Prasanna

We have given detailed NCERT Solutions for Class 10 Sanskrit Shemushi Chapter 1 शुचिपर्यावरणम् Questions and Answers come in handy for quickly completing your homework.

Shemushi Sanskrit Class 10 Solutions Chapter 1 शुचिपर्यावरणम्

Class 10 Sanskrit Shemushi Chapter 1 शुचिपर्यावरणम् Textbook Questions and Answers

अभ्यासः

प्रश्न 1.
एकपदेन उत्तरं लिखत-

(क) अत्र जीवितं कीदृशं जातम्?
उत्तराणि:
दुर्वहमत्र

(ख) अनिशं महानगरमध्ये किं प्रचलति?
उत्तराणि:
कालायासचक्रम्

(ग) कुत्सितवस्तुमिश्रितं किमस्ति?
उत्तराणि:
भक्ष्यम्

(घ) अहं कस्मै जीवनं कामये?
उत्तराणि:
मानवाय

(ङ) केषां माला रमणीया?
उत्तराणि:
ललितलतानां

प्रश्न 2.
अधोलिखितानां प्रश्नानाम् उत्तराणि संस्कृतभाषया लिखत-

(क) कविः किमर्थं प्रकृतेः शरणम् इच्छति?
उत्तराणि:
कविः सुजीवनार्थं प्रकृतेः शरणम् इच्छति।

(ख) कस्मात् कारणात् महानगरेषु संसरणं कठिनं वर्तते?
उत्तराणि:
यानानां हि अनन्ताः पङ्कतयः महानगरेषु सन्ति अतः तत्र संसरणं कठिनं वर्तते।

(ग) अस्माकं पर्यावरणे किं किं दूषितम् अस्ति?
उत्तराणि:
अस्माकं पर्यावरणे वायुमण्डलं जलम्, भक्ष्यम्, धरातलं च सर्व दुषितम् अस्ति।

(घ) कविः कुत्र सञ्चरणं कर्तुम् इच्छति?
उत्तराणि:
कविः एकान्ते कान्तारे सञ्चरणं कर्तुम् इच्छति।

(ङ) स्वस्थजीवनाय कीदृशे वातावरणे भ्रमणीयम्?
उत्तराणि:
स्वस्थजीवनाय खगकुलकलरव गुञ्जिते-कुसुमावलि समीरचालिते वातावरणे भ्रमणीयम्।

(च) अन्तिमे पद्यांशे कवेः का कामना अस्ति?
उत्तराणि:
अन्तिमे पद्यांशे कवेः मानवेभ्यः शान्तिप्रिय-जीवनस्य कामना अस्ति।

प्रश्न 3.
सन्धिं / सन्धिविच्छेदं कुरुत-

(क) प्रकृतिः + _________ = प्रकृतिरेव
उत्तराणि:
एव

(ख) स्यात् + _________ + _________ = स्यान्नैव
उत्तराणि:
न, एव

(ग) _________ + अनन्ता = ह्यनन्ताः
उत्तराणि:
हि

(घ) बहिः + अन्तः + जगति = _________
उत्तराणि:
बहिरन्तर्जगति

(ङ) _________ + नगरात् = अस्मान्नगरात्
उत्तराणि:
अस्मात्

(च) सम् + चरणम् = _________
उत्तराणि:
पञ्चरणम्

(छ) धूमम् + मुञ्चति = _________
उत्तराणि:
धूमंमुञ्चति

प्रश्न 4.
अधोलिखितानाम् अव्ययानां सहायतया रिक्तस्थानानि पूरयत-

भृशम्, यत्र, तत्र, अत्र, अपि, एव, सदा, बहिः

(क) इदानीं वायुमण्डलं ___________ प्रदूषितमस्ति।
उत्तराणि:
भृशम्

(ख) ___________ जीवन दुर्वहम् अस्ति।
उत्तराणि:
अत्र

(ग) प्राकृतिक वातावरणे क्षणं सञ्चरणम् ___________ लाभदायकं भवति।
उत्तराणि:
अपि

(घ) पर्यावरणस्य संरक्षणम् ___________ प्रकृतेः आराधना।
उत्तराणि:
एव

(ङ) ___________ समयस्य सदुपयोगः करणीयः।
उत्तराणि:
सदा

(च) भूकम्पित-समये ___________ गमनमेव उचितं भवति।
उत्तराणि:
बहिः

(छ) ___________ हरीतिमा ___________ शुचि पर्यावरणम्।
उत्तराणि:
यत्र, तत्र

प्रश्न 5(अ).
अधोलिखितानां पदानां पर्यायपदं लिखत-

(क) सलिलम् – _________
(ख) आम्रम् – _________
(ग) वनम् – _________
(घ) शरीरम् – _________
(ङ) कुटिलम् – _________
(च) पाषाणः – _________
उत्तराणि:
(क) जलम्
(ख) रसालम्
(ग) कान्तारम्
(घ) तनुः
(ङ) वक्रम्
(च) प्रस्तर:

प्रश्न 5(आ).
अधोलिखितपदानां विलोमपदानि पाठात् चित्वा लिखत-

(क) सुकरम् – ___________
(ख) दूषितम् – ___________
(ग) गृह्णन्ती – ___________
(घ) निर्मलम् – ___________
(ङ) दानवाय – ___________
(च) सान्ताः – ___________
उत्तराणि:
(क) दुष्करम्
(ख) निर्मलं
(ग) मुञ्चति
(घ) दुषितं
(ङ) मानवाय
(च) ध्वानम्

प्रश्न 6.
उदाहरणमनुसृत्य पाठात् चित्वा च समस्तपदानि समासनाम च लिखत-

उत्तराणि:
(ख) हरिततरूणाम् – कर्मधारय समास
(ग) ललितलतानाम् – कर्मधारय समास
(घ) नवमालिका – कर्मधारय समास
(ङ) धृतसुखसन्देशम् – बहुब्रीहि समास
(च) कज्जलमलिनम् – कर्मधारय समास
(छ) दुर्दान्तैर्दशनै – कर्मधारय समास

प्रश्न 7.
रेखाङ्कित-पदमाधृत्य प्रश्ननिर्माणं कुरुत-

(क) शकटीयानम् कज्जलमलिनं धूम मुञ्चति।
उत्तराणि:
कीदृशम्

(ख) उद्याने पक्षिणां कलरवं चेतः प्रसादयति।
उत्तराणि:
केषाम्

(ग) पाषाणीसभ्यतायां लतातरुगुल्माः प्रस्तरतले पिष्टाः सन्ति।
उत्तराणि:
के

(घ) महानगरेषु वाहनानाम् अनन्ता पङ्क्तयः धावन्ति।
उत्तराणि:
केषु कुत्र

(ङ) प्रकृत्याः सन्निधौ वास्तविक सुखं विद्यते।
उत्तराणि:
कस्याः

योग्यताविस्तारः
यह पाठ आधुनिक संस्कृत कवि हरिदत्त शर्मा के रचना संग्रह ‘लसल्लतिका’ से संकलित है। इसमें कवि ने महानगरों की यांत्रिक-बहुलता से बढ़ते प्रदूषण पर चिन्ता व्यक्त करते हुए कहा है कि यह लौहचक्र तन-मन का शोषक है, जिससे वायुमण्डल और भूमण्डल दोनों मलिन हो रहे हैं। कवि महानगरीय जीवन से दूर, नदी-निर्झर, वृक्षसमूह, लताकुञ्ज एवं पक्षियों से गुञ्जित वनप्रदेशों की ओर चलने की अभिलाषा व्यक्त करता है।

समास-समसनं समासः
समास का शाब्दिक अर्थ होता है-संक्षेप। दो या दो से अधिक शब्दों के मिलने से जो तीसरा नया और संक्षिप्त रूप बनता है वह समास कहलाता है। समास के मुख्यतः चार भेद हैं-

1. अव्ययीभाव
2. तत्पुरुष
3. बहुव्रीहि
4. द्वन्द्व

1. अव्ययीभाव- इस समास में पहला पद अव्यय होता है और वही प्रधान होता है और समस्तपद अव्यय बन जाता है।
यथा-निर्मक्षिकम् मक्षिकाणाम् अभाव:

यहाँ प्रथमपद ‘निर्’ है और द्वितीयपद मक्षिकम् है। यहाँ मक्षिका की प्रधानता न होकर मक्षिका का अभाव प्रधान है, अतः यहाँ अव्ययीभाव समास है। कुछ अन्य उदाहरण देखें-

• उपग्रामम् – ग्रामस्य सपीपे – (समीपता की प्रधानता)
• निर्जनम् – जनानाम् अभाव: – (अभाव की प्रधानता)
• अनुरथम् – रथस्य पश्चात् – (पश्चात् की प्रधानता)
• प्रतिगृहम् – गृहं गृहं प्रतिः – (प्रत्येक की प्रधानता)
• यथाशक्ति – शक्तिम् अनतिक्रम्य – (सीमा की प्रधानता)
• सचक्रम् – सक्रेण सहितम्: – (सहित की प्रधानता)

2. तत्पुरुष- ‘प्रायेण उत्तरपदार्थप्रधानः तत्पुरुषः’ इस समास में प्राय: उत्तरपद की प्रधानता होती है और पूर्वपद उत्तरपद के विशेषण का कार्य करता है। समस्तपद में पूर्वपद की विभक्ति का लोप हो जाता है।
यथा- राजपुरुषः अर्थात् राजा का पुरुष। यहाँ राजा की प्रधानता न होकर पुरुष की प्रधानता है, और राजा शब्द पुरुष के विशेषण का कार्य करता है।

• ग्रामगतः – ग्राम गतः।
• शरणागात – शरणम् आगतः।
• देशभक्तः – देशस्य भक्तः।
• सिंहभीतः – सिंहात् भीतः।
• भयापन्नः – भयम् आपन्नः।
• हरित्रातः – हरिणा त्रातः।

तत्पुरुष समास के दो भेद हैं-कर्मधारय और द्विगु।
1. कर्मधारय- इस समास में एक पद विशेष्य तथा दूसरा पद पहले पद का विशेषण होता है। विशेषण-विशेष्य भाव के अतिरिक्त उपमान-उपमेय भाव भी कर्मधारय समास का लक्षण है।
यथा-

• पीताम्बरम् – पीतं च तत् अम्बरम्।
• महापुरुषः – महान् च असौ पुरुषः।
• कज्जलमलिनम् – कज्जलम् इव मलिनम्।
• नीलकमलम् – नीलं च तत् कमलम्।
• मीननयनम् – मीन इव नयनम्।
• मुखकमलम् – कमलम् इव मुखम्।

2. द्विगु- ‘संख्यापूर्वी द्विगुः’
इस समास में पहला पद संख्यावाची होता है और समाहार (एकत्रीकरण या समूह) अर्थ की प्रधानता होती है।
यथा- त्रिभुजम्-त्रयाणां भुजानां समाहारः।
इसमें पूर्ववद ‘त्रि’ संख्यावाची है।

• पंचपात्रम् – पंचाना पात्राणां समाहारः।
• पंचवटी – पंचानां वटानां समाहारः।
• सप्तर्षिः – सप्तानां ऋषीणां समाहारः।
• चतुर्युगम् – चतुर्णा युगानां समाहारः।

3. बहुब्रीहि- ‘अन्यपदप्रधानः बहुबीहिः’
इस समास में पूर्व तथा उत्तर पदों की प्रधानता न होकर किसी अन्य पद की प्रधानता होती है।
यथा-

• पीताम्बरः – पीतम् अम्बरम् यस्य सः (विष्णुः)। यहाँ न तो पीतम् शब्द की प्रधानता है और न अम्बरम् शब्द की अपितु पीताम्बरधारी किसी अन्य व्यक्ति (विष्णु) की प्रधानता है।
• नीलकण्ठः – नीलः कण्ठः यस्य सः (शिवः)।
• दशाननः – दश आननानि यस्य सः (रावण:)।
• अनेककोटिसारः – अनेककोटिः सारः (धनेम्) यस्य सः।
• विगलितसमृद्धिम् – विगलिता समृद्धिः यस्य तम् (पुरुषम्)।
• प्रक्षालितपादम् – प्रक्षालितौ पादौ यस्य तम् (जनम्)।

4. द्वन्द्व- ‘उभयपदार्थप्रधान: द्वन्द्वः’
इस समास में पूर्वपद और उत्तरपद दोनों की समान रूप से प्रधानता होती है। पदों के बीच में ‘च’ का प्रयोग विग्रह में होता है।
यथा-

• रामलक्ष्मणौ – रामश्च लक्ष्मणश्च।
• पतरौ – माता च पिता च।
• धर्मार्थकाममोक्षाः – धर्मश्च, अर्थश्च, कामश्च, मोक्षश्च।
• वसन्तग्रीष्मशिशिराः – वसन्तश्च ग्रीष्मश्च शिशिरश्च।

कविपरिचय- प्रो० हरिदत्त शर्मा इलाहाबाद केन्द्रीय विश्वविद्यालय में संस्कृत के आचार्य रहे हैं। इनके कई संस्कृत काव्य प्रकाशित हो चुके हैं। जैसे-गीतकंदलिका, त्रिपथगा, उत्कलिका, बालगीताली, आक्रन्दनम्, लसल्लतिका इत्यादि। इनकी रचनाओं में समाज की विसंगतियों के प्रति आक्रोश तथा स्वस्थ वातावरण के प्रति दिशानिर्देश के भाव प्राप्त होते हैं।

भावविस्तारः
पृथिवी, जल, तेजो वायुराकाशश्चेति पञ्चमहाभूतानि प्रकृतेः प्रमुखतत्त्वानि। एतैः तत्त्वैरेव पर्यावरणस्य रचना भवति। आवियते रेतः समन्तात् लोकोऽनेनेति पर्यावरणम्। परिष्कृतं प्रदूषणरहितं च पर्यावरणमस्मभ्यं सर्वविधजीवनसुखं ददाति। अस्माभिः सदैव या प्रयतितव्यं यथा जलं स्थलं गगनञ्च निर्मलं स्यात्। पर्यावरणसम्बद्धाः केचन श्लोकाः अधोलिखिताः सन्ति-
यथा-

पृथिवीं परितो व्याप्य, तामाच्छाद्य स्थितं च चत्।
जगदाधाररूपेण, पर्यावरणमुच्यते।।

दूषणविषये-

सृष्टौ स्थितौ विनाशे च नृविज्ञैर्बहुनाशकम्।
पञ्चतत्वविरुद्ध यत्साधितं तत्प्रदूषणम्।।

युप्रदूषणविषये-

प्रक्षिप्तो वाहनधूमः कृष्णः बह्वपकारकः।
दुष्टैरसायनयुक्तो घातक: श्वासरुग्वहः।।

नप्रदूषणविषये-

यन्त्रशाला परित्यक्तैर्नगरेदूषितद्रवैः।
नदीनदी समुद्राश्च प्रक्षिप्तैर्दूषणं गताः।।

दूषण निवारणाय संरक्षणाय च-

शोधनं रोपणं रक्षावर्धनं वायुवारिणः।
वनानां वन्यवस्तूनां भूमेः संरक्षणं वरम्।।
एते श्लोकाः पर्यावरणकाव्यात् संकलिताः सन्ति।

सम-तद्भव- शब्दानामध्ययनम्-
धोलिखितानां तत्समशब्दानां तदुद्भूतानां च तद्भवशब्दानां परिचयः करणीय:
तत्सम – तद्भव
प्रस्तर – पत्थर
वाष्प – भाप
दुर्वह – दूभर
वक्र – बाँका
कज्जल – काजल
चाकचिक्य – चकाचक, चकाचौंध
धूमः – धुआँ
शतम् – सौ (100)
बहिः – बाहर

दः परिचयः
स्मन् गीते शुचि पर्यावरणम् इति ध्रुवकं (स्थायी) वर्तते। तदतिरिक्त सर्वत्र प्रतिपङ्क्ति 26 मात्राः सन्ति। इदं गीतिकाच्छन्दसः मस्ति।

Class 10 Sanskrit Shemushi Chapter 1 शुचिपर्यावरणम् Additional Important Questions and Answers

अतिरिक्त प्रश्नाः

1. अधोलिखितं पद्यांशं पाठित्वा प्रदत्तान प्रश्नान् उत्तरत-

(क) दुर्वहमत्र जीवितं जातं प्रकृतिरेव शरणम्।
शुचि-पर्यावरणम्॥
महानगरमध्ये चलदनिशं कालायसचक्रम्।
मनः शोषयत् तनुः पेषयद् भ्रमति सदा वक्रम्॥
दुन्तैिर्दशनैरमुना स्यान्नैव जननसनम्। शुचि …॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. चक्रम कीदृशम् चलति?
2. अस्माकम् शरणम् किम्?
3. अत्र जीवितं कथं जातम्?

उत्तराणि:

1. वक्रम्
2. प्रकृतिः
3. दुर्वहम्

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. चक्रम् वक्रम् कदा चलति?
2. अमुना दुर्दान्तैः दशनैः किं न स्यात्?

उत्तराणि:

1. मनः शोषयत् तनुः पेषयद् कालायसचक्रम् सदा वक्रम् चलति।
2. अमुना दुर्दान्तैः दशनैः जनग्रसनं नस्यात

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. “अहर्निशम्” इति पदस्य पर्यायपदं पद्यांशे किम्?
2. “सरलैः” इति पदस्य विपर्यय पदं पद्यांशे किम
3. “पर्यावरणम्” इति पदस्य विशेषणपदं किम्?
4. ‘सदा वक्रम् भ्रमति’ अत्र क्रियापदं किम्?

उत्तराणि:

1. अनिशम्
2. दुर्दान्तः
3. शुचि
4. भ्रमति

(ख) कज्जलमलिनं धूमं मुञ्चति शतशकटीयानम्।
वाष्ययानमाला संधावति वितरन्ती ध्वानम्॥
यानानां पङ्क्तयो ह्यनन्ताः कठिनं संसरणम्। शुचि …।

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. कि कज्जलमलिनं धूमं मुञ्चति?
2. वाष्पयानमाला कि कुर्वन्ती संधावति?
3. वाष्पयानमाला कि वितरन्ती अस्ति?

उत्तराणि:

1. शतशकटीयानम्
2. ध्वानम्
3. ध्वानम्

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. संसरणम् कठिनम् किमर्थम् भवति?
2. शतशकटीयानम् किं करोति?

उत्तराणि:

1. अनन्ताः यानानां पंक्तयः अतः संसरणम् कठिनम् भवति।
2. शतशकटीयानम् कज्जलमलिनं धूमं मुञ्चति।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. ‘शतशकटीयानम्’ कर्तृपदस्य क्रियापदं किम्?
2. अनंताः’ इति पदस्य विशेष्यपदं किम्?
3. ‘चलनम्’ इति क्रियापदस्य पर्यायपदं किम्?
4. श्लोके ‘सरलम्’ इति पदस्य कः विपर्ययः आगतः?

उत्तराणि:

1. मुञ्चति
2. पक्तयः
3. संसरणम्
4. कठिनम्

(ग) वायुमण्डलं भृशं दूषितं न हि निर्मलं जलम्।
कुत्सितवस्तुमिश्रितं भक्ष्यं समलं धरातलम्॥
करणीयं बहिरन्तर्जगति तु बहु शुद्धीकरणम्। शुचि …॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. जलम् कीदृशम् अस्ति?
2. भृशं दूषितम् किम् अस्ति?
3. कुत्सितवस्तु मिश्रितं किम्?

उत्तराणि:

1. निर्मलम्
2. वायुमण्डलम्
3. धरातलम्

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. मनुष्यैः किं करणीयम्?
2. धरातलं कीदृशं जातम्?

उत्तराणि:

1. मनुष्यैः बहिरन्तर्जगति तु बहु शुद्धीकरणम् करणीयम्।
2. समलं धरातलम् जातम्।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. ‘दूषितम्’ इति पदस्य विपर्ययपदं श्लोके किम्?
2. ‘वायुमण्डलम्’ इति पदस्य विशेषणपदं किम्?
3. ‘करणीयम्’ इति क्रियापदस्य कर्तृपदं किम्?
4. श्लोके ‘अत्यधिकम्’ इति पदस्य क: पर्यायः आगतः?

उत्तराणि:

1. निर्मलम्
2. दूषितम्
3. शुद्धीकरणम्
4. भृशम्

(घ) कञ्चित् काल नय मामस्मान्नगराद् बहुदूरम्।
प्रपश्यामि ग्रामान्ते निर्झर-नदी-पयःपूरम्।।
एकान्ते कान्तारे क्षणमपि मे स्यात् सञ्चरणम्। शुचि …॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. कविः कस्मात् दूरं गच्छति?
2. कीदृशे कान्तारे सः क्षणमपि संरचरण कर्तुम् इच्छति।
3. ग्रामान्ते निर्झर-नदी कीदृशं भवति?

उत्तराणि:

1. नगरात्
2. एकान्ते
3. पयः पूरम्

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. ग्रामान्ते सः किम् द्रष्टुम् इच्छति?
2. एकान्ते कान्तारे मे किं स्यात्?

उत्तराणि:

1. ग्रामान्ते सः निर्झर-नदी-पयः पूरम् द्रष्टुम् इच्छति।
2. एकान्ते कान्तारे में क्षणमपि सञ्चरणं स्यात्।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. अतिसमीपम्’ इति पदस्य विपर्ययपदम् किम्?
2. कान्तारे’ इति पदस्य विशेषणपदं किम्?
3. ‘जलम्’ इति पदस्य पर्यायपदं किम्?
4. श्लोके ‘सञ्चरणम्’ इति कर्तृपदस्य क्रियापदं किम्?

उत्तराणि:

1. बहुदूरम
2. एकान्ते
3. पयः
4. स्यात्

(ङ) हरिततरूणां ललितलतानां माला रमणीया।
कुसुमावलिः समीरचालिता स्यान्मे वरणीया।।
नवमालिका रसालं मिलिता रुचिरं संगमनम्। शुचि …॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. कीदृशानां वृक्षाणां माला रमणीया भवेत्?
2. लतानाम् माला कीदृशी भवेत्?
3. संगमनम् कीदृशं भवेत्?

उत्तराणि:

1. हरितानाम्
2. रमणीया
3. रुचिरम्

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. केन चलिता कुसुमावलिः वरणीया स्यात्?
2. नवमालिका संगमनं कीदृशम्?

उत्तराणि:

1. समीरेण चालिता कुसमावलिः वरणीया स्यात्।
2. नवमालिका रुचिरं संगमनम्।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. रमणीया’ इति विशेषणपदस्य विशेष्यपदं श्लोके किम्?
2. श्लोके ‘चालिता’ इति क्रिया पदस्य कर्तृपदं किम्?
3. अत्र ‘आम्रम्’ इति पदस्य पर्यायपदं किम्?
4. श्लोके ‘दुर्गमनम्’ इति कः विपर्ययः?

उत्तराणि:

1. माला
2. समीरेण
3. रसालम्
4. संगमनम्

(च) अयि चल बन्धो! खगकुलकलरव गुञ्जितवनदेशम्।
पुर कलरव सम्भ्रमितजनेभ्यो धृतसुखसन्देशम्।।
चाकचिक्यजालं नो कुर्याज्जीवितरसहरणम्। शुचि … ||

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. कविः कीदृशम् सन्देशम् दातुम् इच्छति?
2. कविः कुत्र चलितुम् इच्छति?
3. कविः कं सम्बोधयति?

उत्तराणि:

1. धृतसुखम्
2. खगकुलकलरवदेशम्
3. बंधुवर

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. किम् जीवितम् रसं न हरेत्?
2. पुर कलरव सम्भ्रमित जनेभ्यः सुखसन्देशं कुत्र प्राप्नोति?

उत्तराणि:

1. नगराणाम् चाकचिक्यजालं जीवितम् रसं न हरेत्।
2. पुर-कलरव सम्भ्रमित जनेभ्यः सुखसन्देशं खगकुलकलरव गुजितवनदेशे प्राप्नोति।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. ‘ग्राम’ इति पदस्य विपर्ययपदं किम्?
2. ‘चल’ इति क्रियापदस्य कर्तृपदं किम्?
3. ‘चाकचिक्यजालं’ इति कर्तृपदस्य क्रियापदं किम्?
4. श्लोके ‘नगर’ इति पदस्य कः पर्यायः आगतः?

उत्तराणि:

1. पुर
2. बन्धो
3. कुर्यात्
4. पुर

(छ) प्रस्तरतले लतातरुगुल्मा नो भवन्तु पिष्टा।
पाषाणी सभ्यता निसर्गे स्यान्न समाविष्टा॥
मानवाय जीवनं कामये नो जीवन्मरणम्। शुचि …॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. लतातरुगुल्माः कुत्र न पिष्टा:भवन्तु?
2. पाषाणी सभ्यता कुत्र समविष्टा न स्यात्?
3. कविः कि न कामयते?

उत्तराणि:

1. प्रस्तरतले
2. निसर्ग
3. जीवन्मरणम्

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. कविः कस्य कामना करोति?
2. प्रस्तरे तले के न भवन्तु?

उत्तराणि:

1. कविः मानवाय जीवनस्य कामना करोति।
2. प्रस्तरतले लतातरुगुल्माः प्रियः न भवन्तु।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. सभ्यता’ इति पदस्य विशेषणपदं किम्?
2. अहम्’ इति कर्तृपदस्य क्रियापदं किम्?
3. ‘जीवन्मरणं’ इति पदस्य विपर्ययपदं किम् प्रयुक्तम्?
4. ‘पाषाणीसभ्यता निसर्गे न स्यात्’। अत्र कर्तृपदं किम्?

उत्तराणि:

1. पाषाणी
2. कामये
3. जीवनम्
4. पाषाणीसभ्यता

2. अधोलिखित कथनेषु रेखांकितपदानि आधृत्य प्रश्ननिर्माणं क्रियाताम्-

(क) मानव जीवनाय शुचि-पर्यावरणं आवश्यकम् भवति।
(i) कीदृशम्
(ii) क:
(iii) कम्
(iv) कुत्र
उत्तराणि:
(i) कीदृशम्

(ख) महानगरमध्ये कालायसचक्रम् अनिशं चलति।
(i) केन
(ii) क:
(iii) कुत्र
(iv) कम्
उत्तराणि:
(iii) कुत्र

(ग) चक्रम् सदा वक्रम भ्रमति।
(i) कः
(ii) केन
(iii) कीदृशम्
(iv) कुत्र
उत्तराणि:
(iii) कीदृशम्

(घ) अमुना दुर्दान्तैः अशनैः जनग्रसनम् न स्यात्।
(i) कः
(ii) केन
(iii) केषाम्
(iv) कुत्र
उत्तराणि:
(ii) केन

(ङ) शतं शकटीयानम् धूम्र मुञ्चति।
(i) केन
(ii) क:
(iii) कति
(iv) केषाम्
उत्तराणि:
(iii) कति

(च) यामानां पक्तयः अनन्ताः कठिनं संसरणम् भवति।
(i) कः
(ii) केषाम्
(iii) कुत्र
(iv) कम्
उत्तराणि:
(ii) केषाम्

(छ) वायुमण्डलं अत्यधिकं दूषितं जातम्।
(i) केषाम्
(ii) कीदृशम्
(iii) कः
(iv) कम्
उत्तराणि:
(ii) कीदृशम्

(ज) प्रकृत्याः सन्निधौ वास्तविकं सुखं विद्यते।
(i) कम्
(ii) कस्याः
(iii) केषाम्
(iv) कः
उत्तराणि:
(ii) कस्याः

(झ) उद्याने पक्षिणां कलरवं चेतः प्रसादयति।
(i) केषाम्
(ii) कः
(iii) कम्
(iv) कुत्र
उत्तराणि:
(i) केषाम्

(त्र) पाषाणीसभ्यतायां लतातरुगुल्माः प्रस्तरतले पिष्टाः सन्ति।
(i) के
(ii) क:
(iii) केषाम्
(iv) कुत्र
उत्तराणि:
(i) के

(ट) शकटीयानम् कज्जलमालिनं धूमं मुञ्चति।
(i) किम्
(ii) कम्
(iii) कीदृशम्
(iv) कथम्
उत्तराणि:
(iii) कीदृशम्

(ठ) प्रस्तरतले लतातरुगुल्मा पिष्टाः न भवन्तु।
(i) के
(ii) कुत्र
(iii) किम्
(iv) केन
उत्तराणि:
(ii) कुत्र

(ड) महानगरेषु, वाहनानाम् अनन्ताः पङक्तयः धावन्ति।
(i) केषु
(ii) कस्मै
(iii) के
(iv) किम्
उत्तराणि:
(i) केषु

(ढ) शकटीयानम् कज्जलमलिन धूम मुञ्चति।
(i) कम्
(ii) किम्
(iii) कथम्
(iv) कानि
उत्तराणि:
(ii) किम्

(ण) कविः मानवस्य जीवनस्य कामनां करोति।
(i) कथम्
(ii) केन
(iii) कस्मै
(iv) कस्य
उत्तराणि:
(iv) कस्य

3. अधोलिखितानां श्लोकानाम् अन्वये रिक्तस्थानपूर्तिः।

(क) दुर्वहमत्र जीवितं जातं प्रकृतिरेव शरणम्।
शुचि-पर्यावरणम्॥
महानगरमध्ये चलदनिशं कालायसचक्रम्।
मनः शोषयत् तनुः पेषयद् भ्रमति सदा वक्रम्॥
दुर्दान्तैर्दशनैरमुना स्यान्नैव जनग्रसनम्॥ शुचि …..॥

अन्वयः
अत्र जीवितं (i) _______ जातं प्रकृतिः एवं शरणम् शुचि-पर्यावरणम् (एव शरणम्) (ii) _______ मध्ये कालाय सचक्रम् अनिशं चलत् मनः (iii) _______ तनुः प्रेक्षेयद् सदा वक्रम भ्र मति अमुना दुर्दान्तैः (iv) _______ जनग्रसनम् न एवं स्यात् ।।1।।

मञ्जूषा- दशनैः, दुर्वहम्, शोषयत्, महानगर
उत्तराणि:
(i) दुर्वहम्
(ii) महानगर
(ii) शोषयत्
(iv) दशनैः

(ख) कज्जलमलिनं धूमं मुञ्चति शतशकटीयानम्।
वाष्पयानमाला संधावति वितरन्ती ध्वानम्॥
यानानां पङ्क्तयो ह्यनन्ताः कठिनं संसरणम्॥ शुचि…॥

अन्वयः
(अद्यत्वे देशे) शतशकटीयानम् (i) ________ धूमं मुञ्चति, वाष्पयानमाला (ii) ________ वितरन्ती संधावति। यानानां (iii) ________ पङ्क्त यः हि (येन) (iv) ________ संसरणम्।

मञ्जूषा- कठिनं, कज्जलमलिनं, ध्वानम्, अनन्ताः
उत्तराणि:
(i) कज्जलमलिनं
(ii) ध्वानम्
(iii) अनन्ताः
(iv) कठिन

(ग) वायुमण्डलं भृशं दूषितं न हि निर्मल जलम्।
कुत्सितवस्तुमिश्रितं भक्ष्यं समलं धरातलम्॥
करणीयं बहिरन्तर्जगति तु बहु शुद्धीकरणम्॥ शुचि…॥

अन्वयः
(अद्य) वायुमण्डलं (i) ________ दूषितं, निर्मल जलम् हि न, (ii) ________ कुत्सित वस्तु मिश्रितं समलं (iii) ________ अन्तजगति तु (iv) ________ करणीयं बहु शुद्धीकरणम्।।

मञ्जूषा- भक्ष्यं, भृशं, धरातलम्, निर्मल
उत्तराणि:
(i) भृशं
(ii) भक्ष्य
(iii) धरातलम्
(iv) निर्मल

(घ) कञ्चित् कालं नय मामस्मान्नगराद् बहुदूरम्।
प्रपश्यामि ग्रामान्ते निर्झर-नदी-पयःपूरम्॥
एकान्ते कान्तारे क्षणमपि मे स्यात् सञ्चरणम्॥ शुचि …॥

अन्वयः
कञ्चित् कालं माम् (i) _________ नगरात् बहु दूरम् नय। ग्रामान्ते (अहम्) (ii) _________ नदी-पयः पूरम् प्रापश्यामि। (iii) _________ कान्तारे में क्षणम् अपि (iv) _________ स्यात्।

मञ्जूषा- सञ्चरणम्, अस्मात्, एकान्ते, निर्झर
उत्तराणि:
(i) अस्मात्
(ii) निर्झर
(iii) एकान्ते
(iv) सञ्चरणम्

(ङ) हरिततरूणां ललितलतानां माला रमणीया।
कुसुमावलिः समीरचालिता स्यान्मे वरणीया॥
नवमालिका रसालं मिलिता रुचिरं संगमनम्। शुचि …॥

अन्वयः
हरित तरूणाम् ललित (i) ________ रमणीया माला (ii) ________ चालिता कुसमावलिः मे वरणीया स्यात् (iii) ________ नवमालिका (iv) ________ संगमनम् मिलिता।

मञ्जूषा- रुचिरं, लतानां, रसालं, समीर
उत्तराणि:
(i) लतानां
(ii) समीर
(iii) रसालं
(iv) रुचिरं

(च) अयि चल बन्धो! खगकुलकलरव गुञ्जितवनदेशम्।
पुर-कलरव सम्भ्रमितजनेभ्यो धृतसुखसन्देशम्॥
चाकचिक्यजालं नो कुर्याज्जीवितरसहरणम्॥ शुचि …॥

अन्वयः
बन्धो! खगकुल (i) ________ गुञ्जितवनदेशम् (ii) ________ पुर-कलरव सम्भ्रमित (iii) ________धृत सुख सन्देशम्। चाकचिक्यजालं जीवित (iv) ________ हरणम् न कुर्यात्।

मञ्जूषा- जनेभ्यः, कलरव, रस, चल
उत्तराणि:
(i) कलरव
(ii) चल
(iii) जनेभ्यः
(iv) रस

(छ) प्रस्तरतले लतातरुगुल्मा नो भवन्तु पिष्टा।
पाषाणी सभ्यता निसर्गे स्यान्न समाविष्टा।
मानवाय जीवनं कामये नो जीवन्मरणम्। शुचि …॥

अन्वयः
लतातरु गुल्मा: (i) ________ न पिष्टाः भवन्तु। (ii) ________ पाषाणी सभ्यता (iii) ________ न स्यात्। (अहम्) मानवाय (iv) ________ कामये जीवन्मरणम् न।
मञ्जूषा- जीवन, प्रस्तरतले, निसर्ग, समाविष्टा
उत्तराणि:
(i) प्रस्तरतले
(ii) निसर्गे
(iii) समाविष्टा
(iv) जीवनं

4. श्लोकक्रमानुसारं वाक्यानि पुनः लेखनीयानि-

प्रश्न (अ).
(क) अमुना दुर्दान्तैः दशनौः जनग्रसनम् न स्यात्।
(ख) महानगरेषु कालायसचक्रम् अहर्निशम् चलति।
(ग) चक्रम सर्वदा वक्रम् भ्रमति।
(घ) अधुना प्रकृतिरेव शरणम् गन्तव्यम्।
(ङ) अस्मिन् संसारे जीवितं कठिन अस्ति।
(च) अस्मभ्यम्-पर्यावरणम् आवश्यकम्।
(छ) चक्रम् मनः शोषयति तनुः च पेषयति।
(ज) शुचि-पर्यावरणम् आवश्यकम् अस्ति।
उत्तराणि:
(क) अस्मिन् संसारे जीवितं कठिनं अस्ति।
(ख) महानगरेषु कालायसचक्रम् अहर्निशम् चलति।
(ग) चक्रम् मनः शोषयति तनुः च पेषयति।
(घ) चक्रम सर्वदा वक्रम् भ्रमति।
(ङ) अमुना दुर्दान्तः दशनैः जनग्रसनम् न स्यात्।
(च) अधुना प्रकृतिरेव शरणम् गन्तव्यम्।
(छ) अस्मभ्यम्-पर्यावरणम् आवश्यकम्।
(ज) शुचि-पर्यावरम् आवश्यकम् अस्ति।

प्रश्न (आ).
(क) शुचि-पर्यावरणम् आवश्यकम् अस्ति।
(ख) वायु-प्रदूषणम् भवति।
(ग) शकटीयानम् धूमं मुञ्चति।
(घ) महानगरेषु वाहनानाम् अन्नताः पङक्तयः धावन्ति।
(ङ) पर्यावरणम् प्रदूषितम् भवति।
(च) शतशकटीयानम् कज्जलमलिनं धूम मुञ्चति।
(छ) महानगरेषु चलनम् कठिनम् भवति।
(ज) वाष्पयानमाला ध्वानम् वितरन्ती सधावति।
उत्तराणि:
(क) शकटीयानम् धूमं मुज्जति।
(ख) शतशकटीयानम् कज्जलमलिनं धूमं मुञ्चति।
(ग) वाष्पयानमाला ध्वानम् वितरन्ती सधावति।
(घ) महानगरेषु वाहनानाम् अन्नताः पङक्तयः धावन्ति।
(ङ) वायु-प्रदूषणम् भवति।
(च) महानगनेषु चलनम् कठिनम् भवति।
(छ) पर्यावरणम् प्रदूषितम् भवति।
(ज) शुचि-पर्यावरणम् आवश्यकम् अस्ति।

5. समुचितानि पर्यायपदानि मेलनं कुरुत-

खण्ड ‘क’ – खण्ड ‘ख’
(क) जीतितम् – सुन्दरी
(ख) दुर्वहम् – पुष्पपक्तिः
(ग) तनुः – वायु:
(घ) वृक्षाणाम् – वरणीया
(ङ) रमणीया – त्यजति
(च) कुसमावलिः – अत्यधिकम्
(छ) चयनीया – कुटिलम्
(ज) समीरः – जीवनम्
(झ) चाकचिक्यजालम् – शरीरम्
(ञ) रसालम् – दुष्करम्
(ट) मुञ्चति – तरुणाम्
(ठ) वक्रम् – कृत्रिमप्रभावपूर्णजगत्
(ड) भक्ष्यम् – आम्रम्
(ढ) संसरणम् – मलेनयुक्तम्
(ण) भृषम् – सञ्चलनम्
(त) समलम् – खाद्यपदार्थ
(थ) शुचिः – मनसि
(द) यानानाम् – इच्छामि
(ध) धरातलम् – अवलोकयामि
(न) अन्तर्जगति – नहि
(प) बहु – वने
(फ) प्रपश्यामि – पवित्रम्
(ब) कान्तारे – मित्र
(भ) बन्धु – वाहनानाम्
(म) नो – पृथ्वी
(य) कामये – अत्यधिकम्
उत्तराणि:
(क) जीवनम्
(ख) दुष्करम्
(ग) शरीरम्
(घ) तरुणाम्
(ङ) सुन्दरी
(च) पुष्पपक्तिः
(छ) वरणीया
(ज) वायुः
(झ) कृत्रिमप्रभावपूर्ण जगत्
(ञ) आम्रम्
(ट) त्यजति
(ठ) कुटिलम्
(ड) खाद्यपदार्थ
(ढ) सञ्चलनम्
(ण) अत्यधिकम्
(त) मलेन युक्तम्
(थ) पवित्रम्
(द) वाहनानाम्
(ध) पृथ्वी
(न) मनसि
(प) अत्यधिकम्
(फ) अवलोकयामि
(ब) वने
(भ) मित्र
(म) नहि
(य) इच्छामि

6. ‘क’ स्तम्भे विशेषणपदं लिखितम् ‘ख’ स्तम्भे पुनः विशेष्यपदम्। तयोः मेलनं कुरुत-

‘क’ स्तम्भ – ‘ख’ स्तम्भ
(क) शुचि – (i) दशनः
(ख) दुर्दान्तः – (ii) वायुमण्डलम्
(ग) दूषितम् – (iii) कान्तरे
(घ) निर्मल – (iv) पर्यावरणम्
(ङ) समलं – (v) धरातलम्
(च) एकान्ते – (vi) धुर्म
(छ) मलिनं – (vii) पक्तयः
(ज) अनंताः – (viii) जलम्
उत्तराणि:
(क) (iv)
(ख) (i)
(ग) (ii)
(घ) (viii)
(ङ) (v)
(च) (iii)
(छ) (vi)
(ज) (vii)

7. निम्न ‘क’ वर्गीय पदार्थ ‘ख’ वर्गीय पदेषु विपर्यायपदानि चीयताम्-

‘क’ पदानि – ‘ख’ विपर्यायपदानि
(क) निर्मलम् – अदुर्दान्तः
(ख) भक्ष्यम् – निर्मलम्
(ग) बहु – ग्रहणति
(घ) करणीयम् – एकम्
(ङ) नगरात् – अन्ताः
(च) बहुदूरम् – सरलम्
(छ) कान्तारे – अकरणीयम्
(ज) रमणीया – ग्रामात्
(झ) रुचिरम् – समीपम्
(ञ) बन्धो! – नगरे
(ट) कुर्यात् – अरमणीया
(ठ) जीवितम् – मलम्
(ड) सभ्यता – अभक्ष्यम्
(ढ) मानवाय – न्यून
(ण) भृषम् – अरुचिरम्
(त) दूषितम् – शत्रों!
(थ) शुचि – अकुर्यात्
(द) वक्रम् – मरणम्
(ध) दुर्दान्तः – असभ्यता
(न) मलिनम् – दानवाय
(प) मुञ्चति – न्यूनम्
(फ) शतम् – अदूषितम्
(ब) अनन्ताः – अशुचि
(भ) कठिनम् – सरलम्
उत्तराणि:
(क) मलम्
(ख) अभक्ष्यम्
(ग) न्यून
(घ) अकरणीयम्
(ङ) ग्रामात्
(च) समीपम्
(छ) नगरे
(ज) अरमणीया
(झ) अरुचिरम्
(ञ) शत्रो!
(ट) अकुर्यात्
(ठ) मरणम्
(ड) असभ्यता
(ढ) दानवाय
(ण) न्यूनम्
(त) अदूषितम्
(थ) अशुचि
(द) सरलम्
(ध) अदुर्दान्तः
(न) निर्मलम्
(प) ग्रहणति
(फ) एकम्
(ब) अन्ताः
(भ) सरलम्

We have given detailed NCERT Solutions for Class 9 Sanskrit Shemushi Chapter 2 स्वर्णकाकः Questions and Answers come in handy for quickly completing your homework.

Shemushi Sanskrit Class 9 Solutions Chapter 2 स्वर्णकाकः

Class 9 Sanskrit Shemushi Chapter 2 स्वर्णकाकः Textbook Questions and Answers

अभ्यासः

प्रश्न 1.
एकपदेन उत्तर लिखत –

(क) माता काम् आदिशत्?
(ख) स्वर्णकाक: कान् अखादत्?
(ग) प्रासादः कीदृशः वर्तते?
(घ) गृहमागत्य तया का समुद्घटिता?
(ङ) लोभाविष्टा बालिका कीदृशीं मञ्जूषां नमति?
उत्तर:
(क) पुत्रीम्
(ख) तण्डुलान्
(ग) स्वर्णमयः
(घ) मञ्जूषा
(ङ) बृहत्तमाम्

प्रश्न 2.
अधोलिखितानां प्रश्नानाम् उत्तराणि संस्कृतभाषया लिखत –

(क) निर्धनायाः वृद्धायाः दुहिता कीदृशी आसीत्?
(ख) बालिकया पूर्व कीदृशः काकः न दृष्टः आसीत्?
(ग) निधनायाः दुहिता मञ्जूषायां कानि अपश्यत?
(घ) बालिका किं दृष्ट्वा आश्चर्यचकिता जाता?
(ङ) गर्विता बालिका कीदृशं सोपानम् अयाचत् कीदृशं च प्राप्नोत्?
उत्तर:
(क) निर्धनायाः वृद्धायाः दुहिता विनम्रा मनोहरा च आसीत्।
(ख) बालिकया पूर्व एतादृशः स्वर्णपक्षः रजतचञ्चुः स्वर्णकाकः न दृष्टः आसीत्।
(ग) निर्धनायाः दुहिता मञ्जूषायां महाहाणि हीरकाणि अपश्यत्।
(घ) बालिका वृक्षस्य उपरि स्वर्णमयं प्रासादं दृष्ट्वा आश्चर्यचकिता जाता।
(ङ) गर्विता बालिका स्वर्णमय सोपानम् अयाचत् ताम्रमयं च प्राप्नोत्।

प्रश्न 3.
(अ) अधोलिखितानां शब्दानां विलोमपदं पाठात् चित्वा लिखतशब्दाः विलोमपदानि –

शब्दाः – विलोमपदानि
(क) पश्चात् – ………..
(ख) हसितुम् – ………
(ग) अधः – ……………
(घ) श्वेतः – ………….
(ङ) सूर्यास्त: – ………….
(च) सुप्तः – …………..
उत्तर:
शब्दाः – विलोमपदानि
(क) पश्चात् x पूर्वम्
(ख) हसितुम् x रोदितुम्
(ग) अधः x उपरि
(घ) श्वेतः x कृष्णः
(ङ) सूर्यास्त: x सूर्योदयः
(च) सुप्तः x प्रबुद्धः

(आ) सन्धिं कुरुत –

(क) नि + अवसत् – ………….
(ख) सूर्य + उदयः – ………….
(ग) वृक्षस्य + उपरि – ………….
(घ) हि + अकारयत् – ………….
(ङ) च + एकाकिनी – ………….
(च) इति + उक्त्वा – ………….
(छ) प्रति + अवदत् – ………….
(ज) प्र + उक्तम् – ………….
(झ) अत्र + एव – ………….
(ञ) तत्र + उपस्थिता – ………….
(ट) यथा + इच्छम् – ………….
उत्तर:
(क) नि + अवसत् – न्यवसत्
(ख) सूर्य + उदयः – सूर्योदयः
(ग) वृक्षस्य + उपरि – वृक्षस्योपरि
(घ) हि + अकारयत् – ह्यकारयत्
(ङ) च + एकाकिनी – चैकाकिनी
(च) इति + उक्त्वा – इत्युक्त्वा
(छ) प्रति + अवदत् – प्रत्यवदत्
(ज) प्र + उक्तम् – प्रोक्तम्
(झ) अत्र + एव – अत्रैव
(ञ) तत्र + उपस्थिता – तत्रोपस्थिता
(ट) यथा + इच्छम् – यथेच्छम्

प्रश्न 4.
स्थूलपदान्यधिकृत्य प्रश्ननिर्माणं कुरुत –

(क) ग्रामे निर्थना स्वी अवसत्।
(ख) स्वर्णकाकं निवारयन्ती बालिका प्रार्थयत्।
(ग) सूर्योदयात् पूर्वमेव बालिका तत्रोपस्थिता।
(घ) बालिका निर्धनमातुः दुहिता आसीत्।
(ङ) लुब्धा वृद्धा स्वर्णकाकस्य रहस्यमभिज्ञातवती।
उत्तर:
(क) ग्रामे का अवसत्?
(ख) कं निवारयन्ती बालिका प्रार्थयत्?
(ग) कस्मात् पूर्वमेव बालिका तत्रोपस्थिता?
(घ) बालिका कस्याः दुहिता आसीत्?
(ङ) लुब्धा वृद्धा कस्य रहस्यमभिज्ञातवती?

प्रश्न 5.
प्रकृति-प्रत्यय-संयोगं कुरुत (पाठात् चित्वा वा लिखत) –

(क) वि + लोक् + ल्यप् – ………….
(ख) नि – क्षिप् + ल्यप् – ………….
(ग) आ + गम् + ल्यप् – ………….
(घ) दृश् + क्त्वा – ………….
(ङ) शी + क्त्वा – ………….
(च) लघु + तमप् – ………….
उत्तर:
(क) वि + लोक् + ल्यप् – विलोक्य
(ख) नि – क्षिप् + ल्यप् – निक्षिप्य
(ग) आ + गम् + ल्यप् – आगम्य
(घ) दृश् + क्त्वा – दृष्ट्वा
(ङ) शी + क्त्वा – शयित्वा
(च) लघु + तमप् – लघुतम (लघुतमम्)

प्रश्न 6.
प्रकृति-प्रत्यय-विभागं कुरुत –

(क) रोदितुम् – ………….
(ख) दृष्ट्वा – ………….
(ग) विलोक्य – ………….
(घ) निक्षिप्य – ………….
(ङ) आगत्य – ………….
(च) शयित्वा – ………….
(छ) लघुतमम् – ………….
उत्तर:
(क) रोदितुम् – रुद् + तुमुन्
(ख) दृष्ट्वा – दृश् + क्त्वा
(ग) विलोक्य –वि + लोक् + ल्यप्
(घ) निक्षिप्य – नि + क्षिप् + ल्यप्
(ङ) आगत्य – आ + गम् + ल्यप्
(च) शयित्वा – शी + क्त्वा
(छ) लघुतमम् – लघु + तमप्

प्रश्न 7.
अधोलिखितानि कथनानि कः/का, कं/कां च कथयति –

कथनानि – क:/का – कं/काम्
(क) पूर्व प्रातराशः क्रियताम्। – …………. – ………….
(ख) सूर्यातपे तण्डुलान् खगेभ्यो रक्ष। – …………. – ………….
(ग) तण्डुलान् मा भक्षय। – …………. – ………….
(घ) अहं तुभ्यं तण्डुलमूल्यं दास्यामि। – …………. – ………….
(ङ) भो नीचकाक! अहमागता, महां तण्डुलमूल्यं प्रयच्छ। – …………. – ………….
उत्तर:
कथनानि – कः/का – कं/काम्
(क) पूर्व प्रातराशः क्रियताम्। – स्वर्णकाक: – बालिकाम्
(ख) सूर्यातपे तण्डुलान् खगेभ्यो रक्ष। – वृद्धा माता – पुत्रीम्
(ग) तण्डुलान् मा भक्षया – बालिका – स्वर्णकाकम्
(घ) अहं तुभ्यं तण्डुलमूल्यं दास्यामि। – स्वर्णकाकः – बालिकाम्
(ङ) भो नीचकाक! अहमागता, मां तण्डुलमूल्यं प्रयच्छ। – लुब्धायाः पुत्री – स्वर्णकाकम्

प्रश्न 8.
उदाहरणमनुसृत्य कोष्ठकगतेषु पदेषु पञ्चमीविभक्तेः प्रयोगं कृत्वा रिक्तस्थानानि पूरयत –

यथा-मूषक: बिलाद् बहिः निर्गच्छति। (बिल)
(क) जनः …………. बहिः आगच्छति। (ग्राम)
(ख) नद्यः …………. निस्सन्ति। (पर्वत)
(ग) …………. पत्राणि पतन्ति। (वृक्ष)
(घ) बालकः …………. बिभेति। (सिंह)
(ङ) ईश्वरः …………. त्रायते। (क्लेश)
(च) प्रभुः भक्तं …………. निवारयति। (पाप)
उत्तर:
(क) ग्रामात्
(ख) पर्वतात्/पर्वतेभ्यः
(ग) वृक्षात्
(घ) सिंहात्
(छ) क्लेशात्
(च) पापात्

Class 9 Sanskrit Shemushi Chapter 2 स्वर्णकाकः Additional Important Questions and Answers

अतिरिक्त कार्यम्

प्रश्न 1.
निम्नलिखितम् अनुच्छेदं पठित्वा तदाधारिताना प्रश्नानाम् उत्तराणि लिखत –

1. पुरा कस्मिंश्चिद् ग्रामे एका निधना वृद्धा स्त्री न्यवसत्। तस्याः च एका दुहिता विनमा मनोहरा चासीत्। एकदा माता स्थाल्या तण्डुलान् निक्षिप्य पुत्रीम् आदिशत्। “सूर्यातपे तण्डुलान् खगेभ्यो रक्षा” किञ्चित् कालादनन्तरम् एको विचित्रः काकः समुड्डीय तस्याः समीपम् अगच्छत् ।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. वृद्धा स्त्री कीदृशी आसीत्?
2. वृद्धायाः कीदृशी दुहिता आसीत्?
3. निर्धनास्त्री कुत्र न्यवसत्?

उत्तर:

1. निर्धना
2. विनम्रा
3. कस्मिंश्चिद्ग्रामे

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. किञ्चित्कालानन्तरम् किम् अभवत्?
2. दुहिता कीदृशी आसीत?

उत्तर:

1. किञ्चित्कालानन्तरम् एको विचित्रः काकः समुड्डीय तस्याः समीपम् अगच्छत्।
2. दुहिता विनम्रा मनोहरा चासीत्।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. अनुच्छेदे ‘न्यवसत्’ इति क्रियापदस्य कर्तृपदं किमस्ति?
2. ‘दुहिता विनम्रा’ अनयोः पदयोः विशेषणपदं किम्?
3. अस्मिन् अनुच्छेदे ‘धनवती’ इति पदस्य कः विपर्ययः आगतोऽस्ति?
4. पुत्री इत्यर्थे किं पदं प्रयुक्तम्?

उत्तर:

1. स्त्री
2. विनम्रा
3. निर्धना
4. दुहिता

2. नैतादृशः स्वर्णपक्षो रजतचञ्चुः स्वर्णकाकस्तया पूर्व दृष्टः। तं तण्डुलान् खावन्तं हसन्तञ्च विलोक्य बालिका रोदितुमारब्धा। तं निवारयन्ती सा प्रार्थयत्-“तण्डुलान् मा भक्षया मदीया माता अतीव निर्धना वर्तते।” स्वर्णपक्षः काकः प्रोवाच, “मा शुचः। सूर्योदयात्प्राग् ग्रामाबहिः पिप्पलवृक्षमनु त्वया आगन्तव्यम्। अहं तुभ्यं तण्डुलमूल्यं दास्यामि।” ङ्केप्रहबिता बालिका निनामपि न लेभा

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. स्वर्णकाकस्य चञ्चुः कीदृशः आसीत्?
2. बालिकायाः माता कीदृशी आसीत्?
3. क: तण्डुलान् अखादत्?

उत्तर:

1. रजतस्य
2. निर्धना
3. स्वर्णकाक:

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. स्वर्णकाकं निवारयन्ती सा कि प्रार्थयत्?
2. स्वर्णकाकः किम् अवदत्?

उत्तर:

1. स्वर्णकाकं निवारयन्ती सा प्रार्थयत्-“तण्डुलान् मा भक्षय। मदीया माता अतीव निर्धना वर्तते।”
2. स्वर्णकाकः प्रोवाच मा शुचः। सूर्योदयात्प्राग् ग्रामाद्बहिः पिप्पवृक्षमनु त्वया आगन्तव्यम्। अहं तुभ्यं तण्डुलमूल्यं दास्यामि।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. अनुच्छेदे ‘हसितुम्’ इत्यस्य पदस्य कः विपर्ययः लिखितः?
2. रोदितुमारब्धा’ इत्यस्याः क्रियायाः अनुच्छेदे कर्तृपदं कि वर्तते?
3. ‘प्रहर्षिता बालिका निद्रामपि न लेभे’। अत्र वाक्ये विशेषणपदं किमस्ति?
4. ‘दृष्ट्वा ‘ इत्यर्थे किं पदं प्रयुक्तम्?

उत्तर:

1. रोदितुम्
2. बालिका
3. प्रहर्षिता
4. विलोक्य

3. सूर्योदयात्पूर्वमेव सा तत्रोपस्थिता। वृक्षस्योपरि विलोक्य सा च आश्चर्यचकिता सजाता यत् तत्र स्वर्णमयः प्रासादो वर्तते। यदा काकः शयित्वा प्रबुद्धस्तदा तेन स्वर्णगवाक्षात्कथितं “हहो बाले! त्वमागता, तिष्ठ, अहं त्वत्कृते सोपानमवतारयामि, तत्कथय स्वर्णमयं रजतमयम् तासमय वा”? कन्या अवदत्-“अहं निर्धनमातुः दुहिता अस्मि। तानसोपानेनैव आगमिष्यामि।” परं स्वर्णसोपानेन सा स्वर्णभवनम् आरोहत् ।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. सूर्योदयात् पूर्वमेव का तत्र उपस्थिता?
2. कन्या कस्याः दुहिता आसी?
3. काकः किं कृत्वा प्रबुद्धः?

उत्तर:

1. बालिका (सा)
2. निर्धनमातुः
3. शयित्वा

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. स्वर्णगवाक्षात् काकेन किं कथितम्?
2. कन्या किं प्रावोचत्?

उत्तर:

1. स्वर्णगवाक्षात् काकेन कथितम्-हंहो बाले! त्वमागता, तिष्ठ, अहं त्वत्कृते सोपानमवतारयामि।
2. कन्या प्रावोचत् अहं निर्धनमातुर्दुहिताऽस्मिा ताम्रसोपानेनैव आगमिष्यामि।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. ‘अधः’ इत्यस्य पदस्य कः विपर्ययः अनुच्छेदे लिखितोऽस्ति?
2. ‘अहं निर्धनमातुः दुहिता अस्मि’। अत्र वाक्ये ‘अहम्’ इति कर्तृपदस्य क्रियापदं किम्?
3. अनुच्छेदे ‘ताम्रमयं सोपानम्’ अनयो: पदयोः विशेषणपदं किमस्ति?
4. अस्ति’ इत्यर्थे कि पदं प्रयुक्तम्?

उत्तर:

1. उपरि
2. अस्मि
3. ताम्रमयं
4. वर्तते

4. चिरकाल भवने चित्रविचित्रवस्तूनि सज्जितानि दृष्ट्वा सा विस्मयं गता। श्रान्तां तां विलोक्य काकः अवदत् “पूर्व लघुप्रातराशः क्रियताम्-वद त्वं स्वर्णस्थाल्यां भोजनं करिष्यसि किं वा रजतस्थाल्याम् उत ताम्रस्थाल्याम्”? बालिका अवदत्-ताप्रस्थाल्याम् एव अहं-“निर्धना भोजनं करिष्यामि।” तदा सा आश्चर्यचकिता सञ्जाता यदा स्वर्णकाकेन स्वर्णस्थाल्यां भोजनं पर्यवेषितम्। न एतादृशम् स्वादु भोजनमद्यावधि बालिका खादितवती। काकोऽववत्-“बालिके! अहमिच्छामि यत् त्वम् सर्वदा अत्रैव तिष्ठ परं तव माता तु एकाकिनी वर्तते। अतः त्वं शीघ्रमेव स्वगृहं गच्छ ।”

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. कीदृशीं तां विलोक्य काकः प्राह?
2. बालिका कीदृशं भोजनम् अद्यावधि न खादितवती?
3. कीदृशानिवस्तूनि दृष्ट्वा सा विस्मयं गता?

उत्तर:

1. श्रान्ताम्
2. स्वादु
3. चित्रविचित्रवस्तूनि

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. सा बालिका कथं विस्मयं गता?
2. काकः किम् अवदत्

उत्तर:

1. चिरकाल भवने चित्रविचित्रवस्तूनि सज्जितानि दृष्ट्वा सा विस्मयं गता।
2. काकः अवदत् बालिके। अहमिच्छामि यत्त्वं सर्वदा अत्रैव तिष्ठ पर तव माता तु एकाकिनी वर्तते। त्वं शीघ्रमेव स्वगृहं गच्छ।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. तदा सा कन्या आश्चर्यचकिता सञ्जाता’। अस्मिन् वाक्ये विशेषणपदं किमस्ति?
2. अनुच्छेदे ‘जननी’ इत्यस्य पदस्य कः पर्यायः लिखितः?
3. ‘त्वम्’ इत्यस्य कर्तृपदस्य अनुच्छेदे क्रियापदं किमस्ति?
4. ‘दृष्ट्वा ‘ इत्यर्थे किम् पदं प्रयुक्तम्?

उत्तर:

1. आश्चर्यचकिता
2. माता
3. तिष्ठ
4. विलोक्य

5. इत्युक्त्वा काकः कक्षाभ्यन्तरात् तिनः मञ्जूषाः निस्सार्य तां प्रत्यवदत् “बालिके! यथेच्छ गृहाण मञ्जूषामेकाम्।” लघुतमा मञ्जूषां प्रगृह्य बालिकया कथितम् इयत् एव मदीयतण्डुलानां मूल्यम्। गृहमागत्य तया मञ्जूषा समुद्घाटिता, तस्यां महार्हाणि हीरकाणि विलोक्य सा प्रहर्षिता तहिनाद्धनिका च सजाता ।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. काकः कक्षाभ्यन्तरात् कति मञ्जूषाः निस्सारयति?
2. कीदृशी मञ्जूषां प्रगृह्य बालिका कथितम्?
3. किं विलोक्य सा प्रहर्षिता सञाता?

उत्तर:

1. तिस्रः
2. लघुतमाम्
3. हीरकाणि

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. मञ्जूषायां कानि विलोक्य सा बालिका प्रहर्षिता अभवत्?
2. काकः किं अवदत्?

उत्तर:

1. मञ्जूषायां महार्हाणि हीरकाणि विलोक्य सा बालिका प्रहर्षिता अभवत्।
2. बालिके! यथेच्छं गृहाण मञ्जूषामेकाम्।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. ‘प्रत्यवदत्’ इति क्रियापदस्य अनुच्छेदे कर्तृपदं किमस्ति?
2. अनुच्छेदे ‘सा’ इति कर्तृपदस्य क्रियापदं किम् आगतम्?
3. “निष्क्रम्य’ इति पदस्य कः विपर्ययः अत्र अनुच्छेदे लिखितः?
4. ‘ग्रहणं कुरु’ इत्यर्थे किं पदं प्रयुक्तम्?

उत्तर:

1. काकः
2. सजाता
3. आगत्य
4. गृहाण

6. तस्मिन्नेव ग्रामे एका अपरा लुब्धा वृद्धा न्यवसत्। तस्या अपि एका पुत्री आसीत्। ईयया सा तस्य स्वर्णकाकस्य रहस्यम् ज्ञातवती। सूर्यातपे तण्डुलान् निक्षिप्य तयापि स्वसुता रक्षार्थं नियुक्ता। तथैव स्वर्णपक्षः काकः तण्डुलान् भक्षयन् तामपि तत्रैवाकारयत्। प्रातस्तत्र गत्वा सा काकं निर्भर्त्सयन्ती प्रावोचत्-“भो नीचकाक! अहमागता, मां तण्डुलमूल्यं प्रयच्छ।” काकोऽब्रवीत्-“अहं त्वत्कृते सोपानम् अवतारयामि। तत्कथय स्वर्णमयं रजतमय तासमयं वा।” गर्वितया बालिकया प्रोक्तम्-“स्वर्णमयेन सोपानेन अहम् आगच्छामि।” परं स्वर्णकाकस्तत्कृते तासमयं सोपानमेव प्रायच्छत्। स्वर्णकाकस्तां भोजनमपि तानभाजने एव अकारयत् ।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. तस्मिन् एव ग्रामे कीदृशी वृद्धा न्यवसत्?
2. स्वर्णकाकः तां बालिका कस्मिन् भोजनम् अकारयत्?
3. लुब्धा वृद्धायाः एका का आसीत्?

उत्तर:

1. लुब्धा
2. ताम्रभाजने
3. पुत्री

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. सा वृद्धा ईर्ष्णया किम् अभिज्ञातवतो?
2. तस्मिन्नेव ग्रामे एकाऽपरा कीदृशी वृद्धा न्यवसत्?

उत्तर:

1. सा वृद्धा ईयया तस्य स्वर्णकाकस्य रहस्यम् अभिज्ञातवती।
2. तस्मिन्नेव ग्रामे एकाऽपरा लुब्धा वृद्धा न्यवसत्।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. अनुच्छेदे ‘अकारयत्’ इति क्रियायाः कर्तृपदं किम्?
2. अनुच्छेदे ‘लुब्धा वृद्धा’ इति कर्तृपदस्य क्रियापदं किमस्ति?
3. ‘प्रशंसयन्ती’ इति पदस्य कः विपर्ययः अनुच्छेदे आगतः?
4. ‘वसति स्म’ इति अर्थ किं पदं प्रयुक्तम्?

उत्तर:

1. स्वर्णकाकः
2. न्यवसत्
3. निर्भर्त्सयन्ती
4. न्यवसत्

प्रश्न 7.
प्रतिनिवृत्तिकाले स्वर्णकाकेन कक्षाभ्यन्तरात् तिम्रः मञ्जूषाः तत्पुरः समुत्क्षिप्ताः। लोभाविष्टा सा बृहत्तमा मञ्जूषां गृहीतवती। गृहमागत्य सा तर्षिता यावद् मञ्जूषामुद्घाटयति तावत् तस्या भीषणः कृष्णसर्पो विलोकितः। लुब्धया बालिकया लोभस्य फलं प्राप्तम्। तदनन्तरं सा लोभ पर्यत्यजत् ।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. केन तिनः मञ्जूषाः कन्यायाः पुरः समुत्क्षिप्ता:?
2. लोभाविष्टा सा कन्या कीदृशीं मञ्जूषां गृहीतवती?
3. मञ्जूषाम् उद्घाट्य सा किम् पश्यति?

उत्तर:

1. स्वर्णकाकेन
2. बृहत्तमाम्
3. कृष्णसर्पः

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. गृहमागत्य यदा सा तर्षिता मञ्जूषाम् उद्घाटयति तदा कं विलोकयति?
2. लुब्धया बालिकया कस्य फल प्राप्तम्?

उत्तर:

1. गृहमागत्य यदा सा तर्षिता मञ्जूषाम् उद्घाटयति तदा तस्यां भीषणं कृष्णसर्प विलोकयति।
2. लुब्धया बालिकया लोभस्य प्राप्तम्।

(iii) भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. अनुच्छेदे ‘समुत्क्षिप्ताः’ इति क्रियायाः कर्तृपदं किम्?
2. अस्मिन् अनुच्छेदे ‘लोभाविष्टा’ इति विशेषणं कस्यै आगतम्?
3. ‘तदनन्तरं सा लोभं पर्यत्यजत्’। अत्र अव्ययपदं किमस्ति?
4. तस्य समक्षे’ इत्यर्थे किं पदं प्रयुक्तम्?

उत्तर:

1. स्वर्णकाकेन
2. बालिकार्य
3. अनन्तरम्
4. तत्पुरः

प्रश्न 2.
निम्नवाक्येषु रेखाङ्कित पदानां स्थानेषु प्रश्नवाचकं पदं लिखत –

प्रश्न 1.
पुरा कस्मिंश्चिद् ग्रामे एका वृद्धा न्यवसत्।
(क) क;
(ख) का
(ग) किम्
(घ) के
उत्तर:
(ख) का

प्रश्न 2.
तस्याः दुहिता विनम्रा मनोहरा च आसीत्।
(क) का
(ख) कीदृशम्
(ग) कीदृशः
(घ) कीदृशी
उत्तर:
(घ) कीदृशी

प्रश्न 3.
सा पुत्रीम् आदिदेश।
(क) काम्
(ख) कम्
(ग) किम्
(घ) कथम्
उत्तर:
(क) काम्

प्रश्न 4.
सूर्यातपे तण्डुलान् खगेभ्यो रक्ष।
(क) कथम्
(ख) काभ्यः
(ग) केभ्यः
(घ) कस्यै
उत्तर:
(ग) केभ्यः

प्रश्न 5.
तदा एक: विचित्रः काकः समुड्डीय ताम् उपाजगाम।
(क) के
(ख) काः
(ग) क:
(घ) कीदृशः
उत्तर:
(ग) क:

प्रश्न 6.
सा प्रार्थयत् – तण्डुलान् मा भक्षय।
(क) का
(ख) कः
(ग) के
(घ) काः
उत्तर:
(क) का

प्रश्न 7.
प्रहर्षिता बालिका निद्राम् अपि न लेभे।
(क) कथम्
(ख) कीदृशी
(ग) कीदृशः
(घ) कीदृशम्
उत्तर:
(ख) कीदृशी

प्रश्न 8.
अहं तुभ्यं तण्डुलमूल्यं प्रदास्यामि।
(क) कम्
(ख) कस्यै
(ग) केभ्यः
(घ) काभ्यः
उत्तर:
(ख) कस्यै

प्रश्न 9.
वृक्षस्य उपरि विलोक्य सा आश्चर्यचकिता सञ्जाता।
(क) का
(ख) काः
(ग) कीदृशः
(घ) कीदृशी
उत्तर:
(घ) कीदृशी

प्रश्न 10.
अहं त्वत्कृते सोपानम् अवतारयामि।
(क) किम्
(ख) कम्
(ग) काम्
(घ) कथम्
उत्तर:
(क) किम्

प्रश्न 11.
अहं निर्धनमातुः दुहिता अस्मि।
(क) कस्याः
(ख) कस्य
(ग) कः
(घ) का
उत्तर:
(क) कस्याः

प्रश्न 12.
श्रान्तां तां विलोक्य काकः प्राह।
(क) के
(ख) कथम्
(ग) काः
(घ) कः
उत्तर:
(घ) कः

प्रश्न 13.
अहं निर्धना ताम्रस्थाल्याम् एव भोजनं करिष्यामि।।
(क) कस्याम्
(ख) के।
(ग) काः
(घ) काभ्यः
उत्तर:
(क) कस्याम्

प्रश्न 14.
तव माता एकाकिनी वर्तते।
(क) किम्
(ख) कस्याः
(ग) के
उत्तर:
(ख) कस्याः

प्रश्न 15.
त्वं शीघ्रमेव स्वगृहं गच्छ।
(क) कीदृशः
(ख) कस्यै
(ग) का
(घ) कुत्र
उत्तर:
(घ) कुत्र

प्रश्न 16.
काकः कक्षाभ्यन्तरात् तिस्रः मञ्जूषाः निस्सार्य अवदत्।
(क) कति
(ख) कथम्
(ग) किम्
(घ) काम्
(घ) का
उत्तर:
(क) कति

प्रश्न 17.
लघुतमा मञ्जूषां प्रगृह्य बालिकया कथितम्।
(क) कम्
(ख) कया
(ग) कस्याः
(घ) काम्
उत्तर:
(ख) कया

प्रश्न 18.
तत्रैव एका अपरा लुब्धा वृद्धा न्यवसत्।
(क) कः
(ख) किम्
(ग) केभ्यः
(घ) का
उत्तर:
(घ) का

प्रश्न 19.
ईय॑या सा स्वर्णकाकस्य रहस्यम् अभिज्ञातवती।
(क) कथम्/कया
(ख) का
(ग) कस्मै
उत्तर:
(क) कथम्/कया

प्रश्न 20.
तथा अपि स्वसुता तण्डुलरक्षार्थं नियुक्ता।
(क) कथम्
(ख) के
(ग) कः
उत्तर:
(घ) का

प्रश्न 21.
महयं तण्डुलमूल्यं प्रयच्छ।
(क) कस्याः
(ख) किम्
उत्तर:
(ख) किम्

प्रश्न 22.
स्वर्णमयेन सोपानेन अहम् आगच्छामि।।
(क) कीदृशेण
(ख) कीदृशी
(ग) काभ्यः
उत्तर:
(क) कीदृशेण

प्रश्न 23.
लोभाविष्टा सा बृहत्तमा मञ्जूषां गृहीतवती।
(क) के
(ख) कस्याः
(ग) का
(घ) कीदृशी
उत्तर:
(घ) कीदृशी

प्रश्न 24.
तस्यां तया भीषणः सर्पः विलोकितः।
(क) कः
(ख) कम्
(ग) का
(घ) कम्
उत्तर:
(क) कः

प्रश्न 25.
लुब्धया बालिकया लोभस्य फलं प्राप्तम्।
(क) क;
(ख) किम्
(ग) केभ्यः
(घ) कस्य
उत्तर:
(घ) कस्य

प्रश्न 3.
निम्नवाक्यानि घटनाक्रमानुसारं पुनर्लिखत –

प्रश्न 1.

1. बालिका सूर्योदयात् पूर्वम् एव तत्र उपस्थिता अभवत्।
2. यदा मञ्जूषां सा उद्घाटितवती तदा तस्यां महाणि हीरकाणि दृष्ट्वा सा प्रसन्ना जाता।
3. यदा काकः शचित्वा प्रबुद्धः जातः तदा सः स्वर्णगवाक्षात् अवदत्।
4. कस्मिंश्चिद् ग्रामे एका धनहीना वृद्धा स्त्री अवसत्।
5. बालिका अकथयत्-अहं ताम्रस्थाल्यामेव भोजनं करिष्यामि।
6. बालिके! त्वम् आगता? तिष्ठ, अहं त्वत्कृते सोपानम् अवतारयामि।
7. सा काकं भर्त्सयन्ती अवदत्।
8. परं स्वर्णसोपानेन सा स्वर्णभवनम् आससाद।

उत्तर:

1. कस्मिंश्चिद् ग्रामे एका धनहीना वृद्धा स्त्री अवसत्।
2. बालिका सूर्योदयात् पूर्वम् एव तत्र उपस्थिता अभवत्।
3. यदा काकः शचित्वा प्रबुद्धः जातः तदा सः स्वर्णगवाक्षात् अवदत्।
4. बालिके। त्वम् आगता? तिष्ठ, अहं त्वत्कृते सोपानम् अवतारयामि।
5. परं स्वर्णसोपानेन सा स्वर्णभवनम् आससाद।
6. बालिका अकथयत् – अहं ताम्रस्थाल्यामेव भोजनं करिष्यामि।
7. यदा मञ्जूषां सा उद्घाटितवती तदा तस्यां महाहाणि हीरकाणि दृष्ट्वा सा प्रसन्ना जाता।
8. सा काकं भर्ल्सयन्ती अवदत्।

प्रश्न 2.

1. इति उक्त्वा काकः कक्षायाः अभ्यन्तरात् तिस्रः मञ्जूषा: निस्सारयत्।
2. लोभाविष्टा सा बृहत्तमा मञ्जूषां गृहीतवती।
3. भो नीचकाक! अहम् आगता अस्मि। मह्यं तण्डुलमूल्यं प्रयच्छ।
4. किञ्चित् कालात् अनन्तरं एकः विचित्रः काकः तत्र तामुपाजगाम।
5. एकस्याः वृद्धायाः एका विनम्रा मनोहरा च कन्या आसीत्।
6. कन्या प्रावोचत्-अहं निर्धन मातुः कन्या अस्मि। अतः ताम्र सोपानेन एव आगमिष्यामि।
7. तस्मिन्नेव ग्रामे एका अपरा लुब्धा वृद्धा स्त्री अवसत्।
8. एकदा सा पुत्रीम् आदिदेश-सूर्यातपे तण्डुलात् खगेभ्यो त्वं रक्ष।

उत्तर:

1. एकस्याः वृद्धायाः एका विनम्रा मनोहरा च कन्या आसीत्।
2. एकदा सा पुत्रीम् आदिदेश-सूर्यातपे तण्डुलात् खगेभ्यो त्वं रक्ष।
3. किञ्चित् कालात् अनन्तरं एक: विचित्रः काकः तत्र तामुपाजगाम।
4. कन्या प्रावोचत्-अहं निर्धन मातुः कन्या अस्मि। अतः ताम्र सोपानेन एव आगमिष्यामि।
5. इति उक्त्वा काकः कक्षायाः अभ्यन्तरात् तिम्रः मञ्जूषाः निस्सारयत्।
6. तस्मिन्नेव ग्रामे एका अपरा लुब्धा वृद्धा स्त्री अवसत्।
7. भो नीचकाक! अहम् आगता अस्मि। मह्यं तण्डुलमूल्यं प्रयच्छ।
8. लोभाविष्टा सा बृहत्तमा मञ्जूषां गृहीतवती।

प्रश्न 3.

1. गृहम् आगत्य यदा सा मञ्जूषाम् उद्घाटयति तदा तस्मिन् एवं कृष्णसर्पम् अपश्यत्।
2. हे नीच काक! अहम् आगता अस्मि, मह्यं तण्डुलानां मूल्यं प्रयच्छ।
3. तदनन्तरं सा लोभम् अत्यजत्।
4. तस्मिन् एवं ग्रामे एका लुब्धा वृद्धा अपि न्यवसत्।
5. गर्विता बालिका अवदत्-अहं स्वर्णमयेन सोपानेव एव आगमिष्यामि।
6. लुब्धया कन्यया लोभस्य फलं प्राप्तम्।
7. तस्याः अपि एका ईर्ष्यालु: कन्या अपि आसीत्।
8. सूर्यातपे सापि तण्डुलान् निक्षिप्य स्वसुता तेषां रक्षार्थ नियुक्तवती।

उत्तर:

1. तस्मिन् एव ग्रामे एका लुब्धा वृद्धा अपि न्यवसत्।
2. तस्याः अपि एका ईर्ष्यालुः कन्या अपि आसीत्।
3. सूर्यातपे सापि तण्डुलान् निक्षिप्य स्वसुतां तेषां रक्षार्थ नियुक्तवती।
4. हे नीच काक! अहम् आगता अस्मि, महयं तण्डुलानां मूल्यं प्रयच्छ।
5. गर्विता बालिका अवदत्-अहं स्वर्णमयेन सोपानेव एव आगमिष्यामि।
6. गृहम् आगत्य यदा सा मञ्जूषाम् उद्घाटयति तदा तस्मिन् एवं कृष्णसर्पम् अपश्यत्।
7. लुब्धया कन्यया लोभस्य फलं प्राप्तम्।
8. तदनन्तर सा लोभम् अत्यजत्।

प्रश्न 4.
निम्न पदानां पर्यायपदानि मेलयत-

पदानि – पर्यायः

1. निर्धना – अवदत्
2. एकदा – तस्याः समक्षे
3. पुरा – शोकं कुरु
4. माता – अतीव प्रसन्ना
5. शुचः – स्वर्णशरीरम्
6. प्रबुद्धः – अद्यपर्यन्तम्
7. स्वर्णमयम् – निक्षिप्ताः
8. चिरकालम् – दृष्ट्वा
9. दुहिता – असन्तुष्टा (उत्कण्ठिता)
10. विस्मयम् – एकवारम् (एकस्मिन् दिवसे)
11. प्राह – खादन्
12. अद्यावधि – प्राचीने काले
13. प्रगृह्य – जागरितः
14. विलोक्य – दीर्धकालम्
15. प्रहर्षिता – जननी
16. भक्षयन् – गृहीत्वा
17. तत्पुरः – धनहीना
18. समुत्क्षिप्ताः – आश्चर्यम्
19. पर्यत्यजत् – तुभ्यम्
20. तर्षिता – सुता
21. त्वत्कृते – त्यक्तवती

उत्तर:

पदानि – पर्यायः

1. निर्धना – धनहीना
2. एकदा – एकवारम् (एकस्मिन् दिवसे)
3. पुरा – प्राचीनेकाले
4. माता – जननी
5. शुचः – शोकं कुरु
6. प्रबुद्धः – जागरितः
7. स्वर्णमयम् – स्वर्णशरीरम्
8. चिरकालम् – दीर्घकालम्
9. दुहिता – सुता
10. विस्मयम् -आश्चर्यम्
11. प्राह – अवदत्
12. अद्यावधि – अद्यपर्यन्तम्
13. प्रगृह्य – गृहीत्वा
14. विलोक्य – दृष्ट्वा
15. प्रहर्षिता – अतीव प्रसन्ना
16. भक्षयन् – खादन्
17. तत्पुरः – तस्याः समक्षे
18. समुत्क्षिप्ताः – निक्षिप्ता:
19. पर्यत्यजत् – त्यक्तवती
20. तर्षिता – असन्तुष्टा (उत्कण्ठिता)
21. त्वत्कृते – तुभ्यम्

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

‘क’ स्तम्भ – ‘ख’ स्तम्भ

1. वृद्धा – स्वर्णकाक:
2. मनोहरा – माता
3. कस्मिंश्चिद् – बालिका
4. स्वर्णपक्ष: – भवनम्
5. निर्धना – मञ्जूषा
6. प्रहर्षिता – सा
7. स्वर्णमयम् – काक:
8. आश्चर्यचकिता – ग्राम
9. प्रबुद्धः – दुहिता
10. एकाकिनी – माता
11. इयत् – स्त्री
12. तिस्रः – मूल्यम्

उत्तर:

‘क’ स्तम्भ – ‘ख’ स्तम्भ

1. वृद्धा – स्त्री
2. मनोहरा – दुहिता
3. कस्मिंश्चिद् – ग्रामे
4. स्वर्णपक्ष: – स्वर्णकाकः
5. निर्धना – माता
6. प्रहर्षिता – बालिका
7. स्वर्णमयम् – भवनम्
8. आश्चर्यचकिता – सा
9. प्रबुद्धः – काकः
10. एकाकिनी – माता
11. इयत् – मूल्यम्
12. तिस्रः – मञ्जूषा

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 13 Amines. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 13 Important Extra Questions Amines

Amines Important Extra Questions Very Short Answer Type

Question 1.
Arrange the following in decreasing order of solubility in water:
(CH₃)₃N, (CH₃)₂H, CH₃NH₂ (CBSE Delhi 2019)
Answer:
CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N

Question 2.
Arrange the following compounds in increasing order of their solubility in water:
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂ (CBSE Delhi 2011, CBSE 2013)
Answer:
C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂

Question 3.
Write the structure of n-methylhexanamine. (CBSE 2013)
Answer:
CH₃ – NH – CH₂CH₃

Question 4.
Write the structure of 2-amino toluene. (CBSE 2013)
Answer:

Question 5.
Write the IUPAC name of the given compound: (CBSE Delhi 2016)

Answer:
2, 4, 6-Tribromoaniline

Question 6.
Arrange the following in increasing order of their basic strength in an aqueous solution:
CH₃NH₂, (CH₃)₃N, (CH₃)₂NH (CBSE Delhi 2013)
Answer:
(CH₃)₃N < CH₃NH₂ < (CH₃)₂NH

Question 7.
Write the IUPAC name of the compound (CBSE Delhi 2014)

Answer:

Question 8.
Arrange the following in increasing order of base strength in the gas phase:
(C₂H₅)₃N, C₂H₅NH₂, (C₂H₅)₂NH (CBSE Delhi 2019)
Answer:
(C₂H₅)₃N > (C₂H5)₂NH > C₂H₅NH₂

Question 9.
Arrange the following in increasing order of boiling points:
(CH₃)₃N, C₂H₅OH, C₂H₅NH₂ (CBSE Delhi 2019)
Answer:
(CH₃)₃N < C₂H₅NH₂ < C₂H₅OH

Question 10.
Arrange the following in the increasing order of their pKb values.
C₆H₅NH₂, C₂H₅NH₂, C₆H₅NHCH₃ (CBSE AI 2018)
Answer:
C₂H₅NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂

Question 11.
Write IUPAC name of the following compound:
CH₃NHCH(CH₃)₂ (CBSE Delhi 2017)
Answer:
N-Methylpropan-2-amine

Question 12.
Write IUPAC name of the following compound: (CH₃CH₂)₂NCH₃ (CBSE Delhi 2017)
Answer:
N-Ethyl-N-methylhexanamine

Question 13.
Give a chemical test to distinguish between ethylamine and aniline. (CBSE AI 2011)
Answer:
Aniline gives azo dye test while ethylamine does not give azo dye test.

Question 14.
Arrange the following in increasing order of their basic strength:
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₆H₅)₂NH, and CH₃NH₂ (CBSE AI 2011)
Answer:
(C₆H₅)₂ NH < C₆H₅NH₂ < C₆H₅N (CH3)₂ < CH₃NH₂

Question 15.
Write the IUPAC name of the compound: (CBSE AI 2016)

Answer:
N-Methyl-2-methyl propanamine

Amines Important Extra Questions Short Answer Type

Question 1.
Give the structures of A, B, and C in the following reactions: (CBSE Delhi 2013)

Answer:

Answer:

Question 2.
Complete the following reaction equations:
(i) C₆H₅N₂CI + H₃PO₂ + H₂0 →
Answer:

(ii) C₆H₅NH₂ + Br₂(oq) → (CBSE 2012)
Answer:

Question 3.
Give the structures of products A, B, and C in the following reactions: (CBSE Delhi 2013)

Answer:

Answer:

Question 4.
Write the structures of A, B, and C in the following reactions: (CBSE 2016)

Answer:

Answer:

Question 5.
Write the structures of A, B, and C In the following: (CBSE Delhi 2016)

Answer:

Answer:

Question 6.
Give a chemical test to distinguish between ethylamine and aniline. (CBSE AI 2011)
Answer:
These can be distinguished by the azo dye test. Dissolve the compound in a cone. HCl and add an ice-cold solution of HNO₂ (NaNO₂ + dil. HCl) and then treat it with an alkaline solution of 2-naphthol. The appearance of brilliant orange or red dye indicates aniline.


Ethylamine does not form a dye. It will give brisk effervescence due to the evolution of N₂ but the solution remains clear.

Question 7.
How will you distinguish between the following? Give one chemical test:
(i) Aniline and benzylamine
Answer:
These can be distinguished by the azo dye test. Aniline reacts with HNO₂ (NaNO₂ + dil. HCl) at 273-278K to form stable benzene diazonium chloride which on treatment with an alkaline solution of 2-naphthol gives an orange dye (as given above).
Benzylamine does not give azo dye tests.

(ii) Aniline and N-methylaniline. (CBSE AI 2010)
Answer:
These can be distinguished by carbylamine test. Aniline being primary amines gives carbylamine test, i.e. when heated with an alcoholic solution of KOH and CHCl₃, it gives the foul smell of phenyl isocyanide.

Question 8.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B, and C. (CBSE Sample Paper 2011)
Answer:
(i) Since the compound C of molecular formula C₆H₇N is formed from B on treatment with Br₂ and KOH (Hoffmann bromamide reaction), therefore, the compound ‘B’ must be an amide and ‘C’ must be an amine. The only aromatic amine having molecular formula C₆H₇N is C₆H₅NH₂ (aniline).

(ii) Since ‘C’ is aniline, the amide from which it is formed must be benzamide (C₆H₅CONH₂).

Thus, B is benzamide.

(iii) Since B is formed from A with aqueous ammonia and heating, therefore, compound ‘A’ must be benzoic acid.

Thus, A = C₆H₅COOH, B = C₆H₅CONH₂, C = C₆H₅NH₂.

Question 9.
Account for the following:
(a) Gabriel phthalimide synthesis is not preferred for preparing aromatic primary amines.
Answer:
Gabriel phthalimide synthesis is not preferred for preparing aryl amines because aryl halides do not undergo nucleophilic substitution reaction with phthalimide.

(b) On reaction with benzene sulphonyl chloride, primary amine yields product soluble in alkali whereas secondary amine yields product insoluble in alkali. (CBSE Al 2019)
Answer:
Answer:
The sulphonamide formed by the reaction of secondary amine and benzene sulphonyl chloride does not contain any hydrogen atom attached to the N atom, so it is not acidic.
Therefore, it is insoluble in alkali.

Due to the presence of ‘H’ on nitrogen, it is soluble in alkali.

Since there is no ‘H’ on nitrogen, it is insoluble in alkali.

Amines Important Extra Questions Long Answer Type

Question 1.
Write equations of the following reactions:
(i) Acetylation of Aniline
Answer:
Acetylation of Aniline:

(ii) Coupling reaction
Answer:
Coupling reaction:

(iii) Carbylamine reaction (CBSE Delhi 2019)
Answer:
Carbylamine reaction:

Question 2.
An aromatic compound ‘A’ on heating with Br₂ and KOH forms a compound ‘B’ of molecular formula C₆H₇N which on reacting with CHCl₃ and alcoholic KOH produces a foul-smelling compound ‘C’.
Write the structures and IUPAC names of compounds A, B, and C. ((SSE Delhi 2019)
Answer:

A: C₆ H₅CONH₂: Benzamide B: C₆H₅NH₂: Benzenamine C: C₆H₅NC: Phenyt isocyanide

Question 3.
Write the structures of main products when benzene diazonium chloride reacts with the following reagents: (CBSE Delhi 2019)
(i) CuCN
Answer:

(ii) CH₃CH₂OH
Answer:

(iii) Kl
Answer:

Question 4.
(a) Write the product formed when
(i) 2-chioropropane is treated with aic. KOH.
Answer:

(ii) Aniline reacts with conc. H₂SO₄ at 453 – 473 K.
Answer:
Sulphonation: Sulphonation of aniline is carried out by heating aniline with sulphuric acid. The product formed is anilinium hydrogen sulfate which on heating gives sulphanilic acid.

The sulphanilic acid exists as a dipolar ion (structure II) which has acidic and basic groups in the same molecule. Such ions are called Zwitter ions or inner salts.

(b) When aniline is heated with CHCl₃ and aic. KOH, a foul-smelling compound is formed. What is this compound? (CBSE 2019C)
Answer:
Pheriyt isocyanide is formed.

Question 5.
(a) Identify ‘A’ and ‘B’ In the following reaction:

Answer:

(b) Why does aniline not undergo Friedel-Crafts reaction? (CBSE 2019C)
Answer:
Aniline being a Lewis base reacts with Lewis acid such as AlCl₃ to form a salt.

As a result, N of aniline acquires +ve charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Hence aniline does not undergo Friedel-Crafts reaction.

Question 6.
Complete the following reactions: (CBSE 2013)
(i) CH₃CH₂NH₂ + CHCl₃ + ale. KOH →
Answer:

(ii) C₆H₅N₂+Cl^– →
Answer:

Answer:

Question 7.
Write the main products of the following reactions: (CBSE 2013, CBSE AI 2013)

Answer:

Answer:

Answer:

Question 8.
Write the main products of the following reactions: (CBSE 2013)

Answer:

Answer:

Answer:

Question 9.
Account for the following:
(i) Primary amines (R-NH₂) have a higher boiling point than tertiary amines (R₃N).
(ii) Aniline does not undergo Friedel — Crafts reaction.
(iii) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.
OR
Give the structures of A, B, and C in the following reactions: (CBSE 2014)

Answer:
(i) Primary amines (RNH₂) have two hydrogen atoms on the N atom and therefore, form intermolecular hydrogen bonding.

Tertiary amines (R₃N) do not have hydrogen atoms on the N atom and therefore, these do not form hydrogen bonds. As a result of hydrogen bonding in primary amines, they have higher boiling points than tertiary amines of comparable molecular mass. For example, b.p. of n-butylamine is 351 K while that of tert-butylamine is 319 K.

(ii) Aniline being a Lewis base reacts with Lewis acid such as AlCl₃ to form a salt.

As a result, N of aniline acquires +ve charge and hence it acts as a strong deactivating group for electrophilic substitution reactions. Hence aniline does not undergo Friedel Crafts reaction.

(iii) Due to the presence of lone pair of electrons on the N atom, amines are basic in nature. The methyl group is the electron releasing group (+I inductive effect) and therefore, it increases the electron density on the N atom, and therefore, basic character increases, so that (CH₃)₃N should be more basic than (CH₃)₂NH. But tertiary ammonium ion formed from tertiary amines is less hydrated than secondary ammonium ion formed from secondary amine. Therefore, (CH₃)₃N has less tendency to form ammonium ion, and consequently, it is less basic than (CH₃)₂NH. Thus, (CH₃)₂NH is more basic than (CH₃)₃N due to the combined effect of inductive effect and hydration effect.
OR

Question 10.
Give the structures of A, B, and C in the following reactions:

OR
How will you convert the following:
(i) Nitrobenzene into aniline
(ii) Ethanoic acid into methanamine
(iii) Aniline into N-phenylethylamine
(Write the chemical equations involved.) (CBSE Delhi 2014)
Answer:

OR

Question 11.
Write chemical equations for the following conversions:
(i) Nitrobenzene to benzoic acid.
Answer:

(ii) Benzyl chloride to 2-phenytethanamine.
Answer:

(iii) Aniline to benzyl alcohol. (CBSE Delhi 2012)
Answer:

Question 12.
(a) Identify ‘A’ and ‘B’ in the following reaction:

Answer:

(b) Why is ethylamine soluble in water whereas aniline is not?
Answer:
Ethylamine dissolves in water due to intermolecular hydrogen bonding as shown below:

However, because of the large hydrophobic part (i.e. hydrocarbon part) of aniline, the extent of hydrogen bonding is less and therefore, aniline is insoluble in water.

Question 13.
(a) Identify X and Y in the following:

Answer:
X =
benzene diazonium ch(orlde,
Y =
Cyanobenzene

(b) Amino group is o, p-directing for aromatic electrophilic substitution reactions. Why does aniline on nitration give m-nitroaniline? (CBSE 2019C)
Answer:
Under strongly acidic conditions of nitration, most of the aniline is converted into anilinium ion having an NH₃⁺ group. This group is an m-directing group, therefore, m-nitro aniline is also obtained along with o- and p-products.

Question 14.
Give the structures of A, B, and C in the following reactions: (CBSE Delhi 2013)

Answer:

Answer:

Question 15.
Do as directed:
(i) Arrange the following compounds in the increasing order of their basic strength in an aqueous solution:
CH₃NH₃, (CH₃)₃N, (CH₃)₂NH.
Answer:
(CH₃)₃N < CH₃NH₂ < (CH₃)₂ NH

(ii) Identify ‘A’ and ‘B’:

Answer:
A: C6H5N2+ Cl- B: C6H5OH

(iii) Write the equation of carbylamine reaction. (CBSE 2018C)
Answer:

Question 16.
(i) Illustrate the following reactions giving a suitable example in each case:
(a) Hoffmann bromamide degradation reaction
(b) Diazotisation
(c) Gabriel phthalimide synthesis
(ii) Distinguish between the following pairs of compounds:
(a) Aniline and N-methylaniline
(b) (CH₃)₂NH and (CH₃)₃N
OR
(i) Write the structures of main products when benzene diazonium chloride (C₆H₅N₂⁺Cl^–) reacts with the following reagents:
(a) CuCN/KCN
(b) H₂0
(c) CH₃CH₂OH
(ii) Arrange the following:
(a) C₂H₅NH₂, C₂H₅OH, (CH₃)₃N – in the increasing order of their boiling point.
(b) Aniline, p-nitroaniline, p-methyl aniline – in the increasing order of their basic strength. (CBSE Delhi 2015)
Answer:
(i) (a) Hoffmann bromamide degradation reaction: Primary amines can be prepared from amides by treatment with Br₂ and KOH solution. The amine formed contains one carbon atom less than the parent amide.

(b) Diazotisation: The reaction of aniline or other aromatic amines, with nitrous acid at 0-5 °C to form diazonium salts is called diazotization. Nitrous acid needed for this reaction is prepared in situ by the action of dil. HCl on NaNO₂.

(c) Gabriel’s phthalimide synthesis. This method is used for preparing only primary amines. In this method, phthalimide is treated with alcoholic KOH to give potassium phthalimide, which is treated with an alkyl halide or benzyl halide to form N-alkyl or aryl phthalimide. The hydrolysis of N-alkyl phthalimide with 20% HCl under pressure or refluxing with NaOH gives primary amine.

Phthalic acid can again be converted into phthalimide and is used again and again. This method is very useful because it gives pure amines. Aryl halides cannot be converted to arylamines by Gabriel synthesis because they do not undergo nucleophilic substitution with potassium phthalimide.

(ii) (a) Add an alcoholic solution of KOH and CHCl3 to the compounds. Aniline gives the foul smell of isocyanide whereas N-methyl aniline does not give a foul smell.

(b) When treated with Hinsberg’s reagent (benzene sulphonyl chloride, C₆H₅SO₂CI), dimethylamine, (CH₃)₂NH gives precipitate which is insoluble in aqueous KOH.

(CH₃)₃N does not react with Hinsberg’s reagent.
Or

(ii) (a) (CH₃)₂ N < C₂H₅NH₂ < C₂H₅OH
(b) p-nitroaniline < aniline < p-methylaniline

Question 17.
Give the IUPAC names of the following compounds:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

(CBSE Delhi 2017)
Answer:

(CBSE Delhi 2017)
Answer:

Answer:

Question 18.
Complete the following reactions:

Answer:

Answer:

Answer:

Answer:

Answer:

Question 19.
Write the main products when benzene diazonium chloride (C₆H₅N₂⁺Cl^–) reacts with the following:
(i) CuCN/KCN
Answer:

(ii) H₂0
Answer:

(iii) CH₃CH₂OH (CBSE AI 2015, 2018)
Answer:

(iv) Copper powder/HCI
Answer:

Question 20.
Complete the following chemical equations:

(CBSE Delhi 2011)
Answer:

(CBSE Delhi 2011)
Answer:

(CBSE Delhi 2011)
Answer:

(CBSE AI 2013)
Answer:

(CBSE AI 2013)
Answer:

(CBSE AI 2013)
Answer:

Answer:

Question 21.
Give the structures of A, B, and C In the following reactions:

(CBSE AI 2017)
Answer:

(CBSE AI 2017)
Answer:

(CBSE Delhi 2016)
Answer:

(CBSE Delhi 2016)
Answer:

(CBSE AI 2016)
Answer:

(CBSE AI 2016)
Answer:

Answer:

Question 22.
An organic compound A’ with molecular formula C₇H₇NO reacts with Br₂/aq KOH to give compound B’, which upon reaction with NaNO₂ and HCI at OC gives C’. Compound C’ on heating with CH₃CH₂OH gives a hydrocarbon D’. Compound B’ on further reaction with Br₂ water gives a white precipitate of compound E’. Identify the compounds A, B, C, D, and E; also justify your answer by giving relevant chemical equations. (CBSE Sample Paper 2019)
OR
(a) How will you convert:
(i) Aniline into Fluorobenzene?
(ii) Benzamide into Benzylamine?
(iii) Ethanamine into N, N-Diethylethanamine?
(b) Write the structures of A and B in the following:

Answer:

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 9 Coordination Compounds.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 9 Important Extra Questions Coordination Compounds

Coordination Compounds Important Extra Questions Very Short Answer Type

Question 1.
Give an example of linkage isomerism. (CBSE Delhi 2010)
Answer:
[Co (NH₃)₅ (N0₂)] Cl₂ and [Co (NH₃)₅ (ONO)]Cl₂

Question 2.
Give an example of coordination isomerism. (CBSE Delhi 2010)
Answer:
[CO(NH₃)₆] [Cr(CN)₆] and [Cr(NH₃)₆] [Co (CN)₆]

Question 3.
Which of the following is more stable complex and why? (CBSE Delhi and Al 2014)
[CO(NH₃)₆]³⁺ and [Co(en)₃]³⁺
Answer:
[Co(en)₃]³⁺ because bidentate ligand (ethylenediamine) forms chelate which is more stable.

Question 4.
A coordination compound with molecular formula CrCl₃.4H₂0 precipitates one mole of AgCl with AgN0₃ solution. Its molar conductivity is found to be equivalent to two ions. What is the structural formula and name of the compound? (CBSE Sample Paper 2017-18)
Answer:
[CrCl₂(H₂0)₄]Cl: Tetraaquadichloridochrom ium(III) chloride.

Question 5.
What is coordination isomerism? Give one example. (CBSE Delhi 2010)
Answer:
Coordination isomerism arises when both the cation and anion are complexes and the ligands are exchanged. For example, [CO(NH₃)₆] [Cr(CN)₆] and [Cr(NH₃)₆] [Co(CN)₆]

Question 6.
Write IUPAC name of the complex [Co(en)₂(H₂0)(CN)]²⁺
OR
Using IUPAC norms, write the formula of Ammonium tetrafluoride cobaltates (II). (CBSE AI 2019)
Answer:
Aquacyanidobis (ethane-1,2-diamine) cobalt (III) ion
OR
(NH₄)₂[Co F₄]

Question 7.
Write the IUPAC name of the complex K₃[Cr(C₂0₄)₃].
OR
Using IUPAC norms write the formula of Hexaamminecobalt(III) sulphate. (CBSE AI 2019)
Answer:
Potassium trioxalatochromate(III)
OR
[Co(NH₃)₆]₂ (S0₄)₃

Question 8.
Write IUPAC name of the complex [Co(en)₂CI₂]
OR
Using IUPAC norms, write the formula of Sodium tetrachloridonickelate(II). (CBSE AI 2019)
Answer:
Dichloridobis (ethylenediamine) cobalt(III) ion.
OR
Na₂[NlCl₄]

Question 9.
Predict the number of unpaired electrons in the square planar [Pt(CN)₄]^2- ion. (CBSE 2019C)
Or
Amongst [Fe(C₂0₄)₃]^3- and [Fe(NH₃)₆]³⁺ which is more stable and why?
Answer:
The Pt (II) ion has 5d⁸ electronic configuration. For square planar geometry, dsp² hybridisation is involved. For this, one empty d-orbital is needed for hybridisation. Therefore, the pairing of electrons takes place in the remaining d-orbitals. Hence, there are no unpaired electrons in [Pt(CN)₄]^2- ion and it is diamagnetic.

OR
[Fe(C₂0₄)₃]^3- is more stable because it is a chelate. C₂0₄^2- is a didentate ligand.

When a didentate or a polydentate ligand uses its two or more donor atoms to bind to the same central metal atom or ion forming a ring structure it is called chelation.

The resulting complex has a ring structure and the ligand coordinating through two or more donor groups is called a chelating ligand. Some common examples of chelating ligands are carbonate ion, oxalate ion (ox^2-), ethylenediamine (en), ortho-phenanthroline (ph) and ethylenediaminetetraacetate ion (EDTA).

Chelate ligands form more stable complexes than similar ordinary complexes in which the ligand acts as monodentate. This is called a chelate effect.

Coordination Compounds Important Extra Questions Short Answer Type

Question 1.
Write IUPAC name of the complex [Pt(en)₂Cl₂]. Draw structures of geometrical isomers for this complex.
OR
Using IUPAC norms write the formulae for the following:
(i) Hexaamminecobalt(III) sulphate
(ii) Potassium trioxalatochromate(III) (CBSE Delhi 2019)
Answer:
Dichlorido bis(ethane-1,2-diamine) platinum (II)
Geometrical isomers.

OR
(i) [CO(NH₃)₆]₂(S0₄)₃
(ii) K₃[Cr(ox)]₃

Question 2.
Out of [COF₆]^3- and [Co(en)₃]³⁺, which one complex is
(i) paramagnetic
Answer:
[CoF₆]^3- is paramagnetic due to the presence of 4 unpaired electrons.

(ii) more stable
Answer:
[Co(en)₃]³⁺ is more stable because of chelation.

(iii) inner orbital complex and
Answer:
[Co(en)₃]³⁺ forms inner orbital complex involving d²sp³ hybridisation.

(iv) high spin complex (Atomic no. of Co = 27) (CBSE Delhi 2019)
Answer:
[COF₆]^3- forms a high spin complex (sp³d² hybridisation).

Question 3.
(i) Write down the IUPAC name of the following complex: (CBSE Delhi 2015)
[Cr(NH₃)₂Cl₂ (en)]Cl
(en = ethylenediamine)
Answer:
Diamminedichlorido(ethane-1,2,- diamine)chromium(III) chloride

(ii) Write the formula for the following complex:
Pentaamminenitrito-O-cobalt (III).
Answer:
[Co(NH₃)₅ (ONO)]²⁺

Question 4.
(i) Write down the IUPAC name of the following complex [Co(NH₃)₅(N0₂)] (N0₃)₂
Answer:
Pentaamminenitrito-N-cobalt(III) nitrate

(ii) Write the formula for the following complex: Potassium tetracyanonickelate (II) (CBSE 2015)
Answer:
K₂[Ni(CN)₄]

Question 5.
When a coordination compound CrCl₃.6H₂0 is mixed with AgN0₃, 2 moles of AgCI are precipitated per mole of the compound. Write (CBSE Delhi 2016)
(i) Structural formula of the complex.
Answer:
For one mole of the compound, two moles of AgCI are precipitated with AgN0₃, which indicates two ionisable chloride ions in the complex. Hence structural formula is [CrCl(H₂0)₅]Cl₂. H₂0

(ii) IUPAC name of the complex.
Answer:
Pentaaquachloridochromium (III) chloride hydrate

Question 6.
(i) For the complex [Fe(CN)₆]^3-, write the hybridisation type, magnetic character and spin nature of the complex. (At. number: Fe = 26).
Answer:
[Fe(CN)₆]^3-:
The oxidation state of Fe = +3
Fe(III)

Hybridisation: d²sp³
Magnetic character: Paramagnetic
Spin type: Low spin complex

(ii) Draw one of the geometrical isomers of the complex [Pt(en)₂Cl₂]²⁺ which is optically active. (CBSE Delhi 2016)
Answer:
cis- form is optically active

Question 7.
When a coordination compound NiCl₂.6H₂0 is mixed with AgNO₃, 2 moles of AgCI are precipitated per mole of the compound. Write (CBSE 2016)
(i) Structural formula of the complex
(ii) IUPAC name of the complex
Answer:
For one mole of the compound, 2 moles of AgCI are precipitated with AgNO₃, which indicates 2 ionisable Cl” ions in the complex.

1. Structural formula: [Ni (H₂0)₆]Cl₂
2. IUPAC name: Hexaaquanickel (II) chloride

Question 8.
Write IUPAC name of the complex [Cr(NH₃)₄Cl₂]⁺. Draw structures of geometrical isomers for this complex.
OR
Using IUPAC norms write the formulae for the following:
(i) Pentaamminenitrito-O-cobalt(III) chloride
(ii) Potassium tetracyanidonickelate(II) (CBSE Delhi 2019)
Answer:
IUPAC name: Tetramminedichloridochromi um(III) ion.
Geometrical isomers:

OR
(i) [CO(NH₃)₅(ONO)]Cl₂
(ii) K₂[Ni(CN)₄]

Question 9.
Write the hybridisation and magnetic character of the following complexes:
(i) [Fe(H₂O)₆]²⁺
(ii) [Fe(CO)₅] (Atomic no. of Fe = 26) (CBSE Delhi 2019)
Answer:
[Fe(H₂0)6]²⁺
Hybridisation: sp³d²
Magnetic character: Paramagnetic due to 4 unpaired electrons.
Fe(C0)₅
Hybridisation: dsp³
Magnetic character: It is diamagnetic.

Question 10.
Draw one of the geometrical isomers of the complex [Pt(en)₂Cl₂]²⁺ which is optically inactive. Also, write the name of this entity according to the IUPAC nomenclature.
Or
Discuss the bonding in the coordination entity [Co(NH₃)₆ ]³⁺ on the basis of valence bond theory. Also, comment on the geometry and spin of the given entity. (Atomic no. of Co= 27) (CBSE Sample Paper 2019)
Answer:

IUPAC Name of the entity:
Dichloridobis(ethane – 1,2-diamine) platinum(IV) ion
Or

Coordination Compounds Important Extra Questions Long Answer Type

Question 1.
(a) Draw the structures of geometrical isomers of [Fe(NH₃)₂ (CN)₄]^–
(b) [NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why? [Atomic number of Ni = 28]
OR
Define the following:
(a) Ambidentate ligands
(b) Spectra chemical series
(C) Heteroleptic complexes
Answer:

(b) In the presence of strong field ligand CO, the unpaired d-electrons of Ni pair up so [Ni(CO)₄] is diamagnetic but Cl^– being a weak ligand is unable to pair up the unpaired electrons, so [Ni(Cl₄)]^2- is paramagnetic.

and

OR
Answer:
(a) Ambidentate ligands: The monodentate ligands which can coordinate with the central atom through more than one site are called ambidentate ligands.
These ligands contain more than one coordinating atoms in their molecules.
For example, NO₂ can coordinate to the metal atom through N or 0 as

Similarly, CN^– can coordinate through C or N as

Thiocyanato (SCN) can coordinate through S or N

(b) The arrangement of ligands in the increasing order of crystal field splitting is called spectrochemical series. This is shown below:
I^– < Br^– < SCN^– < Cl^– < F^– < OH^– < Ox^2- < O^2- < H₂0 < NCS^– < py = NH₃ <en <NO₂^– < CN^– < CO.

Weak field ligands are those which cause less crystal field splitting. These form high spin complexes. For example, Cl^–, F^–, etc.

Strong field ligands are those which cause greater crystal field splitting. These form low spin complexes. For example, CN^–, NO₂^–, CO.

(c) The complexes in which the metal is bound to more than one kind of donor groups (ligands) are called heteroleptic complexes. Some common examples of heteroleptic complexes are [NiCl₂(H₂O)₄], [CoCl₂(NH₃)₄]⁺, etc.

Question 2.
Write the structures and names of all the stereoisomers of the following compounds:
(i) [Co(en)₃]Cl₃
Answer:

(ii) [Pt(NH₃)₂Cl₂]
Answer:

(iii) [Fe(NH₃)4Cl₂]Cl (CBSE 2011)
Answer:

Question 3.
Write the name, stereochemistry and magnetic behaviour of the following:
(i) K₄[Mn(CN)₆]
(ii) [Co(NH₃)₂ C1]Cl₂
(iii) K₂[Ni(CN)₄] (CBSE Delhi 2011)
Answer:

Complex	Name	Stereochemistry	Magnetic behaviour
(i) K4[Mn(CN)6]	Potassium hexacyanomanganate (I)	Octahedral	Paramagnetic
(ii) [Co(NH3)5Cl]Cl2	Pentaamminechtorido cobalt (III) chloride	Octahedral	Diamagnetic
(iii) K2[Ni(CN)4)	Potassium tetracyanonicketate (II)	Square planar	Diamagnetic

Question 4.
For the complex [Fe(en)₂Cl₂]Cl identify the following:
(i) Oxidation number of iron
Answer:
III.

(ii) Hybrid orbitals and shape of the complex
Answer:
d²sp³ hybridisation, octahedral

(iii) Magnetic behaviour of the complex
Answer:
Paramagnetic due to one unpaired electron

(iv) Number of its geometric isomers
Answer:
Two

(v) Whether there may be optical isomer also
Answer:
One optical, an isomer of cls-geometrical isomer.

(vi) Name the complex. (CBSE 2011)
Answer:
Dichloridobis(ethylenediamine) iron (III) chloride.

Question 5.
Give the formula of each of the following coordination entities:
(i) CO³⁺ ion is bound to one Cl^–, one NH₃ molecule and two bidentate ethylene diamine (en) molecules.
(ii) Ni²⁺ ¡on is bound to two water molecules and two oxalate ions. Write the name and magnetic behaviour of each of the above coordination entitles. (At. nos. CO = 27, Ni = 28) (CBSE 2012)
Answer:
(i) [Co(NH₃) (Cl) (en)₂]²⁺ Amminechtoriðo bls-(ethane -1, 2-diamine) cobalt (III) ion Co(27):

Since there are no unpaired electrons,
complex is diamagnetic.

(ii) [Ni(H₂O)₂ (C₂O₄)₂]^2- Diaquadioxatatonickelate (II) ion Ni(28):

The complex has two unpaired electrons, therefore, it will be paramagnetic.

Question 6.
State a reason for each of the following situations:
(i) CO²⁺ is easily oxidised to CO³⁺ in presence of a strong ligand.
Answer:
The configuration of CO²⁺ is t_2g⁶ e_g¹ and for CO³⁺, it is t_2g⁶. The crystal field stabilisation energy is more than compensated for the third ionisation enthalpy. Therefore, CO²⁺ is easily oxidised to CO³⁺ in the presence of a strong ligand.

(ii) CO is a stronger complexing reagent than NH₃.
Answer:
CO is a stronger complexing ligand than NH₃ because it contains both σ and π character and can form a back bond (M → CO) also. Therefore, CO forms a stronger bond with the metal. It is also called a strong field ligand.

(iii) The molecular shape of Ni(CO)₄ is not the same as that of [Ni(CN)₄]^2-. (CBSE Delhi 2012)
Answer:
The molecular shape of [Ni(C0)₄] is tetrahedral because this complex nickel involves sp³ hybridisation. In [Ni(CN)₄]^2-, nickel involves dsp² and its shape is square planer.

Question 7.
Name the following coordination entities and draw the structures of their stereoisomers:
(i) [Co(en)₂Cl₂]⁺(en = ethane-1, 2-diamine)
Answer:
(Co(en)₂Cl₂]⁺
Name: Dichloridobis (ethane-1, 2-diamine) cobalt(III) ion. It shows two geometrical isomers cis- and trAnswer:c/s-shows optical isomerism.


c/s-[Co(en)₂Cl₂]⁺ is optically active.

(ii) [Cr(C₂0₄)₃]^3-
Answer:
[Cr(C₂0₄)₃]^3-
Name: trioxalatochromate(III) ion. This complexion shows optical isomerism and their dextro and laevo forms are shown below:

(iii) [CO(NH₃)₃Cl₃]
(Atomic numbers Cr = 24, CO = 27) (CBSE 2012)
Answer:
[Co(NH₃)₃Cl₃]
Name Triamminetrichloridocobalt (III) It shows two geometrical isomers known as facial (fac) and meridional or merisomer as shown below:

Question 8.
Name the following coordination entities and describe their structures:
(i) [Fe(CN)₆]^4-
(ii) [Cr(NH₃)₄Cl₂]⁺
(iii) [Ni(CN)₄]^2-
(Atomic numbers Fe = 26, Cr = 24, Ni = 28) (CBSE 2012)
Answer:
(i) [Fe(CN)₆]^4-: Hexacyanoferrate (II) ion.
Structure. In this case, iron is in a +2 oxidation state having the outer electronic configuration 3d⁶. To account for the diamagnetic character, all the electrons get paired leaving two vacant orbitals.

The hybridisation scheme and bonding are as shown below:

(ii) [Cr(NH₃)₄Cl₂]⁺: Tetraamminedichlorido chromium (III) ion.

The chromium (Z = 24) has the electronic configuration 3d⁵ 4s¹. The chromium in this complex is in + 3 oxidation state and the ion is formed by the loss of one 4s and two of the 3d-electrons as shown in Fig. The inner d-orbitals are already vacant and two vacant 3d, one 4s and three 4p-orbitals are hybridised to form six d2sp3-hybrid orbitals. Six pairs of electrons one from each NH₃ molecule and Cl- ions (shown by xx) occupy the six vacant hybrid orbitals. The molecule has octahedral geometry.

Since the complex contains three unpaired electrons, it is paramagnetic.
Cr (III):

(iii) [Ni(CN)₄]^2-: Tetracyanonickelate (II) ion.

Question 9.
Write the IUPAC names of the following compounds:
(i) [Cr(NH₃)₃Cl₃]
Answer:
Triamminetrichloridochromium (III)

(ii) K₃[Fe(CN)₆]
Answer:
Potassium hexacyanoferrate (III)

(iii) [CoBr₂(en)₂]⁺, (en = ethylenediamine) (CBSE Delhi 2013)
Answer:
Dibromidobis(ethylenediamine) cobalt (III) ion

Question 10.
Write the types of isomerism exhibited by the following complexes:
(i) [Co(NH₃)₅Cl]S0₄
Answer:
ionisation isomerism

(ii) [Co(en)₃]³⁺
Answer:
optical isomerism

(iii) [Co(NH₃)₆] [Cr(CN)₆] (CBSE Delhi 2013)
Answer:
coordination isomerism

Question 11.
For the complex [NiCl₄]^2-, write
(i) the IUPAC name.
(ii) the hybridisation type.
(iii) the shape of the complex (Atomic no. of Ni = 28)
Or
What is meant by crystal field splitting energy? On the basis of crystal field theory, write the electronic configuration of d4 in terms to f2g and eg in an octahedral field when
(i) Δ_o > P
(ii) Δ_o < P (CBSE 2013)
Answer:
(i) Tetrachloridonickelate(II)ion
(ii) sp³ hybridisation
(iii) tetrahedral
OR
The difference between two sets of energy levels when the five degenerate d-orbitals splits in the presence of the electrical field of Ligands is called crystal field splitting energy. In the octahedral field, it is represented as Δ_o.

Electronic configurations of d⁴

• When Δ_o >P;t_2g⁴
• When Δ_o < P; t_2g³ e_g¹

Question 12.
Write the IUPAC name of the complex [Cr(NH₃)₄Cl₂]. What type of Isomerism does it exhibit? (CBSE Delhi 2014)
Answer:
Tetraammlnedlchloridochromium( III) ion. It shows cis and trans isomerism and cis form shows optical isomerism.

Question 13.
(i) Write the IUPAC name of the complex [Cr(NH₃)₄ Cl₂] Cl.
(ii) What type of Isomerism is exhibited by the complex [Co(en)₃]³⁺?
(en = ethane-1, 2-diamine)
(iii) Why Is [NIC(412 paramagnetic but [Ni(CO)₄] Is diamagnetic?
(At. nos.: Cr = 24, Co = 27, Ni = 28) (CBSE 2014)
Answer:
(i) TetraamminedichLorìdochromium (III) chloride.
(ii) Optical isomerism
(iii) In [NICl₄]^2- complexion, nickel is in +2 oxidation state and the configuration is 3d⁸. Since the molecule is tetrahedral, it involves sp³ hybridisation as shown below:

The molecule is paramagnetic because it contains two unpaired electrons. In [Ni(CO)₄], nickel is in O oxidation state and has the configuration 4s² 3d⁸ or 3d¹⁰. The molecule is tetrahedral and involves sp³-hybridisation as given below:

Each CO donates a pair of electrons forming four Ni—CO bond. The compound is diamagnetic since it contains no unpaired electron.

Question 14.
(i) Draw the geometrical isomers of the complex [Pt(NH₃)₂Cl₂].
Answer:

(ii) On the basis of crystal field theory, write the electronic configuration for d⁴ ion if Δ₀ < P.
Answer:
t_2g³ e_g¹

(iii) Write the hybridization and magnetic behaviour of the complex [Ni(CO)₄]. (At.no. of Ni = 28) (CBSE Delhi 2015)
Answer:
[Ni(CO)₄] involves sp³ hybridization of nickel and the complex is tetrahedral. Magnetic behaviour: Diamagnetic

Question 15.
(i) Draw the geometrical isomers of complex [Co(en)₂Cl₂]⁺.
Answer:
Geometrical isomers of complex [Co(en)₂Cl₂]⁺:

(ii) On the basis of crystal field theory, write the electronic configuration for d⁴ ion if Δ₀ > P.
Answer:
t_2g⁴

(iii) [NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28) (CBSE 2015)
Answer:
In [NiCl₄]^2- complexion, nickel is in +2 oxidation state and the configuration is 3d⁸. Since the molecule is tetrahedral, it involves sp³ hybridisation as shown below:

The molecule is paramagnetic because it contains two unpaired electrons.

In [Ni(CO)₄], nickel is in 0 oxidation state and has the configuration 4s² 3d⁴ or 3d¹⁰. The molecule is tetrahedral and involves sp³-hybridisation as given below :

Each CO donates a pair of electrons forming four Ni-CO bonds. The compound is diamagnetic since it contains no unpaired electron.

Question 16.
What type of isomerism is shown by the complex
(i) [Co (NH₃)₅ (SCN)]²⁺ (CBSE AI 2017)
Answer:
Linkage isomerism

(ii) [Co (NH₃)₆] [Cr (CN)₆] (CBSE Delhi 2017)
Answer:
Coordination isomerism

(iii) [Co (en)₃] Cl₃ (CBSE AI 2014)
Answer:
Optical isomerism

Question 17.
A metal ion Mⁿ⁺ having d⁴ valence electronic configuration combines with three bidentate ligands to form a complex compound. Assuming Δ_o > P:
(i) Write the electronic configuration of the d⁴ ion.
Answer:
t_2g⁴ e_g⁰

(ii) What type of hybridisation will Mn+ ion have?
Answer:
d²sp³

(iii) Name the type of isomerism exhibited by this complex. (CBSE Sample Paper 2017 – 18)
Answer:
Optical isomerism

Question 18.
Write the state of hybridisation, the shape and the magnetic behaviour of the following complex entities :
(a) [Cr(NH₃)₄Cl₂] Cl
(b) [Co(en)₃]Cl₃
(c) K₃[Ni(CN)₄] (CBSE AI 2011)
Answer:

Complex	Hybridisation	Shape	Magnetic behaviour
(a) [Cr(NH3)4Cl2]Cl	d2sp3	Octahedral	Paramagnetic
(b) [Co(en)3]Cl3	d2sp3	Octahedral	Diamagnetic
(c) K2[Ni(CN)4]	dsp2	Square planar	Diamagnetic

Question 19.
(a) Give one chemical test as evidence to show that [Co(NH₃)₅Cl] S0₄ and [Co(NH₃)₅5(S0₄)]Cl are ionisation isomers.
Answer:
When [Co(NH₃)₅(S0₄)]Cl is treated with silver nitrate solution, a white precipitate of AgCl is formed. But [Co(NH₃)₅Cl]S0₄ does not give white ppt with AgN0₃ solution.
[Co(NH₃)₅S0₃ ]Cl+AgN0₃ (Ag) → AgCl White ppt
[Co(NH₃)₅ Cl]S0₄ + AgN0₃ (Ag) → No white ppt

(b) [NiCl₄]^2- is paramagnetic while [Ni(C0)₄] is diamagnetic though both are tetrahedral. Why? (Atomic no. of Ni = 28)
Answer:
In [NiCl₄]^2-, the oxidation state of nickel is +2 and it has the electronic configuration: 3d8

Cl^– ion is a weak field ligand and it causes the pairing of electrons. Since it has two unpaired electrons, it is paramagnetic. It undergoes sp3 hybridisation forming a tetrahedral structure.

In Ni (CO)₄, the oxidation state of nickel is 0 and it has the electronic configuration: 3d⁸ 4s². CO is a strong field ligand and therefore, it causes the pairing of electrons as

Since it has no unpaired electron, it is diamagnetic. It also undergoes sp3 hybridisation resulting in tetrahedral geometry of the complex,

(c) Write the electronic configuration of Fe(III) on the basis of crystal field theory when it forms an octahedral complex in the presence of (i) strong field ligand, and (ii) weak field ligand. (Atomic no. of Fe = 26) (CBSEAI2019)
Answer:
Fe (III): 3d⁵
(i) Strong field ligand: t_2g⁵ e_g⁰
(ii) Weak field ligand: t_2g³ e_g²

Question 20.
A metal complex having the composition Cr (NH₃)₄Cl₂Br has been isolated in two forms A and B. Form A reacts with AgN0₃ to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.
(i) Write the formulae of isomers A and B.
Answer:
Isomer A: [Cr(NH₃)₄ BrCl]Cl
Isomer B: [Cr (NH₃)₄ Cl₂]Br

(ii) State the hybridisation of chromium in each of them.
Answer:
The hybridisation of Cr in isomer A and B is d²sp³.

(iii) Calculate the magnetic moment (spin only value) of the isomer A. (CBSE Sample Paper 2018)
Answer:
The number of unpaired electrons in Cr³⁺(3d³) is 3.
Magnetic moment = \(\sqrt{n(n+2)}\)
= \(\sqrt3(3 + 2)\) = 3.87 BM

Question 21.
(a) Write the formula of the following coordination compound: Iron(III) hexacyanoferrate(II)
Answer:
Fe4[Fe (CN)₆]₃

(b) What type of isomerism is exhibited by the complex [Co(NH₃)₅Cl]S0₄?
Answer:
Ionisation isomerism

(c) Write the hybridisation and number of unpaired electrons in the complex [COF₆]^3-. (Atomic No. of Co = 27) (CBSE 2018)
Answer:
sp³d², 4

Question 22.
Write the IUPAC names of the following :
(a) [Ag(NH₃)₂][Ag(CN)₂]
Answer:
diamminesilver(I)
dicyanidoargentate(I)

(b) K₃[Fe(C₂0₄)₃]
Answer:
potassium trioxalatoferrate(III)

(c) [CoCl₂(en)₂]Cl
Answer:
dichloridobis (ethane-1, 2-diamine) cobalt (III) chloride

(d) K₃[Cr(C₂0₄)₃]
Answer:
potassium trioxalatochromate(III)

(e) K₄[Ni(CN)₄]
Answer:
potassium tetracyanidonickelate (0)

(f) [Pt(NH₃)₂Cl(N0₂)]
Answer:
diamminechloridonitrito-N-platinum(II)

(g) [CoBr₂(en)₂]⁺ (CBSE Delhi 2012)
Answer:
dibromidobis (ethylenediamine) cobalt(III) ion

(h) [Co(NH₃)₅ONO]Cl₂
Answer:
pentaamminenitrito-O-cobalt(III) chloride

(i) [Co(NH₃)₅(N0₂)](N0₃)₂ (CBSE Al 2015)
Answer:
pentaamminenitrito-N-cobalt(III) nitrate

(j) [Cr(NH₃)₂Cl₂ (en)]Cl (CBSE Delhi 2015)
Answer:
diamminedichlorido(ethane-1, 2-diamine)chromium (III) chloride

Question 23.
Write the formulae of the following coordination compounds:
(a) hexaamminecobalt(III) sulphate
Answer:
[CO(NH₃)₆]₂(S0₄)₃

(b) potassium tetrachloridopalladate(II)
Answer:
K₂[PdCl₄]

(c) diamminechloridonitrito -N- platinum(II)
Answer:
[Pt(NH₃)₂Cl(N0₂)]

(d) pentaamminenitrito -N- cobalt(III)
Answer:
[CO(N0₂)(NH₃)₅]²⁺

(e) pentaamminenitrito – 0- cobalt(III) (CBSE Delhi 2015)
Answer:
[CO(ONO)(NH₃)₅]²⁺

(f) sodium dicyanidoaurate(I) (CBSE AI 2017)
Answer:
Na [Au (CN)₂]

(g) tetraamminechloridonitrito-N- platinum(IV) sulphate (CBSE AI 2017)
Answer:
[Pt(NH₃)₄Cl (N0₂)] S0₄

(h) potassium tetracyanidonickelate(II) (CBSE AI 2015)
Answer:
K₂[Ni(CN)₄]

(i) potassium trioxatatachromate(III)
Answer:
K₃[Cr(ox)₃]

(j) tetracarbonylnickel(O)
Answer:
[Ni(C0)₄]

Question 24.
Write IUPAC names of the following :
(i) K₃[AI(C₂0₄)₃]
Answer:
Potassium trioxalatoaluminate (III)

(ii) [Ni(CO)₄]
Answer:
tetracarbonylnickel(O)

(iii) Fe₄[Fe(CN)₆]₃
Answer:
iron hexacyanidoferrate(III)

(iv) [CoCI(NH₃)₅]Cl₂
Answer:
pentaamminechloridocobalt (III) chloride

(v) [PtCl₂(C₅H₅N)(NH₃)]
Answer:
amminedichlorido (pyridine) platinum(II)

(W) Na[PtBrCI(N0₂)(NH₃)]
Answer:
sodium amminebromidochloridonitri- to-N-platinate(II)

(v/i) [Cr(NH₃)₃Cl₃] (CBSE Delhi 2013)
Answer:
triamminetrichloridochromium (III)

(viii) K₃[Fe(CN)₆] (CBSE Delhi 2013)
Answer:
potassium hexacyanoferrate (III)

(ix) Na₃[AlF₆]
Answer:
sodium hexafluoridoaluminate (III).

(x) [Co(en)₃]₂(S04)₃ (CBSE Delhi 2013)
Answer:
tris(ethylenediamine)cobalt(III) sulphate

Question 25.
For the complex [Fe(en)₂Cl₂]Cl, identify the following:
(i) Oxidation number of iron.
Answer:
III.

(ii) Hybrid orbitals and shape of the complex.
Answer:
d²sp² hybridisation, octahedral.

(iii) Magnetic behaviour of the complex.
Answer:
Paramagnetic due to one unpaired electron.

(iv) Number of its geometrical isomers.
Answer:
Two.

(v) Whether there may be optical isomer also?
Answer:
An optical isomer of cis-geometrical isomer.

(vi) Name of the complex. (CBSE Delhi 2011)
Answer:
Dichloridobis (ethane-1, 2-diamine) iron (III) chloride.

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 2 Units and Measurements Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

Class 11 Physics Chapter 2 Important Extra Questions Units and Measurements

Units and Measurements Important Extra Questions Very Short Answer Type

Question 1.
If the size of the atom were enlarged to the tip of the sharp pin, how large would the height of Mount Everest be?
Answer:
10¹⁰ m.

Question 2.
What does the LASER mean?
Answer:
It stands for Light Amplification by Stimulated Emission of Radiation.

Question 3.
If the Universe were shrunk to the size of the Earth, how large would the Earth be on this scale?
Answer:
1o^-11 m (size of an atom.).

Question 4.
A research worker takes 100 careful readings in an experiment. If he repeats the same experiment by taking 400 readings, then by what factor will the probable error be reduced?
Answer:
By a factor of 4.

Question 5.
What is the number of significant figures in 0.06070?
Answer:
4.

Question 6.
Which of the following reading is most accurate?
(a) 7000m,
(b) 7 × 10² m,
(c) 7 × 10³ m
Answer:
(a) i.e. 7000 m.

Question 7.
The density of a cube is calculated by measuring the length of one side and its mass. If the maximum errors in the measurement of mass and length are 3% and 2% respectively, then what is the maximum possible error in the measurement of density?
Answer:
3% + 3 × 2% = 9%.

Question 8.
The mass of a body as measured by two students is given as 1.2 kg and 1.23 kg. Which of the two is more accurate and why?
Answer:
The second measurement is more accurate as it has been made to the second decimal point.

Question 9.
Do the inertial and gravitational masses of ordinary objects differ in magnitude?
Answer:
No.

Question 10.
Are S.I. units Coherent? Why?
Answer:
Yes, because all the derived units in this system can be obtained by multiplying or dividing a certain set of basic units.

Question 11.
Do A.U. And Å represents the same magnitudes of distance?
Answer:
No, 1 A.U. = 1.496 × 10¹¹ m and 1 Å = 10¹⁰ m.

Question 12.
What does SONAR stand for?
Answer:
It stands for Sound Navigation and Ranging.

Question 13.
What is the atomic mass unit (a.m.u.)?
Answer:
It is defined as \(\frac{1}{12}\)th of the mass of one 6C12 atom
i.e. 1 a.m.u. = \(\frac{1}{12} \times \frac{12}{6.023 \times 10^{23}}\) = 1.66 × 10^-27 kg

Question 14.
Which is the most accurate clock?
Answer:
Cesium atomic clock.

Question 15.
Write the S.I. units of the following physical quantities:
(a) Luminous intensity
Answer:
Candela (cd)

(b) Temperature
Answer:
Kelvin (K)

(c) Electric current
Answer:
Ampere (A)

(d) Amount of substance
Answer:
Mole (mol)

(e) Plane angle
Answer:
Radian (rd)

(f) Solid angle
Answer:
Steradian (sr)

(g) Pressure.
Answer:
Nm-2 = pascal (pa).

Question 16.
What is the difference between mN, Nm, and nm?
Answer:

• mN means.milli-newton, 1 mN = 10^-3N.
• Nm means newton-meter, 1 Nm = 1 J
• nm means namometer, 1 nm = 10^-9 m.

Question 17.
If x = a + bt + ct² where x is in meter and t in seconds, what is the unit of c?
Answer:
The unit on the left-hand side is a meter so the units of ct² should also be a meter. Since t² has units of s², so the unit of c is ms^-2

Question 18.
Will the dimensions of a physical quantity be the same, whatever be the units in which it is measured? Why?
Answer:
Yes, the dimensions don’t depend on the system of units chosen.

Question 19.
Write the dimensions of:
(i) gravitational constant
Answer:
[M^-1 L³ T²]

(ii) Plank’s constant
Answer:
[M L² T^-1]

(iii) torque
Answer:
[M L² T²]

(iv) surface tension
Answer:
[M L⁰ T^-2]

(v) angular momentum.
Answer:
[M L² T^-1]

Question 20.
Name at least two physical quantities each having dimensions:
(a) [M L^-1 T^-2]
Answer:
Pressure and stress,

(b) [M L² T^-1]
Answer:
Plank’s constant and angular momentum.

Question 21.
State the principle of homogeneity of dimensions?
Answer:
It states that the dimensions of each term on both sides of an equation are the same.

Question 22.
Which are the main types of errors in a physical measurement?
Answer:
Main errors are systematic error, random error, gross error, relative error, and percentage error.

Question 23.
Which one is large, the number of microseconds in a second or the number of seconds in a year?
The number of seconds in a year = 10⁷s and the number of microseconds in a second = 10⁶μs. So the number of seconds in a year is larger than microseconds in a second.

Question 24.
Do significant figures change if the physical quantity is measured in different systems of units?
Answer:
No, significant figures don’t depend on the system of units. e.s. 250 g = 2.50 × 10^-1 kg.
Both have 3 significant figures.

Question 25.
Suggest a distance corresponding to each of the following order of length:
(a) 10^-4 m
Answer:
Size of the atomic nucleus

(b) 10^-9 m
Answer:
Size of the oil molecule

(c) 10⁴ m
Answer:
Height of Mount Everest

(d) 10⁷ m
Answer:
Radius of Earth

(e) 10⁹ m.
Answer:
The radius of Sun.

Question 26.
What do you understand by the following?
(a) Century
Answer:
It is the largést unit of time, 1 century = 100 years.

(b) Shake
Answer:
It is the smallest unit of time, 1 shake = 10^-8s.

(c) Lunar month
Answer:
It is the time taken by the moon to complete one revolution around the Earth, 1 lunar month = 27.3 days.

(d) Leap year
Answer:
A year that is divisible by four and in which the month of February is of 29 days is called a leap year.

(e) Tropical year.
Answer:
The year in which the total solar eclipse takes place is called a tropical year.

Question 27.
What is the role of power (index) of a measurable in the formula used for calculation of a quantity in regard to the error in the quantity as determined in the given experiment?
Answer:
The error in the quantity becomes power times the error,
i.e. if x = p^a q^b. Then
\(\frac{\Delta x}{x}=a \frac{\Delta p}{p}+b \frac{\Delta q}{q}\)

Question 28.
How will you find the size of a liquid molecule?
Answer:
Using the formula, t = \(\frac{\mathrm{nV}}{500 \mathrm{~A}}\) we can find the size of the liquid molecule.
where V = volume of liquid (say oleic acid),
n = no. of drops in the solution of \(\frac{1}{500}\) concentration
A = area of the film of oleic acid left
t = thickness of the film.

Question 29.
What do you mean by the term measurement?
Answer:
Measurement means the comparison of a physical quantity with its unit to find out how many times the unit is contained in the given physical quantity.

Question 30.
Sort out the incorrect representation of units and write them
(i) m/sec
Answer:
ms^-1

(ii) Newton
Answer:
newton

(iii) kelvin
Answer:
kelvin

(iv) m.m.
Answer:
mm

(v) Jk^-1
Answer:
JK^-1

(vi) kg/m³
Answer:
kgm^-3

(vii) wH
Answer:
Wh

(viii) gms^-2
Answer:
gs2

(ix) length = 5M
Answer:
length = 5 m

(x) B = 4g (B = magnetic field intensity).
Answer:
B = 4G

Question 31.
Define light year.
Answer:
It is defined as the distance traveled by light in one year.
1 L.Y. (ly) = 3 × 10⁸ ms^-1 × 365 × 24 × 60 × 60s ≈ 9.46 × 10¹⁵ m.

Question 32.
Define Astronomical distance.
Answer:
It is defined as the distance between the Earth and Sun.
1 A.U. = 1.496 × 10¹¹ m~ 1.5 × 10¹¹ m.

Question 33.
What is the limit of
(i) accuracy
Answer:
The least count of the measuring instrument is the limit of accuracy with which a physical quantity can be measured.

(ii) error?
Answer:
The error in measurement is taken equal to half the least count.

Question 34.
What do you mean by ‘Order of magnitude’?
Answer:
Order of magnitude is defined as the approximation to the nearest power of 10 used to express the magnitude of a physical quantity under consideration, e.g.

1. The order of magnitude of the time interval of 1.2 × 10^-6 s is -6.
2. The order of magnitude of the distance of 4.5 × 10⁶ is +6.

Question 35.
Find the order of magnitude of a light-year.
Answer:
I light year = 9.46 × 10¹⁵ m ≈ 10¹⁶m
∴ The order of magnitude of light-year is +16.

Question 36.
Derive the dimensional formula of:
(a) Angular velocity
Answer:
Angular velocity = \(\frac{\text { Angle }}{\text { Time }}=\frac{1}{\mathrm{~T}}\) = [M⁰ L⁰ T^-1]

(b) Angular momentum
Answer:
Angular momentum = momentot inertia × Angular velocity
= mass x (radius of rotalion)² (Time)^-1
= [M L² T^-1].

Question 37.
Derive the dimensional formula of:
(a) Impulse
Answer:
Impulse = Force x Time
= [M L T^-2][T]
= [M L² T^-1].

(b) Surface energy
Answer:
Surfäce energy = \(\frac{\text { Energy }}{\text { Surface area }}\)
= \(\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^{2}\right]}\)
= [M L⁰ T^-2].

Question 38.
Derive the dimensional formula of
(a) Specific gravity
Answer:
Specific gravity = \(\frac{\text { Density of substance }}{\text { Density of water at } 4^{\circ} \mathrm{C}}\)
= \(\frac{\left[\mathrm{ML}^{3} \mathrm{~T}^{0}\right]}{\left[\mathrm{ML}^{-3} \mathrm{~T}^{0}\right]}\)
= [M⁰ L⁰ T⁰]

(b) Coefficient of viscosity .
Answer:
Coefficient of viscosity = \(\frac{\text { Force } \times \text { Distance }}{\text { Area } \times \text { velocity }}\)
= \(\frac{\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]}{\left[\mathrm{L}^{2}\right]\left[\mathrm{LT}^{-1}\right]}\)
= [M L^-1 T^-1]

Question 39.
Derive the dimensional formula of:
(a) Universal gas constant
Answer:
Universal gas constant = \(\frac{\text { Pressure } \times \text { Volume }}{\text { Temperature }}\)
= \(\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{3}\right]}{[\mathrm{K}]}\)
= [M L²  T² K^-1]

(b) Specific heat.
Answer:
Specific heat = \(\frac{\text { Heat }}{\text { Mass } \times \text { Temperature }}\)
= \(\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{MK}]}\)
= [M⁰ L² T^-2 K^-1].

Question 40.
Derive the dimensional formula of:
(a) Coefficient of elasticity
Answer:
Coefficient of elasticity = \(\frac{\text { Stress }}{\text { Strain }}\)
= \(\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}{\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]}\)
= [M L^-1 T^-2]

(B) Boltzmann’s constant
Answer:
Boltzmann’s constant = \(\frac{\text { Energy }}{\text { Temperature }}\)
= \(\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{K}]}\)
= [M L² T^-2 K^-1]

Question 41.
Define dimensions of a physical quantity.
Answer:
They are defined as the powers to which the fundamental units of mass, length, and time have to be raised to obtain its units, e.g. derived unit of area is [M⁰ L² T⁰]. Thus its dimensions are 1 zero in mass, 2 in length, and zero in time.

Question 42.
Define the dimensional formula of a physical quantity.
Answer:
It is defined as an expression that shows which of the fundamental units and with what powers appear into the derived unit of a physical quantity.
e.g. dimensional formula of force is [M¹ L¹ T²].

Question 43.
Define dimensional equation of a physical quantity.
Answer:
It is defined as the equation obtained by equating the symbol of a physical quantity with its dimensional formula, e.g. [F] = [M L T^-2] is the dimensional equation of force.

Question 44.
Define one kilogram.
Answer:
It is the mass of platinum-iridium cylinder (90% Pt + 10% Ir) having its diameter equal to its height (both equal to 3.9 cm) kept in the International Bureau of Weights and Measures of Sevres near Paris.

Question 45.
Define one second.
Answer:
It is defined as the time interval occupied by 9, 192, 631, 770 vibrations corresponding to the transition between two hyperfine levels of cesium -133 (Cs¹³³) atom in the ground state.

Question 46.
Define one ampere.
Answer:
It is defined as that constant current which when flowing through two parallel, straight conductors of the infinite length of negligible cross-section held one meter apart in a vacuum produces a force of 2 × 10^-7 N/m between them.

Question 47.
Define Kelvin.
Answer:
It is defined as \(\frac{1}{273.16}\) fraction of the thermodynamic temperature at the triple point of water.

Question 48.
Define radian.
Answer:
It is defined as the angle made at the center of a circle by an arc of length equal to the radius of the circle.

Question 49.
Define steradian.
Answer:
It is defined as the solid angle made at the center of a sphere by an area cut from its surface whose area is equal to the square of the radius of the sphere.

Question 50.
Define one mole.
Answer:
It is defined as the amount of substance that contains the same number of elementary units {i.e. atoms) as there are atoms in 0.012 kg of carbon-12.

Question 51.
Define standard meter.
Answer:
It is defined to be equal to exactly 1650763.73 wavelengths of orange-red light emitted in vacuum by krypton-86 atom i.e. kr⁸⁶.
i. e. 1 metre = 16,50,763.73 wavelengths.
Or
It is also defined as the distance traveled by light in \(\frac{1}{299792458}\) second.

Here 299792458 ms^-1 is the exact value of the velocity of light.

For all practical purposes, c = 2.9 × 10⁸ ms^-1 = 3.0 × 10⁸ ms^-1.

Units and Measurements Important Extra Questions Short Answer Type

Question 1.
If the size of a nucleus is scaled up to the tip of a sharp pin, what roughly is the size of an atom?
Answer:
The size of a nucleus is in the range of 10^-15 m to 10^-14 m. The tip of a sharp pain may be taken to be in the range of 1o^-5 m to 10^-4 m. Thus we are scaling up the size of the nucleus by a factor of 10^-5/10^-15 = 10¹⁰. An atom roughly of size 10^-10 m will be scaled up to a rough size of 10^-10 × 10¹⁰ = 1 m. Thusanucleus in an átom is as small in size as the tip of a sharp pin placed at the center of a sphere of radius about a meter.

Question 2.
(a) What do you mean by physical quantity?
Answer:
It is defined as a quantity that can be measured, e.g. mass, length, time, etc.

(b) What do you understand by:
(i) Fundamental physical quantities?
Answer:
They are defined as those quantities which cannot be expressed in terms of other quantities and are independent of each other, e.g. mass, length, time.

(ii) Derived physical quantities?
Answer:
They are defined as the quantities which can be expressed in terms of fundamental quantities, e.g. velocity, acceleration, density, pressure, etc.

Question 3.
(a) Define the unit of a physical quantity.
Answer:
It is defined as the reference standard used to measure a physical quantity.

(b) Define:
(i) Fundamental units.
Answer:
They are defined as the units of fundamental quantities. They are independent of each other and are expressed by writing the letter of the fundamental quantity in a parenthesis.
e.g. Fundamental units of mass, length and time are [M], [L], [T] respectively.

(ii) Derived units.
Answer:
They are defined as those units which can be derived from fundamental units. They are expressed by writing the symbol of a derived quantity in a parenthesis.
e.g. D.U. of velocity = [u]
acceleration = [a]
pressure = [P]
work = [W] and so on.

Question 4.
Define one Candela.
Answer:
It is defined as the luminous intensity in a perpendicular direction of a surface of \(\frac{1}{600,000}\) square meter area of a black body at a temperature of freezing platinum (1773°C) under a pressure of 101,325 N/m².

It is the S.I. unit of luminous intensity.

Question 5.
What is the advantage of choosing wavelength of light radiation as standard of length?
Answer:

1. It can be easily made available in any standard laboratory as Krypton is available everywhere.
2. It is well defined and does not change with temperature, time, place or pressure, etc.
3. It is invariable.
4. It increases the accuracy of the measurement of length (1 part in 10⁹).

Question 6.
Which type of phenomenon can be used as a measure of time? Give two examples of it.
Answer:
Any phenomenon that repeats itself regularly at equal intervals of time can be used to measure time.

The examples are:

1. Rotation of earth – the time interval for one complete rotation is called a day.
2. Oscillations of a pendulum.

Question 7.
Find the number of times the heart of a human being beats in 10 years. Assume that the heartbeats once in 0.8s.
Answer:
In 0.8 s, the human heart makes one beat.

∴ In 1 s, the human heart makes = \(\frac{1}{0.8}=\frac{10}{8}\) beats.

∴ In 10 years, the human heart makes
= \(\frac{10}{8}\) × 365 × 24 × 60 × 60 beats.
= 3.942 × 10⁸ beats.

Question 8.
Why it is not possible to establish a physical relation involving more than three variables using the method of dimensions?
Answer:
The dimensional analysis fails to derive a relation involving more than three unknown variables. The reason is that there will be more than three unknown factors in that case whose values cannot be determined from the three relations which we get by comparing the powers of M, L, and T.

Question 9.
What is the difference between accurate and precise measurement?
Answer:
A given measurement is said to be accurate in relation to other similar measurements if the error involved in it is least.

A given measurement is said to be precise in relation to other similar measurements if it is taken with an instrument with the minimum least count.

Question 10.
Pick up the most accurate and most precise measurement out of (a) 50.0 m, (b) s.oe m, (e) 5.00 cm, (f) 5.00 mm.
Answer:
The following table gives the relative error in count of the given measurement:

l	Δl	Δl/l
50.00 m	0.01 m	0.01/50 = 0.2 × 10-3
5.00 m	0.01 m	0.01/5 = 0.2 × 10-2
5.00 cm	0.01 cm	0.01/5 = 0.2 × 10-2
5.00 mm	0.01 mm	0.01/5 = 0.2 × 10-2

Clearly, the first measurement is most accurate because the relative error in it is minimum. The fourth measurement is most precise because it is taken with an instrument having the minimum least count among the v given measurements.

Question 11.
Define one parsec.
Answer:
It is defined as the distance at which an arc of length equal to y one astronomical unit subtends an angle of one second at a point.


Note: 1 L.Y. = 6.3 × 10⁴ A.U.
1 parsec = 3.26 L.Y.

Question 12.
Define annual parallax.
Answer:
It is defined as the angle (θ) subtended by the basis at the object (0). It is also called the parallactic angle.

∴ Parallactic angle = \(\frac{\text { Length of arc }}{\text { Radius }}\)
Or
θ = \(\frac{b}{s}\)

∴ s = \(\frac{b}{θ}\)

The parallax method is used to measure the distance of stars.
Here, b = basis = E₁E₂ distance where E₁ and E₂ are the two observation points on earth. θ = angle made by the star at point O. To find θ, let us observe the star 0’ simultaneously and let Φ₁ and Φ₂ be the angles made at E₁ and E₂.
∴ θ = Φ₁ + Φ₂
∴ s = \(\frac{b}{\phi_{1}+\phi_{2}}\)

Question 13.
Give Avogadro’s method to measure distances of the order of 10^-10 m.
Answer:
It is the indirect method of measuring the distances of the order of 10-10 m which is the size of an atom i.e. small distances. An atom is a tiny sphere. When such atoms lie packed in any substance; empty spaces are left in between. According to Avogadro’s hypothesis, the volume of all atoms in one gram of substance is \(\frac{2}{3}\) of the volume occupied by one gram of the substance.

i.e. V’ = \(\frac{2}{3}\)V ….(i)
Where V = actual volume of one gram mass.
V’ = volume occupied by atoms in I gram mass.
ρ = density of the substance.
∴ V = \(\frac{1}{ρ}\) …(ii)

Let m = atomic weight of the substance.
N = Avogadro’s number

∴Number of atoms in I gram of the substance = \(\frac{N}{M}\)
If r be the radius of each atom, then
V’ = no. of atoms in I gram x volume of each atom
Or
V’ = \(\frac{N}{M}\) x \(\frac{4}{3}\) 4πr3 …. (iii)

∴ From (i), (ii), and (iii), we get

Thus, r the radius of an atom can be calculated from equation (iv).

Question 14.
What arè the characteristics of a standard unit?
Answer:
The characteristics of a standard unit are as follows:

1. It should be well defined.
2. It should be of the proper size.
3. It should be easily accessible.
4. It should be reproducible in all places.
5. It should not change with time, place, and physical conditions such as pressure, tèmperature, etc.
6. It should be widely accepted.

Question 15.
What are the advantages of the S.I. system?
Answer:
Following are the main advantages of the S.I. system over other systems of units i.e. (C.G.S, FPS, and MKS).
1. It is a coherent system of units i.e. a system based on a certain set of fundamental units from which all derived units are obtained by multiplication or division without introducing numerical factors i.e. units of a given quantity are related to one another by the power of 10. So the conversions are easy.

2. S.I. is a rational system of units as it assigns only one unit to a particular physical quantity e.g. Joule is the S.I. unit for all types of energies while MKS units of mechanical energy, heat energy, and electrical energy are Joule, calorie, and watt-hour respectively.

3. It is an absolute system of units: There are no gravitational units on the system. The use of factor ‘g’ is thus eliminated.

4. It is a metric system i.e. the multiples and submultiples of units are expressed as powers of 10.

Question 16.
Point out the measurable likely to create the maximum error in the following experimental measurement.
Young’s modulus ‘Y’ of the material of the beam is calculated using the relation Y = mgl³/4bd³δ
When w = mg, δ = depression, I, b, d = length, breadth, thickness.
Answer:

Thus clearly m, l, b, d and 8 introduce the maximum error in the measurement of Y.

Question 17.
Classify the physical quantities on basis of their dimensional formula.
Answer:
They are divided into the following four categories:
1. Dimensional variables: They are defined as the physical quantities which possess dimensions and have variable values e.g. Area, velocity, force, etc.

2. Non-dimensional variables: They are defined as the physical quantities which have no dimensions but have variable values, e.g. Angle, specific gravity, strain, sin0, cos0 (i.e. trigonometric functions).

3. Dimensional constants: They are defined as the physical quantities which have both dimensions and constant values, e.g. Plank’s constant, speed of light in vacuum, gravitational constant.

4. Dimensionless constants: They are defined as the physical quantities which do not have dimensions but have constant values, e.g. n, e, pure numbers 1, 2, 3, …… etc.

Question 18.
What are the limitations of dimensional analysis?
Answer:
Following are the limitations of dimensional analysis:

1. Dimensionless constants involved in the physical relationship can not be determined.
2. It fails to derive the relations involving plus or minus signs like s = ut + \(\frac{1}{2}\)at²,
   v = u + at,
   v² — u² = 2as etc.
3. It fails to derive a relation involving more than three physical quantities.
4. This method does not help to derive the relations containing exponential and trigonometrical functions.
5. This method gives no information on whether a quantity is a scalar or vector.
6. It does not tell about the physical correctness of a relation.

Question 19.
Define significant figures. State the rules for determining the number of significant figures in a given measurement.
Answer:
They are defined as the number of digits up to which we are sure about their accuracy. In other words, they are defined as those digits that are known correctly in an experimental observation plus one more digit that is uncertain.

Following are the rules for determining significant figures:

1. All non-zero digits are significant.
2. All zeros occurring between non-zero digits are significant.
3. All zeros to the right of the last non-zero digit are not significant.
4. All zeros to the right of a decimal point and to the left of a non¬zero digit are not significant, e.g. 0.000879 has (3) significant figures.
5. All zeros to the right of a decimal point and to the right of a non-zero digit are significant, e.g. 0.2370 contains (4) sig-fig.
6. All zeros to the right of a non-zero digit and to the left of an expressed decimal are significant, e.g. 21900. has (5) sig-fig.

Question 20.
Mankind has existed for about 10⁶ years whereas the s. the universe is about 1010 years old. If the age of the universe is taken to be one day, how many seconds has mankind existed?
Answer:
Since

Mankind has existed for 8.64s on the new chosen scale.

Question 21.
State and explain the rule for finding the maximum possible error in a result.
Answer:
The maximum possible error is found in different ways in different types of results as follows:
(a) When the result involves the sum of two quantities,
i. e. if x = p + q
then, maximum possible error = maximum absolute error in first quality + maximum absolute error in the second quality. i.e. Δx = Δp + Δq

(b) When the result involves the difference of the quantities
i.e. if x = p – q
then, maximum possible error in x = maximum absolute error in p + maximum possible error in q
i.e. Δx = Δp + Δq

(c) When the result involves the multiplication of two quantities i.e. x = pq; then maximum relative error in x is given by
\(\frac{\Delta x}{x}=\frac{\Delta p}{p}+\frac{\Delta q}{q}\)

(d) When the result involves the quotient of tw«j observed quantities i.e. when x = p/q, then Maximum relative error in x is given by
\(\frac{\Delta x}{x}=\frac{\Delta p}{p}+\frac{\Delta q}{q}\)

(e) When the result involves some power oFa physical quantity, i.e. when x = pa qb, then maximum absolute e> u r in x is given by
\(\frac{\Delta x}{x}=a \frac{\Delta p}{p}+b \frac{\Delta q}{q}\)

i.e. maximum possible relative error in x = power × relative . error in p + power × relative error in q.

Thus the error is always additive in nature and maximum, permissible error is equal to the sum of maximum possible error in the i. individual quantities of that quantity.

Question 22.
How do you represent very large and very small physical y quantities? Write the prefixes, multiple, submultiple, and their symbols in a tabular form.
Answer:
The very long and very small quantities are written in powers of 10. The prefixes are tabled below:

S.No. Power of 10	Prefix	Symbol
1. 1012	Tera (Trillion)	T
2. 109	Giga (billion)	G
3. 106	mega(million)	M
4. 103	kilo	K
5. 102	Hecta	h
6. 101	Deca	da
7. 10-1	Deci	d
8. 10-2	Centi	c
9. 10-3	mull	m
10. 10-6	micro	u
11. 10-9	nano	n
12. 10-12	pico	p
13. 10-15	femto	f
14. 10-18	Atto	a

Question 23.
Explain the importance of reference frames in measurements.
Answer:
(a) All measurements are made with reference to a point or portion i.e., a frame of reference.
(b) The number of time units contained in a physical quantity gets changed with a change in the reference frame.
(c) The same physical quantity may have different values in different reference frames.

Question 24.
Briefly describe the various techniques to measure time.
Answer:
The various techniques to measure time are:
(a) The synchronous motor run on a.c. of frequency 50 hertz is used to measure time as its rotation provides a time scale.

(b) Electronic oscillations: The semiconductor oscillators produce high-frequency oscillations i.e. of a very small time period. These oscillations thus can be used to measure the small time intervals.

(c) Quartz-crystal clocks: These clocks work on the piezoelectric effect. The oscillations so produced can be used to measure time intervals.

(d) Atomic clock: These are usually Cs-133 atom clocks whose electrons make a definite number of repeated jumps between two energy levels. These are very precise and are used to define second, so time can be measured.

(e) Decay of elementary particles: The study of decay can provide a scale for the measurement of very small intervals of time as unstable elementary particles decay between 10^-16 to 10^-24 seconds.

(f) Radioactive dating: Very long intervals of time can be measured by comparing them with the techniques of radioactive dating. The half-life period of decay of carbon is a standard time frame to determine the age of fossils etc.

Question 25.
Explain the rules for finding significant figures in the sum, difference, product, and quotients of true numbers.
Answer:
The rules for counting significant figures in algebraic operations are given below:
1. Addition and Subtraction: The sum or the difference of two numbers has significant figures only in those places where these are in the least precise amongst the given number. For example, if we subtract 45.7 from 46.9267 the result is 1.2267. But it should be written only 1.2 because the least precise of the two numbers is 45.7 and there is only one digit after the decimal.

Similarly in the sum of numbers 4205, 112.39, 77.93, and 213.2532, the correct result is 824.0 to the significant figures. So, in addition, or subtraction the same number of the decimal. places are retained in the result as are present in the number with the minimum amount of decimal places.

2. Multiplication and Division: The product or quotient of two numbers does not have more significant figures than are present in the least precise of the given numbers. e.g. in the product of two quantities 0.025 with 40, we get 1.000 but the answer is to be written as 1.0 because there are two significant figures in 40 the least of the two numbers. Similarly in a quotient when we divide 16.775 by 2.5, we get 6.71. The result of the significant figure will be 6.7.

Question 26.
Is it possible for an equation to be dimensionally correct still to be wrong? If so indicate the number of ways in which this might happen.
Answer:
It is possible that an equation may be dimensionally correct but physically it is wrong. For example the displacement of a particle moving with velocity u and acceleration ‘a’ after time t is given by
s = ut + 1/2at²

However, s = 1/2at² is dimensionally correct, as [L] = [LT²][T²]² shows dimensions on LHS = dimensions on RHS. Therefore, in certain circumstances, an equation may be dimensionally correct but actually, it is not physically correct. This happens especially in equations involving the sum and difference of two or more terms and in functions involving trigonometric functions. For example, the displacement y maybe y = a where a is the amplitude. This is dimensionally correct but does not give the full picture of the physical phenomenon.

The correct equation is

y = a sin wt or y = a sin \(\frac{2 \pi}{\lambda}\) (vt + x) etc.

Question 27.
Show that bigger is the unit smaller is the numerical value of physical quantity and vice-versa.
Answer:
In any system of unit, the following relations hold good
n₁u₁ = n₂u₂

Where n₁ and n₂ are the numerical values, u₁ and u₂ represent the unit of a physical quantity. This relation is based on the simple thing that the magnitude of a physical quantity remains the same in different systems of units.
Thus nu = constant or n ∝ \(\frac{1}{u}\)

If larger the n smaller will be the unit (u) and smaller the numerical value (n) Larger will be u. E.g. if the length of a rod l be 1 m in the S.I. system. Then it is 100 cm in the C.G.S. system i.e.

l = 1 m = 100 cm. Here clearly 1 < 100 and m > cm. Hence proved.

Question 28.
The difference in the order of magnitude of the longest and shortest distances and most massive and least massive objects are quite different. Write the same and compare them.
Answer:
The size of the nucleus is of the order 10^-15 m and the intergalactic distance is of the order of 10²² m. So, the ratio between the longest and the shortest distance is:
\(\frac{10^{22} \mathrm{~m}}{10^{-15} \mathrm{~m}}\) = 10³⁷

The smallest mass is the mass of an electron of the order of 10^-30 kg and the longest mass is the mass of the galaxy of the order of 10⁴² kg. So, the ratio of maximum to minimum mass is
\(\frac{10^{42} \mathrm{~kg}}{10^{-30} \mathrm{~kg}}\) = 10⁷²

Now the ratio between the mass ratio and distance ratio is
\(\frac{10^{72}}{10^{37}}\) = 10³⁵

The ratio betwe< n minimum mass and minimum distance is
\(\frac{10^{-30} \mathrm{~kg}}{10^{-15} \mathrm{~m}}\) = 10^-15

The ratio between maximum mass and minimum distance is
\(\frac{10^{42} \mathrm{~kg}}{10^{22} \mathrm{~m}}\) = 10²⁰

Question 29.
Define the terms
(i) mean absolute error
Answer:
The arithmetic means of all the absolute errors is known as mean absolute error.

(ii) relative error and
Answer:
The relative error is the ratio of mean absolute error to the mean or true value of the quantity measured.
Relative error = \(\frac{\Delta \mathrm{a}_{\text {mean }}}{\mathrm{a}_{\text {mean }}}\)

(iii) percentage error.
Answer:
Percentage error is the relative error expressed in percent
Percentage error = \(\frac{\Delta \mathrm{a}_{\text {mean }}}{\mathrm{a}_{\text {mean }}}\) × 100

Question 30.
State the rules applied in rounding off measurements.
Answer:
While rounding off measurements, the following rules are applied:
1. If the digit to be dropped is smaller than 5, then the preceding digit should be left unchanged.
e.g. 7.93 is rounded off to 7.9.

2. If the digit to be dropped is greater than 5, then the preceding digit should be raised by 1.
e.g. 17.26 is rounded off to 17.3.

3. If the digit to be dropped in 5 followed by digits other than zero, then the preceding digit should be raised by 1.
e.g. 7.351 on being rounded off to first decimal becomes 7.4.

4. If the digit to be dropped in 5 or 5 followed by zeros, then the preceding digits are not changed if it is even.
e.g. (a) 3.45 on being rounded off becomes 3.4.
(b) 3.450 on being rounded off becomes 3.4.

5. If the digit to be dropped in 5 or 5 followed by zeros, then the preceding digit is raised by one if it is odd.
e.g. (a) 3.35 on being rounded off becomes 3.4.
(b) 3.50 on being rounded off becomes 3.4.

Note: Rules (iv) are based on the convention that the number is to be rounded off to the nearest even number.

Units and Measurements Important Extra Questions Long Answer Type

Question 1.
State the rules for writing the units of physical quantities in the S.I. system.
Answer:
While writing the units of physical quantities following rules are followed with S.L units:
(1) The S.l. units are written in the form of symbols after the number i.e., number of time, the unit is contained in the physical quantity so that physical quantity = nu

With symbols, certain rules are laid down:

• Units in symbols are never written in plural i.e., meters is only m and not ms, years is y.
• The units based on the name of the scientists are written beginning with small letters and with capital letters in symbolic form viz, weber (Wb), newton (N), etc.
• No full stop is used at the end of the symbol.
• Symbols of units not based on the name of scientists are written as small letters viz. kilogram (kg), second (s), etc.

(2) Bigger and smaller number of units are represented with symbols corresponding to the power of 10 viz. 10⁶ is mega (M), 10¹² is Tera (T), 10^-3 is milli (m), 10^-9 is nano (n), etc.

(3) All units are written in numerator viz. kg/m³ is kg m, Nm²c².

(4) The units are written within parenthesis in graphs below the corresponding taxes viz. (ms^-1) and (s) in the velocity-time graph.

(5) Units of a similar physical quantity can be added or subtracted.

Question 2.
Explain the Triangular method.
Answer:
It is used to measure the distance of an accessible or inaccessible hill or a tower by measuring the angle which the object makes at point P (say)

Let x = distancy of point P from the foot of tower = PA .
∴ h = x tan θ

It is also used to measure the distance of an inaccessible object eg. a tree on the other bank of a river.

Let h = height of the inaccessible object.
Let θ₁, θ₂ = be the angle made at P and Q by the object.
Let PA = d, PQ = x.

∴ In ΔPAB and ΔQAB,

Question 3.
What are the uses of dimensional analysis? Explain each of them.
Answer:
Dimensional analysis is used for:
(a) checking the dimensional correctness of the given physical equation or relation.
(b) converting one system of units to another system.
(c) deriving the relationship between various physical quantities.

(a) checking of the dimensional correctness of a physical relationship is done by using the principle of homogeneity of dimensions. If the dimensions of M, L, T of each term on R.H.S. are equal to the dimensions of M, L, T of each term on L.H.S., then the given- physical relation is dimensionally correct, otherwise wrong.

(b) conversion: It is based on the fact that the magnitude of a physical quantity remains the same whatever may be the system of units, i.e. n₁u₁ = n₂u₂.

are the units of M, L, T in the first and second system of units of a physical quantity having dimensions of M, L, T, and a, b, c respectively.

Thus if fundamental units of both systems, dimensions of the quantity, and its numerical value n₁ in one system, are known then we can easily calculate n₂ in another system.

(c) Derivation of a relationship between various physical quantities is based on the principle of homogeneity of dimensions.

Following are the steps used:

1. We must Know the physical quantities (say p, q, r) upon which a physical quantity say x depends.
2. We must know the dimensions of p, q, r say a, b, c respectively.
3. 
4. Now, write the dimensions of each physical quantity on both sides of the equation
5. and compare the powers of M, L, T to find a, b, c. Putting values of a, b, c in the equation
6. we get the required relation.

Numerical Problems:

Question 1.
The average distance from Earth to the Sun is 1.49 × 10¹¹ m. Find out the value of 1 percent in m.
Answer:
According to the definition of parsec,

Question 2.
The parallax of a heavenly body measured from two points: diametrically opposite on the Earth’s equator is 60 seconds. If the radius of the Earth is 6.4 × 10⁶ m, determine the distance of the heavenly body from the center of Earth. Convert this distance in AU.
Answer:
Here

∴ The distance of the heavenly body from Earth is given by

Question 3.
Find the height of a rock mountain if on moving 100 m towards the rock in the horizontal direction through the base of
the rock, the angle of elevation of itš top increases from 300 to 45°.
Answer:
Here d = 100 m, θ₁ = 30°, θ₂ = 45°, h =?

Question 4.
Find the number of air molecules in a room of volume 12 m³. Given 1 mole of air at N.T.P. occupies a volume of 22.4 liters.
Answer:
No. of atoms in I mole of air = N
= 6.023 × 10²³

∴ Also at N.T.P., 1 mole of air occupies 22.4 liters of volume
= 22.4 × 10^-3 m³

∴ No. of molectiles ir 22.4 × 10^-3 m³ volume = 6.023 × 10²³
∴ No. of molecules in 12 m³ volume = \(\frac{6.023 \times 10^{23}}{22.4 \times 10^{-3}}\) × 12
= 3.23 × 10²⁶

Question 5.
(a) Convert ION into dyne using dimensional analysis.
Answer:
Newton (N) and dyne are the S.I. and C.G.S. units of force having dimensional formula [MLT^-2]
∴ a = 1, b = 1, c = – 2

S.I. system
n₁ = 10
M₁ = 1 kg
L₁ = 1 m
T₁ = 1 s

C.G.S. system.
n₂ = ?
m₂ = 1 g
L₂= 1 cm
T₂ = 1 s

(b) Find the units of length, mass, and time of the unit of force, velocity, and energy are 100 dynes, 10 cms^-1, and 500 erg respectively.
Answer:

Question 6.
Check the dimensional correctnesïof the relation
V = \(\sqrt{\frac{2 G M}{R}}\)
Answer:
It is done using the principle of homogeneity of dimensions i.e. if the dimensions of each term of both sides of the equation are the same, Then it is dimensionally correct.
Now dimensions of V = [LT^-1].
G = [M^-1 L³ T²]
M = [M]
R = [L]

∴ Dimensions of L.H.S. = [V] = [LT^-1] …(f)
Dimensions of R.H.S = \(\sqrt{\frac{G M}{R}}\)

Now from equations (i) and (ii), it is clear that the dimensions of L.H.S = R.H.S, so the given relation is dimensionally correct.

Question 7.
Suppose that the oscillatións of a simple pendulum depend on
(i) mass of the bob (m),
(ii) length of the string (1),
(iii) acceleration due to gravity (g) and (iv) angular displacement
(iv) Dimensionally show which of the above factors here an influence upon the period and in what way?
Answer:
Let t ∝ m^a l^b g^c θ^d
or t = k’ m^a l^b g^c q^d ….(i)
here k’ is a dimensionless constant.

Since q is dimensionless, hence equation (i) reduces to
t = K m^a l^b g^c …(ii)

where K. = k¹.θ^d is another dimensionless constant.
Now writing the dimensional formula of each physical quantity on both sides of equation (ii), we get
[M⁰ L⁰ T^-1] = [M]^a [L]^b [LT²]^c = [M^a L^b-c T^-2c]

Comparing dimensions of


Thus from equation (vii), we see that the period of the pendulum is directly proportional to the square root of the length of string, and inversely proportional to the square root of acceleration due to gravity and is independent of the mass of the bob.

Question 8.
Given that the period T of oscillation of a gas bubble from an explosion underwater depends on P, d, and E, where the symbols are pressure, density, and total energy of the explosion. Find dimensionally a relation for T.
Answer:
Let T ∝p^a d^b E^c
or
T = k p^a d^b E^c …(i)
where k is a dimensionless constant.

Writing dimensional formula of each physical quantity on both sides of equation (i), we get

Question 9.
If the velocity of light c, the constant of gravitation G, and Plank’s constant h be chosen as fundamental units, find the value of a gram, a centimeter, and a second in terms of new units of mass, length, and time respectively.
Given c = 3 × 10¹⁰ cms^-1
G = 6.67 × 10^-8 dyne cm² g^-2
h = 6.6 × 10^-27 ergs.
Answer:

Question 10.
Assuming that the mass m of the largest stone that can be moved by a flowing river depends on the velocity y density p and acceleratión due to gravity g. Show that m varies as the sixth power of the velocity of flow.
Answer:
Let, m ∝ v^a
m ∝ρ^bb

where k is a dimensionless constant.

Writing dimensions of each physical quantity on both sides of equation (i), we get
[M¹ L⁰ T⁰] = [LT^-1] [ML^-3]^b [LT^-2]^c
[M^b L ^a-3b+2c T^-a-2c]

Comparing powers of:

Question 11.
The radius of Earth is 6.37 × 106 m and its average density is 5.517 × 10³ kg m^-3. Calculate the mass of Earth to correct significant figures.
Answer:
M = pV
Here, R = 6.37 × 10⁶ m,
ρ = 5.517 × 10³ kg m³

Here R has 3 significant figures and the density has four. Thus the final result should be rounded off to 3 significant figures.

Hence, M = 5.97 × 10²⁴ kg.

Question 12.
The length, breadth, and thickness of a block of metal were measured with the help of Vernier Callipers. The measurements are:
l = (5.250 ± 0.001) cm
b = (3.450 ± 0.001) cm,
t = (1.740 ± 0.001) cm.
Find the percentage error in the volume of the block.
Answer:
The volume of the block is given by:
V = lbt

∴ The relative percentage error in V is given by:

Question 13.
An experiment measures quantities a, b, c, and x is calculated from the relation:
x = ab²/c³
The percentage errors in a, b, c are ± 1%, ± 3%, and ± 2% respectively what is the % error in x?
Answer:

Question 14.
The radius of the proton is about 10^-9 microns and the radius of the universe is about 10²⁸ cm. Name a physical object whose size is approximately halfway between these two extremes on the logarithmic scale.
Answer:
Radius of proton, x₁ = 10^-9 microns = 10^-9 × 10⁶m .
= 10^-15m

Radius of universe, x₂ = 10²⁸ cm = 10²⁶m.

If x be the size halfway between these two extremes on the logarithmic scale is:

Since 10⁶m is the order of the size of the Earth. So the physical object is Earth.

Question 15.
The velocity v (cms^-1) of a particle is given in terms of time t (s) by the equation y = at + \(\frac{b}{t+c}\). What are the dimensions of a, b and c?
Answer:
v = at + \(\frac{b}{t+c}\).

L.H.S. is velocity, so according to principle of homogeneity t, of dimensions, at and must have the dimensions of \(\frac{b}{t+c}\) velocity.

Question 16.
The young’s modulus (Y) of a material is given by the relation Y = \(\frac{\mathbf{M g L}}{\pi \mathbf{r}^{2} \boldsymbol{l}}\). If the percentage error in W(= mg), L, r and l are 0.5%, 1%, 3% and 4% respectively, what is the percentage error in Y? If the calculated value of Y is 18.79 × 10¹¹ dyne cm^-2, to what value should we round off the result?
Answer:
The total percentage error in Y is given by:

Since Y = 18.79 × 10¹¹ dyne cm^-2, it can be rounded off to
Y = 19 × 10¹¹ dyne cm^-2

The total percentage error in this result is:
\(\frac{19 \times 11.5}{100}\) = 2.2

So, the result of Y can be quoted as (19 ± 2.2) × 10¹¹ dyne cm^-2

Question 17.
The refractive index as measured by relation:
μ = \(\frac{\text { real depth }}{\text { apparent depth }}\) was found to have values 1.29, 1.33, 1.34, 1.35, 1.32, 1.36, 1.30, 1.33.
Find the mean value of μ, the mean value of the absolute error, the relative error, and the percentage error.
Answer:
Mean value of μ

Question 18.
Derive the dimensions of a/b in the relation:
F = a\(\sqrt{x}\) + bt²
where F is the force, x is the distance and t is the time.
Answer:
Here, F= a\(\sqrt{x}\) + bt² …(i)

According to the principle of homogeneity of dimensions, a \(\sqrt{x}\) and bt² should have dimensions of F.

Question 19.
Calculate the number of seconds in a:
(i) day
(ii) year
and express them in orders of magnitude.
Answer:
(i) No.of seconds in a day = 24 × 60 × 60s
= 86400s
= 8.64 × 104s
= 0.864 × 105s
∴ order of magnitude = 5

(ii) No.of seconds in a year = 365 days
= 365 × 0.864 × 105s
= 315360 × 105s
= 31536 × 107s
∴ order of magnitude = 7.

Question 20.
A stone is lying in a fluid stream. The force F acting on it depends on the density of the fluid δ, the velocity of flow v, and the maximum area of cross-section A perpendicular to the direction of flow. Find the relation between the force F and the velocity v.
Answer:
Let, F ∝ ρ^a
∝ v^b
∝ A^c
or
F ∝ p^a v^b A^c
or
F = k p^a v^b A^c …(i)

Now writing the dimensional formula of each physical quantity on both sides of equation (i), we get:
[MLT^-2] = [ML^-3]^a [LT^-1]^b [L²]^c
= [M^a L^-3a+b+2c T^-b] …(ii)

Comparing dimensions of M, L, T on both sides of equation (ii), we get:

Question 21.
If we suppose the velocity of light (c), acceleration due to gravity (g), and pressure (p) as the fundamental units, then And the dimensional formula of mass in this system of units.
Answer:

Question 22.
The density p of a piece of metal of a mass m of and volume V is given by the formula, ρ = m/V.
If, m = 325.32 ± 0.0lg
V = 136.41 ± 0.01 cm³.
Find the percentage error in p:
Answer:
Here, m = 325.32g, Δm = 0.01 g
V = 136.41 cm³, ΔV = 0.01 cm³

Question 23.
The specific resistance σ of a circular wire of radius r cm, resistance R Ω and and length L is given by:
σ = \(\frac{\pi r^{2} R}{L}\)
If, r = 0.20 ± 0.02 cm
R = 20 ± 1Ω
L = 80 ± 0.01 cm, then find the % error in σ.
Answer:
% error in σ is given by:

Question 24.
If S² = a t⁴, where S is in meters, t in second, then find the unit of a.
Answer:
Here, S² = a t⁴
According to the principal of homogeneity of dimensions,
at⁴ = m²
∴ a = \(\frac{\mathrm{m}^{2}}{\mathrm{t}^{4}}\) = m² s^-4

Question 25.
Calculate the value of 600m. + 600mm with due regard to the significant digits.
Answer:
Here, because there is no significant digit after the decimal point in 600m, so 600m + 600mm.
= 600m + 600 × 10^-3m
= 600m + 0.600m
= 600.6m = 601m.

Value-Based Type:

Question 1.
The teacher of class XI asked Madan and Prathiva to find the distance of the moon from the Earth. Pratibha said it is impossible to find. But Madan was excited to know. He observed the moon from two diametrically opposite points A and B on Earth. The angle q subtended at the moon by the two directions of observation is 1°54′. Given the diameter of the Earth to be about (1.276 × 10⁷ m).
(i) Which values arc depicted by Madan?
Answer:
The values depicted by Madan are :
(a) Curiosity
(b) Sincerity
(c) Willing to know and implement the scientific ideas.

(ii) Which mathematical concept is used in the above problem?
Answer:
Parallax method

(iii) Compute the distance of the moon from the Earth?
Answer:

We have

Question 2.
Muesli went to London to his uncle who is a doctor over there. He found that the currency is quite different from India. He could not understand the pound and how it is converted into rupees. He asked an English man that how far is central London from here.

He replied that it is about 20 miles. Mukesh was again confused as he never used these units in India. When his uncle came back from his clinic he curiously inquired all about it. His uncle told him about the unit system used in England. He explained that here F.P.S system is used.

It means distance is measured in the foot, Mass in the pound, and time in seconds. But in India, it is an MKS system.
(i) What values are depicted by Mukesh?
Answer:
(a) Curiosity
(b) Willing to know
(c) Intelligence

(ii) How many types of unit systems are there?
Answer:
The unit system is:
(a) CGS (centimeter, gram and second) system
(b) FPS (foot, pound and second) system
(c) MKS (meter, kilogram, and second) system

Question 3.
Two friends Sachin and Dinesh are confused as a book with many printing errors contains four different formulas of the displacement ‘y’ of a particle undergoing a certain periodic motion.
(a) y=asin 2πt/T
(b) y = asinvt

[a = Maximum displacement of the particle,
v = Speed of the particle, T = time-period of motion]

Sachin told that he has no idea whereas Dinesh ruled out the wrong formulas on dimensional grounds.
(i) which values are displayed by Dinesh?
Answer:
the values displayed by Dinesh are:
(a) Sincerity
(b) Curiosity
(c) application of knowledge

(ii) which one of the above is correct? Give justification to support your answer.
Answer:


L.H.S = R.H.S ; Hence the equation is correct.

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 6 Electromagnetic Induction. 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 6 Important Extra Questions Electromagnetic Induction

Electromagnetic Induction Important Extra Questions Very Short Answer Type

Question 1.
What is the function of a step-up transformer? (CBSE AI 2011C)
Answer:
The function of a step-up transformer is to step-up the alternating voltage.

Question 2.
State Lenz’s law. (CBSE AI 2012C)
Answer:
It states that the direction of induced emf is such that it opposes the cause of its production.

Question 3.
How can the self-inductance of a given coil having ‘N’ number of turns, area of cross-section of ‘A’ and length T be increased? (CBSE AI 2012C)
Answer:
By inserting a core of high permeability inside the coil.

Question 4.
How does the mutual inductance of a pair of coils change when
(a) the distance between the coils is increased and
(b) the number of turns in the coils is increased? (CBSE AI 2013)
Answer:
(a) decreases
(b) increases.

Question 5.
The motion of the copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping? (CBSE AI 2013)
Answer:
Production of eddy current.

Question 6.
Why is the core of a transformer laminated? (CBSE Delhi 2013C)
Answer:
To reduce the effects of eddy currents.

Question 7.
A metallic piece gets hot when surrounded by a coil carrying a high-frequency alternating current. Why? (CBSE Delhi 2014C)
Answer:
Due to the production of eddy current which generates heat.

Question 8.
Name any two applications where eddy currents are used to advantage. (CBSE Delhi 2016C)
Answer:

1. Electromagnetic damping
2. Induction furnace.

Question 9.
A long straight current-carrying wire passes normally through the centre of the circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify. (CBSE Delhi 2017)
Answer:
Yes, as there will be a change in magnetic flux.

Question 10.
Predict the polarity of the capacitor in the situation described below. (CBSE AI 2017)

Answer:
The upper plate will be positive with respect to the lower plate in the capacitor.

Question 11.
In the figure given, mark the polarity of plates A and B of a capacitor when the magnets are quickly moved towards the coil. (CBSE AI 2017C)

Answer:
Plate A will be positive with respect to plate B in the capacitor.

Question 12.
A long straight current-carrying wire passes normally through the centre of the circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify. (CBSE Delhi 2017)
Answer:
Yes, as there will be a change in magnetic flux.

Question 13.
An air-cored solenoid has self-inductance 2.8 H. When the core is removed, the self-inductance becomes 2 mH. What is the relative permeability of the core used? (CBSE Delhi 2017C)
Answer:
μ_r = 2.8 / 2 × 10^-3 = 1.4 × 10³

Question 14.
A choke and a bulb are in series to a dc source. The bulb shines brightly. How does its brightness change when an iron core is inserted inside the choke coil?
Answer:
There is no change in the brightness as the inductive reactance is zero for dc.

Question 15.
Why does the acceleration of a magnet falling through a long solenoid decrease?
Answer:
It decreases because of the opposing induced emf produced in the soLenoid due to the rate of change of magnetic flux.

Question 16.
Why is the core of a transformer laminated?
Answer:
It is done to reduce the effect of Eddy Currents.

Question 17.
A vertical metallic pole falls down through the plane of the magnetic meridian. Will any emf be Induced between Its ends?
Answer:
No emf will be induced because the pote neither intercepts the vertical component nor the horizontal component of the earth’s magnetic field.

Question 18.
A magnet Is moved towards a coil and an electric charge is induced in it. If the resistance of the coil is increased, how will the induced charge change?
Answer:
On increasing the resistance of the colt, the magnitude of induced charge decreases.

Question 19.
Can a transformer be used In a dc circuit?
Answer:
No, because there is no change in magnetic flux.

Question 20.
Why does a metallic piece become very hot when it is surrounded by a coil carrying high-frequency alternating current?
Answer:
The high-frequency coil induces eddy currents in the metallic piece. These eddy currents produce heat hence the metalLic piece becomes hot.

Question 21.
The figure shows a horizontal solenoid PQ connected to a battery and a switch. A copper ring R is placed on a frictionless track, the axis of the ring being along the axis of the solenoid. What would happen to the ring as the switch S is closed.

Answer:
The ring WilL be repelled due to opposing induced emf produced in it.

Question 22.
An air-core solenoid is connected to an ac source and a bulb. If an iron core Is inserted in the solenoid, how does the brightness of the bulb change? Give reasons for your answer.
Answer:
Insertion of an iron core In the solenoid increases its inductance. This in turn increases the value of inductive reactance. This decreases the current and hence the brightness of the bulb.

Question 23.
A magnet is moved in the direction indicated by an arrow between two coils AB and CD as shown in the figure. Suggest the direction of current in each coil,

Answer:
In coil AB induced current flows from A to B and in colt CD current flows from C to D.

Question 24.
Predict the directions of induced currents in metal rings 1 and 2 lying in the same place where current I in the wire is increasing steadily. (CBSE Delhi 2012)

Answer:
1 -clockwise, 2-anticlockwise.

Question 25.
The electric current flowing in a wire In the direction from B to A is decreasing. Find out the direction of the induced current in the metallic loop kept above the wire as shown. (CBSE AI 2014)
Answer:
Clockwise.

Question 26.
Predict the polarity of plate A of the capacitor, when a magnet is moved towards it, as is shown In the figure. (CBSE AI 2014C)

Answer:
Positive.

Question 27.
The figure below shows two positions of a loop PQR in a perpendicular uniform magnetic field. In which position of the coil is there induced emf?

Answer:
In position b.

Question 28.
If the self-inductance of an air-core inductor increases from 0.01 mH to 10 mH on introducing an iron core into it, what is the relative permeability of the core used?
Answer:
We know that µ = \(\frac{L}{L_{0}}=\frac{10}{0.01}\) = 1000

Electromagnetic Induction Important Extra Questions Short Answer Type

Question 1.
An induced current has no direction of its own, comment.
Answer:
Yes, it is perfectly correct to say that an induced current has no fixed direction of its own. The direction of induced current depends upon the change in magnetic flux because in accordance with Lenz’s law the induced current always opposes the change in magnetic flux.

Question 2.
How are eddy currents produced? Mention two applications of eddy currents?
Answer:
Eddy currents are the currents induced in the body of a thick conductor when the magnetic flux linked with the conductor changes. When a thick conductor is moved in a magnetic field, magnetic flux linked with it changes. In situations like these, we can have induced currents that circulate throughout the volume of a material.

Because their flow patterns resemble swirling eddies in a river, therefore they are called eddy currents.

• Electromagnetic braking, and
• Induction furnace.

Question 3.
Name and define the unit used for measuring the coefficient of mutual inductance. State the relation of this unit with the units of magnetic flux and electric current.
Answer:
In SI the unit of mutual inductance is henry (H). Now from the expression
ε = – \(\frac{d \phi}{d t}\) = – M \(\frac{d l}{d t}\)
we have M = ε l \(\frac{d l}{d t}\).

Let ε = 1 volt and dl/dt = 1 As^-1, then
M = 1 volt/1 As^-1 = 1 henry.

The mutual-inductance of a coil is said to be 1 henry if a rate of change of current of 1 ampere per sec in the neighbouring coil induces in at an emf of 1 volt.

Question 4.
What are eddy currents? Write any two applications of eddy currents. (CBSE A! 2011)
Answer:
Eddy currents are the currents induced in the body of a thick conductor when the magnetic flux linked with a bulk piece of conductor changes.

1. Dead Beat Galvanometer, and
2. Induction furnace.

Question 5.
(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of the magnetic field B, area A and length l of the solenoid.
(b) How is this magnetic energy per unit volume compared with the electrostatic energy per unit volume stored in a parallel plate capacitor? (CBSE Delhi 2011C)
Answer:
The magnetic field stored in a solenoid is given by the expression U = – \(\frac{1}{2}\)Ll².

But for a solenoid B = μ₀nl
or
l = B / μ₀ n

Substituting in the above expression we have
U = \(\frac{1}{2}\) × (μ₀n²Al)\(\left(\frac{B}{\mu_{0} n}\right)^{2}\) as L = μ₀ n² A l

U = \(\frac{1}{2}\)\(\frac{B^{2} A l}{\mu_{0}}\)

We know that the energy stored per unit volume in a parallel plate capacitor is
U_E = \(\frac{1}{2}\)ε₀E²

It is clear that in both cases the energy stored per unit volume is proportional to the square of the field intensity.

Question 6.
State Lenz’s Law.
A metallic rod held horizontally along the east-west direction is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer. (CBSE Delhi 2013)
Answer:
Lenz’s law states that the polarity of the induced emf is such that it tends to oppose the cause of its production.

Yes, as it will cut the horizontal component of the earth’s magnetic field.

Question 7.
Starting from the expression for the energy W = 1/2Ll², stored in a solenoid of self-inductance L to build up the current l, obtain the expression for the magnetic energy in terms of the magnetic field B, area A and length l of the solenoid having n number of turns per unit length. Hence show that the energy density is given by 8z/2m0. (CBSE Delhi 2013C)
(i) The magnetic energy is
U = \(\frac{1}{2}\)LI² = \(\frac{1}{2}\)L\(\left(\frac{B}{\mu_{0} n}\right)^{2}\) since B = μ₀nl

Now L = μ₀n² Al, therefore we have
U_B = \(\frac{1}{2}\)(μ0n2 Al)\(\left(\frac{B}{\mu_{0} n}\right)^{2}\) = \(\frac{1}{2 \mu_{0}} B^{2} A l\)

(ii) The magnetic energy per unit volume is
U_B = \(\frac{U_{B}}{V}=\frac{U_{B}}{A l}=\frac{B^{2}}{2 \mu_{0}}\)

Question 8.
Define mutual inductance. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? (CBSE Delhi 2016)
Answer:
Mutual inductance is numerically equal to the magnetic flux linked with a coil when the unit current passes through the neighbouring coil.
Given M = 1.5 H, dl = 20 – 0 = 20 A,
dt = 0.5 s, Φ = ?
Φ = – M\(\frac{d l}{d t}\)
or
Φ = – 1.5 × \(\frac{20}{0.5}\) = – 60 Wb

Question 9.
Explain the principle on which the metal detector is used at airports for security reasons.
Answer:
The metal detectors used in airport security checkpoints operate by detecting eddy currents induced in metallic objects. The detector generates an alternating magnetic field. This induces eddy currents in the conduction object carried through the detector. The eddy currents in turn produce an alternating magnetic field. This field induces a current in the detectors receiver coil.

Question 10.
A current is induced in coil C₁, due to the motion of current-carrying coil C₂.
(i) Write any two ways by which a large deflection can be obtained in the galvanometer G.
(ii) Suggest an alternative device to demonstrate the induced current in place of a galvanometer. (CBSE Delhi 2011)

Answer:
(ii) The two ways are
(a) Passing a large current through coil C₂ and
(b) Moving coil C₂ quickly towards the coil.
(ii) A magnetic compass can be placed at the centre of coil C₁. Whenever current will be induced it will show a deflection.

Question 11.
Consider a cube EFGHIJKL of side ‘a’ placed in a magnetic field B acting perpendicular to the face FJKG as shown in the figure. Write magnetic flux through the following faces (a) EFGH (b) EFJI (c) EILH (d) FJKG

Answer:
The magnetic flux depends upon the angle (q) between the magnetic field and area vector.
(a) Here θ = 90°, therefore magnetic flux Φ = BA cos θ = BA cos 90° = 0
(b) Here θ = 90°, therefore magnetic flux Φ = BA cos θ = BA cos 90° = 0
(c) Here θ = 180°, therefore magnetic flux Φ = BA cos 180° = -BA
(d) Here θ = 0°, therefore magnetic flux Φ = BA cos 0° = BA

Question 12.
A metallic rod of ‘L’ length is rotated with an angular frequency of ‘ω’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring. (CBSE Delhi 2012, 2013)
Answer:
Let the rod move from OP to OQ through a small sector of angle θ.

The small area covered by the rod is
dA= πL² × \(\frac{\theta}{2 \pi}=\frac{1}{2}\)L²θ

where L is the radius of the circle in which the rod rotates. Hence the induced emf is
ε = \(\frac{d \phi}{d t}=\frac{d B A}{d t}=\frac{B d A}{d t}=B \frac{d}{d t}\left[\frac{1}{2} L^{2} \theta\right]\)

ε = – \(\frac{1}{2}\)L2B\(\frac{d \theta}{d t}=\frac{1}{2}\)BωL²

Question 13.
A metallic rod of length l length is rotated with a frequency ω, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius R = l, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf induced in the rod. If r is the resistance of the rod and the metallic ring has negligible resistance, obtain the expression for the power generated. (CBSE AI 2013C)
Answer:
For the first part refer above question.
The power generated is given by
P = \(\frac{\varepsilon^{2}}{r}=\frac{B^{2} \omega^{2} l^{4}}{2 r}\)

Question 14.
A rectangular loop PQMN with movable ‘ arm PQ of length 10 cm and resistance 4 Ω is placed in a uniform magnetic field of 0.25 T acting perpendicular to the plane of the loop as is shown in the figure. The resistances of the arms MN, NP and MQare negligible.
Calculate the
(i) emf induced in the arm PQand
(ii) current induced in the loop when arm PQ is moved with velocity 20 m s^-1. (CBSE Delhi 2014C)

Answer:
Given L = 10 cm = 0.1 m, B = 0.25 T, v = 20 m s^-1
(a) ε = BLv = 0.25 × 0.1 × 20 = 0.5 V
(b) l = ε/R = 0.5/4 = 0.125 A

Question 15.
In the given diagram a coil B is connected to a low voltage bulb L and placed parallel to another coil A as shown. Explain the following observations
(i) Bulb lights, and
(ii) Bulb gets dimmer if coil B is moved upwards.

Answer:
(a) When ac is applied across coil A an induced emf is produced in coil B due to mutual induction between the two coils. This makes the lamp light up.
(b) When coil B is moved upwards the mutual induction and hence induced emf in coil B decreases. This makes the lamp dimmer.

Question 16.
A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B.
(a) Write the expression for the instantaneous value of the emf induced in the wire.
Answer:
ε = BLv

(b) What is the direction of the emf?
Answer:
west to east

(c) Which end of the wire is at the higher potential? (CBSEAI2011)
Answer:
East.

Question 17.
Two concentric circular coils one of small radius r₁ and the other of large radius r₂, such that r₁ << r₂, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. (NCERT)
Answer:
Let a current l₂ flow through the outer circular coil. The field at the centre of the coil is B₂ = \(\frac{\mu_{0} I_{2}}{2 r_{2}}\). Since the other coaxially placed coil has a very small radius, B2 may be considered constant over its cross¬sectional area. Hence,
Φ = πr₁²B2 = \(\frac{\mu_{0} \pi r_{1}^{2} l_{2}}{2 r_{2}}\) = M₁₂l₂

Thus M₁₂ = \(\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)

But M₁₂ = M₂₁, therefore M₁₂ = M₂₁ = \(\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)

Note that we calculated M₁₂ from an approximate value of Φ₁ assuming the magnetic field B₂ to be uniform over the area πr₁². However, we can accept this value because r₁ << r₂.

Question 18.
Consider a magnet surrounded by a wire with an on/off switch S (figure). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current flow in the circuit? Explain. (NCERT Exemplar)

Answer:
There is no relative motion between the magnet and the coil. This means that there is no change in magnetic flux, hence no electromotive force is produced and hence no current will flow in the circuit.

Question 19.
A wire in the form of a tightly wound solenoid is connected to a DC source and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will increase. As the wires are pulled apart the flux will leak through the gaps. Lenz’s law demands that induced e.m.f. resist this decrease, which can be done by an increase in current.

Question 20.
A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will decrease. As the iron core is inserted in the solenoid, the magnetic field increases and the flux increases. Lenz’s law implies that induced e.m.f. should resist this increase, which can be achieved by a decrease in current.

Electromagnetic Induction Important Extra Questions Long Answer Type

Question 1.
11 kW of electric power can be transmitted to a distant station at (i) 220 V or (ii) 22,000 V. Which of the two modes of transmission should be preferred and why? Support your answer with possible calculations.
Answer:
1. Consider that 11000 watt of energy has to be transmitted. First at 220 V and then at 22000 V. When the power is transmitted at 220 V then the current flowing through the wires is 11000/220 = 50 A

2. When power is transmitted at 22000 V then the current through the wires is 11000 / 22000 = 0.5 A. If R is the resistance of the line wire then the energy dissipated in the two cases is 2500R joule per sec and 0.25R. joule per sec.

This shows that if energy is transmitted at low voltages there is more loss in energy than when it is transmitted at high voltages. Furthermore, if power is to be transmitted at low voltage then the resistance of the line wire should be low, as such thick wires will be required. To support these thick wires strong poles situated close to each other will be needed. This will increase the cost of transmission. But at high voltages, even thin wires will do.

Question 2.
A coil A is connected to a voltmeter V and the other coil B to an alternating current source D. If a large copper sheet C is placed between the two coils, how does the induced emf in coil A change due to current in coil B. Justify your answer.

Answer:
In the absence of sheet C, an induced emf is set up in coil due to mutual induction phenomenon when an alternating current is passed through coil B.

However, when induced copper sheet C is placed, eddy currents are set up in the sheet due to a change in flux.

Thus, now coil A has a positive effect due to coil B and a negative effect due to eddy currents in C. Consequently, the flux of coil A and hence the induced emf in coil A is decreased, i.e. the reading of voltmeter V is reduced.

Question 3.
A bar magnet is dropped so that it falls vertically through coil C. The graph obtained for the voltage produced across the coil versus time is as shown in figure (b).
(i) Explain the shape of the graph and
(ii) why is the negative peak longer than the positive peak?

Answer:
(a) As the magnet approaches the coil, an emf is induced in it. As the magnet approaches the coil the magnetic flux linked with the coil increases. As a result the induced emf increases. When the magnet enters the coil, the change in magnetic flux linked with the coil begins to decrease and becomes zero when the magnet is completely inside the coil. This starts decreasing the emf and makes it zero.

When the magnet comes out of the coil the direction of induced emf changes direction and begins to increase in the opposite direction. When the magnet moves far away from the coil the induced emf becomes zero.

As the magnet comes out of the coil with a speed greater than the speed at which it approaches the coil, therefore the induced emf is more in the second case. Hence the longer negative peak.

Question 4.
(a) State Lenz’s law: Using this law indicate the direction of the current in a closed loop when a bar magnet with the North Pole is brought close to it. Explain briefly how the direction of the current predicted wrongly results in the violation of the law of conservation of energy.
(b) A rectangular loop and circular loop are moving out of a uniform magnetic field region with a constant velocity v as shown in the figure. In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops. (CBSE AI 2011C)

Answer:
(a) Lenz’s law states that the direction of the induced emf is such that it opposes the cause of its production. When the north pole of the magnet is brought near the coil, the upper end of the coil will acquire north polarity so as to oppose the approaching North Pole (by repelling). This means that the direction of current must be anticlockwise as seen from the side of the magnet.

If the current is wrongly predicted as clockwise, then the upper face will acquire south polarity which will attract the North Pole. This means the current is being produced without doing any work. This leads to the violation of the law of conservation of energy.

(b) It is expected that the induced emf will be constant in the rectangular coil. In the case of the rectangular coil, when pulled out of the magnetic field, the rate of change of magnetic flux will be constant because the rate of change of area is constant. This is not so in the case of the circular coil.

Question 5.
The current through two inductors of self-inductance 15 mH and 25 mH is increasing with time at the same rate. Draw graphs showing the variation of the
(a) emf induced with the rate of change of current
(b) energy stored in each inductor with the current flowing through it. Compare the energy stored in the coils, if the power dissipated in the coils is the same. (CBSE AI 2017C)
Answer:
Given L₁ = 15 mH, L₂ = 25 mH.
(a) The emf induced across an inductor is given by the expression ε = – L\(\frac{d i}{d t}\) since di/dt is the same for both coils therefore induced emf will depend upon the value of inductance. Hence the graphs are as shown

(b) The energy stored in an inductor is given by U = \(\frac{1}{2}\)Ll²

(c) The power dissipated P = l₂X_L
or
l₂ = Pl XL and U = \(\frac{1}{2}\)Ll¹ = \(\frac{1}{2} \frac{L P}{X_{L}}=\frac{1}{2} \frac{P L}{\omega L}=\frac{P}{2 \omega}\)

Since power dissipated is same and co is also same therefore energy stored in the coils will also be the same.

Question 6.
A rectangular frame of wire is placed in a uniform magnetic field directed outwards, normal to the paper. AB is connected to a spring which is stretched to AB and then released at time t = 0. Explain qualitatively how induced e.m.f. in the coil would vary with time. (Neglect damping of oscillations of spring) (CBSE AI 2018C)

Answer:
As the spring is released AB is pulled out of the field. This increases the area of the loop inside the magnetic field. This increases flux and hence an induced emf is produced. The portion AB does not stop at Ab but moves outwards. Now the spring will push AB inwards. This will decrease the area of the loop thereby decreasing the induced emf. This continues and hence the emf increases and decreases periodically.

Question 7.
Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of the same length wound one over the other. (CBSE AI 2017)
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil.

As shown in the figure consider two long co-axial solenoids S₁ and S₂ with S₂ wound over S₁.

Let l = length of each solenoid r₁, r₂ = radii of the two solenoids

A = πr₁² = area of cross-section of inner solenoid S₁
N₁, N₂ = number of turns in the two solenoids First we pass a time-varying current l₂ through S₂. The magnet field set up inside S₂ due to l₂ is
B₂ = μ_o n₂ l₂

where n₂ = N₂/l = the number of turns per unit length of S₂.

Total magnetic flux linked with the inner solenoid S₁ is
Φ₁ = B₂AN₁ = μ0 n₂ l₂ × AN₁

Mutual inductance of coil 1 with respect to coil 2 is
M₁₂ = \(\frac{\phi_{1}}{l_{2}}\) = μ0n2AN1 = \(\frac{\mu_{0} A N_{1} N_{2}}{l}=\frac{\mu_{0} \pi r^{2} N_{1} N_{2}}{l}\)

The mutual inductance of any two coils is always proportional to the product N₁ N₂ of their number of turns. This is termed the reciprocity theorem.

Question 8.
Define mutual inductance and write its SI unit.
A square loop of side ‘a’ carrying a current l₂ is kept at distance x from an infinitely long straight wire carrying a current l1 as shown in the figure. Obtain the expression for the resultant force acting on the loop. (CBSE Delhi 2019)

Answer:
(a) Mutual inductance equals the magnetic flux associated with a coil when unit current flows in its neighbouring coil. (b) Force per unit length between two parallel straight conductors
\(\frac{F}{L}=\frac{\mu_{o} I_{1} I_{2}}{2 \pi r}\)

Force on the part of the loop which is parallel to infinite straight wire and at a distance x from it F = \(\frac{\mu_{o} l_{1} l_{2} a}{2 \pi x}\)(away from the infinitely 2KX long wire)

Force on the part of the loop which is at a distance (x + a) from it
F₂ = \(\frac{\mu_{0} I_{1} I_{2} a}{2 \pi(x+a)}\) (towards the infinite long wire)

Net force F = F₁ – F₂
F = \(\frac{\mu_{o} l_{1} l_{2} a^{2}}{2 \pi(x+a)}\) (away from the infinite straight line)

Question 9.
What are the possible causes of energy loss in a transformer? How are these minimised?
Answer:
The possible causes of energy losses in transformers are:
(a) Flux leakage: There is always some flux leakage. It can be reduced by winding the primary and secondary coils one over the other.
(b) Copper: The copper wires used for the windings have some resistance and hence some energy is lost due to heat produced in the wire. It can be minimised by taking thick wire.
(c) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core, which results in loss of electrical energy. To minimise it we use a laminated iron core.
(d) Hysteresis loss: As the magnetisation cycle of the iron core is repeated again and again some loss of energy takes place due to magnetic hysteresis. To minimise it we prefer a soft iron core for which hysteresis loss is less.

Question 10.
Discuss how Faraday’s law of e.m.f induction is applied in an ac generator for converting mechanical energy into electrical energy. Draw graphs to show the ’phase relationship’ between the instantaneous
(a) magnetic flux (Φ) linked with the coil and
(b) induced emf (ε) in the coil. (CBSE Delhi 2016C)
Answer:
In an ac generator, the change in magnetic flux is brought about by rotating the coil in a magnetic field. According to Faraday’s law, induced emf is set up in the coil on changing the magnetic flux linked with it. Hence mechanical energy, which is supplied to rotate the coil, gets converted into electrical energy.

Let the coil be rotated with a constant angular speed, co, the angle 0 between magnetic field 8, and area vector A of the coil at any instant is θ = ωt

Therefore magnetic flux at any instant is Φ = BA cos ωt

From Faraday’s law, induced emf is
ε = – n\(\frac{d \phi}{d t}\) = -nBA\(\frac{d}{d t}\)cos ωt
ε = nBAa sin ωt

Magnetic flux is given by Φ = BA cos ωt, therefore the graph will be a cosine curve as shown

The induced emf is given by ε = nBA sin ωt, therefore the graph will be a sine curve as shown.

Question 11.
(a) Define self-inductance of a coil. Obtain an expression for the energy stored in a solenoid of self-inductance ‘L’ when the current through it grows from zero to l.
(b) A square loop MNOP of the side 20 cm is placed horizontally in a uniform magnetic field acting vertically downwards as shown in the figure. The loop is pulled with a constant velocity of 20 cm s^-1 till it goes out of the field.

(c) Depict the direction of the induced current in the loop as it goes out of the field. For how long would the current in the loop persist?
(d) Plot a graph showing the variation of magnetic flux and induced emf as a function of time. (CBSE AI 2015)
Answer:
(a) Let i be the current at some instant through a pure inductor. If di / dt is the rate of change of current through the inductor then the voltage between the terminals of the inductor at this instant is
V=L di / dt

Therefore instantaneous power in the inductor is
P = V_ab i = Li \(\frac{d i}{d t}\)

The energy dU supplied to the inductor during an infinitesimal time interval dt is
dU = P dt, so dU = Li di

The total amount of energy supplied while the current increases from zero to a final value l is
U = L ∫_o¹ i di = \(\frac{1}{2}\)L l²

(b) (i) Direction of induced current – clockwise (MNOP), the duration of induced current is 1s.
(ii) The graph is as shown.

Question 12.
(a) Define mutual inductance and write its SI units.
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil. It is measured in henry.

(b) Derive an expression for the mutual inductance of two long co-axial solenoids of the same length wound one over the other.
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil.

As shown in the figure consider two long co-axial solenoids S₁ and S₂ with S₂ wound over S₁.

Let l = length of each solenoid r₁, r₂ = radii of the two solenoids

A = πr₁² = area of cross-section of inner solenoid S₁
N₁, N₂ = number of turns in the two solenoids First we pass a time-varying current l₂ through S₂. The magnet field set up inside S₂ due to l₂ is
B₂ = μ_o n₂ l₂

where n₂ = N₂/l = the number of turns per unit length of S₂.

Total magnetic flux linked with the inner solenoid S₁ is
Φ₁ = B₂AN₁ = μ_o n₂ l₂ × AN₁

Mutual inductance of coil 1 with respect to coil 2 is
M₁₂ = \(\frac{\phi_{1}}{l_{2}}\) = μ0n2AN1 = \(\frac{\mu_{0} A N_{1} N_{2}}{l}=\frac{\mu_{0} \pi r^{2} N_{1} N_{2}}{l}\)

The mutual inductance of any two coils is always proportional to the product N1 N2 of their number of turns. This is termed the reciprocity theorem.

(c) In an experiment, two coils c₁ and c₂ are placed close to each other. Find out the expression for the emf induced in the coil c₁ due to a change in the current through the coil c₂. (CBSE Delhi 2015)
Answer:
Consider the experimental set up as shown.

When the key K is pressed in coil C₁ a magnetic flux linked with C₂ changes.

The magnetic flux linked with coil C₂ is given by
f₂ = M l₁ where M is the mutual inductance between the two coils.

But by Faraday’s flux rule we have
ε₂ = – \(\frac{d \phi}{d t}\) therefore we have
ε₂ = – \(\frac{d M l_{1}}{d t}\) = – M \(\frac{d l_{1}}{d t}\)

Question 13.
(a) When a bar magnet is pushed towards (or away) from the coil connected to a galvanometer, the pointer in the galvanometer deflects. Identify the phenomenon causing this deflection and write the factors on which the amount and direction of the deflection depend. State the laws describing this phenomenon.
(b) Sketch the change In flux, emf and force when a conducting rod PQ of resistance R and length I moves freely to and fro between A and C with speed v on a rectangular conductor placed in the uniform magnetic field as shown in the figure. (CBSE AI 2016)

Answer:
(a) Phenomenon: electromagnetic induction

Factors:

• Strength of the magnetic field of the magnet
• speed of motion of bar magnet.

Direction depends upon

• the motion of magnet whether inward or outward
• the direction of the north/south pole.

Law: The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

(b) The sketch is as shown below.

Question 14.
What is induced emf? Write Faraday’s laws of electromagnetic induction. Express it mathematically. A conducting rod of length ‘L’ with one end pivoted is rotated with a uniform angular velocity ‘ω’ in a vertical plane, normal to a uniform magnetic field ‘B’. Deduce an expression for the emf induced in this rod.
Answer:
It is the emf induced when the magnetic flux linked with a coil changes.

Faraday put forward the following laws called Faraday’s laws of electromagnetic induction.

• Law 1. Whenever the magnetic flux linked with a coil changes an induced emf is produced.
• Law 2. The induced emf lasts as long as the change in magnetic flux continues.

The expression for induced emf in rotating rod: Let the rod move from OP to OQ through a small sector of angle θ.

The small area covered by the rod is
dA = πL² × \(\frac{\theta}{2 \pi}=\frac{1}{2}\)L2θ

where L is the radius of the circle in which the rod rotates. Hence the induced emf is
ε = \(\frac{d \phi}{d t}=\frac{d B A}{d t}=\frac{B d A}{d t}=B \frac{d}{d t}\left[\frac{1}{2} L^{2} \theta\right]\)
Or
ε = \(\frac{1}{2} L^{2} B \frac{d \theta}{d t}=\frac{1}{2}\)BωL²

Question 15.
How is the mutual inductance of a pair of coils affected when:
(a) the separation between the coils is increased?
Answer:
When the separation between the two coils is increased the mutual inductance decreases. It is because the flux linked with the secondary due to a current in the primary decreases.

(b) the number of turns of each coil is increased?
Answer:
When the number of turns of each coil is increased the mutual inductance increases because M₁₂ (= M₂₁) ∝ N₁N₂,

i.e. mutual inductance is directly proportional to the number of turns. The linkage of flux increases with an increase in the number of turns in the coils.

(c) a thin iron sheet is placed between the two coils, other factors remaining the same? Explain your answer in each case.
Answer:
When a thin iron sheet is placed between the coils the permeability of the medium between the coil increases. As mutual inductance is directly proportional to the permeability, therefore the mutual inductance increases.

Question 16.
A rectangular coil of area A, having a number of turns N, is rotated at f revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2π f N B A.
Answer:
Consider an armature of the ac generator having n turns and placed in a uniform magnetic field B.

Suppose at any instant t the normal to the plane of the coil makes an angle 0 with the direction of the magnetic field. If r_o is the uniform angular velocity with which, the coil rotates then θ = ωt.

The flux through the loop equals its area A multiplied by B_⊥ = Bcos θ, the component of magnetic field B perpendicular to the area, hence

Φ = n B A cos Φ = n B A cos ω t …(1)
where is the number of turns in the armature?

By Faraday’s flux rule,
ε = – \(\frac{d \phi}{d t}=-\frac{d}{d t}\) n B A cos ω t
= – n B A \(\frac{d}{d \phi}\)cos ω t = – n B A (- ω sin ω t) ….(2)
Or
ε = n B A ω sin ω t

The induced emf is maximum when sin ωt = maximum = 1, therefore the maximum induced emf is given by
ε₀ = n B A ω sin ω t …(3)

Since ω = 2πf, therefore equation (3) becomes ε₀ = 2πf nBA

Question 17.
State Lenz’s law. Explain by giving examples that Lenz’s law is a consequence of the law of conservation of energy. (CBSE Delhi 2017C)
Answer:
It states that the direction of induced emf in a coil is such that it opposes the cause of its production. Lenz’s law is a consequence of the law of conservation of energy. To show it, let us consider a bar magnet pushed towards a conducting loop. When N-pole moves towards the loop, the face of the loop facing the North Pole develops north polarity as per Lenz’s law so as to oppose the motion of the magnet.

Again, when N-pole moves away from the loop, the nearby face develops south polarity, thus opposing the motion of the magnet away from the loop. It means that the motion of the magnet is automatically opposed every time. Hence, some work is to be done on the magnet to move it and this mechanical work is transformed into electrical energy. Thus, the conservation law of energy is followed.

Question 18.
Define the term ‘self-inductance’ and write its SI unit.
Obtain the expression for the mutual inductance of two long co-axial solenoids S₁ and S₂ wound one over the other, each of length L and radii r₁ and r₂ and n₁ and n₂ number of turns per unit length when a current I is set up in the outer solenoid S₂. (CBSE Delhi 2017)
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil.

As shown in the figure consider two long co-axial solenoids S₁ and S₂ with S₂ wound over S₁.

Let l = length of each solenoid r₁, r₂ = radii of the two solenoids

A = πr₁² = area of cross-section of inner solenoid S₁
N₁, N₂ = number of turns in the two solenoids First we pass a time-varying current l₂ through S₂. The magnet field set up inside S₂ due to l₂ is
B₂ = μ_o n₂ l₂

where n₂ = N₂/l = the number of turns per unit length of S₂.

Total magnetic flux linked with the inner solenoid S₁ is
Φ₁ = B₂AN₁ = μ_o n₂ l₂ × AN₁

Mutual inductance of coil 1 with respect to coil 2 is
M₁₂ = \(\frac{\phi_{1}}{l_{2}}\) = μ_on₂AN₁ = \(\frac{\mu_{0} A N_{1} N_{2}}{l}=\frac{\mu_{0} \pi r^{2} N_{1} N_{2}}{l}\)

The mutual inductance of any two coils is always proportional to the product N₁ N₂ of their number of turns. This is termed the reciprocity theorem.

Question 19.
Obtain the expression for the magnetic energy stored in an ideal inductor of self-inductance L when a current l passes through it. Hence obtain the expression for the energy density of magnetic field B produced in the inductor. (CBSE Delhi 2016C)
Answer:
Let i be the current at some instant through a pure inductor. If di / dt is the rate of change of current through the inductor then the voltage between the terminals of the inductor at this instant is
V=L di / dt

Therefore instantaneous power in the inductor is
P = V_ab i = Li\(\frac{d i}{d t}\)

The energy di supplied to the inductor during an infinitesimal time interval dt is
dU = P dt, so dU = Li di

The total amount of energy supplied while the current increases from zero to a final value l is

U = L ∫₀¹ di = \(\frac{1}{2}\)L l²

This gives the expression for the energy stored in an inductor.

The magnetic energy is

The magnetic energy per unit volume is
U_B = \(\frac{U_{B}}{V}=\frac{U_{B}}{A l}=\frac{B^{2}}{2 \mu_{0}}\)

Question 20.
Derive the expression for the self-inductance of a solenoid.
Answer:
Consider a uniformly wound solenoid with N turns and length l. Let the length of the solenoid be large as compared to its radius and let the core of the solenoid have air. Due to this, we can take the interior field uniform. Therefore the magnetic field in the interior of the solenoid is given by

B = μ₀ n l = μ₀ \(\frac{N}{l}\) l …(1)

where n is the number of turns per unit length. The flux through each turn is given by
Φ_m = B A = μ₀ \(\frac{N A}{l}\) l …(2)

where A Is the area of cross-section of the solenoid. But
L = \(\frac{N \Phi_{\mathrm{m}}}{I}\) …(3)

Therefore from equations (2) and (3) we have
L = \(\frac{N \Phi_{m}}{l}=\frac{\mu_{0} N^{2} A}{l}\) ….(4)

This shows that L depends on the geometric factors and is proportional to the square of the number of turns. Since N N = n l, we can therefore express the above result as
L = µ₀\(\frac{(n l)^{2}}{l}\)A = µ₀ n² A l …(5)

Question 21.
(a) State Faraday’s laws of electromagnetic Induction.
Answer:
Faraday’s Laws of electromagnetic induction
First law: Whenever the magnetic flux (inked with a circuit (or coil) changes, an emf is induced in the circuit. The induced emf Lasts so Long as the change in the magnetic flux continues.

Second law:
The magnitude of induced emf in the circuit (or coil) is directly proportional to the rate of change of magnetic flux linked with the circuit.
i.e. e = – \(\frac{d \phi}{d t}\)

(b) Derive an expression for the emf induced across the ends of a straight conductor of length I moving at right angles to a uniform magnetic field B with a uniform speedy.
Answer:
Let a conductor ab of Length l be placed in a magnetic field \(\vec{B}\) shown by (x) directed towards the reader. When the conductor moves with velocity v perpendicular to B, the force on any free electron of this conductor is given

F = e vB sin 90° = evß

When this electron moves across the Length of the conductor, the work done by the electron
W = F × l = evBl

After some time when all the free electrons are shifted towards the end b, the end b becomes negative and end a becomes positively charged.

Hence the net induced emf in the conductor
e = \(\frac{W}{q}=\frac{e v B l}{e}=v B l\)
e = Blv

(c) Obtain the expression for the magnetic energy stored in a solenoid in terms of the magnetic field B, area A and length I of the solenoid through which a current j is passed. (CBSE 2019C)
Answer:
Energy stored in an inductor Considers an inductor of inductance L. Let I will be instantaneous current in the inductor and dl/dt be the rate of growth of current.

Induced emf, E = L\(\frac{dl}{dt}\) (in magnitude)

If the source sends a charge dq in time dt, then
dq = ldt

Small amount of work done by the source
dW = edq = eldt = L\(\frac{dl}{dt}\). ldt
or
dW = Ll dl

Total work was done during the growth of current in an Inductor by the external source
W = ∫dW =L∫₀¹ ldl
= L \(\left[\frac{l^{2}}{2}\right]_{0}=\frac{1}{2}\)Ll²
W = \(\frac{1}{2}\) Ll² …(1)

This work done is stored in the form of magnetic energy.
In a solenoid,
L = μ₀n²Al
And B = μ₀nl
∴ l = \(\frac{B}{\mu_{0} n}\)

So Eq. (1) becomes
W = \(\frac{1}{2}\)μ0n2Al\(\left(\frac{B}{\mu_{0} n}\right)^{2}=\frac{B^{2} A l}{2 \mu_{0}}\)

Question 22.
(a) A metallic rod of length ‘l’ and resistance ‘R’ is rotated with a frequency ‘v’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius ‘l’, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field ‘B’ parallel to the axis is present everywhere.
(i) Derive the expression for the induced emf and the current in the rod.
(ii) Due to the presence of current in the rod and of the magnetic field, find the expression for the magnitude and direction of the force acting on this rod.
(iii) Hence, obtain an expression for the power required to rotate the rod.
(b) A copper coil l is taken out of a magnetic field with a fixed velocity. Will it be easy to remove it from the same field if its ohmic resistance is increased?
OR
(a) A rectangular coil rotates in a uniform magnetic field. Obtain an expression for induced emf and current at any instant. Also find their peak values. Show the variation of induced emf versus angle of rotation (ωt) on a graph.
(b) An iron bar falling through the hollow region of a thick cylindrical shell made of copper experiences a retarding force. What can you conclude about the nature of the iron bar? Explain. (CBSEAI 2019)
Answer:
(a) (/) Let the rod moves from OP to OQ through a small sector of angle θ.

The small area covered by the rod is
dA = πl² × \(\frac{\theta}{2 \pi}=\frac{1}{2}\)l²θ

where L is the radius of the circle in which the rod rotates. Hence the induced emf is

Now induced current is given by
i = \(\frac{\varepsilon}{R}=\frac{B \omega l^{2}}{2 R}\)

(ii) \(\vec{F}=i(\vec{l} \times \vec{B})\)
F = \(\frac{l^{3} \omega B}{2 R}\)

Direction of \(\vec{F}\) is perpendicular to both \(\vec{i}\) and \(\vec{B}\),

(iii) P = i²R
= \(\left(\frac{B l^{2} \omega}{2 R}\right)^{2}\)R = \(\frac{B^{2} l^{4} \omega^{2}}{4 R}\)

(b) Yes, since induced current will reduce, it will be a little easier to remove the coil.
Or
(a) Consider an armature of the ac generator having n turns placed in a uniform magnetic field B.

Suppose at any instant t the normal to the plane of the coil makes an angle θ with the direction of the magnetic field. If ω is the uniform angular velocity with which the coil rotates then θ = ωt.

The flux through the loop is Φ = n BA cos Φ = n BA cos ω t
where n is the number of turns in the armature.

By Faraday’s flux rule,

Now,i₀ = \(\frac{\varepsilon_{0}}{R}=\frac{n B A \omega}{R}\)

The graph is as shown.

(b) The bar is magnetic in nature. It experiences retardation in accordance with Lenz’s law.

Question 23.
State Lenz’s law. The energy f required to build up a steady current l, in a given coil, varies with l in the manner as shown. Calculate the self-inductance of the coil.

A circular coil of radius r, is placed coaxially with another coil of radius R (R >> r) with the centre of the two coils coinciding with each other. Obtain an expression for the mutual inductance of the two coils.
Answer:
Lenz’s law states that the direction of induced emf is such that it opposes the cause of its production.
The energy stored in an inductor is given by the relation
E = – \(\frac{1}{2}\)Ll²

From the graph for a current of 200 mA the energy is 4 mJ, therefore self-inductance of the coil.
\(\frac{2 E}{l^{2}}\) = L
or
\(\frac{2 \times 4 \times 10^{-3}}{\left(200 \times 10^{-3}\right)^{2}}\) = 0.2 H
or
L = 200 mH

Consider the two coils of radius r and R placed coaxially, as shown in the figure.

Let n₁ and n₂ be the number of turns per unit length in them.

If a current l₂ is passed through the outer coil, there will a magnetic field
B_o = μ_o n₁ l₁ produced in it.

The flux associated with the inner coil will undergo variation as the field B0 grows.
The effective magnetic flux
Φ₁ = B_o × A_eff = \(\frac{\mu_{0} l_{2}}{2 R}\) × πr²

Therefore mutual inductance of the coils is
M = \(\frac{\phi_{1}}{l_{2}}=\frac{\mu_{0} \pi r^{2}}{2 R}\)

Question 24.
(a) Describe a simple experiment (or activity) to show that the polarity of emf induced in a coil is always such that it tends to produce a current which opposes the change of magnetic flux that produces it.
(b) The current flowing through an inductor of self-inductance L is continuously increasing. Plot a graph showing the variation of
(i) Magnetic flux versus the current
(ii) induced emf versus dl/dt
(iii) Magnetic potential energy stored versus the current. (CBSE Delhi 2014)
Answer:
(a) Consider a coil C connected to a galvanometer. When the North Pole of a bar magnet is pushed towards the coil, the pointer in the galvanometer deflects, indicating the presence of electric current in the coil. When the magnet is pulled away from the coil, the galvanometer shows deflection in the opposite direction, which indicates a reversal of the current’s direction.

We see that the North Pole of a bar magnet is being pushed towards the closed coil. As the North Pole of the bar magnet moves towards the coil, the magnetic flux through the coil increases. Hence current is induced in the coil in such a direction that it opposes the increase in flux.

This is possible only if the current in the coil is in a counter-clockwise direction with respect to an observer situated on the side of the magnet. Similarly, if the North Pole of the magnet is being withdrawn from the coil, the magnetic flux through the coil will decrease. To counter this decrease in magnetic flux, the induced current in the coil flows in a clockwise direction and the South Pole faces the receding North Pole of the bar magnet. This would result in an attractive force that opposes the motion of the magnet and the corresponding decrease in flux.

(b) The graphs are as shown.
(i) Φ = Ll
Therefore the graph is

(ii) ε = – L\(\frac{dl}{dt}\)
Therefore the graph is

(iii) U = \(\frac{1}{2}\)Ll²
Therefore the graph is

Numerical Problems:

Formulae for solving numerical problems

• Induced emf ε = –\(\frac{d \phi}{d t}\)
  or
  ε = \(\frac{-\left(\phi_{2}-\phi_{1}\right)}{t_{2}-t_{1}}\)
• Emf induced in a rod of Length L is ε = Blv.
• When a conducting rod of length L kept perpendicular to a uniform magnetic field B is rotated about one of its ends with uniform angular velocity, the emf induced between its ends has a magnitude
  \(\frac{1}{2}\)BωL² = BπvL².
• When a current in a coil changes, it induces a back emf in the same coil. The self-induced emf is given by
  ε = – \(\frac{d \phi}{d t}\) = – L \(\frac{d l}{d t}\) inductance of the coil. It is a measure of the inertia of the coil against the change of current through it. Also Φ = L l
• ε = – M\(\frac{d l}{d t}\) . Also Φ = Ml for mutual induction.
• \(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}=\frac{l_{p}}{l_{s}}\) = k.
• Average power η = \(\frac{\varepsilon_{s} l_{s}}{\varepsilon_{p} l_{p}}\)

Question 1.
A square loop of side 10 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s^-1 in the positive X-direction containing a magnetic field in the positive Z-direction. The field is non-uniform and has a gradient of 10^-3 T cm^-1 along the negative X-direction (i.e. it increases by 10^-3 T cm^-1 as one move in the negative X-direction). Calculate the emf induced. (CBSE 2019C)
Answer:

Question 2.
In a ceiling fan, each blade rotates in a circle of radius 0.5 m. If the fan makes 2 rotations per second and the vertical component of the earth’s magnetic field is 8 × 10^-5 T, calculate the emf induced between the inner and outer ends of each blade. (CBSE2019C)
Answer:

Question 3.
A circular coil of radius 10 cm, 500 turns and resistance 2 Ω are placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the magnitudes of the emf and the current induced in the coil. The horizontal component of the earth’s magnetic field at the place is 3.0 × 10^-5 T. (NCERT)
Answer:
Initial flux through the coil
Φ_in = BA cos θ = 3.0 × 10^-5 × π × 10^-2 × cos 0°
= 3π × 10^-7 Wb

Final flux after the rotation
Φ_f = BA cos θ = 3.0 × 10^-5 × π × 10^-2 × cos 180° = – 3π × 10^-7 Wb

Therefore estimated value of induced emf is
ε = n\(\frac{d \phi}{d t}\) = 500 × (6π × 10^-7)/ 0.25 = 3.8 × 10^-3 V dt

Hence the current induced is
l = ε/R = 3.8 × 10^-3 / 2 = 1.9 × 10^-3 A

Question 4.
Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 turn coil of area 0.10 m². The coil rotates at half a revolution per second and is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil? (NCERT)
Answer:
Given n = 100, A = 0.10 m2, B = 0.01 T, f = 0.5 rps
Using the expression ε₀ = nBAω

we have ε₀ = 100 × 0.01 × 0.10 × 2 × 3.14 × 0.5 = 0.314 V

Question 5.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s^-1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? (NCERT)
Answer:
Given L = 8 cm = 8 × 10^-2 m, b = 2 cm = 2 × 10^-2 m, 8 = 0.3 T, v = 1 cm s^-1 = 0.01 m s^-1 ε = ?
(a) Using the expression ε = B L v
ε = 0.3 × 8 × 10^-2 × 0.01 = 2.4 × 10^-4 V

This emf will last till the loop comes out of the magnetic field. Since the shorter side is moving out, therefore it will take 2 s to cover 2 cm with a velocity of 1 cm s^-1.

(b) Using the expression e = B L v
ε = 0.3 × 2 × 10^-2 × 0.01 = 0.6 × 10^-4 V

This emf will last till the loop comes out of the magnetic field. Since the longer side is moving out, therefore it will take 8 s to cover 8 cm with a velocity of 1 cm s^-1.

Question 6.
A 1.0 m long metallic rod is rotated with an angular frequency 400 rad s-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. (NCERT)
Answer:
Given L = 1.0 m, ω = 400 rad s^-1, B = 0.5 T,

ε = ? Using the relation ε = \(\frac{1}{2}\)BωL²

we have
ε = \(\frac{1}{2}\)0.5 × 400 × 1² =100 V

Question 7.
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s^-1 in a uniform horizontal magnetic field of magnitude 3.0 × 10^-2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from? (NCERT)
Answer:
Given r = 8.0 cm = 8.0 × 10^-2 m, n = 20, ω = 50 rad s-1 B = 3.0 × 10^-2 T, ε₀ = ?, ε_a = ? R= 10 Ω l = ?, P = ?
Using the relation
ε0 = nBAω
= 20 × 3 × 10^-2 × 3.14 × (8 × 10^-2)² × 50 = 0.603 V

The average induced emf is given by
εa = 0 over one cycle.

Also l₀ = ε₀ / R
= 0.603 / l₀
= 0.0603 A

Now P = 1 /2 ε₀ × l₀ = 1 /2 × 0.603 × 0.0603
= 0.018 W

The induced current causes a torque opposing the rotation of the coil. An external agent (rotor) must supply torque to counter this torque in order to keep the coil rotating uniformly. Thus the source of the power dissipated as heat in the coil is the external rotor.

Question 8.
Current in a circuit falls from 5.0 A to 0. 0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit (NCERT)
Answer:
Given dl = 5.0 – 0.0 = 5.0 A, dt = 0.1 s, ε = 200 V, L = ?

Using the relation ε = – L \(\frac{dl}{dt}\)
or
L = \(\frac{\varepsilon}{d l / d t}\)

L = \(\frac{200}{5 / 0.1}=\frac{200 \times 0.1}{5}\) = 4 H

Question 9.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? (NCERT)
Answer:
Given dl = 20- 0.0 = 20 A, dt = 0.5 s, ε = ?,
M = 1.5 H,
Using the relation ε = – \(\frac{d \phi}{d t}\) = -M\(\frac{dl}{dt}\)
or
dΦ = Mdl = 1.5 × 20 = 30 Wb

Question 10.
A jet plane is travelling towards the west at a speed of 1800 km h^-1. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the earth’s magnetic field at the location has a magnitude of 5 × 10^-4 T and the dip angle is 30°? (NCERT)
Answer:
Given v= 1800 km h-1, ε = ?, L = 25 m,
B = 5 × 10^-4 T, δ = 30°
Now vertical component of earth’s magnetic field is
B_V = B × sin 30°
= 5 × 10^-4 × 0.5
= 2.5 × 10^-4T
Now using the expression ε = B L v
ε = 2.5 × 10^-4 × 25 × 500 = 3.125 V

Question 11.
1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of a radius of 0.5 cm for this purpose. Calculate the fraction of ohmic losses to the power transmitted if
(a) power is transmitted at 220 V. Comment on the feasibility of doing this.
(b) a step-up transformer is used to boost the voltage to 11000 V, power transmitted, then a step-down transformer is used to bring the voltage to 220 V. ρ_cu = 1.7 × 10^-8 SI unit. (NCERT Exemplar)
Answer:
Given P = 1 MW, 2L = 20 km = 2 × 104 m, r = 0.5 cm = 0.5 × 10^-2 m
(a) Resistance of the Cu wire used
R = \(\frac{\rho L}{A}=\frac{1.7 \times 10^{-8} \times 2 \times 10^{4}}{\pi \times\left(0.5 \times 10^{-2}\right)^{2}}\) = 4 Ω

Now current at 220 V is

l = \(\frac{P}{V}=\frac{10^{6}}{220}\) = 0.45 × 10⁴ A

Therefore power loss
= l²R = (0.45 × 10⁴)² × 4 = 10⁶W

This is a huge loss; therefore this method is not feasible.

(b) Now P = Vl’ or
l’ = \(\frac{P}{V^{\prime}}=\frac{10^{6}}{11000}=\frac{1}{1.1}\) × 10² A

Hence power loss
= l’²R = \(\frac{1}{1.21}\) × 4 × 10⁴ = 3.3 × 10⁴ W

Hence fraction of power loss
\(\frac{3.3 \times 10^{4}}{10^{6}}\) = 3.3%

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 11 Alcohols, Phenols and Ethers. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 11 Important Extra Questions Alcohols, Phenols and Ethers

Alcohols, Phenols and Ethers Important Extra Questions Very Short Answer Type

Question 1.
Arrange the following in increasing order of their boiling point:
CH₃CH₂OH, CH₃CHO, CH₃—O—CH₃ (CBSE AI 2019)
Answer:
CH₃—O—CH₃ < CH₃CHO < CH₃CH₂OH

Question 2.
Arrange the following in increasing order of their acidic character:
Ethanol, Phenol, Water (CBSE AI 2019)
Answer:
Ethanol <Water < Phenol

Question 3.
How will you convert ethanol to ethene? (CBSE 2011)
Answer:

Question 4.
Draw the structural formula of the 2-methylpropan-2-ol molecule. (CBSE Delhi 2012)
Answer:

Question 5.
Which of the following isomers is more volatile: (CBSE Delhi 2014)
o-nitrophenol or p-nitrophenol?
Answer:
o-nitrophenol

Question 6.
Write the IUPAC name of the given compound: (CBSE 2015)

Answer:
2-Methylpropane-1, 3-diol

Question 7.
Write the IUPAC name of the given compound: (CBSE Delhi 2015)

Answer:
2, 5- Dinitrophenol

Question 8.
Arrange the following in increasing order of their acidic character: Benzoic acid, Phenol, Cresol (CBSE Al 2019)
Answer:
Cresol < Phenol < Benzoic acid

Question 9.
Rearrange the following compounds in the increasing order of their boiling points: (CBSE Al 2013)
CH₃-CHO, CH₃-CH₂-OH, CH₃-CH₂-CH₃
Answer:
CH₃CH₂CH₃ < CH₃CHO < CH₃CH₂OH

Question 10.
An alkoxide is a stronger base than a hydroxide ion. Justify. (CBSE Sample Paper 2012)
Answer:
Due to the presence of electron-donating alkyl group, there is high electron density in alkoxide ion as compared to hydroxide ion. Therefore, the alkoxide ion is more basic than the hydroxide ion.

Question 11.
Why is (±) butan-2-ol optically inactive? (CBSE Delhi 2013, CBSE AI 2013)
Answer:
(±)-Butan-2-ol represents a racemic mixture of (+)-butan-2-ol and (-)-2-butanol which rotate the plane polarized light in different directions but to an equal extent. Therefore, (±) compound is optically inactive.

Question 12.
Write the IUPAC name of the following compound: (CBSE Al 2017)

Answer:
2-Methoxy-2-methylpropane

Question 13.
Write the IUPAC name of the following compound: (CBSE AI 2017)

Answer:
3-Phenylprop-2-en-1-ol

Question 14.
Write the IUPAC name of the following: (CBSEAI 2018)

Answer:
3, 3-Dimethylpentan-2-ol

Alcohols, Phenols and Ethers Important Extra Questions Short Answer Type

Question 1.
Predict the major product obtained when t-butyl bromide reacts with sodium methoxide. Also, give its IUPAC name.
Or
(a) Show the chemical reaction with bond movements and arrows for the nucleophilic attack of water on carbocation in acid catalysed hydration of alkenes.
(b) Give IUPAC name for the following: (CBSE 2019C)

Answer:

OR

Mechanism: hydration of alkenes
The mechanism of the reaction involves the following steps:
Step 1. Alkene gets protonated to form a carbocation by the electrophilic attack of H₃O⁺:

Step 2. Nucleophile (H₂0) attacks the carbocation forming protonated alcohol.

Step 3. Loss of H+ from oxygen (deprotonation) to form alcohol

(b) 2,6-dimethylphenol

Question 2.
How are the following conversions carried out?
(i) Propene to propan-2-ol
Answer:

(ii) Ethyl magnesium chloride to propan-1-ol (CBSE Delhi 2010)
Answer:

Question 3.
How would you obtain:
(i) Picric add (2, 4, 6-trlnitrophenol) from phenol.
Answer:

(ii) 2-Methyl propene from 2-methyl propanol? (CBSE Delhi 2011)
Answer:

Question 4.
Write the chemical equations involved in the following reactions:
(i) Kolbe’s reaction
(ii) Friedel-Crafts acetylation of anisole (CBSE Delhi 2016)
OR
How do you convert:
(i) Phenol to toluene
(ii) Formaldehyde to ethanol?
Answer:
(i) Kolbe’s reaction. Sodium phenoxide reacts with CO₂ under pressure (6-7 atm) at 400 K to form sodium salicylate which upon acidification with HCL gives salicylic acid.

(ii) Friedel-Crafts acetylation of anisole. On reacting anisole with acetyl chloride in the presence of anhyd. AlCl₃, 2-methoxy acetophenone and 4-methoxy acetophenone are formed.

(i) Phenol to toluene

(ii) Formaldehyde to ethanol

Question 5.
Explain the mechanism of the following reaction: (CBSE Delhi 2013, 2016)

Answer:
The mechanism for the formation of diethyl ether from ethanol at 413 K is given below:
(i) Ethyl alcohol gets protonated in the presence of H⁺.

(ii) Nucleophilic attack by another alcohol (unprotonated) molecule occurs on the protonated alcohol with the elimination of a molecule of water.

(iii) Oxonium ion [oses a proton to form an ether.

Question 6.
How will you convert:
(i) Propene to Propan-2-ol?
Answer:

(ii) Phenol to 2, 4, 6 – trinitrophenol? (CBSE Delhi 2013)
Answer:

Question 7.
How will you convert:
(i) Propene to propan-1 -ol?
Answer:

(ii)Ethanal to propan-2-ol?
Answer:

Question 8.
Give chemical tests to distinguish between (CBSE Sample Paper 2011)
(i) Isopropyl alcohol and n-propyl alcohol
Answer:
Isopropyl alcohol gives the iodoform test. On heating with NaOH/l₂ or NaOl, isopropyl alcohol forms a yellow precipitate of iodoform (CHI₃).

(ii) Phenol and alcohol.
Answer:
Phenol reacts with neutral FeCl₃ solution to give red-violet complex, whereas alcohol does not give this test.

Question 9.
(i) Arrange the following compounds in the increasing order of their acid strength:
p-cresol, p-nitrophenol, phenol
Answer:
p-cresol < phenol < p-nitrophenol

(ii) Write the mechanism (using curved arrow notation) of the following reaction:
(a) CH₂ = CH₂ CH₃-CH₂ + H₂O
OR
Write the structures of the products when butan-2-ol reacts with the following: (CBSE AI 2017)
(i) CrO₃
(ii) SOCl₂
Answer:

Or

Question 10.
Alcohols are easily protonated in comparison to phenols. Explain. (CBSE AI 2016)
Answer:
In alcohols, the electron-releasing inductive effect (+ I effect) of the alkyl group attached to the carbon having the -OH group increases the electron density on the oxygen atom. Therefore, alcohols are easily protonated.

On the other hand, in the case of phenol, the oxygen atom acquires a partial positive charge due to resonance.
Thus, it is not protonated.

Question 11.
Explain why ortho nitrophenol is more acidic than ortho methoxy phenol. (CBSE AI 2015)
Answer:
This is because -NO₂ (nitro group) is an electron-withdrawing group and will increase the +ve charge on oxygen to make it more acidic. On the other hand, the -OCH₃ group is an electron-releasing group and will decrease +ve charge on oxygen making it less acidic as the O-H bond will not break easily.

Question 12.
Anisole on reaction with HI gives phenol and CH₃I as main products and not iodobenzene and CH₃OH. Why? (CBSE AI 2016)
Answer:
It this reaction protonated anisole, i.e. methyl phenyl oxonium ion, is first formed and then the halide ion attacks this protonated anisole. Due to steric hindrance of the bulky phenyl group, the attack preferably occurs to the alkyl group forming methyl iodide and phenol.

Question 13.
The boiling points of ethers are lower than their corresponding isomeric alcohols. Explain. (CBSE AI 2012)
Answer:
Ethers have low polarity and therefore, do not show any association by intermolecular hydrogen bonding. On the other hand, their isomeric alcohols have strong intermolecular hydrogen bonding and therefore, their boiling points are high.

Question 14.
Butan-1-ol has a higher boiling point than diethyl ether. Give reason.
Answer:
Butan-1-ol has intermolecular hydrogen bonding between their molecules. Therefore, it exists as associated molecules and a large amount of energy is required to break these bonds and hence its boiling point is high. But diethyl ether does not show any association by intermolecular hydrogen bonding. Hence, its boiling point is low.

Question 15.
Write the mechanism of acid dehydration of ethanol to yield ethene. (CBSE Sample Paper 2018)
Answer:
The reaction is believed to occur as follows:
(i) Formation of protonated alcohol: Alcohol combines with a proton to form a protonated alcohol.

(ii) Formation of the carbocation: The protonated alcohol loses a water molecule to form a carbocation.

(iii) Elimination of a proton to form alkene: The carbocation then eliminates a proton and undergoes rearrangement of electrons to form the alkene.

The acid used in step (i) is released in step (iii). To drive the equilibrium reaction in the forward direction, ethene is removed as soon as it is formed.

Alcohols, Phenols and Ethers Important Extra Questions Long Answer Type

Question 1.
What happens when
(a) Sodium phenoxide is treated with CH₃Cl?
Answer:

(b) CH₂ = CH – CH₂ – OH is oxidised by PCC?
Answer:

(c) Phenol is treated with CH₃COCI/anhydrous AlCl₃?
Write chemical equations in support of your answer.
Answer:

Question 2.
(a) How will you convert the following:
(i) Phenol to benzoquinone
Answer:

(ii) Propanone to 2-methyl propane-2-ol
Answer:

(b) Why does propanol have a higher boiling point than that butane? (CBSE 2019C)
Answer:
The molecules of propanol are held together by intermolecular hydrogen bonding while butane molecules have only weak van der Waals forces of attraction. Since hydrogen bonds are stronger than van der Waals forces, therefore, propanol has a higher boiling point than butane.

Question 3.
Identify the product formed when propan-1 -ol is treated with a cone. H₂S0₄ at 413 K. Write the mechanism involved for the above reaction. (CBSE Sample Paper 2019)
Answer:
(a) 1-Propoxypropane is formed.
Mechanism involved:
Step 1: Formation of protonated alcohol. Propanol gets protonated in the presence of H⁺.

Step 2: Nucleophilic attack. Due to the presence of a +ve charge on the oxygen atom, the carbon of the CH₂ part becomes electron deficient. As a result, a nucleophilic attack by another alcohol molecule (unprotonated) occurs on the protonated alcohol with the elimination of a molecule of water.

Step 3: Deprotonation. Oxonium ion loses a proton to form an ether.

Question 4.
Write the equations involved in the following reactions: (CBSE 2013, 2014)
(i) Reimer-Tiemann reaction
Answer:
Reimer-Tiemann reaction: When phenol is refluxed with chloroform in the presence of aqueous caustic alkali at 340 K, an aldehydic group (CHO) gets introduced in the ring at a position ortho to the phenolic group. Ortho hydroxy benzaldehyde or salicylaldehyde is formed as the product of the reaction.

(ii) Williamson’s ether synthesis
Answer:
Williamson’s ether synthesis. This is used to prepare symmetrical and unsymmetrical ethers by treating alkyl halide with either sodium alkoxide or sodium phenoxide.

Aryl halides cannot be used for the preparation of alkyl-aryl ethers because of their low reactivity.

Question 5.
Draw the structure and name the product formed if the following alcohols are oxidised. Assume that an excess of the oxidising agent is used. (CBSE Delhi 2012)
(i) CH3CH2CH2CH2OH
Answer:

(ii) 2-butenol
Answer:

(iii) 2-methyl-1-propanol
Answer:

Question 6.
Write the main product(s) in each of the following reactions: (CBSE Delhi 2016)

Answer:

Answer:

Answer:

Question 7.
Give reasons for the following: (CBSE Delhi 2016, Outside Delhi)
(i) Protonation of phenols Is difficult whereas ethanol easily undergoes protonation.
Answer:
In phenols, the oxygen atom acquires a partial positive charge due to resonance and therefore, it is not easily protonated. On the other hand, in ethanol, the alkyl group is an electron-releasing group and increases the electron density on 0 atoms. Therefore, ethanol is easily protonated.

(ii) Bolting point of ethanol is higher than that of dimethyl ether.
Answer:
The molecules of ethanol are held together by intermolecular hydrogen bonding while dimethyl molecules have only weak van der Waals forces of attractions. Since hydrogen bonds are stronger than van der Waals forces, ethanol has a higher boiling point than dimethyl ether.

(iii) Anisole on reaction with Hl gives phenol and CH₃-I as main products and not iodobenzene and CH₃OH.
Answer:
This is because, during the reaction, the attack of halide ion occurs to the protonated anisole, i.e. methyl phenyl oxonium ion, which is formed during the protonation of anisole. Due to steric hindrance of the bulky phenyl group, the attack preferably occurs to the alkyl group forming methyl iodide and phenol.

Question 8.
How do you convert the following: (CBSE Delhi 2015, Delhi)
(i) Phenol to anisole
(ii) Propan-2-ol to 2-methylpropan-2-ol
(iii) Aniline to phenol?
OR
(i) Write the mechanism of the following reaction:
2CH₃CH₂OH CH₃CH₂—O—CH₂CH₃
(ii) Write the equation involved In the acetylation of salicylic acid.
Answer:
(i) Phenol to anisole

(iii) Aniline to phenol

Or
(i) The mechanism for the formation of ether from ethanol at 413 K is a nucleophilic bimolecular reaction as given below:
(a) Ethyl alcohol gets protonated in the presence of H⁺

(b) Due to the presence of a +ve charge on the oxygen atom, the carbon of CH₂ part of CH₃CH₂ becomes electron deficient. As a result, nucleophilic attack by another alcohol molecule (unprotonated) occurs on the protonated alcohol with the elimination of a

(c) Oxonium ion loses a proton to form an ether.

Question 9.
Give reasons for the following: (CBSE 2015, Outside Delhi)
(i) o-nitrophenol is more acidic than o-methoxyphenyl.
Answer:
This is because —NO₂ (nitro group) is an electron-withdrawing group and will increase the +ve charge on oxygen to make it more acidic. On the other hand, the -OCH₃ group is an electron¬releasing group and will decrease +ve charge on oxygen making it less acidic as O-H bond will not break easily.

(ii) Butan-1-oi has a higher boiling point than diethyl ether.
Answer:
Butan-1 -ol has intermolecular hydrogen bonding between their molecules. Therefore, it exists as associated molecules and large amount of energy is required to break these bonds and hence, its boiling point is high. But diethyl ether does not show any association by intermolecular hydrogen bonding. Hence, its boiling point is low.

(iii) (CH₃)₃C – O – CH₃ on reaction with HI gives (CH₃)₃C – I and CH₃ – OH as the main products and not (CH₃)₃C – OH and CH₃ – I.
Answer:
The reaction:

gives (CH₃)₃C – I and CH₃OH as the main products and not (CH₃)₃COH and CH₃I. This is because the reaction occurs by the S_N1 mechanism and the formation of products is governed by the stability of carbocation formed from the cleavage of the C-0 bond in the protonated ether. Since tert. butyl carbocation, (CH₃)₃C⁺ is more stable than methyl carbocation, CH₃, the cleavage of C-0 gives a more stable carbocation, [(CH₃)₃C]⁺ and methanol. Then, iodide ion, I^– attacks this tert. butyl carbocation to form tert. butyl iodide.

Question 10.
(i) Write the mechanism of the following reaction: (CBSE 2015, Outside Delhi)

Answer:
HBr → H⁺ + Br^–
(a) H+ attacks oxygen of O-H to form protonated alcohol

(b) Protonated alcohol Loses a molecule of water to form a carbocation.

(C) Br attacks the carbocation to form bromoatkane
CH₃CH₂+ + Br^– → CH₃CH₂Br (Bromoethane)

(ii) Write the equation involved in the Reimer-Tiemann reaction.

Question 11.
(A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C₄H₈0. 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. HCl, give the same product (D).
(i) Write the structures of (A), (B), (C) and (D).
Answer:
A = CH₃CH₂CH₂CHO
B = CH₃COCH₂CH₃
C = (CH₃)₂CHCHO
D = CH₃CH₂CH₂CH₃

(ii) Out of (A), (B) and (C) isomers, which one is least reactive towards the addition of HCN? (CBSE 2018 C)
Answer:
B

Question 12.
(a) How will you convert the following:
(i) Phenol to benzene
Answer:

(ii) Propene to propanol
Answer:

(b) Why is ortho-nitrophenol more acidic than ortho-methoxy phenol? (CBSE 2018C) OH
Answer:
Due to the strong -R and -I effect of the -NO₂ group, the electron density in the O-H bond decreases and hence the loss of a proton becomes easy.

Moreover, the o-nitrophenoxide formed after the Loss of a proton is stabilised by resonance.

o-nitrophenoxide ion is stabilised by resonance and hence o-nitrophenol is a stronger acid. On the other hand, due to the +R effect of the -OCH3 group, the electron density in the O-H bond increases and this makes the loss of proton difficult.

Furthermore, after the Loss of proton o-methoxyphenoxide ion left is destabilised by resonance.

The two negative charges repel each other and therefore, destabilise the o-methoxyphenoxide ion. Thus, o-nitrophenol is more acidic than o-methoxyphenyl.

Question 13.
(a) How will you convert the following:
(i) Ethanal to propan-2-ol
Answer:

(ii) Phenol to 2-hydroxy acetophenone
Answer:

(b) Why is phenol more acidic than ethanol? (CBSE 2019C)
Answer:
It is because phenoxide ion is more stable than ethoxide ion. Phenols are more acidic than alcohols. The greater acidic character of phenoLs as compared to alcohols can be explained on the basis of resonance. Phenol is a resonance hybrid of the following structures:

It is clear that three structures of phenol (III, IV and V) have a +ve charge on the oxygen of the —OH group. This oxygen attracts the electron pair of O—H bond strongly towards itself, weakens the O—H bond and, therefore, facilitates the release of W. Similarly, the phenoxide ion is resonance stabilised as follows:

Thus, we observe that both phenol and phenoxide ion are stabilised by resonance. Now if we carefully observe the resonance structures, we observe that phenoxide ion is more resonance stabilised than phenol. In phenol three contributing structures (III, IV and V) have both positive and negative charges and therefore, these will be unstable. These structures will require energy to separate the charge and therefore, will be unstable.

On the other hand, there is no structure in phenoxide ion which requires charge separation. Thus, the resonance hybrid of phenol is less stable than phenoxide ion and the reaction is very much in favour of the phenoxide ion. Therefore, the phenol is more acidic.

On the other hand, in the case of alcohols, neither the alcohol nor the alkoxide ion is stabilised by resonance.

Thus, phenols are more acidic than alcohols.

Question 13.
(a) How do you convert the following?
(i) Phenol to Anisole
(ii) Ethanol to Propan-2-ol
(b) Write the mechanism of the following reaction:

(c) Why phenol undergoes electrophilic substitution more easily than benzene?
OR
(a) Account for the following:
(i) o-Nitrophenol is more steam volatile than p-nitrophenol.
(ii) Tert-butyl chloride on heating with sodium methoxide gives 2-methylpropene instead of tert-butyl methyl ether.

(b) Write the reaction involved in the following:
(i) Reimer-Tiemann reaction
(ii) Friedel-Crafts Alkylation of Phenol

(c) Give a simple chemical test to distinguish between Ethanol and Phenol. (CBSE Delhi 2019)
Answer:
(i)
(ii) Ethanol to Propan-2-ol

(b) Mechanism CH₃
(i) Ethanol combines with a proton to form protonated alcohol.

(ii) Protonated alcohol loses a water molecule to form a carbocation.

(iii) Carbocation eliminates a proton and undergoes rearrangement of electrons to form an alkene.

(c) In phenol, the -OH group is an activating group and therefore, electrophilic substitution reaction undergoes more easily in phenol than benzene.
OR
(a) (i) o-Nitrophenol is more steam volatile than p-nitrophenol because of chelation due to intramolecular hydrogen bonding.

(ii) Tert-Butyl chloride reacts with sodium methoxide to give 2-methylpropene.

This is because alkoxides are not only nucleophiles but strong bases as well. They react with alkyl halides leading to an elimination reaction.

(b) Reimer-liemann reaction

(C) Phenols react with neutral Fecl₃ to give violet colour, while ethanol does not give any colour.

Question 14.
(i) Write steps to carry out the conversion of phenol to aspirin.
Answer:

(ii) What happens when benzene diazonium chloride Is heated with water?
Answer:

(iii) Write the structures of the Isomers of alcohols with molecular formula C₄H₁₀O. Which of these exhibits optical activity?
Answer:

(ii) is optically active.

Question 15.
How are the following conversions carried out?
(i) Benzyl chloride to benzyl alcohol
Answer:

(ii) Methyl magnesium bromide to 2-methyl propane-2-ol
Answer:

(iii) Propene to propan-2-ol
Answer:

(iv) Ethylmagnesium chloride to propan-1-ol. (CBSE Delhi 2010, 2013)
Answer:

Question 16.
How will you convert:
(i) Phenol to 2, 4, 6-trlnftrophenot
Answer:

(ii) Propan-2-ol to propanone
Answer:

(iii) Phenol to 2, 4, 6-trlbromophenol
Answer:

(iv) Propene to propan-1-ol
Answer:

(v) Ethanal to propan-2-ol? (CBSE DelhI 2013)
Answer:

Question 17.
How will you convert
(i) Propan-2-ol to 2-methylpropan-2-ol (CBSE Delhi 2015)
Answer:

(ii) Aniline to phenol (CBSE Delhi 2015)
Answer:

(iii) Ethanol to propane nitrile (CBSE AI 2015)
Answer:

(iv) Phenol to toluene (CBSE AI 2016)
Answer:

(v) Formaldehyde to ethanol? (CBSE AI 2016)
Answer: