Class 10

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 3 Chemical Reactions and Equations Notes, Textbook Exercise Important Questions and Answers.

Std 10 Science Part 1 Chapter 3 Chemical Reactions and Equations Question Answer Maharashtra Board

Class 10 Science Part 1 Chapter 3 Chemical Reactions and Equations Question Answer Maharashtra Board

Question 1.
Choose the correct option from the bracket and explain the statement giving reasons :
(Oxidation, displacement, electrolysis, reduction, zinc, copper, double diplacement, decomposition)

a. To prevent rusting, a laver of ……… metal is applied on iron sheets.
Answer:
To prevent rusting, a layer of zinc metal is applied on iron sheets.
The rusting of iron is an oxidation process. Due to corrosion of an iron a deposit of reddish substance (Fe₂O₃.H₂O) is formed on it. This substance is called rust. To prevent corrosion, a layer of zinc metal (galvanisation) is applied on iron sheets.

b. The conversion or ferrous sulphate to ferric sulphate is …….. reaction.
Answer:
The conversion of ferrous sulphate to ferric sulphate is an oxidation reaction.
When ferric ion is formed. from ferrous ion, the positive charge is increased by one unit. while this happens the rerrous ion loses one electron. A process in which a metal or its ion loses one or more electrons is called an oxidation.
2FeSO₄ → Fe₂(SO₄)₃
Fe₂ + SO₄^2- → 2Fe³⁺ + SO₄^2-

c. When electric current is passed through acidulated water …….. of water takes place.
Answer:
when electric current is passed through acidulated water decomposition of water takes place. In this reaction. hydrogen and oxygen gas are formed.

This decomposition takes place with the help of an electric current, it is also called electrolytic decomposition.

d. Addition of an aqueous solution of ZnSO₄ to an aqueous solution of BaCl₂ is an example of ……… reaction.
Answer:
Addition of an aqueous solution of ZnSO₄ to an aqueous solution or BaCl₂ is an example or double displacement reaction.

Barium chloride reacts with zinc sulphate to form a white precipitate of barium sulphate. white precipitate is formed by exchange of ions Ba⁺⁺ and SO₄^— between the reactants.

Question 2.
a. What is the reaction called when oxidation and reduction take place simultaneously? Explain with one example.
Answer:
The reaction which involves simultaneous oxidation and reduction is called an oxidation-reduction or redox reaction.
In a redox reaction, one reactant gets oxidised while the other gets reduced during a reaction.
Redox reaction = Reduction + Oxidation

In redox reaction, the reductant is oxidized by the oxidant and the oxidant is reduced by the
reductant.
Example:CuO_(s) + H_2(g) → Cu_(s) + H₂O
In this reaction, oxygen is removed from copper oxide therefore it is a reduction of CuO, while hydrogen accepts oxygen to form water that means oxidation of hydrogen takes place. Thus oxidation and reduction reactions occur simultaneously.

Other examples of redox reactions:

b. How can the rate of the chemical reaction, namely, decomposition of hydrogen peroxide be increased?
Answer:
At room temperature, the decomposition of hydrogen peroxide into water and oxygen takes place slowly. However, the same reaction occurs at a faster rate on adding manganese dioxide (MnO₂),
powder in it.

c. Explain the term reactant product giving examples.
Answer:

1. The substance which undergoes bond breaking while taking part in a chemical reaction is called reactant.
2. The substance formed as a result of a chemical reaction by formation of new bonds is called product.
3. Example: In a chemical reaction, the formation of carbon dioxide gas takes place by combustion of coal in air. In this reaction, coal (carbon) and oxygen (from air) are the reactants while carbon dioxide is the product.

d. Explain the types of reactions with reference to oxygen and hydrogen. Illustrate with examples.
Answer:
With reference to oxygen and hydrogen, there are two types of reaction

1. Oxidation reaction
2. Reduction reaction.

1. Oxidation reaction:
Examples:
(1) When carbon burns in air, it forms carbon dioxide. In this reaction carbon accepts oxygen, therefore, this is an oxidation reaction.
C_(s) + O_2(g) → CO_2(g)

(2) When sodium reacts with ethyl alcohol, sodium ethoxide and hydrogen gas is formed. In this reaction, hydrogen is removed from ethyl alcohol, therefore this is an oxidation reaction.

(3) Acidified potassium dichromate (K₂Cr₂O₇ / H₂SO₄) oxidises ethly alcohol to acetic acid.

2. Reduction reaction:
Examples:
(1) When hydrogen gas is passed over black copper oxide a reddish coloured layer of copper is
formed.
In this reaction an oxygen atom removed from CuO to form copper, hence, this is reduction.

(2) when hydrogen gas is passed over red hot coke, methane is obtained.
Here, hydrogen is added to coke (carbon). Hence, this is reduction.

e. Explain the similarity and difference in two events, namely adding NaOH to water and adding CaO to water.
Answer:
Similarity : Both NaOH and CaO, when dissolved separately in water, solid NaOH dissolves releasing heat, resulting in rise in temperature. This reaction is exothermic reaction. When solid CaO dissolves in water, Ca(OH)₂ is formed, large amount of heat is evolved. This reaction is also exothermic reaction. Both reactions are combination reactions and single product is obtained.
NaOH_(s) + H₂O → NaOH_(aq) + Heat
CaO_(s) + H₂O → Ca(OH)_2(aq) + Heat
Difference:

1. Aqueous solution of NaOH is considered as a strong alkali.
2. Aqueous solution of Ca(OH)₂ is considered as a weak alkali.

Question 3.
Explain the following terms with examples.
a. Endothermic reaction
Answer:
Endothermic reaction: The reaction in which heat is absorbed is called an endothermic
reaction.
when KNO_3(s) dissolves in water, there is absorption of heat during the reaction and the temperature of the solution falls.
KNO_2(s) + H₂O_(l) + Heat → KNO_3(aq)

b. Combination reaction
Answer:
When two or more reactants combine in a reaction to form a single product, it is called a combination reaction.
Examples:
1. The ammonia gas reacts with hydrogen chloride gas to form the salt in gaseous state, immediately it condenses at room temperature and gets transformed into the solid state.

2. Magnesium burns in air to form white powder of magnesium oxide as a single product.

3. Iron reacts with sulphur to form iron sulphide.

c. Balanced equation
Answer:
In a chemical reaction, the number of atoms of the elements in the reactants is same as the number or atoms of those elements in the product, such an equation is called a balanced equation.
Example: AgNO₃ + NaCl → AgCl + NaNO₃
In the above reaction, the number of atoms of the elements in the reactants is same as the number of atoms of elements in the products.

d. Displacement reaction
Answer:
The reaction in which the place of the ion of a less reactive element in a compound is taken by another more reactive element by the formation of its own ions is called displacement reaction.

When zinc granules are added to the blue coloured copper sulphate solution, the zinc ions formed from zinc atoms take the place of Cu²⁺ ions in CuSO₄, and copper atoms, formed from Cu²⁺ ions comes out i.e. the more reactive zinc displaces the less reactive Cu from copper sulphate.

4. Give scientific reason:
a. When the gas formed on heating lime stone is passed through freshly prepared lime water, the lime water turns milky.
Answer:
when lime stone is heated, calcium oxide and carbon dioxide are formed. This carbon dioxide gas is passed through freshly prepared lime water, insoluble calcium carbonate and water are formed. In this reaction, lime water turns milky.

b. It takes time for pieces of Shahabud tile to disappear in HCl, but its powder disappears rapidly.
Answer:
The rate of a reaction depends upon the size of the particles of the reactants taking part in the reaction. The smaller the size of the reactants particles, the more is their total surface area and the faster is the rate of reaction.

In the reaction of dil. HCl with pieces of Shahabad tile, CO₂ effervescence is formed amid the tile disappears slowly. On the other hand. CO₂ effervescence forms at faster rate with Shahabad tile powder and it disappears rapidly.

c. While preparing dilute sulphuric acid from concentrated sulphuric acid in the laboratory, the concentrated sulphuric acid is added slowly to water with constant stirring.
Answer:
(1) The preparation of dilute sulphuric acid falls in the category of extreme exothermic process.

(2) During the preparation of dilute sulphuric acid. large amount of water is taken in a glass container which is surrounded by ice. Cool it for twenty minutes, Now small quantity of conc. H₂SO₄ is added slowly with stirring. Therefore, only a small amount of heat is liberated at a time. In this way dilute sulphuric acid is prepared.

(3) On the other hand, in the process of dilution or conc. sulphuric acid with water, very large amount of heat is liberated. As a result, water gets evaported instantaneously, if it is poured in to conc. H₂SO₄ which may cause an accident.

d. It is recommended to use air tight container for storing oil for long time.
Answer:

1. If edible oil is allowed to stand for a long time, it undergoes air oxidation, it becomes rancid and its smell and taste changes.
2. Rancidity in the rood stuff cooked in oil or ghee is prevented by using antioxidants. The process of oxidation reaction of food stuff can also be slowed down by storing it in air tight container.

Question 5.
Observe the following picture a write down the chemical reaction with explanation.

Answer:
The rusting of iron is an oxidation process. The rust on iron does not form by a simple reaction between oxygen and iron surface. The rust is formed by an electrochemical reaction. Fe oxidises to Fe₂O₃. H₂O on one part of iron surface while oxygen gets reduced to H₂O on another part or surface, Different regions on the surface of iron become anode and cathode.
(1) Fe is oxidised to Fe²⁺ in the anode region.
Fe_(s) → Fe²⁺ _(aq) + 2e^–
(2) O₂ is reduced to form water in the cathode region.
O_2(g) + 4H⁺ _(aq) + 4e— → 2H₂O_(l)

When Fe²⁺ ions migrate from the anode region they react with water and futher get oxidised to form Fe³⁺ ions.
A reddish coloured hydrated oxide is formed from Fe³⁺ ions. It is called rust. It collects on the
surface.
2Fe³⁺ _(aq) + 4H₂O_(l) → Fe₂O₃. H₂O_(s) + 6H⁺ _(aq)…
Because of various components in the atmosphere, oxidation of metals takes place, consequently resulting in their damage. This is called ‘corrosion’. Iron rusts and a reddish coloured layer is formed on it. This is corrosion of iron.

Question 6.
Identify from the following reactions the reactants that undergo oxidation and reduction.
a. Fe + S → FeS
Answer:
Fe + S → FeS
In this reaction, Iron (Fe) undergoes oxidation
and sulphur. (S) undergoes reduction.

b. 2Ag₂O → 4Ag + O₂↑
Answer:
2Ag₂O → 4Ag + O₂↑
In this reaction, reduction of Ag₂O takes place.

c. 2Mg + O₂ → 2MgO
Answer:
2Mg + O₂ → 2MgO
In this reaction, oxidation of Mg takes place.

d. NiO + H₂ → Ni + H₂O
Answer:
NiO + H₂ → Ni + H₂O
In this reaction, reduction of NiO takes place and oxidation of H₂ takes place.

Question 7.
Balance the following equation stepwise.
a. H₂S₂O_7(l) + H₂O_(l) → H₂SO_4(l)
Answer:
Step 1: Rewrite the given equation as it is
H₂S₂O7_(l) + H₂O_(l) → H₂SO_4(l)
Step 2: write the number or atoms of each element in the unbalanced equation on both sides of equations.

Element	Number of atoms in reactant (left side)	Number of atoms in products (right side)
H	4	2
S	2	1
O	8	4

Step 3: To equalise the number of hydrogen atoms, sulphur atoms and oxygen atoms we use 2 as the coemficient or factor in the product.

Element	Number of atoms in reactant (left side)	Number of atoms in products (right side)
H	4	2 × 2
S	2	1 × 2
O	8	4 × 2
Total	14	14

Now the equation becomes H₂S₂O₇ + H₂O → 2H₂SO₄
Now, count the atoms of each element on both sides of the equation. The number of atoms on both sides are equal. Hence, the balanced equation is
H₂S₂O₇ + H₂O → 2H₂SO₄
Now indicate the physical states of the reactants and products.
H₂S₂O_7(l) + H₂O_(l) → 2H₂SO_4(l)

b. SO_2(g) + H₂S_(aq) → S_(s) + H₂O_(l)
Answer:
Step 1:
Rewrite the given equation as it is
SO_2(g) + H₂S_(aq) → S_(s) + H₂O_(l)

Step 2:
Write the number of atoms of each element in the unbalanced equation on both sides of equations.

Element	Number of atoms in reactants (left side)	Number of atoms in products (right side)
S	2	1
O	2	1
H	2	2

The number of hydrogen atoms on both sides of the equation is same, therefore, equalise the number of sulphur atoms and oxygen atoms.

Step 3: To balance the number of sulphur atoms:

Number of atoms of sulphur	In reactants		In products
	S2O	H2S	(S)
Initially	1	1	1
To balance	1	1	1 × 2

To equalise the number of sulphur atoms, we use 2 as the factor in the product, now the equation becomes
SO₂ + H₂S → 2S + H₂O

Step 4:
To equalise the number of oxygen atoms in the unbalanced equation.

Number of atoms of oxygen	In reactants (SO2)	In products H2O
Initially	2	1
To balance	2	1 × 2

To equalise the number of sulphur atoms, we use 2 as the factor in the product i.e. H₂O, now the unbalanced equation becomes
SO₂ + H₂S → 2S + 2H₂O

Step 5:
To equalise the number of hydrogen atoms in unbalanced equation:

Number of atoms of hydrogen	In reactants (H2S)	In products (H2O)
Initially	2	4
To balance	2 × 2	4

To equalise the number of hydrogen atoms we use 2 as the factor in the reactant i.e, H₂S, now the unbalanced equation become
SO₂ + 2H₂S → 2S + 2H₂O
Now, count the atoms of each element on both sides of the equation, there are less number of sulphur atoms in the product. Now equalise the sulphur atoms, the balanced equation becomes,
SO₂ + 2H₂S → 3S + 2H₂O
Now indicate the physical states of reactants and products.
SO_2(g) + 2H₂S_(aq) → 3S_(s) + 2H₂O_(l)

c. Ag_(s) + HCl_(l) → AgCl ↓ + H₂ ↑
Answer:
Step 1:
Rewrite the given equation as it is
Ag_(s) + HCl_(l) → AgCl ↓ + H₂ ↑

Step 2:
write the number of atoms or each element in the unbalanced equation on both sides of equations.

Element	Number of atoms in reactants (left side)	Number of atoms in products (right side)
Ag	1	1
H	1	2
Cl	1	1

The number of silver and chlorine atoms on both sides of the equation are same, therefore, equalise the number of hydrogen atoms.

Step 3:
To balance the number of hydrogen atoms.

Number of atoms of hydrogen	In reactants HCl	In products H2
Initially	1	2
To balance	1 × 2	2

To equalise the number of hydrogen atoms, we use 2 as the factor in the product HCl, now the unbalanced equation become
Ag_(s) + 2HCl → AgCl + H₂

Step 4:
To balance the number of chlorine atoms:

Number of atoms of chlorine	In reactants (2HCl)	In products (AgCl)
Initially	2	1
To balance	2	2 ×1

To equalise the number of chlorine atoms, we use 2 as the factor in the product AgCl. now the unbalanced equation becomes
Ag + 2HCl → 2AgCl + H₂
Now count the atoms of each element on both sides of the equation, there are less number of silver atoms in the reactant. Now equalise the silver atoms, the balanced equation becomes
2Ag + 2HCl → 2AgCl + H₂
Now indicate the physical states of the reactunts and products
2Ag_(s) + 2HCl_(l) → 2AgCl ↓ + H₂ ↑

d. H₂SO_4(aq) + NaOH_(aq) → Na₂SO_4(aq) + H₂O_(l)
Answer:
Step 1:
Rewrite the given equation as it is
H₂SO_4(aq) + NaOH_(aq) → Na₂SO_4(aq) + H₂O_(l)

Step 2:
write the number of atoms of each element in the unbalanced equation on both sides of the equation.

Element	Number of atoms in reactants	Number of atoms in products
Na	1	2
S	1	1
O	5	5
H	3	2

The number of oxygen atoms involved in different compounds on both sides (reactants and products) are equal. Therefore, balance the number of atoms of the second element, sodium.

Step 3:
To balance the number of sodium atoms:

Number of atoms of sodium	In reactants	In products
To begin with	1 (in NaOH)	2 (in Na2SO4)
To balance	1 × 2	2

To equalise the number of sodium atoms, we use 2 as the factor of NaOH in the reactants. Now, the partly balanced equation becomes as follows
H₂SO₄ + 2NaOH → Na₂SO₄ + H₂O

Step 4:
Now, balance the number of hydrogen atoms:

Number of atoms of hydrogen	In reactants	In products
To begin with	(in H2SO4) 2 (in NaOH)	2 (in H2O)
To balance	4	2 × 2

To equalise the number of hydrogen atoms, we use 2 as the factor or H₂O in the products. The equation then becomes
H₂SO₄ + 2NaOH → Na₂SO₄ + H₂O
Now, count the atoms of each element on both sides of the equation. The number of atoms on both sides are equal. Hence, the balanced equation is
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Now indicate the physical states of the reactants and the products.
H₂SO_4(aq) + 2NaOH_(aq) → Na₂SO_4(aq) + 2H₂O_(l)

Question 8.
Identify the endothermic and exothermic reaction.
a. HCl + NaOH → NaCl + H₂O + heat
Answer:
Exothermic reaction.

b. \(2 \mathrm{KClO}_{3}(\mathrm{s}) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2} \uparrow\)
Answer:
Exothermic reaction.

c. CaO + H₂O → Ca(OH)₂ + heat
Answer:
Exothermic reaction.

d. \(\mathrm{CaCO}_{3}(\mathrm{s}) \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2} \uparrow\)
Answer:
Exothermic reaction.

Question 9.
Match the column in the following table:

Reactants	products	Type of chemical reaction
BaCl2(aq) + ZnSO4(aq)	H2CO3(aq)	Displacement
2 AgCl(s)	FeSO4(aq) + Cu(s)	Combination
CuSO4(aq) + Fe(s)	BaSO4↓ + ZnCl2(aq)	Decomposition
H2O(l) + CO2(g)	2Ag(s) + Cl2(g)	Double displacement

Answer:

Reactants	products	Type of chemical reaction
BaCl2(aq) + ZnSO4(aq)	BaSO4↓ + ZnCl2(aq)	Double displacement
2 AgCl(s)	2Ag(s) + Cl2(g)	Decomposition
CuSO4(aq) + Fe(s)	FeSO4(aq) + Cu(s)	Displacement
H2O(l) + CO2(g)	H2CO3(aq)	Combination

Project:
Do it your self:
1. Prepare aqueous solutions or various solid salts available in the laboratory. Observe what happens when aqueous solution of sodium hydroxide is added to these. Prepare a chart of double displacement reactions based on these observation.

2. Observe and note the physical and chemical changes experienced in various incidents in your day to day 1ife.

Can you recall? (Text Book Page No.16)

Question 1.
what are the types of molecules of elements and compounds?
Answer:
Elements are divided into three classes i.e. metals, nonmetals and metalloids. When two or more elements combine chemically in a fixed proportion by weight, a compound is formed. The properties of a compound are altogether different from those of the constitutional elements.

Question 2.
what is meant by valency of element?
Answer:
The number of electrons that an atom of an element gives away or takes up while forming an ionic bond, is called the valency or that element.

Question 3.
What is the requirement for writing molecular formulae of different compounds?
How are the molecular formulae of the compounds written?
Answer:
while writing the molecular formulae of different compounds, the symbol of the radicals and their valence should be known.
The number of the ions is written as subscript on the right of the symbol or the ion. By cross multiplication of valenceies chemical formula is obtained.

Find out (Text Book Page No. 44)
Question1.
How are the blackened silver utensils and patinated (greenish) brass utensils cleaned?
Answer:
The blackened silver utensils and patinated (greenish) brass utensils are cleaned using baking soda, vinegar and lemon mix.

Use your brain power! (Text Book Page No. 35)

Question 1.
write down the physical states of reactants and products in the reaction
SO₂ + 2H₂S → 3S + 2H₂O
Answer:
Reactants : SO_2(g), 2H₂S_(g)
Products : 3S_(s), 2H₂O_(l).

Question 2.
write down the physical states of reactants and products in the reaction
2Ag + 2HCl → 2AgCl + H₂
Answer:
Reactants: 2Ag_(s), 2HCl_(l)
Products: 2AgCl ↓, H₂ ↑

Question 3.
Identify the reactants and products of the following equation.

Answer:
Reactants: vegetable oil, H₂(g)
Product: Vanaspathi ghee

Use your brain power! (Text Book Page No. 42)

Question 1.
Which is the oxidant used for purification of drinking water?
Answer:
The chlorine based oxidants are used in the purification of drinking water.

Question 2.
Why is potassium permanganate used during cleaning water tanks?
Answer:
Potassium permanganate is an oxidising agent. It oxidises dissolved iron, manganese and hydrogen sulphide into solid particles that are filtered out of the water tank. It is used to control iron bacteria growth in tank.

Can you tell? (Text Book Page No. 43)

Question 1.
what is the type of this reaction, in which Vanaspathi ghee is formed from vegetable oil?
Answer:

In the preparation of vanaspathi ghee from vegetable oil hydrogen gas is used. This process is known as hydrogenation. This is reduction reaction.

Find out (Text Book page No. 33)

What are the other uses of silver nitrate in every day life?
Answer:
Silver nitrate is used in the voters-ink. It is used as reactant in the laboratory. Silver nitrate is used to prevent infection in wounds and skin burns.

Use your brain power! (Text Book Page No. 35)

Question 1.
N_2(g) + H_2(g) ⇌ NH_3(g)
Answer:
Step 1:
Rewrite the given equation as it is
N_2(g) + H_2(g) ⇌ NH_3(g)

Step 2:
Write the number of atoms of each element in the unbalanced equation on both sides of equations

Element	Number of atoms in reactants	Number of atoms in products
N	2	1
H	2	3

Step 3:
In the given equation. NH₃ is a compound and it contains hydrogen element. On the left hand side there are two H atoms and on the right side 3H atoms. Equalise H atoms on both sides.

Hydrogen atoms	In reactants	In products
Initially	2	3
To balance	3 × 2	2 × 3

To equalise the number of hydrogen atoms, we use 3 as the factor in the reactant and 2 as the factor in the products. Now the equation becomes
N₂ + 3H₂ → 2NH₃
Now, count the atoms of each element on both sides of the equation. The number of atoms on both sides are equal. Hence, the balanced equation is
N₂ + 3H₂ → 2NH₃
Now indicate the physical states of the reactants and products
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

Question 2.
Calcium chloride + Sulphuric acid → Calcium sulphate + Hydrogen chloride.
Answer:
Step 1:
Write the chemical equation from the given word equation.
CaCl₂ + H₂SO₄ → CaSO₄ + HCl

Step 2:
Write the number of atoms of each element in the unbalanced on both sides of equation.

Element	Number of atoms in reactants	Number of atoms in products
Ca	1	1
Cl	2	1
H	2	1
S	1	1
O	4	4

Step 3:
In the given equation H₂SO₄ is a compound and it contains hygrogen element. On the left hand side there are two hydrogen atoms and on the right side one hydrogen atom. Equalise H atoms on both sides.

Hydrogen atoms	In reactants (H2SO4)	In products (HCl)
Initially	2	1
To balance	2	2  × 1

To equalise the number of hydrogen atoms we use 2 as the factor in the product so that the number of H atoms on both sides are equal. Therefore, the equation becomes
CaCl₂ + H₂SO₄ → CaSO₄ + 2 HCl
Now, count the atoms of each element on both sides of the equation. The number of atoms on both sides are equal hence, the balanced equation is
CaCl₂ + H₂SO₄ → CaSO₄ + 2 HCl
Now, indicate the physical state of the reactants and products.
CaCl_2(s) + H₂SO_4(l) → CaSO_4(s) + HCl_(l)

Can you tell? (Text Book Page No. 39)

Take into account the time required for following processes. Classify them into two groups and give titles to the groups.
(1) Cooking gas starts burning on ignition.
(2) Iron article undergoes rusting.
(3) Erosion of rocks takes place to form soil.
(4) Alcohol is formed on mixing yeast in glucose solution under proper condition.
(5) Effervescence is formed on adding baking soda into a test tube containing dilute acid.
(6) A white precipitate is formed on adding dilute sulphuric acid to barium chloride solution.
Answer:
The above processes are classified into two groups (a) slow speed reactions (b) fast speed reactions.
Slow speed reactions: (2), (3) and (4).
Fast speed reactions: (1),(5) and (6).

Use your brain power! (Text Book Page No. 43)

Question 1.
Some more examples of redox reaction are as follows. Identify the reductants and oxidants from them.
(1) 2H₂S + SO₂ → 3S↓ + 2H₂O
(2) MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂↑
Answer:
Oxidants: SO₂, MnO₂
Reductants: H₂S, HCl

Question 2.
If oxidation means losing electrons, what is meant by reduction.
Answer:
Reduction means gaining one or more electrons.

Question 3.
Write the reaction of formation of Fe²⁺ by reduction Fe³⁺ by making use of the symbol (e^–).
Answer:
Fe³⁺ + e^– → Fe²⁺ (reduction)

Think about it (Text Book Page No. 43)

Question 1.
The luster of the surface of the aluminium utensils in the house is lost after a few days. Why does this happen?
Answer:
The aluminium utensils when kept in the house for a few days, oxidation of aluminium takes place, a thin laver aluminium oxide (Al₂O₃) is deposited on the surface. Hence, aluminium utensils lose their lustre in a few days.

Question 3.
How many products are formed in each of the above reactions?
Answer:
A single product is formed in each of the above reaction.

Use your brain power! (Text Book Page No. 39)

Question 1.
What is the difference in the process of dissolution and a chemical reaction.
Answer:
In the process of dissolution, new substance is not necessarily formed. Whereas in a chemical reaction a new substance is definitely formed.

Question 2.
Does a new substance form when a solute dissolves in a solvent?
Answer:
It is not necessary that a new substance is always formed.

Fill in the blanks:

Question 1.
Organic waste is decomposed by micro-organism and as a result manure and……..are formed.
Answer:
Organic waste is decomposed by micro-organism and as a result manure and bio gas are formed.

Question 2.
……….is formed on mixing yeast in glucose solution under proper condition.
Answer:
Alcohol is formed on mixing yeast in glucose solution under proper condition.

Question 3.
The chemical reaction during which H_2(g) is lost is termed as………
Answer:
The chemical reaction during which H_2(g) s lost is termed as oxidation.

Question 4.
Corrosion can be prevented by using………
Answer:
Corrosion can be prevented by using antirust solution.

Question 5.
The chemical reactions in which heat is liberated are called………..reactions.
Answer:
The chemical reactions in which heat is liberated are called exothermic reactions.

Question 6.
The chemical formula of rust is………
Answer:
The chemical formula of rust is Fe₂O₃.H₂O.

Question 7.
A reaction in which heat is absorbed is called………reaction.
Answer:
A reaction in which heat is absorbed is called endothermic reaction.

Question 8.
The process of rusting or iron is………process.
Answer:
The process of rusting of iron is oxidation process.

Question 9.
when oil and fats are oxidised or even allowed to stand in air for a long time, they become ……….
Answer:
when oil and fats are oxidised or even allowed to stand in air for a long time, they become rancid.

Question 10.
……… are used to prevent oxidation of food.
Answer:
Antioxidants are used to prevent oxidation of food.

Question 11.
Carbon dioxide is passed through water. The reaction is a………reaction.
Answer:
Carbon dioxide is passed through water. The reaction is a combination reaction.

Question 12.
Calcium carbonate is heated. The reaction is a………..reaction.
Answer:
Calcium carbonate is heated. The reaction is a decomposition reaction.

Question 13.
Zinc strip is dipped in a CuSO₄ solution. The reaction is a……….reaction.
Answer:
Zinc strip is dipped in a CuSO₄ solution. The reaction is a displacement reaction.

Question 14.
Silver nitrate solution is added to NaCl solution. The reaction is a……….reaction.
Answer:
Silver nitrate solution is added to NaCl solution. The reaction is a double displacement reaction.

Question 15.
The slow process of decay or destruction of a metal due to effect of air, moisture and acids on it is known as……….
Answer:
The slow process of decay or destruction of a metal due to effect of air, moisture and acids on it is known as corrosion.

Rewrite the following statements by selecting the correct options:

Question 1.
The reaction of iron nail with copper sulphate solution is………reaction. (March 2019)
(a) double displacement
(b) displacement
(c) combination
(d) decomposition
Answer:
(b) displacement

Question 2.
Reddish brown deposit formed on iron nails kept in a solution of copper sulphate is
(a) Cu₂O
(b) Cu
(c) CuO
(d) CuS
Answer:
(b) Cu

Question 3.
The reaction CuSO_4(aq) + Zn_(s) → ZnSO_4(aq) + Cu_(s) is a……..reaction.
(a) displacement
(b) double displacement
(c) decomposition
(d) combination
Answer:
(a) displacement

Question 4.
………is a combination reaction.
(a) Cu + H₂SO₄ → CuSO₄ + H₂
(b) H₂ + Cl₂ → 2HCl
(c) \(2 \mathrm{HgO} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Hg}+\mathrm{O}_{2}\)
(d) \(\mathrm{CaCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}\)
Answer:
(b) H₂ + Cl₂ → 2HCl

Question 5.
………..a decomposition reaction.
(a) \(\mathrm{CaCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}\)
(b) H₂O + CO₂ → H₂CO₃
(c) CaS + 2HCl → CaCl₂ + H₂S
(d) 2H₂ + O₂ → 2H₂O
Answer:
(a) \(\mathrm{CaCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}\)

Question 6.
In a chemical equation the……….are written on the left hand side.
(a) products
(b) reactants
(c) catalysts
(d) elements
Answer:
(b) reactants

Question 7.
The Δ sign written above the arrow indicates………..of the reaction.
(a) reactant
(b) product
(c) heat
(d) direction of the reaction
Answer:
(c) heat

Question 8.
The reaction KNO_3(S) + H₂O_(l) + Heat → KNO_3(aq) is a/an……….reaction.
(a) exothermic
(b) endothermic
(c) oxidation
(d) reduction
Answer:
(b) endothermic

Question 9.
The reaction NaOH_(S) + H₂O_(l) → NaOH_(aq) is a/an……..reaction.
(a) exothermic
(b) endothermic
(c) oxidation
(d) reduction
Answer:
(a) exothermic

Question 10.
A solution of Al₂(SO₄)₃ in water is……….
(a) blue
(b) pink
(c) green
(d) colourless
Answer:
(d) colourless

Question 11.
Carbon dioxide………..
(a) turns lime water milky
(b) is odourless
(c) is colourless
(d) All the three (a), (b) and (c) are correct
Answer:
(d) All the three (a), (b) and (c) are correct

Question 12.
……….is the correct set up to pass CO₂ through lime water.

Answer:
Correct set up D.

Question 13.
when……..is passed through fresh lime water, it turns milky.
(a) H₂
(b) CO
(c) CO₂
(d) SO₂
Answer:
(c) CO₂

Question 14.
Magnesium reacts with con. HCl to form………..salt.
(a) copper chloride
(b) ferrous chloride
(c) calcium chloride
(d) magnesium chloride
Answer:
(d) magnesium chloride

Question 15.
Zinc reacts with hydrochloric acid. The reaction is a reaction.
(a) combination
(b) decomposition
(c) displacement
(d) double decomposition
Answer:
(c) displacement

Question 16.
In a double displacement reaction,………… (Practice Activity Sheet – 1)
(a) ions remain at rest
(b) ions get liberated
(c) ions are exchanged
(d) ions are not created
Answer:
(c) ions are exchanged

State whether the following statements are True or False:

Question 1.
Rusting of iron is a fast reaction.
Answer:
False. (Rusting of iron is a slow reaction.)

Question 2.
Milk is set into curd is a chemical change.
Answer:
True.

Question 3.
The reaction between salt and water is an example of exothermic reaction.
Answer:
False. (The reaction between salt and water is an example of endothermic reaction.)

Question 4.
The speed of a chemical reaction depends on the catalyst used in the chemical reaction.
Answer:
True.

Question 5.
The simple form of representation of a chemical reaction in words is known as word reaction.
Answer:
True.

Question 6.
Nascent oxygen is always denoted by showing the symbol of oxygen.
Answer:
False. (Nascent oxygen is always denoted by showing symbol of oxygen [0] in square brackets.)

Question 7.
Antioxidants are used to prevent oxidation or food containing fats and oils.
Answer:
True.

Question 8.
When oils and fats are allowed to stand for a long time, they become rancid.
Answer:
True.

Question 9.
The chemical formula of rust is Fe₃O₄ .xH₂O.
Answer:
False. (The chemical formula or rust is Fe₂O₃ .xH₂O.)

Question 10.
Glucose combines with oxygen in our body and provides energy. The reaction is an endothermic reaction.
Answer:
False. (Glucose combines with oxygen in our body and provides energy. The reaction is an exothermic reaction.)

Question 11.
Chemical reactions in which reactants gain oxygen are reduction reactions.
Answer:
False. (Chemical reactions in which reactants gain oxygen are oxidation reactions.)

Question 12.
CuSO_4(aq) + Znl_(s) → ZnlSO_4(aq) + Cu_(s) is an example of decomposition reaction.
Answer:
False. (It is an example of displacement reaction.)

Question 13.
The chemical reactions in which heat is liberated are called endothermic reactions.
Answer:
False. (The chemical reactions in which heat is liberated are called exothermic reactions.)

Question 14.
The product or insoluble solid in chemical reaction is indicated by an arrow pointing upwards.
Answer:
False. (The product or insoluble solid in chemical reaction is indicated by an arrow ↑ pointing downwards.)

Question 15.
The rate of a reaction increases on increasing the temperature.
Answer:
True.

Question 16.
The digestion of food is a chemical decomposition process.
Answer:
True.

Question 17.
The reaction between sodium hydroxide and hydrochloric acid is a slow reaction.
Answer:
False (The reaction between sodium hydroxide and hydrochloric acid is a fast reaction.)

Question 18.
When calcium carbonate is heated, it decomposes into calcium oxide and oxygen gas.
Answer:
False (when calcium carbonate is heated. it decomposes into calcium oxide and carbon dioxide gas.

Question 19.
The rate of a chemical reaction changes in presence of catalyst.
Answer:
True.

Question 20.
Chlorines is an oxidant.
Answer:
True.

Taking into consideration the relationship in the first pair, complete the second pair. (OR) Complete the following:

Question 1.
2H₂ + O₂ → 2H₂O Combination reaction :: 2HgO → 2Hg + O₂ :……….
Answer:
Decomposition reaction

Question 2.
NH₃ + HCl → NH₄Cl : Combination reaction :: Fe + CuSO₄ → FeSO₄ + Cu :……..
Answer:
Displacement reaction

Question 3.
2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂ : Oxidation :: CuO + H₂ → Cu + H₂O :……….
Answer:
Reduction

Question 4.
CuCl₂ + 2KI → CuI₂ + 2KCl : Double displacement :: Zn + 2HCl → ZnCl₂ + H₂ :……….
Answer:
Displacement reaction

Question 5.

Answer:
Combination reaction

Question 6.
CuI₂ : Brown :: AgCl :……….
Answer:
White.

Match the column in the following table:

Question 1.

Reactants	products	Type of chemical reaction
Fe + S	NaCl + H2O	Oxidation
CuSO4 + Zn	2CuO	Neutralization
2Cu + O2	ZnSO4 + Cu	Displacement
HCl + NaOH	FeS	Combination

Answer:

Reactants	products	Type of chemical reaction
Fe + S	FeS	Combination
CuSO4 + Zn	ZnSO4 + Cu	Displacement
2Cu + O2	2CuO	Oxidation
HCl + NaOH	NaCl + H2O	Neutralization

Rewrite the second column so as to match the item from first column or Match the following:

Question 1.

Column I	Column II
1. Reduction	(a) Type of a chemical reaction
2. Oxidation	(b) Combination with hydrogen
3. Double displacement	(c) Losing hydrogen
4. Displacement	(d) Exchange of ions

Answer:
(1) Reduction – Combination with hydrogen
(2) Oxidation – Losing hydrogen
(3) Double displacement – Exchange of ions
(4) Displacement – Type of chemical reaction.

Question 2.

Column I	Column II
1. Oils and fats are allowed to stand in air for a long time	(a) Slow  reaction
2. NaOH dissolves in water	(b) Rancid
3. Zinc is added to CuSO4 solution	(c) Exothermic reaction
4. Rusting of water	(d) Colourless Solution

Answer:
(1) Oils and fats are allowed to stand in air for a long time – Rancid
(2) NaOH dissolves in water – Exothermic reaction
(3) Zinc is added to CuSO₂ solution – Colourless solution
(4) Rusting of iron – Slow reaction.

Question 3.

Answer:
(1) Combination reaction – 2Cu +O₄ → 2CuO
(2) Double displacement reaction – AgNO₃ + NaCl → AgCl ↓ + NaNO₃
(3) Decomposition reaction – \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11(\mathrm{s})} \stackrel{\Delta}{\longrightarrow} 12 \mathrm{C}_{(\mathrm{s})}+11 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\)
(4) Displacement reaction – Zn+ 2HCl → ZnCl₂ + H₂ ↑

Classify each of the following reactions as combination, decomposition, displacement or double displacement reactions:

Answer:
Combination reaction

Answer:
Decomposition reaction

Answer:
Displacement reaction

Answer:
double displacement reaction

Answer:
Combination reaction

Answer:
double displacement reaction

Name the following:

Question 1.
The product formed in the thermal decomposition of sugar.
Answer:
Carbon is formed in the thermal decomposition of sugar.

Question 2.
The gas evolved when sorghum metal reacts with ethanol.
Answer:
Hydrogen (H₂) gas is evolved when sodium metal reacts with ethanol.

Question 3.
The precipitate formed when barium sulphide reacts with zinc sulphate.
Answer:
When barium sulphide reacts with zinc sulphide, a precipitate of barium sulphate is formed.
\(\mathrm{BaS}+\mathrm{ZnSO}_{4} \longrightarrow \underset{\text { precipitate }}{\mathrm{BaSO}_{4}}+\mathrm{ZnS}\)

Question 4.
The reducing agent used for the reduction of copper oxide.
Answer:
Hydrogen is used for the reduction of copper oxide.

Question 5.
The catalyst used to accelerate the rate of decomposition of hydrogen peroxide.
Answer:
Manganese dioxide (MnO₂) is used as a catalyst to accelerate the rate of decomposition of hydrogen peroxide.

Question 6.
which oxidising agent is used to oxidise ferrous sulphate.
Answer:
Potassium permanganate (KMnO₄) is used as an oxidising agent to oxidise ferrous sulphate.

Question 7.
The product formed in the oxidation of ethyl alcohol.
Answer:
Acetic acid is formed in the oxidation of ethyl alcohol.

Answer the following questions in one sentence each:

Question 1.
what is meant by a chemical equation?
Answer:
The simple representation or a chemical reaction in a condensed form with the help of chemical formulae is called a chemical equation.

Question 2.
what is meant by a word equation?
Answer:
The simple form or representation or a chemical reaction in words is known as word equation.

Question 3.
what happens in a combination reaction?
Answer:
A single compound (product) is formed from two or more substances during a combination reaction.

Question 4.
what happens in a displacement reaction?
Answer:
In a displacement reaction. a more reactive element displaces another element, having less reactivity, from its compound.

Question 5.
what happens in a decomposition reaction?
Answer:
A single substance is broken down and two or more substances are formed during a decomposition reaction.

Question 6.
what happens in a double displacement reaction?
Answer:
A precipitate is formed by exchange of ions between the reactants during a double displacement reaction.

Question 7.
IdentIry the type of following reaction:
\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \stackrel{\Delta}{\longrightarrow} 12 \mathrm{C}+11 \mathrm{H}_{2} \mathrm{O}\) (Practice Activity Sheet – 2)
Answer:
The above reaction is a decomposition reaction.

Question 8.
what happens in an endothermic reaction?
Answer:
In an endothermic reaction, the reactants absorb heat to form products.

Question 9.
State the use of antioxidants in food containing fats and oils.
Answer:
Antioxidants are used to prevent oxidation of food containing fats and oils.

Question 10.
What are edible oils?
Answer:
Edible oils are compounds of alcohols and organic acids (carboxylic acids). The compounds formed are known as esters of carboxylic acids.

Question 11.
Is rancidity a phenomenon of oxidation or reduction?
Answer:
Rancidity is a phenomenon of oxidation.

Answer the following questions:

Question 1.
What do you understand by a physical change?
OR
Define physical change.
Answer:
The change in which only the physical state of a substance is changed; no new substance is formed. This change is temporary. During this change the composition of the substance does not change.

Question 2.
Explain giving two examples or physical change.
Answer:
(1) Conversion of ice into water is a physical change. On heating, ice melts into water. when the water is cooled, it freezes into ice. Thus, we get ice from water by a simple method and no new substance is formed. Hence, conversion of ice into water is a physical change.

(2) Magnetization of iron nail is a physical change. An iron nail magnetized by induction loses its magnetism as soon as it is detached from the magnet which induces magnetism in it. An iron nail magnetized by some other methods can also be demagnetized by simple means such as hammering or heating it. Thus, the magnetization of an iron nail can be easily reversed to get original nail. Hence, it is a physical change.

Question 3.
what do you understand by a chemical change?
OR
Define Chemical change.
Answer:
The change in which a substance or substances are converted into a new substance or substances, possessing properties altogether different from the original ones, is called a chemcial change. During this change, the original substance cannot be recovered by any simple means. This change is permanent.

Question 4.
Explain giving two examples of chemical change.
Answer:
(1) When carbon is burnt, carbon dioxide is formed. In this process carbon combines with oxygen, therefore carbon and oxygen are reactants, while curbon dioxide is a product. This change is permanent.
\(\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{g})} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{CO}_{2(\mathrm{g})}\)

(2) When a magnesium wire is burnt in air, a white powder of magnesium oxide is formed. We cannot obtain magnesium from magnesium oxide by simple methods. Properties of magnesium oxide are altogether different from those of magnesium. A new substance MgO is formed in the reaction. Hence, this change is a chemical change.
\(\begin{array}{c}
2 \mathrm{Mg} \\
\text { Magnesium }
\end{array}+\mathrm{O}_{2} \stackrel{\Delta}{\longrightarrow} \begin{array}{c}
2 \mathrm{MgO} \\
\text { Magnesium oxide }
\end{array}\)

Question 5.
What is meant by a chemical reaction?
Answer:
A process in which some substances undergo bond breaking and are transformed into new substances by formation of new bonds is called a chemical reaction.

Question 6.
What is the importance of a chemical equation?
Answer:

1. Reactants are converted into products.
2. Mass is conserved.
3. Atoms are conserved.
4. The properties and compositions of the products of a chemical reaction are different from those of its reactants.
5. Generally, energy is either absorbed or evolved.

Question 7.
What are the conventions used in writing a chemical equation?
Answer:
Conventions used in writing a chemical equation:
(1) The reactants are written on the left hand side (LHS), while the products are written on the right hand side (RHS).

(2) Whenever there are two or more reactants, a plus sign (+) is written between each two of them. Similarly, if there are two or more products, a plus sign is written between each two of them.

(3) Reactant side and product side are connected with an arrow (→) pointing from reactants to products. The arrow represents the direction of the reaction. Heat is to be given from outside to the reaction, it is indicated by the sign Δ written above the arrow.

(4) The conditions like temperature, pressure, catalyst, etc., are mentioned above the arrow (→) pointing towards the product side.

(5) The physical states of the reactants and products are also mentioned in a chemical equation. The notations g, l, s, and aq are written in brackets as a subscript along with the symbols / formulae of reactants and products. The symbols g, l, s, and aq stand for gaseous, 1iquid. solid and aqueous respectively.

If the product is gaseous, instead of (g) it can be indicated by an arrow ↑ pointing upwards. If the product formed is insoluble solid, then instead of (s) it can be indicated by an arrow ↓ pointing downwards.

(6) Special information or names of reactants/products are written below their formulae.

Write the balanced equations for the following reactions:

Question 1.
Ba(OH)₂ + HBr → BaBr₂ + H₂O
Answer:
Step 1:
Rewrite the given equation as it is
Ba(OH)₂ + HBr → BaBr₂ + H₂O

Step 2:
write the number of atoms of each element in the unbalanced equation on both sides of the equation.

Element	Number of atoms in reactants	Number of atoms in products
Ba	1	1
Br	1	2
O	2	1
H	3	2

Step 3:
To balance the number of oxygen atoms:

Number of atoms of oxygen	In reactants	In products
To begin with	2 [in Ba(OH)2]	1 (in H2O)
To balance	2	1 × 2

To equalise the number of oxygen atoms, we use 2 as the coefficient of H₂O in the product.
Now, the partly balanced equation become as follows
Ba(OH)₂ + HBr → BaBr₂ + 2H₂O

Step 4:
Now, balance the number of hydrogen atoms.
In the partly balanced equation:

Number of atoms of hydrogen	In reactants	In products
To begin with	2 [in Ba(OH)2] 1 (in HBr)	4 (in 2H2O)
To balance	1 × 2 + 2	4

To equalise the number of hydrogen atoms, we use 2 as the coefficient of HBr in the reactants. Now, the equation becomes
Ba(OH)₂ + 2HBr → BaBr₂ + 2H₂O
Now, count the atoms or each element on both sides of the equation. The number of atoms on both
sides are equal. Hence, the balanced equation is
Ba(OH)₂ + 2HBr → BaBr₂ +2H₂O
Now indicate the physical states of the reactants and products.
Ba(OH)_2(aq) + 2HBr_(aq) → BaBr_2(aq) +2H₂O_(l)

Question 2.
KCN + H₂SO₄ → K₂SO₄ + HCN
Answer:
Step 1:
Rewrite the given equation as it is
KCN + H₂SO₄ → K₂SO₄ + HCN

Step 2:
Write the number of atoms of each element or group in the unbalanced equation on both sides of the equation.

Element	Number of atoms in reactants	Number of atoms in products
K	1	2
CN (group)	1	1
O	4	4
H	2	1
S	1	1

The number of oxygen atoms involved in different compounds on both sides (reactants and products) are equal. Therefore, balance the number of atoms of the second element, potassium.

Step 3:
To balance K atoms:

Number of atoms of Potassium	In reactants	In products
To begin with	2 (KCN)	2 (in K2SO4)
To balance	1 × 2	2

To equalise the number of potassium atoms, we use 2 as the coefficient of KCN in the reactants.
Now, the partly balanced equation becomes
2KCN + H₂SO₄ → K₂SO₄ + 2HCN
Now, count the atoms of each element on both sides of the equation. The number of atoms on both sides are equal. Hence, the balanced equation is
2KCN + H₂SO₄ → KgSO₄ + 2HCN
Now indicate the physical states of the reactants and the products.
2KCN_(aq) + H₂SO_4(aq) → K₂SO_4(aq) + 2HCN_(g)

Question 3.
CH₄ + O₂ → 4CO₂ + H₂O
Answer:
Step 1:
Rewrite the given equation as it is
CH₄ + O₂ → 4CO₂ + H₂O

Step 2:
Write the number of atoms of each element in the unbalanced equation on both sides of the equation.

Element	Number of atoms in reactants	Number of atoms in products
C	1	1
O	2	3
H	4	2

Step 3:
To balance the number of oxygen atoms:

Number of atoms of oxygen	In reactants	In products
To begin with	2 (in O2)	1 (in H2O) 2 (in CO2)
To balance	2 × 2	1 × 2 + 2

To equalise the number of oxygen atoms, we use 2 as the coefficient of O₂ in the reactants and 2 as the coefficient of H₂O in the product.
Now, the partly balanced equation becomes
CH₄ + 2O₂ → CO₂ + 2H₂O
Now, count the atoms of each element on both sides of the equation. The number of atoms on both sides are equal. Hence, the balanced equation is,
CH₄ + 2O₂ → CO₂ + 2H₂O
Now, indicate the physical states of the reactants and the products.
CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(l)

Answer the following questions:

Question 1.
what are the different types of chemical reaction?
Answer:
Types of chemical reaction

1. Combination reaction
2. Decomposition reaction
3. Displacement reaction
4. Double displacement reaction.

Question 2.
What is meant by a combination reaction?
OR
Define: combination reaction.
Answer:
When two or more reactants combine in a reaction to form a single product, it is called a combination reaction.

Question 3.
Give two examples of combination reaction.
Answer:
Examples of combination reaction :
(1) The ammonia gas reacts with hydrogen chloride gas to form the salt in gaseous state, immediately it condenses at room temperature and gets transformed into the solid state.

(2) Magnesium burns in air to form white powder of magnesium oxide as a single product.

(3) Iron reacts with sulphur to form iron sulphide.

Question 4.
What Is meant by a decomposition reaction?
Answer:
The chemical reaction in which two or more products are formed from a single reactant is called decomposition reaction.

Question 5.
what is meant by a thermal decomposition?
Answer:
The reaction in which a compound is decomposed by heating it to a high temperature is called thermal decomposition.

Question 6.
What is meant by a electrolytic decomposition?
Answer:
The reaction in which a compound is decomposed by passing an electric current through its solution or molten mass is called an electrolytic decomposition.

Question 7.
Give two examples of thermal decomposition.
Answer:
(1) At high temperature, calcium carbonate decomposes into calcium oxide and carbon dioxide.

(2) At high temperature sugar decomposes into black mass of carbon and water vapour.

Question 8.
Give an example of electrolytic decomposition.
Answer:
When an electric current is passed through acidified water, it is electrolysed giving hydrogen
and oxygen.

Question 9.
Study the following reaction and answer the questions asked.

(a) State the type of reaction.
(b) Define this reaction. (Practice Activity Sheet – 1)
Answer:
(a) The type of reaction is electrolytic decomposition reaction.
(b) The reaction in which a compound is decomposed by passing an electric current through its solution or molten mass is called an electrolytic decomposition.

Question 10.
what is meant by a displacement reaction?
Answer:
The reaction in which the place of the ion of a less reactive element in a compound is taken by another more reactive element by formation of its own ions, is called displacement reaction.

Question 11.
Give an example of displacement reaction.
Answer:
when zinc granules are added to the blue coloured copper sulphate solution, the zinc ions formed from zinc atoms take the place or Cu²⁺ ions in CuSO₄, and copper atoms, formed from Cu²⁺ ions comes out i.e. the more reactive zinc displaces the less reactive Cu from copper sulphate.

Question 12.
Observe the reaction and answer the following questions.
CuSO_4(aq) + Fe_(s) → FeSO_4(aq) + Cu_(s)
(a) Identify and write the type of chemical reaction.
(b) write the definition of above reaction. (Practice Activity Sheet – 3)
Answer:
(a) When iron powder is added to the blue coloured copper sulphate solution, the iron ions formed from iron atoms take the place or Cu²⁺ ions in CuSO₄, and copper atoms, formed from Cu²⁺ ions comes out i.e. the more reactive iron displaces the less reactive Cu from copper sulphate. Therefore this reaction is a displacement reaction.

(b) The reaction in which the place of the ion of a less reactive element in a compound is taken by another more reactive element by formation of its own ions, is called displacement reaction.

Question 13.
what is meant by a double displacement reaction?
Answer:
The reaction in which the ions in the reactants are exchanged to form a precipitate is called double displacement reaction.

Question 14.
Give two examples of double displacement reaction.
Answer:
(1) Solutions of sodium chloride and silver nitrate react with each other forming a precipitate of silver chloride and a solution of sodium nitrate.

White precipitate of AgCl is formed by exchange of ions Ag⁺ and Cl^– between the reactants.

(2) Barium suiphide reacts with zinc sulphate to form zinc sulphide and a white precipitate of barium sulphate.

white precipitate is formed by exchange of ions Ba⁺⁺ and SO₄^— between the reactants.

Question 15.
Write down what you understand from the following chemical reaction:
AgNO_3(aq) + NaCl_(aq) → AgCl ↓ + NaNO_3(aq)
Answer:
(i) The above reaction is a double displacement reaction.
(ii) AgNO₃ and NaCl are the reactants while AgCl and NaNO₃ are the products.
(iii) The reactants and the product NaNO₃ are in aqueous state. The product AgCl is formed in the form of precipitate.

Question 16.
Study the following chemical reaction and answer the questions given below:
AgNO_3(aq) + NaCl_(aq) → AgCl_(s)↓ + NaNO_3(aq)
(i) Identiry and write the type of chemical reaction.
(ii) write the definition of above type of chemical reaction.
(iii) Write the names of reactants and products of above reaction. (March 2019)
Answer:
(i) The type of chemical reaction: Double displacement reaction.
(ii) The reaction in which the ions in the reactants are exchanged to form a precipitate is called double displacement reaction.
(iii)

1. The above reaction is a double displacement reaction.
2. AgNO₃ and NaCl are the reactants while AgCl and NaNO₃ are the products.
3. The reactants and the product NaNO₃ are in aqueous state. The product AgCl is formed in the form of precipitate.

Question 17.
When sodium chromate solution is mixed with barium sulphate solution, a precipitate is formed.
(i) What is the colour of the precipitate formed?
(ii) Name the precipitate.
(iii) What is the type of chemical reaction?
Answer:
(i) The colour of the precipitate is yellow.
(ii) The yellow precipitate formed is barium chromate.
(iii) The type of chemical reaction is double displacement.

Question 18.
Explain the term Exothermic reaction.
Answer:
Exothermic reaction : The process in which heat is given out is called an exothermic reaction.
When NaOH_(s) dissolves in water, there is evolution of heat leading to a rise in temperature.
NOH_(s) + H₃O_(l) → NaOH_(aq) + Heat

Question 19.
State whether the following reactions are exothermic or endothermic:
(i) 3CaO. Al₂O_3(s) + 6H₂O_(l) → 3CaO. Al₂O₃. 6H₂O_(s) + Heat
Answer:
Exothermic reaction

(ii) 2CaSO₄. H₂O + 3H₂O → 2CaSO₄. 2H₂O + Heat
Answer:
Exothermic reaction

(iii) KNO_3(aq) + H₂O_(l) + Heat → KNO_3(aq)
Answer:
Endothermic reaction

(iv) NaOH_(s) + H₂O_(l) → NaOH_(aq) + Heat
Answer:
Exothermic reaction

(v) Transformation of ice into water.
Answer:
Endothermic reaction

(vi) Water turns into ice.
Answer:
Exothermic reaction

(vii) Cooking of food.
Answer:
Endothermic reaction

(viii) Burning candle.
Answer:
Exothermic reaction

Question 20.
What do you mean by slow speed reaction?
OR
Define: Slow speed reaction.
Answer:
The reaction which requires long time for completion i.e. occurs slowly is called slow speed reaction.

Question 21.
What do you mean by fast speed reaction?
OR
Define: Fast speed reaction.
Answer:
The reaction which is completed in short time i.e. occurs rapidly is called fast speed reaction.

Question 22.
Give two examples of slow speed reactions.
Answer:
(1) On heating potassium chlorate (KClO₃) it decomposes slowly into potassium chloride and oxygen gas.

This reaction requires long time for completion, therefore it is slow speed reaction.

(2) Rusting of iron is a slow speed reaction. In this reaction iron reacts with oxygen from air to form iron oxide.

Question 23.
Give two examples of fast speed reactions.
Answer:
(1) The reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) is a neutralization reaction and it is fast speed reaction.
NaOH_2(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)
This neutralizaton reaction is completed in short time, therefore it is fast speed reaction.

(2) Aqueous solution of sodium chloride reacts with silver nitrate solution to form white precipitate of silver chloride (NaCl) and sodium nitrate.
NaCl_(aq) + AgNO_3(aq) → NaNO_3(aq) + AgCl ↓
This reaction is completed in short time, therefore it is fast reaction.

Question 24.
Write a short note on slow speed and fast speed reactions.
Answer:
Slow speed reaction:
The reaction which requires long time for completion i.e. occurs slowly is called slow speed reaction.
Examples:
(1) On heating potassium chlorate (KClO₃) it decomposes slowly into potassium chloride and oxygen gas.

This reaction requires long time for completion, therefore it is slow speed reaction.

(2) Rusting of iron is a slow speed reaction. In this reaction iron reacts with oxygen from air to form iron oxide.

This reaction requires long time for completion.

Fast speed reaction:
The reaction which is completed in short time i.e., occurs rapidly is called fast speed reaction.
Examples:
(1) The reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) is a neutralization reaction and it is fast speed reaction.
NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)
This neutralizaton reaction is completed in short time, therefore it is fast speed reaction.

(2) Aqueous solution of sodium chloride reacts with silver nitrate solution to form white precipitate of silver chloride (NaCl) and sodium nitrate.

This reaction is completed in short time, therefore it is fast reaction.

Question 25.
State the factors which affect the speed (or rate) of a reaction.
Answer:
The factors which affect the rate of a reaction are

1. Nature of the reactants.
2. Size of the particles of the reactants.
3. Concentration of the reactants.
4. Temperature of the reaction.
5. Catalyst.

Question 26.
How does the rate of reaction depend on the nature of the reactants? Illustrate with suitable example.
Answer:
(1) when the reactant combines with two or more other reactants then the rate of a chemical reaction depends on the nature of the reactants.

(2) Both Al and Zn reacts with dilute hydrochloric acid, H₂ gas is liberated and water soluble salts of these metals are formed. However, aluminium metal reacts faster with dil. HCl as compared to zinc metal.

(3) Al is more reactive than Zn. Therefore, the rate of reaction of Al with hydrochloric acid is higher than that of Zn. Hence, the nature of the reactant affect the rate of a reaction.

Question 27.
How does the rate of a reaction depend on the size of the particles of reactants?
Answer:
(1) In the reaction of dil. HCl and Shahabad tile, CO₂ effervescence is formed slowly. On the other hand, C₂ effervescence forms at faster speed with the powder of Shahabad tile.

(2) The above observation indicates that the rate of a reaction depends upon the size of the particles of the reactants taking part in the reaction. Smaller the size of the reactant particles taking part in a reaction faster will be the rate of reaètion.

Question 28.
How does the rate of a reaction depend upon the concentration of the reactants? Give suitable example.
Answer:
(1) A chemical reaction takes place due to collisions of the reactant molecules. Higher the concentrations of the reactants more will be the frequency of collisions and faster wifi be the rate of the reaction.

(2) In the reaction of dil. HCl and CaCO₃, CaCO₃ disappears slowly and CO₂ also liberates slowly. On the other hand the reaction with concentrated HCl takes place rapidly and CaCO₃ disappears fast.

(3) Concentrated acid reacts faster than dilute acid, that means the rate of a reaction is proportional to the concentration of reactants.

Slow reaction:
CaCO₃ + dli.2HCl → CaCl₂ + CO₂ + H₂O
Fast reaction:

Question 29.
How does the rate of a reaction depend upon the temperature of reactants? Give suitable example.
Answer:
(i) (1) When the temperature of the reactants is increased, the reactant molecules start moving with more velocity and their kinetic energy increases. As a result, the number collisions increases. Hence, the rate of chemical reaction increases.

(2) Lime stone on heating decomposes to give CO₂, which turns 1ime water milky. On the other hand, the lime water does not turn milky before heating the lime stone: because of the zero rate of reaction. The above observation indicates that the rate of a reaction increases on increasing the temperature.

(ii) Solid CaCO₃ does not decompose at room temperature when heated, it decomposes to give CaO and CO₂ that means the rise in temperature increases the rate of reaction. CaCO₃ room temperature No chemical reaction.

Question 30.
How does the rate if a reaction depend upon the catalyst? Give suitable example.
Answer:
(1) The substance in whose presence the rate of a chemical reaction changes, without causing any chemical change to it is called a catalyst.

(2) On heating potassium chlorate (KClO₃) decomposes into potassium chloride and oxygen slowly.
\(2 \mathrm{KClO}_{3} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{KCl}+3 \mathrm{O}_{2}\)
The rate of the above reaction neither increases by reducing the particle size nor by increasing the reaction temperature. However in the presence of manganese dioxide, KClO₃ decomposes at a comparatively lower temperature and oxygen is produced more briskly. No chemical change takes place in MnO₂ in this reaction. It acts as catalyst.

Question 31.
State the Importance of rate in a chemical reaction.
Answer:

1. The use of strong acid and strong base in a chemical reaction increases the rate of reaction.
2. In a chemical reaction, if the smaller size of the reactant particles, the concentrated solution, high temperature and use of catalyst increases the rate of chemical reaction.
3. The rate of chemical reaction is important with respect to environment.
4. If the rate of chemimal reaction is fast it is profitable for the chemical factories.
5. The ozone layer in the earth’s atmosphere protects the life of earth from the ultraviolet radiation of the sun. The process of depletion or maintenance of this layer depends upon the rate of production or destruction of ozone molecules.

Question 32.
Define Oxidation reaction.
Answer:
Oxidation: The chemical reaction in which a reactant combines with oxygen or loses hydrogen to form the product is called oxidation reaction.

Question 33.
Give examples of oxidation.
Answer:
(1) when carbon burns in air, it forms carbon dioxide. In this reaction carbon accepts oxygen, therefore, this is an oxidation reaction.
C_(s) + O_2(g) → CO_2(g)

(2) when sodium reacts with ethyl alcohol, sodium ethoxide and hydrogen gas is formed. In this reaction, hydrogen is removed from ethyl alcohol, therefore this is an oxidation reaction.

(3) Acidified potassium dichromate: (K₂Cr₂O₇/H₂SO₄) oxidises ethly alcohol to acetic acid.

Question 34.
What do you mean by oxidant? Explain with suitable example.
Answer:
The chemical substances which bring about an oxidation reaction by making oxygen available are called oxidants or oxidizing agents.

1. In the combustion of carbon, oxygen is an oxidant.
2. In the oxidation of ethly alcohol, potassium dichromate is used as oxidant.

Question 35.
Name the various oxidants. How nascent oxygen is liberated from these oxidants?
Answer:
K₂Cr₂O₇/H₂SO₄, KMnO₄/H₂SO₄ are the commonly used chemical oxidants. Hydrogen peroxide (H₂O₂) is used as a mild oxidant. Ozone (O₃) is also a chemical oxidant. Nascent oxygen is generated by chemical oxidants and it is used for the oxidation reaction.
O₃ → O₂ + [O].
H₂O₂ → H₂O + [O]
K₂Cr₂O₇ + 4H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O + 3[O]
2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 3H₂O + 5[O]
Nascent oxygen is a state prior to the formation of the O₂ molecule. It is the reactive form of oxygen and is represented by the symbol as [O]

Question 36.
Acidified potassium permanganate (KMnO₄) is a chemical oxidant and explain, how acidified potassium permanganate oxidise ferrous sulphate (FeSO₄). Accordingly write a new definition of oxidation and reduction.
Answer:
Acidified KMnO4 oxidises ferrous sulphate (FeSO₄) to ferric sulphate Fe₂(SO₄)₃ and in addition to above K₂SO₄ and MnSO₄ by-products are formed.
2KMnO₄ + 10 FeSO₄ → 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O
2FeSO₄ → Fe₂(SO₄)₃
Ionic reaction :
2Fe₂ + 2SO₄²⁺ → 2Fe³⁺ + 3SO₂^2-
Net Ionic equation Fe²⁺ → Fe³⁺ + e^–
When ferric ion is formed from ferrous ion the positive change is increased by one unit. While this happens the rerrous ion loses one electron.

When metal or its ion loses electron, it is called an oxidation and gain of electron is called reduction.

Question 37.
Define reduction reaction.
Answer:
The chemical reaction in which a reactant gains hydrogen and loses oxygen to form the product is called the reduction reaction.

Question 38.
Give two examples of reduction.
Answer:
(1) When hydrogen gas is passed over black copper oxide a reddish coloured layer of copper is formed.
In this reaction an oxygen atom removed from CuO to form copper, hence, this is reduction.

(2) When hydrogen gas is passed over red hot coke, methane is obtained.
Here, hydrogen is added to coke (carbon). Hence, this is reduction.

Question 39.
What do you mean by reductant? Explain with suitable example.
Answer:
The chemical substances which bring about reduction by making hydrogen available are called reductant. In the preparation of methane from carbon, hydrogen is a reductant.

Question 40.
What are redox reactions? Identify the substances that are oxidised and the substances that are reduced in the following reactions:
(1) 2H₂S_2(g) + SO_2(g) → 3S_(s) + 2H₂O_(l)
(2) CuO_(s) + H_2(g) → CU_(s) + H₂O_(l)

Answer:
When oxidation and reduction take place simultaneously in a given chemical reaction, it is known as a redox reaction.
(1)

H₂S is oxidised and SO₂ is reduced.

(2)

CuO is reduced and H₂ is oxidised.

(3)

(1) Oxidation: H₂S, H₂O and HCl.
(2) Reduction: SO₂, CuO and MnO₂

Question 41.
Observe the following reaction and answer the questions given below:
BaSO₄ + 4C → BaS + 4CO
(1) what type of reaction is it? Justify.
(2) Give one more example.
Answer:
(1) This is a redox reaction. In this reaction the reduction of BaSO4 and oxidation of carbon take place simultaneously.

(2) Example
CuO+H₂ → Cu + H₂O
2H₂S + SO₂ → 3S + 2H₂O

Question 42.
What is corrosion?
Answer:
The slow process of decay or oxidation of metals due to various components of atmosphere is known as corrosion.
Iron rusts and a reddish coloured layer is collected on it. This is corrosion of iron.

Question 43.
How does rusting of iron occur?
Answer:
Iron when exposed to moist air forms a reddish layer of hydrated ferric oxide.

Question 44.
How can corrosion be prevented?
Answer:

1. Corrosion damages buildings, bridges, automobiles, ships, iron railings and other articles made of iron.
2. It can be prevented by using an anti-rust solution, coating the surface by a paint, processes like galvanising and electroplating with other metals.

Question 45.
What is corrosion? Do gold ornaments corrode? Justify.
Answer:
The slow process of decay or oxidation or metal due to the effect of air, moisture and acids on
it is known as corrosion.
(1) Gold is a noble metal. There is no effect or moist air or action of acid on it at any temperature.
(2) Pure gold is a very soft metal. it breaks and gets bent easily. Hence, in gold ornaments, gold is alloyed with other metals 1ike copper or silver in appropriate proportion to make it hard and resistant to corrosion. Hence gold ornaments do not get corroded.

Question 46.
Complete the process of iron rusting by filling the blanks. Suggest a way to prohibit the process.
The iron rust is formed due to reaction. Different regions on iron surface become anode and cathode.
Reaction on anode region:
Fe_(s) → Fe²⁺ _(aq) + 2e^–
Reaction on cathode region:
O_2(g) + 4H⁺ _(aq) +………→ 2H₂O_(l)
when Fe²⁺ ions migrate from anode region they react with……..to fomm Fe³⁺ ions.
A reddish coloured hytirated oxide is formed from……….ions. It is called rust.
2Fe³⁺ _(aq) + 4H₂O_(l) → +………+ 6H⁺ _(aq)
A way to prevent rusting………..
(Practice Activity Sheer – 2)
Answer:
The iron rust is formed due to electrochemical reaction. Different regions on iron surface become unode and cathode.
Reaction on anode region:
Fe_(s) → Fe²⁺ _(aq) + 2e^–
Reaction on cathode region:
O_2(g) + 4H⁺ _(aq) + 4e^– → 2H₂O_(l)
when Fe²⁺ ions migrate from anode region they react with water to form Fe³⁺ ions.
A reddish coloured hydrated oxide is formed from Fe³⁺ ions. It is called rust.
2Fe³⁺ (aq) + 4H₂O_(l) → + Fe₂O₃. H₂O_(s) + 6H⁺_(aq)
A way to prevent rusting by colouring with acrylic paints, Zn plating, galvanizing, anodizing, alloying, etc.

Question 17.
Deifne: Rancidity.
Answer:
When oil or fat or left over cooking oil for making food stuff undergoes oxidation ir stored for a long time and it is found to have foul odour called rancidity.

Distinguish between the following:

Question 1.
Combination reaction and Decomposition reaction.

Combination reaction	Decomposition reaction
1. In a combination re­action, two or more reactants take part in the chemical reaction.	1. In a decomposition reaction there is only one reactant in the chemical reaction.
1. In the combination reaction, only one product is formed.	2. In a decomposition reaction, two or more products are formed.

Question 2.
Oxidation and reduction
Answer:

Oxidation	Reduction
1. The chemical reaction in which reactants gain oxygen or lose hydrogen is called oxidation.	1. The chemical reaction in which reactants gain hydrogen or lose oxygen is called reduction.
2. A reducing agent undergoes oxidation.	2. An oxidising agent undergoes reduction.

Question 3.
Exothermic and Endothermic reaction.
Answer:

Exothermic reaction	Endothermic reaction
1. The reaction in which heat is evolved is called an exothermic reaction.	1. The reaction in which heat is absorbed is called an endothermic reaction.
2. The evolution of heat leads to a rise in the temperature of the solution.	2. The absorption of heat leads to a fall in the temperature of the solution.

Give scientific reasons:

Question 1.
Grills of doors and windows are always painted before they are used.
Answer:

• Grills of doors and windows are made from iron. Iron has a tendency to undergo corrosion.
• Paint does not allow air or moisture to come in contact with iron surface.
  Therefore, to prevent rusting of iron. grills of doors and windows are always painted before they are used.

Question 2.
Physical states of reactants and products are mentioned while writing a chemical equation.
Answer:
(1) while writing a chemical equation, gaseous, 1iquid and solid states are symbolised as (g), (l) and (s) respectively.

(2) This is done to make it more informative and to emphasise that those reactions occur in that manner only under those conditions. Hence, physical states of reactants and products are mentioned while writing a chemical equation.

Question 3.
Iron articles rust readily whereas steel which is also mainly made of iron does not undergo corrosion.
Answer:
(1) Iron articles rust readily as iron reacts with oxygen and moisture of air to convert into its hydroxide and oxide (Fe₂O₃. x H₂O), while steel is an alloy of iron, carbon and chromium.

(2) The properties of an alloy are different from the properties of its constituents. The added metals increase its resistance to corrosion. It is more durable and clean.

Question 4.
Concentrated hydrochloric acid reacts more vigorously with calcium carbonate than dilute hydrochloric acid.
Answer:

1. The rate of a reaction increases with the concentration of the reactant.
2. As concentrated hydrochloric acid contains more number of HCl molecules than those in an equal volume of dilute HCl, concentrated HCl reacts more vigorously with calcium carbonate.

Question 5.
Zinc powder reacts much faster with dil. H₂SO₄ than does granulated zinc of the Same mass.
Answer:
(1) In a reaction. the rate of the reaction depends upon the particle size of the solid reactant as the reaction takes place on the surface only. Smaller the particles are, the more will be their total surface area and faster will be the rate of the reaction.
(2) Hence, zinc powder reacts much faster with dil. H₂SO₄ than does granulated zinc.

Question 6.
When copper articles exposed to air for a long time, gets corroded.
Answer:
Copper oxidises to form black coloured laver of copper oxide. when copper oxide combines with carbon dioxide from air, copper loses its lustre due to formation of greenish layer of copper carbonate on its surface. Thus, copper articles exposed to air for a long time get corroded.

Question 7.
When silver vessels exposed to air turns blackish after sometime.
Answer:
On exposure to air, silver vessels turns blackish after sometime. This is because of the layer of silver sulphide (Ag₂S) formed by the reaction or silver with hydrogen suphide in air.

Explain the following reactions giving their balanced chemical equations:

Question 1.
Calcium carbonate (Lime stone) is heated.
Answer:
When calcium carbonate (Lime stone) is heated at high temperature it decomposes to form quicklime and carbon dioxide gas.

Question 2.
Copper reacts with dil. nitric acid.
Answer:
When copper reacts with dil. nitric acid, nitric oxide gas is formed.

Question 3.
Copper reacts with conc. nitric acid.
Answer:
When copper reacts with conc. nitric acid, reddish coloured poisonous nitrogen dioxide gas is formed.

Question 4.
Ammonia gas reacts with hydrogen chloride gas.
Answer:
When ammonia gas reacts with hydrogen chloride gas, it forms the salt ammonium chloride in gaseous state, but immediately it got transformed into the solid state.

Question 5.
Magnesium strip is burnt in air.
Answer:
When magnesium strip is burnt in air, a white powder of magnesium oxide is formed.

Question 6.
Calcium oxide is mixed with water.
Answer:
When calcium oxide (slaked lime) is mixed with water, calcium hydroxide is formed with evolution of large amount of heat.

Question 7.
Sugar is heated.
Answer:
When sugar is heated, it decomposes to form carbon (black substance).

Question 8.
Electric current is passed through acidulated water.
Answer:
When an electric current is passed through acidulated water, it decomposes into hydrogen and
oxygen gas.

Question 9.
Zinc powder is added to copper sulphate solution.
Answer:
When zinc powder is added to copper sulphate solution, more reactive zinc displaces less reactive copper from copper sulphate solution. The colourless zinc sulphate is formed with evolution of heat.

Question 10.
Iron powder is added to copper sulphate solution.
Answer:
When iron powder is added to copper sulphate solution, more reactive iron displaces less reactive copper from copper sulphate. The colourless ferrous sulphate solution is formed with evolution of heat.

Question 11.
Lead is added to copper sulphate solution.
Answer:
When lead is added to copper sulphate solution, more reactive lead displaces less reactive copper from copper sulphate. The colourless lead sulphate solution is formed with evolution of heat.

Question 12.
Potassium chromate solution is added to barium sulphate solution.
Answer:
When potassium chromate solution is added to barium sulphate solution, yellow precipitate of barium chromate is formed.

Question 13.
Calcium chloride solution is added to sodium carbonate solution.
Answer:
When calcium chloride solution is added to sodium carbonate solution, white precipitate of calcium carbonate is formed.

Question 14.
Sodium chloride solution is mixed with silver nitrate solution.
Answer:
When sodium chloride solution is mixed with silver nitrate solution, white precipitate or silver chloride is formed.

Question 15.
Dilute sulphuric acid is added to barium chloride solution.
Answer:
When dilute sulphuric acid is added to barium chloride solution, white precipitate of barium sulphate is rormed.

Question 16.
Calcium carbonate (Lime stone) is treated with dil. hydrochloric acid.
Answer:
When calcium carbonate (lime stone) is treated with dil. hydrochloric acid, carbon dioxide gas is formed.

Question 17.
Aluminium is treated with dil. hydrochloric acid.
Answer:
When aluminium is treated with dilute hydrochloric acid, hydrogen gas is liberated.

Question 18.
Magnesium is treated with hydrochloric acid.
Answer:
When magnesium is treated with hydrochloric acid, hydrogen gas is liberated.

Question 19.
Hydrogen peroxide is decomposed in the presence of manganese dioxide (MnO₂).
Answer:
When hvdrogen peroxide is decomposed in the presence of manganese dioxide (MnO₂), water and oxygen are formed.

Question 20.
Ethyl alcohol is treated with acidified potassium dlchromate.
Answer:
When ethly alcohol is treated with acidified potassium dichromate, acetic acid is formed. This is oxidation reaction.

Question 21.
Hydrogen gas is passed over black copper oxide.
Answer:
When hydrogen gas is passed over black copper oxide, a reddish coloured layer of copper is formed.

10th Std Science Part 1 Questions And Answers:

• Gravitation Class 10 Questions And Answers
• Periodic Classification of Element Class 10 Questions And Answers
• Chemical reactions and equations Class 10 Questions And Answers
• Effects of electric current Class 10 Questions And Answers
• Heat Class 10 Questions And Answers
• Refraction of light Class 10 Questions And Answers
• Lenses Class 10 Questions And Answers
• Metallurgy Class 10 Questions And Answers
• Carbon Compounds Class 10 Questions And Answers
• Space Missions Class 10 Questions And Answers

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 6 Animal Classification Notes, Textbook Exercise Important Questions and Answers.

Std 10 Science Part 2 Chapter 6 Animal Classification Question Answer Maharashtra Board

Class 10 Science Part 2 Chapter 6 Animal Classification Question Answer Maharashtra Board

Question 1.
a. I am diploblastic and acoelomate. Which phylum do I belong to ?
Answer:
I am from phylum Cnidaria or Coelenterata.

b. My body is radially symmetrical. Water vascular system is present in my body. I am referred as fish though I am not. What is my name?
Answer:
Starfish. I am from Echinodermata phylum.

c. I live in your small intestine. Pseudocoelom is present in my thread like body. In which phylum will you include me?
Answer:
I am Ascaris. I am included in Aschelminthes.

d. Though I am multicellular, there are no tissues in my body. What is the name of my phylum?
Answer:
Sponge, Porifera.

Question 2.
Write the characters of each of the following animals with the help of classification chart:
a. Bath sponge.
Answer:
Classification:
Kingdom: Animalia
Sub-kingdom: Non-chordata
Phylum: Porifera
Characters:

• Multicellular organisms without cell wall
• Cellular grade organization.
• Asymmetrical body
• Acoelomate

Bath sponge is a marine animal. Blackish in colour and round in shape having porous body. It has spongin fibres and spicules which serve as skeleton. Bath sponges have good water-holding capacity. It is sedentary animal which is fixed to some substratum in the aquatic environment. Reproduction is by budding. It also has a good regeneration capacity.

b. Grasshopper.
Answer:
Classification:
Kingdom: Animalia
Sub-kingdom: Non-chordata
Phylum: Arthropoda
Class: Insecta
Characters:

• Multicellular organisms without cell wall
• Organ-system grade organization
• Bilaterally symmetrical
• Triploblastic and Eucoelomate.

Grasshopper is an insect included under class insecta of phylum arthropoda because it has jointed appendages. There are three pairs of legs and two pairs of wings. It is a terrestrial insect which is well adapted to the surrounding environment by showing camouflage. It has chitinous exoskeleton. The respiration by tracheae.

c. Rohu.
Answer:
Classification:
Kingdom: Animalia
Phylum: Chordata
Class: Pisces
Subclass: Teleostei (Bony fish)
Characters:

• Multicellular organisms without cell wall
• Organ-system grade organization
• Bilaterally symmetrical
• Triploblastic and Eucoelomate.

Rohu is a fresh water bony fish. It is a chordate having a vertebral column, hence included under subphylum vertebrata. The body is well adapted for aquatic mode of life. The shape of the body is streamlined. The exoskeleton is of scales. The gills Eire present which are used for respiration. The endoskeleton is of bones, hence called bony fish. There are paired fins and a impaired caudal fin which is used in steering and changing the direction during swimming.

d. Penguin.
Answer:
Classification:
Kingdom: Animalia
Phylum: Chordata
Class: Aves
Characters:

• Multicellular organisms without cell wall
• Organ-system grade organization
• Bilaterally symmetrical
• Triploblastic and Eucoelomate.

Penguin is a flightless bird inhabitant of cold snow-clad regions. It has exoskeleton of feathers. The body is well adapted to survive in cold regions.

It is a warm-blooded bird. The forelimbs are modified into wings. But due. to excessive body weight, the penguins are not seen flying. It can wade in the water with modified hind limbs.

e. Frog.
Answer:
Classification:
Kingdom: Animalia
Phylum: Chordata
Class: Amphibia
Characters:

• Multicellular organisms without cell wall
• Organ-system grade organization
• Bilaterally symmetrical
• Triploblastic and Eucoelomate.

The frog is a true amphibian that can live in water as well as on land. When on land it respires with the help of lungs while in water it uses its skin for breathing. It does not have exoskeleton. The skin is soft, slimy and moist. It is suitably coloured and hence the frog can camouflage in the surroundings. Body is divisible into head and trunk. Two pairs of limbs are seen. The forelimbs are short and used for support during locomotion. The hind limbs are long and strong, used for jumping when on land and for swimming when in water.

The eyes are large and protruding. Since the neck is absent, such eyes help in looking around. The tympanum is present.

f. Lizard.
Answer:
Classification:
Kingdom: Animalia
Phylum: Chordata
Class: Reptilia
Characters:

• Multicellular organisms without cell wall
• Organ-system grade organization
• Bilaterally symmetrical
• Triploblastic and Eucoelomate.

The lizard is a cold-blooded reptile. The limbs are weak and do not support the body weight, hence lizard is seen creeping. But the feet are provided with pads and suckers due to which lizards are well- adapted to climb on the vertical walls. The exoskeleton has fine scales. The body is divisible into head, neck and trunk. The capacity to regenerate is developed in lizards, hence it can produce the lost tail or limbs. The mode of reproduction is egg laying. It feeds on insects with the help of long and sticky tongue.

g. Elephant.
f. Lizard.
Answer:
Classification:
Kingdom: Animalia
Phylum: Chordata
Class: Mammalia
Characters:

• Multicellular organisms without cell wall
• Organ-system grade organization
• Bilaterally symmetrical
• Triploblastic and Eucoelomate.

Elephant is the terrestrial, herbivorous mammal adapted to survive in hot and humid tropical forests.
It is a mammal and hence shows viviparity and milk secretion. The body is divisible into head, neck, trunk, and tail. The proboscis is a characteristic feature of the elephant which is actually modified nose.

h. Jellyfish.
Answer:
Classification:
Kingdom: Animalia
Sub-kingdom: Non-chordata
Phylum: Cnidaria or Coelenterata
Characters:

• Multicellular organisms without cell wall
• Tissue grade organization
• Radially symmetrical
• Diploplastic and Acoelomate

Jellyfish or Aurelia is a coelenterate. Its body is medusa. It appears as a transparent balloon seen floating in the marine waters. Since it has appearance like a jelly, it is known commonly as jellyfish. There are tentacles provided with cnidoblasts or stinging cells. Tentacles are used for catching the prey. Cnidoblasts are used to secrete a toxin which paralyses the prey.

Question 3.
Write in brief about progressive changes in animal classification.
Answer:
There were different methods of classification of animals.

1. The first classification method was given by the Greek philosopher Aristotle. He took into account the criteria like body size, habits and habitats of the animals. This method was called artificial method of classification.
2. The same artificial method was used by other scientists such as Theophrastus, Pliny, John Ray, Linnaeus, etc.
3. Further due to advances in science the references were changed and there were some new methods of classification proposed.
4. The system of classification called ‘Natural system of classification’ was then proposed. This system of classification was based on criteria such as body organization, types of cells, chromosomes, bio-chemical properties, etc.
5. Later, Dobzhansky and Meyer gave the system of classification based on evolution.
6. In 1977, Carl Woese has also proposed the three domain system of animal classification.

Question 4.
What is the exact difference between grades of organization and symmetry? Explain with examples.
Answer:
I. Grades of organization:
(1) The grades of organization mean the way an organism has different body formation.
(2) Unicellular organisms like amoeba have a single cell in the body and hence the organization in its body is called protoplasmic grade of organization.
(3) Some organisms have only cells in their body which is called cellular grade of organization, e.g. Poriferans.
(4) Some have tissues e.g. Coelenterates. They are said to have tissue grade organization. Some have organs, they are said to have organization-organ grade, e.g. Platyhelminthes. All other higher animals have organ-system grade organization.

II. Symmetry:
(1) Symmetry on the other hand shows the base of the body formation.
(2) The symmetry can be understood by taking an imaginary cut through the animal body.
(3) Based on the symmetry there can be three types.
(4) In asymmetric animals, there is no symmetry in any plane, e.g. Amoeba.
(5) The bilateral symmetry is the one in which an imaginary axis can pass through only one median plane to divide the body into two equal halves. Most of the animals have bilateral symmetry and hence their organs are arranged in symmetric way on both the sides.
(6) The imaginary cut passing through the central axis but any plane of body aan -give more than one equal half. The organs of such animals are arranged in a radius of an imaginary circle, e.g. Cnidarians and some echinoderms.
Both grades of organization and symmetry are the bases for classifying animals into different phyla.

Question 5.
Answer in brief.
a. Give scientific classification of shark upto class.
Answer:
Kingdom: Animalia
Phylum: Chordata
Subphylum: Vertebrata
Class: Pisces
Subclass: Elasmobranchii (Cartilaginous)
Example: Scientific name: Scoliodon sorrakowah.
Common name: Shark

b. Write four distinguishing characters of phylum – Echinodermata.
Answer:
Distinguishing characters of Echinodermata:

1. Marine organisms with skeleton made up of calcareous spines. Calcareous material on the body hence the name is Echiodermata. Some are sedentary while some are free swimming.
2. Body is triploblastic, eucoelomate and radially symmetrical when adult. The larvae are bilateral symmetrical.
3. Locomotion with the help of tube-feet which are also used for capturing the prey.
4. Echinoderms have regeneration capability. Hence they can restore their lost parts.
5. Most of them are unisexual.
6. Examples; Starfish, sea-urchin, brittle star, sea cucumber, etc.

c. Distinguish between butterfly and bat with the help of four distinguish properties.
Answer:
Butterfly:

1. Butterfly is classified as Non-chordate.
2. It is included in class Insecta of phylum Arthropoda.
3. Butterfly has three pairs of legs and two pairs of chitinous wings.
4. Butterfly is a diurnal (active during day) insect.
5. Butterfly lays eggs which hatch into larva. Larva develops into pupa and pupa metamorphoses into an adult.

Bat:

1. Bat is classified as a Chordate.
2. It is included in class Mammalia of subphylum Vertebrata.
3. Bat has one pair of legs and a pair of patagium which are used for flying. Patagium has bones.
4. Bat is a nocturnal (active at night) mammal.
5. Bat is a viviparous animal that gives birth to live young ones. Young ones are fed by milk secreted by mammary glands.

d. To which phylum does Cockroach belong? Justify your answer with scientific reasons.
Answer:
(1) Cockroach belongs to the phylum Arthropoda and class Insecta.
(2) Scientific reasons for placement of Cockroach in the phylum Arthropoda:

• The body is covered by chitinous exoskeleton.
• Jointed appendages present, three pairs of walking legs and two pairs of membranous wings.
• Body is eucoelomate, triploblastic, bilaterally segmented and segmented.
• Respiration by spiracles and tracheal tubes.

Question 6.
Give scientific reasons.
a. Though tortoise lives on land as well as in water, it cannot be included in class-Amphibia.
Answer:

• When tortoise lives on the land, it respires with the help of lungs.
• When in water, it puts out its nares (nasal openings) out of the water and breathes air.
• It cannot take up oxygen dissolved in water. In both the habitats it respires with the help of lungs. In case of true amphibians, this is not the case.
• They can breathe in water with the help of skin and on land with the help of lungs.
• Tortoise also has exo-skeleton which is lacking in Amphibia. Therefore, tortoise cannot be included in class Amphibia.

b. Our body irritates if it comes in contact with jellyfish.
Answer:

• Jellyfish is a coelenterate that has cnidoblasts bearing tentacles.
• These cnidoblasts inject toxins to paralyse the prey at the time of feeding.
• When jellyfish comes in contact with our body, this toxin is released causing reaction to our skin.
• Therefore, our body gets irritation when we come in contact with jellyfish.

c. All vertebrates are chordates but all chordates are not vertebrates.
Answer:

• All chordates possess notochord in some period of their development.
• All vertebrates also have notochord during embryonic life, which is later replaced by vertebral column.
• Therefore all vertebrates are chordates.
• But some chrodate’s like Urochordata and cephalochordata do not possess vertebral column and hence they are not vertebrates.

d. Balanoglossus is connecting link between non-chordates and chordates.
Answer:

• Balanoglossus shows some characters of non-chordates.
• It also has notochord as in case of chordates.
• Since it shares the characters of non-chordates and chordates, from the view point of evolution, it is called connecting link between them.

e. Body temperature of reptiles is not constant. (Board’s Model Activity Sheet)
Answer:

1. Reptiles are cold-blooded animals.
2. The thermoregulatory system is not there in their bodies.
3. Their body temperatures, fluctuate as per the environmental temperatures.
4. Therefore, the body temperature is not maintained at constant level in reptiles.

Question 7.
Answer the following questions by choosing correct option.
a. Which special cells are present in the body of sponges (Porifera)?
1. Collar cells
2. Cnidoblasts
3. Germ cells
4. Ectodermal cells
Answer:
1. Collar cells
Explanation: Porifera animals are attached to the substratum. They do not show locomotion. For gathering and catching the food, they need to produce a current in the water. For this purpose, they have characteristic collar cells in their body. Germ cells and ectodermal cells are seen in all other phyla. Cnidoblasts are characteristic feature of coelenterates.

b. Which of the following animals’ body shows bilateral symmetry?
1. Starfish
2. Jellyfish
3. Earthworm
4. Sponge
Answer:
3. Earthworm
Explanation: When an imaginary plane passing through only one axis can divide the body into two equal halves, then it is called bilateral symmetry. Such symmetry is shown only by earthworm. Sponge body is asymmetrical while starfish and jellyfish are radially symmetrical.

c. Which of the following animals can regenerate it’s broken body part?
1. Cockroach
2. Frog
3. Sparrow
4. Starfish
Answer:
4. Starfish
Explanation: Cockroach, sparrow and frog cannot perform regeneration. Only echinoderms show power of regeneration. So only starfish can regenerate its broken part.

d. Bat is included in which class?
1. Amphibia
2. Reptilia
3. Aves
4. Mammalia
Answer:
4. Mammalia
Explanation: Bat gives birth to young ones and they also possess mammary glands. Amphibia, Reptilia and Aves do not show such features. Therefore, bat is included in Mammalia.

Question 8.

Body cavity	Germ Layer	Phylum
Absent	_____________	Porifera
Absent	Triploblastic	_____________
Pseudocoelom	_____________	Aschelminthes
Present	_____________	Arthropoda

Answer:

Body cavity	Germ Layer	Phylum
Absent	Diploblastic	Porifera
Absent	Triploblastic	Platyhelminthes
Pseudocoelom	Triploblastic	Aschelminthes
Present	Triploblastic	Arthropoda

Question 9.

Type	Character	Examples
Cyclostomata	……………	……………
……………	Gill respiration	……………
Amphibia	……………	……………
……………	……………	Whale, Cat, Man
……………	Poikilotherms	……………

Answer:

Type	Character	Examples
Cyclostomata	Jawless mouth with suckers	Petromyzon, Myxine
Pisces	Gill respiration	Pomfret, Sea horse, Shark
Amphibia	Moist skin without exoskeleton	Frog, Toad, Salamander
Mammalia	Mammary glands	Whale, Cat, Man
Reptilia	Poikilotherms	Tortoise, Lizard, Snake

Question 10.
Sketch, labell and classify.
1. Hydra.
Answer:

Classification:
Kingdom: Animalia
Division: Non-chordata
Phylum: Coelenterata
Example: Hydra

2. Jellyfish
Answer:

Classification:
Kingdom: Animalia
Division: Non-chordata
Phylum: Coelenterata
Example: Jellyfish

3. Planaria
Answer:

Classification:
Kingdom: Animalia
Division: Non-chordata
Phylum: Platyhelminthes
Example: Planaria

4. Roundworm
Answer:

Classification:
Kingdom: Animalia
Division: Non-Chordata
Phylum: Aschelminthes
Example: Ascaris (Roundworm)

5. Butterfly
Answer:

Classification:
Kingdom: Animalia
Division: Non-chordata
Phylum: Arthopoda
Class: Insecta
Example: Butterfly

6. Earthworm
Answer:

Classification:
Kingdom: Animalia
Division: Non-chordata
Phylum: Annelida
Example: Earthworm

7. Octopus
Answer:

Classification:
Kingdom: Animalia
Division: Non-chordata
Phylum: Mollusca
Example: Octopus

8. star fish
Answer:

Classification:
Kingdom: Animalia
Division: Non-chordata
Phylum: Echinodermata
Example: Star fish

9. Shark
Answer:

Classification:
Kingdom: Animalia
Phylum: Chordata
Sub Phylum: Vertebrata
Class: Pisces
Example: Scoliodon (Shark)

10. Frog
Answer:

Classification:
Kingdom: Animalia
Phylum: Chordata
Sub Phylum: Vertebrata
Class: Amphibia
Example: Frog

11. Wall Lizard
Answer:

Classification:
Kingdom: Animalia
Phylum: Chordata
Sub Phylum: Vertebrata
Class: Reptilia
Example: Wall Lizard

12. Pigeon.
Answer:

Classification:
Kingdom: Animalia
Phylum: Chordata
Sub-Phylum: Vertebrata
Class: Aves
Example: Pigeon

Question 11.

Answer:
(1) Jellyfish
(2) Nereis
(3) Flatworm/Planaria
(4) Bony fish.

Project: (Do it your self)

1. In each week, on a specific day of your convenience, observe the animals present around your school and residence. Perform this activity for six months. Keep date-wise record of your observations. After the observation period of six months, analyse your observations with respect to seasons. With the help of your teacher, classify the reported animals.

Can you recall? (Text Book Page No. 61)

Question 1.
Which criteria are used for classification of organisms?
Answer:
The living organisms are classified according to their basic characteristics, such as presence or absence of nucleus, unicellular body or multicellular body, presence or absence of cell wall and the mode of nutrition in them.

Question 2.
How are the plants classified?
Answer:
The plants are classified according to the following basis:

1. Presence or absence of the organs.
2. Presence or absence of separate

Use your brain power: (Text Book Page No. 74)

(A) Animals like gharial and crocodile live in water as well as on land. Are they amphibians or reptiles?
Answer:
Ghariyal and crocodile are reptiles. They can swim in water and crawl on land. But they can respire only with the help of lungs. Their breathing is through nostrils. Even when in water, they have to inhale and exhale by coming up to the surface of water for air. Amphibians can breathe through the skin when in water and by lungs when on land. They also have hard exoskeleton which amphibians do not have. Hence, ghariyal and crocodile are not amphibians, but they are reptiles.

(B) Animals like whale, walrus live in water (ocean). Are they included in Pisces or Mammalia?
Answer:
Whale and walrus are aquatic and marine mammals. They do not belong to class Pisces. They do not have gills to breathe in dissolved oxygen in water. Neither they have scales on the body nor can they lay eggs. Whales and walrus have mammary glands like all other mammals. They give birth to live young one. They breathe only with the help of lungs by putting their nostrils out of the water at surface. Hence they are included in Mammalia.

Choose the correct alternative and write its alphabet against the sub-question number:

Question 1.
System of classification based on evolution was brought into practice by ……….. and …………
(a) Darwin, Mendel
(b) Lamarck, De Vries
(c) Morgan, Mayor
(d) Dobzansky, Meyer
Answer:
(d) Dobzansky, Meyer

Question 2.
Artificial method of animal classification was proposed by ………….
(a) Aristotle
(b) Darwin
(c) Lamarck
(d) Whittaker
Answer:
(a) Aristotle

Question 3.
Animals attached to substratum are called ……….. animals.
(a) sessile
(b) sedentary
(c) lame
(d) motionless
Answer:
(b) sedentary

Question 4.
In coelenterates, ………… are useful for capturing the prey whereas ………. inject the toxin in the body of prey.
(a) tentacles, cnidoblast
(b) hands, legs
(c) flagella, sting
(d) cilia, sting cells
Answer:
(a) tentacles, cnidoblast

Question 5.
Body of annelidan animals is long, cylindrical and …………. segmented.
(a) annular
(b) metamerically
(c) jointed
(d) cuticular
Answer:
(b) metamerically

Question 6.
…………. is second largest phylum in animal kingdom.
(a) Mollusca
(b) Arthropoda
(c) Porifera
(d) Platyhelminthes
Answer:
(a) Mollusca

Question 7.
Endoskeleton of Cyclostomata animals is …………..
(a) bony
(b) bony and cartilaginous
(c) cartilaginous
(d) none of the above
Answer:
(c) cartilaginous

Question 8.
Body cavity between the body and internal organs is called ………….
(a) gastrocoel
(b) enteron
(c) coelom
(d) cave
Answer:
(c) coelom

Question 9.
Larvae of ……….. metamorphose into adults after settling down at bottom of the sea.
(a) Hemichdrdata
(b) Urochordata
(c) Cephalochordata
(d) Cyclostomata
Answer:
(b) Urochordata

Question 10.
The body organization of unicellular organisms is of …………. grade.
(a) cellular
(b) tissue
(c) protoplasmic
(d) organ
Answer:
(c) protoplasmic

Question 11.
………….. is a cold blooded animal. (March 2019)
(a) Bat
(b) Snake
(c) Rabbit
(d) Elephant
Answer:
(b) Snake

Question 12.
Calcareous spines are present on the body of ………… animal. (July 2019)
(a) fish
(b) snail
(c) sponge
(d) starfish
Answer:
(d) starfish

Question 13.
Due to which similar characteristic honey bee and cockroach are included in the same phylum?
(a) Wings
(b) Three pair of legs
(c) Jointed appendages
(d) Antenna
Answer:
(c) Jointed appendages

Write whether the following statements are true or false with proper explanation:

Question 1.
Greek philosopher Linnaeus was the first to perform the animal classification.
Answer:
False. (Greek philosopher Aristotle was the first to perform the animal classification.)

Question 2.
Heart if present in the non-chordates is on dorsal side of body.
Answer:
True.

Question 3.
Arthropoda animals bear numerous pores on their body.
Answer:
False. (Porifera animals bear numerous pores on their body.)

Question 4.
Porifera animals have special type of collar cells.
Answer:
True.

Question 5.
Aschelminthes have acoelomate and bilaterally symmetrical body.
Answer:
False. (Platyhelminthes have acoelomate and bilaterally symmetrical body. OR Aschelminthes have pseudocoelomate and bilaterally symmetrical body.

Question 6.
Planet Earth has highest number of animals from phylum Arthropoda.
Answer:
True.

Question 7.
Animals belonging to phylum Annelida perform locomotion with the help of tube-feet.
Answer:
False. (Animals belonging to phylum Echinodermata perform locomotion with the help of tube-feet.)

Question 8.
Herdmania has notochord in only tail region and hence it is called Urochordate.
Answer:
True.

Question 9.
Mammals breathe with the help of lungs.
Answer:
True.

Question 10.
Amphibians are warm blooded.
Answer:
False. (Amphibians are cold-blooded. OR Mammals are warm blooded.)

Match the columns:

Question 1.

Phylum	Characteristics
(1) Mollusca	(a) Collar cells
(2) Hemichordata	(b) Mantle
	(c) Trunk
	(d) Cnidoblasts

Answer:
(1) Mollusca – Mantle
(2) Hemichordata – Trunk.

Question 2.

Phylum	Characteristics
(1) Porifera	(a) Tunic
(2) Coelenterata	(b) Collar cells
	(c) Tentacles bearing cnidoblasts
	(d) Mantle

Answer:
(1) Porifera – Collar cells
(2) Coelenterata – Tentacles bearing cnidoblasts.

Question 3.

Subphylum/Class	Characteristics
(1) Cyclostomata	(a) Collar cells
(2) Urochordat	(b) Sucker
	(c) Tunic
	(d) Chitinous exoskeleton

Answer:
(1) Cyclostomata – Sucker
(2) Urochordata – Tunic.

Find the odd one out:

Question 1.
Physalia, Hyalonema, Ruplectella, Spongilla
Answer:
Physalia. (Physalia belongs to Coelenterata, all the remaining are poriferans.)

Question 2.
Planaria, Liverfluke, Filarial worm, Tapeworm
Answer:
Filarial worm. (Filarial worm is Aschelminthes remaining are Platyhelminthes.)

Question 3.
Star fish, Sea-urchin, Nereis, Sea-cucumber
Answer:
Nereis. (Nereis belongs to Annelida all the remaining are Echinoderm animals.)

Question 4.
Cockroach, Butterfly, Spider, Honey bee
Answer:
Spider. (Spider is eight-legged Arachnid, remaining are insects.)

Question 5.
Amphioxus, Herdmania, Doliolum,Oikopleura
Answer:
Amphioxus. (Amphioxus is Cepholochordate all the remaining are Urochordates.)

Question 6.
Frog, Tortoise, Toad, Salamander
Answer:
Tortoise. (Tortoise is a reptile, the remaining are amphibians.)

Question 7.
Tube feet, Setae, Parapodia, Sucker
Answer:
Tube feet. (Tube feet are locomotory organs of Echinoderms, the remaining are locomotory organs of Annelids.)

Question 8.
Shark, Sting ray, Electric ray, Pomfret
Answer:
Pomfret. (Pomfret is a bony fish, all the remaining are cartilaginous fish.)

Find the correlation:

Question 1.
Annelida : Earthworm : : Platyhelminthes : …………
Answer:
Annelida : Earthworm : : Platyhelminthes : Planaria/Liverfluke

Question 2.
Horse : Mammal : : Seahorse : ………….
Answer:
Horse : Mammal : : Seahorse : Pisces

Question 3.
Parapodia : Annelida : : Tube feet : ………..
Answer:
Parapodia : Annelida : : Tube feet : Echinodermata

Question 4.
Frog : Amphibia : : Turtle : …………..
Answer:
Frog : Amphibia : : Turtle : Reptilia

Question 5.
Proboscis : Hemichordata : : Suctorial mouth : …………
Answer:
Proboscis : Hemichordata : : Suctorial mouth : Cyclostomata

Question 6.
Bird from very cold regions : Penguin : : Aquatic Mammal from very cold regions : ………..
Answer:
Bird from very cold regions : Penguin : : Aquatic Mammal from very cold regions : Whale.

Distinguish between:

Question 1.
Non-chordates and Chordates.
Answer:
Non-chordates

1. Non-chordates are less evolved animals and are on the lower levels of evolution.
2. Non-chordates do not have notochord.
3. In non-chordates, there are no pharyngeal gill slits.
4. Nerve cord, if present is double and solid.
5. Nerve cord is located on the ventral side of the body.
6. Heart if present is on the dorsal side of the body.

Chordates:

1. Chordates are more evolved animals and are on the higher levels of evolution.
2. Chordates have notochord at least in some stage of development.
3. In chordates, there are pharyngeal gill slits.
4. Nerve cord is single and hollow.
5. Nerve cord is located on the dorsal side of the body.
6. Heart if present is on the ventral side of the body.

Question 2.
Phylum Platyhelminthes and Phylum Aschelminthes. OR Write any two points of differences between flat worms and round worms.
Answer:
Phylum Platyhelminthes (Flat worms):

1. Platyhelminth worms have slender and flat leaf or strip like body hence called flat worms.
2. Platyhelminthes are triploblastic and acoelomate.
3. Most of them are hermaphrodite or bisexual having both male and female reproductive systems in the same body.
4. Examples: Planaria, Liver fluke, Tapeworm, etc.

Phylum Aschelminthes (Round worms):

1. Aschelminthes have long thread-like or Cylindrical body, hence called round worms.
2. Aschelminthes are triploblastic and pseudocoelomate.
3. They are unisexual with male and female sexes separate. There is sexual dimorphism.
4. Examples: Ascaris (Intestinal worm), Filarial worm, Loa loa (Eye worm), etc.

Question 3.
Urochordata and Cephalochordata.
Answer:
Urochordata:

1. Urochordates have notochord in the tail region of the adult body.
2. These animals look like small sacs.
3. Usually urochordates are hermaphrodites.
4. Body of urochordate is covered over by skin-like test or tunic.
5. Examples: Herdmania, Doliolum, Oikopleura, etc.

Cephalochordata:

1. Cephalochrodates have notochord in the entire length of the body.
2. These animals look like small fish.
3. Cephalochordates are unisexual.
4. Body of cephalochordate is not covered in a test.
5. Example: Amphioxus.

Question 4.
Cyclostomata and Pisces.
Answer:
Cyclostomata:

1. Cyclostomata are the poorly evolved first class of vertebrate animals.
2. Cyclostomata have circular jawless mouth with suckers.
3. Paired appendages are absent in cyclostomates.
4. Cyclostomes have soft skin which is without any scales.
5. Endoskeleton is cartilaginous.
6. Examples: Petromyzon, Myxine, etc.

Pisces:

1. Pisces are the better evolved class of vertebrates which is well adapted for aquatic living.
2. Pisces have mouth with upper and lower jaws. Teeth are present in the mouth.
3. Paired and unpaired fins present in all kinds of fishes.
4. Fishes have different types of scales on the body.
5. Endoskeleton may be cartilaginous, or it may be bony.
6. Examples: Shark (Scoliodoh), rays which are cartilaginous fishes and pomfret, makerel, sardines, rohu which are bony fishes.

Question 5.
Amphibia and Reptilia.
Answer:
Amphibia:

1. Amphibians can inhabit both land and water. They can survive on both environments by breathing there.
2. The exoskeleton is absent in amphibians. The skin is soft, slimy and moist.
3. Body is divided into head and trunk. Neck is absent.
4. The digits do not have claws.
5. The respiration is by skin when in water and by lungs when on land. The larvae breathe by gills.
6. There is external fertilization at the time of sexual reproduction.
7. The developmental stages are eggs and tadpole. Metamorphosis is seen in amphibians.
8. Examples : Frog, Toad, Salamander, etc.

Reptilia:

1. Reptilians are terrestrial animals. Though turtle and sea snakes can stay in water, they cannot breathe in water.
2. The exoskeleton in the form of scales. Some animals have plates or scutes (e.g. tortoise and crocodile).
3. Body is divided into head, neck and trunk.
4. The digits have claws.
5. The respiration is only by lungs.
6. There is internal fertilization at the time of sexual reproduction.
7. The developmental stages are eggs and juvenile. Metamorphosis is not seen in reptiles.
8. Examples : Tortoise, Lizard, Snake, etc.

Question 6.
Aves and Mammalia.
Answer:
Aves:

1. Aves are totally adapted for the aerial mode of life.
2. Body is spindle shaped. Body is divisible into head, neck and trunk. There are two pairs of limbs. The forelimbs are modified to form wings for flight.
3. Digits have scales and claws.
4. The exoskeleton is in the form of feathers.
5. Jaws are modified into a beak.
6. Birds are oviparous. The eggs hatch into nestlings.
7. The incubation of eggs and feeding of nestlings is done by both parents.
8. Examples: Crow, Sparrow, Peacock, Parrot, Pigeon, Duck, Penguin, etc.

Mammalia:

1. Mammals are adapted for terrestrial life.
2. Body is not spindle shaped. It is divisible into head, neck, trunk and tail. There are two pairs of limbs. They are adapted for walking or running on the ground.
3. Digits have nails or hoofs. Few have claws.
4. The exoskeleton is in the form of fur, hair, wool, etc.
5. Jaws have teeth and they surround the mouth.
6. Mammals are viviparous. They give birth to live young ones. (Exception: Platypus)
7. Parental care is shown only by mother, who feeds, the babies with milk from mammary glands.
8. Examples: Cat, Dog, Tiger, Lion, Elephant, Human, Kangaroo, Dolphin, Bat, etc.

Classification-based questions:

Question 1.
Identify me:
(1) I am metamerically segmented, blood sucking, ectoparasite. I have suckers. Who am I and to what phylum do I belong to? (OR) Who am I? (July 2019)
I have suckers. I am blood sucking.
Answer:
Leech, Phylum Annelida.

(2) I have chitinous exoskeleton, I have four pairs of walking appendages. I can sting you. Who am I? What phylum do I belong to?
Answer:
Scorpion. Phylum Arthropoda.

Question 2.
Characters of a phylum are given below. Read them carefully and answer the questions:
(a) Spines of calcium carbonate are present on the body, (b) These animals are exclusively marine, (c) They perform the locomotion with the help of tube feet, (d) Their skeleton is made up of calcareous plates or spicules.
(i) Animals of which phylum show the above character?
Answer:
Animals belonging to phylum Echinodermata show the above characters.

(ii) Give an example from that phylum.
Answer:
Starfish, brittlestar, sea urchin.

(iii) These animals can be classified with the help of which criteria of new system of animals classification.
Answer:
Animals are classified on the basis of criteria such as body organization, body symmetry, body cavity, etc.

Question 3.
Identify my class/phylum and give one example of it: (March 2019)
(a) I have mammary glands and exoskeleton in the form of hair.
(b) We form the highest number of animals on the planet. We have bilateral symmetry and our exoskeleton is in the form of chitin.
(c) I live in your small intestine, my body is long and thread like and pseudocoelomate.
Answer:
(a) Class: Mammalia, Example: Cat, Dog, Man.
(b) Phylum: Arthropoda, Example: Prawn, Crab.
(c) Phylum: Aschelminthes, Example: Ascaris or round worm, Filarial worm.

Question 4.
Tell me who am I? What is my class/ phylum?
1. My body is divided into proboscis, collar and trunk. I am marine animal.
Answer:
Balanoglossus; Phylum: Hemichordata.

2. I stay inside two shells. My body is divided into head, foot and visceral mass.
Answer:
Bivalve or Oyster; Phylum: Mollusca.

3. I am male as well as female. I am endoparasite having a coelomate and bilaterally symmetrical and flattened body.
Answer:
Liver fluke or tape worm; Phylum: Platyhelminthes.

4. I am sedentary marine animal drinking water all the time through numerous pores on the body.
Answer:
Sponge; Phylum: Porifera.

5. I am venomous, eight legged creature having chitinous exoskeleton.
Answer:
Scorpion; Phylum: Arthropoda.

6. My body is covered by tunic. As a larva I swim but as an adult I settle down.
Answer:
Doliolum or Salpa; Phylum: Chordata subphylum : Urochordata.

Question 5.
Identify the class of given animals and write one characteristic of each animal:
(1) Kangaroq (2) Penguin (3) Crocodile (4) Frog (5) Sea-horse. (July 2019)
Answer:
(1) Kangaroo: Class Mammalia. It is a marsupial animal with pouch for development of offspring. Long hind limbs for jumping.
(2) Penguin: Class Aves. It is flightless bird. Body covered with thick feathery coat. Oviparous mode.
(3) Crocodile: Class Reptilia. It is a large animal seen near water bodies. Can swim in water but cannot respire in water. Body covered with exoskeleton of scaly plates. Limbs very weak in comparison with huge bodies.
(4) Frog: Class Amphibia. Shows aquatic as well as terrestrial mode. Can breathe with lungs and skin. No exoskeleton and skin is slimy.
(5) Sea-horse: Class Pisces. Bony fish. Highly modified body structure showing brood pouch for development of offspring gills for respiration, fins for swimming.

Answer the following questions:

Question 1.
State any four benefits of animal classification. (March 2019)
Answer:

1. Studying the different animals becomes easy when they are placed under different groups.
2. When few representative animals of the particular group are studied then the idea about other animals belonging to that group also becomes clear.
3. The animal evolution becomes easier to follow after studying classification.
4. The identification of animals can be done accurately.
5. Relationship of the different animals with each other and with other groups can be understood clearly.
6. Habitat of each animal and its role in nature is understood by classification.
7. Various adaptations are understood by learning classification.

Question 2.
Into which phyla is Non-chordata divided? In which three subphyla are Chordates divided?
Answer:
I. The phyla of Non-chordata:

• Protozoa
• Porifera
• Coelenterata or Cnidaria
• Platyhelminthes
• Aschelminthes
• Annelida
• Arthropoda
• Mollusca
• Echinodermata
• Hemichordata

II. The subphyla of Chordata:

• Urochordata
• Cephalochordata
• Vertebrata

Question 3.
Write the characteristics of chordates.
Answer:
Characteristics of Chordates:

1. All chordates possess notochord and pharyngeal gill slits in at least during some developmental stage.
2. Presence of single, tubular and dorsally located spinal cord and ventrally located heart.

Question 4.
Write the characteristics of vertebrates.
Answer:
Characteristics of vertebrates:

• In vertebrates, notochord is replaced by vertebral column.
• Development of head is complete.
• Well-developed cranium which protects the brain.
• Presence of endoskeleton which is either cartilaginous or bony.
• Presence of jaws as in Gnathostomata or absence of jaws as in Agantha.

Write short notes on:

Question 1.
(1) Benefits of classification.
Answer:

• Studying the different animals becomes easy when they are placed under different groups.
• When few representative animals of the particular group are studied then the idea about other animals belonging to that group also becomes clear.
• The animal evolution becomes easier to follow after studying classification.
• The identification of animals can be done accurately.
• Relationship of the different animals with each other and with other groups can be understood clearly.
• Habitat of each animal and its role in nature is understood by classification.
• Various adaptations are understood by learning classification.

Question 2.
Germinal layers.
Answer:

• During the initial embryonic period of any multicellular animal there is formation of germinal layers or germ layer.
• These germ layers give rise to new tissues in the developing animal.
• The primitive animals were diploblastic i.e. they have only two germ layers called ectoderm and endoderm.
• The higher animals are triploblastic, having three germ layers; ectoderm, mesoderm and endoderm.
• Cnidarians are diploblastic while all other animals are triploblastic.

Question 3.
Coelom.
Answer:

• Coelom means body cavity. It is situated between the body wall and the internal organs of the body.
• The coelom is formed during early embryonic life in case of multicellular animals. It is formed from either mesoderm or gut.
• Coelom when present in the body, those animals are called eucoelomate. Phylum Annelida onwards are eucoelomate animals. They are animals with true body cavity.
• Those animals in which coelom are absent are called acoelomate animals. Porifera, Cnidaria and Platyhelminthes are acoelomate animals.
• When coelom is not formed from mesoderm or gut, but formed from other tissues, it is called pseudocoelom. Only Aschelminthes animals have such coelom and hence they are called pseudocoelomate.

Question 4.
Notochord.
Answer:

• Notochord is an important feature of Chordates.
• Notochord is supporting rod like structure.
• This structure is present on the dorsal side of the animals.
• It keeps the nervous tissue separated from the rest of the tissues.
• In Hemichordates, the notochord is present in the proboscis.
• In Urochordates, the notochord is present in the tail region of the free swimming larvae.
• In Cephalochordates, the notochord lies throughout the length of the body.
• In vertebrates, notochord is replaced by the vertebral column.

Complete the paragraph by choosing the appropriate words given in the brackets:

Question 1.
(Linnaeus, Dobzhansky, Carl Woese, Theophrastus, Artificial method, Aristotle, Natural system, Traditional system)
Time to time, different scientists have tried to classify the animals. Greek philosopher ………… was the first to perform the animal classification. Aristotle classified the animals, according to the criteria like body size, habits and habitats. Classification proposed by Aristotle is known as ………… Besides Aristotle, artificial method of classification was followed by ……….., Pliny, John Ray and ……….. Later on,’………… of classification’ was followed. Natural system of classification was based on various other criteria. By the time, system of classification based on evolution was also brought into practice. It was used by …………. and Meyer. Recently, ……….. has also proposed the animal classification.
Answer:
Time to time, different scientists have tried to classify the animals. Greek philosopher Aristotle was the first to perform the animal classification. Aristotle classified the animals, according to the criteria like body size, habits and habitats. Classification proposed by Aristotle is known as ‘Artificial method’. Besides Aristotle, artificial method of classification was followed by Theophrastus, Pliny, John Ray and Linnaeus. Later on, ‘Natural system of classification’ was followed. Natural system of classification was based on various other criteria. By the time, system of classification based on evolution was also brought into practice. It was used by Dobzansky and Meyer. Recently, Carl Woese has also proposed the animal classification.

Question 2.
(neck, lungs, skin, exoskeleton, amphibian, metamorphose, aquatic, gills)
Class Amphibia consist of animals which are strictly ……….. only during their larval stages. At that time they breathe through their …………. Tadpoles are such stages which later ………… to form adult frog. Adult frog respires with the help of ………… when in water and with when on land. Thus, it is a true …………. For performing cutaneous respiration, i.e. respiration through skin, they lack ………. in any form. The skin is also kept moist by staying near the water bodies. Amphibians do not have a ………… but eyes are bulging and prominent, this solves the problems of vision.
Answer:
Class Amphibia consist of animals which are strictly aquatic only during their larval stages. At that time they breathe through their gills. Tadpoles are such stages which later metamorphose to form adult frog. Adult frog respires with the help of skin when in water and with lungs when on land. Thus, it is a true amphibian. For performing cutaneous respiration, i.e. respiration through skin, they lack exoskeleton in any form. The skin is also kept moist by staying near the water bodies. Amphibians do not have a neck but eyes are bulging and prominent, this solves the problems of vision.

Paragraph based questions:

1. Read the paragraph and answer the questions given below:
Locomotion is considered as an important j characteristics of the animals. However, animals belonging to Porifera are said to be sedentary. Every 1 other phylum has typical locomotory organs. E.g. Nereis crawls with the help of parapodia, whereas earthworm buries in soil by setae. Spiders have four pairs of walking legs, crab has five while all insects have three pairs of walking legs. The walking legs are also called appendages. Starfish moves with the help of tube feet. Snails and bivalves use muscular foot for locomotion. Birds flying with their spread out wings and fish swimming with their fins, both have spindle-shaped body tapering at both the ends. While flying or swimming such body offers least resistance during locomotion. Mammals have two pairs of limbs while animals like snakes are limbless. Other animals belonging to the class of snakes also have very weak limbs which make them creep on the ground.

Questions and Answers:

Question 1.
What are the locomotory organs in phylum Annelida?
Answer:
Annelidans have parapodia and setae as the locomotory organs.

Question 2.
Which phylum has a characteristic of jointed appendages?
Answer:
Phylum Arthropoda has a characteristic of jointed appendages.

Question 3.
Which the locomotory organ of animals belong to Phylum Mollusca?
Answer:
Animals belonging to Phylum Mollusca have strong muscular foot which is used for locomotion.

Question 4.
Which class of animals show weak legs?
Answer:
Class Reptilia belonging to subphylum vertebrata show weak legs.

Question 5.
In which class of animals the forelimbs are modified?
Answer:
Class Aves belonging to subphylum vertebrata have wings which are modified forelimbs.

Diagram based questions:

Question 1.
Sketch, label and classify the following organisms:
1. Liverfluke.
Answer:

Classification:
Kingdom: Animalia
Division: Non-chordata
Phylum: Platyhelminthes
Example: Liverfluke

2. Leench.
Answer:

Classification:
Kingdom: Animalia
Division: Non-chordata
Phylum: Annelida
Example: Leech

3. Cockroach:
Answer:

Classification:
Kingdom: Animalia
Division: Non-chordata
Phylum: Arthopoda
Class: Insecta
Example: Cockroach

Question 2.
Identify the animal given in the figure and label the figure:
1.

Answer:
Balanoglossus

2.

Answer:
Herdmania.

3.

Answer:
Amphioxus

4.

Answer:
Petromyzon.

5.

Answer:
bat

Question 3.
Identify the class of the animal shown in the figure and write any two characteristics.
Answer:
(1) The animal shown in the figure is bat.
(2) It belongs to class Mammalia of Subphylum Vertebrata. Phylum Chordata.
(3) Characteristics:
(i) Body is divided into head, neck, torso and tail. Patagium present for the flying mode. Nocturnal in habit. It is warm blooded.
(ii) Gives birth to live young ones. Mammary glands present for nourishing young ones.

Question 4.
Observe the figure and answer the following questions.

(a) To which phylum these organisms belong?
(b) Name the substance with which their body is covered.
(c) Name their organs of locomotion.
Answer:
(a) The starfish and the sea urchin shown in the figure belong to phylum Echinodermata.
(b) The body of echinoderm animal is covered with calcareous spines or ossicles/plates.
This is the substance covering the body is mostly calcium salts and compounds.
(c) Their locomotory organs are tube feet.

Question 5.
Observe the figures given below and answer the given questions: (Board’s Model Activity Sheet)

(a) In which phylum are these animals included?
(b) Which substance forms the outer layer of their exoskeleton?
(c) What are their locomotory organs?
Answer:
(a) These animals are included in phylum Arthropoda.
(b) The outer layer of their exoskeleton is covered by chitinous substance.
(c) Their locomotory organs are jointed paired appendages.

Question 6.
Identify the phylum of the given animal and write any two characteristics of this phylum. (Board’s Model Activity Sheet)
Answer:

This animal is Sycon sponge and its phylum in Porifera.
Characteristics of phylum Porifera
(a) Asymmetrical body.
(b) Many pores on body. Large osculum and smaller ostia.

Question 7.
(a) Identify the animal given here.
(b) Write the phylum to which it belongs.
(c) Identify the pointed parts; p, q, r and s.

Answer:
(a) The given animal is Octopus.
(b) It belongs to the phylum Mollusca.
(c) p = eye, q = sucker, s = siphon and r = tentacle.

Complete the following charts:

Question 1.
Complete the chart by taking into consideration the criteria for classification: (Text Book Page No. 61)

Answer:

Question 2.
Complete the following flow-chart.

Answer:
(A) Eukaryotes
(B) Monera.

Activity based questions:

Question 1.
Observe: (Text Book Page No. 65)
(1) Body organization of human has been shown in the following figure. Use appropriate labels for different organs present in human body.

Answer:
There are different organs in the human body. The liver, pancreas, stomach, intestine, etc. related to the digestive system and a pair of kidney concerned with excretion is present in the abdominal cavity. The cranial cavity shows brain and sense organs. In the thoracic cavity there are lungs and heart. In addition to these organs, there are network of blood capillaries, nerve network, etc. which is spread from head to toes.

Question 2.
Why is earthworm called as friend of farmers? (Get Information: Text Book Page No. 69)
Answer:
Earthworms move through the soil in the farms and fields. They feed on the detritus in the soil. They also help in decomposition of the organic matter. When the soil is loosened due to their activities, the roots of the crops grow well. They enrich the soil by their excreta which act as fertilizers. All these facts make earthworm, a farmer’s friend.

Question 3.
In what way the leech is used in ayurvedic system of treatment? (Get Information: Text Book Page No. 69)
Answer:
Leeches are blood sucking ectoparasite. In Ayurveda leech is used to remove impure blood and blood clots. Such blood is sucked up by leeches and then the patient gets some relief. In the leech body there is. a substance called hirudine which prevent blood clotting as it sucks up the blood. This hirudine is also used for medicinal purpose.

Question 4.
What is chitin? (Find out: Text Book Page No. 70)
Answer:
Chitin is a type of polysaccharide. Its chemical formula is (C8H13O5N)n. It is a long-chain polymer of N-acetylglucosamine, which is actually a derivative of glucose. It is a primary component of cell walls in fungi, the exoskeletons of arthropods, such as crustaceans and insects. In many medicines chitin is used. The industrial processes and the biotechnological experiments also use chitin.

Question 5.
Let’s Think: (Text Book Page No. 70)
(i) What types of benefit and harm occur to human from animals of phylum-Arthropoda?
Answer:
Some insects are very useful for us. We get many products from them. e.g. Honey bee, Lac insect, Silk worm, are the insects that provide us with honey and wax, lac and silk respectively. The culture experiments are done on these insects for large scale production of these substances. Butterflies help in the pollination of crops and are thus helpful for the farmers and gardeners. Lady bug beetle is an insect which acts as a natural pest control as it attacks the other harmful insect pests.

In biological pest control methods it is widely used. Some insects, on the contrary are very harmful. Mosquito, bed bugs, lice are blood sucking parasites which can spread the diseases. Mosquito is a vector for dengue, filariasis and malaria. Some are biting insects that can cause wounds, some cause allergies of various kinds. The grains and crops are destroyed to great extent by the insects. In this way the insects belonging to the phylum Arthropods are harmful to health, wealth and peace of mind too.

(ii) Which are the animals from phylum Arthropoda those have shortest and longest life span?
Answer:
The shortest life span: May fly – About 24 hours. The longest life span : Lobster (Homarus americanus) – About 100 years.

(iii) Why has it been said that only insects directly compete with humans for food?
Answer:
The standing crop in the fields can be totally ruined by insects. The locust can damage the crops when they attack in thousands at a time. The grains are also infested by variety of insects like ants, weevils, beetles, etc. Therefore, we can say that only insects compete with humans for food.

Project: (Do it your self)

Project 1.
How does the infection of tapeworm in man, liver fluke in grazing animals like goat and sheep occur and what are their preventive measures? (Collect the Information, Internet is my friend: Textbook page no. 69)

Project 2.
How does the infection of round worms like Ascaris, filarial worm and plant nematodes occur and what are their preventive measures and treatment? (Collect the Information, Internet is my friend: Textbook page no. 69)

Project 3.
Books are my friend: Collect the information about pearl production from bivalves by reading appropriate books. (Textbook page no. 70)

Project 4.
Book are my friends: The Animal Kingdom: Libbie Hyman and some other similar books.
(Textbook page no. 75)

Project 5.
Use of Information Technology: (Textbook page no. 75)
Prepare the presentation of animal classification using video clips downloaded from internet.

10th Std Science Part 2 Questions And Answers:

• Heredity and Evolution Class 10 Questions And Answers
• Life Processes in Living Organisms Part – 1 Class 10 Questions And Answers
• Life Processes in Living Organisms Part – 2 Class 10 Questions And Answers
• Environmental management Class 10 Questions And Answers
• Towards Green Energy Class 10 Questions And Answers
• Animal Classification Class 10 Questions And Answers
• Introduction to Microbiology Class 10 Questions And Answers
• Cell Biology and Biotechnology Class 10 Questions And Answers
• Social health Class 10 Questions And Answers
• Disaster Management Class 10 Questions And Answers

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 7 Lenses Notes, Textbook Exercise Important Questions and Answers.

Std 10 Science Part 1 Chapter 7 Lenses Question Answer Maharashtra Board

Class 10 Science Part 1 Chapter 7 Lenses Question Answer Maharashtra Board

Question 1.
Match the columns in the following table and explain them:

Column 1	Column 2	Column 3
Farsightedness	Nearby object can be seen clearly	Bifocal  lens
Presbyopia	Faraway object can be seen clearly	Concave  lens
Nearsightedness	Problem of old age	Convex  lens

Answer:

Column 1	Column 2	Column 3
Farsightedness	Faraway object can be seen clearly	Convex  lens
Presbyopia	Problem of old age	Bifocal  lens
Nearsightedness	Nearby object can be seen clearly	Concave  lens

1. Farsightedness:
Hypermetropia or farsightedness is the defect of vision in which a human eye can see distant objects clearly but is unable to see nearby objects clearly.
In this case the image of a nearby object would fall behind the retina instead of on the retina.

Possible reasons for hypermetropia:
(1) Curvature of the cornea and the eye lens decreases. Hence, the converging power of the eye lens becomes less.
(2) The distance between the eye lens and retina decreases (relative to the normal eye) and the focal length of the eye lens becomes very large due to the flattening of the eyeball.

2. Presbyopia:
Presbyopia is the defect of vision in which aged people find it difficult to see nearby objects
comfortably and clearly without spectacles.

Reason of presbyopia: The power of accommodation of eye usually decreases with ageing. The muscles near the eye lens lose their ability to change the focal length of the lens.

Therefore, the near point of the eye lens shifts rarther from the eye, This defect is corrected using a. convex lens of appropriate power. The lens converges light rays before they fall on the eye lens such that the action of the eye lens forms the image on the retina.

3. Nearsightedness:
Myopia or nearsightedness is the defect of vision in which a human eye can see nearby objects distinctly but is unable to see distant objects clearly as they appear indistinct.
In this case the image of a distant object is formed in front of the retina instead of on the retina.
[Figs. 7.29 (a), 7.29 (b)]

Possible reasons for myopia:
(1) The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large. (2) The distance between the eye lens and the retina increases as the eyeball elongates.

Myopia is corrected using a suitable concave lens. Light rays are diverged by the concave lens before they strike the eye lens. A concave lens of proper focal length is chosen to produce the required divergence. Hence, after the converging action of the eye lens, the image is formed on the retina. [Fig. 7.29 (c)]

Question 2.
Draw a figure explaining various terms related to a lens.
Answer:
(1) Centre of curvature (C): The centres of the spheres whose parts form the surfaces of a lens are called the centres of curvature of the lens. A lens has two centres of curvature C₁, and C₂ for its two spherical surfaces.

(2) Radii of curvature (R₁, R₂): The radii of the spheres whose parts form surfaces of a lens are called the radii of curvature of the lens.

(3) Principal axis: The imaginary straight line passing through the two centres of curvature of a lens is called the principal axis of the lens.

(4) Optical centre (O): The point inside a lens on the principal axis, through which light rays pass without changing their path is called the optical centre (O) of the lens.

(5) Principal focus (F): When light rays parallel to the principal- axis are incident on a convex lens, they converge at a point on the principal axis. This point is called the principal focus (F) of the convex lens. Light rays travelling parallel to the principal axis of a concave lens diverge after refraction in such a way that they appear to be coming out of a point on the principal axis. This point is called the principal focus of the concave lens. A lens has two principal foci F₁ and F₂.
[Note: In this chapter, the terms focus and the principal focus are used in the same sense.]

(6) Focal length (f): The distance between the optical centre and the principal focus of a lens is called the focal length (f) of the lens.

C₁, C₂: Centres of curvature, R₁, R₂: Radii of curvature, O: Optical centre.
The cross sections of convex and concave lenses are shown in parts (a) and (b) of Fig. 7.4. The surface marked as 1 is part of sphere S₁ while the surface marked as 2 is part of sphere S₂.

P₁, P₂, P₃: Incident rays of light,
Q₁, Q₂, Q₃: Refracted rays of light, O: Optical centre

F₁, F₂: Principal foci of the lens, f: Focal length of the lens

Question 3.
At which position will you keep an object in front of a convex lens so as to get a real image of the same size as the object? Draw a figure.
Answer:
At 2F₁.

Question 4.
Give scientific reasons:
a. A simple microscope is used for watch repairs.
Answer:
(1) when an object is placed within the focal length of a magnifying glass or simple microscope (convex lens), its larger and erect image is obtained on the same side of the lens as that of the object.

(2) By adjusting the distance between the object and the lens, the image can be obtained at the minimum distance of distinct vision. Thus, a watch repairer can see the minute parts of a watch more clearly with the aid of a magnifying glass (a simple microscope) than with the naked eye, without any stress on the eye. Hence, watch repairers use a magnifying glass (a simple microscope) while repairing the watches.

b. One can sense colours only in bright light.
Answer:
(1) The retina in the eye is made of many light sensitive cells. The rod-shaped cells respond to the intensity of light while the cone-shaped cells j respond to various colours.
(2) The cone-shaped cells do not respond to faint light. They function only in bright light. Hence, one can sense colours only in bright, light.

c. We cannot clearly see an object kept at a distance less than 25 cm from the eye.
Answer:
(1) When we try to see a nearby object, the eye lens becomes more rounded and its focal length decreases. Then a clear image of the object is formed on the retina of the eye.
(2) The focal length of the eye lens cannot be decreased beyond some limit. Therefore we cannot clearly see an object kept at a distance less than 25 cm from the eye.

Question 5.
Explain the working of an astronomical telescope using refraction of light.
Answer:
Construction of a refracting telescope: It consists of two convex lenses called the objective lens (directed towards the object) and the eyepiece (directed towards the eye). The focal length and diameter of the objective lens are respectively greater than the focal length and diameter of the eyepiece. The objective lens is fitted at one end of a long metal tube.

A metal tube of smaller diameter is fitted in this metal tube and the eyepiece is fitted at the outer end of the smaller tube. With the help of a screw it is possible to change the distance between the eyepiece and the objective lens by sliding the tube fitted with the eyepiece. The principal axes of the objective lens and the eyepiece are along the same line. A telescope is usually mounted on a stand.

Working: When the objective lens is pointed towards the distant object to be observed, the rays of light from the distant object, which are almost parallel to each other, pass through the objective lens. The objective lens collects maximum amount of light as it is large in size. It forms a real, inverted and diminished image in the focal plane of the objective lens. Now, the position of the eyepiece is adjusted such that this image falls just within the focal length of the eyepiece and serves as the object for the eyepiece which works as a simple microscope.

The final image is highly magnified, virtual, on the same side as that of the object and inverted with respect to the original object. The final image can be observed by keeping the eye close to the eyepiece. If the image formed by the objective lens lies in the focal plane of the eyepiece, the final image is formed at infinity.

Question 6.
Distinguish between the following:
a. Farsightedness (Hypermetropia) and Nearsightedness (Myopia).
Answer:
Farsightedness:

1. In hypermetropia, a human eye can see distant distinctly but is unable to see nearby objects clearly.
2. In this case, the image of a nearby object would be formed behind the retina.
3. This defect can be corrected using a convex lens of appropriate power.

Nearsightedness:

1. In myopia, a human eye can see near objects distinctly, but is unable to see distant objects clearly.
2. In this case, the image of a distant object is formed in front of the retina.
3. This defect can be corrected using a concave lens of appropriate power.

b. Concave lens and Convex lens.
Answer:
Concave lens:

1. A concave lens has its surfaces curved inwards.
2. It is thicker at the edges than in the middle.
3. It can form only a virtual image.
4. It can form only a diminished image.

Convex lens:

1. A convex lens has its surfaces puffed up outwards.
2. It is thicker in the middle than at the edges.
3. It can form a real image as well as a % virtual image.
4. It can form a magnified, diminished or the same sized image (relative to the object) depending on the position of the object.

Question 7.
What is the function of the iris and the muscles connected to the lens in the human eye?
Answer:
When the incident light is very bright, the muscles of the iris stretch to reduce the size of the pupil. When the incident light is dim, the muscles of the iris relax to increase the size of the pupil. Thus, the iris controls the size of the pupil and thereby regulates the amount of light entering the eye. (Fig. 7.26)

When a distant object is to be observed, the ciliary muscles relax so that the eye lens becomes flat. This increases the focal length of the lens. Therefore, a sharp image of the distant object is formed on the retina.
Thus, we can see a distant object clearly. (Fig. 7.27)

when an object closer to the eye is to be observed. the ciliary muscles contract increasing the curvature. of the eye lens. The eye lens, therefore, becomes rounded. This decreases the focal length of the lens. Therefore, a sharp image of the nearby object is formed on the retina. Thus, we can see a nearby object clearly. (Fig. 7.27)

Question 8.
Solve the following examples:
i. Doctor has prescribed a lens having I power + 1.5 D. What will be the focal length of the lens? What is the type of the lens and what must be the defect of vision?
Solution:
Data: P = + 1.5 D, f = ?
Focal length of the lens, f = \(\frac{1}{P}=\frac{1}{1.5 \mathrm{D}}\)
= \(\frac{10}{15}\) m = 0.6667 m = 0.67 m
P is positive. This shows that the lens is convex. The defect of vision is farsightedness (hypermetropia).

ii. 5 cm high object is placed at a distance of 25 cm from a converging lens of focal length of 10 cm. Determine the position, size and type of the image.
Solution:
Data: Converging lens, f = 10 cm
u = – 25 cm, h₁ = 5 cm, v = ?, h₂ = ?


The height of the image = -3.3 cm (inverted image ∴ minus sign).
(iii) The image is real, inverted and smaller than the object.

iii. Three lenses having powers 2, 2.5 and 1.7 D are kept touching in a row. What is the total power of the lens combination?
Solution:
Data: P₁ = 2 D, P₂ = 2.5 D, P₃ = 1.7 D, P = ?
Total power of the lens combination,
P = P₁ +P₂ + P₃
= 2 D + 2.5 D + 1.7 D
= 6.2 D.

iv. An object kept 60 cm from a lens gives a virtual image 20 cm in front of the lens. What is the focal length of the lens? Is it a converging lens or diverging lens?
Solution:
Data: u = -60 cm, v = -20 cm, f = ?

∴ The focal length of the lens, f = – 30 cm. As f is negative, it is a diverging lens.

Project:

Question 1.
Make a Powerpoint presentation about the construction and use of binoculars. (Do it your self)

Can you recall? (Text Book Page No. 80)

Question 1.
Indicate the following terms related to spherical mirrors in figure 7.1: pole, centre of curvature, radius of curvature, principal focus.

Answer:

Question 2.
How are concave and convex mirrors constructed?
Answer:
The given part of a hollow spherical glass can be converted into a concave mirror by (i) polishing (silvering) its inner side (inner surface or concave surface) to make it reflecting or (ii) coating its outer side with a thin layer of silver and painting it with red colour to protect the silver coating.
[Note: Case (i) corresponds to the front surface silvered concave mirror.]

The given part of a hollow spherical glass can be converted into a convex mirror by (i) polishing (silvering) its outer side (outer surface or convex surface) to make it reflecting or (ii) coating its inner side with a thin layer of silver and painting it with red colour to protect the silver coating.
[Note: Case (i) corresponds to the front surface silvered convex mirror. ]

Use your brain power! (Text Book Page No. 85)

Question 1.
From equations (1) and (2) what is the relation between h₁, h₂, u and v?
Answer:
M = \(\frac{h_{2}}{h_{1}}\) …….(1)
Also, M = \(\frac{v(\text { image distance })}{u \text { (object distance) }}\)……(2)
\(\frac{h_{2}}{h_{1}}\) = \(\frac{v}{u}\)

Try this (Text Book Page No. 88)

(1) Try to read a book keeping it very far from your eyes.
(2) Try to read a book keeping it very close to your eyes.
(3) Try to read a book keeping it at a distance of 25 cm from your eyes.
At which time do you see the alphabets clearly? Why?
Answer:
Minimum distance of distinct vision: Though the focal length of the eye lens is adjustable, it cannot be decreased below a certain limit. Hence, if an object is very close to the eye, it cannot be seen clearly. For a normal human eye, the minimum distance from the eye at which an object is clearly visible without stress on the eye, is called the minimum distance of distinct vision. For the normal human eye, it is 25 cm.

Use your brain power! (Text Book Page No. 89)

Question.
(1) Why do we have to bring a small object near the eyes in order to see it clearly?
(2) If we bring an object closer than 25 cm from the eyes, when can we not see it clearly even though it subtends a bigger angle at the eye?
Answer:
(1) when a small object is brought near the eyes, its apparent size increases. Therefore, it is
seen clearly.

(2) Minimum distance of distinct vision: Though the focal length of the eye lens is adjustable, it cannot be decreased below a certain limit. Hence, if an object is very close to the eye, it cannot be seen clearly. For a normal human eye, the minimum distance from the eye at which an object is clearly visible without stress on the eye, is called the minimum distance of distinct vision. For the normal human eye, it is 25 cm.

Try this (Text Book Page No. 91)

Question 1.
Take a burning incense stick in your hand and rotate it fast along a circle.
Answer:
A circle of red light is seen.

Question 2.
Draw a cage on one side of a cardboard and a bird on the other side. Hang the cardboard with the help of a thread. Twist the thread and leave It. What do you see and why?
Answer:
The bird appears to be inside the cage. This happens due to persistence of vision.
Persistence of vision: We see an object when its image is formed on the retina. The image disappears when the object is removed from our sight. But this is not instantaneous and the image remains imprinted on the retina for about \(\frac{1}{16}\) th of a second after the removal of the object. The sensation on the retina persists for a while. This effect is known as the persistence of vision.

It is due to persistence of vision that we continue to see the object in its position for about \(\frac{1}{16}\) th of a second after it is removed.
Example: When a burning stick of incense is moved fast in a circle, a circle of red light is seen.

Can you tell? (Text Book Page No. 91)

Question 1.
How do we perceive different colours?
Answer:
(1) In nature we field objects of various colours. Perception of colour means to be able to respond to colour.
(2) We can distinguish between various colours due to perception of colour.
(3) The cone-shaped cells on the retina of the eye respond to the various colours when light is bright and communicate to the brain about the colours of the image formed on the retina. This gives us the proper idea about the colours of the object.
(4) If, in the retina of a person, the cone-shaped cells responding to certain specific colours are absent, the person is unable to distinguish between the colours. As a result, he lacks perception of colour.

Fill in the blanks and rewrite the statements:

Question 1.
The focal length of a………..lens is positive.
Answer:
The focal length of a convex lens is positive.

Question 2.
The focal length of a………..lens is negative.
Answer:
The focal length of a concave lens is negative.

Question 3.
The magnification produced by a………..lens is always positive.
Answer:
The magnification produced by a concave lens is always positive.

Question 4.
The power of a………..lens is positive.
Answer:
The power of a convex lens is positive.

Question 5.
The power of a………..lens is negative.
Answer:
The power of a concave lens is negative.

Question 6.
The focal length of a lens with power 2.5 D is………..
Answer:
The focal length of a lens with power 2.5 D is 40 cm (0.4 m).

Question 7.
The power of a lens with focal length 20 cm is………..
Answer:
The power of a lens with focal length 20 cm is 5D.

Question 8.
The minimum distance of distinct vision for a normal human eye is………..
Answer:
The minimum distance of distinct vision for a normal human eye is 25 cm.

Question 9.
If two lenses with focal lengths 10 cm and 20 cm respectively are kept in contact with each other, the effective power of the combination is………..
Answer:
If two lenses with focal lengths 10 cm and 20 cm respectively are kept in contact with each other, the effective power of the combination is 15 D.

Question 10.
A………..lens is used as a simple microscope.
Answer:
A convex lens is used as a simple microscope.

Rewrite the following statements by selecting the correct options:

Question 1.
Inside water, an air bubble behaves………..
(a) like a flat plate
(b) like a concave lens
(c) like a convex lens
(d) like a concave mirror
Answer:
Inside water, an air bubble behaves like a concave lens.

Question 2.
…………represents the lens formula.

Answer:
(b) \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) represents the lens formula.

Question 3.
The power of a convex lens of focal length 25 cm is………..
(a) +4.0 D
(b) 0.25 D
(c) -4.0 D
(d) -0.4D
Answer:
The power of a convex lens of focal length 25 cm is +4.0 D

Question 4.
A lens does not produce any deviation of a ray of light passing through………..
(a) it’s centre of curvature
(b) it’s optical centre
(c) it’s principal focus
(d) an axial point at a distance 2F from its centre
Answer:
A lens does not produce any deviation of a ray of light passing through its optical centre.

Question 5.
The image formed by a concave lens is always………..
(a) virtual and erect
(b) real and erect
(c) virtual and inverted
(d) real and inverted
Answer:
The image formed by a concave lens is always virtual and erect.

Question 6.
A convex lens forms a virtual image of an object placed………..
(a) at infinity
(b) at a distance 2F from the lens
(c) at a distance F from the lens
(d) between the principal focus and the optical centre of the lens.
Answer:
A convex lens forms a virtual image of an object placed between the principal focus and the optical centre of the lens.

Question 7.
When an object is placed at 2F₁ of a convex lens, its image is formed………..
(a) at F₁
(b) at 2F₂
(c) beyond 2F₂
(d) on the same side as the object
Answer:
When an object is placed at 2F₁ of a convex lens, its image is formed at 2F₂

Question 8.
To obtain an image of the same size as that of an object with the help of a convex lens, the object should be placed………..
(a) at infinity
(b) beyond F₁
(c) between F₁ and 2F₁
(d) at 2F₁
Answer:
To obtain an image of the same size as that of an object with the help of a convex lens, the object should be placed at 2F₁

Question 9.
When an object is placed between O and F₁ in front of a convex lens, the image formed is ………..
(a) enlarged and erect
(b) diminished and erect
(c) real and enlarged
(d) diminished and inverted
Answer:
When an object is placed between O and F₁ in front of a convex lens, the image formed is enlarged and erect.

Question 10.
When an object is placed at any finite distance from a concave lens, the image is formed ………..
(a) between F₁ and 2F₁
(b) beyond 2F₁
(c) at F₁
(d) between F₁ and O on the same side as the object.
Answer:
When an object is placed at any finite distance from a concave lens, the image is formed between F₁ and O on the same side as the object.

Question 11.
A student obtained a clear image of window grills on the screen. But the teacher told him to get the image of a tree far away, instead of the window. To get a clear image, the lens must be ……….. (Practice Activity Sheet – 2)
(a) moved towards the screen
(b) moved away from the screen
(c) moved behind the screen
(d) moved far away from the screen
Answer:
A student obtained a clear image of window grills on the screen. But the teacher told him to get the image of a tree far away, instead of the window. To get a clear image, the lens must be moved towards the screen.

Question 12.
The image obtained while finding the focal length of a convex lens is ……….. (Practice Activity Sheet – 3)
(a) real and erect
(b) virtual and erect
(c) real and inverted
(d) virtual and inverted
Answer:
The image obtained while finding the focal length of a convex lens is real and inverted.

Question 13.
Yash found out f₁ and f₂ of a symmetric convex lens experimentally. Then which of the following conclusions is true? (Practice Activity Sheet – 4)
(a) f₁ = f₂
(b) f₁ > f₂
(c) f₁ < f₂
(d) f₁ ≠ f₂
Answer:
(a) f₁ = f₂

State whether the following statements are true or false: (If a statement is false, correct it and rewrite it.)

Group (A)

Question 1.
Power of a lens, P = \(\frac{1}{f}\).
Answer:
True.

Question 2.
If the power of a lens is 2 D, its focal length = 0.5 m.
Answer:
True.

Question 3.
A concave lens is a converging lens. (March 2019)
Answer:
False. (A concave lens is a diverging lens.)

Question 4.
A convex lens is a diverging lens.
Answer:
False. (A convex lens is a converging lens.)

Question 5.
A concave lens always forms a virtual image.
Answer:
True.

Question 6.
A convex lens always forms a virtual image.
Answer:
False. (A convex lens forms a real image or a virtual image depending on the object distance.)

Question 7.
Due to the light sensitive cells in the eye, we get information about the brightness or dimness of the object and the colour of the object.
Answer:
True.

Question 8.
The focal length of a concave lens is negative.
Answer:
True.

Question 9.
The magnification produced by a concave lens is positive or negative depending on the object distance.
Answer:
False. (The magnification produced by a concave lens is always positive.)

Question 10.
The magnification produced by a convex lens is positive or negative depending on the object distance.
Answer:
True.

Question 11.
A concave lens is used as a magnifying glass.
Answer:
False. (A convex lens is used as a magnifying glass.)

Question 12.
A convex lens is used as a simple microscope.
Answer:
True.

Question 13.
A concave lens is used to correct myopia.
Answer:
True.

Question 14.
A convex lens is used to correct hypermetropia.
Answer:
True.

Group (B)

Question 1.
When red light falls on the eyes, the cells responding to red light get excited more than those responding to other colours and we get the sensation of red colour.
Answer:
True.

Question 2.
When an object is placed in front of a concave lens, its image is obtained on the opposite side of the object.
Answer:
False. (When an object is kept in front of a concave lens, its image is obtained on the same side of the lens as the object.)

Question 3.
The image formed by a concave lens is always virtual.
Answer:
True.

Question 4.
The principal focus of a convex lens is virtual.
Answer:
False. (The principal focus of a convex lens is real.)

Question 5.
An object of height 2 cm forms an image of height 3 cm when placed in front of a concave lens.
Answer:
False. (An object of height 2 cm forms an image of height less than 2 cm when placed in front of a concave lens.)

Question 6.
Absence of rod like cells results in colour blindness.
Answer:
False. (Absence of conical cells results in colour-blindness.)

Question 7.
Nearsightedness can be corrected using spectacles having convex lenses.
Answer:
False. (Nearsightedness can be corrected using spectacles having concave lenses.)

Question 8.
Farsightedness can be corrected using spectacles having convex lenses of suitable focal length.
Answer:
True.

Question 9.
As one grows old, ciliary muscles become weak.
Answer:
True.

Question 10.
In a simple microscope, the object is placed within the focal length of the convex lens.
Answer:
True.

Question 11.
A compound microscope forms an erect and real image of a small object.
Answer:
False. (A compound microscope forms an inverted and virtual image of a small object.)

Question 12.
In a compound microscope, a real image acts as an object for the eyepiece.
Answer:
True.

Question 13.
In television, we see a continuous picture due to persistence of vision.
Answer:
True.

Question 14.
The conical cells can respond differently to red, green and blue colours.
Answer:
True.

Question 15.
The rod like cells respond to colours and communicate the presence of colours in the retinal image of the brain.
Answer:
False. (The rod like cells respond to the intensity of light and communicate the degree of brightness and darkness, to the brain.)

Question 16.
The conical cells respond to the intensity of light and communicate the degree of brightness and darkness to the brain.
Answer:
False. (The conical cells respond to colours and communicate the presence of colours in the retinal image to the brain.)

Question 17.
Generally, using the same objective lens, but different eyepieces, different magnification can be obtained.
Answer:
True.

Find the odd one out and give the reason:

Question 1.
Simple microscope, Compound microscope, Telescope, Myopia.
Answer:
Myopia. It is a defect of vision; others are instruments.

Question 2.
Myopia, Presbyopia, Hypermetropia, Spectrometer.
Answer:
Spectrometer. It is an instrument; others are defects of vision.

Question 3.
Presbyopia, Retina, Nearsightedness, Farsightedness.
Answer:
Retina. It is a part of the eye; others are defects of vision.

Question 4.
Compound microscope, Kaleidoscope, Simple microscope, Astronomical telescope.
Answer:
Kaleidoscope. Others are optical instruments.

Question 5.
TV, Motion picture, Complete circle formed by a revolving burning incense stick, Colour blindness.
Answer:
Colour-blindness. Others are examples of persistence of vision.

Question 6.
Planets, Stars, Satellites, Rainbow.
Answer:
Rainbow. Others are celestial bodies.

Considering the correlation between the words of the first pair, pair the third word accordingly with proper answer:

Question 1.
Nearsightedness: Elongated eyeball :: Farsightedness:………
Answer:
Flattened eyeball

Question 2.
Convex lens : Converging :: Concave lens :………..
Answer:
Diverging

Question 3.
Object at 2F₁ of a convex lens : Image at 2F₂ :: Object at F₁ :………..
Answer:
Image on the opposite side at infinity

Question 4.
Magnification positive : Erect image :: Magnification negative :………..
Answer:
Inverted image

Question 5.
Convex lens : Positive power of the lens :: Concave lens:
Answer:
Negative power of the lens.

Question 6.
\(\frac{1}{f(\text { in metre })}\) : Power of the lens (in dioptre) :: \(\frac{\text { Image distance }}{\text { Object distance }}\)
Answer:
Magnification.

Question 7.
Focal length : Metre :: Power of a lens :………..
Answer:
Dioptre.

Question 8.
Iris : Pupil :: Ciliary muscles :……….
Answer:
Eye lens

Question 9.
Nearsightedness : Concave lens :: Farsightedness :………..
Answer:
Convex lens

Question 10.
Nearsightedness : Image in front of the retina :: Farsightedness :………..
Answer:
Image behind the retina

Question 11.
Observation of stars and planets : Telescope :: Repairing a watch :……….
Answer:
Simple microscope

Question 12.
Cinema : Persistence of vision :: Rainbow :………
Answer:
Refraction, dispersion and internal reflection of light.

Match the following:

Question 1.

Column A	Column B
(1) Conical cells	(a) Intensity of light
(2) Rod like cells	(b) Colour of an image
(3) Pupil	(c) Iris
(4) Cornea	(d) Aperture
	(e) Transparent

Answer:
(1) Conical cells – Colour of an image
(2) Rod like cells – Intensity of light
(3) Pupil – Aperture
(4) Cornea – Transparent.

Question 2.

Column A	Column B
(1) Magnification	(a) \(\frac{1}{f}\)
(2) Power of a lens	(b) \(\frac{h_{2}}{h_{1}}\)
(3) Focal length	(c) f
(4) Distance of an object from a lens	(d) u
	(e) \(\frac{h_{1}}{h_{2}}\)

Answer:
(1) Magnification: \(\frac{h_{2}}{h_{1}}\)
(2) Power of a lens: \(\frac{1}{f}\)
(3) Focal length: f
(4) Distance of an object from a lens: u.

Question 3.

Answer:
(1) Lens: \(\frac{1}{f}: \frac{1}{v}-\frac{1}{u}\)
(2) Magnification: \(\frac{h_{2}}{h_{1}}\)
(3) Refractive index: \(\frac{\sin i}{\sin r}\)

Question 4.

Column A	Column B
(Convex lens)	(a) Image virtual, erect and enlarged
(1) Object at 2F1	(b) Image real, inverted and of the same size
(2) Object between F1 and 2F1	(c) Image real, inverted and highly diminished
(3) Object between O and F1	(d) Image real, inverted and highly enlarged
(4) Object at infinity	(e) Image real, inverted and enlarged

Answer:
(1) Object at 2F₁ – Image real, inverted and of the same size
(2) Object between F₁ and 2F₁ – Image real, inverted and enlarged
(3) Object between O and F₁ – Image virtual, erect and enlarged
(4) Object at infinity – Image real, inverted and highly diminished

Question 5.

Column A	Column B
(1) Nearsightedness	(a) Ciliary muscles become weak
(2) Farsightedness	(b) Image in front of the retina
(3) Presbyopia	(c) Colour-blindness
	(d) Image behind the retina

Answer:
(1) Nearsightedness – Image in front of the retina
(2) Farsightedness – Image behind the retina
(3) Presbyopia – Ciliary muscles become weak

Question 6.

Column A	Column B
(1) Convex lens	(a) To see small objects clearly
(2) Astronomical telescope	(b) To observe minute objects
(3) Compound microscope	(c) To observe astronomical objects such as stars, planets, etc.
(4) Simple microscope	(d) Presbyopia
	(e) Power of a lens

Answer:
(1) Convex lens – Presbyopia
(2) Astronomical telescope – To observe astronomical objects such as stars, planets, etc.
(3) Compound microscope – To observe minute objects
(4) Simple microscope – To see small objects clearly.

Question 7.

Column A	Column B
(1) Persistence of vision	(a) Lenses and mirrors are used
(2) Reflecting telescope	(b) To see objects far away from us
(3) Telescope	(c) Motion picture
(4) Compound microscope	(d) To observe blood
	(e) Convex lens

Answer:
(1) Persistence of vision – Motion picture
(2) Reflecting telescope – Lenses and mirrors are used
(3) Telescope – To see objects far away from us
(4) Compound microscope – To observe blood corpuscles.

Name the following:

Question 1.
Name the lens which forms a real image or a virtual image depending on the position of the object.
Answer:
A convex lens.

Question 2.
Name the lens which produces magnification always less than 1.
Answer:
A concave lens.

Question 3.
Name the lens which always forms an image virtual and smaller than the object.
Answer:
A concave lens.

Question 4.
Name the lens used to obtain the image on a screen.
Answer:
A convex lens.

Question 5.
Name the lens for which the image always lies between the object and the lens.
Answer:
A concave lens.

Question 6.
Name the instrument used to observe bacteria.
Answer:
A compound microscope.

Question 7.
Name the instrument used to observe planets.
Answer:
An astronomical telescope.

Answer the following questions in one sentence each:

Question 1.
An object is placed at 60 cm from a convex lens of focal length 20 cm. State the nature and size of the image relative to that of the object.
Answer:
The image is real, inverted and smaller than the object.

Question 2.
If an object is placed at 50 cm from a convex lens of focal length 25 cm, what will be the image distance?
Answer:
The image distance will be.50 cm.

Question 3.
An object is placed at 40 cm from a convex lens of focal length 20 cm. State the nature and the size of the image relative to that of the object.
Answer:
The image is real, inverted and of the same size as that or the object.

Question 4.
An object is placed at 30 cm from a convex lens of focal length 20 cm. State the nature and the size of the image relative to that of the object.
Answer:
The image is real, inverted and larger than the object.

Question 5.
An object is placed at 15 cm from a convex lens of focal length 25 cm. State the nature and size of the image relative to that of the object.
Answer:
The image is virtual, erect and larger than the object.

Question 6.
State the type of lens that can be used to burn paper in sunlight at noon.
Answer:
A convex lens can be used to burn paper in sunlight at noon.

Question 7.
State the type of lens used to correct myopia.
Answer:
A concave lens is used to correct myopia.

Question 8.
State the type of lens used to correct hypermetropia.
Answer:
A convex lens is used to correct hypermetropia.

Question 9.
If two lenses with focal lengths 10 cm and – 20 cm respectively are kept in contact with each other, what will be the effective power of the combination of the lenses?
Answer:
The effective power of the combination of the lenses will be + 5 D.

Question 10.
If two lenses with focal lengths – 10 cm and 40 cm respectively are kept in contact with each other, what can you say about the behaviour of the combination of the lenses?
Answer:
The combination of the lenses will behave as a concave lens.

Answer the following questions:

Question 1.
What is a lens?
Answer:
A lens is a transparent material bound by two surfaces, out of which at least one surface is spherical.
[Note: A lens is normally made of glass or plastic.]

Question 2.
In which instruments have you seen a lens?
Answer:
We have seen a lens in a microscope and a telescope.

Question 3.
How is a lens different from a mirror?
Answer:
A mirror has one reflecting surface. By reflection of light, it forms an image of the object placed in front of it. A mirror is not transparent. A lens has two surfaces that form an image by refraction of light. A lens is transparent.

Question 4.
Make a list of optical devices you know.
Answer:
Microscope, telescope, binoculars, camera, projector.

Question 5.
Do you know which is the natural optical device?
Answer:
Yes. The eye is the natural optical device.

Question 6.
What is a convex lens?
Answer:
A lens having both spherical surfaces puffed up outwards is called a convex lens or double convex lens or biconvex lens. It is thicker in the middle than at the edges.
[Note: A convex lens is also called a converging lens.]

Question 7.
What is a concave lens?
Answer:
A lens having both spherical surfaces curved inwards is called a concave lens or double concave lens or biconcave lens. It is thicker at the edges than in the middle.
[Note: A concave lens is also called a diverging lens.]

Question 8.
Draw neat labelled diagrams: Types of lenses.
Answer:

[Note: Positive meniscus behaves as a convex lens as it is thicker in the middle than at the edges. Negative meniscus behaves as a concave lens as it is thicker at the edges than in the middle.]

Question 9.
In general, when a ray of light passes through a lens, there occurs a change in its direction of propagation. Why?
Answer:
The working of a lens is similar to that of a triangular prism. When a ray of light passes through a lens, it is refracted twice: When entering the lens and when emerging from the lens. There is a change 5 in its direction of propagation every time and as both the changes occur in the same sense, the direction of propagation of the emergent ray is different from that of the incident ray.

Question 10.
State the rules used for drawing ray diagrams for the formation of an image by a convex lens.
Answer:
Rules used for drawing ray diagrams for the formation of an image by a convex lens:
(1) When the incident ray is parallel to the principal axis, the refracted ray passes through the principal focus.

(2) When the incident ray passes through the principal focus, the refracted ray is parallel to the principal axis.

(3) When the incident ray passes through the optical centre of the lens, it passes without changing its direction.

Question 11.
In the case of a convex lens, show the path of the refracted ray when the incident ray of light (1) is parallel to the principal axis of the lens (2) passes through the focus of the lens (3) passes through the optical centre of the lens.
Answer:
1.

2.

3.

Question 12.
Draw neat and well labelled ray diagrams for image formation by convex lens when an object is (1) at infinity (2) beyond 2F₁ (3) at 2F₁ (4) between F₁ and 2F₁ (5) at focus F₁ (6) between focus F₁ and optical centre O. Also, in each case, state the position, nature and size of the image relative to that of the object.
Answer:
(1) Object at infinity:

In this case, the image is formed at focus F₂ of the convex lens. It is real, inverted and highly diminished (point-sized).

(2) Object beyond 2F₁:

In this case, the image is formed between F₂ and 2F₂. It is real, inverted and diminished.

(3) Object at 2F₁:

In this case, the image is formed at 2F₂. It is real, inverted and of the same size as that of the object.

(4) Object between F₁ and 2F₁:

In this case, the image is formed beyond 2F₂. It is real, inverted and magnified (enlarged).

(5) Object at focus F₁:

In this case, the image is formed at infinity. It is real, inverted and infinitely large (highly magnified).

(6) Object between focus F₁ and optical centre O:

In this case, the image is formed on the same side of the lens as the object. It is virtual, erect and larger than the object.
[Note: Here, the image is virtual. Hence, it is shown by a dotted line.]

Question 13.
Observe the following figure and complete the table: (March 2019)

Answer:

Point	Answer
(i) Position of the object	Between F1 and O
(ii) Position of the image	On the same side of the lens as the object
(iii) Size of the image	Very large
(iv) Nature of the image	Virtual and erect

Question 14.
At which position will you keep an object in front of a convex lens to get a real image smaller than the object? Draw a figure.
Answer:
The object should be placed beyond 2F₁.

Question 15.
State the rules used for drawing ray diagrams for the formation of an image by a concave lens.
Answer:

1. When the incident ray is parallel to the principal axis, the refracted ray, when extended backwards, passes through the principal focus.
2. When the incident ray is directed towards the principal focus F₂, the refracted ray is parallel to the principal axis.
3. When the incident ray passes through the optical centre of the lens, it passes without changing its direction.

Question 16.
State the characteristics of an image formed by a concave lens.
Answer:
The image formed by a concave lens is always virtual, erect and smaller than the object. It is on the same side of the lens as the object. Generally, it is formed between the optical centre of the lens and the principal focus F₁. If the object is at infinity, the image is a point image formed at F₁.

Question 17.
In the case of image formation by a concave lens, what can you say about the position, nature and size of the image relative to the size of the object?
Answer:
Image formation by a concave lens :
(1) If the object is at infinity, the image is formed at the focus of the lens, on the same side of the lens as the object. It is virtual, erect and much smaller than the object (point image).

(2) If the object is at any finite distance from the lens, the image is formed on the same side of the lens as the object and between the focus and the optical centre of the lens. It is virtual, erect and smaller than the object. The image distance is less than the object distance.

Question 18.
Draw a ray diagram to show image formation by a concave lens.
Answer:

PQ : Object
P’Q’ : Image (virtual, therefore shown by a dotted line),
O : Optical centre,
F₁ : Principal focus,
f : Focal length of the lens
[Note: If in a Board examination, incomplete diagram (as shown below) is given, students should complete it and label its parts as shown in Figure.]

Question 19.
State the Cartesian sign convention for refraction of light (image formation) by a lens.
Answer:
Cartesian sign convention for refraction of light (image formation) by a lens:
In this case, the optical centre (O) of the lens is taken as the origin and the principal axis of the lens is taken as X-axis of the coordinate system.

(1) The object is always placed at the left of the lens.
All distances parallel to the principal axis are measured from the optical centre of the lens.
(2) All distances measured to the right of the origin are taken as positive while distances measured to the left of the origin are taken as negative.
(3) Distances measured perpendicular to and above the principal axis are taken as positive.
(4) Distances measured perpendicular to and below the principal axis are taken as negative.
(5) The focal length of a convex lens is positive and that of a concave lens is negative. (Fig. 7.21)

Question 20.
(i) What is a lens formula? (ii) State it.
Answer:
(i) The relationship between the object distance (u), image distance (v) and focal length (J) of a lens is called the lens formula.
(ii) It is \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
[Note: The lens formula holds good for all values of u and v and is applicable to a convex lens as well as a concave lens. The sign convention for u, v and f must tie used in solving numerical examples.]

Question 21.
What is meant by the magnification produced by a lens? State the formulae for it.
Answer:
The magnification (M) produced by a lens is the ratio of the height of the image (h₂) to the height of the object (h₁).
M = \(\frac{h_{2}}{h_{1}}\)……….(1)
Also M = \(\frac{v(\text { image distance })}{u \text { (object distance) }}\)……………(2)

Question 22.
When is the magnification produced by a lens (1) positive (2) negative?
Answer:
The magnification produced by a lens is
(1) Positive when the image is virtual (as it is erect)
(2) Negative when the image is real (as it is inverted).

Question 23.
Express the magnification produced by a lens in terms of the focal length of the lens and (1) the object distance (2) the image distance.
Answer:
Magnification (M) produced by a lens

where f is the focal length of the lens:

(2) From eq. (2), we have

Question 24.
An object is kept in front of a lens of focal length + 10 cm. Describe the nature of the image in the following cases: (1) The object” distance is 25 cm. (2) The object distance is 5 cm.
Answer:
Since, the focal length of the lens ( +10 cm) is positive, it is a convex lens.
(1) If an object is kept at 25 cm from the lens, the image will be real, inverted and smaller than the object.
(2) If an object is kept at 5 cm from the lens, the image will be virtual, erect and larger than the object.

Question 25.
Anu and Anand have concave and convex lenses respectively. They took lenses in sunlight and tried to burn two pieces of paper of equal areas and temperature. State which lens will burn the paper. Give the reason. Explain with the help of a diagram, why the other paper did not burn.
Answer:
(1) The convex lens will burn the paper. See Fig. 7.22 for reference. The ray of sunlight will converge at the principal focus of the lens. Hence, if the paper is held at the focus, it will burn due to concentration of heat energy.

(2) The paper held in front of the concave lens, will not burn. For reference, see Fig. 7.23. The concave lens will diverge the rays of sunlight falling on it. Hence, the paper will not burn.

Question 26.
To obtain a magnified real image of a small film strip, which type of lens is used? Where is the film strip placed to obtain the image on the screen?
Answer:
To obtain a magnified real image of a small film strip, a convex lens is used. The film strip is placed between F₁ and 2F₁ and the screen is placed on the other side of the lens.

Question 27.
When an object of height 2 cm is placed in front of a convex lens, the height of the image is found to be 3 cm. State the nature and position of the image giving reason.
Answer:
When an object is placed between the optical centre and the principal focus of a convex lens, the image formed by the lens is virtual and larger than the object. When an object is placed between F₁ and 2F₁ the image formed by the lens is real and larger than the object.

In the above case, if the image is virtual, it will be erect and on the same side of the lens as that of the object. If the image is real, it will be inverted and beyond 2F₂ on the other side of the lens with respect to the object.

Question 28.
You are given a lens which gives a virtual, erect and enlarged image. What type of lens is it?
Answer:
Since the lens gives a virtual, erect and enlarged image, it must be a convex lens.

Question 29.
When an object of height 3 cm is placed in front of a concave lens, the height of the image is found to be 6 cm. State, giving the reason, whether the given statement is true or false.
Answer:
When an object is placed in front of a concave lens, the image formed by the lens is always smaller than the object. In the statement given in the question, the height of the image is reported as greater than that of the object. Hence, the statement given in the question is false.

Question 30.
State two uses of a concave lens.
Answer:

1. A concave lens is used to correct myopia (nearsightedness).
2. In some optical instruments, a combination of a concave lens and a convex lens is used.

Question 31.
State two uses of a convex lens.
Answer:
A convex lens is used (1) to read words in small print (2) to correct hypermetropia (Far-sightedness).

Question 32.
An object is kept in front of a lens of j focal length – 20 cm. Describe the nature of the image when the object distance is 25 cm.
Answer:
Since the focal length of the lens (- 20 cm) is negative, it is a concave lens.
If an object is kept at 25 cm from the lens, the image will be virtual, erect and smaller than the object.
[Note: The nature of the image is independent of the object distance as it is a concave lens.]

Question 33.
An object is placed in front of a convex lens of focal length 20 cm. If the object distance is changed from 60 cm to 40 cm, what can you say about the size of the image relative to that of the object?
Answer:
In this case, the focal length (f) of the lens is 20 cm.
∴ 2f = 40 cm.
When the object distance is 60 cm (which is greater than 2f), the image will be smaller than the object. When the object distance becomes 40 cm (which is equal to 2f), the image will be of the same size as that of the object.

Question 34.
What is the power of a lens?
Answer:
The capacity of a lens to converge or diverge incident rays is called its power. The power (P) of a lens is the inverse of the focal length (f) of the lens.
P = \(\frac{1}{f}\)

Question 35.
What is the unit of power of a lens? Define it.
Answer:
The unit of power of a lens is the dioptre (D).
One dioptre is the power of a lens whose focal length is one metre.
1 dioptre (D) = \(\frac{1}{1 \text { metre }(\mathrm{m})}\)
[Note: The dioptre, the SI unit of power of a lens, is denoted by D.]

Question 36.
What is the sign of the power of (i) a convex lens (ii) a concave lens?
Answer:
The power of a convex lens is positive while that of a concave lens is negative.

Question 37.
If there is an increase or decrease in the focal length of a lens, what will be the effect on the power of the lens?
Answer:
The power of a lens is the inverse of its focal length. Hence, if there is an increase in the focal length of a lens, the power of the lens will decrease accordingly. Similarly, if there is a decrease in the focal length of a lens, the power of the lens will increase accordingly.

Question 38.
If two lenses of focal lengths f₁ and f₂ are kept in contact with each other, state the formula for the focal length of the combination. If P₁ and P₂ are the powers of these lenses, state the formula for the power of the combination.
Answer:
If two lenses of focal lengths f₁ and f₂ are kept in contact with each other, the focal length (f) of the combination is given by \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

[Two lenses kept in contact with each other]
If P₁ and P₂ are the powers of these lenses, the power (P) of the combination is given by P = P₁ + P₂.
[Note: The figures are given only for reference.]

Question 39.
Draw a neat labelled diagram to show the structure of the human eye.
Answer:

Question 40.
What is cornea?
Answer:
The cornea is a thin and transparent cover (membrane) on the human eye through which light enters the eye. Maximum refraction of light rays entering the eye occurs at the cornea.

Question 41.
What is iris?
Answer:
The iris is a dark fleshy screen (muscular diaphragm) behind the cornea in the human eye. Its colours are different for different people.

Question 42.
What is pupil?
Answer:
The pupil is a small circular opening of changing diameter at the centre of the iris in the human eye.

Question 43.
What is the use of the pupil in the human eye?
Answer:
The pupil in the human eye is useful for controlling and regulating the amount of light entering the eye. The pupil contracts in the presence of too much light and dilates when light is insufficient, thus changing the amount of light entering the eye.

Question 44.
With reference to the functioning of the pupil in the human eye, what is adaptation?
Answer:
The tendency of the pupil in the human eye to adjust the opening for light, depending on the intensity of incident light, to control and regulate the amount of light entering the eye is called adaptation.

Question 45.
What is the shape and the size of the human eyeball?
Answer:
The human eyeball is approximately spherical in shape with a diameter of about 2.4 cm.

Question 46.
Name the part of the human eye that forms a transparent bulge on the surface of the eyeball.
Answer:
The cornea forms a transparent bulge on the surface of the eyeball.

Question 47.
Which part of the human eye is located just behind the pupil?
Answer:
A transparent biconvex crystalline lens is located just behind the pupil in the human eye.

Question 48.
What is retina?
Answer:
The retina is a light sensitive screen consisting of a delicate membrane with a large number of light sensitive cells.

Question 49.
What is the nature of the eye lens and what does the eye lens do?
Answer:
The eye lens is a double convex transparent crystalline lens, just behind the pupil. The eye lens provides small adjustment of focal length to form a real and inverted sharp image on the retina.

Question 50.
What happens when light falls on the retina?
Answer:
When light falls on the retina, light sensitive cells of the retina are activated. They generate electrical signals which are passed by optic nerves to the brain. The brain interprets the signals and processes the information such that we perceive the object as it is.

Question 51.
What are ciliary muscles?
Answer:
The muscles which hold the eye lens in its position, and bring about changes in the shape (curvature) of the eye lens, and hence of focal length are known as ciliary muscles.

Question 52.
What is the focal length of the eye lens of a normal eye in relaxed position of eye muscles?
Answer:
The focal length of the eye lens of a normal eye in relaxed position of eye muscles is about 2 cm.

Question 53.
Where does the second focal point of the eye lens of a normal eye in relaxed position of eye muscles lie?
Answer:
The second focal point of the eye lens of a normal eye in relaxed position of eye muscles lies on the retina.

Question 54.
What is meant by power of accommodation of the eye?
Answer:
The ability of the eye lens to adjust its focal length is called the power of accommodation of the eye.

Question 55.
Explain the term power of accommoda¬tion of the eye.
(OR)
Write a short note on the power of accommodation of the eye.
Answer:
Power of accommodation of the eye: The eye lens is held in its position by the ciliary muscles. When we look at a nearby object, the ciliary muscles compress the eye lens so that it becomes rounded. Hence, the focal length of the eye lens decreases. Therefore, the image is formed on the retina of the eye and hence the nearby object is seen clearly.

When we look at a distant object, the ciliary muscles relax so that the eye lens becomes flat. Hence, the focal length of the eye lens increases. Therefore, the image is formed on the retina of the eye and hence the distant object is seen clearly. This ability of the eye lens to adjust its focal length is called the power of accommodation of the eye.

Question 56.
What is meant by accommodation? How is it brought about?
Answer:
The process of focusing the eye on objects at different distances is called accommodation. It is brought about by changing the curvature of the f elastic eye lens making it thinner or thicker.

Question 57.
The human eye is very similar to a photographic camera. The figure given shows the main parts of a photographic camera. Now answer the following questions:

(1) Name the parts of the human eye similar to the following parts of the photographic camera :
(a) Photographic film (b) Aperture.
(2) State one difference between the human eye lens and camera lens.
(3) Name the muscles which adjust the curvature of the eye lens.
(4) Which phenomenon of light is responsible for the working of the eye?
Answer:
(1) (a) The retina in the human eye is similar to the photographic film in a camera.
(b) The pupil in the human eye is similar to the aperture in a camera.
(2) In a photographic camera, the focal length of the lens changes when the position of the lens is changed. In the human eye, the focal length of the eye lens is changed by the ciliary muscles and the distance of the image from the eye lens is fixed.
(3) The ciliary muscles adjust the curvature of the eye lens.
(4) The refraction of light is responsible for the working of the eye.

Question 58.
Have you seen a photographic camera in which a film is used? Compare the human eye with it. State similarities between them. State the points of difference between them.
Answer:
Yes, we have seen a photographic camera in which a film is used.
Cameras, in general, have various shapes and sizes. Some cameras are much bigger than the human eye while some are smaller than the human eye. Here we shall consider a simple camera.

Similarities: In the case of a camera as well as the human eye, it is possible to control the amount of incoming light with the help of a diaphragm and an aperture. Both use a convex lens for focusing. The photographic film in a camera is coated with a photosensitive material. The retina in the eye consists of a large number of light sensitive cells. The photographic film in a camera is processed using chemicals and then prints (photographs) can be obtained using the appropriate paper.

In the human eye, the electrical signals generated by light sensitive cells are passed by optic nerves to the brain which interprets them.

Differences: Cameras come in a variety of sizes and shapes unlike the human eye. Unlike the human eye, a wide variation in exposure time is possible in the case of cameras. The human eye is sensitive in the visible region (red to violet) of the electromagnetic spectrum, while a much wider range of the electromagnetic spectrum can be covered with cameras designed for specific J purposes. In comparison with the human eye, a wider view and range can be covered by a camera.

In comparison with the human eye, a wider intensity (of light) range can be covered with a camera. The retina is indispensable in the human eye, while cameras without a photographic film have been designed with the help of photosensitive materials and are in current use.

[Note: With advances in technology, improved cameras are designed all the time, and the list of differences between the human eye and a camera in general would be practically endless.]

Question 59.
What is meant by the minimum distance of distinct vision?
Answer:
The minimum distance from the normal eye, at which an object is clearly visible without stress on the eye is called the minimum distance of distinct vision.

Question 60.
Explain the term minimum distance of distinct vision.
(OR) Write a short note on distance of distinct vision.
Answer:
Minimum distance of distinct vision; Though the focal length of the eye lens is adjustable, it cannot be decreased below a certain limit. Hence, if an object is very close to the eye, it cannot be seen clearly. For a normal human eye, the minimum distance from the eye at which an object is clearly visible without stress on the eye, is called the minimum distance of distinct vision. For the normal human eye, it is 25 cm.

Question 61.
State four reasons related to problems of vision.
Answer:
Problems of vision are related to (i) weakening of ciliary muscles (ii) change in the size of the eyeball (iii) irregularities on the surface of cornea (iv) formation of a membrane over the eye lens.

Question 62.
What is myopia or nearsightedness? What are the possible reasons of myopia? How is myopia corrected? Explain with diagrams.
Answer:
Myopia or nearsightedness is the defect of vision in which a human eye can see nearby objects distinctly but is unable to see distant objects clearly as they appear indistinct.
In this case the image of a distant object is formed in front of the retina instead of on the retina. [Figs. 7.29 (a), 7.29 (b)]

Possible reasons of myopia: (1) The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large. (2) The distance between the eye lens and the retina increases as the eyeball elongates.

Myopia is corrected using a suitable concave lens. Light rays are diverged by the concave lens before they strike the eye lens. A concave lens of proper focal length is chosen to produce the required divergence. Hence, after the converging action of the eye lens, the image is formed on the retina. [Fig. 7.29 (c)]

Question 63.
Observe the following diagram and answer the questions.
(a) Which eye defect is shown in this diagram?
(b) What are the possible reasons for this eye defect?
(c) How is this defect corrected? Write it in brief.

Answer:
(a) Myopia or Nearsightedness

(b) Possible reasons for the defect:
(i) The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large.
(ii) The eyeball elongates so that the distance between the lens and the retina increases.

(c) Correction of the defect: This defect can be corrected using spectacles with concave lenses.
A concave lens diverges the incident rays and these diverged rays can be converged by the lens in the eye to form an image on the retina.

Question 64.
What is the sign of the power of the lens used to correct myopia?
Answer:
The power of the lens used to correct myopia is negative.
[Note: It is a concave lens. Negative focal length Negative power.]

Question 65.
In a Std. X class, out of 40 students, 10 students use spectacles, 2 students have ( positive power and 8 students have negative power of lenses in their spectacles.
Answer the following questions:
(1) What does the negative power indicate?
(2) What does the positive power indicate?
(3) Generally which type of spectacles do most of the students use?
(4) What defect of eyesight do most of the students suffer from?
(5) Give two possible reasons for the above defect.
Answer:
(1) The negative power indicates a concave lens or myopia.
(2) The positive power indicates a convex lens or hypermetropia.
(3) Generally, most of the students use spectacles with concave lenses.
(4) Most of the students suffer from myopia.
(5) Two possible reasons for myopia:

1. The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large.
2. The distance between the eye lens and the retina increases as the eyeball elongates.

Question 66.
What is hypermetropia or farsightedness? What are the possible reasons of hypermetropia? How is hypermetropia corrected? Explain with figures.
Answer:
Hypermetropia or farsightedness is the defect of vision in which a human eye can see distant objects clearly but is unable to see nearby objects clearly.
In this case the image of a nearby object would fall behind the retina instead of on the retina.
[Figs. 7.31 (a), 7.31 (b)]

Possible reasons of hypermetropia:
(1) Curvature of the cornea and the eye lens decreases. Hence, the converging power of the eye lens becomes less. (2) The distance between the eye lens and retina decreases (relative to the normal eye) and the focal length of the eye lens becomes very large due to the flattening of the eyeball.

Hypermetropia is corrected using a suitable convex lens. Light rays are converged by the convex lens before they strike the eye lens. A convex lens of proper focal length is chosen to produce the required convergence. Hence, after the converging action of the eye lens, the image is formed on the retina. [Fig. 7.31 (c)]

Question 67.
What is the sign of the power of the lens used to correct hypermetropia?
Answer:
The power of the lens used to correct hypermetropia is positive.
[Note: It is a convex lens. Positive focal length ∴ Positive power.]

Question 68.
Given below is a diagram showing a defect of human eye.

Study it and answer the following questions:
(1) Name the defect shown in the figure. (Practice Activity Sheet – 3)
(2) Give two possible reasons for this defect of eye in human beings.
(3) Name the type of lens used to correct the eye defect.
(4) Draw a labelled diagram to show how the defect is rectified by using the lens.
Answer:
A lens having both spherical surfaces puffed up outwards is called a convex lens or double convex lens or biconvex lens. It is thicker in the middle than at the edges.
[Note: A convex lens is also called a converging lens.]

Question 69.
Observe the following figures and complete the table. (Practice Activity Sheet – 1)
Answer:

Question 70.
What is presbyopia? State the reason for this defect. How is presbyopia corrected?
Answer:
Presbyopia is the defect of vision in which aged people find it difficult to see nearby objects comfortably and clearly without spectacles.

Reason of presbyopia: The power of accommodation of eye usually decreases with ageing. The muscles near the eye lens lose their ability to change the focal length of the lens.
Therefore, the near point of the eye lens shifts farther from the eye.

This defect is corrected using a convex lens of appropriate power. The lens converges light rays before they fall on the eye lens such that the action of the eye lens forms the image on the retina.

Question 71.
What is a bifocal lens?
Answer:
A bifocal lens is a lens of which the upper part is a concave lens to correct myopia and the lower part is a convex lens to correct hypermetropia.

[Note: A person suffering from myopia as well as hypermetropia, uses a bifocal lens. Nowadays, the defects of vision such as myopia and hypermetropia can be corrected using contact lenses or by laser surgery.]

Question 72.
(A) Anil cannot see the blackboard writing clearly, but he can see nearby objects clearly.
(i) What is the eye defect he is suffering from?
(ii) How is it corrected?
(B) Anil’s uncle cannot see nearby objects clearly, but he can see distant objects clearly.
(i) What is the eye defect he is suffering from?
(ii) How is it corrected?
Answer:
(A) Anil cannot see the blackboard writing clearly, but he can see nearby objects clearly.
(i) This defect is called myopia (nearsightedness).
(ii) It is corrected using spectacles having concave lenses of appropriate power.

(B) Anil’s uncle cannot see nearby objects clearly, but he can see distant objects clearly.
(i) This defect is called hypermetropia (farsightedness).
(ii) It is corrected using, spectacles having convex lenses of appropriate power.

Question 73.
When are bifocal lenses used in spectacles?
Answer:
When a person cannot see nearby objects as well as distant objects clearly, bifocal lenses are used in spectacles.

Question 74.
Aniket from Std. X uses spectacles. The power of the lenses in his spectacles is -0.5 D. Answer the following questions:
(1) State the type of’ lenses used in his spectacles.
(2) Name the defect of vision Aniket is suffering from.
(3) Find the focal length of the lenses used in his spectacles.
Answer:
(1) Concave lenses are used in the spectacles used by Aniket.
(2) Aniket is suffering from myopia (near-sightedness).
(3) Focal length of the lenses used in his spectacles

= -2 m (Concave lens ∴ Minus sign)

Question 75.
Sunita from Std. X uses spectacles. Her spectacle number is -1.5 D. Answer the following questions:
(1) Name the defect of eye from which she is suffering.
(2) What type of lens is she using?
(3) Find the focal length of the lens.
Answer:
(1) Myopia.
(2) Concave.

(Concave lens ∴ minus sign)
This is the focal length of the lens.

Question 76.
Surabhi from Std. X uses spectacles. The power of the lenses in her spectacles is 0.5 D. Answer the following questions from the given information: (March 2019)
(i) Identify the type of lenses used in her spectacles.
(ii) Identify the defect of vision Surabhi is suffering from.
(iii) Find the focal length of the lenses used in her spectacles.
Answer:
(i) Convex lenses are used in the spectacles used by Surabhi.
(ii) Surabhi is suffering from hypermetropia (farsightedness).
(iii) Focal length of the lenses used in her spectacles

Question 77.
My grandfather uses a bifocal lens in his spectacles. Explain why.
Answer:
In old age, people usually suffer from both myopia and hypermetropia. Therefore, they need spectacles having bifocal lenses.

The upper part of a bifocal lens is a concave lens to correct myopia. The lower part of a bifocal lens is a convex lens to correct hypermetropia.

Question 78.
State uses of concave lens.
Answer:

1. Concave lenses are used for proper working of medical equipment, scanner, CD player – the instruments that employ laser rays.
2. One or more concave lenses are used in a small safety device, fitted in the peep hole in a door, due to which we can see a large area outside the door.
3. Concave lenses are used in spectacles to correct nearsightedness (myopia).
4. A concave lens is used to spread light emitted by the small bulb in a torch over a wide area.
5. A concave lens is used in front of the eyepiece or inside the eyepiece fitted in a camera, telescope and microscope – the instruments employing convex lenses.

Question 79.
State uses of a convex lens.
Answer:
Convex lenses are used in a simple microscope, compound microscope, refracting telescope, camera, projector, spectroscope, spectacles for correcting farsightedness (hypermetropia) and binoculars.

Question 80.
What is meant by the apparent size of an object? With a neat and labelled diagram, explain the relation between the apparent size of an object and the angle subtended by the object at the eye.
Answer:
An object appears small or big depending upon the size of its image formed on the retina of the eye. The size of an object as perceived by the eye is called the apparent size of the object. Consider two objects of the same size, one held near the eye and the other away from the eye as shown in the following figure (Fig. 7.34). The nearby object (PQ) appears larger than the distant object (P₁Q₁).

Also, the angle a subtended by the nearby object at the eye is larger than the angle β subtended by the distant object at the eye. This shows that the apparent size of an object depends upon the angle subtended by the object at the eye. The greater the angle subtended by the object at the eye, the greater is the apparent size of the object. Similarly, the smaller the angle subtended by the object at the eye, the smaller is the apparent size of the object.

Question 81.
With a neat labelled diagram, explain the working of a simple microscope. State uses of a simple microscope.
(OR)
What does a simple microscope consist of? What is the order of magnification obtained by a simple microscope? What is a simple microscope used for?
Answer:
A simple microscope consists of a convex lens of short focal length, usually fixed in a suitable frame with a handle or mounted on a stand.

The object is placed in front of the convex lens of short focal length such that the object distance is less than the focal length. The image is virtual and larger than the object. It is formed on the same side of the lens as the object.

A maximum magnification of about 20 can be obtained by a simple microscope. A simple microscope is used by watch repairers to observe small parts of a watch and by jewellers to examine ornaments. A simple microscope (also called a magnifying glass) is also used to read words in small print.

Question 82.
With a neat labelled diagram, explain the construction and working of a compound microscope.
Answer:
Construction of a compound microscope:

(1) A compound microscope consists of a metal tube fitted with two convex lenses at the two ends. These lenses are called the objective lens (the lens directed towards the object) and the eyepiece (the lens directed towards the eye). Both the lenses are small in size, but the cross section of the objective lens is less than that of the eyepiece. The objective lens has a short focal length. The focal length of the eyepiece is more than that of the objective lens.

(2) The metal tube is mounted on a stand. The principal axes of the objective lens and the eyepiece are along the same line. The distance between the object and the objective lens can be changed with a screw. It is possible to change the distance between the objective lens and the eyepiece.

Working :
(1) The object to be observed is illuminated and placed in front of the objective lens, slightly beyond the focal length of the objective lens. Its real, inverted and enlarged image is formed by the objective lens on the other side.

(2) This intermediate image lies within the focal length of the eyepiece. It serves as an object for the eyepiece. The eyepiece works as a simple microscope. The final image is virtual, highly enlarged and inverted with respect to the original object. It can be formed at the minimum distance of distinct vision from the eyepiece. The final image is observed by keeping the eye close to the eyepiece.

Question 83.
State two uses of a compound microscope.
Answer:
Uses of a compound microscope:

1. It is used to observe blood corpuscles, plant and animals cells, microorganisms like bacteria, etc.
2. It is used in a pathological laboratory to observe blood, urine, etc.
3. It is a part of a travelling microscope used for measurement of very small distance.

Question 84.
What will happen if in a compound microscope, the objective lens is large in size and has a focal length?
Answer:
If the objective lens of a compound microscope is large in size, in addition to the light coming from an object, other unwanted light will be incident on the objective lens. Hence, the image will not be seen clearly. If the objective lens has a large focal length, the magnification produced by it will be less.

Question 85.
(a) In which type of microscope do you find the lens arrangement as shown in the following diagram? (Practice Activity Sheet – 1)

(b) Write in brief, the working of this microscope.
(c) Where is this microscope used?
Answer:
(a) Compound microscope.

(b)
An object appears small or big depending upon the size of its image formed on the retina of the eye. The size of an object as perceived by the eye is called the apparent size of the object. Consider two objects of the same size, one held near the eye and the other away from the eye as shown in the following figure (Fig. 7.34). The nearby object (PQ) appears larger than the distant object (P₁Q₁).

Also, the angle a subtended by the nearby object at the eye is larger than the angle β subtended by the distant object at the eye. This shows that the apparent size of an object depends upon the angle subtended by the object at the eye. The greater the angle subtended by the object at the eye, the greater is the apparent size of the object. Similarly, the smaller the angle subtended by the object at the eye, the smaller is the apparent size of the object.

(c) A simple microscope is used by watch repairers to observe small parts of a watch and by jewellers to
examine ornaments. A simple microscope (also called a magnifying glass) is also used to read words in small print.

Question 86.
(i) Which type of microscope has the arrangement of lenses shown in the following figure?

(ii) Label the figure correctly.
(iii) Write the working of this microscope.
(iv) Where is this microscope used?
(v) Suggest a way to increase the efficiency of this microscope. (Practice Activity Sheet – 2)
Answer:
(i) Compound microscope.

(ii)

(iii) Working:
(1) The object to be observed is illuminated and placed in front of the objective lens, slightly beyond the focal length of the objective lens. Its real, inverted and enlarged image is formed by the objective lens on the other side.

(2) This intermediate image lies within the focal length of the eyepiece. It serves as an object for the eyepiece. The eyepiece works as a simple microscope. The final image is virtual, highly enlarged and inverted with respect to the original object. It can be formed at the minimum distance of distinct vision from the eyepiece. The final image is observed by keeping the eye close to the eyepiece.

(iv)

1. It is used to observe blood corpuscles, plant and animals cells, microorganisms like bacteria. etc.
2. It is used in a pathological laboratory to observe blood, urine, etc.
3. It is a part of a travelling microscope used for measurement of very small distance.

(v) Lenses with appropriate focal lengths should be selected.

Question 87.
State the use of a telescope.
Answer:
A telescope is used to observe a distant object such as mountain, moon, planet, star in the magnified form.

Question 88.
Observe the following figure and answer the questions. (Practice Activity Sheet – 3)
(a) Which optical instrument shows arrangement of lenses as shown in the figure?
(b) Write in brief the working of this optical instrument.

(c) How can we get different magnifications in this optical instrument?
(d) Draw the figure again and labelled it properly.
Answer:
(a) Refracting telescope.

(b) working: When the objective lens is pointed towards the distant object to be observed, the rays
of light from the distant object, which are almost parallel to each other. pass through the objective lens. The objective lens collects maximum amount of light as it is large in size. It forms a real, inverted and diminished image in the focal plane of the objective lens. Now, the position of the eyepiece is adjusted such that this image falls just within the focal length of the eyepiece and serves as the object for the eyepiece which works as a simple microscope.

The final image is highly magnified, virtual, on the same side as that of the object and inverted with respect to the original object. The final image can be observed by keeping the eye close to the eyepiece. If the image formed by the objective lens lies in the focal plane of the eyepiece, the final image is formed at infinity.

(c) We can get different magnifications by using the eyepiece with different focal lengths.

(d)

Question 89.
What is persistence of vision? Give one example of persistence of vision.
Answer:
Persistence of vision: We see an object when its image is formed on the retina. The image disappears when the object is removed from our sight. But this is not instantaneous and the image
remains imprinted on the retina for about \(\frac{1}{16}\) th of a second after the removal of the object.

The sensation on the retina persists for a while. This effect is known as the persistence of vision. It is due to persistence of vision that we continue to see the object in its position for about \(\frac{1}{16}\) th of a second after it is removed.
Example: When a burning stick of incense is moved fast in a circle, a circle of red light is seen.

Question 90.
Name two devices whose working is based on the phenomenon of persistence of vision.
(OR)
Name any two applications based on persistence of vision.
Answer:
The working of a television set and motion picture is based on the phenomenon of persistence of vision.
[Note: These are the examples of persistence of vision in daily life.

Question 91.
How is the phenomenon of persistence of vision used in motion pictures?
Answer:
In motion pictures, photographs of a moving object are taken at the rate of more than sixteen pictures per second. These photographs are projected on the screen at the same rate.
Each picture is slightly different from the other. As a result of persistence of vision, we get the impression of observing the object in continuous motion.

Question 92.
Name the two types of light sensitive cells present in the retina of the human eye. What are their functions?
Answer:
(1) The retina of the human eye contains a large number of light sensitive cells. These cells are of two shapes : (i) rods and (ii) cones.
(2) The rod-like cells respond to the intensity of light.
(3) The conical cells respond to various colours of light. They respond differently to red, green and blue colours. They do not respond to faint light.

Question 93.
When do you say that a person is colour blind?
Answer:
When a person is unable to distinguish between certain colours, he is said to be colour blind.
[Note: (i) Except for being colour-blind, their eyesight is normal, (ii) Rod-shaped cell ≡ rod-like cell, cone-shaped cell = Conical cell.]

Question 94.
Explain the perception of colour in the human eye.
(OR)
Explain in short perception of colour.
(OR)
Write a note on perception of colour.
Answer:
(1) In nature we firld objects of various colours. Perception of colour means to be able to respond to colour.
(2) We can distinguish between various colours due to perception of colour.
(3) The cone-shaped cells on the retina of the eye respond to the various colours when light is bright and communicate to the brain about the colours of the image formed on the retina. This gives us the proper idea about the colours of the object.
(4) If, in the retina of a person, the cone-shaped cells responding to certain specific colours are absent, the person is unable to distinguish between the colours. As a result, he lacks perception of colour.

Question 95.
What is colour-blindness?
Answer:
(1) The retina of the human eye contains a large number of light sensitive cells. These cells are of two shapes : (i) rods and (ii) cones.
(2) The cone-shaped cells respond to various colours of light when light is bright.
(3) Thus, the perception of colour is due to the presence of the cone-shaped cells in the retina.
(4) In the retina of some persons, cone-shaped cells responding to certain specific colours are absent. Hence, these persons are unable to distinguish between certain colours, i.e., they are colour-blind. This defect is known as colour¬blindness.

Question 96.
Why are some persons colour-blind? What is the cause of this defect?
Answer:
In the retina of some persons, cone-shaped cells responding to certain specific colours are absent. Hence, these persons are unable to distinguish between certain colours, i.e., they are colour-blind.

Question 97.
What are the difficulties faced by a colour-blind person?
Answer:
(1) A colour-blind person cannot distinguish between different colours. For example, he cannot distinguish between red and green colours. Also he cannot distinguish between blue and green colours. Red and green, both appear grey. Since a colour¬blind person cannot distinguish between red and green colours, it is difficult for him to cross a road. There is a possibility of an accident while crossing a road.

(2) A colour-blind person cannot distinguish between two objects of different colours, which are otherwise identical, e.g., clothes.

(3) A colour-blind person may have an inferiority complex and hence may find it difficult to mix with other persons.

Give scientific reasons:

Question 1.
A convex lens is known as a converging lens.
Answer:
When rays of light parallel to the principal axis of a convex lens pass through the lens, they converge to a point on the principal axis. Hence, a convex lens is known as a converging lens.

Question 2.
A concave lens is called a diverging lens.
Answer:
When rays of light parallel to the principal axis of a concave lens pass through the lens, they appear to diverge from a point on the principal axis. Hence, a concave lens is called a diverging lens.

Question 3.
In old age, a bifocal lens is necessary for some persons.
Answer:
(1) Some people, in old age, suffer from myopia (nearsightedness) as well as hypermetropia (farsightedness).
(2) Myopia is corrected using a concave lens of appropriate power. Hypermetropia is corrected using a convex lens of appropriate power. Therefore, they need a bifocal lens.

Question 4.
A person suffering from myopia (nearsightedness) uses spectacles of concave lenses.
Answer:
(1) A person suffering from myopia can see nearby objects clearly as the image of a nearby object is formed on the retina, but cannot see distant objects clearly as the image of a distant object is formed in front of the retina instead of on the retina.

(2) A concave lens diverges the rays of light passing through it. When spectacles of concave lenses of appropriate power are used, the parallel rays coming from a distant object are diverged to proper extent before they are incident on the eye lens. Therefore, after the converging action of the eye lens, the image of a distant object is formed on the retina of the eye and hence the distant object can be seen clearly.

Question 5.
A person suffering from hypermetropia (farsightedness) uses spectacles of convex lenses.
Answer:
(1) A person suffering from hypermetropia can see distant objects clearly as the image of a distant object is formed on the retina, but cannot see nearby objects clearly as the image of a nearby object would be formed behind the retina instead of on the retina.

(2) A convex lens converges the rays of light passing through it. When spectacles of convex lenses of appropriate power are used, the rays of light coming from a nearby object are converged to proper extent before they are incident on the eye lens. Therefore after the converging action of the j eye lens, the image of a nearby object is formed on j the retina of the eye and hence the nearby object | can be seen clearly.

Question 6.
You cannot enjoy watching a movie from a very short distance from the screen in a cinema hall.
Answer:
(1) The less the distance between the screen in a cinema hall and the person watching the movie, the more is the intensity of light falling on the eye.
(2) This results in great contraction of the pupil of the eye causing a strain. Hence, you cannot enjoy watching a movie from a very short distance from the screen in a cinema hall.

Question 7.
The rays of light travelling through the optical centre of a lens pass without changing their path.
Answer:
The portion of a lens near the optical centre is like a very thin slab of glass. Hence, the rays of light travelling through the optical centre of a lens pass without changing their path.

Question 8.
A convex lens converges the rays of light falling on it.
Answer:

• A convex lens can be regarded as made of a very large number of portions of triangular prisms. The bases of these prisms are towards the central thicker portion of the lens.
• A ray of light passing through a prism bends towards its base. Hence, a convex lens converges the rays falling on it.

Question 9.
A concave lens diverges the rays of light falling on it.
Answer:

• A concave lens can be regarded as made of a very large number of portions of triangular prisms. The bases of these prisms are towards the edges of the less, i.e, away from the central thinner portion of the lens.
• A ray of light passing through a prism bends towards its base. Hence, a concave lens diverges the rays of light falling on it.

Question 10.
When a burning stick of incense is moved fast in a circle, a circle of red light is seen.
Answer:
The impression of the image on the retina lasts for about \(\frac{1}{16}\) th of a second after the removal of the object. If a burning stick of incense is moved at a rate of more than sixteen revolutions per second, we see a circle of red light due to persistence of vision.

Question 11.
Colour-blind persons are unable to distinguish between different colours.
Answer:
(1) The cone-shaped cells in the retina of a person respond to colours. This makes the perception of colours possible.
(2) In the retina of colour-blind persons, cone-shaped cells responding to certain specific colours are absent. Hence, they are unable to distinguish between different colours.

Question 12.
It is risky to issue a driving license to a person suffering from colour-blindness.
Answer:
A colour-blind person cannot distinguish between different colours. If a driver is colour-blind, he will not be able to distinguish between the colours of the signal and the colours on different sign boards. This will lead to an accident. Hence, it is risky to issue a driving license to a person suffering from colour-blindness.

Distinguish the following:

Question 1.
Real image and Virtual image.
Answer:
Real image:

1. A real image is formed when the light rays starting from an object meet after reflection or refraction.
2. It can be projected on a screen.
3. It is inverted with respect to the object.

Virtual image:

1. A virtual image is formed when the light rays starting from an object (when extended backward) appear to meet after reflection or refraction.
2. It cannot be projected on a screen.
3. It is erect with respect to the object.

Question 2.
Simple microscope and Compound microscope.
Answer:
Simple microscope:

1. In a simple microscope, only one convex lens is used.
2. In this case, the object is placed within the focal length of the convex lens.
3. Its magnifying power is much less than that of a compound microscope.
4. It is used to observe minute parts of a watch, to read words in small print, etc.

Compound microscope:

1. In a compound microscope, two convex lenses, objective and eyepiece, are used.
2. In this case, the object is placed beyond the focal length of the objective lens.
3. Its magnifying power is much greater than that of a simple microscope.
4. It is used to observe blood corpuscles, plant and animal cells, etc.

Question 3.
Compound microscope and Astronomical refracting telescope.
Answer:
Compound microscope:

1. In a compound microscope, the focal length and cross section of the objective lens are respectively smaller than the focal length and cross section of the eyepiece.
2. In this case, to observe the object, the distance between the object and the objective lens is adjusted.
3. It forms a magnified image of a small object.
4. It is used to observe blood corpuscles, plant and animal cells, etc.

Astronomical refracting telescope:

1. In an astronomical refracting telescope, the focal length and cross section of the objective lens are respectively greater than the focal length and cross section of the eyepiece.
2. In this case, to observe the object, the distance between the objective lens and eyepiece is adjusted.
3. It forms a near image of a distant object.
4. It is used to observe sateulites, planets, stars, etc.

Question 4.
Simple microscope and Astronomical refracting telescope.
Answer:
Simple microscope:

1. In a simple microscope, only one convex lens is used.
2. In this case, the object is placed within the focal length of the convex lens.
3. In this case, the image is erect.
4. It is used to observe minute parts of a watch, to read words in small print, etc.

Astronomical refracting telescope:

1. In an astronomical refracting telescope, two convex lenses, objective lens and eyepiece are used.
2. In this case, the object is far away from the objective lens.
3. In this case, the image is inverted.
4. It is used to observe satellites, planets, stars, etc.

Read the following paragraph and answer the questions given below it:

Construction of a compound microscope:
(1) A compound microscope consists of a metal tube fitted with two convex lenses at the two ends. These lenses are called the objective lens (the lens directed towards the object) and the eyepiece (the lens directed towards the eye). Both the lenses are small in size, but the cross section of the objective lens is less than that of the eyepiece. The objective lens has a short focal length. The focal length of the eyepiece is more than that of the objective lens.(2) The metal tube is mounted on a stand. The principal axes of the objective lens and the eyepiece are along the same line. The distance between the object and the objective lens can be changed with a screw. It is possible to change the distance between the objective lens and the eyepiece.
Working:
(1) The object to be observed is illuminated and placed in front of the objective lens, slightly beyond the focal length of the objective lens. Its real, inverted and enlarged image is formed by the, objective lens on the other side.
(2) This intermediate image lies within the focal length of the eyepiece. It serves as an object for the eyepiece. The eyepiece works as a simple microscope. The final image is virtual, highly enlarged and inverted with respect to the original object. It can be formed at the minimum distance of distinct vision from the eyepiece. The final image is observed by keeping the eye close to the eyepiece.
Use: This microscope is used to observe blood cells, microorganisms, etc.

Question 1.
In a compound microscope, which lens has greater focal length?
Answer:
In a compound microscope, the eyepiece has greater focal length.

Question 2.
Where do you place the object to be observed with a compound microscope?
Answer:
In a compound microscope, the object to be observed is placed in front of the objective lens, slightly beyond the focus of the objective lens.

Question 3.
State which distance is adjusted to observe the object with a compound microscope.
Answer:
To observe the object with a compound microscope, the distance between the object and objective lens is adjusted.

Question 4.
State the nature of the final image in compound microscope relative to the object.
Answer:
In a compound microscope, the final image is highly enlarged, inverted and virtual relative to the object.

Question 5.
State the use of a compound microscope.
Answer:
A compound microscope is used to observe blood cells, microorganisms, etc.

Fill in the blanks for a convex lens:

Question 1.

f (m)	0.2	—————–	0.1
P (D)	—————	2	——————

Answer:
[P (D) = \(\frac{1}{f(\mathrm{m})}\)]

f (m)	0.2	0.5	0.1
P (D)	5	2	10

Question 2.

h1 (cm)	—————-	5	10
h2 (cm)	-30	-20	—————-
M	-2	—————–	-0.5

Answer:
[M = \(\frac{h_{2}}{h_{1}}\)]

h1 (cm)	15	5	10
h2 (cm)	-30	-20	-5
M	-2	-4	-0.5

Solve the following numerical problems:

Problem 1.
An object Is kept at 60 cm in front of a convex lens. Its real image is formed at 20 cm from the lens. Find the focal length or the lens.
Solution:
Data: Convex lens, u = -60 cm,

The focal length of the lens = 15 cm.

Problem 2.
The focal length of a convex lens is 20 cm. If an object of height 2 cm is placed at 30 cm from the lens, find (i) the position and nature of the Image (ii) the height of the image (iii) the magnification produced by the lens.
Solution:
Data: Convex lens, f = 20 cm,
u = -30 cm, h₁ = 2 cm, v = ?, h₂ = ?, M = ?

The image will be formed at 60 cm from the lens and on the other side of the lens with respect to the object. It is a real image.

h₂ is negative. This shows that the image is inverted.
The height of the image -4 cm.
(iii) M = \(\frac{h_{2}}{h_{1}}=\frac{-4 \mathrm{cm}}{2 \mathrm{cm}}\) = -2
M is negative, indicating that the image is inverted.
The magnification produced by the lens = -2.

Problem 3.
When a pm of height 3 cm is fixed at 10 cm from a convex lens, the height or the virtual image formed is 12 cm. Find the focal length of the lens.
Solution:
Data: Convex lens, h₁ =3 cm,
h₂ = 12 cm (virtual image), u = -10 cm, f = ?

The focal length of the lens = 13.33 cm [approximately]

Problem 4.
At what distance from a convex lens of focal length 2.5 m should a boy stand so that his image is half his height?
Solution:
Data: Convex lens, f= 2.5 m,
M= –\(\frac{1}{2}\), u=?

∴ u = -3f = -3 × 2.5 m = -7.5 m
This is the object distance.
The boy should stand at 7.5 m from the convex lens so that his image is half his height.

Problem 5.
A convex lens forms a real image or a pencil at a distance of 40 cm from the lens. The image formed is of the same size as the object. Find the focal length and power of the lens. At what distance is the pencil placed from the lens?
Solution:
Data: Convex lens, v = 40 cm,
M = -1, f = ?, h₁ = ?, u = ?
M = = -1 = \(\frac{v}{u}\)
∴ u = -v = -40 cm (object distance)
The pencil is placed at 40 cm from the convex lens.

The focal length of the lens 20 cm.

The power of the lens = 5 D.

Problem 6.
A spherical lens is used to obtain an image on a screen. The size of the image is four times the size of the object. What is the type of lens and at what distance is the screen placed from the lens?
Solution:
Data: M = -4, type of lens? v = ?
As the image formed by the lens is obtained on a screen, it is a real image. The lens is, therefore, a convex 1ens.

The distance of the screen from the lens = 5f.

Problem 7.
An object of height 5 cm Is held 20 cm away from a converging lens of focal length 10 cm. Find the position, nature and size of the image formed.
Solution:
Data: Converging lens (convex lens),
f = 10 cm, h₁ = 5 cm, u = -20 cm, v = ?, h₂ = ?

The image is real and inverted. it is formed at 20 cm from the lens and on the other side of the lens relative to the object.

The height of the image, h₂ = -5 cm
Thus, it is numerically the same as the height of the object.

Problem 8.
An object is placed at 10 cm from a convex lens of focal length 12 m. Find the position and nature of the image.
Solution:
Data: Convex lens, u = -10 cm,
f = 12 cm, v = ?


∴ v = -60 cm
It is negative.
The image is formed at 60 cm from the lens and on the same side of the lens relative to the object. It is virtual, erect, and enlarged.

Problem 9.
An object of height 4 cm is placed in front of a concave lens of focal length 40 cm. If the object distance is 60 cm, find the position and height of the image.
Solution:
Data: f = -40 cm (concave lens),
u = -60 cm, h₁ = 4 cm, v = ?, h₂ = ?

The image Is formed at 24 cm from the lens.
It is on the same side as the object.

The height of the image is 1.6 cm.

Problem 10.
What is the power of a convex lens having focal length 0.5 m?
Solution:
Data: Convex lens, f= 0.5 m, P = ?
P = \(\frac{1}{f}=\frac{1}{0.5 \mathrm{m}}\) = 2D
The power of the lens = 2D.

Problem 11.
The power of a convex lens is 2.5 dioptres. Find its focal length.
(OR)
Calculate the focal length of a corrective lens having power +2.5 D.
Solution:
Data: Convex lens, P = +2.5 D, f = ?
P = \(\frac{1}{f}\)
∴ 2.5 D = \(\frac{1}{f}\)
∴ f = \(\frac{1}{2.5 \mathrm{D}}\) = 0.4 cm = 40 cm
The focal length of the lens = 40 cm.

Problem 12.
Two convex lenses of focal length 20 cm each are kept in contact with each other. Find the power of their combination.
Solution:
Data: f₁ = 20 cm = 0.2 m,
f₂ =20 cm = 0.2 m, P (combination) = ?

∴ Focal length of the combination or the lenses, f = 0.1 m.
P = \(\frac{1}{f}=\frac{1}{0.1 \mathrm{m}}\) = 10 D
The power of the combination of the lenses, P = 10 D.

Problem 13.
Two convex lenses of equal focal lengths are kept in contact with each other. If the power of their combination is 20 D, find the focal length of each convex lens.
Solution:
Data: Convex lens, P = 20 D, f₁ = f₂ = ?
The focal length (f) of the combination of the lenses is given by

This gives the focal length or each convex lens.

Problem 14.
If a convex lens of focal length 10 cm and a concave lens of focal length 50 cm are kept in contact with each other, (i) what will be the focal length of the combination? (ii) what wiil be the power of the combination? (iii) what will be the behaviour of the combination (behaviour as a convex lens/concave lens)?
Solution:
Data: f₁ = +10 cm = +0.1 m (convex lens),
f₂ = -50 cm = -0.5 m (concave lens),
f (combination) = ?, P (combination) = ?

The focal length or the combination of the lenses = 0.125 m = 12.5 cm.

The power of the combination of the lenses 8.D.
(iii) The focal length of the combination of the lenses is positive. This shows that the combination will behave as a convex lens.

Numerical problems for practice:

Problem 1.
Find the focal length of a convex lens which produces a real image at 60 cm from the lens when an object is placed at 40 cm in front of the lens.
Answer:
24 cm

Problem 2.
Find the focal length of a convex lens which produces a virtual image at 10 cm from the lens when an object is placed at 5 cm from the lens.
Answer:
10 cm

Problem 3.
A real image is obtained at 30 cm from a convex lens of focal length 7.5 cm. Find the distance of the object from the lens.
Answer:
u = -10 cm

Problem 4.
An object is kept at 20 cm in front of a convex lens and its real image is formed at 60 cm from the lens. Find (1) the focal length of the lens (2) the height or the image if the height of the object is 6 cm.
Answer:
(1) 15 cm
(2) h₂ = -18 cm]

Problem 5.
An object is kept at 10 cm in front of a convex lens. Its image is formed on the screen at 15 cm from the lens. Calculate (1) the focal length of the lens (2) the magnification produced by the lens.
Answer:
(1) 6 cm
(2)M = -1.5

Problem 6.
An object is kept at 60 cm in front of a convex lens of focal length 15 cm. Find the image distance and the nature of the image. Also find the magnification produced by the lens.
Answer:
v = 20 cm. The image is real, inverted and smaller than the object. M = –\(\frac{1}{3}\)]

Problem 7.
An object of height 2 cm is kept at 30 cm from a convex lens. Its real image is formed at 60 cm from the lens. Find the focal length and power of the lens.
Answer:
f = 20 cm, P = 5 D

Problem 8.
If the power of a lens is 4 dioptres, find its focal length.
Answer:
25 cm

Problem 9.
Find the power of a convex lens of focal length 40 cm.
Answer:
2.5 D

Problem 10.
Find the power of a convex lens of focal length 12.5 cm.
Answer:
8 D

Problem 11.
If for a lens, f = – 20 cm, what is the power of the lens?
Answer:
-5 D

Problem 12.
An object of height 4 cm is kept in front of a concave lens of focal length 20 cm. If the object distance is 30 cm, find the position and the height of the image.
Answer:
v = -12 cm, h₂ = 1.6 cm

Problem 13.
If two convex lenses of focal lengths 10 cm and 5 cm are kept in contact with each other, what is their combined focal length?
Answer:
\(\frac{10}{3}\) cm [approximately 3.33 cm]

Problem 14.
If a convex lens of focal length 20 cm and a concave lens of focal length 30 cm are kept in contact with each other, (i) What will be the focal length of the combination? (ii) What will be the power of the combination? (iii) What will be the behaviour of the combination?
Answer:
(i) f = 60 cm
(ii) P = \(\frac{5}{3}\) D = 1.6667 D (approximately)
(iii) The combination will behave as a convex lens.

Problem 15.
A concave lens of focal length 12 cm and a convex lens of focal length 20 cm are kept in contact with each other, (i) Find the focal length of the combination, (ii) What will be the behaviour of the combination?
Answer:
(i) f = -30 cm
(ii) The combination will behave as a concave lens.

10th Std Science Part 1 Questions And Answers:

• Gravitation Class 10 Questions And Answers
• Periodic Classification of Element Class 10 Questions And Answers
• Chemical reactions and equations Class 10 Questions And Answers
• Effects of electric current Class 10 Questions And Answers
• Heat Class 10 Questions And Answers
• Refraction of light Class 10 Questions And Answers
• Lenses Class 10 Questions And Answers
• Metallurgy Class 10 Questions And Answers
• Carbon Compounds Class 10 Questions And Answers
• Space Missions Class 10 Questions And Answers

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light Notes, Textbook Exercise Important Questions and Answers.

Std 10 Science Part 1 Chapter 6 Refraction of Light Question Answer Maharashtra Board

Class 10 Science Part 1 Chapter 6 Refraction of Light Question Answer Maharashtra Board

Question 1.
Fill in the blanks and explain the completed statements:
a. Refractive index depends on the………….of light.
Answer:
Refractive index depends on the velocity of light.
It is an experimental fact. (There is no question of explanation.)

b. The change in…………of light rays while going from one medium to another is called refraction.
Answer:
The change in the direction of propagation of light rays while going from one medium to another is called refraction. This is definition of refraction. It is assumed that the ray of light passes obliquely from one medium to another. (There is no question of explanation.)

Question 2.
Prove the following statements:
a. If the angle of incidence and angle of emergence of a light ray falling on a glass slab are i and e respectively, prove that i = e. (Practice Activity Sheet – 4)
Answer:
In the following figure, SR || PQ and NM is the refracted ray. Hence, r = i₁.
Now _gn_a = sin i/sin r and _an_g = sin i₁/ sin e.

Also gna = \(\frac{1}{{ }_{\mathrm{a}} n_{\mathrm{g}}}\)
∴ \(\frac{\sin i}{\sin r}=\frac{\sin e}{\sin i_{1}}\)
As r = i₁, it follows that sin i = sin e
∴ i = e.

b. A rainbow is the combined effect (an exhibition) of the refraction, dispersion, and total internal reflection of light (taken together). (Practice Activity Sheet – 1)
(OR)
With a neat labelled diagram, explain how the formation of rainbow occurs.
Answer:
(1) The formation of a rainbow in the sky is a combined result of refraction, dispersion, internal reflection and again refraction of sunlight by water droplets present in the atmosphere after it has rained.

Here, for simplicity only violet and red colours are shown. The remaining five colours lie between these two.

(2) The sunlight is a mixture of seven colours: violet, indigo, blue, green, yellow, orange and red. After it has stopped raining, the atmosphere contains a large number of water droplets. When sunlight is incident on a water droplet, there is (i) refraction and dispersion of light as it passes from air to water (ii) internal reflection of light inside the droplet and (iii) refraction of light as it passes from water to air.

(3) The refractive index of water is different for different colours, being maximum for violet and minimum for red. Hence, there is dispersion of light (separation into different colours) as it passes from air to water. [ See above Figure for reference.]

(4) The combined action of different water droplets, acting like tiny prisms, is to produce a rainbow with red colour at the outer side and violet colour at the inner side. The remaining five colours lie between these two.
The rainbow is seen when the sun is behind the observer and water droplets in the front.

Question 3.
Mark the correct answer in the following questions :
A. What is the reason for the twinkling of stars?
(i) Explosions occurring in stars from time to time
(ii) Absorption of light in the earth’s atmosphere
(iii) Motion of stars
(iv) Changing refractive index of the atmospheric gases
Answer:
Changing refractive index of the atmospheric gases.

B. We can see the Sun even when it is little below the horizon because of
(i) reflection of light
(ii) refraction of light
(iii) dispersion of light
(iv) absorption of light
Answer:
refraction of light

C. If the refractive index of glass with respect to air is 3/2, what is the refractive index of air with respect to glass?
(i) \(\frac{1}{2}\)
(ii) 3
(iii) \(\frac{1}{3}\)
(iv) \(\frac{2}{3}\)
Answer:
(iv) \(\frac{2}{3}\)

Question 4.
Solve the following examples:
a. If the speed of light in a medium is 1.5 × 10⁸ m/s, what is the absolute refractive index of the medium? (Practice Activity Sheet – 1 and 4)
Solution:
Data: v = 1.5 × 10⁸ m/s,
c = 3 × 10⁸ m/s, n = ?
n = \(\frac{c}{v}=\frac{3 \times 10^{8} \mathrm{m} / \mathrm{s}}{1.5 \times 10^{8} \mathrm{m} / \mathrm{s}}\) = 2
This is the absolute refractive index of the medium.

b. If the absolute refractive indices of glass and water are \(\frac{3}{2}\) and \(\frac{4}{3}\) respectively, what is the refractive index of glass with respect to water?
Solution:

This is the refractive index of glass with respect to water.

Project:

Question 1.
Using a laser and soap water. study the refraction of light under the guidance of your teacher. (Do it your self)

Can you recall? (Text Book Page No. 73)

Question 1.
What is meant by reflection of light?
Answer:
Reflection of light: When light is incident on the surface of an object, in general, it is deflected in different directions. This process is called reflection of light.

Question 2.
What are the laws of reflection?
Answer:
Laws of reflection of light:

1. The incident ray and the reflected ray of light are on the opposite sides of the normal to the reflecting surface at the point of incidence and all the three are in the same plane.
2. The angle of incidence j and the angle of reflection are equal in measure.

Can you recall? (Text Book page No. 75)

Question 1.
If the refractive index of the second medium with respect to the first medium is ₂n₁ and that of the third medium with respect to the second medium is ₃n₂, what and how much is ₃n₁.
Answer:
₃n₁ is the refractive index of the third medium with respect to the first medium.

∴ ₃n₁ = ₂n₁ × ₃n₂.

[Suppose medium 1 = air, medium 2 ≡ ice and medium 3 ≡ diamond. Then, ₂n₁ ÷ 1.31, ₃n₂ = 1.847
∴ ₃n₁ = ₂n₁ × ₃n₂ = 1.31 × 1.847 = 2.42 which is the refractive index of diamond with respect to air.]

Can you tell? (Textbook page No. 76)

Question 1.
Have you seen a mirage which is an illusion of the appearance of water on a hot road or in a desert?
Answer:
Due to the changes in refraction of light, the light rays coming from a distant object appear to be coming from the image of the object inside the ground. This is called a mirage. When the earth’s surface is heated by the sun, the temperature of air increases. This produces a layer of hot air of lower density (mass per unit volume) and lower refractive index at the surface.

Hot air works as an optically rarer medium relative to cool air. When the temperature changes rapidly in the vertical direction, as refraction of light takes place, the angle of refraction changes continuously. The rays of light from the top of an object such as a car or tree cross the rays from the bottom of the object on their way to the observer’s eye.

Hence, an inverted image is formed below the object’s true position and downward towards the surface in the direction of air at higher temperature. In this case, some rays of light bend back up into the denser air above figure. Mirage produces an impression of water near the hot ground.

Question 2.
Have you seen that objects beyond and above a holy fire appear to be shaking? Why does this happen?
Answer:
The temperature of air beyond and above a holy fire changes all the time. Hence, the density of air also changes constantly. Hence, the direction of propagation of the rays of light approaching us from the objects beyond and above the holy fire changes constantly. Therefore, those objects appear to be shaking.

Use your brain power! (Text Book Page No. 77)

Question 1.
From incident white light how will you obtain white emergent light by making use of two prisms?
Answer:

• Take a prism. Allow white light to fall on it.
• Obtain a spectrum.
• Take a second identical prism. Place it parallel to the first prism in an upside down position with the first prism [as shown in Figure]
• Allow the colours of the spectrum to pass through the second prism.
• Obtain the beam of light emerging from the other side of the second prism.

The beam of light emerging from the other side of the second prism is a beam of white light.

Explanation: White light is made up of seven colours. The first prism produces dispersion of white light while the second prism combines light of different colours to produce white light again. The net deviation of a ray of light is zero.
[Note: This experiment is due to sir Isaac Newton. It proved that it was not the prism which added colours to the white light but a property of the white light itself.]

Question 2.
You must have seen chandeliers having glass prisms. The light from a tungsten bulb gets dispersed while passing through these prisms and we see coloured spectrum. If we use an LED light instead of a tungsten bulb, will we be able to see the same effect?
Answer:
Light emitted by LED (light-emitting-diode) does not have all wavelengths in the region 400 nm to 700 nm. Hence, its spectrum is not the same as that of light from a tungsten bulb or as that of sunlight.

Fill in the blanks and rewrite the statements:

Question 1.
The phenomenon of change in the………..of light when it passes obliquely from one transparent medium to another is called refraction.
Answer:
The phenomenon of change in the direction of propagation of light when it passes obliquely from one transparent medium to another is called refraction.

Question 2.
The refractive index depends upon the…………of propagation of light in different media.
Answer:
The refractive index depends upon the velocity of propagation of light in different media.

Question 3.
The process of separation of light into its component colours while passing through a medium is called………..
Answer:
The process of separation of light into its component colours while passing through a medium is called dispersion of light.

Question 4.
When a light ray travels obliquely from air to water, it bends………the normal at the point of incidence.
Answer:
When a light ray travels obliquely from air to water, it bends towards the normal at the point of incidence.

Question 5.
When a light ray travels obliquely from benzene to air, it bends…………the normal at the point of incidence.
Answer:
When a light ray travels obliquely from benzene to air, it bends away from the normal at the point of incidence.

Question 6.
In glass, the speed of red ray is……violet ray.
Answer:
In glass, the speed of red ray is greater than that of violet ray.

Question 7.
The speed of light in glass is………in water.
Answer:
The speed of light in glass is less than that in water.

Question 8.
The speed of light in water is…………in benzene.
Answer:
The speed of light in water is greater than that in benzene.

Question 9.
Rainbow occurs due to refraction, dispersion,……….and again refraction of sunlight by water droplets.
Answer:
Rainbow occurs due to refraction, dispersion, internal reflection and again refraction of sunlight by water droplets.

Question 10.
In dispersion of sunlight by a glass prism,………..ray is deviated the least.
Answer:
In dispersion of sunlight by a glass prism, red ray is deviated the least.

Rewrite the following statements by selecting the correct options:

Question 1.
The change in the direction of propagation of light when it passes obliquely from one transparent medium to another is called………
(a) dispersion
(b) scattering
(c) refraction
(d) reflection
Answer:
(c) refraction

Question 2.
When a ray of light travels from air to glass slab and strikes the surface of separation at 90°, then it…………
(a) bends towards the normal
(b) bends away from the normal
(c) passes unbent
(d) passes in zigzag way
Answer:
(c) passes unbent

Question 3.
If a ray of light passes from a denser medium to a rarer medium in a straight line, the angle of incidence must be…………
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Answer:
(a) 0°

Question 4.
A ray of light strikes a glass slab at an angle of 50° with the normal to the surface of the slab. What is the angle of incidence?
(a) 50°
(b) 25°
(c) 40°
(d) 100°
Answer:
(a) 50°

Question 5.
If a ray of light propagating in air strikes a glass slab at an angle of 60° with the surface of the slab, the angle of refraction is…………
(a) more than 30 °
(b) less than 30 °
(c) 60°
(d) 30°
Answer:
(b) less than 30 °

Question 6.
A ray of light gets deviated When it passes obliquely from one medium to another medium because………..
(a) the colour of light changes
(b) the frequency of light changes
(c) the speed of light changes
(d) the intensity of light changes
Answer:
(c) the speed of light changes

Question 7.
The speed of light in turpentine oil is 2 × 10⁸ m/s. The absolute refractive index of turpentine oil is about……..[Speed of light in vacuum ≈ 3 × 10⁸ m/s]
(a) 1.5
(b) 2
(c) 1.3
(d) 0.67
Answer:
(a) 1.5

Question 8.
LASER stands for………..
(a) light amplification by stimulated emission of radiation
(b) light and sound energy radiation
(c) light and simulated energy radiation
(d) light amplification by sound energy radiation
Answer:
(a) light amplification by stimulated emission of radiation

Question 9.
Out of the following……….has the highest absolute refractive index.
(a) fused quartz
(b) diamond
(c) crown glass
(d) ruby
Answer:
(b) diamond

Question 10.
The absolute refractive index…………
(a) is expressed in dioptre
(b) is expressed in m/s
(c) of air is about \(\frac{4}{3}\)
(d) has no unit
Answer:
(d) has no unit

Question 11.
The speed of light in a medium of refractive index n is………., where c is the speed of light
in vacuum.
(a) \(\frac{c}{n}\)
(b) nc
(c) \(\frac{n}{c}\)
(d) \(\sqrt{\frac{c}{n}}\)
Answer:
(a) \(\frac{c}{n}\)

Question 12.
The speed of light in a transparent medium having absolute refractive index 1.25 is……….[Speed of light in vacuum ≈ 3 × 10⁸ m/s]
(a) 1.25 × 10⁸ m/s
(b) 2.4 × 10⁸ m/s
(c) 3.0 × 10⁸ m/s
(d) 1.5 × 10⁸ m/s
Answer:
(b) 2.4 × 10⁸ m/s

Question 13.
…………light is deviated the maximum in the spectrpm of white light obtained with a glass prism.
(a) Red
(b) Yellow
(c) Violet
(d) Blue
Answer:
(c) Violet

Question 14.
………..light is deviated the least in the spectrum of white light obtained with a glass prism.
(a) Red
(b) Yellow
(c) Violet
(d) Blue
Answer:
(a) Red

Question 15.
A ray of light makes an angle of 50° with the surface S1 of the glass slab. Its angle of incidence will be………….(March 2019)
(a) 50°
(b) 40°
(c) 140°
(d) 0°
Answer:
(a) 50°

Question 16.
A glass slab is placed in the path of convergent light. The point of convergence of light:
(a) moves away from the slab
(b) moves towards the slab
(c) remains at the same point
(d) undergoes a lateral shift
Answer:
(a) moves away from the slab

Question 17.
In refraction of light through a glass slab, the directions of the incident ray and the refracted ray are………… (Practice Activity Sheet – 1)
(a) perpendicular to each other
(b) non-parallel to each other
(c) parallel to each other
(d) intersecting each other
Answer:
(c) parallel to each other

Question 18.
If we gradually increase the angle of incidence of a ray of light passing through a prism, then………….. (Practice Activity Sheet – 4)
(a) the angle of deviation goes on decreasing
(b) the angle of deviation decreases but after certain value of incident angle, deviation angle increases
(c) the angle of deviation goes on increasing
(d) the angle of deviation increases but after certain value of incident angle, deviation angle decreases
Answer:
(b) the angle of deviation decreases but after certain value of incident angle, deviation angle increases

State whether the following statements are True or False. (If a statement is false, correct it and rewrite it.):

Question 1.
The incident ray and the refracted ray of light are on the opposite sides of the normal at the point of incidence.
Answer:
True.

Question 2.
The refractive index of a medium (such as glass) does not depend on the wavelength of light.
Answer:
False. (The refractive index of a medium depends on the wavelength of light.)

Question 3.
When a light ray travels obliquely from an optically rarer medium to an optically denser medium, it bends away from the normal.
Answer:
False. (When a light ray travels obliquely from an optically rarer medium to an optically denser medium, it bends towards the normal.)

Question 4.
When a light ray travels obliquely from glass to air, it bends towards the normal.
Answer:
False. (When a light ray travels obliquely from glass to air, it bends away from the normal.)

Question 5.
If the angle of incidence is 0°, the angle of refraction is 90°.
Answer:
False. (If the angle of incidence is 0°, the angle of refraction is also 0°.)

Question 6.
In dispersion of white light by a glass prism, yellow colour is deviated the least.
Answer:
False. (In dispersion of white light by a glass prism, red colour is deviated the least.)

Question 7.
In vacuum, the speed of light does not depend upon the frequency of light.
Answer:
True.

Question 8.
In glass, the speed of violet ray is less than that of red ray.
Answer:
True.

Question 9.
In a material medium, the speed of light depends on the frequency of light.
Answer:
True.

Question 10.
The velocity of light is different in different media.
Answer:
True.

Question 11.
Wavelength of red light is close to 700 nm.
Answer:
True.

Question 12.
Wavelength of orange light is greater than that of blue light.
Answer:
True.

Find the odd one out and give the reason:

Question 1.
Reflection, Neutralization, Refraction, Dispersion.
Answer:
Neutralization. It is associated with a chemical reaction between an acid and an alkali; others are phenomena associated with light.

Answer the following questions in one sentence each:

Question 1.
Mention any two phenomena in nature where refraction of light takes place.
Answer:
Mirage and twinkling of a star.

Question 2.
What is the angle of refraction when the angle of incidence is 0°?
Answer:
When the angle of incidence is 0°, the angle of refraction is also 0°.

Question 3.
In refraction of light, \(\frac{\sin i}{\sin r}\) = constant in sin a particular case. What is this constant called?
Answer:
The constant \(\frac{\sin i}{\sin r}\) (in a particular case) is called the refractive index of the second medium with respect to the first medium.

Question 4.
If the refractive index of medium 2 with respect to medium 1 is 5/3, what is the refractive index of medium 1 with respect to medium 2?
Answer:
The refractive index of medium 1 with respect to medium 2 is 0.6.

Question 5.
In dispersion of sunlight by a glass prism, which colour is deviated the least?
Answer:
In dispersion of sunlight by a glass prism, red colour is deviated the least.

Question 6.
In dispersion of sunlight by a glass prism, which colour is deviated the most?
Answer:
In dispersion of sunlight by a glass prism, violet colour is deviated the most.

Question 7.
What is the wavelength of violet light?
Answer:
The wavelength of violet light is (about) 400 nm.

Question 8.
State the relation between ₂n₁ and critical angle.
Answer:
₂n₁ = sin i, where i is the critical angle.

Answer the following questions:

Question 1.
What is meant by refraction of light?
Answer:
The change in the direction of propagation of light when it passes obliquely from one transparent medium to another is called refraction of light.

Question 2.
Why is there a change in the direction of propagation of light when it passes obliquely from one transparent medium to another?
Answer:
The velocity of light is different in different media. Hence, there is a change in the direction of propagation of light when it passes obliquely from one transparent medium to another.

Question 3.
In the case of refraction of light through a glass slab, the emergent ray is parallel to the incident ray, but it is displaced sideways. Why does this happen?
Answer:

The first refraction takes place as light passes obliquely from air to glass. In this case, the ray of light bends towards the normal at point N. The second refraction takes place as light passes obliquely from glass to air. In this case, the ray of light bends away from the normal at point M. The faces PQ and SR of the glass slab are parallel. Hence, the extent of bending of light at SR is equal in magnitude but opposite in sense relative to the bending of light at PQ. Hence, the emergent ray of light (MD) is parallel to the incident ray of light (AN), but it is displaced sideways as shown in Figure.

Question 4.
Define angle of incidence and angle of refraction.
Answer:
(1) The angle made by the incident ray of light with the normal to the surface at the point of incidence is called the angle of incidence.

(2) The angle made by the refracted ray of light with the normal to the surface at the point of incidence is called the angle of refraction.

[Note: The angle e in Fig. 6.3 is also called the angle of emergence as it is the angle made by the emergent ray with the normal to the surface at the point of emergence. ]

Question 5.
Repeat the activity “Refraction of light passing through a glass sl^b” by replacing the glass slab by a transparent plastic slab.
(i) What similarity do you observe?
(ii) What difference do you notice?
Answer:
(i) Similarity: The emergent ray is parallel to the incident ray, but it is displaced sideways.
(ii) Difference: For a given angle of incidence, the extent of refraction (bending) is different (in general, less) for a transparent plastic slab relative to the glass slab.

Question 6.
State the laws of refraction of light.
Answer:
Laws of refraction of light:
(1) The incident ray and the refracted ray are on the opposite sides of the normal to the surface at the point of incidence and all the three, i.e., the incident ray, the refracted ray and the normal are in the same plane.

(2) For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant (Snell’s law). This constant is called the refractive index of the second medium with respect to the first medium.
[Note: Here, a ray means a ray of light.]

Question 7.
How is refraction of light related to refractive index?
Answer:
When a ray of light travels obliquely from an optically rarer medium (lower refractive index) to an optically denser medium (higher refractive index), the ray bends towards the normal. When a ray of light travels obliquely from an optically denser medium to an optically rarer medium, the ray bends away from the normal. For a given angle of incidence (i ≠ 0), the extent of refraction (bending) of light is different in different media.

If the refractive index of the second medium with respect to the first medium is greater than 1, the greater the refractive index, the greater is the bending of the ray of light towards the normal. If the refractive index of the second medium with respect to the first medium is less than 1, the greater the refractive index, the lesser is the bending of the ray of light away from the normal.

Question 8.
Define the refractive index of the second medium with respect to the first medium.
(OR)
What is meant by refractive index?
Answer:
The refractive index of the second medium with respect to the first medium is defined as the ratio of the sine of the angle of incidence to the sine of the angle of refraction when the ray of light is obliquely incident at the boundary separating the

two media and travels from the first medium to the second medium. (See Fig. 6.4.)
(OR)
The refractive index of the second medium with respect to the first medium is defined as the ratio of (the magnitude of) the velocity of light in the first medium to (the magnitude of) the velocity of light in the second medium.

[Note: Velocity is a vector, i.e., it has magnitude and direction. In definition of refractive index, we consider only the magnitude of velocity of light (speed of light). Velocity of light in a medium depends on the physical condition of the medium as well as the frequency of light. Velocity of light is different in different media. For a given medium, the refractive index depends on the colour of light (frequency of light.)]

Question 9.
State the formulae for the refractive index of the second medium with respect to the first medium.
Answer:
The refractive index of the second medium with respect to the first medium,
₂n₁ = \(\frac{\sin i}{\sin r}=\frac{v_{1}}{v_{2}}\)
where i is the angle of incidence, r is the angle of refraction (as the ray of light passes obliquely from the first medium to the second medium), v₁ is the magnitude of the velocity (speed) of light in the first medium and v₂ is the magnitude of the velocity of light in the second medium.

Question 10.
Define absolute refractive index.
Answer:
The absolute refractive index of a medium is defined as the ratio of the magnitude of the velocity of light in vacuum to the magnitude of the velocity of light in the medium.

[Note: The speed of light is maximum in vacuum, about 3 × 10⁸ m/s. When light travels from one medium to another, there occurs a change in its speed and wavelength (A). But its frequency (v) remain the same.]

Question 11.
Obtain the relation between the refractive index of the second medium with respect to the first medium and the refractive index of the first medium with respect to the second medium.
Answer:
Let v₁ = speed of light in the first medium, v₂ = speed of light in the second medium, ₂n₁ = refractive index of the second medium With respect to the first medium and ₁n₂ = refractive index of the first medium with respect to the second medium.
By definition, ₂n₁ = \(\frac{v_{1}}{v_{2}}\) and ₁n₂ = \(\frac{v_{2}}{v_{1}}\)
Hence,

(OR)
₁n₂ × ₂n₁ = 1.

Question 12.
If the refractive index of a certain material with respect to air is 1.5, what is the refractive index of air with respect to that material?
Answer:
As the refractive index of the given material with respect to air is 1.5, the refractive index of air with respect to the material is
\(\frac{1}{1.5}=\frac{1}{3 / 2}=\frac{2}{3}\) = 0.6667 (approximately)

Question 13.
Explain the terms optically rarer medium and optically denser medium with examples.
Answer:
When we consider two media (such as air and glass), the medium with lower refractive index is called the optically rarer medium (in the present case, air) and the medium with higher refractive index is called the optically denser medium (glass, in the present case).

The higher density does not necessarily mean higher refractive index. For example, the density of water is greater than that of kerosene, but the absolute refractive index of water is less than that of kerosine. Thus, when we consider water and kerosine, water is an optically rarer medium while kerosine is an optically denser medium.

If we consider kerosene and benzene, kerosine is an optically rarer medium while benzene is an optically denser medium.

Question 14.
A ray of light is incident obliquely at a boundary separating two media. What is its behaviour if (1) the refractive index of the second medium is greater than that of the first medium (2) the refractive index of the first medium is greater than that of the second medium? Draw the corresponding neat and labelled diagrams.
Answer:
Consider a ray of light incident obliquely at a boundary separating two media.
(1) If the refractive index of the second medium is greater than that of the first medium, the ray bends towards the normal at the point of incidence as it travels from the first medium (optically rarer medium) to the second medium (optically denser medium). The angle of refraction (r) is less than the angle of incidence (i). (Fig. 6.6)

Fig. 6.6: A ray of light travelling from a rarer medium to a denser medium (Schematic diagram)

Fig. 6.7: A ray of light travelling from a denser medium to a rarer medium (Schematic diagram)

(2) If the refractive index of the first medium is greater than that of the second medium, the ray bends away from the normal at the point of incidence as it travels from the first medium (optically denser medium), to the second medium (optically rarer medium). The angle of refraction (r) is greater than the angle of incidence (i). (Fig. 6.7)

[Note In this chapter, a rarer medium means an optically rarer medium and a denser medium means optically denser medium unless stated otherwise.]

Question 15.
Observe the following figure and write accurate conclusion regarding refraction of light. (Practice Activity Sheet – 2)

Answer:
When a light ray passes obliquely from a rarer medium to a denser medium, it bends towards the normal.

Question 16.
What happens when a ray of light is incident normal to the interface between two media? Draw the corresponding neat and labelled diagram.
Answer:
When a ray of light is incident normal to the interface between two media, the ray propagates undeviated as it travels from the first medium to the second medium irrespective of the refractive indices of the two media. In this case, the angle of incidence (i) is zero and so also the angle of refraction (r).

Fig. 6.9: A ray of light incident normal to the interface between two media propagates without any change in its direction of propagation

Question 17.
Draw a neat and labelled diagram to show the path of a ray of light in air and glass when the ray is incident obliquely on a glass slab. Show the (i) incident ray (ii) refracted ray (iii) emergent ray (iv) angle of incidence (v) angle of refraction (vi) angle of emergence in the diagram.
(OR)
Draw a neat and labelled diagram to show refraction of light through a glass slab.
Answer:

Fig. 6.10: The path of the ray of light in air and glass when the ray is incident obliquely on a glass slab
In Fig. 6.10, i = angle of incidence, r = angle of refraction and e = angle of emergence.

Question 18.
Observe the given figure and name the following rays:

(i) ray AB
(ii) Ray BC
(iii) ray CD
Answer:
(i) The ray AB is the incident ray.
(ii) The ray BC is the refracted ray.
(iii) The ray CD is the emergent ray.

Question 19.
A plane mirror is kept at the bottom of a trough with water in it as shown in the following figure (Fig. 6.12). The ray of light emerging from a source at the point S outside the trough reaches the point A on the surface of water. Draw a neat ray diagram to show the subsequent path of light and complete the ray diagram.

Answer:

Question 20.
Give two examples of the effect of atmospheric refraction on a small scale in local environment.
Answer:

1. The occurrence of a mirage
2. Flickering of an object seen through a turbulent stream of hot air rising above the Holi fire are examples of the effect of atmospheric refraction on a small scale in local environment.

Question 21.
What is a mirage? With a neat labelled diagram, explain the conditions under which it is seen.
Answer:
Due to the changes in refraction of light, the light rays coming from a distant object appear to be coming from the image of the object inside the ground. This is called a mirage.

When the earth’s surface is heated by the sun, the temperature of air increases. This produces a layer of hot air of lower density (mass per unit volume) and lower refractive index at the surface. Hot air works as an optically rarer medium relative to cool air. When the temperature changes rapidly in the vertical direction, as refraction of light takes place, the angle of refraction changes continuously.

The rays of light from the top of an object such as a car or tree cross the rays from the bottom of the object on their way to the observer’s eye. Hence, an inverted image is formed below the object’s true position and downward towards the surface in the direction of air at higher temperature. In this case, some rays of light bend back up into the denser air above figure. Mirage produces an impression of water near the hot ground.

Question 22.
Explain in brief the flickering of an object seen through a turbulent stream of hot air rising above the Holi fire.
Answer:
During the Holi fire, the temperature of the air just above the fire becomes much greater than that of the air further up. The hot air has lower density (mass per unit volume) and lower refractive index. It becomes an optically rarer medium. The cool air has higher density and higher refractive index. It is an optically denser medium relative to hot air. Hence, in refraction of light, the angle of refraction changes continuously due to a continuous variation in refractive index.

As the physical conditions of air change rapidly, the apparent position of an object fluctuates rapidly. This gives rise to the flickering of an object seen through a turbulent stream of hot air rising above the Holi fire.

Question 23.
With a neat labelled diagram, explain twinkling of a star. Also explain why a planet does not twinkle.
Answer:
(1) As a star is far away from the earth, it appears as a point source of light. The density of air decreases with height above the earth’s surface. Hence, the refractive index of air also decreases with height. When starlight enters the earth’s atmosphere, it undergoes refraction continuously in the medium with gradually changing refractive index. The bending of starlight occurs towards the normal as it passes from the optically rarer part of the medium to the optically denser part.

(2) Hence, when a star is observed near the horizon, its apparent position is slightly higher than the actual position (See below figure).

(3) Further, the apparent position varies with time as the medium is not stationary due to mobility of air and change in temperature. When more light is refracted towards the observer the star appears bright. When less light is refracted towards the observer, the star appears dim.

(4) Thus there is fluctuation in the brightness of a star when observed from the earth. This is called twinkling of a star.

(5) Compared to stars, planets are relatively closer to the earth. Hence, a planet appears as a collection of a large number of point sources. Due to the changes in the refractive index of air, there is a change in the position and brightness of these point sources.

There is an increase in intensity of light coming from some point sources while there is a decrease in intensity of light coming from equal number of other point sources, on an average. The average brightness of a planet remain the same. Also, there is no change in the average position of a star. Hence, a planet does not twinkle.

Question 24.
What is the correct reason for blinking/flickering of stars? Explain it.
(a) The blasts in the stars.
(b) Absorption of star light by the atmosphere.
(c) Motion of the stars.
(d) Changing refractive index of gases in the atmosphere. (Practice Activity Sheet – 2)
Answer:
(d) Changing refractive index of the gases in the atmosphere results in blinking/flickering of stars.

Explanation:
(1) As a star is far away from the earth, it appears as a point source of light. The density of air decreases with height above the earth’s surface. Hence, the refractive index of air also decreases with height. When starlight enters the earth’s atmosphere, it undergoes refraction continuously in the medium with gradually changing refractive index. The bending of starlight occurs towards the normal as it passes from the optically rarer part of the medium to the optically denser part.

(2) Hence, when a star is observed near the horizon, its apparent position is slightly higher than the actual position (See below figure).

(3) Further, the apparent position varies with time as the medium is not stationary due to mobility of air and change in temperature. When more light is refracted towards the observer the star appears bright. When less light is refracted towards the observer, the star appears dim.

(4) Thus there is fluctuation in the brightness of a star when observed from the earth. This is called twinkling of a star.

(5) Compared to stars, planets are relatively closer to the earth. Hence, a planet appears as a collection of a large number of point sources. Due to the changes in the refractive index of air, there is a change in the position and brightness of these point sources.

There is an increase in intensity of light coming from some point sources while there is a decrease in intensity of light coming from equal number of other point sources, on an average. The average brightness of a planet remain the same. Also, there is no change in the average position of a star. Hence, a planet does not twinkle.

Question 25.
With a neat labelled diagram, explain advanced sunrise and delayed sunset.
Answer:
(1) The sunrise (the appearance of the sun above the horizon) is advanced due to atmospheric refraction of sunlight. An observer on the earth sees the sun two minutes before the sun reaches the horizon. A ray of sunlight entering the earth’s atmosphere follows a curved path due to atmospheric refraction before reaching the earth. This happens due to a gradual variation in the refractive index of the atmosphere.

For the observer on the earth, the apparent position of the sun is slightly higher than the actual position. Hence, the sun is seen before the sun reaches the horizon.

(2) Increased atmospheric refraction of sunlight occurs also at the sunset (the sun disappearing below the horizon). In this case, the observer on the earth continues to see the setting sun for two minutes after the sun has dipped below the horizon, thus delaying the sunset.

The advanced sunrise and delayed sunset increases the duration of day by four minutes.

Question 26.
Water in a swimming pool or water tank appears shallower than its depth. Why?
Answer:
When light rays travel obliquely from an optically denser medium (water, in this case) to an optically rarer medium (air, in this case), they bend away from the normal at the point of incidence. As a result, the bottom of a swimming pool or water tank appears raised to an observer standing near the edge of the pool or the tank. Therefore, the swimming pool or water tank appears shallower than its depth.

Question 27.
Place a coin at the bottom of a glass jar containing water. Now tilt the jar suitably. When viewed at a suitable angle, the coin appears to be floating. Why?
Answer:
When light rays travel obliquely from an optically denser medium (water, in this case) to an optically rarer medium (air, in this case), they bend away from the normal at the point of incidence. As a result, the coin appears raised. Therefore, when the jar is tilted suitably and observed at a suitable angle, the coin appears to be floating.

Question 28.
State the wavelength range of electromagnetic radiation to which our eyes are sensitive.
Answer:
Our eyes are sensitive to light (electromagnetic radiation). Its wavelength range is 400 nm to 700 nm.
[Note: Wavelength (λ) goes on decreasing and frequency (ν) goes on increasing from red (λ ≃ 700 nm) → orange → yellow → green → blue → indigo → violet (A ≃ 400 nm). c = vλ, where c is the speed of light in vacuum.]

Question 29.
What do you mean by dispersion of light? What is a spectrum of light? Name the different colours of light in the proper sequence in the spectrum of white light.
(OR)
What do you mean by dispersion? Name the different colours of light in the proper sequence in the spectrum of white light.
Answer:
The process of separation of light into its component colours while passing through a medium is called dispersion of light. The band of coloured components of a light beam is called spectrum.
The different colours of light in the spectrum of white light are violet, indigo, blue, green, yellow, orange and red.

Question 30.
What is a prism?
Answer:
A prism is a transparent medium bound by two plane surfaces inclined at an angle. Normally it is made of glass and has triangular cross section.

Question 31.
With a neat labelled diagram, describe the experiment to demonstrate dispersion of sunlight (white light) by a prism.
Answer:
Experiment:
(1) Procedure: Keep a glass prism on a table in a dark room. Hold a plane mirror outside the room so that it reflects a beam of sunlight into the room. Allow this beam to pass through a narrow slit made in cardboard and then fall on the prism. Place a white screen on the other side of the prism as shown in the following figure. [Fig. 6.17]

(2) Observations:

1. A pattern of various colours is observed on the screen. This pattern is called the spectrum.
2. It is found that in dispersion, the ray corresponding to violet colour deviates the most.
3. The ray corresponding to red colour deviates the least.
4. The deviation of rays corresponding to other colours is intermediate.

(3) Conclusion: When sunlight (white light) is incident on a prism, dispersion of light takes place, forming a spectrum.

[Notes: (1) This experiment is due to Sir Isaac Newton (1642 – 1727), English physicist and mathematician. (2) If in a Board examination, incomplete diagram (as shown in Fig. 6.18) is given, students should complete it and label its parts as shown in Fig. 6.17.]

Question 32.
How does the dispersion of white light take place when it passes through a glass prism?
Answer:
When rays of light are incident on a prism, they are refracted twice, while travelling from air to glass and then from glass to air. Even when the incident rays are directed away from the base of the prism, the emergent rays bend towards the base of the prism, as the prism is triangular. Thus, the rays are deviated as they pass through the prism.

The refractive index of glass is different for different colours. Therefore, the rays corresponding to different colours are deviated to different extents. White light is a mixture of seven colours : violet, indigo, blue, green, yellow, orange and red. Hence, when white light is incident on a prism, a spectrum of seven colours is obtained.

The refractive index of glass is maximum for violet light and minimum for red light. Hence, violet light is deviated the most and red light is deviated the least. The deviation of rays corresponding to other colours is intermediate. In this manner, the dispersion of light takes place when it passes through a glass prism. [For reference, see Fig. 6.17.]

Question 33.
What is a spectrum? Why do we get a spectrum of seven colours when while light is dispersed by a prism?
(OR)
Explain how a spectrum is formed.
Answer:
A band of coloured components of a light beam is called a spectrum. When white light is incident on a prism, the rays corresponding to different colours bend through different angles on refraction.

Of the various colours in the visible region, red light bends the least and violet light bends the most. Each colour emerges through the prism along a different path and becomes, distinct. Hence, we get a spectrum of seven colours.

Question 34.
What is partial reflection of light?
Answer:
When light travels from a denser medium to a rarer medium, it is partially reflected, i.e., part of light comes back into the denser medium as per the laws of reflection. This is called partial reflection of light.

[Note: Partial reflection of light occurs even when light travels from a rarer medium to a denser medium. The rest of light is refracted.]

Question 35.
Explain the terms total internal reflection and critical angle.
Answer:
Figure 6.20 shows passage of light from water (denser medium) to air (rarer medium).

The ray of light incident at the boundary separating the two media bends away from the normal on refraction. Here, the angle of refraction r, is greater than the angle of incidence i.
Now _an_w = \(\frac{\sin i}{\sin r}\) < 1. Here, _an_w is the refractive index of sin r air with respect to water.

As _an_w is constant, r increases as i increases. For r = 90°, the ray travels along the boundary. If i is increased further, as r cannot be greater than 90°, light does not enter air. There is no refraction of light and all the light enters water on reflection. This is called total internal reflection.
For r = 90°, _an_w = \(\frac{\sin i}{\sin 90^{\circ}}\) = sin i. This angle i is sin 90° called the critical angle.

Question 36.
Swarali has got the following observations while doing an experiment. Answer her questions with the help of observations. (Practice Activity Sheet – 2)
Swarali observed that the light bent away from the normal, while travelling from a denser medium to a rarer medium. When Swarali increased the values of the angle of incidence (i). the values of the angle of refraction (r) went on increasing. But at a certain angle of incidence, the light rays returned into the denser medium.
So, Swarali has some questions. Answer them.
(a) Name this certain value of i. What is the value of r at that time?
(b) Name this process of returning light in the denser medium. Explain the process.
Answer:
(a) Critical angle r = 90°
(b) Total internal reflection.

As light goes from a denser to rarer medium, if the value of the angle of incidence increases, then the value of the angle of refraction also increases. But after a specific angle of incidence called the critical angle, the light gets reflected back into the denser medium.

The ray of light incident at the boundary separating the two media bends away from the normal on refraction. Here, the angle of refraction r, is greater than the angle of incidence i.
Now _an_w = \(\frac{\sin i}{\sin r}\) < 1. Here, _an_w is the refractive index of sin r air with respect to water. As _an_w is constant, r increases as i increases. For r = 90°, the ray travels along the boundary. If i is increased further, as r cannot be greater than 90°, light does not enter air. There is no refraction of light and all the light enters water on reflection. This is called total internal reflection.

For r = 90°, anw = \(\frac{\sin i}{\sin 90^{\circ}}\) = sin i. This angle i is sin 90° called the critical angle.

Question 37.
The observations made by Swarali while doing the experiment are given below. Based on these write answers to the questions:
Swarali found that the light ray travelling from the denser medium to a rarer medium goes away from the normal. If the angle of incidence (i) is raised by Swarali, the angle of refraction (r) went on increasing. However, after certain value of the angle of incidence, the light ray is seen to return back into the denser medium. (March 2019)
(i) What is the specific value of ∠i called?
(ii) What is the process of reflection of incident ray into a denser medium called?
(iii) Draw the diagrams of three observations made by Swarali.
Answer:
(i) Critical angle
(ii) Total internal reflection
(iii)

Question 38.
Define total internal reflection of light.
Answer:
When light travels from a denser medium to a rarer medium, if the angle of incidence is greater than the critical angle, there is no refraction of light and all the light is reflected in the denser medium. This is called total internal reflection of light.

Question 39.
Define critical angle.
Answer:
When light travels from a denser medium to a rarer medium, the angle of incidence for which the angle of refraction becomes 90°, is called the critical angle.

Question 40.
If the refractive index of a rarer medium with respect to a denser medium is 0.5, what is the critical angle?
Answer:
₂n₁ = 0.5 = sin i
∴ Critical angle i = 30°.

Question 41.
Name the devices in which total internal reflection of light is used.
Answer:

1. Total internal reflecting prisms are used in a camera, binoculars, periscope.
2. Total internal reflection of light is used in optical fibres.

[Note: Total internal reflection of light plays an important role in sparkling brilliance of a diamond.]

Question 42.
Explain why an empty test tube held obliquely in water appears shiny to an observer looking down.
Answer:
When an empty test tube is held obliquely in Water in a beaker, some light rays passing from water to air are incident at an angle greater than the critical angle. They are, thus, totally internally reflected as shown, and the surface of the test tube has a silvery shine.

Question 43.
Observe the given figure and answer the following questions. (Practice Activity Sheet – 3)

(a) Identify and write the natural process shown in the figure.
(b) List the phenomena which are observed in this process.
(c) Redraw the diagram and show the above phenomena in it.
Answer:
(a) The natural process shown in the figure is formation of rainbow.
(b) The phenomena observed in this process are refraction, internal reflection and dispersion of light.
(c)

Write a short note on the following:

Question 1.
Refraction observed in the atmosphere.
Answer:
When a ray of light passes obliquely from an optically rarer medium to an optically denser medium, it bends towards the normal at the point of incidence. If opposite is the case, the ray bends away from the normal.

Atmosphere is never static. Air is mobile and its density and temperature are not uniform. As a result, in general, the path of a ray of light through atmosphere of varying refractive index is a curve. The refractive index of cool air is greater than that of hot air.

Atmospheric refraction of light results in many interesting optical phenomena such as twinkling of a star, advanced sunrise and delayed sunset, mirage and flickering of an object seen through a turbulent stream of hot air rising from a fire.

Question 2.
Dispersion of light.
Answer:
The process of separation of light into its component colours while passing through a medium is called dispersion of light. When white light passes through a glass prism, it spreads out into a band of different colours (components) called the spectrum of light. The colours in the spectrum of white light are violet, indigo, blue, green, yellow, orange and red.

Formation of a rainbow is an example of dispersion of light in nature. In this case, raindrops are responsible for dispersion of sunlight.

Dispersion takes place because the refractive index of a material such as glass or water, is different for different colours. It is maximum for violet colour and minimum for red colour. Hence, in the spectrum of white light (sunlight) obtained with a prism, violet light is deviated the most while red light is deviated the least. The deviation of light corresponding to other colours lies in between.

Give scientific reasons:

Question 1.
A coin kept in a bowl is not visible when seen from one side. But, when water is poured in the bowl, the coin becomes visible.
Answer:
(1) When the bowl is empty, the rays of light coming from the coin are obstructed by the side of the bowl, and hence the coin is not visible when seen from one side of the bowl.

(2) When water is poured in the bowl, the rays of light coming from the coin travel from water (denser medium) to air (rarer medium). Hence, they bend away from the normal on refraction. Therefore, the coin appears to be raised and becomes visible when observed from one side of the bowl.

Question 2.
A pencil dipped in water obliquely appears bent at the surface of water.
(OR)
When a pencil is partly immersed in water and held in a slanting position, it appears to be bent at the boundary separating water and air.
Answer:
(1) When a pencil is partly immersed in water and held in a slanting position, the rays of light coming from the immersed part of the pencil emerge from water (a denser medium) and enter air (a rarer medium). During this propagation, they bend away from the normal on refraction.

The pencil appearing bent at the boundary of water and air (schematic diagram)

(2) As a result, the immersed part of the pencil does not appear straight with respect to the part outside the water, but appears to be raised. Hence, a pencil dipped obliquely in water appears bent at the surface of the water.

Question 3.
The shadow of the edge of an empty vessel is formed due to the slanting rays of the sun. When water is poured in the vessel, the shadow is shifted.
Answer:
(1) When the slanting rays of the sun are obstructed by the edge of the empty vessel, the shadow of the edge is formed.

(2) When water is poured in the vessel, the slanting rays of the sun travel from air (rarer medium) to water (denser medium). During this propagation, they bend towards the normal on refraction. Hence, some part in the region of the shadow is now illuminated and the shadow appears to have shifted.

Question 4.
The bottom of a pond appears raised.
Answer:
(1) The rays of light coming from the bottom of a pond bend away from the normal as they travel from water (denser medium) to air (rarer medium).

(2) Hence, they appear to come from a point above the actual point from which they come.
Therefore, the bottom of the pond appears raised.

Question 5.
While shooting a fish in a lake, the gun is aimed below the apparent position of the fish.
Answer:
(1) The rays of light coming from the fish bend away from the normal as they travel from water (denser medium) to air (rarer medium).
(2) Hence, the position of the fish in water appears to be above Its real position. Therefore, while shooting a fish in a lake, the gun is aimed below the apparent position of the fish.

Question 6.
The sun is seen on the horizon a little before sunrise.
(OR)
The sun is seen on the horizon for sometime even after sunset.
Answer:
(1) The earth is surrounded by an atmosphere which is denser near the surface of the earth. When the rays of light from the sun enter the earth’s atmosphere from outer space, they travel from a rarer medium to a denser medium. Hence, they bend towards the normal on refraction.

(2) Hence, even when the sun is below the horizon while rising or setting, its rays reach us due to refraction and it appears to be on the horizon. Therefore, the sun is seen on the horizon a little before sunrise as well as for some time even after sunset.

Distinguish between:

Question 1.
Reflection of light and Refraction of light:
Answer:

Reflection of light	Refraction of light
1. The rays of light, before and after reflection, travel in the same medium.	1. In refraction of light, the rays travel from one medium to another medium.
2. In reflection, the angle of incidence and the angle of reflection are equal.	2. In refraction, when the rays travel obliquely from one medium to another medium, the angle of incidence and the angle of refraction are not equal.
3. In reflection, there is no change in the speed and wavelength of light.	3. In refraction, there occurs a change in the speed and wavelength of light.
4. In reflection, there is no dispersion of light.	4. Generally, in refraction, there occurs dispersion of light.

[Note: The frequency of light remains the same in reflection and refraction.]

Complete the following or Solve and fill in the blanks :

Question 1.

Speed of light in the first medium (v1)	Speed of light in the second medium (v2)	Refractive index 2n1	Refractive index 2n1
3 × 108 m/s	1.2 × 108 m/s	————————–	————————–
————————–	2.25 × 108 m/s	4/3	————————–
2 × 108 m/s	————————–	————————–	1.5

Answer:

Speed of light in the first medium (v1)	Speed of light in the second medium (v2)	Refractive index 2n1	Refractive index 2n1
3 × 108 m/s	1.2 × 108 m/s	2.5	0.4
3 × 108 m/s	2.25 × 108 m/s	4/3	0.75
2 × 108 m/s	3 × 108 m/s	2/3	1.5

Formulae:
₂n₁ = v₁/v₂, ₁n₂ = v₂/v₁

Solve the following examples/numerical problems:
c = 3 × 10⁸ m/s

Problem 1.
The speed of light in a transparent medium is 2.4 × 10⁸ m/s. Calculate the absolute refractive index of the medium.
Solution:
Data: c = 3 × 10⁸ m/s,
v = 2.4 × 10⁸ m/s, n = ?

The absolute refractive index of the medium = 1.25.

Problem 2.
The velocity of light in a medium is 2 × 10⁸ m/s. What is the refractive index of the medium with respect to air, if the velocity of light in air is 3 × 10⁸ m/s?
Solution:
Data: v₁ = 3 × 10⁸ m/s,
v₂ = 2 × 10⁸ m/s, ₂n₁ = ?
₂n₁ = \(\frac{v_{1}}{v_{2}}\)
\(=\frac{3 \times 10^{8}}{2 \times 10^{8}}\)
= 1.5
The refractive index of the medium with respect to air is 1.5.

Problem 3.
Light travels with a velocity 1.5 × 10⁸ m/s in a medium. On entering second medium its velocity becomes 0.75 × 10⁸ m/s. What is the refractive index of the second medium with respect to the first medium? (Practice Activity Sheet – 3)
Solution:
Given: Velocity of light in the first medium = v₁ = 1.5 × 10⁸ m/s,
velocity of light in the second medium = v₂ = 0.75 × 10⁸ m/s,
refractive index of the second medium with respect to the first medium = ₂n₁ = ?
₂n₁ = \(\frac{v_{1}}{v_{2}}\)
₂n₁ = \(\frac{1.5 \times 10^{8}}{0.75 \times 10^{8}}\) = 2
Hence, the refractive index of the second medium with respect to the first medium is 2.
[Note : The absolute refractive index of the second medium = \(\frac{3 \times 10^{8} \mathrm{m} / \mathrm{s}}{0.75 \times 10^{8} \mathrm{m} / \mathrm{s}}\) = 4 (greater than that of diamond, not likely).]

Problem 4.
The refractive index of water is 4/3 and the speed of light in air is 3 × 10⁸ m/s. Find the speed of light in water.
Solution:
Data: ₂n₁ = 4/3, v₁ = 3 × 10⁸ m/s, v₂ = ?

The speed of light in water = 2.25 × 10⁸ m/s.

Problem 5.
The speed of light in water and glass is 2.2 × 10⁸ m/s and 2 × 10⁸ m/s respectively. What is the refractive index of (i) water with respect to glass (ii) glass with respect to water?
Solution:
Data: u_w = 2.2 × 10⁸ m/s,
v_g= 2 × 10⁸ m/s, _wn_g = ?, _gn_w = ?

The refractive index of water with respect to glass = 0.909 (approximately).

The refractive index of glass with respect to glass = 1.1 (approximately).

Numerical Problems For Practice:
(Given: C = 3 × 10⁸m/s)

Problem 1.
The speed of light in a transparent medium is 2 × 10⁸ m/s. Find the absolute refractive index of the medium.
Solution:
1.5

Problem 2
The absolute refractive index of a transparent medium is 5/3. Find the speed of light in the medium.
Solution:
1.8 × 10⁸ m/s

Problem 3.
The absolute refractive index of a transparent medium is 2.4 and the speed of light in that medium is 1.25 × 10⁸ m/s. Find the speed of light in air.
Solution:
3 × 10⁸ m/s

Problem 4.
The speed of light in water is 2.25 × 10⁸ m/s and that in glass is 2 × 10⁸ m/s. Find the refractive index of (i) the glass with respect to water (ii) water with respect to the glass.
Solution:
(i) 1.125
(ii) 0.889 (approximately)

Problem 5.
If the refractive index of a certain glass with respect to water is 1.25, find the refractive index of water with respect to the glass.
Solution:
0.8

Problem 6.
If the absolute refractive index of glass is 1.5 and that of water is \(\frac{4}{3}\), find the refractive index of water with respect to glass.
Solution:
\(\frac{8}{9}\)

10th Std Science Part 1 Questions And Answers:

• Gravitation Class 10 Questions And Answers
• Periodic Classification of Element Class 10 Questions And Answers
• Chemical reactions and equations Class 10 Questions And Answers
• Effects of electric current Class 10 Questions And Answers
• Heat Class 10 Questions And Answers
• Refraction of light Class 10 Questions And Answers
• Lenses Class 10 Questions And Answers
• Metallurgy Class 10 Questions And Answers
• Carbon Compounds Class 10 Questions And Answers
• Space Missions Class 10 Questions And Answers