Inter 2nd Year

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 6(c) Group-17 Elements Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 6(c) Group-17 Elements

Very Short Answer Questions

Question 1.
Which halogen produces O₂ and O₃ on passing through water?
Answer:
Fluorine produces O₂ and O₃ on passing through water.
3F₂ + 3H₂O → 6HF + O₃
2F₂ + 2H₂O → 4HF + O₂

Question 2.
Interhalogen compounds are more reactive than the constituent halogens except for fluorine – explain.
Answer:
Inter halogen compounds are more reactive than halogens. This is because X – X’ bond in interhalogens is weaker than X – X bond in halogens except F – F bond.

Question 3.
What is the use of ClF₃?
Answer:
ClF₃ is very useful fluorinating agent and it is used for the production of VF₆ in the enrichment
of U²³⁵.
U_(s) + 3ClF_3(l) → UF_6(g) + 3ClF_(g)

Question 4.
Write two uses of ClO₂.
Answer:
Uses of ClO₂:

• ClO₂ is highly reactive oxidising agent.
• It is used as bleaching agent for paper pulp and textiles.
• It is used in water treatment.

Question 5.
Why are halogens coloured?
Answer:
Halogens are coloured due to the absorption of radiations in visible region which results in the excitation of outer electrons to higher energy level. Halogens absorb different quanta of radiation and display different colours.

Question 6.
Write the reactions of F₂ and Cl₂ with water. [Mar. 14]
Answer:
Fluorine produces O₂ and O₃ on passing through water.
3F₂ + 3 H₂O → 6HF + O₃
2F₂ +2 H₂O → 4HF + O₂

Chlorine dissolves in water giving a solution of chlorine water. A freshly prepared solution of chlorine water contains HCl and HOCl .
Cl₂ + H₂O → HCl + HOCl
HOCl is unstable and dissociates to give nascent oxygen
HOCl → HCl + (O)

Question 7.
With which neutral molecule, ClO^– is isoelectronic ? Is that molecule a Lewis base ? (Hint: ClF; Yes)
Answer:

• ClO^– is isoelectronic with the neutral molecule ClF.
• Yes. ClF is a Lewis base (Electron pair donar).

Question 8.
Arrange the following in the order of the property indicated for each set.
a) F₂, Cl₂, Br₂, I₂ – increasing bond dissociation enthalpy.
b) HF, HCl, HBr, HI – increasing acidic strength
c) HF, HCl, HBr, HI – increasing boiling points.
Answer:
a) Increasing bond’ dissociation enthalpy order
I₂ < F₂ < Br₂ < Cl₂

b) Increasing acidic strength order
HF < HCl < HBr < HI

c) Increasing boiling points order
HCl < HBr < HI < HF

Question 9.
Electron gain enthalpy of fluorine is less than that of chlorine – explain.
Answer:
The negative electron gain enthalpy of the fluorine is less them that of chlorine. This is due to small size of fluorine atom which results in strong inter electronic repulsions in the relatively small 2p orbitals of fluorine and thus the incoming electron does not experience much attraction.

Question 10.
HF is a liquid while HCl is a gas – explain.
Answer:
HF is a liquid due to the presence of inter molecular hydrogen bonding where as HCl is a gas due to there is no such type of bonding in it.

Question 11.
Bond dissociation enthalpy of F₂ is less than that of Cl₂ – explain.
Ans. Bond dissociation enthalpy of F₂ is less than that of Cl₂
Explanation:
In F₂ molecule electron repulsions are greater among lone pairs because these lone pairs are much closer to each other than in case of Cl₂.

Question 12.
Write the formulae of the compounds, in which oxygen has positive oxidation states and mention the oxidation states of oxygen in them.
Answer:

• In OF₂ and O₂F₂ oxygen has positive oxidation states.
• In OF₂, oxygen oxidation state is + 2.
• In O₂F₂ oxygen oxidation state is + 1.

Question 13.
What is the use of O₂F₂ and I₂O₅?
Answer:
Use of O₂F₅:
O₂F₂ is a fluorinating agent. O₂F₂ oxidises plutonium to PUF₆ and the reaction is used in removing plutonium as PUF₆ from spent nuclear fuel.
Use of I₂O₅:
I₂O₅ is a good oxidising agent and is used in the estimation of carbon monoxide (CO).

Question 14.
Write two uses of hydrogen chloride.
Answer:
Uses of hydrogen chloride :

• It is used in medicines and as a laboratory reagent.
• It is used in the manufacturing of Cl₂, NH₄Cl and glucose (from com starch).
• It is used in extracting glue from bones-and purifying bone black.

Question 15.
Explain the reactions of Cl₂ with NaOH. [T.S. Mar. 16]
Answer:

1. Reaction with cold dilute NaOH: Chlorine reacts with cold dilute NaOH to give sodium hypochlorite and sodium chloride.
   2 NaOH (cold, dil.) + Cl₂ → NaCl + NaOCl + H₂O
2. Reaction with hot concentrated NaOH : Chlorine reacts with hot concentrated NaOH to give sodium chlorate and sodium chloride.
   3 Cl₂ + 6 NaOH (hot, Conc.) → 5 NaCl + NaClO₃ + 3 H₂O

Question 16.
What happens when Cl₂ reacts with dry slaked lime ? [A.P. Mar. 16, 15]
Answer:
Chlorine reacts with dry slaked lime and forms bleaching powder.
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

Question 17.
Chlorine acts as an oxidizing agent – explain with two examples.
Answer:
Chlorine acts as oxidising agent.
Example – 1: Cl₂ oxidises Iodine to Iodate .
I₂ + 6 H₂O + 5 Cl₂ → 2HIO₃ + 10 HCl
Example – 2 : Cl₂ oxidises Sodium Sulphite to Sodium Sulphate
Cl₂ + Na₂SO₃ + H₂O → Na₂SO₄ + 2 HCl

Question 18.
What is aqua regia ? Write its reaction with gold and platinum.
Answer:
A mixture of 3 parts of Cone. HCl and one part of Cone. HNO₃ constitutes aqua regia. It is used for dissolving noble metals.
It’s reaction with gold :
Au + 4H⁺ + \(\mathrm{NO}_3^{-}\) + 4Cl^– → \(\mathrm{AuCl}_4^{-}\) + NO + 2 H₂O
It’s reaction with Platinum
3Pt + 16H⁺ + 4\(\mathrm{NO}_3^{-}\) + 18 Cl^– → \(3 \mathrm{PtCl}_6^{-2}\) + 4 NO + 8H₂O

Question 19.
How is chlorine manufactured by Deacon’s method ? [T.S. & A.P. Mar. 19 & T.S. Mar. 16]
Answer:
Deacon’s process : In Deacon’s process chlorine is obtained by the oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl₂ (catalyst) at 723 K.

Question 20.
Chlorine acts as a bleaching agent only in the presence of moisture – explain.
Answer:
Moist chlorine is a powerful bleaching agent. This bleaching property is due to oxidation.
Cl₂ + H₂O → 2HCl + (O)
Ex : Coloured substance + (O) → colourless substance.

Question 21.
The decreasing order of acidic character among hypohalogen acids is HClO > HBrO >HIO. Give reason.
Answer:
Given the decreasing order of acidic character among hypohalogen acids is

HaO > HBrO > HIO
From the above mentioned K_a values the given order of hypohalogen acids – acid strength order is
HClO > HBrO > HIO

Question 22.
The acidic nature of the oxoacids of chlorine is
HOCl < HClO₂ < HClO₃ < HClO₄ – explain.
(Hint: HA + H₂O ⇌ H₃O⁺ + A^– conjugate base, greater the stability of A^–, lesser will be its basic strength or greater will be the tendency of HA to release H⁺. In other words, stronger will be the acid HA. Among the conjugate bases of oxoacids of chlorine, the stability order is HOCl^– < \(\mathrm{ClO}_2^{-}\) > \(\mathrm{ClO}_3^{-}\) > \(\mathrm{ClO}_4^{-}\))
Answer:
Given the acidic nature of oxoacids of chlorine is
HOCl^– < H\(\mathrm{ClO}_2^{-}\) > H\(\mathrm{ClO}_3^{-}\) > \(\mathrm{ClO}_4^{-}\)
HA + H₂O ⇌ H₃O⁺ + A^– conjugate base, greater the stability of A^–, lesser will be its basic strength or greater will be the tendency of HA to release H⁺. In other words, stronger will be the acid HA. Among the conjugate bases of oxoacids of chlorine, the stability order is
HOCl^– < H\(\mathrm{ClO}_2^{-}\) > H\(\mathrm{ClO}_3^{-}\) > H\(\mathrm{ClO}_4^{-}\)

Question 23.
What are interhalogen compounds ? Give two examples.
Answer:
The binary diamagnetic compounds of halogens which are formed by the reaction of halogens among themselves are called interhalogen compounds.
E.g.: IF₇, ClF₃, BrF₃, ClF, IF₃ etc.

Question 24.
Explain the structure of ClF₃.
Answer:
Structure of ClF₃:

• Central atom in ClF₃ is ‘Cl’
• Excited state electronic configuration of ‘Cl’ is
  
• Cl atom undergoes sp³d hybridisation
• It is a bent T-shaped molecule (or) trigonal bipyramidal with 2 – positions occupied by lone pairs.

Question 25.
OF₂ should be called oxygen difluolide and not fluorine oxide – Why ?
Answer:
OF₂ should be called oxygen difluoride and riot fluoride oxide.
The binary compotmds of oxygen with fluoride are not pronounced as oxides because the electronegativity of fluorine is greater than oxygen.

Question 26.
Iodine is more soluble in KI than in water – Explain.
Answer:
Iodine is more soluble in KI than in water.

Reason:
Iodine combine with KI forming soluble complex KI₃.
KI + I₂ → KI₃ (Soluble complex)
Iodine does not dissolve in water. This is due to positive free energy change (+ ∆G)

Question 27.
Among the hydrides of halogens
a) Which is most stable ?
b) Which is most acidic ?
c) Which has lowest boiling point ?
Answer:
a) Among hydrides of halogens HF is most stable.
b) Among hydrides of halogens HI is most acidic.
c) Among hydrides of halogens HCl (189K) has lowest boiling point.

Question 28.
Compare the bleaching action of Cl₂ and SO₂.
Answer:

• Cl₂ is a powerful bleaching agent. This bleaching action is due to oxidation.
  Cl₂ + H₂O → 2HCl+ (O)
  Coloured substance + (O) → colourless substance.
• It bleaches vegetable or organic matter in the presence of moisture. Bleaching effect of Cl₂ is permanent.
• SO₂ acts as bleaching agent in presence of moisture.
  SO₂ + 2 H₂O → H₂SO₄ + 2[H]
  
• SO₂ bleaches wool and silk.

Question 29.
Give the oxidation states of halogens in the following :
a) Cl₂O
b) \(\mathrm{ClO}_2^{-}\)
c) KBrO₃
d) NaClO₄
Answer:
a) Cl₂O :
2x – 2 = 0
x = -+ 1
Oxidation state of chlorine in Cl₂O is + 1.

b) \(\mathrm{ClO}_2^{-}\)
x + 2(-2) = -1
x = -1 + 4 = +3
In latex]\mathrm{ClO}_2^{-}[/latex] the oxidation state of ‘Cl’ is + 3

c) KBrO₃ :
1 + x + 3 (-2) = 0
x = + 5
In KBrO₃ the oxidation state of Br is + 5.

d) NaClO₄
1 + x + 4(-2) = 0
x = + 7
In NaClO₄ the oxidation state of Cl is + 7.

Question 30.
Describe the molecular shape of \(\mathrm{I}_3^{-}\) .
(Hint: Central iodine is of sp³d. – linear)
Answer:

• In Tri iodide ion central iodine atom undergo sp³d hybridisation.
• It contains three lone pairs and two bond pairs.
• According to VSEPR theory it has linear shape.
  

Short Answer Questions

Question 1.
How can you prepare Cl₂ from HCl and HCl from Cl₂ ? Write the reactions.
Answer:
Preparation of Cl₂ from HCl:

• On heating MnO₂ with Conc. HCl, Cl₂ gas is liberated
  MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
• By the Õxidation of HCl gas by atmospheric oxygen in presence of CuCl₂ catalyst at 723 K.
  4 HCl + O₂ 2Cl₂ + 2 H₂O

Preparation of HCl from Cl₂:

• Cl₂ when reacted with H₂ to form HCl
  H_2(g) + Cl_2(g) → 2HCl_(g)

Question 2.
Write balanced equations for the following.
a) NaCl is heated with Conc.H₂SO₄ In the presence of MnO₂.
b) Chlorine is passed into a solution of Nal ¡n water.
Answer:
a) NaCl is heated with Conc.H₂SO₄ in presence of MnO₂ liberates Cl₂ gas.
4 NaCl + MnO₂ + 4 H₂SO₄ → MnCl₂ + 4 NaHSO₄ + 2 H₂O + Cl₂.

b) When chlorine water is added to a solution of sodium iodide a brown colour is formed.
Cl₂ + 2 NaI → 2 NaCl + I₂

Question 3.
Explain the structures of
a) BrF₅ and
b) IF₂.
Answer:
a) Structure of BrF₅.

• Central atom in BrF₅ is Br.
• ‘Br’ undergoes sp³d² hybridisation in 2^nd excited state.
  
• Shape of the molecule is octahedral with one position occupied by a lone pair (or) square pyramidal.

b) Structure of IF₇ :

• Central atom in IF₇ is T.
• T undergoes sp³d³ hybridisation in 3^rd excited state.
  
• Shape of the molecule is Penta gonal bipyramid structure
  

Question 4.
Write a short note on the hydrides of halogens.
Answer:
Hydrides of halogens :
Formation :

These compounds dissolve in water to form hydrohalic acids.
Boiling points HF – 293 K
HCl – 189K
HBr – 206 K
HI – 238 K
The decreasing order of stability of these compounds is
HF > HCl > HBr > HI
The increasing order of acidic strength of these compounds
HF < HCl < HBr < HI

Question 5.
How is chlorine obtained in the laboratory ? How does it react with the following ? [T.S. Mar. 15]
a) cold dil. NaOH
b) excess NH₃
c) KI
Answer:
In the laboratory chlorine is prepared by the oxidation of HCl with MnO₂.
4 HCl + MnO₂ → MnCl₂ + 2 H₂O + Cl₂ ↑

a) Chlorine reacts with cold dil. NaOH to form sodium hypochlorite.
Cl₂ + 2 NaOH → NaCl + NaOCl + H₂O

b) Cl₂ reacts with excess of NH3, Nitrogen and ammonium chloride are formed.
8 NH₃ + 3 Cl₂ → 6 NH₄Cl + N₂ ↑

c) Cl₂ reacts with KI to liberate iodine
Cl₂ + 2KI → 2 KCl + I₂ ↑

Question 6.
What are interhalogen compounds ? Give some examples to illustrate the definition. How are they classified ?
Answer:
The binary diamagnetic compounds of halogens which are formed by the reaction of halogens among themselves are called interhalogen compounds.
E.g.: IF₇, ClF₃, BrF₃, ClF, IF₃ etc.
The above examples are binary diamagnetic compounds and formed by* combination of halogens only. .
Inter halogen compounds are classified into four types.

1. AX – Type : Eg : ClF, BrF
2. AX₃ – Type : Eg: ClF₃, IF₃
3. AX₅ – Type : Eg : ClF₅, BrF₅
4. AX₇ – Type : Eg : IF₇

‘A’ is less electronegative halogen.
X is more electronegative halogens.

Long Answer Questions

Question 1.
How is ClF₃ prepared ? How does it react with water ? Explain its structure.
Answer:
Preparation of ClF₃: Chlorine reacts with excess of fluorine to form ClF₃.

Reaction with H₂O: ClF₃ reacts with water explosively and oxidises water to give oxygen or in controlled quantities oxygen diflouride (OF₂) as well as HF and HCl.
ClF₃ + 2H₂O → 3 HF + HCl + O₂
ClF₃ + H₂O → 4 HF + HCl + OF₂
Structure of ClF₃:

• Central atom in ClF₃ is ‘Cl’
• Excited state electronic configuration of ‘Cl’ is
  
• Cl atom undergoes sp³d hybridisation
• It is a bent T-shaped molecule (or) trigonal bipyramidal with 2 – positions occupied by lone pairs.

Question 2.
How is chlorine prepared in the laboratory ? How does it react with the following ? [T.S. Mar. 19, 15]
a) Iron
b) hot, cone. NaOH
c) acidified FeSO₄
d) Iodine
e) H₂S
f) Na₂S₂O₃
Answer:
In the laboratory chlorine is prepared by the oxidation of HCl with MnO₂.
4 HCl + MnO₂ → MnCl₂ + 2 H₂O + Cl₂ ↑

a) Cl₂ reacts with Iron to form FeCl₃
2 Fe + 3Cl₂ → 2 FeCl₃

b) Cl₂ reacts with hot Cone. NaOH to form sodium chloride and sodium chlorate
3Cl₂ + 6 NaOH → 5 NaCl + NaClO₃ + 3 H₂O

c) Cl₂ reacts with acidified FeSO₄ and ferric ions are formed.
2 FeSO₄ + H₂SO₄ + Cl₂ → Fe₂(SO₄)₃ + 2 HCl

d) Cl₂ reacts with Iodine to form ICl

e) Cl₂ reacts with H₂S and forms HCl and ‘S’
Cl₂ + H₂S → 2 HCl + S

f) ‘S’ is precipitated by the reaction of Cl₂ with Na₂S₂O₃.
‘Na₂S₂O₃ + Cl₂ + H₂O → Na₂SO₄ + 2 HCl + S

Question 3.
Discuss the anomalous behaviour of fluorine.
Answer:
Abnormal behaviour of fluorine : Fluorine being the first element in the group differs considerably from other halogens. The reasons may be traced out to one or more of the following, characteristics.
F₂ has

1. a small size
2. The highest electronegativity amongst the known elements.
3. No d-orbital are available in its valence shell.
4. Low enthalpy of F-F bond dissociation and
5. Only two electrons are present in the penultimate shell while other halogens 8 electrons.

The abnormal characteristics of F₂ can be summarised as follows.

1. F₂ exhibits -1 oxidation state because it is the most electronegative element known. Therefore, no higher oxidation state for it in its compounds.
2. In it’s hydride HF, it has hydrogen bonding, but other halides of hydrogen do not show this property. HF can form \(\mathrm{HF}_2^{-}\) ion. No such ions are known with other halogens.
3. It combines with carbon while others do not combine even under drastic conditions.
4. F₂ has a lower EA compared to that of Cl₂ even though F₂ is the most electronegative element.
5. Fluorides have the highest ionic character among the halides.
   Example : AlF₂ is ionic while AlCl₃ is covalent compound.

Question 4.
How is chlorine prepared by electrolytic method ? Explain its reaction with [A.P. Mar. 16, 15]
a) NaOH and
b) NH₃ under different conditions.
Answer:
Preparation of chlorine by electrolytic method : Chlorine is obtained by the electrolysis of brine solutions (Cone. NaCl). Cl₂ gas is liberated at anode.
2 NaCl → 2 Na⁺ + 2Cl^–
2 H₂O + 2e^– → 2 OH^– + H₂ (cathode)
2 Cl^– → Cl₂ + 2e^– (anode)
a)

1. NaOH:
   Chlorine reacts with cold dil. NaOH to form sodium hypochlorite.
   Cl₂ + 2 NaOH → NaCl + NaOCl + H₂O
2. Cl₂ reacts with hot cone. NaOH to form sodium chloride and sodium chlorate.
   3 Cl₂ + 6 NaOH → 5 NaCl + NaClO₃ + 3 H₂O

b)

1. Cl₂ reacts with excess of NH₃, Nitrogen and ammonium chlorate are formed.
   8 NH₃ + 3 Cl₂ → 6 NH₄Cl + N₂ ↑
2. NH₃ reacts with excess of Cl₂ to form NCl₃ and HCl.
   

Question 5.
Write the names and formulae of the oxoacids of chlorine. Explain their structures and relative acidic nature.
Answer:
Four oxyacids of chlorine are known. They are
Hypochlorous acid – HOCl
Chlorous acid – HClO₂
Chloric acid – HClO₃
Perchloric acid – HClO₄
Structure of HClO : In this chlorine atom is sp³ hybridised. Outer electronic configuration of Cl in ClO^– after sp³ hybridisation.

Shape is tetrahedral with 3 lone pairs (or) linear.
No π – bonds.

Chlorous acid : (HClO₂) : Chlorine is in sp³ hybrid state, in first excited state. Shape is tetrahedral with 2 lone pairs (or) angular one π_d-p bond is present.
First excited state

Chloric acid (HClO₃) : The central chlorine atom undergoes sp³ hybridisation in second excited state.

Shape is tetrahedral with one lone pair (or) pyramidal.
Two π_d-p bonds present.

Perchloric acid (HClO₄) : The central chlorine atom undergoes sp³ hybridisation in third excited state.

Shape is perfect tetrahedral. No lone pairs.
Three π_d-p bonds present.

Textual Examples

Question 1.
Halogens have maximum negative electron gain enthalpy in the respective periods of the periodic table. Why ?
Solution:
Halogens have the smallest size in their respective periods and therefore high effective nuclear charge. As a consequence, they readily accept one electron to acquire noble gas electronic configuration.

Question 2.
Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidising agent than chlorine. Why ?
Solution:
It is due to

1. Low enthalpy of dissociation of F-F bond
2. High hydration enthalpy of F^–

Question 3.
Fluorine exhibits only – 1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation states also. Explain.
Solution:
Fluorine is the most electronegative element and cannot exhibit any positive oxidation .state. Other halogens have d orbitals and therefore, can expand their octets and show +1, +3, +5 and +7 oxidation states also.

Question 4.
Write the balanced chemical equation for the reaction of Cl₂ with hot and concentrated NaOH. Is this reaction a disproportionation reaction ? Justify.
Solution:
3 Cl₂ + 6 NaOH → 5NaCl + NaClO₃ + 3H₂O
Yes. Chlorine from zero oxidation state is changed to -1 and +5 oxidation states.

Question 5.
When HCl reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Why ?
Solution:
Its reaction with iron produces H₂.
Fe + 2HCl → FeC₂ + H₂
Liberation of hydrogen prevents the formation of ferric chloride.

Question 6.
Discuss the molecular shape of BrF₃ on the basis of VSEPR theory.
Solution:
The central atom Br has seven electrons in the valence shell. Three of these will form electron pair bonds with three fluorine atoms leaving behind four electrons. Thus, there are three bond pairs and two lone pairs.

According to VSEPR theory, these will occupy the comers of a trigonal bipyramid. The two lone pairs will occupy the equatorial positions to minimise lone pair- lone pair and the bond pair-lone pair repulsions which are greater than the bond pair-bond pair repulsions. In addition, the axial flourine atoms will be bent towards the equatorial fluorine in order to minimise the lone pair-lone pair repulsions. The shape would be that of a slightly bent T.

Intext Questions

Question 1.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and Cl₂.
Oxidizing power is a combined effect of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
Solution:
Comparing F₂ and Cl₂ with the given parameters.

From the data given above, it is clear that the bond dissociation enthalpy and electron gain enthalpy are higher for chlorine but hydration energy is much higher for fluorine. It compensates the effect of other two and thus, makes flourine more oxidizing than chlorine.

The relative oxidizing power of the halogens can be further illustrated by their reactions with water.
2 F₂(g) + 2 H₂O(l) → 4H⁺ (aq) + 4F^– (aq) + O₂(g) ……………. (i)
Cl₂(g) + H₂O(l) → HCl (aq) + HOCl (aq) ……………… (ii)

Question 2.
Give two examples to show the anomalous behaviour of fluorine.
Solution:
The two anomalous behaviour of flourine are as follows :
a) It forms only one oxo-acid while other halogens form a number of oxo-acids.
b) Hydrogen flouride (HF) is a liquid (b.p. 293 K) due to strong hydrogen bonding whereas other hydrogen halides are gases.

Question 3.
Sea is the greatest source of some halogens. Comment.
Solution:
Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and calcium. NaCl is the most abundant (2.5% by mass) among all of these. Certain forms of marine life contain iodine in their systems; Various sea weeds, for example, contain upto . 0.5% of iodine and chile saltpetre contains upto 0.2% of sodium iodate. NaCl and camalite, KCl, MgCl₂. 6H₂O etc are present in the deposits of dried up seas.

Question 4.
Give the reason for bleaching action of Cl₂ ?
Solution:
Bleaching action of chlorine is due to its oxidizing property. When chlorine reacts with water, it gives nascent oxygen which decoloures is the coloured substance.
Cl₂ + H₂O → 2HCl + [O]
Coloured substance + [O] → Colourless substance
Bleaching action of chlorine creates permanent effect. It bleaches the vegetable or organic matter in the presence of moisture.

Question 5.
Name some poisonous gases which can be prepared from chlorine gas.
Solution:

1. Phosgene (COCl₂)
2. Tear gas (CCl₃NO₂)
3. Mustard gas (ClCH₂CH₂SCH₂CH₂Cl)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(e) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(e)

I. Find the I.F. of the following differential equations by transforming them into a linear form.

Question 1.
x\(\frac{dy}{dx}\) – y = 2x² sec² 2x
Solution:
x\(\frac{dy}{dx}\) – y = 2x² sec² 2x
\(\frac{dy}{dx}\) – \(\frac{1}{x}\) . y = 2x. sec² 2x
This is linear in x
I.F = e^{∫-\(\frac{1}{x}\)dx} = e^{-log x} = e^{log 1/x} = \(\frac{1}{x}\)

Question 2.
y\(\frac{dy}{dx}\) – x = 2y³
Solution:
y\(\frac{dy}{dx}\) – x = 2y³
\(\frac{dy}{dx}\) – \(\frac{1}{y}\).x = 2y²
I.F = e^{∫-\(\frac{1}{x}\)dx} = e^{-log y} = e^{log 1/y} = \(\frac{1}{y}\)

II. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}\) + y tan x = cos³ x
Solution:

Solution is 2y = x cos x + sin x. cos² x + c. cos x

Question 2.
\(\frac{dy}{dx}\) + y sec x = tan x
Solution:
I.F. = e^{∫sec x dx} = e^{log(secx + tan x)} = sec x + tan x
y. (sec x + tan x)
= ∫tan x (sec x + tan x)dx
= ∫(sec x. tan x + tan² x)dx
= ∫(sec x tan x + sec² x – 1)dx
Solution is
y(sec x + tan x) = sec x + tan x – x + c

Question 3.
\(\frac{dy}{dx}\) – y tan x = e sec x. dx
Solution:
I.F. = e^{-∫tan x dx} = e^{log cos x} = cos x
y. cos x = ∫e^x.sec x. cos x dx = ∫ e^x dx
= e^x + c

Question 4.
x\(\frac{dy}{dx}\) + 2y = log x.
Solution:
I.F. = e^{∫\(\frac{2}{x}\)dx} = e^{2log x} = e^{log x²} = x²
Solution is

Question 5.
(1 + x²)\(\frac{dy}{dx}\) + y = e^{tan-1 x}
Solution:

Question 6.
\(\frac{dy}{dx}+\frac{2y}{x}\) = 2x².
Solution:
I.F. = e^{∫\(\frac{2}{x}\)dx} = e^{2log x} = e^{log x²} = x²
y. x² = ∫2x⁴ dx = \(\frac{2x^5}{5}\) + c

Question 7.
\(\frac{dy}{dx}+\frac{4x}{1+x^2}y=\frac{1}{(1+x^2)^2}\)
Solution:

Question 8.
x\(\frac{dy}{dx}\) + y = (1 + x)e^x
Solution:

Question 9.
\(\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{1+x^2}{1+x^3}\).
Solution:

Question 10.
\(\frac{dy}{dx}\) – y = -2e^-x.
Solution:
I.F = e^∫-dx = e^-x
y. e^-x = -2∫e^-2x dx = e^-2x + c
y = e^-x + c. e^x

Question 11.
(1 + x²)\(\frac{dy}{dx}\) + y = tan^-1 x.
Solution:

Put t = tan^-1 x so that dt = \(\frac{dx}{1+x^2}\)
R.H.S = ∫t. e^tdt = t. e^t – ∫e^tdt
= t. e^t – e^t = e^t(t – 1)
Solution is y. e^{tan-1 x} = e^{tan-1 x} (tan^-11 x – 1) + c
y = tan^-1 x – 1 + c. e^{-tan-1 x}

Question 12.
\(\frac{dy}{dx}\) + y tan x = sin x.
Solution:
I.F. =e^{∫tan x dx} = e^{log sec x} = sec x
y. sec x = ∫ sin x. sec x dx = ∫tan x dx
= log sec x + c

III. Solve the following differential equations.

Question 1.
cos x. \(\frac{dy}{dx}\) + y sin x = sec² x
Solution:
\(\frac{dy}{dx}\) + tan x. y = sec³ x
I.F. = e^{∫tan x dx} = e^{log sec x} = sec x
y. sec x = ∫sec⁴ x dx = ∫(1 + tan² x) sec² x
dx = tan x + \(\frac{\tan^3x}{3}\) + c

Question 2.
sec x. dy = (y + sin x) dx.
Solution:
\(\frac{dy}{dx}=\frac{y+sin x}{sec x}\) = y cos x + sin x. cos x
\(\frac{dy}{dx}\) – y. cos x = sin x. cos x
I.F. = e^{-∫cos x dx} = e^{– sin x}
= y. e^{-sin x} = ∫ e^{-sin x}. sin x. cos x dx ……….. (1)
Consider ∫e^{-sin x}. sin x. cos x dx
t = – sin x ⇒ dt = -cos x dx
∫e^{-sin x}. sin x. cos x dx = + ∫e^t t dt
= t. e^t – e^t + c.
= e^{-sin x} (- sin x – 1) + c
y. e^{-sin x} = – e^{-sin x} (sin x + 1) + c
or y = – (sin x + 1) + c. e^{sin x}.

Question 3.
x log x.\(\frac{dy}{dx}\) + y = 2 log x.
Solution:

Question 4.
(x + y + 1)\(\frac{dy}{dx}\) = 1.
Solution:
\(\frac{dy}{dx}\) = x + y + 1
\(\frac{dy}{dx}\) – x = y + 1
I.F. = e^∫-dy e^-y
x . e^-y = ∫e^-y (y + 1) dy
= -(y + 1). e^-y + ∫e^-y. dy
= -(y + 1) e^-y – e^-y
= -(y + 2) e^-y+ c
x = -(y + 2) + c. e^-y

Question 5.
x(x – 1)\(\frac{dy}{dx}\) – y = x³(x – 1)³
Solution:
\(\frac{dy}{dx}-\frac{1}{x(x-1)}\)y = x²(x – 1)²

Question 6.
(x + 2y³)\(\frac{dy}{dx}\) = y
Solution:

Solution is x = y(y² + c)

Question 7.
(1 – x²)\(\frac{dy}{dx}\) + 2xy = x\(\sqrt{1-x^2}\)
Solution:

Question 8.
x(x – 1)\(\frac{dy}{dx}\) – (x – 2)y = x³(2x – 1)
Solution:


Solution is y(x – 1) = x²(x² – x + c)

Question 9.
\(\frac{dy}{dx}\)(x²y³ + xy) = 1
Solution:
\(\frac{dy}{dx}\) = xy + x²y³
This is Bernoulli’s equation

Question 10.
\(\frac{dy}{dx}\) + x sin 2y = x³ cos² y
Solution:

Question 11.
y² + (1 – \(\frac{1}{y}\)).\(\frac{dy}{dx}\) = 0
Solution:


Solution xy = 1 + y + cy.e^1/y

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 3(b) Neural Control and Coordination Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 3(b) Neural Control and Coordination

Very Short Answer Questions

Question 1.
Name the cranial meninges covering the brain of a man.
Answer:
The brain is covered by three connective tissue membranes called meninges.

1. Dura mater
2. Arachnoid mater
3. Pia mater.

Question 2.
What is Corpus callosum?
Answer:
Two cerebral hemispheres are internally connected by a transverse, wide, and flat bundle of myelinated fibres beneath the cortex called the corpus callosum.

Question 3.
What do you know about arbor vitae?
Answer:
The white matter of the cerebellum is a branched, tree-like appearance. Hence it is called arbor vitae and is surrounded by a sheath of grey matter.

Question 4.
Why the sympathetic division is called the thoracolumbar division?
Answer:
The preganglionic sympathetic neurons have their cell bodies in the grey matter of the thoracic and lumber regions of the spinal cord. So, the sympathetic division is called the thoracolumbar division.

Question 5.
Why the parasympathetic division is called the Cranio sacral division?
Answer:
The cell bodies of the paraganglionic neurons of the parasympathetic division are located in the brain and in the sacral region of the spinal cord. Hence, the parasympathetic is also known as the Cranio sacral division.

Question 6.
Distinguish between the absolute and relative refractory periods.
Answer:
Absolute refractory period:
During the absolute, refractory period, even a very strong stimulus cannot initiate a second action potential. This period coincides with the period of depolarization and repolarization.

Relative refractory period:
It is the time during which a second action potential can be initiated by a larger than normal stimulus. It coincides with the period of hyperpolarization.

Question 7.
What is all-or-none principle?
Answer:
The action potential occurs in response to a threshold stimulus (or) supra threshold stimulus but does not occur at subthreshold stimuli. It means the nerve impulse is either conducted totally (or) not conducted at all and this called all-or-none principle.

Question 8.
How do rods and cones of human eye differ from each other chemically and functionally?
Answer:
Rods:
Rods contain a purplish red protein called rhodopsin, which contains a derivative of vitamin A. Rods are concerned with dim light.

Cones :
Cones contain a visual pigment called iodopsin, made of a protein called photopsin and they are important in daylight vision and colour vision.

Question 9.
Distinguish between the blind spot and the yellow spot.
Answer:
Blind Spot :
The region of the retina where the optic nerve exists the eyeball and devoid of rods and cones is called blind spot.

Yellow spot:
The centre of the posterior portion of the retina is called yellow spot.

Question 10.
What is organ of corti?
Answer:
The hearing apparatus that is present in the middle canal of the cochlea is called organ of corti. The organ of corti contains hair cells that act as auditory receptors.

Short Answer Questions

Question 1.
Draw a labelled diagram of the T.S. of the spinal cord of man.
Answer:

Question 2.
Distinguish between somatic and autonomic neural systems.
Answer:

Question 3.
Give an account of the retina of human eye.
Answer:
Retina is the inner coat of the eye. It consist of a pigmented epithelium and a neural portion. The pigmented epithelium is a sheet of melanin containing epithelial cells. The neural portion has three layers namely photoreceptor Iaydr, bipolar cell layer and ganglion cell layer.

Photoreceptor layer consist of rods and cones. Rods contain a protein called rhodopsin. Rods are concerned with dim light. Cones contain a visual pigment called iodopsin and they are important in daylight vision and colour vision. There are three types of cones and are response to red, green and blue colours.

The centre of the posterior portion of the retina is called yellow spot. A depression present in the yellow spot is called ‘Forea’ contractile and it contains only cones. Forea is responsible for sharp vision. The region of retina which is devoid of rods and cones is known as blind spot (or) optic disc, which form the optic nerve called 2nd cranial nerve.

Question 4.
Give an account of synaptic transmission.
Answer:
A nerve impulse is transmitted from one neuron to another through junction called synapses.

There are two types of synapses. 1) Electrical synapses 2) Chemical synapses.

Electrical synapses :
These synapses are electrically conductive links between two neurons and are also called “gap junctions”. Impulses transmission across an electrical synapses is always faster than that across a chemical synapses.

Chemical synapses :
Chemicals called neuro transmitters are involved in the transmission of impulses at those synapses. When an impulse arrives at the axon terminal, it depolarizes the membrane opening voltage gated calcium channels. Calcium ions stimulate the release of neurotransmitters in the cleft by exocytosis. The released neurotransmitters bind to their specific receptors, present on the post synaptic membrane.

The post synaptic membrane has ligand gated channels. They are ion channels which respond to chemical signals, rather than to changes in the membrane potential. The entry of ions can generate a new potential in the post synaptic neuron. The new potential developed may be either excitatory (or) inhibitory.

Excitatory post synaptic potentials cause depolarisation, where as inhibitory post synaptic potentials cause hyper polarisation of post synaptic membrane.

Question 5.
List out the differences between Sympathetic and Parasympathetic neural system in man.
Answer:

Long Answer Questions

Question 1.
Give a brief account of the structure and functions of the brain of man.
Answer:
Brain is the site of information, processing and control. It is protected in the cranial cavity and covered by three cranial meninges namely duramater (outer layer), arachnoid mater (thin middle layer) and piamater (inner layer).
The brain can be divided into three major parts called

1. Fore brain
2. Mid brain
3. Hind brain.

1) Fore brain :
The fore brain consists of i) Olfactory bulb ii) Cerebrum and iii) Dience-phalon.

i) Olfactory bulb :
Which receives impulses pertaining to smell from the Olfactory epithe-lium.

ii) Cerebrum :
Cerebrum forms the major part of the human brain. A deep cleft divides the cerebrum longitudinally into two halves, which are termed as the left and right cerebral hemispheres. The hemispheres are connected by a transverse, wide and flat bundle of myelinated fibres beneath the cortex, called corpus callosum. It brings the coordination between the left and right sides of the hemispheres. The surface of the cerebral cortex shows many folds and grooves. The folds are called gyri, the deepest and shallower grooves between folds are called fissures and sulci respectively.

The cerebral cortex contain three functional areas called
a) Sensory areas : receive and interpret the sensory impulses.
b) Motor areas : which control volutntary muscular movements.
c) Association areas : which are neither clearly sensory nor motor in function, they deal integrative functions, such as memory and communications.

The cerebral medulla consist of mostly myelinated axons. Each cerebral hemisphere of the cerebrum is divided into four lobes namely frontal, parietal, temporal and occipital lobes.

iii) Diencephalon :
It contains three main parts namely, a) Epithalamus, b) Thalamus and c) Hypothalamus.
a) Epithalamus :
It is the roof of the diencephalon. It is axon nervous part which is fused with the pia matter to form the anterior choriod plexus. The epithelium of the epithalamus forms a pineal stalk, which ends in a rounded structure called pineal body.

b) Thalamus :
It lies superior to the mid brain. It is the major coordination centre for sensory and motor signalling.

c) Hypothalamus :
It lies at the base of the thalamus. The hypothalamus forms a funnel-shaped downward extension called infundibulum, connecting the hypothalamus with the pituitary gland. It also contains a group of neuro-secretory cells, which secrete hormones called hypothalamic hormones.

Hypothalamus controls and integrates the activities of the autonomous nervous system and it has osmoregulatory, thermoregulatory, thirst, feeding the satiety centres.

Limbic system :
The inner part of cerebral hemisphere and group of associated structures forms limbic system. Limbic system along with hypothalamus is involved in the regulation of sexual behaviour and expression of emotional reactions.

2) Mid brain :
Mid brain is located between the thalamus of the fore brain and pons varolii of hind brain. The ventral portion of mid brain consists of a pair of longitudinal bands of nervous tissues called cerebral peduncles. The dorsal portion of the mid brain consists of four lobes called corpora quadrigmina. The two larger anterior lobes are called superior colliculi, which are concerned with visual function. The smaller posterior lobes are called inferior colliculi and are concerned with auditory functions.

3) Hind brain :
The hind brain comprises of cerebellum, pons varolii and medulla oblongata.
i) Cerebellum :
It is the second largest part of the brain. It consists of two cerebellar hemispheres and a central vermis. Each cerebellar hemisphere consists of three lobes namely anterior, posterior and floccular lobes. It has a branching tree like core of white matter called arbor vitae.

ii) Pons Varolii :
It consists of nerve fibres which form a bridge between the two cerebellar hemispheres. It is a relay station between the cerebellum, spinal cord and the rest of the brain. Pons has the pneumotaxic centre as it regulates the amount of air a person can take in each time.

iii) Medulla Oblongata :
It is the posterior part of brain. It extends from the Pons Varolii above and continuous with the spinal cord below. Medulla includes cardiovasicular, and respiratory centers, the centers for swallowing, vomiting, coughing, sneezing and hiccupping. The mid brain, pons and medulla Oblongata are collectirel referred to as brain stem.

Question 2.
Explain the transmission of nerve impulse through a nerve fibre with the help of suitable diagrams.
Answer:
Nerve impulse is the combination of mechanical, chemical (or) electrical disturbances occur in neuron because of stimulus. The propagation of a impulse along nerve fibre is called transmission. In this process both physical and chemical changes are involved. The entire process is divided into stimulation, excitation, conduction and response.

Resting membrane potential :
The resting membrane potential exists because of a small buildup of negative ions in the axoplasm along the inside of the membrane and an equal buildup of positive ions in the extra cellular fluid along the outer surface of the membrane. Such a Separation of positive and negative electrical chafges is a form of potential energy. In neurons, the resting membrane potential ranges from -40 to -90 mV. A typical value is-70 mV.

At resting phase, the axolemma is polarized. If the inner side becomes less negative, it is said to be depolarized. If the inner side becomes more negative, it is said to be hyperpolarized. During the resting phase the activation gates of sodium are closed, the inactivation gates of sodium are open and the activation gates of potassium are closed.

Sodium-potassium pump : Sodium and potassium ions diffuse inwards and outwards, respectively, down their concentration gradients through leakage channels. Such a movement of ions, if unchecked, would eventually disturb the resting membrane potential. These flows of ions are offset by sodium-potassium pumps (Na⁺/K⁺ ATPases) present in the axonal walls. These pumps expel three Na⁺ ions for each two K⁺ ions imported. As these pumps remove more positive charges from the axoplasm than they bring into it, they contribute to the negativity of the resting membrane potential i.e.,-70mv.

Depolarization (Rising phase):
When a nerve fibre is stimulated, the plasma membrane becomes more permeable to Na⁺ ions than to K⁺ ions as the activation and inactivation voltage gates of sodium open and activation voltage gates of potassium close. As a result the rate of flow of Na+ into the axoplasm exceeds the rate of flow of K⁺ to the ECF. Hence, the axolemma is positively charged inside and negatively charged outside. This reversal of electrical charge is called “depolarization”.

Outer face of the point which is adjacent to the site of depolarization remains positively charged. The electrical potential difference between these two areas is called “action potential”. An action potential occurs in the membrane of the axon of a neuron when depolarization reaches a certain level called ‘threshold potential1 (-55 mV). The particular stimulus which is able to bring the membrane potential to threshold is called ‘threshold stimulus’.

Repolarization (Falling phase) :
As the wave of depolarization passes away from its site of origin to the adjacent point, the activation gates of sodium remain open, inactivation gates of sodium close and activation gates of potassium open at the site of origin of depolarization. As a result the influx of Na⁺ ions into the axoplasm from the ECF is checked and ‘efflux’ of K⁺ ions occurs, which leads to the returning of axolemma to the resting state (exit of potassium ions causes a reversal of membrane potential to negative inside). This is called ‘repolarization’.

Hyperpolarization (Undershoot):
The repolarization typically goes more negative than the resting potential to about -90 mV This is called ‘hyperpolarization’. This occurs because of the increased K⁺ permeability that exists while voltage gated K⁺ channels are open activation and inactivation gates of Na⁺ channels remain closed. The membrane potential returns to its original resting state as the K⁺ channels close completely. As the voltage falls below the -70 mV level of the resting state, it is called ‘undershoot’.

The refractory periods :
Jhe period of time after an action potential begins during which the neuron cannot generate another action potential in response to a normal threshold stimulus is called the ‘refractory period’. There are two kinds of refractory periods, namely absolute refractory period and relative refractory period. During the absolute refractory period, even a very strong stimulus cannot initiate a second action potential. The relative refractory period is the time during which a second action potential can be initiated by a larger than normal stimulus.

Conduction speed:
The conduction speed of a nerve impulse depends on the diameter of the axon : the greater the axon’s diameter, the faster is the conduction. In a myelinated axon, the voltage-gated Na⁺ and K⁺ channels are concentrated at the nodes of Ranvier. As a result the impulse ‘jumps’ from one Ranvier’s node to the next, rather than travelling the entire length pf the nerve fibre. This mechanism of conduction is called Saltatory conduction. Saltatory conduction is faster (in myelinated fibres) than the continuous conduction (in nonmyelinated fibres).

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 11th Lesson Elections and Representation Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 11th Lesson Elections and Representation

Long Answer Questions

Question 1.
Write an essay on the Election System in India.
Answer:
Elections are very important for the political system of modem democratic states. Modem democratic states have representative governments. People participate in the process of government through their elected representatives. The election system is a political device through which the modern state creates among its citizens a sense of involvement and participation in public affairs.

Features of the Indian Election System :
The following features of the Indian Election system highlight its good structure nature.

1. Direct Election of Representatives:
The Constitution provides for a direct election of the representatives of the people. Members of the Lok Sabha, the State Legislative Assemblies, Municipalities, and Village Panchayats are directly elected by the people. These legislative bodies are the real centres of people’s power in the Indian democratic system.

2. Indirect Election for some Institutions :
However, the Constitution also provides for an indirect election in respect of the Rajya Sabha, State Legislative Councils, and the President and the Vice President of India. These are elected indirectly and in accordance with a system of proportional representation vote.

3. Universal Adult Franchise :
The constitution provides for a uniform Franchise to all the citizens. The right to vote was granted to all the citizens of 18 years of age without any discrimination on the basis of caste, religion, gender, education, property and place of birth. All the citizens whose names appear in the electoral lists, are eligible to exercise their vote in elections.

4. Reservation of seats for SCs and STs :
With a view to safeguard the interests of the people belonging to the Scheduled Castes and Scheduled Tribes, the Constitution provided the reservation of seats for them. Article 330 of the Constitution provides the reservation of seats to these classes in the Lok Sabha and Article 332 lays down this provision in respect of elections to every state Assembly. In the reserved constituencies, persons belonging to SC and ST only can contest in the elections.

5. Provision for Nominations :
The Constitution, under its Article 337, lays down that the President may, if he is of the opinion that the Anglo-Indian community is not adequately represented in the Lok Sabha, nominate not more than two members of the community to the Lok Sabha. Likewise, the Governor can also nominate not more than one person from this Anglo – Indian community to the State Legislative Assembly.

6. Regular revision of Electoral Rolls:
The Election Commission revises and prepares the electoral rolls enumerating the names of the eligible voters for every ten years. Besides this, the Election Commission can order the revision of electoral rolls before any election. Provision also exists for a regular annual revision of electoral lists. Only those persons whose names appear in the electoral rolls of the constituency can exercise their franchise on the Election Day.

7. Territorial and Single Member Constituencies :
Indian Election System provides for the creation of single-member territorial constituencies. All the voters living in a particular and defined territory constitute one constituency. Each territorial constituency elects one representative. Each state is divided into as many territorial constituencies as is the number of seats of its Legislative Assemblies and Parliamentary Constituencies and each constituency elects one representative.

8. Delimitation of Constituencies :
After every census the boundaries of the constituencies are delimited. This work is done by a three member Delimitation Commission. This commission can change the boundaries of constituencies, and its decision is final. This cannot be challenged before any court of law.

9. Secret Ballot:
Secret voting enables the voters to exercise their votes in accordance with their wishes and opinions. Special steps are taken in elections for maintaining secrecy and for checking impersonation in voting. This system is essential for making elections free and fair.

10. Introduction of Voting Machines :
The Election Commission has introduced the using of electronic voting machines (EVMs) for casting of votes by people and counting of votes in India.

11. Relative Majority of Votes System :
In the election, the candidate who secures more votes than every other contestant in his constituency is declared elected as representative to the Lok Sabha or the State Legislative Assemblies. In this system, valid votes are taken into consideration for counting.

12. Independent Machinery for the conduct of Elections in India:
According to the Article 324 of the Constitution, the conduct of elections in India is the responsibility of the Election Commission. It is the Constitutional body which is conducting the elections freely, fairly, impartially and independently.

Question 2.
Explain the Functions of Election Commission in India.
Answer:
Article 324(1) of the constitution provides the Election Commission to supervise and conduct the elections to parliament, state legislatures, the offices of the Preside ans the Vice-President of India.

Composition :
The Election Commission of India consists of the Chief Election Commissioner and Two other Commissioners.

Appointment :
The Chief Election Commissioner and other commissioners are appointed by the president of India.

Tenure :
The Chief Election Commissioner and other commissioners hold office for a period of 6 years or until they attain the age of 65 years whichever is earlier.

Removal:
The Chief Election Commissioner and other commissioners can be removed by the president on the basis of a resolution passed to that effect by both the Houses of Parliament with special majority either on the ground of proved misbehaviour or in capacity.

Salary and Allowances :
The Chief Election Commissioner and two other commissioners shall receive salary and Allowances which are similar to that of a Judge of the Supreme Court.

Powers and Functions of Election Commission:
The Constitution of India in its Articles 324-328 enumerates the powers and functions of the Election Commission. These can be mentioned here under.

1. It prepares all periodically revised electoral rolls.
2. It makes every effort to ensure that the voters’ list is free of errors like non-existence of names of registered voters or existence of names of that non-eligible or non¬existent.
3. It notifies the dates and schedules of election and scrutinizes nomination papers.
4. During this entire process, the Election Commission has the power to take decisions to ensure a free and fair poll.
5. It can postpone or cancel the election in the entire country or a specific State or constituency on the grounds that the atmosphere is vitiated and therefore, a free and fair election may not be possible.
6. The Commission also implements a model code of conduct for parties and candidates. It can order a re-poll in a specific constituency.
7. It can also order a recount of votes when it feels that the counting process has not been fully fair and just.
8. The Election Commission accords recognition to political parties and allots symbols to each of them.
9. It advices the President whether elections can be held in a state under President’s rule in order to extend the period of emergency after one year.
10. It advices the Governor on matters relating to the disqualifications of the,members of Stage Legislature.

Question 3.
What is Representation? How many types of representative systems are there in India?
Answer:
Representation:
In Democracy people elect members to the Legislatures. The elected members are the representatives of the people. They represent the people in the Legislature. This process is called representation.

Types of Representative systems in India:
Many methods are in practice for electing the representatives of Legislative bodies in India. They may be explained under the following heads :

i) Direct Method :
This is widely practiced and easy method. The voter takes part in the election directly and casts his vote to the condidate of his choice. The elected representatives serve for a fixed term.
Ex : The members of Lok Sabha and State Legislative Assemblies are being elected through this method.

ii) In Direct Method :
In this method the voters elect their representatives indirectly. Accordingly, first voters (Primary electors) elect the Intermediatoiy electors known as secondary Electors. These secondary electors in turn, elect the Representatives on behalf of the first voters. The sequence is
Voters → Intermediatory voters → Representatives
Or
Primary voters → Secondary voters Representatives
(Members of Electoral College)

Ex: In India, voters elect the members of Legislative Assemblies in the states, who in turn, elect the members of Upper house, Popularly known as the RAJYA SABHA in the centre in accordance with the provisions, Rules and Regulations in force. They also elect the members of Legislative council and form a part and parcel of electoral college for electing the President of India.

iii) Territorial Representation:
Under Territorial Representation, the country will be divided into geographical units called “Constituencies”. Voters in every constituency elect their representatives in periodically held elections, i.e., for every five years.
Ex : a) Representatives of Local bodies.
b) Members of State Legislative Assemblies and
c) Members of Lok Sabha

iv) Functional Representation:
Under this method people are divided into functional groups, such as for example Doctors, Lawyers, Engineers, Teachers, Graduates, Capitalists, Workers etc. representatives from each functional group will be returned to legislative bodies. It is hoped that these representatives will ventilate all the problems related to their occupations.
Ex: In India

• The president of India nominates 12 members to the Rajya Sabha having special knowledge or practical experience in respect of matters such as literature, science, arts and social service.
• 1/3 of the total members of the state legislative council shall be elected by the electorate consisting of the members of local bodies such as Municipalities, Zillaparishads etc.
• 1/12 of the total members of state legislative council are elected by the electorate consisting of university graduates.
• 1/12 of the total members of state legislative council are elected by the electorate consisting of secondary school teachers or those in higher educational institutions.
• 1/6^th of the total members of state legislative council are nominated by the governor on the basis of their special knowledge in literature, science, arts, co-operative movement or social service.

v) Proportional Representation :
Under this system each party gets representation strictly in Accordance with its voting strength, it means majority of electors would have majority of the representatives, but a minority of electors would have minority of representatives. There are two methods in proportional representation. These are
A) Single Transferable Vote System :
The election of President, Vice President and members of Rajya Sabha etc. in India is based on this system.
The main features of this system are

1. Each voter will have only one vote.
2. But the voter has to distribute this vote according to his preference (First preference, second preference etc).
3. The condidate to be declared elected should secure the required QUOTA of votes.
4. Provision for transfer of votes.

QUOTA :
Under this system as stated above, A candidate in order to be declared elected need not secure majority votes. But he requires only quota of votes. For determining quota, two methods are followed.
A) Hare and Andrae method of fixing quota :

Ex:
Number of votes cast – 14000
Number of seats vacant – 14
Quota – 1400 votes

B) Drop method:

Question 4.
Do you think that Indian Election System needs to be Reformed.
Answer:
The overall health of a representative form of Government depends, to a very large extent, on the fairness and effectiveness of the electoral process. In order to strengthen the foundatipns of our democratic polity, several reforms need to be undertaken in the electoral system. The system should provide equal opportunities to all citizens and help to build an egalitarian society. This would call for a careful review and reforms of existing features governing legislative, administrative and institutional dimensions of the electoral system in the country.

Within the broad framework of the existing constituency system of elections, geographies delimitation, multi-party system with right to individual to contest, the main issues that need to be examined include plugging loopholes in the present Representation of People Act and Anti-Defection Act.

The various committees and commissions which were appointed to examine our electoral system, election machinery as well as election process have recommended various reforms that have to be introduced in our electoral system. These can be mentioned briefly here under.

1. The donation of companies to the political parties should be strictly banned.
2. The accounts of the political parties are to be audited by the Election commission periodically.
3. The number of members of the Election commission shall be raised.
4. The limit on election expenditure of the candidates must be proper, practical and realistic.
5. The announcement of new policies, projects and programmes by the party in power during elections should be banned.
6. The members of the election commission should be appointed by the president on the advice of the prime minister, leader of the opposition in the Lok Sabha and the chief justice of India.
7. The government should meet the election expenditure of the candidates.
8. The election commission should be authorised to invalidate the election of a candidate if it was proved that he had used government machinery during elections.
9. Notification should be issued to the voters and electronic voting machines should be introduced after fool-proof arrangements.
10. The candidate who secures 51 percent of the polled votes shall be declared as winner.

Short Answer Questions

Question 1.
Write a Short Note on Electoral Functions.
Answer:
The electoral functions are different for the individual voter and for the political system. For the individual voter, elections may be regarded as a means of political participation and to some extent, of policy choice, although for many voters. Even in democratic societies, elections may be a customary at to which little significance is attached. For the political system, elections are important devices for assuring legitimacy and for system maintenance and support building. Hence, the electoral functions may be considered into four broad categories. They are :
a) Political choice
b) Political participation
c) Support-building and system – maintenance, and
d) Linkage functions.

a) Political choice :
The elections may be used or interpreted as a plebiscite, a referendum, or a mandate. They are the instruments for choosing the leaders and also determining the will of the people. They are devices for controlling political leaders. Control of leaders involves some degree of control of governmental choice and policies. They provide the means for the peaceful and orderly transfer of power.

b) Political participation :
The major electoral function is to provide opportunities and channels for political participation. This function of elections is a central one as political participation is essential in the democratic system. The participation strengthens the democratic system.

c) Support-building and system – maintenance :
The elections support the political system by providing legitimacy, political stability, integration and identification. The elections, are therefore a contributing rather than a controlling factor of a political system. Rosenau uses the term support-building as function of elections and hence, all other functions could be subsumed under it.

d) Linkage functions :
Elections are important agencies of political communication between the people and the government in the sense of political linkages. They provide a chance for the people to have more direct contact with their leaders.

Question 2.
Discuss the Election process in India.
Answr:
The electoral process in India is operationalized in several stages which can be explained as under.
1. Delimitation of Constituencies :
The first step of conducting the elections can be described as the delimitation of constituencies which is done by a Delimitation Commission appointed after every census by the President. This happens for every 10 years. Generally, a constituency which elects a member of the Lok Sabha consists of six or seven State Legislative Assembly constituencies. The decisions of the Delimitation commission are final and cannot be challenged in any court.

2. Recognition of Political Parties :
Political parties have to be registered with the Election Commission. According to certain criteria, set by the Election Commission regarding the length of political activity and success in elections, parties are categorized by the Commission as National or State parties, or simply declared registered-unrecognized parties. National parties are given a symbol that is for their use only, throughout the country. State parties have the sole use of a symbol in the state in which they are recognized as such Registered- unrecognized parties can choose a symbol from a selection of ‘free’ symbols.

3. Photo Identity Cards :
In an attempt to improve the accuracy of the electoral roll and prevent electoral fraud, the Election Commission ordered the making of photo identity cards for all voters in the country in Aug, 1993. To take advantage of latest technological innovations, the Commission issued revised guidelines for Election Photo Identity Card (EPIC) Program in May 2000.

4. Electoral Rolls :
An important step in the conduct of elections is to get prepared constituency-wise electoral rolls which record the names of the eligible voters. The electoral rolls are revised after each census, before every election or after regular intervals.

5. Notification and Appointment of Returning Officers :
When general elections are to be held, the President of India sends a communication to the Election Commission and the latter, after consulting Central and State Governments, announces the poll calendar, i.e., the dates for filing the nomination papers, scrutiny of nomination papers and withdrawal of nominations by the candidates. The Election Commission then appoints Returning Officers for various constituencies. The Regional Election Commissioners help the Election Commission in the smooth conduct of elections.

6. Filing of Nomination Papers :
The candidates who wish to contest in the elections have to file their nomination papers with the Returning Officer in their respective I constituencies. Such candidates must submit their nominations in a given formats prescribed by the Election Commission. If the contestant is a party candidate, the name has to be proposed by a voter and seconded by another voter. In the case of non-party contestant, the candidate has to be subscribed by 10 registered electors of the constituency as proposers.

7. Scrutiny of Nominations :
After the last date for the filing of the nominations, all the nomination papers are duly scrutinized by the Returning Officers in the presence of the candidates or his nominee. The scrutiny is conducted for determining whether the nomination papers have been filed properly, the candidates possess the required qualifications, and they have complied with all rules and regulations. Later, the Returning Officer decides all cases and notifies the names of those candidates whose nomination papers are found in order.

8. Withdraw of Nominations:
After scrutiny, the candidates are allowed to voluntarily withdraw their nominations within a fixed date and time as fixed by the Election Commission. For this purpose a candidate has to apply in writing to the Returning Officers.

9. Election Campaign :
The next stage in the electoral process involves the election campaign. The contestants and parties get engaged in the election campaign. Each party issues an Election manifesto which states its policies, programmes and promises.

10. Electronic Voting Machines (EVMs) :
An electronic voting machine is a simple electronic device used to record votes in place of ballot papers and boxes which were used earlier in conventional voting system. The advantages of the EVM over the traditional ballot paper system are given here.
a) It eliminates the possibility of invalid and doubtful votes.
b) It makes the process of counting of votes much faster than the conventional system.
c) It reduces to a great extent the quantity of paper used, thus saving a large number of trees.
d) It reduces cost printing as only one sheet of Ijallot papers required for each polling station.

11. Polling of Votes:
Polling personnel are appointed and polling booths are set up in different localities. Each polling booth, on an average caters to about a more than thousand voters. The voting is a secret one.

12. Supervising Elections by Observers :
The Election Commission appoints a.lafge number of Election Observers to ensure that the campaign is conducted fairly, and that people are free to vote as they choose. Election Observers keep a check on the amount that each candidate and party spends on the election.

13. Media Coverage:
In order to bring as much transparency as possible to the electoral process, the media are encouraged and provided with facilities to cover the election.

14. Counting of votes and Declaration of Results :
After the process of polling of votes, on a fixed day and time, the Returning Officer and his staff members open the voting machines in the presence of the agents. Then each candidate verifies his votes polled recorded in the EVM. A candidate who gets more valid votes than the other contesting candidate in very constituency is declared elected. These results declared on the basis of relative majority of vote victory system. The Returning Officer makes an announcement of the results in this regard.

15. Election Petitions :
Any elector or candidate can file an election petition if he or she thinks there has been malpractice during the election. An election petition is not an ordinary civil suit, but treated as a contest in which the whole constituency is involved. Election petitions are tried by the High Court of the state concerned and if upheld can even lead to the restaging of the election in that constituency.

Question 3.
Write a short note on composition and Functions of Election Commission. [(AP. Mar. 16) Mar. 18]
Answer:
Composition :
The Election Commission of India consists of the Chief Election Commissioner and two other commissioners. They are appointed by the President of India.

Functions of Election Commission :

1. It prepares all periodically revised electoral rolls.
2. It makes every effort to ensure that the voters’ list is free of errors like non-existence of names of registered voters or existence of names of that non-eligible or non-existent.
3. It notifies the dates and schedules of election and scrutinizes nomination papers.
4. During this entire process, the Election Commission has the power to take decisions to ensure a free and fair poll.
5. It can postpone or cancel the election in the entire country or a specific State or constituency on the grounds that the atmosphere is vitiated and therefore, a free and fair election may not be possible.
6. The Commission also implements a model code of conduct for parties and candidates. It can order a re-poll in a specific constituency.
7. It can also order a recount- of votes when it feels that the counting process has not been fully fair and just.
8. The Election Commission accords recognition to political parties and allots symbols to each of them.
9. It advices the President whether elections can be held in a state under President’s rule in order to extent the period of emergency after one year.
10. It advices the Governor on matters relating to the disqualifications of the members of State Legislature.

Question 4.
What is Representation? What do you know about Territorial Repre-sentation.
Answer:
Representation:
In democracy, people elect members to the Legislatures. The elected members are the representatives of the people. They represent the people in the Legislature. This process is called representation.

Territorial Representation :
In the territorial or geographical representation system, the total electorate of the country is divided into territorial units called constituencies which elect one or more representatives. The constituencies are more or less equal in size and population. All voters living in a particular Constituency take part in the election of representatives. Where one representative is elected from a constituency it is known as a single-member constituency. Where more than one representative is elected, it is known as multi-member constituency. Most of the modem states, including India, have followed single¬members constituencies for the elections to the lower houses of the legislature.

Question 5.
Estimate the pros and cons of FPTP system in India.
Answer:
In our country we have been following a special method of elections. The entire country is dividend into 550 constituencies, each constituency elects one representative and the candidate who secures the highest number of votes in that constituency is declared elected. It is important to note that in this system whoever has more votes than all other candidates is declared elected.

The winning candidate need not secure a majority of votes. This method is called the First Past the Post system (FPTP). In the election race, the candidate who is ahead of others, who crosses the Wining post first of all, is the winner. This method is also- called the Plural System. This is the method of election prescribed by the Constitution.

In India this FPTP System is popular and successful because of its simplicity. The entire election system is extremely simple to understand even for common voters who may have no specialized knowledge about politics and elections. There is also a clear choice presented to the voters at the time of elections. Voters may either give greater importance to the party or the candidate or balance the two.

The FPTP system generally gives the largest party or coalition some extra bonus seats, more than their share of votes would allow. Thus, this system makes it possible for parliamentary government to function smoothly and effectively by facilitating the formation of a stable government. Finally, the FPTP System encourages voters from different social groups to come together to win an election in a locality. Above all, the FPTP System has proved to be simple and familiar to the ordinary voters. It has helped larger parties to win clear majorities at the centre and state levels.

Question 6.
Write a Note on the Electoral Reforms. [Mar. 17]
Answer:
The Electoral Reforms will ensure the free and fair elections in the country. The successful functioning of Indian democracy depends on the electoral reforms.

Some Electoral Reform proposed :

1. The donation of companies to the political parties should be strictly banned.
2. The accounts of the political parties are to be audited by the Election commission periodically.
3. The number of members of the Election commission shall be raised.
4. The limit on election expenditure of the candidates must be proper, practical and realistic.
5. The announcement of new policies, projects and programmes by the party in power during elections should be banned.
6. The membersof the election commission should be appointed by the president on the advice of the prime minister, leader of the opposition in the Lok Sabha and the chief justice of India.
7. The government should meet the election expenditure of the candidates.
8. The election commission should be authorised to invalidate the election of a candidate if it was proved that he had used government machinery during elections.
9. Notification should be issued to the voters and electronic voting machines should be introduced after fool-proof arrangements.
10. The candidate who secure 51 percent of the polled votes shall be declared as winner.

Question 7.
What is meant by proportional representation system.
Answer:
Under this system each party gets representation strictly in accordance with its voting strength. It means majority of electors would have majority of the representatives, but a minority of electors would have minority of representatives.

There are two methods in proportional representation. These are

A) Single Transperable Vote System :
The election of President, Vice President and members of Rajya Sabha etc in India is based on this system.

The main features of this system are.

1. Each voter will have only one vote.
2. But the voter has to distribute this vote according to his preference (First preference, second preference etc.).
3. The candidate to be declared elected should secure the required QUOTA of votes.
4. Provision for transper of votes.

QUOTA :
Under this system as stated above, A candidate in order to be declared elected need not secure majority votes. But he requires only quota of votes. For determining quota, two methods are followed :
A) Hare and Andrae method fixing Quota :

Ex:
Number of votes Cast = 14000
Number of seats-vacant =14
Quota = 1400 votes

B) Drop method :

Very Short Answer Questions

Question 1.
Relations between democracy and elections.
Answer:
Elections are the Central institution of democratic representative governments. In democracy, elections are periodic. Elected officials are accountable to the people, and they must return to the voters at prescribed intervals to seek their mandate to cantinue in office. This means that officials .in a democracy must accept the risk of being voted out of office. In democratic system the elections are also indusive.

The definition of citizen and voter must be large enough to include a large proportion of the adult population. Democratic elections are definitive. They determine the leadership of the government. Subject to the laws and constitution of the country, popularly elected representatives hold the reins of power.

Question 2.
Electronic voting machines. [Mar. 18, 16]
Answer:
An Electronic voting machine is a simple electronic device used to record votes in place of ballot papers and boxes which were used earlier in conventional voting system. The advantages of the EVM over the traditional ballot paper system are give below.

• It eliminates the possibility of invalid and doubtful votes.
• It makes the process of counting of votes much faster that the conventional system.
• It reduces to a great extent the quantity of paper used, thus saving a large number of trees.
• It reduces cost printing as only one sheet of ballot papers required for each polling station.

Question 3.
Territorial Representation.
Answer:
In the territorial or geographical representation system, the total electorate of the Country is divided into territorial units called constituencies which elect one or more representatives. The constituencies are more or less equal in size and populatiqn. All voters living in a particular constituency take, part in the election of representatives. Where one representative is elected from a constituency it is known as a single-member constituency.

Where more than one representative is elected, it is known as multi-member constituency. Most of the modem states, including India, have followed single. Members constituencies for the elections to the lower houses of the legislature.

Question 4.
Functional Representation. [Mar. 17]
Answer:
It is based on occupation. People engaged in the same kind of occupation have more things in common than people living in the same locality. Doctors, farmers, industrial workers, traders, journalists, lawyers, teachers etc., have more in common than those who live as neighbours. One man cannot represent all traders, Hence, representation should be on functional basis. A legislature representing such different occupational groups would be a proper forum where different interests would be projected and pleaded for. But it is not possible to provide representation to each and every occupation which are innumerable in number and the classification of profession is a touch task.

Question 5.
Composition of Election commission of India.
Answer:
Article 324 of the constitution has made the following provisions with regard to the composition of election Commission. Since its inception 1950 and till 15th Oct, 1989, the election commission functioned as a single member body consisting of the Chief Election Commissioner. On 16th October, 1989 the President of India appointed two more election Commissioners to cope up with the increased work of the Election Commission. Thereafter, the Election Commission started functioning as a multimember body consisting of 3 Election Commissioners.

Question 6.
Electoral Reforms.
Answer:
The various commitees and commissions have recommended various reforms that have to be introduced in our electoral system, election machinery and election process. These can be mentioned briefly here under.

1. Lowering of voting age
2. Deputation to Election commission
3. Electronic voting machines
4. Prohibition on the sale of liquor
5. Number of proposers

Question 7.
Election offences.
Answer:
Some of the electoral offences as given below.

1. Promoting enmity between classes or grounds of religion, race, community or language.
2. Convening, holding or attending any public meeting during 48 hours before the end of poll.
3. Causing disturbance at election meetings.
4. Printing of election pamphlets, posters etc., without printers / publishers names and addresses.
5. Violation of maintenance of secrecy of vote.
6. Influencing of voting by officials connected with conduct of elections and police personnel.
7. Disorderly conduct and disturbance in or near a polling station, indulging use of loud speakers etc.
8. Canvassing within 100 meters of a polling station on the day of poll.
9. Misconduct at the polling station or failure to obey the lawful direction of the presiding officer.
10. Illegal hiring or procuring of vehicles for conveying voters to and from polling stations.
11. Unlawful removal of ballot papers / EVMs from the polling stations.
12. Booth capturing.

Question 8.
Corrupt practices in elections.
Answer:
Following are some important corrupt practices during the elections. They are.

1. Briling a person to induce him/her to stand or not to stand as a candidate.
2. Interference with free exercise of anybody’s electoral right.
3. Threat with injury of any kind, including social ostracism, excommunication, divine displeasure or spiritual censure.
4. Appeal on grounds of religion, caste, community or Language or the use of religious or national symbols.
5. Publication of false statement about personal character and conduct of any
6. Hiring or procuring of vehicles for free conveyance of voters.
7. Incurring of election expenditure by a candidate in excess of the prescribed limit.
8. Booth capturing.

Question 9.
Role of the Election Commission in India. [Mar. 17]
Answer:
Over the years, the Election Commission of India has emerged as an independent authority which has asserted its powers to ensure fairness in the election process. It has acted in an importial and unbaised manner in order to protect the sancity of the electoral process. The record of EC also shows that every improvement in the functioning of institutions does not require legal or constitutional change.

It is widely agreed that the Election Commission is more independent and assertive now than it was till twenty five years ago. This is not because the powers and constitutional protection of the Election commission have increased. The Election Commission has started using more effectively the powers it always had in the constitution.

In the past sixty five years, sixteen Lok Sabha Elections have been held. Many more state assembly elections and bye-elections have been conducted by the Election Commis-sion.

Question 10.
Representation.
Answer:
In democracy people elect members to the legislatures. The elected members are the representatives of the people. They represent the people in the legislature. This process is called representation.

Students get through AP Inter 2nd Year Chemistry Important Questions 7th Lesson d and f Block Elements & Coordination Compounds which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 7th Lesson d and f Block Elements & Coordination Compounds

Very Short Answer Questions

Question 1.
What are transition elements? Give examples.
Answer:
Transition elements are the elements which contains partially filled d-subshells in their ionic state (or) in their elementary state. Eg : Mn, Co, Ag etc.

Question 2.
Write the general electronic configuration of transition elements.
Answer:
General electronic configuration of transition elements is (n – 1)d^1-10 ns^1-2.

Question 3.
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state ?
Answer:
Mn⁺² has electronic configuration [Ar] 4s⁰ 3d⁵.
Fe⁺² has electronic configuration [Ar] 4s⁰ 3d⁶.

• Mn⁺² has half filled d-subshell which is more stable.
  Hence Mn⁺² compounds are more stable than Fe⁺² toward oxidation to their +3 state.

Question 4.
Why Zn²⁺ is diamagnetic whereas Mn²⁺ is paramagnetic ?
Answer:

• Zn⁺² electronic configuration is [Ar] 4s⁰3d¹⁰. It has no unpaired electrons. So it is dia magnetic.
• Mn⁺² electronic configuration is [Ar] 4s⁰3d⁵. It has five unpaired electrons so it is paramagnetic.

Question 5.
Write ‘spin only’ formula to calculate the magnetic moment of transition metal ions.
Answer:
Spin only formula to calculate the magnetic moment of transition metal ions is
μ = \(\sqrt{n(n+2)}\) BM BM = Bohr Magneton.

Question 6.
Claculate the ‘spin only’ magnetic moment of \(\mathrm{Fe}^{2+}(\mathrm{aq})\) ion. (Board Model Paper) (AP Mar. ’17)
Answer:
Fe⁺² ion has electronic configuration [Ar] 4s⁰3d⁶.
It has four unpaired electrons n = 4.
Spin only magnetic moment μ = \(\sqrt{n(n+2)}\) BM = \(\sqrt{4(4+2)}\) = \(\sqrt{24}\) BM = 4.89 BM

Question 7.
Aqueous Cu²⁺ ions are blue in colour, where as aqueous Zn²⁺ ions are colourless. Why ?
Answer:

• Electronic configuration of Cu⁺² ion is [Ar] 4s⁰3d⁹. It contains one unpaired electron due to presence of this unpaired electron aq. Cu⁺² ions are blue in colour.
• Electronic configuration of Zn⁺² ion is [Ar] 4s⁰3d¹⁰. It contains no unpaired electrons, due to absence of unpaired electrons aq. Zn⁺² ions are colourless.

Question 8.
What are complex compounds ? Give examples.
Answer:
Complex compounds : Transition metal atoms or ions form a large number of compounds in which anions or neutral groups are bound to metal atom or ion through co-ordinate covalent bonds. Such, compounds are called co-ordination compounds (or) complex compounds.
Eg. : [Fe(CN)₆]^4-, [Co(NH₃)₆]³⁺

Question 9.
What is an alloy? Give example.
Answer:
Alloy : An intimate mixture having physical properties similar to that of the metal formed by a metal with other metals or metalloids or sometimes a non metal is called as an alloy.
Eg.: Invar — 64% Fe, 35% Ni, Mn 8cc in traces
Nichrome — 60% Ni,, 25% Fe, 15% Cr.

Question 10.
What is lanthanoid contraction ? (IPE 2016(TS))
Answer:
The slow decrease of atomic and ionic radii in lanthanides with increase in atomic number is called lanthanide contraction.

Question 11.
What is mischmetall ? Give its composition and uses. (AP Mar. ’16)
Answer:
Mischmetall is an alloy which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al.

• It is used in Mg- based alloy to produce bullets, shell and lighter flint.

Question 12.
What is a coordination polyhedron ?
Answer:
The spatial arrangement of the ligands which are directly bonded to the central atom or ions defines the geometry about the central atom is called co-ordination polyhedron.
Eg.: Octahedral, tetrahedral etc.

Question 13.
What is a ligand ? (TS Mar. ’18) (IPE 2014)
Answer:
Ligand : A co-ordinating entity which is bound to the central atom by donating electron pairs is called a ligand. Eg.: Cl^–, NH₃, CN^– etc.

Question 14.
What is a chelate ligand ? Give example.
Answer:
The ligands which can form two co-ordinate covalent bonds through two donor atoms are called bidentate ligands. These bidentate ligands are also called chelate ligands.
Eg.: C₂\(\mathrm{O}_4^{-2}\), \(\mathrm{CO}_3^{-2}\) etc.

Question 15.
What is an ambidentate ligand ? Give example. (TS Mar. ’18) (IPE 2016(AP))
Answer:
A unidentate ligand containing two possible donor atoms can co-ordinate through either of donor atoms. Such ligands are called ambidentate ligands. Eg : \(\mathrm{NO}_2^{-}\)

Question 16.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Why ?
Answer:

• [Cr(NH₃)₆]³⁺ is paramagnetic due to the presence of three unpaired electrons.
• [Ni(CN)₄]^2- is diamagnetic due to the absence of unpaired electrons.

Question 17.
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why ?
Answer:
Copper exhibits +1 oxidation state most frequently because Cu⁺ has electronic configuration [Ar] 4s⁰3d¹⁰ which has a fulfilled d-subshell which is more stable.

Question 18.
Why do transition elements exhibit more than one Oxidation state (variable oxidation states) ?
Answer:
Transition elements exhibits more than one oxidation state.
Reason :

• The energy difference between (n – 1) d subshell and ns subshell is very low. So both of these subshells complete to lose the electrons.

Question 19.
CuSO₄.5H₂O is blue in colour where as anhydrous CuSO₄ is colourless. Why ? (IPE 2014)
Answer:
CuSO₄.5H₂O is blue in colour whereas anhydrous CuSO₄ is colourless because in the absence of ligand, crystal field splitting doesnot occurs.

Question 20.
Give the formula the following complexes.
a) Potassium hexacyano ferrate (III)
b) Tetra carbonyl nickel (O) (IPE 2016 (TS))
Answer:
a) K₃ [Fe (CN)₆]
b) [Ni(CO)₄]

Question 21.
Scandium is a transition element. But Zinc is not. Why ? (IPE 2014)
Answer:
Scandium has electronic configuration [Ar] 4s²3d¹.
Zinc has electronic configuration [Ar] 4s²3d¹⁰.
Scandium has one unpaired d-electron where as Zinc has zero unpaired d-electrons so Scandium is transition element but Zinc is not.

Question 22.
What are interstitial compounds ? (IPE 2015)
Answer:
Interstitial compounds are non-stichiometric compounds which are formed when non-metals like H, C, N etc., are heated with transition elements. In these compounds the non-metal atoms occupy the interstitial species of metal atoms. Ex : TiH_0.54

Question 23.
In what way is the electronic configuration of transition elements different from non transition elements ?
Answer:

• The general electronic configuration of transition elements is (n – 1) d^1-10 ns^1-2.
• The general electronic configuration of non-transition elements is (n – 1)d¹⁰ ns².

Question 24.
Even though silver has d¹⁰ configuration, it is regarded as transition element. Why ?
Answer:
The outer electronic configuration of silver is – 4d¹⁰5s¹. It is having general electronic configuration of a transition element (n – 1) d^1-10 ns^1-2.
So silver is a transition element.

Question 25.
Though Sc is a transition element, it does not exhibit variable oxidation state. Why ?
Answer:
Scandium has electronic configuration [Ar] 4s²3d¹. It has only one unpaired d-electron. So it does not exhibits variable oxidation state. It exhibits +3 stable oxidation state.

Question 26.
Why is it difficult to obtain M³⁺ oxidation state in Ni, Cu and Zn ?
Answer:

• Ni has electronic configuration [Ar] 4s²3d⁸.
  Ni⁺² has electronic configuration [Ar] 4s⁰3d⁸.
  It is difficult to remove an electron from 3d⁸. So, Ni⁺³ is difficult to obtain. (Ni has high negative enthalpy of hydration).
• Cu has electronic configuration [Ar] 4s¹3d¹⁰.
  Cu⁺ has electronic configuration [Ar] 3d¹⁰.
  It is difficult to remove the electrons from 3d¹⁰ (stable). So, Cu⁺³ is difficult to obtain.
• Zn has electronic configuration [Ar] 4s²3d¹⁰.
  Zn⁺² has electronic configuration [Ar] 4s⁰3d¹⁰.
  It is difficult to remove the electron from 3d¹⁰ (stable). So Zn⁺³ is difficult to obtain.

Question 27.
Cu⁺² forms halides like CuF₂, CuCl₂ and CuBr₂ but not CuI₂. Why ?
Answer:
Cu⁺² forms halides like CuF₂, CuCl₂ and CuBr₂ but not CuI₂ because Cu⁺² oxidises I^– to I₂.
2Cu⁺² + 4I^– → Cu₂I₂ + I₂.

Question 28.
Why is Cr²⁺ reducing and Mn³⁺ oxidizing even though both have the same d⁴ electronic configuration ?
Answer:
Cr⁺² is reducing as its configuration changes from d⁴ to d³, the latter having a half filled t_2g level. On the other hand the change from Mn⁺² to Mn⁺³ results in the half filled (d⁵) configuration which has extra stability.

Question 29.
What is meant by ‘disproportionation’ ? Give an example of disproportionation reaction in aqueous solution.
Answer:
The reactions in which only one element undergo both oxidation as well as reduction are called disproportionation reactions.

• When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation.
  Eg: Cu⁺ ion is not stable in aqueous solution because it undergo disproportionation in aqueous solution.

Question 30.
Why do the transition metals readily form alloys?
Answer:
Because of similar atomic or ionic radii and similar characterstic properties of transition elements alloys are readily formed by these elements.

Question 31.
How do the Ionic character and acidic nature vary among the oxides of first transition series?
Answer:

• As the oxidation number of a metal increases ionic character decreases in case of transition elements. Eg: Mn₂O₇ is a covalent green oil.
• In CrO₃ and V₂O₅ the acidic character is predominant.
• V₂O₅ is however amphoteric though mainlý acidic V₂O₅ reacts with alkali as well as acids to give \(\mathrm{VO}_4^{-3}\) and \(\mathrm{VO}_4^{-3}\).
• CrO is basic.
• Mn₂O₇ gives HMnO₄ and CrO₃ gives H₂CrO₄ and H₂Cr₂O₇.

Question 32.
What are the different oxidation s.tates exhibited by the lanthanoids?
Answer:

• Lanthanoids exhibits +2, +3 states majorly. Sometimes +2 and +4 states exhibited in solid compounds.
• The common oxidation state of lanthanoids is +3.

Question 33.
What is difference between a double salt and a complex compound?
Answer:
Double salts dissociate into simple ions completely when dissolved in water while complex compounds dissociate to give complex ion and the counter ions.

Question 34.
What is meant by coordination number.
Answer:
Co-ordination Number is the number of ligands present around a central metal atom or ion in a complex Ex : The co-ordination number of Co in [Co(NH₃)₃ Cl₃] is six.

Question 35.
Using IUPAC norms write the systematic names of the following.

1. [CO(NH₃)₆]Cl₃
2. [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
3. [Ti(H₂O)₆]³⁺ and
4. [NiCl₄]^2-

Answer:

1. Hexa amine cobalt (III) chloride
2. Diamine chloride (methyl amine) platinum (II) chloride
3. Hexa aquo titanium (III) ion
4. Tetra chloro nickelate (III) ion

Question 36.
Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25. (IPE 2016 (AP)) (AP Mar. ’16)
Solution:
With atomic number 25, the divalent ion in aqueous solution will have d⁵ configuration (five unpaired electrons). The magnetic moment, μ is
μ = \(\sqrt{5(5+2)}\) = 5.92 BM

Short Answer Questions

Question 1.
Write the characteristic properties of transition elements.
Answer:
Transition elements exhibits typical characteristic properties.

• Electronic configurations.
• Para and ferro magnetic properties.
• Alloy forming ability.
• Complex forming ability.
• Interstitial compounds.
• Variable oxidation states.
• Formation of coloured hydrated ions.
• Catalytic property.
• Metallic character.

Question 2.
What is lanthanoid contraction ? What are the consequences of lanthanoid contraction ? (AP Mar. ’17) (IPE Mar. 15 (TS), 14 B.I.E)
Answer:
Lanthanoid contraction : The overall decrease in atomic and ionic radii from lanthanum to leutetium is observed in the lanthanoids. This phenomenon is called lanthanoid contraction. It is due to the fact that with every additional proton in the nucleus, the corresponding electron goes into a 4f-subshell. This is too diffused to screen the nucleus as effectively as the more localised inner shell. Hence, the attraction of the nucleus for the outermost electrons increases steadily with the atomic number.

Consequences of Lanthanoid Contraction : The important consequences of lanthanoid contraction are as follows :

i) Basic character of oxides and hydroxides : Due to the lanthanoid contraction, the covalent nature of La-OH bond increases and thus, the basic character of oxides and hydroxides decreases from La(OH)₃ to Lu(OH)₃.

ii) Similarity in the size of elements of second and third transition series : Because of lanthanoid contraction, elements which follow the third transition series are considerably smaller than would otherwise be expected. The normal size increases from Sc → Y → La and disappears after lanthanides. Thus, pairs of elements such as Zr/Hf, Nb/Ta and Mo/w are almost identical in size.
Due to almost similar size, such pairs have very similar properties which makes their separation difficult.

iii) Separation of lanthanoids : Due to lanthanoid contraction, there is a difference in some properties of lanthanoid like solubility, degree of hydration and complex formation. These difference enable the separation of lanthanoids by ion exchange method.

Question 3.
Explain Werner’s theory of coordination compounds with suitable examples. (Mar. ’14) (TS Mar. ’15) (IPE – May. 2015 (TS), ’14 B.I.E)
Answer:
Werner’s theory :
Postulates :

1) Every complex compound has a central metal atom (or) ion.

2) The central metal shows two types of valencies namely primary valency and secondary valency.

A) Primary valency : The primary valency is numerically equal to the oxidation state of the metal. Species or groups bound by primary valencies undergo complete ionization. These valencies are identical with ionic bonds and are non-directional. These valencies are represented by discontinuous lines (….)
Eg. : CoCl₃ contains Co³⁺ and 3Cl^– ions. There are three Primary Valencies or three ionic bonds.

B) Secondary Valency : Each metal has a characteristic number of Secondary Valencies. They are directed in space around the central metal.
The number of Secondary Valencies is called Coordination number (C.N.) of the metal. These valencies are directional in nature.
For example in CoCl₃. 6NH₃
Three Cl^– ions are held by Primary Valencies and 6NH₃ molecules are held by Secondary Valencies. In CuSO₄.4NH₃ complex \(\mathrm{SO}_4^{2-}\) rion is held by two Primary Valencies and 4NH₃ molecules are held by Secondary Valencies.

3) Some negative ligands, depending upon the complex, may satisfy both primary and secondary valencies. Such ligands, in a complex, which satisfy both primary as well as secondary valencies do not ionize.

4) The primary valency of a metal is known as its outer sphere of attraction or ionizable valency while the Secondary valencies are known as the inner sphere of attraction or coordination sphere. Groups bound by secondary valencies do not undergo ionization in the complex.
Example to Clarify Werner’s Theory

Question 4.
Using IUPAC norms write the formulas for the following.

1. Tetrahydroxozincate (II)
2. Hexaaminecobalt (III) sulphate
3. Potassium tetrachloropalladate (II) and
4. Potassium tri(oxalato) chromate (III) (IPE – May. 2015 (AP, TS), 14)

Answer:

1. Tetrahydroxozincate (II) ion – [Zn(OH)₄]^-2
2. Hexa amine cobalt (III) sulphate – [Co(NH₃)₆]₂ (SO₄)₃
3. Potassium tetrachioro palladate (II) – K₂[PdCl₄]
4. Potassium tri(oxalato) chromate (III) – K₃[Cr(C₂O₄)₃]

Question 5.
What are homoleptic and heteroleptic complexes ? Give one example for each.
Answer:
Homoleptic complexes : These are the complexes in which a metal is bound by only ore kind of ligands, eg. : [Co(NH₃)₆]³⁺
Heteroleptic complexes: These are the complexes in which a metal is bound by more than one kind of ligands, eg. : [Co(NH₃)₄Cl₂]⁺

Question 6.
Give the geometrical shapes of the following complex entities

1. [CO(NH₃)₆]³⁺
2. [Ni(CO)₄]
3. [Pt Cl₄]^2- and
4. [Fe(CN)₆]^4-.

Answer:

1. Geometrical shape of [CO(NH₃)₆]³⁺ is octahedral.
2. Geometrical shape of [Ni(CO)₄] is tetrahedral.
3. Geometrical shape of [PtCl₄]^2- is square planar.
4. Geometrical shape of [Fe(CN)₆]^-4 is octahedral.

Question 7.
Using IUPAC norms write the systematic names of the following.

1. K₄[Fe(CN)₆]
2. [Cu(NH₃)₄]SO₄
3. [Ti (H₂O)₆]³⁺ and
4. [NiCl₄]^2-.

Answer:

1. K₄[Fe(CN)₆] – Potassium hexa cyanao ferrate (II)
2. [Cu(NH₃)₄]SO₄ – Tetra amine copper (II) sulphate
3. [Ti (H₂O)₆]³⁺ and – Hexa aquo titanium (III) ion
4. [NiCl₄]^2- – Tetra chloro nickelate (II) ion

Question 8.
Explain the terms
(i) Ligand
(ii) Coordination number
(iii) Coordination entity
(iv) Central metaf atom/ion.
Answer:
i) Ligand : The ions, or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be

a) simple ions such as Cl^–
b) small molecules such as H₂O or NH₃
c) large molecules such as H₂N.CH₂CH₂NH₂ or N(CH₂CH₂NH₂)₃ or even
d) macro-molecules, such as proteins.

On the basis of the numberof donor atoms available for coordination, the ligands can be classified as :

a) Unidentate : One donor atom, Eg. etc.
b) Bidentate : Two donor atoms, Eg.: H₂NCH₂CH₂NH₂
(ethane-1, 2-diamine or en), C₂\(\mathrm{O}_4^{2-}\) (oxalate), etc.
c) Polydentate : More than two donor atoms, Eg. : N(CH₂CH₂NH₂)₃ EDTA, etc.

ii) Coordination number : The coordination number (CN) of metal ion in a complex can be defined as the number of ligands or donor atoms to which the metal is directly bonded.
Eg : In [Ptcl₆]^2-, CN of Pt = 6, In [Ni(NH₃)₄]²⁺, CN of Ni = 4.

iii) Coordination entity: A central metal atoms or ions bonded to a fixed number of molecules or ions (ligands) is known as coordination entity. For example [CoCl₃(NH₃)₃]. Ni(CO)₄], etc.

iv) Central metal atom/ion : In a coordination entity, the atom/ion to which a fixed no. of ions/groups are bound in a definite geometrical arrangement around it is called central metal atom or ion. Eg : K₄[Fe(CN)₆] ‘Fe’ is central metal.

Question 9.
Explain the terms

1. Unidentate ligand
2. Bidentate ligand
3. Polydentate ligand and
4. Ambidentate ligand giving one example for each.

Answer:

1. Unidentate : The negative ion or neutral molecule having only one donor atom is called unidentate ligand
   Eg : etc.
2. Bidentate (or didentate) : The ions or molecules having two donor atoms are called
   
3. Polydentate ligands : Ligands having more than one donor atom in the coordinating group and are capable of forming two or more coordinate bonds with same central atom simultaneously are called poly dentate ligands. Eg : C₂\(\mathrm{O}_4^{-2}\).
4. Ambidentate: Ligand which can ligate through two different atoms is called ambidentate ligand. Eg : \(\mathrm{NO}_2^{-}\), \(\mathrm{SCN}^{-}\) ions. \(\mathrm{NO}_2^{-}\) ion can coordinate either through nitrogen or through oxygen to a central metal atom/ion. Similarly, \(\mathrm{SCN}^{-}\) ion can coordinate through the sulphur or nitrogen atom.
   

Question 10.
Explain the colour and para magnetic property of transition elements.
Answer:
Transition elements having unpaired electrons in d-orbitals are coloured. The colour is due to d-d transitions of electrons. Electrons are excited from lower set of d-orditals to higher set of d- orbitals by absorbing energy from visible light and emits complementary colour which is the colour of the ion. Transition elements are ions having atleast one unpaired electron in d-orbitals are para magnetic. The extent of para magnetic property is expressed in terms of magnetic moment (μ). μ = \(\sqrt{n(n+2)}\) where n is number of unpaired electrons present in d- orbitals.

Question 11.
Explain different types of isomerism exhibited by Co-ordination compounds, giving suitable examples.
Answer:
Isomerism in Co-ordination compounds : Isomers are compounds that have the same chemical formula but different arrangement of atoms. Two principal types of isomerism are known among co-ordination compounds namely stereo isomerism and structural isomerism.

a) Stereoisomerim : Stereoisomerism is a form of isomerism in which two substances have the same composition and structure but differ in the relative spatial positions of the ligands. This can be sub divided into two classes namely.
i) Geometrical isomerism and
ii) optical isomerism

b) Structural isomerism :
i) Linkage isomerism
ii) Co-ordination isomerism
iii) Ionisation isomerism
iv) Hydrate isomerism

a) (i) Geometrical isomerism :

• Geometrical isomerism arises in Co-ordination complexes due to different possible geometric arrangements of the ligands.
• This isomerism found in complexes with Co-ordination numbers 4 and 6.
• In a square planar complex of formula [Mx₂L₂] the two ligands may be arranged in adjacent to each other in a cis isomer (or) opposite to each other in a trans isomer.
  
• Square planar complex of type [MAB XL] shows three isomers two-cis and one trans. (A, B, X, L are unidentate ligands is square planar complex).
• Geometrical isomerism is not possible in tetrahedral geometry.
• In octahedral complex of formula [MX₂L₄] in which two ligands X may be oriented cis or trans to each other.
  
• Another type of geometrical isomerism occurs in octahedral Co-ordination compounds of type [Ma₃b₃] if three donor atoms of the same ligands occupy adjacent positions at the comers of an octahedral face then it is facial (fac) isomer. When the positions occupied are around the meridian of the octahedran then it is meridonial (mer) isomer.
  

a(ii) Optical isomerism : Optical isomerism arises when two isomers of a compound exist such that one isomer is a mirror image of the other isomer. Such isomers are called optical isomers or enantiomers. The molecules or ions that cannot be superimposed are called chiral.

The two forms are called dextro (d) and laevo (l) depending upon the direction they rotate the plane of polarised light in a polarimeter (d rotates to the right, l to the left). Optical isomerism is common in octahedral complexes involving bidentate ligands.

b(i) Linkage isomerism: Linkage isomerism arises in Co-ordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand-NCS^–, which may bind through the nitrogen to give M-NCS or through sulphur to give M-SCN.
eg. : [Mn(CO)₅SCN] and [Mn(CO)₅NCS]

(ii) Co-ordinate isomerism : This type isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex.
eg. : [CO(NH₃)₆] [Cr(CN)₆] and [Co(CN)₆] [Cr(NH₃)₆]

(iii) Ionisation isomerism: This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion.
eg. : [Co(NH₃)₅SO₄] Br and [Co(NH₃)₅Br]SO₄

(iv) Hydrate isomerism : This form of isomerism is known as ‘hydrate isomerism since water is involved as a solvent. Hydrate isomers differ by whether or not a hydrate molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice.

Question 12.
What is meant by chelate effect ? Give example.
Answer:
When a didentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metql ion, a 5-or 6-membered ring is formed, the effect is known as chelate effect. Example

Question 13.
Discuss the nature of bonding and magnetic behaviour in the following co-ordination entities on the basis of valence bond theory.
(i) [Fe(CN)₆]^4-
(ii) [FeF₆]^3-
(iii) [Co(C₂O₄)₃]^3- and
(iv) [CoF₆]^3-
Answer:
i) [Fe(CN)₆]^4- : In this complex Fe is present as Fe²⁺.
Fe = [Ar] 3d⁶4s²
Outer configuration of Fe²⁺ = 3d⁶4s⁰

CN^– being strong field ligand, pair up the unpaired d electrons Thus, two 3d-orbital are now available for CN^– ions.

Since, all the electrons are paired, the complex is diamagnetic. Moreover (n – 1) d-orbitals are involved in bonding, so, it is an inner orbital or low spin complex.

ii) [FeF₆]^3- : In this complex, the oxidation state of Fe is + 3.

F- is not a strong field ligand. It is a weak field ligand, so no pairing occurs. Thus, 3d- orbitals are not available to take part in bonding.

Because of the presence of five unpaired electrons, the complex is paramagnetic. Moreover, nd-orbitals are involved in bonding, so it is an outer orbital or high spin complex.

iii) [Co(C₂O₄)₃]^3- : In this complex, the oxidation state of Co is + 3.
Outer configuration of Co = 3d⁷ 4s²
Co³⁺ = 3d⁶4s⁰

Oxalate ion being a strong field ligand pair up the 3d electrons, thus two out of the five 3d-orbitals are available for oxalate ions.

Since, all the electrons are paired, this complex is diamagnetic. It is an inner orbital complex because of the involvement of (n – 1) d-orbital for bonding,

iv) [CoF₆]^3- : In this complex, Co is present as Co³⁺
Outer configuration of Co³⁺ = 3d⁶ 4s⁰

Because of the presence of four unpaired electrons, the complex is paramagnetic. Since, nd orbitals take part in bonding, it is an outer orbital complex or high spin complex.

Question 14.
What is spectrochemical series ? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
The arrangement of ligands in order of their increasing field strengths, i.e., increasing crystal field splitting energy (CFSE) values is called spectrochemical series.

The ligands with a small value of CFSE (Δ₀) are called weak field ligands. For such ligands Δ₀ < P where P is the pairing energy. Whereas the ligands with a large value of CFSE are called strong field ligands. In case of such ligands Δ₀ > R

When ligands approach a transition metal ion, the d-orbitals split into two sets (t_2g and e_g), one with lower energy and the other with higher energy. The difference of energy between the two sets of orbitals is called crystal field splitting energy, Δ₀ for octahedral field and Δ_t for tetrahedral field.

If Δ₀ < P (pairing energy), the 4^th electron enters one of the e_g orbitals giving the configuration \(t_{2 g}^3 e_g^1\) thereby forming high spin complex. Such ligands for which Δ₀ < P are known as weak field ligands.
If Δ₀ > P, the 4^th electron pairs up in one of the t_2g orbitals giving the configuration \(t_{2 g}^3 e_g^1\), thus forming low spin complexes. Such ligands for which Δ₀ > P are called strong field ligands.

Question 15.
Explain the terms
(i) Ligand
(ii) Coordination number
(iii) Coordination entity
(iv) Central metal atom/ion.
Answer:
i) Ligand : The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be
a) simple ions such as Cl^–
b) small molecules such as H₂O or NH₃
c) large molecules such as H₂NCH₂CH₂NH₂ or N(CH₂CH₂NH₂)₃ or even
d) macro-molecules, speh as proteins.
On the basis of the number of donor atoms available for coordination, the ligands can be classified as :

a) Unidentate : One donor atom, Eg.: etc.
b) Bidentate : Two donor atoms, Eg.: H₂NCH₂CH₂NH₂
(ethane-1, 2-diamine or en), C₂\(\mathrm{C}_2 \mathrm{O}_4^{2-}\) (oxalate), etc.
c) Polydentate: More than, two donor atoms, Eg.: N(CH₂CH₂NH₂)₃ EDTA etc.

ii) Coordination number: The coordination number (CN) of metal ion in a complex can be defined as thê number of ligands or donor atoms to which the metal is directly bonded.
Eg: In [PtCl₆]^2-, CN of Pt = 6, In [Ni(NH₃)₄]²⁺, CN of Ni = 4.

iii) Coordination entity : A central metal atoms or ion bonded to a fixed number of molecules or ions (ligands) is known as coordination entity. For example [COCl₃(NH₃)₃]. Ni(CO)₄], etc.

iv) Central metal atom/ion: In a coordination entity, the atom/ion to which a fixed no. of ions/groups are bound in a definite geometrical arrangement around it is called central metal atom or ion. Eg: K₄[Fe(CN)₆] ‘Fe’ is central metal.

Students get through AP Inter 2nd Year Chemistry Important Questions 8th Lesson Polymers which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 8th Lesson Polymers

Very Short Answer Questions

Question 1.
What are polymers ? Give example.
Answer:
Polymer : A large molecular weight complex compound which is formed by the repeated combination of simple units (monomers) is called polymer. E.g.: Nylon 6, 6, Buna-S,. rubber etc.

Question 2.
What is polymerization ? Give an example of polymerization reaction. [IPE 2014 Mar. 14]
Answer:
Polymerization: The process of formation of polymers from respective monomers is called polymerization.
(or)
A large molecular weight complex compound which is formed by the repeated combination of simple units is called polymer. This process is called polymerisation.
E.g. : Formation of polyethene from ethene and reaction of hexamethylene diamine and adipic acid leading to the formation of Nylon 6, 6 are examples of two different types of polymerisation reactions.

Question 3.
What is addition polymer ? Give example. [IPE Mar. 2015 (TS)]
Answer:
Addition Polymer: The polymer which is formed by the addition of molecules of monomers of same type (or) different type containing double bonds is called addition polymer.
E.g. : Polyethene, poly acrylonitrile.

Question 4.
What is condensation polymer ? Give example.
Answer:
Condensation polymer : The polymer which is formed by the condensation reaction between molecules having more than one functional group is called condensation polymer.
E.g. : Nylon 6, 6, Poly ethylene terephthalate.

Question 5.
What are homopolymers ? Give example.
Answer:
Homopolymers : The polymers which are formed by the polymerisation of a single monomeric species are known as homopolymers.
E.g.: Polyethene, Poly styrene.

Question 6.
What are copolymers ? Give example. [IPE 2014]
Answer:
Copolymers: A polymer which is formed by the polymerisation of two (or) more chemically different types of monomer units is called copolymer.
E.g.: Butadiene – Styrene polymer (Buna-S)

Question 7.
What are elastomers. Give example.
Answer:
Elastomers : There are rubber like solids with elastic properties. In elastomers the polymer chains are held together by the weak enter molecular forces.
E.g. : Buna – S, Buna – N etc.

Question 8.
What are fibres ? Give example.
Answer:
Fibres : Fibres are the thread forming solids which possess high tensile strength.
E.g. : Nylon 6, 6, polyesters.

Question 9.
What are thermoplastic polymers ? Give example.
Answer:
Thermoplastic Polymers: These are the linear (or) slightly branched long chain molecules capable of softening on heating and hardening on cooling.
E.g. : Polystyrene, polythene.

Question 10.
What are thermosetting polymers ? Give example.
Answer:
Thermo setting polymers: These polymers are cross linked (or) heavily branched molecules which on heating undergo extensive cross linking in moulds and again become infusible.
E.g. : Bakelite, urea-formaldehyde resin etc.

Question 11.
What is Ziegler – Natta catalyst ?
Answer:
A mixture of Tri alkyl aluminium and titanium chloride is called Ziegler – Natta catalyst
E.g. : (C₂H₅)₃ Al + TiCl₄

Question 12.
What are the repeating monomeric units of Nylon 6 and Nylon 6, 6 ?
Answer:
The repeating monomeric units of Nylon – 6 is Capro lactum.

The repeating monomeric units of Nylon 6, 6 are hexamethylene diamine and Adipic acid.
H₂N-(CH₂)₆ – NH₂
Hexamethylene

HOOC – (CH₂)₄ – COOH
Adipic acid

Question 13.
What is the difference between Buna – N and Buna – S ?
Answer:
Buna – N : Buna – N is the copolymer which is formed by the polymerisation of 1, 3 – Butadiene and acrylonitrile.

Buna – S : Buna – S is the copolymer which is formed by the polymerisation of 1, 3 – Butadiene and Styrene.

Question 14.
What is PDI (Poly Dispersity Index) ?
Answer:
Poly Dispersity Index (PDI) : The ratio between weight average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{w}}\)) and the number average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{n}}\)) of a polymer is called Poly Dispersity Index (PDI).

Question 15.
What is vulcanization of rubber ? [A.P. Mar. 17] [IPE – 2016 (TS)]
Answer:
Vulcanization of rubber : The process of heating the raw rubber with sulphur (or) with sulphur compounds to improve it’s physical properties is called vulcanization of rubber.

Question 16.
What is biodegradable polymer ? Give one example of a biodegradable polyester ?
Answer:
Biodegradable polymers : The polymers degradable by enzymatic hydrolysis and to some extent by oxidation are called biodegradable polymers.
Eg.: Nylon – 2, Nylon – 6, PHBV, Polyglycolic acid, Polylactic acid etc.

Question 17.
What is PHBV ? How is it useful to man ? [IPE Mar – 2015 (TS), B.I.E, 2016 (TS)]
Answer:
Poly β – hydroxy butyrate – CO – β – hydroxy Valerate (PHBV) : It is a Copolymer of 3 -hydroxy butanoic acid and 3 – hydroxy pentanoic acid.

Properties & Uses : The properties of PHBV vary according to the ratio of both the acids, 3-hydroxy butanoic acid provides stiffness and 3-hydroxy pentanoic acid imparts flexibility to copolymer.
It is used in medicine for making capsules.
PHBV also undergoes degradation by bacteria.

Short Answer Questions

Question 1.
Write the names and structures of the monomers of the following polymers. [A.P. Mar. 17]
i) Buna – S
ii) Buna – N
iii) Dacron
iv) Neoprene
Ans:
i) Buna – S
Monomers : 1, 3 – Butadiene, Styrene
Structure :

ii) Buna-N
Monomers : 1, 3 – Butadiene, Acrylonitrile
Structure : CH₂ = CH – CH = CH₂, CH₂ = CH – CN

iii) Dacron
Monomers : Ethylene glycol, Terephthalic acid
Structure : HOCH₂ – CH₂OH, HOOC COOH

iv) Neoprene .
Monomers : 2 – chloro – 1, 3 – Butadiene
Structure :

Question 2.
Define thermoplastics and thermosetting polymers with two examples of each.
Answer:
Thermoplastic Polymers: These are the linear (or) slightly branched long chain molecules capable of softening on heating and hardening on cooling. E.g.: Polystyrene, polythene.
Thermo setting polymers: These polymers are cross linked (or) heavily branched molecules which on heating undergo extensive cross linking in moulds and again become infusible.
Eg.: Bakelite, urea-formaldehyde, resin etc.

Question 3.
Distinguish between the terms homo polymer and co polymer. Give one example of each.
Answer:
Homopolymers : The polymers which are formed by the polymerisation of a single monomeric species are known as homopolymers.
E.g.: Polyethene, Poly styrene.
Copolymers: A polymer which is formed by the polymerisation of two (or) more chemically different types of monomer units is called copolymer.
E.g.: Butadiene – Styrene polymer (Buna-S)

Question 4.
Explain the purpose of vulcanization of rubber. [T.S. Mar. 17]
Answer:

1. Natural rubber becomes soft at high temperatures and brittle at low temperatures. It shows high water absorption capacity.
2. Natural rubber is soluble in non-polar solvents and is non-resistant to oxidising agents.
3. To improve these physical properties rubber can be vulcanised.
4. The process of heating the raw rubber with sulphur (or) with sulphur compounds to improve it’s physical properties is called vulcanisation of rubber.
5. The vulcanized rubber has improved properties like elasticity, minimum water absorbing tendency, high resistance to chemical oxidation as well as organic solvents.

Question 5.
Write the names and structures of the monomers used for getting the following polymers [A.P. Mar. 17; Mar. 14]
i) Polyvinyl chloride
ii) Teflon
iii) Bakelite
iv) Polystyrene.
Answer:
i) Polyvinyl chloride .
Monomer: Vinyl chloride
Structure : CH₂ = CH – Cl

ii) Teflon
Monomer: Tetrafluoro ethylene
Structure : CF₂ = CF₂

iii) Bakelite [IPE 2015 (AP), 2014, B.M.P, 2016 (AP)
Monomers : Phenol, Formaldehyde
Structure :

iv) Polystyrene
Monomer : Styrene
Structure :
q

Question 6.
Name the different types of molecular masses of polymers.
Answer:
The different types of important molecular masses of polymers are

1. Number average molecular mass (\(\overline{\mathbf{M}}_{\mathrm{n}}\))
2. Weight average molecular mass (\(\overline{\mathbf{M}}_{\mathrm{w}}\))

Question 7.
What is natural rubber ? How does it exhibit elastic properties ?
Answer:

1. Natural rubber is a polymer and possesses elastic properties.
2. It is an elastomer and it is manufactured from rubber latex. Latex is a colloidal dispersion of rubber in water.
3. Natural rubber may be considered as a linear polymer of isoprene. It is also called as cis-1, 4-Poly isoprene.
4. The cis poly isoprene molecule consists of various chains held together by weak vander waals interactions and has a colloid structure. Thus it stretches like a spring and exhibits elastic properties.

Question 8.
Give the structure of nylon 2 – nylon 6 ?
Answer:
Nylon 2 – Nylon 6 :
It is an alternating polyamide copolymer of glycine and amino caproic acid (H₂N – (CH₂)₅ – COOH). It is a biodegradable polymer.
Structure of Nylon 2 – Nylon – 6

Question 9.
Explain copolymerization with an example. [IPE Mar – 2015 (AP), B.I.E.]
Answer:
Copolymers: A polymer which is formed by the polymerisation of two (or) more chemically different types of monomer units is called copolymer. E.g.: Butadiene-Styrene polymer(Buna-S) The process of formation of copolymer is called copolymerisation.
E.g. : Buna – S : Buna – S is the copolymer which is formed by the polymerisation of 1, 3 – Butadiene and Styrene.

Question 10.
What are LDP and HDP ? How are they formed ?
Answer:
Polythenes are two types :

1. LDP (Low Density Polythene),
2. HDP (High Density Polythene)

1) Low Density Polythene : LDP is formed by the polymerisation of ethene under high pressure of 1000 to 2000 atm. at a pressure of 350 to 570K in the presence of traces of dioxygen (or) a peroxide initiator.
Properties :
a) This is obtained through the free radical additon.
b) LDP is chemically inert and tough.
c) LDP is flexible and a poor conductor of electricity.
Uses:
a) It is used in the insulation of electric cables.
b) It is used in the manufacture of pipes in agriculture irrigation.

2) High Density Polythene : HDP is formed by the polymerisation of ethene in a hydro carbon solvent in presence of Ziegler Natta catalyst at a temperature of 333K to 343 K and under a pressure of 6 – 7 atm.
Properties :
a) HDP consists of linear molecules and has high density due to close packing.
b) It is chemically inert and more tough, hard.
Uses:
a) It is used in manufacture of house hold articles like buckets, dustbins etc.
b) It is used in manufacture of pipes.

Question 11.
What are natural and synthetic polymers ? Give two examples of each type.
Answer:
Natural polymers : The polymers which are obtained from natural sources such as plants and animals are called natural polymers.
E.g. : Starch, cellulose, rubber etc.

Synthetic polymers : The polymers which are artificially prepared i.e., man-made are called synthetic polymers.
These have wide applications in daily life as well as in industry.
E.g. : Plastics, Nylon 6, 6, synthetic rubbers.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(d)

I.

Question 1.
Find the number of ways of arranging the letters of the word.
(i) INDEPENDENCE
(ii) MATHEMATICS
(iii) SINGING
(iv) PERMUTATION
(v) COMBINATION
(vi) INTERMEDIATE
Solution:
(i) The word INDEPENDENCE contains 12 letters in which there are 3 N’s are alike, 2 D’s are alike, 4 E’s are alike and rest are different.
∴ The number of required arrangements = \(\frac{(12) !}{4 ! 3 ! 2 !}\)

(ii) The word MATHEMATICS contains 11 letters in which there are 2 M’s are alike, 2 A’s are alike, 2 T’s are alike and rest are different.
∴ The number of required arrangements = \(\frac{(11) !}{2 ! 2 ! 2 !}\)

(iii) The word SINGING contains 7 letters in which there are 2 I’s are alike, 2 N’s are alike, 2 G’s are alike and rest is different.
∴ The number of required arrangements = \(\frac{7 !}{2 ! 2 ! 2 !}\)

(iv) The word PERMUTATION contains 11 letters in which there are 2 T’s are alike and rest are different.
∴ The number of required arrangements = \(\frac{(11) !}{2 !}\)

(v) The word COMBINATION contains 11 letters in which there are 2 O’s are alike, 2 I’s are alike, 2 N’s are alike and rest are different.
∴ The number of required arrangements = \(\frac{(11) !}{2 ! 2 ! 2 !}\)

(vi) The word INTERMEDIATE contains 12 letters in which there are 2 I’s are alike, 2 Ts are alike, 3 E’s are alike and rest are different.
∴ The number of required arrangements = \(\frac{(12) !}{2 ! 2 ! 3 !}\)

Question 2.
Find the number of 7-digit numbers that can be formed using 2, 2, 2, 3, 3, 4, 4.
Solution:
In the given 7 digits, there are three 2 ‘s, two 3’s and two 4’s.
∴ The number of 7 digited numbers that can be formed using the given digits = \(\frac{7 !}{3 ! 2 ! 2 !}\)

II.

Question 1.
Find the number of 4-letter words that can be formed using the letters of the word RAMANA.
Solution:
The given word RAMANA has 6 letters in which there are 3 A’s alike and the rest are different.
Using these 6 letters, 3 cases arise to form 4 letter words.
Case I: All different letters R, A, M, N
Number of 4 letter words formed = 4! = 24
Case II: Two like letters A, A and two out of R, M, N
The two different letters can be choosen from 3 letters in ³C₂ = 3 ways.
∴ Number of 4 letters word formed = 3 × \(\frac{4 !}{2 !}\)
= 3 × 12
= 36
Case III: Three like letters A, A, A and one out of R, M, N.
One letter can be choosen from 3 different letters in ³C₁ = 3 ways.
∴ Number of 4 letter words formed = 3 × \(\frac{4 !}{3 !}\)
= 3 × 4
= 12
∴ Total number of 4 letter words formed from the word RAMANA = 24 + 36 + 12 = 72

Question 2.
How many numbers can be formed using all the digits 1, 2, 3, 4, 3, 2, 1 such that even digits always occupy even places?
Solution:
In the given 7 digits, there are two 1’s, two 2’s, two 3’s and one 4.
The 3 even places can be occupied by the even digits 2, 4, 2 in \(\frac{3 !}{2 !}\). (Even place is shown by E)

The remained odd places can be occupied by the odd digits 1, 3, 3, 1 in \(\frac{4 !}{2 ! 2 !}\) ways.
∴ The number of required arrangements = \(\frac{3 !}{2 !} \times \frac{4 !}{2 ! \times 2 !}\)
= 3 × 6
= 18

Question 3.
In a library, there are 6 copies of one book, 4 copies each of two different books, 5 copies each of three different books and 3 copies each of two different books. Find the number of ways of arranging all these books in a shelf in a single row.
Solution:
Total number of books in a library = 6 + (4 × 2) + (5 × 3) + (3 × 2) = 35
∴ The number of required arrangements = \(\frac{(35) !}{6 !(4 !)^2(5 !)^3(3 !)^2}\)

Question 4.
A book store has ‘m’ copies each of, ‘n’ different books. Find the number of ways of arranging the books in a shelf in a single row.
Solution:
Total number of books in a book store are = m × n = mn
∴ The number of required arrangements = \(\frac{(m n) !}{(m !)^n}\)

Question 5.
Find the number of 5-digit numbers that can be formed using the digits 0, 1, 1, 2, 3.
Solution:
‘O’ can also be taken as one digit,
the number of 5 digited number formed = \(\frac{5 !}{2 !}\) = 60
Among them, the numer that starts with zero is only 4 digit number.
The number of numbers start with zero = \(\frac{4 !}{2 !}\) = 12
Hence the number of 5 digit numbers that can be formed by using all the given digits = 60 – 12 = 48

Question 6.
In how many ways can the letters of the word CHEESE be arranged so that no two E’s come together?
Solution:
The given word contains 6 letters in which one C, one H, 3 E’s and one S.
Since no two E’s come together, first arrange the remaining 3 letters in 3! ways. Then we can find 4 gaps between them.
The 3 E’s can be arranged in these 4 gaps in \(\frac{{ }^4 P_3}{3 !}\) = 4 ways.
∴ The number of required arrangements = 3! × 4 = 24

III.

Question 1.
Find the number of ways of arranging the letters of the word ASSOCIATIONS. In how many of them
(i) all the three S’s come together
(ii) The two A’s do not come together
Solution:
Hint: The number of linear permutations of ‘n’ things in which ‘p’ alike things of one kind, ‘q’ alike things of 2nd kind, ‘r’ alike things of 3rd kind and the rest are different is \(\frac{n !}{p ! q ! r !}\)
The given word ASSOCIATIONS has 12 letters in which there are 2 A’s are alike, 3 S’s are alike, 2 O’s are alike 2 I’s are alike and rest are different.
∴ They can be arranged = \(\frac{(12) !}{2 ! 3 ! 2 ! 2 !}\)
(i) Treat the 3 S’s as one unit. Then we have 9 + 1 = 10 entities in which there are 2A’s are alike, 20’s are alike, 2 I’s are alike and rest are different.
They can be arranged in \(\frac{(10) !}{2 ! 2 ! 2 !}\) ways.
The 3 S’s among themselves can be arranged in \(\frac{3 !}{3 !}\) = 1 way
∴ The number of required arrangements = \(\frac{(10) !}{2 ! 2 ! 2 !}\)

(ii) Since 2 A’s do not come together, first arrange the remaining 10 letters in which there are 3 S’s are alike, 2 O’s are alike 2 I’s are alike and rest are different in \(\frac{(10) !}{3 ! 2 ! 2 !}\) ways.
Then we can find 11 gaps between them. The 2 A’s can be arranged in these 11 gaps in \(\frac{{ }^{11} P_2}{2 !}\) ways.
∴ The number of required arrangements = \(\frac{(10) !}{3 ! 2 ! 2 !} \times \frac{{ }^{11} P_2}{2 !}\)

Question 2.
Find the number of ways of arranging the letters of the word MISSING so that the two S’s are together and the two I’s are together.
Solution:
In the given word MISSING contains 7 letters in which there are 2 I’s are alike, 2 S’s are alike and the rest are different.
Treat the 2 S’s as one unit and 2 I’s as one unit. Then we have 3 + 1 + 1 = 5 entities. These can be arranged in 5! ways.
The 2 S’s can be arranged among themselves in \(\frac{2 !}{2 !}\) = 1 ways and the 2 I’s can be arranged among themselves in \(\frac{2 !}{2 !}\) = 1 way.
∴ The number of required arrangements = 5! × 1 × 1 = 120

Question 3.
If the letters of the word AJANTA are permuted in all possible ways and the words thus formed are arranged in dictionary order, find the ranks of the words
(i) AJANTA
(ii) JANATA
Solution:
The dictionary order of the letters of the word AJANTA is A A A J N T
(i) In the dictionary order first comes that words begin with the letter A.
If we fill the first place with A, we may set the word AJANTA.
Second place can be filled with A, and the remaining 4 places can be filled in 4! = 24 ways.
On proceeding like this, we get
A A – – – – – = 4! = 24
A J A A – – – = 2! = 2
A J A N A – = 1 = 1
AJANTA = 1 = 1
∴ Rank of the word AJANTA = 24 + 2 + 1 + 1 = 28

(ii) In the dictionary order first comes that words begin with the letter A.
If we fill the first place with A, the remaining 5 letters can be arranged in \(\frac{5 !}{2 !}\) ways (since there 2 A’s remain)
On proceeding like this, we get
A – – – – – – = \(\frac{5 !}{2 !}\) = 60
J A A – – – – = 3! = 6
J A N A A – – = 1 = 1
J A N A T A = 1 = 1
∴ Rank of the word JANATA is = 60 + 6 + 1 + 1 = 68.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(b)

I.

Question 1.
In the experiment of tossing a coin n times, if the variable X denotes the number of heads and P(X = 4), P(X = 5), P(X = 6) are in arithmetic progression then find n.
Solution:
X follows binomial distribution with p = \(\frac{1}{2}\), q = \(\frac{1}{2}\) (∵ a coin is tossed)
Hint: a, b, c are in A.P.
⇒ 2b = a + c (or) b – a = c – a
Given, P(X = 4), P(X = 5), P(X = 6) are in A.P.

⇒ 2 × 30(n – 4) = 5[30 + n² – 9n + 20]
⇒ 12n – 48 = n² – 9n – 50
⇒ n² – 21n + 98 = 0
⇒ n² – 14n – 7n + 98 = 0
⇒ n(n – 14) – 7(n – 14) = 0
⇒ (n – 7) (n – 14) = 0
⇒ n = 7 or 14

Question 2.
Find the minimum number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8.
Solution:
Let n be the number of times a fair coin tossed
X denotes the number of heads getting
X follows binomial distribution with parameters n and p = \(\frac{1}{2}\)
Given P(X ≥ 1) ≥ 0.8
⇒ 1 – P(X = 0) ≥ 0.8
⇒ P(X = 0) ≤ 0.2
⇒ \({ }^n C_o\left(\frac{1}{2}\right)^n \leq 0.2\)
⇒ \(\left(\frac{1}{2}\right)^n \leq \frac{1}{5}\)
The Maximum value of n is 3

Question 3.
The probability of a bomb hitting a bridge is \(\frac{1}{2}\) and three direct hits (not necessarily consecutive) are needed to destroy it. Find the minimum number of bombs required so that the probability of the bridge being destroyed is greater than 0.9.
Solution:
Let n be the minimum number of bombs required and X be the number of bombs that hit the bridge, then
X follows binomial distribution with parameters n and p = \(\frac{1}{2}\)
Now P(X ≥ 3) > 0.9
⇒ 1 – P(X < 3) > 0.9
⇒ P(X < 3) < 0.1
⇒ P(X = 0) + P(X = 1) + P (X = 2) < 0.1

By trial and error, we get n ≥ 9
∴ The least value of n is 9
∴ n = 9

Question 4.
If the difference between the mean and the variance of a binomial variate is \(\frac{5}{9}\) then, find the probability for the event of 2 successes, when the experiment is conducted 5 times.
Solution:
Given n = 5
Let p be the parameters of the Binomial distribution
Mean – Variance = \(\frac{5}{9}\)

∴ Probability of the event of 2 success = \(\frac{80}{243}\)

Question 5.
One in 9 ships is likely to be wrecked when they are set on the sail, when 6 ships are on the sail, find the probability for
(i) Atleast one will arrive safely
(ii) Exactly, 3 will arrive safely
Solution:
p = probability of ship to be wrecked = \(\frac{1}{9}\)
q = 1 – p
= 1 – \(\frac{1}{9}\)
= \(\frac{8}{9}\)
Number of ships = n = 6

Question 6.
If the mean and variance of a binomial variable X are 2.4 and 1.44 respectively, find P(1 < X ≤ 4).
Solution:
Mean = np = 2.4 ………(1)
Variance = npq = 1.44 ……….(2)
Dividing (2) by (1)
\(\frac{n p q}{n p}=\frac{1.44}{2.4}\)
⇒ q = o.6 = \(\frac{3}{5}\)
p = 1 – q
= 1 – 0.6
= 0.4
= \(\frac{2}{5}\)
Substituting in (1)
n(0.4) = 2.4
⇒ n = 6
P(1 < X ≤ 4) = P(X = 2) + P(X = 3) + P(X = 4)
= \({ }^6 C_2 q^4 \cdot p^2+{ }^6 C_3 q^3 \cdot p^3+{ }^6 C_4 q^2 \cdot p^4\)

Question 7.
It is given that 10% of the electric bulbs manufactured by a company are defective. In a sample of 20 bulbs, find the probability that more than 2 are defective.
Solution:
p = probability of defective bulb = \(\frac{1}{10}\)
q = 1 – p
= 1 – \(\frac{1}{10}\)
= \(\frac{9}{10}\)
n = Number of bulbs in the sample = 20
P(X > 2) = 1 – P(X ≤ 2)
= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

Question 8.
On average, rain falls on 12 days every 30 days, find the probability that, the rain will fall on just 3 days of a given week.
Solution:
Given p = \(\frac{12}{30}=\frac{2}{5}\)
q = 1 – p
= 1 – \(\frac{2}{5}\)
= \(\frac{3}{5}\)
n = 7, r = 3
P(X = 3) = ⁿC_r . q^n-r . p^r
= \({ }^7 C_3\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^3\)
= \(\text { 35. }\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^3\)
= \(\frac{35 \times 2^3 \times 3^4}{5^7}\)

Question 9.
For a binomial distribution with mean 6 and variance 2, find the first two terms of the distribution.
Solution:
Let n, p be the parameters of a binomial distribution
Mean (np) = 6 ……..(1)
and variance (npq) = 2 ……..(2)
then \(\frac{n p q}{n p}=\frac{2}{6}\)
⇒ q = \(\frac{1}{3}\)
∴ p = 1 – q
= 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\)
From (1) np = 6
n(\(\frac{2}{3}\)) = 6
∴ n = 9
The first two terms of the distribution are
P(X = 0) = \({ }^9 C_0\left(\frac{1}{3}\right)^9=\frac{1}{3^9}\)
and P(X = 1) = \({ }^9 C_1\left(\frac{1}{3}\right)^8\left(\frac{2}{3}\right)=\frac{2}{3^7}\)

Question 10.
In a city, 10 accidents take place in a span of 50 days. Assuming that the number of accidents follows the Poisson distribution, find the probability that there will be 3 or more accidents in a day.
Solution:
Average number of accidents per day
λ = \(\frac{10}{50}=\frac{1}{5}\) = 0.2
The probability that there win be 3 or more accidents in a day
P(X ≥ 3) = \(\sum_{\mathrm{K}=3}^{\infty} \mathrm{e}^{-\lambda} \cdot \frac{\lambda^{\mathrm{K}}}{\mathrm{K} !}, \lambda=0.2\)

II.

Question 1.
Five coins are tossed 320 times. Find the frequencies of the distribution of a number of heads and tabulate the result.
Solution:
5 coins are tossed 320 times
Probability of getting a head on a coin
p = \(\frac{1}{2}\), n = 5
Probability of having x heads

Question 2.
Find the probability of guessing at least 6 out of 10 answers in (i) True or false type examination (ii) multiple choice with 4 possible answers.
Solution:
(i) Since the answers are in True or false type
probability of success p = \(\frac{1}{2}\), q = \(\frac{1}{2}\)
probability of guessing at least 6 out of 10
P(X = 6) = \({ }^{10} \mathrm{C}_6\left(\frac{1}{2}\right)^{10-6} \cdot\left(\frac{1}{2}\right)^6\)
= \({ }^{10} \mathrm{C}_6\left(\frac{1}{2}\right)^{10}\)

(ii) Since the answers are multiple-choice with 4 possible answers
Probability of success p = \(\frac{1}{4}\), q = \(\frac{3}{4}\)
Probability of guessing at least 6 out of 10
P(X = 6) = \({ }^{10} C_6\left(\frac{3}{4}\right)^{10-6}\left(\frac{1}{4}\right)^6\)
= \({ }^{10} C_6 \cdot \frac{3^4}{4^{10}}\)

Question 3.
The number of persons joining a cinema ticket counter in a minute has Poisson distribution with parameter 6. Find the probability that
(i) no one joins the queue in a particular minute
(ii) two or more persons join the queue in a minute.
Solution:
Here λ = 6
(i) Probability that no one joins the queune in a particular minute
P(X = 0) = \(\frac{\mathrm{e}^{-\lambda} \lambda^0}{0 !}=\mathrm{e}^{-6}\)

(ii) Probability that two or more persons join the queue in a minute

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 5(a) Human Reproductive System Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 5(a) Human Reproductive System

Very Short Answer Questions

Question 1.
Where are the testes located in man? Name the protective coverings of each testis.
Answer:
The testes are male primary sex organs suspended outside the abdominal cavity within a pouch called scrotumr. Each testis is enclosed in a fibrous envelope called ‘tunica albuginea’.

Question 2.
Name the canals that connect the cavities of scrotal sac and abdominal cavity. Name the structures that keep the testes in position.
Answer:
The cavity of scrotal sac is connected to the abdominal cavity through the ‘inguinal canal’. Testes are held in position in the scrotum by the ‘gubemaculum’, a fibrous cord.

Question 3.
What are the functions of Sertoli cells of the seminiferous tubules and the Leydig cells in man?
Answer:
i) Sertoli cells :
Also known as ‘nourishing cells’ helps in the nourishment of spermatozoa and produce a hormone ‘inhibin’, which inhibits the secretion of FSH.

ii) Leydig cells :
Produce Testosterone that controls the secondary sexual characters and spermatogenesis.

Question 4.
Name the copulatory structure of man. What are the three columns of tissues in it?
Answer:
The penis is the copulatory structure of man. It is made up of the three columns of tissue; Two upper ‘Corpora Cavernosa’ on the dorsal aspect and one ‘Corpus Spongiosum’ on the ventral side.

Question 5.
Define Spermiogenesis and Spermiation.
Answer:
Spermiogenesis :
The process in which haploid spermatids are transformed into spermatozoa or sperms.

Spermiation :
The process in which sperm head becomes embedded in the Sertoli cells and finally released from the seminiferous tubules.

Question 6.
Name the yellow mass of cells accumulated in the empty follicle after ovulation. Name the hormone secreted by it and what is its function?
Answer:
The yellow mass of cells accumulated in the empty follicles after ovulation is called ‘Corpus luteum’ (yellow body). It secretes a hormone called Progesterone which has three major functions’.

i) For regular menstrual cycle.
ii) For the formation of thick endometrium in uterus.
iii) Maintenance of pregnancy after fourth month.

Question 7.
Define gestation period. What is the duration of gestation period in the human beings?
Answer:
The period during which embryo development takes place in uterus is called gestation period. In humans the gestation period is 266 days (38 weeks) from the fertilization of egg or 40 weeks from the start of last menstrual cycle.

Question 8.
What is implantation, with reference to embryo?
Answer:
The blastocyte invades the endometrium of uterus and get implanted into the uterine mucosa till the whole of it comes to lie within the thickness of the endometrium, the process is called ‘implantation1. It begins on the 6th day after fertilization.

Question 9.
Distinguish between hypoblast and epiblast.
Answer:

Question 10.
Write two major functions, each of testis and ovary?
Answer:
Functions of Testis :
a) Testis is a cytogenic gland, which produces sperms.
b) Leydig cells of testes produce a male sex hormone called ‘Testosterone’ Which controls the development of secondary sexual characters and spermatogenesis.

Functions of Ovaries :
a) Ovaries are primary female sex organs, producing on ovum during each menstrual cyle.
b) They produce female hormones; Estrogen and Progesterone.

Question 11.
Draw a labelled diagram of a sperm.
Answer:

Question 12.
What are the major components of the seminal fluid?
Answer:
Seminal fluid is an alkaline, viscous fluid that contains fructose, proteins, citric acid, inorganic phosphorus, potassium and prostaglandins. It is produced by seminal vesicles present postero inferior to the urinary bladder in the pelvis.

Question 13.
What is menstrual cycle? Which hormones regulate menstrual cycle?
Answer:
The reproductive cycle in the female primates like monkeys, apes and humans, is called ‘menstrual cycle’. The cycle is regulated by majority four hormones. They are

1. Luteinising hormone (LH),
2. ollicular stimulating hormone (FSH),
3. Estrogen and
4. Progesterone.

Question 14.
What is parturition? Which hormones are involved in inducing parturition?
Answer:
The process of delivery of the foetus, starting from labour (a series of strong, rhythemic uterine contractions that push the foetus the placenta outside the body) is called parturition. The hormone, oxytocin is responsible for the contraction of uterus during parturition.

Question 15.
How many eggs do you think were released by the ovary of a female dog which gave birth to six puppies?
Answer:
The dog might have produced single ovary but it might be fertilised by six different sperms. After fertilization the synkaryon or zygotic nucleus undergoes divisions and forms six euqal or unequal zygotes, those giving birth to six puppies.

Question 16.
What is neurulation?
Answer:
The process of formation of neural tube from neural plate in embryo, as part of organogenesis, is called neurulation.

Question 17.
What is capacitation of sperms?
Answer:
Capacitation of sperm refers to the physiological changes that the spermatozoa undergoes to be able to penetrate and fertilize an egg.

Question 18.
What is compaction in human development?
Answer:
The process in which morula becomes embryo by reducing unequal cleavage smaller and larger blastomeres and forming a superficial flat cell layer trophoblast and inner cell mass embryo proper.

Question 19.
Distinguish between involution and ingression in the human development.
Answer:

Question 20.
What are the four extra embryonic membranes?
Answer:
The four extra embryonic membranes are

1. Chronic membrane
2. Amnionic membrane
3. Allantoic membrane
4. Yolk sac.

Short Answer Questions

Question 1.
Describe microscopic structure of testis of man.
Answer:
The testes or testicles area pair of oval pinkish male primary sex organs suspended outside the abdominal cavity within a pouch called ‘scrotum’. The low temperature (2 – 2.5°C) is maintained in the scrotum to promote spermatogenesis.

Each testis is enclosed in a fibrous envelope, ’tunica albuginea’. The envelope extends inward to form septa that partition the testis into lobules. There are nearly 250 testicular lobules in each testis. Each lobule contains 1 to 3 highly coiled ’seminiferous tubules’. A pouch of serous membrane, called ‘tunica vaginalis’ covers the testis.

Each seminiferous tubule is lined by ‘germinal epithelium’ which consists of undifferentiated male germ cells called ‘spermatogonial mother cells’ and nourishing cells called ‘Sertoli cells’. These cells provide nutrition to spermatozoa and also produce inhibin, that inhibits the secretion of FSH. The region outside the seminiferous tubules contain Leydig cells. These cells produce androgens, the most important is ‘Testosterone’. It controls the development of secondary sexual characters and spermatogenesis. The seminiferous tubules open into ‘vasa efferentia’ through ‘rete testis’. Rete testis is a network of tubules in the testis carrying spermatozoa from seminiferous tubules to vasa efferentia.

Question 2.
Describe the microscopic structure of ovary of woman.
Answer:
Ovaries are the primary female sex organs that produce the ‘female garnets’ or ‘ova’ and several steroid hormones (ovarian hormones). A pair of ovaries is located one on each side of the lower abdomen. The double layered fold of peritoneum connecting the ovary with the wall of abdominal cavity is known as the ‘mesovarium’.

The ovaries are covered on the outside by a’ layer of simple cuboidal epithelium called ‘germinal (ovarian) epithelium. This is actually the visceral peritoneum that envelopes the ovaries. Under this layer there is a dense connective tissue capsule, the ‘tunica albuginea’. The ovarian stroma is distinctly divided into an outer cortex and an inner’ medulla. The cortex is dense and granular due to the presence of numerous ovarian follicles in various stages of development. The medulla is a loose connective tissue with abundant blood vessels, lymphatic vessels and nerve fibers.

Question 3.
Describe the Graafian follicle in woman.
Answer:
During ‘oogenesis’, the formed garnet mother cells or oogonia in each foetal ovary are called primary oocytes. Each primary oocyte gets sorrounded by a flattened layer of follicular cells. It is called ‘primordial follicle’. The follicles become cuboidal and proliferate to produce stratified epithelium made up of cells called granulosa cells. Follicles at this stage of development are called ’primary follicles’. A homogenous membrane, the ‘zona pellucida’ appears between primary oocyte and granulosa cells. The innermost layer of granulosa cells are firmly attached to zona pellucida forming ‘corona radiata’.

A cavity appears in membrane granulosa, it increases in size, wall of follicle becomes thin. As the follicle-expands the stromal cells sorrounding the granulosa become condensed to forma covering called inner theca interna’ and outer ’theca externa’. Now these follicles are called ‘secondary follicles’.

The cells of theca interna secrete a hormone called Oestrogen. At this stage, the primary oocyte within the secondary follicles grows in size and completes ‘meiosi§ I’ forming a large haploid ‘Secondary oocyte’ and a small ‘first polar body’. Then the 2nd meoitic division starts but stops at metaphase. The secondary follicle further changes into the nature follicle called ‘Grafian follicle’. The rupture of graafian follicle by LH results in the release of ovum, a process called ovulation.

Question 4.
Draw a labelled diagram of the male reproductive system.
Answer:

Question 5.
Draw a labelled diagram of the female reproductive system.
Answer:

Question 6.
Describe the structure of seminiferous tubule.
Answer:
Seminiferous tubules are present in the lobules of testes. Each seminiferous tubule is lined by the germinal epithelium which consists of undifferentiated male germ cells called spermatogonial mother cells and nourishing cells’ called ‘Sertoli cells’. The spermatogenia produce primary spermatocytes which undergo meoitic division finally leading to the formation of Spermatozoa or sperms (spermatogenesis).

Sertoli cells provide nutrition to spermatozoa and produce a hormone ‘inhibiri which Sertoli cells provide nutrition to spermatozoa and produce a hormone ‘inhibiri which inhibits the secretion of FSH. The regions outside the seminiferous tubules called interstitial spaces, contain Leydig cells. Leydig cells produce androgens, mainly Testosterone that controls the development of secondary sexual characters and spermatogenesis. The seminiferous tubules open into ‘vasa efferentia’ through the ‘rete testis’.

Question 7.
What is Spermatogenesis? Briefly describe the process of Spermatogenesis in man.
Answer:
During puberty, in the testis the immature male germ cells, spermatogonia produce sperms by spermatogenesis. The spermatogonial stem cells in seminiferous tubules multiply by repeated mitotic divisions and develops> into primary spermatocytes with 46 chromosomes. A primary spermatocyte undergoes first meiotic division to produce 2 equal sized haploid ‘secondary spermatocyte’ which have only 23 chromosomes. These undergo second meotic division to produce four haploid ’spermatidis’ which intum transform into spermatozoa (sperms) by the process called spermoigenesis. After spermiogenesis, sperm heads become embedded is Sertoli cells, and are finally released from the seminiferous tubules by the process ‘spermiation’.

Spermatogenesis starts at the age of puberty due to increase in the secretion of gonadotropin releasing hormone (GnRH), produced from hypothalamus. The increased levels of GnRH intum stimulates pituitary gland to produce two gonadotropins.
a) Lutenising Hormone (LH)
b) Follicular Stimulating Hormone (FSH)

LH acts on Leydig cells and stimulates the secretion of androgens. Androgens inturn stimulate the process of spermato-genesis. FSH acts on Sertoli cells and stimulates secretion of some factors which help in the process of Spermio-genesis.

Question 8.
What is Oogenesis? Give a brief account of Oogenesis in a woman.
Answer:
The process of formation of mature female gamqtds called ‘Oogenesis’. It is initiated during the embryonic development stage when a couple of million garnet mother cells (oogonia) are formed within each foetal ovary and do not multiply thereafter. These cells start division and stop the process at prophase I qf the meiosis-I. At this stage they are called primary oocytes.

Each primary Oocyte is sorrounded by a flattened layer of follicular cells, it is called orimary follicle. These follicular cells become cuboidal and proliferate to produce a stratified epithelium called membrana granulosa. These cells are called granulosa cells. Follicles at this stage are called primary follicles. A homogenous membrane, the ‘zona pellucida’, appears between primary oocyte and granulosa cells.

A cavity appears within the membrana granulosa increases in size and the wall of follicle becomes thin. The oocyte lies eccentrically in the follicle sorrounded by granulosa cells. As the follicle expands the stromal cells sorrounding membrana granulosa become condensed to form a inner covering ‘theca interna’ and outer covering ‘theca externa’. Now these are called ‘secondary follicles’.

The cells of theca interna later secrete a hormone called ’oestrogen1. The primary oocyte within the graafian undergoes two meoitic divisons, finally changing into a Graafian follicle’. The Graafian follicle is at first very small, later it enlarges, becomes so big that it not only reaches the surface of ovary, but also forms a bulging in this situation. Ultimately the follicle ruptures releasing the ovum.

Question 9.
Draw a labelled diagram of Graafian follicle.
Answer:

Question 10.
In our society women are often blamed for giving birth to daughters. Can you explain why this is not correct?
Answer:
The sex of a child depends on the male parent but not on female parent. The sex of the baby has been decided at the time of fertilization itself. The chromosome pattern in human female is XX and that in the male is XY. Therefore all the haploid garnets produced by the female (ova) have the sex chromosome X, whereas the male garnets (sperms) have either X chromosome or Y chromosome. 50 percent of sperms carry the X chromosome while th^ other 50 percent carry the Y chromosome.

After fusion of the male and female garnets the zygote would carry either XX or XY depending on what type of sperm fertilised the ovum. The zygote carrying ‘XX would develop into a female child and that with XY would form a male child. So, the sex of a child depends on the male parent (heterogametic parent) but not on mother.

Question 11.
Describe the accessory glands associated with male reproductive system of man.
Answer:
The male accessory glands are :
a) Seminal vesicles
b) Prostate gland
c) Bulbourethral glands

a) Seminal vesicles :
The seminal vesicles are a pair of simple tubular glands present postero-inferior to the urinary bladder in the pelvis. Each seminal vesicle opens into cprresponding vas deferens thus enters into prostate gland. The secretion of the seminal vesicles constitutes about 60 percent of the volume of seminal fluid. It is an alkaline, viscous fluid that contains fructose, proteins, citric acid, inorganic phosphorous, potassium and prostaglandins. Fructose is main energy source for the sperm and prostaglandins aid fertilization by causing mucous lining of the cervix to be more receptive to sperm as well as by aiding the movement of sperm towards the ovum.

b) Prostate gland :
Prostate gland is located directly beneath the urinary bladder. The gland sorrounds the ‘prostatic urethra’ and sends its secretions through several prostatic ducts. In man, the prostate contributes 15-30 percent of the semen. The prostate secretion activates spermatozoa and provides nutrition.

c) Bulbourethral glands :
They are also called ‘cowper’s glands’, are located beneath the prostate gland at the beginning of the internal portion of the penis. They add an alkaline fluid to semen during the process of ejaculation. The fluid secreted by these glands lubricates the urethra. It also functions as a flushing agent, that washes out the acidic urinary residues that remaindn urethra, before the semen is ejaculated.

Question 12.
Describe the placenta in a women.
Answer:
Placenta is a structural and functional unit of both chorionic villi and uterine tissue and it develops between the embryo (foetus) and the mother. The maternal and foetal blood do not mix with each other. They are seperated by the placental membrane.

The placenta consists of two essential portions : a maternal part of the placenta derived from the endometrium of the uterus and foetal membranes of the foetal part of the placenta.

The maternal components of the Placenta are : .

1. Uterine epithelium
2. terine connective tissue
3. Uterine capillary endothelium.

The foetal components of the Placenta are :

1. Foetal capillary endothelium
2. Foetal connective tissue
3. Foetal chorionic epithelium.

The Placenta of human is called chorioallantoic placenta’as allantois fuse with chorion in the process of vascularisation. Placenta is discoidal as the Villi are restricted to the dorsal surface of blastodisc. Placenta is haemochorial as the maternal blood comes into direct contact with foetal chorion. During parturition the placenta is cast off with the loss of embryonic membranes and the encapsulating maternal tissues (decidua) causing extensive haemorrhage and there by bleeding. So, it is also called deciduate placenta.

Functions of placenta :

1. Supplies Oxygen and nutrients to the embryo.
2. Removes CO₂ and excretory materials produced by embryo.
3. Secretes Progesterone which is essential for maintenance of pregnancy after 4^th month.
4. Secretes Oestrogens (mainly estradiol) that reach maternal blood and promote uterine growth and development of mammary glands.
5. Secretes Human Chorionic Gonadotropin (HCG) that is similar to luteinizing hormone is its action. This hormone is also used as indicator in the detection of pregnancy is early stages.
6. Somatomammotropin secreted by placenta has an anti-insulin effect on the mother leading to increased plasma levels of glucose and aminoacids in the maternal circulation. In this way it increases the availability of these materials to the foetus.

Long Answer Questions

Question 1.
Describe female reproductive system of a woman with the help of a labelled diagram.
Answer:
The female reproductive system consists of a pair of ovaries along with a pair of oviducts, uterus, vagina and the external genetalia located in the pelvic region. These parts of the system along with a pair of mammary glands are integrated structurally and functionally to support the processes of ovulation, fertilization, pregnancy, birth and child care.

1) Ovaries :
Ovaries are the primary female sex organs that produce the female gametes (ova) and several steroid hormones (ovarian hormones). A pair of ovaries is located on each side of lower abdomen. The double layered fold of peritoneum connecting the ovary with the wall of abdominal cavity is known as the meso ovarium.

The Ovaries are covered by a layer of’germinal (ovarian) epithelium. Underneath this layer, there is a dense connective tissue capsule called, ‘tunica albuginea’. The ovarian stroma is distinctly divided into an outer cortex and an inner medulla. The cortex appears more dense and granular due to numerous ovarian follicles. The medulla is a loose connective tissue with abundant blood vessels, lymphatic vessels and nerve fibres.

2) Fallopian tubes (oviducts) :
Each fallopian tube extends from the periphery of each ovary to the uterus and it bears a funnel shaped infundibulum, with finger like projections called ‘fimbriae’, which help in collection of ovum after ‘ovulation’.

The infundibulum leads to a wider part of the oviduct called ‘ampulla’. The last part of the oviduct, ‘isthmus’ has a narrow lumen and it joins the uterus. Fallopian tube is the site of fertilization. It conducts the ovum or zygote towards the uterus by peristalsis. The fallopian tube is attached to the abdominal wall by a peritoneal fold called ‘meso salpinx’.

3) Uterus :
The uterus is a single and it is also called womb. It is a large, muscular, highly vascular and inverted pear-shaped structure present in the pelvis between the bladder and rectum. The lower narrow part through which the uterus opens into the vagina is called ’cervix’. The cavity of the cervix is called ‘cervical canal’ which along with the vagina forms the ‘birth canal’.

The wall of the uterus has three layers of tissue. The external thin membranous ‘perimetrium’, the middle thick layer of ‘myometrium’ and inner glandulas lining layer”chlled ‘endometrium’. The endometrium undergoes cyclic changes during menstrual cycle while myometrium exhibits strong contractions during parturition.

4) Vagina :
The vagina is a large, median, fibromuscular tube that extends from the cervix to the vestibule (the space between labia minora). It is lined by non- keratinised stratified squamous epithelium. It is highly vascular and opens into the vestibule by the vaginal orifice.

5) Vulva:
Vulva or pudendum refers to the external genitals of the female. The vestibule has two apertures – the upper external urethral orifice of the urethra and the lower vaginal orifice of vagina. Vaginal orifice is covered by a mucous membrane ‘hymen’. vestibule is bound by two pairs of fleshly folds of tissue called inner ‘labia minora’ and outer larger ‘labia majora’. Clitoris is a sensitive, erectile structure, that lies at the upper junction of the two labia minora above the urethral opening. There is a cushion of fatty tissue covered by skin and pubic hair present above the labia major, called mons pubis.

Accessory reproductive glands of female : These include; .

a) Bartholin’s glands :
These are two glands located slightly posterior and to the left and right of the opening of the vagina! They secrete mucus to lubricate the vagina and are homologous to the bulbourethral glands of the male reproductive system.

b) Skene’s glands :
These are located on the anterior wall of vagina, around the lower end of the urethra. They secrete a lubricating fluid when stimulated. The skene’s glands are homologous to the prostate gland of the male reproductive system.

c) Mammary glands :
These are paired structures that contain glandular tissue and fat. The alveoli cells present in the mammary lobes of each glandular tissue secrete milk, which is stored in cavities of alveoli. The alveoli open into mammary tubes and then to mammary ducts, from there to mammary ampulla and finally connected to lactiferous duct through which milk is sucked out by the baby.

Question 2.
Describe male reproductive system of a man. Draw a labelled diagram of it.
Answer:
The male reproductive system or male genital system consists of a number of sex organs that are a part of the human reproductive process. The sex organs which are located in the pelvic region include a pair of testes, accessory ducts, glands and external genitalia.

1) Testes:
The testes are a pair of oval pinkish male sex organs suspended in abdominal cavity within a pouch called scrotum. The scrotum helps in maintaining the low temperature of the testes (2-2.5°C) necessary for spermatogenesis. The cavity of- scrotal sac is connected to the abdominal cavity through the ’inguinal canal’ Testes is held in position in the scrotum of the ‘gubemaculum’, a fibrous cord that connects the testis with the bottom of scrotum and a ‘spermatic Cord’, formed by the vas deferens, nerves, blood vessels and other tissues that run from abdomen down to each testicle, through inguinal cartal.

Each testis is enclosed in a fibrous envelope, ’tunica albuginea’, which extends inwards into testis and divide it into lobules. Each lobule contains 1 to 3 highly coiled seminiferous tubules. A pouch of serous membrane ‘tunica vaginalis’ covers the testis.

Miniferous tubules :
Each seminiferous tubule is lined by ‘germinal epithelium’ which consists of undifferentiated male gum cells called ’spermatogonial mother cells’ and it also bears ‘nourishing cells’ called ‘sertoli cells’.

→ Spermatogonial cells (or) primary spermatocytes undergo meiotic division, producing spermatozoa or sperms by a process spermatogenesis.

→ Sertoli cells provide nutrition to spermatozoa and produce a hormone ‘inhibin’, which inhibits secretion of FSH.

The region outside the tubules, contain interstitial cells of ‘Leydig cells’. They produce androgens, the most important in testosterone. It controls the development of secondary sexual characters and spermatogenesis. The seminiferous tubules open into vasa efferntia through the rete testis. Rete testis is a network of tubules is of the testis carrying spermatozoa from the seminiferous tubules to the vasa efferentia,

2) Epididymis :
The vasa efferntia leave the testis and open into a narrow, tightly coiled tube called ‘epididymis’ located along the posterior surface of each testis. The epididymis provides a storage space for sperms and gives them time tofrnature.

It is differentiated into three regions.

1. Caput epididymis
2. Corpus epididymis
3. Cauda epididymis

The caput epididymis receives spermatozoa via the vasa efferntia of the mediastinum testis. It is mass of a connective tissue at the back of the testis that encloses the rete testis. . .

3) Vasa deferentia :
The vas deferens or ductus deferent is a long, narrow mascular tube. The mucosa of the ductus deferens consists of a pseudo stratified columnar epithelium and lamina propia. It starts from the tail of epididymis, passes through the inguinal canal into the abdomen and loops over the urinary bladder. It receives a duct from seminal vesicle.

The vas deferens and the duct of the seminal vesicle units to form a ‘short ejaculatory duct’ or ‘ductus ejaculatorius’ . The two ducts, carrying spermatozoa and the fluid secreted by the seminal vesicles, converge in the centre of prostate and open into urethra, which transports the sperms to outside.

4) Urethra :
In male, Urethra is the shared terminal duct of the reproductive and urinary systems. The urethra originates from urinary bladder and extends through the penis to its external opening called ‘urethral meatus’. The urethra provides an exit for urine as well as semen during ejaculation.

5) Penis :
Urethra opens into the major copulatory organ of male, the ‘penis’. The penis and scrotum constitute the male external genitalia. The penis serves as a urinal duct and intromittent organ the transfers spermatozoa to the vagina of a female.

The penis is made up of three columns of tissue : two upper Corpora cavernosa on the dorsal aspect and one Corpus spongiosum on the ventral side. Skin and a subcutaneous layer encloses all three columns, which consists of special tissue that helps in erection of penis. The enlarged and bulbous end of penis is called ‘glans penis’, which is covered by a loose fold of skin (foreskin) called prepuce.

Male accessory glands : Male accessory glands are :
a) Seminal vesicles (b) Prostate glands (c) Bulbourethral glands

a) Seminal vesicles:
These are a pair of simple tubular glands present postero-inferior to the urinary bladder in the pelvis. Each seminal vesicle enters prostate gland through vas deferens. The vesicles produce seminal fluid rich is fructose, proteins, citric acid, in organic phosphorus, potassium and prostaglandins. All these serve sperm cells.

b) Prostate gland :
It is located directly beneath the urinary bladder. The gland surrounds the ‘Prostatic urethra’, and sends its secretions through prostatic ducts. The prostatic secretion activates spermatozoa and provides nutrition. In man, the prostate contributes 15-30% of the semen.

c) Bulbourethral glands :
These are also called cowper’s glands located beneath the prostate gland at the beginning of the internal portion of the penis. They add an alkaline fluid to semen and the fluid secreted by them lubricates urethra. It acts as flushing agent washing out the acidic urinary residues that remain in the urethra, before the semen is ejaculated.

Question 3.
Write an essay on different events that occur during development of a human.
Answer:
During the development of zygote into a human, a series of events takes place in mother’s womb. They are :

I. Fertilisation
II. Gastrulation
III. Organogenesis
IV. Placenta formation
V. Pregnancy and Parturition.

I. Fertilisation:
It is the formation of zygote by fusing ovum with sperms. Sperm makes it way through corona radiata and zona pellucida. Acrosin released from acrosome of sperm dis¬solves zona pellucida of ovum and easiers penetration of sperm into ovum. The entry of sperm, induces the comple- tion of meiosis of ovum. The nuclear union results in the formation of synkaryon (zygotic nucleus).

After fertilisation the events that follow are :
1) Cleavage :
After the fertilisation, the first phase of embryonic development in cleavage of zygote as it moves through the isthmus of the oviduct towards the uterus. The daughter cells are called blastomeres.

2) Morula :
Morula is a solid mass of cells and is developed in the fallopian tube and reaches the uterus for further development. The morula passes through a process called compaction, after which the embryo has a superficial flat cell layer and inner cell mass. The outer superficial layer becomes the ‘trophoblast’ that serves to attach the embryo to the uterine wall and the inner cell mass constitutes formative cells, which give rise to the ’embryo proper’, also called the ’embryoblast’.

3) Blastocyst:
Some fluid passes into the morula from uterine cavity thus seperating the inner cell mass from trophoblast. As the quantity of fluid increases, the morula acquires a cyst shape. The cells of trophoblast become flattened and the inner cell mass attaches to the innerside of trophoblast on one side only. The morula now becomes a ‘blastocyst’.

The cavity of the blastocyst is the blastocoeV or ‘segmentation cavity’ or ‘primary body cavity’. The side of blastocyst to which the inner cell mass is attached is called ’embryonic’ or animal pole1, while the opposite side is the ‘abembryonic pole’. The cells of the trophoblast above the region of inner cell mass are called ‘cells of rauber’.

4) Implantation :
The zona pellicida around the blastocoel disappears and cells of trophoblast stick to the uterine endometrium. The trophoblast invades endometrium and gets implanted into if, called ‘interstitial implantation’, which .starts on the 6th day after fertilisation. After the implantation, the uterine endometrium is differentiated into ‘decidua’.

a) The portion of the decidua where the placenta is to be formed is called the ‘decidua basalis’.
b) The part of the decidua that seperates the embryo from the uterine lumen is called the ’decidua capsularis’.
c) The part lining the rest of the uterine cavity is called the decidua perietalis.

At the end of pregnancy the decidua is shed off, along with the placenta and membranes.

5) Formation of Bilaminar Embryonic disc :
The inner cell mass of blastocyst forms into a disc called ’embryonic disc’. Then the ‘cells of Rauber’ disapearts, some cells get seperated by delamination and eventually forms a layer of cells, that further develops into the hypoblast, which is the future extra embryonic endoderm. The remaining part of the disc is called ‘epiblast’. Now the embryonic disc is called ‘bilaminar embryonic disc’.

The hypoblast layer below the trophoblast encloses a cavity called ‘yolk sac’ or i ’umblical vesicle’. Gradually the embryonic disc becomes oval.

II. Gastrulation :
Gastrulation involves proliferation, differentiation of movement of cells with in the embryo. Along the longitudinal axis of the embryonic disc, a primitive streak is formed. A longitudinal furrow, ‘primitive groove’ forms along the middle of primitive streak. On either side of it are the ‘primitive folds’. Anteriorly primitive streak has a shallow primitive pit. The thickened part of streak is called ‘primitive knot’ or ‘primitive node’ or ‘Hansen’s node’.

1) Trilaminar Embryo:
The furture epidermal cells from epiblast, forms the endoderm of embryo. The remaining epiblast is known as ‘ectoderm’. The invasion of epiblast cells into space between the epiblast and hypoblast is called gastrulation. The process of gastrulation converts the bilaminar embryonic disc to trilaminar embryonic disc.

2) Extra embryonic membranes :
Now four extra embryonic or foetal membranes are formed. They are chorion, amnion, allantios and yolk sac.

1. Between the amnion and the embryo, there is an ‘amniotic cavity’ filled with aminiotic fluid, that acts as shock absorber and also prevents embryo from shock.
2. Allantios and chorion are fused to form ‘chorio allantoic membrane’ which constitute placenta.
3. Yolk sac encloses a fluid cavity, it has no nutritive value.

III. Organogenesis :
In involves series of stages.
1) Formation of Notochord and Neural tube :
The chora mesodermal cells converge and involute through Hensen’s node and extends forward as ‘notochordal process’. This later transforms into a solid rod – ‘notochord’. The notochord mesoderm induces the overlying endodermal cells to form neural cells which further changes into a neural tube by a process called neurlation.

2) Differentiation of Mesoderm and Formation of Coelom:
The longitudinal column of mesoderm adjacent to neural tube on either side is called ‘epimere’ and the mesoderm around the gut is ‘hypomere’. The mesoderm in between these two is ‘mesomere’. The ‘somites’ of epimere differentiate into myotome, sclerotome and dermatome.

1. Sclerotome – forms the vertebral column
2. Dermatome – forms the dermis of the skin
3. Myotome – forms the voluntary muscles of the body

The hypomere splits into outer somatic a.ij inner splanchnic mesodermal layers.

Intra embryonic coelom is formed in between these two layers, which given rise to oericardial, pleural and peritoneal cavities.

IV. Placenta famation :
The chorionic villi and uterine tissue interdigitate with each, other to form a structural and functional unit called ‘placenta’ between the foetus and the mother. The maternal and foetal blood are seperated by ‘placental membrane.

Functions of Placenta :

1. Supplies O₂ and nutrients to the embryo.
2. Removes CO₂ and waste materials from embryo.
3. Progesterone secreted by it essential for maintenance of pregnancy.
4. Oestrogen secreted by it promotes uterine growth.
5. hCG produced is used as a test to detect pregnancy.
6. Somato mammotropin increases glucose levels of plasma.

V Pregnancy and Parturition :
Pregnancy :
It is the intra uterine development of embryo and foetus. In humans it is 266 days (38 weeks) from the fertilization of egg.

Events during pregnancy:
Human gestation can be divided into 3 trimesters of three months each. The events are :
One month – Embryo’s heart is formed
Second month – Foetus develops limbs and digits
Third month – Major organs are formed
Fifth month – First movements and appearance of hair and head.
Six months – Body is covered with fine hair, eye lids seperate and eye lashes are formed.
Nine months – Foetus is fully developed.

Parturition :
The process of delivery of foetus after labour is called parturition and is favoured by hormone oxytocin that causes stronger uterine contractions. This leads to the expulsion of baby out of the uterus through the birth canal.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(e) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(e)

I.

Question 1.
If ⁿC₄ = 210, find n.
Solution:
ⁿC_r = \(\frac{n !}{(n-r) ! r !}\) = \(\frac{n(n-1)(n-2) \ldots \ldots[n-(r-1)]}{1.2 .3 \ldots \ldots \ldots . . . r}\)
Solution:
ⁿC₄ = 210
⇒ \(\frac{n(n-1)(n-2)(n-3)}{1.2 .3 .4}=10 \times 21^n C_{2 r-1}\)
⇒ n(n – 1) (n – 2) (n – 3) = 10 × 21 × 1 × 2 × 3 × 4
⇒ n(n – 1) (n – 2) (n – 3) = 10 × 7 × 3 × 2 × 3 × 4
⇒ n(n – 1) (n – 2) (n – 3) = 10 × 9 × 8 × 7
⇒ n = 10

Question 2.
If ¹²C_r = 495, find the possible values of r.
Solution:
Hint: ⁿC_r = ⁿC_n-r
¹²C_r = 495
= 5 × 99
= 11 × 9 × 5
= \(\frac{12 \times 11 \times 9 \times 5 \times 2}{12 \times 2}\)
= \(\frac{12 \times 11 \times 10 \times 9}{1.2 .3 .4}\)
= ¹²C₄ or ¹²C₈
∴ r = 4 or 8

Question 3.
If 10 . ⁿC₂ = 3 . ⁿ⁺¹C₃, find n.
Solution:
10 . ⁿC₂ = 3 . ⁿ⁺¹C₃
⇒ 10 × \(\frac{n(n-1)}{1.2}=\frac{3(n+1)(n-1)}{1.2 .3}\)
⇒ 10 = n + 1
⇒ n = 9

Question 4.
If ⁿP_r = 5040 and ⁿC_r = 210, find n and r.
Solution:
Hint: ⁿP_r = r! ⁿC_r and ⁿP_r = n(n – 1) (n – 2) ……. [n – (r – 1)]
ⁿP_r = 5040, ⁿC_r = 210
r! = \(\frac{{ }^n P_r}{{ }^n C_r}=\frac{5040}{210}=\frac{504}{21}\) = 24 = 4!
∴ r = 4
ⁿP_r = 5040
ⁿP₄ = 5040
= 10 × 504
= 10 × 9 × 56
= 10 × 9 × 8 × 7
= ¹⁰P₄
∴ n = 10
∴ n = 10, r = 4

Question 5.
If ⁿC₄ = ⁿC₆, find n.
Solution:
ⁿC_r = ⁿC_s ⇒ r = s or r + s = n
ⁿC₄ = ⁿC₆
∴ n = 4 + 6 = 10, (∵ 4 ≠ 6)

Question 6.
If ¹⁵C_2r-1 = ¹⁵C_2r+4, find r.
Solution:
¹⁵C_2r-1 = ¹⁵C_2r+4
2r – 1 = 2r + 4 or (2r – 1) + (2r + 4) = 15
(2r – 1) + (2r + 4) = 15
⇒ 4r + 3 = 15
⇒ 4r = 12
⇒ r = 3
∴ 2r – 1 = 2r + 4
⇒ -1 = 4 which is impossible
∴ r = 3

Question 7.
If ¹⁷C_2t+1 = ¹⁷C_3t-5, find t.
Solution:
¹⁷C_2t+1 = ¹⁷C_3t-5
2t + 1 = 3t – 5 or (2t + 1) + (3t – 5) = 17
⇒ 1 + 5 = t or 5t = 21
⇒ t = 6 or t = \(\frac{21}{5}\) which is not an integer
∴ t = 6

Question 8.
If ¹²C_r+1 = ¹²C_3r-5, find r.
Solution:
¹²C_r+1 = ¹²C_3r-5
⇒ r + 1 = 3r – 5 or (r + 1) + (3r – 5) = 12
⇒ 1 + 5 = 2r or 4r – 4 = 12
⇒ 2r = 6 or 4r = 16
⇒ r = 3 or r = 4
∴ r = 3 or 4

Question 9.
If ⁹C₃ + ⁹C₅ = ¹⁰C_r then find r.
Solution:
ⁿC_r = ⁿC_n-r
¹⁰C_r = ⁹C₃ + ⁹C₅
∴ ⁹C₃ + ⁹C₅ = ⁹C₃ + ⁹C₅ = ¹⁰C₆ or ¹⁰C₄ = ¹⁰C_r (given)
⇒ r = 4 or 6

Question 10.
Find the number of ways of forming a committee of 5 members from 6 men and 3 ladies.
Solution:
Total number of persons = 6 + 3 = 9
∴ Number of ways of forming a committee of 5 members from 6 men and 3 ladies = ⁹C₅
= \(\frac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1}\)
= 126

Question 11.
In question no. 10, how many committees contain atleast two ladies?
Solution:
Since a committee contains atleast 2 ladies, the members of the committee may be of the following two types.
(i) 3 men, 2 ladies
(ii) 2 men, 3 ladies
The number of selections in the first type = ⁶C₃ × ³C₂
= 20 × 3
= 60
The number of selections in the second type = ⁶C₂ × ³C₃
= 15 × 1
= 15
∴ The required number of ways of selecting the committee containing atleast 2 ladies = 60 + 15 = 75.

Question 12.
If ⁿC₅ = ⁿC₆, then find ¹³C_n.
Solution:
∵ ⁿC₅ = ⁿC₆
⇒ n = 6 + 5 = 11
¹³C_n = ¹³C₁₁ = ¹³C₂
= \(\frac{13 \times 12}{1 \times 2}\)
= 78

II.

Question 1.
Prove that for 3 ≤ r ≤ n, ^(n-3)C_r . ^(n-3)C_(r-1) + 3 . ^(n-3)C_(r-2) + 3 . ^(n-3)C_(r-3) = ⁿC_r
Solution:

Question 2.
Find the value of ¹⁰C₅ + 2 . ¹⁰C₄ + ¹⁰C₃
Solution:
Hint: ⁿC_r + ⁿC_r-1 = ⁽ⁿ⁺¹⁾C_r

Question 3.
Simplify ³⁴C₅ + \(\sum_{r=0}^4{ }^{(38-r)} C_4\)
Solution:

Question 4.
In a class, there are 30 students. If each student plays a chess game with each of the other students then find the total number of chess games played by them.
Solution:
Number of students in a class = 30
Since each student plays a chess game with each of the other students, the total number of chess games played by them = ³⁰C₂ = 435

Question 5.
Find the number of ways of selecting 3 girls and 3 boys out of 7 girls and 6 boys.
Solution:
The number of ways of selecting 3 girls and 3 boys Out of 7 girls and 6 boys = ⁷C₃ × ⁶C₃
= 35 × 20
= 700

Question 6.
Find the number of ways of selecting a committee of 6 members out of 10 members always including a specified member.
Solution:
Since a specified member is always included in a committee, the remaining 5 members can be selected from the remaining 9 members in ⁹C₅ ways.
∴ Required number of ways selecting a committee = ⁹C₅ = 126

Question 7.
Find the number of ways of selecting 5 books from 9 different mathematics books such that a particular book is not included.
Solution:
Since a particular book is not included in the selection, the 5 books can be selected from the remaining 8 books in ⁸C₅ ways.
∴ The required number of ways of selecting 5 books = ⁸C₅ = 56

Question 8.
Find the number of ways of selecting 3 vowels and 2 consonants from the letters of the word EQUATION.
Solution:
The word EQUATION contains 5 vowels and 3 consonants.
The 3 vowels can be selected from 5 vowels in ⁵C₃ = 10 ways.
The 2 consonants can be selected from 3 consonants in ³C₂ = 3 ways.
∴ The required number of ways of selecting 3 vowels and 2 consonants = 10 × 3 = 30

Question 9.
Find the number of diagonals of a polygon with 12 sides.
Solution:
The number of diagonals of a polygon with sides = \(\frac{n(n-3)}{2}\)
= \(\frac{12(12-3)}{2}\)
= 54

Question 10.
If n persons are sitting in a row, find the number of ways of selecting two persons, who are sitting adjacent to each other.
Solution:
The number of ways of selecting 2 persons out of n persons sitting in a row, who are sitting adjacent to each other = n – 1

Question 11.
Find the number of ways of giving away 4 similar coins to 5 boys if each boy can be given any number (less than or equal to 4) of coins.
Solution:
The 4 similar coins can be divided into different groups as follows.
(i) One group containing 4 coins
(ii) Two groups containing 1, 3 coins respectively
(iii) Two groups containing 2, 2 coins respectively
(iv) Two groups containing 3, 1 coins respectively
(v) Three groups containing 1, 1, 2 coins respectively
(vi) Three groups containing 1, 2, 1 coins respectively
(vii) Three groups containing 2, 1, 1 coins respectively
(viii) Four groups containing 1, 1, 1, 1 coins respectively
these groups can given away to 5 boys in = \({ }^5 C_1+2 \times{ }^5 C_2+{ }^5 C_2+{ }^5 C_3 \times \frac{3 !}{2 !}+{ }^5 C_4\)
= 5 + 20 + 10 + 30 + 5
= 70 ways

III.

Question 1.
Prove that \(\frac{{ }^{4 n} C_{2 n}}{{ }^{2 n} C_n}=\frac{1.3 .5 \ldots \ldots(4 n-1)}{\{1.3 .5 \ldots \ldots(2 n-1)\}^2}\)
Solution:

Question 2.
If a set A has 12 elements, find the number of subsets of A having
(i) 4 elements
(ii) Atleast 3 elements
(iii) Atmost 3 elements
Solution:
Number of elements in set A = 12
(i) Number of subsets of A with exactly 4 elements = ¹²C₄ = 495

(ii) The required subset contains atleast 3 elements.
The number of subsets of A with exactly 0 elements is ¹²C₀
The number of subsets of A with exactly 1 element is ¹²C₁
The number of subsets of A with exactly 2 elements is ¹²C₂
Total number of subsets of A formed = 2¹²
∴ Number of subsets of A with atleast 3 elements = (Total number of subsets) – (number of subsets contains 0 or 1 or 2 elements)
= 2¹² – (¹²C₀ + ¹²C₁ + ¹²C₂)
= 4096 – (1 + 12 + 66)
= 4096 – 79
= 4017

(iii) The required subset contains atmost 3 elements
i.e., it may contain 0 or 1 or 2 or 3 elements.
The number of subsets of A with exactly 0 elements is ¹²C₀
The number of subsets of A with exactly 1 element is ¹²C₁
The number of subsets of A with exactly 2 elements is ¹²C₂
The number of subsets of A with exactly 3 elements is ¹²C₃
∴ Number of subsets of A with atmost 3 elements = ¹²C₀ + ¹²C₁ + ¹²C₂ + ¹²C₃
= 1 + 12 + 66 + 220
= 299

Question 3.
Find the number of ways of selecting a cricket team of 11 players from 7 batsmen and 6 bowlers such that there will be atleast 5 bowlers in the team.
Solution:
Since the team consists of at least 5 bowlers, the selection may be of the following types.

The number of selections in the first type = ⁷C₆ × ⁶C₅
= 7 × 6
= 42
The number of selections in the second type = ⁷C₅ × ⁶C₆
= 21 × 1
= 21
∴ The required number of ways selecting the cricket team = 42 + 21 = 63

Question 4.
If 5 vowels and 6 consonants are given, then how many 6-letter words can be formed with 3 vowels and 3 consonants?
Solution:
No. of vowels given = 5
No.of consonants given = 6
We have to form a 6-letter word with 3 vowels and 3 consonants from given letters.
3 vowels can select from 5 in ⁵C₃ ways.
3 consonants can select from 6 in ⁶C₃ ways.
Total No. of words = ⁵C₃ × ⁶C₃ × 6! = 1,44,000

Question 5.
There are 8 railway stations along a railway line. In how many ways can a train be stopped at 3 of these stations such that no two of them are consecutive?
Solution:
Number of ways of selecting 3 stations out of 8 = ⁸C₃ = 56
Number of ways of selecting 3 out of 8 stations such that 3 are consecutive = 6
Number of ways of selecting 3 out of 8 stations such that 2 of them are consecutive = 2 × 5 + 5 × 4
= 10 + 20
= 30
∴ Number of ways for a train to be stopped at 3 of 8 stations such that no two of them are consecutive = 56 – (6 + 30) = 20

Question 6.
Find the number of ways of forming a committee of 5 members out of 6 Indians and 5 Americans so that always the Indians will be in majority in the committee.
Solution:
Since the committee contains the majority of Indians, the members of the committee may be of the following types.

The number of selections in type I = ⁶C₅ × ⁵C₀ = 6 × 1 = 6
The number of selections in type II = ⁶C₄ × ⁵C₁ = 15 × 5 = 75
The number of selections in type III = ⁶C₃ × ⁵C₂ = 20 × 10 = 200
∴ The required number ways of selecting a committee = 6 + 75 + 200 = 281.

Question 7.
A question paper is divided into 3 sections A, B, C Containing 3, 4, 5 questions respectively. Find the number of ways of attempting 6 questions choosing at least one from each section.
Solution:
First Method: The selection of a question may be of the following

Total No. of ways of attempting 6 questions

Second Method:
Required No.of attempting 6 questions = Total no. of arrangements – selection except question from C – selection except Q from A – selection except Q from B
= ¹²C₆ – ⁷C₆ – ⁹C₆ – ⁶C₆
= 805

Question 8.
Find the number of ways in which 12 things be
(i) divided into 4 equal groups
(ii) distributed to 4 persons equally.
Solution:
(i) The number of ways in which 12 things be divided into 4 equal groups = \(\frac{12 !}{3 ! 3 ! 3 ! 3 ! 4 !}\) = \(\frac{12 !}{(3 !)^4 4 !}\)
(ii) The number of ways in which 12 things be distributed to 4 persons equally = \(\frac{12 !}{3 ! 3 ! 3 ! 3 !}\) = \(\frac{12 !}{(3 !)^4}\)

Question 9.
A class contains 4 boys and g girls. Every Sunday, five students with atleast 3boys go for a picnic. A different group is being sent every week. During the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed is 85, find g.
Solution:
No. of boys = 4
No. of girls = g
Since there should be atleast 3 boys it can be done in 2 ways as shown in the table

The number of girls in G₁ = [⁴C₃ × ^gC₂] × 2
Since each group contains 2 girls
The number of girls in G₂ = [⁴C₃ × ^gC₂] × 1
Since each group contains 1 girl.
Given no. of dolls distributed = 85
⇒ [⁴C₃ × ^gC₂] × 2 + [⁴C₄ × ^gC₁] × 1 = 85
⇒ 4 . \(\frac{g(g-1)}{2}\) × 2 + 1 . g . 1 = 85
⇒ 4g² – 4g + g – 85 = 0
⇒ 4g² – 3g – 85 = 0
⇒ 4g² – 20g + 17g – 85 = 0
⇒ 4g(g – 5) + 17(g – 5) = 0
⇒ (g – 5)(4g + 17) = 0
Since g ≠ \(\frac{-17}{4}\)
∴ g = 5
Hence No. of girls = 5

Students get through AP Inter 2nd Year Physics Important Questions 3rd Lesson Wave Optics which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 3rd Lesson Wave Optics

Short Answer Questions

Question 1.
Explain Doppler effect in light. Distinguish between red shift and blue shift. [T.S. Mar. 16]
Answer:
Doppler effect in light : The change in the apparent frequency of light, due to relative motion between source of light and observer. This phenomenon is called Doppler effect.

The apparent frequency of light increases when the distance between observer and source of light is decreasing and the apparent frequency of light decreases, if the distance between source of light and observer increasing.
Doppler shift can be expressed as = \(\frac{\Delta v}{v}=\frac{-v_{\text {radial }}}{c}\)
Applications of Doppler effect in light:

1. It is used in measuring the speed of a star and speed of galaxies.
2. Measuring the speed of rotation of the sun.

Red shift: The apparent increase in wave length in the middle of the visible region of the spectrum moves towards the red end of the spectrum is called red shift.

Blue shift: When waves are received from a source moving towards the observer, there is an apparent decrease in wave length, this is called blue shift.

Question 2.
Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for maximum and zero intensity. [A.P. Mar. 16; T.S. Mar. 15]
Answer:
Let y₁ and y₂ be the displacements of the two waves having same amplitude a and <|> is the phase difference between them.

y₁ = a sin ωt …………… (1)
y₂ = a sin (ωt + Φ) …………….. (2)
The resultant displacement y = y₁ + y₂
y = a sin ωt + a sin (ωt + Φ)
y = a sin ωt + a sin ωt cos Φ + a cos ωt sin Φ
y = a sin ωt [1 + cos Φ] + cos ωt (a sin Φ)
Let R cos θ = a(1+ cos Φ) ……………….. (4)
R sin θ = a sin Φ ………………. (5)
y = R sin ωt. cos θ + R cos ωt. sin θ
y = R sin (ωt + θ) ………………… (6)
where R is the resultant amplitude at P, squaring equations (4) and (5), then adding
R² [cos² θ + sin2 θ] = a²[l + cos² Φ + 2 cos Φ + sin² Φ]
R² [1] = a² [1 + 1 + 2 cos Φ]
I = R² = 2a² [1 + cos Φ] = 2a² × 2 cos² \(\frac{\phi}{2}\);
I = 4a² cos² \(\frac{\phi}{2}\) …………. (7)

i) Minimum intensity (I_max)
cos² \(\frac{\phi}{2}\) = 1
Φ = 2nπ Where n = 0, 1, 2, 3 ……….
Φ = 0, 2π, 4π, 6π
∴ I_max = 4a².

ii) Minimum intensity (I_min)
cos² \(\frac{\theta}{2}\) = 1
Φ = (2n + 1)π Where n = 0, 1, 2, 3 ……….
Φ = π, 3π, 5π, 7π …………..
∴ I_min = 0

Question 3.
Does the principle of conservation of energy hold for interference and diffraction phenomena? Explain briefly. [Mar. 14]
Answer:
Yes, law of conservation of energy is obeyed. In case of constructive interference, intensity becomes maximum. Hence bright fringes are formed on the screen where as in the case of destructive interference, intensity becomes minimum. Hence dark fringes are formed on the screen.

This establishes that in the interference and diffraction pattern, the intensity of light is simply being redistributed i.e., energy is being transferred from dark fringe to bright fringe. No energy is being created (or) destroyed in the process. Hence energy is redistributed.

Thus the principle of conservation of energy is being obeyed in the process of interference and diffraction.

Question 4.
How do you determine the resolving power of your eye ? [A.P. Mar. 17]
Answer:
Make black strips of equal width separated by white strips. All the black strips having same width, while the width of white strips should increase from left to right.

Now watch the pattern with one eye. By moving away (or) closer to the wall, find the position where you can just see some two black strips as separate strips.

All black strips to the left of this strips would merge into one another and would not be distinguishable on the other hand, the black strips to the right of this would be more and more clearly visible.

Note the width d of the white strips and measure the distance D of the wall from eye.
Then resolution of your eye = \(\frac{\mathrm{d}}{\mathrm{D}}\)

Question 5.
Explain polarisation of light by reflection and arrive at Brewster’s law from it.
Answer:
Polarisation of light by reflection : When unpolarized light is incident on the boundary of a denser medium, at a particular angle of incidence the reflected light is completely plane polarised. This incident angle is called Brewster’s angle (i_B).

Brewster’s law:
When light is incident on a transparent surface at Brewster s angle, then reflected and refracted rays are at right angles to each other.
From snell’s law, refractive index,
∴ i_B + r = \(\frac{\pi}{2}\) ⇒ r = \(\frac{\pi}{2}\) – i_B

n = \(\frac{\sin \mathrm{i}_B}{\sin \mathrm{r}}=\frac{\sin \mathrm{i}_B}{\sin \left(\frac{\pi}{2}-\mathrm{i}_B\right)}=\frac{\sin \mathrm{i}_B}{\cos \mathrm{i}_B}\) = tan i_B
n = tani_B, This is known as Brewster’s law.
Brewster’s law – statement: The refractive index of a denser medium is equal to tangent of the polarising angle.

Question 6.
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. [T.S. Mar. 17; Mar. 12]
Answer:
Let I₀ be the intensity of polarised light after passing through the first polariser P₁. Then the intensity of light after passing through second polariser P₂ will be I = I₀cos²θ.
Where θ is the angle between pass axes P₁ and P₂. Since P₁ and P₂ are crossed the angle between the pass axes of P₂ and P₃ will be (\(\frac{\pi}{2}\) – θ)
Hence the intensity of light emerging from P₃ will be
I = I₀cos²θ . cos²(\(\frac{\pi}{2}\) – θ)
= I₀cos²θ . sin²θ
I = \(\frac{\mathrm{I}_0}{4}\) sin² 2θ
∴ The transmitted intensity will be maximum when θ = \(\frac{\pi}{4}\).

Long Answer Questions

Question 1.
Distinguish between Coherent and Incoherent addition of waves. Develop the theory of constructive interferences.
Answer:
Coherent sources : The two sources which maintain zero (or) any constant phase relation between themselves are known as Coherent sources.

Incoherent sources : If the phase difference changes with time, the two sources are known as incoherent sources.
Theory of constructive and destructive interference :
Let the waves of two coherent sources be
y₁ = a sin ωt ………………….. (1)
y₂ = a sin (ωt + Φ) …………….. (2)
where a is amplitude and Φ is the phase difference between two displacements.
According to superposition principle, y = y₁ + y₂
y = a sin ωt + a sin (ωt + Φ) = a sin ωt + a sin ω cos Φ + a cos ωt sin Φ
y = a sin ωt [1 + cos Φ] + cos ωt [a sin Φ] ………………….. (3)
Let A cos θ = a(1 + cos Φ], ………………. (4)
A sin θ = a sin Φ ……………… (5)
Substituting equations (4) and (5) in equation (3)
y = A sin ωt. cos θ + A cos ωt sin θ
y = A sin (ωt + θ) …………………… (6)
Where A is resultant amplitude. Squaring equations (4) and (5), then adding
A²[cos² θ + sin² θ] = a²[1 + cos² Φ + 2 cos Φ + sin² Φ]
A² [1] = a² [1 + 1 + 2 cos Φ]
I = A² = 2a² [1 + cos Φ]
I = 2a² × 2 cos² \(\frac{\phi}{2}\)
I = 4a² cos² \(\frac{\phi}{2}\)
I = 4I₀ cos² \(\frac{\phi}{2}\) …………….. (7) [∵ I₀ = a²]

Case (i) For constructive interference: Intensity should be maximum.
cos \(\frac{\phi}{2}\) = 1 ⇒ Φ = 2nπ
Where n = 0, 1, 2, 3 ⇒ Φ = 0, 2π, 4π, 6π ………….. I_max = 4I₀
Case (ii) For destructive interference: Intensity should be minimum
i.e., cos Φ = 0 ⇒ Φ = (2n + 1) π ; where π = 0, 1, 2, 3 ; Φ = π, 3π, 5π ⇒ I_min = 0.

Question 2.
Describe Young’s experiment for observing interference and hence arrive at the expression for ‘fringe width’.
Answer:
Interference: The modification of intensity obtained by the super position of two (or) more light waves is called interference.
Description :

1. Thomas Young experimentally observed the phenomenon of interference of light using two coherent sources.
2. A small pin hole ‘S’ illuminated by monochromatic source of light which produces a spherical wave.
3. S₁ and S₂ are two narrow pin holes equidistant from S.
4. Screen is placed at a distance D.
5. The points at which any two crests (or) any two troughs are superimposed, constructive interference takes place bright fringe will be observed on the screen.
6. The points at which crest of one wave and trough of another wave are superimposed, destructive interference takes place dark fringe will be observed on the screen.
7. Thus on the screen alternately bright and dark frings are observed.

Expression for fringe width :
1. It is the distance between two successive bright (or) dark fringes, denoted by β.

The path difference (δ) = d sin θ
2. If θ is very small then from figure sin θ ≈ tan θ = \(\frac{x}{D}\)

3. For bright fringes path difference S₂P – S₁P = nλ
∴ d sin θ = nλ
d × \(\frac{x}{D}\) = nλ
x = \(\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) ……………. (1) where n = 0, 1, 2, 3 …………….

This equation represents the position of bright fringe.
When n = 0, x₀ = 0.
n = 1, x₁ = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\) and n = 2, x₂ = \(\frac{2\lambda \mathrm{D}}{\mathrm{d}}\)
The distance between any two consecutive bright fringes is
x₂ – x₁ = \(\frac{2 \lambda D}{d}-\frac{\lambda D}{d} \Rightarrow \beta=\frac{\lambda D}{d}\) ………… (2)

4. For dark fringes path difference S₂P – S₁P = (2n + 1) \(\frac{\lambda}{2}\) ∴ d sin θ = (2n + 1) \(\frac{\lambda}{2}\)
d × \(\frac{x}{D}\) = (2n + 1) \(\frac{\lambda}{2}\) ⇒ x = \(\frac{(2 n+1) \lambda D}{2 d}\) ………….. (3) where n = 0, 1, 2, 3…………
This equation (3) represents, position of dark fringe.
When n = 0, x₀ = \(\frac{\lambda D}{2 d}\) ⇒ n = 1, x₁ = \(\frac{3 \lambda D}{2 d}\); n = 2, x₂ = \(\frac{5 \lambda D}{2 d}\) ………….
The distance between any two consecutive dark fringes is x₂ – x₁ = \(\frac{5 \lambda \mathrm{D}}{2 \mathrm{~d}}-\frac{3 \lambda \mathrm{D}}{2 \mathrm{~d}}=\frac{5 \lambda \mathrm{D}-3 \lambda \mathrm{D}}{2 \mathrm{~d}}\)
β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\) …………………….. (4)
Hence fringe width is same for bright and dark fringes.

Problems

Question 1.
What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm ? .
Solution:
Since vλ = c, \(\frac{\Delta v}{v}=\frac{\Delta \lambda}{\lambda}\)
(for small changes in v and λ). For
∆λ = 589.6 – 589.0 = + 0.6 nm
we get [usmg Equation \(\frac{\Delta v}{v}=\frac{v_{\text {radial }}}{c}\)]
\(\frac{\Delta v}{v}=-\frac{\Delta \lambda}{\lambda}=-\frac{v_{\text {radial }}}{\dot{c}}\)
or, υ_radial ≅ + c(\(\frac{0.6}{589.0}\)) = +3.06 × 10⁵ m s^-1
= 306 km/s
Therefore, the galaxy is moving away from us.

Question 2.
Unpolarised light is incident on a plane glass surface. What should be the angle of the incidence so that the reflected and refracted rays are perpendicular to each other ?
Solution:
For i + r to be equal to π/2, we should have tan i_B = μ = 1.5. This gives i_B = 57°. This is the Brewster’s angle for air to glass interface.

Question 3.
What is the Brewster angle for air to glass transition ? (Refractive index of glass = 1.5.)
Solution:
Here, i_p = ? μ = 1.5; As tan i_p = μ = 1.5
∴ i_p = tan^-1 (1.5); i_p = 56.3

Question 4.
In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3 ?
Solution:
Let I₁ = I₂ = I. If Φ is phase difference between the two light waves, then resultant intensity,
I_R = I₁ + I₂ + \(2 \sqrt{\mathrm{I}_1 \mathrm{I}_2}\) . cos Φ
When path difference = λ, Phase difference Φ = 0°
∴ I_R = I + I + 2\(\sqrt{\text { II }}\) . cos 0° = 4I = k
When path difference = \(\frac{\lambda}{3}\), phase difference Φ = \(\frac{2 \pi}{3}\) rad
∴ I_R = I + I + 2\(\sqrt{\text { II }}\) cos \(\frac{2 \pi}{3}\) ⇒ I’_R = 2I + 2I(\(\frac{-1}{2}\)) = I = \(\frac{\mathrm{k}}{4}\)

Question 5.
Assume that light of wavelength 6000Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch ?
Solution:
A 100 inch telescope implies that 2a = 100 inch = 254 cm. Thus if,
λ = 6000 Å = 6 × 10^-5 cm
then ∆θ ≈ \(\frac{0.61 \times 6 \times 10^{-5}}{127}\)
= 2.9 × 10^-7 radians.

Question 6.
In a Youngs double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Solution:
d = 0.28 mm = 0.28 × 10^-3 m, D = 1.4 m, β = 1.2 × 10^-2 m, n = 4
Since, β = \(\frac{\mathrm{D}}{\mathrm{d}}\) nλ ⇒ \(\frac{\beta}{n} \cdot \frac{d}{D}\) ⇒ λ = \(\frac{1.2 \times 10^{-2} \times 2.8 \times 10^{-2}}{4 \times 1.4}\)
⇒ λ = 600 × 10^-9 m
⇒ λ = 600 nm.

Textual Examples

Question 1.
What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm ?
Solution:
Since vλ = c, \(\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{\Delta \lambda}{\lambda}\)
(for small changes in v and λ). For
∆λ = 589.6 – 589.0 = + 0.6 nm
we get [usmg Equation \(\frac{\Delta v}{v}=\frac{v_{\text {radial }}}{c}\)]
\(\frac{\Delta v}{v}=-\frac{\Delta \lambda}{\lambda}=-\frac{v_{\text {radial }}}{\dot{c}}\)
or, υ_radial ≅ + c(\(\frac{0.6}{589.0}\)) = +3.06 × 10⁵ m s^-1
= 306 km/s
Therefore, the galaxy is moving away from us.

Question 2.
(a) When monochromatic light is incident on a surface separating, two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why ?
(b) When light travels from a rarer to a denser medium, the speed decreases. Does the reduction in speed imply a reduction in the energy carried by the light wave ?
(c) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light.
Solution:
a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) and undergo forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light.

b) No. energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.

c) For a given frequency, intensity of light in the photon picture is determined by the number of photons crossing ah unit area per unit time.

Question 3.
Two slits are made one millimetre apart and the screen is placed one metre away. What is the fringe separation when blue*: green light of wavelength 500 nm is used ?
Solution:
Fringe spacing = \(\frac{\mathrm{D} \lambda}{\mathrm{d}}=\frac{1 \times 5 \times 10^{-7}}{1 \times 10^{-3}}\) m
= 5 × 10^-4 m = 0.5 mm

Question 4.
What is the effect on the interference fringes in a Young’s double-slit experiment due to each of the following operations :
(a) the screen is moved away from the plane of the slits;
(b) the (monochromatic) source is replaced by another (monochro-matic) source of shorter wavelength;
(c) the separation between the two slits is increased;
(d) the source slit is moved Closer to the double-slit plane;
(e) the width of the source slit is increased;
(f) the monochromatic source is replaced by a source of white light ?
(In each operation, take all parameters, other than the one specified, to remain unchanged.)
Solution:
a) Angular separation of the fringes remains constant (= λ/d). The actual separation of the fringes increases in proportion to the distance of the screen from the plane of the two slits.

b) The separation of the fringes (and also angular separation) decreases. See, however, the condition mentioned in (d) below.

c) The separation of the fringes (and also angular separation) decreases. See however, the condition mentioned in (d) below.

d) Let s be the size of the source and S its distance from the plane of the two slits. For interference fringes to be seen, the condition s/S < λ/d should be satisfied; otherwise, interference patterns produced by different parts of the source overlap and no fringes are seen. Thus, as S decreases (i.e., the source slit is brought closer), the interference pattern gets less and less sharp, and when the source is brought too close for this condition to be valid, the fringes disappear. Till this happens, the fringe separation remains fixed.

e) Same as in (d). As the source slit which increases, fringe pattern gets less and less sharp. When the source slit is so wide that the condition s/S ≤ λ/d is not satisfied, the interference pattern disappears.

f) The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are at the same position. Therefore, the central fringe is white. For a point P for which S₂P – S₁P = λ_b/2, where λ_b (≈ 4000 Å) represents the wavelength for the blue colour, the blue component will be absent and the fringe will appear red in colour. Slightly farther away where S₂Q – S₁Q = λ_b = λ_r/2 where λ_r (≈ 8000 Å) is the wavelength for the red colour, the fringe will be predominantly blue.
Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.

Question 5.
In Textual Example 3, what should the width of each slit be to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern ?
Solution:
We want aθ = λ, θ = \(\frac{\lambda}{\mathrm{a}}\)
10 \(\frac{\lambda}{\mathrm{d}}\) = 2\(\frac{\lambda}{\mathrm{a}}\), a = \(\frac{\mathrm{d}}{5}\) = 0.2 mm

Question 6.
Assume that light of wavelength 6000 Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch ?
Solution:
A 100 inch telescope implies that 2a =100 inch = 254 cm.
Thus if,
λ = 6000 Å = 6 × 10^-5 cm
then
∆θ ≈ \(\frac{0.61 \times 6 \times 10^{-5}}{127}\)
≈ 2.9 × 10^-7 radians.

Question 7.
For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm ?
Solution:
Z_F = \(\frac{\mathrm{a}^2}{\lambda}=\frac{\left(3 \times 10^{-3}\right)^2}{5 \times 10^{-7}}\) = 18 m

Question 8.
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids ?
Solution:
Let I₀ be the intensity of polarised light after passing through the first polariser P₁. Then the intensity of light after passing through second polariser P₂ will be I = I₀cos²θ,
where θ is the angle between pass axes of P₁ and P₂. Since P₂ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be
I = I₀cos²θ cos² (\(\frac{\pi}{2}\) – θ)
= I₀ cos² θ sin² θ = (I₀/4) sin² 2θ
Therefore, the ‘transmitted intensity will be maximum when θ = π/4.

Question 9.
Unpolarised light is incident on a plane glass surface. What should be the angle of the incidence so that the reflected and refracted rays are perpendicular to each other ?
Solution:
For i + r to be equal to π/2, we should have tan i_B = μ = 1.5. This gives i_B = 57°. This is the Brewster’s angle for air to glass interface.

Students get through AP Inter 2nd Year Physics Important Questions 14th Lesson Nuclei which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 14th Lesson Nuclei

Very Short Answer Questions

Question 1.
The half-life of ⁵⁸Co is 72 days. Calculate Its average life. [Board model Paper]
Answer:
T_half = 0.693 × T_Mean ⇒ T_Mean = \(\frac{T_{\text {half }}}{0.693}=\frac{72}{0.693}\) = 103.8 days.

Question 2.
Why do all electrons emit during p-decay not have the same energy?
Answer:
When a neutron is converted into a proton, an electron and neutron are emitted along with it.
\({ }_1^1 \mathrm{n} \longrightarrow{ }_1^1 \mathrm{H}+{ }_{-1}^0 \mathrm{e}+\mathrm{v}\)
In β – decay proton remains in the nucleus, but electrons and neutrons are emitted with constant energy.
The energy of the neutron is not constant. So, all electrons do not have the same energy.

Question 3.
Neutrons are the best projectiles to produce nuclear reactions. Why?
Answer:
Neutrons are uncharged particles. So they do not get deflected by the electric and magnetic fields. Hence Neutrons are considered as best projectiles in nuclear reaction.

Question 4.
Neutrons cannot produce ionization. Why ?
Answer:
Because neutrons are uncharged particles and cannot produce ionization.

Question 5.
What are delayed neutrons ?
Answer:
Neutrons are emitted in the fission products after, sometime are called delayed neutrons.

Question 6.
What are thermal neutrons ? What is their importance ?
Answer:
Neutrons having kinetic energies approximately 0.025 eV are called as slow neutrons or thermal neutrons. ²³⁵U undergoes fission only when bombarded with thermal neutrons.

Question 7.
What is the value of neutron multiplication factor in a controlled reaction and in an uncontrolled chain reaction ?
Answer:
In controlled chain reaction K = 1
In uncontrolled chain reaction K > 1

Question 8.
What is the role of controlling rods in a nuclear reactor ?
Answer:
In nuclear reactor, controlling rods are used to absorb the neutrons. Cadmium, boron materials are used in the form of rods in reactor. These control the fission rate.

Question 9.
Why are nuclear fusion reactions called thermo nuclear reactions ?
Answer:
Nuclear fusion occurs at very high temperatures. So it is called as thermo nuclear reaction.

Question 10.
Define Becquerel and Curie.
Answer:
Becquerel: 1 disintegration or decay per second is called Becquerel. It is SI unit of activity.
1 disintegration or decay
i.e., Becquerel = \(\frac{1 \text { disintegration or decay }}{\text { second }}\)
Curie : 3.7 × 10¹⁰ decays per second is called Curie.
1 Curie : 1 Ci = \(\frac{3.7 \times 10^{10} \text { decays }}{\text { second }}\) = 3.7 × 10¹⁰Bq.

Question 11.
What is a chain reaction ?
Answer:
Chain reaction : The neutrons produced in the fission of a nucleus can cause fission in other neighbouring nuclei producing more and more neutrons to continue the fission until the whole fissionable material is disintegrated. This is called chain reaction.

Question 12.
What is the function of moderator in a nuclear reactor ?
Answer:
They are used to slow down the fast moving neutrons produced during the fission process.
e.g.: Heavy water, Beryllium.

Question 13.
What is the energy released in the fusion of four protons to form a helium nucleus ?
Answer:
26.7 MeV energy is released.

Short Answer Questions

Question 1.
Write a short note on the discovery of neutron.
Answer:

1. Bothe and Becker found that when beryllium is bombarded with a – particles of energy 5 MeY which emitted a highly penetrating radiation.
2. The equation for above process can be written as
   \({ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{13} \mathrm{C}\) + γ – (radiation energy)
3. The radiations are not effected by electric and magnetic fields.
4. In 1932, James Chadwick, had subjected nitrogen and argon to the beryllium radiation. He interpreted the experimental results by assuming that the radiation is of a new kind of particles which has no charge and its mass is equal to proton. These neutral particles were named as neutrons. Thus the neutron was discovered.
5. The experimental results can be represented by the following equation.
   \({ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{12} \mathrm{C}+{ }_0^1 \mathrm{n}+\mathrm{Q}\)
   

Question 2.
What are nuclear forces ? Write their properties.
Answer:
The forces which hold the nucleons together in nucleus are called nuclear forces.
Properties of Nuclear forces:

1. Nuclear forces are attractive forces between proton and proton (P – N, proton and neutron (P – N) and neutron and neutron (N – N).
2. Nuclear forces are independent of charge. It was found that force between proton and proton is same as force between neutron and neutron.
3. These forces are short range forces i.e., these forces will act upto a small distance only. Generally the range of nuclear forces is upto few Fermi (10^-15 m).
4. These forces are non central forces, i.e., they do not act along the line joining the two nucleons.
5. These forces are exchange forces. The force between two nucleons is due to exchange of n-mesons.
6. These forces are spin dependent. These forces are strong when the spin of two nucleons are in same direction and they are weak when they are in opposite direction.
7. Nuclear forces are saturated forces i.e., the force between nucleons will extend upto the immediate neighbouring nucleons only.
8. These are the strongest forces in nature. They are nearly 10³⁸ times stronger than gravitational forces and nearly 100 times stronger than Coulombic forces.

Question 3.
Define half life period and decay constant for a radioactive substance. Deduce the relation between them.
Answer:
Half life period (T) : Time taken for the number of radio active nuclei to disintegrate to half of its original number of nuclei is called Half life period.

Decay constant (λ) : The ratio of the rate of radioactive decay to the number of nuclei present at that instant.
It is a proportional constant and is denoted by ‘λ’.
∴ λ = \(\frac{-\left(\frac{d N}{d t}\right)}{N}\)

Relation between half the period and decay constant:

1. The radioactive decay law N = N₀ e^-λt states that the number of radioactive nuclei in a radioactive sample decreases exponentially with time. Here A is called decay constant.
2. If N₀ is the number of nuclei at t = 0 and N is the radioactive nuclei at any instant of time’t’.
3. Substituting N = \(\frac{\mathrm{N}_0}{2}\) at t = T in N = N₀ e^-λt
4. \(\frac{\mathrm{N}_0}{2}\) = N₀ e^-λT
   e^λT = 2
   λT= ln 2
   T = \(\frac{\ln 2}{\lambda}=\frac{2.303 \log _{10}^2}{\lambda}\)
   ∴ T = \(\frac{0.693}{\lambda}\)

Question 4.
What is nuclear fission ? Give an example to illustrate it.
Answer:
Nuclear fission : The process of dividing a heavy nucleus into two or more smaller and stable nuclei due to nuclear reaction is called nuclear fission.
Ex: The fission reaction is \({ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n} \rightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_0^1 \mathrm{n}+\mathrm{Q}\)
Where Q is the energy released.
Q = (Total mass of reactants – Total mass of product) C²
= [(Mass of \({ }_{92}^{235} \mathrm{U}\) + Mass of \({ }_0^1 \mathrm{n}\)) – (Mass of \({ }_{56}^{141} \mathrm{Ba}\) + Mass of \({ }_{36}^{92} \mathrm{Kr}\) + Mass of three neutrons)] C²
= (235.043933 – 140.9177 – 91.895400 – 2 × 1.008665) amu × C².
= 0.2135 × 931.5 MeV = 198.9 MeV = 200 MeV

Question 5.
What is nuclear fusion ? Write the conditions for nuclear fusion to occur.
Answer:
Nuclear fusion : The process of combining lighter nuclei to produce a larger nucleus is known as nuclear fusion.
E.g : Hydrogen nuclei (₁H¹) are fused together to form heavy Helium (₂He⁴) along with 25.71 MeV energy released.
Conditions for nuclear fusion :

1. Nuclear fusion occurs at very high temperatures such as 10⁷ kelvin and very high pressures. These are obtained under the explosion of an atom bomb.
2. Higher density is also desirable so that collisions between light nuclei occur quite frequently.

Question 6.
Distinguish between nuclear fission and nuclear fusion.
Answer:
Nuclear fission

1. In this process heavy nucleus is divided into two fragments along with few neutrons.
2. These reactions will takes place even at room temperature.
3. To start fission atleast one thermal neutron from outside is compulsory.
4. Energy released per unit mass of participants is less.
5. In this process neutrons are liberated.
6. This reaction can be controlled.
   Ex: Nuclear reactor.
7. Atom bomb works on principle of fission reaction.
8. The energy released in fission can be used for peaceful purpose.
   Ex: Nuclear reactor and Atomic power stations.

Nuclear fusion

1. In this process lighter nuclei will join together to produce heavy nucleus.
2. These reactions will takes place at very high temperature such as 107 Kelwin.
3. No necessary of external neutrons.
4. Energy released per unit mass of participants is high. Nearly seven times more than fission reaction.
5. In this process positrons are liberated.
6. There is no control on fusion reaction.
7. Hydrogen bomb works on the principle of fusion reaction.
8. The energy released in fusion cannot be used for peaceful purpose.

Long Answer Questions

Question 1.
Define mass defect and binding energy. How does binding energy per nucleon vary with mass number ? What is its significance ?
Answer:
1. Mass defect (∆M) : The difference in mass of a nucleus and its constituents is called the mass defect. The nuclear mass M is always less than the total mass, Em, of its constituents.
Mass defect, (∆M) = [Zm_p + (A – Z)m_n – M]

2. Binding energy: The energy required to break the nucleus into its constituent nucleons is called the binding energy.
Binding Energy, (E_b) = ∆MC² = [Zm_p + (A – Z)m_n – M] 931.5 MeV
Nuclear binding energy is an indication of the stability of the nucleus.
Nuclear binding energy per nucleon E_bn = \(\frac{E_{\mathrm{b}}}{\mathrm{A}}\)

3. The following graph represents how the binding energy per nucleon varies with the mass number A.

4. From the graph that the binding energy is highest in the range 28 < A < 138. The binding energy of these nuclei is very close to 8.7 MeV

5. With the increase in the mass number the binding energy per nucleon decreases and consequently for the heavy nuclei like Uranium it is 7.6 MeV

6. In the region of smaller mass numbers, the binding energy per nucleon curve shows the characteristic minima and maxima.

7. Minima are associated with nuclei containing an odd number of protons and neutrons such as \({ }_3^6 \mathrm{Li},{ }_5^{10} \mathrm{~B},{ }_7^{14} \mathrm{~N}\) and the maxima are associated with nuclei having an even number of protons and neutrons such as \({ }_2^4 \mathrm{He},{ }_6^{12} \mathrm{C},{ }_8^{16} \mathrm{O}\).

Significance:
8. The curve explains the relationship between binding energy per nucleon and stability of the nuclei.

9. Uranium has a relatively low binding energy per nucleon as 7.6 MeV Hence to attain greater stability Uranium breaks up into intermediate mass nuclei resulting in a phenomenon called fission.

10. On the other hand light nuclei such as hydrogen combine to form heavy nucleus to form helium for greater stability, resulting in a phenomenon called fusion.

11. Iron is the most stable having binding energy per nucleon 8.7 MeV, and it neither undergoes fission per fusion.

Question 2.
What is radioactivity ? State the law of radioactive decay. Show that radioactive decay is exponential in nature. [T.S. Mar. 16]
Answer:
1. Radioactivity : The nuclei of certain elements disintegrate spontaneously by emitting alpha (α), beta (β) and gamma (δ) rays. This phenomenon is called Radioactivity or Natural radioactivity.

2. Law of radioactivity decay : “The rate of radioactive decay \(\left(\frac{d N}{d t}\right)\) (or) the number of
nuclei decaying per unit time at any instant, is directly proportional to the number of nuclei (N) present at that instant is called law of radioactivity decay”.

3. Radioactive decay is exponential in nature : Consider a radioactive substance. Let the number of nuclei present in the sample at t = 0, be N₀ and let N be the radioactive nuclei remain at an instant t.
\(\frac{\mathrm{dN}}{\mathrm{dt}}\) ∝ N ⇒ \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = – λN
dN = – λ Ndt …………………….. (1)
The proportionality constant λ is called decay constant or disintegration constant. The negative sign indicates the decrease in the number of nuclei.

4. From eq. (1) \(\frac{\mathrm{dN}}{\mathrm{N}}\) = – λ dt ……………… (2)

5. Integrating on both sides
\(\int \frac{\mathrm{dN}}{\mathrm{N}}=-\lambda \int \mathrm{dt}\)
ln N = – λt + C …………….. (3)
Where C = Integration constant.

6. At t = O; N = N₀. Substituting in eq. (3), we get ln N₀ = C
∴ ln N = -λt + ln N₀
ln N – ln N₀ = – λt
ln (\(\frac{\mathrm{N}}{\mathrm{N}_0}\)) = – λt
∴ N = N₀ e^-λt
The above equation represents radioactive decay law.

7. It states that the number of radioactive nuclei in a radioactive sample decreases exponentially with time.

Sample Problems

Question 1.
The half life of radium is 1600 years. How much time does lg of radium take to reduce to 0.125g. [IPE 2016 (TS)]
Answer:
Half – life of radium = 1600 years;
\(\frac{N}{N_0}=\frac{1}{2^n} \Rightarrow \frac{0.125}{1}=\frac{1}{2^n} \Rightarrow \frac{125}{1000}=\frac{1}{2^n} \Rightarrow \frac{1}{8}=\frac{1}{2^n} \Rightarrow \frac{1}{2^3}=\frac{1}{2^n} \Rightarrow n=3\)
∴ Time taken = Half life × no. of Half lives = 1600 × 3 = 4800 years

Question 2.
Plutonium decays with a half-life of 24,000 years. If plutonium is stored for 72,000 years, what fraction of it remains ?
Answer:
Half life of plutonium, T = 24000 years; Stored time of plutonium, t = 72000 years
no. of half lives, n = \(\frac{t}{T}=\frac{72000}{24000}\) = 3; Fraction of plutonium remains = \(\frac{N}{N_0}=\frac{1}{2^n}=\frac{1}{2^3}=\frac{1}{8}\)

Question 3.
Explain the principle and working of a nuclear reactor with the help of a labelled diagram. [A.P. & T.S. Mar. 17, A.P. Mar. 16 15, Mar. 14]
Answer:
Principle : A nuclear reactor works on the principle of achieving controlled chain reaction in natural Uranium ²³⁸U enriched with ²³⁵U, consequently generating large amounts of heat.
A nuclear reactor consists of (1) Fuel (2) Moderator (3) Control rods (4) Radiation shielding (5) Coolant.

1. Fuel and clad : In reactor the nuclear fuel is fabricated in the form of thin and long cylindrical rods. These group of rods treated as a fuel assembly. These rods are surrounded by coolant, which is used to transfer of heat produced in them. A part of the nuclear reactor which is used to store the nuclear fuel is called the core of the reactor. Natural uranium, enriched uranium, plutonium and uranium – 233 are used as nuclear fuels.

2. Moderator : The average energy of neutrons released in fission process is 2 MeV. They are used to slow down the velocity of neutrons. Heavy water or graphite are used as moderating materials in reactor.

3. Control Rods : These are used to control the fission rate in reactor by absorbing the neutrons. Cadmium and boron are used as controlling the neutrons, in the form of rods.

4. Shielding : During fission reaction beta and gamma rays are emitted in addition to neutrons. Suitable shielding such as steel, lead, concrete etc., are provided around the reactor to absorb and reduce the intensity of radiations to such low levels that do not harm the operating personnel.

5. Coolant : The heat generated in fuel elements is removed by using a suitable coolant to flow around them. The coolants used are water at high pressures, molten sodium etc.

Working : Uranium fuel rods are placed in the aluminium cylinders. The graphite moderator is placed in between the fuel cylinders. To control the number of neutrons, a number of control rods of cadnium or beryllium or boron are placed in the holes of graphite block. When a few ²³⁵U nuclei undergo fission fast neutrons are liberated. These neutrons pass through the surrounding graphite moderator and loose their energy to become thermal neutrons. These thermal neutrons are captured by ²³⁵U. The heat generated here is used for heating suitable coolants which is turn heat water and produce steam. This steam is made to rotate steam turbine and there by drive a generator of production for electric power.

Question 4.
Explain the source of stellar energy. Explain the carbon – nitrogen cycle, proton – proton cycle occuring in stars.
Answer:
Scientists proposed two types of cyclic processes for the sources of energy in the sun and stars. The first is known as carbon-nitrogen cycle and the second is proton-proton cylce.

1. Carbon-Nitrogen Cycle : According to Bethe carbon-nitrogen cycle is mainly responsible for the production of solar energy. This cycle consists of a chain of nuclear reactions in which hydrogen is converted into Helium, with the help of Carbon and Nitrogen as catalysts. The nuclear reactions are as given below.

2. Proton – Proton Cycle: A star is formed by the condensation of a large amount of matter at a point in space. Its temperature rises to 2,00,000°C as the matter contracts under the influence of gravitational attraction. At this temperature the thermal energy of the protons is sufficient to form a deuteron and a positron. The deuteron then combines with another proton to form lighter nuclei of helium \({ }_2^3 \mathrm{He}\). Two such helium nuclei combine to form a helium nucleus latex]{ }_2^4 \mathrm{He}[/latex] and two protons releasing a total amount of energy 25.71 MeV The nuclear fusion reactions are given below.

Problems

Question 1.
Compare the radii of the nuclei of mass numbers 27 and 64.
Solution:
A₁ = 27; Asub>2 = 64
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left[\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right]^{1 / 3}\) [∵ R = R₀A^1/3]
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left[\frac{27}{64}\right]^{\frac{1}{3}}=\frac{3}{4}\)
∴ R₁ : R₂ = 3 : 4

Question 2.
Find the energy required to split \({ }_8^{16} \mathrm{O}\) nucleus into four α-particles. The mass of an a-particle is 4.002603u and that of oxygen is 15.994915u.
Solution:
The energy required to split O = [Total mass of the products – Total mass of the reactants] c²
Mass of four \({ }_2^4 \mathrm{He}\) – Mass of \({ }_8^{16} \mathrm{O}\)] × c²
= [(4 X 4.002603) – 15.994915] u × c²
= [16.010412 – 15.994915] u × c²
= (0.015497) 931.5 MeV = 14.43 MeV

Question 3.
A certain substance decays to 1/232 of its initial activity in 25 days. Calculate its half-life.
Solution:
Fraction of substance decays
= \(\frac{\text { Quantity remains }}{\text { Initial quantity }}\)
= \(\frac{1}{2^n}=\frac{1}{32}=\frac{1}{2^5}\)
∴ n = 5
Duration of time = 25 days
We know (n) = \(\frac{\text { Duration of time }}{\text { Half life time }}\)
∴ Half life time = \(\frac{\text { Duration of time }}{\mathrm{n}}\)
\(\frac{25}{5}\) = 5 days

Question 4.
One gram of radium is reduced by 2 milli- gram in 5 years by a-decay. Calculate the half¬life of radium.
Solution:
Initial (original) mass (N₀) = 1 gram
Reduced mass – 2 mg = 0.002 grams
Final mass (N)= 1 – 0.002 = 0.998 grams
t = 5 years
e^-λt = \(\frac{\mathrm{N}}{\mathrm{N}_0}\) ⇒ e^λt = \(\frac{\mathrm{N}_0}{\mathrm{~N}}\) ⇒ λt = log_e[latex]\frac{\mathrm{N}_0}{\mathrm{~N}}[/latex]
λt = 2.303 log [latex]\frac{\mathrm{N}_0}{\mathrm{~N}}[/latex]
λt = 2.303 log [latex]\frac{1}{0.998}[/latex]
= 2.303 log (1.002)
= 2.303 × 0.000868
= 0.001999
λ = \(\frac{0.001999}{5}\) = 0.0003998
T = \(\frac{0.693}{\lambda}=\frac{0.693}{0.0003998}\) = 1733.3 years

Question 5.
If one microgram of \({ }_92^{235} \mathrm{U}\) is completely destroyed in an atomhomb, how much energy will be released ?
Solution:
m = 1 μg = 1 × 10^-6 g = 1 × 10^-6 × 10^-3 kg
= 10^-9 kg
c = 3 × 10⁸ m/s
E = mc² = 1 × 10^-9 × 9 × 10⁶ = 9 × 10⁷ J

Question 6.
200 MeV energy is released when one nucleus of ²³⁵U undergoes fission. Find the number of fissions per second required for producing a power of 1 megawatt.
Solution:
E = 200 MeV
P = 1 × 10⁶ W
P = \(\frac{\mathrm{nE}}{\mathrm{t}} \Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=\frac{\mathrm{P}}{\mathrm{E}}=\frac{10^6}{200 \times 10^6 \times 1.6 \times 10^{-19}}\)
= \(\frac{1}{32}\) × 10¹⁸
∴ P = 0.03125 × 10¹⁸
= 3.125 × 10⁶

Textual Examples

Question 1.
Given the mass of iron nucleus as 55.85u and A = 56, find the nuclear density ?
Solution:
m_Fe = 55.85, u = 9.27 × 10^-26 kg
Nuclear density = \(\frac{\text { mass }}{\text { volume }}\)
= \(\frac{9.27 \times 10^{-26}}{(4 \pi / 3)\left(1.2 \times 10^{-15}\right)^3} \times \frac{1}{56}\)
= 2.29 × 10¹⁷ kg m^-3
The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus.

Question 2.
Calculate the energy equivalent of 1 g of substance.
Solution:
Energy, E = 10^-3 × (3 × 10⁸)² J
E = 10^-3 × 9 × 10¹⁶ = 9 × 10¹³ J
Thus, if one gram of matter is converted to energy, there is a release of an enormous amount of energy.

Question 3.
Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of \({ }_8^{16} \mathrm{O}\) in MeV/c².
Solution:
1 u = 1.6605 × 10^-27 kg
To convert it into energy units, we multiply it by c² and find, that energy. equivalent
= 1.6605 × 10^-27 × 2.9979 × 10⁸ kg m²/s²
= 1.4924 × 10^-10 J
= \(\frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}}\) eV
= 0.9315 × 10⁹ eV = 931.5 MeV
or, 1 u = 931.5 MeV/c².
For \({ }_8^{16} \mathrm{O}\), ∆M = 0.13691 u = 0.13691 × 931.5 MeV/c² = 127.5 MeV/c²
The energy needed to separate \({ }_8^{16} \mathrm{O}\) into hs constituents is thus 127.5 MeV/c².

Question 4.
The half-life of \({ }_{92}^{238} \mathrm{U}\) undergoing a – decay is 4.5 × 10⁹ years. What is the activity of 1 g sample of \({ }_{92}^{238} \mathrm{U}\) ?
Solution:
T_1/2 = 4.5 × 10⁹ y
= 4.5 × 10⁹ y × 3.16 × 10⁷ s/y.
= 1.42 × 10¹⁷ s
One k mol of any isotope contains Avogadro’s number of atoms, and so 1 g of \({ }_{92}^{238} \mathrm{U}\) contains
\(\frac{1}{238 \times 10^{-3}}\) k mol × 6.025 × 10²⁶ atoms/kmol
= 25.3 × 10²⁰ atoms.
The decay rate R is
R = λN
= \(\frac{0.693}{\mathrm{~T}_{1 / 2}}\) N = \(\frac{0.693 \times 25.3 \times 10^{20}}{1.42 \times 10^{17}}\) S^-1
= 1.23 × 10⁴ S^-1
= 1.23 × 10⁴ Bq

Question 5.
Tritium has a half-life of 12.5 y undergoing beta decay. What fraction of sample of pure tritium will remain undecayed after 25 y.
Solution:
By definition of half-life, half of the initial i sample will remain undecayed after 12.5 y. In the next 12.5 y, one-half of these nuclei would have decayed. Hence, one fourth of the sample of the initial pure s tritium will remain undecayed.

Question 6.
We are given the following atomic masses:
\({ }_{92}^{238} \mathrm{U}\) = 238.05079 u
\({ }_{2}^{4} \mathrm{He}\) = 4.00260 u
\({ }_{90}^{234} \mathrm{Th}\) = 234.04363 u
\({ }_{1}^{1} \mathrm{H}\) = 1.00783 u
\({ }_{91}^{237} \mathrm{Pa}\) = 237.05121 u
Here the symbol Pa is for the element protactinium (Z = 91).
a) Calculate the energy released during the alpha decay of \({ }_{92}^{238} \mathrm{U}\).
b) Show that \({ }_{92}^{238} \mathrm{U}\) cannot spontaneously emit a proton.
Solution:
a) The alpha decay of \({ }_{92}^{238} \mathrm{U}\) is given by

The energy released in this process is given by
Q = (M_U – M_Th – M_He) c²
Substituting the atomic masses as given in the data, we find
Q = (238.05079 – 234.04363 – 4.00260)u × c²
= (0.00456 u) c²
= (0.00456 u) (931.5 MeV/u)
= 4.25 MeV

b) If \({ }_{92}^{238} \mathrm{U}\) spontaneously emits a proton, the decay process would be
\({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{91}^{237} \mathrm{~Pa}+{ }_1^1 \mathrm{H}\)
The Q for this process to happen is
= (M_U – M_Pa – M_H)c²
(238.05079 – 237.05121 – 1.00783) u × c²
=(- 0.00825 u) c²
= – (0.00825 u) (931.5 MeV/u)
= -7.68 MeV
Thus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply an energy of 7.68 MeV to a \({ }_{92}^{238} \mathrm{U}\) nucleus to make it emit a proton.