Class 10

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 10 Chemical Bonding Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 10th Lesson Chemical Bonding

10th Class Chemistry 10th Lesson Chemical Bonding Textbook Questions and Answers

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Question 1.
List the factors that determine the type of bond that will be formed between two atoms. (AS1)
(OR)
How can you identify the type of bond formation between two atoms?
Answer:

• The strength of attraction or repulsion between atoms.
• Electrons in valence shell (valence electrons).

Question 2.
Explain the difference between the valence electrons and the covalency of an element. (AS1)
(OR)
How are valence electrons different from the covalency of element? Explain with examples.
Answer:
Valence electrons :

• Number of electrons in the outermost orbit or an atom is called its valence electrons.
• Ex: Na (Z = 11). It has 2e^– in I orbit, 8e^– in II orbit and 1e^– in III orbit.
• So number of valence electrons in Na atom are ‘l’.

Covalency of an element:

• Number of valance electrons which are taking part in covalent bond is called covalency.
• The electron configuration of Boran is 1s² 2s² 2p¹.
• It has three valance electrons.
• So its covalency is 3.

Question 3.
A chemical compound has the following Lewis notation : (AS1)
a) How many valence electrons does element Y have?
b) What is the valency of element Y?
c) What is the valency of element X?
d) How many covalent bonds are there in the molecule?
e) Suggest a name for the elements X and Y.

Answer:
a) 6
b) 2
c) 1
d) two
e) X – is hydrogen and Y – is oxygen. The formed molecule is H₂O.

Question 4.
Why do only valence electrons involve in bond formation? Why not electron of inner shells? Explain. (AS1)
(OR)
Which shell electrons involve in bond formation? Explain. What is the reason behind it?
Answer:

• The nucleus and the electrons in the inner shell remain unaffected when atoms come close together.
• But the electrons in the outermost shell (valence shell) of atoms get affected.
• The inner shell electrons are strongly attracted by the nucleus when compared to the valence electrons.
• So electrons in valence shell (valence electrons) are responsible for the formation of bond between atoms.

Question 5.
Explain the formation of sodium chloride and calcium oxide on the basis of the concept of electron transfer from one atom to another atom. (AS1)
(OR)
Explain the formation of any two compounds according to Kossel’s theory.
Answer:
I. Formation of sodium chloride (NaCl) :
1) Sodium chloride is formed from the elements sodium (Na) and chlorine (Cl).

2) Cation formation:
i) When sodium (Na) atom loses one electron to get octet electron configuration, it forms a cation (Na⁺).
ii) Now Na⁺ gets electron configuration that of Neon (Ne) atom.

3) Anion Formation :
i) Chlorine has shortage of one electron to get octet in its valence shell.
ii) So it gains the electron that was lost by Na to form anion and gets electron configuration of Argon (Ar).

4) Formation of NaCl :
i) Transfer of electrons between ‘Na’ and ‘Cl’ atoms, results in the formation of ‘Na⁺‘ and ‘Cl^–’ ions.
ii) These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound sodium chloride (NaCl).
Na⁺_(g) + Cl^–(g) → Na⁺Cl^–_(s) or NaCl

II. Formation of calcium oxide (CaO) :
1. Calcium (Ca) reacts with oxygen (0) to form an ionic compound calcium oxide (CaO).

2. Atomic number of Calcium is 20. Its electronic configuration is 2, 8, 8, 2.
3.

4. Atomic number of Oxygen is 8. Its electronic configuration is 2, 6.

5.

6. These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound calcium oxide (CaO).
Ca²⁺ + O^2- → Ca²⁺O^2- (or) CaO

Question 6.
A, B and C are three elements with atomic number 6, 11 and 17 respectively.
i) Which of these cannot form ionic bond? Why? (AS1)
ii) Which of these cannot form covalent bond? Why? (AS1)
iii) Which of these can form ionic as well as covalent bonds? (AS1)
Answer:
i) ‘A’ cannot form ionic bond. Its valence electrons are 4. It is difficult to lose or gain 4e^– to get octet configuration. So it forms covalent bond [Z of A is 6 so it is carbon (C)].

ii) ‘B’ cannot form covalent bond. Its valence electrons are 1 only. So it is easy to donate for other atom and become an ion. So it can form ionic bond [Z of B is 11, so it is sodium (Na)].

iii) Element C can form ionic as well as covalent bonds. Atomic number of Cl is 17. It is able to participate with Na in ionic bond and with hydrogen in HCl molecule as covalent bond.

Question 7.
How do bond energies and bond lengths of molecule help us in predicting their chemical properties? Explain with examples. (AS1)
(OR)
How can you explain with examples that bond energies and bond lengths are used to recognise chemical properties?
Answer:
1. Bond length :
Bond length or bond distance is the equilibrium distance between the nuclei of two atoms which form a covalent bond.

2. Bond energy :
Bond energy or bond dissociation energy is the energy needed to break a covalent bond between two atoms of a diatomic covalent compound in its gaseous state.

3. If the nature of the bond between the same two atoms changes the bond length also changes. For example, the bond lengths between two carbon atoms are C – C > C = C > C = C.

4. Thus the various bond lengths between the two carbon atoms are in ethane 1.54 Å, ethylene 1.34 Å, acetylene 1.20 Å.

5. The bond lengths between two oxygen atoms are in H₂O₂ (O – O) is 1.48 Å and in O₂ (O = O) is 1.21 Å.
6. Observe the table.

7. When bond length decreases, then bond dissociation energy increases.

8. When bond length increases, then bond dissociation energy decreases.

9. Bond length of H – H in H₂ molecule is 0.74 Å and its bond dissociation energy is 436 KJ/mol, whereas bond length of F – F in F₂ molecule is 1.44 Å and its bond dissociation energy is 159 KJ/mol.

10. Melting and boiling points of substances also can be determined by this bond energies and bond lengths.

Question 8.
Predict the reasons for low melting point for covalent compounds when compared with ionic compounds. (AS2)
(OR)
“Covalent compounds have low melting point.” What Is the reason for this statement? Explain.
Answer:
They are covalent compounds.

• The melting point is low due to the weak Vander Waal’s forces of attractions between the covalent molecules.
• The force of attraction between the molecules of a covalent compound is very weak.
• Only a small amount of heat energy is required to break these weak molecular forces, due to which covalent compounds have low melting points and low boiling points.
• Please note that some of the covalent solids like diamond and graphite have, however very high melting points and boiling points.

Question 9.
Collect the information about properties and uses of covalent compounds and prepare a report. (AS4)
(OR)
Generally these compounds are non-polar in nature. What are those compounds? Explain their properties and uses.
(OR)
Write any two uses and two properties of covalent compounds.
Answer:
The compounds are covalent.
Properties of covalent compounds :

1. Covalent compounds are usually liquids or gases, only some of them are solids.
2. The covalent compounds are usually liquids or gases due to the weak force of attraction between their molecules.
3. Covalent compounds have usually low melting and low boiling points.
4. Covalent compounds are usually in soluble in water but they are soluble in organic solvents.
5. Covalent compounds do not conduct electricity.

Uses of covalent compounds :

1. Covalent compounds form 99% of our body.
2. Water is a covalent compound. We know its many uses.
3. Sugars, food substances, tea and coffee are all covalent compounds.
4. Air we breathe in contains covalent molecules of oxygen and nitrogen.
5. Almost everything on earth other than most simple in organic salts are covalent.

Question 10.
Draw simple diagrams to show how electrons are arranged in the following covalent molecules : (AS5)
a) Calcium oxide (CaO)
b) Water (H₂O)
c) Chlorine (Cl₂)
Answer:
a) Calcium oxide (CaO) :

b) Water (H₂O):

The formation of water molecule can be shown like this also

c) Chlorine (Cl₂):

We can explain the formation of Cl₂ molecule in this way also.

Question 11.
Represent the molecule H₂O using Lewis notation. (AS5)
(OR)
How can you explain the formation of H₂O molecule using dot structure?
Answer:
One atom of oxygen shares its two electrons with two hydrogen atoms to form a water molecule.

Question 12.
Represent each of the following atoms using Lewis notation : (AS5)
a) Beryllium
b) Calcium
c) Lithium
Answer:

Question 13.
Represent each of the following molecules using Lewis notation : (AS5)
a) Bromine gas (Br2)
b) Calcium chloride (CaCl₂)
c) Carbon dioxide (CO₂)
d) Which of the three molecules listed above contains a double bond?
Answer:
a) Bromine gas (Br₂) :

b) Calcium chloride (CaCl₂)

c) Carbon dioxide (CO₂) :

d) CO₂, contains double bond in above list. Its structure is like this : O = C = O.

Question 14.
Two chemical reactions are described below. (AS5)
♦ Nitrogen and hydrogen react to form ammonia (NH₃).
♦ Carbon and hydrogen bond to form a molecule of methane (CH₄).
For each reaction give :
a) The valency of each of the atoms involved in the reaction.
b) The Lewis structure of the product that is formed.
Answer:
a) ♦ Nitrogen and hydrogen react to form ammonia (NH₃):
i) The valency of nitrogen is 3 and hydrogen is 1.
ii) The chemical formula of the product is NH₃

♦ Carbon and hydrogen bond to form a molecule of methane (CH₄):
i) The valency of carbon is 4 and hydrogen is 1.
ii) The chemical formula of the product is CH₄.

b) ♦ The Lewis structure of the product that is formed (: NH₃)

♦ The Lewis structure of the product that is formed (CH₄)

Question 15.
How does Lewis dot structure help in understanding bond formation between atoms? (AS6)
(OR)
What is the use of Lewis dot structure in bond formation? Explain.
Answer:

1. Only the outermost electrons of an atom take part in chemical bonding.
2. They are known as valence electrons.
3. The valence electrons in an atom are represented by putting dots (•) on the symbol of the element, one dot for each valence electron.
4. For example, sodium atom has 1 valence electron in its outermost shell, so we put 1 dot with the symbol of sodium and write Na• for it.
5. Sodium atom loses this 1 electron to form sodium ion.
6. By knowing the valence electrons of two different atoms by Lewis dot structure, we can understand which type of bond is going to establish between them and forms corresponding molecule.

Question 16.
What is octet rule? How do you appreciate role of the ‘octet rule’ in explaining the chemical properties of elements? (AS6)
(OR)
Which rule decides whether given element is chemically stable or not? Appreciate that rule.
Answer:
Octet rule decides whether given element is stable or not.
Octet rule :

• ‘The atoms of elements tend to undergo chemical changes that help to leave their atoms with eight outer shell electrons.”
• It was found that the elements which participate in chemical reaction get octet (or) ns2 np6 configuration similar to that of noble gas elements.

Role of octet in chemical properties of elements :

1. Except He remaining inert gas elements have 8 electrons in their outermost orbit. Since these elements are having stable octet configuration in their outermost orbit, they are very stable.
2. They do not allow the outermost electrons to take part in chemical reactions.
3. So by having octet configuration for these elements we can conclude these are chemically inertial.
4. If any group of elements (take halogens) which contain 7 electrons in their outermost orbit, they require only 1 e^– to get octet configuration.
5. So they try to participate in chemical reaction to get that 1 difference electron for octet configuration.
6. Similarly, Na contains 2, 8, 1 as its electronic configuration.
7. So it loses le^– from its outermost shell; it should have 8e^– in its outer shell and get the octet configuration.
8. Thus the octet rule helps in explaining the chemical properties of elements.

Question 17.
Explain the formation of the following molecules using valence bond theory.
a) N₂ molecule
b) O₂ molecule
(OR)
Write the formation of double bond and triple bond according to valence bond theory.
(OR)
Who proposed Valence Bond Theory? Explain the formation of N₂ molecule by using this theory.
Answer:
Linus Pauling was proposed valence bond theory.
Formation of N₂ molecule :

1. Electronic configuration of Nitrogen is 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹.
2. Suppose that p_x orbital of one Nitrogen atom overlaps the p_x orbital of other ‘N’ atom giving σ p_x – p_x bond along the inter nuclear axis.
3. The p_y and p_z orbitals of one ‘N’ atom overlaps with the p_y and p_z orbital of other ‘N’ atom laterally giving π p_y – p_y and π p_z – p_z bonds.
4. Therefore, N₂ molecule has a triple bond between two Nitrogen atoms.

Formation of O₂ molecule :

1. Electronic configuration of ‘O’ is 1s² 2s² 2p_x² 2p_y¹ 2p_z¹.
2. If the P_y orbital of one ‘O’ atom overlaps the p_y orbital of other ‘O’ atom along inter- nuclear axis, a σ p_y – p_y bond is formed.
3. p_z orbital of oxygen atom overlaps laterally, perpendicular to inter nuclear axis giving a π p_y – p_z bond.
4. So O₂ molecule has a double bond between the two oxygen atoms.

Question 18.
What is hybridisation? Explain the formation of the following molecules using hybridisation.
a) BeCl₂
b) BF₃
Explain the formation of sp and sp² hybridisation using examples.
(OR)
What is the name given to inter mixing of atomic orbitals to form new orbitals. Explain the formation of following molecules by using that process,
a) BeCl₂
b) BF₃
Answer:
This process is called hybridisation.
Hybridisation :
It is a phenomenon of inter mixing of atomic orbitals of almost equal energy which are present in the outer shells of the atom and their reshuffling or redistribution into the same number of orbitals but with equal properties like energy and shape.

a) Formation of BeCl₂ (Beryllium chloride) molecule :

1. ₄Be has electronic configuration 1s² 2s².
2. It has no unpaired electrons.
3. It is expected not to form covalent bonds, but informs two covalent bonds one each with two chlorine atoms. „
4. To explain this, an excited state is suggested for Beryllium in which an electron from ‘2s’ shifts to 2p_x level.
5. Electronic configuration of ₄Be is 1s² 2s¹ 2p_x¹].
6. Electronic configuration of ₁₇Cl is 1s² 2s² 2p⁶ 3s² 3p_x² 3p_y² 3p_z¹.
7. If Be forms two covalent bonds with two chlorine atoms, one bond should be σ 2s-3p due to the overlap of ‘2s’ orbital of Be, the ‘3p_z‘ orbital of one chlorine atom.
8. The other bond should be σ 2p-3p due to the overlap of ‘2p_x’ orbital of Be atom the 3p orbital of the other chlorine atom.
9. As the orbitals overlapping are different, the bond strengths of two Be-Cl must be different.
10. But, both bonds are of same strength and Cl\(\hat{\mathrm{Be}}\) Cl is 180°.

The Hybridisation of BeCl₂ can be explained in this way also :
a) Be atom in its excited state allows its 2s orbital and 2p_x orbital which contain unpaired electrons to intermix and redistribute to two identical orbitals.
b) As per Hund’s rule each orbital gets one electron.
c) The new orbitals based on the types of orbitals that have undergone hybridisation are called sp orbitals.
d) The two sp orbitals of Be get separated by 180°.
e) Now each chlorine atom comes with its 3p_z¹ orbital and overlaps it the sp orbitals of Be forming two identical Be-Cl bonds (σ sp-p bonds).
Cl\(\hat{\mathrm{Be}}\) Cl = 180°.

f) Both the bonds are of same strength.

b) Formation of BF₃ molecule :

1. ₅B has electronic configuration 1s² 2s² 2pxh
2. The excited electronic configuration of ₅B is 1s² 2s¹ 2p_x¹ 2p_y¹
3. As it forms three identical B-F bonds in BF₃.
4. It is suggested that excited ‘B’ atom undergoes hybridisation.
5. There is an intermixing of 2s, 2p_x, 2p_y orbitals and their redistribution into three identical orbitals called sp² hybrid orbitals.
6. For three sp² orbitals to get separated to have minimum repulsion the angle between any two orbitals is 120° at the central atom and each sp² orbital gets one election.
7. Now three fluorine atoms overlap their 2p_z orbitals containing unpaired electrons (F₉ 1s² 2s² 2p_x² 2p_y² 2p_z¹) the three sp² orbitals of ‘B’ that contain unpaired electrons to form three σsp²-p bonds.

Fill in the Blanks

1. Electrons in the outermost orbit are called …………………… .
2. Except …………………… gas all other noble gases have octet in their valence shell.
3. Covalency of elements explains about member of …………………… formed by the atom.
4. Valence bond theory was proposed by …………………… .
5. In …………………… bonding the valence electrons are shared among all the atoms of the metallic elements.
Answer:

1. valence electrons
2. Helium
3. covalent bonds
4. Linus Pauling
5. covalent

Multiple Choice Questions

1. Which of the following elements is electronegative?
A) Sodium
B) Oxygen
C) Magnesium
D) Calcium
Answer:
B) Oxygen

2. An element ₁₁X²³ forms an ionic compound with another element ‘Y’. Then the charge on the ion formed by X is
A) +1
B) +2
C) -l
D) – 2
Answer:
A) +1

3. An element ‘A’ forms a chloride ACl₄. The number of electrons in the valence shell of ‘A’
A) 1
B) 2
C) 3
D) 4
Answer:
D) 4

10th Class Chemistry 10th Lesson Chemical Bonding InText Questions and Answers

10th Class Chemistry Textbook Page No. 153

Question 1.
How do elements usually exist?
Answer:
They may exist as a single atom or as a group of atoms.

Question 2.
Do atoms exist as a single atom or as a group of atoms?
Answer:
Atoms exist as a single atom, sometimes as a group of atoms also.

Question 3.
Are there elements which exist as atoms?
Answer:
Yes. There are elements which exist as atoms.

Question 4.
Why do some elements exist as molecules and some as atoms?
Answer:
By following different laws of chemical combination the chemical compounds take place as a result of combination of atoms of various elements in different ways.

Question 5.
Why do some elements and compounds react vigorously while others are inert?
Answer:
1) Number of electrons in their outermost shell.
2) Bond strength between the atoms in compound.

Question 6.
Why is the chemical formula for water H₂O and for sodium chloride NaCl, why not HO₂ and NaCl₂?
Answee:
Valencies of the atoms participating in the molecules.

Question 7.
Why do some atoms combine dille tl do not?
Answer:
1) Atoms which have 8e“ in their outer shell will not combine.
2) Atoms which have more than or less than 8e“ in their outer shell will combine.

Question 8.
Are elements and compounds simply made up of separate atoms Individually arranged?
Answer:
No. Elements and compounds are not simply made up of separate atoms individually arranged.

Question 9.
Is there any attraction between atoms?
Answer:
Yes. There is some attraction betwen atoms.

Question 10.
What is that holding them together?
Answer:
Force of attraction between them.

10th Class Chemistry Textbook Page No. 155

Question 11.
Why is there absorption of energy in certain chemical reactions and release of energy in other reactions?
Answer:’
Because of bond energy between the atoms in a molecule.

Question 12.
Where does the absorbed energy go?
Answer:
For breaking chemical bonds between atoms in a molecule.

Question 13.
Is there any relation to energy and bond formation between atoms?
Answer:
Yes. There is some relation to energy and bond formation between atoms.

Question 14.
What could be the reason for the change in reactivity of elements?
Answer:
Number of electrons in their outermost orbit.

Question 15.
What could be the reason for this?
Answer:
They have 8 (e^–) electrons in their outermost orbit.

10th Class Chemistry Textbook Page No. 157

Question 16.
What did you notice in Lewis dot structure of noble gases and electronic configurations of the atoms of these elements shown in table – 1?
Answer:
Except He remaining Ne, Ar, Kr have 8 electrons in their outermost orbit.

10th Class Chemistry Textbook Page No. 158

Question 17.
What have you observed from the above conclusions about the main groups?
Answer:

1. Number of gained electrons of non-metals in their valency.
2. Number of lost electrons of metals in their valency.

Question 18.
Why do atoms of elements try to combine and form molecules?
Answer:
To attain stable electronic configuration.

10th Class Chemistry Textbook Page No. 159

Question 19.
Is it accidental that IA to VIIA main group elements during chemical reactions get eight electrons in the outermost shells of their ions, similar to noble gas atoms?
Answer:
No, it cannot be simply accidental.

Question 20.
Explain the formation of ionic compounds NaCl, MgCl₂, Na₂O and AlCl₃ through Lewis electron dot symbols (formulae).
Answer:
1) Lewis electron dot symbol for NaCl:

Formation of sodium chloride (NaCl) :
Sodium chloride is formed from the elements sodium and chlorine. It can be explained as follows.
Na_(s) + ½Cl_2(g) → NaCl₂

Cation formation :
When sodium (Na) atom loses one electron to get octet electron configuration it forms a cation (Na+) and gets electron configuration that of Neon (Ne) atom.

Anion formation :
Chlorine has shortage of one electron to get octet in its valence shell. So it gains the electron from Na atom to form anion and gets electron configuration as that of argon (Ar).

Formation of the compound NaCl from its ions :
Transfer of electrons between ‘Na’ and ‘Cl’ atoms, results in the formation of ‘Na⁺‘ and ‘Cl^–‘ ions. These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound sodium chloride (NaCl).
Na⁺_(g) + Cl^–_(g) → Na⁺Cl^–_(s) or NaCl

2) Lewis electron dot symbol for MgCl₂:
MgCl₂

Formation of magnesium chloride (MgCl₂):
Magnesium chloride is formed from the elements magnesium and chlorine. The bond formation MgCl₂ in brief using chemical equation is as follows :
Mg_(s) + Cl_2(g) → MgCl_2(g)
Cation formation:

Anion formation :

The compound MgCl₂ formation from its ions :
Mg²⁺ gets ‘Ne’ configuration and
Each Cl^– gets ‘Ar’ configuration
Mg²⁺_(g) + 2 Cl^–_(g) → MgCl_2(s)
One ‘Mg’ atom transfers two electrons one each to two ‘Cl’ atoms and so formed Mg²⁺ and 2Cl^– attract to form MgCl₂.

3) Lewis electron dot symbol for (Na₂O) :

Formation of di sodium monoxide (Na₂O):
Di sodium monoxide formation can be explained as follows:
Cation formation (Na⁺ formation):

Two ‘Na’ atoms transfer one electron each to one oxygen atom to form 2 Na⁺ and O^2-
Each Na⁺ gets ‘Ne’ configuration and O^2- gets ‘Ne’ configuration.
These ions (2Na⁺ and O^2-) attract to form Na₂O.

4) Lewis electron dot symbol for (AlCl₃):

Formation of aluminium chloride (AlCl₃):
Aluminium chloride formation can be explained as follows:
Formation of aluminium ion (Al³⁺), the cation:

Each aluminium atom loses three electrons and three chlorine atoms gain them, one electron each.
The compound AlCl₃ is formed from its component ions by the electrostatic forces of attractions.
Al³⁺_(g) + 3 Cl^–_(g) → AlCl_3(s)

10th Class Chemistry Textbook Page No. 163

Question 21.
How do cations and anions of an ionic compound exist in its solid state?
Answer:
Ionic compounds exist in crystalline state.

Question 22.
Do you think that pairs of Na⁺ Cl^– as units would be present in the solid crystal?
Answer:
Yes. I think that pairs of Na⁺ Cl^– as units would be present in the solid crystal.

10th Class Chemistry Textbook Page No. 164

Question 23.
Can you explain the reasons for all these?
Answer:
Ionic bond is formed between atoms of elements with electronegativity, difference equal to or greater than 1.9.

10th Class Chemistry Textbook Page No. 165

Question 24.
Can you say what type of bond exists between atoms of nitrogen molecule?
Answer:
Triple bond exists between atoms of nitrogen molecule.

10th Class Chemistry Textbook Page No. 168

Question 25.
What do you understand from bond lengths and bond energies?
Answer:
Bond lengths and bond energies are not same when the atoms that form the bond are different.

Question 26.
Are the values not different for the bonds between different types of atoms?
Answer:
Yes. The values are not different for the bonds between different types of atoms.

10th Class Chemistry Textbook Page No. 170

Question 27.
What is the bond angle in a molecule?
Answer:
It is the angle subtended by two imaginary lines that pass from the nuclei of two atoms which form the covalent bonds with the central atom through the nucleus of the central atom at the central atom.

10th Class Chemistry Textbook Page No. 172

Question 28.
How is MCI molecule formed?
Answer:
The ‘1s’ orbital of ‘H’ atom containing unpaired electron overlaps the ‘3p’ orbital of chlorine atom containing unpaired electron of opposite spin.

10th Class Chemistry 10th Lesson Chemical Bonding Activities

Activity – 1

1. Write the Lewis structures of the given elements in the table. Also, consult the periodic table and fill in the group number of the element.
Answer:

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 6 Reproduction.

AP State Syllabus SSC 10th Class Biology Important Questions 6th Lesson Reproduction

10th Class Biology 6th Lesson Reproduction 1 Mark Important Questions and Answers

Question 1.
What questions you ask the doctor, who visited your school on World AIDS day?
Answer:

1. How does AIDS disease occurs?
2. How does the AIDS transmit?
3. What are the symptoms of AIDS?
4. What are the precautions to be taken to prevent AIDS?

Question 2.
What is colostrum?
Answer:
The first secretion from the Mammary glands, after giving birth, rich in antibodies.
During the end of pregnancy, a watery yellowish lymph like fluid accumulates in mammary glands. It is known as colostrum.

Question 3.
Name the types of asexual reproduction in the following organisms:
a) Paramoecium b) Yeast
Answer:
a) Paramoecium : Paramoecium reproduces by splitting into two. (Transverse binary fission)
b) Yeast: Yeast reproduces by Budding.

Question 4.
What are the advantages of grafting method in plants.
Answer:

1. Grafting is used to obtain a plant with desirable characters.
2. It can be used to produce varieties in seedless fruits.

Question 5.
What measures can be taken to avoid sexually transmitted diseases?
Answer:

1. Avoid sex with unknown or multiple partners.
2. Sex with life partners only.
3. Follow ethical and healthy life practices because contraceptives always cannot prevent STD’s.
4. In case of any doubt, consult a qualified doctor for early detection if diagonised with disease take complete treatment.

Question 6.
What is parthenogenesis?
Answer:
The process of developing zygote from female gametes without fertilization is known as parthenogenesis.

Question 7.
In flowering plants, I am formed as the result of double fertilization. The cotyledons digest and absorb me. Who am I?
Answer:
Endosperm.

Question 8.
In what way does mitotic division help the living organism?
Answer:

1. growth
2. cell repair
3. healing wounds.

Question 9.
Give any two suggestions to create awareness to stop female foeticide.
Answer:

1. Preparing relevant slogans
2. Organising rallies
3. Awareness campaign by using electronic and print media

Question 10.
Write two precautions you take, while observing Rhizopus in the laboratory.
Answer:

1. Don’t touch the experimental bread with hand.
2. If you touch the bread, thoroughly wash your hands.
3. Leave the bread in the open air for about an hour.
4. Avoid opening of the plastic bag as much as you can.
5. Sprinkle water over bread.
6. Place the bag in a dark and warm place.

Question 11.
Mention two materials you have used to observe Rhizopus on bread mould.
Answer:
Bread mould sample, plain glass slide, cover slip, water, disposable gloves.

Question 12.
What type of reproduction occurs in paramoecium during favourable conditions?
Answer:
During favourable conditions paramoecium reproduce asexually by fission.

Question 13.
What type of reproduction occurs in paramoecium during unfavourable conditions?
Answer:
During unfavourable conditions paramoecium reproduce sexually by conjugation.

Question 14.
Which bacteria is responsible for formation of curd from milk?
Answer:
Lactobacillus bacteria is responsible for formation of curd from milk.

Question 15.
What is asexual reproduction?
Answer:
The reproduction in which a single parent is involved, without formation of gametes is known as asexual reproduction.

Question 16.
What is fission?
Answer:
Splitting of organisms into two or more offsprings in a symmetrical manner is known as fission. Ex: Paramoecium and bacteria.

Question 17.
How budding occurs in yeast?
Answer:

1. A bud develops as an outgrowth due to repeated cell division at specific site.
2. These buds develop when fully mature, detach from the parent body and become new independent individuals.

Question 18.
Which animals reproduce through fragmentation?
Answer:
Fragmentation is a common mode of reproduction in Algae, Fungi and many land plants.

Question 19.
What is Regeneration?
Answer:

1. Many fully differentiated organisms have the ability to give rise to new individual organism from their body parts.
2. If the individual is some how cut or broken up into many pieces, many of these pieces grow into separate individuals. Ex: Hydra and planaria.

Question 20.
In which plant, small plants grow at the edge of leaves?
Answer:
In Bryophyllum, small plants grow at the edge of leaves.

Question 21.
By means of which plants propagate vegetatively through stem?
Answer:
Plants propagate vegetatively through stem by means of stolons, bulbs, corms, tuber etc.

Question 22.
Through which the Vallisneria, Strawberry propagate vegetatively?
Answer:
Vallisneria, Strawberry propagate vegetatively through stolons.

Question 23.
Which plants produce the new plants through roots?
Answer:
New plants are produced from the roots of Dahlia, radish, carrot etc.

Question 24.
What are the artificial propagation methods in plants?
Answer:
Cutting, Layering and Grafting are the artificial propagation methods in plants.

Question 25.
Which method is used to obtain a plant with desirable characters?
Answer:
Grafting is used to obtain a plant with desirable characters.

Question 26.
Which method will you adopt to get two desirable characters from two different plants in a single plant?
Answer:
I will adopt grafting method to get two desirable characters from two different plants in a single plant.

Question 27.
Which fungus is commonly called as bread mould?
Answer:
Rhizopus is commonly called bread mould.

Question 28.
How bread mould appears when you observe it under microspore?
Answer:
The common bread mould consists of fine thread like projections called hyphae and thin knob like structures called Sporangia.

Question 29.
In which plants leaf is known as Sporophyll? Why?
Answer:
In fern plants leaf is known as Sporophyll. Because on the lower surface of the leaf sporangia are present.

Question 30.
What is fertilisation?
Answer:
Union of male and female gametes is known as fertilisation.

Question 31.
What is external fertilisation?
Answer:
If the fertilisation occurs outside the body of the organism then it is known as external fertilisation. Eg : Frog and fish.

Question 32.
What is internal fertilisation?
Answer:
If the fertilisation occurs inside the body of the female organism then it is known as internal fertilisation. Eg : Terrestrial animals (Reptiles, Aves, Mammals).

Question 33.
What are the parts present in male reproductive system of man?
Answer:
A pair of testis, Accessory glands and System of ducts.

Question 34.
The male reproductive organ testis produces?
Answer:
Testis produces male reproductive cells or spermatozoa.

Question 35.
Sperms are temporarily stored in which part of duct system?
Answer:
Sperms are temporarily stored in epididymis of duct system.

Question 36.
What are the accessory glands present in male reproductive system?
Answer:
The accessory glands present in male reproductive system are one prostrate gland . and two cow cowper glands.

Question 37.
The fluid secreted by accessory glands is
Answer:
The fluid secreted by the accessory glands is semen.

Question 38.
What is the function of semen?
Answer:
Semen provide nutrients for sperm to keep alive and helps as a medium for the movement of sperms.

Question 39.
Which hormone regulates the development of the male reproductive organs?
Answer:
The hormone testosterone regulates the development of the male reproductive organs.

Question 40.
How are the secondary sexual characters are controlled in males?
Answer:
Secondary sexual characters in males are controlled by the male sex hormone testosterone.

Question 41.
Men produce sperm from the age of about?
Answer:
Men produce sperm from the age of about 13 or 14 years and can go on doing so most their lives.

Question 42.
Which are capable of changing the sex of the organism in which they grow like wasp?
Answer:
Some bacteria and other micro-organisms have been found capable of changing the sex of the organism of wasp in which they live.

Question 43.
The female gamete ovum is produced by
Answer:
The female gamete ovum is produced by graffian follicles of Ovary.

Question 44.
What is ovulation?
Answer:
The release of ovum from graffian follicle is known as ovulation.

Question 45.
Fertilisation of ovum occurs in which part of female reproductive system?
Answer:
Fertilisation of ovum occurs in fallopian tube or oviduct of female reproductive system.

Question 46.
What is placenta?
Answer:
Placenta is the nourishment tissue formed by the outer membrane of the embryo called chorion and the part of the uterine tissue.

Question 47.
When do placenta is formed during the development of embryo?
Answer:
Placenta is formed at around 12 weeks of pregnancy or during the embryonic development.

Question 48.
What keeps embryo moist and protects it from minor mechanical injury?
Answer:
The embryo develop in amniotic fluid filled cavity which keeps it moist and protects it from minor mechanical injury.

Question 49.
Which membrane forms umbilical cord?
Answer:
Allantois membrane which originates from the digestive canal of the embryo forms the major part of tube like structure called umbilical cord.

Question 50.
What is foetus?
Answer:
From the third month of pregnancy the embryo is called foetus.

Question 51.
What is gestation period?
Answer:
Total time required for the development of embryo and foetus is called gestation period.

Question 52.
What is the gestation period in human beings?
Answer:
The gestation period in human beings is 9 months or 280 days.

Question 53.
Collect the information about gestation periods in different animals.
Answer:
Gestation period in different animals:

Question 54.
What is after birth?
Answer:
The muscular contractions of the uterus continue until they push out the tissues of the placenta, which are commonly called the ‘after birth’.

Question 55.
What are labour pains?
Answer:
The rhythmic contraction and relaxation of muscle layers of the uterus is known as labour pains.

Question 56.
What is colostrum?
Answer:
Colostrum: During the end of pregnancy a watery yellowish lymph like fluid accumulates in the mammary glands. It is known as colostrum.

Question 57.
What is the importance of feeding colostrum to new born baby?
Answer:
It is very important to feed colostrum to the new born baby because it helps in developing the immune system of the child.

Question 58.
What is the need of sexual reproduction?
Answer:
Sexual reproduction help organisms to develop characters that would be help them to adapt better to their surroundings.

Question 59.
In which mountain regions can Sal trees grow?
Answer:
Sal trees grow in the Himalayan mountains.

Question 60.
What are the different parts of a flower?
Answer:
Sepals, petals, stamens and carpels are the different parts of a flower.

Question 61.
What are stamens and carpels?
Answer:
The reproductive parts of a flower which possess the sex cells or germ cells are called stamens and carpels.

Question 62.
What are unisexual flowers? Give examples.
Answer:
Flowers having either stamens or carpels are called unisexual flowers.
Eg: Bottlegourd, papaya.

Question 63.
What are Bisexual flowers? Give some examples.
Answer:
Flowers having both the stamen and carpel are called bisexual flowers. Eg: Datura.

Question 64.
What are the three parts of carpel or gynoecium?
Answer:
The three parts of carpel or gynoecium are ovary, style and stigma.

Question 65.
What is self pollination?
Answer:
Plants having flowers. Where reproductive cells of stamen of the flower fertilise the female reproductive cells of the carpel of the same flower is called self pollination.
Eg: Plants of pea family.

Question 66.
How cross fertilisation occurs?
Answer:
If the male cells of flower of a plant fertilise the female cells of flowers on the same or different plants of the same species, the type of pollination is called cross pollination.

Question 67.
What did Darwin showed regarding fertilization of plants?
Answer:
Darwin in 1876 showed that plants when isolated had the greatest tendency to self fertilize while when surrounded by varieties of the same flower, they readily cross fertilize.

Question 68.
Which cells are composed the embryosac of ovule?
Answer:
The embryosac of ovule composed of gametophyte cells.

Question 69.
How many cells and nuclei does an embryosac consisting in majority of flowering plants?
Answer:
The majority of flowering plants have an embryosac consisting of seven cells and eight nuclei.

Question 70.
What is double fertilisation?
Answer:
Double fertilisation: Union of one male nucleus with an egg and the second male nucleus with the fusion nucleus is called double fertilisation.

Question 71.
What is germination?
Answer:
The seed produced after fertilisation contains the future plant or embryo that develops into a seedling under appropriate conditions. This process is called germination.

Question 72.
Who gave the phrase “omnis cellula de cellula”? What does it mean?
Answer:
The ‘phrase omnis cellula de cellula’ means cells arise from pre-existing cells. It was given by Rudolph Virchow who discovered cell division.

Question 73.
Who stated that the animals can reproduce through binary fission of cells?
Answer:
In 1852 Robert Remak of Germany stated that animals can reproduce through binary fission of cells.

Question 74.
Who discovered the process of mitosis?
Answer:
Mitosis was discovered by Walther Flemming in 1879.

Question 75.
What is the most important discovery of Walther Flemming regarding chromosomes?
Answer:
Walther Flemming’s most important discovery was chromosomes appear double in nature.

Question 76.
Who proposed that chromosomes carried a different set of heritable elements?
Answer:
Wilhelm Roux proposed that chromosomes carried a different set of heritable elements.

Question 77.
What are the hypothesis made by August Weiseman on chromosomes?
Answer:

1. In successive generations, individuals of the same species have the same number of chromosomes.
2. In successive cell division the number of chromosomes always remain constant.

Question 78.
Who confirmed the scheme of mitotic division?
Answer:
The scheme of mitotic division was confirmed in 1904 by Theodor Boveri.

Question 79.
Who discovered the structure of DNA?
Answer:
The structure of (DNA) deoxy ribonucleic acid was discovered in 1953 by James Watson and Francis Crick.

Question 80.
The cells in which organ do not divide?
Answer:
Cells present in organs such as heart and brain of an individual never divide.

Question 81.
What is time required for completion of mitosis?
Answer:
The process of mitosis is completed in 40 to 60 minutes.

Question 82.
What is interphase?
Answer:
The period between two cell divisions is called interphase.

Question 83.
Into how many phases the interphase can be divided?
Answer:
Interphase can be divided into three phases. They are G₁ phase, S phase and G₂ phase.

Question 84.
What is G₁ phase of interphase?
Answer:
G₁ phase is the linking period between the completion of mitosis and the begining of DNA replication (Gap 1 phase).

Question 85.
What is S phase of interphase?
Answer:
S phase is the period of DNA synthesis leading to duplication of chromosomes.

Question 86.
What is G₂ phase of interphase?
Answer:
G₂ phase is the time between the end of DNA replication and the beginning of mitosis (Gap 2 phase).

Question 87.
Who conducted some experiments using the cell fusion technique on phases of interphase?
Answer:
Potu Narasimha Rao and Johnson conducted some experiments using the cell fusion technique to understand the functional relationship between the phases of interphase.

Question 88.
What is cytokinesis?
Answer:
Division of cytoplasm is called cytokinesis.

Question 89.
What are the different stages present in mitosis?
Answer:
Prophase, Metaphase, Anaphase and Telophase are the different stages present in mitosis.

Question 90.
In which phase of the mitosis chromosomes split lengthwise to form chromatids?
Answer:
In prophase of the mitosis chromosomes split lengthwise to form chromatids.

Question 91.
During which phase of mitosis chromatids are pulled towards poles?
Answer:
During anaphase of mitosis chromatids are pulled towards poles.

Question 92.
How many haploid daughter cells are formed after meiosis?
Answer:
Four haploid daughter cells are formed after meiosis.

Question 93.
What are the diseases that can be sexually transmitted?
Answer:
Sexually transmitted diseases include bacterial infections such as Gonorrhoea and Syphilis and Viral infections such as AIDS.

Question 94.
In what way the sexually transmitted diseases spread from person to person?
Answer:
Sexually transmitted diseases spread by unsafe sexual contacts, using infected devices, infected blood transfusion, from an infected mother to child.

Question 95.
Which state has the highest number of HIV patients in the country?
Answer:
Andhra Pradesh and Telangana has the highest number of HIV patients in the country.

Question 96.
Which factors are contributing to the spread of HIV in Andhra Pradesh?
Answer:
Illiteracy, poor health, unemployment, migration, non-traditional sex practise, unethical contacts and trafficking are some of the factors contributing to the spread of HIV in Andhra Pradesh.

Question 97.
Expand “ASHA”.
Answer:
Accredited Social Health Activist.

Question 98.
What is Red ribbon express?
Answer:
Red Ribbon express is an AIDS/HIV awareness campaign train by the Indian Railways. The motto of the Red ribbon express is “Embarking on the Journey of Life”.

Question 99.
What is contraception?
Answer:
The prevention of pregnancy in women by preventing fertilisation is called contraception.

Question 100.
Which device not only prevents fertilisation but also transmitting some sexually transmitted diseases?
Answer:
Condoms and diaphragm (cap) prevents fertilisation and also useful to not transmitting some sexually transmitted diseases like gonorrhoea, syphilis, AIDS.

Question 101.
What are spermicides?
Answer:
Spermicides are the pills used for killing sperms.

Question 102.
What are the surgical methods to birth control in males and females?
Answer:
Vasectomy for males and Tubectomy for female are the surgical birth control methods in human beings.

Question 103.
What is Vasectomy?
Answer:
In males, a small portion of vas deferens is removed by surgical operation ami both ends are tied properly. This method is called vasectomy.

Question 104.
What is Tubectomy?
Answer:
In females, a small portion of oviducts (fallopian tube) is removed by surgical operation and the cut ends are tied. This prevents the ovum from entering into the oviducts. This method is called Tubectomy.

Question 105.
What is the marriage age for girls in India?
Answer:
The marriage age for girls in India is 18 years.

Question 106.
What is foeticide?
Answer:
Foeticide is the act of destruction or aborting a foetus because it is female.

10th Class Biology 6th Lesson Reproduction 2 Marks Important Questions and Answers

Question 1.
What are the questions you asked the doctor who visited your school to know “the ways of transmission of HIV”?
Answer:
I shall ask the following questions to the doctor.

1. What are the ways of transmission of HIV?
2. How can we prevent the spread of HIV?
3. What precautions should we take while doing transfusion of blood:
4. How does HIV transmit from mother to baby?
5. Why should we use disposable syrenges?

Question 2.
The chromosomal number is reduced to half in the daughter cells produced by meiosis. What happens if the number is not reduced to half in daughter cells?
(OR)
In Meiosis, the chromosome number in the daughter cells are reduced to half that of their parent cells. Guess, what would happen, if the reduction of chromosome number is not done.
Answer:

1. If the reduction of chromosomes number is not done, the chromosomal number is doubled in the offsprings.
2. The change in chromosomal number leads to development of abnormal characters in the individual.
3. The offspring differs from parental generation.
4. Abnormal characters will be formed in new generation, which are not useful for the existence of individual.

Question 3.
What questions do you ask a doctor to know about different birth control methods?
Answer:

1. What is family planning?
2. What is meant by contraception?
3. How many types of contraceptive methods are there?
4. What are the contraceptive devices used for female?
5. What are the contraceptive devices used for male?
6. What is tubectomy?
7. What is vasectomy?
8. What are surgical methods of birth controls?

Question 4.
Apparao and Ramulamma are a newly married illiterate couple. They don’t want children for few years. Suggest some birth control methods for them.
(OR)
Mention any four birth control methods.
Answer:
a) condoms
b) diaphragm (Cap)
c) pills
d) copper – T
e) loop

Question 5.
Why is it important for gametes to have half the number of chromosomes?
Answer:

1. If gametes have 2 sets of chromosomes, the number of chromosomes will be 4 sets in zygote after fertilization because of this the chromosomal number will be doubled in each generation. This results in abnormalities in off-spring.
2. Hence, to maintain a constant number of chromosomes, garnets should have half set of chromosomes.

Question 6.
Identify the flower parts a, b, c, d and write their main function.
Answer:
a) Ovary: Female reproductive organ in flower. It produces female gametes called ovules.
b) Style: Ovary has a pipe like structure called style. It allows the pollen tube to enter the ovary for fertilization.
c) Stamen: These are male parts called androecium. It has two parts. They are filament and Anther.
d) Anther : Produces male gametes called pollen grain.

Question 7.
Draw and label the diagram of human sperm cell.
Answer:

Question 8.
How can we get the desired useful triats with the help of two selected triats by using grafting method?
Answer:

1. Two plants are joined together in such a way that two stems join and grow as a single plant.
2. One which is attached to soil is called stock and the cut stem of another plant without roots is called scion.
3. Both stock and scion are tied with the help of a twine thread and covered by a polythene cover.
4. Grafting is used to obtain a plant with desirable characters.
5. This technique is very useful in propagating improved varieties of plants with various flowers and fruits. Ex: Mango, citrus, apple, rose.

Question 9.
Draw the labelled diagram of Embryo-sac A.
Answer:

Question 10.
Observe the diagram and answer the following questions.
i) Which phases take same time duration?
Answer:
G₁ phase and S phase.
ii) In which phase, DMA synthesis takes place?
Answer:
S Phase.

Question 11.
Write the process involved in seedless fruit development with two suitable examples.
Answer:
In some plants ovary directly develops into fruit without the process of fertilization, this phenomenon is called as parthenocarypy.
Ex: Grapes, water melon.

Question 12.
What precautions will you take to keep away from diseases like AIDS and other sexually transmitted diseases?
Answer:

1. Avoid sex with unknown partners or multiple partners.
2. Use condom every time.
3. Use disposable syringes and needles.
4. Transfusion of safe blood to the patients.
5. HIV mother can have child with doctor’s advice only.

Question 13.
Observe the diagram and answer the following questions.
i) Name male and female reproductive parts of the above figure.
Answer:
Male reproductive parts – anther / pollen grain / stamen
Female reproductive parts – ovary / ovule / style / stigma.

ii) Write the names of (1) and (2) in the diagram.
Answer:

1. Sepal or calyx
2. Petal or corolla

Question 14.
When does Parthenogenesis occur? Write names of two animals in which parthenogenesis takes place.
Answer:
a) Parthenogenesis is a process of reproduction where there is a shift from sexual to asexual mode of reproduction.
b) In this process generally the female garnets develops into zygote without fertilization.
c) This strange kind of reproduction occur in bees, ants and wasps.
d) The parthenocarpic zygote develop into male (Monoploid) while the fertilized one developed into female (Diploid)

Question 15.
Draw the figure of metaphase in mitosis, and write about it.
Answer:

1. Chromosomes move to spindle equator, centromeres attached to spindle fibres.
   
2. Centromeres split, separating the chromatids.

Question 16.
Prepare 4 questions on meiosis, to conduct a Quiz programme.
Answer:

1. Where does meiosis occur in?
2. How many daughter cells are produced at the end of meiosis?
3. In which phase of meiosis karyokinesis takes place?
4. Name the scientist who discovered meiosis for the first time.

Question 17.
Write slogans on ‘Child marriages – a social evil’.
Answer:

1. Child marriage, a loosing game.
2. She is a child herself, why burden her with another child?
3. My childhood, my right.
4. A child should call ‘mother’ but a child should not be called mother.
5. Good marriages take place slowly. Go slow with children’s marriage.
6. Say no to child marriage.

Question 18.
Write 5 slogans on the prevention of HIV/AIDS.
Answer:

1. Open your eyes before AIDS closes them.
2. Hate the disease but not the diseased.
3. Spread the knowledge not the virus.
4. Wear protection to prevent infection.
5. AIDS brings pain! Girls please obstain.

Question 19.
What is fission? Give examples.
Answer:

1. Fission is a method of asexual reproduction in which a single-celled organism splits into two or more offsprings.
2. This splitting usually occurs in a symmetrical manner.
3. When an organism is split into two offsprings it is called binary fission.
4. When an organism is split into more offsprings, it is called multiple fission.
5. This is often the only mode of reproduction for single celled organisms.
   Ex : Paramoecium and bacteria.

Question 20.
Write a short notes on fragmentation.
Answer:

1. Fragmentation is a reproductive method in multicellular organisms with relatively simple body organisation.
2. Some can grow from a separate piece of parent organism. This can be from any part of the body.
3. This happens only in the simplest such as some flat-worms, moulds, lichens, spirogyra, etc.
4. Fragmentation is a common mode of reproduction in algae, fungi and many land plants.

Question 21.
What do you know about parthenogenesis? Explain with examples.
Answer:

1. Parthenogenesis is an asexual reproduction in which unfertilized eggs develop into offsprings.
2. In this process generally egg develops into new individual without meiosis and fertilization. So the offsprings are diploid.
3. In some species of animals reproduction occurs only through parthenogenesis. There are no males known in these species. Ex: Rotifers.
4. In another type of parthenogenesis meiosis does occur and the egg can develop whether fertilized or not.
5. The monoploid offsprings develop into males and diploid into females.
   Ex: Bees, Ants and Wasps.
6. Nowadays we are able to develop seed less fruits like watermelon, grapes, pomegranate etc.

Question 22.
Describe the vegetative propagation through the stem with examples.
Answer:

1. Production of new plants from the vegetative parts such as stem, root, leaves of the existing plant is called vegetative propagation.
2. Aerial weak stems like runners and stolons, when they touch the ground, give off adventitious roots.
3. When the connection with the parent plant is broken, the portion with the newly struck roots develops into an independent plant.
4. Some examples for propagation by stem are from stolons, bulbs, corms and tubers as follows.
   a) Stolons – Vallisneria, Strawberry
   b) Bulbs – Alliumcepa or onion
   c) Corms – Colacasia
   d) Tuber – Potato

Question 23.
Write short note on artificial propagation method cutting.
Answer:

1. Cutting is an artificial method of vegetative propagation in which new plants are developed from the cut portion of existing plant.
   
2. Some plants grow individually when a piece of the parent plant having bud is cut from the existing plant.
3. The lower part of this cutting is buried in moist soil.
4. After few days the cut parts having buds grow as an individual plant.
   Ex: Rose, Hibiscus.

Question 24.
What is layering? Explain briefly about it.
Answer:

1. Stems that form roots while still attached to the parent plants are called layers. Propagating the plants in this method is layering.
2. A branch of the plant with at least one node is bent towards the ground and a part of it is covered with moist soil leaving the tip of the branch exposed above the ground.
3. After sometime, new roots develop from the part of the branch hurried in the soil.
4. The branch is then cut off from the parent plant, later it develops roots and grows to become a new plant. Ex: Nerium.

Question 25.
Write a short note on Grafting.
Answer:

1. Grafting is a method of artificial vegetative propagation in which two plants are joined together in such a way that two stems join and grow as a single plant.
   
2. One which is attached to soil is called stock and the cut stem of another plant without roots is called scion.
3. Both stock and scion are tied with help of a twine thread and covered by a polythene cover.
4. After few days both will unite by forming new tissue and grow as a single one.
5. Grafting is used to obtain a plant with desirable characters.
6. Plants in which grafting is done more in mango, apple, citrus, plants.

Question 26.
What are the advantages of grafting?
Answer:

1. Grafting enables us to combine the most desirable characteristics of the two plants (scion and stock) in its flower and fruits.
2. By grafting method, a very young scion can be made to flower and produce fruits quite fast when it is grafted to the stock.
3. Grafting can be used to produce varieties of seedless fruits.

Question 27.
How is tissue culture more beneficial than other traditional methods for the artificial propagation of plants? (OR)
What is tissue culture? What are its uses?
Answer:

1. The traditional methods for the artificial propagation of plants are being replaced by the modern methods of artificial propagation of plants involving tissue culture, as it is more beneficial than the traditional methods.
2. In tissue culture, a few plant cells or plant tissue are placed in a growth medium with plant hormones in it and it grows into new plants.
3. Thousands of plants can be grown in very short interval of time.
4. There will be no climatic impact on the propagation, so multiplication can be done throughout the year.
5. It is possible to obtain plants that are free from pathogens.

Question 28.
How does the Rhizopus propagate?
Answer:

1. Rhizopus propagates by means of spores.
   
2. The Rhizopus parent plant produces hundreds of microscopic reproductive units called spores.
3. When the spore case of the plant bursts, the spores spread into air.
4. These air borne spores fall on food or soil, under favourable conditions like damp and warm conditions, they germinate and produce new plants.

Question 29.
Write a short note on spore formation. (OR)
How spores are produced in sporangia of fungi?
Answer:

1. Spore formation is a method of asexual reproduction which occurs through microscopic reproductive units called spores.
2. Most of the fungi like rhizopus, mucor etc., bacteria and non-flowering plants such as ferns and mosses reproduce by the method of spore formation.
3. In fungi like rhizopus spores are produced in some specialised structures called sporangia which bursts and spreads the spores into air. These spores when fall on food or soil under favourable conditions germinate and produce new plants.
4. In non-flowering plants like fern, the leaves called sporophyll bears clusters of sporangia on their lower side. These sporangia produce the spores which produce the new plant when it falls on ground under favourable conditions.

Question 30.
How is external fertilisation different from internal fertilisation? (OR)
What are the differences between external and internal fertilisation?
Answer:

1. Fertilisation that takes place outside the body of mother is called external fertilisation. This is most common in animals like fishes and amphibians. As the chance of fertilisation is controlled by nature it becomes necessary to give rise to vast number of eggs and sperms by these animals.
2. Fertilisation that takes place inside the body of mother is called internal fertilisation. This is common in most of the land animals. As the chance of fertilisation is not controlled by the nature, these animals generally produce less number of eggs.

Question 31.
Write a short note on ovulation. (OR)
What is ovulation? How it occurs?
Answer:

1. Release of the egg or ovum is called ovulation.
2. The ova develop in tiny cellular structures in ovary called follicles, which at first look like cellular bubbles.
3. As a follicle grows, it develops a cavity filled with fluid.
4. Each follicle contains a single ovum.
5. When an ovum is mature, the follicle ruptures at the surface of the ovary and the tiny ovum is flushed out.
6. This release of ovum is called ovulation.

Question 32.
How does the uterus get adapted to receive the embryo?
Answer:

1. The uterus at the time of fertilization is beautifully adapted to receive the developing embryo, providing it with food and disposing of its wastes.
2. A few days prior to this time, the uterus was small, its tissues were thin, and its supply of blood vessels was poor.
3. When the fertilized egg or zygote is about to enter the uterus become much larger, its inner wall becomes thick, soft and moist with fluid, its blood supply is greatly increased and waiting for an embryonic occupant.

Question 33.
What is colostrum? What is its importance?
Answer:

1. During the end of pregnancy, a watery lymph like fluid accumulates in the mammary glands.
2. This is called colostrum.
3. For the first few days after the baby is born, the mammary glands secrete only colostrum.
4. It is very important to feed the new born baby with colostrum because it helps in developing the immune system of the child.

Question 34.
What is the importance of mitosis in human beings?
Answer:

1. Mitosis is the cell division that transforms a human fertilized egg into a baby in nine months and into an adult in the next 20 years.
2. The bone marrow cells actively divide by mitosis to produce red blood cells.
3. Mitosis helps in replacing the worn out cells in the skin.
4. Mitotic divisions in the cells surrounding the wound helps in cease the wound and healing.

Question 35.
Collect the information about the significance of the experiments done by Dr Potu Narasimha Rao and Johnson.
Answer:

1. Nearly 4 decades back Dr.P.N. Rao and Johnson did some elegant experiments using the cell fusion technique to understand the functional relationship between the phases of cell cycle.
2. These experiments have, for the first time provided evidence that the progression of cells through the cell cycle is sequential and unidirectional and are controlled by a series of chemical signals that can diffuse freely between nucleus and cytoplasm.
3. These experiments revealed for the first time the structure of interphase chromosomes that are not ordinarily visible under the microscope.
4. These experiments are considered to be a ‘milestone’ in the cell cycle studies.

Question 36.
Ramu said that it is very essential to create more awareness in Andhra Pradesh on the risk of HIV infection and AIDS. Do you support him? If so, how can you support his statement?
Answer:
Yes, what Ramu said is right. I support his statement with the following reasons.

1. Andhra Pradesh has the highest number of HIV patients in the country.
2. According to official statistics, the state had 5 lakh of the 24 lakh HIV positive patients
   in the country during 2011-12.
3. While one in every 300 adults is suffering from HIV elsewhere, in Andhra Pradesh one in every 100 adults is a HIV patient, that is almost one per cent.
4. The prevalence of HIV is 1.07 per cent among males and 0.73 among females in the state, which again is higher than in other states.

Question 37.
Briefly explain about the contraception and contraceptive methods.
Answer:
The prevention of pregnancy in the woman by preventing fertilisation is called contraception. Any device or chemical which prevents pregnancy in a woman is called a contraceptive. Contraceptive methods are of various types and used by any of the partners as preferable. Some of the contraceptive methods are:

1. Use of physical devices such as condoms and diaphragm (cap).
2. Use of hormonal pills which stop the ovaries from releasing ovum into oviduct.
   These pills can be induced either orally or inserting into female reproductive organ vagina.
3. Use of spermicides that kills the sperms.
4. Use of intra-uterine device called copper – T, loop, etc.
5. Use of surgical methods such as vasectomy for male and tubectomy for female.

Question 38.
Classify the given organisms basing on the type of reproduction.
Man, Flatworm, Mould, Dog, Bacteria, Frog, Fern, Datura, Hen, Yeast.
Answer:

Question 39.
What will happen if the amnion is ruptured before the foetus is developed completely?
Answer:

1. Amnion is the embryonic membrane that grows around the embryo itself.
2. The cavity within the amnion is filled with a fluid called amniotic fluid, which keeps the growing embryo moist and protects it from minor mechanical injury.
3. If the amnion ruptures by accident before the foetus developed completely, the amniotic fluid is released out through vagina.
4. As there is no protective fluid around the foetus, it starts getting damaged.
5. So if possible delivery must done immediately by surgerical method, otherwise abortion must be done.
6. If baby dies inside the uterus which leads to infections in uterus causing problems
   to mother that leads to death.

Question 40.
How will you appreciate the contribution of August Weiseman to the cell biology?
Answer:

1. Science is not advanced only by the collection of data. Someone must think about and interpret the data. August Weiseman belongs to this category who think and interpret the data.
2. Even though his poor eyesight not allowed him to use a microscope to study cells, he made great contribution to the cell biology making use of his thinking capacity and interpretation skills.
3. He hypothesised that
   a) In successive generations, individuals of the same species have the same number of chromosomes.
   b) In successive cell division, the number of chromosomes remains constant.
4. His hypothesis proved right in case of mitosis.
5. We should take such a great person who overcame his defect with his will as our role model.

Question 41.
How will you appreciate the contribution of Dr. P.N. Rao to the ceil biology?
Answer:

1. Dr. Potu Narasimha Rao, a renowned scholar and eminent cytologist came from a poor family in Muppalla village of Guntur district.
2. He did his research work on the cytogenetics of tobacco plant and cancer cells in culture medium.
3. He conducted research in cell kinematics and triggering factor of cell division i.e., mitosis.
4. He observed the interphase and its three phases.
5. To understand the functional relationship between these phases he did elegant experiments on cell fusion technique along with his research associate Dr.Johnson.
6. His researches revealed that the cell cycle is sequential, unidirectional and controlled by a series of chemical signals.
7. His experiments are considered to be a milestone in the cell cycle.
8. He is an exemplary person who proved that poverty is not a barrier to the talent and wisdom.

Question 42.
Write briefly about natural vegetative propagation in plants.
Answer:

1. In natural vegetative propagation new plants are produced from stem, root, leaves of old plants without the help of any reproductive organs.
2. In bryophyllum small plants grow at the edge of leaves.
3. Aerial weak stems like runners stolons, when they touch the ground give it adventitious roots.
4. When the connection with the parent plant is broken the stem portion with the adventitious roots develops into an independent plant.
5. Some examples for propagation by stem are from stolons, bulbs, corms, tuber etc.
   
6. Stolons – Vallisneria, strawberry.
   Bulbs – Onion (Alliumcepa)
   Corms – Colacasia
   Tuber – Potato

Question 43.
What are sexually transmitted diseases and mention the ways to prevent them?
Answer:

1. A disease which can be transmitted through sexual contact is called sexually transmitted disease or STD.
2. These include bacterial infections such as gonorrhoea, syphilis, Herpis and viral infections such as herpes and AIDS.
3. Lack of hygiene is usually a major factor in providing conditions for spread of STDs.
4. But unprotected sex with multiple and unknown partners is the highest reason for the spread of STDs.
5. Some of the ways to prevent STD are as follows.
   a) Being faithful to one’s life partner.
   b) Avoid sexual contact with unknown person.
   c) Using condom during sexual intercourse.
   d) Maintaining personal hygiene.

Question 44.
Why more complex organisms cannot give rise to new individual through regeneration ?
Answer:

1. Many organisms have the ability to give rise to new individual organisms from their body parts.
2. Regeneration happens through mitosis and a particular type of tissue can give rise to its own kind only.
3. In complex organisms, different tissues and organs have altogether different structures.
4. Regenerating a different kind of tissue from another kind is not possible.
5. Hence complex organisms are not able to give rise to new individuals through regeneration.

Question 45.
How an organism will be benefited if it reproduces through spores?
Answer:

1. Reproduction through spores gives several advantages to an organism like they are produced in very large numbers and it helps in propagation of species.
2. Spores can remain dormant till favourable conditions become available.
3. Spores help an organism to overcome unfavourable conditions.
4. Spores can be spread through water, air or animals and thus is good for the spread of an organism to more places.

Question 46.
What is the role of the placenta in embryo development?
Answer:

1. Placenta is a tissue formed by the cells from the embryo and the mother.
2. It is formed around 12 weeks of pregnancy and becomes an important structure for nourishment of the embryo.
3. Placenta is a disc which is embedded in the uterine wall. It contains villi on the embryo’s side of the tissue.
4. On the other side mother’s blood spaces are present.
5. This provides a large surface area for diffusion of glucose, oxygen and other nutrients from the mother of the embryo.

Question 47.
Why do we practise vegetative propagation for growing some types of plants?
(OR)
Why vegetative propagation is adopted over other types of propagation?
Answer:
Vegetative propagation is practised in some plants because

1. It is the only method of reproduction in seed less plants.
2. We get more number of matured plants in a very short time.
3. Thousands of plants can be grown in very short time.
4. This method can help the breeder in preserving the characters he need.
5. It is very easy and economical method for the multiplication of ornamental plants.

Question 48.
What is Mitosis? Which type of cells it occurs in organisms? Write about the different stages of it.
Answer:

1. Mitosis is a method of cell division, in which the nucleus divides into two daughter nuclei.
2. Each containing the same number of chromosomes as the parent nucleus.
3. Mitosis takes place in all body cells which retains same number of chromosomes.
4. Different stages of mitosis:
   1. Prophase
   2. Metaphase
   3. Anaphase
   4. Telophase

10th Class Biology 6th Lesson Reproduction 4 Marks Important Questions and Answers

Question 1.
Explain the changes involved in the formation of seed from Ovule.
(OR)
Pollen grain reached the stigma of a flower. Explain the changes that occurs up to the formation of seeds in a sequence.
Answer:
Process of double fertilization:

1. At the time of fertilization there will be a total of 7 cells arranged in three groups in a mature embryo sac.
2. They are one egg (female garnet) two synergids, one central cell (secondary or polar nucleus) and three antipodals.
3. While all the cells are in haploid (n) condition only the polar nucleus is diploid (2n). This is due to the fusion of two nuclei.
4. The synergids are also known as helper cells.
5. Fertilization is the process of fusion of male and female gametes. For the fusion pollen grains have to reach the surface of the stigma. This is called pollination.
   
6. Pollen grain received by the stigma, germinate and give rise to pollen tubes. The pollen tube has two male nuclei.
7. Usually the pollen tube enters the ovule through microphyle and discharges the two male gametes into the embryo sac.
8. One male nucleus (garnet) approaches the egg and fuses with it to form diploid (2n) zygote this is called first fertilization.
9. The other male nucleus reaches the secondary nucleus (2n) (polar nucleus) and fuses with it to form endosperm nucleus which will be triploid. This is second fertilization. Thus double fertilization occurs in embryosac.
   Changes after double fertilization:
10. After double fertilization, the ovule increases in size rapidly as a result of formation of endosperm tissue by mitosis and the development of new embryo.
11. The embryo consists of cotyledons an epicotyl and a hypocotyl. The cotyledons become greatly enlarged because of stored food for the seedling.
12. The zygote divides several times to form an embryo within the ovule. The ovule develops a tough coat and is converted into a seed. The ovary grows to form a fruit.

Question 2.
Observe the given diagram and answer the following questions.
i) What are the four main parts of a flower?
Answer:
Calyx, Corolla, Androecium and Gynoecium are the main parts of a flower.

ii) Which parts of the flower produces gametes?
Answer:
Androecium and gynoecium produces gametes.

iii) Which parts of the flower help in pollination?
Answer:
Petals or corolla help flower in pollination.

iv) Which part protect the flower during its bud stage?
Answer:
Sepals or calyx protect flower in bud stage.

v) Which part of the flower will turn into a fruit in the future?
Answer:
Ovary of the flower will change into fruit.

Question 3.
Organisms reproduce asexually in many ways. Some of the organisms are given below. Fill the below table based on the collected information about the organism and mode of asexual reproduction in it.
a) Onion b) Spirogyra c) Strawberry d) Ginger e) Honey-bee f) Paramoecium g) Planaria h) Yeast

Answer:

Question 4.
i) Draw a neat labelled diagram of L.S. of flower.
ii) What are the sexual parts in the flower ?
Answer:
i)
ii) A. Androecium or Stamen
B. Gynoecium or Pistil

Question 5.
Read carefully and answer the following questions.

According to Weismann prediction, every organism undergoes two kinds of cell divisions. In Mitosis, there is no change in chromosomal number (2n) and in Meiosis, chromosomal number is reduced to half (n).

i) What does ‘n’ and ‘2n’ indicate?
Answer:
‘n’ indicates haploid state. ‘2n’ indicates diploid state.

ii) In which cells, Meiosis takes place?
Answer:
Meiosis occurs in sex cells during the formation of gametes.

iii) What happens, if chromosomal number is not reduced in Meiosis?
Answer:
The chromosomal number not constant in successive generations.

iv) Which type of cell division occurs in the skin cells?
Answer:
Mitosis

Question 6.
Observe the diagram and answer the following.

i) Which part produce the female gamete?
Answer:
Ovary

ii) Where does the fertilization takes place in female reproductive system?
Answer:
Fallopian tube

iii) Where does the embryo develops until it is ready to born?
Answer:
Uterus

iv) In some cases doctor’s cut and tie the cut ends of the fallopian tubes. What is the name of surgery?
Answer:
Tubectomy

Question 7.
Briefly explain the stages of cell cycle.
Answer:
The process of cell division is called “mitosis”. The period between two cell divisions is called “Interphase”.
This is actually the period when the genetic material makes it’s copy so that it is equally distributed to the daughter cells during mitosis. Interphase can be devided into three phases.
G₁ Phase: This is the linking period between the completion of mitosis and the beginning of DNA replication (GAP-1 Phase). The cell size increase during this period.
S Phase: This is the period of DNA synthesis (Synthesis phase) leading duplication of chromosomes.
G₂ Phase: This is the time between the end of DNA replication and the beginning of mitosis. Cell organells devide and prepare chromosome for mitosis.

Question 8.
i) Draw a labelled diagram of the human male reproductive system.
ii) What is the function of testosterone?
Answer:
i) Male reproductive system:
ii) The function of testosterone hormone is maintaining of secondary sexual chracters in males.

Question 9.
Describe the life cycle of a flowering plant with a help of neat labelled diagrams. (OR) Draw the life cycle of a flowering plant.
Answer:

1. Adult plant produces flowers:
   When the plant matures and is ready to reproduce, it develops flowers. Flowers are special structures involved in sexual reproduction, which includes pollination and fertilisation.
   
2. Pollination: The transfer of pollen grains from the anther of a stamen to the stigma of a carpel is called pollination.
3. Fertilisation:
   i) After pollen grains falls on the stigma fertilization occurs when the male gamete present in pollen grains joins with the female gametes present in the ovule.
   ii) In the ovary the male nucleus of pollen combines with the nucleus of female gamete or egg present to form zygote.
4. Formation of fruit and seed: After fertilisation, a combined cell i.e. zygote grows into an embryo within a seed formed by the ovule.
5. Each seed contains a tiny plant called an embryo which has root, stem and leaf parts ready to grow into a new plant when conditions are favourable.
6. Another part of the flower (the ovary) grows to form fruit, which protects the seeds and helps them spread away from the parent plant to continue the cycle.

Question 10.
Analyze the following information and answer the following questions.

i) What do you call the given reproduction methods?
Answer:
Given reproduction methods are called ‘vegetative propagation’.

ii) What is the major difference between sexual reproduction and vegetative reproduction in plants?
Answer:
In sexual reproduction gametes form zygote. Plant parts like root, stem and leaf are used in vegetative reproduction. It is one of asexual method.

iii) Potato plants do not produce seeds. How can you propagate this plant?
Answer:
Potato plants propagates through the ‘eyes’.

iv) What are the advantages of propagating plants with the above given methods?
Answer:
In vegetative propagation

1. More plants are produced in less time
2. Characters are not changed.
3. It would be possible to develop new varieties with useful characters.

Question 11.
Explain the methods of artificial propagation in various plants.
Answer:
Artificial propagation:

1. Cutting: Some plants can grow individually when a piece of the parent plant having bud is cut off from the existing plant. The lower part of this cutting is buried in moist soil.
   
   After few days the cut parts having buds grow as an individual plant after developing roots. E.g. Rose, Hibiscus.
2. Layering: A branch of the plant with atleast one node is bent towards the ground and part of it is covered with moist soil. After a few days new roots develop from the part of the branch buried in the soil. The branch is then cut off from the parent plant.
   E.g: Nerium, Jasmine
   
3. Grafting: Two plants are joined together in such a way that two stems join and grow as a single plant. This technique is very useful in propagating improved varieties of various flowers and fruits. Grafting is used to obtain a plant with desirable character. E.g: Mango, citrus, apple, rose.
   

Question 12.
Observe the following figures and find the stages of cell division and explain.
Answer:
In the mitotic cell division, the division of nucleus (karyokinesis) followed by the division of cytoplasm (cytokinesis). Finally brings about the formation of two daughter cells. There are four stages in mitosis division.
They are

1. Prophase
2. Metaphase
3. Anaphase
4. Telophase

Question 13.
Mention the stages of Mitosis with the help of diagrams. Explain the changes that takes place in Prophase.
Answer:
Mitosis is a method of cell division, in which the nucleus divides into two daughter nuclei each containing the same number of chromosomes as the parent nucleus. Mitosis takes place in all body cells which retains same number of chromosomes.
Different stages of mitosis:
1) Prophase 2) Metaphase 3) Anaphase 4) Telophase

1) Prophase

1. In this phase chromosomes condense and get coiled.
2. They become visible even in light microscope.
3. Nucleoli becomes smaller.
4. Chromosome split lengthwise to form chromatids, connected by centromeres.
5. Nuclear membrane breaks down.
6. Centrosome containing rod like centrioles, divide and form ends of spindle.

Question 14.
Describe the process of double fertilization in plants. Explain the uses of endosperm that is formed.
Answer:
Double fertilization:

1. In flowering plant germinated pollen grain forms pollen tube.
2. The end of the pollen tube ruptures and two male garnets are released in the Embryosac.
3. Out of two male garnets one male garnet fuses with female garnet which is called fertilization.
4. Another male garnet fuses with the secondary nucleus and forms endosperm.
5. So in flowering plant fertilization occures twice hence it is called double fertilization.

Uses of Endosperm:

1. Cotyledons develops by utilizing endosperm.
2. The Cotyledons utilizes the stored food in the endosperm.
3. Some of the plants utilizes the endosperm completely and changes in to seed.
4. Because of the stored food the size of the cotyledons increases.

Question 15.
Explain any two natural and two artificial vegetative propagation methods to produce more number of plants in less time period with examples.
Answer:
Natural propagation:
i) Leaves – Small plant grow at the edge of the leaves. Ex: Bryophyllum
ii) Stems:
a) Stolon – Ex: Jasmine, strawberry b) Bulbs – Ex: Onion
c) Corns – Ex: Colocasia d) Rhizome – Ex: Ginger e) Tuber – Ex: Potato
iii) Root – Ex: Roots of murayya, guava
Artificial propagation:
Cutting: Some plants can grow individual when a piece of parent plant having bud is cut off from the existing plants. Ex: Rose, Hibiscus.
Layering: A branch of the plant with at least one node is bent towards the ground and a part of it is covered with moist soil leaving the tip of the branch exposed above the ground. Ex: Nerium, Jasmine.
Grafting: Two plants are joint together in such a way that stems join and grow as a single plant one which is attached to soil is called stock and stem of another plant without roots is called scion. Both stock and scion are tied with a twine thread and cover by a polythene cover. Ex: Mango, citrus, apple, rose.

Question 16.
Read the following table and answer the following questions.

i) Name the structure concerned to the heart.
Answer:
Tricuspid valve

ii) What is the function of acrosome?
Answer:
It helps the sperm in penetrating into ovum.

iii) Name the structures which are helpful for gaseous exchange.
Answer:
Alveoli and guard cells

iv) Name the part performing Excretion.
Answer:
Glomerulus

Question 17.
a) Draw a neat and labelled diagram of Human female reproductive system.
b) What happens when the Fallopian tubes are closed?
Answer:
a) Female reproductive system
b) If fallopian tubes are closed the sperm can not reach the ova, fertilization will not happen and zygote will not form.

Question 18.
Observe the following table.

On the basis of information given in the table write- the answers to the following questions.
i) Write the names of two organisms that show Asexual reproduction.
Answer:
Yeast, Hydra, Bacteria, Paramoecium (any two you may write)

ii) Write two artificial vegetative propagation methods mentioned in the table.
Answer:
Cutting, Grafting

iii) Write the names of two plants, which undergo natural vegetative propagation mentioned in the table.
Answer:
Ginger, Turmeric

iv) In fission, how many organisms can we get from one organism?
Answer:
Two

Question 19.
Among the following organisms can we see asexual reproduction? Write about the method of asexual reproduction in any of the two organisms.
Answer:
а) Paramoecium b) Yeast c) Spirogyra d) Amoeba e) Planaria
Yes, we can see asexual reproduction in all the following organisms.

Method of asexual reproduction – Organism
Binary fission                                – Paramoecium, amoeba
Budding                                        – Yeast
Fragmentation                              – Spirogyra
Regeneration                                – Planaria

1) Binary fission in Paramoecium: A single cell divides into two equal daughter cells. First the cytoplasm divides into two parts followed by nuclear division.
2) Asexual reproduction in Yeast: Budding is the common method of asexual reproduction in yeast. In this method, yeast cell wall at a particular region becomes soft and bulges into an outgrowth called bud. Cytoplasm enters into this bulge and then nucleus divides mitotically into two nuclei, one moves into the bud. Finally bud is detached from the parent cell and grows into an independent yeast cell.

Question 20.
See the adjacent picture. Which type of pollination will occur in this ? Why do you think so?
Answer:

1. Self-pollination occurs if stamens and carpels matures at the same time.
2. If they mature at different times, cross pollination occurs.
3. Cross pollination occurs in this plant.
4. For cross pollination the pollen grains are carried from other plants belonging to the same species.
5. The mechanism of dispersal of pollen grains from one plant to other plant is facilitated mostly by wind and insects.
6. Cross pollination is believed to be advantageous for the plant.
7. The seeds produced by the flower will contain another source of genetic material
8. Which may contain genes which are advantageous to the survival of the seedlings.

Question 21.
What are the consequences if meiosis do not happen in the body cells of the organism?
Answer:

1. Each organism has a fixed number of chromosomes.
2. This number has to be maintained in its offspring.
3. Any sudden change in the number of chromosomes will be harmful to the offspring. Assume parent has 10 chromosomes.
4. In the absence of meiosis during sexual reproduction gametes will also have the same number of chromosomes as parent i.e., 10 chromosomes.
5. Union of female and male gametes occur forming zygote during sexual reproduction. The number of chromosomes doubled in zygote will have 10+10 chromosomes.
6. In the next generation, the offspring will have forty chromosomes. If this continues cells in the offsprings will have thousands of chromosomes within few generation.
7. This results in formation of abnormalities in each generation. Hence by way of meiotic division, the chromosome number is maintained constant from generation to generation.

Question 22.
Describe different artificial vegetative methods to produce large scale production of plants.
Answer:

1. Different artificial vegetative propagation methods are cutting, layering, grafting and tissue culture methods.
2. Cutting: Some plants grow individually when a piece of parent plant having bud is cut from the existing plant. After burying in the soil the cut parts having buds grow as an individual plant after developing roots. E.g. Rose.
3. Layering: A branch of the plant with at least one node is bent towards the ground and part of it is covered with moist soil. After sometime, new roots develop from the part of the branch hurried in the soil. The branch is then cut off from the parent plant. E.g: Nerium.
4. Grafting: Two plants are joined together in such a way that two stems join and grow as a single plant. This technique is very useful in propagating improved vari¬eties of various flower and fruits. Grafting is used to obtain a plant with desirable character. E.g: Mango, citrus, apple, rose.
5. Tissue culture: In this method, few plant cells or plant tissues are placed in a growth medium with plant hormones in it and it grows into new plants. Thousands of plants can be grown in very short interval of time.

Question 23.
i) Labelled parts of A, B, C, D above drawn Human female reproductive system.

ii) In which part fertilization takes place?
iii) Which part is in connection with implantation?
iv) What is ovulation?
Answer:
i) A: Fallopian tube
B: Ovary
C: Uterus
D: Vagina
ii) Fertilization takes place in fallopian tube.
iii) Uterus
iv) Release of ovum from graffian follicle of ovary is known as ovulation.

Question 24.
Write some programmes conducted by you to bring awareness in the people about health and hygeine and family planning?
Answer:

1. Organising Health camps on World Health day to people of the village.
2. Conducting immunisation programs for every three months.
3. Supplying tablets on the deworming day.
4. Organising seminars by expert doctors on individual health and cleanliness programs.
5. Propagating small family norms conducting camps for family planning operations.
6. Educating the masses through pamplets on the needs of taking balanced diet.
7. Need of using toilets and washing hands and legs before and after meals.
8. Educating the people by conducting adult education centres. This is basically required for enlightening the people on health aspects.

Question 25.
Government made an act on determining sex through ultrasound scanning and telling it as crime. What do you do to tell this to others?
Answer:

1. I will educate people knowing the sex of foetus inside mother’s womb is a severe crime as per the act made by government.
2. The purpose of ultrasound tests are to know the growing condition of the foetus and also to see whether it is suffering with severe ailments.
3. By knowing the sex of the foetus, if it is female people are ready for aborting it.
4. This leads to reduction in male female ratio in the country.
5. Children either male or female are equal to parents.
6. We should see proper development of girl child after her birth.

Question 26.
Write about the embryonic membranes that nourish, protect and support to the embryo?
Answer:

1. The growing embryo form two membranes – Chorion and Amnion.
2. Chorion establishes connection with the walls of the uterus and helps in the supply of nutrients to the embryo and in the removal of wastes from the embryo.
3. Amnion forms a sac like structure around the embryo and amniotic fluid is present between layers of Amnion.
4. Amnion and Amniotic fluid give protection to the embryo against mechanical shocks.
5. Placenta is a tissue formed around 12 weeks of pregnancy by the cells from the embryo and mother.
6. Embryo receives all the required nutrients and oxygen for its metabolism from the mother through the blood vessels present in the placenta.
7. Another membrane called allantois, which originates from the digestive canal of the embryo forms the major part of a tube like structure called umbilical cord.
8. Umbilical cord contains very important blood vessels that connect the embryo with the placenta.

Question 27.
Write brief history of cell division.
Answer:

1. In 1852 a German scientist, Robert Remak published his observations on cell division and stated that the binary fission of cells was the means of reproduction of animal cells.
2. This view was widely publicized by Rudolf Virchow who gave the phrase “Omnis cellulade cellula” means all cells arising from pre existing cells.
3. In 1879 Walther Flemming reported that there were string like structures in the nucleus which split longitudinally during cell division. He named the process as mitosis means fine threads as the dividing structures resembled threads.
4. Wilhelm Roux proposed that each chromosome carried a different set of heritable elements and suggested that the longitudinal splitting observed by Flemming ensured the equal division of these elements.
5. Combined with the rediscovery of Gregor Mendel’s 1866 paper on heritable elements in peas, these results highlighted the central role of the chromosomes in carrying heritable material or genetic material.
6. The scheme of mitotic division was confirmed in 1904 by Theodor Boveri.
7. The chemical nature of the genetic material was determined in a series of experiments over the next fifty years.
8. The structure of DNA – the constituent of the genetic material was determined in 1953 by James Watson and Francis Crick.

Question 28.
Explain briefly about child birth. (OR) How child birth occurs after gestation period?
Answer:

1. Total time required for the embryonic and foetal development is about 9 months or 280 days.
2. After this time, foetus is expelled from the uterus by the mother. This is child birth.
3. Child birth is a complicated process and involves the participation of child and mother.
4. The foetal hormones produced inside, stimulate the contraction of the muscles present in the walls of uterus.
5. These contractions called labour pains help in the expulsion of the foetus from the uterus.
6. During this process the amnion ruptures, placenta is separated from the walls of J the uterus.
7. At child birth the head usually comes out first.
8. The foetus is still attached to the mother’s uterus through the umbilical cord, which is later separated by the doctors.

Question 29.
Draw the life history of flowering plant in the form of block diagram.
Answer:
Life history of a flowering plant:

Question 30.
In a flower self fertilization takes place. Write the process, the flower organs which involve in self fertilization.
Answer:

1. Fusion of male and female gametes produced by the same individual is called self fertilization.
2. Self ferlization occurs in bisexual flowering plants.
3. The flower organs which involve in self fertilization are stamens (androecium) and carpels (Gynoecium).
4. Majority of flowering plants have an embryo sac consisting of seven cells and eight nuclei.
   
5. The pollen grains produced by anther of stamen are transferred to the stigma of the same flower by wind or insects.
6. The stigma of the carpel secretes a sticky substance which promotes the growth of pollen grains.
7. Under favourable conditions pollen grains germinate on the stigma and give rise to pollen tubes.
   Only one pollen tube finally reaches the embryo sac.
8. This pollen tube will have two male nuclei, which migrate to the tip of the pollen tube at the time of fertilization. Usually the pollen tube enters the ovule through micropyle and discharges the two male gametes into its embryo-sac.
9. One male nucleus (gamete) approaches the egg and fuses with it to form a diploid zygote. This is first fertilization.
10. The other male nucleus reaches the secondary nucleus (2n) and fuses with it to form the endosperm nucleus which will be triploid. This is second fertilization in the embryo sac.
11. Thus double fertilization occurs in embryo sac which is unique in flowering plants.

Question 31.
Describe the structure of flower with a neatly labelled diagram.
Answer:

1. A typical flower consists of an outer whorl of green sepals (calyx) which protects the parts with in.
2. The second whorl has petals (corolla) which are usually brightly coloured. They sometimes emit fragrance also.
3. Petals are soft and are useful to attract insects to facilitate cross pollination.
   
4. The third whorl of the flower consists of stamens (Androecium) which are the male reproductive organs.
5. Each stamen is made up of a filament and an anther.
6. Each anther usually has two anther lobes. The anther produces pollen grains (microspores).
7. The inner most fourth whorl is gynoecium or pistil. It consists of ovary, style and stigma.
8. Ovary occupies central portion on the thalamus. A swollen ovary is present on the thalamus.
9. Inside the ovary future seeds, known as ovules are present.
10. Ovary has a pipe like extension called style. The tip of the style ends in stigma. The stigma receive the pollen grains.

Question 32.
Write a brief note on male reproductive system of human beings.
Answer:

1. The male reproductive system of human beings consists of a pair of testis, accessory glands and a system of ducts.
2. Testis are male reproductive organs and produces spermotozoa or sperms and also secretes male sex hormone Testosterone.
   
3. Inside each testis several lobules are present. Each lobule has several tubules called seminiferous tubules.
4. Germinal epithelial cells in the seminiferous tubules undergo meiotic division to produce sperms.
5. The accessory glands include one prostrate gland and two cowper glands. Secretion of these glands produce semen.
6. The duct system consists of vasa efferentia.
   They collect spermatozoa from seminiferous tubules.
7. Vasefferentia continue as epididymis where sperms are stored temporarily.
8. From epididymis sperms moved into tubule called vas deference and then into urethra.

Question 33.
Describe the female reproductive system in human beings.
Answer:

1. A pair of ovaries, oviducts, uterus and vagina are the parts present in female reproductive system.
2. Ovaries are present just below the Kidneys in the abdominal cavity.
3. Each ovary has several sac like structures called ovarian follicles or Graffian follicles.
   
4. Every time only one follicle matures and release one ovum into the body cavity.
5. Ovaries secrete two female sex hormones called oestrogen and progesterone which control the development of female reproductive organs, ovulation and menstruation.
6. Just above the ovaries are the tubes called oviducts or fallopian tubes where fertilisation takes place.
7. The two oviducts connect to a bag like organ called uterus at their other ends.
8. The uterus is connected through a narrow opening called cervix to another tube called vagina which opens to the outside of the body.
9. Vagina is a tubular structure and is also called birth canal because it is through this passage that the baby is born after the completion of development inside the uterus of the mother.

Question 34.
Describe briefly about the reduction division or meiosis.
(OR)
Why meiosis is also known as reduction division? Comment on it.
Answer:

1. Meiosis occurs only during the formation of gametes in sexual reproduction.
2. During meiosis only one set of chromosomes are passed on to the daughter cells. Hence daughter cells have half the number of chromosomes of the mother cells.
   
3. In meiosis karyokinesis and cytokinesis occur two times.
4. During first phase of meiosis the parent cell divides twice, though the chromosomes divide only once.
5. The second phase meiosis is similar to normal mitosis, but chromosomes do not duplicate more over the chromosome number distributed equally to each cells.
6. Thus the four daughter cells have just half the number of chromosomes of the parent cells.
7. These are haploid (containing only one set of chromosomes).
8. Thus meiotic division is also called reduction division.

Question 35.
Describe the developmental stages of human embryo after fertilization with the help of neatly labelled diagrams.
Answer:

1. During fertilization, chromosomes of the ovum and the chromosomes of the sperm make up into pairs and the resulting cell is called zygote.
2. Fertilization takes place in the oviduct or fallopian tube.
   
3. The zygote which is diploid travels down the fallopian tube. As it moves it undergoes several mitotic divisions forming the embryonic stage called blastocyst.
4. Blastocyst moves towards the wall of the uterus and finally gets attached and embedded in the wall of the uterus. This is called implantation.
5. The growing embryo forms two membranes Chorion and Amnion.
6. Chorion establishes connection with the walls of the uterus and helps in the supply of nutrients to the embryo and removal of wastes from the embryo.
7. Amnion forms a sac like structure around the embryo. The space between the amnion and embryo is filled with a fluid called amniotic fluid.
8. Amnion and amniotic fluid give protection to the embryo against minor mechanical injury.
9. Placenta is a tissue formed by the cells from the embryo and the mother. It is formed around 12 weeks of pregnancy.
10. Placenta nourishes the growing embryo.
    
11. A tough cord called umbilical cord is also formed by the embryo which is connected to the walls of the uterus through the placenta.
12. From 3 months of pregnancy, the embryo is called foetus.
13. Pregnancy lasts on an average 9 months or 280 days. This period is called gestation period.
14. After this time foetus is expelled from the uterus by the mother – this is child birth.
15. This process is complicated and involves the participation of foetus and mother.

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 12 Electromagnetism Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 12th Lesson Electromagnetism

10th Class Physics 12th Lesson Electromagnetism Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
How do electric appliances work?
Answer:
Electrical appliances work due to the electric force. Electrical force works in displacing the charges. Electric force is independent of the state of rest or the motion of the charged particle. Electric motor, washing machine are some of the examples of electric appliances.

Question 2.
How do electromagnets work?
Answer:
An electromagnet acquires the magnetic properties only when electric current is passed through the solenoid. Once the current is switched off, it almost loses its magnetic properties as retentivity of soft iron is very low. The strength of the electromagnet depends upon number of turns per unit length of the solenoid and the current through the solenoid.

Question 3.
Is there any relation between electricity and magnetism?
Answer:
The first evidence that there exists such a relationship between electricity and magnetism was observed by Oersted. When current carrying conductor was parallel to the axis of the needle, and the needle was deflected. This was much against his expectations. On reversing the direction of the current the needle moved in opposite direction.

Question 4.
Can we produce magnetism from electricity?
Answer:
We can produce magnetism from electronic current. Ampere with his Ampere’s swimming rule explained the direction of electric current and the deflection of magnetic needle.

Improve Your Learning

Question 1.
Are the magnetic field lines closed? Explain. (AS1)
Answer:

• Magnetic field lines are closed.
• If we observe the field lines formed by a current carrying straight wire, circular field lines are formed. They are closed circles.
• If we observe the field lines by a current carrying solenoid the field lines out side the solenoid are continuous with those inside.
• Thus the magnetic field lines are closed loops.

Question 2.
See figure, magnetic lines are shown. What is the direction of the current flowing through the wire? (AS1)

Answer:
If field lines are in anti-clockwise direction as shown in the diagram, the direction of current is vertically upwards. This can be demonstrated with right hand thumb rule.

Question 3.
A bar magnet with north pole facing towards a coil moves as shown in figure. What happens to the magnetic flux passing through the coil? (AS1)

(OR)
Why would induced current be generated in the coil when a north pole of a bar magnet pushed into it ?
Answer:
If north pole of the magnet moves towards the coil, there is a continuous change of magnetic flux linked with closed coil, then current is generated in the coil.

Question 4.
A coil is kept perpendicular to page. At P, current flows into the page and at Q it comes out of the page as shown in figure. What is the direction of magnetic field due to the coil? (AS1)

Answer:
At the top, anti-clockwise direction.
At the bottom, clockwise direction.
Try This:

Take a test tube and wound minimum 50 turns of 24 guage insulated copper wire with 2cms length at the centre of test tube as shown in figure, ‘l Now solenoid is ready. Take 3cms length of iron nail and make it floats on water with appropriate foam (thermocol) on the water. Now connect the j two ends of solenoid two 3-6 volts battery eliminator and switch on the eliminator. You can observe the motion of the nail towards the solenoid. (If not move decrease the water level or increase the potential).
Try to explain motion of the nail into the water using solenoid concept.

Question 5.
The direction of current flowing in a coil is shown in the figure. What type of magnetic pole is formed at the face that has flow of current as shown in the figure? (AS1)

Answer:
North. Since the current in the coil flows in anti-clockwise direction, north pole is formed at the face we are watching. 

Question 6.
Why does the picture appear distorted when a bar magnet is brought close to the screen of a television? Explain. (AS1)
(OR)
Explain magnetic force on moving charge and current carrying wire.
(OR)
What happens when you bring a bar magnet near a picture of TV screen ? What inference do you conclude from this activity?
Answer:
This is due to the fact that magnetic field exerts a force on the moving charge.

TV screen Activity :

• Take a bar magnet and bring it near the TV screen.
• Then the picture on the screen is distorted.
• Here the distortion is due to the motion of the electrons reaching the screen are affected by the magnetic field.
• Now move the bar magnet away from the screen.
• Then the picture on the screen stabilizes.
• This must be due to the fact that the magnetic field exerts a force on the moving charges. This force is called magnetic force.
• The magnitude of the force is F = Bqv where B is magnetic induction, ‘q’ is the charge and v is the velocity of the charged particle.

Question 7.
Symbol ‘X’ indicates the direction of a magnetic held into the page. A straight long wire carrying current along its length is kept perpendicular to the magnetic field. What is the magnitude of force experienced by the wire? In what direction does it act? (AS1)

Answer:
1) Magnetic force (F) experienced by the wire with the magnitude of ILB :
Here I = Current, L = Length of the wire B = Magnetic field

2) Direction of the magnetic force (F) :

1) The direction of force can be find by using Right hand rule.
2) Fore finger → i (North)
Middle finger → B (into the page)
Thumb → F (Towards west parallel to the paper)

Question 8.
Explain the working of electric motor with a neat diagram. (AS1)
(OR)
Which device converts electrical energy into mechanical energy? Explain the working of that device with a neat diagram.
Answer:
Electric motor converts electrical energy into mechanical energy.
Electric motor:
It is a device which converts electrical energy into mechanical energy.

Principle :
It is based on the principle that a current carrying conductor placed perpendicular to the magnetic field experiences a force.

Construction :
a) Armature coil:
It contains a single loop of an insulated copper wire in the form of a rectangle.

b) Strong magnetic field :
Armature coil is placed between two permanent poles (N & S) of a strong magnet.

c) Slip-ring Commutator:
It consists of two halves (C₁ and C₂) of a metallic ring. The two ends of the armature coil are connected to these two halves of the ring. Commutator reverses the direction of current in the armature coil.

d) Brushes:
Two carbon brushes B₁ and B₂ press against the commutator. These brushes act as the contact between the commutator and terminals of the battery.

e) Battery :
A battery is connected across the carbon brushes. The battery supplies the current to the armature coil.

Working and Theory :

1. When current flows through the coil, AB and CD experience magnetic force.
2. In the arm, AB of the coil experiences a force in one direction, similarly, in CD it experiences in opposite direction.
3. These two equal and opposite forces constitute a couple; which rotates the coil.
4. At this position, the supply of current to the coil is cut off because contacts of commutator and brushes break.
5. Hence no force acts on the arms of the coil.
6. The coil will not come to rest because of rotational inertia of motion, till the commutator again comes in contact with the brushes B₁ and B₂.
7. Now the direction of the current in the arms AB and CD is reversed.
8. Then the couple again rotates in opposite direction.
9. The coil of DC motor continues to rotate in the same direction. Hence electrical energy is converted into mechanical energy.
10. The speed of rotation of the motor depends on
    a) current through the armature
    b) number of turns of the coil
    c) area of the coil
    d) magnetic induction.

Question 9.
Derive Faraday’slaw of induction from law of conservation of energy. (AS1)
Answer:
Faraday’s law :
Whenever there is a continuous change in magnetic flux linked with coil closed the current is generated in the coil.

• Consider a pair of parallel bare conductors which are separated by 7′ meters.
• They are placed in uniform magnetic field of induction ‘B’ supplied by ‘N’ and ‘S’ poles of the magnet.
• A galvanometer is connected to the ends of the parallel conductors.
• We can close the circuit by touching the parallel conductor with another bare conductor which is taken in our hand.
• If we move our hand to the left, the galvanometer needle will deflect in one direction.
• If we move our hand to the right, the needle in the galvanometer moves in opposite direction.
• A current will be set up in the circuit only when there is an EMF in the circuit. Let EMF be ‘ε’.
• The principle of conservation energy tells us that this electric energy must come from the work that we have done in moving the cross wire.

Question 10.
The value of magnetic field induction which is uniform is 2T. What is the flux passing through a surface of area 1.5 m2 perpendicular to the field? (AS1)
Answer:
B = 2T ; Φ == ? ; A = 1.5 m²
We know B = \(\frac{\phi}{\mathrm{A}}\)
or Φ = BA = 2 × 1.5 =3 Webers

Question 11.
An 8N force acts on a rectangular conductor 20 cm long placed perpendicular to a magnetic field. Determine the magnetic field induction if the current in the conductor is 40 A. (AS1)
Answer:
F = 8N ; l = 20 cm or 20 × 10^-2 m ; B = ? ; i = 40 Amp

Question 12.
Explain with the help of two activities that current carrying wire produces magne tic field. (AS1)
(OR)
How can you verify that a current carrying wire produces a magnetic field with the help of experiment?
Answer:
Activity – 1

• Take a thermocole sheet and fix two thin sticks ol height 1cm.
• Join the two sticks with the help of copper wire.
• Take a battery, tap key and connect them in series with the copper wire which is thin.
• Keep a marine compass needle beneath the wire.
• If you press the tap key, current flows in the copper wire.
• Immediately the magnetic needle gets deflected.
• This indicates that the magnetic field is increased when current flows through the conductor.

Activity – 2

• Take a wooden plank and make a hole.
• Place the plank on the table.
• Place the retort stand on it.
• Pass copper wire through the hole.
• Connect the two ends of the wire with battery through switch.
• Place some compass needle around the hole.
• When the current flows the magnetic needle deflects.
• We can verify this by changing the direction of current.
• So we can conclude the magnetic field surrounds a current carrying conductor.

Question 13.
How do you verify experimentally that the current carrying conductor experiences a force when it is kept in magnetic field? (AS1)
Answer:

1. A copper wire is passed through splits of wooden sticks.
2. Connect the wire to 3 volts battery.
3. Close the switch of the battery and pass the current.
4. Bring the horse-shoe magnet near the wire.
5. Then a force is experienced on the wire.
6. Reverse the polarities of the magnet, then the direction of the force is also reversed.
7. The right hand rule helps the direction of flow of current and the direction of current.

Question 14.
Explain Faraday’s law of induction with the help of an activity. (AS1)
(OR)
Write an activity which proves changing magnetic flux produces induced current in the circuit.
Answer:

• Connect the terminals of a coil to a sensitive ammeter.
• Push a bar magnet towards the coil, with its north pole facing the coil, the needle in the galvanometer deflects.
• It shows that a current is set up in the coil.
• The galvanometer does not deflect if the magnet is at rest.
• If the magnet is moved away from the coil, the needle in the galvanometer again deflects in opposite direction.
• Further this experiment enables us to understand that the relative motion of the magnet and coil set up a current in the coil. It makes no difference whether the magnet is moved towards the coil. This is one form of Faraday’s law.

Question 15.
Explain the working of AC electric generator with a neat diagram. (AS1)
(OR)
Which device converts mechanical energy into electrical energy? Explain the working of that device with a neat diagram.
Answer:
Generator converts mechanical energy into electrical energy.

• As armature is rotated about an axis, the magnetic flux linked with armature changes. Therefore, an induced current is produced in the armature.
• If the armature rotates in anti-clockwise direction, from Flemming’s right hand rule the direction of current and deflection of the coil are noted.
• Alter armature has turned through 180°, it occupies another position.
• By applying Flemming’s right hand rule we can find the direction of current and deflection of the needle.
• Hence we can conclude the induced current is alternating in nature.

Question 16.
Explain the working of DC generator with a neat diagram. (AS1)
Answer:

• The principle and working of D.C generator is same as that of AC generator except that in place of slip – rings as sliding contacts, we have a slip-ring or a commutator.
• In a slip ring, there are two half rings.
• The ends of armature coil are connected to these rings and these rings rotate the armature.
• By using slip-ring, the direction of induced current does not change in the external circuit throughout the complete rotation of the armature. In other words, the current in the external circuit always flows in the same direction. Hence the induced current is unidirectional.

Question 17.
Rajkumar said to you that the magnetic field lines are open and they start at north pole of bar magnet and end at south pole. What questions do you ask Rajkumar to correct him by saying “field lines are closed”? (AS2)
Answer:

• If the magnetic field lines start at north pole and end at south pole, where do the lines go from south pole?
• What is happening within the bar magnet?
• Are the magnetic field lines passing through bar magnet?
• What is the direction of magnetic field lines inside the bar magnet? (Recall the solenoid activity).
• Can you say now, that the magnetic field lines are open?

Question 18.
As shown in figure, both coil and bar magnet move in the same direction. Your friend is arguing that there is no change in flux. Do you agree with his statement? If not, what doubts do you have? Frame questions about the doubts you have regarding change in flux. (AS2)

Answer:

• What happens if both magnet and coil move in same direction?
• What happens if both magnet and coil move in opposite direction?
• What is the direction of the current in the coil?
• If both move in same direction, is there any linkage of flux with the coil?
• When ‘N’ pole is moved towards the coil what is the direction of current?
• If magnet is reversed, what is the direction of current in the coil?

Question 19.
What experiment do you suggest to understand Faraday’s law? What items are required? What suggestions do you give to get good results of the experiment? Give precautions also. (AS3)
Answer:
Aim :
To understand Faraday’s law of induction.

Materials required :
A coil of copper wire, a bar magnet, Galvanometer, etc.

Procedure :

1. Connect the terminal of a coil to a sensitive galvanometer as shown in the figure.
2. Normally, we would not expect any deflections of needle in the galvanometer because there is to be no electromotive force in this circuit.
3. Now if we push a bar magnet towards the coil, with its north pole facing the coil, we observe the needle in the galvanometer deflects, showing that a current is set up in the coil.
4. The galvanometer does not deflect if the magnet is at rest.
5. If the magnet is moved away from the coil, the needle in the galvanometer again deflects, but in the opposite direction, which means that a current is set up in the coil in the opposite direction.
6. If we use the end of south pole of a magnet instead of north pole in the above activity, the deflections are exactly reversed.
7. This experiment proves “whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil”.

Precautions :

1. The coil should be kept on an insulating surface.
2. Bar magnet should be of good magnetic moment.
3. The centre of the galvanometer scale must be zero.
4. The deflections in the galvanometer must be observed while introducing the bar magnet into the coil and also while withdrawing it.

Question 20.
How can you verify that a current carrying wire produces a magnetic field with the help of an experiment? (AS3)
Answer:
Experiment:

• Take a thermocole sheet and fix two thin wooden sticks of height 1cm.
• These sticks are joined with the help of a copper wire.
• Connect battery and tap key to this copper wire.
• Place a magnetic compass beneath the wire.
• Now press the tap key and allow the current through the wire. It is observed that magnetic needle deflects.
• If you change the direction of the current, the direction of deflection of needle also changes.
• So we can say current carrying conductor produces magnetic field.

Question 21.
Collect information about generation of current by using Faraday’s law. (AS4)
Answer:
Faraday’s law is useful in generation of current.

1. According to this law, the change in magnetic flux induces EMF in the coil.
2. Fie also proposed electromagnetic induction.
3. Electromagnetic induction is a base for generator, which produces electric current.
4. Transformer also works on the principle of electromagnetic induction, which is helpful in transmission of electricity.
5. Hence Faraday’s law is used in the generation and transmission of current.

Question 22.
Collect information about material required and procedure of making a simple electric motor from internet and make a simple ntotor on your own. (AS4)
Answer:
Aim :
Preparation of a simple electric motor.

Material requried :
A wire of nearly 15 cm, 1.5v Battery, Iron nail, strong magnet, paper clip.

Procedure:

1. Attach the magnet to the head of the iron nail.
2. Attach a paper clip to the open end of the magnet.
3. Now attach the other end of the nail (Free end) to the cap (positive terminal) of the battery.
4. Now connect the negative terminal of the battery and the head of the iron nail through a wire.
5. We observe that the paper clip rotates.

Another model:
Materials required :
1.5 m enamelled copper wire (about 25 gauge), 2 safety pins,
1.5 v battery, magnets, rubber bands or bands cut from cycle tube.

Procedure :

1. Wind copper wire on the battery nearly 10 – 15 turns to make a coil.
2. Remove the coil and fix the ends as shown in the figure.
3. Scrape the insulation com¬pletely on one end of the coil.
4. Scrape the insulation on top, left and right of the other end. The bottom should be insulated.
5. Now complete the electric mo¬tor as shown in the figure. “5

Question 23.
Collect information of experiments done by Faraday. (AS4)
Answer:
Experiment – 1

1. Connect the terminals of a coil to a sensitive galvanometer as shown in the figure.
2. Normally, we would not expect any deflection of needle in the galvanometer because there is no EMF in the circuit.
3. Now, if we push a bar magnet towards the coil, with its north pole facing the coil, the needle in the galvanometer deflects, showing that a current has been set up in the coil, the galvanometer does not deflect if the magnet is at rest.
4. If the magnet is moved away from the coil, the needle in the galvanometer again deflects, but in the opposite direction, which means that a current is set up in the coil in the opposite direction.
5. If we use the end of south pole of a magnet instead of north pole, the results i.e., the deflections in galvanometer are exactly opposite to the previous one.
6. This activity proves that the change in magnetic flux linked with a closed coil, produces current.
7. From this Faraday’s law of induction can be stated as “whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil”. This induced EMF is equal to the rate of change of magnetic flux passing through it.

Experiment – 2

1. Prepare a coil of copper wire C₁ and connect the two ends of the coil to a galvanometer.
2. Prepare another coil of copper wire similar to C₂ and connect the two ends of the coil to a battery via switch.
3. Now arrange the two coils C₁ and C₂ nearby as shown in the figure.
4. Now switch on the coil C₂. We observe a deflection in the galvanometer connected to the coil C₁.
5. The steady current in C₂ produces steady magnetic field. As coil C₂ is moved towards the coil C₁ the galvanometer shows a deflection.
6. This indicates that electric current is induced in coil C₁.
7. When C₂ is moved away, the galvanometer shows a deflection again, but this time in the opposite direction.
8. The deflection lasts as long as coil C₂ is in motion.
9. When C₂ is fixed and C₁ is moved, the same effects are observed.
10. This shows the induced EMF due to relative motion between two coils.

Question 24.
Draw a neat diagram of electric motor. Name the parts. (AS5)
Answer:

Question 25.
Draw a neat diagram of an AC generator. (AS5)
(OR)
Draw the diagram of electric generator and label its parts. A.
Answer:

Question 26.
How do you appreciate the Faraday’s law, which is the consequence of conservation of energy? (AS6)
Answer:

• Law of conservation of energy says energy neither be created nor be destroyed, but can be converted from one form to another.
• Faraday’s law says whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil. This induced EMF is equal to the rate of change of magnetic flux passing through it.
• We have to do some work to move the magnet through a coil. This work produces energy.
• This energy is converted into electrical energy in the coil.
• In this way conservation of energy takes place in electromagnetic induction.

Question 27.
How do you appreciate the relation between magnetic field and electricity that changed the lifestyle of mankind? (AS6)
Answer:

• Changing life style of mankind is a result of many inventions, utilising a lot of scientific principles.
• Scientists all ways going on searching for new principles and new applications to make our life more comfortable.
• If you consider electricity, right from amber stone to nuclear power, so many changes have been incorporated.
• The idea of Oersted and Faraday that current carrying wire produces electricity and electromagnetic induction, enable us to use electric motors, generators, fans, mixers, grinders, induction stoves, etc.
• All these appliances makes our life more comfortable. Hence Faraday and Oersted rendered a lot of servies in this field.
• Hence, I appreciate the relation between magnetic field and electricity that changed the life style of mankind.
  So if current is more, induction is also more.

Question 28.
Give a few applications of Faraday’s law of induction in daily life. (AS7)
Answer:
Applications:
The daily life applications of Faraday’s law of induction are

1. Generation of electricity
2. Transmission of electricity
3. Metal detectors in security checking
4. The tape recorder
5. Use of ATM cards
6. Induction stoves
7. Transformers
8. Induction coils (spark plugs in automobiles)
9. Break system in railway wheels
10. AC and DC generators
11. Windmills, etc.

Question 29.
Which of the various methods of current generation protects the nature well? Give examples to support your answer. (AS7)
Answer:
Windmill :

• Electricity is produced when an armature of a generator rotates between two poles of a strong magnet.
• Whereas when wind falls on the wheel of a windmill, it rotates. So the armature of the generator rotates between two poles of a magnet along with the rotation of the wheel of the windmill.
• Thus electric current is produced.
• This is how, KE of the wind is converted into electric energy.

Advantages :
Wind energy produces no smoke and no harmful gases. So this form of energy is pollution free or environment-friendly.

Fill in The Blanks

1. The SI unit of magnetic field induction is ………………….
2. Magnetic flux is the product of magnetic field induction and …………………
3. The charge is moving along the direction of magnetic field. Then force acting on it is ………………..
4. A current carrying wire of length L is placed perpendicular to a uniform magnetic field B. Then the force acting on the wire with current I is ……………..
5. Faraday’s law of induction is the consequence of …………………
Answer:

1. weber/m² (or) Tesla
2. area
3. zero
4. ILB
5. Law of conservation of energy

Multiple Choice Questions

1. Which converts electrical energy into mechanical energy?
A) motor
B) battery
C) generator
D) switch
Answer:
A) motor

2. Waich converts mechanical energy into electrical energy?
(OR)
The device used to convert mechanical energy into electrical energy among the following is
A) motor
B) battery
C) generator
D) switch
Answer:
C) generator

3. The magnetic force on a current carrying wire placed in uniform magnetic field if the wire is oriented perpendicular to magnetic field, is
A) 0
B) ILB
C) 2ILB
D) ILB/2
Answer:
B) ILB

10th Class Physics 12th Lesson Electromagnetism InText Questions and Answers

10th Class Physics Textbook Page No. 221

Question 1.
Why does the needle get deflected by the magnet?
Answer:
Because of strength of the magnetic field of the magnet, the needle gets deflected since it is in the field.

10th Class Physics Textbook Page No. 213

Question 2.
How can we find the strength of the field and direction of the field?
Answer:
We can find the strength of the field with magnetic flux and the direction of the field from the tangent drawn to the line of force.

10th Class Physics Textbook Page No. 214

Question 3.
Can we give certain values to magnitude of the field at every point in the magnetic field?
Answer:
In uniform magnetic field it is same whereas in non-uniform magnetic field it is different.

10th Class Physics Textbook Page No. 215

Question 4.
What is the flux through unit area perpendicular to the field?
Answer:
Flux density or magnetic induction.

Question 5.
Can we generalize the formula of flux for any orientation of the plane taken in the field?
Answer:
Yes, Φ = BA cos θ

Question 6.
What is the flux through the plane taken parallel to the field?
Answer:
Magnetic flux (or) Magnetic field.

Question 7.
What is the use of introducing the ideas of magnetic flux and magnetic flux density?
Answer:
Magnetic flux and flux density help in understanding the concept of electromagnetic induction and relation between electricity and magnetism.

Question 8.
Are there any sources of magnetic field other than magnets?
Answer:
Current carrying straight wires and loops act as sources of magnetic filed.

Question 9.
Do you know how old electric calling bells work?
Answer:
Yes. They work on the principle of magnetic effect of electric currents.

10th Class Physics Textbook Page No. 218

Question 10.
What happens when a current carrying wire is kept in a magnetic field?
Answer:

• Magnetic field applies force on current carrying wire.
• So it gets deflected and the direction of deflection is given by right hand rule.
• Or there will be no force acting on the wire when wire is in the direction of the field.

10th Class Physics Textbook Page No. 219

Question 11.
Do you feel any sensation on your skin?
Answer:
Yes. The hair on my skin rises up when I stand near TV screen.

Question 12.
What could be the reason for that?
Answer:
It is due to the magnetic field produced by electric charges in motion.

Question 13.
Why does the picture get distorted?
Answer:
Due to motion of electrons that form the picture is affected by the magnetic field of bar magnet.

Question 14.
Is the motion of electrons reaching the screen affected by the magnetic field of the bar magnet?
Answer:
Yes. The motion of electrons reaching the screen is affected by the magnetic field of the bar magnet.

Question 15.
Can we calculate the force experienced by a charge moving in a magnetic field?
Answer:
Yes, If the force is F, it is given by the expression F = qvB.

Question 16.
Can we generalize the equation for magnetic force on charge when there is an angle ‘0’ between the directions of field “B” and velocity “v”?
Answer:
No, Then force F is given by the formula F = qvB sin θ.

Question 17.
What is the magnetic force on the charge moving parallel to a magnetic field?
Answer:
When the charge moves parallel to the magnetic field the value of “θ” becomes zero. In the equation F = qvB sin θ, since θ = θ, the value of force F also becomes zero.

Question 18.
What is the direction of magnetic force acting on a moving charge?
Answer:
By applying right hand rule we can guess the direction of magnetic force acting on a moving charge is the “thumb” direction.

10th Class Physics Textbook Page No. 221

Question 19.
Can you determine the magnetic force on a current carrying wire which is placed along a magnetic field?
Answer:
F = BIl sin θ. If the current carrying wire is placed along direction field θ = 0.
∴ F = 0

Question 20.
What is the force on the wire if its length makes an angle ‘θ’ with the magnetic field?
Answer:
F = Bqv sin θ or F = Bil sin θ, where ‘i’ is current. WorhA
Here B = magnetic induction, q = charge, v = velocity of the charge and ‘θ’ is the angle between direction of field and velocity.

10th Class Physics Textbook Page No. 222

Question 21.
How could you find its (current carrying wire) direction?
Answer:
We can find by using right hand rule.

Question 22.
Is the direction of deflection observed experimentally same as that of the theoretically expected one?
Answer:
Yes. But it depends on polarities of the horse shoe magnet.

Question 23.
Does the right hand rule give the explanation for the direction of magnetic force exerted by magnetic field on the wire?
Answer:
The right hand rule does not help us to explain the reason for deflection of wire.

Question 24.
Can you give a reason for it (deflection of wire)?
Answer:
There exists only magnetic field due to external source. When there is a current in the wire, it also produces a magnetic field. These fields overlap and give non-uniform field. This is the reason for it.

10th Class Physics Textbook Page No. 223

Question 25.
Does this deflection fit with the direction of magnetic force found by right hand rule?
Answer:
Yes. This deflection fits with the direction of magnetic force found by right hand rule.

Question 26.
What happens when a current carrying coil is placed in a uniform magnetic field?
Answer:
It gets deflected since magnetic lines of force are perpendicular to the length of the coil.

Question 27.
Can we use this knowledge to construct an electric motor?
Answer:
Yes. This is the principle of electric motor.

Question 28.
What is the angle made by AB and CD with magnetic field?
Answer:

AB and CD are at right angles to the magnetic field.

10th Class Physics Textbook Page No. 224

Question 29.
Can you draw the direction of magnetic force on sides AB and CD?
Answer:
Yes. The direction of magnetic force on sides AB and CD can be determined by applying right hand rule.

Question 30.
What are the directions of forces on BC and DA?
Answer:

ADBC, magnetic force pulls the coil up and at DA magnetic force pulls it down.

Question 31.
What is the net force on the rectangular coil?
Answer:
Net force on the rectangular coil is zero.

Question 32.
Why does the coil rotate?
Answer:

The rectangular coil rotates in clockwise direction because of equal and opposite pan’ of forces acting on the two sides of the coil.

Question 33.
What happens to the rotation of the coil if the direction of current in the coil remains unchanged?
Answer:
The coil comes to halt and rotates in anti-clockwise direction.

Question 34.
How could you make the coil rotate continuously?
Answer:
If the direction of current in coil, after the first half rotation, is reversed, the coil will continue to rotate in the same direction.

10th Class Physics Textbook Page No. 225

Question 35.
How can we achieve this (convertion of electrical energy to mechanical energy)?
Answer:
Brushes B₁ and B₂ are used to achieve this.

Question 36.
What happens when a coil without current is made to rotate in magnetic field?
Answer:
When the coil rotated due to the change in magnetic flux electricity is generated.

Question 37.
How is current produced?
Answer:
The current is produced from the battery to the coil.

Question 38.
Why is there a difference in behaviour in these two cases?
Answer:
The A.C. supply changes its direction a number of times in a second. But D.C. is unidirectional current. So there is a difference in the behaviour of the metal ring in these two cases.

Question 39.
What force supports the ring against gravity when it is being levitated?
Answer:
The magnetic force developed in the coil of copper wire supports the ring against gravity when it is being levitated.

10th Class Physics Textbook Page No. 226

Question 40.
Could the ring be levitated if DC is used?
Answer:
The metal ring is levitated because the net force on it should be zero according to Newton’s second law.

Question 41.
What is this unknown force acting on the metal ring?
Answer:
The change in polarities at certain intervals at the ends of the solenoid causes the unknown force acting on the metal ring.

Question 42.
What is responsible for the current in the metal ring?
Answer:
The field through the metal ring changes so that flux linked with the metal ring changes and this is responsible for the current in metal ring.

Question 43.
If DC is used, the metal ring lifts up and falls down immediately. Why?
Answer:
The flux linked with metal ring is zero. When the switch is on, at that instant there should be a change in the flux linked with ring. So the ring rises up and falls down. If the switch is off, the metal ring again raises up and falls down. There is no change in flux linked with ring when the switch is off.

10th Class Physics Textbook Page No. 227

Question 44.
What could you conclude from the above analysis (metal ring lifts up and falls down)?
Answer:
The relative motion of the magnet and coil sets up a current in the coil.

10th Class Physics Textbook Page No. 228

Question 45.
What is the direction of induced current?
Answer:
The direction of the induced current is such that it opposes the charge that produced it.

Question 46.
Can you apply conservation of energy for electromagnetic induction?
Answer:
Yes, we can apply. The mechanical energy is converted into electrical energy.

10th Class Physics Textbook Page No. 229

Question 47.
Can you guess what could be the direction of induced current in the coil in such case?
Answer:
The direction of the induced current in the coil must be in anti-clockwise direction.

Question 48.
Could we get Faraday’s law of induction from conservation of energy?
Answer:
Yes, we can get. Here we have to ignore the friction everywhere.

10th Class Physics Textbook Page No. 230

Question 49.
Can you derive an expression for the force applied on crosswire by the field “B”?
Answer:
Yes. The force applied F = BIl.

10th Class Physics Textbook Page No. 232

Question 50.
How could we use the principle of electromagnetic induction in the case of using ATM card when its magnetic strip is swiped through a scanner? Discuss with your friend or teacher.
Answer:
If the card is moved through a card reader, then a change in magnetic flux is produced in one direction, which induced potential or EMF. The current received by the pickup coil goes through signal amplification and translated into binary code, so that it can be read by computer.

Question 51.
What happens when a coil is continuously rotated in a uniform magnetic field?
Answer:
An induced current is generated in the coil.

Question 52.
Does it (continuous rotation of coil) help us to generate electric current?
Answer:
Yes. Continuous rotation of coil helps us to generate electric current.

10th Class Physics Textbook Page No. 233

Question 53.
Is the direction of current induced in the coil constant? Does it change?
Answer:
Yes, it changes. When the coil is at rest in vertical position, with side (A) of coil at top position side (B) at bottom position, no current will be induced in it.

Question 54.
Can you guess the reason for variation of current from zero to maximum and vice-versa during the rotation of coil?
Answer:
The reason for variation of current from zero to maximum and vice-versa during the rotation of coil current generated follows the same pattern so that in first half except that the direction of current is reversed.

Question 55.
Can we make use of this current? If so, how?
Answer:
Two carbon brushes are arranged in such a way that they press the slip rings to obtain current from the coil. When these brushes are connected to external devices like TV, Radio we can make them work with current supplied from ends of carbon brushes.

10th Class Physics Textbook Page No. 234

Question 56.
How can we get DC current using a generator?
Answer:
By connecting two half-slip rings instead of a slip ring commutator on either side to the ends of the coil we can get D.C. current.

Question 57.
What changes do we need to make in an AC generator to be converted into a DC generator?
Answer:
Instead of two slip rings, we have to use a slip ring commutator to change A.C. generator into a D.C. generator.

10th Class Physics 12th Lesson Electromagnetism Activities

Activity – 2

Question 1.
Show that the magnetic field around a bar magnet is three dimensional and its strength and direction varies from place to place.
Answer:

• Take a sheet of white paper and place it on the horizontal table.
• Place a bar magnet in the middle of the sheet.
• Place a magnetic compass near the magnet it settles to a certain direction.
• Use a pencil and put dots on the sheet on either side of the needle. Remove the compass. Draw a small line segment connecting the two dots. Draw an arrow on it from south pole of the needle to north pole of the needle.
• Repeat the same by placing the compass needle at various positions on the paper. The compass needle settles in different directions at different positions.
• This shows that the direction of magnetic field due to a bar magnet varies from place to place.
• Now take the compass needle to places far away from magnet, on the sheet and observe the orientation of the compass needle in each case.
• The compass needle shows almost the same direction along north and soiath at places far from the magnet.
• This shows that the strength of the field varies with distance from the bar magnet.
• Now hold the compass a little above the table and at the top of the bar magnet.
• We observe the deflection in compass needle. Hence we can say that the mag¬netic field is three dimensional i.e., magnetic field surrounds its source.
• From the above activities we can generalize that a magnetic field exists in the region surrounding a bar magnet and is characterized by strength and direction.

Activity – 3

Question 2.
Explain how you draw magnetic lines of force in the magnetic field.
(OR)
What is the name given to the imaginary lines joining from north pole to south pole of a bar magnet called? Explain how you can draw those lines around a bar magnet.
Answer;
These lines are called magnetic lines of forces.

Procedure:

1. Take a white drawing sheet.
2. Place a marine compass at the centre of the sheet.
3. Draw a line which shows north and south of the earth on the drawing sheet.
4. Now remove the compass needle and place a bar magnet at the centre of the sheet showing north of the bar magnet pointing north of the earth.
5. Place the magnetic compass near the bar magnet without contact. The needle comes to rest after oscillations.
6. Locate the end of the pointer with pencil. Now place the compass needle at this point and once again notice the end of the pointer.
7. We can repeat the same around the magnet, and draw all the points with the help of the pencil.
8. We can draw the lines taking the needle too far to the magnet and we can observe the orientation of needle of compass.
9. So we can conclude that the strength of field varies with distance from the bar magnet.
10. These lines of force are from north of the bar magnet to south of the bar magnet.

Activity – 4

Question 3.
Explain the direction of magnetic field around the straight conductor carrying current.
(OR)
What field would be formed around straight conductor carrying current? How do you find the direction of that field experimentally?
Answer:
Magnetic field would be formed around current carrying conductor.

Procedure:

1. Take a wooden plank and make a hole and place il on the table.
2. Place a stand on the plants, and suspend a c opper wire from the stand and see that it passes through the hole made to the plank.
3. Connect a battery and switch to this wire in series. Place some magnetic needle at the hole.
4. If the current is passed through the wire, the magnetic needle deflects and it is directed as the tangent to the circle
5. If the current flows in downward direction, the field lines are in anti-clockwise direction and if the current flows in upward direction, the field lines are in clockwise direction.
6. The direction ol the current and magnetic fines of force can be easily explained with the help of right hand thu mb rule. If you hold the current carrying conductor with your right hand grip stretching the thumb, the direction of the I humb shows the direction of the current, the direction of the other four fingers shows direction of magnetic lines of force.

Activity 5

Question 4.
Explain the direction of magnetic field due to circular coil.
Answer:

Procedure :

1. Take a thin wooden plank and cover it with whitepaper.
2. Make two holes to the plank and pass insulated copper wire through the holes and wind the wire 4 to 5 times through the holes such that it looks like a coil.
3. The ends of the wire are connected to the battery terminals.
4. Now place a compass needle at the centre of the coil.
5. Put dots on either side of the compass. Repeat this by keeping at the dots. We can observe that field lines are circular.
6. Here the direction of the field is perpendicular to the plane of the coil.
7. The direction of the magnetic field due to coil points towards you when the current in the coil is in anti-clockwise direction.
8. When you curl your right hand fingers in the direction of current, thumb gives the direction of magnetic field.

Activity – 6

Question 5.
Explain the magnetic field due to solenoid.
(OR)
What is the name given to the device which is a long wire wound in a close pack helix? Find the direction of magnetic field around that device.
Answer:
It is called solenoid.

Procedure :

1. Take a wooden plank covered with white paper.
2. Make holes on its surface.
3. Pass copper wire through the holes.
4. Join the ends of the coil to a battery through a switch.
5. Current passes through the coil, when we switch on the circuit.
6. Now sprinkle iron filings on the surface of the plank, around the coil. Then orderly pattern of iron filings is seen on the paper.
7. The iron filings arrange themselves in orderly way and look like lines of force.
8. The long coil is known as solenoid. The direction of the field due to solenoid is determined by using right hand rule.
9. One end of the solenoid behaves like a north pole and the other behaves like south pole.
10. Outside the solenoid the direction of the field lines of force is from north to south while inside the direction is from south to north. Thus the magnetic field lines are closed loops.
11. Hence electric charges in motion produce magnetic field.

Activity – 8

Question 6.
Explain the field lines due to horse-shoe magnet between its poles.
(OR)
Which field is set up between poles of a horse-shoe magnet? Explain the field lines due to horse magnet between its poles.
Answer:
Non-uniform magnet is set up between poles of a horse shoe magnet.

Procedure :

1. The field in between north and south pole of horse-shoe magnet are straight and parallel.
2. If the wire is passing perpendicular to the paper, the magnetic lines of force are concentric circles, when the current is passed.
3. The direction of field lines due to the wire in upper part coincides with the direction of field lines of horse-shoe magnet.
4. The direction of field lines by the wire in lower part is opposite to the direction of field lines of horse-shoe magnet.
5. Hence the net field in upper part is strong and in the lower part is weak.
6. Hence a non-uniform field is created around the wire.

Activity – 9

Question 7.
Explain electromagnetic induction.
(OR)
Which current will levitate the ring in the following figure? Explain the experimental activity.
Answer:
AC will levitate the ring.

Procedure :

1. Fix a soft iron cylinder on the wooden base vertically.
2. Wind copper wme around the soft iron.
3. Take a metal ring which is slightly greater in radius than the radius of soft iron cylinder and insert it through the soft iron cylinder.
4. Connect the ends of the coil to an AC source and switch on the current.
5. Here metal ring levitates on the coil (appears to rise and floats in the air).
6. In this experiment we can conclude that if AC current is used, the magnetic induction changes in both magnitude and direction in the solenoid and in the ring. The field through the metal ring changes, so that flux linked with the metal ring changes.
7. If DC current is used, the metal ring lifts up and falls down immediately.

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 8 Heredity.

AP State Syllabus SSC 10th Class Biology Important Questions 8th Lesson Heredity

10th Class Biology 8th Lesson Heredity 1 Mark Important Questions and Answers

Question 1.
How did you get the characters from your parents and grandparents?
Answer:
By Genes

Question 2.
Why man is called a moving museum of vestigial organs?
Answer:

1. The organs which are not useful in animals are called ‘vestigial organs’. There are nearly 180 vestigial organs in human beings,
2. Hence, human being is said to be a moving museum of vestigial organs.

Question 3.
Observe the given Flow-chart and answer the question.
Who decides the sex of the baby – mother or father? How?
Answer:
Father decides the sex of the baby.
Mother has XX chromosomes. Father has ‘XY’ chromosomes. Y chromosome is determining factor. So father is responsible.

Question 4.
What examples you will give to prove that Lamarckism is not correct?
Answer:
August Weismann, tested the theory proposed by Lamarck by experiments on rats. He removed tails of parental rats. He observed for twenty two generations but still off springs are normal with tails.

Question 5.
Which chromosomes determine the sex in human beings?
Answer:
Allosomes or Sex chromosomes. They are xx (girls) and xy (boys).

Question 6.
Why do we call appendix as a vestigial organ?
Answer:

1. Vestigial organ is the organ of our body which is smaller and simpler than those in related species they have lost their original function.
2. Appendix is highly developed in ruminants which helps in the digestion of cellulose. But, in human beings the cellulose is eliminated as undigested food. Hence in human beings appendix has no role in cellulose digestion. So, we call it as a vestigial organ.

Question 7.
What are variations?
Answer:
Differences in characters within very closely related groups of organisms are referred to as variations.

Question 8.
Who is known as father of genetics?
Answer:
Gregor Johann Mendel is known as father of genetics.

Question 9.
Why Mendel has chosen garden pea plant as material for his experiments?
Answer:
Pea plant has following advantages.

1. Well defined characters
2. Bisexual flowers
3. Predominently self fertilization
4. Early hybridization
5. Annual plant.

Question 10.
What are the vitamins present in pea plant?
Answer:
The vitamins present in pea plant are ‘A, C, E, K and B’.

Question 11.
What is F₃ generation?
Answer:
F₃ generation represents the offsprings produced from the individuals of F₂ generation.

Question 12.
What is a factor?
Answer:
The determining agent responsible for each trait is called a factor.

Question 13.
WTiat is law of dominance?
Answer:
According to Mendel, among a pair of alleles for a character, only one expresses itself in the first generation as one of the allele is dominant over the other. This is known as law of dominance.

Question 14.
What is phenotype ratio?
Answer:
The characters which can be seen is known as phenotype and their ratio is called phenotype ratio.

Question 15.
What is the phenotype ratio in F₁ generation of monohybrid cross?
Answer:
The phenotype ratio in F₁ generation of monohybrid cross is 3 : 1.

Question 16.
What is genotype ratio?
Answer:
The genetic makeup of an individual with reference to a specific character under consideration is called genotype and their ratio is called genotype ratio.

Question 17.
What is the genotype ratio in F₂ generation of monohybrid cross?
Answer:
The genotype ratio in F₂ generation of monohybrid cross is 1 : 2 : 1.

Question 18.
What are genes?
Answer:
Genes are the factors which are responsible for characters or traits of an organism. These are the units of heredity that are transferred from a parent to offspring. These are small segments of DNA on a chromosome.

Question 19.
What is an allele?
Answer:
The pair of genes which are responsible for character is called allele.

Question 20.
What are homozygous alleles?
Answer:
If an organism has two copies of the same allele for example TT or tt it is homozygous for that trait.

Question 21.
What are heterozygous allele?
Answer:
If an organism has one copy of two different alleles for example Tt, then it is heterozygous.

Question 22.
What is law of independent assortment?
Answer:
In the inheritance of more than one pair of characters (traits), the factors for each pair of characters assorted independently of the other pair. This is known as “Law of independent assortment”.

Question 23.
Wliat is the law of segregation?
Answer:
The law of segregation states that every individual possesses a pair of alleles for any particular trait and that each parent passes a randomly selected copy of only one of these to its offspring.

Question 24.
What are heritable traits?
Answer:
Traits that may be passed on from one generation to the next are called as heritable traits.

Question 25.
What is heredity?
Answer:
The process of acquiring characters or traits from parents is called heredity.

Question 26.
What is inheritance?
Answer:
The process in which traits are passed from one generation to another generation is called inheritance.

Question 27.
What is genetic drift?
Answer:
Change in the frequency of genes in small populations is called genetic drift.

Question 28.
Who was the first person to propose the theory of evolution?
Answer:
Jean Baptist Lamarck was the first person to propose the theory of evolution.

Question 29.
What are acquired characters?
Answer:
The characters developed during the lifetime of an organism are called acquired characters.

Question 30.
What is inheritance of acquired characters?
Answer:
Lamarck proposed that the acquired characters are passed to its offsprings i.e., to next generation. This is known as inheritance of acquired characters.

Question 31.
Who wrote the book “principles of geology” of evolution?
Answer:
The book “principles of geology” of evolution was written by Charles Lyell.

Question 32.
What is micro evolution?
Answer:
Small changes within the species is known as micro evolution.

Question 33.
What is macro evolution or speciation?
Answer:
The process of evolution through which new species are going to be formed is known as speciation or macro evolution.

Question 34.
The double helix structure of DNA was discovered by?
Answer:
James Watson and Francis Crick discovered the double helix structure of DNA.

Question 35.
What are autosomes?
Answer:
Chromosomes whose number and morophology do not differ between males and females of a species are called autosomes.

Question 36.
What are allosomes?
Answer:
The chromosomes that determine sex of the organism are called allosomes.

Question 37.
What is meant by survival of the fittest?
Answer:
According to Darwin’s theory of natural selection, nature only selects or decides which organism should survive or perish in nature. This is the meaning of survival of the fittest.

Question 38.
Write the expanded form of DNA.
Answer:
The expanded form of DNA is Deoxyribo Nucleic Acid.

Question 39.
What is the basis of evolution?
Answer:
Selection of variants by environmental factors forms the basis of evolution.

Question 40.
What are analogous organs?
Answer:
The organs which have similar appearance and functions but have different structure and origin. Ex : Wings of a butterfly, bat and a bird.

Question 41.
What is embryology?
Answer:
Embryology is the study of the development of an organism from egg to adult stage.

Question 42.
What is Palaeontology?
Answer:
The study of fossil is called Palaeontology.

Question 43.
How palaeontologists determine the age of fossil?
Answer:
Palaeontologists determine the age of fossil by using carbon dating method.

Question 44.
Where do scientists collected fossil of dinosaurs?
Answer:
Scientists collected fossils of dinosaurs from Yamanapalli in Adilabad district in Telangana State.

Question 45.
What are connecting links?
A. The organisms which bear the characters of two different groups are called connecting links.

Question 46.
Which organism is recognised as the connecting link between aves and reptiles?
Answer:
Archeopteryx is recognised as the connecting link between aves and reptiles.

Question 47.
What is human evolution?
Answer:
Human evolution is the evolutionary process leading upto the appearance of modern human beings.

Question 48.
How do variations occur?
Answer:
Variations develop during reproduction in organisms. Sexual reproduction and errors in DNA copying lead to variations in offspring in a population.

Question 49.
What is divergent evolution?
Answer:
The evolutionary process through which homologous organs develop is called divergent evolution.

Question 50.
What is convergent evolution?
Answer:
The evolutionary process through which analogous organs develop is called convergent evolution.

Question 51.
What are vestigial organs?
Answer:
Organs which are not useful in animal are called vestigial organs.

Question 52.
Who proposed the theory of inheritance of acquired characters?
Answer:
Jean Baptist Lamarck proposed the theory of inheritance of acquired characters.

Question 53.
Who proved that the bodily changes which may occur due to environment won’t be passed to its offsprings?
Answer:
Augustus Weisemann proved that the bodily changes which may occur due to environment won’t be passed to its offsprings.

Question 54.
How the study of fossil is considered significant?
Answer:
The study of fossil is considered significant because

1. Fossils provide direct evidence of past life and
2. Fossils provide convincing proof of organic evolution.

Question 55.
What are the nitrogen bases present in DNA?
Answer:
The nitrogen bases present in DNA are adenine, guanine, thymine and cytosine.

Question 56.
How do embryological studies provide evidence for evolution?
Answer:
The similarities in embryonic development reinforce the idea of evolution from common ancestors. The sequence of embryonic development in different vertebrates shows similarities.

Question 57.
Why Mendel selected garden pea plant for his experiments? Give a reason.
Answer:
Mendel selected garden pea plant for his experiments as these can be self pollinated.

Question 58.
What is the evolutionary significance of the fossil archaeopteryx?
Answer:
Archaeopteryx serves as a connecting link between birds and reptiles. It is the fossil evidence to show that birds have evolved from reptiles.

Question 59.
How does the creation of variation in a species ensure survival?
Answer:
The creation of variations in a species enable them to adapt according to the changes and the new needs thus they will enable the survival of the species.

Question 60.
Define evolution.
Answer:
Evolution is the sequence of gradual changes which takes place in the primitive organisms over millions of years in which new species are produced.

10th Class Biology 8th Lesson Heredity 2 Marks Important Questions and Answers

Question 1.
Write Phenotypic and Genotypic ratio of table given at side.
(OR)
Write the Phenotypic and Genotypic ratio when heterogygous (Yy) pea plant is hybridised with the same kind of plant.
(OR)
Write the phenotypic and genotypic ratios of Mendel’s cross-pollination experiments in pea plants with heterozygous yellow seeds (Yy) with that of the same type, i.e., Yy.
Answer:
Phenotypic ratio -3:1
Genotypic ratio -1:2:1

Question 2.
Define and explain Variations with examples.
Answer:
Variations: Differences in characters within very closely related groups of organisms are referred to as variations.
(OR)
Differences among living beings are called variations.
Ex:

1. Earlobes in some humans are free and in others attached.
2. Colour of eyes (cornea) in some people are blue and in others black.
3. Colour of skin is black or white.

Question 3.
“Human being is said to be a moving museum of vestigial organs”. How can you support this statement?
Answer:

1. During the course of evolution, some organs remain in organisms. For example, appendix in the digestive system.
2. In human beings it has no role to play in the process of digestion.
3. But in herbivores like rabbit appendix play important role.
4. Such type of organs which are not useful in animal are called vestigial organs.
5. There are nearly 18 vestigial organs in human beings.
6. For example pinna, hair on skin, mammary glands in man, etc.
7. That’s why human being is said to be a moving museum of vestigial organs.

Question 4.
Fill the given table and write the genotypic ratio basing on table.

Answer:

The genotypic ratio is 1: 2: 1

Question 5.
Who decides the sex of the baby, mother or father? Explain with a flow chart.
(OR)
Draw a flow-chart showing the sex determination in human beings.
Answer:
Father decides the sex of the baby.

Question 6.
Define the terms phenotype and genotype.
Answer:
Phenotype: The observable properties of an organism that are produced by the interaction of the genotype and the environment. These characters can be seen.
Genotype: The genotype is the genetic make-up an individual usually with reference to a specific characteristic consideration.

Question 7.
What questions you will ask a palaeontologist about fossils?
Answer:

1. What are fossils?
2. How do they preserve?
3. What can be the actual remains?
4. How do they form?
5. What do we call the study of fossil?
6. Can you tell some examples of fossils?
7. How do the palaeontologists determine the age of fossils?
8. What are dinosaurs and ketosaurs? In which years they belong?
9. Where did they collect the fossil of dinosaurs? What is the length of this fossil?
10. Where did they preserve the fossil of dinosaurs?

Question 8.
How does the embryological evidences support that Evolution has taken place?
Answer:
Evidences:

1. Remarkable similarities in the Embryos of different animals from fish to man.
2. Tadpole of frog resembles the fish more than the frog.
3. Life history of every individual exhibits the structural features of their ancestors.
4. The resemblance is so close at an early stage, it is difficulty to distinguish one embryo from other.

Question 9.
Observe the checker board and answer the following questions.
i) Write phenotypic ratio of monohybrid cross.
ii) How many heterozygous plants are present in the checker board?
Answer:
i) Phenotypic ratio 3 : 1
ii) Two heterozygous plants – (Yy, yY)

Question 10.
What happens if there is no evolution?
Answer:

1. Evolution is a continuous and comprehensive process.
2. If it does not take place, there is no formation of new species.
3. Variations do not take place and hence desirable traits are not developed.
4. All the Earth would be with the primitive species without any changes.

Question 11.
If you cross a plant with pure yellow seeds (YY) with a plant with pure green seeds (yy), what would be the colour of the seeds in F₂ generation? Show in a checker board.
Answer:
All pea plants are yellow in F₁ generation on self pollination in F₁ generation.
In F₂ generation, we can observe that 75% are yellow seed producing pea plants and 25% are green ones.
Hence the phenotypic ratio is 3 : 1 the genotypic ratio is 1 : 2 : 1

Question 12.
If you meet a historian to clarify your doubt on ‘Man has first born in African conti¬nent’, what type of questions will you ask him / her?
Answer:

1. What is human evolution?
2. When were early man like forms appeared on land?
3. When did the fossil of the homosapiens appeared on earth?
4. Where did the early man lived?
5. Where can we trace the earliest members of the human race homosapiens?
6. When do some of our ancestors left Africa?
7. How the residents of Africa migrated to other places?
8. Why did the residents of Africa migrated to other places?
9. Are all humans evolved from single ancestor?

Question 13.
What is the difference between Phenotype and Genotype?
Answer:

Question 14.
What are the differences between homozygous and heterozygous ?
Answer:

Question 15.
How do traits get expressed according to Mendel?
Answer:

1. Mendel hypothesised that each character or trait is expressed due to a pair of factors or alleles.
2. Now these alleles are known as genes.
3. Gene is made up of a segment of DNA which provide information of protein.
4. Protein is needed for biochemical process.
5. If the proteins work efficiently, the traits get expressed in better way.
6. Thus genes control the traits or characters.

Question 16.
What is speciation? How it occurs?
Answer:

1. Origin of new species from existing one is called speciation.
2. It may occur due to a) mutations and b) natural selection.

Question 17.
What is sex chromosome? Name the two types of sex chromosomes. Mention the chromosomes present in male and female?
Answer:

1. The chromosome which determine the sex of a person is called sex chromosome.
2. The two types of sex chromosomes are X chromosome and Y chromosome.
3. Sex chromosomes in male is XY.
4. Sex chromosomes in female is XX.

Question 18.
Write a short notes on the law of “inheritance of acquired characters”.
Answer:

1. Law of inheritance of acquired characters was proposed by Jean Baptist Lamarck.
2. He thought that the characters acquired by an organism in its life time are passed to its offsprings.
3. He thought that at same point of time in the history, the size of giraffe was equal to that of deer.
4. Due to shortage of food material on the ground and lower branches of trees giraffes started stretching their necks.
5. Because of continuous usage of neck, after several generations giraffes obtained longer necks.
6. Such characters that are developed during the lifetime of an organism are called acquired characters.

Question 19.
Write a short notes on the theory of “Natural selection”.
Answer:

1. The theory of natural selection was proposed by Charles Darwin.
2. This theory states that nature only selects or decides which organism should survive or perish in nature.
3. The organisms with useful traits will survive.
4. The organisms having harmful traits are going to be perished or eliminated from its environment.

Question 20.
Write briefly about “Survival of the fittest”.
Answer:

1. Variations which are useful are retained, while those which are harmful are removed.
2. In a population where there is a struggle, the “fittest” will be survived.
3. Nature favours only useful variations.
4. Each species tends to produce large number of offspring.
5. They compete with each other for food, space, mating and other species.
6. In this struggle for existence, only the fittest can survive. This is called survival of the fittest.
7. Over long period of time this leads to formation of new species.

Question 21.
How are new species evolved?
Answer:

1. Sexual reproduction and errors in DNA copying leads to variations in offspring in a population.
2. Organisms contain variations that help to adapt to its environment going to be survived more efficiently.
3. But in the same population, the organisms which contain the trait which may not help to adapt in its environment may be perished or eliminated slowly.
4. These small changes within the population due to variations is called micro evolution.
5. When organisms of the same species with variations are separated by some cause for long years, lot of variations may take place in these species.
6. These accumulated variations make them unable to mate and produce new offsprings.
7. Thus new species form and this is known as speciation or macro evolution.

Question 22.
Write a brief note on homologous organs.
Answer:

1. Homologous organs are the organs which perform different functions but have similar structure and origin.
2. For example forelimb of a whale (swimmer), bat (flyer), horse (runner), mole (digger) and man (grasping).
3. If we carefully observe the anatomy of all these animals, they have a common pattern in the arrangement of bones.
4. Even though their external form and functions are different, they are similar internally.
5. Thus it indicates that all the vertebrates have evolved from a common ancestor.

Question 23.
What are fossils? Write a short note on their formation.
Answer:

1. Fossils are evidence of ancient life forms or ancient habitats which have been preserved by natural processes.
2. Fossil evidence is typically preserved within sediments deposited beneath water and land.
3. They can be actual remains of once lived such as bones or seeds or even traces of past event such as dinosaurs foot print or ripple marks on a pre-historic shore.
4. Usually when organisms die, their bodies will be decomposed and lost.
5. But sometimes the body or at least some parts may be in an environment that does not let it decompose completely.
6. For example if a dead insect get caught in mud, it will not decompose quickly and the mud will eventually harden and retain the impression of the body parts of insect.
7. All such preserved traces of living organisms are called fossils.

Question 24.
How would you appreciate Jean Baptist Lamarck for his contribution to the biology?
Answer:

1. Jean Baptist Lamarck was the first person to propose the theory of evolution.
2. He proposed that the acquired characters are passed to its offspring i.e., to next generation.
3. This is known as inheritance of acquired characters.
4. For example elongation of neck and forelimbs in giraffe.
5. Even though this theory was disproved, his contribution to biology was appreciable because it changed the belief of the people of olden days that the organisms on the earth have not undergone any change.

Question 25.
How did August Weisemann disprove the theory of “Inheritance of acquired characters” proposed by Lamarck? (OR)
What example will you give to prove that Lamarckism is not correct?
Answer:

1. August Weisemann, tested the theory of “Inheritance of acquired characters” proposed by Lamarck by an experiment on rats.
2. He removed tails of parental rats.
3. He observed its offsprings which have normal tails.
4. He has done it again for twenty two generations but still offsprings are normal with tails.
5. He proved that the bodily changes are not inherited. So they won’t be passed to its offspring.
6. Thus he disproved the theory of “Inheritance of acquired characters”.

Question 26.
Some organisms or species adapt better and survive in a community of organisms. Why do you think this may happen?
Answer:

1. This may happen due to the variations posessed by the organisms which are suitable to that habitat.
2. The variations that help the organism to collect food to escape from their enemies, increase the chance of survival for the organism than the other organisms.
3. In general, variations come during sexual reproduction or mutation.
4. If the variations are useful, that organisms can adapt better and survive.
5. These organisms can be selected by the nature.

Question 27.
What do you understand about pure breeds?
Answer:

1. Pure breed is that expresses the selected character over several generations.
2. A pure breed will have both the factors of the same type.
3. It means all the pure breeds are homozygous.
4. All the gamates produced by them will have same type of factor.
5. Pure breed on self pollination will give pure breed again.

Question 28.
What do you understand about F₁ generation?
Answer:

1. F₁ generation or first filial is the offspring of first generation parents.
2. Cross pollination of pure breeds will give F₁ generation.
3. All the individuals produced in F₁ generation are heterozygous.
4. Only the dominant characters are expressed in this generation.

Question 29.
What are the differences between F₁ generation and F₂ generation of mono hybrid cross?
Answer:

Question 30.
What are the differences between monohybrid cross and dihybrid cross ?
Answer:

Question 31.
Write a short note on fossils of dinosaurs, ketosaurs collected in Telangana state.
Answer:

1. A rare and magnificient fossil of the dinosaurs, ketosaurs were collected in Yamanapalli in Adilabad district of Telangana district.
2. They belong to the lower jurassic age going back to about 160 million years.
3. This fossil has 14 meters length and 5 meters height.
4. This fossil is preserved in BM Birla Science Centre. Hyderabad.

Question 32.
Write a short notes on vestigial organs.
Answer:

1. During the course of evolution some organs remain in organisms, even though they don’t have any work to do.
2. For example appendix in the digestive system of human beings has no role to play in the process of digestion.
3. But in herbivores like rabbit appendix plays important role.
4. Such type of organs which are not useful in animal are called “vestigial organs”.
5. There are nearly 180 vestigial organs in human beings such as pinna, hair on skin, mammary glands in human, etc.
6. That’s why human being is said to be a moving museum of vestigial organs.

10th Class Biology 8th Lesson Heredity 4 Marks Important Questions and Answers

Question 1.
Fossils are the precious evidences preserved by the nature to help us knowing about ancient life forms. Write the information you collected about fossils.
Answer:

1. Fossils are the evidences of ancient life forms or ancient habitates which have been preserved by natural processes.
2. Fossils provide information about what lived in the past.
3. Palentologists determine the age of fossils by using carbon-dating method.
4. They convey us about genetic condition, heredity characters through inactive chromosomes which are present in them.
5. They give a detailed information about their diet, life styles, shape of body, etc.
6. Fossils provide the information about how species have changed across long periods of the earth history.

Question 2.
a) If a sperm with ‘X’ chromosome fertilizes with an ovum with ‘X’ chromosome, what will be the gender of the baby?
b) Who determines the sex/gender of the baby, mother or father?
c) Is it correct to blame the mother for giving birth to a baby girl?
d) Are all our characters resembles our parents?
Answer:
a) Female
b) Father
c) Not correct
d) No, some variations occur.

Question 3.
Observe the diagram and answer the following questions. Vamsi and Priya are newly married couple. They want to give birth to a male child.

a) Draw a probable diagram showing transfer of chromosomes from parents to give birth to male child.
Answer:
b) Who determines the sex of the baby? How can you say ?
Answer:
Father (Vamsi) determines the sex of the baby. Because the chromosome ‘Y’ that determines male sex is present in males.

Question 4.

i) What does the given flow chart indicate?
Answer:
Sex determination in human beings

ii) What will happen if the sperm containing ‘X’ chromosomes fertilises the ovum?
Answer:
Baby will be a girl

iii) Who decides the sex of the baby – Mother or Father?
Answer:
Father

iv) How many pairs of chromosomes are present in off-spring?
Answer:
23 pairs

Question 5.
Write a brief note on Homologous and Analogous organs.
Answer:
Homologous organs: Organs which are structurally similar but functionally different are known as “Homologous organs”.
Forelimbs of a whale – swimming
Wings of a bat – flying
Forelegs of cheetah – running

Analogous: Organs which are structurally different but functionally similar are known as “Analogous organs”.
Eg : Wings of a bird – flying
Wings of a bat – flying

Question 6.
Observe the flow – chart and answer the following.
i) What does the flow – chart represent?
Answer:
The flow – chart represents a monohybrid hybridisation between a pure breed Tall (T) and a pure breed dwarf (t) plants resulting first filial generation. On self pollinating with F₁ generation the new breed have any combinations of T, t came in F₂ generation.

ii) What is the phenotype characters in F₁ generation?
Answer:
In the phenotypic characters in F₁ generation all are dominant that is (T) Tall,

iii) What is the Genotype, Phenotype ratio of F₂ generation?
Answer:
Genotype ratio in F₂ generation is 1 : 2 : 1
Phenotype ratio in F₂ generation is 3 : 1

iv) What laws of inheritance did you understand by this flow – chart?
Answer:
understand that i) the law of dominance ii) law of segregation proposed by Mendel.

Question 7.
Write the Darwin’s theory of evolution in a nutshell.
Answer:

1. Any group of population of an organism has variations and all members of group are not identical.
2. Variations maybe passed from parent to offspring through heredity.
3. The natural selection over abundance of offspring leads to a constant struggle for their survival in any population.
4. Individuals with variations that help them to survive and reproduce tend to live longer and have more offsprings than organisms with less useful features.
5. The offsprings of survivors inherit the useful variations, and the same process happens with every new generation until the variation becomes a common feature.
6. As the environment changes, the organism within the environment adapt and changes to the new living conditions.
7. Over a long period of time, each species of organism can accumulate so many changes that it becomes a new species, similar to but distinctly different from the original species. All species on the earth arise in this way.
8. Evolution is a slow and continuous process that involves several thousands of generations.

Question 8.
What are Mendel’s laws of inheritance? What are the reasons to choose pea plant for his experiment?
Answer:
Mendel’s Laws of inheritance:

1. Law of Dominance : Among a pair of closely related ‘alleles’ or factors, only one expresses itself. In the first generation as one of the allele is dominant over the other. This is called as Mendel’s Law of dominance.
2. Law of Segregation : The law of segregation states that every individual possesses a pair of alleles for any particular trait that each parent posses a randomly selected copy only one of these to its off-spring.
3. Law of Independent assortment : In the inheritance of more than one pair of characters (traits), the factors for each pair of characters assort independently of the other pairs. This is known as ‘Law of Independent assortment’.

Mendel has chosen garden pea as material for his experiment because:

1. It has well developed characters.
2. It is a bisexual flower.
3. Predominently self pollinating.
4. Suitable for cross pollination.
5. It is an annual plant.

Question 9.
What is Phenotype and Genotype? Explain them with the help of Mendel’s Monohybrid cross.
Answer:
Phenotype: Expression of visible character of an individual is called phenotype.
Genotype: Genetic constitution of an individual for any character is called Genotype or Probable nature of factors is known as genotype.
Cross Pollinating a pure breed of yellow coloured pea seeds (YY) and green coloured pea seeds (yy) give F₁ generation. All pea seeds were yellow in F₁ generation. So, yellow colour is phenotype. ‘Yy’ is genotype of all pea seeds in F₁ generation.

F₂ Generation : Self pollination of F₁ pea plants (Yy)

Phenotype Ratio is 3:1 Genotype Ratio is 1:2:1.

Question 10.
Explain in brief any two evidences of Evolution.
Answer:
Some of the evidences of evolution are

1. Homologous and analogous organs
2. Evidences from embryology and
3. Evidences from fossils.

I. Homologous and analogous organs :

1. Organs which have common fundamental anatomical plan and similar embryonic origin, whatever varied functions they may perform are regarded as homologous organs.
2. For example forelimb of a whale, wing of bat, leg of leopord, claw of mole and hand of man.
3. They indicate that all the vertebrates are evolved from common ancestor.
4. Organs which are structurally different but functionally similar are known as “Analogous organs”.
5. For example wings of bats and wings of birds.
6. The designs of the two wings, their structure and components are different but they look similar because they have a common use for flying but their origin is not common.

II. Evidences from embryology:

1. There are remarkable similarities in the embryos of different animals from fish to man.
2. The resemblance is so close that at an early stage even an experienced embryologist would find difficulty to distinguish one embryo from the other.
3. This strengthens the view of the existence of a common ancestor from which all these have evolved.

III. Evidences from fossils:

1. Fossils are evidences of ancient life forms or ancient habitats which have been preserved by natural processes.
2. Palaeontologists determine the age of fossils by using carbon dating method.
3. These fossils provide evidences of presence of extinct animals like dinosaurs and how the evolution occurred on the earth, etc.

Question 11.
Observe the given flow-chart and answer the following questions:

i) Name the chromosomes that determine the sex of an individual.
Answer:
Y Chromosome of father.

ii) Show given information in the form of Checker Board.
Answer:

iii) In this situation, which principle of Mendel is applicable?
Answer:
Law of dominance

iv) “Mother determines the sex of the baby”. Is this statement correct or not? Why?
Answer:
The statement is not correct. Because the sperm of father that carries Y chromosome fertilize with ovum of mother that contains X chromosome the resultant will be XY – Male baby.

Question 12.
Keep in mind Mendel’s experiments and write what you know about the following concepts?
a) Pure breed b) Phenotype c) Genotype d) Alleles
Answer:
a) Pure breed: These are the plants that expresses a selected character over several generations. Such plants according to Mendel were pure breed for that character.

b) Phenotype: The characters which can be seen is known as phenotype. We cannot determine the internal factors by phenotype. It tells about only the dominating char-acters which express externally. The phenotype ratio in monohybrid cross is 3 : 1.

c) Genotype: The genetic make up of an individual is known as genotype. Genotype itself is the indication of internal factors. It tells about both dominant and recessive characters present within. The genotype ratio in monohybrid cross is 1 : 2: 1.

d) Alleles: Alleles are corresponding pairs of genes located at specific positions in chromosomes. Together they determine the genotype of their host organism. Every individual possesses a pair of alleles for any particular trait and that each parent passes a rondomly selected copy of only one of these to an offspring. The offspring then receives its own pair of alleles for that trait one each from both parents.

Question 13.
Competition among organisms, variations, natural selection survival of the fittest.
Which theory explains all these aspects? Describe them in a orderly manner.
Answer:

1. Competition: Every living organism in this world reproduces itself. The rate of multiplication and existence of organisms are more or less common for all living things. But the food supply and space remain unchanged, in other words they are limited. Under these conditions, there is a competition among the organisms to fulfil their needs of food and space. This is known as struggle for existence. It may be interspecific, intraspecific and the environment factors.
2. Variations: Every organism has its own specific characteristic and relationship with its environment. The variability caused by variations may be large or small. Large variations are known as macro variation and small variations called micro variations. Variations may be favourable or harmful. Those with useful variations survive while the others perish.
3. Natural selection: The organism with favourable variations are best adapted to the environment in which they live. They have a better chance of survival and perpetuation of race. This principle is called natural selections.
4. Survival of the fittest: Variations which are useful are retained, while those are harmful are lost. In a population where there is a struggle, the fittest will be survived and the less adjusted will be perished. This is known as the elimination of the unfit.

Question 14.
What is genetic drift? Explain how it provides diversity in the population.
Answer:

1. Changes in the frequency of genes in small populations, due to accidents is known as “Genetic drift”.’
2. Let us consider a colour variation occurred in red colour beetles which are living on green coloured leaves of bushes.
3. It results in “blue” colour beetles instead of ‘red’ colour beetles and passed its colour to its progeny.
4. Initially in the population, there are few blue beetles, but most are red.
5. Imagine at this point, an elephant comes by and stamps on the bushes where the beetles live.
6. This kills most of the beetles but by chance a few beetles survived are mostly blue.
7. Again the beetle population slowly increases, but in the beetle population most of them are in blue colour.
8. Thus genetic drift provides diversity in the population.

Question 15.
Have the apparent groups of human beings (races) evolved differently?
Answer:

1. No, there is no biological basis to the notion of human races. All humans are a single species.
2. Regardless of where we have lived for the past few thousand years, we all came from Africa.
3. The earliest members of human species, Homosapiens, can be traced there.
4. Our genetic footprints can be traced back to our African roots.
5. A couple of hundred thousand years ago, some of our ancestors left Africa while others stayed on.
6. While the residents spread across Africa, the migrants slowly spread across the planet.
7. They did not go in a single line. They went forwards and backwards, with groups, sometimes separating from each other, even moving in and out of Africa.
8. Like all other species on planet, they had come into being as an accident of evolution, and were trying to live their lives the best they could.

Question 16.
How would you appreciate Gregor Johann Mendel’s contribution to the genetics?
(OR)
Why Gregor Johann Mendel is considered as the father of genetics?
Answer:

1. Gregor Johann Mendel worked on the problem of how variations were passed from one generation to the other.
2. As he was a monk, he did his experiments with interest in the garden of the monastery.
3. He worked for over seven years after which he presented conclusions from his experimental data in a form of a detailed research paper.
4. Mendel made many careful observations of pea plants and asked himself questions about what he observed and then planned and designed experiments to find the answers.
5. He had worked on nearly 10,000 pea plants of 34 different varieties choosing 7 distinguishing forms of characters.
6. His experimental outcomes gave the idea how the variations were passed on from one generation to another.
7. He was an exemplary person for his observation, planning, patience and experimental skills.
8. His efforts to know the secrets of nature was really appreciable.

Question 17.
What are the hypothesis assumptions and outcomes of Mendel’s experiments with pea plants?
Answer:
Regarding his experiments with pea plants, Mendel hypothesised that

1. Characters were carried as traits and an organism always carried a pair of factors for a character.
2. The distinguishing traits of the same character were present in the population of an organism.
3. The traits shown by the pea plants must be in the seeds that produced them.
4. The seeds must have obtained by the traits from the parent plants.

His assumptions made to explain his observations are:
Assumption 1: Every pea plant has two ‘factors’ which are responsible for producing a particular property or trait.
Assumption 2: During reproduction one ‘factor’ from each parent is taken to form a new pair in the progeny.
Assumption 3: One of these will always dominate the other if mixed together.

Laws made from his experiments:

1. Law of Dominance: Among a pair of alleles for a character, only one expresses itself in the first generation as one of the allele is dominant over the other.
2. Law of Segregation: Every individual possesses a pair of alleles for any particular trait and that each parent passes a randomly selected copy of only one of these to its offspring.
3. Law of Independent Assortment: In the inheritance of more than one pair of characters the factors for each pair of characters assorts independently of the other pairs.

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 9 Classification of Elements- The Periodic Table Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 9th Lesson Structure of Atom

10th Class Chemistry 9th Lesson Structure of Atom Textbook Questions and Answers

Improve Your Learning

Question 1.
Newlands proposed the law of octaves. Mendeleeff suggested eight groups for elements in his table. How do you explain these observations in terms of modem periodic classification? (AS1)
(OR)
Correlate various tables proposed on classification of elements.
Answer:

• According to Newlands, every eighth element starting from a given element jsembles in its properties to that of the starting element, when elements are ranged in ascending order of their atomic weights.
• According to Newlands, the properties of fluorine and chlorine are similar and sodium and potassium are similar. Same aspect is given by modern periodic table.
• Mendeleeff divided it into horizontal rows and vertical columns. He called them peribds and groups respectively. Modem periodic table also gives the same.
• According to Mendeleeff, the elements of same group have similar properties. Modern periodic table also proposed the same thing.
• Mendeleeff gave the general formula for first group elements as R,0, and general formula for second group elements as RO. We can find the same thing in modern periodic table.
• The elements of particular group possess same common valency. Same was proposed by modern periodic table.

Question 2.
What are the limitations of Mendeleeff’s periodic table? How could the modern periodic table overcome the limitations of Mendeleeff’s table? (AS1)
(OR)
How can the limitations of Mendeleeffs table be overcome with the help of modern periodic table?
Answer:
Limitations of Mendeleeffs periodic table :
1) Anomalous pair of elements :
Certain elements of highest atomic weights precede those with lower atomic weights.
Eg : Tellurium (atomic weight 127.6) precedes iodine (atomic weight 126.9).

2) Dissimilar elements placed together :
a) Elements with dissimilar properties were placed in same group as sub-group A and sub-group Bt
Eg : Alkali metals like Li, Na, K, etc. of IA group have little resemblance with coinage metals like Cu, Ag, Au of IB group.

b) Cl of VII A group is a non-metal and Mn of VII B group is a metal.

Method of overcoming the limitations of Mendeleeffs periodic table by modern periodic table :
1. In modern periodic table, elements are arranged in the ascending order of their atomic numbers. So this arrangement eliminated the problem of anomalous series.
Eg : Though Tellurium (Te) has more atomic weight than Iodine (I), its atomic number is one unit less compared to Iodine.

2. The elements with similar outer shell (valence shell) electronic configurations in their atoms are in the same column called group in modern periodic table. So the elements have similar properties overcoming the Mendeleeffs second limitation.

Question 3.
Define the modern periodic law. Discuss the construction of the long form of the periodic table. (AS1)
(OR)
What are the salient features of modern periodic table?
Answer:
Modern periodic law :
‘The physical and chemical properties of elements are the periodic function of the electronic configurations of their atoms”.

Construction of the long form periodic table :

1. Based on the modern periodic law, the modern periodic table is proposed.
2. This periodic table is known as long form of the periodic table.
3. Long form periodic table is the graphical representation of Aufbau principle.
4. The modern periodic table has 18 vertical columns called groups and 7 horizontal rows known as periods.
5. There are 18 groups, represented by using Roman numerals I to VIII, with letters A and B in traditional notation, (or) 1 to 18 by Arabic numerals.
6. There are 7 periods. These periods are represented by Arabic numerals 1 to 7.
7. The number of main shells present in the atom of particular atom decides to which period it belongs.
8. First period consists 2 elements, 2^nd and 3rd periods contains 8 elements each, 4^th and 5^th periods contains 18 elements each, 6 period contains 32 elements and 7^th period is incomplete.
9. The elements are classified into s, p, d and f block elements.
10. Inert gases are placed in 18^th group.

Question 4.
Explain how the elements are classified into s, p, d and f-block elements in the periodic table and give the advantage of this kind of classification. (AS1)
(OR)
How is the periodic table classified based upon the entering of differenciating electron? Explain that classification. What is the advantage of such classification?
Answer:
1) Depending upon which sub-shell the differentiating electron enters, the elements are classified into s, p, d and f-block elements. They are

1. s – block elements,
2. p – block elements,
3. d – block elements,
4. f – block elements.

2) s – block elements :
i) If the differentiating electron enters in s-sub-shell, then the elements are called s-block elements.
ii) IA (1), IIA (2) group elements belong to this block.

3) p – block elements :
i) If the differentiating electron enters in p-sub-shell, then the elements are called p-block elements.
ii) IIIA(13), IV A (14), V A (15), VIA (16), VIIA (17) belong to p-block.

4) d – block elements :
i) If the differentiating electron enters in d-sub-shell, then the elements are called d – block elements.
ii) I B, II B, III B, IV B, V B, VI B, VII B, VIII B belong to d-block elements.
iii) They are also called transition elements.

5) f – block elements :
i) If the differentiating electrons enter in f-sub-shell, then the elements are called f-block elements.
ii) These are divided into two types
a) Lanthanides (41 elements),
b) Actinides (5f elements).
iii) These are also called as inner transition elements.

Advantage of this classification :
1) The systematic grouping of elements into groups made the study simple.
2) Each period begins with the electron entering a new shell and ends with the complete filling of s and p-sub-shells of that shell.

Question 5.
Given below is the electronic configuration of elements A, B, C, D. (AS1)

Answer:
According to electronic configuration
A = Be B = Mg C = P D = Ne
1. Which are the elements coming within the same period?
Answer:
A and D i.e. Be and Ne coming within the same period. [They have same valence shell (n = 2)]

2. Which are the ones coming within the same group?
Answer:
A and B i.e., Be and Mg coming within the same group. [They have same valence subshell with same valency (2s² and 3s²)]

3. Which are the noble gas elements?
Answer:
D, i.e. Ne is the noble gas element. [It has valency as ‘O’ and it has ‘8’ electrons in valence shell].

4. To which group and period does the element ‘C’ belong?
Answer:
Element ‘C’ i.e. ‘P’ belongs to 3^rd period and VA group.

Question 6.
Write down the characteristics of the elements having atomic number 17. (AS1)
1) Electronic configuration ___________
2) Period number _____________
3) Group number _____________
4) Element family ____________
5) No. of valence electrons ___________
6) Valency _____________
7) Metal or non-metal ____________
Answer:

1. 1s² 2s² 2p⁶ 3s² 3p⁵
2. 3
3. VII A or 17
4. Halogen family
5. 7
6. 1
7. Non-metal

Question 7.
a) State the number of valence electrons, the group number and the period number of each element given in the following table : (AS1)

Answer:

b) State whether the following elements belong to a Group (G), Period (P) or neither Group nor Period (N). (AS1)

Answer:

Question 8.
Elements in a group generally possess similar properties, but elements along a period have different properties. How do you explain this statement? (AS1)
(OR)
Elements in a group possess similar properties, but elements along a period have different properties. Explain the reason.
Answer:

• Physical and chemical properties of elements are related to their electronic configurations, particularly the outer shell configurations.
• Therefore, all the elements in a group should have similar chemical properties.
• Similarly, across the table from left to right in any period, elements get an increase in the atomic number by one unit between any two successive elements.
• Therefore, the electronic configuration of valence shell of any two elements in a period is not same. Due to this reason, elements along a period possess different chemical properties.

Question 9.
s – block and p – block elements except 18^th group elements are sometimes called as ‘Representative elements’ based on their abundant availability in the nature. Is it justified? Why? (AS1)
(OR)
Which elements are called representative elements? Why?
Answer:

• s, p – block elements are called representative elements because these are the elements which take part in chemical reactions because of incompletely filled outermost shell.
• These elements undergo chemical reactions to acquire the nearest noble gas configuration by losing or gaining or sharing of electrons.
• So they are called representative elements.

Question 10.
Complete the following table using periodic table. (AS1)

Answer:

Question 11.
Complete the following table using the periodic table. (AS1)

Answer:

Question 12.
The electronic configuration of the elements X, Y, and Z are given below.
a) X = 2
b) Y = 2, 6
c) Z = 2, 8, 2
i) Which element belongs to second period?
Answer:
Y belongs to second period.

ii) Which element belongs to second group?
Answer:
Z belongs to second group,

iii) Which element belongs to 18^th group?
Answer:
X belongs to 18^th group.

Question 13.
Identify the element that has the larger atomic radius in each pair of the following and mark it with a symbol (✓). (AS1)
(i) Mg or Ca
(ii) Li or Cs
(iii) N or P
(iv) B or Al
Answer:

Question 14.
Identify the element that has the lower ionization energy in each pair of the, following and mark it with a symbol (✓). (AS1)
(i) Mg or Na (ii) Li or O (iii) Br or F (iv) K or Br
Answer:

Question 15.
In period 2, element X is to the right of element Y. Then, find which ofitheydements have : (AS1)
i) Low nuclear charge
Answer:
Y has low nuclear charge.

ii) Low atomic size
Answer:
X has lower atomic size,

iii) High ionization energy
Answer:
X has higher ionization energy.

iv) High electronegativity
Answer:
Xhas high electronega^vity.

v) More metallic,character
Answer:
Y has more metallic character.

Question 16.
How does metallic character change when we move
i) Down a group?
ii) Across a period?
Answer:
i) Down a group :
When we move from top to bottom in a group, the metallic character increases.

ii) Across a period:
When we move left to right in a period, the metallic character decreases.

Question 17.
Why was the basis of classification of elements changed from the atomic mass to the atomic number? (AS1)
(OR)
Which atomic property is more suitable for classification of elements? Why?
Answer:

• The first attempt to classify elements was made by Dobereiner.
• Dobereiner’s attempt gave a clue that atomic masses could be correlated with properties of elements:
• Newlands’ law of octaves also followed the same basis for classification but this law is not valid for the elements that had atomic masses higher than calcium.
• Mendeleeff’s classification also based on the atomic masses of elements, but it lead to some limitations like Anomalous pair of elements and Dissimilar elements placed together.
• Moseley by analyzing the X-ray patterns of different elements was able to calculate the number of positive charges in the atoms of respective elements.
• With this analysis, Moseley realized that the atomic number is more fundamental
  characteristic of an element than its atomic weight. ,
• So, he arranged the elements in the periodic table according to the increasing order of their atomic number.
• This arrangement eliminated the problem of anomalous series and dissimilar elements placed together in Mendeleeff’s classification.

Question 18.
What is a periodic property? How do the following properties change in a group and period? Explain. (AS1)
I. a) Atomic radius
b) Ionization energy
c) Electron affinity
d) Electronegativity
II. Explain the ionization energy order in the following sets of elements: (AS1)
a) Na, Al, Cl
b) Li, Be, B
c) C, N, O
d) F, Ne, Na
e) Be, Mg, Ca
Answer:
Periodic property:
The property in which there shall be a regular gradation is called periodic property.

I. a) Atomic radius :
Period :
Atomic radius of elements decreases across a period from left to right because the nuclear charge increases due to increase in atomic number.

Group :
Atomic radius increases from top to bottom in a group due to addition of new shell.

b) Ionization energy:
Period :
When we move from left to right it does not follow a regular trend but generally increases due to increase in atomic number.

Group :
In a group from top to bottom, the ionization energy decreases due to increase in atomic size. –

c) Electron affinity:
Period :
Electron affinity values increase from left to right in a period.

Group :
Electron affinity values decrease from top to bottom in a group.

d) Electronegativity :
Period :
Electronegativity increases from left to right in a period.

Group :
Electronegativity decreases from top to bottom in a group.

II. Ionization energy order :
a) Na, Al, Cl
b) Li, Be, B
c) C, N, O
d) F, Ne, Na
e) Be, Mg, Ca
Answer:
a) In a period ionisation energy increases so the order is Na < kl < Cl.
b) Beryllium has stable configuration 1s² 2s². So it has more ionisation energy. So the order is Li < B < Be.
c) Nitrogen has half-filled p-orbitals. So it has greater ionisation energy. So the order is C < O < N.
d) Ne is inert gas right to F. Whereas Na is a metal ion in third period. So, the order is Na < F < Ne. e) In a group ionisation energy decreases. So the order is Be > Mg > Ca.

Question 19.
Name two elements that you would expect to have chemical properties similar to Mg. What is the basis for your choice? (AS2)
Answer:

• The two elements which have chemical properties similar to Magnesium are Beryllium and Calcium.
• The basis for my expectation is that they belong to same group as we know elements belonging to same group have similar properties.

Question 20.
On the basis of atomic numbers predict to which block the elements with atomic number 9, 37, 46 and 64 belong to? (AS2)
Answer:

1. The element with atomic number 9 belongs to p-block.
2. The element with atomic number 37 belongs to s-block.
3. The element with atomic number 46 belongs to d-block.
4. The element with atomic number 64 belongs to f-block.

Question 21.
Using the periodic table, predict the formula of compound formed between and element X of group 13 and another element Y of group 16. (AS2)
Answer:
The valency of 13^th group elements is 3.
The valency of 16^th group elements is 2.
The formula of compound is X₂Y₃.

Question 22.
An element X belongs to 3rd period and group 2 of the periodic table. State (AS2)
a) The no. of valence electrons
b) The valency.
c) Whether it is metal or a non-metal.
Answer:
a) The number of valence electrons are 2.
b) The valency of element is +2.
c) It is a metal.

Question 23.
An element has atomic number 19. Where would you expect this element in the periodic table and why? (AS2)
Answer:
The clement with atomic number 19 is in 4^th period and first group of the periodic table.
Reason :

1. Electronic configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s or [Ar]4s¹
2. The differentiating electron enters into 4^th shell. Hence it belongs to 4^th period.
3. The differentiating electron is in ‘s’ orbital. So it belongs to ‘s’ block.
4. The outermost orbital has only one electron. Hence it belongs to first group.

Question 24.
Aluminium does not react with water at room temperature but reacts with both dil. HCl and NaOH solutions. Verify these statements experimentally. Write your observations with chemical equations. From these observations, can we conclude that Al is a metalloid? (AS3)
Answer:

• Aluminium reacts with dil. HCl and releases hydrogen gas with formation of Aluminium chloride.
  
• Aluminium reacts with NaOH solution and releases hydrogen gas.
• 
• The above two reactions says that Aluminium is amphoteric.
• Aluminium does not react with water at room temperature.
• This concludes that the properties of Aluminium are in between a metal and non¬metal. So it behaves like a metalloid.

Question 25.
Collect the information about reactivity of VIIIA group elements (noble gases) from internet or from your school library and prepare a report on their special character when compared to other elements of periodic table. (AS4)
Answer:
Reactivity of Noble gases :

• The noble gases show extremely low chemical reactivity.
• He and Ne do not form chemical compounds.
• Xenon, krypton and argon show only minor reactivity.
• The reactivity order follows like this : Ne < He < Ar < Kr < Xe < Rn.
• Xenon can form compounds like XeF₂, XeF₄ and XeF₆, etc.

Reasons for low reactivity :

• The extremely low reactivity of noble gases is due to stable electronic configuration.
• But as we move from top to bottom the reactivity increases. So xenon can form some compounds with high electronegative elements.

Question 26.
Collect information regarding metallic character of elements of IA group and prepare report to support the idea of metallic character increases in a group as we move from top ro bottom. (AS4)
Answer:
Metallic character of IA group elements :

1. Alkali metals exhibit many of the physical properties common to metals but their densities are lower than those of other metals.
2. Alkali metals have one electron in their outer shell which is loosely bound.
3. They have largest atomic radii of the elements in their respective periods.
4. The lower ionization energies result in their metallic properties and high reactivities.
5. An alkali metal can easily lose its valence electron to form positive ion.
6. So they have greater metallic character.
7. The metallic character increases as we move from top to bottom in group due to addition of another shell, it is easy to lose electron.

Question 27.
How do you appreciate the role of electronic configuration of the atoms of elements in periodic classification? (AS6)
(OR)
How does electronic configuration help in the classification of elements in modern periodic table?
Answer:
The quantity is electronic configuration.

1. Modern periodic table is based on electronic configuration. So elements are arranged in ascending order of their atomic numbers.
2. The chemical properties of elements depend on valence electrons. The elements in same group have same number of valence electrons. So the elements belonging to same group have similar properties.
3. So the construction of modern periodic table mainly depends on electronic configuration.
4. Thus electronic configuration plays a major role in the preparation of modern periodic table. So its role is thoroughly appreciated.

Question 28.
Without knowing the electronic configurations of the atoms of elements Mendeleeff still could arrange the elements nearly close to the arrangements in the Modern periodic table. How can you appreciate this? (AS6)
Answer:

• Mendeleeff took consideration about chemical properties while arranging the elements. So the arrangement of elements is close to arrangement of elements in Modern periodic table.
• For this, he violated his periodic law.
• He left some gaps for elements, later those elements are discovered.
• So the efforts of Mendeleeff should be thoroughly appreciated.

Question 29.
Comment on the position of hydrogen in periodic table. (AS7)
Answer:

• Hydrogen is the element which has easier atomic structure than any other element.
• Electron configuration of hydrogen is Is1. It has one proton in its is nucleus and one electron in its is orbital.
• Hydrogen combines with halogens, oxygen and sulphur to form compounds having similar formulae just like alkali metals.
• Similarly, just like halogens, hydrogen also exists as diatomic molecule and combine with metals and non-metals to form covalent compounds.
• As alkali metals hydrogen can lose one electron and accept one electron as halogens.
• So in periodic table, its place may be in IA or VIIA group.
• But based on electronic configuration of hydrogen, it is placed in IA group.

Question 30.
How do the positions of elements in the periodic table help you to predict its chemical properties? Explain with an example. (As7)
Answer:
1) The physical and chemical properties of atoms of the elements depend on their electronic configuration, particularly the outer shell configurations.

2) Elements are placed in the periodic table according to the increasing order of their electronic configuration.

3) The elements in a group possess similar electronic configurations. Therefore all the elements in a group should have similar chemical properties.
Ex : Consider K

• It is the element in 4^th period 1^st group.
• Electron configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹.
• Differentiating electron enters into s-orbital. Hence it belongs to s-block.
• It is on the left side of the periodic table. Hence it is a metal.
• It is ready to lose one electron to get octet configuration. Hence its reactivity is more.
• It is Alkali metal.
• All alkali metals react with both acids and bases and releases H₂ gas.

Fill In The Blanks

1. Lithium, ……………… and potassium constitute a Dobereiner’s triad.
2. ……………… was the basis of the classifications proposed by Dobereiner, Newlands, and Mendeleeff.
3. Noble gases belong to ……………… group of periodic table.
4. The incomplete period of the modern periodic table is
5. The element at the bottom of a group would be expected to show …………….. metallic character than the element at the top

Answer:

1. Sodium
2. Atomic weight
3. VIIIA or 18 group
4. 7^th
5. higher

Multiple Choice Questions

1. Number of elements present in period – 2 of the long form of periodic table …………
A) 2
B) 8
C) 18
D) 32
Answer:
B) 8

2. Nitrogen (Z = 7) is the element of group V of the periodic table. Which of the following is the atomic number of the next element in the group?
A) 9
B) 14
C) 15
D) 17
Answer:
C) 15

3. Electronic configuration of an atom is 2, 8, 7. To which of the following elements would it be chemically similar?
A) Nitrogen (Z = 7)
B) Fluorine (Z = 9)
C) Phosphorous (Z – 15)
D) Argon (Z = 18)
Answer:
B) Fluorine (Z = 9)

4. Which of the following is the most active metal?
A) lithium
B) sodium
C) potassium
D) rubidium
Answer:
D) rubidium

10th Class Chemistry 9th Lesson Classification of Elements-The Periodic Table InText Questions and Answers

10th Class Chemistry Textbook Page No. 129

Question 1.
Can you establish the same relationship with the set of elements given in the remaining rows?
Answer:
Yes, we can establish the approximately same relationship between other elements given in the table.

Question 2.
Find average atomic weights of the first and third elements in each row and compare it with the atomic weight of the middle element. What do you observe?
Answer:
The atomic weight of middle element is arithmetic mean coverage of first and third elements.

10th Class Chemistry Textbook Page No. 135

Question 3.
What is atomic number?
Answer:
The number of positive charges (protons) in the atom of element is the atomic number of element.

10th Class Chemistry Textbook Page No. 142

Question 4.
How does the valency vary in a period on going from left to right?
Answer:
It does not follow a regular trend when we move from left to right in a period. First, it increases and then decreases and finally ‘O’ for inert gases.

Question 5.
How does the valency vary on going down a group?
Answer:
The valency is constant when we move from top to bottom in a group because the number of valence electrons are same for same group elements.

10th Class Chemistry Textbook Page No. 144

Question 6.
Do the atom of an element and its ion have same size?
Answer:
No, the positive ion has smaller size than neutral atom whereas negative ion has greater size than neutral atom.

Question 7.
Which one between Na and Na⁺ would have more size? Why?
Answer:

• The atomic number of Sodium is 1 and it has 11 protons and 11 electrons with outer electron as 3s¹ whereas Na⁺ ion has 11 protons but only 10 electrons.
• The 3s shell of Na⁺ has no electron in it.
• So the outer shell configuration is 2s²2p⁶.
• As proton number is more than electrons, the nucleus of Na⁺ ion attracts outer shell electrons with strong nuclear force.
• As a result the Na⁺ ion shrinks in size.
• Therefore, the size of Na⁺ ion is less than Na atom.

Question 8.
Which one between Cl and Cl^– would have more size? Why?
Answer:

• The electronic configuration of chlorine (Cl) atom is 1s2 2s² 2p⁶ 3s² 3p⁵ and the electronic configuration of chloride (Cl^–) ion is 1s² 2s² 2p⁶ 3s² 3p⁶.
• Both chlorine and chloride ions have 17 protons each but there are 17 electrons in chlorine atom, whereas 18 electrons in chloride ion.
• Therefore, the nuclear attraction is less in Cl^– ion when compared with chlorine atom.
• Therefore the size of the chlorine (Cl) atom is less size than chloride of Cl^– ion.

Question 9.
Which one in each of the following pairs is larger in size? Why?
a) Na, Al
b) Na, Mg⁺²
c) S^2-, Cl^–
d) Fe²⁺, Fe³⁺
e) C^4-, F^–.
Answer:
a) Na has larger size because Sodium and Aluminium are third period elements in which Na is left to Al. As we move from left to right in a period atomic size decreases.

b) Mg²⁺ has smaller size because Mg²⁺ has 10 electrons and 12 protons whereas Na has 11 electrons and 11 protons. So the distance between nucleus and outermost orbital is less in Mg²⁺ due to greater nuclear attraction.

c) S^2- has, larger size because S^2- has 18 electrons and 16 protons and Cl^– has 18 electrons and 17 protons. So nuclear attraction over outermost orbital is more in Cl^– when compared with S^2-. So S^2- has larger size.

d) In Fe²⁺ it has 26 protons and 24 electrons whereas for Fe³⁺ it has 26 protons and 23 electrons. So nuclear attraction over outermost orbital is more in Fe³⁺. So Fe³⁺ has smaller size (or) Fe²⁺ has larger size.

e) C^4- has 6 protons and 10 electrons whereas F^– has 9 protons and 10 electrons. So nuclear attraction is less in C^4-. So size of C^4- is more than F^–.

10th Class Chemistry Textbook Page No. 129

Question 10.
What relation about elements did Dobereiner want to establish?
Answer:
Dobereiner wanted to give a relationship between the properties of elements and their atomic weights.

Question 11.
The densities of calcium (Ca) and barium (Ba) are 1.55 and 3.51 gem-3 respectively. Based on Dobereiner’s law of triads can you give the approximate density of strontium (Sr)?
Answer:
Molecular weight is directly proportional to density.

So density of strontium is mean of calcium and barium according to Dobereiner.
∴ Density of strontium = \(\frac{1.55+3.51}{2}\) = 2.53.

10th Class Chemistry Textbook Page No. 130

Question 12.
Do you know why Newlands proposed the law of octaves? Explain your answer in terms of the modern structure of the atom.
Answer:

• John Newlands found that when elements were arranged in the ascending order of their atomic weights, they appeared to fall into seven groups.
• Each group contained elements with similar properties.
• If we start with hydrogen and move down, the next eighth element is fluorine, and then next eighth element is chlorine and the properties of these elements are similar.
• Similarly, if we start from Lithium their eighth element is Sodium and next eighth element is potassium. These show similar properties.

Question 13.
Do you think that Newlands’ law of octaves is correct? Justify.
Answer:
No, there are some limitations of Newlands’ model:

• There are instances of two elements fitted into the same slot. Eg : Cobalt and Nickel.
• Certain elements, totally dissimilar in their properties, were fitted into the same group.
• Law of octaves holds good only for the elements up to Calcium.
• Newlands’ periodic table was restricted to only 56 elements and did not leave any room for new elements.
• Newlands had taken consideration about active pattern sometimes without caring the similarities.

10th Class Chemistry Textbook Page No. 134

Question 14.
Why did Mendeleeff have to leave certain blank spaces in his periodic table? What is your explanation for this?
Answer:
1) Mendeleeff predicted that some elements which have similar properties with the elements in a group are missing at that time.
2) So he kept some blanks in the periodic table by writing ’eka’ to the name of the element immediately above the empty space.
3) Later these elements are discovered and they are fitted into those empty spaces.

Question 15.
What is your understanding about Ea₂O₃, EsO₂?
Answer:

• Mendeleeff predicted that after aluminium there was another element namely eka- aluminium (Ea) and after silicon, there was another element namely eka-silicon (Es).
• He also gave the formulae of those oxides as Ea203 and Es02.
• Later those elements are discovered namely gallium and germanium and Ea₂O₃ and EsO₂ as Ga₂O₃ and GeO₂.

10th Class Chemistry Textbook Page No. 135

Question 16.
All alkali metals are solids but hydrogen is a gas with diatomic molecules. Do you justify the inclusion of hydrogen in first group with alkali metals?
Answer:
No, hydrogen shows the properties of both alkali metals and halogens. Still the position of hydrogen has some questions. So it was kept just above alkali metals in first group.

10th Class Chemistry Textbook Page No. 141

Question 17.
Why are lanthanoids and actinoids placed separately at the bottom of the periodic table?
Answer:
The properties of these elements do not coincide with other elements because the valence electron enters 4f and 5f orbitals respectively. So they are placed separately at the bottom of the periodic table.

Question 18.
If lanthanoids and actinoids are inserted within the table, imagine how the table would be?
Answer:
It looks very big in size, and it is difficult to identify, as these elements have similar properties.

10th Class Chemistry Textbook Page No. 145

Question 19.
Second ionization energy of an element is higher than its first ionization energy. Why?
Answer:

• The energy required to remove an electron from unipositive ion is called second ionisation energy.
• It is difficult to remove an electron from unipositive ion when compared with neutral atom due to an increase in nuclear attraction.
• So always second ionisation energy is higher than first ionisation energy.

10th Class Chemistry Textbook Page No. 146

Question 20.
The calculated electron gain enthalpy values for alkaline earth metals and noble gases are positive. How can you explain this?
Answer:

• Generally alkaline earth metals having one, two or three valence electrons prefer to lose electrons in order to get inert gas configuration. So it is difficult to add electron to alkaline earth metals. So they have positive electron gain enthalpy values.
• Noble gases are stable. So they do not prefer to take electrons. So they have positive electron gain enthalpy.

Question 21.
The second period element, for example, ‘F’ has less electron gain enthalpy than the third period element of the same group for example ‘Cl’. Why?
Answer:

• Electron gain enthalpy values decrease in a group as we go down and increase from left to right along a period.
• But the size of Fluorine is small compared chlorine.
• So it is difficult to add electron to fluorine.
• So fluorine has less electron gain enthalpy.

10th Class Chemistry 9th Lesson Classification of Elements- The Periodic Table Activities

Activity – 1

Question 1.
Observe the following table. Establish the relationship of other elements given in the table.

Answer:

Activity – 2

Question 2.
Some main group elements of s-block and p-block have family names as given in the following table.
Observe the long form of a periodic table and complete the table with proper information.

Answer:

Activity – 3

Question 3.
Collect valencies of first 20 elements.
Answer:

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 10 Natural Resources.

AP State Syllabus SSC 10th Class Biology Important Questions 10th Lesson Our Environment

10th Class Biology 10th Lesson Natural Resources 1 Mark Important Questions and Answers

Question 1.
Recently a new programme was launched in our state known as “Vanam – Manain”. Prepare any two slogans to promote the programme.
Answer:
a) Save forest, forest will save you.
b) Grow the plants and get the fresh air.

Question 2.
Suggest any two practices suitable to farmers with less water resources.
Answer:

1. Construction of percolation tanks (or) Soak pits
2. Irrigation techniques like drip irrigation and usage of sprinklers.

Question 3.
Why should we conserve forests? Give two reasons.
Answer:
a) Forests serve as lungs for the world. They purify the air and protect the earth from greenhouse effect and global warming,
b) Forests are rich habitats for plants

Question 4.
Ravi observed symbol on the plastic water bottle purchased by him. What does this symbol indicate? and animals.
Answer:
The symbol on the plastic bottle indicates that the bottle is made from recycled plastic and after its use it can be recycled.

Question 5.
Write any two suggestions for the conservation of biodiversity at your village.
Answer:

1. Protecting and preserving the natural habitats of birds and animals.
2. Replace the wood products with alternative sources.
3. Using Recycled products and following the 4’R Principle in day to day life.

Question 6.
Suggest any two activities to check soil erosion in your school.
Answer:

1. Observe the school ground after the rain.
2. Conduct a field project on soil erosion.

Question 7.
To create awareness on “Water conservation” in your locality, what slogan you will suggest?
Prepare two slogans on ‘Save Water’ propaganda.
Answer:
“Don’t Waste Water”.
“Save every drop”.
“Water is life”.

Question 8.
The symbol is there on the item you bought. What it instructs? (OR)
What does the given logo indicate? What does it mean?

Answer:
It is the Recycle logo. It indicates that the item we bought is prepared from recycled materials or the item can be recycled after use.

Question 9.
What happens if the forest area decreases rapidly?
Answer:
If the forest area decreases

1. It destroys wildlife habitat.
2. It increases soil erosion.
3. It releases green house gases into the atmosphere contributing to global warming.
4. It also harms people who relay on forest for their survival, hunting and gathering, harvesting forest products or using timber and firewood.

Question 10.
Write two activities which you are performing to save electricity.
(OR)
Write any two measures vou take in your home to reduce the consumption of electricity.
Answer:

1. We can reduce the consumption of electricity by putting off the fans and lights when there is no need.
2. We can use LED (Lighting Emitting Device) bulbs to save electricity.
3. To shut down laptops and computers when they are not in use.

Question 11.
Prepare two slogans on protecting non-renewable resources.
Answer:

1. Use Biofuel – Reduce Fossil Fuel.
2. Use alternative resources – Save the environment.

Question 12.
Write two examples for non-renewable resources.
Answer:
Examples for non-renewable resources are coal, petroleum and natural gas.

Question 13.
What is sustainable development? Is it needful for us?
Answer:
When we use the environment In ways that ensure we have resources for the future, It Is called sustainable development. It Is needed because development and conservation can coexist In harmony.

Question 14.
What are examples for natural resources?
Answer:
Examples for natural resources are water, soil, forests, flora, fauna, etc.

Question 15.
What are percolation tanks?
Answer:
Percolation tanks are normally earthen dams with masonry structures where water may overflow.

Question 16.
What are Kharif crops?
Answer:
Crops grown In the rainy season are termed as Kharif crops, e.g: Paddy, maize, millet and cotton crops.

Question 17.
What are Rabi crops?
Answer:
The crops that are grown only in winter season are generally called Rabi crops, e.g.: Wheat, Gram and Mustard.

Question 18.
What is the average fall of ground water level in Andhra Pradesh state during the period of 1998 – 2002?
Answer:
The average fall of ground water level In Andhra Pradesh state during the period of 1998 – 2002 Is 3 meters.

Question 19.
Which agency in villages of Warangal district helped in recharging wells that were being dried up?
Answer:
Centre for water solidarity (Secundrabad, T.S.) helped In recharging wells that were drying up In the villages.

Question 20.
Give examples for micro irrigation techniques.
Answer:
Drip irrigation, sprinklers are the examples for micro irrigation techniques.

Question 21.
Mow did the boundaries between the villages were fixed in ancient times?
Answer:
In ancient times village boundaries were decided upon a water shed (Land between water sources usually of two rivers or streams) basis fixed at the common point of the drainage system In between two villages by the expert farmers In the village.

Question 22.
Expand ICRISAT.
Answer:
International Crop Research Institute for Semi-Arid Tropics.

Question 23.
What is the other name for Sri Rama Sagar Project?
Answer:
Sri Rama Sagar Project also known as the Pochampadu project on the Godavari river,

Question 24.
What is the use qf planting Gliricidia on field bunds?
Answer:
Planting Gliricidia on field bunds help In strengthen them and make the soli nitrogen-rich.

Question 25.
What is the micro irrigation system that can reduce water consumption by 70%?
Answer:
Drip irrigation can reduce water consumption by 70%.

Question 26.
Who predicted that by 2025, 1.8 billion people will be living in countries or regions with absolute water scarcity ?
Answer:
The Food and Agriculture Organisation (FAO) of the united nations has predicted that by 2025, 1.8 billion people will be living in countries or regions with absolute water scarcity.

Question 27.
What happens if we use resources wisely?
Answer:
If resources are used wisely and efficiently they will last much longer. Through conservation people can reduce waste and manage natural resources wisely.

Question 28.
Give an example of country where restrictions on water usage were imposed.
Answer:
In Australia restrictions were imposed on activities like, watering lawns by using sprinkler systems, washing vehicles, using house pipes to clean paved areas, and refilling swimming pools.

Question 29.
Why are the natural resources used up quickly?
Answer:
The population of human beings has grown enormously in the past two centuries. Billions of people use up resources quickly as they eat food, build houses, produce goods and burn fuel for transportation and electricity.

Question 30.
What happens if we damage a forest resource?
Answer:
Harm to animals that may be forced to find new habitats. If we damage a forest resource indiscriminately the depletion of resources occur and we may have to face problem for water and timber in future.

Question 31.
What are the results of deforestation?
Answer:
Deforestation destroys wild life habitats and increases soil erosion and also releases green house gases into atmosphere, contributing to global warming.

Question 32.
How the people in China and Mexico recycle paper? (OR)
Give an example of recycling paper by the people. What is the use of recycling paper?
Answer:
People in China and Mexico reuse much of their waste paper, including writing paper, wrapping paper and card board.

Question 33.
How the soil is important for us ? How the soil is importane for us?
Answer:
Soil is vital to food production and also important to plants that grow in the wild.

Question 34.
What are the reasons for depletion of nutrients in soil?
Answer:
Poor farming methods, such as repeatedly planting the same type of crop in the same place cause depletion of nutrients in the soil.

Question 35.
What is biodiversity?
Answer:
Biodiversity is the variety of living things that populate the Earth.

Question 36.
How are people speeding up the loss of biodiversity?
Answer:
Through hunting, pollution, habitat destruction, people are speeding up the loss of biodiversity.

Question 37.
How many plant species are being used by us for medicines world wide?
Answer:
We use between 50,000 to 70,000 plant species for medicines world wide.

Question 38.
What is selective harvesting?
Answer:
The practice or removing individual plants or small groups of plants leaving other plants standing to anchor the soil is called selective harvesting.

Question 39.
How are fossil fuels produced?
Answer:
The fuels that are produced from the remains of ancient plants and animals are called fossil fuels. They include coal, petroleum and natural gas.

Question 40.
What are the alternate sources of energy?
Answer:
The alternate sources of energy are sun, wind and water.

Question 41.
What are the other products made from petroleum?
Answer:
Plastic, synthetic, rubber, fabrics like nylon, medicines, cosmetics, waxes, cleaning products, medical devices, etc., are the other products made from petroleum.

Question 42.
Which plant’s seeds are used for the production of bio-fuel?
Answer:
Seeds from the Jatropa Curcas plant are used for the production of bio-fuel.

Question 43.
How does the mining method, Mountain Top Removal mining (MTR) devastate the environment?
Answer:
The mining method Mountain Top Removal mining devastate the environment. They destroy soil, plants and animal habitats.

Question 44.
In which country car manufacturers recycle many raw materials used in making automobiles?
Answer:
In Japan car manufacturers recycle many raw materials used in making automobiles.

Question 45.
In which country nearly one third of the iron produced comes from recycled automobiles?
A. In the United States, nearly one-third of the iron produced comes from recycled automobiles.

Question 46.
What does the Indian tradition teach us?
Answer:
The Indian tradition teaches us that all forms of life – human, animal and plant are so closely inter linked that disturbance of one gives rise to imbalance in the other.

Question 47.
Expand IUCN.
Answer:
IUCN stands for International Union for the Conservation of Nature.

Question 48.
How is IUCN planning to protect wild life and habitats?
Answer:
IUCN monitors the status of endangered wild life, threatened national parks and preserves.

Question 49.
What are the four R’s to save the environment?
Answer:
Reduce, Reuse, Recycle, Recover are the four R’s to save the environment.

Question 50.
How did Amritha Devi and her daughters protest against the king’s order?
Answer:

1. Amritha Devi and her daughters, followed by villagers, who clung to trees in the forest surrounding their village and laid down their lives to save them.
2. They protested against the king’s order to collect wood for the construction of his palace.

Question 51.
Write a method of soil conservation.
Answer:
One soil conservating method is called contour strip cropping. Several crops such as corn, wheat and clover are planted to alternating strips across a slope or across the path of the prevailing wind.

Question 52.
What is the rate of extinction by the estimation of the scientists?
Answer:

1. Scientists estimate that the current rate of extinction is 1,000 times the natural rate through hunting, pollution, habitat destruction.
2. Based on various estimates of the number of species on Earth, we could be losing anywhere from 200 to 1,00,000 species each year.

Question 53.
What is the need to protect biodiversity?
Answer:
We need to protect biodiversity to ensure plentiful and varied food sources. Biodiversity is important for more than just food because many plant species are being used for medicines.

Question 54.
Mention two ways in which water harvesting can be undertaken?
Answer:
The two ways by which water harvesting can be undertaken are

1. Capturing run off water from, rooftops.
2. Capturing run off water from local catchments.

Question 55.
On the basis of the issues raised in the chapter management of natural resources, what changes you in corporate in your lifestyle in a move towards a sustainable use of our resources?
Answer:
I would incorporate the maximum of four R’s i.e., reduce, recycle, reuse and recover in my lifestyle in a move towards a sustainable use of our resources.

10th Class Biology 10th Lesson Natural Resources 2 Marks Important Questions and Answers

Question 1.
Rahul remarked that different human activities are responsible for global warming.
What might be the reasons for his statement?
Answer:

1. Deforestation.
2. Industrialisation and urbanization.
3. Conversion of agriculture lands into residential areas.
4. Home appliances like A/C, refrigerators, vehicle pollution.
5. Population explosion.

Question 2.
What steps do you take to improve natural resources?
Answer:

1. Motivate the people to conserve water.
2. Try to avoid wastage of water whenever possible.
3. Plantation in the vacant lands.
4. Educating the farmer regarding proper utilization of water for irrigation.
5. Encourage the people to recycle the water wherever possible.

Question 3.
Proper utilisation of natural resources is the way to show gratitude to our nation.
Can you support this statement? Give your argument.
Answer:

1. Natural resources of a nation influence its economical and social development.
2. Natural resources are freely available in nature and help in many activities and development of people.
3. The generation of natural resources take a lot of time.
4. They disappear by indiscriminate usage.
5. So proper utilization of natural resources is the way to show gratitude to our nation.

Question 4.
The humans are utilising natural resources indiscriminately. These resources are decreasing more rapidly. Guess what will be the consequences in future?
Answer:
Indiscriminate usage of natural resources causes the following consequences.

1. Reduction in rainfall
2. Drought will occur.
3. Atmospheric temperature becomes increase.
4. The rare species become extinct.

Question 5.
Write any four slogans on the conservation of natural resources.
Answer:
Slogans:

1. Waste water today – live in desert tomorrow
2. Practice eco-friendly methods.
3. Use natural resources judiciously.
4. Save nature – Save future.

Question 6.
There is an increase in the atmospheric temperature year by year. If it continues, guess and write what would be the consequences?
Answer:
If the temperature on earth increases, the consequences would be as follows.
a) All the glaciers and the frozen ice in the polar region start melting leading to rise in the sea water levels.
b) It results in the submergence of low lying coastal areas throughout the world. Millions of people of those areas would lost their homes.
c) Changes in rainfall patterns take place and it result in the occurance of droughts and decrease in crop production.
d) Global warming results in climate change which cause the breakout of climatic sensitive diseases like Malaria, Dengue, Diarrhoea, etc.

Question 7.
There is water scarcity in Ravi’s village during summer. He wants to conduct a rally to create awareness regarding conservation of water. Write any four slogans required to conduct this rally.
Answer:

1. Water is life.
2. Save water – Save a life.
3. Today’s rain water is tomorrow’s life saver.
4. No matter your occupation, water conservation is your obligation.

Question 8.
What steps you take to conserve the biofuels in your daily life?
Answer:

1. Development and usage of alternative energy resources in place of bio-fuels.
2. Minimise the usage of bio-fuels whenever possible.
3. Use public transport, ride by bicycle and walking regularly.
4. Use and purchase energy efficient appliances to save bio-fuels.

Question 9.
Why do we use fossil fuels judiciously?
Answer:

1. Fossil fuels were produced from the remains of ancient plants and animals.
2. They include coal, petroleum (oil) and natural gas.
3. We need to use fossil fuels judiciously because they are non – renewable resources.
4. We need to conserve fossil fuels so we don’t run out of them.
5. The pollution caused by them when burnt, to limit our fossil fuel use.
6. Future generations may not get these resources.
7. Balance in the nature will be disturbed.
8. Electricity production will be stopped.
9. Vehicles running with fossil fuels become useless.

Question 10.
Write two suggestions to create awareness on groundwater conservation.
Answer:

1. We need to adapt different methods to Improve the quality and increase the quantity of groundwater.
2. We should dig water harvesting pits for every house.
3. We should clean the silt, mud fill in tanks and ponds.
4. We should prohibit the establishment of borewells for extraction of groundwater for agricultural and Industrial use.
5. These measures will improve quality and quantity of groundwater.

Question 11.
What is the importance of 4R’s in achieving the goal of “Swachh Bharat”?
Answer:

1. Reduce the production of garbage.
2. Reuse the garbage for the production of manure and electricity.
3. Recycle the garbage by separating It as dry and wet garbage.
4. Recover the plants.

Question 12.
Suggest four measures to conserve fossil fuels.
Answer:
Measures to conserve fossil fuels:

1. Usage of alternatives to fossil fuel.
2. Minimise the usage of fossil fuel.
3. Walk, ride by bicycle and use public transportation whenever possible.
4. Purchase energy efficient appliances.
5. Turn off light and other electronics when you are not using them.

Question 13.
The indiscriminate digging of Borewells may result in what type of consequences in future?
Answer:

1. Due to over drilling of borewells and pulling out water by electric motors, the ground water level Is decreasing day by day.
2. It Is goes on without recharging, ground water becomes scarce.
3. It shows impact on agriculture and the productivity will decrease.
4. Fluorine levels In ground water will increase.
5. Sometimes, saline water may intrude Into the interior places of land and water becomes unfit for consumption.
6. Farmers have to drill the bore wells to more depths which Increase the losses for them.

Question 14.
Ramaiah made broad bed furrow around his field under employment guarantee scheme. Guess the reasons for if. If all the farmers of your village work together, will their water scarcity meet?
Answer:
The reason for Ramaiah making broad bed furrow around his field was, it is useful to conserve soil and water, fertilizer application weeding operations. It also conserves rain water.

The farmers are over coming the water scarcity by sharing water available in the village. They formed groups of farmer including large and small ones who would use the same water resource. Farmers were also motivated to use irrigation techniques like drip irrigation.

Question 15.
What are renewable sources and non-renewable resources?
Answer:
Renewable resources: Resources that can be replaced after they are used are called renewable resources.
Ex: Air, water and soil.
Non-renewable resources: Some other resources, cannot be replaced at all: Once they are used up they are gone forever and are called non renewable resources.
Ex: Coal, Petroleum, Natural gas (fossil fuels).

Question 16.
How do people waste natural resources?
Answer:

1. People often waste natural resources.
2. Animals are over hunted, forests are cleared, exposing land to wind and water damage.
3. Fertile soil is exhausted and lost to erosion because of poor farming practices.
4. Fuel supplies are depleted.
5. Water and air are polluted.
6. Water resources is indiscriminately used for crop growth.

Question 17.
How do people use the forest resources differently?
Answer:

1. The need to conserve resources often conflicts with other needs.
2. For some people, a forest area may be a good place to put a farm.
3. A timber company may want to harvest the area’s trees for construction materials.
4. A business company may want to build a factory or a shopping mall on the land.

Question 18.
What are die effects of deforestation?
Answer:

1. Deforestation destroys wild life habitats and increases soil erosion.
2. It also releases green house gases into the atmosphere, contributing to global warming,
3. Deforestation accounts for 15 per cent of the world’s green house gas emissions.
4. Deforestation also harms the people who rely on forests for their survival, hunting and gathering, harvesting forest products, or using the timber for firewood.

Question 19.
In your opinion What are the causes for soil erosion?
Answer:

1. Soil erosion is caused by poor farming methods such as repeatedly planting the same type of crop in the same place.
2. These methods deplete nutrients in the soil.
3. Soil erosion Is also caused by water and wind currents.
4. When farmers plough up and down hills, soil erosion occurs.
5. Overgrazing by cattle also causes soil erosion.
6. Natural floods causes the extensive damage to the top layer of the soil.

Question 20.
What is Biodiversity? Explain.
(OR)
What is the importance of biodiversity?
Answer:

1. Biodiversity is the variety of living things that populate the earth.
2. The products and benefits we get from nature rely on biodiversity.
3. We need to protect biodiversity to ensure plentiful and varied food sources.
4. Biodiversity is important for more than just food. For instance we use between 50,000 to 70,000 plant species for medicines world wide.

Question 21.
How can we use the fossil fuels carefully?
Answer:
We can use the fossil fuels carefully by taking the following measures.

1. Turn off lights and other electronics when we are not using them.
2. Purchase energy-efficient appliances.
3. Walk, ride a bicycle, if the distance is less.
4. Use public transportation whenever possible.
5. It is better to prefer public transport system like bus or train, instead of travel in personal vehicles.

Question 22.
Why the prices of aluminium and iron are expensive?
Answer:
Earth’s supply of raw material resources is in danger. Many mineral deposits that have been located and mapped have been depleted. As the ores for minerals like aluminium and iron become harder to find and extract, their prices go up.
This makes tools and machinery more expensive to purchase and operate.

Question 23.
What are the effects of mining?
Answer:

1. Many mining methods such as Mountain Top Removal mining (MTR) devastate the environment.
2. They destroy soil, plants and animal habitats.
3. Many mining methods also pollute water and air, as toxic chemicals leak into the surrounding ecosystem.

Question 24.
What did Smt. Indira Gandhi said, while launching the world conservation strategy in India on 6th March 1980?
Answer:
“The interest in conservation is not a sentimental one but the discovery of a truth well known to our ancient stages. The Indian tradition teaches us that all forms of life- human, animal and plant – are so closely inter-linked that disturbance in one gives rise to imbalance in the other” said by Smt. Indira Gandhi.

Question 25.
What are the steps taken by the government to conserve resources?
Answer:

1. Government enacts laws defining how land should be used and which areas should be set aside as parks and wild life preserves.
2. The government enforces laws designed to protect the environment from pollution, such as requiring factories to install pollution control devices and also provide incentives for conserving resources.

Question 26.
What is the necessity of sustainable management of natural resources? Out of the two methods reuse and recycle which one would you suggest to practice and why?
Answer:

1. Sustainable management of natural resources is necessary to Increase the over all life of natural resources specially non renewable resources and also to control the environmental pollution.
2. Both reuse and recycle are the good choice.
3. Reuse: If we reuse something then the cost of recycle will be saved.
4. Recycle: It is not necessary that each and everything can be reused, so after getting recycled the life of the resource will be enhanced.

Question 27.
“Burning fossil fuels is a cause of global warming”. Justify this statement?
Answer:

1. Fossil fuels are composed of carbon, hydrogen, nitrogen and sulphur.
2. When these are burnt they produce CO₂, H₂O, Oxides of Nitrogen and Sulphur.
3. Incomplete combustion of fossil fuels produces green house gases such as CO₂,
4. If huge amount of fossil fuels are burnt, It would produce high amount of CO₂ resulting intense global warming.

Question 28.
Can you suggest some changes in your school which would make it environment friendly?
Answer:
The changes that would make my school environment friendly are

1. Save energy by turning off lights that we are not using.
2. I will suggest to buy recycled paper for decoration and other purposes.
3. Use writing paper on both the sides.
4. Growing trees and plants all around the play ground.

Question 29.
What is the necessity of replenishment of forest? State four reasons.
Answer:
The replenishment of forest is necessary because of the following reasons.

1. It is used to conserve soil.
2. It provides shelter to wild animals.
3. It reduces atmospheric pollution.
4. It controls flood and increases frequency of rainfall.

10th Class Biology 10th Lesson Natural Resources 4 Marks Important Questions and Answers

Question 1.
Forest is renewable resource. But, each year, the Earth loses about 36 million acres of forest. In this type of situation, what suggestions do you give to save forests from turning into non-renewable resources ?
Answer:

1. Forests are the lungs of the world. So I will suggest the following measures to save forests from turning into non-renewable resources.
2. Sustainable forestry practices for ensuing resources into the future.
3. Low impact logging practices, harvesting with natural regeneration in mind. Prevention of removing all the high value trees or all the largest trees from the forests. Recycling methods should be adopted.
4. Replace wood products with alternative sources.
5. Preventing forest fires.
6. Implementing methods like agro forestry, social forestry crop rotation, green plantation, etc. are essential.

Question 2.
What are four R’s? Explain how they help to conserve the environment?
(OR)
Write about the 4 ‘R’s needed for the protection and conservation of environment.
Answer:
By pursuing the maximum of four R’s i.e., Reduce, Reuse, Recycle and Recover, we can save the environment in an effective way.

1. Reduce : It means to use less, I would save electricity by switching off unnecessary lights and fans, prefer walking or cycling than using a vechicle, turn off the engine of car at red light, repair leaky taps and would not waste food.
2. Recycle: It means to collect used things like plastic, paper, galss and metal items and recycle these materials to make required things instead of synthesising or extracting fresh plastic, paper, glass or metal.
3. Reuse: It refers to use things again and again. For example instead of throwing away used envelops, they can be used by pasting new labels.
4. Recover: We should implement ‘recover’ to prevent environmental threat. For example when we cut trees to construct industries or roads for transportation, it is important to grow trees in another areas.

Question 3.
What steps you would like to follow on your part to conserve bio-diversity?
Answer:

1. Biodiversity is the variety of living things that populate the earth.
2. To conserve biodiversity we should avoid hunting.
3. Sustainable forest conservation methods should be followed.
4. I will actively participate Vana Mahosthavam programmes.
5. I will educate and encourage people and make them participate in conservation programmes.
6. Create awareness programmes in and around school.
7. Writing slogans and also make some posters about conservation of biodiversity
8. Judicious use of electricity wherever possible.
9. Finding out of various alternative sources of energy.
10. Plant the saplings in the habitat.
11. Encouraging of social forestry.

Question 4.
Observe the pie diagram showing water resources available in our state for agriculture and answer the given questions.
A) Which water resource is using more for agriculture?
B) What are the consequences of excess utilization of underground water?
C) Which water resource should be utilized for agriculture?
D) What are the alternative ways to increase the underground water resources?

Answer:
A) Ground water.
B) Underground water table will be depleted and scarcity of drinking water will arise.
C)

1. Tanks should be constructed to harvest with rain water.
2. Projects should be constructed across the rivers to store water that can be utilized for agriculture.

D)

1. Construction of rain water storage structures on large scale.
2. Constructing soaking or percolation pits.
3. Contour field bunding.
4. Recharge of wells by building dykes or barriers in the nalla.
5. Plantation in waste lands.
6. Adapting micro irrigation techniques.
   (Any two points you can write)

Question 5.
Forests are renewable resource. Write four sentences supporting this.
(OR)
“Forest is a renewable resource”. Do you agree? Justify.
Answer:

1. Forests are rich habitat for plants and animals. Forests serve as lungs for the world and a bed of nutrients for new fife to prosper.
2. Forest’s pure air protects the earth from green house effect by removing carbon dioxide from the atmosphere and converts it into oxygen.
3. Many fruits, medicines, dyes, sandle wood and bamboo is obtained from forest by local people.
4. Forest provide employment to large number of people and also help in generating revenue.

Question 6.
Observe the above table and answer the following questions.

1. Which crop is most suitable to cultivate for small farmer in both the villages?
Answer:
Cotton, paddy

2. If you are a large farmer, which crop do you select to cultivate?
Answer:
Cotton, paddy, mirchi

3. What similarities you have identified in village A and village B?
Answer:
Small and large fanners cultivated same type of crop in both villages. Large farmer gets more income per acre on crops than small farmer in both the villages.

4. Which is the lowest income crop ?
Answer:
Mirchi.

5. Is there any relationship between production of crops and income ? How ?
Answer:
Commercial crops are good for income. Income may or may not related to production of crop. It depends upon demand of the market.

Question 7.
Read the given information and answer the following questions.

A survey was conducted in two villages, Vanaparthy and Vaddicherla of Warangal district in Telangana State. The first with no scarcity and the second with scarce groundwater. Well census was carried out in the villages in order to get a complete picture of well irrigation and its status as well as availability of water. There are no alternative sources of supply as against wells in Vaddicherla, where there is an existing tank that has been converted into a percolation tank, so that the water situation is much better in Vanaparthy.

i) Why did they conduct survey?
Answer:
A compartive study on available water resources irigation method in the Vaddicharla and Wanaparthi of Warangal Dist of Telangana State.

ii) What are irrigation resources in Telangana State?
Answer:
Lakes, wells, canals and ground water etc…,

iii) In which village, do you suggest drip irrigation?
Answer:
Vaddicherla.

iv) Why the water situation is much better in Vanaparthy village compared to Vaddicherla?
Answer:
Existing tank has been converted into a percolation tank.

Question 8.
Observe the Pie diagram and answer the following questions.
i) Identify the fossil fuels from the above diagram.
Answer:
Coal, natural gas, oils are fossil fuels.

ii) Why wastes should be considered as primary energy source in future?
Answer:
The fossil fuels may be exhausted in future. So we may be considered that wastes are primary alternative energy resources.

iii) Why can’t we depend on fossil fuels forever?
Answer:
We can’t depend on fossil fuels forever because fossil fuels are non-renewable resources.

iv) What are the alternatives for fossil fuels?
Answer:
Solar energy, wind energy, tide energy, nuclear energy, energy from waste materials.

Question 9.
Explain the importance and implementation of community based interventions and farmer based interventions for water management.
Answer:
Community based interventions:

1. For water harvesting, there is an urgent need to construct earthen and masonry dams. They help us to store rain water during rainy seasons. They are help in increasing the ground water table.
2. Construction of percolation pits and field bunding are very helpful in the harvesting every rain drop.
3. Open dry wells near nalla canal were recharged by building dykes or barriers in the nalla and maintaining the run – off rain water. The ground water is recharged by these community based interventions.
   Farmer based interventions:
4. Broad Bed Furrow (BBF) land form and contour planting methods are very useful to conserve soil, water and fertilizer application and weeding operations.
5. Planting Gliricidia, a leguminous plant adapted to grow in dry areas on field bunds to strengthen them and make the soil nitrogen rich.
6. Farmers were encouraged to use water resource jointly and irrigate land using micro irrigation methods like sprinklers and drip irrigation.

Question 10.
Explain the farmer based and community based interventions to conserve soil and water resources.
Answer:

Question 11.
“The humans who were developed by using the natural resources, today has become the reason for destroying them”. Explain analytically.
Answer:
“The humans who were developed by using the natural resources, today has become the reason for destroying them” – This statement is absolutely true.

1. Primitive man lived in forests and hills. He used the natural resources for his livelyhood. He worshipped nature and used them wisely for his development.
2. After his development, he becomes greedy and using the natural resources indiscriminately and held responsible for their destruction.
3. To meet the needs of growing population, industrialization, urbanization, and huge constructive activities, man utilised natural resources Indiscriminately. At the same time, he did not planned for their revival.
4. But now he realised the importance of natural resources and taken up steps for their conservation. The concept of “Sustainable development” is being implemented in natural resource management.
5. He focussed on development of alternatives for fossil fuels, conservation of water and soil at community level and farmer based interventions.
6. Now he is so keen on conserving forests, wild life and biodiversity.
7. He is so cautious in minimising the utilization of natural resources by following 4’R principle in the day to day life [R – Reduce, R – Reuse, R – Recycle, R – Recover]
8. Now, he is adopting micro-irrigation methods like sprinklers and drip Irrigation to minimise the water usage in low water available areas.
9. He is very interested in following eco-friendly techniques, natural farming methods, using biofertilizers, vermicompost and natural pest control methods in place of toxic chemical pesticides.

Question 12.
The wells and tanks in your village become dry. Ground water levels decreased. Assume the causes for this. Will there be no water scarcity if all the farmers of your village work collectively?
Answer:
Causes for decreasing ground water levels:

1. Varying monsoon behaviour in recent years, there is a pressure on ground water utilization.
2. Indiscriminate tapping of ground water in our village by too much drilling and construction of deep tube wells and bore wells have resulted in over exploitation and depletion of ground water resources.
3. There will be no water scarcity if all the farmers of our village work collectively. Farmers in our village were encouraged to use water resource jointly and irrigate land using micro irrigation techniques. By using micro irrigation techniques farmers in our village obtained more crop yield. Farmers in our village follow the micro irrigation method i.e. drip irrigation and can reduce water consumption by 70% in our village.

Question 13.
Whom do you meet to collect the information of the methods of farmer based, community based water management? Prepare information table to note down your observation.
Answer:
I will meet officials of International Crop Research Institute for Semi – Arid Tropics (ICRISAT) located at Hyderabad to collect information of the methods of farmer based and community based water management.
I also collect information from Central Research Institute for Dry Land Agriculture (CRIDA), National Remote Sensing Agency (NRSA), District Water Management Agency (DWMA) and M Venkatarangaiah Foundation (MVF) and NGO.
The information I gathered from these institutions is summarised below.
Information table:
For Information table See Q.No. 10 in 4 Marks.

Question 14.
Think that there is much scarcity of water for drinking and cultivation in your village. What advice do you give to prevent this?
(OR)
How do you overcome the problem of water scarcity in your village?
Answer:

1. Motivate the people to conserve water.
2. I will educate the people to avoid wastage of water whenever possible.
3. Construction of recharge pits in the house, school and in the open areas to increase the underground water level.
4. Planting trees wherever possible in the village particularly in the vacant lands.
5. Educate the farmers about the micro irrigation system like drip irrigation, sprin¬klers, etc.
6. Encourage the farmers to form groups to share available water among themselves.
7. Construction of percolation tanks in the low lying areas of the village.

Question 15.
What type of fossil fuels are used in your house? What measures do you take to conserve them?
Answer:
Fossil fuels are sources of energy for cooking, heating and burning in our households. Petrol and diesel are being used in our house for transport and running generators and water pumps.
Measures to be taken to conserve fossil fuels in my house :

1. I will put the food material to be cooked on the stove only after arranging all the things which are necessary for cooking.
2. By using pressure cookers 20% gas on rice and 41.5% on meat would be saved when compared to Other cooking means.
3. We must reduce the flame as soon as the boiling process starts in a pressure cooker. This process saves nearly 35% of fuel.
4. I will soak the food material before cooking. It saves 22% of fuel.
5. I will cook food in broad and low depth vessel.
6. I will keep lid on the cooking vessel. If not, it takes more time to cook.
7. For short distances to travel I will go by walk to save fuel for longer distance. I use public transport.
8. Encourage people to use solar water heater and solar cooker.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.3

10th Class Maths 3rd Lesson Polynomials Ex 3.3 Textbook Questions and Answers

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
i) x² – 2x – 8
ii) 4s² – 4s + 1
iii) 6x² – 3 – 7x
iv) 4u² + 8u
v) t² – 15
vi) 3x² – x – 4
Answer:
i) Given polynomial is x² – 2x – 8
We have x² – 2x – 8 = x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x – 4) (x + 2)
So, the value of x² – 2x – 8 is zero
when x – 4 = 0 or x + 2 = 0 i.e.,
when x = 4 or x = -2
So, the zeroes of x² – 2x – 8 are 4 and -2.
Sum of the zeroes = 4 – 2 = 2 Coefficient of ,x -(-2)
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-2)}{1}\) = 2
And product of the zeroes = 4 × (-2) = -8
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-8}{1}\) = -8

ii) Given polynomial is 4s² – 4s + 1
We have, 4s² – 4s + 1
= 4s² – 2s – 2s + 1
= 2s (2s – 1) – 1(2s – 1)
= (2s – 1) (2s – 1)
= (2s – 1)²
So, the value of 4s2 – 4s + 1 is zero
when 2s-1 = 0 or s = \(\frac{1}{2}\)
∴ Zeroes of the polynomial are \(\frac{1}{2}\) and \(\frac{1}{2}\)
∴ Sum of the zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1.
= – \(\frac{\text { Coefficient of } s}{\text { Coefficient of } s^{2}}\) = –\(\frac{-4}{4}\) = 1
And product of the zeroes = \(\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)\) = \(\frac{1}{4}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{1}{4}\)

iii) Given polynomial is 6x² – 3 – 7x
We have, 6x² – 3 – 7x = 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3) (3x + 1)
The value of 6x² – 3 – 7x is zero, when the value of (3x +1) (2x – 3) is 0
i.e., when 3x + 1 = 0 and 2x – 3 = 0
3x = -1 and 2x = 3
x = \(\frac{-1}{3}\) and x = \(\frac{3}{2}\)
∴ The zeroes of 6x² – 3 – 7x = \(\frac{-1}{3}\) and \(\frac{3}{2}\)
∴ Sum of the zeroes = \(\frac{1}{3}\) + \(\frac{3}{2}\) = \(\frac{7}{6}\).
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-7)}{6}\) = \(\frac{7}{6}\)
And product of the zeroes = \(\left(\frac{-1}{3}\right) \times\left(\frac{3}{2}\right)\) = \(\frac{-1}{2}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-3}{6}\) = \(\frac{-1}{2}\)

iv) Given polynomial is 4u² + 8u
We have, 4u² + 8u = 4u (u + 2)
The value of 4u² + 8u is 0,
when the value of 4u(u + 2) = 0, i.e.,
when u = 0 or u + 2 = 0, i.e.,
when u = 0 (or) u = – 2
∴ The zeroes of 4u² + 8u are 0 and – 2.
Therefore, sum of the zeroes = 0 + (-2) = -2
= – \(\frac{\text { Coefficient of } u}{\text { Coefficient of } u^{2}}\) = \(\frac{-8}{4}\) = -2
And product of the zeroes 0 . (-2) = 0
= \(\frac{\text { Constant term }}{\text { Coefficient of } u^{2}}\) = \(\frac{0}{4}\) = 0

v) Given polynomial is t² – 15.
We have, t² – 15 = (t – √15 ) (t + √l5)
The value of t² – 15 is 0,
when the value of (t – √15 ) (t + √l5) = 0, i.e.,
when t – √15 = 0 or t + √15 = 0, i.e.,
when t = √15 (or) t = -√15
∴ The zeroes of t² – 15 are √15 and -√15.
Therefore, sum of the zeroes = √15 + (-√15) = 0
= – \(\frac{\text { Coefficient of } t}{\text { Coefficient of } t^{2}}\) = –\(\frac{0}{1}\) = 0
And product of the zeroes √15 × (-√15) = -15
= \(\frac{\text { Constant term }}{\text { Coefficient of } t^{2}}\) = \(\frac{-15}{1}\) = -15

vi) Given polynomial is 3x² – x – 4
we have, 3x² – x – 4 = 3x² + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1) (3x – 4)
The value of 3x² – x – 4 is 0 when the value of (x + 1) (3x – 4) is 0.
i.e., when x + 1 = 0 or 3x – 4 = 0
i.e., when x = -1 or x = \(\frac{4}{3}\)
∴ The zeroes of 3x² – x – 4 are -1 and \(\frac{4}{3}\)
Therefore, sum of the zeroes = -1 + \(\frac{4}{3}\) = \(\frac{-3+4}{3}\) = \(\frac{1}{3}\)
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-1)}{3}\) = \(\frac{1}{3}\)
And product of the zeroes -1 × \(\frac{4}{3}\) = \(\frac{-4}{3}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-4}{3}\)

Question 2.
Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.
i) \(\frac{1}{4}\), -1
ii) √2, \(\frac{1}{3}\)
iii) 0, √5
iv) 1, 1
v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
vi) 4, 1
Answer:
Let the polynomial be ax² + bx + c
and its zeroes be α and β.
i) Here, α + β = \(\frac{1}{4}\) and αβ = -1
Thus, the polynomial formed = x² – (sum of the zeroes)x + product of the zeroes
= x² – (\(\frac{1}{4}\))x – 1
= x² – \(\frac{x}{4}\) – 1
The other polynomials are (x² – \(\frac{x}{4}\) – 1)
then the polynomial is 4x² – x – 4.

ii) Here, α + β = √2 and αβ = \(\frac{1}{3}\)
Thus, the polynomial formed = x² – (sum of the zeroes)x + product of the zeroes
= x² – (√2)x + \(\frac{1}{3}\)
= x² – √2x + \(\frac{1}{3}\)
The other polynomials are (x² – √2x + \(\frac{1}{3}\))
then the polynomial is 3x² – 3√2x + 1.

iii) Here, α + β = 0 and αβ = √5
Thus, the polynomial formed = x² – (sum of the zeroes)x + product of the zeroes
= x² – (0)x + √5
= x² + √5

iv) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then α + β = 1 = \(\frac{-(-1)}{1}\) = \(\frac{-b}{a}\) and
αβ = 1 = latex]\frac{1}{1}[/latex] = \(\frac{c}{a}\)
If a = 4, then b = 1 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is 4x² + x + 1.

v) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then α + β = \(\frac{-1}{4}\) = \(\frac{-b}{a}\) and
αβ = \(\frac{1}{4}\) = \(\frac{c}{a}\)
If a = 4, then b = 1 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is 4x² + x + 1.

vi) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then α + β = 4 = \(\frac{-(-4)}{1}\) = \(\frac{-b}{a}\) and
αβ = 1 = \(\frac{1}{1}\) = \(\frac{c}{a}\)
If a = 1, then b = -4 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is x² – 4x + 1.

Question 3.
Find the quadratic polynomial, for the zeroes α, β given in each case.
i) 2, -1
ii) √3, -√3
iii) \(\frac{1}{4}\), -1
iv) \(\frac{1}{2}\), \(\frac{3}{2}\)
Answer:
i) Let the polynomial be ax² + bx + c, a ≠ 0 and its zeroes be α and β.
Here α = 2 and β = – 1
Sum of the zeroes = α + β = 2 + (-l) = 1
Product of the zeroes = αβ = 2 × (-1) = -2
Therefore the quadratic polynomial ax² + bx + c is x² – (α + β)x + αβ = [x² – x – 2]
the quadratic polynomial will be x² – x – 2.

ii) Let the zeroes be α = √3 and β = -√3
Sum of the zeroes = α + β
= √3 + (-√3) = 0
Product of the zeroes = αβ
= √3 × (-√3) = -3
∴ The quadratic polynomial
ax² + bx + c is [x² – (α + β)x + αβ]
= [x² – 0.x + (-3)] = [x² – 3]
the quadratic polynomial will be x² – 3.

iii) Let the zeroes be α = \(\frac{1}{4}\) and β = -1
Sum of the zeroes = α + β
= \(\frac{1}{4}\) + (-1) = \(\frac{1+(-4)}{4}\) = \(\frac{-3}{4}\)
Product of the zeroes = αβ
= \(\frac{1}{4}\) × (-1) = \(\frac{-1}{4}\)
∴ The quadratic polynomial
ax² + bx + c is [x² – (α + β)x + αβ]
= [x² – \(\left(\frac{-3}{4}\right)\).x + (\(\frac{-1}{4}\))]
the quadratic polynomial will be 4x² + 3x – 1.

iv) Let the zeroes be α = \(\frac{1}{2}\) and β = \(\frac{3}{2}\)
Sum of the zeroes = α + β
= \(\frac{1}{2}\) + \(\frac{3}{2}\) = \(\frac{1+3}{2}\) = \(\frac{4}{2}\) = 2
Product of the zeroes = αβ
= \(\frac{1}{2}\) × \(\frac{3}{2}\) = \(\frac{3}{4}\)
∴ The quadratic polynomial
ax² + bx + c is [x² – (α + β)x + αβ]
= [x² – 2x + (\(\frac{3}{4}\))]
the quadratic polynomial will be 4x² – 8x + 3.

Question 4.
Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x³ + 3x² – x – 3 and check the relationship between zeroes and the coefficients.
Answer:
Given cubic polynomial
p(x) = x³ + 3x² – x – 3
Comparing the given polynomial with ax³ + bx² + cx + d, we get a = 1, b = 3, c = -1, d = -3
Futher given zeroes are 1,-1 and – 3
p(1) = (1)³ + 3(1)² – 1 – 3
= 1 + 3 – 1 – 3 = 0
p(-1) = (-1)³ + 3(-1)² – 1 – 3
= -1 + 3 + 1 – 3 = 0
p(-3) = (-3)³ + 3(-3)² – (-3) – 3
= -27 + 27 + 3 – 3 = 0
Therefore, 1, -1 and -3 are the zeroes of x³ + 3x² – x – 3.
So, we take α = 1, β = -1 and γ = -3 Now,
α + β + γ = 1 + (-1) + (-3) = -3
αβ + βγ + γα = 1(-l) + (-1) (-3) + (-3)1
= -1 + 3 – 3 = -1
= \(\frac{c}{a}\) = \(\frac{-1}{1}\) = -1
αβγ = 1 (-1) (-3) = 3 = \(\frac{-d}{a}\) = \(\frac{-(-3)}{1}\) = 3

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.1

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.1 Textbook Questions and Answers

Question 1.
Find the distance between the following pair of points,
(i) (2, 3) and (4, 1)
Answer:
Distance = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{(4-2)^{2}+(1-3)^{2}}\)
= \(\sqrt{4+4}\)
= √8 = 2√2 units

ii) (- 5, 7) and (-1, 3)
Answer:
Distance = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{(-1+5)^{2}+(3-7)^{2}}\)
= \(\sqrt{4^{2}+(-4)^{2}}\)
= \(\sqrt{16+16}\)
= √32 = 4√2 units

iii) (- 2, -3) and (3, 2)
Answer:

iv) (a, b) and (- a, – b)
Answer:

Question 2.
Find the distance between the points (0, 0) and (36, 15).
Answer:
Given: Origin O (0, 0) and a point P (36, 15).
Distance between any point and origin = \(\sqrt{x^{2}+y^{2}}\)
∴ Distance = \(\sqrt{36^{2}+15^{2}}\)
= \(\sqrt{1296+225}\)
= \(\sqrt{1521}\)
= 39 units

∴ 1521 = 3² × 13²
\(\sqrt{1521}\) = 3 × 13 = 39

Question 3.
Verify that the points (1, 5), (2, 3) and (-2, -1) are collinear or not.
Answer:
Given: A (1, 5), B (2, 3) and C (- 2, – 1)

Here the sum of no two segments is equal to third segment.
Hence the points are not collinear.
!! Slope of AB, m₁ = \(\frac{3-5}{2-1}\) = -2
Slope of BC, m₂ = \(\frac{-1-3}{-2-2}\) = 1
m₁ ≠ m₂
Hence A, B, C are not collinear.

Question 4.
Check whether (5, -2), (6, 4) and (7,-2) are the vertices of an isosceles triangle.
Answer:
Let A = (5, – 2); B = (6, 4) and C = (7, – 2).

Now we have, AB = BC.
∴ △ABC is an isosceles triangle,
i.e., given points are the vertices of an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in figure. Jarina and Phani walk into the class and after observing for a few minutes Jarina asks Phani “Don’t you think ABCD is a square?” Phani disagrees. Using distance formula, find which of them is correct. Why?

Answer:
Given: Four friends are seated at A, B, C and D where A (3, 4), B (6, 7), C (9, 4) and D (6, 1).
Now distance

BD = \(\sqrt{(6-6)^{2}+(1-7)^{2}}\) = √36 = 6
Hence in □ ABCD four sides are equal
i.e., AB = BC = CD = DA
= 3√2 units
and two diagonals are equal.
i.e., AC = BD = 6 units.
∴ □ ABCD forms a square.
i.e., Jarina is correct.

Question 6.
Show that the following points form an equilateral triangle A(a, 0), B(- a, 0), C(0, a√3).
Answer:
Given: A (a, 0), B (- a, 0), C (0, a√3).

Now, AB = BC = CA.
∴ △ABC is an equilateral triangle.

Question 7.
Prove that the points (-7, -3), (5, 10), (15, 8) and (3, -5) taken in order are the corners of a parallelogram.
Answer:
To show that the given points form a parallelogram.
We have to show that the mid points of each diagonal are same. Since diagonals of a parallelogram bisect each other.
Now let A(-7, -3), B(5, 10), C(15, 8) and D(3, -5)
Then midpoint of diagonal

∴ (1) = (2)
Hence the given are vertices of a parallelogram.

Question 8.
Show that the points (-4, -7), (-1, 2), (8, 5) and (5, -4) taken in order are the vertices of a rhombus. And find its area.
(Hint: Area of rhombus = \(\frac{1}{2}\) × product of its diagonals)
Answer:
Given in ▱ ABCD , A(-4, – 7), B (- 1, 2), C (8, 5) and D (5,-4)


∴  In ▱ ABCD, AB = BC = CD = AD [from sides are equal]
Hence ▱ ABCD is a rhombus.
Area of a rhombus = \(\frac{1}{2}\) d₁d₂
= \(\frac{1}{2}\) × 12√2 × 6√2
= 72 sq. units.

Question 9.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
i) (-1,-2), (1,0), (-1,2), (-3,0)
Answer:
Let A (- 1, -2), B (1, 0), C (- 1, 2), D (- 3, 0) be the given points. Distance formula

In ▱ ABCD, AB = BC = CD = AD – four sides are equal.
AC = BD – diagonals are equal.
Hence, the given points form a square,

ii) (-3, 5), (1, 10), (3, 1), (-1,-4).
Answer:
Let A(-3, 5), B(l,10), C(3, 1), D(-l, -4) then

In ▱ ABCD, \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{CD}}\) and \(\overline{\mathrm{BC}}\) = \(\overline{\mathrm{AD}}\) (i.e., both pairs of opposite sides are equal) and \(\overline{\mathrm{AC}}\) ≠ \(\overline{\mathrm{BD}}\).
Hence ▱ ABCD is a parallelogram,
i.e., The given points form a parallelogram.

iii) (4, 5), (7, 6), (4, 3), (1, 2).
Answer:
Let A (4, 5), B (7, 6), C (4, 3) and D (1, 2) be the given points.
Distance formula

In ▱ ABCD, AB = CD and BC = AD (i.e., both pairs of opposite sides are equal) and AC ≠ BD.
Hence ▱ ABCD is a parallelogram, i.e., The given points form a parallelogram.

Question 10.
Find the point on the X-axis which is equidistant from (2, -5) and (-2,9).
Answer:
Given points, A (2, – 5), B (- 2, 9).
Let P (x, 0) be the point on X – axis which is equidistant from A and B. i.e., PA = PB.
Distance formula = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

But PA = PB.
⇒ \(\sqrt{x^{2}-4 x+29}=\sqrt{x^{2}+4 x+85}\)
Squaring on both sides, we get
x² – 4x + 29 = x² + 4x + 85
⇒ – 4x – 4x = 85 – 29
⇒ – 8x = 56
⇒ x = \(\frac{56}{-8}\) = -7
∴ (x, 0) = (- 7, 0) is the point which is equidistant from the given points.

Question 11.
If the distance between two points (x, 7) and (1, 15) is 10, find the value of x.
Answer:
Formula for distance between two points = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\)
Now distance between (x, 7) and (1,15) is 10.
∴ \(\sqrt{(x-1)^{2}+(7-15)^{2}}\) = 10
∴ (x – l)² + (-8)² = 10²
⇒ (x – l)² = 100 – 64 = 36
∴ x – 1 = √36 = ± 6
∴ x – 1 = 6 or x – 1 = -6
⇒ x = 6 + 1 = 7 or x = -6 + 1 = -5
∴ x = 7 or x = – 5

Question 12.
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Answer:
Given: P (2, – 3), Q (10, y) and
\(\overline{\mathrm{PQ}}\) = 10.
Distance formula = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

⇒ y² + 6y + 73 = 100
⇒ y² + 6y – 27 = 0
⇒ y² + 9y – 3y – 27 = 0
⇒ y (y + 9) – 3 (y + 9) = 0
⇒ (y + 9) (y – 3) = 0
⇒ y + 9 = 0 or y – 3 = 0
⇒ y = -9 or y = 3
⇒ y = – 9 or 3.

Question 13.
Find the radius of the circle whose centre is (3, 2) and passes through (-5,6).
Answer:

Given: A circle with centre A (3, 2) passing through B (- 5, 6).
Radius = AB
[∵ Distance of a point from the centre of the circle]
Distance formula = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

Question 14.
Can you draw a triangle with vertices (1, 5), (5, 8) and (13, 14)? Give reason.
Answer:
Let A (1, 5), B (5, 8) and C (13, 14) be the given points.
Distance formula

Here, AC = AB + BC.
∴ △ABC can’t be formed with the given vertices.
[∵ Sum of the any two sides of a triangle must be greater than the third side].

Question 15.
Find a relation between x and y such that the point (x, y) is equidistant from the points (-2, 8) and (-3, -5).
Answer:
Let A (- 2, 8), B (- 3, – 5) and P (x, y). If P is equidistant from A, B, then PA = PB.
Distance formula =


Squaring on both sides we get, x² + y² + 4x – 16y + 68
= x² + y² + 6x +10y + 34
⇒ 4x – 16y – 6x – 10y = 34-68
⇒ – 2x – 26y = -34
⇒ x + 13y = 17 is the required condition.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.2

10th Class Maths 6th Lesson Progressions Ex 6.2 Textbook Questions and Answers

Question 1.
Fill in the blanks in the following table, given that ‘a’ is the first term, d the common difference and an the nth term of the A.P:

Answer:

Question 2.
Find the i) 30th term of the A.P.: 10, 7, 4,……
ii) 11th term of the A.P.: -3, –\(\frac{1}{2}\), 2,…
Answer:
i) Given A.P. = 10, 7, 4, …….
a₁ = 10; d = a₂ – a₁ = 7 – 10 = – 3
a_n = a + (n – 1) d
a₃₀ = 10 + (30 – 1) (- 3) = 10 + 29 × (- 3) = 10 – 87 = – 77

ii) Given A.P. = – 3, –\(\frac{1}{2}\), 2,…
a₁ = -3; d = a₂ – a₁ = –\(\frac{1}{2}\) – (-3) = – 3
= –\(\frac{1}{2}\) + 3
= \(\frac{-1+6}{2}\)
= \(\frac{5}{2}\)
a_n = a + (n – 1) d
= -3 + (11-1) × \(\frac{5}{2}\)
= -3 + 10 × \(\frac{5}{2}\)
= -3 + 5 × 5
= -3 + 25
= 22

Question 3.
Find the respective terms for the following APs.
i) a₁ = 2; a₃ = 26, find a₂.
Answer:
Given: a₁ = a = 2 …….. (1)
a₃ = a + 2d = 26 …….. (2
Equation (2) – equation (1)
⇒ (a + 2d) – a = 26 – 2
⇒ 2d = 24
d = \(\frac{24}{2}\) = 12
Now a₂ = a + d = 2 + 12 = 14

ii) a₂ = 13; a₄ = 3, find a₁, a₃.
Answer:
Given: a₂ = a + d = 13 ….. (1)
a₄ = a + 3d = 3 ….. (2)
Solving equations (1) and (2);

∴ Substituting d = – 5 in equation (1) we get
a + (-5) = 13
∴ a = 13 + 5 = 18 i.e., a₁ = 18
a₃ = a + 2d = 18 + 2(- 5)
= 18 – 10 = 8

iii) a₁ = 5; a₄ = 9\(\frac{1}{2}\), find a₂, a₃.
Answer:
Given: a₁ = a = 5 ….. (1)
a₄ = a + 3d = 9\(\frac{1}{2}\) ….. (2)
Solving equations (1) and (2);

⇒ 3d = 4\(\frac{1}{2}\)
⇒ 3d = \(\frac{9}{2}\)
⇒ d = \(\frac{9}{2 \times 3}\) = \(\frac{3}{2}\)
∴ a₂ = a + d = 5 + \(\frac{3}{2}\) = \(\frac{13}{2}\)
a₃ = a + 2d = 5 + 2 × \(\frac{3}{2}\) = 5 + 3 = 8

iv) a₁ = -4; a₆ = 6, find a₂, a₃, a₄, a₅.
Answer:
Given: a₁ = a = -4 ….. (1)
a₆ = a + 5d = 6 ….. (2)
Solving equations (1) and (2);
(-4) + 5d = 6
⇒ 5d = 6 + 4
⇒ 5d = 10
⇒ d = \(\frac{10}{5}\)
Now
∴ a₂ = a + d = -4 + 2 = -2
a₃ = a + 2d = -4 + 2 × 2 = -4 + 4 = 0
a₄ = a + 3d = -4 + 3 × 2 = -4 + 6 = 2
a₅ = a + 4d = -4 + 4 × 2 = -4 + 8 = 4

v) a₂ = 38; a₆ = -22, find a₁, a₃, a₄, a₅.
Answer:
Given: a₂ = a + d = 38 ….. (1)
a₆ = a + 5d = -22 ….. (2)
Subtracting (2) from (1) we get

Now substituting, d = – 15 in equation (1), we get
a + (- 15) = 38 ⇒ a = 38 + 15 = 53
Thus,
a₁ = a = 53;
a₃ = a + 2d = 53 + 2 × (- 15) = 53 – 30 = 23;
a₄ = a + 3d = 53 + 3 × (- 15) = 53 – 45 = 8;
a₅ = a + 4d = 53 + 4 × (- 15) = 53 – 60 = – 7

Question 4.
Which term of the AP:
3, 8, 13, 18,…, is 78?
Answer:
Given: 3, 8, 13, 18, ……
Here a = 3; d = a₂ – a₁ = 8 – 3 = 5
Let ‘78’ be the nth term of the given A.P.
∴ a_n = a + (n – 1) d
⇒ 78 = 3 + (n – 1) 5
⇒ 78 = 3 + 5n – 5
⇒ 5n = 78 + 2
⇒ n = \(\frac{80}{2}\) = 16
∴ 78 is the 16th term of the given A.P.

Question 5.
Find the number of terms in each of the following APs:
i) 7, 13, 19, ….., 205
Answer:
Given: A.P: 7, 13, 19, ……….
Here a₁ = a = 7; d = a₂ – a₁ = 13 – 7 = 6
Let 205 be the n^th term of the given A.P.
Then, a_n = a + (n – 1) d
205 = 7 + (n- 1)6
⇒ 205 = 7 + 6n – 6
⇒ 205 = 6n + 1
⇒ 6n = 205 – 1 = 204
∴ n = \(\frac{204}{6}\) = 34
∴ 34 terms are there.

ii) 18, 15\(\frac{1}{2}\), 13, …, -47
Answer:
Given: A.P: 18, 15\(\frac{1}{2}\), 13, …….
Here a₁ = a = 18;
d = a₂ – a₁ = 15\(\frac{1}{2}\) – 18 = -2\(\frac{1}{2}\) = –\(\frac{5}{2}\)
Let ‘-47’ be the nth term of the given A.P.
a_n = a + (n – 1) d

⇒ -94 = 36 – 5n + 5
⇒ 5n = 94 + 41
⇒ n = \(\frac{135}{5}\) = 27
∴ 27 terms are there.

Question 6.
Check whether, -150 is a term of the AP: 11, 8, 5, 2…
Answer:
Given: A.P. = 11, 8, 5, 2…
Here a₁ = a = 11;
d = a₂ – a₁ = 8 – 11 = -3
If possible, take – 150 as the n^th term of the given A.P.
a_n = a + (n – 1) d
⇒ -150 = 11 + (n – 1) × (-3)
⇒ -150 = 11 – 3n + 3
⇒ 14 – 3n = – 150
⇒ 3n= 14 + 150 = 164
∴ n = \(\frac{164}{3}\) = 54\(\frac{2}{3}\)
Here n is not an integer.
∴ -150 is not a term of the given A.P.

Question 7.
Find the 31^st term of an A.P. whose 11^th term is 38 and the 16^th term is 73.
Answer:
Given: An A.P. whose

⇒ -5d = -35
⇒ d = \(\frac{-35}{-5}\) = 7
Substituting d = 7 in the equation (1)
we get,
a + 10 x 7 = 38
⇒ a + 70 = 38
⇒ a = 38 – 70 = -32
Now, the 31^st term = a + 30d
= (-32) + 30 × 7
= -32 + 210 = 178

Question 8.
If the 3rd and the 9^th terms of an A.P are 4 and -8 respectively, which term of this A.P is zero?
Answer:
Given: An A.P. whose

Substituting d = -2 in equation (1) we get
a + 2 × (-2) = 4
⇒ a – 4 = 4
⇒ a = 4 + 4 = 8
Let n^th term of the given A.P be equal to zero.
an = a + (n – 1)d
⇒ 0 = 8 + (n – 1) × (-2)
⇒ 0 = 8 – 2n + 2
⇒ 10 – 2n = 0
⇒ 2n = 10 and n = \(\frac{10}{2}\) = 5
∴ The 5^th term of the given A.P is zero.

Question 9.
The 17^th term of an A.P exceeds its 10 term by 7. Find the common difference.
Answer:
Given an A.P in which a₁₇ = a₁₀ + 7
⇒ a₁₇ – a₁₀ = 7
We know that a_n = a + (n – 1)d

⇒ d = \(\frac{7}{7}\) = 1

Question 10.
Two APs have the same common difference. The difference between their 100^th terms is 100, what is the difference between their 1000^th terms?
Answer:
Let the first A.P be:
a, a + d, a + 2d, ……..
Second A.P be:
b, b + d, b + 2d, b + 3d, ………
Also, general term, a_n = a + (n – 1)d
Given that, a₁₀₀ – b₁₀₀ = 100
⇒ a + 99d – (b + 99d) = 100
⇒ a – b = 100
Now the difference between their 1000^th terms,

∴ The difference between their 1000^th terms is (a – b) = 100.
Note: If the common difference for any two A.Ps are equal then difference between n^th terms of two A.Ps is same for all natural values of n.

Question 11.
How many three-digit numbers are divisible by 7?
Answer:
The least three digit number is 100.

∴ The least 3 digit number divisible by 7 is 100 + (7 – 2) = 105
The greatest 3 digit number is 999

∴ The greatest 3 digit number divisible by 7 is 999 – 5 = 994.
∴ 3 digit numbers divisible by 7 are
105, 112, 119,….., 994.
a₁ = a = 105; d = 7; a_n = 994
a_n = a + (n – 1)d
⇒ 994 = 105 + (n – 1)7
⇒ (n – 1)7 = 994 – 105
⇒ (n – 1)7 = 889
⇒ n – 1 = \(\frac{889}{7}\) = 127
∴ n = 127 + 1 = 128
∴ There are 128, 3 digit numbers which are divisible by 7.
(or)
\(\frac{\text { last number – first number }}{7}\)
\(\frac{999-100}{7}\)
≃ 128.4 = 128 numbers divisible by 7.

Question 12.
How many multiples of 4 lie between 10 and 250?
Answer:
Given numbers: 10 to 250

∴ Multiples of 4 between 10 and 250 are
First term: 10 + (4 – 2) = 12
Last term: 250 – 2 = 248
∴ 12, 16, 20, 24, ….., 248
a = a₁ = 12; d = 4; a_n = 248
a_n = a + (n – 1)d
248 = 12 + (n – 1) × 4
⇒ (n – 1)4 = 248 – 12
⇒ n – 1 = \(\frac{236}{4}\) = 59
∴ n = 59 + 1 = 60
There are 60 numbers between 10 and 250 which are divisible by 4.

Question 13.
For what value of n, are the n^th terms of two APs: 63, 65, 67, ….. and 3, 10, 17,… equal?
Answer:
Given : The first A.P. is 63, 65, 67, ……
where a = 63, d = a₂ – a₁,
⇒ d = 65 – 63 = 2
and the second A.P. is 3, 10, 17, …….
where a = 3; d = a₂ – a₁ = 10 – 3 = 7
Suppose the nth terms of the two A.Ps are equal, where a_n = a + (n – 1)d
⇒ 63 + (n – 1)2 = 3 + (n – 1)7
⇒ 63 + 2n – 2 = 3 + 7n – 7
⇒ 61 + 2n = 7n – 4
⇒ 7n – 2n = 61 + 4
⇒ 5n = 65
⇒ n = \(\frac{65}{5}\) = 13
∴ 13^th terms of the two A.Ps are equal.

Question 14.
Determine the AP whose third term is 16 and the 7^th term exceeds the 5^th term by 12.
Answer:
Given : An A.P in which
a₃ = a + 2d = 16 …… (1)
and a₇ = a₅ + 12
i.e., a + 6d = a + 4d + 12
⇒ 6d – 4d = 12
⇒ 2d = 12
⇒ d = \(\frac{12}{2}\) = 6
Substituting d = 6 in equation (1) we get
a + 2 × 6 = 16
⇒ a = 16 – 12 = 4
∴ The series/A.P is
a, a + d, a + 2d, a + 3d, …….
⇒ 4, 4 + 6, 4 + 12, 4 + 18, ……
⇒ A.P.: 4, 10, 16, 22, …….

Question 15.
Find the 20^th term from the end of the AP: 3, 8, 13,…, 253.
Answer:
Given: An A.P: 3, 8, 13, …… , 253
Here a = a₁ = 3
d = a₂ – a₁ = 8 – 3 = 5
a_n = 253, where 253 is the last term
a_n = a + (n – l)d
∴ 253 = 3 + (n – 1)5
⇒ 253 = 3 + 5n – 5
⇒ 5n = 253 + 2
⇒ n = \(\frac{255}{5}\) = 51
∴ The 20^th term from the other end would be
1 + (51 – 20) = 31 + 1 = 32
∴ a₃₂ = 3 + (32 – 1) × 5
= 3 + 31 × 5
= 3 + 155 = 158

Question 16.
The sum of the 4^th and 8^th terms of an AP is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the AP.
Answer:
Given an A.P in which a₄ + a₈ = 24
⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 ……. (1)
and a₆ + a₁₀ = 44
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 ……. (2)
Also a + 5d = 12
⇒ a + 5(5) = 12
⇒ a + 25 = 12
⇒ a = 12 – 25 = -13
∴ The A.P is a, a + d, a + 2d, ……
i.e., – 13, (- 13 + 5), (-13 + 2 × 5)…
⇒ -13, -8, -3, …….

Question 17.
Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000?
Answer:
Given: Salary of Subba Rao in 1995 = Rs. 5000
Annual increment = Rs. 200
i.e., His salary increases by Rs. 200 every year.

Clearly 5000, 5200, 5400, forms an A.P in which a = 5000 and d = 200.
Now suppose that his salary reached Rs. 7000 after x – years.
i.e., a_n = 7000
But, a_n = a + (n – 1)d
7000 = 5000 + (n – 1)200
⇒ 7000 – 5000 = (n – 1)200
⇒ n – 1 = \(\frac{2000}{200}\) = 10
⇒ n = 10 + 1
∴ In 11^th year his salary reached Rs. 7000.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.3

10th Class Maths 5th Lesson Quadratic Equations Ex 5.3 Textbook Questions and Answers

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
i) 2x² + x – 4 = 0
Answer:
Given: 2x² + x – 4 = 0
⇒ 2x² + x = 4
⇒ (√2x)² + x = 4
⇒ (√2x)² + 2.√2.x.\(\frac{1}{2 \sqrt{2}}\) = 4
Now LHS is in the form a² + 2ab
where b = \(\frac{1}{2 \sqrt{2}}\)
Adding b² = \(\left(\frac{1}{2 \sqrt{2}}\right)^{2}\) on both sides we get

ii) 4x² + 4√3x + 3 = 0
Answer:
Given: 4x² + 4√3x + 3 = 0
⇒ 4x² + 4√3x = -3
⇒ (2x)² + 2(2x)√3 = -3
LHS is of the form a² + 2ab where
where b = √3.
∴ Adding b² = (√3)² = 3 on both sides, we get
(2x)² + 2(2x)(√3) + (√3)² = -3 + (√3)²
⇒ (2x + √3)² = -3 + 3 = 0
∴ (2x + √3)² = 0
⇒ 2x + √3 = 0
⇒ 2x = -√3
⇒ x = \(\frac{-\sqrt{3}}{2}\)
∴ The roots are \(\frac{-\sqrt{3}}{2}\), \(\frac{-\sqrt{3}}{2}\).

iii) 5x² – 7x – 6 = 0
Given quardratic equation = 5x² – 7x – 6 = 0
∴ 5x² – 7x – 6
⇒ x² – \(\frac{7}{5}\)x = \(\frac{6}{5}\), it can be re-written as
x² – 2.\(\frac{7}{10}\)x = \(\frac{6}{5}\) now it is in the form
of a² – 2ab where a = x, and b = \(\frac{7}{10}\)
Now adding b² = \(\left(\frac{7}{10}\right)^{2}\) on both sides, we get

Note: If we take the Q.E. as 5x² – 7x + 6 = 0, then we get the T.B. answer.

iv) x² + 5 = -6x
Answer:
The given Q.E. is x² + 5 = -6x
⇒ x² + 6x = -5
⇒ (x)² + 2.(x).3 = -5
Now L.H.S. is of the form a² + 2ab where b = 3.
Adding b² = 3² on both sides we get
x² + 2(x)(3) + 3² = -5 + 3²
(x + 3)² = -5 + 9 = 4
∴ x + 3 = 74 = ± 2
⇒ x = +2 – 3 or – 2 – 3
= -1 or -5 are the roots of the given Q.E.

Question 2.
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula,
i) 2x² + x – 4 = 0
Answer:
Comparing this Q.E. with ax² + bx + c = 0
a = 2; b = 1; c = -4
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

ii) 4x² + 4√3x + 3 = 0
Answer:
Given: 4x² + 4√3x + 3 = 0
Here a = 4; b = 4√3 ; c = 3

iii) 5x² – 7x – 6 = 0
Answer:
Given: 5x² – 7x – 6 = 0
Here a = 5; b = -7 and c = -6

iv) x² + 5 = -6x
Answer:
Given: x² + 5 = -6x
⇒ x² + 6x + 5 = 0
Here a = 1; b = 6; c = 5

Question 3.
Find the roots of the following equations:
i) x – \(\frac{1}{x}\) = 3, x ≠ 0
Answer:
Given: x – \(\frac{1}{x}\) = 3
⇒ x² + 6x + 5 = 0
⇒ \(\frac{x^{2}-1}{x}\) = 3
⇒ x² – 1 = 3x
⇒ x² – 3x – 1 = 0
Here a = 1; b = -3; c = -1

ii) \(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\), x ≠ -4, 7
Answer:
Given: \(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\)

⇒ x² – 3x – 28 = -30
⇒ x² – 3x – 28 + 30 = 0
⇒ x² – 3x + 2 = 0
⇒ x² – 2x – x + 2 = 0
⇒ x(x – 2) – 1(x – 2) = 0
⇒ (x – 2) (x – 1) = 0
⇒ x – 2 = 0 (or) x – 1 = 0
⇒ x = 2 or x = 1
⇒ x = 2 or 1.

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Answer:
Let the present age of Rehman be x years.
3 years ago Rehman’s age = x – 3 and its reciprocal is \(\frac{1}{x-3}\)
Rehman’s age 5 years from now = x + 5 and its reciprocal is \(\frac{1}{x+5}\)
The sum of the reciprocals


⇒ x² + 2x – 15 = 3(2x + 2)
⇒ x² + 2x – 15 = 6x + 6
⇒ x² + 2x – 15 – 6x – 6 = 0
⇒ x² – 4x – 21 =0
⇒ x² – 7x + 3x – 21 =0
⇒ x(x – 7) + 3(x – 7) 0
⇒ (x – 7) (x + 3) = 0
⇒ x – 7 = 0 or x + 3 = 0
⇒ x = 7 or x = -3
But x can’t be negative, x = 7
i.e., Present age of Rehman = 7 years.

Question 5.
In a class test, the sum of Moulika’s marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.
Answer:
Sum of the marks in Mathematics and English = 30
Let Moulika’s marks in Mathematics be x Then her marks in English = 30 – x
If she got 2 more marks in Mathematics, then her marks would be x + 2.
If she got 3 marks less in English then her marks would be 30 – x – 3 = 27 – x
By problem (x + 2) (27 – x) = 210
⇒ x(27 – x) + 2(27 – x) = 210
⇒ 27x – x² + 54 – 2x = 210
⇒ -x² + 25x + 54 = 210
⇒ x² – 25x – 54 + 210 = 0
⇒ x² – 25x + 156 = 0
⇒ x² – 12x – 13x + 156 = 0
⇒ x(x – 12) – 13(x 12) = 0
⇒ (x – 12) (x – 13) = 0
⇒ x – 12 = 0 or x – 13 = 0
⇒ x = 12 or x = 13
If x = 12, then marks in Mathematics = 12 English = 30 – 12 = 18
If x = 13, then marks in Mathematics = 13 English = 30 – 13 = 17

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Answer:
Let the shorter side of the rectangular field = x m.
Then its longer side = x + 30 m.

The diagonal of a rectangle is also the hypotenuse of the lower triangle Here the diagonal = x + 60
∴ By Pythagoras Theorem
(side)² + (side)² = (hypotenuse)²
⇒ (x + 30)² + x² = (x + 60)²
⇒ x² + 60x + 900 + x² = x² + 120x + 3600
⇒ x² – 60x – 2700 = 0
⇒ x² – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30 (x – 90) = 0
⇒ (x – 90) (x + 30) = 0
⇒ x – 90 = 0 (or) x + 30 = 0
⇒ x = +90 (or) x = -30 But ‘x’ can’t be negative.
∴ x = 90 m
i.e., the shorter side x = 90 m Longer side x + 30 = 90 + 30 = 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Answer:
Let the large number be x.
8 times larger number = Square of the srnall number = 8x
Square of the larger number = x²
By problem, x² – 8x = 180
⇒ x² – 8x – 180 = 0
⇒ x²– 18x + 10x – 180 = 0
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x + 10)(x – 18) = 0
⇒ x + 10 = 0 (or) x – 18 = 0
⇒ x = -10 (or) x = 18
If x = 18, then larger number =18;
(small number)2 = 8 × (+18) = 144
∴ Small number = √144 = 12
The numbers are 18, 12
Note: Discard x = -10.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Answer:
The distance travelled = 360 km.
Let the speed of the train = x kmph.
Time taken to complete a journey = \(\frac{\text { distance }}{\text { speed }}\)

⇒ x² + 5x = 1800
⇒ x² + 5x – 1800 = 0
⇒ x² + 45x – 40x – 1800 = 0
⇒ x(x + 45) – 40(x + 45) = 0
⇒ (x + 45) (x – 40) = 0
x + 45 = 0 or x -40 = 0
x = -45 or x = +40
But x can’t be negative.
∴ The speed of the train = 40 kmph.

Question 9.
Two water taps together can fill a tank in 9\(\frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answer:
Let the time taken to fill the tank by smaller tap = x (hours)
So the part filled by smaller tap in
1 hour = \(\frac{1}{x}\) × \(\frac{75}{8}\) = \(\frac{75}{8x}\) ……. (1)
Again then the time taken to fill the tank by larger tap = (x – 10) hours
∴ the part of tank that can be filled by larger tap alone in one hour of time = \(\frac{1}{x-10}\)
∴ In \(\frac{75}{8}\) hours the part filled by larger tap = \(\frac{75}{8}\left(\frac{1}{x-10}\right)\)
∴ By both taps together

⇒ 150x – 750 = 8x² – 80x
⇒ 8x² – 80x – 150x + 750 = 0
⇒ 8x² – 230x + 750 = 0
⇒ 4x² – 115x + 375 = 0
⇒ 4x² – 100x – 15x + 375 = 0
⇒ 4x(x – 25) – 15(x – 25) = 0
∴ (4x – 15) (x – 25) = 0 15
⇒ 4x = 15, x = \(\frac{15}{4}\) or x = 25
x = 25 hours.
then time taken to fill by larger tap = x – 10 = 25 – 10 = 15 hours
(x cannot be \(\frac{15}{4}\) since we have considered ‘x’ as time taken by smaller tap, which is to be higher one)

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.
Answer:
Let the speed of the passenger train = x kmph.
Then speed of the express train = x + 11 kmph.
Distance travelled = 132 km
We know that time = \(\frac{\text { distance }}{\text { speed }}\)

⇒ x² + 11x = 13 × 11
⇒ x² + 11x – 1452 = 0
⇒ x² + 44x – 33x – 1452 = 0
⇒ x(x + 44) – 33 (x + 44) = 0
⇒ (x + 44) (x – 33) = 0
⇒ x + 44 = 0 (or) x – 33 = 0
⇒ x = -44 (or) x = 33
But x can’t be negative.
∴ Speed of the passenger train = x = 33 kmph.
Speed of the express train = x + 11 = 44 kmph.

Question 11.
Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24m, find the sides of the two squares.
(OR)
If the sum of the areas of two squares is 468 m² and the difference of their perimeters is 24m, then find the measurements of their sides.
Answer:
Let the side of first square = x m say Then perimeter of the first square = 4x [∵ P = 4 . side]
By problem, perimeter of the second square = 4x + 24 (or) 4x – 24
∴ Side of the second square =

Now sum of the areas of the two squares is given as 468 m²
x² + (x + 6)² = 468
⇒ x² + x² + 12x + 36 = 468
⇒ 2x² + 12x + 36 – 468 = 0
⇒ 2x² + 12x – 432 = 0
⇒ x² + 6x – 216 = 0
⇒ x² + 18x – 12x – 216 = 0
⇒ x(x + 18)- 12(x + 18) = 0
⇒ (x + 18) (x – 12) = 0
⇒ x + 18 = 0 (or) x – 12 = 0
⇒ x = -18 (or) 12
But x can’t be negative.
∴ x = 12
i.e., side of the first square = 12
∴ Perimeter = 4 × 12 = 48
∴ Perimeter of the second square = 48 + 24 = 72
∴ Side of the second square = \(\frac{72}{4}\) = 18 m.
(or)
x² + (x – 6)² = 468
⇒ x² + x² – 12x + 36 = 468
⇒ 2x² – 12x – 432 – 0
⇒ x² – 6x – 216 = 0
⇒ x² – 18x + 12x – 216 = 0
⇒ x(x-18) + 12(x-18) = 0
⇒ (x – 18) (x + 12) = 0
⇒ x – 18 = 0 (or) x + 12 = 0
⇒ x = 18 (or) – 12
But x can’t be negative.
∴ x = 18
i.e., side of the first square = 18 m
∴ Perimeter = 4 × 18 = 72
Perimeter of the second square = 72 – 24 = 48
∴ Side of the second square = \(\frac{48}{4}\) = 12 m.
i.e., In any way, the sides of the squares are 12m, 18m.

Question 12.
If a polygon of ‘n’ sides has \(\frac{1}{2}\)n(n – 3) diagonals. How many sides will a polygon having 65 diagonals? Is there a polygon with 50 diagonals?
Answer:
Given: Number of diagonals of a polygon with n-sides = \(\frac{n(n-3)}{2}\)
No. of diagonals of a given polygon = 65
i.e., \(\frac{n(n-3)}{2}\) = 65
where n is number of sides of the polygon
⇒ n² – 3n = 2 × 65
⇒ n² – 3n – 130 = 0
⇒ n² – 13n + 10n – 130 = 0
⇒ n(n – 13) + 10(n – 13) = 0
⇒ (n – 13) (n + 10) = O
⇒ n – 13 = 0 (or) n + 10 = 0
⇒ n = 13 (or) n = -10
But n can’t be negative.
∴ n = 13 (i.e.) number of sides = 13.
Also to check 50 as the number of diagonals of a polygon
∴ \(\frac{n(n-3)}{2}\) = 50
⇒ n² – 3n = 100
⇒ n² – 3n – 100 = 0
There is no real value of n for which the above equation is satisfied.
∴ There can’t be a polygon with 50 diagonals.

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 6th Lesson Refraction of Light at Curved Surfaces

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
Have you ever touched a magnifying glass with your hand?
Answer:
Yes.

Question 2.
Have you touched the glass in the spectacles used for reading with your hand?
A. Yes.

Question 3.
Is it a plane or curved surface?
Answer:
Curved surface.

Question 4.
Is it thicker in the middle or at the edge?
Answeer:
Magnifying glass and some spectacle are thicker in middle whereas some spectacles are thicker at edge.

Improve Your Learning

Question 1.
A man wants to get a picture of a zebra. He photographed a white donkey after fitting a glass, with black stripes, on to the lens of his camera. What photo will he get? Explain. (AS1)
(OR)
A person wants to get a picture of zebra and he photographed a white donkey fitting a glass with black stripes. Does he get photo of zebra? Explain.
Answer:
The person was unable to gel the picture <>l zebra because only two rays are enough to form complete image after convergence. So he will get the image of white donkey but the intensity may be less.
(OR)
He will get a picture of while donkey because e\ery part of lens forms an image so if you cover lens with stripes still it forms a complete image. However, the intensity of the image will be reduced.

Question 2.
Two converging lenses are to be placed in the path of parallel rays so that the rays remain parallel after passing through both lenses. How should the lenses be arranged? Explain with a neat ray diagram. (AS1)
Answer:

• Two lenses are placed in the path of parallel rays as shown in figure.
• The first lens is placed in the direction of parallel lines, which converges at focus.
• The second lens is arranged so that it is the focus of 2nd then emerging rays will be parallel.

Question 3.
The focal length of a converging lens is 20 cm. An object is 60 cm from the lens. Where will the image be formed and what kind of image is it? (AS1)
Answer:
f = 20 cm (by sign conversion f = + 20 cm)
u = 60 cm (by sign conversion u = – 60 cm)

Image will be formed at 30cm in between F₁, and 2F1. Image is real, inverted and diminished.

Question 4.
A double convex lens has two surfaces of equal radii ‘R’ and refractive index n = 1.5. Find the focal length ‘f’. (AS1)
(OR)
What is the focal length ‘f, when its double convex lens has two surfaces of equal radii ‘R’ and refractive index n = 1.5?
Answer:
R₁ = R₂ = R (suppose)
Focal length (f) = ?; Refractive index (n) = 1.5

∴ Focal length of lens = Radius of curvature of surface.

Question 5.
Write the lens maker’s formula and explain the terms in it. (AS1)
(OR)
Ravi wants to make a lens. Which formula he has to follow ? Write the formula and explain the terms in it.
(OR)
Write lens formula.
Answer:
Lens maker’s formula:
\(\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
n = Refractive index of the medium
R₁ = Radius of curvature of 1 st surface
R₂ = Radius of curvature of 2nd surface
f = Focal length

Question 6.
How do you verify experimentally that the focal length of a convex lens is increased when it is kept in water? (AS1)
(OR)
Write an activity to show that the focal length of a lens depends on its surrounding medium.
Answer:
Aim :
To prove focal length of convex lens is increased when it is kept in water.

Apparatus :
Convex lens, water, cylindrical vessel, circular lens holder, stone.

Procedure :

1. Take a cylindrical vessel like glass tumbler.
2. Its height must be greater than the focal length of lens, (the around four times focal length of lens).
3. Keep a black stone inside the vessel at its bottom.
4. Pour the water into the vessel such that the height of the water level from the top of the stone is greater than the focal length of lens.
5. Now dip the lens horizontally using a circular lens holder.
6. Set the distance between stone and lens that is equal to or less than focal length of lens.
7. Now see the stone through the lens.
8. We can see the image of the stone.
9. If we dip the lens to a certain height which is greater than the focal length of lens in air, still we can see the image.
10. This shows that the focal length of convex lens has increased in water.
11. Thus we conclude that the focal length of lens depends upon the surrounding medium.

Note : For convenience, use 5 or 10 cm focal length convex lens.

Question 7.
How do you find the focal length of a lens experimentally? (AS1)
Answer:

• Take the lens (Ex : Convex), which focused towards the distant object.
• A white coated screen (Ex : White paper) is placed on the other side of the lens.
• Adjust the screen till you get a clear image of the object.
• At this position measure the distance between the lens and screen which is equal to the focal length of the lens.

Question 8.
Harsha tells Siddhu that the double convex lens always behaves like a convergent lens. But Siddhu knows that Harsha’s assertion is wrong and corrected Harsha by asking some questions. What are the questions asked by Siddhu? (As2)
Answer:
The questions asked by Siddhu :

1. Is the object placed beyond 2f point?
2. Is the object located at 2f point?
3. Is the object located in between the 2f and the focal point?
4. Is the object located at the focal point?
5. Is the object located in front of the focal point?
6. Is the lens kept in a medium with refractive index less than lens or more than lens?

Question 9.
Assertion (A): A person standing on the land appears taller than his actual height to a fish inside a pond. (AS2)
Reason (R) : Light bends away from the normal as it enters air from water.
Which of the following is correct? Explain.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true and R is not the correct explanation of A.
c) A is true but R is false.
d) Both A and R are false.
e) A is false but R is true.
Answer:
Answer a is correct.
Explanation :
Because the light travelling from water to air it bends away from the normal so the fish observes the apparent image of the person, appears taller than his original.

Question 10.
A convex lens is made up of three different materials as shown in the figure Q-10. How many of images does it form? (AS2)

Answer:

• A lens made of three different materials of refractive indices say n₁, n₂ and n₃.
• These three materials will have three different refractive indices. Thus for a given object it forms three images.

Question 11.
Can a virtual image be photographed by a camera? (AS2)
Answer:
Yes, we can.
Ex : – A plane mirror forms a virtual image, we can able to take photograph of that image in plane mirror.

Question 12.
You have a lens. Suggest an experiment to find out the focal length of the lens. (AS3)
(OR)
Through an experiment, find out the focal length of the lens.
Answer:
Aim :
To find focal length of given lens.

Apparatus :
Object (candle), convex lens, v – stand, screen.

Procedure :

• Take a v-stand and place it on a long table at the middle.
  Place a convex lens on the v-stand. Imagine the principal axis of the lens.
• Light a candle and ask your friend to take the candle far away from the lens along the principal axis.
• Adjust a screen (a sheet of white paper placed perpendicular to the axis) which is on other side of the lens until you get an image on it.
• Measure the distance of the image from the v-stand of lens (image distance V) and also measure the distance between the candle and stand of lens (object distance ‘u’). Record the values in the table.

• Now place the candle at a distance of 60 cm from the lens, try to get an image of the candle flame on the other side on a screen. Adjust the screen till you get a clear image.
• Measure the image distance V and object distance ‘u’ and record the values in table.
• Repeat the experiment lor various object distances like 50 cm, 40 cm, 30 cm, etc. Measure the image distances in all cases and note them in table.
• Using the formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\), find f in all the cases. We will observe the value ‘f is equal in all cases. This value off is the focal length of the given lens.

Question 13.
Let us assume a system that consists of two lenses with focal length f₁, and f₂ respectively. How do you find the focal length of the system experimentally, when
i) two lenses are touching each other
ii) they are separated by a distance ‘d’ with common principal axis? (AS3)
Answer:
Experimental Proof:
i) Two lenses are touching each other :
Aim :
To find focal length of combination of two convex lenses, touching each other. Material required : Convex lenses – 2 (with known focal lengths say f, and f2); V-stands – 2, candle, screen scale.

Procedure:

• Place two V-stands with two convex lenses as they touch each other on a table.
• Place a candle (object) far away from the lenses.
• Adjust a screen, which is placed other side of the lenses until we get a clear image on it.
• At that position, measure the image distance (v) and object distance (u).
• Do this experiment for several object distances and record in the given table.

ii) They are separated by a distance of ‘d’ :
Procedure :

• Now place v-stands along with lenses with distance’d’.
• Do the same procedure again.
• Record the observations in the given table.
• Find the average of the ‘f’_comb.

Question 14.
Collect the information about the lenses available in an optical shop. Find out how the focal length of a lens may be determined by the given power’ of the lens. (AS4)
Answer:
I had collected the information regarding different lenses available at optical shops.
The relationship between power and focal length is power (D) = \(\frac{1}{f}\). f is in meters.

Question 15.
Collect the information about lenses used by Galileo in his telescope. (AS4)
(OR)
What lenses are used by Galileo in his telescope?
Answer:

A Galilean telescope is defined as having one convex lens and one concave lens. The concave lens serves as the ocular lens or the eye piece, while the convex lens serves as the objective. The lens are situated on either side of a tube such that the focal point of the ocular lens is the same as the focal point for the objective lens.

Question 16.
Use the data obtained by activity – 2 in table-1 of this lesson and draw the graphs of u vs v and \(\frac{1}{u}\) vs \(\frac{1}{v}\) (AS5)
(OR)
By obtaining data from activity – 2 in table – 1 of this lesson, draw the graphs of u vs v and \(\frac{1}{u}\) vs \(\frac{1}{v}\)
Answer:
Graph of u – v using data obtained by activity – 2. Take lens with focal length 30 cm.

The graph looks like this

The shape of the graph is rectangular hyperbola.

Graph of \(\frac{1}{u}\) – \(\frac{1}{v}\)

For these values the graph is straight line which touches the axis as shown in figure.

Question 17.
Figure shows ray AB that has passed through a divergent lens. Construct the path of the ray up to the lens if the F position of its foci is known. (AS5)

Answer:

The path of the ray up to the lens if the position of foci is known for ray AB is diverging lens or concave lens path.

Question 18.
Figure shows a point light source and its image produced by a lens with an optical axis N₁, N₂. Find the position of the lens and its foci using a ray diagram. (AS5)

Answer:

1. The object is in between focus and optic centre.
2. The image is virtual, erect and magnified. Nv
3. l is the lens, ‘O’ is the object and T is the image.

Question 19.
Find the focus by drawing a ray diagram using the position of source S and the image S’ given in the figure. (AS5)

Answer:

1.  Image is real.
2. l’ is lens, ‘O’ is object and T is image.
3.  Lens is convex.

(Or)

1. Image is real.
2. l’ is lens, ‘O’ is object and ‘I’ is image.
3. Lens is convex.

Question 20.
A parallel beam of rays is incident on a convergent lens with a focal length of 40 cm. Where should a divergent lens with a focal length of 15 cm be placed for the beam of rays to remain parallel after passing through the two lenses? Draw a ray diagram. (AS5)
Answer:

1. A parallel beam of rays when incident on a convergent lens, after refraction they meet at the focus of the lens.

2. A beam of rays which is incident on a divergent lens, after refraction, pass parallel to the principal axis. If we extend these incident rays, they seems to meet at focus of the lens.

3. Hence the divergent lens should be kept at 25 cm distance from convergent lens (40 – 15 = 25 cm) as shown in the figure.

PF = 40 cm (Focal length of convergent lens)
P’F = 15 cm (Focal length of divergent lens)
PP’ = 40 – 15 = 25 cm (Position of divergent lens)

Question 21.
Draw ray diagrams for the following positions and explain the nature and position of image.
i) Object is placed at 2F₂
ii) Object is placed between F₂ and optic centre P. (AS5)
Answer:
i) Object is placed at 2F₂:

Nature : Real, inverted and diminished.
Position : Image is formed on the principal axis between the points F₁, and 2F₁.

ii) Object is placed between F₂ and optic centre P :

Nature : Virtual, erect and magnified.
Position : Same side of the lens where object is placed.

Question 22.
How do you appreciate the coincidence of the experimental facts with the results obtained by a ray diagram in terms of behaviour of images formed by lenses? (AS6)
Answer:

• Ray diagrams are very useful in optics.
• By the ray diagrams, we can easily find the values of image distance, object distance, focal length, radius of curvature, magnification, etc.
• These results are exactly equal to the result gotten by an experiment.
• For example : In the experiment, with a convex lens, we get clear image of an object, on a screen by adjusting the screen.

Then, we measure the image distane (v) practically. This takes more time and requires equipped lab also.

But, by simply draw a ray diagram on a paper, we can get exact image distance (v) very easily, without lab.

• So, ray diagrams are very useful in the construction of microscopes, telescopes, etc.
• Hence, one can trust and depend on the result of ray diagrams instead of several lab experiments.
• So, I appreciate the ray diagrams.

Question 23.
Find the refractive index of the glass which is a symmetrical convergent lens if its focal length is equal to the radius of curvature of its surface. (AS7)
Answer:
Given that lens is convergent symmetrical

Question 24.
Find the radii of curvature of a convexo – concave convergent lens made of glass with refractive index n = 1.5 having focal length of 24 cm. One of the radii of curvature is double the other. (AS7)
Answer:

Question 25.
The distance between two point sources of light is 24 cm. Where should a convergent lens with a focal length of f = 9 cm be placed between them to obtain the images of both sources at the same point? (AS7)
Answer:
For Source S₁ :

∴ The convex lens may be placed between the two sources, such that a distance of 18 cm from one source, and 6 cm from other source.

Question 26.
Suppose you are inside the water in a swimming pool near an edge. A friend is standing on the edge. Do you find your friend taller or shorter than his usual height? Why? (AS7)
(OR)
If your friend is standing near an edge of the swimming pool and you are in the water, do you find he is taller or shorter than his usual height?
Answer:

1. My friend appears to be taller because the light is travelling from rarer to denser.
2. The rays bend in such away that they seems to be coming from long distance.
3. So it is actually apparent image of my friend which appears to be taller due to refraction.

Fill in the Blanks

1. The rays from the distant object, falling on the convex lens pass through ……………….. .
2. The ray passing through the ……………….. of the lens is not deviated.
3. Lens formula is given by ……………….. .
4. The focal length of the plano-convex lens is 2R where R is the radius of curvature of the surface. Then the refractive index of the material of the lens is ……………….. .
5. The lens which can form real and virtual images is ……………….. .
Answer:

1. Tocus
2. optical centre
3. \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
4. 1.5
5. convex lens

Multiple Choice Questions

1. Which one of the following materials cannot be used to make a lens?
A) water
B) glass
C) plastic
D) clay
Answer:
D) clay