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Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Tamil Book Solutions Guide Pdf Chapter 5.5 தொகைநிலை, தொகாநிலைத் தொடர்கள் Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th Tamil Solutions Chapter 5.5 தொகைநிலை, தொகாநிலைத் தொடர்கள்

கற்பவை கற்றபின்

Question 1.
பாடப்பகுதியில் இடம்பெற்றுள்ள தொகைநிலைத் தொடர், தொகாநிலைத் தொடர்களைக் கண்டறிந்து தனித்தனியே தொகுக்க.
Answer:
தொகைநிலைத் தொடர்கள்

தொகா நிலைத்தொடர்கள்

பாடநூல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
சொற்களுக்கு இடையே வேற்றுமை உருபு மறைந்து வருவது ……………………….
அ) வேற்றுமைத் தொகை
ஆ) உம்மைத் தொகை
இ) உவமைத் தொகை
ஈ) அன்மொழித் தொகை
Answer:
அ) வேற்றுமைத் தொகை

Question 2.
‘செம்மரம்’ என்னும் சொல் …………………. த்தொகை.
அ) வினை
ஆ) பண்பு
இ) அன்மொழி
ஈ) உம்மை
Answer:
ஆ) பண்பு

Question 3.
‘கண்ணா வா!’ – என்பது ……………….. த் தொடர்.
அ) எழுவாய்
ஆ) விளி
இ) வினைமுற்று
ஈ) வேற்றுமை
Answer:
ஆ) விளி

பொருத்துக

1. பெயரெச்சத் தொடர் – அ) கார்குழலி படித்தாள்.
2. வினையெச்சத் தொடர் – ஆ) புலவரே வருக.
3. வினைமுற்றுத் தொடர் – இ) பாடி முடித்தான்.
4. எழுவாய்த் தொடர் – ஈ) எழுதிய பாடல்.
5. விளித் தொடர் – உ) வென்றான் சோழன்.
Answer:
1. ஈ
2. இ
3. உ
4. அ
5. ஆ

சிறுவினா

Question 1.
தொகைநிலைத் தொடர்கள் எத்தனை வகைப்படும்? அவை யாவை?
Answer:
தொகைநிலைத் தொடர்கள் ஆறு வகைப்படும். அவையாவன:

• வேற்றுமைத்தொகை
• உவமைத்தொகை
• வினைத்தொகை
• உம்மைத்தொகை
• பண்புத்தொகை
• அன்மொழித்தொகை

Question 2.
இரவு பகல் என்பது எவ்வகைத் தொடர் என விளக்குக.
Answer:

• ‘இரவு பகல்’ இத்தொடர், ‘இரவும் பகலும்’ என விரிந்து பொருள் தருகின்றது.
• இதில் சொல்லின் இடையிலும் இறுதியிலும் `உம்’ என்னும் இடைச் சொல் மறைந்து நின்று பொருள் தருகிறது.
• இவ்வாறு சொற்களுக்கு இடையிலும் இறுதியிலும் ‘உம்’ என்னும் இடைச் சொல் மறைந்து நின்று பொருள் தருவதை உம்மைத்தொகை என்பர்.

Question 3.
அன்மொழித்தொகையை எடுத்துக்காட்டுடன் விளக்குக.
Answer:
வேற்றுமை, வினை, பண்பு, உவமை, உம்மை ஆகிய தொகைநிலைத் தொடர்களில் அவை அல்லாத வேறு பிற சொற்களும் மறைந்து வருவது அன்மொழித்தொகை எனப்படும்.

சான்று : பொற்றொடி வந்தாள்.

இத்தொடர் ‘பொன்னாலாகிய வளையலை அணிந்த பெண் வந்தாள்’ என்னும் பொருள் தருகிறது. இதில் ‘ஆல்’ என்னும் மூன்றாம் வேற்றுமை உருபும் ‘ஆகிய என்னும் அதன் பயனும் மறைந்து வந்துள்ளது. ‘வந்தாள்’ என்னும் சொல்லால் பெண் என்பதையும் குறிப்பதால், இது மூன்றாம் வேற்றுமைப் புறத்துப் பிறந்த அன்மொழித் தொகை ஆகும்.

மொழியை ஆள்வோம்

கீழ்க்காணும் தலைப்பில் இரண்டு நிமிடம் பேசுக.

1. கைத்தொழில் ஒன்றைக் கற்றுக்கொள்

பெருமை மிகுந்த சான்றோர் சபைக்கு என் முதற்கண் வணக்கத்தைத் தெரிவித்துக் கொள்கின்றேன். இப்பெருமைமிகு சபையில் நான் பேச எடுத்துக் கொண்ட தலைப்பு கைத்தொழில் ஒன்றைக் கற்றுக்கொள் என்பதாகும்.

ஏட்டுச் சுரைக்காய் கறிக்கு உதவாது’ என்பார்கள். அதைப்போல ஏட்டுப் படிப்பு படித்தவர்களுக்கு எந்த வேலையும் கிடைப்பதாகத் தெரியவில்லை.

படித்தாலும், படித்துப் பட்டம் பெற்றாலும் கைத்தொழில் ஒன்றையும் நாம் கூடவே, சேர்த்துக் கற்றுக்கொள்ள வேண்டும். படிப்புக்கேற்ற வேலை கிடைக்கும் வரை அந்த வேலைக்காகக் காத்திராமல் கற்ற கைத்தொழில் நமக்கு மிகவும் பயன்படும்.

தையல், ஓவியம், மரவேலை, மின்னணுச் சாதனங்கள் பழுதுபார்ப்பு, தட்டச்சு, கணிப்பொறி, கூடை பின்னுதல், அலங்காரப் பொருட்கள் செய்தல் இவற்றைப் பொழுதுபோக்கிற்காக நாம் பள்ளியில் கற்றாலும், அங்கு ஆழமாக ஆழ்ந்து கற்க வேண்டும். அதுவேதான், பிற்காலத்தில் படிப்புக்கு ஏற்ற வேலை கிடைக்கும் வரையில் நமக்கு நல்ல சம்பாத்தியத்தைக் கொடுக்கும்.

ஏட்டுக்கல்வி கைவிட்டாலும், கைத்தொழில் கல்வி உன்னைக் கைவிடாது என்பதைப் புரிந்து கொள்ள வேண்டும். எனவே, இளைஞர்களாகிய நாம் கைத்தொழில் ஒன்றைக் கற்றுக் கொள்ள வேண்டும் என்று சொல்லி என்னுடைய உரையை நிறைவு செய்கின்றேன்.

2. இதயம் கவரும் இசை

என்னை ஈன்ற தாய் மொழிக்கும், இந்தச் சான்றோர் பேரவைக்கும் முதற்கண் வணக்கத்தைத் தெரிவித்துக்கொண்டு, இதயம் கவரும் இசை என்ற நல்லதொரு தலைப்பில் சில நிமிடங்கள் பேசுகின்றேன். ஒவ்வொருநாளும் ஒவ்வொரு மனிதனுக்கும் ஒவ்வொரு பிரச்சினைகள் இருக்கத்தான் செய்கின்றன. துன்பங்கள் நம்மைத் துரத்தும் போது மன அமைதி தானாக தேடி வருவதில்லை . இசையின் பக்கம் நாம் தான் ஓடி வர – வேண்டும். புல்லாங்குழல் இசையும், வீணை இசையும், நாத முழக்கமும், மத்தளம் இசையும் மனதைப் பண்படுத்தும். இசைக்கச்சேரி கேட்கும் போது இதயமெல்லாம் மென்மையாகிவிடும்.

சங்க காலத்தில் தலைவன் ஒருவன் கள் உண்ட மயக்கத்தில் படுத்து கிடக்கின்றான். தொலைவில் தலைவிதினையைக் காயவைத்துக் கொண்டிருக்கின்றாள். தலைவன் படுத்து இருந்த இடத்தை நோக்கி மத யானை ஒன்று ஓடி வருகின்றது .ஐயோ! தலைவனுக்கு என்ன ஆகுமோ? என்று கவலைப்படாமல், தலைவி அருகிலிருந்த யாழை எடுத்து மீட்டினாள். மதம் பிடித்த யானை யாழ் இசையில் மயங்கி தலைவனை மிதிக்காமல் தெளிந்து சென்றதாம். இசை உயிரையும் காப்பாற்றும்.

குழந்தை பிறந்ததும் தாலாட்டி இசை செய்தான், அவன் வளர்ந்து திருமணம் ஆகும் போதும் மங்கள இசைதான். இப்படி இசை வாழ்வில் எத்தனையோ இடத்திலும் இடம் பெற்றிருக்கிறது. நம் வாழ்க்கையில் ‘இதயம் கவரும் இசை அனைவரையும் கவரும் தசை’ என்று சொல்லி என் உரையை நிறைவு செய்கிறேன்.

நன்றி! வணக்கம்!!

சொல்லக் கேட்டு எழுதுக.

முல்லை நில மக்களாகிய ஆயர்கள் குழல் ஊதுவதில் வல்லவர்கள். இதனைச் சம்பந்தர் திருப்பதிகத்தில் அமைந்த நிகழ்ச்சி ஒன்று விளக்குகிறது. திருவண்ணாமலைச் சாரலில் ஆயர் ஒருவர் ஆநிரைகளையும் எருமையினங்களையும் மேய்த்துக் கொண்டிருந்தார். மாலையில் அவற்றை எல்லாம் ஒன்று திரட்டினார்.

அப்போது எருமை ஒன்று காணாமல் போனதை அறிந்தார். தன் கையிலிருந்த குழலை எடுத்து இனிய இசையை எழுப்பினார். இன்னிசை கேட்ட எருமை அவரை வந்தடைந்தது. இவ்வாறு ஆயர்களின் இசைத் திறத்தைத் திருப்பதிகம் விளக்குகிறது.

கோடிட்ட இடங்களில் பொருத்தமான சொல்லுருபுகளை இட்டு நிரப்புக.

(கொண்டு, இருந்து, உடைய, காட்டிலும், ஆக, நின்று, உடன், விட, பொருட்டு)

1. இடி ………………….. மழை வந்தது.
2. மலர்விழி தேர்வின் …………………. ஆயத்தமானாள்.
3. அருவி மலையில் …………………. வீழ்ந்தது.
4. தமிழைக் ……………….. சுவையான மொழியுண்டோ !
5. யாழ், தமிழர் …………………….. இசைக் கருவிகளுள் ஒன்று.
Answer:
1. உடன்
2. பொருட்டு
3. இருந்து
4. காட்டிலும்
5. உடைய

பின்வரும் இசைக்கருவிகளின் பெயர்களை அகர வரிசைப்படுத்துக.

படகம், தவில், கணப்பறை, பேரியாழ், உறுமி, உடுக்கை, தவண்டை , பிடில், நாகசுவரம், மகுடி.
Answer:
உடுக்கை, உறுமி, கணப்பறை, தவண்டை , தவில், நாகசுவரம், படகம், பிடில், பேரியாழ், மகுடி.

பின்வரும் இணைச்சொற்களை வகைப்படுத்துக.

உற்றார் உறவினர், விருப்பு வெறுப்பு, காலைமாலை, கன்னங்கரேல், ஆடல்பாடல், வாடிவதங்கி, பட்டிதொட்டி, உள்ளும் புறமும், மேடுபள்ளம், நட்ட நடுவில்.

சரியான இணைச்சொற்களை இட்டு நிரப்புக.

(மேடுபள்ளம், ஈடுஇணை, கல்விகேள்வி, போற்றிப்புகழப்பட, வாழ்வு தாழ்வு, ஆடிஅசைந்து)

1. சான்றோர் எனப்படுபவர் …………………………. களில் சிறந்தவர் ஆவார்.
2. ஆற்று வெள்ளம் …………………… பாராமல் ஓடியது.
3. இசைக்கலைஞர்கள் …………………. வேண்டியவர்கள்.
4. தமிழ் இலக்கியங்களின் பெருமைக்கு …………………….. இல்லை.
5. திருவிழாவில் யானை ………………. வந்தது.
Answer:
1. கல்விகேள்வி
2. மேடுபள்ளம்
3. போற்றிப்புகழப்பட
4. ஈடுஇணை
5. ஆடி அசைந்து

கடிதம் எழுதுக

இருப்பிடச் சான்று வேண்டி வட்டாட்சியருக்கு விண்ணப்பம் எழுதுக.

அனுப்புநர்
சா. சுந்தர்,
த/பெ. ஆ. சங்கர்
34, குறிஞ்சி நகர்,
ஈரோடு – 638 001.

பெறுநர்
உயர்திரு. வட்டாட்சியர் அவர்கள்,
வட்டாட்சியர் அலுவலகம்,
ஈரோடு.

மதிப்புக்குரிய அய்யா,

பொருள் : இருப்பிடச் சான்றிதழ் வேண்டுதல் சார்பாக. வணக்கம்.

ஈரோடு, அரசு உயர்நிலைப் பள்ளியில் எட்டாம் வகுப்பு படிக்கின்றேன். 34, குறிஞ்சி நகர், ஈரோடு – 638 001 என்ற முகவரியில் பத்து ஆண்டுகளாக நாங்கள் வசித்து வருகின்றோம். அரசின் கல்வி உதவித்தொகைக்கு விண்ணப்பிக்க இருப்பிடச் சான்றிதழ் தேவைப்படுகின்றது. இத்துடன் குடும்ப அட்டை நகலும் ஆதார் அட்டை நகலும் இணைத்துள்ளேன். ஆகவே, எனக்கு இருப்பிடச் சான்றிதழ் வழங்கும்படி பணிவுடன் கேட்டுக் கொள்கிறேன்.

நன்றி!

இடம் : ஈரோடு
நாள் : 25.06.2020

இப்படிக்கு,
தங்கள் உண்மையுள்ள,
சா. சுந்தர்.

உறைமேல் முகவரி:
பெறுநர்
உயர்திரு. வட்டாட்சியர் அவர்கள்,
வட்டாட்சியர் அலுவலகம்,
ஈரோடு.

மொழியோடு விளையாடு

குறுக்கெழுத்துப்புதிர்.

இடமிருந்து வலம் :
1. முதற்கருவி எனப் பெயர் பெற்றது.
2. யாழிலிருந்து உருவான பிற்காலக் கருவி
7. இயற்கைக் கருவி
12. விலங்கின் உறுப்பைப் பெயராகக் கொண்ட கருவி

வலமிருந்து இடம் :
4. வட்டமான மணி போன்ற கருவி
8. ஐந்து வாய்களைக் கொண்ட கருவி
9. இசைக்கருவிகளை இசைத்துப் பாடல் பாடுவோர்

மேலிருந்து கீழ் :
1. 19 நரம்புகளைக் கொண்ட யாழ்
3. ஒன்றோடு ஒன்று மோதி இசைக்கப்படுபவை
5. சிறிய வகை உடுக்கை
6. பறை ஒரு ……………… கருவி

கீழிருந்து மேல் :
8. மூங்கிலால் செய்யப்படும் காற்றுக்கருவி
10. வீணையில் உள்ள நரம்புகளின் எண்ணிக்கை
11. திருமணத்தின்போது கொட்டும் முரசு

நிற்க அதற்குத் தக

என் பொறுப்புகள்:

1. கைவினைக்கலைகளுள் ஒன்றைக் கற்றுக்கொள்வேன்.
2. இசைக் கலையை வளர்த்த சான்றோர்களைப் பற்றி அறிந்து போற்றுவேன்.

கலைச்சொல் அறிவோம்

1. கைவினைப் பொருள்கள் – Crafts
2. புல்லாங்குழல் – Flute
3. முரசு – Drum .
4. கூடைமுடைதல் – Basketry
5. பின்னுதல் – Knitting
6. கொம்பு – Horn
7. கைவினைஞர் – Artisan
8. சடங்கு – Rite

இணையத்தில் காண்க

இசையின் வகைப்பாடுகள் பற்றிய செய்திகளை இணையத்தில் தேடி எழுதுக.

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக..

Question 1.
திருவாசகம் படித்தாள் – இதில் மறைந்து வரும் வேற்றுமை உருபு …………………..
அ) இரண்டாம் வேற்றுமை உருபு
ஆ) மூன்றாம் வேற்றுமை உருபு
இ) நான்காம் வேற்றுமை உருபு
ஈ) ஐந்தாம் வேற்றுமை உருபு
Answer:
அ) இரண்டாம் வேற்றுமை உருபு

Question 2.
கம்பர் பாடல் – இதில் இடம்பெற்றுள்ள வேற்றுமையுருபு ……………………
அ) கு
ஆ) இன்
இ) அது
ஈ) கண்
Answer:
இ) அது

Question 3.
காலம் கரந்த பெயரெச்சம் ………………..
அ) வினைத்தொகை
ஆ) பண்புத்தொகை
இ) வேற்றுமைத்தொகை
ஈ) உவமைத்தொகை
Answer:
அ) வினைத்தொகை

Question 4.
ஆடுகொடி, வளர்தமிழ் – ஆகியன …………………. க்குச் சான்றுகள்.
அ) வேற்றுமைத்தொகை
ஆ) உம்மைத்தொகை
இ) உவமைத்தொகை
ஈ) வினைத்தொகை
Answer:
ஈ) வினைத்தொகை

Question 5.
வெண்ணிலவு, கருங்குவளை ஆகியன ………………….. க்குச் சான்றுகளாகும்.
அ) வினைத்தொகை
ஆ) பண்புத்தொகை
இ) உவமைத்தொகை
ஈ) உம்மைத்தொகை
Answer:
ஆ) பண்புத்தொகை

Question 6.
இருபெயரொட்டு பண்புத்தொகைக்குச் சான்றாக அமையும் ஒன்றினைத் தேர்வு செய்க.
அ) மலர்விழி
ஆ) இரவுபகல்
இ) பொற்றொடி வந்தாள்
ஈ) பனைமரம்
Answer:
ஈ) பனைமரம்

Question 7.
இரவுபகல், தாய்தந்தை ஆகியன ………………………. க்குச் சான்றாகும்.
அ) அன்மொழித்தொகை
ஆ) உவமைத்தொகை
இ) உம்மைத்தொகை
ஈ) இருபெயரொட்டுப் பண்புத்தொகை
Answer:
இ) உம்மைத்தொகை

Question 8.
தொகாநிலைத் தொடர் வகைகள்
அ) 6
ஆ) 8
இ) 9
ஈ) 3
Answer:
இ) 9

Question 9.
எழுவாய்த் தொடர் அமையும் சான்றினைத் தேர்ந்தெடுக்க.
அ) மல்லிகை மலர்ந்தது
ஆ) நண்பா படி
இ) சென்றனர் வீரர்
ஈ) வரைந்த ஓவியம்
Answer:
அ) மல்லிகை மலர்ந்தது

Question 10.
‘நண்பா படி’ என்பது ……………………
அ) எழுவாய்த் தொடர்
ஆ) விளித் தொடர்
இ) வினைமுற்றுத்தொடர்
ஈ) பெயரெச்சத் தொடர்
Answer:
ஆ) விளித்தொடர்

குறுவினா

Question 1.
வேற்றுமைத் தொகை என்றால் என்ன? சான்று தருக.
Answer:

•  இருசொற்களுக்கு இடையில் வேற்றுமை உருபு மறைந்துவந்து பொருள் தந்தால், அதனை வேற்றுமைத் தொகை என்பர். 
• சான்று: திருவாசகம் (ஐ)படித்தான். (இரண்டாம் வேற்றுமைத் தொகை)

Question 2.
உருபும் பயனும் உடன்தொக்கத் தொகை என்றால் என்ன? சான்று தருக.
Answer:

• ஒரு தொடரில் வேற்றுமை உருபும் அதன் பொருளை விளக்கும் சொல்லும் மறைந்து வருவது உருபும் பயனும் உடன் தொக்கத் தொகை எனப்படும்.
• சான்று: பால் குடம்.
• பாலைக் கொண்ட குடம்’ இரண்டாம் வேற்றுமை உருபும் பயனும் உடன் தொக்கத் தொகை.

Question 3.
வினைத்தொகையை சான்றுடன் விளக்குக.
Answer:

• காலம் காட்டும் இடைநிலையும், பெயரெச்சவிகுதியும் மறைந்து வரும் பெயரெச்சம் வினைத்தொகை என்பர்.
• சான்று: வளர்தமிழ். இதில் காலம் காட்டும் இடைநிலைகள் தொக்கி வந்துள்ளன.
• வளர்ந்த தமிழ், வளர்கின்ற தமிழ், வளரும் தமிழ் எனவும் முக்காலத்திற்கும் பொருந்தும் படியாகப் பொருள் தருகின்றன. எனவே, இது வினைத்தொகை ஆகும்.

Question 4.
பண்புத்தொகை என்றால் என்ன? சான்று தருக.
Answer:

• பண்புப்பெயருக்கும் அது தழுவி நிற்கும் பெயர்ச்சொல்லுக்கும் இடையே ‘ஆன்’, ‘ஆகிய’ என்னும் பண்பு உருபுகள் மறைந்து வருவது பண்புத்தொகை எனப்படும்.
• சான்று: வெண்ணிலவு, கருங்குவளை.

Question 5.
‘மலர்விழி’ என்னும் சான்று அமையும் தொகை யாது? விளக்கு.
Answer:
‘மலர்விழி’ என்பது உவமைத்தொகை. மலர் போன்ற விழி என்ற பொருளைத் தருகிறது. ‘மலர்’ என்பது உவமை. ‘விழி’ என்பது உவமேயம். ‘போன்ற’ என்பது உவம உருபு. உவமைக்கும் உவமேயத்துக்கும் இடையில் உவம உருபு மறைந்து வந்தால் அது உவமைத்தொகை எனப்படும்.

Question 6.
எண்ணும்மை என்றால் என்ன? சான்று தருக.
Answer:
ஒன்றுக்கு மேற்பட்ட சொற்களில் ‘உம்’ என்னும் உருபு வெளிப்பட வருவது எண்ணும்மை ஆகும். சான்று: இரவும் பகலும், பசுவும் கன்றும்

சிறு வினா

Question 1.
தொகாநிலைத் தொடர் என்றால் என்ன? அதன் வகைகளை எழுதுக.
Answer:
(i) ஒரு தொடரில் இரு சொற்கள் வந்து அவற்றின் இடையில் சொல்லுருபு மறையாமல் நின்று பொருள் தந்தால் அதனை தொகாநிலைத்தொடர் என்பர்.

(ii) தொகாநிலைத்தொடர் ஒன்பது வகைப்படும். எழுவாய்த்தொடர், விளித்தொடர், வினைமுற்றுத் தொடர், பெயரெச்சத் தொடர், வினையெச்சத் தொடர், வேற்றுமைத் தொகாநிலைத் தொடர், இடைச்சொல் தொடர், உரிச்சொல் தொடர், அடுக்குத்தொடர்.

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Samacheer Kalvi 11th Physics Chapter 3 Laws of Motion Textual Questions Solved

Samacheer Kalvi 11th Physics Laws of Motion Multiple Choice Questions

11th Physics Chapter 3 Book Back Answers Question 1.
When a car takes a sudden left turn in the curved road, passengers are pushed towards the right due to
(a) inertia of direction
(b) inertia of motion
(c) inertia of rest
(d) absence of inertia
Answer:
(a) inertia of direction

Samacheer Kalvi 11th Physics Solution Chapter 3 Question 2.
An object of mass m held against a vertical wall by applying horizontal force F as shown in the figure. The minimum value of the force F is [IIT JEE 1994]
(a) Less than mg
(b) Equal to mg
(c) Greater than mg
(d) Cannot determine
Answer:
(c) Greater than mg

11th Physics Lesson 3 Book Back Answers Question 3.
A vehicle is moving along the positive x direction, if sudden brake is applied, then
(a) frictional force acting on the vehicle is along negative x direction
(b) frictional force acting on the vehicle is along positive x direction
(c) no frictional force acts on the vehicle
(d) frictional force acts in downward direction
Answer:
(a) frictional force acting on the vehicle is along negative x direction

11th Physics 3rd Chapter Book Back Answers Question 4.
A book is at rest on the table which exerts a normal force on the book. If this force is considered as reaction force, what is the action force according to Newton’s third law?
(a) Gravitational force exerted by Earth on the book
(b) Gravitational force exerted by the book on Earth
(c) Normal force exerted by the book on the table
(d) None of the above
Answer:
(c) Normal force exerted by the book on the table

Laws Of Motion Class 11 State Board Question 5.
Two masses m₁ and m₂ are experiencing the same force where m₁ < m₂ The ratio of their acceleration \(\frac{a_{1}}{a_{2}}\) is –
(a) 1
(b) less than 1
(c) greater than 1
(d) all the three cases
Answer:
(c) greater than 1

Samacheer Kalvi Guru 11th Physics Question 6.
Choose appropriate free body diagram for the particle experiencing net acceleration along negative y direction. (Each arrow mark represents the force acting on the system).

Answer:

Class 11 Physics Solutions Samacheer Kalvi Question 7.
A particle of mass m sliding on the smooth double inclined plane (shown in figure) will experience –
(a) greater acceleration along the path AB
(b) greater acceleration along the path AC
(c) same acceleration in both the paths
(d) no acceleration in both the paths

Answer:
(a) greater acceleration along the path AC

Samacheer Kalvi 11th Physics Question 8.
Two blocks of masses m and 2m are placed on a smooth horizontal surface as shown. In the first case only a force F₁  is applied from the left. Later only a force F₂  is applied from the right. If the force acting at the interface of the two blocks in the two cases is same, then F₁ : F₂ is [Physics Olympiad 2016]

(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer:
(c) 2 : 1

11th Physics 3rd Lesson Book Back Answers Question 9.
Force acting on the particle moving with constant speed is –
(a) always zero
(b) need not be zero
(c) always non zero
(d) cannot be concluded
Answer:
(b) need not be zero

11th Physics 3rd Chapter Exercise Question 10.
An object of mass m begins to move on the plane inclined at an angle 0. The coefficient of static friction of inclined surface is lay. The maximum static friction experienced by the mass is –
(a) mg
(b) µ_s mg
(c) µ_s mg sin θ
(d) µ_s mg cos θ
Answer:
(d) µ_s mg cos θ

Class 11 Samacheer Physics Solutions Question 11.
When the object is moving at constant velocity on the rough surface –
(a) net force on the object is zero
(b) no force acts on the object
(c) only external force acts on the object
(d) only kinetic friction acts on the object
Answer:
(a) net force on the object is zero

Samacheer Kalvi 11th Physics Book Back Answers Question 12.
When an object is at rest on the inclined rough surface –
(a) static and kinetic frictions acting on the object is zero
(b) static friction is zero but kinetic friction is not zero
(c) static friction is not zero and kinetic friction is zero
(d) static and kinetic frictions are not zero
Answer:
(c) static friction is not zero and kinetic friction is zero

Class 11 Physics Samacheer Kalvi Question 13.
The centrifugal force appears to exist –
(a) only in inertial frames
(b) only in rotating frames
(c) in any accelerated frame
(d) both in inertial and non-inertial frames
Answer:
(b) only in rotating frames

11th Physics Unit 3 Book Back Answers Question 14.
Choose the correct statement from the following –
(a) Centrifugal and centripetal forces are action reaction pairs
(b) Centripetal forces is a natural force
(c) Centrifugal force arises from gravitational force
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion
Answer:
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion

Laws Of Motion Class 11 Numericals With Solutions Pdf Question 15.
If a person moving from pole to equator, the centrifugal force acting on him –
(a) increases
(b) decreases
(c) remains the same
(d) increases and then decreases
Answer:
(a) increases

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions

Question 1.
Explain the concept of inertia. Write two examples each for inertia of motion, inertia of rest and inertia of direction.
Answer:
The inability of objects to move on its own or change its state of motion is called inertia. Inertia means resistance to change its state. There are three types of inertia:
1. Inertia of rest:
The inability of an object to change its state of rest is called inertia of rest.
Example:

• When a stationary bus starts to move, the passengers experience a sudden backward push.
• A book lying on the table will remain at rest until it is moved by some external agencies.

2. Inertia of motion:
The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.
Example:

• When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front seat.
• An athlete running is a race will continue to run even after reaching the finishing point.

3. Inertia of direction:
The inability of an object to change its direction of motion on its own is called inertia of direction.
Example:

• When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle.
• When a bus moving along a straight line takes a turn to the right. The passengers are thrown towards left.

Question 2.
State Newton’s second law.
Answer:
The force acting on an object is equal to the rate of change of its momentum –
\(\overline{\mathrm{F}}\) = \(\frac{d \bar{p}}{d t}\)

Question 3.
Define one newton.
Answer:
One newton is defined as the force which acts on 1 kg of mass to give an acceleration 1 ms^-2 in the direction of the force.

Question 4.
Show that impulse is the change of momentum.
Answer:
According to Newton’s Second Law
F = \(\frac {dp}{dt}\) i.e. dp = Fdt
Integrate it over a time interval from t_i  to t_f

P_i → initial momentum of the object at t_i
P_f → Final momentum of the object at t_f
P_f – P_i = ∆p = change in momentum during the time interval ∆t.
\(\int_{t_{i}}^{t_{f}} \mathrm{F} \cdot d t=\mathrm{J}\) is called the impulse.
If the force is constant over the time interval ∆t, then

Hence the proof.

Question 5.
Using free body diagram, show that it is easy to pull an object than to push it.
Answer:
When a body is pushed at an arbitrary angle θ [0 to \(\frac {π}{2}\)], the applied force F can be resolved into two components as F sin 0 parallel to the surface and F cos 0 perpendicular to the surface as shown in figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is no acceleration along the vertical direction the normal force N is equal to
N_push = mg + F cos θ …………(1)
As a result the maximal static friction also increases and is equal to
\(f_{S}^{\max }\) = \(\mu_{r} \mathrm{N}_{\mathrm{push}}\) = µ_s(mg + F cos θ) ……(2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.

When an object is pulled at an angle θ, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is –
N_pull = mg – F cos θ ………….(3)

Equation (3) shows that the normal force is less than – N_push. From equations (1) and (3), it is easier to pull an object than to push to make it move.

Question 6.
Explain various types of friction. Suggest a few methods to reduce friction.
Answer:
There are two types of Friction:
(1) Static Friction:
Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force f_s lies between
\(0 \leq f_{s} \leq \mu_{s} \mathrm{N}\)
where, µ_s – coefficient of static friction
N – Normal force

(2) Kinetic friction:
The frictional force exerted by the surface when an object slides is called as kinetic friction. Also called as sliding friction or dynamic friction,
f_k – µ_kN
where µ_k – the coefficient of kinetic friction
N – Normal force exerted by the surface on the object

Methods to reduce friction:
Friction can be reduced

• By using lubricants
• By using Ball bearings
• By polishing
• By streamlining

Question 7.
What is the meaning by ‘pseudo force’?
Answer:
Pseudo force is an apparent force which has no origin. It arises due to the non-inertial nature of the frame considered.

Question 8.
State the empirical laws of static and kinetic friction.
Answer:
The empirical laws of friction are:

• Friction is independent of surface of contact.
• Coefficient of kinetic friction is less than coefficient of static friction.
• The direction of frictional force is always opposite to the motion of one body over the other.
• Frictional force always acts on the object parallel to the surface on which the objet is placed,
• The magnitude of frictional force between any two bodies in contact is directly proportional to the normal reaction between them.

Question 9.
State Newton’s third law.
Answer:
Newton’s third law states that for every action there is an equal and opposite reaction.

Question 10.
What are inertial frames?
Answer:
Inertial frame is the one in which if there is no force on the object, the object moves at constant velocity.

Question 11.
Under what condition will a car skid on a leveled circular road?
Answer:
On a leveled circular road, if the static friction is not able to provide enough centripetal ’force to turn, the vehicle will start to skid
\(\mu_{s}<\frac{v^{2}}{r g}\)

Samacheer Kalvi 11th Physics Laws of Motion Long Answer Questions

Question 1.
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it.
Answer:
In nature, conservation laws play a very important role. The dynamics of motion of bodies can be analysed very effectively using conservation laws. There are three conservation laws in mechanics. Conservation of total energy, conservation of total linear momentum, and conservation of angular momentum. By combining Newton’s second and third laws, we can derive the law of conservation of total linear momentum. When two particles interact with each other, they exert equal and opposite forces on each other.

The particle 1 exerts force \(\overrightarrow{\mathrm{F}}_{12}\) on particle 2 and particle 2 exerts an exactly equal and opposite force \(\overrightarrow{\mathrm{F}}_{12}\) on particle 1 according to Newton’s third law.
\(\overrightarrow{\mathrm{F}}_{12}\) = –\(\overrightarrow{\mathrm{F}}_{12}\) ……..(1)
In terms of momentum of particles, the force on each particle (Newton’s second law) can be written as –
\(\overrightarrow{\mathrm{F}}_{12}\) = \(\frac{d \vec{p}_{1}}{d t}\) and \(\overrightarrow{\mathrm{F}}_{21}\) = \(\frac{d \vec{p}_{2}}{d t}\) ………(2)
Here \(\vec{p}_{1}\) is the momentum of particle 1 which changes due to the force \(\overrightarrow{\mathrm{F}}_{12}\) exerted by particle 2. Further Po is the momentum of particle \(\vec{p}_{2}\) This changes due to \(\overrightarrow{\mathrm{F}}_{21}\) exerted by particle 1.
Substitute equation (2) in equation (1)
\(\frac{d \vec{p}_{1}}{d t}\) = – \(\frac{d \vec{p}_{2}}{d t}\) …………(3)
\(\frac{d \vec{p}_{1}}{d t}\) + \(\frac{d \vec{p}_{2}}{d t}\) = 0 ………(4)
\(\frac {d}{dt}\)(\(\vec{p}_{1}\) + \(\vec{p}_{2}\)) = 0
It implies that \(\vec{p}_{1}\) + \(\vec{p}_{2}\) = constant vector (always).
\(\vec{p}_{1}\) + \(\vec{p}_{2}\) is the total linear momentum of the two particles (\(\vec{p}_{tot}\) = \(\vec{p}_{1}\) + \(\vec{p}_{2}\)).It is also called as total linear momentum of the system. Here, the two particles constitute the system. From this result, the law of conservation of linear momentum can be stated as follows.

If there are no external forces acting on the system, then the total linear momentum of the system (\(\vec{p}_{tot}\)) is always a constant vector. In other words, the total linear momentum of the system is conserved in time. Here the word ‘conserve’ means that \(\vec{p}_{1}\) and \(\vec{p}_{2}\) can vary, in such a way that \(\vec{p}_{1}\) + \(\vec{p}_{2}\) is a constant vector.

The forces \(\overrightarrow{\mathrm{F}}_{12}\) and \(\overrightarrow{\mathrm{F}}_{21}\) are called the internal forces of the system, because they act only between the two particles. There is no external force acting on the two particles from outside. In such a case the total linear momentum of the system is a constant vector or is conserved.

Meaning of law of conservation of momentum:
1. The Law of conservation of linear momentum is a vector law. It implies that both the magnitude and direction of total linear momentum are constant. In some cases, this total momentum can also be zero.

2. To analyse the motion of a particle, we can either use Newton’s second law or the law of conservation of linear momentum. Newton’s second law requires us to specify the forces involved in the process. This is difficult to specify in real situations. But conservation of linear momentum does not require any force involved in the process. It is convenient and hence important.

For example, when two particles collide, the forces exerted by these two particles on each other is difficult to specify. But it is easier to apply conservation of linear momentum during the collision process.

Examples:
Consider the firing of a gun. Here the system is Gun+bullet. Initially the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let \(\vec{p}_{1}\) be the momentum of the bullet and \(\vec{p}_{2}\) the momentum of the gun before firing. Since initially both are at rest,

Total momentum before firing the gun is zero, \(\vec{p}_{1}\) + \(\vec{p}_{2}\) = 0.
According to the law of conservation of linear momentum, total linear momentum has to be zero after the firing also.

When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from \(\vec{p}_{1}\) to \(\vec{p}_{1}\) To conserve the total linear momentum of the system, the momentum of the gun must also change from \(\vec{p}_{2}\) to \(\vec{p}_{2}^{\prime}\). Due to the conservation of linear momentum, \(\vec{p}_{1}\) + \(\vec{p}_{2}^{\prime}\) = 0.

It implies that \(\vec{p}_{1}^{\prime}\)= \(\vec{p}_{2}^{\prime}\), the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet. This is the reason after firing, the gun suddenly moves backward with the momentum (-\(\vec{p}_{2}^{\prime}\)). It is called ‘recoil momentum’. Th is is an example of conservation of total linear momentum.

Question 2.
What are concurrent forces? State Lami’s theorem.
Answer:
Concurrent force:
A collection of forces is said to be concurrent, if the lines of forces act at a common point.

Lami’s Theorem:
If a system of three concurrent and coplanar forces is in equilibrium, then Lami’s theorem states that the magnitude of each force of the system is proportional to sine of the angle between the other two forces.
i.e. |\(\overrightarrow{\mathrm{F}}_{1}\)|∝ sin α, |\(\overrightarrow{\mathrm{F}}_{2}\)| ∝ sin β, |\(\overrightarrow{\mathrm{F}}_{2}\)| ∝ sin γ,

Question 3.
Explain the motion of blocks connected by a string in

1. Vertical motion
2. Horizontal motion.

Answer:
When objects are connected by strings and When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Case 1:
Vertical motion:
Consider two blocks of masses m₁ and m₂ (m₁> m₂) connected by a light and in-extensible string that passes over a pulley as shown in Figure.

Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving, m₂ vertically upward and m_k, downward with same acceleration a. The gravitational force m₁g on mass m₁ is used in lifting the mass m₂. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.

Applying Newton’s second law for mass m₂ T \(\hat{j}\) – m₂ g \(\hat{j}\) = m₂ a \(\hat{j}\) The left hand side of the above equation is the total force that acts on m₂ and the right hand side is the product of mass and acceleration of m₂ in y direction.
By comparing the components on both sides, we get
T = m₂ g = m₂ a ……….(1)
Similarly, applying Newton’s second law for mass m₂
T \(\hat{j}\) – m₁ g\(\hat{j}\) = m₁a\(\hat{j}\)
As mass m₁ moves downward (-\(\hat{j}\)), its acceleration is along (-\(\hat{j}\))
By comparing the components on both sides, we get
T = m₁g = -m₁a
m₁g – T = m₁a ………..(2)
Adding equations (1) and (2), we get
m₁g – m₂g = m₁a + m₂a
(m₁ – m₂)g = (m₁ + m₂)a …………(3)
From equation (3), the acceleration of both the masses is –
a = (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ………..(4)
If both the masses are equal (m₁ = m₂), from equation (4)
a = 0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
T = m₂g = m₂(\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))
T = m₂g + m₂ (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ……….(5)
By taking m₂g common in the RHS of equation (5)

Equation (4) gives only magnitude of acceleration.
For mass m₁, the acceleration vector is given by \(\vec{a}\) = –\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\)\(\hat{j}\)
For mass m₂, the acceleration vector is given by \(\vec{a}\) = \(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\) \(\hat{j}\)

Case 2:
Horizontal motion:
In this case, mass m₂ is kept on a horizontal table and mass m₁, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface

As both the blocks are connected to the un stretchable string, if m₁ moves with an acceleration a downward then m₂ also moves with the same acceleration a horizontally.
The forces acting on mass m₂ are

• Downward gravitational force (m₂g)
• Upward normal force (N) exerted by the surface
• Horizontal tension (T) exerted by the string

The forces acting on mass m₁ are

• Downward gravitational force (m₁g)
• Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure.

Applying Newton’s second law for m₁
T\(\hat{i}\) – m₁g\(\hat{j}\) = -m₁a\(\hat{j}\) (alongy direction)
By comparing the components on both sides of the above equation,
T – m₁g = -m₁a …………(1)
Applying Newton’s second law for m₂
Ti = m₁ai (along x direction)
By comparing the components on both sides of above equation,
T = m₂a ………….(2)
There is no acceleration along y direction for m₂.
N\(\hat{j}\) – m₂g\(\hat{j}\) = 0
By comparing the components on both sides of the above equation
N – m₂g = 0
N = m₂g ……….(3)
By substituting equation (2) in equation (1), we can find the tension T
m₂a – m₁g = -m₁a
m₂a + m₁a = m₁g
a = \(\frac{m_{1}}{m_{1}+m_{2}}\)g …………(4)
Tension in the string can be obtained by substituting equation (4) in equation (2)
T = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\)g ………..(5)

Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

Question 4.
Briefly explain the origin of friction. Show that in an inclined plane, angle of friction is pqual to angle of repose.
Answer:
If a very gentle force in the horizontal direction is given to an object at rest on the table it does not move. It is because of the opposing force exerted by the surface on the object which resists its motion. This force is called the frictional force. During the time of Newton and Galileo, frictional force was considered as one of the natural forces like gravitational force. But in the twentieth century, the understanding on atoms, electron and protons has changed the perspective.

The frictional force is actually the electromagnetic force between the atoms on the two surfaces. Even well polished surfaces have irregularities on the surface at the microscopic level. The component of force parallel to the inclined plane (mg sin θ) tries to move the object down. The component of force perpendicular to the inclined plane (mg cos θ) is balanced by the Normal force (N).
N = mg cos θ ………(1)

When the object just begins to move, the static friction attains its maximum value
f_s = \(f_{s}^{\max }\) = µ_s N
This friction also satisfies the relation
\(f_{s}^{\max }\) = µ_s mg sin θ ……….(2)
Equating the right hand side of equations (1) and (2),
(\(f_{s}^{\max }\))/N = sin θ / cos θ
From the definition of angle of friction, we also know that
tan θ = µ_s ………..(3)
in which θ is the angle of friction.
Thus the angle of repose is the same as angle of friction. But the difference is that the angle of repose refers to inclined surfaces and the angle of friction is applicable to any type of surface.

Question 5.
State Newton’s three laws and discuss their significance.
Answer:
First Law:
Every object continues to be in the state of rest or of uniform motion (constant velocity) unless there is external force acting on it.

Second Law:
The force acting on an object is equal to the rate of change of its momentum

Third Law:
For every action there is an equal and opposite reaction.

Significance of Newton’s Laws:
1. Newton’s laws are vector laws. The equation \(\overline{\mathrm{F}}\) = m\(\overline{\mathrm{a}}\) is a vector equation and essentially it is equal to three scalar equations. In Cartesian coordinates, this equation can be written as F_x\(\hat{i}\) + F_y\(\hat{j}\) + F_z\(\hat{j}\) = ma_x\(\hat{i}\) + ma_y\(\hat{j}\) + ma_z\(\hat{j}\)
By comparing both sides, the three scalar equations are

F_x = ma_x The acceleration along the x-direction depends only on the component of force acting along the x – direction.
F_y = ma_y The acceleration along the y direction depends only on the component of force acting along the y – direction.
F_z = ma_z The acceleration along the z direction depends only on the component of force acting along the z – direction.
From the above equations, we can infer that the force acting along y direction cannot alter the acceleration along x direction. In the same way, F_z cannot affect a_y and a_x. This understanding is essential for solving problems.

2. The acceleration experienced by the body at time t depends on the force which acts on the body at that instant of time. It does not depend on the force which acted on the body before the time t. This can be expressed as
\(\overline{\mathrm{F}}\)(t) = m\(\overline{\mathrm{a}}\)(t)
Acceleration of the object does not depend on the previous history of the force. For example, when a spin bowler or a fast bowler throws the ball to the batsman, once the ball leaves the hand of the bowler, it experiences only gravitational force and air frictional force. The acceleration of the ball is independent of how the ball was bowled (with a lower or a higher speed).

3. In general, the direction of a force may be different from the direction of motion. Though in some cases, the object may move in the same direction as the direction of the force, it is not always true. A few examples are given below.

Case 1:
Force and motion in the same direction:
When an apple falls towards the Earth, the direction of motion (direction of velocity) of the apple and that of force are in the same downward direction as shown in the Figure.

Case 2:
Force and motion not in the same direction:
The Moon experiences a force towards the Earth. But it actually moves in elliptical orbit. In this case, the direction of the force is different from the direction of motion as shown in Figure.

Case 3:
Force and motion in opposite direction:
If an object is thrown vertically upward, the direction of motion is upward, but gravitational force is downward as shown in the Figure.

Case 4:
Zero net force, but there is motion:
When a raindrop gets detached from the cloud it experiences both downward gravitational force and upward air drag force. As it descends towards the Earth, the upward air drag force increases and after a certain time, the upward air drag force cancels the downward gravity. From then on the raindrop moves at constant velocity till it touches the surface of the Earth. Hence the raindrop comes with zero net force, therefore with zero acceleration but with non-zero terminal velocity. It is shown in the Figure

4. If multiple forces \(\overrightarrow{\mathrm{F}}_{1}\), \(\overrightarrow{\mathrm{F}}_{2}\), \(\overrightarrow{\mathrm{F}}_{3}\),……. \(\overrightarrow{\mathrm{F}}_{n}\) act on the same body, then the total force (\(\overrightarrow{\mathrm{F}}_{net}\)) is equivalent to the vectorial sum of the individual forces. Their net force provides the acceleration.
\(\overrightarrow{\mathrm{F}}_{net}\) = \(\overrightarrow{\mathrm{F}}_{1}\) + \(\overrightarrow{\mathrm{F}}_{2}\) + \(\overrightarrow{\mathrm{F}}_{3}\) + ……… + \(\overrightarrow{\mathrm{F}}_{n}\)

Newton’s second law for this case is –
\(\overrightarrow{\mathrm{F}}_{net}\) = m\(\overline{\mathrm{a}}\)
In this case the direction of acceleration is in the direction of net force.
Example:
Bow and arrow

5. Newton’s second law can also be written in the following form.
Since the acceleration is the second derivative of position vector of the body \(\left(\vec{a}=\frac{d^{2} \vec{r}}{d t^{2}}\right)\)
the force on the body is –
\(\overline{\mathrm{F}}\) = m\(\frac{d^{2} \vec{r}}{d t^{2}}\)
From this expression, we can infer that Newton’s second law is basically a second order ordinary differential equation and whenever the second derivative of position vector is not zero, there must be a force acting on the body.

6. If no force acts on the body then Newton’s second law, m = \(\frac{d \vec{v}}{d t}\) = 0
It implies that \(\overline{\mathrm{v}}\) = constant. It is essentially Newton’s first law. It implies that the second law is consistent with the first law. However, it should not be thought of as the reduction of second law to the first when no force acts on the object. Newton’s first and second laws are independent laws. They can internally be consistent with each other but cannot be derived from each other.

7. Newton’s second law is cause and effect relation. Force is the cause and acceleration is the effect. Conventionally, the effect should be written on the left and cause on the right hand side of the equation. So the correct way of writing Newton’s second law is –
m\(\overline{\mathrm{a}}\) = \(\overline{\mathrm{F}}\) or \(\frac{d \vec{p}}{d t}\) = \(\overline{\mathrm{F}}\)

Question 6.
Explain the similarities and differences of centripetal and centrifugal forces.
Answer:
Salient features of centripetal and centrifugal forces.
Centripetal Force:

• It is a real force which is exerted on the body by the external agencies like gravitational force, tension in the string, normal force etc.
• Acts in both inertial and non-inertial frames
• It acts towards the axis of rotation or center of the circle in circular motion
  \(\left|\overrightarrow{\mathrm{F}}_{\mathrm{C}_{\mathrm{P}}}\right|\) = mω²r = \(\frac{m v^{2}}{r}\)
• Real force and has real effects
• Origin of centripetal force is interaction between two objects.
• In inertial frames centripetal force has to be included when free body diagrams are drawn.

Centrifugal Force:

• It is a pseudo force or fictitious force which cannot arise from gravitational force, tension force, normal force etc.
• Acts only in rotating frames (non-inertial frame)
• It acts outwards from the axis of rotation or radially outwards from the center of the circular motion
  \(\left|\overrightarrow{\mathrm{I}}_{\mathrm{C}_{\mathrm{f}}}\right|\) = mω²r = \(\frac{m v^{2}}{r}\)
• Pseudo force but has real effects
• Origin of centrifugal force is inertia. It does not arise from interaction. In an inertial frame the object’s inertial motion appears as centrifugal force in the rotating frame.
• In inertial frames there is no centrifugal force. In rotating frames, both centripetal and centrifugal force have to be included when free body diagrams are drawn.

Question 7.
Briefly explain ‘centrifugal force’ with suitable examples.
Answer:
To use Newton’s first and second laws in the rotational frame of reference, we need to include a Pseudo force called centrifugal force. This centrifugal force appears to act on the object with respect to rotating frames.

Circular motion can be analysed from two different frames of reference. One is the inertial frame (which is either at rest or in uniform motion) where Newton’s laws are obeyed. The other is the rotating frame of reference which is a non – inertial frame of reference as it is accelerating.

When we examine the circular motion from these frames of reference the situations are entirely different. To use Newton’s first and second laws in the rotational frame of reference, we need to include a pseudo force called ‘centrifugal force’. This ‘centrifugal force’ appears to act on the object with respect to rotating frames. To understand the concept of centrifugal force, we can take a specific case and discuss as done below.

Free body diagram of a particle including the centrifugal force Consider the case of a whirling motion of a stone tied to a string. Assume that the stone has angular velocity ω in the inertial frame (at rest). If the motion of the stone is observed from a frame which is also rotating along with the stone with same angular velocity ω then, the stone appears to be at rest.

This implies that in addition to the inward centripetal force – mω²r there must be an equal and opposite force that acts on the stone outward with value +mω²r . So the total force acting on the stone in a rotating frame is equal to zero (-mω²r + mω² r = 0). This outward force +mω²r is called the centrifugal force. The word ‘centrifugal’ means ‘flee from center’.

Note that the ‘centrifugal force’ appears to act on the particle, only when we analyse the motion from a rotating frame. With respect to an inertial frame there is only centripetal force which is given by the tension in the rstring. For this reason centrifugal force is called as a ‘pseudo force’. A pseudo force has no origin. It arises due to the non inertial nature of the frame considered. When circular motion problems are solved from a rotating frame of reference, while drawing free body diagram of a particle, the centrifugal force should necessarily be included as shown in the figure.

Question 8.
Briefly explain ‘rolling friction’.
Answer:
The invention of the wheel plays a crucial role in human civilization. One of the important applications is suitcases with rolling on coasters. Rolling wheels makes it easier than carrying luggage. When an object moves on a surface, essentially it is sliding on it. But wheels move on the surface through rolling motion. In rolling motion when a wheel moves on a surface, the point of contact with surface is always at rest. Since Rolling and kinetic friction the point of contact is at rest, there is no relative motion between the wheel and surface.

Hence the frictional fore is very less. At the same time if an object moves without a wheel, there is a relative motion between the object and the surface. As a result frictional force is larger. This makes it difficult to move the object. The figure shows the difference between rolling and kinetic friction. Ideally in pure rolling, motion of the point of contact with the surface should be at rest, but in practice it is not so.

Due to the elastic nature of the surface at the point of contact there will be some deformation on the object at this point on the wheel or surface as shown in figure. Due to this deformation, there will be minimal friction between wheel and surface. It is called ‘rolling friction. In fact, rolling friction’ is much smaller than kinetic friction.

Question 9.
Describe the method of measuring angle of repose.
Answer:
When objects are connected by strings and When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Case 1:
Vertical motion:
Consider two blocks of masses m₁ and m₂ (m₁> m₂) connected by a light and in extensible string that passes over a pulley as shown in Figure.

Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving, m₂ vertically upward and m_k, downward with same acceleration a. The gravitational force m₁g on mass m₁ is used in lifting the mass m₂. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.

Applying Newton’s second law for mass m₂
T \(\hat{j}\) – m₂g\(\hat{j}\) = m₂a\(\hat{j}\) The left hand side of the above equation is the total force that acts on m2 and the right hand side is the product of mass and acceleration of m₂ in y direction.
By comparing the components on both sides, we get
T = m₂g = m₂a ……….(1)

Similarly, applying Newton’s second law for mass m₂
T \(\hat{j}\) – m₁g\(\hat{j}\) = m₁a\(\hat{j}\)
As mass mj moves downward (-\(\hat{j}\)), its acceleration is along (-\(\hat{j}\))
By comparing the components on both sides, we get
T = m₁g = -m₁a
m₁g – T = m₁a ………..(2)

Adding equations (1) and (2), we get
m₁g – m₂g = m₁a + m₂a
(m₁ – m₂)g = (m₁ + m₂)a …………(3)
From equation (3), the acceleration of both the masses is –
a = (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ………..(4)
If both the masses are equal (m₁ = m₂), from equation (4)
a = 0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
T = m₂g = m₂(\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))
T = m₂g + m₂ (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ……….(5)
By taking m2g common in the RHS of equation (5)

Equation (4) gives only magnitude of acceleration
For mass m₁, the acceleration vector is given by \(\vec{a}\) = –\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\)\(\hat{j}\)
For mass m₂, the acceleration vector is given by \(\vec{a}\) = \(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\) \(\hat{j}\)

Case 2:
Horizontal motion:
In this case, mass m₂ is kept on a horizontal table and mass m₁, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface

As both the blocks are connected to the un stretchable string, if m₁ moves with an acceleration a downward then m₂ also moves with the same acceleration a horizontally.
The forces acting on mass m₂ are

1. Downward gravitational force (m₂g)
2. Upward normal force (N) exerted by the surface
3. Horizontal tension (T) exerted by the string

The forces acting on mass m₁ are

1. Downward gravitational force (m₁g)
2. Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure 2.

Applying Newton’s second law for m₁
T\(\hat{i}\) – m₁g\(\hat{j}\) = -m₁a\(\hat{j}\) (alongy direction)
By comparing the components on both sides of the above equation,
T – m₁g = -m₁a …………(1)
Applying Newton’s second law for m₂
T\(\hat{i}\) = m₁a\(\hat{i}\) (along x direction)
By comparing the components on both sides of above equation,
T = m₂a ………….(2)
There is no acceleration along y direction for m₂.
N\(\hat{j}\) – m₂g\(\hat{j}\) = 0
By comparing the components on both sides of the above equation
N – m₂g = 0
N = m₂g ……….(3)
By substituting equation (2) in equation (1), we can find the tension T
m₂a – m₁g = -m₁a
m₂a + m₁a = m₁g
a = \(\frac{m_{1}}{m_{1}+m_{2}}\)g …………(4)
Tension in the string can be obtained by substituting equation (4) in equation (2)
T = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\)g ………..(5)

Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

Question 10.
Explain the need for banking of tracks.
Answer:
In a leveled circular road, skidding mainly depends on the coefficient of static friction py. The coefficient of static friction depends on the nature of the surface which has a maximum limiting value. To avoid this problem, usually the outer edge of the road is slightly raised compared to inner edge. This is called banking of roads or tracks. This introduces an inclination, and the angle is called banking angle. “Let the surface of the road make angle 9 with horizontal surface. Then the normal force makes the same angle 9 with the vertical. When the car takes a turn.

there are two forces acting on the car:
(a) Gravitational force mg (downwards)
(b) Normal force N (perpendicular to surface)
We can resolve the normal force into two components N cos θ and N sin θ. The component balances the downward gravitational force ‘mg’ and component will provide the necessary centripetal acceleration. By using Newton second law.

The banking angle 0 and radius of curvature of the road or track determines the safe speed of the car at the turning. If the speed of car exceeds this safe speed, then it starts to skid outward but frictional force comes into effect and provides an additional centripetal force to prevent the outward skidding.

At the same time, if the speed of the car is little lesser than safe speed, it starts to skid inward and frictional force comes into effect, which reduces centripetal force to prevent inward skidding. However if the speed of the vehicle is sufficiently greater than the correct speed, then frictional force cannot stop the car from skidding.

Question 11.
Calculate the centripetal acceleration of Moon towards the Earth.
Answer:
The centripetal acceleration is given by a = \(\frac{v^{2}}{r}\) This expression explicitly depends on Moon’s speed which is nontrivial. We can work with the formula
ω²R_m = a_m
a_m is centripetal acceleration of the Moon due to Earth’s gravity, ω is angular velocity
R_m is the distance between Earth and the Moon, which is 60 times the radius of the Earth.
R_m = 60R = 60 x 6.4 x 10⁶ = 384 x 10⁶ m
As we know the angular velocity ω = \(\frac { 2π}{ T }\) and T = 27.3 days = 27.3 x 24 x 60 x 60 second = 2.358 x 10⁶ sec.
By substituting these values in the formula for acceleration
a₆ = \(\frac{\left(4 \pi^{2}\right)\left(384 \times 10^{6}\right)}{\left(2.358 \times 10^{8}\right)^{2}}\) = 0.00272 ms^-2

Samacheer Kalvi 11th Physics Laws of Motion Conceptual Questions

Question 1.
Why it is not possible to push a car from inside?
Answer:
While trying to push a car from outside, he pushes the ground backwards at an angle. The ground offers an equal reaction in the opposite direction, so car can be moved. But the person sits inside means car and the person becomes a single system, and the force given will be a internal force. According to Newton’s third law, total internal force acting on the system is zero and it cannot accelerate the system.

Question 2.
There is a limit beyond which the polishing of a surface increases frictional resistance rather than decreasing it why?
Answer:
Friction is a contact force. Friction is directly proportional to area of contact. In the normal surfaces there are bumps to interlock the surfaces in contact. But the surfaces are polished beyond certain limit. The area of contact will be increased and the molecules come closer to each other. It increases electrostatic force between the molecules. As a result it increases friction.

Question 3.
Can a single isolated force exist in nature? Explain your answer.
Answer:
No. According to Newton’s third law, for every action, there is an equal and opposite reaction. So, whatever case we consider, if there is an action there is always a reaction. So it is impossible.

Question 4.
Why does a parachute descend slowly?
Answer:
A parachute descends slowly because the surface area of parachute is large so that air gives more resistance when it descends down.

Question 5.
When walking on ice one should take short steps. Why?
Answer:
Let R represent the reaction offered by the ground. The vertical component R cos θ will balance the weight of the person and the horizontal component R sin θ will help the person to walk forward.
Now, normal reaction = R cos θ
Friction force = R sin θ
Coefficient of friction, µ = \(\frac {R sin θ}{R cos θ }\) = tan θ
In a long step, θ is more. So tan θ is more. But p has a fixed value. So, there is danger of slipping in a long step.

Question 6.
When a person walks on a surface, the frictional force exerted by the surface on the person is opposite to the direction of motion. True or false?
Answer:
False. In frictional force exerted by the surface on the person is in the direction of his motion. Frictional force acts as an external force to move the person. When the person trying to move, he gives a push to ground on the backward direction and by Newton’s third law he is pushed by the ground in the forward direction. Hence frictional force acts along the direction of motion.

Question 7.
Can the coefficient of friction be more than one?
Answer:
Yes. The coefficient of friction can be more than one in some cases such as silicone rubber. Coefficient of friction is the ratio of frictional force to normal force, i.e. F = μR. If p is greater than one means frictional force is greater than normal force. But in general case the value is usually between 0 and 1.

Question 8.
Can we predict the direction of motion of a body from the direction of force on it?
Answer:
Yes. The direction of motion is always opposite to the force of kinetic friction. By using the principle of equilibrium, the direction of force of static friction can be determined. When the object is in equilibrium, the frictional force must point in the direction which results as a net force is zero.

Question 9.
The momentum of a system of particles is always conserved. True or false?
Answer:
True. The total momentum of a system of particles is always constant i.e. conserved. When no external force acts on it.

Samacheer Kalvi 11th Physics Laws of Motion Numerical Problems

Question 1.
A force of 50 N act on the object of mass 20 kg. shown in the figure. Calculate the acceleration of the object in x and y directions.
Answer:
Given F = 50 N and m = 20 kg
(1) component of force along x – direction
F_x = F cos θ
= 50 x cos 30° = 43.30 N
a_x = \(\frac{F_{x}}{m}\) = \(\frac {43.30}{20}\) =2.165 ms^-2

(2) Component of force along y – direction
F_y = F sin θ = 50 sin 30° = 25 N
a_y = \(\frac{F_{y}}{m}\) = \(\frac {25}{20}\) = 1.25 ms^-2

Question 2.
A spider of mass 50 g is hanging on a string of a cob web as shown in the figure. What is the tension in the string?
Answer:
Given m = 50 g, T = ?
Tension in the string T = mg
= 50 x 10^-2 x 9.8 = 0.49 N

Question 3.
What is the reading shown in spring balance?

Answer:
When a spring balance hung on a rigid support and load is attached at its other end, the weight of the load exerts a force on the rigid support in turn support exerts equal and opposite force on that load, so that balance will be stretched. This is the principle of spring balance. Flence the answer is 4 kg.
Given: m = 2 kg, 0 = 30°.
Resolve the weight into its component as mg sin θ and mg cos θ.
Here mg sin θ acts parallel to the surface
∴ W = mg sin θ
= 2 x 9.8 x sin 30° = 2 x 9.8 x \(\frac {1}{2}\) = 9.8 N

Question 4.
The physics books are stacked on each other in the sequence: +1 volumes 1 and 2; +2 volumes 1 and 2 on a table
(a) Identify the forces acting on each book and draw the free body diagram.
(b) Identify the forces exerted by each book on the other.
Answer:
Let m₁, m₂, m₃, m_4, are the masses of +1 volume I and II and +2 volumes I & II

(a) Force on book m₄

• Downward gravitational force acting downward (m₃g)
• Upward normal force (N₃) exerted by book of mass m₃
  |

(b) Force on book m₃

• Downward gravitational force (m₃g)
• Downward force exerted by m₄ (N₄)
• Upward force exerted by m₂ (N₂)
  

(c) Force on book m₂

• Downward gravitational force (m₂g)
• Downward force exerted by m₃ (N₃)
• Upward force exerted by m₁ (CN₁)
  

(d) Force on book m₁

• Downward gravitational force exerted by earth (m₁g)
• Downward force exerted by m₂ (N₂)
• Upward force exerted by the table (N_table)
  

Question 5.
A bob attached to the string oscillates back and forth. Resolve the forces acting on the bob in to components. What is the acceleration experienced by the bob at an angle θ.
Answer:
The gravitational force (mg) acting downward can be resolved into two components as mg cos θ and mg sin θ
T – tension exerted by the string.
Tangential force F_T = ma_T = mg sin θ
∴ Tangential acceleration a_T = g sin θ
Centripetal force Fc = ma_c = T – mg cos θ
a_c = \(\frac { T – mg cos θ }{ m }\)

Question 6.
Two masses m₁ and m₂ are connected with a string passing over a friction-less pulley fixed at the comer of the table as shown in the figure. The coefficient of static friction of mass m₁ with the table is µ_s Calculate the minimum mass m₃ that may be placed on m₁to prevent it from sliding. Check if m₁ = 15 kg, m₂ = 10 kg, m₃ = 25 and µ_s = 0.2.

Solution:
Let m₃ is the mass added on m₁
Maximal static friction
\(f_{s}^{\max }\) = µ_sN = µ_s (m₁ + m₃ )g
Here
N = (m₁ + m₃ )g
Tension acting on string = T = m₂ g
Equate (1) and (2)
µ_s(m₁ + m₃) = m₂g
µ_sm₁ + µ_sm₃ = m₂
m₃ = \(f_{s}^{\max }\) – m₁

(ii) Given,
m₁ = 15 kg, m₂ = 10 kg : m₃ = 25 kg and µ_s = 0.2
m₃ = \(f_{s}^{\max }\) – m₁
m₃ = \(\frac {10}{ 0.2 }\) – 15 = 50 – 15 = 35 kg
The minimum mass m₃ = 35 kg has to be placed on ml to prevent it from sliding. But here m₃ = 25 kg only.
The combined masses (m₁ + m₃) will slide.

Question 7.
Calculate the acceleration of the bicycle of mass 25 kg as shown in Figures 1 and 2.

Answer:
Given:
Mass of bicycle m = 25 kg
Fig. I:
Net force acting in the forward direction, F = 500 – 400 = 100 N
acceleration a = \(\frac { F}{ m }\) = \(\frac { 100 }{25}\) = 4 ms^-2

Fig. II:
Net force acting on bicycle F = 400 – 400 = 0
∴ acceleration a = \(\frac { F}{ m }\) = \(\frac { 0}{25}\) = 0

Question 8.
Apply Lami’s theorem on sling shot and calculate the tension in each string?

Answer:
Given F = 50 N, θ = 30°
Here T is resolved into its components as T sin θ and T cos θ as shown.
According to Lami’s theorem,
\(\frac { F}{sin θ}\) = \(\frac { T}{sin (180 – θ)}\) = \(\frac { T}{sin (180 – θ)}\)
\(\frac { F}{sin θ}\) = \(\frac { T}{sin θ}\)
\(\frac { F}{2 sin θ cos θ}\) = \(\frac { T}{sin θ}\) [T = \(\frac { T}{2 cos θ}\) ]
T = \(\frac {T}{2 cos θ}\) = \(\frac { 50}{ 2 cos 30}\) = 28.868 N

Question 9.
A football player kicks a 0.8 kg ball and imparts it a velocity 12 ms^-1. The contact between the foot and ball is only for one – sixtieth of a second. Find the average kicking force.
Answer:
Given,
Mass of the ball = 0.8 kg
Final velocity (V) =12 ms^-1 and time t =\(\frac {1}{60}\) s
Initial velocity = 0
We know the average kicking force
F = ma = \(\frac {m(v – u)}{t}\) = \(\frac{0.8(12-0)}{\left(\frac{1}{60}\right)}\)
F = 576 N

Question 10.
A stone of mass 2 kg is attached to a string of length 1 meter. The string can withstand maximum tension 200 N. What is the maximum speed that stone can have during the whirling motion?
Solution:
Given,
Mass of a stone = 2 kg,
length of a string = 1 m
Maximum tension = 200 N
The force acting on a stone in the whirling motion is centripetal force. Which is provided by tension of the string.
T_max = F_max = \(\frac{m \mathrm{V}_{\mathrm{max}}^{2}}{r}\)
200 = \(v_{\max }^{2}\) = 100
\(v_{\max }^{2}\) = 10 ms^-1

Question 11.
Imagine that the gravitational force between Earth and Moon is provided by an invisible string that exists between the Moon and Earth. What is the tension that exists in this invisible string due’ to Earth’s centripetal force? (Mass of the Moon = 7.34 x 1022 kg, Distance between Moon and Earth = 3.84 x 108 m).

Solution:
Given,
Mass of the moon = 7.34 x 1022 kg
Distance between moon and earth = 3.84 x 108 m
Centripetal force = F = \(\frac{m \mathrm{V}^{2}}{r}\) = \(\frac{7.34 \times 10^{22} \times\left(1.023 \times 10^{3}\right)^{2}}{3.84 \times 10^{8}}\) = 2 x 10²⁰

Question 12.
Two bodies of masses 15 kg and 10 kg are connected with light string kept on a smooth surface. A horizontal force F = 500 N is applied to a 15 kg as shown in the figure. Calculate the tension acting in the string.

Answer:
Given,
m₁  = 15 kg, m₂  = 10 kg, F = 500 N
Tension acting in the string T = \(\frac{m_{2}}{m_{1}+m_{2}}\) F
T = \(\frac {10}{25}\) x 500 = 200 N

Question 13.
People often say “For every action there is an equivalent opposite reaction”. Here they meant ‘action of a human’. Is it correct to apply Newton’s third law to human actions? What is meant by ‘action’ in Newton third law? Give your arguments based on Newton’s laws.
Answer:
Newton’s third law is applicable to only human’s physical actions which involves physical force. Third law is not applicable to human’s psychological actions or thoughts.

Question 14.
A car takes a turn with velocity 50 ms^-1 on the circular road of radius of curvature To m. Calculate the centrifugal force experienced by a person of mass 60 kg inside the car?
Answer:
Given,
Mass of a person = 60 kg
Velocity of the car = 50 ms^-1
Radius of curvature = 10 m
Centrifugal force F = \(\frac{m \mathrm{V}^{2}}{r}\) = \(\frac{60 \times(50)^{2}}{10}\) = 15,000 N

Question 15.
A long stick rests on the surface. A person standing 10 m away from the stick. With what minimum speed an object of mass 0.5 kg should he thrown so that it hits the stick. (Assume the coefficient of kinetic friction is 0.7).
Answer:
Given,
Distance (s) = 10 m
Mass of the object (m) = 0.5 kg
Coefficient of kinetic friction (µ) = 0.7
Work done in moving a body in horizontal surface ω = µ_R x s = µmg x s
This work done is equal to initial kinetic energy of the object
\(\frac{1}{2} m v^{2}\) = µ mg s
\(\left|v^{2}\right|\) = 2 µgs = 2 x 0.7 x 9.8 x 10
v² = 14 x 9.8 = 137. 2
v = 11. 71 ms^-1

Samacheer Kalvi 11th Physics Laws of Motion Additional Questions Solved

Samacheer Kalvi 11th Physics Laws of Motion Multiple Choice Questions

Question 1.
The concept “force causes motion” was given by –
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(b) Aristotle

Question 2.
Who decoupled the motion and force?
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(a) Galileo

Question 3.
The inability of objects to move on its own or change its state of motion is called as –
(a) force
(b) momentum
(c) inertia
(d) impulse
Answer:
(c) inertia

Question 4.
Inertia means –
(a) inability
(b) resistance to change its state
(c) movement
(d) inertial frame
Answer:
(b) resistance to change its state

Question 5.
When a bus starts to move from rest, the passengers experience a sudden backward push is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(c) Inertia of rest

Question 6.
If the brake is applied in the moving bus suddenly, passengers move forward is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(a) Inertia of motion

Question 7.
In whirling motion, if the string is cut suddenly, the stone moves tangential to circle is an –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(b) Inertia of direction

Question 8.
Newtons laws are applicable in –
(a) Inertial frame
(b) non inertial frame
(c) in any frame
(d) none
Answer:
(a) Inertial frame

Question 9.
The accelerated train is an example for –
(a) inertial frame
(b) non-inertial frame
(c) both (a) and (b)
(d) none of the above
Answer:
(b) non-inertial frame

Question 10.
Rate of change of momentum of an object is equal to –
(a) acceleration
(b) work done
(c) force
(d) impulse
Answer:
(c) force

Question 11.
The product of mass and velocity is –
(a) force
(b) impulse
(c) momentum
(d) acceleration
Answer:
(c) momentum

Question 12.
Unit of momentum –
(a) kg ms^-2
(b) kg ms^-1
(c) MLT^-2
(d) MLT^-1
Answer:
(b) kg ms^-1

Question 13.
According to Newton’s third law –
(a) F₁₂ = F₂₁
(*) F₁₂ = -F₂₁
(c) F₁₂ + F₂₁ = 0
(d) F₁₂ x F₂₁ = 0
Answer:
(a) F₁₂ = F₂₁

Question 14.
According to Newton’s third law –
(a) \(\overrightarrow{\mathrm{F}_{12}}=\overrightarrow{\mathrm{F}_{21}}\)
(b) \(\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}\)
(c) \(\mathrm{F}_{12}+\mathrm{F}_{21}\) = 0
(d) \(\mathrm{F}_{12}x\mathrm{F}_{21}\) = 0
Answer:
(b) \(\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}\)

Question 15.
The law which is valid in both inertial and non-inertial frame is –
(a) Newton’s first law
(b) Newton’s second law
(c) Newton’s third law
(d) none
Answer:
(c) Newton’s third law

Question 16.
When a force is applied on a body, it can change –
(a) velocity
(b) momentum
(c) direction of motion
(d) all the above
Answer:
(d) all the above

Question 17.
The rate of change of velocity is 1 ms^-2 when a force is applied on the body of mass 75 gm the force is –
(a) 75 N
(b) 0.75 N
(c) 0.075 N
(d) 0.75 x 10^-3 N
Answer:
(c) Force is given by
F = m a
= 75 gm x 1 cm s^-2 = 75 x 10^-3 x 1 = 75 x 10^-3 = 0.075 N

Question 18.
The action and reaction forces acting on –
(a) same body
(b) different bodies
(c) either same or different bodies
(d) none of the above
Answer:
(b) different bodies

Question 19.
Newton’s first law of motion gives the concept of –
(a) velocity
(b) energy
(c) momentum
(d) Inertia
Answer:
(d) Inertia

Question 20.
Inertia of a body has direct dependence on –
(a) velocity
(b) area
(c) mass
(d) volume
Answer:
(c) mass

Question 21.
If a car and a scooter have the same momentum, then which one is having greater speed?
(a) scooter
(b) car
(c) both have same velocity
(d) data insufficient
Answer:
(a) scooter

Question 22.
Newton’s second law gives –
(a) \(\overrightarrow{\mathrm{F}} \propto \frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(b) \(\overrightarrow{\mathrm{F}}=\frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(c) \(\overrightarrow{\mathrm{F}}=m \vec{a}\)
(d) all the above
Answer:
(d) all the above

Question 23.
1 dyne is –
(a) 10⁵N
(b) 10^-5N
(c) 1N
(d) 10^-3N
Answer:
(b) 10^-5N

Question 24.
If same force is acting on two masses m₁ and m₂, and the accelerations of two bodies are a₁ and a₂ respectively, then –
(a) \(\frac{a_{2}}{a_{1}}=\frac{m_{2}}{m_{1}}\)
(b) \(\frac{a_{1}}{a_{2}}=\frac{m_{1}}{m_{2}}\)
(c) \(\frac{a_{1}}{a_{2}}=\frac{m_{2}}{m_{1}}\)
(d) m₁ a₁ + m₂a₂ = 0
Answer:
(c) \(\frac{a_{1}}{a_{2}}=\frac{m_{2}}{m_{1}}\)

Question 25.
If a force \(\overline{\mathrm{F}}\) = 3\(\hat{i}\) – 4\(\hat{j}\) N produces an acceleration of 10 ms^-2 on a body, then the mass of a body is –
(a) 10 kg
(b) 9 kg
(c) 0.9 kg
(d) 0.5 kg
Answer:
\(\overline{\mathrm{F}}\) = 3\(\hat{i}\) – 4\(\hat{j}\)
Magnitude:
|\(\overline{\mathrm{F}}\)| = \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5N
F = ma
⇒ m = \(\frac{|\mathrm{F}|}{a}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\) = 0.5 kg

Question 26.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop?
(a) 0.75 s
(b) 1.33 s
(c) 6 s
(d) 35 s
Answer:
Acceleration a = \(\frac{-F}{m}\) = \(\frac{50}{20}\) = – 2.5 ms^-2
u = l5 ms^-1
v = 0
t = ?
v = u + at
0 = 15 – 2.5t
t = \(\frac{15}{2.5}\) = 6s

Question 27.
Rain drops come down with –
(a) zero acceleration and non zero velocity
(b) zero velocity with non zero acceleration
(c) zero acceleration and non zero net force
(d) none
Answer:
(a) zero acceleration and non zero velocity

Question 28.
If force is the cause then the effect is –
(a) mass
(b) potential energy
(c) acceleration
(d) Inertia
Answer:
(c) acceleration

Question 29.
In free body diagram, the object is represented by a –
(a) line
(b) arrow
(c) circle
(d) point
Answer:
(d) point

Question 30.
When an object of mass m slides on a friction less surface inclined at an angle 0, then normal force exerted by the surface is –
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg tan θ
Answer:
(b) mg cos θ

Question 31.
The acceleration of the sliding object in an inclined plane –
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg sin θ
Answer:
(c) g sin θ

Question 32.
The speed of an object sliding in an inclined plane at the bottom is –
(a) mg cos θ
(b) \(\sqrt{2 s g sin θ}\)
(c) \(\sqrt{2 s g cos θ}\)
(d) \(\sqrt{2 s g tan θ}\)
Answer:
(b) \(\sqrt{2 s g sin θ}\)

Question 33.
The acceleration of two bodies of mass m₁ and m₂ in contact on a horizontal surface is –
(a) \(a=\frac{\mathbf{F}}{m_{1}}\)
(b) \(a=\frac{F}{m_{2}}\)
(c)  \(a=\frac{\mathrm{F}}{m_{1}+m_{2}}\)
(d) \(a=\frac{\mathrm{F}}{m_{1} m_{2}}\)
Answer:
(c)  \(a=\frac{\mathrm{F}}{m_{1}+m_{2}}\)

Question 34.
Two blocks of masses m₁ and m₂ (m₁ > m₂) in contact with each other on frictionless, horizontal surface. If a horizontal force F is given on m₁, set into motion with acceleration a, then reaction force on mass m₁ by m₂, is –
(a) \(\frac{\mathrm{F} m_{1}}{m_{1}+m_{2}}\)
(b) \(\frac{m_{1} m_{2}}{\mathrm{F} m_{1}}\)
(c) \(\frac{m_{1} m_{2}}{\mathrm{F} m_{2}}\)
(d) \(\frac{\mathrm{F} m_{2}}{m_{1}+m_{2}}\)
Answer:
(d) \(\frac{\mathrm{F} m_{2}}{m_{1}+m_{2}}\)

Question 35.
If two masses m₁ and m₂ (m₁ > m₂) tied to string moving over a frictionless pulley, then acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 36.
if two masses m₁ and m₂ (m₁ > m₂)tied to string moving over a frictionless pulley, then acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 37.
Three massses is in contact as shown. If force F is applied to mass m₁, the acceleration of three masses is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Answer:
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Question 38.
Three masses in contact is as shown above. If force F is applied to mass m₁ then the contact force acting on mass m₂ is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)

Question 39.
Three masses is contact as shown. It force F is applied to mass m₁, then the contact force acting on mass m₃ is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Question 40.
Two masses connected with a string. When a force F is applied on mass m₂. The acceleration produced is –

(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}}\)
(b) \(\frac{\mathbf{F}}{m_{1}-m_{2}}\)
(c) \(\frac{m_{1}+m_{2}}{\mathrm{F}}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}}\)

Question 41.
Two masses connected with a string. When a force F is applied on mass m₂. The force acting on m₁ is –
(a) \(\frac{m_{1} \mathrm{F}}{m_{1}+m_{2}}\)
(b) \(\frac{m_{2} \mathrm{F}}{m_{1}+m_{2}}\)
(c) \(\frac{m_{1}+m_{2}}{m_{1}} \mathbf{F}\)
(d) \(\frac{m_{1}+m_{2}}{m_{2}} \mathbf{F}\)
Answer:
(b) \(\frac{m_{2} \mathrm{F}}{m_{1}+m_{2}}\)

Question 42.
If a block of mass m lying on a frictionless inclined plane of length L height h and angle of inclination θ, then the velocity at its bottom is –
(a) g sin θ
(b) g cos θ
(c) \(\sqrt{2 g h}\)
(d) \(\sqrt{2 a sin θ}\)
Answer:
(c) \(\sqrt{2 g h}\)

Question 43.
If a block of mass m lying on a frictionless inclined plane of length L, height h and angle of inclination θ, then the time take taken to reach the bottom is –
(a) g sing θ
(b) sin θ \(\sqrt{\frac{2 h}{g}}\)
(c) sin θ \(\sqrt{\frac{g}{h}}\)
(d) \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)
Answer:
(d) \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)

Question 44.
A rocket works on the principle of conservation of –
(a) energy
(b) mass
(c) angular momentum
(d) linear momentum
Answer:
(b) mass

Question 45.
A bomb at rest explodes. The total momentum of all its fragments is –
(a) zero
(b) infinity
(c) always 1
(d) always greater then 1
Answer:
(a) zero

Question 46.
A block of mass m₁ is pulled along a horizontal friction-less surface by a rope of mass m₂ If a force F is given at its free end. The net force acting on the block is –
(a) \(\frac{m_{1} \mathrm{F}}{m_{1}-m_{2}}\)
(b) F
(c) \(\frac{m_{2} \mathrm{F}}{\left(m_{1}+m_{2}\right)}\)
(d) \(\frac{m_{1} \mathrm{F}}{\left(m_{1}+m_{2}\right)}\)
Answer:
(b) F

Question 47.
A block of mass m is pulled along a horizontal surface by a rope. The tension in the rope will be same at all the points –
(a) if the rope is accelerated
(b) if the rope is mass less
(c) always
(d) none of the above
Answer:
(b) if the rope is mass less

Question 48.
The lines of forces act at a common point is called as –
(a) concurrent forces
(b) co-planar forces
(c) equilibrium
(d) resultant
Answer:
(a) concurrent forces

Question 49.
If the lines of forces act in the same plane, they can be –
(a) concurrent forces
(b) coplanar forces
(c) either concurrent force or coplanar forces
(d) Lami’s force
Answer:
(d) concurrent forces

Question 50.
Lami’s theorem is applicable only when the system of forces are is –
(a) same plane
(b) different plane
(c) equilibrium
(d) none of the above
Answer:
(c) equilibrium

Question 51.
Due to the action of internal forces of the system, the total linear momentum of the system is –
(a) a variable
(b) a constant
(c) always zero
(d) always infinity
Answer:
(c) always zero

Question 52.
The velocity with which a gun suddenly moves backward after firing is –
(a) linear velocity
(b) positive velocity
(c) recoil velocity
(d) v₁ + v₂
Answer:
(c) recoil velocity

Question 53.
If a very large force acts on an object for a very short duration, then the force is called as –
(a) Newtonian force
(b) impulsive force
(c) concurrent force
(d) coplanar force
Answer:
(A) impulsive force

Question 54.
The unit of impulse is –
(a) Nm
(b) Ns
(c) Nm²
(d) Ns^-2
Answer:
(b) Ns

Question 55.
The force which always opposes the relative motion between an object and the surface where it is placed is –
(a) concurrent force
(b) frictional force
(c) impulsive force
(d) coplanar force
Answer:
(b) frictional force

Question 56.
The force which opposes the initiation of motion of an object on the surface is –
(a) static friction
(b) kinetic friction
(c) friction
(d) zero
Answer:
(d) static friction

Question 57.
When the object is at rest, the resultant of gravitational force and upward normal force is –
(a) Static force
(b) zero
(c) one
(d) infinity
Answer:
(b) zero

Question 58.
The magnitude of static frictional force d lies between –
(a) 0 ≤ f ≤ µ_sN
(b) 0 ≥f ≥ µ_sN
(c) 0 and 1
(d) 0 and minimal static frictional force.
Answer:
(a) 0 ≤ f ≤ µ_sN

Question 59.
The unit of co-efficient of static friction is –
(a) N
(b) N m
(c) N s
(d) no unit
Answer:
(d) no unit

Question 60.
If the object is at rest and no external force is applied on the object, the static friction acting on the object is –
(a) µ_sN
(b) zero
(c) one
(d) infinity
Answer:
(d) no unit

Question 61.
When object begins to slide, the static friction acting on the object attains –
(a) zero
(b) minimum
(c) maximum
(d) infinity
Answer:
(c) maximum

Question 62.
The static friction does not depend upon –
(a) the area of contact
(b) normal force
(c) the magnitude of applied force
(d) none of the above
Answer:
(a) the area of contact

Question 63.
Which of the following pairs of materials has minimum amount of coefficient of static friction is –
(a) Glass and glass
(b) wood and wood
(c) ice and ice
(d) steel and steel
Answer:
(c) ice and ice

Question 64.
Kinetic friction is also called as –
(a) sliding friction
(b) dynamic friction
(c) both (a) and (b)
(d) static friction
Answer:
(c) both (a) and (b)

Question 65.
The unit of coefficient of kinetic friction is/has –
(a) Nm
(b) Ns
(c) Nm²
(d) no unit
Answer:
(d) no unit

Question 66.
The nature of materials in mutual contact decides –
(a) µs
(b) µk
(c) µs or µk
(d) none
Answer:
(c) µs or µk

Question 67.
Coefficient of kinetic friction is less than –
(a) O
(b) one
(c) µs
(d) µsN
Answer:
(c) µs

Question 68.
The static friction –
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(a) increases linearly

Question 69.
The kinetic friction –
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(b) is constant

Question 70.
Kinetic friction is independent of –
(a) nature of materials
(b) temperature of the surface
(c) applied force
(d) none of the above
Answer:
(c) applied force

Question 71.
The angle between the normal force and the resultant force of normal force and maximum frictional force is –
(a) angle of friction
(b) angle of repose
(c) angle of inclination
(d) none of the above
Answer:
(a) angle of friction

Question 72.
The angle friction θ is given by –
(a) tan µ_s
(b) tan^-1 µ_s
(c) \(\frac{f S^{\mathrm{max}}}{N}\)
(d) sin^-1 µ_s
Answer:
(b) tan^-1 µ_s

Question 73.
The angle of inclined plane with the horizontal such that an object placed on it begins to slide is –
(a) angle of friction
(b) angle of repose
(c) angle of response
(d) angle of retardation
Answer:
(b) angle of repose

Question 74.
Comparatively, which of the following has lesser value than others?
(a) static friction
(b) kinetic friction
(c) Rolling friction
(d) skiping friction
Answer:
(c) Rolling friction

Question 75.
The origin of friction is –
(a) electrostatic interaction
(b) electromagnetic interaction magnetic
(c) photon interaction
(d) interaction
Answer:
(b) electromagnetic interaction

Question 76.
Friction can be reduced by –
(a) polishing
(b) lubricating
(c) using ball bearings
(d) all the above
Answer:
(c) using ball bearings

Question 77.
For a particle revolving in a circular path, the acceleration of the particle is –
(a) along the tangent
(b) along the radius
(c) along the circumference of the circle
(d) zero
Answer:
(b) along the radius

Question 78.
A particle moves along a circular path under the action of a force. The work done by the force is –
(a) Positive and non zero
(b) zero
(c) Negative and non zero
(d) none of the above
Answer:
(b) zero

Question 79.
A bullet hits and gets embedded in a solid block resting on a horizontal frictionless table. Which of the following is conserved?
(a) Momentum and kinetic energy
(b) kinetic energy alone
(c) Momentum alone
(d) potential energy alone
Answer:
(c) Momentum alone

Question 80.
The origin of the centripetal force can be –
(a) gravitational force
(b) frictional force
(c) coulomb force
(d) all the above
Answer:
(d) all the above

Question 81.
Centripetal acceleration is –
(a) \(\frac{m v^{2}}{r}\)
(b) \(\frac{v^{2}}{r}\)
(c) r v²
(d) rω
Answer:
(b) \(\frac{v^{2}}{r}\)

Question 82.
Centripetal acceleration is –
(a) \(\frac{m v^{2}}{r}\)
(b) r ω²
(c) rv²
(d) rω
Answer:
(c) rω²

Question 83.
The centripetal force is –
(a) \(\frac{m v^{2}}{r}\)
(b) rω²
(c) both (a) and (b)
(d) none
Answer:
(c) both (a) and (b)

Question 84.
When a car is moving on a circular track the centripetal force is due to –
(a) gravitational force
(b) frictional force
(c) magnetic force
(d) elastic force
Answer:
(b) frictional force

Question 85.
If the road is horizontal then the normal force and gravitational force are –
(a) equal and along the same direction
(b) equal and opposite
(c) unequal and along the same direction
(d) unequal and opposite
Answer:
(b) equal and opposite

Question 86.
The velocity of a car for safe turn on leveled circular road –
(a) \(v \leq \sqrt{\mu_{s} r g}\)
(b) \(v \geq \sqrt{\mu_{s} r g}\)
(c) \(v=\sqrt{\mu_{s} rg}\)
(d) \(v \leq \mu_{s} rg\)
Answer:
(a) \(v \leq \sqrt{\mu_{s} r g}\)

Question 87.
In a leveled circular road, skidding mainly depends on –
(a) µ_s
(b) µ_k
(c) acceleration
(d) none
Answer:
(a) µ_s

Question 88.
The speed of a car to move on the banked road so that it will have safe turn is –
(a) µ_srg
(b) \(\sqrt{r g \tan \theta}\)
(c) rg tan θ
(d) r²g tan θ
Answer:
(b) \(\sqrt{r g \tan \theta}\)

Question 89.
Centrifugal force is a –
(a) pseudo force
(b) real force
(c) forced acting towards center
(d) none of the above
Answer:
(a) pseudo force

Question 90.
Origin of centrifugal force is due to –
(a) interaction between two
(b) inertia
(c) electromagnetic interaction
(d) inertial frame
Answer:
(b) inertia

Question 91.
Centripetal force acts in –
(a) inertial frame
(b) non inertial frame
(c) both (a) and (h)
(d) linear motion
Answer:
(c) both (a) and (b)

Question 92.
Centrifugal force acts in –
(a) inertial frame
(b) non inertial frame
(c) both (a) and (b)
(d) linear motion
Answer:
(b) non inertial frame

Question 93.
A cricket ball of mass loo g moving with a velocity of 20 ms-¹ is brought to rest by a player in 0.05s the impulse of the ball is –
(a) 5 Ns
(b) – 2 Ns
(c) – 2.5 Ns
(d) zero
Answer:
(b) – 2 Ns
mass = 0.1 kg
Initial velocity t = 20 ms^-1
Final velocity y = 0
Change in momentum in impulse = m(v – u) = 0.1(0 – 20) = – 2 Ns

Question 94.
If a stone tied at the one end of a string of length 0.5 m is whirled in a horizontal circle with a constant speed 6 ms^-1  then the acceleration of the shone is –
(a) 12 ms^-2
(b) 36 ms^-2
(c) 2π² ms^-2
(d) 72 ms^-2
Answer:
(d) Centripetal acceleration = \(\frac{v^{2}}{r}\) = \(\frac{6^{2}}{0.5}\) = \(\frac{36}{0.5}\) = 72 ms^-2

Question 95.
A block of mass 3 kg is at rest on a rough inclined plane with angle of inclination 30° with horizontal. If .is 0.7, then the frictional force is –
(a) 17.82 N
(b) 1.81 N
(c) 3.63 N
(d) 2.1 N
Answer:
(a) Frictional force = µmg cos θ = 0.7 x 3 x 9.8 cos 30° = 17.82 N

Question 96.
Two masses 2 kg and 4 kg are tied at the ends of a mass less string and which is passing over a friction-less pulley. The tension in the string is –
(a) 3.68 N
(b) 78.4 N
(c) 26 N
(d) 13.26 N
Answer:
(c) Tension in the string T = \(\frac{2 m_{1} m_{2}}{\left(m_{1}+m_{2}\right)}\)g
T = \(\frac{2 x 2 x 4}{2 + 4}\) x 9.8 = \(\frac{16}{6}\) x 9.8 = 26.13 N

Question 97.
A bomb of 10 kg at rest explodes into two pieces of mass 4 kg and 6 kg. if the velocity of 4 kg mass is 6 ms-¹ then the velocity of 6 kg is –
(a) – 4 ms^-1
(b) – 6 ms^-1
(c) – 24 ms^-1
(d) – 2.2 ms^-1
Answer:
(a) According to law of conservation of momentum
m₁v₁ + m₂v₂ = 0
v₂ = –\(\frac{m_{1} v_{1}}{m_{2}}\) = \(\frac{4 x 6}{6}\) = -4ms^-1

Question 98.
A body is subjected under three concurrent forces and it is in equilibrium. The resultant of any two forces is –
(a) coplanar with the third force
(b) is equal and opposite to third force
(c) both (a) and (b)
(d) none of the above
Answer:
(c) both (a) and (b)

Question 99.
An impulse is applied to a moving object with the force at an angle of 20° with respect to velocity vector. The angle between the impulse vector and the change in momentum vector is –
(a) 0°
(b) 30°
(c) 60°
(d) 120°
Answer:
(a) Impulse and change in momentum are in same direction. So the angle is zero.

Question 100.
A bullet of mass m and velocity v₁ is fired into a large block of wood of mass M. The final velocity of the system is-
(a) \(\frac{v_{1}}{m+\mathrm{M}}\)
(b) \(\frac{m v_{1}}{m+\mathrm{M}}\)
(c) \(\frac{m+m}{m} v_{1}\)
(d) \(\frac{m+m}{m-M} v_{1}\)
Answer:
(b) \(\frac{m v_{1}}{m+\mathrm{M}}\)

Question 101.
A block of mass 2 kg is placed on the floor. The co – efficient of static friction is 0.4. The force of friction between the block and floor is –
(a) 2.8 N
(b) 7.8 N
(c) 2 N
(d) zero
Answer:
(b) The force required to move = = µR = µmg = 0.4 x 2 x 9.8 = 7.84 N

Question 102.
A truck weighing 1000 kg is moving with velocity of 50 km/h on smooth horizontal roads. A mass of 250 kg is dropped into it. The velocity with which it moves now is –
(a) 12.5 km/h
(b) 20 km/h
(c) 40 km/h
(d) 50 km/h
Answer:
(c) According to law of conservation of linear momentum
m₂ v₂ = (m₁ + m₂)v₂
v₂ = \(\frac{m_{1} v_{1}}{m_{1}+m_{2}}\) = \(\frac{1000 \times 50}{1250}\) = 40 km/h

Question 103.
A body of mass loo g is sliding from an inclined plane of inclination 30°. if u = 1.7, then the frictional force experienced is –
(a) \(\frac{3.4}{\sqrt{3}}\)N
(b) 1.47 N
(c) \(\frac{\sqrt{3}}{3.4}\)N
(d) 1.38 N
Answer:
(b) Frictional force F = µ mg cos θ = 1.7 x 0.1 x 10 cos 30°= \(\frac{1.7}{2}\) x \(\sqrt{3}\) = 1.47 N

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (1 Mark)

Question 1.
A passenger sitting in a car at rest, pushes the car from within. The car doesn’t move, why?
Answer:
For motion, there should be external force.

Question 2.
Give the magnitude and directions of the net force acting on a rain drop falling with a constant speed.
Answer:
as \(\overline{\mathrm{a}}\) = 0 so \(\overline{\mathrm{F}}\) = 0.

Question 3.
Why the passengers in a moving car are thrown outwards when it suddenly takes a turn?
Answer:
Due to inertia of direction.

Question 4.
You accelerate your car forward. What is the direction of the frictional force on a package resting on the floor of the car?
Answer:
The package in the accelerated car (a non inertial frame) experiences a Pseudo force in a direction opposite to that of the motion of the car. The frictional force on the package which acts opposite to this pseudo force is thus in the same direction (forward) as that of the car.

Question 5.
What is the purpose of using shockers in a car?
Answer:
To decrease the impact of force by increasing the time for which force acts.

Question 6.
Why are types made of rubber not of steel?
Answer:
Since coefficient of friction between rubber and road is less than the coefficient of friction between steel and road.

Question 7.
Wheels are made circular. Why?
Answer:
Rolling friction is less than sliding friction.

Question 8.
If a ball is thrown up in a moving train, it comes back to the thrower’s hands. Why?
Answer:
Both during its upward and downward motion, the ball continues to move inertia of motion with the same horizontal velocity as the train. In this period, the ball covers the same horizontal distance as the train and so it comes back to the thrower’s hand.

Question 9.
Calculate the force acting on a body which changes the momentum of the body at the rate of 1 kg-m/s² .
Answer:
As F = rate change of momentum
F = 1 kg-m/s² = 1N

Question 10.
On a rainy day skidding takes place along a curved path. Why?
Answer:
As the friction between the types and road reduces on a rainy day.

Question 11.
Why does a gun recoils when a bullet is being fired?
Answer:
To conserve momentum.

Question 12.
Why is it difficult to catch a cricket ball than a tennis ball even when both are moving with the same velocity?
Answer:
Being heavier, cricket ball has higher rate of change of momentum during motion so more force sumed.

Question 13.
The distance travelled by a moving body is directly proportional to time. Is any external force acting on it?
Answer:
As s ∝ t, so acceleration a = 0, therefore, no external force is acting on the body.

Question 14.
Calculate the impulse necessary to stop a 1500 kg car moving at a speed of 25 ms^-1.
Answer:
Use formula I = change in momentum = m(v – u) (Impulse – 37500 Ns)

Question 15.
Lubricants are used between the two parts of a machine. Why?
Answer:
To reduce friction and so to reduce wear and tear.

Question 16.
What provides the centripetal force to a car taking a turn on a level road?
Answer:
Force of friction between the type and road provides centripetal force.

Question 17.
A body is acted upon by a number of external forces. Can it remain at rest?
Answer:
Yes, if the external forces acting on the body can be represented in magnitude and direction by the sides of a closed polygon taken in the same order.

Question 18.
Bodies of larger mass need greater initial effort to put them in motion. Why?
Answer:
As F = ma so for given a, more force will be required to put a large mass in motion.

Question 19.
An athlete runs a certain distance before taking a long jump Why?
Answer:
So that inertia of motion may help him in his muscular efforts to take a longer jump.

Question 20.
Action and reaction forces do not balance each other. Why?
Answer:
As they acts on different bodies.

Question 21.
The wheels of vehicles are provided with mudguards. Why?
Answer:
When the wheel rotates at a high speed, the mud sticking to the wheel flies off tangentially, this is due to inertia of direction. If order that the flying mud does not spoil the clothes of passer by the wheels are provided with mudguards.

Question 22.
China wares are wrapped in straw paper before packing. Why?
Answer:
The straw paper between the China ware increases the Time of experiencing the jerk during transportation. Hence impact of force reduces on China wares.

Question 23.
Why is it difficult to walk on a sand?
Answer:
Less reaction force.

Question 24.
The outer edge of a curved road is generally raised over the inner edge Why?
Answer:
In addition to the frictional force, a component of reaction force also provides centripetal force.

Question 25.
Explain why the water doesn’t fall even at the top of the circle when the bucket full of water is upside down rotating in a vertical circle?
Answer:
Weight of the water and bucket is used up in providing the necessary centripetal force at the top of the circle.

Question 26.
Why does a speedy motor cyclist bends towards the center of a circular path while taking a turn on it?
Answer:
So that in addition of the frictional force, the horizontal component of the normal reaction also provides the necessary centripetal forces.

Question 27.
An impulse is applied to a moving object with a force at an angle of 20° wr.t. velocity vector, what is the angle between the impulse vector and change in momentum vector ?
Answer:
Impulse and change in momentum are along the same direction. Therefore angle between these two vectors is zero.

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (2 Marks)

Question 28.
A man getting out of a moving bus runs in the same direction for a certain distance. Comment.
Answer:
Due to inertia of motion.

Question 29.
If the net force acting upon the particle is zero, show that its linear momentum remains constant.
Answer:
As F x \(\frac {dp}{dt}\)
when F = 0, \(\frac {dp}{dt}\) = 0 so P = constant

Question 30.
A force of 36 dynes is inclined to the horizontal at an angle of 60°. Find the acceleration in a mass of 18 g that moves in a horizontal direction.
Answer:
F = 36 dyne at an angle of 60°
F_x = F cos 60° = 18 dyne
F_x = ma_x
So a_x = \(\frac{F_{x}}{m}\) = 1 cm /s²

Question 31.
The motion of a particle of mass m is described by h = ut + \(\frac {1}{2}\) gt². Find the force acting on particle.
Answer:
a = ut + \(\frac {1}{2}\) gt²
find a by differentiating h twice w.r.t.
a = g
As F = ma so F = mg (answer)

Question 32.
A particle of mass 0.3 kg is subjected to a force of F = -kx with k= 15 Nmr^-1. What will be its initial acceleration if it is released from a point 20 cm away from the origin?
Answer:
As F = ma so F = -kx = ma
a = \(\frac {-kx}{m}\)
for x = 20 cm, ⇒ a = -10 m/s².

Question 33.
A 50 g bullet is fired from a 10 kg gun with a speed of 500 ms-¹. What is the speed of the recoil of the gun?
Answer:
Initial momentum = 0
Using conservation of linear momentum mv + MV = 0
V = \(\frac {-mv}{M}\) ⇒ V = 2.5 m/s

Question 34.
Smooth block is released at rest on a 45° incline and then slides a distance d. If the time taken of slide on rough incline is n times as large as that to slide than on a smooth incline. Show that coefficient of friction, µ = \(\left(1-\frac{1}{n^{2}}\right)\)
Answer:
When there is no friction, the block slides down the inclined plane with acceleration. a = g sin θ
when there is friction, the downward acceleration of the block is a’ = g (sin θ – µ cos θ)
As the block Slides a distance d in each case so
d = \(\frac {1}{2}\) at² = \(\frac {1}{2}\) a’t’²
\(\frac{a}{a^{\prime}}=\frac{t^{\prime 2}}{t^{2}}=\frac{(n t)^{2}}{t^{2}}\) = n²
or \(\frac {g sin θ}{g(sin θ – µ cos θ)}\) = n²
Solving, we get (Using θ = 45°)
µ = 1 – \(\frac{1}{n^{2}}\)

Question 35.
A spring balance is attached to the ceiling of a lift. When the lift is at rest spring balance reads 49 N of a body hang on it. If the lift moves:

1. Downward
2. upward, with an acceleration of 5 ms²
3. with a constant velocity.

What will be the reading of the balance in each case?
Answer:
1.  R = m(g – a) = 49 N
so = m = \(\frac {49}{9.8}\) = 5 kg
R = 5 (9.8 – 5)
R = 24 N

2. R = m(g + a)
R = 5 (9.8 + 5)
R = 74 N

3.  as a = 0 so R = mg = 49 N

Question 36.
A bob of mass 0.1 kg hung from the ceiling of room by a string 2 m long is oscillating. At its mean position the speed of a bob is 1 ms^-1. What is the trajectory of the ‘oscillating bob if the string is cut when the bob is –

1. At the mean position
2. At its extreme position.

Answer:

1. Parabolic
2. vertically downwards

Question 37.
A block placed on a rough horizontal surface is pulled by a horizontal force F. Let f be the force applied by the rough surface on the block. Plot a graph of f versus F.
Answer:

Unto point A, f = F (50 Long as block is stationary) beyond A, when F increases, block starts moving f remains constant.

Question 38.
A mass of 2 kg is suspended with thread AB. Thread CD of the same type is attached to the other end of 2 kg mass.

• Lower end of the lower thread is pulled gradually, hander and hander is the downward direction so as to apply force on AB Which of the thread will break & why?
• If the lower thread is pulled with a jerk, what happens?

Answer:

• Thread AB breaks down
• CD will break.

Question 39.
A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is p and the acceleration due to gravity is g, calculate the minimum force required to be applied by the finger to held the block against the wall?
Answer:

For the block not to fall f = Mg
But f = µR = µF so
µF = Mg
F = \(\frac {Mg}{µ}\)

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (3 Marks) & Numericals

Question 40.
A block of mass 500 g is at rest on a horizontal table. What steady force is required to give the block a velocity of 200 cm s^-2 in 4 s?
Answer:
Use F – ma
a = \(\frac {v – u}{ t }\) = \(\frac {200- 0}{ 4 }\) = 50 cm/s²
F = 500 x 50 = 25,000 dyne.

Question 41.
A force of 98 N is just required to move a mass of 45 kg on a rough horizontal surface. Find the coefficient of friction and angle of friction?
Answer:
F = 48 N,R = 45 x 9.8 = 441 N
µ = \(\frac {F’}{ R}\) = 0.22
Angle of friction θ = tan^-1 0.22 = 12°24′

Question 42.
Calculate the force required to move a train of 2000 quintal up on an incline plane of 1 in 50 with an acceleration of 2 ms^-2. The force of friction per quintal is 0.5 N.
Answer:
Force of friction = 0.5 N per quintal
f = 0.5 x 2000 = 1000 N
m = 2000 quintals = 2000 x 100 kg
sin θ = \(\frac {1}{50}\), a – 2 m/s²
In moving up an inclined plane, force required against gravity
mg sin θ = 39200 N
And force required to produce acceleration = ma
= 2000 x 100 x 2 = 40,0000 N
Total force required = 1000 + 39,200 + 40,0000 = 440200 N.

Question 43.
A force of 100 N gives a mass m₁, an acceleration of 10 ms^-2 and of 20 ms^-2 to a mass m₂.
What acceleration must be given to it if both the masses are tied together?
Answer:
Suppose, a = acceleration produced if m₁ and m₂ are tied together,
F = 100 N
Let a₁ and a₂ be the acceleration produced in m₁ and m₂ respectively.
∴ a₁ and a₂ = 20ms^-2 (given)
Again m₁ = \(\frac{\mathrm{F}}{a_{1}}\) and m₂ = \(\frac{\mathrm{F}}{a_{2}}\)
⇒ m₁ = \(\frac {100}{10}\) = 10kg
and m₂ = \(\frac {100}{20}\) = 5kg
∴ m₁ + m₂ = 10 + 5 = 15
so, a = \(\frac{\mathbf{F}}{m_{1}+m_{2}}\) = \(\frac {100}{15}\) = \(\frac {20}{3}\) = 6.67 ms²

Question 44.
The pulley arrangement of figure are identical. The mass of the rope is negligible. In (a) mass m is lifted up by attaching a mass (2m) to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2 mg. In which case, the acceleration of m is more?

Answer:
Case (a):
a = \(\frac {2m – m}{2m + m}\) g = a = \(\frac {g}{3}\)
Case (b):
FBD of mass m
ma’ = T – mg
ma’ = 2 mg – mg
⇒ ma’ = mg
a’ = g
So in case (b) acceleration of m is more.

Question 45.
Figure shows the position-time graph of a particle of mass 4 kg. What is the

(a) Force on the particle for t < 0, t > 4s, 0 < t < 4s?
(b) Impulse at t = 0 and t = 4s?
(Consider one dimensional motion only)
Answer:
(a) For t < 0. No force as Particles is at rest. For t > 4s, No force again particle comes at rest.
For 0 < t < 4s, as slope of OA is constant so velocity constant i.e., a = 0, so force must be zero.

(b) Impulse at t = 0
Impulse = change in momentum
I = m(v – w) = 4(0 – 0.75) = 3 kg ms^-1
Impulse at t = 4s
1 = m(v – u) = 4 (0 – 0.75) = -3 kg ms^-1

Question 46.
What is the acceleration of the block and trolley system as the figure, if the coefficient of kinetic friction between the trolley and the surface is 0.04? Also Calculate friction in the string: Take g = 10 m/s², mass of the string is negligible.

Answer:
Free body diagram of the block
30 – T = 3a
Free body diagram of the trolley

T – f_k = 20 a ………….(2)
where f_k = µ_k= 0.04 x 20 x 10 = 8 N
Solving (i) & (ii), a = 0.96 m/s² and T = 27.2 N

Question 47.
Three blocks of masses ml = 10 kg, m² = 20 kg are connected by strings on smooth horizontal surface and pulled by a force of 60 N. Find the acceleration of the system and frictions in the string.

Solution:
All the blocks more with common acceleration a under the force F = 60 N.
F = (m₁ + m₂ + m₃)a
a = \(\frac{\mathrm{F}}{\left(m_{1}+m_{2}+m_{3}\right)}\) = 1 m/s²
to determine, T₁ →Free body diagram of m₁.

T₁ = m₁a = 10 x 1 = 10 N
to determine, T₂ →Free body diagram of m₃

F – T₂ = m₃a
Solving, we get T₂ = 30 N

Question 48.
The rear side of a truck is open and a box of 40 kg mass is placed 5m away from the open end. The coefficient of friction between the box and the surface below it is 0.15 on a straight road, the truck starts from rest and accelerates with 2 m/s². At what distance from the starting point does the box fall off the truck ? (ignore the size of the box)
Answer:
Force on the box due to accelerated motion of the truck
F = ma = 40 x 2 = 80 N (in forward direction)
Reaction on the box, F’ = F = 80 N (in backward direction)
Force of limiting friction, f = µR = 0 .15 x 40 x 10 = 60 N
Net force on the box in backward direction is P = F’ f = 80 – 60 = 20 N
Backward acceleration in the box = a= \(\frac {p}{m}\) = \(\frac {20}{40}\) = 0.5 ms^-2
t = time taken by the box to travel s = 5 m and falls off the truck, then from
s = ut + \(\frac {1}{2}\) at²
5 = 0 x t + \(\frac {1}{2}\) x 0.5 x t²
t = 4.47
If the truck travels a distance x during this time
then x = 0 x 4.34 +\(\frac {1}{2}\) x 2 x (4.471)²
x = 19.98 m

Question 49.
A block slides down as incline of 30° with the horizontal. Starting from rest, it covers 8 m in the first 2 seconds. Find the coefficient of static friction.
Use s = ut + \(\frac {1}{2}\) at²
a = \(\frac{2 s}{t^{2}}\) at² as u = 0
µ = \(\frac{g sin θ – a}{g Cos θ }\)
Putting the value and solving, µ = 0.11

Question 50.
A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m/s² . The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the:
(a) Force on the floor of the helicopter by the crew and passenger.
(b) Action of the rotor of the helicopter on the surrounding air
(c) Force on the helicopter due to the surrounding air (g = 10 m/s² )
Answer:
(a) Force on the floor of the helicopter by the crew and passengers
= apparent weight of crew and passengers
= 500(10+ 15)
=12500 N

(b) Action of rotor of helicopter on surrounding air is Obviously vertically downwards, because helicopter rises on account of reaction of this force. Thus force of action
= (2000 + 500) (10 + 15)
= 2500 x 25
= 62,500 N

(c) Force on the helicopter due to surrounding air is obviously a reaction. As action and reaction are equal and opposite, therefore
Force of reaction F’ = 62,500 vertically upwards.

Question 51.
A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is (µ). Let the mass of the box be m.

1.  At what angle of inclination θ of the plane to the horizontal will the box just start to slide down the plane ?
2. What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to a > θ.
3. What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed ?
4. What is the force needed to be applied upwards along the plane to pk kg f make the box move up the plane with acceleration a ?

Answer:
1. When the box just starts sliding
µ = tanθ
or 0 = tan^-1 µ

2. Force acting on the box down the plane
= mg (sin a – µ cos a)

3. Force needed mg (sin a + µ cos a)

4. Force needed = mg (sin a + µ cos a) + ma.

Question 52.
Two masses of 5 kg and 3 kg are suspended with help of mass less in extensible string as shown. Calculate T₁ and T₂ when system is going upwards with acceleration m/s². (Use g 9.8 m/s²)
Answer:
According Newton’s second law of motion
(1) T₁ – (m₁ + m₂)g = (m₁ + m₂)a
T₁ = (m₁ + m₂)(a + g) = (5 + 3) (2 + 9.8)
T₁ = 94.4 N

(2) T₂ – m₂g = m₂a
T₂ = m₂ (a + g)
T₂ = 3(2 + 9.8)
T₂ = 35.4 N

Question 53.
There are few forces acting at a Point P produced by strings as shown, which is at rest. Find the forces F₁ & F₁

Answer:
Using Resolution of forces IN and 2N and then applying laws of vector addition. Calculate for F₁ & F₁.
F₁ = \(\frac{1}{\sqrt{2}}\) N, F₂ = \(\frac{3}{\sqrt{2}}\)N

Question 54.
A hunter has a machine gun that can fire 50g bullets with a velocity of 150 ms A 60 kg tiger springs at him with a velocity of 10 ms^-1. How many bullets must the hunter fire into the target so as to stop him in his track?
Answer:
Given m = mass of bullet = 50 gm = 0.50 kg
M = mass of tiger = 60 kg
v = Velocity of bullet – 150 m/s
V = Velocity of tiger = – 10 m/s
(v It is coming from opposite direction n = no. of bullets fired per second at the tiger so as to stop it.)
P_i = 0, before firing ……..(i)
P_f = n (mv) + MV …………(ii)
∴ From the law of conservation of momentum,
P_i = P_f
⇒ 0 = n (mv) + MV
n = \(\frac{MV}{mv}\) = \(\frac{-60 \times(-10)}{0.05 \times 150}\) = 80

Question 55.
Two blocks of mass 2 kg and 5 kg are connected by an ideal string passing over a pulley. The block of mass 2 kg is free to slide on a surface inclined at an angle of 30° with the horizontal whereas 5 kg block hangs freely. Find the acceleration of the system and the tension in the string.

Let a be the acceleration of the system and T be the Tension in the string. Equations of motions for 5 kg and 2 kg blocks are
5g – T = 5a
T – 2g sin θ – f = 2a
where f = force of limiting friction
= µR = µ mg cos θ = 0.3 x 2 g x cos 30°
Solving (1) & (2)
a = 4.87 m/s²

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Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Tamil Book Solutions Guide Pdf Chapter 5.3 நாட்டுப்புறக் கைவினைக் கலைகள் Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th Tamil Solutions Chapter 5.3 நாட்டுப்புறக் கைவினைக் கலைகள்

கற்பவை கற்றபின்

Question 1.
உங்கள் பகுதியில் கிடைக்கும் களிமண், பனையோலை போன்ற பொருள்களைப் பயன்படுத்திக் கைவினைப் பொருள்களைச் செய்து காட்சிப்படுத்துக.
Answer:

Question 2.
பனையோலையால் செய்யப்படும் பல்வேறு கைவினைப்பொருள்களின் படங்களைச் சேகரித்து படத்தொகுப்பு உருவாக்குக.
Answer:

பாடநூல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
பழந்தமிழ் இலக்கியங்களைப் பாதுகாத்து வைத்தவை ……………………
அ) கல்வெட்டுகள்
ஆ) செப்பேடுகள்
இ) பனையோலைகள்
ஈ) மண்பாண்டங்கள்
Answer:
இ) பனையோலைகள்

Question 2.
பானை ………………….. ஒரு சிறந்த கலையாகும்.
அ) செய்தல்
ஆ) வனைதல்
இ) முடைதல்
ஈ) சுடுதல்
Answer:
ஆ) வனைதல்

Question 3.
‘மட்டுமல்ல’ எனும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ……………………
அ) மட்டு + மல்ல
ஆ) மட்டம் + அல்ல
இ) மட்டு + அல்ல
ஈ) மட்டும் + அல்ல
Answer:
ஈ) மட்டும்+அல்ல

Question 4.
கயிறு + கட்டில் என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் …………………….
அ) கயிற்றுக்கட்டில்
ஆ) கயிர்க்கட்டில்
இ) கயிறுக்கட்டில்
ஈ) கயிற்றுகட்டில்
Answer:
அ) கயிற்றுக்கட்டில்

பின்வரும் சொற்களைச் சொற்றொடரில் அமைத்து எழுதுக.

1. முழுவதும் – பாடநூல் முழுவதும் வாசித்தால்தான் தெளிவு கிடைக்கும்.
2. மட்டுமல்லாமல் – ஏட்டுக் கல்வி மட்டுமல்லாமல் தொழில் கல்வியும் கற்க வேண்டும்.
3. அழகுக்காக – பல அரங்குகளில் சுடுமண் சிற்பங்களை அழகுக்காக வைத்திருப்பார்கள்.
4. முன்பெல்லாம் – முன்பெல்லாம் மண்பாண்டங்களை அதிகம் பயன்படுத்தினார்கள்.

குறுவினா

Question 1.
எவற்றையெல்லாம் கைவினைக்கலைகள் எனக் கூறுகிறோம்?
Answer:
மண் பொம்மைகள் செய்தல், மரப்பொம்மைகள் செய்தல், காகிதப்பொம்மைகள் செய்தல், தஞ்சாவூர்த்தட்டு செய்தல், சந்தன மாலையும் ஏலக்காய் மாலையும் செய்தல், மாட்டுக்கொம்பினால் கலைப் பொருட்கள் செய்தல் ஆகியவற்றை எல்லாம் கைவினைக் கலைகள் எனக் கூறுகிறோம்.

Question 2.
மண்பாண்டம், சுடுமண் சிற்பம் – ஒப்பிடுக.
Answer:

Question 3.
பனையோலையால் உருவாக்கப்படும் பொருள்கள் யாவை?
Answer:
குழந்தைகளுக்கான கிளுகிளுப்பை பொம்மைகள், பொருள்களை வைத்துக்கொள்ள உதவும் சிறிய கொட்டான், பெரிய கூடை, சுளகு, விசிறி, தொப்பி, ஓலைப்பாய், பனை மட்டை நாரிலிருந்து கயிறு, கட்டில், கூடை போன்றவை செய்யப்படுகின்றன.

சிறுவினா

Question 1.
பிரம்பினால் பொருள்கள் செய்யும் முறையைக் கூறுக.
Answer:
(i) பிரம்பு என்பது ஒரு தாவரம். முதலில் பிரம்புகளை நெருப்பில் காட்டிச் சூடுபடுத்த வேண்டும்.

(ii) சூடான பிரம்பை நட்டு வைத்திருக்கும் இரண்டு பாறைகளுக்கு இடையே செலுத்தி வளைக்க வேண்டும். அது வேண்டிய வடிவத்தில் கம்பி போல வளையும்.

(iii) பின்னர் அதனை தண்ணீரில் நனைத்து வைத்து விட்டால், அப்படியே நிலைத்து விடும். பிறகு அவற்றை இணைத்துச் சிறு ஆணிகளை அறைந்தும், சிறு பிரம்பு இழைகளைக் கொண்டும் தேவையான பொருட்களாக மாற்ற வேண்டும்.

Question 2.
மூங்கிலால் செய்யப்படும் பொருள்கள் குறித்து எழுதுக.
Answer:
மட்டக்கூடை, தட்டுக்கூடை, கொட்டுக்கூடை, முறம், ஏணி, சதுரத்தட்டி, கூரைத்தட்டி, தெருக்கூட்டும் துடைப்பம், மாடுகளுக்கான மூஞ்சிப்பெட்டி, பழக்கூடை, பூக்கூடை, பூத்தட்டு, கட்டில், புல்லாங்குழல், புட்டுக்குழாய், கால்நடைகளுக்கு மருந்து புகட்டும் குழாய், தொட்டில், பாடை ஆகியவை அனைத்தும் மூங்கிலால் செய்யப்படும் பொருள்கள் ஆகும்.

நெடுவினா

Question 1.
தமிழகக் கைவினைக் கலைகளைப் பற்றிய செய்திகளைத் தொகுத்து எழுதுக.
Answer:
மண்பாண்டக் கலை :

• குடம், தோண்டி, கலயம், கடம், மூடி, உழக்கு, அகல், உண்டியல், அடுப்பு, தொட்டி ஆகிய அனைத்துப் பொருட்களும் சுத்தமான களிமண்ணால் செய்யப்பட்டவை.
• பக்குவப்படுத்தப்பட்ட களிமண், மெல்லிய மணல் சாம்பல் ஆகியவற்றைக் கலந்து எடுத்துக் கொள்ள வேண்டும்.
• சக்கரத்தின் நடுவே வைத்து உரிய வடிவத்தால் அதைக் கொண்டு வர வேண்டும். பிறகு அடிப்பகுதியில் நூல் அல்லது ஊசியால் அறுத்து எடுத்து காயவைக்க வேண்டும். பிறகு உரிய மண்பாண்டம் தயாராகிவிடும்.
• மண்பாண்டங்களில் சமைத்த உணவு உடலுக்கு நல்லது.
• திருவிழாக் காலங்களிலும் சமயச் சடங்குகளிலும் மண்பானைகள் இன்றுவரை பயன்படுத்தப்பட்டு வருகின்றன.

மூங்கில் கலை :
(i) மூங்கில் கொண்டு பல கைவினைப் பொருட்கள் செய்யப்படுகின்றன.

(ii) மட்டக்கூடை, தட்டுக்கூடை, கொட்டுக்கூடை, முறம், ஏணி, சதுரத்தட்டி, கூரைத்தட்டி, தெருக்கூட்டும் துடைப்பம், மாடுகளுக்கான மூஞ்சிப்பெட்டி, பழக்கூடை, பூக்கூடை, பூத்தட்டு, கட்டில், புல்லாங்குழல், புட்டுக்குழாய், கால்நடைகளுக்கு மருந்து புகட்டும் குழாய், தொட்டில், பாடை ஆகிய அனைத்தும் மூங்கிலால் செய்யப்படும் பொருள்கள் ஆகும்.

(iii) முன்பு எல்லாம் திருமணத்தின் போது சீர்த்தட்டுகளாகப் பயன்படுத்தினர்

பனையோலைக் கலை :
(i) பனையோலையில் பல கைவினைப் பொருட்கள் உருவாக்கப்படுகின்றன.

(ii) குழந்தைகளுக்கான கிளுகிளுப்பை பொம்மைகள், பொருள்களை வைத்துக் கொள்ள உதவும் சிறிய கொட்டான், பெரிய கூடை, சுளகு, விசிறி, தொப்பி, ஓலைப்பாய், பனை மட்டை நாரிலிருந்து கயிறு, கட்டில், கூடை போன்றவை செய்யப்படுகின்றன.

பிரம்புக் கலை :
(i) பிரம்பு என்பது ஒரு தாவரம்.

(ii) முதலில் பிரம்புகளை நெருப்பில் காட்டி சூடுபடுத்த வேண்டும். சூடான பிரம்பை நட்டு வைத்திருக்கும் இரண்டு பாறைகளுக்கு இடையே செலுத்தி வளைக்க வேண்டும்.

(iii) அது வேண்டிய வடிவத்தில் கம்பி போல வளையும். பின்னர் அதனைத் தண்ணீரில் நனைத்து வைத்து விட்டால், அப்படியே நிலைத்து விடும். பிறகு அவற்றை இணைத்துச் சிறு ஆணிகளை அறைந்தும், சிறு பிரம்பு இழைகளைக் கொண்டு கூட்டியும் தேவையான பொருட்களாக மாற்ற வேண்டும்.

(iv) பிரம்பு மிகவும் குளிர்ச்சியானது. எனவே அதில் அமர்வது உடல்நலத்துக்கு நல்லது.
(v) மேலும் பிரம்புப்பொருள் வீட்டுக்கு அழகையும் கொடுக்கும்.

சிந்தனை வினா

Question 1.
கைவினைக் கலைகளுக்கும் சுற்றுச்சூழல் பாதுகாப்பிற்கும் இடையேயுள்ள தொடர்பு குறித்து எழுதுக.
Answer:
(i) கைவினைப் பொருட்கள் அனைத்தும் இயற்கையான பொருளால் தயாரிக்கப் படுபவை.

(ii) செயற்கையான பொருளோ தீங்கு விளைவிக்கும் இரசாயனமோ இதில் பயன்படுத்தப்படுவது இல்லை.

(iii) இயற்கையாகக் கிடைக்கும் களிமண், பனை ஓலை, மூங்கில், பிரம்பு ஆகியவற்றை முதன்மைப் பொருளாகக் கொண்டு கைவினைப் பொருட்கள் செய்யப்படுகின்றன.

(iv) கைவினைக் கலைகளுக்குப் பயன்படுத்தப்படக்கூடிய பொருட்களின் மீதத்தைப் பூமியில் புதைத்தாலும், அவை மக்கி விடும். இதனால் சுற்றுப்புறத்திற்கு எந்தத் தீங்கும் ஏற்படாது.

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
பானை ஓடுகள் கிடைத்துள்ள இடம் …………………..
அ) சிந்துசமவெளி
ஆ) ஆதிச்சநல்லூர்
இ) செம்பியன் கண்டியூர்
ஈ) கீழடி
Answer:
அ) சிந்துசமவெளி

Question 2.
முதுமக்கள் தாழிகள் கிடைத்துள்ள தமிழக இடம் ……………………..
அ) சிந்துசமவெளி
ஆ) ஆதிச்சநல்லூர்
இ) செம்பியன் கண்டியூர்
ஈ) கீழடி
Answer:
ஆ) ஆதிச்சநல்லூர்

Question 3.
மிகவும் பழமையான கைவினைக் கலைகளில் ஒன்று ………………….
அ) பிரம்புக் கலை
ஆ) மண்பாண்டக் கலை
இ) பனை ஓலைக்கலை
ஈ) மூங்கில் கலை
Answer:
ஆ) மண்பாண்டக் கலை

Question 4.
குளங்கள், ஆற்றங்கரைகள், வயல்வெளிகள் ஆகிய இடங்களில் கிடைக்கும் ……………………
அ) செம்மண்
ஆ) களிமண்
இ) வண்டல் மண்
ஈ) உவர்மன்
Answer:
ஆ) களிமண்

Question 5.
பானை செய்யும் சக்கரத்தின் வேறு பெயர் ………………..
அ) ஊசி
ஆ) நூல்
இ) திருவை
ஈ) சுழல்
Answer:
இ) திருவை

Question 6.
குழந்தைகளைப் படுக்க வைக்க உதவும் பாய் …………………..
அ) பந்திப்பாய்
ஆ) திண்ணைப் பாய்
இ) பட்டுப் பாய்
ஈ) தடுக்குப் பாய்
Answer:
ஈ) தடுக்குப் பாய்

Question 7.
உட்காரவும் படுக்கவும் உதவும் ………………… பாய்.
அ) பந்திப்பாய்
ஆ) திண்ணைப் பாய்
இ) பட்டுப் பாய்
ஈ) தடுக்குப் பாய்
Answer:
ஆ) திண்ணைப் பாய்

Question 8.
“முற்காலத்தில் பாய்மரக் கப்பல்களில் பயன்படுத்தியது கூட பாய் தான்” என்பதைக் கூறும் நூல் ………………….
அ) அகநானூறு
ஆ) பரிபாடல்
இ) புறநானூறு
இ) பதிற்றுப்பத்து
Answer:
இ) புறநானூறு

Question 9.
தமிழ்நாட்டின் மாநில மரம் ……………………
அ) தென்னை மரம்
ஆ) மாமரம்
இ) பனை மரம்
ஈ) மூங்கில் மரம்
Answer:
இ) பனை மரம்

Question 10.
பிரம்பு என்பது …………………. வகையைச் சேர்ந்த தாவரம்.
அ) செடி
ஆ) கொடி
இ) மரம்
ஈ) நீர்நிலை
Answer:
ஆ) கொடி

Question 11.
கலாமஸ் ரொடாங் என்னும் தாவரவியல் பெயர் கொண்டது …………………
அ) பிரம்பு
ஆ) மூங்கில்
இ) மண்பாண்டம்
ஈ) பனையோலை
Answer:
அ) பிரம்பு

குறுவினா

Question 1.
கைவினைக் கலைகள் என்றால் என்ன?
Answer:
அன்றாடப் பயன்பாட்டுக்காக அழகிய பொருள்களைத் தொழில் முறையில் உருவாக்கும் கலையைக் கைவினைக் கலை என்பர்.

Question 2.
மூங்கில் வகைகள் யாவை?
Answer:

• கல் மூங்கில்
• மலை மூங்கில்
• கூட்டு மூங்கில்

என மூங்கில் மூன்று வகைப்படும்.

Question 3.
கைவினைப் பொருட்கள் செய்யப் பயன்படும் மூங்கில் எவை?
Answer:
கூட்டு மூங்கிலே கைவினைப் பொருட்கள் செய்யப் பயன்படும் மூங்கில் ஆகும்.

Question 4.
மண்பாண்டப் பொருட்கள் யாவை?
Answer:
பானை, சட்டி, குடம், தோண்டி, கலயம், கடம், மூடி, உழக்கு, அகல், உண்டியல், அடுப்பு, தொட்டி ஆகியன மண்பாண்டப் பொருட்கள் ஆகும்.

Question 5.
களிமண் கிடைக்கும் இடங்கள் யாவை?
Answer:
குளங்கள், ஆற்றங்கரைகள் மற்றும் வயல்வெளிகள் ஆகியவற்றில் களிமண் கிடைக்கும்.

Question 6.
திருவை என்றால் என்ன?
Answer:
மண் பானை செய்யும் சக்கரத்தையே திருவை என்பர்.

Question 7.
மண்பாண்டங்களால் கிடைக்கும் நன்மைகள் யாவை?
Answer:

• மண்பாண்டங்களில் சமைத்த உணவு நல்ல சுவையுடன் இருக்கும்.
• மண்பாண்டத்தில் செய்யப்படும் உணவே உடல் நலத்திற்கும் நல்லது.
• மண் பானையில் வைத்த தண்ணீர் குளிர்ச்சியாக இருக்கும்.

Question 8.
சுடுமண் சிற்பங்கள் என்றால் என்ன?
Answer:
மண்பாண்டங்களைப் போன்றே களிமண்ணால் செய்யப்பட்டுச் சூளையில் சுட்டு எடுக்கப்படுபவை சுடுமண் சிற்பங்கள் ஆகும்.

Question 9.
பாயின் வகைகள் யாவை?
Answer:

• படுக்கும் பாய்
• பந்திப் பாய்
• திண்ணைப் பாய்
• பட்டுப்பாய்
• தொழுகைப் பாய்

Question 10.
பிரம்பு தருவிக்கப்படும் இடங்கள் யாவை?
Answer:
அஸ்ஸாம், அந்தமான், மலேசியா ஆகிய இடங்களில் இருந்து பிரம்பு தருவிக்கப் படுகிறது.

Question 11.
பிரம்புப் பொருட்களால் கிடைக்கும் நன்மைகள் யாவை?
Answer:

• பிரம்பு மிகவும் குளிர்ச்சியானது. எனவே, அதில் அமர்வது உடல்நலத்துக்கு நல்லது.
• மேலும், பிரம்புப் பொருள் வீட்டுக்கு அழகையும் கொடுக்கும்.

சிறுவினா

Question 1.
தமிழருக்கும் மண்பாண்டக் கலைக்கும் உள்ள தொடர்பைக் காட்டும் சான்றுகள் யாவை?
Answer:

• சிந்து சமவெளி அகழாய்வில் பானை ஓடுகள் கிடைத்துள்ளன.
• தமிழ்நாட்டில் ஆதிச்சநல்லூரில் முதுமக்கள் தாழிகள் கிடைத்துள்ளன.
• நாகை மாவட்டம் செம்பியன் கண்டியூரில் கலையழகு மிகுந்த மண்கலங்கள் கண்டுபிடிக்கப்பட்டுள்ளன.
• மதுரைக்கு அருகில் உள்ள கீழடியில் ஏராளமான சுடுமண் பொருட்கள் கிடைத்துள்ளன.
• இவையெல்லாம் தமிழருக்கும் மண்பாண்டக் கலைக்கும் உள்ள தொடர்பைக் காட்டும் சான்றுகளாகும்.

Question 2.
பாயின் வகைகளை விளக்குக.
Answer:

• குழந்தைகளைப் படுக்க வைப்பது தடுக்குப் பாய்.
• உட்கார்ந்து உண்ண உதவுவது பந்திப் பாய்.
• உட்காரவும் படிக்கவும் உதவுவது திண்ணைப் பாய்.
• திருமணத்துக்குப் பயன்படுவது பட்டுப் பாய்.
• இஸ்லாமியர் தொழுகைக்குப் பயன்படுத்துவது தொழுகைப் பாய்.

Question 3.
பாய்களில் எவையெவை இடம்பெற்றிருந்தன?
Answer:
(i) திருமணத்திற்குப் பயன்படுத்தும் பாய்களில் மணமக்கள் பெயர்கள் இடம் -5 பெற்றிருந்தன.

(ii) குத்துவிளக்கு, மயில், பூக்கள், வழிபாட்டுச் சின்னங்கள் ஆகியவையும் பாய்களில் இடம்பெற்றிருந்தன.

Question 4.
பிரம்பினால் செய்யப்பட்ட பலவகைப் பொருட்கள் யாவை?
Answer:

• கட்டில்
• ஊஞ்சல்
• நாற்காலி
• மேசை
•  பூக்கூடை
• பழக்கூடை
• இடியாப்பத் தட்டு
• அர்ச்சனைத் தட்டு
• வெற்றிலைப் பெட்டி

Students can Download Maths Chapter 1 Rational Numbers Ex 1.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th 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 8th Maths Solutions Term 1 Chapter 1 Rational Numbers Ex 1.1

8th Maths Exercise 1.1 Samacheer Kalvi Question 1.
Fill in the blanks:
(i) \(\frac{-19}{5}\) lies between the integers _____ and _____
(ii) The rational number that is represented by 0.44 is ______.
(iii) The standard form of \(\frac{+58}{-78}\) is _____.
(iv) The value of \(\frac{-5}{12}+\frac{7}{15}\) = ______
(v) The value of \(\left(\frac{-15}{23}\right) \div\left(\frac{+30}{-46}\right)\) is ______
Solution:
(i) -4 and -3
(ii) \(\frac{11}{25}\)
(iii) \(\frac{-29}{39}\)
(iv) \(\frac{1}{20}\)
(v) 1

8th Maths Exercise 1.1 In Tamil Question 2.
Say True or False.
(i) 0 is the smallest rational number.
(ii) There are an unlimited rationals between 0 and 1.
(iii) The rational number that does not have a reciprocal is 0.
(iv) The only rational number which is its own reciprocal is -1.
(v) The rational numbers that are equal to their additive inverses are 0 and -1.
Solution:
(i) False
(ii) True
(iii) True
(iv) False
(v) False

8th Maths Exercise 1.1 Question 3.
List five rational numbers between 2 and 0
(i) -2 and 0
(ii) \(\frac{-1}{2}\) and \(\frac{3}{5}\)
(iii) 0.25 and 0.35
(iv) -1.2 and -2.3
Solution:
(i) -2 and 0

Samacheer Kalvi 8th Maths Solutions Term 1 Pdf Question 4.
Write four rational numbers equivalent to -3 7
(i) \(\frac{-3}{5}\)
(ii) \(\frac{7}{-6}\)
(iii) \(\frac{8}{9}\)
Solution:

8th Standard Maths Exercise 1.1 Answers Question 5.
Draw the number line and represent the following rational numbers on it.

Solution:

8th Maths Book Example Sums Question 6.
Find the rational numbers for the points marked on the number line.

Solution:
(i) The number lies between -3 and -4. The unit part between -3 and -4 is divided into 3 equal parts and the second part is asked.
∴ The required number is \(-3 \frac{2}{3}=-\frac{11}{3}\).
(ii) The required number lies between 0 and -1. The unit part between 0 and -1 is divided into 5 equal parts, and the second part is taken.
∴The required number is \(-\frac{2}{5}\)
(iii) The required number lies between 1 and 2. The unit part between 1 and 2 is divided into 4 equal parts and the third part is taken.
∴ The required number is \(1 \frac{3}{4}=\frac{7}{4}\)

Samacheer Kalvi.Guru 8th Maths Question 7.
Using average, write 3 rational numbers between \(\frac{14}{5}\) and \(\frac{16}{3}\)
Solution:

Maths 8th Guide Samacheer Kalvi Question 8.
Verify that -(-x) is the same x for:
(i) x = \(\frac{11}{15}\)
(ii) x = \(\frac{-31}{45}\)
Solution:

Samacheer Kalvi 8th Maths Solutions Question 9.
Re-arrange suitable and add :
\(\frac{-3}{7}+\frac{5}{6}+\frac{4}{7}+\frac{1}{3}+\frac{13}{-6}\)
Solution:

8th Maths 1.1 Question 10.
What should be added to \(\frac{-8}{9}\) to get \(\frac{2}{5}\).
Solution:
Let the number to be added = x

8th Std Maths Exercise 1.1 Question 11.
Subtract \(\frac{-8}{44}\) from \(\frac{-17}{11}\)
Solution:

Maths Term 1 Question 12.
Evaluate:
(i) \(\frac{9}{2} \times \frac{-11}{3}\)
(ii) \(\frac{-7}{27} \times \frac{24}{-35}\)
Solution:

Maths 8th Class Chapter 1 Exercise 1.1 Question 13.
Divide
(i) \(\frac{-21}{5}\) by \(\frac{-7}{-10}\)
(ii) \(\frac{-3}{13}\) by -3
(iii) -2 by \(\frac{-6}{15}\)
Solution:

8th Maths In Tamil Question 14.
Simplify \(\left(\frac{2}{5}+\frac{3}{2}\right)+\frac{3}{10}\) as a rational number and show that it is between 6 and 7.
Solution:

8th Standard Maths Book Exercise 1.1 Question 15.
Write the five rational numbers which are less than -2.
Solution:
All the integers are rational numbers
∴ Rational numbers less than -2 are -10, -15, -20, -25, -30

8th Maths Guide Question 16.
Compare the following pairs of rational numbers

Solution:


\(\frac{10}{15}<\frac{12}{15}\)
∴ \(\frac{2}{3}<\frac{4}{5}\)

8th Maths Book Answer Question 17.
Arrange the following rational numbers is ascending and descending order.

Solution:

Objective Type Questions

8th Std Maths Guide Question 18.
The number which is subtracted from \(\frac{-6}{11}\) to get \(\frac{8}{9}\) is

Solution:
(B) \(\frac{-142}{99}\)
Hint:
Let x be the number be subtracted

Samacheer Kalvi Guru 8th Maths Book Solutions Question 19.
Which of the following rational numbers is the greatest?

Solution:
(A) \(\frac{-17}{24}\)
Hint:
LCM of 24, 16, 8, 32 = 8 × 2 × 3 × 2 = 96


∴ \(\frac{-17}{24}\) is the greatest number.

8th Std Maths Book Answers Question 20.
\(\frac{-5}{4}\) is a rational number which lies between
(A) 0 and \(\frac{-5}{4}\)
(B) -1 and 0
(C) -1 and -2
(D) -4 and -5
Solution:
(C) -1 and -2
Hint:
\(\frac{-5}{4}\) = \(-1 \frac{1}{4}\)
∴ \(\frac{-5}{4}\) lies between -1 and -2.

Samacheer Kalvi 8th Maths Book Question 21.
The standard form of \(\frac{3}{4}+\frac{5}{6}+\left(\frac{-7}{12}\right)\) is

Solution:
(D) 1
Hint:

8th Maths Book Samacheer Kalvi Question 22.
The sum of the digits of the denominator in the simplest form of \(\frac{112}{528}\)
(A) 4
(B) 5
(C) 6
(D) 7
Solution:
(C) 6
Hint:

Sum of digits in the denominator = 3 + 3 = 6

Question 23.
The rational number (numbers) which has (have) additive inverses is (are)
(A) 7
(B) \(\frac{-5}{7}\)
(C) 0
(D) all of these
Solution:
(D) all of these
Hint:
Additive inverse of 7 is -7
Additive inverse of \(\frac{-5}{7}\) is \(\frac{-5}{7}\)
Additive inverse of 0 is 0.

Question 24.
Which of the following pairs is equivalent?

Solution:
(B) \(\frac{16}{-30}, \frac{-8}{15}\)
Hint:

∴ \(\frac{16}{-30}\) and \(\frac{8}{15}\) are equivalent fraction.

Question 25.
\(\frac{3}{4} \div\left(\frac{5}{8}+\frac{1}{2}\right)\) =

Solution:
(C) \(\frac{2}{3}\)
Hint:

Guys who are planning to learn and understand the topics of 10th Social Science Geography can grab this Tamilnadu State board solutions for Chapter 1 India: Location, Relief and Drainage 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.

Students who feel tough to learn concepts can take help from this Samacheer Kalvi 10th Social Science Book Solutions Guide Pdf, all the Questions and Answers can easily refer in the exams. Go to the below sections and get 10th Social Science Geography Chapter 1 India: Location, Relief and Drainage Tamilnadu State Board Solutions PDF.

Tamilnadu Samacheer Kalvi 10th Social Science Geography Solutions Chapter 1 India: Location, Relief and Drainage

Do you feel scoring more marks in the 10th Social Science Geography 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 Geography Board solutions for Chapter 1 India: Location, Relief and Drainage 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 Geography Chapter 1 textbook solutions for Tamilnadu State Board.

India: Location, Relief and Drainage TEXTUAL EXERCISE

I. Choose the correct answer.

10th Social Geography 1st Lesson Question 1.
The north-south extent of India is
(a) 2,500 km
(b) 2,933 km
(c) 3,214 km
(d) 2,814 km
Answer:
(c) 3,214 km

India Location Relief And Drainage Question 2.
The Southern most point of India is:
(a) Andaman
(b) Kanyakumari
(c) Indira Point
(d) Kavaratti
Answer:
(c) Indira Point

India – Location, Relief And Drainage Questions And Answers Question 3.
The extent of Himalayas in the east-west is about
(a) 2,500 km
(b) 2,400 km
(c) 800 km
(d) 2,200 km
Answer:
(a) 2,500 km

India Location Relief And Drainage Pdf Question 4.
……………. River is known as ‘Sorrow of Bihar’.
(a) Narmada
(b) Godavari
(c) Kosi
(d) Damodar
Answer:
(c) Kosi

India Location Relief And Drainage Questions And Answers Question 5.
Deccan Plateau covers an area of about ……. sq.km.
(a) 8 lakh
(b) 6 lakh
(c) 5 lakh
(d) 7 lakh
Answer:
(d) 7 lakh

India Location, Relief And Drainage Notes Question 6.
A landmass bounded by sea on three sides is referred to as:
(a) Coast
(b) Island
(c) Peninsula
(d) Strait
Answer:
(c) Peninsula

South Indian Rivers Are East Flowing Give Reason Question 7.
The Palk Strait and Gulf of Mannar separates India from ………
(a) Goa
(b) West Bengal
(c) Sri Lanka
(d) Maldives
Answer:
(c) Sri Lanka

India Location, Relief And Drainage Pdf Question 8.
The highest peak in South India is:
(a) Ooty
(b) Kodaikanal
(c) Anaimudi
(d) Jindhagada
Answer:
(c) Anaimudi

10th Geography Samacheer Kalvi Question 9.
………… Plains are formed by the older alluviums.
(a) Bhabar
(b) Tarai
(c) Bhangar
(d) Khadar
Answer:
(c) Bhangar

Samacheer Kalvi 10th Geography Book Question 10.
Pulicat Lake is located between the states of:
(a) West Bengal and Odisha
(b) Karnataka and Kerala
(c) Odisha and Andhra Pradesh
(d) Tamil Nadu and Andhra Pradesh
Answer:
(d) Tamil Nadu and Andhra Pradesh

II. Match the following.

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

III. Give Reasons.

10th Social Geography Unit 1 Question 1.
The Himalayas are called young fold mountains.
Answer:
They were formed by earth movements which affected the relief of the earth in the last phase of its geological history. Because of the young age which is evident from the striking contrast in relief, Himalayan ranges are called young fold mountains.

10th Samacheer Kalvi Social Question 2.
North Indian Rivers are perennial.
Answer:
North Indian Rivers have their origin from the snow-covered Himalayas. As these rivers have water throughout the year they are referred to as perennial rivers.

Question 3.
Chottanagpur Plateau is rich in mineral resources.
Answer:
Chottanagpur Plateau is a store house of mineral resources such as mica, bauxite, copper, limestone, iron ore and coal.

Question 4.
The great Indian desert is called Marusthali.
Answer:
In Sanskrit Marusthali means ‘dead land’. It is the region of the actual desert with severe arid climate and very low vegetation. This region has different types of sand dunes.

Question 5.
The Eastern states are called seven sisters.
Answer:
Mizoram and Manipur are connected to the rest of India through Basak Village in Assam. And
it is due to this interdependence, they were given the sobriquet. It is a well known fact that the
states of Arunachal Pradesh, Assam, Meghalaya, Manipur, Mizoram, Nagaland and Tripura were named the seven sisters in 1972.

Question 6.
The river Godavari is often referred as Vridha Ganga.
Answer:
“Vridha” means old. It originates and flow through the Peninsular plateau, the oldest among the physiographic divisions of India. Like River Ganga,Godavari is the longest with an area of 3.13 lakh km2, among the peninsular rivers. Hence the river is often referred as “Vridha Ganga”.

IV. Distinguish between the following

Question 1.
Himalayan rivers and Peninsular rivers.
Answer:

S.No.	Himalayan Rivers	Peninsular Rivers
1.	Originate from Himalayas	Originate from Western Ghats
2.	Long and wide Perennial in nature	Short and arrow
3.	Perennial in nature	Non Perennial in nature
4.	Unsuitable for hydropower generation	Suitable for hydropower generation
5.	Middle and lower courses are navigable	Not useful for navigation

Question 2.
Western Ghats and Eastern Ghats.
Answer:

Question 3.
Himadri and Himachal.
Answer:

Question 4.
Western Coastal Plains and Eastern Coastal Plains.
Answer:

V. Answer in brief.

Question 1.
Name the neighbouring countries of India.
Answer:
India shares its land boundaries with Pakistan in the west, Afghanistan in the north-west, China, Nepal and Bangladesh in the east.
Our Southern neighbours across the sea consists of the two island countries, namely Sri Lanka and Maldives.

Question 2.
Give the importance of lST.
Answer:
The longitudinal extent of India between the West and the East is about 30° from Gujarat to Arunachal Pradesh. This longitudinal difference make a difference of nearly 2 hours in local time between Gujarat in the West and Arunachal Pradesh in the East. In order to avoid the time difference between the places 1ST is calculated. The Indian Standard Time is calculated based on 82°30′ East longitude.

Question 3.
Write a short note on Deccan Plateau.
Answer:

• The Deccan Plateau is a triangular landmass that lies to the south of the river Narmada.
• This is the largest physiographic division of our country.
• It covers an area of about 16 lakh sq.km (about half of the total area of the country)
• It is an old rocky plateau region.
• The topography consists of a series of plateaus and hill ranges interspersed with river valleys.
• It is higher in the west and slopes gently eastwards.
• The Western Ghats and the Eastern Ghats mark the western and the eastern edges of the Deccan plateau respectively.
• Aravalli hills mark the north-western boundary of the plateau region.
• Its northern and north-eastern boundaries are marked by the Bundelkhand upland, Kaimur and Rajmahal hills.
• Western Ghats lie parallel to the Western coast. They are continuous and can be crossed through passes only like Thai, Bhor and the Pal Ghats. It is higher then the Eastern Ghats. It cause orographic rain by facing the rain-bearing moist winds to rise along the western slopes of the Ghats. The height of the Western Ghats progressively increases from north to south. The highest peak include the Anaimudi (2,695 metres) and the DodaBetta (2,637 metres).
• The Eastern Ghats stretch from the Mahanadi valley to the Nilgiris in the south. The Eastern Ghats are discontinuous and irregular and dissected by rivers draining into the Bay of Bengal. Mahendragiri (1,501 metres) is the highest peak in the Eastern Ghats. Shevroy Hills and the Javadi Hills are located to the southeast of the Eastern Ghats. The famous hill stations of Udagamandalam, popularly Known as Ooty and the Kodaikanal are located here.
• One of the distinct features of the peninsular plateau is the black soil area known as Deccan Trap. This is of volcanic region hence the rocks are igneous. These rocks have denuded over time and are responsible for the formation of black soil. ‘

Question 4.
State the west-flowing rivers of India.
Answer:
Narmadha, Tapti and Mahi are the west-flowing rivers of India. They flow into the Arabian Sea through Gulf of Cambay.

Question 5.
Write a brief note on the island group of Lakshadweep.
Answer:

• It lies close to the Malabar coast of Kerala.
• This group of islands is composed of small coral islands.
• Earlier they were known as Laccadive, Minicoy and Amindive. In 1973, these were named as Lakshadweep.
• It covers small area of 32 sq.km. Kavaratti island is the administrative headquarters of Lakshadweep.
• This island group has great diversity of flora and fauna. The Pitti island which is uninhabited, has a bird sanctuary.

VI. Answer in a paragraph.

Question 1.
Explain the divisions of the Northern Mountains and its importance to India.
Answer:

1. Northern mountains are the youngest and the loftiest mountain chains in the world.
2. It stretches for a distance of2500km from the Indus gorge in the West to Brahmaputra gorge in the East.
3. The major divisions of the Northern mountains are:
   • The Trans Himalayas
   • Himalayas
   • Eastern or Purvanchal hills

(i) The Trans Himalayas:

• It lies in Jammu and Kashmir and Tibetian plateau.
• It is also known as Western Himalayas. As its areal extent is more in Tibet, it is also known as Tibetean Himalayas.
• The rocks of this region are of Thethys sediments and contain fossils bearing marine sediments.
• The prominent ranges of this division are Zaskar, Ladakh, Kailash and Karakoram.

(ii) The Himalayas:

• It is formed by the uplifted compression of the Thethys sea due to tectonic forces.
• It has three parallel ranges.

(a) The Greater Himalayas (Himadri)
(b) The Lesser Himalayas (Himachal)
(c) The Outer Himalayas (Siwaliks)

(a) The Greater Himalayas or Himadri:

• The most continuous range.
• Almost all the lofty peaks of the Himalayas are located in this range.
  Eg: Mt. Everest (8,848m) and Kanchenjunga (8,586m).
• It is the region of permanent snow cover.
• It has many glaciers like Gangothri, Yamunothri and Siachen.

(b) The Lesser Himalayas or Himachal:

• It is the middle range of Himalayas.
• Made up of rocks like slate, limestone and quartzite.
• Important ranges pir panjal, Dhauladhar and Mahabharat.
• Familiar for hill stations – Shimla, Mussourie Nainital, Almora, Ranikhet and Daijeeling.

(c) The Outer Himalayas or Siwaliks:

• It is the most discontinuous range dissected by the Himalayan rivers.
• The longitudinal valleys found between the Siwaliks and the Himachal are called Duns in the West and Duars in the East. Eg: Dehradun.
• This range is ideal for the development of settlements.

(iii) The Eastern Himalayas or Purvanchal Hills:

• They are the Eastern off shoots of Himalayas.
• Most of these hills are located along the border of India and Myanmar.
• Some of the important hills are Patkai Bum, Naga hills, Manipur hills, Mizo hills, Garo hills, Khasi hills and Jaintia hills.
• Collectively known as Purvanchal hills.

Importance:

• Forms as the natural barrier to the Sub continent.
• Source for many perennial rivers such as Indus, Ganges and Brahmaputra.
• Paradise of tourists due to its natural beauty.
• Renowned for rich bio-diversity.
• Many hill stations and Pilgrim centres like Amamath, Kedamath, Badrinath and Vaishnavidevi temple are located.
• Natural climatic barrier prevents the cold winds from Central Asia. Blocks the South west monsoon winds and causes heavy rainfall to North India.

Question 2.
Give an account on the major peninsular rivers of India.
Answer:

• The rivers in South India are called the Peninsular rivers.
• Most of the rivers originate from the Western Ghats.
• These are seasonal rivers, (non-perennial)
• They have a large seasonal fluctuation in volume of water as they are solely fed by rain.
• These rivers flow in valleys with steep gradients.
• Based on the direction of flow, the Peninsular rivers are divided into the west flowing and east flowing rivers.

East Flowing Rivers:

1. Mahanadi:

1. It originates near Sihawa in Raipur district of Chattisgarh and flows through Odisha.
2. Its length is 851 km.
3. Seonath, Telen, Sandur and lb are its major tributaries.
4. The main stream of Mahanadi gets divided into several distributaries such as Paika, Birupa, Chitartala, Genguti and Nun.
5. The Mahanadi empties its water in Bay of Bengal.

2. Godavari:

1. Godavari is the longest river (1,465 km) with an area of 3.13 lakh km2 among the Peninsular rivers.
2. It is also called Vridha Ganga.
3. It originates in Nasik district of Maharashtra, a portion of Western Ghats.
4. It flows through the states of Telangana and Andhra Pradesh before joining Bay of Bengal.
5. Puma, Penganga, Pranitha, Indravati, Tal and Salami are its major tributaries.
6. The river near Rajahmundry gets divided into two Channels called Vasistha and Gautami and forms one of the largest deltas in India.
7. Kolleru, a fresh water lake is located in the deltaic region of the Godavari.

3. Krishna:

1. The river Krishna originates from a spring at a place called Mahabaleshwar in the Western Ghats of Maharashtra.
2. Its length is 1,400 km and an area of 2.58 lakh sq km.
3. It is the second longest Peninsular river Bhima, Peddavagu, Musi, Koyna and Thungab hadra are the major tributaries of this river.
4. It also flows through Andhra Pradesh and joins in Bay of Bengal, at Hamasaladevi.

4. Kaveri:

1. The river Kaveri originates at Talakaveri, Kudagu hills of Karnataka.
2. Its length is 800 km.
3. The river Kaveri is called Dhakshin Ganga or Ganga of south.
4. Harangi, Hemavati, Kabini, Bhavani, Arkavathy, Noyyal, Amaravathi etc are the main tributaries of the river Kaveri.
5. In Karnataka the river bifurcates twice, forming the sacred islands of Srirangapatnam and Sivasamudram.
6. While entering Tamil Nadu, the Kaveri continues through a series of twisted wild gorges until it reaches Hogenakkal Falls and flows through a straight, narrow gorge near Salem.
7. The Kaveri breaks at Srirangam Island with two channels, river Coleroon and Kaveri.
8. At last, it empties into the Bay of Bengal at Poompuhar.

West Flowing Rivers:

1. Narmada:

1. It rises in Amarkantak Plateau in Madhya Pradesh at an elevation of about 1057 m and flows for a distance of about 1,312 km.
2. It covers an area of 98,796 sq km and forms 27 km long estuary before outfalling into the Arabian Sea through the Gulf of Cambay.
3. It is the largest among the west flowing rivers of Peninsular India.
4. Its principal tributaries are Burhner, Halon, Heran, Banjar, Dudhi, Shakkar, Tawa, Bama and Kolar.

2. Tapti:

1. The Tapti is one of the major rivers of Peninsular India with the length of about 724 km.
2. It covers an area of 65,145 sq km.
3. Tapti river rises near Multai in the Betul district of Madhya Pradesh at an elevation of about 752 m.
4. It is one of only the three rivers in Peninsular India that run from east to west – the others being the Narmada and the Mahi.
5. The major tributaries are Vaki, Gomai, Arunavati, Aner, Nesu, Buray, Panjhra and Bori.
6. It out falls into the Arabian Sea through the Gulf of Cambay.

3. Give a detailed account on the basin of the Ganga.

1. The Ganga River system is the largest drainage system of India.
2. The river Ganga is 2,480 km long.
3. It rises in the Gangotri glacier in the Himalayas at a height of 6000 metres.
4. It cuts deep gorges through the Siwalik range and enters into the plain at Haridwar.
5. The Ganga plain occupies an area of about 3,37,000 covering the states of Uttar Pradesh, Bihar and West Bengal.
6. The river Yamuna rises in Yamunotri glacier. After flowing for a distance of about 1300 km, it joins Ganga on its right bank at Allahabad.
7. The rivers Chambal, Betwa, Son and Ken rise in the Deccan Plateau and join Ganga on its right bank.
8. The Ghandak, the Gomati, the Ghaghara and the Kosi join the Ganga on its left bank.
9. It is covered by thick alluvial sediments.
10. The Ganga plains slopes gently from Haryana and drains into Bay of Bengal.
11. It is covered by thick alluvial sediments.
12. The largest distributary of Ganga is Hooghly.
13. Most of the Ganga Delta lies in Bangladesh.
14. The Seaward of the Ganga Delta has tidal estuaries, sand banks and islands known as Sunderbans.

VII. Map exercises: Mark the following in the outline map of India

Question 1.
Major mountain ranges – Karakoram, Ladakh, Zaskar, Aravalli, Western Ghats, Eastern Ghats.
Answer:

Question 2.
Major rivers — Indus, Ganga, Brahmaputra, Narmada, Tapti, Mahanadi, Godavari, Krishna & Kaverl.
Answer:

Question 3.
Major plateaus – Malwa, Chotanagpur, Deccan.
Answer:

VIII. Activities

Question 1.
Observe the Peninsular Plateau map of India and mark the major plateau divisions of India.
Answer:

Question 2.
Prepare a table showing the major West flowing and East flowing rivers of peninsular India.
Answer:

Question 3.
Assume that you are travelling from West Bengal to Gujarat along the beautiful coasts of India. Find out the states which you would pass through.
I will pass through the states of Odisha, Andhra Pradesh, Tamil Nadu, Kerala, Karnataka and Maharashtra.
Answer:

Question 4.
Find out the states through which the river Ganga flows.
The river Ganga flows through the states of Uttarakhand, Uttar Pradesh, Bihar, Jharkhand, and West Bengal.
Answer:

Question 5.
Prepare a table showing the major rivers in India and find out it’s tributaries, origin, length and area.
Answer:

Find Out:

Question 1.
The number of Union Territories along the Western Coast and Eastern Coast?
Answer:

1. Along Western coast – Four – Diu, Daman, Dadra and Nagar Haveli, Lakshadweep islands.
2. Along the Eastern coast -Two – Puducherry and Andaman and the Nicobar Islands.

Question 2.
Area wise which is the smallest and largest state?
Answer:
Largest state: Rajasthan
Smallest State: Goa

Question 3.
The states which do not have an international border or lie on the coast.
Answer:
The States that do not share an International boundary or lie on the coast are Haryana, Madhya Pradesh, Jharkhand and Chattisgarh and Telangana.
Union Territories – Chandigarh and Delhi.

Question 4.
Classify into four groups each having common frontiers with
i) Pakistan
ii) China
iii) Myanmar and
iv) Bangladesh
Answer:
Pakistan: Jammu and Kashmir, Punjab, Rajasthan and Gujarat.
China: Jammu and Kashmir, Himachal Pradesh, Sikkim and Arunachal Pradesh.
Myanmar: Arunachal Pradesh, Nagaland, Manipur, Mizoram.
Bangladesh: Bihar, West Bengal, Jharkhand, Assam, Meghalaya and Tripura.

Question 5.
Find the Hill stations which are located in Himalayan Mountains.
Answer:
Shimla, Mussourie, Nainital, Almora. Ranikhet, Darjeeling and Kulu Manali are some of the hill stations in Himalayan mountains.

Question 6.
In which river the Gerosappa (jog) fall is found?
Answer:
Sharavathi River

India: Location, Relief and Drainage Additional Questions

I. Choose the correct answer.

Question 1.
India covers an area of ………. million sq.kms.
(a) 3.2
(b) 3.5
(c) 3.8
Answer:
(a) 3.2

Question 2.
India’s longest border is with:
(a) Bangladesh
(b) Srilanka
(c) Bhutan
(d) Afghanistan
Answer:
(a) Bangladesh

Question 3.
In India, there is a vast plain in ………
(a) north
(b) south
(c) east
Answer:
(a) north

Question 4.
Till 2024 the ………………… will be the capital for both the States of Andhra pradesh and Telangana.
(a) Hyderabad
(b) Secunderabad
(c) Nellore
(d) Amaravati
Answer:
(a) Hyderabad

Question 5.
India is situated into southern part of ……….
(a) Indo-Myanmar
(b) Asia
(c) Sri Lanka
Answer:
(b) Asia

Question 6.
The ………………… ranges of the Himalayas is the most continuous of all the ranges.
(a) Himachal
(b) Aravalli
(c) Siwaliks
(d) Himadri
Answer:
(d) Himadri

Question 7.
India occupies ………. % of the world’s land area.
(a) 3.5
(b) 2.4
(c) 7.5
Answer:
(b) 2.4

Question 8.
The deltaic region of lower Ganga, the uplands are known as:
(a) Chars
(b) Bhangar
(c) Terai
(d) Bils
Answer:
(a) Chars

Question 9.
India is the ……….. largest country regarding its area.
(a) seventh
(b) third
(c) fourth
Answer:
(a) seventh

Question 10.
The ………………… river system is the largest drainage system of India.
(a) Yamuna
(b) Godavari
(c) Ganga
(d) Kaveri
Answer:
(c) Ganga

Question 11.
India is ………….. times smaller than USA.
(a) ten
(b) seven
(c) three
Answer:
(c) three

Question 12.
India has a predominant position in the ……. realm.
(a) Bay of Bengal
(b) Indian Ocean
(c) Arabian Sea
Answer:
(b) Indian Ocean

Question 13.
No other country has such a large ……….. line.
(a) mountain
(b) plateau
(c) coastal
Answer:
(c) coastal

Question 14.
India is connected with China, Japan and Australia through ………..
(a) Malaccan Strait
(b) Suez canal
(c) Palk Strait
Answer:
(a) Malaccan Strait

Question 15.
The 0° Meridian passes through Greenwich in ……………
(a) New york
(b) England
(c) Los Angeles
Answer:
(b) England

Question 16.
Dehra Dun is the capital of ………..
(a) Uttar Pradesh
(b) Sikkim
(c) Uttranchal
Answer:
(c) Uttranchal

Question 17.
India is a ……….. country with total freedom of worship.
(a) republic
(b) democratic
(c) secular
Answer:
(c) secular

Question 18.
According to gaseous mass theory, the earth was separated from the ………….
(a) Universe
(b) Sun
(c) Milky way
Answer:
(b) Sun

Question 19.
The ……… force on Tethys sea gave rise to “Himalayas”.
(a) depressional
(b) compressed
(c) longitudinal
Answer:
(b) compressed

Question 20.
The Himalayas started growing up due to ………. of the agents of denudation.
(a) deposition
(b) transportation
(c) erosion
Answer:
(c) erosion

Question 21.
The Gangetic plain was formed due to ………..
(a) erosion
(b) transportation
(c) deposition
Answer:
(c) deposition

Question 22.
The major physiographical units of India are ………
(a) six
(b) five
(c) four
Answer:
(a) six

Question 23.
The Himalayas are made of ………… rocks.
(a) sedimentary
(b) igneous
(c) volcanic
Answer:
(a) sedimentary

Question 24.
Pamir knot is in the ………… part of India.
(a) north-east
(b) north-west
(c) south-east
Answer:
(b) north-west

Question 25.
Ladakh is in the north-west of ……….
(a) Himachal Pradesh
(b) Kashmir
(c) Kerala
Answer:
(b) Kashmir

Question 26.
The northernmost range of Himalayas is ………
(a) Himachal
(b) Siwaliks
(c) Himadri
Answer:
(c) Himadri

Question 27.
Himachal is highly ………. topography.
(a) smooth
(b) rugged
(c) narrow
Answer:
(b) rugged

Question 28.
Pirpanjal lies in ……… state.
(a) Kashmir
(b) Himachal Pradesh
(c) Arunachal Pradesh
Answer:
(a) Kashmir

Question 29.
Amamath, Kedranath and Badrinath are the …….. of Himachal range.
(a) valleys
(b) hill resorts
(c) pilgrimages
Answer:
(c) pilgrimages

Question 30.
Siwaliks is the ………. range, made up of mud and soft rocks.
(a) discontinuous
(b) broad
(c) continuous
Answer:
(a) discontinuous

Question 31.
An example of longitudinal valley of Siwaliks is ………..
(a) Nainital
(b) Mussouri
(c) Dehra Dun
Answer:
(c) Dehra Dun

Question 32.
……… supports the growth of thick forest in siwaliks.
(a) Pebbles and gravels
(b) Muds and soft rock
(c) fine silt
Answer:
(c) fine silt

Question 33.
The Purvachal mountains are of ……… height.
(a) medium
(b) short
(c) very tall
Answer:
(a) medium

Question 34.
The Himalayas act as the ……. barrier protecting from the foreign invasions.
(a) physical
(b) natural
(c) climate
Answer:
(a) physical

Question 35.
…… soil helps the cultivation of crops in great plain.
(a) block
(b) alluvium
(c) fertile
Answer:
(b) alluvium

Question 36.
There are dense forests on the ……….. of the Himalayan.
(a) peak
(b) valleys
(c) slopes
Answer:
(c) slopes

Question 37.
The ……… attract the tourists by its scenic beauty and pleasant climate.
(a) hill stations
(b) pilgrimages
(c) valleys
Answer:
(a) hill stations

Question 38.
The great Plain was formed due to …….. force.
(a) compression
(b) depression
(c) longitudinal
Answer:
(b) depression

Question 39.
The depression was caused in the plains out of ………..
(a) erosion
(b) transportation
(c) deposition
Answer:
(c) deposition

Question 40.
The rivers passed their ways by eroding their ……. when Himalayas lifted themselves high.
(a) passes
(b) mountains
(c) valleys
Answers:
(c) valley

II. Fill in the blanks.

1. The land in India gets abundance sunshine from the ………. sun.
2. India is the ……….. largest country with respect to area.
3. The shallow sea divided the Angara and Gondwana land was ………
4. The south eastern part of the Deccan plateau is known as ……….
5. The description of the physical relief features of a country is known as ………
6. The Pamir Knot, popularly known as the ……….
7. Mount Everest is the highest peak of the world with an altitude of ………..
8. The largest tributary of the Ganga is ……..
9. The largest plateau of the peninsular region is ………
10. The largest and longest of the Peninsular river is …………
11. The Cauvery rises in hills of …….. district in Karnataka.
12. The ………. river rises in Agasthiar hills.
13. India is connected with Europe through ………….
14. Himalayas and Karakoram provide a ……… boundary in the north.
15. Godwin Austin known as ………. is the highest peak of India.
16. The water body around the compact land mass was called …………
17. Karakoram stretches out eastern from ………..
18. …………. is the longest glacier.
19. The gaps providing natural routes across the mountains are called …………
20. The beautiful Kashmir valley is found in ……… ranges.
21. The narrow longitudinal valleys of siwaliks are called ………..
22. A flat low lying land of the Great plain is made up of ……….
23. The ……….. is a narrow porous zone along the foothills of Siwaliks.
24. The ………. is a zone of dampers and marshes covered with forests.
25. The Indus river flows into Pakistan through ………..
26. Ganga river reaches plain at …………..
27. Dams across the rivers help in generating ……….. power.
28. The Sub division of the peninsular plateau are the …….. and the ………..
29. At the apex of the Deccan plateau is ……….
30. The other name of the Deccan plateau is ………. plateau.
31. The Passes in the Western Ghats are ………., ………. and ………
32. The rivers of the Peninsular India originate on the ……… ghats.
33. ………, ……….. are the distributaries of the cauvery.
34. ……… and ……….. of the special markages of the fast coast plain.
35. River Ganga is known as the river ………. in Bangladesh.
Answers
1. tropical
2. seventh
3. tethys
4. Telengana
5. physiography
6. Roof of the World
7. 8,848 metres
8. Hooghly
9. Deccan plateau
10. Godavari
11.coorg
12. Thamiraparani
13. suez canal
14. natural
15. Mount K2
16. Panthalasa
17. Pamir Knot
18. Siachin
19. Passes
20. dhauladhar
21. duns
22. alluvium
23. Bhabar
24. Terail
25. Kashmir
26. Haridwar
27. hydroelectric
28. Central highland, Deccan plateau
29. Kanyakumari
30. lava
31. Thai ghat, Bohr ghat, Pal ghat
32. western
33. vennar, vettar
34. Lagoons, delta
35. Padma

III. Match the following.

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

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

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

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

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

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

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

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

IV. Distinguish Between:

Question 1.
Konkan Coast and Malabar Coast.
Answer:

S.No.	Konkan Coast	Malabar Coast
1.	It extent from Gujarat to Goa.	It is located between Mangalore and Kanyakumari.
2.	It has some features of marine erosions like cliffs and reefs.	It contains a large number of long and narrow lagoons.
3.	The Northern part is sandy and the southern part is rocky and rugged.	The lagoons and backwaters serve as inland waterways.

Question 2.
Andaman and Nicobar Islands and Lakshadweep Islands.
Answer:

S.No.	Andaman and Nicobar Islands	Lakshadweep Islands
1.	They are located in the Bay of Bengal.	They are located in the Arabian Sea.
2.	They are far off from India.	They are of coral origin.
3.	Port Blair is the capital.	Kavaratti is the administration headquaters.
4.	They are about 572 islands.	They are 27 islands here.
5.	Only 38 are inhabited.	Only 11 islands are inhabited.

Question 3.
The Bundelkhand and The Baghelkhand.
Answer:

S.No.	The Bundalkhand	The Baghelkhand
1.	It is located towards the south of the Yamuna river.	It lies to the east of “Maikala Range”.
2.	It is composed of igneous and metamorphic rocks.	It is made up of sand stone and limestone.
3.	In the northern part of it, there is rich deposit of alluvium.	The central part of the plateau acts as water divide between the Son and the Mahanadhi drainage basins.

Question 4.
GMT and IST.
Answer:

S.No.	GMT	IST
1.	It means Greenwich Mean Time.	It means Indian Standard Time
2.	GMT is calculated at 0° longitude.	India’s central meridian is 82° 30′ E longitude.
3.	It passes through Greenwich.	It passes through Mirzapur.
4.	It is the World Standard Time.	It is the Indian Standard Time.

Question 5.
The Bhabar Plain and The Khadar Plain.

S.No.	The Bhabar Plain	The Khadar Plain
1.	It lies to the south of the Siwalik from west to east.	It is the new alluvium of the plains.
2.	It is a pebble studded zone of porous beds.	They form a continuous belt of fresh deposits of silt every year during rainy season.
3.	This plain is not suitable for cultivation. Only big trees with large roots thrive in this region	The Khadar land consists of sand, silt, clay and mud. It is highly fertile soil.

V. Answer in One Word.

Question 1.
The tropic of Cancer does not pass through which state?
Answer:
Odisha

Question 2.
What is the easternmost longitude of India?
Answer:
97° 25′ E

Question 3.
Which latitude divides India into two equal halves?
Answer:
Tropic of Cancer (23° 30′ N)

Question 4.
Name the island group which lies towards east of India.
Answer:
Andaman and Nicobar

Question 5.
What is the percentage of land area that India occupies with respect to the world?
Answer:
2.4%

Question 6.
What is the distance between east and weat extremity of India in Kilometres?
Answer:
2933 km

Question 7.
What is the time lag from Gujarat to Arunachal Pradesh?
Answer:
2 hrs

Question 8.
The Standard Meridian of India passes through Mirzapur, it is located on which state of India.
Answer:
Uttar Pradesh

Question 9.
How much distance has been reduced between India and Europe by the construction of the Suez Canal?
Answer:
7000 km distance

Question 10.
Which is the smallest state of India area wise?
Answer:
Goa

Question 11.
What is the southernmost part of India?
Answer:
Indira Point

Question 12.
What is southernmost part of Indian mainland?
Answer:
Kanyakumari

Question 13.
Which states of India shares their border with Myanmar?
Answer:
Tripura, Mizoram, Manipur, and Nagaland.

Question 14.
Name the Eastern Coastal States of India.
Answer:
Odisha, West Bengal, Tamil Nadu, Andhra Pradesh.

Question 15.
Name the landmass bounded by sea on three sides.
Answer:
Peninsular

Question 16.
Which is the most of the volcanoes and earthquakes in the world located?
Answer:
The Peninsular Plateau

Question 17.
Name the mountain ranges in the eastern part of India forming its boundary with Myanmar.
Answer:
Purvanchal

Question 18.
What is the name given to longitudinal valley that lies between lesser Himalayas and Shivaliks?
Answer:
Duns

Question 19.
Name the part of the Himalayas lying between the Kali and Tista rivers.
Answer:
Nepal Himalayas

Question 20.
Which river has the largest inhabited riverine islands in the world?
Answer:
Brahmaputra

Question 21.
What is the soil of Bangar region that contains calcareous deposits locally known as?
Answer:
Kankar

Question 22.
What is the swampy and marshy region of Northern plains called as?
Answer:
Terai

Question 23.
What is the highest peak in the Eastern Ghats?
Answer:
Mahendragiri (1502 mt)

Question 24.
Name the second highest peak of Western Ghats.
Answer:
Doda Betta

Question 25.
Which type of soil is found in Deccan Trap region?
Answer:
Black soil

Question 26.
Name the major west flowing rivers in the peninsular plateau.
Answer:
Narmada and Tapti

Question 27.
Name the largest river of the Indian Ocean.
Answer:
Luni

Question 28.
What is the name given to the southern part of eastern coastal plains?
Answer:
Coromandel

Question 29.
Which is the largest salt water lake of India?
Answer:
Chilika

Question 30.
Which physiographic division is the storehouse of minerals?
Answer:
Peninsular Plateau

VI. Answer in brief.

Question 1.
India is a Sub-continent.
Answer:
India possesses distinct continental characteristics in physiography, climate, natural vegetation, minerals, human resources etc., Hence India is known as “Subcontinent”.

Question 2.
Mention the tributaries and distributaries of Cauvery.
Answer:
Tributaries: Amaravathi, Noyyal and Bhavani.
Distributaries: Vennar, vettar and Kudamurutti

Question 3.
Write the latitudinal and longitudinal extent of India.
Answer:
India extends from 8°4’N to 37°6 ‘N latitudes and 68°7 ‘E to 97°25 ‘E longitude. It is located in the North-Eastern hemisphere. The Tropic of cancer (23°30’N) passes through the middle of the country dividing it into two halves.

Question 4.
When did the Suez Canal start functioning and how did it benefit India?
Answer:
The Suez Canal started functioning in 1869.
Benefits of India.

1. It reduced the distance between India and Europe by 7,000 km.
2. The Canal in the boom for trade as it had reduced the transportation cost and number of days involved.

Question 5.
Name the major physiographic divisions of India.
Answer:
Based on the geological structure, India is divided into 6 major physiographic divisions namely.

1. The Himalayan Mountains (Northern mountains)
2. The Great Northern plains
3. The Peninsular plateau
4. The Great Indian desert
5. The Coastal plains
6. The Islands

Question 6.
Which features modified the relief features of India?
Answer:
Besides geological functions, a lot of processes such as weathering erosion and deposition have created and modified the relief to its present form.

Question 7.
Name the prominent ranges of Trans-Himalayas.
Answer:
The prominent ranges of Trans-Himalayas are Karakoram, Zaskar, Ladak and Kailash.

Question 8.
What is the extent of Northern plains of India?
Answer:
It spreads over an area of 7 lakh sq.km. The plain being about 2400 km long and 240 to 320 km broad.

Question 9.
What are ‘Duns’?
Answer:
The longitudinal valley found between the Siwaliks and Himachal range are called Duns in the West and Duars in the East. These are the ideal sites for human settlements and agriculture.

Question 10.
What is doab?
Answer:
A tract of land between two rivers is called a doab. It made up of two words ‘do’ meaning ‘two’ and ab meaning ‘water’.

Question 11.
To which parts are Ganga plains extended?
Answer:
The Ganga plain extends between Ghagger and Teesta rivers. It is spread over Haryana, Delhi, Uttar Pradesh, Jharkhand and West Bengal.

Question 12.
How are riverine islands formed?
Answer:
Rivers moving from mountains carry alluvium with them and do the depositional work. In the lower course because of gentle slope, the river velocity decrease, and therefore islands are formed.

Question 13.
Where is -Deccan Trap located and what does it composed of?
Answer:
The black soil area in the peninsular plateau regions is known as Deccan trap. They extend from Gujarat to Delhi in a southwest – northeast direction. This is of volcanic origin hence the rocks are igneous.

Actually these rocks have descended over time and are responsible for the formation of black soil. The Aravalli hills lie on the western and northwestern margins of the peninsular plateau. These are highly eroded hills and are found as broken hills.

Question 14.
What are Barchans?
Answer:
They are crescent shaped dunes.

Question 15.
What are the climate characteristics of Indian desert?
Answer:
The region recevies very low rainfall below 150 mm per year. Streams appear during the rainy season. Soon after they disappear into the sand as they do not have enough water to reach the sea.

VII. Answer the following in paragraph.

Question 1.
What do you mean by drainage system? Write a note on it.
Answer:

1. A drainage system is an integrated system of tributaries and trunk stream which collects and drains surface water into the sea, lake or some other water body.
2. The total area drained by a river and its tributaries is known as a drainage basin.
3. The drainage pattern of an area is the result of the geological structure of the respective areas.
4. The river system provides irrigation, drinking water, power generation and livelihood for a large number of population.
5. The drainage system of India is broadly divided into two major groups on the basis of their location. They are
   • The Himalayan rivers
   • Peninsular rivers

Question 2.
“Unity in Diversity”. Explain.
Answer:
India is a vast country with a diversity of Physical characteristics.

Diversity in Natural Phenomena

1. India has unique landforms ranging from the highest peaks to the lowest plains. In the north India, Mount Godwin Austin, otherwise known as Mt. K2 is the highest peak of India and coastal plains are the lowest in the South India.
2. The Climate varies from the tropical to the temperate zone. Mawsynram in Meghalaya receives the highest amount of rainfall whereas the thar desert receives very low rainfall.
3. We have wet dense tropical forest on the western Ghats, Mangrove trees in the sundarbans of West Bengal and the shrubs and sparse vegetation in the Thar desert.
4. The diversity of the physical environment and climate has made India an ideal habitat for varieties of flora and fauna.

Diversity in Natural Phenomena
India is a secular country with total freedom of worship. People follow Hinduism, Christianity, Islam, Sikkism, Buddhism, Jainism and Zoroastrianism with cultural diversities.

Unity in Diversity
Inspite of its physical, religion and racial varities, the ‘Indian Culture’ unites all people. Hence India is known for her “Unity in diversity”.

People shed all their differences and stand together when is a crisis. The best examples are kargil invasions and natural calamities like floods and Tsunami. Even if we differ in many factors, we are all one in the fact we are Indians.

Question 3.
Give an account on Punjab-Haryana plains.
Answer:

1. Punjab Haryana plains are one among the subdivisions of the Northern plains.
2. They are formed by the deposition of the rivers Sutlej, Beas and Ravi, the tributaries of River Indus.
3. They lie to the North-east of the Great Indian Desert.
4. These plains are found over an area of about 1.75 lakh sq.km.
5. These plains act as water-divide (Doab) to the rivers Yamuna-Sutlej and Ganga-Yamuna.

Question 4.
Explain the origin of the Himalayas.
Answer:

1. Millions of years ago there was only one landmass on the surface of the earth. It was surrounded by ocean on all sides.
2. The landmass was called “ Pangea”. It was surrounded by a waterbody Known as “Panthalasa”.
3. This large landmass split up into two parts. The northern part was known as “Angaraland” and the southern part was known as “Gondwana land”.
4. The Sea seperating these two was called the Tethys Sea.
5. This Sea stretched along an east-west direction.
6. The rivers from Angara and Gondwana deposited the silts along the Tethys Sea.
7. After a long period, due to tectonic forces the deposits up lifted to forms fold mountains called the Himalayan range.
8. It is an young fold mountain.
9. The Himalayas is the home of several high peak. However, it holds the record of having the maximum number of highest peaks among any mountains range in world. Out of the heights peaks in this world, Himalayas holds nine.

Question 5.
Give an account of the four divisions of Himalayas from west to east along with Purvanchal hills respectively.
Answer:
Divisions of the Himalayas of the basis of regions from west to east. These division have been demarcated by river valleys.

(i) Punjab Himalayas:
The part of the Himalayas lying between Indus and Sutlej has been traditionally known as Punjab Himalayas but it is also known as regionally as Kashmir and Himachal Himalaya from west to east respectively.

(ii) Kumaon Himalayas:
The part of the Himalayas lying between Satluj and Kali rivers is known as Kumaun Himalayas.

(iii) Assam Himalayas:
The Kali and Tista rivers demarcate the Nepal Himalayas and the past lying between Tista and Dihang rivers is known as Assam Himalayas.

Purvachal:

1. The Brahmaputra marks the eastern most boundary of the Himalayas. Beyond the Dihang gorge, the Himalayas bend sharply to the south and spread along the eastern boundary of India. They are known as Purvanchal or the Eastern hills and mountains.
2. These hills running through the north-eastern states are mostly composed of strong sandstones which are sedimentary rocks.
3. It covered with dense forests, they mostly runs as parallel ranges and valleys.
4. The Purvanchal comprises the Patkai hills, the Naga hills, Manipur hills and the Mizo hills.

Question 6.
Give a detailed account on the Great Northern Plains.
Answer:
Formation:
The Northern Plain has been formed by the interplay of the three major river systems namely – The Indus, The Ganga, and The Brahmaputra along with their tributaries. This plain is formed of alluvial soil. The deposition of alluvium in a vast basis lying of the foothills of the Himalayas over millions of years formed this fertile plain.

Extension:

It spread over an area of 7 lakh sq.km. The plain being about 2400 km long and 240 to 320 km broad, is a densely populated physiographic division.
Importance:
With a rich soil caves combined with adequate water supply and favourable climate it is agriculturally a very production past of India.

Important Features:

1. In the lower course, due to gentle slope, the velocity of the river decreases which results in the formation of riverine islands. Majuli, in the Brahmaputra River is the largest inhabited riverine island in the world.
2. The rivers in their lower course split into numerous channels due to the deposition of silt. These channels are known as tributaries.
3. The Northern Plain is broadly divided into three sections. Punjab plains, Ganga plains and Brahmaputra plains.
4. According to the variations in relief features, the Northern plains can be divided into four regions – Bhabar, Terai, Bhangar and Khadar.

Punjab Plains:
The Western part of the Northern plain is referred to as the Punjab plains formed by the Indus and its tributaries. The larger part of this plain lies in Pakistan Indus and its tributaries – The Jhelum, the Chenab, the Ravi, the Beas and the Satluj originate in the Himalayas. This section of the plain is dominated by the doabs.

Ganga Plains:
They extend between Ghagger and Teesta rivers. It is spread over the states of North India like Haryana, Delhi, U.P., Bihar, Partly Jharkhand and West Bengal.

Brahmaputra plains:
In the east of Ganga plains lies the Brahmaputra plains. They cover the areas of Assam and Arunachal Pradesh.
According to the variations in relief features, the Northern plains can be divided into four regions.

Bhabar:
The rivers after descending from the mountains deposit pebbles in a narrow belt of about 8 to 16 km in width lying parallel to the slopes of the Shivaliks. It is known as Bhabar. All the streams disappear in this Bhabar belt.

Terai:
South of this belt, the streams and rivers re-emerge and create a wet swampy and marshy region known as ‘Terai’. This was a thickly forested region full of wildlife.

The Terai is wider in the eastern parts of the Great plains, especially in Brahmaputra valley due to heavy rainfall. In many states, the Terai forest have been cleaned for cultivation.

Khadar:
The newer younger deposits of the flood plains are called Khadar. They are renamed almost every year and so are fertile. Thus ideal for intensive agriculture. The Khadar tracts are enriched by fresh deposits of silt every year during rainy seasons. The Khadar land consists of sand, silt, clay and mud. It is highly fertile soil.

Delta Plains:
The deitaic plains is an extension of the Khadar land. It covers about 1.9 sq.km in the lower reaches of the Ganga River. It is an area of deposition as the river flows in this tract sluggishly. The deltaic plain consists mainly of old mud, new mud and marsh. In the delta region, the uplands are called “chars”. While the marshy areas are called ‘Bils’

Question 7.
Write in brief about Peninsular Plateau.
Answer:
Location:

1. The Peninsular plateau is located to the south of northern great plains.
2. It covers an area of about 16 lakh sq.km.
3. Boundaries of the Peninsular plateau.
   • North – Aravalli, Vindhya, Satpura and Rajmahal ranges
   • West – Western Ghats
   • East – Eastern Ghats
4. The average height varies between 600 – 900 mts above the mean sea level.
5. The general slope is from west to east broad divisions, namely the Central Highlands and the Deccan Plateau.

Central Highlands:

• The part of the Peninsular plateau lying to the north of the Narmada river covering a major area of the Malwa Plateau is known as the Central Highlands

Malwa Plateau:

• Malwa Plateau is bounded by the Aravalli range, the Vindhya range and Bundelthand.
• The Chambal river and its tributaries have created ravines in the northern part of the plateau.

The Bundelkhand:

• It is located towards the south of the Yamuna river.
• It is composed of igneous and metamorphic rocks.
• In the northern part, the Ganga and Yamuna rivers have deposited alluvium.
• The hilly areas are made up of sandstone and granite.
• Betwa and Ken rivers have carved out deep gorges.

The Baghelkhand:

• It lies to the east of “Maikala Range”.
• It is made up of sandstone and limestone in the west and granite in the east.
• The central part of the plateau acts as water divide between the Son and the Mahanadhi drainage system.

The Chotanagpur Plateau:

• It is located towards the northeast.
• It is drained by Damodar, Sabamarekha, Koel and Barakar rivers.
• The Damodar river flows from west to east through the middle of this region.
• This region has a series of plateau and hills.

Deccan Plateau:

1. It is made up of lava and is covered with black soil.
   Boundaries of Deccan Plateau.
   • North-east – Vindhya and Satpura mountains
   • North – Mahadev and Maikala range
   • West – Western Ghats
   • East – Eastern Ghats
2. It is a tableland composed of the old crystalline, igneous and metamorphic rocks.
3. It is formed due to the breaking and drifting of the Gondwana land.
4. It has broad and shallow valleys and rounded hills.
5. It is a triangular landmass that lies to the south of the river Narmada.
6. The Satpura range flanks its broad base in the north while the Mahadev, the Kaimur hills and the Maikal range form its eastern extensions.
7. The Deccan plateau is higher in the weat and slopes gently eastwards.
8. Rivers like Mahanadhi, Godavari, Krishna and Kaveri flow eastwards and join the Bay of Bengal.
9. It is separated by a fault from the Chotanagpur plateau.
10. It made up of lava rocks and has black regur soils.

Question 8.
Write about the Indian Desert.
Answer:

1. The Indian desert lies towards the western margins of the Aravalli Hills.
2. It is an undulating sandy plain covered with sand dunes.
3. This regions receives very low rainfall below 150 mm per year.
4. It has arid climate with low vegetation cover.
5. Streams appear during the rainy season, soon after they disappear into the sand as they do not have enough water to reach the sea. Luni is the only large rivers in this region.
6. Barchans (Crescent-shaped dunes) covers large areas more prominent in Jaisalmer but longitudinal dunes become more prominent near the Indo-Pakistan boundary.
7. It is the world 7th largest desert and world 9th largest sub tropical desert located in western part of the India.
8. There are two major division on the Thar desert. They are known as the Actual desert region (Marusthali) and the semi desert region (Bhangar).

Question 9.
Differences between Himalayan rivers and Peninsular rivers.
Answer:

We think the data given here clarify all your queries of Chapter 1 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 Geography Chapter 1 India: Location, Relief and Drainage Questions and Answers & score well. Need any information regarding this then ask us through comments & we’ll give the best possible answers very soon.

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.4

9th Maths Exercise 3.4 Question 1.
Expand the following :
(i) (2x + 3y + 4z)²
(ii) (-p + 2q + 3r)²
(iii) (2p + 3)(2p – 4)(2p – 5)
(iv) (3a + 1)(3a – 2)(3a + 4)
Solution:
(i) (2x + 3y + 4z)²
(a + b + c)² ≡ a² + b² + c² + 2ab + 2bc + 2ca
∴ (2x + 3y + 4z)² = (2x)² + (3y)² + (4z)² + 2 (2x) (3y) + 2 (3y) (4z) + 2 (4z) (2x)
= 4x² + 9y² + 16z² + 12xy + 24yz + 16zx

(ii) (-p + 2q + 3r)²
(a + b + c)² ≡ a² + b² + c² + 2ab + 2bc + 2ca
(-p + 2q + 3r)² = (-p)² + (2q)² + (3r)² + 2(-p) (2q) + 2(2q) (3r) + 2 (-p) (3r)
= p² + 4q² + 9r² – 4pq +12qr – 6rp

(iii) (2p + 3)(2p – 4)(2p – 5)
(x + a) (x + b) (x + c) ≡ x³ + (a + b + c) x² + (ab + be + ca) x + abc
(2p + 3)(2p – 4)(2p – 5) = (2p)³ + (3 – 4 – 5) (2p)² + [(3 × – 4)² + (-4 × -5) + (-5 × 3)] 2p + 3 × – 4 × – 5
= 8p³ + (-6) (4p²) + [-12 + 20 + (-15)] 2p + 60
= 8p³ – 24p² + (-7)2p + 60
(2p + 3)(2p – 4)(2p – 5) = 8p³ – 24p² – 14p + 60

(iv) (3a + 1)(3a – 2)(3a + 4)
(x + a) (x + b) (x + c) ≡ x³ + (a + b + c) x² + (ab + bc + ca) x + abc
(3a + 1)(3a – 2)(3a + 4) = (3a)³ + (1 – 2 + 4) (3a)² + [1 × (- 2) + (-2 × 4) + 4 × 1] (3a) + 1 × -2 × 4
= 27a³ + 3 (9a²) + (-2 – 8 + 4)3a – 8
= 27a³ + 27a² – 8a – 8

9th Maths Algebra Exercise 3.4 Question 2.
Using algebraic identity, find the coefficients of x2, x and constant term without actual expansion.
(i) (x + 5)(x + 6)(x + 7)
(ii) (2x + 3)(2x – 5)(2x – 6)
Solution:
(i) (x + 5)(x + 6)(x + 7)
(x + a) (x + b) (x + c) ≡ x³ + (a + b + c) x² + (ab + bc + ca) x + abc
Co-efficient of x² = a + b + c = 5 + 6 + 7 = 18
Co-efficient of x² = ab + bc + ca = (5 × 6) + (6 × 7) + (7 × 5)
= 30 + 42 + 35 = 107
Constant term = abc = 5 × 6 × 7
Co-efficient of constant term = 210

(ii) (2x + 3)(2x – 5)(2x — 6)
∴ Co-efficient of x² = 4 (a + b + c) = 4 (3 + (-5) + (-6))
= 4 × (-8) = -32
Co-efficient of x = 2 (ab + bc + ca)
= 2 [3 × (-5) + (-5) (-6) + (-6) (3)]
= 2[-15 + 30- 18] = 2 × (-3) = -6
Constant term = abc = 3 × (-5) × (-6) = 90

9th Maths 3.4 Question 3.
If (x + a)(x + b)(x + c) = x³ + 14x² + 59x + 70, find the value of
(i) a + b + c
(ii) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
(iii) a² + b² + c²
(iv) \(\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}\)
Solution:
(x + a)(x + b)(x + c) = x³ + 14x² + 59x + 70 …………….. (1)
(x + a) (x + b) (x + c) ≡ x³ + (a + b + c) x² + (ab + bc + ca) x + abc …………. (2)
(i) Comparing (1) & (2)
We get, a + b + c = 14
(ii) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{b c+a c+a b}{a b c}=\frac{59}{70}\)
(iii) (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
= 14² – 2 (59) = 196 – 118 = 78
(iv) \(\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}=\frac{a^{2}+b^{2}+c^{2}}{a b c}=\frac{78}{70}=\frac{39}{35}\)

Exercise 3.4 Class 9 Question 4.
Expand
(i) (3a – 4b)³
(ii) (x + \(\frac{1}{y}\))³
Solution:
(i) (3a – 4b)³ We know that
(x – y)³ = x³ – 3x²y + 3xy² – y³
(3a-4b)³ = (3a)³ – 3 (3a)² (4b) + 3 (3a) (4b)² – (4b)³
= 27a³ – 108a²b + 144 ab² – 64b³

(ii) (x + \(\frac{1}{y}\))³
(x + y)³ ≡ x³ + 3x²y + 3xy² + y³
\(\left(x+\frac{1}{y}\right)^{3}=x^{3}+\frac{3 x^{2}}{y}+\frac{3 x}{y^{2}}+\frac{1}{y^{3}}\)

9th Maths Exercise 3.4 In Tamil Question 5.
Evaluate the following by using identities:
(i) 98³
(ii) 1001³
Solution:
(i) 98² = (100 – 2)³
(a – b)³ ≡ a³ – 3a²b + 3ab² – b³
98³ = (100 – 2)³ = 100³ – 3 × 100² × + 3 × 100 × 2² – 2³
= 1000000 – 3 × 10000 × 2 + 300 × 4 – 8
= 1000000 – 60000 + 1200 – 8 = 1001200 – 60008 = 941192

(ii) 1001³ = (1000 + 1)³
(a + b)³ ≡ a³ + 3a²b + 3ab² + b³
(1000 + 1)³ = 1000³ + 3(1000²) × 1 + 3 × 1000 × 1² + 1³
= 1000,000,000 + 3,000,000 + 3000 + 1 = 1,003,003,001

Ex 3.4 Class 9 Question 6.
If (x + y + z) = 9 and (xy + yz + zx) = 26 then find the value of x² + y² + z².
Solution:
(x + y + z) = 9 and(xy + yz + zx) = 26
x² + y² + z² = (x + y + z)² – 2 (xy + yz + zx)
= 9² – 2 × 26 = 81 – 52 = 29

9th Standard Maths Exercise 3.4 Question 7.
Find 27a³ + 64b³, if 3a + 4b = 10 and ab = 2.
Solution:
3a + 4b = 10, ab = 2
(3a + 4b)³ = (3a)³ + 3 (3a)² (4b) + 3 (3a) (4b)² + (4b)³
(27a³ + 64b³) = (3a + 4b)³ – 3 (3a) (4b) (3a + 4b)
∵ x³ + y³ = (x + y)³ – 3xy – (x + y)
= 10³ – 36 ab (10)= 1000 – 36 × 2 × 10
= 1000 – 720 = 280

Exercise 3.4 Class 9 Maths Solution Question 8.
Find x³ – y³, if x – y = 5 and xy = 14.
Solution:
x – y = 5, xy = 14
x³ – y³ = (x – y)³ + 3xy (x – y) = 5³ + 3 × 14 × 5
= 125 + 210 = 335

Class 9 Maths Exercise 3.4 Solutions Question 9.
If a + \(\frac{1}{a}\) = 6, then find the value of a³ + \(\frac{1}{a^{3}}\)
Solution:
a³ + b³ = (a + b)³ – 3ab (a + b)

9th Math 3.4 Solution Question 10.
If x² + \(\frac{1}{x^{2}}\) = 23, then find the value of x + \(\frac{1}{x}\) and x³ + \(\frac{1}{x^{3}}\)
Solution:

Maths Exercise 3.4 Class 9 Question 11.
If (y – \(\frac{1}{y}\))³ = 27, then find the value of y³ – \(\frac{1}{y^{3}}\)
Solution:

9th Class Maths Exercise 3.4 Solution Question 12.
Simplify : (i) (2a + 36 + 4c)(4a² + 9b² + 16c² – 6ab – 12bc – 12bc – 8ca)
(ii) (x – 2y + 3z) (x² + 4y² + 9z² + 2xy + 6yz – 3xz)
Solution:
(i) (2a + 36 + 4c)(4a² + 9b² + 16c² – 6ab – 12bc – 12bc – 8ca)
We know that
(a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³ + c³ × 3 abc
∴ (2a + 36 + 4c) (4a² + 9b² + 16 c² – 6 ab – 12 bc – 8 ca)
= (2a)³ + (3b)³ + (4c)³ – 3 × 2a × 36 × 4c
= 8a³ + 27b³ + 64c³ – 72 abc

(ii) (x – 2,y + 3z) (x² + 4y² + 9z² + 2xy + 6yz – 3xz)
(a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³ + c³ – 3abc .
∴ (x – 2y + 3z) (x² + 4y² + 9z² + 2xy + 6yz – 3xz)
= x³ + (-2y)³ + (3z)³ – 3 × x × (-2y) (3z)
= x³ – 8y³ + 27z³ + 18 xyz

9th Maths Exercise 3.4 Samacheer Kalvi Question 13.
By using identity evaluate the following:
(i) 7³ – 10³ + 3³
(ii) 1 + \(\frac{1}{8}-\frac{27}{8}\)
Solution:
(i) 7³ – 10³ + 3³
(a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³ + c³ – 3 abc
If a + b + c = 0, then a³ + b³ + c³ = 3 abc
∴ 7 – 10 + 3 = 0
⇒ 7³ – 10³ + 3³ = 3 × 7 × – 10 × 3
= 9 × -70 = -630

9th Class Math Exercise 3.4 Solution Question 14.
If 2x – 3y – 4z = 0, then find 8x³ – 27y³ – 64z³.
Solution:
If 2x – 3y – 4z = 0 then 8x³ – 27y³ – 64z³ = ?
If x + y + z = 0 then x³ + y³ + z³ = 3xyz
8x³ – 21 y³ – 64z³ = (2x)³ + (-3y)³ + (-4z)³
= 3 × 2x × -3y × – 4z = 72 xyz

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Tamilnadu Samacheer Kalvi 10th Social Science History Solutions Chapter 6 Early Revolts against British Rule in Tamil Nadu

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 6 Early Revolts against British Rule in Tamil Nadu 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 6 textbook solutions for Tamilnadu State Board.

Early Revolts against British Rule in Tamil Nadu Textual Exercise

I. Choose the correct answer.

Early Revolts Against British Rule In Tamil Nadu Question 1.
Who was the first Palayakkarar to resist the East India Company’s policy of territorial aggrandizement?
(a) Marudhu brothers
(b) Puli Thevar
(c) Velunachiyar
(d) Veera Pandya Kattabomman
Answer:
(b) Puli Thevar

Early Revolts Against British Rule In Tamil Nadu Book Back Answers Question 2.
Who had borrowed money from the East India Company to meet the expenses he had incurred during the Carnatic wars?
(a) Velunachiyar
(b) Puli Thevar
(c) Nawab of Arcot
(d) Raja of Travancore
Answer:
(c) Nawab of Arcot

Early Revolts Against British Rule Question 3.
Who had established close relationship with the three agents of Chanda Sahib?
(a) Velunachiyar
(b) Kattabomman
(c) Puli Thevar
(d) Oomai thurai
Answer:
(c) Puli Thevar

Early Revolts Against British Rule In Tamilnadu Question 4.
Where was Sivasubramanianar executed?
(a) Kayathar
(b) Nagalapuram
(c) Virupachi
(d) Panchalamkurichi
Answer:
(b) Nagalapuram

Question 5.
Who issued the Tiruchirappalli proclamation of Independence?
(a) Marudhu brothers
(b) Puli Thevar
(c) Veera Pandya Kattabomman
(d) Gopala Nayak
Answer:
(a) Marudhu brothers

Question 6.
When did the Vellore Revolt breakout?
(a) 24 May 1805
(b) 10 July 1805
(c) 10 July 1806
(d) 10 September 1806
Answer:
(c) 10 July 1806

Question 7.
Who was the Commander-in-Chief responsible for the new military regulations in Vellore Fort?
(a) Col. Fancourt
(b) Major Armstrong
(c) Sir John Cradock
(d) Colonel Agnew
Answer:
(c) Sir John Cradock

Question 8.
Where were the sons of Tipu Sultan sent after the Vellore Revolt?
(a) Calcutta
(b) Mumbai
(c) Delhi
(d) Mysore
Answer:
(a) Calcutta

II. Fill in the blanks.

1. The Palayakkarar system was put in place in Tamil Nadu by ……………..
2. Except the Palayakkarars of ……………., all other western Palayakkarars supported Puli-Thevar.
3. Velunachiyar and her daughter were under the protection of …………… for eight years.
4. Bennerman deputed ……………… to convey his message, asking Kattabomman to surrender.
5. Kattabomman was hanged to death at ……………..
6. The Rebellion of Marudhu Brothers was categorized in the British records as the ……………….
7. …………. was declared the new Sultan by the rebels in Vellore Fort.
8. ……………. suppressed the revolt in Vellore Fort.
Answers:
1. Viswanatha Nayaka
2. Sivagiri
3. GopalaNayaker
4. Ramalinga Mudaliar
5. Kayathar
6. South Indian Rebellion
7. Fateh Hyder
8. Col. Gillespie

III. Choose the correct statement.

Question 1.
(i) The Palayakkarar system was in practice in the Kakatiya Kingdom.
(ii) Puli Thevar recaptured Nerkattumseval in 1764 after the death of Khan Sahib.
(iii) Yusuf Khan who was negotiating with the Palayakkarars, without informing the Company administration was charged with treachery and hanged in 1764.
(iv) Ondiveeran led one of the army units of Kattabomman.
(a) (i) (ii) and (iv) are correct
(b) (i) (ii) and (iii) are correct
(c) (iii) and (iv) are correct
(d) (i) and (iv) are correct
Answer:
(b) (i) (ii) and (iii) are correct

Question 2.
(i) Under Colonel Campbell, the English Army went along with Mafuzkhan’s army.
(ii) After Muthu Vadugar’s death in Kalaiyar Kovil battle, Marudhu Brothers assisted Velunachiyar in restoring the throne to her.
(iii) Gopala Nayak spearheaded the famous Dindigul League.
(iv) In May 1799 Cornwallis ordered the advance of Company armies to Tirunelveli.
(a) (i) and (ii) are correct
(b) (ii) and (iii) are correct
(c) (ii) (iii) and (iv) are correct
(d) (i) and (iv) are correct.
Answer:
(b) (ii) and (iii) are correct

Question 3.
Assertion (A): Puli Thevar tried to get the support of Hyder Ali and the French.
Reason (R): Hyder Ali could not help Puli Thevar as he was already in a serious conflict with the Marathas.
(a) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
(b) Both (A) and (R) are wrong
(c) Both (A) and (R) are correct and R is the correct explanation of (A)
(d) (A) is wrong and (R) is correct
Answer:
(d) (A) is wrong and (R) is correct

Question 4.
Assertion (A): Apart from the new military Regulations the most objectionable was the addition of a leather cockade in the turban.
Reason (R): The leather cockade was made of animal skin.
(a) (A) is wrong and (R) is correct
(b) Both. (A) and (R) are correct and (R) is the correct explanation of (A)
(c) Both (A) and (R) are wrong
(d) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
Answer:
(b) Both. (A) and (R) are correct and (R) is the correct explanation of (A)

IV. Match the following.

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

V. Answer the questions briefly.

Question 1.
What were the duties of the Palayakkarars?
Answer:

1. The Palayakkarars were free to collect revenue.
2. Administer the territory.
3. Settle disputes and maintain law and order also.
4. Helped the Nayak rulers to restore the kingdom.

Question 2.
Identify the Palayams based on the division of east and west.
Answer;
Among the 72 Palayakkarars, there were two blocks namely the eastern and the western Palayams.

• The eastern Palayams were – Sattur, Nagalapuram, Ettayapuram and Panchalam Kurichi.
• The western Palayams were – Uttrumalai, Thalavankottai, Naduvakurichi, Singampatti and Seithur.

Question 3.
Why was Heron dismissed from service?
Answer:

1. Puli Thevar continued were defy the authority of the English East India Company.
2. Col. Heron was urged by the company to deal the issue of Puli Thevar.
3. For want of canon and of supplies and pay to soldiers, colonel Heron abandoned the plan. Hence he was dismissed from service.

Question 4.
What was the significance of the Battle of Kalakadu?
Answer:
In the Battle of Kalakadu, Mahfuzkhan’s troops were routed by the huge forces of Puli Thevar.

Question 5.
What was the bone of contention between the Company and Kattabomman?
Answer:

1. The company gained the right to collect taxes from Panchalamkurichi which was under Kattabomman.
2. The company appointed collectors to collect taxes from all the Palayams.
3. The collectors humiliated the Palayakkarars and adopted force to collect the taxes.
4. This was the bone of contention between the company and Kattabomman.

Question 6.
Highlight the essence of the Tiruchirappalli Proclamation of 1801.
Answer:
The Tiruchirappalli Proclamation of 1801 was the first call to the Indians to unite against the British, cutting across region, caste, creed and religion. The Proclamation was pasted on the walls of the Nawab’s palace in Tiruchirappalli Fort and on the walls of the Srirangam Temple. As a result, many Palayakkarars of Tamil country rallied together to fight against the English.

Question 7.
Point out the importance of the Treaty of 1801.
Answer:

1. The Treaty of 1801 was known as “Carnatic Treaty”. Under the terms of the camatic Treaty of 31st July 1801.
2. The British assumed the direct control over Tamilagam.
3. The Palayakkarar system came to an end with the demolition of all forts and disbandment of their army.

VI. Answer all the questions given under each caption.

Question 1.
Velunachiyar
(a) Who was the military chief of Velunachiyar?
Answer:
Gopala Nayaker

(b) What were the martial arts in which she was trained?
Answer:
The martial arts in which she was trained were valari, stick fighting and to wield weapons.

(c) Whom did she marry?
Answer:
She was married to Muthu Vadugar, the Raja of Sivagangai.

(d) What was the name of her daughter?
Answer:
Her Daughter’s name was Vellachinachiar.

Question 2.

Dheeran Chinnamalai

(a) When was Dheeran Chinnamalai bom?
Answer:
Dheeran Chinnamalai was bom in 1756.

(b) How did he earn the title “Chinnamalai”?
Answer:
The tax money collected by Tipu’s Diwan was confiscated by Theerthagiri (original name of Dheeran Chinnamalai) While returning to Mysore. He let Diwan to go by instructing him to tell his sultan that “Chinnamalai” who is between Sivamalai and Chennimalai was the one who took away taxes. Thus he gained the name “Dheeran Chinnamalai”.

(c) Name the Diwan of Tipu Sultan.
Answer:
Diwan of Tipu Sultan was Mohammed Ali.

(d) Why and where was he hanged to death?
Answer:
Dheeran Chinnamalai refused to accept the overlordship of the British. So he was captured and imprisoned. In 31st July 1805 he was hanged to death at the top of the Sankagiri Fort.

VII. Answer the following in detail.

Question 1.
Attempt an essay of the heroic fight Veerapandya Kattabomman conducted against the East India Company.
Answer:
(i) Veera Pandya Kattabomman became the Palayakkarar of Panchalamkurichi after the death of his father, Jagavira Pandya Kattabomman. The company administrators did not give him much importance. But soon several events led to the conflict between him and the East India Company.

(ii) The Company had gained the right to collect taxes from Panchalamkurichi. The Collectors adopted force to collect the taxes. This was the bone of contention between the English and Kattabomman.

(iii) When Kattabomman refused to pay the land revenue, he was asked to meet Jackson, the company official in Ramanathapuram where he had to stand for hours before the official. Kattabomman somehow escaped from there.

(iv) On his return to Panchalamkuruchi, Kattabomman represented to the Madras council about how he was ill-treated by the collector Jackson. The council asked Kattabomman to appear before a Committee with William Brown, William Oram and John Casamajor as members. The committee found Kattabomman innocent and Jackson was dismissed from service.

(v) Thereafter Kattabomman along with Marudhu Brothers confronted with the English. Kattabomman was asked to surrender. But his ‘evasive reply’ prompted Major Bannerman to attack his fort. The major soon seiged Panchalamkuruchi.

(vi) Kattabomman escaped to Pudukottai where he was captured. During the trail Kattabomman bravely admitted all the charges levelled against him. He was hanged from a tamarind tree in the old fort of Kayathar.

Question 2.
Highlight the tragic fall of Sivagangai and its outcome.
Answer:
Tragic fall of Sivagangai: In May 1801, the English attacked the rebles in Thanjaviir and Tiruchirapaili.

1. The rebels went to piranmalai and Kalaiyar kovil.
2. They were again defeated by the able commanders and superior military strength of the English company.
3. The rebellion failed and Sivagangai was annexed in 1801.
4. In 24th October 1801 the Marudhu brothers were executed in the Fort of Thirupathur near Ramanathapuram.

Outcomes:

1. The exploits and sacrifices of the Palayakkarars inspired later generations.
2. The Rebellion of Murudhu brothers called as “South Indian Rebellion is a landmark event in the history of Tamil Nadu.
3. The rebellion resulted in the liquidation of all the chieftains of Tamil Nadu.
4. The British assumed direct control over Tamilagam.
5. The Palayakkarar system came to an end with the demolition of all Forts and disbandment of their army.

Question 3.
Account for the outbreak of Vellore Revolt in 1806.
Answer:
After the suppression of resistance of Kattabomman and Marudhu Brothers in 1801, the British charged the Nawab of Arcot with disloyalty and forced a treaty on him. According to this treaty, the Nawab was forced to cede the districts of North Arcot, South Arcot, Tiruchirappalli, Madurai and Tirunelveli to the Company and transfer all the administrative powers to it.

But the resistance did not die down. The dispossessed little kings and feudal chieftains were continuously deliberating on the future course of action against the company. This finally resulted in the Vellore Revolt of 1806. The sepoys of the British Indian army nursed a strong sense of resentment over low salary and poor prospects of promotion. The English army officers gave little respect for the social and religious sentiments of the Indian sepoys also angered them.

The immediate cause of the revolt came in the form of a new military regulation according to which the Indian soldiers were asked not to wear caste marks or ear rings when in uniform. They were to be cleanly shaven on the chin and maintain uniformity how their moustache looked. The new turban added fuel to fire. The new turban had the leather cockade made of animal skin. The sepoys refused to wear it.

On 10 July 1806, in the early hours, guns began to boom. The Indian sepoys revolted against the company rule. However, the revolt was suppressed brutally.

VIII. Activity

Question 1.
Teacher can ask the students to prepare an album of patriotic leaders of early revolts against the British rule in Tamil Nadu. Using their imagination they can also draw pictures of different battles in which they attained martyrdom.
Answer:
Do it yourself.

Question 2.
Stage play visualizing the conversation between Jackson and Kattabomman be attempted by students with the help of teachers.
Answer:
Do it yourself.

Question 3.
A comparative study of Vellore Revolt and 1857 Revolt by students be tried enabling them to find out to what extent Vellore Revolt had all the forebodings of the latter.
Answer:
The Vellore Mutiny took place on July 10, 1806. This major act of defiance happened even before the famous Rebellion of 1857. Though lasting only for a day, the Vellore Mutiny marked the first ever large-scale and violent mutiny by Indian sepoys against the East India Company. It was triggered by the English disregard to the religious sensitivities of the Hindu and Muslim Indian sepoys. The Revolt of 1857 was a rather large revolt which went on for days. It was fed by resentments bom of diverse perceptions, including invasive British-style social reforms, harsh land taxes, as well as scepticism about the improvements brought about by British rule.
Students can make a comparative study under the guidance of their teacher.

Early Revolts against British Rule in Tamil Nadu Additional Questions

I. Choose the correct answer.

Question 1.
Palayakkarar system was in practice during the rule of ………………. of Warangal.
(a) Rajendra Chola
(b) Prataba Rudhra
(c) Ashoka
Answer:
(b) Prataba Rudhra

Question 2.
Puli Thevar’s three major ports came under the control of Yusuf khan on
(a) 18^th August 1798
(b) 16^th May 1761
(c) 19^th September 1798
(d) 16^th October 1799
Answer:
(b) 16^th May 1761

Question 3.
On many occasions the Palayakarars helped the ………….. rulers to restore the kingdom.
(a) Nayak
(b) Pallava
(c) Pandya
Answer:
(a) Nayak

Question 4.
Muthu vadagar died in the battle of :
(a) Kalakad
(b) Tiruchirapalli
(c) Kalaiyar kovil
(d) Palayamkottai
Answer:
(c) Kalaiyar kovil

Question 5.
The Proclamation of ………………. was the first call to the Indians to unite against the British.
(a) 1805
(b) 1809
(c) 1801
Answer:
(c) 1801

Question 6.
As per this treaty of 1801 Nawab of Arcot transferred all the administrative powers to the company:
(a) Treaty of Mangalore
(b) Treaty of Madras
(c) Treaty of Carnatic
(d) None of the above
Answer:
(c) Treaty of Carnatic

Question 7.
Hyder Ali could not help Puli Thevar because ……………..
(a) he was not well
(b) he was in a serious conflict with the Marathas
(c) he did not like Puli Thevar
(d) he was hostile to Puli Thevar
Answer:
(b) he was in a serious conflict with the Marathas

Question 8.
Puli Thevar was defeated by:
(a) Colonel Heron
(b) Captain campbell
(c) Major cootes
(d) Colonel Fancourt
Answer:
(b) Captain campbell

Question 9.
Dheeran was well trained in ……………..
(a) modem warfare
(b) archery
(c) horse riding
(d) all of the above
Answer:
(d) all of the above

Question 10.
………………….. ordered the release of Sivasubramanianar and the suspension of the collector Jackson.
(a) Lord Wellesley
(b) John casamajor
(c) Governor edward clive
(d) General Bannerman
Answer:
(c) Governor edward clive

II. Fill in the blanks :

1. Chinna Marudhu collected nearly ……………. men to challenge the English army.
2. ……………. tried to get the support of Hyder Ali and the French.
3. Puli Thevar wielded much influence over the ………………
4. …………… was the military chief of Velunachiyar.
5. The leather cockade was made of ……………..
6. The Marudhu brothers were executed in the Fort of Tirupathur near …………… on 24 October 1801.
7. ………….. policy of the English split the forces of the Palayakkarars.
8. ………….. wielded much influence over the western Palayakkarars.
9. Trained by the French Dheeran mobilised the …………….. youth in thousands and fought the British together with Tipu.
10. Tipu was killed at the end of the Anglo-Mysore war in …………….
Answers :
1. 1.20,000
2. Puli Thevar
3. Western Palayakkarars
4. Gopala Nayakar
5. animal skin
6. Ramanathapuram
7. Divide and Rule
8. Puli Thevar
9. Kongu 10.1799

III. Choose the correct statement.

Question 1.
(i) Kuyili was a faithful friend of Velunachiyar.
(ii) She led the unit of women soldiers named after Udaiyaal.
(iii) Udaiyaal was a timid girl who divulged information on Kuyili.
(iv) Kuyili is said to have walked into the British arsenal after setting herself on fire.
(a) (i), (ii) and (iii) are correct
(b) (i), (iii) and (iv) are correct
(c) (i), (ii) and (iv) are correct
(d) (ii),(iii) and (iv) are correct
Answer:
(c) (i), (ii) and (iv) are correct

Question 2.
(i) Velunachiyar was crowned as Queen with the help of Puli Thevar.
(ii) She was the first female ruler or queen to resist the British colonial power in India.
(iii) Gopala Nayak drew inspiration from Tipu Sultan.
(iv) The Carnatic Treaty took place in the year 1801.
(a) (i), (ii) and (iv) are correct
(b) (ii), (iii) and (iv) are correct
(c) (i) and (iii) are correct
(d) (i), (iii) and (iv) are correct
Answer:
(b) (ii), (iii) and (iv) are correct

IV. Match the following

Question 1.

1.	Khan Sahib	(a)	Friend of Velunachiyar
2.	Velunachiyar	(b)	Kattabomman’s brother
3.	Kuyili	(c)	Tipu’s diwan
4.	Oomathurai	(d)	Marudhanayagam Pillai
5.	Mohammed Ali	(e)	Bom in 1730

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

Question 2.

1.	Yusuf Khan	(a)	Military chief
2.	Dalavay	(b)	guerrilla attack
3.	Pagodas	(c)	Khan Sahib
4.	Dheeran Chinnamalai	(d)	Raja of Sivagangai
5.	Muthu Vadugar Periya Udaya Thevar	(e)	Forced labour

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

V. Answer briefly:

Question 1.
Name the Palayakkarars who revolted against the British rule in Tamil Nadu.
Answer:
Puli Thevar, Velunachiyar, Dheeran chinnamalai, Marudhu brothers and Veerapandiya Kattabomman were some of the prominent Palayakkarars who revolted against the British rule in Tamil Nadu.

Question 2.
Explain about Ondiveeran’s bravery.
Answer:
Ondiveeran led one of the army units of Puli Thevar. Fighting by the side of Puli Thevar, he caused much damage to the company’s army. According to oral tradition, in one battle, Ondiveeran’s hand was chopped off and Puli Thevar was saddened. But Ondiveeran said it was a reward for his penetration into enemy’s fort causing many heads to roll.

Question 3.
Who supported the British and who did not join the confederacy of Puli Thevar?
Answer:
The English succeeded in getting the support of the Raj as of Ramanathapuram and Pudukkottai. The Palayakkarars of Ettayapuram, Panchalamkurichi and Sivagiri did not join the confederacy of Puli Thevar.

Question 4.
Who was Muthu Vadugar? How was he killed?
Answer:
Muthu Vadugar was the Raja of Sivagangai. He was married to Velunachiyar. In 1772, the Nawab of Arcot and the Company troops under the command of Lt. col. Bon Jour stormed the Kalaiyar Kovil Palace. In the ensuing battle Muthu Vadugar was killed.

Question 5.
Write a short note on Velunachiyar.
Answer:
Velunachiyar was bom in 1730 she was the only daughter of Raja Sellamuthu sethupathy of Ramanathapuram. She was trained in martial arts like valari, stick fighting and wield weapons. She was expert in horse riding and archery and had proficiency in English, French and Urdu.

Question 6.
What status did the Palayakkarars avail during the seventeenth and eighteenth centuries?
Answer:
During the seventeenth and eighteenth centuries the Palayakkarars dominated the politics of Tamil country. They functioned as independent sovereign authorities within their respective Palayams.

Question 7.
Write a brief note on Marudhu brothers.
Answer:

1. Periya Marudhu or Vella Marudhu and his younger brother Chinna Marudhu were the able generals of Muthuvadugar of Sivagangai.
2. After Muthuvadugar’s death in the Kalaiyur kovil battle, they assisted in restoring the throne to Velunachiyar.
3. In the last years of the eighteenth century Marudhu brothers organised resistance against the British.

Question 8.
Give an estimate of the Vellore Revolt of 1806.
Answer:
The Vellore Revolt failed because there was no immediate help from outside. According to recent studies, the organizing part of the revolt was done perfectly by Subedars Sheik Adam and Sheik Hamid and Jamedar Sheik Hussain of the 2nd Battalion of 23^rd regiment. Vellore revolt had all the forebodings of the Great Rebellion of 1857. The only difference was that there was no civil rebellion following the mutiny.

VI. Answer all the questions given under each caption:

Question 1.
Vellore Revolt:

(a) When did the outbreak of Vellore revolt occur?
Answer:
Out break of Vellore revolt occurred on 10th July 1806.

(b) Who commanded the garrison?
Answer:
Colonel Fancourt commanded the garrison.

(c) By whom was the organising part of the revolt done?
Answer:
The organising part of the revolt was done perfectly by Subedars Sheik Adam and Sheik Hamid and Jamedar Shiek Hussain of 2nd battalion of 23rd regiment and subedar Jamedar Sheik Kasim of the 1st battalion of 1st regiment.

(d) Name the places where the Vellore revolt of 1806 echoed?
Answer:
Vellore Revolt of 1806 had its echoes in Bellary, Walajabad, Hyderabad, Bengaluru, Nandydurg and Sankaridurg.

Question 2.
Appearance Before Madras Council

(a) Whom did the Madras council ask to appear before the committee?
Answer:
Kattabomman.

(b) Who were the committee members of Madras council?
Answer:
William Brown, William oram, John Casamajor.

(c) When did Kattabomman appear before the committee?
Answer:
Kattabomman appeared before the committee on 15th December 1798.

(d) What did Kattabomman report to the committee? What was the result?
Answer:
Kattabomman reported on what transpired in Ramanathapuram. The committee found Kattabomman was not guilty.

VII. Answer the following in detail.

Question 1.
Give an estimation of Revolt at Vellore (1806).
Answer:

1. General Gillespie from Arcot along with the captain young cavalry commander crushed the revolt. Nearly eight hundred soldiers were found dead. The organising part of the revolt was done perfectly by the subedars of 1st battalion of the 1st regiment and the 2nd battalion of 23rd regiment.
2. The Vellore revolt failed because there was no immediate help from outside.
3. It was the 1st open uprising of the Indian soldiers under British army.
4. This revolt was not confined to Vellore Fort alone but echoed outside regions also.
5. Vellore Revolt had all the forebodings of the Great Rebellion of 1857.
6. This was also called as Vellore mutiny as it arose only from the soldiers.

Question 2.
Give on assessment of Velunachiyar’s resistance to the British colonial power in India,
Answer:
Velunachiyar belonged to the royal family. She was brought up as a princess and was trained in marital arts. She was also adept in horse riding and archery. She was married to Muthu Vadugar, the Raja of Sivangangai. When the Raja was killed in the battle, Velunachiyar had to live under the protection of Gopala Nayakar at Virupachi near Dindigul for eight years. During her period in hiding, Velunachiyar organised an army and succeeded in securing an alliance with not only Gopala Nayakar but Hyder Ali as well. Dalavay (military chief) Thandavarayanar wrote a letter to Sultan Flyder Ali on behalf of Velunachiyar asking for 5000 infantry and 5000 cavalry to defeat the English. Velunachiyar explained in detail in Urdu all the problems she had with East India Company.

She conveyed her strong determination to fight the English. Impressed by her courage, Hyder Ali ordered his commandant Syed in Dindigul Fort to provide the required military assistance. Velunachiyar employed agents for gathering intelligence to find where the British had stored their ammunition. With military assistance from Gopala Nayakar and Hyder Ali, She recaptured Sivagangai. She was crowned as Queen with the help of Marudhu Brothers. ” She became the first female ruler to resist the British colonial power in India.

Impotant Events And Years:

Years	Events
1730	Velunachiyar born
1801	Sivagangai annexed, Carnatic Treaty
1799	Anglo – Mysore war
1806	Vellore (Revolt) Mutiny

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Tamilnadu Samacheer Kalvi 7th Tamil Solutions Term 2 Chapter 2.5 ஒரெழுத்து ஒருமொழி, பகுபதம், பகாப்பதம்

மதிப்பீடு

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
நன்னூலின்படி தமிழிலுள்ள ஓரெழுத்து ஒரு மொழிகளின் எண்ணிக்கை …………..
அ) 40
ஆ) 42
இ) 44
ஈ) 46
Answer:
ஆ) 42

Question 2.
எழுதினான்’ என்பது ………………..
அ) பெயர்ப் பகுபதம்
ஆ) வினைப் பகுபதம் இ) பெயர்ப் பகாப்பதம்
ஈ) வினைப் பகாப்பதம்
Answer:
ஆ) வினைப் பகுபதம்

Question 3:
பெயர்ப்ப குபதம் ……………… வகைப்படும்.
அ) நான்கு
ஆ) ஐந்து
இ) ஆறு
ஈ) ஏழு
Answer:
இ) ஆறு

Question 4.
காலத்தைக் காட்டும் பகுபத உறுப்பு ……………….
அ) பகுதி
ஆ) விகுதி
இ) இடைநிலை
ஈ) சந்தி
Answer:
இ) இடைநிலை

பொருத்துக

1. பெயர்ப் பகுபதம் – வாழ்ந்தான்
2. வினைப் பகுபதம் – மன்
3. இடைப் பகாப்பதம் – நனி
4. உரிப் பகாப்பதம் – பெரியார்
Answers:
1. பெயர்ப் பகுபதம் – பெரியார்
2. வினைப் பகுபதம் – வாழ்ந்தான்
3. இடைப் பகாப்பதம் – மன்
4. உரிப் பகாப்பதம் – நனி

சரியான பகுபத உறுப்பை எழுதுக

1. போவாள் – போ + வ் + ஆள்
போ – பகுதி
வ் – எதிர்கால இடைநிலை
ஆள் – படர்க்கைப் பெண்பால் வினைமுற்று விகுதி

2. நடக்கின்றான் – நட + க் + கின்று + ஆன்
நட – பகுதி
க் – சந்தி
கின்று – நிகழ்கால இடைநிலை
ஆன் – படர்க்கை ஆண்பால் வினைமுற்று விகுதி

பின்வரும் சொற்களைப் பிரித்துப் பகுபத உறுப்புகளை எழுதுக

1. பார்த்தான் – பார் + த் + த் + ஆன்
பார் – பகுதி
த் – சந்தி
த் – இறந்தகால இடைநிலை
ஆன்- படர்க்கை ஆண்பால் வினைமுற்று விகுதி

2. பாடுவார் – பாடு + வ் + ஆர்
பாடு – பகுதி
வ் – எதிர்கால இடைநிலை
ஆர் – படர்க்கைப் பலர்பால் வினைமுற்று விகுதி

குறுவினா

Question 1.
ஓரெழுத்து ஒருமொழி என்றால் என்ன?
Answer:
(i) ஓரெழுத்து தனித்து நின்று பொருள் தரும் சொல்லாக அமைவதே ஓரெழுத்து ஒரு மொழி ஆகும்.
(ii) எ.கா. (தீ, நீ, வா, போ).

Question 2.
பதத்தின் இரு வகைகள் யாவை?
Answer:
பதம் இரண்டு வகைப்படும். அவை, பகுபதம், பகாப்பதம்.

Question 3.
பகுபத உறுப்புகள் எத்தனை வகைப்படும்?
Answer:
அவை யாவை? பகுபத உறுப்புகள் ஆறு வகைப்படும். அவை, பகுதி, விகுதி, இடைநிலை, சந்தி, சாரியை, விகாரம்.

சிறுவினா

Question 1.
விகுதி எவற்றைக் காட்டும்?
Answer:
சொல்லின் இறுதியில் நிற்கும் உறுப்பே விகுதி ஆகும். இது திணை, பால், எண், இடம்,
முற்று, எச்சம் போன்றவற்றைக் காட்டும். (எ.கா.) படித்தான் = ஆன் – விகுதி

Question 2.
விகாரம் என்பது யாது? எடுத்துக்காட்டுடன் விளக்குக.
Answer:
பகுதி, விகுதி, சந்தி, இடைநிலை இவற்றில் ஏற்படும் மாற்றமே விகாரம் எனப்படும்.
(எ. கா.) வந்தான் = வா – பகுதி வா ‘வ’ எனக் குறுகியது விகாரம்.

Question 3.
பெயர்ப்பகுபதம் எத்தனை வகைப்படும்? அவை யாவை?
Answer:
பெயர்ப் பகுபதம் ஆறு வகைப்படும். அவை
(i) பொருள் – பொன்னன் (பொன் + அன்)
(ii) இடம் – நாடன் (நாடு + அன்)
(iii) காலம் சித்திரையான் (சித்திரை + ஆன்)
(iv) சினை கண்ண ன் (கண் + அன்)
(v) பண்பு இனியன் (இனிமை + அன்)
(vi) தொழில் – உழவன் (உழவு + அன்)

கற்பவை கற்றபின்

Question 1.
பாடப்பகுதியில் இடம்பெற்ற சொற்களில் பகுபதம், பகாப்பதம் ஆகியவற்றைக் கண்டறிந்து தனித்தனியே தொகுக்க.
Answer:
பகுபதம்
பெயர்ப்பகுபதம் :
பொருள் – பொன்னன் (பொன் + அன்)
இடம் – நாடன் (நாடு + அன்)
காலம் – சித்திரையான் (சித்திரை + ஆன்)
சினை – கண்ண ன் (கண் + அன் )
பண்பு – இனியன் (இனிமை + அன்)
தொழில் – உழவன் (உழவு + அன்)
வினைப்பகுபதம் : உண்கின்றான் – உண் + கின்று + ஆன்

பகாப்பதம் :
பெயர்ப் பகாப்பதம் – நிலம், நீர், நெருப்பு, காற்று
வினைப் பகாப்பதம் – நட, வா, படி, வாழ்.
இடைப் பகாப்பதம் – மன், கொல், தில், போல்
உரிப் பகாப்பதம் – உறு, தவ, நனி, கழி.

Question 2.
உங்கள் வகுப்பு மாணவ – மாணவிகளின் பெயர்களைப் பகுபதம், பகாப்பதம் என வகைப்படுத்துக.
Answer:
மாணவர்கள் தாங்களாகவே செய்ய வேண்டியவை.

கூடுதல் வினாக்கள்

நிரப்புக.

Question 1.
தே என்பதன் பொருள் ………….. எனப்படும்.
Answer:
கடவுள்

Question 2.
நன்னூல் என்னும் இலக்கண நூலை எழுதியர் ………….
Answer:
பவணந்தி முனிவர்

ஓரெழுத்து ஒரு மொழிகளும் அவற்றின் பொருளும்

மொழியை ஆள்வோம்

கேட்க.

Question 1.
சிறந்த கல்வியாளர்களின் சொற்பொழிவுகளை இணையத்தில் கேட்டு மகிழ்க.
Answer:
மாணவர்கள் தாங்களாகவே செய்ய வேண்டியவை

கீழ்க்காணும் தலைப்பில் இரண்டு நிமிடம் பேசுக

Question 1.
கல்வியின் சிறப்பு
Answer:
அனைவருக்கும் வணக்கம்!
நான் கல்வியின் சிறப்பு என்ற தலைப்பில் பேசவிருக்கிறேன். இன்றைய உலகின் தே இன்றியமையாத ஒன்றாகத் திகழ்வது யாதெனில் கல்வியே ஆகும். அனைவருக்கும் பகுத்தறிவு தேவைப்படுகிறது. அவர்களுக்குப் பகுத்தறியும் சக்தியைக் கொடுப்பதே கல்வியின் சிறப்பாகும்.

கல்வியின் சிறப்பு என்றாலே கல்வி கற்றவன் எங்குச் சென்றாலும் சிறப்பிக்கப்படுபவன் – என்பதே நினைவுக்கு வரும். அவனுக்கு எல்லா நாட்டினரும் உறவினர்கள் ஆவார்கள். எல்லா நாடும் சொந்த நாடாகும். கல்வி கற்கவில்லையெனில் வாழ்நாள் முழுவதும் அனைவராலும் அவமதிக்கப்படுவான். இதனையே வள்ளுவர்.

“யாதானும் நாடாமல் ஊர்ஆமால் என்னொருவன் சாந்துணையும் கல்லாத வாறு. என்று கூறுகிறார்.
அதுமட்டுமா? கற்றவரைக் கண்ணுடையார்’ என்றும் கல்லாதவரை முகத்தில் இரண்டு புண்ணுடையவர் என்றும் இடித்துரைக்கிறார் வள்ளுவர்.

கற்றோர்க்கு அணிகலன் கல்வியே; கற்றோரே கண்ணுடையவர்; கற்றாரே தேவர் எனப் போற்றப்படத்தக்கவர்; கற்றோரே மேலானவர் என்பதை அனைவரும் உணர வேண்டும்.
“கல்வி வந்தது எனில் கடைத்தேறிற்று உலகே!” என்று புரட்சிக்கவி கூறுகின்றார். கல்வியால் எல்லா வளங்களும் கிடைக்கும் என்பதே இதன் பொருளாகும்.

கல்வி உடையவர் எல்லா மக்களிடமும், நன்றாக பழகிக் கொள்வதோடு மட்டுமல்லாமல் அவர்களுடன் மகிழ்ச்சியாக சேர்ந்து வாழ்வதையே விரும்புவர்.

மனிதன் வாழ்நாள் முழுவதும் கற்றுக் கொண்டே இருக்க வேண்டும். கற்க மறுப்பவன் வாழ மறுப்பவன் ஆகின்றான். கல்வி என்னும் விளக்கால் வாழ்க்கையில் எதிர்ப்படும் இருள்களையெல்லாம் நீக்க முடியும். கல்வி போல மனப்பயத்தைப் போக்கும் மருந்து வேறொன்றுமில்லை. கல்வித் துணை வறுமையில் கை கொடுக்கும். கல்வியின் பயனே மனித வாழ்வின் பெரும்பேறாகும்.

கல்வி, தொழிலுக்கு வழி காட்டும். கல்வி என்பது வாழ்வதற்கு உதவும் கருவியாகும். வாழ்க்கையின் வெற்றிக்குக் கல்வி மிகவும் இன்றியமையாததாகிறது. வாழ்க்கையை நெறிப்படுத்தவும் மேம்படுத்தவும் கல்வி பயன்படுகிறது. கல்வி கற்ற பண்பு, நீதி, நேர்மை இவைகள் அனைத்தும் ஒருங்கே அமைந்து காணப்படும்.

கல்வியினால் மட்டுமே உலக அறிவினை வளர்த்துக் கொள்ள முடியும். உலகை முழுமையாகப் படிக்கவும் முடியும். கல்வி மனிதனுக்கு ஓர் உன்னதமான தேவையாகும்”
“கற்கை நன்றே! கற்றை நன்றே! பிச்சைப் புகினும் கற்கை நன்றே!” என்ற கூற்றினை மனதில் நிறுத்தி அள்ள அள்ளக் குறையாதக் கல்வியை அள்ளிப் பருகுவோம்.
கல்வி என்பது பலமே !
கற்றல் என்பது சுகமே!

Question 2.
குழந்தைத் தொழிலாளர் முறை ஒழிப்பு
Answer:
ஒரு குழந்தை கூலிக்காக வேலை பார்ப்பது மிகவும் தவறு. இளமைக் காலம் கல்வி கற்பதற்கே. குழந்தைத் தொழிலாளர்கள் இல்லாத நிலை உருவாக வேண்டும்.
பள்ளி செல்லாத குழந்தைகள், குழந்தைத் தொழிலாளர் ஆகிறார்கள். இது ஒரு பக்கம் இருக்க; வேலைவாய்ப்புக்காக இடம்பெயர்ந்து மாநிலம் விட்டு மாநிலம், மாவட்டம் விட்டு மாவட்டம் என்று செல்கின்றனர். இவ்விரண்டு நிலைகளில் ஒரு சிலர் தங்கள் பெற்றோருடன் சேர்ந்து வேலை செய்வார்கள் அல்லது தனியாகச் செல்வார்கள்.

மூன்றாவது நிலையில் உள்ளவர்கள் மிகவும் பரிதாபத்துக்குரியவர்கள். கொத்தடிமை முறையில் வேலைக்குச் செல்கிறார்கள். பெற்றோர் வாங்கிய கடனை ஈடுகட்டுவதற்காக பிள்ளைகள் வேலைக்குச் செல்கிறார்கள்.

குழந்தைத் தொழிலாளர்கள் உடல் ரீதியாகப் பாதிக்கப்படுகிறார்கள். உளவியல் ரீதியான பாதிப்பு, உணர்வு மற்றும் சமூக ரீதியான பாதிப்பு ஏற்படுகிறது.

கொடிய வறுமை, ஊட்டச்சத்துக் குறைவு, கல்வியறிவு பெற முடியாத நிலை, உடல் நலனைப் பாதிக்கக் கூடிய ஆபத்தான சூழல், காற்றோட்டம் இல்லாத குறுகிய அறை போன்றவை குழந்தைகளின் உடல் நலனைப் பெரிதும் பாதிப்பதால் ஆஸ்துமா, காசநோய் போன்ற நோய்கள் தாக்குகிறது. இதனைத் தடுக்க வேண்டும்.

குழந்தைகளுக்குத் தொடக்கக் கல்வி கட்டாயம் ஆக்கப்பட வேண்டும். பள்ளி மற்றும் கல்லூரிகளில் குழந்தைத் தொழிலாளர் ஒழிப்பு பற்றிய விழிப்புணர்வு முகாம்களை நடத்தலாம்.

சொல்லக் கேட்டு எழுதுக

1. இளமைப் பருவத்திலேயே கல்வி கற்க வேண்டும்.
2. கல்வியே அழியாத செல்வம்.
3. கல்வி இல்லாத நாடு விளக்கு இல்லாத வீடு.
4. பள்ளித்தலம் அனைத்தும் கோயில் செய்குவோம்.
5. நூல்களை ஆராய்ந்து ஆழ்ந்து படிக்க வேண்டும்.

கீழ்க்காணும் சொற்களை அறுவகைப் பெயர்களாக வகைப்படுத்துக

நல்லூர், வடை, கேட்டல், முகம், அன்னம், செம்மை, காலை, வருதல், தோகை, பாரதிதாசன், பள்ளி, இறக்கை, பெரியது, சோலை, ஐந்து மணி, விளையாட்டு, புதன்

Answer:

அறிந்து பயன்படுத்துவோம்

மூவிடம் :

இடம் மூன்று வகைப்படும். அவை 1. தன்மை, 2. முன்னிலை, 3. படர்க்கை
தன்னைக் குறிப்பது தன்மை.
(எ.கா.) நான், நாம், நாங்கள், என், எம், எங்கள்.
முன்னால் இருப்பவரைக் குறிப்பது முன்னிலை.
(எ.கா.) நீ , நீங்கள், நீர், நீவிர், உன், உங்கள்.
தன்னையும், முன்னால் இருப்பவரையும் அல்லாமல் மூன்றாமவரைக் குறிப்பது – படர்க்கை .
(எ.கா.) அவன், அவள், அவர், அவர்கள், அது, அவை, இவன், இவள், இவை.

சரியான சொல்லைக் கொண்டு நிரப்புக

(அது, நீ, அவர்கள், அவைகள், அவை, நாம், உன்)

Question 1.
……… பெயர் என்ன ?
Answer:
உன்

Question 2.
ஏழாம் வகுப்பு மாணவர்கள்.
Answer:
நாம்

Question 3.
………….. எப்படி ஓடும்?
Answer:
அது

Question 4.
………………என்ன செய்து கொண்டிருக்கிறாய்?
Answer:
நீ

Question 5.
…. வந்து கொண்டு இருக்கிறார்கள்.
Answer:
அவர்கள்

பின்வரும் தொடர்களில் மூவிடப் பெயர்களை அடிக்கோடிடுக. அவற்றை வகைப்படுத்துக.

1. எங்கள் வீட்டு நாய்க்குட்டி ஓடியது.
2. இவர்தான் உங்கள் ஆசிரியர்.
3. நீர் கூறுவது எனக்குப் புரியவில்லை
4. எனக்கு, அது வந்ததா என்று தெரியவில்லை , நீயே கூறு.
5. உங்களோடு நானும் உணவு உண்ணலாமா?

Answer:

கடிதம் எழுதுக

உங்கள் பகுதியில் நூலகம் ஒன்று அமைத்துத் தர வேண்டி நூலக ஆணையருக்குக் கடிதம் எழுதுக.
Answer:
அனுப்புநர் :
ஊர்ப் பொதுமக்கள்,
மறைமலை நகர்,
காஞ்சிபுரம் மாவட்டம்

பெறுநர் :
நூலக ஆணையர்,
பொதுநூலகத் துறை,
சென்னை – 600 002.

மதிப்பிற்குரிய ஐயா,
பொருள் : நூலகம் அமைத்துத் தர வேண்டுதல் தொடர்பாக

எங்கள் ஊர் மறைமலைநகர். இங்கு இரண்டாயிரம் பேருக்கு மேல் வாழ்கிறோம். பள்ளி மாணவர்கள், கல்லூரி மாணவர்கள், போட்டித் தேர்வுக்குப் படிக்கும் இளைஞர்கள், அன்றாடச் செய்தியை அறிந்து கொள்ளும் ஆர்வலர், பணி ஓய்வு பெற்றவர்கள் எனப் பலரும் உள்ளனர்.

அவரவர்களுக்குத் தேவையான நூல்கள், செய்தித்தாள்கள், இதழ்கள் போன்றவை இங்குக் கிடைப்பதற்கரிதாக உள்ளது. இவர்கள் அனைவரும் பயன்பெறும் வகையில் நூலகம் ஒன்றை எங்கள் ஊரில் அமைத்துத்தர ஆவன செய்யுமாறு கேட்டுக் கொள்கிறோம்.

நன்றி!
இடம் : மறைமலை நகர்,
தேதி : 5-2-2020

இப்படிக்கு ,
தங்கள் உண்மையுள்ள,
ஊர்ப் பொதுமக்கள்

மொழியோடு விளையாடு

கீழே உள்ள குறிப்புகளைப் பயன்படுத்திக் கட்டத்தில் எழுத்துகளை நிரப்புக.
1. காலையில் பள்ளி மணி……………………
2. திரைப்படங்களில் விலங்குகள் ……………………… காட்சி குழந்ை தகளுக்குப் பிடிக்கும்.
3. கதிரவன் காலையில் கிழக்கே ………………
4. நாள்தோறும் செய்தித்தாள் ………………. வழக்கம் இருக்க வேண்டும்.

Answer:

1. காலையில் பள்ளி மணி அடிக்கும்.
2. திரைப்படங்களில் விலங்குகள் நடிக்கும் காட்சி குழந்ை தகளுக்குப் பிடிக்கும்.
3. கதிரவன் காலையில் கிழக்கே உதிக்கும்.
4. நாள்தோறும் செய்தித்தாள் படிக்கும் வழக்கம் இருக்க வேண்டும்.

ஓர் எழுத்துச் சொற்களால் நிரப்புக

Question 1.
…………………… புல்லை மேயும்.
Answer:
ஆ

Question 2.
……………………… சுடும்.
Answer:
தீ

Question 3.
……………….. பேசும்.
Answer:
நா

Question 4.
…………………… பறக்கும்.
Answer:
ஈ

Question 5.
…………….. மணம் வீசும்
Answer:
பூ

பின்வரும் எழுத்துகளுக்குப் பொருள் எழுதுக

(எ.கா.) தா – கொடு

Answer:

1. தீ – நெருப்பு
2. பா – பாடல்
3. தை – தை மாதம்
4. வை – புல், வைக்கோல்
5. மை – அஞ்சனம்

பின்வரும் சொற்களை இருபொருள் தருமாறு தொடரில் அமைத்து எழுதுக

(ஆறு, விளக்கு, படி, சொல், கல், மாலை, இடி)
(எ.கா.) ஆறு – ஈ ஆறு கால்களை உடையது.
தஞ்சாவூரில் காவிரி ஆறு பாய்கிறது.

விளக்கு : இலக்கணப் பாடத்தை விளக்கிக் கூறு.
அறியாமை என்னும் இருளைப் போக்குவது கல்வி என்னும் விளக்கு.

படி : காலையில் தினமும் படி.
மாடிப்படி ஏறி வா.

சொல் : சொற்கள் சேர்ந்தால் பாமாலை.
பெரியோர் சொல் கேட்டு சிறியோர் நடக்க வேண்டும்.

கல் : கற்களால் ஆனது கோபுரம்.
இளமையில் கல்.

மாலை : நேற்று மாலை பூங்காவிற்குச் சென்றேன்.
பூ மாலை நல்ல மணம் வீசியது.

இடி இடிக்கும் சப்தம் கேட்டது.
தவறுகளைக் கண்டால் இடித்துரைத்தல் வேண்டும்.

நிற்க அதற்குத் தக

என் பொறுப்புகள்

1. பாடப்புத்தகங்கள் மட்டுமன்றிப் பிற புத்தகங்களையும் படிப்பேன்.
2. பெற்றோர், ஆசிரியர், மூத்தோர் இவர்களை எப்போதும் மதித்து நடப்பேன்.

கலைச்சொல் அறிவோம்

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

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Samacheer Kalvi 11th Chemistry Chapter 3 Periodic Classification of Elements Textual Evaluation Solved

Choose The Correct Answer from The Following

11th Chemistry Chapter 3 Book Back Answers Question 1.
What would be the IUPAC name for an element with atomic number 222?
(a) bibibiium
(b) bididium
(c) didibium
(d) bibibium
Answer:
(d) bibibium

11th Chemistry Lesson 3 Book Back Answers Question 2.
The electronic configuration of the elements A and B are 1s², 2s², 2p⁶, 3s² and 1s², 2s², 2p⁵, respectively. The formula of the ionic compound that can be formed between these elements is ……….
(a) AB
(b) AB₂
(c) A₂B
(d) none of the above.
Answer:
(a) AB₂

11th Chemistry Unit 3 Book Back Answers Question 3.
The group of elements in which the differentiating electron enters the anti-penultimate shell of atoms are called –
(a) p-block elements
(b) d-block elements
(c) s-block elements
(d) f-block elements
Answer:
(d) f-block elements

11th Chemistry 3rd Lesson Answers Question 4.
In which of the following options the order of arrangement does not agree with the variation of property indicated against it? (NEET 2016 Phase 1)
(a) I < Br < Cl < F (increasing electron gain enthalpy)
(b) Li < Na < K < Rb (increasing metallic radius)
(c) Al³⁺< Mg²⁺< Na⁺ < F^– (increasing ionic size)
(d)  B < C < O < N (increasing first ionization enthalpy)
Answer:
(a) I < Br < Cl < F (increasing electron gain enthalpy)

11th Chemistry 3rd Lesson Book Back Answers Question 5.
Which of the following elements will have the highest electro negativity?
(a) Chlorine
(b) Nitrogen
(c) Cesium
(d) Fluorine
Answer:
(d) Fluorine

Question 6.
Various successive ionization enthalpies (in kJ mol^-1) of an element are given below. The element is ………….

(a) phosphorus
(b) sodium
(c) aluminium
(d) silicon table
Answer:
(c) aluminium

11th Chemistry 3rd Lesson Question 7.
In the third period, the first ionization potential is of the order …………..
(a) Na > Al > Mg > Si > P
(b) Na < Al < Mg < Si < P
(c) Mg > Na > Si > P > Al
(d) Na< Al < Mg < Si < P
Answer:
(b) Na < Al < Mg < Si < P

11th Chemistry Chapter 3 Question 8.
Identify the wrong statement ……………..
(a) Among st the iso electronic species, smaller the positive charge on cation, smaller is the ionic radius
(b) Among-st iso electric species greater the negative charge on the anion, larger is the ionic radius
(c) Atomic radius of the elements increases as one moves down the first group of the periodic table
(d) Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table.
Answer:
(a) Among-st the iso electronic species, smaller the positive charge on cation, smaller is the ionic radius

Samacheer Kalvi Guru 11th Chemistry Question 9.
Which one of the following arrangements represent the correct order of least negative to most negative electron gain enthalpy?
(a) Al< O<C< Ca< F
(b) Al < Ca<O< C< F
(c) C < F < O < Al < Ca
(d) Ca < Al < C < O < F
Answer:
(d) Ca < Al < C < O < F

Samacheer Kalvi Class 11 Chemistry Solutions Question 10.
The correct order of electron gain enthalpy with negative sign of F, Cl, Br and I having atomic number 9, 17, 35 and 53, respectively is ………..
(a) J > Br > Cl >F
(b) F > Cl > Br >I
(c) Cl > F > Br >I
(d) Br > I > Cl > F
Answer:
(c) Cl > F > Br > I

Samacheer Kalvi 11 Chemistry Solutions Question 11.
Which one of the following is the least electro negative element?
(a) Bromine
(b) Chlorine
(c) Iodine
(d) Hydrogen
Answer:
(d) Hydrogen.
Solution:
Hydrogen is the least electro negative element. Since electro negativity increases across the period from left to right. Hydrogen is the first element and it has less electro negativity and down the group electro negativity decreases.

Samacheer Kalvi 11th Chemistry Chapter 1 Solutions Question 12.
The element with positive electron gain enthalpy is ……….
(a) hydrogen
(b) sodium
(c) argon
(d) fluorine
Answer:
(c) argon
Solution:
Argon has completely filled configuration. So addition of the electron is not possible and has positive electron gain enthalpy.

11th Chemistry 3rd Chapter Question 13.
The correct order of decreasing electro negativity values among the elements X, Y, Z and A with atomic numbers 4, 8, 7 and 12 respectively –
(a) Y > Z > X > A
(b) Z > A > Y > X
(c) X > Y > Z > A
(d) X > Y >A >Z
Answer:
(a) Y > Z > X > A

Periodic Classification Of Elements Class 11 Question 14.
Assertion : Helium has the highest value of ionization energy among all the elements known Reason: Helium has the highest value of electron affinity among all the elements known –
(a) Both assertion and reason are true and reason is correct explanation for the assertion
(b) Both assertion and reason are true but the reason is not the correct explanation for the assertion
(c) Assertion is true and the reason is false
(d) Both assertion and the reason are false
Answer:
(c) Assertion is true and the reason is false

Periodic Classification Of Elements Class 11 Notes Pdf Question 15.
The electronic configuration of the atom having maximum difference in first and second ionization energies is ……….
(a) 1s², 2s², 2p⁶, 3s¹
(b) 1s², 2s², 2p⁶, 3s²
(c) 1s², 2s², 2p⁶, 3s², 3s², 3p⁶, 4s¹
(d) 1s², 2s², 2p⁶, 3s², 3p¹
Answer:
(a) 1s², 2s², 2p⁶, 3s¹

Chemistry Chapter 3 Answers Question 16.
Which of the following is second most electro negative element?
(a) Chlorine
(b) Fluorine
(c) Oxygen
(d) Sulphur
Answer:
(a) Chlorine

Periodic Classification Of Elements Class 11 Questions And Answers Question 17.
IE₁  and IE₂ of Mg are 179 and 348 k cal mol^-1 respectively. The energy required for the reaction
Mg → Mg²⁺ + 2e^– is ……..
(a) +169 kcal mol^-1
(b) -169 kcal mol^-1
(c) +527 kcal mol^-1
(d) -527 kcal mol^-1
Answer:
(c) +527 kcal mol^-1

Class 11 Chemistry Chapter 3 Important Questions With Answers Question 18.
In a given shell the order of screening effect is …………..
(a) s > p > d > f
(b) s > p > f > d
(c) f > d > p > s
(d) f > p > s > d
Answer:
(a) s > p > d > f

Samacheerkalvi.Guru 11th Chemistry Question 19.
Which of the following orders of ionic radii is correct?
(a) H^– > H⁺ > H
(b) Na⁺ > F“ > O^–
(c) F > O^2- > Na⁺
(d) None of these
Answer:
(d) None of these

Samacheer Kalvi.Guru 11th Chemistry Question 20.
The first ionization potential of Na, Mg and Si are 496, 737 and 786 kJ mol^-1 respectively. The ionization potential of Al will be closer to
(a) 760 kJ mol^-1
(b) 575 kJ mol^-1
(c) 801 kJ mol^-1
(d) 419 kJ mol^-1
Answer:
(b) 575 kJ mol^-1

Class 11 Chemistry Samacheer Solutions Question 21.
Which one of the following is true about metallic character when we move from left to right in a period and top to bottom in a group?
(a) Decreases in a period and increases along the group
(b) Increases in a period and decreases in a group
(c) Increases both in the period and the group
(d) Decreases both in the period and in the group
Answer:
(a) Decreases in a period and increases along the group

11th Chemistry Solutions Samacheer Kalvi Question 22.
How does electron affinity change when we move from left to right in a period in the periodic table?
(a) Generally increases
(b) Generally decreases
(c) Remains unchanged
(d) First increases and then decreases
Answer:
(a) Generally increases.

Samacheer Kalvi Guru 11 Chemistry Question 23.
Which of the following pairs of elements exhibit diagonal relationship?
(a) Be and Mg
(b) Li and Mg
(c) Be and B
(d) Be and Al
Answer:
(d) Be and Al

II. Write brief answer to the following questions

Samacheer Kalvi 11th Chemistry Solution Question 24.
Define modern periodic law.
Answer:
The modem periodic law states that, “The physical and chemical properties of the elements are periodic function of their atomic numbers.”

Question 25.
What are isoelectronic ions? Give examples.
Answer:
There are some ions of different elements having the same number of electrons are called isoelectronic ions.
Example:
Na⁺, Mg²⁺, Al³⁺, F^– , O^2- and N^3-

Question 26.
What is effective nuclear charge?
Answer:
The net nuclear charge experienced by valence electrons in the outermost shell is called the effective nuclear charge.
Z_eff = Z – S
Where,
Z = Atomic number
S = Screening constant calculated by using Slater’s rules.

Question 27.
Is the definition given below for ionization enthalpy is correct?
“Ionization enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom”
Answer:
No. It is not correct. The accurate and absolute definition is as follows:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 28.
Magnesium loses electrons successively to form Mg⁺, Mg²⁺ and Mg³⁺ ions. Which step will have the highest ionization energy and why?
Answer:

• The third step will have the highest ionization energy. I.E₃>I.E₂>I.E₁
• Because from a neutral gaseous atom, the electron removal is easy and less amount of energy is required. But from a di positive cation, there will be more number of protons than the electrons and there is more forces of attraction between the nucleus and electron. So the removal of electron in a di positive cation, becomes highly difficult and more energy is required.

Question 29.
Define electro negativity.
Answer:
Electro negativity is the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

Question 30.
How would you explain the fact that the second ionization potential is always higher than first ionization potential?
Answer:

• Second ionization potential is always higher than first ionization potential.
• Removal of one electron from the valence orbit of a neutral gaseous atom is easy so first ionization energy is less. But from a uni positive ion, removal of one more electron becomes difficult due to the more forces of attraction between the excess of protons and less number of electrons.
• Due to greater nuclear attraction, second ionization energy is higher than first ionization energy.

Question 31.
Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10^-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of kJ mol^-1.
Answer:
Energy of an electron in the ground state of the hydrogen atom = -2.18 x 10^-18 J
H → H⁺ + e^–
Energy required to ionize 1 mole of hydrogen atoms, we multiply by the Avogadro constant.
E = 2.18 x 10^-18 x 6.023 x 10²³
= 13.123 x 10⁵ J mol^-1
I.E = +1312 K J mol^-1

Question 32.
The electronic configuration of an atom is one of the important factor which affects the value of ionization potential and electron gain enthalpy. Explain.
Answer:

• Electronic configuration of an atom affects the value of ionization potential and electron gain enthalpy.
• Half filled valence shell electronic configuration and completely filled valence shell electronic configuration are more stable than partially filled electronic configuration.
• For e.g. Beryllium (Z = 4) 1s² 2s² (completely filled electronic configuration)
  Nitrogen (Z = 7) 1s²  2s²  2p_x¹  2p_y¹ 2p_z¹ (half filled electronic configuration) Both beryllium and nitrogen have high ionization energy due to more stable nature.
• In the case of beryllium (1s² 2s²), nitrogen (1s² 2s² 2p³) the addition of extra electron will disturb their stable electronic configuration and they have almost zero electron affinity.
• Noble gases have stable ns² np⁶ configuration and the addition of further electron is unfavorable and they have zero electron affinity.

Question 33.
In what period and group will an element with Z = 118 will be present?
Answer:
The element with atomic number Z = 118 is present in 7^th period and 18^th group.

Question 34.
Justify that the fifth period of the periodic table should have 18 elements on the basis of quantum numbers.
Answer:
Fifth period of the periodic table have 18 elements. 5^th period starts from Rb to Xe (18 elements). 5^th period starts with principal quantum number n = 5 and 1 = 0, 1,2,3 and 4. When n = 5, the number of orbitals = 9.
1 for 5s
5 for 4d
3 for 5p
Total number of orbitals = 9.
Total number of electrons that can be accommodated in 9
orbitals = 9 x 2 = 18. Hence the number of elements in 5th period is 18.

Question 35.
Elements a, b, c and d have the following electronic configurations:
a : 1s², 2s², 2p⁶
b : 1s², 2s², 2p⁶, 3s², 3p¹
c : 1s², 2s², 2p⁶ 3s²,3p⁶
d : 1s², 2s², 2p¹
Which elements among these will belong to the same group of periodic table?
Answer:

1. 
2. In the above elements, Ne and Ar belong to same group (Noble gases – 18th group).
3. Al and B belong to the same group (13^th group).

Question 36.
Give the general electronic configuration of lanthanides and actinides?
Answer:

• The electronic configuration of lanthanides is 4f^1-14 5d^0-16s².
• The electronic configuration of actinides is 5f^1-14 6d^0-1 7s².

Question 37.
Why halogens act as oxidizing agents?
Answer:
Halogens act as oxidizing agents. Their electronic configuration is ns² np⁵. So all the halogens are ready to gain one electron to attain the nearest inert gas configuration. An oxidizing agent is the one which is ready to gain an electron. So all the halogens act as oxidizing agents. Also halogens are highly electro negative with low dissociation energy and high negative electron gain enthalpies. Therefore, the halogens have a high tendency to gain an electron. Hence they act as oxidizing agents.

Question 38.
Mention any two anomalous properties of second period elements.
Answer:

• In the 1^st group, lithium forms compounds with more covalent character while the other elements of this group form only ionic compounds.
• In the 2^nd group, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.

Question 39.
Explain the Pauling’s method for the determination of ionic radius.
Answer:
1. Ionic radius is defined as the distance from the center of the nucleus of the ion up-to which it exerts its influence on the electron cloud of the ion.
2. Ionic radius of uni-univalent crystal can be calculated from the inter-ionic distance between the nuclei of the cation and anion.
3. Pauling assumed that ions present in a crystal lattice are perfect spheres and they are in contact with each other, therefore
d = r_C⁺ + r_A^– ………(1)
Where, d = distance between the center of the nucleus of cation C+ and the anion A-
r_C⁺ = radius of cation
r_A^– = radius of anion.
4. Pauling assumed that the radius of the ion having noble gas configuration (Na⁺ and F^– having 1s², 25², 2p⁶ configuration) is inversely proportional to the effective nuclear charge felt at the periphery of the ion.

Where Z_eff is the effective nuclear charge
Z_eff = Z – S
5. Dividing the equation (2) by (3)

On solving equation (1) and (4), the values of r_C⁺ and r_A^– can be obtained.

Question 40.
Explain the periodic trend of ionization potential.
Answer:
(a) The energy required to remove the most loosely held electron from an isolated gaseous atom is called as ionization energy.
(b) Variation in a period:
Ionization energy is a periodic property. On moving across a period from left to right, the ionization enthalpy value increases. This is due to the following reasons.

• Increase of nuclear charge in a period
• Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus. Therefore, ionization enthalpy increases.

(c) Variation in a group:
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons.

• A gradual increase in atomic size
• Increase of screening effect on the outermost electrons due to the increase of number of inner electrons.

Hence, ionization enthalpy is a periodic property.

Question 41.
Explain the diagonal relationship.
Answer:

• On moving diagonally across the periodic table, the second and the third period elements show certain similarities.
• Even though the similarity is not same as we see in a group, it is quite pronounced in the following pair of elements.
• 
• The similarity in properties existing between the diagonally placed elements is called “diagonal relationship”.

Question 42.
Why the first ionization enthalpy of sodium is lower than that of magnesium while its second ionization enthalpy is higher than that of magnesium?
Answer:
The 1st ionization enthalpy of magnesium is higher than that of Na due to higher nuclear charge and slightly smaller atomic radius of Mg than Na. After the loss of first electron, Na⁺ formed has the electronic configuration of neon (2,8). The higher stability of the completely filled noble gas configuration leads to very high second ionization enthalpy for sodium. On the other hand, Mg⁺ formed after losing first electron still has one more electron in its outermost (3 s) orbital. As a result, the second ionization enthalpy of magnesium is much smaller than that of sodium.

Question 43.
By using Pauling’s method calculate the ionic radii of K⁺ and Cl^– ions in the potassium chloride crystal. Given that \(\mathrm{d}_{\mathrm{K}}+_{-} \mathrm{cl}^{-}\) = 3.14 Å
Answer:
Given

We know that,

(Z_eff)_Cl^– = Z – S
= 17 – [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 17 – 11.25 = 5.75
(Z_eff)_K⁺ = Z – S
= 19 – [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 19 – 11.25 = 7.75

r_(K⁺) = 0.74 r_(Cl^–)
Substitute the value of r_(K⁺) in equation (1)
0.74 r_(Cl^–) + r_(Cl^–) = 3.14 Å
1.74 r_(Cl^–) = \(\frac {3.14 Å}{1.74}\) = 1.81 Å.

Question 44.
Explain the following, give appropriate reasons.

1. Ionization potential of N is greater than that of O
2. First ionization potential of C-atom is greater than that of B-atom, where as the reverse is true for second ionization potential.
3. The electron affinity values of Be, Mg and noble gases are zero and those of N (0.02 eV) and P (0.80 eV) are very low
4. The formation of F^– (g) from F(g) is exothermic while that of O^2-(g) from O (g) is endothermic.

Answer:
1. N (Z = 7) 1s² 2s² 2p_x¹ 12p_y¹ 2p_z¹. It has exactly half filled electronic configuration and it is more stable. Due to stability, ionization energy of nitrogen is high.
O (Z = 8) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹. It has incomplete electronic configuration and it requires less ionization energy.
I.E₁ N > I.E₁O

2. C (Z = 6) 1s² 2s² 2p_x¹ 2p_y¹. The electron removal from p orbital is very difficult. So carbon has highest first ionization potential. B (Z = 5) 1s² 2s² 2p¹. In boron nuclear charge is less than that of carbon, so boron has lowest first ionization potential.
I.E₁ C > I.E₁ B
But it is reverse in the case of second ionization energy. Because in case of B+ the electronic configuration is 1s² 2s², which is completely filled and it has high ionization energy. But in C+ the electronic configuration is 1s² 2s² 2p¹, one electron removal is easy so it has low ionization energy.
I.E₂ B > I.E₂ C

3. Be (Z = 4) 1s² 2s²
Mg (Z = 12) 1s² 2s² 2p⁶ 3s²
Noble gases has the electronic configuration of ns² np⁶. All these are completely filled and are more stable. For all these elements Be, Mg and noble gases, addition of electron is unfavorable and so they have zero electron affinity.

Nitrogen (Z = 7) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹. It has half filled electronic configuration. So addition of electron is unfavorable and it has very low electron affinity value of 0.02 eV. Phosphorus (Z = 15) 1s² 2s² 2p⁶ 3s² 3p_x¹ 3p_y¹ 3p_z¹. It also has half filled electronic configuration. Due to the symmetry and more stability, it has very low electron affinity value of 0.80 eV.

4. F_(g) + e^– → F_(g)^– exothermic
F (Z = 9) 1s² 2s² 2p⁵. It is ready to gain one electron to attain the nearest inert gas configuration. By gaining one electron, energy is released, so it is an exothermic reaction.
O_(g) + 2e^– → O^2-_(g) endothermic
O (Z = 8) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹. It is the small atom with high electron density. The first electron affinity is negative because energy is released in the process of adding one electron to the neutral oxygen atom. Second electron affinity is always endothermic (positive) because the electron is added to an ion which is already negative, therefore it must overcome the repulsion.

Question 45.
What is screening effect? Briefly give the basis for Pauling’s scale of electro negativity. Screening effect:
Answer:
The repulsive force between inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect . is called shielding effect (or) screening effect.

Pauling’s scale:

• Electro negativity is the relative tendency of an element present in a covalently bonded molecule to attract the shared pair of electrons towards itself.
• Pauling assigned arbitrary value of electronegativities for hydrogen and fluorine as 2.2 and 4, respectively.
  • Based on this the electronegativity values for other elements can be calculated using the following expression.
    (X_A-X_B) = 0.182 √E_AB – (E_AA E_BB)
    Where E_AB , E_AA and E_BB are the bond dissociation energies of AB, A₂ and B₂ molecules respectively.
    X_A and X_B are electronegativity values of A and B.

Question 46.
State the trends in the variation of electro negativity in period and group.
Answer:
Variation of electron negativity in a period:
The electro negativity increases across a period from left to right. Since the atomic radius decreases in a period, the attraction between the valence electron and the nucleus increases. Hence the tendency to attract shared pair of electrons increases. Therefore, electro negativity increases in a period.

Variation of electro negativity in a group:
The electro negativity decreases down a group. As we move down a group, the atomic radius increases and the nuclear attractive force on the valence electron decreases. Hence electro negativity decreases in a group.

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements  In-Text Question – Evaluate your self

Question 1.
What is the basic difference in approach between Mendeleev’s periodic table and modern periodic table?
Answer:
The main basic difference between Mendeleev’s periodic table and modem periodic table is that first one is constructed on the basis of atomic weight and the later is constructed on the basis of atomic number.

Question 2.
The element with atomic number 120 has not been discovered so far. What would be the IUPAC name and the symbol for this element? Predict the possible electronic configuration of this element.
Answer:
Atomic number : 120
IUPAC temporary symbol : Unbinilium
IUPAC temporary symbol : Ubn
Possible electronic configuration : [Og] 8s²

Question 3.
Predict the position of the element in periodic table satisfying the electronic configuration (n – 1 )d² ns² where n = 5?
Answer:
Electronic Configuration : (n – 1 )d² ns²
for n = 5, the electronic configuration is,
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d² 5s²
Atomic number : 40
4th group 5th period (d block element) = Zirconium

Question 4.
Using Slater’s rule calculate the effective nuclear charge on a 3p electron in aluminium and chlorine. Explain how these results relate to the atomic radii of the two atoms.
Answer:
Electronic Configuration of Aluminium

Effective nuclear charge = Z – S = 13 – 9.5
(Z_eff)_Al = 3.5
Electronic Configuration of chlorine

Effective nuclear charge = Z- S = 17 – 10.9
(Z_eff)_Cl = 6.1
(Z_eff)_Cl > (Z_eff)_Cl and hence r_Cl< r_Al

Question 5.
A student reported the ionic radii of iso electronic species X³⁺ , Y²⁺ and Z^– as 136 pm, 64 pm and 49 pm respectively. Is that order correct? Comment.
Answer:
X³⁺, Y²⁺, Z^– are iso electronic.
∴ Effective nuclear charge is in the order
(Z_eff)_Cl^– < (Z_eff)Y_Y²⁺ < (Z_eff)_X³⁺ and hcnce, ionic radii should be in the order r_Z^– > r_Y²⁺ > r_X³⁺
∴ The correct values are:

Question 6.
The first ionisation energy (IE₁) and second ionisation energy (IE₂) of elements X, Y and Z are given below.

Which one of the above elements is the most reactive metal, the least reactive metal and a noble gas?
Answer:
Noble gases:
Ioniation energy ranging from 2372 KJmol^-1 to 1037 kJ mol^-1. For element X, the IE₁ value is in the range of noble gas, moreover for this element both IE₁ and IE₂ are higher and hence X is the noble gas. For Y, the first ionisation energy is low and second ionisation energy is very high and hence Y is most reactive metal.
For Z, both IE₁ and IE₂ are higher and hence it is least reactive.

Question 7.
The electron gain enthalpy of chlorine is 348 kJ mol^-1. How much energy in kJ is released when 17.5 g of chlorine is completely converted into Cl^– ions in the gaseous state?
Cl_(g)+ e^– → Cl^–_(g)
∆H = 348 kJ mol^-1
For one mole (35.5g) 348 kJ is released.
∴ For 17.5g chlorine,  energy leased.
∴ The amount of energy released = \(\frac {348}{2}\) = 174 kJ.

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements  Additional Questions

Question 1.
The chemical symbol of carbon and cobalt are
(a) Ca and CO
(b) Ca and Cl
(c) C and CO
(d) Cr and Cb
Answer:
(c) C and CO

Question 2.
Consider the following statements.
(i) The chemical symbol of nickel is N.
(ii) An element is a material made up of different kind of atoms.
(iii) The physical state of bromine is liquid.
Which of the above statement is/are not correct?
(a) (i) and (iiii)
(b) (iii) only
(c) (ii) and (iii)
(d) (i) and (ii)
Answer:
(d) (i) and (ii)

Question 3.
Match the list-I and list-II using the correct code given below the list.
List – I
A. Jewels
B. Bolts and cot
C. Table salt
D. Utensils

List – II
1. Sodium chloride
2. Copper
3. Gold
4. Iron

Answer:

Question 4.
The law of triads is not obeyed by
(a) Ca, Sr, Ba
(b) Cl, Br, I
(c) Li, Na, K
(d) Be, B, C
Answer:
(d) Be, B, C

Question 5.
The law of triads is obeyed by
(a) Fe, CO, Ni
(b) C, N, O
(c) He, Ne, Ar
(d) Al, Si, P
Answer:
(a) Fe, CO, Ni

Question 6.
Match the list-I and list-II using the code given below the list.
List-I
A. Law of triads
B. Law of octaves
C. First periodic law
D. Modem periodic law

List-II
1. Chancourtois
2. Henry Moseley
3. Newland
4. Johann Dobereiner

Answer:

Question 7.
Consider the following statements.
(i) In Chancourtois classification, elements differed from each other in atomic weight by 16 or multiples of 16 fell very nearly on the same vertical line.
(ii) Mendeleev’s periodic law is based on atomic weight.
(iii) Mendeleev listed the 117 elements known at that time and are arranged in the order of atomic numbers.
Which of the following statement is/are not correct?
(a) (i) only
(b) (ii) and (iii)
(c) (iii) only
(d) (i),(ii), (iii)
Answer:
(c) (iii) only

Question 8.
Which of the following elements were unknown at that time of Mendeleev?
(a) Na, Mg
(b) Fe, CO
(c) K, Cu
(d) Ga, Ge
Answer:
(d) Ga, Ge

Question 9.
Consider the following statements.
(i) Position of hydrogen could not be made clear.
(ii) Isotopes find correct place in Mendeleev’s periodic table.
(iii) Mendeleev’s periodic table could not explain the variable valencies of elements.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) and (iii)
(c) (ii) only
(d) (i), (ii), (iii)
Answer:
(c) (ii) only

Question 10.
According to modem periodic law, the physical and chemical properties of the elements are periodic functions of their
(a) atomic volume
(b) atomic numbers
(c) atomic weights
(d) valency
Answer:
(B) atomic numbers

Question 11.
Which period contain 32 elements?
(a) Period 1
(b) Period 4
(c) Period 5
(d) Period 6
Answer:
(d) Period 6

Question 12.
There are horizontal rows of the periodic table known as
(a) groups
(b) periods
(c) families
(d) chalcogens
Answer:
(b) periods

Question 13.
The shortest period contains elements.
(a) H, He
(b) Li, Be
(c) B, C
Answer:
(a) H, He

Question 14.
The longest form of periodic table was constructed by
(a) Dmitri Mendeleev
(b) Henry Moseley
(c) Lothar Meyer
(d) New lands
Answer:
(b) Henry Moseley

Question 15.
Match the list-I and list-II using the correct code given below the list.
List – I
Z = 100
Z = 101
Z = 102
Z = 103

List – II
1. Mendelevium
2. Lawrencium
3. Fermium
4. Nobelium

Answer:

Question 16.
Which one of the following is the first transition series?
(a) Sc
(b) Zn
(c) Ti
(d) Cu
Answer:
(a) Sc

Question 17.
Which period mostly include man made radioactive elements?
(a) 4^th period
(b) 7^th period
(c) 6^th period
(d) 31 period
Answer:
(b) 7^th period

Question 18.
Which one of the following is called halogen family?
(a) Group 17
(b) Group 16
(c) Group 1
(d) Group 2
Answer:
(a) Group 17

Question 19.
Group 16 constitutes family.
(a) halogen
(b) nobel gas
(c) chalcogen
(d) alkali metals
Answer:
(c) chalcogen

Question 20.
Consider the following statements.
(i) The valency of the elements increases from left to right in a period.
(ii) Valency decreases from 7 to I with respect to oxygen.
(iii) The metallic character of the elements decreases across a period.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) only
(c) (i) and (iii)
(d) (i) (ii) and (iii)
Answer:
(b) (ii) only

Question 21
Match the list – I and list -II using the correct code given below the list.
List – I
A. Li
B. Na
C. K
D. Cs

List – II
1. 2,8,8,1
2. 2,1
3. 2,8, 18, 18, 8, 1
4. 2,8, 1

Answer:

Question 22.
What will be the change in valency down the group in the periodic table?
(a) increases
(b) decreases
(c) remains same
(d) zero
Answer:
(c) remains same

Question 23.
Which one of the following is a metalloid?
(a) N
(b) P
(c) Bi
(d) Sb
Answer:
(d) Sb

Question 24.
Which one of the following is a metal?
(a) N
(b) Br
(c) Bi
(d) As
Answer:
(c) Bi

Question 25.
Match the list – I and list – II using the correct code given below the list.
List – I
A. Alkali metal
B. Alkaline earth metals
C. d-block elements
D. p-block elements

List – II
1. ns² np^1-6
2. ns¹
3. ns²
4. (n – 1)d^1-10 ns^0-2

Answer:

Question 26.
Consider the following statements.
(i) Oxidation character increases from left to right in a period.
(ii) Reducing character increases from left to right in a period.
(iii) Metallic character increases from left to right in a period.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (i), (ii), (iii)
Answer:
(c) (ii) and (iii)

Question 27.
The general electronic configuration of d-block elements is
(a) ns² nd^1-10 10
(b) (n-1)d^1-10 ns^0-2
(c) (n-2)d^1-10 (n – 1)^0-2
(d) ns²nd⁵
Answer:
(b) (n-1)d^1-10 ns^0-2

Question 28.
Consider the flowing statements.
(i) d-block elements show variable oxidate states.
(ii) Mostly d-block elements form colourless compounds.
(iii) Mostly d-block elements are diamagnetic due to paired electrons.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) only
(c) (i) and (ii)
(d) (ii) and (iii)
Answer:
(d) (ii) and (iii)

Question 29.
All the s – block and p-block elements excluding l8 group are called elements.
(a) representative
(b) transition
(c) inner – transition
(d) trans uranium
Answer:
(a) representative

Question 30.
Which of the following is the correct electronic configuration of noble gases?
(a) ns² np⁶ nd¹⁰
(b) ns² np⁵
(c) ns² np⁶
(d) ns² np³
Answer:
(c) ns² np⁶

Question 31.
Group numbers 13 to 12 in the periodic table are called …………..
(a) inner transition elements
(b) Representative elements
(c) synthetic elements
(d) transition elements
Answer:
(d) transition elements

Question 32.
Which one of the following is in solid state at room temperature?
(a) Bromine
(b) Mercury
(c) Bismuth
(d) Gallium
Answer:
(c) Bismuth

Question 33.
Which of the following is not a metalloid (or) semi-metal?
(a) Silicon
(b) Arsenic
(c) Germanium
(d) Sodium
Answer:
(d) Sodium

Question 34.
Which of the following metal is not in liquid state?
(a) Gallium
(b) Aluminium
(c) Mercury
(d) Calsium
Answer:
(b) Aluminium

Question 35.
Which of the following is not a periodic property?
(a) Atomic radius
(b) Ionization enthaphy
(c) Electron affinity
(d) Oxidation number
Answer:
(d)Oxidation number

Question 36.
Which of the following property increases as we go down the group in the periodic property?
(a) ionization energy
(b) Electro negativity
(c) Atomic radius
(d) Electron affinity
Answer:
(c) Atomic radius

Question 37.
The metallic radius of copper is ………………
(a) 0.99 Å
(b) 1.28 Å
(c) 1.98 Å
(d) 2.56 Å
Answer:
(5) 1.28 Å

Question 38.
Consider the following statements………………
(i) Atomic radius of elements increases with increase in atomic number as we go down the group.
(ii) Atomic radius of elements increases with increase in atomic number as we go across the period.
(iii) Atomic radius of elements decreases as we go from left to right in a period.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) and (ii)
(c) (ii) only
(d) (i) and (iii)
Answer:
(c) (ii) only

Question 39.
Which one of the following is not an iso electronic ion?
(a) Na⁺
(b) Mg²⁺
(c) Cl^–
(d) O^2-
Answer:
(c) Cl^–

Question 40.
Which one of the following is not an isoelectronic ion?
(a) Al³⁺
(b) N^3-
(c) Mg²⁺
(d) K⁺
Answer:
(d) K⁺

Question 41.
Which of the following possess almost same properties due to lanthanide contraction?
(a) Zr, HF
(b) Na, K
(c) Zn, Cd
(d) Ag. Au
Answer:
(a) Zr, HF

Question 42.
Consider the following statements.
(i) Ionization is always an exothermic process.
(ii) Ionization energies always increase in the order I.E₁> IE₂>I.E₃.
(iii) Ionization energy measurements are carried out with atoms in the solid state.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) and (ii)
(c) (iii) only
(d) (i), (ii) and (iii)
Answer:
(d) (i), (ii) and (iii)

Question 43.
Statement-I : Ionization enthalpy of Be is greater than that of 13.
Statement-II : The nuclear charge of B is greater than that of Be.
(a) Statement-I and II are correct and statement-II is the correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not the correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(b) Statement-I and II are correct but statement-II is not the correct explanation of statement-I.

Question 44.
Statement-I: Ionization enthalpy of nitrogen is greater than that of oxygen.
Statement-lI: Nitrogen has exactly half filled electronic configuration which is more stable than electronic configuration of oxygen.
(a) Statement-I is wrong but statement-II is correct.
(b) Statement-I is correct but statement-II is wrong.
(c) Statement-I and II are correct and statement-TI is the correct explanation of statement-I.
(d) Statement-I and II are correct but statement-II is not the correct explanation of statement-I.
Answer:
(c) Statement-I and II are correct and statement-II is the correct explanation of statement-I.

Question 45.
Which of the following does not have zero electron gain enthalpy?
(a) Be
(b) Cl
(c) Mg
(d) N
Answer:
(b) Cl

Question 46.
Which of the following have zero electron gain enthalpy?
(a) Halogens
(b) Noble gases
(c) Chalcogens
(d) Gold
Answer:
(b) Noble gases

Question 47.
Which of the following have the highest value of electronegativity?
(a) Halogens
(b) Alkali metals
(c) Alkaline earth metals
(d) Transition metals
Answer:
(a) Halogens

Question 48.
Among all the elements which one has the highest value of electronegativity?
(a) Chlorine
(b) Bromine
(c) Fluorine
(d) Iodine
Answer:
(c) Fluorine

Question 49.
Among the alkali metals which one form compounds with more covalent character?
(a) Sodium
(b) Potassium
(c) Rubidium
(d) Lithium
Answer:
(d) Lithium

Question 50.
Which of the following pair is not diagonally related?
(a) Li, Mg
(b) Li, Na
(c) Be, Al
(d) B, Si
Answer:
(b) Li, Na

Question 51.
In the modern periodic table, the period indicates the value of
(a) atomic number
(b) mass number
(c) principal quantum number
(d) azimuthal quantum number
Answer:
(c) principal quantum number
Hint:
In the modern periodic table, each period begins with the filling of a new shell. Therefore, the period indicates the value of principal quantum number. Thus, option (c) is correct.

Question 52.
Which of the following statements related to the modern periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitais in a p-subshell.
(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitais in a d-subsheli.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (1) for the last subshell that received electrons in building up the electronic configuration.
Answer:
(b)The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitais in a d-subshell.

Question 53.
The size of isoelectronic species- F^–, Ne and Na⁺ is affected by
(a) nuclear charge (Z)
(b) valence principal quantum number (n)
(c) electron-electron interaction in the outer orbitais
(d) none of the factors because their size is the same
Answer:
(a) nuclear charge (Z).

Question 54.
Which of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electrons from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitais bearing lower n value is easier than from orbital having high n value.
Answer:
(d)Removal of electron from orbitais bearing lower n value Is easier than from orbital having high n value.

Question 55.
Considering the elements B, Al, Mg and K, the correct order of their metallic character is:
(a) B > Al >Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Answer:
(d) K > Mg > Al> B
Hint:
In a period, metallic character decreases as we move from left to right. Therefore, metallic character of I< Mg and Al decreases in the order: K> Mg > Al. However, within a group, the metallic character, increases from top to bottom. Thus, Al is more metallic than B.Therefore, the correct sequence of decreasing metallic character is K> Mg >Al > B, i.e,
option (d) is correct.

Question 56.
Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is
(a) B>C>Si>N>F
(b) Si>C>B>N>F
(c) F>N>C>B>Si
(d) F>N>C>Si>B
Answer:
(c) F>N>C>B> Si
Hint:
In a period, the non-metallic character increases from left to right. Thus, among B, C, N and F, non-metallic character decreases in the order: F > N > C> B. However, within a group, non-metallic character decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correct sequence of decreasing non-metallic character is: F> N > C > B > Si, i.e., option (c) is correct.

Question 57.
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is
(a) F>Cl>O>N
(b) F>O>Cl>N
(c) Cl>F>O>N
(d) O>F>N>Cl
Answer:
(b) F>O>Cl>N.
Hint:
Within a period, the oxidizing character increases from left to right. Therefore, among F, O and N,oxidizing power decreases in the order: F> O> N. However, within a group, oxidizing power decreases from top to bottom. Thus, F is a stronger oxidizing agent than Cl. Further because O is more electronegative than Cl, therefore, O is a stronger oxidizing agent than Cl. Thus, overall decreasing order of oxidizing power is: F > O > Cl > N, i.e.,
option (b) is correct.

Question 58.
The highest ionization energy is exhibited by ………………
(a) halogens
(b) alkaline earth metals
(c) transition metals
(d) noble gases
Answer:
(b) alkaline earth metals

Question 59.
Which of the following is arranged in order of increasing radius?
(a) K⁺_(aq) < Na⁺_(aq) <Li⁺_(aq)
(b) K⁺_(aq)> Na⁺_(aq)> Zn²⁺_(aq)
(c) K⁺_(aq)> Li⁺_(aq) > Na⁺_(aq)
(d) Li⁺_(aq)< Na⁺_(aq) < K⁺_(aq)
Answer:
(d) Li⁺_(aq)< Na⁺_(aq) < K⁺_(aq)

Question 60.
Among the following elements, which has the least electron affinity?
(a) Phosphorous
(b) Oxygen
(c) Sulphur
(d) Nitrogen
Answer:
(d) Nitrogen

Question 61.
Which one of the following is isoelectronic with Ne?
(a) N^3-
(b) Mg²⁺
(c) Al³⁺
(d) All the above
Answer:
(d) All the above

Question 62.
Which clement has smallest size?
(a) B
(b) N
(c) Al
(d) P
Answer:
(b) N

Question 63.
In halogens, which of the following decreases from iodine to fluorine?
(a) Bond length
(b) Electronegativity
(c) ionization energy
(d) Oxidizing power
Answer:
(a) Bond length

Question 64.
What is the electronic configuration of the elements of group 14?
(a) ns² np⁴
(b) ns² np⁶
(c) ns² np²
(d) ns²
Answer:
(c) ns² np²

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements  2-Mark Questions

Write brief answer to the following questions:

Question 1.
State Johann Dobereiner’s law of triads.
Answer:
Johann Dobereiner noted that elements with similar properties occur in groups of three which he called triads. It was seen that invariably, the atomic weight of the middle number of the triad was nearly equal to the arithmetic mean of the weights of the other two numbers of the triad.
For e.g.,

Question 2.
Write a note about Chancourtois classification.
Answer:
in this system, elements that differed from each other in atomic weight by 16 or multiples of 16 fell very nearly on the same vertical line. Elements lying directly under each other showed a definite similarity. This was the first periodic law.

Question 3.
State the New land’s law of octaves.
The Law of octaves states that, “when elements are arranged in the order of increasing atomic weights, the properties of the eighth element are a repetition of the properties of the first element.

Question 4.
State Mendeleev’s periodic law.
Answer:
This law states that “The physical and chemical properties of elements are a periodic function of their atomic weights.”

Question 5.
Explain about the relationship between the atomic number of an element and frequency of the X-ray emitted from the elements.
Answer:
Henry Moseley studied the X-ray spectra of several elements and determined their atomic numbers (Z). He noticed that the frequencies of X-ray emitted from the elements concerned could be correlated by the equation
\(\sqrt{υ}\) = a(Z – b)
Where, υ Frequency of the X-ray emitted by the element.
a and b = Constants and have same values for all the elements.
Z = Atomic number of the element.

Question 6.
State modern periodic law.
Answer:
The modem periodic law states that, “The physical and chemical properties of the elements are periodic function of their atomic numbers.”

Question 7.
What are the anomalies of the long form of periodic table?
Answer:
The long form of periodic table need clarification about the following:

• Position of hydrogen is not defined till now.
• Lanthanides and actinides still find place in the bottom of the table.

Question 8.
Mention the names of the elements with atomic number 101, 102, 109 and 110.
Z = 101  IUPAC  name : Mendelevium
Z = 102  IUPAC  name : Nobelium
Z = 109  IUPAC  name : Meitnerium
Z = 110  IUPAC  name : Darmstadtium

Question 9.
Write a note about the electronic configuration of elements in groups.
Answer:
A vertical column of the periodic table is called a group. A group consists of a series of elements having similar configuration of the outermost shell. There are 18 groups in the periodic table. it may be noted that the elements belonging to the same group are said to constitute a family. For example, elements of group 17 are called halogen family.

Question 10.
Give the name and electronic configuration of elements of group and 2 group.
Answer:

• Elements of 1^st group are called alkali metals. Their electronic configuration is ns¹.
• Elements of 2^nd group are called alkaline earth metals. Their electronic configuration is ns².

Question 11.
Write any two characteristic properties of alkali metals.
Answer:

• Alkali metals readily lose their outermost electron to form +1 ion.
• Alkali metals are soft metals with low melting and boiling points.

Question 12.
Write any two characteristic properties of alkaline earth metals.
Answer:

• Alkaline earth metals readily lose their outermost electrons to form +2 ion.
• As we go down the group. their metallic character and reactivity are increased.

Question 13.
Groups from 13 to 1 in the periodic table are called p-block elements. Give reason.
Answer:

• The elements whose last electron enters into the p-orbital of the outermost shell are having similar properties and thus form a group. The ‘np’ orbital of these elements is being progressively tilled. Hence, these elements are named as p-block elements.
• The groups of 3^th to 18^th in the periodic table belongs to p-block.

Question 14.
Why noble gases do not show much of chemical reactivity?
Answer:
Noble gases having closed valence shell configuration as ns² np⁶. The valence shell orbitais of noble gases are completely filled by electrons and it is very difficult to alter this stable arrangement by the addition or removal of electrons. Because of these reasons noble gases do not show much of chemical reactivity.

Question 15.
Halogens and chalcogens have highly negative electron gain enthalpies. Why?
Answer:

• Group 16 (chaicogens) and Group 17 (halogens) are interested to add two or one electrons respectively to attain stable noble gas configuration.
• Because of this interest these elements have highly negative electron gain enthalpies.

Question 16.
What are d-biock elements? Why are they called so?
Answer:

• The elements whose last electron enters into the d-orbitalof the penultimate shell (n-1) are having similar properties and called as d-block elements.
• The groups of 3 to 12 in the center of the periodic table belongs to d-block.

Question 17.
Elements Zn, Cd and Hg with electronic configuration (n-1)d¹⁰ ns² do not show most of transition elements properties. Give reason.
Answer:

• Zn, Cd and Hg are having completely filled d-orbitais (d¹⁰ electronic configuration).
• They do not have partially filled d-orbitais Like other transition elements. So they do not show much of the transition elements properties.

Question 18.
Why Zn, Cd and Hg are considered as soft metals?
Answer:

• Zinc, cadmium and mercury are metals with low melting points. This is because they have an especially stable electronic configuration.
• Mercury is so poor at forming metallic bonds that it is liquid at room temperature.
• Zinc and cadmium arc soft metals that oxidize to the +2 oxidation states.

Question 19.
Why d-block elements are called as transition elements?
Answer:
d-block elements form a bridge between the chemically active metals of s-block elements and the less active elements of groups of 13^th and 14^th and thus take their familiar name transition elements.

Question 20.
What are f-block elements? how many series are there? Why they are called f-block elements?
Answer:

• The elements, whose last electron enters into the f-orbital of the ante-penultimate shell (n-2) are having similar properties are called f-block elements. In these elernents (n-2)f orbitais are being filled progressively.
• The two rows of elements placed at the bottom of the periodic table. They are lanthanides and actinides.

Question 21.
Write the electronic configuration of lanthanides and actinides?
Answer:

• The electronic configuration of lanthanides is 4f^1-144 5d^0-11 6s².
• The electronic configuration of actinides is 5f^1-14 6d^0-17s².

Question 22.
What are lanthanides and actinides? ,
Answer:

• In 4f senes, 4f’orbitals arc being progressively filled with electrons, 4f^1-14 5d^0-1 6s². These elements lie in 6^th period and are called rare earths or lanthanides or lanthanones.
• In 5f series, 5f orbitais are being progressively filled with electrons, 5f¹6d^0-1 7s². These elements lie in 7^th period and are called actinides or actonones.

Question 23.
What are semi-metals? Give example.
Answer:

• Some elements in the periodic table show properties that are characteristic of both metals and non-metals. These elements are called semi-metals or metalloids.
• Example: Silicon. germanium, arsenic, antimony and tellurium.

Question 24.
What are periodic properties? Give example.
Answer:
The term periodicity of properties indicates that the elements with similar properties reappear at certain regular intervals of atomic number in the periodic table.
Example:

• Atomic radii
• Ionisation energy
• Electron affinity
• Electronegativity.

Question 25.
Define ionic radius.
Answer:
The ionic radius of an ion is the distance between the center of the ion and the outermost point of its electron cloud.

 

Question 26.
Cationic radius is smaller than its corresponding neutral atom. Justify this statement.
Answer:

• When an neutral atom lose one or more electrons it forms cation.
  Na → Na⁺ + e^–
• The radius of this cation (r_Na⁺)is decreased than its parent atom (r_Na).
• When an atom is charged to cation, the number of nuclear charge becomes greater than the number of orbital electrons. Hence the remaining electrons are more strongly attracted by the nucleus. Hence the cationic radius is smaller than its corresponding neutral atom.

Question 27.
Anionic radius is higher than the corresponding neutral atom. Give reason.
Answer:
When an atom gain one or more electrons it forms anion. During the formation of anion, the number of orbital electrons become greater than the nuclear charge. Hence, the electrons are not strongly attracted by the lesser number of nuclear charges. Hence anionic radius is higher than the corresponding neutral atom.

Question 28.
What are iso electronic ions? Give example.
Answer:
There are some ions of different elements having the same number of electrons are called isoelectronic ions.
Example: Na⁺ Mg²⁺, Al³⁺, F^–, O^2-, N^3-

Question 29.
Define ionization energy. Give its unit.
Answer:
The energy required to remove the most loosely held electron from an isolated gaseous atom is called ioniiation energy.
M_(g) + energy M⁺_(g) + electron
The unit of ionization energy is KJ mole^-1.

Question 30.
Ionization energy of beryllium is greater than the ionization energy of boron. Why?
Answer:
Be (Z= 4) 1s² 2s². it has completely filled valence electrons, which requires high IE₁.
B (Z =5) 1s² 2s² 2p¹. It has incompletely filled valence electrons, which requires comparatively
less IE₁ Hence I.E₁ Be > I.E₁ B.

Question 31.
Ionization energy of nitrogen is greater than the ionization energy of oxygen. Give reason.
Answer:
₇N 1s² 2s¹ 2p³ (or) . Nitrogen has exactly half filled valence electrons, which requires high I.E₁.
₈O 1s² 2s² 2p⁴ (or). Oxygen has incomplete valence shell electrons, which requires comparatively less I.E₁.
I.E₁N>I.E₁O

Question 32.
Define electron gain enthalpy or electron affinity. Give its unit.
Answer:
The electron gain enthalpy of an element is the amount of energy released when an electron is added to the neutral gaseous atom.
A + electron → A^– + energy (E.A)
Unit of electron affinity is KJ mole^–.

 

Question 33.
Electron gain enthalpy of F ¡s less negative than Cl. Why?
Answer:
When an electron is added to F, the added electron goes to the L shell (n = 2). As the ‘L’ shell possess smaller region of space, the added electron feels significant repulsion from the other electrons present in this level.
E.A of F = -328 KJ mole^-1
E.A of Cl = -349 KJ mole^-1

Question 34.
Electron affinity of oxygen is less negative than sulphur. Justify this statement.
Answer:
When an electron is added to oxygen, the added electron goes to the ‘L’ shell (n 2). As the ‘L’ shell possess smaller region of space, the added electron feels significant repulsion from the other electrons present in this level.
E.A of O = – 141 KJ mole^–.
E.A of S = – 200 KJ mole^–.

Question 35.
Explain about the factors that affect electro negativity.
Answer:

• Effective nuclear charge:
  As the nuclear charge increases, electro negativity also increases along the periods.
• Atomic radius:
  The atoms in smaller size will have larger electronegativity.

Question 36.
Explain about periodic variation of electro negativity across a period.
Answer:
As we move from left to right in a period, electro negativity increases. This is due to the following reasons:

• Nuclear charge increases in a period
• Atomic size decrease in a period

Halogens have the highest value of electro negativity in their respective periods.

Question 37.
Explain about the period variation of electro negativity along a group.
Answer:
As we move down from top to bottom in a group, electro negativity decreases due to increased atomic radius. Fluorine has the highest value of clectro negativity among all the elements.

Question 38.
Define valency. How is it determined?
Answer:
The valency of an element may be defined as the combining capacities of elements. The electrons present in the outermost shell are called valence electrons and these electrons determine the valency of the atom.

Question 39.
What is the basic difference in approach between Mendeleev’s periodic law and the modern periodic law?
Answer:
The basic difference in approach between Mendeleev’s periodic law and modern periodic law is the change in basis of classification of elements from atomic weight to atomic number.

Question 40.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer:
The sixth period corresponds to sixth shell. The orbitais present in this shell are 6s, 4f. 5p and 6d. The maximum number of electrons which can be present in these sub-shell is 2 + 14 + 6 + 10 = 32. Since the number of elements in a period corresponds to the number of electrons in the shells, the sixth period should have a maximum of 32 elements.

 

Question 41.
Why do elements in the same group have similar physical and chemical properties?
Answer:
The elements in a group have same valence shell electronic configuration and hence have similar physical and chemical properties.

Question 42.
How do atomic radius vary in a period and in a group? How do you explain the variation.
Answer:
Within a group atomic radius increases down the group Reason :
This is due to continuous increases in the number of electronic shells or orbit numbers in the structure of atoms of the elements down a group.

Variation across period:
Atomic radii:
From left to right across a period atomic radii generally decreases due to increase in effective nuclear charge from left to right across a period.

Question 43.
Explain why cation are smaller and anions are larger ¡n radii than their parent atoms?
Answer:
A cation is smaller than the parent atom because it has fewer electrons while its nuclear charge remains the same. The size of anion will be larger than that of parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

Question 44.
What is basic difference between the terms electron gain enthalpy and electronegativity?
Answer:
Electron gain enthalpy refers to tendency of an isolated gaseous atom to accept an additional electron to form a negative ion. Whereas electronegativity refers to tendency of the atom of an clement to attract shared pair of electrons towards it in a covalent bond.

Question 45.
Would you expect the first ionization enthalpies of two isotopes of the same element to be the same or different? Justify your answer.
Answer:
Ionization enthalpy, among other things, depends upon the electronic configuration (number of electrons) and nuclear charge (number of protons). Since isotopes of an element have the same electronic configuration and same nuclear charge, they have same ionization enthalpy.

 

Question 46.
Write the general electronic configuration of s-, p-, d-, and f-block elements?
Answer:

• s-block elements : ns^1-2 where n 2 – 7.
• p-block elements : ns² np^1-6 where n = 2 – 6.
• d-block elements : (n – 1) d^1-0 ns^0-2 where n = 4 – 7.
• f-block elements : (n – 2) f^0-14 (n – 1) d^0-1 ns² where n = 6 – 7.

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements 3-Mark Questions

Question 1.
Why there is a need for classification of elements?
Answer:

• Classification is a fundamental and essential process in our day-to-day life for the effective utilization of resources, daily events and materials.
• In such a way, for the effective utilization of discovered elements becomes fundamentally essential process.
• The periodic classification of the elements is one of the outstanding contributions to the progress of chemistry.

Question 2.
Prove that the halogens, chlorine, bromine and iodine follow the law of triads.
Answer:
When the halogens, chlorine, bromine and iodine are placed on below the others, they had similar properties. The atomic weight of bromine was close to the average of the atomic weights of chlorine and iodine.

Question 3.
What are the salient features of New land’s law of octaves?
Answer:

1. This law is quite well for lighter elements but not supported to heavier elements.
2. Elements were arranged in increasing atomic masses without taking an account on the properties of elements.
3. This law was seemed to be applicable only for elements upto calcium.

Question 4.
How the properties of Eka – silicon was related to germanium?
Answer:

Question 5.
Compare the properties of Eka – aluminium and gallium.
Answer:

Question 6.
How Moseley determined the atomic number of an element using X-rays?
Answer:

• Henry Moseley studied the X-ray spectra of several elements and determined their atomic numbers (Z).
• He discovered a correlation between atomic number and the frequency of X-rays generated by bombarding an clement with high energy of electrons.
• Moseley correlated the frequency of the X-ray emitted by an equation as,
  \(\sqrt{v}\) = a (Z – b)
  Where υ = Frequency of the X-rays emitted by the elements.
  a and b = Constants.
• From the square root of the measured frequency of the X-rays emitted, he determined the atomic number of the element.

 

Question 7.
What are the reasons behind the Moseley’s attempt in finding atomic number?
Answer:

• The number of electrons increases by the same number as the increase in the atomic number.
• As the number of electrons increases, the electronic structure of the atom changes.
• Electrons in the out cannost shell of an atom (valence shell electrons) determine the chemical properties of the elements.

Question 8.
Draw a simplified form of periods and elements present in modern period table.
Answer:

Question 9.
Write the electronic configuration of alkali metals ₂Li,₁₁Na, ₁₉K, ₃₇Rb, ₅₅Cs and ₈₇Fr.
Answer:

Question 10.
Explain about the classification of elements based on electronic configuration.
Answer:

• The distribution of electrons into orbitais, s, p, d and f of an atom is called its electronic configuration. The electronic configuration of an atom is characterized by a set of four quantum numbers, n, l, m and s. of these the principal quantum number (n) defines the main energy level known as shells.
• The position of an element in the periodic table is related to the configuration of that element and thus reflects the quantum numbers of the last orbital filled.
• The electronic configuration of elements in the periodic table can be studied along the periods and groups separately for the best classification of elements.
• Elements placed in a horizontal row of a periodic table is called a period. There are seven periods.
• A vertical column of the periodic table is called a group. A group consists of a series of elements having similar configuration of the outermost shell. There are 18 groups in periodic table.

 

Question 11.
Write about the electronic configuration of 1^st and 2^nd period.
Answer:
Electronic configuration of t period:
In 1 period only two elements are present. This period starts with the filling of electrons in first energy level, n¹. This level has only one orbital as is. Therefore it can accommodate two electrons maximum.

Electronic configuration of 2nd period:
In the 2’ period 8 elements are present. This period starts with filling of electrons in the second energy level, n = 2. In this level four orbitais (one 2s and three 2p) are present. Hence the second energy level can accommodate 8 electrons. Thus, second period has eight elements.

Question 12.
How many elements are there in 4^th period? Prove it.
Answer:
In fourth period, 18 elements are present. In this period electrons are entering into fourth energy level, i.e., n = 4. it starts with the filling of 4s-orbitals. However, after the 4s, but before the 4p orbitais, there are five 3d-orbitais also to be filled. Thus, nine orbitais (one 4s, five 3d and three 4p) have to be filled. These nine orbitais can accommodate (9 x 2 = 18)18 electrons. Hence, period contain 18 elements in it.

Question 13.
How many elements are there in 6th period? Prove it.
Answer:
In sixth period, 32 elements are present. This period starts with the filling of 6th energy shell, n = 6. There are sixteen orbitais (one 6s, seven 4f, five 4d and three 6p) to be filled. These sixteen orbitais can accommodate 32 (16 × 2 = 32) electrons. Hence, 32 elements are present in sixth period.

Question 14.
What are the two exceptions of block division in the periodic table?
Answer:
1. Helium has two electrons. Its electronic configuration is 1s². As per the configuration, it is supposed to be placed in ‘s’ block, but actually placed in s group which belongs to ‘p’ block. Because it has a completely filled valence shell as the other elements present in 18^th group. It also resembles with 18^th group elements in other properties. Hence helium is placed with other noble gases.

2. The other exception is hydrogen. it has only one s-electron and hence can be placed in group 1. It can also gain an electron to achieve a noble gas arrangement and hence it can behave as halogens (17th group elements). Because of these assumptions, position of hydrogen becomes a special case. Finally, it is placed separately at the top of the periodic table.

 

Question 15.
Explain about the salient features of metals.
Answer:

• Metals comprise more than 78% of all known elements. They are present on the left side of the periodic table.
• They are usually solids at room temperature. [Mercury is an exception (Hg-liquid), gallium (303 K) and cesium (302 K) also have very low melting points].
• Metals usually have high melting and boiling points.
• They are good conductors of heat and electricity.
• They are malleable and ductile, and also can be flattened into thin sheets by hammering and drawn into thin wires.

Question 16.
Explain about the characteristics of non-metals.
Answer:

• Non-metals are located at the top right hand side of the periodic table.
• In a period, as we move from left to right the non-metallic character increases while the metallic character increases as we go down a group.
• Non-metals are usually solids or liquids or gases at room temperature with low melting and low boiling points (Exceptions : boron and carbon).
• They are poor conductors of heat and electricity.
• Most of the non-metallic solids are brittle and are neither malleable nor ductile.

Question 17.
Periodic change in electronic configuration is responsible for the physical and chemical properties of element. Justify this statement.
Answer:

• The electronic configuration of the elements changes periodically in a period and group as well.
• We could find a pattern in the physical and chemical properties as we go down in a group or move across a period.
• For example : The chemical reactivity is high at the beginning, lower at the middle and increases to a maximum at group 17 in a period.
• The reactivity increases on moving down the group of alkali metals. But the reactivity decreases on moving down the group of halogens.
• Atomic radii increases down the group and decreases across the period.

Question 18.
What is covalent radius? How would you determine the covalent radius of chlorine atom?
Answer:
The distance between the nuclei of two covalent bonded atoms is known as covaLent distance or inter-nuclear distance. The one – half of this inter-nuclear distance is called covalent radius. The covalent distance (Cl – Cl) of Cl₂ molecule is experimentally found as 198 pm (1.98 A). Its covalent radius is 99 pm (0.99 Å).
Cl – Cl Inter nuclear distance = 1.98 Å
∴ r_Cl = 1.98 / 2=0.99 Å.

 

Question 19.
Write a note about metallic radius.
Answer:

• It is defined as one half of the distance between the centers of nuclei of the two adjacent atoms in the metallic crystal.
• The metallic radius is always larger than its covalent radius.
• The distance between two adjacent copper atoms in solid copper is 2.56 A. Hence, the metallic radius of copper is 1.28 A.

Question 20.
Arrange Na⁺, Mg²⁺ and Al³⁺ in the increasing order of ionic radii. Give reason.
Answer:
Na⁺, Mg²⁺ and Al³⁺ are iso electronic cations.

The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is,
r_Na⁺ > r_Mg²⁺ > r_Al³⁺

Question 21.
Arrange the ions F^–, O^2- and N^3- ¡n the increasing order of their ionic radii. Give reason.
Answer:
F^–, O^2- and N^3- are isoelectronic species.

The anion with the greater negative charge will have a larger radius because of the lesser attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is,
r_Na^3– > r_O^2– > r_F^–

Question 22.
Mention some characteristics of ionization energy.
Answer:

• Ionization is always an endothermic process. It absorbs energy.
• Ionization energies always increase in the order, I.E₁< I.E₂<IE₃.
• Ionization energy measurements are carried out with atoms in the gaseous state.

Question 23.
Why ionization energy and electron affinity are calculated in gaseous state?
Answer:

• Inter molecular force can affect the value of ionization energy and electron affinity.
• In gaseous state, there is little inter molecular force in a substance and it can be considered negligible In some cases. So the value of I.E and E.A are almost unaffected if they are calculated for gaseous atoms.
• When we arc talking about ionization energy and electron affinity, we need to consider
  atoms and we can find free atoms only when the substance is in gaseous state.

Question 24.
How is screening effect of inner electrons affect ionization energy?
Answer:

• The electrons of inner shells form a cloud of negative charge. This cloud shields the outer electron from the nucleus and also acts as a screen between the nucleus and the outer electrons.
• This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. Therefore, the outer electrons do not feel the full effect of the positive charge of the nucleus.
• If screening effect increases, ionization energy decreases.

 

Question 24.
How is screening effect of inner electrons affect ionization energy?
Answer:

• The electrons of inner shells form a cloud of negative charge. This cloud shields the outer electron from the nucleus and also acts as a screen between the nucleus and the outer electrons.
• This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. Therefore, the outer electrons do not feel the full effect of the positive charge of the nucleus.
• If screening effect increases, ionization energy decreases.

Question 25.
Ionization energy of Mg is greater than that of Al. Why?
Answer:
Mg (Z = 12) 1s² 2s² 2p⁶ 3s².
Al (Z = 13) 1s² 2s² 2p⁶ 3s² 3p¹.
Although the nuclear charge of aluminium is greater than that of magnesium, I.E of Mg is greater than that of Al. It is because Mg atom has more stable configuration than Al atom. IE₁ of Mg > IE₁ of Al.

Question 26.
What are all the factors that influences electron gain enthalpy?
Answer:
1. Size of the atom:
The new electron which was added experiences stronger attraction to its nucleus if the atoms are smaller in size.
Atomic size α \(\frac {1}{ Electron gain enthalpy }\)

2. Nuclear charge:
The new electron which was added experiences stronger attraction to its nucleus if the atom possess greater nuclear charge. Nuclear charge α Electron gain enthalpy

3. Electronic configuration:
An atom with stable electronic configuration has no tendency to gain an electron. Such atoms have zero or almost zero electron gain enthalpy.

Question 27.
Explain about the periodic variation of electron gain enthalpy in a period and in a group.
Answer:
1. The electron gain enthalpy increases as we move from left to right in a period due to the increase of nuclear charge. However, Be, Mg, N and noble gases have almost zero value of electron gain enthalpy due to extra stability of completely and half filled orbitais.

2. When we move in a group of periodic table, the size and nuclear charge increase. But the effect of increase in atomic size is much more pronounced than that of nuclear charge and thus the additional electron feels less attraction by the large atom. Consequently, electron gain enthalpy decreases.

 

Question 28.
Explain about the electro negativity and non-metallic character across the period and down the group.
Answer:
Eletro negativity α Non-metallic character:

• As the electro negativity is directly proportional to the non-metallic character, thus across the period, with an increase in electro negativity. the non-metallic character also increases.
• As we move down the group. decrease in electro negativity is accompanied by a decrease in non-metallic character.

Question 29.
Prove that valency is a periodic property.
Answer:
Variation in period:
The number of valence electrons increases from I to 8 on moving across a period. The valency of the elements with respect to hydrogen and chlorine increases from 1 to 4 and then decreases from 4 to zero.

Variation in group:
On moving down a group, the number of valence electrons remains same. All the elements in a group exhibit same valency. For example, all the elements of group I have valency equal to 1. Hence. valency is a periodic property.

Question 30.
Write a note about periodic trends and chemical reactivity.
Answer:

• The group 1 elements are extremely reactive because these elements can lose one electron to form cation. Their ionization enthalpy is also least.
• The high reactivity of halogens is due to the ease with which these elements can gain an electron to form anion. Their electron gain enthalpy is most negative.
• The elements at the extreme left (alkali) exhibit strong reducing behavior, whereas the elements at the extreme right (halogens) exhibit strong oxidizing behaviour.
• The reactivity of elements at the center of the periodic table becomes low when compared with extreme right and left.

Question 31.
How would you explain the fact that the first ionization enthalpy of sodium ¡s lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer:
Electronic configuration of Na and Mg are
Na = 1s² 2s² 2p⁶ 3s¹
Mg = 1s² 2s² 2p⁶ 3s²
First electron in both cases has to be removed from 3s-orbital but the nuclear charge of Na (+11) is lower than that of Mg (+12) therefore first ionization energy of sodium is lower than that of magnesium.

After the loss of first electron, the electronic configuration of
Na = 1s² 2s² 2p⁶
Mg = 1s²2s²2p⁶3s¹
Here electron is to be removed from inert (neon) gas configuration which is very stable and hence removal of second electron requires more energy when compared to Mg. Therefore, second ionization enthalpy of sodium is higher than that of magnesium.

 

Question 32.
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down the group?
Answer:
Atomic size:
With the increase in atomic size, the number of electron shells increase. Therefore, the force that binds the electrons with the nucleus decreases. The ionization enthalpy thus decreases with the increase in atomic size.

Screening or shielding effect of inner shell electron:
With the addition of new shells, the number of inner electron shells which shield the valence electrons increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionization enthalpy decreases.

Question 33.
Which of the following pairs of elements would have more negative electron gain enthalpy?

1. O or F
2. F or Cl.

Answer:
1. O or F. Both O and F lie in 2e” period. As we move from O to F the atomic size decreases. Due to smaller size off nuclear charge increases. Further, gain of one electron by
F → F^–
F^– ion has inert gas configuration, While the gain of one electron by
O → O^–
gives O ion which does not have stable inert gas configuration. Consequently, the energy released is much higher in going from
F → F^–
than going from O → O^–. In other words electron gain enthalpy off is much more negative than that of oxygen. –

2. The negative electron gain enthalpy of Cl (e.g. ∆H = – 349 mol^-1) is more than that of F (e.g. ∆H = -328 U mol^-1).

The reason for the deviation is due to the smaller size off. Due to its small size, the electron repulsion in the relatively compact 2p-subshell are comparatively large and hence the attraction for incoming electron is less as in the case of Cl.

Question 34.
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
For oxygen atom:
O_(g) + e^– → O^-1_(g) (e.g. ∆H = -141 Id mor^-1)
O^-1_(g) + e^– → O^2-_(g) (e.g. ∆H = + 780 kJ mol^-1)
The first electron gain enthalpy of oxygen is negative because energy is released when a gaseous atom accepts an electron to form monovalent anion. The second electron gain enthalpy is positive because energy is needed to overcome the force of repulsion between monovalent anion and second incoming electron.

Question 35.
What are major differences between metals and non-metals?
Metals:

• Have a strong tendency to lose electrons to form cations.
• Metals are strong reducing agents.
• Metals have low ionization enthalpies.
• Metals form basic oxides and ionic compounds.

Non-Metals:

• Non-metals have a strong tendency to accept electrons to form anions.
• Non-metals are strong oxidizing agents.
• Non-metals have high ionization enthalpies.
• Non-metals form acidic oxides and covalent compounds.

Question 36.
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb Cl> Br>. Explain.
Answer:
The elements of group 1 have only one electron in their respective valence shells and thus have a strong tendency to lose this electron. The tendency to lose electrons in turn, depends upon the ionization enthalpy. Since the ionization enthalpy decreases down the group therefore, the reactivity of group 1 elements increases in the same order Li

 

Question 37.
Arrange the following as stated:

1. N₂ O₂, F₂, Cl₂ (Increasing order of bond dissociation energy)
2. F, Cl, Br, I (Increasing order of electron gain enthalpy)
3. F₂, N₂, Cl₂, O₂ (Ipcreasing order of bond length).

Answer:

1. F₂, N₂, Cl₂, O₂
2. I< Br < F < Cl
3. N₂ O₂, F₂, Cl₂

Question 38.
The first ionization enthalpy of magnesium is higher than that of sodium. On the other hand, the second ionization enthalpy of sodium is very much higher than that of magnesium. Explain.
Answer:
The 1^st ionization enthalpy of magnesium is higher than that of Na⁺ due to higher nuclear charge and slightly smaller atomic radius of Mg than Na. After the loss of first electron, N& formed has the electronic configuration of neon (2, 8). The higher stability of the completely filled noble gas configuration leads to very high second ionization enthalpy for sodium. On the other hand. Mg⁺ formed after losing first electron still has one more electron in its outermost (3s) orbital. As a result, the second ionization enthalpy of magnesium is much smaller than that of sodium.

Question 39.
Give reasons:

1. lE₁ of sodium is lower than that of magnesium whereas IE₂ of sodium is higher than that of magnesium.
2. Noble gases have positive value of electron gain enthalpy.

Answer:
1. The effective nuclear charge of magnesium is higher than that of sodium. For these reasons, the energy required to remove an electron from magnesium is more than the energy required in sodium. Hence, the first ionization enthalpy of sodium is lower than that of magnesium.

2. Noble gases have completely filled electronic configuration and they are more stable. So in Noble gases addition of electron is not possible. Electron gain enthalpy is always the amount of energy released (-ve sign) when an electron is added to an atom. – Butin noble gases, if an electron is added, they have positive value of electron gain enthalpy.

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements 5-Mark Questions

Question 1.
(a) State Mendeleev’s periodic law.
(b) Describe about the merits of Mendeleev’s periodic table.
Answer:
(a) Mendeleev’s periodic law:
Mendeleev’s periodic law states that the physical and chemical properties of elements are a periodic function of their atomic weights.

(b) Merits of Mendeleev’s periodic table:

• The comparative studies of elements were made easier.
• The table shóws the relationship in properties of elements in a group.
• The table helped to correct the atomic masses of some elements later on, At the time of Mendeleev, the atomic weight of Au and Pt were known as 196.2 and 196.7 respectively. However, Mendeleev placed Au (196.2) after Pt (196.7) saying that atomic weight of Au is incorrect, which was later on found to be 197.
• At the time of MendeLeev, about 70 elements were known and thus blank spaces were left for unknown elements which helped further discoveries.
• Both Gallium (Ga) in III group and Germanium (Ge) in IV group, were unknown at that time by Mendeleev predicted their existence and properties. He referred the predicted elements as eka-aluminium and eka-silicon. After discovery of the actual elements, their properties were found to match closely to those predicted by Mendeleev.

Question 2.
Explain about the anomalies of Mendeleev’s periodic table. Anomalies of Mendeleev’s periodic table
Answer:

1. Some elements with similar properties were placed in different groups whereas some elements having dissimilar properties were placed in same group, but iodine (127) was placed in VII group.
   Example: Tellurium (127.6) was placed in VI group.
2. Some elements with higher atomic weights were placed before lower atomic masses in order to maintain the similar chemical nature of elements. This concept was called inverted pair of elements concept.
   Example : ⁵⁹₂₇CO and ^58.7₂₈Ni
3. Isotopes did not find any place in Mendeleev’s periodic table.
4. Position of hydrogen could not be made clear.
5. He did not leave any space for lanthanides and actinides which were discovered later on.
6. Elements with different nature were placed in one group,
   Example: Alkali metals and coinage metals were placed together.
7. Diagonal and horizontal relationships were not explained.

Question 3.
Explain about the structural features of Moseley’s long form of periodic table.
Answer:

• The long form of periodic table of the elements is constructed on the basis of modem periodic law. The arrangement resulted in repeating electronic configurations of atoms at regular intervals.
• The elements placed in horizontal rows are called periods and in vertical columns are called groups.
• According to IUPAC, the groups are numbered from I to 18.
• There are 18 vertical columns which constitute 18 groups or families. All the members of a particular group have similar outer shell electronic configuration.
• There are 7 horizontal rows called periods.
  
  The elements are shown in the above table along with its atomic numbe.
• The atomic number also indicates the number of electrons in the atoms of an element.
• This periodic table is important and useful because we can predict the properties of any element using periodic trend, even though that element may be unfamiliar to us.

Question 4.
Explain the merits of Moseley’s long form of periodic table.
Answer:
Merits of Moseley’s long form of periodic table:

•  As this classification is based on atomic number, it relates the position of an element to its electronic configuration.
• The elements having similar electronic configuration fall in a group. They also have similar physical and chemical properties.
• The completion of each periõd is more logical. In a period as the atomic number increases, the energy shells are gradually filled up until an inert gas configuration is reached.
• The position of zero group is also justified in the table as group 18.
• The table completely separates metals and non-metals.
• The table separates two sub groups. lanthanides and actinides, dissimilar elements do not fall together.
• The greatest advantage of this periodic table is that this can be divided into four blocks namely s, p. d and f-block elements.
• This arrangement of elements is easier to remember, understand and reproduce.

 

Question 5.
Explain about the general characteristics of periods.
Answer:
1. Number of electrons in outermost shell:
The number of electrons present in the outermost shell increases from 1 to 8 as we proceed in a period.

2. Number of shells:
As we move from left to right in a period the shells remains the same. The number of shells present in the elements corresponds to the period number. For example : all the elements of 2 period have on 2 shells (K, L)

3. Valency:
The valency of the elements increases from left to right in a period. With respect to hydrogen. the valency of period elements increases from 1 to 4 and then falls to one. With respect to oxygen, the valency increases from1 to 7.

4. Metallic character:
The metallic character of the elements decreases across a period.
For example:

Question 6.
Explain about the salient features of groups.
Answer:
1. Number of electrons in outermost shell:
The number of electrons present in the outermost shells does not change on moving down in a group, i.e remains the same. Hence, the valency also remains same within a group.

2. Number of shells:
In going down a group the number of shells increases by one at each step and ultimately becomes equal to the period number to which the element belongs.

3. Valency:
The valencies of all the elements of the same group are the same. The valency of an element with respect to oxygen is same in a group.

4. Metallic character:
The metallic character of the elements increases in moving from top to bottom in a group.

Question 7.
Explain the classification of elements based on chemical behaviour and on physical properties.
Answer:

Based of chemical behavior:

1. Main group elements:
   All s-block and p-block elements excluding group elements are called representative elements.
2. Noble gases:
   The 18th group elements are exclusively called noble gases. They have completely filled electronic configuration as ns² np⁶. These elements are highly stable.
3. Transition elements:
   The elements of d-block are called transition elements. These include elements of groups from 3 to 12 lying between s-block and p-block elements.
4. Inner transition elements.
   The elements of f-block are called inner-transition elements. These consist of lanthanides and actinides, with 14 elements in each.

Based of physical properties:

1. Metals:
   Metals comprise more than 78% of all known elements. They are usually solids at room temperature (except Hg, Ga and Cs). They have high melting and boiling points. They are good conductors of heat and electricity.
2. Non-metals:
   Non-metals are usually solids or liquids or gases at room temperature with low melting and low boiling points (except boron and carbon). They are poor conductors of heat and electricity. Most of the non-metallic solids are brittle and are neither malleable nor ductile.
3.  Metalloids or Semi-metals:
   Some elements in the periodic tables show properties that are characteristic of both metals and non-metals. They are called metalloids. Example : Silicon, germanium, arsenic, antimony and tellurium.

 

Question 8.
(a) Define atomic radius.
(b) What are the difficulties in determining atomic radius?
Answer:
(a) Atomic radius is the distance between the center of its nucleus and the outermost shell
containing the electron.
(b) Difficulties in determining atomic radius

• The size of an atom is very small (∼ 1.2Å i.e 1.2 × 10¹⁰)
• The atom is not a rigid sphere; it is more like a spherical cotton ball rather than like a cricket ball.
•  It is not possible to isolate an atom and measure its radius.
• The size of an atom depends upon the type of atoms in its neighborhood and also the nature of bonding between them.

Question 9.
Prove that the atomic radii is a periodic property.
Answer:
Atomic radius is the distance between the center of its nucleus and the outermost shell containing the electron.

Atomic radius is a periodic property.
1. Variation in periods:
The atomic radius decreases while going from left to right in a period. As we move from left to right in a period, the nuclear charge increases by one unit in each succeeding element. But the number of the shell remains same. Hence, the electrons are attracted strongly by the nucleus. Hence the atomic radius decreases along the period. In 2nd period r_Li>r_Be>r_B>r_C>r_N>r_O>r_F

2. Variation in a group:
The atomic radius of elements increases with increase in atomic number as we move from top to bottom in a group. The attraction of the nucleus for the electrons decreases as shell number increases. Hence atomic radius increases along the group. In 1 group r_Li < r_Na < r_K <r_Rb < r_Cs
Hence, atomic radii is a periodic property.

Question 10.
Explain about the factors that influence the ionization enthalpy. Factors influencing ionization enthalpy:
Answer:
1. Size of the atom:
If the size of an atom is larger, the outermost electron shell from the nucleus is also larger and hence the outermost electrons experience lesser force of attraction. Hence it would be more easy to remove an electron from the outermost shell. Thus, ionization energy decreases with increasing atomic sizes.

2. Magnitude of nuclear charge:
As the nuclear charge increases, the force of attraction between the nucleus and valence electrons also increases. So, more energy is required to remove a valence electron. Hence I.E increases with increase in nuclear charge.
Ionization enthalpy α nuclear charge

3. Screening or shielding effect of the inner electrons:
The electrons of inner shells form a cloud of negative charge and this shields the outer electron from the nucleus. This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. 1f screening effect increases, ionization energy decreases.

4. Penetrating power of sub shells s, p, d and f:
The s-orbital penetrate more closely to the nucleus as compared to p-orbitais. Thus, electrons in s-orbitals are more tightly held by the nucleus than electrons in p-orbitais. Due to this, more energy is required to remove a electron from an s-orbital as compared to a p-orbital. For the same value of ‘n’, the penetration power decreases in a given shell in the order.
s > p > d > f.

5. Electronic configuration:
If the atoms of elements have either completely filled or exactly half filled electronic configuration, then the ionization energy increases.

 

Question 11.
Define ionization energy.
Prove that ionization energy is a periodic property.
Answer:
(a) The energy required to remove the most loosely held electron from an isolated gaseous atom is called as ionization energy.
(b) (i) Variation in a period:
On moving across a period from left to right, the ionization enthalpy value increases. This is due to the following reasons:

• Increase of nuclear charge in a period
• Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus, thus ionization enthalpy increases. Hence, ionization energy is a periodic property.

(ii) Variation in a group:
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons:

• A gradual increase in atomic size
• Increase of screening effect on the outermost electrons due to the increase of number of inner electrons.
  Hence, ionization enthalpy is a periodic property.

Question 12.
Distinguish between electron affinity and electron negativity.
Answer:
Electron affinity:

• It is the tendency of an isolated gaseous atom to gain an electron.
• Ills the property of an isolated atom.
• It does not change regularly in a period or a group.
• It is measured in electron volts/atom or kcal/mole or kJ/mole.

Electron negativity:

• It is the tendency of an atom in a molecule to attract the shared pair of electrons.
• It is the property of bonded atom.
• It changes regularly in a period or a group.
• It is a number and has no units.

Question 13.
What are the anomalous properties of second period elements?
Answer:

1. In the Pt group, lithium differs in many aspects from its own family elements. Similarly, in the 2rd group, beryllium differs in many aspects from its own family.
2. For example. lithium forms compounds with more covalent character. But other alkali metals of this group form only ionic compounds.
3. Similarly, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.
4. Lithium and beryllium resemble more with the elements lying at their right hand side in the 3t1 period than with the other members of their own family.
5. These kinds of anomalies are also observed from 13th to 17th groups.
6. This sort of similarity is commonly referred to as diagonal relationship in the periodic properties.
7. The anomalous behaviors are attributed to the following factors:
   • Smaller atomic size
   • Higher ionization enthalpy
   • High electronegativity

Activity 3.1

Covalent radii (in Å) for sonic elements of different groups and periods are listed below. Plot these values against atomic number. From the plot, explain the variation along a period and a group.
2^nd group elements : Be (0.89), Mg (1.36), Ca (1.74), Sr (1.91) Ba( 1.98)
17^th group elements : F (0.72), Cl (0.99), Br (l.14),I (1.33)
3^nd Period elements : Na (1.57), Mg(1.36),AI (1.25), Si (1.17), P(1.10), S (1.04), Cl (0.99)
4^thperiod elements : K (2.03), Ca (1.74), Sc (l.44), Ti(1.32), V (1.22), Cr (1.17), Mn (1.17), Fe( 1.17), CO (1.16), Ni (1.15), Cu (1.17), Zn(1.25), Ga(1.25), Ge(1.22), As(1.21), Se(1.14), Br( 1. 14)
Solution:
2^nd group elements:

17^th group elements:

As we move down the group, atomic radii increases with the increase in atomic number . As we move down the group, the number energy levels increases, as the number of electrons Periodic.

 

17^th group elements: F (4.i), Cl (3.0), Br (2.8), I (2.5)
3^rd Period elements : Na(0.9), Mg(l.2), Al (1.5), Si(l.8), P(2.1), S(2.5), Cl(3.0)
4^th period elements : K(O.8), Ca (1.0), Sc (1.3), Ti (l.5), V(1.6), Cr(1.6), Mn(1.5), Fe(l.8), CO(1.9), Ni(1.9), Cu(1.9), Zn(1.6), Ga(1.6), Ge(1.8), As(2.0), Se(2.4), Br(2.8)
Solution:
2^nd group elements:

17^th group elements:

As we go down the group, the electro negativity value decreases. Moving down the group, the electro negativity decreases due to the longer distance between the nucleus and the valence electron shell thereby decreasing the attraction making the atom have less attraction for electrons or protons.
3^rd period:

4^th period:

The positively charged protons in the nucleus attract the negatively charged electrons. As the number of protons in the nucleus increases, the electro negativity or attraction will increase. Therefore electro negativity increases from left to right across the period. This occurs due to the greater charge on the nucleus, causing the electron bonding pairs to be very attracted to atoms placed further right on the periodic table.

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