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Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Tamil Book Solutions Guide Pdf Chapter 3.3 தமிழ்ர் மருந்தும் Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th Tamil Solutions Chapter 3.3 தமிழ்ர் மருந்தும்

கற்பவை கற்றபின்

Question 1.
நீங்கள் மருத்துவரிடம் கேட்க விரும்பும் ஐந்து வினாக்களை எழுதுக.
Answer:
(i) பல மருந்துகளின் பெயர்களை மருத்துவ நூல்களில் படிக்கின்றோம். ஆனால், அந்த மருந்துகளைப் பார்த்ததே இல்லை. மற்றவர்களுக்கும் தெரிவது இல்லை. அதைத் தெரிந்து கொண்டால் அந்த மருந்துகளின் பயன்பாட்டை நாங்கள் பயன்படுத்த வழி உண்டு. அதற்கு மருத்துவராக விளங்கும் நீங்கள் வழிவகை செய்ய முடியுமா?

(ii) வேதிக்கலப்பு இல்லாத உணவு இன்று குறைவு. அப்படி இருக்கும் போது நோய்கள் விரைவாகவே வந்து விடுகின்றன. இதிலிருந்து மீண்டுவர தாங்கள் கூறும் அறிவுரை யாது?

(iii) பழைய மருத்துவ தாவரங்களை மீட்டுருவாக்கம் செய்ய வழிவகை உள்ளதா?

(iv) நவீன மருத்துவத்தைத் தவிர்த்து நாட்டு மருத்துவத்திற்கு நுழைய அரசு மருத்துவமனையில் பழைய மருத்துவமுறைக்கு வழி உள்ளதா?

(v) தமிழர் மருத்துவத்தைப் பெரும்பாலான தமிழர்களே ஏற்றுக்கொள்ளாத போது, தமிழர் மருத்துவத்தை உலக அளவில் பறைசாற்றுவது எப்படி?

Question 2.
உங்கள் பகுதிகளில் கிடைக்கும் மூலிகைகளின் மாதிரிகளைத் திரட்டி அவற்றின் பயன்களை எழுதிக் காட்சிப்படுத்துக.
Answer:
மாணவர் செயல்பாடு.

பாடநூல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
தொடக்க காலத்தில் மனிதர்கள் மருத்துவத்திற்கு ………………………… பயன்படுத்தினர்.
அ) தாவரங்களை
ஆ) விலங்குகளை
இ) உலோகங்களை
ஈ) மருந்துகளை
Answer:
அ) தாவரங்களை

Question 2.
தமிழர் மருத்துவத்தில் மருந்து என்பது ……………………… நீட்சியாகவே உள்ளது.
அ) மருந்தின்
ஆ) உடற்பயிற்சியின்
இ) உணவின்
ஈ) வாழ்வின்
Answer:
இ) உணவின்

Question 3.
உடல் எடை அதிகரிப்பதால் ஏற்படும் நோய்களுள் ஒன்று ……………………..
அ) தலைவலி
ஆ) காய்ச்ச ல்
இ) புற்றுநோய்
ஈ) இரத்தக்கொதிப்பு
Answer:
ஈ) இரத்தக்கொதிப்பு

Question 4.
சமையலறையில் செலவிடும் நேரம் ……………………… செலவிடும் நேரமாகும்.
அ) சுவைக்காக
ஆ) சிக்கனத்திற்காக
இ) நல்வாழ்வுக்காக
ஈ) உணவுக்காக
Answer:
இ) நல்வாழ்வுக்காக

குறுவினா

Question 1.
மருத்துவம் எப்போது தொடங்கியது?
Answer:
தொடக்க காலத்தில் மனிதனுக்கு நோய் வந்தபோது இயற்கையாக வளர்ந்த தாவரங்களைக் கொண்டும், அவனுக்கு அருகில் கிடைத்த தாவர, கனிம, சீவப் பொருள்களைக் கொண்டும் நோயைத் தீர்க்க முயன்றிருப்பான். அப்போதே மருத்துவம் தொடங்கியது.

Question 2.
நல்வாழ்விற்கு நாம் நாள்தோறும் செய்ய வேண்டியவை யாவை?
Answer:

• 45 நிமிடத்தில் 3 கி.மீ. நடைப்பயணம்.
• 15 நிமிடம் யோகா, தியானம், மூச்சுப்பயிற்சி.
• 7 மணி நேர தூக்கம்.
• 3 லிட்டர் தண்ணீ ர் அருந்துதல்.

Question 3.
தமிழர் மருத்துவத்தில் மருந்துகளாகப் பயன்படுவன யாவை?
Answer:
மூலிகை, தாவர இலை, தாவர வேர், உலோகங்கள், பாஷாணங்கள், தாதுப்பொருட்கள் ஆகியன தமிழர் மருத்துவத்தில் மருந்துப்பொருட்களாகப் பயன்படுகின்றனவாகும்.

சிறுவினா

Question 1.
நோய்கள் பெருகக் காரணம் என்ன?
Answer:

• மனிதன் இயற்கையை விட்டு விலகி வந்ததுதான் முதன்மைக் காரணம்.
• மாறிப்போன உணவு, மாசு நிறைந்த சுற்றுச்சூழல், மன அழுத்தம் இவை மூன்றும் குறிப்பிடத்தக்க காரணங்கள்.
• தன் உணவுக்காக வேறு எதைப்பற்றியும் கவலை கொள்ளாமல், நிலத்தை உரங்களாலும், பூச்சிக்கொல்லிகளாலும் நச்சுப்படுத்தலாம் என்ற
• அலட்சியமான எண்ணம், மன அழுத்தம், எது கேளிக்கை? எது குதூகலம்? எது படிப்பு? எது சிந்தனை? என்ற புரிதல் இல்லாமை ஆகியவற்றைக் கூடுதல் காரணங்களாகச் சொல்லலாம்.
• நம்முடைய வாழ்வியலைச் செம்மைப்படுத்துவதற்காக நாம் அறிவியல் அறிவை, மேம்பட்ட அறிவை வளர்த்தோம். ஆனால் நுண்ணறிவைத் தொலைத்துவிட்டோம்.
• இயற்கையோடு இயைந்து வாழலாம் என்கிற அறிவை நாம் மறந்துவிட்டோம். இதுவே இன்றைக்குப் பல நோய்கள் பெருக மிக முக்கியமான காரணம் ஆகும்.

Question 2.
பள்ளிக் குழந்தைகளுக்கு மருத்துவர் கூறம் அறிவுரைகள் யாவை?
Answer:

• நோய் வந்த பின்பு மருத்துவமனைக்குச் செல்வதை விட வருமுன் காக்கும் வாழ்க்கையை வாழக் கற்றுக் கொள்ளுங்கள்.
• சரியான உணவு, சரியான உடற்பயிற்சி, சரியான தூக்கம் ஆகிய மூன்றும் உங்களை நலமாக வாழவைக்கும்.
• விலை உயர்ந்த உணவுதான் சரியான உணவு என்று எண்ணாதீர்கள்.
• எளிமையாகக் கிடைக்கக்கூடிய காய்கறிகள், கீரைகள், பழங்கள், சிறுதானியங்களை உணவில் சேர்த்துக் கொள்ளுங்கள்.
• கணினித் திரையிலும், கைபேசியிலும் விளையாடுவதைத் தவிர்த்து நாள்தோறும் ஓடியாடி விளையாடுங்கள்.
• இரவுத் தூக்கம் மிகவும் இன்றியமையாதது. உரிய நேரத்தில் உறங்கச் செல்லுங்கள். அதிகாலையில் விழித்தெழுங்கள் உங்களை எந்த நோயும் அண்டாது.

நெடுவினா

Question 1.
தமிழர் மருத்துவத்தின் சிறப்புகளாக மருத்துவர் கூறும் செய்திகளைத் தொகுத்து எழுதுக.
Answer:
(i) வேர்பாரு; தழைபாரு மிஞ்சினக்கால் பற்பசெந்தூரம் பாரே’ என்றனர் சித்தர்கள்.

(ii) வேர், தழையால் குணம் அடையாதபோது சில நாட்பட்ட நோய்களுக்கு, தாவரங்கள் மட்டும் அல்லாமல் உலோகங்களையும் பாஷாணங்களையும் சித்த மருந்துகளாக நம் முன்னோர்கள் பயன்படுத்தியிருக்கிறார்கள்.

(iii) அந்தக் காலத்தில் எப்படி மூலிகைகளை மருந்தாகப் பார்த்தார்களோ அப்படியே தாதுப்பொருட்களையும், உலோகத்தையும் பார்த்தார்கள்.

(iv) அவற்றை மருந்துகளாக மாற்றும் வல்லமை சித்தமருத்துவத்தில் இருந்திருக்கிறது.

(v) ஒரு மருந்தை எடுத்துக்கொண்டால் அதற்கு விளைவும் இருக்கும்; பக்கவிளைவும் இருக்கும். ஆனால் தமிழர் மருத்துவத்தில் பக்க விளைவுகள் இல்லை. அதற்குக் காரணம் மருந்து என்பதே உணவின் நீட்சியாக இருக்கிறது.

(vi) ஒரு கவளம் சோற்றை உடல் எப்படி எடுத்துக்கொள்கிறதோ, அப்படியே தான் சித்த மருத்துவத்தின் லேகியத்தையும், சூரணத்தையும் உடல் எடுத்துக்கொள்ளும்.

(vii) அதனால் உணவு எப்படி பக்கவிளைவுகளைத் தருவதில்லையோ அதே போலச் சித்த மருந்துகளும் பக்கவிளைவுகளை ஏற்படுத்துவதில்லை.

(viii) தமிழர் மருத்துவத்தின் சிறப்பு என்னவென்றால், தனித்துவமான பார்வை இதன் முதல் சிறப்பு; இரண்டாவது சூழலுக்கு இசைந்த மருத்துவம் இது. இந்த மருத்துவத்தின் பயன்பாடோ, மூலக்கூறுகளோ, மருந்துகளோ சுற்றுச்சூழலைச் சிதைக்காது.

(ix) மிக முக்கியமான சிறப்பு என்னவென்றால், நோய்க்கான சிகிச்சையை மட்டும் சொல்லாமல், நோய் மீண்டும் வராமலிருப்பதற்கான வாழ்வியலையும் சொல்கிறது.

(x) அதாவது நோய்நாடி நோய் முதல் நாடி’ என்ற திருக்குறளின்படி நோயை மட்டுமன்றி, அதன் காரணிகளையும் கண்டறிந்து ஒருவரை நோயில்லாத மனிதராக்குகிறது.

சிந்தனை வினா

Question 1.
நோயின்றி வாழ நாம் என்னென்ன வழிகளைக் கையாளலாம்?
Answer:
இயற்கையோடு இணைந்து உண்ணல்:
மனிதனின் அடிப்படைத் தேவைகளுள் முதன்மையானது உணவு. மக்கள் உண்ணும் உணவும் உணவுப் பழக்கவழக்கங்களுமே அவர்களது உடல் நலத்தையும் உள நலத்தையும் தீர்மானிக்கின்றன. தமிழர் மருத்துவத்தில் உணவு என்பது அனைத்து நோய்களையும் தீர்க்கக் கூடிய சஞ்சீவி மருந்தாகக் கருதப்படுகிறது.

உண்ணும் முறை :
எளிதில் செரிக்கக் கூடிய பழம், காய், பருப்பு, அரிசி, கோதுமை, பால் இவற்றையே குடல் ஏற்றுக் கொள்கிறது. நாச்சுவை கருதி உண்ணாமல், உடல் நலங்கருதி உண்ணுதலே நல்லது. உணவை விரைவாக விழுங்கக்கூடாது; நன்றாக மென்று விழுங்குதல் வேண்டும்.

பயிற்சிகள் :
தினமும் நாற்பத்தைந்து நிமிடத்தில் மூன்று கி.மீ. நடைப்பயணம், பதினைந்து நிமிடம் யோகா, தியானம், மூச்சுப்பயிற்சி, ஏழு மணிநேர தூக்கம், மூன்று லிட்டர் தண்ணீ ர் அருந்துவது அவசியம்.

தவிர்க்க வேண்டியன:
நோய்க்கு முதல் காரணம் உப்பு. இதனைக் குறைவாக சேர்த்தல் நன்று. உப்பு நிறைந்த பொருள்களான ஊறுகாய், அப்பளம், வடகம், கருவாடு, வறுத்த முந்திரிபருப்பு, வறுத்த உருளைச் சீவல், வாழைக்காய்ச் சீவல், புளித்த மோர் முதலியனவற்றை முழுவதுமாகத் தவிர்த்தல் வேண்டும். கொழுப்பு நிறைந்த இறைச்சிகள், முட்டையின் மஞ்சள் கரு, தயிர், நெய், வெண்ணெய், பாலாடை, பனிக்கூழ், இனிப்புக்கட்டி ஆகியவற்றை நீக்குதல் வேண்டும்.

சமச்சீர் உணவு :
‘உணவே மருந்து மருந்தே உணவு’ என்று வாழ்ந்தவர்கள் நம் முன்னோர்கள். ஒருவர் உட்கொள்ளும் உணவில் புரதம், கொழுப்பு, மாச்சத்து, கனிமங்கள், நுண்ணூட்டச் சத்துகள் சேர்ந்ததே சமச்சீர் உணவு. எனவே, அளவறிந்து உண்ண வேண்டியது அவசியமாகும்.

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
அருந்தும் உணவே அருமருந்தென அறிந்தவர் ………………………… மக்கள்.
அ) கேரள
ஆ) தெலுங்கு
இ) தமிழ்
ஈ) கர்நாடகம்
Answer:
இ) தமிழ்

Question 2.
………………….. மக்கள் உடற்கூறு பற்றிய அறிவிலும், மருத்துவம் பற்றிய புரிதலிலும் சிறந்து விளங்கினர்.
அ) தமிழ்
ஆ) சீனா
இ) ஆங்கிலேயர்
ஈ) கேரளம்
Answer:
அ) தமிழ்

Question 3.
தமிழர் மருத்துவம் ………………………. என்று அழைக்கப்படுகிறது.
அ) ஹோமியோபதி
ஆ) அலோபதி
இ) அக்குபஞ்சர்
ஈ) சித்த மருத்துவம்
Answer:
ஈ) சித்த மருத்துவம்

Question 4.
உடலை வளப்படுத்தி உள்ளத்தைச் சீராக்குவது …………………… கலை.
அ) நாட்டியம்
ஆ) ஓவியம்
இ) உடற்பயிற்சி
ஈ) யோகனம்
Answer:
ஈ) யோகனம்

Question 5.
தமிழர் ………………………. சாங்கியம், ஆசீவகம் ஆகும்.
அ) மெய்யியல்
ஆ) வரலாறு
இ) அறிவியல்
ஈ) தத்துவங்கள்
Answer:
ஈ) தத்துவங்கள்

Question 6.
…………………. மற்றும் ……………………. காலத்தில் அந்தந்த மதங்களின் கூறுகள் நம் மருத்துவத்தில் இருந்தன.
அ) இந்து, முஸ்லிம்
ஆ) சைவம், வைணவம்
இ) இந்து, சமணம்
ஈ) சமணம், பௌத்தம்
Answer:
ஈ) சமணம், பௌத்தம்

Question 7.
………………………….. களுடைய நவீன அறிவியல் பார்வை நம்மீது தாக்கத்தை ஏற்படுத்தியது.
அ) பாரசீகர்கள்
ஆ) ஆங்கிலேயர்கள்
இ) ஆரியர்கள்
ஈ) முகலாயர்கள்
Answer:
ஆ) ஆங்கிலேயர்கள்

Question 8.
‘வேர்பாரு; தழைபாரு மிஞ்சினக்கால் பற்பசெந்தூரம் பாரே’ என்றனர் ……………………….
அ) தமிழர்கள்
ஆ) சித்தர்கள்
இ) சைவர்கள்
ஈ) முன்னோர்கள்
Answer:
ஆ) சித்தர்கள்

Question 9.
நாட்பட்ட நோய்களுக்கு, தாவரங்கள் மட்டும் அல்லாமல் ……………………….. களையும் பயன்படுத்தினர்.
அ) உலோகங்கள்
ஆ) அலோகங்கள்
இ) சாயங்கள்
ஈ) வேதியல்
Answer:
அ) உலோகங்கள்

Question 10.
தாதுப்பொருட்களையும், உலோகத்தையும் மருந்துகளாக மாற்றும் வல்லமை …………………….. மருத்துவத்தில் இருந்திருக்கிறது.
அ) நவீன
ஆ) மூலிகை
இ) சித்த
ஈ) சீன
Answer:
இ) சித்த

Question 11.
‘நோய்நாடி நோய் முதல்நாடி’ என்று கூறியவர் ………………………
அ) கம்பர்
ஆ) வள்ளுவர்
இ) ஔவையார்
ஈ) திருமூலர்
Answer:
ஆ) வள்ளுவர்

Question 12.
தினமும் ………………………… நிமிடத்தில் மூன்று கி.மீ. நடைப்பயணம் மேற்கொள்ள வேண்டும்.
அ) 25
ஆ) 30
இ) 40
ஈ) 45
Answer:
ஈ) 45

Question 13.
தினமும் …………………… நிமிடம் யோகா செய்ய வேண்டும்.
அ) 10
ஆ) 15
இ) 20
ஈ) 25
Answer:
ஆ) 15

Question 14.
தினமும் …………………… மணிநேரம் தூங்குவது அவசியம்.
அ) 7
ஆ) 8
இ) 6
ஈ) 5
Answer:
அ) 7

Question 15.
தினமும் ………………… லிட்டர் தண்ணீ ர் அருந்துவது அவசியம்.
அ) 5
ஆ) 4
இ) 2
ஈ) 3
Answer:
ஈ) 3

Question 16.
கணினித்திரையிலும் கைபேசியிலும் விளையாடுவதைத் தவிர்த்து நாள்தோறும் ………………………….. விளையாடுங்கள்.
அ) கபடி
ஆ) நொண்டி
இ) ஓடியாடி
ஈ) செஸ்
Answer:
இ) ஓடியாடி

பிரித்து எழுதுக

மருந்தென = மருந்து + என
உடற்கூறுகள் = உடல் + கூறுகள்
தங்களுக்கென = தங்களுக்கு + என
வந்துள்ளோம் = வந்து + உள்ளோம்
முயன்றிருப்பான் = முயன்று + இருப்பான்

அறிந்திருப்பர் = அறிந்து + இருப்பர்
பழந்தமிழர் = பழமை + தமிழர்
வளப்படுத்தி = வளம் + படுத்தி
மருந்தில்லா = மருந்து + இல்லா
இவற்றுக்கெல்லாம் = இவற்றுக்கு + எல்லாம்

கண்டறிந்து = கண்டு + அறிந்து
வாழ்வியலுடன் = வாழ்வியல் + உடன்
துரிதமாக = துரிதம் + ஆக
மாறிப்போனது = மாறி + போனது
ஒளிந்திருக்கும் = ஒளிந்து + இருக்கும்

சேர்த்தெழுதுக

இரத்தம் + கொதிப்பு = இரத்தக்கொதிப்பு
உடல் + பயிற்சி = உடற்பயிற்சி
அவசியம் + ஆயிற்று = அவசியமாயிற்று
நாள் + பட்ட = நாட்பட்ட
மீண்டு + எழுந்து = மீண்டெழுந்து

என்று + எல்லாம் = என்றெல்லாம்
பட்டியல் + இட்டு = பட்டியலிட்டு
சுற்று + சூழல் = சுற்றுச்சூழல்
மட்டும் + அன்றி = மட்டுமன்றி
மற்று + ஒரு = மற்றொரு

உணவுக்கு + ஆக = உணவுக்காக
நுண் + அறிவு = நுண்ணறிவு
சரி + அன்று = சரியன்று
சமையல் + அறை = சமையலறை
விழித்து + எழுங்கள் = விழித்தெழுங்கள்

குறுவினா

Question 1.
பாடப்பகுதியில் வள்ளுவர் மருந்து பற்றிய குறளாக குறிப்பிடுவது எது?
Answer:
‘மருந்தென வேண்டாவாம் யாக்கைக்கு அருந்தியது
அற்றது போற்றி உணின்’ என்கிறார் வள்ளுவர்.

Question 2.
பாரம்பரிய மருத்துவ முறையில் தமிழ் மக்கள் எவ்வாறு சிறந்திருந்தனர்?
Answer:

• தமிழ்மக்கள் உடற்கூறுகள் பற்றிய அறிவிலும், மருத்துவம் பற்றிய புரிதலிலும் சிறந்து விளங்கினர்.
• உலகில் பல்வேறு மருத்துவ முறைகள் இருந்தாலும் தங்களுக்கெனப் பாரம்பரிய மருத்துவ முறைகளையும் உருவாக்கிப் பின்பற்றி வந்தனர்.

Question 3.
பழந்தமிழரின் மருத்துவம் பற்றிய குறிப்புகள் எவற்றில் கிடைக்கின்றன?
Answer:
பழந்தமிழர்கள் மருத்துவத்தை அறிந்தது மட்டுமன்றி மருத்துவத்தில் சிறந்தும் விளங்கினார்கள் என்பதற்கான குறிப்புகள் பழந்தமிழ் இலக்கியங்களில் கிடைக்கின்றன.

Question 4.
பழந்தமிழர்கள் அறிந்திருந்த மருத்துவ முறைகள் யாவை?
Answer:
பழந்தமிழர்கள், மூலிகை மருத்துவம், அறுவை மருத்துவம், மருந்தில்லா மருத்துவம் போன்றவற்றையும் உடலை வளப்படுத்தி உள்ளத்தைச் சீராக்கும் யோகம் முதலிய
கலைகளையும் அறிந்திருந்தார்கள்.

Question 5.
தமிழர் எவ்வுண்மையை விளக்கினர்?
Answer:
நோயை இயற்கையில் கிடைக்கும் பொருட்கள், அப்பொருளின் தன்மை அதன் சுவை இவற்றைக் கொண்டே குணப்படுத்த முடியும் என்ற உண்மையை மிகத்தெளிவாக விளக்கினர்.

Question 6.
தமிழர் மருத்துவத்தின் முக்கியச் சிறப்பு என்ன?
Answer:
(i) தமிழர் மருத்துவத்தின் மிக முக்கியமான சிறப்பு என்னவென்றால், நோய்க்கான சிகிச்சையை மட்டும் சொல்லாமல், நோய் மீண்டும் வராமலிருப்பதற்கான வாழ்வியலையும் சொல்கிறது.

(ii) அதாவது நோய்நாடி நோய் முதல் நாடி’ என்ற குறளின்படி நோயை மட்டுமன்றி அதன் காரணிகளையும் கண்டறிந்து ஒருவரை நோயில்லாத மனிதராக்குகிறது.

Question 7.
இன்று நோய்கள் பெருகுவதற்கான மூன்று காரணங்கள் யாவை?
Answer:

• மனிதன் இயற்கையை விட்டு விலகி வந்ததுதான் முதன்மைக் காரணம்.
• மாறிப்போன உணவு, மாசு நிறைந்த சுற்றுச்சூழல், மன அழுத்தம் இவை மூன்றும் குறிப்பிடத்தக்க காரணங்கள்.

சிறுவினா

Question 1.
தமிழர் மருத்துவ முறைகள் எவ்வாறு பரிணாம வளர்ச்சி அடைந்தது?
Answer:
(i) தமிழரது நிலம், நிறைந்த பண்பாடுகளும் தத்துவங்களும் அடங்கியது.

(ii) நோய்கள் எல்லாம் பேய், பிசாசுகளால் வருகின்றன; பாவ, புண்ணியத்தால் வருகின்றன என்று உலகத்தின் பல பகுதிகளில் சொல்லிக் கொண்டிருந்த காலத்தில், தமிழர் தத்துவங்களான சாங்கியம், ஆசீவகம் போன்றவை உடலுக்கும் ஏற்படும் மாற்றங்களை விளங்கின.

(iii) நோயை இயற்கையில் கிடைக்கும் பொருட்கள், அப்பொருள்களின் தன்மை அதன் சுவை இவற்றைக் கொண்டே குணப்படுத்த முடியும் என்ற உண்மையை மிகத் தெளிவாக விளக்கினர்.

(iv) தமிழர் மருத்துவம் பண்பாட்டுக் கூறாக ஆகும்போது நாட்டு வைத்தியமாகவும், பாட்டி வைத்தியமாகவும் மரபு சார்ந்த சித்த வைத்தியமாகவும், உணவு சார்ந்த
மருத்துவமாகவும், பண்பாடு சார்ந்த மருத்துவமாகவும் விரிந்திருக்கிறது.

Question 2.
உயர்வாக இருந்த தமிழர் மருத்துவமுறை பிறகு பின்தங்கிப் போனதற்குக் காரணம் என்ன ?
Answer:

• நம்மீது நிகழ்ந்த படையெடுப்புகள் தாக்கத்தை ஏற்படுத்தின.
• தமிழர் மருத்துவம் அவரவர்வாழ்வியலுடனும், தத்துவங்களுடனும் பிணைந்துதான் வந்து கொண்டிருந்தது.
• சமணர் மற்றும் பௌத்தர் காலத்தில் அந்தந்த மதங்களின் கூறுகள் நம் மருத்துவத்தில் இருந்தன. பிறகு சைவம் ஓங்கிய போது சைவ சித்தாந்தத்தின் கூறுகள் கலந்தன.
• இறுதியில் ஆங்கிலேயர்கள் வந்தனர். அவர்களுடைய நவீன அறிவியல் பார்வை நம்மீது தாக்கத்தை ஏற்படுத்தியது.
• சித்த மருத்துவம் என்பது மரபுவழி மருத்துவமாகவும், நாட்டு மருத்துவமாகவும் சுருங்கியது. இறுதியில் கிராமம் சார்ந்த மருத்துவமாக மாறிப்போனது.
• நவீன மருத்துவத்தில், துரிதமாகச் சில நோய்களுக்குக் கிடைத்த தீர்வுகள் பெரும்பயனையும் வரவேற்பையும் கொடுத்தன.

Question 3.
தமிழர் மருத்துவம் மறுமலர்ச்சி அடைவதற்கான காரணங்கள் யாவை?
Answer:
(i) பல ஆண்டுகளுக்குப் பின்னால் தான் பாரம்பரிய மருத்துவம் மிகப்பெரிய அனுபவத்தின் நீட்சி என்பதும் மிகப்பெரிய பட்டறிவில் ஒரு பெரிய அறிவியல் கண்டிப்பாக ஒளிந்திருக்கும் என்பதும் புரியத் தொடங்கின. குறிப்பாகச் சர்க்கரை, இரத்தக்கொதிப்பு, புற்றுநோய், மாரடைப்பு முதலிய வாழ்வியல் நோய்கள் பெருகிய நிலையைச் சொல்லலாம்.

(ii) இவற்றைத் தீர்க்க வெறும் இரசாயன மருந்துகள் போதா. கூடவே உணவு, வாழ்வியல், உடற்பயிற்சி, யோகம் இவையும் கூட்டாக மேற்கொள்ளப்பட வேண்டும்.

(iii) தொடர் சிகிச்சைக்குப் பிற பக்க விளைவுகள் இல்லா மருந்துகளின் தேவை அவசியமாயிற்று. அதன் பிறகுதான் எல்லா நாடுகளிலும் இருக்கும் பாரம்பரிய மருத்துவ முறைகளின் மீது, நவீன அறிவியல் பார்வை விழத் தொடங்கியது.

(iv) அதனால் சித்த மருத்துவத்தின் தொன்மையும், தமிழர்களின் தொன்மையும் புரிய ஆரம்பித்தன் ஆய்வுகள் மேற்கொள்ளப்பட்டன.

(v) நாட்பட்ட நோய்களுக்கு மட்டுமல்லாது புதிய தொற்றுநோய் மாதிரியான சவால்களுக்கும் இது சிறந்த மாற்றாக இருப்பது தெரிய வந்தது.

(vi) இன்றைக்குப் பெருவாரியாக இது மீண்டெழுந்து வந்து கொண்டிருக்கிறது.

Question 4.
மருத்துவத்தில் பக்கவிளைவுகள் பற்றி மருத்துவர் கூறிய கருத்துகள் யாவை?
Answer:
(i) ஒரு மருந்தை எடுத்துக்கொண்டால் அதற்கு விளைவும் இருக்கும்; பக்க விளைவும் இருக்கும். ஆனால், தமிழர் மருத்துவத்தில் பக்க விளைவுகள் இல்லை . அதற்குக் காரணம் மருந்து என்பதே உணவின் நீட்சியாக இருக்கிறது.

(ii) ஒரு கவளம் சோற்றை உடல் எப்படி எடுத்துக்கொள்கிறதோ, அப்படியே தான் சித்த மருத்துவத்தின் இலேகியத்தையும், சூரணத்தையும் உடல் எடுத்துக்கொள்ளும்.

(iii) அதனால், உணவு எப்படிப் பக்க விளைவுகளைத் தருவதில்லையோ அதே போலச் சித்த மருந்துகளும் பக்கவிளைவுகளை ஏற்படுத்துவதில்லை.

(iv) இருந்த போதிலும் சித்த மருத்துவத்தின் மீது தற்போது நடக்கும் பல அறிவியல் ஆய்வுகள் மூலம் அவற்றைத் தர நிர்ணயம் செய்து யாருக்கு எந்த மருந்து, எந்த அளவில், எந்தத் துணை மருந்து கொடுத்தால் பக்க விளைவு இருக்காது என்று பட்டியலிட்டுள்ளனர்.

Question 5.
உணவைக் குறைப்பதுதான் எடையைக் குறைக்கும் வழியா விளக்குக.
Answer:

• இன்றைக்குப் பல உணவுக் கட்டுப்பாட்டு முறைகள் உள்ளன. எல்லாம் நல்ல முயற்சிகள் தாம். ஆனால், அவற்றைப் பின்பற்றுவதற்குத் தம் உடல் ஏற்றதாக இருக்கிறதா என்பதைக் குடும்ப மருத்துவரிடம் கேட்டு முடிவெடுக்க வேண்டும்.
• ஏனென்றால், இன்றைக்குப் பலரும் இணையத்தைப் பார்த்து அம்முறைகளுக்குச் செல்கிறார்கள்.
• சிலருக்கு அவை கேடு விளைவிக்கக்கூடும். தடாலடியாக எடையைக் குறைப்பது சரியன்று .
• ஒரு குறிப்பிட்ட கால இடைவெளிகளில் ஒன்று, ஒன்றரை ஆண்டுகளில் எடையைக் கட்டுக்குள் கொண்டு வர முயற்சிக்க வேண்டும்.
• அவசர யுகம் என்றாலும் உணவு உண்பதில், சில ஒழுக்கங்களைக் கடைப்பிடித்தாக வேண்டும்.
• உணவுக்காக சமையலறையில் செலவிடும் நேரத்தை, நல்வாழ்விற்காகச் செலவிடும் நேரம் என நினைக்க வேண்டும்.

நெடுவினா

Question 1.
மருத்துவத்தின் தொடக்கம் குறித்தும், நாள்தோறும் நாம் செய்ய வேண்டியன குறித்தும், மாணவர்கள் கடைப்பிடிப்பன குறித்தும் எழுதுக.
Answer:
மருத்துவத்தின் தொடக்கம் :
தொடக்க காலத்தில் மனிதனுக்கு நோய் வந்தபோது இயற்கையாக வளர்ந்த தாவரங்களைக் கொண்டும் அவனுக்கு அருகில் கிடைத்தத் தாவர, கனிம, சீவப் பொருள்களைக் கொண்டும் நோயைத் தீர்க்க முயன்றிருப்பான். தாவரங்களின் வேர், பட்டை, இலை, பூ, கனி முதலியவற்றை மருந்தாகப் பயன்படுத்தியிருப்பான். இவ்வாறுதான் மனிதர்களுக்கும் மருத்துவத்திற்குமான தொடர்பு தொடங்கியது.

நாள்தோறும் செய்ய வேண்டியன :
தினமும் நாற்பத்தைந்து நிமிடத்தில் மூன்று கி.மீ. நடைப்பயணம், பதினைந்து நிமிடம் யோகா, தியானம், மூச்சுப்பயிற்சி, ஏழுமணி நேர தூக்கம், மூன்று லிட்டர் தண்ணீர் அருந்துவது அவசியம். எங்கோ விளையும் ஆப்பிளைச் சாப்பிடுவதை விட, நமது ஊரில் விளையும் கொய்யா, இலந்தை, நாவல், பப்பாளி, நெல்லி, வாழைப்பழங்கள் ஆகியவற்றைக் காலை உணவுக்கு முன்பு சாப்பிடலாம்.

மாணவர்கள் கடைப்பிடிக்க வேண்டியன:
நோய் வந்த பின்பு மருத்துவமனைக்குச் செல்வதை விட வருமுன் காக்கும் வாழ்க்கையை வாழக் கற்றுக் கொள்ளுங்கள். சரியான உணவு, சரியான உடற்பயிற்சி, சரியான தூக்கம் ஆகிய மூன்றும் உங்களை நலமாக வாழ வைக்கும். விலை உயர்ந்த உணவுதான் சரியான உணவு என்று எண்ணாதீர்கள்.

எளிமையாகக் கிடைக்கக்கூடிய காய்கறிகள், கீரைகள், பழங்கள், சிறுதானியங்களை உணவில் சேர்த்துக் கொள்ளுங்கள். கணினித்திரையிலும் கைபேசியிலும் விளையாடுவதைத் தவிர்த்து நாள்தோறும் ஓடியாடி விளையாடுங்கள். இரவுத்தூக்கம் இன்றியமையாதது. உரிய நேரத்தில் உறங்கச் செல்லுங்கள்; அதிகாலையில் விழித்தெழுங்கள் உங்களை எந்த நோயும் அண்டாது.

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.10

9th Maths Exercise 3.10 Solutions Question 1.
Draw the graph for the following
(i) y = 2x
(ii) y = 4x – 1
(iii) y = \(\left(\frac{3}{2}\right)\) x + 3
(iv) 3x + 2y = 14
Solution:
(i) Put x = -1, y = 2 × -1 = -2
When x = 0, y = 2 × 0 = 0
When x = 1, y = 2 × 1 = 2

The points (x, y) to be plotted: (-1, -2), (0, 0), (1, 2)

(ii) When x = -1 ⇒ y = 4 (-1) -1
y = – 4 – 1 = – 5
x = 0 ⇒ y = 4 × 0 – 1 = -1
x = 1 ⇒ y = 4 × 1 – 1 = 3

The points (x, y) to be plotted: (-1, -5), (0, -1), (1, 3)

The points to be plotted: (-2, 0), (0, 3), (2, 6)

The points to be plotted: (-2, 10), (0, 7), (2, 4)

9th Maths Graph Question 2.
Solve graphically
(i) x + y = 7; x – y = 3
(ii) 3x + 2y = 4; 9x + 6y – 12 =0
(iii) \(\frac{x}{2}+\frac{y}{4}=1 ; \frac{x}{2}+\frac{y}{4}=2\)
(iv) x – y = 0; y + 3 = 0
(v) y = 2x + 1; y + 3x – 6 = 0
(vi) x = -3, y = 3
Solution:
(i) We can find x and y intericepts and thus of the two points on the lines (1), (2)
x + y = 7 ……… (1), x – y = 3 …………. (2)
To draw the graph of (1)
Put x = 0 in (1)
0 + y = 7 ⇒ y = 7
Thus A (0, 7) is a point on the line
Put y = 0 in (1)
x + 0 = 7 ⇒ x = 7
Thus B (7, 0) is another point on the line
Plot A and B. Join them to produce the line (1).
To draw the graph of (2), we can adopt the same procedure.

When x = 0,(2) ⇒ x – y = 3
0 – y = 3 ⇒ y = -3
P (0, -3) is a point on the line.
Put y = 0 in (2); x – 0 = 3
x = 3
∴ Q (3, 0) is another point on the line (2)
Plot P, Q
1 The point of intersection (5, 2) of lines (1), (2) is a solution

(ii) 3x + 2y = 4 ……. (1)
9x + 6y= 12 ………. (2)
To draw the graph of (1)
Put x = 0 in (1) ⇒ 3 (0) + 2y = 4
2y = 4
y =2
∴ A (0, 2) is a point on the line (1)
Put y = 0 in (1) ⇒ 3x + 2(0) = 4
3x = 4
x = \(\frac{4}{3}\) = 1.3
∴ B (1.3, 0) is another point on the line (1)
Plot the points A, B. Join them to produce the line (1)
To draw the graph of (2)
Put x = 0 in (2) ⇒ 9 (0) + 6y = 12
6y = 12
y = 2
∴ P (0, 2) is a point on the line (2)
Put y = 0 in (2) ⇒ 9x + 6(0) = 12
9x = 12
x = \(\frac{12}{9}=\frac{4}{3}\)
x = 1.3

Q (1.3, 0) is another point on the line (2)
Plot P, Q. Join them to produce the line (2).
The point of intersection of the lines (1), (2) is a solution.
[A, B],[P, Q] represent the same line.
∴ All the points on one line are also on the other.
This means we have an infinite number of solutions.

∴ Comparing (1), (2) we can conclude their slopes are equal
∴ The lines are parallel and will not meet at any point and hence no solution exists.
Let us draw the graphs of (1) & (2)
(1) ⇒ 2x + y = 4, Put x = 0 in (1) ⇒ y = 4
∴ A (0, 4) is a point on (1)
Put y = 0 in (1) ⇒ 2x = 4
x = 2
∴ B (2, 0) is another point on (1)
Plot A, B ; Join them to produce the line (1)
(2) ⇒ 2x + y = 8, Put x = 0 in (2)
2(0) + y = 8
y = 8, P (0,8) is a point
Put y = 0 in (2) ⇒ 2x + 0 = 8
2x = 8
x = 4, Q (4, 0) is another point on the line (2)

Plot P, Q: Join them to produce the line (2)

(iv) x – y = 0 ………….. (1) ⇒ -y = -x ⇒ y = x

Put x = 0 in (1), 0 – y = 0
-y = 0 ⇒ y = 0
A (0, 0) is a point on the line (1)
Put y = 0 in (1) ⇒ x – 0 = 0 ⇒ x = 0
B (-3, -3) is also the same point as A
(2) ⇒ y + 3 = 0
y = -3
∴ from (1) y = -3 = x
∴ B (-3, -3) is the solution

(v) y = 2x + 1 …………… (1)


The points of intersection (1, 3) of the lines (1) and (2) is a solution. The solution is the point that is common to both the lines.
∴ The solution is as x = 1, y = 3.

(vi) The point of intersection (-3, 3) is a solution.

x + y = 7 ……… (1), x – y = 3 …………. (2)
To draw the graph of (1)
Put x= 0 in (1)
0 + y = 7 ⇒ y =7
Thus A (0, 7) is a point on the line
Put y = 0 in (1)
x + 0 = 7 ⇒ x =7
Thus B (7, 0) is another point on the line
Plot A and B. Join them to produce the line (1).
To draw the graph of (2), we can adopt the same procedure.
When x = 0, …….. (2) ⇒ x – y = 3
0 – y = 3 ⇒ y = -3
P (0, -3) is a point on the line.
Put y = 0 in (2) ; x – 0 = 3
x = 3
∴ Q (3, 0) is another point on the line (2) Plot P, Q

9th Graph Exercise 3.10 Question 3.
Two cars are 100 miles apart. If they drive towards each other they will meet in 1 hour. If they drive in the same direction they will meet in 2 hours. Find their speed by using graphical method.
Solution:
Let x, y be the speed of the two cars. If the two cars travel towards each other they will meet in 1 hr. The distance between them d = 100; \(\frac{d}{s}\) = t
i.e., \(\frac{100}{x+y}\) = 1 ⇒ x + y = 100 ………. (1)
If the two cars travel in the same direction they will meet in 2 hrs.
x + y = \(\frac{100}{2}\) ⇒ x – y = 50 ………….. (2)

x + y = 10 ………….. (1)
Put x = 0 in (1), then 0 + y = 100 ⇒ y =100
A (0, 100) is a point on (1)
Put y = 0 in (1), then x + 0 = 100 ⇒ x =100
B (100, 0) is another point on (1)
Plot A & B. Join them to produce the line (1)
Similarly by x – y = 50
Put x = 0 in (2), then 0 – y = 50 ⇒ y = -50
P (0, -50) is a point on (2)
Put y = 0 in (2), then x – 0 = 50 ⇒ x = 50
Q (50, 0) is another point on (2)

Plot P & Q. Join them to produce the line (2)
The point of intersection (75, 25) of the two lines (1) & (2) is the solution.
∴ The solution i.e., the speed of the two cars x and y is given by x = 75 km and y = 25 km

Guys who are planning to learn and understand the topics of 10th Social Science History can grab this Tamilnadu State board solutions for Chapter 3 World War II Questions and Answers from this page for free of cost. Make sure you use them as reference material at the time of preparation & score good grades in the final exams.

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Tamilnadu Samacheer Kalvi 10th Social Science History Solutions Chapter 3 World War II

Do you feel scoring more marks in the 10th Social Science History Grammar sections and passage sections are so difficult? Then, you have the simplest way to understand the question from each concept & answer it in the examination. This can be only possible by reading the passages and topics involved in the 10th Social Science History Board solutions for Chapter 3 World War II Questions and Answers. All the Solutions are covered as per the latest syllabus guidelines. Check out the links available here and download 10th Social Science History Chapter 3 textbook solutions for Tamilnadu State Board.

World War II Textual Exercise

I. Choose the correct answer.

10th Social World Map Question 1.
When did the Japanese formally sign of their surrender?
(a) 2 September, 1945
(b) 2 October, 1945
(c) 15 August, 1945
(d) 12 October, 1945
Answer:
(a) 2 September, 1945

10th World Outline Map Question 2.
Who initiated the formation of League of Nations?
(a) Roosevelt
(b) Chamberlain
(c) Woodrow Wilson
(d) Baldwin
Answer:
(c) Woodrow Wilson

Samacheer Kalvi 10th Social Solutions Question 3.
Where was the Japanese Navy defeated by the US Navy?
(a) Battle of Guadalcanal
(b) Battle of Midway
(c) Battle of Leningrad
(d) Battle of El Alamein
Answer:
(b) Battle of Midway

10th World Map Question 4.
Where did the US drop its first atomic bomb?
(a) Kavashaki
(b) Innoshima
(c) Hiroshima
(d) Nagasaki
Answer:
(c) Hiroshima

History Chapter 3 Question 5.
Who were mainly persecuted by Hitler?
(a) Russians
(b) Arabs
(c) Turks
(d) Jews
Answer:
(d) Jews

Meaning In Tamil Question 6.
Which Prime Minister of England who signed the Munich Pact with Germany?
(a) Chamberlain
(b) Winston Churchill
(c) Lloyd George
(d) Stanley Baldwin
Answer:
(a) Chamberlain

Social Science World Map Question 7.
When was the Charter of the UN signed?
(a) June 26, 1942
(b) June 26, 1945
(c) January 1, 1942
(d) January 1, 1945
Answer:
(b) June 26, 1945

World Map For 10th Class Question 8.
Where is the headquarters of the International Court of Justice located?
(a) New York
(b) Chicago
(c) London
(d) The Hague
Answer:
(d) The Hague

II. Fill in the blanks.

1. Hitler attacked ……… which was a demilitarised zone.
2. The alliance between Italy, Germany and Japan is known as ……..
3. ……… started the Lend Lease programme.
4. Britain Prime Minister …….. resigned in 1940.
5. Saluting the bravery of the ………. Churchill said that “Never was so much owed by so many to so few”.
6. is a device used to find out the enemy aircraft from a distance.
7. ………. The Universal Declaration of Human Rights set forth fundamental human rights in articles.
8. After the World War II ……… was voted into power in Great Britain.
Answers:
1. Rhineland
2. Axis power
3. US – President Roosevelt
4. Chamberlain
5. British Royal airforce
6. Radar
7. 30
8. Labour party

III. Choose the correct statement.

World War Question 1.
(i) Banking was a major business activity among Jews.
(ii) Hitler persecuted the Jews.
(iii) In the concentration camps Jews were killed.
(iv) The United Nations has currently 129 member countries in it.
(a) (i) and (ii) are correct
(b) (i) and (ii) is correct
(c) (ii) and (iv) are correct
(d) (i) is correct and (ii), (iii) and (iv) are correct
Answer:
(a) (i) and (ii) are correct 2.

History Chapter 3 Class 10 Question 2.
Assertion: President Roosevelt realised that the United States had to change its policy of isolation.
Reason: He started a programme of Lend Lease in 1941.
(a) Both A and R are correct
(b) A is right but R is not the correct reason
(c) Both A and R are wrong
(d) R is right but it has no relevance to A
Answer:
(b) A is right but R is not the correct reason

IV. Match the following.

Answers:
1. (e)
2. (d)
3. (a)
4. (b)
5. (c)

V. Answer the questions briefly.

History Class 10 Chapter 3 Question 1.
Mention the important clauses of the Treaty of Versailles relating to Germany.
Answer:
Here are the important clauses of the Treaty of Versailles relating to Germany:

1. Germany was forced to give up territories to the west, north and east of the German border;
2. Germany had to disarm and was allowed to retain a very restricted armed force (army, navy and air force);
3. As reparations for the War, Germany was expected to pay for the military and civilian cost of the War to the Allied nations.

World War 2 In Tamil Question 2.
Who were the three prominent dictators of the post World War I ?
Answer:
Mussolini of Italy, Hitler from Germany and Franco from Spain were the three prominent dictators of post-world war I.

Question 3.
How did Hitler get support from the people of Germany?
Answer:
Hitler was well aware of the discontent among the Germans. He used his oratorical skills to sway the common people and promised them to return the glorious military past of Germany. He founded the National Socialist Party, generally known as “the Nazis”. The fundamental platform on which Hitler built his support was the notion of the racial superiority of the Germans as a pure. ‘Aryan’ race and a deep-seated hatred of the Jews.

Hitler came to power in 1933 and ruled Germany for twelve long years.

Question 4.
Describe the Pearl Harbour incident.
Answer:
America had installed strong naval base in Pearl harbour, Hawaii on December 1941, Japan attacked America’s pacific fleet without any warning. Japan’s idea was to cripple America so that they can freely move towards South-east Asian countries. But, when many battle ships and numerous fighter planes were destroyed, the U.S.A. declared war on Japan.

Question 5.
What do you know of Beveridge Report?
Answer:
In 1942, a report commonly known as the Beveridge Report was published in the United Kingdom. The report proposed a series of measures which the government should adopt to provide citizens with adequate income, health care, education, housing and employment to overcome poverty and disease which were the major impediments to general welfare.

Question 6.
Name the Bretton Woods Twins.
Answer:
The World Bank and the International Monetary Fund (IMF) are referred to as the Bretton Woods Twins. The International Bank for Reconstruction and Development (IBRD) and International Development Agency (IDA) together often referred as the World Bank.

Question 7.
What are the objectives of IMF?
Answer:
The objectives of the IMF are to:

1. foster global monetary cooperation,
2. secure financial stability,
3. facilitate international trade,
4. promote high employment and sustainable economic growth, and
5. reduce poverty around the world.

VI. Answer the questions given under each caption.

Question 1.
Battle of Stalingrad

(a) When did Germany attack Stalingrad?
Answer:
The Germans attack Stalingrad in August 1942.

(b) What were the main manufactures of Stalingrad?
Answer:
The main manufactures of Stalingrad were armaments and tractors.

(c) What was the name of the plan formulated by Hitler to attack Stalingrad?
Answer:
“Operation Blue” or “Fall Blau” was the name of the plan formulated by ‘ Hitler to attack Stalingrad.

(d) What is the significance of the Battle of Stalingrad?
Answer:
The people of Russia successfully defended the city of Stalingrad. It stopped the German advance into the Soviet Union and favoured the allies. It was a great patriotic war by the Russians.

Question 2.
Japanese Aggression in South-east Asia

(а) Name the South-east Asian countries which fell to the Japanese.
Answer:
Guam, the Philippines, Hong Kong, Singapore, Malaya, the Dutch East Indies (Indonesia) and Burma.

(b) Account for the setback of Allies in the Pacific region?
Answer:
The Allies faced many reverses in the Pacific region because of their inadequate preparations. The colonial rulers, especially the British, withdrew from their territories, leaving the local people to face the atrocities of the Japanese.

(c) What is the significance of Battle of Midway?
Answer:
The US navy defeated the Japanese navy in the Battle of Midway.

(d) What happened to the Indians living in Burma?
Answer:
Many Indians walked all the way from Burma to the Indian border, facing hardships. Many died of disease and exhaustion. Those who remained suffered under the Japanese.

Question 3.
General Assembly and Security Council

(a) List the permanent member countries of the Security Council.
Answer:
The United States, Britain, France, Russia and China are permanent member countries of the Security Council.

(b) What is the Holocaust?
Answer:
Holocaust refers to World’s worst ever genocide of killing nearly six million Jews by the Germans during the Second World War.

(c) Who was the Chairperson of the UN Commission on Human Rights?
Answer:
The widow of the American President Franklin Roosevelt was the chairperson of the UN commission on Human Rights.

(d) What is meant by veto?
Answer:
Veto is an constitutional right to reject a decision or proposal made by a law making body. Each of the permanent members of the UNO has the right of veto.

VII. Answer in detail.

Question 1.
Attempt an essay on the rise and fall of Adolf Hitler.
Answer:
Hitler was great orator. He swayed the people by his impassioned speeches, promising a return to the glorious military past of Germany. He founded the National Socialist Party, known as ‘the Nazis’. He came to power in 1933 and ruled Germany till 1945, with a small group of fanatic followers. He rearmed Germany. He made huge expenditure on the recruitment of armed forces and the manufacture of armaments and machinery for the army, navy and air force. Soon the economic condition of Germany got strengthened and the problem of unemployment came to an end. In 1938, Hitler invaded Austria and Czechoslovakia. Sudetenland in Czechoslovakia was German-speaking, and Hitler’s claim was that the German-speaking people should be united in one nation. Though Hitler gave an assurance in the Munich Pact that Germany would not attack any other country, but this was broken immediately.

In 1939, he invaded Czechoslovakia. Poland was attacked next, and this was the final act which resulted in declaration of war by Britain and France against Germany. In June 1940, Italy joined Germany, and in September 1940, Japan also joined the Axis Powers. The German army followed a tactic of ‘lightning strike’ to storm into various countries and overrun them. In June 1941, German army invaded Russia and remained successful in the initial years. But ultimately got defeated due to the resistance by Soviet army, and the fierce Russian winter. In the Battle of Alamein 1942, the Allied forces counter-attacked and defeated the German and Italian forces in North Africa. The German army was chased across the desert, out of North Africa. The war continued till Hitler’s suicide in April 1945.

Question 2.
Analyse the effects of World War II.
Answer:
The long term effects of II World War were mostly unpleasant but a little platform of cherishment to the colonies by way of extending them independence. The radiation from the atom bombs were not clearly estimated at the time of war.

New geo-political power alignment: The world got itself split into two new political super powers. One led by the United States with anti-communist ideas and the other one led by Soviet Russia with communist ideas.

Nuclear proliferation: The United States and the Soviet Union built large stock of nuclear powered weapons. This was just like a race between them. Britain and France also joined in the race. More and more resources were spent by the countries for destructive purposes.

International agencies: International agencies like the United Nations Organisation, the World Bank and the International Monetary Fund came into existence.

Independence of the colonies: Colonial powers were forced to give independence.

Empowerment of Women: Women power entered into the labour force in great numbers. There was a change in social relations and more and more women become economically independent.

Question 3.
Assess the structure and activities of the UN.
Answer:
The United Nations came into existence in the year 1945 to achieve lasting peace among all nations which were inter-dependent. It functions like any government, through its principal organs which are similar to the legislative, executive and judicial wings of a state.

1. The General Assembly is the body in which each member state is represented. It meets once a year and issues of interest and points of conflict are discussed in the Assembly.
2. The Security Council has fifteen members, five of them (the USA, Britain, France, Russia and China) are permanent members. The other ten temporary members are elected in rotation from different parts of the world. Each of the permanent members has the right to veto any decision by the other members of the Security Council.
3. The UN Secretariat is headed by the Secretary General, who is elected by the General Assembly on the recommendation of the Security Council.
4. The International Court of Justice is the Judicial wing of the United Nations. Its headquarter is at The Hague.
5. The fifth organ of the UN is the Economic and Social Council (ECOSOC). It is responsible for coordinating all the economic and social work of the United Nations.

Activities of the United Nations

Human rights, the problems of refugees, climate change, gender equality are all within the ambit of the activities of the United Nations. The UN Peace-keeping force has acted in many areas of conflict all over the world.

VIII. Students Activity

Question 1.
Group project involving students to prepare an album with pictures on different phases of the World War II.
Answer:
You can do this activity under the guidance of your teacher.

Question 2.
A debate in the class on the success or failure of the UN in preserving World Peace.
Answer:
You can do this activity by discussing the points in the class under the guidance of your teacher.

Question 3.
Marking the Allies and Axis countries, as well as important battlefields of World War II in a world map.
Answer:

IX. Map Work

Mark the following on the world map:
1. Axis Power Countries
2. Allied Power Countries
3. Hiroshima, Nagasaki, Hawai Island, Moscow, San Fransico

2.

World War II Additional Questions

I. Choose the correct answer:

Question 1.
This treaty contained the seeds of the second world war ………
(a) Treaty of London
(b) Treaty of Rome
(c) Treaty of Versailles
Answer:
(c) Treaty of Versailles

Question 2.
The Dictator of Spain was called:
(a) Mussolini
(b) Franco
(c) Hitler
(d) Edi-Amin
Answer:
(b) Franco

Question 3.
The country emerged as a World power after the First World war was ………
(a) China
(b) India
(c) Japan
Answer:
(c) Japan

Question 4.
Hitler came to power in ……………….. and ruled till 1945.
(a) 1932
(b) 1936
(c) 1934
(d) 1933
Answer:
(d) 1933

Question 5.
In September 1938, Hitler threatened a war on ………
(a) Czechoslovakia
(b) Poland
(c) Finland
Answer:
(a) Czechoslovakia

Question 6.
Italy invaded Ethiopia in ……………….. when they asked help from League of nations, got no help.
(a) 1931
(b) 1935
(c) 1937
(d) 1939
Answer:
(b) 1935

Question 7.
Blitzkrieg means a ………
(a) Lightning war
(b) Trench warfare
(c) Guerilla warfare
Answer:
(a) Lightning war

Question 8.
World War II was ……………….. war.
(a) Traditional
(b) Modem
(c) Scientific
(d) Alarming
Answer:
(b) Modem

Question 9.
Hitler signed the Non-Aggression pact with ……………
(a) Gorbachev
(b) Stalin
(c) Lenin
Answer:
(b) Stalin

Question 10.
In the year ……………….. Italy and Japan joined the axis powers.
(a) 1939
(b) 1937
(c) 1935
(d) 1940
Answer:
(d) 1940

II. Fill in the blanks:

1. The great powers of the world split up into two opposing groups namely the ……… and the ………
2. ……… was humiliated by the Treaty of Versailles.
3. The dictators ……… and ……… started Nazism and Fascism.
4. The Japanese attacked American fleet stationed at ……… on December 7, 1941.
5. America dropped atom bombs on the cities of ……… and ……… on August 6th and
9th of 1945.
6. The Second World War came to an end in ………
7. The German Air force was known as ………
8. In 1941, Prime Minister Winston Churchill and president F.D.Roosevelt concluded the ………
9. The Munich agreement was signed between Britain and ………
10. In 1945, at the end of the Second World War Hiroshima and Nagasaki were destroyed by ………
11. Hitler sent his airforce to drop bombs on ………
12. The UNO was established in ………
13. The UN charter was signed at ………
14. UNO’s main deliberative body is ………
15. The seat of International Court of Justice is at ………
Answers:
1. Allies and Axis
2. Germany
3. Hitler and Mussolini
4. Pearl Harbour
5. Hiroshima and Nagasaki
6. 1945
7. Luftwaffe
8. Atlantic Charter
9. Germany
10. America
11. Russia
12. 1945
13. San Francisco
14. The General Assembly
15. The Hague

III. Match the following:

Answers:
1.(c)
2. (e)
3. (a)
4. (b)
5. (d)

Answers:
1. (c)
2. (d)
3. (a)
4. (e)
5. (b)

Answers:
1. (d)
2. (e)
3. (b)
4. (c)
5. (a)

IV. Answer briefly:

Question 1.
How did Japan sowed the seeds for Second World War?
Answer:

1. Japan emerged as a world power after the First World War.
2. The industrial development and economic growth forced Japan to follow the policy of imperialism.
3. It signed Rome – Berlin – Tokyo Axis with Italy and Germany and sowed the seeds for the second world war.

Question 2.
List down the important organs of the UN.
Answer:

• FAO – Food and Agricultural organisation
• WHO – World Health Organisation
• UNESCO – United Nations Educational, Scientific and Cultural organisation
• UNICEF – United Nations Children’s Fund
• UNDP – United Nations Development programme

Question 3.
What was the immediate cause of the Second World War?
Answer:

1. In 1939, Hitler demanded from Poland and right to construct a military road connecting East Prussia and Germany through Polish corridor.
2. He also demanded the surrender of Danzig.
3. When Poland refused, Hitler made a lightning attack on Poland known as Blitzkrieg on 1st September 1939. It was the immediate cause of the second world war.
4. Britain and France declared war on Germany in support of Poland and Second World War started.

Question 4.
What are the principal organs of the UNO?
Answer:

1. The General Assembly
2. The Security Council
3. The Secretariat
4. The International Court of Justice
5. The Economic and Social Council.

Question 5.
Write a short note on Atlantic charter.
Answer:

1. In August 1941, the British prime minister Winston Churchill and U.S. president F.D.Roosevelt met on the board of the battleship Augusta.
2. They signed the Atlantic charter and agreed to launch a massive attack against the common enemy Germany.

Question 6.
What is meant by “Lend-Lease”?
Answer:
President Roosevelt of U.S.A. brought the Lend-Lease programme in 1941. Arms, food, military equipment, other supplies were sent to power in the name of loan which would be returned after use.

Question 7.
What were the main objectives of the UNO?
Answer:

1. To maintain international peace and security.
2. To develop friendly relations among nations.
3. To settle International disputes by peaceful means.
4. To be a centre for helping nations to achieve these goals.

Question 8.
What is the difference between a ‘Vote’ and a ‘Veto’?
Answer:
Veto: It is the right to block a decision or can be called as Negative vote by law.
Vote: It is the right law that is done infavour of a person/party. This can be called as positive vote.

Question 9.
Mention any two major achievements of the UNO.
Answer:

1. The UNO settled disputes between Israel and Palestine, Iran and Iraq, and withdrawal of Soviet troops from Afghanistan.
2. It has signed many Nuclear Test Ban Treaties like NTBT in 1963 and CTBT in 1996.

Question 10.
What is meant by the Holocaust?
Answer:
The word ‘Holocaust’ refers to the mass killing of nearly six million jews by the Germans during II World War. This was done because the Nazis completely want to exterminate jews completely from Germany and all other countries where they are.

Question 11.
Name the five permanent members of the security council.
Answer:
The USA, the UK, the France, the Russian Federation and the China.

Question 12.
What are the official languages of the UNO?
Answer:
The official languages of the UNO are Arabic, Chinese, English, French, Russian and Spanish. However its working languages are English and French.

Question 13.
How does the UNO get its financial aid?
Answer:
UNO gets its financial aid mainly from USA and from other member nations.

V. Answer all the questions given under each caption:

Question 1.
Aggressive Act of Hitler

(a) When did Hitler came to power?
Answer:
Hitler came to power in 1933.

(b) What did he do to bring back glorious Germany?
Answer:
He began to re-arm Germany. The recruitment of the armed forces, new armaments and machinery for the army, navy and air force.

(c) What was the name of the alliances he had with Japan and Italy?
Answer:
Rome-Berlin-Tokyo axis.

(d) What was his claim for Sudetenland?
Answer:
He claimed that the German-speaking people should be United into one nation and Sudetenland people were German-speaking people.

Question 2.
End of War

(a) Where did a big American and British force land?
Answer:
The American and British force landed in Normandy.

(b) With whom did they join?
Answer:
They joined with the secret underground French forces.

(c) What did Hitler do?
Answer:
He committed suicide.

(d) When did America drop atom bomb on Hiroshima and Nagasaki?
Answer:
America dropped atom bombs on the cities of Hiroshima and Nagasaki on August 6th and 9th of 1945.

Question 3.
Outbreak Of II World War.

(a) What are the axis powers?
Answer:
Germany, Italy and Japan.

(b) What is the significance of Blitzkrieg?
Answer:
Blitzkrieg means lightning strike which was the tactic followed by Germany to storm into various countries and over run them.

(c) Which country’s naval force was very powerful among the european naval forces?
Answer:
The British Royal Navy was very powerful among the European naval forces.

(d) How did British and French Saved their soldiers at the Battle of Dunkrik?
Answer:
British saved their men by calling all those who had boats/small ship and put them into use.The French formed the nucleus of their army.

VI. Answer the following in detail:

Question 1.
Write and explain any two International agencies of the UNO.
Answer:
The need for international co-operation and achieving peace among the world nations led to the formation of the international agencies of the UNO.

These agencies are:

1. The World Bank
2. The International Monetary Fund
3. The International Labour Office (ILO)
4. The International Finance Corporation.

The World Bank:
(i) The World Bank was established in 1945. It is located in Washington, USA. All the members of IMF are also the members of the World Bank.

(ii) The two main organs of the World Bank are the International Bank for Reconstruction and Development (IBRD) and the International Development Agency (IDA). Together they are called “World Bank”.

(iii) Earlier, the IBRD was funding only for the reconstruction activities of the European countries after World War II. Later, its work expanded to promote economic development for poor countries and development projects for developing countries.

(iv) The Bank also works on the key areas like poverty alleviation, and other basic issues of the Societies of the World. The International Development Agency (IDA): It lends money to the governments for developmental activities.

The International Development Agency (IDA): It lends money to the governments for developmental activities.

These loans are given at low rate of interest and for 50 long years. Therefore they are called as soft loans.

The International Finance Corporation (IFC): It mainly functions with private enterprises in developing countries.

The International Labour Office: Is located in Geneva, Switzerland. It works on issues related to Labour and employment.

Question 2.
What are the major achievements of the UNO?
Answer:

1. The UNO has rendered a great service in establishing peace, and security by solving various problems.
2. Political disputes are solved by security council, legal disputes by International court of Justice and others by special agencies.
3. UNO has solved many International disputes.
4. It preserves peace in the world through peaceful negotiations.
5. It settled disputes between Israel and Palestine, Iran and Iraq and withdrawal of Soviet troops from Afghanistan.
6. In the conference, all countries adopted “Agenda 21”, a blueprint to promote economic development and protect natural resources.
7. The UNO established the International Research and Training Institute for the Advancement of Women.
8. It has supported many programmes and projects to improve the quality of life for women in over 100 countries.
9. The UNO played a vital role in the Suez canal crisis of 1956. It made France, Britain, and Israel to withdraw their troops from Egypt.
10. The UNO also settled the Korean war and Vietnam war.

Question 3.
Describe the International Monetary Fund (IMF).
Answer:

1. The International Monetary Fund was primarily the brainchild of Harry Dexter White and John Maynard Keynes, the famous economist.
2. It was formally organized in 1945 with 29 member countries.
3. Now it has a membership of 189 countries.
4. Its primary objective is to ensure financial stability and development across the world,
5. The main agenda is to promote international monetary co-operation, expansion of international trade and exchange stability.
6. The Fund lends money from its resources to countries facing balance of payments problems.
7. It imposes stringent conditions on the borrowing nations to tighten their budgets, practice fiscal prudence and reduce their expenditure.
8.  This is often unpopular, especially among the developing countries which may have to cut down on various programmes which provide subsidies to the people.

We think the data given here clarify all your queries of Chapter 3 and make you feel confident to attempt all questions in the examination. So, practice more & more from Tamilnadu State Board solutions for 10th Social Science History Chapter 3 World War II Questions and Answers & score well. Need any information regarding this then ask us through comments & we’ll give the best possible answers very soon.

Students can Download Maths Chapter 1 Number System Ex 1.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 1 Number System Ex 1.1

7th Standard Maths Number System Exercise 1.1 Question 1.
Fill in the blanks:
(i) (-30) + ____ = 60
(ii) (-5) + ___ = -100
(iii) (-52) + (-52) = ____
(iv) ____ + (-22) = 0
(v) ____ + (-70) = 70
(vi) 20 + 80 + ___ = 0
(vii) 75 + (-25) = ____
(viii) 171 + ___ = 0
(ix) [(-3) + (-12)] + (-77) = _____ + [(12) + (-77)]
(x) (-42) + [____ + (-23)] = [____ + 15] + ____
Solution:
(i) 90
(ii) -95
(iii) -104
(iv) 22
(v) 140
(vi) -100
(vii) 50
(viii) -171
(ix) -3
(x) +15; -42; -23

7th Maths Guide Exercise 1.1 Question 2.
Say True or False.
(i) The additive inverse of (-32) is -32
(ii) (-90) + (-30) = 60
(iii) (-125) + 25 = -100
Solution:
(i) False
(ii) False
(iii) True

7th Maths Exercise 1.1 Samacheer Kalvi Question 3.
Add the following.
(i) 8 and-12 using number line.
Solution:
Starting at zero on the number line facing positive direction and move 8 steps forward reaching 8.

Then we move 12 steps
backward to represent -12 –
and reach at -4.
∴ 8 + (-12) = -4

(ii) (-3) and (-5) using number line.
Solution:
Starting at zero on the number line facing positive direction and move 3 steps backward reaching-3.

Then we move 5 steps backward to represent -5 and reach -8.
∴ (-3) + (-5) = -8

(iii) (-100) + (-10)
Solution:
(-100) + (-10) = -100 – 10 = -110

(iv) 20 + (-72)
Solution:
20 + (-72) = 20 – 72 = -52

(v) 82 + (-75)
Solution:
82 + (-75) = 82 – 75 = 7

(vi) -48 + (-15)
Solution:
-48 + (-15) = -48 – 15 = -63

(vii) -225 + (-63)
Solution:
-225 + (-63) = -225 – 63 = -288

Samacheer Kalvi 7th Maths Book Answers Question 4.
Thenmalar appeared for competitive exam which has negative scoring of 1 mark for each incorrect answers. In paper I she answered 25 question incorrectly and in paper II13 questions incorrectly. Find the total reduction of marks.
Solution:
For each incorrect question the score = -1
In paper I, score for 25 incorrect questions – 25 × (-1) = -25
In paper II, for 13 incorrect question the score = 13 × (-1) = -13
The total marks get reduced = (-25) + (-13) = -38
-38 marks will be reduced.

Samacheer Kalvi 7th Maths Book Solutions Question 5.
In a quiz competition, Team A scored +30, -20, 0 and team B scored -20, 0,+30 in three successive rounds. Which team will win? Can we say that we can add integers in any order?
Solution:
Total score of team A = [(+30) + (-20)] + 0 = (+10) + 0 = 10
Total score of team B = [(-20) + 0] + (+30)
= -20 + 30 = +10
Score of team A = Score of team B.
Yes, we say that we can add integers in any order.

7th Standard Maths Exercise 1.1 Question 6.
Are (11 + 7) +10 and 11 + (7 + 10) equal? Mention the property.
Solution:
First we take (11 + 7) + 10 = 18 + 10 = 28
Now 11 + (7 + 10) = 11 + 17 = 28
In both the cases the sum is 28. ∴ (11 + 7) + 10 = 11 + (7 + 10)
This property is known as associative property of integers under addition.

7th Maths Exercise 1.1 Question 7.
Find 5 pairs of integers that added to 2.
Solution:
0 + 2 = 2
1 + 1 = 2
-1+3 = 2
-2 + 4 = 2
-3 + 5 =2 (and many more.)

Objective Type Questions

Samacheer Kalvi Guru 7th Maths Question 8.
The temperature at 12 noon at a certain place was 18° above zero. If it decreases at the rate of 3° per hour at what time if would be 12° below zero?
(i) 12 mid night
(ii) 12 noon
(iii) 10 am
(iv) 10 pm
Solution:
(iv) 10 pm
: ’Temperature at 12 noon = 18° above zero = +18°
Rate of decrease per hour = -3°
Temperature 12° below zero = -12°
-12 is 30 units to the left of+18°
Time at which it reach -12° = \(\frac{30}{3}\) = 10 h
10 hrs after 12 noon = 10 pm

Samacheer Kalvi 7th Maths Question 9.
Identify the problem with negative numbers as its answer.
(i) -9 +(-5) + 6
(ii) 8 + (-12) – 6
(iii) -4 + 2 + 10
(iv) 10 + (-4) + 8
Solution:
(i) -9 +(-5) + 6
(i) -9 + (-5) + 6 = -14 + 6 = -8
(ii) 8 + (-12) + 6 = -4 + 6 = – 2
(iii) -4 + 2 + 10 = -2 + 10 = 8
(iv) 10 + (-4) + 8 = 6 + 8 = 14

Samacheer Kalvi 7th Maths Book Question 10.
(-10) + (+7) = ____
(i) +3
(ii) -3
(iii) -17
(iv) +17
Solution:
(ii) -3

Samacheer Kalvi Guru 7th Standard Maths Question 11.
(-8) + 10 + (-2) = ____
(i) 2
(ii) 8
(iii) 0
(iv) 20
Solution:
(iii) 0

Maths 7th Samacheer Kalvi Question 12.
20 + (-9) + 9 = ____
(i) 20
(ii) 29
(iii) 11
(iv) 38
Solution:
(i) 20

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All these concepts of Chapter 18 Heredity are explained very conceptually by the subject teachers in Tamilnadu State Board Solutions PDF as per the prescribed Syllabus & guidelines. You can download Samacheer Kalvi 10th Science Book Solutions Chapter 18 Heredity State Board Pdf for free from the available links. Go ahead and get Tamilnadu State Board Class 10th Science Solutions of Chapter 1 Heredity.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity

Kickstart your preparation by using this Tamilnadu State Board Solutions for Class 18th Science Chapter 18 Heredity Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 18 easily after studying the Tamilnadu State Board Class 18th Science Textbook solutions pdf. Download the Tamilnadu State Board Science Chapter 18 Heredity solutions of Class 18th by accessing the links provided here and ace up your preparation.

Samacheer Kalvi 10th Science Heredity Textual Evaluation Solved

I. Choose the Correct Answer.

Question 1.
According to Mendel, alleles have the following character ______.
(a) Pair of genes
(b) Responsible for character
(c) Production of gametes
(d) Recessive factors.
Answer:
(a) Pair of genes

Question 2.
9 : 3 : 3 : 1 ratio is due to:
(a) Segregation
(b) Crossing over
(c) Independent assortment
(d) Recessiveness
Answer:
(c) Independent assortment

Question 3.
The region of the chromosome where the spindle fibres get attached during cell division ______.
(a) Chromomere
(b) Centrosome
(c) Centromere
(d) Chromonema.
Answer:
(c) Centromere

Question 4.
The centromere is found at the centre of the ……… chromosome.
(a) Telocentric
(b) Metacentric
(c) Sub-metacentric
(d) Acrocentric
Answer:
(b) Metacentric

Question 5.
The ______ units form the backbone of the DNA.
(a) 5 carbon sugar
(b) Phosphate
(c) Nitrogenous bases
(d) Sugar phosphate.
Answer:
(c) Nitrogenous bases

Question 6.
Okasaki fragments are joined together by:
(a) Helicase
(b) DNA polymerase
(c) RNA primer
(d) DNA ligase
Answer:
(d) DNA ligase

Question 7.
The number of chromosomes found in human beings are ______.
(a) 22 pairs of autosomes and 1 pair of allosomes.
(b) 22 autosomes and 1 allosome.
(c) 46 autosomes.
(d) 46 pairs of autosomes and 1 pair of allosomes.
Answer:
(a) 22 pairs of autosomes and 1 pair of allosomes.

Question 8.
The loss of one or more chromosome in a ploidy is called:
(a) Tetraploidy
(b) Aneuploidy
(c) Euploidy
(d) polyploidy
Answer:
(b) Aneuploidy

II. Fill in the blanks.

Question 1.
The pairs of contrasting character (traits) of Mendel are called ______.
Answer:
Alleles or allelomorphs.

Question 2.
The physical expression of a gene is called ______.
Answer:
Phenotype.

Question 3.
The thin thread-like structures found in the nucleus of each cell are called ______.
Answer:
Chromosomes.

Question 4.
DNA consists of two _______ chains.
Answer:
Polynucleotide.

Question 5.
An inheritable change in the amount or the structure of a gene or a chromosome is called ______.
Answer:
Mutation.

III. Identify whether the statement is True or False. Correct the False Statement.

Question 1.
A typical Mendelian dihybrid ratio of F₂ generation is 3 : 1.
Answer:
False.
Correct statement: A typical Mendelian dihybrid ratio of F₂ generation is 9 : 3 : 3 : 1.

Question 2.
A recessive factor is altered by the presence of a dominant factor.
Answer:
True.

Question 3.
Each gamete has only one allele of a gene.
Answer:
True.

Question 4.
hybrid is an offspring from a cross between genetically different parent.
Answer:
True.

Question 5.
Some of the chromosomes have an elongated knob-like appendage known as a telomere.
Answer:
False.
Correct statement: Some of the chromosomes have an elongated knob – like appendage known as the satellite.

Question 6.
New nucleotides are added and a new complementary strand of DNA is formed with the help of enzyme DNA polymerase.
Answer:
True.

Question 7.
Down’s syndrome is a genetic condition with 45 chromosomes.
Answer:
False.
Correct statement: Down’s syndrome is the genetic condition with 21 chromosomes.

IV. Match the following.

Question 1.

Answer:
1. (c) 22 pair of chromosome
2. (d) 2n
3. (e) 23rd pair of chromosome
4. (a) Trisomy 21
5. (b) 9 : 3 : 3 : 1.

V. Answer in a Sentence.

Question 1.
What is a cross in which inheritance of two pairs of contrasting characters is studied?
Answer:
Dihybrid cross.

Question 2.
Name the conditions when both the alleles are identical.
Answer:
Homozygous

Question 3.
A garden pea plant produces axial white flowers. Another of the same species produced terminal violet flowers. Identify the dominant trait?
Answer:

• Axial position – a dominant trait
• White flowers – a recessive trait
• Terminal position – a recessive trait
• Violet flower – dominant trait.

Question 4.
What is the name given to the segments of DNA, which are responsible for the inheritance of a particular character?
Answer:
Genes

Question 5.
Name the bond which binds the nucleotides in a DNA.
Answer:
The hydrogen bonds bind the nucleotides in a DNA.

VI. Short Answer Questions.

Question 1.
Why did Mendel select pea plant for his experiments?
Answer:

• The pea plant is self-pollinating and so it is very easy to raise pure breeding individuals.
• It has a short life span, as it is an annual.
• It is easy to cross-pollinate.
• It has deeply defined contrasting characters.
• The flowers are bisexual.

Question 2.
What do you understand by the term phenotype and genotype?
Answer:
Phenotype is the outward appearance or morphological character of an organism. The expression of gene or the genetic make up of an individual for a particular trait is called genotype.

Question 3.
What are allosomes?
Answer:
Out of 23 pairs of chromosomes, 22 pairs are autosomes and the 23rd pair is the allosome or sex chromosome.

Question 4.
What are Okazaki fragments?
Answer:
For the synthesis of new DNA, two things are required one is RNA primer and the enzyme primase. The DNA polymerase moves along the newly formed RNA primer nucleotides, which leads to the elongation of DNA. In the other strand, DNA is synthesis in small fragments called okazaki fragments. These fragments are linked by the enzyme called ligase.

Question 5.
Why is euploidy considered to be advantageous to both plants and animals?
Answer:
In euploidy condition, the individual bears more than the usual number of diploid chromosomes. The triploid plants and animals are sterile. The tetraploid plants, often result in increased fruit and flower size.

Question 6.
A pure tall plant (TT) is crossed with pure dwarf plant (tt), what would be the F₁ and F₂ generations? Explain.
Answer:
When a pure tall plant (TT) is crossed with pure dwarf plant (tt), In F₁ generation all the plants will be tall. In F₂ generation by selling the F₁ monohybrid the tall and dwarf plant will be in the ratio of 3 : 1. In F₂ generation phenotypic ratio will be 3 : 1. genotypic ratio will be 1 : 2 : 1

Question 7.
Explain the structure of a chromosome.
Answer:
The chromosomes are thin long and thread-like structures, consisting of two identical strands called sister chromatids. They are held together by a centromere. Each chromatid is made up of a spirally coiled thin structure called chromonema. The bead-like structures along the length are called chromomeres. The chromosomes are made up of DNA, RNA, chromosomal proteins (histones and non-histones) and certain metallic ions. These proteins give structural support to the chromosome.

Question 8.
Label the parts of the DNA in the diagram given below. Explain the structure briefly. Structure of DNA.
Answer:
DNA is the hereditary material as it contains the genetic information. DNA is a large molecule, consisting of millions of nucleotides. Each nucleotide consists of three compounds.

(a) A sugar molecule – Deoxy Ribose sugar

(b) A nitrogenous base [Purines and Pyrimidines]

• Purines (Adenine and Guanine)
• Pyrimidines (Cytosine and Thymine)

(c) A phosphate group Nucleoside and Nucleotide:
Nucleoside = Nitrogen base + sugar
Nucleotide = Nucleoside + Phosphate
The nucleotides are formed according to the purines and pyrimidines present in them.

VII. Long Answer Questions.

Question 1.
Explain with an example of the inheritance of the dihybrid cross. How is it different from a monohybrid cross?
Answer:
The dihybrid cross involves the inheritance of two pairs of contrast characteristics, round – yellow seeds and wrinkled – green seeds. When pea plants having round – yellow seeds cross – bred with pea plants having wrinkled – green seeds, in the first generation (F₁), only round yellow seeds were produced.
No wrinkled – green seeds were obtained. Round yellow colour seeds were dominant and wrinkled-green seeds were recessive.

When round – yellow seeds were cross-bred by self-pollination, four types of seeds having different combinations of shape and colour were obtained in the F₂ generation. They were round- yellow, round-green, wrinkled-yellow and wrinkled – green seeds.

A dihybrid cross produced four types of F₂ offsprings in the ratio of 9 with two dominant traits, 3 with one dominant trait and one recessive trait, 3 with another dominant trait and another recessive trait and one with two recessive traits. The new combinations of traits with round green and wrinkled yellow had appeared in the dihybrid cross (F₂ generation). The ratio of each phenotype of seeds in the F₂ generation is 9 : 3 : 3 : 1. This is known as the Dihybrid ratio.

Difference between a monohybrid cross and dihybrid cross:
Monohybrid cross:
Monohybrid cross is a genetic cross, that involves a single pair of genes, which is responsible for one trait.
Parents differ by a single trait.
Monohybrid ratio in F₂ generation is 3 : 1.

Dihybrid cross:
Dihybrid cross is a genetic cross, that involves two pairs of genes, which are responsible for two traits,
The parents have two different independent traits.
The dihybrid ratio in the F₂ generation is 9 : 3 : 3 : 1.

Question 2.
How is the structure of DNA organised? What is the biological significance of DNA?
Answer:
DNA is the hereditary material, as it contains the genetic information. It is a large molecule consisting of millions of nucleotides, so it is called a polynucleotide. Each nucleotide consists of three components.
(a) A sugar molecule – Deoxyribose sugar

(b) A nitrogenous base – There are two types of the nitrogenous base in DNA they are

• Purines (Adenine and Guanine)
• Pyrimidines (Cytosine and Thymine)

(c) A phosphate group – The polynucleotide chains from a double helix. Nitrogenous bases in the centre are linked to sugar – phosphate units, which form the backbone of the DNA. Pairing between the nitrogenous bases is very specific and is always between purine and pyrimidine, linked by hydrogen bonds.

Adenine (A) links Thymine (T) with two hydrogen bonds [A=T]. Cytosine (C) links Guanine (G) with three hydrogen bonds (C = G). The hydrogen bonds between the nitrogenous bases make the DNA molecule stable. The nucleotides in a helix are joined together by phosphodiester bonds.

The biological significance of DNA:

• It is responsible for the transmission of heredity information from one generation to the next generation.
• It contains the information required for the formation of proteins.
• It controls the developmental process and life activities of an organism.

Question 3.
The sex of the newborn child is a matter of chance and neither of the parents may be considered responsible for it. What would be the possible fusion of gametes to determine the sex of the child?
Answer:
Out of 23 pairs of chromosomes, 22 pairs are autosomes and one pair (23rd pair) is the sex chromosome. The female gametes or the eggs formed are similar in their chromosome type [22 + XX], So human females are homogametic. The male gametes or sperms produced are of two types. They are produced in equal proportions. The sperm bearing [22 + X] chromosomes and the sperm bearing (22 + Y) chromosomes. So human males are called heterogametic.

It is a chance, as to which category of sperm fuses with the egg. If the egg [X] is fused by the X – bearing sperm an [XX] individual (female) is produced. If the egg [X] is fused by the Y – bearing sperm an [XY] individual (male) is produced. The sperm produced by the father determines the sex of the child. The mother is not responsible for determining the sex of the child.

VIII. Higher Order Thinking Skills (HOTS) Questions

Question 1.
Flowers of the garden pea are bisexual and self-pollinated. Therefore, it is difficult to perform hybridization experiment by crossing a particular pistil with the specific pollen grains. How Mendel made it possible in his monohybrid and dihybrid crosses?
Answer:
As the garden pea is self-pollinating plant the parent plant were emasculated to prevent self pollination.
The anthers were collected from male parent and dusted on the female parent and the stigma was bagged.

Question 2.
Pure-bred tall pea plants are first crossed with pure-bred dwarf pea plants. The pea – plants obtained in the F₁ generation are then cross-bred to produce F₂ generation of pea plants.

1. What do the plants of the F₁ generation look like?
2. What is the ratio of tall plants to dwarf plants in the F₂ generation?
3. Which type of plants was missing in F₁, generation but reappeared in the F₂ generation?

Answer:

1. Tall
2. 1 : 2 : 1
3. Dwarf plants.

Question 3.
Kavitha gave birth to a female baby. Her family members say that she can give birth to only female babies because of her family history. Is the statement given by her family members true. Justify your answer.
Answer:
No, the statement is not true.
Sex determination is a chance of probability as to which category of sperm fuses with the eggs. If the egg(x) is fused by the x-bearing sperm, then
individual is female. If the egg (x) is fused by the y-bearing sperm then the individual is male. The sperm produced by the father only determines the sex of the child.

IX. Value-Based Questions

Question 1.
Under which conditions does the law of independent assortment hold good and why?
Answer:
Mendel gave this law based on his dihybrid cross experiment. Here the total number of individuals is F₂ will be sixteen which occur in a ratio of 9 : 3 : 3 : 1 where two parental classes and two new combination will be produced.

Samacheer Kalvi 10th Science Heredity Additional Questions Solved

I. Fill in the blanks.

Question 1.
The branch of biology that deals with the genes, genetic variation and heredity of living organisms is called ______.
Answer:
Genetics.

Question 2.
The chromatids are held together by ______.
Answer:
Centromere.

Question 3.
______ is the enzyme, which separates the double helix of DNA, above the replication fork.
Answer:
Topoisomerase.

Question 4.
_____ mutation is the changes occurring in the nucleotide sequence of a gene.
Answer:
Gene or point.

Question 5.
The end of the chromosome is called ______.
Answer:
Telomere.

Question 6.
The constriction of the chromosome at any point is called ______.
Answer:
Nuclear Zone.

Question 7.
The nucleotides in a helix are joined together by ______ bonds.
Answer:
Phosphodiester.

II. Match the following:

Question 1.

Answer:

1. (c) occurs in pairs and alike
2. (d) segments of DNA
3. (a) polynucleotide
4. (f) occurs in pairs and unlike
5. (b) Human male
6. (g) somatic characters
7. (e) bind DNA to the origin of replication site.

III. Choose the correct answer.

Question 1.
The thin long thread-like structures consisting of two identical strands ______.
(a) Hybrid
(b) Chromosomes
(c) Genes
(d) DNA and RNA.
Answer:
(b) Chromosomes

Question 2.
A cross between a tall plant (TT) and short pea plant (tt) resulted in progeny that were all tall plants because:
(a) Tallness is the dominant trait
(b) Shortness in the dominant trait
(c) Tallness in the recessive trait
(d) Height of pea plant is not governed by gene ‘T’ or ‘t’
Answer:
(a) Tallness is the dominant trait

Question 3.
The number, size and shape of chromosomes in the cell nucleus of an organism ______
(a) Karyotype
(b) Heterozygous
(c) Autosome
(d) Nucleotide
Answer:
(a) Karyotype

Question 4.
In human males all the chromosomes are paired perfectly except one. The unpaired chromosomes are:
(a) large chromosome
(b) small chromosome
(c) Y- chromosome
(d) X- chromosome
Answer:
(c) Y- chromosome

Question 5.
The addition or deletion in the number of chromosomes present in a cell is called _______
(a) Homogametic
(b) Polymerase
(c) Ploidy
(d) Nucleoside
Answer:
(c) Ploidy

IV. Write true or false for the statements. Correct the false statement.

Question 1.
The character which expresses itself is called the dominant condition and that which is masked is called recessive condition. Answer:
True

Question 2.
The chromosomes of body cells of an organism are haploid in condition and the single set of chromosomes in gametes are diploid in condition.
Answer:
False
Correct statement: The chromosomes of body cells of an organism are diploid in condition and the single set of chromosomes in gametes are haploid in condition.

Question 3.
The mother is responsible for determining the sex of the child. The sperm produced by the father does not determine the sex of the child.
Answer:
False
Correct statement: The mother is not responsible for determining the sex of the child. The sperm produced by the father determines the sex of the child.

Question 4.
The human female is called Homogametic. The human males are called heterogametic.
Answer:
True

Question 5.
Adenine always links with Guanine with three hydrogen bonds and cytosine always links with thymine with two hydrogen bonds.
Answer:
False
Correct statement: Adenine always links with thymine with two hydrogen bonds. Cytosine always links with Guanine with three hydrogen bonds.

V. Match the following persons with the correct statements.

Question 1.

Answer:

1. (d) chromosomes
2. (f) Model of DNA
3. (e) Mutation
4. (c) Down’s syndrome
5. (a) Role of chromosome in heredity
6. (b) Father of genetics

VI. Answer the following in a word or with a sentence.

Question 1.
Who is the father of genetics?
Answer:
Gregor Johann Mendel.

Question 2.
Name the plant on which Mendel performed his experiment.
Answer:
Pisum sativum (Garden Pea).

Question 3.
What is locus?
Answer:
Each gene is present at a specific position on a chromosome called its locus.

Question 4.
Who proposed the double helical model of DNA?
Answer:
Watson and Crick

Question 5.
What does the dihybrid cross involve?
Answer:
The dihybrid cross involves the inheritance of two pairs of contrasting characteristics.

Question 6.
What are the components of chromosome?
Answer:
The components of chromosomes are DNA, RNA, chromosomal proteins like histones and non histones and certain metallic ions.

Question 7.
What are Nucleoside and Nucleotide?
Answer:
Nucleoside = Nitrogen base + sugar
Nucleotide = Nucleoside + phosphate

VII. Answer the following briefly.

Question 1.
Write the Law of purity of gametes.
Answer:
When a pair of contrasting alleles are brought together in a heterozygote, the two members of the allelic pair remain together without mixing and when gametes are formed, the two separate out, so that only one enters each gamete.

Question 2.
What are chromonema and chromomeres?
Answer:
Each chromatid is made up of a spirally coiled thin structure called chromonema. The chromonema has a number of bead – like structures along its length which are called chromomeres.

Question 3.
Explain the types of chromosome-based on the position of the centromere.
Answer:

Based on the position of the centromere, the chromosomes are classified as follows:

• Telocentric: The centromere is at the proximal end. The chromosomes are rod – shaped.
• Acrocentric: The centromere is found at the end with a short arm and a long arm. They are rod-shaped chromosomes.
• Sub metacentric: The centromere is found near the centre of the – chromosome. Thus forming two unequal arms. They are J shaped or L shaped chromosomes.
• Metacentric: The centromere occurs in the centre of the chromosome and forms two equal arms. They are V – shaped chromosomes.

Question 4.
Give reason for the appearance of new combination of characters in F₂ progene.
Answer:
At the time of gamete formation in the F₁ Hybrid genes character asserted independently resulted in the appearance of new combination of character in the F₂ progene.

Question 5.
What is sickle cell anaemia?
Answer:
Sickle cell anaemia is caused by the mutation of a single gene. This alteration in gene brings a change in the structure of protein part of Haemoglobin molecule. The red blood cells, that carry the haemoglobin is sickle-shaped and oxygen-carrying capacity reduces, causing sickle cell anaemia.

Question 6.
Represent the phenotypic, genotypic ratio of both monohybrid cross and dihybrid cross in 72 generations of pea plants.
Answer:
Monohybrid cross:
Phenotypic ratio – 3 : 1
Genotypic ratio – 1 : 2 : 1

Dihybrid cross:
Phenotypic ratio – 9 : 3 : 3 : 1
Genotypic ratio – 1 : 2 : 2 : 1 : 4 : 1 : 2 : 2 : 1.

Question 7.
How do chromosomes take part in the formation of the male and female child?
Answer:
Fertilization of the egg (22 + X) with a sperm [22 + X] will produce a female child [44 + XX] The fertilization of the egg (22 + X) with a sperm [22 + Y] will give rise to a male child [44 + XY].

VIII. Answer the following in detail.

Question 1.
Explain the regions of chromosomes with a neat labelled diagram.
Answer:
The chromosomes are thin, long and thread-like structures, with two identical strands called chromatids. They are held together by a centromere. The chromatid is made up of spirally coiled, a thin structure called chromonema, which has a number of bead-like structures along its length called chromomeres.

A chromosome consists of the following regions:

1. Primary constriction: The two arms of a chromosome meet at a point called primary constriction or centromere. The centromere is the region, where spindle fibres attach to the chromosome during cell division.
2. Secondary constriction: Some chromosomes have a secondary constriction at any point of the chromosome, called the nuclear zone or nucleolar organizer. (Formation of the nucleolus in the nucleus).
3. Telomere: The end of the chromosome is called telomere. Each extremity of the chromosome has a polarity and prevents it from joining the adjacent chromosome. It maintains and provides stability to the chromosomes.
4. Satellite: Some of the chromosomes have an elongated knob-like appendage at one end of the chromosome, known as a satellite. The chromosomes with satellites are called as sat- chromosomes.
   

Question 2.
Given an account of the Laws of Mendel.
Answer:
Mendel proposed three important laws which are now called as Mendel’s Laws of Heredity.
(i) Law of Dominance : “When two homozygous individuals with one or more sets of contrasting characters are crossed, the characters that appear in the F₁ hybrid are dominant and those that do not appear in F₁ are recessive characters”.

(ii) Law of Segregation or Law of purity of gametes : “When a pair of contrasting factors or genes or allelomorphs are brought together in a heterozygote or hybrid, the two members of the allelic pair remain together without mixing and when gametes are formed, the two separate out, so that only one enters each gamete.”

(iii) Law of independent assortment : “In case of inheritance of two or more pairs of characters simultaneously, the factors or genes of one pair assort out independently of the other pair.”

Question 3.
Explain the monohybrid cross, with a diagram and describe the interpretation of Mendel oh monohybrid cross.
Answer:
Crosses involving the inheritance of only one pair of contrasting characters are called monohybrid cross. It is a cross between two forms of a single trait like a cross between tall and dwarf plant. In a monohybrid cross, a pure breeding tall plant and a pure breeding dwarf plant, results, tall and monohybrids in the F₁ generation.

In the F₂ generation, selfing of the F₁ monohybrids resulted in tall and dwarf plants in the ratio of 3 : 1. The external expression of a particular trait is known as the phenotype. So the phenotypic ratio is 3 : 1.
In the F₂ generations, 3 different types were obtained.
Tall Homozygous – TT (pure) – 1
Tall Heterozygous – Tt – 2
Dwarf Homozygous – tt – 1
So the genotypic ratio 1 : 2 : 1.
A genotype is the genetic expression of an organism.

Mendel’s interpretation on monohybrid cross:
Factors are passed on from one generation to another, factors are referred to as genes. Tallness and dwarfism are determined by a pair of contrasting factors tall plant T, a dominant character and a plant is dwarf’s’, recessive character in pairs. Pure breeding tall plants (TT) and pure dwarf plants (tt) are called homozygous. If they are unlike, (Tt) they are referred to as heterozygous.

(i) Two factors make up a pair of contrasting characters are called alleles or allelomorphs. One member of each pair is contributed by one parent.

(ii) When two factors of a trait are brought together, by fertilization, only one expresses itself (tallness), masking the expression of the other (dwarfness). The character which expresses itself is called dominant and the character, which is masked is called recessive.

(iii) The factors for tallness [T] and dwarfness (t) are separate entities, and in a gamete either T or t is present. When F₁ hybrids are self crossed the two entities separate and then unite independently, forming tall and dwarf plants.

Question 4.
Explain Mendel’s laws of heredity with the results of a dihybrid cross.
Answer:
Mendel crossed pea plants having round yellow seeds (dominant) with pea plants having wrinkled green seeds. In the F₁ generation, round shape, yellow colour of the seeds were dominant over the wrinkled green colour seeds. When the hybrids of the F₁ generation were cross-bred by self-pollination, the dihybrid cross, produced four types of F₂ offsprings in the ratio of 9, with two dominant traits, 3 with one dominant trait and one recessive trait, 3 with another dominant trait and another recessive trait and 1 with two recessive traits.

Two new combinations of traits with round green and wrinkled yellow had appeared in the dihybrid cross. [F₂ generation]
Mendel’s law of heredity:
(i) Law of dominance: When two homozygous individuals with contrasting characters are crossed, the characters, that appear in the F₁ hybrid are dominant and those do not appear in F₁ are recessive characters.

(ii) Law of segregation or Law of purity of gametes: When a pair of contrasting factors or genes are brought together, in a heterozygote the contrasting pair remain together without, mixing and when gametes are formed, the two separate out so that only one enters each gamete.

(iii) Law of Independent Assortment: In case of inheritance of two or more pairs of characters simultaneously, the factors or genes of one pair assort out independently of the other pair.

Question 5.
Explain the Watson and Crick model of DNA.
Answer:
Watson and Crick model of DNA:
DNA molecules consist of two polynucleotide chains. These chains form a double helix. Structure, with two strands, which run anti-parallel to one another. Nitrogenous bases in the centre are linked to sugar – phosphate units, which form the backbone of the DNA. Pairing between the nitrogenous bases is very specific and is always between purine and pyrimidine linked by hydrogen bonds.

Adenine [A] links Thymine [T] with two hydrogen bonds (A = T)
Cytosine [C] links Guanine [G] with three hydrogen bonds (C = G). This is called complementary base pairing.
Hydrogen bonds between the nitrogenous bases make the DNA molecule stable.
Each turn of the double helix is 34 A° (3.4 nm). There are ten base pairs in a complete turn.
Nucleotides in a helix are joined together by phosphodiester bonds.

Question 6.
Explain in detail, the various steps of DNA replication.
Answer:
Replication of DNA:
Replication of DNA occurs within a cell. DNA molecule produces exact copies of its own structure during replication.
The two strands of a DNA molecule have complementary base pairs, the nucleotides of each strand, provide the information needed to produce its new strand. The two resulting daughter cells contain exactly the same genetic information as the parent cell.

DNA replication involves the following steps:
(a) Origin of replication: The points on the DNA, where replication begins, is the site of origin of replication. The two strands open and separate at this point, forming the replication fork.

(b) An unwinding of DNA molecule: The enzyme, helicase, bind to the origin of a replication site. Helicase separates the two strands of DNA. The enzyme called topoisomerase separates the double helix above the replication fork and removes the twists formed during the unwinding process.

(c) Formation of RNA primer: An RNA primer is a short segment of RNA nucleotides. The primer is synthesized by the DNA template, close to the origin of a replication site.

(d) Synthesis of the new complementary strand from the parent strand: After the formation of RNA primer, nucleotides are added with the help of an enzyme DNA polymerase, a new complementary strand of DNA is formed from each of the parent strands. The daughter strand is synthesized as a continuous strand, which is called the leading strand.

The short segments of DNA are synthesized, in the other strand and called lagging strand. The short segments of DNA are called Okazaki fragments. The enzyme DNA ligase joins the fragments. The replication fork of the two sides meets at a site called terminus, which is stimulated opposite to origin of a replication site.

IX. Higher Order Thinking Skills [HOTS] Questions

Question 1.
What do you understand about DNA, gene and chromosome in short?
Answer:
Deoxy ribonucleic acid [DNA] is the material located in the cells and nucleus, that makes up the chromosomes and genes. Its molecule is in the shape of a double helix.

A gene is a segment of DNA that is passed down from parents to children and confers a trait to the offspring. Genes are organised and packaged in units called chromosomes.

Chromosomes are thin, long thread-like structures contain two identical strands, called chromatids, held together by a centromere. Humans have 23 pairs of chromosomes.

Question 2.
What is the difference between a gene mutation and a chromosomal mutation?
Answer:

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You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.6

Exercise 1.6 Class 9 Maths Samacheer Question 1.
(i) If n(A) = 25, n(B) = 40, n(A ∪ B) = 50 and n(B’) = 25 , find n(A ∩ B) and ii(U).
(ii) If n(A) = 300, n(A ∪ B) = 500, n(A ∩ B) = 50 and n(B’) = 350, find n(B) and n(U).
Solution:
(i) n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
n(A ∩ B) = 25 + 40 – 50 = 65 – 50 = 15
n(U) = n(B) + n(B’) = 40 + 25 = 65

(ii) n(U) = n(B) + n(B’)
n(A ∩ B) = n(A) + n(B) – n(A B)
n(B) = n(A ∪ B) + n(A ∩ B) – n(A) = 500 + 50 – 300 = 250
n(U) = 250 + 350 = 600

9th Maths Exercise 1.6 Question 2.
If U = {x : x ∈ N, x ≤ 10}, A = {2, 3, 4, 8, 10) and b = {1, 2, 5, 8, 10}, then verify that n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Solution:
n(A) = 5, n(B) = 5
A ∪ B = {1, 2, 3, 4, 5, 8, 10}, A ∩ B= {2, 8, 10}
n(A ∪ B) = 7, n(A ∩ B) = 3
L.H.S n(A ∪ B) = 7
R.H.S = n(A) + n(B) – n(A ∩ B) = 5 + 5 – 3 = 7
∴ L.H.S = R.H.S proved.

9th Standard Maths Exercise 1.6 Question 3.
Verify n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) for the following sets.
(i) A = {a, c, e, f, h}, B = {c, d, e, f} and C = {a, b, c, f}
(ii) A = {1, 3, 5} B = {2, 3, 5, 6} and C = {1, 5, 6, 7}.
Solution:
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)

(i) A = {a, c, e, f, h}, B = {c, d, e, f}, C = {a, b, c, f}
n (A) = 5, n (B) = 4, n (C) = 4
n( A ∩ B) =3
n(B ∩ C) = 2
n( A ∩ C) =3
n( A ∩ B ∩ C) = 2
A ∩ B = {c, e, f}
B ∩ C = {c, f}
A ∩ C = {a, c, f}
A ∩ B ∩ C = {c, f}
A ∪ B ∪ C = {a, c, d, e, f, b, h}
∴ n(A ∪ B ∪ C) = 7 ……………. (1)
n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
= 5 + 4 + 4 – 3 – 2 – 3 + 2 = 15 – 8 = 7 ……………… (2)
∴ (1) =(2)
⇒ n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
Hence it is verified.

(ii) A = {1, 3, 5}, B = {2, 3, 5, 6 }, C = {1, 5, 6,7} = 3, n (B) = 4, n (C) = 4
n(A ∩ B) = 2
n(B ∩ C) = 2
n(C ∩ A) = 2
n(A ∩ B ∩ C) = 1
n(A ∪ B ∪ C) = 6
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
6 = 3 + 4 + 4 – 2 – 2 – 2 + 1 = 12 – 6 = 6
Hence it is verified.

9th Maths 1.6 Question 4.
In a class, all students take part in either music or drama or both. 25 students take part in music, 30 students take part in drama and 8 students take part in both music and drama. Find
(i) The number of students who take part in only music.
(ii) The number of students who take part in only drama.
(iii) The total number of students in the class.
Solution:
Let the number of students take part in music is M.
Let the number of students take part in drama is D.
By using venn diagram

(i) The number of students take part in only music is 17.
(ii) The number of students take part in only drama is 22.
(iii) The total number of students in the class is 17 + 8 + 22 = 47.

Samacheer Kalvi Guru 9th Maths Question 5.
In a party of 45 people, each one likes tea or coffee or both. 35 people like tea and 20 people like coffee. Find the number of people who
(i) like both tea and coffee.
(ii) do not like tea.
(iii) do not like coffee.
Solution:
Let the people who like tea be T.
Let the people who like coffee be C
By using formula
n( A ∪ B) = n(A) + n(B) – n(A ∩ B)

(i) n(T ∩ C) = n(T) + n(C) – n(T ∪ C) = 35 + 20 – 45 = 55 – 45 = 10
The number of people who like both coffee and tea = 10.

(ii) The number of people who do not like Tea
n(T) = n(U) – n(T) = 45 – 35 = 10

(iii) The number of people who do not like coffee
n(C’) = n(U) – n(C) = 45 – 20 = 25.

Samacheer Kalvi 9th Standard Maths Question 6.
In an examination 50% of the students passed in Mathematics and 70% of students passed in Science while 10% students failed in both subjects. 300 students passed in at least one subjects. Find the total number of students who appeared in the examination, if they took examination in only two subjects.
Solution:
Let the students who appeared in the examination be 100%.
Let the percentage of students who failed in mathematics be M.
Let the percentage of students who failed in science be S.
Failed in Maths = 100 % – Pass% = 100% – 50% = 50%
Failed in Science% 100% – 70% = 30%
Failed in both% = 10%
n(M ∪ S) = n(M) + n(S) – n(M ∩ S)
= 50% + 30% – 10% = 70%
% of students failed in atleast one subject = 70%
∴ The % of students who have passed in atleast one subject = 100% – 70% = 30%
30% = 300
∴ \(100 \%=\frac{100 \times 300}{30}=1000\)
∴ The total number of students who appeared in the examination = 1000 students.

Samacheer Kalvi 9th Maths Solutions Pdf Question 7.
A and B are two sets such that n(A – B) = 32 + x, n(B – A) = 5x and n(A ∩ B) = x. Illustrate the information by means of a venn diagram. Given that n(A) = n(B), calculate the value of x.
Solution:

n (A – B) = 32 +x
n(B – A) = 5x
n(A ∩ B) = x
n( A) = n(B)
32 + x + x = 5x + x
32 + 2x = 6x
4x = 32
x = 8

Samacheer Kalvi 9th Maths Question 8.
Out of 500 car owners investigated, 400 owned car A and 200 owned car B, 50 owned both A and B cars. Is this data correct?
Solution:
n( A ∪ B) = n(A) + n(B) – n( A ∩ B)
n(A ∪ B) = 500 (given) …………. (1).
n( A) = 400
n(B) = 200
n( A ∩ B) =50
∴ n(A ∩ B) = 400 + 200 – 50 = 550 …………… (2)
1 ≠ 2
∴ This data is incorrect.

9th Maths Samacheer Kalvi Question 9.
In a colony, 275 families buy Tamil newspaper, 150 families buy English newspaper, 45 families buy Hindi newspaper, 125 families buy Tamil and English newspapers, 17 families buy English and Hindi newspapers, 5 families buy Tamil and Hindi newspapers and 3 families buy all the three newspapers. If each family buy atleast one of these newspapers then find
(i) Number of families buy only one newspaper
(ii) Number of families buy atleast two newspapers
(iii) Total number of families in the colony.
Solution:
(i) Tamil Newspaper buyers n(A) = 275
English Newspaper buyers n(B) = 150
Hindi Newspaper buyers n(C) = 45
Tamil and English Newspaper buyers n(A ∩ B) = 125
English and Hindi Newspaper buyers n(B ∩ C) = 17
Hindi and Tamil Newspaper buyers n(C ∩ A) = 5
All the three Newspaper buyers n(A ∩ B ∩ C) = 3

(i) Number of families buy only one newspaper = 148 + 11 + 26 = 185
(ii) Number of families buy atleast two news papers = 122 + 14 + 2 + 3 = 141
(iii) Total number of families in the colony = 148 + 11 + 26 + 122 + 14 + 2 + 3 = 326

Question 10.
A survey of 1000 farmers found that 600 grew paddy, 350 grew ragi, 280 grew corn, 120 grew paddy and ragi, 100 grew ragi and corn, 80 grew paddy and corn. If each farmer grew atleast any one of the above three, then find the number of farmers who grew all the three.
Solution:

a = 600 – (120 – x + x + 80 – x)
= 600 – (200 – x)
= 600 – 200 + x
= 400 + x
b = 350 – (120 – x + x + 100 – x)
= 350 – (220 – x)
= 350 – 230 + x
= 130 + x
c = 280 – (80 – x + x + 100 – x)
= 2800 – (180 – x) = 280 – 180 + x = 100 + x
Each farmer grew atleast one of the above three, the number of farmers who grew all the three is x.
= a + b + c + 120 – x + 100 – x + 80 – x + x = 1000
400 + x + 130 + x + 100 + x + 120 – x + 100 – x + 80 – x + x= 1000
∴ 930 + x = 1000
x = 1000 – 930 = 70
∴ 70 farmers grew all the three crops

9th Standard Maths Samacheer Kalvi Question 11.
In the adjacent diagram, if n(U) = 125,y is two times of x and z is 10 more than x, then find the value of x, y and z.

Solution:
n(U) = 125
y = 2x
z = x + 10
∴ x + y + z + 4 + 17 + 6 + 3 + 5 = 125
x + 2x + x + 10 + 35 = 125
4x + 45 = 125
4x = 125 – 45
4x = 80
x = 20
∴ y = 2x = 2 × 20 = 40
z = x + 10 = 20 + 10 = 30
Hence x = 20 ; y = 40; z = 30

Samacheer Kalvi Guru Maths 9th Question 12.
Each student in a class of 35 plays atleast one game among chess, carrom and table tennis. 22 play chess, 21 play carrom, 15 play table tennis, 10 play chess and table tennis, 8 play carrom and table tennis and 6 play all the three games. Find the number of students who play (i) chess and carrom but not table tennis (ii) only chess (iii) only carrom (Hint: Use Venn diagram)
Solution:

A – Chess
B – Carrom
C – Table Tennis
n(A) = 22
n(B) = 21
n(C) = 15
n(A ∩ C) = 10
n(B ∩ C) = 8
n(A ∩ B ∩ C) = 6
(i) y = 22 – (x + 6 + 4) = 22 – (x + 10)
= 22 – x – 10
= 12 – x
z = 21 – (x + 6 + 2) = 21 – (8 + x)
21 – 8 – x = 13 – x
y + z + 3 + x + 2 + 4 + 6 = 35
12 – x + 13 – x + 15 + x = 35
40 – x = 35
x = 40 – 35 = 5
(i) Number of students who pay only chess and Carrom but not table tennis = 5
(ii) Number of students who play only chess = 12 – x = 12 – 5 = 7
(iii) Number of students who play only carrom = 13 – x = 13 – 5 = 8

9th Class Maths Exercise 1.6 Solution Question 13.
In a class of 50 students, each one come to school by bus or by bicycle or on foot. 25 by bus, 20 by bicycle, 30 on foot and 10 students by all the three. Now how many students come to school exactly by two modes of transport?
Solution:

A – by bus
B – by bicycle
C – on foot
n(A) = 25
n(B) = 20
n(C) = 30
n(A ∩ B ∩ C ) = 10
n(A ∪ B ∪ C ) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)
50 = 25 + 20 + 30 – (10 + x) – (10 + y) – (10 + z) + 10
50 = 75 – 10 – x – 10 – y – 10 – z + 10
= 75 – 20 – (x + y + z)
= 55 – (x + y + z)
x + y + z = 55 – 50 = 5
∴ The number of students who come to school exactly by two modes of transport = 5

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Tamilnadu Samacheer Kalvi 10th Social Science Economics Solutions Chapter 1 Gross Domestic Product and its Growth: an Introduction

Do you feel scoring more marks in the 10th Social ScienceEconomics Grammar sections and passage sections are so difficult? Then, you have the simplest way to understand the question from each concept & answer it in the examination. This can be only possible by reading the passages and topics involved in the 10th Social ScienceEconomics Board solutions for Chapter 1 Gross Domestic Product and its Growth: an Introduction Questions and Answers. All the Solutions are covered as per the latest syllabus guidelines. Check out the links available here and download 10th Social ScienceEconomics Chapter 1 textbook solutions for Tamilnadu State Board.

Gross Domestic Product and its Growth: an Introduction Textual Exercise

I. Choose the correct answer.

Gross Domestic Product And Its Growth An Introduction Question 1.
GNP equals …………
(a) NNP adjusted for inflation
(b) GDP adjusted for inflation
(c) GDP plus net property income from abroad
(d) NNP plus net property income or abroad
Answer:
(c) GDP plus net property income from abroad

10th Social Economics 1st Lesson Question 2.
National Income is a measure of:
(a) Total value of money
(b) Total value of producer goods
(c) Total value of consumption goods
(d) Total value of goods and services
Answer:
(d) Total value of goods and services

10th Economics Book Samacheer Kalvi Question 3.
Primary sector consist of ………………
(a) Agriculture
(b) Automobiles
(c) Trade
(d) Banking
Answer:
(a) Agriculture

Samacheer Kalvi Economics Question 4
………………….. approach is the value added by each intermediate good is summed to estimate the value of the final good.
(a) Expenditure approach
(b) Value-added approach
(c) Income approach
(d) National Income
Answer:
(b) Value-added approach

Class 10 Social Science Economics Chapter 1 Question 5.
Which one sector is highest employment in the GDP?
(a) Agricultural sector
(b) Industrial sector
(c) Service sector
(d) None of the above.
Answer:
(c) Service sector

Growth Of Gross Domestic Product Match The Following Question 6.
Gross value added at current prices for services sector is estimated at …………………..
(a) 91.06
(b) 92.26
(c) 80.07
(d) 98.29
Answer:
(b) 92.26

10th Social Science Economics Question 7.
India is …………… larger producer in agricultural product.
(a) 1^st
(b) 3^rd
(c) 4^th
(d) 2^nd
Answer:
(d) 2^nd

10th Economics Question 8.
India’s life expectancy at birth is ………………. years.
(a) 65
(b) 60
(c) 70
(d) 55
Answer:
(a) 65

Gross Meaning In Tamil Question 9.
Which one is a trade policy?
(a) irrigation policy
(b) import and export policy
(c) land-reform policy
(d) wage policy
Answer:
(b) import and export policy

Question 10.
Indian economy is:
(a) Developing Economy
(b) Emerging Economy
(c) Dual Economy
(d) All the above
Answer:
(d) All the above

II. Fill in the Blanks.

1. …………….. sector is largest sector in India.
2. GDP is the indicator of ………………. economy.
3. Secondary sector otherwise called as …………..
4. ……………… sector is the growth engine of Indian economy.
5. India is …………….. largest economy of the world.
6. India is ………………. fastest growing nation of the world.
7. ………………. policy envisages rapid industrialization with modernization for attaining rapid economic growth of GDP.
Answers:
1. Service
2. a country’s
3. industry sector
4. Service
5. sixth
6. fifth
7. Industrial

III. Choose the Correct Statement.

Question 1.
The rate of saving is low in India for the following reason
(i) Low per capita income.
(ii) Poor performance and less contribution of public sector.
(iii) Poor contribution of household sector.
(iv) Savings potential of the rural sector not tapped fully.

(a) i, ii and iv are correct
(b) i, ii and iii are correct
(c) i, ii, iii and iv are correct
(d) i, iii and iv are correct
Answer:
(a) i, ii and iv are correct

IV. Match the Following.

Answers:
1. (c)
2. (d)
3. (e)
4. (a)
5 .(b)

V. Give Short Answers.

Question 1.
Define National Income.
Answer:
National Income is a measure of the total value of goods and services produced by an economy over a period of time, normally a year. Commonly National Income is called a Gross National product (or) National Dividend.

Question 2.
What is meant by Gross domestic product?
Answer:
The GDP is the market value of all the final goods and services produced in the country during a time period.

Question 3.
Write the importance of Gross Domestic product.
Answer:

1. It helps in the study of economic growth of an economy.
2. To understand to unequal distribution of wealth in an economy.
3. To analyse the problem of inflation and deflation.
4. To compare domestic country with the developed countries of the world.
5. To estimate the purchasing power of the economy.
6. Helps public sector to frame suitable policies for development.
7. It acts as a guide to economic planning.

Question 4.
What is per capita income?
Answer:
Per capita income or PCI is an indicator to show the living standard of people in a country. It is obtained by dividing the National Income by the population of a country.

Question 5.
Define the value added approach with example.
Answer:
The value of each intermediate good is added together to estimate the value of the final good. It is called as value added approach. Eg: In order to find the value of a cup of tea, we need to add the value of tea powder plus milk plus sugar. Here Tea powder, milk and sugar are intermediate goods, whereas tea is the final good. Eg: Tea powder – ₹ 3 /-, Milk – ₹ 5 /-, Sugar – ₹ 2 /- Market price of one cup of tea is ₹ 10 /- (3 + 5 + 2).

Question 6.
Name the sectors contribute to the GDP with examples.
Answer:
Name of the sectors are: (i) Primary sector, (ii) Secondary sector and (Hi) Tertiary sector.

Question 7.
Write the sector wise Indian GDP composition in 2017.
Answer:
Sector wise contribution in GDP of India for the year 2016 – 2017
Primary Sector – 17.09%
Secondary Sector – 29.03%
Tertiary Sector – 52.08%

Question 8.
What are the factors supporting to develop the Indian Economy?
Answer:

1. A fast growing working age population
2. Strong Legal system
3. Many English language speakers
4. Low wage cost
5. Highly advanced space technology
6. External economics of scale.

Question 9.
Write the name of economic policies in India.
Answer:
Name of economic policies in India are:

• Agriculture Policy
• Industrial Policy
• New Economic Policy
• Trade Policy
• Employment Policy
• Currency and Banking Policy
• Fiscal and Monetary Policy
• Wage Policy
• Population Policy

Question 10.
Write a short note on
(i) Gross National Happiness (GNH)
(ii) Human Development Index (HDI).
Answer:
(i) Gross National Happiness: This term was introduced in 1972, by the King of Bhutan Jigme Wang chuck. It is an index which is used to measure the collective happiness and well-being of a population.
(ii) Human Development Index: This term was introduced in 1990 by a Pakistani Economist at the United Nations. It is a composite index of life expectancy at birth, adult literacy rate and standard of living measured in terms of GDP adjusted to purchasing power parity.

VI. Write in detail answer.

Question 1.
Briefly explain various terms associated with measuring of national income.
Answer:
Various terms associated with measuring of national income.
(i) Gross National Product or GNP is the total value of goods and services produced and income received in a year by domestic residents of a country. It excludes profits earned from capital invested abroad.

(ii) Gross Domestic Product or GDP is the total value of output of goods and services produced by the factors of production within the geographical boundaries of the country.

(iii) Net National Product or NNP refers to gross national product, i.e., the total market value of all final goods and services produced by the factors of productions of a country or other polity during a given time period, minus depreciation.

(iv) Net Domestic Product or NDP is a part of Gross Domestic Product. It is obtained from the Gross Domestic Productby deducting the Quantum of ten wear and tear expenses (depreciation).
NDP = GDP – Depreciation

(v) Per Capita Income or PCI is an indicator to show the living standard of people in a country. It is obtained by dividing the National Income by the population of a country.
Per Capita Income = \(\frac{\text { National Income }}{\text { Population }}\)

(vi) Personal Income or PI is the total money income received by individuals and households of a country from all possible services before direct taxes. Personal income can be expressed as follows:
PI = NI Corporate Income Texes – Undistributed Corporate Profits – Social Security Contributions + Transfer payment.

(vii) Disposable Income or DI means actual income which can be spent on consumption by . individuals and families. It can expressed as DPI = PI – Direct Taxes.

Question 2.
What are the methods of calculating Gross Domestic Product?
Answer:
Explain it.
There are three methods to calculate Gross Domestic Product.
(i) Expenditure Approach
(ii) Income Approach
(iii) Value-Added Approach

(i) Expenditure Approach: According to this method, the expenditure on all the final goods and services produced in the country during a specific period are added together, to get the GDP.
Y = C + I + G + (X – M)
Y – National Income
C – Consumption Expenditure,
I – Investment Expenditure
G – Government Expenditure,
X – Exports
M – Imports.

(ii) The Income Approach: In this method, the earnings of all the men and women who are involved in producing goods and services are added together to measure GDP.
Y = W + R + I + π
Y = National Income, W- Wages,
R – rent, I – interest, π – profit

(iii) Value-Added Approach: In this method, the value of each intermediate good is added together to estimate the value of final goods. The sum of the value of all the final goods gives us the total value of the final goods produced in the economy, which is Measured as GDP.

GDP = Sum of final goods produced in the economy S_f
S_f = Sum of Intermediate goods produced in the economy.

Question 3.
Write about the composition of GDP in India.
Answer:
Indian economy is broadly divided into three sectors which contribute to the GDP –
(i) Primary Sector – It includes agriculture-based allied activities, production of raw materials such as cattle farm, fishing, mining, forestry etc. It is also called agricultural sector.

(ii) Secondary Sector – It includes industries that produce a finished, usable product or are involved in construction. This sector generally takes the output of the primary sector and manufactures finished goods. It is also called industrial sector.

(iii) Tertiary Sector – It is known as service sector and includes transport, insurance, banking, trade, education, health care etc. ’

Question 4.
Write the differences between the growth and development.
Answer:
Differences between the Economic growth and Economic development

Question 5.
Explain the development path based on GDP and employment.
Answer:

1. In the earlier stages of Indian Independence, India remained as closed economy and the interaction with the outside world remained limited.
2. The reason for closed trade was to give importance to domestic industries and reduce the dependence on foreign products and companies.
3. Later in the year 1991, India adopted free trade policy, and liberalised the economy.
4. It has given permission for the foreign companies to enter into the Indian economy.
5. To give employment to the increasing size of work force, a thrust was given to employment generation under the Five year plans.
6. Rural Development was also given special importance.
7. Eradication of poverty became a very important part in context of Rural Development.
8. The private companies and Industries were subject to strict rules and regulations.
9. It was believed that the social welfare of the people could be possible only by the government, therefore it gained importance.
10. India’s Per Capita Income have doubled in 12 years.
11. India falls under Middle Income country category.
12. There is reduction in poverty percentage and the life expectancy at birth is 65 years.
13. 44% of children under 5 are malnourished.
14. The literacy rate for the population of 15 years of age is only 63% compared to 71% of other middle income countries.
15. India has followed a different path of development by moving from agricultural sector to service sector very quickly. This help India to expect emerging Industrialists in Indian Economy.

Question 6.
Explain the following the economic policies
1. Agricultural Policy
2. Industrial policy
3. New economic policy.
Answer:
Many economic policies have been framed by the Government of India since independence for increasing rate of economic growth and economic development. The important economic policies are :
(i) Agricultural Policy – This policy is the set of Government decisions and actions relating to domestic agriculture and imports of foreign agricultural products. Governments usually implement agricultural policies with the goal of achieving a specific outcome in the domestic agricultural product markets. Some agricultural policies are price policy, land reform policy, irrigation policy, food policy etc.

(ii) Industrial Policy – It is a very important aspect of any economy. It create employment, promotes research and development, leads to modernisation and ultimately make the economy self sufficient. Several industrial policies since 1948 have come into existence – textile industrial policy, sugar industry policy, price policy of industrial growth etc.

(iii) New Economic Policy – The economy of India had undergone a significant policy shifts in the beginning of the 1990s. This new model of economic reforms is commonly known as the LPG or Liberalisation, Privatisation and Globalisation model. The primary objective of this model was to make the economy of India the fastest developing economy in the globe with capabilities that help at match up with the biggest economies of the world. These economic reforms influenced the overall economic growth of the country in a significant manner.

Gross Domestic Product and its Growth: an Introduction Additional Questions

I. Choose the correct answer.

Question 1.
What is ‘H’ in GNH?
(a) Holislic
(b) Happiness
(c) Human
(d) Hazardous
Answer:
(b) Happiness

Question 2.
Goods are …………………
(a) Tangible
(b) Intangible
(c) Both (a) and (b)
(d) services
Answer:
(a) Tangible

Question 3.
What is India’s world rank in industrial sector?
(a) 6
(b) 7
(c) 8
(d) 9
Answer:
(a) 6

Question 4.
…………………. is the market value of all goods and services produced in the country.
(a) GNP
(b) GDP
(c) NNP
(d) NDP
Answer:
(b) GDP

Question 5.
India finally decided to liberalise its economy in the year …………..
(a) 1991
(b) 1995
(c) 2000
(d) 2001
Answer:
(a) 1991

Question 6.
The Goods and Services are measured in terms of ………………… of that country.
(a) wealth
(b) currency
(c) type
(d) size
Answer:
(b) currency

Question 7.
Income method sums all forms of …………
(a) Expenditure
(b) Income
(c) Savings
Answer:
(b) Income

Question 8.
Only those goods and services that has a market value are included in the …………………
(a) GNP
(b) GDP
(c) NNP
(d) NDP
Answer:
(b) GDP

Question 9.
The primary function of the Government is …………….
(a) to maintain law and order
(b) Military defence
(c) Social Security measures
Answer:
(a) to maintain law and order

Question 10.
According to Economists Tyler and Alex, Final goods and services a part of other goods and services ………………… a part of other goods and services
(a) will be
(b) will not be
(c) fully
(d) None
Answer:
(b) will not be

Question 11.
Economic development is the process as well as an increase in real …………. income.
(a) individual
(b) family
(c) national
Answers:
(c) national

II. Fill in the Blanks:

1. Economic development focuses on balanced and ……….. distribution of wealth among all individuals and tries to uplift the downgrade society.
2. The GDP of the United States of America is 19.3 trillion USD and ranked ………….
3. Human development Index is apt tool to measure the real development in an …………….
4 ………….. has emerged as a hub of global software business.
5. In India the GDP is measured both annually and …………….
6. Economic growth means an increase in ……………. and …………. in an economy.
7. Human Resources are ……………. for economic development.
8. Per capita income is calculated by dividing National Income by ……………
9. Per capita income is an indicator of …………… in a country.
10. Economic development is economic growth and allocation of resources from primary sector ………… to sector.
11. Tertiary sector is known as ……………
Answers:
1. equitable
2. one
3. economy
4. Bangalore
5. quarterly
6. production of good, services
7. instruments
8. population
9. living standard of people
10. tertiary
11. service sector

III. Match the Following.

Answers:
1. (d)
2. (a)
3. (e)
4. (c)
5. (b)

IV. Answer in brief.

Question 1.
What are the four pillars of Gross National Happiness Index?
Answer:
The four pillars of GNHI are

1. Sustainable and equitable socio-economic development
2. Environmental conservation
3. Preservation and Promotion of culture
4. Good Governance

Question 2.
“India has followed a different path of development from many other countries”. Explain.
Answer:
India has gone more quickly from agriculture to services that tend to be less tightly regulated than heavy industry. There are some emerging manufacturing giants in the Indian economy.

Question 3.
What do you understand by the term ‘double counting’?
Answer:
The value of final goods are included in the calculation of GDP, but not the value of Intermediate goods. This is because the value of the Intermediate goods is already included in the final good. So, if it is added again it will result in double counting.

Question 4.
What are the nine domains of Gross National Happiness or GNH?
Answer:
The nine domains of GNH are – psychological well-being, health, time use, education, cultural diversity and resilience, good governance, community vitality, ecological diversity and resilience, and living standards.

Question 5.
What is meant by Market value?
Answer:
The price at which the goods and services are sold in the market is called as Market value.

Question 6.
Write a note on Income method.
Answer:

• Income method is one of the methods of calculating National Income.
• In this method, the income and payments received by all the people in the country are calculated.

Question 7.
What are the key parameters of Economic Growth?
Answer:
The key parameters of Economic Growth in an economy are its Gross Domestic Product (GDP) and Gross National Product (GNP) which helps in measuring the actual size of an economy.

Question 8.
What is Net Domestic Product?
Answer:
Net Domestic Product is obtained from the Gross Domestic Product by deducting the Quantum of tear and wear expenses (depreciation).
NDP = GDP (-) Depreciation.

Question 9.
What are the nine domains of GNHI?
Answer:

1. Psychological well-being
2. health
3. time-use
4. Education
5. Cultural diversity and Resilience
6. Good governance
7. Community vitality
8. Ecological diversity
9. Living standards.

Question 10.
What are the basic concepts of National Income?
Answer:

1. Gross National Product (GNP)
2. Gross Domestic Product (GDP)
3. Net National Product (NNP)
4. Net Domestic Product (NDP)
5. Per Capita Income (PCI)
6. Personal Income (PI)
7. Disposable Income (DI)

Question 11.
What is meant by GNP Deflator?
Answer.
It means the change in the Gross National Product (GNP) with the change in the price levels symbolically.\(\frac{\Delta \mathrm{GNP}}{\Delta \mathrm{P}}\)

Question 12.
What are called sectors?
Answer:
Sectors are the groups of various economic activities that produced goods and services.

Question 13.
Why is the primary sector also called agricultural and related sector?
Answer:
Since most of the natural products we get are from Agriculture based allied activities. Production of raw materials such as cattle farm, dairy, fishing, mining, forestry, com and coal. The primary sector is also called agriculture and related sector.

Question 14.
Why is the tertiary sector also called the ‘Service Sector’?
Answer:
Since the activities that fall under the tertiary sector generate services rather than goods, this sector is also called the service sector.

V. Answer in detail.

Question 1.
What are the limitations of the Gross Domestic Product?
Answer:
The GDP is the most widely used measure of the state of the economy. While appreciating its usefulness, we should be aware of some of its limitations –
(i) Several important goods and services are left out of the GDP – The GDP includes only the goods and services sold in the market. Clean air. which is vital for a healthy life, has no market value and is left out of the GDP.

(ii) GDP measures only quantity but not quality – In the 1970s schools and banks were not allowed to use ball point pens r because of their poor quality. Since then, not only has these been a substantial increase in the quantity of ballpoint pens produced in India but their quality has also improved a lot. The improvement in quality of goods is very important but it is not captured by the GDP.

(iii) GDP does not tell us about the way income is distributed in the country – The GDP of a country’ may be growing rapidly but income may be distributed to unequally that only a small percentage of people may be benefiting from it.

(iv) GDP does not tell us about the kind of life people are living – A high level of per capita real GDP can go hand – in – hand with very low health condition of people, an undemocratic political system, high pollution and high suicide rate.

Question 2.
Write a detailed note on the measurement of GDP.
Answer:
(i) The GDP Gross Domestic Product of a country measures the market value of goods and services produced during a particular period of time.

(ii) In the measurement of GDP, the GDP of the previous years are not included. Eg: the GDP of 2018 will include the market value of goods and services produced only during 2018.

(iii) In India, GDP is measured both annually and in quarterly.

(iv) The Annual GDP financial year (2017-18) is from 1st April 17 to 31st March 18.

(v) The quarterly GDP for any financial year is calculated by dividing the year into 4 quarters with 3 months each.

1^st Quarter – April, May, June (Q₁)
2^nd Quarter – July, Aug, Sept (Q₂)
3^rd Quarter – Oct, Nov ,Dec (Q₃)
4^th Quarter – Jan, Feb, March (Q₄)
(vi) The GDP of Q₂ will not include GDP of Q₁ and vice versa. This means, only the goods and services produced in that quarter is included for the measurement of GDP.

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.2

10th Maths Exercise 6.2 Samacheer Kalvi Question 1.
Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10\(\sqrt{3}\) m.
Solution:

Ex 6.2 Class 10 Samacheer Question 2.
A road is flanked on either side by continuous rows of houses of height 4\(\sqrt{3}\) m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30°. Find the width of the road.
Solution:

Exercise 6.2 Class 10 Samacheer Kalvi Question 3.
To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? (\(\sqrt{3}\) = 1.732)
Solution:

Let ‘H’ be the fit of the window. Given that elevation of top of the window is 60°.

Given that elevation of bottom of the window is 45°.

∴ Height of the window = 3.66 m

10th Maths Trigonometry Exercise 6.2 Question 4.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 40°. Find the height of the pedestal. (tan 40° = 0.8391, \(\sqrt{3}\) = 1.732)
Solution:

Let ‘p’ be the fit of the pedestal and d be the distance of statue from point of cabs, on the ground.
Given the elevation of top of the statue from p^f on ground is 60°.

10th Maths Exercise 6.2 Question 5.
A flag pole ‘h’ metres is on the top of the hemispherical dome of radius V metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30°. Find
(i) the height of the pole
(ii) radius of the dome.
Solution:

10th Maths Ex 6.2 Question 6.
The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Solution:

Let BD be tower of height = 15 m
AE be pole of height = ‘p’

10th Maths 6.2 Question 7.
A vertical pole fixed to the ground is divided in the ratio 1 : 9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?
Solution:

10th Samacheer Kalvi Maths Trigonometry Question 8.
A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8°. What is the height of the peak if the distance between consecutive milestones is 1 mile, (tan 4° = 0.0699, tan 8° = 0.1405).
Solution:

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Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

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Samacheer Kalvi 12th Physics Electrostatics Textual Evaluation Solved

Samacheer Kalvi 12th Physics Electrostatics Multiple Choice Questions

Physics Class 12 Samacheer Kalvi Question 1.
Two identical point charges of magnitude -q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge +q is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q be stable with respect to the displacement?

(a) A₁ and A₂
(b) B₁ and B₂
(c) both directions
(d) No stable
Answer:
(b) B₁ and B₂

Samacheer Kalvi 12th Physics Chapter 1 Question 2.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) infinite uniform line charge
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
(c) uniformly charged infinite plane

Samacheer Kalvi Guru 12th Physics Question 3.
What is the ratio of the charges \(\left|\frac{q_{1}}{q_{2}}\right|\) for the following electric field line pattern?

(a) \(\frac { 1 }{ 5 }\)
(b) \(\frac { 25 }{ 11 }\)
(c) 5
(d) \(\frac { 12 }{ 25}\)
Answer:
(d) \(\frac { 12 }{ 25}\)

Samacheer Kalvi 12th Physics Question 4.
An electric dipole is placed at an alignment angle of 30° with an electric field of 2 x 10⁵ N C^-1. It experiences a torque equal to 8 N m. The charge on the dipole if the dipole length is 1 cm is-
(a) 4 mC
(b) 8 mC
(c) 5 mC
(d) 1 mC
Answer:
(b) 8 mC

12 Physics Samacheer Kalvi Question 5.
Four Gaussian surfaces are given below with charges inside each Gaussian surface. Rank the electric flux through each Gaussian surface in increasing order-

(a) D < C < B < A
(b) A < B = C < D
(c) C < A = B < D
(d)D > C > B > A
Answer:
(a) D < C < B < A

Class 12 Physics Samacheer Kalvi Question 6.
The total electric flux for the following closed surface which is kept inside water-

(a) \(\frac { 80q }{{ ε }_{0}}\)
(b) \(\frac { q }{{ 40ε }_{0}}\)
(c) \(\frac { q }{{ 80ε }_{0}}\)
(d) \(\frac { q }{{ 40ε }_{0}}\)
Answer:
(b) \(\frac { q }{{ 40ε }_{0}}\)

12th Physics 1st Chapter Question 7.
Two identical conducting balls having positive charges q₁ and q₂ are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be- (NSEP 04-05)
(a) less than before
(b) same as before
(c) more than before
(d) zero
Answer:
(c) more than before

12th Physics Samacheer Kalvi Question 8.
Rank the electrostatic potential energies for the given system of charges in increasing order

(a) 1 = 4 < 2 < 3
(b) 2 = 4 < 3 < 1
(c) 2 = 3 < 1 < 4
(d) 3 < 1 < 2 < 4
Answer:
(a) 1 = 4 < 2 < 3

Samacheerkalvi.Guru 12th Physics Question 9.
An electric field \(\vec { E } \) = 10x\(\hat{i} \) exists in a certain region of space. Then the potential difference V = V₀ – V_A, Where V₀ is the potential at the origin and V_A is the potential at x = 2 m is-
(a) 10 J
(b) -20 J
(c) + 20 J
(d) – 10 J
Answer:
(a) 10 J

Electrostatics Notes Class 12 State Board Question 10.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. The correct plot for electrostatic potential due to this spherical shell is-

Answer:

12th Physics Lesson 1-Electrostatics Question 11.
Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is-
(a) 8.80 x 10^-17 J
(b) -8.80 x 10^-17 J
(c) 4.40 x 10^-17 J
(d) 5.80 x 10^-17 J
Answer:
(a) 8.80 x 10^-17 J

Electrostatics Class 12 Questions And Answers Pdf Question 12.
If voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion.
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains same, Q doubled
(d) Both Q and C remain same
Answer:
(c) C remains same, Q doubled

Electrostatic Problems And Solutions Pdf Question 13.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
(a) Capacitance
(b) Charge
(c) Voltage
(d) Energy density
Answer:
(d) Energy density

Samacheer Kalvi Physics Question 14.
Three capacitors are connected in triangle as shown in the figure. The equivalent capacitance between the points A and C is

(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) \(\frac { 1 }{ 4 }\) μF
Answer:
(b) 2 μF

Physics Solution Class 12 Samacheer Kalvi Question 15.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 x 10^-2 C and 5 x 10^-2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is (AIIPMT 2012)
(a) 3 x 10^-2 C
(b) 4 x 10^-2 C
(c) 1 x 10^-2 C
(d) 2 x 10^-2 C
Answer:
(a) 3 x 10^-2 C

Samacheer Kalvi 12th Physics Electrostatics Short Answer Questions

Samacheer Kalvi Class 12 Physics Solutions Question 1.
What is meant by quantisation of charges?
Answer:
The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.
q = ne
Here n is any integer (0, ±1, ±2, ±3, ±4). This is called quantisation of electric charge.

Samacheer Kalvi Guru Physics Question 2.
Write down Coulomb’s law in vector form and mention what each term represents.
Answer:
The force on a charge q₁ exerted by a point charge q₁ is given by
\(\vec { F } \)₁₂ = \(\frac { 1 }{{ 4πε }{0}}\) \(\frac {{ q }_{1}{ q }_{2}}{{ r }^{2}}\) \(\hat{r} \)₂₁
Here \(\hat{r} \)₂₁ is the unit vector from charge q₁ to q₁.
But \(\hat{r} \)₂₁ = –\(\hat{r} \)₁₂,

Therefore, the electrostatic force obeys Newton’s third law.

Samacheer Kalvi 12th Physics Solutions Question 3.
What are the differences between Coulomb force and gravitational force?
Answer:

• The gravitational force between two masses is always attractive but Coulomb force between two charges can be attractive or repulsive, depending on the nature of charges.
• The value of the gravitational constant G = 6.626 x 10^-11 N m² kg^-2. The value of the constant k in Coulomb law is k = 9 x 10⁹ N m² C².
• The gravitational force between two masses is independent of the medium. The electrostatic force between the two charges depends on nature of the medium in which the two charges are kept at rest.
• The gravitational force between two point masses is the same whether two masses are at rest or in motion. If the charges are in motion, yet another force (Lorentz force) comes into play in addition to coulomb force.

Samacheer Kalvi.Guru 12th Physics Question 4.
Write a short note on superposition principle.
Answer:
According to this superposition principle, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
\({ \vec { F } }_{ 1 }^{ tot }\) = \(\vec { F } \)₁₂ + \(\vec { F } \)₁₃ + \(\vec { F } \)₁₄ + \(\vec { F } \)_1n

Samacheer Kalvi 12 Physics Question 5.
Define ‘Electric field’.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced
by a unit charge and is given by
\(\vec { E } \) = \(\frac { \vec { F } }{ { q }_{ 0 } } \)
The electric field is a vector quantity and its SI unit is Newton per Coulomb (NC^-1).

Question 6.
What is mean by ‘Electric field lines’?
Answer:
Electric field vectors are visualized by the concept of electric field lines. They form a set of continuous lines which are the visual representation of the electric field in some region of space.

Question 7.
The electric field lines never intersect. Justify.
Answer:
As a consequence, if some charge is placed in the intersection point, then it has to move in two different directions at the same time, which is physically impossible. Hence, electric field lines do not intersect.

Question 8.
Define ‘Electric dipole’
Answer:
Two equal and opposite charges separated by a small distance constitute an electric dipole.

Question 9.
What is the general definition of electric dipole moment?
Answer:
The electric dipole moment vector lies along the line joining two charges and is directed from -q to + q. The SI unit of dipole moment is coulomb meter (Cm).
\(\vec { P } \) = qa\(\hat{i} \) -qa(\(\hat{-i} \)) = 2 qa\(\hat{i} \)

Question 10.
Define “electrostatic potential”.
Answer:
The electric potential at a point P is equal to the work done by an external force to bring a unit positive charge with constant velocity from infinity to the point P in the region of the external
electric field \(\vec { E } \).

Question 11.
What is an equipotential surface?
Answer:
An equipotential surface is a surface on which all the points are at the same potential.

Question 12.
What are the properties of an equipotential surface?
Answer:
Properties of equipotential surfaces
(i) The work done to move a charge q between any two points A and B,
W = q (V_B – V_A). If the points A and B lie on the same equipotential surface, work done is zero because V_A = V_B.

(ii) The electric field is normal to an equipotential surface. If it is not normal, then there is a component of the field parallel to the surface. Then work must be done to move a charge between two points on the same surface. This is a contradiction. Therefore the electric field must always be normal to equipotential surface.

Question 13.
Give the relation between electric field and electric potential.
Answer:
Consider a positive charge q kept fixed at the origin. To move a unit positive charge by a small distance dx in the electric field E, the work done is given by dW = -E dx. The minus sign implies that work is done against the electric field. This work done is equal to electric potential difference. Therefore,
dW = dV.
(or) dV = -Edx
Hence E = \(\frac { dV }{ dx }\)
The electric field is the negative gradient of the electric potential.

Question 14.
Define electrostatic potential energy?
Answer:
The potential energy of a system of point charges may be defined as the amount of work done in assembling the charges at their locations by bringing them in from infinity.

Question 15.
Define ‘electric flux’.
Answer:
The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux. Its unit is N m² C^-1. Electric flux is a scalar quantity.

Question 16.
What is meant by electrostatic energy density?
Answer:
The energy stored per unit volume of space is defined as energy density u_E = \(\frac { U }{ Volume }\)
From equation u_E = \(\frac { 1 }{ 2 }\) \(\frac{\left(\varepsilon_{0} A\right)}{d}\) (Ed)² = \(\frac { 1 }{ 2 }\) ε₀ (Ad) E² or u_E = \(\frac { 1 }{ 2 }\) ε₀E²

Question 17.
Write a short note on ‘electrostatic shielding’.
Answer:
Consider a cavity inside the conductor. Whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be protected from external electrical disturbance is kept inside this cavity. This is called electrostatic shielding.

Question 18.
What is Polarisation?
Answer:
Polarisation \(\vec { P } \) is defined as the total dipole moment per unit volume of the dielectric.
\(\vec { P } \) = X_e \(\vec { P } \)_ext

Question 19.
What is dielectric strength?
Answer:
The maximum electric field the dielectric can withstand before it breakdowns is called dielectric strength.

Question 20.
Define ‘capacitance’. Give its unit.
Answer:
The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors.
C = \(\frac { q }{ V }\) or Q ∝ V.
The SI unit of capacitance is coulomb per volt or farad (F).

Question 21.
What is corona discharge?
Answer:
The electric field near the edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge. This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

Samacheer Kalvi 12th Physics Electrostatics Long Answer Questions

Question 1.
Discuss the basic properties of electric charges.
Answer:
Basic properties of charges
(i) Electric charge:
Most objects in the universe are made up of atoms, which in turn are made up of protons, neutrons and electrons. These particles have mass, an inherent property of particles. Similarly, the electric charge is another intrinsic and fundamental property of particles. The SI unit of charge is coulomb.

(ii) Conservation of charges:
Benjamin Franklin argued that when one object is rubbed with another object, charges get transferred from one to the other. Before rubbing, both objects are electrically neutral and rubbing simply transfers the charges from one object to the other. (For example, when a glass rod is rubbed against silk cloth, some negative charge are transferred from glass to silk. As a result, the glass rod is positively charged and silk cloth becomes negatively charged).

From these observations, he concluded that charges are neither created or nor destroyed but can only be transferred from one object to other. This is called conservation of total charges and is one of the fundamental conservation laws in physics. It is stated more generally in the following way. The total electric charge in the universe is constant and charge can neither be created nor be destroyed. In any physical process, the net change in charge will always be zero.

(iii) Quantisation of charges:
The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.
q = ne
Here n is any integer (0, ±1, ±2, ±3, ± ….). This is called quantisation of electric charge. Robert Millikan in his famous experiment found that the value of e = 1.6 x 10^-19C. The charge of an electron is -1.6 x 10^-19 C and the charge of the proton is +1.6 x 10^-19C. When a glass rod is rubbed with silk cloth, the number of charges transferred is usually very large, typically of the order of 10¹⁰. So the charge quantisation is not appreciable at the macroscopic level. Hence the charges are treated to be continuous (not discrete). But at the microscopic level, quantisation of charge plays a vital role.

Question 2.
Explain in detail Coulomb’s law and its various aspects.
Answer:
Consider two point charges q₁ and q₂ at rest in vacuum, and separated by a distance of r. According to Coulomb, the force on the point charge q₂ exerted by another point charge q₁ is
\(\vec { F } \) ₂₁ = K\(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)₁₂,
where [/latex] \(\hat{r} \)₁₂ is the unit vector directed from charge q₁ to charge q₂ and k is the proportionality constant.

Important aspects of Coulomb’s law:
(i) Coulomb’s law states that the electrostatic force is directly proportional to the product of the magnitude of the two point charges and is inversely proportional to the square of the distance between the two point charges.

(ii) The force on the charge q₂exerted by the charge q₁ always lies along the line joining the two charges. \(\hat{r} \)₂₁is the unit vector pointing from charge q₁ to q₂ Likewise, the force on the charge q₁ exerted by q₂ is along – (i.e., in the direction opposite to \(\hat{r} \)₂₁).

(iii) In SI units, k = \(\frac { 1 }{{ 4πε }_{0}}\) and its value is 9 x 10⁹ Nm²C^-2. Here e0 is the permittivity of free space or vacuum and the value of ε₀ = \(\frac { 1 }{{ 4πε }_{0}}\) = 8.85 x 10^-12 C² N^-1 m^-2

(iv) The magnitude of the electrostatic force between two charges each of one coulomb and separated by a distance of 1 m is calculated as follows:
[F] = \(\frac{9 \times 10^{9} \times 1 \times 1}{1^{2}}\) = 9 x 10⁹N. This is a huge quantity, almost equivalent to the weight of one million ton. We never come across 1 coulomb of charge in practice. Most of the electrical phenomena in day-to-day life involve electrical charges of the order of pC (micro coulomb) or nC (nano coulomb).

(v) In SI units, Coulomb’s law in vacuum takes the form \(\vec { F } \) ₂₁ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)₁₂. sin Since ε > ε₀, the force between two point charges in a medium other than vacuum is always less than that in vacuum. We define the relative permittivity for a given medium as ε = \(\frac { ε }{{ ε }_{0}}\) .For vacuum or air, ε_r = 1 and for all other media ε_r > 1

(vi) Coulomb’s law has same structure as Newton’s law of gravitation. Both are inversely proportional to the square of the distance between the particles. The electrostatic force is directly proportional to the product of the magnitude of two point charges and gravitational force is directly proportional to the product of two masses.

(vii) The force on a charge q₁ exerted by a point charge q₂ is given by \(\vec { F } \)₁₂ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)₁₂ Here \(\hat{r} \)₂₁ is sthe unit vector from charge q₂ to q₁.

Therefore, the electrostatic force obeys Newton’s third law.

(viii) The expression for Coulomb force is true only for point charges. But the point charge is an ideal concept. However we can apply Coulomb’s law for two charged objects whose sizes are very much smaller than the distance between them. In fact, Coulomb discovered his law by considering the charged spheres in the torsion balance as point charges. The distance between the two charged spheres is much greater than the radii of the spheres.

Question 3.
Define ‘Electric field’ and discuss its various aspects.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by

Here \(\hat{r} \) is the unit vector pointing from q to the point of interest P. The electric field is a vector quantity and its SI unit is Newton per Coulomb (NC^-1).

Important aspects of Electric field:
(i) If the charge q is positive then the electric field points away from the source charge and if q is negative, the electric field points towards the source charge q.

(ii) If the electric field at a point P is \(\vec { E } \), then the force experienced by the test charge qo placed at the point P is \(\vec { F } \) = q₀ \(\vec { E } \). This is Coulomb’s law in terms of electric field. This is shown in the below Figure.

(iii) The equation implies that the electric field is independent of the test charge q0 and it depends only on the source charge q.

(iv) Since the electric field is a vector quantity, at every point in space, this field has unique direction and magnitude as shown in Figures (a) and (b). From equation, we can infer that as distance increases, the electric field decreases in magnitude. Note that in Figures (a) and (b) the length of the electric field vector is shown for three different points. The strength or magnitude of the electric field at point P is stronger than at the point Q and R because the point P is closer to the source charge.

(v) In the definition of electric field, it is assumed that the test charge q₀ is taken sufficiently small, so that bringing this test charge will not move the source charge. In other words, the test charge is made sufficiently small such that it will not modify the electric field of the source charge.

(vi) The expression is valid only for point charges. For continuous and finite size charge distributions, integration techniques must be used. However, this expression can be used as an approximation for a finite-sized charge if the test point is very far away from the finite sized source charge.

(vii) There are two kinds of the electric field: uniform (constant) electric field and non-uniform electric field. Uniform electric field will have the same direction and constant magnitude at all points in space. Non-uniform electric field will have different directions or different magnitudes or both at different points in space. The electric field created by a point charge is basically a non-uniform electric field. This non-uniformity arises, both in direction and magnitude, with the direction being radially outward (or inward) and the magnitude changes as distance increases.

Question 4.
How do we determine the electric field due to a continuous charge distribution? Explain. Electric field due to continuous charge distribution.
Answer:
The electric charge is quantized microscopically. The expressions of Coulomb’s Law, superposition principle force and electric field are applicable to only point charges. While dealing with the electric field due to a charged sphere or a charged wire etc., it is very difficult to look at individual charges in these charged bodies. Therefore, it is assumed that charge is distributed continuously on the charged bodies and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus.

Consider the following charged object of irregular shape. The entire charged object is divided into a large number of charge elements ∆q₁, ∆q₂, ∆q₃ ……..∆q_n,…… and each charge element Δq is taken as a point charge.
The electric field at a point P due to a charged object is approximately given by the sum of the fields at P due to all such charge elements.

Here ∆ qi is the i^th charge element, r_ip is the distance of the point P frome the i^th charge element, r_ip is the unit vector from ith charge element to the pont P.
However the equation is only an approximation. To incorporate the continuous distribution of charge, we take the limit ∆q → 0(= dq). In this limit, the summation in the equation becomes an integration and takes the following form

Here r is the distance of the point P from the infinitesimal charge dq and \(\hat{r} \) is the unit vector from dq to point P. Even though the electric field for a continuous charge distribution is difficult to evaluate, the force experienced by some test charge q in this electric field is still given by \(\vec { F } \) = q\(\vec { E } \).

(a) Line charge distribution: If the charge Q is uniformly distributed along the wire of length L, then linear charge density (charge per unit length) is λ = \(\frac { Q }{ L }\). Its unit is colomb per meter (Cm^-1). The charge present in the infinitestimal length dl is dq = λdl.

The electric field due to the line of total charge Q is given by

(b) Surface charge distribution: If the charge Q is uniformly distributed on a surface of area A, then surface charge density (charge per unit area) is σ = \(\frac { Q }{ A }\). Its unit is coulomb per square meter (C m^-2). The charge present in the infinitesimal area dA is dq = σdA. The electric field due to a of total charge Q is given by

(c) Volume charge distribution: If the charge Q is uniformly distributed in a volume V, then volume charge density (charge per unit volume) is given by ρ = \(\frac { Q }{ V }\). Its unit is coulomb per cubic meter (Cm^-3) The charge present in the infinitesimal volume element dV is dq = ρdV. The electric field due to a volume of total charge Q is given by

Question 5.
Calculate the electric field due to a dipole on its axial line and equatorial plane.
Case (I) :
Electric field due to an electric dipole at points on the axial line. Consider an electric dipole placed on the x-ax is as shown in figure. A point C is located at a distance of r from the midpoint O of the dipole along the axial line. Axial line

The electric field at a point C due to +q is

Since the electric dipole moment vector \(\vec { P } \) is from -q to +q and is directed along BC, the above equation is rewritten as

where \(\hat{p} \) is the electric dipole moment unit vector from -q to +q. The electric field at a point C due to -q is

Since +q is located closer to the point C than -q, \(\vec { E } \) _. \(\vec { E } \) + us stronger than \(\vec { E } \) _. Therefore, the length of the E + vector is drawn large than that of \(\vec { E } \) _vector.
The total electric field at point C is calculated using the superposition principle of the electric field.

Note that the total electric field is along \(\vec { E } \)₊, since +q is closer to C than -q.

The direction of \(\vec { E } \)_tot is shown in Figure
If the point C is very far away from the dipole then (r >> a). Under this limit the term(r² – a²)² ≈ r⁴ Substituting this into equation, we get

If the point C is chosen on the left side of the dipole, the total electric field is still in the

Case (II) :
Electric field due to an electric dipole at a point on the equatorial plane
Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure. Since the point C is equi-distant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of E+ is along BC and the direction of E is along CA. E+ and E_ are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it.

The perpendicular components \(\left|\vec{E}_{+}\right|\) sin θ and \(\left|\vec{E}_{-}\right|\) sin θ are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the paralle component of \(\vec { E } \)₊ and \(\vec { E } \)_– and its direction is along \(\hat{-p} \).


The magnitudes \(\vec { E } \)₊ and \(\vec { E } \)_– are the same and are given by

By substituting equation (1) into equation (2), we get

At very large distances (r >> a), the equation becomes
\(\vec { E } \)_tot \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{ r }^{3}}\) (r >>) …… (4)

Question 6.
Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
Torque experienced by an electric dipole in the uniform electric field:
Consider an electric dipole of dipole moment \(\vec { p } \) placed in a uniform electric field E whose field lines are equally spaced and point in the same direction. The charge +q will experience a force q\(\vec { E } \) in the direction of the field and charge -q will experience a force -q\(\vec { E } \) in a direction opposite to the field.

Since the external field \(\vec { E } \) is uniform, the total force acting on the dipole is zero. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole. (Note that electric field lines of a uniform field are equally spaced and point in the same direction). The total torque on the dipole about the point O
\(\vec { τ } \) = \(\overrightarrow{\mathrm{OA}}\) × (-q\(\vec { E } \)) + \(\overrightarrow{\mathrm{OB}}\) × q\(\vec { E } \)
Using right-hand corkscrew rule, it is found that total torque is perpendicular to the plane of the paper and is directed into it.

The magnitude of the total torque
\(\vec { τ } \) = \(|\overrightarrow{\mathrm{OA}}|\)(-q\(\vec { E } \)) sin θ + \(|\overrightarrow{\mathrm{OB}}|\) \(|q \overrightarrow{\mathrm{E}}|\) sin θ
where θ is the angle made by \(\vec { P } \) with \(\vec { E } \). Since p = 2aq, the torque is written in terms of the vector product as
\(\vec { τ } \) = \(\vec { p } \) x \(\vec { E } \)
The magnitude of this torque is τ = pE sin θ and is maximum Torque on dipole
when θ =90°.
This torque tends to rotate the dipole and align it with the electric field \(\vec { E } \). Once \(\vec { E } \) is aligned with \(\vec { E } \), the total torque on the dipole becomes zero.

Question 7.
Derive an expression for electrostatic potential due to a point charge.
Answer:
Electric potential due to a point charge:
Consider a positive charge q kept fixed at the origin. Let P be a point at distance r from the charge q.
The electric potential at the point P is


Electric field due to positive point charge q is

The infinitesimal displacement vector, d\(\vec { r } \) = dr\(\hat{r} \) and using \(\hat{r} \) . \(\hat{r} \) = 1, we have

After the integration,

Hence the electric potential due to a point charge q at a distance r is
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{ r }\) …… (2)
Important points (If asked in exam)
(i) If the source charge q is positive, V > 0. If q is negative, then V is negative and equal to
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{ r }\)
(ii) The description of motion of objects using the concept of potential or potential energy is simpler than that using the concept of field.
(i) From expression (2), it is clear that the potential due to positive charge decreases as the distance increases, but for a negative charge the potential increases as the distance is increased. At infinity (r = ∞) electrostatic potential is zero (V = 0).
(iv) The electric potential at a point P due to a collection of charges q₁,q₂,q₃… q_n is equal to sum of the electric potentials due to individual charges.

Where r₁, r₂,r₃,…..r_n are the distances of q₁,q₂,q₃… q_n
respectively from P

Question 8.
Derive an expression for electrostatic potential due to an electric dipole.
Answer:
Electrostatic potential at a point due to an electric dipole:
Consider two equal and opposite charges separated by a small distance 2a. The point P is located at a distance r from the midpoint of the dipole. Let 0 be the angle between the line OP and dipole axis AB.
Let r₁ be the distance of point P from +q and r₁ be the distance of point P from -q.
Potential at P due to charge +q = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }_{1}}\)
Potential at P due to charge -q = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }_{2}}\)
Total Potential at the point P,
V = \(\frac { 1 }{{ 4πε }_{0}}\)q \(\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)\) ….. (1)
Suppose if the point P is far away from the dipole, such that r >> a, then equation can be expressed in terms of r. By the cosine law for triangle BOP,

\(r_{1}^{2}\) = r² + a² – 2ra cos θ = r² \(\left(1+\frac{a^{2}}{r^{2}}-\frac{2 a}{r} \cos \theta\right)\)
Since the point P is very far from dipole, then r >> a. As a result the term \(\frac {{ a }^{ 2 }}{{ r }^{ 2 }}\) is very small and can be neglected. Therefore

since \(\frac { a }{ r }\) << 1, we can use binominal theorem and retain the terms up to first order
\(\frac { 1 }{{ r_{1}} }\) = \(\left(1+\frac{a}{r} \cos \theta\right)\) ……. (2)
Similarly applying the cosine law for triangle AOP,
\(r_{2}^{2}\) = r² + a² – 2ra cos (180 – θ)
Since cos (180 – θ) = cos θ we get
\(r_{2}^{2}\) = r² + a² + 2ra cos θ
Neglecting the term \(\frac {{ a }^{ 2 }}{{ r }^{ 2 }}\) (because r >> a)
\(r_{2}^{2}\) = r² \(\left(1+\frac{2 a \cos \theta}{r}\right)\) (or) r₂ = r \(\left(1+\frac{2 a \cos \theta}{r}\right)^{\frac{1}{2}}\)
Using Binomial theorem, we get
\(\frac { 1 }{{ r_{2}} }\) = \(\frac { 1 }{ r }\) \(\left(1-a \frac{\cos \theta}{r}\right)\)
Substituting equations (3) and (2) in equation (1)

But the electric dipole moment p = 2qa and we get,
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\left(\frac{p \cos \theta}{r^{2}}\right)\)
Now we can write p cos θ = \(\vec { P } \), \(\hat{r} \) where \(\hat{r} \) is the unit vector from the point O to point P. Hence the electric potential at a point P due to an electric dipole is given by
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{\vec{p} \cdot \hat{r}}{r^{2}}\) (r >> a) ….. (4)
Equation (4) is valid for distances very large compared to the size of the dipole. But for a point dipole, the equation (4) is valid for any distance.
Special cases:

Case (I):
If the point P lies on the axial line of the dipole on the side of +q, then θ = 0. Then the electric potential becomes
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{r}^{ 2 }}\)

Case (II):
If the point P lies on the axial line of the dipole on the side of -q, then θ = 180°, then
V = – \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{r}^{ 2 }}\)

Case (III):
If the point P lies on the equatorial line of the dipole, then θ = 90°. Hence, V = 0.

Question 9.
Obtain an expression for potential energy due to a collection of three point charges which are separated by finite distances.
Answer:
Electrostatic potential energy for collection of point charges:
The electric potential at a point at a distance r from point charge ql is given by
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 1 }}{r}\) …… (1)
This potential V is the work done to bring a unit positive charge from infinity to the point. Now if the charge q₂ is brought from infinity to that point at a distance r from qp the work done is the product of q₂ and the electric potential at that point. Thus we have W = q₂V …… (2)
This work done is stored as the electrostatic potential energy U of a system of charges q₁ and q₂ separated by a distance r. Thus we have
U = q₂ V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r}\) …… (3)
The electrostatic potential energy depends only on the distance between the two point charges. In fact, the expression (3) is derived by assuming that q₁ is fixed and q₂ is brought from infinity. The equation (3) holds true when q₂ is fixed and q₁ is brought from infinity or both q₂and q₂ are simultaneously brought from infinity to a distance r between them.
Three charges are arranged in the following configuration as shown in Figure.

To calculate the total electrostatic potential energy, we use the following procedure. We bring all the charges one by one and arrange them according to the configuration.
(i) Bringing a charge q₁ from infinity to the point A requires no work, because there are no other charges already present in the vicinity of charge q₁

(ii) To bring the second charge q₂ to the point B, work must be done against the electric field created by the charge q₁ So the work done on the charge q₁ is W = q₂V_1B. Here V_1B is the electrostatic potential due to the charge q₁ at point B.
U = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{{r}_{12}}\) ….. (4)
Note that the expression is same when q₂ is brought first and then q₁ later.

(iii) Similarly to bring the charge q3 to the point C, work has to be done against the total electric field due to both charges q₁ and q₂. So the work done to bring the charge q₃ is = q₃ (V_1C + V_2C). Here V_1C is the electrostatic potential due to charge q₁ at point C and V_2C is the electrostatic potential due to charge q₂ at point C. The electrostatic potential is

(iv) Adding equations (4) and (5), the total electrostatic potential energy for the system of three charges q₁,q₂ and q₃ is

Note that this stored potential energy U is equal to the total external work done to assemble the three charges at the given locations. The expression (6) is same if the charges are brought to their positions in any other order. Since the Coulomb force is a conservative force, the electrostatic potential energy is independent of the manner in which the configuration of charges is arrived at.

Question 10.
Derive an expression for electrostatic potential energy of the dipole in a uniform electric field.
Answer:
Electrostatic potential energy of a dipole in a uniform electric field:
Consider a dipole placed a torque when kept in an uniform electric field \(\vec { E } \). A dipole experiences a torque when kept in an uniform electric field \(\vec { E } \). This torque rotates the dipole to align it with the direction of the electric field. To rotate the dipole (at constant angular velocity) from its initial angle θ’ to another angle θ against the torque exerted by the electric field, an equal and opposite external torque must be applied on the dipole.

The work done by the external torque to rotate the dipole from angle θ’ to θ at constant angular velocity is

Since τ_ext is equal and opposite to τ_E = \(\vec { P } \) x \(\vec { E } \), we have
\(\left|\overrightarrow{\mathrm{r}}_{\mathrm{ext}}\right|\) = \(\left|\overrightarrow{\mathrm{r}}_{\mathrm{E}}\right|\)= \(|\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{E}}|\) …. (2)
Substituting equation (2) in equation (1) We get,

This work done is equal to the potential energy difference between the angular positions θ and θ’.
U(θ) – (Uθ’) = AU = -pE cos θ +PE cos θ’.
If the initial angle is = θ’ = 90° and is taken as reference point, then U(θ’) + pE cos θ’ = θ.
The potential energy stored in the system of dipole kept in the uniform electric field is given by El = -pE cos θ = –\(\vec { P } \) . \(\vec { E } \) ….. (3)
In addition to p and E, the potential energy also depends on the orientation θ of the electric dipole with respect to the external electric field.
The potential energy is maximum when the dipole is aligned anti-parallel (θ = π) to the external electric field and minimum when the dipole is aligned parallel (θ = 0) to the external electric field.

Question 11.
Obtain Gauss law from Coulomb’s law.
Answer:
Gauss law: Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux Φ_E through the closed surface is
Φ_E = \(\oint { \vec { E } } \) .d \(\vec { A } \) = \(\frac{\mathrm{Q}_{\mathrm{end}}}{\varepsilon_{0}}\)
A positive point charge Q is surrounded by an imaginary sphere of radius r as shown in figure. We can calculate the total electric flux through the closed surface of the sphere using the equation.

Φ_E = \(\oint { \vec { E } } \) .d \(\vec { A } \) = \(\oint { EdA } \) cos θ …… (1)
The electric field of the point charge is directed radially outward at all points on the surface of the sphere. Therefore, the direction of the area element d \(\vec { A } \) is along the electric field \(\vec { E } \) and θ = 0°.
Φ_E = \(\oint { EdA } \) since cos 0° = 1 ….. (2)
E is uniform on the surface of the sphere,
Φ_E = \(\oint { EdA } \) ….. (3)
Substituting for
\(\oint { dA } \) = 4π² and E = \(\frac { 1 }{{ 4πε }_{0}}\) Q in equation 3, we get
Φ_E = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }^{2}}\) × 4π² = 4π \(\frac { 1 }{{ 4πε }_{0}}\) = \(\frac { q }{{ ε }_{0}}\) ……. (4)
The equation (4) is called as Gauss’s law. The remarkable point about this result is that the equation (4) is equally true for any arbitrary shaped surface which encloses the charge Q.

Question 12.
Obtain the expression for electric field due to an infinitely long charged wire.
Answer:
Electric field due to an infinitely long charged wire:
Consider an infinitely long straight wire having uniform linear charge density λ. Let P be a point located at a perpendicular distance r from the wire. The electric field at the point P can be found using Gauss law. We choose two small charge elements A₁ and A₁ on the wire which are at equal distances from the point P.

The resultant electric field due to these two charge elements points radially away from the charged wire and the magnitude of electric field is same at all points on the circle of radius r. From this property, we can infer that the charged wire possesses a cylindrical symmetry.

Let us choose a cylindrical Gaussian surface of radius r and length L. The total electric flux in this closed surface is

It is seen that for the curved surface, \(\vec { E } \) is parallel to \(\vec { A } \) and \(\vec { E } \).d \(\vec { A } \) = EdA. For the top and bottom surface, \(\vec { E } \) is perpendicular to \(\vec { A } \) and \(\vec { E } \).d\(\vec { A } \) = 0
Substituting these values in the equation (2) and applying Gauss law


Since the magnitude of the electric field for the entire curved surface is constant, E is taken out of the integration and Q_encl is given by Q_encl = λL.

Here,

dA = total area of the curved surface = 2πrL. Substituting this in
equation (4), We get

The electric field due to the infinite charged wire depends on \(\frac { 1 }{ r }\) rather than \(\frac { 1 }{{r}^{ 2 }}\) for a point charge.
Equation (6) indicates that the electric field is always along the perpendicular direction (\(\hat{r} \) ) to wire. In fact, if λ > 0 then E points perpendicular outward (\(\hat{r} \) ) from the wire and if λ < 0, then E points perpendicular inward (- \(\hat{r} \) ).

Question 13.
Obtain the expression for electric field due to an charged infinite plane sheet.
Answer:
Electric field due to charged infinite plane sheet: Consider an infinite plane sheet of charges with uniform surface charge density o. Let P be a point at a distance of r from the sheet. Since the plane is infinitely large, the electric field should be same at all points equidistant from the plane and radially directed at all points. A cylindrical shaped Gaussian surface of length 2r and area A of the flat surfaces is chosen such that the infinite plane sheet passes perpendicularly through the middle part of the Gaussian surface.

Applying Gauss law for this cylindrical surface,

The electric field is perpendicular to the are a element at all points on the curved surface and is parallel to the surface areas at P and P’. Then,

Since the magnitude of the electric field at these two equal surfaces is uniform, E is taken out of the integration and Q_encl is given by Q_encl = σA, we get

The total area of surface either at P or P’

Hence 2EA = \(\frac { σA }{{ ε }_{0}}\) or E = \(\frac { σ }{{ 2ε }_{0}}\) …… (3)
In vector from, E = \(\frac { σ }{{ 2ε }_{0}}\) \(\hat{n} \) ….. (4)
Hence \(\hat{n} \) is the outward unit vector normal to the plane. Note that the electric field due to an infinite plane sheet of charge depends on the surface charge density and is independent of the distance r.

The electric field will be the same at any point farther away from the charged plane. Equation (4) implies that if o > 0 the electric field at any point P is outward perpendicular n to the plane and if σ < 0 the electric field points inward perpendicularly (\(\hat{n} \) ) to the plane. For a finite charged plane sheet, equation (4) is approximately true only in the middle region of the plane and at points far away from both ends.

Question 14.
Obtain the expression for electric field due to an uniformly charged spherical shell.
Answer:
Electric field due to a uniformly charged spherical shell:
Consider a uniformly charged spherical shell of radius R and total charge Q. The electric field at points outside and inside the sphere is found using Gauss law.

Case (a):
At a point outside the shell (r > R): Let us choose a point P outside the shell at a distance r from the center as shown in figure (a). The charge is uniformly distributed on the surface of the sphere (spherical symmetry). Hence the electric field must point radially outward if Q > 0 and point radially inward if Q < 0. So we choose a spherical Gaussian surface of radius r and the total charge enclosed by this Gaussian surface is Q. Applying Gauss law,
\(\oint { \vec { E } } .d\vec { A } \) = \(\frac { Q }{{ ε }_{0}}\) …….(1)
The electric field \(\vec { E } \) and d\(\vec { A } \) point in the same direction (outward normal) at all the points on the Gaussian surface. The magnitude of \(\vec { E } \) is also the same at all points due to the spherical symmetry of the charge distribution.

But

dA = total area of Gaussian surface = 4πr². Substituting this value in equation (2).

The electric field is radially outward if Q > 0 and radially inward if Q < 0. From equation (3), we infer that the electric field at a point outside the shell will be same as if the entire charge Q is concentrated at the center of the spherical shell. (A similar result is observed in gravitation, for gravitational force due to a spherical shell with mass M)

Case (b):
At a point on the surface of the spherical shell (r = R): The electrical field at points on the spherical shell (r = R) is given by
\(\vec { E } \) = \(\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\) \(\hat{r} \) …… (4)

Case (c):
At a point inside the spherical shell (r < R): Consider a point P inside the shell at a distance r from the center. A Gaussian sphere of radius r is constructed as shown in the figure (b). Applying Gauss law

Since Gaussian surface encloses no charge, So Q = 0. The equation (5) becomes E = 0 (r < R) …(6)
The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.

Question 15.
Discuss the various properties of conductors in electrostatic equilibrium.
Answer:
Properties of conductors in electrostatic equilibrium:
(i) The electric field is zero everywhere inside the conductor. This is true regardless of whether the conductor is solid or hollow. This is an experimental fact. Suppose the electric field is not zero inside the metal, then there will be a force on the mobile charge carriers due to this electric field.

As a result, there will be a net motion of the mobile charges, which contradicts the conductors being in electrostatic equilibrium. Thus the electric field is zero everywhere inside – the conductor. We can also understand this fact by applying an external uniform electric field on the conductor.

Before applying the external electric field, the free electrons in the conductor are uniformly distributed in the conductor. When an electric field is applied, the free electrons accelerate to the left causing the left plate to be negatively charged and the right plate to be positively charged.

Due to this realignment of free electrons, there will be an internal electric field created inside the conductor which increases until it nullifies the external electric field. Once the external electric field is nullified the conductor is said to be in electrostatic equilibrium. The time taken by a conductor to reach electrostatic equilibrium is in the order of 10^-6s, which can be taken as almost instantaneous.

(ii) There is no net charge inside the conductors. The charges must reside only on the surface of the conductors. We can prove this property using Gauss law. Consider an arbitarily shaped conductor. A Gaussian surface is drawn inside the conductor such that it is very close to the surface of the conductor.

Since the electric field is zero everywhere inside the conductor, the net electric flux is also zero over this Gaussian surface. From Gauss’s law, this implies that there is no net charge inside the conductor. Even if some charge is introduced inside the conductor, it immediately reaches the surface of the conductor.

(iii) The electric field outside the conductor is perpendicular to the surface of the conductor and has a magnitude of \(\frac { σ }{{ ε }{0}}\) where a is the surface charge density at that point. If the electric field has components parallel to the surface of the conductor, then free electrons on the surface of the conductor would experience acceleration. This means that the conductor is not in equilibrium. Therefore at electrostatic equilibrium, the electric field must be perpendicular to the surface of the conductor.

We now prove that the electric field has magnitude \(\frac { σ }{{ ε }{0}}\) just outside the conductor’s surface. Consider a small cylindrical Gaussian surface. One half of this cylinder is embedded inside the conductor. Since electric field is normal to the surface of the conductor, the curved part of the cylinder has zero electric flux. Also inside the conductor, the electric field is zero. Hence the bottom flat part of the Gaussian surface has no electric flux. Therefore the top flat surface alone contributes to the electric flux. The electric field is parallel to the area vector and the total charge inside the surface is σA. By applying Gauss’s law,
EA = \(\frac { σA }{{ ε }{0}}\)
In vector from, \(\vec { E } \) = \(\frac { σ }{{ ε }{0}}\) \(\hat{n} \)
Here n represents the unit vector outward normal to the surface of the conductor. Suppose σ < 0, then electric field points inward perpendicular to the surface.

(iv) The electrostatic potential has the same value on the surface and inside of the conductor. We know that the conductor has no parallel electric component on the surface which means that charges can be moved on the surface without doing any work. This is possible only if the electrostatic potential is constant at all points on the surface and there is no potential difference between any two points on the surface. Since the electric field is zero inside the conductor, the potential is the same as the surface of the conductor. Thus at electrostatic equilibrium, the conductor is always at equipotential.

Question 16.
Explain the process of electrostatic induction.
Answer:
Whenever a charged rod is touched by another conductor, charges start to flow from charged rod to the conductor. This type of charging without
actual contact is called electrostatic induction:
(i) Consider an uncharged (neutral) conducting sphere at rest on an insulating stand. Suppose a negatively charged rod is brought near the conductor without touching it, as shown in figure (a). The negative charge of the rod repels the electrons in the conductor to the opposite side.

Various steps in electrostatic induction
As a result, positive charges are induced near the region of the charged rod while negative charges on the farther side. Before introducing the charged rod, the free electrons were distributed uniformly on the surface
of the conductor and the net charge is zero. Once the charged rod is brought near the conductor, the distribution is no longer uniform with more electrons located on the farther side of the rod and positive charges are located closer to the rod. But the total charge is zero.

(ii) Now the conducting sphere is connected to the ground through a conducting wire. This is called grounding. Since the ground can always receive any amount of electrons, grounding removes the electron from the conducting sphere. Note that positive charges will not flow to the ground because they are attracted by the negative charges of the rod (figure (b)).

(iii) When the grounding wire is removed from the conductor, the positive charges remain near the charged rod (figure (c)).

(iv) Now the charged rod is taken away from the conductor. As soon as the charged rod is removed, the positive charge gets distributed uniformly on the surface of the conductor (figure (d)). By this process, the neutral conducting sphere becomes positively charged.

Question 17.
Explain dielectrics in detail and how an electric field is induced inside a dielectric.
Answer:
Induced Electric field inside the dielectric:
When an external electric field is applied on a conductor, the charges are aligned in such a way that an internal electric field is created which cancels the external electric field. But in the case of a dielectric, which has no free electrons, the external electric field only realigns the charges so that an internal electric field is produced.

The magnitude of the internal electric field is smaller than that of external electric field. Therefore the net electric field inside the dielectric is not zero but is parallel to an external electric field with magnitude less than that of the external electric field. For example, let us consider a rectangular dielectric slab placed between two oppositely charged plates (capacitor) as shown in the figure.

The uniform electric field between the plates Induced electric field lines inside the dielectric acts as an external electric field \(\vec { E } \)_ext which polarizes the dielectric placed between plates. The positive charges are induced on one side surface and negative charges are induced on the other side of surface But inside the dielectric, the net charge is zero even in a small volume. So the dielectric in the external field is equivalent to two oppositely charged sheets with the surface charge densities +σb and -σb. These charges are called bound charges. They are not free to move like free electrons in conductors. This is shown in the figure.

(a) Balloon sticks to the wall
(b) Polarisation of wall due to the electric field created by the balloon
For example, the charged balloon after rubbing sticks onto a wall. The reason is that the negatively charged balloon is brought near the wall, it polarizes opposite charges on the surface of the wall, which attracts the balloon.

Question 18.
Obtain the expression for capacitance for a parallel plate capacitor.
Answer:
Capacitance of a parallel plate capacitor:
Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d. The electric field between two infinite parallel plates is uniform and is given by E = \(\frac { σ }{{ ε }{0}}\) where σ is the surface charge density on the plates σ = \(\frac { Q }{ A }\) .If the separation distance d is very much smaller than the size of the plate (d² << A), then the above result is used even for finite-sized
parallel plate capacitor.

Capacitance of a parallel plate capacitor
The electric field between the plates is
E = \(\frac { Q }{{ Aε }{0}}\) ….. (1)
Since the electric field is unifonn, the electric potential between the plates having separation d is given by
V = Ed = \(\frac { Qd }{{ Aε }{0}}\) ….. (2)
Therefore the capacitance of the capacitor is given by
C = \(\frac { Q }{ V }\) = \(\frac{\mathrm{Q}}{\left(\frac{\mathrm{Q} d}{\mathrm{A} \varepsilon_{0}}\right)}\) = \(\frac{\varepsilon_{0} \mathrm{A}}{d}\) ….. (3)
From equation (3), it is evident that capacitance is directly
proportional to the area of cross section and is inversely proportional to the distance between the plates. This can be understood from the following.
(i) If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.

(ii) If the distance d between the two plates is reduced, the potential difference between the plates (V = Ed) decreases with E constant.

Question 19.
Obtain the expression for energy stored in the parallel plate capacitor.
Answer:
Energy stored in the capacitor:
Capacitor not only stores the charge but also it stores energy. When a battery is connected to the capacitor, electrons of total charge -Q are transferred from one plate to the other plate. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor. To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
dW = VdQ ….. (1)
Where V = \(\frac { Q }{ C }\)
The total work done to charge a capacitor is

This work done is stored as electrostatic potential energy (U_E) in the capacitor.
U_E = \(\frac {{ Q }^{2}}{ 2C }\) = \(\frac { 1 }{ 2 }\) CV² (∴ Q = CV) ….. (3)
where Q = CV is used. This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor. But where is this energy stored in the capacitor? To understand this question, the equation (3) is rewritten as follows using the results
C = \(\frac{\varepsilon_{0} \mathrm{A}}{d}\) and V = Ed
U_E = \(\frac { 1 }{ 2 }\) \(\left(\frac{\varepsilon_{0} \mathrm{A}}{d}\right)\) (Ed)² = \(\frac { 1 }{ 2 }\) ε₀(Ad)² …… (4)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy density \(\overline { Volume } \). Frome equation (4) we get
u_E = \(\frac { 1 }{ 2 }\) ε₀E²
From equation (5), we infer that the energy is stored in the electric field existing between the plates of the capacitor. Once the capacitor is allowed to discharge, the energy is retrieved.

Question 20.
Explain in detail the effect of a dielectric placed in a parallel plate capacitor.
Answer:
(i) When the capacitor is disconnected from the battery:
Consider a capacitor with two parallel plates each of cross-sectional area A and are separated by a distance d. The capacitor is charged by a battery of voltage V₀ and the charge stored is Q₀. The capacitance of the capacitor without the dielectric is
C₀ = \(\frac {{ Q }_{0}}{{ V }_{0}}\) ….. (1)
The battery is then disconnected from the capacitor and the dielectric is inserted between the plates. The introduction of dielectric between the plates will decrease the electric field. Experimentally it is found that the modified electric field is given by

(a) Capacitor is charged with a battery
(b) Dielectric is inserted after the battery is disconnected
E = \(\frac {{ E }_{0}}{{ ε }_{r}}\) …… (2)
Here E₀ is the electric field inside the capacitors when there is no dielectric and ε_r is the relative permeability of the dielectric or simply known as the dielectric constant. Since ε_r > 1, the electric field E < E₀. As a result, the electrostatic potential difference between the plates (V = Ed) is also reduced. But at the same time, the charge Q₀ will remain constant once the battery is disconnected. Hence the new potential difference is
V = Ed = \(\frac {{ E }_{0}}{{ ε }_{r}}\)d = \(\frac {{ V }_{0}}{{ ε }_{r}}\) ….. (3)
We know that capacitance is inversely proportional to the potential difference. Therefore as V decreases, C increases. Thus new capacitance in the presence of a dielectric is
C = \(\frac {{ Q }_{0}}{ V }\) = ε_r \(\frac {{ Q }_{0}}{{ V }_{0}}\) = ε_r C₀ …… (4)
Since ε_r > 1, we have C > C₀. Thus insertion of the dielectric constant ε_r increases the capacitance. Using equation,
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } \)
C = \(\frac{\varepsilon_{r} \varepsilon_{o} A}{d}\) = \(\frac { εA }{ d }\) …… (5)
where ε = ε_rε₀ is the permittivity of the dielectric medium. The energy stored in the capacitor before the insertion of a dielectric is given by U₀ = \(\frac { 1 }{ 2 }\) \(\frac{\mathrm{Q}_{0}^{2}}{\mathrm{C}_{0}}\) ….. (6)
After the dielectric is inserted, the charge Q₀ remains constant but the capacitance is increased. As a result, the stored energy is decreased.

Since ε_r> 1 we get U < U₀. There is a decrease in energy because, when the dielectric is inserted, the capacitor spends some energy in pulling the dielectric inside.

(ii) When the battery remains connected to the capacitor: Let us now consider what happens when the battery of voltage V₀ remains connected to the capacitor when the dielectric is inserted into the capacitor.
The potential difference V₀ across the plates remains constant. But it is found experimentally (first shown by Faraday) that when dielectric is inserted, the charge stored in the capacitor is increased by a factor ε_r.

(a) Capacitor is charged through a battery
(b) Dielectric is inserted when the battery is connected.
Q = ε_rQ₀ ….. (1)
Due to this increased charge, the capacitance is also increased. The new capacitance is
C = \(\frac {{ Q }_{0}}{ V }\) = ε_r \(\frac {{ Q }_{0}}{{ V }_{0}}\) = ε_r C₀ …… (2)
However the reason for the increase in capacitance in this case when the battery remains connected is different from the case when the battery is disconnected before introducing the dielectric.
Now, C₀ = \(\frac {{ ε }{_0}A}{ d }\) and, C = \(\frac { εA }{ d }\) …… (3)
U₀ = \(\frac { 1 }{ 2 }\) C₀ \({ V }_{ 0 }^{ 2 }\) ….. (4)
Note that here we have not used the expression
U₀ = \(\frac { 1 }{ 2 }\)\({{ V }_{ 0 }^{ 2 }}{{C}_{0}}\)
because here, both charge and capacitance are changed, whereas in equation 4, V₀ remains constant. After the dielectric is inserted, the capacitance is increased; hence the stored energy is also increased.
U = \(\frac { 1 }{ 2 }\) \({ CV }_{ 0 }^{ 2 }\) = \(\frac { 1 }{ 2 }\) ε_r \({ CV }_{ 0 }^{ 2 }\) = ε_r U₀
Since e_r > 1 we have U > U₀
It may be noted here that since voltage between the capacitor V₀ is constant, the electric field between the plates also remains constant.

Question 21.
Derive the expression for resultant capacitance, when capacitors are connected in series and in parallel.
Answer:
Capacitor in series and parallel:
(i) Capacitor in series:
Consider three capacitors of capacitance C₁, C₂ and C₃ connected in series with a battery of voltage V as shown in the figure (a).
As soon as the battery is connected to the capacitors in series, the electrons of charge -Q are transferred from negative terminal to the right plate of C₃which pushes the electrons of same amount -Q from left plate of C₃ to the right plate of C₂ due to electrostatic induction. Similarly, the left plate of C₂ pushes the charges of Q to the right plate of which induces the positive charge +Q on the left plate of C₁ At the same time, electrons of charge -Q are transferred from left plate of C₁ to positive terminal of the battery.

By these processes, each capacitor stores the same amount of charge Q. The capacitances of the capacitors are in general different, so that the voltage across each capacitor is also different and are denoted as V₁, V₂ and V₃ respectively.
The total voltage across each capacitor must be equal to the voltage of the a battery.
V = V₁ + V₂ + V₃ ….. (1)
Since Q = CV, we have V = \(\frac { Q }{{ C }_{1}}\) + \(\frac { Q }{{ C }_{2}}\) + \(\frac { Q }{{ C }_{3}}\)
Q = \(\left( \frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \right) \) ….. (2)
If three capacitors in series are considered to form an equivalent single capacitor Cs shown in figure (b), then we have V = \(\frac { Q }{{ C }_{s}}\)
Substituting this expression into equation (2) we get
V = \(\frac { Q }{{ C }_{s}}\) = Q\(\left( \frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \right) \)
\(\frac { 1 }{{ C }_{s}}\) = \(\frac { 1 }{{ C }_{1}}\) + \(\frac { 1 }{{ C }_{2}}\) + \(\frac { 1 }{{ C }_{3}}\) ….. (3)
Thus, the inverse of the equivalent capacitance C_s of three capacitors connected in series is equal to the sum of the inverses of each capacitance. This equivalent capacitance C_s is always less than the smallest individual capacitance in the series.

(ii) Capacitance in parallel:
Consider three capacitors of capacitance C₁,C₂ and C₃ connected in parallel with a battery of voltage V as shown in figure (a).

Since corresponding sides of the capacitors are connected to the same positive and negative terminals of the battery, the voltage across each capacitor is equal to the battery’s voltage. Since capacitance of the capacitors is different, the charge stored in each capacitor is not the same. Let the charge stored in the three capacitors be Q₁,Q₂, and Q₂ respectively. According to the law of conservation of total charge, the sum of these three charges is equal to the charge Q transferred by the battery,
Q = Q₁ + Q₂ + Q₃ ….. (1)
Now, since Q = CV, we have
Q = C₁V + C₂ V + C₃ V ….. (2)
If these three capacitors are considered to form a single capacitance C_P which stores the total charge Q as shown in the figure (b), then we can write Q = CPV. Substituting this in equation (2), we get
C_p V = C₁ V + C₂ V + C₃ V
C_p = C₁ + C₂ + C₃
Thus, the equivalent capacitance of capacitors connected in parallel is equal to the sum of the individual capacitance. The equivalent capacitance C_p in a parallel connection is always greater than the largest individual capacitance. In a parallel connection, it is equivalent as area of each capacitance adds to give more effective area such that total capacitance increases.

Question 22.
Explain in detail how charges are distributed in a conductor, and the principle behind the lightning conductor.
Answer:
Distribution of charges in a conductor: Consider two conducting spheres A and B of radii r₁ and r₂ respectively connected to each other by a thin conducting wire as shown in the figure. The distance between the spheres is much greater than the radii of either spheres.

If a charge Q is introduced into any one of the spheres, this charge Q is redistributed into both the spheres such that the electrostatic potential is same in both the spheres. They are now uniformly charged and attain electrostatic equilibrium. Let q₁ be the charge residing on the surface of sphere A and q₂ is the charge residing on the surface of sphere B such that Q = q₁ + q₂ The charges are distributed only on the surface and there is no net charge inside the conductor. The electrostatic potential at the surface of the sphere A is given by
V_A = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) …. (1)
The electrostatic potential at the surface of the sphere B is given by
V_B = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) ….. 2)
The surface of the conductor is an equipotential. Since the spheres are connected by the conducting wire, the surfaces of both the spheres together form an equipotential surface. This implies that
V_A = V_B or \(\frac {{ q }_{ 1 }}{{ r }_{ 1 }}\) = \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) ….. (3)
Let us take the charge density on the surface of sphere A is σ₁ and charge density on the surface of sphere B is σ₁. This implies that q₁ = \({ 4\pi r }_{ 1 }^{ 2 }\)σ₁ and q₁ = \({ 4\pi r }_{ 1 }^{ 2 }\)σ₂. Substituting these values into equation (3), we get
σ₁ r₁ = σ₂r₂ ….. (4)
from which we conclude that
σr = constant …. (5)
Thus the surface charge density o is inversely proportional to the radius of the sphere. For a smaller radius, the charge density will be larger and vice versa.

Lightning arrester or lightning conductor:
This is a device used to protect tall buildings from lightning strikes. It works on the principle of action at points or corona discharge. The device consists of a long thick copper rod passing from top of the building to the ground. The upper end of the rod has a sharp spike or a sharp needle.

The lower end of the rod is connected to the copper plate which is buried deep into the ground. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike. Since the induced charge density on thin sharp spke is large, it results in a corona discharge.

This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud. The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the Earth. The lightning arrester does not stop the •lightning; rather it divers the lightning to the ground safety.

Question 23.
Explain in detail the construction and working of a Van de Graaff generator.
Answer:
Principle: Electrostatic induction and action at points.
Construction:
A large hollow spherical conductor is fixed on the insulating stand. A pulley B is mounted at the center of the hollow sphere and another pulley C is fixed at the bottom. A belt made up of insulating materials like silk or rubber runs over both pulleys. The pulley C is driven continuously by the electric motor.

Two comb shaped metallic conductors E and D are fixed near the pulleys. The comb D is maintained at a positive potential of 104 V by a power supply. The upper comb E is connected to the inner side of the hollow metal sphere.

Working:
Due to the high electric field near comb D, air between the belt and comb D gets ionized. The positive charges are pushed towards the belt and negative charges are attracted towards the comb D. The positive charges stick to the belt and move up. When the positive charges reach the comb E, a large amount of negative and positive charges are induced on either side of comb E due to electrostatic induction. As a result, the positive charges are pushed away from the comb E and they reach the outer surface of the sphere. Since the sphere is a conductor, the positive charges are distributed uniformly on the outer surface of the hollow sphere. At the same time, the negative charges nullify the positive charges in the belt due to corona discharge before it passes over the pulley.

When the belt descends, it has almost no net charge. At the bottom, it again gains a large positive charge. The belt goes up and delivers the positive charges to the outer surface of the sphere. This process continues until the outer surface produces the potential difference of the order of 10⁷ which is the limiting value. We cannot store charges beyond this limit since the extra charge starts leaking to the surroundings due to ionization of air. The leakage of charges can be reduced by enclosing the machine in a gas filled steel chamber at very high pressure. Uses: The high voltage produced in this Van de Graaff generator is used to accelerate positive ions (protons and deuterons) for nuclear disintegrations and other applications.

Samacheer Kalvi 12th Physics Electrostatics Numarical Problems

Question 1.
When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.
Solution:
Charge produced in each object q = 50 nC
q = 50 x 10^-9 C
Charge of electron (e) = 1.6 x 10^-9 C
Number of electron transferred, n = \(\frac { q }{ e }\) = \(\frac {{ 50 × 10 }^{-9}}{{ 1.6 × 10 }^{-19}}\)
=31. 25 × 10^-9 × 10¹⁹
n = 31.25 x 10¹⁰ electrons
Ans. n = 31.25 x 10¹⁰ electrons

Question 2.
The total number of electrons in the human body is typically in the order of 10²⁸. Suppose, due to some reason, you and your friend lost 1% of this number of electrons. Calculate the electrostatic force between you and your friend separated at a distance of 1 m. Compare this with your weight. Assume mass of each person is 60 kg and use point charge approximation.
Solution:
Number of electrons in the human body = 10²⁸
Number of electrons in me and my friend after lost of 1% = 10²⁸ x 1%
= 10²⁸ x \(\frac { 1 }{ 100 }\)
n = 10²⁶ electrons
Separation distance d = 1 m
Charge of each person q = 10²⁶ x 1.6 x 10^-19
q = 1.6 x 10⁷ C
Electrostatic force, F = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r^{2}}\) = \(\frac{9 \times 10^{9} \times 1.6 \times 10^{7} \times 1.6 \times 10^{7}}{1^{2}}\)
F = 2.304 x 10²⁴N
Mass of the person, M = 60 kg
Acceleration due to gravity, g = 9.8 ms^-2
Weight (W) = mg
= 60 x 9.8
W = 588 N
Comparison: Electrostatic force is equal to 3.92 x 10²¹ times of weight of the person.

Question 3.
Five identical charges Q are placed equidistant on a semicircle as shown in the figure. Another point charge q is kept at the center of the circle of radius R. Calculate the electrostatic force experienced by the charge q.

Solution:
Force acting on q due to Q₁ and Q₅ are opposite direction, so cancel to each other.
Force acting on q due to Q₃ is F₃ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ qQ }_{3}}{{ R }^{2}}\)
Force acting on q due to Q₂ and Q₄

Resolving in two component method:
(i) Vertical Component:
Q₂ Sin θ and Q₄ Sinθ are equal and opposite direction, so they are cancel to each other.

(ii) Horizontal Component:
Q₂ Sin θ and Q₄ cos θ are equal and same direction, so they can get added.
F₂₄ = F_2q + F_4q = F₂ cos 55° + F₄ cos 45°
F₂₄ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ qQ }_{2}}{{ R }^{2}}\) cos 45° + \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { qQ 4}{{ R }^{2}}\) cos 45°
Resultant net force F

Question 4.
Suppose a charge +q on Earth’s surface and another +q charge is placed on the surface of the Moon, (a) Calculate the value of q required to balance the gravitational attraction between Earth and Moon (b) Suppose the distance between the Moon and Earth is halved, would the charge q change? (Take m_E = 5.9 x 10²⁴ kg, m_M = 7.348 x 10²² kg)
Solution:
Mass of the Earth, M_E = 5.9 x 10²⁴ kg
Mass of the Moon, M_M = 7.348 x 10²² kg
Charge placed on the surface of Earth and Moon = q
(a) Required charge to balance the F_G between Earth and Moon
F_C = F_G (or) \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }^{2}}{{ r }^{2}}\) = \(\frac{\mathrm{G} \mathrm{M}_{\mathrm{E}} \times \mathrm{M}_{\mathrm{M}}}{r^{2}}\)
q² = G × M_E × M_M × 4πε₀ = 320.97 × 10²⁵
q = \(\sqrt { 320.97\times { 10 }^{ 25 } } \) = 5.665 x 10¹³ = 5.67 x 10¹³ C

(b) The distance between Moon and Earth is

so q = 5.67 x 10¹³ C
There is no change.

Question 5.
Draw the free body diagram for the following charges as shown in the figure (a), (b) and (c).

Solution:

Question 6.
Consider an electron travelling with a speed V_Φ and entering into a uniform electric field \(\vec { E } \) which is perpendicular to \(\overrightarrow{\mathrm{V}_{0}}\) as shown in the Figure. Ignoring gravity, obtain the electron’s acceleration, velocity and position as functions of time.
Solution:

Speed of an electron = V₀
Uniform electric field = \(\vec { E } \)
(а) Electron’s acceleration:
Force on electron due to uniform electric field, F = Ee
Downward acceleration of electron due to electric field, a = \(\frac { F }{ m }\) = – \(\frac { eE }{ M }\)
Vector from, \(\vec { a } \) = – \(\frac { eE }{ M }\) \(\hat{j} \)

(b) Electron’s velocity:
Speed of electron in horizontal direction, u = V₀ From the equation of motion, V = u + at
V = V₀ \(\frac { eE }{ M }\) t
Vector from \(\vec { V } \) = V₀ \(\hat{j} \) – \(\frac { eE }{ M }\) t \(\hat{j} \)

(c) Electron’s position:
Position of electron, s = r
From equation of motion, r = V₀ t + \(\frac { 1 }{ 2 }\) \(\left(-\frac{e \mathrm{E}}{\mathrm{M}}\right)\) t²
r = V₀ t + \(\frac { 1 }{ 2 }\) \(\frac { eE }{ M }\) t² \(\hat{j} \)
Vector from,
\(\vec { r } \) = V₀ t \(\hat{j} \) \(\frac { 1 }{ 2 }\) \(\frac { eE }{ M }\) t² \(\hat{j} \)

Question 7.
A closed triangular box is kept in an electric field of magnitude E = 2 × 10³ N C^-1 as shown in the figure.

Calculate the electric flux through the (a) vertical rectangular surface (b) slanted surface and (c) entire surface.
Answer:
Electric field of magnitude E = 2 × 10³ NC^-1
(a) Vertical rectangular surface:
Rectangular area A= 5 × 10^-2 × 15 × 10^-2
A= 75 × 10^-24 m²
θ =180°
⇒ cos 180° = -1

Electric flux, Φ_v.s = EA cos θ
= 2 × 10³ × 75 × 10^-4 × cos 180°
= -150 × 10^-1
Φ_v.s = -15 Nm² C^-1

(b) Slanted surface:
cos θ = cos 60° = 0.5
sin θ = sin 30° = \(\frac { Opposite }{ hyp }\)

hyp = \(\frac {{ 5 × 10 }^{2}}{ 0.5 }\)
hyp = 0.1m
Area of slanted surface A₂ = (0.1 × 15 × 10^-2)
A₂ = 0.015 M²
Electric flux, Φ_v.s = EA = cos θ
= 2 × 10³ × 0.015 × cos 60°
= 2 × 10³ × 0.015 × 10³
= 0.015 × 10³
Φ_v.s = 15 Nm² C^-1
Horizontal surface
θ = 90° ; cos 90° = 0
Electric flux, Φ_H.S = E. A₃ Cos 90° = 0

(c) Entire surface:
Φ_Total = Φ_V.S + Φ_S.S + Φ_H.S = -15 + 15 + 0
Φ_Total = 0

Question 8.
The electrostatic potential is given as a function of x in figure (a) and (b). Calculate the corresponding electric fields in regions A, B, C and D. Plot the electric field as a function of x for the figure (b).

Answer:
The relation between electric field and potential
E = – \(\frac { dv }{ dx }\)

(a) Region A :
dv = -3V ; dx = 0.2 m
Electric field, E_A = \(\frac { (-3) }{ 0.2 }\) = 15 V m^-1
Region B:
dv = 0V ; dx = 0.2 m
Electric field, E_B = \(\frac { 0 }{ 0.2 }\) = 0
Region C:
dv = 2V ; dx = 0.2 m
Electric field, E_C = \(\frac { -2 }{ 0.2 }\) = 10 V m^-1
Region D:
dv = -6V ; dx = 0.2 m
Electric field, E_D = \(-\left(\frac{-6}{0.2}\right)\) = 10 V m^-1 = 30 V m^-1

Electric field, E_A = 15 V m^-1
Electric field, E_B = 0
Electric field, E_C = \(\frac { (-3) }{ 0.2 }\) = 10 V m^-1
Electric field, E_D = 30 V m^-1

(b)

Question 9.
A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure.

To create the spark, an electric field of magnitude 3 x 10⁶Vm^-1 is required, (a) What potential difference must be applied to produce the spark? (b) If the gap is increased, does the potential difference increase, decrease or remains the same? (c) find the potential difference if the gap is 1 mm.
Answer:
Separation gap between two electrodes, d = 0.6 mm
d = 0.6 × 10^-3 m
Magnetude of electric field Electric field = E = 3 × 10⁶ V m^-1
Electric field E = \(\frac { V }{ d }\)
(a) Applied potential difference, V = E . d
= 3 × 10⁶ × 0.6 10^-13 = 1.8 × 10³
V = 1800 V

(b) From equation, V = E . d
If the gap (distance) between the electrodes increased, the potential difference also increases.

(c) Gap between the electrodes, d = 1mm = 1 x 10^-3 m
Potential difference, V = E.d
= 3 × 10⁶ × 1 × 10^-3 = 3 × 103
V = 3000 V

Question 10.
A point charge of +10 μC is placed at a distance of 20 cm from another identical point charge of +10 μC. A point charge of -2 μC is moved from point a to b as shown in the figure. Calculate the
change in potential energy of the system? Interpret your result.

q₁ = 10μC = 10 x 10^-6 C
q₂ = 2μC = -2 x 10^-6 C
distance, r = 5cm = 5 x 10^-2 m
Answer:
Change in potential energy,

= -36 × 1 × 10⁹ × 10^-12 × 10² = -36 × 10^-1
∆ U = -3.6 J

Negative sign implies that to move the charge -2pC no external work is required. System spends its stored energy to move the charge from point a to point b.
Ans:
∆ U = -3.6 J, negative sign implies that to move the charge -2μC no external work is required. System spends its stored energy to move the charge from point a to point b.

Question 11.
Calculate the resultant capacitances for each of the following combinations of capacitors.

Answer:

Parallel combination of capacitor 1 and 2
C_p = C₀ + C₀ = 2C₀
Series combination of capacitor C_p and 3
\(\frac { 1 }{{ C }_{S}}\) = \(\frac { 1 }{{ C }_{p}}\) + \(\frac { 1 }{{ C }_{3}}\) = \(\frac { 1 }{{ 2C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = (or) \(\frac { 1 }{{ C }_{S}}\) = \(\frac { 3 }{ 2 }\) C₀  (or)C_S = \(\frac { 2 }{ 3 }\) C₀ 

\(\frac { 1 }{ { C }_{ { S }_{ 1 } } } \) = \(\frac { 1 }{{ C }_{1}}\) + \(\frac { 1 }{{ C }_{2}}\) = \(\frac { 1 }{{ C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = \(\frac { 1 }{{ C }_{0}}\) (or)
\(\frac { 1 }{ { C }_{ { S }_{ 1 } } } \) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 1 } }\) = \(\frac {{ C }_{0}}{ 2 }\)
Similarly 3 and 4 are series combination
\(\frac { 1 }{ { C }_{ { S }_{ 2 } } } \) = \(\frac { 1 }{{ C }_{3}}\) + \(\frac { 1 }{{ C }_{4}}\) = \(\frac { 1 }{{ C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }{0}}{ 2 }\)

\({ C }_{ { S }_{ 1 } }\) and \({ C }_{ { S }_{ 2 } }\) are in parallel combination
C_p = \({ C }_{ { S }_{ 1 } }\) + \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }_{0}}{ 2 }\) + \(\frac {{ C }_{0}}{ 2 }\) (or) C_p = \(\frac {{ 2C }_{0}}{ 2 }\) C_p = C₀

(c) Capacitor 1, 2 and 3 are in parallel combination
C_p = C₀ + C₀ + C₀ = 3C₀
C_p = 3C₀

(d) Capacitar C₁ and C₂ are in combination

Similarly C₃ and C₄ are in series combination


\({ C }_{ { S }_{ 1 } }\) and \({ C }_{ { S }_{ 2 } }\) are in parallel combination across RS:

(e) Capacitor 1 and 2 are series combination

Similarly 3 and 4 are series combination
\(\frac { 1 }{ { C }_{ { S }_{ 2 } } } \) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }_{0}}{ 2 }\)
Three capacitors are in parallel combination

Question 12.
An electron and a proton are allowed to fall through the separation between the plates of a parallel plate capacitor of voltage 5 V and separation distance h = 1 mm as shown in the figure.

(a) Calculate the time of flight for both electron and proton
(b) Suppose if a neutron is allowed to fall, what is the time of flight?
(c) Among the three, which one will reach the bottom first?
(Take m_p = 1.6 x 10^-27 kg, m_e= 9.1 x 10^-31 kg and g = 10 m s^-2)
Answer:
Potential difference between the parallel plates V = 5 V
Separation distance, h = 1 mm =1 x 10^-3 m
Mass of proton, mp = 1.6 x 10^-27 kg
Mass of proton, m =9.1 x 10^-31 kg
Charge of an a proton (or) electron, e— 1.6 x 10^-19 C
[u = 0; s = h]
From equation of motion, S = ut + \(\frac { 1 }{ 2 }\) at²
From equation of motion, h = \(\frac { 1 }{ 2 }\) at²
t = \(\sqrt { \frac { 2h }{ a } } \)
Acceleration of an electron due to electric field, a = \(\frac { F }{ m }\) = \(\frac { eE }{ m }\)
[E = \(\frac { V }{ d }\)]

(a) Time of flight for both electron and proton,

t_p = 63 ns……. (2)

(b) time of flight of neutron t_n = \(\sqrt { \frac { 2h }{ g } } \) = \(\sqrt{\frac{2 \times 1 \times 10^{3}}{10}}\) = \(\sqrt{0.2 \times 10^{-3}}\)
t_n = 0.0141 s = 14.1 x 10^-3 s
t_n = 14.1 x 10^-3 ms ……. (3)
(c) Compairision of values 1,2 and 3. The electron will reach the bottom first.

Question 13.
During a thunder storm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 x 10⁶ Vm^-1 ), lightning will occur.

(a) If the bottom part of the cloud is 1000 m above the ground, determine the electric potential difference that exists between the cloud and ground.
(b) In a typical lightning phenomenon, around 25C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?
Answer:
(a) Electric field between the cloud and ground,
V = E.d
V= 3 x 10⁶ x 1000 = 3 x 10⁹V
(a) Electrons transfered from cloud to ground,
q = 25 C
Electron static potential energy,
U = \(\frac { 1 }{ 2 }\) CV²
[C = \(\frac { q }{ V }\)]
= \(\frac { 1 }{ 2 }\) qV = \(\frac { 1 }{ 2 }\) x 25 x 3 x 10⁹
U = 37.5 x 10⁹ J

Question 14.
For the given capacitor configuration
(a) Find the charges on each capacitor
(b) potential difference across them
(c) energy stored in each capacitor.

Answer:
Capacitor b and c in parallel combination
C_p = C_b + C_c = (6 + 2) μF = 8 μF
Capacitor a, c_p and d are in series combination, so the resulatant copacitance
\(\frac { 1 }{{ C }_{s}}\) = \(\frac { 1 }{{ C }_{a}}\) + \(\frac { 1 }{{ C }_{cp}}\) + \(\frac { 1 }{{ C }_{d}}\) = \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 8 }\) = \(\frac { 3 }{ 8 }\)
C_s  = \(\frac { 8 }{ 3 }\) μF

(a) Charge on each capacitor,
Charge on capacitor a, Q_a = C_s V = \(\frac { 8 }{ 3 }\) x 9
Q_a = 24 μC
Charge on capacitor, d, Q_d = C_s V = \(\frac { 8 }{ 3 }\) x 9
Q_d = 24 μC
Capacitor b and c in parallel
Charge on capacitor, b, Q_b = \(\frac { 6 }{ 3 }\) x 9 = 18
Q_b = 18 μC
Charge on capacitor, c, Q_c = \(\frac { 2 }{ 3 }\) x 9 = 6
Q_c = 6 μC

(b) Potential difference across each capacitor, V = \(\frac { q }{ C }\)
Capacitor C_a, V_a = \(\frac{ { q }_{a}}{{ C }_{a}}\) = \(\frac {{ 24 × 10 }^{6}}{{ 8 × 10 }^{6}}\) = 3 V
Capacitor C_b, V_b = \(\frac{ { q }_{b}}{{ C }_{b}}\) = \(\frac {{ 18 × 10 }^{6}}{{ 6 × 10 }^{6}}\) = 3 V
Capacitor C_c, V_c = \(\frac{ { q }_{c}}{{ C }_{c}}\) = \(\frac {{ 6 × 10 }^{6}}{{ 2 × 10 }^{6}}\) = 3 V
Capacitor C_d, V_d = \(\frac{ { q }_{d}}{{ C }_{d}}\) = \(\frac {{ 24 × 10 }^{6}}{{ 8 × 10 }^{6}}\) = 3 V

(c) Energy stores in a capacitor, U = \(\frac { 1 }{ 2 }\) CV²
Energy in capacitor C_a, U_a = \(\frac { 1 }{ 2 }\) C_a \({ V }_{ a }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 8 x 10^-6 x (3)²
U_a = 36 μJ
Capacitor C_b, U_b = \(\frac { 1 }{ 2 }\) C_b \({ V }_{ b }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 6 x 10^-6 x (3)²
U_a = 27 μJ
C_c, U_c = \(\frac { 1 }{ 2 }\) C_c \({ V }_{ c }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 2 x 10^-6 x (3)²
U_a = 9 μJ
C_d, U_d = \(\frac { 1 }{ 2 }\) C_d \({ V }_{ d }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 8 x 10^-6 x (3)²
U_a = 36 μJ

Question 15.
Capacitors P and Q have identical cross sectional areas A and separation d. The space between the capacitors is filled with a dielectric of dielectric constant er as shown in the figure. Calculate the capacitance of capacitors P and Q.

Answer:
Cross-sectional area of parallel plate capacitor = A
Each area of different medium between parallel plate capacitor = \(\frac { A }{ 2 }\)
Separation distance = d
Capacitance of parallel plate capacitor, C = \(\frac { εA }{ d }\)
Air medium of dielectric constant, ε_r = 1
dielectric medium of dielectric constant = ε_r

Case 1:
Capacitance of air filled capacitor

Capacitance of dielectric filled capacitor

Capacitance of parallel plate capacitor

Case 2:
Each distance of different medium between the parallel plate capacitor = \(\frac { d }{ 2 }\)
Capacitance of dielectric filled capacitor

Capacitance of air filled capacitor,

Capacitance of parallel plate capacitor,

Samacheer Kalvi 12th Physics Electrostatics Additional Questions Solved

I. Multiple Choice Questions

Question 1.
When a solid body is negatively charged by friction, it means that the body has
(a) acquired excess of electrons
(b) lost some, problems
(c) acquired some electrons and lost a lesser number of protons
(d) lost some positive ions
Answer:
(a) acquired excess of electrons

Question 2.
A force of 0.01 N is exerted on a charge of 1.2 x 10^-5 G at a certain point. The electric field at that point is
(a) 5.3 x 10⁴ NC^-1
(b) 8.3 x 10^-4 NC^-1
(c) 5.3 x 10² NC^-1
(d) 8.3 x 10⁴ NC^-1
Answer:
(d) 8.3 x 10⁴ NC^-1
Hint:
E = \(\frac { F }{ q }\) = \(\frac { 0.01 }{{ 1.2 × 10 }^{-5}}\) = 8.3 x 10² NC^-1

Question 3.
The electric field intensity at a point 20 cm away from a charge of 2 x 10 5 C is
(a) 4.5 x 10⁶ NC^-1
(b) 3.5 x 10⁵ NC^-1
(c) 3.5 x 10⁶ NC^-1
(d) 4.5 x 10⁵ NC^-1
Answer:
(a) 4.5 x 10⁶ NC^-1
Hint:
E = \(\frac{q}{4 \pi \varepsilon_{0} r^{2}}\) = \(\frac{9 \times 10^{9} \times 2 \times 10^{-5}}{(0.2)^{2}}\) = 4.5 x 10⁶ NC^-1

Question 4.
How many electrons will have a charge of one coulomb?
(a) 6.25 x 10¹⁸
(b) 6.25 x 10¹⁹
(c) 1.6 x 10¹⁸
(d) 1.6 x 10¹⁹
Answer:
(a) 6.25 x 10¹⁸
Hint:
Number of electron, n = \(\frac { q }{ e }\) = \(\frac { 1 }{{ 1.6 × 10 }^{-19}}\) = 6.25 × 10¹⁸

Question 5.
The ratio of the force between two charges in air and that in a medium of dielectric constant K is
(a) K : 1
(b) 1 : K
(c) K² : 1
(d) 1 : K²
Answer:
(a) K : 1

Question 6.
The work done in moving a positive charge on an equipotential surface is
(a) finite and positive
(b) infinite
(c) finite and negative
(d) zero
Answer:
(d) zero

Question 7.
If a charge is moved against the coulomb force of an electric field.
(a) work is done by the electric field
(b) energy is used from some outside source
(c) the strength of the field is decreased
(d) the energy of the system is decreased
Answer:
(b) energy is used from some outside source

Question 8.
No current flows between two charged bodies when connected
(a) if they have the same capacitance
(b) if they have same quantity of charge
(c) if they have the same potential
(d) if they have the same charge density
Answer:
(c) if they have the same potential

Question 9.
Electric field lines about a negative point charge are
(a) circular, anticlockwise
(b) circular, clockwise
(c) radial, inwards
(d) radial, outwards
Answer:
(c) radial, inwards

Question 10.
Two plates are 1 cm apart and the potential difference between them is 10 V. The electric field between the plates is
(a) 10 NC^-1
(b) 250 NC^-1
(c) 500 N^-1
(d) 1000 NC^-1
Answer:
(d) 1000 NC^-1
Hint:
E = \(\frac { V }{ d }\) = \(\frac { 10 }{{ 1 × 10 }^{-2}}\) = 8.3 x 10² NC^-1

Question 11.
At a large distance (r), the electric field due to a dipole varies as
(a) \(\frac { 1 }{ r }\)
(b) \(\frac { 1 }{{ r }^{2}}\)
(c) \(\frac { 1 }{{ r }^{3}}\)
(d) \(\frac { 1 }{{ r }^{4}}\)
Answer:
(c) \(\frac { 1 }{{ r }^{3}}\)

Question 12.
Two thin infinite parallel plates have uniform charge densities +c and -σ. The electric field in the space between then is
(a) \(\frac { σ }{{ 2ε }_{0}}\)
(b) \(\frac { σ }{{ ε }_{0}}\)
(c) \(\frac { 2σ }{{ 2ε }_{0}}\)
(d) Zero
Answer:
(b) \(\frac { σ }{{ ε }_{0}}\)

Question 13.
Two isolated, charged conducting spheres of radii R₁, and R₂ produce the same electric field near their surfaces. The ratio of electric potentials on their surfaces is-
(a) \(\frac {{ R }_{1}}{{ R }_{2}}\)
(b) \(\frac {{ R }_{2}}{{ R }_{1}}\)
(c) \(\frac { { R }_{ 1 }^{ 2 } }{ { R }_{ 2 }^{ 2 } } \)
(d) \(\frac { { R }_{ 2 }^{ 2 } }{ { R }_{ 1 }^{ 2 } } \)
Answer:
(b) \(\frac {{ R }_{2}}{{ R }_{1}}\)

Question 14.
A 100 μF capacitor is to have an energy content of 50 J in order to operator a flash lamp. The voltage required to charge the capacitor is
(a) 500 V
(b) 1000 V
(c) 1500 V
(d) 2000 V
Answer:
(b) 1000 V
Hint:

Question 15.
A 1 μF capacitor is placed in parallel with a 2 μF capacitor across a 100 V supply. The total charge on the system is
(a) \(\frac { 100 }{ 3 }\) μC
(b) 100 μC
(c) 150 μC
(d) 300 μC
Answer:
(d) 300 μC
Hint:
Equivalent capacitor = 1 + 2 = 3 μF
Total charge, q = CV = 3 x 100 = 300 μF

Question 16.
A parallel plate capacitor of capacitance 100 μF is charged to 500 V. The plate separation is then reduce to half its original value. Then the potential on the capacitor becomes
(a) 250 V
(b) 500 V
(c) 1000V
(d) 2000 V
Answer:
(a) 250 V
Hint:
Here, C’ = 2C, since the charge remains the same.
q = C’V’ = CV ⇒ V = \(\frac { CV }{ 2C }\) = \(\frac { 500 }{ 2 }\) = 250 V

Question 17.
A point charge q is placed at the midpoint of a cube of side L. The electric flux emerging from the cube is ‘
(a) \(\frac { q }{{ ε }_{0}}\)
(b) \(\frac { q }{{ 6Lε }_{0}}\)
(c) \(\frac { 6Lq }{{ ε }_{0}}\)
(d) zero
Answer:
(a) \(\frac { q }{{ ε }_{0}}\)

Question 18.
The capacitor C of a spherical conductor of radius R is proportional to
(a) R²
(b) R
(c) R^-1
(d) R⁰
Answer:
(b) R

Question 19.
Energy of a capacitor of capacitance C, when subjected to a potential V, is given by
(a) \(\frac { 1 }{ 2 }\) CV²
(b) \(\frac { 1 }{ 2 }\) C²V
(c) \(\frac { 1 }{ 2 }\) CV
(d) \(\frac { 1 }{ 2 }\) \(\frac { C }{ V }\)
Answer:
(a) \(\frac { 1 }{ 2 }\) CV²

Question 20.
The electric field due to a dipole at a distance r from its centre is proportional to
(a) \(\frac { 1 }{{ r }^{3/2}}\)
(b) \(\frac { 1 }{{ r }^{3}}\)
(c) \(\frac { 1 }{ r }\)
(d) \(\frac { 1 }{{ r }^{3}}\)
Answer:
(b) \(\frac { 1 }{{ r }^{3}}\)

Question 21.
A point charge q is rotating around a charge Q in a circle of radius r. The workdone on it by the coulomb force is
(a) 2πrq
(b) 2πQq
(c) \(\frac { Q }{{ 2ε }^{0}r}\)
(d) zero
Answer:
(d) zero

Question 22.
The workdone in rotating an electric dipole of moment P in an electric field E through an angle 0 from the direction of the field is
(a) pE (1 – cos θ)
(b) 2pE
(c) zero
(d) -pE cos θ
Answer:
(a) pE (1 – cos θ)
Hint:
W = pE(cos θ₀ – cos θ)
[θ₀ = cos 0, cos 0 = 1]
W = pE(1 – cos θ)

Question 23.
Capacitance of a parallel plate capacitor can be increased by
(a) increasing the distance between the plates
(b) increasing the thickness of the plates
(c) decreasing the thickness of the plates
(d) decreasing the distance between the plates
Answer:
(d) decreasing the distance between the plates

Question 24.
Two charges are placed in vacuum at a distance d apart. The force between them is F. If a medium of dielectric constant 2 is introduced between them, the force will now be
(a) 4F
(b) 2F
(c) F/2
(d) F/4
Answer:
(d) F/4

Question 25.
An electric charge is placed at the centre of a cube of side a. The electric flux through one of its faces will be
(a) \(\frac { q }{{ 6ε }^{0}}\)
(b) \(\frac { q }{ { ε }_{ 0 }{ a }^{ 2 } } \)
(c) \(\frac { q }{ { 4πε }_{ 0 }{ a }^{ 2 } } \)
(a) \(\frac { q }{{ ε }^{0}}\)
Answer:
(a) \(\frac { q }{{ 6ε }^{0}}\)
Hint:
According to Gauss’s law, the electric flux through the cube is \(\frac { q }{{ ε }^{0}}\). Since there are six faces, the flux through one face is \(\frac { q }{{ 6ε }^{0}}\).

Question 26.
The electric field in the region between two concentric charged spherical shells-
(a) is zero
(b) increases with distance from centre
(c) is constant
(d) decreases with distance from centre
Answer:
(d) decreases with distance from centre

Question 27.
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80 V. The potential at the centre of the sphere is-
(a) 800 V
(b) zero
(c) 8 V
(d) 80 V
Answer:
(d) 80 V

Question 28.
A 4 μF capacitor is charged to 400 V and then its plates are joined through a resistance of 1 K Ω. The heat produced in the resistance is-
(a) 0.16 J
(b) 0.32 J
(c) 0.64 J
(d) 1.28 J
Answer:
(b) 0.32 J
Hint:
The energy stored in capacitor is converted into heat
U = H = \(\frac { 1 }{ 2 }\) CV² = \(\frac { 1 }{ 2 }\) x 4 x 10^-6 x (400)² = 0.32 J

Question 29.
The workdone in carrying a charge Q, once round a circle of radius R with a charge Q₂ at the centre is-
(a) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\)
(b) zero
(c) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}}\)
(d) infinite
Answer:
(b) zero
Hint:
The electric field is conservative. Therefore, no work is done in moving a charge around a closed path in a electric field.

Question 30.
Two plates are 2 cm apart. If a potential difference of 10 V is applied between them. The electric field between the plates will be
(a) 20 NC^-1
(b) 500 NC^-1
(c) 5 NC^-1
(d) 250 NC^-1
Answer:
(b) 500 NC^-1
Hint:
\(\frac { V }{ d }\) = \(\frac { 10 }{{ 2 ×10 }^{-2}}\) 500 NC^-1

Question 31.
The capacitance of a parallel plate capacitor does not depend on
(a) area of the plates
(b) metal of the plates
(c) medium between the plates
(d) distance between the plates
Answer:
(b) metal of the plates

Question 32.
A capacitor of 50 μF is charged to 10 volts. Its energy in joules is
(a) 2.5 x 10^-3
(b) 5 x 10^-3
(c) 10 x 10^-4
(d) 2.5 x 10^-4
Answer:
(a) 2.5 x 10^-3
Hint:
U = \(\frac { 1 }{ 2 }\) CV² = \(\frac { 1 }{ 2 }\) x 50 x 10^-6 x (10)² = 2.5 x 10^-3 J

Question 33.
A cube of side b has a charge q at each of its vertices. The electric field due to this charge distribution at the centre of the cube is
(a) \(\frac { q }{{b}^{ 2 }}\)
(b) \(\frac { q }{{2b}^{ 2 }}\)
(c) \(\frac { 32q }{{b}^{ 2 }}\)
(d) zero
Answer:(d) zero
Hint:
There
is an equal charge at diagonally opposite comer. The fields due the these at the centre cancel out. Therefore, the net field at the centre is zero.

Question 34.
Total electric fulx coming out of a unit positive charge put in air is
(a) ε₀
(b) \({ \varepsilon }_{ 0 }^{ -1 }\)
(c) (4πε₀)^-1
(d) 4πε₀
Answer:
(b) \({ \varepsilon }_{ 0 }^{ -1 }\)

Question 35.
Electron volt (eV) is a unit of
(a) energy
(b) potential
(c) current
(d) charge
Answer:
(a) energy

Question 36.
A point Q lies on the perpendicular bisector of an electric dipole of dipole moment P. If the distance of Q from the dipole is r, then the electric field at Q is proportional to-
(a) p^-1 and r^-2
(b) p and r^-2
(c) p and r^-3
(d) p² and r^-3
Answer:
(c) p and r^-3

Question 37.
A hollow insulated conducting sphere is given a positive charge of 10 μC. What will be the electric field at the centre of the sphere is its radius is 2 metres?
(a) zero
(b) 8 μCm^-2
(c) 20 μCm^-2
(d) 5 μCm^-2
Answer:
(d) zero

Question 38.
A particle of charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is-
(a) qE²y
(b) q²Ey
(c) qEy²
(d) qEy
Answer:
(d) qEy
Hint:
Force on the particle = qE
KE = Workdone by the force = F.y = qEy

Question 39.
Dielectric constant of metals is-
(a) 1
(b) greater then 1
(c) zero
(d) infinite
Answer:
(d) infinite

Question 40.
When a positively charged conductor is earth connected
(a) protons flow from the conductor to the earth
(b) electrons flow from the earth to the conductor
(c) electrons flow from the conductor to the earth
(d) no charge flow occurs
Answer:
(b) electrons flow from the earth to the conductor

Question 41.
The SI unit of electric flux is
(a) volt metre²
(b) newton per coulomb
(c) volt metre
(d) joule per coulomb
Answer:
(c) volt metre

Question 42.
Twenty seven water drops of the same size are charged to the same potential. If they are combined to form a big drop, the ratio of the potential of the big drop to that of a small drop is-
(a) 3
(b) 6
(c) 9
(d) 27
Answer:
(c) 9
Hint:
V’ = n^2/3 V
⇒ \(\frac { V’ }{ V }\) = (27)^2/3 = 9

Question 43.
A point charge +q is placed at the midpoint of a cube of side l. The electric flux emerging ’ from the cube is-
(a) \(\frac { q }{{ ε }^{0}}\)
(b) \(\frac {{ 6ql }^{2}}{{ ε }^{0}}\)
(c) \(\frac { q }{ { 6l }^{ 2 }{ { ε }^{ 0 } } } \)
(d) \(\frac { { C }^{ 2 }{ V }^{ 2 } }{ 2 } \)
Answer:
(a) \(\frac { q }{{ ε }^{0}}\)

Question 44.
The energy stored in a capacitor of capacitance C, having a potential difference V between the plates, is-
Answer:
(c)

Question 45.
The electric potential at the centre of a charged conductor is-
(a) zero
(b) twice that on the surface
(c) half that on the surface
(d) same as that on the surface
Answer:
(d) same as that on the surface

Question 46.
The energy stored in a capacitor is given by
(a) qV
(b) \(\frac { 1 }{ 2 }\)qV
(c) \(\frac { 1 }{ 2 }\) CV
(d) \(\frac { q }{ 2C }\)
Answer:
(b) \(\frac { 1 }{ 2 }\)qV

Question 47.
The unit of permitivity of free space so is
(a) coulomb/newton-metre
(b) newton-metre²/coulomb²
(c) coulomb²/newton-metre²
(d) coulomb/(newton-metre)²
Answer:
(c) coulomb²/newton-metre²

Question 48.
An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. It its dipole moment is along the direction of the field, the force on it and its potential energy are, respectively.
(a) 2qE and minimum
(b) qE and pE
(c) zero and minimum
(d) qE and maximum
Answer:
(c) zero and minimum
Hint:
Potential energy, U = -pE cos θ
For q = 0°; U = -pE, which is minimum.

Question 49.
An electric dipole of moment \(\vec { P } \) is lying along a uniform electric field \(\vec { E } \) . The workdone in rotating the dipole by 90° is
(a) \(\frac { pE }{ 2 }\)
(b) 2pE
(c) pE
(d) √2pE
Answer:
(c) pE

Question 50.
Aparallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates
(a) does not charge
(b) becomes zero
(c) increases
(d) decreases
Answer:
(c) increases

Question 51.
When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance
(a) increases K times
(b) increases K^-1 times
(c) decreases K times
(d) remains constant
Answer:
(c) decreases K times

Question 52.
A comb run through one’s dry hair attracts small bits of paper. This is due to the fact that
(a) comb is a good conductor
(b) paper is a good conductor
(c) the atoms in the paper gets polarised by the charged comb
(d) the comb posseses magnetic properties
Answer:
(c) the atoms in the paper gets polarised by the charged comb

Question 53.
Which of the following is not a property of equipotential surfaces?
(a) they do not cross each other
(b) they are concentric spheres for uniform electric field
(c) the rate of change of potential with distance on them is zero
(d) they can be imaginary spheres.
Answer:
(b) they are concentric spheres for uniform electric field

Question 54.
A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will be
(a) reduced to half
(b) doubled
(c) becomes 4 times
(d) remains the same
Answer:
(d) remains the same

Question 55.
If the electric field in a region is given by \(\vec { E } \) = 5\(\hat{j} \) + 4\(\hat{j} \) + 9\(\hat{k} \) , then the electric flux through a surface of area 20 units lying in the y-z plane will be-
(a) 20 units
(b) 80 units
(c) 100 units
(d) 180 units
Answer:
(c) 100 units
Hints:
The area vector \(\vec { A } \) = 20\(\hat{j} \); \(\vec { E } \) = (5\(\hat{j} \) + 4\(\hat{j} \) + 9\(\hat{k} \))
Flux (Φ) = \(\vec { E } \) – \(\vec { A } \) = 5 x 20 =100 units

Question 56.
A, B and C are three points in a uniform electric field. The electric potential is-

(a) maximum at A
(b) maximum at B
(c) maximum at B
(d) same at all the three points A, B and C
Answer:
(b) maximum at B
Hint:
The potential decreases in the direction of the field. Therefore V_B > V_C>C_A.

Question 57.
A conducting sphere of radius R is give a charge Q. The electric potential and the electric field at the centre of the sphere are, respectively-
(a) zero, \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R }^{ 2 } } \)
(b) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \)
(c) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \), zero
(d) zero,zero
Answer:
(c) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \), zero.

II. Fill in the blanks

Question 1.
A dipole is placed in a uniform electric field with its axis parallel to the field. It experiences …………………
Answer:
neither a net force nor a torque

Question 2.
The unit of permittivity is…………………
Answer:
C²N^-1m^-2

Question 3.
The branch of physics which deals with static electric charges or charges at rest is …………………
Answer:
electrostatics

Question 4.
The charges in a electrostatics field are analogous to ………………… in a gravitational field.
Answer:
mass

Question 5.
The substances which acquire charges on rubbing are said to be …………………
Answer:
electrified

Question 6.
Electron means …………………
Answer:
amber

Question 7.
A glass rod rubbed with a silk cloth. Glass rod and silk cloth acquires …………………
Answer:
positive and negative charge respectvely .

Question 8.
When ebonite rod is rubbed with fur, ebonite rod and fur acquires …………………
Answer:
negative and positive charge respectively

Question 9.
………………… termed the classification of positive and negative charges.
Answer:
Franklin

Question 10.
Applications such as electrostatic point spraying and powder coating, are based on the property of ………………… between charged bodies.
Answer:
attraction and repulsion

Question 11.
Bodies which allow the charge to pass through them are called …………………
Answer:
conductor

Question 12.
Bodies which do not allow the charge to pass through them are called …………………
Answer:
insulators

Question 13.
The unit of electric charge is …………………
Answer:
coulomb

Question 14.
Total charge in an isolated sysem …………………
Answer:
remains a constant

Question 15.
The force between two charged bodies was studied by …………………
Answer:
coulomb

Question 16.
The unit of permittivity in free space (s0) is …………………
Answer:
C²N^-1m^-2

Question 17.
The value of s, for air or vacuum is …………………
Answer:1

Question 18.
Charges can neither be created nor be destroyed is the statement of law of conservation of …………………
Answer:
charge

Question 19.
The space around the test charge, in which it experiences a force is known as field …………………
Answer:
electric

Question 20.
Electric field at a point is measued in terms of …………………
Answer:
electric field intensity

Question 21.
The unit of electric field in tensity is …………………
Answer:
NC^-1.

Question 22.
The lines of force are far apart, when electric field E is …………………
Answer:
small

Question 23.
The lines of force are close together, when electric field E is …………………
Answer:
large

Question 24.
Electric dipole moment …………………
Answer:
P = 2qd

Question 25.
Torque experienced by electric dipole is …………………
Answer:
x = PE sin θ

Question 26.
An electric dipole placed in a non-uniform electric field at an angle 0 experiences …………………
Answer:
both torque and force

Question 27.
When thee dipole is aligned parallel to the field, its electric potential energy is …………………
Answer:
u = -PE

Question 28.
Change of potential with distance is known as …………………
Answer:
potential distance

Question 29.
The number of electric lines of force crossing through the given area is …………………
Answer:
electric flux

Question 30.
The process of isolating a certain region of space from external field is called …………………
Answer:
electrostatic shielding

Question 31.
Capacitor is a device to store …………………
Answer:
charge

Question 32.
The charge density in maximum at …………………
Answer:
pointed

Question 33.
The principle made use of in lightning arrestor is …………………
Answer:
action of points

Question 34.
Van de Graaff generator producers large electrostatic potential difference of the order of …………………
Answer:
10⁷ V

III. Match the following

Question 1.

Answer:
(i) → (d)
(ii) → (a)
(iii) → (b)
(iv) → (c)

Question 2.

Answer:
(i) → (c)
(ii) → (d)
(iii) → (a)
(iv) → (b)

Question 3.

Answer:
(i) → (b)
(ii) → (d)
(iii) → (a)
(iv) → (c)

Question 4.

Answer:
(i) → (b)
(ii) → (d)
(iii) → (a)
(iv) → (c)

IV. Assertion and reason type

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but the reason is not correct explanation of the assertion.
(c) If assertion is true but reason is false.
(d) If the assertion and reason both are false.
(e) If assertion is false but reason is true.

Question 1.
Assertion: Electric lines of force cross each other.
Reason: Electric field at a point supermpose to give one resultant electric field.
Answer:
(e) Both assertion and reason are true but the reason is not correct explanation of the assertion.
Explanation: If electric lines of forces cross each other, then the electric field at the point of intersection will have two direction simultaneously which is not possible physically.

Question 2.
Assertion: Charge is quantized.
Reason: Charge, which is less than 1 C is not possible.
Answer:
(c) If assertion is true but reason is false.
Explanation: Q = ±ne and charge lesser than 1 C is possible.

Question 3.
Assertion:
A point charge is brought in an electric field. The field at a nearby point will increase, whatever be the nature of the charge.
Reason: The electric field is independent of the nature of charge.
(d) If the assertion and reason both are false.
Explanation: Electric field at the nearby-point will be resultant of existing field and field due to the charge brought. It may increase or decrease if the charge is positive or negative depending on the position of the point with respect to the charge brought.

Question 4.
Assertion: The tyre’s of aircraft’s are slightly conducting.
Reason: If a conductor is connected to ground, the extra charge induced on conductor will flow to ground.
Answer:
(b) Both assertion and reason are true but the reason is not correct explanation of the assertion.
Explanation: During take off and landing, the friction between treys and the run way may cause electrification of treys. Due to conducting to a ground and election sparking is avoided.

Question 5.
Assertion: The lightening conductor at the top of a high building has sharp ends.
Reason: The surface density of charge at sharp points is very high, resulting in setting up of electric wind.
Answer:
(a) Both assertion and reason are true and the reason is the correct explanation of the assertion.

Samacheer Kalvi 12th Physics Electrostatics Short Answer Questions

Question 1.
What is meant by triboelectric charging?
Answer:
Charging the objects through rubbing is called triboelectric charging.

Question 2.
What is meant by conservation of total charges?
Answer:
The total electric charge in the universe is constant and charge can neither be created nor be destroyed. In any physical process, the net change in charge will always be zero.

Question 3.
State Gauss’s Law?
Answer:
Definition:
Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux OE through the closed surface is
Φ_E = \(\oint { \vec { E } } \) .d\(\vec { A } \) = \(\frac {{ q }_{encl}}{{ ε }_{0}}\)

Question 4.
What is meant by electrostatic shielding?
During lightning accompanied by a thunderstorm, it is always safer to sit inside a bus than in open ground or under a tree. The metal body of the bus provides electrostatic shielding, since the electric field inside is zero. During lightning, the charges flow through the body of the conductor to the ground with no effect on the person inside that bus.

Question 5.
What is meant by dielectric?
Answer:
A dielectric is a non-conducting material and has no free electrons. The electrons in a dielectric are bound within the atoms. Ebonite, glass and mica are some examples of dielectrics.

Question 6.
What are non-polar molecules? Give examples.
A non-polar molecule is one in which centers of positive and negative charges coincide. As a result, it has no permanent dipole moment. Examples of non-polar molecules are hydrogen (H₂), oxygen (O₂), and carbon dioxide (CO₂) etc.

Question 7.
What are polar molecules? Give examples.
Answer:
In polar molecules, the centers of the positive and negative charges are separated even in the absence of an external electric field. They have a permanent dipole moment.
The net dipole moment is zero in the absence of an external electric field. Examples of polar molecules are H₂O, N₂O, HCl, NH₃.

Question 8.
What is a capacitors?
Answer:
Capacitor is a device used to store electric charge and electrical energy. Capacitors are widely used in many electronic circuits and have applications in many areas of science and technology.

Samacheer Kalvi 12th Physics Electrostatics Long Answer Questions

Question 1.
Derive an expression for electric field due to the system of point charges?
Answer:
Electric field due to the system of point charges:
Suppose a number of point charges are distributed in space. To find the electric field at some point P due to this collection of point charges, superposition principle is used. The electric field at an arbitrary point due to a collection of point charges is simply equal to the vector sum of the electric fields created by the individual point charges. This is called superposition of electric fields.
Consider a collection of point charges q₁, q₂, q₃,…., q_n located at various points in space. The ‘ total electric field at some point P due to all these n charges is given by

Here r_1p, r_2p, r_3p,…., r_np, are the distance of the charges ₁, q₂, q₃,…., q_n from the point respectively. Also \(\hat{r} \)_1p + \(\hat{r} \)_2p + \(\hat{r} \)_3p,…., \(\hat{r} \)_np are the corresponding unit vectors directed from q₁, q₂, q₃,…., q_n tpo P.

Equation (2) can be re-written as,

For example in figure, the resultant electric field due to three point charges q₁, q₂, q₃ at point P is shown. Note that the relative lengths of the electric field vectors for the charges depend on relative distantes of the charges to the point P.

Question 2.
Derive an expression for electric flux of rectangular area placed in uniform electric field.
Answer:
(i) Electric flux for uniform Electric field:
Consider a uniform electric field in a region of space. Let uschoose an area A normal to the electric field lines as shown in figure (a). The electric flux for this case is
Φ_E = EA ….. (1)
Suppose the same area A is kept parallel to the uniform electric field, then no electric field lines pierce through the area A, as shown in figure (b). The electric flux for this case is zero.
Φ_E = 0 ….. (2)
If the area is inclined at an angle θ with the field, then the component of the electric field perpendicular to the area alone contributes to the electric flux. The electric field component parallel to the surface area will not contribute to the electric flux. This is shown in figure (c). For this case, the electric flux
ΦE = (E cosθ) A …(3)
Further, θ is also the angle between the electric field and the direction normal to the area. Hence in general, for uniform electric field, the electric flux is defined as
Φ_E= \(\vec { E } \).\(\vec { A } \) = EA cos θ …(4)

Here, note that \(\vec { A } \) is the area vector \(\vec { A } \) = A\(\hat{n} \). Its magnitude is simply the area A and the direction is along the unit vector h perpendicular to the area. Using this definition for flux, Φ_E= \(\vec { E } \).\(\vec { A } \), equations (2) and (3) can be obtained as special cases.
In figure (a), θ = 0° so Φ_E= \(\vec { E } \).\(\vec { A } \) = EA
In figure (b), θ = 90° so Φ_E= \(\vec { E } \).\(\vec { A } \) = 0

(ii) Electric flux in a non uniform electric field and an arbitrarily shaped area: Suppose the electric field’is not uniform and the area A is not flat, then the entire area is divided
into n small area segments ∆\(\vec { A } \)₁ ∆\(\vec { A } \)₂, ∆\(\vec { A } \)₃,…..∆\(\vec { A } \)_n, such that each area element is almost flat and the electric field over each area element is considered to be uniform.
The electric flux for the entire area A is approximately written as

By taking the limit ∆\(\vec { A } \)₁ → 0 (for all i) the summation in equation (5) becomes integration. The total electric flux for the entire area is given by
Φ_E = ∫\(\vec { E } \).d\(\vec { A } \) ….. (6)
From Equation (6), it is clear that the electric flux for a given surface depends on both the electric field pattern on the surface area and orientation of the surface with respect to the electric field.

(iii) Electric flux for closed surfaces: In the previous section, the electric flux for any arbitrary curved surface is discussed. Suppose a closed surface is present in the region of the non-uniform electric field as shown in figure (a).

The total electric flux over this closed surface is written as
Φ_E = \(\oint { \vec { E } } \).d\(\vec { A } \) …… (7)
Note the difference between equations (6) and (7). The integration in equation (7) is a closed surface integration and for each areal element, the outward normal is the direction of d\(\vec { A } \) as shown in the figure (b).
The total electric flux over a closed surface can be negative,
positive or zero. In the figure (b), it is shown that in one area element, the angle between d\(\vec { A } \) and \(\vec { E } \) is less than 90°, then the electric flux is positive and in another areal element, the angle between dA and E is greater than 90°, then the electric flux is negative. In general, the electric flux is negative if the electric field lines enter the closed surface and positive if the electric field lines leave the closed surface.

Samacheer Kalvi 12th Physics Electrostatics Numerical Problems

Question 1.
Electrons are caused to fall through a potential difference of 1500 volts. If they were initially at rest. Then calculate their final speed.
Solution:
The electrical potential energy is converted into kinetic energy. If v is the final speed then

Question 2.
Small mercury drops of the same size are charged to the same potential V. If n such drops coalesce to form a single large drop, then calculate its potential.
Solution:
Let r be the radius of a small drop and R that of the large drop. Then, since the volume remains conserved,
\(\frac { 1 }{ 2 }\) πR² = \(\frac { 4 }{ 3 }\) πR³n
⇒ R³ = r³n
R = r³(n)^1/3
Further, since the total charge remains conserved, we have, using Q = CV
C_large V = n C_small v
Where V is the potential of the large drop.
4πε₀ RV = n (4πε₀r)v
V = \(\frac { nrv }{ R }\) = \(\frac { nrv }{{ r(n) }^{1/3}}\)
V = vn^2/3

Question 3.
Two particles having charges Q₁ and Q₂ when kept at a certain distance, exert a force F on each other. If the distance between the two particles is reduced to half and the charge on each particle is doubled. Find the force between the particles.
Solution:
F = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{r^{2}}\)
If the distance is educed by half and two particles of charges are doubled.

Question 4.
Two charged spheres, separated by a distance d, exert a force F on each other. If they are immersed in a liquid of dielectric constant 2, then what is the force.
Solution:
Force between the charges (vacuum)

Force between the charges (medium)

Question 5.
Find the force of attraction between the plates of a parallel plate capacitor.
Solution:
Let d be the distance between the plates. Then the capacitor is
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } \)
Energy stored in a capacitor,

Energy magnitude of the force is,

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You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.17

10th Maths Exercise 3.17 Answers Question 1.
If A = \(\left[\begin{array}{cc}{1} & {9} \\ {3} & {4} \\ {8} & {-3}\end{array}\right]=\left[\begin{array}{ll}{5} & {7} \\ {3} & {3} \\ {1} & {0}\end{array}\right]\) then verify that
(i) A + B = B + A
(ii) A + (-A) = (-A) + A = 0
Solution:

Exercise 3.17 Class 10 Question 2.

Solution:

Ex 3.17 Class 10 Question 3.
Find X and Y if X + Y = \(\left[\begin{array}{ll}{7} & {0} \\ {3} & {5}\end{array}\right]\) and X – Y = \(\left[\begin{array}{ll}{3} & {0} \\ {0} & {4}\end{array}\right]\)
Solution:

10th Maths Exercise 3.17 Question 4.
If A = \(\left[\begin{array}{lll}{0} & {4} & {9} \\ {8} & {3} & {7}\end{array}\right]\), B = \(\left[\begin{array}{lll}{7} & {3} & {8} \\ {1} & {4} & {9}\end{array}\right]\) find the value of
(i) B – 5A
(ii) 3A – 9B
Solution:

10th New Syllabus Maths Exercise 3.17 Question 5.
Find the values of x, y, z if

Solution:
(i) \(\left(\begin{array}{cc}{x-3} & {3 x-z} \\ {x+y+7} & {x+y+z}\end{array}\right)=\left(\begin{array}{ll}{1} & {0} \\ {1} & {6}\end{array}\right)\)
x – 3 = 1 ⇒ x = 4
3x – z = 0
3(4) – z = 0
-z = -12 ⇒ z = 12
x + y + 7 = 1
x + y = -6
4 + y = -6
y = -10
x = 4, y = -10, z = 12

(ii) \(\left[\begin{array}{ccc}{x} & {y-z} & {z+3}\end{array}\right]+\left[\begin{array}{lll}{y} & {4} & {3}\end{array}\right]=\left[\begin{array}{lll}{4} & {8} & {16}\end{array}\right]\)
x + y = 4 ……………. (1)
y – z + 4 = 8 ………….. (2)
z + 3 + 3 = 16 ………….. (3)
From (3), we get z = 10
From (2), we get y – 10 + 4 = 8
From (2), we get y = 14
From (1) we get x + 14 = 4
x = -10
x = -10, y = 14, z = 10

10th Maths Exercise 3.17 Samacheer Kalvi Question 6.

Solution:

10th Samacheer Maths Question 7.
Find the non-zero values of x satisfying the matrix equation

Solution:

Samacheer Kalvi 10th Guide Maths Question 8.

Solution:
x² – 4x = 5
y² – 2y = 8
y² – 2y – 8 = 0
(y – 4)(y + 2) = 0
y = 4, -2
x² – 4x – 5 = 0
(x – 5)(x + 1) = 0
x = 5, -1
x = -1, 5, y = 4, -2