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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 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 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 Syllabus AP Board 9th Class Biology Solutions Chapter 1 Cell its Structure and Functions Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 1st Lesson Cell its Structure and Functions

9th Class Biology 1st Lesson Cell its Structure and Functions Textbook Questions and Answers

Improve Your Learning

Question 1.
Differentiate between a) Plant cell and animal cell b) Prokaryotic and Eucaryotic cells. (AS 1)
Answer:
a)

b)

Question 2.
What happens if plasma membrane ruptures or breaks? (AS 2)
Answer:

• Cell membrane or plasma membrane is covering of the animal cell.
• It separate cytoplasm from the external environment.
• It defined the shape and size of the cell.
• It plays a crucial role in maintaining a balance of various substances inside the cell.
• It controls the exchange of substances between the cell and its external environment.
• If it ruptured or broke, then the above activities will stop, the cell will die.

Question 3.
Prepare a model of plant cell or animal cell with locally available materials. (AS 5)
Answer:

Question 4.
What would happen to the life of cell if there was no golgi complex? (AS 2)
Answer:

• The golgi apparatus packed various substances before they are transported to other parts of the cell.
• If there was no golgi complex in the cell the proteins and other substances are not altered and packed.
• Substances transport will not occur.
• Regeneration or repair of the membrane will not takes place.

Question 5.
What happen to the cell if nucleus is removed? Give two reasons to support your answer. (AS 1)
Answer:

• If nucleus is removed from a cell, there would be no control on the functions of a cell.
• Cells are not involved in the process of cell division.
• The cell will not live for more time.
• E.g.: Red blood cells, not having nucleus live less time than the other cells, which are having nucleus.

Question 6.
Lysosomes are known as suicidal bags of the cell. Why? (AS 1)
Answer:

• Lysosomes contained the destructive enzymes.
• Thus the enzymes normally do not come in contact with the rest of the cell.
• The materials that need to be destroyed are transported to the lysosomes.
• At times, the lysosomes burst and the enzymes are released to digest the cell.
• Hence, lysosomes are known as suicidal bags of the cell.

Question 7.
Why do plant cell possess large sized vacuole? (AS 1)
Answer:

• Vacuoles are fluid filled sac-like structures.
• In a newly formed plant cell, the vacuoles are small.
• As the cell becomes old, these vacuoles, fuse to form a single large vacuole.
• In mature plant cells, they might occupy almost the entire cell space.

Question 8.
Prepare a temporary mount of any leaf peel observe the stomata draw their picture. Write a short note on the same. (AS 5)
Answer:

• A fresh leaf of Rheo is taken.
• Making a slit in the pith material and keep the leaf inside the slit.
• To get the T.S a leaf, section cutting with a blade should be done.
• The thin section with brush and keep the section on the slide.
• Putting a drop of water, glycer¬ine on it.
• Staining the section with saffronin.
• Cover the section with a cover slip.
• By observing under the microscope of the leaf. We can see stomata in the lower epidermis.
• They are enclosed by two kidney shaped cells, called guard cells.
• In between two guard cells a pore formed stomata.

Question 9.
“Cell is the basic unit of life” – explain the statement. (AS1)
Answer:

• The fundamental organizational unit of life is the cell.
• All living organisms are composed of cells.
• In unicellular organisms, a single cell performs all the the functions.
• In multicellular organisms, a no. of cells together performs different functions.
• So, we can say that “Cell is the basic unit of life”.

Question 10.
How do you appreciate about the organisation of cell in the living body? (AS 6)
Answer:

• Cell is the basic unit in the structural organisation of all living organisms.
• Cell carry physiological functions like oxidise food materials to derive energy.
• Excrete the waste materials.
• Increase in number by dividing into two identical cells.
• Defend itself against the attack of foreign organisms.
• Try to adjust to the conditions in its surroundings.
• Function of an organism depends on the functions carried out by the cell.

Question 11.
If the organisation of cell is destroyed due to physical and chemical influence, what will happen? (AS 6)
Answer:

• If the organisation of cell is destroyed due to physical and chemical influence, the cell will die.
• Sometimes it also effects the functions of other cells nearby.

Question 12.
Read the chapter carefully collect the information about the functions of different cell organelles and make a table which contains serial number. Cell organelle, function. Don’t forget to write your specific findings below the table. (AS 4)
Answer:

Question 13.
How could you appreciate the function of tiny cell in a large body of an organism? (AS 6)
Answer:

• Cell is the basic unit in the structural organisation of ail living organisms.
• It is the functional and structural unit of the organism.
• Functions essential for survival of the organism are carried out at the level of a cell only.
• Each cell acts as an individual unit.
• In each cell excretion, generation of energy, defending itself, adjust to the conditions, production of new cells etc. functions are carried out.
• So we must appreciate the function of a tiny cell in a large body of an organism.

Question 14.
Look at the following cartoon of a cell. Find out the functions of cell organelles. (AS 5)

Answer:

Question 15.
Who and when was “The cell theory” proposed? When did they prepare it? What are its salient features? (AS 1)
Answer:
M.J. Schleiden and Theodar Schwann proposed “The cell theory”. They prepared it in 1838 – 39.

Statements of modern form of cell theory :

1. All the living organisms are made up of cells and their products.
2. All the cells are formed from pre-existing cells.
3. All the cells are made up of similar chemicals and show similar metabolic activities.
4. Functioning of an organism depends on the functions carried out and the interac-tion of different cells present in the organism.

Question 16.
When you observing the nucleus of cheek cell in laboratory, what precautions do you take?
Answer:
While observing the nucleus of cheek cell in laboratory the following precautions are to be taken.

Precautions:

1. Do not scrap the cheek too hard as it may injure the buccal mucosa.
2. Scrapped material should be spread uniformly on the slide.
3. Excess stains should be drained off.
4. There should be no air bubbles under the coverslip.

Question 17.
Draw the typical animal cell and lable its parts.
Answer:

9th Class Biology 1st Lesson Cell its Structure and Functions InText Questions and Answers

9th Class Biology Textbook Page No. 1& 2

Question 1.
Observe the following figures.

a) What common features do you see in both the cells?
Answer:
We can observe some common features in plant and animal cells. They both are having plasma membrane, mitochondria, golgi apparatus, endoplasmic reticulum, nucleus etc.

b) Which cell organelles are found exclusively in plant cell?
Answer:
Chloroplasts and big vacuoles are the cell organelles exclusively found in plant cell.

9th Class Biology Textbook Page No. 3

Question 2.
What is the role of the cell wall in plant cells?
Answer:
It exerts an inward wall pressure to resist the outward pressure exerted by the cell sap.

9th Class Biology 1st Lesson Cell its Structure and Functions Activities

Activity – 1

Question 1.
How do you observe cell membrane in a peel of Rheo leaf under microscope? Draw the diagram of it. Write your observations.
Answer:
Take Rheo leaf, tear the leaf in single stroke take a small piece of leaf peel with light coloured (transparent) portion. Put it on slide and put a drop of water on it. Cover it with cover slip and observe the light portion of leaf under the microscope.

Observations:

1. Cells are arranged in rows.
2. Cell membrane is clearly seen.
3. Nucleus is present in the cell.

Lab Activity

Question 2.
To observe the nucleus in cheek cells.
Answer:
Aim :
To observe the nucleus in cheek cells.

Material:
A tooth pick or ice-cream spoon or spatula, glass slide, coverslip, watch glass, needle, blotting paper, 1% methylene blue, normal saline, glycerine, microscope etc.

Procedure:

1. Wash your mouth and scrap a little of the internal living of your cheek inside your mouth with a clean tooth pick or spatula or ice-cream spoon.
2. Place the scrap in a watch glass containing a very small quantity of normal saline.
3. Then place the material on a glass slide.
4. Put a drop of methylene blue and wait for a couple of minutes.
5. Wipe off the extra stain with a fine cloth of blotting paper.
6. Put a drop of glycerine over it.
7. Place a coverslip. Tap the coverslip with the blunt end of needle so as to spread the cells.

Precautions :

1. Do not scrap the cheek too hard as it may injure the buccal mucosa.
2. Scrapped material should be spread uni¬formly on the slide.
3. Excess stains should be drained off.
4. There should be no air bubbles under the coverslip.

Observations :

1. The shape of the cells are circular in shape.
2. These cells are not similar to the structure in onion peel cell.
3. Near the centre of the cell there is a darkly coloured oval dot like structure present.

Activity – 2

Question 3.
How do you observe mitochondria in onion peel ? Observe and make a sketch of mitochondria.
Answer:
Observing mitochondria :

1. Make a fresh solution of Janus Green-B in a beaker.
2. Mix 200 mg Janus Green-B in 100 ml of water.
3. Take a watch glass pour some solution. Put the onion peel in this solution and keep it about half an hour.
4. Keep a piece of onion peel on the slide and wash thoroughly with water. Mitochondria in onion peel ceil
5. Cover the slide with a cover slip and observe it under microscope at high magnification.

Observations :
Green oval or cylindrical grains scattered in the cytoplasm. They are mitochondria.

Activity – 3

Question 4.
Observe a chSoroplast in Rheo leaf under microscope ? Draw the diagram of it and write your observation.
Answer:
Observing chloroplast:

1. Take the peel of Rheo leaf and mount it in water on a slide.
2. Observe it under high power microscope.

Observations :

1. Small green granules called chloroplasts are present in the cells of Rheo leaf.
2. Chloroplasts mainly contain green substance called chlorophyll.

Activity – 4

Question 5.
How do you observe chloroplast in Algae under microscope? Draw the diagram and write your findings.
Answer:
Observing chloroplast:

1. Collect some algae from pond and separate out thin filaments of them.
2. Place a few filaments on slide. Observe it under microscope.

Observations:

1. In algae the chloroplasts are found as ladders, stars, spirals or reticulate.
2. The primary function of chloroplasts is to trap the energy of sunlight and transform it to chemical energy in photosynthesis.

Activity – 5

Question 6.
How do you observe under microscope the vacuoles of succulent plant like cactus?
Write small note on them.
Answer:
Observing vacuoles :

1. Take the leaf or stem of any succulent plant like cactus.
2. Take thin cross section of stem of cactus in a watch glass containing water.
3. Stain it with dilute safranine solution.
4. Observe it under low and high power microscope.

Observations :

1. The large empty spaces visible in the cell are vacuoles.
2. These are fluid filled sac like structures.

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

2. Which of the following is true?
A) The distance of virtual image is always greater than the object distance for convex lens.
B) The distance of virtual image is not greater than the object distance for convex lens.
C) Convex lens always forms a real image.
D) Convex lens always forms a virtual image.
Answer:
B) The distance of virtual image is not greater than the object distance for convex lens.

3. Focal length of the plano-convex lens is when its radius of curvature of the surface is R and n is the refractive index of the lens.

4. The value of the focal length of the lens is equal to the value of the image distance when the rays are
A) passing through the optic centre
B) parallel to the principal axis
C) passing through the focus
D) in all the cases
Answer:
D) in all the cases

5. Which of the following is the lens maker’s formula?

Answer:
C

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces Additional Questions and Answers

Question 1.
Derive a relation between refractive indices of two media (n₁, n₂), object distance (u), image distance (v) and radius of curvature (R) for a curved surface.
(OR)
Derive \(\frac{\mathbf{n}_{2}}{\mathbf{v}}-\frac{\mathbf{n}_{1}}{\mathbf{u}}=\frac{\mathbf{n}_{2}-\mathbf{n}_{1}}{\mathbf{R}}\)
(OR)
Derive curved surface formula.
Answer:

• Consider a curved surface separating two media of refractive indices n₁, and n₂.
• A point object is placed on the principal axis at point ‘O’.
• The ray which travels along the principal axis passes through the pole undeviated.
• The second ray, which forms an angle with a princi¬pal axis, meets the interface at A. The angle of incidence is Q₁. The ray bends and passes through the second medium along the line AI. The angle of refraction is Q₂.
• The two refracted rays meet at I and the image is formed there.
• 6) Let the angle made by the second refracted ray with principal axis be γ and the angle between the normal and principal axis be β.
• From figure,

PO = u (object distance), PI = v (Image distance),
PC = R (radius of curvature) and n₁, n₂ are refractive indices of the media.
From ∆ACO, θ₁ = α + β
∆ACI, β = θ₂ + γ
⇒ θ₂ = β – γ
According to Snell’s law, n₁sin θ₁ = n₂ sin θ₂.
∴ n₁ sin (α + β) = n₂ sin (β – γ) …………….. (1)
As per paraxial approximation,
sin (α + β) = α + β and sin (β – γ) = β – γ.
∴ (1) ⇒ n₁(α + β) = n₂ (β – γ)
⇒ n₁ α + n₁β = n₂ β – n₂ γ — (2)
Since all angles are small, we can write

∴ This is the required relation for curved surfaces.

Question 2.
Derive expression for lens maker’s formula.
(OR)
Prove \(\frac{1}{\mathbf{f}}=(\mathbf{n}-\mathbf{1})\left(\frac{1}{\mathbf{R}_{1}}-\frac{1}{\mathbf{R}_{2}}\right)\).
Answer:

Procedure :

• Imagine a point object ‘O’ placed on the principal axis of the thin lens
• Let this lens be placed in a medium of refractive index n_a and let refractive index of lens be n_b.
• Consider a ray, from ‘O’ which is incident on the convex surface of the lens with radius of curvature R₁ at A.
• The incident ray refracts at A.
• It forms image at Q, if there were no concave surface.
• From figure Object distance PO = – u;

Image distance PQ = v = x
Radius of curvature R = R₁
n₁ = n_a and n₂ = n_b.

• But the ray that has refracted at A suffers another refraction at B on the concave surface with radius of curvature (R₂).
• At B the ray is refracted and reaches I.
• The image Q of the object due to the convex surface. So I is the image of Q for concave surface.
• Object distance u = PQ = + x
  Image distance PI = v
  Radius of curvature R = – R₂
• The refraction of the concave surface of lens is medium -1 and surrounding is medium – 2.
  ∴ n₁ = n_b and n₂ = n_a

Question 3.
Derive the lens formula.
Answer:
1. Consider an object 00′ placed on the principal axis in front of a convex lens as shown in the figure. Let II’ be the real image formed by the lens, i.e. the other side of it.

2. From the figure : PO, PI, PFt are the object distance, image distance and focal length respectively.

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces InText Questions and Answers

10th Class Physics Textbook Page No. 64

Question 1.
What happens to a ray that is incident on a curved interface separating the two media? Are the laws of refraction still valid?
Answer:
It undergoes deviation from its path. Yes, the laws of reflection are still valid.

Question 2.
How do rays betid when they are incident on a curved surface?
Answer:
A ray will bend towards the normal when it travels from rarer to denser medium and bends away from the normal when it travels from denser to a rarer medium.

10th Class Physics Textbook Page No. 65

Question 3.
What happens to ray that travels along the principal axis? Similarly, a ray that travels through the centre of curvature?

Answer:
According to Snell’s law the ray which travels along the normal drawn to the surface does not deviate from its path. Hence both rays in the given condition travel along normal, so they do not deviate.

Question 4.
What difference do you notice in the refracted rays in 4 (a) and 4 (b)? What could be the reason for that difference?

Answer:

• In figure 4 (a) ray travelling parallel to the principal axis strikes a convex surface and passes from a rarer medium to a denser medium.
• In figure 4 (b) a ray travelling parallel to the principal axis strikes a convex surface passes from a denser medium to a rarer medium.
• Figure 4 (a) : The refracted ray moves towards the normal.
• Figure 4 (b) : The refracted ray moves away from the normal.
  Reason : The main reason is that light passes through different media.

10th Class Physics Textbook Page No. 66

Question 5.
What difference do you notice in refracted rays in 4 (c) and 4 (d)? What could be the reasons for that difference?
(OR)
Draw the ray diagrams when the incident ray passes through the curved surfaces.
a) Rarer medium to denser medium.
b) Denser medium to rarer medium.

Answer:

• In figure 4 (c) a ray travelling parallel to the principal axis strikes a concave surface and passes from a denser medium to a rarer medium.
• In figure 4 (d) a ray travelling parallel to the principal axis strikes a concave surface and passes from a rarer medium to a denser medium.

Reasons :

• Figure 4 (c) :The refracted ray reaches a particular point on the principal axis.
• Figure 4 (d) : The refracted ray moves away from the principal axis.
• The main reason is that light passes through different media.

Question 6.
You might have observed that a lemon in the water of a glass tumbler appears bigger than its actual size, when viewed from the sides of tumbler.
1) How can you explain this (appeared) change in size of lemon?
Answer:
It can be explained by using refraction. When light travels from one medium to another medium it undergoes refraction.

2) Is the lemon that appears bigger in size an image of lemon or is it the real lemon?
Answer:
That is image of lemon.

3) Can you draw a ray diagram to explain this phenomenon?
Answer:

10th Class Physics Textbook Page No. 70

Question 7.
What happens to the light ray when a transparent material with two curved surfaces is placed in its path?
Answer:
The light ray undergoes refraction.

Question 8.
Have you heard about lenses?
Answer:
Yes, we have heard about lenses. A transparent material bounded by two spherical v surfaces is called lens.

Question 9.
How does a light ray behave when it is passed through a lens?
Answer:
A light ray will deviate from its path in some cases and does not deviate in some other cases.

10th Class Physics Textbook Page No. 72

Question 10.
How does the lens form an image?
Answer:
Lens forms an image through converging light rays or diverging light rays.

Question 11.
If we allow a light ray to pass through the focus, which path does it take?
Answer:
The ray passing through the focus takes a parallel path to principal axis after refraction.

10th Class Physics Textbook Page No. 73

Question 12.
What happens when parallel rays of light fall on a lens making some angle with the principal axis?
Answer:’
The rays converge at a point (or) appear to diverge from a point lying on the focal plane.

Question 13.
What do you mean by an object at infinity? What type of rays fall on the lens?
Answer:
The distance between the lens and the object is very much greater than when compared to object size is known as object at infinity. Parallel rays fall on the lens.
The object at infinity means distant object. The rays falling on the lens from an object at infinity are parallel to principal axis.

10th Class Physics Textbook Page No. 77

Question 14.
Could you get an image on the screen for every object distance with a convex lens?
Answer:
No, when the object is placed between pole and focus we will get virtual, erect and enlarged image on the other side of the- object.

Question 15.
Why don’t you get an image for certain object distances?
Answer:
Because at those distances the light rays diverge each other.

Question 16.
Can you find the minimum limiting object distance for obtaining a real image? What do you call this minimum limiting object distance?
Answer:
Yes, this minimum limiting object distance is called focal length.

Question 17.
When you do not get an image on the screen, try to see the image with your eye directly from the place of the screen. Could you see the image? What type of image do you see?
Answer:
Yes, we can see the image. This is a virtual image which we cannot capture on screen.

Question 18.
Can you find the image distance of a virtual image? How could you do it?
Answer:
We can find the image distance of virtual image by using lens formula \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\) (if we know the focal length of lens and object distance.)

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces Activities

Activity – 1

Question 1.
Write an activity to observe the light refraction at curved surface.
Answer:
Procedure and observation :

• Draw an arrow of length 4 cm usfng a black sketch pen on a thick sheet of paper.
• Take an empty cylindrical-shaped transparent vessel.
• Keep it on the table.
• Ask your friend to bring the sheet of paper on which arrow was drawn behind the vessel while you look at it from the other side.
• We will see a diminished image of the arrow.
• Ask your friend to fill vessel with water.
• Look at the arrow from the same position as before.
• We can observe an inverted image.

Explanation :

• In the first case, when the vessel is empty, light from the arrow refracts at the curved interface, moves through the glass, enters in to air then it again undergoes refraction on the opposite curved surface of vessel and comes out into the air.
• In this way light travels through two media, comes out of the vessel and forms a diminished image.
• In the second case, light enters the curved surface, moves through water, comes out of the glass and forms an inverted image.

Lab Activity

Question 2.
Write an activity to know the characteristics of image due to convex lens at various distances.
Answer:
Aim:
Determination of focal length of bi-convex lens using UV method.

Material Required :
V Stand, convex lens, light source, screen, meter scale. Take a V-stand and place it on a long (nearly 2m) 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.

Procedure :

1. Take a V-stand and place a convex lens on this stand.
2. Imagine the principal axis of the lens.
3. Light a candle and ask your friend to take the candle far away from the lens along the principal axis.
4. We use a screen because it forms a real image generally which will form on a screen. Real images cannot be seen with an eye.
5. Adjust the screen, on other side of lens until clear image forms on it.
6. Measure the distance of the image from the stand and also measure the distance between the candle and stand of lens.
7. Now place the candle at a distance of 60 cm from the lens such as the flame of the candle lies on the principal axis of the lens.
8. Try to get an image of candle flame on the other side on a screen.
9. Adjust the screen till you get a clear image.
10. Measure the distance of image (v) from lens and record the value of’u’ and V in the table.
11. Repeat this for various distances of images; in all cases note them in the table.

Observation :

Conclusion : From this we conclude that a convex lens forms both real and virtual images when object is placed at various positions.

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 8 Structure of Atom Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 8th Lesson Structure of Atom

10th Class Chemistry 8th Lesson Structure of Atom Textbook Questions and Answers

Improve Your Learning

Question 1.
What information does the electronic configuration of an atom provide? (AS1)
Answer:

• The distribution of electrons in shells, sub-shells and orbital in an atom is known as electronic configuration.
• It provides the information of position of an electron in the space of atom.
• The distribution of electrons in various atomic orbitals provides an understanding of the electronic behaviour of the atom and in turn its reactivity.
• The short hand notation is as shown below.

Question 2.
a) How many maximum number of electrons that can be accommodated in a principal energy shell?
Answer:
The maximum number of electrons that can be accommodated in a principal energy shell is 2n². Here n is principal quantum number.

b) How many maximum number of electrons that can be accommodated in a sub-shell?
Answer:
The maximum number of electrons that can be accommodated in a sub-shell is 2(2l +1) (where l is orbital quantum number).

c) How many maximum number of electrons can that be accommodated in an orbital?
Answer:
The maximum number of electrons that can be accommodated in an orbital is 2.

d) How many sub-shells are present in a principal energy shell?
Answer:
The number of sub-shells in a principal energy shell is n (n is principal quantum number).

e) How many spin orientations are possible for an electron in an orbital?
Answeer:
The spin orientations possible for an electron in an orbital are 2.

Question 3.
In an atom the number of electrons in M-shell is equal to the number of electrons in the K and L-shell. Answer the following questions. (AS1)
a) Which is the outermost shell?
Answer:
The outermost shell is N shell.

b) How many electrons are there in its outermost shell?
Answer:
Two electrons are there in outermost shell.

c) What is the atomic number of element?
Answer:
The atomic number of element is 22.

d) Write the electronic configuration of the element.
Answer:
The element is Ti (Titanium). Its electronic configuration is 1s²2s²2p⁶3s²3p⁶4s²3d².

Reason :

• Electrons enter M shell after completion of K and L.
• So the number of electrons in M shell is 10.
• But after completion of 3p orbital electron enters 4s before entering to 3d.
• So outermost orbit or shell is N shell.
• So the atomic number of element is 22.
• Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d².

Question 4.
Rainbow is an example for continuous spectrum – explain. (AS1)
(OR)
Which is naturally occurring continuous spectrum ? Explain.
Answer:

• Rainbow is a spectrum of different colours (VIBGYOR) with different wavelengths.
• These colours are continuously distributed.
• There is no fixed boundary for each colour.
• Hence, rainbow is a continuous spectrum.

Question 5.
How many elliptical orbits are added by Sommerfeld in third Bohr’s orbit ? What was the purpose of adding these elliptical orbits? (AS1)
Answer:
Sommerfeld added two elliptical orbits to Bohr’s third orbit.

Purpose of adding elliptical orbits :

• Bohr’s model failed to account for splitting of line spectra and line spectrum.
• In an attempt to account for the structure of line spectrum, Sommerfeld modified Bohr’s atomic model by adding elliptical orbits.

Question 6.
What is absorption spectrum?
Answer:
Absorption spectrum: The spectrum formed by the absorption of energy when electron jumps from lower energy level to higher energy level is called absorption spectrum. It contains dark lines on bright background.

Question 7.
What is an orbital? How it is different from Bohr’s orbit? (AS1)
(OR)
Comparison between orbit and orbital.
Answer:
The region of space around the nucleus where the probability of finding electron is maximum is called orbital. Whereas orbit is the path of the electron around the nucleus.

These two are differentiated like this.

Question 8.
Explain the significance of three quantum numbers in predicting the positions of an electron in an orbit. (AS1)
(OR)
How are quantum numbers helpful to understand the atomic structure?
Answer:
Significance of three quantum numbers in predicting the positions of an electron in an orbit.

1) Principal quantum number (n) :
The principal quantum number explains about the size and energy of shells (or) orbitals. It is denoted by n.

As ‘n’ increases, the orbitals become larger and the electrons in those orbitals are farther from the nucleus.

It takes values 1, 2, 3, 4, ……………. for that the shells are represented by letters K, L, M, N, ……….

The number of electrons in a shell is limited to 2n².

2) The Angular – momentum quantum number (l) :
The angular momentum quantum number defines the shape of the orbital occupied by the electron and the orbital angular momentum of the electron, is in motion.

l takes values from 0 to n – 1 for these values the orbitals are designated by letters s, p, d, f, ………….. etc.

l also governs the degree with which the electron is attached to nucleus. The larger the value of l, the smaller is the bond with which it is maintained with the nucleus.

3) Magnetic orbital quantum number (m_l) :
The orientation of orbital with external magnetic field determines magnetic orbital quantum number.

m_l has integer values between – l and l including zero.

The number of values for m, are 2l + l, which give the number of orbitals per sub-shell. The maximum number of electrons in orbitals in the sub-shell is 2 (2l + l).

Question 9.
What is nl^x method? How is it useful? (AS1)
(OR)
What is nl^x method? How is it useful in electronic configuration?
Answer:
The shorthand notation consists of the principal energy level (n value) the letter representing sub – level (l value), and the number of electrons (x) in the sub-shell is written as superscript nl^x.

It is useful in writing electron configuration of elements. For example, in Hydrogen (H), the set of quantum numbers is n = 1, l = 0, m_l = 0, m_s = ½ or – ½. The electronic configuration is

Question 10.
Following orbital diagram shows the electronic configuration of nitrogen atom. Which rule does not support this? (AS1)

(OR)
Write the correct electronic configuration of the given nitrogen atom with the help of Hund’s rule.
Answer:

• This electron configuration does not support Hund’s rule.
• According to Hund’s rule, the orbitals of equal energy are occupied with one elec-tron each before pairing of electrons starts.
• Here, pairing of electrons in 2p_x orbital was taken place without filling of an elec-tron in 2p_z orbital.
• Hence the correct electron configuration is as follows.

Question 11.
Which rule is violated in the electronic configuration 1s⁰ 2s² 2p⁴?
Answer:

• Aufbau principle is violated in this electronic configuration because according to Aufbau principle, electron enters orbital of lowest energy.
• Among 1s, 2s and 2p, Is has least energy.
• So Is orbital must be filled before the electron should enter 2s.

Question 12.
Write the four quantum numbers for the differentiating electron of sodium (Na) atom. (AS1)
Answer:
The electronic configuration of sodium (Na) is 1s² 2s² 2p⁶ 3s¹. So the differentiating electron enters 3s. Therefore the four quantum numbers are

Question 13.
What is emission spectrum?
(OR)
When radiation is emitted what is the name given to such spectrum? Explain such spectrum.
Answer:

• The spectrum produced by the emitted radiation is known as emission spectrum.
• This spectrum corresponds to liberation of energy when an excited electron returns back to ground state.

Emission spectrum is of two types :

1) Continuous spectrum :
When white light passes through a prism it dissociates into seven colours. This spectrum is called continuous spectrum.

2) Discontinuous spectrum :
Discontinuous spectrum is of two types.

a) Line spectrum :
The spectrum with sharp and distinct lines. It is given by gaseous atoms.

b) Band spectrum :
The spectrum very closely spaced lines is known as band spectrum. It is given by molecule.

Question 14.
i) An electron in an atom has the following set of four quantum numbers to which orbital it belong to : (AS2)
Answer:

This electron belongs to 2s orbital.
Spin is in clockwise direction. ⇒ 2s¹

ii) Write the fojur quantum numbers for Is1 electron. (AS1)
Answer:
The four quantum numbers for Is1 electron are

Question 15.
Which electronic shell is at a higher energy level K or L? (AS2)
Answer:
L – shell is at higher energy level, because it is far from nucleus than K shell.

Question 16.
Collect the information regarding wavelengths and corresponding frequencies of three primary colours red, blue and green. (AS4)
Answer:
The wavelengths and corresponding frequencies of three primary colours red, blue and green are given below.

Question 17.
The wavelength of a radio wave is 1.0 m. Find its frequency. (AS7)
Answer:
c = 3 × 10⁸ m/s ; λ = 1m ; c = vλ ⇒ v = \(\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^{8}}{1}\) = 3 × 10⁸ Hz.

Question 18.
Why are there exemptions in writing the electronic configurations of Chromium and Copper?
Answer:
1. Elements which have half-filled or completely filled orbitals have greater stability.

2. So in chromium and copper the electrons in 4s and 3d redistributes their energies to attain stability by acquiring half-filled and completely filled d-orbitals.

3. Hence the actual electronic configuration of chromium and copper are as follows.

Fill In The Blanks

1. If n = 1, then angular momention quantum number (l) = …………………
2. If a sub-shell is denoted as 2p, then its magnetic quantum number values are …………………, …………………, …………………
3. Maximum number of electrons that an M-shell contain is / are …………………
4. For ‘n’, the minimum value is ………………… and the maximum value is …………………
5. For?, the minimum value is ………………… and the maximum value is …………………
6. For’m/ the minimum value is ………………… and the maximum value is …………………
7. The value of ‘ms’ for an electron spinning in clockwise direction is ………………… and for anti-clockwise direction is …………………
Answer:

1. 0
2. – 1, 0, + 1
3. 18
4. 1, – ∞
5. 0, (n – 1)
6. – l, + l
7. + ½, – ½

Multiple Choice Questions

1. An emission spectrum consists of bright spectral lines on a dark back ground. Which one of the following does not correspond to the bright spectral lines?
A) Frequency of emitted radiation
B) Wavelength of emitted radiation
C) Energy of emitted radiations
D) Velocity of light
Answer:
D) Velocity of light

2. The maximum number of electrons that can be accommodated in the L-shell of an atom is
A) 2
B) 4
C) 8
D) 16
Answer:
C) 8

3. If l = 1 for an atom, then the number of orbitals in its sub-shell is
A) 1
B) 2
C) 3
D) 0
Answer:
C) 3

4. The quantum number which explains about size and energy of the orbit or shell is
A) n
B) l
C) m_l
D) m_s
Answer:
A) n

10th Class Chemistry 8th Lesson Structure of Atom InText Questions and Answers

10th Class Chemistry Textbook Page No. 112

Question 1.
How many colours are there in a rainbow? What are they?
Answer:
There are seven colours in a rainbow. They are Violet, Indigo, Blue, Green, Yellow,Orange and Red.

Question 2.
What are the characteristics of electromagnetic waves?
A.nswer:
Electromagnetic energy is characterised by wavelength (l) and frequency (u).

10th Class Chemistry Textbook Page No. 113

Question 3.
Can we apply this equation c = uA, to a sound wave?
Answer:
Yes. It is a universal relationship and applies to all waves.

10th Class Chemistry Textbook Page No. 114

Question 4.
What happens when you heat an iron rod on a flame? Do you find any change in colour while heating an iron rod?
Answer:

• When we heat an iron rod, some amount of heat energy that was absorbed by iron rod is emitted as light.
• First iron turns into red (lower energy corresponding to higher wavelength) and as the temperature rises it glows and turns into orange, yellow, blue or even white respectively (higher energy and lower wavelength).
• If we go on heating the rod, it turns into white light which includes all visible wavelengths.
• So we find some changes in colour while heating an iron rod.

Question 5.
Do you observe any other colour at the same time when one colour is emitted?
Answer:
While heating the rod if the temperature is high enough, other colours will also be emitted, but due to higher intensity of one particular emitted colour (eg.: red), others cannot be observed.

Question 6.
How do various colours come from fire works?
(OR)
Do you enjoy Deepavali fire works? Variety of colours is seen from fire works. How do these colours come from fire works?
Answer:
Yes. The electrons present in atoms of elements absorb energy and move to excited states and they return to ground state with emission of energy in visible spectrum. So the colours observed during fire works are the emitted energy by various elements in different fire works.

10th Class Chemistry Textbook Page No. 115

Question 7.
Do you observe yellow light in street lamps? Which will produce yellow light?
Answer:
Yes, sodium vapours produce yellow light in street lamps.

Question 8.
Why do different elements emit different flame colours when heated by the same non-luminous flame?
Answer:

• All the materials are made up of atoms and molecules. These atoms and molecules possess certain fixed energy.
• An atom or molecule having lowest possible energy is said to be in ground state.
• When we heat the materials the electrons of these atoms gain energy and move to excited states (higher energy state).
• An atom of molecule in excited state can emit light to lower its energy in order to get stability and come back to ground state.
• Light emitted in such process has certain fixed wavelength for one kind of atoms.
• The light emitted by different kinds of atoms is different because the excited states electrons will go are different. So different elements produce different flame colours.

Question 9.
What happens when an electron gains energy?
Answer:
The electron moves to higher energy level called the excited state.

10th Class Chemistry Textbook Page No. 116

Question 10.
Does the electron retain the energy forever?
Answer:
The electron loses the energy and comes back to its ground state. The energy emitted by the electron is seen in the form of electromagnetic energy.

Question 11.
Did Bohr’s model account for the splitting of line spectra of a hydrogen atom into finer lines?
Answer:
No, Bohr’s model failed to account for splitting of line spectra.

Question 12.
Why is the electron in an atom restricted to revolve around the nucleus at certain fixed distances?
Answer:
In order to explain the atomic spectra, Bohr-Sommerfeld model proposed that the electrons are restricted to revolve around the nucleus at certain fixed distances.

10th Class Chemistry Textbook Page No. 117

Question 13.
Do the electrons follow defined paths around the nucleus?
Answer:
No, they revolve around the nucleus in a region called orbital.

Question 14.
What is the velocity of the electron?
Answer:
It is very close to light.

Question 15.
Is it possible to find exact position of electron? How do you find the position and velocity of an electron?
Answer:
No, as the electrons are very small, light of very short wavelength is required for this task.

This short wavelength light interacts with the electron and disturbs the motion of electron. So it is not possible to find exact position and velocity of electron simultaneously. Whereas we can find the region where the probability of finding electron is more.

Question 16.
Do atoms have a definite boundary, as suggested by Bohr’s model?
Answer:
Yes, atoms have definite boundary.

Question 17.
What do we call the region of space where the electron might be, at a given time?
Answer:
The region of space around the nucleus where the probability of finding an electron is maximum, called an orbital.

10th Class Chemistry Textbook Page No. 118

Question 18.
What information do the quantum numbers provide?
Answer:
The quantum numbers describe the space around the nucleus where the electrons are found and also their energies.

Question 19.
What does each quantum number signify?
Answer:
The quantum numbers signify the probability of finding electron in the space around nucleus.

10th Class Chemistry Textbook Page No. 119

Question 20.
What is the maximum value of/for n = 4?
Answer:
The maximum value of / for n = 4 is 3.

Question 21.
How many values can l have for n = 4?
Answer:
l takes values from 0 to n – 1. So l has 4 values for n = 4. Those values are 0, 1,2, 3.

Question 22.
Do all the p-orbitals have the same energy? A. Orbitals in the sub-shell belonging to same shell possess same energy but they differ in their orientations.

10th Class Chemistry Textbook Page No. 121

Question 23.
How are two electrons in the Helium atom arranged?
Answer:
They are arranged in pair in Is orbital and the electronic configuration is 1s².

10th Class Chemistry Textbook Page No. 122

Question 24.
What are the spins of two electrons in an orbital?
Answer:
The two electrons in an orbital have opposite spins. If one is clockwise spin, then other electron has anti-clockwise spin.

Question 25.
How many electrons can occupy an orbital?
Answer:
An orbital can hold only two electrons.

10th Class Chemistry 8th Lesson Structure of Atom Activities

Activity – 1

Question 1.
Explain the wave nature of light.
(OR)
How does light behave? Explain.
Answer:

• Light is an electromagnetic wave.
• Electromagnetic waves are produced when an electric charge vibrates.
• This vibrating electric charge creates a change in the electric field. The changing electric field creates a changing magnetic field.
• This process continues with both the created fields being perpendicular to each other and at right angles to the direction of propagation of the wave.
• This electromagnetic wave is produced.

Activity – 2

Question 2.
Write an activity which shows metal produces colour in flame.
(OR)
‘Metal produces colour in a flame.’ Prove the statement by giving examples.
Answer:
A)

• Take a pinch of cupric chloride in a watch glass and make a paste with concentrated hydrochloric acid.
• Take this paste on a platinum loop and introduce it into a non-luminous flame.
• Cupric chloride produces a green colour flame.

B)

• Take a pinch of strontium chloride in a watch glass and make a paste with concentrated hydrochloric acid.
• Take this paste on a platinum loop and introduce it into a non-luminous flame.
• Strontium chloride produces a crimson red flame.

Activity – 3

Question 3.
Complete the electronic configuration of the following elements.

Answer:

AP State Syllabus AP Board 9th Class Social Studies Important Questions Chapter 24 Disaster Management.

AP State Syllabus 9th Class Social Studies Important Questions 24th Lesson Disaster Management

9th Class Social 24th Traffic Education 1 Mark Important Questions and Answers

Question 1.
Write any two causes for road accidents. (SA-II 2018-19)
Answer:

1. Alcohol & drunk & driving
2. Street racing
3. Speeding
4. Weather related
5. Cell phones, etc.

9th Class Social 24th Traffic Education 2 Marks Important Questions and Answers

Question 1.
The Pie chart below shows the Age profile of road crash victims for 2016. Read the chart and answer the following question.
Q : What is the best conclusion that can be drawn from the information provided in the “Pie” chart? (SA-II 2018-19)

Answer:
According to the chart, the main thing is

1. In road accidents all kinds of age people are there.
2. Majority of accident victims are between 25-35 years of age.
3. Majority of victims are adolescence stage.
4. The main reason behind is lack of traffic education.
5. So create awareness about traffic education.

9th Class Social 24th Traffic Education 4 Marks Important Questions and Answers

Question 1.
Analyse the information given above and based on that write the need and significance of Traffic Education. (SA-III : 2016-17)

Answer:

1. In the given year 2012, 2577 accidents took place.
2. Two wheelers (30%), Four wheelers (28%), caused more accidents.
3. Over speed, no awareness about traffic rules are main reasons for the accidents.
4. Youth are the main victims of the accidents.

Significance of traffic education :

1. Traffic education is essential for equipping children sufficiently to become safe road users.
2. Traffic education is heeded to encourage safe traffic activities.
3. Traffic education is very essential for every age group.
4. Traffic education for educated road users can be easily done through media, T.V. and radio.

9th Class Social 24th Traffic Education Important Questions and Answers

Question 1.
What is traffic?
Answer:
Movement of any object from one place to another is traffic like wire movement of any vehicle from one place to another place on the road is road traffic.

Question 2.
What is Traffic Education?
Answer:
Traffic Education is the education which describes the traffic rules and regulations in a clear and simple way.

Question 3.
What is the need and significance of Traffic Education?
Answer:
As young people become more independent, they are exposed to increase risks. Especially teenagers are an important group of road users. Many are unaware that road incidents are the biggest cause for serious accidents and deaths. It is necessary to teach all the road safety measures clearly to avoid accidents.

Question 4.
How does breath analyser work?
Answer:
When a person drinks alcohol it is absorbed in to the blood and is circulated through out the body. As this blood reaches the lungs, the air we exhale carries traces of alcohol which is measured by the gadget. In a way the exhaled air would contain alcohol traces along with Carboh-di-oxide. These machines can pick up even the slightest traces of alcohol. Police Officer cannot delete the record in breath analyser even though he wants to help the victim.

Question 5.
Why does the RTA cancel the driving licenses?
Answer:
The Regional Transport Authority can disqualify persons from holding driving license or revoke the same if the person:

1. is a habitual drunkard.
2. is an addict to any narcotic drug
3. is using a vehicle in the commission of cognizable offence
4. is driving dangerously
5. is using the vehicle without registration
6. is not giving any information required to the police
7. is not shifting the victim of the accident in which his or her vehicle is involved to the nearest hospital
8. does not produce the following certificates on demand by police- Certificate of Insurance, – Certificate of Registration- Driving License- Pollution Certificate

Question 6.
Explain the division of Roads.
Answer:
Division of roads:
1. Footpath:
It is laid on either side of the road for the use of pedestrians. It is built with of about 2 meters

2. Road divider :
The road is divided into two halves with cement slabs

3. Zebra Crossings :
Zebra Crossing is the place where the pedestrians cross the road these are laid at places where traffic is heavy.

Question 7.
What are the hurdles of road safety?
Answer:
Hurdles of road safety :

1. Negligence of civilians
2. Pathetic conditions of roads
3. Unsafe vehicle design
4. Under implementation of road safety standards
5. Lack of proper enforcement of laws
6. Lack of emergency services.

Question 8.
Read the following rules and answer the given questions.
Rules for pedestrians :

1. Walk on the footpath. If footpath is not available and the road is narrow, walk on the right side of the road watching the oncoming traffic.
2. Must use reflective clothing at night when walking outside built up area.
3. Always carry a torch while walking at night time.
4. Do look for safe place to cross and look left and right listen for traffic.
5. Let any traffic coming in either direction.
6. Walk briskly straight across the road when it is clear and continue to watch and listen for walking.
7. Use zebra Lines for crossing roads.
8. Don’t use the mobile in any form while walking/ crossing i.e. either listening to songs.
9. Take the help of traffic police while crossing the road.
10. Pedestrians do not walk on roads in inebriated condition.

1) Where should we walk if footpath is not available?
Answer:
If footpath is not available and the road is narrow, walk on the right side of the road watching the on coming traffic.

2) What should we carry at night time while walking?
Answer:
Torch light.

3) What should we use for crossing roads?
Answer:
Zebra Lines.

4) Can we use mobile phones while walking on footpath?
Answer:
No, we cannot.

5) Pedestrians must use ……………… clothing at night when walking outside built up area.
Answer:
reflective

Question 9.
Read the following passage and answer the given questions.
Traffic Signals :
A traffic Light, traffic signal or a stop light is a signatory device position at a road intersection to indicate when it is safe to drive through. Follow traffic signals at the junction, i.e. Red- stop before line; Orange- get ready to go; Green- move the vehicle.

1) Write the names of signatory devices.
Answer:
Traffic light, traffic signal or stop light.

2) What does it indicate?
Answer:
It indicates when it is safe to drive through.

3) Where can we find signals?
Answer:
At the junctions.

4) Match the following :
1. Red ( ) a. Move the vehicles
2. Orange ( ) b. Get ready to go
3. Green ( ) c. Stop before line
Answer:
1 – c,
2 – b,
3 – a

AP State Syllabus AP Board 9th Class Social Studies Important Questions Chapter 23 Disaster Management.

AP State Syllabus 9th Class Social Studies Important Questions 23rd Lesson Women Protection Acts

9th Class Social 23rd Lesson Disaster Management 1 Mark Important Questions and Answers

Question 1.
Give examples of human induced disasters. (SA-III : 2016-17)
Answer:

1. Road accidents
2. Rail accidents
3. Air accidents
4. Fire accidents
5. Terrorism

9th Class Social 23rd Lesson Disaster Management Important Questions and Answers

Question 1.
Write about the air accidents.
Answer:
Across the world including India, air accidents have increased. Many factors govern the safety of the passengers in the aircraft like increase in the number of aircrafts, technical problems, fire, landing and take off condition, the environment that the airline operates in (mountainous terrains or frequent storms), factors like airport security in cases of hijackings, bomb attacks, etc.

Question 2.
Who should not drive vehicles?
Answer:
The following people should not be driving vehicles :

1. Having consumed alcohol.
2. Have been taking any medicine or drug that affects their responses.
3. Are tired, as tiredness affects the driving skills and reaction time.
4. Are sick or injured.
5. Are angry or upset.

Question 3.
What safety measures should be taken while air travel?
Answer:
Some safety measures that can be taken up are given below :

1. Pay attention to the flight crew safety demonstration.
2. Carefully read the safety briefing card, that is there in the seat pocket.
3. Know where the nearest safety emergency exit is and know how to open in case of emergency.
4. Always keep your seat belt fastened when you are in seat.

Question 4.
Collect the information about this.
Answer:
Date : 26-11-2008
Time : 20.00 hours
Attack type : Bombings, shootings, hostage crisis, siege.
Deaths : 166 including attackers
Injured : More than 600
Perpetrators : ISI, Lashkar-e-Toiba
Note : One was caught and executed later.

Question 5.
Read the following suggestions and answer the given questions.

i) Why should we inform police?
Answer:
As it could contain explosive device.

ii) What is the police control room number?
Answer:
100

iii) What should we do after informing the police?
Answer:
We should keep a watch on the suspected object.

iv) ………………. of any suspicious behaviour, parked vehicle.
Answer:’
Be alert

Question 5.
Read the following and answer the questions given below.
The 2004 fire in a school in Kumbakonam, Tamil Nadu sparked off debates and arguments on the safety of schools in the country. 93 innocent lives were charred to death. The main causes of this fire were lack of awareness amongst children and teachers as to what to do in case of a fire, lack of preparedness, kitchen close to the classroom etc. However, incidents like these are not new.
1) When was the fire sparked off?
Answer:
In 2004.

2) Where is Kumbakonam?
Answer:
It is in Tamil Nadu district.

3) What were the main causes of this fire?
Answer:
The main causes of this fire were lack of awareness amongst children and teachers.

4) Where is the kitchen?
Answer:
The kitchen is close to the classroom.

5) Who died in the accident?
Answer:
93 innocent lives were charred to death.

AP State Syllabus AP Board 9th Class Social Studies Important Questions Chapter 22 Women Protection Acts.

AP State Syllabus 9th Class Social Studies Important Questions 22nd Lesson Women Protection Acts

9th Class Social 22nd Lesson Women Protection Acts 1 Mark Important Questions and Answers

Question 1.
Many acts are made by the government of India to protect the rights of women and girls. Mention any two such acts enforced by the government of India. (SA-II : 2018-19)
Answer:

• Child marriage Act – 2006
• The Dowry Prohibition Act -1961
• Sexual Assault and Torturing Act – 2013
• The Immoral Trafficking (prevention) Act -1956 and Amendment – 2006.

9th Class Social 22nd Lesson Women Protection Acts 2 Marks Important Questions and Answers

Question 1.
Even after the ‘Prohibition of child marriage Act1, child marriages are still going on here and there. Suggest some measures to stop this dogma. (SA-III : 2016-17)
Answer:

1. Empower girls with information, skills and support networks.
2. Provide economic support and incentives to girsl and their families.
3. Educate and rally parents and community members.
4. Enhance girls.

Question 2.
The Government made many programmes and acts for girls, women protection and development. What is your suggestion for the better implementation?
Answer:
Govt made so many acts for women and girld child protection and developments.
Suggestions for better implementation :

1. To raise awareness on women’s right issues.
2. Enhancement of the political and legal status of women through political empowerment.
3. Protesting when violations against women, children take place.
4. Wider publicity, propaganda about acts to be made and educate people.
5. Campaign against injustice by coming on to the streets, approaching the courts and by sharing information.

9th Class Social 22nd Lesson Women Protection Acts Important Questions and Answers

Question 1.
What are children rights?
Answer:

• All the people below 18 years age are all children without gender discrimination.
• Protection from government for children rights.
• Right to live.
• Right to live with parents as far as possible.
• Right to gain knowledge and get awareness about the world through media like radio, newspapers, TV, etc.
• Right to protection from violence and harmful incidents.
• Right to get special care to live and for development in case of disabled children.
• Right to have good health and get medical care.
• Right to use mother tongue, follow their religion and tradition.
• Right to play.
• Right to avoid the harm for children from getting education and health.
• Right to get protection from using harmful medicines also manufacturing and purchasing.
• Right to get help when the children are neglected and being suffered.

Question 2.
What are the forms of domestic violence?
Answer:
Forms of Domestic Violence :

• Sexual abuse (forcible sexual intercourse, watching obscene films )
• Physical abuse (behaviour caused to bodily pain or harm, beating, torturing)
• Verbal and emotional abuse (humiliation, name calling or ridicule not speaking to them, ignoring.)
• Mental abuse (threaten to harm, not to send the children for study, stopping from jobs, force to marry whom she doesn’t like)
• Economic abuse (deprivation of economic or financial resources, use the women wealth, sell their property)

Question 3.
Write some of the important features of the ordinance on ‘Sexual Assault and Torturing.
Answer:

• Minimum 20 years of imprisonment.
• Recruitment of women police to address the complaints from the victims.
• It is not necessary for the victim to attend before the police officers individually.
• There will be no punishment if the accused dies in struggle at the time of attack with acid on women.
• There is a scope to shoot video at the time of complaining and trial on the request of the victims.

Question 4.
Who are eligible for judicial assistance?
Answer:

• Citizens belong to scheduled castes and scheduled tribes.
• Victims of immoral human trafficking, beggars, women, children, mentally imbalanced and physically challenged. „
• Victims of natural disasters, agricultural and industrial labourers and victims of Domestic Violence and Caste enmity.
• Citizens earning annual income less than Rs. 50,000.

Question 5.
What are the methods of Judicial Assistance?
Answer:
Methods of Judicial Assistance :

• Judicial advice is given free of cost by the advocate.
• After verification of the cases and if it is found necessary, the advocates may be appointed on behalf of the complaint and take up the cases in the court.
• To bear the expenditures of court and court fee.
• The xerox copies of the judgment will be given at free of cost to those who sought judicial assistance.

Question 6.
Comment on this picture.

Answer:
They both are couple. The man is angry and ready to kick his wife. She is cheerful and warning him with the section 498 A of I PC.
Note: Section 498 A of IPC, 1860: Husband or relative of husband of women subjecting her to cruelty – whoever beingthe husband orthe relative ofthe husband of a woman, subjects such woman to cruelty shall be punished with the imprisonment for a term which may extend to three years and shall also be liable to fine.

Question 7.
What about the Immoral Trafficking Act (1956) and amendment 2006? What are the different forms of trafficking?
Answer:
The Immoral Trafficking (Prevention) Act 1956 and amendment 2006 :
Girls and women are being lured in the name of providing jobs, bright future, offering roles in cinema etc., and take them to towns and cities and sell them for prostitution. Even they beat and, cause physical violence to make them obey. Certain times they remove body parts.

Trafficking and selling of them for prostitution or encouraging them to enter into prostitution is a punishable crime. Though they come on their will and interest is also treated a crime.

Different forms of trafficking :
Sexual assault

1. Forced prostitution
2. Social and religious prostitution (Devadasi, Jogini, Mathangi etc.)
3. Sexual assault in tourism
4. Reading rustic writings and watching obscene pictures

Activities against the law

1. It is a crime to make the trafficking children involve in the begging for money and beating them to obey. Certain times they make them?physically handicapped by removing their body parts and make them as beggars.
2. Removing body parts and doing business.
3. Selling narcotics with the trafficked children.

Labourers
1. Vetti :
Taking work without making proper payments, not providing nutrious food, clothes and even not treat them as human beings.

2. Domestic labour:
Washing clothes, cleaning the houses and utensils without proper wages, entrusting work in neighbours’ houses.

3. Agriculture Labour:
Involving in the agriculture work and exploiting them for more physical work.

4. Construction work :
Labour involved in construction of buildings, roads, etc. with nominal payments and providing them with substandard food.

Sadistic pleasure

1. Tying up the children on camels and make them run for entertainment
2. Marriages and adoption without their consent

Question 8.
Read the following passage and answer the given questions.
Lok Adalat means people’s court. All are equal before law. The Article 39-A of the Constitution of India, contains various provisions for settlement of disputes through Lok Adalat. It is an Act to constitute legal services authorities to provide free and competent legal services to the weaker sections of the society. It is to ensure that opportunities for securing justice are not denied to any citizen by reason of economic or other disabilities. And to organize Lok Adalats to secure that the operation of the legal system promotes justice on a basis of equal opportunity.

In addition to above, an Act was formulated which is called “Act of Judicial Services Authority.” This is a central law. The State government and High Court jointly formulated certain principles as per this Act.
1) What is Lok Adalat?
Answer:
It is people’s court.

2) Write about the Article 39 – A.
Answer:
The Article 39 – A of the Constitution of India, contains various provisions for settlement of disputes through Lok Adalat.

3) To whom does it help?
Answer:
It helps the weaker sections of the society.

4) What is the name of the Act?
Answer:
Act of Judicial Services Authority

5) This is a _______ law.
Answer:
central

Question 9.
Read the following passage and answer the given questions.

1) What is punishment mentioned here?
Answer:
Five years of imprisonment and a fine.

2) What is observed?
Answer:
It is observed that after the marriage the bride is teased, scolded, abused, beaten and sometimes killed or forced to commit suicide.

3) What are all these called?
Answer:
These are all called ‘Offences’.

4) What is the law for offerings? (Write any two)
Answer:
1) Enlisting the offerings
2) Offerings cannot be demanded.

5) The value of offerings shall not be _____ and ______ the parents’ economic status.
Answer:
Over, above

Question 10.
Write some misconceptions and realities of domestic violence.
Answer:
Domestic Violence – Misconceptions and Realities :
Misconception:
Parents punishing their children now and then is not a major issue to be considered.

Reality:
Domestic violence starts slowly and it becomes a routine, if nobody opposes it. The victims certain times are hospitalised for treatment and even lose their life and it adversely affects their dignity and self-respect.

Misconception :
Addiction to the drinking is the major cause for violence.

Reality :
The alcoholism for domestic violence is only an assumption but the statistics reveals that 40% of the men who beat their wives are not addicted to alcoholism.

Misconception :
Women shall bear the violence for the sake of children.

Reality :
Violence leads to more and more violence. Children that grow in the violent atmosphere will become more violent in future.

Misconception :
There is no way to get out of domestic violence.

Reality :
The violence free environment is the birth right of every woman. The Women Protecting Acts and laws are available and one should take their support.

AP State Syllabus AP Board 9th Class Social Studies Important Questions Chapter 21 Human Rights and Fundamental Rights.

AP State Syllabus 9th Class Social Studies Important Questions 21st Lesson Human Rights and Fundamental Rights

9th Class Social 21st Lesson Human Rights and Fundamental Rights 1 Mark Important Questions and Answers

Question 1.
What is a writ?
(OR)
What do you understand about ‘writt1? (SA-II : 2017-18)
Answer:

• Writ is an authority given to the court to issue directions to the government to protect and enforce any constitutional right.
• It is an inherent power given to the court.
• The court can “Suo motu” (on its own motion) take congnizance of a violation of fundamental rights.

Question 2.
Write any two slogans about abolition of child labour. (SA-II : 2017-18)
Answer:
1. Child is your heart not born to push your cart.
2. Children are the future heroes, they need to learn before they can work.

9th Class Social 21st Lesson Human Rights and Fundamental Rights 2 Marks Important Questions and Answers

Question 1.
Read the following situations. Which Fundamental Right is being violated in each case and how? (SA-II : 2018-19)
i. A director who makes a documentary film that criticizes the policies of the government is arrested and sent to prison.
ii. A ten years old boy is not allotted to go to school and is forced to work in a firecracker factory.
Answer:
i) In first case Right to freedom of speech and expression is violated. Everyone has a right to express their opinion on the administration of govt. But some limitations are there.
ii) In the second case, Right against exploitation is violated.
The Constitution says “no child below the age of 14 shall be employed engaged in any other hazardous employment.

Question 2.
Read the following concepts and write under the table with relevant fundamental right. (SA-II : 2017-18)

• Prohibition of employment for children in factories.
• All minorities shall have the right to establish and administer educational institutions of their choice.
• Abolition of titles.
• Right to life.

Answer:

9th Class Social 21st Lesson Human Rights and Fundamental Rights Important Questions and Answers

Question 1.
Write a short note about UDHR.
Answer:

• A group of people from 9 countries around the world formed by UNO drafted a list of 30 articles.
• This became the Universal Declaration of Human Rights.
• It was passed by the UN General Assembly in 1948.

Question 2.
What are the functions of NHRC?
Answer:

• NHRC is vested with the authority to make an inquiry, Suo motu or on a petition presented to it by a victim.
• It works for protecting human rights.
• It intervenes in any case involving human rights in the court or outside the court.

Question 3.
What are our fundamental duties?
Answer:
The fundamental duties to be discharged by the citizens were not there in our Constitution earlier. They were included through 42nd amendment. They are :

1. Respecting the constitution, National Flag and National Anthem.
2. Cherishing the noble ideals of the freedom struggle.
3. Defending the country and rendering national service when called for.
4. Upholding and protecting the sovereignty, unity and integrity of India.
5. Promoting harmony and the spirit of common brotherhood amongst all the people of India and renouncing any practice derogatory to the dignity of women.
6. Preserving the rich heritage of the Nation’s composite culture.
7. Protecting and improving the natural environment including forests, lakes, rivers and wild life and having compassion for living creatures.
8. Developing scientific temper, humanism and spirit of inquiry or reform.
9. Safeguarding public property and abjuring violence.
10. Striving for the excellence in all individual and collective activities.

Question 4.
What are the differences between the fundamental rights and fundamental duties?
Answer:
1) Fundamental rights :
a) Fundamental rights are definite.
b) No one can deprive us of these rights.
c) Even Government cannot ignore them.
d) These rights are safeguarded through court of law.

2) Fundamental duties :
a) No one can force us to abide to them.
b) The discharge of fundamental duties is left to our discretion.
c) No court can force us to discharge these duties.

Question 5.
Read the passage and answer the following questions.
Human Rights Commission :

The United Nations has been working to protect the basic human rights of people and in this effort they have encouraged their member nations to do the same. In 1993,the Government of India passed an act of legislature to protect human rights. A commission for human rights was set up under this law called the National Human Rights Commission (NHRC). The commission has been given powers of a civil court extendable to the entire country. The Government has also established other institutions to monitor and inquire into situations in which rights are violated. Thus we have different institutions such as the National Commission for Women, National Commission for Protection of Child Rights, National Minority Commission and State Human Rights Commissions.
i) When did the Government of India pass an act of legislature to protect human rights?
Answer:
In 1993 the Government of India passed a legislature.

ii) Expand NHRC.
Answer:
NHRC means National Human Rights Commission.

iii) What are the institutions that work on Human Rights?
Answer:
1) National Human Rights Commission.
2) National Commission for Women.
3) National Minority Commission.
4) State Human Rights Commission.

iv) What does the above Logo stand for?
Answer:
It is the logo of Human Rights designed by the United Nations Organization.

v) Which organization is behind the establishment of Human Rights Commission?
Answer:
The UNO is instrumental in establishing the Human Rights Commission.

Question 6.
Gather certain information about violation of fundamental rights and how the court protected them.
Answer:
I.

1. One factory was emitting dangerous gases and health of the people was effected.
2. A public interest litigation is filed in the court.
3. The factory was forced to close down. As it is against the fundamental right of Right to life.

II.

1. One child was working as a domestic servant in a family.
2. The family members went on a tour for two days.
3. The child (servant) was kept in the house and the door was locked from outside.
4. The neighbours recognized it. Given a complaint to the police.
5. The police broke open the doors and a case was filed against the head of the family.

Question 7.
Which of the following rights is available under the Indian Constitution?
a) Right to work
b) Right to adequate livelihood
c) Right to protect one’s culture
d) Right to privacy
Answer:
c) Right to protect one’s culture

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 7 Human Eye and Colourful World Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 7th Lesson Human Eye and Colourful World

10th Class Physics 7th Lesson Human Eye and Colourful World Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
What is the function of lens in human eye?
Answer:
The eye lens forms a real image on retina then we can see the objects.

Question 2.
How does it help to see objects at long distances and short distances?
Answer:
Eye lens can adjust itself in shape, so that it helps to see the objects at long and short distances.

Question 3.
How is it possible to get the image at the same distance on the retina?
Answer:
When the eye is focussed at distant object, the ciliary muscles are relaxed. So the focal length of eye lens adjusted itself which is equal to the distance of object from the retina. Then we can see the object clearly.
When the eye is focussed on a closer object the ciliary muscles adjust the focal length in such a way that the image is formed on retina and we see the object clearly.

Question 4.
Are we able to see all objects in front of our eye clearly?
Answer:
Yes. By maintaining the 25 cm distance of the object from our eye, we can see the objects in front of our eye.

Question 5.
How do the lenses used in spectacles correct defects of vision?
Answer:
To form image on retina.
To correct
myopia- Concave lens
hypermetropia i- Convex lens

Improve Your Learning

Question 1.
How do you correct the eye defect Myopia? (AS1)
Answer:

• Some people cannot see objects at long distances but can see nearby objects clearly. This type of defect in vision is called ‘Myopia’ or ‘Near sightedness’.
• Myopia is corrected by using a con-cave lens of focal length equal to the distance of the far point F from the eye.
• This lens diverges the parallel rays from distant object as if they are coming from the far point.
• Finally the eye lens forms a clear im-age at the retina.

Question 2.
Explain the correction of the eye defect Hypermetropia. (AS1)
Answer:

• A person with hypermetropia can see distant objects clearly but cannot see objects at near distances. This is also known as ‘far sightedness’.
• Eye lens can form a clear image on the retina when any object is placed beyond near point.
• To correct the defect of hypermetropia, we need to use a lens which forms an image of an object beyond near point at H, when the object is between H and L. This is possible only when a double con¬vex lens is used.
• The image acts like an object for the eye lens. Hence final image due to eye is formed at retina.

Question 3.
How do you find experimentally the refractive index of material of a prism? (AS1)
(OR)
Write the experimental procedure in finding the refractive index of a prism.
(OR)
Which quantity will decide whether a given medium is denser or rarer? How do you find that quantity of prism experimentally?
Answer:
Refractive index decides whether a medium is denser or rarer.
Aim :
To find the refractive index of a prism.

Materials required :
Glass prism, white chart of size 20 x 20 cm, pencil, pins, scale and protractor.

Procedure :

1. Keep a prism on white chart.
2. Draw the triangular base of the prism with pencil.
3. Remove the prism.
4. The shape of the outline drawn prism is triangle and name its vertices as P, Q and R.
5. PQ and PR be the refracting surfaces.
6. Find the angle between PQ and PR. This is the angle of the prism (A) (or) Refracting angle.
7. Mark M on PQ and draw a perpendicular line to PQ at M.
8. Place the centre of the protractor at M, along the normal and mark an angle of 30° and then draw a line up to M. This line denotes incident ray. This angle is called angle of incidence.
9. Place the prism in its position (triangle) again.
10. Now fix two pins vertically on the line at points A and B .
11. See the images of pins through the 2nd refracting side (PR).
12. Fix another two pins at points C and D such that all the four pins appear to lie along the same line.
13. Remove the pins and prism, join the pin-holes. Draw the incident and emergent rays.
14. The angle between the normal and the emergent ray at N is the angle of emergence.
15. The line passing through the points A, B, M, N, C and D represents the path of light when it suffers refraction through prism.

The angle of deviation :

• Extend incident and emergent rays are intercept at a point ‘O’.
• The angle between these two rays is the angle of deviation (d).
• Note the emergent deviation angles for different values of i, in the given table.

• Draw the graph between angle of incidence on X – axis and the angle of deviation on Y – axis.
• We notice that the angle of deviation decreases first and then increases with increase of the angle of incidence.
• Mark points on a graph paper and join the points to obtain a graph (smooth curve).
• Draw a tangent line to the curve, which parallel to X-axis, at the lowest point of the graph.

• The point where this line cuts the Y- axis gives the angle of minimum deviation. It is denoted by D.
• From the graph, we notice that, at angle of minimum deviation, the angle of incidence is equal to the angle of emergence.
• By finding A and D we can find refractive index of prism by using formula .

Question 4.
Explain the formation of rainbow. (AS1)
(OR)
What is the natural spectrum occuring in sky? Explain the formation of that spectrum.
Answer:

• A rainbow is a natural spectrum of sunlight in the form of bows appearing in the sky when the sun shines on rain drops.
• It is combined result of reflection, refraction and dispersion of sunlight from water droplets, in atmosphere.
• It always forms in the direction opposite to the sun.
• To see a rainbow, the sun be must behind us and the water droplets falls in front of us.
• When a sunlight enters a spherical raindrop, it is refracted and dispersed. The different colours of light are bent in different angles.
• When different colours of light falls on the back inner surface of drop, it (Water drop) reflects (different colours of light) interwnally (total internal reflection).
• The water drops again refract the different colours, when it comes out from the raindrop.
• After leaving this different colours from the raindrop as rainbow, reach our eye. Thus, we see a rainbow.

Question 5.
Explain briefly the reason for the blue of the sky. (AS1)
Answer:

• The reason for blue sky is due to the molecules N₂ and O₂.
• The size of these molecules are comparable to the wavelength of blue light.
• These molecules act as scattering centres for scattering of blue light.
• So scattering of blue light by molecules of N₂ and O₂ is responsible for blue of the sky.

Question 6.
Explain two activities for the formation of artificial rainbow. (AS1)
(OR)
Give two activities for the formation of the artificial rainbow? And explain it.
(OR)
Suggest an experiment to create a rainbow in your classroom and explain the procedure.
Answer:
Activity -1 :

• Select a white coated wall on which the sun rays fdll.
• Stand in front of a wall in such a way that the sun rays fall on your back.
• Hold a tube through which water is flowing.
• Place your finger in the tube to obstruct the flow of water.
• Water comes out from small gaps between the tube and finger like a fountain.
• Observe the changes on wall while showering the water.

Activity – II : (Activity – 4)

• Take a metal tray and fill with water.
• Place a mirror in water such that it makes an angle to the water surface.
• Now focus white light on the mirror through the water.
• Keep a white card board sheet above the water surface.
• We may observe the colours VIBGYOR on the board.
• The splitting of white light into different colours (VIBGYOR) is called dispersion.
• So consider a white light is a collection of waves with different wavelengths.
• Violet has shortest wavelength, and red has longest wavelength.

Question 7.
Derive an expression for the refractive index of the material of a prism. (AS1)
Answer:
Derivation of formula for refractive index of a prism :
PQ and PR are refracting surfaces of prism, i, is an angle of incidence, i2 is an angle of emergence, is an angle of refraction of first surface and r2 is an angle of incidence on second surface.

1. From triangle OMN, we get
d = i₁ – r₁ + i₂ – r₂
∴ d = (i₁ + i₂)-(r₁+ r₂) ……… (1)
2. From triangle PMN, we have
∠A + (90° – r₁) + (90° – r₂) = 180°
⇒ r₁ + r₂ = A ……… (2)
3. From (1) and (2),
⇒ d = (i₁ + i₂) – A
⇒ A + d = i₁ + i₂
4. This is the relation between angle of incidence, angle of emergence, angle of deviation and angle of prism.

5. From Snell’s law, we know that
n₁ sin i = n₂ sin r

6. Let ‘n’ be the refractive index of the prism.
7. Using Snell’s law at M, refractive index of air n₁ = 1; i = i₁; n₂ = n; r = r₁.
⇒ sin i₁ = n sin r₁ …………. (4)
8. Similarly, at N, n₁ = n; i = r₂; n₂ = 1; i₁ = i₂
⇒ n sin r₂ = n sin i₂ ………… (5)
9. We know that at the angle of minimum deviation position (D), i.e. i₁ = i₂
10. We will notice that MN is parallel to the base of prism QR.

14. This is the formula for the refractive index of the prism.

Question 8.
Light of wavelength λ₁ enters a medium with refractive index n₂ from a medium with refractive index n₁ What is the wavelength of light in second medium? (AS1)
Answer:
Wavelength of first medium =λ₁
Refractive index of first medium = n₁
Let the wavelength of second medium = λ₂,
Refractive index of second medium = n₂
Light enters from first medium to second medium = \(\Rightarrow \frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\lambda_{1}}{\lambda_{2}}\) (∵ υ = const)
\(\Rightarrow \lambda_{2}=\frac{\lambda_{1} \mathrm{n}_{1}}{\mathrm{n}_{2}}\)

Note:
For the below questions the following options are given. Choose the correct option by making hypothesis based on given assertion and reason. Give an explanation.
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.

Question 9.
Assertion (A) : The refractive index of a prism depends only on the kind of glass of which it is made of and the colour of light. (AS2)
Reason (R) : The refractive index of a prism depends on the refracting angle of the prism and the angle of minimum deviation.
Answer:
(b) Both A and R are true and R is not the correct explanation of A.

Question 10.
Assertion (A) : Blue colour of sky appears due to scattering of light.
Reason (R) : Blue colour has shortest wavelength among all colours of white light. (AS2)
Answer:
(C) A is true but R is false.

Question 11.
Suggest an experiment to produce a rainbow in your classroom and explain the procedure. (AS3)
Answer:
Activity to produce a rainbow in classroom :

• Take a prism and place it on the table near a vertical white wall.
• Take a thin wooden plank, make a small hole in it and fix it vertically on the table.
• Place the prism between the wooden plank and wall.
• Place a white light source behind the hole of the wooden plank. Switch on the light.
• Adjust the height of the prism such that the light falls on one of the lateral surfaces.
• Observe the changes in the emerged ray of the prism.
• Adjust the prism by slightly rotating it till you get an image on the wall.
• We observe a band of different colours on the wall.
• These colours are nearly equal to the colours of the rainbow, i.e., VIBGYOR.

Question 12.
Prisms are used in binoculars. Collect information why prisms are used in binoculars. (AS4)
Answer:

• Binoculars consists of a pair of identical or mirror symmetrical telescope mounted side by side and aligned to point accurately in the same direction, allowing the viewer to use both eyes, when viewing distant objects.
• The size of binoculars is reduced by using prisms.
• We get good image with more brightness.
• Objective size and optical quality should be increased by using prisms in binoculars.

Question 13.
Incident ray on one of the face (AB) of prism and emergent ray from the face AC are given in figure. Complete the ray diagram. (AS5)

Answer:

Question 14.
How do you appreciate the role of molecules in the atmosphere for the blue colour of the sky? (AS6)
(OR)
How do you appreciate the nature of molecules, responsible for the blue of the sky?
Answer:

• The sky appear blue due to atmospheric refraction and scattering of light through molecules.
• Molecules are scattering centres.
• The reason to blue sky is due to the molecules N2 and 02.
• The sizes of these molecules are comparable to the wavelength of blue light.
• In the absence of molecules there will be no scattering of sunlight and the sky will appear dark.
• We should appreciate the molecules which are scattering centres.

Question 15.
Eye is the only organ to visualise the colourful world around us. This is possible due to accommodation of eye lens. Prepare a six line stanza expressing your wonderful feelings. (AS6)
Answer:
“Many people simply see
The world in black and white
But through my eyes the world’s alive
So colourful and bright
Don’t close your mind to the sights
That light up the night and day
There’s so much to see here on this earth
And not a rupee do you have to pay”
The most obvious things aren’t the ones
There is beauty in the unknown
with willing eyes and an open mind
The true wonders you will be shown

Question 16.
How do you appreciate the working of ciliary muscles in the eye? (AS6)
(OR)
Which muscles are helpful in changing the focal length of eye lens? How do you appreciate those muscles?
Answer:

• Ciliary muscles are helpful in changing focal length of eye lens.
• The ciliary muscles which attached with eye lens help to change the focal length of eye lens.
• When the eye is focussed on a distant object, these are relaxed. So the focal length of eye lens increases to its maximum value.
• The parallel rays coming into the eye are focussed on the retina then we can see the object clearly.
• When the eye is focussed on a nearer object the muscles are strained so the focal length of eye-lens decreases. The ciliary muscles adjust the focal length and the image is formed on retina then we can see the object.
• Their process of adjusting focal length of eve lens is called accomodation.
• So ciliary muscles must be appreciated for its accomodation of eye lens.

Question 17.
Why does the sky sometimes appear white? (AS7)
(OR)
The sky sometimes appears white. What is the reason behind it?
Answer:

• Our atmosphere contains atoms and molecules of different sizes.
• According to their sizes, they are able to scatter different wavelengths of light.
• For example, the size of the water molecule is greater than the size of the N₂ or O₂.
• It acts as a scattering centre for light other frequencies which are lower than the frequency of blue light.
• On a hot day due to rise in the temperature, water vapour enters atmosphere which leads to abundant presence of water molecules in atmosphere. These water molecules scatter the colours of other frequencies (other than blue).
• All such colours of other frequencies reach our eye and white colour is appeared to us.

Question 18.
Glass is known to be transparent material. But ground glass is opaque and white in colour. Why? (AS7)
Answer:

• Ground glass is glass whose surface has been ground that produces a flat but rough finish.
• Ground glass has the effect of rendering the glass translucent by scattering of light during transmission thus blurring visibility while still transmitting light.
• To get more permanent frost, the glass may be ground by rubbing with some gritty substance.

Question 19.
If a white sheet of paper is stained with oil, the paper turns transparent. Why? (AS7)
(OR)
What is the reason behind the paper turns transparent when it is immersed in oil?
Answer:

• The paper fibres have higher index of refraction probably much greater than 1.5.
• The oil or fat also has a high index of refraction so that it nearly matches the index of refraction of the paper fibres and it reduces the scattering significantly.
• The fat adhering to the cellulose fibers lowers the index of refraction of the cellulose and also fills in air voids, so that visible light passes through the bag with significantly less scattering.
• The oil connects the fibres in the paper with a liquid which can transmit by refraction (rather than scatter) light that falls upon it. As a result, the paper stained with oil is turned transparent.

Question 20.
A light ray falls on one of the faces of a prism at an angle 40° so that it suffers angle of minimum deviation of 30°. Find the angle of prism and angle of refraction at the given surface. (AS7)
Answer:
Given that, incident ray on one of the prisms (i₁) = 40°
Angle of minimum deviation (D) = 30°
Angle of prism (A) = ?
A+D = 2i₁
⇒ A = 2i₁ – D
⇒ A = 2(40°) -30°
= 80° – 30°
= 50°
∴ Angle of prims (A) = 50°
Angle of refraction (r₁) = ?

Question 21.
The focal length of a lens suggested to a person with Hypermetropia is 100 cm. Find the distance of near point and power of the lens. (AS7)
Answer:
i) The distance of near point :
If the distance of near point is ‘d’ and focal length is ‘f’ then the relation between

Question 22.
A person is viewing an extended object. If a converging lens is placed in front of his eye, will he feel that the size of object has increased? Why? (AS7)
Answer:

• A simple magnifier allows us to put the object closer to the eye than we could normally focus and forms an enlarged virtual images.
• The principle behind this is angular magnification.
• The magnification is Ma = \(\frac{25}{f}\) for the close focus point, but since that causes eye strain, it is usually desirable to put the images at infinity giving Ma = \(\frac{25}{f}\)
• So he fees the size of object is increased. The reason is mentioned.
• Angular magnification: ‘The ratio of the angle substended at the eye by the image formed by an optical instrument to the angle substended at the eye by the object being viewed.”

Fill In The Blanks

1. The value of least distance of distinct vision is about ……………………… .
2. The distance between the eye lens and retina is about ……………………… .
3. The maximum focal length of the eye lens is about ……………………… .
4. The eye lens can change its focal length due to working of ……………………… muscles.
5. The power of lens is ID then focal length is ……………………… .
6. Myopia can be corrected by using ……………………… lens.
7. Hypermetropia can be corrected by using ……………………… lens.
8. In minimum deviation position of prism, the angle of incidence is equal to angle of ……………………… .
9. The splitting of white light into different colours (VIBGYOR) is called ……………………… .
10. During refraction of light, the character of light which does not change is ……………………… .
Answer:

1. 25 cm
2. 2.5 cm
3. 2.5 cm
4. ciliary
5. 100 cm
6. bi-concave
7. bi-convex
8. emergence
9. dispersion of light
10. frequency

Multiple Choice Questions

1. The size of an object as perceived by an eye depends primarily on
A) actual size of the object
B) distance of the object from the eye
C) aperture of the pupil
D) size if the image formed on the retina
Answer:
B) distance of the object from the eye

2. When objects at different distances are seen by the eye which of the following remains constant?
A) focal length of eye-lens
B) object distance from eye-lens
C) the radii of curvature of eye-lens
D) image distance from eye-lens
Answer:
D) image distance from eye-lens

3. During refraction, …………….. will not change.
A) wavelength
B) frequency
C) speed of light
D) all the above
Answer:
B) frequency

4. A ray of light falls on one of the lateral surfaces of an equilateral glass prism placed on the horizontal surface of a table as shown in figure. For minimum deviation of ray, which of the following is true?

A) PQ is horizontal
B) ‘QR’ is horizontal
C) ‘RS’ is horizontal
D) Either ‘PQ’ or ‘RS’ is horizontal
Answer:
B) ‘QR’ is horizontal

5. Far point of a person is 5 m. In order that he has normal vision what ki nd of spectacles should he use?
A) concave lens with focal length 5 m
B) concave lens with focal length 10 m
C) convex lens with focal length 5 m
D) convex lens with focal length 2.5 mm
Answer:
A) concave lens with focal length 5 m

6. The process of re-emission of absorbed light in all directions with different intensi¬ties by the atom or molecule is called ……………
A) scattering of light
B) dispersion of light
C) reflection of light
D) refraction of light
Answer:
A) scattering of light

10th Class Physics 7th Lesson Human Eye and Colourful World InText Questions and Answers

10th Class Physics Textbook Page No. 104

Question 1.
Can you imagine the shape of rainbow when observed during travel in an airplane?
Answer:

The shape of rainbow, when observed during travel in an airplane is circular as shown in the following figure.

10th Class Physics Textbook Page No. 88

Question 2.
Why do the values of least distance of distinct vision and angle of vision change with person and age?
Answer:

• The ciliary muscle helps the eye lens to change its focal length by changing radii of curvature of eye lens.
• When the eye is focussed on a distant object, the ciliary muscles are relaxed so that the focal length of eye lens has its maximum value which is equal to its distance from the retina.
• The working of ciliary muscle in eye changes from person to person.
• So, the values of least distance of distinct vision and angle of vision changes with person and age.

10th Class Physics Textbook Page No. 89

Question 3.
How can we get same image distance for various positions of objects?
Answer:
For different positions of object, the image distance remains constant only when focal length of lens changes.

Question 4.
Can you answer above question using concepts of refraction through lenses?
Answer:
The focal length of a lens depends on the material by which it has been made and radii of curvatures of lens. We need to change focal length of eye lens to get same image distance for various positions of object.

10th Class Physics Textbook Page No. 90

Question 5.
How does eye lens change its focal length?
(OR)
What is the role of ciliary muscles in the eye ? Write the answer in one or two sentences only.
Answer:
The ciliary muscles to which eye lens is attached help the eye lens to change its focal length by changing the radii of curvature of eye lens.

Question 6.
How does this change (eye lens changes its focal lengths) take place in eye ball?
Answer:
When the eye is focussed on a distant object the ciliary muscles are relaxed. So the focal length of eye lens has its maximum value which is equal to its distance from the retina. The parallel rays coming into the eye are then focussed on to the retina and we see the object clearly.

Question 7.
Does eye lens form a real image or virtual image?
Answer:
The eye lens forms a real and inverted image of an object on the retina.

Question 8.
How does the image formed on retina help us to perceive the object without change in its shape, size and colour?
Answer:

• The eye lens forms a real and inverted image of an object on the retina.
• This retina is a delicate membrane, which contains large number of receptors (125 million) called ‘rods’ and ‘cones’.
• They receive the light signals and identify the colour, and the intensity of light.
• These signals are transmitted to the brain through optic-nerve fibres.
• The brain interprets these signals and finally processes the information so that we perceive the object in terms of its shape, size and colour.

Question 9.
Is there any limit to change of focal length of eye lens?
Answer:
Yes. When the object is at infinity, the parallel rays from the object falling on the eye lens are refracted. They form a point-sized image on retina. In this situation, eye lens has a maximum focal length.

Question 10.
What are the maximum and minimum focal lengths of the eye lens?
Answer:
Maximum focal length is 2.5 cm and minimum focal length is 2.27 cm.

Question 11.
How can we find the maximum and minimum focal lengths of the eye lens?
(OR)
Calculate the maximum and minimum focal lengths of the eye lens.
Answer:’
a) When the object is at infinity,
u = – ∞ ; v = 2.5 cm (image distance which is equal to distance between eye lens and retina)

b) Object at a distance of 25 cm from eye. In this case, eye has minimum focal length.
Here u = -25 cm ; v 2.5 cm

10th Class Physics Textbook Page No. 91

Question 12.
What happens if eye lens is not able to adjust its focal length?
Answer:
In this case the person cannot see the object clearly and comfortably.

Question 13.
What happens if the focal length of eye lens is beyond the range of 2.5 cm to 2.27 cm?
Answer:
The vision (image) becomes blurred due to defects of eye lens.

10th Class Physics Textbook Page No. 92

Question 14.
What can we do to correct myopia?
Answer:
To correct myopia, we use concave lens.

10th Class Physics Textbook Page No. 93

Question 15.
How can you decide the focal length of the lens to be used to correct myopia?
Answer:
Assume that the object distance (u) is infinity and image distance (v) is equal to distance of far point.
u = – ∞ ; v = distance of far point = – D
Let ‘f be the focal length of bi-concave lens.

Here ‘f is negative showing that it is a concave lens.

Question 16.
What happens when the eye has a minimum focal length greater than 2.27 cm?
Answer:
In this case, the rays coming from the nearby object after refraction at eye lens, forms image beyond the retina.

10th Class Physics Textbook Page No. 94

Question 17.
How can you decide the focal length of convex lens to be used?
Answer:
Here object distance (u) = -25 cm
Image distance (v) = – d (distance of near point)
Let ‘f be the focal length of bi-convex lens.

If d > 25 cm, then ‘f’ becomes positive then we need to use bi-convex lens to correct hypermetropia.

10th Class Physics Textbook Page No. 95

Question 18.
Have you ever observed details in the prescription?
Answer:
A prescription that contains some information regarding type of lens to be used to correct vision.

Question 19.
What does it (“my sight is increased or decreased”) mean?
Answer:
Usually doctors, after testing the defects of vision prescribe corrective lenses indicating
their power which determines the type of lens to be used and its focal length.

Question 20.
What do you mean by power of lens ? flftorH)
Answer:
The reciprocal of focal length is called power of lens.

Question 21.
Doctor advised to use 2D lens. What is its focal length?
Answer:
Given power of lens P = 2D

∴ Focal length of lens (f) = 50 cm.

10th Class Physics Textbook Page No. 96

Question 22.
How could the white light of the sun give us various colours of the rainbow?
Answer:
Due to reflection, refraction and dispersion of sunlight.

Question 23.
What happens to a light ray when it passes through a transparent medium bounded by plane surfaces which are inclined to each other?
Answer:
When light incident on one of the plane surfaces and emerges from the other.

Question 24.
What is a prism?
Answer:
A prism is a transparent medium separated from the surrounding medium by consisting two refracting plane surfaces which are inclined.

10th Class Physics Textbook Page No. 97

Question 25.
What is the shape of the outline drawn for a prism?
Answer:
The outline drawn for a prism is in a triangle shape.

10th Class Physics Textbook Page No. 98

Question 26.
How do you find the angle of deviation?
Answer:
The angle between the extended incident and emergent rays is called angle of deviation.
(OR)
Extend both incident and emergent rays till they meet at a point ‘O’. Measure the angle between these two rays. This is the angle of deviation.

Question 27.
What do you notice from the angle of deviation?
Answer:
The angle of deviation decreases first and then increases with increase in the angle of incidence.

Question 28.
Can you draw a graph between angle of incidence and angle of deviation?
Answer:

Yes, we can draw the graph between angle of incidence and angle of deviation.

Question 29.
From the graph, can you find the minimum of the angle of deviation?
Answer:
Yes, we can. Draw a tangent line to the curve, parallel to X – axis, at the lowest point of the graph. The point where this line cuts Y – axis gives the angle of minimum deviation.

Question 30.
Is there any relation between the angle of incidence (i) angle of emergence (r) and piangle of deviation (d)?
Answer:
(i₁ + i₂)= A + D
i + r = A + D

10th Class Physics Textbook Page No. 101

Question 31.
In activity-3,we noticed that light has chosen different paths. Does this mean that the refractive index of the prism varies from colour to colour?
Answer:
Yes, refractive index of the prism varies from colour to colour.

Question 32.
Is the speed of light of each colour different?
Answer:
In vacuum – No, the speed of each colour is constant.
In medium – Speed is different for different colours.

Question 33.
Can you guess now, why light splits into different colours when it passes through a prism?
Answer:
Due to dispersion of light and different wavelength of colours in medium.

10th Class Physics Textbook Page No. 102

Question 34.
Does it (light passing through a prism) split into more colours? Why?
Answer:
We know the frequency of light is the properly of the source and it is equal to number of waves leaving the source per second. This cannot be changed by any medium. Hence frequency doesn’t change due to refraction. The coloured light passing through any transparent medium retains its colour.

Question 35.
Can you give an example in nature, where you observe colours as seen in activity 3?
Answer:
Yes, in rainbow. It is a good example of dispersion of light.

Question 36.
When do you see a rainbow in the sky?
Answer:
Dut to the refraction, reflection and dispersion of sunlight, when the sunlight passes through the rain drops, we can see the rainbow in the sky.

Question 37.
Can we create a rainbow artificially?
Answer:
Yes, we can create a rainbow artificially.

10th Class Physics Textbook Page No. 104

Question 38.
Why does the light dispersed by the raindrops appear as a bow?
Answer:

• A rainbow is not the flat two dimensional arc as it appears to us.
• The rainbow we see is actually a three dimensional cone with the tip of our eye.
• All the drops that disperse the light towards us lie in the shape of the cone – a cone of different layers.
• The drops that disperse red colour to our eye are on the outermost layer of the cone, similarly, the drops that disperse orange colour to our eye are on the layer of the cone beneath the red colour cone.
• In this way the cone responsible for yellow lies beneath orange and so on it till the violet colour cone becomes the innermost cone.

Question 39.
Why is the sky blue?
Answer:
A clear cloudless day – time sky is blue because molecules in the air scatter blue light from the sun more than thev scatter red liuht.

Question 40.
What is scattering?
A. Atoms or molecules which are exposed to absorb light energy and emit some part of the light energy in different directions is called scattering of light.

10th Class Physics Textbook Page No. 106

Question 41.
Why is that the sky appears white sometimes when you view it in certain direction on hot days?
Answer:

• On a hot day, due to rise in the temperature, water vapour enters atmosphere which leads to abundant presence of water molecules in atmosphere.
• These water molecules scatter the colours of other frequencies (other than blue).
• All such colours of other frequencies reach our eye and the sky appears white.

Question 42.
Do you know the reasons for appearance the red colour of sun during sunrise and at sunset?
(OR)
Sun appears red in colour during sunrise and sunset. Give reason.
(OR)
Why does sky appear red at Sunshine and Sunset?
Answer:

• The light rays from the sun travel more distance in atmosphere to reach our eye in morning and evening times.
• During sunrise and sunset except red light all colours scatter more and vanish before they reach us.
• Since scattering of red light is very less, it reaches us.
• As a result sun appears red in colour during sunrise and sunset.

Question 43.
Can you guess the reason why sun does not appear red during noon hours?
Answer:
During noon hours, the distance to be travelled by the sun rays in atmosphere is less than when compared to morning and evening hours. Therefore all colours reach our eye without scattering. Hence the sunlight appears white in noon hours.

10th Class Physics 7th Lesson Human Eye and Colourful World Activities

Activity – 1

Question 1.
How do you find least distance of distinct vision?
(OR)
What is the least distance a person can see an object comfortably and distinctly known as ? Write an activity to find that (least, distance of distinct vision) distance.
Answer:

• The least distance a person can see an object comfortably and distinctly is known as least distance of distinct vision.
• Hold the textbook at certain distance with your hands.
• Try to read the contents on the page.
• Gradually move the book towards eye, till it reaches very close to your eyes.
• You may see that printed letters on the page appear blurred or you feel strain to read.
• Now move the book backwards to a position where you can see clear printed letters without strain.
• Ask your friend to measure distance between your eye and textbook at this position.
• Note down its value.
• Repeat this activity with other friends and note down the distances for distinct vision in each case.
• Find the average of all these distances of clear vision.
• You will notice that to see an object comfortably and distinctly, you must hold it at a distance about 25 cm from your eyes.
• This 25 cm distance is called least distance of distinct vision.
• This value varies from person to person and with age.

Activity – 2

Question 2.
How can you find angle of vision?
(OR)
What is maximun angle a person is able to see whole object? Write an activity to find that angle.
Answer:

• The maximum angle, at which we are able to see the whole object is called angle of vision.
• Arrange a retort stand.
• Collect a few wooden sticks (or) PVC pipes that are used for electric wiring.
• Prepare sticks or pipes of 20 cm, 30 cm, 35 cm, 40 cm, 50 cm from them.
• Keep the retort stand on a table and stand near the table such that vour head is beside the vertical stand.
• Adjust the clamp on horizontal rod and fix at a distance of 25 cm from the eyes.
• Ask one of your friends to fix a wooden stick of 30 cm height to the clamp in vertical position.
• Now keeping your vision parallel to horizontal rod of the stand, try to see the top and bottom of wooden stick kept in vertical position.
• If you are not able to see both ends of stick at this distance (25 cm), adjust the vertical stick on the horizontal rod till you are able to see both ends of the stick at nearest possible distance from your eye.
• Fix the clamp to the vertical stick at this position.
• Without changing the position of the clamp on horizontal rod, replace this stick of 30 cm length.
• Try to see the top and bottom of the stick simultaneously without any change in the position of eye.
• Try the same activity with various lengths of the sticks.
• You can see the whole object AB which is at a distance of 25 cm (least distance of clear vision) because the rays coming from the ends A and B of the object will enter the eye.
• Similarly we can also see complete object CD with eye as explained above.
• Let us assume that AB is moving closer to eye to a position A B .
• You notice that you will be able to see only the part (EF) of the object A B because the rays coming from E and F enter your eye.

• The rays coming from A and B cannot enter your eye.
• The rays coming from the extreme ends of an object forms an angle at the eye.
• If this angle is below 60°, we can see whole object.
• If this angle is above 60°, then we see only the part of the object.
• This maximum angle, at which we are able to see the whole object is called angle of vision.
• The angle of vision for a healthy human being is about 60°.
• It varies from person to person and with age.

Activity – 3

Question 3.
Describe an activity for dispersion of light.
(OR)
What is the name given to process when white light passes through a prism it splits into different colours ? Explain the process with an activity.
Answer:

• The splitting of white light into different colours is called despersion of light.
• Do this experiment in the dark room.
• Take a prism and place it on the table near a vertical white wall.
• Take a thin wooden plank.
• Make a small hole in it and fix it vertically on the table.
• Place the prism between the wooden plank and wall.
• Place a white light source behind the hole of wooden plank.
• Switch on the light.
• The rays coming out of the hole of plank become a narrow beam of light.
• Adjust the height of the prism such that the light falls on one of the lateral surfaces.
• Observe the changes in emerged rays of the prism.
• Adjust the prism by slightly rotating it till you get an image on the wall.
• You will observe that white light is splitting into certain different colours.

Activity – 6

Question 4.
Describe an experiment for scattering of light.
(OR)
What is the principle involved in blue of the sky ? Explain the principle with an experiment?
Answer:

• Take the solution of sodium-thio-Sulphate (hypo) with sulphuric acid in a glass beaker.
• Place the beaker in which reaction is taking place in an open place where abundant sunlight is available.
• Watch the formation of grains of sulphur and observe changes in beaker.
• You will notice that sulphur precipitates as the reaction is in progress.
• At the beginning, the grains of sulphur are smaller in size as the reaction progresses, their size increases due to precipitation.
• Sulphur grains appear blue in colour at beginning and slowly their colour becomes white as their size increases.
• The reason for this is scattering of light.
• At the beginning, the size of grains is small and almost comparable to the wavelength of blue light.
• Hence they appear blue in the beginning.
• As the size of grains increases, their size becomes comparable to the wavelengths of other colours.
• As a result of this, they act as scattering centres for other colours.
• The combination of all these colours appears as white.

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 5th Lesson Refraction of Light at Plane Surfaces

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Textbook Questions and Answers

Improve Your Learning

Question 1.
Why is it difficult to shoot a fish swimming in water? (AS1)
(OR)
If the fish is swimming in water, why it is difficult to shoot?
(OR)
A shooter finds it difficult to shoot a fish swimming in water. Why?
Answer:
Due to refraction of light, it is difficult to shoot a fish swimming in water.

Reason :
The light rays coming from the fish towards shooter, bend at water-air interface. So, shooter sees only image of the fish, but not actual fish.

Question 2.
The speed of light in a diamond is 1,24,000 km/s. Find the refractive index of diamond if the speed of light in air is 3,00,000 km/s. (AS1)
Answer:
Speed of light in diamond = 1,24,000 km/s
Speed of light in vacuum = 3,00,000 km/s

Question 3.
Refractive index of glass relative to water is 9/8. What is the refractive index of water relative to glass? (AS1)
Answer:
Refractive index of glass relative to water = \(\frac{n_{g}}{n_{w}}=\frac{9}{8}\)
∴ Refractive index of water relative to glass = \(\frac{\mathrm{n}_{\mathrm{w}}}{\mathrm{n}_{\mathrm{g}}}=\frac{8}{9} \cdot\left[\because \mathrm{n}_{12}=\frac{1}{\mathrm{n}_{21}}\right]\)

Question 4.
The absolute refractive index of water is 4/3. What is the critical angle ? (AS1)
Answer:
Absolute refractive index of water = 4/3

Critical angle of water = C = 48°5′ = 48.5°.

Question 5.
Determine the refractive index of benzene if the critical angle is 42°. (AS1)
Answer:
Critical angle of benzene = 42°.

Question 6.
Explain the formation of mirage. (AS1)
(OR)
How is the mirage formed? Explain.
(OR)
A person walking on a road observed some water being present on the road but when he went there actually he did not find water. What is that actually formed called? Explain that process.
(OR)
Sometimes during the hot summer at noon time on tar roads, it appears that there is water on the road, but there would really be no water. What do you call this phenomenon? Explain why it happens.
(OR)
Why do you see a mirage the road on a hot summer day?
Answer:

• During hot summer day, air just above the road surface is very hot and the air at higher altitudes is cool.
• We know that refractive index of air increases with density.
• So, the cooler air at the top has greater refractive index than hotter air just above the road.

• Light travels faster through the thinner hot air than the denser cool air above it.
• On hot days, the temperature decreases with height.
• Thus the refractive index of air increases with height.
• When the light from a tall object such as tree or from the sky passes, through a medium just above the road whose refractive index decreases towards ground, suffers refraction and takes a curved path because of total internal reflection.
• This refracted light reaches the observer in a direction shown as in second figure.
• This appears to the observer that the ray is reflected from ground.
• Hence we will see water on road, which is the virtual image of sky and an inverted image of tree on the road.
• Such virtual images of distant high objects cause the optical illusion called ‘mirage’.

Question 7.
How do you verify experimentally that \(\frac{\sin i}{\sin r}\) is a constant? (AS1)
(OR)
Explain the experiment that shows the relation between angle of incidence and angle of refraction through figure.
(OR)
Write an experiment to obtain the relation between angle of incidence and angle of refraction.
Answer:
Aim:
Identifying relation between angle of incidence and angle of refraction.

Materials required :
A plank, white chart, protractor, scale, small black painted plank, a semi-circular glass disc of thickness nearly 2 cm pencil and laser light.

Procedure :

1. Take a wooden plank which is covered with white chart.
2. Draw two perpendicular lines, passing through the middle of the paper as shown in the figure (a).
3. Let the intersecting point be O.
4. Mark one line as NN which is normal to the another line marked as MM.
5. Here MM represents the line drawn along the interface of two media and NN represents the normal drawn to this line at ‘O’.
6. Take a protractor and place it along NN in such a way that its centre coincides with ‘O’ as shown in figure (b).
7. Then mark the angles from 0° to 90° on both sides of the line NN.
8. Repeat the same on the other side of the line NN.
9. The angles should be represented on circular line.
10. Now place semi circular glass disc so that its diameter coincides with the interface line (MM) and its centre coincides with the point O.
11. Take the laser light and send it along NN in such a way that the laser propagates from air to glass through the interface at point O and observe the way of laser light coming from other side of disc.
12. There is no deviation.
13. Send Laser light along a line which makes 15° (angle of incidence) with NN and see that it must pass through point O.
14. Measure its corresponding angle of refraction.
15. Repeat the experiment with angle of incidences of 20°, 30°, 40°, 50° and 60° and note the corresponding angles of refraction.

Observation :

• Find sin i, sin r for every i and r note down the values in table.

• Evaluate \(\frac{\sin i}{\sin r}\) for every incident angle i.
• We get \(\frac{\sin i}{\sin r}\) as constant.
• That is the relationship between angle of incidence and angle of refraction.
• The ratio of sin i and sin r is called refractive index.

Question 8.
Explain the phenomenon of total internal reflection with one or two activities. (AS1)
Answer:
Procedure :

1. Place the semi-circular glass disc in such a way that its diameter coincides with interlace line MM and its center coincides with point O’.
2. Now send light from the curved side of the semicircular glass disc.
3. The light travels from denser medium to rarer medium.
4. Start with angle of incidence (i), equals to 0° and observe for refracted on other side of the disc.
5. It does not deviate into its path when entering rarer medium.
6. Send laser light along angles of incidence 5°, 10°, 15°, etc. and measure the angle of refraction.
7. And tabulate the results in the given table.

Observation :

• Make a table shown below and note the values ‘i’ and ‘r’.

• At particular angle of incidence, the refracted ray does not come out but grazes the interface separating the air and glass. This angle is called critical angle.
• When the angle of incidence is greater than critical angle, the light ray gets reflected into denser medium at the interface, i.e. light never enters rarer medium. This phenomenon is called total internal reflection.

Question 9.
How do you verify experimentally that the angle of refraction is more than angle of incidence when light rays travel from denser to rarer medium? (AS1)
(OR)
When the light rays travel from denser to rarer medium, how can you explain, the angle of refraction is more than angle of incidence experimentally?
Answer:
Procedure :

• Take a metal disc. Use a protractor and mark angles along its edge as shown in the figure.
• Arrange two straws at the centre of the disc, in such a way that they can be rotated freely about the centre of the disc.
• Adjust one of the straws to make an angle 10°.
• Immerse half of the disc vertically into the water, filled in a transparent vessel. While dipping, verify that the straw at 10° must be inside the water.
• From the top of the vessel, try to view the straw which is inside the water as shown in the figure.
• Then adjust the other straw which is outside of the water until both straws look like they are in a single straight line.
• Then take the disc out of the water and observe the two straws on it. You will find that they are not in a single straight line.
• Measure the angle between the normal and second straw. Note the values in the following table.

• Do the same for various angles and find the corresponding angles of refraction and note them in the table.

Observation :
We will find the angle of refraction is more than angle of incidence.
i. e., r > i.

Conclusion :
When light travels from denser (water) to rarer (air) it bends away from the normal.

Question 10.
Take a bright metal ball and make it black with soot in a candle flame. Immerse it in water. How does it appear and why? (Make hypothesis and do the above experiment) (AS2)
Answer:

• The black metallic .ball appears to be raised up in the water because the path of the ray changes its direction at the interface, separating the two media, i.e., water and air.
• This path is chosen by light ray so as to minimize time of travel between ball and eye.
• This can be possible only when the speed of light changes at interface of two media.
• In another way the speed of light is different in different media.

Hypothesis :
Speed of light changes when it travels from one medium to another medium.

Question 11.
Take a glass vessel and pour some glycerine into it and then pour water up to the brim. Take a quartz glass rod. Keep it in the vessel. Observe the glass rod from the sides of the glass vessel.
1) What changes do you notice?
2) What could be the reasons for these changes? (AS2)
Answer:

1. We cannot see the glass rod in glycerine but we can see the rod in water.
2. We can also observe an apparent image of glass rod in water.
3. Reasons:
   i) Glycerine has essentially same refractive index as glass.
   ii) So, any light passing through these is bent equally.
   iii) Since both are transparent, it is not possible for our eye to distinguish the boundary by a change in the angle of reflection, and the glass seems to vanish.
   iv) But, the refractive index of glass and water are different.
   v) So the glass rod is visible to our eye in water. .

Question 12.
Do Activity-7 again. How can you find critical angle of water? Explain your steps briefly. (AS3)
Answer:

Procedure:

1. Take a cylindrical transparent vessel.
2. Place a coin at the bottom of the vessel.
3. Now pour water until you get the image of the coin on the water surface.
4. This is due to total internal reflection.

Critical angle of water :

1. Refractive index of water = 1.33
2. The sine of critical angle of water = \(\frac{1}{\text { Refractive index }}\)
3. Sin C = \(\frac{1}{\text { 1.33 }}\) ⇒ sin C = 0.7518.
   ∴ C = 8.7°
4. ∴ The critical angle of water = 48.7°.

Question 13.
Collect the values of refractive index of the following media. (AS4)

Answer:

Question 14.
Collect information on working of optical fibres. Prepare a report about various uses of optical fibres in our daily life. (AS4)
(OR)
What do you know about the working of optical fibres and make a report of various uses of optical fibres in our daily life?
(OR)
How are the optical fibres working? What are the various uses of optical fibres in our daily life?
Answer:
1) An optical fibre is very thin fibre made of glass or plastic having radius about a micrometer (10-6 m).
2) A bunch of such thin fibres form a light pipe.

Working :
1. Optical fibre having three parts-namely core (n = 1, 7), clading (n = 1, 6) and shielding.
2. The ray of light AB gets refracted at point ‘B’ into core and incident at ‘C’ with angle of incidence i (i > c).

3. The angle of incidence is greater than the critical angle and hence total internal reflection takes place.
4. The light is thus transmitted along the fibre.
5. The optical fibre is also based on ‘Fermat’s principle.

Uses :

1. Optical fibres are used in ‘endoscopy’ to see the internal organs like throat, stomach, intestines, etc.
2. Optical fibres are used in transmitting communication signals through light pipes.
3. Optical fibres are used in international telephone cables laid under the sea, in large computer networks, etc.
4. In optical fibre about 2000 telephone signals appropriately mixed with light waves may be simultaneously transmitted through a typical optical fibre.

Question 15.
Take a thin thermocol sheet. Cut it in circular discs of different radii like 2 cm, 3 cm, 4 cm, 4.5 cm, 5 cm etc. and mark centers with sketch pen. Now take needles of length nearly 6 cm. Pin a needle to each disc at its centre vertically. Take water in a large opaque tray and place the disc with 2 cm radius in such a way that the needle is inside the water as shown in figure. Now try to view the free end (head) of the needle from surface of the water.

1) Are you able to see the head of the needle?
Now do the same with other discs of different radii. Try to see the head of the needle, each time.
Note : The position of your eye and the position of the disc on water surface should not be changed while repeating the Activity with’other discs.
2) At what maximum radius of disc, were you not able to see the free end of the needle ?
3) Why were you not able to view the head of the nail for certain radii of the discs ?
4) Does this Activity help you to find the critical angle of the medium (water) ?
5) Draw a diagram to show the passage of light ray from the head of the nail in different situations. (AS4)
Answer:

1. Yes, we can see head of the needle.

2. Height of the pin = 6 cm
Radius of the disc = x cm
Critical angle of water = C = 48.7°
Tan C = 48.7°
\(\frac{x}{6}\) = 1.138 ⇒ x = 6.828 cm
So, at radius of 6.8 cm we cannot see the free end of the needle.

3. Because the light rays coming from object undergoing total internal reflection by touching the surface of disc.

4. Yes, we can find critical angle.
Refractive index of air (n₂) = 1.003 ; Refractive index of water (n₁) = 1.33
Sin C = \(\frac{n_{2}}{n_{1}}=\frac{1.003}{1.33}\) = 0.7541 ⇒ C = 48.7°

5.

Question 16.
Explain the refraction of light through the glass slab with a neat ray diagram. (AS5)
(OR)
Draw a glass slab diagram and explain the refraction of light through glass slab.
(OR)
Write the procedure of a lab Activity to understand lateral shift of light rays through a glass slab.
(OR)
How can you find lateral shift using glass slab?
Answer:
Aim :
A) Determination of position and nature of image formed by a glass slab.
B) Understanding lateral and vertical shift.
C) Determination of refractive index of given glass slab.

Materials required :
Plank, chart paper, clamps, scale, pencil, thin glass slab and pins.

Procedure :

1. Place a piece of chart on a plank. Clamp it. Place a glass slab in the middle of the paper.
2. Draw border line along the edges of the slab by using a pencil. Remove it. You will get a figure of a rectangle.
3. Name the vertices of the rectangle as A, B, C and D.
4. Draw a perpendicular at a point on the longer sides (AB) of the rectangle.
5. Now draw a line, from the point of intersection where side AB of rectangle and perpendicular meet, in such a way that it makes 30° angle with the normal.
6. This line represents the incident ray falling on the slab and the angle it makes with normal represents angle of incidence.
7. Now place the slab on the paper in such a way that it fits in the rectangle drawn. Fix two identical pins on the line making 30° angle with normal, such that they stand vertically with same height.
8. By looking at the two pins from the other side of the slab, fix two pins in such a way that all pins appear to be along a straight line.
9. Remove the slab and take out pins. Draw a straight line by joining the dots formed by the pins up to the edge CD of the rectangle.
10. This line represents emergent ray of the light.
11. Draw a perpendicular to the line CD where our last line drawn meets the line CD.
12. Measure the angle between emergent ray and normal.
13. This is called angle of emergence.
14. The angle of incidence and angle of emergence are equal.
15. Incident emergent rays are parallel.
16. The distance between the parallel rays is called shift.

Question 17.
Place an object on the table. Look at the object through the transparent glass slab. You will observe that it will appear closer to you. Draw a ray diagram to show the passage of light ray in this situation. (AS5)
Answer:

Question 18.
What is the reason behind the shining of diamond and how do you appreciate it? (AS6)
(OR)
For which reason is the diamond shining and how is it appreciable?
Answer:

• The critical angle of diamonds is very low, i.e., 24.4°.
• So if a light ray enters diamond, it undergoes total internal reflection.
• It makes the diamond shine brilliant.
• So total internal reflection is main cause of brilliance of diamonds.
• Majority of people are attracted towards diamonds due to this property.
• So we have to thoroughly appreciate total internal reflection for brilliance of diamonds.

Question 19.
How do you appreciate the role of Fermat’s principle in drawing ray diagrams? (AS6)
(OR)
How do you admire the role of Fermat’s principle in drawing ray diagrams? Fermat’s principle: The light ray always travels in a path which needs shortest possible time to cover distance between two points.
This principle has lot of importance on optics. This is used in

1. Laws of reflection (i.e., angle of incidence = angle of reflection)
2. Laws of refraction (Snell’s law)
3. To derive refractive index of a medium.
4. To derive refractive index of glass slab.
   So, I appreciate the Fermat’s principle.

Question 20.
A light ray is incident on air-liquid interface at 45° and is refracted at 30°. What is the refractive index of the liquid? For what angle of incidence will the angle between reflected ray and refracted ray be 90°? (AS7)
Answer:
i) Given that angle of incidence (i) = 4S°
angle of refraction (r) = 30°

ii) Given that angle between reflected and refracted ray is 90°.
We know angle of incidence = angle of reflection
∴ Angle of refraction (r) = 90 – angle of incidence
= 90 – i

Critical angle = 54.7°. This angle is also known as polarising angle.

Question 21.
Explain why a test tube immersed at a certain angle in a tumbler of water appears to have a mirror surface from a certain viewing position. (AS7)
Answer:
When a test tube is immersed at a certain angle in a tumbler of water appears to have
a mirror surface from a certain viewing positions due to total internal reflection.

Explanation :

• The critical angle for glass is 42°.
• The glass and air in test tube works as denser and rarer mediums.
• The rays of light while travelling through water strike glass – air interface of test tube at an angle of more than 42° (i > c) they get totally internal reflected as shown figure.
• When these reflected rays reach the eye, they appear to come from the surface of test tube itself.
• Now the test tube appears like silvary.

Question 23.
In what cases does a light ray not deviate at the interface of two media? (AS7)
Answer:

1. When a light ray incident is perpendicular to the interface of surface, it does not undergo deviation.
2. When a light ray incident is more than critical angle, it does not undergo deviation (refraction) but it undergoes reflection to come back into the original medium.

Question 25.
When we sit at camp fire, objects beyond the fire seen swaying. Give the reason for it. (AS7)
(OR)
What are the reasons for the objects beyond the fire seen swaying, when we sit at camp fire?
Answer:

• The temperature of the surrounding air changes due to convection of heat by the camp fire.
• This leads to chang in density and refractive index of air, continuously.
• The continuous change in refractive index of air changes the refracted path of the light ray.
• This is the cause for swaying of an object.

Question 26.
Why do stars appear twinkling? (AS7)
(OR)
What is the reason for the appearance of stars like twinkling?
Answer:

• The twinkling of a star is due to atmospheric refraction of star light.
• The atmosphere consists of a number of layers of varying densities.
• When light rays coming from a star pass through this layers and undergo refraction for several times.
• Thats why stars appear twinkling.

Question 27.
Why does a diamond shine more than a glass piece cut to the same shape? (AS7)
(OR)
What is the reason for shining of diamond brightly as compared to glass piece cut?
Answer:

• The critical angle of a diamond is very low (i.e., 24.4°).
• So if a light ray enters a diamond it definitely undergoes total internal reflection.
• Whereas it is not possible with glass piece cut to the same shape.
• So diamond shines more than a glass piece.

Fill In The Blanks

1. At critical angle of incidence, the angle of refraction is ……………… .
2. n₁ sin i = n₂ sin r, is called ……………… .
3. Speed of light in vacuum is ……………… .
4. Total internal reflection takes place when a light ray propagates from …………. to …………… medium.
5. The refractive index of a transparent material is 3/2. The speed of the light in that medium is …………… .
6. Mirage is an example of ……………… .
Answer:

1. 90°
2. Snell’s law
3. 3 × 10⁸ m/s
4. denser, rarer
5. 2 × 10⁸ m/s
6. optical illusion / total internal reflection

Multiple Choice Questions

1. Which of the following is Snell’s law?

Answer:
B)

2. The refractive index of glass with respect to air is 2. Then the critical angle of glass air interface is ………………….
A) 0°
B) 45°
C) 30°
D) 60°
Answer:
C) 30°

3. Total internal reflection takes place when the light ray travels from …………….. .
A) rarer to denser medium
B) rarer to rarer medium
C) denser to rarer medium
D) denser to denser medium
Answer:
C) denser to rarer medium

4. The angle of deviation produced by the glass slab is …………… .
A) 0°
B) 20°
C) 90°
D) depends on the angle formed by the light ray and normal to the slab
Answer:
D) depends on the angle formed by the light ray and normal to the slab

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Additional Questions and Answers

Question 1.
Derive Snell’s law.
(OR)
Prove n₁ sin i = n₂ sin r.
(OR)
Derive the Snell’s formula from Fermat’s principle.
(OR)
Derive the formula in realtion with and of incidence and angle of refraction.
Answer:
Let X be the path and A be the point above X and B be the point below X.
Now, we have to find the way from A to B.

1. Let us try to calculate how long it would take to go from A to B by the two paths through point D and another through point C.
2. If we draw a perpendicular DE, between two paths at D, we see that the path on line is shortened by the amount EC.
3. On the other hand, in the water, by drawing corresponding perpendicular CF we find that we have to go to the extra distance DF in water. These times must be equal since we assumed there was no change in time between two paths.
4. Let the time taken by the man to travel from E to C and D to F be ∆t and v₁ and v₂ be the speeds of the running and swimming. From figure we get,
   EC = v₁ ∆t and DF = v₂ ∆t
   ⇒ \(\frac{\mathrm{EC}}{\mathrm{DF}}=\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\) ………….. (1)
5. Let i and r be the angles measured between the path ACB and normal NN, perpendicular to shore line X.

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces InText Questions and Answers

10th Class Physics Textbook Page No. 56

Question 1.
Why should you see a mirage as a flowing water?
Answer:

• A mirage is a naturally occuring optical phenomenon, in which light rays are bent to produce a displaced image of distant objects or the sky.
• As light passes from colder air (higher place) to warmer air (lower place), the light ray bends away from the direction of the temperature gradient.
• Once the rays reach the viewer’s eye, the visual cortex interprets it as if it traces back along a perfectly straight “line of sight”. However this line is at a tangent to the path the ray takes at the point it reaches the eye.
• The result is that an “inferior image ” of the sky above appears on the ground.
• The viewer may incorrectly interprets this sight as water that is reflecting the sky, which is to the brain, a more reasonable and common occurrence.

Question 2.
Can you take a photo of a mirage?
Answer:

• Yes, I can take a photo of a mirage.
• Our eye can catches the total internal reflected rays.
• So, camera lens also catches the same.

10th Class Physics Textbook Page No. 46

Question 3.
What difference do you notice in fig 2(a) and Fig 2(b) with the respect to refracted rays?
(OR)
Draw the ray diagram of refraction in between denser and rarer medium.

Answer:
In figure 2(a) the light ray bends towards normal whereas in 2(b) the light ray bends away from the normal.

Question 4.
Is there any relation between behaviour of refracted rays and speed of the light?
Answer:
Yes. The speed of light changes when it travels from one medium to another medium. So the light may bend towards normal or away from normal.

10th Class Physics Textbook Page No. 47

Question 5.
Why db different material media possess different values of refractive indices?
Answer:
Refractive index depends on nature of material. So different media have different values of refractive indices.

10th Class Physics Textbook Page No. 48

Question 6.
On what factors does the refractive index of a medium depend?
Answer:
Refractive index depends on (1) Nature of material and (2) Wavelength of light used.

10th Class Physics Textbook Page No. 49

Question 7.
Can we derive the relation between the angle of incidence and the angles of refraction theoretically?
Answer:
Yes, we can derive the relation between angle of incidence and angles of refraction theoretically. We can get nt sin i = n2 sin r.

10th Class Physics Textbook Page No. 53

Question 8.
Is there any chance that angle of refraction is equal to 90° ? When does this happen?
Answer:
Yes, when angle of incidence is equal to critical angle then angle of refraction is equal to 90°.

10th Class Physics Textbook Page No. 54

Question 9.
What happens to light when the angle of incidence is greater than critical angle?
Answer:
When the angle of incidence is greater than critical angle, the light ray gets reflected into denser medium at the interface, i.e., light never enters rarer medium. This phenomenon is called total internal reflection.

10th Class Physics Textbook Page No. 57

Question 10.
How does light behave when a glass slab is introduced in its path?
Answer:
The light ray undergoes refraction two times.

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Activities

Activity – 1

Question 1.
Procedure :
Take some water in a glass tumbler. Keep a pencil in it. See the pencil from one side of glass and also from the top of the glass.
Observation:

1. How does it look?
Answer:
From the side it appears to be bent. From the top it appears as it is raising.

2. Do you find any difference between two views?
Answer:
Yes, the position of pencil is different.

Activity – 2

Question 2.
Procedure :

1. Go to a long wall (of length of 30 feet) facing the Sun. Go to the one end of a wall and ask someone to bring a bright metal object near the other end of the wall.
2. When the object is few inches from the wall, it will distort and we will see a reflected image on the wall as though the wall were a mirror.

Observation:

Why is there an image of the object on the wall?
Answer:
The image is due to refraction of light.

Activity-3 Refraction

Question 3.
Procedure: –

1. Take a shallow vessel with opaque walls such as a mug, a tin or a pan.
2. Place a coin at the bottom of the vessel.
3. Move away from the vessel until we cannot see the coin (fig. 2). Ask someone to fill the vessel with water. When the vessel is filled with water the coin comes back into view (fig. 3).

1. Why are you able to see the coin when the vessel is filled with water?
Answer:
The ray of light originated from the coin does not reach your eye when the vessel is empty. Hence you are not able to see the coin. But the coin becomes visible after the vessel is filled with water.

2. How is it possible? Do you think that the ray reaches your eye when the vessel is filled with water?
Answer:
Yes, it reaches the second instance.

3. What happens to the light ray at interface between water and air?
Answer:
It bends towards the normal.

4. What could be the reason for this bending of the light ray in the second instance?
Answer:
It is based on Fermat’s principle, which states that the light ray always travels in a path which needs shortest possible time to cover the distance between the two points.

Activity – 4

Question 4.
Prove that when light ray travels from denser to rarer medium it bends away from the normal.
Answer:


Procedure :

1. Take a metal disc. Use protractor and mark angles along its edge as shown in the figure.
2. Arrange two straws from the centre of the disk.
3. Adjust one of the straws to the angle 10°.
4. Immerse half of the disc vertically into the water, filled in a transparent vessel.
5. Inside the water the angle of straw should be at 10°.
6. From the top of the vessel try to view the straw which is inside the water.
7. Then adjust the other straw which is outside the water until both straws are in a single straight line.
8. Then take the disc out of the water and observe the two straws on it.
9. We will find that they are not in a single straight line.
10. It could be seen from the side view while half of the disc is inside the water.
11. Measure the angle between the normal and second straw. Draw table for various angles and corresponding angles of refraction.

Observation :
We observe that ‘r’ is greater than ‘i’ in all cases and when light travels from
denser to rarer medium it bends away from the normal.

Activity – 6

Question 5.
Why can we not see a coin placed in water from the side of glass?
Answer:
Procedure :

1. Take a transparent glass tumbler and coin.
2. Place a coin on a table and place glass on the coin.
3. Observe the coin from the side of the glass. We can see the coin.
4. Now fill the glass with water and observe the coin from the side of the glass tumber.
5. Now we cannot see the coin because the coin rises up due to refraction.

Activity – 7

Question 7.
Why can we see the coin in water from top? What is the phenomenon behind that?
Answer:

Procedure :

1. Take a cylindrical transparent vessel. Place a coin at the bottom of the vessel.
2. Now pour water until we will get the image of the coin on the water surface.
3. This is due to total internal reflection.

Activity – 8

Question 8.
Write an Activity to find refractive index of glass slab by calculating vertical shift.
(OR)
Explain the experiment with glass slab in determination of refraction through vertical shift.
Answer:

Procedure :

1. Take a glass slab and measure the thickness of the slab.
2. Take a white chart and fix it on the table.
3. Place the slab in the middle of the chart.
4. Draw line around it.
5. Remove the slab from its place.
6. The lines form a rectangle. Name the vertices of it as A. B, C and D. ‘
7. Draw a perpendicular to the longer line AB of the rectangle at any point on it.
8. Place slab again in the rectangle ABCD.
9. Take a needle. Place at a point P in such a way that its length is parallel to the AB on the perpendicular line at a distance of 20 cm from the slab.
10. Now take another needle and by seeing at the first needle from the other side of the slab, try to keep the needle so that it forms a straight line with the first needle. 1 1)
11. Remove the slab and observe the positions of the needles.
12. They are not in same line.
13. Draw a perpendicular line from the second needle to the line on which the first needle is placed.
14. Take the intersection point as Q.
15. The distance between P and Q is vertical shift.
16. We will get the same vertical shift placing needle at different distances.
17.