Previous Year Paper

UPPSC Polytechnic Lecturer Electrical 22 Nov 2021 Official Paper I (Previous Year Paper)

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General Hindi (25 questions)

Question 1

'कपिश-कपीश' का सही शब्द-युग्म चुनें?

  1. AOption (a)

    तपन-वानर

  2. BOption (b)

    जलता हुआ-हनुमान्

  3. COption (c)

    मटमैला-सुग्रीव

  4. DOption (d)

    मलिन-खजाना 

View answer & explanation

Correct answer

Answer: (c)C.

मटमैला-सुग्रीव

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "मटमैला-सुग्रीव" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • कपिश-कपीश :- मटमैला-सुग्रीव
  • कपिश
  • विशेषण
  • भूरे रंग का।
  • सुग्रीव के पर्याय
  • दाढ़वाला, जबड़ेवाला, संकटमोचन, प्लवगद्र, रामदूत, अंगद, हनुमंत, पवनपुत्र, मारुति
  • भीम, कपींद्र, कपीश, हरीश, सुरपति, इंद्र, मरुत्वान, पवनसुत, बजरंग।

Additional Information

  • प्रमुख शब्द युग्म:-
  • अध्ययन :- पढ़ना
  • अध्यापन :- पढ़ाना
  • अधम :- नीच
  • अधर्म :- पाप
  • अली :- सखी
  • अलि :- भौंरा
  • अन्त :- समाप्ति
  • अन्त्य :- नीच, अन्तिम
  • अम्बुज :- कमल
  • अम्बुधि :- सागर
  • असन :- भोजन
  • आसन :- बैठने की वस्तु
  • कटिबद्ध :- तैयार, कमर बाँधे
  • कटिबन्ध :- कमरबन्द, करधनी
  • कृशानु :- आग
  • कृषाण :- किसान
  • कटीली :- तीक्ष्ण, धारदार
  • कँटीली :- काँटेदार
  • कोष :- खजाना
  • कोश :- शब्द-संग्रह (डिक्शनरी)
  • कदन :- हिंसा
  • कदन्न :- खराब अन्न
  • कुच :- स्तन    
  • कूच :- प्रस्थान
  • गण :- समूह
  • गण्य :- गिनने योग्य
  • गुड़ :-  शक्कर
  • गुड़ :- गम्भीर

Question 2

अनेकार्थी शब्द 'अर्क' का इनमें से एक अर्थ नहीं है।

  1. AOption (a)

    पारा

  2. BOption (b)

    मदार

  3. COption (c)

    ताॅंबा 

  4. DOption (d)

    स्फटिक

View answer & explanation

Correct answer

Answer: (a)A.

पारा

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "पारा" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • पारा अर्क का अनेकार्थी शब्द नहीं है।
  • पारा शब्द ईश्वर का अनेकार्थी है।
  • ईश्वर के अनेकार्थी
  • परमात्मा, स्वामी, शिव, पारा, पीतल।
  • अर्क के अनेकार्थी
  • सूर्य, सत्त्व, ताँबा, बिजली की चमक, स्फटिक, मदार, क्वाथ (काढ़ा) रविवार।

Additional Information

  • अन्य अनेकार्थी शब्द
  • गुरु :- शिक्षक, ग्रहविशेष, श्रेष्ठ, बृहस्पति, भारी, बड़ा, भार।
  • गो :- बाण, आँख, वज्र, गाय, स्वर्ग, पृथ्वी, सरस्वती, सूर्य, बैल, इत्यादि।
  • गुण :- कौशल, शील, रस्सी, स्वभाव, धनुष की डोरी।
  • गति :- चाल, दशा, मोक्ष, हालत।
  • कनक :- सोना, धतूरा, गेंहूँ।
  • कुशल :- खैरियत, चतुर ।
  • अवकाश :- छुटटी, अवसर, अंतराल
  • आम :- आम का फल, सर्वसाधारण, मामूली, सामान्य।
  • अन्तर :-  शेष, दूरी, हृदय, भेद।
  • आराम :- विश्राम, निरोग होना।
  • उत्तर :- उत्तर दिशा, जवाब, हल, अतीत, पिछला, बाद का इत्यादि।

Question 3

'परीक्षा' शब्द का सन्धि-विच्छेद है

  1. AOption (a)

    पर + ईक्षा

  2. BOption (b)

    परि + ईक्षा

  3. COption (c)

    परि + इक्षा

  4. DOption (d)

    परः + ईक्षा

View answer & explanation

Correct answer

Answer: (b)B.

परि + ईक्षा

Why this answer is correct

परि + ईक्षा = 'परीक्षा' शब्द का सन्धि-विच्छेद है।

  • रि (इ) + (ई) ई = ई

Key Points

  • दीर्घ संधि : 
  • ​जब दो शब्दों की संधि करते समय (अ, आ) के साथ (अ, आ) हो तो 'आ' बनता है
  • जब (इ, ई) के साथ (इ, ई) हो तो 'ई' बनता है
  • जब (उ, ऊ) के साथ (उ, ऊ) हो तो 'ऊ' बनता है।
  • इस संधि को हम ह्रस्व संधि भी कह सकते हैं।
  • जैसे: पुस्तक + आलय : पुस्तकालय बनता है।
  • यहाँ अ+आ मिलकर आ बनाते हैं।

Important Points​ 

  • ​(ई + इ = ई) मही + इंद्र = महींद्र
  • नारी + इंदु = नारींदु ।
  • (ई + ई = ई) नदी + ईश = नदीश
  • मही + ईश = महीश । (
  • उ + उ = ऊ) भानु + उदय = भानूदय
  • विधु + उदय = विधूदय ।
  • (उ + ऊ = ऊ) लघु + ऊर्मि = लघूर्मि
  • सिधु + ऊर्मि = सिंधूर्मि ।

Question 4

'सहर-शहर' का सही शब्द-युग्म चुनें?

  1. AOption (a)

    सवेरा-नगर

  2. BOption (b)

    नगर-प्रातःकाल

  3. COption (c)

    शैर-जीत

  4. DOption (d)

    सरल-नगर

View answer & explanation

Correct answer

Answer: (a)A.

सवेरा-नगर

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "सवेरा-नगर" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • 'सहर' शब्द का अर्थ 'सवेरा' तथा 'शहर' शब्द का अर्थ 'नगर' होता है।
  • अतः सहर-शहर युग्म का अर्थ सवेरा-नगर होगा।
  • सहर
  • पुल्लिंग
  • प्रातःकाल, सवेरा।
  • शहर
  • पुल्लिंग
  • नगर।

Additional Information

  • प्रमुख शब्द युग्म
  • अंचल :- साड़ी का छोर
  • अंचला :- साधुओँ का एक वस्त्र
  • अंचित :- गूंथा हुआ, पूजित
  • अचित् :- जड़, चेतन रहित
  • आकर :- खजाना
  • आकार :- बनावट
  • आभरण :- गहना
  • आवरण :- ढकना
  • इतर :- दूसरा
  • इत्र :- पुष्पसार
  • इति :- समाप्ति
  • ईति :- दैव आपत्ति
  • उद्योत :- प्रकाश
  • उद्योग :- उपाय
  • उतर :- नीचे आना
  • उत्तर :- जवाब
  • ऋत :- सत्य
  • ऋतु :- मौसम
  • ओर :- तरफ
  • और :- दूसरा, तथा
  • कंथा :- गुदड़ी
  • कथा :- कहानी

Question 5

इनम��ं से 'आकाश' का पर्यायवाची शब्द है

  1. AOption (a)

    पुष्कर

  2. BOption (b)

    अपरिमित

  3. COption (c)

    प्रपंच

  4. DOption (d)

    अवलम्ब

View answer & explanation

Correct answer

Answer: (a)A.

पुष्कर

Why this answer is correct

'आकाश' का पर्यायवाची शब्द है: पुष्कर

आकाश के अन्य: अंबर, अनंत, अंतरिक्ष, अभ्र, आसमान, गगन, दिव, नभ

Confusion Points

  • पुष्कर अनेकार्थी शब्द है, इसलिए आकाश का पर्यायवाची शब्द माना जाएगा।
  • पुष्कर के एक से अधिक अर्थ – तालाब, कमल, आकाश, तलवार।

Key Points

  • अपरिमित : जो बहुत अधिक हो, ज्यादा हो
  • प्रपंच : स्वांग, ढोंग, आडंबर, दिखावा, ढ़कोसला
  • अवलम्ब : भरोसा , पनाह , प्रश्रय आश्रम , आसरा

Additional Information

  • फूल- पुष्प, कुसुम,पुहुप, सुमन, प्रसून।
  • कमल– पंकज, राजीव, पद्म, सरोज, नलिन, जलज।
  • पेड़– वृक्ष, पादप, शाखी, तरु, विटप।
  • पानी-  नीर, सलिल, जीवन, तोय, उदक, पय,अंबु, अंभ, रस, आप, आब, वारि ।
  • राजा – क्षत्र, क्षत्री, द्विजलिंगी, नाभि, नृप, पार्थिव, बाहुज, मूर्द्धक, मूद्र्धाभिषिक्त, राजन्य, वर्म, विराज, विराट, वीर, सार्वभौम।
  • हवा– वायु, अनिल, समीर, पवन।
  • आम – अतिसौरभ, रसाल, फलराज, आम्र, सहकार, पिकबंधु, च्युतफल।
  • दिन-  अह:, दिवस, वासर, दिवा, वार।
  • हाथी – गज, हस्ती, मतंग, द्विरद, गयंद, सिंधुर, दंती, कुंभी, वितुण्ड

Question 6

तत्सम शब्दों की दृष्टि से एक शब्द अशुद्ध है, वह है

  1. AOption (a)

    अंगरक्षक

  2. BOption (b)

    अमृत

  3. COption (c)

    आर्द्रक

  4. DOption (d)

    ग्यारह 

View answer & explanation

Correct answer

Answer: (d)D.

ग्यारह 

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "ग्यारह" सही है, क्योंकि यह अशुद्ध है। तथा अन्य विकल्प तत्सम शब्द के हैं।

Key Points

  • तत्सम शब्द : एकादश
  • तद्भव शब्द : ग्यारह

Additional Information

  • तत्सम - तद्भव

अंगरक्षक :- अंगरखा

अमृत - अमिय

अक्षोट :- अखरोट

ग्रन्थि :- गाँठ

गर्गर :- गागर

एकादश :- ग्यारह

गोपाल :- ग्वाला

Question 7

निम्नलिखित वाक्यों में से एक में सकर्मक क्रिया है, वह वाक्य है

  1. AOption (a)

    श्याम सोता है।

  2. BOption (b)

    विद्यार्थी कहानी लिखता है।

  3. COption (c)

    चिड़िया उड़ती है।

  4. DOption (d)

    बच्चा रोता है।

View answer & explanation

Correct answer

Answer: (b)B.

विद्यार्थी कहानी लिखता है।

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "विद्यार्थी कहानी लिखता है।" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • "विद्यार्थी कहानी लिखता है।" वाक्य में सकर्मक क्रिया है।
  • विद्यार्थी क्या लिखता है? उत्तर में हमें प्राप्त होता है कहानी।
  • अतः किसी वाक्य में क्या, किसे तथा किसको लगाकर उत्तर प्राप्त होता है तो वहाँ सकर्मक क्रिया होगी।

Additional Information

  • अकर्मक क्रिया
  • जिस क्रिया का फल कर्ता पर ही पड़ता है वह क्रिया अकर्मक क्रिया कहलाती हैं।
  • इस क्रिया में कर्म का अभाव होता है।
  • उदाहरण
  • राजेश दौड़ता है।
  • सांप रेंगता है।
  • पूजा हंसती है।
  • श्याम सोता है।
  • बच्चा रोता है।

Question 8

'अनवधान' शब्द का विलोम है

  1. AOption (a)

    वरदान

  2. BOption (b)

    सावधानी

  3. COption (c)

    निरभिमान

  4. DOption (d)

    अननुमत

View answer & explanation

Correct answer

Answer: (b)B.

सावधानी

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "सावधानी" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • अनवधान शब्द का विलोम सावधानी होगा।
  • अनवधान का विलोम 
  • अवधान या सावधानी।
  • अनवधान
  • पुल्लिंग 
  • अवधान का अभाव,
  • असावधानी।
  • अवधान
  • पुल्लिंग
  • ध्यान।
  • मनोयोग।

Additional Information

  • प्रमुख विलोम शब्द
  • अदोष :- सदोष
  • अधम :-उत्तम
  • अधर्म :-सध्दर्म
  • अधुनातन :- पुरातन
  • उपाय :- निरुपाय
  • उषा :- संध्या
  • एकता :- अनेकता
  • एकांगी :- सर्वांगीण
  • कनिष्ठ :- ज्येष्ठ
  • कृतज्ञ :- कृतघ्न
  • कृष :- स्थूल
  • कृष्ण :- शुक्ल
  • क्रय :- विक्रय
  • क्षणिक :- शाश्वत
  • खेद :- प्रसन्नता
  • गुप्त :- प्रकट
  • ग्रामीण :- शहरी
  • घात :- प्रतिघात
  • घृणा :- प्रेम
  • छली :- निश्चल
  • छूत :- अछूत
  • जंगली :- पालतू
  • दयालु :- निर्दयी
  • दाता :- याचक
  • निंदा :- स्तुति
  • निरक्षर :- साक्षर
  • नूतन :- पुरातन
  • प्रत्यक्ष :- परोक्ष
  • बंधन :- मुक्ति
  • मूक :- वाचाल
  • लगभग :- पूरा
  • लगाव :- हटाना
  • लहू :-पसीना

Question 9

इनमें से तद्भव-तत्सम का एक युग्म अशुद्ध है

  1. AOption (a)

    ओस - अवश्याय

  2. BOption (b)

    काढ़ा - क्वाथ

  3. COption (c)

    उबटन - उद्वर्तन

  4. DOption (d)

    रहट - अरहट्ट

View answer & explanation

Correct answer

Answer: (d)D.

रहट - अरहट्ट

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "रहट-अरहट्ट" एक अशुद्ध युग्म है तथा अन्य विकल्प असंगत हैं।

Key Points

  • रहट का तत्सम अरघट्ट है।
  • रहँट
  • पुल्लिंग
  • कुएँ से पानी निकालने का यंत्र विशेष।

Additional Information

  • अन्य उदाहरण
  • अम्बा :-अम्मा
  • अग्नि :- आग
  • अक्षी :- आंख
  • क्लेश :- कलेश
  • ग्रीष्म :- गर्मी
  • घृत :- घी
  • चूर्ण :- चूरन
  • उलूक :- उल्लू
  • कृपा :- किरपा
  • कुंभकार :- कुम्हार
  • कादलीपाली :- केला

Question 10

निम्नलिखित में से 'गुफा' का पर्यायवाची शब्द नहीं है

  1. AOption (a)

    गुहा

  2. BOption (b)

    दरी

  3. COption (c)

    गहवर

  4. DOption (d)

    विजन

View answer & explanation

Correct answer

Answer: (d)D.

विजन

Why this answer is correct

विजन : 'गुफा' का पर्यायवाची शब्द नहीं है।

Key Points

  • 'गुफा' का पर्यायवाची : गोह, खोह, गुहा, कंदरा, सुरंग, गार, गड्ढा, कंदरा, गुहा, दरी, गह्वर/गहवर
  • विजन के पर्याय : सूनापन, एकांत, सुनसान, जनशून्य।

Additional Information

  • पति-- भर्ता, वल्लभ, स्वामी, आर्यपुत्र
  • पक्षी-- विहंग, विहग, खग, पखेरू, परिंदा, चिड़िया, शकुंत, अंडज, पतंग, द्विज
  • पंडित-- सुधी, विद्वान, कोविद, बुध, धीर, मनीषी, प्राज्ञ, विचक्षण
  • पुष्प-- फूल, सुमन, कुसुम, प्रसून
  • बाण-- तीर, शर, विशिख, आशुग, शिलीमुख, इषु, नाराच
  • ब्रह्मा-- आत्मभू, स्वयंभू, चतुरानन, पितामह, हिरण्यगर्भ, लोकेश, विधि, विधाता
  • वृक्ष-- तरु, द्रुम, पादप, विटप, अगम, पेड़, गाछ
  • मछली-- मत्स्य, झख, मीन, जलजीवन, सफरी, झष, जलीय जीव
  • महादेव-- शंभु, ईश, पशुपति, शिव, महेश्वर, शंकर, चंद्रशेखर, भव, भूतेश, गिरीश, हर, त्रिलोचन
  • मेघ-- घन, जलधर, वारिद, बादल, नीरद, वारिधर, पयोद, अंबुद, पयोधर
  • मुनि-- यती, अवधूत, संन्यासी, वैरागी, तापस, संत, भिक्षु, महात्मा, साधु, मुक्तपुरुष

Question 11

'विष्णु' शब्द में इनमें से कौन-सा प्रत्यय लगकर 'वैष्णव' शब्द बनता है?

  1. AOption (a)

    एव

  2. BOption (b)

    अव

  3. COption (c)

    आ

  4. DOption (d)

    अ

View answer & explanation

Correct answer

Answer: (d)D.

अ

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "अ" सही है तथा अन्य विकल्प असंगत हैं।

Confusion Points

  • यहाँ आप अव , विष्णु + अव, नहीं समझ लेना क्योंकि अव  कोई प्रत्यय ही नही होता है।

Key Points

  • वैष्णव शब्द में "अ" प्रत्यय है।
  • अ :- शैव, वैष्णव, तैल, पार्थिव।
  • इसमें सम्बन्धवाचक तद्धित प्रत्यय है।
  • सम्बन्धवाचक तद्धित प्रत्यय
  • सम्बन्धवाचक तद्धित प्रत्यय से सम्बन्ध का बोध होता है। इसमें भी कहीं-कहीं पर आदि-स्वर की वृद्धि हो जाती है।

Important Points

  • सम्बन्धवाचक तद्धित प्रत्यय के उदाहरण
  • इक :- लौकिक, धार्मिक, वार्षिक, ऐतिहासिक।
  • इत :- पीड़ित, प्रचलित, दुःखित, मोहित।
  • इम :-  स्वर्णिम, अन्तिम, रक्तिम।
  • इल :- जटिल, फेनिल, बोझिल, पंकिल।
  • ईय :- भारतीय, प्रान्तीय, नाटकीय, भवदीय।
  • य :- ग्राम्य, काम्य, हास्य, भव्य।

Additional Information

  • विष्णु
  • विष् + नु
  • व्यंजन संधि
  • विष्णु के पर्यायवाची
  • नारायण, दामोदर, पीताम्बर, चक्रपाणी, जनार्दन, मुकुंद, गोविन्द, कमलाकांत, लक्ष्मीपति, उपेन्द्र।

Question 12

'युद्ध करने की इच्छा रखनेवाला' - इस वाक्यांश के लिए एक उपयुक्त शब्द है

  1. AOption (a)

    मुमुक्षु

  2. BOption (b)

    विजिगीषु

  3. COption (c)

    युयुत्सु

  4. DOption (d)

    उत्साही

View answer & explanation

Correct answer

Answer: (c)C.

युयुत्सु

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "युयुत्सु" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • युद्ध करने की इच्छा रखने वाला वाक्य के लिए एक शब्द युयुत्सु है।
  • युयुत्सु
  • पुल्लिंग
  • योद्धा।

Additional Information

  • मुमुक्षु
  • विशेषण
  • मोक्ष की कामना करनेवाला।
  • विजिगीषु
  • जिसे विजय पाने की इच्छा हो।
  • उत्साही (जो उत्साह रखता हो)
  • विशेषण
  • उत्साह युक्त।
  • उद्यमी।

Important Points

  • अनेक शब्दों के लिए एक शब्द
  • जो उपासना करता हो :- उपासक
  • देखने योग्य :- दर्शनीय
  • युग परिवर्तन की चेतना :- युगान्तकारी
  • सुनने योग्य :- श्रव्य
  • जंगल की आग :- दावानल
  • मर्म को भेदने वाला :- मर्मभेदी
  • श्रम पर जीने वाला    :- श्रमजीवी
  • जिसका कोई उपाय नही हो :- निरुपाय
  • प्रवास करने वाला :- प्रवासी
  • पेट की आग :- जठरानल
  • जिसकी चिंता संभव नही :- अचिंत्य
  • जिसकी कल्पना नही की जा सके :- अकल्पनीय
  • इन्द्रियों को जितने वाला  :- जितेन्द्रिय
  • एक समय में विद्यमान :- समसमायिक
  • जितने की इच्छा :- जिगीषा
  • पूछने योग्य :- प्रष्टव्य

Question 13

निम्नलिखित में से एक शुद्ध वाक्य है

  1. AOption (a)

    तुम्हारे सब काम गलत होते हैं।

  2. BOption (b)

    मैं अपनी स्वेच्छा से आया हूँ।

  3. COption (c)

    आपके साथ उचित न्याय किया जाएगा।

  4. DOption (d)

    आज वर्षा होने की संभावना की जा रही है।

View answer & explanation

Correct answer

Answer: (a)A.

तुम्हारे सब काम गलत होते हैं।

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "तुम्हारे सब काम गलत होते हैं।" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • "तुम्हारे सब काम गलत होते हैं।" वाक्य सही है।
  • वाक्य के अंत में "हैं" में लगने वाली अनुस्वार बिंदु बहुवचन के लिए प्रयुक्त की जाती है।
  • किसी व्यक्ति को सम्मान देने अथवा किसी बड़े व्यक्ति को इज्जत देने के लिए भी "है" में अनुस्वार बिंदु का प्रयोग किया जाता है।

Additional Information

  • अशुद्ध वाक्य :- मैं अपनी स्वेच्छा से आया हूँ।
  • शुद्ध वाक्य :- मैं स्वेच्छा से आया हूँ।
  • 'अपनी' शब्द अनावश्यक प्रयुक्त हुआ है।
  • अशुद्ध वाक्य :- आपके साथ उचित न्याय किया जाएगा।
  • शुद्ध वाक्य :- आपके साथ न्याय किया जाएगा।
  • (न्याय कभी अनुचित नहीं हो सकता है। अतः उचित शब्द अनावश्यक प्रयुक्त हुआ है।)
  • अशुद्ध वाक्य :- आज वर्षा होने की संभावना की जा रही है।
  • शुद्ध वाक्य :- आज वर्षा होने की संभावना है।

Question 14

निम्नलिखित में से 'अपादान तत्पुरुष' क��� उदाहरण है

  1. AOption (a)

    यथाशक्ति

  2. BOption (b)

    शरणागति

  3. COption (c)

    वज्राहत

  4. DOption (d)

    चोर-भयम् 

View answer & explanation

Correct answer

Answer: (d)D.

चोर-भयम् 

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "चोर-भयम्" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • चोरभयम् :- चोर से भय
  • पंचमी तत्पुरुष समास अथवा अपादान तत्पुरुष होगा।
  • परिभाषा:
  • ऐसे समास शब्द जिनमें अपादान कारक चिन्ह “से” का लोप हो , उन्हें अपादान तत्पुरुष समास कहा जाता है।

Important Points

  • अपादान तत्पुरुष समास के अन्य उदाहरण
  • अन्नहीन :- अन्न से हीन
  • ऋणमुक्त :- ऋण से मुक्त
  • कर्जमुक्त :- कर्ज से मुक्त।
  • कर्मरहित :- कर्म से रहित
  • कर्महीन :- कर्म से हीन
  • जीवनमुक्त :- जीवन से मुक्त
  • दूरागत :- दूर से आगत
  • देशनिकाला :- देश से  निकाला
  • देशनिर्वासित :- देश से निर्वासित
  • धनहीन :- धन से हीन

Additional Information

  • शरणागति
  • शरण को आया हुआ
  • कर्म तत्पुरुष समास
  • वज्राहत
  • वज्र से आहत
  • करण तत्पुरुष समास
  • यथाशक्ति 
  • शक्ति के अनुसार
  • अव्ययी भाव समास

Question 15

इनमें से किस वाक्य में 'अच्छा' शब्द विशेषण के रूप में प्रयुक्त हुआ है?

  1. AOption (a)

    आपने अच्छा किया, जो वहाँ नहीं गये।

  2. BOption (b)

    अच्छा, तुम अब घर जाओ।

  3. COption (c)

    अच्छा है, वह आज नहीं आया।

  4. DOption (d)

    यह काम बहुत अच्छा है।

View answer & explanation

Correct answer

Answer: (d)D.

यह काम बहुत अच्छा है।

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "यह काम बहुत अच्छा हुआ।" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • "यह काम बहुत अच्छा हुआ।" वाक्य में काम के अच्छा होने की विशेषता बताई गई है।
  • अतः यहाँ 'अच्छा' शब्द विशेषण के रूप में प्रयुक्त हुआ है।

Important Points

  • विशेषण की परिभाषा
  • जो संज्ञा अथवा सर्वनाम की विशेषता बताए उस शब्द को 'विशेषण' कहते है।
  • जैसे :- 'कालाकोट' और 'अच्छा लड़का' में 'काला' तथा 'अच्छा' शब्द विशेषण है।

Additional Information

  • विशेषण के भेद 
  • संख्यावाचक विशेषण (संख्या के रूप में विशेषता बतलाने वाले)
  • परिमाणवाचक विशेषण (मात्रा के रूप में विशेषता)
  • गुणवाचक विशेषण (गुण के रूप में)
  • सार्वनामिक विशेषण (सर्वनाम शब्द के रूप में, जैसे :- यह , वह , कौन , क्या , कोई , कुछ आदि।)

Question 16

कोष्ठक में अंकित (^) विराम चिन्ह का इनमें से एक नाम गलत है, वह है:-

  1. AOption (a)

    योजक - चिन्ह

  2. BOption (b)

    विस्मरण - चिन्ह

  3. COption (c)

    हंसपद चिन्ह

  4. DOption (d)

    काकपद चिन्ह

View answer & explanation

Correct answer

Answer: (a)A.

योजक - चिन्ह

Why this answer is correct

कोष्ठक में अंकित (^) विराम चिन्ह का इनमें से एक नाम गलत है, वह है: योजक - चिन्ह

उपर्युक्त विकल्पों में से विकल्प "योजक चिह्न" सही है, क्योंकि यह अलग अथवा गलत चिह्न है और प्रश्न में पूछा हुआ है की कौनसा चिह्न अलग है।

Key Points 

  • योजक चिह्न (-)
  • इस चिह्न को विभाजक या समास बोधक चिह्न भी कहते हैं।
  • इसका प्रयोग प्राय: द्वंद्व समास के बीच किया जाता है।
  • जैसे
  • माता-पिता दोनों खुश थे।

Additional Information

  • त्रुटिबोधक/हंसपद/काकपद/विस्मरण - चिन्ह (^)
  • वाक्य में जब कोई शब्द बीच में छूट जाता है तो इस चिह्न का प्रयोग करके छूटे हुए शब्द को लिखते हैं।
  • जैसे
  • 7
  • वह रोज^ बजे स्कूल जाता है।

Question 17

विलोम शब्दों की दृष्टि से इनमें से एक युग्म अशुद्ध है, वह है

  1. AOption (a)

    आतप - निरातप

  2. BOption (b)

    उदय - अस्त

  3. COption (c)

    कुटिल - जटिल

  4. DOption (d)

    गहरा - छिछला

View answer & explanation

Correct answer

Answer: (c)C.

कुटिल - जटिल

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "कुटिल-जटिल" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • कुटिल व जटिल दोनों का विलोम "सरल" होगा।
  • कुटिल
  • विशेषण
  • टेढ़ा।
  • छली, चालबाज।
  • जटिल
  • विशेषण
  • कठिन दुर्बोध
  • उलझा हुआ, पेचीदा।

Additional Information

  • प्रमुख विलोम शब्द
  • आध्यात्मिक :- आधिभौतिक
  • आम :- ऱवास
  • आकुँचन    :- प्रसारण
  • आलस्य :- स्फूर्ति
  • आदत्त :- प्रदत्त
  • अनृत :- ऋत, तथ्य
  • अभिमुख    :- विमुख
  • उदीची :- प्रतीची
  • उत्साह :- निरुत्साह, अनुत्साह
  • उपार्जित :- स्वयंप्राप्त
  • उत्पतन :- निपतन, अवतरण
  • उदात्त :- अनुदात्त
  • औद्धत्य    :- अनौद्धत्य
  • औदार्य :- अनौदार्य
  • अंतर  :- सतत
  • औपचारिक :- अनौपचारिक
  • गमन :- आगमन
  • गरल :-सुधा, अमृत

Question 18

इनमें से एक वाक्य में 'अपादान कारक' नहीं है, वह

  1. AOption (a)

    उसके हाथ से कलम गिरा।

  2. BOption (b)

    पतझड़ में पीपल के पेड़ों से पत्ते झड़ने लगते हैं।

  3. COption (c)

    वह कलम से लिखता है।

  4. DOption (d)

    नूतन को गंदगी से बहुत घृणा है।

View answer & explanation

Correct answer

Answer: (c)C.

वह कलम से लिखता है।

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "वह कलम से लिखता है।" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • वह कलम से लिखता है। इस वाक्य में 'कलम' करण है, क्योंकि लिखने का काम कलम से किया गया है।
  • करण कारक
  • इसके हिन्दी पर्याय 'से' के 'द्वारा' है।
  • जिसकी सहायता से कोई कार्य किया जाए, उसे करण कारक कहते हैं।

Additional Information

  • अपादान कारक
  • वाक्य में जिस स्थान या वस्तु से किसी व्यक्ति या वस्तु की पृथकता अथवा तुलना का बोध होता है, वहाँ अपादान कारक होता है।
  • प्रेम, घृणा, लज्जा, ईर्ष्या, भय और सीखने आदि भावों की अभिव्यक्ति के लिए अपादान कारक का ही प्रयोग किया जाता है।
  • क्योंकि उक्त कारणों से अलग होने की क्रिया किसी-न-किसी रूप में जरूर होती है।

Question 19

'किसी के पास रखी हुई दुसरे की वस्तु' - इस वाक्यांश के लिए इनमें से एक सही शब्द नहीं है

  1. AOption (a)

    रिक्थ

  2. BOption (b)

    थाती

  3. COption (c)

    अमानत

  4. DOption (d)

    धरोहर

View answer & explanation

Correct answer

Answer: (a)A.

रिक्थ

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "रिक्थ" सही है,क्योंकि यह सही विकल्प नहीं है। तथा अन्य विकल्प असंगत हैं।

Key Points

  • 'किसी के पास रखी हुई दूसरे की वस्तु' के लिए एक शब्द रिक्थ नहीं है।
  • 'वह जिसे उत्तराधिकार में धन या सम्पत्ति मिले या मिलने को हो' के लिए एक शब्द - रिक्थ होगा।
  • 'किसी के पास रखी हुई दूसरे की वस्तु' के लिए एक शब्द  : थाती/अमानत/धरोहर होगा।

Additional Information

  • रिक्थ
  • पुल्लिंग
  • वह सम्पत्ति जो उत्तराधिकारी को दी जाय।
  • वह धन-संपत्ति या ऐसी ही कोई और चीज जो किसी को उत्तराधिकारी के रूप में मिली हो या मिले।

Confusion Points

  • रिक्त का पर्यायवाची
  • ख़ाली, शून्य, कोतल, खाली, शून्यगर्भ, सारहीन।

Additional Information

  • अमानत का पर्यायवाची
  • थाती, धरोहर, उपनिधि, न्यास।
  • धरोहर शब्द स्त्रीलिंग शब्द है।

Question 20

'दुर्व्यवहार' शब्द में कितने उपसर्ग हैं?

  1. AOption (a)

    एक

  2. BOption (b)

    दो

  3. COption (c)

    तीन

  4. DOption (d)

    चार

View answer & explanation

Correct answer

Answer: (c)C.

तीन

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "तीन" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • दुर्व्यवहार शब्द में तीन उपसर्ग है।
  • दुर्व्यवहार :- दुर् + वि + अव + हार।
  • दुर्व्यवहार 
  • किसी शक्ति या अधिकार के अन्यायपूर्ण या अनौचित्यपूर्ण प्रयोग से किसी अन्य व्यक्ति, समूह, प्राणी, जीव या अन्य चीज़ के साथ करे गये बुरे या हानिकारक व्यवहार को कहते हैं।

Additional Information

  • एक से अधिक उपसर्गों का प्रयोग
  • कुछ शब्द एक से अधिक उपसर्गों के प्रयोग से बने होते हैं।
  • जैसे 
  • परोपकार :- पर + उप + कार
  • स्वावलम्बन :- स्व + अव + लम्बन
  • समालोचना :- सम् + आ + लोचना
  • समाचार :- सम् + आ + चार
  • अव्यवस्था :- अ + वि + अव + स्था

Question 21

निम्नलिखित में से शुद्ध वर्तनी का शब्द है

  1. AOption (a)

    संग्रहीत

  2. BOption (b)

    हिरण्यकश्यपु

  3. COption (c)

    अन्तर्धान

  4. DOption (d)

    अपन्हुति

View answer & explanation

Correct answer

Answer: (c)C.

अन्तर्धान

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "अंतर्धान" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • अंतर्धान शब्द शुद्ध शब्द है।
  • अंतर्धान शब्द अंतर्ध्यान शब्द का शुद्ध रूप है।
  • अंतर्धान
  • पुल्लिंग
  • लोप हो जाना।

Additional Information

  • अशुद्ध शब्द :- संग्रहित 
  • शुद्ध शब्द :- संगृहीत
  • संगृहीत
  • संग्रह किया हुआ, एकत्र किया हुआ; जमा किया हुआ, संकलित
  • ग्रस्त, जकड़ा हुआ
  • निगृहीत या संयत किया हुआ, शासित , आगत, प्राप्त, स्वीकृत।
  • अशुद्ध शब्द :- हिरण्यकश्यपु
  • शुद्ध शब्द :- हिरण्यकशिपु
  • अशुद्ध शब्द :- अपन्हुति
  • शुद्ध शब्द :- अपह्नुति
  • (अर्थ :- अलंकार का एक भाग)

Question 22

'अण्डे का शहजादा' इस मुहावरे का सही अर्थ है

  1. AOption (a)

    बहुत चालाक व्यक्ति

  2. BOption (b)

    अनुभवी व्यक्ति

  3. COption (c)

    अनुभवहीन व्यक्ति

  4. DOption (d)

    महामूर्ख व्यक्ति

View answer & explanation

Correct answer

Answer: (c)C.

अनुभवहीन व्यक्ति

Why this answer is correct

'अंडे का शहजादा’ मुहावरे का सही अर्थ है – अनुभवहीन व्यक्ति। अन्य विकल्प असंगत हैं। 

Key Points

स्पष्टीकरण:- 

मुहावरेअर्थवाक्य प्रयोग
अंडे का शहजादाअनुभवहीन व्यक्ति‌‌‌अजीत सिंह अपने पिता की सिफ़ारिश पर प्रमोशन तो पा गए लेकिन काम ढंग का नहीं आता, बस समझो कि वो अंडे का शहजादा है।

अन्य:-

  • अनुभवी व्यक्ति : घाट घाट का पानी पिया हुआ
  • महामूर्ख व्यक्ति : काठ का उल्लू

Additional Information

मुहावरा परिभाषाउदाहरण
किसी भाषा की अभिव्यंजना के विशिष्ट रूप को 'मुहावरा' कहते हैं। एक अन्य पक्ष है कि विशिष्ट शब्दों विचित्र प्रयोगों एवं प्रयोग-सिद्ध विशिष्ट वाक्यांशों वाक्य-पद्धति को 'मुहावरा' कहते हैं।मुहावरा - अन्धे के हाथ बटेर लगना अर्थ - इच्छित वस्तु का अचानक प्राप्त होना वाक्य - रहीम की लटरी खुलते ही ऐसा लगा मानो अन्धे के हाथ बटेर लग गई हो।

Question 23

अनेकार्थक शब्द 'आराम' का इनमें से एक अर्थ है

  1. AOption (a)

    सरोवर

  2. BOption (b)

    बगीचा

  3. COption (c)

    समीर

  4. DOption (d)

    लहर

View answer & explanation

Correct answer

Answer: (b)B.

बगीचा

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "बगीचा" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • बगीचा शब्द आराम का अनेकार्थी शब्द है।
  • आराम के अनेकार्थी
  • बाग, विश्राम, रोग का दूर होना।

Important Points

  • प्रमुख अनेकार्थी शब्द
  • कुशल :- खैरियत, चतुर ।
  • कल :- बीता हुआ दिन, आने वाला दिन, मशीन।
  • खग :- पक्षी, तारा, गन्धर्व, बाण।
  • कर्ण :- कर्ण (नाम), कान।
  • काम :- वासना, कामदेव, कार्य, पेशा, धंधा।
  • खल :- दुष्ट, धतूरा, दवा कूटने का खरल।
  • गण :- समूह, मनुष्य, भूतप्रेतादि, शिव के गण, पिंगल के गण।
  • अधर :- धरती (आकाश के बीच का स्थान), पाताल, नीचा, होंठ।
  • अंक :- भाग्य, गिनती के अंक, नाटक के अंक, चिन्ह संख्या, गोद।

Additional Information

  • सर के अनेकार्थी
  • तालाब, जलाशय, सरोवर, तड़ाग।
  • वायु के पर्यायवाची
  • पवन , हवा , वात , मारुत , अनिल , पवमान , समीर, स्पर्शन।
  • तरंग का अनेकार्थी शब्द
  • स्वर लहरी, लहर, उमंग।

Question 24

निम्नलिखित में से अन्तस्थ व्यंजन हैं

  1. AOption (a)

    श्, ष्, स्, ह्

  2. BOption (b)

    अ्, ब्, स्, द

  3. COption (c)

    ञ्, ण्, न्, म्

  4. DOption (d)

    य्, र्, ल्, व्

View answer & explanation

Correct answer

Answer: (d)D.

य्, र्, ल्, व्

Why this answer is correct

य्, र्, ल्, व्... अन्तस्थ व्यंजन हैं।

  • अंत करण के माध्यम से उच्चारित होने वाले व्यंजनों को अंतस्थ व्यंजन कहा जाता है।
  • अंतस्थ व्यंजन के उदाहरण- य', 'र', 'ल' और 'व'।
  • अंतर शब्द के अर्थ अंदर का, बीच में स्थित, भीतर रहने वाला, भीतरी, अन्त:स्थ, मध्य में रहने वाला इत्यादि होते हैं।

Important Points

उच्चारण स्थान (ध्वनि वर्ग)उच्चरित ध्वनि
द्वयोष्ठ्यप, फ, ब, भ, म
दन्त्योष्ठ्यफ़
दन्त्यत, थ, द, ध
वर्त्स्यन, स, ज़, र, ल, ळ
मूर्धन्यट, ठ, ड, ढ, ण, ड़, ढ़, र, ष
कठोर तालव्यश, च, छ, ज, झ
कोमल तालव्यक, ख, ग, घ, ञ, ख़, ग़
पश्च-कोमल-तालव्यक़
स्वरयंत्रामुखीह

Question 25

'दान की बछिया के दाॅंत नहीं देखे जाते' - इस लोकोक्ति का सही अर्थ है

  1. AOption (a)

    उपकार करते समय पैसे का लोभ उचित नहीं।

  2. BOption (b)

    मुफ्त के माल में दोष नहीं देखे जाते।

  3. COption (c)

    दान उपयुक्त वस्तु का होना चाहिए।

  4. DOption (d)

    दान की बछिया से मोह नहीं करना चाहिए।

View answer & explanation

Correct answer

Answer: (b)B.

मुफ्त के माल में दोष नहीं देखे जाते।

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प "मुफ़्त के माल में दोष नहीं देखे जाते।" सही है तथा अन्य विकल्प असंगत हैं।

Key Points

  • दान की बछिया के दाँत नहीं देखे जाते
  • अर्थ :- मुफ़्त के माल में दोष नहीं देखे जाते।
  • वाक्य प्रयोग
  • जब परिवार के सदस्य शादी की विदाई में मिले सूट के कपड़े की कीमत आंकने लगे तो पिताजी बोले कि दान की बछिया के दांत नहीं देखे जाते।

Additional Information

  • प्रमुख लोकोक्तियाँ
  • बुढि़या मरी तो आगरा तो देखा :- प्रत्येक घटना के दो पहलू होते हैं - अच्छा और बुरा।
  • लिखे ईसा पढ़े मूसा :- गंदी लिखावट।
  • अंधा सिपाही कानी घोड़ी, विधि ने खूब मिलाई जोड़ी :- जहां दो व्यक्ति हों और दोनों ही एक समान मूर्ख, दुष्ट या अवगुणी हों वहां ऐसा कहते हैं।
  • अपना रख पराया चख :- निजी वस्तु की रक्षा एवं अन्य वस्तु का उपभोग।
  • अच्छी मति जो चाहो बूढ़े पूछन जाओ :- बड़े बूढ़ों की सलाह से कार्य सिद्ध हो सकते हैं।
  • अब की अब, जब की जब के साथ :- सदा वर्तमान की ही चिन्ता करनी चाहिए।
  • छप्पर पर फूंस नहीं, ड्योढ़ी पर नाच :- दिखावटी ठाट-वाट परन्तु वास्तविकता में कुछ भी नहीं।
  • छछूंदर के सर पर चमेली का तेल :- अयोग्य के पास योग्य वस्तु का होना।
  • जिसके हाथ डोई, उसका सब कोई :- धनी व्यक्ति के सब मित्र होते हैं।
  • योगी था सो उठ गया आसन रहा भभूत :- पुराण गौरव समाप्त।

Technical (100 questions)

Question 26

Two identical R.C. coupled amplifiers each having a lower cut-off frequency (fc) are cascaded with negligible loading. The lower cut-off frequency of the overall amplifier is

  1. AOption (a)

    fc(2−1)\frac{f_c}{\sqrt{(\sqrt{2}-1)}}

  2. BOption (b)

    fc(2−1)f_c{\sqrt{(\sqrt{2}-1)}}

  3. COption (c)

    fc(2+1)\frac{f_c}{\sqrt{(\sqrt{2}+1)}}

  4. DOption (d)

    fc(2+1)f_c{\sqrt{(\sqrt{2}+1)}}

View answer & explanation

Correct answer

Answer: (a)A.

fc(2−1)\frac{f_c}{\sqrt{(\sqrt{2}-1)}}

Why this answer is correct

Concept:

Effect of cascading on the cutoff frequency and Bandwidth:

If N-Identical non-interacting amplifiers are cascaded and each amplifier has an individual cutoff frequency fL1 and fH1.

Then the overall lower cut-off frequency of the cascaded amplifier will be:

fL=fL121/N−1{f_L} = \frac{{{f_{{L_1}}}}}{{\sqrt {{2^{1/N}} - 1} }}        

Now, the overall upper cut-off frequency of the cascaded amplifier will be:

fH=fH1×21/N−1{f_H} = {f_{{H_1}}} \times \sqrt {{2^{1/N}} - 1}

For, 21/N−1<1\sqrt {{2^{1/N}} - 1} < 1 

{f_L} > {f_{{L_1}}};;& ;;{f_H} < {f_{{H_1}}}

Note: Cascading Increases lower cutoff and decreases higher cutoff frequency hence bandwidth decreases.

Calculation:

Given:

fL1= fL2 = fc

N = 2

The overall lower cutoff frequency (fL) is given as,

fL=fL121/N−1{f_L} = \frac{{{f_{{L_1}}}}}{{\sqrt {{2^{1/N}} - 1} }}

fL=fc2(1/2)−1{f_L} = \frac{{{f_{{c}}}}}{{\sqrt {{2^{(1/2)}} - 1} }} = fc(2−1)\frac{f_c}{\sqrt{(\sqrt{2}-1)}}

For a cascade of interacting amplifiers, net higher cutoff freq. is calculated as:

1fH=1.1×1f12+1f22+…\frac{1}{{{f_H}}} = 1.1 \times \sqrt {\frac{1}{{f_1^2}} + \frac{1}{{f_2^2}} + \ldots }

‘OR’

1fH=1f1+1f2+…\frac{1}{{{f_H}}} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} + \ldots

Where, f1, f2, f3, etc are pole frequencies, which are calculated from the Transfer function of a cascaded amplifier.

Net or overall lower cutoff is calculated as:

fL=1.1×fL12+fL22+…{f_L} = 1.1 \times \sqrt {f_{{L_1}}^2 + f_{{L_2}}^2 + \ldots }

Question 27

Super-position theorem based on

  1. Only linear principle
  2. Only homogeneity principle
  3. Both linearity and homogeneity principle 

Which one of the following is/are correct answer?

  1. AOption (a)

    1 and 2

  2. BOption (b)

    2 and 3

  3. COption (c)

    Only 3

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

Only 3

Why this answer is correct

Superposition Theorem:

  • It is stated that in any linear, active, bilateral network having more than one source, the response across any element is the Algebraic sum of the response obtained from each source considered separately and all other sources are replaced by their internal resistance.
  • The principle of the superposition theorem is based on linearity and homogeneity.
  • Voltage Source  →   short
  • Current source   →  open
  • Do not disturb the dependent source present in the network.

 

Limitations of SPT:

  • This theorem cannot be used to measure power.
  • This theorem is not applicable to unbalanced bridge circuits.
  • Applicable only to linear circuits not for nonlinear circuits.
  • Applicable only for the circuits having more than one source.

 

Application of SPT:

 

Superposition Theorem is applied to determine the current in one particular branch of a network containing several voltage source and/or current source.

Question 28

Which of the following statement is correct?

  1. AOption (a)

    The flow of current in electric circuit involves discontinuous expenditure of energy

  2. BOption (b)

    The flow of current in electric circuit requires energy for creating current but not to maintain it

  3. COption (c)

    In the magnetic circuit energy is needed continuously to maintain the flux

  4. DOption (d)

    In the magnetic circuit energy is needed for creating flux initially but not to maintain it

View answer & explanation

Correct answer

Answer: (d)D.

In the magnetic circuit energy is needed for creating flux initially but not to maintain it

Why this answer is correct

  • Flux is the line of force through a medium from N pole to S pole. It is measured in weber.
  • Though the electrical and magnetic circuits are alike in many ways they exhibit various dissimilarities as well.
  • In a magnetic circuit, the current is the actual flow of electrons whereas in a magnetic circuit flux is the direction of force- nothing flows between the N pole and S pole.
  • It is the same reason why energy is required to produce current and to maintain it in an electrical circuit whereas energy is required to produce flux but not for its maintenance.

Question 29

The values of current i and voltage V2 in the following circuit are respectively

  1. AOption (a)

    2 A, 4 V

  2. BOption (b)

    0.5 A, 6 V

  3. COption (c)

    1 A, 8 V

  4. DOption (d)

    2 A, 6 V

View answer & explanation

Correct answer

Answer: (c)C.

1 A, 8 V

Why this answer is correct

Concept:

Kirchhoff’s Laws:

  • Ohm’s law only applies to the simple circuit.
  • When we have to use Kirchhoff’s law along with Ohm’s law then we have sufficient powerful tools for analyzing a large variety of electric circuits.
  • These laws are formally known as Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL)

Kirchhoff’s Current Law (KCL): It states that the algebraic sum of currents entering a node or closed boundary is zero.

Mathematically, KCL implies that

\mathop \sum \limits_{n = 1}^N {i_n} = 0

Where N is the number of branches connected to the node

And in is the nth current entering or leaving the node.

By this law current entering a node may be regarded as positive and current leaving a node may be regarded as negative or vice-versa.

Considered the node in the figure shown below, in which i1, i3, and i5 are incoming current and i2, i4, and i6 are outgoing current.

Applying KCL gives,

i1 + i3 + i5 = i2 + i4 + i6

Hence, the sum of currents entering a node is equal to the sum of the currents leaving the node.

Note: KCL is based on the conservation of charge.

Kirchhoff’s Voltage Law (KVL): It states that the algebraic sum of all voltage around a close path or loop is zero.

Mathematically, KVL implies that

\mathop \sum \limits_{m = 1}^M {v_m} = 0

Where M is the number of voltages in a loop or number of branches in a loop

And, vm is mth voltage.

Consider a circuit shown below in which R1 and R2 are two resistance, v1, v2 are two voltage source that causes current (I) flow in the loop.

The sign of voltage drop across the passive element is such a way that the current enters from the positive terminal.

Applying KVL on this circuit

V1 + V4 – IR1 – IR2 = 0

Or, V1 + V4 = IR1 + IR2

Hence, the Sum of Voltage drops = Sum of Voltage rises

Note: KVL deals with the conservation of energy.

Calculation:

By applying KVL,

6 - 6i + 8i -8i = 0

⇒ I = 1 A

V2 = 8i = 8 V

Question 30

A measuring system consisting of a sensor, an amplifier and an oscilloscope, the sensitivity of each equipment is as follows, sensitivity of sensor; 0.4 mV/°C, gain of Amplifier; 5.0 V/mV and sensitivity of oscilloscope: 10 mV/V, the sensitivity of complete setup will be

  1. AOption (a)

    15.0 mV/°C

  2. BOption (b)

    15.4 mV/°C

  3. COption (c)

    20.0 mV/°C

  4. DOption (d)

    20.4 mV/°C

View answer & explanation

Correct answer

Answer: (c)C.

20.0 mV/°C

Why this answer is correct

Given that,

Sensitivity of sensor SA = 0.4 mV/°C

Sensitivity of amplifier gain SB = 5 V/mV

Sensitivity of Oscilloscope SC = 10 mv/V

Overall sensitivity = Product of all sensitivities = SA × SB × SC = .4 × 5 × 10 = 20 mV/°C

Question 31

Two point charges Q1 = 10 μC and Q2 = 20 μC are placed at co-ordinates (1, 1, 0) and (-1, -1, 0) respectively. The total electric flux passing through a plane z = 20 will be

  1. AOption (a)

    7.5 μC

  2. BOption (b)

    13.5 μC

  3. COption (c)

    15.0 μC

  4. DOption (d)

    22.5 μC

View answer & explanation

Correct answer

Answer: (c)C.

15.0 μC

Why this answer is correct

Concept:

  • As per gauss's law, total electric flux passing through a closed surface is equal to the total change enclosed in it.

ψ =Qene = ∮⃗D.⃗dS\oint⃗ D.⃗{dS}

  • The surface is chosen such that D⃗\vec D is normal or tangential to the gaussion surface.
  • When D⃗\vec D is normal to the surface, D⃗.dS⃗=Dds\vec D.\vec{dS}=D ds & when D⃗\vec D is tangential to the surface D⃗.dS⃗=0\vec D.\vec{dS}=0 Thus we must chosen a surface that has some of the symmatry exhibited by the charge distribution.

Calculation:

For the given configuration; for a cylindrical surface with z = 20 & z = -20 as top & bottom

Half of the flux will pass through z = 20 & other half with z = -20.

ψ = Total flux = Qenc = 20 + 10 μ c = 30 μ C

Flux passing through z = 20 ⇒ ψ2=302=15\frac{\psi}{2}=\frac{30}{2}=15  μ C

Question 32

A current of -8 + 6√2 sin (wt + 30) is passed through three ammeters. They are a centre zero PMMC meter, a true RMS meter and a moving iron meter. The respective reading (in A) will be

  1. AOption (a)

    8, 6, 6

  2. BOption (b)

    8, 10, 10

  3. COption (c)

    -8, 6, 6

  4. DOption (d)

    -8, 10, 10

View answer & explanation

Correct answer

Answer: (d)D.

-8, 10, 10

Why this answer is correct

Concept:

PMMC meterTrue RMS meterMoving iron meter (MI)
Measured valueAverage valueRMS valueRMS value
Working sourceDCACAC and DC
Working principleElectromagnetic inductionSee beck effectChange in self-inductance

 

Calculation:

Given current is

i(t) = −8+6;√2;(sin⁡ωt+30∘) - 8 + 6;√ 2 ;\left( {\sin \omega t + 30^\circ } \right) A

PMMC meter measures only DC value

∴ Reading of PMMC meter = -8 A

RMS value of current is

IRMS=Io2+IRMS12+…{I_{RMS}} = \sqrt {I_o^2 + I_{RMS1}^2 + \ldots }

Here, Io = -8 A

IRMS1 = (6√2 / √2) = 6 A

IRMS = √((-8)2 + 62) 

⇒ IRMS = 10 A.

RMS value of True RMS meter = 10 A

RMS value of MI meter = 10 A.

Points to remember:

  • PMMC with rectifier works in both AC and DC
  • MI with CT (Current transformer) works in AC only.

Question 33

A moving coil instrument of resistance 4 ohm, gives a full-scale deflection of 20 mA. It is desired to convert this instrument into a voltmeter to read voltage upto 30 V. The value of series resistance needed is

  1. AOption (a)

    896 ohm

  2. BOption (b)

    1296 ohm

  3. COption (c)

    1496 ohm

  4. DOption (d)

    1996 ohm

View answer & explanation

Correct answer

Answer: (c)C.

1496 ohm

Why this answer is correct

Concept:

To increase the range of a voltmeter, we need to the series resistance and it is given by

Rse=Rm(VVm−1);{R_{se}} = {R_m}\left( {\frac{V}{{{V_m}}} - 1} \right);

Where V is the required voltmeter range

Vm is the voltmeter range

Rm­ is the meter internal resistance

Calculation:

Given that, internal resistance (Rm) = 4 Ω full scale deflection (IFSD) = 20 mA

Voltage full scale deflection (VFSD) = IFSD × Rm = 80 mV

⇒ Vm = 80 mV and Rm = 4 Ω

Required potential difference (V) = 30 V

Series resistance (Rs)=(VVm−1)Rm;\left( {{R_s}} \right) = \left( {\frac{V}{{{V_m}}} - 1} \right){R_m};

=(3080×10−3−1)(4);= \left( {\frac{{30}}{{80 \times {{10}^{ - 3}}}} - 1} \right)\left( {4} \right);

= (374) (4) = 1496 Ω

Question 34

Two parallel conductors carry the same current in the same direction. What kind of mutual force they will experience?

  1. AOption (a)

    Repulsion

  2. BOption (b)

    Attraction

  3. COption (c)

    Zero

  4. DOption (d)

    Either A or B

View answer & explanation

Correct answer

Answer: (b)B.

Attraction

Why this answer is correct

CONCEPT:

Magnetic field:

  • The space or region around the current-carrying wire/moving electric charge or around the magnetic material in which force of magnetism can be experienced by other magnetic material is called as magnetic field/magnetic induction by that material/current.
  • It is denoted by B.
  • The magnetic force per unit length between two parallel wires is given by;

Fl=μo2πI1I2d\frac{F}{l} = \frac{{{\mu _o}}}{{2\pi }}\frac{{{I_1}{I_2}}}{d}

Where μ0 = permittivity of free space, I1 = current in a first wire, I2 = current in a second wire,d = distance between two wires, and l = length of current-carrying wire.

EXPLANATION:​ 

  • The magnetic force per unit length between two parallel wires is given by;

Fl=μo2πI1I2d\frac{F}{l} = \frac{{{\mu _o}}}{{2\pi }}\frac{{{I_1}{I_2}}}{d}

  • If the conductor carries the current in the same direction, then the force between them will be attractive.
  • If the conductor carries the current in the opposite direction, then the force between them will be repulsive. Therefore option 2 is correct.

Question 35

The sphere of radius 0.1 m has a charge of 0.04 μC. Which of the following value of potential inside the sphere is valid?

  1. AOption (a)

    720 volt

  2. BOption (b)

    3600 volt

  3. COption (c)

    0 volt

  4. DOption (d)

    3240 volt

View answer & explanation

Correct answer

Answer: (b)B.

3600 volt

Why this answer is correct

CONCEPT:

Electric potential:

  • The electric potential at any point in the electric field is defined as the amount of work done in moving a unit positive test charge from infinity to that point without acceleration.

⇒V=kqr⇒ V=\frac{kq}{r}

Where V = electric potential at a point, k = 9×109 N-m2/C2, q = charge, and r = distance of a point from the center of the charge

CALCULATION:

Given, 

Q = 0.04 × 10-6 C

r = .1 m

⇒ V = (9× 109 × 0.04 × 10-6) /.1 = 3600V.

As electric field intensity inside the metal sphere zero, the portion inside the sphere will become an equipotential surface means inside the sphere at any point the potential is equal to the potential at the surface.

Question 36

A current transformer has double turn primary and a 100 turn secondary winding. The secondary draws current of 6 A and magnetising ampere-turns is 60 A. What is actual transformation ratio?

  1. AOption (a)

    100.5

  2. BOption (b)

    101

  3. COption (c)

    50.24

  4. DOption (d)

    25.12

View answer & explanation

Correct answer

Answer: (c)C.

50.24

Why this answer is correct

Magnetizing ampere-turns = 60 A = NI

Magnetizing current (Im) = 30 A

Number of primary turns (N1) = 2

Number of secondary turns (N2) = 100

Nominal ratio, n=N2N1=1002=50n = \frac{{{N_2}}}{{{N_1}}} = \frac{{100}}{2} = 50

Exciting current (I0) = Im = 30 A

The rated secondary current (I2) = 6 A

Reflected secondary current = nI2 = 50 × 6 = 300 A

Primary current, I1=(nI2)2+Im2{I_1} = \sqrt {{{\left( {n{I_2}} \right)}^2} + I_m^2}

=3002+302=301.49;A= \sqrt {{{300}^2} + {{30}^2}} = 301.49;A

Actual ratio =I1I2=301.496=50.249= \frac{{{I_1}}}{{{I_2}}} = \frac{{301.49}}{6} = 50.249

Question 37

In which bias condition, zener diode is used as a voltage regulator?

  1. AOption (a)

    Forward bias

  2. BOption (b)

    Reverse bias

  3. COption (c)

    Both forward and reverse bias

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

Reverse bias

Why this answer is correct

CONCEPT:

Zener diode:

  • A Zener diode is a semiconductor device that allows current to flow either in a forward or reverse direction.
  • Since it is highly doped, the depletion layer is very thin and has a high voltage gradient and electric field.

​

EXPLANATION:

  • The Zener diode will be operated in the reverse breakdown region so that the voltage will remain constant. Therefore option 3 is correct.
  • In that region, the voltage will be equal to the breakdown voltage and this will make the Zener diode work as a voltage regulator.

Additional Information

Zener Breakdown:

  • If the reverse-biased voltage increases, the electrons from the covalent bonds within the depletion region come out and make the depletion region conductive.
  • This is the Zener breakdown and the voltage at this point is the Zener voltage.
  • After the breakdown, there is a sharp increase in the reverse current irrespective of the external voltage supplied.

Question 38

A D.C. commulatively compound motor delivers rated load torque at rated speed. If the series field is short-circuited, then armature current and speed will

  1. AOption (a)

    Both increase

  2. BOption (b)

    Bothe decrease

  3. COption (c)

    Increase and decrease respectively

  4. DOption (d)

    Decrease and increase respectively

View answer & explanation

Correct answer

Answer: (a)A.

Both increase

Why this answer is correct

Circuit diagram of a cumulative compound motor:

Where,

Ish is the current through the shunt field

Ia is the armature current

IL is the load current

Rsh is thew shunt field resistance

Ra is the armature resistance

Rse series field resistance

Eb is the induced EMF

V is the terminal voltage

For a cumulative compound motor

When series field winding is short-circuited:

  • Then The motor will behave like a DC shunt motor.
  • As the series field is short-circuited, only the shunt field will provide flux.
  • Therefore flux of the motor has been decreased, the speed will increase slightly and would remain almost constant only if the shunt current remains constant.

When shunt winding is short-circuited: 

  • The load current will pass completely through the shunt winding
  • Then there won't be any current through the armature of the motor
  • Therefore it won't produce any induced EMF even though field excitation is given.
  • Hence, the motor will stop rotating.

When series field winding is open-circuited:

  • Then there won't be any current through the armature of the motor
  • Therefore it won't produce any induced EMF even though field excitation is given.
  • Hence, the motor will stop rotating.

​

When shunt field winding is open-circuited:

  • Mainly flux will be produced by shunt winding in the cumulative compound motor.
  • As more flux is reduced the motor will rotate at dangerously high speed.

Question 39

A series resonant circuit has L = 1 mH and C = 10 F, the required R (in Ω) for BW = 15.9 Hz is

  1. AOption (a)

    0.1

  2. BOption (b)

    0.2

  3. COption (c)

    0.0159

  4. DOption (d)

    500

View answer & explanation

Correct answer

Answer: (a)A.

0.1

Why this answer is correct

Concept:

In a series resonant circuit,

Bandwidth (BW) =frQ=fH−fL=R2πL= \frac{{{f_r}}}{Q} = {f_H} - {f_L} = \frac{R}{2π L}

Where fr is the resonant frequency

Q is the quality factor

fH is the higher half-power frequency

fL is the lower half-power frequency

Resonant frequency (fr) is defined as:

fr=12πLC=fHfLf_r= \frac{1}{{2π \sqrt {LC} }} = \sqrt {{f_H}{f_L}}

Calculation:

Given that,

Resistance = R Ω

Inductance L = 1 mH

Bandwidth BW = 15.9 Hz

And we know that,

BW = (R/2πL)

15.9 = (R/2π× 1 × 10-3)

⇒ R = .1 Ω

Question 40

The 'ampere second' could be the unit of

  1. AOption (a)

    Conductances

  2. BOption (b)

    Power

  3. COption (c)

    Energy

  4. DOption (d)

    Charge

View answer & explanation

Correct answer

Answer: (d)D.

Charge

Why this answer is correct

Current is defined as the charge flowing through a particular cross section in unit time 

i.e. I = Q/t

in terms of units 

Ampere = coulombs/sec

Ampere- Sec = Coulombs = unit of charge 

Regarding Other units in options

Power P=I2R=watt{\rm{P}} = {{\rm{I}}^2}{\rm{R}} = {\rm{watt}}

Energy E=P.t=Joules{\rm{E}} = {\rm{P}}.{\rm{t}} = {\rm{Joules}}

Question 41

Two long parallel wires A and B, 12 cm apart carry current of 750 A and 500 A respectively in opposite direction. The flux density at the mid point of the perpendicular line between wire will be

  1. AOption (a)

    0.833 × 10-3 Wb

  2. BOption (b)

    4.167 × 10-3 T

  3. COption (c)

    0.833 × 10-3 T

  4. DOption (d)

    0 Wb

View answer & explanation

Correct answer

Answer: (b)B.

4.167 × 10-3 T

Why this answer is correct

Flux density due to infinite long line = H = I2πρaϕ^\frac{I}{2π \rho}\hat{a_{\phi}}

HA=7502π(0.06)H_A=\frac{750}{2π (0.06)} (inside the paper)

HB=5002π(0.06)H_B=\frac{500}{2π (0.06)} (outside the paper)

Total flux density = H = HA + HB = 12502π(0.06)=3315.72 Am\frac{1250}{2π (0.06)}=3315.72 \ \frac{A}{m}

B = μ H

= 4 π × 10-7 × 3315.72 = 4.16 × 10-3

Question 42

For depletion type MOSFET : ID = 4.5 mA at VGS = -2 V. What is the value of IDSS if VP = -5V ?

  1. AOption (a)

    5.5 mA

  2. BOption (b)

    12.5 mA

  3. COption (c)

    5.0 mA

  4. DOption (d)

    15 mA

View answer & explanation

Correct answer

Answer: (b)B.

12.5 mA

Why this answer is correct

Formula Used:

ID=IDSS(1−VGSVP)2I_D=I_{DSS}(1-\frac{V_{GS}}{V_P})^2

Application:

Given,

ID = 4.5 mA

VGS = -2 V

VP = -5V

From above concept,

4.5 × 10-3 = IDSS (1−−2−5)2(1-\frac{-2}{-5})^2

or, IDSS = 4.5×10−3(1−0.4)2=12.5 mA\frac{4.5\times 10^{-3}}{(1-0.4)^2}=12.5\ mA

Question 43

The percentage liming error, in the case of an instrument reading of 10 A with a 0 to 200 A ammeter having a guaranteed accuracy of 1% full scale reading is

  1. AOption (a)

    10%

  2. BOption (b)

    20%

  3. COption (c)

    30%

  4. DOption (d)

    50%

View answer & explanation

Correct answer

Answer: (b)B.

20%

Why this answer is correct

Formula: 

%Guaranteed;accuracy;error;=Limiting;Error;(L.E.)Full;scale;value×100{\rm{\% Guaranteed;accuracy;error;}} = \frac{{Limiting;Error;\left( {L.E.} \right)}}{{Full;scale;value}} \times 100

%Limiting;Error;(%L.E.);=Limiting;Error;(L.E.)True;value×100{\rm{\% L}}imiting;Error;\left( {\% L.E.} \right){\rm{;}} = \frac{{Limiting;Error;\left( {L.E.} \right)}}{{True;value}} \times 100

Calculation:

Given-

Full scale reading = 200 A

Guaranteed accuracy = 1% of full-scale reading = 0.01 × 200 = 2 A

Measured value = 10 A

Limiting error =210×100=20% = \frac{{2}}{{10}} \times 100 = 20\%

Question 44

A change in 700 mV in base emitter voltage causes a change of 200 μA in the base current. Determine dynamic input resistance. 

  1. AOption (a)

    2 kΩ

  2. BOption (b)

    10 kΩ

  3. COption (c)

    3 kΩ

  4. DOption (d)

    3.5 kΩ

View answer & explanation

Correct answer

Answer: (d)D.

3.5 kΩ

Why this answer is correct

Concept:

Dynamic input resistance is defined as the ratio of change in input voltage or base voltage (VBE) to the corresponding change in input current or base current (IB), with the output voltage or collector  voltage (VCE) kept at constant.  rd = VBE / IB 

Calculation: 

For the given question: VBE = 700 mV, IB = 200µA

⇒ rd = 0.7/0.0002

= 3500 ohm = 3.5 kΩ

Question 45

The Q – meter works on the principle of

  1. AOption (a)

    A mutual inductance

  2. BOption (b)

    Self inductance

  3. COption (c)

    Series resonance

  4. DOption (d)

    Parallel resonance

View answer & explanation

Correct answer

Answer: (c)C.

Series resonance

Why this answer is correct

Q meter is an instrument which is designed to measure the value of quality factor directly and it is useful in measuring the characteristics of coils and capacitors.

Important:

  • Q meter works on the principle of series resonance.
  • At series resonance, the voltage across capacitor is equal to Q times of applied input voltage. Thus, we can measure the value of Q directly by connecting a voltmeter across capacitor.
  • It is used for the measurement of quality factor, inductance, self-capacitance, inductance, bandwidth and capacitance of a coil.

Question 46

If temperature increases then conductivity of a semiconductor is

  1. AOption (a)

    Decreases

  2. BOption (b)

    Increases

  3. COption (c)

    Constant

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

Increases

Why this answer is correct

The electrical conductivity of a semiconductor material increases with increasing temperature. Due to increase in temperature more electron-hole pair are generated and hence conductivity increases.

Question 47

The point form of Ampere's Law is given by

  1. AOption (a)

    Curl (B) = I

  2. BOption (b)

    Curl (H) = J

  3. COption (c)

    Curl (D) = J

  4. DOption (d)

    Curl (V) = I

View answer & explanation

Correct answer

Answer: (b)B.

Curl (H) = J

Why this answer is correct

Maxwell equations:

  • are an extension of the works of Gauss, Faraday, and Ampere
  • help to study the application of both electrostatic and magnetic fields
  • can be written in integral form and point form
  • need to be modified depending upon the media involved in the problem.

 

Important Points:

Maxwell’s equation for static electromagnetic fields are as shown:

Differential (or point form)Integral formEquation Name
∇⃗⋅D⃗=ρv\vec \nabla \cdot \vec D = {\rho _v}\mathop \oint \nolimits_s \vec D \cdot d\vec s = \mathop \smallint \nolimits_v {\rho _v}dGauss’s law for the static electric field
∇⃗⋅B⃗=0\vec \nabla \cdot \vec B = 0∮sB⃗⋅dS⃗=0\mathop \oint \nolimits_s \vec B \cdot d\vec S = 0Gauss’s law for the static magnetic field.
∇⃗×E⃗=0\vec \nabla \times \vec E = 0∮LE⃗⋅dℓ⃗=0\mathop \oint \nolimits_L \vec E \cdot d\vec \ell = 0Conservative nature of the electrostatic field
∇⃗×H⃗=J⃗\vec \nabla \times \vec H = \vec J\mathop \oint \nolimits_L \vec H \cdot d\vec \ell = \mathop \smallint \nolimits_s \vec J \cdot d\vec sAmpere’s law

Question 48

The open circuit test in a 100 kVA, 2000/200 V, 50 Hz, single phase transformer gives 200 V, 90 A, 5 kW on low voltage side. Core loss and magnetizing components of the current approximately are respectively

  1. AOption (a)

    9.0 A, 89.5 A

  2. BOption (b)

    12.5 A, 88.7 A

  3. COption (c)

    25.0 A, 86.5 A

  4. DOption (d)

    30.0 A, 84.9 A

View answer & explanation

Correct answer

Answer: (c)C.

25.0 A, 86.5 A

Why this answer is correct

Concept:

  • Open circuit test is used to determine the no-load current and losses of the transformer because of which their no-load parameters are determined.
  • This test is performed on the primary winding of the transformer.
  • The wattmeter, ammeter, and voltage are connected to their primary winding
  • The nominal rated voltage is supplied to their primary winding with the help of the ac source.

 

 

  • The secondary winding of the transformer is kept open, and the voltmeter is connected to their terminal.
  • This voltmeter measures the secondary induced voltage.
  • As the secondary of the transformer is open, thus no-load current flows through the primary winding.
  • The value of no-load current is very small as compared to the full rated current.
  • The copper loss occurs only on the primary winding of the transformer because the secondary winding is open.
  • The reading of the wattmeter only represents the core and iron losses.
  • The core losses of the transformer are the same for all types of loads.

Formula:

W0 = wattmeter reading

V1 = voltmeter reading

I0 = Ammeter reading

∴ Iron loss of the transformer, Pi = W0

W0 = V1 I0 cos ϕ0          ---(1)

No-load power factor =cos⁡ϕ0=W0V1I0= \cos {ϕ _0} = \frac{{{W_0}}}{{{V_1}{I_0}}} 

Working component or core loss component, IW=W0V1{I_W} = \frac{{{W_0}}}{{{V_1}}}        ---(2)

From (1) and (2)

IW = I0 cos ϕ0

Calculation:

Given:

W0 = 5000 W

V1 = 200 V

I­0 = 90 A

Core loss component ⇒IW=W0V1⇒ {I_W} = \frac{{{W_0}}}{{{V_1}}} 

IW=5000200{I_W} = \frac{{5000}}{{200}}

IW = 25 A

And we know that

I02 = Im2 + IW2

⇒ 902 = Im2 + 252

⇒ Magnetizing current Im = 86.46 A

Question 49

Spherical charge density distribution is given by

ρv=ρv0(1−r2a2)ρ_v=ρ_{v_0}\left(1-\frac{r^2}{a^2}\right); r < a

ρv = 0; r > a

The value of total charge and maximum value of E will be

i) 0, 415πρv0a\frac{4}{15}\pi\rho_{v_0}a

ii) 0, 0.745 a

iii) 815πρv0a3\frac{8}{15}\pi\rho_{v_0}a^3, 0.745 a

iv) 415πρv0a3\frac{4}{15}\pi\rho_{v_0}a^3, 0.745 a2

  1. AOption (a)

    i, ii are correct

  2. BOption (b)

    ii, iii are correct

  3. COption (c)

    only ii correct

  4. DOption (d)

    only iii correct

View answer & explanation

Correct answer

Answer: (d)D.

only iii correct

Why this answer is correct

ρv = ρa (1−r2a2)\left(1-\frac{r^2}{a^2}\right) r < a

ρv = 0 r > a

We know that Qenclosed = ∫ρv dV\int \rho_v \ dV

For spherical distribution dV = r2 sin θ dr dθ dϕ

∴ Qenc=∫vρa(1−r2a2)r2sin⁡θ dr dθ dϕ\rm Q_{enc}=\displaystyle\int_v \rho_a\left(1-\frac{r^2}{a^2}\right)r^2\sin \theta \ dr\ d\theta\ d\phi

=∫02π∫0π∫0aρa(1−r2a2)r2sin⁡θ dr dθ dϕ=\displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi}\displaystyle\int_0^{a} \rho_a\left(1-\frac{r^2}{a^2}\right)r^2\sin \theta \ dr\ d\theta\ d\phi

=4π∫0aρa(1−r2a2)r2sin⁡θ dr=4\pi\displaystyle\int_0^{a} \rho_a\left(1-\frac{r^2}{a^2}\right)r^2\sin \theta \ dr

=4πρ0[r33−r55a2]0a=4\pi \rho_0 \left[\frac{r^3}{3}-\frac{r^5}{5a^2}\right]_0^a     ......(1)

Qenc=815πρ0a3\rm Q_{enc}=\frac{8}{15}\pi \rho_0 a^3

∫D.ds=Qenc\int D.ds=Q_{enc}

∮E.ds=Qencϵ0\oint E.ds=\frac{Q_{enc}}{\epsilon_0}

For the closed spherical surface

E.4πr2=1ϵ0∫0rρdvE.4\pi r^2=\frac{1}{\epsilon_0}\displaystyle\int_0^r\rho dv

E.4πr2=1ϵ04πρ0[r33−r55a2]E.4\pi r^2=\frac{1}{\epsilon_0}4 \pi\rho_0 \left[\frac{r^3}{3}-\frac{r^5}{5a^2}\right]

E=ρ0ϵ0[r3−r35a2]E=\frac{\rho_0}{\epsilon_0} \left[\frac{r}{3}-\frac{r^3}{5a^2}\right]

For max value of E ⇒ dEdr=0\frac{dE}{dr}=0

dEdr=ρ0ϵ0[13−3r25a2]=0\frac{dE}{dr}=\frac{\rho_0}{\epsilon_0}\left[\frac{1}{3}-\frac{3r^2}{5a^2}\right]=0

On solving we get r = 0.74 a

Question 50

For dynamic characteristics of any instruments out of the following which qualities are desirable?

i. Speed of response

ii. Lag

iii. Fidelity

iv. Dynamic error

  1. AOption (a)

    i, ii only correct

  2. BOption (b)

    ii, iii only correct

  3. COption (c)

    iii, iv only correct

  4. DOption (d)

    i, iii only correct

View answer & explanation

Correct answer

Answer: (d)D.

i, iii only correct

Why this answer is correct

The performance characteristics of an instrument are mainly divided into two categories: 

i) Dynamic characteristics 

ii) Static characteristics 

Dynamic characteristics: 

The set of criteria defined for the instruments, which are changes rapidly with time, is called ‘dynamic characteristics’. The various dynamic characteristics are:

 i) Speed of response

 ii) Measuring lag

 iii) Fidelity

iv) Dynamic error

Static characteristics: 

The set of criteria defined for the instruments, which are used to measure the quantities which are slowly varying with time or mostly constant, i.e., do not vary with time, is called ‘static characteristics’. 

The various static characteristics are: 

i) Accuracy  

ii) Precision  

iii) Sensitivity  

iv) Linearity  

v) Reproducibility  

vi) Repeatability  

viii) Threshold  

ix) Drift  

x) Stability  

xi) Tolerance  

xii) Range or span  

(i) Fidelity: It is defined as the degree to which a measuring instrument is capable of faithfully reproducing the changes in input, without any dynamic error.  

(ii) Resolution: It is the smallest quantity being measured which can be detected with certainty by an instrument.

(iii) Threshold: If the instrument input is increased very gradually from zero there will be some minimum value below which no output change can be detected. This minimum value defines the threshold of the instrument. 

(iv) Hysteresis:  Hysteresis is a phenomenon that illustrates the different output effects when loading and unloading.

Question 51

The equivalent circuit of a zener diode consists of

  1. AOption (a)

    A resistance in parallel with DC battery

  2. BOption (b)

    A capacitance in series with DC battery

  3. COption (c)

    A capacitance in series with current source

  4. DOption (d)

    ‘A small dynamic resistance in series with DC battery

View answer & explanation

Correct answer

Answer: (d)D.

‘A small dynamic resistance in series with DC battery

Why this answer is correct

CONCEPT:

Zener diode:

  • A Zener diode is a semiconductor device that allows current to flow either in a forward or reverse direction.
  • Since it is highly doped, the depletion layer is very thin and has a high voltage gradient and electric field.

​

  • Generally, the equivalent circuit of the ideal Zener diode (RB) is represented by a constant voltage source or dc battery.
  • But, in the case of practical Zener diode, the equivalent is represented by a small dynamic resistance in series with DC battery

Additional Information

  • The Zener diode will be operated in the reverse breakdown region so that the voltage will remain constant. Therefore option 3 is correct.
  • In that region, the voltage will be equal to the breakdown voltage and this will make the Zener diode work as a voltage regulator.

Zener Breakdown:

  • If the reverse-biased voltage increases, the electrons from the covalent bonds within the depletion region come out and make the depletion region conductive.
  • This is the Zener breakdown and the voltage at this point is the Zener voltage.
  • After the breakdown, there is a sharp increase in the reverse current irrespective of the external voltage supplied.

Question 52

Consider the following bridges, used for measurement of inductance:

  1. Maxwell bridge
  2. Owen bridge
  3. Hay's bridge
  4. Anderson bridge

Bridge used for measurement of inductance over a wide range are

  1. AOption (a)

    1 and 2 only

  2. BOption (b)

    2 and 3 only

  3. COption (c)

    3 and 4 only

  4. DOption (d)

    2 and 4 only

View answer & explanation

Correct answer

Answer: (c)C.

3 and 4 only

Why this answer is correct

  • A bridge circuit that is used to measure the resistance & inductance of coils with high Q-factor is known as Hays Bridge.
  • This is the modification of Maxwell’s bridge. So this bridge is used to determine the high-quality factor in the circuit.
  • The Anderson’s bridge gives an accurate measurement of the self-inductance of the circuit.
  • The bridge is the advanced form of Maxwell’s inductance capacitance bridge.
  • In Anderson bridge, the unknown inductance is compared with the standard fixed capacitance which is connected between the two arms of the bridge.

Mistake Points All four bridges can be used for the measurement of inductance, Here question is about the measurement of inductance in a wide range,  Anderson and Hays bridges will be used for a wide range of inductance measurements. So this answer is correct.

Question 53

If Vx=xx2+y2,Vy=yx2+y2,Vz=0\rm V_x=\frac{x}{x^2+y^2},V_y=\frac{y}{x^2+y^2},V_z=0, the divergence and curl in cylindrical co-ordinates will be

i. 1r2+1r,0\frac{1}{r^2}+\frac{1}{r},0

ii. r2+1r2,0r^2+\frac{1}{r^2}, 0

iii. 0, 1r2+1r\frac{1}{r^2}+\frac{1}{r}

iv. 0, r2+1r2r^2+\frac{1}{r^2}

v. 0, 0

  1. AOption (a)

    only iii

  2. BOption (b)

    i, iii

  3. COption (c)

    only v

  4. DOption (d)

    iii, v

View answer & explanation

Correct answer

Answer: (c)C.

only v

Why this answer is correct

Concept:

Divergence in rectangular coordinates ∇⃗.A⃗=∂Ax∂x+∂Ay∂y+∂Az∂z\vec {\nabla}.\vec A=\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}

Divergence in cylindrical coordinates ∇⃗.A⃗=1ρ∂∂ρ(ρAρ)+1ρ∂Aϕ∂ϕ+∂Az∂z\vec {\nabla}.\vec A=\frac{1}{\rho}\frac{\partial}{\partial\rho}(\rho A_{\rho})+\frac{1}{\rho}\frac{\partial A_{\phi}}{\partial \phi}+\frac{\partial A_{z}}{\partial z}

Curl in the rectangular coordinate ∇⃗×A⃗=∣ax^ay^az^∂∂x∂∂y∂∂zAxAyAz∣\vec {\nabla}\times\vec A=\begin{vmatrix}\hat{a_x}&\hat{a_y}&\hat{a_z}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ A_x&A_y&A_z\end{vmatrix}

Curl in cylindrical coordinates ∇⃗×A⃗=1ρ∣aρρaϕ^az^∂∂ρ∂∂ϕ∂∂zAρρAϕAz∣\vec {\nabla}\times\vec A=\frac{1}{\rho}\begin{vmatrix}a_{\rho}&\rho \hat{a_\phi}&\hat{a_z}\\ \frac{\partial}{\partial \rho}&\frac{\partial}{\partial \phi}&\frac{\partial}{\partial z}\\ A_\rho&\rho_{A\phi}&A_z\end{vmatrix}

Calculation:

First divergence & curl calculated in rectangular coordinates then connected to cylindrical coordinates

given V⃗=Vxax^+Vyay^+Vzaz^\vec V=V_x\hat{a_x}+V_y\hat{a_y}+V_z\hat{a_z}

Vx=xx2+y2V_x=\frac{x}{x^2+y^2}

Vy=yx2+y2V_y=\frac{y}{x^2+y^2}

Vz = 0

Divergence:- 

∇⃗.V⃗=∂∂x[xx2+y2]+∂∂y[yx2+y2]\vec {\nabla}.\vec V=\frac{\partial}{\partial x}\left[\frac{x}{x^2+y^2}\right]+\frac{\partial}{\partial y}\left[\frac{y}{x^2+y^2}\right]

∇⃗.V⃗=x2+y2−2x2(x2+y2)2+x2+y2−2y2(x2+y2)2\vec {\nabla}.\vec V=\frac{x^2+y^2-2x^2}{(x^2+y^2)^2}+\frac{x^2+y^2-2y^2}{(x^2+y^2)^2}

∇⃗.V⃗=0\vec {\nabla}.\vec V=0

Divergence = 0

Curl:

∇⃗×V⃗=∣ax^ay^az^∂∂x∂∂y∂∂zxx2+y2yx2+y20∣\vec {\nabla}\times\vec V=\begin{vmatrix}\hat{a_x}&\hat{a_y}&\hat{a_z}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ \frac{x}{x^2+y^2}&\frac{y}{x^2+y^2}&0\end{vmatrix}

∇⃗×V⃗=ax^(−∂∂z(yx2+y2))−ay^(∂∂z(xx2+y2))\vec {\nabla}\times\vec V=\hat{a_x}\left(-\frac{\partial}{\partial z}\left(\frac{y}{x^2+y^2}\right)\right)-\hat{a_y}\left(\frac{\partial}{\partial z}\left(\frac{x}{x^2+y^2}\right)\right)

+az^(∂∂x(yx2+y2)−∂∂y(xx2+y2))+\hat{a_z}\left(\frac{\partial}{\partial x}\left(\frac{y}{x^2+y^2}\right)-\frac{\partial}{\partial y}\left(\frac{x}{x^2+y^2}\right)\right)

∇⃗×V⃗=0\vec {\nabla}\times\vec V=0

curl = 0

Question 54

A series Circuit Containing a circuit element has following current and applied voltage

V = 200 Sin (2000t + 50°) V

i = 4 cos (2000t + 13.2°) A

The element Comprising the circuit is

  1. AOption (a)

    A pure inductor

  2. BOption (b)

    A pure capacitor

  3. COption (c)

    A practical capacitor

  4. DOption (d)

    A practical inductor

View answer & explanation

Correct answer

Answer: (c)C.

A practical capacitor

Why this answer is correct

Given that,

V = 200 Sin (2000t + 50°) V

i = 4 cos (2000t + 13.2°) A

We know that,

sin (90 + ϕ) = cos ϕ 

⇒ V = 200 sin (2000t + 50°) V  --------- (i)

⇒ i = 4 sin (2000t + 13.2°+ 90°) A

⇒ i = 4 sin (2000t + 103.2°) A    ------- (ii)

From, (i) and (ii) we can observe that,

Current 'i' is leading voltage V by 53.2° 

Hence, the element is practical capacitor

Question 55

Which of the following is NOT a sine wave oscillator?

  1. AOption (a)

    R-C phase shift oscillator

  2. BOption (b)

    Wein bridge oscillator

  3. COption (c)

    Calpits oscillator

  4. DOption (d)

    Astable oscillator

View answer & explanation

Correct answer

Answer: (d)D.

Astable oscillator

Why this answer is correct

Phase shift oscillator:

  • The phase shift oscillator is a linear electronic circuit that produces a sine wave output.
  • It consists of an inverting amplifier element such as a transistor or op-amp with its output feedback to its input through a phase shift network consisting of resistors and capacitors in a ladder network.
  • The feedback network shifts the phase of the amplifier output by 1800 at the oscillation frequency to give positive feedback.

 

Wein bridge oscillator :

  • The Wein bridge oscillator uses two RC networks connected together to produce a sinusoidal oscillator.
  • The Wein bridge oscillator uses a feedback circuit consisting of a series RC circuit connected with a parallel RC of the same component values producing a phase delay or phase advance depending upon the circuit frequency
  • At the resonant frequency, the phase shift is 00.

 

Clapp oscillator:

  • The Clapp oscillator is an LC oscillator that uses a particular combination of an inductor and three capacitors to set the oscillator frequency.
  • The Clapp is often drawn as a Colpitts oscillator that has an additional capacitor placed in series with the inductor.
  • This comes under linear or harmonic oscillators, which produce a sine wave output.
  • Clapp oscillator is also one kind of phase shift oscillator containing L, C elements, and a transistor or op-amp along with feedback, so it provides phase shift.

 

​Relaxation oscillator:

  • A relaxation oscillator is a nonlinear electronic oscillator circuit that produces nonsinusoidal repetitive output signals such as a triangle wave or square wave.
  • Relaxation oscillators are generally used to produce low-frequency signals.
  • These oscillators will not provide any phase shift in their output
  • Examples of Relaxation oscillators are Astable multivibrator, flyback or sweep oscillator, etc.

Question 56

An energy meter having a meter constant of 1200 revolutions per kWh is found to make 5 revolutions in 75 seconds. The load power is

  1. AOption (a)

    100 W

  2. BOption (b)

    200 W

  3. COption (c)

    250 W

  4. DOption (d)

    300 W

View answer & explanation

Correct answer

Answer: (b)B.

200 W

Why this answer is correct

Concept:

  • The number of revolutions made by the energy meter per kilowatt-hour is known as the meter constant of an energy meter.
  • Unit of meter constant is revolution per kilowatt-hour (rev/kWh)
  • It is constant for a particular energy meter.

 

Meter constant (K) = No. of revolution by meter/Energy consumed (E)

Calculation:

Number of revolution = K × E = K × P × t

∴5=1200×P×753600\therefore 5 = 1200 × P × \frac{{75}}{{3600}}

⇒P=0.2 kW=200W\Rightarrow P = 0.2 \ kW= 200W

Question 57

Select incorrect statement for N-P-N transistor.

  1. AOption (a)

    In a N-P-N transistor, Majority charge carrier are holes

  2. BOption (b)

    In a N-P-N transistor majority charge carrier are electrons

  3. COption (c)

    Mobility of electrons is higher than that of holes

  4. DOption (d)

    Emitter arrow shows direction of flow of conventional current

View answer & explanation

Correct answer

Answer: (a)A.

In a N-P-N transistor, Majority charge carrier are holes

Why this answer is correct

CONCEPT:

  • The material which is not a good conductor or a good insulator is called as a semiconductor.
  • For example: Silicon
  • The charge carriers which are present in more quantity in a semiconductor compared to other particles are called the majority charge carrier.

There are two types of semiconductors:

P-type semiconductor:

  • The semiconductor having holes as majority charge carriers and electrons as a minority charge carrier is called as P-type semiconductor.

N-type semiconductor:

  • The semiconductor having electrons as majority charge carriers and holes as a minority charge carrier are called as N-type semiconductor.

EXPLANATION:

NPN Transistor:

An N-P-N transistor is a type of bipolar junction transistor (BJT) and consists of three layers of semiconductor material: the emitter, base, and collector. These layers are typically made from silicon.

In a N-P-N transistor:

  1. The emitter layer is made of N-type semiconductor material, which means it has an excess of electrons as majority carriers. Electrons are negatively charged, and they are the majority charge carriers in the N-type material.

  2. The base layer is made of P-type semiconductor material, which has an excess of holes as majority carriers. Holes are positively charged vacancies left when electrons move from the P-type material to the N-type material.

  3. The collector layer is also made of N-type semiconductor material, similar to the emitter.

  4. The arrow in the transistor symbol typically points from the emitter to the base, and it indicates the direction of conventional current flow (from positive to negative). Conventional current flow is opposite to the actual flow of electrons, so it points in the direction of hole movement.

So, let's break down each statement:

  1. In a N-P-N transistor, Majority charge carriers are holes. (This is incorrect, as the majority charge carriers in the N-P-N transistor are electrons in the N-type regions and holes in the P-type base region.)

  2. In a N-P-N transistor, majority charge carriers are electrons. (This is correct. Electrons are the majority charge carriers in the N-type emitter and collector regions.)

  3. Mobility of electrons is higher than that of holes. (This is generally correct. In most semiconductor materials, electrons have higher mobility than holes.)

  4. Emitter arrow shows the direction of flow of conventional current. (This is correct. The arrow in the transistor symbol indicates the direction of conventional current flow, which is opposite to the direction of electron flow.)

So, the only incorrect statement is statement 1, where incorrectly states that the majority charge carriers in an N-P-N transistor are holes. In reality, the majority charge carriers in the emitter and collector regions are electrons, and in the base region, they are holes.

Question 58

The value of point charge will be, if it is required to hold four equal point charges +q each in equilibrium at the corners of the square, the point charge is placed at center of square

  1. AOption (a)

    Q=22+14qQ=\frac{2\sqrt{2}+1}{4}q

  2. BOption (b)

    Q=32+14qQ=\frac{3\sqrt{2}+1}{4}q

  3. COption (c)

    Q=33+14qQ=\frac{3\sqrt{3}+1}{4}q

  4. DOption (d)

    Q=42+14qQ=\frac{4\sqrt{2}+1}{4}q

View answer & explanation

Correct answer

Answer: (a)A.

Q=22+14qQ=\frac{2\sqrt{2}+1}{4}q

Why this answer is correct

Consider a square of side '2a' and q charge at the corner, Q charge at the center.

The diagram can be drawn as,

Consider, FR is the resultant force at point A due to corner point charge q.

It is required to hold four equal point charges +q each in equilibrium at the corners of the square.

In that case,

FR⃗=FQ⃗\vec{F_R}=\vec{F_Q} .... (1)

FR⃗=FAB⃗+FCA⃗+FAD⃗\vec{F_R}=\vec{F_{AB}}+\vec{F_{CA}}+\vec{F_{AD}}

Since FAB⃗\vec{F_{AB}}= FCA⃗\vec{F_{CA}} and apart from each other by 90°.

FR⃗=FAD⃗+2FAB⃗\vec{F_{R}}=\vec{F_{AD}}+\sqrt 2 \vec{F_{AB}} .... (2)

By using Coulomb law,

|FAB| = |FCA| = kq24a2\frac{kq^2}{4a^2}

|FAD| = kq28a2\frac{kq^2}{8a^2}

And, |FQ| = kqQ2a2\frac{kqQ}{2a^2}

From equations (1) and (2),

kqQ2a2\frac{kqQ}{2a^2} = kq28a2\frac{kq^2}{8a^2} + 2×kq24a2\sqrt2 \times \frac{kq^2}{4a^2}

⇒ Q=q4+2a2qQ=\frac{q}{4}+\frac{\sqrt 2a}{2q}

⇒ Q=q+22q4Q=\frac{q+2\sqrt2 q}{4}

Hence,​

Q=22+14qQ=\frac{2\sqrt{2}+1}{4}q

Question 59

A moving coil galvanometer is made into a d. c. ammeter by connecting

  1. AOption (a)

    A high resistance in series with the meter

  2. BOption (b)

    A low resistance across the meter

  3. COption (c)

    A pure inductance across the meter

  4. DOption (d)

    A capacitor in series with the meter

View answer & explanation

Correct answer

Answer: (b)B.

A low resistance across the meter

Why this answer is correct

  • A galvanometer can be converted into an ammeter by connecting a shunt resistance parallel to it
  • A shunt is basically a wire of very small resistance
  • A galvanometer can be converted into a voltmeter by connecting a very high resistance in the series
  • A voltmeter is used to measure the potential difference whereas an ammeter is used to measure the electric current
  • A galvanometer is used to measure a small electric current in a circuit

 

Points to remember:

A galvanometer can be converted into a voltmeter by connecting a resistance in series with it

Question 60

Which operating condition is satisfied by the transistor if is supposed to function in cut-off region ?

  1. AOption (a)

    VCE > 0

  2. BOption (b)

    VCE = 0

  3. COption (c)

    VCE < 0

  4. DOption (d)

    VCE = VCC

View answer & explanation

Correct answer

Answer: (d)D.

VCE = VCC

Why this answer is correct

Operating Regions of Transistor:

RegionsEmitter JunctionCollector JunctionApplication
ActiveForward BiasedReverse BiasedAmplifier
SaturationForward BiasedForward BiasedOn switch
Cut offReverse BiasedReverse BiasedOff switch
Reverse ActiveReverse BiasedForward BiasedAttenuator

 

Cut-off: ​

  • The point where the load line intersects the IB = 0 curve is known as a cut-off.
  • At this point, IB = 0, and only a small collector current (i.e. collector leakage current ICEO) exists.
  • At cut-off, the base-emitter junction no longer remains forward biased, and normal transistor action is lost.
  • The collector-emitter voltage is nearly equal to VCC i.e. VCE (cut off) = VCC

Additional Information 

  • The transistor mainly works as either amplifier, ON switch, or OFF switch.
  • Working regions corresponding to these applications of the transistor are plotted on a graph between collector current (IC) and collector to emitter voltage (VCE).
  • This graph is obtained when a line is drawn joining saturation and cutoff points. The line obtained is known as the load line.
  • The point where the load line intersects with the output characteristic curve is known as the Quiescent point or Q-point.

 

The following graph shows the representation of the Q-point.

  • When the transistor operates as an amplifier, its signal handling capacity is maximum because amplification increases the strength and bandwidth of the input signal.
  • Amplification of signal is achieved in the active region only which is present in the middle of the load line graph.

Question 61

For Stokes Theorem which of the following is correct?

  1. AOption (a)

    ∮SA.dS=∮VdivA dV\rm \displaystyle\oint_SA.dS=\displaystyle\oint_VdivA\ dV

  2. BOption (b)

    ∫S(∇×A).dS=∮SA. dL\rm \displaystyle\int_S(\nabla\times A).dS=\displaystyle\oint_SA.\ dL

  3. COption (c)

    ∫V(∇.A)dV=∮A.dl\rm \displaystyle\int_V(\nabla. A)dV=\displaystyle\oint A.dl

  4. DOption (d)

    ∫S(∇×A).dS=∮A.dS\rm \displaystyle\int_S(\nabla\times A).dS=\displaystyle\oint A.dS

View answer & explanation

Correct answer

Answer: (b)B.

∫S(∇×A).dS=∮SA. dL\rm \displaystyle\int_S(\nabla\times A).dS=\displaystyle\oint_SA.\ dL

Why this answer is correct

CONCEPT:

Stokes theorem:

  • Stoke's theorem enables us to transform the surface integral of the curl of the vector field A into the line integral of that vector field A over the boundary C of that surface and vice-versa. The theorem states.
  • The flux of the curl of a vector function A over any surface S of any shape is equal to the line integral of the vector field A over the boundary C of that surface i.e.

∬S,curl,,A.ds=∮C,A.dl\underset{S}{\mathop \iint },curl,,A.ds=\underset{C}{\mathop \oint },A.dl

  • It states that the line integral of a vector field F⃗\vec F around any closed surface C is equal to the surface integral of the normal component of curl of vector F⃗\vec F over an unclosed surface ‘S’.

Gauss divergence theorem:

  • It states that the surface integral of the normal component of a vector function F⃗\vec F taken over a closed surface ‘S’ is equal to the volume integral of the divergence of that vector function F⃗\vec F taken over a volume enclosed by the closed surface ‘S’.
  • Mathematically, it can be written as:

⇒∬S,F⃗.n^ds=∭V∇.F⃗dv\Rightarrow \underset{S}{\mathop \iint },\vec{F}.\hat{n}ds=\iiint\limits_{V}{\nabla .\vec{F}dv}

Question 62

For an n-channel silicon FET with a = 3 × 10-4 cm and ND = 1015 electron/cm3, what is the pinch - off voltage if the dielectric constant of silicon is ϵ = 12ϵ0 and ϵ0=136π×10−9\rm \epsilon_0 = \frac{1}{36 \pi} \times 10^{-9}  ?

  1. AOption (a)

    6.8 V

  2. BOption (b)

    5.2 V

  3. COption (c)

    8.8 V

  4. DOption (d)

    9.2 V

View answer & explanation

Correct answer

Answer: (a)A.

6.8 V

Why this answer is correct

Formula Used:

Pinch off voltage (VP) = a2qND2ϵ\frac{a^2qN_D}{2\epsilon}

Application:

Given,

a = 3 × 10-4 cm

ND = 1015 electron/cm3

ϵ = 12ϵ0

q = 1.6 × 10-19 C

From above concept,

VP = (3×10−4)2×1.6×10−19×10152×12ϵ0\frac{(3\times 10^{-4})^2\times 1.6\times 10^{-19}\times 10^{15}}{2\times 12 \epsilon_0}

Where, ϵ0=136π×10−9\rm \epsilon_0 = \frac{1}{36 \pi} \times 10^{-9}

After solving it,

VP = 6.8 V

Question 63

What is the value of shunt resistance of an ammeter, if the range is extended from 1 mA to 10 mA and metre has a resistance of 27 Ω ?

  1. AOption (a)

    3 Ω

  2. BOption (b)

    4 Ω

  3. COption (c)

    2.5 Ω

  4. DOption (d)

    3.5 Ω

View answer & explanation

Correct answer

Answer: (a)A.

3 Ω

Why this answer is correct

Concept:

We can extend the range of the ammeter by keeping a shunt resistance.

Here Rm = internal resistance of the coil

Rsh = Shunt resistance

I = Required full-scale range

Im = Full scale deflection of current

As the two resistances, Rm and Rsh are in parallel, the voltage drop across the resistance is equal.

ImRm=(I−Im)Rsh{I_m}{R_m} = \left( {I - {I_m}} \right){R_{sh}}

Rm=(IIm−1)Rsh{R_m} = \left( {\frac{I}{{{I_m}}} - 1} \right){R_{sh}}

⇒Rsh=Rm(IIm−1)⇒ {R_{sh}} = \frac{{{R_m}}}{{\left( {\frac{I}{{{I_m}}} - 1} \right)}}

⇒Rsh=Rm(m−1)⇒ {R_{sh}} = \frac{{{R_m}}}{{\left( {m - 1} \right)}}

Where m=IImm = \frac{I}{{{I_m}}}

‘m’ is called multiplying power

Calculation:

Given that,

Rsh = Shunt resistance

I = Required full-scale range = 10 mA

Im = Full scale deflection of current = 1 mA

And meter resistance Rm = 27 Ω  

⇒ m = 10 / 1 = 10

Rsh=Rm(m−1) {R_{sh}} = \frac{{{R_m}}}{{\left( {m - 1} \right)}}

⇒ Rsh=27(10−1)=3Ω {R_{sh}} = \frac{{{27}}}{{\left( {10- 1} \right)}} = 3 \Omega

Question 64

In a common base configuration, the alpha of the transistor is 0.99, its collector current is 1 mA and the collector to base current with emitter open is 1 μA. The value of base current is  

  1. AOption (a)

    29 μA

  2. BOption (b)

    19 μA

  3. COption (c)

    9 μA

  4. DOption (d)

    39 μA

View answer & explanation

Correct answer

Answer: (c)C.

9 μA

Why this answer is correct

Concept:

In a common base connection,

IE = IB + IC

And IC = α IE + ICBO

Where α is the current amplification factor

IC=α1−αIB+11−αICBOI_C= \frac{α}{1-α} I_B+\frac {1}{1-α}I_{CBO}

Calculation:

Given that,

α = .99

ICBO = 1 μA

IC = 1 mA

⇒ IC=α1−αIB+11−αICBOI_C= \frac{α}{1-α} I_B+\frac {1}{1-α}I_{CBO}

⇒ 1mA=.991−.99IB+11−.99×1μA1 mA= \frac{.99}{1-.99} I_B+\frac {1}{1-.99} \times 1 μ A

⇒ IB ≈ 9 μA

Question 65

A shunt of 200 Ω resistance used with a galvanometer of 1000 Ω resistance, determine the value of shunt resistance to give a multiplying power of 50

  1. AOption (a)

    20.4 Ω

  2. BOption (b)

    18 Ω

  3. COption (c)

    64 Ω

  4. DOption (d)

    21 Ω

View answer & explanation

Correct answer

Answer: (a)A.

20.4 Ω

Why this answer is correct

Concept:  

  • Shunts are used for the range extension of ammeters.
  • A shunt is a low-value resistance having minimum temperature co-efficient and is made up of manganin. Because manganin has a very low value of temperature coefficient.
  • It is connected in parallel with the ammeter whose range is to be extended. The combination is connected in series with the circuit whose current is to be measured.
  • Shunt provides a path for extra current as it is connected across (in parallel with) the instrument.
  • These shunted instruments can be used to measure currents many times greater than their normal full-scale deflection currents.
  • The ratio of maximum current (with shunt) to the full-scale deflection current (without shunt) is known as the ‘multiplying power’ or ‘multiplying factor’ of the shunt.

In the figure

I is total current flowing in the circuit

Ish is the current through the shunt resistor

Rm is the ammeter resistance

Formula:

Rse = Rm(M – 1)

Rm=VmIm{R_m} = \frac{{{V_m}}}{{{I_m}}}

Rsh=RmM−1{R_{sh}} = \frac{{{R_m}}}{{M - 1}}

M= Multiplying factor = (Required full scale deflection) / (Initial full scale deflection)

Where,

Rsh = Series resistance

Rm = Meter resistance

Vm = Potential difference across the meter

Im = Meter current

Calculation:

Given 

Rm = 1000 Ω

M=50M = 50 

Rsh=100050−1=20.4;ΩR_{sh} = \frac{{1000}}{{50 - 1}} = 20.4 ;Ω

Rsh =  20.4 Ω

Question 66

Find the current density of a material with resistivity 20 units and electric field intensity 2000 units.

  1. AOption (a)

    400

  2. BOption (b)

    300

  3. COption (c)

    200

  4. DOption (d)

    100

View answer & explanation

Correct answer

Answer: (d)D.

100

Why this answer is correct

Concept:  

The current density (D) with respect to resistivity and electric field intensity(E) is given by the formula D=EresistivityD = {E \over resistivity}

Calculation:

For the given question;

Resistivity = 20 units, E = 2000 units

 D=200020D = {2000 \over 20} = 100 units.

Question 67

When Moving iron instrument is used to measure AC quantities which types of error occurs?

i. Friction

ii. Stray magnetic field

iii. Waveform

iv. frequency

  1. AOption (a)

    i, ii, iv are correct

  2. BOption (b)

    ii, iii, iv are correct

  3. COption (c)

    i, ii, iii are correct

  4. DOption (d)

    i, ii, iii, iv all are correct

View answer & explanation

Correct answer

Answer: (d)D.

i, ii, iii, iv all are correct

Why this answer is correct

When moving iron instrument is used to measure AC quantities following errors occur:

  • Hysteresis error is a serious type of error in MI instruments. The effect of this error is that the readings are higher when the current increases than when it decreases.
  • Stray magnetic field error may be serious because of the weakness of the operating magnetic field. This error can be reduced by enclosing the movement in an iron case (or) shielding the instrument with the help of a steel case.
  • Temperature error is due to a change of temperature that affects the instrument resistance and stiffness of the control spring.
  • Friction error is due to the friction of moving parts this can be avoided by making the torque-weight ratio high
  • Frequency Error: Change in frequency may because errors due to changes in the reactance of the operating coil and also to the changes in the magnitude of eddy current set up in the metal parts of the instrument near the operating coil.
  • Error due to reactance of the instrument coil: The change in instrument coil reactance owing to change in frequency causes serious error in the case of voltmeters.
  • Eddy current error affects both ammeters and voltmeters equally. The frequency error may be compensated for by connecting a suitable capacitor C in parallel with swamping resistance R (C = L/R2, if the frequency is not too high.
  • MI instruments may be seriously affected by waveform both on account of the change in the form of flux waveform and in the case of voltmeter, the effect of harmonic upon the inductance.

Hence, all statements are correct.

Question 68

A loss less transmission line having 50 Ω characteristic impedance and length λ4\frac{\lambda}{4} is short circuited at one end and connected to an ideal voltage source of 1 V at the other end. The current drawn from the voltage source is

  1. AOption (a)

    zero

  2. BOption (b)

    0.02

  3. COption (c)

    50

  4. DOption (d)

    infinity

View answer & explanation

Correct answer

Answer: (a)A.

zero

Why this answer is correct

Concept: 

The input impedance for a loss less transmission line for a quater wave(λ4\lambda \over 4 )  Zin = Zo2 / ZL 

Where,

Zo is characteristic impedance, ZL  is load impedance

Calculation: 

For the given question,

Zo = 50 ohm, ZL = 0 (short-circuited)

Therefore, Zin = 502 / 0 = infinite

Current drawn = I = Vs / Zin = 1/infinite = 0

Question 69

For the following Waveform, match the correct option.

1.A.Form factor = 23\frac{2}{\sqrt{3}}
2.B.Form factor = Unity
3.C.Form factor = 1−4α3π1−απ\frac{\sqrt{1-\frac{4\alpha}{3\pi}}}{1-\frac{\alpha}{\pi}}
4.D.Form factor = π22\frac{\pi}{2\sqrt{2}}
  1. AOption (a)

    A - 2, B - 4 , C - 3, D - 1

  2. BOption (b)

    A - 2, B - 4 , C - 1, D - 3

  3. COption (c)

    A - 3, B - 4 , C - 2, D - 1

  4. DOption (d)

    A - 3, B - 4 , C - 1, D - 2

View answer & explanation

Correct answer

Answer: (c)C.

A - 3, B - 4 , C - 2, D - 1

Why this answer is correct

Explanation:

Form Factor:

The form factor of an alternating current waveform is the ratio of the RMS value to the average value.

\(FormFactor=\frac{RMSValue}{Average~Value}\)

Peak factor

  • The peak factor is the peak amplitude of the waveform divided by the RMS value of the waveform.

\(PeakFactor=\frac{PeakAmplitude}{RMS~Value}\)

  • For square waveform form factor and peak factor is the same and value is one.

IMPORTANT EVALUATIONS:

WAVEFORMSHAPEMAX. VALUEAVERAGE VALUERMS VALUEFORM FACTORCREST FACTOR
SINUSOIDAL WAVEAm{A_m}2Amπ\frac{{2{A_m}}}{\pi }Am2\frac{{{A_m}}}{{\sqrt 2 }}Am22Amπ=1.11\frac{{\frac{{{A_m}}}{{\sqrt 2 }}}}{{\frac{{2{A_m}}}{\pi }}} = 1.11AmAm2=2\frac{{{A_m}}}{{\frac{{{A_m}}}{{\sqrt 2 }}}} = \sqrt 2
SQUARE WAVEAm{A_m}Am{A_m}Am{A_m}AmAm=1\frac{{{A_m}}}{{{A_m}}} = 1AmAm=1\frac{{{A_m}}}{{{A_m}}} = 1
TRIANGULAR WAVEAm{A_m}Am2\frac{{{A_m}}}{2}Am3\frac{{{A_m}}}{{\sqrt 3 }}Am3Am2=23\frac{{\frac{{{A_m}}}{{\sqrt 3 }}}}{{\frac{{{A_m}}}{2}}} = \frac{2}{{\sqrt 3 }}AmAm3=3\frac{{{A_m}}}{{\frac{{{A_m}}}{{\sqrt 3 }}}} = \sqrt 3
HALF-WAVE RECTIFIED WAVEAm{A_m}Amπ\frac{{{A_m}}}{\pi }Am2\frac{{{A_m}}}{2}Am2Amπ=π2\frac{{\frac{{{A_m}}}{2}}}{{\frac{{{A_m}}}{\pi }}} = \frac{\pi }{2}22

Question 70

Three resistance each of R ohm are connected in delta, its equivalent star will comprise three resistance each of value

  1. AOption (a)

    3R ohm

  2. BOption (b)

    2R ohm

  3. COption (c)

    R3\frac{R}{3} ohm

  4. DOption (d)

    R2\frac{R}{2} ohm

View answer & explanation

Correct answer

Answer: (c)C.

R3\frac{R}{3} ohm

Why this answer is correct

Concept:

For interconversions of delta and star, when all resistances are the same we can directly conclude two points

  1. From Delta to Star resistance will be divided by 3
  2. From Star to Delta resistance is multiplied by 3

Shortcuts for conversion:

Delta connection is shown below

Star connection of resistors is shown as

To convert Star from Delta

Branch resistance = product of connected resistance/sum of all resistances

For branch resistance PS, Ra and Rb are connected

RA=Ra×RbRa+Rb+Rc{R_A} = \frac{{{R_a} \times {R_b}}}{{{R_a} + {R_b} + {R_c}}}

For branch resistance QS, Ra and Rc are connected

RB=Ra×RcRa+Rb+Rc{R_B} = \frac{{{R_a} \times {R_c}}}{{{R_a} + {R_b} + {R_c}}}

For branch resistance, SR, Rb and Rc are connected

RC=(Rc×Rb)Ra+Rb+Rc{R_C} = \frac{{\left( {{R_c} \times {R_b}} \right)}}{{{R_a} + {R_b} + {R_c}}}

To convert Delta from Star

Branch resistance = ∑ connected resistances + product;of;connected;resistanceresistance;which;is;not;connected\frac{{product;of;connected;resistance}}{{resistance;which;is;not;connected}}

For branch resistance PQ, RA and RB are connected

Ra=RA+RB+RA×RBRC{R_a} = {R_A} + {R_B} + \frac{{{R_A} \times {R_B}}}{{{R_C}}}

For branch resistance PR, RB and RC are connected

Rb=RC+RB+RC×RBRC{R_b} = {R_C} + {R_B} + \frac{{{R_C} \times {R_B}}}{{{R_C}}}

For branch resistance QR, RA and RC are connected

Rc=RC+RA+RA×RCRB{R_c} = {R_C} + {R_A} + \frac{{{R_A} \times {R_C}}}{{{R_B}}}

Question 71

If a coil has a resistance of 10 Ω and an inductance of 1 H, what will be value of current 0.1 sec. after switching on a 500 V d. c. supply?

  1. AOption (a)

    6.32 A

  2. BOption (b)

    31.6 A

  3. COption (c)

    3.7 A

  4. DOption (d)

    4.0 A

View answer & explanation

Correct answer

Answer: (b)B.

31.6 A

Why this answer is correct

Concept: 

For a source included RL circuit the current values at any instance of time is given by the formula.

i(t) = i(∞) + [i(0) - i(∞)]e-t/τ

where i(∞) = value of current after a long time

i(0) = value of current at t = 0, instance of switch close/open.

Calculation:

for the give circuit

τ=LR=110=0.1 sτ = \frac{L}{R} = \frac{1}{10}= 0.1\ s; i(0) = 0; i(∞) = 50

i(t) = i(∞) + (i(0) - i(∞))e-t/τ 

at t = 0.1 sec

i(0.1) = 50 + (-50)e-0.1/0.1

i(0) = 50 - 50e\frac{50}{e} = 31.605 Amp

Question 72

Above pinch off voltage in a JFET the drain current

  1. AOption (a)

    Decreases

  2. BOption (b)

    Increases sharply

  3. COption (c)

    Remains constant

  4. DOption (d)

    Both 1 and 2

View answer & explanation

Correct answer

Answer: (c)C.

Remains constant

Why this answer is correct

In JFET after pinch off the drain current becomes almost constant.

Explanation:

Shockley's equation is given as:

ID=IDSS(1−VGSVP)2{I_D} = {I_{DSS}}{\left( {1-\frac{{{V_{GS}}}}{{{V_P}}}} \right)^2}

Where,

VGS = Gate to source voltage

IDSS = Drain to source saturation current

VP = Pinch-off voltage

According to Shockley's equation and characteristics curves for a P-channel JFET, the Drain current ID decreases with an increasing positive Gate-Source voltage (VGS).

The Drain current is zero when VGS = VP. For normal operation, VGS is biased to be somewhere between VP and 0. 

The characteristics curves for a P-channel junction field-effect transistor are shown below

Case I: If VDS = 0 and VGS = 0, the device will be idle with no current i.e. IDS = 0.

Case II:

  • Now consider VDS to be negative while VGS is 0.
  • At this state, the current flows from the source to the drain (as per conventional direction) as the holes within the p-substrate move towards the drain while being repelled from the source.
  • The value of this current is restricted only by the channel-resistance and is seen to increase with a decrease in VDS (Ohmic region).
  • However once the pinch-off occurs (VDS = VP), the current IDS saturates at a particular level IDSS, during which the device acts as a constant current source.

 

Case III:

  • Next, let VGS = positive while VDS is negative.
  • Here the effect exhibited is similar to that in Case II with the fact that the saturation occurs at a faster rate as the VGS becomes more and more positive.
  • Here the current stop or ceases to flow as the value of VDS becomes equal to VP, turning the device into OFF state.

Question 73

The current wave form as shown in figure is passed through a resistance of 5 Ω. The average power will be

  1. AOption (a)

    167 W

  2. BOption (b)

    500 W

  3. COption (c)

    100 W

  4. DOption (d)

    33.33 W

View answer & explanation

Correct answer

Answer: (a)A.

167 W

Why this answer is correct

Concept:

Average Value:

The average value can be calculated by dividing the area by time period for once cycle

Vavg=12VmTT=VM2=0.5;Vm{V_{avg}} = \frac{{\frac{1}{2}{V_m}T}}{T} = \frac{{{V_M}}}{2} = 0.5;{V_m}

Rms value:

{V_{rms}} = \sqrt {\frac{1}{T}\mathop \smallint \limits_0^T {{\left( {\frac{{{V_m}}}{T}t} \right)}^2}dt;} ;

Vrms = Vm / √3 = 0.577 Vm

Note: The result of RMS value for standard triangular waveform and sawtooth waveform should be remembered as Vm/√3

For square wave RMS value is Vmax

Calculation:

The rms current of the given waveform is given by

Irms = .577 Im = .577 × 10 = 5.77 A

Average Power =(Irms)2R = 5.772 × 5 = 166.67 ≈ 167 W

Question 74

A three-phase 500 V motor load has a p.f. of 0.4. Two watt meter are Connected to measure, read 20 kW and 10 KW. Find the reactive power

  1. AOption (a)

    10 kVar

  2. BOption (b)

    17.32 kVar

  3. COption (c)

    30 kVar

  4. DOption (d)

    51.92 kVar

View answer & explanation

Correct answer

Answer: (b)B.

17.32 kVar

Why this answer is correct

Concept:

Power in a three-phase circuit is,

P=√3VLILcos⁡ϕP = √ 3 {V_L}{I_L}\cos \phi

Two wattmeter method:

The connection diagram using wattmeters as shown below.

W1=IRVRBcos⁡(IR ∧VRB){W_1} = {I_R}{V_{RB}}\cos \left( {{I_R} \ ^\wedge{V_{RB}}} \right)

W2=IYVYBcos⁡(IY∧VYB){W_2} = {I_Y}{V_{YB}}\cos \left( {{I_Y}^\wedge{V_{YB}}} \right)

From the phasor diagram,

IR∧VRB=30−ϕ{I_R}^\wedge{V_{RB}} = 30 - \phi

IY∧VYB=30+ϕ{I_Y}^\wedge{V_{YB}} = 30 + \phi

W1=IRVRBcos⁡(30−ϕ){W_1} = {I_R}{V_{RB}}\cos \left( {30 - \phi } \right)

⇒W1=VLILcos⁡(30−ϕ)\Rightarrow {W_1} = {V_L}{I_L}\cos \left( {30 - \phi } \right)

W2=IYVYBcos⁡(30+ϕ){W_2} = {I_Y}{V_{YB}}\cos \left( {30 + \phi } \right)

⇒W2=VLILcos⁡(30+ϕ)\Rightarrow {W_2} = {V_L}{I_L}\cos \left( {30 + \phi } \right)

W1+W2=VLIL[cos⁡(30−ϕ)+cos⁡(30+ϕ)]{W_1} + {W_2} = {V_L}{I_L}\left[ {\cos \left( {30 - \phi } \right) + \cos \left( {30 + \phi } \right)} \right]

=√3VLILcos⁡ϕ= √ 3 {V_L}{I_L}\cos \phi

⇒ Total three-phase power =;W1+W2=√3VLILcos⁡ϕ = {\rm{;}}{{\rm{W}}_1} + {{\rm{W}}_2} = √ 3 {V_L}{I_L}\cos \phi

Total three-phase power is the sum of two wattmeters.

W1=VLILcos⁡(30−ϕ){W_1} = {V_L}{I_L}\cos \left( {30 - \phi } \right)

W2=VLILcos⁡(30+ϕ){W_2} = {V_L}{I_L}\cos \left( {30 + \phi } \right)

W1−W2=√3VphIphsin⁡ϕ{W_1} - {W_2} = √ 3 {V_{ph}}{I_{ph}}\sin \phi

√3(W1−W2)=3VphIphsin⁡ϕ√ 3 \left( {{W_1} - {W_2}} \right) = 3{V_{ph}}{I_{ph}}\sin \phi

Reactive power =√3;(W1−W2)= \surd 3;\left( {{W_1}-{W_2}} \right)

Reactive power is equal to √3 times the difference between the readings of the two wattmeters.

W1+W2=3VphIphcos⁡ϕ{W_1} + {W_2} = 3{V_{ph}}{I_{ph}}\cos \phi

√3(W1−W2)=3VphIphcos⁡ϕ√ 3 \left( {{W_1} - {W_2}} \right) = 3{V_{ph}}{I_{ph}}\cos \phi

⇒ϕ=tan⁡−1(√3(W1−W2)W1+W2)\Rightarrow \phi = {\tan ^{ - 1}}\left( {\frac{{√ 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)

Power factor =cos;ϕ= cos;\phi

cosϕ=cos⁡[tan⁡−1(√3(W1−W2)W1+W2)]cos\phi = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{√ 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]

Calculation:

Reactive power =√3;(W1−W2)= \surd 3;\left( {{W_1}-{W_2}} \right)

= √3 (20 - 10) = 17.32 kVar

Question 75

A 0 to 200 V voltmeter has guaranteed accuracy of 1% of full-scale reading. The voltage measured by this instrument is 50 V the limiting error is

  1. AOption (a)

    0.25%

  2. BOption (b)

    1.0%

  3. COption (c)

    2.0%

  4. DOption (d)

    4.0%

View answer & explanation

Correct answer

Answer: (d)D.

4.0%

Why this answer is correct

Concept:

Limiting error: The maximum allowable error in the measurement is specified in terms of true value, which is known as limiting error. It will give a range of errors. It is always with respect to the true value, so it is a variable error.

%Limiting;Error;(%L.E.);=Limiting;Error;(L.E.)True;value×100{\rm{\% L}}imiting;Error;\left( {\% L.E.} \right){\rm{;}} = \frac{{Limiting;Error;\left( {L.E.} \right)}}{{True;value}} \times 100

Guaranteed accuracy error: The allowable error in measurement is specified in terms of full-scale value is known as guaranteed accuracy error. It is a variable error seen by the instrument since it is with respect to full-scale value.

%Guaranteed;accuracy;error;=Limiting;Error;(L.E.)Full;scale;value×100{\rm{\% Guaranteed;accuracy;error;}} = \frac{{Limiting;Error;\left( {L.E.} \right)}}{{Full;scale;value}} \times 100

Calculation:

Full scale voltage = 200 V

Accuracy = 1%

Voltage to be measured = 50 V

The limiting error is given as :-

= 1100×200×150\frac{1}{{100}} \times 200\times \frac{1}{{50}}

=4/100

= 4100×100=4%\frac{4}{{100}} \times 100 = 4\%

Question 76

An AM signal is represented by

x(t) = (20 + 4 sin 500πt) cos (2πt × 105)) V.

What is total signal power?

  1. AOption (a)

    208 W

  2. BOption (b)

    204 W

  3. COption (c)

    408 W

  4. DOption (d)

    416 W

View answer & explanation

Correct answer

Answer: (b)B.

204 W

Why this answer is correct

Concept:

The general form of an amplitude modulation signal is given by:

x(t) = [Ac + m(t)] cos (ωct)  

The total transmitted power for an AM system is given by:

Pt=Pc(1+μ22){P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)   ---(1)

Pc = Carrier Power

μ = Modulation Index

The modulation index is defined as:

μa=max⁡∣m(t)∣Ac{\mu _a} = \frac{{\max \left| {m\left( t \right)} \right|}}{{{A_c}}}

Equation (1) can be written as:

Pt=Pc+Pcμ22{P_t} = {P_c}+\frac{P_c\mu ^2}{2}

Pt = PC + Psb

Calculation:

Given, the amplitude-modulated signal as:

x(t) = (20 + 4 sin 500 πt) cos (2π × 105 t)       ---(2)

The general form of an amplitude modulation signal is given by:

x(t) = [Ac + m(t)] cos (ωct)     ---(3)

Now, comparing equation (2) and (3), we can write:

Ac = 20

m(t) = 4 sin 500 πt

So, the modulation will be:

μa=max⁡∣m(t)∣Ac{\mu _a} = \frac{{\max \left| {m\left( t \right)} \right|}}{{{A_c}}}

μa=max⁡∣4sin⁡500πt∣Ac{\mu _a} = \frac{{\max \left| {4\sin 500\pi t} \right|}}{{{A_c}}}

μa=420=0.2{\mu _a} = \frac{4}{{20}} = 0.2

Now, the total power of the amplitude-modulated signal is given by:

Pt=12Ac2(1+μa22){P_t} = \frac{1}{2}A_c^2\left( {1 + \frac{{\mu _a^2}}{2}} \right)

=12×(20)2[1+(0.2)22] = \frac{1}{2} \times {\left( {20} \right)^2}\left[ {1 + \frac{{{{\left( {0.2} \right)}^2}}}{2}} \right]

Pt = 204 W

Question 77

A resistance wire strain guage with a guage factor of 2, is bonded to steel structural member subjected to a stress of 100 MN/m2, the modulus of elasticity of a steel is 200 GN/m2. The percentage change in value of guage resistance due to the applied stress will be

  1. AOption (a)

    0.3%

  2. BOption (b)

    0.2%

  3. COption (c)

    0.4%

  4. DOption (d)

    0.1%

View answer & explanation

Correct answer

Answer: (d)D.

0.1%

Why this answer is correct

Concept:

Gauge factor= GF=dR/RdL/LGF = {{dR/R} \over {dL/L}} 

Where,

 dLL {dL \over L} is strain

Modulus of elasticity  =stressstrain = {stress \over strain}

Therefore;

 dRR=GF.stressModulusOfElasticity {dR \over R} = GF.{stress \over ModulusOfElasticity}

Calculation: 

For the given question,

GF =2, stress = 100 GN/m2, Modulus of elasticity = 200 GN/m2 

dRR=2.100M200G {dR \over R} = 2.{100 M \over 200 G}

solving we get 

dRR=0.001 {dR \over R} = 0.001

Thus, percentage change in value of gauge resistance = 0.001*100 = 0.1%

Question 78

The primary current in a CT is

  1. AOption (a)

    Independent of secondary current

  2. BOption (b)

    Dependent of secondary current

  3. COption (c)

    Depends upon transformation ratio

  4. DOption (d)

    Depends on the nominal ratio

View answer & explanation

Correct answer

Answer: (a)A.

Independent of secondary current

Why this answer is correct

Concept:

  • A current transformer(CT) is a device used to produce an alternating current in its secondary, which is proportional to the AC current in its primary.
  • This is primarily used when a current or voltage is too high to measure directly.
  • The induced secondary current is then suitable for measuring instruments or processing in electronic equipment, which typically needs isolation between the primary and secondary circuits.

Explanation:

Primarily, a CT is a measuring device where the primary current is the current to be measured which is reduced in proportion to the turns ratio which allows for a convenient way of safely monitoring the actual electrical current flowing in an AC transmission line using a standard ammeter. Hence, the primary current in a CT is independent of the secondary current.

Question 79

The negative feedback in an amplifier leads to which one of the following?

  1. AOption (a)

    Increase in current gain  

  2. BOption (b)

    Increase in voltage gain

  3. COption (c)

    Decrease in voltage gain

  4. DOption (d)

    Decrease in bandwidth

View answer & explanation

Correct answer

Answer: (c)C.

Decrease in voltage gain

Why this answer is correct

Concept:

Negative feedback circuit:

 

The feedback amplification factor is given by:

 Af = Ao1+Aoβ\frac{A_o}{1 + A_o β }

Where,

Ao is open-loop gain

Aoβ is the loop gain.

Explanation:

The negative feedback in amplifiers causes:

  • Reduced the gain and increases the stability in G.
  • Increases the bandwidth to maintain constant gain-bandwidth product
  • Reduces the distortion and noise in the amplifier
  • The signal-to-noise ratio is not affected.
  • The voltage gain (Av) of an amplifier is defined as the ratio of output voltage to the input voltage.

Av = Vo/Vin

Here, Vo is the output voltage of an amplifier and Vin is the input voltage of an amplifier.

  • In a negative feedback amplifier, closed-loop voltage gain is given

            Av = Vo / Vin = 1/(1+Aoβ)

Here, β = feedback factor,

Ao = open-loop gain of the amplifier.

  • This expression clearly shows that closed-loop voltage gain has reduced by introducing negative feedback.
  • We know that a product of gain and bandwidth is inversely proportional so here bandwidth of amplifier will increase ;

            (gain × bandwidth = 0.35)

  • A negative feedback amplifier decreases the current gain.

Question 80

The value of current I in the following circuit is shown in fig.

  1. AOption (a)

    2.45 Amp.

  2. BOption (b)

    1.45 Amp.

  3. COption (c)

    2.40 Amp.

  4. DOption (d)

    1.40 Amp.

View answer & explanation

Correct answer

Answer: (d)D.

1.40 Amp.

Why this answer is correct

Concept:

 

Nodal Analysis:

  • Nodal Voltage Analysis uses the “Nodal” equations of Kirchhoff’s first law to find the voltage potentials around the circuit. So by adding together all these nodal voltages the net result will be equal to zero.
  • If there are “n” nodes in the circuit there will be “n-1” independent nodal equations and these alone are sufficient to describe and hence solve the circuit.
  • At each node point write down Kirchhoff’s first law equation(KCL), that is: “the currents entering a node are exactly equal in value to the currents leaving the node” then express each current in terms of the voltage across the branch using Ohm's law.
  • So the nodal analysis is primarily based on the application of KCL and Ohm's law.
  • For “n” nodes, one node will be used as the reference node and all the other voltages will be referenced or measured with respect to this common node

Calculation:

From the above circuit,

V = 20 - 2 I        ------- (i) 

Apply KCL at node,

I + 2 = (20 - 2 I - 5I) / 3

3I + 6 = 20 - 7 I

⇒ 10 I = 14

⇒ I = 1.4 A

Question 81

The following equation ∂2u∂t2−c2(∂2u∂x2+∂2u∂y2)=0\rm \frac{\partial ^2 u}{\partial t^2} - c^2 \left( \frac{\partial ^2 u}{\partial x^2}+ \frac{\partial ^2 u}{\partial y^2}\right) = 0 represents

i. Poisson equation

ii. Wave equation

iii. Laplace equation

iv. Heat equation

  1. AOption (a)

    i, ii, iv are correct

  2. BOption (b)

    i, ii are correct

  3. COption (c)

    ​only iv is correct

  4. DOption (d)

    ​only ii is correct

View answer & explanation

Correct answer

Answer: (d)D.

​only ii is correct

Why this answer is correct

Explanation:

3-D heat equation is given as below

(∂2Tdx2+∂2T∂y2+∂2T∂z2)+Q(x,t)K=1α∂T∂t\left( {\frac{{{\partial ^2}T}}{{d{x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} + \frac{{{\partial ^2}T}}{{\partial {z^2}}}} \right) + \frac{{Q\left( {x,t} \right)}}{K} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}

For 1 – D & without heat generation:

∂2T∂x2=1α∂T∂t\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}

Where α ÷ thermal diffusivity.

Wave equation is given by:

∂2u∂t2=c2(∂2u∂x2+∂2u∂y2)\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {c^2}\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}}} \right)      (2-D)

Laplace equation:

∂2u∂x2+∂2u∂y2+∂2u∂z2=0\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} + \frac{{{\partial ^2}u}}{{\partial {z^2}}} = 0     (3-D)

∇2u=0{\nabla ^2}u = 0

​Poisson’s equation: 

∇2V=−ρvϵ{\nabla ^2}V = - \frac{{{\rho _v}}}{\epsilon}

Question 82

A 0-150 V, voltmeter has a guaranteed accuracy of 1% of full-scale reading. The voltage measured by this instrument is 75 V, what will be percentage of limiting error ?

  1. AOption (a)

    2%

  2. BOption (b)

    1%

  3. COption (c)

    4%

  4. DOption (d)

    3%

View answer & explanation

Correct answer

Answer: (a)A.

2%

Why this answer is correct

Full scale voltage = 150 V

Accuracy = 1%

Voltage to be measured = 75 V

The limiting error is given as :-

= 1100×150×175\frac{1}{{100}} \times 150 \times \frac{1}{{75}}

=2/100

= 2100×100=2%\frac{2}{{100}} \times 100 = 2\%

Question 83

The current gain of a bipolar transistor drops at high frequency because of

  1. AOption (a)

    transistor internal capacitance

  2. BOption (b)

    High current effect in base

  3. COption (c)

    Parasitic inductive elements

  4. DOption (d)

    The early effect

View answer & explanation

Correct answer

Answer: (a)A.

transistor internal capacitance

Why this answer is correct

In amplifier circuits coupling and bypass, capacitors look short to ac at mid-band frequencies. For less frequency capacitive reactance of these capacitors disturbs the gain and phase shift of signals therefore they are taken into consideration. The variation in gain or phase shift for a certain value of input signal frequency is known as frequency response.

  • For large frequencies, the coupling and bypass capacitors behave like ac shorts and have no effect on the amplifier’s response.
  • Inner transistor junction capacitance, though, does come into play, dropping a gain of the amplifier and causing phase shift as signal frequency rises.
  • For less frequency, the inner capacitance has large reactance because of their less capacitance value and less frequency value.
  • So, they appear like open circuits and have no effect on the working of transistors.
  • With the increment in frequency, the inner capacitive decrease and at a certain point have an effect on the gain of the transistor.
  • When reactance Cbe is less enough a certain quantity of output signal voltage is fed back out of phase with the input. hence, efficiently dropping the gain

Question 84

Transform the vector B̅ =yâx + (x + z)ây located at point (-2, 6, 3) into cylindrical co- ordinate

  1. AOption (a)

    6.325âρ - 71.57 âϕ 

  2. BOption (b)

    6.325âρ + 71.57 âϕ  + 3âz

  3. COption (c)

    6.325âρ​ + 73.57 âϕ

  4. DOption (d)

    −0.9487aρ−6.008aϕ-0.9487a_{\rho}-6.008a_{\phi}

View answer & explanation

Correct answer

Answer: (d)D.

−0.9487aρ−6.008aϕ-0.9487a_{\rho}-6.008a_{\phi}

Why this answer is correct

Calculation:

At Point P: x = -2, y = 6, z = 3. Hence,

rho=x2+y2=4+36=6.32\rm rho=\sqrt{x^2+y^2}=\sqrt{4+36}=6.32

ϕ=tan−1yx=tan⁡−16−2=108.43∘\rm ϕ=tan^{-1}\frac{y}{x}=\tan^{-1}\frac{6}{-2}=108.43^{\circ}

z = 3

r=x2+y2+z2=4+36+9=7\rm r=\sqrt{x^2+y^2+z^2}=\sqrt{4+36+9}=7

θ=tan⁡−1x2+y2z=tan⁡−1403=64.62∘\rm \theta=\tan^{-1}\frac{\sqrt{x^2+y^2}}{z}=\tan^{-1}\frac{\sqrt {40}}3{}=64.62^\circ

Thus,

P(-2, 6, 3) = P(6.32, 108.43°, 3) = P(7, 64.62°, 108.43°)

In the Cartesian system, A at P is

A = 6ax + ay

For vector A, Ax = y, Ay = x + z, Az = 0. Hence, in the cylindrical system.

[AρAϕAz]=[cos⁡ϕsin⁡ϕ0−sin⁡ϕcos⁡ϕ0001][yx+z0]\rm \begin{bmatrix}A_{ρ}\\ A_{ϕ}\\ A_z\end{bmatrix}=\rm \begin{bmatrix}\cos ϕ&\sin ϕ&0\\ -\sin ϕ&\cos ϕ&0\\ 0&0&1\end{bmatrix}\rm \begin{bmatrix}y\\ x+z\\ 0\end{bmatrix}

or

Aρ = y cos ϕ + (x + z) sin ϕ

Aϕ = -y sin ϕ + (x + z) cos ϕ

Az = 0

But x = ρ cos ϕ, y = sin ϕ, and substituting these yields

A = (Aρ, Aϕ, Az) = [ρ cos ϕ sin ϕ + (ρ cos ϕ + z) sin ϕ]aρ + [-ρ sin2 ϕ + (ρ cos ϕ + z)cos ϕ]aϕ

At P

ρ = 40,tan⁡ϕ=6−2\sqrt{40}, \tan \phi=\frac{6}{-2}

Hence, 

cosϕ=−240,sin⁡ϕ=640\rm cos \phi=\frac{-2}{\sqrt{40}}, \sin \phi=\frac{6}{\sqrt{40}}

A=[40.−240.640+(40.−240+3).640]aρ+[−40.3640+(40.−240+3).−240]aϕ\rm A=\left[\sqrt{40}.\frac{-2}{\sqrt{40}}.\frac{6}{\sqrt{40}}+\left(\sqrt{40}.\frac{-2}{\sqrt{40}}+3\right).\frac{6}{\sqrt{40}}\right]a_{\rho}+\rm \left[-\sqrt{40}.\frac{36}{{40}}+\left(\sqrt{40}.\frac{-2}{\sqrt{40}}+3\right).\frac{-2}{\sqrt{40}}\right]a_{\phi}

−640aρ−3840aϕ=−0.9487aρ−6.008aϕ\rm \frac{-6}{\sqrt{40}}a_{\rho}-\frac{38}{\sqrt{40}}a_{\phi}=-0.9487a_{\rho}-6.008a_{\phi}

Question 85

The impedance of a parallel circuit is (10 – j 30) Ω at 1 MHz. The values of circuit elements will be

  1. AOption (a)

    10 Ω and 6.4 mH

  2. BOption (b)

    100 Ω and 4.7 nF

  3. COption (c)

    10 Ω and 4.7 mH

  4. DOption (d)

    100 Ω and 6.4 nF

View answer & explanation

Correct answer

Answer: (b)B.

100 Ω and 4.7 nF

Why this answer is correct

Given, Z = (10 – j 30) Ω

Let’s assume R and jXL are in parallel.

Y=;1Z=1R+1jXL=1(10;−;j;30)Y = ;\frac{1}{Z} = \frac{1}{R} + \frac{1}{{{\rm{j}}{{\rm{X}}_{\rm{L}}}}} = \frac{1}{{\left( {10{\rm{;}}-{\rm{;j;}}30} \right)}}

Y=10;+;j30100;+;900=1100+j301000Y = \frac{{10; + ;j30}}{{100; + ;900}} = \frac{1}{{100}} + \frac{{j30}}{{1000}}

Comparing, R=100;Ω,;XL=−;100030;ΩR = 100;\Omega ,;{X_L} = - ;\frac{{1000}}{{30}};\Omega

As the reactance is negative, it is not an inductive reactance. The element must be a capacitor.

1/(ωC) = 1000/30

⇒C=3100;×;2π;×;106=4.7;nF\Rightarrow C = \frac{3}{{100; \times ;2\pi ; \times ;{{10}^6}}} = 4.7;nF

Question 86

A point charge of 10-9 C is placed at a point ‘A’ in the free space. The potential difference between two points 30 cm and 20 cm away from the charge at ‘A’ will be

  1. AOption (a)

    45 volt

  2. BOption (b)

    15 volt

  3. COption (c)

    90 volt

  4. DOption (d)

    30 volt

View answer & explanation

Correct answer

Answer: (b)B.

15 volt

Why this answer is correct

Concept:

The potential at point B due to change Q at point A at distance d can be given as –

V=Q4πε0dV = \frac{Q}{{4\pi {\varepsilon _0}d}}

Q is the charge

d is the distance

ε0 is the absolute permittivity

Calculation:

The potential at point B from the charge at point A is,

VB=10−94π×8.854×10−12×20×0.01=45;V{V_B} = \frac{{{{10}^{ - 9}}}}{{4\pi \times 8.854 \times {{10}^{ - 12}} \times 20 \times 0.01}} = 45;V

The potential at point C from the charge at point A is,

VC=10−94π×8.854×10−12×30×0.01=30;V{V_C} = \frac{{{{10}^{ - 9}}}}{{4\pi \times 8.854 \times {{10}^{ - 12}} \times 30 \times 0.01}} = 30;V

Therefore, the potential difference is given by,

VBC = VB – VC = 45 – 30 = 15 V

Question 87

Calculate the value of RL to be connected across A - B for maximum power transfer.

  1. AOption (a)

    2.5 Ω

  2. BOption (b)

    10 Ω

  3. COption (c)

    0 Ω

  4. DOption (d)

    15 Ω

View answer & explanation

Correct answer

Answer: (a)A.

2.5 Ω

Why this answer is correct

Concept:

Maximum power transfer for DC circuit:

According to the MPT the maximum power transfer to the load when the load resistance is equal to the source resistance or Thevenin resistance.

RL = Rth 

RL = load resistance

Rth = Thevenin or source resistance

The power at maximum power transfer (Pmax) = Vth2 / 4Rth

The maximum power transfer theorem is used in electrical circuits.

Calculation:

Rth = RL

= ( 5 × 5 )  / 10

= 2.5 Ω

Question 88

The inductance of a moving iron instruments is given by:

L = (10 + 5θ - θ2) μH, when θ is the deflection in radian from Zero position. The spring Constant is 12 × 10-6 N.m/rad. Estimate the deflection for a current of 5A

  1. AOption (a)

    100°

  2. BOption (b)

    96.8°

  3. COption (c)

    66°

  4. DOption (d)

    57.8°

View answer & explanation

Correct answer

Answer: (b)B.

96.8°

Why this answer is correct

Concept:

For a MI instrument Td(Deflection Torque) = 12I.I.dLdθ {1 \over 2}I.I.{dL \over d\theta}

Tc(Controlling Torque) = Kcθ\theta

During equilibrium: Tc = Td

On solving; θ\theta  = 12I.I.dLdθ1Kc {1 \over 2}I.I.{dL \over d\theta}{1 \over Kc}

Calculation: 

For the given question,

L = (10+5θ−θ2)(10 + 5\theta - \theta^2) µH, Kc = 12 x 10-6 Nm/rad, I = 5A

θ=12I.I.dLdθ1Kc\theta = {1 \over 2}I.I.{dL \over d\theta}{1 \over Kc}

Substituting the values in the above formula,

We get,

 θ=12.25.(5−2θ).112\theta = {1 \over 2}.25.{(5-2\theta)}.{1 \over 12}

on solving θ=\theta =  1.689 rad = 96.8o

Question 89

Which of the following points locate the quiescent point ?

  1. AOption (a)

    (IC, VCB)

  2. BOption (b)

    (IE, VCE)

  3. COption (c)

    (IE, VCB)

  4. DOption (d)

    (IC, VCE​)

View answer & explanation

Correct answer

Answer: (d)D.

(IC, VCE​)

Why this answer is correct

The operating point is also called the quiescent point and for a transistor, it is given as \(\left( {{{\rm{I}}{\rm{C}}},{\rm{;}}{{\rm{V}}{{\rm{CE}}}}} \right)\)

  • The ac load line of a transistor circuit is steeper than its dc load line but the two intersect at Q point
  • When AC and DC Load lines are represented in a graph, it can be understood that they are not identical
  • Both lines intersect at the Q-point or quiescent point; The endpoints of the AC load line are saturation and cut off points; This is understood from the figure below

Question 90

Which of the following instrument is free from hysteresis and eddy current loss?

  1. AOption (a)

    Moving iron instruments

  2. BOption (b)

    PMMC instruments

  3. COption (c)

    Electrostatic instruments

  4. DOption (d)

    Dynamometer type instruments

View answer & explanation

Correct answer

Answer: (c)C.

Electrostatic instruments

Why this answer is correct

  • In dynamometer type instruments air core is used for both fixed coil and moving coil. Hence this instrument is almost free from or has minimum hysteresis and eddy current error.
  • Hysteresis and eddy current errors are present in MC, MI instruments.
  • These errors are not present in electrostatic instruments because the flux is constant in electrostatic instruments.

Question 91

For the circuit shown below the Thevenin's voltage and equivalent resistance at terminal ab are respectively:

  1. AOption (a)

    8 V, 12 Ω

  2. BOption (b)

    -8 V, 12 Ω

  3. COption (c)

    6 V, 8 Ω

  4. DOption (d)

    -6 V, 8 Ω

View answer & explanation

Correct answer

Answer: (b)B.

-8 V, 12 Ω

Why this answer is correct

Thevenin voltage(Vth)

Let's say current i flows through the voltage source 10 V

Applying KVL in loop ⊥ 

10 = 10 i + 40(i - 2)

Solving we get i = 1.8 A

Vab = Vth = 40(i - 2) = -8 V

Thevenin Resistance (Rth)

For Rth; SC voltage source and OC current source

Rab = Rth = (10 || 40) + 4

= 12 Ω

Question 92

Which of the following instrument is used for measuring frequency ?

  1. AOption (a)

    Owen's bridge

  2. BOption (b)

    Schering bridge

  3. COption (c)

    Anderson's bridge

  4. DOption (d)

    Wien's bridge

View answer & explanation

Correct answer

Answer: (d)D.

Wien's bridge

Why this answer is correct

Explanation:

  • The Wien’s bridge is primarily known as a frequency-determining bridge.
  • It is suitable for the measurement of frequencies from 100 Hz to 100 kHz.
  • It finds its application in various other circuits other than ac bridge circuits.
  • It is employed in harmonic distortion analyser, where it is used as notch filter, discriminating against one specific frequency.
  • It also finds applications in audio and HF oscillators as the frequency determining device.

Important Points

Type of BridgeName of BridgeUsed to measureImportant
DC BridgesWheatstone bridgeMedium resistance
Corey foster’s bridgeMedium resistance
Kelvin double bridgeVery low resistance
Loss of charge methodHigh resistance
MeggerHigh insulation resistanceResistance of cables
AC BridgesMaxwell’s inductance bridgeInductanceNot suitable to measure Q
Maxwell’s inductance capacitance bridgeInductanceSuitable for medium Q coil (1 < Q < 10)
Hay’s bridgeInductanceSuitable for high Q coil (Q > 10), slowest bridge
Anderson’s bridgeInductance5-point bridge, accurate and fastest bridge (Q < 1)
Owen’s bridgeInductanceUsed for measuring low Q coils
Heaviside mutual inductance bridgeMutual inductance
Campbell’s modification of Heaviside bridgeMutual inductance
De-Sauty’s bridgeCapacitanceSuitable for perfect capacitor
Schering bridgeCapacitanceUsed to measure relative permittivity
Wein’s bridgeCapacitance and frequencyHarmonic distortion analyzer, used as a notch filter, used in audio and high-frequency applications

Question 93

Find the volume charge density that is associated with the field D = ar C/m2

  1. AOption (a)

    1 C/m3

  2. BOption (b)

    0 C/m3

  3. COption (c)

    2r\frac{2}{r} C/m3

  4. DOption (d)

    r C/m3

View answer & explanation

Correct answer

Answer: (c)C.

2r\frac{2}{r} C/m3

Why this answer is correct

Concept:

Volume charge density is given by the formula Pv = ∇⃗⋅D⃗\vec{\nabla} \cdot \vec{D}

Calculation:

For spherical coordinates ∇⃗⋅D⃗\vec{\nabla} \cdot \vec{D} = 1r2.δδr(r2Ar)\frac{1}{r^2}.\frac{\delta}{\delta r}(r^2Ar) having only ar^\hat{a_r} component

D⃗=ar^\vec{D} = \hat{a_r}

∴ 1r2δδr(r2)=2rr2=2r\frac{1}{r^2}\frac{\delta}{\delta r}(r^2) =\frac{2r}{r^2}=\frac{2}{r}

Question 94

A moving iron instrument gives full-scale deflection with 100 V, it has a coil of 10,000 turns and resistance of 1000 Ω. It is to be used as an ammeter to give full-scale deflection at 40A. How many turns in coil would be. required ?

  1. AOption (a)

    500

  2. BOption (b)

    1000

  3. COption (c)

    25

  4. DOption (d)

    50

View answer & explanation

Correct answer

Answer: (c)C.

25

Why this answer is correct

⇒ current across the resistance coil while

full scale deflection = i = 100 V1000 Ω\frac{100 \ V}{1000 \ Ω} = 0.1 A

∴ current through shunt coil = ish = (40 - 0.1) = 39.9 A

Shunt resistance = V39.9\frac{V}{39.9} = 10039.9\frac{100}{39.9} = 2.50 Ω

⇒ A 1000 Ω resistor has 10000 coils turn

∴ a 2.50 Ω resistor wil have = 2.50 × 100001000\frac{10000}{1000}

= 25 coils turns

Question 95

Consider a unity negative feedback control system, here G(s)=1(s+1)\rm G(s) = \frac{1}{(s + 1)} and H(s)=Ks(s+2)\rm H(s) = \frac{K}{s(s + 2)} The closed loop system is stable for

  1. AOption (a)

    K > 20

  2. BOption (b)

    15 < K < 20

  3. COption (c)

    8 ≤ K ≤ 14

  4. DOption (d)

    K < 6

View answer & explanation

Correct answer

Answer: (d)D.

K < 6

Why this answer is correct

Given that characteristic equation is,

C.E = 1+ G(s) H(s) = 0

⇒  1+ (1/s+1)(K/s(s+2)) = 0

⇒ C.E = s3 + 3s2 + 2s + K = 0

\(\left. {\begin{array}{{20}{c}} {{s^3}}\ {{s^2}}\ {{s^1}}\ {{s^0}} \end{array}} \right|\begin{array}{{20}{c}} 1&{\left( { 2} \right)}\ 3&k\ {\frac{{6 - K}}{3}}&0\ k&{} \end{array}\)

For system to be stable,

K > 0, K (6 - K)/3 > 0

⇒ K > 0, 6 - K > 0

⇒ K > 0, K < 6

Question 96

The Trans conductance of a JFET is computed at constant VDS by:

  1. AOption (a)

    Ratio of change in ld to change of Vgs

  2. BOption (b)

    Ratio of change in Vgs to change of ld

  3. COption (c)

    Product of change in Vgs to change of Id

  4. DOption (d)

    Ratio of change in Vds to change of Id

View answer & explanation

Correct answer

Answer: (a)A.

Ratio of change in ld to change of Vgs

Why this answer is correct

In a JFET transconductance is defined as gm=ΔIdΔVgs;at;constant;VDS{g_m} = \frac{{{\rm{\Delta }}{I_d}}}{{{\rm{\Delta }}{V_{gs}}}};at;constant;{V_{DS}}.

Hence it is computed at constant VDS by ratio of change in ld to change of Vgs.

Question 97

The magnetizing force at the centre of a circular coil varies

i) Directly as the number of its turns

ii) Directly as the current

iii) Inversely proportional to its radius

iv) Directly proportional to its radius

  1. AOption (a)

    only i, ii, iv are correct

  2. BOption (b)

    only i, ii, iii are correct

  3. COption (c)

    only ii, iii, iv are correct

  4. DOption (d)

    only i, iii are correct

View answer & explanation

Correct answer

Answer: (b)B.

only i, ii, iii are correct

Why this answer is correct

Electric field intensity at (0, 0, h) due to current I flow in a circular loop of radius ‘a’, where the center of loop coincides with z-axis is given by

H⃗(0,0,h)=Ia22(h2+a2)32az^\vec H\left( {0,0,h} \right) = \frac{{I{a^2}}}{{2{{\left( {{h^2} + {a^2}} \right)}^{\frac{3}{2}}}}}\widehat {{a_z}}

Calculation:

For the given question, a = r, h = 0

H⃗(0,0,h)=Ir22(0+r2)32=I2rAt/m\vec H\left( {0,0,h} \right) = \frac{{I{r^2}}}{{2{{\left( {0 + {r^2}} \right)}^{\frac{3}{2}}}}} = \frac{I}{{2r}}At/m

For the n-turn circular coil, H=NI2rH = \frac{{NI}}{{2r}}

From the above expression,

The magnetizing force at the center of a circular coil varies

  • Directly as the number of its turns
  • Directly as the current
  • Inversely as its radius

Question 98

Equipotential surface is a

  1. AOption (a)

    Real surface

  2. BOption (b)

    Complex surface

  3. COption (c)

    Imaginary surface

  4. DOption (d)

    Not existing surface

View answer & explanation

Correct answer

Answer: (c)C.

Imaginary surface

Why this answer is correct

Concept:

Equipotential surface:

 

  • The locus of all the points on the surface which has the same electric potential is called the equipotential surface. Since the electric field lines are imaginary lines points over the field lines would also be imaginary.
  • The electric field is always perpendicular to the equipotential surface.
  • Two equipotential surfaces can never intersect.
  • The direction of the equipotential surface is from high potential to low potential.
  • The potential difference between any two points on the equipotential surface is always zero.
  • So, zero work is to be done to move a charge from one point to another point on the equipotential surface. For example surface of the charged conductor is an equipotential surface.

For point charge, the equipotential surface would be a concentric circle.

Question 99

A phase shift oscillator uses 5 pF capacitors. Find the value of R to produce a frequency of 800 kHz

  1. AOption (a)

    39.8 kΩ

  2. BOption (b)

    28.1 kΩ

  3. COption (c)

    16.2 kΩ

  4. DOption (d)

    22.9 kΩ

View answer & explanation

Correct answer

Answer: (c)C.

16.2 kΩ

Why this answer is correct

Concept:

  • An oscillator consists of an amplifier with gain A & a feedback network with gain β.
  • The feedback network is used for phase shifting.
  • The amplifier receives the output of the phase-shifting network.
  • The amplifier then amplifies it and adds a 180° phase shift. This phase-shifted output of the amplifier is applied to the input of feedback.
  • Feedback networks shift the amplifier output to 180°.
  • Thus, due to the total 360° phase shift, the feedback becomes positive feedback.
  • This gives rise to the oscillation if the Barkhausen criteria is satisfied.

 

Explanation:

 

  • Three RC sections are used which are called ladder networks.
  • The phase shift provided by three RC networks is 180°.
  • Thus the total phase shift in the circuit is 360°.

 

→ The gain of the amplifier must be sufficient so that,

|Aβ| ≥ 1

Then, the Barkhausen criterion is satisfied and Barkhausen oscillation are produced.

Calculation:

The frequency of oscillation is given by

f0=12πRC6\rm f_0 = \frac{1}{2 \pi RC \sqrt 6}

Given, C = 5 pF, f0 = 800 kHz

R=12πf0C6\rm R = \frac{1}{2 \pi f_0 C \sqrt 6}

R=12π×800×103×5×10−12×6R = \frac{1}{2 \pi \times 800 \times 10^3 \times 5 \times 10^{-12} \times \sqrt 6}

∴ R = 16.2 kΩ

Hence, correct option is (3)

Question 100

The correct sequence of the following types of ammeters with increasing accuracy is

  1. Moving iron type
  2. Moving coil permanent magnet type
  3. Induction type

Select the Correct answer using the code given below:

  1. AOption (a)

    1, 3, 2

  2. BOption (b)

    2, 1, 3

  3. COption (c)

    1, 2, 3

  4. DOption (d)

    3, 1, 2

View answer & explanation

Correct answer

Answer: (d)D.

3, 1, 2

Why this answer is correct

Induction type instruments are least accurate among the given ones. Moving coil instruments has more accuracy than moving iron instruments.

Important Points

Moving Iron (MI) type instruments:

  • Moving Iron (MI) type instruments can be used for both AC & DC measurements. But Moving Iron connected through the current Transformer block DC supply. So that Moving Iron connected through the current Transformer is only used for AC measurements.
  • The moving iron meter has a large magnetic reluctance as compared to the PMMC meter. That’s why more power is required to operate the moving iron meter.

 

PMMC instruments:

  • The scale is uniform (Linear scale)
  • High torque to weight ratio compared to all the other type of instruments
  • Frequency error or reactance error is absent
  • High sensitivity
  • Hysteresis loss is lesser
  • No internal heating problem
  • High accuracy
  • Very less power consumption

Question 101

According to Gauss theorem, the surface integral of the normal component of electric field intensity E̅ over closed surface containing charge Q is

  1. AOption (a)

    Q

  2. BOption (b)

    Q∈0\frac{Q}{\in_0}

  3. COption (c)

    ∈0Q

  4. DOption (d)

    Q2∈0\frac{Q^2}{\in_0}

View answer & explanation

Correct answer

Answer: (b)B.

Q∈0\frac{Q}{\in_0}

Why this answer is correct

CONCEPT:

Gauss law:

  • According to this law, the total flux linked with a closed surface called Gaussian surface is 1/ϵo times the charge enclosed by the closed surface.

⇒ϕ=∮E⃗.dA⃗=Qϵo\Rightarrow \phi=\oint \vec{E}.\vec{dA}=\frac{Q}{\epsilon_o}

  • This law is applicable to all the closed surfaces.
  • According to this law, the total flux linked with a closed surface called Gaussian surface is 1/ϵo times the charge enclosed by the closed surface.
  • So this law is applicable to all the closed surfaces. Hence, option 2 is correct.

Question 102

Which meter has the highest accuracy in the prescribed limit of frequency range ?

  1. AOption (a)

    PMMC

  2. BOption (b)

    Moving iron

  3. COption (c)

    Electrodynamometer

  4. DOption (d)

    Rectifier

View answer & explanation

Correct answer

Answer: (d)D.

Rectifier

Why this answer is correct

The correct answer is option Rectifier.

Concept:

Rectifier Meter:

  • The Rectifier meter consists of a rectifier and PMMC. Hence it can measure AC quantity with accuracy.
  • It has PMMC so it measures average value.
  • To get the RMS value of AC quantity to be measured, the measured value is being multiplied with the form factor of the signal. Generally, for AC signal measurement the meter reading is already multiplied by 1.11 while calibrating.

Hence the correct answer is Rectifier meter has the highest accuracy in the prescribed limit of the frequency range.

 

Additional Information 

PMMC meter:

  • It uses a permanent magnet. So it can't be used in an AC circuit.
  • t is the most accurate instrument and gives a linear reading.
  • it has a higher cost.

 

Dynamometer:

  • It is a modified form of permanent-magnet type.
  • It consists of two air-cored fixed coils placed on either side of the moving coil which produces the required magnetic field.
  • It can be used for AC as well DC circuits as voltmeter and ammeter.
  • It uses spring for controlling torque. Hence it gives a more accurate result than moving iron instruments.

 

Moving Iron:

  • Moving-iron instruments are generally used to measure alternating voltages and currents.
  • In moving-iron instruments, the movable system consists of one or more pieces of specially-shaped soft iron.
  • It measures true RMS value.

 

Induction Wattmeter:

  • For AC power measurement only.
  • Works on the principle of electromagnetic induction.

Question 103

Consider the following statements associated with moving iron instruments.

i. These can be used in d.c. as well as a.c. circuits.

ii. The scale is non-uniform.

iii. The moving iron is placed in a field of permanent magnet.

Which of these statements are correct?

  1. AOption (a)

    i, ii, iii

  2. BOption (b)

    i, ii

  3. COption (c)

    ii, iii

  4. DOption (d)

    i, iii

View answer & explanation

Correct answer

Answer: (b)B.

i, ii

Why this answer is correct

Moving Iron (MI) type instruments:

  • Moving Iron (MI) type instruments can be used for both AC & DC measurements. But Moving Iron connected through the current Transformer block DC supply. So that Moving Iron connected through the current Transformer is only used for AC measurements.
  • The moving iron meter has a large magnetic reluctance as compared to the PMMC meter. That’s why more power is required to operate the moving iron meter.

 

Advantages of moving iron:

  • It is a universal instrument which can be used for the measurement of AC and DC quantities
  • These instruments can withstand large loads and are not damaged even under severe overload conditions
  • It is very cheap due to the simple construction

 

Disadvantages of moving iron:

  • These instruments suffer from error due to hysteresis, frequency change, and stray losses
  • The reading of the instrument is affected by temperature variation

 

In the moving iron instruments, the vane is made up of soft iron and high permeability steel because it can move in the magnetic field of the stationary coil carrying the current.

Question 104

What would be the computational value of feedback voltage in a negative feedback amplifier with A = 100, β = 0.03 and input signal voltage = 30 mV?

  1. AOption (a)

    0.06 V

  2. BOption (b)

    0.03 V

  3. COption (c)

    0.09 V

  4. DOption (d)

    0.15 V

View answer & explanation

Correct answer

Answer: (c)C.

0.09 V

Why this answer is correct

Concept:

Output voltage signal is given by

V0 = A Vi

Feedback voltage is given by

Vf = β V0 = Aβ Vi

Calculation:

Given A = 100

β = 0.03

Vi = 30 mV = 0.03 V

Vf = 100 × 0.03 × 0.03 = 0.09 V

Question 105

For the following circuit, the value of ic(t) and time constant will be, if S is closed at t = 0 (Consider C = 0.5 F)

  1. AOption (a)

    ic(t) = e-2t; τ = 2 sec

  2. BOption (b)

    ic(t) = e-2t; τ = .5 sec

  3. COption (c)

    ic(t) = e-t; τ = 1 sec

  4. DOption (d)

    ic(t) = e-t; τ = 2 sec

View answer & explanation

Correct answer

Answer: (b)B.

ic(t) = e-2t; τ = .5 sec

Why this answer is correct

Concept:

A capacitor doesn’t allow a sudden change in voltage, i.e.

VC(0+) = VC(0-)

Also, the transient voltage across a capacitor is given by:

Vc(t) = V(∞) – [V(∞) – V(0)] e-t/τ  (τ is the Time constant for the RC Circuit)       ----1)

Calculation:

For t >0

Vc(∞) = 1 V

Vc(0) = 0 V

τ = Req C = 1 × .5 = .5 sec

Vc(t) = 1 + (0-1) e-t/.5 = 1- e-2t

Ic(t) = C. dVc(t) / dt = .5 × 2e-2t = e-2t

Question 106

Total energy associated in a system of four identical charges of Q = 3 nC at corners of 1 m on a side, is Approximately equal to

  1. AOption (a)

    292 nJ

  2. BOption (b)

    486 nJ

  3. COption (c)

    438 nJ

  4. DOption (d)

    1876 nJ

View answer & explanation

Correct answer

Answer: (c)C.

438 nJ

Why this answer is correct

The correct option is 438 nJ.

Concept:

The potential energy of n- charge system is given as-

U=14πϵo∑qiqjrirjU=\frac{1}{4\pi\epsilon_o}\sum \frac{q_iq_j}{r_ir_j}

The total energy of the 4-charge system can be given as-

U=14πϵo[q1q2r12+q1q3r13+q1q4r14+q2q3r23+q2q4r24+q3q4r34]U=\frac{1}{4\pi\epsilon_o}[\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_1q_4}{r_{14}}+\frac{q_2q_3}{r_{23}} +\frac{q_2q_4}{r_{24}} +\frac{q_3q_4}{r_{34}}]

Calculation:

q1=q2=q3=q4=3nCq_1=q_2=q_3=q_4=3nC

r12=r23=r34=r14=1mr_{12}=r_{23}=r_{34}=r_{14}=1m

r13=r24=2mr_{13}=r_{24}=\sqrt 2m  

q1q2=q1q3=q1q4=q2q3=q2q4=q3q4=3nC×3nC=9×10−18C{q_1q_2}={q_1q_3}={q_1q_4}={q_2q_3}={q_2q_4}= {q_3q_4}=3nC\times3nC=9\times10^{-18}C

14πϵo=9×109\frac{1}{4\pi\epsilon_o}=9\times10^{9}

Hence, Total energy is -

U=14πϵo[q1q2r12+q1q3r13+q1q4r14+q2q3r23+q2q4r24+q3q4r34]U=\frac{1}{4\pi\epsilon_o}[\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_1q_4}{r_{14}}+\frac{q_2q_3}{r_{23}} +\frac{q_2q_4}{r_{24}} +\frac{q_3q_4}{r_{34}}]

U=9×109×9×10−18[1+12+1+1+12+1]U=9\times10^9\times9\times10^{-18}[1+\frac{1}{\sqrt2}+ 1+1+\frac{1}{\sqrt2}+1]

U=438.534nJU= 438.534nJ

Hence the correct answer is  438nJ.

Question 107

Which of the following statements are correct

(i) Accuracy is the closeness with which an instrument approaches the true value as the quantity being measured.

(ii) Precision is a measure of reproducibility of the measurement.

(iii) Precision of an instrument can be improved upon by calibration.

(iv) Accuracy may be specified in terms of limiting error.

  1. AOption (a)

    i, ii, iv only

  2. BOption (b)

    ii, iii, iv only

  3. COption (c)

    i, iii, iv only

  4. DOption (d)

    i, ii, iii only

View answer & explanation

Correct answer

Answer: (a)A.

i, ii, iv only

Why this answer is correct

Accuracy: It is the closeness with which an instrument reading approaches the true value of the quantity being measured. It means conformity to truth.

It can be expressed in the following ways.

Point accuracy: This is the accuracy of the instrument only at a point on its scale. It doesn’t give information about the accuracy at any other point on the scale.

Accuracy as a percentage of scale range: When an instrument has a uniform scale, its accuracy can be expressed in terms of the scale range.

Example: The accuracy of a voltmeter having a range of 200 V may be expressed as ±1 percent of the scale range

±1 percent of scale range = 0.01 × 200 = 2 V, i.e. the reading will have ±2 V error.

Accuracy as a percentage of true value: While measuring we can express the accuracy of reading as a percentage of true value.

Example: The accuracy of a voltmeter having a full-scale range of 200 V and it is used to measure 20 V and the accuracy is ±1 percent of true value

True value = 20 V

±1 percent of true value = 0.01 × 20 = 0.2 V

Precision: It is a measure of the reproducibility of the measurements, i.e. given a fixed value of quantity. It is a measure of the degree of agreement within a group of measurements.

Question 108

For following Fig. if C(s) is Laplace Transform of output and R(s) is Laplace transform of input, the equivalent transfer function T(s) will be

  1. AOption (a)

    T(s)=s3+12s4+s2+2s\rm T(s) = \frac{s^3 + 1}{2s^4 + s^2 + 2s}

  2. BOption (b)

    T(s)=s3+12s4+s2+1\rm T(s) = \frac{s^3 + 1}{2s^4 + s^2 + 1}

  3. COption (c)

    T(s)=s3−12s4+s2+2s\rm T(s) = \frac{s^3 - 1}{2s^4 + s^2 + 2s}

  4. DOption (d)

    T(s)=s3+12s4+s2−1\rm T(s) = \frac{s^3 + 1}{2s^4 + s^2 -1}

View answer & explanation

Correct answer

Answer: (a)A.

T(s)=s3+12s4+s2+2s\rm T(s) = \frac{s^3 + 1}{2s^4 + s^2 + 2s}

Why this answer is correct

Concept:

According to Mason’s gain formula, the transfer function is given by

TF = \frac{{\mathop \sum \nolimits_{k = 1}^n {M_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}

Where, n = No of forward paths

Mk = kth forward path gain

Δk = the value of Δ which is not touching the kth forward path

Δ = 1 – (sum of the loop gains) + (sum of the gain product of two non-touching loops) – (sum of the gain product of three non-touching loops)

Application:

Given signal flow graph is

Forward paths

M1 = s × s × 1/s = s

M2 = 1/s × 1/s = 1/s2

Loops:

L1 = s × s (-1) = - s2

L2 = -1/s 

L3 =  s × s × 1/s × (-s) = -s2

L4= 1/s × 1/s × (-s) = - 1/s

Δ = 1 - ( - s2  - s2  - 1/s  - 1/s) = 1 + 2s2 + 2/s

TF=s+1s21+2s2+2s=s3+12s4+s2+2sTF = \frac{s+\frac{1}{s^2}}{1+2s^2+\frac{2}{s}}= \frac{s^3+1}{2s^4+s^2+2s}

Question 109

An electrodynamometer instrument can be used as

  1. Wattmeter and VAR meter
  2. Power factor meter and Frequency meter
  3. Transfer instrument

Which of the above statements are correct?

  1. AOption (a)

    1 and 2 only

  2. BOption (b)

    1 and 3 only

  3. COption (c)

    2 and 3 only

  4. DOption (d)

    1, 2 and 3

View answer & explanation

Correct answer

Answer: (d)D.

1, 2 and 3

Why this answer is correct

Precision type electrodynamometer is used in the standardization process of the potentiometer.

It is a transfer instrument. A transfer instrument is one that is calibrated with a d.c. source and used without any modifications for a.c. measurements. It has the same accuracy for both D.C. and A.C. measurements.

Hence an electrodynamometer type of instrument is used as both standard and transfer instruments.

This instrument can be used as:

  • Ammeter
  • Voltmeter
  • Wattmeter
  • Reactive power meter (VAR meter)
  • Power factor meter
  • Frequency meter

Question 110

For an electric field E = E0 sinωt, what is the phase difference between the conduction current and displacement current?

  1. AOption (a)

    0°

  2. BOption (b)

    45°

  3. COption (c)

    90°

  4. DOption (d)

    180°

View answer & explanation

Correct answer

Answer: (c)C.

90°

Why this answer is correct

Explanation:

An electric current can be classified as,

Conduction current: It is a movement of charges (flow of charger) in the presence of electric field

Convection current: It is a movement of charger or flow of current in the absence of electric field (it flows in response to mechanical forces)

Displacement current: It is defined as the rate of change of the electric displacement field. It follows that a time-varying electric field produces a magnetic field and vice-versa.

The ratio of conduction current to displacement current,

\(\frac{{{{\rm{J}}{\rm{C}}}}}{{{{\rm{J}}{\rm{D}}}}} = \frac{{{\rm{\sigma E}}}}{{{\rm{\omega }}\epsilon{\rm{E}}}} = \frac{{\rm{\sigma }}}{{{\rm{\omega \varepsilon }}}}\)

The phase difference between the conduction current and displacement current is 90°

Question 111

A voltmeter has a resistance of 300 Ω and inductance of 0.12 H. This instrument reads correctly on DC. What is the reading on AC at 100 V when the frequency is 25 Hz ? 

  1. AOption (a)

    99.80 V 

  2. BOption (b)

    120.36 V

  3. COption (c)

    142.00 V

  4. DOption (d)

    151.00 V

View answer & explanation

Correct answer

Answer: (a)A.

99.80 V 

Why this answer is correct

With DC Supply:

Z = R = 300 Ω

IDC = 100300=0.3333 A\frac{100}{300}=0.3333 \ A

With AC Supply:

Z = R2+XL2\sqrt{R^2+X_L^2} = 3002+(2π×25×0.12)2=300.59 Ω\sqrt{300^2+(2\pi\times 25 \times 0.12)^2} = 300.59\ \Omega

IAC = 100300.59=0.3326 A\frac{100}{300.59}=0.3326 \ A

Since, in DC 0.3333 A produced by 100 V

It means, in AC 0.3326 A produced by X V.

Hence,

100 ∝ 0.3333 ... (1)

X ∝ 0.3326 .... (2)

From equation (1) & (2),

Reading (X) = 100×0.33260.3333\frac{100\times 0.3326}{0.3333} = 98.80 V

Question 112

The system with open loop transfer function G(s)H(s) = 1s(s+1)\frac{1}{s(s + 1)} is

  1. AOption (a)

    Type 0 and order 0

  2. BOption (b)

    Type 1 and order 1

  3. COption (c)

    Type 1 and order 2

  4. DOption (d)

    Type 2 and order 1

View answer & explanation

Correct answer

Answer: (c)C.

Type 1 and order 2

Why this answer is correct

Concept:

Transfer Function:

The transfer function of a control system is defined as the ratio of the Laplace transform of the output variable to the Laplace transform of the input variable assuming all initial conditions to be zero.

G(s) = C(s) / R(s)

Where G(s) = Transfer Function

C(s) = Laplace Transform of output

R(s) = Laplace Transform of input

  • Order of the system can be defined as the value of the highest exponent that appears in the denominator of the closed-loop transfer function.
  • The total number of poles of the closed-loop system gives the order of the system.
  • The type of the system can be defined as the number of poles located exactly at s=0 of the open-loop transfer function.

Calculation:

Given, Transfer Function is G(s)H(s) = K / s(s+1)

∴ The Type of the given system is 1 and the order of the system is 2

Question 113

Which of these is NOT valid at point P(0, 4, 0) ?

  1. AOption (a)

    aϕ = -ax

  2. BOption (b)

    aθ = -az

  3. COption (c)

    ar = 4ay

  4. DOption (d)

    aρ = aγ

View answer & explanation

Correct answer

Answer: (c)C.

ar = 4ay

Why this answer is correct

Given that

P (0, 4, 0) = P (x, y, z)

x = 0, y = 4, z = 0

In cylindrical system ρ=x2+y2=4ρ =\sqrt{x^2+y^2} = 4

ϕ=tan⁡−1(yx)=tan⁡−1(40)ϕ=\tan^{-1}\left(\frac{y}{x}\right)=\tan^{-1}\left(\frac{4}{0}\right)= 90°

And z = 0

In spherical system r=x2+y2+z2r=\sqrt{x^2+y^2+z^2}

= 4

θ=tan⁡−1x2+y2zθ=\tan^{-1}\frac{\sqrt{x^2+y^2}}{z} = 90°

ϕ = 90°

we know that

[aρaϕaz]=[cos⁡ϕsin⁡ϕ0sin⁡ϕcos⁡ϕ0001][axayaz]\begin{bmatrix}a_{ρ}\\ a_{ϕ}\\ a_{z}\end{bmatrix}=\begin{bmatrix}\cos ϕ&\sin ϕ&0\\ \sin ϕ&\cos ϕ&0\\ 0&0&1\end{bmatrix}\begin{bmatrix}a_{x}\\ a_{y}\\ a_{z}\end{bmatrix}

[aρaϕaz]=[010100001][axayaz]\begin{bmatrix}a_{ρ}\\ a_{ϕ}\\ a_{z}\end{bmatrix}=\begin{bmatrix}0&1&0\\ 1&0&0\\ 0&0&1\end{bmatrix}\begin{bmatrix}a_{x}\\ a_{y}\\ a_{z}\end{bmatrix}

aρ = ay

aϕ = ax

az = az

similarly we know that

⇒[araθaϕ]=[sin⁡θcos⁡ϕsin⁡θsin⁡ϕcos⁡θcos⁡θcos⁡ϕcos⁡θsin⁡ϕsin⁡θsin⁡θcos⁡ϕ0][axayaz]\Rightarrow \begin{bmatrix}a_{r}\\ a_{θ}\\ a_{ϕ}\end{bmatrix}=\begin{bmatrix}\sin θ\cos ϕ&\sin θ\sin ϕ&\cos θ\\ \cos θ\cos ϕ&\cos θ \sin ϕ&\sin θ\\ \sin θ&\cos ϕ&0\end{bmatrix}\begin{bmatrix}a_{x}\\ a_{y}\\ a_{z}\end{bmatrix}

⇒[araθaϕ]=[010001100][axayaz]\Rightarrow \begin{bmatrix}a_{r}\\ a_{θ}\\ a_{ϕ}\end{bmatrix}=\begin{bmatrix}0&1&0\\ 0&0&1\\ 1&0&0\end{bmatrix}\begin{bmatrix}a_{x}\\ a_{y}\\ a_{z}\end{bmatrix}

ar = ay

aθ = az

aϕ = ax

Question 114

For salient - pole rotor, alternator check the following statements and choose the correct

answer.

i. The machine has non-uniform air-gap.

ii. The machine has uniform air-gap.

iii. The machine runs at high speed.

iv. The machine runs at low speed.

  1. AOption (a)

    i, ii are correct

  2. BOption (b)

    i, iv are correct

  3. COption (c)

    ii, iii are correct

  4. DOption (d)

    ii, iv are correct

View answer & explanation

Correct answer

Answer: (b)B.

i, iv are correct

Why this answer is correct

The correct option is i, iv are correct.

Concept:

Salient pole rotor properties:

  • Salient pole rotors are large in diameter and shorter in length
  • It is used in lower-speed electrical machines (100 RPM to 1500 RPM) because of the projected pole. At high speed, these projected poles will suffer a large amount of centrifugal force and noise will be more.
  • Since the poles are not uniformly spread over the rotor surface it has no uniform air gap.
  • Salient pole rotors normally require damper windings to avoid oscillations of rotors during operation.
  • They are mostly used in hydropower plants.

 

Hence i and iv statements are correct.

Additional Information Non- salient pole rotor-

  • It has a uniform air gap.
  • Can rotate at high speed (normally 1500 or 3000 rpm).
  • Doesn't require damper winding.

Question 115

A long straight wire carries a current I = 100 A. At what distance the magnetic field is 1 A/m ?

  1. AOption (a)

    1.59 m

  2. BOption (b)

    0.159 m

  3. COption (c)

    0.0159 m

  4. DOption (d)

    15.9 m

View answer & explanation

Correct answer

Answer: (d)D.

15.9 m

Why this answer is correct

We know that

Magnetic field intensity is given as:

H = I/2πr

Here I = current, r = distance

H=I2πr=1002π×r=1;Am−1H = \frac{I}{{2\pi r}} = \frac{{100}}{{2\pi \times r}} = 1;A{m^{ - 1}}

⇒ r = 15.9 m

Question 116

Maxwell's equation in ______ form gives information at points of discontinuity is electromagnetic field.

  1. AOption (a)

    Differential

  2. BOption (b)

    Integral

  3. COption (c)

    Algebraic

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (b)B.

Integral

Why this answer is correct

The correct option is integral form.

Concept:

Maxwell’s equations:

These are basic equations of electromagnetism which are a collection of

  • Gauss’s law for electricity,
  • Gauss’s law for magnetism,
  • Faraday’s law of electromagnetic induction, and
  • Ampere’s law for currents in conductors.
  • Maxwell equations give a mathematical model for electric, optical, and radio technologies, like power generation, electric motors, etc.

The integral form is used with the finite volume method. These are equivalent in uniform grids.

The differential form does not have a solution in presence of discontinuities (eg. compressible flows with shocks).

hence Maxwell's equation in integral form gives information at points of discontinuity is an electromagnetic field.

 

Maxwell's 4 laws in differential and integral form are as follows:

Differential formIntegral formName
∇×E=−∂B∂t\nabla \times E = - \frac{{\partial B}}{{\partial t}}\mathop \oint \nolimits_L^{} E.dl = - \frac{\partial }{{\partial t}}\mathop \smallint \nolimits_S^{} B.d SFaraday’s law of electromagnetic induction
∇×H=J+∂D∂t\nabla \times H =J+ \frac{{\partial D}}{{\partial t}}\mathop \oint \nolimits_L^{} H.dl = \mathop \smallint \nolimits_S^{} (J+\frac{{\partial D}}{{\partial t}})dSAmpere’s circuital law
∇ . D = ρv\mathop \oint \nolimits_S^{} D.dS = \mathop \smallint \nolimits_v^{} \rho_v.dVGauss’ law
∇ . B = 0∮SB.dS=0\mathop \oint \nolimits_S^{} B.dS = 0Gauss’ law of Magnetostatics (non-existence of magnetic monopole)

Question 117

In respect of Laplacian operator, which of the following equations are correct ?

i. ∇2V=∂2V∂x2+∂2V∂y2+∂2V∂z2\rm \nabla^2 V = \frac{\partial ^2 V}{\partial x^2} + \frac{\partial ^2 V}{\partial y^2} + \frac{\partial ^2 V}{\partial z^2}

ii. ∇2V=1ρ∂∂ρ(∂V∂ρ)+1ρ2∂2V∂ϕ2∂2V∂z2\nabla^2 V = \frac{1}{\rho} \frac{\partial }{\partial \rho} \left( \frac{\partial V}{\partial \rho} \right)+ \frac{1}{\rho^2} \frac{\partial^2 V }{\partial \phi^2} \frac{\partial^2 V}{\partial z^2}

iii. ∇2V=1ρ∂∂ρ(ρ∂V∂ρ)+1ρ2∂2V∂ϕ2+∂2V∂z2\rm \nabla^2 V = \frac{1}{\rho} \frac{\partial }{\partial \rho} \left( \rho \frac{\partial V}{\partial \rho} \right)+ \frac{1}{\rho^2} \frac{\partial^2 V }{\partial \phi^2}+ \frac{\partial^2 V}{\partial z^2}

iv. ∇2V=1r2∂∂r(r2∂V∂r)+1r2sin⁡θ∂∂θ(∂V∂θ)+1r2sin⁡2θ∂2V∂ϕ2\rm \nabla^2 V = \frac{1}{r^2} \frac{\partial }{\partial r} \left( r^2 \frac{\partial V}{\partial r} \right)+ \frac{1}{r^2 \sin \theta} \frac{\partial }{\partial \theta} \left( \frac{\partial V}{\partial \theta }\right)+ \frac{1}{r^2 \sin^2 \theta } \frac{\partial^2 V}{\partial \phi^2}

  1. AOption (a)

    i, ii are correct

  2. BOption (b)

    i, iii, iv are correct

  3. COption (c)

    i, ii, iv are correct

  4. DOption (d)

    i, iii are correct

View answer & explanation

Correct answer

Answer: (d)D.

i, iii are correct

Why this answer is correct

The correct option is option 2.

Concept-

Laplacian operator-

The Laplacian operator is the divergence of the gradient of a function.

The laplacian function can be written in the different coordinate systems as follows

Cartesian coordinate system:

∇2V=∂2V∂x2+∂2V∂y2+∂2V∂z2\rm \nabla^2 V = \frac{\partial ^2 V}{\partial x^2} + \frac{\partial ^2 V}{\partial y^2} + \frac{\partial ^2 V}{\partial z^2}

Cylindrical-coordinate system:

∇2V=1ρ∂∂ρ(ρ∂V∂ρ)+1ρ2∂2V∂ϕ2+∂2V∂z2\rm \nabla^2 V = \frac{1}{\rho} \frac{\partial }{\partial \rho} \left( \rho \frac{\partial V}{\partial \rho} \right)+ \frac{1}{\rho^2} \frac{\partial^2 V }{\partial \phi^2}+ \frac{\partial^2 V}{\partial z^2}

Spherical Coordinate system:

∇2V=1r2∂∂r(r2∂V∂r)+1r2sin⁡θ∂∂θ(sinθ∂V∂θ)+1r2sin⁡2θ∂2V∂ϕ2\rm \nabla^2 V = \frac{1}{r^2} \frac{\partial }{\partial r} \left( r^2 \frac{\partial V}{\partial r} \right)+ \frac{1}{r^2 \sin \theta} \frac{\partial }{\partial \theta} \left( sin\theta \frac{\partial V}{\partial \theta }\right)+ \frac{1}{r^2 \sin^2 \theta } \frac{\partial^2 V}{\partial \phi^2}

Question 118

In a transformer, the primary and secondary voltages are

  1. AOption (a)

    60° out of phase

  2. BOption (b)

    90° out of phase

  3. COption (c)

    180° out of phase

  4. DOption (d)

    Always in phase

View answer & explanation

Correct answer

Answer: (c)C.

180° out of phase

Why this answer is correct

Transformer:

  • A transformer is a static device that transfers power from one circuit to another circuit without a change in the frequency.
  • Generally, the winding which is connected to the supply is called as primary winding and the winding on which load connected is called as secondary winding.
  • In the transformer, two windings are electrically isolated but magnetically coupled. Hence the transformer is a coupled circuit.
  • The transformer is a singly excited device since it requires only one external voltage source to energize any number of windings placed on its core.
  • The transformer can be treated as a phase-shifting device since it offers a displacement of approx 1800 between two circuits.
  • As the amount of flux in the transformer core is constant irrespective of power transfer, it can be treated as a "Constant flux device". Hence primary and secondary fluxes of a transformer are equal.
  • The transformer is a negative feedback circuit because it satisfies 'Lenz's law'.

Question 119

For the vector identity (A.B)2 + (A × B)2 is

  1. AOption (a)

    (AB)2

  2. BOption (b)

    A2B

  3. COption (c)

    AB2

  4. DOption (d)

    AB

View answer & explanation

Correct answer

Answer: (a)A.

(AB)2

Why this answer is correct

The correct option is (AB)2.

Concept:

Dot product(Scalar product):

  • The dot product of two vectors is given by the product of the magnitude of the two vectors and cosine of the angle between the vector.

 A.B=∣A∣∣B∣cosθA.B=|A||B|cos\theta

  • The dot product essentially tells us how much of the force vector is applied in the direction of the motion vector.
  • The dot product can also help us measure the angle formed by a pair of vectors and the position of a vector relative to the coordinate axes.

Cross product (Vector Product):

  • The Cross product of two vectors is given by the product of the magnitude of the two vectors and sine of the angle between the vector.

 A×B=∣A∣∣B∣sinθA\times B=|A||B|sin\theta

  • The cross product essentially gives us a vector that is perpendicular to both the vector.

Calculation:

(A.B)2+(A×B)2(A.B)^2 + (A\times B)^2

=(ABcosθ)2+(ABsinθ)= (ABcos\theta)^2+(ABsin\theta)

=A2B2cos2θ+A2B2sin2θ=A^2B^2cos^2\theta+A^2B^2sin^2\theta

=A2B2(cos2θ+sin2θ)=A^2B^2(cos^2\theta+sin^2\theta)

=A2B2=(AB)2=A^2B^2 =(AB)^2

Hence, the correct answer is (AB)2.

Question 120

For R.C circuit shown in following Fig. the impulse response value of i(t) will be

  1. AOption (a)

    i(t)=[1R−1RCe−tRC]u(t)\rm i(t) = \left[ \frac{1}{R} - \frac{1}{RC}e^{\frac{-t}{RC}} \right] u(t)

  2. BOption (b)

    i(t)=1R[δ(t)−1RCe−tRC]u(t)\rm i(t) = \frac{1}{R}\left[ \delta (t) - \frac{1}{RC}e^{\frac{-t}{RC}} \right] u(t)

  3. COption (c)

    i(t) = [R - RCe-t/RC]u(t)

  4. DOption (d)

    i(t) = [Rδ(t) - RCe-t/RC​]u(t)

View answer & explanation

Correct answer

Answer: (b)B.

i(t)=1R[δ(t)−1RCe−tRC]u(t)\rm i(t) = \frac{1}{R}\left[ \delta (t) - \frac{1}{RC}e^{\frac{-t}{RC}} \right] u(t)

Why this answer is correct

The correct option is i(t)=[1Rδ(t)−1RCe−t/RC]u(t)i(t)=[\frac{1}{R}δ(t)-\frac{1}{RC}e^{-t/RC}]u(t)

The impedance of the given circuit is given by,

 Z=R+1CS=1+RCSCSZ= R+\frac{1}{CS}=\frac{1+RCS}{CS}

i(t)=V(t)Zi(t)=\frac{V(t)}{Z}

Taking Laplace to transform,

I(s)=V(s)Z(s)I(s)=\frac{V(s)}{Z(s)}

=1(1+RCS)/CS=\frac{1}{(1+RCS)/CS}

=CS1+RCS=1RSS+1/RC=\frac{CS}{1+RCS}=\frac{1}{R}\frac{S}{S+1/RC}

=1R[1−1/RCS+1/RC]=\frac{1}{R}[1-\frac{1/RC}{S+1/RC}]

Taking inverse laplace transform-

i(t)=1R[δ(t)−1RCe−tRC]u(t)\rm i(t) = \frac{1}{R}\left[ \delta (t) - \frac{1}{RC}e^{\frac{-t}{RC}} \right] u(t)

Question 121

The state-space representation of a system is given below

ẋ = [−100−2]x+[10]u\rm \begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix} x + \begin{bmatrix} 1 \\ 0 \end{bmatrix} u and y = [1 1]x\rm \begin{bmatrix} 1 \ 1 \end{bmatrix} x 

The transfer function of the system is

  1. AOption (a)

    1s2+3s+2\rm \frac{1}{s^2 + 3s + 2}

  2. BOption (b)

    1s+2\rm \frac{1}{s + 2}

  3. COption (c)

    ss2+3s+2\rm \frac{s}{s^2 + 3s + 2}

  4. DOption (d)

    1s+1\rm \frac{1}{s + 1}

View answer & explanation

Correct answer

Answer: (d)D.

1s+1\rm \frac{1}{s + 1}

Why this answer is correct

Concept:

State gives the future behavior of the system with present input and history which is nothing but the initial condition described by the state variables.

NOTE: Resistive circuit doesn’t have any State variable, called a “Memoryless system”

The number of State variables:

  1. The number of capacitors and inductors for the RLC circuit.

  2. For the differential equation, it is the order.

A standard form of State-space matrix:

X˙=AX+BU\dot X = AX + BU

It is called as “State equation or Dynamic equation”

Y = CX + DU 

It is known as “Output equation”

X˙\dot X: Differential State Vector

Y: Output vector

U: Input vector

A: State matrix

B: Input matrix

C: Output matrix

D: Transition matrix

Transfer Function:

For the state model defined by we have to write the Transfer function for the analysis.

It is given by

TF=C[sI−A]−1B+DTF = C{\left[ {sI - A} \right]^{ - 1}}B + D

TF=Cadj[sI−A]det⁡[sI−A]B+DTF = C\frac{{adj\left[ {sI - A} \right]}}{{\det \left[ {sI - A} \right]}}B + D

The characteristic equation is defined by

|sI – A| = 0

Calculation:

Transfer function = C (sI - A)-1 B + D

In the given question,

\(A = \left[ {\begin{array}{{20}{c}} -1&0\ { 0}&-2 \end{array}} \right],B = \left[ {\begin{array}{{20}{c}} 1\ 0 \end{array}} \right]\)

C = \left[ {\begin{array}{*{20}{c}} 1&1 \end{array}} \right],~D = 0

\(sI - A = \left[ {\begin{array}{{20}{c}} s&0\ 0&s \end{array}} \right] - \left[ {\begin{array}{{20}{c}} -1&0\ { 0}&-2 \end{array}} \right] \)

sI-A= \left[ {\begin{array}{*{20}{c}} {s +1}&0\ 0&{s +2} \end{array}} \right]

\ {\left( {sI - A} \right)^{ - 1}} = \frac{1}{{\left( {s + 2} \right)\left( {s +1} \right)}}\left[ {\begin{array}{*{20}{c}} {s +2}&0\ { 0}&{s +1 } \end{array}} \right]

\(\frac{T}{F} = \left[ {\begin{array}{{20}{c}} 1&1 \end{array}} \right]\frac{1}{{\left( {s + 2} \right)\left( {s +1} \right)}}\left[ {\begin{array}{{20}{c}} {s +2}&0\ { 0}&{s + 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\ 0 \end{array}} \right]\)

=1(s+1)= \frac{1}{{\left( {s + 1} \right)}}

Question 122

Calculate the electric field at a distance of 2 m from a Vander Graff generator carrying a charge of 1 × 10-6 C

  1. AOption (a)

    2.25 × 103 Vm-1

  2. BOption (b)

    225 × 103Vm-1

  3. COption (c)

    0.225 × 103Vm-1

  4. DOption (d)

    22.5 × 103 Vm-1

View answer & explanation

Correct answer

Answer: (a)A.

2.25 × 103 Vm-1

Why this answer is correct

CONCEPT:

Electric field intensity: 

  • It is defined as the force experienced by a unit positive test charge in the electric field at any point.

⇒E=Fqo⇒ E=\dfrac{F}{q_{o}}

Where E = electric field intensity, qo = charge on the particle

We know that the electric field intensity at a point due to a point charge Q is given as,

⇒E=kQr2⇒ E=\dfrac{kQ}{r^2}     ---(1)

⇒ E ∝ Q

⇒E∝1r2⇒ E\propto\dfrac{1}{r^2}

Where k = electrostatic force constant, r = distance of the point from the charge

Calculation:

Given that,

Q = 1 μC

r = 2 m

And, we know that k = 9 × 109

E =  (9 × 109 × 1 × 10-6) / (22) = 2.25 × 103 Vm-1

Question 123

A 100 V, 60 W lamp is connected in series with a 100 V, 100 W lamp across 200 V supply. What will be the power consumed by each lamp in watt ?

  1. AOption (a)

    33.75 W, 26 W

  2. BOption (b)

    56.25 W, 63.18 W

  3. COption (c)

    93.75 W, 56.25 W

  4. DOption (d)

    40 W, 93.75 W

View answer & explanation

Correct answer

Answer: (c)C.

93.75 W, 56.25 W

Why this answer is correct

Series connection of Bulbs:

Two bulbs of rating W1 and W2 are connected in a series connection across a voltage source of V volts as shown below figure,

  • In series connection current through each bulb is always the same.
  • The source voltage will divide between the bulbs according to their individual resistance values.
  • The total power consumed by bulbs is the product of source voltage (V) and current through each bulb (I).
  • Individual powers can be calculated by either P = I2 R or P = V2/R.

 

Calculation:

Given that 

Source voltage V = 200 V

Power rating bulb1 P1 = 60 W

Power rating of bulb2 P2 = 100 W

Voltage rating of bulb1 V1 = 100 V 

Voltage rating of bulb2 V2 = 100 V

Resiatnce of Bulb1 R1 = V12/P1 = 1002/60 = 166.67 Ω

Resiatnce of Bulb2 R2 = V22/P2 = 1002/100 = 100 Ω 

As bulbs are connected in series in series Req = R1 + R2 = 266.67 Ω 

Current through the circuit for 200V supply is given by I = V/ Req = 200 / 266.67 = 0.75 A

Then, the power consumed by lamp1 = I2R1 = (.75)2 × 166.67 = 93.75 W

The power consumed by lamp2 = I2 R2 = (.75)2 × 100 = 56.25 W

Question 124

Two small identical conducting spheres have charges 2 × 10-9 C and -0.5 × 10-9 C respectively, if they are brought into contact and then separated by 4 cm, the force between them will be

  1. AOption (a)

    -56.25 × 10-7 N

  2. BOption (b)

    +56.25 × 10-7 N

  3. COption (c)

    -0.316 × 10-5 N

  4. DOption (d)

    +0.316 × 10-5 N

View answer & explanation

Correct answer

Answer: (d)D.

+0.316 × 10-5 N

Why this answer is correct

The correct option is **+**0.316 × 10-5 N.

Concept:

  • When two charged metal spheres are connected, the charge will flow from one to the other until their potentials are equal.
  • Two small identical conducting spheres are brought together to touch then a total charge of the two spheres spreads out across the whole surface of the spheres. When the spheres are moved apart again, each one is left with half of the total original charge.

 

Force between two charge particles is given by-

F=kq1q2r2F= k\frac{q_1q_2}{r^2}

Where, q1 ,q2 are charge.

k = Couloumb's constant = k = 9 ×109 

r = distance between the charge particles.

Calculation:

Given the sphere with charges-

q1 = 2 × 10-9 C

q2 = -0.5 × 10-9 C

Total charge,

Q = q1 + q2

    = 1.5 ×10-9 C

When both the spheres are separated then the total charge is equally distributed over both spheres.

Hence new charge is given as -

q1new = q2new = Q/2 = 0.75 × 10-9 C

Hence force between both spheres is given by-

F=kq1q2r2F= k\frac{q_1q_2}{r^2}

Where k = 9 ×109 

 r = 4 cm = 0.04 m

Hence,

F=9×109×0.75×10−9×0.75×10−90.042F= \frac{9×10^9×0.75× 10^{-9}× 0.75×10^{-9}}{0.04^2}

F=3.16×10−6NF=3.16×10^{-6}N

Hence, the correct answer is +0.316 × 10-5 N.

Question 125

The main reason for generation of harmonics in a transformer could be

  1. AOption (a)

    Fluctuating load

  2. BOption (b)

    Poor insulation

  3. COption (c)

    Mechanical vibrations

  4. DOption (d)

    Saturation of core

View answer & explanation

Correct answer

Answer: (d)D.

Saturation of core

Why this answer is correct

The harmonic is the distortion in the waveform of the voltage and current. It is the integral multiple of some reference waves. The harmonic wave increases the core and copper loss of the transformer and hence reduces their efficiency. It also increases the dielectric stress on the insulation of the transformer. These harmonics are produced by saturation of core.

Saturation of core means the thermal limit after which a core or magnet loose its magnetic strength or stops production of magnetic lines of force.

125 questions in the full paper