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You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.4

11th Maths Exercise 3.4 Samacheer Kalvi Question 1.

Solution:

11th Maths Exercise 3.4 Question 2.

Solution:

11th Maths Exercise 3.4 Answers Question 3.

Solution:


since x is in III quadrant
Both sin x and cos x are negative

11 Maths Samacheer Solution Question 4.

Solution:

Class 11 Maths Ex 3.4 Solutions Question 5.

Solution:

Samacheer Kalvi 11 Maths Solutions Question 6.

Solution:

(ii) L.HS = cos(π + θ) = cos(180° + θ)
= cos 180° cos θ – sin 180° sin θ
= (-1) cos θ – (0) sin θ
= – cos θ = RHS

(iii) LHS = sin (π + θ) = sin π cos θ + cos π sin θ
= (0) cos θ + (-1) sin θ
= -sin θ = RHS

Samacheer Kalvi Guru 11th Maths Solution Question 7.
Find a quadratic equation whose roots are sin 15° and cos 15°
Solution:

11th Maths Samacheer Solutions Question 8.
Expand cos(A + B + C). Hence prove that cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B, if A + B + C = \(\frac{\pi}{2}\)
Solution:
Taking A + B = X and C = Y
We get cos (X + Y) = cos X cos Y – sin X sin Y
(i.e) cos (A + B + C) = cos (A + B) cos C – sin (A + B) sin C
= (cos A cos B – sin A sin B) cos C – [sin A cos B + cos A sin B] sin C
cos (A + B + C) = cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C If (A + B + C) = π/2 then cos (A + B + C) = 0
⇒ cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C = 0
⇒ cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B

Trigonometry 3.4 Class 11 Question 9.
Prove that
(i) sin(45° + θ) – sin(45° – θ) = \(\sqrt{2}\)sin θ.
(ii) sin(30° + θ) + cos(60° + θ) = cos θ.
Solution:

Samacheer Kalvi Guru 11th Maths Question 10.
If a cos(x + y) = b cos(x – y), show that (a + b) tan x = (a – b) cot y.
Solution:
a cos (x + y) = b cos (x – y)
a[cos x cos y – sin x sin y] = 6[cos x cos y + sin x sin y]
(i.e) a cos x cos y – a sin x sin y = b cos x cos y + b sin x sin y
a cos x cos y – b sin x sin y = a sin x sin y + b cos x cos y

⇒ a cot y – b tan x = a tan x + b cot y
a cot y – b cot y = a tan x + b tan x
⇒ (a + b) tan x = (a – b) cot y.

Samacheer Kalvi Class 11 Maths Solutions Question 11.
Prove that sin 105° + cos 105° = cos 45°.
Solution:
sin 105° = sin (60°+ 45°)
= sin 60° cos 45° + cos 60° cos 45°

= cos 45° = RHS

Samacheer Kalvi Maths 11th Question 12.
Prove that sin 75° – sin 15° = cos 105° + cos 15°.
Solution:
RHS = cos 105° + cos 15° = cos (90° + 15°) + cos (90° – 75°)
= – sin 15° + sin 75°
= sin 75° – sin 15° = LHS

Samacheer Kalvi 11th Maths Solutions Question 13.
Show that tan 75° + cot 75° = 4
Solution:

Samacheer Kalvi Guru 11 Maths Question 14.
Prove that cos(A + B) cos C – cos(B + C) cos A = sin B sin(C – A).
Solution:
LHS = (cos A cos B – sin A sin B) cos C – (cos B cos C – sin B sin C) cos A
= cos A cos B cos C – sin A sin B cos C – cos A cos B cos C + cos A sin B sin C
= cos A sin B sin C – sin A sin B cos C
= sin B [sin C cos A – cos C sin A]
= sin B [sin (C – A)] = RHS

Samacheerkalvi.Guru 11th Maths Question 15.
Prove that sin(n + 1) θ sin(n – 1) θ + cos(n + 1) θ cos(n – 1)θ = cos 2θ, n ∈ Z.
Solution:
Taking (n + 1) θ = A and (n – 1) θ = B
LHS = sin A sin B + cos A cos B
= cos (A – B)
= cos[(n + 1) – (n – 1)]θ
= cos (n + 1 – n + 1)θ = cos 2θ = RHS

Samacheer Kalvi.Guru 11th Maths Question 16.

Solution:

Samacheer Kalvi 11th Maths Question 17.
Prove that
(i) sin(A + B) sin(A – B) = sin² A – sin² B
(ii) cos(A + B) cos(A – B) = cos² A – sin² B = cos² B – sin² A
(iii) sin²(A + B) – sin²(A – B) = sin2A sin2B
(iv) cos 8θ cos 2θ = cos² 5θ – sin² 3θ
Solution:
(i) LHS = sin (A + B) sin (A – B)
= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)
= sin²A cos² B – cos² A sin² B
= sin² A (1 – sin² B) – (1 – sin² A) sin² B
= sin² A – sin² A sin² B – sin² B + sin² A sin² B
= sin² A – sin² B = RHS

(ii) LHS = cos (A + B) cos (A – B) = (cos A cos B – sin A sin B) (cos A cos B + sin (A sin B)
= cos² A cos² B – sin² A sin² B
= cos² A (1 – sin² B) – (1 – cos² A) sin² B
= cos² A – cos² A sin² B – sin² B + cos² A sin² B
= cos² A – sin² B = RHS
Now cos² A – sin² B = (1 – sin² A) – (1 – cos² B)
= 1 – sin² A – 1 + cos² B
= cos² B – sin² A

(iii) sin² A – sin² B = sin (A + B) sin (A – B)
LHS = sin² (A + B) – sin² (A – B) = sin [(A + B) + (A – B)] [sin (A + B) – (A – B)]
= sin 2A sin 2B = RHS

(iv) LHS = cos 8θ cos 2θ
= cos (5θ + 3θ) cos (5θ – 3θ) .
We know cos (A + B) cos (A – B) = cos² A – sin² B
∴ cos (5θ + 3θ) cos (5θ – 3θ) = cos² 5θ – sin² 3θ = RHS

Samacheer Kalvi 11th Maths Book Solutions Question 18.
Show that cos² A + cos² B – 2 cos A cos B cos(A + B) = sin²(A + B).
Solution:
LHS = cos² A + cos² B – 2 cos A cos B [cos A cos B – sin A sin B]
= cos² A + cos² B – 2 cos² A cos² B + 2 sin A cos A sin B cos B
= (cos² A – cos² A cos² B) + (cos² B – cos² A cos² B) + 2 sin A cos A sin B cos B
= cos² A (1 – cos² B) + cos² B (1 – cos² A) + 2 sin A cos A sin B cos B
= cos² A sin² B + cos² B sin² A + 2 sin A cos B sin B cos A
= (sin A cos B + cos A sin B)²
= sin² (A + B) = RHS

Samacheer Kalvi 11th Maths Solution Question 19.
If cos(α – β) + cos(β – γ) + cos(γ – α) = \(-\frac{3}{2}\), then prove that cos α + cos β + cos γ = sin α + sin β + sin γ
Solution:

2 cos (α – β) + 2cos (β – γ) + 2cos (γ – α) = -3
2cos(α – β) + 2cos(β – γ) + 2cos (γ – α) + 3 = 0
[2 cos α cos β + 2 sin α sin β] + [2 cos β cos γ + 2 sin β sin γ] + [2 cos γ cos α + sin γ sin α] + 3 = 0
= [2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α] + [2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α] + (sin² α + cos² α) + (sin² β + cos² β) + (sin² γ + cos² γ) = 0
⇒ (cos² α + cos² β + cos² γ + 2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α) + (sin² α + sin² β) + (sin² γ + 2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α) = 0
(cos α + cos β + cos γ)² + (sin α + sin β + sin γ)² = 0
=(cos α + cos β + cos γ) = 0 and sin α + sin β + sin γ = 0
Hence proved

Samacheer 11 Maths Solutions Question 20.

Solution:

Samacheer Kalvi 11th Maths Book Question 21.

Solution:

11th Maths Solutions Samacheer Kalvi Question 22.

Solution:

11th Samacheer Kalvi Maths Question 23.

Solution:

Question 24.

Solution:

Maths Samacheer Kalvi 11th Question 25.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.4 Additinal Questions

10th Maths Exercise 3.4 Samacheer Kalvi Question 1.
Prove that sin (A + B) sin (A – B) = cos² B – cos² A
Solution:
LHS = sin (A + B) sin (A – B)
= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)
= sin² A cos² B – cos² A sin² B
= (1 – cos² A) cos² B – cos² A (1 – cos² B)
= cos² B – cos² A cos² B – cos² A + cos² A cos² B
= cos² B – cos² A = RHS

Question 2.
Prove that
(i) sin A + sin(120° + A) + sin (240° + A) = 0
(ii) cos A + cos (120° +A) + cos (120° – A) = 0
Solution:
(i) sin A + sin(120° + A) + sin (240° + A)
= sin A+ sin 120° cos A + cos 120° sin A + sin 240° cos A + cos 240° sin A …… (1)

By substituting these values in (1), we get,

(ii) cos 120° = cos (180° – 60°) = – cos 60° = \(\frac{-1}{2}\)
LHS = cos A + cos (120° + A) + cos (120° – A)
= cos A + cos 120° cos A – sin 120° sin A + cos 120° cos A + sin 120° sin A
= cos A + 2 cos 120° cos A

Question 3.

Solution:

tan (A + B) = 1 ⇒ A+ B = 45°

Question 4.
If A + B = 45°, show that (1 + tan A) (1 + tan B) = 2 and hence deduce the value of tan \(22 \frac{1^{0}}{2}\).
Solution:
Given A + B = 45° ⇒ tan (A + B) = tan 45°

(i.e.) tan A + tan B = 1 – tan A.tan B
(i.e.) 1 + tan A + tan B = 2 – tan A tan B (add 1 on both sides)
1 + tan A + tan B + tan A tan B = 2
(i.e.) (1 + tan A) (1 + tan B) = 2

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Tamil Book Solutions Guide Pdf Chapter 3.1 நோயும் மருந்தும் Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th Tamil Solutions Chapter 3.1 நோயும் மருந்தும்

கற்பவை கற்றபின்

Question 1.
ஐம்பெருங்காப்பியங்கள், ஐஞ்சிறுகாப்பியங்கள் ஆகியவற்றின் பெயர்களைத் தொகுத்து எழுதுக.
Answer:

மாடநூல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
உடல்நலம் என்பது ……………………… இல்லாமல் வாழ்தல் ஆகும்.
அ) அணி
ஆ) பணி
இ) பிணி
ஈ) மணி
Answer:
இ) பிணி

Question 2.
நீலகேசி கூறும் நோயின் வகைகள் ……………………
அ) இரண்டு
ஆ) மூன்று
இ) நான்கு
ஈ) ஐந்து
Answer:
ஆ) மூன்று

Question 3.
‘இவையுண்டார்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது_ ………………………
அ) இ + யுண்டார்
ஆ) இவ் + உண்டார்
இ) இவை + உண்டார்
ஈ) இவை + யுண்டார்
Answer:
இ) இவை + உண்டார்

Question 4.
தாம் + இனி என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ……………………
அ) தாம் இனி
ஆ) தாம்மினி
இ) தாமினி
ஈ) தாமனி
Answer:
இ) தாமினி

குறுவினா

Question 1.
நோயின் மூன்று வகைகள் யாவை?
Answer:

• மருந்தினால் நீங்கும் நோய்.
• எதனாலும் தீராத தன்மையுடைய நோய் மற்றொரு வகை.
• வெளியில் ஆறி உள்ளுக்குள் இருந்து துன்பம் தரும் நோய்.

Question 2.
நீலகேசியில் பிறவித் துன்பத்தைத் தீர்க்கும் மருந்துகளாகக் கூறப்படுவன யாவை?
Answer:
நல்லறிவு, நற்காட்சி, நல்லொழுக்கம் என்பவையே பிறவித் துன்பத்தைத் தீர்க்கும் மருந்துகளாக நீலகேசி கூறுகின்றது.

சிறு வினா

Question 1.
நோயின் வகைகள் அவற்றைத் தீர்க்கும் வழிகள் பற்றி நீலகேசி கூறுவன யாவை?
Answer:

• ஒளிபொருந்திய அணிகலன்களை அணிந்த பெண்ணே! நோயின் தன்மை பற்றி யார் வினவினாலும் அது மூன்று வகைப்படும் என அறிவாயாக.
• மருந்தினால் நீங்கும் நோய்.
• எதனாலும் தீராத தன்மையுடைய நோய் மற்றொரு வகை.
• வெளியில் ஆறி உள்ளுக்குள் இருந்து துன்பம் தரும் நோய்.
• அகற்றுவதற்கு அரியவை பிறவித் துன்பங்கள் ஆகும்.
• இவற்றைத் தீர்க்கும் மருந்துகள் மூன்று. நல்லறிவு, நற்காட்சி, நல்லொழுக்கம் என்பவையே அம்மருந்துகள்.
• இவற்றை ஏற்றோர் பிறவித்துன்பத்திலிருந்து நீங்கி உயரிய இன்பத்தை அடைவர்.

சிந்தனை வினா

Question 1.
துன்பமின்றி வாழ நாம் கைக்கொள்ள வேண்டிய நற்பண்புகள் யாவை?
Answer:
தருமம் செய்தல், கோபத்தைத் தணித்தல், முயற்சி செய்தல், கல்வி கற்றல், உலக நடையை அறிந்து நடத்தல், நல்ல நூல்களைப் படித்தல், பொறாமை படாமல் இருத்தல், பொய்சாட்சி சொல்லாமல் இருத்தல், இனிமையாகப் பேசுதல், பேராசையைத் தவிர்த்தல், நட்புடன் பழகுதல், பெரியோர்களை மதித்தல், ஒழுக்கம் தவறாமல் இருத்தல், நன்றியை மறவாமல் இருத்தல், காலத்தைக் கடைபிடித்தல், களவு செய்யாதிருத்தல், இழிவானதைச் செய்யாதிருத்தல், இரக்கம் கொள்ளுதல், பொய் சொல்லாதிருத்தல், ஆணவம் கொள்ளாதிருத்தல், சுறுசுறுப்புடன் இருத்தல், உடற்பயிற்சி செய்தல், அதிகாலையில் எழுந்திருத்தல் போன்றவை கைக்கொள்ள வேண்டிய நற்பண்புகள் ஆகும்.

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
மக்களின் உடலுக்கும் உள்ளத்திற்கும் துன்பம் தருவன …………………….
அ) நாய்கள்
ஆ) நோய்கள்
இ) பேய்கள்
ஈ) மனிதர்கள்
Answer:
ஆ) நோய்கள்

Question 2.
உள்ளத்தில் தோன்றும் தீய எண்ணங்களால் ஏற்படும் துன்பங்களையும் ……………………… என்றே நம் முன்னோர்கள் குறிப்பிட்டனர்.
அ) கவலை
ஆ) துன்பம்
இ) நோய்கள்
ஈ) பொறுமை
Answer:
இ) நோய்கள்

Question 3.
நோய்களை நீக்கும் மருந்துகளாக விளங்கும் அறக்கருத்துகளை விளக்குபவை …………………..
அ) இலக்கியங்கள்
ஆ) இலக்கணங்கள்
இ) படைப்புகள்
ஈ) முன்னோர்கள்
Answer:
அ) இலக்கியங்கள்

Question 4.
நோயைத் தீர்க்கும் மருந்துகள் ………………….
அ) இரண்டு
ஆ) மூன்று
இ) நான்கு
ஈ) ஐந்து
Answer:
ஆ) மூன்று

Question 5.
………………… ஐஞ்சிறு காப்பியங்களுள் ஒன்று.
அ) சிலப்பதிகாரம்
ஆ) நீலகேசி
இ) குண்டலகேசி
ஈ) வளையாபதி
Answer:
ஆ) நீலகேசி

Question 6.
நீலகேசி ………………… சமயக் கருத்துகளைக் கூறுகிறது.
அ) சமணம்
ஆ) புத்தம்
இ) கிறித்தவம்
ஈ) இந்து
Answer:
அ) சமணம்

Question 7.
நீலகேசி, கடவுள் வாழ்த்து நீங்கலாக ……………….. சருக்கங்களைக் கொண்டது.
அ) எட்டு
ஆ) ஒன்பது
இ) ஏழு
ஈ) பத்து
Answer:
ஈ) பத்து

Question 8.
‘போலாதும்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ………………………..
அ) போ + தும்
ஆ) போல் + ஆதும்
இ) போல் + அனதும்
ஈ) போலா + தும்
Answer:
ஆ) போல்+ஆதும்

Question 9.
‘உய்ப்பனவும்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………
அ) உய் + பனவும்
ஆ) உய்ப் + பனவும்
இ) உய்ப்ப ன + உம்
ஈ) உய்ப்ப ன + அம்
Answer:
இ) உய்ப்பன+உம்

Question 10.
‘கூற்றவா’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………….
அ) கூ + அவா
ஆ) கூற்று + அவா
இ) கூற் + அவா
ஈ) கூற்று + ஆவா
Answer:
ஆ) கூற்று+அவா

Question 11.
‘ஐம்பெருங்காப்பியம்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………..
அ) ஐந்து + காப்பியம்
ஆ) ஐந்து + பெரு + காப்பியம்
இ) ஐம்பெருங் + காப்பியம்
ஈ) ஐந்து + பெருமை + காப்பியம்
Answer:
ஈ) ஐந்து + பெருமை + காப்பியம்

Question 12.
‘அரும்பிணி’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ………………………….
அ) அரும் + பிணி
ஆ) அரு + பிணி
இ) அருமை + பிணி
ஈ) அரும் + பணி
Answer:
இ) அருமை + பிணி

Question 13.
‘அ + பிணி’ என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ……………………….
அ) அபிணி
ஆ) அப்பிணி
இ) அப்பிணி
ஈ) அதுபிணி
Answer:
இ) அப்பிணி .

Question 14.
‘தெளிவு + ஓடு’ என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் …………………..
அ) தெளிவுவூடு
ஆ) தெளிவோடு
இ) தெளிவுஓடு
ஈ) தெளிவாடு
Answer:
ஆ) தெளிவோடு

Question 15.
‘பிணி + உள்’ என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ……………….
அ) பிணியுள்
ஆ) பிணினள்
இ) பிணியாள்
ஈ) பிணிபுள்
Answer:
அ) பிணியுள்

Question 16.
‘இன்பம் + உற்றே’ என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ………………….
அ) இன்பமுற்று
ஆ) இன்பமற்றே
இ) இன்பம் உற்றோ
ஈ) இன்பமுற்றே
Answer:
ஈ) இன்பமுற்றே

குறுவினா

Question 1.
நோய்கள் எவற்றிற்கெல்லாம் துன்பம் தருவன?
Answer:
மக்களின் உடலுக்கும் உள்ளத்திற்கும் துன்பம் தருவன நோய்கள்.

Question 2.
நம் முன்னோர்கள் எவற்றையும் நோய்கள் என்று கூறினர்?
Answer:
உள்ளத்தில் தோன்றும் தீய எண்ணங்களால் ஏற்படும் துன்பங்களையும் நோய்கள் என்றே நம் முன்னோர் கூறினர்.

Question 3.
இலக்கியங்கள் விளக்குவன யாவை?
Answer:
நோய்களை நீக்கும் மருந்துகளாக விளங்கும் அறக்கருத்துகளை இலக்கியங்கள் விளக்குகின்றன.

Question 4.
ஐம்பெருங்காப்பியங்கள் யாவை?
Answer:
சிலப்பதிகாரம், மணிமேகலை, சீவகசிந்தாமணி, வளையாபதி, குண்டலகேசி.

Question 5.
ஐஞ்சிறுகாப்பியங்கள் யாவை?
Answer:
சூளாமணி, நீலகேசி, உதயண குமார காவியம், யசோதர காவியம், நாககுமார காவியம்.

சிறுவினா

Question 1.
நீலகேசி குறித்து எழுதுக.
Answer:

• நீலகேசி என்றால் கருத்த கூந்தலை உடையவள் என்று பொருள்.
• ஆசிரியர் பெயர் அறியப்படவில்லை .
• கடவுள் வாழ்த்து நீங்கலாகப் பத்துச் சருக்கங்களைக் கொண்டது.
• 894 பாடல்களைக் கொண்டது.
• நீலகேசி தெருட்டு என்ற வேறு பெயரும் உண்டு.

சொல்லும் பொருளும்

தீர்வன – நீங்குபவை
உவசமம் – அடங்கி இருத்தல்
நிழல் இகழும் – ஒளிபொருந்திய
பேர்தற்கு அகற்றுவதற்கு
திரியோகமருந்து – மூன்று யோகமருந்து
தெளிவு – நற்காட்சி
திறத்தன – தன்மை யுடையன
கூற்றவா – பிரிவுகளாக
பூணாய் – அணிகலன்களை அணிந்தவளே
பிணி – துன்பம்
ஓர்தல் – நல்லறிவு
பிறவார் – பிறக்கமாட்டார்

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Tamilnadu Samacheer Kalvi 10th Social Science Civics Solutions Chapter 4 India’s Foreign Policy

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India’s Foreign Policy Textual Exercise

I. Choose the correct answer.

Indian Foreign Policy Class 10 Questions And Answers Question 1.
Which Minister plays a vital role in molding the foreign policy of our country?
(a) Defense Minister
(b) Prime Minister
(c) External Affairs Minister
(d) Home Minister
Answer:
(c) External Affairs Minister

Indian Foreign Policy 10th Standard Notes Question 2.
The Panchseel treaty has been signed between:
(a) India and Nepal
(b) India and Pakistan
(c) India and China
(d) India and Sri Lanka
Answer:
(c) India and China

10th Social Indian Foreign Policy Notes Question 3.
Which article of Indian constitution directs to adopt foreign policy?
(a) Article 50
(b) Article 51
(c) Article 52
(d) Article 53
Answer:
(b) Article 51

Foreign Policy Of India 10th Notes Question 4.
Apartheid is:
(a) An international association
(b) Energy diplomacy
(c) A policy of racial discrimination
(d) None of these
Answer:
(d) None of these

Indian Foreign Policy Notes 10th Class Question 5.
The Agreement signed by India and China in 1954 related to …………..
(a) Trade and Commerce
(b) Restoration of normal relations
(c) Cultural exchange programmes
(d) The Five Principles of Co-existence
Answer:
(d) The Five Principles of Co-existence

Indian Foreign Policy Class 10 Notes Question 6.
Which is not related to our foreign policy?
(a) World co-operation
(b) World peace
(c) Racial equality
(d) Colonialism
Answer:
(d) Colonialism

Indian Foreign Policy Class 10 Questions And Answers Pdf Question 7.
Which of the following country is not the founder member of NAM?
(a) Yugoslavia
(b) Indonesia
(c) Egypt
(d) Pakistan
Answer:
(d) Pakistan

Indian Foreign Policy 10th Notes Question 8.
Find the odd one:
(a) Social welfare
(b) Health care
(c) Diplomacy
(d) Domestic affairs
Answer:
(c) Diplomacy

Question 9.
Non-Alliance means ………..
(a) being neutral
(b) freedom to decide on issues independently
(c) demilitarisation
(d) none of the above
Answer:
(b) freedom to decide on issues independently

Question 10.
Non – military issues are:
(a) Energy security
(b) Water security
(c) Pandemics
(d) All the above.
Answer:
(d) All the above.

II. Fill in the blanks.

1. India conducted its first nuclear test at …………
2. At present our foreign policy acts as a means to generate ……….. for domestic growth and development.
3. ……….. is the instrument for implementing foreign policy of a state.
4. …………. was India’s policy in the face of the bipolar order of the cold war.
5. Our tradition and national ethos is to practice …………..
Answers:
1. Pokhran
2. inward investment, business and technology
3. Diplomacy
4. Non-Alignment
5. disarmament

III. Consider the following statement and tick the appropriate answer.

Question 1.
Arrange the following in the correct chronological order and choose the correct answer from the code given below.
(i) Panchsheel
(ii) Nuclear test at Pokhran
(iii) Twenty-year Treaty
(iv) First Nuclear test
(a) i, iii, iv, ii
(b) i, ii, iii, iv
(c) i, ii, iv, iii
(d) i, iii, ii, iv
Answer:
(c) i, ii, iv, iii

Question 2.
Which of the following is not about NAM?
(i) The term Non-Alignment was coined by V. Krishna Menon
(ii) It aimed to maintain national independence in foreign affairs by joining any military alliance
(iii) At present it has 120 member countries
(iv) It has transformed to an economical movement
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (ii) only
(d) (iv) only
Answer:
(c) (ii) only

Question 3.
Write true or false against each of the statement.
(a) During Cold War India tried to form a third bloc of nations in the international affairs.
(b) The Ministry of Home Affairs is responsible for the conduct of the country’s foreign relations.
(c) The nuclear test at Pokhran was done under Subterranean Nuclear Explosions Project.
Answer:
(a) True
(b) False
(c) True

Question 4.
Assertion (A): India aligned with Soviet Union by the Indo-Soviet treaty on 1971.
Reason (B): This began with a disastrous Indo-China war of 1962.
(a) A is correct and R explains A
(b) A is correct and R does not explain A
(c) A is correct and R is Wrong
(d) Both A and R are wrong
Answer:
(b) A is correct and R does not explain A

Question 5.
Assertion (A): India has formal diplomatic relations with most of the nations.
Reason (R): India is the World’s second-most populous country.
(a) A is correct and R explains A
(b) A is correct and R does not explain A
(c) A is wrong and R is correct
(d) Both are wrong
Answer:
(b) A is correct and R does not explain A

Question 6.
Avoidance of military blocs was a necessity for India after political freedom. Because India had to redeemed from
(a) acute poverty
(b) illiteracy
(c) chaotic socio-economic conditions
(d) all the above
Answer:
(d) all the above

IV. Match the following.

1.	Indian Ocean island	(a)	1955
2.	Landbridge to ASEAN	(b)	1954
3.	Panchsheel	(c)	Maldives
4.	Afro Asian Conference	(d)	Foreign Policy
5.	World Peace	(e)	Myanmar

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

V. Give Short Answers.

Question 1.
What is foreign policy?
Answer:
Foreign policy can be defined as a country’s policy that is conceived, designed and formulated to safeguard and promote her national interests in her external affairs in the conduct of relationships with other countries, both bilaterally and multilaterally.

Question 2.
Explain India’s nuclear policy.
Answer:
Indian nuclear programme in 1974 and 1998 is only done for strategic purposes. The two themes of India’s nuclear doctrine are
• No first use
• Credible minimum deterrence
It has decided not to use nuclear power for ‘offensive purposes’ and would never use against any non-nuclear state.

Question 3.
Highlight the contribution by Nehru to India’s foreign policy.
Answer:

1. The most idealistic phase of India’s foreign policy under the guidance of India’s first Prime Minister, Jawaharlal Nehru.
2. The new nations that got independence after the long period of colonial struggle found themselves in a very difficult situation with respect to economic development.
3. So it was necessary to align with either of the blocs – United States of America U.S.A (or) Union Soviet Socialist Republic (U.S.S.R).

Question 4.
Differentiate: Domestic policy and Foreign policy
Answer:

Domestic Policy	Foreign Policy
Domestic policy is the nation’s plan for dealing issues within its own nation.	Foreign policy is the nation’s plan for dealing with other nations.
It includes laws focusing on domestic affairs, social welfare, health care, education, civil rights, economic issues and social issues.	Trade, diplomacy, sanctions, defence, intelligence and global environments are the types of foreign policy.

Question 5.
List any four guiding principles of Panchsheel.
Answer:

1. Mutual non – aggression
2. Mutual non – interference
3. Equality and co-operation for mutual benefit
4. Peaceful co-existence

Question 6.
What was the reason for India to choose the path of Non-Alignment?
Answer:
The new nations that got independence after the long period of colonial struggle found themselves in a very difficult situation with respect to economic development. So it was necessary to align with either of the blocs – United States of America (USA) or United Soviet Socialist Republic (USSR). Nehru, India’s first Prime Minister, was opposed to the rivalry of the two superpowers (America and Russia). So he chose the path of Non-Alignment.

Question 7.
In what ways are India’s global security concerns reflected?
Answer:
India’s global security concerns are reflected in its military modernisation, maritime security and nuclear policies.

Question 8.
List out the member countries of SAARC.
Answer:
The member countries are Afghanistan, Bangladesh, Bhutan, India, Nepal, Maldives, Pakistan and Sri Lanka.

Question 9.
Name the architects of the Non-Aligned movement.
Answer:
Jawaharlal Nehru of India, Tito of Yugoslavia, Nasser of Egypt, Sukarno of Indonesia, and KwameNkumarah of Ghana were the architects of Non Aligned Movement.

Question 10.
Mention the main tools of foreign policy.
Answer:
The main tools of foreign policy are treaties and executive agreements, appointing ambassadors, foreign aid, international trade and armed forces.

VI. Answer in detail.

Question 1.
Write a detailed note on Non-alignment.
Answer:

• The term ‘Non-Alignment’ was coined by V. Krishna Menon.
• Non-alignment has been regarded as the most important feature of India’s foreign policy.
• It aimed to maintain national independence in foreign affairs by not joining any military alliance.
• The Non-Aligned Movement (NAM) was formed with a membership of 120 countries and 17 states as observers and 10 international organisations.

The founding fathers of Non-Aligned Movement:
Jawaharlal Nehru of India, Tito of Yugoslavia, Nasser of Egypt, Sukarno of Indonesia, and Kwame Nkumarah of Ghana were the founding fathers of NAM.

• • Non-aligned countries have been successful in establishing a foundation of economic co-operation among underdeveloped countries.

Question 2.
Discuss the core determinants of India’s foreign policy?
Answer:

1. Geographical position and size of territory.
2. Nation’s history, traditions and philosophical basis.
3. Natural resources.
4. The compulsion of economic development.
5. Political stability and structure of Government.
6. The necessity of peace, disarmament and non – proliferation of nuclear weapons.
7. Military strength
8. International milieu.

Question 3.
Make a list on basic concepts followed by India to maintain friendly relations with its neighbours.
Answer:
(i) Indian foreign policy has always regarded the concept of neighbourhood as one of widening concentric circles, around the central axis of historical and cultural commonalties.

(ii) India gives political and diplomatic priority to her immediate neighbours and the Indian Ocean Island states such as Maldives.

(iii) India provides neighbours with support as needed in this form of resources equipment and training.

VII. Project and activity

Question 1.
Identify any two aspects of India’s foreign policy that you would like to retain and two that you would like to change if you were the decision maker.
Answer:
Do it yourself.

India’s Foreign Policy Additional Questions

I. Choose the correct answer.

Question 1.
India is a country with an unbounded faith in …………
(a) War
(b) Peace
(c) Love
Answer:
(b) Peace

Question 2.
Find out the main tools of the foreign policy of the following.
(a) Treaties
(b) International trade
(c) Foreign Aid
(d) All the above
Answer:
(d) All the above

Question 3.
Apartheid was abolished on …………
(a) 1990
(b) 1991
(c) 1890
Answer:
(a) 1990

Question 4.
In which place the Foreign Service Training Institute was established?
(a) New Delhi
(b) Mumbai
(c) Calcutta
(d) None of the above
Answer:
(a) New Delhi

Question 5.
Apartheid was abolished by …………..
(a) Jawaharlal Nehru
(b) Nelson Mandela
(c) Gandhi
Answer:
(b) Nelson Mandela

Question 6.
In which year Panchsheel was signed?
(a) 1951
(b) 1952
(c) 1954
(d) 1956
Answer:
(c) 1954

Question 7.
SAARC’s first meeting was held at ………….
(a) Colombo
(b) Cairo
(c) Dacca
Answer:
(c) Dacca

Question 8.
Who was opposed to the rivalry of the two super powers? (America and Russia)
(a) V. Krishna Menon
(b) Nasser
(c) Jawaharlal Nehru
(d) None
Answer:
(c) Jawaharlal Nehru

Question 9.
The first SAARC’s meeting was held at Dacca in the year …………..
(a) 1985
(b) 1965
(c) 1995
Answer:
(c) 1995

Question 10.
Which year did India conduct its first nuclear test at Pokhran?
(a) 1971
(b) 1972
(c) 1974
(d) 1976
Answer:
(c) 1974

II. Fill in the blanks :

1. India followed the policy of …………..
2. India has rendered whole-hearted support to the ………….. to bring World Peace.
3. ………….. is an economic and geopolitical organization of eight countries are particularly located in South Asia.
4. SAARC Disaster Management Centre was set up at ………….
5. …………policy is the nation’s plan for dealing issues within its own nation.
Answers:
1. Non-alignment
2. UNO
3. SAARC
4. New Delhi
5. Domestic

III. Match the following.

1.	Jawaharlal Nehru	(a)	Indonesia
2.	Chou-En-Lai	(b)	Egypt          .
3.	Nelson Mandela	(c)	India
4.	Nasser	(d)	China
5.	Sukarno	(e)	South Africa

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

IV. Answer briefly:

Question 1.
Mention any three objectives of our Foreign Policy.
Answer:

1. National security
2. National prosperity
3. Achieving world peace and enable every nation to peacefully co-exist.

Question 2.
Why is world peace an essential one?
Answer:
The economic development of the nations can be achieved only through world peace. So world peace is very essential not only for the economic development of India but also for all the developing countries of the world.

Question 3.
Explain the foreign policy stance of India?
Answer:
The foreign policy stance of India was:

1. Supporting the cause of decolonisation.
2. Staunch opponent of the apartheid regime in South Africa.
3. Accepted the importance of defence preparedness.

Question 4.
Name the areas identified by the SAARC countries for mutual co-operations.
Answer:
The SAARC countries identified mutual co-operation in the following areas. They are transportation, postal service, tourism, shipping meteorology, health, agriculture, rural construction and telecommunications.

Question 5.
What are the two themes of India’s nuclear doctrine?
Answer:
The two themes of India’s nuclear doctrine are:

1. No first use
2. Credible minimum deterrence.

V. Detail.

Question 1.
Write a paragraph about Panchsheel and the policy of Non-alignment.
Answer:
Panchsheel:
India is called by the name of “A Great Peace Maker”. It followed five principles which are popularly known as ‘Panchsheel’. Jawaharlal Nehru said, stress on these five principles. They are:

1. Each country should respect the territorial integrity and sovereignty of others.
2. No country should attack any other country.
3. No one should try to interference in the internal affairs of others.
4. All countries shall strive for equality and mutual benefit.
5. Every country should try to follow the policy of peaceful co-existence.

These Panchsheel greatly added to the international status of India.
Policy of Non-allignments:

1. After the Second World War the world was divided into two hostile blocs. The
   American Bloc and The Russian Bloc.
2. Both of them are trying to increase their influence at the cost of the other.
3. But India has not joined either of these two blocs.
4. Whenever any difference arises between there blocs, India tries to remove that difference. Thus India has contributed substantially towards world peace.

Question 2.
Write about policy of Disarmament.
Answer:

1. Our tradition and national ethos is to practice disarmament.
2. Since Independence, global non-proliferation has been a dominant theme of India’s nuclear policy.
3. So India supported UN disarmament progrmme.
4. Indian nuclear programme in 1974 and 1998 is only done for strategic purposes.

Question 3.
Point out the basic concepts of India’s foreign policy:
Answer:

• Preservation of national interest.
• Achievement of world peace.
• Disarmament
• Fostering cordial relationship with other countries.
• Solving conflicts by peaceful means.
• Independence of thought and action as per the principle of NAM.
• Equality in conducting international relations.
• Anti-Colonialism, anti-imperialism anti- racism.

We think the data given here clarify all your queries of Chapter 4 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 Civics Chapter 4 India’s Foreign Policy 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 10th Maths Solutions Chapter 3 Algebra Ex 3.11

10th Maths Exercise 3.11 Samacheer Kalvi Question 1.
Solve the following quadratic equations by completing the square method
(i) 9x² – 12x + 4 = 0
(ii) \(\frac{5 x+7}{x-1}\) = 3x + 2
Solution:

Ex 3.11 Class 10 Samacheer Question 2.
Solve the following quadratic equations by formula method
(i) 2x² – 5x + 2 = 0
(ii) \(\sqrt{2} f^{2}\) – 6f + \(3 \sqrt{2}\)
(iii) 3y² – 20y – 3 = 0
(iv) 36y² – 12 ay + (a² – b²) = 0
Solution:
(i) 2x² – 5x + 2 = 0
The formula for finding roots of a quadratic equation ax² + bx + c = 0 is

Exercise 3.11 Class 10 Samacheer Question 3.
A ball rolls down a slope and travels a distance d = t² – 0.75t feet in t seconds. Find the time when the distance travelled by the ball is 11.25 feet.
Solution:
Distance d = t² – 0.75 t,
Given that d = 11.25 = t² – 0.75 t.
t² – 0.75t – 11.25 = 0

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Samacheer Kalvi 6th Maths Book Solutions Term 2 Question 1.
Say the time in two ways:

Solution:
(i) 10 : 15 hours; quarter past 10; 45 minutes to 11
(ii) 6 : 45 hours; quarter to 7; 45 minutes past 6
(iii) 4 : 10 hours; 10 minutes past 4; 50 minutes to 5
(iv) 3 : 30 hours; half-past 3; 30 minutes to 4
(v) 9 : 40 hours; 20 minutes to 10; 40 minutes past 9.

Samacheer Kalvi 6th Maths Term 2 Question 2.
Match the following:
(i) 9.55 – (a) 20 minutes past 2
(ii) 11.50 – (b) quarter past 4
(iii) 4.15 – (c) quarter to 8
(iv) 7.45 – (d) 5 minutes to 10
(v) 2.20 – (e) 10 minutes to 1?
Solution:
(i) 9.55 – (d) 5 minutes to 10
(ii) 11.50 – (e) 10 minutes to 12
(iii) 4.15 – (b) quarter past 4
(iv) 7.45 – (c) quarter to 8
(v) 2.20 – (a) 20 minutes past 2

6th Maths Guide Term 2 Question 3.
Convert the following :
(i) 20 minutes into seconds
(ii) 5 hours 35 minutes 40 seconds into seconds
(iii) 3\(\frac { 1 }{ 2 }\) hours into minutes
(iv) 580 minutes into hours
(v) 25200 seconds into hours
Solution:
(i) 20 minutes into seconds
20 minutes = 20 × 60 seconds = 1200 seconds
(ii) 5 hours 35 minutes 40 seconds into seconds

∴ 5 hours 35 minutes 40 seconds = 20,140 seconds
(iii) 3\(\frac { 1 }{ 2 }\) hours into minutes
3\(\frac { 1 }{ 2 }\) hours = 3 hours 30 minutes
= 3 × 60 minutes + 30 minutes
= 180 minutes + 30 minutes
= 210 minutes
∴ 3\(\frac { 1 }{ 2 }\) hours = 210 minutes
(iv) 5580 minutes into hours
580 minutes = \(\frac{580}{60}\) hours = 9 hours 40 minutes
∴ 580 minutes = 9 hours 40 minutes
(v) 25200 seconds into hours
25200 seconds = \(\frac{25200}{60}\) minutes = 420 minutes = \(\frac{420}{60}\) hours = 7 hours
∴ 25200 seconds = 7 hours

Samacheer Kalvi Term 2 Question 4.
The duration of electricity consumed by the farmer for his pumpset on Monday and Tuesday was 7 hours 20 minutes 35 seconds and 3 hours 44 minutes 50 seconds respectively. Find the total duration of consumption of electricity.
Solution:
Total duration of electricity consumed on both the days
= 7 hours 20 min 35 sec + 3 hours 44 min 50 sec
= (7 + 3) hours (20 + 44) min (35 + 50) sec
= 10 hours 64 min 85 seconds
= 11 hours 5 min 25 seconds

Samacheer Kalvi 6th Maths Book Solutions Question 5.
Subtract 10 hrs 20 min 35 sec from 12 hrs 18 min 40 sec.
Solution:

1 hour 58 minutes 05 seconds

6th Standard Maths Exercise 2.2 Question 6.
Change the following into 12 hour format:
(i) 02 : 00 hours
(ii) 08 : 45 hours
(iii) 21 : 10 hours
(iv) 11 : 20 hours
(v) 00 : 00 hours
Solution:

Term 2 Samacheer Kalvi Question 7.
Change the following into 24 hour format.
(i) 3.15 a.m.
(ii) 12.35 p.m.
(iii) 12.00 noon
(iv) 12.00 midnight.
Solution:

Samacheer Kalvi 6th Maths Guide Term 2 Question 8.
Calculate the duration of time.
(i) from 5.30 a.m to 12.40 p.m
(ii) from 1.30 p.m to 10.25 p.m
(iii) from 20:00 hours to 4:00 hours
(iv) from 17:00 hours to 5 :15 hours
Solution:
(i) from 5.30 a.m. to 12 .40 p.m.
Duration of time from 5.30 a.m. to noon = 12 : 00 – 5 : 30 = 6 : 30 i.e 6 hours 30 minutes
From noon to 12.40 p.m the duration = 00 hours 40 minutes
Total duration = 6 hours 30 minutes + 00 hours 40 minutes
= 6 hours 70 minutes
= 6 hours + (60 + 10) minutes
= 6 hours + 1 hr 10 minutes
= 7 hours 10 minutes
∴ Duration of time from 5.30 am to 12.40 pm = 7 hours 10 minutes
(ii) from 1.30 pm to 10.25 p.m.

(iii) from 20 : 00 hrs to 4 : 00 hrs.

(iv) from 17 : 00 hours to 5 : 15 hours.

Samacheer Kalvi 6th Maths Question 9.
The departure and arrival timing of the Vaigai Superfast Express (No. 12635) from Chennai Egmore to Madurai Junction are given. Read the details and answer the following.

(i) At what time does the Vaigai Express start from Chennai and arrive at Madurai?
(ii) How many halts are between there Chennai and Madurai?
(iii) How long does the train halt at the Villupuram junction?
(iv) At what time does the train come to Sholavandan?
(v) Find the journey time from Chennai Egmore to Madurai?
Solution:
(i) It starts from Chennai at 13 : 40 hrs and arrive at Madurai at 21 : 20 hrs.
(ii) There are 8 halts.
(iii) Departure from Villupuram = 15 hours 55 minutes
Arrival at Villupuram = 15 hours 50 minutes
The train halt at Villupuram for = 05 minutes
(iv) At 20 : 34 hours the train come to Sholavandan
(v) Arrival time at Madurai = 20 hours 80 (20 + 60) minutes
Departure time from Chennai Egmore = 13 hours 40 minutes
Journey Time = 07 hours 40 minutes

Samacheer Kalvi 6th Standard Maths Question 10.
Manickam joined a chess class on 20.02.2017 and due to exam, he left practice after 20 days. Again he continued practice from 10.07.2017 to 31.03.2018. Calculate how many days did he practice?
Solution:
From the date of joining = 20 days From 10.07.2017 to 31.03.2018
July – 22
Aug – 31
Sep – 30
Oct – 31
Nov – 30
Dec – 31
Jan – 31
Feb – 28
Mar – 31
Total – 265
Total no of practice days = 265 + 20 = 285 days

Class 6 Maths Chapter 2 Exercise 2.2 Solutions Question 11.
A clock gains 3 minutes every hour. If the clock is set correctly at 5 a.m, find the time shown by the clock at 7 p.m?
Solution:
5 a.m. = 5.00 hours
7 p.m = 19.00 hours
Difference = 19.00 – 5.00 = 14.00 hours
Time gain = 14 × 3 = 42 minutes
Time shown by the clock = 7.42 p.m

Samacheer Kalvi 6th Maths Solutions Question 12.
Find the number of days between the republic day and kalvi valarchi day in 2020.
Solution:
2020 is a leap year republic day – 26.01.2020
Kalvi valarchi day – 15.07.2020
Jan – 5
Feb – 29
Mar – 31
April – 30
May – 31
June – 30
July – 14
Total – 170 days

6th Class Maths Exercise 2.2 Question 13.
If 11th of January 2018 is Thursday, what is the day on 20th July of the same year?
Solution:
January – 21 Days (31 – 10)
February – 28 Days
March – 31 Days
April – 30 Days
May – 31 Days
June – 30 Days
July – 19 Days
Total – 190 Days
190 days ÷ 7
190 days = 27 weeks + 1 day
Required day is the first day after Thursday.
∴ 20th of July is Friday.

Samacheer Kalvi Maths 6th Question 14.
(i) Convert 480 days into years
(ii) Convert 38 months into years
Solution:
(i) 480 days = \(\frac{480}{365}\)
= 1 year 115 days
= 1 year 3 months 25 days

(ii) 38 months = \(\frac{38}{12}\)
= 3 years 2 months

Samacheer Kalvi Guru 6th Maths Term 2 Question 15.
Calculate your age as on 01.06.2018
Solution:

Age is 12 years 2 months

Objective Type Questions

Samacheer Kalvi 6th Maths Solution Question 16.
2 days = …….. hours
(i) 38
(ii) 48
(iii) 28
(iv) 40
Solution:
(ii) 48

Samacheer Kalvi Class 6 Maths Solutions Question 17.
3 weeks = _____ days.
(a) 21
(b) 7
(c) 14
(d) 28
Solution:
(a) 21

Question 18.
Number of ordinary years between two consecutive leap years is
(i) 4 years
(ii) 2 years
(iii) 1 year
(iv) 3 years
Solution:
(iv) 3 years

Question 19.
What time will it 5 hours after 22 : 35 hours?
(a) 2 : 30 hrs
(b) 3 : 35 hrs
(c) 4 : 35 hrs
(d) 5 : 35 hrs
Solution:
(b) 3 : 35 hrs

Question 20.
2 \(\frac{1}{2}\) years is equal to months.
(i) 25
(ii) 30
(iii) 24
(iv) 5
Solution:
(ii) 30

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Tamilnadu Samacheer Kalvi 10th Social Science History Solutions Chapter 7 Anti-Colonial Movements and the British of Nationalism

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 7 Anti-Colonial Movements and the British of Nationalism 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 7 textbook solutions for Tamilnadu State Board.

Anti-Colonial Movements and the British of Nationalism Textual Exercise

I. Choose the correct answer.

Anti Colonial Movements And The Birth Of Nationalism Question 1.
Which one of the following was launched by Haji Shariatullah in 1818 in East Bengal?
(a) Wahhabi Rebellion
(b) Farazi Movement
(c) Tribal uprising
(d) Kol Revolt
Answer:
(b) Farazi Movement

Question 2.
Who declared that “Land belongs to God” and collecting rent or tax on it was against divine law?
(a) Titu Mir
(b) Sidhu
(c) Dudu Mian
(d) Shariatullah
Answer:
(c) Dudu Mian

Question 3.
Who were driven out of their homeland during the process of creation of Zamins under Permanent Settlement?
(a) Santhals
(b) Titu Mir
(c) Munda
(d) Kol
Answer:
(a) Santhals

Question 4.
Find out the militant nationalist from the following.
(a) Dadabhai Naoroji
(b) Justice Govind Ranade
(c) Bipin Chandra pal
(d) Romesh Chandra
Answer:
(c) Bipin Chandra pal

Question 5.
When did the Partition of Bengal come into effect?
(a) 19 June 1905
(b) 18 July 1906
(c) 19 August 1907
(d) 16 October 1905
Answer:
(d) 16 October 1905

Question 6.
What was the context in which the Chota Nagpur Tenancy Act was passed?
(a) Kol Revolt
(b) Indigo Revolt
(c) Munda Rebellion
(d) Deccan Riots
Answer:
(c) Munda Rebellion

Question 7.
Who set up the first Home Rule League in April 1916?
(a) Annie Basant
(b) Bipin Chandra Pal
(c) Lala Lajpat Rai
(d) Tilak
Answer:
(d) Tilak

Question 8.
Who drew the attention of the British to the suffering of Indigo cultivation through his play Nil darpan?
(a) Dina Bandhu Mitra
(b) Romesh Chandra Dutt
(c) Dadabhai Naoroji
(d) Birsa Munda
Answer:
(a) Dina Bandhu Mitra

II. Fill in the blanks.

1. In 1757, Robert Clive was financially supported by…………….., the moneylenders of Bengal.
2. …………….. was an anti-imperial and anti-landlord movement originated in and around 1827.
3. The major tribal revolt took place in Cbotanagpur region was ……………….
4. The ………….. Act, restricted the entry of non-tribal people into the tribal land.
5. Around 1854 activities of social banditry were led by ………………
6. The British Commander of Kanpur killed by the rebels during the 1857 Rebellion was ……………..
7. Chota Nagpur Act was passed in the year ……………..
8. W.C. Bannerjee was elected the president of Indian National Congress in the year …………..
Answers:
1. Jagat Seths
2. Wahhabi rebellion
3. Kol revolt
4. Chotanagpur Tenancy Act
5. Bir Singh
6. Major General High Wheelor
7. 1908
8. 1885

III. Choose the correct statement.

Question 1.
(i) The company received ₹ 22.5 million from Mir Jafar and invested it to propel the industrial revolution in Britain.
(ii) Kols organized an insurrection in 1831-1832, which was directed against Government officers and money lenders.
(iii) In 1855, two Santhal brothers Sidhu and Kanu led the Santhal Rebellion.
(iv) In 1879, an Act was passed to regulate the territories occupied by the Santhals.
(a) (i) (ii) and (iii) are correct
(b) (ii) and (iii) are correct
(c) (iii) and (iv) are correct
(d) (i) and (iv) are correct
Answer:
(a) (i) (ii) and (iii) are correct

Question 2.
(i) Dudu Mian laid emphasis on the egalitarian nature of Islam and declared that “Land belongs to God” .
(ii) According to the Doctrine of Lapse, new territories under the corrupt Indian rulers were to be annexed.
(iii) The British officials after the suppression of 1857 Revolt were given power to judge and take the lives of Indians without due process of law.
(iv) One of the causes of the failure of the Revolt of 1857 was many of the Indian princes and zamindars remained loyal to the British.
(a) (ii) (iii) and (iv) are correct
(b) (i) (ii) and (iv) are correct
(c) (i) (iii) and (iv) are correct
(d) (i), (ii) and (iii) are correct.
Answer:
(c) (i) (iii) and (iv) are correct

Question 3.
(i) One of the most significant contributions of the early Indian Nationalists was the formulation of an economic critique of colonialism.
(ii) The early Congress leaders stated that the religious exploitation in India was the primary reason for the growing poverty.
(iii) One of the goals of the moderate Congress leaders w as to achieve Swaraj or self-rule.
(iv) The objective of partition was to curtail the Bengali influence and weaken the nationalist movement.
(a) (i) and (iii) are correct
(b) (i) (iii) and (iv) are correct
(c) (ii) and (iii) are correct
(d) (iii) and (iv) are correct
Answer:
(b) (i) (iii) and (iv) are correct

Question 4.
Assertion (A): Under colonial rule, for the first time in Indian history, Government claimed a direct proprietary right over forests.
Reason (R): Planters used intimidation and violence to compel farmers to grow indigo.
(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:
(a) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)

Question 5.
Assertion (A): The Revolt of 1857 was brutally suppressed by the British army.
Reason (R): The failure of the rebellion was due to the absence of Central authority.
(a) Both (A) and (R) are wrong
(b) (A) is wrong and (R) is correct
(c) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(d) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
Answer:
(b) (A) is wrong and (R) is correct

IV. Match the following.

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

V. Answer the following questions briefly.

Question 1.
How are the peasant uprisings in British India classified?
Answer:
The peasants uprising in British India are classified in to four categories:

1. Restorative Rebellions – To attempt to restore old order and old social relations.
2. Religious movements – Led by religions leaders to liberate the locals by restructuring society on certain religious principles.
3. Social banditry – Looked upon by their people as heroes or champions of their cause.
4. Mass Insurrection – Leaderless and spontaneous uprisings.

Question 2.
Write about the Kanpur Massacre of 1857.
Answer:
The Siege of Cawnpur was a key episode in the Indian rebellion of 1857. The besieged Company forces and civilians in Cawnpur (now Kanpur) were unprepared for an extended siege and surrendered to rebel forces under Nana Sahib, in return for a safe passage to Allahabad. However under ambiguous circumstances, their evacuation from Cawnpur turned into a massacre, and most of the men were killed.

Question 3.
Name the territories annexed by the British under the Doctrine of Lapse.
Answer:
Satara, Sambalpur, parts of Punjab, Jhansi, and Nagpur were the territories annexed under the Doctrine of Lapse.

Question 4.
What do you mean by drain of wealth?
Answer:
The colonial economy was a continuous transfer of resources from India to Britain without any favourable returns back to India. This is called the drain to wealth.

Question 5.
Explain the concept of constructive Swadeshi.
Answer:
Constructive Swadeshi: It largely stress upon self – help.

1. It focused on building alternative-institutions of self governance that would operate free of British control.
2. It also laid emphasis on the need of self strengthening of the people.
3. In turn it would help them in creating a worthy citizen for political agitation.
4. Swadeshi shops spread all over the place selling textiles, handlooms, soaps, earthenwares match and leather goods.

Question 6.
Highlight the objectives of Home Rule Movement.
Answer:
The following were the objectives of the Home Rule Movements in India –

• To attain self-government within the British Empire by rising constitutional means.
• To obtain the status of dominion, apolitical position accorded later to Australia, Canada, South Africa and New Zealand.
• To use non-violent constitutional methods to achieve goals.

Question 7.
Summarise the essence of Lucknow Pact.
Answer:
Lucknow Pact was signed in the year 1916. Under the provision of this pact.

1. Congress and Muslim League agreed to work together to attain self Government in India.
2. A separate electorate for Muslim was accepted by the Congress leadership.

VI. Answer all the questions under each caption.

Question 1.
Deccan Riots
(а) When and where did the first recorded incident of rioting against the moneylenders in the Deccan appear?
Answer:
The first recorded incident of rioting against the moneylenders in the Deccan appeared in May 1875, in Supa, a village near Poona.

(b What was the right given to moneylenders under a new law of the British?
Answer:
Under a new law, the British moneylenders were allowed to attach the mortgaged land of the defaulters and auction it off.

(c) What did it result in?
Answer:
It resulted in transfer of lands from the cultivators to the non-cultivating classes.

(d) Against whom was the violence directed in the Deccan riots.
Answer:
The violence was directed mostly at the Gujarat moneylenders.

Question 2.
The Revolt of 1857.

(a) Who assaulted his officer, an incident that led to the outbreak of 1857 Revolt?
Answer:
Mangal Pandey assaulted his officer, this was an incident that led to the outbreak of 1857 Revolt.

(b) Who was proclaimed the Sahhensha – e- Hindustan in Delhi?
Answer:
Bahadhur Shah – II was proclaimed as Shahensha – e – Hindustan in Delhi.

(c) Who was the correspondent of London Times who reported on the brutality of the 1857 revolt?
Answer:
William Howard Russell was the correspondent of London Times who reported on the brutality of the 1857 Revolt.

(d) What did the Queen’s proclamation say on matters relating to religion?
Answer:
The Queen’s proclamation said that the British would not interfere in traditional institutions and religious matters.

Question 3.
Indian National Congress
(a) What were the techniques adopted by the Congress to get its grievances redressed?
Answer:
The techniques adopted by the Congress to get its grievances redressed included appeals, petition and delegation to Britain.

(b) What do you know of Lal-Bal-Pal triumvirate?
Answer:
Lala Lajpat Rai of Punjab, Bal Gangadhar Tilak of Maharashtra and Bipin Chandra Pal of Bengal were three prominent leaders during the Swadeshi period. They are often referred to as Lal-Bal-Pal triumvirate.

(c) Where was the first session of Indian National Congress held?
Answer:
The first session of Indian National Congress was held at Bombay.

(d) How did the British respond to the Swadeshi Movement?
Answer:
The British brutally crushed the Swadeshi movement by arresting prominent leaders and putting them into the prison.

VII. Answer the following in detail.

Question 1.
Discuss the causes and consequences of the Revolt of 1857.
Answer:
Political Cause: The annexation policy of British India: Two major policies were followed to bring more territories under British rule.

(a) Doctrine of Paramountcy:
(i) New territories were annexed on the grounds that the native rulers were corrupt and inefficient.
(ii) British claimed themselves as paramount, supreme authority.

(b) Doctrine of Lapse:
(i) If a native ruler does not have biological male heir of their own to the throne after his death the territory would ‘lapse’ in to British India.
(ii) Sambalpur, Satara, Jhansi, parts of Punjab and Nagpur were annexed by the British through Doctrine of Lapse.

Religious Cause:
(i) Indian Sepoys were prohibited from wearing religious marks on their forehead.
(ii) To render their services overseas. Crossing sea meant the loss of their caste. So Sepoys at Barrackpore refused to go to Burma by sea.

Economic Cause: Discrimination in salary and promotion. They felt humiliated by racial abusement.

Immediate Cause:

1. Introduction of cartridges to the new Enfield Rifles.
2. There was a strong suspicion that the new cartridge was greased with cow and pig fat
3. The cartridge had to be bitten off before loading (pork is forbidden to the Muslims and cow is sacred to the Hindus) so the Sepoys refused to use it.
4. In 29th March a Sepoy named Mangal Pandey assaulted his European officer Along with some other Sepoys and Mangal Pandey were court martialled and hanged.

This fuelled the anger among the Sepoys and started revolt.

Consequences of the Revolt: The Revolt was suppressed: Army had been restored to British control under Martial law.

1. The British officers were given power to judge and take the life of Indians without dae process of law.
2. Bahadhur shah II was captured and sent to Burma.
3. The British parliament passed a law proclaiming the transfer of power from the East India Company to the British Crown.
4. The structure of the Indian army was revised army was reduced and Indians were restrained from holding important ranks and positions.
5. Queen Victoria proclaimed that the British would not interfere in traditional institutions and religious matters of Indians.
6. It was promised that more Indians would be absorbed in Government services.
7. Secretary of state was appointed to review the Indian affairs.

Question 2.
How did the people of Bengal respond to the partition of Bengal (1905)?
Answer:

1. The partition of Bengal that took place in 1905, resulted in widespread protests all across India. It started a new phase of the Indian national movement.
2. The partition instead of dividing the Bengali people along the religious line united the leaders of both the groups – extremist and moderate.
3. The people of Bengal adopted new techniques of protest. The boycott of British goods was one such method.
4. The day Bengal was officially partitioned was declared as a day of mourning. Thousand’s of people took bath in River Ganga and marched on the streets of Calcutta singing Vande Mataram.
5. The people of Bengal were outraged at what they saw as a ‘divide and rule’ policy of the British government.

Question 3.
Attempt a narrative account of how Tilak and Annie Besant by launching Home Rule Movement sustained the Indian freedom struggle after 1916?
Answer:
The Indian National Movement was revived and also radicalised during the Home Rule Movement (1916- 1918), led by Lokamanya Tilak and Annie Besant.

1. It emerged as a mass movement against the British betrayal to the Indian cause of self Government after the I World War in return for the support rendered.
2. Tilak first set up the Home Rule League in April 1916.
3. In September 1916 Annie Besant by the demand of her followers started the Home Rule Movement in Madras (Adyar) without the support of the congress.
4. Both the leagues worked independently.
5. They succeeded in enrolling young people in large numbers and extending the movement to the areas,
6. The Home Rule league was utilised to carry propaganda through press, speeches, public meetings, lectures, discussions and touring in favour of self Government.
7. Both declared their strong view of self Government as right based on the national principle of self determination.
8. The Home Rule Movement led to the reunion of moderate and the extremists leaders of Congress at the Lucknow session.
9. Opened the fresh talks with the leaders of the Muslim League and decided to work together for self Government.
10. The demand for Swaraj was raised by Tilak and Annie Besant gained popularity.
11. This strong hold of both the leaders made the British parliament to announce Montagau – Chelmsford reforms which promised gradual progress of India towards self Government.

VIII. Activity

Question 1.
Identify the Acts passed in British India from 1858 to 1919, with a brief note on each. The Government of India Act, 1858.
Answer:

1. The Government of India Act 1858, marked the beginning of new chapter in the constitutional history of India.
2. Liquidation of East India Company, and the powers of government, territories and revenues ‘ were transfer to the British Crown.

The Indian Council Act, 1861

1. It enabled the Governor-General to associate the people of the land with work of legislation.
2. However, the legislative councils were merely talk shops with no power to criticize the administration or ask for some information. Their scope was fixed in legislation purpose alone; they had no right to move some kind of vote of no confidence.

The Indian Council Act, 1892

1. This act marks the beginning of representative form of Government in India.
2. Indian National Congress had adopted some resolutions in its sessions in 1885 and 1889 and put its demand. The major demands placed were as follows:
   • A simultaneous examination of ICS to be held in England and India
   • Reforms of the legislative council and adoption of the principle of election in place of nomination
   • Opposition to the annexation of Upper Burma
   • Reduction in the Military expenditure.

The Indian Councils Act, 1909 (The Morely-Minto Reforms)
(i) The British viceroy of India (1905-10)—was able to introduce several important innovations into the legislative and administrative machinery of the British Indian government.

(ii) The act also increased the maximum additional membership of the Imperial Legislative Council from 16 to 60.

The Government of India Act, 1919 (The Montague-Chelmsford Reforms)
(i) Subjects of administration were divided into two categories – ‘Central’ and ‘Provincial’. All important subjects (like Railways and Finance) were brought under the category of Central, while matters relating to the administration of the Provinces were classified as Provincial.

(ii) The provincial subjects were divided into two groups viz. reserved and transferred.

• The reserved subjects were kept with the Governor and transferred subjects were kept with the Indian Ministers.
• This division of subjects was basically what they meant by introducing the Diarchy.
• The reserved subjects were the essential areas of law enforcement such as justice, police, and revenue. The transferred subjects were such as public health, public works, education etc.

Question 2.
Mark the important Centres of 1857 Revolt in an outline map.
Answer:

Question 3.
Prepare album with pictures of frontline leaders of all the ant-colonial Struggles launched against the British.
Answer:
Do it yourself.

Anti-Colonial Movements and the British of Nationalism Additional Questions

I. Choose the correct answer.

Question 1.
The British historians call the Revolt of 1857 as ……………….
(a) Military revolt
(b) The great revolt
(c) Freedom struggle
Answer:
(a) Military revolt

Question 2.
The battle of plassey was orchestrated by:
(a) Robert Clive
(b) Hugh Rose
(c) Birsa Munda
(d) Major Munro
Answer:
(a) Robert Clive

Question 3.
During the great revolt of 1857 the Governor General of India was ……………….
(a) Lord Lytton
(b) Lord Ripon
(c) Lord Canning
Answer:
(c) Lord Canning

Question 4.
The British favoured the ……………….. to suppress the peasant uprising.
(a) religious leaders
(b) tribal leaders
(c) local population
(d) zamindars
Answer:
(d) zamindars

Question 5.
The first sign of unrest appeared at ………………
(a) Meerut
(b) Bareilly
(c) Barrackpore
Answer:
(c) Barrackpore

Question 6.
In ……………….. the sepoys at Barrackpur near Calcutta refused to go to Burma by sea, as it meant the loss of their caste.
(a) 1814
(b) 1824
(c) 1815
(d) 1826
Answer:
(b) 1824

Question 7.
The wife of Nawab of Oudh was ……………..
(a) Mumtaj Mahal
(b) Fathima Begum
(c) Begum Hazrat Mahal
Answer:
(c) Begum Hazrat Mahal

Question 8.
After the sepoy revolt, the first civil rebellion broke out in parts of:
(a) North-Western Provinces and Oudh
(b) Bengal and Bihar
(c) North-Eastern Provinces
(d) Central Provinces
Answer:
(a) North-Western Provinces and Oudh

Question 9.
Viceroy means ………………
(a) The representative of the crown
(b) General of an army
(c) Religious leader
Answer:
(a) The representative of the crown

Question 10.
The British Parliament adopted the ……………….. in November 1858.
(a) Lucknow Pact
(b) Swadeshi Movement
(c) Indian Government Act
(d) Rowlatt Act
Answer:
(c) Indian Government Act

Question 11.
The last Emperor of the Mughals …………..
(a) Babur
(b) Akbar
(c) Bahadur Shah – II
Answer:
(c) Bahadur Shah – II

Question 12.
The second half of the ……………….. saw the emergence of national political consciousness.
(a) 16th Century
(b) 17th Century
(c) 18th Century
(d) 19th Century
Answer:
(d) 19th Century

Question 13.
Queen Victoria’s proclamation of 1858 was described as ……………
(a) Human Rights
(b) Magna Carta
(c) citizenship
Answer:
(b) Magna Carta

Question 14.
Partition of Bengal was introduced by:
(a) Lord Lytton
(b) Lord Dalhousie
(c) Lord Rippon
(d) Lord Curzon
Answer:
(d) Lord Curzon

Question 15.
The Nawab of Bengal after Alivardi Khan was ……………..
(a) Tipu Sultan
(b) Siraj-ud-daulah
(c) Mir Questionaim
Answer:
(b) Siraj-ud-daulah

Question 16.
The British who lead the company’s army against Siraj-ud-daulah at plassey was …………..
(a) Robert Clive
(b) Lord Hastings
(c) Edmund Burke
Answer:
(a) Robert Clive

Question 17.
One-third of the population was wiped out from Bengal because ……………
(a) a terrible famine occurred there
(b) a civil war broke out
(c) an epidemic broke out
Answer:
(a) a terrible famine occurred there

Question 18.
The riots were ……………….
(a) cultivators
(b) zamindars
(c) traders
Answer:
(a) cultivators

Question 19.
The Indigo Commission was set up to enquire into the system of indigo production. Whom did the commission hold guilty?
(a) The riots
(b) The government
(c) The planters
Answer:
(c) The planters

Question 20.
After the indigo production collapsed in Bengal, the planters shifted their operation to ……………….
(a) Gujarat
(b) Bihar
(c) Odissa
Answer:
(b) Bihar

Question 21.
At which battle was the Nawab of Bengal Siraj-ud-daulah defeated by the East India Company?
(a) Battle of Plassey
(b) Battle of Buxar
(c) Battle of Panipat
(d) None of the above
Answer:
(a) Battle of Plassey

Question 22.
Who led the Farazi Movement after the death of Dudu Mian?
(a) Titu Mir
(b) Birsa Munda
(c) Sidhu
(d) Noah Mian
Answer:
(d) Noah Mian

Question 23.
Who provided leadership of the 1857 revolt in Bareilly?
(a) Begum Hazrat Mahal
(b) Khan Bahadur
(c) Nana Saheb
(d) Rani Lakshmi Bai
Answer:
(b) Khan Bahadur

Question 24.
The Indigo Revolt began in the year ……………
(a) 1857
(b) 1858
(c) 1859
(d) 1875
Answer:
(c) 1859

Question 25.
Who among the following was not an extremist leader?
(a) Lala Lajpat Rai
(b) Vallabhbhai Patel
(c) Bipin Chandra Pal
(d) Bai Gangadhar Tilak
Answer:
(b) Vallabhbhai Patel

II. Fill in the blanks :

1. The British territories were broadly divided into administrative units called …………..
2. The Royal charter could not prevent other European powers from entering the ………….. Markets.
3. The Bengal Nawabs asserted their power and autonomy after the death of …………..
4. ………….. was made the Nawab of Bengal after the defeat of Siraj-ud-daulah at Plassey.
5. The first Anglo-Maratha war ended with the treaty of …………..
6. Indigo cultivation was done under two main systems known as …………… and …………..
7. By the terms of the Permanent Settlement, the rajas and taluqdars were recognized as …………………
8. ……………… developed Ryotwari System which gradually extended all over South India.
9. The indigo villages were usually around indigo factories owned by ……………….
10. The planters at times pressurised the village between to sign the …………… on behalf of the ryots.
11. The British described the tribal people as …………….
12. Tribals went to work in the ………….. and the ………….. in Bihar.
13. The new law passed in 1850 made …………… into ………….. easier.
14. The Revolt of 1857 began from …………….
15. Mangal Pandey a young soldier was hanged to death for ………….. his officer in ……………
16. The Mughal Emperor Bahadur Shah Zafar died in the …………… jail.
17. In the countryside peasants and zamindars resented the …………….. and the rigid methods of ……………. collection.
18. Nana Saheb, the adopted son of the late Peshwa Baji Rao, gathered armed forces and expelled the……………….. from the city.
19. …………… fought a guerrilla war against the British with the support of several tribal and peasant leaders.
20. The section of Indian nationalists who had little patience for the non-political constructive programmes was called ……………..
21. After the Battle of Plassey, the British adopted the policy of ……………..
22. Many of the peasant revolts were led by …………….. leaders.
23. The Kol uprising took place in Chotanagpur and Singbhum region of Jharkhand and Orissa, under the leadership of ……………. and …………
24. After the 1857 revolt, the British captured Bahadur Shah and transported him to …………….
Answers:
1. Presidencies
2. Eastern
3. Aurangzeb
4. Mir Jafar
5. Salbai
6. Nij, ryots
7. Zamindars
8. Thomas Munro
9. Planters
10. contract
11. Savage
12. tea plantations, Coal mines
14. Meerut
15. Attacking, Barrackpore
16. Rangoon
17. High takes, revenue
18. British garrison
19. Nana Saheb
20. Militant nationalists
21. territorial expansion
22. religious
23. Bindrai, Singhrai
24. Burma

III. True or false

1. Siraj-ud-daulah got help from his commander Mir Jafar and finally won victory in the Battle of Plassey.
2. Birsa urged his followers to purify themselves, give up drinking liquor and stop believing in witchcraft and sorcery.
3. The British wanted to preserve the tribal way of life.
4. The traders and moneylenders never deceived the tribal people.
5. Many tribal groups did not like the colonial forest laws and therefore revolted.
Answers:
1. False
2. True
3. False
4. False
5. True

IV. Choose the correct statement.

Question 1.
(i) The Munda rebellion prompted the British to formulate a policy on tribal land.
(ii) The early objectives of the Congress were to develop and consolidate sentiments of national unity.
(iii) The Indian princes and zamindars were neither loyal to the British power.
(iv) The leaders of the 1857 revolt were defeated due to lack of weapons, organisation, discipline and betrayal by their aides.
(a) (i), (ii) and (iii) are correct
(b) (i), (ii) and (iv) are correct
(c) (i), (iii) and (iv) are correct
(d) (iii) and (iv) are correct
Answer:
(b) (i), (ii) and (iv) are correct

Question 2.
(i)The boycott and Swadeshi were liked to each other.
(ii) Pheroze Shah Mehta and Gokhale were the two important extremist leaders.
(iii) The methods of moderate leaders failed to yield any substantive change in the British attitude
(iv) The Poona Sarvajanik Sabha came into existence in 1870.
(a) (i) and (ii) are correct
(b) (ii), (iii) and (iv) are correct
(c) (i), (iii) and (iv) are correct
(d) (iii) and (iv) are correct
Answer:
(d) (iii) and (iv) are correct

Question 3.
Assertion (A): The plight of the indigo growers war miserable.
Reason (R): They were trapped in debts which were often passed from father to son.
(a) Both A and R are correct
(b) Both A and R are wrong
(c) Both A and R are correct and R is the correct explanation of A
(d) Both A and R are correct but R .is not the correct explanation of A
Answer:
(c) Both A and R are correct and R is the correct explanation of A

V. Match the following.
a.

1.	Mangal Pandey	(a)	Kanpur
2.	Bahadur Shah II	(b)	Lucknow
3.	Nana Saheb	(c)	Central India
4.	Begum Hazarat Mahal	(d)	Barrackpore
5.	Rani Lakshmi Bai	(e)	Delhi

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

b.

1.	Rani Lakshmi Bai	(a)	Mughal Emperor
2.	Bahadur Shah II	(b)	Colin Campbell
3.	The Great Revolt	(c)	Jhansi
4.	Lucknow	(d)	Magna Carta
5.	Queen Victoria’s Proclamation	(e)	1857

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

c.

1.	Birjis Qadr	(a)	Bihar
2.	Rani lakshmi bai	(b)	Faizabad
3.	Kunwar Singh	(c)	Jhansi
4.	Bakht Khan	(d)	Lucknow
5.	Ahmadullah Shan	(e)	Bareilly

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

d.

1.	Khunt Katti	(a)	Mirror of the Indigo
2.	Bethbehari	(b)	Joint holding
3.	Niladarpan	(c)	Self rule
4.	Swaraj	(d)	Great tumult
5.	Ulugulan rebellion	(e)	Forced labour

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

VI. Answer briefly:

Question 1.
How did the East India Company restructure the Mughal revenue system?
Answer:

1. British – The East India Company introduced land tenures and altered the agrarian relations.
2. Zamindars and other who collected the revenue never given the possession right on land.
3. No wide spread system of private ownership. In this way East India Company restructured the Mughal revenue system.

Question 2.
What was the main reason for the defeat of Siraj-ud-daulah at Plassey?
Answer:
Mir Jafar, one of Siraj-ud-daulah’s commanders did not fight the battle.

Question 3.
“Tribes” who are they?
Answer:

1. Tribes in India were and are very much part of the Indian society.
2. Infact they are the Indian peasantry subsisting through shifting cultivation.
3. The modem usage of word tribe in India restricts the definition to distinguish them from the rest of the Indian society, a stratified system based on caste.

Question 4.
How did the company purchase Indian goods?
Answer:
It purchased Indian goods with gold and silver imported from Britain.

Question 5.
What were the impacts of the transfer of power from East India Company to the Crown?
Answer:

1. After the transfer of power to the Crown.
2. De-industrialisation led to massive unemployment workers were forced out of the land.
3. Heavy taxation mined agriculture.
4. Famine deaths increased.
5. Peasants were forced to pay revenue directly to the Government.
6. Mortgaged land of the defaulters were auctioned by the moneylenders.
7. Peasants were trapped in the vicious-cycle of debt and forced to abandon cultivation.

Question 6.
When was Robert Clive appointed the Governor of Bengal?
Answer:
He was appointed the Governor of Bengal in 1764.

Question 7.
What were the two elements of early Indian response to colonialism?
Answer:
First response: Tribal uprisings and peasant rebellions made an attempts to restore the old order in the late 18th and early 19th century. It was restorative in nature.

Second response: In the second half of 19th-century response was in the ‘ form of Indian Nationalism emphasising on consciousness of unity and National aspiration.

Question 8.
What is indigo?
Answer:
It is a plant that produces rich blue colour used as a dye.

Question 9.
Why did cloth dyers prefer indigo to woad?
Answer:
Cloth dyers preferred indigo as a dye because it produced a rich blue colour whereas the dye from woad was pale and dull.

Question 10.
Mention different types of activities of tribal people.
Answer:
(a) Some practised Jhum cultivation
(b) Some were hunter-gatherers
(c) Some herded animals
(d) Some took to settled cultivation

Question 11.
Who was Birsa?
Answer:
Birsa belonged to a family of Mundas, a tribal group that lived in Chottanagpur.

Question 12.
What did people say about Birsa?
Answer:
People said that he had miraculous powers. He could cure all diseases and multiply grain.

Question 13.
What problems did Birsa set out to resolve?
Answer:
(a) The familiar ways of tribals seemed to be disappearing.
(b) Their livelihoods were under threat.
(c) The religion appeared to be in danger.

Question 14.
On what charges was Birsa convicted?
Answer:
Birsa was convicted on the charges of rioting.

Question 15.
When did Birsa die and how?
Answer:
He died of cholera in 1900 in jail.

Question 16.
Name any two smaller rulers who acknowledged the suzerainty of Bahadur Shah Zafar.
Answer:
Nana Saheb and Birjs Qadar.

Question 17.
Who was Tantia Tope?
Answer:
He was the general of Nana Saheb.

Question 18.
Name the important leaders and centres of the mutiny.
Answer:
Important leaders:

1. Rani Lakshmi Bai of Jhansi
2. Tantia Tope
3. Nana Saheb and Kunwar Singh.

Important centres:

1. Kanpur
2. Delhi
3. Lucknow

Question 19.
Why was not the revolt widespread?
Answer:

• There was no unity among the rebels.
• It did not extend beyond north India.
• South India, Punjab, Sind and Rajasthan kept quiet.
• The rulers of the Indian States who did not support the movement remained neutral.
• A large number of rulers of the Indian States and the big zamindars did not join the movement
• The educated Indians did not support the movement.

Question 20.
The early Indian response to colonial exploitation and the colonial political and economic domination consisted of two elements. Mention them.
Answer:
The two responses:
(i) The first response in the late eighteenth and early nineteenth centuries was restorative in nature. Both tribal uprising and peasant rebellion made an attempt to restore the older order.

(ii) The second response appeared in the second half of the nineteenth century in the form of Indian nationalism that progressively imagined India as a nation emphasising on a consciousness of unity, and national aspiration.

Question 21.
Mention some of the issues that added to a sense of resentment among the Indian peasants and tribals against the British.
Answer:
Some of the issues are – the concept of private property rights in land, rigorous collection of land revenue, encroachment of tribal land by the non-tribal people, the interference of Christian missionaries in the socio-religious life of the local people.

Question 22.
What is the significance of the peasant rebellion that erupted in the first half of the nineteenth century India?
(OR)
What were the characteristics of the peasant uprisings that began except in India in the first half of the nineteenth century ?
Answer:
(i) The peasant uprisings clearly showed an awareness of the power structure in rural society and a strong will to restructure the authority.

(ii) The rebels were quite familiar with the political source of oppression and demonstrated in their actions against the Zamindar houses, grain stocks, the money lenders, and the merchants.

(iii) At times, the British state machinery, which came forward to protect these local agents of oppression, was also attacked.

Question 23.
Highlight the reasons that make the Great Rebellion of 1857 so significant.
(OR)
Why are the events of 1857- 58 significant?
Answer:
The event of 1857-58 is significant for the following reasons,

1. This was the first major revolt of armed forces accompanied by civilian rebellions.
2. The revolt witnessed the unprecedented violence, peipetrated by both sides.
3. The revolt brought an end to the East India Company and the governance of the Indian subcontinent was taken over by the British Crown.

Question 24.
Mention some of the key demands of the Indian National Congress.
Answer:
Here are some of the key demands made by the Indian National Congress:

1. Creation of legislative councils at provincial and central level.
2. Increasing the number of elected members in the legislative council.
3. Separating judicial and executive functions.
4. Reducing military expenditure.
5. Reduction of home charges.

VII. Answer all the questions given under each caption:

Question 1.
Farazi – Movement

(a) When and by whom was Farazi movement launched?
Answer:
In 1818 Farazi movement was lunched by Shariatullah.

(b) What did he advocate?
Answer:
He advocated the participants to abstain from un islamic activities.

(c) What was the result of it?
Answer:
This resulted in conflict directly with zamindars and in turn with the British who favoured the zamindars.

(d) Who revived this movement in the 1870?
Answer:
This movement was revived by Noahmian in the 1870s.

Question 2.
Revolt at Kanpur.
(a) Who joined the rebels at Kanpur and with whom?
Answer:
Nana Saheb joined the rebels with his commander Tantia Tope.

(b) What happened to the English?
Answer:
The English surrendered to the rebel force. The English men, women and children were mercilessly massacred.

(c) Who defeated Nana Saheb?
Answer:
Sir Colin Campbell defeated Nana Saheb.

(d) When was Cawnpur brought under British control?
Answer:
Kanpur was brought under the control of the British by the middle November 1857.

Question 3.
State of indian After the Battle of Plassey

(a) What was the policy adopted by the British after Plassey battle?
Answer:
British adopted the policy of territorial expansion.

(b) Name the areas that have undergone changes?
Answer:
Army, police, judicial system, land revenue administration and other institutions of Governance had systematic changes.

(c) What was the early Indians response to?
Answer:
The early Indians response to Colonial exploitation, political and Economic Domination.

(d) What was the nature of the response?
Answer:
The response in the. late 18th and early 19th Century was restorative in nature.

Question 4.
Munda Rebellion.
(a) Mention two reasons that aggravated the miseries of Munda people.
Answer:

• The corrupt police
• Lack of access to justice and the disillusionment with Christian missionaries.

(b) Describe briefly the role of Birsa Munda in this rebellion.
Answer:
Birsa Munda gave an impetus to the movement by declaring himself as the messenger of God. He claimed that he had a prophecy and promised supernatural solution to the problem of Munda people. The Munda leaders utilised the cult of Birsa Munda to recruit more people to their cause. A serious of night meetings were held and subsequently a revolt was planned.

(c) When did the revolt take place? How can you say that it was a violent revolt?
Answer:
The revolt took place on the Christmas day in 1889. The rebels resorted to violence. They burnt down buildings and shot arrows at the Christian missionaries, and Munda Christian converts. They also made attacks on police stations and government officials.

(d) How was resistance crushed?
Answer:
Their resistance was crushed and Birsa Munda was arrested who later died in jail.

VIII. Answer the following in detail.

Question 1.
Discuss the four major trends discerned during the Swadeshi Movement in Bengal.
Answer:
Boycott and Swadeshi were the part of the plan to make India self-sufficient. Four major trends can be discerned during the Swadeshi Movement in Bengal.

The Moderate Trend:

1. Had faith in British rule and their sense of justice and democratic practice.
2. The moderate leaders were not ready to wrest power from British in one single movement.
3. Boycott and Swadeshi Movement was of limited significance to them.

Constructive Swadeshi:

1. Focused on self-help through Swadeshi industries national schools arbitration courts and constructive programmes in the villages.
2. Rejected the self-defeating modest approach of moderates.
3. It remained non-political in nature.

Militiant Nationalism:

1. Focused more on a relentless boycott of foreign goods.
2. Had little patience for the non – political constructive programmes.
3. They ridiculed the idea of self-help.

Revolutionary terrorism:

1. A radical response to the British rule with violent methods.
2. It marked the shift from the mass – based movement to individual action.
3. British officials who were anti – Swadeshi or repressive towards native population were targeted.

Question 2.
Mention the results of the Great Revolt of 1857.
Answer:

1. It put an end to the company’s rule in India in 1858. The administration of India was directly taken over by the British Crown.
2. By a special Act, both the Board of Control and the Court of Directors were abolished.
3. The office of the Secretary of State for India was created. He was assisted by an Indian council of 15 members.
4. The Governor General of India was designated as viceroy of India.
5. The policy of ruthless conquest in India was given up.
6. The Indian princess were given the right of adoption.

Social and religious changes:

1. Full religious freedom was guaranteed to Indians.
2. Indian were also given assurance that high posts would be given to them without any discrimination.

Military changes:

1. The Indian army was thoroughly re-organized.
2. The number of the European forces were increased.
3. The artillery was put under the charge of the British.

Queen Victoria’s Proclamation of 1858:

1. Queen Victoria’s Proclamation was issued in 1858.
2. According to Queen Victoria’s Proclamation the English East India. The administration of India was taken over by the British Crown. The revolt paved the way for the rise of the national movement and it served as a source of inspiration in the later struggle for freedom.

Question 3.
What do you know about Militant Nationalism?
Answer:

1. The following three were the well known extremist leaders of the Indian National Congress. Lala Lajpat Rai of Punjab, Bala Gangadhar Tilak of Maharashtra and Bipin Chandrapal of Bengal during the period of Swadeshi Movement.
2. They were referred to as Lai – Bal – Pal Triumvirate.
3. Punjab, Bengal and Maharashtra emerged as the hotbed of militant nationalism during the Swadeshi Movement.
4. In south India Tuticorin became the most important location of Swadeshi activity.
5. V.O.Chidambaranar launched Swadeshi Navigation Company as a part of it.
6. One of the common goals of the extremist leaders was to achieve Swaraj or Self-rule.
7. Political murders and individual acts of terrorism were not approved by the Militant leaders.

Question 4.
Describe the circumstances that led to the Indigo Revolt in 1859.
Answer:
(i) Natural Indigo dye was highly valued by cloth makers around the World. Many Europeans sought to make their fortunes by becoming Indigo planters in India. They employed Indian peasants to grow the Indigo which was processed into dye at the planters factories. The dye was then exported to Europe.

(ii) The Indian peasants were not interested in growing Indigo. Instead they wanted to grow food crops. But they were forced to grow indigo. The British planters gave them cash advances to help pay for the rent of the land or other costs. These advances needed to be repaid with interest.

(iii) At the end of the season, the planters paid the cultivators low prices for their indigo. Moreover, the small amount they earned was not enough to pay back the cash advance with interest. So, they fell into debt.

(iv) The peasants would be forced to enter into another contract to grow indigo. And same thing would be repeated once again. Thus, the peasants were never able to clear their debts. Debts were often passed from father to son.
Finally, they revolted against the oppressive system in 1859 which came to be known as the Indigo Revolt. The peasants took a collective decision and refused to grow indigo. The movement quickly spread far and wide.

IMPORTANT EVENTS AND YEARS:
Year – Events
1855 – Birth of Indian National Congress
1856 – The General Services Enlistment Act
1857 – The Great Revolt
1858 – Queen Victoria’s Proclamation
1858 – Darbar at Allahabad
1858 – Vernacular Press Act
1905 – Partition of Bengal, Swadeshi Movement
1906 – Formation of the Muslim league
1907 – Surat Split
1909 – Minto – Morley Reforms Act

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Tamilnadu Samacheer Kalvi 7th Tamil Solutions Term 2 Chapter 3.1 ஒரு வேண்டுகோள்

மதிப்பீடு

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
மயிலும் மானும் வனத்திற்கு ………………………… தருகின்ற ன.
அ) களைப்பு
ஆ) வனப்பு
இ) மலைப்பு
ஈ) உழைப்பு
Answer:
ஆ) வனப்பு

Question 2.
மிளகாய் வற்றலின் ……………… தும்மலை வரவழைக்கும்.
அ) நெடி
ஆ) காட்சி
இ) மணம்
ஈ) ஓசை
Answer:
அ) நெடி

Question 3.
அன்னை தான் பெற்ற ………….. ….. சிரிப்பில் மகிழ்ச்சி அடைகிறார்.
அ) தங்கையின்
ஆ) தம்பியின்
இ) மழலையின்
ஈ) கணவனின்
Answer:
இ) மழலையின்

Question 4.
வனப்பில்லை ‘ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………
அ) வனம் + இல்லை
ஆ) வனப்பு + இல்லை
இ) வனப்பு + யில்லை
ஈ) வனப் + பில்லை
Answer:
ஆ) வனப்பு + இல்லை

Question 5.
வார்ப்பு + எனில்’ என்பதனைச் சேர்தெழுதக் கிடைக்கும் சொல்
அ) வார்ப்எனில்
ஆ) வார்ப்பினில்
இ) வார்ப்பெனில்
ஈ) வார்ப்பு எனில்
Answer:
இ) வார்ப்பெனில்

நயம் அறிக

ஒரே எழுத்திலோ ஓசையிலோ முடியும் இயைபுச் சொற்களைப் பாடலில் இருந்து எடுத்து எழுதுக.

குறுவினா

Question 1.
தாய்மையின் ஓவியத்தில் நிறைந்திருக்க வேண்டியவை யாவை?
Answer:
அன்பும் பாசமும் தாய்மையின் ஓவியத்தில் நிறைந்திருக்க வேண்டும்.

Question 2.
ஒரு கலை எப்பொழுது உயிர்ப்புடையதாக அமையும்?
Answer:
மானுடப் பண்பு நிறைந்திருந்தால் ஒரு கலை உயிர்ப்புடையதாக அமையும்.

சிறுவினா

சிற்பங்களும் ஓவியங்களும் எவ்வாறு அமைய வேண்டும் என்று கவிஞர் கூறுகிறார்?
Answer:
(i) சிற்பங்கள் : ஒரு சிற்பி, பாறை உடைப்பவரின் சிலையைச் செதுக்கினால் அதிக வியர்வை நாற்றம் வீச வேண்டும். உழவனின் உருவச் சிலையாக இருந்தால் ஈரமண் வாசம் வீச வேண்டும்.

(ii) ஓவியங்கள் : ஓர் ஓவியன், தாயின் உருவத்தைத் தீட்டினால் அன்பும் பண்பும் மேலோங்கிட வேண்டும். சிறு குழந்தையின் சித்திரமானால் உடலெங்கும் பால் மணம் கமழ வேண்டும். சிற்பங்களும் ஓவியங்களும் இவ்வாறு அமைவதே சிறப்பு என்று கவிஞர் கூறுகிறார்.

சிந்தனை வினா

நீங்கள் ஒரு கலைஞராக இருந்தால் எத்தகைய படைப்புகளை உருவாக்குவீர்கள்?
Answer:
நான் ஒரு கலைஞராக இருந்தால் பச்சைப்பசேல் என விளங்கும் மலைகள், அங்கு விழும் அருவிகள், பயமறியாமல் பறக்கும் பறவைகள், புலி, மான், சிங்கம் என அனைத்து வனவாழ் விலங்குகளும் அச்சமின்றி அருவியில் நீர் அருந்துதல், ஒன்றையொன்று நட்புடன் நோக்குதல் இவற்றை உருவாக்குவேன்.

கற்பவை கற்றபின்

Question 1.
உங்களுக்குப் பிடித்த ஏதேனும் ஒரு கலை பற்றிய தகவல்களைத் திரட்டுக.
Answer:
மயிலாட்டம்: மயில் வடிவுள்ள கூட்டுக்குள் ஒருவர், தன் உருவத்தை மறைத்துக் கொண்டு, நையாண்டி மேளத்திற்கேற்ப ஆடும் ஆட்டமே மயிலாட்டமாகும். நையாண்டி மேளம் இசைக்க, காலில் கட்டப்பட்டுள்ள சலங்கை ஒலிக்க மயிலின் அசைவுகளை ஆடிக்காட்டுவர்.

கரகாட்டத்தின் துணையாட்டமாகவும் மயிலாட்டம் ஆடப்படுகிறது. ஊர்ந்து ஆடுதல், மிதந்து ஆடுதல், சுற்றி ஆடுதல், இறகை விரித்தாடுதல், தலையைச் சாய்த்தாடுதல், தாவியாடுதல், இருபுறமும் சுற்றியாடுதல், அகவுதல், தண்ணீ ர் குடித்துக் கொண்டே ஆடுதல் ஆகிய அடவுகளைக் கலைஞர்கள் இவ்வாட்டத்தில் ஆடிக் காட்டுவர்.

Question 2.
உழைப்பாளர்களின் பெருமையைக் கூறும் கவிதைகளைத் தொகுத்து வந்து வகுப்பறையில் பகிர்க.
Answer:
உழைப்பு இல்லையேல்
உணவும், மகிழ்ச்சியும்
இல்லை இன்று
உழைத்தால் வயதான போது
உட்கார்ந்து உண்ணலாம்
உழைக்க வேண்டிய காலத்தில்
உழைக்கவில்லை என்றால்
ஓய்வு எடுக்க வேண்டிய
காலத்தில் உழைக்க வேண்டும்
உயிரினங்கள் கூட உழைக்கின்றன
கையிருந்தும் உழைக்காமல்
இருந்தால் வாழ்க்கை இல்லாமலாகும்.

கூடுதல் வினாக்கள்

சொல்லும் பொருளும் :

1. பிரும்மாக்கள் – படைப்பாளர்கள்
2. நெடி – நாற்றம்
3. மழலை – குழந்தை
4. வனப்பு – அழகு
5. பூரிப்பு – மகிழ்ச்சி
6. மேனி – உடல்

நிரப்புக.

Question 1.
பூரிப்பு என்பதன் பொருள்
Answer:
மகிழ்ச்சி

Question 2.
தேனரசன் …………….. பணியாற்றியவர்.
Answer:
தமிழாசிரியராகப்

Question 3.
ஒரு வேண்டுகோள் கவிதை ……………. என்னும் நூலிலிருந்து எடுத்துத் தரப்பட்டுள்ளது.
Answer:
பெய்து பழகிய மேகம்

விடையளி :

Question 1.
தேனரசன் எழுதிய நூல்கள் யாவை?
Answer:

• மண்வாசல்
• வெள்ளை ரோஜா
• பெய்து பழகிய மேகம்.

Question 2.
தேனரசன் எந்த இதழ்களில் கவிதைகள் எழுதினார்?
Answer:

• வானம்பாடி
• குயில்
• தென்றல்

பாடலின் பொருள்

கலையுலகப் படைப்பாளர்களே! மண்ணின் அழகுக்கு அழகு சேர்ப்பவர்களே! உங்களுக்கு ஒரு மனித சமுதாயத்தின் வேண்டுகோள்!

நீங்கள் பாறை உடைப்பவரின் சிலையைச் செதுக்கினால், அதில் வியர்வை நாற்றம் வீசவேண்டும். உழவரின் உருவ வார்ப்பாக இருந்தால், அதில் ஈரமண்ணின் மணம் வீச வேண்டும்.

தாயின் மகிழ்ச்சியான உருவத்தை ஓவியமாக வரைந்தால், அவரின் முகத்தில் அன்பும் பாசமும் நிறைந்திருக்க வேண்டும். சிறு குழந்தையின் சித்திரத்தைத் தீட்டினால் அதன் உடலில் பால் மணம் கமழ வேண்டும்.

ஆல்ப்ஸ் மலைச் சிகரங்கள், அட்லாண்டிக் பெருங்கடல் அலைகள், அமேசான் காடுகள், பனிபடர் பள்ளத்தாக்குகள், தொங்கும் தோட்டங்கள் என இயற்கையின் விந்தைத் தோற்றங்கள் எவையும் கலைவடிவம் பெறலாம். ஆனால் அதில் மானுடப் பண்பு கட்டாயமாக இருக்க வேண்டும். மானுடம் இல்லாத எந்த அழகும் அழகன்று. மனிதன் கலக்காத எதிலும் உயிர்ப்பில்லை

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

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Samacheer Kalvi 11th Chemistry Chapter 2 Quantum Mechanical Model of Atom Textual Evaluation Solved

I. Choose the correct answer
Quantum Mechanical Model Of Atom Class 11 Book Back Answers Question 1.
Electronic configuration of species M²⁺ is 1s² 2s² 2p⁶3s² 3p⁶ 3d⁶ and its atomic weight is 56. The number of neutrons in the nucleus of species M is ………..
(a) 26
(b) 22
(c) 30
(d) 24
Answer:
(c) 30
Solution:
M²⁺ : 1s² 2s² 2p⁶3s² 3p⁶ 3d⁶
M : 1s² 2s² 2p⁶3s² 3p⁶ 3d⁸
Atomic number = 26
Mass number = 56
No. of neutrons = 56 – 26 = 30.

11th Chemistry Quantum Mechanical Model Of Atom Question 2.
The energy of light of wavelength 45 nm is
(a) 6.67 x 10¹⁵ J
(b) 6.67 x 10¹¹ J
(c) 4.42 .x 10¹⁸ J
(d) 4.42 x 10^-15 J
Answer:
(c) 4.42 .x 10¹⁸ J
Solution:
E = hv = hc / λ
\(\frac{6.626 \times 10^{-34} \mathrm{J} \mathrm{s} \times 3 \times 10^{8} \mathrm{ms}^{-1}}{45 \times 10^{-9} \mathrm{m}}\) = 4.42 .x 10¹⁸ J.

11th Chemistry Chapter 2 Book Back Answers Question 3.
The energies E₁ and E₂ of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e. λ₁ and λ₂ will be …………
(a) \(\frac{\lambda_{1}}{\lambda_{2}}=1\)
(b) λ₁ = 2 λ₂
(c) λ₁ = \(\sqrt{25 \times 50} \lambda_{2}\)
(d) 2 λ₁ = λ₂
Answer:
(b) λ₁ = 2 λ₂
Solution:
\(\frac{E l}{E 2}\) = \(\frac{25eV}{50eV}\) = \(\frac{1}{2}\)
\(\frac{\mathrm{hc}}{\lambda_{1}} \times \frac{\lambda_{2}}{\mathrm{hc}}\) = \(\frac{1}{2}\)
2λ₂ = λ_1.

Samacheer Kalvi Guru 11th Chemistry Question 4.
Splitting of spectral lines in an electric field is called …………….
(a) Zeeman effect
(b) Shielding effect
(c) Compton effect
(d) Stark effect
Answer:
(d) Stark effect
Solution:
Splitting of spectral lines in magnetic field is called Zeeman effect and splitting of spectral lines in electric field, is called Stark effect.

Samacheerkalvi.Guru 11th Chemistry Question 5.
Based on equation E = -2.178 x 10¹⁸ J \(\left(\frac{z^{2}}{n^{2}}\right)\) certain conclusions are written. Which of them is not correct ? (NEET)
(a) Equation can be used to calculate the change in energy when the electron changes orbit
(b) For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit
(c) The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.
(d) Larger the value of n, the larger is the orbit radius.
Answer:
(b) For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit
Solution:
Correct statement:
For n = 6, the electron has more negative energy than it does for n = 6 which means that the electron is strongly bound in the smallest allowed orbit.

11th Chemistry Samacheer Kalvi Question 6.
According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?
(a) n = 6 to n = 1
(b) n = 5 to n = 4
(c) n = 5 to n = 3
(d) n = 6 to n = 5
Answer:
(d) n = 6 to n = 5
Solution:
n = 6 to n = 5
E₆ = -13.6 / 6² ; E₅ = – 13.6 / 5²
E₆ – E₅ = (-13.6 / 6²) – (-13.6 / 5²)
= 0.166 eV atom^-1
E₅ – E₄ = (-13.6 / 5²) – (-13.6 / 4²)
= 0.306 eV atom^-1

Samacheer Kalvi.Guru 11th Chemistry Question 7.
Assertion : The spectrum of He⁺ is expected to be similar to that of hydrogen
Reason : He⁺ is also one electron system,
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false
(d) If both assertion and reason are false
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.

Samacheer Kalvi Guru 11 Chemistry Question 8.
Which of the following pairs of d-orbitals will have electron density along the axes ? (NEET Phase – II)
(a) d_z², d_xz
(b) d_xz, d_yz
(c) d_z², \(d_{x^{2}-y^{2}}\)
(d) d_xy, \(d_{x^{2}-y^{2}}\)
Answer:
(c) d_z², \(d_{x^{2}-y^{2}}\)

Samacheer Kalvi 11th Chemistry Question 9.
Two electrons occupying the same orbital are distinguished by …………
(a) azimuthal quantum number
(b) spin quantum number
(c) magnetic quantum number
(d) orbital quantum number
Answer:
(b) spin quantum number
Solution:
Spin quantum number For the first electron ms = +\(\frac {1}{2}\)
For the second electron ms = –\(\frac {1}{2}\).

Quantum Mechanical Model Of Atom Question 10.
The electronic configuration of Eu (atomic no. 63) Gd (atomic no. 64) and Tb (atomic no. 65) are (NEET – Phase II)
(a) [Xe] 4f⁷ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁸ 5d¹ 6s²
(b) [Xe] 4f⁷, 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
(c) [Xe] 4f⁷ , 6s², [Xe] 4f⁸ 6s² and [Xe] 4f⁸ 5d¹ 6s²
(d) [Xe] 4f⁸ 5d¹ 6s²[Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
Answer:
(b) [Xe] 4f⁷, 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
Solution:
Eu : [Xe] 4f⁷, 5d⁰, 6s²
Gd : [Xe] 4f⁷, 5d¹, 6s²
Tb : [Xe] 4f⁹, 5d⁰,6s²

Chemistry Class 11 Samacheer Kalvi Question 11.
The maximum number of electrons in a sub shell is given by the expression …………..
(a) 2n²
(b) 21 + 1
(c) 41 + 2
(d) none of these
Answer:
(c) 41 + 2
Solution:
2 (21 + 1) = 41 + 2.

11th Chemistry 2nd Lesson Question 12.
For d-electron, the orbital angular momentum is ………….
(a) \(\frac{\sqrt{2} h}{2 \pi}\)
(b) \(\frac{\sqrt{2 \mathrm{h}}}{2 \pi}\)
(c) \(\sqrt{2 \times 4}\)
(d) \(\frac{\sqrt{6} h}{2 \pi}\)
Answer:
(d) \(\frac{\sqrt{6} h}{2 \pi}\)
Solution:
Orbital angular momentum
= \(\sqrt{(1(1+1)}) \mathrm{h} / 2 \pi\)
For d orbital = \(\sqrt{(2 × 3)} \mathrm{h} / 2 \pi\) = \(\sqrt{6} \mathrm{h} / 2 \pi\).

Quantum Mechanical Model Of Atom Class 11 Question 13.
What is the maximum numbers of electrons that can be associated with the following set of quantum numbers ? n = 3,l = 1 and m = -1
(a) 4
(b) 6
(c) 2
(d) 10
Answer:
(c) 2
Solution:
n = 3; l = 1; m = -1 either 3p_x or 3p_y

Samacheer Kalvi Guru Chemistry Question 14.
Assertion: Number of radial and angular nodes for 3p orbital are l, l respectively. Reason: Number of radial and angular nodes depends only on principal quantum number.
(a) both assertion and reason are true and reason is the correct explanation of assertion.
(b) both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(c) assertion is true but reason is false
Solution:
No. of radial node = n- l – 1
No. of angular node = l for 3p orbital
No. of angular node = l =1
No. of radial node = n- l – 1 = 3 – 1 – 1 = 1.

Question 15.
The total number of orbitals associated with the principal quantum number n = 3 is ………..
(a) 9
(b) 8
(c) 5
(d) 7
Answer:
(a) 9
Solution:
n = 3; l = 0; m₁ = 0 – one s orbital n = 3; l = 1; m₁ = -1, 0, 1 – three p orbitals n = 3; l = 2; m₁ = -2, -1, 0, 1, 2 – five d orbitals, overall nine orbitals are possible.

Question 16.
If n = 6, the correct sequence for filling of electrons will be, …………
(a) ns → (n – 2) f → (n – 1)d → np
(b) ns → (n – 1) d → (n – 2) f → np
(c) ns → (n – 2) f → np → (n – 1) d
(d) none of these are correct
Answer:
(a) ns → (n – 2)f → (n – l)d → np
Solution:
n = 6 According Aufbau principle,
6s → 4f → 5d → 6p
ns → (n – 1)f → (n – 2)d → np.

Question 17.
Consider the following sets of quantum numbers:

Which of the following sets of quantum number is not possible ?
(a) (i), (ii), (iii) and (iv)
(b) (ii), (iv) and (v)
(c) (z) and (iii)
(d) (ii), (iii) and (iv)
Answer:
(b) (ii), (iv) and (v)
Solution:
(ii) l can have the values from 0 to n – 1 n = 2; possible l values are 0, 1 hence l = 2 is not possible.
(iv) for l = 0; m = -1 not possible
(v) for n = 3 l = 4 and m = 3 not possible.

Question 18.
How many electrons in an atom with atomic number 105 can have (n + 1) = 8 ?
(a) 30
(6) 17
(c) 15
(d) unpredictable
Answer:
(b) 17
Solution:
n + 1 = 8
Electronic configuration of atom with atomic number 105 is [Rn] 5f¹⁴ 6d³ 7s²

Orbital	(n+1)	No. of electrons
5f	5 + 3 = 8	14
6d	6 + 2 = 8	3
7s	7 + 0 = 0	2
No. of electrons = 14 + 3 = 17

Question 19.
Electron density in the yz plane of 3 d_x²_-y² orbital is …………….
(a) zero
(b) 0.50
(c) 0.75
(d) 0.90
Answer:
(a) zero

Question 20.
If uncertainty in position and momentum are equal, then minimum uncertainty in velocity is ……….
(a) \(\frac{1}{m} \sqrt{\frac{h}{\pi}}\)
(b) \(\sqrt{\frac{\mathrm{h}}{\pi}}\)
(c) \(\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)
(d) \(\frac{\mathrm{h}}{4 \pi}\)
Answer:
(c) \(\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)
Solution:

Question 21.
A macroscopic particle of mass 100 g and moving at a velocity of 100 cm s^-1d will have a de Broglie wavelength of ………….
(a) 6.6 x 10^-29 cm
(b) 6.6 x 10^-30 cm
(c) 6.6 x 10^-31 cm
(d) 6.6 x 10^-32 cm
Answer:
(c) 6.6 x 10^-31 cm
Solution:
m = 100 g = 100 x 10^-3 kg
v = 100 cm s^-1 = 100 x 10^-2 m s^-1
λ = \(\frac{h}{mv}\) =
= 6.626 x 10^-31 ms^-1
= 6.626 x 10^-31 cm s^-1

Question 22.
The ratio of de Broglie wavelengths of a deuterium atom to that of an a – particle, when the velocity of the former is five times greater than that of later, is ……………
(a) 4
(b) 0.2
(c) 2.5
(d) 0.4
Answer:
(d) 0.4

Question 23.
The energy of an electron in the 3rd orbit of hydrogen atom is -E. The energy of an electron in the first orbit will be ……………..
(a) – 3E
(b) – E /3
(c) – E / 9
(d) – 9E
Answer:
(c) – E / 9
Solution:
E_n  = \(\frac{-13.6}{n^{2}}\) eV atom^-1
E₁ = \(\frac{-13.6}{1^{2}}\)13.6 = \(\frac{-13.6}{9}\)
Given that,
E₃ = – E
\(\frac{-13.6}{9}\) = -E
13.6 = – 9E = E₁ = – 9E
E₁ = – 9E

Question 24.
Time independent Schnodinger wave equation is ………….

Answer:
(a) \(\widehat{\mathrm{H}} \psi=\mathrm{E} \psi\).

Question 25.
Which of the following does not represent the mathematical expression for the Heisenberg uncertainty principle?
(a) ∆E.∆p ≥ h/4π
(b) ∆E.∆v ≥ h/4πm
(c) ∆E.∆t ≥ h/4π
(d) ∆E.∆x ≥ h/4π
Answer:
(d) ∆E.∆x ≥ h/4π.

II. Write brief answer to the following questions

Question 26.
Which quantum number reveal information about the shape, energy, orientation and size of orbitals?
Answer:
Magnetic quantum number reveal information about the shape, energy, orientation and size of orbitals.

Question 27.
How many orbitals are possible for n =4?
Answer:
If n = 4, the possible number of orbitals are calculated as follows –
n = 4, main shell = N
If n = 4, l values are 0, 1, 2, 3
If l = 0,  4s orbital = 1 orbital
If l = 1,  m = -1,0, +1 = 3 orbitals
If l = 2,  m = -2,-1,0, +1,+2 = 5 orbitals
If l = 3,  m = -3,-2,-1,0, +1,+2,+3 = 7 orbitals
∴ Total number of orbitals = 16 orbitals

Question 28.
How many radial nodes for 2s, 4p, 5d and 4f orbitals exhibit? How many angular nodes?
Answer:
Formula for total number of nodes = n – 1

1. For 2s orbital: Number of radial nodes =1.

2. For 4p orbital: Number of radial nodes = n – l – 1. = 4 – 1 – 1 = 2
Number of angular nodes = l
∴ Number of angular nodes = 1
So, 4p orbital has 2 radial nodes and 1 angular node.

3. For 5d orbital:
Total number of nodes = n – 1 = 5 – 1 = 4 nodes
Number of radial nodes = n – l – 1 = 5 – 2 – 1 = 2 radial nodes.
Number of angular nodes = l = 2
∴ 5d orbital have 2 radial nodes and 2 angular nodes.

4. For 4f orbital:
Total number of nodes = n – 1 = 4 – 1 = 3 nodes
Number of radial nodes = n – 7 – 1 = 4 – 3 – 1 = 0 node.
Number of angular nodes = l = 3 nodes
∴ 4f orbital have 0 radial node and 3 angular nodes.

Question 29.
The stabilization of a half filled d-orbital is more pronounced than that of the p-orbital why?
Answer:
The exactly half filled orbitals have greater stability. The reason for their stability are –

1. symmetry
2. exchange energy.

(1) Symmetry:
The half filled orbitals are more symmetrical than partially filled orbitals and this symmetry leads to greater stability.

(2) Exchange energy:
The electrons with same spin in the different orbitals of the same sub shell can exchange their position. Each such exchange release energy and this is known as exchange energy. Greater the number of exchanges, greater the exchange energy and hence greater the stability. In d-orbital, 10 exchanges are possible but in p-orbital 6 exchanges are possible. So, d – orbital with 5 unpaired electrons (10 exchanges)n i.e. half filled is more stable than p – orbital with 3 unpaired electrons (6 exchanges).

Question 30.
Consider the following electronic arrangements for the d5 configuration.

(1) Which of these represents the ground state
(2) Which configuration has the maximum exchange energy.
Answer:
(1)  – This represents the ground state.
(2)   – This represents the maximum exchange energy.

Question 31.
State and explain Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle states that “No two electrons in an atom can have the same set of values of all four quantum numbers”.
Illustration: H(Z = 1) 1s¹.
One electron is present in hydrogen atom, the four quantum numbers are n = 1, l = 0, m = 0 and s = + \(\frac {1}{2}\). For helium Z = 2. He: 1s². In this one electron has the quantum number same as that of hydrogen, n = 1,l = 0, m = 0 and s = +½ For other electron, fourth quantum number is different, i.e. n = 1, l = 0, m = 0 and s = – ½.

Question 32.
Define orbital? What are the n and l values for 3p_x and 4 d_x²_-y² electron?
Answer:
(i) Orbital is a three dimensional space which the probability of finding the electron is maximum.

Question 33.
Explain briefly the time independent Schrodinger wave equation?
Answer:
The time independent Schrodinger equation can be expressed as
\(\widehat{\mathrm{H}}\) Ψ = EΨ ……………(1)
Where \(\widehat{\mathrm{H}}\) is called Hamiltonian operator.
Ψ is the wave function.
E is the energy of the system.

Since Ψ is a function of position coordinates of the particle and is denoted by Ψ (x, y, z)
∴ Equation (1) can be written as,

Multiply the equation (3) by \(\widehat{\mathrm{H}}\) and rearranging

The above equation (4) Schrodinger wave equation does not contain time as a variable and is referred to as time independent Schrodinger wave equation.

Question 34.
Calculate the uncertainty in position of an electron, if ∆v = 0.1% and n = 2.2 x 10⁶ ms^-1.
Answer:
Mass of an electron = m = 9.1 x 10^-31 kg.
∆v = Uncertainty in velocity = \(\frac {0.1}{100}\) x 2.2 x 10³ ms^-1 .
∆v = 0.22 x 10⁴ = 2.2 x 10³ ms^-1
∆x . ∆v . m = \(\frac {h}{4π}\)
∆x = \(\frac {h}{∆v . m x 4π}\)

= 0.02635 x 10^-6
∆x = 2.635 x 10^-8
Uncertainty in position = 2.635 x 10^-8.

Question 35.
Determine the values of all the four quantum numbers of the 8th electron in O – atom and 15^th electron in Cl atom and the last electron in chromium.
Answer:
(1) O (Z = 8) 1s² 2s² 2p_x² 2p_y¹ 2p_z¹
Four quantum numbers for 2p_x¹ electron in oxygen atom:
n = principal quantum number = 2
l = azimuthal quantum number =1
m = magnetic quantum number =+1
s = spin quantum number = +\(\frac {1}{2}\)

(2) Cl (Z = 17) 1s² 2s² 2p⁶ 3s² 3p_x² 3p_y² 3p_z¹
Four quantum numbers for 15^th electron in chlorine atom:
n = 3, l = 1, m = 0, s = + ½

(3) Cr (Z = 24) 1s² 2s² 2p² 3s² 3p² 3d² 4s¹
n = 3, l = 2, m = +2, s = + ½

Question 36.
The quantum mechanical treatment of the hydrogen atom gives the energy value:
E_n = \(\frac{-13.6}{n^{2}}\) eV atom^-1

1. use this expression to find ∆E between n = 3 and n = 4
2. Calculate the wavelength corresponding to the above transition.

Answer:
(1) When n = 3
E₃ = \(\frac{-13.6}{3^{2}}\) = \(\frac {-13.6}{9}\) = – 1.511 eV atom^-1
When n = 4 E₄ = \(\frac{-13.6}{4^{2}}\) = – 0.85 eV atom^-1
∆E = E₄ – E₃ = – 0.85 – (-1.511) = + 0.661 eV atom
∆E = E₃ – E₄
= – 1.511 – (-0.85)
= – 0.661 eV atom^-1

(2) Wave length = λ
∆E = \(\frac {hc}{ λ}\)
λ = \(\frac {hc}{∆E}\)
h = Planck’s constant = 6.626 x 10^-34 Js^-1
c = 3 x 10⁸ m/s
λ = \(\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{0.661}\)
= 10.02 x 10^-34 x 3 x 10⁸
= 30 x 10^-26
λ = 3 x 10^-25 m.

Question 37.
How fast must a 54 g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green light 5400 Å?
Answer:
m = mass of tennis ball = 54 g = 5.4 x 10^-2 kg.
λ = de Broglie wavelength = 5400 Å. = 5400 x 10^-10 m.
V = velocity of the ball = ?
λ = \(\frac {h}{mV}\)
V = \(\frac {h}{λ.m}\)

= 0.2238 x 10^-24
= 2.238 x 10^-25 m.

Question 38.
For each of the following, give the sub level designation, the allowable m values and the number of orbitals.

1. n = 4, l = 2,
2. n = 5, l = 3
3. n = 7, l = 0

Answer:
1. n = 4, l = 2
If l = 2, ‘m’ values are -2, -1, 0, +1, +2
So, 5 orbitals such as d_xy,d_yz,d_xz,\(d_{x^{2}-y^{2}}\) and d^_z

2. n = 5 , l = 3
If l = 3, ‘m’ values are -3, -2, -1, 0, +1, +2, +3
So, 7 orbitals such as f z, fxz, fyz, fxyz, fz(x2 y2)’ ^x(x2-3y2)’ ^y(3×2 ??y

3. n = 7 , l = 0
If l = 0, ‘m’ values are 0. Only one value.
So, 1 orbital such as 7s orbital.

Question 39.
Give the electronic configuration of Mn²⁺ and Cr³⁺
Answer:
1. Mn (Z = 25)
Mn → Mn²⁺ + 2e^–
Mn²⁺ electronic configuration is 1s 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

2. Cr (Z = 24)
Cr → Cr³⁺ + 3e^–
Cr³⁺ electronic configuration is Is² 2s² 2p⁶ 3s²3p⁶ 3d³

Question 40.
Describe the Aufbau principle.
Answer:
In the ground state of the atoms, the orbitals are filled in the order of their increasing energies. That is the electrons first occupy the lowest energy orbital available to them. Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals.
The order of filling of various orbitals as per Aufbau principle is –
1 s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d ………..
For e.g., K (Z =19)
The electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹.
After filling 4s orbital only we have to fill up 3d orbital.

Question 41.
A n atom of an element contains 35 electrons and 45 neutrons. Deduce

1. the number of protons
2. the electronic configuration for the element
3. All the four quantum numbers for the last electron

Answer:
An element X contains 35 electrons, 45 neutrons

1. The number of protons must be equal to the number of electrons. So the number of protons = 35.
2. Number of electrons = 35. So the electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵.
3. The last electron i.e. 5^th electron in 4p orbital has the following quantum numbers. n = 4, l = 1, m =+1, s = + \(\frac {1}{2}\)

Question 42.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wave length associated with the electron revolving around the nucleus.
Answer:
In order for the electron wave to exist in phase, the circumference of the orbit should be an integral multiple of the wavelength of the electron wave. Otherwise, the electron wave is out of phase.
mvr = nh / 2π, 2πr = nλ,
where mvr = angular momentum
where 2πr = circumference of the orbit

n = 3, n = 4

Question 43.
Calculate the energy required for the process.
He⁺_(g) → He²⁺_(g) + e^–
The ionization energy for the H atom in its ground state is – 13.6 eV atom^-1.
Answer:
The ionization energy for the H atom in its ground state =-13.6 eV atom^-1.
Ionization energy = \(\frac{13.6 z^{2}}{n^{2}}\) eV
Z = atomic number
n = principal quantum number or shell number
For He, n = 1, z = 2
IE = \(\frac{-13.6 \times 2^{2}}{1^{2}}\)eV.

Question 44.
An ion with mass number 37 possesses unit negative charge. If the ion contains 11.1% more neutrons than electrons. Find the symbol of the ion.
Answer:
Let the number of electrons in an ion = x
number of neutrons = n = x + \(\frac{11.1}{100}\) eV = 1.111 x
(As the number of neutrons are 11.1% more than the number of electrons)
In the neutral of atom, number of electron.
e^– = x – 1 (as the ion carries -1 charge)
Similarly number of protons = P = x – 1 Number of protons + number of neutrons = mass number = 37
(x – 1) + 1.111 x = 37 .
2.111 x = 37 +1
2.111 x = 38
x = \(\frac{38}{2.111}\) = 18.009 = 18
∴ Number of protons = atomic number – 1 = 18-1 = 17
∴ The symbol of the ion = \(_{17}^{37} \mathrm{Cl}\).

Question 45.
The Li²⁺ ion is a hydrogen like ion that can be described by the Bohr model. Calculate the Bohr radius of the third orbit and calculate the energy of an electron in 4^th orbit.
Answer:
Li²⁺ hydrogen like ion.
Bohr radius of the third orbit = r₃ = ?
r₃ = \(\frac{(0.529) n^{2}}{Z}\) A
Where n = shell number, Z = atomic number.
r₃ = \(\frac{(0.529) 3^{2}}{3}\) A [∴for lithium Z = 3, n = 3]
= \(\frac{0.529 x 9}{3}\)
r₃ = l.587Å
E_n = \(\frac{(-13.6) Z^{2}}{n^{2}}\) eV atom^-1.
E₄ = Energy of the fourth orbit = ?
E₄ = \(\frac{(-13.6) \times 3^{2}}{4^{2}}\) = \(\frac{-13.6 \times 9}{16}\) = -7.65 eV atom^-1
E₄ = – 7.65 eV atom^-1

Question 46.
Protons can be accelerated in particle accelerators. Calculate the wavelength (in Å)of such accelerated proton moving at 2.85 × 108 ms^-1 (the mass of proton is 1.673 x 10^-27 Kg).
Answer:
m = mass of the proton = 1.673 x 10^-27 Kg
v = velocity of the proton = 2.85 x 10⁸ ms^-1
λ = \(\frac{h}{mv}\)
h = Planck’s constant = 6.626 x 10³⁴ Kg m² s^-1

Wavelength of proton = λ = 1.389 x 10^-15 m.

Question 47.
What is the de Broglie wavelength (in cm) of a 160g cricket ball travelling at 140 Km hr^-1.
Answer:
m = mass of the cricket ball = 160g = 0.16 kg.
v = velocity of the cricket ball =140 Km h^-1
= \(\frac {140 x 5}{18}\) = 38.88 ms^-1
de Broglie equation = λ = \(\frac{h}{mv}\)
h = Planck’s constant = 6.626 x 10^-34 kg m² s^-1

λ = 1.065 x 10^-34m
Wave length in cm = 1.065 x 10^-34 x 100
= 1.065 x 10^-32 cm.

Question 48.
Suppose that the uncertainty in determining the position of an electron in an orbit is 0.6 A. What is the uncertainty in its momentum?.
Answer:
∆x = uncertainty in position of an electron = 0.6 Å = 0.6 x 10^-10 m.
∆p = uncertainty in momentum = ?
Heisenberg’s uncertainty principle states that,
∆x.∆p ≥ \(\frac{h}{4π}\)
∆p = \(\frac{h}{4π.∆x}\)
h = Planck’s constant = 6.626 x 10^-34 kg m² s^-1

Uncertainty in momentum = 0.8792 x 10^-24 kg ms^-1 (or) = 8.792 x 10^-25 kg ms^-1

Question 49.
Show that if the measurement of the uncertainty in .the location of the particle is equal to its de Broglie wavelength, the minimum uncertainty in its velocity is equal to its velocity /4π
Answer:
If, uncertainty in position = ∆x = λ , the value of uncertainty in velocity = \(\frac{v}{4π}\)
Heisenberg’s principle states that
∆x.∆v. m = \(\frac{h}{4π}\) …………(1)
de Broglie equation states that
λ = \(\frac{h}{mv}\) ………….(2)
∴ h = λ .m.v …………(3)
∆x = \(\frac{h}{∆v.4π}\) ………….(4)
Substituting the value of h in equation (4)
∆x = \(\frac{λ x m. v}{∆v.4π.m}\)
if ∆x = λ
∆v =  = \(\frac{v}{4π}\)

Question 50.
What is the de Broglie wave length of an electron, which is accelerated from the rest, through a potential difference of 100V?
Answer:
Potential difference = V = 100 V
Potential energy = eV = 1.609 x 10^-19c x 100V
\(\frac{v}{4π}\) m v² = 1.609 x 10^-19 x 100
\(\frac{v}{4π}\) m v² = 1.609 x 10^-19 J
v² = \(\frac{2 \times 1.609 \times 10^{-17}}{m}\)
m = mass of electron = 9.1 x 10^-31 Kg

v = 5.93 x 10⁶ m/s
λ = \(\frac{h}{mv}\) where h = 6.62 x 10^-34 JS
= \(\frac{6.62 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.93 \times 10^{6}}\)
= 1.2x 10^-10m
A= 1.2 Å.

Question 51.
Identify the missing quantum numbers and the sub energy level

Answer:

Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom In-Text Questions – Evaluate Yourself

Question 1.
Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 k eV.
Answer:
λ = \(\frac{h}{mv}\)
Potential difference of an electron = V = 1 keV.
Potential energy = \(\frac{1}{2}\) mv² = eV
e = charge of an electron = 1.609 x 10^-19c
l k V = 1000 V
:. Potential energy = 1.609 x 10^-19 x 1000 = 1.609 x 10^-19
\(\frac{1}{2}\) mv² = 1.609 x 10^-16V
m = 9.1 x 10^-31 kg
λ = \(\frac{h}{mv}\)

= 1.2 x 10^-11 m
λ = 1.2 x 10^-11 m.

Question 2.
Calculate the uncertainty in the position of an electron, if the uncertainty in its velocity is 5.7 x 10 s ms^-1.
Answer:
Uncertainty in velocity = Av = 5.7 x 10⁵ ms^-1
Mass of an electron = m = 9.1 x 10^-31 kg.
Uncertainty in position = ∆x = ?
∆x.m.∆v = \(\frac{h}{4π}\)

= 1 x 10^-10m.
Uncertainty in position = 1 x 10^-10 m.

Question 3.
How many orbitals are possible in the 4th energy level? (n = 4)
Answer:
n = 4
Number of orbitals in 4^th energy level = ?
When n = 4, l = 0,1,2,3
If l = 0 orbital = 4s =1
If l = 1 orbital = 4p_x, 4p_y, 4p_z = 2
If l = 2 orbital = \(4 \mathrm{d}_{\mathrm{xy}}, 4 \mathrm{d}_{\mathrm{yz}}, 4 \mathrm{d}_{\mathrm{zx}}, 4 \mathrm{d}_{\mathrm{x}} 2_{\mathrm{y}}, 2,4 \mathrm{d}_{\mathrm{z}^{2}}\) = 5
If l = 3 orbital = -3,-2, -1, 0, +1, +2, +3 = 7
Number of orbitals in 4th energy level = 16.

Question 4.
Calculate the total number of angular nodes and radial nodes present in 3d and 4f orbitals.
Answer:
Number of angular nodes in 3d orbital = ?
Number of radial nodes in 3d orbital = ?
Number of angular nodes = l
Number of radial nodes = n – l – 1

1. For 3d orbital:
Number of angular nodes = 2 because l = 2
Number of radial nodes = 3 – 2 -1 = 0
Total number of nodes in 3d orbital = 2

2. For 4f orbital:
Number of angular nodes = 3 because l = 3
Number of radial nodes = n – l – l =4 – 3 – 1 = 0
Total number of nodes in 4f orbital = 3.

Question 5.
Energy of an electron in hydrogen atom in ground state is -13.6 eV. What is the energy of the electron in the second excited state?
Answer:
Energy of an electron in ground state = -13.6 eV.
∴ Energy of an electron in the second excited state = E₂.
n = 2
E₂ = \(\frac{-13.6 \mathrm{eV}}{\mathrm{n}^{2}}\) = \(\frac{-13.6}{2^{2}}\) = \(\frac{-13.6}{4}\) = -3.4 eV.

Question 6.
How many unpaired electrons are present in the ground state of Fe³⁺ (z = 26), Mn2+ (z = 25) and argon (z=18)?
Answer:

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 3d⁵ for Fe³⁺ ion.
So, it contain 5 unpaired electrons.
Mn (Z = 25). Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 3d⁵
Mn → Mn²⁺ + 2e^–
Number of unpaired electrons in Mn²⁺ = 5
Ar (Z = 18). Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶.
All orbitals are completely filled. So, no unpaired electrons in it.

Question 7.
Explain the meaning of the symbol 4f². Write all the four quantum numbers for these electrons.
Answer:
4f² : It means that the element has 2 electrons in outermost 4f shell.
Quantum number values are,
n = principal quantum number = 4
l = azimuthal quantum number = 3
m = magnetic quantum number = – 3, -2
s = spin quantum number = +\(\frac {1}{2}\) – \(\frac {1}{2}\).

Question 8.
Which has the stable electronic configuration? Ni²⁺ or Fe³⁺
Answer:
Ni (Z = 28). 1s² 2s² 2p⁶ 3s² 3p⁶4s²3d⁸
Ni²⁺ electronic configuration = Is² 2s²2p⁶ 3s² 3p⁶ 3d⁸
Fe (Z = 26). 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
Fe³⁺ Is² 2s² 2p⁶ 3s²3p⁶ 3d⁵
If d orbital is half filled, according to Aufbau principle, it is more stable. So Fe³⁺ is more stable than Ni²⁺.

Samacheer Kalvi 11th Chemistry Solutions Quantum Mechanical Model of Atom Additional Questions Solved

I. Choose the correct answer

Question 1.
Which of the following experiment proves the presence of an electron in an atom?
(a) Rutherford’s α-ray scattering experiment
(b) Davisson and Germer experiment
(c) J.J. Thomson cathode ray experiment
(d) G.R Thomson gold foil experiment
Answer:
(c) J.J. Thomson cathode ray experiment.

Question 2.
Consider the following statements regarding Rutherford’s α-ray scattering experiment.
i. Most of the α-particles were deflected through a small angle.
ii. Some of α-particles passed through the foil.
iii. Very few α-particles were reflected back by 180°.
Which of the above statements is/are not correct.
(a) i and ii
(b) ii and iii
(c) i and iii
(d) i ii and iii
Answer:
(a) i and ii.

Question 3.
Considering Bohr’s model which of the following statements is correct?
(a) The energies of electrons are continuously reduced in the form of radiation.
(b) The electron is revolving around the nucleus in a dynamic orbital.
(c) Electrons can revolve only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of h/2 π.
(d) In an atom, electrons are embedded like seeds in watermelon.
Answer:
(c) Electrons can revolve only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of h/2π.

Question 4.
The energy of an electron of hydrogen atom in 2nd main shell is equal to
(a) – 13.6 eV atom^-1
(b) – 6.8 eV atom^-1
(c) – 0.34 eV atom^-1
(d) – 3.4 eV atom^-1
Answer:
(d) -3.4 eV atom^-1
Hints:
Energy of an electron in 2nd main shell = \(\frac{(-13.6) Z^{2}}{n^{2}}\); Z = 1, n = 2
E = \(\frac{-13.6}{2^{2}}\) = \(\frac{-13.6}{4}\) = -3.4 atom^-1.

Question 5.
The energy of an electron of Li²⁺ in the 3^rd main shell is …………..
(a) – 1.51 eV atom^-1
(b) – 6.8 eV atom^-1
(c) + 1.51 eV atom^-1
(d) – 3.4 eV atom^-1
Answer:
(a) -1.51 eV atom^-1
Hints:
E = \(\frac{(-13.6) Z^{2}}{n^{2}}\) eV atom^-1
Li²⁺= H atom. So Z = 1, n = 3.
E = \(\frac{(-13.6) 1^{2}}{3^{2}}\) = \(\frac{-13.6}{9}\) = -1.51 eV atom^-1

Question 6.
The energy of an electron of hydrogen atom in main shell in terms of U mold is
(a) – 1312.8 k J mol^-1
(b) – 82.05 k J mol^-1
(c) – 328.2 kJ mol^-1
(d) – 656.4 k J mol^-1
Answer:
(b) – 82.05 k J mol^-1
Hints:
E = \(\frac{(-13.6) Z^{2}}{n^{2}}\) kJ mol^-1 , Z = 1, n = 4
∴ E = \(\frac{-1312.8}{16}\) = -82.50 kJ mol^-1

Question 7.
The Bohr’s radius of Li2 0f21d orbit is
(a) 0.529 Å
(b) 0.0753 Å
(c) 0.7053 Å
(d) 0.0529 Å
Answer:
(c) 0.7053 Å
Hints:
r_n = \(\frac{(0.529) Z^{2}}{n^{2}}\)Å, n = 2, Z = 3(for Li²⁺)
r = \(\frac{(0.529) 3^{2}}{2^{2}}\) = \(\frac{0.529 x 4}{3}\) = 0.7053 Å.

Question 8.
The formula used to calculate the Boh’s radius is ………..
(a) r_n = \(\frac{(-13.6) Z^{2}}{n^{2}}\) eV atom^-1
(b) r_n = \(\frac{(0.529) Z^{2}}{n^{2}}\) A
(c) r_n= \(\frac{(-1312.8) Z^{2}}{n^{2}}\) kJ mol^-1
(d) r_n = \(\frac{(+1312.8) Z^{2}}{n^{2}}\) kJ mol^-1
Answer:
(b) r_n = \(\frac{(0.529) Z^{2}}{n^{2}}\) A.

Question 9.
Who proposed the dual nature of light to all forms of matter?
(a) John Dalton
(b) Neils Bohr
(c) Albert Einstein
(d) J.J. Thomson
Answer:
(c) Albert Einstein

Question 10.
dc Brogue equation is ………..
(a) E = h γ
(b) E = mc²
(c) γ = \(\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{h}}\)
(d) λ = \(\frac{h}{mv}\)
Answer:
(d) λ = \(\frac{h}{mv}\).

Question 11.
The crystal used in Davison and Germer experiment is …………….
(a) nickel
(b) zinc suiphide
(c) gold foil
(d) NaCl
Answer:
(a) nickel.

Question 12.
Which one of the following is the time independent Schrodinger wave equation?

Answer:

Question 13.
Match the list-I and list-II correctly using the code given below the list.
List – I
A. Principal quantum number
B. Azimuthal quantum number
C. Magnetic quantum number
D. Spin quantum number

List – II
1. represents the directional orientation of orbital
2. represents the spin of the electron
3. represents the main shell
4. represents the sub shell

Answer:

Question 14.
The maximum number of electrons that can be accommodated in N shell is …………..
(a) 8
(b) 18
(c) 32
(d) 36
Answer:
(c) 32
Hints:
Number of electrons in the main shell = 2n² n = 4, for N shell.
∴ Maximum number of electrons in N shell = 2(4)² = 32.

Question 15.
The maximum number of electrons that can be accommodated in f orbital is ………….
(a) 10
(b) 14
(c) 16
(d) 6
Answer:
(6) 14
Hints:
forbital – l = 3.
Maximum number of electrons in sub shell = 2(2l + 1)
∴ For ‘f’ orbital, the maximum number of electrons = 2(2 x 3 + l) = 14.

Question 16.
When l = 0, the number of electrons that can be accommodated in the sub shell is ……………..
(a) 0
(b) 2
(c) 6
(d) 8
Answer:
(b) 2
Hints:
If l = 0, number of electrons = (2l + 1)
= 2 (2 x 0 + 1) = 2.

Question 17.
Which one of the quantum number is used to calculate the angular momentum of an atom?
(a) n
(b) m
(c) l
(d) s
Answer:
(c) l

Question 18.
What is the formula used to calculate the angular momentum?
(a) \(\sqrt{l^{(l+1)}} \frac{h}{2 \pi}\)
(b) \(\frac{\mathrm{mvr}}{2 \pi}\)
(c) \(\frac{mvr}{2}\)
(d) m . ∆v
Answer:
(a) \(\sqrt{l^{(l+1)}} \frac{h}{2 \pi}\).

Question 19.
Which of the following provides the experimental justification of magnetic quantum number?
(a) Zeeman effect
(b) Stark effect
(c) Uncertainty principle
(d) Quantum condition
Answer:
(a) Zeeman effect.

Question 20.
What are the values of n, l, m and s for 3p_x electron?
(a) 3, 2, 1, 0
(b) 3, 1,-l, +½
(c) 3, 2, +1, -½
(d) 3, 0, 0, +½]
Answer:
(b) 3, 1, -1, +½
Hint:
3p_x electron ; n = 3 (main shell)
for p_x orbitaI, l = 1, m = -1, s = \(\frac {1}{2}\).

Question 21.
Identify the quantum number for \(4 d_{x^{2}-y^{2}}\) electron.
(a) 4, 2, -2, +½
(b) 4, 0, 0, +½
(c) 4, 3, 2, +½
(d) 4, 3, 2, -½
Answer:
(a) 4, 2, -2, +½.

Question 22.
How many orbitals are possible in 3^rd energy level?
(a) 16
(6) 9
(c) 3
(d) 27
Answer:
(b) 9
Hints:
3^rd energy level Number of orbitals = ?
n = 3 main shell = m
l = 0, 1,2 m = 0, -1,0, +1
Total = 9 orbitals.

Question 23.
The region where the probability density function of electron reduces to zero is called
(a) orbit
(b) orbital
(c) nodal surface
(d) sub shell
Answer:
(c) nodal surface.

Question 24.
Consider the following statements.
(i) The region where the probability density of electron is zero, called nodal surface.
(ii) The probability of finding the electron is independent of the direction of the nucleus.
(iii) The number of radial nodes is equal to n + l + 1 Which of the above statements is/are correct?
(a) (i) and (iii)
(b) (i) and (ii)
(c) (iii) only
(d) (ii) and (iii)
Answer:
(b) (i) and (ii).

Question 25.
Match the list-I and list-II correctly using the code given below the list.
List-I
A. s – orbital
B. p – orbital
C. d – orbital
D. f – orbital

List-II
1. complex three-dimensional shape
2. symmetrical sphere
3. dumb-bell shape
4. clover leaf shape

Answer:

Question 26.
Which one of the following is the correct increasing order of effective nuclear charge felt by an electron?
(a) s>p>d>f
(b) s<p<d<f
(c) s>p>f>d
(d) f<p<d<s
Answer:
(a) s>p>d>f.

Question 27.
The value of n, l, m and s of 8th electron in an oxygen atom are respectively
(a) 1, 0, 0, + ½
(b) 2, 1, +1, – ½
(c) 2, 1, -1, – ½
(d) 2, 1, 0, +½
Answer:
(a) 2, 1, +1, – ½.

Question 28.
The number of impaired electrons in carbon atom in the gaseous state is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(c) 2

Question 29.
Which of the following is not used in writing electronic configuration of an atom?
(a) Aufbau principle
(b) Hund’s rule
(c) Pauli’s exclusion principle
(d) Heisenberg’s uncertainty principle
Answer:
(d) Heisenberg’s uncertainty principle.

Question 30.
Which of the following is the expected configuration of Cr (Z = 24)?
(a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²
(b) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
(c) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
(d) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s³
Answer:
(a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²

Question 31.
Which of the following is the actual configuration of Cr (Z = 24)?
(a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²
(b) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
(c) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
(d) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s³
Answer:
(b) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

Question 32.
Assertion (A) : Cr with electronic configuration [Ar] 3d⁵ 4s¹ is more stable than [Ar] 3d⁴ 4s¹.
Reason(R ): Half filled orbitals have been found to have extra stability than partially filled orbitals.
(a) A and R are correct and R is the correct explanation of A.
(b) A and R are correct but R is not the correct the explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) A and R are correct and R is the correct explanation of A.

Question 33.
Assertion (A): Copper (Z = 29) with electronic configuration [Ar] 4s¹ 3d¹⁰ is more stable than [Ar] 4s¹ 3d¹⁰.
Reason(R): Copper with [Ar] 4s² 3d⁹ is more stable due to symmetrical distribution and exchange energies of d electrons.
(a) A and R are correct and R is the correct explanation of A.
(b) A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) A and R are correct and R is the correct explanation of A.

Question 34.
In a sodium atom (atomic number = 11 and mass number = 23) and the number of neutrons is …………..
(a) equal to the number of protons
(b) less than the number of protons
(c) greater than the number of protons
(d) none of these
Answer:
(c) greater than the number of protons.

Question 35.
The idea of stationary orbits was first given by …………
(a) Rutherford
(b) J.J. Thomson
(c) Nails Bohr
(d) Max Planck
Answer:
(c) Niels Bohr.

Question 36.
de Broglie equation is ……………
(a) λ = \(\frac {h}{mv}\)
(b) λ = \(\frac {hv}{m}\)
(c) λ = \(\frac {mv}{h}\)
(d) λ = hmv
Answer:
(a) λ = \(\frac {h}{mv}\).

Question 37.
The orbital with n = 3 and l = 2 is …………..
(a) 3s
(b) 3p
(c) 3d
(d) 3J
Answer:
(c) 3d

Question 38.
The outermost electronic configuration of manganese (at. no. = 25) is …………
(a) 3d⁵ 4s²
(b) 3d⁶ 4s¹
(c) 3d⁷ 4s⁰
(d) 3d⁶ 4s²
Answer:
(a) 3d⁵ 4s²

Question 39.
The maximum number of electrons in a sub-shell is given by the equation
(a) n²
(b) 2 n²
(c) 2 l – l
(d) 2 l + 1
Answer:
(d) 2 l + 1

Question 40.
Which of the following statements is correct for an electron that has the quantum numbers n = 4 and m = -2.
(a) The electron may be in 2 p orbital
(b) The electron may be in 4 d orbital
(c) The electron is in the second main shell
(d) The electron must have spin quantum number as +\(\frac {1}{2}\).
Answer:
(b) The electron may be in 4d orbital.

Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom 2 – Marks Questions

Question 1.
Write a note about J.J. Thomson’s atomic model.
Answer:

• J.J. Thomson’s cathode ray experiment revealed that atoms consist of negatively charged particles called electrons.
• He proposed that atom is a positively charged sphere in which the electrons are embedded like the seeds in the watermelon.

Question 2.
Explain about theory of electromagnetic radiation.
Answer:

• The theory of electromagnetic radiation states that a moving charged particle should continuously loose its energy in the form of radiation.
• O Therefore, the moving electron in an atom should continuously loose its energy and finally collide with nucleus resulting in the collapse of the atom.

Question 3.
Explain how matter has dual character?
Answer:

• Albert Einstein proposed that light has dual nature, i.e. like photons behave both like a particle and as a wave.
• Louis de Broglie extended this concept and proposed that all forms of matter showed dual character.
• He combined the following two equations of energy of which one represents wave character (hυ) and the other represents the particle nature (mc²).

Question 4.
Explain about the significance of de Broglie equation.
Answer:

• X = \(\frac {h}{mv}\). This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle.
• For a particle with high linear momentum (mv) the wavelength will be so small and cannot be observed.
• For a microscopic particle such as an electron, the mass is of the order of 10^-31 kg, hence the wavelength is much larger than the size of atom and it becomes significant.
• For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant.

Question 5.
How many electrons can be accommodated in the main shell l, m and n?
Answer:

Question 6.
How many electrons that can be accommodated in the sub shell s, p, d, f ?
Answer:

Question 7.
What are quantum numbers?
Answer:

• The electron in an atom can be characterized by a set of four quantum numbers, namely principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s).
• When Schrodinger equation is solved for a wave function T, the solution contains the first three quantum numbers n, l
• and m.
• The fourth quantum number arises due to the spinning of the electron about its own axis.

Question 8.
How many orbitals are possible in the 3rd energy level?
Answer:
n = 3, main shell is m.
Total number of orbitals in 3rd energy level = ?

Total number of orbitals = 9.

Question 9.
What are Ψ and Ψ² ?
Answer:

• Ψ itself has no physical meaning but it represents an atomic orbital.
• Ψ² is related to the probability of finding the electrons within a given volume of space.

Question 10.
What is meant by nodal surface?
Answer:

• The. region where there is probability density function reduces to zero is called nodal surface or a radial node.
• For ns orbital, (n-1) nodes are found in it.

Question 11.
Mention the shape of s, p, d orbitals.
Answer:

• Shape of s – orbital – sphere
• Shape of p – orbital – dumb bell
• Shape of d – orbital – clover leaf

Question 12.
Calculate the total number of angular nodes and radial nodes present in 4p and 4d orbitals.
Answer:
1. For 4p orbital:
Number of angular nodes = l
For 4p orbital 7 = l
Number of angular nodes = l
Number of radial nodes = n – l – 1 = 4 -1 -1 = 2
Total number of nodes = n -1 = 4 – 1 = 3
1 angular node and 2 radial nodes.

2. For 4d orbital:
Number of angular nodes = l
For 4d orbital l = 2
Number of angular nodes = 2
Number of radial nodes = n – l – 1 = 4 – 2 – 1 = 1
Total number of nodes = n – l = 4 – l = 3
1 radial nodes and 2 angular node.

Question 13.
Write the equation to calculate the energy of nth orbit.
Answer:
E_n = \(\frac{(-1312.8) Z^{2}}{n^{2}}\) kJ mol^-1
Where Z = atomic number, n = principal quantum number.

Question 14.
what are degenerate orbitals?
Answer:

• Three different orientations in space that are possible for a p-orbital. All the three p-orbitals, namely p_x, p_y and p_z have same energies and are called degenerate orbitals.
• In the presence of magnetic or electric field, the degeneracy is lost.

Question 15.
Energy of an electron in hydrogen atom in ground state is -13.6 eV. What is the energy of the electron in the third excited state?
Answer:
E₁ = – 13.6 eV
E₃ = \(\frac{-13.6}{n^{2}}\) Where n = 3
E₃ = \(\frac{-13.6}{9}\) = 1.511 eV
Energy of the electron in the third excited state = 1.511 eV.

Question 16.
The energies of the same orbital decreases with an increase in the atomic number. Justify this statement.
Answer:
The energy of the 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium and that of lithium is greater than that of sodium and so on because H (Z =1), Li (Z = 3) and Na (Z = 11). When atomic number increases, the energies of the same orbital decreases. E_2s(H) > E_2s(Li) > E_2s(Na) > E_2s(K) ………….

Question 17.
State Hund’s rule of maximum multiplicity.
Answer:
It states that electron pairing in the degenerate orbitals does not take place until all the available orbitals contain one electron each.

Question 18.
How many unpaired electrons are present in the ground state of –
1. Cr³⁺ (Z = 24)
2. Ne (Z = 10)
Answer:
1. Cr³⁺ (Z = 24) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
Cr³⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴.
It contains 4 unpaired electrons.

2. Ne (Z = 10) 1s²2s²2p⁶. No unpaired electrons in it.

Question 19.
What is meant by electronic configuration? Write the electronic configuration of N (Z = 7).
Answer:
The distribution of electrons into various orbitals of an atom is called its electronic configuration.

Question 20.
Which is the actual configuration of Cr (Z = 24) Why?
Answer:
Cr (Z = 24) 1s²2s²2p⁶.
The reason for this is, Cr with 3d⁵ configuration is half filled and it will be more stable. Chromium has [Ar] 3d⁵ 4s¹ and not [Ar] 3d⁴ 4s² due to the symmetrical distribution and exchange energies of d electrons.

Question 21.
What is the actual configuration of copper (Z = 29)? Explain about its stability.
Answer:
Copper (Z = 29)
Expected configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s²
Actual configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
The reason is that fully filled orbitals have been found to have extra stability. Copper has the electronic configuration [Ar] 3d¹⁰ 4s¹ and not [Ar] 3d⁹ 4s² due the symmetrical distribution and exchange energies of d electrons. Symmetry leads to stability. The full filled configuration have symmetrical distribution of electrons and hence they are more stable than unsymmetrical configuration.

Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom 3 – Mark Questions

Question 1.
What are the conclusions of Rutherford’s α – rays scattering experiment?
Answer:

• Rutherford bombarded a thin gold foil with a stream of fast moving α – particles.
• It was observed that most of the a-particles passed through the foil.
• Some of them were deflected through a small angle.
• Very few α- particles were reflected back by 180°.
• Based on these observations, he proposed that in an atom, there is a tiny positively charged nucleus and the electrons are moving around the nucleus with high speed.

Question 2.
What are the limitations of Bohr’s atom model?
Answer:

• The Bohr’s atom model is applicable only to species having one electron such as hydrogen, Li²⁺ etc and not applicable to multi – electron atoms.
• It was unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric field (Stark effect).
• Bohr’s theory was unable to explain why the electron is restricted to revolve around the nucleus in a fixed orbit in which the angular momentum of the electron is equal to nh / 2π.

Question 3.
Illustrate the significance of de Broglie equation with an iron ball and an electron.

1. 6.626 kg iron ball moving with 10 ms^-1.
2. An electron moving at 72.73 ms^-1.

Answer:
1. λ_{iron ball} = \(\frac{h}{mv}\)

= 1 x 10^-35m

2. λ_{iron ball} = \(\frac{h}{mv}\)

= \(\frac{6.626}{662.6}\) x 10^-3m = 1 x 10⁵m
For an electron, the de Broglie wavelength is significant and measurable while for iron ball it is too small to measure, hence it becomes insignificant.

Question 4.
Explain Davisson and Germer experiment.
Answer:

• The wave nature of electron was experimentally confirmed by Davisson and Germer.
• They allowed the accelerated beam of electrons to fall on a nickel crystal and recorded the diffraction pattern.
• The resultant diffraction pattern is similar to the X – ray diffraction pattern.
• The finding of wave nature of electron leads to the development of various experimental ’ techniques such as electron microscope, low energy electron diffraction etc.

Question 5.
Bohr radius of 1st orbit of hydrogen atom is 0.529 Å. Assuming that the position of an electron in this orbit is determined with the accuracy of 0.5% of the radius, calculate the uncertainty in the velocity of the electron in hydrogen atom.
Answer:
Uncertainty in position = ∆x
= \(\frac{0.5}{100}\) x 0.529 Å
= \(\frac{0.5}{100}\) x 10^-10 x 0.529 m
∆x = 2.645 x 10^-13 m
From Heisenberg’s uncertainty principle,
∆x.∆p ≥ \(\frac{h}{4π}\)
∆x.m.∆p ≥ \(\frac{h}{4π}\)
∆v ≥ \(\frac{h}{∆x.m.4π}\)
∆v =
∆v = 2.189 x 10⁸m.

Question 6.
Write a note about principal quantum number.
Answer:

• The principal quantum number represents the energy level in which electron revolves around the nucleus and is denoted by the symbol ‘n’.
• The ‘n’ can have the values 1, 2, 3,… n = 1 represents K shell; n=2 represents L shell and n = 3, 4, 5 represent the M, N, O shells, respectively.
• The maximum number of electrons that can be accommodated in a given shell is 2n².
• ‘n’ gives the energy of the electron,

E_n = \(\frac{(-1312.8) Z^{2}}{n^{2}}\) KJ mol^-1 and the distance of the electron from the nucleus is given by r_n = \(\frac{(-0.529) n^{2}}{Z}\) A.

Question 7.
Explain about azimuthal quantum number.
Answer:

• It is represented by the letter 7′ and can take integral values from zero to n – 1, where n is the principal quantum number.
• Each l value represents a subshell (orbital). l = 0, 1, 2, 3 and 4 represents the s, p, d, f and g orbitals respectively.
• The maximum number of electrons that can be accommodated in a given subshell (orbital) is 2(2l + 1).
  It is used to calculate the orbital angular momentum using the expression Angular momentum = \(\sqrt{l(l+1)} \frac{h}{2 \pi}\).

Question 8.
Draw the shapes of 1s, 2s and 3s orbitals
Answer:

Question 9.
Explain how effective nuclear charge is related with stability of the orbital.
Answer:

• In a multi-electron atom, in addition to the electrostatic attractive force between the electron and nucleus, there exists a repulsive force among the electrons.
• These two forces are operating in the opposite direction. This results in the decrease in the nuclear force of attraction on electron.
• The net charge experienced by the electron is called effective nuclear charge.
• The effective nuclear charge depends on the shape of the orbitals and it decreases with increase in azimuthal quantum number l.
• The order of the effective nuclear charge felt by a electron in an orbital within the given shell is s > p > d > f.
• Greater the effective nuclear charge, greater is the stability of the orbital. Hence, within a given energy level, the energy of the orbitals are in the following order s < p < d < f.

Question 10.
Calculate the wavelength of an electron moving with a velocity of 2.05 x 10⁷ ms^-1.
Answer:
According to de Broglie’s equation, λ = \(\frac {h}{mv}\)
Mass of electron (m) = 9.1 x 10^-31 kg
Velocity of electron (υ) = 2.05 x 10⁷ ms^-1
Planck’s constant (h) = 6.626 x 10^-34 kg m² s^-1
λ =  =355 x 10^-4m.

Question 11.
The mass of an electron is 9.1 x 10^-31 kg. If its kinetic energy is 3.0 x 10^-25 J, calculate its wavelength.
Answer:
Step I.
Calculation of the velocity of electron
Kinetic energy = 1 / 2 mυ² = 3.0 x 10^-25 kg m² s^-2
υ² = 65.9 x 10⁴ m² s^-2
υ = (65.9 x 10⁴ m s^-2) = 8.12 x 10² ms^-1

Step II.
Calculation of wavelength of the electron
According to de Broglie’s equation,
λ = \(\frac {h}{mv}\) =
=  0.08967 x l0^-5 m = 8967 x 10^-10 m = 8967 Å (∴1Å = 10^-10m).

Question 12.
From the following sets of quantum numbers, state which are possible. Explain why the others are not possible.

1. n = 0, l = 0, m_l = 0, m_s = + \(\frac {1}{2}\)
2. n = 1, l = 0, m_l = 0, m_s = – \(\frac {1}{2}\)
3. n = 1, l = 1, m_l = 0, m_s = + \(\frac {1}{2}\)
4. n = 1, l = 0, m_l = +1, m_s= +\(\frac {1}{2}\)
5. n = 3, l = 3, m_l = -3, m_s = + \(\frac {1}{2}\)
6. n = 3, l = 1, m_l = 0, m_s = +\(\frac {1}{2}\)

Answer:

1. The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
2. The set of quantum numbers is possible.
3. The set of quantum numbers is not possible because, for n = 1,1 cannot be equal to 1. It can have 0 value.
4. The set of quantum numbers is not possible because for l = 0, m_l; cannot be +1. It must be zero.
5. The set of quantum numbers is not possible because, for n = 3, l = 3.
6. The set of quantum numbers is possible.

Question 13.
How many electrons in an atom may have the following quantum numbers?
(a) n = 4; m_s = – ½
(b) n = 3, l = 0.
Answer:
(a) For n = 4
1 Total number of electrons = 2n² = 2 x 16 = 32
Half out of these will have ms = – \(\frac {1}{2}\)
Total electrons with m_s (-½) = 16.

(b) For n = 3
l = 0; m₁ = 0, m_s = + ½ – ½ (two e^–).

Question 14.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:
According to Bohr’s theory,
mυr = \(\frac {nh}{2π}\) (n = 1,2,3, …… so on)
or 2πr = \(\frac {nh}{mυ}\) or mυ = \(\frac {nh}{2πr}\) ………..(i)
According to de Brogue equation,
λ = \(\frac {h}{mυ}\) or mυ = \(\frac {h}{λ}\) ……….(ii)
Comparing (i) and (ii),
\(\frac {nh}{2πr}\) = \(\frac {h}{λ}\) or 2πr = nλ
Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an into the de Broglie wave length.

Question 15.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion.
Answer:
Let the no. of electrons in the ion = x
the no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + \(\frac {30.4x }{100}\) = x + 0.304 x
Now, mass number of ion = Number of protons + Number of neutrons
= (x + 3) + (x + 0.304 x)
∴ 56 = (x + 3) + (x + 0.304 x) or 2.304 x = 56 – 3 = 53
x = \(\frac {53}{2.304}\) = 23
Atomic number of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

Question 16.
The uncertainty in the position of a moving bullet of mass 10 g is 10 s m. Calculate the uncertainty in its velocity?
Answer:
According to uncertainty principle,
∆x.m∆υ = \(\frac {h}{4π}\) or ∆υ = \(\frac {h}{4πm∆x}\);
h = 6.626 x 1o^-34 kg m₂ s^-1; m = 10 g = 10^-2 kg
∆x = 10^-5m; ∆v =  = 5.27 x 10^-28mv
= 1.6 x 10^-15 kg m² s^-15
Or
\(\frac {1}{2}\) mv² = 1.6 x 10^-15kg m²s^-2
v =  = 5.93 x 10⁷m^-1

Question 17.
The uncertainty in the position and velocity of a particle are 10^-10 m and 5.27 × 10^-24 ms^-1 respectively. Calculate the mass of the particle.
Answer:
According to uncertainty principle.
∆x. m∆υ = \(\frac {h}{4π}\)
or
m = \(\frac {h}{4π∆x.∆υ}\);
h = 6.626 x 10^-34 kg m² s^-1
∆x = 10^-10 m; ∆x = 5.27 x 10^-24ms^-1
m  = 0.1 kg.

Question 18.
With what velocity must an electron travel so that its momentum Is equal to that of a photon of wave length = 5200 A?
Answer:
According to de Brogue equation, λ = \(\frac {h}{mv}\)
Momentum of electron, mv = \(\frac {h}{λ}\)
= 1.274 x 10^-27 kg ms^-1 ………(i)
The momentum of electron can also be calculated as = mv = (9.1 x 10^-31kg) x v ………(ii)
Comparing (i) and (ii)
(9.1 X 10^-31kg) v = (1.274 x 10^-27 kg ms^-1)
v   = 1.4 x 10³ ms^-1

Question 19.
Using Aufbau principle, write the ground state electronic configuration of following atoms.

1. Boron (Z = 5)
2. Neon (Z = 10)
3. Aluminium (Z = 13)
4. Chlorine (Z = 17)
5. Calcium (Z = 20)
6. Rubidium (Z = 37)

Answer:

1. Boron (Z = 5) ; 1s² 2s² 2p¹
2. Neon (Z = 10) ; 1s² 2s² 2p⁶
3. Aluminium (Z = 13) ; 1s² 2s² 2p⁶ 3s² 3p¹
4. Chlorine(Z = 17) ; 1s² 2s² 2p⁶ 3s² 3p⁵
5. Calcium (Z = 20) ; 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
6. Rubidium (Z = 37) ; 1s² 2s²2p⁶ 3s² 3p⁶3d¹⁰ 4s² 4p⁶ 5s¹

Question 20.
Calculate the de Broglie wavelength of an electron moving with 1 % of the speed of light?
Answer:
According to de Brogue equation, A = \(\frac {h}{mv}\)
Mass of electron = 9.1 x 10^-31 kg; Planck’s constant 6.626 x 10^-34 kg m² s^-1
Velocity of electron = 1% of speed of light = 3.0 x 10⁸ x 0.01 = 3 10⁶ ms^-1
Wavelength of electron (λ) = \(\frac {h}{mv}\)
= 2.43 x 10^-10m.

Question 21.
What is the wavelength for the electron accelerated by 1.0 X i0 volts?
Answer:
Step I.
Calculation of the velocity of electron
Energy (kinetic energy) of electron = 1.0 x 10⁴ volts.
= 1.0 x 10⁴ x 1.6 x 10^-19 J = 1.6 x 10^-15J.

Step II.
Calculation of wavelength of the electron
According to de Broglie equation,
λ = \(\frac {h}{mυ}\); λ 
= 1.22 x 10^-11m .

Question 22.
In a hydrogen atom, the energy of an electron in first Bohr’s orbit is 13.12 x 10⁵ J mol^-1. What is the energy required for its excitation to Bohr’s second orbit?
Answer:

The energy required for the excitation is:
∆E = E₂ – E₁ = (-3.28 x l0⁵) – (- 13.12 x 10⁵) = 9.84 x 10⁵ J mol^-1

Question 23.
Dual behavior of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 x 10⁶ ms^-1, calculate de Broglie wavelength associated with this electron.
Answer:
λ = \(\frac {h}{mυ}\); λ =
= 0.455 x 10^{-34 + 25} m = 0.455 nm = 455 pm.

Question 24.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the symbol to the element.
Answer:
An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
∴ Number of neutrons = x + \(\frac {x × 31.7}{100}\) = (x + 0.317 x)
Now, Mass no. of element = No. of protons + No. of neutrons
81 = x + x + 0.317 x = 2.317 x
Or
x = \(\frac {81}{2.317}\) = 35
∴ No. of protons = 35, No. of neutrons = 81 – 35 = 46
Atomic number of element (Z) = Number of protons = 35
The element with atomic number (Z) 35 is bromine ⁸¹₃₅Br.

Question 25.
The electron energy in hydrogen atom is given by E_n = (- 2.18 × 10^-18) / n² J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Answer:
Step I.
Calculation of energy required
The energy required is the difference in the energy when the electron jumps from orbit with
n = ∞ to orbit with n = 2.
The energy required (∆E) = E_∞ – E₂
= 0 – \(\left(-\frac{2.18 \times 10^{-18}}{4} \mathrm{J}\right)\) = 5.45 x 10^-19 J.

Step II.
Calculation of the longest wavelength of light in cm used to cause the transition
∆E = hv = hc / λ.
λ = \(\frac {hc}{∆E}\) = n= 3.644 x 10^-7
m = 3.644 x 10^-7 x 10² = 3.645 x 10^-5 cm.

Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom 5-Mark Questions

Question 1.
Describe about Bohr atom model.
Answer:
Assumptions of Bohr atom model.
1. The energies of electrons are quarantined

2. The electron is revolving around the nucleus in a certain fixed circular path called the stationary orbit.

3. Electron can revolve only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of h/2π
mvr = \(\frac {nh}{2π}\) where n = 1,2,3,…etc.,

4. As long as an electron revolves in a fixed stationary orbit, it doesn’t lose its energy. But if an electron jumps from a higher energy state (E₂) to a lower energy state (E₁), the excess energy is emitted as radiation. The frequency of the emitted radiation is E₂ – E₁= hv.
∴ v = \(\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{h}}\)
Conversely, when suitable energy is supplied to an electron, it will jump from lower energy orbit to a higher energy orbit.

5. Bohr’s postulates are applied to a hydrogen like atom (H, He⁺ and Li²⁺ etc..) the radius of the nth orbit and the energy of the electron revolving in the th orbit were derived.
r_n = \(\frac{(0.529) n^{2}}{Z}\) A(0.529) n²
E_n = \(\frac{(-1 3.6) Z^{2}}{n}\) eV atom^-1
E_n = \(\frac{(1312.8) Z^{2}}{n}\) kJ mol^-1

Question 2.
Derive de Brogue equation and give its significance.
Answer:
1. Louis de Brogue extended the concept of dual nature of light to all forms of matter. To quantify this relation, he derived an equation for the wavelength of a matter-wave.

2. He combined the following two equations of the energy of which one represents wave character (hu) and the other represents the particle nature (mc²).
Planck’s quantum hypothesis:
E = hv ……….(1)
Einsteins mass-energy relationship:
E = mc² ………(2)
From (1) and (2)
hv = mc²
hc/λ = mc²
∴ λ = \(\frac{h}{mc}\) ………(3)
The equation (3) represents the wavelength of photons whose momentum is given by mc. (Photons have zero rest mass).

3. For a particle of matter with mass m and moving with a velocity y, the equation (3) can be written as λ = \(\frac{h}{mc}\) ………(4)

4. This is valid only when the particle travels at speed much less than the speed of Light.

5. This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle (i.e momentum).

6. Significance of de Brogue equation:
For a particle with high linear momentum, the wavelength will be too small and cannot be observed. For a microscopic particle such as an electron, the mass is 9.1 x 10^-31 kg. Hence the wavelength is much larger than the size of atom and it becomes significant.

Question 3.
What are the main features of the quantum mechanical model of an atom.
Answer:
1. The energy of electrons in an atom is quarantined.

2. The existence of quarantined electronic energy levels is a direct result of the wave-like properties of electrons. The solutions of the Schrodinger wave equation gives the allowed energy levels (orbits).

3. According to Heisenberg’s uncertainty principle, the exact position and momentum of an electron cannot be determined with absolute accuracy. As a consequence, quantum mechanics introduced the concept of orbital. Orbital is a three-dimensional space in which the probability of finding the electron is maximum.

4. The solution of the Schrodinger wave equation for the allowed energies of an atom gives the wave function Ψ, which represents an atomic orbital. The wave nature of the electron present in an orbital can be well defined by the wave function Ψ.

5. The wave function Ψ itself has no physical meaning. However, the probability of finding the electron in a small volume dx, dy, dz around a point (x,y,z) is proportional to |Ψ (x,y,z)|² dx dy dz |Ψ (x,y,z)|² is known as probability density and is always positive.

Question 4.
Explain about –
(1) Magnetic quantum number
(2) Spin quantum number
Answer:
(1) Magnetic quantum number:

• It is denoted by the letter m_l. It takes integral values ranging from – l to +l through 0.
  i.e. if l = 1; m = -1, 0 and +1.
• The Zeeman Effect (the splitting of spectral lines in a magnetic field) provides the experimental justification for this quantum number.
• The magnitude of the angular momentum is determined by the quantum number l while its direction is given by magnetic quantum number.

(2) Spin quantum number:

• The spin quantum number represents the spin of the electron and is denoted by the letter ‘m_s‘.
• The electron in an atom revolves not only around the nucleus but also spins. It is usual to write this as electron spins about its own axis either in a clockwise direction or in anti-clockwise direction.
• Corresponding to the clockwise and anti-clockwise spinning of the electron, maximum two values are possible for this quantum number.
• The values of ‘m_s‘ is equal to –\(\frac {1}{2}\) and +\(\frac {1}{2}\).

Question 5.
Explain about the shape of orbitals.
Answer:
Orbital: The solution to Schrodinger equation gives the permitted energy values called eigen values and the wave functions corresponding to the eigen values are called atomic orbitals.

Shape of orbital:
s – orbital:
For Is orbital, l = 0, m = 0, f(θ) = 1√2 and g(φ) = 1/√2π. Therefore, the angular distribution function is equal to 1/√2π. i.e. it is independent of the angle θ and φ. Hence, the probability of finding the electron is independent of the direction from the nucleus. So, the shape of the s orbital is spherical.

p – orbital:
For p orbitals l = 1 and the corresponding m values are -1, 0 and +1. The three different m values indicates that there are three different orientations possible for p orbitals. These orbitals are designated as p_x, p_y and p_z. The shape of p orbitals are dumb bell shape.

d – orbital:
For ‘d’ orbital 1 = 2 and the corresponding m values are -2, -1, 0, +l,+2. The shape of the d orbital looks like a clover leaf. The five m values give rise to five d orbitals namely d_xy, d_yz, d_zx, d_x²_-y² and d_z² The 3d orbitals contain two nodal planes.

f – orbital
For f orbital, 1 = 3 and the m values are -3, -2,-1, 0, +1, +2, +3 corresponding to seven f orbitals, \(\mathrm{f}_{\mathrm{z}^{3}}, \mathrm{f}_{\mathrm{xz}^{2}}, \mathrm{f}_{\mathrm{yz}^{2}}, \mathrm{f}_{\mathrm{xyz}}, \mathrm{f}_{\mathrm{z}\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{x}\left(\mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{y}\left(3 \mathrm{x}^{2}-\mathrm{y}^{2}\right)}\). They contain 3 nodal planes.

Question 6.
What is exchange energy? How it is related with stability of atoms? Explain with suitable examples.
Answer:
1. If two or more electrons with the same spin are present in degenerate orbitals, there is a possibility for exchanging their positions. During exchange process, the energy is released and the released energy is called exchange energy.

2. If more number of exchanges are possible, more exchange energy is released. More number of exchanges are possible only in the case of half filled and fully filled configurations.

3. For example, in chromium, the electronic configuration is [Ar]3d⁵ 4s¹. The 3d orbital is half filled and there are ten possible exchanges.

4. On the other hand only six exchanges are possible for [Ar] 3d⁴ 4s² configuration.

5. Hence, exchange energy for the half filled configuration is more This increases the stability of half filled 3d orbitals.

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.2

11th Maths Exercise 1.2 Solutions Question 1.
Discuss the following relations for reflexivity, symmetricity and transitivity:

(i) The relation R defined on the set of all positive integers by “mRn if m divides n”.
Solution:
S = {set of all positive integers}

(a) mRm ⇒ ‘m’ divides’m’ ⇒ reflexive

(b) mRn ⇒ m divides n but
nRm ⇒ n does not divide m
(i.e.,) mRn ≠ nRm
It is not symmetric

(c) mRn ⇒ nRr as n divides r
It is transitive

(ii) Let P denote the set of all straight lines in a plane. The relation R defined by “lRm if l is perpendicular to m”.
Solution:
P = {set of all straight lines in a plane}
lRm ⇒ l is perpendicular to m

(a) lRl ⇒ l is not perpendicular to l
⇒ It is not reflexive

(b) lRm ⇒ l is perpendicular to m
mRl ⇒ m is perpendicular to l
It is symmetric

(c) l perpendicular to m ⇒ m perpendicular to n ⇒ l is parallel to n It is not transitive

(iii) Let A be the set consisting of all the members of a family. The relation R defined by “aRb if a is not a sister of b”.
Solution:
A = {set of all members of the family}
aRb is a is not a sister of b

(a) aRa ⇒ a is not a sister of a It is reflexive

(b) aRb ⇒ a is not a sister of b.
bRa ⇒ b is not a sister of a.
It is symmetric

(c) aRb ⇒ a is not a sister of b.
bRc ⇒ b is not a sister of c.
⇒ aRc ⇒ a can be a sister of c
It is not transitive.

(iv) Let A be the set consisting of all the female members of a family. The relation R
defined by “aRb if a is not a sister of b”.
Solution:
A = {set of all female members of a family}

(a) aRa ⇒ a is a sister of a
It is reflexive

(b) aRb ⇒ a is a sister of b
bRa ⇒ b is a sister of a
⇒ It is symmetric

(c) aRb ⇒ a is a sister of b bRc ⇒ b is a sister of c aRc ⇒ a can be sister of c It is not transitive.

(v) On the set of natural numbers the relation R defined by “xRy if x + 2y = 1”.
Solution:
N= {1, 2, 3, 4, 5,….}
xRy if x + 2y = 1 R is an empty set

(a) xRx ⇒ x + 2x = 1 ⇒ x = \(\frac{1}{3}\) ∉ N. It is not reflexive
xRy = yRx ⇒ x + 2y = 1 It does not imply that y + 2x = 1 as y = \(\frac{1-x}{2}\) It is not symmetric.

(b) -x = y ⇒ (-1, 1) ∉ N
It is not transitive.

11th Maths Exercise 1.2 Answers Question 2.
Let X = {a, b, c, d} and R = {(a, a), {b, b), (a, c)}. Write down the minimum number
of ordered pairs to be included to R to make it
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
X = {a, b, c, d}
R = {(a, a), (b, b), (a, c)}
(i) To make R reflexive we need to include (c, c) and (d, d)
(ii) To make R symmetric we need to include (c, a)
(iii) R is transitive
(iv) To make R reflexive we need to include (c, c)
To make R symmetric we need to include (c, c) and (c, a) for transitive
∴ The relation now becomes
R = {(a, a), (b, b), (a, c), (c, c), (c, a)}
∴ R is equivalence relation.

Exercise 1.2 Class 11 Maths State Board Question 3.
Let A = {a, b, c} and R = {(a, a), (b, b), (a, c)}. Write down the minimum number of ordered pairs to be included to R to make it
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
(i) (c, c)
(ii) (c, a)
(iii) nothing
(iv) (c, c) and (c, a)

11th Maths Exercise 1.2 Question 4.
Let P be the set of all triangles in a plane and R be the relation defined on P as aRb if a is
similar to b. Prove that R is an equivalence relation.
Solution:
P = {set of all triangles in a plane}
aRb ⇒ a similar to b

(a) aRa ⇒ every triangle is similar to itself
∴ aRa is reflexive

(b) aRb ⇒ if a is similar to b ⇒ b is also similar to a.
⇒ It is symmetric

(c) aRb ⇒ bRc ⇒ aRc
a is similar to b and b is similar to c
⇒ a is similar to a
⇒ It is transitive
∴ R is an equivalence relation

11 Maths Exercise 1.2 Question 5.
On the set of natural numbers let R be the relation defined by aRb if 2a + 3b = 30. Write down the relation by listing all the pairs. Check whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
N = {set of natural numbers}
R ={(3, 8), (6, 6), (9, 4), (12, 2)}

(a) (3, 3) ∉ R ⇒ R is not reflexive
2a + 3b = 30
3b = 30 – 2a
b = \(\frac{30-2 a}{3}\)

(b) (3, 8) ∈ R(8, 3) ∉ R
⇒ R is not symmetric

(c) (a, b) (b, c) ∉ R ⇒ R is transitive
∴ It is not equivalence relation.

11th Maths Exercise 1.2 Answers In Tamil Question 6.
Prove that the relation “friendship” is not an equivalence relation on the set of all people in Chennai.
Solution:
(a) S = aRa (i.e. ) a person can be a friend to himself or herself.
So it is reflextive.

(b) aRb ⇒ bRa so it is symmetric

(c) aRb, bRc does not ⇒ aRc so it is not transitive
⇒ It is not an equivalence relation

11th Maths Chapter 1 Exercise 1.2 Question 7.
On the set of natural numbers let R be the relation defined by aRb if a + b ≤ 6. Write down the relation by listing all the pairs. Check whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
Set of all natural numbers aRb if a + b ≤ 6
R= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)}
(i) (5, 1) ∈ R but(5, 5) ∉ R
It is not reflexive

(ii) aRb ⇒ bRa ⇒ It is symmetric

(iii) (4, 2), (2, 3) ∈ R ⇒ (4, 3) ∉ R
∴ It is not transitive

(iv) ∴ It is not an equivalence relation

11th Std Maths Exercise 1.2 Answers Question 8.
Let A = {a, b, c}. What is the equivalence relation of smallest cardinality on A? What is the equivalence relation of largest cardinality on A?
Solution:
R = {{a, a), (b, b), (c, c)} is this smallest cardinality of A to make it equivalence relation n(R) = 3

(i) R = {(a, a), {a, b), (a, c), (b, c), (b, b), {b, c), (c, a), (c, b), (c, c)}
n(R) = 9 is the largest cardinality of R to make it equivalence.

11 Maths Samacheer Kalvi Question 9.
In the set Z of integers, define mRn if m – n is divisible by 7. Prove that R is an equivalence relation.
Solution:
mRn if m – n is divisible by 7
(a) mRm = m – m = 0
0 is divisible by 7
∴ It is reflexive

(b) mRn = {m – n) is divisible by 7
nRm = (n – m) = – {m – n) is also divisible by 7
It is symmetric

It is transitive
mRn if m – n is divisible by 7
∴ R is an equivalence relation.

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.2 Additional Questions

Samacheer Kalvi 11th Maths Guide Question 1.
Find the range of the function.
f = {(1, x), (1, y), (2, x), (2, y), (3, z)}
Solution:
The range of the function is {x, y, z}.

Samacheer Kalvi 11th Maths Example Sums Question 2.
For n, m ∈ N, nln means that tt is a factor of n&m. Then find whether the given relation is an equivalence relation.
Solution:
Since n is a factor of n. So the relation is reflexive.
When n is a factor of m (where m ≠ n) then m cannot be a factor of n.
So the relation is not symmetric when n is a factor of m and m is a factor of p then n will be a factor of p. So the given relation is transitive. So it is not an equivalence relation.

11th Maths Solution Samacheer Question 3.
Verify whether the relation “is greater than” is an equivalence relation.
Solution:
You can do it yourself.

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.6

9th Maths Exercise 3.6 Question 1.
Factorise the following:
(i) x² + 10x + 24
(ii) z² + 4z – 12
(iii) p² – 6p – 16
(iv) t² + 72 – 17t
(v) y² – 16y – 80
(vi) a² + 10a – 600
Solution:
(i) x² + 10x + 24
x² + 10x + 24 = x² + 6x + 4x + 24

= x(x + 6) + 4 (x + 6)
= (x + 6) (x + 4)

(ii) z² + 4z – 12
z² + 4z – 12 = z² + 6z – 2z- 12

= z (z + 6) – 2 (z + 6)
= (z + 6) (z – 2)

(iii) p² – 6p – 16
p² – 6p – 16 = p² – 8p + 2p – 16

= p(p – 8) + 2(p – 8)
= (p – 8)(p + 2)

(iv) t² + 72 – 17t
t² + 72 – 17t = t² – 17t + 72
= t² – 9t – 8t + 72

= t(t – 9) – 8 (t – 9)
= (t – 9) (t – 8)

(v) y² – 16y – 80
y² – 16y – 80 = y² – 20y + 4y – 80

= y(y – 20) + 4 (y – 20)
= (y – 20) (y + 4)

(vi) a² + 10a – 600
a² + 10a – 600 = a² + 30a – 20a – 600

= a(a + 30) -20 (a + 30)
= (a + 30) (a – 20)

Exercise 3.6 Class 9 Question 2.
Factorise the following
(i) 2a² + 9a + 10
(ii) 5x² – 29xy – 42y²
(iii) 9 – 18x + 18x²
(iv) 6x² + 16xy + 8y²
(v) 12x² + 36x²y + 27y²x²
(vi) (a + b)² + 9 (a + b) + 18
Solution:
(i) 2a² + 9a + 10
2a² + 9a + 10 = 2a² + 4a + 5a + 10

= 2a(a + 2) + 5 (a + 2)
= (a+ 2) (2a+ 5)

(ii) 5x² – 29xy – 42y²
5x² – 35xy + 6xy – 42y²

= 5x (x – 7) + 6y (x – 7)
= (x – 7) (5x + 6y)

(iii) 9 – 18x + 8x²
= 8x² – 18x + 9
= 8x² – 6x – 12x + 9

= 2x (4 x – 3) – 3 (4x – 3)
= (4x – 3) (2x – 3)

(iv) 6x² + 16xy + 8y²
= 2 (3x² + 8xy + 4y²)
= 2 (3x² + 8xy + 4y²)

= 2 (3x² + 6xy + 2xy + 4y²)
= 2 (3x (x + 2y) + 2y (x + 2y))
= 2 (x + 2y) (3x + 2y)

(v) 12x² + 36x²y + 27y²x²
= 27y²x² + 36x²y + 12x² = 3x²(9y² + 12y + 4)

= 3x² (9y² + 6y + 6y + 4) = 3x² (3y (3y + 2) + 2 (3y + 2))
= 3x² (3y + 2) (3y + 2) = 3x² (3y + 2) (3y + 2)

(vi) (a + b)² + 9 (a + 6) + 18
= (a + b)² + 6 (a + b) + 3 (a + b) + 18

= (a + b) ((a + b) + 6) + 3 ((a + b) + 6)
= ((a + 6) + 6) ((a + b) + 3) = (a + b + 6) (a + b + 3)

9th Maths 3.6 Question 3.
Factorise the following:
(i) (p – q)² – 6(p – q) – 16
(ii) m² + 2mn – 24n²
(iii) \(\sqrt{5} a^{2}\) + 2a – \(3 \sqrt{5}\)
(iv) a⁴ – 3a² + 2
(v) 8m³ – 2m²n – 15mn²
(vi) \(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{2}{x y}\)
Solution:
(i) (p – q)² – 6 (p – q) – 16
= (p – q)² – 8(p – q) + 2(p – q) – 16

= (p – q)((p – q) – 8) + 2((p – q) – 8)
= (p – q – 8)(p – q + 2)

(ii) m² + 2mn – 24n²
= m² + 6mn – 4mn – 24n²

= m(m + 6n) – 4n(m + 6n)
= (m + 6n)(m – 4n)