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Tamilnadu State Board New Syllabus Samacheer Kalvi 9th Science Guide Pdf Chapter 14 Acids, Bases and Salts Text Book Back Questions and Answers, Notes.

Samacheer Kalvi 9th Science Solutions Chapter 14 Acids, Bases and Salts

9th Science Guide Acids, Bases and Salts Text Book Back Questions and Answers

I. Choose the correct answer :

Question 1.
Zn + 2HCl → ZnCl₂ + ….↑ (H₂, O₂, CO₂)
Answer:
H₂

Question 2.
Apple contains malic acid. Orange contains …………………(citric acid, ascorbic acid).
Answer:
ascorbic acid

Question 3.
Acids in plants and animals are organic acids. Whereas Acids in rocks and minerals are …………………. (Inorganic acids, Weak acids).
Answer:
Inorganic acids

Question 4.
Acids turn blue litmus paper to …………….. (green, red, orange).
Answer:
Red

Question 5.
Since metal carbonate and metal bicarbonate are basic, they react with acids to give salt and water with the liberation of ……………….. (NO₂, SO₂, CO₂).
Answer:
CO₂

Question 6.
The hydrated salt of copper sulphate has …………….colour (red, white, blue).
Answer:
Blue

II. Answer in briefly :

Question 1.
Classify the various types of Acids based on their sources.
Answer:
The acids are classified based on their sources and organic and inorganic acids.
Organic acids – acids present in plants and animals.
Inorganic acids – acids prepared from rocks and minerals.

Question 2.
Write any four uses of acids.
Answer:

• Sulphuric acid is called King of Chemicals because it is used in the preparation of many other compounds. It is used in car batteries also.
• Hydrochloric acid is used as a cleansing agent in toilets.
• Citric acid is used in the preparation of effervescent salts and as a food preservative.
• Nitric acid is used in the manufacture of fertilizers, dyes, paints and drugs.
• Oxalic acid is used to clean iron and manganese deposits from quartz crystals. It is also used as bleach for wood and removing black stains.
• Carbonic acid is used in aerated drinks.
• Tartaric acid is a constituent of baking powder.

Question 3.
Give the significance of p^H of .soil in agriculture.
Answer:
In agriculture, the p^H of soil is very important. Citrus fruits require slightly alkaline soil, while rice requires acidic soil and sugarcane requires neutral soil.

Question 4.
What are the various uses of Aquaregia.
Answer:

• It is used chiefly to dissolve metals such as gold and platinum.
• It is used for cleaning and refining gold.

Question 5.
What are the uses of Plaster of Paris?
Answer:

• It is used for plastering bones.
• It is used for making casts for statues.

Question 6.
Two acids ‘A’ and ‘B’ are given. Acid A gives one hydrogen ion per molecule of the acid in solution. Acid B gives two hydrogen ions per molecule of the acid in solution.
(i) Find out acid A and acid B.
(ii) Which acid is called the King of Chemicals?
Answer:
(i) Acid A – HCl – Hydrochloric acid. Acid B – H₂SO₄ – Sulphuric acid.
(ii)Sulphuric acid – H₂SO₄.

Question 7.
Define aquaregia.
Answer:
It is the mixture of hydrochloric acid and nitric acid prepared optimally in a molar ratio of 3 : 1.

Question 8.
Correct the mistakes :
(a) Washing soda is used for making cakes and bread soft, spongy.
(b) Calcium sulphate hemihydrate is used in textile industry.
Answer:
(a) Baking soda (Sodium bicarbonate – NaHCO₃) is used for making cakes and bread soft spongy, (or) Washing soda is used for softening hard water.
(b) Calcium sulphate hemihydrate (CaSO₄, 1/2 H₂O)is used for plastering bones (or) Bleaching powder (Calcium oxy Chloride – CaOC₁₂) is used in textile industry.

Question 9.
What is neutralization reaction? Give an example.
Answer:
Neutralization reaction is a reaction in which an acid reacts with a base to form salt and water and H⁺ ion and OH^– ion combines to generate water. The neutralization of a strong acid and strong base has a p^Hequal 7.

III. Answer in detail :

Question 1.
Differentiate hydrate and anhydrous salts with examples.
Answer:

Question 2.
Give the tests to identify Acids and Bases.
Answer:

• Acids turn blue litmus. Red bases turn red litmus blue.
• In acid phenolphthalein is colourless. In base Phenolphthalein is pink in colour.
• In acid methyl orange is pink. In bases methyl orange is yellow

Question 3.
Write any four uses of bases.
Answer:

• Sodium hydroxide is used in the manufacture of soap.
• Calcium hydroxide is used in the whitewashing of buildings.
• Magnesium hydroxide is used as a medicine for a stomach disorder.
• Ammonium hydroxide is used to remove grease stains from cloth.

Question 4.
Write any five uses of salts.
Answer:
Common Salt (NaCl) :
It is used in our daily food and used as a preservative
Washing Soda (Sodium Carbonate – Na₂CO₃) :

• It is used in softening hard water.
• It is used in glass, soap and paper industries.

Baking Soda (Sodium bicarbonate -NaHCO₃):

• It is used in making of baking powder which is a mixture of baking soda and tartaric acid.
• It is used in soda-acid fire extinguishers.
• Baking powder is used to make cakes and bread, soft and spongy.
• It neutralizes excess acid in the stomach and provides relief.

Bleaching powder (Calcium Oxychloride – CaOCl₂):

• It is used as a disinfectant.
• It is used in textile industry for bleaching cotton and linen.

Plaster of Paris (Calcium Sulphate Hemihydrate – CaSO₄ .1/2 H₂O):

• It is used for plastering bones
• It is used for making casts for statues.

Question 5.
Sulphuric acid is called King of Chemicals. Why is it called so?
Answer:
Sulphuric acid is called King of Chemicals because it is used in the preparation of many other compounds

Intext Activities

ACTIVITY – 2

Take solutions of hydrochloric acid or sulphuric acid. Fix two nails on a cork and place the cork in a 100 ml beaker.
Connect the nails to the two terminals of a 6 V battery through a bulb and a switch as shown in Figure. Now pour some dilute HCl in the beaker and switch on the current. Repeat the activity with dilute sulphuric acid, glucose and alcohol solutions. What do you observe now? Does the bulb glow in all cases?

Answer:

• The bulb glows when sulphuric acid and hydrochloric acid are used.
• The bulb does not glow when the activity is done with alcohol and glucose solution.

ACTIVITY – 3

Collect the following samples from the science laboratory – Hydrochloric acid, Sulphuric acid and Nitric acid, Sodium hydroxide, Potassium hydroxide. Take 2 ml of each solution in a test tube and test with a litmus paper and indicators phenolphthalein and Methyl orange. Tabulate your observations.

Answer:

ACTIVITY-4

Answer:

ACTIVITY – 5
Boil about 100 ml of groundwater in a vessel to dryness. After all the water get evaporated observe the inner wall of the vessel. Can you observe any deposits? This is the deposit of dissolved salts present in water.
Answer:
This is the deposit of dissolved salts present in water.

9th Science Guide Acids, Bases and Salts Additional Important Questions and Answers

I. Choose the correct answer :

Question 1.
Acid secreted in our stomach is ……………….
(a) hydrochloric acid
(b) sulphuric acid
(c) nitric acid
(d) carbonic acid
Answer:
(a) hydrochloric acid

Question 2.
Hydrochloric Acid reacts with metal bicarbonates to give ……………
(a) metal chloride
(b) water
(c) carbon di – oxide
(d) all the above
Answer:
(d) all the above

Question 3.
…………… & ………….. metals do not react with HCl or HNO₃.
(a) Gold & Magnesium
(b) Silver & Magnesium
(c) Gold & Silver
(d) Zinc & Silver
Answer:
(c) Gold & Silver

Question 4.
The molar ratio of hydrochloric acid and nitric acid in aquaregia is ………………..
(a) 1 : 3
(b) 6 : 3
(c) 2 : 3
(d) 3 : 1
Answer:
(d) 3 : 1

Question 5.
Bases ionise in water to form ……………………. ions.
(a) H⁺
(b) H₃O⁺
(c) OH^–
(d) O^2-
Answer:
(c) OH ^–

Question 6.
Which of the following pairs are weak base?
(a) NH₄OH & NaOH ‘
(b) Ca(OH)₂ & KOH
(c) NH₄OH & Ca(OH)₂
(d) NaOH&KOH
Answer:
(c) NH₄OH & Ca(OH)₂

Question 7.
NaOH & KOH are …………….
(a) strong bases
(b) metal Oxides
(c) weak bases
(d) diacidic bases
Answer:
(a) strong bases

Question 8.
Which of the following solution is soapy to touch?
(a) Acidic
(b) Basic
(c) Salt
(a(d) Aquaregia
Answer:
(b) Basic

Question 9.
Which of the following solutions do not conduct electricity?
(a) alcohol
(b) glucose
(c) sulphuric acid
(d) both a & b
Answer:
(d) both a & b

Question 10.
The p^H value of neutral solution is ………………
(a) = 7
(b) <7
(c) none of the above
(d) -7
Answer:
(a) = 7

Question 11.
The p^H of stomach fluid is ……………….
(a) 4
(b) 2
(c) 6
(d) 7
Answer:
(b) 2

Question 12.
A salt which is formed by complete neutralization of an acid and a base is called ………………. salt.
(a) basic
(b) acid
(c) double
(d) normal
Answer:
(d) normal

Question 13.
The number of water molecules present in one molecule of copper sulphate is
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) 5

Question 14.
The molecular formula of copper sulphate pentahydrate is ………………..
(a) CuSO₄. H₂O
(b) CuSO₄. 2H₂O
(c) CuSO₄. 5H₂O
(d) CuSO₄. 3H₂O
Answer:
(c) CuSO₄ . 5H₂O

Question 15.
……………is the gas produced when HCl is added with carbonate salt.
(a) H₂
(b) N₂
(C) CO₂
(d) O₂
Answer:
(c)CO₂

Question 16.
The formula of bleaching powder is ……………..
(a) CaCl₂
(b) CaOCl₂
(c) Ca(OH)₂
(d) CaO
Answer:
(b) CaOCl₂

Question 17.
The chemical name of plaster of paris is ……………………..
(a) Calcium sulphate hemihydrate
(b) Calcium sulphate monohydrate
(c) Calcium sulphate dihydrate
(d) Calcium sulphate trihydrate
Answer:
(a) Calcium sulphate hemihydrate

Question 18.
Which of the following metal does not react with sodium hydroxides?
(a) Cu
(b) Ag
(c) Cr
(d) All the above
Answer:
(d) All the above

Question 19.
Curd contains …………….acid.
(a) malic ‘
(b) formic
(c) lactic
(d) ascorbic
Answer:
(c) lactic

Question 20.
Which one of the following acids undergoes complete ionisation?.
(a) HCl
(b) CH₃COOH
(c) H₂SO₄
(d) all the above
Answer:
(a) HCl

II. Fill in the blanks :

1. Acid reacts with base to form a neutral product called ………………
Answer:
Salt

2. The taste of acid is ………………..
Answer:
Sour

3. …………….. contain one or more replaceable hydrogen atoms.
Answer:
Acids

4. ………………..acids have a relatively smaller amount of acids dissolved in a solvent.
Answer:
Dilute

5. Acids react with metallic oxides to produce ……………….
Answer:
salt and water

6. _________ acid is used in aerated drinks.
Answer:
Carbonic acid

7. Chemical formula of aquaregia is ________
Answer:
3HCl + HNO₃

8. Water soluble bases are called ________
Answer:
alkali

9. Non-metallic oxides are ___________ in nature.
Answer:
acid

10. __________ are bitter in taste.
Answer:
Bases

11. __________ alkali has a relatively high percentage of alkali in its aqueous solution.
Answer:
Concentrated

12. Acids turn blue litmus to ………….
Answer:
red

13. Phenolphthalein and methyl Orange are ………………
Answer:
indicators

14. p^H stands for …………….. in a solution.
Answer:
power of hydrogen ion concentration

15. The p^H value of acids are …………….. than 7.
Answer:
lesser

16. White enamel coating of our teeth is…………….
Answer:
Calcium phosphate

17. Salt is ………….. in nature.
Answer:
hygroscopic

18. Salt which is formed by the partial replacement of hydrogen ions of an acid by a metal is called ………………………..
Answer:
acid salt

19 Salts contaíning water of crystallisation are called …………. salts.
Answer:
hydrated

20. Salts that do not contain water of crystallisation is called …………..
Answer:
anhydrous salt

21. p^H value of human blood is ……………..(7.0, 7.4, 7.6).
Answer:
7.4

22. The nature of the toothpaste commonly used is ……………. in nature (acidic, basic, neutral)
Answer:
basic

23. You are given pure water to test the p^H value using p^H paper. It shows colour (White, black, green)
Answer:
green

III. To Match:

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

IV. Complete the following equations.

Question 1.
HCl + H₂O → ? + ?
Answer:
HCl + H₂O → H₃O⁺ + Cl^–

Question 2.
H⁺ + H₂O → ?
Answer:
H⁺ + H₂O → H₃O⁺

Question 3.
Mg+ ? → ? + H₂
Answer:
Mg + H₂SO₄ → MgSO₄ + H₂↑

Question 4.
Na₂CO₃ + 2HCl → ? + ? + CO₂↑
Answer: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂↑

Question 5.
ZnO + 2HCl → ? + ?
Answer:
ZnO + 2HCl → ZnCl₂ + H₂O↑

Question 6.
Zn + ? → Na₂ZnO₂ + H₂ ↑
Answer:
Zn + 2NaOH → Na₂ZnO₂+ H₂↑

Question 7.
CaO + H₂SO₄ → ? + H₂O
Answer:
CaO + H₂SO₄ → CaSO₄ + H₂O

Question 8.
HCl + ? → NaCl + H₂O
Answer:
HCl + NaOH → NaCl + H₂O

Question 9.
Ca(OH)₂ + ? → ? + H₂O
Answer:
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Question 10.

Answer:
A = NH₄Cl, B = NH₃↑
C = HCl
D = NaHSO₄,E = H₂O

V. Answer in briefly :

Question 1.
What are organic acids? Given examples.
Answer:
Acids present in plants and animals (living things) are organic acids.
Example: HCOOH, CH₃COOH

Question 2.
How are acids classified based on ionisation? Give examples.
Answer:
Acids get ionise in water (produce H⁺ ions) completely or partially. Based on the extent of ionisation, acids are classified as follows:
Strong Acids:
These are acids that ionise completely in water.
Example: HCl

Weak Acids :
These are acids that ionise partially in the water.
Example: CH₃COOH.

Question 3.
What is Aquaregia? Mention its uses.
Answer:
It is a mixture of hydrochloric acid and nitric acid prepared optimally in a molar ratio of 3:1. It is a yellow-orange fuming liquid. It is a highly corrosive liquid, able to attack gold and other resistant substances.

• Aquaregia is used tp dissolve noble metals such as gold, platinum, and palladium.
• It is used Tor cleaning and refining gold.

Question 4.
What does the acidity of base mean?
Answer:
It means the number of replaceable hydroxyl groups present in one molecule of a base.

Question 5.
What is Potash alum? Write its formula.
Answer:
Potash alum is a mixture of potassium sulphate and aluminium sulphate. KAl(SO₄)₂.12H₂O.

Question 6.
What are double salts? Give an example?
Answer:
Double salts are the salts formed by the combination of the saturated solution of two simple salts in equimolar ratio followed by crystallization. For example : Potash alum.

Question 7.
What are basic salts? Give suitable reaction for this.
Answer:
Basic salts are the product formed by the partial replacement of hydroxide ions of a diacidic or triacidic base with an acid radical.
Pb(OH)₂ + HCl → Pb(OH)Cl + H₂O

Question 8.
Why are tooth pastes basic?
Answer:
Toothpastes which are generally basic and used for cleaning the teeth can neutralise the excess acid and prevent tooth decay.

Question 9.
What is water of crystallisation?
Answer:
Many salts are found as crystals with water molecules they contain. These water molecules are known as water of crystallisation.

Question 10.
Why do blue colour copper sulphate becomes white on heating?
Answer:
On heating, blue colour copper sulphate loses its water molecules and becomes white.

Question 11.
Acidic or basic solutions are good conductors of electricity. Justify your answer.
Answer:
Acidic and basic solutions in water conduct electricity because they produce hydrogen and hydroxide ions respectively.

Question 12.
What are hygroscopic substances?
Answer:
Substance which absorbs water from the surroundings are called hygroscopic substances.

Question 13.
Define indicator. Give examples
Answer:
Chemical substances used to find out whether the given solution is acid or base are called indicators. Eg: Phenolphthalein, methyl orange.

Question 14.
Define ionisation.
Answer:
Ionisation is the condition of being dissociated into ions by heat or radiation or chemical reactions or electrical discharge.

Question 15.
How is normal salt obtained? Give a suitable reaction.
Answer:
A normal salt is obtained by complete neutralization of an acid by a base.
NaOH + HCl → NaCl + H₂O

Question 16.
How are bases classified based on ionisation?
Answer:
Based on Ionisation
(a) Strong Bases :
These are bases which ionise completely in aqueous solution.
Example: NaOH, KOH

(b) Weak Bases:
These are bases that ionise partially in aqueous solution.
Example: NH₄OH, Ca(OH)₂

VI. To Interpret:

Question 1.
CH₄ and NH₃, are not acids.
Answer:
CH₄ and NH₃ do not produce hydrogen ion (H⁺) in its aqueous solution.

Question 2.
Acetic acid (CH₃COOH) is a mono basic acid.
Answer:
Though acetic acid contains four hydrogen atoms only one hydrogen can be replaced in its aqueous solution. So CH₃COOH is a mono basic acid.
CH₃COOH → H⁺+ + CH₃COO^–

Question 3.
Al(OH)₃ & Zn(OH)₂ are not alkalis.
Answer:
Al(OH)₃ & Zn(OH)₂ are water insoluble bases. So they are bases not alkalies.

Question 4.
NaOH & KOH are strong bases.
Answer:
These are bases which ionise completely in aqueous solution.

Question 5.
NaHSO₄ is an acid salt.
Answer:
It is formed by the partial replacement of hydrogen ion of sulphuric acid (H₂ SO₄ ) by a metal present in sodium hydroxide (NaOH).
NaOH + H₂ SO₄ → NaHSO₄ + H₂O

Question 6.
Non-metallic oxides are acidic is nature.
Answer:
When non – metallic oxides react with bases, they can form a salt and water similar to the reaction of base with acids. So non-metallic oxides are acidic in nature.

VII. Complete the following table:

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:
1. – d, 2. – f, 3. – a, 4. – e, 5. – c

Question 4.

Answer:
l.-e,2.-f,3.-d,4-b,5.-c

VIII. Find odd one out & give reason :

Question 1.
HCl, HNO₃, HCOOH, H₃PO₄
Answer:
H₃PO₄,
It is a tribasic acid whereas other three are monobasic acids.

Question 2.
Acetic acid, formic acid, tartaric acid, sulphuric acid.
Answer:
Sulphuric acid.
It is a mineral acid whereas other three are organic acids.

Question 3.
CaO, Na₂O, ZnO, NaOH
Answer:
NaOH.
It is a base containing replaceable hydroxyl ion. whereas other three bases contain replaceable oxide ions.

Question 4.
Bitter taste, soupy to touch, turns red litmus to blue, produce pink colour with methyl orange.
Answer:
Produce pink colour with methyl orange.
It is the property of acids whereas other three are properties of bases.

Question 5.
Litmus paper, phenolphthalein, methyl orange, Aquaregia.
Answer:
Aquaregia.
It is a mixture of two acids namely, hydrochloric acid and nitric acid, used to dissolve metals such as gold and platinum whereas other three are indicators used to identify the nature of the solution.

IX. Spot the error / Correct the wrong statement given below :

Question 1.
An acid is the compound which are capable of forming hydroxyl ions (OH^– ) in aqueous solution. .
Answer:
An acid is the compound which are capable of forming hydrogen ions (H+) in aqueous , solution, (or) A base is the compound which are capable of forming hydroxyl ion (OH-) in aqueous solution.

Question 2.
Nitric Acid is a constituent of baking powder.
Answer:
Tartaric acid is a constituent of baking powder.

Question 3.
The p^H value of the base in lesser than 7.
Answer:
The p^H value of the base is greater than 7. (or) p^H value of an acid is lesser than 7.

Question 4.
Ca(OH)₂ is a triacidic base.
Answer:
Ca(OH)₂ is a diacidic base (or) Al(OH)₃ is a triacidic base.

Question 5.
Magnesium hydroxide is used in whitewashing of buildings.
Answer:
Magnesium hydroxide is used as an antacid. (or) Calcium hydroxide is used in whitewashing of buildings.

X. Answer in detail :

Question 1.
Explain the classification of acids based on their basicity.
Answer:
Monobasic Acid :
Acid that contain only one replaceable hydrogen atom per molecule is called monobasic acid. It gives one hydrogen ion per molecule of the acid in solutions.
Example: HCl, HNO₃
.
Dibasic Acid :
An acid which gives two hydrogen ions per molecule of the acid in solution.
Example : H₂SO₄, H₂CO₃
‘
Tribasic Acid :
An acid which gives three hydrogen ions per molecule of the acid in solution.
Example: H₃PO₄.

Question 2.
Write notes on the properties of acids.
Answer:
(a) They have sour taste.
(b) Their aqueous solutions conduct electricity since they contain ions
(c) Acids turns blue litmus red.
(d) Acids react with active metals to give hydrogen gas.
Mg + H₂SO₄ → MgSO₄ + H ₂↑
Zn + 2HCl → ZnCl₂ + H₂↑

(e) Acids react with metal carbonate and metal hydrogen carbonate to give carbon dioxide.
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂↑
NaHCO₃ + HCl → NaCl + H₂O + CO₂↑

(f) Acids react with metallic oxides to give salt and water.
CaO + H₂SO₄ → CaSO₄+ H₂O

(g) Acids react with bases to give salt and water.
HCl + NaOH → NaCl + H₂O.

Question 3.
Write notes on the properties of bases. .
Answer:
(a) They have bitter taste.
(b) Their aqueous solutions have soapy touch.
(c) They turn red litmus blue.
(d)Their aqueous solutions conduct electricity.
(e) Bases react with metals to form salt with the liberation of hydrogen gas.
Zn + 2 NaOH → Na₂ZnO₂ + H₂↑

(f) Bases react with non-metallic oxides to produce salt and water. Since this is similar to the reaction between a base and an acid, we can conclude that non-metallic oxides are acidic in nature,
Ca(OH)₂ + CO₂ → CaCO₃+ H₂O

(g) Bases react with acids to form salt and water.
KOH + HCl → KCl + H₂O
The above reaction between a base and an acid is known as Neutralisation reaction.

(h) On heating with ammonium salts, bases give ammonia gas.
NaOH + NH₄Cl → NaCl + H₂O+ NH₃↑

Question 4.
Describe the classification of bases based on their acidity.
Answer:
(a) Monoacidic Base :
It is a base that ionises in water to give one hydroxide ion per molecule.
Example: NaOH, KOH

(b) Diacidic Base:
It is a base that ionises in water to give two hydroxide ions per molecule.
Example: Ca(OH)₂, Mg(OH)₂

(c) Triacidic Base :
It is a base that ionises in water to give three hydroxide ions per molecule.
Example: Al(OH)₃, Fe(OH)₃

Question 5.
Write notes on importance of pH in everyday life.
Answer:
p^H in our digestive system :
Hydrochloric acid produced in our stomach helps in the digestion of food without harming the stomach. During indigestion, the stomach produces too much acid and this causes pain and irritation. pH of stomach fluid is approximately 2.0

p^H changes is the cause of tooth decay :
White enamel coating of our teeth is calcium phosphate, the hardest substance in our body. Toothpaste which are generally basic and used for cleaning the teeth can neutralise the excess acid and prevent tooth decay.

p^H of soil:
In agriculture, the pH of soil is very important. Citrus fruits require slightly alkaline soil, while rice requires acidic soil and sugarcane requires neutral soil.

p^H of rain water:
The pH of rainwater is approximately 7 which means that it is neutral and also represents its high purity. If the atmospheric air is polluted with oxides of sulphur and nitrogen, they get dissolved in rainwater and make its pH less than 7. Thus, if the pH of rainwater is less than 7, then it is called acid rain. When acid rain flows into the rivers it lowers the pH of the river water. The survival of aquatic life in such rivers becomes difficult.

Question 6.
List the properties of salts.
Answer:

•  Salts are mostly solids which melt as well as boil at high temperature.
• Most of the salts are soluble in water.
  For example, chloride salts of potassium and sodium are soluble in water. But silver chloride is insoluble in water.
• They are odourless, mostly white, cubic crystals or crystalline powder with a salty taste.
• Salt is hygroscopic in nature.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5

Choose the correct or the most suitable answer:

Question 1.
The value of 2 + 4 + 6 + …………… + 2n is
(1) \(\frac{n(n-1)}{2}\)
(2) \(\frac{n(n+1)}{2}\)
(3) \(\frac{2 n(2 n+1)}{2}\)
(4) n(n + 1)
Answer:
(4) n(n + 1)

Explaination:
2 + 4 + 6 + ……… + 2n = 2(1 + 2 + 3 + ………….. + n)
= 2 × \(\frac{n(n+1)}{2}\)
= n(n + 1)

Question 2.
The coefficient of x⁶ in (2 + 2x)¹⁰ is
(1) 10C₆
(2) 2₆
(3) 10C₆2⁶
(4) 10C₆2¹⁰
Answer:
(4) 10C₆2¹⁰

Explaination:

Question 3.
The coefficient of x⁸ y¹² in the expansion of (2x + 3y)²⁰ is
(1) 0
(2) 2⁸ 3¹²
(3) 2⁸ 3¹² + 2¹² 3⁸
(4) 20C₈ 2⁸ 3¹²
Answer:
(4) 20C₈ 2⁸ 3¹²

Explaination:

Question 4.
If nC₁₀ > nC_r for all possible r then a value of n is
(1) 10
(2) 21
(3) 19
(4) 20
Answer:
(4) 20

Explaination:
Out of ¹⁰C₁₀, ²¹C₁₀, ¹⁹C₁₀ and ²⁰C₁₀, ²⁰C₁₀ is larger.

Question 5.
If a is the Arithmetic mean and g is the Geometric mean of two numbers then
(1) a ≤ g
(2) a ≥ g
(3) a = g
(4) a > g
Answer:
(2) a ≥ g

Explaination:
Given Arithmetic mean = a,
Geometric mean = g
We have A. M ≥ G. M
∴ a ≥ g

Question 6.
If (1 + x²)² (1 + x)ⁿ = a₀ + a₁x + a₂x² + ………… + x^{n + 4} and if a₀, a₁, a₂ are in A. P then n is
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(3) 3

Explaination:

n² – 5n + 6 = 0
(n – 2) (n – 3) = 0
n = 2 or n = 3

Question 7.
If a, 8, b are in A .P , a, 4 , b are in G. P and if a, x ,b are in H . P then x is
(1) 2
(2) 1
(3) 4
(4) 16
Answer:
(1) 2

Explaination:
Given a, 8, b are in A. P ∴ 2 × 8 = a + b ⇒ a + b = 16 ——— (1)
Also a, 4, b are in G.P ∴ 4² = a . b ⇒ ab = 16 ——- (2)
Also a, x, b are in H.P. ∴ \(\frac{1}{a}, \frac{1}{x}, \frac{1}{b}\) are in A.P

Question 8.

(1) A. P
(2) G.P
(3) H.P
(4) AGP
Answer:
(3) H.P

Explaination:

Question 9.
The H.M of two positive numbers whose A.M and G.M are 16,8 respectively is
(1) 10
(2) 6
(3) 5
(4) 4
Answer:
(4) 4

Explaination:
Let a, b be the two numbers. Given A. M = \(\frac{a+b}{2}\) = 16
G.M = \(\sqrt{a b}\) = 8
a+b = 16 × 2 = 32
ab = 8² = 64
H.M = \(\frac{2 a b}{a+b}\)
= \(\frac{2 \times 64}{32}\) = 2 × 2 = 4

Question 10.
If S denote the sum of n terms of an A. P whose common difference is d, the value of S_n – 2S_{n- 1} + S_{n – 2} is
(1) d
(2) 2d
(3) 4d
(4) d²
Answer:
(1) d

Explaination:

= a + (n – 1) d – (a + (n – 2)d)
= a + (n – 1) d – a – (n – 2)d
= a + nd – d – a – nd + 2d = d

Question 11.
The remainder when 38¹⁵ is divided by 13 is
(1) 12
(2) 1
(3) 11
(4) 5
Answer:
(1) 12

Explaination:
38¹⁵ = (39 – 1)¹⁵ = 39¹⁵ – 15C₁ 39¹⁴(1) + 15C₂ (39)¹³(1)² – 15C₃ (39)¹²(1)³ ….. + 15C₁₄ (39)¹(1) – 15C₁₅(1)
Except -1 all other terms are divisible by 13.
∴ When 1 is added to it the number is divisible by 13. So the remainder is 13 – 1 = 12.

Question 12.
The nth term of the sequence 1, 2, 4, 7, 11, ………….. is
(1) n² + 3n² + 2n
(2) n³ – 3n² + 3n
(3) \(\frac{n(n+1)(n+2)}{3}\)
(4) \(\frac{n^{2}-n+2}{2}\)
Answer:
(4) \(\frac{n^{2}-n+2}{2}\)

Explaination:

Question 13.
The sum up to n terms of the series

(1) \(\sqrt{2 n+1}\)
(2) \(\frac{\sqrt{2 n+1}}{2}\)
(3) \(\sqrt{2 n+1}-1\)
(4) \(\frac{\sqrt{2 n+1}-1}{2}\)
Answer:
(4) \(\frac{\sqrt{2 n+1}-1}{2}\)

Explaination:

Question 14.
The n^th term of the sequence

(1) 2ⁿ – n – 1
(2) 1 – 2^-n
(3) 2^-n + n – 1
(4) 2^n-1
Answer:
(2) 1 – 2^-n

Explaination:

Question 15.
The sum up to n terms of the series

(1) \(\frac{\mathbf{n}(\mathbf{n}+1)}{2}\)
(2) 2n (n + 1)
(3) \(\frac{\mathbf{n}(\mathbf{n}+1)}{\sqrt{2}}\)
(4) 1
Answer:
(3) \(\frac{\mathbf{n}(\mathbf{n}+1)}{\sqrt{2}}\)

Explaination:

Question 16.
The value of the series

(1) 14
(2) 7
(3) 4
(4) 6
Answer:
(1) 14

Explaination:

Question 17.
The sum of an infinite G.P is 18. If the first term is 6 the common ratio is
(1) \(\frac{1}{3}\)
(2) \(\frac{2}{3}\)
(3) \(\frac{1}{6}\)
(4) \(\frac{3}{4}\)
Answer:
(2) \(\frac{2}{3}\)

Explaination:
Let the geometric series be a, ar, ar², …………… ar^n-1
Given a = 6, S_∞ = 18

Question 18.
The coefficient of x⁵ in the series e^-2x is
(1) \(\frac{2}{3}\)
(2) \(\frac{3}{2}\)
(3) \(-\frac{4}{15}\)
(4) \(\frac{4}{15}\)
Answer:
(3) \(-\frac{4}{15}\)

Explaination:

Question 19.
The value of

(1) \(\frac{e^{2}+1}{2 e}\)
(2) \(\frac{(e+1)^{2}}{2 e}\)
(3) \(\frac{(e-1)^{2}}{2 e}\)
(4) \(\frac{e^{2}+1}{2 e}\)
Answer:
(3) \(\frac{(e-1)^{2}}{2 e}\)

Explaination:

Question 20.
The value of

(1) log \(\left(\frac{5}{3}\right)\)
(2) \(\frac{3}{2}\) log \(\left(\frac{5}{3}\right)\)
(3) \(\frac{5}{3}\) log \(\left(\frac{5}{3}\right)\)
(4) \(\frac{2}{3}\) log \(\left(\frac{2}{3}\right)\)
Answer:
(2) \(\frac{3}{2}\) log \(\left(\frac{5}{3}\right)\)

Explaination:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 1.
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.
(i) \(\frac{1}{5+x}\)
Answer:

(ii) \(\frac{2}{(3+4 x)^{2}}\)
Answer:

(iii) (5 + x²)^2/3
Answer:

(iv) \((x+2)^{-\frac{2}{3}}\)
Answer:

Question 2.
Find \(\sqrt[3]{1001}\) approximately. (two decimal places)
Answer:

Question 3.
Prove that \(\sqrt[3]{x^{3}+6}-\sqrt[3]{x^{3}+3}\) is approximately equal to \(\frac{1}{x^{2}}\) when x is sufficiently large.
Answer:

Question 4.
Prove that \(\sqrt{\frac{1-x}{1+x}}\) is approximately equal to 1 – x + \(\frac{x^{2}}{2}\) when x is very small.
Answer:

Question 5.
Write the first 6 terms of the exponential series
(i) e^5x
Answer:

(ii) e^-2x
Answer:

(iii) e^x/2
Answer:

Question 6.
Write the first 4 terms of the logarithmic series.
(i) log (1 + 4x)
(ii) log (1 – 2x)
(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)
(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)
Find the intervals on which the expansions are valid.
Answer:
(i) log ( 1 + 4x )

(ii) log (1 – 2x)

(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)

(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)

Question 7.

Answer:

Question 8.

Answer:

Question 9.
Find the coefficient of x⁴ in the expansion of \(\frac{3-4 x+x^{2}}{e^{2 x}}\)
Answer:

Question 10.
Find the value of
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3

Question 1.
Find the sum of the first 20 terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77 .
Answer:
Sum of the first n terms of an Arithmetic progression is S_n = \(\) [2a + (n – 1)d]
Given S₁₀ = 52
52 = \(\frac{10}{2}\) [ 2a + (10 – 1) d ]
52 = 5 [2a + 9d]
52 = 10a + 45d ……….. (1)
Also given S₁₅ = 77
77 = \(\frac{15}{2}\) [2a + (15 – 1)d]
77 = \(\frac{15}{2}\) [2a + 14d]
77 = 15 [a + 7d]
77 = 15a + 105d ………….. (2)

Substituting the value of d in (1)

Question 2.
Find the sum up to the 17^th term of the series
Answer:

Question 3.
Compute the sum of first n terms of the following series.
(i) 8 + 88 + 888 + 8888 + . . . . . . .
Answer:

(ii) 6 + 66 + 666 + 6666 + . . . . . . .
Answer:

Question 4.
Compute the sum of first n terms of 1 + (1 + 4) + (1 + 4 + 4²) + (1 + 4 + 4² + 4³) + …………..
Answer:
The given series is 1 + (1 + 4) + (1 + 4 + 4²) + (1 + 4 + 4² + 4³) + …………..
n^th term of the series is T_n = 1 + 4 + 4² + 4³ + ………………. + 4^n-1

Sum to n terms of the series
S_n = T₁ + T₂ + ………….. + T_n

Question 5.
Find the general terms and sum to n terms of the sequence 1, \(\frac{4}{3}\), \(\frac{7}{9}\), \(\frac{10}{27}\), ………….
Answer:
The given sequence as 1, \(\frac{4}{3}\), \(\frac{7}{9}\), \(\frac{10}{27}\), ………….
Consider the terms in the numerator 1, 4, 7, 10, ……………….. which is an Arithmetic progression
with first term a = 1, common difference d = 4 – 1 = 3
a_n = a + (n – 1)d = 1 + (n – 1)3 = 1 + 3n – 3 = 3n – 2
The given sequence can be written as 1, (1 + 3)\(\left(\frac{1}{3}\right)\), (1 + 2 × 3)\(\left(\frac{1}{3}\right)^{2}\) (1 + 3 × 3) \(\left(\frac{1}{3}\right)^{3}\), …………….
where \(1, \frac{1}{3}, \frac{1}{3^{2}}, \frac{1}{3^{3}}, \ldots \ldots\) is a G. P with first term a = 1 , common ratio r = \(\frac{1}{3}\)
∴ The given sequence is an Arithmetico – Geometric sequence.


which is an arithmetic – geometric sequence.
∴ The sum of first n terms of the arithmetico – geometric sequence is given by

Question 6.
Find the value of n if the sum to n terms of the series \(\sqrt{3}+\sqrt{75}+\sqrt{243}+\ldots \ldots . . . \text { is } 435 \sqrt{3}\)
Answer:

Question 7.
Show that the sum of ( m + n)^th and ( m – n)^th term of an A.P is equal to twice the m^th term.
Answer:
Let the A.P. be a, a + d, a + 2d, ……..
t_{m + n} = a + (m + n – 1)d
t_{m – n} = a + (m – n – 1)d
t_m = a + (m – 1)d
2t_m = 2[a + (m – 1)d]
To prove t_{m + n} + t_{m – n} = 2t_m
LHS t_{m + n} + t_{m – n} = a + (m + n – 1)d + a + (m – n – 1)d

= 2a + d [2m – 2]
= 2[a + (m – 1)d) = 2 t_m = RHS

Question 8.
A man repays an amount of Rs.3250 by paying Rs.20 in the first month and then increases the payment by Rs. 15 per month. How long will it take him to clear the amount?
Answer:
Amount of loan = 3250
Let n be the number of months taken to clear the loan
Amount paid in the first month a = 20
Increased payment in every month d = 15
∴ Amount paid in the second month = 20 + 15 = 35
Amount paid in the third month = 35 + 15 = 50
∴ The sequence of amount paid in every month is 20, 35, 50, …………. which is an A.P with first term a = 20 and common difference = 15
Given S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S_n = 3250
3250 = \(\frac{n}{2}\) [2 × 20 + (n – 1) 15]
6500 = n[40 + 15n – 15]
6500 = n[25 + 15n]
6500 = 25n + 15n²
1300 = 5n + 3n²
3n² + 5n – 1300 = 0
3n² + 65n – 60n – 1300 = 0
n(3n + 65) – 20 (3n + 65) = 0
(n – 20) (3n + 65) = 0
n = 20 = 0 or 3n + 65 = 0
n = 20 or n = \(-\frac{65}{3}\)
n = \(-\frac{65}{3}\) is not possible ∴ n = 20
Thus, in 20 months the loan is cleared.

Question 9.
In a race, 20 balls are placed in a line at intervals of 4 meters with the first ball, 24 meters away from the starting point. A contestant is required to bring the balls back to the starting place one at a time. How far would the contestant run to bring back all balls?
Answer:
Number of balls placed in a line = 20
Let A be the starting point.
The distance of the first ball from A = 24 m
The distance of the second ball from the first ball = 4 m

A contestant starts from point A, travels a distance of 24 m and picks the first ball and brings it back to starting point A. This is continued for each ball.
Distance travelled by the contestant to bring the first ball = 24 + 24 = 2 × 24 = 48 m
Distance travelled to bring the second ball = 2 ( 24 + 4 ) = 2 × 28 = 56 m
Distance travelled to bring the third ball = 2 × (24 + 4 + 4) = 64 m

∴ The sequence of distances travelled are 48, 56, 64 ……………
This is an Arithmetic progression with first term a = 48 ,
common difference d = 56 – 48 = 8 ,
number of terms n = 20.
The sum of 20 terms of this A.P gives the total distance travelled by the contestant in bringing all balls to the starting place
S_n = \(\frac{\mathrm{n}}{2}\) [2a + (n – 1)d]
S₂₀ = \(\frac{\mathrm{20}}{2}\) [2 × 48 + (20 – 1)8] = 10 [96 + 19 × 8]
= 10 [ 96 + 152] = 10 × 248 = 2480
Total distance traveled = 2480 m

Question 10.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally how many bacteria will be present at the end of the 2nd hour, 4th hour, and nth hour?
Answer:
Number of bacteria at the beginning = 30
Number of bacteria after 1 hour = 30 × 2 = 60
Number of bacteria after 2 hours = 30 × 2² = 120
Number of bacteria after 4 hours = 30 × 24 = 30 × 16 = 480
∴ Number of bacteria after n^th hour = 30 × 2ⁿ

Question 11.
What will Rs. 500 amounts to in 10 years after its deposit in a bank which pays an annual interest rate of 10 % compounded annually?
Answer:

Question 12.
In a certain town, a viral disease caused severe health hazards, upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infected virus particle gets doubled each day, being 5 particles on the first day. Find the day when the infections virus particles just grow over 1,50,000 units?
Answer:
The number of viruses present at the beginning = 5, Given virus, gets doubled each day
∴ The sequence of a number of viruses in each day is 5, 10, 20, 40, 80, ………………. which is a G. P with First-term a = 5, Common ratio r = \(\frac{10}{5}\) = 2
n^th term t_n = ar^n-1

∴ On the 15^th day, the infectious virus grows over 1,50,000 units.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2

Question 1.
Write the first 6 terms of the sequence whose nth terms are given below and classify them as Arithmetic progression Geometric progression, Arithmetic – geometric progression, Harmonic progression and none of them.
(i) \(\frac{1}{2^{n+1}}\)
Answer:

(ii)
Answer:

(iii) \(4\left(\frac{1}{2}\right)^{n}\)
Answer:

(iv) \(\frac{(-1)^{n}}{n}\)
Answer:

(v) \(\frac{2 n+3}{3 n+4}\)
Answer:

(vi) 2018
Answer:
The n^th term a_n = 2018
a₁ = 2018,
a₂ = 2018,
a₃ = 2018,
a₄ = 2018,
a₅ = 2018,
a₆ = 2018,
∴ The given sequence is 2018, 2018, 2018, 2018, 2018, 2018, ………….
This is a œnstant sequence which has same common ratio and common difference.
Hence this is an A. P, G . P and AGP.

(vii) \(\frac{3 n-2}{3^{n-1}}\)
Answer:

Question 2.
Write the first 6 terms of the sequences whose n^th term a_n is given below
(i)
n = 1, a_n = n + 1, a₁ = 1 + 1 = 2
n = 2, a_n = n, a₂ = 2
n = 3, a_n = n + 1, a₃ = 3 + 1 = 4
n = 4, a_n = n, a₄ = 4
n = 5, a_n = n + 1, a₅ = 5 + 1 = 6
n = 6, a_n = n, a₆ = 6
∴ The first six terms are 2, 2, 4, 4, 6, 6

(ii)
n = 1, a₁ = 1, n = 2, a₂ = 2
n = 3, a_n = a_n+1 + a_n-2, a₃ = a_{3 – 1} + a_{3 – 2} = a₂ + a₁ = 2 + 1 = 3

n = 4, a_n = a_n+1 + a_n-2, a₄ = a_{4 – 1} + a_{4 – 2} = a₃ + a₂ = 3 + 2 = 5

n = 5, a_n = a_n+1 + a_n-2, a₅ = a_{5 – 1} + a_{5 – 2} = a₄ + a₃ = 5 + 3 = 8

n = 6, a_n = a_n+1 + a_n-2, a₆ = a_{6 – 1} + a_{6 – 2} = a₅ + a₄ = 8 + 5 = 13
∴ The first six terms are 1, 2, 3, 5, 8, 13

(iii)
Answer:
n = 1, a_n = n, a₁ = 1
n = 2, a_n = n, a₂ = 1
n = 3, a_n = n, a₃ = 1

n = 4, a_n = a_n-1 + a_n-2 + a_n-3
a₄ = a_4-1 + a_4-2 + a_4-3
a₄ = a₃ + a₂ + a₁
a₄ = 3 + 2 + 1 = 6

n = 5, a_n = a_n-1 + a_n-2 + a_n-3
a₅ = a_5-1 + a_5-2 + a_5-3
a₅ = a₄ + a₃ + a₂
a₅ = 6 + 3 + 2 = 11

n = 6, a_n = a_n-1 + a_n-2 + a_n-3
a₆ = a_6-1 + a_6-2 + a_6-3
a₆ = a₅ + a₄ + a₃
a₆ = 11 + 6 + 3 = 20
∴ The first six terms are 1, 2, 3, 6, 11, 20

Question 3.
Write the n^th term of the following sequences.
(i) 2, 2, 4, 4, 6, 6, ……………..
Answer:
The odd terms are a₁ = 2, a₃ = 4, a₅ = 6
The even terms are a₂ = 2, a₄ = 4, a₆ = 6

(ii)
Answer:

The terms in the numerator are 1 , 2 , 3, 4
a = 1 , d = 2 – 1 = 1
a_n = a + (n – 1) d
a_n = 1 + (n – 1)(1) = 1 + n – 1 = n
a_n = n
The terms in the denominator are 2 , 3 , 4 , 5 , 6 .
a = 2, d = 3 – 2 = 1
a_n = a + (n – 1) d
a_n = 2 + (n – 1) (1) = 2 + n – 1 = n + 1
a_n = n + 1
∴ The n^th term of the given sequence is a_n = \(\frac{n}{n+1}\) for all n ∈ N

(iii)
Answer:

The terms in the numerator are 1, 3, 5, 7, 9, ………….
a = 1 , d = 3 – 1 = 2
a_n = a + (n – 1) d
a_n = 1 + (n – 1)2
a_n = 1 + 2n – 2 = 2n – 1
The terms in the denominator are 2, 4, 6, 8, 10, …………..
a = 2, d = 4 – 2 = 2
a_n = a + (n – 1) d
a_n = 2 + (n – 1)(2)
a_n = 2 + 2n – 2 = 2n
∴ The n^th term of the given sequence is a_n = \(\frac{2 n-1}{2 n}\) for all n ∈ N

(iv) 6, 10, 4, 12, 2, 14, 0, 16, – 2 …………………
Answer:
The given sequence is 6, 10, 4, 12, 2, 14, 0, 16, – 2 ……………..
The odd terms are a₁ = 6, a₃ = 4 , a₅ = 2 , a₇ = 0, a₉ = – 2
∴ a_n = n – 7, n is odd
The even terms are a₂ = 10, a₄ = 12 , a₆ = 14 , a₈ = 16
∴ a_n = 8 + n, n is even.

Question 4.
The product of three increasing numbers in a G.P is 5832 . If we add 6 to the second number and 9 to the third number, then resulting numbers form an A.P. Find the numbers in G.P.
Answer:
Let the increasing numbers in G.P be \(\), a, ar.
Given \(\frac{a}{r}\) × a × ar = 5832 ⇒ a³ = 5832 = 18³ ⇒ a = 18
Also given \(\frac{a}{r}\), a + 6, ar + 9 form an A.P.
∴ 2(a + b) = \(\frac{a}{r}\) + (ar + 9)
⇒ (a + 6) + (a + 6) = \(\frac{a}{r}\) + (ar + 9)
⇒ (a + 6) – \(\frac{a}{r}\) = (ar + 9) – (a + 6)
⇒ a + 6 – \(\frac{a}{r}\) = ar + 9 – a – 6
⇒ a + 6 – \(\frac{a}{r}\) = ar – a + 3
Substituting the value of a = 18, we get


39r = 18r² + 18
18r² – 39r + 18 = 0
(2r – 3)(3r -2) = 0
2r – 3 = 0 or 3r – 2 = 0
r = \(\frac{3}{2}\) or r = \(\frac{2}{3}\)

Case (i) When a = 18, r = \(\frac{3}{2}\) the numbers in G.P are

Case (ii) When a = 18, r = \(\frac{2}{3}\)

Question 5.
Write the n^th term of the sequence \(\frac{3}{1^{2} \cdot 2^{2}}, \frac{5}{2^{2} \cdot 3^{2}}, \frac{7}{3^{2} \cdot 4^{2}}\), …………….. as a difference of two terms.
Answer:

The terms in the numerator are 3,5, 7
which forms an A. P with first term a = 3 and common difference d = 5 – 3 = 2
n^th term t_n = a + (n – 1) d
= 3 + (n – 1)(2)
= 3 + 2n – 2 = 2n + 1
t_n = 2n + 1
The terms in the denominator are 1² . 2², 2² . 3², 3² . 4² ……………….
n^th term t_n = n² . (n + 1)²

Question 6.
If t_k is the k^th term of a G.P then show that t_{n – k}, t_k, t_{n + k} also form a G.F for any positive integer k.
Answer:
Given t^k is the k^th term of a G.P. We have n^th term of a G.P is t_n = ar^n-1

Question 7.
If a, b, c are in geometric progression and if a^1/x = b^1/y = c^1/z are in Arithmetic progression.
Answer:

Question 8.
The A.M of two numbers exceeds their G.M by 10 and H.M by 16. Find the numbers.
Answer:
Let the numbers be a and b

Question 9.
If the roots of the equation (q – r)x² + (r – p)x + (p – q) = 0 are equal then show that p , q and r are in A. P.
Answer:
The roots are equal ⇒ ∆ = 0
(i.e.) b² – 4ac = 0
Hence, a = q – r ; b = r – p ; c = p – q
b² – 4ac = 0
⇒ (r – p)² – 4(q – r)(p – q) = 0
r² + p² – 2pr – 4[qr – q² – pr + pq] = 0
r² + p² – 2pr – 4qr + 4q² + 4pr – 4pq = 0
(i.e.) p² + 4q² + r² – 4pq – 4qr + 2pr = 0
(i.e.) (p – 2q + r)² = 0
⇒ p – 2q + r = 0
⇒ p + r = 2q
⇒ p, q, r are in A.P.

Question 10.
If a , b , c are respectively the p^th, q^th and r^th terms of a G . P show that (q – r) log a + (r – p) log b + (p – q) log c = 0
Answer:
Let A be first term and R be the jmnon ratio of the G.P.
Given a = p^th term of the G.P
General term of a G. P with first term A and common ratio R is t_n = AR^{n – 1}
∴ a = t_p = AR^{P – 1}
log a = log AR^p-1 = log A + log R^p-1 = log A + (p – 1) log R

b = q^th term of the G.P
b = t_q = AR^q-1
log b = log AR^q-1 = log A + log R^q-1 = log A + (q – r)log R

c = r^th term of the G.P
c = t_r = AR^r-1
log c = log AR^r-1 = log A + log R^r-1 = log A + (r – 1) log R

(q – r) log a + (r – p) log b + (p – q) log c
= (q – r) [log A + (p – 1) log R] + (r – p) [ log A + (q – 1) log R ] + (P – q) [ log A + (r – 1) log R]
= (q – r) log A + (q – r) (p – 1 ) log R + (r – p) log A + (r – p) (q – 1) log R + (P – q) log A + (p – q) (r – 1 ) log R
= [ q – r + r – p + p – q ] log A + [ (q – r) (p – 1) + (r – p) (q – 1) + (p – q)(r – 1)] log R
= 0 × log A + [pq – q – rp + r + rq – r – pq + p + pr – p – rq + q] log R
= 0 × log R = 0
∴ (q – r) log a + (r – p) log b + (p – q) log c = 0

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th English Guide Pdf Prose Chapter 4 The Summit Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 12th English Solutions Prose Chapter 4 The Summit

12th English Guide The Summit Text Book Back Questions and Answers

Textual Questions:

1. Based on your reading of the text, answer the following questions in one or two sentences each. (Text Book Page No. 116) (Note: IQ → Important Questions)

Question a.
What did Hillary do with his wet boots? (IQ)
Answer:
Hillary cooked his wet boots over the fierce flame of the Primus and managed to soften them.

Question b.
Name equipment and a tool carried by the climbers during their expedition. (IQ)
Answer:
They carried oxygen bottles, ice-ax, and crampons during their expedition.

Question c.
Why did Hillary become clumsy-fingered and slow-moving?
Answer:
After reaching the peak, Hillary ran out of oxygen, he was becoming clumsy-fingered and slow-moving. So, he quickly replaced his oxygen set.

Question d.
What did Hillary find in a tiny hollow? (IQ)
Answer:
Hillary found two oxygen bottles left by Evan and Bourdillon, who had their earlier attempt to reach Mt. Everest.

Question e.
When did Hillary feel a sense of freedom and well being?
Answer:
Their first partly-full bottle of oxygen got exhausted. They had only one oxygen bottle to cany. With reduced load of 20 litre bottle, Hillary cut steps down off the South Summit. So, he felt a sense of freedom and well-being.

Question f.
What did Hillary mean by saying “We had had enough to do the job, but by no means too much”?
Answer:

1. Hillary meant that they carried an exact number of oxygen bottles needed for their expedition.
2. They didn’t even carry a single bottle more.

2. Answer the following questions in two or three sentences each:(Text Book Page No. 116)

Question a.
How did the mountaineers belay?
Answer:
Belaying refers to a variety of techniques used by mountaineers to exert tension on a climbing rope so that a climber does not fall very far. A climbing partner typically applies tension at the other end of the rope whenever the climber is not climbing and removes the tension from the rope when the climber needs a rope to continue to climb.

Question b.
Why was the original zest fading away?
Answer:
Their original zest was fading away as they had to struggle a lot to find the top of Mt. Everest.

Question c.
What did Edmund Hillary do to escape the ’ large overhanging ice cornices?
Answer:
In a number of places, the overhanging ice cornices were very large. In order to escape them, Hillary cut a line of steps down to where the snow met the rocks on the west.

Question d.
What did Tenzing and Edmond Hillary gift to the God of Softy Summit? How did they do it? (IQ)
Answer:

1. Tenzing gifted a bar of chocolate, a packet of biscuits and a handful of lollies whereas Hillary placed a small crucifix given by Colonel Hunt, besides Tenzing’s gifts.
2. They did so by making a little hole in the snow.

Question e.
What did the photograph portray?
Answer:
The photograph portrayed North Col and the old route which had been made famous by the struggles of those great climbers in 1920’s and 1930’s.

Question f.
The soft snow was difficult and dangerous. Why?
Answer:
The soft snow was difficult and dangerous as it sometimes held Hillary’s weight but often fell down suddenly.

Question g.
How did the firm snow at the higher regions fill them with hope?
Answer:
They were a little perturbed by slippery soft snow. But as they reached firmer snow higher up, they felt better. As one bottle of oxygen got exhausted, their load was now less. As Hillary’s ax bit into the first steep slope of the ridge, his high hopes were realized. The snow was crystalline and firm. They were able to make comfortable belays to haul themselves up slowly.

3. Based on the text, answer the following questions in a paragraph of about 100 – 150 words each: (Text Book Page No. 116)

Question a.
How did Hillary and Tenzing prepare themselves before they set off to the summit?
Answer:
They started up their cookers and drank lots of lemon juice and sugar. Then they took sardines and biscuits. Hillary cleaned the ice off the oxygen sets. He rechecked and tested them. He had removed his boots which had become wet the day before. They were now frozen solid. It would be very challenging to start climbing the ice-cold Himalayas with such wet and chilling boots. So, he cooked them over the fierce flame of Primus and managed to soften them up. They were also conscious of the probability of braving snowstorms during the ascent. They fortified their clothing with windproof and also pulled three pairs of gloves silk, woolen, and windproof on to their hands. At 6.30 am they crawled out of their tent into the snow. They hoisted their 30 lb. of oxygen gear on their backs. Connecting their oxygen masks they turned on the valves to bring life-giving oxygen into their lungs. Taking a few deep breaths, They got ready to go.

Question b.
Give an account of the journey to the South Coi from 28,000 feet.
Answer:
They reached a crest of a ridge at about 28000 feet from where the ridge narrowed to a knife-edge. Hillary took over the lead. The route on top of the ridge was filled with soft snow which was both difficult and dangerous. After several hundred feet they reached a tiny hollow where they found two oxygen bottles left by the earlier expeditors. They took it for their use. Hillary took the lead for nearly 400 feet. They made frequent changes of the lead. When Hillary was stamping a trail in deep soft snow, a portion of it gave way and he slipped back. After this, they had a dilemma to proceed further. Still, without losing hope they proceeded further and climbed on to the south peak.

Question c.
Describe the feelings of Edmund Hillary and Tenzing as they reached the top of the Summit.
Answer:
Hillary’s first feelings were of relief on reaching Everest. There were no more ridges to traverse and no more humps to tease them off with the hope of success. He looked at Tenzing. In spite of the balaclava helmet, goggles, and oxygen mask, all encrusted with long icicles, that concealed his face, his delight was visible. He looked around with a grin of delight. They shook hands. Tenzing hugged Hillary and thumped each other till they were breathless. It was 1 .30 am. The ridge had taken them two and a half hours but it seemed like a lifetime.

Question d.
The ridge had taken us two and a half hours, but it seemed like a lifetime. Why?
Answer:
Tenzing and Hillary struggled hard to reach the top of the ridge. They helped each other belaying. After crossing the crack with great struggle, Hillary signaled to Tenzing to come on up who wriggled his way up the crack and collapsed at the top like a giant fish hailing from the sea. Again the ridge curved away to the right and they were unable to find its top. They felt that the ridge seemed never-ending. At this juncture, they lost their original enthusiasm but didn’t give up their try.

Hillary felt at one spot that the ridge seemed to be dropped sharply away instead of rising. Later he found a narrow snow ridge running up to a snowy summit. With the help of the ice-axe they paved the way in the firm snow and reached the top. Though it took two and a half hours for them to reach the top of the ridge, they felt that it took their whole lifetime just because of the struggle they had to reach it.

Question e.
Describe the view from the top. What was the most important photograph?
Answer:
On reaching the peak, Tenzing and Hillary felt a great relief. To the east was their giant neighbour Makalu, unexplored and unclimbed. Far away across the clouds, the great bulk of Kanchenjunga loomed on the horizon. To the west, they could see the unexplored ranges of Nepal stretching off into the distance. The most important photograph was a shot down at the north ridge showed the North Col and the old route. It had been made famous by the famous climbers of 1920’s and 1930’s. It was the breath-taking view of the snow’ clad peak all round.

Question f.
‘There is no height, no depth that the spirit of man, guided by higher Spirit cannot attain’. Discuss the above statement in the context of the achievement of Edmund Hillary and Tenzing.
Answer:
Tenzing and Hillary started with their ascent from 28,000 feet on 29th May. They had undergone many hardships before they reach the summit. The ridge was narrow from 28,000 feet and they had to make their way through soft snow which was both difficult and dangerous. The soft snow gave way suddenly when Hillary had his trail in the deep snow. In spite of his slipping back, he did not give up his effort to move forward. They also found it very difficult to cross the ridge when the snow was firm.

Crossing the narrow crack between the rock and the cornice was really thrilling. It was done by them with fierce determination. Though their energy level went down when they found the ridge to be never-ending, they did not lose hope and proceeded further. It was this strong determination and undying spirit that made them reach the summit successfully. Thus the statement, ‘There is no height, no depth that the spirit of man, guided by the higher spirit cannot attain gets proved.

Paragraph:

Introduction:
Edmund Hillary in his “The Summit” tells how the summit of Everest was reached.

Preparation:
Hillary and his friend Tenzing got ready for the climb from camp-8 on May 29. They drank large quantities of lemon juice and had some biscuits. He softened his boots which became frozen solid. They also wore three pairs of gloves.

Journey:
They reached a crest of a ridge at about 28000 feet which narrowed to a knife-edge. The route on top of the ridge was both difficult and dangerous as it was filled with soft snow without losing hope they proceeded further and crampons on to the south peak.

The struggle:
The snow was crystalline and firm. With the help of his ice-axe he made away and they proceeded further belaying each other. The overhanging ice cornices threatened them. They struggled hard to cross a great cornice which had a forty feet narrow crack between the cornice and the rock. Hillary found a narrow snow ridge running up to a snowy summit. Thus they reached the top. It took nearly two and a half hours for them to/each the top of the ridge.

Sense of fulfillment:
They thumped each other on their back till they became breathless. As they had undergone a huge risk they felt that they had risked their whole lifetime to reach the top. The most important photograph was that of the North ridge. They made a little hole in the snow and placed some offerings to the Gods. From there they safely started descending to the South Col.

Conclusion:
Thus their ascent of Everest depicts the value of teamwork which helps one to overcome any difficulties. The adventure of Hillary and Tenzing enriches the fact that there is no height, no depth, that the spirit of man, guided by a higher spirit cannot attain.

Vocabulary:

Idioms:

I. Given below are some Idiomatic expressions with their meanings. Understand the meaning: (Text Book Page No.117)

1. wait for the dust to settle – to wait for a situation to become clear or certain.
2. get/have all your ducks in a row – to have made all the preparations needed to do something / to be well-organized
3. fetch and carry (for somebody) – to do a lot of little jobs for somebody as if you were their servant
4. do the math – to think carefully about something before doing it, so that you know all the relevant facts or figures
5. round the corner – very near

II. Fill in the blanks with the right idioms. Choose from the above-given idioms: (Text Book Page No.117)

Question a.
The Sherpas are cheerful, gallant men, who________________ tents, oxygen, food etc., for climbers during their ascent of the summit.
Answer:
fetch and carry

Question b.
The team _________________ carefully so as to reach the summit successfully.
Answer:
does the math

Question c.
When they had to climb through deep new snow the party sometimes had to _________________.
Answer:
wait for the dust to settle

Question d.
Each member of the team had all their _________________.
Answer:
get/have all your ducks in a row

Question e.
We could not believe that with a few more whacks of the ice axe in the firm snow we were _______________ to the top.
Answer:
round the corner

III. Understand the meaning of the given idiomatic expression and choose the right one to complete the sentence:
(Text Book Page No.117)

• the icing on the cake — something extra and not essential, but is added to make it even better
• break the ice — to make people more relaxed, especially at the beginning of the meeting

Question a.
The conference room was silent though packed. The chairman introduced an interactive session to _________.
Answer:
break the ice

Question b.
Our headmistress not only promised us to take us for an excursion but also announced that on return we would get a holiday. It was like _________.
Answer:
the icing on the cake

Phrasal Verbs: (Text Book Page No.118)

i. Given below are the phrasal verbs with their meanings. Use the given phrasal verbs in sentences of your own (Text Book Page No.118)

1. I want to turn on the television.
2. Mr. Prem took over charge as manager.
3. The children set off for school.
4.  The meeting was put off to next week.

ii. Given below are some Phrasal Verbs which are frequently used in connection with travelling. Guess the meaning and match: (Text Book Page No.118)

Question a.

Answer:

Question b.

Answer:

Question c.

Answer:

Compound words: (Text Book Page No. 118)

i. Here are some compound words chosen from the text:

I. Match the following with their right field, choosing appropriately from the box given: (Text Book Page No. 119)

Question 1.
snow-board
Answer:
Sports

Question 2.
snow-mobile
Answer:
transportation

Question 3.
snow-chains
Answer:
machinery

Question 4.
snow-storm
Answer:
weather

Question 5.
snow-bird
Answer:
travel

Question 6.
snow-belt
Answer:
geography

Listening:

First, read the following statements. Then, listen to the passage read aloud by your teacher or played on the recorder and complete the statements. You may listen to it again if required. Complete the following: (Text Book Page No. 119)

Question a.
List any three aspects which contributed to the success of the ascent of the summit.

1. ___________
2. __________
3. __________.

Answer:

1. Knowledge gained from other climbers
2. Careful planning
3. Excellence of equipment

Question b.
Without the help of ____________ nothing would have been possible.
Answer:
Teamwork / Sherpas

Question c.
The main idea of the passage is _________
Answer:
reasons for the success of the summit

Question d.
The biggest thing of all is _________.
Answer:
togetherness

Question e.
_________ were cheerful and gallant men.
Answer:
Sherpas

Speaking:

a) Group Activity: (Text Book Page No. 119)

Question i.
Have you ever been on an adventurous trip? If so, share your success story with your friends.
Answer:
I with my brother went on a trip from Tirunelveli to Chennai on a bike. When we were just nearing Chennai, the tire of the bike burst, we did not know what to do. Luckily we managed not to hurt ourselves. We were waiting for someone to help us. Finally, a car came that way. We pleaded with them to help us. They promised to send a mechanic to that place. We waited for nearly two hours. Then the mechanic came and repaired our bike. Later we reached our uncle’s home in Chennai safely.

Question ii.
How will you organize or plan for a trip or an event? Do you have the habit of preparing a check-list? Discuss.
Answer:
I organize a trip with my family to Mysore for four days to visit all the important places. I planned to go by train. So I will book the to and fro tickets to Mysore. I will book a room to stay there in advance, I will prepare a checklist a day before my travel one toothbrush, towel, soap, money, my Id card, medicines, first aid kit. A tourist guide book of Mysore. I will plan everything properly to make my trip safe and pleasant.

b) Individual Activity: (Text Book Page No. 119)

Given below are a few proverbs. Prepare a short speech of two minutes on one of the proverbs:

a) Nothing is impossible:
The word ‘impossible’ means it is possible. You can do anything. “The word impossible is found only in the dictionary of fools,” says Napoleon Bonaparte. In this word, everything is possible on the basis of will power, determination, and sacrifice. To accomplish the most difficult tasks, you need to put in a lot of hard work, extra perseverance, and concentration on a simple objective. You should have determination, dedication, and devotion to attain success. Never bother about the results keep on going even if the pace is slow. Just ensure it remains steady.

b) Where there is a will there’s away.
“Where there is a will there’s a way” is a message of hope and encouragement. Perseverance leads to success. Most of us are endowed with a fair share of intelligence, and all that is needed for success is diligence and perseverance. Some may not be blessed with good health. Some may not be blessed with a sharp intelligence or a retentive memory. Some may be dull. Yet, it is in the will power of all to attain to a certain standard of success. Thus, we may conclude with another saying “Strong Determination is the key to success”.

c) Together we can achieve more.
This means Unity is strength. If people work together, the work is easier and is completed more quickly. This proverb means that you should take advantage of a favourable situation before it changes. It is best to always be honest and tell the truth. By doing so, you will win the trust and respect of others. A country is strong whose citizens are united, a family is strong whose members stay together. Thus unity is very important in each and every sphere of our life.

Giving instructions: (Text Book Page No. 119)
Here are a few instructions are given by a Health Inspector to a group of students, in order to prevent malaria and dengue. Complete the series by adding some more important instructions. (Text Book Page No. 120)

1. Do not allow water to stagnate in and around your house.
2. Keep your surroundings clean.
3. Wear long-sleeved shirts/blouses and long pants/skirts that cover your arms and legs.
4. Do not litter the place with wastes
5. Divide the waste into perishable and non¬perishable and put them in the proper place

Now, write a set of 8 to 10 instructions for the following situations:

1. A doctor instructing a patient regarding a healthy diet and proper care after surgery.

• Eat food at regular intervals.
• Do not take fatty food
• Eat more green vegetables
• Go for a walk daily
• Do not take more sugar and salt

2. A traffic police personnel to the public, as to how to move around in safety, in crowded public places during festival seasons.

• Do not wear lots of jewels
• Take care of personal belongings.
• Do not allow children to stray away.
• Always follow the queue

3. A mother to her children, on safety measures to be taken before leaving home on vacation.

• Do not go alone
• Carry your belongings safely
• Do not eat from roadside shops
• Do not waste time unnecessarily

Reading:

On the basis of your understanding of the given passage, make notes in any appropriate format:

The Sherpas were nomadic people who first migrated from Tibet approximately 600 years ago, through the Nangpa La pass and settled in the Solukhumbu District, Nepal. These nomadic people then gradually moved westward along salt trade routes. During the 14th century, Sherpa ancestors migrated from Kham. The group of people from the Kham region, east of Tibet, was called “ShyarKhamba”.

The inhabitants of ShyarKhumbu were called Sherpa. Sherpa migrants travelled through O and Tsang, before crossing the Himalayas. According to Sherpa’s oral history, four groups migrated out of Solukhumbu at different times, giving rise to the four fundamental Sherpa clAnswer: Minyagpa, Thimmi, Sertawa, and Chawa. These four groups have since split into the more than 20 different clans that exist today.

Sherpas had little contact with the world beyond the mountains and they spoke their own language. Ang Dawa, a 76-year-old former mountaineer recalled “My first expedition was to Makalu [the world’s fifth highest mountain] with Sir Edmund Hillary’ We were not allowed to got really heavy when wet, and we only got a little salary, but we danced the Sherpa dance, and we were able to buy firewood and make campfires, and we spent a lot of the time dancing and singing and drinking. Today Sherpas get good pay and good equipment, but they don’t have good entertainment. My one regret is that I never got to the top of Everest. I got to the South Summit, but I never got a chance to go for the top.

The transformation began when the Sherpa Tenzing Norgay and the New in 1953. Edmund Hillary took efforts to
build schools and health clinics to raise the living standards of the Sherpas. Thus life in Khumbu improved due to the efforts taken by Edmund Hillary and hence he was known as ‘Sherpa King’.

Sherpas working on the Everest generally tend to perish one by one, casualties of crevasse falls, avalanches, and altitude sickness. Some have simply disappeared on the mountain, never to be seen again. Apart from the bad seasons in 1922, 1970, and 2014 they do not die en masse. Sherpas carry the heaviest loads and pay the highest prices on the world’s tallest mountain. In some ways, Sherpas have benefited from the commercialization of Everest more than any group, earning income from thousands of climbers and trekkers drawn to the mountain. While interest in climbing Everest grew gradually over the decades after the first ascent, it wasn’t until the 1990s that the economic motives of commercial guiding on Everest began.

This leads to an eclipse of the amateur impetus of traditional mountaineering. Climbers looked after each other for the love of adventure and “the brotherhood of the rope” now are tending to mountain businesses. Sherpas have taken up jobs as guides to look after clients for a salary. Commercial guiding agencies promised any reasonably fit person a shot at Everest.

Note Making:

Notes:
Life of Sherpas

I. Nomadic Sherpas migrated from Tibet to Nepal
a) 600 years ago
b) Shyarkhamba
c) 4 Sherpa clans – Minyaagpa, Thimmi, Sertawa, Chawa.

II. Little contact beyond mountains
a) have their own language
b) have no opportunity to reach the top
c) carry things for others
d) Edmund Hillary the ‘Sherpa King’

III. Sherpas die in mountain casualties:
a) carry heaviest loads
b) earn money from climbers
c) 1990s commercial guiding on Everest

Grammar:

Kinds of sentences — Simple, Complex, and Compound:

a) Simple sentence: (Text Book Page No.121)

Task 1:
Pick out the finite verbs in the following sentences: (Text Book Page No. 121)

Question a.
You can solve this problem in different ways.
Answer:
can solve

Question b.
The professor has been working on the last chapter of the book since March.
Answer:
as been working

Question c.
Despite being a celebrity, Ravi mingles easily with everyone.
Answer:
mingles

Question d.
You must speak clearly to make yourself understood.
Answer:
must speak

Question e.
The chairman being away, the clerk is unable to approve the proposal.
Answer:
is

Question f.
Getting down from the car, the Chief Guest walked towards the dais amidst applause.
Answer:
walked

Question g.
The old man struggled to walk without support.
Answer:
struggled

Question h.
In case of emergency, please contact this number.
Answer:
contact

Question i.
The sun having set, the temperature fell rapidly.
Answer:
fell

Question j.
But for your help, I could not have completed the assignment.
Answer:
completed

Task 2:

II. Read the following passage and identify the simple sentences: (Text Book Page No. 121)

Question  1.
Sunflowers turn according to the position of the sun. In other words, they ‘chase the light’. Have you ever wondered what happens on cloudy, rainy days when the sun is completely covered by clouds? If you think the sunflower withers or turns its head towards the ground, you are completely mistaken. Do you know what happens? Sunflowers turn to each other to share their energy. Learning from Nature, we too should support and empower each other.
Answer:
Sunflowers turn according to the position of the sun. In other words, they ‘chase the light’. Have you ever wondered what happens on cloudy, rainy days when the sun is completely covered by clouds? If you think the sunflower withers or turns its head towards the ground, you are completely mistaken. Do you know what happens? Sunflowers turn to each other to share their energy. Learning from Nature, we too should support and empower each other.

b) Complex sentence: (Text Book Page No.122)

Task 1:

Look at the following complex sentences. Circle the Main clauses and underline the Subordinate clauses: (Text Book Page No. I23) 

Question a.
Nobody knows when the power supply will resume.
Answer:
Nobody knows when the power supply will resume.
M.C                                              S.C

Question b.
Please tell me what the time is.
Answer:
Please tell me             what the time is.
M.C                                   S.C

Question c.
The man who directed the film was my schoolmate.
Answer:
The man             who directed the film          was my schoolmate.
M.C                             S.C                                          M.C

Question d.
I believe that all men are basically good.
Answer:
I believe                  that all men are basically good.
M.C                                           S.C

Question e.
No one knows when he will return.
Answer:
No one knows               when he will return.
M.C                                           S.C

Task 2:

Pick out the complex sentences in the following passage: (Text Book Page No. 123)

Question  1.
A man saw a lion in the bush, as he was walking through the forest. He did not know what to do. He was helpless. He was too scared to turn around and run. He just knelt down as if he were getting ready to pray. He closed his eyes, thinking that the lion would pounce on him anytime. Out of the corner of his eye, he saw the lion on its knees too. Shocked, he asked the lion what it was doing. The lion replied that he was praying before he started his meal.
Answer:
A man saw a lion in the bush, as he was walking through the forest. He did not know what to do. He was helpless. He was too scared to turn around and run. He just knelt down as if he were getting ready to pray. He closed his eyes, thinking that the lion would pounce on him anytime. Out of the corner of his eye, he saw the lion on its knees too. Shocked, he asked the lion what it was doing. The lion replied that he was praying before he started his meal.

c) Compound sentence: (Text Book Page No. 123)
Identify the two Main clauses and conjunction in each of the following sentences:

Question a.
It started raining suddenly and people ran for shelter.
Answer:
It started raining suddenly            and                              people ran for shelter.
M.C                                            conjunction                                    M.C

Question b.
Understand the concept well, otherwise, you cannot solve the problem.
Answer:
Understand the concept well,              otherwise,                          you cannot solve the problem.
M.C                                                      conjunction                                            M.C

Question c.
Fifty candidates appeared for the interview, but only five were selected.
Answer:
Fifty cT’rlidates appeared for the interview,                 but                   only five were selected.
M.C                                                                         conjunction                          M.C

Question d.
Ramesh did not know Spanish, so he wanted a translator.
Answer:
Ramesh did not know Spanish,                                so                                        he wanted a translator.
M.C                                                                     conjunction                                           M.C

Question e.
He is a good actor, still, he is not popular.
Answer:
He is a good actor,                   Still,                         he is not popular.
M.C                                       conjunction                          M.C

Task 2:

Pick out the compound sentences in the following passage: (Text Book Page No. 123) 

Question  1.
The food we eat has to be digested and then thrown out of the body. The air we breathe in, has to be thrown out, to help us survive. But we hold negative emotions like insecurity, anger, and jealousy within ourselves for years. If these negative emotions are not eliminated, the mind grows corrupt and diseased. Let us do away with hatred and lead a healthy life filled with peace and joy.
Answer:
The food we eat has to be digested and then thrown out of the body. The air we breathe in, has to be thrown out, to help us survive. But we hold negative emotions like insecurity, anger, and jealousy within ourselves for years. If these negative emotions are not eliminated, the mind grows corrupt and diseased. Let us do away with hatred and lead a healthy life filled with peace and joy.

Task 3:

Complete the sentences choosing the right endings: (Text Book Page No. 124)


Answer:

Conditional clause:

Task 1:

Read the following sentences and fill in the blanks: (Text Book Page No. 124)

Question a.
If I _______ (be) a spider, I would ______ (weave) webs.
Answer:
were, weave

Question b.
If Raj ______ (be) a sculptor, he would ______ (make) beautiful idols.
Answer:
were, make

Question c.
If Mary had an umbrella, she would ______ (lend) it to me.
Answer:
lend

Question d.
Rex would have played with me if he ______ (has) time.
Answer:
had

Question e.
If I were you, I would ______ (accept) this offer.
Answer:
accept

Question f.
We will ______ (select) storybooks for kids if we allow time for storytelling.
Answer:
select

Question g.
The Education Minister will ______ (visit) our school tomorrow if he goes by this way.
Answer:
visit

Question h.
You will be rewarded by the wise if you ______ (stand) for truth.
Answer:
stand

Question i.
If my mother ______ (know) of my poor performance in the exam, she will not allow me to watch a movie.
Answer:
knows

Question j.
If I had won the lottery, I would have ______ (donate) relief materials for the flood victims.
Answer:
donated

Task 2:

Rewrite the following sentences using ‘If’ without changing the meaning: (Text Book Page No. 125)

e.g. Unless you go for a walk regularly, you cannot reduce your weight. (Use ‘If’)
If you do not go for a walk regularly, you cannot reduce your weight.

Question a.
Sindhu would not have won the world championship unless she had single-minded devotion.
Answer:
If Sindhu had not had single-minded devotion, she would not have won the world championship.

Question b.
You will not reach your goal unless you chase your dream.
Answer:
If you do not chase your dream, you will not reach your goal.

Question c.
Unless we plant more trees, we cannot save our planet.
Answer:
If we plant more trees, we can save our planet.

Question d.
The rescue team would not have saved the victims unless they had received the call in time.
Answer:
If the rescue team had received the call in time, they would have saved the victims.

Question e.
The palace cannot be kept clean unless we appoint more people.
Answer:
If we do not appoint more people, the palace cannot be kept clean.

Question f.
The portraits would not have been so natural unless the artist had given his best.
Answer:
If the artist had not given his best, the portraits would not have been so natural.

Question g.
The manager would not have selected Nithiksha unless she exhibited good accounting skills.
Answer:
If she had not exhibited good accounting skills, the manager would not have selected Nithiksha.

Question h.
The policeman would not have arrested the man unless he had violated the rules.
Answer:
If he had not violated the rules, the policeman would not have arrested the man.

Question i.
Mr. Kunaal would not sponsor my higher education unless I studied well.
Answer:
If I studied well, Mr. Kunaal would sponsor my higher education.

Question j.
Kavin will not stop flying kites unless he understands the risk involved in it.
Answer:
If kavin understands the risk involved in it, he will stop flying kites.

Question k.
Tanya would not know the answer unless she referred to the answer key.
Answer:
If she referred to the answer key, Tanya would know the answer.

Question l.
My village cannot achieve a 100 % literacy rate unless the elders of the village cooperate with the education department.
Answer:
If the elders of the village do not cooperate with the education department, my village cannot achieve 100% literacy.

Writing:

Summarizing is to briefly sum up the various points from the notes made from the original passage.

Refer to the reading passage (Text Book Page No. 120). You must have completed reading. Now go through the passage once again and refer to the notes made and do the summarizing.

Rough Copy:
The Sherpas were nomadic people who migrated from Tibet about 600 years ago. They moved to the west. The ancestors from the Kham region were called Shayar khamba. The inhabitants were called Sherpas. There were fundamental clans, sherpas lived beyond the mountains. An old Sherpa Ang Dawa accompanied Edwin Hillary – He was called the Sherpa king, because he took efforts to build schools and health clinics to raise their living standards, Sherpas working on the Everest got perished due to Crevasse falls, Avalanches, and altitude sickness. They can carry heavy loads. They have benefited from the commercialization. Sherpas have taken up to look after clients for a salary. Sherpas’ life is entangled with the mountains.

Fair Copy:

The Sherpas

The Sherpas were nomadic people who migrated from Tibet about 600 years ago. They moved to the west. The ancestors from the Kham region were called Shayar Khamba. The inhabitants were called Sherpas. There were fundamental clans. Sherpas lived beyond the mountains. An old Sherpa Ang Dawa accompanied Edwin Hillary – He was called the Sherpa king because he took efforts to build schools and health clinics to raise their living standards. Sherpas working on the Everest got perished due to Crevasse falls, Avalanches, and altitude sickness. They can cany heavy loads. They have benefited by the commercialization. Sherpas have taken up jobs as guides to look after clients for a salary. Sherpas’ life is entangled with the mountains.

Interpreting non-verbal presentation:

Reading a map:

Let us together scale the summit. Here is a drawing of Everest showing the way to the summit, and the position of the camps with their heights. Trace the trekking trail to reach the summit with the given detail and write an interesting paragraph in about 100 words. (Text Book Page No. 126)

Paragraph about trekking trail to reach the summit:

We started our journey at the base camp. We have sufficient oxygen with us. At the 3rd camp, we suffered a lot due to ice fail. We didn’t come out till the weather was clear. Only three of our team could reach the 8th camp. From here the ridge narrowed to a knife-edged and it was a steep slope. So we had to cut steps and then moved. Since there was a lot of firm snow, we could do it. With great efforts. We reached the 9th camp. One of my friends became sick, I started my challenging journey with my other friend. Having fierce determination, nothing could stop us from reaching the top. When we realized that the ridge ahead of us dropped sharply away, we stood on the summit. It was a splendid joy we shared with each other and shouted, at the height of 28,700 feet.

ஆசிரியரைப் பற்றி:

சர் எட்மெண்ட் பெர்சிவல் ஹிலாரி (Sir Edmund Percival Hillary) (20 ஜீலை 1919 – 11 ஜனவரி 2008) நியூசிலாந்து நாட்டைச் சேர்ந்த மலை ஏறுபவர் (mountaineer), கண்டுபிடிப்பாளர் (explorer) மற்றும் தாராளமான நன்கொடையாளர் (philanthropist). இவர் இரண்டாம் உலகப்போரின் போது “நியூசிலாந்து ராயல் விமானப் படையில்” மாலுமியாக (navigator) பணியாற்றினார்.

எவரெஸ்ட் மலையில் ஏறியதற்கு பின்பு, அவர் தான் உருவாக்கிய “ஹிமாலயன் டிரஸ்ட்” மூலம் நேபாள நாட்டில் உள்ள “ஷெர்பா” (Sherpa) மக்களுக்கு சேவை செய்வதில் தன்னை ஈடுபடுத்திக் கொண்டார். “High Adventure, No Latitude for Error, Nothing Venture, Nothing Win. View from the Summit, The Remarkable Memoir by the First Person to Conquer Everest” ஆகியவை அவருடைய பிரபலமான (famous) புத்தகங்களாகும்.

பாடத்தைப் பற்றி:

இப்பாடத்தில் எவரெஸ்ட் சிகரத்தை தொட்ட உலக புகழ் பெற்ற மலைஏறும் வீரர் எட்மண்ட் ஹிலாரி தன் அனுபவத்தை கொடுத்திருக்கிறார். குறிப்பாக மே மாதம் 29ம் தேதி மலையேறும் இறுதி நாளில் நடந்த மெய் சிலிர்க்கும் நிகழ்வை பற்றியும் காலை 4.30 மணி முதல் பிற்பகல் 2 மணி வரை டென்சிங்குடன் சேர்ந்து அவர் எதிர்கொண்ட அபாயங்கள், புத்திசாலித்தனங்கள் பற்றியும் விரிவாக சொல்லப்பட்டு இருக்கிறது. அவர் புரிந்த இந்த அபாயகரமான நிகழ்வை, பாடத்தைப் படித்து விரிவாகக் காண்போம். சாகசங்கள், பயணங்கள் மற்றும் கண்டுபிடிப்புகள் எப்போதும் உற்சாகம் தரக்கூடியவை.

குறிப்பாக அவை எப்பொழுதும் மெய்யானதாகவும் மற்றும் முதன்முதலில் நடப்பதாகவும் இருந்தால் மெய்சிலிப்பேற்றக்கூடியவை. ஒருவர் மனதில் எழும் ஒரே கேள்வி என்னவென்றால், இவ்வாறு அதிகப்படியான ஆபத்துக்கள் அடங்கிய காரியங்களை ஒருவரை செய்யத் தூண்டுவது எது என்பதே ஆகும். இது வலிமைமிக்க துணிகரச் செயலை புரியக்கூடிய மனநிலை மற்றும் தனிச்சிறப்புமிக்க பண்புகளே அவர்களை அத்தகைய உயரங்களை தொடச் செய்கிறது.

The Summit Summary in Tamil

“எவரெஸ்ட்டிற்கு ஏறும் வழி” (The Ascent of Everest) என்னும் ஜான் ஹன்ட் (John Hunt) எழுதிய நூலை சற்று தழுவி இப்பாடப்பகுதி எடுக்கப்பட்டது. எவரெஸ்ட் சிகரம் எவ்வாறு அடையப்பட்டது என்பது சர் எட்மெண்ட் ஹிலாரியின் (Sir Edmund Hillary) சொந்த வார்த்தைகளால் சொல்லப்படுகிறது.

மே 28ல் தெற்கு கணவாயில் போடப்பட்ட முகாமில் 8 பேர் இருந்தனர், முறையே எட்மெண்ட் ஹிலாரி (Edmund Hillary), டென்சிங் (Tenzing), ஜார்ஜ் லோ (George Lowe), அல்பிரட் கிரிகோரி (Albert Gregory) மற்றும் இரண்டு செர்பாக்கள் (Sherpas), பெம்பா (Pemba), அங்க் நியிமா (Ang Nyima). ஆனால் பெம்பா ஏற முடியாத அளவிற்கு உடல்நிலை சரியில்லாமல் இருந்தார். மற்றவர்கள் அதிக சுமையுடன் அன்று 27900 அடி ஏறினர். அங்கே ஹிலாரியும், டென்சிங்கும் சிறிய கூடாரத்தில் இருந்து அவர்களுடன் வந்த மூன்று நபர்கள் முகடுகளில் சறுக்கி தெற்கு கணவாய்க்கு திரும்புவதை கண்டார்கள்.

சூரியன் மறையும் போது, ஹிலாரி மற்றும் டென்சிங் தங்கள் கூடாரத்திற்குள் சென்று, வெப்பம் தரும் (warm clothing) ஆடைகளை அணிந்து, தூங்கும் பைகளுக்குள் நெளிந்து சென்றார்கள். மறுநாள், மே 29ல் காலை 4 மணிக்கு ஏறுவதற்கு அவர்கள் தயாரானார்கள் (ready).

நாங்கள் எங்களது சூட்டடுப்பை (cooker) பற்ற வைத்து, அதிக அளவு எலுமிச்சை சாறையும் (lemon juice), சர்க்கரையும் (sugar) கலந்து குடித்தோம், அதன்பின் எங்களது கடைசி தகரக்குவளையில் உள்ள சார்டின் (sardines) எனப்படும் கடல் உணவையும், பிஸ்கெட்டையும் எடுத்துக் கொண்டோம். நான் எனது ஆக்சிஜன் தொகுப்பு பொருட்களை கூடாரத்தின் உள்ளே இழுத்து, அதன் மேலிருந்த பனிக்கட்டிகளை சுத்தம் செய்து அதனை பரிசோதனை செய்து சோதித்து பார்த்தேன்.

முந்தைய நாள் ஈரமாயிருந்து. இன்று உறைந்து போன (frozen) எனது காலணிகளை கழற்றினேன். எனவே அவைகளை நான் அடுப்பில் இருந்து வரும் புகையினால் சூடேற்றி மென்மையாக்கினேன். எங்களது கீழ் ஆடைக்கு மேலே நாங்கள் காற்று சீராக்கும் ஆடையை அணிந்தும், எங்கள் கைகளில் பட்டு, கம்பளி மற்றும் காற்று சீராக்கும், கையுறைகளையும் (gloves) அணிந்துக் கொண்டோம்.

காலை 6.30 மணியளவில் நாங்கள் எங்கள் கூடராத்தை விட்டு பனிக்குள் (snow) நகர்ந்து வந்து, 30 அளவுடைய ஆக்சிஜன் (oxygen) உருளையை (gear) முதுகில் சுமந்து, முகமூடியை அணிந்து, ஆக்சிஜன் உருளையில் குழாயைத் திறந்து எங்களது நுரையீரலுக்குள் ஆக்சிஜன் வாழ்வை அனுப்பினோம். ஒரு சில சிறந்த ஆழமான சுவாசத்திற்குப் பிறகு நாங்கள் புறப்பட தயாரானோம். எனது உறைந்து போன பாதங்களைப் பற்றி நான் சிறிது கவலைப்பட்டாலும், நான் டென்சிங்கை நகரச் சொன்னேன்.

டென்சிங் நீண்ட மலைமுகட்டை நோக்கி (traverse) செல்லக் கூடிய குறுக்கு பாதையில் நடந்தார். நாங்கள் 28000 அடி பனிக்குவியலை உண்டாக்கிய மலைமுகட்டின் உச்சியை அடைந்தோம். அதன் பிறகு அந்த முகடானது கத்தி போல் கூர்மையான பாதையாக காணப்பட்டது. என் காலகள் இப்போது வெப்பமானதால், நான் டென்சிங்கை முந்தினேன்.

கடினமானதாகவும், ஆபத்தாகவும் இருக்கக்கூடிய மலை முகட்டிற்கு செல்லக்கூடிய பாதையானது மென்பனியால் உருவாகயிருந்தது. அது சில நேரங்களில் என் நிலையிலேயே இருக்க செய்தாலும், அடிக்கடி மேலே செல்ல வழி விட்டது. பல நூறு அடிகளுக்கு அப்புறம் நாங்கள் சிறிய காலியான இடத்திற்கு வந்தோம். அங்கே எவான் மற்றும் போர்டில்லான் ஆகியோரின் முந்தைய முயற்சியில் விட்டுச் சென்ற இரண்டு ஆக்சிஜன் குடுவைகளை கண்டோம்.

மதிப்பு அளவீட்டிலிருந்த பனியை நான் விலக்கி, நாங்கள் அதனை சிக்கனமாக உபயோகித்தால் திரும்ப தெற்கு கணவாய் வரை செல்லத் தேவையான பல நூறு லிட்டர் ஆக்சிஜனை அவை இன்னும் கொண்டுள்ளன என்பதை கண்டறிந்து நிம்மதியானேன்.

நான் தெற்கு உச்சியின் கடைசி 400 அடிக்கு முன்னால் உள்ள மலைமுகட்டிற்கு (ridge) செல்லக்கூடிய பாதையில் தொடர்ந்து பயணமானோம்.

முன்னேறி செல்வதில் நாங்கள் அடிக்கடி (frequent) மாற்றங்களை கொண்டு வந்தோம். நான் எனது அடியை ஆழமான பாதையின் மீதிருந்த ஆழமான பனியில் வைத்த போது, ஒரு பகுதி விலகி சென்றதனால், நான் மூன்று நான்கு அடிகள் பின்னோக்கி வழுக்கினேன். நாங்கள் முன்னேறி செல்வது குறித்து டென்சிங்கிடம் நான் அறிவுரைகளை (advisability) கேட்ட போது, அவர் பனியைக் குறித்து மகிழ்ச்சியற்றுக் காணப்பட்டாலும் அவருடைய வழக்கமான கூற்றான “நீங்கள் விரும்பிய படியே” என்று உரையை முடித்தார்.

நான் முன்னேறிச் செல்வதற்கு முடிவெடுத்தேன் (decided). கடைசியாக நாங்கள் மேலே இருந்த உறுதியான பனியை அடைந்த போது நாங்கள் கடைசியா இருந்த செங்குத்தான சரிவினில் பனியை உடைக்கும் கம்பி கொண்டு பனியை வெட்டி படிகளை உருவாக்கினோம். அப்பொழுது நேரம் காலை 9 மணி.

தெற்கு சிகரத்திற்கு (south summit) கீழே நாங்கள் அமர்வதற்கு ஒரு இருக்கையை வெட்டினோம், பின்பு எங்களது பாதி நிரப்பப்பட்ட முதல் ஆக்சிஜன் குடுவை தீர்ந்திருந்த போது எங்களிடம் ஒரே ஒரு ஆக்சிஜன் நிரம்பிய குடுவை மட்டுமே மீதம் இருந்தது. இப்போது எங்களது உபகரணங்கள் சுமப்பதற்கு இலகுவாகவும், சுமார் 20 க்கு சற்று அதிகமாகவும் தான் எடை கொண்டிருந்தன.

என்னுடைய பனிக் கோடாரி (ice-axe) அந்த முகட்டின் முதல் செங்குத்தான சாய்வினை வெட்டிய போது, என்னுடைய உயர்ந்த நம்பிக்கைகளை (high hopes) நான் உணர்ந்தேன். பனியாகவும் படிகமாகவும் (crystalline), உறுதியானதாகவும் இருந்தது. பனிக்கோடாரியின் இரண்டு அல்லது மூன்று அடிகள் எங்களது பெரிய படிகட்டுகளை (large steps) உண்டு பண்ணியது. பனிக்கோடாரியின் ஒரு உறுதியான அடியானது கைப்பிடியை பாதி வரை மூழ்கச் செய்து, உறுதியான வசதியான மலை ஏற்றத்துக்கு வழி வகுக்கும்.

நாங்கள் ஒரு நேரத்தில் ஒருவர் மட்டுமே நகர்ந்தோம். என்னால் 40 படிகட்டுகளை வெட்ட முடியும், நான் அவ்வாறு வெட்டும் போது என் கயிரை டென்சிங் பிடித்துக் கொண்டு என் எடையைத் தாங்குவார் (belaying). அதன்பின் நான் என் கைப்பிடியை மூழ்கச் செய்து, அதனைச் சுற்றி கயிரால் கண்ணிகள் (loops) போட்டேன். டென்சிங்கோ , உடையும் படிக்கட்டுகளுக்கு எதிராக என்னை நோக்கி நகர்ந்து வந்தார். பின்னர் மறுபடியும் அவர் என் கயிரை பிடித்துக் கொண்டு என் எடையத்ை தாங்கும் போது, நான் வெட்டினேன்.

அதிக இடங்களில் தொங்குகின்ற பனிப்பாறைகள் மிகப் பெரியதாக இருந்ததால், அவற்றிலிருந்து தப்பிக்க நான் பனியின் மேற்பரப்பில் இருந்து பாறை வரையிலும் படிக்கட்டுகளை வெட்டினேன். இந்தப் பெரிய பாறையிலிருந்து நேராக 8000 அடிக்கு கீழாக மேற்கு பள்ளத்தாக்கில் உள்ள முகாம் 4ன் சிறிய கூடாரங்களைப் (tiny tents) பார்த்து சிலிர்ப்பேற்றியது. பாறைகளில் மேலே தவழ்ந்ததாலும் (crawl) பனியில் கைப்பிடிகளை வெட்டியதாலும் எங்களால் இந்த கடினமான பகுதிகளை வேகமாக கடந்து செல்ல முடிந்தது.

அதன் கிழக்குப் பகுதியில் (east side) இன்னொரு பெரிய பனிப்பாதை இருந்தது. அது 40-அடி உயர படியில் உருண்டோடி வந்தது, அங்கே பனிப்பாறைக்கும் பாறைக்கும் இடையே ஒரு விரிசல் (crack) இருந்தது. டென்சிங்கை எவ்வளவு சிறப்பாக என் எடையைத் தாங்க முடியுமோ, அவ்வாறு செய்து விட்டு, நான் அந்த விரிசலுக்குள் ஏறினேன். பின்னர், பின்னோக்கி உதைத்தவாறு என் மலையேறும் கருவியில் இருந்த ஆணிகளை உறைந்த பனியில் முழுவதுமாக மூழ்கச் செய்து நான் தரையில் இருந்து நெம்பியவாறு மேலே எழுந்தேன்.

அங்கிருந்து அனைத்து சிறிய பாறைப்பிடிகளின் துணையோடு, என் மூட்டு, தோள்பட்டை, கைகளால் முடிந்த அளவு பலத்தை திரட்டிக்கொண்டு, பனிப்பாறையுடன் ஒட்டிக்கொள்ள வேண்டும் என்று ஜெபித்துக் (praying) கொண்டே, நான் விரிசலில் மலையேறும் கருவியின் மூலம் பின்புறமாக நகர்ந்தேன்.

எனது முன்னேற்றம் (progress) மெதுவாக இருந்தாலும், நிலையானதாக இருந்தது. டென்சிங் கயிரை விட்டுவிட்டதால், நான் மேல் நோக்கி மெதுமெதுவாக பாறையின் உச்சியைத் தொடும் வரை நகர்ந்து, விரிசலிலிருந்து வெளிவந்து நான் தெற்கு சிகரத்திற்கு படிகளை வெட்டிக் கொண்டிருந்த போது, நான் சுதந்திரமாகவும், நன்றாக இருப்பது போன்றும் உணர்ந்தேன்.

ஒரு சில நிமிடங்களுக்கு நான் மூச்சுவிடுவதற்காக படுத்திருந்தேன். முதன் முறையாக நாங்கள் மலைச் சிகரத்தை அடைவதை எதுவும் தடுக்க முடியாது என்று மிக உறுதியாக உணர்ந்தோம், நான் அந்த தொங்கும் பாறையில் நிலையாக நின்று கொண்டு டென்சிங்கை நோக்கி மேலே வருமாறு கையசைத்தேன் (signed).

நான் அந்த கயிறில் நிம்மதி பெருமூச்சு விட்டபோது, டென்சிங்கோ அந்த விரிசலை நோக்கி மேலே நெளிந்து (wriggled) வந்து கடைசியாக உச்சத்தை அடைந்த போது, மிகப்பெரிய போராட்டத்திற்கு பின், கடலில் இருந்து வெளியே இழுத்து வரப்பட்ட பெரிய இராட்சச மீன் போல் நிலைகுலைந்தார்.

மலை முகடுகள் முன்பு போலவே தொடர்ந்தன, பெரிய பனிப்பாறையின் (giant cornices) வலது புறமும், கூர்மையான மலைச் சரிவுகள் இடது புறமும் இருந்தன. அந்த மலைமுகடானது வலது புறமாக வளைந்து (curved) சென்றதால் எங்களுக்கு மலை உச்சி எங்கு இருக்கிறது என்று தெரியாமல் போனது. நான் (கூனல்) திமிலின் பின்புறத்தை வெட்டும் போது அடுத்த திமில் கண்ணில் தென்படும். நேரம் செலவழிந்து கொண்டே இருந்தது, மலைமுகடுகளும் முடிவது போலத் தோன்றவில்லை.

எங்களது உண்மையான பேரார்வம் சிறிது விலகியது. இது ஒரு கடுமையான போராட்டமாகவே மாறியது. எங்களுக்கு முன்னே இருந்த பனிப்பாறை (snowy summit) எழுவதற்குப் பதிலாக, கிழே விழ ஆரம்பித்ததை நான் அதன்பின் உணர்ந்தேன். ஒரு குறுகலான பனிமுகடு (ice-axe) ஒரு மலை உச்சியை நோக்கி ஓடுவதைப் பார்க்க நான் மேல் நோக்கி பார்த்தேன்.

இதிலிருந்து விடுபட வேண்டும் (relief) என்பதே என் முதல் உணர்வாக இருந்தது. மேலும் இனிமேல் வெட்டுவதற்கு படிகட்டுகளோ, கடப்பதற்கு மலை முகடுகளோ, வென்றுவிடலாம் என்ற நம்பிக்கைக்கு ஏமாற்றம் அளிக்கும் மலை அங்கே காணப்படவில்லை. நான் டென்சிங்கை நோக்கினேன்.

மறைக்கக்கூடிய தலைகவசத்தையும் (balaclavahelmet), பனிக் கண்ணாடியையும், ஆக்சிஜன் வாயு முகமூடியையும் அணிந்திருந்தால் அனைத்தும் ஈட்டி போன்ற கூர்மையான வடிவுடைய பனிக்கட்டியும் சேர்த்து அவருடைய முகத்தை மறைத்தாலும், அவரைச் சுற்றிலும் அவர் பார்த்து விட்டு சந்தோஷத்தினால் சிரித்ததை எதையும் மறைக்கவில்லை.

நாங்கள் இருவரும் கை குலுக்கிக் கொண்டோம். டென்சிங் அவரது கையை என் தோள்மேல் போட்டுக் கொண்டார், கிட்டத்தட்ட மூச்சு முட்டும் வரை நாங்கள் இருவரும் முதுகில் தட்டிக் கொண்டோம். அப்போது நேரம் காலை 11,30 மணியாய் இருந்தது. இந்த மலை முகடு 2 1/2 மணி நேரத்தை எடுத்துக் கொண்டது, அது எங்களுக்கு ஒரு ஆயுட்காலமாகவே தோன்றியது.

எங்களுக்கு கிழக்கில் இதுவரை கண்டறியப்படாத ஏற்படாத பெரிய இராட்சச (giant) அண்டையரான மக்காலு (makalu) இருந்தது. மேகங்களுக்கிடையில் மிகத் தூரத்தில் கஞ்சன்சங்காவின் (Kanchenjunga) மிகப்பெரும் பகுதி அடிவானத்தில் நிழல் போல அச்சுறுத்தும் விதமாக காணப்பட்டது. மேற்கில் இதுவரை கண்டறியப்படாத நேபாளத்தின் மலைப்பகுதிகள் தூரத்திற்கு நீண்டு கிடந்ததை எங்களால் பார்க்க முடிந்தது.

வடக்கு மலைமுகட்டுக்கு கிழே உள்ள வடக்கு கணவாயையும் (North Col), 1920 மற்றும் 1930 களில் மலை ஏறுபவர்களால் போராடி பிரபலமாக்கப்பட்ட பழைய பாதையையும் (old route) காட்டும் மிக முக்கியமான புகைப்படமாக (photograph) நான் கருதும் புகைப்படம் அப்போது எடுக்கப்பட்டது. பத்து மணித்துளிகளுக்கும் பின், நான் மெதுவாக நகர்பவனாகவும் என் விரல்கள் வளைவுத் தன்மையற்றும் மாறின அதனால் நான் உடனே என் ஆக்சிஜன் உபகரணத்தை மாற்றினேன்.

இந்த வேளையில் (meanwhile), டென்சிங் பனியின் நடுவே ஒரு சிறிய துளையிட்டு (hole), அதனுள் – பலதரப்பட்ட சிறிய உணவுப் பொருட்களான (articles), சாக்லேட், பிஸ்கட், குச்சிமிட்டாய் (lollies) போன்றவற்றை வைத்தார். உண்மையில் இவை சிறிய படைப்புகள் தான் ஆனாலும் இந்த உயரமான மலைச்சிகரத்தை இருப்பிடமாக கொண்டுள்ளதாக நம்பும் அனைத்து பக்திமயமான புத்தமதத்தவர்களின் கடவுள்களுக்கு (devout Buddhists) ஒரு சிறு அடையாளப் பரிசாக இருக்கும்.

2 நாட்கள் முன்னதாக நாங்கள் தெற்கு கணவாயில் ஒன்றாக இருந்த போது, கர்னல் ஹன்ட் (Colonel Hunt) என்னிடம் ஒரு சிறிய சிலுவையை (crucifix) கொடுத்து, அதனை மலை உச்சிக்கு எடுத்துச் செல்லுமாறு கேட்டுக் கொண்டார். நானும் பனியில் சிறிய துளையிட்டு, டென்சிங்கின் பரிசுப்பொருட்களுக்கு அருகில் அந்த சிலுவையை வைத்தேன்.

15 நிமிடங்களுக்கு பிறகு, நான் உச்சியிலிருந்து எங்களது படிக்கட்டுகளை நோக்கி கீழே இறங்கினேன். ஆக்ஸிஜன் ‘ குறைந்து கொண்டிருப்பதன் அவசரத்தால் தூண்டப்பட்ட நாங்கள் நேரத்தை வீணாக்காமல் எங்கள் பாதையில் முன்னேறி சென்றோம் (cramponed). நாங்கள் மிக கவனமாக பாறையின் குறுக்கே தவழ்ந்து கடந்து, வழுக்கக்கூடிய பனிப்பகுதியில் ஒருவர் பின் ஒருவராக நகர்ந்து, கடைசியாக எங்களின் படிக்கட்டுகளில் மலையேறினோம் மேலும் திரும்பவும் தெற்கு வந்தோம்.

அப்போது நாங்கள் மிகவும் களைப்பாக இருந்தோம், ஆனாலும், நாங்கள் முகட்டில் சேமித்து வைத்திருந்த உருளைகளை (cylinders) நோக்கி கீழே நகர்ந்தோம். நாங்கள் எங்கள் முகாமை விட்டு மிக குறுகிய தூரத்தில் இருந்தாலும், எங்களது சொந்த குடுவைகளில் மிகச் சிறிய அளவே ஆக்சிஜன் மீதமிருந்ததாலும், நாங்கள் கூடுதல் உருளைகளை சுமந்து எதிர்பார்க்க முடியாத தரைத்தளத்தை (crazy platform) உடைய எங்களது கூடாரத்தை மதியம் 2 மணி அளவில் அடைந்தோம்.

எங்களுக்கு இதுவரை எல்லாவித தேவைகளுக்கும் நன்றாக உதவிய முகாமை கடைசியாக ஒருமுறை பார்த்த படி, நாங்கள் கீழ்நோக்கி திரும்பி, எங்கள் கால்களை இழுத்த படியே, மலை முகட்டிலிருந்து தெற்கு கணவாய்க்கு பாதுகாப்பாக கீழறங்கும் வேலையில் எங்களை ஈடுபடுத்திக் கொண்டோம்.

கனவைப் (dream) போலவே நேரம் கடந்தது முகாமிற்கு 200 அடிக்கு மேலே எங்களை நோக்கி வந்த இருவர் எங்களை சந்தித்தன. அவர்கள் சூடான சூப்பையும், நெருக்கடி கால ஆக்சிஜனையும் சுமந்து கொண்டிருந்த ஜரர்ஜ் லோ மற்றும் வில்பிரட் நாய்ஸ் ஆவார்கள். உயரத்திற்கு மிக அருகில் வந்த போது எனது ஆக்ஸிஜன் தீர்ந்தது. எங்களது வேலையை முடிப்பதற்கு போதுமான அளவு ஆக்ஸிஜனை நாங்கள் வைத்திருந்தோம், ஆனால் கொஞ்சம் கூட தேவைக்கு அதிகமாக இல்லை.

நாங்கள் எங்கள் கூடாரத்திற்குள் தவழ்ந்து வந்து, மகிழ்ச்சிப் பெருமூச்சு விட்டு (delight), எங்கள் தூங்கும் பைகளின் (sleeping bags) மேல் நிலை குலைந்து விழுந்த போது எங்களின் கூடாரங்கள் முடிவில்லாத தெற்கு கணவாயில் காற்றினால் சிறகடித்து (flapped) விழுந்து எங்களை பொட்டலம் போல் பொதிந்தது.

எங்களது பிரயாணத்தை வழிநடத்திய ஜான் ஹண்ட் என்பார் இவ்வாறு கூறுகிறார், “இது ஒரு மறக்க முடியாத நாளாக இருந்தது”. அவர்கள் உச்சியை ஏறி அடைந்தார்கள். அங்கே சந்தோஷ கூச்சலும், கைகலக்கல்களும், அரவணைப்புகள் இந்த இரண்டு கதாநாயகர்களுக்காக காணப்பட்டன. அவர்களது சந்தோஷமும் பெருமிதமும் டென்சிங் மற்றும் ஹிலாரியால் மிகச் கச்சிதமாக முடிக்கப்பட்ட இந்த சாலையை அவர்கள் எவ்வாறு பகிர்ந்து கொண்டார்கள், என்பதை காண்பித்தது. சாகசமானது (adventure) நிறைவுக்கு (concluded) வந்தது.

ஆபத்து மற்றும் கடினமான சூழ்நிலைகளில் உருவாக்கப்பட்ட குழுப்பணியும் (teamwork), தோழமை உணர்வும், ஒன்றையொன்று சந்தித்து, ஒன்றாக வென்றது தான் எவரெஸ்ட் சிகரத்தை ஏறிய கதையாகும், அவர்கள் இந்த பெரிய மலையின் உச்சியில் ஏறிய பிறகும், மற்றவர்களுக்கு அவர்களுடைய சொந்த “எவரெஸ்ட்களை” கண்டறியும் வண்ணம்.

சாகசத்திற்கான நிறைய சந்தர்ப்பங்கள் இன்னனமும் உள்ளன. அவற்றுள் சில, கைக்கு மிக அருகாமையிலும், மற்றவை வேறு நிலங்களில் வெகு தூரத்திலும் உள்ளன. அனைத்து சாகசங்களும் உற்சாகமளிப்பதில்லை. அதுபோல மலையில், மட்டுமே சாகசங்கள் புரியப்படுவதில்லை. நம்முடைய அன்றாட வாழ்க்கையிலும், கடவுளின் உன்னத சக்தியால் மேலே ஏறுவதற்கு எவெரெஸ்ட்கள் உள்ளன.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1

Question 1.
Expand
(i) \(\left(2 x^{2}-\frac{3}{x}\right)^{3}\)
(ii) \(\left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4}\)
Answer:
(i) \(\left(2 x^{2}-\frac{3}{x}\right)^{3}\)

(ii) \(\left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4}\)

Question 2.
Compute
(i) 102⁴
(ii) 99⁴
(iii) 9⁷
Answer:
(i) 102⁴
102⁴ = (100 + 2)⁴

= 100000000 + 8 × 1000000 + 24 × 10000 + 3200 + 16
= 100000000 + 8000000 + 240000 + 3216
= 108243216

(ii) 99⁴
99⁴ = (100 – 1)⁴

= 100000000 – 4 × 1000000 + 6 × 10000 – 400 + 1
= 100000000 – 4000000 + 60000 -400+1
= 96059601

(iii) 9⁷
9⁷ = (10 – 1)⁷


= 10³ (10⁴ – 7 × 10³ + 21 × 10² – 35 × 10 + 35) – 21 × 100 + 70 – 1
= 10³ (10000 – 7000 + 2100 – 350 + 35 ) – 2100 + 70 – 1
= 10³ (12135 – 7350) – 2031
= 10³ × 4785 – 2031
= 4785000 – 2031
= 4782969

Question 3.
Using binomial theorem, indicate which of the following two number is larger. (l.Ol)1000000, 10000 Answer:

= 1000000C₀ + 1000000C₁( . 01) + …………………….
= 1 + 1000000 × 0 . 01 + ……………………..
= 1 + 10000 + other positive terms
= 10001 + other positive terms
(1.01)^1000000 – 10000 = 10001 + other positive terms – 10000 > 0
∴ (1.01)^1000000 > 10000 ⇒ (1.01)^1000000 is larger.

Question 4.
Find the coefficient of x¹⁵ in \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\)
Answer:
General term T_r+1 = nC_r x^n-r . a^r
∴ In the expansion \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\)

To find the coefficient of x¹⁵, Put 20 – 5r = 15
20 – 15 = 5r ⇒ 5r = 5 ⇒ r = 1
∴ T_{1 + 1} = 10C₁x^{20 – 5} ⇒ T₂ = 10 . x¹⁵
∴ The coefficient of x¹⁵ is 10

Question 5.
Find the coefficient of x6 and the coefficient of x² in \(\left(x^{2}-\frac{1}{x^{3}}\right)^{6}\)
Answer:
The general terms is T_r+1 = nC_ra^n-r . a^r
∴ The general term in the expansion of \(\left(x^{2}-\frac{1}{x^{3}}\right)^{6}\) is

To find the coefficient of x⁶ , Put 12 – 5r = 6
12 – 6 = 5r ⇒ 5r = 6 ⇒ r = \(\frac{6}{5}\) which is impossible.
∴ There is no x⁶ term in the expansion.
To find the coefficient of x²,
Put 12 – 5r = 2
⇒ 12 – 2 = 5r
⇒ 10 = 5r
⇒ r = \(\frac{10}{5}\) = 5
Substituting in (1) we have

T₃ = 15
∴ The coefficient of x² is 15

Question 6.
Find the coefficient of x⁴ in the expansion of (1 + x³)⁵⁰ \(\left(x^{2}+\frac{1}{x}\right)^{5}\)
Answer:


= 10x⁴ × 1 + 5 × 1225 x⁴ + 19600 x⁴ + 500 x⁴
= x⁴ (10 + 6125 + 19600 + 500) = 26235 . x⁴
∴ The coefficient of x⁴ is 26235.

Question 7.
Find the constant term of \(\left(x^{3}-\frac{1}{3 x^{2}}\right)^{5}\)
Answer:


To get the constant term,
Put 15 – 5r = 0
⇒ 5r = 15
⇒ r = 3

Question 8.
Find the last two digits of the number 3⁶⁰⁰.
Answer:
Consider 3⁶⁰⁰
3⁶⁰⁰ = (3²)³⁰⁰ = 9³⁰⁰ = (1o – 1)³⁰⁰

= 10³⁰⁰ – 300 (10)²⁹⁹ + ………………. + 300 C₁ × 10 × – 1 + 1 × 1 × 1
= 10³⁰⁰ – 300 (1o)²⁹⁹ + …………….. – 300 × 10 + 1
= 10³⁰⁰ – 300 × 10²⁹⁹ + ……………… – 3000 + 1
All the terms except the last are multiples of 100 and hence divisible by 100.
∴ The last two digits will be 01.

Question 9.
If n is a positive integer show that 9ⁿ⁺¹ – 8n – 9 is always divisible by 64.
Answer:

which is divisible by 64 for all positive integer n.
∴ 9ⁿ – 8n – 1 is divisible by 64 for all positive integer n.
Put n = n + 1 we get
9^{n + 1} – 8 (n + 1) – 1 is divisible by 64 for all possible integer n
(9^{n + 1} – 8n – 8 – 1) is divisible by 64
∴ 9^{n + 1} – 8n – 9 is always divisible by 64

Question 10.
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)ⁿ are equal.
Answer:
Given (x + y)ⁿ
If n is odd the middle term in the expansion of (x + y)ⁿ are \(\frac{T_{n-1}}{2}+1\) and \(\mathrm{T}_{\frac{\mathrm{n}+1}{2}+1}\)

Question 11.
If n is a positive integer and r is a non-negative integer, prove that the coefficients of x^r and x^n-r in the expansion of (1 + x)ⁿ are equal.
Answer:
Given (1 + x)ⁿ.
General term T_r+1 = nC_r x^{n – r} . a^r
∴ The general term in the expansion of (1 + x)ⁿ is T_r+1 = nC_r . (1)^{n – r} . x^r
T_r+1 = nC_r . x^r ……….. (1)
∴ Coefficient of x^r is nC_r,
Put r = n – r in (1)
T_{n – r + 1} = nC_{n – r} . x^n-r
∴ The coefficient of x^n-r is nC_n-r …………. (2)
we know nC_r = nC^n-r
∴ The coefficient of x^r and coefficient of x^n-r are equal, (by (1) & (2))

Question 12.
If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.
Answer:
Let a = a + b – b
= b + (a – b)
aⁿ = [b + (a – b)]ⁿ
Using binomial expansion

Question 13.
In the binomial expansion of (a + b )n , the coefficients of the 4th and 13th terms are equal to each other , find n.
Answer:
In (a + b)ⁿ general term is t_{r + 1} = ⁿC_r a^{n – r} b^r
So, t₄ = t_{3 + 1} = ⁿC₃ = ⁿC₁₂
⇒ n = 12 + 3 = 15
We are given that their coefficients are equal ⇒ ⁿC₃ = ⁿC₁₂ ⇒ n = 12 + 3 = 15
[ⁿC_x = ⁿC_y ⇒ x = y (or) x + y = n]

Question 14.
If the binomial co – efficients of three consecutive terms in the expansion of (a + x)ⁿ are in the ratio 1 : 7 : 42, then find n.
Answer:
The general term in the expansion of (a + b)ⁿ is T_r+1 = nC_r . a^n-r . b^r
∴ The general term in the expansion of (a + x)ⁿ is T_r+1 = nC_r . a^n-r . x^r
Let the three consecutive terms be r^th term, (r + 1 )^th term, (r + 2 )^th term.

Question 15.
In the binomial coefficients of (1 + x)ⁿ, the coefficients of the 5^th, 6^th, and 7^th terms are in A.P. Find all values of n.
Answer:
Given (1 + x)ⁿ

12(n – 4) = 30 + n² – 5n – 4n + 20 ⇒ 12n – 48 = 30 + n² – 9n + 20
⇒ n² – 9n – 12n + 50 + 48 = 0 ⇒ n² – 21n + 98 = 0
⇒ n² – 14n – 7n + 98 = 0 ⇒ n(n – 14) – 7(n – 14) = 0
⇒ (n – 7) (n – 14) = 0 ⇒ n = 7 or n = 14

Question 16.
prove that

Answer:

This relation is true for all values of n. Equating coefficient of xⁿ on both sides, we have
The general term in the expansion of (1 + x)²ⁿ is T_{r + 1} = 2nC_r (1)^{2n – r} . x^r
Put r = n we get, T_n+1 = 2nC_n . xⁿ
∴ The coefficient of xⁿ in the expansion of (1 + x)²ⁿ is 2nC_n

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 4 Combinatorics and Mathematical Induction Ex 4.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.5

Choose the correct or the most suitable answer:

Question 1.
The sum if the digits at the 10^th place of all numbers formed with the help of 2, 4, 5, 7 taken all at a time is
(1) 432
(2) 108
(3) 36
(4) 18
Answer:
(2) 108

Explaination:

Number of digits given = 2, 4, 5, 7
Number of 4 digit numbers formed = 4! =24
So each digit occur \(\frac{24}{4}\) = 6 times
Sum of the digits = 2 + 4 + 5 + 7 = 18
So sum of the digits in each place = 18 × 6 = 108

Question 2.
In an examination, there are three multiple-choice questions and each question has 5 choices. The number of ways in which a student can fail to get all answer correct is
(1) 125
(2) 124
(3) 64
(4) 63
Answer:
(2) 124

Explanation:
Number of multiple-choice questions = 3
Number of choices in each question = 5
Number of ways of answering = 5 × 5 × 5 = 5³ = 125
Number of ways of getting all answer correct = 1
∴ Number of ways of getting incorrect answer = 125 – 1 = 124

Question 3.
The number of ways in which the following prize be given to a class of 30 boys first and second in mathematics, first and second in Physics, first in Chemistry and first in English is
(1) 30⁴ × 29²
(2) 30² × 29³
(3) 30² × 29⁴
(4) 30 × 29⁵
Answer:
(1) 30⁴ × 29²

Explanation:
I and II in maths can be given in 30 × 29 ways.
I and II in physics can be given in 30 × 29 ways.
I and chemistry can be given in 30 ways.
I in English can be given in 30 ways.
So total number of ways = 30 × 29 × 30 × 29 × 30 × 30 = 30⁴ × 29²

Question 4.
The number of 5 digit numbers all digits of which are odd is
(1) 25
(2) 5⁵
(3) 5⁶
(4) 625
Answer:
(2) 5⁵

Explanation:
The odd digit numbers = 1, 3, 5, 7, 9
Since repetition is allowed, each digit can be filled by 1, 3, 5, 7, 9
Unit’s place can be filled in 5 ways
Ten’s place can be filled in 5 ways etc.
∴ The number of 5 digit numbers with an odd digit in each place = 5 × 5 × 5 × 5 × 5 = 5⁵ ways

Question 5.
In 3 fingers, the number of ways four rings can be worn is ————- ways.
(1) 4³ – 1
(2) 3⁴
(3) 68
(4) 64
Answer:
(4) 64

Explanation:
Each letter can be ported in 3 ways
∴ 4 letter is 3⁴ ways

Question 6.

then the value of n are
(1) 7 and 11
(2) 6 and 7
(3) 2 and 11
(4) 2 and 6
Answer:
(2) 6 and 7

Explaination:

( n + 5 ) ( n + 4) = 2 × 11 × (n – 1)
n² + 5n + 4n + 20 = 22n – 22
n² + 9n + 20 – 22n + 22 = 0
n² – 13n + 42 = 0
n (n – 6) – 7 (n – 6) = 0
(n – 7) (n – 6) = 0
n – 7 = 0 or n – 6 = 0
n = 7 and n = 6

Question 7.
The product of r consecutive positive integers is divisible by
(1) r !
(2) (r – 1) !
(3) ( r + 1 ) !
(4) r^r
Answer:
(1) r !

Explanation:
1(2) (3) ….. (r) = r! which is ÷ by r!

Question 8.
The number of five-digit telephone numbers having at least one of their digits repeated is
(1) 90000
(2) 10000
(3) 30240
(4) 69760
Answer:
(4) 69760

Explaination:
Case (i) When zero is allowed in the first place.
The number of five-digit telephone numbers which can
be formed using the digits 0, 1, 2, …………,9 is 10⁵.
The number of five-digit telephone numbers which have none of their digits repeated is 10P⁵
= 10 × 9 × 8 × 7 × 6 = 30240
∴ The required number of telephone number
= 10⁵ – 30240
= 100000 – 30240
= 69,760

Case (ii) When zero is not allowed in the first place,
(a) Repetition allowed:

The number of ways of filling the first box, second box, …………….. , fifth box
= 9 × 10 × 10 × 10 × 10
= 90000

(b) Repetition riot allowed:
Number of ways of filling 1 box = 9
Number of ways of filling 2^nd box = 9
Number of ways of filling 3^rd box = 8
Number of ways of filling 4^th box = 7
Number of ways of filling 5^th box = 6
∴ Number of numbers with no digit is repeated = 9 × 9 × 8 × 7 × 6 = 37216
∴ Number of numbers having atleast one of the digits repeated = 90000 – 37216 = 52,784

Question 9.
If (a² – a ) C₂ = ( a² – a) C₄ then the value of a is
(1) 2
(2) 3
(3) 4
(4) 5
Answer:
(2) 3

Explaination:
(a² – a)C₂ = (a² – a) C₄
(a² – a)C₂ = (a² – a) C_{(a² – a) – 4} [nC_r = nC_n-r]
(a² – a)C₂ = (a² – a) C_{a² – a – 4}
∴ 2 = a² – a – 4
a² – a – 4 – 2 = 0 ⇒ a² – a – 6 = 0
a² – 3a + 2a – 6 = 0
a (a – 3) + 2 (a – 3) = 0
⇒ (a + 2)(a – 3) = 0 ⇒ a = – 2 or 3

Question 10.
There are 10 points in a plane and 4 of them are collinear. The number of straight lines joining any two points is
(1) 45
(2) 40
(3) 39
(4) 38
Answer:
(2) 40

Explaination:
Total number of points = 10
The number of straight lines joining any two points is same as the number ways of selecting 2 points from 10 points.
∴ The number of straight lines = 10C₂
Given 4 points are collinear.
∴ The number of straight lines that these 4 points contribute = 4C₂
Since 4 points are collinear they lie on a line.
∴ Required number of lines = 10C₂ – 4C₂ + 1

= 5 × 9 – 2 × 3 + 1
= 45 – 6 + 1 = 40

Question 11.
The number of ways in which a host lady invite for a party of 8 out of 12 people of whom two do not want to attend the party together is
(1) 2 × 11C₇ + 10C₈
(2) 11C₇ + 10C₈
(3) 12C₈ – 10C₆
(4) 10C₆ + 2!
Answer:
(3) 12C₈ – 10C₆

Explaination:
Number of people = 12
Number of people invited for the party = 8
Number of people not willing to attend the party = 2
Number of ways of selecting 8 people from 12 people is 12C₈
Two people do not attend party out of 10, remaining people 8 can attend in 10C₆ ways.
∴ Number of ways in which two of them do not attend together = 12C₈ – 10C₆

Question 12.
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines.
(1) 6
(2) 9
(3) 12
(4) 18
Answer:
(4) 18

Explaination:
Number of first set of parallel lines = 4
Number of second set of parallel lines = 3
Number of parallelograms = 4C₂ × 3C₂
= \(\frac{4 \times 3}{1 \times 2} \times \frac{3 \times 2}{1 \times 2}\)
= 6 × 3
= 18

Question 13.
Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is
(1) 11
(2) 12
(3) 10
(4) 6
Answer:
(2) 12

Explaination:
Number of shake hands = 66
Let the number of persons = n
First person shakes hands with the remaining n – 1 persons.
Second person shakes hands with the remaining n – 2 persons.

∴ Total number of shake hands
= (n – 1) + (n – 2) + …………. + 2 + 1
= \(\frac{(\mathrm{n}-1)(\mathrm{n}-1+1)}{2}\)
Given 66 = \(\frac{(\mathrm{n}-1) \mathrm{n}}{2}\)
n² – n = 132
n² – n – 132 = 0
(n – 12) (n + 11) = 0
n = 12 or n = – 11
n = – 11 is not possible.
∴ n = 12

Question 14.
Number of sides of a polygon having 44 diagonals is
(1) 4
(2) 4!
(3) 11
(4) 22
Answer:
(3) 11

Explaination:
Number of diagonals of a polygon with n sides
= nC₂ – n
= \(\frac{n(n-1)}{2}\)
Given \(\frac{n(n-1)}{2}\) – n = 44
n² – n + 2n = 88
n² – 3n – 88 = 0
(n – 11 ) (n + 8) = 0
n = 11 and n = – 8
n = – 8 is not possible.
∴ n = 11

Question 15.
If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then the total number of point of intersection are
(1) 45
(2) 40
(3) 10!
(4) 210
Answer:
(1) 45

Explaination:
Number of non-parallel lines = 10
Two lines will interest.
∴ Number of points of intersection
= 10C₂ = \(\frac{10 \times 9}{1 \times 2}\) = 45

Question 16.
In a plane there are 10 points are there out of which 4 points are collinear, then the number of triangles formed is
(1) 110
(2) 10C₃
(3) 120
(4) 116
Answer:
(4) 116

Explaination:
Number of points = 10
Number of triangles formed by using 10 points is same as number of ways of choosing 3 points from 10 points = 10C₃
Also given 4 points are collinear.
∴ These 4 points do not contribute triangles.
Thus, total number of triangles = 10C₃ – 4C₃
= \(\frac{10 \times 9 \times 8}{1 \times 2 \times 3}\) – 4
= 5 × 3 × 8 – 4
= 120 – 4
= 116

Question 17.
In 2nC₃ : nC₃ = 11 : 1 then
(1) 5
(2) 6
(3) 11
(4) 7
Answer:
(2) 6

Explaination:
Given 2nC₃ : nC₃ = 11 :

4(2n – 1) = 11 (n – 2)
8n – 4 = 11 n – 22
-4 + 22 = 11 n – 8n
18 = 3n
⇒ n = 6

Question 18.
(n – 1) C_r + ( n – 1 ) C_{r – 1} is
(1) (n + 1) C_r
(2) (n – 1) C_r
(3) nC_r
(4) nC_{r – 1}
Answer:
(3) nC_r

Explaination:
[nC_r + nC_{r – 1} = (n + 1 ) C_r
(n – 1)C_r + (n – 1)C_{r – 1} = (n – 1 + 1) C_r
= nC_r

Question 19.
The number of ways of choosing 5 cards out of a deck of 52 cards which include atleast one king is
(1) 52C₅
(2) 48 C₅
(3) 52C₅ + 48C₅
(4) 52C₅ – 48C₅
Answer:
(4) 52C₅ – 48C₅

Explaination:
Number of cards = 52,
Number of kings = 4
Number of ways of choosing 5 cards out of 52 cards = 52 C₅
Number of cards without kings = 48
Number of ways of choosing 5 cards with no kings from 48 cards = 48 C₅
∴ Number of ways of choosing 5 cards out of 52 cards
= 52C₅ – 48C₅

Question 20.
The number of rectangles that a chessboard has
(1) 81
(2) 9⁹
(3) 1296
(4) 6561
Answer:
(3) 1296

Explaination:
Number of rectangles in a chessboard is

= 36 × 36
= 1296

Question 21.
The number of 10 digit number that can be written by using digits 2 and 3 is
(1) 10C₂ + 9C₂
(2) 2¹⁰
(3) 2¹⁰ – 2
(4) 10!
Answer:
(2) 2¹⁰

Explaination:

Each box can be filled in two ways using digits 2 and 3.
∴ A number of 10 digit number that can be written by using the digits 2 and 3 is 2 × 2 × 2 × …………. 10 times. = 210

Question 22.
If Pr stands for rPr, then the sum of the series 1 + P₁ + 2P₃ + 3P₃ + ……… + nP_n is
(1) P_{n + 1}
(2) P_{n + 1} – 1
(3) P_{n + 1} + 1
(4) (n + 1)P_{n – 1}
Answer:
(1) P_{n + 1}

Explaination:
Given P, = rP_r = r!
1 + 1 × P₁! + 2 × P₂ + 3 × P₃ + …………… + n × P_n!
= 1 + (1 × 1! + 2 × 2! + 3 × 3! + …………. + n × n!)
= (n + 1 )!
= P_{n + 1}
[(1 × 1 !) + (2 × 2!) + (3× 3 !) + ….. + (n × n !) = (n + 1)! – 1]

Question 23.
The product of first n odd natural numbers equals
(1) 2nC_n × nP_n
(2) \(\left(\frac{1}{2}\right)^{n}\) 2nC_n × nP_n
(3) \(\left(\frac{1}{4}\right)^{n}\) × 2nC_n × 2nP_n
(4) nC_n × nP_n
Answer:
(2) \(\left(\frac{1}{2}\right)^{n}\) 2nC_n × nP_n

Explaination:
1 . 3 . 5 …………. (2n – 1)

Question 24.
If nC₄, nC₅, nC₆ are in A. P. the value of n can be
(1) 14
(2) 11
(3) 9
(4) 5
Answer:
(1) 14

Explaination:
Given nC_n, nC₅, nC₆ are in A. P.
∴ 2 × nC₅ = nC₄ + nC₆

12(n – 4) = 30 + (n – 4) (n – 5)
12n – 48 = 30 + n² – 5n – 4n + 20
12n – 48 = 50 + n² – 9n
n² – 9n – 50 – 12n + 48 = 0
n² – 21n + 98 = 0
n² – 14n – 7n + 98 = 0
n(n – 14) – 7(n – 14) = 0
(n – 7) (n – 14) = 0
n – 7 = 0 or n – 14 = 0
n = 7 or n = 14

Question 25.
1 + 3 + 5 + 7 + …………. + 17 is equal to
(1) 101
(2) 81
(3) 71
(4) 61
Answer:
(2) 81

Explaination:
Given 1 + 3 + 5 + 7 + ………… + 17
This is an A.P. with first term a = 1
Common difference d = 2
Last term t_n = 17
a + (n – 1)d = 17
1 + (n – 1)2 = 17
(n – 1)2 = 17 – 1
(n – 1)2 = 16
n – 1 = \(\frac{16}{2}\) = 8
n = 9

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.4

Question 1.
By the principle of mathematical induction, prove that, for n ≥ 1
1² + 2² + 3² + ……….. + n³ = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)
Answer:
Let P(n) = 1² + 2² + 3² + ……….. + n³
p(n) = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)

Step 1:
Let us verify the statement for n = 1
p(1) = 1³ = \(\left(\frac{1(1+1)}{2}\right)^{2}\)

P(1) = 1
∴ The given statement is true for n = 1

Step 2 :
Let us assume that the given statement is true for n = k
P(k) = 1³ + 2³ + 3³ + …………… + k³

Step 3 :
Let us prove the statement is true for n = k + 1
P(k + 1 ) = 1³ + 2³ + 3³ + …………… + k³ + (k + 1)³

This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
1² + 2² + 3² + ……….. + n³ = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)
is true for all natural numbers n.

Question 2.
By the principle of mathematical induction, prove that, for n ≥ 1

Answer:

Step 1 :
Let us verify the statement for n = 1

∴ The given statement is true for n = 1

Step 2 :
Let us assume that the given statement is true for n = k

Step 3:
Let us prove the statement is true for n = k + 1


This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.

is true for all natural numbers n.

Question 3.
Prove that the sum of the first ’n’ non-zero even numbers is n² + n.
Answer:
Let P(n) : 2 + 4 + 6 +…+2n = n² + n, ∀ n ∈ N

Step 1:
P( 1) : 2 = 1² + 1 = 2 which is true for P( 1)

Step 2:
P(k): 2 + 4 + 6+ …+ 2k = k² + k. Let it be true.

Step 3:
P(k + 1) : 2 + 4 + 6 + … + 2k + (2k + 2)
= k²+ k + (2k + 2) = k² + 3k + 2
= k² + 2k + k + 1 + 1
= (k+ 1)²+ (k + 1)
Which is true for P(k + 1)
So, P(k + 1) is true whenever P(k) is true.

Question 4.
By the principle of mathematical induction, prove that, for n ≥ 1, 1 . 2 + 2 . 3 + 3 . 4 + …………. + n . (n + 1) = \(\frac{\mathbf{n}(\mathbf{n}+1)(\mathbf{n}+2)}{3}\)
Answer:
Let P(n) = 1 . 2 + 2 . 3 + 3 . 4 + ………… + n . (n + 1) = \(\frac{\mathbf{n}(\mathbf{n}+1)(\mathbf{n}+2)}{3}\)

Step 1:
Let us verify the statement for n = 1
P(1) = 1 . 2 = \(\frac{1(1+1)(1+2)}{3}\)
= 1 . 2 = \(\frac{1 \cdot 2 \cdot 3}{3}\)
∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k
P (k) = 1-2 + 2-3 + 3- 4 + + k (k + 1 ) =

Step 3:
Let us prove the result for n = k + 1
P(k + 1) = 1 . 2 + 2 . 3 + 3 . 4 + …………… + k (k + 1) (k + 1) (k + 1 + 1)
P (k + 1) = 1 . 2 + 2 . 3 + 3 . 4 + ………….. + k (k + 1) + (k + 1 ) (k + 2)
P (k + 1 ) = P(k) + (k + 1) (k + 2)

This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1, Hence by the principle of mathematical induction, the result is true for all natural numbers n.
1 . 2 + 2 . 3 + 3 . 4 + ………. + n (n + 1) = \(\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3}\)
is true for all natural numbers n ≥ 1

Question 5.
Using the Mathematical induction, show that for any natural number n ≥ 2,

Answer:
Let

The first stage is n = 2

Step 1:
Let us verify the result for n = 2

∴ The given result is true for n = 2.

Step 2:
Let us assume that the result is true for n = k.

Step 3:
Let us prove the result for n = k + 1


∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n ≥ 2.

is true for all natural numbers n ≥ 2.

Question 6.
Using the Mathematical induction, show that for any natural number n ≥ 2,

Answer:

Step 1:
Since n ≥ 2 the first stage is n = 2
Let us verify the result for n = 2

∴ P (2 ) is true.
The result is true for n = 2.

Step 2:
Let us assume that the result is true for n = k.

Step 3:
Let us prove the result for n = k + 1


This implies p(k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n ≥ 2.

is true for all natural numbers n ≥ 2

Question 7.
Using the mathematical induction, show that for any natural number n,

Answer:

Step 1:
First, let us verify the result for n = 1

∴ The given result is true for n = 1

Step 2:
Let us assume the result for n = k

Step 3:
Let us prove the result for n = k + 1

Factorizing k³ + 6k² + 9k + 4
f (k) = k³ + 6k² + 9k + 4
f(- 1) = (- 1)³ + 6(- 1)² + 9(- 1) + 4
f(- 1) = – 1 + 6 – 9 + 4 = 0
∴ (k + 1) is a factor of f(k)

k³ + 6k² + 9k + 4 = (k + 1) (k² + 5k + 4).
= (k + 1) (k² + 4k + k + 4)
= (k + 1) [k(k + 4) + 1(k + 4)]
= (k + 1) (k + 4) (k + 1)

This implies P(k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.

is true for all natural numbers n

Question 8.
Using the Mathematical induction, show that for any natural number n,

Answer:

Step 1:
First, let us verify the result for n = 1


∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k

Step 3:
Let us prove the result for n = k + 1


This implies P(k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.

Question 9.
Prove by mathematical induction that
1! + (2 × 2 !) + (3 × 3!) +…. .+ (n × n!) = (n + 1)! – 1
Answer:
P(n) is the statement
1! + (2 × 2!) + (3 × 3!) + ….. + (n × n!) = (n + 1)! – 1
To prove for n = 1
LHS = 1! = 1
RHS = (1 + 1)! – 1 = 2! – 1 = 2 – 1 = 1
LHS = RHS ⇒ P(1) is true
Assume that the given statement is true for n = k
(i.e.) 1! + (2 × 2!) + (3 × 3!) + … + (k × k!) = (k + 1)! – 1 is true
To prove P(k + 1) is true
p(k + 1) = p(k) + t_{(k + 1)}
P(k + 1) = (k + 1)! – 1 + (k + 1) × (k + 1)!
= (k + 1)! + (k + 1) (k + 1)! – 1
= (k + 1)! [1 + k + 1] – 1
= (k + 1)! (k + 2) – 1
= (k + 2)! – 1
= (k + 1 + 1)! – 1
∴ P(k + 1) is true
⇒ P(k) is true, So by the principle of mathematical induction P(n) is true.

Question 10.
Using the Mathematical induction, show that for any natural number n, x²ⁿ – y²ⁿ is divisible by x + y.
Answer:
Let P(n) = x²ⁿ – y²ⁿ is divisible by (x + y)
For n = 1
P(1) = x^{2 × 1} – y^{2 × 1} is divisible by (x + y)
⇒ (x + y) (x – y) is divisible by (x + y)
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = x^2k – y^2k is divisible by (x + y)
⇒ x^2k – y^2k = λ(x + y) …… (i)
For n = k + 1
⇒ P(k + 1) = x^{2(k + 1)} – y^{2(k + 1)} is divisible by (x + y)
Now x^{2(k + 2)} – y^{2(k + 2)}
= x^{2k + 2} – x^2ky² + x^2ky² – y^{2k + 2}
= x^2k.x² – x^2ky² + x^2ky² – y^2ky²
= x^2k (x² – y²) + y²λ (x + y) [Using (i)]
⇒ x^{2k + 2} – y^{2k + 2} is divisible by (x + y)
∴ P(k + 1) is true.
Thus P(k) is true ⇒ P(k + 1) is true.
Hence by the principle of mathematical induction, P(n) is true for all n ∈ N

Question 11.
By the principle of mathematical induction, prove that, for n ≥ 1, 1² + 2² + 3² + ………….. + n² > \(\frac{n^{3}}{3}\)
Answer:
Let P (n) = 1² + 2² + 3² + ………….. + n² > \(\frac{n^{3}}{3}\)

Step 1:
Let us first verify the result for n = 1

∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k.
P(k) = 1² + 2² + 3² + …………… + k² > \(\frac{\mathrm{k}^{3}}{3}\)

Step 3:
Let us prove the result for n = k + 1
P(k + 1) = 1² + 2² + 3² + …………… + k² + (k + 1)²
P(k + 1) = P(k) + ( k + 1)²

This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
1² + 2² + 3² + ………….. + n² > \(\frac{n^{3}}{3}\)

Question 12.
Use induction to prove that n³ – 7n + 3, is divisible by 3, for all natural numbers n.
Answer:
Let P(n) : n³ – 7n + 3
Step 1:
P(1) = (1)³ – 7(1) + 3
= 1 – 7 + 3 = -3 which is divisible by 3
So, it is true for P(1).

Step 2:
P(k) : k³ – 7k + 3 = 3λ. Let it be true
⇒ k³ = 3λ + 7k – 3

Step 3:
P(k + 1) = (k + 1)³ – 7(k + 1) + 3
= k³ + 1 + 3k² + 3k – 7k – 7 + 3
= k³ + 3k² – 4k – 3
= (3λ + 7k – 3) + 3k² – 4k – 3 (from Step 2)
= 3k² + 3k + 3λ – 6
= 3(k² + k + λ – 2) which is divisible by 3.
So it is true for P(k + 1).
Hence, P(k + 1) is true whenever it is true for P(k).

Question 13.
Use induction to prove that 5^{n + 1} + 4 × 6ⁿ when divided by 20 leaves a remainder 9 , for all natural numbers n.
Answer:
To prove 5^{n + 1} + 4 × 6ⁿ when divided by 20 leaves a remainder 9
That is to prove 5^{n + 1} + 4 × 6ⁿ – 9 is divisible by 20
Let P (n) be the statement 5^{n + 1} + 4 × 6ⁿ
which is divisible by 20
P (n) = 5^{n + 1} + 4 × 6ⁿ – 9 = 20λ

Step 1 :
Let us verify the statement for n = 1
P (1) = 5^{i + 1} + 4 × 6¹ – 9
= 5² + 4 × 6 – 9
= 25 + 24 – 9
= 49 – 9 = 40
which is divisible by 20
∴ The statement is true for n = 1

Step 2:
Let us assume that the statement is true for n = k
P(k) = 5^{k + 1} + 4 × 6^k – 9 is divisible by 20
P(k) = 5^{k + 1} + 4 × 6^k – 9 = 20λ …………… (1)

Step 3:
Let us prove the result for n = k + 1

Hence, P(k + 1) is divisible by 20.
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
Thus, 5^{n + 1} + 4 × 6ⁿ when divided by 20 leaves a remainder 9, for all natural numbers n.

Question 14.
Use induction to prove that 10ⁿ + 3 × 4^{n + 2} + 5 is divisible by 9, for all natural numbers n.
Answer:
Let P( n) = 10ⁿ + 3 × 4^{n + 2} + 5 is divisible by 9

Step 1:
Let us verify the result for n = 1
P ( 1 ) = 10¹ + 3 × 4^{1 + 2} + 5
= 15 + 3 × 4³
= 15 + 3 × 64
= 15 + 192 = 207
which is divisible by 9
Thus we have verified the result for n = 1

Step 2:
Let us assume the result is true for n = k
P(k) = 10^k + 3 × 4^{k + 2} + 5 is divisible by 9
∴ 10^k + 3 × 4^{k + 2} + 5 = 9m for some m
10^k = 9m – 5 – 3 × 4^{k + 2} for some m

Step 3:
Let us prove the result for n = k + 1

which is divisible by 9
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
10ⁿ + 3 × 4^{n + 2} + 5 is divisible by 9,
for all natural numbers n.

Question 15.
Prove that using mathematical induction

Answer:

This implies P (k + 1) is true. Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.

Tamilnadu State Board New Syllabus Samacheer Kalvi 9th Science Guide Pdf Chapter 13 Chemical Bonding Text Book Back Questions and Answers, Notes.

Samacheer Kalvi 9th Science Solutions Chapter 13 Chemical Bonding

9th Science Guide Chemical Bonding Text Book Back Questions and Answers

I. Choose the correct answer:

Question 1.
Number of valence electrons in carbon is
(a) 2
(b) 4
(c) 3
(d) 5
Answer:
(b) 4

Question 2.
Sodium having atomic number 11, is ready to ______________ electron/electrons to attain the nearest noble gas electronic configuration.
(a) gain one
(b) gain two
(c) lose one
(d) lose two
Answer:
(c) lose one

Question 3.
The element that would form anion by gaining electrons in a chemical reaction is ………………..
(a) potassium
(b) calcium
(e) fluorine
(d) iron
Answer:
(c) fluorine

Question 4.
Bond formed between a metal and non metal atom is usually …………………..
(a) ionic bond
(b) covalent bond
(e) co-ordinate bond
Answer:
(a) ionic bond

Question 5.
______________ compounds have high melting and boiling points.
(a) Covalent
(b) Coordinate
(e) Ionic
Answer:
(c) Ionic

Question 6.
Covalent bond is formed by …………………
(a) transfer of electrons
(b) sharing of electrons
(c) sharing a pair of electrons
Answer:
(b) sharing of electrons

Question 7.
Oxidising agents are also called as …………………. because they remove eletrons form other substances.
(a) electron donors
(b) electron acceptors
Answer:
(b) electron acceptors

Question 8.
Elements with stable electronic configurations have eight electrons in their valence shell. They are ……………….
(a) halogens
(b) metals
(c) nobel gases
(d) non metals
Answer:
(c) noble gases

II. Answer briefly :

Question 1.
How do atoms attain Noble gas electronic configuration?
Answer:
Atoms of all elements, other than inert gases, combine to form molecules because they have incomplete valence shell and tend to attain a stable electronic configuration similar to noble gases. Atoms can combine either by transfer of valence electrons from one atom to another or by sharing of valence electrons in order to achieve the stable outer shell of eight electrons.

Question 2.
NaCl is insoluble in carbon tetrachloride but soluble in water. Give reason.
Answer:
NaCl is an ionic compound, it is soluble in polar solvent (water). Whereas CCl₄ is a covalent compound. So.it is insoluble in polar solvent (water). But it is soluble in nonpolar solvents.

Question 3.
Explain the Octet rule with an example.
Answer:
The tendency of atoms to have eight electrons in the valence shell is known as the ‘Octet rule’ or the ‘Rule of eight’ For example, Sodium with atomic number 11 will readily loose one electron to attain Neon’s stable electronic configuration. Similarly, chlorine has electronic configuration 2,8,7. To get the nearest noble gas (i.e. Argon) configuration, it needs one more electron. So chlorine readily gains one electron from another atom and obtains stable electronic configuration. Thus elements tend to have stable valence shell (eight electrons) either by losing or gaining electrons.

Question 4.
Write a note on different types of bonds.
Answer:
There are different types of chemical bonding possible between atoms which make the molecules. Depending on the type of bond they show different characteristics or properties.

Question 5.
Correct the wrong statements.
(a) Ionic compounds dissolve in non-polar solvents.
(b) Covalent compounds conduct electricity in molten or solution state.
Answer:
(a) Covalent compounds dissolve in non-polar solvents. (or) Ionic compounds dissolve in polar solvents.
(b) Ionic compounds conduct electricity in molten or solution state.

Question 6.
Complete the table given below.

Answer:

Question 7.
Draw the electron distribution diagram for the formation of Carbon dioxide (CO₂) molecule.
Answer:

Question 8.
Fill in the following table according to the type of bonds formed in the given molecule.
CaCl₂, H₂O, CaO, CO, KBr, HCl, CCl₄, HF, CO₂, Al₂Cl₆

Answer:

Question 9.
The property which is characteristic of an Ionic compound is that
(a) it often exists as a gas at room temperature.
(b) it is hard and brittle.
(c) it undergoes molecular reactions
(d) it has a low melting point.
Answer:
(b) it is hard and brittle

Question 10.
Identify the following reactions as oxidation or reduction.
(a) Na → Na⁺ + e^–
(b) Fe³⁺ + 2 e^– → Fe⁺
Answer:
(a) oxidation
(b) reduction

Question 11.
Identify the compounds as Ionic/Covalent/Coordinate based on the given characteristics.
(a) Soluble in non-polar solvents
(b) Undergoes faster/instantaneous reactions
(c) Non-conductors of electricity
(d) Solids at room temperature
Answer:
(a) Co-ordinate Covalent compound.
(b) Ionic compound.
(c) Covalent compound.
(d) Ionic compound.

Question 12.
An atom X with atomic number 20 combines with atom Y with atomic number 8. Draw the dot structure for the formation of the molecule XY
Answer:

Question 13.
Considering MgCl₂ as ionic compound and CH₄ as covalent compound give any two differences between these two compounds.
Answer:

Question 14.
Why are Noble gases inert in nature?
Answer:
Noble gases are inert in nature due to the completely filled subshells and thus have stable electronic structures which is very difficult to change. The elements Helium, Neon, Argon, Krypton, Xenon and Radon of group 18 in the periodic table are Noble gases.

III. Answer in detail :

Question 1.
List down the differences between Ionic and Covalent compounds.
Answer:

Question 2.
Give an example for each of the following statements.
(a) A compound in which two Covalent bonds are formed.
(b) A compound in which one ionic bond is formed.
(c) A compound in which two Covalent and one Coordinate bond are formed.
(d) A Compound in which three covalent bonds are formed.
(e) A compound in which a coordinate bond is formed.
Answer:
(a) Oxygen molecule (O₂) (O = O)
(b) Sodium’Chloride (NaCl)
(c) Carbon monoxide?

(d) Nitrogen molecule (N₂) (N ≡N)
(e) NH₃ → BF₃

Question 3.
Identify the incorrect statement and correct them.
(a) Like covalent compounds, coordinate compounds also contain charged particles (ions). So they; are good conductors of electricity.
(b) Ionic bond is a weak bond when compared to Hydrogen bond.
(c) Ionic or electrovalent bonds are formed by the mutual sharing of electrons between atoms.
(d) Loss of electrons is called Oxidation and gain of the electron is called Reduction.
(e) The electrons which are not involved in bonding are called valence electrons.
Answer:
(a) Incorrect statement. Like covalent compounds, co-ordinate compounds also do not contain charged particles (ions), so they are bad conductors of electricity.
(b) Incorrect statement. An ionic bond is a strong bond when compared to a hydrogen bond.
(c) Incorrect statement. Covalent bonds are formed by the mutual sharing of electrons between atoms, (or) Ionic or electrovalent bonds are formed by the transfer of electrons between atoms.
(d) Correct statement
(e) Incorrect statement. The electrons which are not involved in bonding are called lone pair of electrons.

Question 4.
Discuss in brief the properties of coordinate covalent compounds.
Answer:
The compounds containing coordinate covalent bonds are called coordinate compounds.
(a) Physical state – These compounds exist as gases, liquids or solids.
(b) Electrical conductivity-Like covalent compounds, co-ordinate compounds also do not contain charged particles (ions), so they are bad conductors of electricity.
(c) Melting point – These compounds have to melt and boiling points higher than those of purely covalent compounds but lower than those of purely ionic compounds.
(d) Solubility – Insoluble in polar solvents like water but are soluble in non-polar solvents like benzene, CCl₄, and toluene.
(e) Reactions – Co-ordinate covalent compounds undergo molecular reactions which are slow.

Question 5.
Find the oxidation number of the elements in the following compounds.
(a) C in CO₂
(b) Mn in MnSO₄
(c) N in HNO₃
Answer:
(a) C in CO₂
1(C) + 2(0) = 0
1x + 2(-2) = 0
x-4 = 0 ‘
x = +4 .
ON of C in CO₂ is +4

(b) Mn in MnSO₄
1 (Mn) + 1 (S) + 4(0) = 0
x + 1(+6) + 4(-2) = 0
x + 6 – 8 = 0
x – 2 = 0
x = +2
ON of Mn in MnSO₄ is +2,

(c) N in HNO₃
1(H) + 1 (N) +3(0) =0
1 (+1) + 1 (x) + 3 (-2) =0
+ 1 + x – 6 =0
x – 5 =0
x = +5
ON of N in HNO₃ is + 5.

9th Science Guide Chemical Bonding Additional Important Questions and Answers

I. Choose the correct answer:

Question 1.
Which of the following atom can exist independently?
(a) Magnesium
(b) Chlorine
(c) Hydrogen
(d) Neon
Answer:
(d) Neon

Question 2.
Alkali and alkaline earth metals form ………… compound when they react with non-metals.
(a) ionic
(b) covalent
(c) co-ordinate covalent
(d) all the above
Answer:
(a) ionic

Question 3.
……………..compounds are highly brittle.
(a) Ionic
(b) Covalent
(c) Co-ordinate covalent
Answer:
(a) Ionic

Question 4.
The bond which is formed by mutual sharing of electrons is called ……………….. bond.
(a) ionic
(b) covalent
(c) co-ordinate covalent bond
(d) all the above
Answer:
(b) covalent

Question 5.
………………. is an example of a covalent compound having a high melting point.
(a) Magnesium oxide
(b) Silicon carbide
(c) Ammonia
(d) All the above
Answer:
(b) Silicon carbide

Question 6.
Which of the following compound(s) possesses a high melting point?
(a) NaCl
(b) MgCl₂
(c) CCl₄
(d) Both a & b
Answer:
(d) Both a & b

Question 7.
The element that would form cation due to the loss of electron during the chemical reaction is ………………..
(a) calcium
(b) Fluorine
(c) Chlorine
(d) all the above
Answer:
(a) calcium

Question 8.
Fajan’s rule is formulated by considering …………… the cation and ………………..of the cation and anion.
(a) charge
(b) size
(c) charge & size
(d) none
Answer:
(c) charge & size

Question 9.
The formation of brown colour on the freshly unit surface of vegetables and fruits is because ………………. of organic compounds present in them.
(a) oxidation
(b) reduction
(c) both a & b
(d) none
Answer:
(a) oxidation

Question 10.
Which of the following compounds has melting and boiling points higher than covalent compounds but lower than ionic compounds?
(a) NaCl
(b) MgCl₂
(c) H₂O
(d) NH₃→BF₃
Answer:
(d) NH₃→BF₃

Question 11.
Atoms having 1,2 or 3 electrons in their valence shell will readily form ……………..
(a) cation
(b) anion
Answer:
(a) cation

II. Fill in the blanks:

1. ……………theory explains the formation of molecules.
Answer:
Kossel – Lewis theory

The valency of noble gases is ……………..
Answer:
zero

3. …………….. is the only noble gas which does not have eight electrons in their valence shell.
Answer:
Helium

4. The atom that loses electrons will from a ________
Answer:
cation

5. ______ compounds have high density.
Answer:
Ionic

6. In covalent bond formation, the sharing of ………….. electrons takes place in their outermost shell.
Answer:
unpaired

7. Polar solvents contain bond between atoms with ………………
Answer:
different electronegativities

8. ……………….. & ………….. atoms have similar electronegativities.
Answer:
Carbon & hydrogen

9. Molecular reactions are ……………… in the covalent compound.
Answer:
slow

10. Ionic compounds are ……………..in nature.
Answer:
solid

11. The tendency of atoms to have eight electrons in the outer shell is known as ……………….
Answer:
Octet rule

12. As per Fajan’s rule, A1I₃is ……………..
Answer:
covalent

13. Oxidising agents are otherwise called as ………………..
Answer:
electron acceptors

14. The tarnishing of metals is due to the formation of ……………….
Answer:
metal oxide

15. The tarnishing of metals is an example of ……………. reaction.
Answer:
oxidation

16. The sum of oxidation number of all atoms in a compound is ………………..
Answer:
Zero

17. The ……………….. is a metal that has a high resistance to corrosion.
Answer:
Gold

III. Spot the error / correct the wrong statement:

Question 1.
In the formation of compounds, the inner shell electrons of an atom involved in bonding.
Answer:
In the formation of compounds, the valence electrons of an atom involved in bonding.

Question 2.
The atom that gains electrons will form a cation.
Answer:
The atom that gains electrons will form an anion.

Question 3.
Ionic compounds have low melting and boiling point.
Answer:
Ionic compounds have high melting and boiling point, (or) Covalent compounds have low melting and boiling point.

Question 4.
Non-polar solvents contain bonds between atoms with different electronegativities.
Answer:
Non-polar solvents contains bonds between atoms with similar electronegativities.

Question 5.
Covalent compounds are soluble in polar solvents.
Answer:
Covalent compounds are readily soluble in non-polar solvents, (or) Ionic compounds are soluble in polar solvents.

Question 6.
Greater the charge of the cation greater will be the ionic character.
Answer:
Greater the charge of the cation greater will be the covalent character.

IV. Match the following :

Question 1.

Answer:
1. – e, 2. – d, 3. – a, 4. – b, 5. – c

Question 2.

Answer:
1. – c, 2. – a, 3. – b, 4. – e, 5, – d.

V. Find the odd one out and write the reason:

Question 1.
Water, acetone, benzene, toluene turpentine.
Answer:
Water. It is a polar solvent where a^ others air non-polar solvents.

Question 2.
Addition of oxygen, removal of hydrogen, loss of an electron, gain of electron.
Answer:
Gain of electron. It is a reduction reaction whereas other three are oxidation reactions.

Question 3.
Platinum, palladium NaBH₄, CrO₃.
Answer:
CrO₃ It is an oxidising agent whereas the other three are reducing agents.

Question 4.
Ionic bond, metallic bond, Coordinate covalent bond, Hydrogen bond.
Answer:
Hydrogen bond. It is a weak bond whereas the other three are strong bonds.

Question 5.
Soft & waxy, a bad conductor of electricity, low boiling point, solid at room temperature.
Answer:
Solid at room temperature.
It is the property of ionic compounds whereas the other three are the properties of covalent compounds.

.
VI. Answer in brief :

Question 1.
What is a chemical bond?
Answer:
A chemical bond may be defined as the force of attraction between the two atoms that bind them together as a unit called a molecule.

Question 2.
Write the basic concept of Kossel – Lewis theory.
Answer:
Kossle – Lewis theory is based on the concept of electronic configuration of noble gases.

Question 3.
Define the ionic bond.
Answer:
An ionic bond is a chemical bond formed by the electrostatic attraction between positive and negative ions. The bond is formed between two atoms when one or more electrons are transferred from the valence shell of one atom to the valence shell of the other atom.

Question 4.
The following shows the electronic distribution diagram for the formation of MgCl₂ molecule. Based on this answer the following questions.

(a) Which of the above atom loses electrons to form a cation?
(b) Which of the above atom gain electrons to form an anion?
(c) How many electrons are transferred from Mg to Cl?
(d) Write the name of the anion formed.
(e) Which noble gas configuration do these ions resemble?
(f) Write the electronic configuration of Mg²⁺ & Cl^–
Answer:
(a) Magnesium atom loses electrons to form a cation.
(b) Chlorine atom gains 1 electron to from anion.
(c) Two electrons are transferred from Mg – atom to 2 Cl – atoms (each Cl – atom gains 1e^– from Mg – atom).
(d) Chloride anion (Cl^– )
(e) Mg²⁺ ion resembles noble gas configuration of Neon Cl^– ion resembles noble gas configuration of Argon.
(f) Electronic configuration of Mg²⁺ is 2, 8. and Electronic configuration of Cl^– is 2, 8, 8.

Question 5.
What is covalent bond?
Answer:
Bond which is formed between atoms by the mutual sharing of electrons is known as a covalent bond.

Question 6.
Name of the following:
(a) An element which obtains the noble gas configuration of neon by losing three electrons.
(b) An element which gains two electrons to obtain the noble gas configuration of Neon.
Answer:
(a) Aluminium (Al → Al³⁺ + 3e^–)
(b) Oxygen atom (O + 2e^– → 0^2-)

Question 7.
Identify the following reactions as oxidation/reduction/redox reaction.
(a) Zn + CuSO₄ → Cu + ZnSO₄
(b) CuO+H₂ → Cu+H₂O
(c) 2Mg +O₂  → 2MgO
Answer:
(a) Redox reaction
(b) Reduction
(e) Oxidation

Question 8.
What are oxidising agents? Give an example.
Answer:
Substances which have the ability to oxidise other substances are called oxidizing agents. These are also called electron acceptors because they remove electrons from other substances. ’
Example:

Question 9.
What are reducing agents? Give examples.
Answer:
Substances which have the ability to reduce other substances are called reducing agents. These are also called electron donors because they donate electrons to other substances.
Example: NaBH₄, LiAlH₄ and metals like Palladium, Platinum.

Question 10.
What are redox reactions? Give examples.
Answer:
Both the oxidation and reduction occurs in the same reaction simultaneously is known as a redox reaction. If one reactant gets oxidised, the other gets reduced. Such reactions are called oxidation-reduction reactions or redox reactions.
Ex. 1 : 2 PbO + C → 2 Pb + CO₂
Ex. 2 : Zn + CuSO₄  → Cu + ZnSO₄

Question 11.
Define (a) oxidation (b) reduction reactions : Give examples.
Answer:

Question 12.
What is rancidity?
Answer:
The oxidation reaction in food materials that were left open for a long period is responsible for spoiling of food. This is called rancidity.

Question 13.
Define oxidation number.
Answer:
Oxidation number of an element is defined as the formal charge which an atom of that element appears to have when electrons are counted.

Question 14.
Identify the type of bond in \(\mathrm{NH}_{4}^{+}\)

Answer:

VII. To interpret:

Question 1.
Ionic bond is also called electrostatic bond.
Answer:
In ionic bond formation the bond is formed between the oppositely charged ions and these ions come closer to each other due to electrostatic force of attraction. So ionic bond is also called an electrostatic bond.

Question 2.
Ionic compounds are crystalline solids at room temperature.
Answer:
Ionic compounds are formed because of the strong electrostatic force between cations and anions which are arranged in a well-defined geometrical pattern. Thus ionic compounds are crystalline solids at room temperature.

Question 3.
Covalent compounds have a low melting point.
Answer:
In covalent compounds, atoms are held by a weak force of attraction. When heat is applied, the molecules are readily pulled out and get free movement.

VIII. Assertion and Reason type questions :

Question 1.
Statement (A) : Ionic compounds do not conduct electricity in a solid-state.
Reason (B) : The ions in ionic compounds are tightly held together by a strong electrostatic force of attraction and they can not move freely.
(a) B explains A
(b) B does not explain A
(c) B is wrong A
(d) A is right B is wrong
Answer:
(a) B explains A

Question 2.
Statement (A) : Covalent compounds are bad conductors of electricity.
Reason (B) : Covalent compounds contain charged particles (ions)
a) B explains A
b) B does not explain A
c) Both A & B are right
d) Both A & B are wrong
Answer:
(b) B does not explain A
Reason: Since covalent compounds do not have charged particles (ions), they are bad conductors of electricity.

IX. Find the oxidation number of the elements in the following compounds.

(1) Zn in ZnSO₄
(2) Ca in CaH₂
(3) Mg in MgO
(4) N in NH₃
(5) A1 in AlCl₃
Answer:
(1) ZnSO₄
1(Zn) + 1 (S) + 4 (0) = 0
x + 1(+6) + 4(-2) = 0
x + 6 – 8 = 0
x = + 2
Oxidation number of Zn in ZnSO₄ is +2

(2) CaH₂
1(Ca) + 2 (H) = 0
x + 2 (-1) = 0
x -2 = 0
x = + 2
Oxidation number of Ca in CaH₂ is +2

(3) MgO
1(Mg) + 1 (O) = 0
x + 1 (-2) = 0
x -2 = 0
x = + 2
Oxidation number of Mg in MgO is +2

(4) NH₃
1(N) + 3 (H) = 0
x + 3 (-1) = 0
x – 3 = 0
x = +3
Oxidation number of N in NH₃ is +3
(5) AlCl₃
1(Al) + 3(Cl) = 0
x + 3 (-1) = 0
x – 3 = 0
x = + 3
Oxidation number of Al in AlCl₃ is +3

X. Complete the following table:

Question 1.

Answer:

Question 2.

Answer:

XI. To Match:

Answer:

XII. Answer in detail :

Question 1.
Explain the ionic bond formation in sodium chloride with electron distribution diagram.
Answer:
Formation of Sodium Chloride (NaCl)

(i) The atomic number of Sodium is 11 and its electronic configuration is 2, 8, 1. It has one electron excess to the nearest stable electronic configuration of a noble gas – Neon.

(ii) So sodium has a tendency to lose one electron from its outermost shell and acquire a stable electronic configuration forming sodium cation (Na⁺).

(iii) The atomic number of chlorine is 17 and its electronic configuration is 2, 8, 7. It has one electron less to The nearest stable electronic configuration of a noble gas – Argon.

(iv) So chlorine has a tendency to gain one electron to acquire a stable electronic
configuration forming chloride anion (Cl^–)

(v) When an atom of sodium combines with an atom of chlorine, an electron is transferred from the sodium atom to chlorine atom forming sodium chloride molecule. Thus both the atoms achieve stable octet electronic configuration.

Question 2.
Explain the covalent bond formation in the following molecules.
(a) Chlorine
(b) Nitrogen
(c) Hydrogen
(d) Oxygen
Answer:
(a) Chlorine
Chlorine molecule is formed by two chlorine atoms. Each chlorine atom has seven valence electrons (2,8,7). These two atoms achieve a stable completely filled electronic configuration (octet) by sharing a pair of electrons.

(b) Nitrogen
Nitrogen molecule is formed by two nitrogen atoms. Each nitrogen atom has five valence electrons (2, 5). These two atoms achieve a stable completely filled electronic configuration (octet) by sharing three pair of electrons. Hence a triple bond is formed in between the two atoms.

(c)Hydrogen
Hydrogen molecule is formed by two hydrogen atoms. Both hydrogen atoms contributes one electron each to the shared pair and both atoms acquire stable completely filled electronic configuration.

(d) Oxygen
Oxygen molecule is formed by two oxygen atoms. Each oxygen atom has six valence electrons (2,6). These two atoms achieve a stable electronic configuration (octet) by sharing two pair of electrons. Hence a double bond is formed in between
the two atoms.

Question 3.
Explain the co-ordinate covalent bond formation in between NH₃ → BF₃ molecules.
Answer:
The ammonia molecule gives a lone pair of electrons to boron trifluoride (BF₃) molecule which is electron deficient. Thus a coordinate covalent bond is formed between NH₃ (donor molecule) and BF₃ (acceptor molecule) and is represented by

Question 4.
Write notes on the characteristics of covalent compounds.
Answer:
a. Physical state: Depending on the force of attraction between covalent molecules the bond may be weaker or stronger. Thus covalent compounds exist in gaseous, liquid, and solid forms. Eg. Oxygen-gas; Water-liquid: Diamond-solid.

b. Electrical conductivity: Covalent compounds do not contain charged particles (ions), so they are bad conductors of electricity.

c. Melting point: Except few covalent compounds (Diamond, Silicon carbide), they have relatively low melting points compared to Ionic compounds.

d. Solubility: Covalent compounds are readily soluble in non-polar solvents like benzene (C₆H₆), carbon tetrachloride (CCl₄). They are insoluble in polar solvents like water.

e. Hardness and brittleness: Covalent compounds are neither hard nor brittle. But they are soft and waxy.

f. Reactions: Covalent compounds undergo molecular reactions in solutions and these reactions are slow.

Question 5.
Write notes on the characteristics of ionic compounds.
Answer:
a. Physical state: These compounds are formed because of the strong electrostatic force between cations and anions which are arranged in a well-defined geometrical pattern. Thus ionic compounds are crystalline solids at room temperature.

b. Electrical conductivity: Ionic compounds are crystalline solids and so their ions are tightly held together. The ions, therefore, cannot move freely, so they do not conduct electricity in a solid-state. However, in a molten state and their aqueous solutions conduct electricity.

c. Melting point: the strong electrostatic force between the cations and anions hold the ions tightly together, so very high energy is required to separate them. Hence ionic compounds have high melting and boiling points.

d. Solubility: Ionic compounds are soluble in polar solvents like water. They are insoluble in non-polar solvents like benzene (C₆H₆), carbon tetrachloride (CCl₄).

e. Density, hardness, and brittleness: Ionic compounds have high density and they are quite hard because of the strong electrostatic force between the ions. But they are highly brittle.

f. Reactions: Ionic compounds undergo ionic reactions which are practically rapid and instantaneous.