal. Electrons, however, heated, find it difficult or practically impossible to jump this gap and thus never reach the conduction band. Thus electrical conduction is not possible through an insulator figure (iii).

Semiconductors: There is a separation between the valence band and the conduction band. The energy gap is of the order of 1 eV (0.67 eV for germanium and 1.12 eV for silicon). At absolute zero the electrons cannot gain this energy. But at room temperature, these electrons gain energy and move into the conduction band where they are free to move even under the effect of a weak electric field figure (i).

Question 16.
What is a Zener diode? How is it symbolically represented? With the help of a circuit diagram, explain the use of the Zener diode as a voltage stabilizer.
Answer:
It is a special diode made to work only in the reverse breakdown region.
Symbol:

The figure below shows the use of the Zener diode in providing a constant voltage supply.

This use of the Zener diode is based on the fact that in the reverse breakdown (or Zener) region, a very small change in voltage across the Zener diode produces a very large change in the current through the circuit. If voltage is increased beyond Zener voltage, the resistance of the Zener diode drops considerably. Consider that the Zener diode and a resistor R, called dropping resistor, are connected to a fluctuating voltage supply, such that the Zener diode is reverse biased.

Whenever voltage across the diode tends to increase, the current through the diode rises out of proportion and causes a sufficient increase in the voltage drop across the dropping resistor. As a result, the output voltage lowers back to the normal value. Similarly, when the voltage across the diode tends to decrease, the current through the diode goes down out of proportion, so that the voltage drop across the dropping resistor is much less and now the output voltage is raised to normal.

Question 17.
Explain briefly with the help of a circuit diagram how V-l characteristics of a p-n junction diode are obtained in (i) forward bias and (ii) reverse bias.
Answer:
Forward biased characteristics: A p-n junction is said to be forward-biased if its p-type is connected to the positive terminal and its n-type is connected to the negative terminal of a battery shows a circuit diagram that is used to study the forward characteristics of a p-n junction. The p-n junction is forward biased. Different readings are taken by changing the voltage and noting the corresponding milliammeter current.

Practically no current is obtained till the applied voltage becomes greater than the barrier potential. Above the potential barrier voltage, even a small change in potential causes a large change in current.

Reverse biased characteristics: In reverse biased characteristics, instead of a milliammeter, a microammeter is used. The voltage across the p-n junction is increased and the corresponding current is noted.

In the reverse bias, the diode current is very small. As the voltage has increased the current also increases. At a certain voltage, the current at once becomes very large. This voltage is called Breakdown voltage or Zener voltage. At this voltage, a large number of covalent bonds break releasing a large number of electrons and holes. Hence a large current is obtained. The characteristics are as shown.

Question 18.
Explain how the heavy doping of the p and n sides of a p-n junction diode helps in internal field emission (or Zener breakdown), even with a reverse bias voltage of a few volts only. Draw the general shape of the V-I characteristics of a Zener diode. Discuss how the nature of these characteristics led to the use of a Zener diode as a voltage regulator.
Answer:
Consider a p-n junction where both p- and n-sides are heavily doped. Due to the high dopant densities, the depletion layer junction width is small and the junction field will be high. Under large reverse bias, the energy bands near the junction and the junction width decrease. Since the junction width is < 10^-7 m, even a small voltage (say 4 V) may give a field as large as 4 × 10^-7 Vm^-1. The high junction field may strip an electron from the valence band which can tunnel to the n-side through the thin depletion layer. Such a mechanism of emission of electrons after a certain critical field or applied voltage V is termed as internal field emission which gives rise to a high reverse, current, or breakdown current.

The general shape of the V-l characteristics of a Zener diode is as shown in the figure.

Suppose an unregulated dc input voltage (V_i) is applied to the Zener diode (whose breakdown voltage is V_z as shown in the figure. If the applied voltage V_i > V_z, then the Zener diode is in the breakdown condition. As a result of a wide range of values of load (R_L), the current in the circuit or through the Zener diode may change but the voltage across it remains unaffected by the load. Thus, the output voltage across the Zener is a regulated voltage.

Question 19.
(i) Describe briefly with the help of a necessary circuit diagram, the working principle of a solar cell.
Answer:
Solar Cell: A solar cell is a junction diode that converts solar energy into electrical energy. In a solar cell, the n-region is very thin and transparent so that most of the incident light reaches the junction. The thin region is called the emitter and the other base. When light is incident on it, it passes through the crystal onto the junction. The electrons and holes are generated due to light (with h_v > E_g). The electrons are kicked to the n-side and holes to the p-side due to the electric field of the depletion region. Thus p-side becomes positive and the n-side becomes negative giving rise to a photo-voltage. Thus it behaves as a cell.

(ii) Why are Si and GaAs preferred materials for solar cells? Explain. (CBSE AI 2011C)
Answer:
Si and GaAs are preferred for solar cell fabrication due to the fact that their bandgap is ideal. Further, they have high electrical conductivity and high optical absorption.

Question 20.
Explain, with the help of a circuit diagram, the working of a photo-diode. Write briefly how it is used to detect optical signals. (CBSE Delhi 2013)
Answer:
When the photodiode is illuminated with light (photons) with energy (h_v) greater than the energy gap (E_g) of the semiconductor, then – electron-hole pairs are generated due to the absorption of photons. The diode is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode. Due to the electric field of the junction, electrons and holes are separated before they recombine. The direction of the electric field is such that electrons reach the n-side and holes reach the p-side. Electrons are collected on the n-side and holes are collected on the p-side giving rise to an emf. When an external load is connected, current flows.

The magnitude of the photocurrent depends on the intensity of incident light incident on it. This helps in detecting optical signals.

Question 21.
(i) Explain with the help of a diagram, how depletion region and potential barrier are formed in a junction diode.
(ii) If a small voltage is applied to a p-n junction diode, how will the barrier potential be affected when it is (i) forward biased and (ii) reverse biased? (CBSE AI2015)
Answer:
(i) Since there is an excess of electrons in the n-type and excess of holes in the p-type, on the formation of a p-n junction the electrons from the n-type diffuse into the p-region, and the holes in the p-type diffuse into the n-region.

The accumulation of electric charge of opposite polarities in the two regions across the junction establishes a potential difference between the two regions. This is called the potential barrier or junction barrier. The potential barrier developed across the junction opposes the further diffusion of the charge carriers from p to n and vice versa. There is a region on either side of the junction where there is a depletion of mobile charges and has only immobile charges. The region around the junction, which is devoid of any mobile charge carriers, is called the depletion layer or region.

(ii) (a) In forwarding bias the potential barrier decreases.
(b) In reverse bias the potential barrier increases.

Question 22.
Write the two processes that take place in the formation of a p-n junction. Explain with the help of a diagram, the formation of depletion region and barrier potential in a p-n junction. (CBSE Delhi 2017)
Answer:

1. Drift and
2. diffusion.

The n-type has an excess of electrons and the p-type has an excess of holes. When a p-n junction has formed the electrons from the n-type diffuse into the p-region and the holes in the p-type diffuse into the n-region. These diffusing electrons and holes combine near the junction. Each combination eliminates an electron and a hole. This results in the n-region near the junction becoming positively charged by losing its electrons and the p-region near the junction becoming negatively charged by losing its holes.

This accumulation of electric charge of opposite polarities in the two regions across the junction establishes a potential difference between the two regions. This is called the potential barrier or junction barrier.
The potential barrier developed across the junction opposes the further diffusion of the charge carriers from p to n and vice versa. As a result, a region develops on either side of the junction where there is a depletion of mobile charges and has only immobile charges. The region around the junction which is devoid of any mobile charge carriers is called the depletion layer or region.

Question 23.
(i) State briefly the processes involved in the formation of the p-n junction explaining clearly how the depletion region is formed.
Answer:
As we know that n-type semi-conductor has more concentration of electrons than that of a hole and a p-type semi-conductor has more concentration of holes than an electron. Due to the difference in concentration of charge carriers in the two regions of the p-n junction, the holes diffuse from p-side to n-side, and electrons diffuse from n-side to p-side. When an electron diffuses from n to p, it leaves behind it an ionized donor on the n-side. The ionized donor (+ve charge) is immobile as it is bound by the surrounding atoms. Therefore, a layer of positive charge is developed on the n-side of the junction. Similarly, a layer of negative charge is developed on the p-side.

Hence, a space-charge region is formed on either side of the junction, which has immobile ions and is devoid of any charge carrier, called depletion layer or depletion region.

The potential barrier is the fictitious battery which seems to be connected across the junction with its positive end on the n-type and the negative end on the p-type.

(ii) Using the necessary circuit diagrams, show how the V – l characteristics of a p-n junction are obtained in
(a) Forward biasing
(b) Reverse biasing
How these characteristics are made use of in rectification? (CBSE Delhi 2014)
Answer:
(a) p-n junction diode under forwarding bias: The p-side is connected to the positive terminal and the n-side to the negative terminal. Applied voltage drops across the depletion region. Electron in n-region moves towards the p-n junction and holes in the p-region move towards the junction. The width of the depletion layer decreases and hence, it offers less resistance. Diffusion of majority carriers takes place across the junction. This leads to the forward current.

(b) p-n junction diode under reverse bias: Positive terminal of the battery is connected to the n-side and negative terminal to p-side. Reverse bias supports the potential barrier. Therefore, the barrier height increases and the width of the depletion region also increases. Due to the majority of carriers, there is no conduction across the junction. A few minority carriers across the junction after being accelerated by the high reverse bias voltage. This constitutes a current that flows in opposite direction, which is called reverse current.

For V-l curves

A p-n junction diode is used as a half-wave rectifier. Its work is based on the fact that the resistance of the p-n junction becomes low when forward biased and becomes high when reversing biased. These characteristics of the diode are used in rectification.

Question 24.
(i) Explain with the help of a suitable diagram, the two processes which occur during the formations of a p-n junction diode. Hence define the terms (i) depletion region and (ii) potential barrier.
(ii) Draw a circuit diagram of a p-n junction diode under forwarding bias ‘ and explain its working. (CBSE 2018C)
Answer:
(i) The two important processes are diffusion and drift.
Due to the concentration gradient, the electrons diffuse from the n-side to the p side, and holes diffuse from the n-side to the n side.

Due to the diffusion, an electric field develops across the junction. Due to the field, an electron moves from the p-side to the n-side; a hole moves from the n-side to the p-side. The flow of the charge carriers due to the electric field is called drift.

(a) Depletion region: It is the space charge region on either side of the junction that gets depleted of free charges is known as the depletion region.
(b) Potential Barrier: The potential difference that gets developed across the junction and opposes the diffusion of charge carriers and brings about a condition of equilibrium, which is known as the barrier potential.

(ii) The circuit diagram is as shown

Working:
In the forward bias condition, the direction of the applied voltage is opposite to the barrier potential. This reduces the width of the depletion layer as well as the height of the barrier. A current can, therefore, flow through the circuit. This current increases (non¬linearly) with the increase in the applied voltage.

Numerical Problems

Question 1.
(i) Three photodiodes D₁, D₂, and D₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV, and 3 eV respectively. Which of them will not be able to detect light of wavelength 600 nm? (CBSE Delhi 2019)
Answer:
λ = 600 nm
The energy of a photon of wavelength λ
E = \(\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{600 \times 10^{-9} \times 1.6 \times 10^{-19}}\) = 2.08 eV

The bandgap energy of diode D₂ (= 2eV) is less than the energy of the photon. Hence diode D₂ will not be able to detect light of wavelength 600 nm.

(ii) Why photodiodes are required to operate in reverse bias? Explain.
Answer:
A photodiode when operated in reverse bias can measure the fractional change in minority carrier dominated reverse bias current with greater ease than when forward biased.

Question 2.
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz? What is the output frequency of a full-wave rectifier for the same input frequency? (NCERT)
Answer:
The output frequency of the half-wave rectifier is the same as the input frequency, while that of the full-wave rectifier is double that of the input. Therefore the frequency is 50 Hz for half-wave and 100 Hz for full-wave.

Question 3.
A p-n photodiode is fabricated from a semiconductor with a bandgap of 2.8 eV. Can it detect a wavelength of 6000 nm? (NCERT)
Answer:
Given λ = 6000 nm,
Energy of photon

Since the bandwidth is greater than this energy, it will not be able to detect the wavelength.

Question 4.
Three photodiodes D₁, D_2, and D₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV, and 3 eV, respectively. Which ones will be able to detect light of wavelength 600 nm? (NCERT Exemplar)
Answer:
The energy of an incident light photon is given by

For the incident radiation to be detected by the photodiode, the energy of the incident radiation photon should be greater than the bandgap. This is true only for D₂. Therefore, only D₂ will detect this radiation.

Question 5.
If each diode in the figure has a forward bias resistance of 25 Ω and infinite resistance in reverse bias, what will be the values of the current l₁ l₂, l₃, and l₄? (NCERT Exemplar)

nswer:
Current l₃ is zero as the diode in that branch is reverse biased.

Resistance in the branches AB and EF is each (125 +25) Ω = 150 Ω
As AB and EF are identical paraLleL branches, their effective resistance is 150/2 = 75 Ω

Therefore net resistance in the circuit Is
Rnet =75 + 25 = 100 Ω

Therefore current l1 is
l₁ = 5/100 = o.05 A

As resistance of branches AB and EF is equal, the current l1 will be equally shared by the two, hence
l₂ = l₄ = 0.05/2 = 0.025A
Hence l₁ = 0.05 A, l₂ = l₄ = 0.025 A, l₃ = 0

Question 6.
Assuming the ideal diode, draw the output waveform for the circuit given in the figure. ExplaIn the waveform. (NCERT Exemplar)

Answer:
When input voltage Will is greater than 5 V, the diode wilt becomes forward biased and will conduct. When the input is Less than 5 V, the diode will be reverse biased and will not conduct, i.e. open circuit, hence the output is as shown.

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 10 Wave Optics. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 10 Important Extra Questions Wave Optics

Wave Optics Important Extra Questions Very Short Answer Type

Question 1.
Sketch the refracted wavefront emerging from convex tens, If a plane wavefront is an incident normally on it.
Answer:
The figure is as shown.

Question 2.
How would you explain the propagation of light on the basis of Huygen’s wave theory?
Answer:
To explain the propagation of light we have to draw a wavefront at a later instant when a wavefront at an earlier instant is known. This can be drawn by the use of Huygen’s principle.

Question 3.
Draw the shape of the reflected wavefront when a plane wavefront is an incident on a concave mirror.
Answer:
The reflected wavefront is as shown.

Question 4.
Draw the shape of the refracted wavefront when a plane wavefront is an incident on a prism.
Answer:
The shape of the wavefront is as shown.

Question 5.
Draw the type of wavefront that corresponds to a beam of light diverging from a point source.
Answer:
The wavefront formed by the light coming from a very far off source is a plane and for a beam of light diverging from a point, a wavefront is a number of concentric circles.

Question 6.
Draw the type of wavefront that corresponds to a beam of light coming from a very far off source.
Answer:
The wavefront is as shown.

Question 7.
Name two phenomena that establish the wave nature of light.
Answer:
Interference and diffraction of light.

Question 8.
State the conditions which must be satisfied for two light sources to be coherent.
Answer:
(a) Two sources must emit light of the same wavelength (or frequency).
(b) The two light sources must be either in-phase or have a constant phase difference.

Question 9.
Draw an intensity distribution graph for diffraction due to a single-slit.
Answer:
The intensity distribution for a single-slit diffraction pattern is as shown.

Question 10.
Name one device for producing plane polarised light. Draw the graph showing the variation of intensity of polarised light transmitted by an analyser.
Answer:
Nicol prism can be used to produce plane polarised light. The graph is as shown.

Question 11.
State Huygens’ principle of diffraction of light. (CBSE AI 2011C)
Answer:
Huygens principle states that
(a) Each point on a wavefront is a source of secondary waves which travel out with the same velocity as the original waves.
(b) The new wavefront is given by the forward locus of the secondary wavelets.

Question 12.
In what way is a plane polarised tight different from an unpolarised light? (CBSE AI 2012C)
Answer:
Plane polarized light vibrates 1n only one plane.

Question 13.
Which of the following waves can be polarised: (i) Heatwaves (ii) Sound waves? Give a reason to support your answer. (CBSE Delhi 2013)
Answer:
Heatwaves are transverse In nature.

Question 14.
Define the term ‘wavefront’. (CBSE AI 2014C)
Answer:
It Is defined as the locus of all points In a medium vibrating in the same phase.

Question 15.
Define the term ‘coherent sources’ which are required to produce interference pattern in Young’s double-slit experiment. (CBSE Delhi 2014C)
Answer:
Two sources that are In phase or have a constant phase difference are called coherent sources.

Question 16.
What change would you expect if the whole of Young’s double-slit apparatus were dipped into the water?
Answer:
The wavelength λ, of light In water, is less than that in air. Since the fringe width β is directly proportional to the wavelength of light, therefore, the fringe width will decrease.

Question 17.
When light travels from a rarer to a denser medium, it loses some speed. Does the reduction in speed Imply a reduction in the energy carried by the light wave?
Answer:
No, the energy carried by a wave depends upon the amplitude of the wave and not on Its speed of propagation.

Question 18.
If one of the slits say S₁, is covered then what changes occur in the Intensity of light at the centre of the screen?
Answer:
The intensity 1s decreased four times because l ∝ 4a² where a is the amplitude of each wave.

Question 19.
How does the angular separation between fringes in a single-slit diffraction experiment change when the distance of separation between the slit and screen is doubled? (CBSE AI 2012)
Answer:
No change.

Question 20.
What is the effect on the interference fringes in Young’s double-slit experiment If the separation between the screen and slits Is Increased?
Answer:
The fringe width Increases.

Question 21.
How does the Intensity of the central maximum change If the width of the slit Is halved in a single-slit diffraction experiment?
Answer:
The width of the central maxima is doubled and the intensity is reduced to one-fourth of Its original value.

Question 22.
The polarising angle of a medium Is 60°, what is the refractive index of the medium? (CBSE Delhi 2019)
Answer:
Using the expression
μ = tan i_p = tan 60° = 1.732.

Question 23.
In the wave picture of light intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light?
Answer:
For a given frequency intensity of light in the photon, the picture is determined by the number of photons crossing a unit area per unit time.

Question 24.
How does the Intensity of the central maximum change if the width of the slit is halved in the single-slit diffraction experiment?
Answer:
The width of the central maximum is doubled and the intensity is reduced to one-fourth of its original value.

Question 25.
What would happen If the path difference between the interfering beams that is S₂P – S₁P became very large?
Answer:
If the path difference becomes very large it may exceed the coherent length. Thus the coherence of the waves reaching P is lost and no interference takes place.

Question 26.
In Young’s double-slit experiment, what would happen to the intensity of the maxima and the minima if the size of the hole illuminating the two coherent holes were gradually Increased?
Answer:
The fringe width will decrease and finally, there will be general illumination on the screen.

Question 27.
What is the Brewster angle for air to glass transition? (Refractive index of glass =1.5.) (NCERT)
Answer:
Given μ = 1.5, i_P = ?
Using the relation μ = tan i_P we have
l_p = tan^-1 (1.5) = 56.3°

Question 28.
Is Huygen’s principle valid for longitudinal sound waves? (NCERT Exemplar)
Answer:
Yes

Question 29.
Consider a point at the focal point of a convergent lens. Another convergent lens of the short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image? (NCERT Exemplar)
Answer:
Spherical.

Question 30.
What is the shape of the wavefront on earth for sunlight? (NCERTExemplar)
Answer:
Spherical with a huge radius as compared to the earth’s radius so that it is almost a plane.

Question 31.
Draw a graph showing the intensity distribution of fringes due to diffraction at a single-slit. (CBSE 2018C)
Answer:

Wave Optics Important Extra Questions Short Answer Type

Question 1.
How can one distinguish between an unpolarised and linearly polarised light beam using polaroid? (CBSE Delhi 2019)
Answer:
The two lights will be allowed to pass through a polariser. When the polarizer is rotated in the path of these two light beams, the intensity of light remains the same in all the orientations of the polariser, then the light is unpolarised. But if the intensity of light varies from maximum to minimum then the light beam is a polarised light beam.

Question 2.
What is meant by plane polarised light? What type of waves shows the property of polarisation? Describe a method of producing a beam of plane polarised light?
Answer:

1. The light that has its vibrations restricted in only one plane is called plane polarised light.
2. Transverse waves show the phenomenon of polarization.
3. Light is allowed to pass through a polaroid. The polaroid absorbs those vibrations which are not parallel to its axis and allows only those vibrations to pass which are parallel to its axis.

Question 3.
Write the Important characteristic features by which the Interference can be distinguished from the observed diffraction pattern. (CBSE AI 2015)
Answer:
(a) In the interference pattern the bright fringes are of the same width, whereas in the diffraction pattern they are not of the same width.
(b) In interference all bright fringes are equally bright while in diffraction they are not equally bright.

Question 4.
State Brewster’s law. The value of Brewster’s angle for the transparent medium is different for the light of different colours. Give reason. (CBSE Delhi 2016)
Answer:
When the reflected ray and the refracted ray are perpendicular then μ = tani_p where i_p is the polarising angle or Brewster angle.

Brewster’s angle depends upon the refractive index of the two media in contact. The refractive index in turn depends upon the wavelength of light used (different colours) hence Brewster’s angle is different for different colours.

Question 5.
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids.
Answer:
Let l_o be the intensity of polarised light after passing through the first polarizer P₁. Then the intensity of light after passing through the second polarizer P2 will be l = l_ocos 2θ, where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be
l = l_o cos² θ cos² (90° – θ) = l_o cos² θ sin² θ = (l_o /4)sin² 2θ

Therefore, the transmitted intensity will be maximum when θ = π/4

Question 6.
Is energy conserved in interference? Explain.
Answer:
Yes, energy is conserved in interference. Energy from the dark fringes is accumulated in the bright fringes. If we take
l = 4a²cos²\(\frac{\phi}{2}\), then intensity at bright points is l_max = 4a² and intensity at the minima l_min = 0. Hence average intensity in the pattern of the fringes produced due to interference is given by
Ī = \(\frac{I_{\max }+I_{\min }}{2}=\frac{4 a^{2}+0}{2}\) = 2a²

But if there is no interference then total intensity at every point on the screen will be l = a² + a² = 2a², which is the same as the average intensity in the interference pattern.

Question 7.
An incident beam of light of intensity lo is made to fall on a polaroid A. Another polaroid B is so oriented with respect to A that there is no light emerging out of B. A third polaroid C is now introduced midway between A and B and is so oriented that its axis bisects the angle between the axes of A and B. What is the intensity of light now between (i) A and C (ii) C and B? Give reasons for your answers.
Answer:
Polaroids A and B are oriented at an angle of 90°, so no light is emerging out of B. On placing polaroid C between A and B such that its axis bisects the angle between axes of A and B, then the angle between axes of polaroids A and B is 45° and that of C and B also 45°.
(a) Intensity of light on passing through Polaroid A or between A and C is l₁ = \(\frac{l_{0}}{2}\)
(b) On passing through polaroid C, intensity of light between C and B becomes
l₂ = l₁ cos² θ = \(\frac{l_{0}}{2}\) × cos² 45° = \(\frac{l_{0}}{4}\)

Question 8.
One of the slits of Young’s double-slit experiment is covered with a semi¬transparent paper so that it transmits lesser light. What will be the effect on the interference pattern?
Answer:
There will be an interference pattern whose fringe width is the same as that of the original. But there will be a decrease in the contrast between the maxima and the minima, i.e. the maxima will become less bright and the minima will become brighter.

Question 9.
Light from a sodium lamp is passed through two polaroid sheets P₁ and P₂ kept one after the other. Keeping P₁, fixed, P₂ is rotated so that its ‘pass axis can be at different angles, θ, with respect to the pass-axis of P₁.
An experimentalist records the following data for the intensity of light coming out of P₂ as a function of the angle θ.

I = Intensity of beam falling on P₁
(a) One of these observations is not in agreement with the expected theoretical variation of I, identify this observation and write the correct expression.
(b) Define the Brewster angle and write the expression for It in terms of the refractive index of the medium.
Answer:
(a) The observation \(\frac{1}{\sqrt{2}}\) is not correct. It should be 1/2.
(b) It is the angle of incidence at which the refracted and the reflected rays are perpendicular to each other. It is related to the refractive index as tan i_p = μ

Question 10.
How will the interference pattern in Young’s double-slit experiment get affected, when
(a) distance between the slits S₁, and S₂ reduced and
(b) the entire set-up is immersed in water? Justify your answer in each case. (CBSE Delhi 2011C)
Answer:
We know that fringe width β of the dark or bright fringes is given by β = \(\frac{D \lambda}{d}\) where d is the distance between the slits.
(a) When the distance between the slits, i. e. d is reduced then p will increase. The interference pattern will thus become broader.
(b) When the entire set up is immersed in water, the pattern will become narrow due to the decrease in the wavelength of light. The new wavelength λ’ = λ/n, hence β’= β/n

Question 11.
Discuss the intensity of transmitted light when a Polaroid sheet is rotated between two crossed polaroids? (NCERT)
Answer:
Let l0 be the intensity of polarised light after passing through the first polarizer P₁. Then the intensity of light after passing through the second polarizer P₂ will be l = l_o cos² θ

where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be

Therefore, the transmitted intensity will be maximum when θ = π/4

Question 12.
A Polaroid (I) is placed In front of a monochromatic source. Another Polaroid (II) is placed in front of this Polaroid (I) and rotated till no light passes. A third Polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II)? Explain. (NCERT Exemplar)
Answer:
Only in the special case when the pass axis of (III) is parallel to fill or (II) there shall be no light emerging. In all other cases, there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).

Question 13.
(a) Good quality sunglasses made of polaroids are preferred over ordinary coloured glasses. Explain why.
(CBSE AI2019)
Answer:
Polaroid sunglasses are preferred over coloured sunglasses because they reduce the intensity of light.

(b) How is plane polarised light defined?
Answer:
Plane polarised light: Light in which vibrations of electric field vector are restricted to one plane containing the direction of propagation.

(c) A beam of plane polarised light is passed through a polaroid. Show graphically, a variation of the intensity of the transmitted light with the angle of rotation of the Polaroid.
Answer:
The graphical variation is as shown.

Wave Optics Important Extra Questions Long Answer Type

Question 1.
Define the term wavefront. Using Huygen’s wave theory, verify the law of reflection.
Or
Define the term, “refractive index” of a medium. Verify Snell’s law of refraction when a plane wavefront is propagating from a denser to a rarer medium. (CBSE Delhi 2019)
Answer:
The wavefront is a locus of points that oscillate in the same phase.

Consider a plane wavefront AB incident obliquely on a plane reflecting surface MM^–. Let us consider the situation when one end A of was front strikes the mirror at an angle i but the other end B has still to cover distance BC. The time required for this will be t = BC/c.

According to Huygen’s principle, point A starts emitting secondary wavelets and in time t, these will cover a distance c t = BC and spread. Hence, with point A as centre and BC as radius, draw a circular arc. Draw tangent CD on this arc from point C. Obviously, the CD is the reflected wavefront inclined at an angle ‘r’. As incident wavefront and reflected wavefront, both are in the plane of the paper, the 1^st law of reflection is proved.

To prove the second law of reflection, consider ΔABC and ΔADC. BC = AD (by construction),
∠ABC = ∠ADC = 90° and AC is common.

Therefore, the two triangles are congruent and, hence, ∠BAC = ∠DCA or ∠i = ∠r, i.e.the angle of reflection is equal to the angle of incidence, which is the second law of reflection.
Or
The refractive index of medium 2, w.r.t. medium 1 equals the ratio of the sine of the angle of incidence (in medium 1) to the sine of the angle of refraction (in medium 2), The diagram is as shown.

From the diagram

Question 2.
(a) Sketch the refracted wavefront for the incident plane wavefront of the light from a distant object passing through a convex lens.
Answer:

(b) Using Huygens’s principle, verify the laws of refraction when light from a denser medium is incident on a rarer medium.
Answer:
Refraction from denser to the rarer medium: Let XY be plane refracting surface separating two media of refractive index μ₁ and μ₂ (μ₁ > μ₂)

Let a plane wavefront AB incident at an angle i. According to Huygen’s principle, each point on the wavefront becomes a source of secondary wavelets and

Time is taken by wavelets from B to C = Time taken by wavelets from A to D


(c) For yellow light of wavelength 590 nm incident on a glass slab, the refractive index of glass Is 1.5. Estimate the speed and wavelength of yellow light Inside the glass slab. (CBSE 2019C)
Answer:
Given λ = 590 nm, μ = 1.5
Velocity of light inside glass slab.
∴ v = \(\frac{C}{\mu}=\frac{3 \times 10^{8}}{1.5}\) = 2 × 10⁸ ms^-1

Wavelength of yellow light inside the glass slab.
λ₁ = \(\frac{\lambda}{\mu}=\frac{290}{1.5}\) = 393.33 nm

Question 3.
(a) State the postulates of Huygen’s wave theory.
Answer:
The postulates are
All points on a given wavefront are taken as point sources for the production of spherical secondary waves, called wavelets, which propagate outward with speed characteristic of waves in that medium.

(b) Draw the type of wavefront that corresponds to a beam of light (i) coming from a very far off the source and (ii) diverging from a point source.
Answer:
After some time has elapsed, the new position of the wavefront is the surface tangent to the wavelets or the envelope of the wavelets in the forward direction.


Question 4.
What is meant by the diffraction of light? Obtain an expression for the first minimum of diffraction.
Answer:
The divergence of light from its initial line of travel when it passes through an opening or an obstacle is called diffraction or the phenomenon of bending of light around the sharp corners and spreading into the regions of the geometrical shadow is called diffraction.

Consider that a monochromatic source of light S, emitting light waves of wavelength λ, is placed at the principal focus of the convex lens L₁. A parallel beam of light, i.e. a plane wavefront, gets incident on a narrow slit AB of width ‘a’ as shown in the figure.

The diffraction pattern is obtained on a screen Lying at a distance D from the slit and at the focal plane of the convex lens L₂.

Consider a point P on the screen at which wavelets travelLing in a direction making angle O with CO are brought to focus by the lens. The wavelets from different parts of the slit do not reach point P in phase, although they are initially in phase. It is because they cover unequal distances in reaching point R The waveLets from points A and B will have a path difference equal to BN.

From the right angLed ANB, we have
BN = AB sin θ or BN = a sin θ …(1)

Suppose that the point P on the screen is at such a distance from the centre of the screen that BN = λ. and the angle θ = θ₁.

Then, equation 1 gives
λ = a sin θ₁ or sin θ₁ = \(\frac{λ}{a}\)

Such a point one screen will be the position of the first secondary minimum.

Question 5.
Describe an experiment to show that light waves are transverse in nature.
Answer:
Light is a transverse wave. This can be shown with the help of this simple experiment. The figure shows an unpolarized light beam incident on the first polarising sheet, called the polariser where the transmission axis is indicated by the straight line on the polariser.

The polariser can be a thin sheet of tourmaline (a complex boro-silicate). The Light, which is passing through this sheet, is polarised vertically as shown, where the transmitted electric vector is E_o. A second polarising sheet called the analyser intercepts this beam with its transmission axis at an angle θ to the axis of the polariser. As the axis of the analyser is rotated slowly, the intensity of Light received beyond It goes on decreasing.

When the transmitting axis of the analyser becomes perpendicular to the transmission axis of the polariser, no beam is obtained beyond the analyser. This means that the anaLyser has further polarised the beam coming from the polariser. Since only transverse waves can be polarised, therefore, this shows that light waves are transverse In nature.

Question 6.
Derive an expression for the width of the central maxima for diffraction of light at a single-slit. How does this width change with an increase in the width of the slit?
Answer:
The width of the central maxima for diffraction of light at a single-slit is the distance from the 1st diffraction minima on one side of the central maxima to the 1st diffraction minima on another side of the central maxima.

If a be the width of slit then for 1st diffraction minima, we have
sin θ = ± λ
or
sin θ = θ = ± \(\frac{λ}{a}\)

Angular width of central maxima
= 2θ = ± \(\frac{2λ}{a}\)

If D be the distance between the slit and the screen, then linear width ‘x’ of the central maxima is given by
x = d × 2θ = D × \(\frac{2λ}{a}\) = \(\frac{2Dλ}{a}\)

As the width (a) of the slit is increased, the linear width of central maxima goes on decreasing because x ∝ \(\frac{1}{a}\)

Question 7.
(a) Sketchthegraphshowingthevariation of the intensity of transmitted light on the angle of rotation between a polarizer and an analyser.
Answer:
The graph is as shown below

(b) A ray of light is incident at an angle of incidence i_p on the surface of separation between air and a medium of refractive index µ, such that the angle between the reflected and refracted ray is 90°. Obtain the relation between i_p and µ.
Answer:
Suppose an unpolarised light beam is an incident on a surface as shown in the figure below. The beam can be described by two electric field components, one parallel to the surface (the dots) and the other perpendicular to the first and to the direction of propagation (the arrows). It is found that the parallel components reflect more strongly than the other component, this results in a partially polarised beam. Furthermore, the refracted ray is also partially polarized.

Now suppose the angle of incidence, i is varied until the angle between the reflected and the refracted beam is 90°. At this particular angle of incidence, the reflected beam is completely polarised with its electric vector parallel to the surface, while the refracted beam is partially polarised. The angle of incidence at which this occurs is called the polarising angle i_p.

An expression can be obtained relating the polarising angle to the index of refraction n, of the reflecting surface. From the figure, we see that at the polarising angle
i_p + 90° + r = 180°
or
r = 90° – i_p.

Using Snell’s law we have n = \(\frac{\sin i}{\sin r}=\frac{\sin i_{p}}{\sin r}\)

Now sin r = sin (90° – i_p) = cos i_p, therefore the above expression becomes
n = \(\frac{\sin i_{p}}{\cos i_{p}}\) = tan ip

Question 8.
Describe Young’s double-slit experiment to produce an interference pattern due to a monochromatic source of light. Deduce the expression for the fringe width. (CBSE Delhi 2011)
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S₂ and S₁ at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r1 and r2 are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,
y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe
y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,
y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, a width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

Question 9.
(a) Explain, with the help of a diagram, how plane polarised light Is obtained by scattering.
Answer:
It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

(b) Between two polaroids placed In crossed position a third Polaroid is introduced. The axis of the third Polaroid makes an angle of 30° with the axis of the first polaroid. Find the intensity of transmitted light from the system assuming / to be the intensity of polarised light obtained from the first polaroid. (CBSE A! 2011C)
Answer:
Let l_o be the intensity of light passing through the first polaroid.

The intensity of light passing through the middle polaroid whose axes are inclined at 30° to the first polaroid by Malus law is
l’ = l_o cos² 30° = l_o × 3/4 = 3l_o/4

The intensity of light passing through the system is, therefore, (for the second crystal θ = 60°)
l” = l’cos² 60° = 3l_o/4 × 1 /4
or
l” = 3l_o/16

Question 10.
(a) Why are coherent sources necessary to produce a sustained interference pattern?
Answer:
Interference will be sustained if there is a constant phase difference between the two interfering waves. This is possible if the two waves are coherent.

(b) In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference Is λ, is K units. Find out the intensity of light at a point where the path difference is λ/3. (CBSE Delhi 2012)
Answer:
Intensity at any point on the screen
l = l₁ + l₂ + 2\(\sqrt{1_{1}1_{2}}\) cos Φ

Let l0 be the intensity of either source, then l₁= l₂ = l_o
When p = λ, Φ = 2π

Question 11.
Use Huygen’s principle to explain the formation of the diffraction pattern due to a single-slit illuminated by a monochromatic source of light.
When the width of the slit is made double the original width, how would this affect the size and intensity of the central diffraction band? (CBSE Delhi 2012)
Answer:
(a) The arrangement is shown below in the figure.

When a plane wavefront WW’ falls on a single-slit AB, each point on the unblocked portion ADB of wavefront sends out secondary, wavelets in all the directions. For secondary waves meeting at point O, the path difference between waves is zero and hence the secondary waves reinforce each other giving rise to the central maximum at the symmetrical point O.

Consider secondary waves travelling in a direction making an angle θ with DO and reaching the screen at point P. Obviously, path difference between the extreme secondary waves reaching point P from A and B.
= BC = AB sin θ = a sin θ.

If this path difference a sin θ = λ, then point P will be minimum Intensity. In this situation, wavefront may be supposed to consist of two equal halves AD and BD and for every point on AD, there will be a corresponding point on DB having a path difference λ/2. Consequently, they nullify the effect of each other and point P behaves as the first secondary minimum. In general, if path difference a sin θ = nλ where n = 1, 2, 3, …. then we have secondary minima corresponding to that angle of diffraction θ_n.

However, if for some point P₁ on the screen secondary waves BP₁ and AP₁ differ In path by 3λ/2 then point P₁ will be the position of the first secondary maxima. Because in this situation, wavefront AB may be divided into three equal parts such that path difference between corresponding points on the first and second will be λ/2 and they will nullify. But secondary waves from the third part remain as such and give rise to the first secondary maxima, whose Intensity will be much less than that of central maxima.

In general, if path difference a sin θ = (2n + 1)θ/2, where n = 1, 2, 3, …… then we have nth secondary maxima corresponding to these angles.

The width of the central maxima is the distance between the first secondary minima on either side of the centre of the screen. The width of the central maxima is twice the angle θ subtended by the first minima on either side of the central maxima. Now sin θ = \(\frac{\lambda}{a}\). Since θ is small a therefore it can be replaced by tan θ, hence sin θ = \(\frac{\lambda}{a}=\frac{y}{L} or y = [latex]\frac{L \lambda}{\mathrm{a}}\).

This gives the distance of the first secondary minima on both sides of the centre of the screen. Therefore, the width of the central maxima is 2y, hence
2y = \(\frac{2L \lambda}{\mathrm{a}}\)

(b) The size reduces by half according to the relation: size = X/d. Intensity becomes twice the original intensity.

Question 12.
(a) Using the phenomenon of polarization, show how the transverse nature of light can be demonstrated.
Answer:
Light is a transverse wave. This can be shown with the help of this simple experiment. The figure shows an unpolarized light beam incident on the first polarising sheet, called the polariser where the transmission axis is indicated by the straight line on the polariser.

The polariser can be a thin sheet of tourmaline (a complex boro-silicate). The Light, which is passing through this sheet, is polarised vertically as shown, where the transmitted electric vector is E_o. A second polarising sheet called the analyser intercepts this beam with its transmission axis at an angle θ to the axis of the polariser. As the axis of the analyser is rotated slowly, the intensity of Light received beyond It goes on decreasing.

When the transmitting axis of the analyser becomes perpendicular to the transmission axis of the polariser, no beam is obtained beyond the analyser. This means that the anaLyser has further polarised the beam coming from the polariser. Since only transverse waves can be polarised, therefore, this shows that light waves are transverse in nature.

(b) Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other. Unpolarised light of intensity l0 is Incident on P₁. A third polaroid P₃ is kept in between P₁ and P₂ such that its pass axis makes an angle of 30° with that of P₁. Determine the intensity of light transmitted through P₁ P₂ and P₃. (CBSE AI 2014)
Answer:
The intensity of light passing through P₁ is half of the light falling on it. Therefore, the light coming out of P₁ is l0_o/2 Now light coming out of P₃ is

and light coming out of polariser P₂ is

Question 13.
What does a Polaroid consist of? Show using a simple Polaroid that light waves are transverse in nature. The intensity of light coming out of a Polaroid does not change irrespective of the orientation of the pass axis of the polaroid. Explain why. (CBSE AI 2015)
Answer:
A polaroid consists of long-chain molecules aligned in a particular direction.

Let the light from an ordinary source (like a sodium lamps pass through a polaroid sheet P₁ it is observed that its intensity is reduced by half. Now, let an identical piece of polaroid P₂ be placed before P₁ As expected, the light from the lamp is reduced in intensity on passing through P₂2 alone. But now rotating P₁ has a dramatic effect on the light coming from P₂. In one position, the intensity transmitted by P₂ followed by P₁ is nearly zero. When turned by 90° from this position, P₁ transmits nearly the full intensity emerging from P₂ shown in the figure.

Since only transverse waves can be polarised, therefore, this experiment shows that light waves are transverse in nature. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarized light wave is an incident on such a polaroid then the light wave will get linearly polarized with the electric vector oscillating along a direction perpendicular to the aligned molecules; this direction is known as the pass-axis of the Polaroid.

Thus, if the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet P it is observed that its intensity is reduced by half. Rotating P has no effect on the transmitted beam and transmitted intensity remains constant as there is always an electric vector that oscillates in a direction perpendicular to the direction of the aligned molecules.

Question 14.
Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted Intensity be maximum? (CBSE Delhi 2015)
Answer:
Let l0 be the intensity of polarised light after passing through the first polarizer P₁. Then the intensity of light after passing through the second polarizer P₂ will be
l = l_o cos² θ

where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be

The transmitted intensity will be maximum when 2θ = π/2 or θ = π/4

Question 15.
Distinguish between unpolarised and linearly polarised light. Describe with the help of a diagram how unpolarised light gets linearly polarised by scattering. (CBSE Delhi 2015)
Answer:
The two lights will be allowed to pass through a polariser. When the polarizer is rotated in the path of these two light beams, the intensity of light remains the same in all the orientations of the polariser, then the light is unpolarised. But if the intensity of light varies from maximum to minimum then the light beam is a polarised light beam.

It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

Question 16.
(a) Explain how a diffraction pattern is formed due to interference of secondary wavelets of light waves from a slit.
Answer:
(a) Diffraction at single slit: Consider monochromatic plane wavefront WW’ incident on the slit AB as shown in Fig. Imagine the slit to be divided into a large number of very narrow strips of equal width parallel to the slit. When the wavefront reaches the slit, each narrow strip parallel to the slit can be considered to be a source of Huygen’s secondary wavelets.

Central Maxima. Let us first consider the effect of all the wavelets at point O. All wavelets cover the same distance and reach 0 in the same phase. The wavelets superimpose constructively and give maximum intensity at O.

Position of secondary minima. Consider the intensity at a point P at an angle θ with the original direction.
Draw AL ⊥ BP.
In ΔABL


or
sin θ = \(\frac{λ}{a}\)

For second minima, BL = 2A
or a sin θ = 2λ

Similarly, for nth minima a sin θ = nλ
or sin θ = \(\frac{nλ}{a}\) , where n = 1, 2, ……

Position of secondary maxima: If path difference BL = \(\frac{3λ}{2}\), then slit AB can be divided into three equal parts and the path difference between wavelets from corresponding points in the first two parts will be \(\frac{λ}{2}\) and hence cancels each other’s effect and produces destructive Interference and the third path will produce its maxima at P called first secondary maxima.

Similarly, if BL = \(\frac{5λ}{2}\), we get second secondary maxima and so on.

For nth maxima
a sin θ = (2n + 1)\(\frac{λ}{a}\)
or
sin θ = \(\frac{(2 n+1) \lambda}{2 a}\)

Thus we find that the intensity of the central fringe is maximum whereas that of other fringes fall off rapidly in either direction from the centre of the fringe pattern.

(b) Sodium light consists of two wavelengths, 5900 Å and 5960 Å. If a slit of width 2 × 10^-4 m is Illuminated by sodium light, find the separation between the first secondary maxima of the diffraction pattern of the two wavelengths on a screen placed 1.5m away. (CBSE 2019C)
Answer:
The separation between the first secondary maxima of the diffraction pattern of two wavelengths is:

Question 17.
(a) Derive Snell’s law on the basis of Huygen’s wave theory when light Is travelling from a denser to a rarer medium.
Answer:
The ray diagram is as shown.


(b) Draw the sketches to differentiate between plane wavefront and spherical wavefront. (CBSE AI 2016)
Answer:
Spherical wavefront and plane wavefront

Question 18.
The figure drawn here shows the geometry of path differences for diffraction by a single-slit of width a.

Give appropriate ‘reasoning’ to explain why the intensity of light is
(a) Maximum at the central point C on the screen.
(b) (Nearly) zero for point P on the screen when θ = λ / a.
Hence write an expression for the total linear width of the central maxima on a screen kept at a distance D from the plane of the slit. (CBSE Delhi 2016C)
Answer:
(a) At central point C, the angle is zero, all path differences are zero. Hence all the parts of the slit contribute to the same phase. This gives the maximum intensity at point C.
(b)

From figure
NP – LP = NQ= a sin θ = aθ When θ = λ/a

Then path difference NP – LP = aθ = λ.
Hence MP – LP = NP – MP = λ/2

It implies that the contribution from corresponding points in two halves of the slit has a phase difference of π. Therefore, contributions from two halves cancel each other in pairs, resulting in a zero net intensity at point P on the screen. Half angular width of central maxima = λ/a Half linear width = λD/a

Linear width of central maxima = 2λD/a

Question 19.
Two polaroids, P₁ and P₂, are ‘set-up’ so that their ‘pass axis is ‘crossed’ with respect to each other. AQ third Polaroid, P₃ is now introduced between these two so that its ‘pass axis makes an angle with the ‘pass axis of P_1. A beam of unpolarized light, of Intensity I, Is Incident on P₁ If the Intensity of light that gets transmitted through this combination of three polaroids, Is I’, find the ratio \(\left(\frac{I^{\prime}}{I}\right)\) when θ equals: (i) 30°, (ii) 45° (CBSE Delhi 2016C)
Answer:
The diagram is as shown.

The intensity of unpolarized light is given as l.
It becomes half after passing through Polaroid P₁ (l/2)
Using Malus law for the intensity of light passing through P3 we have
l₁ = \(\left(\frac{l}{2}\right)\) cos²θ

This intensity l₁ is Incident on P₂, hence the intensity of light coming out of P₂ will be

Question 20.
What is the effect on the Interference pattern observed In Young’s double-slit experiment In the following cases:
(a) Screen is moved away from the plane of the slits,
(b) Separation between the slits is Increased and
(c) Widths of the slits are doubled. Give the reason for your answer.
Answer:
The fringe width is given by the expression
β = \(\frac{Dλ}{d}\)

(a) When D Is Increased, the fringe width Increases.
(b) When d 1s Increased the fringe width decreases.
(c) If the width w of the slits Is changed then Interference occurs only If \(\frac{1}{w}\) > \(\frac{1}{d}\) remains satisfied, where d is the distance between the slits.

Question 21.
Two slits In Young’s double-slit experiment are illuminated by two different lamps emitting light? Will you observe the Interference pattern? Justify your answer. Find the ratio of Intensities at two points on a screen In Young’s double-slit experiment, when waves from two slits have a path difference of (i) 0 and (ii) λ/4.
Answer:
The two sources act as Independent sources of light and hence can never be coherent. In a light, source light is produced by billions of atoms under proper excitation condition and each atom acts independently of the other atoms. Thus there Is no coherence ‘ between these two Independent sources hence no interference.

The phase difference corresponding to the two paths are Φ = 0 and Φ = π/2

Now intensity at the screen when the phase difference is Φ = 0 is
l_A = l₁ +l₂ + 2\(\sqrt{l_{1} l_{2}}\)cos Φ
or
l_A = l + l + 2\(\sqrt{l l}\)cos Φ = 4l

Now intensity at the screen when the phase difference is Φ = 90° Is
l_B = l₁ +l₂ + 2\(\sqrt{l_{1} l_{2}}\)cos Φ
or
l_B = l + l + 2\(\sqrt{l l}\)cos 90° = 2l
Therefore, ratio of intensities is
\(\frac{l_{A}}{l_{B}}=\frac{4l}{2l}\) = 2

Question 22.
(a) In a single-slit diffraction pattern, how does the angular width of the central maximum vary, when
(i) the aperture of the slit Is Increased?
(ii) distance between the slit and the screen is decreased?
Justify your answer In each case.
Answer:
The angular width of the central maxima in a single-slit diffraction pattern is given by 2θ = \(\frac{2λ}{a}\) where λ is the wavelength of light and ‘a’ the slit width.
(i) When the aperture of the slit is increased the angular width decreases.
(ii) When the distance between the slit and the screen is decreased, the angular width will remain the same but the linear width will increase.

(b) How Is the diffraction pattern different from the interference pattern obtained In Young’s double¬slit experiment? (CBSE Delhi 2011C)
Answer:
The difference is shown in the table:

Diffraction	Interference
1. It is due to interference between the wavelets starting from two parts of the same wavefront.	1. It is a superposition of the two waves starting from two coherent sources.
2. The intensity of the consecutive bands goes on decreasing.	2. All bright fringes have the same intensity.
3. Fringes have poor contrast.	3. Fringes have good contrast.
4. Diffraction fringes are not of the same width.	4. Fringes may or may not be of the same width.

Question 23.
(a) Can two independent monochromatic light sources be used to obtain a steady interference pattern? Justify your answer. (CBSE 2019C)
Answer:
No, because the phase difference between the light waves from two independent sources keeps on changing continuously and for a steady interference pattern, the phase difference between the waves should remain constant with time. Hence two independent monochromatic light sources cannot produce a steady interference pattern.

(b) In Young’s double-slit experiment, explain the formation of interference fringes and obtain an expression for the fringe width.
Answer:
Interference of light

Let S₁, S₂ be the two fine slits illuminated by a monochromatic source S of wavelength λ.

The intensity of light at any point P on the screen at a distance D from the slit depends upon the path difference between S₂P and S₁P.
∴ Path difference = S₂P – S₁P
= (S₂A + AP) – S₁P

Path difference = S₂A = d sin θ
Since θ is small, sin θ can be replaced by tan θ.
∴ Path difference = d tan θ = d\(\frac{y}{D}\)

Constructive interference [Bright Fringes]. For bright fringes, the path difference should be equal to integral multiple of A.
∴ \(\frac{dy}{D}\) = nλ
or
y = n\(\frac{λD}{d}\)

For nth fringe, let us write y as y_n
so y_n = n\(\frac{λD}{d}\)

The spacing between two consecutive bright fringes is equal to the width of a dark fringe.
∴ Width of the dark fringe.
β = y_n – y_n-1 = \(\frac{nλD}{d}\) – (n – 1)\(\frac{λD}{d}\) = \(\frac{λD}{d}\) …(i)

Destructive interference. [Dark Fringes].
For dark fringes, the path difference should be an odd multiple of λ/2.
∴ \(\frac{d}{D}\)y = (2n + 1)λ/2
or
y’ = \(\frac{D}{d}\)(2n + 1)\(\frac{λ}{2}\) = \(\frac{(2 n+1) \lambda D}{2 d}\)

For nth fringe, writing y as yn, we get
y’_n = \(\frac{(2 n+1) \lambda D}{2 d}\).

The spacing between two consecutive dark fringes is equal to the width of a bright fringe.
∴ Width of the bright fringe.

From Eqs. (i) and (ii), we find that dark and bright fringes are of same width given by
β = \(\frac{λD}{d}\).

(c) In an interference experiment using monochromatic light of wavelength A, the intensity of light of point, where the path difference is X, on the screen is K units. Find out the Intensity of light at a point when path difference is λ/4. (CBSE 2019C)
Answer:
The intensity of light on the screen where the waves meet having phase difference Φ is:


Question 24.
(a) Two monochromatic waves emanating from two coherent sources have the displacements represented by y₁ = a cos ωt and y₂ = a cos (ωt + Φ), where Φ is the phase difference between the two displacements. Show that the resultant intensity at a point due to their superposition is given by l = 4l_o cos² Φ/2, where l_o = a².
Answer:
Let the displacements of the waves from the sources S₁ and S₂ at a point on the screen at any time t be given by y₁ = a cos ωt and y₂ = a cos (ωt + Φ), where Φ is the constant phase difference between the two waves. By the superposition principle, the resultant displacement at point P is given by

Thus the amplitude of the resultant displacement is A = 2a cos (Φ/2)

Therefore the intensity at that point is
l = A² = 4a² cos² \(\frac{Φ}{2}\)

(b) Hence obtain the conditions for constructive and destructive interference. (CBSE AI 2014C)
Answer:
Bright fringes: For bright fringes I = max, therefore Φ = 0° or cos Φ = +1
or
Φ = 2 n π
Dark fringes: For dark fringes l = 0,
therefore Φ = π or cos Φ = -1
or
Φ = (2n + 1) π

Question 25.
A parallel beam of monochromatic light falls normally on a narrow slit of width ‘a’ to produce a diffraction pattern on the screen placed parallel to the plane of the slit. Use Huygens’ principle to explain that
(a) the central bright maxima is twice as wide as the other maxima.
Answer:
The width of the central maxima is the distance between the first secondary minima on either side of the centre of the screen. The width of the central maxima is twice the angle 6 subtended by the first minima on either side of the central maxima.

Now sin θ = \(\frac{λ}{a}\). Since θ is small there, a fore it can be replaced by tan 0, hence
tan θ = \(\frac{λ}{a}\) = \(\frac{y}{L}\)
or
y = \(\frac{Lλ}{a}\). This gives the distance of the first secondary min¬ima on both sides of the centre of the screen. Therefore, the width of the central maxima is 2y, hence
2y = \(\frac{2Lλ}{a}\)

(b) the Intensity falls as we move to successive maxima away from the centre on either side. (CBSE Delhi 2014C)
Answer:
The maxima in diffraction pattern are formed at (n +1 /2) λ/a, with n = 2, 3, etc. These become weaker with increasing n, since only one third, one- fifth, one-seventh, etc., of the slit contributes in these cases.

Question 26.
Answer the following questions:
(a) In a double-slit experiment using the light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.
Answer:

(b) Light of wavelength 5000 A propagating 1n air gets partly reflected from the surface of the water. How will the wavelengths and frequencies of the reflected and refracted light be affected? (CBSE Delhi 2015)
Answer:
No change in the wavelength and frequency of reflected light. In the case of refracted light, there is no change in frequency but the wavelength becomes 1/1.33 times the original wavelength.

Question 27.
(a) Good quality sun-glasses made of polaroids are preferred over ordinary coloured glasses. Justify your answer,
Answer:
A Polaroid sun-glass limits the light entering the eye, thus providing a soothing effect.

(b) Two polaroids and P₂ are placed In crossed positions. A third Polaroid P₃ is kept between P₁ and P₂ such that the pass axis of P₃ is parallel to that of P₁. How would the Intensity of light l₂ transmitted through P₂ vary as P₃ Is rotated? Draw a plot of Intensity l₂ Vs the angle ‘θ’, between pass axes of P₁ and P₃. (CBSE AI 2015C)
Answer:
Let l0 be the intensity of polarised light after passing through the first polarizer P₁ Then the intensity of light after passing through the second polarizer P₃ will be
l = l_o cos² θ
where θ is the angle between pass axes of P₁ and P₃. Since P₁ and P₂ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ).

Hence the intensity of light emerging from P2 will be

Therefore, the transmitted intensity will be maximum when θ = π/4

For graph

Question 28.
In a single-slit diffraction pattern, how does the angular width of central maximum change, when
(a) slit width is decreased,
(b) distance between the slit and screen is increased, and
(c) light of smaller visible wavelength is used? Justify your answer in each case.
Answer:
We know that the angular width of the central maximum of the diffraction pattern of a single-slit is given by
w = \(\frac{2Dλ}{a}\).

(a) If slit width ‘a’ is decreased, the angular width will increase because
x ∝ \(\frac{1}{a}\)

(b) Increase in distance between the slit and the screen does not affect the angular width of diffraction maxima. However, linear width of the maxima
w = \(\frac{2Dλ}{a}\) will increase.

(c) If the light of a smaller visible wavelength is used, the angular width is decreased because x ∝ λ.

Question 29.
Light, from a sodium lamp, is passed through two polaroid sheets P₁ and P₂ kept one after the other. Keeping P₁ fixed, P₂ is rotated so that its ‘pass axis can be at different angles, θ, with respect to the pass-axis of P₁
An experimentalist records the following data for the Intensity of light coming out of P₂ as a function of the angle θ.

(a) l_o = Intensity of beam falling on P₁ One of these observations is not in agreement with the expected theoretical variation of l, Identify this observation and write the correct expression.
Answer:
The observation \(\frac{l_{0}}{2 \sqrt{2}}\) is not correct. It should be l_o/4.

(b) Define the Brewster angle and write the expression for it in terms of the refractive index of the medium.
Answer:
It is the angle of incidence at which the refracted and the reflected rays are perpendicular to each other. It is related to the refractive index as tan i_p = n

Question 30.
(a) Light, from a monochromatic source, is made to fall on a single-slit of variable width. An experimentalist records the following data for the linear width of the principal maxima ‘ on a screen kept at a distance of 1 m from the plane of the slit.

Use any two observations from this data to estimate the value of the wavelength of light used.
Answer:
The width of the central maxima is given by the expression β = \(\frac{2Lλ}{a}\)
or
λ₁ = \(\frac{βa}{2L}\)

Using the values of the first observation we have
λ₁ = \(\frac{6 \times 10^{-3} \times 0.1 \times 10^{-3}}{2 \times 1}\) = 0.3 × 10^-6

Using the values of the second observation we have
λ₁ = \frac{3 \times 10^{-3} \times 0.2 \times 10^{-3}}{2 \times 1} = 0.3 × 10^-6

Thus the wavelength of light used is λ = 0.3 × 10^-6 m

(b) Show that the Brewster angle iB for a given pair of transparent media is related to their critical angle ic through the relation i_c = sin^-1 (ωt i_B)
Answer:
We know that

Question 31.
For a single-slit of width, “a” the first minimum of the interference pattern of monochromatic light of wavelength λ occurs at an angle of λ/a. At the same angle λ/a, we get a maximum for two narrow slits separated by a distance ‘a’. Explain. (CBSE Delhi 2014)
Answer:
The path difference between two secondary wavelets is given by nλ = a sin θ. Since θ is a very small sin θ = 0. So, for the first-order diffraction n = 1, the angle is λ/a. Now we know that θ must be very small θ = 0 (nearly) because of which the diffraction pattern is minimum.

Now for interference case, for two interfering waves of intensity l1 and l2 we must have two slits separated by a distance. We have the resultant intensity,
l = l₁ + l₂ + 2\(\sqrt{l_{1} l_{2}}\) cos θ

Since θ = 0° (nearly) corresponding to angle λ/a so cos θ = 1 (nearly)
So,
l = l₁ + l₂ + 2\(\sqrt{l_{1} l_{2}}\) cos θ
l = l₁ + l₂ + 2\(\sqrt{l_{1} l_{2}}\)

We see the resultant intensity is the sum of the two intensities, so there is a maxima corresponding to the angle λ/a.

This is why, at the same angle of λ/a, we get a maximum for two narrow slits separated by a distance “a”.

Question 32.
Show using a proper diagram of how unpolarised light can be linearly polarized by reflection from a transparent glass surface. (CBSE AI 2018, Delhi 2018)
Answer:
An ordinary beam of light, on reflection from a transparent medium, becomes partially polarised. The degree of polarisation increases as the angle of incidence is increased. At a particular value of the angle of incidence, the reflected beam becomes completely polarised. This angle of incidence is called the polarising angle (i_p).

Question 33.
Answer the following questions:
(a) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
Answer:
The size reduces by half according to the relation: size = λ/d. Intensity increases fourfold.

(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
Answer:
The intensity of interference fringes in a double-slit arrangement is modulated by the diffraction pattern of each slit.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
Answer:
Waves diffracted from the edge of the circular obstacle interfere constructively at the centre of the shadow producing a bright spot.

(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily? (NCERT)
Answer:
For diffraction or bending of waves by obstacles/apertures by a large angle, the size of the latter should be comparable to the wavelength. If the size of the obstacle/aperture is much too large compared to the wavelength, diffraction is by a small angle. Here the size is of the order of a few metres. The wavelength of light is about 5 × 10^-7 m, while sound waves of, say, 1 kHz frequency have a wavelength of about 0.3 m. Thus, sound waves can bend around the partition while light waves cannot.

Question 34.
(a) When an unpolarized light of intensity l0 is passed through a polaroid, what is the intensity of the linearly polarised light? Does it depend on the orientation of the polaroid? Explain your answer.
Answer:
(a) The intensity of the linearly polarised light would be l_o/2.
No, it does not depend on the orientation.
Explanation: The polaroid will let the component of the unpolarized light, parallel to its pass axis, to pass through it irrespective of its orientation.

(b) A plane polarised beam of light is passed through a polaroid. Show graphically the variation of the intensity of the transmitted light with an angle of rotation of the polaroid incomplete one rotation. (CBSE Delhi 2018C)
Answer:
We have l = l_o cos² θ
∴ The graph is as shown below

Question 35.
(a) If one of two identical slits producing interference in Young’s experiment is covered with glass so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the Interference pattern.
Answer:
As intensity is directly proportional to the square of the amplitude.


(b) What kind of fringes do you expect to observe if white light is used Instead of monochromatic light?
Answer:
The central fringe will be white and the remaining will be coloured fringes of different width in the VIBGYOR sequence.

Question 36.
Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum? (CBSE Delhi 2015)
Answer:
Let l0 be the intensity of polarised light after passing through the first polarizer P₁ Then the intensity of light after passing through the second polarizer P₂ will be
l = l_o cos² θ
where 0 is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P3 will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be

The transmitted intensity will be maximum when 2θ = π/2 or θ = π/4

Question 37.
Derive the expression for the fringe width In Young’s double-slit experiment.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S₂ and S₁ at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r₁ and r₂ are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o0 = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,

y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe

y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,

y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, a width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

Question 38.
(a) Using Huygens’ principle, draw a diagram to show the propagation of a wave-front originating from a monochromatic point source.
Answer:

(b) Describe diffraction of light due to a single-slit. Explain the formation of a pattern of fringes obtained on the screen and plot showing a variation of intensity with path difference in single-slit diffraction.
Answer:
Let us discuss the nature of the Fraunhofer diffraction pattern produced by a single-slit. Let us examine the waves coming from the various portions of the slit, as shown in the figure. According to Huygens principle, each portion of the slit acts as a source of waves. Hence, light from one portion of the slit can interfere with the light of another portion, and the resultant intensity on the screen will depend on the direction of θ. The secondary waves coming from the different parts of the slit interfere to produce either maxima or minima, thereby giving rise to a diffraction pattern.

For the diagram

2y = \(\frac{2Lλ}{a}\)

Question 39.
What is the effect on the interference fringes in Young’s double-slit experiment due to each of the following operations:
(a) the screen is moved away from the plane of the slits;
Answer:
The angular separation of the fringes remains constant (= λ/d). The actual separation of the fringes increases in proportion to the distance of the screen from the plane of the slits

(b) the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength;
Answer:
The separation of the fringes (and also angular separation) decreases.

(c) the separation between the two slits is Increased
Answer:
The separation of the fringes (and also angular separation) decreases.

(d) the source slit is moved closer to the double-slit plane;
Answer:
Let s be the size of the source and S its distance from the plane of the two slits. For interference, fringes to be seen the
condition \(\frac{s}{S}<\frac{\lambda}{d}\) should be satisfied otherwise interference patterns produced by different parts of the source overlap and no fringes are seen. Thus as S decreases (i.e. the source slit is brought closer), the interference pattern gets less and less sharp and when the source is brought too close for this condition to be valid the fringes – disappear. Till this happens, the fringe separation remains fixed.

(e) the width of the source slit is Increased;
Answer:
Same as in (d). As the source slit width increases fringe pattern gets less and less sharp. When the source slit is so wide that the condition \(\frac{s}{S}<\frac{\lambda}{d}\) is not satisfied the interference pattern disappears.

(f) the monochromatic source is replaced by a source of white light? (In each operation take all parameters other than the one specified to remain unchanged.)
Answer:
The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are in the same position. Therefore, the central fringe is white. For a point P for which S₂P – S₁P= λ_b /2, where (λ_b = 400 nm) represents the wavelength for the blue colour, the blue component will be absent and the fringe will appear red in colour. Slightly farther away from where S₂Q – S₁Q = λ_b = λ_r/2 where λ_r (800 nm) is the wavelength for the red colour, the fringe will be predominantly blue. Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.

Question 40.
What is the diffraction of light? Draw a graph showing the variation of intensity with the angle in a single-slit diffraction experiment. Write one feature which distinguishes the observed pattern from the double-slit interference pattern. How would the diffraction pattern of a single-slit be affected when:
(i) the width of the slit is decreased?
(ii) the monochromatic source of light is replaced by a source of white light?
Answer:
Diffraction is the bending of light around obstacles or openings. It is a consequence of the wave nature of light. For diffraction to take place the obstacle should be of the order of the wavelength of light.

The intensity distribution for the single-slit diffraction pattern is as shown.

The intensity of all bright fringes is the same in Young’s interference pattern, but in diffraction, the intensity of bright fringes falls off on both sides of the central fringe.
We know that the angular width of the central maximum of the diffraction pattern of a single-slit is given by = \(\frac{2Dλ}{a}\)
(i) If slit width ‘a’ is decreased, the angular width will increase because
x ∝ \(\frac{1}{a}\)

(ii) When monochromatic light is replaced by white light, all the seven wavelengths form their own diffraction pattern, so coloured fringes are formed. The first few fringes are visible but due to overlapping the clarity of fringes decreases as the order increases.

Question 41.
(a) Define a wavefront. Using Huygens’ geometrical construction, explain with the help of a diagram how the plane wavefront travels from the instant t₁ to t₂ in the air.
Answer:
(a) Wavefront: It is the locus of the medium or points of a medium that is in the same phase of disturbance
First, consider a plane wave moving through free space as shown in the figure. At t = 0, the wavefront is indicated by the plane labelled AA’. In Huygens’s construction, each point on this wavefront is considered a point source. For clarity, only a few points on AA’ are shown. With these points as sources for the wavelets, we draw circles each of radius cΔt, where c is the speed of light in free space and Δt is the time of propagation from one wavefront to the next. The surface drawn tangent to these wavelets is the plane BB’, (Wavefront at a later time t). Here A₁ A₂ = B₁ B₂ = C₁C₂ = C_τ

(b) A plane wavefront is incident on a convex lens. Explain, with the help of the diagram, the shape of the refracted wavefront formed. (CBSE AI 2019)
Answer:
Spherical wavefront

Question 42.
(a) In Young’s double-slit experiment, derive the condition for
(i) constructive Interference and
(ii) destructive Interference at a point on the screen.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a wavelength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S₂ and S₁ at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r₁ and r₂ are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,

y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe

y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,

y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, the width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

(b) A beam of light consisting of two wavelengths, 800 nm and 600 nm, is used to obtain the interference fringes in Young’s double-slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide. (CBSE Al 2012)
Answer:
Let at a distance y from central maxima the bright fringes due to wavelengths l1 and l2 coincide first time. For this to happen, nl₁ = (n + 1)l₂, where n is an integer.
or
\(\frac{n+1}{n}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{800 \times 10^{-9}}{600 \times 10^{-9}}=\frac{4}{3}\) solving for n we have n = 3

It means that at distance nth (3^th) maxima for wavelength λ₁ is just coinciding with (n + 1)^th (4^th) maxima for wavelength l₂

Question 43.
(a) How does an unpolarized light incident on a Polaroid get polarised? Describe briefly, with the help of a necessary diagram, the polarisation of light by reflection from a transparent medium.
(b) Two polaroids ‘A’ and ‘B’ are kept in a crossed position. How should a third Polaroid ‘C’ be placed between them so that the intensity of polarised light transmitted by Polaroid B reduces to 1/8th of the intensity of unpolarized light incident on A? (CBSE AI 2012)
Answer:
(a) The polariser allows only those vibrations of light to pass which are parallel to the pass axis of the polariser and blocks the remaining vibrations. Thus the light vibrations are restricted to only one plane.

Suppose an unpolarised light beam is an incident on a surface as shown in (figure a). The beam can be described by two electric field components, one parallel to the surface (the dots) and the other perpendicular to the first and to the direction of propagation (the arrows). It is found that the parallel components reflect more strongly than the other component, this results in a partially polarised beam. Furthermore, the refracted ray is also partially. polarised. Now suppose the angle of incidence 6, is varied until the angle between the reflected and the refracted beam is 90° (fig b). At this particular angle of incidence, the reflected beam is completely polarised with its electric vector parallel to the surface, while the refracted beam is partially polarised.

(b) Let l0 be the intensity of unpolarized light incident on Polaroid A. Then l_o/2 will be the intensity of polarised light after passing through the first polariser A. Then the intensity of light after passing through second polariser B will be l = (l_o/2)cos²θ

where θ is the angle between pass axes of A and B. Since A and B are crossed the angle between the pass axes of B and C will be (π/2 – θ). Hence the intensity of light emerging from C will be

Therefore, we have
\(\frac{l_{0}}{8}=\frac{l_{0}}{8}\) sin² 2θ or sin 2θ = 1 or 2θ = 90°
i.e. θ = π/4

Therefore, the transmitted intensity will be maximum when θ = π/4

Question 44.
(a) Describe any two characteristic features which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young’s double-slit experiment.
(b) In the diffraction due to a single slit experiment, the aperture of the slit is 3 mm. If monochromatic light of wavelength 620 nm is incident normally on the slit, calculate the separation between the first order minima and the 3^rd order maxima on one side of the screen. The distance between the slit and the screen is 1.5 m. (CBSE Delhi 2019)
Answer:
(a) (i) Interference pattern has a number of equally spaced bright and dark bands, while the diffraction pattern has a central bright maximum which is twice as wide as the other maxima.

(ii) In interference pattern, the intensity of all bright fringes is the same, while in diffraction pattern intensity of bright fringes goes on decreasing with the increasing order of the maxima.

Let the displacements of the waves from the sources S₁ and S₂ at a point P on the screen at any time t be given by y₁ = a cos cot and y₂ = a cos (ωt + Φ), where Φ is the constant phase difference between the two waves.

By the superposition principle, the resultant displacement at point P is given by

Thus the amplitude of the resultant displacement is
A = 2a cos (Φ/2) …(4)

Therefore the intensity at that point is
l = A² = 4a² cos²\(\frac{Φ}{2}\) …(5)

(b) Given a = 3mm = 3 × 10^-3m, λ = 620nm = 620 × 10^-9 m, D = 1.5 m Distance of first-order minima from centre of the central maxima = X_D1 = λ_D/a

Distance of third order maxima from centre of the central maxima X_D3 = 7λ_D/2a

Distance between first order minima and third order maxima = X_D3 – X_D1

Question 45.
(a) In Young’s double-slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence deduce the expression for the fringe width.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S_D1 and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S2 and S1 at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r1 and r2 are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. Therefore, the condition for dark fringes, or destructive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,

y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe

y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,

y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, the width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

(b) Show that the fringe pattern on the screen is actually a superposition of single-slit diffraction from each slit.
Answer:
It is a broader diffraction peak in which there appear several fringes of smaller width due to the double-slit interference pattern. This is shown below:

(c) What should be the width of each slit to obtain 10 maxima of the double-slit pattern within the central maximum of the single-slit pattern, for the green light of wavelength 500 nm, if the separation between two slits is 1 mm? (CBSE AI 2015)
Answer:
Given 10 b = width of central maxima in diffraction pattern λ = 500 nm, d= 1 mm,
Now 10\(\frac{Dλ}{d}\) = \(\frac{Dλ}{a}\)
or
a = \(\frac{d}{5}=\frac{1}{5}\) = 0.2 mm

Question 46.
(a) Define a wavefront. How is it different from a ray?
Answer:
A wavefront is defined as the locus of all adjacent points at which the phase of vibration of a physical quantity associated with the wave is the same.

It is in two dimensions while a ray is in one dimension.

(b) Depict the shape of a wavefront in each of the following cases.
(i) Light diverging from a point source,
Answer:

(ii) Light emerging out of a convex lens when a point source is placed at its focus.
Answer:

(c) Using Huygen’s construction of secondary wavelets, draw a diagram showing the passage of a plane wavefront from a denser into a rarer medium. (CBSE AI 2015C)
Answer:
The diagram is as shown

Question 47.
(a) Why does unpolarised light from a source show a variation in intensity when viewed through a Polaroid which is rotated? Show with the help of a diagram how unpolarised light from sun got linearly polarised by scattering.
(b) Three identical Polaroid sheets P1, P2 and P3 are oriented so that the pass axis of P2 and P3 are inclined at angles of 60° and 90° respectively with the pass axis of P1. A monochromatic source S of unpolarised light of intensity l0 is kept in front of the polaroid sheet P! as shown in the figure. Determine the intensities of light as observed by the observer at O when Polaroid P3 is rotated with respect to P2 at angles θ = 30° and 60°. (CBSE AI 2016)

Answer:
(a) When light passes through a Polaroid it absorbs all vibrations which are not parallel to its pass axis.
It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

(b) Let l0 be the intensity of light before passing P₁ Intensity of light coming out of P₁ will be l_o/2.

Then the intensity of light after passing through second poloroid (analyser) is
l₂ = \(\frac{l_{0}}{2}\)cos²60° = \(\frac{l_{0}}{8}\)

When P₃ is rotated with respect to P₂ at an angle of 30°, then angle be¬tween the pass axis of P₂ and P₃ will be θ = 30°+ 30° = 60°.
Hence l3 = l2 cos²60° = \(\frac{l_{0}}{8} \times \frac{1}{4}=\frac{l_{0}}{32}\)
or
θ = 30° – 30° = 0°

Therefore, l₃ = l₂ cos² 0° = l₂ = l_o/8
When P₃ is rotated with respect to P₂ at an angle of 60°, then the angle between the pass axis of P₂ and P₃ will be
θ = 30° + 60° = 90°

Question 48.
(a) Derive an expression for path difference in Young’s double-slit experiment and obtain the conditions for constructive and destructive interference at a point on the screen.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S₂ and S₁ at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r₁ and r₂ are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,
y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe
y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,
y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, the width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

(b) The Intensity at the central maxima in Young’s double-slit experiment is l0. Find out the intensity at a point where the path difference is λ/6, λ/4 and λ/3. (CBSE AI 2016)
Answer:
Let l be the intensity of light coming out of each slit. Since waves at the central maxima are in phase, therefore, we have

Question 49.
(a) Define a wavefront. Using Huygens’ Principle, verify the laws of reflection at a plane surface.
Answer:
The wavefront is a locus of points that oscillate in the same phase.

Consider a plane wavefront AB incident obliquely on a plane reflecting surface MM^–. Let us consider the situation when one end A of wavefront strikes the mirror at an angle i but the other end B has still to cover distance BC. The time required for this will be t = BC/c.

According to Huygen’s principle, point A starts emitting secondary wavelets and in time t, these will cover a distance c t = BC and spread. Hence, with point A as centre and BC as radius, draw a circular arc. Draw tangent CD on this arc from point C. Obviously, the CD is the reflected wavefront inclined at an angle ‘r’. As incident wavefront and reflected wavefront, both are in the plane of the paper, the 1^st law of reflection is proved.

To prove the second law of reflection, consider ΔABC and ΔADC. BC = AD (by construction),
∠ABC = ∠ADC = 90° and AC is common.

Therefore, the two triangles are congruent and, hence, ∠BAC = ∠DCA or ∠i = ∠r, i.e.the angle of reflection is equal to the angle of incidence, which is the second law of reflection.
Or
The refractive index of medium 2, w.r.t. medium 1 equals the ratio of the sine of the angle of incidence (in medium 1) to the sine of the angle of refraction (in medium 2), The diagram is as shown.

From the diagram

(b) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? Explain.
Answer:
Size of central maxima reduces to half, (Size of central maxima = 2λD/d) Intensity increases.
This is because the amount of light, entering the slit, has increased and the area, over which it falls, decreases.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the obstacle. Explain why. (CBSE AI 2018, Delhi 2018)
Answer:
This is because of the diffraction of light. Light gets diffracted by the tiny circular obstacle and reaches the centre of the shadow of the obstacle.
β = \(\frac{Dλ}{d}\).

(i) When D is decreased, the fringe width decreases.
(ii) When d is Increased, the fringe width decreases.

Question 50.
The following figure shows an experiment set up similar to Young’s double-slit experiment to observe interference of light.

Here SS₂ – SS₁ = λ/4
Write the condition of (i) constructive,
(ii) destructive Interference at any point P in terms of path difference Δ = S₂P – S₁P.

Does the central fringe observed in the above set up tie above or below O? Give reason in support of your answer.
Yellow light of wavelength 6000°A produces fringes of width 08 mm in Young’s double-slit experiment. What will be the fringe width if the light source is replaced by another monochromatic source of wavelength 7500° A and separation between the slits is doubled?
Answer:
Given SS₂ – SS₁ = λ/4
Now path difference between the two waves from slits S₁ and S₂ on reaching point P on screen is

(a) For constructive interference at point P both difference Δx = nλ.
Therefore

where n = 0, 1, 2, 3, …

(b) For destructive interference at point P path difference

For central bright fringe, putting n = 0 in equation (1), we get
\(\frac{y d}{D}=-\frac{\lambda}{4}\)
or
y = – \(\frac{\lambda D}{4 d}\)

The -ve sign indicates that the central bright fringe will be observed below centre O of the screen, at distance below it.
Given λ₁ = 6000°A, β₁ = 0.8 mm,
λ₂ = 7500°A, β₂ = ?, d₂ = 2d₁

Using the expression
β = \(\frac{Dλ}{d}\) we have
\(\frac{\beta_{2}}{\beta_{1}}=\frac{\lambda_{2} d_{1}}{\lambda_{1} d_{2}}=\frac{7500}{6000} \times \frac{d_{1}}{2 d_{1}}\)

Solving we have β₂ = 0.5 mm

Question 51.
(a) There are two sets of apparatus of Young’s double-slit experiment. Inset A, the phase difference between the two waves emanating from the slits does not change with time, whereas in set B, the phase difference between the two waves from the slits changes rapidly with time. What difference will be observed in the pattern obtained on the screen in the two setups?
(b) Deduce the expression for the resultant intensity in both the above- mentioned setups (A and B), assuming that the waves emanating from the two slits have the same amplitude A and same wavelength λ.
Or
(a) The two polaroids, in a given setup, are kept ‘crossed’ with respect to each other. A third polaroid, now put in between these two polaroids, can be rotated. Find an expression for the dependence of the intensity of light I, transmitted by the system, on the angle between the pass axis of the first and the third polaroid. Draw a graph showing the dependence of l on θ.
(b) When an unpolarized light is an incident on a plane glass surface, find the expression for the angle of incidence so that the reflected and refracted light rays are perpendicular to each other. What is the state of polarization, of reflected and refracted light, under this condition?
Answer:
(a) Set A: Stable interference pattern, the positions of maxima and minima, does not change with time.
Set B: Positions of maxima and minima will change rapidly with time and an average uniform intensity distribution will be observed on the screen.

(b) Let the displacements of the waves from the sources S1 and S2 at a point on the screen at any time t be given by y₁ = a cos ωt and y₂ = a cos (ωt + Φ), where Φ is the constant phase difference between the two waves. By the superposition principle, the resultant displacement at point P is given by

Thus the amplitude of the resultant displacement is A = 2a cos (Φ/2)

Therefore, the intensity at that point is l = A² = 4a² cos²\(\frac{Φ}{2}\) – 4l_o cos²\(\frac{Φ}{2}\)
Since Φ = 0, l.e. there is no phase difference, hence l = 4 l_o

Inset B, the intensity will be given by the average intensity
l = 4l_o\(\left(\cos ^{2} \frac{\phi}{2}\right)\) = 2l_o
Or
(a) Let P₁ and P₂ be the two crossed Polaroids and P₃ be the polaroid kept between the two. Let l_o, be the intensity of polarised light after passing through the first Polaroid P₁. Then the Intensity of light after passing through the second Polaroid P₂ will be,
l = l_o cos² θ
where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₂ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – 0). Hence the intensity of light emerging from P2₂ will be

The transmitted intensity will be maximum when π = π/4. The graph is as shown

(b) The ray diagram is as shown.

When light Is incident at a certain angle called Brewster angle, the reflected beam is completely polarised with its electric vector parallel to the surface, while the refracted beam is partially polarised. The angle of incidence at which this occurs is called the polarising angle i_B.

From figure we see that at the polaris-ing angle
i_B + 90° + r = 180°
or
r = 90° – i_B.

Using Snell’s law we have
μ = \(\frac{\sin i_{\mathrm{B}}}{\sin r}\)

Now sin r = (90° – i_B), therefore the above expression becomes

Nature of polarisation: Reflected light and linearly polarised

Numerical Problems:
Formulae for solving numerical problems

• Fringe width is given by β = \(\frac{Dλ}{d}\)
• Brewster’s Law: μ = \(\frac{\sin \theta_{p}}{\cos \theta_{p}}\) = tan θ_p
• Intensity of light coming out of a polariser l = l_o cos² θ
• If l₁ and l₂ are the intensities of light emitted by the two sources, w₁ and w₂ be the widths of the two slits, a₁ and a₂ be the amplitudes of the waves from the two slits then,
  \(\frac{l_{1}}{l_{2}}=\frac{w_{1}}{w_{2}}=\frac{a_{1}^{2}}{a_{2}^{2}}\)
• lf l_max and lmin be the intensities of light at the maxima and the minima the
  \(\frac{l_{\max }}{l_{\min }}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}\)

Question 1.
Laser light of wavelength 630 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another source of laser light that produces interference fringes separated by 8.1 mm using the same pair of slits. (CBSE Al 2011C)
Answer:
Given λ₁ = 630 nm, β₁ = 7.2 mm, β₂ = 8.1 mm, λ₂ =?

We know that β = \(\frac{Dλ}{d}\), for the same value of D and d we have β₂ ∝ λ,
Therefore, we have

Question 2.
What is the speed of light in a denser medium of polarising angle 30°? (CBSE Delhi 2019)
Answer:

Question 3.
Laser light of wavelength 640 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another source of light that produces interference fringes separated by 8.1 mm using the same arrangement. Also, find the minimum value of the order (n) of the bright fringe of the shorter wavelength which coincides with that of the longer wavelength. (CBSE AI 2012C)
Answer:
Given λ₁ = 640 nm, λ₂ = ?, b₁ = 7.2 mm and b₂ = 8.1 mm

Using the expression β = \(\frac{Dλ}{a}\) we have

Now n fringes of shorter wavelength will coincide with (n – 1) fringes of Longer wavelength
n₁λ₁ = n₂λ₂
or
n × 640 = (n – 1) × 720
Solving for n we have n =9

Question 4.
Yellow light (λ = 6000 Å) illuminates a single-slit of width 1 × 10 m. Calculate
(a) the distance between the two dark tines on either side of the central maximum, when the diffraction pattern is viewed on a screen kept 1.5 m away from the slit;
(b) the angular spread of the first diffraction minimum. (CBSE AI 2012C)
Answer:

Question 5.
Find the ratio of Intensities of two points P and Q on the screen in Young’s double-slit experiment when the waves from sources S₁ and S₂ have a phase difference of (i) π/3 and (11) π/2.
Answer:
Intensity at the screen when the phase difference is Φ = π/3 is

Now intensity at the screen when the phase difference is Φ = 90° is

Therefore, ratio of intensities is
\(\frac{l_{p}}{l_{Q}}=\frac{3l}{2l}=\frac{3}{2}\)

Question 6.
In young’s double-slit experiment, two sifts are separated by 3 mm distance and illuminated by the light of wavelength 480 nm. The screen is at 2 m from the plane of the slits. Calculate the separation between the 8th bright fringe and the 3N dark fringes observed with respect to the central bright fringe.
Answer:
Using the formuLa

Question 7.
Two coherent sources have Intensities In the ratio 25: 16. Find the ratio of the Intensities of maxima to minima, after the interference of light occurs.
Answer:
In the given problem

Question 8.
The ratio of intensities of maxima and minima in an interference pattern is found to be 25: 9. Calculate the ratio of light intensities of the sources producing this pattern.
Answer:
Given

Question 9.
In Young’s double-slit experiment using the light of wavelength 400 nm, Interference fringes of width X are obtained. Th. the wavelength of light is Increased to 600 nm and the separation between the slits Is halved. If on. wants the observed fringe width on the screen to b. the same in the two cases, find the ratio of the distance between the screen and the plan. of the Interfering sources In the two arrangements.
Answer:
Let D₁ be the distance between the screen and the sources when a tight of wavelength 400 nm Is used.

β = \(\frac{Dλ}{a}\)
or
X = \(\frac{D_{1} \times 400 \times 10^{-9}}{d}\) …(i)

Let D₂ be the distance between the screen and the sources to obtain the same fringe width, when light of wavelength 600 nm is used. Then,
X = \(\frac{D_{2} \times 600 \times 10^{-9}}{d}\) ….(ii)

From the equations (i) and (ii), we have
\(\frac{D_{1}}{D_{2}}=\frac{600 \times 10^{-9}}{400 \times 10^{-9}}\) = 1.5

Question 10.
In Young’s slit experiment, Interference fringes are observed on a screen, kept at D from the slits. If the screen Is moved towards the slits by 5 × 10^-2 m, the change in fringe width Is found to be 3 × 10^-5 m. If the separation between the slits is 10-3 m, calculate the wavelength of the light used.
Answer:
Given ΔD = 5 × 10^-2-2 m, Δβ = 3 × 10^-5-5 m, d = 10^-3 m,

Using the relation β = \(\frac{Dλ}{a}\) we have for the two cases
β₁ = \(\frac{D_{1} \lambda}{d}\) and
β₂ = \(\frac{D_{2} \lambda}{d}\) subtracting we have
β₁ – β₂ = \(\frac{λ}{d}\)(D1 – D2)
or

Question 11.
A slit of width ‘a’ is illuminated by monochromatic light of wavelength 700 nm at Normal Incidence. Calculate the value of ‘a’ for the position of
(a) the first minimum at an angle of diffraction of 30°.
(b) first maximum at an angle of diffraction of 30°.
Answer:
Given X = 700 nm = 7 × 10^-7 m

Question 12.
Estimate the angular separation between . first order maximum and third order minimum of the diffraction pattern due to a single-slit of width 1 mm, when light of wavelength 600 nm is incident normal on it. (CBSE AI 2015C)
Answer:
Given d = 1 mm = 10^-3 m,
λ = 600 nm = 6 × 10^-7 m,
For first order maxima
d sin θ = n λ
or
sin θ = \(\frac{n \lambda}{d}=\frac{6 \times 10^{-7}}{10^{-3}}\) = 6 × 10^-4
or
θ_max = 1.047°
Now for minima we have
d sin θ = (2n+1 )\(\frac{λ}{2}\)

For third-order minima we have n = 3
Therefore we have

sin θ = (2 × 3 + 1)\(\frac{λ}{2d}\) = \(\frac{7λ}{2d}\)
= \(\frac{7 \times 6 \times 10^{-7}}{2 \times 10^{-3}}\)
= 21 × 10^-4
or
θ_min = 6.397°

Therefore, angular separation is
θ_min – θ_max = 6.397 – 1.047 = 5.35°

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 15 Polymers. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 15 Important Extra Questions Polymers

Polymers Important Extra Questions Very Short Answer Type

Question 1.
Define the term ‘homopolymerisation’ giving an example. (CBSE Delhi 2012)
Answer:
The polymers formed from one type of monomers and having same repeating units are called homopolymers. For example polyvinyl chloride (PVC), polythene.

Question 2.
Give one example of condensation polymer. (CBSE 2013)
Answer:
Nylon-6, 6

Question 3.
Which of the following is a natural polymer? (CBSE 2014)
Buna-S, Proteins, PVC
Answer:
Proteins.

Question 4.
Based on molecular forces what type of polymer is neoprene? (CBSE 2014)
Answer:
Elastomer.

Question 5.
Which of the following is a fibre?
Nylon, Neoprene, PVC (CBSE 2014)
Answer:
Nylon

Question 6.
Draw structure for the polymer used for the manufacture of non-stick utensils. (CBSE2019C)
Answer:

Question 7.
What is the primary structural feature necessary for a molecule to make it useful in a condensation polymerisation reaction? (CBSE AI 2010)
Answer:
The monomers must be bifunctional i.e. contain two functional groups.

Question 8.
Arrange the following polymers in the increasing order of tensile strength. Nylon 6, Buna-S, Polythene. (CBSE Sample Paper 2010)
Answer:
Buna – S < Polythene < Nylon 6.

Question 9.
What type of reaction occurs in the formation of Nylon 6,6 polymer? (CBSE Sample Paper 2019)
Answer:
Condensation

Question 10.

Is a homopolymer or R copolymer? (CBSE 2019C)
Answer:
Homopolymer

Question 11.
Identify the Monomers of the following polymer:

Answer:
Buta-1,3-diene, styrene.
It is obtained by the polymerisation of buta-1, 3-diene and styrene in the ratio of 3: 1 in the presence of sodium.

Polymers Important Extra Questions Short Answer Type

Question 1.
Draw the molecular structures of the monomers of
(i) PVC
Answer:
PVC : CH₂ = CH – Cl (Vinyl chloride)

(ii) Teflon (CBSE 2010)
Answer:
Teflon : F₂C = CF₂ (Tetrafluoroethene)

Question 2.
Mention two important uses of each of the following : (CBSE Delhi 2011)
(i) Bakelite
Answer:
Uses of bakelite:

• Used for making combs, fountain pens, barrels, electrical switches and handles of utensils.
• Soft bakelites are used for making glue for binding wooden laminated planks in varnishes.

(ii) Nylon 6
Answer:
Uses of Nylon 6:

• It is used for making tyre cords.
• It is used for making fabrics and ropes.

Question 3.
Draw the structure of the monomer for each of the following polymers:
(i) Nylon 6
Answer:
The monomer of Nylon-6:

(ii) Polypropene (CBSE Delhi 2012)
Answer:
The monomer of polypropene:

Question 4.
Differentiate between thermoplastic and thermosetting polymers. Give one example of each. (CBSE Delhi 2010, 2012) Answer:
Thermoplastics when heated become soft and more or less fluid. These can be moulded into any desired shape. The thermoplastics ‘ can be processed again and again.

The intermolecular forces in thermoplastics are intermediate between those of elastomers and fibres.

On the other hand, thermosetting plastics on heating become hard and insoluble masses. These cannot be moulded into the desired shape and cannot be reprocessed. The intermolecular forces in thermosetting are strong and there are cross-links that hold the molecules in place so that heating does not allow them to move freely.

The common examples are:

1. Thermoplastics: Polythene, polystyrene and polyvinyl chloride.
2. Thermosetting: Bakelite and melamine formaldehyde.

Question 5.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Bakelite
Answer:
Phenol and formaldehyde

(ii) Neoprene
Answer:
2-Chloro-1,3-butadiene(chloroprene)

Question 6.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Terylene
Answer:
Ethylene glycol, terephthalic acid

(ii) Nylon-6, 6
Answer:
Hexamethylenediamine and adipic acid.

Question 7.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Teflon
Answer:
Tetrafluoroethylene

(ii) Buna-N
Answer:
1, 3-butadiene and acrylonitrile

Polymers Important Extra Questions Long Answer Type

Question 1.
Write the structures of monomers used for getting the following polymers:
(i) Nylon-6,6
(ii) Glyptal
(iii) Buna-S
OR
(i) Is it a homopolymer or copolymer? Give reason.
(ii) Write the monomers of the following polymer:

(iii) What is the role of the Sulphur in the vulcanization of rubber? (CBSE Delhi 2019)
Answer:
(i) Nylon-6,6
Hexamethylenediamine: NH₂ – (CH₂)₆ – NH₂ and Adipic acid

(ii) Glyptal

(iii) Buna – S
CH₂ = CH — CH = CH₂
1, 3 Butadiene

OR
(i) It is a homopolymer because it has only one monomer.

(iii) Sulphur makes the rubber hard, tough with greater tensile strength and non-sticky by forming sulphur crosslinks.

Question 2.
(a) Write the names of monomers of the following polymer:

Answer:
Ethylene glycol and terephthalic acid or Ethane-1,2- diol and Benzene 1, 4- dicarboxylic acid.

Terylene: It is a polymer of ethylene glycol (ethane-1,2-diol) and terephthalic acid (benzene-1, 4-dicarboxylic acid). It is obtained by heating a mixture of ethylene glycol and terephthalic acid at 420 to 460 K in the presence of zinc acetate-antimony trioxide, [Zn(OOCCH₃)₂ + Sb₂O₃] catalyst. It is known as terylene or dacron.

(b) What does part 6,6 mean in the polymer Nylon-6,6?
Answer:
It represents 6 carbon atoms present in both the monomer units.

Nylon-6,6: The monomer units of nylon-6,6 are hexamethylenediamine and adipic acid. It is prepared by condensation of hexamethylene diamine with adipic acid under high pressure and high temperature.

(c) Give an example of a Biodegradable polymer. (CBSE 2019C)
Answer:
Poly-p-hydroxybutyrate-co-p-hydroxy valerate (PHBV).

It is a copolymer of 3-hydroxybutyric acid and 3-hydroxypentanoic add, in which the monomer units are joined by ester linkages.

Question 3.
Explain the following terms giving a suitable example for each :
(i) Elastomers
Answer:
Elastomers: These are polymers in which the polymer chains are held together by weak intermolecular forces. Because of the presence of weak forces, the polymer can be easily stretched. However, a few cross-linked are also introduced in the chains which impart the property of regaining the original positions after the stretching force is released. A common example is vulcanised rubber.

(ii) Condensation polymers
Answer:
Condensation polymers: A polymer formed by the condensation of two or more than two monomers is called condensation polymer. In this type, each monomer generally contains two functional groups, and condensation takes place by the loss of molecules such as H₂0, HCl, etc. Common examples are nylon-6, nylon-6,6, bakelite, terylene and alkyl resins.

(iii) Addition polymers (CBSE 2012)
Answer:
Addition polymers: A polymer formed by the direct addition of repeated monomers is called an addition polymer. In this type, the monomers are unsaturated compounds and are generally derivatives of ethene. In addition, polymers have the same empirical formula as their monomers. For example, the addition of polymers polyethene or polypropylene are obtained as:

Question 4.
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
Answer:
Buna-S:
1, 3-Butadiene: CH₂ = CH – CH == CH₂
Styrene: C₆H₅CH = CH₂

(ii) Neoprene
Answer:
Neoprene:
Chloroprene: CH₂ = C – CH = CH₂

(iii) Nyton-6, 6 (CBSE Delhi 2013)
Answer:
Nylon 6, 6:

• Adipic acid: HOOC (CH₂)₄ COOH
• Hexamethylenediamine: NH₂(CH₂)₆NH₂

Question 5.
Write the names and structures of the monomers of the following polymers:
(i) Bakelite
Answer:
Bakelite:

• Phenol:
  
• Formaldehyde: HCHO

(ii) Nylon-6
Answer:
Nylon-6:
Caprolactam

(iii) Polythene (CBSE Delhi 2013)
Answer:
Polythene: Ethene: CH₂ = CH₂.

Question 6.
(i) What is the role of benzoyl peroxide in the polymerisation of ethene?
(ii) Identify the monomers in the following polymer:

(iii) Arrange the following polymers in the increasing order of their intermolecular forces: Nylon-6, 6, Polythene, Buna-S
OR
Write the mechanism of free radical polymerisation of ethene. (CBSE 2016)
Answer:
(i) Benzoyl peroxide acts as a free radical generator. In the presence of benzoyl peroxide, phenyl free radical is formed which initiates the chain initiation step.

(iii) Buna-S < Polythene < Nylon 6, 6
OR
Mechanism
(i) Chain initiation step:

(ii) Chain propagating step:

(iii) Chain termination step:

Question 7.
Write the names and structures of the monomers of the following polymers:
(a) Neoprene
Answer:
Chloroprene CH₂ = C — CH = CH₂

(b) Bakelite
Answer:
C₆H₅OH(Phenol) and HCHO (formaldehyde)

(c) PVC (CBSE 2019C)
Answer:
Polyvinyl chloride CH₂ = CH – Cl

Question 8.
Write the structures of monomers used to obtain the following polymers:
(a) Natural rubber
(b) PVC
(c) Nylon-6,6
OR
Write the mechanism of free radical polymerisation of ethene. (CBSE AI 2019)
Answer:
(a) Natural rubber: 2-methyl-1,3-butadiene

(b) PVC: vinyl chloride
CH₂ = CH – Cl

(c) Nylon-6,6:
HOOC-(CH₂)₄ – COOH (Adipic acid)

NH₂ — (CH₂)₆ — NH₂
(Hexamethylenedlamine)
OR
Chain Initiation Step:

Chain propagation:

Chain termination step

Question 9.
Write the structures of monomers used to obtain the following polymers:
(a) Buna-S
(b) Glyptai
(c) Nylon-6
OR
(a) Arrange the following polymers in increasing order of their intermolecular forces: Polyvinylchloride, Neoprene, Terylene
(b) Write one example of each of
(i) Natural polymer
(ii) Thermosetting polymer
(c) What is the significance of numbers 6,6 in the polymer Nylon-6,6? (CBSE Al 2019)
Answer:
(a) Buna-S
CH₂ = CH – CH = CH₂ (1,3-Butadiene) and C₆H₅CH = CH₂ (Styrene)

(b) Glyptal
HO – CH₂ — CH₂ — OH (Ethylene glycol) and

(c) Nylon-6

OR
(a) Neoprene < Polyvinyl chloride (PVC) < Terylene
(b) (i) Isoprene
(ii) Bakelite
(c) The numbers ‘6, 6’ signifies that both the monomers of nylon-6, 6 have six carbon atoms.

Question 10.
What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester. (CBSE AI 2013)
Answer:
Polymers that are degraded by micro-organisms within a suitable period so that the polymers and their degraded products do not cause any serious effects on the environment are called biodegradable polymers. For example,

Poly β – hydroxybutyrate – co – β – hydroxy valerate (PHBV)

Question 11.
Write the names and structures of the monomers of the following polymers:
(i) Polystyrene
Answer:
Polystyrene: Styrene:

(ii) Dacron
Answer:
Dacron:

• Ethylene glycol: HOCH₂CH₂OH
• Terephthalic acid:
  

(iii) Teflon
Answer:
Teflon:
Tetrafluoroethene: F₂C = CF₂

Question 12.
Write the names and structures of the monomers of the following polymers:
(a) Teflon
Answer:
Tetrafluoroethylene, CF₂ = CF₂

(b) Terylene
Answer:

(c) Buna-N
Answer:

Question 13.
Write the structures of monomers used to obtain the following polymers:
(a) Neoprene
(b) PHBV
(c) Bakelite
OR
(a) Arrange the following polymers in decreasing order of their intermolecular forces:
Bakelite, Polythene, Buna-S, Nylon-6,6
(b) Write the monomers of the following polymer:

(c) What is the structural difference between high-density polythene (HDP) and low-density polythene (LDP)? (CBSE AI 2019)
Answer:
(a) Neoprene:
Chloroprene

(b) PHBV

(c) Bakelite

OR
(a) Buna-S < potythene < Bakelite < Nylon-6,6

(b) HO — CH₂ — CH₂ — OH and

(c) Low-density polythene (LDP) consists of highly branched chain molecules. Due to branching, the polythene molecules do not pack well and therefore, it has low density.

High-density polythene (HDP) consists of linear chains and therefore, the molecules can be closely packed. Hence, it has a high density.

Question 14.
What are high density and low-density polythene?
Answer:
High-density polythene is obtained by heating ethene at about 333-343 K under a pressure of 6-7 atm in the presence of a catalyst such as triethylaluminium and titanium tetrachloride (known as Zeigler-Natta catalyst)

This polymer consists of linear chains and therefore the molecules can be closely packed in space. It, therefore, has a high density (0.97g/cm3) and a higher melting point (403K). It is harder, tougher and has greater tensile strength than low-density polythene.

Low-density polythene (LDP) is prepared by free radical addition and hydrogen atom abstraction. It consists of highly branched chain molecules. Due to branching, the polythene molecules do not pack well and therefore it has low density (0.92 g/cm3) and low melting point (384 K). Low-density polythene is transparent. It has moderate tensile strength and high toughness. It is chemically inert.

Question 15.
Discuss the mechanism of free radical addition polymerisation.
Answer:
A large number of unsaturated compounds such as alkenes or dienes and their derivatives are polymerised by this process. The polymerisation takes place through the generation of an initiator, which is a molecule that decomposes to form free radicals. The commonly used initiator is t-butyl peroxide.

The addition occurs as

For example, most of the commercial addition polymers are obtained from alkenes and their derivatives,

The general model of polymerisation is:

Chain initiation step:

Chain Propagating Step:

Chain terminating step

Question 16.
List some important differences between natural rubber and vulcanised rubber.
Answer:
Differences between natural rubber and vulcanised rubber:

Natural rubber	Vulcanized rubber
1. Natural rubber is soft and sticky.	1. Vulcanized rubber is hard and non-sticky.
2. It has low tensile strength.	2. It has a high tensile strength.
3. It has low elasticity.	3. It has high elasticity.
4. It can be used over a narrow range of temperature (from 10°C to 60°C).	4. It can be used over a wide range of temperature (-40°C to 100°C).
5. It has low wear and tears resistance.	5. It has high wear and tears resistance.
6. It is soluble in solvents like ether, carbon- tetrachloride, petrol etc.	6. It is insoluble in all the common solvents.

Question 17.
(a) What is the difference between two notations: nylon-6 and nylon-6,6? Give their synthesis.
Answer:
Nylon-6 has only one compound having 6-carbon atoms while nylon-6,6 refers to a polymer obtained from 6-carbon atoms of dicarboxylic acid (adipic acid) and 6-carbon atoms of diamine (hexamethylene diamine).

Nylon-6 is synthesised from caprolactam as

Nylon-6,6 is synthesised from hexamethylenediamine and adipic acid.

(b) How is Buna-S prepared?
Answer:
Buna-S is obtained by copolymerisation of styrene and 1, 3-butadiene.

Question 18.
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
Answer:
1, 3-Butadiene: CH₂ = CH — CH = CH₂
Styrene: C₆H₅CH = CH₂

(ii) Buna-N
Answer:
1, 3-Butadiene: CH₂ = CH – CH = CH₂
Acrytonitnie: CH₂ = CH — CN

(iii) Dacron
Answer:

• Ethylene glycol:
  
• Terephthalic acid:
  

(iv) Neoprene.
Answer:
Chloroprene:

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 12 Aldehydes, Ketones and Carboxylic Acids.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 12 Important Extra Questions Aldehydes, Ketones and Carboxylic Acids

Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Very Short Answer Type

Question 1.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions: ethanal, propanal, propanone, butanone. (CBSE Delhi 2012)
Answer:
butanone < propanone < propanal < ethanal

Question 2.
Which of the following compounds would undergo the Cannizzaro reaction? Benzaldehyde, Cyclohexanone, 2- Methylpentanal. (CBSE Sample Paper 2019)
Answer:
Benzaldehyde

Question 3.
Draw the structure of semicarbazone of cyclopentanone. (CBSE 2019C)
OR
Draw the structure of the product formed when propanal is treated with zinc amalgam and concentrated hydrochloric acid.
Answer:

OR

Question 4.
Write the IUPAC name of the following: (CBSE AI 2012)

Answer:
Pent-2-enal.

Question 5.
Write the structure of 3-methylbutanal. (CBSE Delhi 2013)
Answer:

Question 6.
Write the structure of the p-Methylbenzal- dehyde molecule. (CBSE Delhi 2013, 2014)
Answer:

Question 7.
Draw the structure of the compound whose IUPAC name is 4-chloropentan 2-one. (CBSE Delhi 2008, CBSE AI 2011, 2013, 2014)
Answer:

Question 8.
Draw the structural formula of a 1-phenyl propane-2-one molecule. (CBSE 2010)
Answer:

Question 9.
Draw the structure of 3-methylbutanal. (CBSE 2011, CBSE Delhi 2013)
Answer:

Question 10.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions:
ethanal, propanal, propanone, butanone. (CBSE 2012)
Answer:
butanone < propanone < propanal < ethanal

Question 11.
Write the IUPAC name of (CBSE 2012)
Ph — CH = CH — CHO.
Answer:
3-Phenyl prop-2-en-al.

Question 12.
Rearrange the following compounds in the increasing order of their boiling points:
CH₃ – CHO, CH₃ – CH₂ – OH, CH₃ – CH₂ – CH₃ (CBSE 2013)
Answer:
CH₃CH₂CH₃ < CH₃CHO < CH₃CH₂OH

Question 13.
Ethanol is soluble in water. Why? (CBSE 2013)
Answer:
Ethanol is soluble in water because of hydrogen bonding between the polar carbonyl group of ethanal and water molecules:

Question 14.
Write the IUPAC name of the compound (CBSE Delhi 2014)

Answer:
3-Hydroxybutanoic acid

Question 15.
Write the IUPAC name of the compound (CBSE Delhi 2014)

Answer:

Question 16.
Write the IUPAC name of the compound (CBSE Delhi 2014)

Answer:
3-Aminobutanal

Question 17.
Write the IUPAC name of the following compound: (CBSE AI2019)

Answer:
But-3-en-2-one

Question 18.
Write the IUPAC name of the following compound: (CBSE AI 2019)

Answer:
1-Phenylbutan-2-one.

Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Short Answer Type

Question 1.
Write structures of compounds A and B in each of the following reactions: (CBSE Delhi 2019)

Answer:

Answer:

Question 2.
What is Tollen’s reagent? Write one use of this reagent. (CBSE 2010)
Answer:
The ammoniacal solution of silver nitrate is called Tollen’s reagent. It is used as an oxidizing reagent and helps to distinguish between aldehydes and ketones. Aldehydes are oxidized by Tollen’s reagent whereas ketones are not.

Question 3.
Name the reagents used in the following reactions: (CBSE Delhi 2015)

Answer:
Lithium aluminium hydride, LiAlH₄

Answer
Alkaline potassium permanganate (KMnO₄), KOH

Question 4.
Distinguish between
C₆H₅CH = CH – COCH₃ and C₆H₅CH = CH CO CH₂ CH₃ (CBSE AI 2016)
Answer:
Heat both the compounds with NaOH and l₂. C₆H₅CH = CHCOCH₃ gives yellow ppt of iodoform. C₆H₅CH = CHCOCH₂CH₃ does not give yellow ppt. of iodoform.

Question 5.
Arrange the following in the increasing order of their reactivity in nucleophilic addition reactions.
CH₃CHO, C₆H₅CHO, HCHO (CBSE AI 2016)
Answer:
C₆H₅CHO < CH₃CHO < HCHO

Question 6.
Write the product in the following reaction: (CBSE AI 2017)

Answer:

Question 7.
Write the structure of the products formed: (CBSE Sample Paper 2019)
(a) CH₃CH₂COOH
Answer:
CH₃ – CH(Cl) – C00H

(b) C₆H₅COCl
Answer:
C₆H₅CHO

Question 8.
Name the reagents in the following reactions: (CBSE AI 2015)

Answer:
Lithium aluminium hydride, LiAlH₄

Answer:
Alkaline potassium permanganate (KMnO₄), KOH

Answer:
CH₃ Mg Br, H₃O⁺

Answer:
Cl₂, P (Hell-Volhard-Zelinsky reaction)

Question 9.
Predict the organic products of the following reactions:

Answer:

Answer:

Answer:

Answer:

Question 10.
Describe how the following conversions are carried out:
(i) Toluene to benzoic acid
Answer:

(ii) Bromobenzene to benzoic acid (CBSE AI 2013)
Answer:

(iii) Ethylcyanide to ethanoic acid
Answer:

(iv) Butan-1-ol to butanoic acid (CBSE Delhi 2010)
Answer:

Question 11.
Write the structures of A, B, C, and D in the following reaction: (CBSE AI 2016)

Answer:

Question 12.
Aromatic carboxylic acids do not undergo Friedel Crafts reaction. Explain. (CBSE Al 2018)
Answer:
Aromatic carboxylic acids do not undergo Friedel Crafts reaction because -COOH group is deactivating and the catalyst aluminum chloride (Lewis acid) gets bonded to the carboxyl group.

Question 13.
pK_a value of 4-nitrobenzoic acid is lower than that of benzoic acid. Explain. (CBSE AI 2018)
Answer:
Due to the presence of a strong electron-withdrawing (-NO₂) group in 4-nitrobenzoic acid, it stabilizes the carboxylate anion and hence strengthens the acid. Therefore, 4-nitrobenzoic acid is more acidic than benzoic acid and its pK_a value is lower.

Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Long Answer Type

Question 1.
Complete the following reactions:

Or
Write chemical equations for the following reactions:
(i) Propanone is treated with dilute Ba(OH)₂.
(ii) Acetophenone is treated with Zn(Hg)/Conc. HCI
(iii) Benzoyl chloride is hydrogenated in presence of Pd/BaSO₄. (CBSE Delhi 2019)
Answer:



Or

Question 2.
Predict the products of the following B reactions: (CBSE Delhi 2015)

Answer:

Answer:

Answer:

Question 3.
Predict the products of the following reactions: (CBSE 2015)

Answer:

Answer:

Answer:

Question 4.
What happens when
(a) Salicylic acid Is treated with (CH₃CO)₂O/H⁺?
Write a chemical equation in support of your answer. (CBSE AI 2019)
Answer:
Salicylic acid is treated with acetic anhydride In the presence of H⁺ to give aspirin.

(b) Phenol is oxidized with Na₂Cr₂O₇/H⁺?
Answer:
Phenol in the presence of acidified sodium dichromate gives benzoquinone.

(C) Anisole Is treated with CH₃Cl/anhydrous AlCl₃?
Answer:

Question 5.
Draw the structures of the following compounds:
(i) 3-Methylbutanal
Answer:

(ii) 4-Chlocopentan-2-one
Answer:

(iii) 4-Methylpent-3-en-2-one
Answer:

(iv) p-Methyl benzaldehyde (CBSE AI 2014)
Answer:

Question 6.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to propene (CBSE Delhi 2017)
Answer:
Propanone to propene

(ii) Propanal to butanone
Answer:
Propanal to butanone

(iii) Benzaldehyde to benzophenone
Answer:
Benzaldehyde to benzophenone

(iv) Benzaldehyde to 3-phenyl propan – 1-ol
Answer:
Benzaldehyde to 3- phenyl propan -1 -ol

(v) Benzaldehyde to α-hydroxyphenyl acetic acid
Answer:
Benzaldehyde to α-hydroxyphenyl acetic acid

(vi) Ethanol to 3-hydroxybutanal
Answer:
Ethanol to 3-hydroxybutanal

Question 7.
Convert the following:
(i) Ethanal to propanone. (CBSE AI 2018)
Answer:
Ethanal to propanone

(ii) Ethanal to lactic acid.
Answer:
Ethanal to lactic acid

(iii) Ethanal to 2-hydroxy-3-butenoic acid.
Answer:
Ethanal to 2-hydroxy-3-butenoic acid

(iv) Acetaldehyde to formaldehyde.
Answer:
Acetaldehyde to formaldehyde

(v) Formaldehyde to acetaldehyde.
Answer:
Formaldehyde to acetaldehyde

(vi) Acetaldehyde to crotonic acid.
Answer:
Acetaldehyde to crotonic acid

Question 8.
A compound ‘X’ (C₂H₄O) on oxidation gives ‘Y’ (C₂H₄O₂). ‘X’ undergoes haloform reaction. On treatment with HCN ‘X’ forms a product ‘V which on hydrolysis gives 2-hydroxy propanoic acid.
(i) Write down structures of ‘X’ and ‘Y’.
(ii) Name the product when ‘X’ reacts with dil NaOH.
(iii) Write down the equations for the reactions involved. (CBSE Sample Paper 2007)
Answer:
Compound ‘X’ (C₂H₄O) is oxidized to ‘Y’ (C₂H₄O₂). Since it undergoes a haloform reaction, it must be acetaldehyde.
(i) X = CHCHO Y = CH₃COOH
On treatment with HCN, X gives cyanohydrin which on hydrolysis gives 2-hydroxypropanoic acid.

(ii) When ‘X’ reacts with dil. NaOH, undergoes an aldol condensation reaction forming aldol which on heating gives but-2-enal.

(iii) Equations for reactions

Other equations are given above.

Question 9.
Complete the following reactions:

Answer:

Answer:

Answer:

Answer:

Answer:

(CBSE Sample Paper 2017 – 2018)
Answer:

(CBSE Delhi 2013)
Answer:

(CBSE AI 2013)
Answer:

Question 10.
(i) An organic compound (A) has a characteristic odor. On treatment with NaOH, it forms two compounds (B) and (C). Compound (B) has molecular formula C₇H₈0 which on oxidation gives back (A). The compound (C) is a sodium salt of an acid. When (C) is treated with soda lime it yields an aromatic hydrocarbon (D). Deduce the structures of (A), (B), (C) and (D). Write the sequence of reactions involved.
Answer:
The compound A is C₆H₅CHO, benzaldehyde having a characteristic odor. The reactions are:

(ii) Complete each synthesis by filling the missing starting materials, reagents, or products. (X and Z).
(a) C₆H₅CHO + CH₃CH₂CHO NaOH X
Answer:

(b) CH₃(CH₂)₉ COOC₂H₅ CH₃(CH₂),CHO
Answer:

(iii) How will you bring about the following conversions in not more than two steps?
(a) Toluene to benzaldehyde
Answer:

(b) Ethylcyanide to 1-phenylpropanoid. (CBSE Sample Paper 2011)
Answer:

Question 11.
(i) How do you convert the following?
(a) Ethanal to propanone
(b) Toluene to benzoic acid
OR
(ii) (A), (B), and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C₄H₈O. isomers (A) and (C) give positive Tollens’ test whereas isomer (B) does not give Tollens’ test but gives positive iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/conc. HCI gives the same product (D).
(a) Write the structures of (A), (B), (C), and (D).
(b) Out of (A), (B), and (C) isomers, which one is least reactive towards the addition of HCN? (CBSEAI 2018)
Answer:

OR
(ii) (a) Out of A, B, and C, (A) and (C) give positive Tollens’ test, and therefore, these are aldehydes. (B) does not give Tollens’ test and therefore, it is a ketone, with a group because it gives positive iodoform test. Thus, the three isomers are:

Question 12.
Predict the products of the following reactions:

(CBSE Delhi 2015)
Answer:

(CBSE Delhi 2015)
Answer:

(CBSE AI 2015)
Answer:

(CBSE AI 2015)
Answer:

(CBSE AI 2017)
Answer:

(CBSE AI 2017)
Answer:

Question 13.
Write the structures of A, B, C, D, and E in the following reactions: (C8SE Delhi 2016)

Answer:

Question 14.
(a) Give chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanone.
(ii) Pentan-2-one and Pentan-3-one.
(b) Arrange the following compounds in increasing order of their acid strength: Benzoic acid, 4- Nitrobenzoic acid, 3,4 -Dinitrobenzoic acid, 4- Methoxybenzoic acid.
OR
Compare the reactivity of benzaldehyde and ethanal towards nucleophilic addition reactions. Write the cross aldol condensation product between benzaldehyde and ethanal. (CBSE Sample Paper 2019)
Answer:
(a) (i)

Experiments	Ethanal	Propanone
1. Tollen’s Test: Warm the organic compound with freshly prepared ammoniacal silver nitrate solution (Tollen’s reagent).	A bright silver mirror is produced.	No silver mirror is formed.
2. Fehling’s Test: Heat the organic compound with Fehling’s reagent.	A reddish-brown precipitate is obtained.	No precipitate is obtained.
Anyone test

Acetaldehyde (Ethanal) and Acetone (Propanone)
(i) Acetaldehyde gives silver mirror with Tollen’s reagent.

Acetone does not give this test.

(ii) Acetaldehyde gives red ppt with Fehling solution.

(ii) Pentan-2-one and Pentan-3-one

Experiment	Pentan-2-one	Pentan-3-one
Iodoform Test: The organic compound is heated with iodine in presence of sodium hydroxide solution.	A yellow precipitate is obtained.	No yellow precipitate is obtained.

Pentan-3-one and Pentan-2-one
(i) Pentan-2-one forms yellow ppt with an alkaline solution of iodine (iodoform test), but pentane-3-one does not give iodoform test.
CH₃COCH₂CH₂CH₃ + 3I₂+ 4NaOH → CH₃CH₂CH₂COONa + CHI₃ + 3H₂0 + 3NaI

(ii) Pentan-2-one gives white ppt with sodium bisulphite while pentan-3-one does not.

Or any other suitable test.

(b) 4- Methoxybenzoic add < Benzoic acid < 4- Nitrobenzoic acid < 3,4-Dinitrobenzoic acid Effect of substituents on acidic strength of acids. The substituents have a marked effect on the acidic strength of carboxylic acids. The nature of substituents affects the stability of the conjugate base (carboxylate ion) and hence affects the acidity of the carboxylic acids.

In general, electron-withdrawing groups (EWG) increase the stability of the carboxylate ion by delocalizing the negative charge and hence increase the acidity of the carboxylic acid. Conversely, electron-donating groups (EDG) decrease the stability of the carboxylate ion by intensifying the negative charge and hence decrease the acidity of the carboxylic acid.
OR
The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than the carbon atom of the carbonyl group present in ethanol. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance hence less reactive than ethanal.

Aromatic Aldehydes and Ketones
In general, aromatic aldehydes and ketones are less reactive than the corresponding aliphatic analogs. For example, benzaldehyde is less reactive than aliphatic aldehydes. This can be easily understood from the resonating structures of benzaldehyde as shown below:

It is clear from the resonating structures that due to the electron releasing (+I effect) of the benzene ring, the magnitude of the positive charge on the carbonyl group decreases, and consequently it becomes less susceptible to the nucleophilic attack. Thus, aromatic aldehydes and ketones are less reactive than the corresponding aliphatic aldehydes and ketones.

However, amongst aromatic aldehydes and ketones, aromatic aldehydes are more reactive than alkyl aryl ketones which in turn are more reactive than diaryl ketones. Thus, the order of reactivity of aromatic aldehydes and ketones is:

Question 15.
(i) Illustrate the following name reaction giving a suitable example:
(a) Clemmensen reduction
(b) Hell-Volhard-Zelinsky reaction

(ii) How are the following conversions carried out?
(a) Ethylcyanide to ethanoic acid
(b) Butan-1-ol to butanoic acid
(c) Benzoic acid to m-bromobenzoic acid .
OR
(i) Illustrate the following reactions giving a suitable example for each.
(a) Cross aldol condensation
(b) Decarboxylation

(ii) Give simple tests to distinguish between the following pairs of compounds:
(a) Pentan-2-one and Pentan-3-one
(b) Benzaldehyde and Acetophenone
(c) Phenol and Benzoic acid (CBSE Delhi 2012)
Answer:
(i) (a) Clemmensen reduction

(b) Hell-Volhard-Zelinsky reaction

(ii)

OR
(i) (a) Cross Aldol condensation

(b) Decarboxylation

(ii) (a) Pentan-2-one and pentan-3-one can be distinguished by iodoform test

Pentan-3-one does not give this test.

(b) Benzaldehyde and acetophenone: Benzaldehyde forms a silver mirror with ammoniacal silver nitrate solution (Tollen’s reagent).

Acetophenone does not react with Tollen’s reagent.

(c) Phenol and Benzoic acid: Benzoic acid reacts with NaHC03 to give effervescence due to the evolution of CO₂.

Phenol does not give effervescence.

Question 16.
(i) Write a suitable chemical equation to complete each of the following transformations:
(a) Butan-1-ol to butanoic acid
(b) 4-Methylacetophenone to benzene-1, 4-dicarboxylic acid.

(ii) An organic compound with molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Toilen’s reagent, and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid. Identify the compound.
OR
(i) Give chemical tests to distinguish between (a) Propanol and propanone
(b) Benzaldehyde and acetophenone

(ii) Arrange the following compounds in increasing order of their property as indicated:
(a) Acetaldehyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN)
(b) Benzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Answer:

(ii) The given compound forms a 2,4-DNP derivative. Therefore, it is an aldehyde or ketone. Since it reduces Tollen’s reagent, it must be an aldehyde. The compound undergoes Cannizzaro’s reaction, so it does not contain hydrogen. On vigorous oxidation, it gives 1,2-benzene dicarboxylic acid, which means that it must be containing an alkyl group at 2-position with respect to the CHO group on the benzene ring.
The molecular formula suggests it should be
(2- Ethytbenza(dehyde)

Answer:
(i) (a) Propanone gives iodoform test but propanol does not. When propanone is heated with aqueous sodium carbonate and iodine solution, a yellow precipitate of iodoform is obtained.

Propanol does not give this test.

(b) Acetophenone forms yellow ppt. of iodoform with an alkaline solution of iodine (iodoform test). Benzaldehyde does not react.

(ii) (a) Methyl tert-butyl ketone < Acetone < Acetaldehyde.
(b) 4-Methoxybenzoic acid < Benzoic add < 3, 4-Dinitrobenzoic acid.
(c) (CH₃)₂CHCOOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH

Question 17.
(i) Although phenoxide ion has more resonating structures than carboxylate ion, the carboxylic acid is a stronger acid than phenol. Give two reasons.
(ii) How will you bring about the following conversions?
(a) Propanone to propane
(b) Benzoyl chloride to benzaldehyde
(c) Ethanal to but-2-enal
OR
(i) Complete the following reactions:

(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Ethanal and Propanal
(b) Benzoic acid and Phenol (CBSE AI 2013)
Answer:
(i) The carboxylic acids are stronger acids than phenols. The difference in the relative acidic strengths can be understood if we compare the resonance hybrid structures of carboxylate ion and phenoxide ion.

The resonance hybrids may be represented as:

(a) The electron charge in the carboxylate ion is more dispersed in comparison to the phenoxide ion since there are two electronegative oxygen atoms in the carboxylate ion as compared to only one oxygen atom in the phenoxide ion.

(b) Carboxylate ion is stabilized by two equivalent resonance structures in which negative charge is on more electronegative 0 atoms. But phenoxide ion has non-equivalent resonance structures in which the negative charge is also on a less electronegative carbon atom.

Therefore, the carboxylate ion is relatively more stable as compared to the phenoxide ion. Thus, the release of H⁺ ions from carboxylic acid is comparatively easier or it behaves as a stronger acid than phenol.

OR

(ii) (a) Ethanal gives a yellow ppt. of iodoform on heating with iodine and NaOH solution whereas propanal does not give.

(b) Benzoic acid reacts with NaHC03 to give effervescence due to the liberation of CO₂.
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂0 + CO₂
Phenol does not give effervescence.
Phenol gives violet color with FeCl₃ solution but benzoic acid does not give such color.

Question 18.
(i) How will you convert the following:
(a) Propanone to Propan-2-ol
(b) Ethanal to 2-hydroxypropanoic acid
(c) Toluene to benzoic acid

(ii) Give a simple chemical test to distinguish between:
(a) Pentan-2-one and Pentan-3-one
(b) Ethanal and Propanal
OR
(i) Write the products of the following reactions:

(ii) Which acid of each pair shown here would you expect to be stronger? (CBSE AI 2013)
(a) F – CH₂ – COOH or Cl – CH₂ – COOH OH

Answer:

(ii) (a) Pentan-2-one forms yellow ppt. with an alkaline solution of iodine (iodoform test), but pentane-3-one does not give iodoform test.

(b) Ethanal gives a yellow ppt. of iodoform on heating with iodine and NaOH solution whereas propanal does not give.

OR

(ii) (a) FCH₂COOH
(b) CH₃COOH

Question 19.
(a) Predict the main product of the following reactions:

(b) Give a simple chemical test to distinguish between

(c) Why is alpha (a) hydrogen of carbonyl compounds acidic in nature?
OR
(a) Write the main product formed when propanal reacts with the following reagents:
(i) 2 moles of CH₃OH in presence of dry HCI
(ii) Dilute NaOH
(iii) H₂N – NH₂ followed by heating with KOH in ethylene glycol

(b) Arrange the following compounds in increasing order of their property as indicated:
(i) F – CH₂COOH, O₂N – CH₂COOH, CH₃COOH, HCOOH – acid character
(ii) Acetone, Acetaldehyde, Benzaldehyde, Acetophenone — reactivity towards the addition of HCN (CBSE AI 2019)
Answer:

(b) On adding NaOH/l2 and heat, acetophenone gives yellow ppt of iodoform but benzophenone does not.

(c) α – hydrogen of carbonyl compounds is acidic because of the electron-withdrawing nature of the carbonyl group.
OR

(b) (i) CH₃COOH < HCOOH < FCH₂COOH < NO₂-CH₃ COOH
(ii) Acetophenone < Benzaldehyde < Acetone < Acetaldehyde

Question 20.
(a) An organic compound with the molecular formula C₇H₆0 forms a 2,4-DNP derivative, reduces Tollen’s reagent, and undergoes the Cannizzaro reaction. On oxidation, it gives benzoic acid. Identify the compound and state the reactions involved.
(b) Give chemical tests to distinguish between the following pair of compounds:
(i) Phenol and propanol
(ii) Benzoic acid and benzene
OR
(a) Predict the products of the following:

(b) Arrange the following in increasing order of acidic character: (CBSE 2019C)
HCOOH, CF₃COOH, ClCH₂COOH, CCl₃COOH
Answer:
(a) Compound = Benzaldehyde or C₆H₅CHO
Reaction
Reaction with 2,4-DNP

With Tollens reagent
RCHO + 2[Ag(NH₃)₂ ]⁺ + 30H^– → RCOO^– + 2 Ag + 2H₂O + 4NH₃ (where R = – C₆H₅)
Cannizzaro

(b) (i) FeCl₃ test: Add 2 drops of neutral FeCl₃ to both the compounds separately. Phenol gives violet color.

(ii) Sodium bicarbonate test: Add NaHCO₃ to both the compounds: Benzoic acid reacts with NaHCO₃ to give effervescence due to liberation of CO₂.
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂
OR
(a) A = CH₃COOH
B = CH₃COCl
C = CH₃CONH₂
D = CH₃NH₂

(b) HCOOH < ClCH₂COOH < CCl₃COOH < CF₃COOH

• the electron-withdrawing substituents disperse the negative charge on carboxylate ion and stabilize it and thus, increase acidity.
• the electron-releasing substituents intensify the negative charge of the carboxylate ion, destabilize it and thus, decrease the acidity.

Question 21.
(i) Give chemical tests to distinguish between:
(a) Propanal and propanone
(b) Benzaldehyde and acetophenone

(ii) How would you obtain:
(a) But-2-enal from the ethanal
(b) Butanoic acid from butanol
(c) Benzoic acid from ethylbenzene? (CBSE 2011)
Answer:
(i) (a) Propanal gives a silver mirror with Tollen’s reagent.

Propanone does not give this test.

(b) Benzaldehyde forms a silver mirror with ammoniacal silver nitrate solution (Tollen’s reagent). Acetophenone does not react.

Question 22.
(i) Describe the following giving linked chemical equations:
(a) Cannizzaro reaction
Answer:
Aldehydes that do not contain any a-hydrogen atom (e.g. benzaldehyde, formaldehyde) undergo self- oxidation and reduction reaction on treatment with cone, solution of caustic alkali. In this reaction, one molecule is oxidized to acid while another molecule is reduced to an alcohol.

(b) Decarboxylation
Answer:
The process of removal of a molecule of C02 from a carboxylic acid is called decarboxylation. It is usually carried out by heating a mixture of carboxylic acid or its sodium salt with soda lime (NaOH + CaO).

(ii) Complete the following chemical equations: (CBSE Delhi 2011)

Answer:

Answer:

Answer:

Question 23.
(i) Give chemical tests to distinguish between the following:
Benzoic acid and ethyl benzoate.
Answer:
(a) When treated with NaHCO₃ solution, benzoic acid gives brisk effervescence while ethyl benzoate does not

(b) Ethyl benzoate on boiling with an excess of NaOH gives ethyl alcohol which on heating with iodine gives yellow ppt. of iodoform.

(ii) Complete each synthesis by giving missing reagents or products in the following:

Answer:

Answer:

(CBSE 2011)
Answer:

Question 24.
(i) Write the products formed when CH₃CHO reacts with the following reagents:
(a) HCN
(b) H2N – OH
(C) CH₃CHO in the presence of dilute NaOH
(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Benzoic acid and Phenol
(b) Propanal and Propanone
OR
(i) Account for the following:
(a) Cl – CH₂COOH is a stronger acid than CH₃COOH.
(b) Carboxylic acids do not give reactions of the carbonyl group.
(ii) Write the chemical equations to illustrate the following name reactions:
Rosenmund reduction
(iii) Out of CH₃CH₂—CO—CH₂ – CH₃ and CH₃CH₂ — CH₂ — CO — CH₃, which gives iodoform test? (CBSE 2014)
Answer:

(ii) (a) Add neutral FeCl₃ in both the solutions, phenol gives violet color with FeCl₃ solution, but benzoic acid does not give such color.
(b) Add an ammoniacal solution of silver nitrate (Tollen’s reagent) in both the solutions, propanal gives silver mirror whereas propanone does not.

OR
(0) Chlorine is an electron-withdrawing atom (-I effect). It withdraws the electrons from carbon to which it is attached and therefore, this effect is transmitted throughout the chain. As a result, electrons are withdrawn more strongly towards oxygen of 0—H bond and promotes the release of the proton.

Consequently, ClCH₂COOH is a stronger acid than CH₃COOH.

(b) Carboxylic acids do not give the characteristic reactions of the carbonyl group (>C = 0) as given by aldehydes and ketones. In carboxylic acids, the carbonyl group is involved in resonance, as follows:

Therefore, it is not a free group. But no resonance is possible in aldehydes and ketones. They give the characteristic reactions of the group.

(ii) Rosenmund reduction. Acid chlorides are converted to corresponding aldehydes by catalytic reduction. The reaction is carried out by passing through a hot solution of the acid chloride in the presence of palladium deposited over barium sulfate (partially poisoned with sulfur or quinoline).

The poisoning of palladium catalyst decreases its activity and it does not allow the further reduction of an aldehyde into alcohol.

(iii) CH₃CH₂CH₂COCH₃ gives iodoform test

Question 25.
(i) Write the structures of A, B, C, and D in the following reactions:

(ii) Distinguish between:
(a) C₆H₅ – CH = CH – COCH₃ and C₆H₅ – CH = CH – CO CH₂CH₃
(b) CH₃CH₂COOH and HCOOH

(iii) Arrange the following in the increasing order of their boiling points:
CH₃CH₂OH, CH₃COCH₃, CH₃COOH
OR
(i) Write the chemical reaction involved in the Etard reaction.
(ii) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction:
CH₃ – CHO, C₆H₅COCH₃, HCHO
(iii) Why pKa of Cl-CH₂-COOH is lower than the pK_a of CH₃COOH?
(iv) Write the product in the following reaction.

(v) A and B are two functional isomers of compound C₃H₆O. On heating with NaOH and l₂, isomer A forms a yellow precipitate of iodoform whereas isomer B does not form any precipitate. Write the formulae of A and B. (CBSE 2016)
Answer:

(ii) (a) Heat both the compounds with NaOH and l₂, C₆H₅ CH = CHCOCH₃ gives yellow ppt. of iodoform. C₆H₅CH = CHCOCH₂CH₃ does not give yellow ppt. of iodoform.
(b) Add ammoniacal AgNO₃ solution (Tollen’s reagent), HCOOH gives silver mirror while CH₃CH₂COOH does not.

(iii) CH₃COCH₃ < CH₃CH₂OH < CH₃COOH
OR
(i) Etard’s reaction

(ii) C₆H₅COCH₃ < CH₃CHO < HCHO

(iii) Chlorine is an electron-withdrawing group (-I inductive effect).
It withdraws the electrons from the C to which it is attached and this effect is transmitted throughout the chain. As a result, the electrons are withdrawn more strongly towards 0 of the O-H bond and promote the release of the proton. Consequently, Cl — CH₂COOH is more acidic than acetic acid.

Therefore, pKa of ClCH₂COOH is less than that of CH₃COOH.

(iv) CH₃ CH₂ CH = CH CH₂ CN CH₃ CH₂ CH = CH CH₂ CHO

(v) Two isomers are CH₃COCH₃ and CH₃CH₂CHO
Since A forms yellow ppt. of iodoform on heating with NaOH and l₂, it is CH₃COCH₃.
CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃COONa + 3Nal + 3H₂O
B does not give iodoform, it is CH₃CH₂CHO.
A: CH₃COCH₃ B: CH₃CH₂CHO

Question 26.
Write the structures of A, B, C, D, and E in the following reactions:

OR
(i) Write the chemical equation for the reaction involved in the Cannizzaro reaction.
(ii) Draw the structure of the semicarbazone of ethanal.
(iii) Why pKa of F-CH₂-COOH is lower than that of Cl – CH₂ – COOH?
(iv) Write the product in the following reaction:
CH₃ – CH = CH – CH₂CN
(v) How can you distinguish between propanal and propanone? (CBSE Delhi 2016)
Answer:

OR
(i) Cannizzaro’s reaction: Aldehydes that do not contain a-hydrogen undergo self-oxidation and reduction on treatment with cone. KOH.


(iii) Fluorine is more electronegative than Cl and therefore, will have a greater electron-withdrawing effect (-I effect). As a result, the carboxylate ion will be stabilized to a larger extent and therefore fluoro acetic acid will be a stronger acid than chloroacetic acid. Hence, the pK_a of FCH₂COOH will be lower than that of ClCH₂COOH.

(iv) CH₃ — CH = CH — CH₂ CN CH₃ CH = CH CH₂ CHO

(v) Propanal gives red ppt with Fehling solution but propanone does not.

Question 27.
(i) Give reasons :
(a) HCHO is more reactive than CH₃-CHO towards the addition of HCN. (CBSE 2018C)
(b) pK_a of O₂N—CH₂—COOH is lower than that of CH₃—COOH.
(c) Alpha hydrogen of aldehydes and ketones is acidic in nature.

(ii) Give simple chemical tests to distinguish between the following pairs of compounds :
(a) Ethanal and Propanal
(b) Pentan-2-one and Pentan-3-one
OR
(i) Write the structure of the product(s) formed :
(a) CH₃ — CH₂ — COOH
(b) C₆H₅COCl
(c) 2HCHO

(ii) How will you bring the following conversions in not more than two steps :
(a) Propanone to propene
(b) Benzyl chloride to phenyl ethanoic acid
Answer:
(i) (a) Due to + I effect of the methyl group is CH₃CHO.
(b) Due to – I effect of the nitro group in nitroacetic acid.
(c) Due to the strong electron-withdrawing effect of the carbonyl group and resonance stabilization of the conjugate base.

(ii) (a) Add NaOH and l₂ to both the compounds and heat, ethanal gives yellow ppt of iodoform.
(b) Add NaOH and l₂ to both the compounds and heat, pentan-2-one gives yellow ppt of iodoform.
OR