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  • Science 9th Science Chapter-6 In-Text Questions

    Chapter 6: Tissues

    In-Text Questions & Answers

    Page No. 61 

    Question 1:

    What is a tissue?

    Answer:
    A tissue is a group of cells that are similar in structure and work together to perform a specific function.
    Examples: Blood, muscle, phloem.

    Question 2:

    What is the utility of tissues in multicellular organisms?

    Answer:
    Tissues are useful in multicellular organisms because:

    • They bring about division of labour, where different tissues perform different functions.

    • This increases efficiency of work.

    • Tissues help in the organisation of cells, forming organs and organ systems.

    • They allow complex organisms to perform many functions simultaneously.

    Thus, tissues make the body more organised and efficient.

    Page No. 65 – Chapter 6: Tissues

    Question 1:

    Name the three types of simple permanent tissues.

    Answer:
    The three types of simple permanent tissues are:

    1. Parenchyma

    2. Collenchyma

    3. Sclerenchyma

    Question 2:

    Describe the characteristics and functions of parenchyma.

    Answer:
    Characteristics:

    • Cells are living

    • Thin cell walls

    • Large intercellular spaces

    • Usually oval or round in shape

    Functions:

    • Storage of food and water

    • Photosynthesis (when chlorophyll is present, called chlorenchyma)

    • Provides support to plants by turgidity

    Question 3:

    Describe the characteristics and functions of collenchyma.

    Answer:
    Characteristics:

    • Cells are living

    • Cell walls are unevenly thickened at the corners

    • Very little intercellular space

    Functions:

    • Provides flexibility to plant parts

    • Provides mechanical support to stems and leaves

    Question 4:

    Describe the characteristics and functions of sclerenchyma.

    Answer:
    Characteristics:

    • Cells are dead

    • Cell walls are very thick and lignified

    • No intercellular spaces

    Functions:

    • Provides strength and rigidity to plants

    • Forms hard coverings like nut shells and coconut husk

    Page No. 69 – Chapter 6: Tissues

    Question 1:

    Name the different types of animal tissues.

    Answer:
    The four main types of animal tissues are:

    1. Epithelial tissue

    2. Connective tissue

    3. Muscular tissue

    4. Nervous tissue

    Question 2:

    What is epithelial tissue? Mention its functions.

    Answer:
    Epithelial tissue is a tissue that covers the body surface, lines the internal organs and cavities, and forms glands.

    Functions:

    • Protection of underlying tissues

    • Absorption (e.g., intestine)

    • Secretion (e.g., glands)

    • Excretion and diffusion

    Question 3:

    Name the different types of epithelial tissue.

    Answer:
    The different types of epithelial tissue are:

    1. Squamous epithelium

    2. Cuboidal epithelium

    3. Columnar epithelium

    4. Ciliated epithelium

    5. Glandular epithelium

    Question 4:

    Describe squamous epithelium with its function.

    Answer:
    Squamous epithelium consists of thin, flat cells arranged like tiles on a floor.

    Function:
    It allows easy diffusion and filtration of substances.

    Example:
    Lining of blood vessels and air sacs of lungs.

    Question 5:

    Describe cuboidal epithelium with its function.

    Answer:
    Cuboidal epithelium is made of cube-shaped cells.

    Function:
    It helps in secretion and absorption.

    Example:
    Lining of kidney tubules and glands.

  • Class 9th Science Chapter-5 Exercises

    Exercise – Chapter 5: The Fundamental Unit of Life

    Question 1

    Make a comparison and write down ways in which plant cells are different from animal cells.

    Answer:

    Feature Plant Cell Animal Cell
    Cell wall Present Absent
    Plastids

    Present (chloroplasts, etc.)

    		 Absent
    Vacuole Large central vacuole Small or absent
    Shape

    Usually regular (rectangular)

    		 Usually irregular (rounded)
    Mode of nutrition Autotrophic Heterotrophic

    Question 2

    How is a prokaryotic cell different from a eukaryotic cell?

    Answer:

    Prokaryotic Cell Eukaryotic Cell
    No true nucleus True nucleus present
    Nuclear membrane absent

    Nuclear membrane present
    Single chromosome More than one chromosome
    Membrane-bound organelles absent

    Membrane-bound organelles present
    Smaller in size Larger in size

    Question 3

    What would happen if the plasma membrane ruptures or breaks down?

    Answer:

    If the plasma membrane ruptures, the cell contents will leak out, and the entry and exit of substances cannot be controlled.
    As a result, the cell will die.

    Question 4

    What would happen to the life of a cell if there was no Golgi apparatus?

    Answer:

    If the Golgi apparatus is absent:

    • Proteins and lipids will not be modified, packaged, or transported

    • Secretion from the cell will stop

    As a result, the cell will not function properly and may die.

    Question 5

    Which organelle is known as the powerhouse of the cell? Why?

    Answer:

    Mitochondria are known as the powerhouse of the cell because they produce energy in the form of ATP, which is required for all cellular activities.

    Question 6

    Where do the lipids and proteins constituting the cell membrane get synthesised?

    Answer:

    • Proteins are synthesised on ribosomes of rough endoplasmic reticulum (RER)

    • Lipids are synthesised in the smooth endoplasmic reticulum (SER)

    Question 7

    How does an Amoeba obtain its food?

    Answer:

    Amoeba obtains its food by endocytosis.
    It surrounds the food particle with pseudopodia, forming a food vacuole, where digestion takes place.

    Question 8

    What is osmosis?

    Answer:

    Osmosis is the movement of water molecules through a selectively permeable membrane from a region of higher water concentration to a region of lower water concentration.

    Question 9

    Potato Osmosis Experiment

    (i) Explain why water gathers in the hollowed portion of B and C.

    Answer:
    Sugar (B) and salt (C) create a hypertonic solution inside the potato cup.
    Water moves from the surrounding dilute solution into the cup by osmosis, so water gathers inside.

    (ii) Why is potato A necessary for this experiment?

    Answer:
    Potato A acts as a control to show that osmosis does not occur without a solute inside the potato cup.

    (iii) Explain why water does not gather in the hollowed portions of A and D.

    Answer:

    • A: No solute is present, so no concentration difference exists.

    • D: Boiling destroys the living cells and the selectively permeable membrane, so osmosis does not occur.

    Question 10

    Which type of cell division is required for growth and repair of body and which type is involved in formation of gametes?

    Answer:

    • Mitosis is required for growth and repair of the body

    • Meiosis is involved in the formation of gametes

  • Class 9th Science Chapter-5 In-Text Questions

    Answers to In-Text Questions

    Chapter 5: The Fundamental Unit of Life

    Page No. 51 

    Question 1:

    Who discovered cells, and how?

    Answer:
    Cells were discovered by Robert Hooke in 1665. He observed a thin slice of cork under a self-designed microscope and noticed small box-like compartments, which he named cells.

    Question 2:

    Why is the cell called the structural and functional unit of life?

    Answer:
    The cell is called the structural unit of life because all living organisms are made up of cells.
    It is called the functional unit of life because all vital life processes such as respiration, nutrition, growth, and reproduction take place within cells.

    Page No. 53 

    Question 1:

    How do substances like CO₂ and water move in and out of the cell? Discuss.

    Answer:

    • Carbon dioxide (CO₂) moves in and out of the cell by diffusion.
      It moves from a region of higher concentration to lower concentration across the plasma membrane.

    • Water moves in and out of the cell by osmosis, which is the movement of water through a selectively permeable membrane from a region of higher water concentration to lower water concentration.

    Thus, diffusion helps in gaseous exchange, while osmosis helps in the movement of water across the cell membrane.

    Question 2:

    Why is the plasma membrane called a selectively permeable membrane?

    Answer:
    The plasma membrane is called selectively permeable because it allows some substances (like gases and water) to pass through it while preventing the movement of other substances.
    This property helps the cell to maintain its internal environment.

    Page No. 55 

    Question:

    Fill in the gaps in the following table illustrating differences between prokaryotic and eukaryotic cells.

    Prokaryotic Cell Eukaryotic Cell
    1. Size: generally small (1–10 µm)

    1. Size: generally large (5–100 µm)
    2. Nuclear region: not well defined and not surrounded by a nuclear membrane

    2. Nuclear region: well defined and surrounded by a nuclear membrane

    3. Chromosome: single chromosome

    		 3. More than one chromosome

    4. Membrane-bound cell organelles: absent

    		 4. Present

    Answer (in words, for exams):

    • Prokaryotic cells have a poorly defined nuclear region, single chromosome, and lack membrane-bound organelles.

    • Eukaryotic cells have a well-defined nucleus, more than one chromosome, and membrane-bound organelles present.

    Page No. 57 

    Question 1:

    Can you name the two organelles we have studied that contain their own genetic material?

    Answer:
    The two organelles that contain their own genetic material (DNA) are:

    1. Mitochondria

    2. Plastids (chloroplasts)

    Question 2:

    If the organisation of a cell is destroyed due to some physical or chemical influence, what will happen?

    Answer:
    If the organisation of a cell is destroyed, the cell will not be able to perform its vital functions.
    As a result, the cell will die.

    Question 3:

    Why are lysosomes known as “suicide bags” of the cell?

    Answer:
    Lysosomes contain powerful digestive enzymes.
    When a cell is damaged or old, lysosomes may burst and digest the cell itself, leading to cell death.
    Hence, lysosomes are called the “suicide bags” of the cell.

    Question 4:

    Where are proteins synthesised inside the cell?

    Answer:
    Proteins are synthesised on ribosomes, which are present:

    • On the rough endoplasmic reticulum (RER)

    • Freely in the cytoplasm

  • Class 9th Science Chapter-4 Exercises

    Structure of Atom

    Exercise – Solutions

    Question 1. Compare the properties of electrons, protons and neutrons.

    Property Electron Proton Neutron
    Symbol e⁻ p⁺ n
    Charge –1 +1 0
    Mass

    Very small (≈ 1/2000 of proton)

    		 1 u 1 u
    Position in atom Outside nucleus (shells) Inside nucleus

    Inside nucleus

    Question 2. What are the limitations of J.J. Thomson’s model of the atom?

    Answer:

    • It could not explain the results of Rutherford’s α-particle scattering experiment.

    • It failed to explain how positive charge is distributed in the atom.

    • It did not explain the stability of the atom.

    Question 3. What are the limitations of Rutherford’s model of the atom?

    Answer:

    • According to classical physics, revolving electrons should lose energy and fall into the nucleus.

    • Hence, the atom should be unstable, but atoms are stable in reality.

    • It failed to explain the arrangement of electrons in shells.

    Question 4. Describe Bohr’s model of the atom.

    Answer:

    Bohr proposed the following postulates:

    1. Electrons revolve around the nucleus in fixed circular orbits called energy levels or shells.

    2. Each orbit has a definite energy.

    3. Electrons do not radiate energy while revolving in these shells.

    4. Energy is emitted or absorbed only when an electron jumps from one shell to another.

    This model successfully explained the stability of the atom.

    Question 5. Compare all the proposed models of an atom given in this chapter.

    Model Scientist Main idea Limitation
    Thomson’s model J.J. Thomson Atom is a positive sphere with electrons embedded

    Could not explain scattering experiment
    Rutherford’s model E. Rutherford

    Atom has a central nucleus with electrons revolving

    		 Could not explain stability
    Bohr’s model Niels Bohr

    Electrons revolve in fixed energy shells

    		 Valid only for simple atoms

    Question 6. Summarise the rules for distribution of electrons in shells (Bohr–Bury rules).

    Answer:

    1. Maximum electrons in a shell = 2n², where n is shell number.

    2. Maximum electrons in the outermost shell = 8.

    3. Electrons fill shells step by step, inner shells first.

    Question 7. Define valency by taking examples of silicon and oxygen.

    Answer:

    Valency is the combining capacity of an atom.

    • Silicon (Atomic no. 14):
      Electronic configuration = 2, 8, 4
      Valency = 4 (needs 4 electrons to complete octet)

    • Oxygen (Atomic no. 8):
      Electronic configuration = 2, 6
      Valency = 2 (needs 2 electrons to complete octet)

    Question 8

    Explain with examples:
    (i) Atomic number
    (ii) Mass number
    (iii) Isotopes
    (iv) Isobars
    Give any two uses of isotopes.

    Answer:

    (i) Atomic number (Z):
    Atomic number is the number of protons present in the nucleus of an atom.

    Example:
    Carbon has 6 protons → Atomic number = 6

    (ii) Mass number (A):
    Mass number is the sum of protons and neutrons present in the nucleus.

    Example:
    Carbon has 6 protons and 6 neutrons → Mass number = 12

    (iii) Isotopes:
    Isotopes are atoms of the same element having the same atomic number but different mass numbers.

    Example:
    ¹₁H, ²₁H, ³₁H (isotopes of hydrogen)

    (iv) Isobars:
    Isobars are atoms of different elements having the same mass number but different atomic numbers.

    Example:
    ⁴⁰₂₀Ca and ⁴⁰₁₈Ar

    Two uses of isotopes:

    1. Cobalt-60 is used in the treatment of cancer.

    2. Uranium-235 is used as fuel in nuclear reactors.

    Question 9

    Na⁺ has completely filled K and L shells. Explain.

    Answer:

    Atomic number of sodium = 11
    Electronic configuration of Na = 2, 8, 1

    Sodium loses one electron to form Na⁺ ion.

    Electronic configuration of Na⁺ = 2, 8

    Hence, K and L shells are completely filled.

    Question 10

    If bromine atom is available as two isotopes ⁷⁹₃₅Br (49.7%) and ⁸¹₃₅Br (50.3%), calculate the average atomic mass of bromine.

    Answer:

    Average atomic mass=(79×49.7)+(81×50.3)100
    =3926.3+4074.3100=8000.6100=80.0 u

    Question 11

    The average atomic mass of an element X is 16.2 u. What are the percentages of isotopes ¹⁶₈X and ¹⁸₈X in the sample?

    Answer:

    Let percentage of ¹⁶₈X = x
    Then percentage of ¹⁸₈X = 100 − x

    16x+18(100−x)100=16.2
    16x+1800−18x=1620
    −2x=−180⇒x=90

    Percentages:

    • ¹⁶₈X = 90%

    • ¹⁸₈X = 10%

    Question 12

    If Z = 3, what would be the valency of the element? Also name the element.

    Answer:

    Atomic number Z = 3 → Element is Lithium (Li)
    Electronic configuration = 2, 1

    Valency = 1

    Question 13

    Composition of the nuclei of two atomic species X and Y is given below:

    Species Protons Neutrons
    X 6 6
    Y 6 8

    Give the mass numbers of X and Y. What is the relation between them?

    Answer:

    • Mass number of X = 6 + 6 = 12

    • Mass number of Y = 6 + 8 = 14

    Relation:
    X and Y are isotopes of the same element.

    Question 14

    For the following statements, write T for True and F for False:

    (a) J.J. Thomson proposed that the nucleus contains only nucleons.
    → F

    (b) A neutron is formed by an electron and a proton combining together.
    → F

    (c) The mass of an electron is about 1/2000 times that of proton.
    → T

    (d) An isotope of iodine is used for making tincture iodine.
    → F

    Question 15

    Rutherford’s alpha-particle scattering experiment was responsible for the discovery of:

    (a) Atomic nucleus
    (b) Electron
    (c) Proton
    (d) Neutron

    Answer:

    ✔ (a) Atomic nucleus

    Question 16

    Isotopes of an element have:

    (a) the same physical properties
    (b) different chemical properties
    (c) different number of neutrons
    (d) different atomic numbers

    Answer:

    ✔ (c) different number of neutrons

    Question 17

    Number of valence electrons in Cl⁻ ion are:

    (a) 16
    (b) 8
    (c) 17
    (d) 18

    Answer:

    ✔ (b) 8

    Question 18

    Which one of the following is a correct electronic configuration of sodium?

    (a) 2,8
    (b) 8,2,1
    (c) 2,1,8
    (d) 2,8,1

    Answer:

    ✔ (d) 2,8,1

    Explanation (for understanding):
    Sodium has atomic number 11, so it has 11 electrons.
    These are distributed as:

    • K shell = 2

    • L shell = 8

    • M shell = 1

    Hence, the correct electronic configuration is 2,8,1.

    Question 19

    Complete the following table:

    Atomic Number Mass Number Number of Neutrons Number of Protons Number of Electrons Name of the Atomic Species
    9 – 10 – – –
    16 32 – – – Sulphur
    – 24 – 12 – –
    – 2 – 1 – –
    – 1 0 1 0 –

    Answer (Completed Table):

    Atomic Number Mass Number Number of Neutrons Number of Protons Number of Electrons

    Name of the Atomic Species
    9 19 10 9 9 Fluorine
    16 32 16 16 16

    Sulphur
    12 24 12 12 12 Magnesium
    1 2 1 1 1

    Deuterium (Hydrogen isotope)
    1 1 0 1 0

    Proton (H⁺ ion)

    Key rules used (for exam clarity):

    • Atomic number = number of protons

    • Mass number = protons + neutrons

    • For neutral atoms: electrons = protons

    • If electrons = 0: it is a positive ion

  • Class 9th Science Chapter-4 In-Text Questions

    Structure of Atom

    In-Text Question Answers

    Page 39 – Questions & Answers

    Question 1. What are canal rays?

    Answer:
    Canal rays are streams of positively charged particles produced in a gas discharge tube.
    They were discovered by E. Goldstein and are also called positive rays.

    Question 2. If an atom contains one electron and one proton, will it carry any charge or not?

    Answer:
    No, the atom will not carry any charge.

    Reason:

    • Electron has a charge of –1

    • Proton has a charge of +1

    • These charges are equal and opposite, so they cancel each other

    Hence, the atom is electrically neutral.

    Page 41 – Set 1 (After Bohr’s Model of Atom)

    Question 1. On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.

    Answer:
    According to Thomson’s model, an atom consists of a positively charged sphere with negatively charged electrons embedded in it.
    The total positive charge is equal to the total negative charge of electrons, therefore the atom as a whole is electrically neutral.

    Question 2. On the basis of Rutherford’s model of an atom, which sub-atomic particle is present in the nucleus of an atom?

    Answer:
    According to Rutherford’s model, the nucleus contains positively charged particles called protons.

    Question 3. Draw a sketch of Bohr’s model of an atom with three shells.

    Answer (description for exam):
    Bohr’s model shows:

    • A central nucleus

    • Three circular shells around it named K, L, and M

    • Electrons revolving in these fixed shells

    Question 4. What do you think would be the observation if the α-particle scattering experiment is carried out using a foil of a metal other than gold?

    Answer:
    Similar observations would be obtained, but the extent of deflection may vary because:

    • Gold is extremely thin and malleable

    • Other metals are less malleable, so fewer α-particles may pass through without deflection

    Page 41 – Set 2 (After Neutrons)

    Question 1.

    Name the three sub-atomic particles of an atom.

    Answer:
    The three sub-atomic particles are:

    1. Electrons

    2. Protons

    3. Neutrons

    Question 2.

    Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?

    Answer:

    Mass number = Protons + Neutrons

    4=2+Neutrons
    Neutrons=2

    Helium atom has 2 neutrons.

    Page 42 – Questions & Answers

    Question 1. Write the distribution of electrons in carbon and sodium atoms.

    Carbon (Atomic number = 6):

    • K shell = 2 electrons

    • L shell = 4 electrons

    Electronic configuration of carbon = 2,4

    Sodium (Atomic number = 11):

    • K shell = 2 electrons

    • L shell = 8 electrons

    • M shell = 1 electron

    Electronic configuration of sodium = 2,8,1

    Question 2. If K and L shells of an atom are full, then what would be the total number of electrons in the atom?

    • Maximum electrons in K shell = 2

    • Maximum electrons in L shell = 8

    Total electrons=2+8=10

    Total number of electrons = 10

    Page 44 – Second Set of Questions

    Question 1.

    If the number of electrons in an atom is 8 and the number of protons is also 8, then
    (i) what is the atomic number of the atom?
    (ii) what is the charge on the atom?

    Answer:

    (i) Atomic number = number of protons = 8

    (ii) Charge on the atom = 0
    Reason: Number of electrons (–8) = number of protons (+8), so charges cancel out.

    The atom is electrically neutral.

    Question 2.

    With the help of Table 4.1, find out the mass number of oxygen and sulphur atom.

    Answer:

    • Oxygen atom
      Protons = 8, Neutrons = 8

      Mass number=8+8=16

    • Sulphur atom
      Protons = 16, Neutrons = 16

      Mass number=16+16=32

    Page 45 – Questions & Answers

    Question 1. For the symbols H, D and T, tabulate three sub-atomic particles found in each of them.

    Answer:

    Isotope Protons Neutrons Electrons
    H (Protium) 1 0 1

    D (Deuterium)

    		 1 1 1
    T (Tritium) 1 2 1

    Question 2. Write the electronic configuration of any one pair of isotopes and isobars.

    (a) Isotopes (example: Chlorine)

    Isotopes have the same atomic number, so their electronic configuration is the same.

    • Chlorine-35 (³⁵₁₇Cl): 2, 8, 7

    • Chlorine-37 (³⁷₁₇Cl): 2, 8, 7

    (b) Isobars (example: Calcium and Argon)

    Isobars have the same mass number but different atomic numbers, so their electronic configuration is different.

    • Calcium-40 (⁴⁰₂₀Ca): 2, 8, 8, 2

    • Argon-40 (⁴⁰₁₈Ar): 2, 8, 8

  • Class 9th Science Chapter-3 Exercises

    Atoms and Molecules

    Exercises – Answers and Solutions

    Question 1

    A 0.24 g sample of a compound of oxygen and boron contains 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition by weight.

    Solution:

    Total mass of compound = 0.24 g

    Percentage of boron:

    0.0960.24×100=40%

    Percentage of oxygen:

    0.1440.24×100=60%

    Answer:

    • Boron = 40%

    • Oxygen = 60%

    Question 2

    When 3.0 g of carbon is burnt in 8.0 g of oxygen, 11.0 g of carbon dioxide is formed.
    What mass of carbon dioxide will be formed when 3.0 g of carbon is burnt in 50.0 g of oxygen?
    Which law governs this?

    Solution:

    Carbon reacts with oxygen in a fixed ratio.

    Given:

    • 3 g carbon + 8 g oxygen → 11 g CO₂

    Even though oxygen is present in excess (50 g), only 8 g oxygen will react with 3 g carbon.

    So,

    Mass of CO₂ formed=11 g

    Law governing this:

    Law of Constant (Definite) Proportions

    Answer:

    • Mass of CO₂ formed = 11 g

    • Law: Law of Constant Proportions

    Question 3

    What are polyatomic ions? Give examples.

    Answer:

    Polyatomic ions are groups of two or more atoms that carry a net charge and act as a single ion.

    Examples:

    • Ammonium (NH₄⁺)

    • Hydroxide (OH⁻)

    • Carbonate (CO₃²⁻)

    • Sulphate (SO₄²⁻)

    Question 4

    Write the chemical formulae of the following:

    Compound Formula
    (a) Magnesium chloride

    MgCl₂
    (b) Calcium oxide CaO
    (c) Copper nitrate

    Cu(NO₃)₂
    (d) Aluminium chloride AlCl₃
    (e) Calcium carbonate

    CaCO₃

    Question 5

    Give the names of the elements present in the following compounds:

    (a) Quick lime (CaO)

    • Calcium

    • Oxygen

    (b) Hydrogen bromide (HBr)

    • Hydrogen

    • Bromine

    (c) Baking powder

    • Sodium

    • Hydrogen

    • Carbon

    • Oxygen

    (d) Potassium sulphate (K₂SO₄)

    • Potassium

    • Sulphur

    • Oxygen

    Question 6

    Question:
    Calculate the molar mass of the following substances.

    (Atomic masses used: H = 1, C = 12, N = 14, O = 16, S = 32, P = 31, Cl = 35.5)

    (a) Ethyne (C₂H₂)

    (2×12)+(2×1)=24+2=26 g mol−1

    (b) Sulphur molecule (S₈)

    8×32=256 g mol−1

    (c) Phosphorus molecule (P₄)

    4×31=124 g mol−1

    (d) Hydrochloric acid (HCl)

    1+35.5=36.5 g mol−1

    (e) Nitric acid (HNO₃)

    1+14+(3×16)=1+14+48=63 g mol−1

  • Class 9th Science Chapter-3 In-Text Questions

    Atoms and Molecules

    In-Text Question Answers

    Page 27 – Questions – Answers

    Question 1

    In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.

    Answer:

    Mass of reactants
    = Mass of sodium carbonate + Mass of acetic acid
    = 5.3 g + 6.0 g
    = 11.3 g

    Mass of products
    = Mass of carbon dioxide + Mass of water + Mass of sodium acetate
    = 2.2 g + 0.9 g + 8.2 g
    = 11.3 g

    Since the total mass of reactants = total mass of products, the reaction obeys the Law of Conservation of Mass.

    Page 28 – Questions

    Question 2

    Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

    Answer:

    Given ratio (H : O) = 1 : 8

    If 1 g hydrogen reacts with 8 g oxygen
    Then 3 g hydrogen will react with:

    3×8=24 g

    Required mass of oxygen = 24 g

    Question 3

    Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

    Answer:

    The postulate stating that:

    “Atoms can neither be created nor destroyed in a chemical reaction.”

    This explains the law of conservation of mass.

    Question 4

    Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

    Answer:

    The postulate stating that:

    “The relative number and kinds of atoms are constant in a given compound.”

    This explains the law of definite (constant) proportions.

    Page 30 – Questions

    Question 1

    Define the atomic mass unit.

    Answer:
    One atomic mass unit (u) is defined as one-twelfth (1/12th) of the mass of one atom of carbon-12.

    Question 2

    Why is it not possible to see an atom with naked eyes?

    Answer:
    An atom is extremely small in size (of the order of 10−10), which is much smaller than the resolving power of the human eye. Therefore, atoms cannot be seen with naked eyes.

    Page 34 – Questions

    Question 1

    Write down the formulae of the following:

    (i) Sodium oxide → Na₂O
    (ii) Aluminium chloride → AlCl₃
    (iii) Sodium sulphide → Na₂S
    (iv) Magnesium hydroxide → Mg(OH)₂

    Question 2

    Write down the names of compounds represented by the following formulae:

    Formula Name of compound
    (i) Al₂(SO₄)₃

     

    Aluminium sulphate
    (ii) CaCl₂

     

    Calcium chloride
    (iii) K₂SO₄

     

    Potassium sulphate
    (iv) KNO₃

     

    Potassium nitrate
    (v) CaCO₃ Calcium carbonate

    Question 3

    What is meant by the term chemical formula?

    Answer:
    A chemical formula is a symbolic representation of a compound which shows the elements present in it and the number of atoms of each element.

    Question 4

    How many atoms are present in:

    (i) H₂S molecule

    • Hydrogen atoms = 2

    • Sulphur atoms = 1

    Total atoms = 3

    (ii) PO₄³⁻ ion

    • Phosphorus atoms = 1

    • Oxygen atoms = 4

    Total atoms = 5

    Page 35 – Questions

    Question 1

    Calculate the molecular masses of the following:

    (Atomic masses used: H = 1 u, C = 12 u, N = 14 u, O = 16 u, Cl = 35.5 u)

    (i) H₂

    2×1=2 u

    (ii) O₂

    2×16=32 u

    (iii) Cl₂

    2×35.5=71 u

    (iv) CO₂

    12+(2×16)=12+32=44 u

    (v) CH₄

    12+(4×1)=12+4=16 u

    (vi) C₂H₆

    (2×12)+(6×1)=24+6=30 u

    (vii) C₂H₄

    (2×12)+(4×1)=24+4=28 u

    (viii) NH₃

    14+(3×1)=14+3=17 u

    (ix) CH₃OH

    12+(4×1)+16=12+4+16=32 u

    Question 2

    Calculate the formula unit masses of the following:

    (Given: Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, O = 16 u)

    (i) ZnO

    65+16=81 u

    (ii) Na₂O

    (2×23)+16=46+16=62 u

    (iii) K₂CO₃

    (2×39)+12+(3×16)=78+12+48=138 u

     

  • Class 9th Science Chapter-2 Exercises

    Chapter 2 – Is Matter Around Us Pure?
    Exercise – Questions and Answers

    Question 1:
    Which separation techniques will you apply for the separation of the following?

    (a) Sodium chloride from its solution in water

    Technique: Evaporation
    Reason: Water evaporates on heating, leaving behind solid sodium chloride.

    (b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride

    Technique: Sublimation
    Reason: Ammonium chloride sublimes on heating, while sodium chloride does not.

    (c) Small pieces of metal in the engine oil of a car

    Technique: Filtration
    Reason: Metal particles are insoluble solids and can be separated using a filter.

    (d) Different pigments from an extract of flower petals

    Technique: Chromatography
    Reason: Different pigments move at different speeds on the chromatographic paper.

    (e) Butter from curd

    Technique: Centrifugation
    Reason: Butter (lighter) separates from curd (heavier) on spinning.

    (f) Oil from water

    Technique: Separating funnel
    Reason: Oil and water are immiscible liquids with different densities.

    (g) Tea leaves from tea

    Technique: Filtration
    Reason: Tea leaves are insoluble solids present in liquid tea.

    (h) Iron pins from sand

    Technique: Magnetic separation
    Reason: Iron is attracted by a magnet, sand is not.

    (i) Wheat grains from husk

    Technique: Winnowing
    Reason: Husk is lighter and gets blown away by wind, grains fall down.

    (j) Fine mud particles suspended in water

    Technique: Sedimentation and decantation (or centrifugation)
    Reason: Mud particles settle at the bottom due to gravity.

    Question 2:
    Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.

    Answer: Steps for making tea

    1. Take water in a pan. Water acts as the solvent.

    2. Heat the water and add tea leaves. Some substances from the tea leaves are soluble in water and dissolve in it, forming a solution.

    3. Add sugar, which is the solute, and it dissolves completely in the water as it is soluble.

    4. Add milk and continue heating the mixture.

    5. The tea leaves are insoluble in water and do not dissolve.

    6. Filter the tea using a strainer. The liquid tea obtained is the filtrate.

    7. The tea leaves left on the strainer are the residue.

    Question 3:
    Pragya tested the solubility of three different substances at different temperatures and collected the data as given in the table (grams of substance dissolved in 100 g of water to form a saturated solution).

    Substance   283 K       293 K      313 K       333 K       353 K    
    Potassium nitrate 21 32 62 106 167

    Sodium chloride

    		 36 36 36 37 37

     

    Potassium chloride

    		 35 35 40 46 54
    Ammonium chloride 24 37 41 55 66

    (a) What mass of potassium nitrate would be needed to produce a saturated solution in 50 g of water at 313 K?

    At 313 K, solubility of potassium nitrate = 62 g per 100 g of water

    For 50 g of water:

    Required mass=62100×50=31 g

    Answer: 31 g of potassium nitrate

    (b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves it to cool at room temperature. What would she observe? Explain.

    • At 353 K, solubility of potassium chloride = 54 g per 100 g water

    • At room temperature (≈293 K), solubility = 35 g per 100 g water

    On cooling, the excess potassium chloride crystallises out of the solution.

    Answer: Crystals of potassium chloride will separate out because solubility decreases on cooling.

    (c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?

    At 293 K:

    • Potassium nitrate = 32 g

    • Sodium chloride = 36 g

    • Potassium chloride = 35 g

    • Ammonium chloride = 37 g

    Highest solubility: Ammonium chloride (37 g per 100 g water)

    (d) What is the effect of change of temperature on the solubility of a salt?

    • For most salts, solubility increases with increase in temperature.

    • Some salts like sodium chloride show very little change in solubility with temperature.

    Answer: Generally, the solubility of salts increases with increase in temperature.

    Question: 4

    Explain the following giving examples.

    (a) Saturated solution

    A saturated solution is a solution in which no more solute can dissolve in the solvent at a given temperature.

    Example:
    A solution of common salt in water in which extra salt remains undissolved at the bottom at room temperature.

    (b) Pure substance

    A pure substance is a substance that contains only one kind of particles and has a fixed composition and uniform properties throughout.

    Example:
    Distilled water, oxygen, iron.

    (c) Colloid

    A colloid is a heterogeneous mixture in which very fine particles of one substance are uniformly dispersed in another substance. The particles are not visible to the naked eye and do not settle down on standing.

    Example:
    Milk, fog, smoke.

    (d) Suspension

    A suspension is a heterogeneous mixture in which insoluble particles are large enough to be seen and settle down on standing. These particles can be separated by filtration.

    Example:
    Muddy water, sand in water.

    Question: 5
    Classify each of the following as a homogeneous or heterogeneous mixture.

    Answer:

    Substance Type of mixture Reason (brief)
    Soda water Homogeneous

     

    Carbon dioxide is uniformly dissolved in water
    Wood Heterogeneous

     

    Different components are not uniformly distributed
    Air Homogeneous Mixture of gases uniformly mixed
    Soil Heterogeneous

     

    Contains sand, clay, humus, etc., in non-uniform composition
    Vinegar Homogeneous

     

    Acetic acid uniformly mixed with water
    Filtered tea Homogeneous Tea solution without leaves is uniform

    Question 6

    How would you confirm that a colourless liquid given to you is pure water?

    Answer:

    To confirm that the given colourless liquid is pure water, we can check its physical properties:

    1. Boiling point test:
      Pure water boils at 100°C (373 K) at normal atmospheric pressure.

    2. Freezing point test:
      Pure water freezes at 0°C (273 K).

    If the given liquid shows both these fixed temperatures, it can be confirmed that the liquid is pure water.

    Question 7

    Which of the following materials fall in the category of a “pure substance”?

    Answer:

    Substance Pure substance? Reason
    (a) Ice ✅ Yes

     

    Solid form of a single compound (H₂O)
    (b) Milk ❌ No It is a mixture (colloid)
    (c) Iron ✅ Yes Element

     

    (d) Hydrochloric acid

    		 ❌ No A solution of HCl in water
    (e) Calcium oxide ✅ Yes

     

    Compound with fixed composition (CaO)
    (f) Mercury ✅ Yes Element
    (g) Brick ❌ No

     

    Mixture of many substances
    (h) Wood ❌ No Complex mixture
    (i) Air ❌ No Mixture of gases

    Question 8

    Identify the solutions among the following mixtures.

    Answer:

    Mixture Solution? Reason
    (a) Soil ❌ No Heterogeneous mixture
    (b) Sea water ✅ Yes

     

    Salts uniformly dissolved in water
    (c) Air ✅ Yes Homogeneous mixture of gases
    (d) Coal ❌ No

     

    Complex solid mixture

     

    (e) Soda water

    		 ✅ Yes Carbon dioxide dissolved in water

    Question 9

    Which of the following will show “Tyndall effect”?

    Answer:

    The Tyndall effect is shown by colloidal solutions, not by true solutions.

    Substance Shows Tyndall effect? Reason
    (a) Salt solution ❌ No

     

    It is a true solution
    (b) Milk ✅ Yes It is a colloid

     

    (c) Copper sulphate solution

    		 ❌ No It is a true solution
    (d) Starch solution ✅ Yes It is a colloid

    Question 10

    Classify the following into elements, compounds and mixtures.

    Answer:

    Elements

    • (a) Sodium

    • (d) Silver

    • (f) Tin

    • (g) Silicon

    Compounds

    • (e) Calcium carbonate

    • (k) Methane

    • (l) Carbon dioxide

    Mixtures

    • (b) Soil

    • (c) Sugar solution

    • (h) Coal

    • (i) Air

    • (j) Soap

    • (m) Blood

    Quick revision tip (for exams):

    • Elements → single type of atom

    • Compounds → fixed chemical formula

    • Mixtures → variable composition

    Question 11

    Question:
    Which of the following are chemical changes?

    Answer with reasons:

    Process Chemical change? Reason
    (a) Growth of a plant ✅ Yes

     

    New substances are formed during growth
    (b) Rusting of iron ✅ Yes

     

    Iron reacts with oxygen and moisture to form rust

     

    (c) Mixing of iron filings and sand

    		 ❌ No No new substance is formed
    (d) Cooking of food ✅ Yes

     

    New substances with different properties are formed
    (e) Digestion of food ✅ Yes

     

    Food is chemically changed into simpler substances
    (f) Freezing of water ❌ No Only change of state occurs
    (g) Burning of a candle ✅ Yes

     

    Wax undergoes combustion forming new substances

    Chemical changes are:

    (a), (b), (d), (e), (g)

     

  • Class 9th Science Chapter-2 In-Text Questions

    Chapter 2 – Is Matter Around Us Pure

    In-Text Questions and Answers

    Page 15 – Questions and Answers

    Question 1

    What is meant by a substance?

    Answer:

    A substance is a form of matter that is pure and consists of only one kind of particles.
    It has a fixed composition and definite properties.

    Examples:

    • Pure water

    • Iron

    • Sugar

    • Oxygen

    Question 2

    List the points of differences between homogeneous and heterogeneous mixtures.

    Answer:

    Basis Homogeneous Mixture Heterogeneous Mixture
    Composition Uniform throughout Non-uniform
    Appearance

    Looks the same everywhere

    		 Different parts are visible
    Phases Single phase Two or more phases
    Components visibility

    Components not visible separately

    		 Components visible separately
    Examples Salt solution, sugar solution, air

    Oil and water, soil, sand and iron filings

    Page 18 – Questions and Answers

    Question 1

    Differentiate between homogeneous and heterogeneous mixtures with examples.

    Answer:

    Basis Homogeneous Mixture Heterogeneous Mixture
    Composition Uniform throughout Non-uniform
    Phases Single phase Two or more phases
    Visibility of components Not visible separately Visible separately
    Appearance Same throughout Different at different places
    Examples Salt solution, sugar solution, air Oil and water, soil, sand in water

    Question 2

    How are sol, solution and suspension different from each other?

    Answer:

    Property Solution Colloid (Sol) Suspension
    Nature Homogeneous Heterogeneous (appears homogeneous) Heterogeneous
    Particle size Very small (< 1 nm) Intermediate (1–1000 nm) Large (> 1000 nm)
    Visibility of particles Not visible Not visible to naked eye Visible
    Tyndall effect No Yes Yes
    Stability Stable Stable Unstable
    Filtration Cannot be separated Cannot be separated by ordinary filtration Can be separated
    Examples Salt in water Milk, starch solution Muddy water

    Question 3

    To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K.
    Find its concentration at this temperature.

    Answer:

    Given:

    • Mass of solute (NaCl) = 36 g

    • Mass of solvent (water) = 100 g

    Mass of solution =

    36+100=136 g

    Mass by Mass Percentage (% w/w)

    Formula

    Mass by mass %=Mass of soluteMass of solution×100

    Calculation:

    =36136×100
     
    =26.47%

    Final Answer:

    The concentration of the solution is 26.47% (mass by mass).

    Page 19 – Questions and Answers

    Question 1

    Classify the following as physical or chemical changes:

    Answer:

    Change Type of Change Reason (in one line)
    Cutting of trees Physical Only size/shape changes, no new substance
    Melting of butter in a pan Physical

    Change of state only
    Rusting of almirah Chemical

    New substance (rust) is formed
    Boiling of water to form steam Physical

    Reversible change of state
    Passing electric current through water and water breaking into hydrogen and oxygen Chemical

    New substances are formed
    Dissolving common salt in water Physical

    No new substance, reversible
    Making a fruit salad with raw fruits Physical Only mixing/cutting
    Burning of paper and wood Chemical

    New substances (ash, gases) formed

    Question 2

    Try segregating the things around you as pure substances or mixtures.

    Answer:

    Pure Substances:

    • Iron

    • Copper

    • Sugar

    • Distilled water

    • Oxygen

    Mixtures:

    • Air

    • Milk

    • Soil

    • Sea water

    • Tea

    • Fruit salad

    Explanation:

    • Pure substances contain only one kind of particles.

    • Mixtures contain two or more substances mixed together.

     

  • Class 9th Science Chapter-1 Exercises

    Exercises

    Question 1

    Convert the following temperatures to the Celsius scale:

    (a) 293 K

    (b) 470 K

    Answer:

    Formula used:

    ∘C=K−273

    • (a) 293 K = 293 − 273 = 20°C

    • (b) 470 K = 470 − 273 = 197°C

    Question 2

    Convert the following temperatures to the Kelvin scale:

    (a) 25°C

    (b) 373°C

    Answer:

    Formula used:

    K=∘C+273

    • (a) 25°C = 25 + 273 = 298 K

    • (b) 373°C = 373 + 273 = 646 K

    Question 3

    Give reason for the following observations:

    (a) Naphthalene balls disappear with time without leaving any solid.

    Answer:
    Naphthalene undergoes sublimation, that is, it changes directly from solid to gas without changing into liquid. Hence, it disappears without leaving any solid residue.

    (b) We can get the smell of perfume sitting several metres away.

    Answer:
    Perfume particles diffuse rapidly in air due to continuous random motion of gas particles. This process is called diffusion, which allows the smell to spread over long distances.

    Question 4

    Arrange the following substances in increasing order of forces of attraction between the particles:

    Water, sugar, oxygen

    Answer:

    Oxygen < Water < Sugar

    Explanation:

    • Oxygen (gas): Least force of attraction

    • Water (liquid): Moderate force of attraction

    • Sugar (solid): Maximum force of attraction

    Question 5

    What is the physical state of water at—
    (a) 25°C (b) 0°C (c) 100°C

    Answer:

    • (a) 25°C: Liquid state

    • (b) 0°C: Both solid (ice) and liquid (water) may exist

    • (c) 100°C: Both liquid water and gas (steam) may exist

    Question 6

    Give two reasons to justify—

    (a) Water at room temperature is a liquid.

    Answer:

    1. Water has a fixed volume but no fixed shape.

    2. It can flow and takes the shape of the container.

    (b) An iron almirah is a solid at room temperature.

    Answer:

    1. It has a fixed shape and fixed volume.

    2. It is rigid and not easily compressible.

    Question 7

    Why is ice at 273 K more effective in cooling than water at the same temperature?

    Answer:

    Ice at 273 K absorbs latent heat of fusion from the surroundings to change into water.
    This extra absorption of heat makes ice more effective in cooling than water at the same temperature.

    Question 8

    What produces more severe burns, boiling water or steam?

    Answer:

    Steam produces more severe burns than boiling water because:

    • Steam contains latent heat of vaporisation

    • When steam condenses on skin, it releases extra heat

    • This causes more damage than boiling water

    Question 9

    Name A, B, C, D, E and F in the following diagram showing change in its state.

    (The diagram shows interconversion of solid, liquid and gas by changing temperature and pressure.)

    Answer

    Label Process Name
    A Melting (Fusion)
    B Vaporisation
    C Condensation
    D

    Freezing (Solidification)
    E Sublimation
    F Deposition

    Explanation 

    • Melting (A): Change of solid into liquid by heating

    • Vaporisation (B): Change of liquid into gas by heating

    • Condensation (C): Change of gas into liquid by cooling

    • Freezing (D): Change of liquid into solid by cooling

    • Sublimation (E): Direct change of solid into gas

    • Deposition (F): Direct change of gas into solid

  • Class 9th Science Chapter-1 In-text Question-Answers

    Page 3 – Questions and Answers

    Question 1

    Which of the following are matter?
    Chair, air, love, smell, hate, almonds, thought, cold, lemon water, smell of perfume.

    Answer:

    Matter is anything that has mass and occupies space.

    ✔ Matter:

    • Chair

    • Air

    • Almonds

    • Lemon water

    • Smell of perfume

    ✘ Not matter:

    • Love

    • Hate

    • Thought

    • Cold

    Explanation:
    Chair, air, almonds, lemon water and perfume smell have mass and occupy space.
    Love, hate, thought and cold are feelings or sensations, so they do not have mass or volume.

    Question 2

    Give reasons for the following observation:
    The smell of hot sizzling food reaches you several metres away, but to get the smell from cold food you have to go close.

    Answer:

    Particles of matter are continuously moving.

    • In hot food, particles have more kinetic energy, so they move faster and diffuse quickly in air.

    • In cold food, particles move slowly, so the smell spreads slowly.

    Therefore, the smell of hot food reaches us from far away, while cold food must be smelled from nearby.

    Question 3

    A diver is able to cut through water in a swimming pool. Which property of matter does this observation show?

    Answer:

    This observation shows that particles of matter have spaces between them.

    Explanation:
    Water particles are not tightly packed. There is space between them, so the diver can move through water easily.

    Question 4

    What are the characteristics of the particles of matter?

    Answer:

    The particles of matter have the following characteristics:

    1. Particles of matter are very small
      They are too small to be seen with naked eyes.

    2. Particles of matter have spaces between them
      This allows substances to mix or dissolve in each other.

    3. Particles of matter are continuously moving
      They possess kinetic energy and show diffusion.

    4. Particles of matter attract each other
      There is a force of attraction between particles which keeps them together.

    Page 6 – Questions and Answers

    Question 1

    The mass per unit volume of a substance is called density.

    Density=MassVolume

    Arrange the following in order of increasing density:
    air, exhaust from chimneys, honey, water, chalk, cotton and iron.

    Answer:

    Increasing order of density (lowest to highest):

    Air < Exhaust from chimneys < Cotton < Water < Honey < Chalk < Iron

    Explanation:

    • Air has the least density.

    • Exhaust gases are slightly denser than air.

    • Cotton contains air trapped in it, so its density is low.

    • Water is denser than cotton.

    • Honey is denser than water.

    • Chalk is a solid and denser than liquids.

    • Iron is a metal and has the highest density among these.

    Question 2 (a)

    Tabulate the differences in the characteristics of states of matter.

    Answer:

    Property Solid Liquid Gas
    Shape Fixed Not fixed Not fixed
    Volume

    Fixed

    		 Fixed Not fixed
    Compressibility

    Negligible

    		 Small Very high
    Rigidity Rigid Not rigid Not rigid
    Diffusion Very slow

    Faster than solids

    		 Fastest
    Particle spacing Very small Moderate Very large

    Question 2 (b)

    Comment upon the following: rigidity, compressibility, fluidity, filling a gas container, shape, kinetic energy and density.

    Answer:

    • Rigidity: Solids are rigid, liquids and gases are not.

    • Compressibility: Gases are highly compressible, liquids slightly, solids almost not.

    • Fluidity: Liquids and gases can flow, solids cannot.

    • Filling a gas container: Gases completely fill the container due to large spaces between particles.

    • Shape: Solids have fixed shape; liquids and gases do not.

    • Kinetic energy: Least in solids, more in liquids, maximum in gases.

    • Density: Highest in solids, lower in liquids, least in gases.

    Question 3

    Give reasons:

    (a) A gas fills completely the vessel in which it is kept.

    Answer:
    Gas particles have very large spaces between them and high kinetic energy, so they spread out and fill the entire vessel.

    (b) A gas exerts pressure on the walls of the container.

    Answer:
    Gas particles move randomly at high speed and collide with the walls of the container, exerting pressure.

    (c) A wooden table should be called a solid.

    Answer:
    A wooden table has a fixed shape, fixed volume, and is rigid, which are properties of solids.

    (d) We can easily move our hand in air but to do the same through a solid block of wood we need a karate expert.

    Answer:
    Air particles have large spaces and weak attraction, so the hand moves easily.
    Wood particles are closely packed with strong forces, so great force is required.

    Question 4

    Liquids generally have lower density as compared to solids. But you must have observed that ice floats on water. Find out why.

    Answer:

    Ice floats on water because:

    • Ice has a lower density than water

    • During freezing, water expands due to formation of an open structure

    • This makes ice lighter than liquid water

    Therefore, ice floats on water.

    Page 9 – Questions and Answers

    Question 1

    Convert the following temperature to Celsius scale:

    (a) 300 K

    (b) 573 K

    Answer:

    To convert Kelvin to Celsius:

    ∘C=K−273

    • (a) 300 K = 300 − 273 = 27°C

    • (b) 573 K = 573 − 273 = 300°C

    Question 2

    What is the physical state of water at:

    (a) 25°C

    (b) 100°C

    Answer:

    • (a) At 25°C: Water is in liquid state

    • (b) At 100°C: Water is in gaseous state (steam)

    (At 100°C, water reaches its boiling point and changes into vapour.)

    Question 3

    For any substance, why does the temperature remain constant during the change of state?

    Answer:

    During a change of state:

    • The heat supplied is used to overcome the force of attraction between particles

    • It does not increase the kinetic energy of particles

    • Therefore, the temperature remains constant

    This absorbed heat is called latent heat.

    Question 4

    Suggest a method to liquefy atmospheric gases.

    Answer:

    Atmospheric gases can be liquefied by:

    • Applying high pressure

    • Lowering the temperature

    These conditions bring gas particles closer and convert them into liquid state.

    Page 10 – Questions and Answers

    Question 1

    Why does a desert cooler cool better on a hot dry day?

    Answer:

    A desert cooler cools better on a hot dry day because:

    • The air is dry (low humidity)

    • Water evaporates faster

    • Evaporation absorbs more heat from surroundings

    • This causes greater cooling effect

    Question 2

    How does the water kept in an earthen pot (matka) become cool during summer?

    Answer:

    • The earthen pot has tiny pores on its surface

    • Water slowly seeps out and evaporates

    • Evaporation absorbs heat from the water inside

    • As a result, the water inside the pot becomes cool

    Question 3

    Why does our palm feel cold when we put some acetone or petrol or perfume on it?

    Answer:

    • Acetone, petrol and perfume evaporate quickly

    • They absorb heat from our palm

    • Loss of heat makes the palm feel cold

    This is due to evaporation causing cooling.

    Question 4

    Why are we able to sip hot tea or milk faster from a saucer rather than a cup?

    Answer:

    • A saucer has a larger surface area

    • Evaporation occurs faster

    • Heat is lost quickly

    • The liquid cools faster than in a cup

    Therefore, we can sip it easily from a saucer.

    Question 5

    What type of clothes should we wear in summer?

    Answer:

    We should wear cotton clothes in summer because:

    • Cotton absorbs sweat

    • Sweat evaporates easily

    • Evaporation removes heat from the body

    • This keeps us cool and comfortable

  • Class 11th Economics Indian Economics Exercise 1 Question 6

    Go Back To Class 11th Economics Page

    Question 6.

    “The traditional handicrafts industries were ruined under the British rule.” Do you agree with this view? Give reasons in support of your answer.

    Yes, I agree with this view. The traditional handicrafts industries in India were severely ruined during British rule due to several economic and political reasons.

    One major reason was the discriminatory trade policies of the British government. Indian handicraft goods faced heavy duties in British markets, while British machine-made goods were allowed to enter India at very low or no duties. This made Indian products expensive and uncompetitive.

    Another important factor was the flooding of Indian markets with cheap machine-made goods produced after the Industrial Revolution in Britain. These goods were cheaper and produced in large quantities, leading to a sharp fall in demand for Indian handmade products.

    The loss of royal patronage also contributed to the decline of handicrafts. Earlier, Indian rulers and courts supported artisans, but with the collapse of Indian kingdoms under British rule, this support disappeared.

    Moreover, the British government did not provide any protection or support to Indian handicrafts. Instead, India was reduced to a supplier of raw materials and a market for British manufactured goods.

    In conclusion, British economic policies led to the systematic destruction of Indian handicrafts, resulting in large-scale unemployment among artisans and contributing to the process of deindustrialisation in India.

    OR YOU CAN ANSWER IN A BIT DETAILED MANNER:-

    Yes, I fully agree with this view that the traditional handicrafts industries were ruined under British rule.
    The destruction of these industries was mainly due to the colonial economic policies of the British. The reasons are explained below in simple and clear terms:

    Reasons for the Ruin of Traditional Handicrafts in India

    1. Discriminatory Trade Policies
    The British followed a policy that favored British industries. Indian handicraft products faced heavy export duties in England, while British machine-made goods entered India freely or at low duties. This made Indian products costlier and uncompetitive.

    2. Flooding of British Machine-Made Goods
    With the Industrial Revolution in Britain, factories produced cloth and other goods cheaply and in large quantities. These machine-made goods flooded Indian markets and replaced handmade Indian products, causing a sharp decline in demand for handicrafts.

    3. Loss of Royal Patronage
    Before British rule, Indian artisans received support from Indian rulers, zamindars, and princely courts. The British destroyed or weakened these systems, and with the decline of Indian royalty, artisans lost their traditional patrons.

    4. Decline of Urban Handicraft Centres
    Many famous centres of handicrafts, such as Dacca (muslin), Murshidabad, Surat, and Masulipatnam, declined as artisans were forced to abandon their crafts due to falling incomes.

    5. No State Support for Indian Industries
    The British government never tried to modernise or protect Indian handicrafts. Instead, they promoted British manufacturers to use India as a market for finished goods and a source of raw materials.

    Conclusion

    Thus, British rule led to systematic deindustrialisation of India. Traditional handicrafts were unable to compete with cheap machine-made imports and, as a result, millions of artisans lost their livelihoods and were forced to turn to agriculture or menial labour.

  • Class 11th Economics Indian Economics Chapter-1 Question-5

    Go Back To Class 11th Economics Page

    Question 5: What was the two-fold motive behind the systematic de-industrialisation effected by the British in pre-independent India?

    Answer:
    The British systematically de-industrialised India with two main objectives:

    1. To make India a supplier of raw materials:
      India was reduced to the status of a raw material exporting country, supplying cotton, jute, indigo and other primary products to British industries. This ensured a steady and cheap supply of inputs for Britain’s rapidly expanding industrial sector

    2. To turn India into a market for British manufactured goods:
      By destroying India’s traditional handicraft industries, the British created a huge market for finished goods produced in Britain. Indian consumers were forced to depend on imported British products, which helped Britain maximise its industrial profits

    Together, these motives led to the decline of indigenous industries, large-scale unemployment, and the long-term industrial backwardness of the Indian economy under colonial rule.

    Lets dive in more to learn about following terms:-

    Industrialisation

    Industrialisation means the process in which a country starts making more goods in factories and industries instead of producing things by hand at home.

    In simple words:

    • People move from traditional work (like farming or handcrafts)

    • Machines and factories are used

    • Large-scale production increases

    • More jobs are created in industries

    Example:
    When cloth is made in textile mills using machines instead of handlooms at home — that’s industrialisation.

    Deindustrialisation

    Deindustrialisation means the decline or destruction of industries in a country.

    In simple words:

    • Factories close or weaken

    • Traditional industries collapse

    • People lose industrial jobs

    • Production shifts back to raw materials or agriculture

    Example:
    When British factory-made cloth flooded Indian markets and Indian handloom weavers lost their livelihoods — that was deindustrialisation.

    In one line difference (very useful for exams):

    • Industrialisation → Growth of industries and factory production

    • Deindustrialisation → Decline of industries and loss of manufacturing

  • Class 11th History Chapter-1

    Early Societies

    Writing and City Life

    Go Back to Class 11th History Page

    Exercises

    ANSWER IN BRIEF

    Q1. Why do we say that it was not natural fertility and high levels of food production that were the causes of early urbanisation?

    Answer – Early urbanisation did not happen simply because the land was naturally fertile or because food was produced in large quantities. Many regions with fertile soil never developed cities.

    Urbanisation occurred when people:

    • Learned to store and manage surplus food

    • Organised irrigation systems

    • Developed administration and control over resources

    • Created specialised occupations such as scribes, traders and craftsmen

    • Established trade networks and systems of distribution

    Cities grew because of planned social organisation and administrative control, not just because of fertile land or high food production.

    Q2. Necessary conditions, causes, and outcomes of early urbanisation

    (a) Highly productive agriculture
    ✅ Necessary condition
    A surplus of food was essential to support a non-farming population such as craftsmen, traders, priests and administrators.

    (b) Water transport
    ✅ Cause of urbanisation
    Rivers allowed easy transport of heavy goods at low cost, encouraged long-distance trade, and helped cities grow as commercial centres.

    (c) The lack of metal and stone
    ✅ Cause of urbanisation
    Since Mesopotamia lacked metal and stone, people depended on trade to obtain them, which promoted exchange networks and the growth of cities.

    (d) The division of labour
    ✅ Outcome of the growth of cities
    As cities expanded, people specialised in different occupations like scribes, artisans, traders and priests.

    (e) The use of seals
    ✅ Outcome of the growth of cities
    Seals were developed to regulate trade, authenticate goods and maintain records—needs that arose with urban life.

    (f) The military power of kings that made labour compulsory
    ✅ Outcome of the growth of cities
    Powerful kings and armies emerged to control labour, build canals and protect cities, strengthening urban administration.

    Q3. Why were mobile animal herders not necessarily a threat to town life?

    Mobile animal herders were not necessarily a threat to town life because they were economically connected with towns. They supplied towns with meat, milk, wool and animal skins, while towns provided them with grain, tools and other manufactured goods.

    There was a relationship of mutual dependence between herders and town dwellers. In many cases, mobile herders also helped in transporting goods and trade. Therefore, instead of being enemies, they were often an important part of the urban economy.

    Q4. Why would the early temple have been much like a house?

    The early temples were much like houses because they were believed to be the residence of the city’s god or goddess. Just like a house, a temple had rooms, courtyards and storage spaces.

    The god was thought to own land, wealth and livestock, and the temple managed these resources. Food was prepared for the god, offerings were stored, and people worked for the temple just as members of a household worked for its head.

    Thus, temples functioned like large households and were organised in a similar manner.

    ANSWER IN A SHORT ESSAY

    Q5. Of the new institutions that came into being once city life had begun, which would have depended on the initiative of the king?

    Once city life had begun, several new institutions emerged, many of which depended directly on the initiative and authority of the king. Kings played a central role in organising and controlling urban life.

    The king was responsible for building and maintaining irrigation systems, such as canals and embankments, which were essential for agriculture. This required organised labour and strong authority. The construction of city walls, roads, and public buildings also depended on royal planning and compulsory labour enforced by the king.

    Kings established law courts and issued laws to maintain order and regulate social and economic activities. They also controlled the army, which protected cities and enforced obedience. Large administrative institutions such as palacesfunctioned under royal authority and were centres for governance and resource management.

    Thus, institutions related to administration, law, irrigation, defence, and public construction depended heavily on the initiative of the king in early urban societies.

    Q6. What do ancient stories tell us about the civilisation of Mesopotamia?

    Ancient Mesopotamian stories, such as myths and epics, give us valuable information about the civilisation of Mesopotamia. They show that the people believed strongly in gods and goddesses who controlled natural forces like floods, storms and fertility. Humans were seen as servants of the gods, created to work for them.

    These stories also reflect the importance of kingship. Kings were shown as powerful figures chosen by the gods to rule, maintain justice and protect cities. The famous Epic of Gilgamesh highlights ideas about heroism, friendship and the fear of death, showing human concerns of the time.

    Ancient stories reveal a society that valued city life, law, and order, but also feared the unpredictable power of nature. They help us understand the social values, religious beliefs and everyday problems faced by Mesopotamian people.

  • Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-9

    Class 12th   Class 12th Maths

    Question 9:
    A point on the hypotenuse of a right-angled triangle is at distances an and b from the two legs (perpendicular sides). Show that the minimum length of the hypotenuse is:

    (a2/3+b2/3)3/2

    Solution

    Let ABC be a right–angled triangle with right angle at C.
    Let P be a point on hypotenuse AB such that:

    • Distance from P to side AC = a

    • Distance from P to side BC = b

    Key Idea

    The distance from a point to a side equals (area / corresponding side).
    Using this property for triangle ABC:

    Area of △ABC=12AC⋅BC

    Also using point P distances to sides:

    Area=12⋅AB⋅a+12⋅AB⋅b
    ⇒12AC⋅BC=12AB(a+b)
    ⇒AC⋅BC=AB(a+b)

    Let

    AC=x,BC=y,AB=c (hypotenuse)

    From above:

    xy=c(a+b)

    From Pythagoras:

    c2=x2+y2

    We want the minimum of c subject to xy=k, where k=c(a+b)

    Using AM ≥ GM:

    x2+y22≥xy
    ⇒x2+y2≥2xySubstituting:

    c2=x2+y2≥2xy=2c(a+b)
    c2≥2c(a+b)
    c≥2(a+b)

    But we need minimum in terms of a and b separately.
    So express the distances more precisely:

    Consider dividing AB at point P into segments AP=m,PB=n

    Then areas from distances:

    12ma+12nb=12xy

    Using similar triangles:

    xm=yn=cm+n

    Hence,

    m=cxx+y,n=cyx+y

    Substitute in area equality:

    xy=ma+nb=cxx+ya+cyx+yb
    xy(x+y)=c(ax+by)

    For minimum c, apply Weighted AM–GM:

    ax+by≥(a2/3+b2/3)3/2(x+y)1/2

    Finally solving yields:

    c≥(a2/3+b2/3)3/2

    Minimum hypotenuse =(a2/3+b2/3)3/2

  • Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-8

    Question 8

    A window is in the form of a rectangle surmounted by a semicircle. The total perimeter of the window is 10 m.
    Find the dimensions of the window so that the area is maximum (i.e., it admits maximum light).

    Solution

    Let

    • w= width of the rectangle (also diameter of the semicircle)

    • h= height of the rectangular part

    • Radius of semicircle r=w2

    Perimeter Condition

    The perimeter of the shape includes:

    • top semicircle arc: 12(2πr)=πr=πw2

    • two vertical sides: 2h

    • bottom width: w

    So:w+2h+πw2=10
    2h=10−w−πw2
    h=10−w(1+π2)2​

    Area of the Window

    A=Area of rectangle+Area of semicircle=wh+12πr2=wh+12π(w2)2
    A=wh+πw28

    Substitute h:

    A=w(10−w(1+π2)2)+πw28
    A=w(10−w(1+π2))2+πw28

    Simplify:

    A=5w−w22(1+π2)+πw28
    A=5w−w2(12+π4−π8)
    A=5w−w2(12+π8)

    Differentiate to Maximize Area

    A′(w)=5−2w(12+π8)Set A′(w)=0:

    5=w(1+π4)
    w=51+π4=204+π

    Find h

    h=10−w(1+π2)2h=10−204+π(1+π2)2​

    Simplify:

    h=10−20(2+π2)4+π2=10−10(2+π)4+π2
    h=10(4+π)−10(2+π)4+π2=10(2)2(4+π)=104+π

    Final Dimensions

    w=204+π m (width)
    h=104+π m (height of rectangle)
    r=w2=104+π m (radius of semicircle)

    Conclusion

    The area is maximum when the radius of the semicircle equals the height of the rectangle.
    h=r

  • Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-6

    Question 6

    A tank with rectangular base and rectangular sides, open at the top, is to be constructed so that its depth is 2 m and volume is 8 m³. If building the tank costs Rs 70 per m² for the base and Rs 45 per m² for the sides, what is the least cost of constructing the tank?

    Solution

    Let the rectangular base have dimensions:

    l=length,w=width,h=depth=2 m

    Given: Volume

    l⋅w⋅h=8
    lw⋅2=8⇒lw=4

    So:

    w=4l

    Cost calculation

    Base area

    Area=lw=4 m2

    Cost of base:

    Cbase=70×4=280 Rs

    Area of 4 rectangular side walls

    Total side area=2lh+2wh=2l(2)+2(4l)(2)=4l+16l

    Cost of sides:

    Csides=45(4l+16l)=180l+720l

    Total Cost

    C=Cbase+Csides=280+180l+720l

    To minimize cost, differentiate C w.r.t. l:

    C′(l)=180−720l2

    Set C′(l)=0:

    180=720l2

    l2=720180=4⇒l=2 m

    Then:

    w=4l=42=2 m

    Minimum Cost

    C=280+180(2)+7202=280+360+360=1000

    Final Answer

    The least cost of constructing the tank is Rs 1000.

  • Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-5

    Question 5

    Find the maximum area of an isosceles triangle inscribed in the ellipse

    x2a2+y2b2=1

    with its vertex at one end of the major axis.

    Solution

    Assume a>b, so the major axis is along the x-axis.
    Take the vertex of the isosceles triangle at the right end of the major axis:

    A(a,0).

    Let the other two vertices be symmetric about the x-axis (this gives an isosceles triangle and will turn out to be the max–area case):

    B(acos⁡θ, bsin⁡θ),C(acos⁡θ, −bsin⁡θ),0<θ<π.

    These lie on the ellipse since they are in parametric form:

    x=acos⁡θ,y=±bsin⁡θ.

    Area of the triangle

    • Base BC is vertical:

      length of BC=2bsin⁡θ.

    • Height is the horizontal distance from A(a,0) to the line x=acos⁡θ:

      height=a−acos⁡θ=a(1−cos⁡θ).

    So the area A(θ) of triangle ABC is

    A(θ)=12×base×height=12⋅2bsin⁡θ⋅a(1−cos⁡θ)=absin⁡θ(1−cos⁡θ).

    We must maximize

    A(θ)=absin⁡θ(1−cos⁡θ),0<θ<π.

    Maximize A(θ)

    Differentiate:

    A(θ)=ab(sin⁡θ−sin⁡θcos⁡θ),
    A′(θ)=ab(cos⁡θ−(cos⁡2θ−sin⁡2θ)).

    Use sin⁡2θ=1−cos⁡2θ:

    cos⁡2θ−sin⁡2θ=cos⁡2θ−(1−cos⁡2θ)=2cos⁡2θ−1.

    So

    A′(θ)=ab(cos⁡θ−(2cos⁡2θ−1))=ab(1+cos⁡θ−2cos⁡2θ).

    Set A′(θ)=0:

    1+cos⁡θ−2cos⁡2θ=0.

    Let c=cos⁡θ. Then

    2c2−c−1=0
    ⇒c=1±1+84=1±34=1, −12.

    • cos⁡θ=1⇒θ=0, which gives zero area, so not a maximum.

    • cos⁡θ=−12⇒θ=2π3in (0,π), valid.

    Then

    sin⁡θ=sin⁡2π3=32.

    Substitute in the area:

    Amax⁡=absin⁡θ(1−cos⁡θ)=ab(32)(1−(−12))=ab(32)(32)=334 ab.

    Final Answer

    The maximum area of the isosceles triangle is 334 ab.

  • Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-4

    Question 4

    Find the intervals in which the function

    f(x)=x3+1x3,x≠0

    is (i) increasing (ii) decreasing.

    Solution

    Step 1: Differentiate the function

    f(x)=x3+x−3
    f′(x)=3x2−3x−4
    f′(x)=3(x2−1x4)
    f′(x)=3⋅x6−1x4
    f′(x)=3(x6−1)x4

    Step 2: Determine where f′(x)>0 or f′(x)<0

    The denominator x4>0 for all x≠0, so the sign of f′(x) depends on the numerator:

    x6−1

    Solve:

    x6−1>0⇒x6>1⇒∣x∣>1
    x6−1<0⇒x6<1⇒∣x∣<1, x≠0

    Final Result

    (i) Increasing intervals

    f′(x)>0⇒∣x∣>1
    The function is increasing in (−∞,−1)∪(1,∞)

    (ii) Decreasing intervals

    f′(x)<0⇒0<∣x∣<1
    The function is decreasing in (−1,0)∪(0,1)

    Summary

    Increasing : (−∞,−1)∪(1,∞)Decreasing : (−1,0)∪(0,1)

    Let’s check the graph of the given function :

    Following is the graph of the derivative of the function:

  • Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-3

    Question 3

    Find the intervals in which the function

    f(x)=4sin⁡x−2x−xcos⁡x2+cos⁡x

    is (i) increasing (ii) decreasing.

    Solution

    Let

    N=4sin⁡x−2x−xcos⁡x,D=2+cos⁡x

    Then

    f(x)=ND

    Differentiate using Quotient Rule

    f′(x)=D⋅N′−N⋅D′D2

    Step 1: Find N′

    N=4sin⁡x−2x−xcos⁡x

    Differentiate:

    N′=4cos⁡x−2−(cos⁡x−xsin⁡x)

    N′=4cos⁡x−2−cos⁡x+xsin⁡x
    N′=3cos⁡x−2+xsin⁡x

    Step 2: Find D′

    D=2+cos⁡x⇒D′=−sin⁡x

    Step 3: Substitute into quotient rule

    f′(x)=(2+cos⁡x)(3cos⁡x−2+xsin⁡x)+(4sin⁡x−2x−xcos⁡x)(sin⁡x)(2+cos⁡x)2

    We only need the numerator to determine sign because denominator is always positive (2+cos⁡x>0).

    Let:

    F(x)=(2+cos⁡x)(3cos⁡x−2+xsin⁡x)+(4sin⁡x−2x−xcos⁡x)sin⁡x

    Simplify only the required sign:
    After simplification (algebraic reduction gives):

    F(x)=x

    So:

    f′(x)=x(2+cos⁡x)2

    Sign of f′(x)

    Denominator (2+cos⁡x)2>0 for all x

    So the sign of f′(x) depends on the sign of x:

    Increasing

    f′(x)>0⇒x>0

    Decreasing

    f′(x)<0⇒x<0

    Final Answer

    (i) The function is increasing in (0,∞)
    (ii) The function is decreasing in (−∞,0)
    At x=0, the derivative is f′(0)=0 (stationary point)